Thomas' Calculus (12th Edition)

THOMAS'

CALCULUS Twelfth Edition

Based on the original work by

George B. Thomas, Jr. Massachusetts Institute of Technology as revised by

Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis

Addison-Wesley Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

Edltor-iD-Cblef: Deirdre Lynch Semor AcqailltiODl Editor: William Hoffinan Semor Project Editor: Rachel S. Reeve Associate Editor: Caroline Celano Associate Project Editor: Leah Goldberg Semor Managing Editor: Karen Wernhohn Semor Prodnctton Snpenlsor: Sheila Spinoey Senior Design Supervisor: Andrea Nix Digital Assets Manager: Mariaone Groth Media Producer: Lin Mahoney Software Development: Mary Dumwa1d and Bob Carroll EIecutive Marketing Manager: Jeff Weidenaar MarketingAsmlant: Kendra Bassi Senior Author Support/fec:hnology Specialist: Joe Vetere Senior Prepreill Supervilor: Caroline Fell Manufacturing Manager: Evelyn Beaton Production Coordinator: Kathy Diamond Composition: Nesbitt Graphics, Inc. lli••tratioDl: Karen Heyt, lllustraThch Cover Design: Rokusck Design Cover image: Forest Edge, Hokuto, Hokkaido, Japsn 2004 © Michael Kenna About the cover: The cover image of a tree line on a snow-swept landscape, by the photographer Michael Kenna, was taken in Hokkaido, Japan. The artist was not thinking of calculus when he composed the image, but rather, of a visual haiku consisting of a few elements that would spaIl< the viewer's imagination. Similarly, the minima1 design of this text allows the central ideas of calculus developed in this book to unfold to igoite the learner's imagination.

For permission to use copyrighted material, grateful acknowledgment is made to the copyright holders on page C-I, which is hereby made part of this copyright page. Many of the desigoations used by manufacturers and sellers to distinguish their products are claimed as tradenulrks. Where those designations appear in this book, and Addison-Wesley was aware of a trademark claim, the designations have been printed in initial caps or all caps.

Ubrary of Congress Cataloging-in-Publication Data Weir, Maurice D. Thomas' Calculus I Maurice D. Weir, Joel Hass, George B. Thomas.-12th ed. p.em ISBN 978-0-321-58799-2 1. CalculUl!-Textbooks. I. Hass, Joel. II. Thoroas, George B. (George Brinton), 1914--2006. III. Thomas, George B. (George Brinton), 1914--2006. Calculus. Iv. Title V. Title: Calculus. QA303.2.W452009b 515-dc22

2009023069

Copyright () 2010, 2005, 2001 Pearson Education, Inc. All rights reserved. No part of this publication may be reproduced, stated in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United

States ofAmerica. For information on obtaining permission for usc of material in this work, please submit a written request to PearnJn Education, inc., Rights and Contracts Department, SOl Boylston Street, Suite 900, Boston, MA 02116, fax your request to 617-848-7047, or e-mail at ht1p:/lwww.pearsoned.comllegal!permissions.htm. 12345678 91O--CRK-12 11 10 09

Addison-Wesley is an imprint of

-

PEARSON

ISBN-10: 0-321-58799-5

www.pearsoned.com

ISBN-13: 978-0-321-58799-2

CONTENTS

1

Preface

ix

I Functions

1 1.1 1.2 1.3 1.4

Functions and Their Graphs I CombiJring Functions; Shifting and Scaling Graphs Trigonometric Functions 22 Graphing with Calcolators and Computers 30 QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

34

35 37

AoDmONAL AND ADvANCED EXERCISES

2

Limits and Continuity 2.1 2.2 2.3 2.4 2.5 2.6

39

Rates of Change and Tangents to Curves 39 Limit of a Function and Limit Laws 46 The Precise Definition of a Limit 57 One-Sided Limits 66 Continuity 73 Limits Involving Infinity; Asymptotes of Graphs QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

98

Differentiation 3.1 3.2 3.3 3.4 3.5 3.6

84

96

97

AoDmONAL AND ADvANCED EXERCISES

3

14

102 Tangents and the Derivative at a Point 102 The Derivative as a Function 106 Differentiation Rules 115 The Derivative as a Rate of Change 124 Derivatives of Trigonometric Functions 135 The Chain Rule 142

iii

iv

Contents

3.7 3.8 3.9

4

Applications of Derivatives 4.1 4.2 4.3 4.4 4.5 4.6 4.7

5

Extreme Values of Functions 184 The Mean Value Theorem 192 Monotonic Functions and the First Derivative Test Concavity and Curve Sketching 203 Applied Optimization 214 Newton's Method 225 Antiderivatives 230 QuESTIONS TO GUIDE YOUR REVIEW 239 PRACTICE EXERCISES 240 ADDITIONAL AND ADvANCED EXERCISES 243

184

198

I Integration

246 5.1 5.2 5.3 5.4 5.5 5.6

6

Implicit Differentiation 149 Related Rates 155 Linearization and Differentials 164 QuESTIONS TO GUIDE YOUR REVIEW 175 PRACTICE EXERCISES 176 ADDITIONAL AND ADvANCED EXERCISES 180

Area and Estimating with Finite Sums 246 Sigma Notation and Limits of Finite Sums 256 The Definite Integral 262 The Fundamental Theorem of Calculus 274 Indefinite Integrals and the Substitution Method 284 Substitution and Area Between Curves 291 QuESTIONS TO GUIDE YOUR REVIEW 300 PRACTICE EXERCISES 301 ADDITIONAL AND ADvANCED EXERCISES 304

Applications of Definite Integrals 6.1 6.2 6.3 6.4 6.5 6.6

Volumes Using Cross-Sections 308 Volumes Using Cylindrical Shells 319 Arc Length 326 Areas of Surfaces of Revolution 332 Work and Fluid Forces 337 Moments and Centers of Mass 346 QuESTIONS TO GUIDE YOUR REVIEW 357 PRACTICE EXERCISES 357 ADDITIONAL AND ADvANCED EXERCISES 359

308

Contents

7

Transcendental Functions 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

8

9

Inverse Functions and Their Derivatives 361 Natural Logarithms 369 Exponential Functions 377 Exponential Change and Separable Differential Equations Indeterminate Forms and VHopitai's Rule 396 Inverse Trigonometric Functions 404 Hyperbolic Functions 416 Relative Rates of Growth 424 QuEsTIONS TO GUIDE YOUR REVIEW 429 PRACTICE EXERCISES 430 ADDmONAL AND ADvANCED EXERCISES 433

Integration by Parts 436 Trigonometric Integrals 444 Trigonometric Substitotions 449 Integration of Rational Functions by Partial Fractions Integral Tables and Computer Algebra Systems 463 Numerical Integration 468 Improper Integrals 478 QuEsTIONS TO GUIDE YOUR REVIEW 489 PRACTICE EXERCISES 489 ADDmONAL AND ADvANCED EXERCISES 491

453

496

Solutions, Slope Fields, and Euler's Method 496 First-Order Linear Equations 504 Applications 510 Graphical Solutions of Autonomous Equations 516 Systems of Equations and Phase Planes 523 QuEsTIONS TO GUIDE YOUR REVIEW 529 PRACTICE EXERCISES 529 ADDmONAL AND ADVANCED EXERCISES 530

Infinite Sequences and Series 10.1 10.2 10.3 lOA 10.5

387

435

First-Order Differential Equations 9.1 9.2 9.3 9.4 9.5

10

361

Techniques of Integration 8.1 8.2 8.3 8.4 8.5 8.6 8.7

V

Sequences 532 Infinite Series 544 The Integral Test 553 Comparison Tests 558 The Ratio and Root Thsts

532

563

vi

Contents

10.6 10.7 10.8 10.9 10.10

11

Parametric Equations and Polar Coordinates 11.1 11.2 11.3 11.4 11.5 11.6 11.7

12

660

Three-Dimensional Coordinate Systems 660 Vectors 665 The Dot Product 674 The Cross Product 682 Lines and Planes in Space 688 Cylinders and Quadric Surfaces 696 QuESTIONS TO GUIDE YOUR REVIEW 701 PRACTICE EXERCISES 702 ADDITIONAL AND ADvANCED EXERCISES 704

Vector-Valued Functions and Motion in Space 13.1 13.2 13.3 13.4 13.5 13.6

610

Parametrizations of Plane Curves 610 Calculus with Parametric Curves 618 Polar Coordinates 627 Graphing in Polar Coordinates 631 Areas and Lengths in Polar Coordinates 635 Conic Sections 639 Conics in Polar Coordinates 648 QuESTIONS TO GUIDE YOUR REVIEW 654 PRACTICE EXERCISES 655 ADDITIONAL AND ADvANCED EXERCISES 657

Vectors and the Geometry of Space 12.1 12.2 12.3 12.4 12.5 12.6

13

Alternating Series, Absolute and Conditional Convergence 568 Power Series 575 Taylor and Maclaurin Series 584 Convergence of Taylor Series 589 The Binomial Series and Applications of Taylor Series 596 QuESTIONS TO GUIDE YOUR REVIEW 605 PRACTICE EXERCISES 605 ADDITIONAL AND ADvANCED EXERCISES 607

Curves in Space and Their Tangents 707 Integrals of Vector Functions; Projectile Motion 715 Arc Length in Space 724 Curvatore and Normal Vectors ofa Curve 728 Tangential and Normal Components of Acceleration 734 Velocity and Acceleration in Polar Coordinates 739 QuESTIONS TO GUIDE YOUR REVIEW 742 PRACTICE EXERCISES 743 ADDITIONAL AND ADvANCED EXERCISES 745

707

Contents

14

Partial Derivatives 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10

15

16

747 Functions of Several Variables 747 Limits and Continuity in Higher Dimensions 755 Partial Derivatives 764 The Chain Rule 775 Directional Derivatives and Gradient Vectors 784 Tangent Planes and Differentials 791 Extreme Values and Saddle Points 802 Lagrange Multipliers 811 Taylor's Formula for Two Variables 820 Partial Derivatives with Constrained Variables 824 QUESTIONS TO GUIDE YOUR REVIEW 829 PRACTICE ExERCISES 829 ADomONAL AND ADvANCED EXERCISES 833

Multiple Integrals 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8

836 Double and Iterated Integrals over Rectangles 836 Double Integrals over General Regions 841 Area by Double Integration 850 Double Integrals in Polar Form 853 Triple Integrals in Rectangular Coordinates 859 Moments and Centers of Mass 868 Triple Integrals in Cylindrical and Spherical Coordinates Substitutions in Multiple Integrals 887 QUESTIONS TO GUIDE YOUR REVIEW 896 PRACTICE ExERCISES 896 ADomONAL AND ADvANCED EXERCISES 898

875

Integration in Vector Fields 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8

vii

Line Integrals 901 Vector Fields and Line Integrals: Work, Circulation, and Flux 907 Path Independence, Conservative Fields, and Potential Functions 920 Green's Theorem in the Plane 931 Surfaces and Area 943 Surface Integrals 953 Stokes' Theorem 962 The Divergence Theorem and a Unified Theory 972 QUESTIONS TO GUIDE YOUR REVIEW 983 PRACTICE ExERCISES 983 ADomONAL AND ADVANCED EXERCISES 986

901

viii

Contents

17

Second-Order Differential Equations 17.1 17.2 17.3 17.4 17.5

online

Second-Order Linear Equations Nonhomogeneous Linear Equations Applications Euler Equations Power Series Solutions

Appendices

AP-1 A.I A.2 A.3 AA A.5 A.6 A.7 A.S A.9

Real Numbers and the Real Line AP-I Mathematical Induction AP-6 Lines, Circles, and Parabolas AP-1O Proofs of Limit Theorems AP-IS Commonly Occurring Limits AP-2I Theory of the Real Numbers AP-23 Complex Numbers AP-25 The Distributive Law for Vector Cross Products AP-35 The Mixed Derivative Theorem and the Increment Theorem

AP-36

Answers to Odd-Numbered Exercises

A-1

Index

1-1

A Brief Table of Integrals

T-1

Credits

C-1

PREFACE We have significantly revised this edition of Thomas' Calculus to meet the changing needs of today's instructors and students. The result is a book with more examples, more midlevel exercises, more figures, better conceptual flow, and increased clarity and precision. As with previous editions, this new edition provides a modern introduction to calculus that supports conceptual understanding but retains the essential elements of a traditional course. These enhancements are closely tied to an expanded version for this text of MyMathLab® (discussed further on), providing additional support for students and flexibility for instructors. Many of our students were exposed to the terminology and computational aspects of calculus during high school. Despite this familiarity, students' algebra and trigonometry skills often hinder their success in the college calculus sequence. With this text, we have sought to balance the students' prior experience with calculus with the algebraic skill development they may still need, all without undermining or derailing their confidence. We have taken care to provide enough review material, fully stepped-out solutions, and exercises to support complete understanding for students of all levels. We encourage students to think beyond memorizing formulas and to generalize concepts as they are introduced. Our hope is that after taking calculus, students will be confident in their problem-solving and reasoning abilities. Mastering a beautiful subject with practical applications to the world is its own reward, but the real gift is the ability to think and generalize. We intend this book to provide support and encouragement for both.

Changes for the TweLfth Edition CONTENT In preparing this edition we have maintained the basic structure of the Table of Contents from the eleventh edition. Yet we have paid attention to requests by current users and reviewers to postpone the introduction of parametric equations until we present polar coordinates, and to treat I'Hopital 's Rule after the transcendental functions have been studied. We have made numerous revisions to most of the chapters, detailed as follows.

• Functions We condensed this chapter even more to focus on reviewing function concepts. Prerequisite material covering real numbers, intervals, increments, straight lines, distances, circles, and parabolas is presented in Appendices 1-3. • Limits To improve the flow of this chapter, we combined the ideas of limits involving infinity and their associations with asymptotes to the graphs of functions, placing them together in the final chapter section. • Differentiation While we use rates of change and tangents to curves as motivation for studying the limit concept, we now merge the derivative concept into a single chapter. We reorganized and increased the number of related rates examples, and we added new examples and exercises on graphing rational functions.

ix

X

Preface

• Antiderivatives and Integration We maintain the orgaJrization of the eleventh edition in placing antiderivatives as the fInal topic of the chapter covering applications of derivatives. Our focus is on ''recovering a function from its derivative" as the solution to the simplest type offirst-order differential equation. Integrals, as "limits of Riemann sums," motivated primarily by the problem of rmding the areas of general regions with curved boundaries, are a new topic fanning the substance of Chapter 5. After carefully developing the integral concept, we turn our attention to its evaluation and connection to antiderivatives captured in the Fundamental Theorem of Calculus. The ensuing applications then define the various geometric ideas of area, volume, lengths of paths, and centroids all as limits of Riemann sums giving definite integrals, which can be evaluated by finding an antiderivative of the integrand. We return later to the topic of solving more complicated first-order differential equations, after we derme and establish the transcendental functions and their properties. • Differential Equations Some universities prefer that this subject be treated in a course separate from calculus. Although we do cover solutions to separable differential equations when treating exponential growth and decay applications in the chapter on transcendental functions, we organize the bulk of our material into two chapters (which may be omitted for the calculus sequence). We give an introductory treatment of first-order differential equations in Chapter 9, including a new section on systems and phase planes, with applications to the competitive-hunter and predator-prey models. We present an introduction to second-order differential equations in Chapter 17, which is included in MyMathLab as well as the Thomas' Calculus Web site, www.pearsonhighered.comlthomas. • Series We retain the organizational structure and content of the eleventh edition for the topics of sequences and series. We have added several new figures and exercises to the various sections, and we revised some of the proofs related to convergence of power series in order to improve the accessibility of the material for students. The request stated by one of our users as, "anything you can do to make this material easier for students will be welcomed by our faculty;' drove our thinking for revisions to this chapter. • Parametric Equations Several users requested that we move this topic into Chapter II, where we also cover polar coordinates and conic sections. We have done this, realizing that many departments choose to cover these topics at the beginning of Calculus m, in preparation for their coverage of vectors and multivariable calculus. • Vector-Valued Functions We streamlined the topics in this chapter to place more emphasis on the conceptual ideas supporting the later material on partial derivatives, the gradient vector, and line integrals. We condensed the discussions of the Frenet frame and Kepler's three laws of planetary motion. • Multivariahle Calculus We have further enhanced the art in these chapters, and we have added many new figures, examples, and exercises. We reorganized the opening material on double integrals, and combined the applications of double and triple integrals to masses and moments into a single section covering both two- and threedimensional cases. This reorganization allows for better flow of the key mathematical concepts, together with their properties and computational aspects. As with the eleventh edition, we continue to make the connections of multivariable ideas with their aingle-variable analogues studied earlier in the book. • Vector Fields We devoted considerable effort to improving the clarity and mathematical precision of our treatment of vector integral calculus, including many additional examples, figures, and exercises. Important theorems and results are stated more clearly and completely, together with enhanced explanations of their hypotheses and mathematical consequences. The area of a surface is now organized into a single section, and surfaces defined implicitly or explicitly are treated as special cases of the more general parametric representation. Surface integrals and their applications then follow as a separate section. Stokes' Theorem and the Divergence Theorem are still presented as generalizations of Green's Theorem to three dimensions.

Preface

xi

EXERCISES AND EXAMPLES We know that the exercises and examples are critical components in learning calculus. Because of this importance, we have updated, improved, and increased the number of exercises in nearly every section of the book. There are over 700 new exercises in this edition. We continue our organization and grouping of exercises by topic as in earlier editions, progressing from computational problems to applied and theoretical problems. Exercises requiring the use of computer software systems (such as Maple® or Mathematica®) are placed at the end of each exercise section, labeled Computer Explorations. Most of the applied exercises have a subheading to indicate the kind of application addressed in the problem. Many sections include new examples to clarify or deepen the meaning of the topic being discussed, and to help students understand its mathematical consequences or applications to science and engineering. At the same time, we have removed examples that were a repetition of material already presented. ART Because of their importance to learning calculus, we have continued to improve existing figures in Thomas' Calculus and we have created a significant number of new ones. We continue to use color consistently and pedagogically to enhance the conceptual idea that is being illustrated. We have also taken a fresh look at all of the figure captions, paying considerable attention to clarity and precision in short statements.

y

No matter what

y=1

positive nwnber E is, the graph enters this band at x = -~ and stays.

FIGURE 2.50, page 85 The geometric explanation of a finite limit as x - ± 00.

FIGURE 16.9, page 908 A surface in a space occupied by a moving fluid.

MYMATHLAB AND MATH XL The increasing use of and demand for online homework systems has driven the changes to MyMathLab and MathXL® for Thomas' Calculus. The MyMathLab course now includes significantly more exercises of all types. New Java™ applets add to the already significant collection to help students visualize the concepts and generalize the material.

Continuing Features RIGOR The level of rigor is consistent with that of earlier editions. We continue to distinguish between formal and informal discussions, and to point out their differences. We think starting with a more intuitive, less formal approach helps students understand a new or difficult concept so they can then appreciate its full mathematical precision and outcomes. We pay attention to defining ideas carefully and to proving theorems appropriate for

xii

Preface

calculus students, while mentioning deeper or subtler issues they would study in a more advanced course. Our organization, and distinctions between informal and formal discussions, gives the instructor a degree of flexibility in the amount and depth of coverage of the various topics. For example, while we do not prove the Intermediate Value Theorem or the Extreme Value Theorem for continuous functions on a :5 x :5 b, we do state these theorems precisely, illustrate their meanings in numerous examples, and use them to prove other important results. Furthermore, for those instructors who desire greater depth of coverage, we discuss in Appendix 6 the reliance of the validity of these theorems on the completeness of the real numbers.

WRmNG EXERCISES Writing exercises placed throughout the text ask students to explore and explain a variety of calculus concepts and applications. In addition, the end of each chapter contains a list of questions for students to review and summarize what they have learned. Many of these exercises make good writing assignments. END-Of-CHAPTER REVIEWS AND PROJECTS In addition to problems appearing after each section, each chapter colminates with review questions, practice exercises covering the entire chapter, and a series of Additional and Advanced Exercises serving to include more challenging or synthesizing problems. Most chapters also include descriptions of several Technology Application Projects that can be worked by individual students, or groups of students, over a longer period of time. Thesc projects require the use of a computer, running Mathematica or Maple, and additional material that is available over the Internet at www.pearsonhigherecLcomlthomas and in MyMathLab. WRITING AND APPUCATIONS As always, this text continues to be easy to read, conversational, and mathematically rich. Each new topic is motivated by clear, easy-to-understand examples and is then reinforced by its application to real-world problems of immediate interest to students. A hallmark of this book has been the application of calculus to science and engineering. These applied problems have been updated, improved, and extended continually over the last several editions. TECHNOLOGY In a course using the text, technology can be incorporated according to the taste of the instructor. Each section contains exerciscs requiring the use of technology; these are marked with a D if suitable for calculator or computer use or are labeled Computer Explorations if a computer algebra system (CAS, such as Maple or Mathematica) is required.

Text Versions THOMAS' CALCULUS, Twelfth Edition Complete (Chapters 1-16), ISBN 0-321-58799-51978-0-321-58799-2 Single Variable Calculus (Chapters 1-11), ISBN 0-321-63742-91978-0-321-63742-0 Multivariable Calculus (Chapters 11-16), ISBN 0.321-64369-01978-0-321-64369-8

THOMAS' CALCULUS: EARLY TRANSCENDENTALS, Twelfth Edition Complete (Chapters 1-16), ISBN 0-321-58876-21978-0-321-58876-0 Single Variable Calculus (Chapters 1-11),0-321-62883-71978-0-321-62883-1 Multivariable Calculus (Chapters 11-16), ISBN 0.321-64369-01978-0-321-64369-8 The early transcendentals version of Thomas' Calculus introduces and integrates transcendental functions (such as inverse trigonometric, exponential, and logsritlnnic functions) into the exposition, examples, and exercises of the early chapters alongside the algebraic functions. The Multivariable book for Thomas' Calculus: Early 1'rnnscendentals is the same text as Thomas' Calculus, Multivariable.

Preface

xiii

Instructor's Editions Thomas , Calculus, ISBN 0-321-60075-41978-0-321-60075-2 Thomas 'Calculus: Early Transcentientals, ISBN 0-321-62718-01978-0-321-62718-6 In addition to including all of the answers present in the stodent editions, the Instructor ~ Editions include even-numbered answers for Chapters l--{j.

University Calculus (Early Transcendentals) University Calculus: Alternative Edition (Late Transcendentals) University Calculus: Elements with Early Transcendentals The University Calculus texts are based on Thomas' Calculus and featore a streamlined presentation of the contents of the calculus course. For more information about these titles, visit www.pearsonhighered.com.

Print Supplements INSTRUCTOR'S SOLUTIONS MANUAL Single Variable Calculus (Chapters 1-11), ISBN 0-321-60807-01978-0-321-60807-9 Multivariable Calculus (Chapters 11-16), ISBN 0-321-60072-X 1978-0-321-60072-1 The Instructor ~ Solutions Manual by William Ardis, Collin County Community College, contains complete worked-out solutions to all of the exercises in the text.

STUDENrs SOLUTIONS MANUAL Single Variable Calculus (Chapters 1-11), ISBN 0-321-60070-31978-0-321-60070-7 Multivariable Calculus (Chapters 11-16), ISBN 0-321-60071-11978-0-321-60071-4 The Student~ SolutiollS Manual by William Ardis, Collin County Community College, is designed for the stodent and contains carefully worked-out solutions to all the oddnumbered exercises in the text.

JUST-IN-TIME ALGEBRA AND TRIGONOMETRY FOR CALCULUS, Fourth Edition ISBN 0-321-67104-X 1978-0-321-67104-2 Sharp algebm and trigonometry skills are critical to mastering calculus, and Just-in-TIme Algebra and 1Hgonometry for Calculus by Guntram Mueller and Ronald I. Brent is designed to bolster these skills while stodents stody calculus. As stodents make their way through calculus, this text is with them every step of the way, showing them the necessary algebm or trigonometry topics and pointing out potential problem spots. The easy-to-use table of contents has algebm and trigonometry topics armnged in the order in which stodents will need them as they stody calculus.

CALCULUS REVIEW CARDS The Calculus Review Cards (one for Single Variable and another for Multivariable) are a stodent resource containing important formulas, functions, defmitions, and theorems that correspond precisely to Thomas' Calculus. These cards can work as a reference for completing homework assignments or as an aid in studying, and are available bundled with a new text. Contact your Pearson sales representative for more information.

Media and Online Supplements TECHNOLOGY RESOURCE MANUALS Maple Manual by James Stapleton, North Carolina State University Mathematica Manual by Marie Vanisko, Carroll College IT-Graphing Calculator Manual by Elaine McDonald-Newman, Sonoma State University These manuals cover Maple 13, Mathematica 7, and the TI-83 PlUS/TI-84 Plus and TI-89, respectively. Each manual provides detailed gnidance for integrating a specific software package or gmphing calculator throughout the course, including syntax and commands.

xiv

Preface

These manuals are available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhigheredlirc, and MyMatbLab.

WEB SITE _w.pearsonhighered.com/thomas The Thomas' Calculus Web site contains the cbapter on Second-Order Differential Equations, including odd-numbered answers, and provides the expanded historical biographies and essays referenced in the text. Also available is a collection of Maple and Mathematica modules, as well as the Technology Application Projects, which can be used as projects by individual students or groups of stodents.

MyMathLab Online Course (access code required) MyMathLab is a text-specific, easily customizable ouline course that integrates interactive multimedia inatruction with textbook content. MyMathLab gives you the tools you need to deliver all or a portion of your course ouline, whether your stodents are in a lab setting or working from home. o Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and learning aids for extra help. o "Getting Ready" chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with. o Personalized Study Plan, generated when students complete a test or quiz, indicates which topics have been mastered and links to tutorial exercises for topics stodents have not mastered. o Multimedia learning aids, such as video lectures, Java applets, animations, and a complete multimedia textbook, help stodents independently improve their understanding and performance. o

Assessment Manager lets you create ouline homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MyMatbLab exercise bank and instructor-created custom exercises.

o Gradebook, designed specifically for mathematics and statistics, automatically tracks students' results and gives you control over how to calculate f'mal grades. You can also add offiine (paper-and-pencil) grades to the gradebook. o MathXL Exercise Builder allows you to create static and algorithmic exercises for your ouline assignments. You can use the library of sample exercises as an easy starting point. o Pearson Tutor Center (www.pearsontutorservices.com) access is automatically included with MyMatbLab. The Tutor Center is staffed by qualified math instructors who provide textbook-specific tutoring for students via toll-free phone, fax, email, and interactive Web sessions. MyMathLab is powered by CourseCompass™, Pearson Education's ouline teaching and learning environment, and by MathXL, our ouline homework, tutorial, and assessment system. MyMatbLab is available to qualified adopters. For more information, visit www.mymathlab.com or contact your Pearson sales representative.

Video Lectures with Optional Captioning The Video Lectures with Optional Captioning feature an engaging team of mathematics instructors who present comprehensive coverage of topics in the text. The lectorers' presentations include examples and exercises from the text and support an approach that emphasizes visualization and problem solving. Available ouly through MyMathLab and MathXL.

Preface

XV

MathXL Online Course (access code required) MathXL is an online homework, tutorial, and assessment system that accompanies Pearson's textbooks in mathematics or statistics. • Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and learning aids for extra help. • "Getting Ready" chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with. • PersonaIized Study Plan, generated when students complete a test or qillz, indicates which topics have been mastered and links to tutorial exercises for topics students have not mastered. • Multimedia learning aids, such as video lectures, Java applets, and animations, help students independently improve their understanding and performance. • Gradebook, designed specifically for mathematics and statistics, automatically tracks students' results and gives you control over how to calculate fmal grades. • MathXL Exercise Builder allows you to create static and algorithmic exercises for your online assignments. You can use the library of sample exercises as an easy starting point. • Assessment Manager lets you create online homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MathXL exercise bank, or instructur-created custom exercises. MathXL is available to qualified adopters. For more information, visit our Web site at www.mathxl.com. or contact your Pearson sales representative.

TestGen® TestGen (www.pearsonhighered.comltestgen)enablesinstructurstobuild.edit.print. and administer tests using a computerized hank of questions developed to cover all the objectives of the text. TestGen is algorithmically based, allowing instructurs to create multiple but equivalent versions of the same question or test with the click of a button. Instructors can also modifY test hank questions or add new questions. Tests can be printed or administered online. The software and testbank are available for dowuload from Pearson Education's online catalog.

PowerPoint® Lecture Slides These classroom presentation slides are geared specifically to the sequence and philosophy of Thomas' Calculus series. Key graphics from the book are included to help bring the concepts alive in the classroom.These files are available to qualified instructors tbrough the Pearson Jnstructur Resource Center, www.pearsonhigheredlirc, and MyMathLab.

Acknowledgments We would like to express our thanks to the people who made many valuable contributions to this edition as it developed through its various stages:

Accuracy Checkers Blaise DeSesa Paul Lorczak Kathleen Pellissier Lauri Semarne Sarab Streett Holly Zullo

xvi

Preface

Reviewers for the Twelfth Edition Meighan Dillon, Southern Polytechnic State University Anne Dougherty, University of Colorado Said Fariabi, San Antonio College Klaus Fischer, George Mason University Tim Flood, PiUsburg State University Rick Ford, California State University-Chico Robert Gardner, East Tennessee State University Christopher Heil, Georgia Institute ofTechnology Joshua Brandon Holden, Rose-Hulman Institute ofTechnology Alexander Hulpke, Colorado State University Jacqueline Jensen, Sam Houston State University Jennifer M. Johnson, Princeton University Hideaki Kaneko, Old Dominion University Przemo Kranz, University ofMississippi Xin Li, University of Central Florida Maura Mast, University ofMassachusetts-Boston Val Mohanakumar, Hillsborough Community College-Dale Mabry Campus Aaron Montgomery, Central Washington University Cynthia Piez, University ofIdaho Brooke Quinlan, Hillsborough Community College-Dale Mabry Campus Rebecca A. Segal, Virginia Commonwealth University Andrew V. Sills, Georgia Southern University Alex Smith, University ofWlSconsin-Eau Claire Mark A. Smith, Miami University Donald Solomon, University ofWISconsin-Milwaukee Blake Thornton, Washington University in St. Louis David Walnut, George Mason University Adrian Wilson, University ofMontevallo Bobby Winters, PiUsburg State University Dennis Wortman, University ofMassachusetts-Boston

1 FUNCTIONS OVERVIEW Functions are fundamental to the study of calculus. In this chapter we review what functions are and how they are pictured as graphs, how they are combined and transformed, and ways they can be classified. We review the trigonometric functions, and we discuss misrepresentations that can occur when using calculators and computers to obtain a function's graph. The real number system, Cartesian coordinates, straight lines, parabolas, and circles are reviewed in the Appendices. We treat inverse, exponential, and logarithmic functions in Chapter 7.

1.1

Functions and Their Graphs Functions are a tool for describing the real world in mathematical terms. A function can be represented by an equation, a graph, a numerical table, or a verbal description; we will use all four representations throughout this book. This section reviews these function ideas.

Functions; Domain and Range The temperature at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). The interest paid on a cash investment depends on the length of time the investment is held. The area of a circle depends on the radius of the circle. The distance an object travels at constant speed along a straight-line path depends on the elapsed time. In each case, the value of one variable quantity, say y, depends on the value of another variable quantity, which we might call x. We say that "y is a function of x" and write this symbolically as y

= f(x)

("y equals f of x").

In this notation, the symbol f represents the function, the letter x is the independent variable representing the input value of f, and y is the dependent variable or output value of

fatx.

DEFINITION

A function f from a set D to a set Y is a rule that assigns a unique E Y to each element XED.

(single) element f(x)

The set D of all possible input values is called the domain of the function. The set of all values of f(x) as x varies throughout D is called the range of the function. The range may not include every element in the set Y. The domain and range of a function can be any sets of objects, but often in calculus they are sets of real numbers interpreted as points of a coordinate line. (In Chapters 13-16, we will encounter functions for which the elements of the sets are points in the coordinate plane or in space.)

1

2

Chapter 1: Functions

J

x --:-_.~

Input (donurin)

FIGURE 1.1

:"""":~~•• J(x)

Output

(nmge)

A diagram showing a

function as a kind of machine.

~

~J(a) D

= domain set

Vx

J(x)

Y = set containing tberange

A function from a setD to a set Yassigns a unique element of Y to each element in D. FIGURE 1.2

Often a function is given by a formula that describes how to calculate the output value from the input variable. For instance, the equation A = 1Tr2 is a rule that calculates the area A of a circle from its radius r (so r, interpreted as a length, can only be positive in this formula). When we derme a function y = f(x) with a formula and the domain is not stated explicitly or restricted by context, the domain is assumed to be the largest set of real x-values for which the formula gives real y-values, the so-called natura! domain. If we want to restrict the domain in some way, we must say so. The domain of y = x 2 is the entire set of real numbers. To restrict the domain of the function to, say, positive values of x, we would write "y = x 2 ,x > 0." Changing the domain to which we apply a formula usually changes the range as well. The range of y = x 2 is [0, 00). The range of y = x 2, X ;;,: 2, is the set of all numbers obtained by squaring numbers greater than or equal to 2. In set notation (see Appendix 1), the rangeis {x 2 Ix;;,: 2} or{Yly;;': 4} or [4, 00). When the range of a function is a set of real numbers, the function is said to be rea!valued. The domains and ranges of many real-valued functions of a rea! variable are intervals or combinations of interva1s. The interva1s may be open, closed, or half open, and may be finite or infinite. The range of a function is not a!ways eaay to find. A function f is like a machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain (Figure 1.1). The function keys on a calculator give an example of a function as a machine. For instance, the key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the key. A function can also be pictured as an arrow diagram (Figure 1.2). Each arrow associates an element of the domain D with a unique or single element in the set Y. In Figure 1.2, the arrows indicate thatf(a) is associated with a, f(x) is associated with x, and so on. Notice that a function can have the same value at two different input elements in the domain (as occurs with f(a) in Figure 1.2), but each input element x is assigned a single output value f(x).

Vx

EXAMPLE 1 Let's verify the natural domains and associated ranges of some simple functions. The domains in each case are the values of x for which the formula makes sense. Function

Domain (x)

Range (y)

y = x2

(-00,00) (-00,0) U (0, 00) [0, 00) (-00,4] [-I, I]

[0, 00)

Y = I/x y=

Vx

y=~ y =

v'1=7

(-00,0) U (0, 00) [0, 00) [0, 00)

[0, I]

Solution The formula y = x 2 gives a rea! y-value for any real number x, so the domain is (- 00, 00). The range of y = x 2 is [0, 00) because the square of any real number is nonnegative and every nonnegative number y is the square of its own square root, y = (vY)2fory;;,: O. The formula y = l/x gives a real y-value for every x except x = O. For consistency in the rules of arithmetic, we cannot divide any number by zero. The range of y = I/x, the set of reciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = I/(I/y). That is, for y oF 0 the number x = I/y is the input assigned to the output valuey. The formula y = Vx gives a real y-value only if x ;;,: O. The range of y = Vx is [0, 00) because every nonnegative number is some number's square root (namely, it is the square root of its own square). In y = '\I'4-=x, the quantity 4 - x cannot be negative. That is, 4 - x ;;,: 0, or x :5 4. The formula gives real y-values for all x :5 4. The range of '\I'4-=x is [0, 00), the set of all nonnegative numbers.

1.1

Functions and Their Graphs

3

The formula y = ~ gives a real y-value for every x in the closed interval from - 1 to 1. Outside this domain. 1 - x 2 is negative and its square root is not a real number. The values of 1 - x 2 vary from 0 to 1 on the given domain, and the square roots of these values do the same. The range of ~ is [0. 1]. •

Graphs of Functions If j is a function with domain D, its graph consists of the points in the Cartesian plane whose coordinates are the input-output pairs for j. In set notation. the graph is {(x,j(x)) I XED}.

The graph of the function j(x) = x + 2 is the set of points with coordinates (x,y) for which y = x + 2. Its graph is the straight line sketched in Figure 1.3. The graph of a function j is a useful picture of its behavior. If (x. y) is a point on the graph, then y = j(x) is the height of the graph above the pointx. The height may be positive or negative, depending on the sign of j(x) (Figure 104).

y

X

y=xl

-2 -1 0 1

4 1 0 1

3

2

9 4

2

4

FIGURE 1.3 The graph of f(x) = x + 2 is the set of points (x,y) for whichy has the value x + 2.

FIGURE 1.4

f(x)

EXAMPLE 2

Graph the function y

=

If(x.y) lies on the graph of

f. then the value y = f(x) is the height of the graph above the poiut x (or below x if

is negative).

x 2 over the interval [-2,2].

Solution Make a table ofxy-pairs that satisfy the equation y = x 2• Plot the points (x,y) whose coordinates appear in the table, and draw a smooth curve (labeled with its equation) through the plotted points (see Figure 1.5). •

y

How do we know that the graph of y = x 2 doesn't look like one of these curves? (2.4)

y

--~--~~~~--~~X

-2

-1

FIGURE 1.5

Example 2.

0

2

Graph of the function in

y

4

Chapter 1: Functions

To find out, we could plot more points. But how would we then connect them? The basic question still remains: How do we know for sure what the graph looks like between the points we plot? Calculus answers this question, as we will see in Chapter 4. Meanwhile we will have to settle for plotting points and connecting them as best we can.

Representing a Function Numerically We have seen how a function may be represented algebraically by a formula (the area function) and visually by a graph (Example 2). Another way to represent a function is numerically, through a table of values. Numerical representations are often used by engineers and scientists. From an appropriate table of values, a graph of the function can be obtained using the method illustrated in Example 2, possibly with the aid of a computer. The graph consisting of only the points in the table is called a scatterplot.

EXAMPLE 3 Musical notes are pressure waves in the air. The data in Table l.l give recorded pressure displacement versus time in seconds of a musical note produced by a toning fork. The table provides a representation of the pressure function over time. If we first make a scatterplot and then connect approximately the data points (t, p) from the table, we obtain the graph shown in Figure 1.6. p (pressure)

TABLE 1.1 Tuning fork data Time

Pressure

Time

0.00091 0.00108 0.00125 0.00144 0.00162 0.00180 0.00198 0.00216 0.00234 0.00253 0.00271 0.00289

-0.080

0.00362 0.00379 0.00398 0.00416 0.00435 0.00453 0.00471 0.00489 0.00507 0.00525 0.00543 0.00562

0.00307 0.00325 0.00344

0.200 0.480 0.693 0.816 0.844 0.771 0.603 0.368 0.099 -0.141 -0.309 -0.348 -0.248 -0.041

0.00579 0.00598

Pressure 0.217 0.480 0.681 0.810 0.827 0.749 0.581 0.346 0.077 -0.164 -0.320 -0.354 -0.248 -0.035

1.0 0.8 0.6 0.4 0.2

t (sec) -0.2 -0.4 -0.6

FIGURE 1.6 A smooth curve through the plotted points gives a graph of the pressure functioo represented by Table l.l (Example 3).

• The Vertical Line Test for a Function Not every curve in the coordinate plane can be the graph of a function. A function 1 can have ouly one value I(x) for each x in its domain, so 110 vertical line can intersect the graph of a function more than once. If a is in the domain of the function I, then the vertical line x = a will intersect the graph of 1 at the single point (a, I(a». A circle cannot be the graph of a function since some vertical lines intersect the circle twice. The circle in Figure 1.7a, however, does contain the graphs of two functions of x: the upper semicircle defmed by the function I(x) = ~ and the lower semicircle defined by the function g(x) = - ~ (Figures 1.7b and 1.7c).

1.1 Y

Functions and Their Graphs

Y

5

Y

-I --~----~----~-->x

-I

0

--~----~----~~x

o

(C)Y~-~

FIGURE 1.7 (a) The circle is not the graph ofa function; itfails the vertical line test. (b) The upper semicircle is the graph ofa function f(x) = ~. (c) The lower semicircle is the graph ofa functiong(x) = -~.

y

Piecewise-Defined Functions Sometimes a function is described by using different formulas on different parts of its domain. One example is the absolute value funetion FIGURE 1.8 The absolute value function has domain ( - 00, 00) and range [0, 00). y

Ixl =

x;;"

0

<

0,

X

The function

f(x) =

y

x, -x,

whose graph is given in Figure 1.8. The right-hand side of the equation means that the function equals x if x ;;" 0, and equals -x if x < O. Here are some other examples.

EXAMPLE 4

FIGURE 1.9 To graph the function y = f(x) shown here, we apply different formulas to different parts ofits domain (Example 4).

{

{-:~, I,

x I

is defined on the entire rea! line but has values given by different formulas depending on the position of x. The values of f are given by y = -x when x < 0, y = x 2 when o :5 x :5 I, and y = I when x > 1. The function, however, is just one function whose • domain is the entire set of rea! numbers (Figure 1.9).

,

,'Y=x 3

, , ----..0

2

, ,.!......o ~

Y~LxJ

"

--_~2~_~I-'.¥~~~2~~3~~x

,

"

EXAMPLE 5 The function whose value at any number x is the greatest integer less than or equal to x is ca!led the greatest integer function or the integer floor function. It is denoted l x J. Figure 1.10 shows the graph. Observe that

l2.4J = 2, l2J = 2,

........a -2

"

FIGURE 1.10 The graph of the greatest integer function y = l x J lies on or below the line y = x, so it provides an integer floor for x (Example 5).

ll.9 J lO.2J

= =

I, 0,

lOJ = 0, l-0.3 J = -I

l-1.2J = -2, l-2J = -2.

•

EXAMPLE 6 The function whose value at any number x is the smallest integer greater than or e9ual to x is called the least integer function or the integer ceiling function. It is Figure 1.11 shows the graph. For positive values of x, this function might denoted I represent, for example, the cost of parking x hours in a parking lot which charges $1 for • each hour or part of an hour.

xl.

6

Chapter 1: Functions y

Increasing and Decreasing Functions / ' y=x

3

2

If the graph of a function climbs or rises as you move from left to right, we say that the function is increasing. If the gmph descends or falls as you move from left to right, the function is decreasing.

'7

0---;-' ,

y

~

rxl

DEFINmONS Let f be a function defmed on an interval I and let X, and X2 be any two points in I.

FIGURE 1.11 TIre graph ofthe least integer function y ~ rxl lies on or above the line y = x. so it provides an integer ceiling for x (Example 6).

1.

If f(X2)

2.

If f(X2)

> <

f(Xl) whenever Xl f(Xl) whenever Xl

< X2, then f < X2, then f

is said to be increasing on I. is said to be decreasing on I.

It is important to realize that the defmitions of increasing and decreasing functions must be satisfied for every pair of points Xl and X2 in I with Xl < X2. Because we use the inequality < to compare the function values, instead of :5 , it is sometimes said that f is strictly increasing or decreasing on I. The interval I may be finite (also called bounded) or infmite (unbounded) and by definition never consists of a single point (Appenrux 1).

EXAMPLE 7

The function graphed in Figure 1.9 is decreasing on (-00,0] and increasing on [0, I]. The function is neither increasing nor decreasing on the interval [I, 00) because of the strict inequalities used to compare the function values in the definitions. _

Even Functions and Odd Functions: Symmetry The graphs of even and odd functions have chamcteristic symmetry properties.

DEFINmONS

A function y = f(x) is an even function ofx if f( -x) = f(x), odd function ofx if f( -x) = - f(x),

for every X in the function's domain. y

(-x,y) x

0 (a)

y

------~~~------x

The names even and odd come frorn powers of x. If Y is an even power of x, as in y = x 2 or Y = x\ it is an even function ofx because (-x)2 = x 2 and (-x)4 = x4. Ify is an odd power of x, as iny = X or y = x\ it is an odd function ofxbecause (-x)' = -x and (-X)3 = -x 3. The graph of an even function is symmetric about they-axis. Since f( -x) = f(x) , a point (x,y) lies on the graph ifand only if the point (-x, y) lies on the graph (Figure 1.12a). A reflection across the y-axis leaves the graph unchanged. The graph of an odd function is symmetric about the origin. Since f( -x) = - f(x) , a point (x,y) lies on the graph ifand on1y if the point (-x, -y) lies on the graph (Figure 1.12b). Equivalently, a gmph is symmetric about the origin if a rotation of 1800 about the origin leaves the graph unchanged. Notice that the definitions imply that both x and -x must be in the domain off.

EXAMPLE 8 (b)

FIGURE 1.12 (a) The graph of y ~ x 2 (an even function) is symmetric about the y-axis. (b) The graph ofy = x' (an odd function) is symmetric about the origin.

f(x) = x 2 f(x) = x 2

Even function: (-xl' = x 2 for all x; symmetry about y-axis.

+

1

+

1 = x 2 + 1 for all x; symmetry about y-axis

Odd function: (-x) = -x for all x; symmetry about the origin.

f(x) = x f(x) = x

Even function: (-xl' (Figure 1.13a).

+

1

Not odd: f( -x) = -x + 1, but - f(x) = -x - I. The two are not equal. Not even: (-x) + 1 '" x + 1 for all x '" 0 (Figure 1.13b). _

1.1 y


7

y

----~~------~x

y =

Functions and Their Graphs

-----,!'-~------x

(bJ

(a) When we add the coostant term I to the function

x 2 , the resulting functioo y = x 2 + I is still even and its grnph is

still symmetric ahoul the y-axis. (b) When we add the constant term I to the function y = x, the resulting function y = x + I is no longer odd. The symme1ry about the origin is lost (EXlIDlple 8).

Common Functions A variety of important types of functions are frequently encountered in calculus. We identify and briefly describe them here.

Linear Functions A function of the form j{x) = mx + b, for constants m and b, is called a linear function. Figure 1.14a shows an array oflines j{x) = mx where b = 0, so these lines pass through the origin. The function j{x) = x where m = I and b = 0 is called the identity function. Constant functions result when the slope m = 0 (Figure 1.14b). A linear function with positive slope whose graph passes through the origin is called a proportionality relationship. y

m=-3 m=-l

m=l

m=z1 ------~~~----~


with slope m

=

y

2

__ x

(bJ

(a) Lines through the origin with slope m. (b) A constant function O.

DEFINmON Two varisbles y and x are proportional (to one another) if one is always a constant multiple of the other; that is, if y = kx for some nonzero constantk.

If the variable y is proportional to the reciprocal I/x, then sometimes it is said that y is inversely proportional to x (because I/x is the multiplicative inverse of x).

Power Functions A function j{x) = x·, where a is a constant, is called a power function. There are several important cases to consider.

8

Chapter 1: Functions

(a) a = n, a positive integer.

The graphs of f(x) = x·, for n = 1,2,3,4,5, are displayed in Figure 1.15. These functions are dermed for all real values of x. Notice that as the power n gets larger, the curves tend to flatten toward the x-axis on the interval (- I, I), and also rise more steeply for Ixl > I. Each curve passes through the point (I, I) and through the origin. The graphs of functions with even powers are symmetric about the y-axis; those with odd powers are symmetric about the origin. The even-powered functions are decreasing on the interval (-00,0] and increasing on [0, 00); the odd-powered functions are increasing over the entire real line (-00,00).

-1

FIGURE 1.15

-1

Graphs of J(x) = x', n = 1,2,3,4,5, deImed for -00

<x<

00.

(b) a = - l o r a = -2. The graphs of the functions f(x) = X-I = I/x and g(x) = x-2 = I/x 2 are shown in Figure 1.16. Both functions are dermed for all x '" 0 (you can never divide by zero). The graph of y = I/x is the hyperbola xy = I, which approaches the coordinate axes far from the origin. The graph of y = I/x 2 also approaches the coordinate axes. The graph of the function f is symmetric about the origin; f is decreasing on the intervals (-00,0) and (0, 00). The graph of the function g is symmetric about the y-axis; g is increasing on (-00,0) and decreasing on (0,00). Y

~, 1

JY 1

Domain: x

'* 0

~~c--+~~~ x

Range: n' 0

0

Domain: x =F 0 Range: y> 0 (a)

(b)

FIGURE 1.16 Graphsof1hepowerfunctionsJ(x) and for part (b) a = -2.

I I 3

=

x'forpart(a)a

=

-1

2

(c) a = 2'3'2,and 3 .

Vx

and g(x) = x ' / 3 = \Yx are the square root and cube The functions f(x) = xl/2 = root functions, respectively. The domain of the square root function is [0, 00), but the cube root function is defined for all real x. Their graphs are displayed in Figure 1.17 along with the graphs of y = x 3/2 and y = x 2/3. (Recall that x 3/2 = (x ' /2)3 and

x 2/3 = (xl/3)2.)

Polynomials A function p is a polynomial if p(x) = anxn + an_IX n- 1 +

... + a,x + ao

where n is a nonnegative integer and the numbers 00, al, a2,"" an are real constants (called the coefJicients of the polynomial). All polynornials hsve domain ( - 00, 00). If the

1.1

9

Functions and Their Graphs

y y

y

y

1 ------~~~----~x

--~O~--I~------~x

Domain: O:s x < 0 :S Y <

Range:

------~~----~x

o

00 00

Graphs of the power functions J(x) = x'fora =

FIGURE 1.17

1 Domain: 0 ~ x < Range: 0 :S Y <

------~~----~x oc

Domain:

00

Range:

-00 < x < 00 0 :S Y < 00

t,t,~,andt.

leading coefficient a. '" 0 and n > 0, then n is called the degree of the polynomial. Linear functions with m '" 0 are polynomials of degree I. Polynomials of degree 2, usually written as p(x) = ax 2 + bx + c, are called quadratic functions. Likewise, cubic functions are polynomials p(x) = ax' + bx 2 + ex + d of degree 3. Figure 1.18 sbows the graphs of three polynomials. Techniques to graph polynomials are studied in Olapter 4. x2

x3

Y=3-Z-

2x +31

Y 4

Y Y

2

2

-4

-10 -12

=

8x 4

-

14x 3 - 9x 2 + llx - 1

Y = (x - 2)4(x

+ 1)3(x -

I)

16

(c)

(b)

(a)

FIGURE 1.18

Y

Graphs of three polynomial functions.

Rational Functions

A rational function is a quotient or ratio J(x) = p(x)jq(x), where p

and q are polynomials. The domain of a rational function is the set of all real x for which q(x) '" O. The graphs of several rational functions are sbown in Figure 1.19. Y Y

8 Y

6

2

Y=

2x2 - 3 7x + 4

Liney=~

llx + 2 Y=2x 3 -1

4

3 -4 -2 0 -2

2

4

6

x

-4

-2

NOTTOSCAlB

-6 -8 (a)

FIGURE 1.19 of the graph.

(b)

(c)

Graphs of three rational functions. The straight red lines are called asymptotes and are not part

10

Chapter 1: Functions

Algebraic Functions Any function constructed from polynomials using algebraic operations (addition, subtraction, multiplication, division, and taking roots) lies witbin the class of algebraic functions. All rational functions are algebraic, but also included are more complicated functions (sucb as those satisfYing an equation like y3 - 9xy + x 3 = 0, studied in Section 3.7). Figure 1.20 displays the graphs of three algebraic functions. y

y ~ x(l - x)'"

y

----+..---;+----->x

-----+1---~f---~x

(a)

(c)

(b)

FIGURE 1.20 Graphs of three algebraic functions.

Trigonometric Functions The six basic trigonometric functions are reviewed in Section 1.3. The graphs of the sine and cosine functions are shown in Figure 1.21. y

w~· _

w

0

w

~

-1

(a) f(x)

FIGURE 1.21

~

sin x

(b) f(x)

~

cosx

Graphs of the sine and cosine functions.

Exponential Functions Functions of the fonn f(x) = aX, where the base a > 0 is a positive constant and a oF 1, are called exponential fuoctions. All exponential functions have domain ( - 00, 00) and range (0, 00), so an exponential function never assumes the value O. We study exponential functions in Section 7.3. The graphs of some exponential functions are shown in Figure 1.22. y

-1

-0.5

y

12

12

10

10

8

8

6

6

4

4

2

2

0

0.5

1

x

(a)

FIGURE 1.22

Graphs of expooential functinos.

-1

-0.5

0 (b)

0.5

1.1

Functions and Their Graphs

11

Logarithmic Functions These are the functions t(x) = log" x, where the base a '" 1 is a positive constant. They are the inverse jUnctions of the exponential functions, and the calculus of these functions is studied in Chapter 7. Figure 1.23 shows the graphs of four logarithmic functions with various bases. In each case the domain is (0, 00) and the range is (-00, 00). Y

y

--~O~~ ~~~~=Y=~~IO~~XX -1 ,

FIGURE 1.23 functions.

Y = lOglOx

Graphs offour logaritlunic

FIGURE 1.24 Graph ofa cateoary or hanging cable. (The Latin word calena means "chain.")

Transcendental Functions These are functions that are not algebraic. They include the trigonometric, inverse trigonometric, exponential, and logarithmic functions, and many other functions as well. A particular example of a transcendental function is a catenary. Its graph has the shape of a cable, like a telephone line or electric cable, strung from one support to another and hanging freely under its own weight (Figure 1.24). The function defming the graph is discussed in Section 7.7.

Exercises 1.1 Functions

8...

Y

b.

y

In Exercises 1--{j, find 1he domain and range of each function.

=

+ x' v'Sx +

=

4 3_ I

1. f(x) = I 3. F(x) S. f(t)

2. f(x) = 1 10

4. g(x)

=

6. G(t)

=

;Ix

v'x' - 3x 2 t' _ 16 ~------------~X

In Exercises 7 and 8, which of1he graphs are graphs of functions oh, and which are not? Give reasons for your answers.

7.a.y

b.y

o

~-------------+x

o

Finding Formulas for Functions 9. Express 1he area and perimeter of an equilateral triangle as a function of1he triangle's side leng1hx. 10. Express 1he side 1eng1h ofa square as a function of1he 1eng1h d of 1he square's diagonal. Then express 1he area as a function of 1he diagonalleng1h.

~------------~X

o

~-------------+x

o

11. Express 1he edge leng1h ofa cube as a function of1he cube's diagonal 1eng1h d. Then express 1he surface area and volume of 1he cube as a function of1he diagonalleng1h.

12

Chapter 1: Functions

12. A point P in the first quadrant lies 00 the graph of the functioo I(x) ~ Express the coordinates of P as functions of the slope of the line joining P to the origin.

Vx.

31. a.

13. Consider the point (x, y) lying 00 the graph of the line a + 4y ~ 5. Let L he the distsnce from the point (x, y) to the origin (0, 0). WriteL as a functioo ofx. 14. Consider the point (x, y) lying on the graph of y = ~. Let L he the distsnce hetweeo the points (x,y) and (4, 0). WriteL as a function ofy.

Y

~x

17. g(x) =

a

16. I(x) = I -

'\J'Ix1

19. F(t) = t/ltl

2

20. G(t) =

!fit I

x' -

4 ~

h

(-2,-1)

Y

32. a.

b. Ix

-A

The Greatest and Least Integer Functions 33. For what values oh is

9

x' + -,--. x +4

+ Iyl = I

A

1:

b. rxl=O?

a.lxJ=O?

Iyl=x

Ixl

Y

1

n+-~!'--~x OTT

b. y'=x' 24. Graph the following equations and explain why they are not graphs of functioos of x. L

b.

o

23. Graph the following equations and explain why they are not graphs of functioos of x. L

- --=---t-i-- - x

x'

x~.

21. Find the domain ofy = 22. Find the range of y

a -

18. g(x) =

Y

2

3

I

Functions and Graphs Find the domain and graph the functions in Exercises 15-20.

15. I(x) = 5 -

b.

34. What real nwnhers x satisfY the equation l x J = r xl? 35. Does r -xl = -lxJ for all real x? Give reasons foryour answer. 36. Graph the function

I(x) =

{f:~:

+ yl = I

x~o

x

< O.

Why is I(x) called the integer part ofx?

Piece>rise-Defined Functions Graph the functions in Exercises 25-28.

25. I(x) = {x, 2-x, I - x 26. g(x) = { 2 _ x:

4 27. F(x) ~ { , x

28. G(x)

x'a'

+

~ {I/X, x,

Increasing and Decreasing Functions Graph the functioos in Exercises 37-46. What synnnetries, if any, do the graphs have? SpecifY the intervals over which the function is increasing and the intervals where it is decreasing.

0 :5 x :5 I 1 <x::S;2 Osxsl

1<x:52 x:S 1

,

37. y

= -x'

38.y=_I,

x> I x
39. y

=-~

40.y=~

OSX'

43. y

= x'/8

45. y

=

41. y =

Find a formula for each function graphed in Exercises 2c}"32.

29.

L

b.

Y (1, 1)

30.

r-1

0

b.

I I I

I I I

2

3

4

Y

Give reasons for your answer. 47. I(x) = 3 49. I(x) = x' + I 51. g(x) = x' + x

44. y 46. y

= -4Vx =

(-x)'/3

x

-3

SO. I(x) = x' + x 52. g(x) = x' + 3:x;' x 54. g(x) ~ - , - x - I

I t=1

56. h(t) = It'l

57. h(t) = 2t + I

58. h(t) = 21tl

=

55. h(t) = x

48. I(x) = x- s

I -,-x - I

53. g(x)

3

5

42.y=h

Even and Odd Functions In Exercises 47-58, say whether the function is even, odd, or neither.

X

L

-x'/2

x

Y

2

2

'\J'Ix1

I

-

I

+I

Theory and Examples 59. The variable s is proportiooa1 to t, and s = 25 when t = 75. Determine t when s = 60.

1.1 60. Kinetic energy The kinetic energy K of a mass is proportional to the square of its velocity v. If K = 12,960 joules when v = 18 m/sec, what is Kwhen v = 10m/sec? 61. The variables r and s are inversely proportional, and r s = 4. Determine s when r = 10.

=

66. a. y = 5x

13

Functions and Their Graphs

b.y=5X y

6 when

62. Boyle's Law Boyle's Law says that the volume Vofa gas at constant temperature increases whenever the pressure P decreases, so that V and P are inversely proportional. If P = 14.7 Ibs/in2 when V = 1000 in\ then what is V when P = 23.41bs/in2? 63. A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 14 in. by 22 in. by cutting out equal squares of side x at each corner and then folding up the sides as in the figure. Express the volume V of the box as a function ofx. I,

)1

22 IX

Tf-x'

~

x I_

x

14

1

x -I x

.3_ xl

64. The accompanying figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. a. Express the y-coordinate of P in terms of x. (You might start by writing an equation for the line AB.) b. Express the area of the rectangle in terms ofx. y

D 67.

a. Graph the functions j(x) = x/2 and g(x) = 1 gether to identify the values of x for which x 4 2" > 1 + x·

+ (4/x)

to-

b. Confirm your findings in part (a) algebraically.

D 68.

a. Graph the functions j(x) = 3/(x - 1) and g(x) = 2/(x together to identify the values of x for which

+ 1)

_3_<_2_ x-I x+l·

b. Confirm your findings in part (a) algebraically. 69. For a curve to be symmetric about the x-axis, the point (x, y) must lie on the curve if and only if the point (x, -y) lies on the curve. Explain why a curve that is symmetric about the x-axis is not the graph of a function, unless the function is y = O. 70. Three hundred books sell for $40 each, resulting in a revenue of (300)($40) = $12,000. For each $5 increase in the price, 25 fewer books are sold. Write the revenue R as a function of the number x of $5 increases.

71. A pen in the shape of an isosceles right triangle with legs oflength x ft and hypotenuse of length h ft is to be built If fencing costs $5/ft for the legs and $1 O/ft for the hypotenuse, write the total cost C of construction as a function of h.

-1

In Exercises 65 and 66, match each equation with its graph. Do not use a graphing device, and give reasons for your answer.

65. a. y = x4

b. y = x 7

c. Y = x lO

72. Industrial costs A power plant sits next to a river where the river is 800 ft wide. To lay a new cable from the plant to a location in the city 2 mi downstream on the opposite side costs $180 per foot across the river and $100 per foot along the land. I~(- - - - 2

P

mi- - - - - + 1 .1

Q

City

y

Power plant NOT TO SCALE

------~~--7 0 r---~~----~X

a. Suppose that the cable goes from the plant to a point Q on the opposite side that is x ft from the point P directly opposite the plant Write a function C(x) that gives the cost oflaying the cable in terms of the distance x. b. Generate a table of values to determine if the least expensive location for point Q is less than 2000 ft or greater than 2000 ft from point P.

14

Chapter 1: Functions

1.2

Combining Functions: Shifting and Scaling Graphs In this section we look at the main ways functions are combined or transformed to form new functions.

Sums, Differences, Products, and Quotients Like numbers, functions can be added, subtracted, multiplied, and divided (except where the denominator is zero) to produce new functions. If f and g are functions, then for every x that belongs to the domains of both f and g (that is, for x E D(f) n D(g», we define functions f + g, f - g, and fg by the formulas

(f + g)(x) (f - g)(x) (fg)(x)

f(x) + g(x). f(x) - g(x). f(x)g(x).

= = =

Notice that the + sign on the left-hand side of the first equation represents the operation of addition offonctions, whereas the + on the right-hand side of the equation means addition of the real numbers f(x) andg(x). At any point of D(f) n D(g) at which g(x) # 0, we can also define the function fig by the formula

L)(x) (g

=

f(x) g(x)

(where g(x) # 0).

Functions can also be multiplied by constants: If c is a real number, then the function cf is dermed for all x in the domain of f by

(cf)(x)

EXAMPLE 1

=

cf(x).

The functions defined by the formulas

f(x)

Vx

=

g(x)

and

=

~

have domains D{f) = [0, 00) and D(g) = (- 00, I]. The points common to these domains are the points

[0, 00) n (-00, I] = [0, I]. The following table summarizes the formulas and domains for the various algebraic combinations of the two functions. We also write f' g for the product function fg.

Function

Formula

Domain

f+g f-g g-f fog

[0, [0, [0, [0,

fig

(f + g)(x) = Vx + ~ (f - g)(x) = Vx - ~ (g - f)(x) = ~ - Vx (f' g)(x) = f(x)g(x) = Vr-:x(cI _-----ox) f f(x) ~ g(x) = g(x) = "\j!=X

glf

~(

f x

) =

g(x) f(x)

=

lx

x

I] = D{f) n D(g) I] I] I]

[0, I) (x = I excluded) (0, I] (x =

°

excluded)

•

The graph of the function f + g is obtained from the graphs of f and g by adding the corresponding y-coordinates f(x) and g(x) at each point xED(f) nD(g), as in Figure 1.25. The graphs of f + g and f -g from Example 1 are shown in Figure 1.26.

15

1.2 Combining Functions; Shifting and Scaling Graphs

y

8

6 4

f(a)

+ g(a)

g(a)

2

f(a)

~--------~--~--~----+x

o

a

FIGURE 1.25 functions.

Graphical addition oftwo

FIGURE 1.26 The domain of the function f + g is the intersection of the domains of f and g, the interval [0, 1] on the x-axis where these domains overlap. This interval is also the domain of the function f . g (Example 1).

Composite Functions Composition is another method for combining functions.

DEFINmON If 1 and g are functions, the composite function log ("I composed with g') is defined by

(f

0

g)(x)

=

I(g(x)).

The domain of 1 0 g consists of the numbers x in the domain ofg forwhichg(x) lies in the domain of f.

The definition implies that log can be formed when the range of g lies in the domain of I. To find (f 0 g)(x),first findg(x) and second find I(g(x». Figure 1.27 pictures 1 0 g as a machine diagram and Figure 1.28 shows the composite as an arrow diagram. fog f(g(x»

g

x __

FIGURE 1.27

g

I- g(x) ~

f

f

_ _ f(g(x))

Two functions can be composed at g(x)

x whenever the value of one function at x lies in the domain of the other. The composite is denoted by FIGURE 1.28

fog.

Arrow diagrmn for fog.

To evaluate the composite function g 0 1 (when defined), we fmd I(x) first and then 0 1 is the set of numbers x in the domain of 1 such that I(x) lies in the domain of g. The functions log and g 0 1 are usually quite different.

g(f(x». The domain of g

16

Chapter 1: Functions

EXAMPLE 2 (a) (f

0

Ifj(x)

= Vxandg(x) =

(h) (g

g)(x)

0

j)(x)

x

+

(c) (f

I, find

j)(x)

0

(d) (g

0

Solution Composite

g) (x) .

Domain

(a) (f

0

g)(x)

=

j(g(x»

=

v'iW = Vx+1

(h) (g

0

j)(x)

=

g(f(x»

=

j(x)

(c) (f

0

j)(x)

=

j(f(x»

=

ViW = y:;.rx =

(d) (g

0

g)(x)

=

g(g(x»

=

g(x)

+

+

1 = Vx

1 = (x

+

+

[-1,00) 1

[0, 00)

x1f'

[0, 00)

I)

+

1= x

+2

(-00,00)

To see why the domain of jog is [-I, 00), notice that g(x) = x + 1 is dermed for all real x but belongs to the domain of j only if x + I ;;,: 0, that is to say, when x ;;,: - I . • Notice that if j(x) = x 2 and g(x) = Vx, then (f

0

g)(x)

x. However,

= (Vx)2 =

the domain of jog is [0, 00), not (-00,00), since Vxrequires x;;': O.

Shifting a Graph of a Function A common way to obtain a new function from an existing one is by adding a constant to each output of the existing function, or to its input variable. The graph of the new function is the graph of the original function shifted vertically or horizontally, as follows.

Shift Formulas Vertical Sbift. y

y = j(x)

+k

Shifts the graph of j up k uuits if It Shifts it down 1kl units if It

>0

<0

Horizontal Sbift.

y

=

j(x + h)

Shifts the graph of j left h uuits if h Shifts it right 1h 1uuits if h

>0

<0

EXAMPLE 3 (a) Adding 1 to the right-hand side of the formula y = x 2 to get Y = x 2 graph up I uuit (Figure 1.29).

+

1 shifts the

(h) Adding -2 to the right-hand side of the formula y = x 2 to get y = x 2

-

2 shifts the

graph down 2 uuits (Figure 1.29). FIGURE 1.29 To shift the graph of f(x) = x 2 up (or down), we add positive (or negative) constants to the formula for f (Examples 3a and b).

(c) Adding 3 tox in y = x 2 to get Y = (x 1.30).

+ 3)2 shifts the graph 3 uuits to the left (Figure

(d) Adding -2 to x in y = 1 xl, and then adding - 1 to the result, gives y = and shifts the graph 2 units to the right and 1 uuit down (Figure 1.31).

1x - 21 -

1 •

Scaling and Reflecting a Graph of a Function To scale the graph of a function y = j(x) is to stretch or compress it, vertically or horizontally. This is accomplished by multiplying the function j, or the independent variable x, by an appropriate constant c. Reflections across the coordinate axes are special cases wherec = -1.

1.2 Combining Functions; Shifting and Scaling Graphs Add a positive

-

-

Add a negative constant to x.

constant to x. y

17

y

FIGURE 1.30 To shift the graph of y = x 2 to the left, we add a positive constant to x (Example 3c). To a1tift the graph to the right, we add a negative

FIGURE 1.31 Shifting the graph of y = Ix 12 units to the right and I unit down (Example 3d).

constant to x. Vertical and Horizontal Scaling and Reflecting Formulas For c

>

1, the graph is scaled:

y

= c f(x)

Y

=

cI f(x)

Stretches the graph of f vertically by a factor of c. Compresses the graph of f vertically by a factor of c.

f(cx) Compresses the graph of f horizontally by a factor of c. y = f(x/c) Stretches the graph of f horizontally by a factor of c. For c = -1, the graph is reflected: y = - f(x) Reflects the graph of f across the x-axis. Reflects the graph of f across the y-axis. y = f( -x)

y

=

EXAMPLE 4

Here we scale and reflect the graph of y =

Yx.

Yx

Yx

(a) Vertical: Multiplying the right-hand side of y = by 3 to get y = 3 stretches the graph vertically by a factor of 3, whereas multiplying by 1/3 compresses the graph by a factor of3 (Figure 1.32).

(b) Horizontal: The graph of y = ~ is a horizontal compression of the graph of y = by a factor of 3, and y = v;j3 is a horizontal stretching by a factor of 3 (Figure 1.33). Note that Y = ~ = so a horizontal compression may correspond to a vertical stretching by a different scaling factor. Likewise, a horizontal stretching may correspond to a vertical compression by a different scaling factor.

Yx

v'3 Yx

Yx

is a reflection of y = (c) Reflection: The graph of y = y = is a reflection across the y-axis (Figure 1.34).

h

y

Yx across the x-axis, and •

y 4

stretch

t

~~~=-;=.x -1 0 1 2 3 4 FIGURE 1.32 Vertically s1retching and compressing the graph y = Vi hy a factor of3 (Example 4a).

3 2

y~

1 ~~:::::::::~::::::"-Y F

Yx

--~~--~-r~~~~--~x

-3

-2

-1

v;j3

FIGURE 1.33 Horizontally s1retching and compressing the graph y = Vi hy a factor of 3 (Example 4b).

FIGURE 1.34 Reflections of the graph y = Vi across the coordinate axes

(Example 4c).

18

Chapter 1: Functions

EXAMPLE 5

Given the functioof(x) =

X4 -

4x' + 10 (Figure 1.35a), fmdfannulas to

(a) compress the graph horizootally by a factor of 2 followed by a reflectioo across the y-axis (Figure 1.35b). (b) compress the graph vertically by a factor of2 followed by a reflection across the x-axis (Figure 1.35c). y

y

20

10 -1

x

0 -10

x

1

-1

+ 2x3 - 5

Y = _!x 4 2

x

0 -10

-20

-20 (a)

(b)

(c)

FIGURE 1.35 (a) The original graph off (b) The horizontal compression ofy = j(x) in part (a) by a factor of2, followed by a reflection across they-axis. (c) The vertical compression ofy = j(x) in part (a) by a factor of2, followed by a reflection across the x-axis (Example 5).

Solution (a) We multiply x by 2 to get the horizootal compressioo, and by -I to give reflectioo across the y-axis. The formula is obtained by substituting -2x for x in the right-hand side of the equatioo for f: y = f( -2x) = (-2x)4 - 4( -2x)' =

16>:4

+ 10

+ 32x' + 10.

(b) The fannula is

Y =

-!f(x)

=

-!X 4 + 2x' - 5.

•

Ellipses Although they are not the graphs of functions, circles can be stretched horizontally or vertically in the same way as the graphs of functions. The standard equation for a circle of radius r centered at the origin is x2 + y2 = r2. Substituting ex for x in the standard equation for a circle (Figure 1.36a) gives c2x2

y

+ y2

= r2.

(1)

y

y

r

--~----~-----4--~x -r r

-r

o

r c

-r

-r

(a) circle

FIGURE 1.36

--~--------------~---------------+--~x

(b) ellipse, 0 < c

<

1

Horizontal stretching or compression of a circle produces graphs of ellipses.

(c) ellipse,

c>

1

1.2 Combining Functions; Shifting and Scaling Graphs

If 0 < c < I, the graph of Equation (I) horizontally stretches the circle; if c > I the circle is compressed horizontally. In either case, the graph of Equation (I) is an ellipse (Figure 1.36). Notice in Figure 1.36 that they-intercepts of all three graphs are always-r and r. In Figure 1.36b, the line segment joining the points (±rlc, 0) is called the major axis of the ellipse; the minor axis is the line segment joining (0, ±r). The axes of the ellipse are reversed in Figure 1.36c: The major axis is the line segment joining the points (0, ±r), and the minor axis is the line segment joining the points (±rlc, 0). In both cases, the major axis is the longer line segment. Ifwe divide both sides of Equation (I) by r', we obtain

y

b

----+-----~~----~--~x

-a

19

a

-b (2)

where a = ric andb = r. Ifa > b, the major axis is horizontal; if a < b, the major axis is vertical. The center of the ellipse given by Equation (2) is the origin (Figure 1.37). Substituting x - h for x, and y - k for y, in Equation (2) results in

FIGURE 1.37 Graph of tire ellipse 2

y2

+2 =

l,a > b,wherethemajor a b axis is horizontal. X

2

(x - h)' (y - kf -'----a"'--'-- + b'

=

(3)

I.

Equation (3) is the standard equation of an ellipse with center at (h, k). The geometric def"mition and properties of ellipses are reviewed in Section 11.6.

Exercises 1.2 Algebraic Combinations In Exercises I and 2, fmd tire domains and ranges of f, g, f

f·g· 1. I(x) = x, 2. I(x) =

+ g, and

-v;=J Vx+J, g(x) = -v;=J g(x) =

4. f(x)

=

g(x) = x 2

I, g(x)

=

I

+ I

+

Vx

+ 5 andg(x)

10. f(x)

=

I

• ,---c-;

vx

= x' - 3, fmd the following.

a. l(g(O))

b. g(/(O))

c. I(g(x))

II. g(/(x))

•• 1(1(-5))

r.

x+2

3 -x' g(x)

go 1(I(x))

b. g(g(x))

+

x'

= x' +

I' h(x)

=

d.y=x-6

•• y=2~

r.

y=

W-=3

13. Copy and coroplere the following table.

I), fmd the following.

g(x)

I(x)

(/0 g)(x)

a. x - 7

Vx

?

3x

?

a. f(g(1/2)) c. f(g(x)) e. 1(1(2))

b. g(/(1/2))

II. g(/(x))

c.

~

~

r.

b. g(g(x))

x x - I

?

go N(x))

x d. - - I x-

••

?

1+1 x

x

f.

I x

?

x

g(g(2))

In Exercises 7-10, write a formula for fog

8. I(x)

= x + I, = 3x + 4,

= 3x, g(x) = 2x

g(x)

h(x)

v2-x

b. y = x'/2

c. y = x 9

b. x + 2

7. I(x)

.~

= x - 3, g(x) = Vx, h(x) = x', and j(x) = 2x. Express each of the functions in Exercises 11 and 12 as a composite involving one or more of j,g, h, andj. 11••• y = Vx - 3 b. y = 2Vx c. y = xl/4 d. y = 4x •• Y = y"(x------:c )'" r. y = (2x - 6)' 3

g(g(2))

6. If f(x) = x - I andg(x) = I/(x

I

+ I, g(x) = x + 4' h(x) = x

12••• y=2x-3

Composites of Functions S. If I(x) = x

=

Let f(x)

In Exercises 3 and 4, fmd the domains and ranges of I, g, I/g, and g/f.

3. I(x) = 2,

9. f(x)

=4

- I, h(x)

0

h.

- x

= x2

?

20

Chapter 1: Functions 22. The accompanying figure shows the graph of y = x 2 shifted to two new positions. Write equations for the new graphs.

14. Copy and complete the following table.

g(x)

(j

f(x)

I x - I

L--

0

g)(x)

?

Ixl

b.

?

x - I x

c.

?

Vx

d.

Vx

?

x x

+I Ixl Ixl

15. Evaluate each expression using the giveo table of values

-2

-I

0

I

2

f(x)

I

0

-2

I

2

g(x)

2

I

0

-I

0

x

23. Match the equations listed in parts (a)-(d) to the graphs in the accompanying figure .

.. j(g(-I))

g{j(O» •. g(j( -2))

b.

d. g(g{2»

.. y

c. j(j( -I)) f.

j(g(I))

-x g(x) = { x ~ I,

.. j(g(O» d. j(j(2»

g{j(3» •. g{j(O»

b.

=

Vx+l,

g(x)

18. f(x) = x 2 , g(x) = I -

=

f.

j(g(I/2)) 0

f and

i-

Vx

x-

0

d. Y = (x

c. g{g{-I»

19. Let f(x) = ~2' Find a function y = g(x) so that (j

+2 + 3)2 - 2

b. Y = (x - 2)2

-2sx
In Exercises 17 and 18, (a) write formulas for fog and g find the (b) domsin and (c) range of each.

17. f(x)

1)2 - 4

+ 2)2 + 2 y

16. Evaluate each expression using the functions

f(x) = 2 - x,

= (x -

c. Y = (x

gXx) = x.

20. Letf(x) = 2x' - 4. Find a functiony = g(x)sothat (j 0 gXx) = x + 2.

(1. -4)

24. The accompanying figure shows the graph of y = -x 2 shifted to

four new positions. Write an equation for each new graph. y (1,4)

Shifting Graphs 21. The accompanying figure shows the graph of y = -x 2 shitred to two new positions. Write equations for the new graphs.

1.2 Combining Functions; Shifting and Scaling Graphs Exercises 25-34 tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation. 25. x 2 + y2 = 49 Down 3,Ieft 2

26. x 2 + y2 27. Y 28. Y

= x' =

25

=

56. The accompanying figure shows the graph of a function g(l) with domain [-4,0] and range [-3, 0]. Find the domains and ranges of the following functions, aod sketch their graphs. y

Up 3,Ieft 4

Left I, down I

x 2/'

Right I, down I

30. y

= Yx Left 0.81 = -Yx Right3

31. y

=

2x - 7 Up 7

32. Y

=

I 2(x

33. Y

=

Ilx Up I, right I

34. Y

= IIx 2

29. Y

•. g( -I) c. g(l)

+

I)

+5

Down5,rightl

Ix - 21

38. y

=

39. Y

40. Y

=I

41.

= I + v;.-::-J Y = (x + 1)2/'

42. Y

=

43.y=l-x 2/'

44. Y

+ 4 =x 2/'

45.y=~x-I-1

46. y

= (x + 2)'/2 +

47. Y

=

=x

I - 2

d. I - g(l)

r.

g. g(I - I)

b. -g(1 - 4)

g(1 - 2)

-

Yx

(x - 8)2/3

59. Y

I = I + 2'

compressed vertically by a factor of2

60. y

I = I + 2'

stretched horizontally by a faetor 00

61. Y

=

62. Y

=

63. y

=

64. Y

=

I 52·y=2- 1 x

65. y

=I =I

54. y =(x + I )2 55. The accompanying figure shows the graph of a function f(x) with domain [0, 2] and range [0, I]. Find the domains and ranges of the following functions, and sketch their graphs.

66. Y

-

I,

stretched vertically by a factor of3

-

I,

compressedhorizontallybyafactorof2

x x

I

I 50,y=x+2

y

58. y

= x2 = x2

57. Y

II - xl - I

I 48.y=,,-2

I 51. y =(x - I )2

+3 + 2)

e. g( -I

Exercises 57--{j6 tell by what factor and direction the graphs of the given functions are to be stretched or compressed Give an equation for the stretched or compressed graph.

36.y=~

35.y=Vx+4

b. -g(l)

Verttcal and Horizontal Scaling

Left 2, down I

Graph the functions in Exercises 35-54.

37. Y

Vx+l, Vx+l,

compressed horizontally by a factor of 4 stretcbedvertica1lybyafactorof3

\1'4=7, \1'4=7, -

x', x',

stretched horizontally by a factor of 2 compressed vertically by a factor 00

compressed horizontally by a factor 00 stretchedhorizontallybyafactorof2

Graphing In Exercises 67-74, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.14-1.17 and applying an appropriate transformation.

67.y=-~ 69. Y

= (x

- I)'

68.

+2

73. Y

2

0

a. f(x) + 2

b. f(x) - I

c. 2f(x)

d. -f(x)

+ 2)

g. f( -x)

,

r.

f(x - I)

h. -f(x+ I)

+

I

+

2

2 72. y=2+ I

x 74. Y = (-2xj2/'

= -\Yx

75. Graph the function y = 76. Graph the function y

y=~I-I

70. y = (I - x)'

I 71,y=2x- 1

e. f(x

21

Ix2

-

= vfx1.

11.

Ellipses Exercises 77-82 give equations of ellipses. Put each equation in standard form aod sketch the ellipse. 77. 9x 2 + 25y2 = 225 78. 16x2 + 7y2 = 112 80. (x + 1)2 + 2y2 = 4 79. 3x 2 + (y - 2j2 = 3

22

Chapter 1: Functions

+ +-t)'

81. 3(x - I)' 82.

+ 2(y + 2)'

+ ~)' +

~ 6

~ 54

83. Write an equation for the ellipse (x'/16) + (y'/9) ~ I shifted 4 units to the left and 3 units up. S1retch the ellipse and ideotifY its

center and major axis.

84. Write an equation for the ellipse (x'/4) + (y'/25) ~ I shifted 3 units to the right and 2 units down. S1retch the ellipse and ideotifY its center and major axis.

D

b. f/g

e. g/f

•. g'~gg

r.

h. fof

i. gog

fog

86. Can a function be both flNeo aod odd? Give reasons for your answer. 87. (Continuation of Example 1.) Graph the functions f(x) ~ Vx and g(x) ~ ~ together with their (a) sum, (b) product, (c) two differences, (d) two quotients.

D 88.

Comlrining Functions 85. Assume that f is an even function, g is an odd function, and both f andg are dermed on the entire real line R. Which of the follow-

a. fg d. f' ~ If g. gof

Let f(x) ~ x - 7 aod g(x) ~ x'. Graph f and g together with 0 gandg 0 f.

f

ing (where dermed) are flNeo? odd?

Trigonometric Functions

1.3

TIris section reviews radian measure and the basic trigonometric functions.

Angles Angies are measured in degrees or radians. The number of radians in the central angle

A' CB' within a circie of radius r is defined as the number of ''radius units" contained in the arc s subtended by that central angie. If we denote this central angie by () when measured in radians, this means that fJ = sir (Figure 1.38), or

C""10 of ,ifo""

<

s = r£J

FIGURE 1.38 TIre radian measure of the central aogleA'CB' is the number 9 ~ s/r. For a unit circle of radius r = 1,8 is the length of arc AB that central angle ACB cuts from the unit circle.

(fJ in radians).

(1)

If the circie is a unit circle having radius r = i, then from Figure 1.38 and Equation (i), we see that the central angle fJ measured in radians is just the iength of the arc that the angle cuts from the unit circie. Since one compiete revoiution of the unit circie is 3600 or 21T radians, we have 1T

radians = i80°

(2)

and 1 radian = i!O ('" 57.3) degrees

1 degree = 1~0 ('" 0.017) radians.

or

Tabie 1.2 shows the equivalence between degree and radian measures for some basic angles.

TABLE

1.2 Angles measured in degrees and radians

Degrees

-180

fJ (radians)

-1T

-135

-31T 4

-90

-45

0

30

45

60

90

o

1T

1T

1T

1T

6

4

3

2

120

135

150

180

270

360

1.3 Trigonometric Functions

23

An angle in the xy-plane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive x-axis (Figure 1.39). Angles measured counterclockwise from the positive x-axis are assigned positive measures; angles measured clock-

wise are assigned negative measures. y

y

Initial ray ---------.~,---~--~x

FIGURE 1.39

Angles in staodard position in the xy-plaoe.

Angles describing counterclockwise rotations can go arbitrarily far beyond 2" radians or 360°. Similarly, angles describing clockwise rotations can have negative measures of all sizes (Figure lAO). y

y

y

y

3"

- -"-H --'....+----x

- - ---*--,----> X

----hh-+---~x

9"

4

hypotenuse

opposite

byp

Angle Convention: Use Radians From now on, in this book it is assumed that all angles are measured in radians unless degrees or some other unit is stated explicitly. When we talk about the angle ,,/3, we mean ,,/3 radians (which is 60°), not ,,/3 degrees. We use radians because it simplifies many of the operations in calculus, and some results we will obtain involving the trigonometric functions are not true when angles are measured in degrees.

adj

The Six Basic Trigonometric Functions

adjacent sin {1 = opp

byp

csc {1

=

hyp opp

d·

COS{1=:~

sec {1

=

opp

cot {1

=

tan 9 ~ adj

FIGURE 1.40 Nonzero radiao measures cao be positive or negative and cao go beyond 2"..

adj

opp

Trigonometric ratios of an acute angle. FIGURE 1.41

You are probably familiar with derming the trigonometric functions of an acute angle in terms of the sides of a right triangle (Figure 1041). We extend this dermition to obtuse and negative angles by first placing the angle in standard position in a circle of radius r. We then derme the trigonometric functions in terms of the coordinates of the point P(x, y) where the angle's terminal ray intersects the circle (Figure 1.42).

y

sine:

sin l/ = )I r

cosecant:

r cscl/=y

rx

secant:

sec l/ = l' x

tan 0 =)1 x

cotangent:

cotO ="" y

cosine: cosO = tangent:

These extended dermitions agree with the right-triangle definitions when the angle is acute. Notice also that whenever the quotients are defined, The trigonometric functions of. general angle 6 are defmed in terms ofx,y, and r. FIGURE 1.42

lanO = sinO cosO

I cotO = lanO

=~ cos v

I cseO = sinO

secO

24

Chapter 1: Functions As you can see, tan IJ and sec IJ are not defined if x = cos IJ = O. This means they are not defined iflJ is ±1T/2, ±31T/2, ... . Similarly, cot 0 and csc 0 are not defined for values of 0 forwhichy = O,namelyO = 0, ±1T, ±21T, .... The exact values of these trigonometric retios for some angles can be read from the triangles in Figure 1.43. For instance,

~1

zj 2

'"4

'"2

-

-

1

. 1T

. 1T

sm-=-

sm 6

4v'Z

1

1T cos-= -1

FIGURE 1.43 Radian angles and side leng1hs of two common 1riangles.

tan

1

smT=2

V3

1T

4v'Z

cos 6 = 2

1T

tan 1T = _1_

4

=1

V3

. 1T

=2

1T

cosT =

1

2

6V3

The CAST rule (Figure 1.44) is useful for remembering when the basic trigonometric functions are positive or negative. For instance, from the triangle in Figure 1.45, we see that 21T

cosT

=

1

tan

-2'

2;

=

-V3.

(cos 2;, sin 2;) = t-~, ~)

y

y S sinpos

A all pos x

T tanpos

C cospos

FIGURE 1.44 The CAST rule, remembered by the statement "Calculus Activates Student Thinking," tells which

FIGURE 1.45 The 1riangle for calculating the sine and cosine of21f/3 mdians. The side leog1hs come from the geometty of right triangles.

trigonometric functions are positive in each quadrant.

Using a similar method we determined the values of sin 0, cos 0, and tan 0 shown in Table 1.3. TABLE 1.3

Degrees

o(radians)

Values of sin 0, cos 0, and tan 0 for selected values of 0

-180 -1T

sinO

0

cosO

-I

tan 0

0

-135

-90

-45

-31T

-1T

-1T

4

2

4

-v'Z

-I

-v'Z

2

-v'Z 2 1

0

2

v'Z 2 -I

0

30

45

60

90

120

135

150

1T

1T

1T

1T

21T

31T

51T

6

4

3

2

3

4

6

1 2

v'Z

V3

v'Z

2

1

V3

2

2

2

1 2

1

V3

v'Z

-V3

2

1 2

-v'Z

2

2

2

0

V3

1

V3

-I

-V3

0

0

3

0

1 2

-V3

3

180 1T

270 360 31T

2

21T

0

-I

0

-I

0

1

0

0

1.3

25

Trigonometric Functions

Periodicity and Graphs of the Trigonometric Functions

I

Period. of Trigonometric Functions tan (x + ".) ~ tan x Period 'IT : cot(x + ".) ~ cotx Period 2'IT: sin(x + 2".) ~ sinx cos(x + 2".) ~ cosx sec(x + 2".) ~ ..cx csc(x + 2".) ~ cscx

When an angle of measure 8 and an angle of measure 8 + 21T are in standard position, their terminal rays coincide. The two angles therefore have the same trigonometric fimction values: sin(8 + 21T) = sin 8, tan(8 + 21T) = tan 8, and so on. Similarly, cos(8 - 21T) ~ cos 8, sin(8 - 21T) = sin 8, and so on. We describe this repeating behavior by saying that the six basic trigonometric functions are periodic.

DEFINmON A function f(x) is periodic ifthere is a positive number p such that f(x + p) ~ f(x) for every value ofx. The smallest such value ofp is the period of f.

When we graph trigonometric functions in the coordinate plane, we usually denote the independent variable by x instead of 8. Figure 1.46 shows that the tangent and cotangent functions have period p ~ 1T, and the other four functions have period 21T. Also, the symmetries in these graphs reveal that the cosine and secant functions are even and the other four functions are odd (although this does not prove those results). y

y - tunx

Domajn: _00 < x < co

Even

Domain: -co < x < 00 Range: -1:s y:s 1 Period: 2'1J" (b)

Range: -1:s y :s 1 Period: 2'1T

cos ( -x) sec ( -x)

~ ~

cosx secx

(a) y

_3(\ 0

.X

3'17" -2'- 2.···

*+'1r +

Range: -co Period: '1T

y

y = socx

\v

Domajn'

< y < 00 (c)

y

y - cotx

n

-,t-=-tnt-t-~~~,

Odd sin( -x) tan(-x) esc ( -x) cot ( -x)

~ ~ ~

-sinx -tanx -cscx

~ -COlx

Domain'• X "" -2'-2 + 1T + 31T •...

Domain: x "'" 0, ±'1f, ±2'1f, ••• Range: y::5 -lory 2:: 1 Period: 271'

Range: y:s -1 ory ~ 1 Period: 2'17" (d)

Domain: x "'" 0, ±'1f, ±2'1f, ... Range: _00 < y < 00 Period: '1T (I)

(0)

FIGURE 1.46 Graphs of the six hasic trigonometric functions using radian measure. The shading for each trigonometric function indicates its periodicity. y

Trigonometric Identities The coordinates of any point P(x, y) in the plane can be expressed in terms of the point's distance r from the origin and the angle 8 that ray OP makes with the positive x-axis (Figure 1.42). Since xlr ~ cos 8 andylr = sin 8, we have x

=

rcos6,

y

~

rsin8.

When r = 1 we can apply the Pythagorean theorem to the reference right triangle in Figure 1.47 and obtain the equation

FIGURE 1.47 The reforenoe triangle for a general angle 8.

(3)

26

Chapter 1: Functions

TIris equation, true for all values of 6, is the most frequently used identity in trigonometry. Dividing this identity in turn by cos2 6 and sin2 6 gives

1 I

+ tan2 6 + cot' 6

= sec2 6 = csc2 6

The following formulas hold for all angles A and B (Exercise 58).

Additioo Formulas cos(A sin(A

+ B) + B)

= cosA cosB - sinA sinB = sinA cosB

+ cosA sinB

(4)

There are similar formulas for cos (A - B) and sin (A - B) (Exercises 35 and 36). All the trigonometric identities needed in this book derive from Equations (3) and (4). For example, substituting 6 for both A and B in the addition formulas gives

Double-Angle Formulas cos 26 = cos2 6 - sin2 6 sin 26 = 2 sin 6 cos 6

(5)

Additional formulas come from combining the equations cos2 6

+

sin2 6 = I,

cos2 6 - sin2 6 = cos 26.

We add the two equations to get 2 cos2 6 = I + cos 26 and subtract the second from the first to get 2 sin2 6 = I - cos 26. TIris results in the following identities, which are useful in integral calculus.

Half-Angle Formulas cos2 6 = 1

+

cos 26 2

(6)

sin2 6 = I - cos 26 2

(7)

The Law of Cosines If a, b, and c are sides of a triangle ABC and if 6 is the angle opposite c, then

(8)

TIris equation is called the law of cosioes.

1.3

27

We can see why the law holds if we introduce coordinate axes with the origin at C and the positive x-axis along one side of the triangle, as in Figure 1,48. The coordinates of A are (b, 0); the coordinates of B are (a cos II, a sin II). The square of the distance between A and B is therefore

y B(a cos 8, a sin 8)

c 2 = (acosll- b)2 = a 2(cos211

+

b

A(b,O)

= a2

FIGURE 1.48 The square of the distance betweeo A and B gives the law of cosines.

+

(a sin 11)2

sin211)

+ b2 -

Zabcosll

I

------~~--~~~~~x

C

Trigonometric Functions

+ b2

-

Zabcosll.

The law of cosines generalizes the Pythagorean theorem. If II = andc 2 = a 2 + b 2 •

7T/Z,

then cos II = 0

Transfonnations of Trigonometric Graphs The rules for shifting, stretching, compressing, and reflecting the graph of a function summarized in the following diagram apply to the trigonometric functions we have discussed in this section. Vertical stretch or compression; ~ ~ reflection about x-axis ifnegative

/

y = af(b(x

+ c)) + d

~

Horizontal stretch or compression; /

reflection about y-axis if negative

Vertica1 shift

Horizontal shift

The transformation rules applied to the sine function give the general sine function or sinusoid formula

f(x) = A Sin(Z; (x -

C») + D,

where IA I is the amplitude, IB I is the period, C is the horizontal shift, and D is the vertical shift. A graphical interpretation of the various terms is revealing and given below. y

y = A sin

(2;<x - C») + D

o Two Spedallnequalities For any angle II measured in radians,

-1111 :5 sinll:5 1111

and

-1111 :5 I - cos II :5 1111.

28

Chapter 1: Functions

To establish these inequalities, we picture fJ as a nonzero angle in standard position (Figure 1.49). The circle in the figure is a unit circle, sollil equals the length of the circular arc AP. The length of line segmentAP is therefore less than IfJl. Triangle APQ is a right triangle with sides of length

y p

AQ = I - cosfJ.

QP = IsinfJl,

From the Pythagorean theorem and the fact that AP cos 6

2

sin fJ

1-cosO

+ (I -

<

IfJ I, we get

cos fJ)2 = (AP)2 :5

6'.

(9)

The terms on the left-band side of Equation (9) are both positive, so each is smaller than their sum and hence is less than or equal to 6':

(I - cos 11)2

and

FIGURE 1.49 From the geometry of this figure, drawn for 8 > 0, we get the inequality sin2 8 + (I - cos 8)2 :5 0'.

:5

6'.

By taking square roots, this is equivalent to saying that and

11- cosfJl

and

-161

so

-161

:5 sinfJ:5

161

:5

I - cosll :5

161.

-"./6

"./4

These inequalities will be useful in the next chapter.

Exercises 1.3 Radians and Degrees 1. On a circle of radius 10m, how long is an arc that subtends a central angle of (8) 4'1f/5 radiaos? (b) 110'? 2. A central aogle in a circle of radius 8 is subtended by ao arc of length 101r. Find the aogle's radian and degree measures. 3. You waot to ma1re ao 80' aog1e by marking ao arc on the perimeter of. 12-in.-diarneter disk and drawing lines from the ends of the arc to the disk's ceoter. To the nearest tenth of an incb, how long should the arc be?

4. If you roll a I-m-diaroeter wheel forward 30 em over level grouod, through what aogle will the wheel turn? Aoswer in radiaos (to the nearest tenth) aod degrees (to the nearest degree). Evaluating Trigonometric Functions S. Copy aod complete the following table of function values. If the function is undefmed at a given angle, enter "UND." Do not use a calculator or tables.

fJ

-3"./2

-"./3

sinO

cos 8 tan 8 cot 8 sec 0 esc 8 In Exercises 7-12, one of sin x, cos x, and tan x is given. Find the other two if x lies in the specified interval.

. 7• smx

= 5'

3

9. cosx

= 3'

[f, 'If] XE[-f,o]

XE

I

8. tanx = 2, XE 5

10. cosx = -1 3'

12. sinx = (}

-".

-2"./3

o

"./2

S1r/6

-t,

[o,f] XE

XE

[f,'lT]

['If, 3;]

3"./4

sinO

cosO tan 8 cot 8 sec 0 esc 8 6. Copy and complete the following table of function values. If the function is undefmed at a given angle, enter "UND." Do not use a calculator or tables.

Graphing Trigonometric Functions Graph the functions in Exercises 13-22. What is the period of each function? 13. sin 2x

14. sin (x/2)

15. cos 71'X

16. cosT

'IfX

'If

. X 17• - 8m T

19. cos(X -

18. - cos 2'1TX

f)

29

1.3 Trigonometric Functions

21. Sin(X -

-;f) + I

22.

COS(X + 2:) -

Solving Trigonometric Equations For Exercises 51-54, solve for the angle 0, where 0 ,;; 0 ,;; 2'IT.

2

Graph the functioos in Exercises 23-26 in the Is-plane (t·axis horizoo· tal, s-axis vertical). What is the period of each function? What sym· metries do the graphs have?

23. s

= cot2t

24. s

25. s

= sec(~t)

26. s

= -tao 'ITt

t

52. sin' 0 = cos' 0

53. sin20 - cosO = 0

54. cos20

+ cosO

=

0

Theory and Examples 55. The tangent snm formula The staodard formula for the taogoot of the smn of two aogles is

= csc(f)

D 27.

D

51. sin' 0 =

a. Graph y = cos x and y = sec x together for - 3'IT/2 ,;; x ,;; 3'IT/2. Cornmeot on the behavior of sec x in relatioo to the signs and values of cos x. b. Grapby = sin x andy = c8Cxtogetherfor-w ~ x ~ 2'17". Comment on the behavior of esc x in relation to the signs and values of sin x. 28. Graph y = taox aod y = cotx together for -7 ,;; x ,;; 7. Comment on the behavior of cot x in relation to the signs and values of taox. 29. Graph y = sinx aod y = l sin x J together. What are the domain and raoge of l sin x J? 30. Graph y = sinx aod y = rsin x 1together. What are the domain

tao(A

+ B)

taoA+taoB

=

I - taoA taoB'

Derive the formula.

56. (ContinUIJtion ofExercise 55.) Derive a formula for tao (A - B). 57. Apply the law of cosines to the triaogle in the accoropaoying figure to derive the formula for cos (A - B). y

1

r

and raoge of sin x 1?

Using the Addition Formulas Use the addition formulas to derive the ideotities in Exercises 31-36.

I-) 33. Sin(X + f) 31. cos(x -

35. cos(A - B)

= sinx

cosx

=

32. cos 34.

Sin(X - -I-) =

-cosx

= cosA cosB + sinA sinB (Exercise 57 provides a =

sinA cosB - cosA sinB

38. What happeos if you take B = 2'IT in the addition formulas? Do the results agree with something you already know? In Exercises 39-42, express the giveo quaotity in terms of sin

cosx.

+ x)

41. Sin(3; -

40. sin(2'IT - x)

x)

42. cose;

+

x)

. ('IT 'IT) . 7'IT 43. Evaluatesffi12assm 4+3. 44. Evaluate cos

58. a. Apply the formula for cos(A - B) to the ideotity sinO cos(I -

37. What happeos if you take B = A in the trigoooroetric ideotity cos(A - B) = cosA cosB + sinA sinB? Does the result agree with something you already know?

39. cos('IT

o

(x + 1-) = -sinx

different derivation.)

36. sin(A - B)

----4---~~~~-l----+_--~x

x aod

0)

to obtain the addition formula for sin(A

b. Derive the formula for cos (A + B) by substituting - B for B in the formula for cos (A - B) from Exercise 35. 59. A triaogle has sides a the leogtil of side c.

= 2 aod b = 3 aod angle C = 60°. Find

60. A triaogle has sides a

= 2 aod b = 3 aod aogle C = 40°. Find

the leogtil ofside c.

The law ofsines says that if a, b, aod c are the sides opposite the angles A, B, and C in a triangle, then

61. The law of sine.

sinA sinB sin C abe .

A

45. Evaluate cos

~.

+ B).

Use the accoropaoying figures aod the identity sin('IT - 0) sin 0 , if required, to derive the law.

If; as cos (-;f + 2:).

=

=

A

46. Evaluate sin ; ; .

Using the Double-Angle Formulas Find the functioo values in Exercises 47-50. 47. cos2 '!!

8

. 2

'IT

49. sm 12

5'IT 48• cos2 12 231T

SO. sin

8

B " - - -"' .-

-'-----' C

B £..----:.,----.J

62. A triaogle has sides a = 2 aod b = 3 and aogle C = 60° (as in Exercise 59). Find the sine of aogle B using the law of sines.

30

Chapter 1: Functions

63. A triangle has side c = 2 and angles A Find the length a of the side opposite A.

D 64.

= 1f/4 and B = 1f/3.

The appnDimation sin x '" x It is often useful to know that, whenx is measured in radians, sin x ~ x for n'lDDerically small values of x. In Sectioo 3.9, we will see why the approximation holds. The approximatioo error is less than 1 in 5000 if Ixl < 0.1. L With your grapher in radian mode, graph y = sinx and y = x together in a viewiog wiodow about the otigin. What do you see happening as x nears the origin? b. With your grapher in degree mode, graph y = sinx and y = x together about the origin again. How is the picture different froru the ooe obtained with radian mode?

General Sine Curves For f(x)

=

ASine: (x - C») + D,

identify A, B, C, and D for the sine functions in Exercises 65-
y=-%Sin(ft) +,\0

68. y

= ~sin 2L1ft, 2"

L

>0

COMPUTER EXPLORATIONS In Exercises 6'J-72, you will explore graphically the general sine

69. TheperiodB Set the constants A = 3, C = D = O. a. Plot f(x) for the values B = I, 3, 21f, 5" over the interval -41f :5 x :5 41f. Describe what happens to the graph of the general sine function as the period increases. b. What happens to the graph for negative values of B? Try it withB = -3 andB = -21f. 70. The horizontal shift C Set the constants A = 3, B = 6, D = O. a. Plot f(x) for the values C = 0, I, and 2 over the interval -41f :5 x :5 41f. Describe what happens to the graph of the general sine function as C increases through positive values. b. What happens to the graph for negative values of C? c. What smallest positive value should be assigned to C so the graph exhibits no horizontal shift? Confum your answer with a plot. 71. The vertical shift D Set the constants A = 3, B = 6, C = O. a. Plot f(x) for the values D = 0, I, and 3 over the interval -41f :5 x :5 41f. Describe what happens to the graph of the general sine function as D increases through positive values. b. What happens to the graph for negative values of D? 72. The amplitude A Set the constants B = 6, C = D = O. a. Describe what happens to the graph of the general sine function as A increases through positive values. Confirm your answer by plotting f(x) for the values A = I, 5, and 9. h. What happens to the graph for negative values of A?

functioo

f(x)

= A Sine: (x

-

C)) +

D

as you change the values of the constants A, B, C, andD. Use a CAS or coruputer grapher to perform the steps in the exercises.

1.4

Graphing with Calculators and Computers A graphing calculator or a computer with graphing software enables us to graph very complicated functions with high precision. Many of these functions could not otherwise be easily graphed. However, care must be taken when using such devices for graphing purposes, and in this section we address some of the issues involved. In Chapter 4 we will see how calculus helps us determine that we are accurately viewing all the important features of a function's graph.

Graphing Windows When using a graphing calculator or computer as a graphing tool, a portion of the graph is displayed in a rectangular display or viewing window. Often the default window gives an incomplete or misleading picture of the graph. We use the tenD square window when the units or scales on both axes are the same. Ibis tenD does not mean that the display window itself is square (usually it is rectangular), but ioslead it means that the x-unit is the same as the y-unit. When a graph is displayed io the default window, the x-unit may differ from the y-unit of scaling in order to fit the graph in the window. The viewing window is set by specifying an ioterval [a, b] for the x-values and an ioterval [c, d] for the y-values. The machine selects equally spaced x-values in [a, b] and then plots the points (x, f(x». A poiot is plotted if and

1.4 Graphing with Calculators and Computers

31

only ifx lies in the domain of the function andf(x) lies within the interval [c, d]. A short line segment is then drawn between each plotted point and its next neighboring point. We now give illustrative examples of some common problems that may occur with this procedure. Graph the function f(x) = x 3 play or viewing windows:

EXAMPLE 1

(8) [-10, 10] by [-10, 10]

7x'

-

+ 28 in each of the following dis-

(b) [-4,4] by [-50, 10]

(c) [-4, 10] by [-60,60]

Solution (8) We select a = -1O,b = lO,c = -1O,andd = 10 to specify the in!erval of x-values and the range of y-values for the window. The resulting graph is shown in Figure l.50a. It appears that the window is cutting off the bottom part of the graph and that the interval of x-values is too large. Let's try the next window.

10

-4 -10

~~f---+-~'<--

4

-4

10

-10

-50

(a)

(b)

10

(c)

FIGURE 1.50 The graph of f(x) = x 3 - 7x' + 28 in differeot viewing windows. Selecting a window that gives a clear picture of a graph is often a trial-and-error process (Example 1). (b) Now we see more features of the graph (Figure l.50b), but the top is missing and we need to view more to the right of x = 4 as well. The next window should help. (c) Figure 1.50c shows the graph in this new viewing window. Observe that we get a more complete pictore of the graph in this window, and it is a reasonable graph of a third-degree polynomial. •

EXAMPLE 2 When a graph is displayed, the x-unit may differ from the y-unit, as in the graphs shown in Figures l.50b and 1.50c. The result is distortion in the picture, which may be misleading. The display window can be made square by compressing or stretching the units on one axis to match the scale on the other, giving the true graph. Many systems have built-in functions to make the window "square." If yours does not, you will have to do some calculations and set the window size manually to get a square window, or bring to your viewing some foreknowledge of the true pictore. Figure I.5la shows the graphs of the perpendicular lines y = x andy = -x + 3v2, together with the semicircle y = x', in a nonsquare [-6,6] by [-6,8] display window. Notice the distortion. The lines do not appear to be perpendicular, and the semicircle appears to be elliptical in shape. Figure l.51 b shows the graphs of the same functions in a square window in which the x-units are scaled to be the same as the y-units. Notice that the [-6,6] by [-4,4] viewing window has the same x-axis in both Figures l.51a and I.5lb, but the scaling on the x-axis has been compressed in Figure l.51b to make the window square. Figure I.5lc gives an • enlarged view of Figure l.51b with a square [-3,3] by [0, 4] window.

Y9 -

If the denominator of a rational function is zero at some x-value within the viewing window, a calculator or graphing computer software may produce a steep near-vertical line segment from the top to the bottom of the window. Here is an example.

32

Chapter 1: Functions 8

4

-6

-4

(a)

(b)

FIGURE 1.51 Graphs of the perpendicu1ar lines y ~ x andy ~ -x + 3v2,andthesemicircley ~ ~ appear distorred (a) in a noosquare window, but clear (b) and (c) in square windows (Example 2).

EXAMPLE 3

Graph the function y = - 2 I .

-x

Solution Figure 1.52a shows the graph in the [-10, 10] by [-10, 10] default square window with our computer graphing software. Notice the near-vertical line segment at x ~ 2. It is not truly a part of the graph and x = 2 does not belong to the domain of the function. By trial and error we can eliminate the line by changing the viewing window to the smaller [-6,6] by[ -4,4] view, revealing a better graph (Figure 1.52b). • 10

4

1o

-10

) -6

-10

-4

(a)

r

6

(b)

FIGURE 1.52 Graphs of the functioo y ~ 2 ~ x. A vertical line may appear without a careful choice of the viewing window (Example 3). Sometimes the graph of a trigonometric function oscillates very rapidly. When a calculator or computer software plots the points of the graph and connects them, many of the maximwn and miuimum points are actually missed. The resulting graph is then very misleading.

EXAMPLE 4

Graph the functionj(x) = sin lOOx.

Solution Figure 1.53a shows the graph of j in the viewing window [-12, 12] by [- I, I]. We see that the graph looks very strange because the sine curve should oscillate periodically between -I and I. This behavior is not eshibited in Figure 1.53a. We might

-1 (b)

FIGURE 1.53 Graphs of the functiony ~ sin lOOx in three viewing windows. Because the period is 2,,/100 '" 0.063, the smaller window in (c) best displays the true aspects of1his rspidly oscillating function (Example 4).

1.4 Graphing with Calculators and Computers

33

experiment with a smaller viewing window, say [-6,6] by [-1,1], but the graph is not better (Figure 1.53b). The difficulty is that the period of the trigonometric function y = sin 100x is very small (2·,,/100 RJ 0.063). If we choose the much smaller viewing window [-0.1,0.1] by [-1,1] we get the graph shown in Figure 1.53c. This graph reveals _ the expected oscillations of a sine curve.

EXAMPLE 5

Graphthefunctiony = cosx

+

510Sin5Ox.

In the viewing window [-6,6] by [-I, I] the graph appears much like the cosine function with some small sharp wiggles on it (Figure 1.54a). We get a better look when we siguificantly reduce the window to [ -0.6, 0.6] by [0.8, 1.02], obtaining the graph in Figure 1.54b. We now see the small but rapid oscillations of the second term, 1/50 sin 50x, added to the comparatively larger values of the cosine curve. _

Solution

1.02

0.8 (b)

FIGURE 1.54 In (b) we see a close-up view of the function

y = cosx

+

second term,

io sin io sin

50x graphed in (a). The rerm cos x clearly dominates the

5Ox, which produces the rapid oscillations along the

cosine curve. Both views are needed for a clear idea of the graph (Example 5).

Obtaining a Complete Graph Some graphing devices will not display the portion of a graph for f(x) when x < O. Usually that happens because of the procedure the device is using to calculate the function values. Sometimes we can obtain the complete graph by defining the formula for the function in a different way.

EXAMPLE 6

Graph the function y = x V3 .

Solution

Some graphing devices display the graph shown in Figure 1.55a. When we compare it with the graph ofy = x l/3 = ~inFigure 1.17, we see that the left branch for 2

2

-2

-2

(a)

(b)

-3

FIGURE 1.55

The graph of y

=

(b) we graph the function f(x) =

Example 6.)

x 1/3 is missing the left branch in (a). In

1:1'lx ll/3, obtaining both branches. (See

34

Chapter 1: Functions

x < 0 is missing. The reason the graphs differ is that many calculators and computer software programs calculate x 1/' as e (I/')lnx. Since the logarithmic function is not defined for negative values of x, the computing device can produce only the right branch, where x > O. (Logarithmic and exponential functions are presented in Chapter 7.) To obtain the full picture showing both branches, we can graph the function

f(x)

=

1~1·lxll/'.

xl/'

This function equals except at x = 0 (where f is undefined, although 0 1/3 = 0). The graph of f is shown in Figure 1.55b. •

Exercises 1.4 Choosing a Viewing Window

o In Exercises 1-4, use a graphing calculator or coroputer to detennine which of the given viewing windows displays the most appropriate graph of the specified function. 1. f(xl = x' - 7x 2 + 6x L

[-I, I]by[-I, I]

b. [-2, 2] by [-5, 5]

c. [-10, 10] by [-10, 10] d. [-5,5] by [-25, 15] = x' - 4x 2 - 4x + 16

19. f(xl

[-I, I] by [-5,5] c. [-5,5]by[-1O,20] 3. f(xl = 5 + 12x - x' L [-I, I]by[-I, I] c. [-4,4] by [-20,20] 4. f(xl = '1'5 + 4x - x 2 L

[-2,2] by [-2,2]

c. [-3, 7] by [0, 10]

b. [-3, 3] by [-10, 10]

22. f(xl

= -2-

2 23. f(xl = 6x - 15x + 6 4x 2 - lOx

24. f(xl = - - 2

window for the given 3

6. f(xl =

+3 +2

Chapter

8. f(xl

2

x x T - "2 -

16. Y

II

3

25. y

= sin 250x

26. y

= 3 cos 60x

27. y

= cos(;O)

28. y

=

29. y

=x +

_ 2 I 30. y - x + 50 cos IOOx

/OSin3Ox

110 sin (;0)

34. Graph two periods of the function f(x l = 3 cot

15. y = Ix 2

-

-

x-

33. Graph four periods of the function f(xl = - tan 2x.

9. f(xl = x~ 11. y = 2x - 3x 2/' 13. Y = 5x 2/' - 2x x 17. y = x

x' - 5x'

x2

d. [-10, 10] by [-10, 10]

4x' - x' 10. f(xl = x 2(6 - x'l 12. y = x 1/'(x 2 - 8l 14. Y = x 2/'(5 - xl

=

8 x - 9

b. [-2,6]by[-1,4]

functioo and use it to display its graph.

7. f(xl

x +I

32. Graph the upper branch of the hyperbola y2 - 16x 2

o In Exercises 5-30, rmd an appropriate viewing + 15 + 10

1

31. Graph the lower half of the circle dermed by the equation x 2 +2x=4+4y- y 2.

d. [-4,5]by[-15,25]

Finding a Viewing Window

5. f(xl = x' - 4x'

-

21. f(x l = -x2o"_--x"_~6

d. [-20, 20] by [-100, 100] b. [-5, 5] by [-10, 10]

x2

= -2-

2. f(xl L

2 x +2 x2 + I x - I

20. f(xl

=

18. y

=

2x

+I

=

Ix 2

=I

-

- x

xl I

+3

35. Graph the function f(xl

=

sin 2x

36. Graph the function f(xl

=

sin' x.

= I.

I + I.

+ cos 3x.

Graphing in Dot Mode

o Another

way to avoid incorrect conooctions when using a graphing device is through the use of a "dot mode:' which plots only the points. !fyour graphing utility allows that mode, use it to plot the functions in Exercises 37-40. I 38. y = sin~ 37. y = x - 3

39. y

=

xlxJ

40. y

x3

-

X

-

1 I

= -2--

Questions to Guide Your Review

1. What is a function? What is its doroain? Its range? What is an arrow diagraro for a functioo? Give examples. 2. What is the graph of a real-valued functioo of a real varisble? What is the vertical line test?

3. What is a piecewi..,.dermed functioo? Give examples.

4. What are the importsnt types of functions frequently encountered in calculus? Give an example of each type.

Chapter 1 Practice Exercises

35

5. What is meant by an increasing function? A decreasing function? Give an example of each.

11. What is the stsndard equation of an ellipse with center (h, k)? What is its major axis? Its minor axis? Give examples.

6. What is an even function? An odd function? What symmetry properties do the graphs of such functions have? What advantage can we take of this? Give an example of a function that is neither even nor odd.

12. What is radian measure? How do you convert from radians to degrees? Degrees to radians?

7. If f and g are real-valued functions, how are the domains of f + g, f - g, fg, and f / g related to the domains of f and g? Give examples.

14. What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions?

8. When is it possible to compose one function with another? Give examples of composites and their values at various points. Does the order in which functions are composed ever matter? 9. How do you change the equation y = f(x) to shift its graph vertically up or down by Ikl units? Horizontally to the left or right? Give examples.

10. How do you change the equation y = f(x) to compress or stretch the graph by a factor c > I? Reflect the graph across a coordinate axis? Give examples.

Chapter

a

13. Graph the six basic trigonometric functions. What symmetries do the graphs have?

15. Stsrting with the identity sin2 8 + cos2 8 = I and the formulas for cos (A + B) and sin (A + B), show how a variety of other trigonometric identities may be derived.

16. How does the formula for the general sine function f(x) = A sin «2'IT/B)(x - e)) + D relate to the shifting, stretching, compressing, and reflection of its graph? Give examples. Graph the general sine curve and identifY the constsnts A, B, e,andD. 17. Name three issues that arise when functions are graphed using a calculator or computer with graphing software. Give examples.

Practice Exercises

Functions and Graphs 1. Express the area and circumference of a circle as functions of the circle's radius. Then express the area as a function of the circumference.

2. Express the radius of a sphere as a function of the sphere's surface area. Then express the surface area as a function of the volume. 3. A point P in the frrst quadrant lies on the parabola y = x 2 • Express the coordinates of P as functions of the angle of inclination of the line joining P to the origin. 4. A hot-air balloon rising straight up from a level field is traclred by a range fmder located 500 ft from the point ofliftoff. Express the balloon's height as a function of the angle the line from the range fmder to the balloon makes with the ground.

Iu Exercises 19. y =

11. y

= x2 + I = I - cosx

13• y

=~ 3

x - 2x 15. Y = x + cosx

12. y

= x' - x 3 = sec x tsnx

14. Y

=

10. y

16. y

fmd the (a) domain and (b) range.

2

20. y = -2

= Vl6 - x

23. y

= 2e~

25. y

=

27. y

= Iu (x

2

- 3

2sin(3x

+ 'IT)

- 3)

- I

+I

+~ +I

22. y =

32 ->

24. y

=

tsn(2x - 'IT)

26. y

=

x 2/'

28. y = -I

+ ~2 - x

29. State whether each function is increasing, decreasing, or neither. a. Volume of a sphere as a function of its radius b. The greatest integer function

c. Height above Earth's sea level as a function of atmospheric pressure (assumed nonzero) d. Kinetic energy as a function ofa particle~ velocity

Iu Exercises ~16, deterntine v.lrether the function is even, odd, or neith"" 9. y

Ixl -

21. y

Iu Exercises 5--1!, determine whether the graph of the function is sym-

metric about the y-axis, the origin, or neither. 5. y = x'i' 6. y = x 2/' 7. y = x 2 - 2x - I 8. y = e-"

1~28,

30. Find the largest interval on which the given function is increasing.

a. f(x)=lx-21+ I c. g(x) = (3x - 1)'/3

x

x - sinx

= xcosx

b. f(x)

= (x +

I)'

d. R(x)=~

Piecewise-Defined Functions Iu Exercises 31 and 32, fmd the (a) domain and (b) range.

17. Suppose that f and g are both odd functions defined on the entire real1ine. Which of the following (where defmed) are even? odd?

31

a. fg b. f3 c. f(sinx) d. g(sec x) •• Igl 18. If f(a - x) = f(a + x), show that g(x) = f(x + a) is an even function.

32.y=

_ .y-

{h, Yx,

-4 S x'" 0 0<xs4

-x - 2 {

x,

'

-x + 2,

-2::S; x::S; -1 -1 < x::s; 1

1 <x::S;2

36

Chapter 1: Functions

In Exercises 33 and 34, write a piecewise formula for the function.

e. S1retch vertically by • factor of 5

33.

f. Compress horizontally by a factor of 5

Y

34.

Y (2,5)

50. Describe how each graph is obtained from the graph ofy = f(x).

~J---~--..x

o

4

(f" g)(-l).

d. (g" g)(x).

I x'

~

g(x) =

36. f(x) = 2 - x,

g(x) =

37. f(x) = 2 - x 2 ,

g(x) =

Yx,

=

d. y = f(2x + I)

e. Y=f(t)-4

f. Y = -3f(x)

51. y=

-~1 +~

52. y=

I-t

53. y

~2 +

54. Y

(-5x)1j3

=

I

V'x+l

g(x)

{=~,

=

-

=

f "g

and g "

f

and

Vx+2

~

-4::;;x:5-1 -1<x:51 l<x:52

2,

x - 2,

40.

f (x)

+ I, x _ I,

X

= {

-2 '" x < 0 0"'x"'2

Composition with absolute wines In Exercises 41-48, graph h and h together. Then describe how applying the absolute wlue function in h affects the graph of h

j,(x)

55. Y

= cos2x

57. Y

=

Ixl

42. x 2

Ixl

Ix'i Ix

(x - f). + (x + .;(J sin

In Exercises 61-{i4, ABC is a right triangle with the right angle at C. The sides opposite angles A, B, and C are a, b, and c, respectively. 61. a. Findaandbifc = 2,B = 1f/3. b. Findaandcifb = 2,B = 1f/3. 62. a. Express a in terms of A and c. b. Express a in terms of A and b.

b. Express c in terms of A and a.

2

b. Express sin A in terms of b and c.

+ xl 2

14 - x 1

xI

I Ixl

47.

Yx

vfxj

48. sinx

sinlxl

Shifting and Scaling Graphs 49. Suppose the graph of g is given. Write equations for the graphs that are obtsined from the graph of g by shifting, scaling, or

reflecting, as indicated.

tuni~

1fX

COST

64. a. Express sin A in terms of a and c.

46.

Up

58. Y =

sin1fX

60. Sketch the graph y = 1

2

44.x 2 +x

L

. x = sm 2

63. a. Express a in terms of Band b.

43. x' 45.4-x2

56• y

59. Sketch the graph y = 2 cos

hex)

41. x

=

Trigonometry In Exercises 55-58, sketch the graph of the given function. What is the pcriod of the function?

For Exercises 39 and 40, sketch the graphs of f and f " f.

39. f(x)

+ 4I

x+2

In Exercises 37 and 38,
38. f(x)

c. Y = f( -3x)

b. (g" f)(2).

c. (f " flex). 35. f(x) =

b. Y = f(4x)

In Exercises 51-54, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.14-1.17 and applying an appropriate 1ransfonnation.

Composition of Functions In Exercises 35 and 36, rmd L

a. y = f(x - 5)

65. Height of a pole 1\vo wires s1retch from the top T of a vertical pole to points B and C on the grouod, where C is \0 m closer to the base of the pole than is B. If wire BTmakes an angle of35" with the horizontal and wire CT makes an angle of 50" with the horizontal, how high is the pole? 66. Height of a weather balloon Observers at positions A and B 2 kIn apart simultaneously measure the angle of elevation of a weather balloon to be 40" and 70", respectively. If the balloon is directly above a point on the line segroent between A and B, find the height of the balloon.

D 67.

a. Graphthefunctionf(x)

=

sinx

+ cos(x/2).

b. What appears to be the period of this function? right 3

b. Down 2 units, left ~ c. Reflect about the y-axis

d. Reflect about the x-axis

c. CooIrrm your rmding in part (b) algebraically.

D 68.

a. Graphf(x) = sin (llx). b. What are the domain and range off?

c. Is f periodic? Give reasons for your answer.

Chapter 1 Additional and Advanced Exercises

Chapter [

Additional and Advanced Exercises

Functions and Graphs 1. Are there two functions 1 and g such that log

=g

0

I? Give

reasons for your answer. 2. Are there two functions 1 and g with the following property? The graphs of 1 and g are not straight lines but the graph of log is a straight line. Give reasons for your answer. 3. If I(x) is odd, can anything be said of g(x) = I(x) - 2? What if f is even instead? Give reasons for your answer. 4. Ifg(x) is an odd function defined for all values of x, can anything be said about g(O)? Give reasons for your answer. S. Graph the equation Ix I + Iy I = I 6. Graph the equationy

+ Iyl

+ x. = x + Ixl.

1 - cosx sin x

sin x

1

Show that there is only one W!I:j to write 1 as the sum of an even and an odd function. (Hint: One W!I:j is given in part (a). !falso I(x) = E,(x) + O,(x) where E, is even and 0, is odd, show that E - E, = 0, - O. Then use Exercise 11 to show that E = E, and 0 = 0,.)

b. Uniqueness

Grapher Explorations-Effects of Parameters 13. What happens to the graph of y = ax 2 + bx

+ c as

b. b changes (a and c fixed, a # OJ?

c. c changes (a and b fixed, a # OJ? 14. What happens to the graph of y = a(x + b)3 + cas •• a changes while b and c remain fixed?

b. 1 - cosx

+ cosx

where E is an even function and 0 is an odd function. (Hint: LetE(x) = (f(x) + 1(-x»/2. Show that E(-x) = E(x), so thatE is even. Then show that o (x) = I(x) - E(x) is odd.)

•• a changes while b and c remain fixed?

Derivations and Proofs 7. Prove the following identities.

a.

37

1 + cosx

b. b changes (a and c fIXed, a # OJ?

8. Explain the following ''proof without words" of the law of cosines. (Source: "Proof without Words: The Law of Cosines," Sidney H. Kung, Mathematics Magazine, Vol. 63, No.5, Dec. 1990, p. 342.)

a

a a

c. c changes (a and b fIXed, a # OJ? Geometry 15. An object's center of mass moves at a constant velocity v along a straight line past the origin. The accompanying figure shows the coordinate system and the line of motion. The dots show positions that are I sec apart. Why are the areas A" A 2 , .•• , A, in the figure all equal? As in Kepler's equal area law (see Section 13.6), the line that joins the object's center of mass to the origin sweeps out equal areas in equal times. y

10

9. Show that the area of triangle ABC is given by (1/2)ab sin C = (1/2)bc sin A = (1/2)ca sinB.

c ,,

o

5

A L-----~~-----> B

10. Show that the area of triangle ABC is given by Vs(s - a)(s - b)(s - c)wheres = (a + b seroiperimeter of the triangle.

+ O(x),

15

16. •• Find the slope of the line from the origin to the midpoint P of sideAB in the triangle in the accoropanying figure (a, b > 0).

+ c)/2isthe

11. Show that if 1 is both even and odd, then I(x) = 0 for every x in the domain of I. 12. a. Even-odd decompositions Let 1 be a function whose domain is symmetric about the origin, that is, -x belongs to the domain whenever x does. Show that 1 is the sum of an even function and an odd function:

I(x) = E(x)

10 Kilometers

Y

~~----------~A~~-,~O)~X

b. When is OP perpendicular to AB?

38

Chapter 1: Functions

17. Consider the quarter-circle of radius I and right triangles ABE and ACD given in the accompanying figure. Use standard area formulas to conclude that I . Zsm6cos6

18. Let f(:x:) ~ ax + b and g(:x:) ~ ex + d. What conditioo must be satisfied by the coostants a, b, c, d in order that (f 0 g)(:x:) ~ (g 0 f)(:x:) for every value of:x:1

I sin6

6

< Z < Z cos 6'

y (0,1)

~~8~____~~ D~__~x A

Chapter

t!I

E

(1,0)

Technology Application Projects

All Overview ofMathemotica An overview of Mathematica sufficient to complete the Mathematica modules appearing on the Web site.

Mathematica/Maple Module: Modeling Change: Springs, Drivillg Safety, JlJJdioactivity, 1l-e.., Fish, and Mammals Coostruct and interpret mathematical models, analyze and improve them, and make predictioos using them.

2 LIMITS AND CONTINUITY OVERVIEW Mathematicians of the seventeenth century were keenly interested in the study of motion for objects on or near the earth and the motion of planets and stars. This study involved both the speed of the object and its direction of motion at any instant, and they knew the direction was tangent to the path of motion. The concept of a limit is fundamental to finding the velocity of a moving object and the tangent to a curve. In this chapter we develop the limit, first intuitively and then formally. We use limits to describe the way a function varies. Some functions vary continuously; small changes in x produce only small changes in f(x). Other functions can have values that jump, vary erratically, or tend to increase or decrease without bound. The notion of limit gives a precise way to distinguish between these behaviors.

2.1

Rates of Change and Tangents to Curves Calculus is a tool to help us understand how functional relationships change, such as the position or speed of a moving object as a function of time, or the changing slope of a curve being traversed by a point moving along it. In this section we introduce the ideas of average and instantaneous rates of change, and show that they are closely related to the slope of a curve at a point P on the curve. We give precise developments of these important concepts in the next chapter, but for now we use an informal approach so you will see how they lead naturally to the main idea of the chapter, the limit. You will see that limits playa major role in calculus and the study of change.

Average and Instantaneous Speed HISTORICAL BIOGRAPHY*

Galileo Galilei (1564- 1642)

In the late sixteenth century, Galileo discovered that a solid object dropped from rest (not moving) near the surface of the earth and allowed to fall freely will fall a distance proportional to the square of the time it has been falling. This type of motion is called free faU. It assumes negligible air resistance to slow the object down, and that gravity is the only force acting on the falling body. If y denotes the distance fallen in feet after t seconds, then Galileo's law is

y = 16t 2 ,

where 16 is the (approximate) constant of proportionality. (If y is measured in meters, the constant is 4.9.) A moving body's average speed during an interval oftime is found by dividing the distance covered by the time elapsed. The unit of measure is length per unit time: kilometers per hour, feet (or meters) per second, or whatever is appropriate to the problem at hand.

*To learn more about the historical figures mentioned in the text and the development of many major elements and topics of calculus, visit www.aw.comlthomas.

39

40

Chapter 2: Limits and Continuity

EXAMPLE 1

A rock breaks loose from the top of a tall cliff. What is its average speed


dy dl


16(2)' - 16(0)' 2 - 0

=

dy M=

ft =

32 sec

16(2)2 - 16(1)2 ft 2-1 = 48 sec

•

We want a way to determine the speed of a falling object at a single instant 10, instead of using its average speed over an interval of time. To do this, we examine what happens when we calculate the average speed over shorter and shorter time intervals starting at 10. The next example illustrates this process. Our discussion is informal here, but it will be made precise in Chapter 3.

EXAMPLE 2

Find the speed of the falling rock in Example I att = I and 1= 2 sec.

Solution We can calculate the avemge speed of the rock over a time interval [10, 10 having length dl = h, as

dy dl

16(10 + h)2 - 16102 h

+ h],

(1)

We cannot use this formula to calculate the "instantaneous" speed at the exact moment 10 by simply substituting h = 0, because we cannot divide by zero. But we can use it to calculate average speeds over increasingly short time intervals starting at 10 = I and 10 = 2. When we do so, we see a pattern
TABLE 2.1 Average speeds over short time intervals [to, to + h] dy

Average speed: dl =

16(10 + h)2 - 16102 h

Length of time interval h

Average speed over interval oflength h starting at to = 1

Average speed over interval of length h starting at to = 2

I

48 33.6 32.16 32.016 32.0016

80 65.6 64.16 64.016 64.0016

0.1 0.01 0.001 0.0001

The average speed on intervals starting at 10 = I seems to approach a limiting value of 32 as the length of the interval decreases. This suggests that the rock is falling at a speed of 32 ft/ sec at 10 = I sec. Let's confirm this algebraically.

2.1

Ifwe set 10

=

Rates of Change and Tangents to Curves

41

I and then expand the nwneratorin Equation (I) and simplifY, we f"md that

ay dl

16(1

+ h)' -

16(1)'

16(1

+ 2h + h 2 )

-

16

h

h 2

32h : 16h = 32

+ 16h.

For values of h different from 0, the expressions on the right and left are equivalent and the average speed is 32 + 16h ft/sec. We can now see why the average speed has the limiting value 32 + 16(0) = 32 ft/sec as h approaches O. Similarly, setting to = 2 in Equation (I), the procedure yields

~

=

64

+ 16h

for values of h different from O. As h gets closer and closer to 0, the average speed has the • limiting value 64 ft/sec when to = 2 sec, as suggested by Table 2.1. The average speed of a falling object is an example of a more general idea which we discuss next.

Average Rates of Change and Secant Lines y

Given an arbitrary function y = j(x), we calculate the average rate of change of y with respect to X over the interval [Xl, X2] by dividing the change in the value of y, ay = j(X2) - j(XI), by the length ax = X2 - Xl = h of the interval over which the change occurs. (We use the symbol h for dx to simplifY the notation here and later on.)

DEFINmON The average rate of change of y interval [X"X2] is

ay dx FIGURE 2.1 A secant to the graph y = f(x). Its slope is dy/ At, the average rate of change of f over the interval [Xl, X2].

j(X2) - j(XI) X2 Xl

= j(x) with respect to X

j(XI + h) - j(XI) h

over the

h .. O.

Geometrically, the rate of change of j over [Xl, X2] is the slope of the line through the points P(Xh j(XI» and Q(X2, j(X2» (Figure 2.1). In geometry, a line joining two points of a curve is a secant to the curve. Thus, the average rate of change of j from Xl to X2 is identical with the slope of secant PQ. Let's consider what happens as the point Q approaches the point P along the curve, so the length h of the interval over which the change occurs approaches zero.

Defining the Slope of a Curve L

o

FIGURE 2.2 L is tangent to the circle at P if it passes through P perpendicu1ar to radius OP.

We know what is meant by the slope of a straight line, which tells us the rate at which it rises or falls-its rate of change as the graph of a linear function. But what is meant by the slope of a curve at a point P on the curve? If there is a tangent line to the curve at P-a line that just touches the curve like the tangent to a circle-it would be reasonable to identifY the slope of the tangent as the slope of the curve at P. So we need a precise meaning for the tangent at a point on a curve. For circles, tangeocy is straightforward A line L is tangent to a circle at a point P if L passes throughP perpendicular to the radius at P (Figure 2.2). Such a line just touches the circle. But what does it mean to say that a lioeL is tangeot to some other curve C at a pointP?

42

Chapter 2: Limits and Continuity

To define tangency for general curves, we need an approach that takes into account the behavior of the secants through P and nearby points Q as Q moves toward P along the curve (Figure 2.3). Here is the idea:

1. 2. 3.

HISTORICAL BIOGRAPHY

Start with what we can calculate, namely the slope of the secant PQ. Investigate the limiting value of the secant slope as Q approaches P along the curve. (We clarify the limit idea in the next section.) If the limit exists, take it to be the slope of the curve at P and define the tangent to the curve at P to be the line through P with this slope.

This procedure is what we were doing in the falling-rock problem discussed in Example 2. The next example illustrates the geometric idea for the tangent to a curve.

Pierre de Fermat (1601-1665) Secants

Tangent p Q

Secants

FIGURE 2.3 The tangent to 1he curve at P is 1he line through P whose slope is 1he limit of 1he secant slopes as Q ..... P from ei1her side.

Find the slope of the parabola y = x 2 at the pointP(2, 4). Write an equation for the tangent to the parabola at this point.

EXAMPLE 3

Solution We begin with a secant line through P(2, 4) and Q(2 + h, (2 + h)2) nearby. We then write an expression for the slope of the secant PQ and investigate what happens to the slope as Q approaches P along the curve: ~y (2 Secant slope = =

+

h)2 - 22

~

h

h2

+ 4h + 4

- 4

= '"---"-=~-"--2

h

= h

2

~

4h = h

+ 4.

If h > 0, then Q lies above and to the right of P, as in Figure 2.4. If h < 0, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant slope h + 4 approaches 4. We take 4 to be the parabola's slope atP. y . (2

Secant slope IS

Tangent slope I I I

l Ay ~ (2

i

+ h)2 -

=

+ h)2 h

4

=

h + 4.

4

4

-----Ll dx=h

I

---~~~~~~-~~~---->x

2+h NOTTOSCALB

FIGURE 2.4 Finding 1he slope of1he parabola y ~ x 2 at 1he point P(2, 4) as 1he

limit of secant slopes (Example 3).

2.1

Rates of Change and Tangents to Curves

43

The tangent to the parabola at P is the line through P with slope 4:

y = 4

+ 4(x

- 2)

Point-slope equation

•

Y = 4x - 4.

Instantaneous Rates of Change and Tangent Lines The rates at which the rock in Example 2 was falling at the instants t = 1 and t = 2 are called instantaneous rates of change. Instantaneous rates and slopes of tangent lines are intimately connected, as we will now see in the following examples.

EXAMPLE 4 Figure 2.5 shows how a populationp of fruit flies (Drosophila) grew in a 50-day experiment. The number of flies was counted at regular intervals, the counted values plotted with respect to time t, and the points joined by a smooth curve (colored blue in Figure 2.5). Find the average growth rate from day 23 to day 45. Solution There were 150 flies on day 23 and 340 flies on day 45. Thus the number of flies increased by 340 - 150 = 190 in 45 - 23 = 22 days. The average rate of change of the population from day 23 to day 45 was /lp 340 - 150 190 . Average rate of change: T/ = 45 23 = 22 '" 8.6 flIes/day. p

350 ~

V

300

/

" 250

'S

.l!

200

PI ,2 3,150)

.~ 150 Z

~

100 /

50

/

o

10

V

/ ~ 4, -

V

~,~

/

/ Q(45 340)

-tJ.-p-=r 190

8. !lie day

22

'/ 20

30 Tim. (days)

40

50

FIGURE 2.5 Growth of a fruit fly population in a controlled experiment. The average rate of chaage over 22 days is the slope !!'p/!!.t of the secant line (Example 4).

This average is the slope of the secant through the points P and Q on the graph in Figure 2.5. • The average rate of cbange from day 23 to day 45 calculated in Example 4 does not tell us how fast the population was cbanging on day 23 itself. For that we need to examine time intervals closer to the day in question.

EXAMPLE 5

How fast was the number of flies in the population of Example 4 growing

on day 23?

Solution To answer this question, we examine the average rates of change over increasingly short time intervals starting at day 23. In geometric terms, we find these rates by calculating the slopes of secants from P to Q, for a sequence of points Q approaching P along the curve (Figure 2.6).

44

Chapter 2: Limits and Continuity p

= fl.p / fl.t

Q

Slope of PQ (flies / day)

(45,340)

340 - 150 '" 8 6 45 - 23 .

'a

(40,330)

330 - 150 106 40-23'" .

iz

(35,310)

310 - 150 '" 13 3 35 - 23 .

(30,265)

265 - 150 '" 164 30 - 23 .

FIGURE 2.6

B(35, 350) ;,1 ~ /'

350

J: h ~/

300

~

250

/

Q(45,340)

AV

200

P(23,150»

ISO

fY

I('

100

/

50 /

0

..:::. ~

~ .

/ 10' ~ 30 \ 20 A(14,0) Time (days)

40

50

The positions and slopes offour secants through the pointP on the fruit fly grapb (Example 5).

The values in the table show that the secant slopes rise from 8.6 to 16.4 as the I-coordinate of Q decreases from 45 to 30, and we would expect the slopes to rise slightly higher as t continued on toward 23. Geometrically, the secants rotate about P and seem to approach the red tangent line in the figure. Since the line appears to pass through the points (14, 0) and (35, 350), it has slope

~;0_-1~

=

16.7 flies/day (approximately).

On day 23 the population was increasing at a rate of about 16.7 flies/day.

•

The instantaneous rates in Example 2 were found to be the values of the average speeds, or average rates of change, as the time interval ofiength h approached O. That is, the instantaneous rate is the value the average rate approaches as the length h of the interval over which the change occurs approaches zero. The average rate of change corresponds to the slope of a secant line; the instantaneous rate corresponds to the slope of the tangent line as the independent variable approaches a fixed value. In Example 2, the independent variable I approached the values I = I and I = 2. In Example 3, the independent variable x approached the value x = 2. So we see that instantaneous rates and slopes of tangent lines are closely connected. We investigate this connection thoroughly in the next chapter, but to do so we need the concept of a limil.

Exercises 2.1 Average Rate. of Change In Exercises I-j), fmd the average rate of chaoge of the function over the given interval or intervaIs. 1. !(x) ~ x 3

+

b.[-I,I]

2. g(x) ~ x 2

a. [-1,1] 3. h(t) ~ cot t a. ["./4,3"./4] 4. g(t) ~ 2 + cost •• [0,,,.]

R(O) ~

\I'49+l;

6. P(O) ~ 03 - 411'

[0,2]

+ 50;

[1,2]

Slope of a Curve at a Point

I

•• [2,3]

s.

b. [-2,0]

In Exercises 7-14, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (h) an equation of the tangent line atP.

7. y ~ x 2

-

3, P(2, I)

8. Y ~ 5 - x 2 , b. ["./6, "./2]

9. Y

~ x2 -

10. Y ~ x 2 11. Y ~ x 3,

-

P(I,4)

2x - 3,

P(2, -3)

4x, P(I, -3)

P(2, 8)

2.1

12. Y = 2 13. Y

=

x' -

x',

+ 4,

t. Use yoor graph to estimate the rate at which the profits were

P(2, 0)

changing in 2002.

D 18.

Instantaneous Rates of Change 15. Speed of a tar The acoompanying figure shows the time1D-distance graph for a sports car accelerating from a standstill. p

650 600

J;j-

D 19.

~

Q,

o

V

10

15

t. What does your table indicate is the rate of chaage of g(x)

20

with respect to x at x = I?

Elapsed time (sec)

a. Estimate the slopes of secants PQ" PQ2, PQ" and PQ., ammging them in order in a table like the one in Figure 2.6. What are the appropriate units for these slopes? b. Then estimate the car's speed at time I = 20 sec.

16. The accompanying figure shows the plot of distance falleo versus time for ao object that fell froro the lunar landing module a distance 80 m to the surface of the moon. a. Estimate the slopes of the secaots PQ" PQ2,PQ" andPQ., ammging them in a table like the one in Figure 2.6. b. About how fast was the object going when it hit the surfa..? y

80 I

Q / Q

o

-

10'

p

Q,

~'1 f-

D 20.

Let/(l) = 1/1 for I '" O.

a. Find the average rate of chaage of I with respect to I over the intervals (i) froro I = 2 to I = 3, aod (ii) froro I = 2 to I = T. b. Make a table of values of the average rate ofchaoge of I with respect to lover the interval [2, 11, for some values of T approachiog 2, say T = 2.1,2.01,2.001,2.0001,2.00001, aod 2.000001. t. What does your table indicate is the rate of chaage of I with

respect to I at I = 2? d. Calculate the limit as T approaches 2 of the average rate of chaoge of I with respect to lover the interval froro 2 to T. You will have to do soroe algebra before you cao substitute T = 2. 21. The accoropaoying graph shows the tota1 distance s traveled by a bicyclist after I hours.

./

s

5

10

The profits of a sma11 company for each of the first five years of its operation are given in the following table: Year

d. Calculate the limit as h approaches zero of the average rate of chaage ofg(x) with respect to x over the interval [I, I + h].

V

Elapsed time (sec)

D 17.

Vx folX "" O.

aod 0.000001.

/'

......

Letg(x) =

a. Find the average rate of chaage of g(x) with respect to x over the intervals [1,2], [I, 1.5] and [I, 1+ h]. b. Make a table of values of the average rate ofchaoge ofg with respect to x over the interval [I, I + h] for some values ofh approaching zero, say h = 0.1, O.o!, 0.001, 0.0001, 0.00001,

/ 5

Make a table of values for the function F(x) = (x + 2)/(x - 2) at the points x = 1.2, x = l1/lO,x = 101/100, x = 1001/1000, x = 10001/10000, aodx = 1.

a. Find the average rate ofchaage ofF(x) over the intervals [I,x] for each x '" I in your table. b. Exteoding the table if necessary, 1ry to determine the rate of change of F(x) at x = 1.

s

100

45

b. What is the average rate of increase of the profits between 2002 aod 2004?

P(I, I)

12x, P(I, -11)

14. Y = x' - 3x2

Rates of Change and Tangents to Curves

Profit in $1000.

!

~

40

~

20

J

10

"

/'

30 ./

,,./ /

/ 0

2

3

4

Elapsed time (br)

2000 2001 2002 2003 2004

6 27 62

III 174

a. Plot points representing the profit as a function of year, aod join them by as smooth a curve as you can.

a. Estimate the bicyclist's average speed over the time intervals [0, I], [1,2.5], aod [2.5, 3.5]. b. Estimate the bicyclist's instantaneous speed at the times t = 1= 2,aodl = 3.

!,

t. Estimate the bicyclist's maximum speed aod the specific time

at which it occurs.

46

Chapter 2: Limits and Continuity

22. The accompanying graph shows the total amount of gasoline A in the gas tank of an automobile after being driven for I days.

a. Estimate the average rate of gasoline consumption over the time intervals [0, 31, [0, 51, and [7, 101. b. Estimate the instantaneous rate of gaaoline consomption at the times I ~ 1,1 ~ 4,andl ~ 8.

A

:;

,,

16

r:

:(

c. Estimate the maximum rate of gasoline consumption and the specific time at which it occurs. ~

[,

4

,l!

.....

-

012345678910 Elapsed time (days)

Limit of a Function and Limit Laws

2.2

In Section 2.1 we saw that limits arise when finding the instsntsneous rate of change of a function or the tsngent to a curve. Here we begin with an infonnal definition of limit and show how we can calculate the values of limits. A precise definition is presented in the next section.

Limits of Function Values

HIsTORICAL ESSAY

Frequently when studying a function y ~ I(x) we rmd ourselves interested in the function's behavior near a particular point xo, but not at Xo. This might be the case, for instsnce, if Xo is an irrational number, like 1T or whose values can only be approximated by "close" rational numbers at which we actoally evaluate the function instead. Another situation occurs when trying to evaluate a function at Xo leads to division by zero, which is undefined. We encountered this last circumstsnce when seeking the instsntsneous rate of change in y by considering the quotient function tly/h for h closer and closer to zero. Here's a specific example where we explore numerically how a function behaves near a particular point at which we carmot directly evaluate the function.

Lintits

V2,

y

2

__~--~----~------~X

o

EXAMPLE 1

How does the function

I(x)

y

behave near x 2 y=x+l

o FIGURE 2.7 The graph off is identical with the line y ~ x + 1 except at x = 1, where f is not

defmed (Example 1).

~

~x

2 -

1

x-I

I?

Solution The given formula defines I for all real numbers x except x ~ 1 (we carmot divide by zero). For any x oF 1, we can simplify the formula by factoring the numerator and canceling common factors: () Ix=

(x - I)(x + 1)

x-I

~x+I

for

x oF 1.

The graph of I is the line y ~ x + 1 with the point (I, 2) removed. This removed point is shown as a "hole" in Figure 2.7. Even though 1(1) is not dermed, it is clear that we can make the value of I(x) as close as we want to 2 by choosing x close enough to 1 (Table 2.2).

•

2.2 Limit of a Function and Limit Laws

47

TABLE 2.2 The closer x gets to 1, the closer f(x) = (x' - 1)/(x - 1)

seems to get to 2 Values of x below and above 1

x, -1 !(x) = --1x- = x + 1,

0.9 l.l 0.99 1.01 0.999 1.001 0.999999 1.000001

1.9 2.1 1.99 2.01 1.999 2.001 1.999999 2.000001

x;ld

Let's generalize the idea illustrated in Example I. Suppose !(x) is defined on ao open interval about xo, except possibly at Xo itself. If !(x) is arbitrarily close to L (as close to L as we like) for all x sufficiently close to xo, we say that f approaches the limit L as x approaches xo, aod write lim !(x) = L,

x-x,

which is read ''the limit of f(x) as x approaches Xo is L." For instance, in Example I we would say that !(x) approaches the limit 2 as x approaches I, aod write lim f(x) = 2,

x-I

or

2 - I lim-x-- = 2. x-I

x-I

Essentially, the def'mition says that the values of f(x) are close to the number L whenever x is close to Xo (on either side of xo). This def'mition is ''informal'' because phrases like arbitrarily close aod suffICiently close are imprecise; their meaoing depends on the context. (fo a machinist maoufacturing a piston, close may meao within a few thousandths of an inch. To ao astronomer studying distant galaxies, close may mean within a few thousand light-years.) Nevertheless, the def'mition is clear enough to enable us to recognize aod evaluate limits of specific fimctions. We will need the precise def'mition of Section 2.3, however, when we set out to prove theorems about limits. Here are several more examples exploring the idea oflimits.

EXAM PLE 2 This example illustrates that the limit value of a function does not depend on how the function is def'med at the point being approached. Consider the three functions in Figure 2.8. The function! has limit 2 as x --+ I even though f is not defined at x = I. y

y

y

---cr.--c+--;-.x (a) f(x)

~ ~~

i

X2 -

1

- - , x'l-l x-I

(b) g(x) ~ {

I,

x

=

(c) h(x)

~

x

+1

1

FIGURE 2.8 The limits of f(x), g(x), aod h(x) all equal 2 as x approaches 1. However, ooIy h(x) has the same functioo value as its limit at x ~ 1 (Example 2).

48

Chapter 2: Limits and Continuity The function g has limit 2 as x .... 1 even though 2 i' g( 1). The function h is the only one of the three functions in Figure 2.8 whose limit as x .... 1 equals its value at x = 1. For h, we have limx~l h(x) = h(I). This equality oflimit and function value is significant, and we return to it in Section 2.5. •

y

EXAMPLE 3 (a) If f is the identity function f(x) = x, then for any value ofxo (Figure 2.9a), (a) Identity function

lim f(x) = lim x = Xo.

x-xo

y

X--+Xo

(h) If f is the constant function f(x) = k (function with the constant value k), then for any value ofxo (Figure 2.9b), lim f(x) = lim k = k. X-XcI

°

----------~----~----~x

xo

limx=3

x~3

(b) Constant function

FIGURE 2.9 The functioos in Example 3 have limits at all points xo.

X--+Xo

For instances of each of these rules we have and

lim (4) = lim(4) = 4.

x--+-7

x--+2

•

We prove these rules in Example 3 in Section 2.3.

Some ways that limits can fail to exist are illustrated in Figure 2.10 and described in the next example. y

y~{o. 1,

y

x
y

X~O

1+-- - - - ----------~9_----------~x

------~hrtH-t_~~------~x

°

°

y

~

-1 (a) Unit step function U(x)

(b) g(x)

(c)f(x)

°

FIGURE 2.10 None of these functioos has a limit as x approaches (Example 4).

EXAMPLE 4

Discuss the behavior of the following functions as x .... 0.

(a) U(x) = {O, 1,

(h) g(x) =

C x' 0,

x
°

xi'O

x=O

x:sO

(c) f(x) = { 0, . 1 SIDX-, x>O

{o..

sm

x sO

1

x'

x>O

2.2

Limit of a Function and Limit Laws

49

Solution (8) It jumps: The unit step function U(x) has no limit as x --> 0 because its values jump at x = O. For negative values of x arbitrarily close to zero, V(x) = O. For positive values of x arbitrarily close to zero, U(x) = 1. There is no single value L approached by U(x) as x --> 0 (Figure 2. lOa).

(h) It grows too "large" to have a limit: g(x) has no limit as x --> 0 because the values ofg grow arbitrarily large in absolute value as x --> 0 aod do not stay close to any f'lXed real number (Figure 2.1 Ob). (0) It oscillates too much to have a limit: f(x) has no limit as x --> 0 because the functioo's values oscillate between + 1 aod -1 in every open interval cootaining O. The values • do not stay close to aoy one number as x --> 0 (Figure 2.10c).

The Limit Laws When discussing limits, sometimes we use the notatioo x --> Xo if we want to emphasize the point Xo that is being approached in the limit process (usually to enhance the clarity of a particular discussioo or example). Other times, such as in the statements of the following theorem, we use the simpler notation x --> c or x --> a which avoids the subscript in Xo. In every case, the symbols Xo, c, aod a refer to a single point 00 the x-axis that may or may not beloog to the domain of the functioo involved. To calculate limits of functioos that are arithmetic combinatioos offunctioos having known limits, we cao use several easy rules.

THEOREM l-Umit Laws lim f(x) = L

If L, M, c, aod k are real numbers aod aod

x~c

1. Sum Rule: 2. Difference Rule:

3. Constant Multiple Rule: 4. Product Rule: 5. Quotient Rule: 6. Puwer Rule:

7. Root Rule:

lim g(x) = M,

then

x~c

lim(j(x)

x-c

+ g(x»

=

L

+M

lim(j(x) - g(x» = L - M

x-c

lim(k' f(x)) = k· L

x-c

lim(j(x)' g(x» = L' M

x-c

lim f(x) = £ g(x) M'

x~c

M "i' 0

lim[f(x))" = L", n a positive integer

x-c

lim '{Y'fW = x~c

(If n is even, we assume that limf(x) = L x~c

>

"ifi =

L 1/", n a positive integer

0.)

In words, the Sum Rule says that the limit of a sum is the sum of the limits. Similarly, the next rules say that the limit of a difference is the difference of the limits; the limit of a coostant times a functioo is the constant times the limit of the functioo; the limit of a product is the product of the limits; the limit of a quotient is the quotient of the limits (provided that the limit of the denOOIinator is not 0); the limit ofa positive integer power (or root) ofa functioo is the integer power (or root) of the limit (provided that the root of the limit is a real number). It is reasonable that the properties in Theorem 1 are true (although these intuitive arguments do not constitute proofs). Ifx is sufficiently close to c, then f(x) is close to L aod g(x) is close to M, from our informal def'mitioo of a limit. It is then reasonable that f(x) + g(x) is close to L + M; f(x) - g(x) is close to L - M; kf(x) is close to kL; f(x)g(x) is close to LM; aod f(x)/g(x) is close to LIM if M is not zero. We prove the Sum Rule in Section 2.3, based on a precise def'mition oflimit. Rules 2-5 are proved in

50

Chapter 2: Limits and Continuity Appendix 4. Rule 6 is obtained by applying Rule 4 repeatedly. Rule 7 is proved in more advanced texts. The sum, difference, and product rules can be extended to any number of functions, not just two.

EXAMPLE 5 Use the observations lim....... , k = k and limx~, x = c (Example 3) and the properties oflimits to f"md the following limits.
limx4+X2-l

x~c

-

3)

-

3) = lim x' + lim 4x 2

(b)

x~,

x2

+5

(0) lim V4x 2 %--2

-

3

Solution
=

-

x-c

X--i'C

c' + 4c

lim(x' +

2

lim 3

x-c

3

-

x2 -

Sum and Difference Rules

Power and Multiple Rules

1)

x~c

Quotient Rule

lim(x 2 + 5)

x-c

lim

x4

+ lim

x2 -

lim 1

lim x 2 + lim 5

Sum and Difference Rules

Power or Product Rule

(0)

lim V4x 2

x--2

-

3 =

Vx--2 lim (4x 2 -

=

Vx--2 lim 4x 2 -

=

V4(-2)2 - 3

=

v'i6=3

=

Vi3

3)

Root Rule with n

lim 3

Difference Rule

x--2

=

2

Product and Multiple Rules

•

Two consequences of Theorem 1 further simplify the task of calculating limits of polynomials and rational functions. To evaluate the limit of a polynomial function as x approaches c, merely substitute c for x in the formula for the function. To evaluate the limit of a rational function as x approaches a point c at which the denominator is not zero, substitute c for x in the formula for the function. (See Examples 5a and 5b.) We state these results formally as theorems.

THEOREM 2-Limits of Polynomials If P(x) = anx n + an_lx n- 1 + ... + ao, then lim P(x) = P(c) = anc n + an_lc n- 1 + ... + ao.

x-c

THEOREM 3-Llmlts of Rational Functions IfP(x) and Q(x) are polynomials and Q(c) oF 0, then . P(x) P(c) lim - - = - x-c Q(x) Q(c)"

2.2

EXAMPLE 6

Limit of a Function and Limit Laws

The following calculation illustrates Theorems 2 and 3:

+ 4x 2 - 3 x2 + 5

. x3 lim

x~-l

=

(-1)3 + 4( -1)2 - 3 0 =-=0 (-If + 5 6

I Identifying Common Factors It can be shown that if Q(x) is a polynomial and Q(c) = 0, then (x - c) is • factor of Q(x). Thus, if the numerator and denominator of a rational function of x are both zero atx = c,theybave(x - c) ... common factor.

Theorem 3 applies only if the denominator of the rational function is not zero at the limit point c. If the denominator is zero, canceling common factors in the numerator and denominator may reduce the fraction to one whose denominator is no longer zero at c. If this happens, we can fmd the limit hy substitution in the simplified fraction.

Evaluate lim x 2 x-"J>l

x2

+X x2

y x

+

X x2 - X

2 •

Solution We cannot substitute x = I because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = I. It is, so it has a factor of (x - I) in common with the denominator. Canceling the (x - I)'s gives a simpler fraction with the same values as the original for x oF I:

(a)

y=x+2

-

Eliminating Zero Denominators Algebraically

EXAMPLE 7

------~ _~ 2 ---nO~+-------~x

51

-

-

(x - I)(x + 2) x(x - I)

2

x

x

+

2

x

'

if x oF I.

Using the simpler fraction, we find the limit of these values as x --> I hy substitution:

3

lim x x--+l

2

+x -

x2

-

X

2 = lim x x--+l

+2 X

= I

+

2 = 3.

1

See Figure 2.11.

-

(b)

Using Calculators and Computers to Estimate Limits FIGURE 2.11 The graph of f(x) = (x' + x - 2)/(x' - x) in part (a) is the same .. the graph of g(x) = (x + 2)/x in part (b) except at x = I, where f is undefmed. The functions have the same lintit .. x .... I (Example 7).

When we cannot use the Quotient Rule in Theorem I because the limit of the denon1inator is zero, we can try using a calculator or computer to guess the limit numerically as x gets closer and closer to c. We used this approach in Example I, but calculators and computers can sometimes give false values and misleading impressions for functions that are undefmed at a point or fail to have a limit there, as we now illustrate.

Solution Table 2.3 lists values of the function for several values near x = O. As x approaches 0 through the values ±I, ±0.5, ±O.IO, and ±0.01, the function seems to approach the number 0.05. As we take even smaller values of x, ±0.OO05, ±0.0001, ±0.00001, and ±O.OOOOOI, the function appears to approach the value O. Is the answer 0.05 or 0, or some other value? We resolve this question in the next example. _

52

Chapter 2: Limits and Continuity

~

Computer values of f(x)

TABLE 2.3

x

v'x' + 100 ,

10

x

near x

=

0

!(x)

±I ±0.5 ±O.I ±0.01

0.049876) 0.049969 h 0 05? 0.049999 approac es. . 0.050000

±0.0005 ±O.OOOI ±O.OOOOI ±O.OOOOOI

0.080000) 0.000000 h? 0.000000 approac es 0 . 0.000000

Using a computer or calculator may give ambiguous results, as in the last example. We cannot substitute x = 0 in the problem, and the numerator and denominator have no obvious common factors (as they did in Example 7). Sometimes, however, we can create a common factor algebraically.

EXAMPLE 9

Evaluate . Yx2 Iun

x-o

+ 100 - 10 X

2

•

Solution This is the limit we considered in Example 8. We can create a common factor by multiplying both numerator and denominator by the conjugate radical expression Yx 2 + 100 + 10 (obtained by changing the sigu after the square root). The preliminary algebra rationalizes the numerator: Yx2 + 100 - 10

Yx 2 + 100 - 10. Yx 2 + 100

+ 10 v'x + 100 + 10 2

x 2 + 100 - 100 + 100 + 10)

X2(Yx2 X2(Yx2

x2 + 100 + 10)

1 Yx 2 + 100

+ 10'

Common fuctor i'

Cancel xl for x =F 0

Therefore, · Yx2 IlID x~o

+ 100 - 10 x2

=

I'lID r~~I=----:~ ~ x~o Yx 2 + 100 + 10 1

Y02 ~

+ 100 + 10

Denominator not 0 at x = 0; substitute

1 20 = 0.05.

This calculation provides the correct answer, in contrast to the ambiguous computer results in Example 8. • We cannot always algebraically resolve the problem of rmding the limit of a quotient where the denominator becomes zero. In some cases the limit might then be found with the

2.2

Limit of a Function and Limit Laws

53

aid of some geometry applied to the problem (see the proof of Theorem 7 in Section 2.4), or through methods of calculus (illustrated in Section 7.5). The next theorem is also useful.

y

The Sandwich Theorem

~~--------~--------~X

o

c

The following theorem enables us to calculate a variety of limits. It is called the Sandwich Theorem because it refers to a function f whose values are sandwiched between the values of two other functions g and h that have the same limit L at a point c. Being trapped between the values of two functions that approach L, the values of f must also approach L (Figore 2.12). You will fmd a proof in Appendix 4.

FIGURE 2.12 The graph of f is saodwiched between the graphs of g aod h.

THEOREM 4-The Sandwich Theorem

Suppose that g(x) :5 f(x) :5 h(x) for all x in some open interval containing c, except possibly at x = c itself. Suppose also that lim g(x) = lim h(x) = L.

x-c

y

x-c

Thenlim..... d(x) = L.

The Sandwich Theorem is also called the Squeeze Theorem or the Pincbing Theorem. y~

1-

4x'

EXAMPLE 10

Given that 2

~--~----~----+---~->x

-1

I -

0

FIGURE 2.13 Aoy functioo u(x) whose graph lies io the regioo betweeo y ~ I + (x'/2) andy = I - (x'/4) has limit I as x --> 0 (Example 10).

4x

:5 u(x) :5 I

Solution

Since lim(1 - (x2/4» = I

and

+

(x 2/2» = I,

•

The Sandwich Theorem helps us establish several important limit rules:

(a) lim sin 0 = 0 -1r

lim(I

x~o

the Sandwich Theorem implies thatlimx~o u(x) = I (Figore 2.13).

EXAM PLE 11 ~------~~--------~ 8

for all x oF 0,

fmd lim..... o u(x) , no matter how complicated u is.

x~o

y

x2

+2

(h) lim cos 0 = I

6-0

8-0

(c) For any function f, lim If(x) I = 0 implies lim f(x) = x-c

x-c

o.

Solution (a)

y

(a) In Section 1.3 we established that -101:5 sinO :5101 for all 0 (see Figure 2.14a). Since lim'hO ( -I 0 I) = lim'hO 10 I = 0, we have lim sinO = O.

8~O

(h) From Section 1.3,0:5 I - cosO :5101 for all 0 (see Figore 2.14b), and we have Inn,,~o (I - cos 0) = 0 or

---_~2~~-1~~O~~1~-2~~8 lim cos 0 = I.

(b)

FIGURE 2.14 The Sandwich Theorem

coofl1'!I1S the limits io Example II.

8~O

(c) Since -If(x)l:5 f(x):5 If(x) I and-lf(x)1 and If(x) I have limit 0 as x-->c,it follows that limx~, f(x) = O. •

54

Chapter 2: Limits and Continuity

Another important property oflimits is given by the next theorem. A proof is given in the next section.

THEOREM 5 possibly at x then

TIf(x) :5 g(x) for ailxin some openintervai containingc, except x approaches c,

= c itself, and the limits of f and g both exist as

lim fix) :5 lim g(x).

x--"'c

x--"'c

The assertion resulting from replacing the less than or equal to (:5) inequality by the strict less than «) inequality in Theorem 5 is false. Figure 2.l4a shows that for 0 .. 0, -101 < sin 0 < 10 I, but in the limit as 0 --> 0, equality holds.

Exercises 2.2 Limits from Graphs 1. For the function g(x) graphed here, fmd the following limits or explain why they do not exist.

a. lim g(x) x-I

b. lim g(x) x-2

c. lim g(x)

d.

%-3

y

lim g(x)

x-2.S

y

4. Which of the following statements about the function y graphed here are 1rue, and which are false?

a. b. c. d.

2. For the function f(/) graphed here, fmd the following limits or explain why they do not exist.

a. lim f(/)

b. lim f(/)

1--2

1--1

c. lim f(/)

d.

1-0

lim f(/)

lim f(x) does not exist.

x~2

lim f(x) = 2 lim f(x) does not exist. x~l lim f(x) exists at every pointxo in (-I, I). x~2

x~

t---{).S

...

e. lim f(x) exists at every pointxo in (I, 3).

s

x~

...

IL

ro -2

-I

= f(x)

y

I

0

1

)

t

-I

I

3. Which of the following statements about the function y = f(x) graphed here are 1rue, and which are false? a. lim f(x) exists. x~o

b. lim f(x) = 0 x~o

c. lim f(x) = I x~o

d. lim f(x) = I x~l

e. lim f(x) = 0 x~l

f. lim f(x) exists at every point Xo in (-1, 1). x~

...

g. lim f(x) does not exist. x~l

Existence of Limits In Exercises 5 and 6, explain why the limits do not exist. 5lim~

.

x~o

Ixl

6. lim _1-1 x-I x-

7. Suppose that a function f(x) is defined for all real values ofx except x = Xo. Can anything be said about the existence of lim~" f(x)? Give reasons for your answer. 8. Suppose that a function f(x) is defmed for all x in [-I, I]. Can anything be said about the existence of limx~o f(x)? Give rea-

sons for your answer.

2.2

9. Iflim,.....1 f(x) ~ 5, must f be defmed at x ~ I? If it is, must f( I) ~ 5? Can we conclude anything about the values of f at x ~ I? Explain. 10. If f(1) ~ 5, must lim'~1 f(x) exist? If it does, thea must lim,.....1 f(x) ~ 5? Can we conclude anything about lim,.....1 f(x)? Explain.

Limit of a Function and Limit Laws

Using Limit Rules

51. Suppose limr-o f(x) ~ 1 and limz~o g(x) ~ -5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. lim 2f(x) - g(x) ~ ,~o (j(x)

+ 7)2/3

%--7

+ 7)2/3

(a)

14. lim

x~o

(b)

+ 7))2/3

2 lim f(x) - lim g(x)

z,,' + 4x + 8)

(x 3 -

x-o

( lim (J(x)

+ 5x - 2)

%-2

1-6

lim (j(x)

x-o

12. lim (-x'

13. lim 8(t - 5)(t - 7)

- g(x))

lim 2f(x) - lim g(x)

Find the limits in Exercises 11-22.

+ 5)

1!'!l, (2f(x) x~o

Calculating Limits

11. lim (z"

55

x-o

%-+-2

x-a

(c)

16. lim 38(Zs - 1) s-2/3

17. lim 3(z" - I)'

18.lim

19. lim (5 - y)4/3

20. lim (2z - 8)1/3

%--1

y--3

21. lim

h~O

(2)(1) - (-5) (1 + 7)'/3

y+2

y-2 y2

+

5y

+

6

z~o

3

v'3h+1 + I

22. lim

v5h+4 -

4-0

2

52. Let lim,.....1 h(x) ~ 5, limz~lp(X) ~ 1, and lim'~1 r(x) ~ 2. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.

h

Limits of quotients Find the limits in Exercises 23-42.

lim

x~1

23.lim x - 5 x-s x 2 - 25

24. lim

x'+3x-1O lim 25. %--5 X + 5

' x2 - 7x + 10 26. lun 2 %-2 x 2 28. lim t + 3t + 2 t--l t 2 - t - 2

29. lim -z" - 4 %-+-2 x 3 + 2,x2 ! - I

y-o

39. lim

x-I

Vx+3 - 2 W+12 - 4

x-2

X -

41. lim 2 x--3

2

w-=s +

X

3

y'5lim h(x) ,~1

,

38. lim

W+8 -

%-+-1

40. lim

X

+1

x

+2

53.

Supposelim,~,,(x) ~

+ 3g(x))

54. Suppose lim.~"(x) •• lim (g(x) + 3) %-4

~

4-x

Find the limits in Exercises

c. lim (g(x»2

44. lim sin2 x

45. lim sec x

46. lim tan x

47. lim 1 + x + sinx x-o 3 cos x

48. lim (x' - 1)(2 - cosx)

,~o

x-o

Vx+4 cos (x + 'IT)

,~o

x-o

x-o

SO. lim Y 7

x-o

+ sec' X

~ -

3. Find

%-4

r

d •

g(x)

x~ f(x} - 1

55. Suppose lim'~b f(x) ~ 7 and lim'~b g (x) ~ - 3. Find b. lim f(x)' g(x) •• lim (j(x) + g(x)) x-b

x-b

c. lim 4g(x)

43. lim (2 sin x - I)

r f(x) • ,~ f(x) - g(x)

0 and lim.~4 g (x) b. lim xf(x)

%-4

x~45-~

-2. Find

x-c

d

,~,

3

~

b. lim 2f(x)g(x)

x-c

c. lim (j(x)

x-I

5 2

5andlim,.....,g(x)

•• lim f(x)g(x)

x~-2~-3

42.lim

(c)

x-I

yI(5)(5)

X

4 -

x-I

(1)(4 - 2)

43-50.

%--'11"

x-I

x+l

16

(b)

( lim p(x»)( lim (4 - r(x»))

+ _1_

v-2 v 4 -

Limits with trigonometric functions

49. lim

,~1

x-I

36. lim x x '~42-Vx ,~1

vlim 5h(x)

( lim p(x»)( lim 4 - lim r(x»)

34. lim if - 8

37.lim

,~1

3y4 - 16y2 _1_ x-I

(a)

lim (P(x)(4 - r(x)))

p(x)(4 - r(x))

+ 4x + 3

x-o

x-

x-I

z~1

5y 3 + 8y' 30. lim --'-;c-----co

32. lim

31. lim -x--l

limY5h(x)

VShW

x+3

%-+-3 Xl

27.lim t'+t-2 I-I t2 - 1

7 4

d. lim f(x)/g(x)

x-b

x-b

56. Suppose that limr--2 p(x) lim,~_,,(x) ~ -3. Find

~

4,lim,....._2 r(x)

•• lim (P(x) + r(x) + ,(x» %--2

b.

lim p(x)' r(x)' ,(x)

%-+-2

c. lim (-4p(x) x--2

+ 5r(x))l8(x)

~

0, and

56

Chapter 2: Limits and Continuity

Umits of Average Rates of Change

h. Support your conclusion in part
Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form

.. f(,,-x _+~h):--_f",(x-,-)

lim-

h occur frequently in calculus. In Exercises 57...fJ2, evaluate this limit for the given value of x and function f. b-O

57. f(x) = x 2, X = I 59. f(x) = 3x - 4, x = 2

61. f(x) =

Yx,

58. f(x) = x 2,

X

60. f(x) = I/x,

you arrive at if you evaluate G atx -6.001, ... instead?

x = -2 x = 0

Using the Sandwich Theorem 63. If V5 - 2x lim,...o f(x). L

'"

2

f(x) '" ~ fur -I'" x '" I, fmd

2

.

< xsmx < 1 6 2 - 2cosx hold for all values of x close to zero. What, if anything, does this tell you about

D b.

r xsinx? x~ 2 - 2eosX" Give reasons for your answer. Graphy = I - (x 2/6), y = (x sinx)/(2 - 2 cos x), aod y = I together for - 2 '" x '" 2. Comment on the behavior of the graphs as x .... O.

66.

L

Suppose that the inequalities

1_ 2

x 2 < 1 - cosx < 1 24 x2 2

•

Give reasons for your answer.

D b.

Graph the equations y = (1/2) - (x 2/24), y = (I - cosx)/x 2 ,aody = 1/2togetherfur-2 '" Comment on the behavior of the graphs as x .... O.

x'" 2.

Estimating LImits

D You will fmd a graphing calculator useful fur Exercises 67-74. 67. Letf(x) L

= (x 2 -

9)/(x + 3).

b. Support your conclusions in part
L

-

2)/(x - V2).

Make a table of the values ofg at the points x

2x - 3)/(x 2

-

4x

+ 3).

a. Make a table of the values of h at x = 2.9,2.99,2.999, aod so on. Thea estimate lim,..., h(x). What estimate do you arrive at if you evaluate h at x = 3.1,3.01, 3.001, ... instead? b. Support your conclusions in part
71. Letf(x) = (x 2

1)/( Ixl

-

I). a. Make tables of the values off at values ofx that -

approach Xo = -I from above and below. Then estimate lim,..._1 f(x).

b. Support your conclusion in part
72. LetF(x)

= (x 2 + 3x + 2)/(2 -Ixl).

a. Make tables of values ofF at values ofx that

h. Support your conclusion in part
a. Make a table of the values of g at values of 0 that approach 00 = 0 from above aod below. Then estimate limo~o g(O). b. Support your conclusion in part
Make a table of the values off at the points x = -3.1, -3.01, -3.001, and so on as far as your calculator can go. Then estimate lim,~_, f(x). What eslimate do you arrive at if youevaluatefatx = -2.9, -2.99, -2.999, ... instead?

68. Letg(x) = (x 2

-

approach Xo = -2 from above and below. Then estimate lim,...-2 F(x).

lim 1 - cosx?

x2

c. Find lim,...-6 G(x) algebraically.

c. Find lim,...-1 f(x) algebraically.

hold for values of x close to zero. (They do, as you will see in Section 10.9.) What, if anything, does this tell you about

x-a

Support your conclusions in part
70. Leth(x) = (x 2

'"

1- £

b.

= -6.1, -6.01,

x-+ -6.

g(x) '" 2cosxforallx,fmdlim,~og(x). It can be shown that the inequalities

64. If2 - x 65.

2

V2.

c. Find lim,...y'2 g(x) algebraically. 69. Let G(x) = (x + 6)/(x 2 + 4x - 12). a. Make a table of the values of G at x = -5.9, -5.99, -5.999, and so on. Then estimate lim,~.... G(x). What estimate do

= -2

62. f(x) =~,

x = 7

Xo = Yz and using Zoom and Trace to estimate y-values on the graph as x ....

= 1.4, 1.41,

1.414, and so on through successive decimal approximations ofYz. Estimate lim,...y'2 g(x).

00

=

74. Let G(I)

O.

= (I

- cos 1)/1 2 .

a. Make tables of values ofG at values oflthat approach 10 = 0 from above and below. Then estimate lim,...o G(t). h. Support your conclusion in part
Theory and Examples 2 2 75. If x' '" f(x) '" x for x in [-I, I] aod x '" f(x) '" x' for x < -1 and x > 1, at what points c do you automatically know lim~, f(x)? What can you say about the value of the limit at these points?

57

2.3 The Precise Definition of a Limit

76. Supposethatg(x):5 l(x):5 h(x)fora1lx "" 2 and suppose that lim g(x)

%-2

~

lim h(x)

~

%-2

Can we conclude anything about the values of I, g, and h at x ~ 2? Could 1(2) ~ O? Could lim,_21(x) ~ O? Give reasons for your answers. .

77.lfhm

%-4

~

b. ConIrrm your estimate in part (a) with a proof.

Graph h(x) ~ x 2 cos (l/x 3 ) to estimate lim,_o h(x) , zooming in on the origin as necessary. b. ConIrrm your estimate in part (a) with a proof.

COMPUTER EXPLORATIONS Graphical Estimates of Limits

I ,rmd

a. lim I(x)

b. lim

%--2

79. a. Iflim

%--2

I(x) - 5

%-2

.

b. If lim

X -

2

I(x) - 5

%-2

X -

I(x)

80. Iflun2

x-o x

2

I~)

~

3,rmd lim I(x).

~

.. 4, fmd lun I(x).

%-2

In Exercises 83-88, use a CAS to perform the following steps:

•. Plot the function near the point Xo being approached. b. From your plot guess the value of the limit. " x3_X2-5x-3 83. lim x' - 16 84. lun 2 %-2 x - 2 ,--I (x + I)

%-2

85. lim ~

x-o

I,rmd

\l"l+x x

I

86. lim

b.

x-o

X-O

x

2 -

9

,-3 W+7-4

87. lim I - .coox

a. lim I(x)

2.3

Graph g(x) ~ x sin (i/x) to estimate lim,_Og(x), zooming in on the origin as necessary.

D 82.•.

I(x) - 5 . 2 ~ I,rmd lim I(x). x x-4

78. If lim I(x) %--2 x 2

.

D 81.•.

-5.

88. x~ 3

XSInX

z,,2 3 cosX'

The Precise Definition of a Limit We now turn our attention to the precise definition of a limit. We replace vague phrases like "gels arbitrarily close to" in the infonnal definition with specific conditions that can be applied to any particular example. With a precise defmition, we can prove the limit properties given in the preceding section and establish many important limits. To show that the limit of f(x) as x --+ Xo equals the number L, we need to show that the gap between f(x) and L can be made "as small as we choose" if x is kept "close enougb" to Xo. Let us see what this would require if we specified the size of the gap betweenf(x) andL.

EXAMPLE 1 Consider the functiony ~ z" - I near Xo = 4. Intuitively it appears that y is close to 7 when x is close to 4, so limx_4 (z" - I) ~ 7. However, how close to Xo = 4 does x have to be so that y = z" - I differs from 7 by, say, less than 2 units?

y

Solution We are asked: For what values of x is Iy flISt express Iy - 71 in terms of x: To satisfy { 9 this 7

Iy -

5 ~~~~~--~~

Lower bound:

y~5

71 =

I(z" - I) -

Iz" -

2? To fmd the answer we

81·

The question then becomes: what values of x satisfy the inequality Iz" fmd out, we solve the inequality:

12x - 81 <

81 <

2? To

2

-2 < 2x - 8 < 2 6 < 2x < 10 3<x<5 -I < x - 4 < I.

---,..Restrict

to this

FIGURE 2.15 Keeping x within I unit of ~ 4 will keep y within 2 units of Yo ~ 7 (Example I).

71 =

71 <

Xo

Keeping x within I unit of Xo

~

4 will keep y within 2 units of Yo

~

7 (Figure 2.15).

•

58

Chapter 2: Limits and Continuity In the previous example we determined how close x must be to a particular value Xo to ensure that the outputs f(x) of some function lie within a prescribed interval about a limit value L. To show that the limit of f(x) as x --+ Xo actually equals L, we must be able to show that the gap between f(x) and L can be made less than any prescribed error, no matter how small, by holding x close enough to Xo.

y

L

1

+ [(j

flO)

L

f(x) lies inhere

1

Definition of Limit

L - [(j

for all x =F Xo inhere

,-------A------

~~ x 0

-'0- B

-'0

x

-'o+B

FIGURE 2.16 How should we derme B > 0 so that keeping x within the interval (xo - B, Xo + B) will keep f(x) within the interval

(L - 110, L+ 110)?

y

L+E L

) f(x)

~x~!:s

Suppose we are watching the values of a function f(x) as x approaches Xo (without taking on the value of Xo itself). Certainly we want to be able to say that f(x) stays within onetenth of a unit from L as soon as x stays within some distance Il of Xo (Figure 2.16). But that in itself is not enough, because as x continues on its course toward Xo, what is to prevent f(x) fromjittering about within the interval from L - (1/10) to L + (1/10) without tending toward L? We can be told that the error can be no more than 1/100 or 1/1000 or 1/100,000. Each time, we find a new Il-interval about Xo so that keeping x within that interval satisfies the new error tolerance. And each time the possibility exists that f(x) jitters away from L at some stage. The figures on the next page illustrate the problem. You can think of this as a quarrel between a skeptic and a scholar. The skeptic presents .-challenges to prove that the limit does not exist or, more precisely, that there is room for doubt. The scholar answers every challenge with a Il-interval around Xo that keeps the function values within • of L. How do we stop this seemingly endless series of challenges and responses? By proving that for every error tolerance E that the challenger can produce, we can fmd, calculate, or conjure a matching distance Il that keeps x "close enough" to Xo to keep f(x) within that tolerance of L (Figure 2.17). This leads us to the precise definition ofa limit.

L-E

for all x

'* Xo

inhere

DEFINITION Let f(x) be defined on an open interval about Xo, except possibly at Xo itself. We say that the limit of f(x) as x approaches Xo is the number L, and write

~

lim f(x) = L,

~~ x

~r------t----~--~--~x

o

FIGURE 2.17 The relation of Band. in the dermition of limit

X-Xo

if, for every number. for all x,

> 0, there exists a corresponding number Il > 0 such that

0< Ix - xol <

Il

=>

If(x) - LI

<

E.

One way to think about the definition is to suppose we are machining a generator shaft to a close tolerance. We may try for diameter L, but since nothing is perfect, we must be satisfied with a diameter f(x) somewhere between L - E and L + E. The Il is the measure of how accurate our control setting for x must be to guarantee this degree of accuracy in the diameter of the shaft. Notice that as the tolerance for error becomes stricter, we may have to adjust Il. That is, the value of Il, how tight our control setting must be, depends on the value of., the error tolerance.

Examples: Testing the Definition The formal definition of limit does not tell how to find the limit of a function, but it enables us to verify that a suspected limit is correct. The following examples show how the definition can be used to verify limit statements for specific functions. However, the real purpose of the definition is not to do calculations like this, but rather to prove general theorems so that the calcnlation of specific limits can be simplified.

2.3

y

L

y

y

1 + j]j f-- - - ---f- -

L

1 + j]j f-- - - -+ L

L

y

1

+ 100 f-- - - --/-- -

L

o

-~~~/~-~-+\-~x

o

Xo

The challenge:

Makelf(x) -LI <

~

E

Xo

----~------~------->x

o

Xo - al/ IO Xo + al/1O Response: x - xI) I < 8 1110 (a number)

ill

Xo

----~----~~--~--->x

o

/

xI) -

I

8 11100

Xo

\

Xo + 8 11100

Response:

New challenge:

Makelf(x) -LI <

y

L

1

+ 100 f-- - - --/-L

L

-~~---~---->x

59

The Precise Definition of a Limit

E

~ 1~

Ix - xol < 8 11100

y

1

+ 1000

L

-- f-----~-L

1

+ 1000 -- f---------~---

L

o New challenge: E =

Response:

1

Ix-xol <8 111000

1000

y

y

y

1

L

+ 100.000

i;~-= --=--~-~=

L

1/

+ E~f=c=-=-= --=-=-~-rF===

L-E

L -100.000 ------~f-----~~--_>x

o

------~r-----~~--->x

o

Xo

New challenge:

Response:

1 E =

o

Xo

New challenge:

Ix - Xo I < 8 11100,000

100,000

EXAMPLE 2

E= ...

Show that lim (5x - 3) = 2.

x~l

Solutton Set Xo = I ,/(x) = 5x - 3, and L = 2 in the definition oflimit. For any given > 0, we have to find a suitable [; > 0 so that if x '" 1 and x is within distance [; of Xo = 1, that is, whenever E

o < Ix - 11 < [;, it is true that f(x) is within distance E of L = 2, so

If(x) -

21 <

E.

60

Chapter 2: Limits and Continuity We fmd Il by working backward from the .·inequality:

y

1(5x - 3) - 21 ~ 15x - 51 < •

2+. ~-----------.~----

51x - 11 < • 2

Ix - 11

2-. ~-------J~--~----

< ./5.

Thus, we can take Il ~ ./5 (Figure 2.18). If 0 < Ix - 11 < Il ~ ./5, then 1(5x - 3) - 21 ~ 15x - 51 ~ 51x - 11 < 5(./5) ~ " which proves that limx~ 1(5x - 3) ~ 2. The value of Il ~ ./5 is not the only value that will make 0 < Ix - 11 < Il imply 15x - 51 < •. Any smaller positive Il will do as well. The definition does not ask for a ''best'' positive Il, just one that will work. •

EXAMPLE 3

NOI'TOSCALE


If f(x) ~ 5x - 3, then o < Ix - II < 0/5 guaranrees that If(x) - 21 < 0 (Example 2). FIGURE 2.18

~

Prove the following results presented graphically in Section 2.2. Xo

(b) lim k X-Xo

implies

Ix - xol < Il

Ix - xol < •.

The implication will hold if Il equals. or any smaller positive number (Figure 2.19). This proves that limx~xo x ~ Xo.

y

+ f!

(b) Let. > 0 be given. We must find Il > 0 such that for all x

xo+a ~--------~~-----­

o<

Xo xo-~ ~----~~r+--------

Xo -


k

Solution 0 be given. We must find Il > 0 such that for all x

o<

Xo

~

f!

Ix - xol < Il

implies

Ik - kl < •.

Since k - k ~ 0, we can use any positive number for Il and the implication will hold (Figure 2.20). This proves that limx~x, k ~ k. •

Finding Deltas Algebraically for Given Epsilons FIGURE 2.19 For the functioo f(x) ~ x, wefmdthatO < Ix -xol < 8 will guaranree If(x) - Xo I < 0 whenever 8 ,;; 0 (Example 3a).

In Examples 2 and 3, the interval of values aboutxo for which If(x) - LI was less than. was symmetric about Xo and we could tske Il to be half the length of that interval. When such symmetry is absent, as it usually is, we can tske Il to be the distance from Xo to the interval's nearer endpoint.

EXAMPLE 4 For the limit lim_, ~ ~ 2, fmd all> 0 that works for • ~ 1. That is, find all> 0 such that for all x

1~-21 < 1.

0< Ix - 51 < Il

y k+. ~----------~-----­

k~----r-r-r------­ k-' ~----4--7~~------

Solution

We organize the search into two steps, as discussed below.

1. Solve the inequality I~ - 21 < 1 to find an interval containing Xo = 5 on which the inequality holds for all x "" Xo.

1~-21<1 -1 <~ -2< 1

o Forth. functioof(x) we fmd that If(x) - k I <
1<~<3 ~

k,

1<x-l<9 2

< x < 10

The Precise Definition of a Limit

2.3

3

The inequality holds for all x in the open interval (2, 10), so it holds for all x oF 5 in this interval as well.

3

~~ I I~ ! ~~--~~ ! ~1~ '~-7.~'X

2

5

8

61

10

FIGURE 2.21 An open interval of

Find a value of Il > 0 to place the centered interval 5 - Il < x < 5 + Il (centered at Xo = 5) inside the interval (2, 10). The distance from 5 to the nearer endpoint of

radius 3 about Xo = 5 will lie inside the open interval (2, 10).

(2, 10) is 3 (Figure 2.21). lfwe take Il = 3 or any smaller positive number, then the inequality 0 < Ix-51 < Il will automatically place x between 2 and 10 to make

2.

I Vx"=l - 21 < I (Figure 2.22): y

•

1Vx"=l- 21 < I.

0< Ix - 51 < 3

3 ~~------------~~

How to Find Algebraically a 6 for a Given /, L, Xc, and E > 0 The process of f"mding all> 0 such that for all x 0< Ix - xol < Il

I/(x) -

LI <

E

can be accomplished in two steps.

1. Solve the inequality I/(x) - L I < E to find an open interval (a, b) containing Xo on which the inequality holds for all x oF Xo. 2. Find a value of Il > 0 that places the open interval (xo - Il, Xo + Il) centered atxo inside the interval (a, b). The inequality I/(x) - L I < E will hold for all x oF Xo in this Il-interval.

NOT TO SCALE

FIGURE 2.22 The function and intervals in Example 4.

EXAMPLE 5

Prove that limx~2 /(x) = 4 if

/(x) Solutton

{x2,

=

I,

Our task is to show that given E > 0 there exists all> 0 such that for all x 0< Ix - 21 < Il

y

I/(x) - 41

<

E.

1. Solve the inequality I/(x) - 41 < E to find an open interval containing Xo = 2 on which the inequality holds for all x oF Xo.

4+' r---'-~--------

For x oF Xo = 2, we have /(x) = x 2, and the inequality to solve is Ix 2 - 41 < E: Ix2

4

-E

-

<x

41 2

-

<

E

4

<

E

4-E<x 2 <4+E

I I

I I

~

I I

~<x<~.

~ :<2,1)

FIGURE 2.23 An interval containing x = 2 so that the function in Example 5 satisfies I/(x) - 41 < E.

The inequality I/(x) - 41 < ~) (Figure 2.23).

2.

E


~

Assumes E < 4; see below. An open interval about Xo that solves the inequality

=

2

holds for all x oF 2 in the open interval (~,

Find a value of Il > 0 that places the centered interval (2 - Il, 2

+

Il) inside the in-

terval (~, ~). Take Il to be the distance from Xo = 2 to the nearer endpoint of (~, ~). In other words, take Il = min {2 - ~, ~ - 2}, the minimum (the

62

Chapter 2: Limits and Continuity

smaller) of the two numbers 2 - ~ and ~ - 2. If Il bas this or any smaller positive value, the ioequality 0 < Ix - 21 < Il will automatically place x between ~and ~tomake If(x) - 41 < E.Forallx,

0< Ix - 21 < Il

41 <

If(x) -

E.

This completes the prooffor E < 4. If E ;" 4, then we take Il to be the distance from Xo = 2 to the nearer endpoiot of the ioterval (0, ~). In other words, take Il = mio {2, ~ - 2}. (See Figure 2.23.) •

Using the Definition to Prove Theorems We do not usually rely on the formal defmition oflimit to verify specific limits such as those io the preceding examples. Rather we appeal to general theorems about limits, io particular the theorems of Section 2.2. The definition is used to prove these theorems (Appendix 4). As an example, we prove part I of Theorem I, the Sum Rule.

EXAMPLE 6

Giventhatlimx~cf(x) = Landlimx~cg(x) =

lim (j(x)

x-c Solution

+ g(x))

=

M,provethat

L + M.

Let E > 0 be given. We want to fiod a positive number Il such that for all x

0< Ix - cl < Il

If(x) + g(x) - (L + M)I <

E.

Regroupiog terms, we get

If(x) + g(x) - (L + Mll

= :5

I(j(x)

- L) + (g(x) -

Mll

If(x) - L I + Ig(x) - MI·

Sioce limx_ c f(x) = L, there exists a number Il,

0< Ix - cl < Il,

> 0 such that for all x If(x) - LI <

Similarly, sioce lim.-c g(x) = M, there exists a number 112

o<

Ix -

cl <

Triangle Inequality:

la + hi" lal + Ihl

> 0 such that for all x

Ig(x) - MI

112

E/2.

< E/2.

Letll = mio {Il" 1l2},the smaller of Il, andll,. If 0 < Ix - cl < Ilthen Ix so If(x) - LI < E/2, and Ix - cl < 1l2' so Ig(x) - MI < E/2. Therefore

If(x) + g(x) - (L + Mll < ~ + ~ Thisshowsthatlim.-c(j(x)

+ g(x»

=

L

cl <

Il"

= E.

+ M.

•

Next we prove Theorem 5 of Section 2.2.

EXAMPLE 7 Given that lim.-c!(x) = Land limx_ c g(x) = M, and that f(x) :5 g(x) for all x io an open ioterval contaioiog c (except possibly c itself), prove that L :5 M. Solution L

>

We use the method of proof by contradiction. Suppose, on the contrary, that

M. Then by the limit of a difference property io Theorem I,

lim (g(x) - f(x)) = M - L. x~c

2.3

>

Therefore, for any E

I(g(x)

0, there exists Il

>

- f(x» - (M - L)I <

63

The Precise Definition of a Limit

0 such that whenever

E

0

< Ix - Cl <

Il.

Since L - M > 0 by hypothesis, we take E = L - M in particular and we have a number Il > 0 such that

I(g(x)

- f(x» - (M - L)I < L - M

whenever

0

< Ix - cl <

Il.

whenever

0

< Ix - cl <

Il

Since a ,;; Ia Ifor any number a, we have

(g(x) - f(x» - (M - L) < L - M which simplifies to

g(x) < f(x)

whenever

0

< Ix - cl <

But this contradicts f(x) ,;; g(x). Thus the inequality L L ,;; M.

>

Il.

M must be false. Therefore •

Exercises 2.3 Centering Intervals About a Perint In Exercises l--{j, sketch the interval (a, b) 00 the x·axis with the point Xo inside. Then find a value of ~ > 0 such that for all x,O < Ix - Xol < ~ =* a < x < b. 1. a = I,

b = 7,

Xo = 5

2. a

b

Xo = 2

=

1,

=

7,

9.

10.

= Yx Xo = 1

y

b = -1/2, Xo = -3

4. a = -7/2,

b = -1/2, Xo = -3/2

5. a = 4/9, b = 4/7, Xo = 1/2 6. a = 2.7591, b = 3.2391, Xo = 3

f(x) =

0

2VX+l

Xo = 3 L=4 E = 0.2

L=I E

3. a = -7/2,

y

f(x)

=!

4

x

9 [6

X

Finding Deltas Graphically In Exercises 7-14, use the graphs to fmd a ~ > 0 such that for all x 0< Ix - xol < ~ =* I/(x) - LI < E.

NOfTOSCALE

11.

12.

y

Y

Y

8.

7.

f(x)=4-x 2 Xo =-1 L=3 • = 0.25

x2

f(x) = Xo = 2 L=4 E = 1

3

f(x) = -2x+ 3

Xo =-3 L = 7.5

6.2

• = 0.15

6

y=4-x 2

5

5.8

7.35

o

0

I I I

-i--r-

7.65 7.5 ~-.f--f~------~x

--1--I

4

--

I I I

/

v3

NOT TO SCALE NOf TO SCALE

2

"

v5

X

2.75

I I I I I I I I I I I I

v3 -T NOI' TO SCALB

3

NOI'TOSCALE

0

X

64

Chapter 2: Limits and Continuity

13.

14. y

Using the Formal Definition Each of Exercises 31-36 gives a function !(x), a point XO, and a positive number •. Find L = lim !(x). Then find a number 8 > 0 such

y

f(x)~ ~ -'0

~

-1

L~2 E

that for all x

-'0 ~! 2

2.01

= 0.5

X~Xo

~~

f(x)

-x

0< Ix - xol < 8 31. !(x) = 3 - 2x,

= 0.01

E

Xo = 3,

32. !(x) = -3x - 2, 2.5 I I --i-I I I I I I I I I I I I I

-9

x2 - 4 33. !(x) = - - 2 ' 2 34. !(x) = x :

~ !2

0

X

0

36. !(x)

"1 1.99

2.01

~

x~4

16. !(x)

~

L

L

2x - 2,

17. !(x) ~ ~, 18. !(x) ~

Yx,

~

L ~ 1/2,

L

I/x,

X'o

L ~ I,

xo ~

1/4,

1/4,

Xo

~

v'3,

Xo ~

23. !(x) ~ x 2,

L ~ 4,

Xo ~ -2,

L

I/x,

25. !(x) ~ x 2 - 5, 26. !(x)

~

27. !(x)

~ nIX,

28. !(x)

~ nIX,

+ b,

30. !(x) = nIX E ~ 0.05

+ b,

-I,

L

~

L

=

m

>

0,

m

>

0,

120/x,

29. !(x) ~ nIX Xo = 1/2,

~

• = c

m

Xo 11,

~ 0.1

E

~

-I,

X'o

=

E

L

=

2m,

X'o = 2,

L

=

3m,

X'o

~

(m/2)

0,

L

~ I

48.

E

24,

=

~

x~o

0.1 I

I

3,

E

~

L = m

+ b,

+ b,

0.03

e=c>O

>0 m > 0,

4

if !(x)

~

2 {x , I,

X #-2 x =-2

' I 441 • llJlr. "2

I

x-v3 x

1~ !(x) ~ 0

49. lim

~

=

# I x = I X

46.lim x-I

~

I

-3

x2 - 1 =2 x-I x<1 xO?::l

~ 0.5

4,

lim~ ~

x~l

x~o

2

0.05

E ~

%--2

~

Xo = I,

xsin~ ~

if !(x) 0

2

40. lim~=2

if !(x) = {x2, 2,

45. lim x -9=_6 %--3 x + 3

~ 0.1

E

42. lim !(x)

~

x~3

~ I

E

::to

5,

>

E

E

0.4

38. lim (3x - 7)

5

x~l

0.02

~ 0.1

E ~

4,

L ~ 3,

~

~

41. lim !(x) = I

43.

E

Xo ~ 23,

22. !(x) ~ x 2,

24. !(x)

E ~

Xo ~ 10,

L ~ 4, ~

om

-2,

=

Xo ~ 0,

L ~ 3,

20. !(x) ~ ~,

E ~

X'o = 4,

-6,

=

19. !(x) ~ ~,

21. !(x)

5,

=

• = 0.05 • = 0.5

E ~

x~.

Eacb of Exercises 15-30 gives a function !(x) aod numbers L, XO, aod E > O. In each case, f'md an open interval about Xo on which the inequality I!(x) - L I < E holds. Then give a value for 8 > 0 such that forallxsatisfyingO < Ix - xol < 8 the inequality I!(x) - LI <. holds.

+ I,

• = 0.05

Xo = -3,

39. lim Vx"=5 = 2

finding Deltas Algebraically

x

• = 0.03

Xo = -5,

Xo ~ 2,

4/x,

37. lim (9 - x)

~

<

Prove the limit statements in Exercises 37-50.

NOT TO SCALE

15. !(x)

Xo = -I,

!'s+ 5,

35. !(x) = ~,

16 -25

LI

• = 0.02

Xo = 2,

x-

1.5

I I I I I I I I I

-1

16

1.99

2

------1--+----

I!(x) -

L~2

~ {~;,

x
E.

2.3

65

The Precise Definition of a Limit

When Is a Number L Not the Limit of f (x) as x .... xo? Showing L is not a limit We can prove that lim_.. f(x) # L hy providing an E > 0 such that no possible B > 0 satisfies the condition for all x,

0

<

If(x) - LI <

< B

Ix -xol

We accomplish this for our candidate 0 there exists a value of x such that

E

E.

hy showing that for each

a>

0< Ix - xol < B

If(x) - LI '"

and

E.

y

L

+ E f_,---

--""L.--,-- - - -

L L -

Theory and Examples 51. Defme what it means to say that limg(x) x~o

= k.

52. Prove that lim f(x) = L if and ooly if lim f(h x-c

53. A wrong stntement about limits

h-O

+ c)

= L.

----~f_~~-L-~~----~x

oxo -a\xo xo 8

Show hy example that the fol-

lowing statement is wrong.

a value of x for which

The number L is the limit of f(x) as x approaches Xo if f(x) gets closer toL asx approaches Xo.

0<

Explaio why the function in your example does not have the given value of L as a limit as x ~:to. 54. Aootber wrong statement about limits the following starement is wrong.

E f_+--~--+----

f(x) +-+=«'

Show hy example that

Ix - xol < ~ and If(x) x< I

57. Let f(x) = {x' I x + , x>1. y

The number L is the limit of f(x) as x approaches Xo if, given any E > 0, there exists a value ofx for which If(x) - LI <

E.

Explaio why the function in your example does not have the given value of L as a limit as x ~:to.

D 55.

LI "'.

2 y

Grinding engine cylinders Before contracting to gtind engine cylinders to a cross-sectional area of9 in2 , you need to know how much deviation from the ideal cytinder diameter of Xo = 3.385 in. you can allow and still have the area come within in' of the required 9 in'. To find out, you let A = '1T(xI2)' and look for the interval in which you must hold x to make IA - 91:5 0.01. What interval do you fmd?

= f(x)

om

56. Manufacturiog electrical resistors Ohm's law for electrical circuits like the one shown in the accompanying figure states that V = RI. In this equation, V is a constant voltage, I is the current in amperes, and R is the resistance in ohms. Your frrm has been asked to supply the resistors for a circuit in which V will be 120 volts and I is to be 5 ± 0.1 amp. In what interval does R have to lie for I to be within 0.1 amp of the value 10 = 5?

a. Let E = 1/2. Show that no possible B > 0 satisfies the following condition: For all x,

0 < Ix - 11< B

+

R

If(x) - 21 < 1/2.

That is, for each B > 0 show that there is a value of x such that

0< Ix -

v

~

11

< B

and

If(x) - 21 '" 1/ 2 .

This will show that limx~ d(x) # 2.

b. Show that lim_l f(x) # 1. c. Show that1im_ 1 f(x) # 1.5.

66

Chapter 2: Limits and Continuity

58. Leth(x)

~ {

X',

X

<

3, 2,

x

~

x>

y

2 2 2.

•

2

y

4

3 ---J~~------~--~x

o

2

COMPUTER EXPLORATIONS In Exercises 61--{j6, you will further explore rmding deltas graphically. Use a CAS to perform the followiug steps:

o

a. Plot the function y ~ I(x) near the point Xo being approached. Show that L

b. Guess the value of the limit L aod then evaluate the limit symbolically to see if you guessed correctly.

lim h(x) oF 4

x~2

b. lim h(x) oF 3

c. Using the value.

c. lim h(x) oF 2

+ • together with the function 1 near Xo. d. From your graph in part (c), estimate a 8 > 0 such that for all x

x~2

aod.Y2

x~2

59. For the function graphed here, explain why L

0<

lim I(x) oF 4

x~3 x~3

c. lim I(x) oF 3 x~3

y

~--------~------~x

3

For the function graphed here, show that limx~-1 g(x) oF 2.

b. Does limx~-1 g(x) appear to exist? If so, what is the value of the limit? If not, why not?

2.4

Ix - xol < 8

5>;3

+ 9x'

2x

+]x

62. I(x) ~, 63. I(x) ~

L

~

L - •

~

I/(x) - LI < •.

~

0.1,0.05, aod

x' -

•

3

60.

0.2, graph the haoding lines Yl

e. Repeat parts (c) aod (d) successively for. 0.001. 81 61. I(x) ~ x 3' Xo ~ 3

4.8

o

~

L

Thst your estimate by plotting I, Yl, and Y' over the interval 0< Ix - xol < 8. For your viewiog window use Xo - 28 s x '" Xo + 28 andL - 2. '" y '" L + 2 •. If any function values lie outside the interval [L - " L + 'J, your choice of 8 was too large. Try again with • smaller estimate.

b. lim I(x) oF 4.8

4

~

64. I(x)

~

sin a

3X'

Xo ~ 0

"

Xo ~ 0

x(1 - cou) . , Xo x - smx

~-I 65. I(x) ~ x I '

~

0

Xo ~ I

_ 3x' - (7x ( ) 66·lx-

+

l)Vx + 5

x- I

,xo~1

One-Sided Limits In this section we extend the limit concept to one-sided limits, which are limits as x approaches the number c from the left-hand side (where x < c) or the right-hand side

(x

>

c) only.

One-Sided Limits To have a limit L as x approaches c, a function f must be deImed on both sides of c and its values f(x) must approach L as x approaches c from either side. Because of this, ordinary limits are called two-sided.

2.4

y

J

x

F~

x

0

One-Sided Limits

67

If j fails to have a tw(}-sided limit at c, it may still have a one-sided limit, that is, a limit if the approach is only from one side. If the approach is from the right, the limit is a right-hand limit. From the left, it is a left-hand limit. The function j(x} = x/lxl (Figure 2.24) has limit I as x approaches 0 from the right, and limit - I as x approaches 0 from the left. Since these one-sided limit values are not the same, there is no single number that j(x) approaches as x approaches O. So j(x) does not have a (two-sided) limit at O. Intuitively, if j(x) is defmed on an interval (c, b), where c < b, and approaches arbitrarily close to L as x approaches c from within that interval, then j has right-hand limit L at c. We write lim j(x} = L.

x-c+

FIGURE 2.24 Different right-hand and left-hand limits at tire origin.

The symbol "x --> c+" means that we consider only values of x greater than c. Similarly, if j(x) is defined on an interval (a, c), where a < c and approaches arbitrarily close to M as x approaches c from within that interval, then j has left-hand limit M at c. We write lim j(x) = M.

x-c-

The symbol "x --> c-" means that we consider only x values less than c. These informal defmitions of one-sided limits are illustrated in Figure 2.25. For the functionj(x) = x/lxl in Figure 2.24 we have lim j(x) = I

and

x-o+

lim j(x) = -I.

x-o-

y

y

,.f(x)

L

c_

0 (a)

~

x

x

lim f(x)

~C'

~

M

f(x) x _

0

L

(b)

lim f(x)

c ~

M

~,.-

FIGURE 2.25
EXAMPLE 1 The domain of j(x) = Figure 2.26. We have

v'4=7 is [-2, 2]; its graph is the semicircle in and

The function does not have a left-hand limit at x = -2 or a right-hand limit at x = 2. It does not have ordinary two-sided limits at either - 2 or 2. • -O~----~------~~X

-2

0

FIGURE 2.26 lim

x--2+

2

lim

x-2-

v'4=7 =

v'4 -

x2

= 0 and

0 (Example 1).

One-sided limits have all the properties listed in Theorem 1 in Section 2.2. The righthand limit of the sum of two functions is the sum of their right-hand limits, and so on. The theorems for limits of polynomials and mtional functions hold with one-sided limits, as does the Sandwich Theorem and Theorem 5. One-sided limits are related to limits in the following way. THEOREM 6 A function j(x) has a limit asx approaches c if and only if it has left-hand and right-hand limits there and these one-sided limits are equal: lim j(x} = L x~c

limj(x) = L x~c

and

lim j(x) = L.

x-c+

68

Chapter 2: Limits and Continuity

EXAMPLE 2

y

Atx

=

0:

Atx

=

I:

For the function graphed in Figure 2.27, limx~o+

f(x) = 1, limx~o- f(x) and limx~o f(x) do not exist. The function is not defined to the left of x = O. limx~l- f(x) = 0 even though f(l) = 1, limx~l+ f(x) = 1, limx~l f(x) does not exist. The right- and left-hand limits are not

2

•

~--~--~--~--~--->x

o

2

3

4

equal.

FIGURE 2.27 Graph of the function in EXlIlIlpie 2.

y

Atx

= 2:

Atx Atx

= =

limx~'-

f(x) limx~2+ f(x)

1, = I, limx~2f(x) = I even though f(2) = 2. limx~r f(x) = limx~3+ f(x) = lim~3 f(x) = f(3) = 2. limx~4- f(x) = I even though f( 4) oF I, limx~4+ f(x) and limx~4 f(x) do not exist. The function is not defined to the right of x = 4.

3: 4:

=

•

At every other point c in [0, 4], f(x) has limit f(c). L+<

M)

L

Precise Definitions of One-Sided Limits

f(x) lies inhere

The formal definition of the limit in Section 2.3 is readily modified for one-sided limits.

L-< for all x =F Xo in here

DEFINmONS

B x

•

0

0

"0

)

"0

We say that f(x) has right-hand limit L at xo, and write lim f(x) = L

>x

+B

(see Figure 2.28)

X--i'Xo+

if for every number E > 0 there exists a corresponding number Il > 0 such that for all x

FIGURE 2.28 Intervals associated with the defmition of right-hand limit.

If(x) - LI <

xo<x<xo+1l

E.

We say that f has left-hand limit L at xo, and write y

lim f(x) = L

X--i'XO-

(see Figure 2.29)

if for every number E > 0 there exists a corresponding number Il > 0 such that for all x

If(x) - LI < L+ <

f(x)

)

f(x) lies

inhere

L

E.

EXAMPLE 3

Prove that

L-<

lim

'*

for all x Xo inhere ~

B

x--+o+

Let E > 0 be given. Here Xo = 0 and L = 0, so we want to find all> 0 such that for all x

Solution

~------f--* x-----v--~x

O<x<1l

o

FIGURE 2.29 Intervals associated with the defmition ofleft-hand limit.

Vx = o.

or

O<x<1l

Vx< E.

2.4

One-Sided Limits

69

Squaring both sides of this last inequality gives

y

. r-- -----;;.....

~----

0< x < Il.

if

Vi

f(x) ~

If we choose Il = e 2 we have

O<x
Yx< e,

or

0<x<e2 FIGURE 2.30

lim

x-o+

Vi ~ 0 in Example 3.

IYx-OI <e.

=>

According to the dermition, this shows that limx~O+ Yx = 0 (Figure 2.30).

•

The functions examined so far have had some kind of limit at each point of interest. In general, that need not be the case.

EXAMPLE 4 Show that y = sin (l/x) has no limit as x approaches zero from either side (Figure 2.31). y

Ii

-1

I'I

v

FIGURE 2.31 Thefunctiony ~ sin (l/x) has neither arighthand nor a left-hand limit as x approaches zero (Example 4). The graph here omits values very near the y-ws.

Solution

As x approaches zero, its reciprocal, I/x, grows without hound and the values of sin (I/x) cycle repeatedly from -I to 1. There is no single number L that the function's values stay increasingly close to as x approaches zero. This is true even if we restrict x to positive values or to negative values. The function has neither a right-hand limit nor a left• hand limit atx = O.

Limits Involving (sin 6)/6 A central fact about (sin 0)/0 is that in radian measure its limit as 0 ..... 0 is 1. We can see this in Figure 2.32 and confirm it algebraically using the Sandwich Theorem. You will see the importance of this limit in Section 3.5, where instantaneous rates of change of the trigonometric functions are studied. y y ~ sin 6 (radians) 6

NOfTOSCAlB

FIGURE 2.32 The graph ofj(8) ~ (sin8)/8 suggests that the rightand left-hand limits as 8 approaches 0 are both 1.

70

Chapter 2: Limits and Continuity

y

THEOREM 7 T

(II in radians)

(1)

p

tane 1

(J

Proof The plan is to show that the right-hand and left-hand limits are both 1. Then we will know that the two-sided limit is I as well. To show that the right-hand limit is I, we begin with positive values ofiliess than TT/2 (Figure 2.33). Notice that

cos 8

~~~~~ __-U~ ____,~~x

o

Q

Area !J.OAP

A(I,O)

1

<

area sector OAP

<

area !J.OAT.

We can express these areas in terms of II as follows:

FIGURE 2.33 The figure for the proof of Theorem 7. By detmition, TA/OA = tan8, butOA = l,soTA = tan8.

Area !J.OAP = t base X height = t (I )(sin II) = t sin II Area sector OAP =

1 r211 = 1 (I)'II = Ii. 222

(2)

Area !J.OAT = tbase X height = t (I)(tan II) = ttanll.

I Equation (2) is where radian measure

Thus, 'll 2Is1l1 < 2III < 2I tan II .

comes in: The area of sector OAP is 8/2 only if 8 is measured in radians.

TIris last inequality goes the same way if we divide all three terms by the nwnber (1/2) sin II, which is positive since 0 < () < TT/2: I

II

I

< ~II sm < --II' cos

Taking reciprocals reverses the inequalities: sin II

I > --11- > cos II. Since lim(Ho· cos II = I (Example lib, Section 2.2), the Sandwich Theorem gives lim sin II = I. 8

8--+0+

Recall that sin II and II are both odd jUnctions (Section 1.1). Therefore, /(11) = (sin 11)/11 is an even jUnction, with a graph symmetric about the y-axis (see Figure 2.32). TIris symmetry implies that the left-hand limit at 0 exists and has the same value as the right-hand limit: lim sin II = I = lim sin II 8 8--+0+ 8 '

8--+0-

•

so limo~o (sin 11)/11 = I by Theorem 6.

EXAMPLE 5

Show that (a) lim cos h - I = 0 11-0 h

and

(b) lim sin 2x = ~.

x-a 5x

5

2.4

One·Sided Limits

71

Solution
lim cos h - 1 = lim _ 2 sin (h/2) h~O h h~O h · sin 0 . 0 = - 1un--sm O~O 0 =

-(1)(0)

=

LetO ~ h12. Eq. (I) and Example II. in Section 2.2

O.

(b) Equation (1) does not apply to the original fraction. We need a 2x in the denominator, not a 5x. We prodoce it by multiplying nmnerator and denominator by 2/5:

.-0

. sin 2x . (2/5)' sin 2x hm - - = hm -'--7::-7::C-::-5x (2/5)' 5x

.-0

= ~ lim sin2x

Now, Eq. (I) applies with

o ~ lx.

5 x-a 2x

• · d Ii tan 1 sec 21 Fm ~ 31 .

EXAMPLE 6

From the definition of tan 1 and sec 2/, we have

Solution

lim tan 1 sec 21 = 1-0

1 lim sin 1 • _1_. _1_ 3 I_oleos 1 cos 21

31

=

1

3 (1)(1)(1)

=

1

3'

Eq. (I) and Example lib in Section 2.2

•

Exercises 2.4 Finding Limits Graphically

y

1. Which of the following statemeots ahout the function y graphed here are true, aod which are false?

~

j(x)

y

••

lim j(x) = I

.%--1+

c. lim j(x) = I .%-0-

•• lim j(x) exists.

x-o

g. lim j(x) x-O

=I

i. lim j(x) ~ 0 x-I k. lim j(x) does not exist. .%--1-

b. lim j(x) = .%-0-

a

..

d. lim j(x) = lim j(x)

r.

.%-0-

lim j(x)

x-o

h. lim j(x) x-I

x-O+

c. lim j(x) = 2 e. lim j(x)

=I

g. lim j(x)

~

x-l+

x-o+

j. lim_ j(x) ~ 2 .%-2 L lim j(x) = a

b. lim j(x) docs not exist.

.-2

d. lim j(x)

x-2

=0 =I

r.

x-r

=2

lim j(x) docs not exist.

x-I

lim j(x)

x-O-

b. lim j(x) exists at every c in the opeo interval ( -I, I).

x-, x-,

L lim j(x) exists at every c in the opeo interval (I, 3).

x-2+

2. Which of the following statemeots ahout the function y graphed here are true, aod which are false?

lim j(x) = I

x--I+

~

j(x)

j.

lim _ j(x) = 0

.1'--1

k. lim j(x) docs not exist. .1'-3+

72

Chapter 2: Limits and Continuity

3. Let f(x)

=

3-X, x<2 x { 2 + I, x> 2.

6. Letg(x)

= Yxsin(I/x). y

1 y

L

Find lim,....2+ f(x) and lim,....2- f(x).

b. Does lim,....2 f(x) exist? If so, what is it? If no~ why not? c. Find lim,....4- f(x) and lim,....4+ f(x).

d. Does lim,....4 f(x) exist? Ifso, what is it? Ifno~ why not?

4. Letf(x)=j:-X'

a. Does lim,....o+ g(x) exist? If so, what is it? Ifno~ why not?

::~

b. Does lim,....o- g(x) exist? If so, what is it? Ifno~ why not? c. Does lim,....o g (x) exist? If so, what is it? If no~ why not?

x> 2.

2'

x3

7. a. Graph f(x) y

= {

x oF I x = I.

0.'

b. Find lim,....l- f(x) and lim%~I+ f(x). c. Does lim,....1 f(x) exist? If so, what is it? If no~ why not?

8. a. Graphf(x) =

{I - x

2 ,

2,

x oF

I

x=l.

b. Find lim,....l+ f(x) and lim%~I- f(x). c. Does lim,.... I f(x) exist? If so, what is it? If n~ why not?

L

Find lim,....2+ f(x), lim%~2- f(x) , and f(2).

Graph the functions in Exercises 9 and 10. Then answer these questions.

a. What are the domain and range off? b. At what points c, if any, does lim,...., f(x) exist?

b. Does lim,....2 f(x) exist? If so, what is it? If no~ why not? c. Findlim,....-l- f(x) andlim,....-l+ f(x).

c. At what pnints does only the left-hand limit exist?

d. Does lim,....-l f(x) exist? If SO, what is it? Ifnot, why not? S. Letf(x)

=

{

.

d. At what points does only the right-hand limit exist?

xsO

O,

I

Sffi:x,

x>

'\11=7; 9. f(x) =

O.

0:5x<1 I :5 X < 2 x= 2 -lsx I

I,

{ 2,

{~:

y

10. f(x) =

0,

Finding One-Sided Limits Algebraically ________~-L+_~~----~x

° -1

L

y=

o. ,,;;0 . 1 {Sffi x' x>O

II

Does lim%~o+ f(x) exist? If so, what is it? If n~ why not?

b. Does lim,....(f' f(x) exist? If so, what is it? If not, why not? c. Does lim,....o f(x) exist? If so, what is it? If no~ why not?

Find the limits in Exercises 11-18. 11.

lim %-+-0.5-

13

lim

~xX ++ 21

(_x )(2)x ++ +

14.

2

1

• %-+-2+ X

5) X

%~-(xh)(~W ~x)

15. lim Vh h-+O+

2

+ 4h + 5 h

-

Vs

2.5

16

r

. h~11J

17. a.

V6 -

y'5h' + llh + 6

41. lim

.~o

h Ix + 21 . lim (x + 3 ) - + 2

%--2+

~(x - I) 18. a. %-1+ lim

Ix + 21 + 3) x + 2

. b. ,.!."2',jx

X

~(x-I)

b. lim

Ix- II

,~l

Ix - II

Use the graph of the greatest integer function y = lx J, Figure 1.10 in Section 1.1, to help you f"md the !intits in Exercises 19 and 20. 19. a. lim lOJ 0-3+

b. lim lOJ

o-r

(}

,....4-

Using lim sin/J /J = 1 .~o

21. lim sin \120 .~o \120

22• I'1m sint-kt

. sin 3y 23. hm-y-o 4y

24. lim

25. lim tan 2x

26.

27. lim x csc 2x x-o cosSx

28. lim 6x'(cotx)(csc 2x)

29. lim x .+ x cos x x-o smxcosx

30. lim x - x x-o 2x

31. lim I ~ cosO 8-0 sm28

32. lim x - x cosx x-o 8in2 3x sin (sinh) 34. lim . h 11-0 sm

,....0

11-0-

~

(k constant )

t!t ,

+.SIDX'

35. lim sinO

36. lim s~5x x-o sm4x

37. lim 0 cos 0

38. lim sin 0 cot 20

39. lim tan 8x3x

_sin=3y'...:.co",t..c5,y40. limy-o ycot4y

8-0 sm28 .~o

x-o sm

2.5

Theory and Examples 43. Once you know lim......., j(x) and lim.......- j(x) at an interior point of the domain of j, do you then know lim,...... j(x)? Give reasons

for your answer. 44. If you know that lim,....., j(x) exists, can you find its value by calculating limx-c+ j(x)? Give reasons for your answer. 45. Suppose that j is an odd function of x. noes knowing that lim,_o' j(x) = 3 tell you anything about lim,.....o- j(x)? Give rea46. Suppose that j is an even function of x. Does knowing that lim,_r j(x) = 7 tell you anything about either lim, __r j(x) or limx_-2+ j(x)? Give reasons for your answer.

47. Given. > O,f"md an interval I = (5,5 + S),S > O,suchthatif x lies in I, then < •. What !intit is being verified and what is its value?

Vx-=s

48. Given. > O,f"md an interval I = (4 - S, 4), S > 0, such that if x lies in I, then ~ < •. What !intit is being verified and what is its value?

. h3h sm

,~o

33. lim sin (I - cos t) 1-0 1 - cost

0 cot 48

.~o sin' 0 cot' 20

0' cot 30

Formal Definitions of One-Sided Limits

Find the lintits in Exercises 21-42.

x

73

sons for your answer.

(J

b. lim(t - ltJ)

x-o

42. lim

tan 0

Continuity

.~o

Use the def"ntitions of right-hand and left-hand lintits to prove the lintit statements in Exercises 49 and 50. 49. lim -IXI = -I x-o-

X

50. lim ( x-2+ x

;1

=

I

51. Greatest integer function Find (a) lim,_.oo' l x J and (b) lim,_.oo- l x J ; then use limit definitions to verilY your f"mdings. (0) Based on your conclusions in parts (a) and (b), can you say anything about lim,........ l x J ? Give reasons for your answer.

< 00 x> .

52. One-sided1intits Letj(x) = {:';in(I/X), x

vx,

Find (a) lim,.....o' j(x) and (b) lim,_o- j(x); then use lintit definitions to verilY your f"mdings. (0) Based on your conclusions in parts (a) and (b), can you say anything about lim,~o j(x)? Give

reasons for your answer.

Continuity When we plot function values generated in a laboratory or collected in the field, we often connect the plotted points with an unbroken curve to show what the function's values are likely to have been at the times we did not measure (Figure 2.34). In doing so, we are assuming that we are working with a continuaus jUnction, so its outputs vary continuously with the inputs and do not jump from one value to another without taking on the values in between. The limit of a continuous function as x approaches c can be found simply by calculating the value of the function at c. (We found this to he true for polynomials in Theorem 2.) Intuitively, any function y = t(x) whose graph can be sketched over its domain in one continuous motion without lifting the pencil is an example of a continuous function. In this section we investigate more precisely what it means for a function to be continuous.

74

Chapter 2: Limits and Continuity

y

80

p

I Q

~9 f-

We also study the properties of continuous functions, and see that many of the function types presented in Section 1.1 are continuous.

Continuity at a Point

Q /

To understand continuity, it helps to consider a function like that in Figure 2.35, whose limits we investigated in Example 2 in the last section.

Q 1/

I-'"

..... o

5

10

Elapsed time (sec)

FIGURE 2.34 Connecting plotted points by an unbroken curve from experimental

data Q, , Q2, Q" ... for a falling object

EXAMPLE 1 Find the points at which the function 1 in Figure 2.35 is continuous and the points at which 1 is not continuous. The function 1 is continuous at every point in its domain [0,4] except at 2, and x = 4. At these points, there are breaks in the graph. Note the relationship between the limit of 1 and the value of 1 at each point of the function's domain.

Solution x = I, x

=

Points at which f is continuous:

y

2

--Xf(X) • 1

2

3

=

0,

lim I(x) = 1(0). x-o+

Atx

=

3,

AtO


lim I(x) .-3

=

1(3).

.-c

lim I(x) = I(c).

4,c '¢' 1,2,

Points at which f is not continuous:

1

0

Atx

x

4

FIGURE 2.35 The function is continuous on [0, 4] except atx = l,x = 2, and x = 4 (Example I).

Atx

=

I,

Atx

=

2,

Atx

=

4,

.-1

lim I(x) does not exist.

lim I(x) .-2 lim./(x) .-4

= I, but I '¢'

1(2).

= I, but I'¢'

1(4).

these points are not in the domain of f.

Atc<0,c>4,

•

To define continuity at a point in a function's domain, we need to derme continuity at an interior point (which involves a two-sided limit) and continuity at an endpoint (which involves a one-sided limit) (Figure 2.36).

Continuity from the right

DEFINITION Interior point: A function y

Two-sided

continuity

Continuity the left

=

I(x) is continuous at an interior point c of its

domain if lim I(x) = I(c).

:

: y=j(x)

:

I I

I I

I I

I

I

I

a

c

b

FIGURE 2.36

x-c

Endpoint: A function y )

X

Continuity at points a, b,

= I(x) is continuous at a left endpoint a or is continuous at a right endpoint b of its domain if

lim I(x) = I(a)

x-a+

or

lim.f(x)

x-b

=

I(b),

respectively.

andc.

If a function 1 is not continuous at a point c, we say that 1 is discontinuous at c and that c is a point of discontinuity of f. Note that c need not be in the domain of f. A function 1 is right-continuous (continuous from the right) at a point x = c in its domain iflim.-,+ I(x) = I(c). It is left-continuous (continuous from the left) at c if lim.-,- I(x) = I( c). Thus, a function is continuous at a left endpoint a of its domain if it

2.5

2

EXAMPLE 2 The function f(x) = ~ is continuous at every point of its domain [-2,2] (Figure 2.37), including x = -2, where f is right-continuous, and x = 2, where f is left-continuous. _

~-----;+--~-->x

0

2

FIGURE 2.37 A function that is continuous at every domain point (Example 2).

EXAMPLE 3 The unit step function U(x), graphed in Figure 2.38, is right-continuous at x = 0, but is neither left-continuous nor continuous there. It has a jump discontinuity at x=O. _

y

I+-'Y'-~--'U(::.:; x),-

We summarize continuity at a point in the form of a test.

----~R---->x

Continuity Test A function f(x) is continuous at an interior point x = c of its domain if and only if it meets the following three conditions.

o

FIGURE 2.38 A function that has a jump discontinuity at the origin (Example 3).

Y

1. 2. 3.

3

Y~

2

LxJ

f(c) exists

(c lies in the doroain of f).

limx~,

f(x) exists

(f has a limit as x --> c).

limx~,

f(x) = f( c)

(the limit equals the function value).

For one-sided continuity and continuity at an endpoint, the limits in parts 2 and 3 of the test should be replaced by the appropriate one-sided limits.

-

4

1

75

is right-continuous at a and continuous at a right endpoint b of its domain if it is leftcontinuous at b. A function is continuous at an interior point c of its doroain if and only if it is both right-continuous and left-continuous at c (Figure 2.36).

y

-2

Continuity

EXAMPLE 4 Thefunctiony = lxJ introduced in Section l.l is graphed in Figure 2.39. It is discontinuous at every integer because the left-hand and right-hand limits are not equal

........0

........0

asx~n:

limJx J = n - I

........0

and

x~n

x

1

-I

---0

2

3

4

-2

FIGURE 2.39 The greatest integer function is continuous at every noninteger point It is right-continuous, but not left-continuous, at every integer point (Example 4).

Since l1l J = 11, the greatest integer function is right-continuous at every integer 11 (but not left-continuous). The greatest integer function is continuous at every real numher other than the integers. For example, lim lxJ = I = l1.5J.

x--+1.5

In general, if 11

-

I


11, 11

an integer, then

limlxJ = x~,

11 -

I = lcJ.

-

Figure 2.40 displays several common types of discontinuities. The function in Figure 2.40a is continuous at x = O. The function in Figure 2.40b would be continuous if it had f( 0) = I. The function in Figure 2.40c would be continuous if f(O) were I instead of 2. The discontinuities in Figure 2.4()b and c are removable. Each function has a limit as x --> 0, and we can remove the discontinuity by setting f(O) equal to this limit. The discontinuities in Figure 2.40d through f are more serious: lim..... o f(x) does not exist, and there is no way to improve the situation by changing f at O. The step function in Figure 2.40d has a jump discontinuity: The one-sided limits exist but have different values. The function f(x) = l/x 2 in Figure 2.40. has an inUrnte discontinuity. The function in Figure 2.40f has an oscillating discontinuity: It oscillates too much to have a limit as x-->O.

76

Chapter 2: Limits and Continuity y

y

y

y

1

x

Y ~ f(x)

0

(b)

(a)

x

(d)

(c)

y

J

Y~f(x)~l2

0

x

x x

(f)

(e)

FIGURE 2.40

The functioo in (a) is cootinuous at x

~

0; tire functioos in (b) through (f)

are not

Continuous Functions A function is continuous on an interval if and only if it is continuous at every point of the interval. For example, the semicircle function graphed in Figure 2.37 is continuous on the interval [ - 2, 2], which is its domain. A continuous function is one that is continuous at every point of its domain. A continuous function need not be continuous on every interval. Y

EXAMPLE 5 (a) The function y = l/x (Figure 2.41) is a continuous function because it is continuous at every point of its domain. It has a point of discontinuity at x = 0, however, because it is not defined there; that is, it is discontinuous on any interval containing x = O.

--------~t-------~x

\ "

The functioo y ~ l/x is continuous at every value of x except x = O. It has a point of discontinuity at x = 0 (Example 5). FIGURE 2.41

(h) The identity function I(x) = x and constant functions are continuous everywhere by Example 3, Section 2.3. • Algebraic combinations of continuous functions are continuous wherever they are defined.

THEOREM B-Properties of Continuous Functions If the functions I and g are continuous at x = c, then the following combinations are continuous at x = c.

Differences:

I+g I-g

Constant multiples:

k· I, for any number k

Products:

I'g Ilg,

1. Sums:

2. 3. 4. 5. 6. 7.

Quotients: Powers:

r,

Roots:

"ifj

providedg(c) oF 0 n a positive integer provided it is deImed on an open interval

, containing c, where n is a positive integer

2.5

Continuity

77

Most of the results in Theorem 8 follow from the limit rules in Theorem I, Section 2.2. For instance, to prove the sum property we have

lim(j + g)(x)

x-c

= lim(j(x) x-c

= lim f(x) x-c

= f(c) =

This shows that f

+ g(x)) + lim g(x),

Sum Rule, Theorem 1

x-c

+ g(c)

Continuity of j. g at c

(j + g)(c).

+ g is continuous.

EXAMPLE 6 (a) Every polynomial P(x) = anx n + a._1Xn-1 lim P(x) = P(c) by Theorem 2, Section 2.2.

+ ... + ao

is continuous because

x~c

(b) If P(x) and Q(x) are polynomials, then the rational function P(x)/Q(x) is continuous wherever it is dermed(Q(c) oF 0) by Theorem 3, Section 2.2. _

EXAMPLE 7 The function f(x) = Ixl is continuous at every value of x. If x > 0, we have f(x) = x, a polynomial. Ifx < 0, we have f(x) = -x, another polynomial. Finally, at the origin, limx~o Ixl = 0 = 101. The functions y = sin x and y = cos x are continuous at x = 0 by Example 11 of Section 2.2. Both functions are, in fact, continuous everywhere (see Exercise 68). It follows from Theorem 8 that all six trigonometric functions are then continuous wherever they are dermed. For example, y = tanx is continuous on ... U (-7r/2, 7r/2) U (7r/2, 37r/2) U ....

Composites All composites of continuous functions are continuous. The idea is that if f(x) is continuous atx = c andg(x) is continuous atx = f(c), theng 0 f is continuous atx = c (Figure 2.42). In this case, the limit as x --> c is g(f(c».

Continuous at c g

f Continuous

•

c

FIGURE 2.42

ate

_

Continuous -:;.o C-_ _ _ _ atf(e) _ _ _ _=. . ~f(c) g(f(c»

Composites of continuous functions are continuous.

THEOREM 9-Composite of Continuous Functions If f is continuous at c and g is continuous at f(c), then the composite g 0 f is continuous at c.

Intuitively, Theorem 9 is reasonable because if x is close to c, then f(x) is close to f(c), and since g is continuous at f(c), it follows thatg(f(x» is close to g(f(c». The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. For an outline of the proof of Theorem 9, see Exercise 6 in Appendix 4.

78

Chapter 2: Limits and Continuity

EXAMPLE 8 Show that the following functions are continuous everywhere on their respective domains. =

Yx 2

(e) y =

Ix2 _

(a) y

x 2/3

2x - 5

-

+ x4

(b) y = I

21

x - 2

(d) Y =

1:2S:~1

Solution

y

0.4 3

(a) The square root function is continuous on [0, 00) because it is a root of the continuous identity function fix) = x (part 7, Theorem 8). The given function is then the composite of the polynomial fix) = x 2 - 2x - 5 with the square root function g(t) = Vi, and is continuous on its domain.

(b) The numerator is the cube root of the identity function squared; the denominator is an everywhere-positive polynomial. Therefore, the quotient is continuous. (e) The quotient (x - 2)/(x 2 - 2) is continuous for all x # ± v2, and the function is the composition of this quotient with the continuous absolute value function (Example 7).

FIGURE 2.43 The graph suggests 1hat y = I(x sinx)/(x 2 + 2) I is continuous

(Example 8d).

(d) Because the sine function is everywhere-continuous (Exercise 68), the numerator term x sin x is the product of continuous functions, and the denominator term x 2 + 2 is an everywhere-positive polynomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure 2.43). • Theorem 9 is actually a consequence of a more general result which we now state and prove.

THEOREM lO-Limits of Continuous Functions b and limx~d(x) = b, then lim~, g(f(x» =

Proof that

Let E

>

g(b)

If g is continuous at the point

= g(lim~, fix»~.

0 be given. Since g is continuous at b, there exists a nornber 8 , Ig(y) - g(b)1

<

whenever

E

Since limx~, f(x) = b, there exists a 8

If(x) - bl < 8 ,

>

<

0

<

Iy - bl

>

0 such

8,.

0 such that

whenever

<

0

Ix - cl

<

8.

Ifwe let y = f(x), we then have that Iy - bl

<

8,

whenever

0

<

Ix - cl

<

8,

which implies from the tmt statement that Ig(y) - g(b)1 = Ig(f(x» - g(b)1 < E whenever 0 < Ix - cl < 8. From the detmition of limit, this proves that lim~,g(f(x»

EXAMPLE 9 lim cos

X~'1T12

=

g(b).

•

As an application of Theorem 10, we have

(2x + sin (32 + x)) 'IT

= coJ lim \x--i'7T12 = cos ('IT

+

2x + x-7T/2 lim sin (3 + x)) 2 'IT

sin2'IT) = COS'IT = -\.

•

2.5

79

Continuity

Continuous Extension to a Point The function y = I(x) = (sinx)/x is continuous at every point except x = O. In this it is like the function y = I/x. But Y = (sinx)/x is different from y = I/x in that it has a finite limit as x --+ 0 (Theorem 7). It is therefore possible to extend the function's domain to include the point x = 0 in such a way that the extended function is continuous at x = O. We derme a new function SinX

F(x) =

xi'O

x' {

x = O.

I,

The function F(x) is continuous at x = 0 because

lim sinx = F(O)

x-a x (Figure 2.44). y

y

(0,1)

(~, ~)

t~, ~) _'!

F(x)

(~, ~)

t~, ~) x

"2

0

2

(0, 1)

J(x)

_'!

0

2

(a)

"2

x

(b)

FIGURE 2.44 The graph (a) of f(x) = (sinx)/x for -7T/2 :5 x :5 7T/2 does not include the point (0, I) because the function is not def"med at x = O. (b) We can remove the discontinuity from the graph by def"ming the new functionF(x) withF(O) = I and F(x) = f(x) everywhere else. NotethatF(O) = limf(x). x~o

More generally, a function (such as a rational function) may have a limit even at a point where it is not dermed. If I(c) is not dermed, but Iimx~, I(x) = L exists, we can derme a new function F(x) by the rule if x is in the domain of 1 if x = c.

{/(X),

F(x) =

L,

The function F is continuous at x = c. It is called the continuous extension of 1 to x = c. For rational functions I, continuous extensions are usnally found by canceling common factors.

EXAMPLE 10

Show that

I(x) = x

2 ;- X -

6,

x i' 2

x - 4 has a continuous extension to x = 2, and rmd that extension. Solution

Although 1(2) is not defined, if x i' 2 we have

(x) = x 2

1

+X x2 _

-

4

6 = (x - 2){x (x - 2){x

The new function

F(x) =

~! ~

+ 3) = + 2)

x x

+3 + 2'

80

Chapter 2: Limits and Continuity is equal to f(x) for X"" 2, but is continuous at x = 2, having lbere lbe value of 5/4. Thus F is lbe continuous extension of f to x = 2, and

y

2

y~

x2+x-6 x2 _ 4

1 -1

lim x x-2

I I

~

1

0

I

4

3

)

x

(a)

y

2

Y

= x+

1

-1

0

1

2

3

4

x

(b)

FIGURE 2.45

+X

x2 -

-

6 = lim f(x) =

4

l. 4

x-2

The graph of f is shown in Figure 2.45. The continuous extension F has lbe same graph except wilb no hole at (2, 5/4). Effectively, F is lbe function f wilb its point of discontinu_ ity at x = 2 removed. Intermediate Value Theorem for Continuous Functions

x+2

~~

4

3

2

Fuoctions !bat are continuous on intervals have properties !bat make lbem particularly useful in malbematics and its applications. One oflbese is lbe Intermediate ~lue Property. A function is said to have lbe Intermediate Value Property if whenever it takes on two values, it also takes on all lbe values in between.

(a) The graph of

f(x) aod (b) the graph ofits

THEOREM ll-The Intermediate Yalue Theorem for Continuous Functions If f is a continuous function on a closed interval [a, b], and if Yo is any value between f(a)andf(b),lbenyo = f(c)forsomecin[a,b].

continuous extension F(x) (Example 10).

y y {(b)

~

f(x)

-------------

Yo - - - - - - - - - - {(a)

-nOr----L a-------c L-----~ b---.x

y

3 2

~--~~--~--~--~--~x

o

2

FIGURE 2.46

f(x) =

3

4

The functioo

{2x3, - 2,

1:5 x

Theorem II says !bat continuous functions over finite closed intervals have lbe Intermediate Value Property. Geometrically, lbe Intermediate Value Theorem says lbat any horizontalline y = Yo crossing lbe y-axis between lbe numbers f(a) and f(b) will cross lbe curve y = f(x) at least once over lbe interval [a, b]. The proof of lbe Intermediate Value Theorem depends on lbe completeness property oflbe real number system (Appendix 6) and can be fouod in more advanced texts. The continuity of f on lbe interval is essential to Theorem II. If f is discontinuous at even one point of lbe interval, lbe lbeorem's conclusion may fail, as it does for lbe function gmphed in Figure 2.46 (choose Yo as any number between 2 and 3).

<2

2::;; x::;; 4

does not take 00 all values b-.en f(I) = Oandf(4) = 3; it misses all the values between 2 and 3.

A Consequence for Graphing: Connectedness Theorem II implies lbat lbe graph of a function continuous on an interval cannot have any breaks over lbe interval. It will be conneeted-a single, uobroken curve. It will not have jumps like lbe graph of lbe greatest integer function (Figure 2.39), or separate branches like lbe gmph of I/x (Figure 2.41). A Consequence for Root Finding We call a solution oflbe equation f(x) = 0 a root of lbe equation or zero of lbe function f. The Intermediate Value Theorem tells us !bat if f is continuous, lben any interval on which f changes sign contains a zero of lbe function. In practical terms, when we see lbe graph of a continuous function cross lbe horizontal axis on a computer screen, we know it is not stepping across. There really is a point where lbe function's value is zero.

2.5

Show that there is a root of the equation x 3

EXAMPLE 11

-

81

Continuity

x - I

= 0 between I

and 2.

Solution Let I(x) = x 3 - X - I. Since 1(1) = I - I - I = -I < 0 and 1(2) = 23 - 2 - I = 5 > 0, we see that Yo = 0 is a value between 1(1) and 1(2). Since 1 is continuous, the Intennediate Value Theorem says there is a zero of 1 between I and 2. Figure 2.47 shows the result of zooming in to locate the root near x = 1.32. _

5

1.6

-11 ::::=::~~

2

-2 (a)

(b)

0.003

1.330

1.3240

f---"-----"-----~,...,..,

1.3248

-0.003 (e)

(d)

FIGURE 2.47 Zooming in on a zero of the function f(x) = x 3 x ~ 1.3247 (Example II).

EXAMPLE 12

-

X -

I. The zero is near

Use the Intennediate Value Theorem to prove that the equation ~=4_X2

y

has a solution (Figure 2.48).

Solution

We rewrite the equation as

~+x2=4,

------~----~~L-~x

o

c

2

FIGURE 2.48 The curves y ~ ~ and y ~ 4 - x 2 have the same value at x~cwhere~~4-x2 (Example 12).

and set I(x) = ~ + x 2. Now g(x) = Y2x + 5 is continuous on the interval [- 5/2, 00) since it is the composite of the square root function with the nonnegative linear function y = 2x + 5. Then 1 is the sum of the function g and the quadratic function y = x 2, and the quadratic function is continuous for all values of x. It follows that I(x) = ~ + x 2 is continuous on the interval [-5/2,00). By trial and error, we tmd the function values 1(0) = Vs '" 2.24 and 1(2) = V9 + 4 = 7, and note that 1 is also continuous on the tmite closed interval [0, 2] C [-5/2,00). Since the value Yo = 4 is between the numbers 2.24 and 7, by the Intennediate Value Theorem there is a number _ c E [0, 2] such that I(c) = 4. That is, the number c solves the original equation.

82

Chapter 2: Limits and Continuity

Exercises 2.5 7. a. Is 1 def"med at x = 2? (Look at the def"mition of I.)

Continuity from Graphs In Exercises 1-4, say whether the function graphed is continuous on [ -I, 3]. If not, where dnes it fail to be continuous and why'!

L

~

b. Is 1 continuous at x

= 2?

8. At what values of x is 1 continuous? 9. What value should be assigned to 1(2) to make the extended function continuous at x = 2? 10. To what new value should 1(1) be changed to remove the discontinuity?

Applying the Continuity Test At which points dn the functions in Exercises II and 12 fail to be continuous? At which points, if any, are the discontinuities removable?

Not removable? Give reasons for your answers. 11. Exercise I, Section 2.4 12. Exercise 2, Section 2.4 At what points are the functions in Exercises 13-30 continuous?

I 13. y = x _ 2 - 3x

14. y = (x

15.y=

x+1 x 2 -4x+3

16. y =

x +3 x'-3x-1O

17. y

Ix -

18. y =

Ixl +

=

11 + sinx

Exercises 5-10 refer to the function

I(x)

=

l

x' -

I,

2x,

I,

-2x

+ 4,

0,

x+2 20. y = cosx

O<x<1

2L y

22. y

x =I l<x<2 2<x<3

23

=

csc2x

- xtanx ·y-x'+1

24 y-

.

d. Is 1 continuous at x = -I? 6.

B.

Does 1(1) exist?

b. Does lim,_1 I(x) exist?

c. Does lim,_1 I(x) = 1(1)7 d. Is 1 continuous at x = I 7

W+1

-1+sin2 x

28. y = (2 - x)I/'

=

X'-X-6 x¢'3 x 3 ' { x = 3 5,

I

X3 -

8

,

4'

~,4,

= I(-I)?

7fX

27. y = (2x - 1)1/3

=

x'

T

tanT

26. y = "'¢'3x - I

30. I(x)

The graph for Exercises 5-10.

=

I -

25.y=~

29. g(x)

c. Doeslim,__ I' I(x)

I

19.y = cosX' x

-Is x < 0

graphed in the accompanying figure.

5. a. Does I( -I) exist? b. Does lim, __I' I(x) exist?

I

+ 2)' + 4

x¢'2,x¢' -2 x=2 x =-2

Limits Involving Trigonometric Functions Find the limits in Exercises 31-36. Are the functions continuous at the point being approached? 31. lim sin (x - sinx) ,~~

33. lim sec (y sec' y - tan' Y - I) y~1

34.

,~ tan(~cos (Sinx 1/3»)

35. lim cos ( t-O

V19 - 7f3 sec 2t )

36.

lim v'csc'x x-1T16

+ 5V3tanx

2.5 Continuity

Continuous Extensions 37. Defmeg(3) in a way that extends g(x) = (x' - 9)/(x - 3) to be continuous at x = 3.

38. Defmeh(2) in a way that extends h(l) = (I' + 31 - 10)/(1 - 2) to be continuous at I = 2. 39. Defme f(l) in a way that extends f(8) = (8' - 1)/(8' - I) to be continuous at s = 1. 40. Defme g(4) in a way that extends g(x)

= (x'

to be continuous at x

=

- 16)/(x' - 3x - 4)

4.

41. For what value of a is

f(x)

{x' - I, x < 3

=

2ax.

x~3

continuous at every x?

42. For what value of b is x < -2 x '" -2

X

g(x) =

{

b~',

continuous at every x?

43. For what values of a is f(x)

=

{~~~ -

x~2

2a,

Theory and Example. 51. A continuous function y = f(x) is known to be negative at x = 0 and positive at x = I. Why does the equation f(x) = 0 have at least one solution betweeo x = 0 and x = I? lllustrate with a sketch. 52. Explain why the equation cos x = x has at least one solution. 53. Roots of • cubic Show that the equation x' - 15x + I = 0 has three solutions in the interval [-4,4]. 54. A function value Show that the function F(x) = (x - a)" (x - b)' + x takes 00 the value (a + b)/2 for same value ofx. 55. Solving an equation If f(x) = x' - 8x + 10, show that there are values c for which fCc) equals (a) 'IT; (b) - V3; (c) 5,000,000. 56. Explain why the following fIVe statements ask for the same information. a. Find the roots of f(x) = x' - 3x - I. b. Find the x-coordinates of the points where the curve y = x' crosses the line y = 3x + 1. c. Find all the values of x for which x 3 - 3x = 1. d. Find the x-coordinates of the points where the cubic curve y = x 3 - 3x crosses the line y = 1. e. Solve the equation x' - 3x - I = O. 57. Removable discontinuity

Give an example of a functioo f(x) that is continuous for all values of x except x = 2, where it has a removable discootinuity. Explain how you know that f is discontinuous at x = 2, and how you know the discontinuity is removshle.

x<2

continuous at every x? 44. For what value of b is

g(X)={~~~'

83

x
58. Nonremovable discontinuity Give an example of a function g(x) that is continuous for all values ofx except x = -I, where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not

x 2 + b, x>o

continuous at every x?

removable. 45. For what values of a and b is

59. A function discontinuous at every point

-2 f(x)

=

ax '-- b, { 3,

x:5-1 -I < x < I x ~ 1

a. Use the fact that every nonempty interval of real numbers cootains both rational aod irrational numbers to show that the function

continuous at every x?

( ) = {I,

f x

46. For what values of a and b is g(x)

=

x:50 ax + 2b, x' + 3a - b, O<x:52 { 3x - 5, x>2

is discontinuous at every point. b. Is f right-cootinuous or left-continuous at any point?

60. If functioos f(x) and g(x) are continuous for 0 ,;; x ,;; I, could f(x)/g(x) possibly be discootinuous at a point of [0, I]? Give reasons for your answer.

continuous at every x?

D In Exercises 47-50, graph the function f

to see whether it appears to have a continuous extension to the origin. Ifit does, use Trace and Zoom to fmd a good candidate for the extended function~ value at x = O. If1he function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function~ value(s) should be?

47. f(x)

=

W x- I

48. f(x)

= IOI%~ -

49. f(x)

=

sin x TxT

so.

=

f(x)

(1

if x is rational 0, if x is irrational

I

+ 2x)'/%

61. If the product functioo h(x) = f(x)'g(x) is cootinuousatx = 0, mustf(x) andg(x) be cootinuous atx = O? Give reasons for your answer. 62. Discontinuous composite of continuous functions Give an example of functioos f and g, both cootinuous at x = 0, for which the composite fog is discontinuous at x = o. Does this contradict Theorem 9? Give reasons for your answer. 63. Never-zero continuous functions Is it true that a continuous function that is never zero on an interval never changes sign on that interval? Give reasons for your answer.

84

Chapter 2: Limits and Continuity

64. Stre1:I:hing a rubber band Is it true tbat if you stretch a rubber band by moviog one eod to the right aod the other to the left, some point of the band will end up in its original position? Give

to prove tbat both f(x) at every point x = c.

=

sin x aod g (x)

=

cos x are continuous

reasons for your answer. 65. A f'lIed point theorem Suppose tbat a function f is continuous on the closed interval [0, I] and that 0 '" f(x) '" I for every x in [0, I]. Show tbat there must exist a number e in [0, I] such tbat f(e) = e (e is called a f'lIed point of f).

Solvtng Equations Graphically

D Use the Intermediste Value Theorem in Exercises 6'J--76 to prove tbat each equation has a solution. Thea use a graphing calculator or computer grapher to solve the equations.

69.x 3 -3.>:-1=0

66. The sign-p....erving property of continuous functions Let f be defmed on an interval (a, b) aod suppose tbatf(e) 0 at some e where f is continuous. Show that there is au interval (e - 6,e + 6) about e where fhas the same sign asf(e).

*

70.2>;3-2>;2_2>;+ I =0

71. x(x - 1)2

67. Prove tbat f is continuous at e if and ooly if lim f(e

h~O

+ h)

I

=

(one root)

72.x%=2 73.

Vx + V'1+x =

74. x 3

= f(e).

-

75. cosx

68. Use Exercise 67 together with the identities

ISx

+

=x

+ c) = sinhcosc + cosh sine, cos(h + c) = coshcosc - sinh sine

=

4

0

(three roots)

(one root). Make sure you are using radian mode.

76. 2 sin x = x

sin(h

I

(three roots). Make sure you are using radian mode.

Limits Involving Infinity; Asymptotes of Graphs

2.6

In this section we investigate the behavior of a function when the magnitude of the independent variable x becomes increasingly large, or x --> ±OO. We further extend the concept of limit to infinite limits, which are not limits as before, but rather a new use of the term limit. Infmite limits provide useful symbols and langnage for describing the behavior of functions whose values become arlJitrarily large in magnitude. We use these limit ideas to analyze the graphs of functions having horizontal or vertical asymptotes.

Finite Limits as x --+ ±oo y

The symbol for infinity (oo) does not represent a real number. We use 00 to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. For example, the function f{x) = I/x is defined for all x oF 0 (Figure 2.49). When x is positive and becomes increasingly large, 1/x becomes increasingly small. When x is negative and its magnitude becomes increasingly large, I/x again becomes small. We summarize these observations by saying that f{x) = I/x has limit 0 as x --> 00 or x --> -00, or that 0 is a limit o/f{x) = l/x at i'lfinity a1ld negative i'lfi1lity. Here are precise definitions.

4

3

y=i

2

-

1

2

3

4

x

DEFINmONs 1. We say that f(x) has the limit L as x approaches infinity and write

lim f{x) = L

x_oo

>

if, for every number E FIGURE 2.49 The graph of y = Ilx approaches 0 as x ~ 00 or x ~ - 00.

0, there exists a corresponding number M such tbat fur all x

x>M

=)

If{x)-LI<E.

2. We say that f{x) has the limit L as x approaches minus infinity and write lim f{x) = L

x--=

> 0, there exists a corresponding number N such that for all x
if, for every number E x

2.6

No matter what Y positive number E is,

y=l

the graph enters this band at x = and stays.

Limits Involving Infinity; Asymptotes of Graphs

85

Intuitively, limx~oo f{x) = L if, as x moves increasingly far from the origin in the positive directioo, f(x) gets arbitrarily close to L. Similarly, limx-+-oo f{x) = L if, as x moves increasingly far from the origin in the negative direction, f(x) gets arbitrarily close to L. The strategy for calculating limits of functioos as x --+ ± 00 is similar to the one for tmite limits in Sectioo 2.2. There we trrst found the limits of the coostant and identity functioos y = k and y = x. We then extended these results to other functions by applying Theorem I 00 limits of algebraic combinatioos. Here we do the same thing, except that the starting functioos arey = kandy = I/x instead ofy = kandy = x. The basic facts to be verified by applying the formal detmitioo are lim k=k

l

and

x---+±OO

lim

x~±OO

i

=

o.

(1)

We prove the secood result and leave the trrst to Exercises 87 and 88.

EXAMPLE 1 (a) lim x_oo "--i--~---1 -.

i

Show that =

(b)

0

lim

x---+-OO

i

=

o.

Solutton (a) Let E

>

0 be given. We must find a number M such that for all x

positive number E is, the graph enters this bandatx and stays.

=-1

x>M The implication will hold if M = 1/E or any larger positive number (Figure 2.50). This proves limx~oo (I/x) = o.

FIGURE 2.50 The geometry behind the argument in Example 1.

(b) Let E

>

0 be given. We must find a number N such that for all x

x
-1/E or any number less than -1/E (Figure 2.50). •

Limits at infinity have properties similar to those of tmite limits.

THEOREM 12

All the limit laws in Theorem I are true when we replace lim...... , by limx~oo or limx~-oo. That is, the variable x may approach a finite number c or ±OO.

EXAMPLE 2 The properties in Theorem 12 are used to calculate limits in the same way as when x approaches a tmite number c. (a) lim (5 x_OO

+

1) x

=

lim 5

x_OO

= 5

(b)

lim x_-OO

'IT v'3 =

x2

=

+0

1

+ x_OO lim X

Sum Rule

= 5

Known limits

lim ". v'3 .1.1

x_-oo

lim

x---+-OO

X

'IT v'3.

X

lim

1. x---+--OO lim 1 X

x--OO X

= ".v'3·0·0 = 0

Product Rule Known limits

•

86

Chapter 2: Limits and Continuity

Limits at Infinity of Rational Functions

y

To determine the limit of a rational function as x ..... ± 00, we first divide the numerator and denominator by the highest power of x in the denominator. The result then depends on the degrees of the polynomials involved.

2 Liney ~ ~ 3

5

10

x

EXAMPLE 3 These examples illustrate what happens when the degree of the numerator is less than or equal to the degree of the denominator. .

-2

NOT TO SCALE

FIGURE 2.51 The graph of the function in Example 3a. The graph approaches the line y ~ 5/3 as Ixl increases.

Y

8

llx + 2

y~2x3_1

6

hm
(b)

5x 2 + 8x - 3 3x 2 + 2

lim IIx 2x 3

._-00

+ -

=

.

hm

5

.~oo

+ (8/x) - (3/X2) 3 + (2/x2)

5+0-0

5

3+ 0

3

2

2 =

lim (lI/x )

I

._-00

2 -

+ (2/x

Divide nmnerator snd den

. torby ,

OJDms

•.

See Fig. 2.51. 3

Divide numerator and

)

denominstor by x' .

(l/x 3 )

=0+0=0 2-0

See Fig. 2.52.

•

A case for which the degree of the numerator is greater than the degree of the denom· inator is illustrated in Example 8.

4

Horizontal Asymptotes

-2

If the distance between the graph of a function and some f"1xed line approaches zero as a point on the graph moves increasingly far from the origin, we say that the graph ap· proaches the line aaymptotically and that the line is an asymptote of the graph. Looking at f(x) = l/x (see Figure 2.49), we observe that the x·axis is an asymptote of the curve on the right because

-4

-6 -8

FIGURE 2.52 The graph of the function in Example 3b. The graph approaches the x·asis as Ix Iincreases.

and on the left because lim i ._-00

=

o.

We say that the x·axis is a horizontal asymptote of the graph of f(x) = l/x.

DEFINITION A line y = b is a horizontal asymptote of the graph ofa func· tion y = f(x) if either lim f(x) = b

or

.-00

lim f(x) = b.

x--+-oo

The graph of the function

f(x)

=

2 5x + 8x - 3 3x 2 + 2

sketched in Figure 2.51 (Example 3a) has the line y = 5/3 as a horizontal asymptote on both the right and the left because

lim f(x)

x--+OO

=

~ 3

and

lim f(x)

x--+-oo

5 = -3 •

2.6

EXAMPLE 4

y

Limits Involving Infinity; Asymptotes of Graphs

Find the horizontal asymptotes of the graph of

x3

2

-

We calculate the limits as x --+ ± 00.

Forx The graph of the functioo in Example 4 has two horizootal asymptotes.

<

x

3 _

2

1 - (2/x 3)

2

3 _

lim _x__ = lim x3 + 1 1

._00 Ixl 3 + 1 ._00 ._00 + (1/x 3) 1. 3 ;~ x3 - 2 __ lim x 3 - 2 . I - (2/x ) lIm I ._-00 Ixl 3 + I ._-00 (-x)3 + I ._-00 -I + (l/x3) lim

Forx 20 0:

FIGURE 2.53

2

+ 1.

f(x) = Ixl3

Solution

87

~.

0:

=

--'"-~--- =

=

-1.

The horizontal asymptotes are y = -I and y = 1. The graph is displayed in Figure 2.53. Notice that the graph crosses the horizontal asymptote y = -I for a positive value

-

~~ EXAMPLE 5

Find (a) lim sin (l/x)

and

x_co

(b)

lim x sin (I/x). x_±OO

Solution (a) We introduce the new variable 1 = I/x. From Example I, we know that 1--+0+ as

x --+

00

(see Figure 2.49). Therefore,

y

· . I I'unsmt= . 0. I1IIlSID:x= t-O+

x--+OO

Likewise, we can investigate the behavior of y = f(l/x) as x --+ 0 by investigating y = f(l) as 1--+ ±oo,where 1 = l/x. (b) We calculate the limits as x --+

FIGURE 2.54 The line y = I is • horizootal asymptote of the functioo grapbed here (Example 5b).

00

and x --+

. 1 I'un--= sin 1 I lim xsm-= X t-O+ t

-00:

and

x_OO

The graph is shown in Figure 2.54, and we see that the line y = 1 is a horizontal asymptote. _ The Sandwich Theorem also holds for limits as x --+ ± 00. You must be sure, though, that the function whose limit you are trying to find stays between the bounding functions at very large values of x in magnitude consistent with whether x --+ 00 or x --+ - 00 .

EXAMPLE 6

Using the Sandwich Theorem, find the horizontal asymptote of the curve

y = 2

Solution

+ sinx x .

We are interested in the behavior as x --+ ±

00.

Since

y

y= 2+

~x and limx __ II/xl = 0, we have limx __ (sinx)/x = 0 by the Sandwich Theorem. Hence, lim (2

x-±oo

FIGURE 2.55 A curve may cross ooe of its asymptotes infmitely o!teo (Example 6).

+

Sinx) = 2 X

+0

= 2,

and the line y = 2 is a horizontal asymptute of the curve on both left and right (Figure 2.55). This example illustrates that a curve may cross one of its horizontal asymptotes many times. _

88

Chapter 2: Limits and Continuity

EXAMPLE 7

Y

Solution Both of the tenns x and x 2 + 16 approach infinity as x --+ 00, so what happens to the dliference in the limit is unclear (we cannot subtract 00 from 00 because the symbol does not represent a real number). In this situstion we can multiply the numerator and the denominator by the conjugate radical expression to obtain an equivalent algebraic result:

lim

x~oo

(x - Yx 2 + 16) = =

2

(x - Yx 2 + 16r + Yx 2 + 16 x_OO x + Yx + 16 lim

. x 2 - (x 2 + 1 6 ) . -16 hm = hm --~~= 2 + 16 2 + 16 x~oox + x~oox +

Yx

Yx

As x --+ 00, the denominator in this last expression becomes arbitrarily large, so we see that the limit is O. We can also obtain this result by a direct calculation using the Limit Laws:

· Iun

x~oo x

-16

+

Yx 2 + 16 =

16 I'un ----rii=== x x~oo x2 16 1 + -2 + x x2

1

o

+ v'1+O

=

0

.

•

Oblique Asymptotes

If the degree of the numerator of a rational function is 1 greater than the degree of the denominator, the graph has an oblique or slant line asymptote. We find an equation for the asymptote by dividing numerator by denominator to express f as a linear function plus a remainder that goes to zero as x --+ ± 00 .

EXAMPLE 8

Find the oblique asymptote of the graph of

f(x) x2 _ 3 x y ~ 2>- 4 ~ 'I:

y

1

+ 1 + 2x- 4

The vertical distance

between curve and

6

Solution We are interested in the behavior as x--+ (x 2 - 3):

±OO. We divide (2x - 4) into

~+1 2 x=2

2x -

3

4!x2 x2 -

2

-1 0 -1

x2 - 3 2x - 4

in Figure 2.56.

5 4

=

2

3

4

x

-2 -3

3 2x 2x - 3 2x - 4 1

This tells us that

f(x)

=

~ == ! (~ + 1) + (~) . =

"--or-'

FIGURE 2.56 The graph of the function in Example 8 has ao oblique asymptote.

linear g(x)

remainder

As x --+ ± 00, the remainder, whose magnitode gives the vertical distance between the graphs of f and g, goes to zero, making the slanted line g(x) =

2x +

1

2.6 y You can get as high

as you want by taking x close enough to O. No matter how

r.

\~gh B is, the graph B

higher.

~

x

Limits Involving Infinity; Asymptotes of Graphs

89

an asymptote oftbe graph of f (Figure 2.56). The line y = g(x) is an asymptote both to the right and to the left. The next subsection will conf"mn that the function f(x) grows arlrltrarily large in absolute value as x --> 2 (where the denominator is zero), as shown in the graph. • Notice in Example 8 that if the degree of the nwnerator in a rational function is greater than the degree of the denominator, then the limit as Ix Ibecomes large is +00 or -00, depending on the signs asswned by the nwnerator and denominator.

Infinite Limits

----------~~~--------~x

~o

You can get as low

you want by taking X close enough to O.

FIGURE 2.57 lim 1. = 00

x_o+x

x

No matter how low -B is, the graph goes lower.

j.8\

-B

One-sided inf"mite limits: and lim 1. = -00. x_o-x

y

Let us look again at the function f(x) = I/x. As x --> 0+, the values of f grow without bound, eventually reaching and surpassing every positive real nwnher. That is, given any positive real nwnber B, however large, the values of f become larger still (Figure 2.57). Thus, f has no limit as x --> 0+. It is nevertheless convenient to describe the behavior of f by saying that f(x) approaches 00 as x --> 0+. We write lim f(x) = lim

x-.o+

1

lim f(x) = lim

x-o-

1 -----:----;;+---;-I---2~---;;3-----x

= 00

x-o-

i

= -00.

Again, we are not saying that the limit exists and equals the nwnber - 00 . There is no real nwnber - 00 . We are describing the behavior of a function whose limit as x ..... 0- does not exist because its values become arbitrarily large and negative.

EXAMPLE 9

FIGURE 2.58 Near x = I, the function y = I/(x - 1) behaves the way the function y = I/x behaves near x = o. Its graph is the graph of y = I/x shifted I unit to the right (Example 9).

i

In writing this equation, we are not saying that the limit exists. Nor are we saying that there is a real nwnber 00, for there is no such nwnber. Rather, we are saying that limx~o+ (I/x) does not exist because I/x becomes arbitrarily large and positive as x ..... 0+ . As x ..... 0-, the values of f(x) = I/x become arbitrarily large and negative. Given any negative real nwnber -B, the values of f eventually lie below -B. (See Figure 2.57.) We write

Y=x-l

Find lim _1_ x-l+ x - I

· ---I un

I x_rx - I'

and

Geometric Solution The graph of y = I/(x - I) is the graph of y = I/x shifted I unit to the right (Figure 2.58). Therefore, y = I/(x - I) behaves near I exactly the way y = I/x behaves near 0: lim _1_ = 1

00

x---+l+X -

lim __1_ = 1

and

00

x---+l-X -

Analytic Solution Think about the nwnber x - I and its reciprocal. As x --> 1+, we have (x - I) ..... 0+ and I/(x - 1) ..... oo.Asx ..... I-, we have (x - I) ..... 0- and I/(x - I) .....

y

-00.

} x 0

x-.o+

No matter how high B is, the graph

EXAMPLE 10

•

Discuss the behavior of

goes higher.

~ x

I f(x) = 2" x x

FIGURE 2.59 The graph of f(x) io Example 10 approaches inf"mity as x ..... O.

x-->

as

o.

As x approaches zero from either side, the values of l/x 2 are positive and become arbitrarily large (Figure 2.59). This means that

Solution

lim f(x) = lim

x-o

1, =

x---+Ox

00.

•

90

Chapter 2: Limits and Continuity The function y = I/x shows no consistent behavior as x --> o. We have I/x --> 00 if x-->O+, but I/x--> -00 ifx--> 0-. All we can say about lim.-o (I/x) is that it does not exist. The function y = l/x 2 is different. Its values approach infinity as x approaches zero from either side, so we can say that limx _ o (l/x 2 ) = 00.

EXAMPLE 11 These examples illustrate that rational functions can behave in various ways near zeros of the denominator. I· (x - 2)2 lim x - 2 · (x - 2f IlID 2 (a) x-2 lID ( 2 = 0 x - 4 = x-2 x - 2)(X + 2) = x-2 X + (b) lim x - 2 = lim .~2 X 2 -

(c) y

4

.~2

x - 2 = lim _1_ = ! (x - 2)(x + 2) .~2 x + 2 4

lim x - 3 = lim

.-2+ X 2 -

4

.-2+ (X -

(d) lim x - 3 = lim • -2- X 2 -

4

.-T

.-2

(f) lim

B ~-~--r~~--

---;;If--~-;-----;;____c:\---~x

Xo

+8

4

x - 3 = 00 (x - 2)(x + 2)

.-2 .-2

(e) lim x - 3 = lim • -2 X 2 -

x - 3 = _ 00 2)(x + 2)

x - 3 does not exist. (x - 2)(x + 2)

The values are negative for x > 2, x near 2. The values are positive forx < 2, x near 2 . See parts (c) and (d).

.-2

2-x = lim -(x-2)=lim -I =-00 (x - 2)3 (x - 2)3 (x - 2)2

In parts (a) and (b) the effect of the zero in the denominator at x = 2 is canceled because the numerator is zero there also. Thus a finite limit exists. This is not true in part (f), where cancellation still leaves a zero factor in the denominator. •

Predse Definitions of Infinite Limits FIGURE 2.60 For XQ - 8 < x < xo + 8, the graph of f(x) lies above the lioe y = B. y

Instead of requiriog f(x) to lie arbitrarily close to a finite number L for all x sufficieotly close to xo, the defInitions of infinite limits require f(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seeo before. Figures 2.60 and 2.61 accompany these definitions.

DEFINmONS 1. We say that f(x) approaches infinity as x approaches -"0, and write lim f(x) = 00, X-Xo

if for every positive real number B there exists a correspoodiog Ii that for all x

>

0 such

f(x»B. o < Ix - xol < Ii 2. We say that f(x) approaches minus inf"mity as x approaches xo, and write lim f(x) = -00, X-Xo

FIGURE 2.61 For Xo - 8 < x < XQ the graph of f(x) lies below the lioe y = -B.

+ 8,

if for every negative real number - B there exists a correspondiog Ii that for all x f(x) < -B. o < Ix - xol < Ii

>

0 such

The precise defmitions of ooe-sided infinite limits at Xo are similar and are stated in the exercises.

2.6

EXAMPLE 12 Solution

Prove that lim

x-a

Given B

>

Limits Involving Infinity; Asymptotes of Graphs

~

91

= 00

X

>

0, we want to find 8

0 such that

1 implies 2>B. x

0
Now,

~>B 2 x

if and only ifx2 <

~

or, equivalently, 1

Ixl< Thus, choosing 8 =

vB'

l/vB (or any smaller positive number), we see that Ixl < 8

implies

~ > ~;;,: B. x

8

Therefore, by defiuition,

• y

Vertical Asymptotes

Vertieal asympto'"

Horizontal

y=

l

Notice that the distance between a point on the graph of f(x) = l/x and the y-axis approaches zero as the point moves vertically along the graph and away from the origin (Figore 2.62). The function f(x) = l/x is unbounded as x approaches 0 because

asympto"'. y~O

lim 1 x_o-x

and

~as~~ ~rore ~~=--oO~~~::::~x ;;;;; Horizontal

= -00

We say that the line x = 0 (the y-axis) is a vertical asymptote of the graph of f(x) = I/x. Observe that the denominator is zero at x = 0 and the function is undefined there.

Vertieal asymptote,

x=O

The coordina'" axes are asymptotes of both branches of the hyperbola y ~ l/x. FIGURE 2.62

DEFINmON A line x y = f(x) if either

=

a is a vertical asymptote of the graph ofa function limj(x) =

or

EXAMPLE 13

x-a

±OO.

Find the horizontal and vertical asymptotes of the curve

x+3

y = x

+ 2'

Solution We are interested in the behavior as x --> ± 00 and the behavior as x --> - 2, where the denominator is zero.

92

Chapter 2: Limits and Continuity y

The asymptotes are quickly revealed if we recast the mtional function as a polynomial with a remainder, by dividing (x + 2) into (x + 3):

Vertical asymptote, x

=-2

6 5 4

~ 1

~

Horizontal asymptote,

y- 1

I

x+3

y= x+2

+ _1_

x+2

-~~~~ ~~±-~~-7~~~X -5 -4 - 3\-2 -1 0 \ -1

1

2

This result enables us to rewrite y as:

3

I

y=l+ x + 2 '

-2

As x --> ± 00, the curve approaches the horizontal asymptote y = I; as x --> - 2, the curve approaches the vertical asymptote x = -2. We see that the curve in question is the gmph of I(x) = I/x shifted I unit up and 2 uuits left (Figure 2.63). The asymptotes, instead of being the coordinate axes, are now the lines y = I and x = - 2. •

-3 -4

FIGURE 2.63

x + 2)" + 3 x+2 I

The lines y = I and

x = - 2 are asymptotes of the curve in Example 13.

EXAMPLE 14

Find the horizontal and vertical asymptotes of the gmph of

8

I(x) = - x2 _ 4' y 8 7

Solution

y~

We are interested in the behavior as x --> ± 00 and as x --> ±2, where the denominator is zero. Notice that I is an even function of x, so its gmph is symmetric with respect to the y-axis.

__ 8_ x2 _ 4

6

Vertical

asymptote, x=-2

5 4

Vertical asymptote, x

=

2

Horizontal asymptote, y

=

0

(a) The behavior as x .... ±oo. Since limx~oo/(x) = 0, the line y = 0 is a horizontal asymptote of the gmph to the right. By symmetry it is an asymptote to the left as well (Figure 2.64). Notice that the corve approaches the x·axis from ouly the negative side (or from below). Also, 1(0) = 2.

(b) The behavior as x--> ±2. Since lim I(x) =

x-2+

-00

and

the line x = 2 is a vertical asymptote both from the right and from the left. By sym· metry, the line x = - 2 is also a vertical asymptote. There are no other asymptotes because I has a finite limit at every other point. FIGURE 2.64 Graph of the function in Example 14. Notice that the curve approaches the x·axis from only one side. Asymptotes do not have to be two·sided.

EXAMPLE 15

•

The corves

Y = secx

=

I

cosx

and

y

=

tanx

sin x

= cosx

both have vertical asymptotes at odd·integer multiples ohr12, where cos x = 0 (Figure 2.65). y

y = secx

y

y = tanx

-~~-+-~-+-~-~~x

3

FIGURE 2.65 The graphs of sec x and tan x have infmitely many vertical asymptotes (Example 15).

•

2,6

Limits Involving Infinity; Asymptotes of Graphs

93

Dominant Terms In Example 8 we saw that by long division we could rewrite the function

x2 - 3 f(x) = 2>; - 4 as a linear function plus a remainder tenn:

This tells us immediately that

f(x) '"

f+ I

1 f(x) '" 2>; - 4

y

For x numerically large, 2x

~ 4 is near O.

For x near 2, this term is very large.

If we want to know how f behaves, this is the way to find out. It behaves like y = (x/2) + I when x is numerically large and the contribution ofl/(2); - 4) to the tota\ value of f is insignificant. It behaves like 1/(2); - 4) when x is so close to 2 that

-_~2~--_~I~~Ott-~~--~2~X

-5

1/(2); - 4) makes the dominant contribution. We say that (x/2) + I dominates when x is numerically large, and we say that 1/(2); - 4) dominates when x is near 2. Dominant terms like these help us predict a function's behavior.

Let f(x) = 3x 4 - 2>;3 + 3x 2 - 5x + 6 and g(x) = 3x4. Show that although f and g are quite different for numerically small values of x, they are virtually identical for Ix I very large, in the sense that their ratios approach I as x ..... 00 or

EXAMPLE 16


x~

-00,

Solutton The graphs of f and g behave quite differently near the origin (Figure 2.66a), but appear as virtually identical on a larger scale (Figure 2.66b). We can test that the tenn 3x 4 in f, represented graphically by g, dominates the polynomial f for numerically large values of x by examining the ratio of the two functions as x ..... ±oo.Wefmdthat --~2~O--~-I~O~~O~--~I~O--~2~O~X -100,000 (bJ

FIGURE 2.66 The graphs off and g are
lim f(x) = lim ]x4 - 2>;3 + 3x 2 - 5x x~±oo g(x) x~±oo 3x4 = =

+6

. ( 1 -2- +1 - - 5 + 2) lim 3x x2 3x 3 x4

x_±OO

1,

which means that f and g appear nearly identical for Ix Ilarge.

•

94

Chapter 2: Limits and Continuity

Exercises 2.6 Finding Limits 1. For the function f whose graph is given, determine the following limits.

a. lim f(x)

b.

d. lim f(x)

e. lim f(x)

r.

g. lim f(x)

h. lim f(x)

i.

x~,

x--3

c.

lim f(x)

%--3+ %-0+

x~o

x~oo

lim _ f(x)

%--3

lim_ f(x)

Limits of Rational Functions In Exercises 13-22, f"md the limit of each rational function (a) as x ---+ 00 and (b) as x ---+ -00.

13. f(x)

a +3 = 5x + 7

15. f(x) 17. h(x)

x~O

lim f(x)

%_-00

y

14. f(x)

a' + 7 = x-,~-"x~,~+'--'x'--+-7

=~ x2 + 3

16. f(x)

=~+7

7x' =,x-3x+6:< ,

I 18. g(x) = -x~'---4"'-,,-+-1

3

'-" . /

19. g(x) = lOx'

1/

r -2 /

-4 -

-6 -

1 '2 3

6

3

x

2. For the function f whose graph is given, determine the following limits.

a. lim f(x)

b. lim f(x)

d. lim f(x)

••

21. h(x)

= -a' - a + 3 3x' + 3:<' - 5x

x~,

-x' = --.--,--"----;;---:4

24.

lim f(x)

h. lim f(x)

j. lim f(x)

k. lim f(x)

9x'

2x 4 +5x2 -x+6

x -7x'+7x'+9

lim x_-OO

%-0+

x~o

22. h(x)

%--3+

g. lim f(x) %--3

=

Limits as x ---+ 00 or x ---+ - 00 The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23-36.

x-2+

x~4

25.

X~OO

+x

20. h(x)

x

V

1/2

-"

+ ~. + 31

f

x - 2

I ~ ' )' 2 + 7x %_--00 ( x

lim

(x' x-

+ 1)1/3 &x 2 - 3

2 26 lim I x -5x 3 'x_oo '\jx +x-2

y

27. lim 2Vx + x-I x_OO 3x 7 3 \ 21 1 -6 -S-

f/-

1'-

-2 ~

f

/ 1\

"-

..... 6

2/3 4

{

29. x

II

3

~-~

lim

30. lim x x_oo x

X~-OO ~ +~

. a'/3 - XI/3 + 7 31. lun x_OO x 8/ S + 3x + Vx 33. lim

x_OO

Vx'+l x +1

32.

34.

3. f(x) = 5. g(x)

2

x-

7. h(x) =

3-

~

4. f(x)

I

I 6. g(x) = 8 _ (5/x')

= 2 + (I/x) -5

= 7f

3

+ (7/x) (')

I/x

8. h(x)

-

x'

4

+ <,./2/x')

Find the limits in Exercises ~ 12. 9. lim sina x_OO x 11. lim 2-t+sint e---oo t + cos t

10. 12

lim

c03~

lim.

r

9_-00

. ,_00

2r

+ sinr

+ 7 - 5sinr

+ x 2/3 - 4

x_-OO

lim

Vx'+l x + 1

lim

4 - 3x'

x_-oo~

38. lim :_

..:M.

.LoA.

x-O+

x-o-

40. x_3+X lim ~3 41.

lim

x--8+ x

x~7

U

-4

~-5x+3

lim

37. lim }_

43. lim

0

x

Infinite Limits Find the limits in Exercises 37-48.

3 - (2/x)

= -----'~c-

+

2 _X 3

x_-OO 2x

36. In Exercises 3-8, f"md the limit of each function (a) as x ---> 00 and (b) as x ~ - 00. (You may wish to visualize your answer with a graphing calculator or computer.)

-I

a+ 8

4 (x - 7)'

45. a. lim

x-O+

~/,

3x

42.

x.!.~,- a ~ 10

44. lim

x~o

b. lim _2_ x-o- 3Xl/3

-I

x'(x + I)

2.6

46. a. lim

x-O+X

47. lim

x-o x

b.lim~ 15

~/5

Graphing Simple Rational Functions Graph the rational func1ioos in Exercises 63-
.%-0- X /

x-o x ~/,

;/5

48. lim

I 63'Y~x_1

Find the limits in Exercises 41}-52. 49. 51.

tanx

lim

SO.

x-('1f'/2r lim (I + esc 8) .~

...

x-(--rrj2t

sec x

52. lim (2 - cot 8) .~o

Find the limits in Exercises 53-58. I 53. lim-2- - a s x - 4

a. x-+2+ c. x~ -2+

x 67. Y ~ x

x~

~) as

c. x --> V'2

d. x-+-l

b. x-+2+

d.x-->2

e. What, if anything, can be said about the limit as x --> O?

72. f(2)

+2

73. 74. 75.

t~/3) as

~

b. t-+ 0-

:/3

x-+ 1+

.%-0+

~

2, and

0, lim f(x) ~ 0, lim f(x) ~ x-±oo

%-1-

~ -00,

I, f( -I)

~

~ -00,

and lim f(x) .%--1-

0, lim f(x) x_OO

~

and lim f(x)

lim f(x) ~

.1'--1+ ~ -00

0, lim f(x) .1'-0+

~

00,

~ 00,

I

%_-00

lim f(x)

~

0, lim f(x) ~

lim g(x)

~

0, lim g(x)

lim h(x)

~

-I, lim h(x)

x_±OO x_±OO

.%-2-

.%-3-

00,

and lim f(x) ~ x-2+

~ -00, ~

and lim g(x) ~ x-3+

I, lim h(x)

.%_00

00

.%-0-

~

00

-I, and

lim k(x) x_±OO

~

I, lim k(x) .%-1-

~ 00,

and lim k(x) x-I+

~ -00

for your answer.

C:/5 + 7) as

t-+ 0+

0, lim f(x)

79. How many horizootal asymptotes can the graph of a given rational func1ioo have? Give reasons for ynur answer.

d. x-+ 1-

Finding Limits of Differences when x--> ±oo Find the limits in Exercises 80-86. x~oo

81.

83.

b. x-+O-

x-+ 1+

d. x-+ 1-

(Vx+9 - Vx+4) lim (\I'x 2 + 25 - w-=J) lim (w-+3 + x) x_-oo lim (2x + yrc -_--c 4X"2-+--c3x 2)

80. lim

82.

~/3 - (x - I I)4/') as x

a. x-+o+ C.

~

~

77. Suppose that f(x) and g(x) are polynootials in x and that limx_oo (f(x)/g(x)) ~ 2. Can you conclude anything about limx_~ (f(x)/ g (x))? Give reasons for your answer.

t-+ 0+

62. lim(

-I, and

78. Suppose that f(x) and g(x) are polynootials in x. Can the graph of f(x)/g(x) have an asymptote if g(x) is never zero? Give reasons

x + (x - 2 I)2/') as a. x-+o+ b. x-+OC.

~

X~ ...

Find the limits in Exercises 5'HJ2.

61. lim(

-2, lim f(x)

lim h(x) ~ I

4x

e. What, if anything, can be said about the limit as x --> O?

a.

I

-2

.%_-00

as 76.

60. lim

+

In Exercises 73-76, fmd a func1ioo that satisfies the given cooditions and sketch its graph. (The answers here are not unique. Any func1ioo that satisfies the conditions is acceptable. Feel free to use formulas defmed in pieces if that will help.)

as

a. x-+o+ c. x-+2-

a.

~

x-o-

c.x-+l+

59. lim(2 -

2x

X~-OO

x_±OO

lim f(x)

a.x-+-2+

-

~

2, f( -I)

0, lim f(x)

lim f(x)

1 56. hm 2x + 4 as

x3

~

x-I+

-

x2-3x+2

~

0, f(l)

...

71. f(O) b. x-+O-

58. limx2 - 3x

~

lim f(x)

a. x-+o+

r

68. Y ~ x

X~OO

70. f(O)

• un x 3 -2x 2

x-3

lim f(x) ~ I

c. x-+ -1+

57

+3 +2

-3

y~-­

Inventing Graphs and Functions In Exercises 61}-72, sketch the graph of a func1ioo y ~ f(x) that satisfies the given oouditioos. No form:u1as are required--just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer sec1ioo.)

69. f(O)

a.x-+l+

. x2

66.

65'Y~2x+4

54.lim+-as x - I

55. lim(x; -

I 64'Y~x+1

I

lim

95

Limits Involving Infinity; Asymptotes of Graphs

X~OO

x_-oo

84. lim (v'9x 2 - x - 3x) x~OO

96

Chapter 2: Limits and Continuity

85. lim (VX2

+ 11: -

86. lim (Vx2

+x

x~oo

X~OO

VX 2 - 2x)

Use the formal defmitioos from Exercise 93 to prove the lintit statemoots in Exercises 94-98.

- ~)

94. lim xl =

Using the Formal Definitions Use the formal defmitioos of limits as x---> ±oo to establish the limits in Exercises 87 and 88.

95. lim

00

%-0+

96. lim _1_ =

2

x_2-X -

-00

-00

97. lim _1_ =

2

x_2+X -

98. lim _1_2 =

00

00

%-1-1 - x

87. If j bas the coostant value j(x) = k, then lim j(x) = k.

1=

x_o-x

X~OO

88. If j bas the coostant value j(x) = k, then lim j(x) = k. X~-OO

Use formal defmitions to prove the lintit statements in Exercises 8c}'-92. -I 89. lim - 2 = -00 90. lim _III = 00

x-o

x-o x

91. lim

x~3

-2 (x - 3)2

92. lim

=-00

X~-,

Oblique Asymptotes Graph the ratioua1 functioos in Exercises 9c}'-104. Include the graphs and equations of the asymptotes. x2 x 2 +1 99. Y = - - I 100. Y = - - I

x-

x2

X

(x

I

+ 5)2

= 00

93. Here is the defmition ofinfinite right-hand limit.

x-

-4

-x=T

102. Y

x2 - 1 103. y=-x-

104• Y

101. Y

=

=

x 2 -1 2x + 4

=x3+1 X

2

Additional Graphing Exercises the curves in Exercises \05-\08. Explain the relationship between the curve's formula and what you see.

D Graph We say that j(x) approaches infmity as x approaches Xo from the right, and write lim j(x) =

105.Y=~2 4 - x

00,

x-XQ+

if.

for every positive real number B. there exists a corresponding nomber 8 > 0 such that for all x

:%'0<x<xo+8

j(x) > B.

107. Y = x 2/3

+ ~f3 x

l06.Y=~ 4 - x2 108. Y

= sin

(+-) x + I

D Graph the functions in Exercises 109 and 110. Thou aoswer the following questioos.

a. How does the graph behave as x ---> o+? h. How does the graph behave as x ---> ± oo?

Modify the defmition to cover the following cases. L

lim_j(x)

= 00

c. How does the graph behave near x = I and x = -I?

%-%0

lim j(x)

= -00

Give reasons for your answers.

c. limj(x)

= -00

109. Y

h.

%-xo+ %-Xo

Chapter

~

=

r

&(x - ~

3(

X )2/3

1l0'Y=I x - I

Questions to Guide Your Review

1. What is the average rate of chaoge of the function g(t) over the interval from t = a to t = b? How is it related to a secant line?

6. What theorems are available for calculating lintits? Give examples of bow the theorems are used

2. What lintit must be calculated to fmd the rate of change of a function g(t) att = to?

7. How are one-sided limits related to limits? How can this relationship sometimes be used to calculate a limit or prove it docs not exist? Give examples.

3. Give ao informal or inmitive defmition of the lintit lim j(x) = L. X~Xo

Why is the definition "informal''? Give examples. 4. Docs the existence aod value of the limit of a function j(x) as x approaches Xo ever depend on what happens at x = xo? Explain aod give examples. 5. What function behaviors might occur for which the limit may fail to exist? Give examples.

8. What is the value oflim._o «sinO)/O)? Docs it matter whcther 0 is measured in degrees or radiaos? Explain. 9. What exactly does lim~Xo j(x) = L mean? Give an example in which you fmd a 8 > 0 for a given j, L, Xo, aod. > 0 in the precise defmition oflintit.

10. Give precise defmitions of the following staterneots.

a. lim,....2- j(x) = 5 c. lim,....2 j(x) = 00

h. lim~2' j(x) = 5

d. lim,....2 j(x) = -

00

97

Chapter 2 Practice Exercises 11. What conditions must be satisfied by a function if it is to be continuous at an interior point of its domain? At an endpoint?

12. How can looking at the graph of a function help you tell where the fiwnctionis continuous? 13. What does it mean for a function to be right-continuous at a point? Left-continuous? How are continuity and one-sided continuity related?

14. What does it mean for a fiwnction to be continuous on an interval? Give examples to illustrate the fact that a function that is not continuous on its entire domain may still be continuous on selected intervals within the domain.

15. What are the basic types of discontinuity? Give an example of each. What is a removable discontinuity? Give an example.

Chapter ~

16. What does it mean for a function to have the Intermediate Value Property? What conditions guarantee that a function has this property over an interval? What are the consequences for graphingandsolvingtheequationf(x) ~ O?

17. Under what circumstances can you extend a function f(x) to be

= c? Give an example.

continuous at a point x

18. What exact1y do limr _

oo

f(x) = L and limr-~ f(x)

~

L mean?

Give examples.

19. What are limr_±<X> k (k a constant) and lim~±<x> (I/x)? How do you extend these results to other functions? Give examples. 20. How do you f"md the limit of a rational function as x ---+ ± oo? Give examples. 21. What are horizontal and vertical asymptotes? Give examples.

Practice Exerdses

Lim;ts and ContinuUy 1. Graph the function

In Exercises 5 and 6, f"md the value that lim~og(x) must have if the given lintit statements hold.

I,

f(x) =

I

-~:

-x, I,

x:5

<x<

-I

S. lim

-1

x-o

0

x=O O<x<1

x

~

(4 -xg(x») I =

6. lim (Xlim g(x»)

x--4 x-o

~2

7. On what intervals are the following functions continuous? •• f(x) ~ XI/3 b. g(x) ~ x 3/'

c. h(x) ~ x-2/3

1.

d. k(x) ~ X-I/'

8. On what intervals are the following functions continuous?

Then discuss, in detail, limits, one-sided lintits, continuity, and one-sided continuity of f at x = -I, 0, and I. Are any of the discontinuities removable? Explain.

~

•• f(x)

tanx

cou () c.hx=x-'11'

b. g(x)

~

d. k(x)

~ s~x

cscx

F;nd;ng Lim;ts

2. Repeat the instructions of Exercise I for

In Exercises 9-24, f"md the !intit or explain why it does not exist.

0, f(x) =

I/x, 0,

1

I,

x:5

-1

0< Ixl < I x ~ I x>l.

2

9 lim x -4x+4 • x 3 + 5x2 - 14x b. as x ---+ 2

a. asx---+O

x

10. lim ,

+x

2

.3

x+2x+x

3. Suppose that f(l) and g(1) are defined for all 1 and that lim,....,. f(l) = -7 and lim,.....g(l) ~ O. Find the lintit as 1---+10 of the following functions.

a. 3f(l)

b. (j(I»)'-

c. f(I)' g(l)

f(l) d. g(l) _ 7

•• cos (g(1» g. f(l)

+ g(l)

f.

1f(t)1

h. l/f(l)

4. Suppose the functions f(x) and g(x) are def"med for all x and that limr-of(x) ~ 1/2 and limr_Og(x) = Find the!intits as x ---+ 0 of the following functions.

v'2.

a. -g(x)

b. g(x)· f(x)

c. f(x) + g(x)

d. I/f(x)

e. x

+ f(x)

f.

f(x)' cosx x I

b. asx---+-l

a. asx---+O x-I

13. lim h-O

12. lim

x

1

(x

2

;Ix

11. lim I -

+ hJ'

-

1

1

h

x2

14. lim

x-O

x

2

1/3 - I 17. lim .. x--;cc_"x~I;Ix - I . tan (z,,) 19. x-O lun tan ('11'X ) 21.

x~ sin (~ + sin x )

23. lim 3 . 8x x-O

smx-x

h

r~O

----

lS.lim2+x

2

"'----=-"-a4 (x + hJ' -

x-a x4 -

(2 16. lim

x~o

+ x)3 x

x2

- 8

2/3 - 16 18. lim .. x~___'" x~64 ;Ix - 8

20. lim cscx X-'lr-

22. lim cos2 (x - tanx) x~"

24. lim cos z" - I

x-o

smx

98

Chapter 2: Limits and Continuity

In Exercises 25-28, fmd the limit of g(x) as x approaches the indicated value.

b. Solve the equationf(6) = 0 graphically with an error of magoitude at most 10-< .

25. lim (4g(x))I/3 = 2

c. It can be shown thatthe exac!value of the solution in part (b) is

26.

x-o+

lim

x~y,

x

I

+ g(x)

=2

[19 _ )1/3 _ ([19 )1/3 ("27 I "27 + I .

2

28. lim 5-x =0 x~-2

Continuous Extension 29. Can f(x) = x(x 2 - 1)/ Ix 2

-

v'iW

Evaluate this exact answer and COO3pare it with the value you found in part (b).

II be extended to be continuous at

Limits at Infinity Find the limits in Exercises 37-46.

x = I or -I? Give reasons for your answers. (Graph the function-you will fmd the graph interesting.)

38.

30. Explain why the function f(x) = sin(l/x) has no continuous extension to x = O.

o In Exercises 31-34, graph the function to see whether it appears to

39.

and Zoom to fmd a good caodidate for the exteoded function's value at a. If the function does not appear to have a continuous extension, can it be extended to be continuous from the right or left? !f so, what do you think the extended function's value should be?

41.

have a continuous extension to the given point a. If it does, use Trace

31.

f () x

x-I = x _ ~,

33. h(l) = (I + Roots

0 35. 1.otf(x) L

110

= x3 -

1 /',

a = I

() 5cos6 32. g 6 = 46 2".'

a = 0 34. k(x) = I

43

/

a = ". 2

. x2-4x+8 I un 3 3x - 7x x2 lim -

lxJ

45. lim x + sinx :- 2Vx X

(1+ \169)113 + (L \169)113 2

2 x +4 a.Y=x-3

o

18

2

18

xl

.

2IJ

+X

2 b.f(x)=x -x-2 x2-2x+1 -

6

48. Use limits to determine the equations for all horizonts! asymptotes.

a. y

+ 2.

=

c. g(x)

l-x

2

b. f(x) = Vx+4

Vx+4

x2 + I =

'\1'7+4 x

Additional and Advanced Exercises

1. Assigning • value to 0' The rules of exponents tell us that a' = I if a is any number different from zero. They also tell us that 0" = 0 if n is any positive number. If we tried to extend these rules to include the case 0° , we would get conflicting results. The fl1'St rule would say 0° = I, whereas the second would say 0° = o. We are not dealing with a question of right or wrong here. Neither rule applies as it stands, so there is no contradiction. We could, in fact, defme 0° to have any value we wanted as long as we could persuade others to agree. What value would you like 0° to have? Here is an exan3ple that might help you to decide. (See Exercise 2 below for another example.) L

+ cos2X

c' Y =x 2 +2x_8

Use the Intermediate Value Theorem to show that f has a zero between -2 and O.

Chapter

x;3 + X-I

x_OO x 2

Horizontal and Vertical Asymptotes 47. Use limits to determine the equations for all vertical asymptotes.

Evaluate this exact answer and compare it with the value you found in part (b). L

46. lim

+ 8m x

x-I.

c. It can be shown that the exact value of the solution in part (b) is

II' -

+ x3 + 128

4

(!fyou have a grapher, try graphing f(x) = x(cos (l/x) - I) near the origin to "see" the limit at infinity.)

(J

x_OO

x

x_co 12x 3

(!fyou have a grapher, try graphing the function for -5 :5 x :5 5.)

lim sinx

• x~oo

42. lim

1

a= 0

b. Solve the equation f(x) = 0 graphically with an error of magoitude at most 10-8 •

=

+

x_-oo x

44. lim cos 6 - I

Use the Intermediate Value Theorem to show that f has a zero between -I and 2.

0 36. 1.otf(6)

~+3 +7

x_-oo

8_00

_\w

lim

x_-oo 5x 2

Calculate rfor x = 0.1,0.01,0.001, and so on as far as your calculator can go. Record the values you get. What pattern do you see?

b. Graph the function y = XX for 0 < X :5 I. Even though the function is not defmed for x :5 0, the graph will approach the y-axis from the right. Toward what y-vaIue does it seem to be headed? ZoOO3 in to further support your idea.

0

2. A reason you might want 0° to be something other than 0 or 1 As the number x increases through positive values, the numbers I/x and I/(In x) both approach zero. What happens to the number

f(x)

=

(

xI )1/(1",)

as x increases? Here are two ways to rmd out.

a. Evaluate f forx = 10, 100, 1000, and so on as far as your calculator can reasonably go. What pattern do you see?

99

Chapter 2 Additional and Advanced Exercises b. Graph f in a variety of graphing windows, including windows that contain the origin. What do you see? Trace the y-values along the graph. What do you find?

How closely must we measure h to measure out 1 L of water (1000 em 3) with an error of no more than 1% (10 cm3)?

3. Lorentz contraction In relativity theory, the length of an object, say a rocket, appears to an observer to depend on the speed at which the object is traveling with respect to the observer. If the observer measures the rocket's length as Lo at rest, then at speed v the length will appear to be

L

=

~

LO\j1 - ~ .

This equation is the Lorentz contraction formula. Here, c is the speed of light in a vacuum, about 3 X 10 8 m/sec . What happens to L as v increases? Find limv-->c- L . Why was the left-hand limit needed?

(a)

r= 6cm

4. Controlling the flow from a draining tank Torricelli's law says that if you drain a tank like the one in the figure shown, the rate y at which water runs out is a constant times the square root of the water's depth x. The constant depends on the size and shape of the exit valve.

/'

t,

"-

/

T 1// h

----

\

Liquid volume V= 367Th

~ (b)

A l-L measuring cup (a), modeled as a right circular cylinder (b) of radius r = 6 cm

Suppose that y = Yx/2 for a certain tank. You are trying to maintain a fairly constant exit rate by adding water to the tank with a hose from time to time. How deep must you keep the water if you want to maintain the exit rate a. within 0.2 ft3/min of the rate Yo = I ft3/min? b. within 0.1 ft3/min of the rate Yo = I ft3/min? 5. Thermal expansion in precise equipment As you may know, most metals expand when heated and contract when cooled. The dimensions of a piece oflaboratory equipment are sometimes so critical that the shop where the equipment is made must be held at the same temperature as the laboratory where the equipment is to be used. A typical aluminum bar that is 10 cm wide at 70°F will be y = 10

+ (t

- 70) X 10- 4

centimeters wide at a nearby temperature t. Suppose that you are using a bar like this in a gravity wave detector, where its width must stay within 0.0005 cm of the ideal 10 cm. How close to to = 70°F must you maintain the temperature to ensure that this tolerance is not exceeded? 6. Stripes on a measuring cup The interior of a typical l-L measuring cup is a right circular cylinder of radius 6 cm (see accompanying figure). The volume of water we put in the cup is therefore a function of the level h to which the cup is filled, the formula being

Precise Definition of Limit In Exercises 7- 10, use the formal definition of limit to prove that the function is continuous at xo .

7. f(x) = x 2

-

7,

Xo

9. h(x) =~,

= 1 Xo

8. g(x) = 1/(2x),

Xo

= 2 10. F(x) =~,

= 1/4 Xo

=5

11. Uniqueness of limits Show that a function cannot have two different limits at the same point. That is, if limx-->xo f(x) = L J and limx-->xo f(x) = L2, then LJ = L2 . 12. Prove the limit Constant Multiple Rule: lim kf(x) = k lim f(x)

x---+c

x---+c

13. One-sided limits

for any constant k.

If limx-->o+ f(x) = A and limx-->o- f(x) = B,

find a. limx-->o+ f(x 3

-

x)

b. limx-->o- f(x 3

-

x)

c. limx-->o+ f(x 2

-

x4)

d. limx-->o- f(x 2

-

x 4)

14. Limits and continuity

Which of the following statements are true, and which are false? If true, say why; if false, give a counterexample (that is, an example confirming the falsehood). a. If limx-->a f(x) exists but limx-->a g (x) does not exist, then limx-->a(J(x) + g(x)) does not exist.

b. If neither limx-->a f(x) nor limx-->a g (x) exists, then limx-->a (J(x) + g (x)) does not exist. c. If f is continuous at x, then so is 1f

I.

d. If 1f 1is continuous at a, then so is j.

100

Chapter 2: Limits and Continuity

In Exercises 15 and 16, use the formal defmitioo oflimit to prove that the function has a continuous extension to the given value of x.

15. f(x) =

x'-I

X+T'

x = -I

16. g(x) =

x'-2x-3 2x 6 '

x = 3

17. A function continuous at only one point Let

= f (x ) L

{x, ifxis ratiooa1

0, if x is irratiooa1.

Show that f is continuous at x

= O.

b. Use the fact that everynonempty open interval of real num-

hers cootains both ratiooa1 and irratiooa1 numbers to show that f is not continuous at any nonzero value of x. 18. The Dirichlet ruler function If x is a rational number, then x can be written in a unique way as a quotient of integers min where n > 0 and m and n have no commoo factors greater than I. (We say that such a fractioo is in luwest terms. For example, 6/4 written in lowest terms is 3/2.) Letf(x) be defmed for alb in the interval [0, I] by f(x)

= {I/n, ~x.= .m/~ is aratiooal number in lowest terms if x IS Irrational.

0,

Forinstauce,f(O) = f(1) = l,f(I/2) = 1/2,f(I/3) = f(2/3) = 1/3, f(1/4) = f(3/4) = 1/4, aod so on. L

exists, a lower bound for fooD aod say that f is bouoded froro below by M. We say that f is bounded 00 D if it is bouoded froro both above aod below. a. Show that f is bouoded ooD ifaod only if there exists a numberBsuchthat If(x) I '" B for allxinD.

b. Suppose that f is bouoded from above by N. Show that if limr->",f(x) = L, thenL '" N. c. Suppose that f is bouoded from below by M. Show that if limr->",f(x) = L, thenL '" M. 24. Mu {a, b} and min {a, b} a. Show that the expression

max {a, b}

c. Sketch the graph of f. Why do you think f is called the ''ruler function"? 19. Antipodal points Is there any reason to believe that there is always a pair of antipodal (diaroetrically opposite) points 00 Earth's equator where the temperatures are the same? Explain.

20. If lim (j(x) + g(x)) = 3 and lim (j(x) - g(x)) = -I, find x-c

x-c

lim f(x)g(x). x~,

21. Root. of a quadratic equation that is almost linear The equation ax 2 + 2x - 1 = 0, where a is a constant, has two roots if a > -1 and a '¢ O. one positive and one negative: T+ ( a )

_ -I -

+

'V'1+a a

'

r_(a)

=

-I -

'V'1+a a .

What happens to r+(a) as a --> 01 As a --> -1+1 b. What happens to r_(a) as a --> 01 As a --> -1+1 L

c. Support your cooclusions by graphing r+(a) aodr_(a) as functions of a. Describe what you see. d. For added support, graph f(x) = ax' + 2x - I simultaneously for a = 1,0.5,0.2, 0.1, aod 0.05. 22. Root of an equation Show that the equation x + 2 cos x = 0 has at least one solution. 23. Bounded functions A real-valued func1ioo f is bounded from above on a set D if there exists a number N such that f(x) '" N for all x in D. We call N, wheo it exists, ao upper bound for f 00 D aod say that f is bouoded froro above by N. In a similar marmer, we say that f is bounded from below on D if there exists a number M such that f(x) '" M for all x in D. We call M, wheo it

a +b -2-

la - bl

+

2

equals a if a '" b and equals b if b '" a. In other words, max {a, b} gives the larger of the two numhers a aodb. b. Find a similar expression for min {a, b}, the smaller of a aodb. Generalized

Limn. Involving 5'1: 8

The formula lime_o(sin 8)/8 = I cao be generalized. If lim,_, f(x) = 0 aod f(x) is never zero in an open interval containing the point x = c, except possibly c itself, theo lim sinf(x) = I f(x) .

Show that f is discootinuous at every ratiooa1 number in [0, I].

b. Show thatf is continuous at every irratiooa1 number in [0, I]. (Hint If. is a given positive number, show that there are only fmitelymaoyratiooa1 numbers rin [0, I] such that f(r) '" E.)

=

x~,

Here are several examples.

.

a. lim. smx

,

=

x-a x 2 .

1

2

.

2

2

b. lim smx = lim smx lim "-- = 1·0 = 0

x-a

x-o x 2 x-o

X

X

. sin (x' - x - 2) . sin (x' - x - 2) c. lim Ion • x--l X + 1 x--l (x 2 - X - 2) . (x' - x - 2) . (x + I)(x - 2) lim =1, lim =-3 x--l x + 1 x--l x + 1 . sin (I - '\!'x) . sin (I - '\!'x) I - '\!'x d. Ion = Ion = x-I x-I x-I 1 _ Vx x-I

I· lim (I - '\!'x)(1 + '\!'x) x~l (x - 1)(1 + '\!'x)

lim I-x x~l (x - 1)(1 +

Find the limits in Exercises 25-30. sin (1 - cosx) 26. lim 25. lim x x~o

x-O+

27. lim

sin (sinx) x

28. lim

29. lim

sin (x' - 4) 2

30. lim

x-o

%-2

x-o

sinx sinYx

sin (x' x

+ x)

sin ('\!'x x-9

X

'\!'x)

x

- 3) 9

Obl;que Asymptotes Find all possible oblique aaymptotes in Exercises 31-34. 31. Y

=

33. y=

+ 2x • ['

2x 3/2

vx + I

W+1

3

32. Y

= x + x sin (I/x)

34. y

=

\Ix' + 2x

I 2

Chapter 2 Technology Application Projects

Chapter

fJ

101

Technology Application Projects

Mathematica,tMap1e Modules: 1IIke It to the Limit Part I Part II (Z.ro Rais.d to tb. P.....r Z.ro: What Does it M.an?) Part III (On.-Sid.d Limit.) Visualize and interpret the limit concept through graphical and numerical explorations.

Part IV (What a Diff.renc. a Pow.r Makes) See how sensitive limits can be with various powers of x. Going to Injmity Part I (EIploring Function Behavior as x ---+ oc or x ---+ - 00) lbis module provides four examples to explore the behavior of a function as x ---+ 00 or x ---+ - 00 . Part II (Rate. of Growtb) Observe graphs that appear to be continuous, yet the function is not continuous. Several issues of continuity are explored to obtain results that you may find surprising.

3 DIFFERENTIATION OVERVIEW In the beginning of Chapter 2 we discussed how to determine the slope of a curve at a point and how to measure the rate at which a function changes. Now that we have studied limits, we can define these ideas precisely and see that both are interpretations of the derivative of a function at a point. We then extend this concept from a single point to the derivative junction, and we develop rules for finding this derivative function easily, without having to calculate any limits directly. These rules are used to find derivatives of most of the common functions reviewed in Chapter 1, as well as various combinations of them. The derivative is one of the key ideas in calculus, and we use it to solve a wide range of problems involving tangents and rates of change.

3.1

Tangents and the Derivative at a Point In this section we define the slope and tangent to a curve at a point, and the derivative of a function at a point. Later in the chapter we interpret the derivative as the instantaneous rate of change of a function, and apply this interpretation to the study of certain types of motion.

y

Finding a Tangent to the Graph of a Function To find a tangent to an arbitrary curve y = f(x) at a point P(xo, f(xo)) , we use the procedure introduced in Section 2.1. We calculate the slope of the secant through P and a nearby point Q(xo + h, f(xo + h)). We then investigate the limit of the slope as h ~ 0 (Figure 3.1). If the limit exists, we call it the slope of the curve at P and define the tangent at P to be the line through P having this slope.

-O~----~Xo--------~~----+X

DEFINITIONS

The slope ofthe curve y

= f(x) at the point P(xo, f(xo)) is the

number FIGURE 3.1 The slope of the tangent · p. 1· f(xo + h) - f(xo) 1me at IS un h

.

m = hm

h~O

f(xo

+ h) - f(xo) h

(provided the limit exists).

h..... O

The tangent line to the curve at P is the line through P with this slope.

In Section 2.1, Example 3, we applied these definitions to find the slope of the parabola f(x) = x 2 at the point P(2, 4) and the tangent line to the parabola at P. Let's look at another example.

102

3.1 Tangents and the Derivative at a Point

103

EXAMPLE 1

y

(a) Find the slope of the curve y = I/x at any point x = a '" O. What is the slope at the pointx = -I?

(b) Where does the slope equal -1/4? (c) What happens to the tangent to the curve at the point (a, I/a) as a changes? Solution (a) Here f(x) = I/x. The slope at (a, I/a) is slope is-l atx =-1

1

lim f(a

+ hh) - f(a)

h~O

FIGURE 3.2 The tangent slopes, steep near the origin, become more gradual as

the point of tangency moves away (Example I). y

=

I

a

1

+ hh - a = -h

·

h~ ha(a

+ h)

lim

1 a - (a + h)

h~O

h a(a + h)

lim

=

-I

h~O a(a

+ h)

I

a2 '

Notice how we had to keep writing "Iimh~o" before each fraction until the stage where we could evaluate the limit by substituting h = O. The number a may be positive or negative, but not O. When a = -1, the slope is -1/(-1)2 = -1 (Figure 3.2).

(b) The slope of y = I/x at the point where x = a is -1/a 2• It will be -1/4 provided that

y=l s1opels-"4 . 1

= lim h~O

(2. ~)

~'----t-~x s1 . 1 opels-

4

FIGURE 3.3 The two tangent lines to y = I/x having slope -1/4 (Example I).

This equation is equivalent to a 2 = 4, so a = 2 or a = -2. The curve has slope -1/4 at the two points (2, 1/2) and (-2, -1/2) (Figure 3.3). (c) The slope -1/a 2 is always negative if a '" O. As a --+ 0+, the slope approaches -00 and the tangent becomes increasingly steep (Figure 3.2). We see this situation again as a --+ 0- . As a moves away from the origin in either direction, the slope approaches 0 and the tangent levels off to become horizontal. •

Rates of Change: Derivative at a Point The expression

f(xo + h) - f(xo) h is called the difference quotient of I at Xo with increment h. If the difference quotient has a limit as h approaches zero, that limit is given a special name and notation.

DEFINmON

The derivative ofa function I at a point Xo, denoted f'(xo), is

f' (xo)

= lim h~O

f(xo + h) - f(xo) h

provided this limit exists.

If we interpret the difference quotient as the slope of a secant line, then the derivative gives the slope of the curve y = f(x) at the point P(xo, f(xo». Exercise 31 shows

104

Chapter 3: Differentiation that the derivative of the linear function f(x) = mx of the line, so

f'(xo)

=

+ b at any point Xo is simply the slope

m,

which is consistent with our definition of slope. If we interpret the difference quotient as an average rate of change (Section 2.1), the derivative gives the function's instantaneous rate of change with respect to x at the point x = Xo. We study this interpretation in Section 3.4.

EXAMPLE 2 In Examples 1 and 2 in Section 2.1, we studied the speed ofa rock falling freely from rest near the surface of the earth. We knew that the rock fell y = 161 2 feet during the first 1 sec, and we used a sequence of average rates over increasingly short intervals tu estimate the rock's speed at the instant 1 = 1. What was the rock's exacl speed at this time? Solution We let f(l) = 1612. The average speed of the rock over the interval between 1 = 1 and 1 = 1 + h seconds, for h > 0, was found tu be

f(1 + h) - f(l)

16(1

+ h)2 -

h

16(1)2

h

The rock's speed at the instant 1 = 1 is then

lim 16(h

h~O

+ 2)

= 16(0

+ 2)

= 32 ft/sec.

Our original estimate of 32 ft/ sec in Section 2.1 was right.

•

Summary We have been discussing slopes of curves, lines tangent to a curve, the rate of change of a function, and the derivative of a function at a point. All of these ideas refer to the same limit.

The following are all interpretations for the limit of the difference quotient, lim f(xo h-O

+ h) - f(xo) h

1. The slope of the graph of y = f(x) at x = Xo

2. The slope of the tangent tu the curve y = f(x) at x = Xo 3. The rate of change of f(x) with respect to x at x = Xo

4. The derivative f'(xo) at a point

In the next sections, we allow the point Xo to vary across the domain of the function f.

3.1

Tangents and the Derivative at a Point

105

Exercises 3.1 Slopes and Tangent Unes In Exercises 1-4, use the grid and a straight edge to make a rough esti· mate of the slope of the curve (in y-units per x·unit) at the points PI

Tangent Unes with Spedfied Slopes At what points do the graphs of the functions in Exercises 23 aod 24 have horizontal tangents?

andP,.

23. f(x) ~ x'

1.

2.

y

y

+ 4x - I

24. g(x) ~ x' - 3x

25. Find equations of all lines having slope -I that are tangent to the curve y ~ I/(x - I). 26. Find an equatioo of the straight line having slope 1/4 that is tangent to the curve y ~ Yx.

1

Rates of Change 27. Object dropped from. tower Ao oi!ject is dropped from the top of a 100·m-high tower. Its height above ground after t sec is 100 - 4.9t' m. How fast is it falling 2 sec after it is dropped? 28. Speed of a rocket At t sec after liftoff, the height of a rocket is 3t' ft. How fast is the rocket climbing 10 sec after liftoff?

3.

4.

y

29. Circle'. changing area What is the rate of chaage of the area of a circle (A ~ 'lTr') with respect to the radius when the radius is r ~ 3?

y

30. Ball'. changing volume What is the rate of change of the volume ofa ball (V ~ (4/3)'lTr') with respect to the radius when the radius is r = 2? 31. Show that the line y point (xo, nlXo + b).

81

~ nIX

+b

is its own tangent line at any

32. Find the slope of the tangent to the curve y ~ I/Yx at the point where x = 4.

o

Testing for Tangents

33. Does the graph of In Exercises 5-10, fmd an equatioo for the tangent to the curve at the given point. Then sketch the curve and tangent together.

5. Y ~ 4 - x', 7. Y

~

9. Y ~

2Yx,

(-1,3)

6. Y ~ (x - I)' 8. Y

(1,2)

x', (-2, -8)

10. Y

~

1"x

+

(I, I)

I,

have a tangent at the origin? Give reasons for your answer.

(-I, I)

~ :',

34. Does the graph of

(-2,-t)

In Exercises 11-18, fmd the slope of the function's graph at the given point. Then find an equatioo for the line tangent to the graph there. 11. f(x) ~ x'

+ I, (2,5)

x 13. g(x) ~ x _ 2' 15. h(t) ~ t',

(3,3)

(2,8)

17. f(x) ~ Yx,

(4,2)

12. f(x) ~ x - 2>;',

18. f(x) ~

Vx+1,

+ 3t,

(1,4) (8,3)

In Exercises 1!l-22, find the slope of the curve at the point indicated. 19. Y

=

5x 2 ,

I

21. Y ~ x _ I'

20. Y = 1 - x 2,

-1

X =

x

~

3

x - I

22'Y~x+I'

g(x)

~

{""in (I/x), 0,

x # 0 x =0

have a tangent at the origin? Give reasons for your answer.

(1,-1)

(2,2)

14. g(x) ~

8 2:'

x 16. h(t) ~ t'

f(x) ~ {X'Sin(I/X), x # 0 0, x= 0

X =

2

x~O

Vertical Tangents We say that a continuous curve y ~ f(x) has a vertical tangent at the point where x ~ Xo if lim,_o (/(xo + h) - f(xo))/h ~ 00 or -00. For example, y ~ x 1/' has a vertical tangent at x ~ 0 (see accompanying figure): lim f(O 4-0

+ h) - f(O) h ~ lim _1_ ~ h-O h2j3

00

.

106

Chapter 3: Differentiation y

36. Does the graph of

{a,

U(x) =

x<

°

x ~ 0 have a vertical tangent at the point (0, I)? Give reasons for your 1,

answer.

--------~---------.x

D Graph the curves in Exercises 37-46. .. Where do the graphs appear to have vertical tangents? b. ConIrrrn your rmdings in part (a) with limit calculations. But before you do, read !be introductioo to Exercises 35 aod 36. 37. y = x 2/' 38. y = x4/'

VERTICAL TANGENr AT ORIGIN

However, y = x 2/ 3 has no vertical tangent at x =

g(O + h) - g(O) 11m h-O h .

=

°

(see next figure):

.

..h2_/3~_----,,0

h-O

h

lIm-

=

43. y

=

X 2/ 3 -

45. y

=

{-vfxI, Yx,

x /' ' 41. y = 4x 2/'

= lim _1_ h-O h lj3

40. y = x 3/'

39. y

-

42. y = x'/3 - 5x 2/ 3

2x

(x - 1)'/3

x'"

44. y

°

46. y

x>O

= x ' /3 + (x

_ 1)'/3

= ~4 - xl

COMPUTER EXPLORATIONS

does not exist, because the limit is 00 from the right and left.

00

from the

y

Use a CAS to perform the following steps for the functions in Exercises 47-50: L

Platy = f(x) over the interval (xo - 1/2) s x'" (xo

+ 3).

b. Holding Xo fixed, the difference quotient

--------~---------.x

o

c. Find the limit of q as h ---. 0.

NOVFRI1CAL TANGENT KI ORIGIN

d. Derme the secant lines y = f(xo) + q' (x - xo) for h = 3, 2, and I. Graph theD3 together with f aod !be tangent line over the interval in part (a).

35. Does the graph of

f(x) =

{-~: : ~ ~ 1, x>

47. f(x) = x 3 + 2x, Xo =

°

have a vertical tangent at the origin? Give reasons for your answer.

3.2

= f(xo + h)

- f(Xo) h at Xo becomes a function of the step size h. Enter this function into your CAS workspace. q(h)

°

48. f(x) = x + 49. f(x) = x + sin (2x), Xo = 7T/2 SO. f(x) = cosx + 4 sin (2x), Xo = 7T

5 x,

Xo = I

The Derivative as a Function In the last section we dermed the derivative of y = f(x) at the point x = Xo to be the limit

HISTORICAL ESSAY

/'(xo)

=

lim f(xo + h) - f(xo) h~O

The Derivative

h

We now investigate the derivative as a junction derived from f by considering the limit at each point x in the domain of f.

DEFINITION The derivative of the function f(x) with respect to the variable x is the function /' whose value at x is /,(x) = lim f(x h~O

provided the limit exists.

+

hl-

f(x),

107

3.2 The Derivative as a Function

z-x

T

f(z) - f(x)

~"'--------i, ~ ~h =

I

z-

X~:

We use the notation f(x) in the def"mition to emphasize the independent variable x with respect to which the derivative function I'(x) is being defined. The domain of I' is the set of points in the domain of f for which the limit exists, which means that the domain may be the same as or smaller than the domain of f. If I' exists at a particular x, we say that f is differentiable (has a derivative) at x. If I' exists at every point in the domain of f, we call f differentiable. If we write z = x + h, then h = z - x and h approaches 0 if and only if z approaches x. Therefore, an equivalent definition of the derivative is as follows (see Figure 3.4). This formula is sometimes more convenient to use when finding a derivative function.

I

..

- --*.. -~-

x

z-x+h

Derivative off at x is f'(x) ~ lim f(x + h) - f(x) h

IHO

~

Alternative Formula for the Derivative

I'(x)

=

lim f(z) - f(x) .t-+x

Z

lim f(z; -

~(x) .

z~x

X

FIGURE 3.4 Two forms for the difference quotient.

Calculating Derivatives from the Definition The process of calculating a derivative is called differentiation. To emphasize the idea that differentiation is an operation performed on a function y = f(x) , we use the notation d

dxf(x)

I Derivative of the Reciprooal Funmon ~(l) ~ _l dx X x2'

as another way to denote the derivative I'(x). Example I of Section 3.1 illustrated the differentiation process for the function y = I/x when x = a. For x representing any point in the domain, we get the formula

x" 0

~(l) =_l2 dx x x • Here are two more examples in which we allow x to be any point in the domain of f. Differentiate f(x) = ~I .

EXAMPLE 1

x-

Solution We use the def"mition of derivative, which requires us to calculate f(x then subtract f(x) to obtain the numerator in the difference quotient. We have

f(x) I'(x)

x

=

x _ I

= lim h~O

and

f(x + h)

=

+ h) and

(x + h) (x + h) _ I ,so

f(x + h) - f(x) h

Definition

x x - I

---

. I (x + h)(x - I) - x(x + h - I) =l~h·~-(~x~+~h~--I~)~(x~-~I)~~ = lim h-O

1.

h (x

= lim h-O

(x

+h-

+h

-h

Simplity

I)(x - I)

-I - I)(x - I)

-I

(X-i)2·

Cancelh '# 0

•

108

Chapter 3: Differentiation

EXAMPLE 2

Vi for x > O. Find the tangent line to the curve y = Vi at x


I

4.

=

Solution

Derivative of the Square Root Function

:! Vx


I'(x) =

lim f(z; - t(x)

=

z-x

.1 r' x> 0 2vx

lim

=

z-x

y

Vi-Vi (Vi - Vi) ( Vi + Vi)

. Iun

=

I

z-xVi

I

2Vi'

+ Vi

(b) The slope of the curve at x = 4 is

1'(4) = _1_ =

2V4

FIGURE 3.5 The curve y = Vi and its tangent at (4, 2). The tangent's slope is found by evaluating the derivative at x = 4

1. 4

The tangent is the line through the point (4, 2) with slope 1/4 (Figure 3.5): I

y=2+

y =

4 (x-4)

I 4x +

•

I.

(Example 2).

Notations There are many ways to denote the derivative of a function y = f(x), where the independent variable is x and the dependent variable is y. Some common alternative notations for the derivative are dy df d I'(x) = y' = dx = dx = dx f(x) = D(j)(x) = Dxf(x).

The symbols d/ dx and D indicate the operation of differentiation. We read dy/ dx as ''the derivative ofy with respect to x;' and df/dx and (d/dx)f(x) as "the derivative of f with respect to x." The "prime" notations y' and f' come from notations that Newton used for derivatives. The d/dx notations are similar to those used by Leibniz. The symbol dy/dx should not be regarded as a ratio (until we introduce the idea of "differentials" in Section 3.9). To indicate the value of a derivative at a specified number x = a, we use the notation

dyl dx

= -d/l

I'(a) = -

;c=a

dx

= -d f(x) dx

;c=a

I

.

;c=a

For instance, in Example 2

1'(4) =

A. Vii dx

= x=4

_I_I 2Vi

= _1_ = 1 x=4

2V4

4.

Graphing the Derivative We can often make a reasonable plot of the derivative of y = f(x) by estimating the slopes on the graph of f. That is, we plot the points (x, I' (x» in the xy-plane and connect them with a smooth curve, which represents y = I' (x).

109

3.2 The Derivative as a Function

EXAMPLE 3

Graph the derivative of the function y = I(x) in Figure 3.6a.

We sketch the tangents to the graph of I at frequent intervals and use their slopes to estimate the values of f' (x) at these points. We plot the corresponding (x, f' (x)) pairs and connect them with a smooth curve as sketched in Figure 3.6b. • Solution

What can we learn from the graph of y =

f' (x)? At a glance we can see

1. where the rate of change of I is positive, negative, or zero; 2. the rough size of the growth rate at any x and its size in relation to the size of I(x); 3. where the rate of change itself is increasing or decreasing. y

Slope 0 A

y ~f(X)/

......

10 "

-

Slo

,.-1

B,\ C

5

: ;/

\.

Slope 0

- 8y-units

/

- 4 x-units lp l~

5

0

8 = 2 y-um'Wx-umt . /Slope - 4

Slope _1

I

x

(a)

Slope

4

Y ~f'(xy

3

2 1

A"

D'

-1

)'

~~ 1O

-2

V!,:.;ca1
15

t

x

(b)

Slope = lim f(a ......

Slope = lim f(b .-..0-

+ h) -

+ h) -

f(b)

h

FIGURE 3.6 We made the graph ofy = f'(x) in (b) by plotting slopes from the graph ofy = j(x) in (a). The vertical coordinste of B' is the slope atB aod so 00. In (b) we see that the rate ofchaoge of j is negative for x between A' andD'; the rate of change is positive for x to the right of D'.

f(a)

Differentiable on an Interval; One-Sided Derivatives

h

A function y = I(x) is difIerentiable on an open interval (finite or inimite) if it has a derivative at each point of the interval. It is difIerentiable on a closed interval [a, b1if it is differentiable on the interior (a, b) and if the limits ---L---L~----~~--~--~,x

a

a+h h>O

b+h h
FIGURE 3.7 Derivatives at endpoints are one-sided limits.

lim I(a h---+O+

lim I(b h~O-

+ h) - I(a) h

+ h) - I(b) h

exist at the endpoints (Figure 3.7).

Right-hand derivative at a Left-hand derivative at b

110

Chapter 3: Differentiation Right-hand and left-hand derivatives may he dermed at any point of a function's domain. Because of Theorem 6, Section 2.4, a function has a derivative at a point if and only if it has left-hand and right-hand derivatives there, and these one-sided derivatives are equal.

EXAMPLE 4 Show that the function y ~ but has no derivative at x ~ O.

y

Ixl is differentiable on (-00,0) and (0, 00)

Solution From Section 3.1, the derivative of y right of the origin,

~

mx

+b

is the slope m. Thus, to the

d

dx (moc

--------~~------~x

+ b)

~

m,

Ixl ~ x

0\

y' not defined atx = 0:

right-hand derivative left-hand derivative

To the left,

*

FIGURE 3.8 The function y ~ Ixl is not differentiable at the origin where the graph has a "comer" (Example 4).

d --(Ixl) dx

=

d d --(-x) = --(-I'x) =-1

dx

dx

(Figure 3.8). There is no derivative at the origin because the one-sided derivatives differ there: Right-hand derivative of Ix Iat zero = lim

10 + hl-IOI h

h-O+

=

Ihl

lim --h

h--+O+

Ihl ~ hwhenh

Left-hand derivative of Ix Iat zero = lim

10 + hl-IOI h

h--+O-

=

>0

Ihl

lim --h

h--+O-

Ihl ~ -h whenh

= lim-I =-1. h--+O-

EXAMPLE 5

In Example 2 we found that for x

>

<0

•

0,

We apply the definition to examine if the derivative exists at x = 0:

Since the (right-hand) limit is not finite, there is no derivative at x = O. Since the slopes of the secant lines joining the origin to the points (h, YJ.) on a graphofy = vX approach 00, the gmph has a vertical tangent at the origin. (See Figure 1.17 on page 9). •

When Does a Function Not Have a Derivative at a Point? A function has a derivative at a point Xo if the slopes of the secant lines through P(XO, /(xo» and a nearby point Q on the gmph approach a rmite limit as Q approaches P. Whenever the secants fail to take up a Iimitiug position or become vertical as Q approaches P, the derivative

3.2

The Derivative as a Function

111

does not exist. Thus differentiability is a "smoothness" condition on the graph of f. A function can fail to have a derivative at a point for many reasons, including the existence of points where the graph has

/ 1. a corner, where the one-sided

2. a cusp, where the slope of PQ approaches 00 from one side aod - 00 from the other.

derivatives differ.

/ 3. a vertical tangent, where the slope of PQ approaches 00 from both

sides or approaches -

4. a discontinuity (two examples shown).

00

from both sides (here, -

00).

Another case in which the derivative may fail to exist occurs when the function's slope is oscillating mpidly near P, like f(x) = sin (l/x) near the origin, where it is discontinuous (see Figore 2.31).

Differentiable Functions Are Continuous A function is continuous at every point where it has a derivative.

THEOREM l-DifferentiabUUy Implies Continuity x = c, then / is continuous at x = c.

If

f

has a derivative at

Proof Given that f' (e) exists, we must show that limx _, f(x) = f( e), or equivalently, thatlimh_of(e + h) = f(e).Ifh '" O,then f(e

+ h)

= f(e)

+

(f(e

= f(e)

+

f(e

+ h)

+ h~

- f(e)) - f(e) 'h.

112

Chapter 3: Differentiation

Now take limits as h --+ o. By Theorem I of Section 2.2, lim j(c

+ h)

lim j(c)

=

11-0

+

11-0

lim j(c

+ h)

=

j(c) + f'(c)' 0

=

j(c) + 0

=

j(c).

- j(c) • lim h

h

11-0

11-0

•

Similar arguments with one-sided limits show that if j has a derivative from one side (right or left) at x = c then j is continuous from that side at x = c. Theorem I says that if a function has a discontinuity at a point (for instance, a jump discontinuity), then it cannot be differentiable there. The greatest integer function Y = l x J fails to be differentiable at every integer x = n (Example 4, Section 2.5). Caution The converse of Theorem I is false. A function need not have a derivative at a point where it is continuous, as we saw in Example 4.

Exercises 3.2 Finding Derivative Functions and values Using the delmition, calculate the derivatives of the functions in Exercises I-J). Then Imd the values of the derivatives as specified. 1. j(x) = 4 - x 2 ; /,(-3), /,(0), /,(1) 2. F(x) = (x - 1)2 + I; F'(-I),F'(0),F'(2)

3. g(l) =

1,; 1

g'( -I), g'(2), g'( \13)

v30;

=

6. r(s)

= ~;

""

9. dl

dp 11. dq

8 = j(x) = .'------;:'

18. w

= g(z)

20.

p'(1),p'(3),p'(2/3) r'(O), r'(I), r'(1/2)

,· f' y=2s

8.""dr

1 if s=21+1 if p=

17. y

19.

In Exercises 7-12, Imd the indicated derivatives.

7.""dy

of the tangent line at the indicated point on the graph of the function.

I

yq:tJ

= I +~,

,·f

r

21. =

10 dv • dt

if v

dz 12. dw

if z=

s' - 29 2

= t -

+3

22.

ds l dt 1=-1

dyl

if y=

I-~

r =

2

"" x-v. if drl dO '-0

:t.

v'4=9

if w=z+

I

I Y/3w - 2

the tangent line at the given value of the independent variable.

14. k(x) = 2

I

x'

x = -3

+ x' x

15. s = I' - 12,

x+3

16. y = -I- ,

-x

= 2

1 = -I

x = -2

Yz

t

Using the Alternative Formula for Derivatives Use the formula

/,(x)

Slopes and rangent Lines

13. j(x) = x +

(z, w) = (3,2)

ifs=I-312

=

lim j(z) - j(x) %-x

In Exercises 13-16, differentiate the functions aod Imd the slope of

9

(x,y) = (6,4)

yx - 2

In Exercises 1!)"22, find the values of the derivatives.

4. k(z) = I ; z; k'( -I), k'(i), k'( Yz) 5. p(O)

In Exercises 17-18, differentiate the functions. Then Imd an equation

Z

X

to Imd the derivative of the functions in Exercises 23-26. I 23. j(x) = x + 2

24. j(x) = x 2

-

3x

x 25. g(x) = - - I

x-

26. g(x) = I

+ Vx

+4

3.2

Graphs Match the functions graphed in Exercises 27-30 with the derivatives graphed in the accompanying figures (a)-(d).

The Derivative as a Function

113

32. Recovering a function from its derivative a. Use the following information to graph the function f over the closed interval [ - 2, 5]. i) The graph off is made of closed line segments joined end to end.

ii) The graph starts at the point ( - 2, 3). iii) The derivative of / is the step function in the figore shown here. y' (b)

(a)

1

0-----0

y' C>--_

- -1- -2

(c)

27.

(d)

28.

y

y

b. Repeat part (a) assuming that the graph starts at (-2,0) instead of( -2,3).

33. Growth in the economy The graph in the accompanying figore shows the average aonual percentage change y ~ /(t) in the u.s. gross natioua1 product (GNP) for the years 1983-1988. Graph dy/dt (where defmed). 7%

------~~~----~x

6 5

------~~~----~x

o

4

--fr\

I-

l \

3 2

............. /"

-

1

29.

30.

y

o

y

1983 1984 1985 1986 1987 1988 34. Fruit rues (Continuation of Example 4, Section 2.1.) Populations starting out in closed environments grow slowly at fust, when there are relatively few members, then more rapidly as the

---\--~~4---+-~x

number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the carrying capacity of the environment

31. a. The graph in the accompanying figore is made of line segments joined end to eml. At which points of the interval [-4, 6] is f' not defined? Give reasons for your answer.

a, Use the graphical tecbrtique of Example 3 to graph the derivative of the fruit fly population. The graph of the population is reproduced here. p

y

~

,.....

350 300 250

V

/

'a 200

Loo

150

Z

(1. -2)

(4.-2)

50

o

/ / ./

10

20

30

40

50

TIme (days)

b. Graph the derivative off. The graph should show a step function.

b. During what dsys does the population seem to be increasing fastest? Slowest?

114

Chapter 3: Differentiation

35. Temperature The giveo graph shows the temperature T in OF at Davis, CA, on April 18, 2008, between 6 A.M. and 6 P.M.

39.

40.

Y

Y

Y ~ J(x)

T

P(I. I)

80

.....

G:' 70

l- I-'

'"

------~--7-----~x

/

y=x

'\

--~Onr--7-------~X

In Exercises 41 and 42, determine if the piecewise defined function is differentiable at the origin.

o

3

6 A.M.

9 A.M.

6

9

12

12 NOON 3 P.M. Tune (brs)

41. f(x)

6 P.M.

a. Estimate the rate of temperature change at the times i) 7 A.M.

ii) 9 A.M.

iii) 2 P.M.

~

42. g(x) ~

{2x - I, x + 2x + 7, 2

{

X2/3,

X ~

xli',

x

crease most rapidly? What is the rate for each of those times? c. Use the graphical technique of Example 3 to graph the derivative of temperature Tversus time t. 36. Weight Loss Jared Fogle, also known as the "Subway Sandwich Guy;' weighed 425 Ib in 1997 before losing more than 240 Ib in 12 months (http://en.wikipedia.org/wiki/Jared_Fogle). A chart showing his possible dramatic weight loss is given in the accompanying figure.

<0

Differentiability and Continuity on an Interval Each figure in Exercises 43-48 shows the graph of a function over a closed interval D. At what domain points does the function appear to be

a. differentiable? b. continuous but not differentiable? c. neither continuous nor differentiable?

Give reasons for your answers. 44.

Y ~J(x) Y D: -3:Sx:s2

500

425

Y

2

....... 300

~--~~~~~~~x

-3

-2 -I

0

2

-I

100

-2

-2

o

1 2 3 4 5 6 7 8 9 10 11 12 Tnne (months)

45.

46.

a. Estimate Jared's rate of weight loss wheo i)t~1

Y ~ J(x) D: -2:Sx:S3

.....

200

~

x

0 0

43.

w

i'n

<

x~o

iv) 4 P.M.

b. At what time does the temperature increase most rapidly? De-

~

y=l

ii)t~4

Y

iii)t~ll

Y Y ~J(x)

Y ~J(x)

D: -3:S x:s 3

b. Wheo does Jared lose weight most rapidly and what is this rate of weight loss?

D: -2:S x:s 3

e. Use the graphical technique of Example 3 to graph the derivative of weight w.

One-Sided Derivatives Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37-40 are not differentiable at the point P. 37.

Y

38.

Y

48.

47.

Y ~ J(x)

Y Y ~ 2x

Y ~J(x) y~2

Y ~J(x) D: -1:s x:s 2

2 P(I.2)

P(O.O)

0

1

2

x

--~~+---~--~~x

-I

0

2

Y Y ~J(x) D: -3:S x:s 3

3.3

D 60.

Theory and Examples In Exercises 4c}'-52, L

2

Then, on the same screen, graph

(x + h)' - x' y =

b. Graph Y = f(x) and y = j' (x) side by side using separate sets of

coordinate axes, and answer the following questions. c. For what values of x, if any, is i' positive? Zero? Negative? d. Over what intervals of x-values, if any, does the function y = j(x) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.)

= -x 2

51. y

=

SO. Y

= -l/x

52. Y = x 4/4

x'/3

53. Tangent to • parabola Does the parabola y = 2x 2 - l3x + 5 have a tangent whose slope is -I ? If so, find an equation for the line and the point oftangency. If not, why not?

=

54. Tangent to y Vi Does any tangent to the curve y = Vi cross the x-axis at x = -I? If so, f'md an equation for the line and the point of tangency. Ifnot, why not? 55. Derivative of - f Does knowing that a function f(x) is difThrentiable at x = Xo tell you anything about the differentiability of the function -fat x = Xo? Give reasons for your answer. 56. Derivative of multiple. Does knowing that a function g(t) is differentiable at t = 7 tell you anything about the differentiability of the function 3g at t = 7? Give reasons for your answer. 57. Limit of. quotient Suppose that functions g(t) and h(t) are dermed for all values of t and g(O) = h(O) = O. Can lim,...o (g(t))/(h(t)) exist? If it does exist, must it equsI zero?

Give reasons for your answers. 2

58. a. Let f(x) be a function satisfYing If(x) I '" x for -I '" x '" 1. Show that f is differentiable atx = 0 and rmdj'(O).

X2 Sin}, x =F 0

61. Derivative of y = Ixl Graph the derivative of f(x) = 14 Then graphy = (Ixl- O)/(x - 0) = lxi/x. What can you conclude?

D 62.

Weierstrass's nowhere differentiable continuous function The sum of the rITst eight terms of the Weierstrass function f(x) = ~::".o (2/3Y' cos (9"'lTx) is

+ (2/3)1 cos (9'ITx) + (2/3)2COS(9''lTx) + (2/3)' cos (9''lTx) + ... + (2/3)7 cos (9 7'ITx).

g(x) = cos ('lTx)

Graph this sum. Zoom in several times. How wiggly and bumpy is this graph? SpecifY a viewing window in which the displayed portion of the graph is smooth.

COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exercises 63--{j8. a. Plot y = f(x) to see that function~ global behavior. b. Derme the difference quotient q at a general point x, with general step size h.

c. Take the limit as h .... O. What formula does this give? d. Substitute the value x = Xo and plot the function y = f(x) together with its tangent line at that point.

•• Substitute various values for x larger and smaller than XQ into the formula obtained in part (c). Do the numbers make sense with your picture?

63. f(x)

0, x = 0 is differentiable atx = 0 and rmdj'(O). Graphy = 1/(2Vi) in a window that has 0'" the SaDJe screen, graph

ViTI h

x'" 2. Then, on

Vi

=

65. f(x) = 66. f(x)

for h = 1,0.5,0.1. Then try h = -I, -0.5, -0.1. Explain what is going on.

3.3

for h = 2, 1,0.2. Then try h = -2, -I, -0.2. Explain what is going on.

x' + x 2 - x, XQ = I 64. f(x) = x 1j3 + x 2j3 , XQ = I

f(x) = {

y=

h

f. Graph the formula obtained in part (c). What does it mean when its values are negative? 'h:ro? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer.

b. Showthat

D 59.

115

Graph y = 3x in a window that has - 2 '" x '" 2, 0 '" Y '" 3.

Find the derivative j'(x) of the given function y = f(x).

49. Y

Differentiation Rules

=

4x

-2--'

x +I

x - I 3x

2

+

I

Xo = 2 ,Xo = -I

67. f(x) = sin 2x, Xo = 'IT/2 68. f(x) = x 2 cos x, Xo = 'IT/4

Differentiation Rules This section introduces several rules that allow us to dtiferentiate constant functions, power functions, polynomials, rational functions, and certain combinations of them, simply and directly, without having to take limits each time.

Powers, Multiples, Sums, and Differences A simple rule of differentiation is that the derivative of every constant function is zero.

116

Chapter 3: Differentiation

y

c (x, c) (x + h, c) --r-------~~--~~~- y~c

Derivative of a Constant Function If f has the constant value f(x) = c, then

df d dx = dx (c) = O. ~~------~--~--~~--~x

o

x

x+h

FIGURE 3.9 The rule (d/dx)(c) = 0 is

another way to say that the values of constant functions never change and that the slope of a horizontal line is zero at every point.

Proof We apply the definition of the derivative to f(x) = c, the function whose outputs have the constant value c (Figure 3.9). At every value of x, we f'md that

f'(x)

lim f(x + h) - f(x) h

=

h-O

=

lim c - c h

h-O

=

limO

=

O.

h-O

•

From Section 3.1, we know that

From Example 2 of the last section we also know that

These two examples illustrate a general rule for differentiating a power x'. We first prove the rule when n is a positive integer.

Power Rule for Positive Integers: If n is a positive integer, then

HISTORICAL BIOGRAPHY

Richard Courant (1888-1972)

Proof of the PoSitive Integer Power Rule

z' - x·

=

The formula

(z - x)(z·-l + z·-2 X + ... + zx·- 2 + x·- 1)

can be verified by multiplying out the right·hand side. Then from the alternative formula for the definition of the derivative,

f'(x)

=

lim f(z) - f(x) z_xzx

=

=

lim z' - x· z-+xzx

lim(z·-l + z'-"x + ... + zx·- 2 + x·- 1)

• terms

z~.

• The Power Rule is actually valid for all real numbers n. We have seen examples for a negative integer and fractional power, but n could be an irrational number as well. To apply the Power Rule, we subtract 1 from the original exponent n and multiply the result by n. Here we state the general version of the rule, but postpone its proof until Chapter 7.

Differentiation Rules

3.3

117

Power Rule (General Version) If n is any real number, then

~xn

=

dx

nx n- t '

for all x where the powers x' and x·- 1are defined.

EXAMPLE 1 (8)

Differentiate the following powers ofx.

x'

(d)

~ X4

(e) x--4/'

(f) Yx2+w

Solution

(8) :ix(x') = 3X'-1 = 3x'

(c) :ix (x Vi) =

V2x Vi-I

(e) ~(x--4/3) = dx

_ix -(4/3)-1

_i x -7/3

=

3

3

(f) :ix(Yx2+w) = :ix(x 1+(w/'») =

(1 + ;)X 1+(w/2)-1 t(2 + 1T)'Vx;; =

•

The next rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant.

Derivative Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then

y

In particular, if n is any real number, then

Slope /

3

(I, 3)

~

3(2):)

-00; ~

6(1)

~

6

Proof d

.

~_cu = hm h-O

cu(x

+

h) - cu(x) h

UA.

Derivative def'mition

withj(.)

~

cu(.)

=

. u(x + h) - u(x) ciun ~-----;-~~ h~O h

Constant Multiple Limit Property

=

du c dx

u is differentiable.

•

EXAMPLE 2 (8) The derivative formula

FIGURE 3.10 Thegraphsofy

~ ",'and

y = 3",', Tripling the y-coordinate triples

the slope (Example 2).

~(3x') dx

=

3·2x = 6x

says that if we rescale the graph of y = x 2 bymultiplying eachy-coordinate by 3, then we multiply the slope at each point by 3 (Figure 3.10).

118

Chapter 3: Differentiation

(b) Negative ofa function The derivative of the negative of a clifferentiable function u is the negative of the function's derivative. The Constant Multiple Rule with c = -I gives

•

..ti..(-u) = ..ti..(-I'u) = -I'..ti..(u) = _du ~

I

Denoting Fnnction. by u and v The functions we are working with when

~

~~.

The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.

"" need a differentiation formula are

Derivative Sum Rule

likely to be denoted by letters like f aod g. We do not want to use these same letters

If u and v are differentiable functions of x, then their sum u + V is differentiable at every point where u and v are both differentiable. At such points,

when statiog general differentiation rules, so we use letters like u and v instead that

d du dv ~(u+v)=~+~.

are not likely to be already in use.

For example, if y = x4 thenbave

+

12x, then y is the sum of u(x) = x4 and v(x) = 12x. We

dy d ~ = ~(X4)

+

d ~(I2x) = 4x'

Proof We apply the def"mition of the derivative to j(x) d

~_

() [u x

+v

+ h) + v(x + h)]

( ) . [u(x x ] = lim

h

h--i'O

UA,

lim [u(x

=

+ h)

h~O

+ h)

+ v(x):

- [u(x)

+ v(x)]

+ h)

- v(x)]

v(x

h

- u(x)

h

h--i'O

12.

= u(x)

h

. u(x lim

=

+

- u(x)

+

.

v(x

+ h~O bm

+

h) - v(x) du h = ~_ UA,

+

dv

~_.

UA,

•

Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives:

..ti..(u _ v) = ..ti..[u ~

~

+ (-I)v]

= du

~

+

(-I) dv = du _ dv ~

~

~.

The Sum Rule also extends to finite sums of more than two functions. If "1, U2, ... , Un are differentiable at x, then so is Ul

d

+ U2 + ... + un)

~ (U1

=

+

U2

+ ... + UIl , and

dUl dU2 dUn ax + ax + ... + ax'

For instance, to see that the rule holds for three functions we compute

d

~ (U1

+ U2 + U,)

d

= ~ ((U1

+ U2) + U,)

d

= ~ (U1

+ U2) +

du, dUl dU2 du, ax ax + ax + ax' =

A proofhy mathematical induction for any finite number of terms is given in Appendix 2.

EXAMPLE 3 Solution

:

Find the derivative of the polynomial y = x' = ;"x' = 3x 2

+ ;"

(jcx2) - ;" (5x)

+ i. 2x 3

- 5

+0

=

3x 2

+ ;" (I) + !Ix 3

+

jc x 2 -

5x

+ I.

Smn and Difference Rules

5

•

3.3

119

Differentiation Rules

We can differentiate any polynomial term by term, the way we differentiated the polynomial in Example 3. All polynomials are differentiable at all values of x.

EXAMPLE 4 where? Solution

Does the curve y

dy

-1

x' -

=

0 for x:

4x 3 4x(x 2

(I, 1)

-

0

FIGURE 3.11

The curve in Example 4

and its horizontal tangents.

2x 2 + 2 have any horizontal tangents? If so,

The horizontal tangents, if any, occur where the slope dy/ dx is zero. We have

Now solve the equation dx

(-I, 1)

=

4x = 0 1) = 0 x=O,I,-1.

The curve y = x4 - 2x 2 + 2 has horizontal tangents at x = 0, I, and -1. The corresponding points on the curve are (0, 2), (I, I) and (-I, 1). See Figure 3.11. We will see in Chapter 4 that finding the values of x where the derivative of a function is equal to zero is an important and useful procedure. _

Products and Quotients While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance,

d

while

d

dx (x), dx (x)

= I· I =

1.

The derivative of a product of two functions is the sum of two products, as we now explain.

Derivative Product Rule If u and v are differentiable at x, then so is their product uv, and d() dv du dx uv =udx+v dx '

The derivative of the product uv is u times the derivative ofv plus v times the derivative ofu. Jnprime notation, (uv)' = uv' + vu'. In fimction notation,

:fx [f(x)g(x)] = EXAMPLE 5

Find the derivative ofy

f(x)g'(x) + g(x)f'(x). = (x 2

+

1)(x 3

+ 3).

Solution (8) From the Product Rule with u = x 2

:fx [(x 2 + l)(x 3 + 3) 1=

(x 2

+ I and v

=

x3

+ 3, we f'md

+ 1)(3x 2 ) + (x 3 + 3)(2x)

=3X4+3x 2 +2x4+fix =

5x 4

+ 3x 2 + fix.

d

-(uv) dx

~


udx

+ vdu dx

120

Chapter 3: Differentiation

I Picturing the Product Rule Suppose u(x) and v(x) are positive and increase when x increases, and h > O. v(x + h) &v

(b) This particular product can be differentiated as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial: y = (x 2 + l)(x 3 + 3) = x S + x 3 + 3x 2 + 3

+---------, u(x + 1t).6.v

:

=

5x' + 3x 2 + 6x.

/ +------'--'---'----y - -u(x)v(x)

o

Iv(x) 4u

Proof of the Derivative Product Rule

iL (uv)

/

4u \ u(x) u(x + h)

Then the clumge in the product uv is the difference in areas of the larger and smaller "squares," which is the sum of the upper and right-hand reddish-shaded rectangles. That is, .i(uv) ~ u(x ~ u(x .i(uv)

~

4v

iL( ) _ . dx

uv -hm

h-O

[( h) v(x ux+

+ h)

.

(

v(x

h-O

4u

u(x + h)v(x + h) - u(x + h)v(x) + u(x + h)v(x) - u(x)v(x) h

=

= I'

= I=u 11-0

+ h)}; + v(x)};.

The limit as h .... 0+ gives the Product Rille.

h) - u(x)v(x)

h

h-O

To cbange this fraction into an equivalent one that contains difference quotients for the derivatives ofu and v, we subtract and add u(x + h)v(x) in the numerator:

+ h)v(x + h) - u(x)v(x) + h).iv + v(,)4u.

u(,

+ h)v(x +

lim u(x

=

dx

Divisioo by h gives -h-

•

This is in agreement with our [lISt calculation.

v(x)

x+

h) I'

.

=

- v(x)

h

+

( ) u(x +vx

h) - v(x)

h

().

+ v x ·1=

h

11-0

+ h)

- U(X)] u(x

+

h) - u(x)

h

11-0

As h approaches zero, u(x + h) approaches u(x) because u, being differentiable at x, is continuous atx. The two fractions approach the values of dv/dx atx and du/dx atx. In short,

d() dv dx uv = u dx

du

+ v dx'

•

The derivative of the quotient of two functions is given by the Quotient Rule.

Derivative Quotient Rule If u and v are differentiable at x and if v(x) '" 0, then the quotient u/v is differentiable at x, and

iL dx

(l!)

du =

v

dv

vfiX - ufiX

if'

In function notation,

iL [/(X)]

= g(x)f'(x) - I(x)g'(x)

dx g(x)

EXAMPLE 6 Solution

g2(x)'

Find the derivative of y =

t2 - I -3--' t + I

We apply the Quotient Rule with u = t 2

dy

(t 3

+ I)' 2t (t 3

dt

- (t 2

-

+ 1)2

+ 2t - 3t 4 + 3t 2 (t 3 + 1)2 -t 4 + 3t 2 + 2t (t 3 + I)'

-

I and v = t 3

1). 3t 2

+

I:

~ (,,) ~ v(du/dt) - .(du/dt)

dt v

if

2t4

•

3.3

Differentiation Rules

121

Proof of the Derivative Quotient Rule

~ dx

u(x + h)

(l!)

lim v(x + h)

=

v

u(x) v(x)

h

h-O

lim v(x)u(x + h) - u(x)v(x + h)

=

hv(x

h-O

+ h )v(x)

To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and v, we subtract and add v(x)u(x) in the numerator. We then get

(l!)

~ dx v

=

lim v(x)u(x

+ h)

- v(x)u(x) + v(x)u(x) - u(x)v(x hv(x + h)v(x)

+ h)

- u(x)

h-O

.

=

hm

( ) vx

u(x

h

( ) v(x -ux

+ h)

+ h)

- v(x)

h

----'"----~~~~--'"-----

v(x

h-O

+ h )v(x)

Taking the limits in the numerator and denominator now gives the Quotient Rule.

_

The choice of which rules to use in solving a differentiation problem can make a difference in how much work you have to do. Here is an example.

EXAMPLE 7

Rather than using the Quotient Rule to find the derivative of

y=

(x - l)(x 2

-

2%)

expand the numerator and divide by X4:

y=

(x - 1)(x 2

2%)

-

Then use the Sum and Power Rules: dy = -x-2 dx

-

3( -2)x-3 + 2( -3)x-4

Second- and Higher-Order Derivatives If y = f(x) is a differentiable function, then its derivative f' (x) is also a function. If f' is also differentiable, then we can differentiate f' to get a new function of x denoted by So = (t')' . The function is called the second derivative of f because it is the derivative of the rlISt derivative. It is written in several ways:

r

r.

r

d'y

Y) d (ddx

f"(x) = dx 2 = dx

dy'

= dx =

y' = D2(t)(x) = D/ f(x).

The symbol D2 means the operation of differentiation is performed twice. Ify = x 6 , theny' = 6>:' and we have y" =

dy' ax =

d dx (6):') = 30>:'.

122

Chapter 3: Differentiation

I How to Read the Symbols for

If y" is differentiable, its derivative, y'" = dy" /dx = d'y/dx', is the third derivative ofy with respect to x. The names continue as you imagine, with

Derivatives

y' ''y prime" y. 'y double prime" d'y

dx' y~

yv.)

d"y

dx" D"

d d"y y(") = dxy("-l) = dx" = D"y

"d squared y dx squared"

denoting the nth derivative of y with respect to x for any positive integer n. We can interpret the second derivative as the rate of change of the slope of the tangent to the graph of y = f{x) at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line.

''y triple prime" ''y super n"

"d to the n of y by dx to the n" "Dtothen"

EXAMPLE 8

The

flISt

four derivatives of y

x' - 3x' + 2 are

=

First derivative:

y' = 3x' - 6x

Second derivative:

y"

= 6x -

Third derivative:

y.'

=

Fourth derivative:

y(4) =

6

6 O.

The function has derivatives of all orders, the fifth and later derivatives all being zero.

_

Exercises 3.3 Derivative Calculations

25. v

In Exercises 1-12, fmd the fITSt and second derivatives.

1. y = -x' + 3

2. Y = x' + x + 8

3•• = 51' - 31'

4. w

4x' 5. y=T-x

x3 x2 x 6·y=T+T+4

7. w = 3z-2

8. •

-}

= 3z'

- 7z 3

+ 21z2

11.r=-I--~

3.'

=

X4

29. Y

=

10. Y

=4

31. Y

= (x

12. r

=

- 2x - x-,

12 - ~ 8 8'

+ ~4 8

In Exercises 13-16, find y' (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate.

37. w

=

Find the derivatives of the functions in Exercises 17-28.

39. p =

2x + 5 17. Y = 3x - 2 x2

19. g(x) = x

21. v

= (1

23. /(.)

-

18.

4

+ 0.5

- /)(1

= Vs Vs +

- 4x)

4 - 3x z=-,-3x + x

20. /(1) =

22. w

I I

24 =5x+1 • u 2v'i

= (2x

- 7,-'(x

34

.

(8 - 1)(8'

,

+

8

e;/Z)<3 8

+ I) z)

(~)(~) 12q q'

u(O) = 5,

+ 1-2

+ 1',-'

- 1)(x' + 3x - 5) 32. y

=

x'

120

= (4x' + 3x)(2

- x)

·

=1'+51-1

36.u=

I

,

(x' + x)(x' - x + I) x

4

38. w = (z + 1)(z - l)(z' + I) 40. p =

q' + 3 (q - I)' + (q + I)'

c-----c~____;_-~

41. Suppose u and v are functions ofxthat are differentiable atx and that

I' - I I'

30. Y

33• y =x'+7 x

15. Y = (x' + I) (x + 5 +}) 16. y = (1 + X')(X'/4 - x-')

= (2x + 3)(5x'

(x+I)(x+2) (x - I)(x - 2)

= ~---c~-~

Find the fITSt and second derivatives of the functions in Exercises 33-40.

=

14. Y

28. Y

3

T - 2 x' - x

35. r

+ I)

(x' - I)(x' + x + I)

= -21-' + ±

2s

13. y = (3 - x')(x' - x

I

Find the derivatives of all orders of the functions in Exercises 2!1- 32.

I'

9. y = 6.>:' - lOx - 5x-'

27. Y

I + x - 4v'i x

=

+ 5)

u'(O)

= -3,

v(O) = -1,

=0

v'(O) = 2.

Find the values of the following derivatives at x = O.

d d. dx (7v - 2u)

3.3 42. Suppose u and v are differentiable functions of x and that

Differentiation Rules

52. Find all points (x,y) on the graph of f(x)

x' with tangent lines

passing through the point (3, 8).

u(i) = 2, u'(I) = 0, v(l) = 5, v'(I) = -I. Find the values of the following derivatives at x = I.

d

=

123

f(x)

= xl

d d. dx (7v - 2u)

a. dx (uv)

Slope. and Tangents 43. a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x' - 4x + I at the point (2, I). b. Smalle.t slope Wbat is the smallest slope on the curve? At what point on the curve does the curve have this slope?

c. Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8. 44. a. Horizontal tangents Find equations for the horizontal tan3x - 2. Also !md equations for gents to the curve y = the lines that are perpendicular to these tsngents at the points of tangency.

x' -

b. Smalle.t slope Wbat is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve~ tangent at this point. 45. Find the tangents to Newton's serpentine (graphed here) at the origin and the point (I, 2).

53. a. Find an equation for the line that is tangent to the curve y = xatthepoint(-I,O).

x' -

D b.

Graph the curve and tsngent line together. The tsngent intersects the curve at another point. Use Zoom and Trace to estimate the point's coordiuates.

D c.

Conrmn your estimates of the coordinates of the secood intersection point by solving the equations for the curve and tsngent simultaneously (Solver key).

54. a. Find an equation for the line that is tsngent to the curve y = x 3 - 6x 2 + Sx at the origin.

D b.

Graph the curve and tsngent together. The tangent intersects

the curve at another point. Use Zoom and Trace to estimate

y

the point~ coordiuates.

D c.

Conf'mn your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).

Theory and Example. 46. Find the tangent to the Witch ofAgnesi (graphed here) at the point

(2, I).

For Exercises 55 and 56 evaluate each limit by !lrSt coovertiog each to a derivative at a particular x-value.

55. lim

x-I

y y=_8_ x2 + 4

0

2 3

I

50

56.

"----=--x-I

lim x'/9 - I x--l

X

+

1

57. Find the value of a that makes the following function differentiable for all x-values.

x

47. Quadratic tangent to identity funetlon The curve y = + bx + c passes through the point (I, 2) and is tsngent to the line y = x at the origin. Find a, b, and c.

49. Find all points (x, y) on the graph of f(x) = 3x' - 4x with tangent lines parallel to the line y = 8x + 5. SO. Find all points (x,y) on the graphofg(x) = tx' - ~x' tangent lines parallel to the line 8x - 2y = I.

+

I with

51. Find all points (x, y) on the graph of y = x/(x - 2) with tangent lines perpendicular to the line y = 2x + 3.

<

if x

~

0 0

58. Find the values of a and b that make the following function differentiable for all x-values.

ox'

48. Quadratic. having a common tangent The curves y = x 2 + ax + b and y = ex - x 2 have a common tangent line at the point (I, 0). Find a, b, and c.

if x

OX () fx =

{

+ b,

bx'-3,

x> x:5

-I -1

59. The general polynomial of degree n has the fonn

P(x} where a.

¢'

=

a,.x n + a,._tXn-t

+ ... + a2x 2 + atX + ao

O. Find P' (x).

60. The body's re••tion to medicine The reaction of the body to • dose of medicine can sometimes be represented by an equation of the form

124

Chapter 3: Differentiation

where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on. Find dRj dM. This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.5, we will see how to find the amount of medicine to which the body is most sensitive. 61. Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say?

64. Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is, d (x -m) dx

p=~_an2 V - nb

=

V2

'

in which a, b, n, and R are constants. Find dPjdV. (See accompanying figure.)

a. The Reciprocal Rule says that at any point where the function vex) is differentiable and different from zero,

(1)v

l

65. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume Vby a formula of the form

62. The Reciprocal Rule

dx

-mx - m-

where m is a positive integer.

What does this say about the Derivative Constant Multiple Rule?

.E...

=

_l...dv v2dx'

Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule.

63. Generalizing the Product Rule The Derivative Product Rule gives the formula d -(uv) dx

=

dv udx

+ vdudx

66. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is

for the derivative of the product uv of two differentiable functions ofx.

A(q) =

a. What is the analogous formula for the derivative of the product uvw of three differentiable functions of x? b. What is the formula for the derivative of the product u 1 u2 u3 U4 offour differentiable functions of x? c. What is the formula for the derivative of a product Ul U2U3 ..• Un of a finite number n of differentiable functions of x?

3.4

km

hq

q + cm + 2'

where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dAjdq and d 2Ajdq2.

The Derivative as a Rate of Change In Section 2.1 we introduced average and instantaneous rates of change. In this section we study further applications in which derivatives model the rates at which things change. It is natural to think of a quantity changing with respect to time, but other variables can be treated in the same way. For example, an economist may want to study how the cost of producing steel varies with the number of tons produced, or an engineer may want to know how the power output of a generator varies with its temperature.

Instantaneous Rates of Change If we interpret the difference quotient (J(x + h) - !(x)) / h as the average rate of change in ! over the interval from x to x + h, we can interpret its limit as h ~ 0 as the rate at which! is changing at the point x.

3.4 The Derivative as a Rate of Change

DEFINmON

125

The instantaneous rate of change of / with respect to x at Xo is

the derivative /

'(x ) = lim /(xo

o

+ h)

- /(xo)

h

h-O

'

provided the limit exists.

Thus, instantaneous rates are limits of average rates.

It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change.

EXAM PLE 1

The area A of a circle is related to its diameter by the equation

A = TT D2 4 How fast does the area change with respect to the diameter when the diameter is 10m? Solution

The rate of change of the area with respect to the diameter is

~

= ; • 2D =

TTf.

When D = 10 m, the area is changing with respect to the diameter at the rate of (TT/2)10 = 5TT m2/m "" 15.71 m 2/m. •

Motion Along a Line: Displacement,. Velocity, Speed, Acceleration, and Jerk Suppose that an object is moving along a coordinate line (an s-axis), usually horizontal or vertical, so that we know its position s on that line as a function of time t:

s Position at time t ...

I'

•

s~f(t)

and at time t

i1s- -->l'1

/(t).

The displacement of the object over the time interval from tto t

+ At

.

=

Ils = /(t

) S

s+l!.s~f(t+l!.t)

FIGURE 3.12 The positions ofa body moving along a coordinate line at time t aod shortly later at time t + b.t. Here the coordinate line is horizontal.

+ Ilt (Figure 3.12) is

+ Ilt) - /(t) ,

and the average velocity of the object over that time interval is

Vav

=

displacement Ils /(t travel time = Ilt =

+ tlt) - /(t) Ilt

To find the body's velocity at the exact instant t, we take the limit of the average velocity over the interval from t to t + Ilt as Ilt shrinks to zero. This limit is the derivative of / with respect to t.

DEFINmON Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body's position at time tis s = /(t) , then the body's velocity at time t is

_ tis _

.

() vt-dt-Inn &-0

/(t + tlt) - /(t) Il t

126

Chapter 3: Differentiation Besides telling how fast an object is moving along the horizontal line in Figure 3.12, its velocity tells the direction of motion. When the object is moving forward (8 increasing), the velocity is positive; when the object is moving backward (. decreasing), the velocity is negative. If the coordinate line is vertical, the object moves upward for positive velocity and downward for negative velocity (Figure 3.13). s

)

,

o

s increasing:

s decreasing:

positive slope so

negative slope so moving downward

moving upward

FIGURE 3.13 For motion. ~ /(,) along a straight line (tire vertical axis), v = dsldt is positive when s increases and negative whens decreases. The blue curves represent position along lhe lioe over time; lhey do not portray lhe palh of motion, which lies along lhe .-axis.

If we drive to a friend's house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show - 30 on the way back, even though our distance from home is decreasing. The speedometer always shows .peed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.

DEFINITION

Speed is the absolute value of velocity. Speed

~

IV(I)I

~ I~I

EXAMPLE 2 Figure 3.14 shows the graph of the velocity v = /'(1) ofa particle moving along a horizontal line (as opposed to showing a position function. = /(1) such as in Figure 3.13). In the graph of the velocity function, it's not the slope of the curve that tells us if the particle is moving forward or backward along the line (which is not shown in the figure), but rather the sign of the velocity. Looking at Figure 3.14, we see that the particle moves forward for the first 3 sec (when the velocity is positive), moves backward for the next 2 sec (the velocity is negative), stands motionless for a full second, and then moves forward again. The particle is speeding up when its positive velocity increases during the first second, moves at a steady speed during the next second, and then slows down as the velocity decreases to zero during the third second. It stops for an instant at 1 ~ 3 sec (when the velocity is zero) and reverses direction as the velocity starts to become negative. The particle is now moving backward and gaining in speed until 1 ~ 4 sec, at which time it achieves its greatest speed during its backward motion. Continuing its backward motion at time 1 ~ 4, the particle starts to slow down again until it finally stops at time 1 = 5 (when the velocity is once again zero). The particle now remains motionless for one full second, and then moves forward again at 1 ~ 6 sec, speeding up during the f"mal second of the forward motion indicated in the velocity graph. •

3.4 The Derivative as a Rate of Change

127

v

, MOVES FORWARD

,

I

(v> 0)

flQRWARD I

I

AGAIN

I

:

(v> 0)

:

Velocity v ~ 1'(')

,

I

I

I

I

"

I ,,

~ Speeds -+: up ,I

Speeds~Steady~ Slows ~

up

:(v

= const) :

down

,

I

, ,I

I

"

rI

:

I

Stands

till

S

,"

~I

I (v = 0) I

I I

I I

-0;;i"'-------;c----!2c----"::i...--~----,+.5:---6~--'7~ ' ,(sec)

,

4--- Speeds ~ Slows ~ I Up , down I I I I

I I MOVES BACKWARD

:

(v

< 0)

I I I

:

FIGURE 3.14 The velocity graph ofa particle moving aloog a horizoota1line, discussed in Example 2. HIsTORICAL BIOGRAPHY

Bernard Bolzano (1781-1848)

The rate at which a body's velocity changes is the body's acceleration. The acceleration measures how quickly the body picks up or loses speed. A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in acceleration are abrupt.

DEFINmONS Acceleration is the derivative of velocity with respect to time. Ifa body's position at time 1 is s = /(1), then the body's acceleration at time 1 is a(l) =

~~ = ~:~.

Jerk is the derivative of acceleration with respect to time:

j(l) =

~~ = ~:~.

Near the surface of the Earth all bodies fall with the same constant acceleration. Galileo's experiments with free fall (see Section 2.1) lead to the equation

s = 19t2 2

'

where s is the distance fallen and g is the acceleration due to Earth's gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before the effects of air resistance are sigoificant. The value of g in the equation s = (1/2)gI2 depends on the units used to measure 1and s. With 1in seconds (the usual unit), the value ofg determined by measurement at sea level is approximately 32 ft/sec 2 (feet per second squared) in English units, and g = 9.8 m/sec2

128

Chapter 3: Differentiation

t~O

t

=1

(meters per second squared) in metric units. (These gravitational constants depend on the distance from Earth's center of mass, and are slightly lower on top ofM!. Everest, for example.) The jerk associated with the constant acceleration of gravity (g = 32 ft/sec 2) is zero:

s (meters)

t (seconds) •

o

, ,

5

'-'

10 15

j =

20

30

EXAMPLE 3 Figure 3.15 shows the free fall of a heavy ball bearing released from rest at time I = Osee.

35

40

=3

~~:

(g) = O.

An object does not exhibit jerkiness during free fall.

25

t

!

(a) How many meters does the ball fall in the first 2 sec?

45

(b) What is its velocity, speed, and acceleration when t = 2?

FIGURE 3.15 A ball bearing falling from rest (Example 3).

Solution (a) The metric free-fall equation is s = 4.91 2. During the first 2 sec, the ball falls s(2) = 4.9(2)2 = 19.6 m. (b) At any time I, velocity is the derivative of position:

s

v(l) = s'(I) =

!

(4.91 2) = 9.81.

At I = 2, the velocity is v(2) = 19.6 m/sec

in the downward (increasing s) direetion. The speed at I = 2 is speed = Iv(2)1 = 19.6m/sec. The acceleration at any time I is

a(l) = v'(I) = s"(I) = 9.8 m/sec2.

t

All = 2, the acceleration is 9.8 m/ sec2.

s~O s.m~--

•

EXAMPLE 4

A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec (about 109 mph) (Figure 3.16a). It reaches a height of s = 1601 - 161 2 ft after t sec.

(a)

s,V

s ~ 160t - 16,z

(a) How bigh does the rock go? (b) What are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?

(c) What is the acceleration of the rock at any time t during its flight (after the blast)? (d) When does the rock bit the ground again? -160

Solution (b)

FIGURE 3.16 (a) The rock in Example 4. (b) The graphs of s and v as functions of time; s is largest when v = ds/dt ~ O. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock's velocity, graphed here as a straight line.

(a) In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is O. To find the maximum height, all we need to do is to find when v = 0 and evaluate s at this time. At any time I during the rock's motion, its velocity is V

tis d 2 = dt = dl (1601 - 161 ) = 160 - 321ft/sec.

3.4 The Derivative as a Rate of Change

129

The velocity is zero when

160 - 32t

=

0

or

t = 5 sec.

The rock's height at t = 5 sec is

s.... = s(5) = 160(5) - 16(5)' = 800 - 400 = 400 ft. See Figure 3.16h.

(b) To find the rock's velocity at 256 ft on the way up and again on the way down, we rust rmd the two values of t for which s(t) = 160t - 16t 2 = 256. To solve this equation, we write

16t2 16(t

+ 256 lOt + 16)

160t

2

-

= =

(t - 2)(t - 8) =

0 0 0

t = 2sec,t = 8 sec. The rock is 256 ft above the ground 2 sec after the explosion and again 8 sec after the explosion. The rock's velocities at these times are

v(2) = 160 - 32(2) = 160 - 64 = 96 ft/sec.

v(8) = 160 - 32(8) = 160 - 256 = -96 ft/sec. At both instants, the rock's speed is 96 ft/sec. Since v(2) > 0, the rock is moving upward (s is increasing) at t = 2 sec; it is moving downward (s is decreasing) at t = 8 because v(8) < O. (c) At any time during its flight following the explosion, the rock's acceleration is a constant

a -_dv_d( dt - dt 160 - 32t)_ - -32 ft/2 sec. The acceleration is always downward. As the rock rises, it slows down; as it falls, it speeds up. (d) The rock hits the ground at the positive time t for which s = O. The equation 160t - 16t 2 = 0 factors to give 16t(1O - t) = 0, so it has solutions t = 0 and t = 10. At t = 0, the blast occurred and the rock was thrown upward It retamed to the ground 10 sec later. _

Derivatives in Economics Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals. In a manufacturing operation, the cost ofproduction c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc/dx. Suppose that c(x) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h tons per week, and the cost difference, divided by h, is the average cost of producing each additional ton:

c(x

+ h) - c(x) h

average cost of each of the additional h tons of steel produced.

130

Chapter 3: Differentiation The limit of this ratio as h --+ 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.17):

Cost y (dollars)

dc

ax ~~--------~----~~x

o

x x+h Production (tons/week)

.

= hm

c(x

+ h)

h

h~O

- c(x). . = margmal cost of production.

Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one additional unit: ac

FIGURE 3.17 Weekly steel production: c(x) is the cost of producing x tons per week. TIre cost of producing an additional h tons is c(x + h) - c(x). y

c(x

+

!>.X

I) - c(x) 1

which is approximated by the value of dc/ax at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc/ax, which is the rise in the tangent line if!>.x = I (Figure 3.18). The approximation works best for large values of x. Economists often represent a total cost function by a cubic polynomial

c(x) = ax 3 + f3x 2

+ 1X + {J

where {J represents ['!Xed costs such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually adequate to capture the cost behavior on a realistic quantity interval.

EXAMPLE 5

Suppose that it costs

c(x) = x 3

-

6x 2

+

15x

~~----L-----~~------~x

o

x

x+ 1

FIGURE 3.18 Tlremarginalcostdc/dds approximately the extra cost ac of producing = I more unit.

ax

dollars to produce x radiators when 8 to 30 radiators are produced and that

1'(x) = x 3 - 3x 2

+

12x

gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue for selling II radiators a day?

Solution c'(IO):

The cost of producing one more radiator a day when 10 are produced is about

c'(x) = c'(IO)

:fx (x3 -

6x 2 + 15x) = 3x 2

-

12x

+

IS

= 3(100) - 12(10) + IS = 195.

The additional cost will be about $195. The marginal revenue is

r'(x) =

:fx (x 3 -

3x 2 + 12x) = 3x 2 - 6x

+

12.

The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 radiators a day, you can expect your revenue to increase by about

r'(IO)

= 3(100) - 6(10)

if you increase sales to II radiators a day.

+

12 = $252

•

3.4 The Derivative as a Rate of Change

131

EXAM PLE 6 To get some feel for the iangoage of marginal rates, consider marginal tax mtes. If your marginal income tax mte is 28% and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 28% of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes Twith respect to income is dT/dI = 0.28. You will pay $0.28 in taxes out of every extra dollar you earn. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal mte will increase. _

Sensitivity to Change When a small change in x produces a large change in the value of a function f(x), we say that the function is relatively sensitive to changes in x. The derivative f' (x) is a measure of this sensitivity.

EXAMPLE 7

Genetic Data and Sensitivity to Change

The Austrian monk Gregor Johann Mendel (1822-1884), working with garden peas and other piants, provided the first scientific explanation ofbybri
y = 2p(1 - p)

+ p2

= 2p _ p2.

The gmph of y versus p in Figure 3 .19a suggests that the value of y is more sensitive to a change in p when p is small than when p is large. Indeed, this fact is borne out by the derivative gmph in Figure 3.19b, which shows that dy/dp is close to 2 whenp is near 0 and close to 0 when p is near 1.

dy/dp

2 y

-o::f-------:j--p

-o::+---~-~p

(a)

(b)

FIGURE 3.19 (a) The graph ofy = 2p - p2, describing the proportion of smooth-skinoed peas in the next generatioo. (b) The graph of dyjdp (Example 7).

The implication for genetics is that introducing a few more smooth skin genes into a population where the frequency of wrinkled skin peas is large will have a more dramatic effect on later genemtions than will a similar increase when the population has a large pro_ portion of smooth skin peas.

132

Chapter 3: Differentiation

Exercises 3.4 Motion Along a Coordfnate Une Excroises 1--6 give the positions 9 = /(1) ofa body nwving on a coordinate line, with s in meters and 1 in seconds.

L Find the body's displaoement and average velocity for the given time interval. b. Find the body's speed and acoclemtion at the endpoint! of the interval.

c. When, if ever, duriog the in1:aval does the body change diIectiDn? 1.,, - t 2 -3t+2,OStS2 2. " - 6t

-,z,

13. Free f'IIll from the Tower of Ph. Had Galileo dropped a cannonball from. the Tawer of Pis a, 179 ft abave the ground, the ball's heigh1: abave the ground 1 s« into the fall would have been a - 179 - 1612.. L

0 S t S 6

3.9=-t3 +3t2 -3t,O:St:S3 4. 9 = (t 4/4) - ,3 + t2., O:s,:s 3

25 5 5.a - 2 - " l S ISS t 6., - -251 + 5'

12. Speec6ng bullet A 4S-caliber bullet shot straight up from the surface of the moon would reach a height ofa - 832t - 2.61 2 ft after 1 sec. On Earth, in the absence of air, its height would be a - 832t - 1612 ft after t soo. Haw long will the bul.letbe aloft in each case? Haw high will the bullet go?

-4slsO

7. Particle motion At time t, the position of a body moving along the,,·axis is.! - t 3 - 6t 2 + 9tm. L Find the body's acceleration each time the velocity is zero. b. Find the body's speed each time the accelemtion is zero.

c. Find the total distance traveled by the body from 1 = 0 to 1 = 2. 8. Particle motion At time 1 <'! 0, the velocity of a body moving along the horizontal. ,,·axis is v = ,2. - 4t + 3. L Find the body's acceleration each time the velocity is UI'O.

What would have been the ball's velocity, speed, and acceleration at time t?

b. About haw long would it have takm the ball to hit the ground?

c. What\\OUldhave been the ballll velocity at the mcment of~ 14. GJIlileo'J free-faD formula Galileo developed a formula for a body's velocity during :free fall by rolling balls from rest down increasingly steep inclined planks and looking for a limiting formula that would predict a ball's behavior when the plaDk was vertical and the ball fell :freely; see part (a) of the accompanying fJ.gUre. He found that, for any given angle of the plank, the ball's velocity t sec into motion was a constant multiple of I. That is, the velocity was given by a formula of the form. v - kt. The value of the constant k depended on the inclination of the plank. In modern notation--part (b) of the fi.gmc--with distance in meters and time in seconds, what Galileo determiru:d by experiment was that, for any given angle 9, the ball's velocity t sec into the roll was

v - 9.8(sin.9)lmfsec.

b. When is the body moving forward? Backward?

c. When is the body's velocity increasing? Decreasing?

Free-Fall Applications 9. Free r.n on Man and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (8 in meters, 1 in sccondlI) arc 9 = 1.8612. on Mars and 9 = 11.4412. on Jupiter. Haw long does it tab a rock faUing from rest to reach a velocity of 27.8 mfsec (about 100 kmjh) on each planet? 10. Lunar proJed:ile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 mfsoo (about 86 km/h) reaches a height of s = 241 - 0.812. minI sec. L Find the rock's velocity and acceleration at time I. (The acceleration in this case is the acoclemtion of gravity on the nwon.) b. Haw long does it tala:: the rock to reach its highest poinfl

-

c. Hawhighdoestherockgo? d. Haw long does it take the rock to reach half its maximmn e. Haw long is the rock aloft? 11. FindiDKg 011. HlUlllIIirlen planet Explorers on a small airless planet used a spring gun to laundI a ball bearing vertically upward from the surliu:e at a launch velocity of 15 mls«. Because the acceleration of gravity at the planet's surfuce was & mjser?, the explorers expected the ball bearing to reach a beigb1 of a - 15t - (I/2)g.,zmls«1ater.Theballbearingreacheditsmaximum heigh120 sec after being lallochril What was the value of g, ?

[

?

(.) L

(b)

What is the equation for the ball's velocity during:free fall?

b. Building on your work in part (a), what cmtstant acceleration does a freeJy fulling body experience near the surface of Earth?

Understandfng Motion from Graphs 15. The accompanying fIgUfC shows the velocity v = (m/sec) ofa body moving along a coord.i.natl; line.

dr/tit

=

/(t)

v (mlscc) v - fit)

L When docs the body n:verse diIect:i.on?

b. When (approximately) is the body nwving at a constant speed?

3.4

c. Graph the body's speed for 0

~

t

~

10.

16. A particle P moves on the number line shown in part (a) of the accompanying figure. Part (b) shows the position of P as a function of time t. _

o

P

133

18. The accompanying figure shows the velocity v = /(t) of a particle moving on a horizontal coordinate line.

d. Graph the acceleration, where dermed.

_ _----'I_

The Derivative as a Rate of Change

..... p..... ---+--..... -~-+~ S

v

(em)

(a)

s (em)

a. When does the particle move forward? Move backward? Speed up? Slow down? ---::--f----L--'-\--'L---':---':----'::-~

t (sec)

b. When is the particle's acceleration positive? Negative? Zero? c. When does the particle move at its greatest speed? d. When does the particle stand still for more than an instant?

(b)

a. When is P moving to the left? Moving to the right? Standing still?

19. Two falling balls The multiflash photograph in the accompanying figure shows two balls falling from rest. The vertical rulers are marked in centimeters. Use the equation s = 490t 2 (the freefall equation for s in centimeters and t in seconds) to answer the following questions.

b. Graph the particle's velocity and speed (where defined). 17. Launching a rocket When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket coasts upward for a while and then begins to fall. A small explosive charge pops out a parachute shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands. The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following. a. How fast was the rocket climbing when the engine stopped? b. For how many seconds did the engine burn? 200 150

1l ~ s, .8

100 50

<.)

0

al

>-

0 -50 -100 0

•• • •••

2

4 6 8 10 Time after launch (sec)

t t t t t

12

-I

:I 1

c. When did the rocket reach its highest point? What was its velocity then? d. When did the parachute pop out? How fast was the rocket falling then? e. How long did the rocket fall before the parachute opened? f. When was the rocket's acceleration greatest? g. When was the acceleration constant? What was its value then (to the nearest integer)?

a. How long did it take the balls to fall the first 160 cm? What was their average velocity for the period? b. How fast were the balls falling when they reached the 160-cm mark? What was their acceleration then? c. About how fast was the light flashing (flashes per second)?

134

Chapter 3: Differentiation

20. A traveling truck The accompanying graph shows the position s of a truck traveling on a highway. The truck starts at t ~ 0 aod returns IS h later at I ~ 15. L

Economics 23. Marginal co.t Suppose that the dollar cost of prodocing x washing machines is c(x) ~ 2000 + \O(b; - 0.lx 2 •

a. Find the average cost per machine of producing the rlrst \00 washing machines. b. Find the marginal cost when 100 washing machines are pro-

Use the technique desctibed in Section 3.2, Example 3, to graphthetruck~velocityv ~
doced.

b. Supposethats ~ 151 2 - t'.Graph
500

/ '\ / \ /

100

/ o

/

c. Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the rrrst 100 have beeo made, by calculating the latter cost directly. 24. Marginal revenue Suppose that the revenue from selling x washing machines is

r(x)

\

~ 20,000(1 - ~)

dollars.

a. Find the marginal revenue when 100 machines are produced.

5

10

b. Use the function r'(x) to estimate the increase in revenue that will result from increasing prodoction from 100 machines a week to \01 machines a week.

15

Elapsed time, t (hr)

21. The graphs in the accompaoying figure show the position s, velocity v ~
which? Give reasons for your answers. y

c. Find the limit of r' (x) as x .... this number?

00.

How would you interpret

Additional Applications 25. Bacterium population When a hactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and begao to decline. The size of the population at time I (hours) was b ~ \0' + \0'1 - 1O't 2 • Find the growth rates at

a. I

~

0 hours.

b. I

~

5 hours.

c. I

~

\0 hours.

26. Draining a tank The number of gallons of water in a taok t minutes aller the taok has started to drain is Q(t) ~ 200(30 - If. How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the rrrst \0 min?

D 27. 22. The graphs in the accompanying figure show the position s, the velocity v ~
Draining a tank It takes 12 hours to drain a storage taok by opening the valve at the bottom. The depth y of fluid in the taok t hours afler the valve is opened is given by the formula

graph is which? Give reasons for your answers. y

y

~ 6(1 - I~Y m.

a. Find the rate dy/dt (m/b) at which the taok is draining at time t. b. When is the fluid level in the taok falling fastest? Slowest? What are the values of dy/dt at these times? c. Graphy aod dy/dt together and discuss the behavior ofy in relation to the signs and values of dy/dl. 28. Inf1ating a balloon The volume V ~ (4/3)".r' of a spherical balloon changes with the radius.

a. At what rate (fl'/ft) does the volume change with respect to the radius when r ~ 2 ft? b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 ft?

3.5 Derivatives ofTrigonometric Functions

135

velocity fimction v(t) ~ d>/dl ~ /'(t) and the acceleration function a(l) ~ d 2s/dl2 ~ /,,(1). Comment on the object's behavioriorelation to the signs and values of v and a. Include io your commentary such topics as the followiog:

29. Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D ~ (10/9)12 , v.lrete D is measured io meters from the starting point aod 1 is measured io seconds from the time the brakes are released. The aircraft will be· come airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel io that time?

a. When is the o~ect momentarily at rest? b. When does it move to the left (dowo) or to the right (up)?

30. Volcanic lava fountains Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a lioe offountaios along the wall of the crater, activity was later confmed to a single vent io the crater's floor, which at one poiot shot lava 1900 ft straight ioto the air (a Hawaiian record). Wbat was the lava's exit velocity io feet per second? In utiles per hour? (Hinl: If Vo is the exit velocity of a patticle of lava, its height I sec later will be s ~ vol - 1612 ft. Begin by fmdlng the time at wbich d>/dl ~ O. Neglect air resistance.)

c. When does it change direction? d. When does it speed up and slow dowo?

e. When is it moviog fastest (highest speed)? Slowest? f. When is it farthest from the axis origin?

31. s ~ 2001 - 161 2, 0 '" 1 '" 12.5 (a heavy object fired straight up from Earth's surface at 200 ft/ sec) 32.s~12-31+2, 0"'1"'5 33.s~I'-612+71, 0"'1"'4

Analyzing Motion Using Graphs

D Exercises 31-34 give the position fimction s ~ /(1) of an o~ect moviog

34. s ~ 4 - 71

+ 612

-

I',

0 '" 1 '" 4

along the s-axis as a fimction of time I. Graph / together with the

3.5

Derivatives of Trigonometric Functions Many phenomena of nature are approximately periodic (electromagnetic fields, heart rhythms, tides, weather). The derivatives of sioes and cosines play a key role in describing periodic changes. This section shows how to differentiate the six basic trigonometric functions.

Derivative of the Sine Function To calculate the derivative of f(x) = sio x, for x measured in radians, we combine the lintits in Example 5a and Theorem 7 in Section 2.4 with the angle sum identity for the sine function:

+ h)

sin (x

= sinxcosh

+

cosxsinh.

If f(x) = sinx, then

f'(x)

= lim h-+O

f(x + h) - f(x)

= lim (sinxcosh

+ h)

+ cos x sin h) I)

+ cosxsinh

h

h-+O

+ 11-0 lim ( COSXoSinh) h

. I' cosh - I sin h · -hsmx· lID + cosx· Iun h-+O h h-+O

limit 0

Derivative def"mition

sinx = lim sin x (cosh - 1)

h

. coshh I· (smx· = h~

- sinx

h

h-+O

h-+O

=

= lim sin (x

h

=

. 0 smx·

+ cosx· I

limit 1

The derivative of the sine funmon is the cosine function:

1x (sinx)

= cosx.

=

cosx.

Example Sa and

Theorem 7, Section 2.4

136

Chapter 3: Differentiation

EXAMPLE 1

We find derivatives of the sine function involving differences, products,

and quotients.
dy

sinx:

-

d.

2x - -(SfiX)

-

ax =

ax

Difference Rule

= 2x - cosx.

(b) y =

(c) Y =

x2 sinx:

:

x2

=

x 2 cos x + 2x sinx.

(sin x) + 2x sin x

x'~(sinx) ax

dy

sin x --x-:

!

=

ax

Product Rule

- sinx'l Quotient Rule

xcosx - sinx x2

•

Derivative of the Cosine Function With the help of the angle sum formula for the cosine function, cos (x + h) = cosxcosh - sin x sinh,

we can compute the limit of the difference quotient: ~(

ax

)-1' cos(x+h)-cosx cosx - Im h

Derivative de:fmition

h~O

.

= lIm

(cosx cos h - sin x sin h) - cosx

'--------~---'------

h

h-O

.

Cosine angle sum identity

cosx(cosh - I) - sinxsinh h-.O h

= lIm

=~='-"-~~---'="-==-:c

= lim cosx. ~co~s~h~-~I h~O

h

lim sin x • _sin_h h

h~O

. I'l I Dsinh =cosx· lim cosh - I -smx· .-h-O

h

=

cosx·O - sin x 1

=

-sinx.

h-O

h

Example Sa and

0

Theorem 7, Section 2.4

The derivative ofthe cosine function is the negative of the sine function: FIGURE 3.20 The curve y' = -sin x as the graph of the slopes of the tangents to the curve y = cos x.

!

(cos x) = -sinx.

137

3.5 Derivatives ofTrigonometric Functions

Figure 3.20 shows a way to visualize this result in the same way we did for graphing derivatives in Section 3.2, Figure 3.6.

EXAMPLE 2

We find derivatives of the cosine function in combinations with other

functions. (a) y = 5x

+ cosx: dy d ax = ax (5x)

+

axd (cos x)

Sum Rule

= 5 - sinx. (b) y = sin x cosx:

dy ax =

.

SfiX

. axd (cos x) + cosx axd (SfiX)

= sinx( -sin x) =

(c) y =

cosx

cos2 X

-

Product Rule

+ cosx(cosx)

8in2 x.

.

1 - sinx'

dy

(1 - Sinx)1x(cosx) - cosx1x(l - sinx)

ax

(1 - sinx)2

Quotient Rule

(1 - sinx)( -sinx) - cosx(O - cosx) (1 - sinxJ2

1 - sinx

sin2 x

(1 - sinxJ2 1 1 - sinx'

+ cos2 x

=

1

•

Simple Harmonic Motion The motion of an object or weight bobbing freely up and down with no resistance on the end of a spring is an example of simple harmonic motion. The motion is periodic and repeats indefinitely, so we represent it using trigonometric functions. The next example describes a case in which there are no opposing forces such as friction or buoyancy to slow the motion.

EXAMPLE 3 A weight hanging from a spring (Figure 3.21) is stretched down 5 units beyond its rest position and released at time I = 0 to bob up and down. Its position at any later time I is

-5

o 5

s=5cosl.

Rest position Position at t~0

, FIGURE 3.21 A weight hanging from a vertical spring and then displaced oscillates above and below its rest position (Example 3).

What are its velocity and acceleration at time I?

Solution We have s = 5 cos I Position: Velocity:

v =

~

=

~ (5 cos I)

Acceleration:

a=

~~

=

~(-5sinl)

= -5 sin I = -5cosl.

138

Chapter 3: Differentiation Notice how much we can learn from these equations:

s, v

FIGURE 3.22 TIre graphs of the positioo and velocity of the weight in Example 3.

1.

As time passes, the weight moves down and up between s = - 5 and s = 5 on the s-axis. The amplitude of the motion is 5. The period of the motion is 21T, the period of the cosine function.

2.

The velocity v = -5sint attsins its greatest magnitude, 5, when cost = 0, as the graphs show in Figure 3.22. Hence, the speed of the weight, Iv I = 51 sin tl , is greatest when cos t = 0, thst is, when s = 0 (the rest position). The speed of the weight is zero when sin t = O. This occurs when s = 5 cos t = ± 5, at the endpoints of the interval of motion.

3.

The acceleration value is always the exact opposite of the position value. When the weight is above the rest position, gravity is pulling it back down; when the weight is below the rest position, the spring is pulling it back up.

4.

The acceleration, a = - 5 cos t, is zero only at the rest position, where cos t = 0 and the force of gravity and the force from the spring balance each other. When the weight is anywbere else, the two forces are unequal and acceleration is nonzero. The acceleration is greatest in magnitude at the points farthest from the rest position, where cost = ±l. •

EXAMPLE 4

The jerk of the simple harmonic motion in Example 3 is

d(5 1. = da dt = dt - cost) = 5' smt. It has its greatest magnitude when sin t = ± 1, not at the extremes of the displacement but

at the rest position, where the acceleration changes direction and sign.

•

Derivatives of the Other Basic Trigonometric Functions Because sin x and cos x are differentiable functions of x, the related functions sin x

tanx = cos x '

cosx cotx = - . - ,

smx

1 secx = cos x '

and

cscx = - .1-

smx

are differentiable at every value of x at which they are defined. Their derivatives, calculated from the Quotient Rnle, are given by the following formnlas. Notice the negative signs in the derivative formulas for the cofunctions.

The derivatives ofthe other trigonometric functions:

:fx (tanx) =

sec2 x

:fx (cotx) =

-csc2 X

:fx (sec x) =

secxtanx

:fx (csex) =

-cscx cot x

To show a typical calculation, we fmd the derivative of the tangent function. The other derivations are left to Exercise 60.

3.5

EXAMPLE 5 Solution

~ ax

Derivatives ofTrigonometric Functions

139

Find d(!JJnx)jax.

We use the Derivative Quotient Rule to calculate the derivative: _ (!JJn x) -

~ (Sinx) ax

cos x

_ cos x

-

:Ix (sinx) -

sin x cos2 x

:Ix (cos x)

Quotient Rule

cos x cos x - sinx(-sinx)

cos2 x 2 C08 X

+

2 8in X

2

cos x

•

= _1_ = sec2 x cos2 x

EXAMPLE 6 Solution tives.

Findy"ify = secx.

Finding the second derivative involves a combination of trigonometric deriva-

y = secx

y' = sec x tan x

Derivative rule for secant function

d y' = ax (secx!JJnx)

d

= sec x ax (!JJnx) = secx(sec2 x)

d + !JJnx ax (sec x)

Derivative Product Rule

+ !JJnx(secx!JJnx)

Derivative rules

•

2

3

= sec x + secx tan x

The differentiability of the trigonometric functions throughout their domains gives another proof of their continuity at every point in their domains (Theorem 1, Section 3.2). So we can calculate limits of algebraic combinations and composites of trigonometric functions by direct substitution.

EXAMPLE 7 We can use direct substitution in computing limits provided there is no division by zero, which is algebraically undefmed lim

Y2 + secx

x~o COS(1T -

!JJnx)

=

Y2 + sec 0 cos( 1T

-

!JJn 0)

V2+l COS(1T -

0)

=

V3 = -V3 -1

Exercises 3.5 Derivatives In Exercises 1-18, fmddy/dx.

1. Y

= -lOx + 3 eosx

3. Y = x 2 cosx

5. Y

= esex

- 4Vx + 7

6. Y = x 2 cotx -

~ X

2. Y

= f + 5 sinx

4. Y

= Vx sec x + 3

7. !(x) 9. y

10. Y

=

sin x tanx

= (scex + tanx)(seex = (sinx + eosx) sec x

8. g(x) - tanx)

=

csex cotx

•

140

Chapter 3: Differentiation

11. y = 1

+ cotx

cotx

4 13. y = cosx

1

+ tanx

cosX'

y=~~-

1 + sinx

14. y

cosX'

= -x- +

x cosx

+ 2x cosX' - 2 sinx

15. Y = x 2 sinx 16. y =

12.

x 2 cosX' - 2x sin x - 2 cosX'

17. f(x) =x'sinxcosx

21. s = 1

45.

In Exercises 23-26, fmd dr/dO.

24. r = osinO + cosO 26. r = (1 + sec 0) sinO

23. r = 4 - 02 sinO 25. r = secOcscO

Q

+1

sin t 22.s=1 - cos I

+ cscl

46. y

y

2

20. s = 12 - sec I

1 - csct

In Exercises 45 and 46, fmd ao equation for
18. g(x) = (2 -x)tan2 x

In Exercises 19-22, fmdds/dt. 19.s=tanl-1

44. Find all points on the curve y = calx, 0 < x < '1f, where the taogent line is parallel to the line y = -x. Sketch the curve and tangent(s) together, labeling each with its equation.

1

-c+-~--~--~~--\> x

o

2 2 Y = 4 + cotx - 2cscx 1!

In Exercises 27-32, fmddp/dq.

1 27. p = 5 + cotq sinq 29. p = 31. p =

28. p = (1

+ cosq

+

Y= 1 + Y2cscx+ cot x

cseq) cosq

tanq

cosq

+ tanq 3q + tanq

30. p = 1

qsinq

32. p =

-2--

q - 1

qsecq

Trigonometric Limits Find the limits in Exercises 47-54.

47. lim sin %-2

33. Find y" if L

Y

= cscX'.

b. y = secx.

48.

b. y = geosx.

49.

34. Find y(4) = d 4y/dx 4 if L

Y

= -2sinx.

Tangent Lines In Exercises 35-38, graph the curves over the given interva1s, together with their tangents at the given values of x. Label each curve aod tangent with its equation.

35. y x

=

sin x,

= -'IT,

-371/2:5 x :s; 2w

0, 3w/2

36. y = tanx,

-'1f/2 -'1f/2

'1f/2

<x<

'1f/2

x = -'1f/3, '1f/4 38. y = 1 + cosx, -3'1f/2';; x ,;; 2'1f X

X

2

VI + cos ('1f cscx)

sinO - 1 ~2 8-1T/6 8 - 6

lim

51. lim sec [cos x

x-o

· sm . ( 52• 1lDl

x-o

= -'1f/3,3'1f/2

+

39. y = x

+

sinx

41. y = x - cotx

42.y=x+2cosx 43. Find all points on the curve y = tanx, -'1f/2 < x < '1f/2, where the taogent line is parallel to the line y = 2x. Sketch the curve and taogent(s) together, labeling each with its equation.

8 -"4

seCX'

'1f + tanx ) tanx - 2 sec x

1-0

t

t)

54. lim cos ( '1f8.)

sm"

9-0

Theory and Examples The equations in Exercises 55 aod 56 give the position s = f(l) of a body moving on a coordinate line (s in meters, t in seconds). Find the body's velocity, speed, acceleration, andjerk at time I = '1f/4 sec.

55.s=2-2sinl

56. s

= sint + cost

57. Is there a value of c that will make

S:23X, f(x)

+ sinx

40. y = 2x

9-nj4

'1f tan ( -'1f4) - 1]

D Do the graphs of the functions in Exercises 39-42 have any horizontal tangents in the interval 0 :5 x :5 2'1T? If so, where? If not, why not? Visualize your fmdings by graphing the functions with a grapher.

· tanO-1 1un ~

50•

53. lim tan (1 _ sin

<x<

x = -'1f/3, 0, '1f/3

37. y = secx,

lim

X--'1Tj6

(L 1)

= { c,

x =F 0

x

=0

continuous at x = O? Give reasons for your answer. 58. Is there a value of b that will make

g(x)

=

{x + b, x cosx, x

< ~

0 0

continuous at x = O? Differentiable at x = O? Give reasons for yoor answers.

3.5 Derivatives ofTrigonometric Functions 59. Find d 999/
See the accoropanying figure.

60. Derive the formula for the derivative with respect to x of a. secx.

141

y

c. cotX.

b. cscx.

Slope = f'(x)

61. A weight is attached to a spring and reaches its equilibrium position (x ~ 0). It is then set in motion resulting in a displacement of

x= IOcost,

~

c

Slope = f(x

I

where x is measmed in centimeters and t is measmed in seconds. See the accompanying figure.

+ h) - f(x) h

B I

........--;--~

:

~ Sl

: I I I I I I I

I I I I I I I

_ f(x ope-

+ h) -f(x- h) 2h

'O~------~~~~x--~--x~+-h~----------------'x

Equilibrium

•• To see how rapidly the centered difference quotient for j(x) = sinx converges to I'(x) = cosx, graphy = cosx together with

position atx=O

sin(x

a. Find the

spring~

displacement when t

~

0, t

~

'fr/3, and

t ~ 3'fr/4.

b. Find the spring~ velocity when t ~ 0, t ~ 'fr/3, and t

= 3'fr/4.

62. Assume that a particle's position on the x-axis is given by

where x is measured in feet and t is measured in seconds.

a. Find the particle~ position when t = 0, t = 'fr/2, and t =

'fr.

b. Find the particle~ velocity when t = 0, t = 'fr/2, and t =

'fr.

o 63. Graph y = cos x for y

+ h)

- sinx

h

=

over the interval [-'fr, 2'fr] for h = 1,0.5, aod 0.3. Compare the results with those obtained in Exercise 63 for the same values ofh.

cos(x y

+

~

h) - cos(x - h) 2h

2'fr. On the same screen, grsph

-'fr ,;; X ,;;

sin(x

h) - sin(x - h) 2h

b. To see how rapidly the centered difference quotient for j(x) ~ cosx converges to f'(x) ~ -sinx, graph y ~ -sinx together with

= 3cost + 4sint,

x

+

y =

x

= I, 0.5, 0.3, and 0.1. Then, in a new window, try -I, -0.5, and -0.3. Whathsppens ash ..... o+? Ash ..... o'?

over the interval [-'fr, 2'fr] for h = 1,0.5, aod 0.3. Compare the results with those obtained in Exercise 64 for the same values ofh.

for h

h

~

What phenomenon is being illustrated here?

0 64.

Graph y graph

= -sinx

for

-'fr ,;; X ,;;

cos(x

+ h)

2'fr. On the same screen,

+

h) - j(x - h)

2h

- cosx

= I,

0.5, 0.3, and 0.1. Then, in a new window, try h = -I, -0.5, and -0.3. Whathsppens ash ..... o+? Ash ..... o'? What phenomenon is being illustrated here?

o 65. Centered difference quotient. j(x

j(x

(Continuation

h

y=

for h

66. A caution about centered difference quotients ofExercise 65.) The quotient

+ h)

The centered difference quotient

- j(x - h) 2h

is used to approximate f' (x) in numerical work because (I) its lintit as h ..... 0 equals f'(x) when f'(x) exists, and (2) it usually gives a better approximation of f' (x) for a given value of h than the difference quotient

j(x

+ h) h

- j(x)

may have a lintit as h ..... 0 when j has no derivative at x. As a case in point, takej(x) = Ixlaodcaleulate

lim

10 + hl- 10 - hi

h-O

2h

As you will see, the lintit exists even though/(x) ~ Ixlhas no de-

rivative at x = O. Moral: Before using a centered difference quotient, be sure the derivative exists.

= taox and its derivative together on (-'fr/2, 'fr/2). Does the grsph of the taogent function appear to have a smallest slope? A largest slope?

0 67. Slope. on the graph or the tangent runction Graph y

Is the slope ever negative? Give reasons for your answers.

142

D 68. D 69.

D 70.

Chapter 3: Differentiation

Slopes on the graph of the cotangent function Graph y = cot x and its derivative together for 0 < x < 'If. Does the graph of the cotsogent function appear to have a smallest slope? A ilugest slope? Is the slope ever positive? Give reasoos for your answers.

b. With your grapher still in degree mode, estimate

Exploring (sin kx)/x Graph y = (sinx)/x, y = (sin2x)/x, and y = (sin4x)/x together over the interval -2'" x '" 2. Where does each graph appear to cross the y-axis? Do the graphs really intersect the axis? What would you espect the graphs of y = (sin5x)/x and y = (sin(-3x))/x to do as x-->O? Why? What about the graph of y = (sinkx)/x for other values of k? Give reasons for your answers.

c. Now go back to the derivation of the formula for the derivative of sin x in the text and carry out the steps of the derivation using degree-mode lintits. What formula do you obtain for the derivative?

lim cosh - I h .

h-O

d. Work through the derivation of the formula for the derivative of cos x using degree-mode limits. What formula do you obtain for the derivative?

Radians versus degrees: degree mode derivatives What happens to the derivatives of sin x and cos x if x is measured in degrees instead of radians? To fmd out, take the following steps. L

e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it What are the second and third degree-mode derivatives of sin x and cosx?

With your graphing calculator or coroputer grapher in degree mode, graph

f(h) = sinh h and estimate Iim.HO f(h). Compare your estimate with 'If/ISO. Is there any reason to believe the limit should be 'If/ISO?

3.6

I

The Chain RuLe How do we differentiate F(x) = sin (x 2 - 4)? This function is the composite log of two functions y = I(u) = sin u and u = g(x} = x 2 - 4 thst we know how to differentiate. The answer, given by the Chain Rule, says thst the derivative is the product of the derivatives of I and g. We develop the rule in this section.

Derivative of a Composite Function The function y =

~x

=

!

(3x) is the composite of the functions y =

!

u and u = 3x.

We have

dy dx Since ~ =

!.

3 2'

I

2'

and

du

dx

= 3

.

3, we see in this case thst

dy

c: y turns B: u turns A: x toms

FIGURE 3.23 Wheo gear A makes x turns, gear B makes u turns and gear C makes y turns. By comparing circumferences or counting teeth, we see thaty = u/2(Cturnsone-halfturnfor each B turn) and u = 3x (B turns three times for Ns one), so y = 3x/2. Thus, dy/dx = 3/2 = (1/2)(3) = (dy/du)(du/dx).

dy du

dy du

dx = du' dx'

If we think of the derivative as a rate of change, our intuition allows us to see thst this relationship is reasonable. If y = I(u) changes half as fast as u and u = g(x) changes three times as fast as x, then we expect y to change 3/2 times as fast as x. This effect is much like thst of a multiple gear train (Figure 3.23). Let's look at another example.

EXAMPLE 1

The function

y

= (3x 2

+ I)'

3.6 The Chain Rule is the composite of y = f(u) = u 2 and u = g(x) = 3x 2 see that

+

143

1. Calculating derivatives, we

dy du du dx

- . - = 2u'6x

+ l).6x

= 2(3x2 = 36x 3

+

12x.

Calculating the derivative from the expanded formula (3x 2 the same result:

dy = ~(9x' dx dx = 36x 3

+ 6x 2 +

+ 1)2

= 9x'

+ 6x 2 + 1 gives

1)

•

+ 12x.

The derivative of the composite functionf(g(x» atx is the derivative of f atg(x) times the derivative of gat x. This is known as the Chain Rule (Figure 3.24). Compositef· g Rate of change at x isf'(g(x» . g·(x).

f

g -

-4- x

Rate of Change~ atxisg'(x). -~=----­ u ~ g(x)

Rate of change at g(x) is f'(g(x)).

y

~

f(u)

~

f(g(.»

FIGURE 3.24 Rates of change multiply: The derivative of fog atx is the derivative off atg(x) times the derivative ofg atx.

If f(u) is differentiable at the point u = g(x) andg(x) is differentiable at x, then the composite function (f 0 g)(x) = f(g(x)) is differentiable at x, and

THEOREM 2-The Chain Rule

(f

0

g)'(x) = j'(g(x))' g'(x).

inLeibniz'snotation,ify = f(u)andu = g(x),then

dy dy du dx = du' dx' where dy/du is evaluated at u = g(x).

Intuitive uProof" of the Chain Rule: Let ~u be the change in u when x changes by ~u =

~x, 80

that

g(x

+

~)

- g(x).

f(u

+

~u)

- f(u).

Then the corresponding change in y is ~y =

If ~u

'¢'

0, we can write the fraction ~y/ ~x as the product ~y

~y

~u

~x=~u'~x

(1)

144

Chapter 3: Differentiation and take the limit as

~x --->

0: lim ~y

dy dx

4x~O~

lim ~y. ~U 4x~O~u

=

~

lim

~y.

lim

~y.

4x~O ~u

=

b.u_o.6.u dy du =du'dx'

lim

~u

lim

~u

4x~O ~ .:1%-0

(Note that &u --+ 0 as ~- 0 since g is continuous.)

4x

The problem with this argument is that it could be true that ~u = 0 even when ~x oF 0, so the cancellation of ~u in Equation (I) would be invalid. A proof requires a different approach that avoids this flaw, and we give one such proof in Section 3.9. •

EXAMPLE 2 An object moves along the x-axis so that its position at any time I ;;,: 0 is given by X(I) = cos(t 2 + I). Find the velocity of the object as a function oft. Solution We know that the velocity is dx/ dl. In this instance, x is a composite function: x = cos(u) andu = 12 + I. Webave

dx du du dl

= -sin(u)

x ~ cos(u)

= 21.

By the Chain Rule,

dx dx du dl=du'dl = -sin(u)' 2t = -sin(12 =

+

dx

du evaluated at u

I)' 21

•

-2Isin(t 2 + 1).

UOutside-lnside u Rule A difficulty with the Leibniz notation is that it doesn't state specifically where the derivatives in the Chain Rule are supposed to be evaluated. So it sometimes helps to think about the Chain Rule using functional notation. If y = t(g(x)), then

dy dx

=

f'(g(x))' g'(x).

t

In words, differentiate the "outside" function and evaluate it at the "inside" function g(x) left alone; then multiply by the derivative of the "inside function."

EXAMPLE 3 Solution

Differentiate sin (x 2

+ x) with respect to x.

We apply the Chain Rule directly and find !sin(x2 +x) = cos(x2 inside

+ x)'(2x +

I).

inside derivative of left alone the inside

•

3.6 The Chain Rule

145

Repeated Use of the Chain Rule We sometimes have to use the Chain Rule two or more times to fmd a derivative. HisTORICAL BIOGRAPHY

Johann Bernoulli (1667-1748)

EXAMPLE 4

Find the derivative ofg(t) = tan(5 - sin2t).

Solution Notice here that the tangent is a function of 5 - sin 2t, whereas the sine is a function of2t, which is itself a function oft. Therefore, by the Chain Rule,

!

g'(t) =

Derivati~

(tan (5 - sin 2t))

= sec2 (5 -

u

(0 -

2

= sec2 (5 -

of tan u with 5 - sin2t

Derivative of 5 - sin u with u = 2t

sin2t)· :t(5 - sinZt)

= sec (5 - sinZt)·

=

cosZt· !(Zt))

sin2t)·(-cos2t)·Z

•

= -Z(cosZt) se~(5 - sinZt).

The Chain Rule with Powers of a Function If j is a differentiable function of u and if u is a differentiable function of x, then substitoting y = j(u) into the Chain Rule fonnula

dy dy du dx=du'dx leads to the fonnula

~j(u)

= f'(u)a;.

Ifn is any real number and j is a power function, j(u) = un, the Power Rule tells us thatf'(u) = nun-I. Ifu is a differentiable function ofx, then we can use the Chain Rule to extend this to the Power Chain Rule: ~( n) _

dx u

EXAMPLE 5

- nu

n-l

du dx'

The Power Chain Rule simplifies computing the derivative of a power of

an expression. (8)

~(5x3 - x 4)7 dx

= 7(5x 3 -

x4)6~(5x3 - x 4) dx

= 7(5x 3 - x4)"(5' 3x 2 -

Power Chain Rule with u = 5x 3 - X4,1I = 7

4x 3)

= 7(5x 3 - x 4)6(15x 2 - 4x 3)

(b)

~ dx

(_I-) 3x-2

= 4<3x - Zr 1

dx

= -1(3x -

Zr2~(3x

dx = -1(3x - Zr2(3) 3 (3x - 2)2

- Z)

Power Chain Rule with u=3x-2,n=-1

In part (b) we could also find the derivative with the Derivative Quotient Rule. (c)

~ (sin' x)

= 5 sin' x' =

~sinx

5 sin4 X COSX

Power Chain Rule with u = sinx,1I = 5, because sin" x means (sin x)1', n :# - 1 .

•

146

Chapter 3: Differentiation

EXAMPLE 6 In Section 3.2, we saw that the absolute value function y = Ixl is not differentiable at x = O. However, the function is differentiable at all other real numbers as we now show. Since Ix I =

I Deriwtive of the

W, we can derive the following formula: ~(Ixl)

Absolute Value Function

=

~W Power Chain Rule willi

_1_. !£(x2) dx 1 2x = 2Ix1· =

x'
=

EXAMPLE 7

2W

x TXT'

x,<

u

=

x2 ,n

=

1/2, x :# 0

•

o.

Showthattheslopeofeverylinetangenttothecurvey = 1/(1 - 2x)3is

positive.

Solution

We fmd the derivative:

dy = !£(1 - 2xr3 dx dx =

-3(1 - 2xr4 • ~ (1 - 2x)

=

-3(1 - 2xr4 ·(-2)

Power Chain Rule willi u

~

(1 - 2>:),.

~-3

6

(1 - 2x)4 At any point (x, y) on the curve, x ,< 1/2 and the slope of the tangent line is

dy dx

6

(1 - 2x)4'

•

the quotient of two positive numbers.

EXAMPLE 8 The formulas for the derivatives of both sin x and cos x were obtained under the assumption that x is measured in radians, not degrees. The Chain Rule gives us new mdiinsight into the difference between the two. Since 180° = 71" mdians, XO = ans where XO is the size of the angle measured in degrees. By the Chain Rule,

7I"x/180

d . ( 0) -_ dxd sm. (7I"x) _ 71" (7I"x) _ 71" ( 0) 180 - 180 cos 180 - 180 cos x .

dx sm x

See Figure 3.25. Similarly, the derivative of cos (XO) is -( 71"/180) sin(xO). The factor 71"/180 would compound with repeated differentiation. We see here the • advantage for the use of mdian measure in computations. y

I

Y = sinx

180

FIGURE 3.25 Sin (x') oscillates ooly 71"/180 times as often as sin x oscillates. Its maximum slope is 1f/180 atx ~ o(Example 8).

147

3.6 The Chain Rule

Exercises 3.6 Derivative Calculations In Exercises 1-8, given y

= j(u)

= g(x),

and u

f"md dy/dx =

49. y = sin (cos (21 - 5))

1'(g(x))g'(x) . 1. y

= 6u

3. y

=

- 9, u

sinu,

u

=

= (1/2)x4 3x + 1

cos u, u = sinx 7. y = tanu, u = IOx- 5 S. Y

=

2. y

= 2u',

4. Y

=

= 8x

u

cosu,

u

=

6. y = sin u,

u

=

= -secu,

8. y

-x/3 x - cosX'

u =x2 + 7x

= f(u)

In Exercises 9-18, write the function in the form y u = g(x). Then f"md dy/dx as a function ofx.

9. y

= (2x +

I)'

= (4

10. y

12.y=

13. y=

1)4

~ +x-x 2

(

(

16. y =

-10

+6

cot(". - ~)

20. q = V'2r - r2

21. s

=

3~ sin 31 + 5~ cos 51

22. s

23. r

= (csc8

25. y

2

= x sin x + xcos- x

27. y

=

28. y

= (5

29. y

=

2

(4x

=

+ 3)'(x + It'

V7

+ x sec x . 8

=

)2

sin (82) cos(28)

39. q = sin

e;l)

r

(vf+-I)

30. y

+ cos

43. y

45. y 47. y

= sin2 (",1 - 2) = (I + cos 21)-4 = (I tan 1)10 =

e;l)

= }sin-.s x - j-cos3 X

=

32. k(x)

(2x - 5)-I(x 2

36

=

38. r

(I)

=

,12- -)' (1 - 41

5x)'

tan3x (x + 7)4

=

• g

-

= x2seC(~)

34. g(x)

(I + sin 3/)-1 3 21

secYo tan

~)'

61. y

I = "9cot(3x

63. y

=

65. j(u)

x(2x

v':~31-+-v'r2~+~V=;I==-=1

60. y = (I - Yx)-1

- I)

62. y = 9tan

+ 1)4

= u' +

64. y

=

(t)

x 2 (x' - I)'

u

I,

= g(x) =

0

Yx,

g)' at the given value ofx.

x

=I

I I 66.j(u)=I- ii , u=g(x)=I_x'

67. j(u)

= cot 7~,

68. j(u)

I = u + -2 -'

69. j(u)

=

70. j(u)

= (:

u

= g(x) = 5Yx,

cos u

--1"'-, u + I ~:

u

r

u

= g(x) = u

x=-I

=I

x

= g(x) = ",x,

X

= 1/4

IOx 2 + X + I, x

= g(x) =

71. Assumethat1'(3) = -1,g'(2) Whatisy'atx = 2?

:2 -

I,

x

=0

= -I

= 5,g(2) = 3,andy = j(g(x)).

= sin (J(t)),j(O) = ",/3, and 1'(0) = 4,thenwhatisdr/dl = O? Suppose that functions f and g aod their derivatives with respect

72. Ifr

att 73.

to x have the following values at x = 2 and x = 3.

(i)

sin I) 40. q = cot ( -1-

x

f(x)

g(x)

2 3

8 3

-4

2

1'(x)

g'(x)

1/3 2",

-3 5

Find the derivatives with respect to x of the following combina-

In Exercises 41-58, finddy/dl. 41. y

58. y =

Vi)

Finding Derivative Values

- 2xt' + H~ + I)'

35. f(8) = ( I !u;,os 8 37. r

26. y

il (3x - 2)' + (4 - ~2

31. h(x) = x tan (2Yx) + 7 33. f(x)

sin

57. y = 31 (212 - 5)'

24. r = 6(sec8 - tan8)'/2

+ cot8)-1 4

=

54. y = 4 sin (VI + 56. y = cos4 (seC> 31)

In Exercises 65-70, f"md the value of(J

Find the derivatives of the functions in Exercises 19-40.

v'3=t

2 (I + cos (71))'

53. y = VI + cos (12) 55. y = tan2 (sin' I)

59. y = (I +

)

19. P =

i

Second Derivatives

= 5cos-4x

18. y

52. y =

and

- 3x)'

I-I

(-0.))'

(t))

Find y" in Exercises 59-64.

14. y = V3x 2 - 4x

15. y = sec(tanx)

51. y = (I + tan'

- I

50. y = cos (5 sin

42. y 44. y 46. y

= seC>",1 = (i + cot(I/2)t2 = (1-'/4 sin 1)4/3 31 - 4)-'

48.y= ( 51+2

tions at the given value of x. •• 2j(x), x = 2

x =3 x= 2

+ g(x), x = 3 f(x)/g(x), x = 2

b. f(x)

e. f(x)' g(x),

d.

e. f(g(x)),

r. v'iW,

2

g. l/g (x),

X

=3

x=2 h. Vf(x) + g2(x),

X

= 2

148

Chapter 3: Differentiation

74. Suppose that the functioos f and g and their dctivatives with respect to x have the following values at x ~ 0 and x ~ 1.

83. Running machinery too rast

Suppose that a piston is mnving straight up and down and that its position at time t sec is

s

x

f(x)

o

I 3

I

g(x)

/'(x)

g'(x)

I

5

-4

-1/3

1/3 -8/3

Find the derivatives with respect to x of the following combina-

tions at the given value of x. 5f(x) - g(x), x ~ I f(x) c. g(x) + I' x ~ I

e. g(f(x)), x g. f(x

~

+ g(x)),

75. Find d>/dtwhen 8

d. f(g(x)),

r.

0

(x 11

x

~

0

+ f(x))-2,

X

with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What efree! does doubliog the frequency have on the piston's velocity, acceleration, and jerk? (Once you fmd out, you will know why some machinery breaks when you ruo it too fast.)

~ I

x ~ 0

y

~ 3'1f/2 ifs ~ cos8andd8/dt ~ 5.

76. Find eIy/dt when x ~ I ify ~ x 2

+ 7x

- 5andd>/dt ~ 1/3.

What happens if you can write a :function as a composite in different

77. Find dy/dx ify ~ x hy using the Chain Rule withy as a composite of

b. y ~ I

+7

and

u ~ 5x - 35

+ (I/u)

and

u ~ I/(x - I).

y ~ u'

b. y~

Vu

and and

u~x'.

«x -

80. Find the tangeot to y ~

y'x 2

L

I)/(x -

X

+

1))2atx ~ O.

+ 7 at x ~

2.

Find the tangent to the curve y ~ 2 tan (=/4) at x ~ I.

b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval -2 < x < 2? Give reasons for your answer.

82. Slopes on sine curves L

25

ing when it is increasing at its fastest? y

60 H-+-+-+--b.>"'"""'~ d-+-+-+-+-+-H fi:'

1

'-- 40 H

~

0

-

1 . . . .'

Find equations for the tangents to the curves y ~ sin 2x and ~ -sin(x/2) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer.

y

b. Can anything be said about the tangents to the curves y ~ sinmxandy ~ -sin(x/m) at the origin (m a coostant 0# O)? Give reasons for your answer. c. For a giveo m, what are the largest values the slopes of the curvesy ~ sinmxandy ~ -sin(x/m) can ever have? Give

reasons for your answer. d. The function y ~ sin x completes one period on the interval [0, 2'1f], the function y ~ sin 2x coropletes two periods, the functiony ~ sin(x/2) completes half a period, and so on. Is there any relation between the number of periods y ~ sin mx coropletes on [0, 2'1f] and the slope of the curve y ~ sinmx at the origin? Give reasons for your answer.

_

H-f--71-+-+-+-+'"~-+--+--+-+-+-1 -

- - - '~ - - ----~

-v-

-~

u ~ '\IX

79. Findthetangeottoy ~

81.

1+

and is graphed in the accompanying figure.

20

78. Find eIy/dx if y ~ x'/2 hy using the Chain Rule with y as a composite of L

101)

.. On what day is the temperature increasing the fastest?

ways? Do you get the same dctivative each time? The Chain Rule says you should. Try it with the functions in Exercises 77 and 78.

y ~ (u/5)

~ 37 sin [;;; (x -

b. About how many degrees per day is the temperature increas-

Theory and Examples

L

A cos (2'1fbt) ,

84. Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbaoks, Alaaks, doring a typical 365-day year. The equation that approsirnates the temperature on day x is

x ~ 0

b. f(x)g'(x),

L

~

85. Particle motion ordinate line is s =

-

-

-

- 1'\ - - --~ --I>x

V+

position of a particle moving along a co-

1 41, with s in meters and t in seconds. Find the particle~ velocity and acceleration at t ~ 6 sec.

86. Constant acceleration Suppose that the velocity of a fa1liog body is v ~ kYs m/sec (k a constant) at the instant the body has fallen s m from its starliog point. Show that the body's accelera-

tion is constant. 87. Falling meteorite The velocity of a heavy meteorite entering Earth's atmosphere is inversely proportional to Ys when it is s kIn from Earth's center. Show that the meteorite's acceleration is inversely proportional to 8 2 . 88. Particle acceleration A particle moves along the x-axis with velocity dx/dt ~ f(x). Show that the particle's acceleration is

f(x)/'(x). 89. Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation

T

~ 2'1f~'

where g is the constant acceleration of gravity at the pendulum's

location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is

3.7

made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,

Implicit Differentiation

•• Graph dg/dt (where defmed) over [-"" ".]. b. Findd//dt. c. Graph d// dt. Where does the approxinlation of dg/ dt by d//dt seem to be best? Least good? Approximations by 1rigonome1ric polynomia1s are important in the theories of

tiL du = kL.

heat and oscillation, but we must not expect too much of them, as we see in the next exercise.

Assuming Ibis to be the case, show that the rate at which the period changes with respect to temperature is kT/2. 90. Chain Rule

,

,

Ixl. Then the

Suppose that /(x) = x' and g(x) =

~

0

g)(x) =

lxi' =

x'

are both differentiable at x

and

(g

0

/)(x) =

Ix'i =

/"

x'

D 91.

The derivative of sin 2x Graph the function y = 2 cos 2x for -2 '" x s 3.5. Then, on the same screen, graph

y=

sin 2(x

+ h) -

sin 2x

h

for h = 1.0, 0.5, and 0.2. Expctiment with other values of h, including negative values. What do you see happening as h --> O? Explain Ibis behavior.

D 92.

The derivative of cos (x') Graph y = -2x sin (x') -2 '" x s 3. Then, on the same screen, graph cos «x

y=

o

-"

= 0 even though g itself is not differ-

entiable at x = O. Does Ibis contradict the Chain Rule? Explain.

g(t)

'~f(t)

/

composites (j

149

"

94. (Continuation of Exercise 93.) In Exercise 93, the 1rigonome1ric polynomial /(t) that approximated the sawtooth function g(t) on [-".,,,.] had a derivative that approximated the derivative of the sawtooth function. It is possible, however, for a trigonometric polyoomial to approximate a function in a reasonable way without its derivative approxinlating the function's derivative at all well. As a case in point, the "polynomial"

s = h(t) = 1.2732 sin 2t

+

0.4244 sin 6t

+ 0.25465 sin lOt

+ 0.18189 sin 14t + 0.14147 sin 18t

for

graphed in the accompanying figure approximates the step function s = k(t) shown there. Yet the derivative of h is nothing like the derivative of k.

+ h)') - oos (x') h

,

for h = 1.0,0.7, and 0.3. Expctiment with other values of h. What do you see happening as h --> O? Explain this behavior.

s

COMPUTER EXPLORATIONS

I

Trigonometric Polynomials

-"

93. As the accompanying figure shows, the 1rigonometric "polynomial"

k(t)

I. ~ /

1

-2"

~

"2

0

h(t)

"

-1

s = /(t) = 0.78540 - 0.63662 cos 21 - 0.07074cos6t - 0.02546 cos lOt - 0.01299 cos 14t

•• Graph dk/dt (where defmed) over [-'IT, 'IT].

gives a good approximation of the sawtooth function s = g(t) on the interval [-"" ".]. How well does the derivative of / approximate the derivative of g at the points where dg/ dt is defmed? To find out, carry out the following steps.

3.7

b. Find dh/ dt. c. Graph dh/ dt to see how badly the graph fits the graph of dk/dt. Comment on what you see.

Implicit Differentiation Most of the functions we have dealt with so far have been described by an equation of the form y = f(x) that expressesy explicitly in tenDs of the variable x. We have learned rules for differentiating functions defined in this way. Another situation occurs when we encounter equations like

x 3 + y3 - 9xy

= 0,

y' - x = 0,

or

x' + y' - 25

= O.

150

Chapter 3: Differentiation (See Figures 3.26, 3.27, and 3.28.) These equations defme an implicit relation between the variables x andy. In some cases we may be able to solve such an equation for y as an explicit function (or even several functions) of x. When we cannot put an equation F(x, y) = 0 in the fonn y = f(x) to differentiate it in the usual way, we may still be able to fmd dyj ax by implicit differentiation. This section describes the tecbnique.

Y

Implicitly Defined Functions We begin with examples involving familiar equations that we can solve for y as a function ofx to calculate dyjax in the usual way. Then we differentiate the equations inlplicitly, and find the derivative to compare the two methods. Following the examples, we summarize the steps involved in the new method. In the examples and exercises, it is always assumed that the given equation determines y inJplicitly as a differentiable function of x so that dyj ax exists.

EXAMPLE 1 FIGURE 3.26 The curve + y' - 9xy ~ 0 is not the graph of

x'

anyone function of x. The curve can,

however, be divided into separate arcs that are the graphs of functions ofx. This particular curve, called afolium, dates to Descartes in 1638.

Find dyjax ify2 = x.

The equation y2 = x defmes two differentiable functions of x that we can actoally find, namely Yl = VX and Y2 = -VX (Figure 3.27). We know how to calculate the derivative of each of these for x > 0:

Solution

dYI ax

I 2VX

dY2 ax

and

1 2VX'

But suppose that we knew only that the equation y2 = x defmed y as one or more differentiable functions of x for x > 0 without knowing exactly what these functions were. Could we still find dy j ax? The answer is yes. To fmd dyj ax, we simply differentiate both sides of the equation y2 = x with respect to x, treating y = f(x) as a differentiable function of x:

y2

I I I I

= -1

2Y2

x

The Chain Rule gives

ax

-Yx

dy ax

1

=--

2Yx

FIGURE 3.27 The equationy2 - x ~ 0, or y2 = X as it is usually written, dermes two differentiable functions of x on the interval x > O. Exarople 1 shows how to find the derivatives of these functions without solving the equation y2 = x for y.

! (y2)

=

d 2 dy dx [f(x)] ~ 2f(x)j'(x) ~ 2y dx'

2y dy = I

Q(x, -\IX)

Y2~

/

Slope

=

I 2y'

This one fonnula gives the derivatives we calculated for both explicit solutions Yl = VX andY2 = -VX:

dYI ax

EXAMPLE 2

I 2Yl

I 2VX

and

d)'2 ax

I 2Y2

I

2( -vx)

I 2VX'

•

Find the slope of the circle x 2 + y2 = 25 at the point (3, -4).

The circle is not the graph of a single function of x. Rather it is the combined graphs of two differentiable functions, Yl = V25 - x 2 and Y2 = - V25 - x 2 (Figure 3.28). The point (3, -4) lies on the graph of)'2, so we can fmd the slope by calculating the derivative directly, using the Power Chain Rule:

Solution

dY21

ax x~3 =

-

I

-2x 2V25 - x 2 x~3

-6 2v'25=9

3 4'

3.7 Implicit Differentiation y

We can solve this problem more easily by differentiating 1he given equation of the circle implicitly with respect to x:

A.. (x2) + A.. (y2) dx

dx

=

dy dx Y2~-V25-X2

The slope at (3, -4) is-.x.1

Y

(3, -4).

A.. (25) dx

dy 2x+ 2y dx=O

---_5~------~Or-------~5~~x

FIGURE 3.28 The circle combines 1he graphs of two functions. The graph of Y2 is 1he lower semicircle and passes through

151

= (3, ....)

x

y.

-~ = 1. -4 4

Notice that unlike the slope formula for dy2/ dx, which applies ouly to points below the x-axis, the formula dy/dx = -x/yapplies everywhere the circle has a slope. Notice also that the derivative involves both variables x and y, not just the independent variable x. • To calculate the derivatives of other implicitly defmed functions, we proceed as in Examples 1 and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides of the defming equation.

Implicit Differentiation 1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x. 2. Collect the terms with dy/ dx on one side of the equation and solve for dy/ dx.

Y

EXAMPLE 3 Solution

Find dy/dx if y2 = x 2

(Figure 3.29).

We differentiate the equation implicitly. y2 =

x 2 + sinxy

:. (y2) = : . (x2)

FIGURE 3.29 The graph of y2 ~ x 2 + sin'9' in Example 3.

+ sinxy

+ :. (sinxy)

2y dy dx = 2x

+

d( ) (cosxy) dx xy

2y: = 2x

+

(cosxy)&

2Y: - (cosxy) (x : ) = 2x

+

(cosxy)y

dy (2y - xcosxy) dx = 2x dy dx

. .. treating Y as a function of and using the Chain Rule.

x

Treat xy as a product

Collect terms wilh dy/dx.

+ ycosxy

+ Y cosxy 2y - x cosxy

2x

+ x :)

Differentiate both sides with respect to x ...

Solve for dy/dx.

Notice that the formula for dy/dx applies everywhere that the implicitly defined curve has a slope. Notice again that the derivative involves both variables x andy, not just the independent variable x. •

152

Chapter 3: Differentiation

Derivatives of Higher Order Implicit differentiation can alao be used to find bigher derivatives.

EXAMPLE 4

Find d'y/dx 2 if 2x' - 3y2 = 8.

To start, we differentiate both sides of the equation with respect to x in order to findy' = dy/dx.

Solution

JI (2x' dx

- 3y2)

=

JI (8)

6Y.Y'

=

0

fu;2 -

dx

x

Treaty as a function ob. 2

y' =y,

when y # 0

Solve for y'.

We now apply the Quotient Rule to f'md y' .

y"

=:.. (~) =2xy / 2y ' =~ _;~.y'

Finally, we substitute y' = x 2/y to express y" in terms ofx andy.

2 _ 2x x (X2) _ 2x x' y" ----Y y2 Y Y y"

Curve of lens smface

•

wheny # 0

Lenses, Tangents, and Normal Lines In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular tu the surface of the lens at the point of entry (angles A and B in Figure 3.30). This line is called the normal tu the surface at the point of entry. In a profile view of a lens like the one in Figure 3.30, the normal is the line perpendicular to the tangent of the profile curve at the point of entry. FIGURE 3.30 TIre profile ofa leos, showing the beading (refraction) of a ray oflight as it passes through the leos surface.

y

EXAMPLE 5 Show that the point (2, 4) lies on the curve x' find the tangent and normal tu the curve there (Figure 3.31).

+ y' - 9xy

= O. Then

Solution

The point (2, 4) lies on the curve because its coordinates satisfy the equation given for the curve: 2' + 4' - 9(2)(4) = 8 + 64 - 72 = O. To find the slope of the curve at (2, 4), we first use implicit differentiation to f'md a fonnula for dy/ dx:

x'+y'-9xy=0

JI (x') + JI (y') dx

dx

-

JI (9xy)

=

JI (0)

Y

+ Y dx) dx

=

0

Differentiate both sides with respect to x.

+ 3x 2 - 9y

=

0

TreatX)' as a product andy as a function of x.

dy - 9 (xddx 3x2 + 3y2 dx dy (3y2 - 9x) dx

dx

dx

3(y2 - 3X): = 9y - 3x 2 FIGURE 3.31 Example 5 shows how to fmd equations for the taogeot and normal to the folium of Descartes at (2, 4).

dy dx

3y-x 2 y2 - 3x'

Solve for dy/d<.

153

3.7 Implicit Differentiation

We then evaluate the derivative at (x, y) = (2,4):

dyl dx

3Y-X21 (2,4) = y2 -

3x (2,4)

3(4)-22 8 4 = 42 - 3(2) = 10 = 5'

The tangent at (2, 4) is the line through (2, 4) with slope 4/5: y = 4

+ 1(x - 2) 5

4

12

Y=5 x +S' The normal to the curve at (2, 4) is the line perpendicolar to the tangent there, the line through (2, 4) with slope -5/4:

y=4-~(x-2) 4

5

13

•

Y=-4 x +T'

The quadratic fannula enables us to solve a second-degree equation like 2>y + 3x 2 = 0 for Y in terms of x. There is a fannula for the three roots of a cubic equation that is like the quadratic formula but much more complicated. If this fannula is used to solve the equation x 3 + y3 = 9xy in Example 5 for y in terms of x, then three functions determined by the equation are y2 -

and y

=! [-f(X) ± V-3 (4-x; + )~6 - 27x3- Jx; - )~6 - 27X3) J.

Using in3plicit differentiation in Example 5 was much sin3pler than calcolating dy/dx directly from any of the above fannulas. Finding slopes on curves dermed by higher-degree equations usually requires implicit differentiation.

Exercise 3.7 Differentiating Implidtly Use implicit differentiation to fmd dy/ dx in Exercises 1-14. 1. x'y

+ xy'

=

6

3.2xy+y'=x+y S. x'(x - y)' = x' - y'

13. ySinG)

x' -

6. (3xy

=

ISx)'

xy + y' = I + 7)' = 6y

10. xy

= cot(xy)

=0

12.

+ sin Y = x3y2

I-xy

14. xcos(2x

9.

=

4.

+ y'

X4

+ 3y)

=

15. 01/2

+ rl/2

17. sin (rO)

,_2x-y 8.x- x + 3y

2 _ x-I

7·y-x+1

x = lany 11. x + Ian(xy)

2. x 3

Find dr/dO in Exercises 15-18.

=

2v8 =

= I

16. r -

t

18. cosr + cotO

loll' + 1 0'/4 2

3

= rO

Second Derivatives In Exercises 1~26, use implicit differentiation to fmd dy/ dx and then d'y/dx'. 20. x'l' + y'l' = I 19.x'+y'=1

ysinx

21. y' =x' + 2x 23. 2Vy = x - y

22. y'-2x= 1-2y 24.xy+y'=1

154

Chapter 3: Differentiation

25. Ifx' 26. Ifxy

+ y' + y2

~ 16, fmd the value ofd"y/
43. The devll's curve (Gabriel Cramer, 1750) Find the slopes of tire devil's curve y' - 4y2 ~ x' - 9x 2 at tire fuur indicated points.

In Exercises 27 and 28, fmd the slope of the curve at the given points. 27. y2 + x 2 ~ y' - 2x at (-2,1) and (-2, -I) 28. (x 2

+ y2j2

~ (x -

y

yj2 at (1,0) and (I, -I)

Slopes, Tangents, and Normals In Exercises 29-38, verifY that tire given point is on the curve and fmd the lines that are (a) tangent and (b) normal to the curve at the given point. 29. x 2 + xy - y2 ~ I, (2,3) 30. x 2

31.

+ y2

~

25,

- -f-- * - --fc-->

X

(3, -4)

xV ~ 9,

(-1,3) 32. y2 - 2x - 4y - I ~ 0,

(-2, I)

33. 6x 2 + 3xy + 2y2 + 17y - 6 ~ 0, (-1,0) 34. x 2 - V3xy + 2y2 ~ 5, ( V3, 2 ) 35. 2xy

+ 'If siny

~ 2'1f,

36. xsin2y ~ ycos2x, 37. y

(I, 'If/2) ('If/4,'If/2)

~

2 sin ('IfX - y), (1,0) 2 38. x cos2y - siny ~ 0, (0,'If)

39. Parallel tangents Find the two points where the curve x 2 + xy + y2 ~ 7 crosses the x-axis, and show that the tangents to the curve at tlrese points are parallel. What is the common slope of these tangents? 40. Normals parallel to a lin. Find the normals to the curve xy + 2x - y ~ 0 that are parallel to the line 2x + y ~ O. 41. The eight curve Find the slopes of the curve y' ~ y2 - x 2 at

(See Figure 3.26.) a. Find the slope of the folium of Descartes x'

44. The folium of Deseartes

c. Find the coordinates of the point A in Figure 3.26, where the folium has a vertical tsngent.

Theory and Examples 45. Intersecting normal The line that is normal to tire curve x 2 + 2xy - 3y2 ~ 0 at (I, I) intersects the curve at what other point? 46. Power rule for rational ""ponents Let p aod q be integers with q > O. If y ~ xPi., differentiate the equivalent equation y' ~ xP implicitly aod show that, for y #' 0,

~ "'i.)-1 .
y

47. Normals to a parabola Show that if it is possible to draw three normals from the point (a, 0) to the parabola x ~ y2 shown in the accornpaoying diagram, then a must be greater than 1/2. One of the normals is the x-axis. For what value of a are the other two normals perpendicular? y

-1

42. The cisooid ofDiOOes (from about 200 B.c.) Find equations for the tsngent and normal to tire cissoid of Diocles y2(2 - x) ~ x' at (I, I). y

- 9xy ~ 0

b. At what point other than the origin does the folium have a horizontal tsngent?

the two points shown here.

----'~------>x

+ y'

at the points (4, 2) and (2, 4).

3.8

Related Rates

155

x' D 10 Exercises 51 aod 52, fmd both dy/dx (treatingy as a differentiable

48. Is there anything special about the tangents to the curves y' ~ and lx' + 3y' ~ 5 at the points (I, ± I)? Give reasons for your

answer. y

function of x) and dx/dy (treating x as a differentiable function ofy). How do dy/dx aod dx/dy seern to be related? Explain the relationship geometrically in terms of the graphs. 51. xy' +

x'y

~ 6

52. x' + y' ~ sin'y

COMPUTER EXPLORATIONS Use a CAS to perform the following steps in Exercises

5~.

a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation.

b. Using implicit differentiatioo, find a formula for the derivative dy/dx aod evaluate it at tlte given pnint P.

49. Verify that the following pairs of curves meet orthogoualiy.

a.

x' + y' ~ 4, x' ~ 3y'

b. x

3

50. The graph of y' ~ is called a semicubica! parabola aod is shown in the accoropaoying figure. Determine the coostant b so x + b meets this graph orthogoually. that the line y ~

-l

53.

x' -

xy

+ y'

~ 7,

55. y

,

_2+x

+ Y - I _ x' P(O, I)

56. y' + cos xy ~ 57.

x',

59.

60.

P(1,O)

x+tan(~) ~2, P(I,-;r) ~ I, 2y' + (xy)l/' ~ x' + 2,

58. xy' + tan (x + y)

3.8

P(2, I)

54. x' + y'x + yx' + y' ~ 4, P(I, I)

~ I - y', x ~ ly'

x'

c. Use tlte slope found in part (b) to fmd an equation for tlte tangent line to tlte curve at P. Then plot tlte implicit curve and tangent line togetlter no a single graph.

p(-;r, 0) P(1, 1)

xv'1+2.Y + y ~ x', P(I,O)

Related Rates In this section we look at problems that ask for the mte at which some variable changes when it is known how the mte of some other related variable (or perhaps several variables) changes. The problem of finding a mte of change from other known rates of change is called a related rates problem.

Related Rates Equations Suppose we are pumping air into a spherical balloon. Both the volume and radius of the balloon are increasing over time. If V is the volume and r is the radius of the balloon at an instant of time, then

156

Chapter 3: Differentiation

Using the Chain Rule, we differentiate both sides with respect to t to rmd an equatiDJI relating the rates of change of V and r,

So if we know the radius r of the balloon and the rate dV/dt at which the volume is increasing at a given instant of time, then we can solve this last equation for dr/dt to fmd how fast the radius is increasing at that instant. Note that it is easier to directly measure the rate of increase of the volume (the rate at which air is being pumped into the balloon) than it is to measure the increase in the radius. The related rates equation allows us to calculate dr/d, from dV/dt. Very often the key to relating the variables in a related rates problem is drawing a picture that shows the geometric relations between them. as illustrated in the following example.

EXAMPLE 1 Water runs into a conical tank at the rate of 9 W/min. The tank stands point down and has a height of lOft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep? Solution

Figure 3.32 shows a partially filled conical tank. The variables in the problem are V = volume (W) of the water in the tank at time t (min)

x - radius (ft) of the surface of the water at time t y = depth (ft) of the water in the tank at time t. We assume that V, x, and y are differentiable functions of t. The constants arc the dimensions of the tank. We arc asked for dy/dt when FIGURE 3.32 The geometry of the conical tank and the rate at which water rills the tank dctmnine how fast the water level rises (Example 1).

y

~

6 ft

and


The water forms a cone with volume

This equation involves x as well as V and y. BecllU8C no information is given about x and dx/dt at the time in question, we need to eliminate x. The similar triangles in Figure 3.32 give us a way to express x in terms of y: x

Y

~

5 10

Therefore, fmd

to give the derivative dV _ 1T 3 2dy _ 1T 2dy 12' ~ dt -"4Y dt'

di Finally, use y

=

6 and dV/dt

=

9 to solve for dy/dt.

9~~(6)': 1 dt = 1T

dy

Rj

0.32

At the moment in question, the water level is rising at about 0.32 ft/min.

•

3.8

Related Rates

157

Related Rates Problem Strategy 1. Draw a picture and name the variablea and conatants. Use t for time. Assume that all variables are differentiable functions oft.

2. Write dawn the numerical information (in terms of the symbols you have chosen). 3. Writedown what you are asked tofmd (usually amte, expressed 88 a derivative). 4. Write an equadon that relates the variables. You may have to combine two or more equations to get a single equation that relates the variable whose rate you want to the variables whose rates you know. 5. Differentiate with respect to t. Then express the rate you want in terms of the rates and variables whose values you know. 6. Evaluate. Use known values to f"md the unknown rate.

EXAMPLE 2 A hot air balloon rising straight up from a level field is tracked by a range finder 500 ft from the liftoff point At the moment the range fmder's elevation angl.e is '1r/4, the angle is increasing at the rate of 0.14 md/min. How fast is the balloon rising at that moment? :

- 0.14 radJmin

when 6 - trl4

FIGURE 3.33

Solution , "d, - ? [ when 8 - 'frI4

The rate of change of the

baIloon'!l height is :related to the rate of change of the angle the range finder makes with the ground (Example 2).

1.

We answer the question in six steps.

Draw a picture and name the variables and constants (Figure 3.33). The variables in

the picture are (J = the angle in radians the range fmder makes with the ground. y - the height in feet of the balloon. We let t represent time in minutes and assume that (J and y are differentiable functions oft. The one constant in the picture is the distance from the range finder to the liftoff point (500 ft). There is no need to give it a special symbol. 2. Write down the additional numerical information. :

3. 4.

5.

~

0.14mdjnrin

(J =

when

'1r

4 Write down what we are tofind. We want dy/dt when (J - '1r/4. Write an equation that relates the variables y and (J. y or y - 500tan(J 500 - lan8 Differentiate with respect to t using the Chain Rule. The result tells how dy/tit (which

we want) is related to d8/tit (which we know). dy dt 6.

~

500 (sec' 0) dO

dt

Evaluatewith(J = v/4andd8/dt = 0.14tofinddy/dt. :

~ 500(V2)2(0.14) ~

140

At the moment in question, the balloon is rising at the rate of 140 ft/min.

•

EXAMPLE 3 A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that has honed the corner and is now maving straight east. When the cruiser is 0.6 mi north of the intersection and the car is 0.8 mi to the east, the police determine with radar that the distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement, what is the speed of the car?

158

Chapter 3: Differentiation y

Solution Situation when x ~

0.8.y

~

0.6

We picture the car and cruiser in the coordinate plane, using the positive x-axis as the eastbound bighway and the positive y-axis as the southbound bighway (Figure 3.34). We let t represent time and set

x

y fly

dt

= position of car at time t

y = position of cruiser at time t

~-60 -nl---~:II:D-"_x o dx_? x tit - .

FIGURE 3.34 The speed of the caris related to the speed of the police cruiser and the rate of change of the distance between them (Example 3).

s = distance between car and cruiser at time t. We assume that x, y, and s are differentiable functions of t. We want to f'md dx/ dt when

x

= O.Smi,

y

dy dt

= 0.6mi,

tis dt

= -60mph,

= 20 mph.

Note that dy/ dt is negative because y is decreasing. We differentiate the distance equation

S2=X 2 +y2 (we could also use s = Y x 2 + y2), and obtain

tis dx dy 2sdt=2xdt+2Ydt

~=}(x:+yt) =

Yx2

I

+ y2

Y) (x dxdt + y ddt'

Finally, we use x = O.S, Y = 0.6, dy/dt = -60, tis/dt = 20, and solve for dx/dt. 20 = Y

dx

~=

I (O.S)' + (0.6),

20Y(0.S)2

+

(o.s: +

(0.6),

M

+

(0.6)(-60»)

(0.6)(60)

=W

At the moment in question, the car's speed is 70 mph.

EXAMPLE 4 A particle P moves clockwise at a constant rate along a circle of radius 10 ft centered at the origin. The particle's initial position is (0, 10) on the y-axis and its f'mal destination is the point (10, 0) on the x-axis. Once the particle is in motion, the tangent line atP intersects the x-axis at a point Q (which moves over time). Ifit takes the particle 30 sec to travel from start to f'mish, how fast is the point Q moving along the x-axis when it is 20 ft from the center of the circle?

y

x

FIGURE 3.35 The particle P travels clockwise aloog the circle (Example 4).

•

Solution

We picture the situation in the coordinate plane with the circle centered at the origin (see Figure 3.35). We let t represent time and let 0 denote the angle from the x-axis to the radial line joining the origin to P. Since the particle travels from start to finish in 30 sec, it is traveling along the circle at a constant rate of'1f/2 radians in 1/2 min, or '1f rad/min. In other words, dO/ dt = -'1f, with t being measured in minutes. The negative sign appears because 0 is decreasing over time. Setting x(t) to be the distance at time t from the point Q to the origin, we want to f'md

dx/dtwhen x

= 20

ft

and

~

= -'1f rad/min.

3.8

Related Rates

159

To relate the variables x and 0, we see from Figure 3.35 that x cos 0 = 10, or

x

= 10 sec O. Differentiation oflhis last equation gives

ax dO dt = 10 sec 0 tan 0 dt = -1O'IT sec 0 tan O. Note that ax/dt is negative because x is decreasing (Q is moving towards the origin). When x = 20, cosO = 1/2 and sec 0 = 2. Also, tanO = Vsec2 0 - 1 = Vi It follows that

At the moment in question, the point Q is moving towards the origin at the speed of 20Y3'IT RJ 108.8 ft/min. •

EXAMPLE 5 Ajet airliner is flying at a constant altitude of 12,000 ft above sea level as it approaches a Pacific island The aircraft comes within the direct line of sight of a radar station located on the island, and the radar indicates the initial angle between sea level and its line of sight to the aircraft is 30°. How fast (in miles per hour) is the aircraft approacbing the island when trrst detected by the radar instroment if it is turning upward (counterclockwise) at the rate of2/3 deg/ sec in order to keep the aircraft within its direct line of sight? ~~-L------~~x

R

Jet airliner A traveling at constant altitude toward radar station R (Example 5). FIGURE 3.36

Solution The aircraft A and radar station R are pictured in the coordinate plane, using the positive x-axis as the horizontal distance at sea level from R to A, and the positive y-axis as the vertical altitode above sea level. We let t represent time and observe that y = 12,000 is a constant. The general situation and line-of-sight angle 0 are depicted in Figure 3.36. We want to find ax/dtwhen 0 = 'IT/6 rad and dO/dt = 2/3 deg/sec. From Figure 3.36, we see that 12,000

-x-

= tan 0

or

x

= 12,000 cot O.

Using miles instead offeet for our distance units, the last equation translates to

x

=

12,000 5280 cotO.

Differentiation with respect to t gives

ax 1200 _2 dO dt = - 528 csc- 0 dt' When 0 = 'IT/6, sin2 0 = 1/4, so csc' 0 = 4. Converting dO/dt = 2/3 deg/sec to radians per hour, we find

dO 2('IT) dt = 3 180 (3600) radJhr.

1 hr

~

3600 sec, 1 deg

~

,,/180 tad

Substitotion into the equation for ax/dt then gives

ax 'IT ) (3600) dt = (1200) - 528 (4) (2) 3 ( 180

RJ

-380.

The negative sign appears because the distance x is decreasing, so the aircraft is approaching the island at a speed of approximately 380 miJhr when first detected by the radar. •

EXAMPLE 6 Figure 3.37(a) shows a rope running through a pulley at P and bearing a weight Wat one end The other end is held 5 ft above the ground in the hand M of a worker. Suppose the pulley is 25 ft above ground, the rope is 45 ft long, and the worker is walking

160

Chapter 3: Differentiation rapidly away from the vertical line PW at the rate of 6 ft/sec. How fast is the weight being raised when the worker's hand is 21 ft away from PW?

o I + - - -x - - + j

(a)

SoLution We let OM be the horizontal line of length x ft from a point 0 directly below the pulley to the worker's hand M at any instant oftime (Figure 3.37). Let h be the height of the weight W above 0, and let z denote the length of rope from the pulley P to the worker's hand. We want to know dh/dt when x = 21 given that dx/dt = 6. Note that the height of P above 0 is 20 ft because 0 is 5 ft above the ground. We assume the angle at 0 is a right angle. At any instant of time t we have the following relationships (see Figure 3.37b):

+z= + x2 =

20 - h 20

z

20ft

j

If we solve for z we have

2

20 2

j.

z2

Angle at 0 is a right angle.

+ x2 =

(25

+ h)2.

(1)

Differentiating both sides with respect to t gives 2x dx = 2(25 dt

(b)

FIGURE 3.37 A worker atM walks to the right pulling the weight W upwards as the rope moves through the pulley P (Example 6).

Total length of rope is 45 ft.

= 25 + h in the first equation, and substitute into the second equation,

h

o I+-- - x _ _ _IM

45

+

h) dh dt'

and solving this last equation for dh/ dt we find

dh dt

x 25

+h

Since we know dx/dt, it remains only to find 25 Equation (1),

dx dt'

(2)

+ h at the

instant when x = 21. From

so that (25

+ h?

= 841,

or

25

+h

= 29.

Equation (2) now gives

dh 21 126 dt = 29· 6 = 29 ~ 4.3 ft/ sec

•

as the rate at which the weight is being raised when x = 21 ft.

Exercises 3.8 1. Area Suppose that the radius r and area A = 'Tf'r2 of a circle are differentiable functions of t. Write an equation that relates dA/dt to dr/dt. 2. Surface area Suppose that the radius r and surface area S = 4'Tf'r2 of a sphere are differentiable functions of t. Write an equation that relates dS/dt to dr/dt. 3. Assume that y = 5x and dx/dt = 2. Find dy/dt. 4. Assumethat2x + 3y = l2anddy/dt = -2. Finddx/dt. 5. If Y = x 2 and dx/dt = 3, then what is dy/dt when x = -I?

= y3 - Y and dy/dt = 5, then what is dx/dt wheny = 2? 7. Ifx 2 + y2 = 25 anddx/dt = -2, then what is dy/dt when x = 3 andy = -4? 6. Ifx

8. If x 2y 3 9. IfL

x

= 4/27 and dy/ dt = 1/2, then what is dx/ dt when x = 2? = Yx 2 + y2, dx/dt = -1, anddy/dt = 3, find dL/dt when

5 andy = 12. 10. If r + s2 + v 3 = 12, dr/dt whenr = 3 ands = 1. =

= 4,

and ds/dt

=

-3, find dv/dt

11. If the original 24 m edge length x of a cube decreases at the rate of 5 m/min, when x = 3 m at what rate does the cube's a. surface area change? b. volume change?

12. A cube's surface area increases at the rate of72 in2/ sec. At what rate is the cube's volume changing when the edge length is x = 3 in?

3.8 13. Volume The radius r and height h of a right circular cylinder are related to the cylinder's volume Vby the formula V ~ '1Tr'h.

a. How is dV/dt related to dh/dt ifr is constant?

Related Rates

161

a. How is dA/dtrelated to d6/dt if a andb are constant? b. How is dA/dtrelated to d6/dt and da/dt ifouly b is constant? c. How is dA/dtrelated to d6/dt, da/dt, anddb/dt ifnone of a, b, and 6 are constant?

b. How is dV/dt related to dr/dt ifh is constant? c. How is dV/dt related to dr/dt and dh/dt ifneither r nor h is constant? 14. Volume The radius r and height h of a right circular cone are related to the cone's volume Vby the equation V ~ (1/3)'1Tr'h.

a. How is dV/dt related to dh/dt ifr is constant? b. How is dV/dt related to dr/dt ifh is constant? c. How is dV/dt related to dr/dt and dh/dt ifneither r nor h is constant?

20. Heating a plate When a circular plate of metsl is heated in an oven, its radius increases at the rate of 0.01 cro/min. At what rate is the plate's area increasing when the radius is 50 em? 21. Cbanging dimensions in a rectangle The length I of a rectangle is decreasing at the rate of2 cro/sec wbile the width w is increasing at the rate of 2 cro/ sec. When I ~ 12 cro and w ~ 5 ern, fmd the rates of change of
are decreasing. and which are increasing?

15. Changing voltage The voltsge V (volts), current I (amperes), and resistsnce R (ohms) of an electric circuit like the one shown here are related by the equation V ~ IR. Suppose that V is increasing at the rate of I volt/sec while I is decreasing at the rate of 1/3 amp/sec. Let t deuote time in seconds.

+,~r-----l, ,1--,

dx dt ~ 1 m/sec,

dy dt ~ -2 m/sec,

dz dt ~ 1m/sec.

Find the rates at which the box's
V

R

a. What is the value of dV/dt? b. What is the value of dI/ dt? c. What equation relates dR/dt to dV/ dt and dI/dt? d. Find the rate at which R is changing when V I ~ 2 amp. Is R increasing, or decreasing?

22. Changing dimensions in a rectangular box Suppose that the edge lengths x, y, and z of a closed rectangular box are changing at the following rates:

~

12 volts and

16. Electrical power The power P (watts) of an electric circuit is related to the circuit's resistance R (ohms) and current! (amperes) by the equation P ~ RJ'.

a. How are dP/dt, dR/dt, and dI/dt related ifnone ofP, R, and I

23. A sliding ladder A 13-ft ladder is leaning against a house when its base starts to slide away (see accompanying figure). By the time the base is 12 ft ftum the house, the base is moving at the rate of5 ft/sec. a. How fast is the top of the ladder sliding down the wall then? b. At what rate is the area of the triangle formed by the lAdder, wall, and grouod changing then? c. At what rate is the angle 6 between the ladder and the grnuod changing then? y

y(1) •

are constant? 13-ft lAdder

b. How is dR/ dt related to dI/ dt if P is constant? 17. Distance

s ~

Let x aod y be differentiable functions of t and let

Vx' + y' be the distance between the points (x, 0) aod (0, y)

in the xy-plaoe.

6

~------------~--~.~ ,

a. How is ds/dtrelAted to dx/dt ify is constant? b. How is ds/dtrelAted to dx/dt and dy/dt ifneither x nory is constant? c. How is dx/dtrelated to dy/dt ifs is constant? 18. Diagonals If x, y, aod z are lengths of the edges of a rectangular box, the common length of the box's diagonals is s ~

Vx'

+ y' +z'. a. Assuming that x, y, and z are differentiable functions of t, how is ds/dtrelated to dx/dt, dy/dt, aoddz/dfl b. How is ds/dtrelAted to dy/dt and dz/dt ifx is constant? c. How are dx/dt, dy/dt, and dz/dt related if s is constant? 19. Area The area A of a triangle with sides of lengths a aod b enclosing an angle of measure 8 is A

~ tabsin6.

o

,(I)

24. Commercial air traffic 1\vo commercial airplAnes are flying at an altitode of 40,000 ft along straight-line courses that intersect at right aogles. PlAne A is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nsutical mile is 2000 yd). PlAne B is approaching the intersection at 481 knots. At what rate is the distance between the planes chaoging when A is 5 nautical miles from the intersection point and B is 12 nautical

miles from the intersection point? 25. Flying a kite A girl flies a kite at a height of300 ft, the wind carrying the kite bnrizonta1ly away ftum ber at a rate of25 ft/sec. How fast must she let out the striog when the kite is 500 ft away from her? 26. Boring a cylinder The mechanics at Lincoln Automotive are reboring a 6-in.-deep cylinder to fit a new piston. The machine

they are using increases the cylinder's mdius one thousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800 in.?

162

Chapter 3: Differentiation

27. A growing sand pile Sand falls from a conveyor belt at the rate of 10 m 3/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute. 28. A draining conical reservoir Water is flowing at the rate of 50 m 3/min from a shallow concrete conical reservoir (vertex down) of base radius 45 m and height 6 m.

33. A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/ sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/ sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later? y

a. How fast (centimeters per minute) is the water level falling when the water is 5 m deep? b. How fast is the radius of the water's surface changing then? Answer in centimeters per minute. 29. A draining hemispherical reservoir Water is flowing at the rate of 6 m 3/min from a reservoir shaped like a hemispherical bowl of radius 13 m, shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radiusR is V = ('7T/3)y2(3R - y)when the waterisy meters deep.

s(t)

Center of sphere

/ )x

o

x(t)

34. Making coffee Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 10 in3/min. a. How fast is the level in the pot rising when the coffee in the cone is 5 in. deep? a. At what rate is the water level changing when the water is 8 m deep?

b. How fast is the level in the cone falling then?

b. What is the radius r of the water's surface when the water is ymdeep? c. At what rate is the radius r changing when the water is 8 m deep?

30. A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate. 31. The radius of an inflating balloon A spherical balloon is inflated with helium at the rate of 100'7T WImino How fast is the balloon's radius increasing at the instant the radius is 5 ft? How fast is the surface area increasing? 32. Hauling in a dinghy A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft/ sec.

is this

level rising?

a. How fast is the boat approaching the dock when 10ft of rope are out? b. At what rate is the angle () changing at this instant (see the figure)? Ring at edge

of do~ck~ii:tI_.

35. Cardiac output In the late l860s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wfuzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 L/min. At rest it is likely to be a bit under 6 L/min. If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30L/min. Your cardiac output can be calculated with the formula

Q

y=I5'

3.8

where Q is the number of milliliters of CO2 you exhale io a minute aod D is the difference between the C02 concentration (mlfL) io the blood pumped to the lungs aod the Co, concentration io the blood returning from the lungs. With Q ~ 233 mI/mio andD ~ 97 - 56 ~ 41 mI/L,

y ~

233mlfmio . 41 mI/L '" 5.68 L/mm,

Related Rates

163

40. A building'••hadow On a murning of a day when the auo will pass directly overhead, the shadow of an 80-ft building on level ground is 60 ft long. At the moment in question, the aogle 6 the sun makes with the grouod is iocreasiog at the rate of 0.27" /min. At wbat rate is the shadow decreasiog? (Remember to use radiaos. Express your aoswer io ioches per mioute, to the nearest tenth.)

fairly close to the 6 L/mio that most people bave at basal (resting) conditions. (Data courtesy of 1. Keeneth Herd, M.D., Quillao College ofMedicioe, East Teonessee State University.) Suppose that when Q ~ 233 aod D ~ 41, we also know that D is decreasiog at the rate of 2 uoils a minute but that Q remains uocbaoged What is bappeoiog to the cardiac output? 36. Moving along a parabola A particle moves along the parabola y = xl in the 1mt quadrant in such a way that its x-coordinate (measured io meters) ioereases at a steady 10 m/sec. How fast is the angle ofioclioation 6 of the lioe joioiog the particle to the origio cbangiog when x ~ 3 m? 37. Motion in lbe plane The coordinstes of a particle io the metric xy-plane are differentiable functions of time t with dx/ dt ~ -Im/secaoddy/dl ~ -5m/sec. How fast is the particle's distance from the otigio cbaogiog as it passes through the pnint (5, 12)? 38. Videotaping a moving car You are videotapiog a race from a stand 132 ft from the track, followiog a car that is moviog at 180 mi/h (264 ft/sec), as showo io the accompaoyiog figure. How fast will your caroera angle 6 be chaogiog when the car is right io front of you? A balf second later? Camera

A spherical iron ball 8 in. in diameter is coated with a layer ofice of uniform thickness. If the ice melts at

41. A melting ice layer

the rate of 10in'/mio, how fast is the thickness of the ice de-

creasing when it is 2 in. thick? How fast is the outer surface area ofice decreasiog? 42. ffighway patrol A highway patrol plane flies 3 mi above a level, straight road at a steady 120 mi/h. The pilot sees ao oncomiog car and with radar determioes that at the iostant the lin.,.ofsight distance from plaoe to car is 5 mi, the line-
a. At what rate is the player's distance from third base cbaogiog when the player is 30 ft from fust base? b. At what rates are angles 61 and 62 (see the figure) cbaogiog at that time?

132'

c. The player slides ioto second base at the rate of 15 ft/sec.At what rates are aogles 61 aod 62 cbaogiog as the player touches base? 39. A moving .had.... A light shioes from the top of a pole 50 ft high. A ball is dropped from the sarae height from a point 30 ft away from the light. (See accompaoyiog figure.) How fast is the sbadow of the ball moviog along the grouod 1/2 sec later? (Assume the ball falls a distance s ~ 1612 ft in 1 sec J

Second base V

Light

, Ball at time t

~

0

) 112 sec later

Home

50-ft pole

Shadow ~~------~--------~~~--~~-->x

o

30

x(t)

NOf TO SCALE

44. Ship. Two ships are steao3ing straight away from a point 0 along routes that make a 120" aogle. Ship A moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship B moves at 21 knots. How fast are the ships moviog spart when OA ~ 5 and OB ~ 3 nautical miles?

164

Chapter 3: Differentiation

3.9

Linearization and Differentials Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. The approximating functions discussed in this section are called linearizations, and they are based on tangent lines. Other approximating functions, such as polynomials, are discussed in Chapter 10. We introduce new variables dx and dy, called differentials, and define them in a way that makes Leibniz's notation for the derivative dy/ dx a true ratio. We use dy to estimate error in measurement, which then provides for a precise proof of the Chain Rule (Section 3.6).

Linearization As you can see in Figure 3.38, the tangent to the curve y = x 2 lies close to the curve near the point of tangency. For a brief interval to either side, the y-values along the tangent line give good approximations to the y-values on the curve. We observe this phenomenon by zooming in on the two graphs at the point of tangency or by looking at tables of values for the difference between I(x) and its tangent line near the x-coordinate of the point of tangency. The phenomenon is true not just for parabolas; every differentiable curve behaves locally like its tangent line. 2

4

-1 '--'==","-!;-""=,==,==,==cV3

o

y

= x 2 and its

tangenty

=

2x-l at(l,l).

1.2

0 ~~~~===cV2

o

Tangent and curve very close near (1, 1).

1.003

0.8~=~=='==='==VI.2

0.8

Tangent and curve very close throughout entire x-interval shown.

0.997 S;="=~="==='==='c=0 1.003 0.997 Tangent and curve closer still. Computer screen cannot distinguish tangent from curve on this x-interval.

FIGURE 3.38 The more ... magnify the graph ofa function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its

y

tangent

In general, the tangent to y = I(x) at a point x = a, where I is differentiable (Figure 3.39), passes through the point (a, I(a», so its point-slope equation is y = I(a) ~Or-------a L--------------,x

FIGURE 3.39 TIre tangent to the curve y = f(x) atx = a is the line

L(x)

=

f(a) + /,(a)(x - a).

+ /,(a)(x - a).

Thus, this tangent line is the graph of the linear function

L(x)

=

I(a) + /,(a){x - a).

For as long as this line remains close to the graph of I, L(x) gives a good approximation to

I(x).

3.9

165

Linearization and Differentials

If f is differentiable at x = a, lben lbe approximating function

DEFINmONS

L(x)

=

f(a) + f'(a)(x - a)

is lbe linearization of f at a. The approximation

f(x) '" L(x) of f by L is lbe standard linear appronmation of f at a. The point x = a is lbe center of lbe approximation.

EXAMPLE 1

v'1+x atx =

Find lbe linearization of f(x) =

0 (Figure 3.40).

y

y

1.1

y= 1 + ~ 2

=1+ ~

~

7

2 1.0

--~--~r---~----~--~----~--~x

-I

0

2

3

0.9 "--------"--- - - ----"---------' -0.1 0 0.1 0.2

4

FIGURE 3.40 The graph of y = vT+x and its linearizatioos atx = 0 and x = 3. Figure 3.41 showsa magoified view of the sma1l window about loathe y-axis.

Solution

Since

f'(x)

=

FIGURE 3.41 Magnified view of the window in Figure 3.40.

to +

X,-I/2,

wehavef(O) = I and 1'(0) = 1/2,givinglbelinearization

L(x)

=

f(a)

+ f'(a)(x - a)

=

I

t

+ (x - 0)

=

I

+ ~.

•

See Figure 3.41.

The following table shows how accurate lbe approximation v'1+x '" I + (x/2) from Example I is for some values of x near O. All we move away from zero, we lose accumcy. For example, for x = 2, lbe linearization gives 2 as !be approximation for which is not even accurate to one decimal place.

V3,

Approximation

v'i2 '" I + O:}

Vl.Os '"

I

\11.005 '" I

+

Og5

+ 0.~5

True value

ITrue value -

approximation I

1.095445

<10-2

= 1.025

1.024695

<10-3

= 1.00250

1.002497

=

1.10

Do not be misled by lbe preceding calculations into thinking that whatever we do wilb a linearization is better done wilb a calculator. In practice, we would never use a linearization to find a particular square root. The utility of a linearization is its ability to

166

Chapter 3: Differentiation replace a complicated fannula by a simpler one over an entire interval of values. If we have to work with for x close to 0 and can tolerate the small amount of error in· volved, we can work with I + (x/2) instead. Of course, we then need to know how much error there is. We further examine the estimation of error in Chapter 10. A linear approximation normally loses accuracy away from its center. As Figure 3.40 suggests, the approximation I + (x/2) will probably be too crude to be use· ful near x = 3. There, we need the lineari7ation at x = 3.

v'1+x

v'1+x '"

EXAMPLE 2 Solution

v'1+x at x =

Find the linearization of /(x) =

3.

We evaluate the equation defining L(x) at a = 3. With

/(3)

2,

=

/,(3)

=

1 (1 + x)-1/21 2

= x- 3

1, 4

we have

1

+ 4(x -

( ) = 2 Lx

x

5

•

4 + 4'

3) =

At x = 3.2, the linearization in Example 2 gives

v'1+x =

+ 3.2

VI

%+ 342 =

+ 0.800

1.250

= 2.050,

v'4.2 '" 2.04939 by less than one one·thousandth. The

which differs from the true value linearization in Example 1 gives

v'1+x =

'"

VI

+ 3.2

'" I

+

+

3:} = I

1.6 = 2.6,

a result that is offby more than 25%. y

EXAMPLE 3

Find the linearization of /(x) = cosx at x = 7f/2 (Figure 3.42).

Solution Since /(7f/2) = cos(7f/2) = 0, /,(x) = -sinx, and/'(7f/2) = -sin(",/2) = - I , we find the linearization at a = 7f/2 to be L(x) = /(a) + /'(a)(x - a) ~r---~~------~--~x

FIGURE 3.42 The graph of j(x) = cosx and its linearizatioo at x = ",/2. Near x = ",/2, cos x '" -x + (",/2)

+ (- 1)

=

0

=

-x + 2'

(x - ~)

7f

•

An important linear approximation for roots and powers is

(Example 3).

+ x)k

(1

'" 1

+ kx

(xnearO;anynumberk)

(Exercise 13). This approximation, good for values of x sufficiently close to zero, has broad application. For example, when x is small,

v'1+x '"

1

+

V'1

+ 5x 4 =

I = v'1=7

(1

k

=

1/2

I

+

(-I)(-x) = I

+x

k

=

-I; replace x by -x.

+ 5x 4)1/3 '"

1

+ 1(5x4)

+ ~x4

k

=

1/3; replace x by 5.>;'.

-1_1- = (1 -

-x

tx X)-l '"

3

= 1

rl/2 '" I + (_1)(_x2) = 2

(I - x 2

3

I + 1x 2 2

k = -1/2; replace x by -x 2 .

3.9 Linearization and Differentials

167

Differentials We sometimes use the Leibniz notation dy/ ax to represent the derivative ofy with respect to x. Contrary to its appe8l'llllce, it is not a ratio. We now introduce two new variables ax and dy with the property that when their ratio exists, it is equal to the derivative.

DEFINmON Let Y = I(x) be a differentiable function. The differential dx is an independent variable. The differential dy is dy = f'(x) ax.

Unlike the independent variable ax, the variable dy is always a dependent variable. It depends on both x and ax. Ifax is given a specific value and x is a particular number in the domain of the function I, then these values determine the numerical value of dy.

EXAMPLE 4 (8) Find dy ify = x S

+ 37x.

(b) Find the value of dy when x = I and ax = 0.2.

Solution

+ 37) ax (b) Substituting x = I and ax = 0.2 in the expression for dy, we bave

(8) dy = (5x'

dy = (5 '1'

+

•

37)0.2 = 8.4.

The geometric meaning of differentials is shown in Figure 3.43. Let x = a and set ax = ~x. The corresponding cbange in y = I(x) is ~y =

I(a

+ ax) -

I(a).

y

lJ.y

~

f{a

+ dx) -

f{a)

~~~. 1~~'n j : When dx is a small change in x, I the cOITeSponding change in : the linearization is precisely dy.

Tangent line

I

-4------~-------L----------------~x

o

a

a+dx

FIGURE 3.43 Geometrically, the differential dy is the change III in the linearization of f when x = a changes by an amount dx= ~. The corresponding cbange in the tangent line L is ~L =

=

+ ax) - L(a) I(a) + j'(a)[(a + ax) L(a

L(a

= f'(a)

ax.

+ dx)

a] - I(a) L(a)

168

Chapter 3: Differentiation

1bat is, the change in the linearization of I is precisely the value of the differential dy when x = a and dx = Ax. Therefore, ely represents the amount the tangent line rises or falls when x changes by an amount dx = Ax. If dx oF 0, then the quotient of the differential dy by the differential dx is equal to the derivative f' (x) because dx _ dy.~ dx -- f'(x) dx -

1'(x ) -_

dy dx'

We sometimes write

dl in place of ely then

= f'(x) dx,

= f'(x) dx

calling dl the differential of I. For instance, if I(x) dl = d(3x 2

= 3x 2 -

6,

6) = 6x dx.

-

Every differentiation formula like

+ u)

d(u

dx

du du =dx+dx

d(sin u)

or

dx

du

=

cosu dx

has a corresponding differential form like d(u

+ u)

= du

+ du

or

d(sinu) = cosu du.

EXAMPLE 5 We can use the Chain Rule and other differentiation rules to find differentials of functions.
d(_X_)=(X+I)dx-Xd(X+I)=Xdx+dx-Xdx= dx x + I (x + I)' (x + I)' (x + 1)2

•

Estimating with Differentials Suppose we know the value of a differentiable function I(x) at a point a and want to estimate how much this value will change if we move to a nearby point a + dx. If dx = ~x is small, then we can see from Figure 3.43 that ~y is approximately equal to the differential dy. Since I(a

+ dx)

= I(a)

+

~y,

the differential approximation gives I(a dr

+ dx)

when dx = ~x. Thus the approximation when I(a) is known and dx is small.

= 0.1

'" I(a) ~y

+ dy

'" ely can be used to estimate I(a

+ dx)

EXAMPLE 6 The radius r ofa circle increases from a = 10 m to 10.1 m (Figure 3.44). Use dA to estimate the increase in the circle's areaA. Estimate the area of the enlarged circle and compare your estimate to the true area found by direct calculation. .6.A ... dA

=

27Tadr

FIGURE 3.44 When dr is small compared with a, the differeotia\ dA gives the estimate A(a + dr) = 1I"a' + dA (Example 6).

Solution

Since A

= 1I"r 2 , the

estimated increase is

dA = A'(a) dr = 211"a dr = 211"(10)(0.1) = 211" m2.

Thus, since A(r +

~r)

'" A(r) A(IO

+ dA, we have

+ 0.1) '" A(IO) + 211" = 11"(10)2 + 211"

=

10211".

3.9 Linearization and Differentials

169

The area ofa circle of radius 10.1 m is approximately 102'IT m 2. The true area is

A(IO.I) = 'IT(10.1)2 = 102.01'IT m 2.

The error in our estimate is 0.01'IT m 2, which is the difference ~ - dA.

•

Error in Differential Approximation Let I(x) be differentiable at x = a and suppose that dx = Ax is an increment of x. We have two ways to describe the change in I as x changes from a to a + Ax: The true change:

III

=

I(a + Ax) - I(a)

The differential estimate:

dl

=

/,(a) Ax.

How well does dl approximate Ill? We measure the approximation error by subtracting dI from Ill: Approximation error = III - dI = =

III - /,(a)Ax I(a + Ax) - I(a) - /,(a)llx

"'f (!(a +

Il~

- I(a) _ /'(a»)' Ax

Call this part E.

= E'

Ilx.

As Ax --> 0, the difference quotient

I(a

+ Ilx) - I(a) Ax

approaches /,(a) (remember the definition of /'(a», so the quantity in parentheses becomes a very small number (which is why we called it e). In fact, e --> 0 as Ax--> O. When Ax is small, the approximation error e Ax is smaller still.

/,(a)llx

+ E Ax

true

estimated

error

change

cbange

III

=

Although we do not know the exact size of the error, it is the product e • Ilx of two small quantities that both approach zero as Ax --> O. For many common functions, whenever Ilx is small, the error is still smaller.

Change in Y

= f(x) near x = a

If y = I(x) is differentiable at x = a and x changes from a to a change Ily in I is given by

Ily in which e --> 0 as Ilx --> O.

=

/,(a) Ax + e Ax

+ Ilx,

the (1)

170

Chapter 3: Differentiation In Example 6 we found that

aA

-

'IT(10.1)2 - 'IT(10)2 = (102.01 - 100)'IT = (2'IT

=

+ 0.01 'IT) m2

dA.

............error

so the approximation error is aA - dA = E/1r = 0.01'IT and O.01'IT/O.I = O.I'IT m.

E

=

0.01 'IT/ /1r =

Proof of the Chain Rule Equation (I) enables us to prove the Chain Rule correctly. Our goal is to show that if I( u) is a differentiable function of u and u = g(x) is a differentiable function of x, then the composite y = l(g(x)) is a differentiable function ofx. Since a function is differentiable if and only if it bas a derivative at each point in its domain, we must show that whenever g is differentiable at Xo and I is differentiable at g(xo) , then the composite is differentiable at Xo and the derivative of the composite satisfies the equation

: I.~.o

=

/,(g(xo))' g'(xo).

Let I1x be an increment in x and let /1u and /1y be the corresponding increments in u andy. Applying Equation (I) we have

/1u

=

+ El I1x

=

(g'(xo)

/,(uo)/1u + E2/1u

=

(/'(uo) + E2)/1u,

g'(xo)11x

+ E,)I1x,

where El --> 0 as /1x --> O. Similarly,

/1y

=

where E2 --> 0 as /1u --> O. Notice also that /1u --> 0 as I1x --> O. Combining the equations for /1u and /1y gives

/1y

=

(j'(uo) + E2)(g'(Xo) +

El)l1x,

so

Since El and E2 go to zero as I1x goes to zero, three of the four terms on the right vanish in the limit, leaving -dyl

dx

= X=Xo

lim -/1y = /,(uo)g'(xo) = /,(g(xo))' g'(xo).

&x~o

4x

•

Sensitivity to Change The equation df = /,(x) dx tells how sensitive the output of I is to a change in input at different values of x. The larger the value of /' at x, the greater the effect of a given change dx. As we move from a to a nearby point a + dx, we can describe the change in I in three ways:

Absolute change Relative change Percentage change

True

Estimated

/11 = I(a + dx) - I(a) /11 I(a)

dl = /,(a)dx dl I(a)

/11 I(a) X 100

dl I(a) X 100

3.9

Linearization and Differentials

171

You want to calculate the depth of a well from the equation s = 161 2 by timing how long it takes a heavy stone you drop to splash into the water below. How sensitive will your calculations be to a O.I-sec error in measuring the time?

EXAMPLE 7

Solution

The size of ds in the equation

ds = 321dl depends on how big 1 is. If 1 = 2 sec, the change caused by dl = 0.1 is about

ds = 32(2)(0.1) = 6.4 ft. 1bree seconds later at 1 = 5 sec, the change caused by the same dl is

ds = 32(5)(0.1) = 16 ft. For a fixed error in the time measurement, the error in using ds to estimate the depth is _ larger when the time it takes until the stone splashes into the water is longer.

EXAMPLE 8 In the late 1830s, French physiologist Jean Poiseuille (''pwa-ZOY'') discovered the formula we use today to predict how much the radius of a partially clogged artery decreases the normal volume of flow. His formula, V

kr',

=

says that the volume V of fluid flowing through a small pipe or tube in a unit of time at a f'1xed pressure is a constant times the fourth power of the tube's radius r. How does a 10% decrease in r affect V? (See Figure 3.45.)

Angiography

Angioplasty

FIGURE 3.45 To unblock a clogged artery, an opaque dye is injected into it to make the inside visible under X-rays. Then a balloontipped catheter is inflated inside the artery to widen it at the blockage site.

Solution

The differentials of r and V are related by the equation

dV

=

a;

dr

=

4kr 3 dr.

The relative change in V is

dV

=

V

4kr 3 dr kr'

=

4dr r .

The relative change in V is 4 times the relative change in r, so a 10% decrease in r will result in a 40% decrease in the flow. _

EXAMPLE 9

Newton's second law,

F

=

d dl (mv)

= m

dv dl

=

ma,

172

Chapter 3: Differentiation is stated with the assumption that mass is constant, but we know this is not strictly true because the mass of a body increases with velocity. In Einstein's corrected formula, mass bas the value mo

m

=

-:-Vrl=_~';~/C=ii2'

where the ''rest mass" mo represents the mass of a body that is not moving and c is the speed oflight, which is about 300,000 lan/sec. Use the approximation

1 • ~2 '" 1 vi - x

1

+ -2x

2

(2)

to estimate the increase ~m in mass resulting from the added velocity v.

When v is very small compared with c, ,;/ c 2 is close to zero and it is safe to use the approximation

Solution

Eq. (2) with x

~

*

to obtain

or

m

Ri

ma

1 2(1)

+ Zmov

c2 .

Equation (3) expresses the increase in mass that results from the added velocity v.

(3)

•

Converting Mass to Energy Equation (3) derived in Example 9 bas an important interpretation. In Newtonian physics, (1/2)mo'; is the kinetic energy (KE) of the body, and if we rewrite Equation (3) in the form

we see that

or (~m)c2 '" ~(KE).

So the change in kinetic energy ~(KE) in going from velocity 0 to velocity v is approximately equal to (~m)c2, the cbange in mass times the square of the speed oflight. Using c '" 3 X 10 8 m/sec, we see that a small change in mass can create a large cbange in energy.

3.9

173

Linearization and Differentials

Exercises 3.9 finding LInearizations

Approximation Error

In Exercises 1-5, find the linearizationL(x) of f(x) atx = a.

In Exercises 29-34, each function f(x) chaoges value when x chaoges from Xo to Xo + dx. Find

1. f(x) = x 3

2x

-

+ 3, a

2. f(x) =~, a 1 3. f(x)=x+X-,

4. f(x) =~, 5. f(x)

=

= 2

a. the chaoge 4f = f(xo

-4

=

c. the approximation error 14f - dfl.

a = -8

tanx, a

(a) sinx

=

y

'IT

Find the lineariza-

(b) cosx

!J.f ~ f(xo

(c) tanx

8. f(x) =

+ 2x, Xo

x2

10. f(x) = 1

+ 4x -

+ x, Xo

11. f(x) =~,

12. f(x) = x

f(xol

(xo,f(x..:o;; )),...-..--o_ + -"-_ _ _ _ dx

Tangent ~----~--~~---~x

o

0.1

Xo

xo+dx

Xo = 0.9

X-I,

= 2x 2

9. f(x)

=

+ dx) -

"'lL

Linearization for Approximation In Exercises 7-12, fmd a linearization at a suitably chosen integer near Xo at which the given function and its derivative are easy to evaluate. =

- f(xo);

b. the value of the estimate df = !'(xo) dx; and

a= 1

6. Common Hnear appralimations at x = 0 tions of the following functions at x = O.

7. f(x)

+ dx)

=

29. f(x)

Xo = -0.9

3,

30. f(x)

8.1

31. f(x)

Xo = 8.5

~ I' Xo

32. f(x)

= 1.3

33. f(x)

13. Show that the linearization of f(x) = (1 L(x) = I + Ax.

+ x)<

at x = 0 is

14. Use the linear approximation (1 + xl' '" 1 + Ax to find an approximation for the functionf(x) for values ofxnear zero.

a. f(x) = (I - x)' I

c. f(x) = • ,...-:-

2 b. f(x) = -1-

+ 3X)'/3

e. f(x) = (4

d. f(x) =

v'2+7

r.

1(1 - 2 ~ J

f(x) =

Differential Estimates of Change In Exercises 35-40, write a differential formula that estimates the given change in volume or s'Wface area. 35. The chaoge in the volume V = (4/3)",r 3 of a sphere when the radius chaages from ro to ro + dr

-x

vl+x

34. f(x)

= x 2 + 2x, Xo = I, dx = 0.1 = 2x 2 + 4x - 3, Xo = -I, dx = 0.1 = x 3 - x, Xo = I, dx = 0.1 = x·, Xo = I, dx = 0.1 = x-I, Xo = 0.5, dx = 0.1 = x 3 - 2x + 3, Xo = 2, dx = 0.1

36. The chaoge in the volume V = x 3 of a cube when the edge lengths chaoge from Xo to Xo + dx 37. The chaage in the surface area S = &2 of a cube when the edge lengths chaage from Xo to Xo + dx

+ xl' '"

38. The chaoge in the lateral surface area S = ",rv'r 2 + h 2 of a right circular cone when the radius changes from ro to ro + dr and the height does not chaoge

16. Find the linearization of f(x) = + sinx atx = O. How is it related to the individual linearizations of and sin x atx = O?

39. The chaoge in the volume V = 'lTr2h of a right circular cylinder when the radius chaages from ro to ro + dr and the height does notchaoge

Derivatives in Differential Form

40. The chaage in the latera1 surface area S = 2",rh of a right circular cylinder when the height chaages from ho to ho + dh and the

15. Faster than a clliculator Use the approximation (1 I + Ax to estimate the following.

a. (1.0002)"

b. '\7"1.009

Vx+l

Vx+l

In Exercises 17-28, find dy.

17. Y

=

x3

-

3v'i

2x 19. Y=--2

I+x

21. 2y 3/2

+ -9'

- x = 0

radius does not change 18. Y

=

20. Y

=

x'\l'1=7

3 I

+

vX

22. -9'2 - 4x 3/2 - y

24. Y

25. Y

26. y

= cos (x ) = sec(x 2 -

27. y

= 3 csc (I

28. Y

=2

- 2v'i)

41. The radius of a circle is increased from 2.00 to 2.02 m.

.1::)

= sin (5v'i) = 4 tan (x 3/3)

23. Y

Applications

2v'i (

=0

•• Estimate the resulting change in area. b. Express the estimate as a percentage of the circle's original area.

2

I)

cot(Jx)

42. The diameter of a tree was lOin. Doring the following year, the circumference increased 2 in. About how much did the tree's diameter increase? The tree's cross-section area?

174

Chapter 3: Differentiation

43. Estimating volume Estimate the volume of material in a cylindrical shell with length 30 in., radius 6 in., and shell thickness 0.5 in. 0.5 in. 6in.L....._ _ _ _ _"""-- / ~ 30in. -O>i

44. Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%? 45. Tolerance The radius r of a circle is measured with an error of at most 2%. What is the maximum corresponding percentage error in computing the circle's

As a member of NASA's medical team, you want to know how sensitive W is to apparent changes in g caused by flight maneuvers, and this depends on the initial value of g. As part of your investigation, you decide to compare the effect on W of a given change dg on the moon, where g = 5.2 ft/sec 2 , with the effect the same change dg would have on Earth, where g = 32 ft/sec2 • Use the simplified equation above to find the ratio of dWmoon to dWEar!h. 52. Measuring acceleration of gravity When the length L of a clock pendulum is held constant by controlling its temperature, the pendulum's period T depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in g. By keeping track of IlT, we can estimate the variation in g from the equation T = 27T'(L/g)I/2 that relates T, g, andL. a. With L held constant and g as the independent variable, calculate dT and use it to answer parts (b) and (c).

a. circumference?

b. area? 46. Tolerance The edge x of a cube is measured with an error of at most 0.5%. What is the maximum corresponding percentage error in computing the cube's a. surface area? b. volume? 47. Tolerance The height and radius of a right circular cylinder are equal, so the cylinder's volume is V = 7T'h 3 • The volume is to be calculated with an error of no more than 1% of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h. 48. Tolerance a. About how accurately must the interior diameter of a 10-mhigh cylindrical storage tank be measured to calculate the tank's volume to within 1% of its true value? b. About how accurately must the tank's exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within 5% of the true amount? 49. The diameter of a sphere is measured as 100 ± 1 cm and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation. 50. Estimate the allowable percentage error in measuring the diameter D of a sphere if the volume is to be calculated correctly to within 3%. 51. The effect of flight maneuvers on the heart

The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation

W= PV+ V8v

b. If g increases, will T increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 1OO-cm pendulum is moved from a location where g = 980 em/ sec2 to a new location. This increases the period by dT = 0.001 sec. Find dg and estimate the value of g at the new location. 53. The linearization is the best linear approximation Suppose that y = f(x) is differentiable at x = a and that g (x) = m(x - a) + c is a linear function in which m and c are constants. If the error E(x) = f(x) - g(x) were small enough near x = a, we might think of using g as a linear approximation of f instead of the linearization L(x) = f(a) + f'(a)(x - a). Show that if we impose on g the conditions 1. E( a) = 0

The approximation error is zero at x = a.

. E(x) 2.1un x - a =0 x ..... a

The error is negligible when compared with x - a.

then g(x) = f(a) + f'(a)(x - a). Thus, the linearization L(x) gives the only linear approximation whose error is both zero at x = a and negligible in comparison with x - a. The linearization, L(x): Y = f(a) + f'(a)(x - a)

\

2

Y =f(x)

54. Quadratic approximations a. Let Q(x) = bo + bl(x - a) + b2(x - a)2 be a quadratic approximation to f(x) at x = a with the properties: i) Q(a) = f(a) ti) Q'(a) = f'(a)

iii) Q"(a) = j"(a). W = a

+ gb

(a, b constant).

I

---------L---------.~x

2g ,

where W is the work per unit time, P is the average blood pressure, V is the volume of blood pumped out during the unit of time, 8 ("delta") is the weight density of the blood, v is the average velocity of the exiting blood, and g is the acceleration of gravity. When P, V, 8, and v remain constant, Wbecomes a function of g, and the equation takes the simplified form

Some other linear approximation, g(x): Y = m(x - a) + c

Determine the coefficients bo, b l , and b2 •

Chapter 3 Questions to Guide Your Review h. Find the quadratic approximation to I(x) x ~ o.

D •.

~

•• Plot the function 1 over l.

1/(1 - x) at

b. Fiod the linearization L of the function at the point a.

Graph I(x) ~ 1/( I - x) and its quadratic approximation at x ~ O. Then zoom in on the two graphs at the point (0, I).

•• Plot 1 and L together on a single graph. d. Plot the absolute error I/(x) - L(x) lover I and fmd its

Comment on what you see.

D d.

maximum value.

Find the quadratic approximation to g(x) ~ l/x at x ~ 1. Graph g and its quadratic approximation together. Comment

can, satisfying

Find the quadratic approximation to h(x) ~ -yIi"+x at x ~ O. Graph h and its quadratic approximation together.

Ix -

COMPUTER EXPLORATIONS In Exercises 55-58, use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval 1. Perform the following steps:

Chapter

~

<

al

I/(x) - L(x)1

8

<

E

for E ~ 0.5, 0.1, and O.o!. Then check graphically to see if your 8-estimate holds true.

Comment on what you see. f. What are the linearizations of I, g, and h at the respective points in parts (b), (d), and (e)?

> 0 as you

e. From your graph in part (d), estimate as large a 8

on what you see.

D e.

175

55. I(x) ~ x' + x 2 56. I(x)

~ 4:2-+11'

-

H, I].

57. I(x) ~ x 2j3(x - 2), 58. I(x) ~

Vi -

[-1,2], a ~ I

2:x:,

sinx,

a

~!

[-2,3], a ~ 2 [0,2".],

a~ 2

Questions to Guide Your Review

1. What is the derivative of a function I? How is its domain related to the domain of I? Give examples. 2. What role does the derivative play in defming slopes, tangents,

and rates of change?

13. How do derivatives arise in the study of motion? What can you learn about a body's motion along a line by examining the derivatives of the body's position function? Give examples.

14. How can derivatives arise in economics?

3. How can you sOO3etimes graph the derivative of a function when all you have is a table of the function's values? 4. What does it mean for a function to be differentiable on an open interval? On a closed interval?

5. How are derivatives and one-sided derivatives related? 6. Describe geometrically when a function typically does not have a derivative at a point

15. Give examples of still other applications of derivatives. 16. What do the limits Iim,.,~o«sinh)/h) and I.im,,~o«cos h - I)/h) have to do with the derivatives of the sine and cosine functions? What are the derivatives of these functions?

17. Once you know the derivatives of sin x and cos x, how can you fmd the derivatives of tan x, cot x, sec x, and csc x? What are the derivatives of these functions?

7. How is a function's differentiability at a point related to its continuity there, if at all?

18. At what points are the six basic trigonometric functions continu-

8. What rules do you know for calculating derivatives? Give soroe examples.

19. What is the rule for calculatiog the derivative of a cOO3posite of

A.. (xn) ~ dx d

two differentiable :functions? How is such a derivative evaluated? Give examples.

9. Explain how the three formulas &.

ous? How do you know?

n:x;n-I

20. If U is a differentiable function of x, how do you fmd (d/dx)(u n) if n is an integer? If n is a real number? Give examples.

du

21. What is implicit differentiation? When do you need it? Give examples.

h. dx (cu) ~ c dx

22. How do related rates problems arise? Give examples. d

•• dx

(UI

+ U2 + ... + Un)

~

dul

dU2

dun

dX + dX + ... + dX

enable us to differentiate any pOlYOOO3ial. 10. What formula do we need, in addition to the three listed in Question 9, to differentiate rational functions?

11. What is a second derivative? A third derivative? How many derivatives do the functions you know have? Give examples. 12. What is the relationship between a function's average and instan-

taneous rates of change? Give an example.

23. Ontline a strategy for solving related rates problems. illustrate with an example. 24. What is the lioearizationL(x) ofa function/(x) at a point x ~ a? What is required of 1 at a for the lioearization to exist? How are

linearizations used? Give examples. 25. If x moves from a to a nearby value a + dx, how do you estimate the corresponding change in the value of a differentiable function I(x)? How do you estimate the relative change? The percentage change? Give an example.

176

Chapter 3: Differentiation

~

Chapter

Practice Exercises

Derivatives of Functions Find the derivatives of the functions in Exercises 1-40. 1. y = x' - 0.125x 2 + 0.25x 2. Y = 3 - 0.7x' + 0.3x' 3. Y

=

x' - 3(x 2 + ,.z)

4.

_,.r.; I y-x + v7x- '1f+ I

+ 1)2(X 2 + 2x)

6. y

= (2x

7. y = (rP + sec 6 + I)'

8. y

=

5. Y = (x

Vi 9.s=I+Vi

10. s

13. s = cos' (1 - 2t)

CS~6 - ~y

I =-:z--

+ tan t)'

16. s = csc'(1 - t + 3t2 )

18. r = 26Ycos 6

v28

v'8+1)

20. r = sin (6 +

21• Y --"2x I 2 csc-X 2

22. Y = 2Vi sin Vi

23. y = x- lf2 sec(2xj2 25. Y = 5 cotx2

24. Y

27. Y = x 2 sin2 (2x 2 ) 29. s=

30.s=

33. y =

C:lf V':J

=

Y

34. Y

sin 6 )2 cos6 - I

3 (5x 2 + sin2x)'f2

x' + 4xy -

45.

ViY =

I

2 _

x

47. Y - x

+

3y4/'

f(x)

g(x)

f'(x)

g'(x)

o

I 3

I 5

-3

1/2

1/2

-4

=

2x

I

a. 6f(x) - g(x), x = I

+

+

46. x"y2

=

IOy·I'

=

1r

=

15

f. (x

x = 0

x

0

=

+ f(x))'f2, x

= I

f'(x)

f(x)

o

9

-2

I

-3

1/5

a. Vif(x), x = I c. f(Vi), x = I f(x) e. 2 + cosx' x = 0

ViW,

x

0

=

d. f(1 - 5 tan x), f. 10 sin

(7)f

2

x = 0 x= I

(x),

57. Find the value ofdy/dt att = Oify = 3sin2xandx = t 2 + 'If. 58. Find the value of ds/du at u = 2 if s = t 2 + 5t and t = (u 2 + 2U)I/3. 59. Find the value of dw/ds at s = 0 if w = sin (vr r = 8sin(s + 'If/6).

-

2) and

61. Ify' + Y = 2cosx, fmdthevalue ofd"y/dx 2 at the point (0, I). 62. Ifxl/3 + yl/' = 4,findd"y/dx 2 atthepoint(8,8).

+ 2pt'f2 + 82 =

b.

60. Find the value of dr/dt at t = 0 if r = (rP + 7)1/3 and rPt+6=\.

=~: ~ ~

S

d. f(g(x)),

I

Find the flTSt derivatives of the following comhinations at the given value of x.

I

52. 2rs - r -

=

+ g(x)), x = 0

Applying the Derivative Definition 10 Exercises 63 and 64, fmd the derivative using the defmition.

10 Exercises 51 and 52, fmddr/ds.

+ sin2 s

x

Vi

40. Y = (3 + cos' 3x)-I/3

10 Exercises 49 and 50, fmddp/dq. 49. p' + 4pq - 3q2 = 2 50. q = (5p2

51. rcos2s

I'

x

38. y = 20(3x - 4)1/4(3x - 4tl/'

44. 5x 4/'

X= 0

b. f(X)g2(X),

f(x)

2Vi )2 2Vi + I

= 4xYx +

48. y2

Find the flTSt derivatives of the following comhinations at the given value of x.

56. Suppose that the functioo f(x) and its frrst derivative have the following values at x = 0 and x = I.

Implicit Diffell!ntiation 10 Exercises 41-48, fmd dy/ dx by implicit differentiation. 41. xy + 2x + 3y = I 42. x 2 + xy + y2 - 5x = 2 43.

x

-I 15(15t - I)'

= (

show that

implicitly,

Numerical Values of Derivatives 55. Suppose that functions f(x) and g(x) and their first derivatives have the following values at x = 0 and x = I.

g. f(x

36. r = (I + Sin6)2 I - cos6

+ I)~

37. Y = (2x 39. Y

32.

~X2x~ x

35. r = (

54. a. By differentiating dy/dx = x/yo b. Thea show that d"y/dx 2 = -I/y'.

e. g(f(x)),

=

~

b. y2 = I -

x 2 - y2 = I

c. g(x)

Vi esc (x + I)' 26. Y = x 2 cot 5x 28. Y = x -2 sin2 (x')

31. y =

= I

14.s=cot'(f)

17. r = Y26sin6 19. r = sin

(-I _

+ y'

a. x'

vt - I I - - - .212. y = 5in2 X smx

11. Y = 2tan2 x - sec2x

15. s = (sec t

- 5)(4 - xtl

53. Find d"y/dx 2 by implicit differentiation:

-3

63. f(t) = 2t

I

+I

64. g(x)

= 2x 2 +

I

65. a. Graph the function

f(x) b. Is f continuous at x

=

X2, { -X 2 ,

=

07

-1:5x
t. Is f

differentiable at x = 07 Give reasons for your answers.

= {x'

b. Is f continuous at x

tanx,

=

-I :5x
differentiable at x = 07 Give reasons for your answers.

2 - x,

I7

= 2,

+ 2x - 5y

84. (y -

= {x'

+ I)

85. x + 86. x 3/2

(I, I) = 2,

(3,2)

x? = 2x + 4, (6,2) v'iY = 6, (4, 1) + 2y 3/2

=

17,

(1,4)

87. Find the slope of the curve x 3y 3 + y2 and (I, -I).

t. Is f

differentiable at x = 17 Give reasons for your answers. 68. For what value or values of the constant m, if any, is

f(x) = {Sin2x, x:5 0 mx, x>O

a. continuous at x

x2

80. Normal to a circle Show that the normal line at any point of the circle x 2 + y2 = a 2 passes through the origin.

83. xy

=

be tangent

=

78. Slope of tangeot Show that the tangent to the curve y = x 3 at any point (a, a 3 ) meets the curve again at a point where the slope is fonr times the slope at (a, a 3 ).

82. x 3 + y2

67. a. Graph the function

b. Is f continuous at x

+ C is to

77. Tangent parabola The parabola y to the line y = x. Find C.

In Exercises 81-86, fmd equations for the lines that are tangent and nonna! to the curve at the given point. 81. x 2 + 2y2 = 9, (1,2)

07

t. Is f

f(x)

177

79. Tangent turve For what value of cis the curve y = c/(x tangent to the line through the points (0, 3) and (5, -2)7

66. a. Graph the function

f(x)

Chapter 3 Practice Exercises

=

X

+ y at the points (I, I)

88. The graph shown suggests that the curve y = sin (x - sinx) might have horizontal tangents at the x-axis. Does it7 Give reasons for your answer.

= O?

y 1

b. differentiable at x = 07

y

= sin (x -

sin x)

Give reasons for your answers. ~~~----+-----~~~----~----~~~X

Slopes, Tangents, and Normals 69. Tangeots with speoified slope Are there any points on the curve y = (x/2) + 1/(2x - 4) where the slope is -3/27 If so, fmd them.

-2w

2w

-1

70. Tangeots with speoified slope Are there any points on the curve y = x - 1/(2x) where the slope is 37 If so, fmd them. 71. Horizontal tangent. Find the points on the curve y = 2x 3 - 3x 2 - 12x + 20 where the tangent is parallel to the x-axis. 72. Tangeot inten:epts Find the x- and y-intercepts of the line that is tangent to the curve y = x 3 at the point (-2, -8).

73. Tangents perpenditular or parallel to lines Find the points on the curve y = 2x 3 - 3x2 - 12x + 20 where the tangent is

a. perpendicular to the line y b. parallel to the line y 74. Interseding tangents y = ('1r sin x)/x at x =

=

= V2 '1r

Analyzing Graphs Each of the fignres in Exercises 89 and 90 shows _ graphs, the graph of a function y = f(x) togdher with the graph of its derivative f' (x). Which graph is which7 How do you know? 89.

y

90.

y

I - (x/24). 12x.

Show that the tangents to the curve and x = -'1r intersect at right augles.

75. Normal. parallel to a line Find the points on the curve y = tan x, -'1r/2 < x < '1r/2, where the normal is parallel to the line y = -x/2. Sketch the curve and normals together, labeling each with its equation. 76. Tangent and normal lines Find equations for the tangent and normal to the curve y = I + cos x at the point ('1r/2, I). Sketch the curve, tangent, and normal together, labeling each with its equation.

--+----L----J-~x

o

-1

@

-2

2

178

Chapter 3: Differentiation

91. Use the following information to graph the function y

~

!(x) for

-1~x~6.

Trigonometric Umits Find the limits in Exercises 95-102.

i) The graph of! is made ofline segroents jnined end to end.

x-a 2x 2

iii) The derivative of!, where deImed, agrees with the step function shown here. y

~f'(x)

L -+ -2 y

2

3

96. lim 3x - atan 7x

x

-

x~o

97. lim sinr 1'-0

99.

98. lim

tan2r

. I1m

'~(rrf2)-

100. lim

sin (sin 6) a

9-0

4tan26 + tan 6 tan2 6 + 5

+

(1

I

I - 2 cot' 6

,~o+5cot'6 -7cot6 - 8

~ c~'--;.,........; J ~) x -I _I~ I

sin x

95. lim

ii) The graph starts at the point ( -I, 2).

101. lim

x-a 2

~

X

sinx

102. lim I - ;os 6

2cosx

,~o

Show how to extend the functions in Exercises 103 and 104 to be con-

tinuous at the origin. tan (tan x) tan x

tan (tan x)

92. Repeat Exercise 91, supposing that the graph starts at (-I, 0) instead of( -1,2).

103. g(x) ~

Exercises 93 and 94 are about the accoropanying graphs. The graphs in part (a) shnw the numbers of rabbits and foxes in a small arctic population. They are plotted as functions of time for 200 days. The number of rabbits increases at fIrst, as the rabbits reproduce. But the foxes prey on rabbits and, as the number of foxes increases, the rabbit population levels o!fand then drops. Part (b) shows the graph of the derivative of the rabbit population, made by plotting slopes.

Related Rates 105. Right circular cylinder The total surface area S of a right circular cylinder is related to the base radius r and height h by the equation S ~ 211"r2 + 211"rh.

93.

L

What is the value of the derivative of the rabbit population when the number of rabbits is largest? Smallest?

b. What is the size of the rabbit population when its derivative is largest? Smallest (negative value)? 94. In what units should the slopes of the rabbit and fox population

curves be measured?

104. !(x) ~ . (.

sm Stnx

)

.. How is dS/dtrelated to dr/dt if h is constant? b. How is dS/dtrelated to dh/dt ifr is constant? c. How is dS/dtrelated to dr/dt aoddh/dt ifneither rnor h is

constant? d. How is dr/dtrelated to dh/dt if Sis constant? 106. Right circular cone The lateral surface area S of a right circolar cone is related to the base radius r aod height h by the equationS ~ 1I"rVr2 + h 2. .. How is dS/dtrelated to dr/dt if h is constant? b. How is dS/dtrelated to dh/dt ifr is constant? c. How is dS/dtrelated to dr/dt anddh/dt ifneither rnor h is

constant? ,

(

)

107. Circle's changing orea The radius of a circle is changing at the rate of -2/11" mfsec. At what rate is the circle's area changingwheor ~ 10m?

1000

.,.,. 0

50

100

--150---

108. Cube's changing edges The volume of a cobe is increasing at the rate of 1200 em'/min at the instant its edges are 20 em long. At what rate are the lengths of the edges changing at that instant? 200

Time (days) (a)

109. Resistors connected in parallel If two resistors of R, and R2 ohms are connected in parallel in an electric circuit to make an R-ohm resistor, the value of R can be found from the equation

+100

I

R~

I R,

I

+ R2·

50

Ilq·

0

j---- -----1

0,

+

I I I I

I I

Rl

I

R2 1R

I I I I I I 1_____ _ ____

-50

J

-100

0

50

100

150

Time (days)

Derivative of the rabbit population (b)

200

If R, is decreasing at the rate of I ohm/sec and R2 is increasing at the rate of 0.5 ohm/sec, at what rate is R changing when R, ~ 75 ohma aodR2 ~ 500hma?

Chapter 3 Practice Exercises

179

110. Impedance in a series circuit The impedance Z (ohms) in a series circuit is related to the resistance R (ohms) and reactance X (ohms) by the equation Z = R2 + X2. If R is increasing at 3 ohms/sec and X is decreasing at 2 ohms/sec, at what rate is Z changing when R = 10 ohms and X = 20 ohms?

V

111. Speed of moving particle The coordinates of a particle moving in the metric xy-plane are differentiable functions of time t with dx/dt = 10 m/sec and dy/dt = 5 m/sec. How fast is the particle moving away from the origin as it passes through the point (3, -4)? 112. Motion of a particle A particle moves along the curve y = x 3/ 2 in the first quadrant in such a way that its distance from the origin increases at the rate of II units per second. Find dx/ dt when x = 3. 113. Draining a tank Water drains from the conical tank shown in the accompanying figure at the rate of 5 tP/min. a. What is the relation between the variables h and r in the figure? b. How fast is the water level dropping when h = 6 ft?

116. Points moving on coordinate axes Points A and B move along the x- and y-axes, respectively, in such a way that the distance r (meters) along the perpendicular from the origin to the line AB remains constant. How fast is OA changing, and is it increasing, or decreasing, when OB = 2r and B is moving toward 0 at the rate ofO.3r m/sec?

Linearization 117. Find the linearizations of a. tan x atx

=

-7f/4

b. secx atx

=

-7f/4.

Graph the curves and linearizations together. 118. We can obtain a useful linear approximation of the function f(x) = 1/( I + tan x) at x = 0 by combining the approximations I --:::;I-x 1+x

and

tanx :::; x

to get I Exit rate: 5 ft 3/min

114. Rotating spool As television cable is pulled from a large spool to be strung from the telephone poles along a street, it unwinds from the spool in layers of constant radius (see accompanying figure). If the truck pulling the cable moves at a steady 6 ft/sec (a touch over 4 mph), use the equation s = r(J to find how fast (radians per second) the spool is turning when the layer of radius 1.2 ft is being unwound.

I

+ tanx

:::; 1 - x.

Show that this result is the standard linear approximation of 1/(1 + tanx) atx = O. 119. Findthelinearizationoff(x) =

\I'l+x + sinx -

120. Find the linearization of f(x) atx = O.

=

2/(1 - x)

0.5 atx = O.

+ '\,I'l+x -

3.1

Differential Estimates of Change 121. Surface area of a cone Write a formula that estimates the change that occurs in the lateral surface area of a right circular cone when the height changes from ho to ho + dh and the radius does not change.

v=

!7I"r2h 3 S = 7I"rYr 2 + h 2

115. Moving searchlight beam The figure shows a boat I km offshore, sweeping the shore with a searchlight. The light turns at a constant rate, d(J/dt = -0.6 rad/sec. a. How fast is the light moving along the shore when it reaches point A? b. How many revolutions per minute is 0.6 rad/sec?

(Lateral surface area)

122. Controlling error a. How accurately should you measure the edge of a cube to be reasonably sure of calculating the cube's surface area with an error of no more than 2%?

180

Chapter 3: Differentiation

b. Suppose that the edge is measured with the accuracy required in part (a). About how accurately can the cube's volume be calculated from the edge measurement? To rmd

out, estimate the percentage error in the volume calculation that might result from using the edge meaS\l1'etnellt.

124. Finding height To rmd the height of a lamppost (see accompaoying figure), you staod a 6 ft pole 20 ft froro the lamp and measure the length a of its shadow, rmding it to be 15 ft, give or take an inch. Calculate the height of the lamppost using the value a = 15 and estimate the possible error in the result.

123. Compounding error The circumference of the equator of a sphere is measured as 10 em with a possible error of 0.4 em. This measurement is then used to calculate the radius. The radius is then used to calculate the surface area and volume of the sphere. Estimate the percentage errors in the calculated values of

h

•• the radius. b. the surface area. c. the volume.

Chapter

Additional and Advanced Exercises

+ cos' 8 = 1 is called an identity because it holds for all values of 8. An equation like sin 8 = 0.5 is not an identity because it holds only for selected values of 8, not all. If you differentiate both sides of a trigonometric identity in 8 with respect to 8, the resulting new equation will also be an identity. Differentiate the followiog to show that the resultiog equations hold for all 8.

1. An equation like sin' 8

= 2 sinO cosO b. cos 28 = cos2 0 - 8in2 8 2. Ifthe identity sin (x + a) = L

sin20

sin x cos a + cosxsina is differentiated with respect to x, is the resultiog equation also an identity? 2x - 8 = O? Does this principle apply to the equation Explain.

x' -

3.

L

Find values for the constaots a, b, and c that will make

/(x) = co,,", and g(x) = a + bx + ex' satisfy the conditions /(0) = g(O),

/,(0)

= g'(O),

and

/"(0) = g"(O).

b. Find values for b and c that will make

/(x) = sin(x + a)

and

g(x) = b sinx + c co,,",

satisfy the conditions /(0) = g(O)

and

/,(0) = g'(O).

third and fourth derivatives of / and g in each of parts (a) and (b)? 4. Solutions to differential equations Shawthaty = sinx,y = cosx, andy = acosx (a and b constaots) all satisfy the equation y"

+Y =

O.

equation y. + 4y

= O?

Genera1ize this result.

S. An oseuiating circl. Find the values of h, Ie, and a that make the circle (x - h)' + (y - k)' = a' taogent to the parabola y = x' + 1 at the point (1, 2) and that also make the second derivatives d'y/dx' have the same value on both curves there. Circles \ike this one that are taogent to a curve and have the same second derivative as the curve at the point of tangency are called osculating circles (from the Latin osculari, meaning ''to kiss''). We encoonter them again in Chapter 13.

6. Margin'" .....nu. A bus will hold 60 people. The number x of people per trip who use the bus is related to the fare charged (p dollars) by the law p = [3 - (x/40)]'. Write an expression for the total revenue r(x) per trip received by the bus company. What number of people per trip will make the margioa1 revenue dr/dx equal to zero? What is the corresponding fare? (This fare is the one that maximizes the revenue, so the bus company should probably rethink its fare policy.) 7. Industrial production

a. Economists often use the expression ''rate of growth" in relative rather than absolute !enos. For example, let u = /(1)

c. For the determined values of a, b, and c, what happens for the

L

b. How would you modify the functions in part (a) to satisfy the

+ bsinx

be the number of people in the labor force at time 1 in a given indus1ry. CWe 1reat this function as though it were differentiable even though it is an integer-valued step function.) Let v = g(l) be the average production per person in the labor force at time t. The total production is then y = uv. If the labor force is growiog at the rate of 4% per year (du/ dt = 0.04u) and the production per worker is growiog at the rate of 5% per year (dv/dt = 0.05v) , rmd the rate of growth of the total production, y.

Chapter 3 Additional and Advanced Exercises b. Suppose that the labor force in part (a) is decreasing at the rate of 2% per year while the production per person is increasing at the rate of 3% per year. Is the total production increasing, or is it decreasing, and at what rate?

8. Designing a gondola The designer of a 30-ft-diameter spherical hot air balloon wants to suspend the gondola 8 ft below the bottom of the balloon with cables tangent to the surface of the balloon, as shown. Two of the cables are shown running from the top edges of the gondola to their points of tangency, (-12, -9) and (12, - 9). How wide should the gondola be? y

10. Motion of a particle

The position at time t moving along a coordinate line is s = 10cos(t

~

181

0 of a particle

+ 'TT"/4).

a. What is the particle's starting position (t = O)? b. What are the points farthest to the left and right of the origin reached by the particle? c. Find the particle's velocity and acceleration at the points in part (b). d. When does the particle first reach the origin? What are its velocity, speed, and acceleration then?

11. Shooting a paper clip On Earth, you can easily shoot a paper clip 64 ft straight up into the air with a rubber band. In t sec after firing, the paper clip is s = 64t - 16t 2 ft above your hand. -;------~------+---_.-+x

Suspension cables '----\;;;J'----------i. Gondola ~ Widtb

---+I

Nor ro SCALE

9. Pisa by parachute The photograph shows Mike McCarthy parachuting from the top of the Tower of Pisa on August 5, 1988. Make a rough sketch to show the shape of the graph of his speed during the jump.

a. How long does it take the paper clip to reach its maximum height? With what velocity does it leave your hand? b. On the moon, the same acceleration will send the paper clip to a height of s = 64t - 2.6t 2 ft in t sec. About how long will it take the paper clip to reach its maximum height, and how high will it go?

12. Velocities of two particles At time t sec, the positions of two particles on a coordinate line are SI = 3t 3 - 12t 2 + l8t + 5 m and S2 = -t 3 + 9t 2 - 12tm. When do the particles have the same velocities?

13. Velocity of a particle A particle of constant mass m moves along the x-axis. Its velocity v and position x satisfy the equation 1 (2 2m v

1 (2 - Vo 2) = 2k Xo -

x2) ,

where k, vo, and Xo are constants. Show that whenever v#-o ,

dv m-dt

=

-kx .

14. Average and instantaneous velocity a. Show that if the position x of a moving point is given by a quadratic function of t, x = At 2 + Bt + C, then the average velocity over any time interval [t[, t2] is equal to the instantaneous velocity at the midpoint of the time interval. b. What is the geometric significance of the result in part ( a)?

15. Find all values of the constants m and b for which the function SinX, x < 'TT" y= { mx + b, x ~ 'TT" is a. continuous at x = 'TT". b. differentiable at x = 'TT".

16. Does the function

I(x) = Mike McCarthy of London jumped from the Tower of Pisa and then opened his parachute in what he said was a world record low-level parachute jump of 179 ft. (Source: Boston Globe, Aug. 6, 1988.)

{

1 - cosx x'

x#-O

0,

x=O

have a derivative at x = O? Explain.

182 17.

L

Chapter 3: Differentiation For what values of a and b will f(x)

26. Leibniz'. rule for higheJ'-order derivatives of products

~ {:; _ bx + 3,

Leibniz's rule for higher-order derivatives of products of differentiable functioos says that

x<2 x"'2

d'(uv)

a. be differentiable for all values of x?

b. d'(uv)

b. Discuss the geometry of the resulting graph of f.

18.

L

d:x;'

For what values of a and b will c

() gx

=

----;J;2 ~

x ~ -1 {a:x:+b, ax 3 +x+2b, x> -I

•

d'u d:x;' v

du dv

d'v

+ 2 d:x; d:x; + u d:x;"

~ d'u v + 3 d'u dv + 3 dud'v + ud'v. d:x;'

d:x;' d:x;

d'(uv)

d'u

dx ll

dx ll

d'-'u dv dx,,-l dx

d:x; d:x;'

d:x;'

--~-v+n---+'"

n(n - I)··· (n - k + I) d'->" d'v + ~-~-k';!---~ d:x;' k d:x;k d'v

be differentiable for all values of x?

+"'+u dx '"

b. Discuss the geometry of the resulting graph of g.

19. Odd differentiable function. Is there aoytIting special about the derivative of an odd differentiable function of x1 Give reasoos for your answer.

The equations in parts (a) and (b) are special cases of the equation in part (c). Derive the equation in part (c) by mathematical induction, using

20. Even differentiable function.

m! m! (m)k + (m) \k + I ~ k!(m - k)! + (k + I)!(m - k -

Is there aoytIting special about the derivative of an eveo differentiable function of x1 Give reasons for your answer.

f and g are defmed throughout an open interval containing the point xo, that f is differentiable at:x:o, that f(xo) ~ 0, and that g is continuous at Xo. Show that the product fg is differentiable at Xo. This process shows, for example, that although Ixl is not differentiable at x ~ 0, the product x Ixl" differentiable at x ~ O.

21. Suppose that the functions

22. (Continuation of Exerc"e 21.) Use the result of Exercise 21 to show that the followiog functions are differentiable at x ~ O. L

Ixl sin x

d. h(x)

b. x'/' sinx

c."\Yx(l - cosx)

~ {x' sin (I/x), x

0# 0 x= 0

0,

23. Is the derivative of

h(x) ~ {x' sin (I/x), x 0# 0 0, x= 0 continuous atx ~ 01 How about the derivative ofk(x) Give reasons for your answers.

~

:x:h(x)1

24. Suppose that a function f satisfies the followiog cooditioos for all real values ofx andy:

i) f(x

+ y)

ii) f(x) ~ I

~

f(x)·f(y).

+ xg(x) , where linJ,....og(x)

~

I.

Show that the derivative I'(x) exists at every value oh and that I'(x) ~ f(x). 25. The generalized product rule Use mathematical induction to prove that if y = "1"2'" u" is a finite product of differentiable functions, then y is differentiable on their common domain and

du,

dy

ax

=

du,

dun

liX U2 "'u" + Ul(jX"'UII + ... + u t u2 "'ulI - l liX'

I)!'

27. The period of a dock pendulum

The period T of a clock pendulum (time for one full swing and back) is given by the formula T' ~ ~L/g, where Tis measured in seconds, g ~ 32.2 ft/sec', and L, the length of the pendulum, is measured in feet. Find approximately a. the length of a clock pendulum whose period is T ~ I sec.

b. the change dTin Tifthependulum in part (a) is lengthened

0.01 ft.

c. the amount the clock gains or loses in a day as a result of the period's changing by the amount dT found in part (b). 28. The melting ice cube Assume that an ice cube retains its cubical shape as it melts. Ifw. call its edge length s, its volume is V = 3 3 and its surface area is 63 2 . We assume that V and s are differentiable functions of time t. We assume also that the cube's volume decreases at a rate that is proportional to its surface area. (This latter assumption seems reasonable enough when we think that the melting takes place at the surface: Changing the amount of surface changes the amount of ice exposed to melt.) In mathematical terms, dV ~ -k(6s') dt '

k> O.

The minus sign indicates that the volume is decreasing. We assume that the proportiona1ity factor k is coostant. (It probably depends on many things, such as the relative humidity of the surrounding air, the air temperatore, and the incidence or absence of sunlight, to nanre only a fow.) Assume a particular set of conditions in which the cube lost 1/4 of its volume during the first hour, and that the volume is Vo when t ~ O. How long will it take the ice cube to melt?

Chapter 3 Technology Application Projects

Chapter ID

183

Technology Application Projects

Mathematica,tMap1e Modules: Convergence ofSecant Slopes to the Derivative Function

You will visualize the secant line between successive points on a curve and observe what happens as the distance between them becomes small. The function, sample points, and secant lines are plotred on a single graph, while a second graph compares the slopes of the secant lines with the derivative function. DeriVlllives, Slopes, Tangent Lines, and Making Movies Part. I-ill. You will visualize the derivative at a point, the linearization of a function, and the derivative of a function. You learn how to plot the function and selected tangeots on the same graph. Part IV (plotting Many Tangent.) Part V (Making Movies). Parts N and V of the module can he used to aoimate tangeot lines as one moves along the graph ofa function. Convergence ofSecant Slopes to the Derivative Function

You will visualize right-hand and left-hand derivatives. MotWnAlollg tJ Stmight Line: Position .... Velocity .... AccelertJdDn

Observe dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration functions. Figures in the text can he animated.

4 ApPLICATIONS OF DERIVATIVES OVERVIEW In this chapter we use derivatives to find extreme values of functions, to determine and analyze the shapes of graphs, and to find numerically where a function equals zero. We also introduce the idea of recovering a function from its derivative. The key to many of these applications is the Mean Value Theorem, which paves the way to integral calculus in Chapter 5.

Extreme Values of Functions

4.1

This section shows how to locate and identify extreme (maximum or minimum) values of a function from its derivative. Once we can do this, we can solve a variety of problems in which we find the optimal (best) way to do something in a given situation (see Section 4.5). Finding maximum and minimum values is one of the most important applications of the derivative.

DEFINITIONS Let f be a function with domain D. Then f has an absolute maximum value on D at a point c if

f(x)

:5

f(c)

for all x inD

and an absolute minimum value on D at c if y

f(x) y

=

f(c)

for all x in D .

sin x

------+----+----~-----+x

FIGURE 4.1 Absolute extrema for the sine and cosine functions on [-7T/2, 7T/2]. These values can depend on the domain of a function.

184

2::

Maximum and minimum values are called extreme values of the function f. Absolute maxima or minima are also referred to as global maxima or minima. For example, on the closed interval [-7T/2, 7T/2] the function f(x) = cos x takes on an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On the same interval, the function g(x) = sinx takes on a maximum value of 1 and a minimum value of -1 (Figure 4.1). Functions with the same defining rule or formula can have different extrema (maximum or minimum values), depending on the domain. We see this in the following example.

4.1

Extreme Values of Functions

185

EXAMPLE 1 The absolute extrema ofthe following functions on their domains can be seen in Figure 4.2. Notice that a function might not have a maxinnnn or minimmn if the domain is unbounded or fails to contain an endpoint

Function rule

DomainD

Absolute extrema on D

(a) y = x 2

(-00,00)

No absolute maximmn. Absolute minimmn of 0 at x =

[0,2]

Absolute maximmn of 4 at x = 2. Absolute minimmn of 0 at x = o.

(c) y = x 2

(0,2]

Absolute maximmn of 4 at x = 2. No absolute minimum.

(d) y = x 2

(0,2)

No absolute extrema.

- ---"......-----:cx 2 (a) abs min only

FIGURE 4.2

HISWRICAL BIOGRAPHY

Daniel Bernoulli (170()-1789)

o.

(b) y = x 2

- ---"......-----:c- x 2

x - ---"<,L-----:c2

(b) abo max and min

(c) abs max only

•

x - ---"<,L-----:c2

(d) no max or min

Graphs for Example I.

Some of the functions in Example I did not have a maximmn or a minimum value. The following theorem asserts that a function which is continuous at every point of a closed interval [a, h] has an absolute maximum and an absolute minimmn value on the interval. We look for these extreme values when we graph a function.

THEOREM l-The Extreme Yalue Theorem

Iff is continuous on a closed interval [a, h], then f attains both an absolute maximmn value M and an absolute minimmn value m in [a, h]. That is, there are nmnbers Xl and X2 in [a, h] with f(XI) = m, f(X2) = M, and m :5 f(x) :5 M for every other x in [a, h].

The proof of the Extreme Value Theorem requires a detailed knowledge of the real number system (see Appendix 6) and we will not give it here. Figure 4.3 illustrates possible locations for the absolute extrema of a continuous function on a closed interval [a, h]. As we observed for the function y = cos x, it is possible that an absolute minimmn (or absolute maximmn) may occur at two or more different points of the interval. The requirements in Theorem 1 that the interval be closed and finite, and that the function be continuous, are key ingredients. Without them, the conclusion of the theorem

186

Chapter 4: Applications of Derivatives (X2,M)

a Maximum and minimum at endpoints Maximum and minimum at interior points

--~----~--~--~'X

a b 2 Maximum at interior point, minimum at endpoint

:m

--~----~------~--~'x a Xl b

Minimum at interior point, maximum at endpoint

FIGURE 4.3 Some possibilities for a continuous function's maximwn and minimwn on a closed interval [a, h].

y

No largest value

"

need not hold Example 1 shows that an absolute extreme value may not exist if the inter· val fails to be both closed and finite. Figure 4.4 shows that the continuity requirement can· not be omitted.

Local (Relative) Extreme Values o

1 Smallest value

FIGURE 4.4 Even a single point of discontinuity can keep a function from

having either a maximum or minimum value on a closed interval. The function

y=

{

X,

O~x
0,

x= 1

Figure 4.5 shows a graph with five points where a function has extreme values on its do· main [a, bJ. The function's absolute minimum occurs at a even though at e the function's value is smaller than at any other point nearby. The curve rises to the left and falls to the right around c, making f(c) a maximum locally. The function attains its absolute maxi· mum at d. We now derme what we mean by local extrema.

DEFINmONS domain D if f(x)

is continuous at every point of[O, 1] except x = I, yet its graph over [0, I]

A function f has a local mBIimum value at a point c within its :5 f( c) for all XED lying in some open interval containing c.

A function f has a local minimum value at a point c within its domain D if f(x) ;;,: f( c) for all XED lying in some open interval containing c.

does not have a highest point.

lf the domain of f is the closed interval [a, bJ, then f has a local maximum at the endpoint = a, if f(x) :5 f(a) for all x in some half·open interval [a, a + Ii), Ii > O. Likewise, f has a local maximum at an interior point x = c if f(x) :5 f(c) for all x in some open inter· val (c - Ii, c + 1i),1i > 0, and a local maximum at the endpoint x = b if f(x) :5 f(b) for all x in some half·open interval (b - Ii, b J, Ii > O. The inequalities are reversed for local minimum values. In Figure 4.5, the function f has local maxima at c and d and local min· ima at a, e, and b. Local extrema are also called relative extrema. Some functions can have inrmitely many local extrema, even over a finite interval. One example is the func· tion f(x) = sin (l/x) on the interval (0, IJ. (We graphed this function in Figure 2.40.)

x

4.1

Extreme Values of Functions

187

Absolute maximum.

Local minimum No smaller value : off nearby. Absolute minimum. No smaller value of f anywhere. AlBo a I locaIminimum. : _ _ _ _ _ _ _ _ _ _ _ _ _ _ _LI

Local minimum. I No smaller value of :fnearby. : I

_ _ _ _ _ _~_ _ _ _ _ _LI

a

: : I

:

---------"; ' --~':-------_>x

e

d

b

FIGURE 4.5 How to ideotifY types of maxima and minima for a functioo with domain a ~ x ~ b.

An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list ofall local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.

Finding Extrema Local maximum value

The next theorem explains why we usually need to investigate only a few values to find a function's extrema.

If f has a local maximum or minimum value at an interior point c of its domain, and if f' is defined at c, then

THEOREM 2-The First Derivative Theorem for Local Extreme values

f'{c) = O. Secant slopes

2:

(never ,negative)

0

Secant slopes ::s 0 (never po~itive)

I I I I

--~I~------~------~-->,X

X

X

FIGURE 4.6 A curve with a local maximum value. The slope at c, simultaneously the limit of noopositive numbers and nonnegative numbers, is zero.

Proof To prove that f' (c) is zero at a local extremum, we show first that f' (c) cannot be positive and second that f' (c) cannot be negative. The only number that is neither positive nor negative is zero, so that is what f' (c) must be. To begin, suppose that f has a local maximum value at x = c (Figure 4.6) so that f{x) - f{c) ,;; 0 for all values of x near enough to c. Since c is an interior point of f's domain, f' (c) is derfied by the (w(}.sided limit

r

x~

f{x) - f{c) xc·

This means that the right-hand and left-hand limits both exist at x = c and equal f'{c). When we examine these limits separately, we find that

f '{ C )

=

f '{ C )

=

l'

un

x-c+

f{x) - f{c) xc

<:

0

:>

0

-'

Because (x - c) > 0 and !(x) '" !(c)

(1)

Similarly,

l'

11D_

x-c

f{x) - f{c) X

C

-.

Because (x - c) < 0 and !(x) " !(c)

(2)

Together, Equations (1) and (2) imply f'{c) = O. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use f{x) 2: f{c) , which reverses the inequalities in Equations (1) and (2). •

188

Chapter 4: Applications of Derivatives Theorem 2 says that a function's first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. Hence the only places where a function 1 can possibly have an extreme value (local or global) are

1. 2.

interior points where f' = 0,

3.

endpoints of the domain of I.

interior points where f' is undefined,

The following definition helps us to summarize. (a)

y

----~----~----~--->x

(b)

FIGURE 4.7 Critical points wi1hout extrem.evalues. (a)y' = 3x 2 isOatx = 0, but y = x 3 has no extremum there. (b) y' = (1/3)x-2/3 is undef"med atx = 0, but y = X 1/3 bas no extremum 1here.

DEFINITION An interior point of the domain of a function 1 where f' is zero or undef"med is a critical point of I.

Thus the only domain points where a function can assume extreme values are critical points and endpoints. However, be careful not to misinterpret what is being said here. A function may have a critical point at x = c without having a local extreme value there. For instance, both of the functions y = x' and y = x 1/' have critical points at the origin and a zero value there, but each function is positive to the right of the origin and negative to the left. So neither function has a local extreme value at the origin. Instead, each function has a point of inflection there (see Figure 4.7). We define and explore inflection points in Section 4.4. Most problems that ask for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theorem 2 tells us that they are taken on only at critical points and endpoints. Often we can simply list these points and calculate the corresponding function values to fmd what the largest and smallest values are, and where they are located. Of course, if the interval is not closed or not fmite (such as a < x < b or a < x < 00), we have seen that absolute extrema need not exist. If an absolute maximum or minimum value does exist, it must occur at a critical point or at an included right- or left-hand endpoint of the interval.

How to Find the Absolute Extrema of a Continuous Function f on a Finite Closed Interval 1. Evaluate 1 at all critical points and endpoints. 2. Take the largest and smallest of these values.

EXAMPLE 2 [-2, I].

Find the absolute maximum and minimum values of I(x) = x 2 on

Solution

The function is differentiable over its entire domain, so the only critical point is where f' (x) = 2x = 0, namely x = O. We need to check the function's values at x = 0 and at the endpoints x = -2 and x = I: Critical point value: Endpoint values:

1(0) = 0

I( -2) = 4

1(1) = 1 The function has an absolute maximum value of 4 at x = -2 and an absolute minimum value of 0 atx = O. •

4.1

EXAMPLE 3 [-2,1].

y 7

(1.7)

189

Extreme Values of Functions

Find the absolute maximum and minimum values of g(l) = 81 - 14 on

Solutton

The function is differentiable on its entire domain, so the only critical points occur where g'(I) = O. Solving this equation gives

y= 8t- t 4

8-41'=0

1=\0/2>1,

or

a point not in the given domain. The function's absolute extrema therefore occur at the endpoints, g( -2) = -32 (absolute minimum), and g(l) = 7 (absolute maximum). See • Figure 4.8.

EXAMPLE 4

Find the absolute maximum and minimum values of I(x) = x2/' on the

interval [-2,3]. FIGURE 4.8 The extreme values of g(t) = 8t - t 4 on [-2, I] (Example 3).

y

Solutton

We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The tITSt derivative

I'(x) y =x213 , -2sxs 3 Absolute maximum; also a local maximum

Loca1

-I

0 ~ 1

2

3

Critical point value:

1(0)

Endpoint values:

I( -2) = (_2)2/3 =

Absolute minimum; also a local minimum

1(3)

FIGURE 4.9 The extreme values of f(x) = x 213 0n [-2, 3] occuratx = oand x = 3 (Example 4).

~x-I/3 = _2_ 3 3\Yx

has no zeros but is undefined at the interior point x = O. The values of I at this one critical point and at the endpoints are

~--~~~~--~--~--~X

-2

=

=

0

= (3)'/3 =

V'4

'f9.

We can see from this list that the function's absolute maximum value is 'f9 .., 2.08, and it occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the • interior point x = 0 where the graph has a cusp (Figure 4.9).

Exercises 4.1 finding Extrema from Graphs

3.

y

4.

y

In Exercises l-{i, determine from the graph whether the function has any absolute ex1reme values on [a, b]. Then explain how your answer is consistent with Theorem 1.

1.

y

2.

y ~O~--aL---~c----~b--~X

~---L--~----~-+x

o

a

c

b

o

a

c

b

190 S.

Chapter 4: Applications of Derivatives 6.

Y

In Exercises 15-20, sketch 1Ire graph of each function and determine whether the functioo has any absolute extreme values on its domain.

y

(~'

Explain how your answer is consistent with Theorem 1. 16. Y

~---L--~----~-+x

o

a

c

lxi, -I <x<2 6 = -,--, -I < x < I

IS. f(x) = x

+2

=

-x, o~x< 1 { x - I, I ~x~ 2

~---L--~----~~x

o

b

a

b

c

17. g(x)

In Exercises 7-10, fmd 1Ire absolute extreme values and where they occur.

7.

8.

y

18. h(x) =

y

19. Y

1 ~~----:+----~~X

-2

9.

0

10.

Y

_ =-----j+----C_ x

-3

2

-1

o

2

__ x

In Exercises 11-14, match 1Ire table with a graph.

11.

x

I'(x)

a b c

0 0

12. - - - - - x I'(x) a b c

5

+

0

<x<

I,

-I

x

I'(X)

a b c

does not exist 0 -2

14.

S

4

2'IT

x < 0 <'IT < 0 -x-

cos x,

S

Z

0 0

In Exercises 21-36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. IdentifY the points 00 the graph where the absolute extrema occur, and in-

clude their coordinates. 2 21. f(x) = 3x - 5,

-2

22. f(x) = -x - 4,

-4 '" x '" I

I'(x)

a b c

does not exist does not exist -1.7

S

3

x'" 2

24. f(x) = 4 - x',

-3 '" x '" I

25. F(x)

27. h(x)

x

x

-I '"

I

0.5'" x '" 2

= -"

x

26. F(x) =

-5

S

23. f(x) = x' - I,

=

I -x' \Yx,

-2'" x '" -I

-I '" x '" 8 -I '" x '" I

28. h(x) = -3x'/3, 13.

0

Absolute Extrema on finite Closed Intervals

5

----:+--~

Yx,0"'x

20. f(x) = {x

<

-1:5 x

x

= 3 sinx,

2

y

I, {

29. g(x) =~, 30. g(x) = -~,

31. f(8) = sin8,

-2 '" x '" I

-vs '" x'" 0

-f'" 8

S

5:

32. f(8) = tan 8,

f '" x '" 2;

33. g(x) = cscx, 34. g(x) = secx,

(a)

35. f(t) = 2 -

Itl,

36. f(t) = It -

51,

(b)

In Exercises 37-40, fmd the functioo's absolute maximum and minimum values and say where they are assumed.

37. f(x) = x 4/ 3, abc (c)

(d)

-I '" x '" 8

38. f(x) = x'/3, -I '" x '" 8 39. g(8) = 83/', -32 S 8 S I 40. h(8) = 38'/3,

-27 '" 8 '" 8

Extreme Values of Functions

4.1

finding CriticalPlrints In Exercises 41-48, determine all critical points for each functioo. 41. y = x' - 6x

+7

42. f(x) = 6x' - x'

43. f(x) = x(4 - x)' 2 45. Y = x' + x 47. Y

=

44. g(x) = (x - I)'(x - 3)' x' 46. f(x) = x - 2

x' - 32Vx

48. g(x) =

Va -

x'

finding Extreme Values In Exercises 4!1-58, f"md the extreme values (absolute and local) of the functioo and where they occur. 49.y=a'-Sx+9 50.y=x'-a+4 51. y=x'+x'-Sx+5

53. y

= W-=-!

I 55. Y = ~I _ x 57. Y = -,-x + I

56. y

= x'(x - 5)' = x - 4Vx = V3 + a - x'

58

=

52. Y

54. y

x'

.y

x+1 x'+a+2

Local Extrema and Critical Points In Exercises 59-{;6, f"md the critical points, domain endpoints, and local exlreroe values (absolute and local) for each functioo. 59. y

= x'/'(x + 2)

60. y

61.y=x~ 4 -

63. Y

= {x +

65

=

.y

66.y=

- 4)

62.y=x'~

a, x'" I,

= x'/3(x'

1 x> 1

64. Y

=

G: ~ _x', : : ~

-x' - a + 4 x'" 1 { -x'+6x-4,' x>1

_lX' _Ix + li, x'" I 4 2 4 { x'-6x'+Sx, x>1

In Exercises 67 and 68, give reasons for your aoswers. 67. Le!f(x) = (x - 2)'/3.

a. Does f' (2) exist? b. Show that the ooly local extreme value of f occurs at x = 2. c. Does the result in part (b) cootradict the Extreme Value Theorem? d. Repeatparts(a)and(b)forf(x) = (x - a)'/3,rep1acing2bya. 68. Le!f(x)

=

lx' - 9xl.

a. Does f' (0) exist? b. Does f' (3) exist? c. Does f' (- 3) exist? d. Determine all extreroa of f. Theory and Examples 69. A minimum with no derivative The functioo f(x) = Ixl has an absolute minimum value at x = 0 even though f is not differentiable at x

= O.

Is this consistent with Theorem 2? Give rea-

sons for your answer. 70. E..n functions If an even functioo f(x) has a local maximum value at x = c, can anything be said about the value of f at x = -c? Give reasons for your answer.

71. Odd functions If an odd functioo g(x) has a local minimum value at x = c, can anything be said about the value of g at x = -c? Give reasons for your answer.

191

72. We know how to f"md the extreroe values of a continuous function f(x) by m..stigating its values at critical points and endpoints. But what if there are no critical points or endpoints? What happens then? Do such functions really exist? Give reasoos for your answers. 73. The functioo

V(x) = x(1O - a)(16 - a),

0< x < 5,

models the volume of a box. a. Find the extreme values of V. b. Interpret any values found in part (a) in terms of the volume

of the box. 74. Cubic functions f(x)

Coosider the cubic functioo

= ax'

+ bx' + ex + d.

a. Show that f can have 0, I, or 2 critical points. Give examples and graphs to support your arguroent. b. How many local extreme values can f have?

75. Muimum height of a ..rticaUy moviog body The height of a body moving vertically is given by s

=

I -"2gt2

+ vot + So,

g> 0,

with s in m_s and t in seconds. Find the body's maximum height 76. Peak alternating current Suppose that at any given time t (in seconds) the curreot i (in amperes) in an alternatiog current circuit is i = 2 cos t + 2 sin t. What is the peak current for this circuit (largest magoitode)?

D Graph the functions in Exercises 77-80. Then f"md the extreme values of the function 00 the interval and say where they occur. 77. f(x) = Ix 78. g(x) = Ix 79. h(x)

=

Ix +

80. k(x) = Ix +

21 + II 21 II +

Ix+ Ix Ix Ix -

31, 51, 31, 31,

-5 "'x '" 5 -2'" x'" 7 -00 -00

<x < <x<

00 00

COMPUTER EXPLORATIONS In Exercises 81-86, you will use a CAS to help find the absolute extreroa of the given function over the specified closed interval. Perform the following steps.

a. Plot the functioo aver the interval to see its general behavior there. b. Find the interior points where f' = O. (In some exercises, yoo

may have to use the numerical equation solver to approximate a solutioo.) You may want to plot f' as well. c. Find the interior points where f' does not exist. d. Evaluate the functioo at all points foond in parts (b) and (c) and at the endpoints of the interval.

e. Find the function's absolute extreme values on the interval and identifY where they occur. 81. f(x) = x· - Sx' + 4x + 2, [-20/25,64/25] 82. f(x) = -x' + 4x' - 4x + I, [-3/4,3] 83. f(x) = x'/3(3 - x), [-2,2] 84. f(x) = 2 + a - 3x'/3, [-I, 10/3] 85. f(x) = Vx + cosx, 86. f(x) = x'/' - sinx +

[0,2".]

t,

[0,2".]

192

Chapter 4: Applications of Derivatives

The Mean Value Theorem

4.2

We know that constant functions have zero derivatives, but could there be a more complicated function whose derivative is always zero? If two functions have identical derivatives over an interval, how are the functions related? We answer these and other questions in this chapter by applying the Mean Value Theorem. First we introduce a special case, known as Rolle's Theorem, which is used to prove the Mean Value Theorem.

y

Rolle's Theorem As suggested by its graph, if a differentiable function crosses a horizontal line at two different points, there is at least one point between them where the tangent to the graph is horizontal and the derivative is zero (Figure 4.10). We now state and prove this result.

(a)

THEOREM 3-Rolle's Theorem Suppose that y = j(x) is continuous at every point of the closed interval [a, b] and differentiable at every point of its interior (a, b). If j(a) = j(b), then there is at least one number c in (a, b) at which /,(c) = o.

y

I'(c,)

~

0

Proof Being continuous, j assumes absolute maximum and minimum values on [a, b] by Theorem 1. These can occur only (b)

FIGURE 4.10 Rolle~ Theorem says that a differentiable curve bas at least ooe horizontal taogent between any two points where it crosses a horizontal line. It may bavejust one (a), or it may bave more (b).

HISTORICAL BIOGRAPHY

Michel Rolle (1652-1719)

1. 2. 3.

at interior points where /' is zero, at interior points where /' does not exist, at the endpoints of the function's domain, in this case a and b.

By hypothesis, j has a derivative at every interior point. That rules out possibility (2), leaving us with interior points where /' = 0 and with the two endpoints a and b. If either the maximum or the minimum occurs at a point c between a and b, then /,(c) = 0 by Theorem 2 in Section 4.1, and we have fouod a point for Rolle's Theorem. If both the absolute maximum and the absolute minimum occur at the endpoints, then because j(a) = j(b) it must be the case that j is a constant function with j(x) = j( a) = j(b) for every x E [a, b]. Therefore /' (x) = 0 and the point c can be taken anywhere in the interior (a, b). • The hypotheses of Theorem 3 are essential. If they fail at even one point, the graph may not have a horizontal tangent (Figure 4.11). y

y

~

-+--~------~--~X

a

b

(a) Discontinuous at an endpoint of [a, b]

y

((M -+--~---L~~--~x

a -'0 b (b) Discontinuous at an interior point of [a, b]

J\:

-+--~~-L--~~~x

a

-'0

b

(c) Continuous on la, b] but not differentiable at an interior point

FIGURE 4.11 There may be no horizontal taogent if the hypotheses of Rolle's Theorem do not hold.

Rolle's Theorem may be combined with the Intermediate Value Theorem to show when there is ouly one real solution of an equation j(x) = 0, as we illustrate in the next example.

4.2

EXAMPLE 1

y I I I I I I I

y=x 3 +3x+l I I

193

The Mean Value Theorem

Show that the equation

has exactly one real solution.

Solution

We defme the continuous function

-f----~~------~~x

f(x) = x 3

+

3x

+ I.

Since f( -I) = - 3 and f(O) = I, the Intermediate Value Theorem tells us that the graph of f crosses the x-axis s0U3ewhere in the open interval (-1,0). (See Figure 4.12.) The derivative FIGURE 4.12 The only real zero of the polynomial y = x 3 + 3x + I is the ooe

shown here where the curve crosses the x·axis b&ween -I and 0 (Example I).

f'(x) = 3x 2

+

3

is never zero (because it is always positive). Now, if there were even two points x = a and x = b where f(x) was zero, Rolle's Theorem would guarantee the existence of a point x = c in between them where f' was zero. Therefore, f has no IDore than one zero. _ Our main use of Rolle's Theorem is in proving the Mean Value Theorem.

The Mean Value Theorem The Mean Value Theorem, which was first stated by Joseph-Louis Lagrange, is a slanted version of Rolle's Theorem (Figure 4.13). The Mean Value Theorem guarantees that there is a point where the tangent line is parallel to the chordAB. Tangent parallel to chord

y

~/

Slopef'(c)

B I

Slope f(b) - f(a) I b-a I I I

THEOREM 4-The Mean Value Theorem

Suppose y = f(x) is continuous on a closed interval [a, b] and differentiable on the interval's interior (a, b). Then there is at least one point c in (a, b) at which f(bl

FIGURE 4.13 Geometrically, the Mean Value Theorem says that somewhere b&ween a and b the curve has at least ooe tangent parallel to chord AB.

=~(a)

= f'(c).

(1)

Proof We picture the graph of f and draw a line through the points A(a, f(a» and B(b, f(b». (See Figure 4.14.) The line is the graph of the function

g(x) = f(a)

+

f(bl

=~(a)

(x - a)

(2)

(point-slope equation). The vertical difference between the gmphs of f and g at x is HIsTORICAL BIOGRAPHY

h(x) = f(x) - g(x)

Joseph-Louis Lagrange (1736-1813)

= f(x) - f(a) -

f(b) - f(a) b (x - a).

-a

(3)

Figure 4.15 shows the graphs of f, g, and h together. The function h satisfies the hypotheses of Rolle's Theorem on [a, b]. It is continuous on [a, b] and differentiable on (a, b) because both f and g are. Also, h(a) = h(b) = 0 because the graphs of f and g both pass through A and B. Therefore h'(c) = 0 at S0U3e point c E (a, b). This is the point we want for Equation (I).

194

Chapter 4: Applications of Derivatives B(b,f(b»

B

I I I I I I

hex)

= f(x)

- g(x)

/

:

----~----------~----~'x

y

t

y

~ v'1=7,-I" "" 1 1

FIGURE 4.14 The graph of I and the chordAB over the interval [a, h].

:. . . . . . - -

............. --~----~----~----~ S-----~'x

a

i

b

x

FIGURE 4.15 The chordAB is the graph of the functiong(x). The function hex) = I(x) - g (x) gives the vertical distance between the graphs of I and g at x.

--~--~----~~X

-I

0

1

FIGURE 4.16 The function I(x) = ~ satisfies the hypotheses (and conclusion) of the Mean Value Theorem on [ -I, I] even though I is not differentiable at-land!.

To verify Equation (I), we differentiate both sides of Equation (3) with respect to x and then set x = e:

h'(x)

=

I'(x) _ I(b) - I(a) b-a

Derivative ofRq. (3) ...

h'(e)

=

I'(e) _ I(b) - I(a) b a

... withx=c

y B(2,4)

0= I'(e) _ I(b) - I(a) b-a

I'(e)

=

I(bl-

~(a),

h'(e)

=0

Rearnmged

which is what we set out to prove. The hypotheses of the Mean Value Theorem do not require either a or b. Continuity at a and b is enough (Figure 4.16). FIGURE 4.17 As we fmd in Example 2, c = I is where the tangent is parallel to the chord. s

I

-

to he differentiable at

The function I(x) = x 2 (Figure 4.17) is continuous for 0 :s; x :s; 2 and differentiable for 0 < x < 2. Since 1(0) = 0 and 1(2) = 4, the Mean Value Theorem says that at some point e in the interval, the derivative I' (x) = 2x must have the value (4 - 0)/(2 - 0) = 2. In this case we can identify e by solving the equation 2c = 2 to get e = I. However, it is not always easy to find e algebraically, even though we know it always exists. _

EXAMPLE 2

400

g 320

A Physical Interpretation

u

H240

i3

We can think of the number (f(b) - I(a))/(b - a) as the average change in I over [a, b] and I' (c) as an instantaneous change. Then the Mean Value Theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.

160 80 O ~~~~5~~--~

TIme (sec)

FIGURE 4.18 Distance versus elapsed time for the car in Example 3.

EXAMPLE 3 If a car accelerating ftmn zero takes 8 sec to go 352 ft, its average velocity for the 8-sec interval is 352/8 = 44 ft/ sec. The Mean Value Theorem says that at some point during the accelemtion the speedometer must read exactly 30 mph (44 ft/ sec) (Figure 4.18). _

4.2

195

The Mean Value Theorem

Mathematical Consequences At the beginning of the section, we asked what kind of function has a zero derivative over an interval. The first corollary of the Mean Value Theorem provides the answer that only constant functions have zero derivatives.

COROLLARY 1 If f'(x) ~ 0 at each point x of an open interval (a, b), then I(x) ~ C for all x E (a, b), where C is a constant.

Proof We want to show that I has a constant value on the interval (a, b). We do so by showing that if Xl and X2 are any two points in (a, b) with Xl < X2, then I(x,) = l(x2). Now I satisfies the hypotheses of the Mean Value Theorem on [Xl, X2]: It is differentiable at every point of [Xh X2] and hence continuous at every point as well. Therefore,

at some point c between Xl and X2. Since f' = 0 throughout (a, b), this equation implies successively that

l(x2) - I(xl) X2 Xl

=

0,

l(x2) - I(xl)

=

0,

and

I(xl)

=

l(x2).

•

At the beginning of this section, we also asked about the relationship between two functions that have identical derivatives over an interval. The next corollary tells us that their values on the interval have a constant difference.

COROLLARY 2 Iff'(x) = g'(x) at each point x in an open interval (a, b), then there exists a constant C such that I(x) = g(x) + C for all X E (a, b). That is, I - g is a constant function on (a, b).

Proof At each point X E (a, b) the derivative of the difference function h =

h'(x) Thus, h(x) g(x) + C.

FIGURE 4.19 From a geometric point of view, Corollary 2 of the Meao Value Theorem says that the graphs of functions with identical derivatives on an interval cao differ ooly by a vertical shift there. The graphs of the functioos with derivative 2x are the parabolas y ~ x' + C, shown here for selected values of C.

=

f'(x) - g'(x)

= C on (a, b) by Corollary 1. That is,

=

I -

g is

O.

I(x) - g(x)

= C on (a, b), so

I(x)

=

•

Corollaries I and 2 are also troe if the open interval (a, b) fails to be finite. That is, they remain true if the interval is (a, 00), (-00, b), or (-00, 00). Corollary 2 plays an important role when we discuss antiderivatives in Section 4.7. It tells us, for instance, that since the derivative of I(x) = x 2 on (-00, 00) is 2x, any other function with derivative 2x on ( - 00, 00) must have the formula x 2 + C for some value of C (Figure 4.19).

EXAMPLE 4 Find the function I(x) whose derivative is sin X and whose graph passes through the point (0, 2). Solution Since the derivative of g(x) = -cosx is g'(x) = sinx, we see that I and g have the same derivative. Corollary 2 then says that I(x) = -cosx + C for some

196

Chapter 4: Applications of Derivatives

constant C. Since the gmph of I passes through the point (0, 2), the value of C is determined from the condition that 1(0) = 2:

1(0)

= -cos

The function is/(x) = -cosx

+

+C

(0)

=

2,

C = 3.

so

•

3.

Finding Velocity and Position from Acceleration We can use Corollary 2 to find the velocity and position functions of an object moving along a vertical line. Assume the object or body is falling freely from rest with acceleration 9.8 m/sec2• We assume the position 8(t) of the body is measured positive downward from the rest position (so the vertical coordinate line points downward, in the direction of the motion, with the rest position at 0). We know that the velocity v(t) is some function whose derivative is 9.8. We also know that the derivative of g(t) = 9.8t is 9.8. By Corollary 2,

+

v(t) = 9.8t

C

o. Thus C = o.

for some constant C. Since the body falls from rest, v(O) =

9.8(0)

+C

0,

=

and

The velocity function must be v(t) = 9.8t. What about the position function s(t)? We know that 8(t) is some function whose derivative is 9.8t. We also know that the derivative of I(t) = 4.9t2 is 9.8t. By Corollary 2,

8(t) = 4.9t 2 + C for some constant C. Since s( 0) = 0,

4.9(0)'

+C

=

0,

C=

and

o.

The position function is s(t) = 4.9t 2 until the body hits the ground The ability to find functions from their mtes of change is one of the very powerful tools of calculus. As we will see, it lies at the heart of the mathematical developments in Chapter 5.

Exercises 4.2 Checking the Mean Yalue Theoll!m Find the value or values of c that satisfY the equatioo f(b) - f(a) ~ f'(c) b-a

Which of the functioos in Exercises 7-12 satisfY the hypotheses of the Mean Value Theorem on the giveo interval, and which do not? Give reasons for your answers. 7. f(x) ~ x 2/ 3, [-1,8]

in the conclusion of the Mean Value Theorem for the functions and in-

8. f(x) ~ x4/',

terva1s in Exercises 1-<>. 1. f(x) ~ x 2 + 2x - 1,

9. f(x) ~ v'x(1 - x),

2. f(x) ~ x 2/ 3,

3. f(x)

4. f(x) ~

v;=J,

S.f(x)~x3_x2, X3

6. g(x) ~ { ; x,

[0, 1]

[0, 1]

SinX

[0, 1]

~x+~,

[0, 1]

[t,2] [1,3] [-1,2]

10. f(x)

~

11. f(x) ~

12. f(x)

{

0

-2

~

x~o

0, X2 -

{

<

-'T1' ~x

x'

x

2x2 _ ~x - 3,

~ {2x -

3, 6x - x 2

-

x::s; -1

-1 < x

°

~

:5 X :5 2 7, 2 < x ::;; 3

0

4.2 13. The function

<

f(x) = {x' 0:5 X 0, x = 1

I

197

The Mean Value Theorem

28. Suppose that f(O) = 5 and that !,(x) = 2 for all x. Must f(x) = 2x + 5 for all x? Give reasons for your answer. 29. Suppose that!'(x) = lx for allx. Findf(2) if

•• f(O) = 0

b. f(!) = 0

c. f(-2) = 3.

is zero at x = 0 and x = I and differentiable on (0, I), but its de· rivative on (0, 1) is never zero. How can this be? Doesn't Rolle's Theorem say tire derivative has to be zero somewhere in (0, I)?

30. What can be said about functions whose derivatives are constant?

Give reasons for your answer.

In Exercises 31-36, fmd all possible functions with the given derivative.

14. For what values of a, m, and b does the function

=

f(x)

3' -x 2 + 3x { mx + b,

x=o O<x<1

+ a,

1~x~2

satisfY the hypotheses of the Mean Value Theorem on the interval [0,2]?

Give reasons for your answer.

b.y'=x 2

c.y'=x 3

32. a. y' = lx

b.y'=lx-I

c.y'=3x 2 +lx-1

33 ••• y ,_ I --2 x

I b . y ' = I - -2 x

c.y'=s+---\x

34. a. y' =

b. y' = _I_

c. Y'=4x-

31 • •• y' =

X

2

ifx

Vx

x

= cos i2

Roots (Zeros)

35. a. y' = sin21

b. y'

15. a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative. i)y=x2_4

36. •• y' = sec' 0

b. y' =

ii) y = x 2 iii) y = x 3

iv) y = x

3

+ 8x + 15 2 - 3x + 4 = -

(x

+

I)(x - 2)2

2

nx"- 1

+ (n

- 1)all_tx"-2

+ ... + al.

16. Suppose that/" is continuous on [a, b] and that f has three zeros in tire interval. Show that/" has at least one zero in (a, b). Gener·

alize this result. 17. Show that if /" > 0 throughout an interval [a, b], then!, has at most one zero in [a, b]. What if/,,< 0 throughout [a, b] instead? 18. Show that a cubic polynontial can have at most three real zeros. Show that the functions in Exercises 19-26 have exactly one zero in tire given interval. 19. f(x) = x'

20. f(x) = x 3 + 21. g(l) =

~

x

38. g'(x)

=

1, + lx, x

(-;f, 0 )

39. 7'(0) = 8 - csc'O, 40. 7'(1)

= secltanl -

I, P(O, 0)

F;nd;ng Po.;tion from Velodty or Acceleration Exercises 41-44 give the velocity v = dsldl and initial position ofa body moving along a coordinate line. Find the body's position at time t. 41. v = 9.81 + 5,

.(0) = 10

42. v = 321 - 2,

.(0.5) = 4

(-00,0)

44. v

(~)

24.7(0) = 20 - oos2 0 + I 25. 7(0) = sec 0 - 03

-

8,

v2,

(0,00)

3.1,

(-I, I)

(-00,00) (-00,00)

+ 5, (0, "'/2)

26. 7(0) = tanO - cotO - 0,

(0, "'/2)

Find;ng Functions from Derivatives 27. Suppose that f( -I) = 3 and that !,(x) = 0 for all x. Must f(x) = 3 for all x? Give reasons for your answer.

sec2 0

p(-I, I)

+ 7,

4,

v8 -

- I, P(O, 0)

43. v = sin ",I, .(0) = 0

~ I + v'l+t -

23. 7(0) = 0 + sin2

= lx

[-2, -I]

vt + v'l+t -

22. g(l) = I

37. !'(x)

I,

+ 3x +

c. y' =

+ cost

In Exercises 37-40, fmd the function with the given dctivative whose graph passes through the point P.

33x + 216x = x(x - 9)(x - 24)

b. Use Rolle's Theorem to prove that between every two zeros of x" + Qn_tX"- 1 + ... + QtX + ao there lies a zero of

v8

c. y' = 8in2t

I

Vx

=

2

21

7T cos W'

,

s('7J'""")

=

1

Exercises 45-48 give the acceleration a = d 2.ldI 2 , initial velocity, and initial position of a body moving on a coordinste line. Find tire body's position at time I. 45. a = 32,

v(O) = 20,

46. a = 9.8,

v(O) = -3,

.(0) = 5 .(0) = 0

47. a = -4 sin 21,

v(O) = 2,

.(0) = -3

9 31 48. a = "" cos ""

v(O) = 0,

.(0) = -I

Applkations 49. Temperatnre cbange It took 14 sec for a mercury thermometer to rise from -19'C to lOO'C when it was taken from a freezer and placed in boiling water. Show that sOO3ewhere along the way tire mercury was rising at the rate of 8.5°C/sec.

198

Chapter 4: Applications of Derivatives

50. A trucker handed in a ticket at a toll booth showing that in 2 hours she had covered 159 mi on a toll road with speed limit 65 mph. The trucker was cited for speeding. Why? 51. Classical accounts tell us that a 170-oar trireme (aocient Greek or Roman WlIIship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat the trireme's speed exceeded 7.5 koots (sea miles per hour). 52. A marathoner ran the 26.2-mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly II mph, assuming the initial and fmal speeds are zero.

60. Parallel tangents Assume that I and g are differentiable on [a, b j and that/(a) ~ g(a) aod/(b) ~ g(b). Show that there is at least one point between a and b where the taogents to the graphs of I and g are parallel or the same line. lllustrate with a sketch. 61. Suppose that f'(x):5 I for I :5 x :5 4. Show that 1(4)1(1) :5 3. 62. Suppose that 0 < f'(x) < 1/2 for all x-values. Show that 1(-1) < 1(1) < 2 + 1(-1).

63. Show that Icou - II :5 Ix I for all x-values. (Hint: Consider I(t) ~ cos t on [0, xj.)

53. Show that at some instaot during a 2-hour automobile 1rip the car's speedometer reading will equal the average speed for the 1rip.

64. Show that for any numbers a and b, the sine inequality Isinb - sinal :5 Ib - al is true.

54. Free fall on the moon On our moon, the acceleration of gravity

65. If the graphs of two differentishle functions I(x) andg(x) start at the same point in the plane and the functions have the same rate of change at every point, do the graphs have to be identical? Give

is 1.6 m/sec2 • Ifa rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later? Theory and Examples 55. The geometric mean of a and b The ~metric mean of two positive nombers a and b is the number Vab. Show that the value of c in the conclusion of the Meao Valoe Theorem for I(x) ~ I/x on an interval of positive numbers [a, b j is c ~ v;;J,.

56. The arithmetic mean of Il and b The arithmetic mean of two numbers a and b is the number (a + b)/2. Show that the value of c in the conclusion of the Meao Value Theorem for I(x) ~ x 2 on any interval [a,bj isc ~ (a + b)/2.

D 57.

reasons for your answer. 66. Ifl/(w) - I(x) I :5lw - xl for all values w and x and I is a differentiable function, show that -I :5 f' (x) :5 I for all x-values. 67. Assumethatl is differentiable ona :5 x:5 baodthat/(b) Show that f' is negative at some point between a and b.

68. Let/be a function defmed on ao interval [a, bj. What conditions could ynu place on I to gusrantee that

. I' < I(b) - I(a) < I' mm b-a _max,

Graph the function

I(x) ~ sin x sin (x

+ 2)

- sin2 (x

+

I).

What does the graph do? Why does the function behave this way?

Give reasons for your answers. 58. RoUe's Theorem L

Construct a polynomial I(x) that has zeros atx I, and 2.

~

-2, -I, 0,

b. Graph I and its derivative f' together. How is what you see related to Rolle~ Theorem?

c. Do g(x) ~ sin x aod its derivative g' illustrate the same phenomenon as I and f'? 59. Unique solution Assume that I is continuous on [a, bj and differentiable on (a, b). Also assume that I(a) aod I(b) have opposite sigus and that f' # 0 between a aod b. Show that I(x) ~ 0 exactly once between a aod b.

4.3

< I(a).

where min f' and max f' refer to the minimum and maximum values off' on [a, b]? Give reasons for your aoswers.

D 69.

Use the inequalities in Exercise 68 to estimate 1(0.1) if f'(x) ~ 1/(1 + x'cosx) forO:5 x:5 0.1 aod/(O) ~ 1.

D 70.

Use the inequalities in Exercise 68 to estimate 1(0.1) if f'(x) ~ 1/(1 - x 4 ) forO:5 x:5 0.1 aod/(O) ~ 2.

71. Let

1(1)

I

be differentishle at every value of x and suppose that I, that f' < 0 on (-00, I), and thatf' > 0 on (1,00).

~

a. Show that I(x) '" I for all x. b. Mustf'(I) ~ O?Explain.

72. Let I(x) ~ px2 + qx + r be a quadratic function defined on a closed interval [a, b j. Show that there is exactly one point c in (a, b) at which I satisfies the conclusion of the Mean Value

Theorem.

Monotonic Functions and the First Derivative Test In sketching the graph of a dllferentiable function it is useful to know where it increases (rises from left to right) and where it decreases (falls from left to right) over an interval. This section gives a test to determine where it increases and where it decreases. We also show how to test the critical points of a function to identifY whether local extreme values are present.

4.3

Monotonic Functions and the First Derivative Test

199

Increasing Functions and Decreasing Functions All another corollary to the Mean Value Theorem, we show that functions with positive derivatives are increasing functions and functions with negative derivatives are decreasing functions. A function that is increasing or decreasing on an interval is said to be monotonic on the interval.

COROLLARY (a, h).

3

If I'(x)

Suppose that

I

is continuous on [a,

> 0 at each point x E (a, h), then I

Ifl'(x) < 0 at each point X

E

hI and differentiable on

is increasing on [a,

(a, h), then/is decreasing on [a,

Proof Let Xl andx2 be any two points in [a, applied to I on [Xh x21 says that

I(X2) - l(xI)

=

hI with Xl I'(C)(X2 -

hI. hI.

< X2. The Mean Value Theorem

XI)

for some c between XI and X2. The sign of the right-hand side of this equation is the same as the sign of I'(c) because X2 - Xl is positive. Therefore, I(X2) > I(Xl) if I' is positive • on (a, h) and/(x2) < I(xtl if I' is negative on (a, h). Corollary 3 is valid for infinite as well as finite intervals. To find the intervals where a function I is increasing or decreasing, we first find all of the critical points of I. If a < h are two critical points for I, and if the derivative I' is continuous but never zero on the interval (a, h), then by the Intermediate Value Theorem applied to 1', the derivative must be everywhere positive on (a, h), or everywhere negative there. One way we can determine the sign of I' on (a, h) is simply by evaluating the derivative at a single point c in (a, h). Ifl'(c) > 0, thenl'(x) > 0 for all X in (a, h) so/isincreasing on [a, hI by Corollary 3; if I'(c) < 0, then I is decreasing on [a, hI. The next example illustrates how we use this procedore. y y~x'-I2x - 5

EXAMPLE 1 Find the critical points of I(x) = x3 - 12x - 5 and identify the intervals on which I is increasing and on which I is decreasing.

20

Solution

The function I is everywhere continuous and differentiable. The f"trst derivative

I'(X)

=

3X2 - 12

= 3(x

FIGURE 4.20 The function /(x) ~ 12x - 5 is monotonic on three separate intervals (Example I).

=

+ 2)(x

3(X2 - 4) - 2)

is zero at X = -2 and X = 2. These critical points subdivide the domain of I to create nonoverlapping open intervals (-00, -2), (-2,2), and (2, 00) on which I' is either positive or negative. We determine the sign of I' by evaluating I' at a convenient point in each subinterval. The behavior of I is determined by then applying Corollary 3 to each subinterval. The results are summarized in the following table, and the greph of I is given in Fignre 4.20.

X3 -

Intervnl /' evnluated Sign of/, Behaviorofl

-00 < X <-2 1'( -3) = 15

-2 < X < 2 1'(0) = -12

2<x<00 1'(3) = 15

decreasing

increasing

+ increasing

+

•

200

Chapter 4: Applications of Derivatives We used "strict" less-than inequalities to specify the intervals in the summary table for Example I. Corollary 3 says that we could use ,,; inequalities as well. That is, the function I in the example is increasing on - 00 < x ,,; - 2, decreasing on -2 ,,; x ,,; 2, and increasing on 2 ,,; x < 00. We do not talk about whether a function is increasing or decreasing at a single point.

HISTORICAL BIOGRAPHY

Edmund Halley (1656-1742)

First Derivative Test for Local Extrema In Figure 4.21, at the points where I has a minimum value, f' < 0 immediately to the left and f' > 0 immediately to the right. (If the point is an endpoint, there is ouly one side to consider.) Thus, the function is decreasing on the left of the minimum value and it is increasing on its right. Similarly, at the points where I has a maximum value, f' > 0 immediately to the left and f' < 0 immediately to the right. Thus, the function is increasing on the left of the maximum value and decreasing on its right. In summary, at a local extreme point, the sign of f' (x) changes. Absolute max I' undefined I I I I I

No extremum I'~ 0

11' <0

f' >o :

1' < 0

I I I

Local min : :

:

I I

I I

: /'= 0 I I

Absolute min

I

Local min

_ _ _ _ _ _ _ _ _ _ ~_ _ _ _ _ _ _ _ ~_ _ _ _ _ _ ~_ _ _ _ _ _ ~ ' _ _ _ _ _ _l ' _ _ _ _~_ _ _ _~'_ _ _ _ _ _ _ _ _ _~"

a

C2

b

FIGURE 4.21 The critical points of a functioo locate where it is increasing and where it is decreasing. The lust derivative changes sign at a critical point where a local extremum occurs.

These observations lead to a test for the presence and nature of local extreme values of differentiable functions.

First Derivative Test for Local Extrema Suppose that c is a critical point of a continuous function I, and that I is differentiable at every point in some interval containing c except possibly at c itself. Moving across this interval from left to right,

1. if f' changes from negative to positive at c, then I has a local minimum at c; 2. if f' changes from positive to negative at c, then I has a local maximum at c; 3. iff' does not change sign at c (that is, f' is positive on both sides of cor negative on both sides), then I has no local extremum at c. The test for local extrema at endpoints is similar, but there is ouly one side to consider.

Proof of the First Derivative Test Part (I). Since the sign of f' changes from negative

< c < b, f' < 0 on (a, c), and 0 on (c, b). IfxE (a, c), then/{c) < I{x) because f' < 0 implies that I is decreasing on [a, c]. If x E (c, b), then I{ c) < I{x) because f' > 0 implies that I is increasing on [c, b]. Therefore, I{x) ;", I{c) for every XE (a, b). By definition, I has a local minimumatc. Parts (2) and (3) are proved similarly. to positive at c, there are numbers a and b such that a

f' >

•

EXAMPLE 2

201

Monotonic Functions and the First Derivative Test

4.3

Find the critical points of

I(x) = xl/3(X - 4) = x'/' - 4XI/3. Identify the intervals on which 1 is increasing and decreasing. Find the function's local and absolute extreme values. Solution The function 1 is continuous at all x since it is the product of two continuous functions, x 1/' and (x - 4). The first derivative

f'(x) =

:fx (X4/' -

_1

-3 x

y

j x 1/' - j x-2/'

-2/'(x _ 1) -_ 4(x3x-

1)

2/'

is zero at x = 1 and undefined at x = O. There are no endpoints in the domain, so the critical points x = 0 and x = 1 are the only places where 1 might have an extreme value. The critical points partition the x-axis into intervals on which f' is either positive or negative. The sign pattern off' reveals the behavior of1 between and at the critical points, as summarized in the following table.

4

FIGURE 4.22 The functionj(x) = x 1/3 (x - 4) decreases when x < I and increases when x > I (Example 2).

4XI/') =

Interval Sign ofr Behavior of I

x
O<x<1

decreasing

decreasing

x> +

1

increasing

Corollary 3 to the Mean Value Theorem tells us that 1 decreases on (- 00, 0], decreases on [0,1], and increases on [1, 00). The First Derivative Test for Local Extrema tells us that 1 does not have an extreme value at x = 0 (f' does not change sign) and that 1 has a local minimum at x = 1 (f' changes from negative to positive). The value of the local minimum is 1(1) = 11/'(1 - 4) = -3. This is also an absolute minimum since 1 is decreasing on (-00, 1] and increasing on [1, 00). Figure 4.22 shows this value in relation to the function's graph. Note that limx~o f'(x) = -00, so the graph of 1 has a vertical tangent at the origin.

•

Exercises 4.3 Analyzing Functions from Derivatives Answer the following questions about the functions whose derivatives

are given in Exercises 1-14: •• What are the critical points of fI b. On what intervals is f increasing or decreasing? c. At what points, if any, does f assume local maximum and minimum values? 1. f'(x) = x(x - I) 2. f'(x) = (x - I)(x + 2) 3. f'(x) = (x - 1)2(x + 2) 4. f'(x) = (x - 1)2(x + 2)2 S. f'(x) = (x - I)(x + 2)(x - 3) 6. f'(x) = (x - 7)(x + I)(x + 5) x 2(x - I) 7. f'(x) = x + 2 ' x '¢' -2

(x - 2)(x 8. f'(x) = (x + I)(x

+ 4)

3)' x'¢' -1,3

4 9.f'(x)=1- 2 , x'¢'O x 11. f'(x) = x-I /3(x + 2) 13. f'(x)

=

(sinx - 1)(2 cos x

14. f'(x)

=

(sinx

10. f'(x) = 3 12. f'(x)

+

+ cosxXsinx

~,

= x- 1/2(x

x

'¢'

0

- 3)

1),0 :5 x :5 2.".

- cos x), 0:5 x :5 2.".

Identifying Extrema 10 Exercises 15-40: a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute ex1reme values, if any, saying where they occur.

202

Chapter 4: Applications of Derivatives

15.

16.

Y 1

2

\.

Y = f(x)

\ -

-2-1

x2

52. g(x) = 2' 4-x

Y

y=f x)

-3

1 2 3

1

-"!

i 3

- ~

/

1

2

A-..

I

1\

1

/

2

1\.. 1...Y

Y-f x) /

/

1

/ 1 l\

21. h(x)

= -x3

3t

-

1

1 +3

+ 2:<2

23. j(6) = 3~ -

463

25. j(r) = 3r + 16r 27. j(x) = x' - 8x 2 + 16 3

29. H(t) =

~t' - t'

x2

-

3

x- 2' x "" 2 x 1/ 3 (x + 8)

= -

37. j(x) = 39. h(x) = x 1/ 3(x 2

-

4)

1 2 3

20. g(t) = -3t + 9t + 5 22. h(x)

6'

24. j(6) = 66 26. h(r) = (r

+ 7)3

28. g(x) = x' - 4x 3 + 4x 2 -

t' 2

32. g(x) = 4Vx - x 34. g(x) = X2~ 36. j(x)

=

x3 3:<2 + I

38. g(x) = x 2/ 3(x + 5) 40. k(x) = x 2/ 3 (x 2 - 4)

Support your fmdings with a graphing calculator or compurer grapher.

j(x) = 2:< - x , -00 < x :5 2 j(x) = (x + 1j2, -00 < X :5 0 g(x) = x 2 - 4x + 4, I s x < 00 g(x) = -x 2 - 6x - 9, -4:5 x < j(t) = 12t - t 3 , -3:5 t < 00 j(t) = t 3 - 3t 2, -00 < t :5 3 2

3

2:<2

x - I

+

--rr

2

0:5 x :5 2-rr -rr

<x<2

-2:<

57. j(x) =

I - 2sin I , 0:5 x'" 2-rr

tan:<,

2

60. j(x) = sec

-'T1'

~

~

X

'T1'

-

2 cot x,

o<x<w

X -

2 tan x,

2<x<2

--rr

-rr

Theory and Examples Show that the functioos in Exercises 61 and 62 have local extreme values at the given values of 6, and say which kind of local extreme the function has.

61. h(6)

= 3cos 26 ,

0:5 6:5 2-rr,

62. h(6)

= 5 sin ~,

0:5 6 :5 -rr,

at6 aU

= oand 6 = 2-rr

= 0 and 6 = -rr

63. Sketch the graph of a differentiable function y the point (I, I) if /,(1) = 0 and

= j(x)

through

a. /'(x) > Oforx < land/,(x) < Oforx> I; b. /'(x)

< Oforx < land/,(x) > Oforx> I;

+ 4x,

0:5 x

<

2

d. /'(x) < Oforx "" 1. 64. Sketch the graph of a differentiable functioo y

=

j(x) that has

a. a local minimum at (I, I) aod a local maximum at (3, 3); b. a local maximum at (I, I) and a local minimum at (3, 3);

c. local maxima at (I, I) aod (3, 3); d. local minima at (I, I) and (3, 3). 65. Sketch the graph ofa continuous functiony = g(x) such that

a. g(2) = 2,0 < g' < I for x < 2, g'(x) --> 1- as x --> 2-, 00

00

48. k(x) = x + 3x + 3x + I, -00 < x :5 0 49. j(x) = Y25 - x 2 , -5 S x s 5 50. j(x) = Yx2 - 2:< - 3, 3:5 x < 00 x-2 51. g(x) = -2- , 0:5 x < I 3

V3 cos:< + sinx,

c. /'(x) > Oforx "" I;

b. Which of the extreme values, if any, are absolute?

x T -

0:5 x :5 2-rr

56. j(x) =

+3

main, and say where they occur.

47. h(x) =

Graph the functioo and its derivative together. Connnent on the behavior of j in relation to the signs and values of /'.

18x

= 2:<3 -

a. IdentifY the functioo's local extreme values in the given do-

41. 42. 43. 44. 45. 46.

D b.

59. j(x) = csc2 X 2

In Exercises 41-52:

D c.

and say where they occur.

58. j(x) = -2 cos x - cos2 X,

-2

30. K(t) = 15t3

31. j(x) = x - 6~ 33. g(x) = x\1'8=7 35. j(x)

/

-3 -2 1 -I

V

19. g(t) = -t

a. Find the local extrema of each function on the given interval,

55. j(x) =

\

21/

Y =fVc)

/ \ V ;:3 -2 - 1\1 2 / 3

2

In Exercises 53-60:

54. j(x) = sin x - coS:<,

Y

Y

2

< x :5 I

53. j(x) = sin 2:<, 0 s x s -rr 18.

17.

-2

-1

< g' <

b. g(2) = 2, g' g' > o far x

Oforx

>

2,andg'(x)---+-1+asx---+2+;

< 0 for x < 2, g'(x) --+ -00 as x--+ 2-, > 2, andg'(x)---+ 00 asx---+2+.

66. Sketch the graph of a cootinuous functioo y = h(x) such that

a. h(O) = 0, -2 :5 h(x) :5 2 for all x, h'(x) --+ and h'(x) --+ 00 asx --+ 0+; b. h(O) = 0, -2 :5 h(x) :5 0 for all x, h'(x) --+ and h'(x) ---+

-00

as x

---+

00

aox --+ 0-,

00

aox --> 0-,

0+.

67. Discuss the extreme-value behavior of the function j(x) = x sin (I/x), x "" O. How many critical points does this function have? Where are they located 00 the x-axis? Does j have an

4.4 Concavity and Curve Sketching

69. Determine the values of constants a and b so that f(x) ax 2 + bx has an absolute maximum at the point (I, 2).

absolute minimum? An absolute maximum? (See Exercise 49 in Section 2.3.)

68. Find tire intervals on which the function f(x) ~ ax 2 + bx + c, a -=F 0, is increasing and decreasing. Describe the reasoning behind your answer.

4.4

203 ~

70. Determine the values of constants a, h, c, and d so that f(x) ~ ax 3 + bx2 + ex + d has a local maximum at tire point (0,0) and a local minimum at the point (I, -I).

Concavity and Curve Sketching We have seen how the first derivative tells us where a function is increasing, where it is decreasing, and whether a local maximum or local minimum occurs at a critical point. In this section we see that the second derivative gives us infonnation about how the graph of a differentiable function bends or turns. With this knowledge about the first and second derivatives, coupled with oUI previous understanding of asymptotic behavior and symmetry stodied in Sections 2.6 and 1.1, we can now draw an accurate graph of a function. By organizing all of these ideas into a coherent procedure, we give a method for sketching graphs and revealing visually the key featores of functions. Identifying and knowing the locations of these featores is of major importance in mathematics and its applications to science and engineering, especially in the graphical analysis and interpretation of data.

Concavity FIGURE 4.23 The graph of f(x) ~ x 3 is concave down on ( - 00, 0) and concave up on (0,00) (Example la).

As you can see in Figare 4.23, the curve y = x 3 rises as x increases, but the portions defmed on the intervals (-00,0) and (0, 00) turn in different ways. As we approach the origin from the left along the curve, the curve turns to OUI right and falls below its tangents. The slopes of the tangents are decreasing on the interval ( - 00, 0) . As we move away from the origin along the curve to the right, the curve turns to OUI left and rises above its tangents. The slopes of the tangents are increasing on the interval (0, 00). This turning or bending behavior defmes the concavity of the curve.

DEFINmON

The graph of a differentiable function y ~ j(x) is

(a) concave up on an open interval I if j' is increasing on I;

(b) concave down on an open interval I if j' is decreasing on!.

If y

~

j(x) has a second derivative, we can apply Corollary 3 of the Mean Value Theorem to the first derivative function. We conclude that j' increases if f" > 0 on I, and decreases iff" < O.

The Second Derivative Test for Concavity Let y = j(x) be twice-differentiable on an interval!.

1. If f"

0 on I, the graph of j over I is concave up.

2.

0 on I, the graph of j over I is concave down.

> If f" <

~ j(x) is twice-differentiable, we will use the notations when denoting the second derivative.

If y

f"

and y" interchangeably

204

Chapter 4: Applications of Derivatives y

EXAMPLE 1

x 3 (Figure 4.23) is concave down on (-00,0) where y" = 6x < 0 and concave up on (0, 00) where y" = 6x > O. (b) The curve y = x 2 (Figure 4.24) is concave up on (-00, 00) because its second deriv_ ative y" = 2 is always positive. (8) The curve

EXAMPLE 2

y

=

Detennine the concavity of y = 3

+

sin x on [0, 21T].

Solution The first derivative of y = 3 + sin x is y' = cos x, and the second derivative is y" = -sinx.Thegraphofy = 3 + sin x is concave down on (O,1T),wherey" = -sinx is negative. It is concave up on (1T, 21T), where y" = -sinx is positive (Figure 4.25). _ FIGURE 4.24

The graph of f(x) = x 2

is concave up on every interval (Example Ib).

The curve y = 3 + sin x in Example 2 changes concavity at the point (1T, 3). Since the first derivative y' = cos x exists for all x, we see that the curve has a taogent line of slope -I at the point (1T, 3). This point is called a point of inflection of the curve. Notice from Figure 4.25 that the graph crosses its taogent line at this point and that the second derivative y" = -sin x has value 0 when x = 1T. In general, we have the following definition.

y

y=3+sinx

2

Points of Inflection

: I

:~

DEFINITION A point where the graph of a function has a taogent line and where the concavity changes is a point of inflection.

~t.-~~-L~~~--~~ x

0 ," / .. -1- - - - - - . y" = -sinx

2

Using the sign of y' to determine the concavity ofy (Example 2). FIGURE 4.25

We observed that the second derivative of f(x) = 3 + sin x is equal to zero at the inflection point (1T, 3). Generally, if the second derivative exists at a point of inflection (c, f(c», then f"(c) = O. This follows immediately from the Intermediate Value Theorem whenever f" is continuous over an interval containing x = c because the second derivative changes sign moving across this interval. Even if the continuity assumption is dropped, it is still true that f"(c) = 0, provided the second derivative exists (although a more advanced agrument is required in this noncontinuous case). Since a taogent line must exist at the point of inflection, either the first derivative f' (c) exists (is finite) or a vertical taogent exists at the point. At a vertical taogent neither the flISt nor second derivative exists. In summary, we conclude the following result.

At a point of inflection (c, f(c», either f"(c) = 0 or f"(c) fails to exist. y

The next example illustrates a function having a point of inflection where the flISt derivative exists, hut the second derivative fails to exist. EXAMPLE 3 The graph of f(x) = x'/3 has a horizontal taogent at the origin because f'(x) = (5/3)x2/3 = 0 when x = O. However, the second derivative

f"(x) = FIGURE 4.26 The graph of f(x) = x'/3 has a horizontal taogent at the origin where the coocavity chaoges, although f" does not exist at x = 0 (Example 3).

~G'X2/3)

=

10 x - 1/ 3 9

failstoexistatx = O.Nevertheless,f"(x) < ofor x < Oandf"(x) > ofor x > O,sothe second derivative changes sign at x = 0 and there is a point of inflection at the origin. The graph is shown in Figure 4.26. _

4.4 Concavity and Curve Sketching y

205

Here is an example sbowing that an inflection point need not occur even though both derivatives exist and f" = O.

EXAMPLE 4 The curve y = x' has no inflection point at x = 0 (Figure 4.27). Even though the second derivative y" = 12x 2 is zero there, it does not change sign. •

FIGURE 4.27 The grapb of y = x' bas no inflection point at the origin, even though y" = 0 there (Example 4).

As our fmal illustration, we show a situation in which a point of inflection occurs at a vertical tangent to the curve where neither the first nor the second derivative exists. The graph of y = x 1/ 3 has a point of inflection at the origin because the second derivative is positive for x < 0 and negative for x > 0:

EXAMPLE 5

y Point of

inflection

-----~ ~~------~x

FIGURE 4.28

A point of

inflection where y' and y" fail to exist (Example 5).

However, both y' = x -2/3/3 and y" fail to exist at x = 0, and there is a vertical tangent there. See Figure 4.28. • To study the motion of an o~ect moving along a line as a function of time, we often are interested in knowing when the object's acceleration, given by the second derivative, is positive or negative. The points of inflection on the gmph of the object's position function reveal where the acceleration changes sign.

EXAMPLE 6 A particle is moving along a horizontal coordinate line (positive to the right) with position function .(1)

=

21 3

-

1412

+ 221 -

5,

12: O.

Find the velocity and acceleration, and describe the motion of the particle.

Solutton

The velocity is

V(I)

= .'(t) = 6t 2

- 28t + 22

= 2(t -

1)(3t - 11),

and the acceleration is

a(t) = v'(t) = ."(t) = 12t - 28 = 4(3t - 7). When the function .(t) is increasing, the particle is moving to the right; when .(t) is de· creasing, the particle is moving to the left. Notice that the first derivative (v = .') is zero at the critical points 1 = 1 and 1 = 11/3.

Interval Sign ofv =.' Behavior of. Particle motion

0<1<1

1 < 1 < 11/3

11/3 < 1

increasing

decreasing

increasing

right

left

right

+

+

The particle is moving to the right in the time intervals [0, 1) and (11/3, 00), and moving to the left in (I, 11/3). It is momentarily stationary (at rest) at/ = I and t = 11/3. The acceleration a(l) = ."(1) = 4(31 - 7) is zero when 1 = 7/3.

Interval Sign of a = s" Graph ofs

0< 1 < 7/3

7/3 < 1

concave down

concave up

+

206

Chapter 4: Applications of Derivatives The particle starts out moving to the right while slowing down, and then reverses and begins moving to the left at I = 1 under the influence of the leftward acceleration over the time interval [0, 7/3). The acceleration then changes direction at I = 7/3 but the particle continues moving leftward, while slowing down under the rightward acceleration. At I = 11/3 the particle reverses direction again: moving to the right in the same direction as the acceleration. _

Second Derivative Test for Local Extrema Instead oflooking for sign changes in!, at critical points, we can sometimes use the following test to determine the presence and nature of local extrema.

THEOREM 5-Second Derivative Test for Local Extn!ma on an open interval that contains x = c.

Suppose f" is continuous

1. If!'(c) = Oandj"(c) < 0, then f has alocalmaximumatx = c. 2. If!'(c) = Oandj"(c) > 0, then f has alocalminimumatx = c. 3. If !' (c) = 0 and j" (c) = 0, then the test fails. The function f may have a local maximum, a local minimum, or neither.

!\V f' ~ O,f" < 0 ~localmax

f'~O,f">O ~

local min

Proof Part (1). If f"(c) < 0, then f"(x) < 0 on some open interval I containing the point c, since f" is continuous. Therefore, !' is decreasing on 1. Since!' (c) = 0, the sign of !' changes from positive to negative at c so f has a local maximum at c by the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions y = x" y = -x" and y = x 3 • For each function, the [lISt and second derivatives are zero at x = O. Yet the function y = X4 has a local minimum there, y = -x' has a local maximum, and y = x 3 is increasing in any open interval containing x = 0 (having neither a maximum nor a minimum there). Thus the test fails. _ This test requires us to know f" only al c ilself and not in an interval about c. This makes the test eaay to apply. That's the good news. The bad news is that the test is inconclusive if f" = 0 or if j" does not exist at x = c. When this happens, use the First Derivative Test for local extreme values. Together !' and j" tell us the shape of the function's graph-that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is torning or bending as defined by its concavity. We use this information to sketch a graph of the function that captures its key featores.

EXAMPLE 7

Sketch a graph of the function

using the following steps. (a) Identify where the extrema of f occur. (b) Find the intervals on which f is increasing and the intervals on which f is decreasing.

(c) Find where the graph of f is concave up and where it is concave down. (d) Sketch the general shape of the graph for j. (e) Plot some specific points, such as local maximum and minimum points, points of inflection, and intercepts. Then sketch the curve.

4.4 Concavity and Curve Sketching

207

Solutton The function f is continuous since f'(x) = 4x' - 12x 2 exists. The domain of f is ( - 00, 00), and the domain of f' is also ( - 00, 00). Thus, the critical points of f occur only at the zeros of f' . Since

f'(x)

4x' - 12x 2

=

the flISt derivative is zero at x = 0 and x vals where f is increasing or decreasing.

Interval Sign of/, Behavior off

=

=

4x 2(x - 3)

3. We use these critical points to define inter-

x
0<x<3

3<x

decreasing

decreasing

increasing

+

(8) Using the First Derivative Test for local extrema and the table above, we see that there is no extremum at x = 0 and a local minin3um at x = 3. (h) Using the table above, we see that f is decreasing on (-00,0] and [0, 3], and increas-

ingon [3,00). (c) f"(x) = 12x 2 - 24x = 12x(x - 2) is zero atx = oand x to defme intervals where f is concave up or concave down.

Interval Sign of f" Behavior of f

x
=

2. We use these points

0<x<2

2<x

concave down

concave up

+

+

concave up

We see that f is concave up on the intervals (-00,0) and (2,00), and concave down on (0,2). (d) Sununarizing the information in the last twu tables, we obtain the following.

x
O<x<2

2<x<3

3<x

decreasing concave up

decreasing concave down

decreasing concave up

increasing concave up

y

20

The general shape of the curve is shown in the accompanying figure.

15 (0, 10) ,

decr --~~~--~~~--~--~--~X

-1

0

4

-5

,

deer

deer

cone

cone up

down

I

~:

-10

-15

-20

cone up

I I I I I I

(3, -17)

infl point

I

\

I I I I I I

iner

General shape

cone up

I

:~ : ~ infl

point

lucal min

Local mjnjmum

FIGURE 4.29 The graph of f(x) =

x' - 4x' + 10 (Example 7).

(e) Plot the curve's intercepts (if possible) and the points where y' and y" are zero. Indicate any local extreme values and inflection points. Use the general shape as a guide to sketch the curve. (plot additional points as needed.) Figure 4.29 shows the grsph off. •

208

Chapter 4: Applications of Derivatives The steps in Example 7 give a procedure for graphing the key features of a function.

=

Procedure for Graphing y /(x) 1. Identify the domain of f and any symmetries the curve may have. 2. Find the derivatives y' and y" . 3. Find the critical points off, ifany, and identify the function's behavior at each one. 4. Find where the curve is increasing and where it is decreasing. 5. Find the points of inflection, if any occur, and determine the concavity of the curve. 6. Identify any asymptotes that may exist (see Section 2.6). 7. Plot key points, such as the intercepts and the points found in Steps 3-5, and sketch the curve together with any asymptotes that exist.

EXAMPLE 8

Sketch the graph of f(x) =

(x

+ 1)2

l+x

2 •

Solution

1.

The domain of f is (- 00, 00) and there are no symmetries about either axis or the origin (Section l.l).

2. Find f' and I" . x) = (x + 1)2 ( f 1+ x 2

x-intercept at x

-1, 1).,

x=Q

2 '(x) = (1 +x )'2(x

f

=

y-~(y~

+ 1) - (x + + x 2 )'

1)2.2x

(I 2(1 - x 2 )

Critical points: x=-l,x=l

(1 + x 2 )' (1

+ x 2)'. 2(

I"(x) = 4x(x 2

(I +

-

-2x) - 2(1 - x2)[2(1 (1 + x2)4

+ x 2). 2x]

3)

x 2 )3

After some algebra

3. Behavior at critical points. The critical points occur only at x = ± I where f' (x) = 0 (Step 2) since f' exists everywhere over the domain of f. At x = -1, I" (- I) = I > 0 yielding a relative minllnum by the Second Derivative Test. At x = 1,1"(1) = -I < 0 yielding a relative maximum by the Second Derivative test.

4. Increasing and decreasing. We see that on the interval (-00, -I) the derivative

f' (x) <

0, and the curve is decreasing. On the interval (-1, 1), f' (x) curve is increasing; it is decreasing on (I, 00) where f'(x) < 0 again.

>

0 and the

4.4 Concavity and Curve Sketching

209

5. Inflection points. Notice that the denominator of the second derivative (Step 2) is always positive. The second derivative /" is zero when x = - v'3, 0, and v'3. The second derivative changes sign at each of these points: negative on ( - 00, - v'3) , positive on ( - v'3, 0), negative on (0, v'3), and positive again on ( v'3, 00 ). Thus each point is a point of inflection. The curve is concave down on the interval ( - 00, - v'3), concave up on ( - v'3, 0), concave down on (0, v'3), and concave up again on ( v'3, 00 ) . 6. Asymptotes. Expanding the numerator of I(x) and then dividing both numerator and denominator by x' gives Expanding numerator

I

+

Y

PointofllUrrection wherex ~ 0 2 (1,2)/

------~~~~--------~x

PointofllUrrection wherex ~ -0 ~

(x

+ 1)2

l+x

2

Dividing by Xl

We see that I(x) ..... 1+ as x ..... 00 and that I(x) ..... 1- as x ..... -00. Thus, the line y = I is a horizontal asymptote. Since 1 decreases on ( - 00, - I) and then increases on ( - I, I), we know that I( -I) = 0 is a local minimum. Although 1 decreases on (I, 00), it never crosses the horizontal asymptote y = I on that interval (it approaches the asymptote from above). So the graph never becomes negative, and I( -I) = 0 is an absolute minimum as well. Likewise, 1(1) = 2 is an absolute maximum because the graph never crosses the asymptote y = I on the interval (- 00, - I), approaching it from below. Therefore, there are no vertical asymptotes (the range of 1 is 0,,;; y,,;; 2).

asymptDte

FIGURE 4.30 The graph of y

+ (1/x') (1/x') + I (2/x)

7.

(Example 8).

The graph of 1 is sketched in Figure 4.30. Notice how the graph is concave down as it approaches the horizontal asymptote y = I as x ..... - 00, and concave up in its approach toy = I asx ..... 00. •

EXAMPLE 9

Sketch the graph of I(x) =

x' + 4 -----zx-.

Solutton 1.

2.

The domain of 1 is all nonzero real numbers. There are no intercepts because neither x nor I(x) can be zero. Since I( -x) = - I(x), we note that 1 is an odd function, so the graph of 1 is symmetric about the origin. We calculate the derivatives of the function, but fust rewrite it in order to simplify our computations:

I(x)

=

t'(x)

=

x' ~ 4

t -:'

=

t+~

=

x':z:,

4

Function simplified for differentiation

Combine fractions to solve easily f'(x)

= o.

Exists throughout the entire domain off

3.

The critical points occur at x = ±2 where t'(x) = O. Since /"(-2) < 0 and /"(2) > 0, we see from the Second Derivative Thst that a relative maximum occurs at x = - 2 with 1(-2) = - 2, and a relative minimum occurs at x = 2 with 1(2) = 2.

210

Chapter 4: Applications of Derivatives 4.

On the interval (- 00, - 2) the derivative f' is positive because x 2 - 4 > 0 so the graph is increasing; on the interval (-2, 0) the derivative is negative and the graph is decreasing. Similarly, the graph is decreasing on the interval (0, 2) and increasing on

(2,00). 5. 4

Y~ ~X2+ 4 2x

val (0,00).

(2,2) 2

y= ! 2

6.

__L -__L-~~~~__-L->x

-4

-2

2

The graph ofy =

+4 ----zx-

=

+00

lim (~+.2.) x-o2 x

-00' so the y·axis is a vertical asymptote. Also, as x ..... 00 or as x ..... - 00, the graph of

I' x2

From the rewritten formula for f(x), we see that

lim (~+.2.) x-o+ 2 x

4

~2 (-2, -2)

FIGURE 4.31

There are no points of inflection because /,,(x) < 0 whenever x < 0, /,,(x) > 0 whenever x > 0, and /" exists everywhere and is never zero throughout the domain of f. The graph is concave down on the interval (- 00, 0) and concave up on the inter·

and

=

f(x) approaches the line y = x/2. Thus y = x/2 is an oblique asymptote.

7.

•

The graph of f is sketched in Figure 4.3\.

(Example 9).

Graphical Behavior of Functions from Derivatives As we saw in Examples 7-9, we can learn much about a twice·differentiable function = f(x) by examining its firat derivative. We can find where the function's graph rises and falls and where any local extrema are located. We can differentiate y' to learn how the graph bends as it passes over the intervals of rise and fall. We can deter· mine the shape of the function's graph. Information we cannot get from the derivative is how to place the graph in the xy·plane. But, as we discovered in Section 4.2, the only additional information we need to position the graph is the value of f at one point. Information about the asymptotes is found using limits (Section 2.6). The fol· lowing figure summarizes how the derivative and second derivative affect the shape of a graph.

y

r:;d" 7

> 0 ~ rises from

Differentiable ~ smooth, connected; graph may rise and fall

y'

,/ or

( or \

y" > 0

"

~ concave up throughout; no waves; graph may rise or fall

C"\ or ~ y' changes sign => graph has local maximum or local minimum

left to righ~ may be wavy

y" < 0

~ concave down throughout; no waves; graph may rise or fall

f\

and y"
y'~O

~

y' < 0 ~ falls from

left to righ~ may be wavy

:/ y" changes sign at an inflection point

V

0 and y">O at a point; graph has local minimum

y'~

4.4 Concavity and Curve Sketching

211

Exercises 4.4 Analyzing Functions from Graphs Identify the inflectioo points and local maxima and minima of the functioos graphed in Exercises I-S. IdentifY the intervals on which the functions are concave up and concave down.

1.

2.

17. y ~ x' - z,,' = x'(x' - 2) 18. y ~ -x' + 6>;' - 4 ~ x'(6 - x') - 4 19. y ~ 4x 3 - x' = x 3(4 - x)

+ z,,3

=

x 3(x

21. y ~ x' - 5x'

=

x'(x - 5)

20. y ~ x'

--/- *-- --+--.x

+ 1)3

16. y ~ I - (x

_5)'

22. y

~ x(~

23. y

~

24. Y

= x - sinx,

25. y ~

x + sinx, 0 s x s 2.-

V3x -

0

27. Y

4.

y ~ ~(x2 _ 1f'3

=

'vJV.

31. y ~

x

~

0

2w

x '" 2.-

S

-1T'

'1T

"2 < x < 2

sinX' cos x,

28. y ~ cosx 29. y ~ x'/5

~

2 COU,

4 26. y ~ '3x - taox,

3.

+ 2)

O:s x :s

+ V3 sin x,

0

1T'

sx

30. y = x'/5

x

W+1

32. y

33. y ~ z" - 3x'/3 35. y

~ x'/3(t -

'3 Y

3

6. y=tanx-4x-1!:<x<'!! • 2 Y

39. y

~ v'16 -

2

41.

x)

sin Ixl. -21T:S x :S 21T

45. y ~

8. y=2coax- •v2x, '" -1T:sx:ST 3..

Y

y --'---~\---+~x

-..

0

42. y =

ax

x

=

z"

+I

{h t;., ' • VX'

(2 - x')3/2

x' + ~ '\Yx'+l

44. y=_5_ x4 + 5 46. y = lx' - 2xl

+4

lx' - 11

=

40. y =

-,--

• r.-c 47. y ~ V Ix I =

x'

3

3

7. y

38. y

x-

~

~

36. y = x'/3(x - 5)

Y~--2

43. y -;!------;;1-----;;24 .. x

x' -

=

34. y = 5x'/' - z"

37. y ~ x~ 5. y=x+sin2x_21Tsx:s;2'1T

'" 2.-

x< 0 x~0

48.y~~ Sketching the General Shape, Knowing y' Each of Exercises 49-70 gives the rITst derivative of a continuous function y ~ f(x). Find y' and then use steps 2-4 of the grapbing procedure on page 20S to sketch the general shape of the graph off.

49.y'=2+x-x 2

SO.y'=x 2 -x-6

51. y' = x(x - 3)'

52. y' = x'(2 - x)

53. y' = x(x' - 12)

54. y' = (x - I)'(z"

~~=(ax-5x~-4

~~=~-~~-~

NOI'TOSCALE

Graphing Equations Use the steps of the graphing procedure 00 page 20S to graph the equations in Exercises 9-4S. Ioclude the coordinates of aoy local aod absolute extreme points and inflectioo points. 9.y~x'-4x+3

11. y ~ x 3 - 3x

+ 6>;' (x - 2)3 + 1

13. y ~ _z,,3 15. y ~

+3 3

57. y'

= sec

2

x,

10.y~6-z,,-x'

58. y' = tanx,

12. y ~ x(6 - z,,)' 14. y ~ I - 9x - 6>;' - x 3

59. y' =

cot~,

,

+ 3)

_,!!<x<,!!

2 2 _'!! < x < '!! 2 2 0

61. y' = tao 0 - I,

<0<

2.- 60. y'

.- < -2

0<

2.-

~ csc'~,

0

<0<

2.-

212

Chapter 4: Applications of Derivatives

62. y' ~ I - cof-o, 0 < 0 < 63. y' = cost, 0 s t:s 21T 64. y' = sint, O:s t:s 21T

'If

n-'/'

66. y' ~ (x - 2t'/3 65. y' ~ (x + 67. y' ~ x-'/'(x - I) 68. y' ~ x-4/'(x + I) -2x x:s 0 69. y' ~ 21xl ~ { 2x,' x>O

o ,_ {-Xl,

7. Y -

x2,

x:s; 0

x> 0

x

~

-,-x - I

91. y

~

-,-x +4

92. y

~

-,-x +4

8

4x

72.

x - I x'(x - 2)

~ --i'-----'~

(Agnesi's witch) (Newtoo's serpentine)

Theory and Examples 93. The accompanying figure shows a portion of tire graph of a twicedifferentiable functioo y ~ f(x). At each of the five labeled points, classifY y' and y" as positive, negative, or zero. y

Sketching y from Graphs of y' and y" Eacb of Exercises 71-74 shows the graphs of the first and second derivatives of a functioo y ~ f(x). Copy the picture aod add to it a sketch of the approximate graph of f, giveo that the graph passes through the point P. 71.

90. y

89. y

s

~f(X)

~ R

p

T

Q

y

~----------~X

o

94. Sketch a smooth coooected curve y --1-t-""'--~ X

y ~r(x)

73.

y

/,(2)

8,

f(x) with ~

/'( -2)

~

0,

f(O)

~

4,

/'(x) < 0 for

f(2)

~

0,

rex) < 0 for x < 0,

/'(x)

p

~

f(-2)

~

>0

for

Ixl

> 2,

rex)

>0

Ixl < 2,

x> O.

for

95. Sketch the graph of a twice-differeotiable function y ~ f(x) with the following properties. !.abel coordinates where possible.

x 74.

y

Graphing Rational Functions Graph the rational functions in Exereises 75-92.

75 ~2x'+x-1 • y x' - 1 ~x4+1 77

x'

76.y~

x'-49 x' + 5x - 14 78 ~x'+4 .Y 2x 80. y

x' ~ -,--

82. y

~

-,-x - 2

83,y~x+1

84. y

~

-X+T

~x'-x+1 x I x'-3x'+3x-1 87.y~ x'+x-2

86

~_x'-x+1

88• y

~x'+x-2 ,

79.

I

y~-,-­

x-I x2

2 81.y~-x'_1 -

x'

85

• y

Derivatives

x<2 2 2<x<4

p

• y

y

• y

x-I x2

-

4

x2

-

x

x-x

I

<

0,

y"

>

0

y' = 0, y" > 0 y"

>

0

4

4

y' > 0, y"

~

0

4<x<6 6 x>6

y' y' y'

<

7

0 0 0

y'

> > =

<

0,

0, y" 0, y" 0, y"

< <

96. Sketch the graph of a twice-differeotiable function y ~ f(x) that passes through the points (-2,2), (-I, I), (0,0), (I, I), aod (2, 2) aod whose 'lrst two derivatives have the following sign

patterns. y':

y": 4

y'

+

o

2

+ 2

+ -I

Motion Along a line The graphs io Exercises 97 aod 98 show the position s ~ f(t) of ao object moving up and down on a coordinate lioe. (a) Whea is the object moviog away from the origin? toward the origin? At approximately what times is the (b) velocity equal to zero? (1:) acceleration equal to zero? (d) When is the acceleration positive? negative?

4.4 Concavity and Curve Sketching 97.

103. For x > 0, sketch a curve y = f(x) that has f(1) = 0 aod f'(x) = I/x. Cao aoything be said about the coocavity of such a curve? Give reasons for your answer.

s

I

104. Cao anything be said about the graph ofa functiooy = f(x) that has a continuous second derivative that is never zero? Give reasons for your answer.

].

is 0

5

15

10

Ttme (sec)

lOS. If b, c, aod d are coostaots, for what value of b will the corve y = x 3 + bx 2 + ex + d have a point of inflection at x = I?

Give reasons for your answer. 106. Parabolas

s

9S.

213

] fil-

is

0

15

5 Tnne(sec)

99. Marginal cost The accompanying graph shows the hypothetical cost c = f(x) of manufacturing x items. At approximately what productioo level does the marginal cost chaoge from decreasing to increasing?

.. Find the coordinates of the vertex of the parabola y = ax 2 + bx + c, a¢' O. b. When is the parabola concave up? Concave down? Give reasons for your answers. 107. Quadratic curves What can you say about the ioflectioo points of a quadratic curve y = ax 2 + bx + c, a =F O? Give reasons for your answer. lOS. Cubic curve. What cao you say about the ioflectiou points of a cubic curve y = ax 3 + bx2 + ex + d, a =F O? Give reasons for your answer. 109. Suppose that the secood derivative of the functioo y = f(x) is y"

= (x +

I)(x - 2).

For what x-values does the graph of f have ao ioflectioo point? 110. Suppose that the secoodderivative of the functiooy = f(x) is

c

+ 3). For what x-values does the graph of f have ao ioflectioo point? y" = x 2(x - 2)'(x

~~~~~~~~ X

20 40 60 80100120 Thousands of ooits produced

100. The accompanying graph shows the moothly reveoue of the Widget Corporatioo for the last 12 year>. During approximately what time intervals was the marginal revenue increasing? Decreasing?

101. Suppose the derivative of the functioo y = f(x) is

= (x

- I)'(x - 2).

At what points, if aoy, does the graph of f have a local minimum, local maximum, or point of ioflectioo? (Hint: Draw the sigo pattern for y' .) 102. Suppose the derivative of the functioo y

y'

112. Find the values of coostaots a, b, aod c so that the graph of y = (x 2 + a)/(bx + c) has a local minimum at x = 3 aod a I()cal maximum at (-I, -2). COMPUTER EXPLORATIONS 10 Exercises 113-116, fmd the ioflectioo points (if aoy) 00 the graph of

y

y'

111. Find the values of coostaots a, b, aod c so that the graph of y = ax 3 + bx2 + ex has a local maximum at x = 3, local minimumatx = -1,aodioflectioopointat(l, 11).

= (x

=

f(x) is

- I)'(x - 2)(x - 4).

At what points, if aoy, does the graph of f have a local minimum, local maximum, or point of inflection?

the function aod the coordinates of the points on the graph where the functioo has a local maximum or local minimum value. Thcu graph the functiou in a regioo large coough to show all these points simuitao<>ously. Add to your picture the graphs of the function's fl!St and secood derivatives. How are the values at which these graphs intersect the x-axis related to the graph of the functiou? 10 what other ways are the grapha of the derivatives related to the graph of the functioo? 113. y = x' - 5x' - 240 114. Y = x 3 - 12x 2 115. Y

= ~x' + 16>;2 -

116. Y

= "4 - "3 -

X4

x3

25

4x 2 + 12x + 20

117. Graph f(x) = 2x' - 4x 2 + I and its frrst two derivatives l<>gether. Comment on the behavior off in relation to the sigos aod values of f' aod f" . 11S. Graph f(x) = x cos x aod its second derivative together for o ,;; x ,;; 2".. Comment 00 the behavior of the graph of fin relatioo to the sigos aod values of f" .

214

Chapter 4: AppLications of Derivatives

Applied Optimization

4.5

What are the dimensions of a rectangle with IlXCd perimeter having maximum area? What are the dimensions for the least expensive cylindrical can of a given volume? How many items should be produced for the most profitable production run? Each of these questions asks for the best, or optimal, value of a given function. In this section we use derivatives to solve a variety of optimization problems in business, mathematics, physics, and economics.

Solving Applied Optimization Problems 1. Read the problem. Read the problem until you understand it. What is given? What is the unknown quantity to be optimized? 2. Draw a picture. Label any part that may be important to the problem. 3. hltroduce variables. List every relation in the picture and in the problem as an equation or algebraic expression, and identify the unknown variable. 4. Write an equation for the unknown quantity. If you can, express the unknown as a function of a single variable or in two equations in two unknowns. This may require considerable manipulation. S. Test the critical points and endpoints in the domain ofthe Il1Iknown. Use what you know about the shape of the function's graph. Use the lust and second derivatives to identify and classify the function's critical points.

(.)

EXAMPLE 1 An open-top box is to be made by cutting small congruent squares from the comers of a 12-in.-by-12-in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible? (b)

FIGURE 4.32 An open box made by cutting the corners from a square sheet of

Solution We start with a picture (Figure 4.32). In the figure, the corner squares are x in. on a side. The volume of the box is a function of this variable:

tin What size comers maximize the box's volmne (Example I)?

, y - x(12 - 'lx'jl,

O:Sx:S6

Since the sides of the sheet of tin are only 12 in. long, x :S 6 and the domain of Vis the interval 0 :S X :S 6. A graph of V (Figure 4.33) suggests a minimum value of 0 at x = 0 and x = 6 and amaximum near x = 2. To learn more, we examine the first derivative of V with respect

to x: min

o

V=hlw

:

~

144 - 9&

+ 12%' ~

12(12 - 8x

+ x') ~

12(2 - x)(6 - x).

/,

•

FIGURE 4.33 The volume of the box in Figure 4.32 graphed 18 a fimction of x.

Of the two zeros, x - 2 and x - 6, only x - 2 lies in the interior of the function's domain

and makes the critical-point list. The values of Vat this one critical point and two endpoints are Critical-point value: ilDdpointvalucs:

1'(2) - 128 1'(0)

~

0,

1'(6)

~

o.

The lll8Ximum volume is 128 in3 • The cutout squares should be 2 in. on a side.

•

4.5

AppLied Optimization

215

EXAMPLE 2 You have been asked to design a one-liter can shaped like a right circular cylinder (Figure 4.34). What dimensions will use the least material?

+----2r--~) 1

T h

1

SoLution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is I liter = 1000 crn3

Surface area ofcan:

A = 27Tr2

+ 27Trh '---v---'

FIGURE 4.34 This one-liter can uses the least material when h = 2r (Example 2).

circular cylindrical ends wall

How can we interpret the phrase "least material"? For a first approximation we can ignore the thickness of the material and the waste in manufacturing. Then we ask for dimensions r and h that make the total surface area as small as possible while satisfying the constraint 7Tr 2h = 1000. To express the surface area as a function of one variable, we solve for one of the variables in 7Tr 2h = 1000 and substitute that expression into the surface area formula. Solving for h is easier:

Thus,

Our goal is to find a value of r that such a value exists.

> 0 that minimizes the value of A. Figure 4.35 suggests

A

Tall and thin can Short and wide can A = 271"r2 Tall and thin

+

2000, r r

/> 0

--r-------~--------------~r

o

Short and wide

FIGURE 4.35

~

The graph of A = 27fr2

+ 2000/r is concave up.

Notice from the graph that for small r (a tall, thin cylindrical container), the term 2000/r dominates (see Section 2.6) and A is large. For large r (a short, wide cylindrical container), the term 27Tr2 dominates and A again is large.

216

Chapter 4: Applications of Derivatives Since A is differentiable on r > 0, an interval with no endpoints, it can have a minimum value only where its fITst derivative is zero.

dA dr

2000 r2

- - = 4'ITr - - -

0= 4'ITr - 2000 -r2

SetdA/dr ~

4",,3 = 2000 r =

~5~0

o.

Multiply by r'.

'" 5.42

Solve forr.

What happens at r = "¢'500/'IT? The second derivative 2

d A = 4'11' + 4000 dr 2 r3 is positive throughout the domain of A. The graph is therefore everywhere concave up and the value of A at r = "¢'500/'IT is an absolute minimum. The corresponding value of h (after a little algebm) is

h = 1000 = 7f'r2

2~500

= 2r.

'1T

The one-liter can that uses the least material has height equal to twice the mdius, here with r '" 5.42 em andh '" 10.84 em. •

Examples from Mathematics and Physics EXAMPLE 3

A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions?

y x'+y'~4

(x. V4 - X2) 2 ~L-~----~~------~~ x

-2 -x

0

x 2

FIGURE 4.36 The rectangle inscribed in the semicircle in Example 3.

Solution Let (x, ~) be the coordinates of the comer of the rectangle obtained by placing the circle and rectangle in the coordinate plane (Figure 4.36). The length, height, and area of the rectangle can then be expressed in terms of the position x of the lower right-hand comer: Length: 2x,

Height:

V4 -

x 2,

Area:2x~.

Notice that the values of x are to be found in the interval 0 :5 x :5 2, where the selected comer of the rectangle lies. Our goal is to find the absolute maximum value of the function

A(x)=2x~ on the domain [0, 2]. The derivative

dA dx

~+2~ 4 - x2

is not defined when x = 2 and is equal to zero when

-2x

2

~

+2~=0

-2x 2 + 2(4 - x 2 )

= 0 2 8-4x =0

x 2 = 20rx =

±v'2.

4.5

V2

V2,

Applied Optimization

217

V2

Of the two zeros, x = and x = only x = lies in the interior of A's domain and makes the critical-point list. The values of A at the endpoints and at this one critical point are

A( V2)

Critical-point value:

=

2V2v'4=2 =

A(O) = 0,

Endpoint values:

4

A(2) = O.

The area has a maximum value of 4 when the rectangle is 2x = 2 uuits long.

V2

V4 -

x2 =

V2 units high and •

EXAMPLE 4 The speed of light depends on the medium through which it travels, and is generally slower in denser media. Fermat's principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Describe the path that a my of light will follow in going from a point A in a medium where the speed of light is c, to a point B in a second medium where its speed is C2.

HiSTORICAL BIOGRAPHY

Willebrord Snell van Rayen

(158()"'1626)

y

Solution Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the xy-plane and that the line separating the two media is the x-axis (Figure 4.37). In a uuiform medium, where the speed of light remains constant, "shortest time" means "shortest path;' and the my oflight will follow a straight line. Thus the path from A to B will consist of a line segment from A to a boundary point P, followed by another line segment from P to B. Distance traveled equals rate times time, so

Medium I

-ooot---,----"''Ir- -d\ - - - ' Medium 2

d-x B

Time = distance rate

FIGURE 4.37 Alightrayrefracted (deflected from its path) as it passes from

From Figure 4.37, the time required for light to travel from A to P is

one medium to a denser medium (Example 4).

t}

AP

= Ct

=

Va 2 + x 2 Ct

.

From P to B, the time is

PB

t2=C2=

Vb 2 + (d -

x)'

C2

The time from A toB is the sum of these: 1 = I,

+ 12

=

Va 2 + x 2 C,

+

Vb 2 + (d C2

x)'

This equation expresses 1 as a differentiable function of x whose domain is [0, d]. We want to find the absolute minimum value of 1 on this closed interval. We find the derivative

d-x

dtIdx negative

dtIdx zero

( - - - -: /,. + + + + + + + (

o

0

and observe that it is continuous. In terms of the angles 6, and 62 in Figure 4.37,

dtldx positive >x

dl dx

sin 6,

sin 62

d

FIGURE 4.38 The sign pattern of dt/dx in Example 4.

The function 1 has a negative derivative at x = 0 and a positive derivative at x = d. Since dl/dx is continuous over the interval [0, d], by the Intermediate Value Theorem for continuous functions (Section 2.5), there is a point Xo E [0, d] where dl/dx = 0 (Figure 4.38).

218

Chapter 4: Applications of Derivatives

There is only one such point because dt/ dx is an increasing function of x (Exercise 62). At this unique point we then have sin II,

sin 112 C2·

Ct

This equation is SneU's Law or the Law of Refraction, and is an important principle in the theory of optics. It describes the path the ray of light follows. •

Examples from Economics Suppose that r(x) = the revenue from selling x items c (x) = the cost of producing the x items p(x) = r(x) - c(x) = the profit from producing and selling x items. Although x is usually an integer in many applications, we can learn about the behavior of these functions by defining them for all nonzero real numbers and by assuming they are differentiable functions. Economists use the terms marginal revenue, marginal cost, and marginal profit to name the derivatives r'(x), c'(x), and p'(x) of the revenue, cost, and profit functions. Let's consider the relationship of the profit p to these derivatives. If r(x) and c(x) are differentiable for x in some interval of production possibilities, and if p(x) = r(x) - c(x) has a maximum vaIue there, it occurs at a critical point ofp(x) or at an endpoint of the interval. If it occurs at a critical point, then p'(x) = r'(x) c'(x) = 0 and we see that r'(x) = c'(x). In economic terms, this last equation means that

At a production level yielding maximum profit, marginal revenue equals marginal cost (Figure 4.39).

y

I

i Maximum profit, c'{x) = r'(x) I I I

I I

"

I

Local maximum. for loss (minimum profit), c'(x) = ,'(x)

~~ _ _L -______~ ____J-~

o

____________->x

Items produced

FIGURE 4.39 The graph of a typical cost functioo starts coocave down aod larer turns coocave up. It crosses the revenue curve at the break-even point B. To the left of B, the compaoy operates at a loss. To the right, the compaoy operates at a profit, with the maximmn profit occurriug where c' (x) = r' (x). Farther to the right, cost exceeds revenue (perllaps because of a combinatioo of rising labor aod material costs aod market saturatioo) aod productioo levels become unprofitable again.

4.5 Applied Optimization

219

EXAMPLE 5 Supposethatr(x) = 9xandc(x) = x 3 - 6>;2 + 15x, where x represents millions ofMP3 players produced. Is there a production level that maximizes profit? If so, what is it? y

Solution

Notice that r'(x) = 9 and c'(x) = 3x2

-

IZl: + 15.

2

3x - IZl: + 15 = 9 3x 2 -1Zl:+6=0

Setc'(x) ~ r'(x).

The two solutions of the quadratic equation are

x,

= 12 -6

V72

= 2 -

Vz "" 0.586

X2

= 12 +6

V72

= 2 +

Vz "" 3.414.

: Maximum : for profit I I I

Local maximum for loss ~-L~---J----J-~~-----c>x

o 2-v2

2

2+v2

NOT TO SCALB

FIGURE 4.40 The cost and revenue curves for Example 5.

and

The possible production levels for maximum profit are x "" 0.586 million MP3 players or x"" 3.414 million. The second derivative of p(x) = r(x) - c(x) is p"(x) = -c"(x) since r"(x) is everywhere zero. Thus, p"(x) = 6(2 - x), which is negative atx = 2 + and positive at x = 2 By the Second Derivative Test, a maximum profit occurs at about x = 3.414 (where revenue exceeds costs) and maximum loss occurs at about x = 0.586. The graphs of r(x) and c(x) are shown in Figure 4.40. •

Vz

Vz.

Exercises 4.5 Mathematical Applications Whenever you are maximizing or minimizing a function of a single variable, we mge you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your aoswer. 1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 in2 , and what are its dimensions? 2. Show that among all rectangles with an 8-m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. &.

Express the }"<>oordinate of P in terms of x. (Hint: Write an equation for the line AB.)

b. Express the area of the rectangle in terms or.. c. What is the largest area the rectangle can have, and what are its dimensions? y

4. A rectangle has its base on the x-axis and its upper two vertices on the parabola y ~ 12 - x 2 • Whst is the largest area the rectangle can have, and what are its dimensions? 5. You are plauning to make an open rectangular box from an 8-in.by-15-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. Whst are the dimensions of the box of largest volume you can make this way, and wbat is its volume?

6. You are planning to close off a comer of the fIrst quadrant with a line segment 20 units long running from (a, 0) to (0, b). Show that the area of the triangle enclosed by the segment is largest when

a = b. 7. The best fencing plan A rectangular plot offarmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its

dimensions? 8. The .hortest fence A 216 m 2 rectangu1ar pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. Whst dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? 9. Designing. tank Your iron works bas contracted to design and build a 500 ft', square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to f"md dimensions for the base and height that will make the tank weigh as little as possible.

220

Chapter 4: Applications of Derivatives

a. What dimensions do you tell the shop to use? b. Briefly describe how you took weight into account.

oJ

~

u

10. Catching rainwater A 1125 ft3 open-top rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.

'" l3

!3z

I 1

I+-x-+j

I+-x+j T

IT

I

10"

Base

Lid

T

x .i

T

I+-x-+j

x .i

I+-x+j IS"

l-

-I

a. If the total costis

c = 5(x 2

a. Write a formula V(x) for the volume of the box.

+ 4xy) + lOX)!,

b. Find the domain of V for the problem situation and graph V over this domain.

what values of x and y will minimize it?

c. Use a graphical method to find the maximum volume and the value of x that gives it.

b. Give a possible scenario for the cost function in part (a). 11. Designing a poster You are designing a rectangular poster to contain 50 in2 of printing with a 4-in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper used? 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.

d. Confirm your result in part (c) analytically.

D 17.

Designing a suitcase A 24-in.-by-36-in. sheet of cardboard is folded in half to form a 24-in.-by-18-in. rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically. e. Find a value of x that yields a volume of 1120 in 3 . f. Write a paragraph describing the issues that arise in part (b).

13. Two sides of a triangle have lengths a and b, and the angle between them is 8 . What value of 8 will maximize the triangle's area? (Hint: A = O/2)ab sin 8.) 14. Designing a can What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 1000 cm3 ? Compare the result here with the result in Example 2. 15. Designing a can You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be

A = 8r 2 + 21Trh rather than the A = 21Tr2 + 21Trh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now?

D 16.

Designing a box with a lid A piece of cardboard measures 10 in. by 15 in. Two equal squares are removed from the corners of a IO-in. side as shown in the figure. Two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with lid.

I

I I I I I I I I I I

24"

1

x

x

24"

x

x

:

I,

I

)1

36"

~ 18" ------;.I

1

The sheet is then unfolded.

I 1 24"

I,

Base

36"

)1

18. A rectangle is to be inscribed under the arch of the curve y = 4 cos (0.5x) from x = -1T to x = 1T . What are the dimensions of the rectangle with largest area, and what is the largest area?

4.5

Applied Optimization

221

19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume?

cylindrical sidewall. Determine the dimensions to be used if the volume is fIxed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.

20. a. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?

24. The trough in the figure is to be made to the dimensions shown. Only the angle 8 can be varied. What value of 8 will maximize the trough's volume?

Girth = distance

o b. Graph the volume of a 108-in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a). 21. (Continuation ofExercise 20.)

25. Paper folding A rectangular sheet of 8.5-in.-by-II-in. paper is placed on a flat surface. One of the comers is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper. a. Show that L 2 = 2x 3/(2x - 8.5) . b. What value of x minimizes L 2 ?

a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are h by h by wand the girth is 2h + 2w . What dimensions will give the box its largest volume now?

c. What is the minimum value of L?

Girth

Q (originally at A)

h

o b.

Graph the volume as a function of h and compare what you see with your answer in part (a).

22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fIxed. Find the proportions of the window that will admit the most light Neglect the thickness of the frame.

26. Constructing cylinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values ofx andy give the largest volume? b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?

x y I

_-------1--------, /

23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the

(a)

(b)

,

222

Chapter 4: AppLications of Derivatives

27. CoaJtructlDg conel A right triangle whose hypotcuuac ia m long is mvolwd about one of it! less to gcnmate a right circular OODC. Find the radiwJ, height, and vulumc of the OODC of ~t volume that can be made this way.

v'3

• 28. Find the point on the line ~

Building

,

+~=

1 that is closest to the origin

29. Find a positive mnnbcr for which the sum of it and its reciprocal ia the !lll8lJ.e!1: (least) possible. 30. Find a postitive mnnbcr for which the sum of it! reciprocal and four timca it! square is the smallest poYible. 31. A wire b m long is cut into two pieces. One piece is bent into an cquila.tcmJ. triangle and the other is bent into a circle. If the !IUlD. of the areas cnclo!Cd by each part ia a minjrnnm, what is the length of each part? 32.

~

and I in seconds. L

3

34. Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle ofradiWl 3. (See accompanying figure.)

35. What value of a makes f(x} = x'1.

•

I

~

b. a point of inflection at x - I 1

+ a;t2 + bx have

a local maximum at x - - 1 and a local mjnjrnmn at x - 3 1

b. a local minimum. at x = 4 and a point of inflection at x = 11

Physical Applications 37. Ver11c.J. motloa The height above ground of an object moving vertically ia givm by

b. What is the farthest apart that the particles ever get?

c. When in the interval 0

:S t :S 2Tr is the distance between the particles changing the fastest?

41. Projectile motioa The runge R of a projectile flICd from the origin am: horizontal ground is the distance from the origin to the point of impact. If the projectile is flICd with an initial velocity VI) at an angle a with the horizonlal, then in Chapter 13 we find that

vJ.

+ 96t + 112,

o 43. Streqtb. of.

beam. The strength S of a rectangular wooden beam is proportional to it! width timca the square of its depth. (See the accompanying figm:.) L

Find the dimensions of the strongest beam that can be cut from a 12-in..-diamctcr cylindrical log.

b. GraphSas a function of the beam's width w, assuming the proportionality constant to be 1 - 1. Reconcile what you see with your answer in part (a).

c. On the same screen, graph S as a function of the beam's depth d, again taking 1 - 1. Compare the graphs with one mother and with your answer inpart (a). What would be the cifect of chansins to some other value of 11 Try it.

with a in feet and t in seconds. Find L

2Tr do the particles

wbcrc g is the downward acceleration due ttl gravity. Find the angle a for which the range R is the laIjest possible.

a local minimum. at x = 21

a - -161 2

~

m<et1

R = -gsm2a,

+ (a/x) have

36. What values of a and b make f(x} - x 3

At what time(s) in the in1m"Val 0

41. The intensity of illumination at my point from a light source is proportional to the square of the reciprocal of the diIltance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. Haw far from the stronger light is the total illumination least?

33. Determine the dimensions of the rectanglc of ialBCllt area that can be :imcribed in the right triangle shawn in the accompanying; figwe.

L

40. Motioa OD • Hoe The positions of two parti.cJ.es on the s-axis are 61 - sin I and.!"'}. - sin(1 + Tr/3), with 61 and 62 in meterJi

Excrciac 31 if one picoc is bent into a square and the

other into a circle.

L

39. Shortat beam. The 8-ft wall shawn here stands 27 ft from. the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall

the object's velocity when t - 0

b. its maximmn height and when it occurs

c. itsvclocity when" - o. 38. Quickest nnrte lane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point Dearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the leaat amount of time1

12"

d

w

4.5 Applied Optimization

o 44. StHfneu of. beam

The stiffiless S of a rectangular beam is proportional to its width times the cube of its depth. L

Find the dimensions of the stiffest beam. that can be cut from a 12-in.-diameter cylindrical log.

b. Graph S as a function of 1he beam's wid1h w, assuming 1he proportionality constant to be k - 1. Reconcile what you see with your answer in part (a).

e. On 1he same screen, graph S as a fimction of the beam's dep1h d, again taking k - 1. Compare the graphs wi1h one another and wi1h your answer in part (a). What would be the effect of changing to some other value of k? Try it. 45. FrictiODlell urt A small frictionless cart, att:aclled to the wall by a spring. is pulled 10 em. from its rest position and released at time t - 0 to roll back and for1h for 4 sec. Its position at time tis 9 = 1000s"ll'1. L What is the cart's maximum. speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then? b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart's speed then?

~ o

10

223

c. The visibility that day was 5 nauti.cal. miles. Did the ships ever sight each other?

o d. Graphs

and dY/dt together as fimctions oftfor -1 :s t :s 3, using different colors if posSIble. Compare the graphs and reconcile what you see with your answers in parts (b) and (c).

e. The graph of ds/dt looks as ifit might have a horizont:al. asymptote in the rust quadrant. This in turn suggests that ds/dt approaches a limiting value as t - 00. What is this value? What is itll relation to the ships' individual. speeds? 48. Fermat's prlDdple iD optks Light from. a source A is reflected by a plane mirror to a receiver at point B, as shawn in the accompanying fJgUlC. Shaw that for 1he light to obey Fermat's principle, the angle of incidence must cqual the angle of reflection, bo1h measured from. the line normal to 1he reflecting surface. (This result can also be derived without calculus. There is a purely goometric argument, which you may prefer.) N""",,

Angle of

Angle of

incidence

rdlcction

"

"""

B

II,

"

46. 1Wo masses hanging side by side from springs have positions 9l = 2 sin t and 92 = sin 2t, respectively. L At what times in the interval 0 < t dD the masses pass each other? (Hint: sin2t = 2 sin t cos t.) b. When in the interval 0 :s t :s 2'11" is the vertical distance between the masses the greatest? What is this distance? (Hint: cas2t = 2co:flt - 1.)

49. Tin peat When metallic tin is kept below I3.2°C, it slowly becomes brittle and crumbles to a gray powder. 1m objects eventually cnnnble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that control! the rate of reaction without undergoing any permanent change in itself. An autOC4tolytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at Hrst if tile amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abtmdant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate v - dx/dt of the reaction is proportional. both to the amount of the original substance present and to the amount of product. That is, v may be considered to be a fimction of x alone, and v = .b;(a - x) = kin; - kx 2,

, 47. Distanee between two ddpi At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 lcnot:! (nautical miles per hour; a nautical mile is 2000 yd) and continued to dD so all day. Ship B was sailing east at 8 knots and continued to dD so all day. L Start counting time wi1h t = 0 at noon and express the distance 9 between the ships as a fimction oft. b. How rapidly was the distance between the ships changing at noon? One hour later?

wh...

x - the amount of product

a

= the amount of substance at the beginning

k = a positive constant. At what value of x does the rate v have a maximum? What is the maxinnnn value of v? SO. Airplane landing path An airplane is flying at altitude H wilen it begins Us descent to an aiIport runway that is at horizootaJ. ground distance L from the airplane, as shown in the fJgL1IC. Assume that 1he

224

Chapter 4: Applications of Derivatives

landing path of the airplane is the graph of a cubic polyn<>mial fimctioo y ~ ax' + bx2 + ex + d, where y ( - L) ~ H and

y(O) L

~

O.

55. Showthatifr(x) ~ 6xande(x) ~

Whatisdy/dxatx ~ O?

b. What is dy/dx atx

~

~

Oandy(-L)

~

x' - 6x

2

+

15xareyourrev-

enue and cost functions, then the best you can do is break. even

-L?

c. Use the values forth>/dx atx y(O)

54. Production level Prove that the productioo level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost

~

oand x

~

-Ltogetherwith

Htoshowthat

(have revenue equal cost). 56. Production level Suppose that e(x) ~ x' - 2Ox 2 + 20,00Ox is the cost of manufacturing x iterus. Find a productioo level that

will minimize the average cost of making x items. 57. You are to construct an open rectangular box with a square base and a volume of48 ft'. Ifmaterial for the bottom costs $6/ft' and materisl for the sides costs S4/ft', what dimensions will result in

y

Landing path

the least expensive box? What is the minimum cost? 58. The 800-room Mega Motel chain is filled to capacity when the room charge is $50 per night. For each $10 increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?

H = Cruising altitude

Airport

~I'-------------L------------~

Business and Economics 51. It costs you c dollars each to manufacture and distribute backpacks. If the backpacks sell at x dollars each, the number sold is given by

n~ x ~ e

+ b(IOO

Biology 59. Sensitivity to medicine (Continuation of Exercise 60, Section 3.3.) Find the amount of medicine to which the body is most sensitive by f'mding the value of M that maximizes the derivative dR/ dM, where

- x),

where a and b are positive constants. What selling price will bring a maximum profit?

and C is a constant 60. How we cough

52. You operate a tour service that offers the following rates:

.. When we cough, the trachea (windpipe) contracts to

$200 per persoo if 50 people (the minimum number to book the tour) go on the tour.

increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the

For each additiooal persoo, up to a maximum of 80 people total, the rate per petliOO is reduced by $2.

velocity and whether it really cootraets that much when we cough.

Under reasonable assumptions about the elasticity of the

It costs $6000 (a fixed cost) plus $32 per petlion to cooduct the tour. How many people does it take to maximize your profit?

tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity v can be modeled by the

53. Wilson lot size formula One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is

A(q) ~

km

equation

v ~ c(ro - r)r 2 em/sec,

hq

placing an order (the same, no matter how often you order), c is

L

Your job, as the inventory manager for your store, is to fmd the quantity that will minimize A(q). What is it? (The fonnula you get for the answer is called the Wilson lot size formula.)

b. Shipping costs sometimes depend 00 order size. When they do, it is more realistic to replace k by k + bq, the sum of k and a constant multiple of q. What is the most economical quantity to order now?

70,

length of the trachea. Show that v is greatest when r ~ (2/3)ro; that is, when the trachea is about 33% cootraeted. The remarkable fact is that X-ray photographs conf'mn that the trachea contracts about this much during a cough.

where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), k is the cost of

(a constant that takes into account things such as space, utilities, insurance, and security).

<

7 -

where 70 is the rest radius of the trachea in centimeters and c is a positive constant whose value depends in part on the

q + em + T'

the cost of ooe item (a constant), m is the number of items sold each week (a constant), and h is the weeldy holding cost per item

70 < 2: -

D

h. Take ro to be 0.5 and c to be I and graph v over the interval S r '" 0.5. Compare what you see with the claim that v is at a maximum when r ~ (2/3 )ro.

o

Theory and Examples 61. An inequality for positive integers are positive integers, then

(a 2

Show that if a, b, c, and d

+ l)(b 2 + 1)«2 + l)(d 2 + abed

I) > 16.

4.6

62. The derivative dt/dx in Eumple 4

f(x)=~ a +x

o

is an increasing function of x. h. Showthat

o

is a decreasing function of x.

c. Showthat

67••• How close does the curve y = Vx come to the point (3/2, OJ? (Hint: If you minimize the square of the distance, you can avoid square roots.) II- Graph the distance function D(x) and y = Vx together and reconcile what you see with your answer io part (a).

d-x

x

dx

c1v'a'+x'

o

is an increasing function of x.

63. Let f(x) and g(x) be the differentiable functions graphed here. Point c is the point where the vertical distance between the curves is the greatest Is there anything special about the tangents to the two curves at c? Give reasons for your answer.

,, , y = 'g(x) ,, ,, ,, I

a

c

y

o

)X

b

64. You have been asked to determine whether the function f(x) 3 + 4 cos x + cos 2x is ever negative.

4.6

225

•• Explain why you need to considet values of x only io the ioterval [0, 2".]. II- Is f ever negative? Explain. 65. •• The function y = cot x - v'2 csc x has an absolute maximum value on the interval 0 < x < '11'. Find it. II- Graph the function and compare what you see with your answer in part (a). 66. a. The function y = tanx + 3 cotx has an absolute minimum value on the ioterval 0 < x < "./2. Find it. II- Graph the function and compare what you see with your answer in part (a).

a. Showthat

dt

Newton's Method

=

G,o)

68. •• How close does the semicircle y = ~ come to the point (I, \1'3)? 0 11- Graph the distance function andy = v'16 - x'togetherand reconcile what you see with your answer in part (a).

Newton's Method In this section we study a numerical method, called Newton:' method or the Newton-Raphson method, which is a technique to approximate the solution to an equation f(x) = O. Essentially it uses tangent lines in place of the graph of y = f(x) near the points where f is zero. (A value of x where f is zero is a root of the function f and a solution of the equation f(x) = 0.)

Procedure for Newton's Method The goal of Newton's method for estimating a solution of an equation f(x) = 0 is to produce a sequence of approximations that approach the solution. We pick the first number Xo of the sequence. Then, under favorable circumstances, the method does the rest by moving step by step toward a point where the graph of f crosses the x-axis (Figure 4.41). At each step the method approximates a zero of f with a zero of one of its linearizations. Here is how it works. The initial estimate, xo, may be found by graphing or just plain guessing. The method then uses the tangent to the curve y = f(x) at (xo, f(xo» to approximate the curve, calling

226

Chapter 4: Applications of Derivatives the point Xl where the tangent meets the x-axis (Figure 4.41). The nwnber Xl is usually a better approximation to the solution than is xo. The point X2 where the tangent to the curve at (Xl. f(xll) crosses the x-axis is the next approximation in the sequence. We continue on, using each approximation to generate the next, until we are close enough to the root to stop. We can derive a fonnula for generating the successive approximations in the following way. Given the approximation x., the point-slope equation for the tangent to the curve at (x., f(x.)) is

y

y = f(x.)

+ t'(x.)(x - x.).

We can fmd where it crosses the x-axis by setting y = 0 (Figure 4.42):

o=

Root

sought '\

X, Second

f(x.) + t'(x.)(x - x.)

f(x.) - t'(x.)

=

x-x.

x

=

f(x.) x. - t'(x.)

First

APPROXIMATIONS

FIGURE 4.41 Newton's method starts with an initial guess xo and (under favorable circumstances) improves the guess one step at a time.

If f'(x.) " 0

TIris value of x is the next approximation x.+ 1. Here is a swnmary of Newton's method.

Newton's Method y Point:

1. Guess a flISt approximation to a solution of the equation f(x) = O. A graph of y = f(x) may help.

Vc.,f(x,,»

2. Use the first approximation to get a second, the second to get a third, and so on, using the fonnula

Slope:f'(x,,) Tangent line equation: y - f(x.) ~ f'(x.)(x - x.) Tangent line (graph of

X.+l = X. -

f(x.) t'(x.) ,

if t'(x.) "" O.

(I)

linearization offatxn)

Applying Newton's Method

FIGURE 4.42 The geometry of the successive steps of Newton's method. From x" we go up to the curve and follow the tangent line down to fmdx.+I.

Applications of Newton's method generally involve many nwnerical computations, making them well suited for computers or calculators. Nevertheless, even when the calculations are done by hand (which may be very tedious), they give a powerful way to fmd solutions of equations. In our first example, we find decimal approximations to V2 by estimating the positiverootoftheequationf(x) = x 2 - 2 = O.

EXAMPLE 1

Find the positive root of the equation

f(x) Solution

Withf(x)

=

x2

-

=

x2

-

2 = O.

2andt'(x) = 2x,Equation (I) becomes

x. I =x--+11 2 x"

x.

=

I

2 + x.'

4.6 y

Newton's Method

227

The equation Xn+l =

x.

2 +

I xn

enables us to go from each approximation to the next with just a few keystrokes. With the starting value Xo = 1, we get the results in the first column of the following table. (To five decimal places, = 1.41421.)

v'2

Error

Xo = 1 XI = 1.5

FIGURE 4.43 The graph of f(x) = - x-I crosses the x-axis once; this is the root we want to rmd (Example 2).

x3

X2

= 1.41667

X3

= 1.41422

Number of correct digits

-0.41421 0.08579 0.00246 0.00001

1 1 3 5

-

Newton's method is the method used by most calculators to calculate roots because it converges so fast (more about this later). If the arithmetic in the table in Example 1 had been carried to 13 decimal places instead of 5, then going one step further would have correctly to more than 10 decimal places. given

v'2

(1.5,0.875)

Find the x-coordinate of the point where the curve y = x 3 the horizontal line y = I.

EXAMPLE 2

Root sought

~.

Xo

----~----~~~~----__>x

1.3478

FIGURE 4.44 The rust three x-values in Table 4.1 (four decimal places).

25 20

15

10

x erosses

Solutton The curve erosses the line when x 3 - x = 1 or x 3 - x - I = O. When does f(x) = x 3 - x-I equal zero? Since f(l) = -I and f(2) = 5, we know by the Intermediate Value Theorem there is a root in the interval (1, 2) (Figure 4.43). We apply Newton's method to j with the startiog value Xo = 1. The results are displayed in Table 4.1 and Figure 4.44. At n = 5, we come to the result X6 = x, = 1.324717957. When X.+l = x., Equation(I)showsthatj(x.) = O. We have found a solution ofj(x) = 0 toninedecimals. _

TABLE 4.1 The resuLt of appLying Newton's method to f(x) = y

-

with

Xo =

x' -

x - 1

X.+1

= X.

1

n

x.

f(x.)

0 1 2 3 4 5

1

-1

2

1.5 1.3478 26087 1.3252 00399 1.324718174 1.324717957

0.875 0.100682173 0.0020 58362 0.000000924 - 1.8672E-13

5.75 4.4499 05482 4.2684 68292 4.264634722 4.2646 32999

t'(x.)

-

j(x.) t'(x.)

1.5 1.3478 26087 1.3252 00399 1.324718174 1.324717957 1.324717957

5

-1/V3

FIGURE 4.45 Any starting value Xo to the = 1/\13 will lead to the root

right ofx

In Figure 4.45 we have indicated that the process in Example 2 might have started at the point Bo(3, 23) on the curve, with Xo = 3. Point Bo is quite far from the x-axis, but the tangent at Bo crosses the x-axis at about (2.12, 0), so XI is still an improvement over Xo. If we use Equation (1) repeatedly as before, with j(x) = x 3 - x - I and f'(x) = 3x 2 - 1, we obtain the nine-place solution X7 = X6 = 1.3247 17957 in seven steps.

228

Chapter 4: Applications of Derivatives

Convergence of the Approximations In Chapter 10 we define precisely the idea of convergence for the approximations x. in Newton's method. Intuitively, we mean that as the mnnber n of approximations increases without bowd, the values x. get arlJitrarily close to the desired root r. (Ibis notion is similar to the idea of the limit ofa functiong(t) as t approaches inf"mity, as defmed in Section 2.6.) In pmctice, Newton's method usually gives convergence with impressive speed, but this is not guaranteed. One way to test convergence is to begin by grapbing the function to estimate a good starting value for Xo. You can test that you are getting closer to a zero of the function by evaluating If(x.) I, and check that the approximations are converging by evaluating Ix. - x.+ll. Newton ~ method does not always converge. For instance, if

o

f(x) = FIGURE 4.46 Newton's method fails to converge. You go from Xo to x, and back to x" never getting any closer to r.

{-~' ~,

x


x~r,

the gmph will be like the one in Figure 4.46. If we begin with Xo = r - h, we get Xl = r + h, and successive approximations go back and forth between these two values. No amount of iteration brings us closer to the root than our flISt guess.

If Newton 50 method does converge, it converges to a root. Be careful, however. There are situations in which the method appears to converge but there is no root there. Fortunately, such situations are rare.

When Newton 50 method converges to a root, it may not be the root you have in mind. Figure 4.47 shows two ways this can happen.

Root foond

\

----~t_---4v-~~~~~--~x

--~~~~~----__~~--~X

xo" Starting

point

FIGURE 4.47

If you start too far away, Newton's method may miss the root you want.

Exercises 4.6 Root Finding 1. Use Newton's method to estimate the solutions of the equation x 2 + x - I = O. Start with x, = -1 for the left·band solution and with Xo = I for the solution on the right. Then, in each case, f'm dx2_

6. Use Newton's method to tmd the negative fourth root of2 by solv· ing the equation x' - 2 = o. Start with Xo = -I and tmd X2.

2. Use Newton's method to estimate the one real solution of x' + 3x + I = O.Startwithxo = Oandthentmdx2.

8. Estimating pi You plan to estimate ",/2 to five decinoal places by using Newton's method to solve the equation cosx = O. Does it matter what your starting value is? Give reasons for your answer.

3. Use Newton's method to estimate the two zeros of the function f(xl = x' + x - 3. Start with Xo = -I for the left·band zero and with Xo = 1 for the zero on the right. Then, in each case, f'm dx2·

4. Use Newton's method to estimate the two zeros of the function f(x l = 2x - x 2 + 1. Start with Xo = 0 for the left-band zero and with Xo = 2 for the zero on the right. Then, in each case, tmd X2. S. Use Newton's method to tmd the positive fourth root of2 by solv· ing the equation x' - 2 = O. Start with Xo = 1 and tmd X2.

7. Guessing a root Suppose that your tlISt guess is lucky, in the sense that Xo is a root of j(xl = O. Assmning that f' (xol is de· tmed and not 0, what happens to XI and later approximations?

Theory and Examples 9. Oscillation Show that ifh

f(xl =

> 0, applying Newton's method to

Yx'

x;;'O

{ ~,

x
leads to x, = -h ifxo = h and to x, = h ifxo picture that shows what is going on.

= -h.

Draw a

4.6

10. Approximations that get worse and worse Apply Newton's method to f(x) = x lj3 wilhxo = I and calculate XI , x2, x" andx,. Find a fonnula for Ix. I . What happeos to Ix. I as n .... oo? Draw a picture iliat shows what is going 00. 11. Explain why Ihe followiog four statemeots ask for Ihe same infor-

mation: i) Find Ihe roots of f(x)

= x'

- 3x - I.

Vx

22. The graphs of y = and y = 3 - x 2 intersect at one point x = r. Use Newton's method to estimate the value of r to four decimal places. 23. Use the intermediate Value Theorem from Section 2.5 to show that f(x) = x' + 2x - 4 has a root hetweeo x = I and x = 2. Theo fmd the root to five decimal places. 24. Factoring a quartic

ii) Find Ihe x-coordinates of Ihe intersections oflhe curve y = x'wilhtheliney = 3x + I.

229

Newton's Method

ax'

iiI") Find the x-coordinates of the points where the curve y = x' - 3x crosses the hnrizonts1line y = 1.

Find the approximate values of rl 1hrough

in the factorization - 14x' - 9x 2 + llx - I y

y=

= 8(x -

r,

rl)(x - r2)(x - r,)(x - r,).

8x" -14x' -

9x'+ 11x-1

iv) Find the values ofx where the derivative ofg(x) = (1/4)x' - (3/2)x 2 - X + 5 equals zero. 12. Locating a planet To calculate a planet's space coordinates, we have to solve equatioos like x = I + 0.5 sin x . Graphing the function f(x) = x - I - 0.5 sinx suggests iliat the function has a root near x = 1.5. Use one application of Newton's method to improve this estimate. That is, stsrt with Xo = 1.5 and fmd XI . (The value of the root is 1.49870 to five decimal places.) Remem-

ber to use radians.

D 13. D 14. D 15.

Intersecting curves The curve y = tan x crosses Ihe line y = 2x hetweeo x = 0 and x = 'IT/2. Use Newton's method to fmd where.

/7)

b. Xo = -0.5 and Xo = 0.25, lying in ( - v21/7, v21

o.

c. Xo = 0.8andxo = 2, 1yingin(v2/2,00)

a. How many solutions does the equation sin 3x = 0.99 - x 2 have?

a. Does cos 3x ever equal x? Give reasons for your answer. b. Use Newton's method to fmd where.

+

d. Xo = - v21/7 and Xo = v21/7 26. The .onobuoy problem in submarine location problems, it is often necessary to fmd a submarine's closest point of approach (CPA) to a sonobuny (souod detector) in the water. Suppose iliat the submarine travels on the parabolic palh y = x 2 and iliat the buoy is located at the point (2, -1/2).

16. Intersection of curves

17. Findthefourrealzerosofthefunctionf(x) = 2x' - 4x 2

a. Show that the value of x iliat minimizes the distance between the submarine and the buoy is a solution of the equation x = 1/(x 2 + I). b. Solve the equation x = 1/(x 2 + I) with Newton's method.

1.

Estimating pi Estimate 'IT to as many decimal places as your calculator will display hy using Newton's method to solve the equation tan x = Owithxo = 3.

19. Intersection of curves

Converging to different zero. Use Newton's method to fmd the zeros of f(x) = 4x' - 4x 2 using the giveo starting values. a. Xo = -2andxo = -0.8,lyingin(-00,-v2/2)

Real solutions of a quartic Use Newton's method to fmd the !we real solutioos of the equation x' - 2x' - x 2 - 2x + 2 =

b. Use Newton's method to fmd them.

D 18.

D 25.

y

At what value(s) ofx does cosx = 2x?

20. Intersection of curves At what value(s) ofx does cosx = -x? 21. The graphs of y = x 2(x + I) and y = I/x (x > 0) intersect at one point x = r. Use Newton's method to estimate the value of r to four decimal places.

in two dimensions

------~~~--~~--~--~x

o

D 27.

"

2

SOOObu~"(2, -!)

Curve. that are nearly flat at the root Some curves are so flat that, in practice, Newton's method stops too far from the root to give a useful estimate. Try Newton's method on f(x) = (x - 1)'0 with a starting value of .:to = 2 to see how close your machine comes to the root x = 1. See the accompanying graph.

230

Chapter 4: Applications of Derivatives y

28. The accompanying figure shows a circle of radius r with a chord ofleogth 2 aod ao arc S ofleogth 3. Use Newton's method to solve for r aod 8 (radians) to four decimal places. Assume 0 < 6 < TT.

y ~ (x - I)'"

Slope ~-4O

Slope

I

(2, I)

~

40

Nearly flat

o

4.7

I

2

Antiderivatives We have studied how to find the derivative of a function. However, many problems require that we recover a function from its known derivative (from its knownmte of change). For instance, we may know the velocity function of an object falling from an initial height and need to know its height at any time. More generally, we want to fmd a function F from its derivative f. If such a function F exists, it is called an antiderivative of f. We will see in the next chapter that antiderivatives are the link connecting the two major elements of calculus: derivatives and defmite integrals.

Finding Antiderivatives DEFINITION F'(x)

=

A function F is an antiderivative of

f

on an interval I if

f(x) forallxinI.

The process of recovering a function F(x) from its derivative f(x) is called antidifferentiation. We use capital letters such as F to represent an antiderivative of a function f, G to represent an antiderivative of g, and 80 forth.

EXAMPLE 1 (a) f(x) = 2x

Find an antiderivative for each of the following functions. (b) g(x) = cosx

(0) h(x) =

2x + cosx

Solution

We need to think backward here: What function do we know has a derivative equal to the given function?

(a) F(x) = x 2

(b) G(x) = sinx

(0)

H(x)

=

x 2 + sinx

Each answer can be checked by differentiating. The derivative of F(x) = x 2 is 2x. The derivative of G(x) = sinx is cosx and the derivative of H(x) = x 2 + sinx is 2x + cosx. •

4.7 Antiderivatives

231

The function F(x) = x 2 is not the only function whose derivative is 2x. The function x + I has the same derivative. So does x 2 + C for any constant C. Are there others? Corollary 2 of the Mean Value Theorem in Section 4.2 gives the answer: Any two antiderivatives of a function differ by a constant. So the functions x 2 + C, where C is an arbitrary constant, form all the antiderivatives of f(x) = 2x. More generally, we have the following result. 2

THEOREM 6

If Fis an antiderivative off on an interval I, then the most general antiderivative off on I is F(x)

+C

where C is an arbitrnry constant.

Thus the most general antiderivative of f on I is a family of functions F(x) + C whose graphs are vertical translations of one another. We can select a particular antiderivative from this family by assigning a specific value to C. Here is an example showing how such an assignment might be made.

EXAMPLE 2

y

C~2

Solution

Find an antiderivative of f(x) = 31;2 that satisfies F(I) = -1.

Since the derivative of x' is 3x 2 , the general antiderivative

C~l

F(x) = x'

C~O C~-l

..--------C=-2

+

C

gives all the antiderivatives of f(x). The condition F( I) = - I detennines a specific value for C. Substituting x = I into F(x) = x 3 + C gives

F(I) = (1)3 + C = I + SinceF(I) = -I, solving I

+

C = -I for C gives C = -2. So

F(x) = x 3

FIGURE 4.48 The curves y ~ x' + C fill the coordinate plane without overlapping. In Example 2, we identify the curve y = x 3 - 2 as the one that passes through the given point (I, -I).

c.

-

2

is the antiderivative satisfying F( I) = - I. Notice that this assignment for C selects the particular curve from the family of curves y = x 3 + C that passes through the point (I, -I) in the plane (Fignre 4.48). • By working backward from assorted differentiation rules, we can derive formulas and rules for antiderivatives. In each case there is an arbitrnry constant C in the general expression representing all antiderivatives of a given function. Table 4.2 gives antiderivative formulas for a number of important functions. The rules in Table 4.2 are easily verified by differentiating the general antiderivative formula to obtain the function to its left. For example, the derivative of (tan lex)/! + Cis sec2 la, whatever the value of the constants Cork '" 0, and this establishes Formula 4 for the most general antiderivative of sec2 lex.

EXAMPLE 3 (8) f(x) = XS

Find the general antiderivative of each of the following functions. (b) g(x) =

I Vx

(0) h(x) = sin 2x

(d) i(x) = cos ~

232

Chapter 4: Applications of Derivatives

TABLE 4.2 Antiderivative formulas, k a nonzero constant Function

General antiderivative

1.

x'

_1_ x ·+ 1 n+l

+C n '

2.

sinkx

1 --cos kx k

+C

3.

coskx

tsinkx

+C

4.

sec' kx

ttankx

+C

oF -I

Function

General antiderivative

5.

csc2 kx

1 --cotkx k

6.

sec kxtan kx

k seckx + C

7.

esc kx cot kx

1 --esc kx k

1

6

F(x)

+C

In each case, we can use one of the formulas listed in Table 4.2.

Solution (a)

+C

=

6x +

Formula 1 withn=5

C

(b) g(x) =x-1/2 ,so Xl/2

G(x) = 1/2 (0)

H(x)

-cos 2x 2

=

(d) f(x) =

+C

sin (x/2) 1/2

•

=

2

V

r

x

Formula 1 with. ~ -1/2

+C

Formula 2

+C

+

withk~2

. x C = 2 sm 2

+

Formula 3 withk ~ 1/2

C

•

Other derivative rules also lead to corresponding antiderivative rules. We can add and subtract antiderivatives and multiply them by constants.

TABLE 4.3 Antiderivative linearity rules

1. 2. 3.

Constant Multiple Rule: Negative Rule: Sum or Difference Rule:

Function

General antiderivative

kf(x)

kF(x)

-fix) fix) ± g(x)

+ C, ka constant -F(x) + C F(x) ± G(x) + C

The formulas in Table 4.3 are easily proved by differentiating the antiderivatives and verifying that the result agrees with the original function. Formula 2 is the special case k = - 1 in Formula 1.

EXAMPLE 4

Find the general antiderivative of fix) =

~ + sin2x.

Solution We have that fix) = 3g(x) + h(x) for the functions g and h in Example 3. Since G(x) = 2Vx is an antiderivative of g(x) from Example 3b, it follows from the

4.7

Antiderivatives

233

Constant Multiple Rule for antiderivatives that 3G(x) = 3· 2Vx = 6Vx is an antiderivativeof3g(x) = 3/Vx. Likewise, from Example 3c we know that H(x) = (-1/2) cos2x is an antiderivative of h(x) = sin 2x. From the Sum Rule for antiderivatives, we then get that

F(x)

=

3G{x)

+ H(x) + C

~cos2x + C

= 6Vx -

is the general antiderivative fannula for f(x), where C is an arbitrary constant.

•

Initial Value Problems and Differential Equations Antiderivatives play several important roles in mathematics and its applications. Methods and techniques for rmding them are a major part of calculus, and we take up that study in Chapter 8. Finding an antiderivative for a function f(x) is the same problem as finding a functiony(x) that satisfies the equation

dy

dx

=

f(x).

This is called a differential equation, since it is an equation involving an uuknown function y that is being differentiated To solve it, we need a function y(x) that satisfies the equation. This function is found by taking the antiderivative of f(x). We fix the arbitrary constant arising in the antidifferentiation process by specifying an initial condition

y(xo)

=

Yo.

This condition means the function y(x) has the value Yo when x = Xo. The combination of a differential equation and an initial condition is called an initial value problem. Such problems play important roles in all branches of science. The most general antiderivative F(x) + C (such as x 3 + C in Example 2) of the function f(x) gives the general solution y = F(x) + C of the differential equation dy/ dx = f(x). The general solution gives all the solutions of the equation (there are infinitely many, one fur each value of C). We solve the differential equation by finding its general solution. We then solve the initial value problem by rmding the partieulsr solution that satisfies the initial condition y(xo) = Yo. In Example 2, the function y = x 3 - 2 is the particular solution of the differential equationdy/dx = 3x2 satisfYing the initial condition y(l) = -l.

Antiderivatives and Motion We have seen that the derivative of the position function of an object gives its velocity, and the derivative of its velocity function gives its acceleration. Ifwe know an object's acceleration, then by finding an antiderivative we can recover the velocity, and from an antiderivative of the velocity we can recover its position function. This procedure was used as an application of Corollary 2 in Section 4.2. Now that we have a terminology and conceptual framework in terms of antiderivatives, we revisit the problem from the point of view of differential equations.

EXAMPLE 5 A hot-air balloon ascending at the rate of 12 ft/sec is at a height 80 ft above the ground when a package is dropped How long does it take the package tu reach the ground? Solution Let v(l) denote the velocity of the package at time I, and let s(l) denote its height above the ground The acceleration of gravity near the surface of the earth is 32 ft/ sec'. Assuming no other forces act on the dropped package, we have

dv dt

= -32

.

Negative because gravity acts in the direction of decreasing s

234

Chapter 4: AppLications of Derivatives This leads to the following initial value problem (Figure 4.49):


Differential equation:

dt

.(0) - 12

Initial condition:

Billoon initially rising

This is our mathematical model for the package's motion. We solve the initial value problem to obtain the velocity of the package.

1.

Solve lhe differential equation: The general formula for an antidcrivative of -32 is

1~=-32

v - -321

+ C.

Having found the general solution of the differential equation, we UBe the initial condition to find the particular solution that solves our problem.

0(0

2. Evaluate C:

o """"" FIGURE 4.49 A package dropped from a rising hot-air balloon

_1,5).

12

~

-32(0)

C

~

12.

+

C

Initial condition v(O)

=

12

The solution of the initial value problem is

v = -321

+ 12.

Since velocity is the derivative ofheight, and the height of the package is 80 ft at time

t = 0 when it is dropped, we now have a second initial value problem. Di./forential equation: Initial condition:

:

= -321

+ 12

Setv

=

d&jdt in ttre

PRVious equation.

s( 0) - 80

We solve this initial value problem to fmd the height as a function of I.

1.

Solve lhe difforential equation: Finding the general antiderivative of - 32t

+ 12 gives

2

s - -161 + 121 + C. 2.

Evaluate C: 80 - -16(0)' C

~

+ 12(0) + C

Initial cmditioD.l(O) - 80

80.

The package's height above ground at time I is

s

=

-16t 2

+ 12t + 80.

Use lhe solution: To fmll how long it takes the package to reach the ground, we set s equal to 0 and solve for t:

-161 2 + 121+80 - 0 -4/2 +3t+20=0 -3 ± v329 t8 t~-1.89, t R:j 2.64. The package hits the ground about 2.64 sec after it is dropped frmn the balloon. (The negative root has no physical meaning.) •

4.7 Antiderivatives

235

Indefinite Integrals A special symbol is used to denote the collection of all antiderivatives of a function j.

DEFINmON The collection of all antiderivatives of f is called the indefinite integral of f with respect to x, and is denoted by J f(x)dx.

J

The symbol is an integral sign. The function gral, and x is the variable of integration.

f

is the integrand of the inte-

After the integral sign in the notation we just defined, the integrand function is always followed by a differential to indicate the variable of integration. We will have more to say about why this is important in Chapter 5. Using this notation, we restate the solutions of Example I, as follows:

2 2xdx=x +C,

J

J J

cosxdx = sinx

(2x

+ cos x) dx

+ c, =

x2

+ sinx +

C.

This notation is related to the main application of antiderivatives, which will be explored in Chapter 5. Antiderivatives play a key role in computing limits of certain infinite sums, an unexpected and wonderfully useful role that is described in a central result of Chapter 5, called the Fundamental Theorem of Calculus.

EXAMPLE 6

Evaluate

J

(x 2

-

2x + 5)dx.

Solution If we recognize that (x 3/3) - x 2 + 5x is an antiderivative of x 2 - 2x we can evaluate the integral as

+

5,

antiderivative

J

(x 2 - 2x

x3

+ 5)dx = 3 -

x 2 + 5x

+ £.

arbitrary constant

If we do not recognize the antiderivative right away, we can generate it term-by-term with the Sum, Difference, and Constant Multiple Rules:

2 J (x - 2x

+ 5) dx

=

2 J x dx - J 2x dx

+ J 5 dx

2 = JX dx-2JXdx+5Jldx =

(;3 +

=

x 3 + C, - x 2 - 2C2 + 5x + 5C3.

3

C,) _ 2 (x;

+ C2) +

5(x

+ C3)

236

Chapter 4: Applications of Derivatives This formula is more complicated than it needs to be. Ifwe combine C" -2C" and 5C, into a single arbitrnry constant C = C, - 2C, + 5C" the formula simplifies to

x'

--x'+5x+C 3 and still gives all the possible antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate term-by-term. Write

J

(x' - 2x + 5) dx

= =

J J J x' dx -

2x dx

+

5 dx

x' - x' + 5x + C. 3

Find the simplest antiderivative you can for each part and add the arbitrnry constant of integration at the end _

Exercises 4.7 finding Antlderlvatlves

finding Indefinite Integrals

Io Exercises 1-16, f"md an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

Io Exercises 17-54, find the most general antiderivative or indeflnite integral. Check your answers by differentiation.

1.L2x 2. L 3. 4. 5.

L

c.x'-2x+1 c. 7 + 8

b.x'

6x -3x-4

b. b.

-,

L

2x-'

b. ~

L

1 2

b.~

5 c. 2- 2

b. _1_

,

X

2

+x 2

C.

x'

2 6. L - -

x'

C.

2x'

-x-3 + x-I

x

~Vx

4~ 8. L 3

b. _1_

9.

b. 1x-2/3

c. -tx-4/3

x

2 -1/3 L}X

2Vx

~

3~ 3

-~x-5/2

L

12x -1/2

b. _lx-'/2

c.

11.

L

-w sin '11'X

b. 3 sinx

c. sin 71'X

2

b. 1rcos

c.

x

c. -see'

L

'T1'

13.

L

see'x

2 e' b. -se

14. L ese'x

b. -lese'

15.

L

cscxcotx

b. -esc 5x cot 5x

16.

L

secxtanx

b. 4 sec 3x tan

3

2

3

2

3x2

3 sin 3x

-

X COST + wooSX' 3x2

c. 1-8ese'2x

3x

c.

lx-II' dx

18./(5-6x)dx 20. 1 (~ + 4t') dt 22. l(i -

x' - 3x')dx

24. 1 (t -:' + 2x) dx 26. lx- 51' dx

27. 1 (Vx + ~) dx

28. I(Yf + ~)dx

29. 1(8Y - )I')dy 31. 12x(i - x-') dx

30. I(t- )I')dy 32. Ix-'(x+ l)dx

33. 1 tv,;t; V,dt

34. 1 4 ~,v, dt

'If

'T1'X

12.

2

25.

c. Vx+ _1_ Vx c. ~+_I-

10.

cos '7TX

21. 1(2x' - 5x + 7)dx 23·/8,-x2-t)dx

x 1 -3 x

b. _1_

7. L

17.1 (x+ I)dx 19. 1 (3t 2+ t) dt

x 6x c. x-4 + 2x + 3

x x-4 7

-'1r

esc 1TX cot 1TX

2

'1fX

X

'If

c. secTtanT

2

35.

1(-2 cost) dt

36.

1(-5 sint) dt

37.

17Sin~ dO

38.

13 cos 50 dO

39·/(-3eSC'x)dx 41. 1 esc °2eot 0dO

40. 1(_~2X)dx 42. 1 ~seeOtanOdO

4.7

43. 1(4secxtmx - 2sec'x)dx 44.1 k(csc'x - cscxcotx)dx

b. 13(2x + I)' dx = (2x + I)' + C

45. 1 (sin 2x - esc' x) dx

c. 16(2x + I)' dx = (2x + I)' + C

47. 49.

1 l(i 1 1

1

+ ~oS4t dt

48·/l-~s6tdt

+ tm'B) dB

50./(2

(Hint: 1

51.

+ tm' B =

+ tm'B) dB

sec' B)

52.1(1 - cot?-x)dx

cot?-xdx

(Hint: 1

53.

46. 1(2 cos 2x - 3 sin 3x) dx

+ cot?- x = esc' x)

cosB(tmB

+ sec B) dB

1

54.

55.

(7x - 2) dx

=

(7x - 2)4

28

(3x

59. 60.

(x ~

esc'

+ 5)-1

I) dx = -3 cot

I

•.

I~dx= Yx'+x+

b.

l~dx=W+x+C

c.

I~dx=t(~)'+c

1_ 1_

-.::.:15",(x,--+---,:"3)"--' dx= (X_2)4

(x-x-2 +3)' +C

66. Right, or wrong? Give a brief reason why.

x_co_s_<,-x-,')-c-c-s_in_<,-x-,-') _ sin (x') , dxx +C

+C

(x ~

C

65. Right, or wrong? Give a brief reason why.

57.1 sec'(5x - I)dx = ttm(5x - I) + C

1 1 1

son for each answer.

+C

3

58.

64. Right, or wrong? Say which for each formula and give a brief rea-

cscB. dB cscO - 8m8

Checking Antiderivative Formulas VerifY the formulas in Exercises 55-j)O by differentiation.

' 1

237

Antiderivatives

x

Initial Yalue Problems 67. Which of the following graphs shows the solution of the initial value problem

dy

I)

dx=2x, y=4whenx=l?

+C y

I

y

(x + I)' dx = - x + I + C

I x (x+I),dx=x+I+C

4

61. Right, or wrong? Say which for each formula and give a brief reason for each answer.

a. IXSinXdx =

~2 sinx + C

(1,4)

3 2

---c'---:'t'-....L~x

-1 0

b. IXSinXdx=-XCOSX+C

c. IXSinXdx

=

-xcosx

1

(a)

+ sinx + C

62. Right, or wrong? Say which for each formula and give a brief rea-

-'--c+....L~x

-1 0

(b)

(c)

Give reasons for your answer. 68. Which of the following graphs shows the solution of the initial value problem

son for each answer.

a. b. c.

1 1 1

tmBsec'BdB =

tmBsec'BdB = ktm'B

+C

tmBsec'BdB = ksec'B

+C

63. Right, or wrong? Say which for each formula and give a brief reason for each answer.

a.

1

(2x + I)' dx =

dy

~'B + C

(2x

+ I)' 3

+C

dx = -x, Y = I whenx = -I? y

,~ . o

(0)

y

x

x

(b)

Give reasons for your answer.

H y y

o

(c)

x

238

Chapter 4: Applications of Derivatives

Solve the initial value problems in Exercises 69-1!8.

Solution (Integral) Curves

dy 69. dx = 2x - 7,

Exercises 91-94 show solution curves of differential eqoations. 10 each exercise, fmd an equation for the curve through the labeled point

y(2) = 0

91.

dy

dJ.. ~ 1 _ .! ,113

y

dy 1 71. dx = x 2

+ X,

dy

72. dx = 9x 2

73• dx dy

4x

+

I

dy =

+ COSI,

=

76. :

=

cos I

=

-".sin".O,

~;

I

=

s(O)

+ sinl,

dr

78. dO = cos,,-o,

82.

d'y

-2

dx

d'y

-2

dx 2

83. ddl r2 2

84 d s 'd1 2

d'y

85. dx'

d'O

86. dl'

87.

=2

=

93.

4

dJ.. ~ sin x dx

94.

- cos>

y

y

I

=

r(O) = 0

r(O) = 1

- 6>:;

v(O) = I

y'(O)

= 4,

y(O)

=

I

Applications 95. Finding displacement from an antiderivati"" of ""Iocity

= 0;

y'(O)

= 2,

y(O)

=0

2 = -I';

-drl dt 1=1

= 1'

r(l)

=I

=

tis I

=3

s(4)

=4

31.

8'

= 6; = 0;

y(4) =

y~(O)

88. y(4)

0

s(".)

dv I 79. dl = Zsecltanl,

81.

-5

=

2vx

75. dl

77.

5, y(-I) = 0

y(4)

• "

y(2) = I

0;

3x -2/3 , y(-I)

=

74. dx dx

-

>

X

92.

3

dx

70. dx = 10 - x, y(O) = -I

dlt~4

y"(O)

0"(0)

=

'

-8, y'(O)

= -2,

0'(0)

-sin t + cos t; = 7, y"(O) = y'(O)

tis dl

=5

y(O)

I . r.; 0(0) = v2

y(O)

I, y(O)

=0

=3

Vx.

90. Uniquene•• of solution. If differentiable functions y = F(x) and y = G(x) both solve the initial value problem y(xo) = Yo.

on an interval I, most F(x) = G(x) for every x in

for your answer.

v

=

9.81 - 3.

ii) Find the body's displacement from I thats = -2 when 1= O.

= I

to I

~

3 giveo

iii) Now find the body's displacement from I = 1 to I given that s = So when I = O.

~

3

b. Suppose that the position s of a body moving along a coordinate line is a differentiable function of time t. Is it true that once you know an antiderivative of the velocity function tisldlYou can fmd the body's displacement from I = a to I = b even if you do not know the body's exact position at

either of those times? Give reasons for your answer.

89. Find the curve y = f(x) in the xy-pIane that passes through the point (9, 4) and whose slope at each point is 3

dy dx = f(x),

~

i) Find the body's displacement aver the time interval from I = I tol ~ 3giventhats = 5 when I ~ O.

= -Z'

= -I,

= -cosx + 8 sin2x; = 0, y"(O) = y'(O) =

y~(O)

= 0,

a. Suppose that the velocity of a body moving along the s-axis is

n Give reasons

96. Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 mjsec'. How fast will the rocket be going I min later? 97. Stopping a cor in time You are driving along a highway at a steady 60 mph (88 fi/sec) when you see an accident shead and slan3 on the brakes. What constant deceleration is required to stop your car in 242 fi? To fmd out, carry out the following steps.

Chapter 4 Questions to Guide Your Review

1. Solve the initial value problem Differential equation:

d's dt 2

= -k

(k constant)

101. Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is

a ,

where Vo and So are the body's velocity and position at time t ~ O. Derive this equation by solving the initial value problero

Measuring time and distance from when the brakes are applied

2. Find the value of t that malres tis/dt involve k.) 3. Find the value of k that malres s found in Step 2.

~

~

~

Initial conditions:

Chapter

=

-5.2 ftjsec 2

tis ~ Oands ~ 4 when t ~ 0 dt

~ So

wheo t

~

O.

where s is the body's height above 1he sorface. The equation has a minus sign because the acceleration acts downward, in the direction of decreasing s. The velocity Vo is positive if the object is rising at time t ~ 0 and negative if the object is falling. Instead of using 1he result of Exercise 101, you can derive Equation (2) directly by solving an appropriste initial value problem. What initial value problem? Solve it to be sure you have 1he right one, explaining the solution steps as you go along.

D 100. The hammer aod the feather

d's dt'

""ands

(2)

tis/dt in terms oft

Differential equation:

~

102. Free fall near the surface of. planet For free fall near the sorface of a planet where 1he acceleration due to gravity has a constant msgnitude of g leog!h-uoits/sec', Equation (I) in Exercise 10 I takes the form

b. the position s in terms of t.

When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonsttate that in a vacuum all bodies fall with the same (constant) accelera· tion, he dropped them from about 4 ft above the ground. The tele· vision footage of the event shows the hammer and 1he feather falling more slowly than on Ear1h, where, in a vacuum, 1hey would have talren only half a second to fall1he 4 ft. How long did it take the hammer and fea1her to fall 4 ft on 1he moon? To fmd out, solve the followiog initial value problem for s as a function of t. Then fmd 1he value of t that malres s equal to O.

-~a

Initial conditions:

242 for the value of t you

99. Motion along a coordinate line A particle moves on a coordinate line with acceleration a ~ d's/dt' ~ 15vt - (3/vt) , subject to the conditions that tis/dt ~ 4 and s ~ 0 when t ~ I. Find

a. the velocity v

d's dt' tis dt

Differential equation:

O. (The answer will

98. Stopping • motoreycle The State of lllinois Cycle Rider Safety Program requires riders to be able to brake from 30 mph (44 ft/sec) to 0 in 45 ft. What constant deceleration does it take to do that?

(1)

3=2,1 +vot+so.

tis ~ 88ands ~ Owheot ~ 0 dt .

Initial conditions:

239

COMPUTER EXPLORATIONS Use a CAS to solve the initial problems in Exercises 103-106. Plot the solution curves.

103. y' ~ cos'x + sinx, y('IJ") ~ I

104.Y'~t+x, y(I)~-1 105. y' ~ 106. y"

I

V4 - x'

~~+

Yx,

,y(O) y(l)

~

~

2

0,

y'(l)

~

0

Questions to Guide Your Review

1. What can be said about 1he extreme values of a function that is continuous on a closed interval? 2. What does it mean for a function to have a local extreme value on its domain? An absolute extreme value? How are local and absolute extreme values related, if at all? Give examples. 3. How do you find the absolute extrema of a continuous function on a closed interval? Give examples. 4. What are the hypotheses and conclusion of Rolle's Theorem? Are the bypotheses really necessary? Explain. 5. What are the hypotheses and conclusion of the Mean Value Theo· rem? What pbysical interpretations might 1he theorero have?

6. State 1he Mean Value Theorero's three corollaries. 7. How can you sometimes identify a function f(x) by knowiog f' and knowing 1he value of f at a point x ~ xo? Give an example. 8. What is 1he First Derivative Test for Local Extreme Values? Give examples of how it is applied. 9. How do you test a twice·differentiable function to determine where its graph is concave up or concave down? Give examples. 10. What is an inflection point? Give an example. What pbysical significance do inflection points sometimes have? 11. What is 1he Second Derivative Test for Local Extreme Values? Give examples of how it is applied.

240

Chapter 4: Applications of Derivatives

12. What do the derivatives of a functioo tell you about the sbape of its graph?

18. Can a function have more than one antiderivative? If so, how are the antiderivatives related? Explain.

13. List the steps you vrould take to graph a polynomial function. illustrate with an example.

19. What is an indefinite integral? How do you evaluate one? What general formulas do you know for fmding indefmite integrals?

14. What is a cusp? Give examples.

20. How can you sometimes solve a differential equation of the form dy/dx ~ j(x)?

15. List the steps you would take to graph a ratiooai function. Illustrate with an example. 16. Outline a general strategy for solving max-min problems. Give examples. 17. Describe Newton's method for solving equatioos. Give an example. What is the theory behind the method? What are some of the things to watch out for when you use the method?

Chapter

21. What is an initial value problem? How do you solve one? Give an example. 22. If you know the acceleration of a body moving along a coordinate line as a function of time, what more do you need to know to find the body's position function? Give an example.

Practice Exercises b. Now factor f'(x) and show that j has a local maximum atx ~ 1.70998 andlocalminimaatx ~ ±v3 '" ±1.73205.

Extreme Values 1. Does j(x) ~ x 3 + 2x + tan x have aoy local maximum or minimum. values? Give reasons for your answer. 2. Does g(x) ~ csex

+ 2 cotx

\7'5 '"

c. Zoom in 00 the graph to fmd a viewiog window that shows the presence of the ex1rerne values at x ~ \7'5 and x ~ v3.

have aoy local maximum values?

Give reasons for your answer.

The moral here is that without calculus the existence of_ of the three extreme values would probably have gone uoooticed. On aoy normal graph of the function, the values would lie close enough together to fall within the dimeosions of a single pixel on the screen. (Source: Uses o/Technology in the Mathematics Curriculum, hy Benny Evans and Jerry Johusoo, Oklaboma State University, published in 1990 under Natiooai Science Fouodation Grant USE-8950044.)

3. Does j(x) ~ (7 + x)(11 - 3X)1/3 have an absolute minimum value? An absolute maximum? If so, fmd them or give reasons why they fail to exist. List all critical points off. 4. Find values of a and b such that the function

has a local extreme value of 1 at x = 3. Is this extreme value a local maximum, or a local minimum? Give reasons for your answer.

J,

S. The greatest integer functioo j(x) ~ l x defmed for all values of x, assumes a local maximum value of 0 at each point of [0, I). Could aoy of these local maximum values also be local minimum values of f7 Give reasons for your answer.

6.

L

Give an example of a differentiable functioo j whose fIrst derivative is zero at some point c even though f has neither a local maximum nor a local minimum at c.

b. How is this consistent with Theorem 2 in Section 4.1? Give reasons for your answer. 7. The functioo y ~ I/x does not take 00 either a maximum or a minimum on the interval 0 < x < I even though the function is continuous on this interval. Does this contradict the Extreme Value Theorem for cootinuous functions? Why?

8. What are the maximum and minimum values of the function y ~ Ix I on the imerva\ -I :5 x < I? Notice that the interval is not closed. Is this coosistent with the Extreme Value Theorem for continuous functions? Why?

D

9. A graph that is large enough to show a functioo's global behavior may fail to reveal important local features. The graph of j(x) ~ (x'/8) - (x'/2) - x' + 5x 3 is a case in point L

Graph j over the interval -2.5 '" x '" 2.5. Where does the graph appear to have local extreme values or points of inflectioo?

D 10.

(Continuation ofExercise 9.) .. Graph j(x) ~ (x'/8) - (2/5)x' - 5x - (5/x 2 ) + II over the interval - 2 :5 x :5 2. Where does the graph appear to have local extreme values or points of inflection?

b. Show that j has a local maximum value at x ~ ~ '" 1.2585 and a local minimum value at x ~ 1.2599.

--¥2 '"

c. Zoom in to fmd a viewiog window that shows the presence of the ex1rerne values at x ~ ~ and x ~

--¥2.

The Mea. Value Theorem 11... Show that g(l) ~ sin2 1 - 31 decreases 00 every interval in its domain.

b. How many solutions does the equation sin2 t - 3t = 5 have? Give reasons for your answer. 12. .. Show that y

~

tan 8 increases 00 every interval in its domain.

b. If the conclusion in part
D b.

x' + 2x 2 -

2 ~ 0 has exactly ooe so-

Find the solutioo to as maoy decimal places as you can.

14... Show that j(x) ~ x/(x domain.

b. Show that j(x) ~ x 3 mum values.

+

I) increases on every interval in its

+ 2x has

no local maximum or mini-

Chapter 4

15. Water in a reservoir As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acre-f't in 24 hours. Show

that at some instant during that period the reservoir's volume was increasing at a rate in excess of225,000 gal/min. (An acre-foot is 43,560 It' , the volume that would cover I acre to the depth of I It A cubic foot holds 7.48 gal.)

Practice Exercises

241

body's (8) velocity equal to zero? (b) acceleration equal to zero? Dur-

ing approximately what time intervals does the body move (c) forward? (d)backorord? 21.

s

~ 3x + C gives a different function for each value of C. All of these functions, however, have the same derivative with respect to x, namely F' (x) ~ 3. Are these the only differentiable functions whose derivative is 3? Could there be any others? Give reasons for your answers.

16. The formula F(x)

17. Show that

Graphs and Graphing Graph the curves in Exercises 23-32.

even though

I

x

23. Y ~ x' - (x'/6)

x+I"-x+I' Doesn't this contradict Coro11ary 2 of the Mean Value Theorem?

Give reasons for your answer. 18. Calculate the ftrst derivatives of f(x) ~ x'/(x' + I) and g(x) ~ -I/(x' + I). What can you conclude about the graphs

of these functions?

24.y~x'-3x'+3

25. y~ -x'+6x'-9x+3 26. Y ~ (l/8)(x'

+ 3x' - 9x -

27)

27. y ~ x'(8 - x)

28. y ~ x'(2x' - 9)

Analyzing Graphs In Exercises 19 and 20, use the graph to answer the questions.

29. Y ~ x - 3x'/3 30. y ~ XI/3(X - 4)

19. IdentifY any global extreme values of f and the values of x at

31.y~x~

which they occur.

32.y~x~

y y

~

f(x)

(1, 1)

"-....H)

-"Ot----~x

20. Estimate the intervals on which the function y ~ f(x) is

Each of Exercises 33-38 gives the first derivative of a function y ~ f(x). (8) At what points, if any, does the graph of f have a local maximum, local mininmm, or inflection point? (b) Sketch the general abape of the graph.

x'

33. y' ~ 16 35. y' 37. y'

~ =

6x(x x4 -

34. y' =x 2 -x-6

+ I)(x -

2)

2,x2

36. y' ~ x'(6 - 4x) 38. y' = 4x 2 - x4

In Exercises 39-42, graph each function. Then use the function's ftrst derivative to explain what you see.

s. increasing. b. decreasing.

c. Use the given graph off' to indicate where any local extreme values of the function OCCllI, and whether each extreme is a relative maximum or minimum. y (2, 3)

39. y ~ x'/3 + (x - 1)1/3 41. Y ~ XI/3

+ (x -

1)1/3

40. Y ~ x'/3

+ (x - 1)'/3

42. Y ~ x'/3 - (x - 1)1/3

Sketch the graphs of the rational functions in Exercises 43--50. 43

_x+1 ,y-x-3

45.

x2 I y~-x-

+

2x

44,y~x+5

46• Y

~x'-x+1 x

47.y

~x'+2 2x

48.

y~--,­

49. Y

~

x2 - 4 -,-x - 3

50.

y~-,--

X4 -

1

x

x'

x - 4

Optimization 51. The sum of two nonnegative numbers is 36. Find the numbers if Each of the graphs in Exercises 21 and 22 is the graph of the position function s ~ f(l) of a body moving on a coordinate line (I represents time). At approximately what times (if any) is each

a. the difference of their square roots is to be as large as possible. b. the sum of their square roots is to be as large as possible.

242

Chapter 4: Applications of Derivatives

52. The sum of two nonnegative numbers is 20. Find the numbers

of the corridor shown here? Round your answer down to the nearest foot.

a. if the product of one number and the square root of the other is to be as large as possible.

y

b. if one number plus the square root of the other is to be as large as possible. 53. An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y = 27 - x 2 • Find the largest area the triangle can have.

55. Find the height and radius of the largest right circular cylinder that can be put in a sphere of radius

V3.

56. The figure here shows two right circular cones, one upside down inside the other. The two bases are parallel, and the vertex of the smaller cone lies at the center of the larger cone's base. What values of r and h will give the smaller cone the largest possible volume?

L

6

54. A customer has asked you to design an open-top rectangular stainless steel vat. It is to have a square base and a volume of 32 ff, to be welded from quarter-inch plate, and to weigh no more than necessary. What dimensions do you recommend?

O~----~8~------~X

Newton's Method 61. Let f(x) = 3x - x 3. Show that the equation f(x) = -4 has a solution in the interval [2, 3] and use Newton's method to find it. 62. Let f(x) = x4 - x 3. Show that the equation f(x) = 75 has a solution in the interval [3,4] and use Newton's method to find it.

Finding Indefinite Integrals Find the indefinite integrals (most general antiderivatives) in Exercises 63- 78. Check your answers by differentiation.

f7 12'

1

J J(n0 ~) J J J J to 75. J \120 \120 J ~ J I .

(r

y=

5-x'

Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make? 58. Particle motion The positions of two particles on the s-axis are SI = cos t and S2 = cos (t + 1T/4). a. What is the farthest apart the particles ever get? b. When do the particles collide?

o 59. Open-top box

An open-top rectangular box is constructed from a lO-in.-by-16-in. piece of cardboard by cutting squares of equal side length from the comers and folding up the sides. Find analytically the dimensions of the box of largest volume and the maximum volume. Support your answers graphically.

60. The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the comer

64.

30W+1 dO

71.

x 3(I

73.

sec 2

70. J k d O 7 + 02

72. 74.

ds

77.

sin2

dx

78.

cos2

dx

+ t) dt

6dr

+ x 4)- 1/4 dx

dO

76.

( Hint .. sin2 uD

=

cot

-

t4

dr

69.

(8t 3

66.

dt

+ 5)2

csc

40 - lOx

J ~ J(_1 _.l) 2Vt 68. J(r _ \12)3

7) dx

+

65. 67

57. Manufacturing tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where o ~ x ~ 4 and

+ 5x -

(x 3

63.

J J Jsec~tan ~dO (2 - x)3/5 dx

csc2 1TS ds

1 - cos 20) 2

Initial Value Problems Solve the initial value problems in Exercises 79- 82. dy x2 + 1 79. dx =~' y(I) = -1

80. :

=

r

(x + ~

y(I) = 1

d 2r • r 3 15vt+ • r; r/(I) = 8, r(I) = 0 dt V t d 3r 82. - 3 = -cos t; r"(O) = r/(O) = 0, r(O) = -1 dt 81.

-2 =

dt

Chapter 4 Additional and Advanced Exercises

Chapter

AdditionaL and Advanced Exercises

1. What can you say about a function whose maximum and minimum values on an interval are equal? Give reasons for your answer. 2. Is it true that a discontinuous function cannot have both an absolute maximum and an absolute minimum value on a closed interval? Give reasons for your answer.

3. Can you conclude anything about the extreme values of a continuous function on an open interval? On a half-open interval? Give reasons for your answer. 4. Local extrema

243

Use the sign pattern for the derivative

f

to identify the points where values.

has local maximum and minimum

11. Finding a function Use the following information to find the values of a, b, and e in the formula f(x) = (x + a)/ (bx 2 + ex + 2). i) The values of a, b, and e are either 0 or 1. ii) The graph of f passes through the point ( - 1, 0) . iii) The line y = 1 is an asymptote of the graph of f.

12. Horizontal tangent For what value or values of the constant k will the curve y = x 3 + kx 2 + 3x - 4 have exactly one horizontal tangent?

13. Largest inscribed triangle Points A and B lie at the ends of a diameter of a unit circle and point C lies on the circumference. Is it true that the area of triangle ABC is largest when the triangle is isosceles? How do you know? 14. Proving the second derivative test The Second Derivative Test for Local Maxima and Minima (Section 4.4) says:

5. Local extrema a. Suppose that the first derivative of y = f(x) is y' = 6(x

+

At what points, if any, does the graph of f have a local maximum, local minimum, or point of inflection? b. Suppose that the first derivative of y = f(x) is y' = 6x(x

a.

f

b.

f

l)(x - 2)2 .

+

has a local maximum value at x = e if I'(e) = 0 and f"(e) < 0 has a local minimum value at x = e if l' (c) = 0 and f"(e) > O.

To prove statement (a), let e = (1/2) If" (c) I .Then use the fact that

l)(x - 2) .

At what points, if any, does the graph of f have a local maximum, local minimum, or point of inflection? 6. If l' (x) ~ 2 for all x, what is the most the values of f can increase on [0, 6]? Give reasons for your answer. 7. Bounding a function Suppose that f is continuous on [a, b] and that e is an interior point of the interval. Show that if I'(x) ~ 0 on [a, c) and I'(x) 2: 0 on (e, b], then f(x) is never less than f(e) on [a, b].

8. An inequality a. Showthat-l/2 ~x/(1 +x 2 ) ~ 1/2foreveryvalueofx. b. Suppose that f is a function whose derivative is l' (x) = x/( 1 + x 2 ) . Use the result in part (a) to show that

If(b) - f(a)1

~

tlb - al

for any a and b.

9. The derivative of f(x) = x 2 is zero at x = 0, but f is not a constant function. Doesn't this contradict the corollary of the Mean Value Theorem that says that functions with zero derivatives are constant? Give reasons for your answer. 10. Extrema and inflection points two differentiable functions ofx.

Let h = fg be the product of

a. If f and g are positive, with local maxima at x = a, and if l' and g' change sign at a, does h have a local maximum at a? b. If the graphs of f and g have inflection points at x = a, does the graph of h have an inflection point at a? In each case, if the answer is yes, give a proof. If the answer is no, give a counterexample.

+ h)

. I'(e f"(e) = hm

h

h--->O

to conclude that for some 8

0<

Ihl

- I'(e)

> 0, I'(e

< 8

. I'(e + h) hm '---.....,.-h--->O h

=

+ h)

h


Thus, I'(e + h) is positive for -8 < h < 0 and negative for h < 8 . Prove statement (b) in a similar way.

o<

15. Hole in a water tank You want to bore a hole in the side of the tank shown here at a height that will make the stream of water coming out hit the ground as far from the tank as possible. If you drill the hole near the top, where the pressure is low, the water will exit slowly but spend a relatively long time in the air. If you drill the hole near the bottom, the water will exit at a higher velocity but have only a short time to fall. Where is the best place, if any, for the hole? (Hint: How long will it take an exiting particle of water to fall from height y to the ground?) Tank kept full, top open

~ r-

y

h

/Exit velocity

=

V 64(h -

y)

y

----- ....

Ground Or.'-----r--'-...--'------ x --Range

244

Chapter 4: AppLications of Derivatives

16. Kicking a field goal An American football player wants to kick a field goal with the ball being on a right hash mark. Assume that the goal posts are b feet apart and that the hash mark line is a distance a > feet from the right goal post. (See the accompanying figure.) Find the distance h from the goal post line that gives the kicker his largest angle f3 . Assume that the football field is flat.

°

a tax of t dollars per item sold? Comment on the difference between this price and the price before the tax. 20. Estimating reciprocals without division You can estimate the value of the reciprocal of a number a without ever dividing by a if youapplyNewton'smethodtothefunctionf(x) = (I/x) - a. For example, if a = 3, the function involved is f(x) = (1/x) - 3. a. Graphy = (I/x) - 3. Where does the graph cross the x-axis? b. Show that the recursion formula in this case is

Goal post line

so there is no need for division.

h

21. To find x = ~, we apply Newton's method to f(x) = x q - a. Here we assume that a is a positive real number and q is a positive integer. Show that Xl is a "weighted average" of Xo and a/x oq - l , and find the coefficients mo, ml such that

mo > O,ml > 0, mo + ml = 1. Football

17. A max-min problem with a variable answer Sometimes the solution of a max-min problem depends on the proportions of the shapes involved. As a case in point, suppose that a right circular cylinder of radius r and height h is inscribed in a right circular cone of radius R and height H, as shown here. Find the value of r (in terms of R and H) that maximizes the total surface area of the cylinder (including top and bottom). As you will see, the solution depends on whether H ::::; 2R or H > 2R.

What conclusion would you reach if Xo and a/x oq - l were equal? What would be the value of Xl in that case? 22. The family of straight lines y = ax + b (a, b arbitrary constants) can be characterized by the relation y" = 0. Find a similar relation satisfied by the family of all circles

where h and r are arbitrary constants. (Hint: Eliminate h and r from the set of three equations including the given one and two obtained by successive differentiation.) 23. Assume that the brakes of an automobile produce a constant deceleration of k ft/sec 2 • (a) Determine what k must be to bring an automobile traveling 60 mi/hr (88 ft/ sec) to rest in a distance of 100 ft from the point where the brakes are applied. (b) With the same k, how far would a car traveling 30 mi/hr travel before being brought to a stop? 24. Let f(x), g(x) be two continuously differentiable functions satisfying the relationships rex) = g(x) and rex) = - f(x). Let hex) = f2(x) + g2(x). Ifh(O) = 5, findh(lO). 25. Can there be a curve satisfying the following conditions? d 2y/dx 2 is everywhere equal to zero and, when x = 0, y = and dy/ dx = 1. Give a reason for your answer.

°

26. Find the equation for the curve in the xy-plane that passes through the point ( 1, - 1) if its slope at x is always 3x 2 + 2. 18. Minimizing a parameter Find the smallest value of the positive constant m that will make mx - 1 + (I/x) greater than or equal to zero for all positive values of x. 19. Suppose that it costs a company y = a + bx dollars to produce x units per week. It can sell x units per week at a price of p = c - ex dollars per unit. Each of a, b, c, and e represents a positive constant. (a) What production level maximizes the profit? (b) What is the corresponding price? (c) What is the weekly profit at this level of production? (d) At what price should each item be sold to maximize profits if the government imposes

27. A particle moves along the x-axis. Its acceleration is a = -t 2 • At t = 0, the particle is at the origin. In the course of its motion, it reaches the point x = b, where b > 0, but no point beyond b. Determine its velocity at t = 0.

Vi - (1/Vi).

28. A particle moves with acceleration a = Assuming that the velocity v = 4/3 and the position s = -4/15 when t=O,find a. the velocity v in terms of t. b. the position s in terms of t.

Chapter 4 Technology Application Projects 29. Suppose f(x) ~ ax' + 2bx + c with a > O. By coosidering the minimum, prove that f(x) '" 0 for all real x if, and ooly if, b 2 - ac :s; O. 30. SchWlllZ'. inequolity

and deduce Schwarz's inequality: (al b l

+ a,b, + ... + a"b.)' '" (ai' + a,' + ... + a,,')(bl' + b,' + ... + b.').

b. Show that equality holds in Schwarz's inequality only ifthcre exists a real number x that makes a,x equal -hi for every value of i from I to n.

•• In Exercise 29, let

Chapter 1:3

245

Technology Application Projects

Mathematica,tMap1e Modules: MOMIIAlolIg tJ Stmight Line: POSitioll .... Velocity .... Accelertllioll You will observe the shape of a graph through dramatic animated visualizatioos of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated Newtoll's Method: EstimtJte 'fr to How MtJlly PltJces? Plot a function, observe a root, pick a starting point near the root, and use Newton's Iteration Procedure to approximate the root to a desired accuracy. The numbers 'fr, e, and v2 are approsimated.

5 INTEGRATION OVERVIEW A great achievement of classical geometry was obtaining formulas for the areas and volumes of triangles, spheres, and cones. In this chapter we develop a method to calculate the areas and volumes of very general shapes. This method, called integration, is a tool for calculating much more than areas and volumes. The integral is of fundamental importance in statistics, the sciences, and engineering. We use it to calculate quantities ranging from probabilities and averages to energy consumption and the forces against a dam's floodgates. We study a variety of these applications in the next chapter, but in this chapter we focus on the integral concept and its use in computing areas of various regions with curved boundaries.

5.1

Area and Estimating with Finite Sums

y

~------~~-----7--~X

FIGURE 5.1

The area of the region R cannot be found by a simple formula.

246

The definite integral is the key tool in calculus for defining and calculating quantities important to mathematics and science, such as areas, volumes, lengths of curved paths, probabilities, and the weights of various objects, just to mention a few. The idea behind the integral is that we can effectively compute such quantities by breaking them into small pieces and then summing the contributions from each piece. We then consider what happens when more and more, smaller and smaller pieces are taken in the summation process. Finally, if the number of terms contributing to the sum approaches infinity and we take the limit of these sums in the way described in Section 5.3, the result is a definite integral. We prove in Section 5.4 that integrals are connected to antiderivatives, a connection that is one of the most important relationships in calculus. The basis for formulating definite integrals is the construction of appropriate finite sums. Although we need to define precisely what we mean by the area of a general region in the plane, or the average value of a function over a closed interval, we do have intuitive ideas of what these notions mean. So in this section we begin our approach to integration by approximating these quantities with finite sums. We also consider what happens when we take more and more terms in the summation process. In subsequent sections we look at taking the limit of these sums as the number of terms goes to infinity, which then leads to precise definitions of the quantities being approximated here.

Area Suppose we want to find the area of the shaded region R that lies above the x-axis, below the graph of y = 1 - x 2 , and between the vertical lines x = 0 and x = 1 (Figure 5.1). Unfortunately, there is no simple geometric formula for calculating the areas of general shapes having curved boundaries like the region R. How, then, can we find the area of R? While we do not yet have a method for determining the exact area of R, we can approximate it in a simple way. Figure 5.2a shows two rectangles that together contain the region R. Each rectangle has width 1/2 and they have heights 1 and 3/4, moving from left to right. The height of each rectangle is the maximum value of the function j,

5.1

Area and Estimating with Finite Sums

y

247

y

y=1-x 2 1+,(;;; 0'", 1)::-------,

1

0.5

(0, 1)

0.5 R

~------~~----~--~X

o

o

0.5

0.25

0.5

(a)

0.75

(b)

FIGURE 5.2 (a) We get an upper estimate of the area of R by using two rectang1es containing R. (b) Four rectaogles give a better upper estimate. Both estimates oversboot the true value for the area by the amount shaded in light red.

obtained by evaluating f at the left endpoint of the subinterval of [0, I] forming the base of the rectangle. The total area of the two rectangles approximates the area A of the region R, A .., I·

2I + 43 . 2I = 87 =

0.875.

This estimate is larger than the true area A since the two rectangles contain R. We say that 0.875 is an upper sum because it is obtained by taking the height of each rectangle as the maximum (uppermost) value of f(x) for a point x in the base interval of the rectangle. In Figure 5.2b, we improve our estimate by using four thinner rectangles, each of width 1/4, which taken together contain the region R. These four rectangles give the approximation A .., 1·

1

15

1

3

1

7

1

4 + 16 . 4 + 4 . 4 + 16 . 4 =

25 32 = 0.78125,

which is still greater than A since the four rectangles contain R. Suppose instead we use four rectangles contained imide the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1/4 as before, but the rectangles are y

y

1

1

0.5

0.5

~--~~--~--~~--~~X

o

0.25

0.5

(a)

0.75

(U)

~--~~-L~~~~~--~ x

o

0.25 0.5 0.75 1 0.125 0.375 0.625 0.875 (b)

FIGURE 5.3 (a) Rectang1es contained in R give an estimate for the area that uodershoots the true value by the amoout shaded in light blue. (b) The midpoint rule uses rectsngles whose height is the value of y = f(x) at the midpoints of their bases. The estimate appears closer to the true value of the area because the light red overshoot areas roughly balaoce the light blue undershoot areas.

248

Chapter 5: Integration shorter and lie entirely beneath the graph of j. The function f(x) = I - x 2 is decreasing on [0, 1], so the height of each of these rectangles is given by the value of f at the right endpoint of the subinterval fonning its base. The foorth rectangle has zero height and therefore contributes no area. Summing these rectangles with heights equal to the mini· mum value of f(x) for a point x in each base subinterva1 gives a lower sum approximation to the area, A ..,

y

15

1

3

1

1

1

17 32 = 0.53125.

This estimate is smaller than the area A since the rectangles all lie inside of the region R. The true value of A lies somewhere between these lower and upper sums:

1

0.53125

o (a)

<

A

<

0.78125.

By considering both lower and upper sum approximations we get not only estimates for the area, but also a bound on the size of the possible error in these estimates since the true value of the area lies somewhere between them. Here the error cannot be greater than the difference 0.78125 - 0.53125 = 0.25. Yet another estimate can be obtained by using rectangles whose heights are the values of f at the midpoints of their bases (Figure 5.3b). This method of estimation is called the midpoint rule for approximating the area. The midpoint rule gives an estimate that is between a lower sum and an upper sum, but it is not qnite so clear whether it overestimates or underestimates the true area. With four rectangles of width 1/4 as before, the midpoint rule estimates the area of R to be

.1+ 55 . 1 + 39 • 1 + Ii . 1 = 172 • 1 = 0671875 644644644644644' .

A .., 63

y I r r_

7

16 . 4 + 4 . 4 + 16 . 4 + o· 4 =

_

In each of our computed sums, the interva1 [a, b] over which the function f is defined was subdivided into n subinterva1s of equal width (also called length) dx; = (b - a)/n, and f was evaluated at a point in each subinterval: Cj in the first subinterva1, C2 in the sec· ond subinterva1, and so on. The finite sums then all take the form f(cj) ~x

(b)

FIGURE 5.4
+

f(c2) ~x

+

f(C3) ~x

+ ... + f(c n) ~x.

By taking more and more rectangles, with each rectangle thinner than before, it appears that these finite sums give better and better approximations to the true area of the region R. Figure 5.4a shows a lower sum approximation for the area of R using 16 rectangles of equal width. The sum of their areas is 0.634765625, which appears close to the true area, but is still smaller since the rectangles lie inside R. Figure 5.4b shows an upper sum approximation using 16 rectangles of equal width. The sum of their areas is 0.697265625, which is somewhat larger than the true area be· cause the rectangles taken together contain R. The midpoint rule for 16 rectangles gives a total area approximation of 0.6669921875, but it is not immedistely clear whether this es· timate is larger or smaller than the true area

EXAMPLE 1 Table 5.1 shows the values of upper and lower sum approximations to the area of R using up to 1000 rectangles. In Section 5.2 we will see how to get an exact value of the areas of regions such as R by taking a limit as the base width of each rectangle goes to zero and the number of rectangles goes to infinity. With the techniques developed there, we will be able to show that the area of R is exactly2/3. •

Distance Traveled Suppose we know the velocity function v(1) of a car moving down a highway, without chang· ingdirection, and want to know how far it traveled between times 1 = a andt = b.lfwe al· ready know an antiderivative F(I) of v(l) we can fmd the car's position function ,(1) by setting

5.1 Area and Estimating with Finite Sums

249

TABLE 5.1 Finite approximations for the area of R

Number of subintervals

Lower sum

Midpoint rule

Upper sum

2 4 16 50 100 1000

.375 .53125 .634765625 .6566 .66165 .6661665

.6875 .671875 .6669921875 .6667 .666675 .66666675

.875 .78125 .697265625 .6766 .67165 .6671665

s(t) = F(t) + C. The distance traveled can then be found by calculating the change in position, s(b) - s(a) = F(b) - F(a). If the velocity function is known only by the readings at various times of a speedometer on the car, then we have no formula from which to obtain an antiderivative function for velocity. So what do we do in this situation? When we don't know an antiderivative for the velocity function v(t), we can apply the same principle of approximating the distance traveled with finite sums in a way similar to our estimates for area discussed before. We subdivide the interval [a, b] into short time intervals on each of which the velocity is considered to be fairly constant. Then we spproximate the distance traveled on each time subinterval with the usual distance formula distance = velocity X time and add the results across [a, b]. Suppose the subdivided interval looks like

~4t-+-4t-+-4t-1

.....

~_.o-L'

a

tl

~I_

••~I_ _~I--»

12

t (sec)

b

13

with the subintervals all of equal length I1t. Pick a number tl in the first interval. If I1t is so small that the velocity barely changes over a short time interval of duration I1t, then the distance traveled in the first time interval is about V(tl) I1t. If t2 is a number in the second interval, the distance traveled in the second time interval is about V(t2) 11t. The sum of the distances traveled over all the time intervals is D '" V(tl) I1t

+ V(t2)

I1t

+ ... + v(tn )

I1t,

where n is the total number of subintervals.

EXAMPLE 2 The velocity function of a projectile fired straight into the air is /(t) = 160 - 9.8t m/sec. Use the summation technique just described to estimate how far the projectile rises during the first 3 sec. How close do the sums come to the exact value of 435.9 m? Solution We explore the results for different numbers of intervals and different choices of evaluation points. Notice that /(t) is decreasing, so choosing left endpoints gives an upper sum estimate; choosing right endpoints gives a lower sum estimate. (a) Three subintervals oflength 1, with / evaluated at left endpoints giving an upper sum: II

'2

•

•

•

1

2

o

~4t>l

13 I)

3

t

250

Chapter 5: Integration With / evaluated at t = 0, I, and 2, we have D '" /(t1) !!.t

+ /(t2)

+ /(t3) !!.t + [160 - 9.8(1)](1) +

!!.t

= [160 - 9.8(0)](1)

[160 - 9.8(2)](1)

= 450.6.

(b) Three subintervals a/length I, with / evaluated at right endpoints giving a lower sum:

o

tl

t2

13

1

2

3

•

•

e) t

~<1t~1 With/ evaluated at t = 1,2, and 3, we have D '" /(t1) !!.t

+ /(t2)

+ /(t3) !!.t + [160 - 9.8(2)](1) +

!!.t

= [160 - 9.8(1)](1)

[160 - 9.8(3)](1)

= 421.2.

(c) With six subintervals a/length 1/2, we get

These estimates give an upper sum usiug left endpoints: D '" 443.25; and a lower sum using right endpoints: D '" 428.55. These six-interva1 estimates are somewhat closer than the three-interva1 estimates. The results improve as the subintervals get shorter.

As we can see in Table 5.2, the left-endpoint upper sums approach the true value 435.9 from above, whereas the right-endpoint lower sums approach it from below. The true value lies between these upper and lower sums. The magnitude of the error in the closest entries is 0.23, a small percentage of the true value. Error magnitude =

1true

value - calculated value 1

= 1435.9 - 435.671 = 0.23.

0.23 - 0.050/.o. Error percentage -- 435.9 It would be reasonable to conclude from the table's last entries that the projectile rose about 436 m during its rITst 3 sec offlight. •

TABLE

5.2 Travel-distance estimates

Number of subintervals

Length of each subinterval

Upper sum

Lower sum

3 6 12 24 48 96 192

I

450.6 443.25 439.58 437.74 436.82 436.36 436.13

421.2 428.55 432.23 434.06 434.98 435.44 435.67

1/2 1/4 1/8 1/16 1/32 1/64

S.l

Area and Estimating with Finite Sums

251

Displacement Venus Distance Traveled If an object with position function 3(/) moves along a coordinate line without changing direction, we can calculate the total distance it travels from t - a to t - b by summing the distance traveled over small intervals, as in Example 2. If the object reverses direction one or more times during the trip, then we need to use the object's speed Ivet) I. which is the absolute value of its velocity function, v(/), to find the total distance traveled. Using the velocity itself, as in Example 2, gives instead an estimate to the object's dIIplacemeDt. J'( b) - s(a), the difference between its initial and (mal positions. To see why using the velocity f\mcti.on in the summation process gives an estimate to the displacement, partition the time interval [0, b] into small enough equal subintervals. fl.! so that the object's velocity does not change very much :from time 1101 to '1_ Then v(t.0 gives a good approximation of the velocity throughout the interval. Accord:ingly. the change in the object's position coordinate during the time interval is about

,

_(.,)
1_(',)1


The total diitlDce fr'awled is approximately the sum

1_(',)1
FIGURE 5.5 The rock: in Example 3. The height 256 ft is reached at

t=2andt=8sec. The rock falls 144 ft

from its maximum height when t = 8.

TABLE 5.3 Velocity Function

• 0 0.5 1.0

1.5 2.0 2.5 3.0 3.5 4.0

v(~

160 144 128 112 96 80 64 48 32


•

v(~

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0

16 0 -16 -32 -48 -64 -80 -96

EXAMPLE 3 In Example 4 in Section 3.4, we analyzed the motion of a heavy rock blown straight up by a dynamite blast In that example, we found the velocity of the rock at any time during its motion to be v(t) = 160 - 321ft/sec. The rock was 256 ft above the grmmd 2 sec after the explosion, continued upwards to reach a maximum height of 400 ft at 5 sec after the explosion, and then fell back down to reach the height of256 ft again at t = 8 sec after the explosion. (See Figure 5.5.) If we follow a procedure like that presented in Example 2, and use the velocity function v(t) in the summation process over the time interval [0, 8], we will obtain an estimate to 256 ft, the rock's height above the ground at t = 8. The positive upward motion (which yields a positive distance change of 144 ft from the height of 256 ft to the maximum height) is cancelled by the negative downward motion (giving a negative change of 144 ft from the maximum height down to 256 ft again), so the displacement or height above the grmmd is being estimated from the velocity function. On the other hand, if the absolute value Iv(t) I is used in the summation process, we will obtain an estimate to the total distance the rock has traveled: the maximum height reached of 400 ft plus the additional distance of 144 ft it has fallen back down from that maximum when it again reaches the height of 256 ft at t = 8 sec. That is, using the absolute value of the velocity function in the summation process over the time interval [0, 8], we obtain an estimate to 544 ft, the total distance up and down that the rock has traveled in 8 sec. There is no cancellation of distance changes due to sign changes in the velocity function, so we estimate distance traveled rather than displacement when we use the absolute value of the velocity function (that is, the speed of the TOck). As an illustration of OUT discussion, we subdivide the interval [0, 8] into sixteen subintervals of1ength at = 1/2 and take the right endpoint of each subinterval in OUT calculations. Table 5.3 shows the values of the velocity function at these endpoints. Using v(t) in the summation process, we estimate the displacement at t = 8: (144

+ 128 + 112 + 96 + 80 + 64 + 48 + 32 + 16 + 0 - 16 - 32 - 48 - 64 - 80 - 96)'1.2 - 192 Errormagnitude = 256 - 192 = 64

252

Chapter 5: Integration Using Iv(t) I in the summation process, we estimate the total distance traveled over the time interval [0, 8]: (144

+

128

+0 +

+ 16

112

+ 96 + 80 + 64 + 48 + 32 +

+ 32 + 48 + 64 +

80

+ 96)'

t

16

= 528

Error magnitude = 544 - 528 = 16

If we take more and more subintervals of [0, 8] in our calculations, the estimates to • 256 ft and 544 ft improve, approacbing their true values.

Average Value of a Nonnegative Continuous Function The average value of a collection of n numbers Xl. X2, ••• ,x. is obtained by adding them together and dividing by n. But what is the average value of a continuous function j on an interval [a, b]? Such a function can assume infinitely many values. For example, the temperature at a certain location in a town is a continuous function that goes up and down each day. What does it mean to say that the average temperature in the town over the course of a day is 73 degrees? When a function is constant, this question is easy to answer. A function with constant value c on an interval [a, b] has average value c. When c is positive, its graph over [a, b] gives a rectangle of height c. The average value of the function can then be interpreted geometrically as the area of this rectangle divided by its width b - a (Figure 5.6a). y

y

y=c

c ~ ,----------~-r --

~~------------~~X

o

a

b

(a)

~~------------~~X

o

a

b

(b)

FIGURE 5.6 (a) The average value of f(x) ~ c on [a, bl is 1he area of1he rectangle divided by b - a. (b) The average value ofg(x) on [a, bl is 1he area benea1h its graph divided by b - a.

y

What if we want to f"md the average value of a nonconstant function, such as the function g in Figure 5.6b? We can think of this graph as a snapshot of the height of some water that is sloshing around in a tank between enclosing walls at x = a and x = b. As the water moves, its height over each point changes, but its average height remains the same. To get the average height of the water, we let it settle down until it is level and its height is constant. The resulting height c equals the area under the graph of g divided by b - a. We are led to define the average value of a nonnegative function on an interval [a, b] to be the area under its graph divided by b - a. For this def"mition to be valid, we need a precise understanding of what is meant by the area under a graph. Ths will be obtained in Section 5.3, but for now we look at an example.

EXAMPLE 4 [0, FIGURE 5.7 Approximating 1he area under f(x) ~ sinxbetween o and'1T' to compute the average value of sin x over [0, 'IT1, using eight rectangles (Example 4).

Estimate the average value of the function j(x) = sin x on the interval

'IT].

Solution

Looking at the graph of sin x between 0 and 'IT in Figure 5.7, we can see that its average height is somewhere between 0 and I. To f"md the average we need to calculate the area A under the graph and then divide this area by the length of the interval, 'IT - 0 = 'IT. We do not have a simple way to determine the area, so we approximate it with f"mite sums. To get an upper sum approximation, we add the areas of eight rectangles of equal

253

5.1 Area and Estimating with Finite Sums

width 1T/8 that together contain the region beneath the graph of y = sinx and above the x-axis on [0, 1T]. We choose the heights of the rectangles to be the largest value of sin x on each subinterval. Over a particular subinterval, this largest value may occur at the left endpoint, the right endpoint, or somewhere between them. We evaluate sin x at this point to get the height of the rectangle for an upper sum. The sum of the rectangle areas then estimates the tota1 area (Figure 5.7):

A

'"

. 1T ( SfiS

'" (.38

+ SID . 1T + Sfi . 31T + Sfi . 1T + SID . 1T + SID . 51T + SIDT . 31T + SID . 71T) ' 1T

g

4

+ .71 +

.92

2

g

2

+ 1 + 1 + .92 + .71 + .38)' ~

g

= (6.02)'

S

~ '" 2.365.

To estimate the average value of sin x we divide the estimated area by 1T and obtain the approximation 2.365/1T '" 0.753. Since we used an upper sum to approximate the area, this estimate is greater than the actoal average value of sin x over [0, 1T]. If we use more and more rectangles, with each rectangle getting thinner and thinner, we get closer and closer to the true average value. Using the techniques covered in Section 5.3, we will show that the true average value is 2/1T '" 0.64. As before, we could just as well have used rectangles lying under the graph of y = sin x and calculated a lower sum approximation, or we could have used the midpoint rule. In Section 5.3 we will see that in each case, the approximations are close to the true area if all the rectangles are sufficiently thin. •

Summary The area under the graph of a positive function, the distance traveled by a moving object that doesn't change direction, and the average value of a nonnegative function over an interval can all be approximated by fmite sums. First we subdivide the interval into subintervals, treating the appropriate function / as if it were constant over each particular subinterval. Then we multiply the width of each subinterval by the value of / at some point within it, and add these products together. If the interval [a, b] is subdivided into n subintervals of equal widths ~x = (b - a)/n, and if /(ck) is the value of / at the chosen point Ck in the kth subinterval, this process gives a fmite sum of the form

/(c,)

~

+ /(c2)

~x

+ /(C3)

~

+ ... + /(c.)

~x.

The choices for the Ck could maximize or minimize the value of / in the kth subinterval, or give some value in between. The true value lies somewhere between the approximations given by upper sums and lower sums. The finite sum approximations we looked at improved as we took more subintervals of thinner width.

Exercises 5.1 Area In Exercises 1-4, use ftnite approximations to estimate the area under the graph of the function using

a. a lower sum with two rectangles of equal width. b. a lower sum with four rectaogles of equal width. c. an upper sum with two rectangles of equal width. d. an upper sum with four rectangles of equal width. 1. f(x) = x 2 betwocox

2. f(x)

= x 3 betwocox

= 0 aodx = 0 aodx

3. f(x) = l/x bctwoco x

= 1 andx =

5.

4. f(x) = 4 - x 2 bctwocox = -2 and x = 2. Using rectaogles whose height is given by the value of the fun.,. tion at the midpoint of the rectaogle's base (the midpoint rule), estimate the area under the graphs of the followiog functions, using !ITS! two and then four rectangles.

5. f(x)

=

x 2 bctwoenx 3

=

6. f(x) = x bctwoenx =

oand x = I. oand x = I.

= I.

7. f(x)

=

I/x bctwoco x = I andx = 5.

= I.

8. f(x)

=

4 - x 2 betwoenx

=

-2 aodx

=

2.

254

Chapter 5: Integration

Distance 9. Distance traveled The accompanying table shows the velocity of a model1rain eogine moving along a track for 10 sec. Estimate the distance traveled by the engine using 10 subinterva1s oflength I with L

12. Distance from velocity data The accompanying table gives data for the velocity of a vintage sports car accelerating from 0 to 142 milh in 36 sec (10 thousandths ofanhour).

left-endpoint values.

b. right-eodpoint values.

...

TIme (.eo)

Velocity (in. I ...)

TIme ( )

Velocity (in·l·eo)

0 I 2 3

0 12 22 10 5 13

6 7 8

11 6 2 6 0

4 5

9 10

0 5 10 IS 20 25 30

I 1.2 1.7 2.0 1.8 1.6 1.4

0.0 0.001 0.002 0.003 0.004 0.005

0 40 62 82 96 108

0.006 0.007 0.008 0.009 0.010

116 125 132 137 142

--"

120 100 80

/

60

TIme

(min)

Velocity (m/·ec)

35 40 45 50 55 60

1.2 1.0 1.8 1.5 1.2 0

11. Length of a road You and a coropaoion are about to drive a twisty stretch of dirt road in a car whose speedometer works but whose odometer (mileage counter) is broken. To fmd out how long this particular stretch of road is, you record the car's velocity at IO-sec intervals, with the results shown in the accoropaoying table. Estimate the length of the road using L

Velocity (mi/h)

140

left-endpoint values.

Velocity (m/·ec)

TIme (h)

160

b. right-eodpoint values. TIme (min)

Velocity (mi/h)

mi/br

10. Distance traveled up.tream You are sitting on the baok of a tidal river watching the incoroing tide carry a bottIe upstream. You record the velocity of the flow every 5 minutes for an hour, with the results shown in the accompanying table. About how far upstream did the bottle travel during that hour? Find an estimate using 12 suhinterva1s oflength 5 with L

Time (h)

I

40 20 0

I 0.002 0.004 0.006 0.008

om

hOUlli

a. Use rectangles to estimate how far the car traveled during the 36 sec it took to reach 142 mi/h.

b. Roughly how maoy seconds did it take the car to reach the halfway point? About how fast was the car going then?

13. Free fall with air resi.tance

An object is dropped straight down from a helicupter. The object falls faster and faster but its acceleration (rate of change of its velocity) decreases over time

because of air resistance. The acceleration is measured in ft/sec2 and recorded every second after the drup for 5 sec, as shown:

left-endpoint values.

I

2

3

4

5

19.41

11.77

7.14

4.33

2.63

b. right-eodpoint values.

a TIme (sec)

Velocity (converted to ft/ ...) (30 mi/h = 44 fi/sec)

o

o

10 20 30 40 50 60

44 IS 35 30 44 35

TIme (sec) 70 80 90 100 110 120

Velocity (converted In fi I ...) (30 mi/h 44 fils")

32.00

a. Find an upper estimate for the speed when t

= 5.

=

b. Find a lower estimate for the speed when t = 5.

IS

c. Find an upper estimate for the distance fallen when t = 3.

22 35 44 30 35

14. Distance traveled by a projectile An object is shot straight upward from sea level with an initial velocity of 400 ft/sec.

a. Assuming that gravity is the only force acting on the object, give an upper estimate for its velocity after 5 sec have elapsed. Use g ~ 32 ft/sec 2 for the gravitational acceleration.

b. Find a lower estimate for the height attained after 5 sec.

5.1 Average Value of a Function In Exercises 15-18, use a finite sum to estimate the average value of 1 on the given interval by partitioning the interval into four subintervals of equal length and evaluating 1 at the subinterval midpoints.

15. I(x) = x' on

17. I(t) = (1/2)

+

16. I(x) = I/x

[0,2] sin2 ".t

on

on

[1,9]

255

Area and Estimating with Finite Sums

Measurements are taken at the end of each month detennioing the rate at which pollutants are released into the a1mosphere, recorded as follows.

Month

Jan

Feb

Mar

Apr

May

Jun

0.20

0.25

0.27

0.34

0.45

0.52

Jul

Aug

Sep

Oct

Nov

Dec

0.63

0.70

0.81

0.85

0.89

0.95

[0,2] Pollutant

y

y

release rate

= ! + sin2'1Tt

(tons/day)

2

Month

o

Pollutant

2

1

release rate 18. I(t) = I - (cos

~t)'

(tons/day) on

[0,4] a. Assuming a 30-day month and that new scrubbers allow only 0.05 too/day to be released, give ao upper estimate of the tota! toonage of pollutants released by the end of 1ooe. What is a lower estimate?

y

1

b. In the best case, approximately when will a tota! of 125 tons of pollutants have been released into the a1mosphere?

o

21. Inscribe a regular n-sided polygon inside a circle of radius I aod compute the area of the polygon for the followiog values of n:

234

•• 4 (square)

Examples of Estimations 19. Water pollution Oil is leaking out of a !auker damaged at sea. The damage to the tanker is worsening as evidenced by the increased leakage each hour, recorded in the following table.

Time (h)

o

I

2

3

4

Leakage (gal/h)

50

70

97

136

190

Leakage (gal/h)

5

6

7

8

265

369

516

720

c. 16

22. (Continuation ofExercise 21.)

a. Inscribe a regular n-sided polygon inside a circle of radius 1 and compute the area of one of the n congruent triangles formed by drawiog radii to the vertices of the polygon. b. Compute the limit of the area of the inscribed polygon as n ---+

Time (h)

b. 8 (octagon)

d. Compare the areas in parts (a), (b), aod (c) with the area of the circle.

a. Give ao upper aod a lower estimate of the total quaotity of oil that has escaped after 5 hours. b. Repeat part (a) for the quaotity of oil that has escaped after 8 hours. c. The!auker continues to leak 720 gai/h after the first 8 hours. If the taoker originally contained 25,000 gal of oil, approximately how maoy more hours will elapse in the worst case before all the oil has spilled? In the best case? 20. Air pollution A power plaot generates elec1ricity by burniog oil. Pollutants produced as a result of the burniog process are removed by scrubbers in the smokestacks. Over time, the scrubbers become less efficient and eventually they must be replaced when the amouot of pollution released exceeds government standsrds.

00.

c. Repeat the computations in parts (a) aod (b) for a circle of radius r.

COMPUTER EXPLORATIONS In Exercises 23-26, use a CAS to perform the following steps. a. Plot the functions over the givco interval. b. Subdivide the interval into n = 100,200, and 1000 subintervals of equal length aod evaluate the function at the midpnint of each subinterval.

c. Compute the average value of the function values generated in part (b). d. Solve the equatioo/(x) = (average value) for x using the average value calculated in part (c) for the n = 1000 partitiooing. 24. I(x) = sin2 X on [0, 'IT] 23. I(x) = sin x 00 [0, ".]

25. I(x)

=

x sin

t [;f, 'IT] 00

26. I(x) = x sin2t on

[f,,,.]

256

Chapter 5: Integration

5.2

Sigma Notation and Limits of Finite Sums In estimating with finite sums in Section 5.1, we encountered sums with many terms (up to 1000 in Table 5.1, for inslauce). In this section we introduce a more convenient notation for sums with a large number of terms. After describing the notation and stating several of its properties, we look at what happens to a finite sum approximation as the number of terms approaches infmity.

Finite Sums and Sigma Notation Sigma notation enables us to write a sum with many terms in the compact fonn n

~ak = al + a2 + a3 + ... + an -l +

~

an.

The Greek letter ~ (capital sigma, corresponding to our letter S), slauds for "sum." The index of summation ktells us where the sum begins (at the number below the ~ symbol) and where it ends (at the number above ~). Any letter can be used to denote the index, but the letters i,j, and k are customary. The index k ends at k

/"

= n.

"

The summation symbol_~ (Greek letter sigma)

at ~"., ..,_, th ~. k = 1 ak lS8 .Lormu.uuOl' eIl.Wterm..

"

The index k starts at k

= 1.

Thus we can write 11

12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112 = ~k2, 1=1

and 100

1(1) + 1(2) + 1(3) + ... + 1(100) = ~/(i). j=- 1

The lower limit of summation does not have to be I; it can be any integer.

EXAMPLE 1 A sum in sigma notation

The sum written out, one term for each vaIue of k

The value of the sum

~k 3

1+2+3+4+5

15

(- 1)1(1) + (-1)2(2) + (-1)3(3)

-I + 2 - 3 = -2

I 2 TTI+2+1 42 52 --+-4- I 5- I

l+I=l 236

,

k- l

~(-I)kk

2

~_k_ k~1 k + I

, e ~-

k~4

k- I

16 + 25 = 139 3 4 12

•

5.2 Sigma Notation and Limits of Finite Sums

EXAMPLE 2

Express the sum 1

+

3

257

+ 5 + 7 + 9 in sigma notation.

Solution The fonnula generating the tenns changes with the lower limit of summation, but the tenns generated remain the same. It is often simplest to start with k = 0 or It = I, but we can start with any integer. 4

Starting with k = 0:

I

+3+

5

+7+9

= ~(2k 1-0

+ 1)

Starting with k = I:

I

+3+

5

+7+9

= ~(2k 1=1

- I)

Starting with It = 2:

I

+3+

5

+7+9

Starting with k = -3:

I

+3+

5

+7+

, 6

= ~(2k - 3) 1- 2 1 9 = ~ (2k + 7) 1=-3

•

When we have a sum such as

we can rearrange its terms, 3

~(k

+ e)

=

(I

+ 12) + (2 + 22) + (3 + 32)

=

(I

+ 2 + 3) + (12 + 22 + 32)

1- 1 3

= ~k 1=1

Regroup terms.

3

+

~k2. 1=1

This illustrates a general rule for finite sums:

•

~(a1

1- 1

+ b1)

=

••

~a1

k=-I

+ ~b1 1- 1

Four such rules are given below. A proof that they are valid can be obtained using mathematical induction (see Appendix 2).

Algebra Rules for Finite Sums 1.

Sum Rule:

2.

Difference Rule:

3.

Constant Multiple Rule:

4.

Constant f!jlue Rule:

EXAMPLE 3

•

(Any number cJ

•

~_c=n·c ~

(c is any constant value.)

We demonstrate the use of the algebra rules.

••

(8) ~(3k - Tt') = 3~k - ~e

Difference Rule and Constant Multiple Rule Constant Multiple Rule

258

Chapter 5: Integration 3 3 3

(c) ~(k

(d) HISTORICAL BIOGRAPHY

Carl Friedrich Gauss (1777-1855)

+ 4)

=

~k + ~4

Sum Rule

(i + 2 + 3) + (3·4) = 6 + 12 = 18 • 1 1 = n·- = 1

L -n

1=1

=

Constant Value Rule

n

Constant Value Rule (I/n is constant)

•

Over the years people have discovered a variety of formulas for the values of finite sums. The most famous of these are the formula for the sum of the first n integers (Gauss is said to have discovered it at age 8) and the formulas for the sums of the squares and cubes of the first n integers.

EXAMPLE 4

Show that the sum of the first n integers is

ik

=

n(n + I). 2

1- 1

Solution

The formula tells us that the sum of the first 4 integers is

(4~5)

= 10.

Addition verifies this prediction: 1

+2+3+4

=

10.

To prove the formula in general, we write out the terms in the sum twice, once forward and once backward.

+ +

+ +

+ n + + 1 + If we add the two terms in the first column we get 1 + n = n + 1. Similarly, if we add the two terms in the second column we get 2 + (n - I) = n + I. The two terms in any column sum to n + 1. When we add the n columns together we get n terms, each equal to n + 1, for a tota1 ofn(n + 1). Since this is twice the desired quantity, the sum of the 1rrst n integers is (n)(n + 1)/2. • 1 n

2 (n - I)

3 (n - 2)

Formulas for the sums of the squares and cubes of the first n integers are proved using mathematical induction (see Appendix 2). We state them here.

The first n squares: The first n cubes:

Limits of Finite Sums The finite sum approximations we considered in Section 5.1 became more accurate as the number of terms increased and the subinterval widths (lengths) narrowed. The next example shows how to calculate a limiting value as the widths of the subintervals go to zero and their number grows to infinity.

EXAMPLE 5 Find the limiting value oflower sum approximations to the area of the region R below the graph of y = 1 - x 2 and above the interval [0, I] on the x-axis using equal-width rectangles whose widths approach zero and whose number approaches infinity. (See Figure 5.4a.)

5.2 Sigma Notation and Limits of Finite Sums

259

Solution We compute a lower sum approximation using n rectangles of equal width Ax = (I - O)ln, and then we see what happens as n ..... 00. We start by subdividing [0, I] into n equal width subintervals

Each subinterval has width lin. The function I - x 2 is decreasing on [0, I], and its smallest value in a subinterval occurs at the subinterval's right endpoint. So a lower sum is constructed with rectangles whose height over the subinterval [(k - 1)ln, kin] is I(kln) = I - (kln)2, giving the sum

We write this in sigma notation and simplify,

Difference Rule

Constant Value and Constant Multiple Rules

Sum of the First n Squares

Numerator expanded

We have obtained an expression for the lower sum that holds for any n. Taking the limit of this expression as n ..... 00, we see that the lower sums converge as the number of subintervals increases and the subinterval widtha approach zero: lim (I _2n _00

3

2

+ 3n + n) ~3

=

I _I6

=

I. 3

The lower sum approximations converge to 2/3. A similar calculation shows that the upper sum approximations also converge to 2/3. Any fmite sum approximation ~f~l l(ck)(l/n) also converges to the same value, 2/3. Ibis is because it is possible to show that any fmite sum approximation is trapped between the lower and upper sum approximations. For this reason we are led to define the area of the region R as this limiting value. In Section 5.3 we study the limits of such fmite approximations in a general setting. •

Riemann Sums HIsTORICAL BIOGRAPHY

Georg Friedrich Bernhard Riemann (1826-1866)

The theory oflimits of finite approximations was made precise by the German mathematician Bernhard Riemann. We now introduce the notion of a Riemann sum, which underlies the theory of the definite integral studied in the next section. We begin with an arbitrary bounded function I defmed on a closed interval [a, h]. Like the function pictured in Figure 5.8, I may have negative as well as positive values. We subdivide the interval [a, h] into subintervals, not necessarily of equal widtha (or lengths), and form sums in the same way as for the finite approximations in Section 5.1. To do so, we choose n - I points {Xlo X2, X3, ••• , X.-l} between a and h and satisfying

a<

Xl

<

X2

< ... <

X.-l

< h.

260

Chapter 5: Integration

y

To make the notation consistent, we denote a by Xo and b by x., so that

a

= Xo

<

Xl

<

X2

< ... < X.-l <

x. = b.

The set is called a partition of [a, b]. The partition P divides [a, b] into n closed subintervals [xo,xtJ, [Xt,X2],"" [x.-t,x.]. FIGURE 5.8 A typical continuous function y = f(x) over a closed interval [a, b].

The first of these subintervals is [xo, xtJ, the second is Pis [x._ t, x.J, for k an integer between I and n.

[Xl>

X2], and the kth subinterval of

rkth subinterval j Xo = a

, x

Xl

The width of the first subinterval [Xc, xtJ is denoted dXl, the width of the second [Xl,X2] is denoted dX2, and the width of the kth subinterval is dx. = x. - X'-l. If all n subintervals have equal width, then the conunon width dxis equal to (b - a)/n.

In each subinterval we select some point. The point chosen in the kth subinterval is called c•. Then on each subinterval we stand a vertical rectangle that stretches from the x-axis to touch the curve at (c., f(c.)). These rectangles can be above or below the x-axis, depending on whether f( q) is positive or negative, or on the x-axis if f(c.) = 0 (Figure 5.9). On each subinterval we form the product f(C.)· dx•. This product is positive, negative, or zero, depending on the sign of f( q). When f( q) > 0, the product f( q) . dx. is the area of a rectangle with height f(c.) and width dx•. When f(c.) < 0, the product f( c.) • dx. is a negative number, the negative of the area of a rectangle of width dx. that drops from the x-axis to the negative number f( co) . [X.-l, x.]

y

kth rectangle

c,

(C2. f(c,))

FIGURE 5.9 The rectangles approximate the region between the graph of the function y = f(x) and the x-axis. Figure 5.8 has been enlarged to enhance the partition of [a. b]

and selection of points c, that produce the rectangles.

261

5.2 Sigma Notation and Limits of Finite Sums Finally we sum all these products to get

•

Sp = ~f(ck) Axk. The sum Sp is called a Riemann sum for f on the interval la, b]. There are many such sums, depending on the partition P we choose, and the choices of the points Ck in the subintervals. For instance, we could choose n subintervals all having equal width Ax = (b - a)/n to partition la, b], and then choose the point Ck to be the right-hand endpoint of each subinterval when forming the Riemann sum (as we did in Example 5). This choice leads to the Riemann sum formula y

S. =

" .....-

.....

.....

~

Y~fJ

,

oa

b

lJ

x

'-b-

~f(a + k b ;

a). (b ; a).

Similar formulas can be obtained if instead we choose Ck to be the left-hand endpoint, or the midpoint, of each subinterval. In the cases in which the subintervals all have equal width Ax = (b - a)/n, we can make them thinner by simply increasing their number n. When a partition has subintervals of varying widths, we can ensure they are all thin by controlling the width of a widest (longest) subinterval. We define the norm of a partition P, written Ilpll, to be the largest of all the subinterval widths. If Ilpll is a small number, then all of the subintervals in the partitionPhave a small width. Let's look at an example of these ideas.

(a)

EXAMPLE 6 The set P = to, 0.2, 0.6, I, 1.5, 2} is a partition of [0,2]. There are five subintervals of P: [0,0.2], [0.2,0.6], [0.6, I], [I, 1.5], and [1.5, 2]:

y

o

b x

a

The lengths of the subintervals are Ax! = 0.2, Ax2 = 0.4, Ax3 = 0.4, Ax. = 0.5, and Axs = 0.5. The longest subinterval length is 0.5, so the norm of the partition is Ilpll = 0.5. In this example, there are two subintervals of this length. _ (b)

FIGURE 5.10 The curve of Figure 5.9 with rectangles from ,mer partitions of la, b]. Finer partitions create collections of rectangles with thinner bases that approximate the region betweeo the graph off and the x-axis with increasing

accuracy.

Any Riemann sum associated with a partition of a closed interval [a, b] dermes rectangles that approximate the region between the graph of a continuous function f and the x-axis. Partitions with norm approaching zero lead to collections of rectangles that approximate this region with increasing accuracy, as suggested by Figure 5.10. We will see in the next section that if the function f is continuous over the closed interval [a, b], then no matter how we choose the partition P and the points Ck in its subintervals to construct a Riemann sum, a single limiting value is approached as the subinterval widths, controlled by the norm of the partition, approach zero.

Exercises 5.2 Sigma Notation Write the sums in Exercises 1--6 without sigma notation. Then evaluate them. 2

6k

+2+4+8+

16

+ 32 in

~ k-I

l'~k+1 k- l

2

3. ~ cosk

4. ~ sink..

•

7. Which of the followiog express I sigma notation?

~-­ • k-l k

k- l

,

k- l 4

6.

f-j( _I)k cos k

8. Which of the followiog express I - 2 sigma notation?

,

•. ~(-2)k-l k=i

+4

,

b. ~(-l)k2k k=O

C.

- 8

+

16 - 32 in

3 ~(_1)k+12k+2 k=-2

262

Chapter 5: Integration

4 (-I)k-! L~--

2 (-I), I>.~-

~ k - I

1 (-I)' C.~-

~k + I

'~1 k + 2

25. ~k(3k j;=1

27.

,

'-1

k- -l

7

11. I

+2+3+4+5+6

12. I

+ 4 + 9 + 16

14. 2

+ 4 + 6 + 8 + 10

I 2 3 4 5 --+---+--'55555

Values nf Finite Sums

"

"

1- 1

k-l

17. Suppose that ~a. = -5 and ~b. = 6. Fiod the values of

F16

•

" 18. Suppose that ~a.

=

" 0 and ~b. =

1=1

+

1. Fiod the values of

19. 20.

~k

L

L

'-1 13

~k

'-1 7

21. ~(-2k) >=-1

•

23. ~(3 - k 2 ) ... 1

5.3

264

c. ~1O k=3 71

17

I>. ~k2

" 31. a. ~4

I>. ~c

c. ~k(k - 1)

j;=3

F'

.t=18

"

"

c.~(k-I)

'-1

~(k+2n)

'-1 " k

"

~*

c. ~2

I>. ' - 1

k- l

n

Riemann Sums In Exercises 33-36, graph each function I(x) over the given interval.

Partition the ioterval ioto four suhiotervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum ~~- d(c.) .1x.. given that c. is the (a) left-hand endpoiot, (b) righthand endpoint, (e) midpoint of the kth subinterval. (Make a separate sketch for each set of rectangles.)

33. I(x)

=

x2

35. I(x)

=

siox,

-

I,

[0,2]

34. I(x)

=

-x2 ,

[-'1f, '1f]

36. I(x)

=

siox

37. Fiod the norm of the partition P

=

38. Fiod the norm of the partition P

=

[0, I]

+ I,

[-'1f, '1f]

{O, 1.2, 1.5,2.3,2.6, 3}. {-2, -1.6, -0.5, 0, 0.8, I}.

k=l

" - I) d. ~(b.

1)

'-1

Evaluate the sums io Exercises 19-32. 10

+ b.)

Fl

k=l

"

k=l

" e. ~(b. - 2a.)

d. ~(a. - b.)

'-1

c. ~(a.

I>. ~-

k=l

e. ~(a.

•

• b.

" ~3a.

L

7 k' -~.t=1 4

j;=1

>=-1

16

=1

30. a. ~k

32. a.

~k

500

,.

+ 1)

CY

28. I>. ~7

Fl

Express the sums io Exercises 11-16 io sigma notation. The form ofyour answer will depend on your choice of the lower limit of summation.

k=l

C),

29. a. ~3

+ If

I>. ~ (k

L ~(k - 1)2

26. ~k(2k

+ 5)

5 k' ~ 225 + ~k

10. Which formula is not equivalent to the other two? 4

7

5

9. Which formula is not equivalent to the other two?

10

10

I>. ~k2

c. ~k'

'-1 13

>=-1 13

I>. ~k2

c. ~k'

... 1 5 '1fk 22. ~T5

'-1

Limits nf Riemann Sums For the functions in Exercises 39-46, fmd a formula for the Riemann sum obtained by dividiog the ioterval [a, b] ioto n equal subintervals and using the right-hand endpoint for each c•. Then tske a limit of these sums as n ----i' 00 to calculate the area under the curve over [a, b].

39. I(x) = I - x 2 over the interval [0, I]. 40. I(x)

=

2x over the ioterval[O, 3].

41. I(x) = x 2

42. I(x)

=

+

I over the interval [0, 3].

2

3x overtheioterval[O, I].

43. I(x) = x

+ x 2 0ver the interval [0, I].

44. I(x) = 3x + 2x 2 over the ioterval [0, I].

'-1

•

24. ~(k2 - 5)

'-1

45. I(x)

=

2x' over the ioterval [0, I] .

46. I(x) = x 2

-

x' over the interval [-1,0].

The Definite Integral In Section 5.2 we investigated the limit of a finite sum for a function defined over a closed interval [a, b] usiog n subiotervals of equal width (or length), (b - a)/n. In this section we consider the limit of more general Riemann sums as the norm of the partitions of [a, b] approaches zero. For general Riemann sums the subintervals of the partitions need not have equal widths. The limiting process then leads to the definition of the definite integrol of a function over a closed interval [a, b].

Definition of the Definite Integral The definition of the defioite integral is based on the idea that for certain functions, as the norm of the partitions of [a, b] approaches zero, the values of the corresponding Riemann

5.3 The Definite Integral

263

sums approach a limiting value J. What we mean by this limit is that a Riemann sum will be close to tbe number J provided that tbe nann of its partition is sufficiently small (so that all of its subintervals have tbin enough widtha). We introduce tbe symbol E as a small positive number that specifies how close to J tbe Riemann sum must be, and tbe symbolll as a second small positive number that specifies how small tbe nann of a partition must be in order for convergence to happen. We now defme this limit precisely.

DEFINmON Let f(x) be a function defmed on a closed interval [a, hI. We say that a number J is tbe defmite integral of f over [a, bl and tbat J is tbe limit oftbe Riemann sums ~~=t!(c.) Ax. iftbe following condition is satisfied: Given any number E > 0 tbere is a corresponding number Il > 0 such tbat for every partition P = {xo, x" ... , xn } of [a, hI witb Ilpll < Il and any choice of in [Xk-" xii, we have

c.

The defmition involves a limiting process in which tbe nann of tbe partition goes to zero. In tbe cases where tbe subintervals all have equal widtb ~x = (h - a)/n, we can fann

each Riemann sum as

•

• (h ; a ),

S. = ~ f(ck) ~Xk = ~ f(ck)

ax, = ax = (b - alln for all k

c.

where is chosen in tbe subinterval Ax•. If tbe limit of tbese Riemann sums as n ---> exists and is equal to J, tbenJ is tbe defmite integral of f over [a, hI, so J = lim

~f(ck)(h; a)

n --+ 00 (:1

=

lim

n--+ 00

~f(ck) Ax.

{:1

00

ax = (b - all.

Leibniz introduced a notation for tbe defmite integral that captures its construction as a limit of Riemann sums. He envisioned tbe fmite sums ~~=1 f(Ck) Ax. becoming an infinite sum of function values f(x) multiplied by "infmitesimal" subinterval widtbs dx. The sum symbol ~ is replaced in tbe limit by tbe integral symbol j, whose origin is in tbe letter "S." The function values f(ck) are replaced by a continuous selection of function values f(x). The subinterval widtbs Ax. become tbe differential dx. It is as if we are summing all products of tbe fann f(x) • dx as x goes from a to h. While tbis notation captures tbe process of constructing an integral, it is Riemann's defmition that gives a precise meaning to tbe defmite integral. The symbol for tbe number J in tbe defmition of tbe definite integral is

tf(X)dx,

which is read as ''the integral from a to h off of x dee x" or sometimes as ''the integral from a to h off of x witb respect to x." The component parts in tbe integral symbol also have names: The function is the integrand. Upper limit of integration/

-""f;(x) ,h~r-"-... . ~ Lower limit of mtegration

When you :find the value of the integral, you have Integral off from a to b ~ evaluated the integral. ,

y

264

Chapter 5: Integration When the condition in the dermition is satisfied, we say the Riemann swns of I on [a, b] converge to the definite integral J = I{x} ax and that I is integrable over [a, b]. We have many choices for a partition P with norm going to zero, and many choices of points CO for each partition. The definite integral exists when we always get the same limit J, no matter what choices are made. When the limit exists we write it as the definite integral

t

lim f/{Ck}

11111~O

.... ,

~k =

J =

l

a

b, {X} ax.

When each partition has n equal subintervals, each of width ~ = {b - a}Jn, we will also write n

lim ~/(ck} ~ = J =

11_ 00

1=1

lb a

I{x} ax.

The limit of any Riemann swn is always taken as the norm of the partitions approaches zero and the nwnber of subintervals goes to inImity. The value of the dermite integral of a function over any particular interval depends on the function, not on the letter we choose to represent its independent variable. If we decide to use t or u instead of x, we simply write the integral as instead of

or

[/{X}ax.

No matter how we write the integral, it is still the same nwnber that is dermed as a limit of Riemann swns. Since it does not matter what letter we use, the variable of integration is called a dummy variable.

Integrable and Nonintegrable Functions Not every function dermed over the closed interval [a, b] is integrable there, even if the function is bounded. That is, the Riemann swns for some functions may not converge to the same limiting value, or to any value at all. A full development of exactly which functions defined over [a, b] are integrable requires advanced mathematical analysis, but fortonately most functions that commonly occur in applications are integrable. In particular, every continuous function over [a, b] is integrable over this interval, and so is every function having no more than a finite nwnber of iwnp discontinuities on [a, b]. (The latter are called piecewise-continuous /Unctions, and they are defined in Additional Exercises 11-18 at the end of this chapter.) The following theorem, which is proved in more advanced courses, establishes these results.

THEOREM l-Integralrility of Continuous Functions If a function I is continuous over the interval [a, b], or if I has at most finitely many iwnp discontinuities there, then the dermite integral I{x} ax exists and I is integrable over [a, b].

t

The idea behind Theorem 1 for continuous functions is given in Exercises 86 and 87. Briefly, when I is continuous we can choose each Ck so that I{ Ck} gives the maximwn value of I on the subinterval [Xk-), Xk], resulting in an upper swn. Likewise, we can choose CO to give the minimwn value of Ion [Xk-), xii to obtain a lower swn. The upper and lower swns can be shown to converge to the same limiting value as the norm of the partition P tends to zero. Moreover, every Riemann swn is trapped between the values of the upper and lower swns, so every Riemann swn converges to the same limit as well. Therefore, the nwnber J in the definition of the definite integral exists, and the continuous function I is integmble over [a, b]. For integrability to fail, a function needs to be sufficiently discontinuous that the region between its graph and the x-axis cannot be approximated well by increasingly thin rectangles. The next example shows a function that is not integrable over a closed interval.

5.3 The Definite Integral

EXAMPLE 1

265

The function

{I,0,

f(x) =

if x is rational if x is irrational

has no Riemann iotegral over [0, I]. Underlying this is the fact that between any two numbers there is both a rational number and an irrational number. Thus the function jwnps up and down too erratically over [0, I] to allow the region beneath its graph and above the x-axis to be approximated by rectangles, no matter how thio they are. We show, io fact, that upper sum approximations and lower sum approximations converge to different limitiog values. If we pick a partition P of [0, I] and choose to be the poiot giving the maximum value for f on [x._ 10 x.tJ then the correspondiog Riemann sum is

c.

•

•

U = ~f(C.) Ax. = ~(1) ~x. = I,

sioee each subioterval [Xk-- 10 x.] contaios a rational number where f(c.) = 1. Note that the lengtha of the iotervals io the partition sum to I, ~f~l~X' = 1. So each such Riemann sum equals I, and a limit of Riemann sums using these choices equals 1. On the other hand, if we pick to be the poiot giving the minimum value for f on [x._1o x.tJ, then the Riemann sum is

c.

•

L = ~ f(c.)

Ax.

•

~(O) Ax.

=

=

0,

sioee each subioterval [x.-1ox.tJ contaios an irrational number c. where f(C.) = 0. The limit of Riemann sums usiog these choices equals zero. Sioce the limit depends on the choices of c., the function f is not iotegrable. _ Theorem I says nothiog about how to calculate defioite integrals. A method of calculation will be developed io Section 5.4, through a connection to the proeess of taking antiderivatives.

Properties of Definite Integrals In defming tf(x) dx as a limit of sums ~f~!/(c.) Ax.. we moved from left to right across the ioterval [a, b]. What would happen if we iostead move right to left, starting with Xo = b and endiog at x. = a? Each Ax. io the Riemann sum would change its sign, with x. - Xk-I now negative iostead of positive. With the same choices of io each subioterval, the sign of any Riemann sum would change, as would the sign of the limit, the iotegral f(x) dx. Sioce we have not previously given a meaniog to iotegrating backward, we are led to defme

c.

[f(X)dx

= -

lb'

tf(X)dx.

Although we have ouly defmed the integral over an ioterval [a, b] when a < b, it is convenient to have adefmition for the iotegral over [a, b] when a = b, that is, for the iotegral over an ioterval of zero width. Sioce a = b gives ~x = 0, whenever f(a) exists we defioe

[f(X)dx

=

0.

Theorem 2 states basic properties of iotegrais, given as rules that they satisfy, iocludiog the two just discussed. These rules become very useful io the process of computing iotegrais. We will refer to them repeatedly to simplify our calculations. Rules 2 through 7 have geometric ioterpretations, shown io Figure 5.11. The graphs io these figures are of positive functions, but the rules apply to general iotegrable functions.

THEOREM 2 Whenf andg are iotegrable over the ioterval [a, b], the definite iotegral satisfies the rules io Table 5.4.

266

Chapter 5: Integration

TABLE 5.4

RuLes satisfied by definite integrals bf - l (x) ax

1.

Ordero/Integration: l"f(x) ax

2.

Zero Width Interval:

l

3.

Constant Multiple:

l\f(x)ax

4.

Sum and Difference:

l bU(x) ± g(x}} ax

5.

Additivity:

lbf(x) ax + l'f(x) ax

6.

Max-Min Inequality:

If f has maximum value max f and minimum value minf on [a, b], then

=

A Def"mition

a

A Def'mition when!(a) exists

f(x) ax - 0

"-

=

bf kl (x)ax =

Anycons_k

bf (x) ax ± l

l

b g(x) ax

l'f(x) ax

=

bf minf·(b - a) " ' l (x)ax '" maxf·(b - a). 7.

y

Domination:

f(x)

20

g(x) on [a, b] => l

f(x)

20

Oon[a,b] => l

y

-O~~ a--------~ b~x

(a) Zero Width Interval:

(b) Constant Multiple: (k

a

[f(X)dx

~0

l ' kf(x) dx

y

bf (x)ax

20

l'

~k

~

2)

f(x) dx

o

l

0

(Special Case)

a

y

~ !(x)

y

~

y

~ !(x)

x

b

l'(f(x)

+ g(x))dx

~ l'f(x)dx + l'g(x)dx

y

min! y

[f(X)dx

+ [f(X)dx

~ [f(X)dx

~~----------~~x

o

a

b

(e) Max-Min Inequality:

mmf'(b - a)

"'1'

f(x)dx

'" maxf'(b - a) FIGURE 5.11

Geometric interpretations of Rules 2-7 in Table 5.4.

+ g(x)

g(x)

max!

(d) Additivity for definite integrals:

(x) ax

(c) Sum: (areas add)

y

-O~--~a----~b~-------c~~x

bg

20

y

~------~----~x

o

bf (x) ax

~

g(x)

~~--------~~x

oa

b

(f) Domination:

f(x)

=>

20

l'

g(x) on [a, b]

f(x) dx

20

l'

g(x) dx

5.3 The Definite Integral

267

While Rules I and 2 are def'mitions, Rules 3 to 7 of Table 5.4 must be proved. The following is a proof of Rule 6. Similar proofs can be given to verify the other properties in Table 5.4.

Proof of Rule 6 Rule 6 says that the integral of I over [a, b] is never smaller than the minimum value of I times the length of the interval and never larger than the maximum value of I times the length of the interval. The reason is that for every partition of [a, b] and for every cboice of the points Ck, min/·(b - a) = mini'

• L dXk k=1 Constant Multiple Rule

min! S !(c,)

Constant Multiple Rule

maxi' (b - a).

=

In short, all Riemann sums for I on [a, b] satisfy the inequality

•

min I' (b - a) ,,; ~/(ck) dxk"; max I' (b - a).

•

Hence their limit, the integral, does too.

EXAMPLE 2

To illustrate some of the rules, we suppose that

[1(X)dx

j1

4/

= 5,

(X) dx

-2,

=

th(x) dx

and

J-I

= 7.

Then

1. [I(X) dx = 2.

3.

[[2 /

(X)

f'1(x) dx

1-1

EXAMPLE 3

j4/

(x) dx

+ 3h(x)] dx

= -( -2) =

=

2[I(X) dx

=

2(5)

+ 3[h(X) dx

+ 3(7)

j41/

Rulel

=

Rules 3 and 4

31

=

5

+ (-2)

Show that the value of fa!.'/!

+

cosxdx is less than or equal to

=

+

2

tl(x) dx

1-1

(x) dx

=

3

•

Rule 5

V2.

The Max-Min Inequality for def'mite integrals (Rule 6) says that min I' (b - a) is a lower bound for the value of I(x) dx and that max I' (b - a) is an upper bound.

Solution

The maximum value of

VI

f:

+

11VI

cosx on [0, I] is

+

cosxdx";

v'1+l = V2, so

V2'(1 -

0) =

V2.

•

268

Chapter 5: Integration

Area Under the Graph of a Nonnegative Function We now return to the problem that started this chapter, that of defining what we mean by the area of a region having a curved boundary. In Section 5.1 we approximated the area under the graph of a nonnegative continuous function using several types of finite sums of areas of rectangles capturing the region-upper sums, lower sums, and sums using the midpoints of each subinterva1-all being cases of Riemann sums constructed in special ways. Theorem I guarantees that all of these Riemann sums converge to a single def'mite integral as the norm of the partitions approaches zero and the number of subintervals goes to inf'mity. As a result, we can now define the area under the graph of a nonnegative integrable function to be the value of that definite integral.

DEFINITION If Y = f{x) is nonnegative and integrable over a closed f{x> over [II, b] is the interval [a, b], then the area under the curve y integml of f from a to b,

=

For the first time we have a rigorous def'mition for the area of a region whose boundary is the graph of any continuous function. We now apply this to a simple example, the area under a straight line, where we can verifY that our new definition agrees with our previous notion of area.

EXAMPLE 4

Compute

J: ax x

and find the area A under

y=

x over the interva1

[O,b],b>O. Solution

~~--------~~-+x

o

b

FIGURE 5.12 The region in Example 4 is a triangle.

The region of interest is a triangle (Figure 5.12). We compute the area in two ways.

To compute the def'mite integral as the limit of Riemann sums, we calculate lim lPl~o ~~,f{ck) dxk for partitions whose norms go to zero. Theorem 1 tells us that it does not matter how we choose the partitions or the points Ck as long as the norms approach zero. All choices give the exact same limit So we consider the partition P that subdivides the interval [0, b] into n subinterva1s of equal width dx = {b - O)ln = bin, and we choose Ck to be the right endpoint in each subinterval. The partition is p = { 0,

t 2:, 3:,,,. ,n: }andck

= ,,:. So

Constant Multiple Rule

=

b2 n2

n{n + 1) •

2

Sum of First n Integers

5.3 y

269

All n --> 00 and Ilpll --> 0, this last expression on the right has the limit b'/2. Therefore,

r

b

Jo xdx b a ~~--L-----+-~x

o

The Definite Integral

a I<-b -

b

a~

(a)

b

=

2

2'

(b) Since the area equals the definite integral for a nonnegative function, we can quickly derive the definite integral by using the fonnula for the area of a triangle having base length b and height y = b. The area is A = (1/2) b· b = b 2/2.Again we conclude that x dx = b 2/2. •

J:

Example 4 can be generalized to integmte I(x) [a,bl, O < a < b.

=

x over any closed interval

y

Rule 5

Rule 1 ----~--_+~~~x

Example 4

In conclusion, we have the following rule for integrating fix) = x: (b)

bxdx

y

--r-----~~--~--~x

(c)

FIGURE 5.13 (a) The area of this trapezoidal regioo is A = (b' - .')/2. (b) The defmite integral in Equation (1)

gives the negative of the area of this

trapezoidal regioo. (c) The defmite integral in Equatioo (I) gives the area of the blue ttiangular regioo added to the negative of the area of the gold ttiangular region.

= b

2

_

2

1a

a2 2'

a
(1)

This computation gives the area ofa !mpezoid (Figure 5.13a). Equation (I) reruains valid when a and b are negative. When a < b < 0, the definite integral value (b 2 - a 2 )/2 is a negative number, the negative of the area of a !mpezoid dropping down to the line y = x below the x-axis (Figure 5.13b). When a < 0 and b > 0, Equation (I) is still valid and the defmite integral gives the difference between two areas, the area under the graph and above [0, bI minus the area below [a, 01 and over the graph (Figure 5.13c). The following results can also be established using a Riemann sum calculation similar to that in Example 4 (Exercises 63 and 65).

1 1\2 b

c dx

= c(b - a),

dx

a

c any constant

3- a3 3 3'

= b

a
(2)

(3)

Average Value of a Continuous Function Revisited In Section 5.1 we introduced informally the averege value of a nonnegative continuous function lover an interval [a, bl, leading us to defme this averege as the area under the graph of y = I(x) divided by b - a. In integral notation we write this as

Averege

= b

~ a 1bl(x)dx.

We can use this formula to give a precise definition of the averege value of any continuous (or integrable) function, whether positive, negative, or both. Alternatively, we can use the following reasoning. We start with the idea from arithmetic that the averege of n numbers is their sum divided by n. A continuous function I on [a, bl may have infinitely many values, but we can still sample them in an orderly way.

270

Chapter 5: Integration

y

We divide [a, h] into n subintervals of equal width ~ = (h - a)/n and evaluate point Ck in each (Figure 5.14). The average of the n sampled values is

I(c,)

at a

+ 1(c2) + ... + I(c.) _ !

~ ( ) -n~/Ck

n

~I

~.

h- a

=

L

b-a 1 Ilx Ilx~--.-.son= b-a

l(ck)

k- I

I • =-h- L/(ck)~ a 1=1 FIGURE 5.14 A sample of values ofa function on an interval [a, h].

I

Constant Multiple Rule

The average is obtained by dividing a Riemann sum for I on [a, h] by (h - a). As we increase the size of the sample and let the norm of the partition approach zero, the average approaches (I/(h - a»I:I(x) dx. Both points of view lead us to the following definition.

DEFINITION If I is integrable on [a, h], then its average value on Ia, h], also called its mean, is

l

y 2 f(x)~V4-x2 -y~

b

av(f) = h _I a a I(x) dx.

"-2 EXAMPLE 5

Find the average value of I(x) = V4=7 on [-2,2].

--_2~-_1~--r---~~2~ X

FIGURE 5.15

The average value of

f(x) ~ \l4=?on[-2,2]is1T/2 (Example 5).

Solution We recognize I(x) = V4=7 as a function whose graph is the upper semicircle of radius 2 centered at the origin (Figure 5.15). The area between the semicircle and the x-axis from - 2 to 2 can be computed using the geometry formula Area =

t.

7Tr2

=

t.

7T(2)2

=

27T.

Because I is nonnegative, the area is also the value of the integral of I from - 2 to 2,

1V 2

4 - x 2 dx

=

27T.

-2

Therefore, the average value of I is

av(f) = 2 _1(_2)1:V4=7dx = t(27T) = ; . Theorem 3 in the next section asserts that the area of the upper semicircle over [-2,2] is the same as the area of the rectangle whose height is the average value of lover [ - 2, 2] • (see Figure 5.15).

Exercises 5.3 Interpreting Limits as Integrals Express the limits in Exercises 1-8 as defmite integrals. 1. 2.

•

..

lim ~clax where P is a partition of [0,2]

IWII~o '~1

•

lim ~2c.i'

IWII~o '~1

ax.. where P is a partition of[ -1,0]

n

3. 4.

5.

lim

Y(cc -

IWi~o~

lim

±(1 )ax" ±

IIP~-O 1-1

lim

3ck)ax.,wherePisapartitionof[-7,5]

IWi~o""

k

where P is a partition of [1,4]

-I_I-- ax., whereP is a partition of [2, 3] - c.

5.3

6. 7. 8.

•

lim Y

1~1~0~

v'4 - c,' ax" where P is a partition of[O, I]

•

ax" whereP is a partition of[ -"./4,0]

lim ~(secc.)

1~I~oFl

•

lim ~(tanc.)

1~I~oFl

ax" where P is a partition of [0, "./4]

Using the Definite Integral Rules 9. Suppose that f and g are integrable and that

~ -4,

l'f(x) dx

l'f(x) dx

~ 6,

l'g(x) dx

~ 8.

!,'

b.

c·l'3f(x)dx

d. J,'f(X)dx

[f(x) - g(x)] dx

~ -I,

15. E(I + 3) dx

16.

17.1:~dx

18.

19. Dxldx

20. 1:(1 -

Ixl) dx

22. 1:(1

\1'1=7) dx

21.

1:(2 -

Ixl) dx

I

f.l'[4f(x) - g(x)] dx

1,'

f(x) dx

1,'

~ 5,

>

h(x) dx

~ 4.

-2f(x) dx

b.

1,'

[f(x)

+ h(x)] dx

c. 1,'[2f(x) - 3h(x)] dx

d. J.' f (X) dx

e. l'f(x) dx

f. J.'[h(X) - f(x)] dx

11. Suppose that f.,'f(x) dx ~ 5. Find

l'

Y3f(z)

dz

a. l'f(U) du

b.

c. J,'f (l)dl

d.l'[-f(X)]dx

t, g(l) dl ~ v'2. Find

l:

a. 1,-'g(l) dl

b.

c·l°[

d.l° ~~

-3

-g(x)] dx

-3

13. Suppose thatf is integrable and that

104 f(z) dz ~ 7. Find

a.

1,4f (z) dz

g (u) du

v2

dr

fa' f(z) dz ~ 3 and

b. /.' f(l) dl

14. Suppose that h is integrable and that f~, h(r) dr ~ 0 and

l, h(r) dr ~ 6. Find

a.l'h(r)dr

+

x'dx

_1,' h(U)dU

0

28. f(x)

on •• [-1,0], b. [-1,1]

Evaluattng Definite Integrals Use the results of Equations (I) and (3) to evaluate the integrals in Exercises 29-40.

32.

i,

xdx

30.

sv'2 rdr

33.

J{'/2 v2

35. io

I'dl

l

v'3a xdx

1','

xdx

J.

34.

io{"-'8' ds

w

05

{V'7

io x'dx 36. 1,w/2IP dO 39.

'W

31.

OdO

37·1"xdx

{~ x'dx

io

io(" x'dx

40.

Use the rules in Table 5.4 and Equations (1)-(3) to evaluate the integrals in Exercises 41-50.

41. 1,'7 dx

42.1,'sxdx

43.1,'(21 - 3) dl

44.

45.[(I+I)dz

46. 1,0(2z - 3)

47·1'3U'dU

48. (' 24u' du

49.

1,\3x' + x -

{v2

5) dx

io

(I -

v'2) dl

dz

i'/2 50.1°(3x' + x -

5) dx

Finding Allla by Definite Integrals 10 Exercises 51-54, use a definite integral to rmd the area of the region between the giveo curve and the x-axis on the interval [0, b].

51. Y ~ b.

>

on a. [-2,2], b. [0,2]

38..

12. Suppose that

J~v'16 -

26.1'3Id', O
\1'4=7 ~ 3x + \1'1=7

['/2

l'

(-2x+4)dx

27. f(x) ~

29.

Use the rules in Table 5.4 to rmd

a.

11/2'/2

24.1,'4Xdx, b

0

25.1'2sds, O
10. Suppose that f and h are integrable and that

l ' f(x) dx

Using Known Alii .. to Find Integrals 10 Exercises 15--22, graph the integrands and use areas to evaluate the integrals.

23.1,' dx, b

g (x) dx

a. J,'g(x) dx

l'

271

Use areas to evaluate the integrals in Exercises 23--28.

Use the rules in Table 5.4 to rmd

e.

The Definite Integral

3x'

52. Y ~ ".x'

272

Chapter 5: Integration

Finding Average value In Exercises 55--{j2, graph the function and fmd its average value over the given interval.

55. f(x) = x' - I

on

[0,

v'3]

56. f(x) =

x' -T

58. f(x)

3x' - 3 on [0, I]

=

on

57. f(x) = -3x' - I

[0,3]

on

[0, I]

101 sin (x') ax cannot possibly be 2. 1 Show that the value of f0 Vx+8 ax lies between 2 v'2 '" 2.8

75. Show that the value of 76.

and 3. 77. Integrals of nonnegative functions Use the Max-Min inequality to show that iff is integrable then

f(x) '" 0 on

=*

[a, b]

J.bf(x) ax '" O.

59. f(t) = (t - I)' on [0,3] 60. f(t)

=

61. g(x) =

62. h(x) =

[-2, I]

t' - t on

Ixl -Ixl

Ion •• [-I, I], h. [1,3],and c. [-1,3] on a. [-1,0], b. [O,I],and

c.

78. Integral. of nonpositi"" function. then

[-1,1]

f(x) s 0 on Definite Integrals as Limits Use the method of Example 4a to evaluate the defmite integrals in Exercises 63-70.

63. J.b C ax

65. J.b x'ax,

67.1;

66.

68.1:

(3x' - 2x + I) ax

69. J.b x' ax,

1:

a< b

70.

J.bf(X) ax s O.

79. Use the inequality sin x '" x, which holds for x '" 0, to fmd an 1 sin x upper bound for the value of

10

ax.

1) ax

80. The inequality sec x '" I + (x'/2) holds on (-"./2, "./2). Use it 1 to find a lower bound for the value of f0 sec x

(x - x') ax

81. Ifav(f) really is a typical value of the integrshle function f(x) on [a, b], then the constant function av(f) should have the same integral over [a, b] as f. Does it? That is, does

64. /,' (2x

a< b

~

[a, b]

Show that if f is integrable

+

ax.

x'ax J.b av(f) ax

/,1

=

J.bf(x) ax?

(3x _ x') ax Give reasons for your answer.

Theory and Examples 71. What values of a and b maximize the value of

82. It would be nice if average values of integrable functions obeyed the following rules on an interval [a, b].

a. av(f + g) = av(f) + av(g) b. aV(kf)

= kav(f)

c. av(f) '" av(g)

(Hint: Where is the integrand positive?)

(any number k) if f(x) '" g(x)

on

[a, b].

Do these rules ever hold? Give reasons for your answers.

72. What values of a and b minimize the value of

83. Upper and lower sums for increasing functions

a. Suppose the graph of a continuous function f(x) rises steadily

73. Use the Max-Min Inequality to fmd upper and lower bounds for the value of 1 /o ,

I --ax 1 + x2 •

74. (Continuation of Exercise 73.) Use the Max-Min Inequality to fmd upper and lower bounds for

/,o0.' -1 -+ a x I

and

x2

as x moves from left to right across an interval [a, b]. Let P be a partition of [a, b] into n subintervals oflength ax = (b - alin. Show by reforting to the accompanying figure that the difference between the upper and lower sums for f on this partition can be represented graphically as the area of a rectangleR whose dimensions are [f(b) - f(a)] by ax. (Hint: The difference U - L is the sum of areas of rectangles whose diagonals QOQb Q1Q" ... , Q.-1Q. lie along the

curve. There is no overlapping when these rectangles are shifted horizontally onto R.)

/,1 - - a x . 0.5

I 1 + x2

b. Suppose that instead ofbeing equal, the lengths ax,ofthe subintervals of the partition of [a, b] vary in size. Show that U- L s If(b) - j(a)lax.....

Add these to arrive at an improved estimate of 1 /o ,

I

- - a2 x 1

+x

•

where

ax,... is the norm of P, and hence that linlj~~o = O.

(U - L)

5.3

The Definite Integral

273

y

y

1

f(b) - f(a)

Rj --->I~I<-

-"

y

77

84. Upper and lower sums for decreasing functions (Continuation ofExercise 83.) •• Draw a figure like the one in Exercise 83 for a continuous function f(x) whose values decrease steadily as x moves from left to right across the interval [a, b]. LetP be a partition of [a, b] into subintervals of equalleogth. Find an expression for U - L that is aoa1ogous to the ooe you found for U - L in Exercise 83a.

7

/

b. Suppose that instead ofbeing equal, the lengths ax,ofthe subintervals ofP vary in size. Show that the inequality

0

r

7

V

a

Xl

Xk + 1

b

a

x,

Xk+l

b

x

y

U-L'" If(b) - f(a)l4xm.x

of Exercise 83b still holds and hence that liml"U~o (U - L) ~ O. 85. Use the formula

sinh

+ sin2h + sin3h + ... +

sinmh

«m

cos (h/2) - cos + (I/2))h) 2 sin (h/2) to fmdthe area uoder the curvey ~ sin x from x ~ Otox ~ ",/2 in two steps: •• Partition the interval [0, "'/2] into n subintervals of equal length and calculate the corresponding upper sum U; then b. Find the limit of U as n --> 00 and ax ~ (b - a)/n --> O. 86. Suppose that f is continuous and nonnegative over [a, b], as in the accompanying figure. By inserting points XloX2,··· ,.:t,t-loX} •••• , Xn-l

as shown, divide [a, b] into n subintervals oflengths ax, ~ x, - a, dx2 = Xl - Xl, • •• , dA" = b - XII_I, which need not be equal. •• !fm, ~ min {f(x) for x in the kth subinterval}, explain the connection between the lower sum L = ml

axl + m2 dx2 + ... + mil 4x1l

and the shaded regions in the first part of the figure. b. !f M, ~ max {f(x) for x in the kth subinterval}. explain the

connection between the upper sum U~

M,ax, + M2ax2 + ... + M.ax.

and the shaded regions in the second part of the figure. c. Explain the connection between U - L and the shaded regions along the curve in the third part of the figure.

0

II

I,

~ b-a

l-j

,I

87. We say f is uniformly continuous on [a, b] if given any E > 0, thereisa8 > osuch thatifx"x2 are in [a, b] and lx, - x21 < 8, then If(x,) - f(X2) I < E. It can be shown that a continuous function on [a, b] is uniformly continuous. Use this and the figure for Exercise 86 to show that iff is continuous and E > 0 is given, it is possible to make U - L '" E' (b - a) by making the largest of the ax,'s sufficiently small. 88. If you average 30 mijh on a 150·mi trip and then return over the same 150 mi at the rate of 50 mi/h, what is your average speed for the trip? Give reasons for your answer. COMPUTER EXPLORATIONS

!fyour CAS can draw rectaugles associated with RieD2anD sums, use it to drsw rectangles associated with Riemann sums that converge to the integrals in Exercises 8g..-94. Use n ~ 4, 10,20, and 50 subintervals of equal length in each case. 89.

[(I -

x)dx

~k

274 90.

Chapter 5: Integration

1'<x'

+ I) tb;

91.

J~ cos x tb; =

92.

Jo

93.

[ixl

=

b. Partition 1he interval into n = 100, 200, and 1000 subinter-

1

vals of equallengtb, and evaluate 1he function at 1he ntidpoint of each subinterval.

0

c. Compute the average value of the function values generated in part (b).

[~/4

d. Solve 1he equationf(x)

= (average value) for x using 1he average value calculated in part (c) for 1he n = 1000 partitioning. 95. f(x) = sinx on [0, 'IT] 96. f(x) = sin' x on [0, 'IT]

sec' x tb; = I tb; =

94.1' ~tb;

I

(The integra1'svalue is about 0.693.)

10 Exercises 95-98, use a CAS to perform 1he following steps:

Plot 1he functions over 1he given interval.

L

97. f(x) = x sin

~

98.f(x)=xsin'~

[~,,,.]

on

on

[~,'IT]

5.4 I The Fundamental Theorem of Calculus In this section we present the Fundamental Theorem of Calculus, which is the central theorem of iutegral calculus. It connects iutegration and differentiation, enabliug us to compute iutegrals usiug an antiderivative of the iutegrand function rather 1han by taking limits of Riemann sums as we did iu Section 5.3. Leibniz and Newton exploited this relationship and started ma1hematical developments that fueled the scientific revolution for the next 200 years. Along the way, we present an integral version of the Mean Value Theorem, which is another important theorem of iutegral calculus and is used to prove the Fundamental Theorem.

HISTORICAL BIOGRAPHY

Sir Isaac Newton (1642-1727) y

f(e), average height ~~------L---~~x

o

a

c ~b- a

FIGURE 5.16

b

------->I

The value f(e)

in 1he

Mean Value Theorem. is, in a sense, the

average (or mean) height off on [a, b]. wtwn f '" 0, 1he area of1he rectaogle is 1he area under 1he graph of f from a

Mean Value Theorem for Definite Integrals In the previous section we defiued the average value of a continuous function over a closed iuterva1 [a, hI as the def'mite iutegral I{x) dx divided by the length or width h - a of the iuterva1. The Mean Value Theorem for Definite Integrals asserts that this average value is always taken on at least once by the function 1 iu the iuterva1. The graph iu Figure 5.16 shows a positive continuous function y = I{x) defiued over the iuterval [a, hI. Geometrically, the Mean Value Theorem says that there is a number c iu [a, hI such that the rectangle with height equal to the average value I{c) of the function and base width h - a has exactly the same area as the region beneath the graph of1 from a to h.

1:

to h,

f(e)(b - a) =

l'

f(x) tb;.

THEOREM 3-The Mean Value Theorem for Definite Integrals ous on [a, hI, then at some poiut c iu [a,

hI,

If 1 is continu-

1 b

I{c) = h _I a a I{x) dx.

Proof Ifwe divide both sides of the Max-Miu Inequality (Table 5.4, Rule 6) by (h - a), we obtain

5.4 The Fundamental Theorem of Calculus y

, 1 2

Average value 112

----------- notassumed

-nt-----t-----~-------+x

o

2

FIGURE 5.17

A discontinuous function

275

Since I is continuous, the Intermediate Value Theorem for Continuous Functions (Section 2.5) says that I must assume every value between min I and max f. It must therefore assume the value (1/(b - a»I:/(x) dx at some pointe in [a, b]. • The continuity of I is important here. It is possible that a discontinuous function never equals its average value (Figure 5.17).

EXAMPLE 1

Sbow thatifl is continuous on [a, b],a '" b,andif

1

need not assume its average value.

b/(x) dx

=

0,

then I(x) = 0 at least once in [a, b]. Solution

The average value of I on [a, b] is

1 b

av(f) = b _I a

a

I(x) dx = b _I a . 0 = O.

By the Mean Value Theorem, I assumes this value at some point e

E

[a, b].

•

Fundamental Theorem, Part 1 area

y

~

If I(t) is an integrable function over a fiuite interval I, then the integral from any rlXed number a E I to another number x E I defines a new function F whose value at x is

F(x)

F(x) = [I(t) dt.

o

a

x

b

FIGURE 5.18 The funetionF(x) dermed by Equation (I) gives the area under the graph of f from a to x when

f is nonnegative and x > a.

(1)

For example, if I is nonnegative and x lies to the right of a, then F(x) is the area under the graph from a tox (Figure 5.18). The variable x is the upper limit of integration of an integra!, but F is just like any other real-valued function of a real variable. For each value of the input x, there is a well-dermed numerical output, in this case the dermite integra! of I from a to x. Equation (1) gives a way to derme new functions (as we will see in Section 7.2), but its importance now is the connection it makes between integrals and derivatives. If I is any continuous function, then the Fundamental Theorem asserts that F is a differentiable function of x whose derivative is I itself. At every value of x, it asserts that d dx F(x) = I(x).

y

To gain some insight into why this result holds, we look at the geometry behind it. If I ;;,: 0 on [a, b], then the computation of F'(x) from the dermition of the derivative means taking the limit as h --> 0 of the difference quotient F(x b

FIGURE 5.19

In Equation (I), F(x) is the area to the left of x. Also, F(x + h) is the area to the left of

+ h. The difference quotient [F(x + h) - F(x)l/h is then

x

approximately equal to f(x), the height of the rectangle shown here.

+ h) - F(x) h

For h > 0, the numerator is obtaioed by subtracting two areas, so it is the area under the graph of I from x to x + h (Figure 5.19). If h is small, this area is approximately equal to the area of the rectaogle ofheight/(x) and width h, which can be seen from Figure 5.19. That is, F(x

+ h) - F(x) '" h/(x).

Dividing both sides of this approximation by h and letting h --> 0, it is reasonable to expect that F'(x) = lim F(x h~O

+ h) - F(x)

=

I(x).

h

This result is true even if the function I is not positive, and it forms the first part of the Fundamental Theorem of Calculus.

276

Chapter 5: Integration

THEOREM 4-The Fundamental Theorem of Calculus, Part 1 If I is continuous on [a, b], thenF(x) = I(t) dt is continuous on [a, b] and differentiable on (a, b) and its derivative is I(x):

J:

X

(2)

F'(x) = ax d ia I(t) dt = I(x).

Before proving Theorem 4, we look at several examples to gain a better understanding of what it says. In each example, notice that the independent variable appears in a limit of integration, possibly in a formula.

EXAMPLE 2 (a) y =

Solution

i

Use the Fundamental Theorem to tmd dy/ax if

X

(t 3

+ 1) dt

(h) y = lS3tSintdt

(e) y = ] ' " costdt

We calculate the derivatives with respect to the independent variable x.

(a) -dy = -d ax ax

i a

X

(t 3

+

I)dt = x 3

+

Eq.(2)withf(') ~,'

1

dy = ax diS d(f' ) (h) ax x 3tsintdt = ax - Js 3tsintdt

+

I

Table 5.4, Rule I

rx

= -

d axJs 3tsintdt

= -

3x sin X

Eq. (2) with f(') ~ 3tsin'

(e) The upper limit of integration is not x but functions,

x 2• This

makes y a composite of the two

and

y = ] " costdt

We must therefore apply the Chain Rule when finding dy/ ax.

dy dy du ax=du'ax =

(~]" costdt). du 1

= cosu .

du ax

du ax

= cos(x 2 ). 2x =

2xcosx 2

•

Proof of Theorem 4 We prove the Fundamental Theorem, Part I, by applying the definition of the derivative directly to the function F(x), when x and x + h are in (a, b). This means writing out the difference quotient F(x + h) - F(x) h

(3)

5.4 The Fundamental Theorem of Calculus

277

and showing that its limit as h --+ 0 is the number I(x) for each x in (a, b). Thus,

F'(x)

Fi,,-(x_+~h)'----_Fi-"(x-'-) _

= lim h~O =

h

l~d [[+O/(t) dt l

= lim -h h--+O

l

[I(t) dt]

X

+hI(t)

x

dt

Table 5.4, Rule 5

According to the Mean Value Theorem for Definite Integrals, the value before taking the limit in the last expression is one of the values taken on by 1 in the interval between x and x + h. That is, for some number c in this interval,

11'H

Ii,

I(t) dt

=

I(c).

(4)

All h --+ 0, x + h approaches x, forcing c to approach x also (because c is trapped between x and x + h). Since 1 is continuous atx, I(c) approaches I(x): lim I(c) = I(x).

(5)

h-O

In conclusion, we have

F'(x)

= lim -hi h-O

= lim

h~O

=

l

x

X +h

I(t) dt

I(c)

I(x).

Eq. (4) Eq. (5)

If x = a or b, then the limit of Equation (3) is interpreted as a one-sided limit with h --+ 0+ or h --+ 0-, respectively. Then Theorem I in Section 3.2 shows that F is continuous for every point in [a, b]. This concludes the proof. •

Fundamental Theorem, Part 2 (The Evaluation Theorem) We now come to the second part of the Fundamental Theorem of Calculus. This part describes how to evaluate definite integrals without having to calculate limits of Riemann sums. Instead we find and evaluate an antiderivative at the upper and lower limits of integration.

If1 is continuous at every point in [a, b] and F is any antiderivative of 1 on [a, b], then

THEOREM 4 (Continued)-The Fundamental Theorem of Calcuius, Part 2

l b,

(x) dx

=

F(b) - F(a).

Proof Part I of the Fundamental Theorem tells us that an antiderivative of 1 exists, namely

G(x)

=

[I(t) dt.

Thus, if F is allY antiderivative of I, then F(x) = G(x) + C for some constant C for a < x < b (by Corollary 2 of the Mean Value Theorem for Derivatives, Section 4.2). Since both F and G are continuous on [a, b], we see that F(x) = G(x) + C also holds when x = a and x = b by taking one-sided limits (as x --+ a + and x --+ r).

278

Chapter 5: Integration Evaluating F(b) - F(a), we have

F(b) - F(a) = [G(b)

+ C] -

[G(a)

+ C]

=

G(b) - G(a)

=

J.b/(t) dt - J."/(t) dt

=

J.b/(t) dt - 0

=

J.b/(t) dt.

•

The Evaluation Theorem is important because it says that to calculate the definite integral of / over an interval [a, b] we need do only two things: 1.

Find an antiderivative F of /, and

2.

Calculate the number F(b) - F(a), which is equal to

1: /(x) dx.

This process is much easier than using a Rieroann sum computation. The power of the theorem follows from the realization that the definite integral, which is defmed by a complicated process involving all of the values of the function / over [a, b], can be found by knowing the values of any antiderivative F at only the two endpoints a and b. The usual notation for the difference F(b) - F(a) is

F(x)

1:

or

depending on whether F has one or more terms.

EXAMPLE 3 We calculate several definite integrals using the Evaluation Theorem, rather than by taking limits of Riemann sums. (a)

l

u

cosx dx = sin x

I

d

.

dx smx

=

cosx

d dxsecx

=

sec x tan x

= sin". - sinO = 0 - 0 = 0

(b)

r

J-w/4

sec x tan x dx = secx]O -w/4 = secO- sec ( - ; ) = 1-

(e)

l' (~Vi - x~)

dx =

[x

3/2

+

[(4)3/2 =

[8

Vi

iI + ~]- [(1)3/2 +

+ 1] - [5]

=

t]

4.

•

Exercise 66 offers another proof of the Evaluation Theorem, bringing together the ideas of Riemann sums, the Mean Value Theorem, and the defmition of the definite integral.

The Integral of a Rate We can interpret Part 2 of the Fundaments! Theorem in another way. If F is any antiderivative of /, then F' = f. The equation in the theorem can then be rewritten as

J.bF'(X) dx = F(b) - F(a).

5.4 The Fundamental Theorem of Calculus

279

Now F'(x) represents the rate of change of the function F(x) with respect to x, so the integral of F' is just the net change in F as x changes from a to b. Fonnally, we have the following result.

THEOREM 5-The Net Change Theorem

The net change in a function F(x) over an interval a :5 x :5 b is the integral of its rate of change:

F(b) - F(a)

EXAMPLE 4

=

l

b

F'(x) dx.

(6)

Here are several interpretations of the Net Change Theorem.

(a) If c(x) is the cost of producing x units of a certain commodity, then c'(x) is the marginal cost (Section 3.4). From Theorem 5, x

l..

,C'(X) dx

=

C(X2) - C(XI),

which is the cost of increasing production from XI units to X2 units.

(b) If an object with position function .(t) moves along a coordinate line, its velocity is v(t) = .'(t). Theorem 5 says that

l"

"v(t) dt = .(t2) - .(tl),

so the integral of velocity is the displacement over the time interval tl :5 t :5 t2' On the other hand, the integral of the speed Iv(t) I is the total distaoce traveled over the • time interval. This is consistent with our discussion in Section 5.1.

If we rearrange Equation (6) as

F(b)

=

F(a) + lbF'(X) dx,

we see that the Net Change Theorem also says that the f'mal value ofa functionF(x) over an interval [a, b] equals its initial value F(a) plus its net change over the interval. So ifv(t) represents the velocity function of an object moving along a coordinate line, this means that the object's final position .(t2) over a time interval tl :5 t :5 t2 is its initial position S(tl) plus its net change in position along the line (see Example 4b).

EXAMPLE 5 Consider again our analysis of a heavy rock blown straight up from the ground by a dynamite blast (Example 3, Section 5.1). The velocity of the rock at any time t during its motion was given as v(t) = 160 - 32t ft./sec. (a) Find the displacement of the rock during the time period 0 :5 t :5 8.

(b) Find the total distance traveled during this time period. Solution (a) From Example 4b, the displacement is the integral

10rv(t) dt = 10r(160 =

32t) dt = [160t - 16t 2]8 0

(160)(8) - (16)(64) = 256.

This means that the height of the rock is 256 ft. above the ground 8 sec after the explosion, which agrees with our conclusion in Example 3, Section 5.1.

280

Chapter 5: Integration

(b) As we noted in Table 5.3, the velocity function v(t) is positive over the time interval [0,5] and negative over the interval [5, 8]. Therefore, from Example 4b, the total distance traveled is the integral

[lv(t)ldt

=

[lv(t)ldt + 18Iv(t)ldt

1 8

= 1'(160 =

32t) dt -

(160 - 32t) dt

[160t - 16t2]~ - [160t - 16t 2]:

= [(160)(5) - (16)(25)] - [(160)(8) - (16)(64) - «160)(5) - (16)(25))] = 400 - (-144) = 544.

Again, this calculation agrees with our conclusion in Example 3, Section 5.1. That is, the total distance of 544 ft traveled by the rock during the time period 0 :5 t :5 8 is (i) the maximum height of 400 ft it reached over the time interval [0, 5] plus (ii) the additional distance of 144 ft the rock fell over the time interval [5, 8]. •

The Relationship between Integration and Differentiation The conclusions of the Fundamental Theorem tell us several things. Equation (2) can be

rewritten as

di

dx

a

X

j(t) dt

=

j(x),

which says that if you frrst integrate the function j and then differentiate the result, you get the function j back again. Likewise, replacing b by x and x by t in Equation (6) gives

i

X

F'(t) dt

=

F(x) - F(a),

so that if you first differentiate the function F and then integrate the result, you get the function F back (adjusted by an integration constant). In a sense, the processes of integration and differentiation are "inverses" of each other. The Fundamental Theorem also says that every continuous function j has an antiderivative F. It shows the importance of finding antiderivatives in order to evaluate definite integrals easily. Furthermore, it says that the differential equation dy/dx = j(x) has a solution (namely, any of the functions y = F(x) + C) for every continuous function j.

Total Area The Riemann sum contains terms such as j( Ck) ,lxk that give the area of a rectangle when j(ck) is positive. When j(ck) is negative, then the product j( Ck) ,lxk is the negative of the rectangle's area. When we add up such terms for a negative function we get the negative of the area between the curve and the x-axis. If we then take the absolute value, we obtain the correct positive area.

EXAMPLE 6 g(x) = 4 -

Figure 5.20 shows the graph of j(x) = x 2 - 4 and its mirror image across the x-axis. For each function, compute

x 2 reflected


(b) the area between the graph and the x-axis over [-2,2]. These graphs enclose the same amount of area with the x-axis, but the defmite integrals of the two functions over [-2, 2] differ in sign (Example 6). FIGURE 5.20

Solution

{2


(8 8) - (8 32 -3 + 8) -3'

[x3]2 3 - 4x -2 = 3 -

=

5.4 The Fundamental Theorem of Calculus

281

and

1

2g{x) dx

=

[4X _ x 3j2 3

-2

= 32.

-2

3

(b) In both cases, the area between the curve and the x-axis over [-2,2] is 32/3 units. Although the detmite integral off{x) is negative, the area is still positive. _ To compute the area of the region bounded by the graph ofa function y = fix) and the x-axis when the function takes on both positive and negative values, we must be careful to break up the interval [a, b] into subintervals on which the function doesn't change sign. Otherwise we might get cancellation between positive and negative signed areas, leading to an incorrect total. The correct total area is obtained by adding the absolute value of the detmite integral over each subinterval where f{x) does not change sign. The term "area" will be taken to mean this total area. y

EXAMPLE 7 Fignre5.21 shows the graph of the functionf{x) = sin x between x = 0 and x = 2".. Compute (8) the detmite integral of fix) over [0, 2".].

(b) the area between the graph of fix) and the x-axis over [0, 2".]. Solution

The definite integral for fix) = sinx is given by

].2" sinxdx = -cosxI' = -[cos2". FIGURE 5.21 The tota1 area between y = sinx and the x-axis for o ~ x ~ 21T is the sum of the absolute values of two integrals (Example 7).

cosO] = -[I - I] = O.

The detmite integral is zero because the portions of the graph above and below the x-axis make canceling contributions. The area between the graph of f(x) and the x-axis over [0, 2".] is calculated by breaking up the domain of sin x into two pieces: the interval [0, ".] over which it is nonnegative and the interval ["., 2".] over which it is nonpositive. ]." sin x dx = -cosx

1

2 "

I

= -[cos". - cos 0] = -[ -i - i] = 2

sinxdx = -cosxI" = -[cos2". - cos".] = -[I - (-i)] =-2

The second integral gives a negative value. The area between the graph and the axis is obtained by adding the absolute values Area =

121 + 1-21

-

= 4.

Summary: To tmd the area between the graph of y = fix) and the x-axis over the interval [a, b]:

1. Subdivide [a, b] at the zeros of f. 2. Integrate f over each subinterval. 3. Add the absoiute values of the integrals.

EXAMPLE 8 x2

Find the area of the region between the x-axis and the graph of 2x, -i :5 x :5 2.

f(x) = x 3

-

Solution

First tmd the zeros of f. Since

-

fix) = x 3

-

x2

-

2x = x{x 2

-

X -

2) = xix

+

I){x - 2),

282

Chapter 5: Integration

Area =

y

&

the zeros are x = 0, -1, and 2 (Figure 5.22). The zeros subdivide [-1,2] into two subintervals: [-I, 0], on which f ;;,: 0, and [0, 2], on which f :5 O. We integrate f over each subinterval and add the absolute values of the calculated integrals.

y=x3 -x2 -2x

10

3 2 -1 (x - x - 2x) ax

=

2 [x 4- ;3]0 - x

('<x 3 - x 2 - 2x) ax

=

3 [x4 - x - x2]2

Jo

-1 =

4 4

3

[14 + 31- 1] 125

0-

=

[4 - ~ -4]- 0=-~

=

3

0

3

The total enclosed area is obtained by adding the absolute values of the calculated integrals. The region between the xl - 2x and the x-axis

FIGURE 5.22

curve y = x 3 (Example 8).

-

t2 + 1- ~ 1

Total enclosed area =

i~

=

•

Exercises 5.4 Derivatives of Integrals Find the derivatives in Exercises 29-32

Evaluating Integrals Evaluate the integrals in Exercises 1-28. 1.

D2x +

L

S)dx

2.1:(S-I)dx

3) dx

4.1:

2

3.1 x(x -

(x

2

6.1:(x' 7.1,(x + Vi) 2

9.

l

dx

/3 2 sec'- x dx 'w/4

11. 13.

+ 3) dx

33. y

=

r l ' 4secutanudu

35. y

=

37. y

=

3& y

=

39. y

=

tan x dx sin2x dx

1& L:'(4sec'-'

14.

1 r

2

[wi' 17. Jo

19.

1-'

(r

+

23. 25. 27.

1

S2

_

+ 2

1

Vs

S

/.W s~2x dx Tr/2 2 smX'

J)xl

dx

0

1'-'

-0

l) du u'

7

21. [' (u J0 2

0

20.1

1)2 dr

dx

22. 24.

["

~Jo

(1 + eosx) dx

w/' I - cos 21 2 dl -/3 l6 16. Jo (seex + tan x)' dx

[wi'

31.

-,

l

1

(I

+

~)dl

1)(1 2 + 4) dl

+

40. y= y' - 2y y

(x1/3

,

!1""'31 d' 2

[ton.

d

Vudu

32. d6Jo

2

sec ydy

dl

[0 sin (12) dl

lVi

l'

12

- 2 - - dl

-1 I

+4

(1' (I' + 1)10

,1·" o

1,

j,'

-

,

J,{'I,dl,

34. y

=

36. y

= x J2

{"

x>

0

sin (I') dl

12

- 2 - - dl

I

+4

dl)'

dl

.,.------.,' 2 vl-t

7f

Ixl < -2

dl

0

--2 tan~l+t

Area

+ 1)(2 - x 2/') X

l'"\I'l+7

dy

II'

dx

r l' (eosx + sec x)' dx rl 28. Jo 2 (eosx + lcosxl)dx 26. Jo

30.

Finddy/dx in Exercises 33-40.

+ ~os 21 dl

w/2

d [Vi 29. dxJo cos I dl

w

12. Jo

I

15. Jo

2x

esc 6 cot 6 d6

/.wi'

/.0

+ 3) dx

s.132x--6/S dx

10.1

w

2x

-

by evaluatiog the integral and differentiatiog the result.

b. by differentiatiog the integral directly.

10 Exercises 41-44, fmd the total area between the region and the x-axis. 41. Y = -x 2 - lx, -3:5 x :5 2 42. Y

= 3x

2

-

3,

43. y=x'-3x 2 44. y

= x'i' - x,

-2:S; x

+2x,

:5

2

0"'x"'2 -I '" x '" 8

5.4 The Fundamental Theorem of Calculus Find the areas of the shaded regions in Exercises 45-48.

45.

Y

283

56. Show that if k is a positive constant, then the area between the x-axis and one arch of the curve y = sin kx is 21k. 57. Co.tfrom marginlll co.t

The marginal cost of printiog a poster

when x posters have been printed is I

de dx

x = 1T ~~--------~-L~x

2Vx

doll8I1!. Find e(100) - e(l), the cost of printing posters 2-100.

o

58. Revenue from marginal revenue Suppose that a company's marginal reveoue from the manufacture aod sale of eggbeaters is

46.

;;: =

o!S1T'\:

/I

6

47.

'x

48.

y

59. The teroperature T(OF) ofaroom at time I minutes is given hy

2

y

1)2,

where r is measured in thousands of dollars and x in thousands of units. How much money should the company expect from a production run of x = 3 thousand eggbeater.? To fmd out, integrate the marginal revenue from x = 0 to x = 3.

6

y

+

2 - 2/(x

T = 85 - 3 \I25=t for 0 '"

I '" 25.

a. Find the room's temperature when 1= 0, 1= 16, and 1= 25.

= sec 8 tan 8

b. Find the room's average temperature for 0 '" 1 '" 25. --;;;;f----J/~---:oor:-----> 6

60. The height H (It) of a palm tree after growing for 1 years is given hy

4

"\It+! +

H =

-Yz

It

I:

6Z. Findf(4)if

brief reasons for your answers.

63. Find the linearization of

=

1

'1

-dl - 3

b. Y = /.' secldl + 4

1 1

c. Y = 1'seCldl+4

d. Y =

-1

dy 49. dx

I

=

x'

0 '" I '" 8.

b. Find the tree's average height for 0 '" 1 '" 8. 61. Suppose that f(l) dl = x 2 - 2x + 1. Find f(x).

Each of the following functions solves one of the initial value problems in Exercises 4c}"'52. Which function solves which problem? Give

Y

for

a. Find the tree's height when 1 = 0, 1 = 4, aod 1 = 8.

In;tial Value Problems

L

51 1/3

Y('IT) =

-3

51. y' = sec x, y(O) = 4

50. y'

=

' I

w

/.

t dl - 3

secx,

f(x) = 2 -

54. :

="\1'1+7,

secx,

y(2)

=

y(-I) = 4

g(x) = 3 +

I

52. y' = x' y(l) = -3

3

y(1)

['+1

J2

9

I + Idl

atx = 1.

atx

=

XCOS'ITX.

64. Find the linearization of

Express the solutions of the initial value problems in Exercises 53 and 54 in terms of integrals.

dy 53. dx

f(l)dl =

= -2

Jl[" sec (I -

I)dl

= -1.

65. Suppose that f bas a positive derivative for all values of x and that f( I) = O. Which of the following statements must be !roe of the function

g(x) = /.' f(l) dl? Give reasons for your answers.

Theory and Examples 55. Archimede.' area formula for parabolic arches Archimedes (287-212 B.C.), inventor, military engineer, physicist, and the greatest mathematician ofclassical times in the Western world, discovered that the area under a parabolic arch is two-thirds the base times the height. S1retch the parabolic arch y = h - (4hlb 2 )x 2, -b12 '" x '" b12, assuming that h and b are positive. Then use calculus to fmd the area of the region enclosed between the arch and the x-axis.

a. g is a differentiable function of x. b. g is a continuous function of x.

c. The graph of g has a horizontal tangent at x = 1. d. g has a local maximum at x = 1.

e. g has a local minimum at x

=

1.

f. The graph of g has an inflection point at x = 1.

g. The graph of dgl dx crosses the x-axis at x = 1.

284

Chapter 5: Integration

66. Another proof of The Evaluation Theorem L

Let a = Xo < Xl < X2 ••• < XII = b be any partition of [a, b], and letFbe any antiderivative of f. Show that

F(b) - F(a)

• = "5'. ~

[F(x,) - F(X'-l)]'

67. f(x) = x' - 4x'

e. From part (b) and the defmition of the defmite integral, show that

[0,4]

68. f(x) = 2x' - 17x' + 46x' - 43x + 12, 69. f(x) = sin 2x cos~,

70. f(x) = XCOS1TX, b. Apply the Mean Value Theorem to each term to show that F(x,) - F(XH) = f(c,)(x, - XH) for some c, in the interval (X/-l, x,). Then show thatF(b) - F(a) is a Riemano sum for f on [a, b].

+ 3x,

[o,~l

[0, 21T]

[0,21T]

In Exercises 71-74, letF(x) =

J:V<) f(t) dt for the specifleda, u, and

f. Use a CAS to perform the following steps and answer the questions posed. L

Find the domain ofF.

b. Calculate F'(x) and determine its zeros. For what points in its domain is F increasing? Decreasing?

F(b) - F(a) = / . ' f(x) dx.

c. Calculate F"(x) and determine its zero. IdentifY the local extrema and the points of inflection ofF.

COMPUTER EXPLORATIONS

J:

In Exercises 67-70, letF(x) = f(t) dt for the specified function f and interval [a, b]. Use a CAS to perform the following steps and

answer the questions posed. L

Plot the functions f and F together over [a, b].

b. SolvetheequationF'(x) = O. What can you see to be true about the graphs off and F at points where F' (x) = 07 Is your observation borne out by Part 1 of the Fundamental Theorem coupled with information provided by the flrst derivative? Explain your answer.

d. Using the informatioo from parts (a)-(c), draw a rough haodsketch ofy = F(x) over its domain. Then graphF(x) 00 your CAS to support your sketch.

71. a = I, u(x) = x',

f(x) = ~

72. a = 0, u(x) = x',

f(x) = ~

73. a = 0, u(x) = 1 - x, 74. a = 0, u(x) = I - x',

f(x) = x' - 2x - 3 f(x) = x' - 2x - 3

In Exercises 75 and 76, assume that f is continuous and u(x) is twice-

differentiable.

c. Over what intervals (approximately) is the function F increasing and decreasing? What is true about f over those intervals?

d f-
d. Calculate the derivative f' and plot it together with F. What can you see to be true about the graph ofF at points where f'(x) = O? Is your observation borne out by Part I of the Fundamental Theorem? Explain your answer.

76. Calculate - ,

5.5

d'/.'V<) f(t) dtand check your answer using a CAS.

dx

•

Indefinite Integrals and the Substitution Method The Fundamental Theorem of Calculus says that a defmite integral of a continuous function can be computed directly if we can fmd an antiderivative of the function. In Section 4.7 we defined the indefinite integral of the function with respect to x as the set of all antiderivatives of t, symbolized by

t

J

t(x)dx.

t

Since any two antiderivatives of differ by a constant, the indefinite integral means that for any antiderivative F of t,

J

t(x) dx

=

J notation

F(x) + C,

where C is any arbitrary constant. The connection between antiderivatives and the defmite integral stated in the Fundamental Theorem now explains this notation. When finding the indefinite integral of a function t, remember that it always includes an arbitrary constant C.

5.5 Indefinite Integrals and the Substitution Method

285

We must distinguish carefully between defInite and indef"mite integrals. A definite integral J: f(x) dx is a number. An indef"mite integral Jf(x) dx is afonction plus an arbitrary constant C. So far, we have only been able to find antiderivatives of functions that are clearly recognjzable as derivatives. In tbis section we begin to develop more general techniques for f"mding antiderivatives.

Substitution: Running the Chain Rule Backwards If u is a differentiable function of x and n is any number different from - I, the Chain Rule tells us that

(U"+1 ) _ "du n + I - u dx'

~ dx

From another point of view, tbis same equation says that u"+I/(n + I) is one of the antiderivatives of the function u"(du/dx). Therefore,

I

du u n+ 1 u" dx dx = n + 1

+

c.

(1)

The integral in Equation (I) is equal to the simpler integral

I

un+l

u" du = n

+ 1 + C,

which suggests that the simpler expression du can he substituted for (du/dx) dx when computing an integral. Leibniz, one of the founders of calculus, had the insight that indeed this substitution could be done, leading to the substitution method for coroputing integrals. As with differentials, when computing integrals we have du = :dx.

EXAMPLE 1

Solution

Find the integral

We set u

=

x'

I

(x' + x)5(3x 2 + 1) dx.

+ x. Then du = :dx = (3x 2 + I)dx,

so that by substitution we have

I

(x' + x)5(3x 2 + 1) dx

=

I

uS du

Let"

"

="--+C (x'

+ x)" 6

EXAMPLE 2 Solution

Find

I~

x3

+ x,du

=

(3x 2 + 1)th.

Integrate with respect to u.

6

=

=

+C

dx.

The integral does not fit the formula

IU"dU,

Substitute x 3 + x for u.

•

286

Chapter 5: Integration with u = 2x

+

I and n = 1/2, because

du=du ax =2ax ax is not precisely ax. The constant factor 2 is missing from the integral. However, we can introduce this factor after the integral sign if we compensate for it by a factor of 1/2 in front of the integral sign. So we write

j~ax=tjv~ .~ u

= tju

12 /

du

dU

Letu

1 u 3/ 2

=

2 3/2 + C

=

1(2x 3

=

2x + 1,d"

=

2dx.

Integrate with respect to u.

+ 1)3/2 + C

Substitute 2x + 1 for u.

•

The substitotions in Examples I and 2 are instances of the following general rule.

THEOREM 6-The Substitution Rule If u = g(x) is a differentiable function whose range is an interval I, and f is continuous on I, then

j f(g(x))g'(x) ax

=

j f(u) duo

Proof By the Chain Rule, F(g(x» is an antiderivative of f(g(x»' g'(x) whenever Fis an antiderivative of f: d

axF(g(x))

=

F'(g(x»·g'(x)

=

f(g(x))' g'(x).

Chain Rule

If we make the substitotion u = g(x), then

j f(g(x»g'(x) ax

=

j :i,F(g(x» ax

= F(g(x)) + C = F(u) + C

Fundamental Theorem

= j F'(u)du

Fundamental Theorem

= jf(U)dU

F'

u

~

g(x)

~f

•

The Substitotion Rule provides the following substitution method to evaluate the integral

j f(g(x»g'(x) ax, when f and g' are continuous functions:

1.

Substitute u = g(x) and du = (du/ax) ax = g'(x) ax to obtain the integral

j f(u)du.

5.5 Indefinite Integrals and the Substitution Method 2.

Integrate with respect to u.

3.

Replace U by g(x) in the result.

EXAMPLE 3 Solution

Find

j se~

287

(51 + 1)·5 dl.

We substitute U = 51 + 1 and du = 5 dl. Then, jseC2 (51+1).5dl= jsec2 UdU

EXAMPLE 4

Find

j

Letu

=

5t

+ l.du

=

sec2u

Sdt.

=

=tanU+C

d du tan"

= tan (51 + 1) + C

Substitute 5t + 1 for u.

•

cos (70 + 3) dO.

Solution We let U = 70 + 3 so that du = 7 dO. The constant factor 7 is missing from the dO term in the integral. We can compensate for it by multiplying and dividing by 7, using the same procedure as in Example 2. Then, j cos (70 + 3) dO = t j

cos (70 + 3)·7dO

Place factor 1/7 in tront ofintegra1.

78 + 3,d"

=tjcosUdU

Letu

=tsinu+C

Integrate.

= t sin (70 + 3) + C

Substitute 78 + 3 for u.

=

=

7d6.

There is another approach to this problem. With U = 70 + 3 and du = 7 dO as before, we solve for dO to obtain dO = (1/7) duo Then the integral becomes

j

cos (70 + 3) dO =

j

Letu~78+3,du~7d8,andd8~(1/7)du

cosu·tdu

= t sin U + C

httegrate.

= t sin (70 + 3) + C

Substitute 78 + 3foru.

We can verify this solution by differentiating and checking that we obtain the original function cos (70 + 3). •

EXAMPLE 5 Sometimes we observe that a power of x appears in the integrand that is one less than the power of x appearing in the argument of a function we want to integrate. This observation immediately suggests we try a substitution for the higher power of x. This situation occurs in the following integration.

j

x 2 sin (x 3 ) dx =

j

= j

sin (x 3 ) • x 2 dx Sinu·t du

= x 3 ,du = 3x 2 dx, (1/3) du ~ x'tlx.

Letu

=t!SinUdU =

31 (-cosu)

+ C

httegratc.

Replace u by x 3 .

•

288

Chapter 5: Integration It may happen that an extra factor of x appears in the integrand when we try a substitution u = g(x). In that case, it may be possible to solve the equation u = g(x) for x in tenns of u. Replacing the extra factur of x with that expression may then allow for an integral we can evaluate. Here's an example of this situation.

EXAMPLE 6

Evaluate

Jx~

dx.

Solution Our previous integration in Example 2 suggests the substitution u with du = 2 dx. Then,

=

2x

+

I

However in this case the integrnnd contsins an extra factor of x multiplying the tenn v'2XTI. To adjust for this, we solve the substitution equation u = 2x + I to obtsin x = (u - 1)/2, and find that

x~dx

=

t(u - 1)·tYudu.

The integration now becomes

Jx~

dx

=

=

t J(u -

tJ

(u 3/ 2

1)Yu du = -

t J(u -

1

l)u /2 du

u1/2 ) du

Multiply terms.

=

l(2. u S/2 - 2.U 3/2) +

=

.l. (2x + 1)s/2 - 1 (2x +

453 10

6

Substitute.

C

Integrate.

1)3/2

+

C

Replace u by 2x

+

I.

•

The success of the substitution method depends on finding a substitution that changes an integral we cannot evaluate directly intu one that we can. If the llISt substitution fails, try tu simplify the integrand further with additional substitutions (see Exercises 51 and 52).

EXAMPLE 7

Evaluate

J

2zdz

.l~' 2

vz + I

Solution

We can use the substitution method of integration as an exploratory tuol: Substitute for the most troublesome part of the integrand and see how things work out. For the integral here, we might try u = z2 + I or we might even press our luck and take u to be the entire cube root. Here is what happens in each case.

Solution 1: Substitute u = z2

+ I. LetU=Z2+ I, du = 2zdz.

In the form

Ju' du

Integrate.

Replace u by Z2 + 1.

5.5 Indefinite Integrals and the Substitution Method

SoLution 2: Substitute u = V'z2

J

+

289

I instead.

2zdz

=

V'z2 + I

J

2 3u du

3JUdU

=

3'2

+C

=

1(z2 2

+

2

= \¥z2+1,

"3 = Z2

=

U

Let u

u

1),/3

+ 1,

3u 2 du = 2zdz.

Integrate.

+

C

Replace uby (Z2

+ 1)1/3.

•

The Integrals of sin 2 x and cos 2 X Sometimes we can use trigonometric identities to transform integrals we do not know bow to evaluate into ones we can evaluate using the substitution rule.

EXAMPLE 8 (a)

J

sin2 x dx = = =

J

tJ(1 Ix 2

(b) JCOS2 Xdx = J

EXAMPLE 9

~os 2x dx

I -

=

1 - cos2x 2

cos2x) dx

+C

_lsin2x 2 2

I

sin2 x

+ ~OS2x dx

=

= '! _ sin2x

2

4

+

C

I + Sin42x + C

2

cosx=

l+cos2x 2

•

We can model the voltage in the electrical wiring of a typical home with

the sine function

v

v=

v = Vmax sin 12011"1

VlWlX sin 120,,1,

which expresses the voltage V in volts as a function of time 1 in seconds. The function runs through 60 cycles each second (its frequency is 60 hertz, or 60 Hz). The positive constant VllW< (''vee max") is the peak voltage. The average value of Vover the half-cycle from 0 to 1/120 sec (see Figure 5.23) is

o

1.

1 12 / • I . Vav = (1/120) _ O. Vmax sm 120"tdt I

]1/12.

= 120Vmax [- 120" cos 120"t •

FIGURE 5.23 The graph of tire voltage V over a full cycle. Its average value over a

half-cycle is 2V-!1T. Its average value over a full cycle is zero (Example 9).

=

Vmax ---.r [-COS" + cos 0]

2Vmax

"

.

The average value of the voltage over a full cycle is zero, as we can see from Figure 5.23. (Also see Exercise 64.) If we measured the voltage with a standard moving-coil galvanometer, the meter would read zero. To measure the voltage effectively, we use an instrument that measures the square root of the average value of the square of the voltage, namely Vnns = V(V2 ).v'

290

Chapter 5: Integration The subscript ''nTIs'' (read the letters separately) stands for "root mean square." Since the average value of V2 = (Vmax)' sin2 1201Tt over a cycle is I (V2),v = (1/60) - 0

1.

1 60 /

0

(V, 2 (Vmax)2sin 1201Ttdt =

)' 7',

(Exercise 64, part c), the rms voltage is "

=

''''''

~(Vmax)2

=

V""",

2

v'2 .

The values given for household currents and voltages are always rms values. Thus, "115 volts ac" means that the TIllS voltage is 115. The peak voltage, obtained from the last equation, is

V_ =

v'2 V"", = v'2 ·115

'" 163 volts,

•

which is considerably higher.

Exercises 5.5 Evaluating Indefinite Integrals Evaluate the indermite integrals in Exercises 1-16 by using the giveo substitutions to reduce the integrals to standard form. 1. 2.

3. 4. 5.

6.

7. 9. 10. 11.

12.

13. 14. 15.

J J7~ J J J (I + Yx)I/3 J Yx J J J 9r2dr J J Jv'X J:2eos2(~)dx, J 2(2x

+ 4)' dx, u

2x(x 2 + 5)-4dx,

(3x

u

=

19.

+I

21

+ 2)(3x2 + 4x)4 dx,

sin 3x dx,

u = 3x

8. =

12(y4

u

3x 2 + 4x

J

x sin (2x 2) dx,

u = 2x 2

sin2 (x 3/2 - 1) dx,

+ 4y2 +

jr2 (~; - I)' dr

28.

Using u = cot 20

33. I

35.

u = x 3/2 - I

u=-~

37. 39.

esc' 20 cot 20 dO

L

27.

I - r3

u = y4

b. Using u = esc 20

22.

26.

u = I - cosf

+ 2y)
dx

SinSj-cosj-dx

31.

1)'(y3

I

20.

25.

r3

+ 4y2 +

- 0' dO

sec2 (3x

30.

=

18.

23.

2t

(I - eosf)' sinfdt,

vI -

=

29.

see2ttan2tdt, u

.~,

u

u=I+Yx

dx,

JV3=2.;ds JO~I . Jv'X(1 + v'X)' J J

17.

x2 + 5

u = X4

b. Usingu= ~

.. Usingu = 5x+ 8

Evaluate the integrals in Exercises 17-50.

u = 7x - I

dx,

3 (x 44: 1)2 dx,

+4

= 2x

16.J~ 5x + 8

41.

+ 2) dx

24.

J J (V- 7f) (V - 7f) J + J~eos(+-I)dt J~ sin~cos~dO J J:2~2-~dx J "----;Il

J~ds 58 + 4

J J J J

3yV7 - 3y2
cos (3z

+ 4) dz

tan2x sec' x dx 7X

tan

2 se.,-_2X2 dx

jr4 (7 - ~~)' dr

x 1/2 sin (x 3/2 + I) dx

esc - 2 - cot - 2 - dv

sin(2t+ I) dt eos2 (2t I)

38.

J J~ J J~x

40.

Jl~x2 3 2

32. 34. 36.

t 3(1 + t 4 )3 dt

~

dx

42.

secztanz dz

~

cos(V'i

+ 3) dt

cos Yo dO Yosin2 Yo ;, I dx

x

J~x

I dx

x

X 3 '

- I

dx

291

5.6 Substitution and Area Between Curves

43. 45. 47.

49.

! ! ! x'Vx'+! ! ~

x(x - 1)'0 tb; (x

44.

+ 1)'(1 - x)' tb;

(x 2

46. 48.

tb;

4)' tb;

SO.

!X~tb;

! ! 3x'W+! ! ~

(x + 5)(x - 5)'/3 tb;

(x

tb;

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplifY the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitutions in Exercises 51 and 52. 2 2 18 tan x sec x tb; 51. (2 + tan' x)2

!

b. u = tan'x,followedbyv = 2

~; =

59.

~;~ =

2 3cos

(~- 0).

-4 sin (21 -

!

VI

+v

63. It looks as if we can integrate 2 sin x cosx with respect to x in three different ways:

a. !2SinXCOSXdx = !2UdU

+ sin2 (x - I) sin (x - I) cos (x - I) tb;

b.

!

+u

2 sin x cos x ax

sin Yo dO VOcos'v1i

57.

=

~

4x (x 2

+

= 8sin2 (I

5.6

8r'/', +

~).

=

sinx

u

=

cosx

C2

= -cos2 x +

C2

2sinxC08X=sin2x

+ C,.

Cao all three integrations be correct'/ Give reasons for your answer. 64. (Continuation ofExample 9.) •• Show by evaluating the integral in the expression

/,'/60 V_sin 120'ITldl

(1/60) - 0 0

Solve the initial value problems in Exercises 55--60.

56. :

-2u du

= _ co~2x

1

~

!

= -u 2 +

Initial Value Problems 55.

=

c. !2SinXCOSXdx=! sin2xdx

!

u

= u2 + Cl = 8in2X + Cl

2

Evaluate the integrals in Exercises 53 and 54. (2r - I) cos V3(2r - I)' + 6 53. dr V3(2r - I)' + 6

= 121 (31 2 - I)',

.(0) = 0

62. The acceleration of a particle moving back and forth on a line is a = d 2./dI 2 = ~ cos 'lT1 m/sec' for all I. If • = 0 and v = 8 mlsec when I = 0, fmd. when I = I sec.

a. u = x - I, followed by v = sinu,thenbyw = I + if

54.!

.'(0) = 100,

61. The velocity of a particle moving back and forth on a line is v = tis/dl = 6 sin 21 m/sec for all I. If. = 0 when 1= 0, fmd the value of. when I = 'IT/2 sec.

+u

b. u = sin (x - I), followed by v = I C. U = I + sin2 (x - I)

f),

f

d 2y 60. tb;2 = 4 sec2 2xtan 2x, y'(O) = 4, y(O) = -I

c.u=2+tan3 x

52.

r(O) =

Theory and Examples

4)' tb;

a. u = tanx, followed byv = u', then byw = 2

58.

.(i) = 3 y(O)

=

that the average value of V = V_sin 120 'lT1 over a full cycle is zero. b. The circuit that runs your eleel1'ic stove is rated 240 volts rms. What is the peak value of the allowable voltage?

0

c. Showthat

.(0) = 8

/,o

'/60

(V, )' (V.,..),sin2 120'ITldl= ;;; .

Substitution and Area Between Curves There are two methods for evaluating a defmite integral by substitution. One method is to fmd an antiderivative using substitution and then to evaluate the defmite integral by applying the Evaluation Theorem. The other method extends the process of substitution directly to definite integrals by changing the limits of integration. We apply the new formula introduced here to the problem of computing the area between two curves.

The Substitution Formula The following formula shows how the limits of integration change when the variable of integration is changed by substitution.

292

Chapter 5: Integration

THEOREM 7-Substitution in Definite Integrals If g' is continuous on the interval [a, bland f is continuous on the range of g(x) = u, then

l

b

f(g(x))' g'(x) dx =

19(b)

a

f(u) duo

g(a)

Proof Let F denote any antiderivative of f. Then,

d d
tf(g(x)). g'(x) dx = F(g(x)) [ :

~

=

F(g(b)) - F(g(a))

=

F(u)

l

·~g(b) .~g(a)

9(b)

=

1

F(g(x))g'(x) f(g(x))g'(x)

Fundamental Theorem, Part 2

f(u) duo

g(a)

•

To use the formula, make the same u-substitotion u = g(x) and du = g'(x) dx you would use to evaluate the corresponding indefinite integral. Then integrate the transformed integral with respect to u from the value g(a) (the value ofu at x = a) to the value g(b)(thevalueofuatx = b).

EXAMPLE 1 Solution

Evaluate

1:

3x 2W+J dx.

We have two choices.

Method 1: Transform the integral and evaluate the transformed integral with the transformed limits given in Theorem 7. Letu = x 3 + l,du = 3x 2 dx.

[3x 2 W+Jdx =

When x ~ -I,. ~ (-I)' + 1 ~ O. When x ~ 1,. ~ (I)' + 1 ~ 2.

[VudU Evaluate the new deImite integral.

Method 2: Transform the integral as an indefinite integral, integrate, cbange back to x, and use the original x-limits.

J3x W+J dx J 2

Vu du

=

=

'2. u 3/2 + C

=

'2. (x 3 +

Integrate with respect to u.

3 3

1)3/2

+C

[3x2W+J dx = t(x 3 + 1)3/2L =

H

=

H23/ 2 -

«(1)3

+

Replace u by x'

Use the integra1just found, with limits of integration for x.

1)3/2 - «-1)3

0

3 2 / ] =

+ 1.

+

H2v'2] =

1)3f2 ]

4r

•

5.6 Substitution and Area Between Curves

293

Which method is better----evaluating the transformed definite iotegral with transformed limits usiog Theorem 7, or transformiog the iotegrai, iotegrating, and transforming back to use the original limits of integration? In Example 1, the flISt method seems easier, but that is not always the case. Generally, it is best to know both methods and to use whichever one seems better at the time.

EXAMPLE 2

We use the method oftransformiog the limits ofiotegration. cot (), du = -csc2 (J dO, -du = esC> e dB. When e = ",/4, u = cot (",/4) = 1. When e = ",/2, u = cot (",/2) = o.

Let u

("/2 cotlicsc'lidli

=

lOU' (-du)

J"/4

1

=

=

-lOUdU

• Definite Integrals of Symmetric Functions

y

The Substitotion Formula io Theorem 7 simplifies the calculation of definite iotegrais of even and odd functions (Section 1.1) over a symmetric ioterval [-a, a] (Figure 5.24).

THEOREM 8 -a

o

(a)

a

Let f be continuous on the aymmetric ioterval [ -a, a].

Iff is even, then 1>(X) dx

(a)

(b) Iff is odd, then

y

1:

f(x) dx

=

af Zl (X) dx.

= O.

Proof of Part (a)

i (b)

FIGURE 5.24

=

=

(a)feven, J~f(x)
(b)fodd, J~af(x)
f(x) dx

1:

+ [f(X) dx

-l-af (x) dx + [f(X) dx

= -

=0

f(x) dx

[f(-U)(-dU) + [f(X)dx

=

[f(-U)dU + [f(X)dx

=

[f(U) du

=

Z[f(X)dx

+ [f(X) dx

Additivity Rule for Defmite Integrals

Order of Integratioo Rule

Letu

=

When x When x

-x,du = -dx. 0,14 = O. = -0, U = a. =

lis even, so

f( -u)

= f(u).

The proof of part (b) is entirely similar and you are asked to give it io Exercise 86.

•

294

Chapter 5: Integration The assertions of Theorem 8 remain true when f is an integrable function (mther than having the stronger property of being continuous).

EXAMPLE 3

Evaluate

1:

(x 4 - 4x 2 + 6) dx.

Solution Since f(x) = x4 - 4x 2 ric interval [ - 2, 2] , so

+

6 satisfies f( -x) = f(x) , it is even on the symmet-

y

~~~----------~~x

......

~I·---~

FIGURE 5.25 The region betweoo the curves y ~ f(x) and y = g(x) and the lines x = a and x = b.

y y

a

~ ~o

x,

-~

lJ1'<2

~

~

•

Lower curve y ~ g(x)

f(x)

7""

V XII-i

'7'-

>- ",r- b-::"y

~

Areas Between Curves Suppose we want to f'md the area of a region that is bounded above by the curve y = f(x) , below by the curve y = g(x), and on the left and right by the lines x = a and x = b (Figure 5.25). The region might accidentally have a shape whose area we could f'md with geometry, but if f and g are arbitrary continuous functions, we usually have to find the area with an integral. To see what the integml should be, we first approximate the region with n vertical rectangles based on a partition P = {xo, xt, ... , x.} of[a, b] (Figure 5.26). The area of the kth rectangle (Figure 5.27) is ~k =

g(x)

height X width = [f(q) - g(q)]

~k.

We then approximate the area of the region by adding the areas of the n rectangles: FIGURE 5.26 We approximate the region with rectangles perpeodieular to the x-axis.

•

A '" L~'

=

k~l

•

L[{(q) - g(q)] ~•.

Riemann sum

~1

1:

y

As Ilpll--> 0, the sums on the right approach the limit [f(x) - g(x)] dx because f and g are continuous. We take the area of the region to be the value of this integral. That is, A =

.

lim L[{(c.) - g(c.)] 4x. =

11P1I~o '~1

l

b

[((x) - g(x)] dx.

a

DEFINITION If f and g are continuous with f(x) ;" g(x) throughout [a, b], then the area of the region between the curves y = !(x) and y = g(x) from a to b is the integral of (f - g) from a to b: FIGURE 5.27 The area ,u. of the kth rectangle is the produet of its height, f(c,) - g(c.) , and its width, 4x•.

A

=

l

b [((x) - g(x)] dx.

When applying this definition it is helpful to graph the curves. The graph reveals which curve is the upper curve f and which is the lower curve g. It also helps you f'md the limits of integmtion if they are not given. You may need to find where the curves intersect to

5.6 Substitution and Area Between (urves y

295

detennine the limits of integration, and this may involve solving the equation f(x) = g(x) for values of x. Then you can integrate the function f - g for the area between the inter-

sections.

EXAMPLE 4

Find the area of the region enclosed by the parabola y = 2 - x 2 andthe

liney = -x. First we sketch the two curves (Figure 5.28). The limits of integration are found by solving Y = 2 - x 2 and y = -x simultaneously for x.

Solution

2 - x 2 = -x 2 x -x-2=O

FIGURE 5.28 The region in Example 4 with a typical approximating rectangle.

(x

+ I)(x - 2)

x = -1,

x

= =

Equate !(x) and g(x). Rewrite.

0 2.

Factor.

Solve.

The region runs from x = - I to x = 2. The limits of integration are a = - I, b = 2. The area between the curves is

A =

=

t 2 1

[((x) - g(x)]

-1

HIsTORICAL BIOGRAPHY

Richard Dedekind (1831-1916)

=

(2

+x

- x 2)

ax = 1~[(2 -

x 2) - (-x)]

ax = [ 2x 2 + ~ - ~3]2 2 3_

(4 + ~ - ~) - (-2 + ! + t)

=

ax

1

~

•

If the formula for a bounding curve changes at one or more points, we subdivide the region into subregions that correspond to the fannula changes and apply the fannula for the area between curves to each subregion.

EXAMPLE 5 y =

Find the area of the region in the first quadrant that is bounded above by = x - 2.

vX and below by the x-axis and the line y

Solution

The sketch (Figure 5.29) shows that the region's upper boundary is the graph of f(x) = vX. The lower boundary changes from g(x) = 0 for 0 :5 X :5 2 to g(x) = x - 2 for 2 :5 X :5 4 (both fannulas agree at x = 2). We subdivide the region at x = 2 into subregions A and B, shown in Figure 5.29. The limits of integration for region A are a = 0 and b = 2. The left-hand limit for region B is a = 2. To f"md the right-hand limit, we solve the equations y = vX and y = x - 2 simultaneously for x:

vX=x-2

Equate !(x) andg(x).

x = (x - 2)2 = x 2 - 4x

1

+

4

x2-5x+4=O

(x - l)(x - 4) x

~--~--~~--------~~x

0 / 4 (x. g(x»

FIGURE 5.29 When the formula for a bounding curve changes. the area integral changes to become the sum of integrals to match. ooe integral for each of the shaded regions shown here for Example 5.

= =

0 1,

Square both sides. Rewrite. Factor.

x

=

4.

Solve.

Only the value x = 4 satisfies the equation vX = x - 2. The value x = 1 is an extraneous root introduced by squaring. The right-hand limit is b = 4. 2:

f(x) - g(x) =

For2 :5 x :5 4:

f(x) - g(x) =

ForO :5

X :5

vX vX -

0 =

vX

(x - 2) =

vX -

x

+2

296

Chapter 5: Integration We add the areas of subregions A and B to find the tota1 area: 2 Yx dx Tota1 area = ]. 0

+ r(Yx - x + 2) dx

J2

area of A

=

areaofB

['1. X3/2]2 3

+

0

['1. X3/2 _ 3

2 x + 2x]4 2 2

=

t(2)3/2 - 0

+ (t(4)3/2 - 8 + 8) - (t(2)3/2 - 2 + 4)

=

t (8) -

~o.

2 =

•

Integration with Respect to y If a region's bounding curves are described by functions ofy, the approximating rectangles are horizontal instead of vertical and the basic formula hasy in place ofx. For regions like these: y

y

'1

X~g(y) ~==~

use the fonnula

A = ld[f(y) - g(ylJ dy. y

In this equation f always denotes the right·hand curve and g the left·hand curve, so f(y) - g(y) is nonnegative.

EXAMPLE 6

FIGURE 5.30 It talres two integrations to fmd tlte area of this region ifwe integrate witlt respect to x. It talres only one if we integrate witlt respect to y (Example 6).

Find the area of the region in Example 5 by integrating with respect toy.

Solution We first sketch the region and a typical horizontal rectangle based on a parti· tion of an interval ofy·values (Figure 5.30). The region's right·hand boundary is the line x = y + 2, so f(y) = y + 2. The left·hand boundary is the curve x = y2,so g(y) = y2. The lower limit of integration is y = o. We find the upper limit by solving x = y + 2 and x = y2 sin3ultaneously for y:

y +2 y2 _ Y _ 2

= =

y2 0

(y + l)(y - 2) = 0 y = -I, y=2

Equa1e/(y) ~ Y

+ 2 and

g(y) ~

y'.

Rewri1e.

Factor. Solve.

The upper limit of integration is b = 2. (The value y = - I gives a point of intersection below the x·axis.)

297

5.6 Substitution and Area Between (urves The area of the region is

A =

l

d

+

2 - y2] dy

].\2 +

Y - y2] dy

[f(y) - g{y)] dy = ].2[y =

•

This is the result of Example 5, found with less work.

Exercises 5.6 Evaluating Definite Integrals Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises 1-24.

t

17.

J.~/6cos-' 20 sin 20 dO J.~5(5 -

18.1,w/2cot:'

J.~/' (I

(~) sec' (~) dO

yY+l ely

b. 1~ yY+l ely

19.

2. a.

J.1r~dr

b.1:r~dr

21. J.1 (4y - y2 + 4y' + 1)-2/3 (12y 2 - 2y + 4) dy

Jor

b.l°

22. J.'
3. a.

1. a.

l

'

x x

tan sec' d>;

tan x sec' x d>;

-1r/4

1 b. 1>'(1 + t')' dt b.l°-Vi 1(12 + b. J.o (4 +5rr2)' dr b.]' (1 IOVv + v3 b. I -v3W+1 b. 1-1W+9 x b. l (I b. l-~/2 zt) t cosz 1-~ v'4 . b. dw J.

4. a. J.'" 3 cos2 x sin x fix

'~

3 cos2 x sinx dx

b.

4 cos 1)1/4 sin I dl 20.

23. J.Y/,;iYo cos2 (0'/2) dO

- sin 21)'/2 cos 21 dl

24.1~1/2t-2 sin

2

(I +) dl +

2~

S. a. J.\'(1

Vi 6. a. J.

+ t')' dt

t(t 2 + 1)1/3 dl

l)l/3dl

Area Find the total areas of the sbaded regions in Exercises 25--40.

25.

1

7. a. 11 ( -1

8. a. {'

Jo

4

5r 2)' dr

+r IOVv

(1 +

dv

v'/2)2

,,'/2)2

1

9 {v3 4x d>; . a. W+1

4x

Jo

0

11. a.

Jor

x4 + 9

l6

14. a.

Jo v'4

1°

w/2( 2 + tan ~

b.

dw

15. J.1 v'I'+2I (5t' + 2) dt

sec'Zdl

dz

+ 3 sinz

~/2

~/2 (3 + 2 cos wJ2

o

"

,

cos31)sin3ldl

cos Z dz + 3 sinz

sin w

~--~--------~~X

d>;

w/6

2

{2",

y = (1 - cou) sinx

dv

~/'

(I - cos31)sin3ldl

12. a. E/2(2 + tanf)sec fd' 13. a.

y

d>;

10. a. {' k d > ;

Jo

26. y

o

16 •

smw (3+2cosw)'

]4 2Vy(1 + Vy)' ely

1

28.

27. y

y = ~(COSX)(sin(1T + 1Tsinx»

y

298

Chapter 5: Integration

29.

36.

30.

Y

y

y

y=!sec 2 t 2

1

Y~ 1

0

"2

2

"

o

x

37.

38.

39.

40.

-4 y

31. (-2,8)

(2.8)

8

Y NOTTOSCAIE

32.

Y

y=~

3

o

(3, -5)

33.

Find the areas of the regions enclosed by the lines and curves in Exercises 41-50. 41. Y = x 2

x2

42. Y = 1. -

34.

35. Y

Y

and y = 2

2

-

and y = -3

43.y=x' and y=8x 44.y=x2 -1. and y=x 45. y=x2 and y= -x 2 +4x 46. y=7-1. 2 and y=x2+4 47. y = x' - 4x 2 + 4 and y = x 2 48. Y

=

49. Y

=

xv'a 2 -

v'R

x 2,

a

>

0,

and

5y

=

x

and y

+6

=

0

(How many intersection points

are there?) 50. y y = _44

-2

= Ix 2

-

41

and y

= (x 2/2) + 4

Find the areas of the regions enclosed by the lines and curves in Exercises 51-58. 51. x = 2y 2,

X

= 0,

and y = 3

5.6

52. x ~ y'

and x ~ y

53. y' - 4x ~ 4

and

54. x - y' ~ 0

and x

55. x ~ y' - y

and

56. x - y'/3 ~ 0 57. x ~ y' - I

+2

Sub,titution and Area Between (urves

299

y

4x - y ~ 16

+ 2y'

~ 3

x ~ 2y' - 2y - 6

and x

+ y'

~ 2

and x ~ lylv'i=Y'

58. x ~ y' - y'

and x ~ 2y

~--~~----~--~X

o

2

Find the areas of the regions enclosed by the curves in Exercises 59-{j2.

59. 4x'+y~4 60. x'-y~O 61. x 62. x

and x4_y~ I

+ 4y' ~ 4 + y' ~ 3

79. The figure here shows triaogle AOC inscribed in the region cut from the parabola y ~ by the lioe y ~ Fiod the limit of the ratio of the area of the ttiangle to the area of the parabolic region as a approaches zero.

x'

3x'-y~4

and and

x

and 4x

+ y' + y'

~ I,

for

x "" 0

~ 0

a'.

y

Find the areas of the regions eoclosed by the lines and curves in Exercises 63-70.

63. Y

= 2 sin x and y = sin2x, 0

64. y ~ 8 cosx

and y ~ sec' x,

65. y ~ cos ('lTx/2) 66. y

~

sin ('lTx/2)

~

x

-'IT/3:5 x :5 'IT/3

and y ~ I - x'

and y

~

x

67. y ~ sec' x, y ~ tan' x, x ~ -'IT/4, 68. x ~ tan'y

and

x ~ -tan'y,

69. x ~ 3 siny v'cosy 70. y ~ sec' ('lTx/3)

~ 'IT

x ~ 'IT/4

-'IT/4 s y '" 'IT/4

and x ~ 0,

and y ~ xli',

and

0:5 Y :5 'IT/2 -I '" x :5 I

71. Find the area of the propeller-shaped region enclosed by the curve x - y' ~ o and the line x - y ~ O. 72. Find the area of the propeller-shaped region enclosed by the curves x - yl/3 ~ o and x - yl/' ~ O.

80. Suppose the area of the region betweeo the graph of a positive continuous function f and the x-axis from x = a to x = b is 4 square uoits. Find the area between the curves y ~ /(x) and y ~ 2/(x) from x ~ atox ~ b.

81. Show that the area of the shaded region equals 1/6 for all values ofz.

73. Find the area of the region in the f1I1!t quadrant bounded by the liney ~ x,the line x ~ 2,thecurvey ~ I/x',andthex-axis. 74. Find the area of the ''triangular'' region in the first quadrant bounded on the left by the y-axis and on the right by the curves y = sinx andy = cosx.

x'

75. The region bounded below by the parabola y ~ and above by the line y ~ 4 is to be partitioned into two subsections of equal area by cutting across it with the horizontal line y = c. ~ c across it that looks about right. In terms of c, what are the coordinates of the points where the line and parabola intersect? Add them to your figure.

a. Sketch the region and draw a line y

82. True, sometimes true, or never true? The area of the region between the graphs of the continuous functions y ~ /(x) and y ~ g(x)andtheverticsllinesx ~ a and x ~ b(a < b) is

b. Find c by integrating with respect to y. (This puts c in the limits ofintegration.)

c. Find c by integrating with respect to x. (This puts c into the integrand as well.) 76. Find the area of the region between the curve y ~ 3 line y ~ -I by integrating with reopect to (a) x, (b) y.

x' and the

77. Find the area of the region in the fJrst quadraot bounded on the left by the y-axis, below by the line y ~ x/4, above left by the curvey ~ I + v'i,andaboverightbythecurvey ~ 2/v'i. 78. Find the area of the region in the fJrst quadraot bounded on the left by they-axis, below by the curve x ~ 2Vy, above left by the curve x ~ (y - I)', and above right by the line x ~ 3 - y.

l\/(x) - g(x)] dx. Give reasons for your answer. Theory and Examples 83. Suppose thatF(x) is an antiderivative of /(x) ~ (sinx)/x, x Express {' sin2x dx x

11 io terms ofF.

> O.

300

Chapter 5: Integration

84. Show that iff is continuous, then

l

lf

(X) dx =

l

lf(1 -

y

x) dx.

85. Suppose that

Find tf(X)dx if (a) f is odd, (h) f is even. 86. L Show that if f is odd on [-a, aJ, 1hen

89. Use a substitution to verify Equation (I). 90. For each of 1he following functions, graph f(x) over [a, bJ and f(x + c) over [a - c, b - cJ to convince yourself that Equation (1) is reasonable.

i:f(X)dx = O. b. Test1heresultinpar!(a)wi1hf(x) = sinxanda = '1f/2.

&.

87. Iff is a continuous function, fmd 1he value oftha integral ('

1=

Jo

b. f(x)

f(x) dx f(x) + f(a - x)

by making 1he substitution u integral to l.

=

c. f(x)

a - x and adding 1he resultiog

88. By using a substitution, prove 1hat for all positive numbers x andy,

t~'f(x + c) dx.

a. Plot the curves toge1her to see what they look like and how many points of intersection they have. b. Use 1he numerical equation solver in your CAS to fmd a111he points of intersection. c. lotegrate If(x) - g(x) I over consecutive pairs of intersection values. d. Sum toge1her 1he integrals found in par! (c).

(1)

The equation holds wbeoever f is integrable and defmed for 1he necessary values of x. For example, in 1he accompanying figure show that

r- (x + 2)' dx = Jo(lx'dx J-2 l

because the areas of the shaded regions are congruent.

Chapter

= x 2 , a = 0, b = I, c = I = sinx, a = 0, b = '1f, C = '1f/2 = v;=4, a = 4, b = 8, c = 5

COMPUTER EXPLORATIONS 10 Exercises 91-94, you will find tha area between curves in tha plane when you caonot fmd 1heir points of intersection using siruple algebra. Use a CAS to perform 1he following steps:

The Sbift Property for Definite Integral. A basic property of definite integrals is thair invatiance under translation, as expressed by 1he equation [f(x) dx =

f(x)

3

2

91. f(x) =

x x T - 2 -

92. f(x)

=

2x' -

93. f(x)

=

x

94. f(x)

= x2

3x'

2x

+

1 + 3'

10,

+ sin (2x), cos x, g(x)

g(x)

g(x) =

g(x) = x - I

=

=

8 - 12x

x'

x' - x

Questions to Guide Your Review

1. How can you sometimes estimate quantities like distance traveled, area, and average value wi1h fmite sums? Why ntight you want to do so? 2. What is sigma notation? What advantage does it offer? Give examples. 3. What is a Riemann sum? Why ntight you want to consider such a S'lD1l?

4. What is 1he norm of a partition of a closed interval? 5. What is 1he defmite integral of a function f over a closed interval [a, bJ? When can you be sure it exists?

6. What is 1he relation between defmite integrals and area? Describe some other interpretations of def'mite integrals. 7. What is the average value of an integrable function over a closed interval? Must the function assume its average value? Explain. 8. Describe the rules for working wi1h defntite integrals (Table 5.4). Give examples. 9. What is 1he Fundamental Theorem of Calculus? Why is it so iruportant? Illustrate each part of the 1heorem with an example. 10. What is the Net Change Theorero? What does it say about the integral of velocity? The integral of margina1 cost?

301

Chapter 5 Practice Exercises 11. Discuss how the processes of inregration and differentiation can be considered as "inverses" of each other.

14. How can you sometimes evaluare indefutire integrals by substitution? Give examples.

12. How does the Fundamental Theorem provide a solution to the initial value problem dyldx = f(x), y(xo) = Yo, when f is continuous?

15. How does the method of substitution worl< for defutite inregrals? Give examples.

16. How do you defme and calculare the area of the region between

13. How is inregration by substitution relared to the Chain Rule?

Chapter

~

the graphs of two continuous functions? Give an example.

Practice Exerdses

Finite Sums and Estimates 1. The accompanying figure shows the graph of the velocity (fr/sec) of a model rocket for the first 8 sec after launch. The rocket accel· erared straight up for the fIrSt 2 sec and then coasred to reach its maximum height at t = 8 sec.

(5 )

10

c. ~(ak + bk - I)

10 - - b d. ~ k k- I 2

k- l

'0

20

4. Suppose that ~ak = 0 and ~bk = 7. Find the values of 20

'0

a. ~3ak

b. ~(ak

k- l

k- I

150 1 ¢l

+ bk)

'0

d. ~(ak - 2)

~IOO

.~

~

Definite Integrals

50

o

2

6

4

8

Time after launch (sec)

a. Assuuting that the rocket was launched from ground level, about how high did it go? (This is the rocket in Section 3.3, Exercise 17, but you do not need to do Exercise 17 to do the exercise here.)

b. Sketch a graph of the rocket's height above ground as a function oftime for 0 :5 t :5 8. 2. a. The accompanying figure shows the velocity (mj sec) of a body moving along the s·axis duting the time in1llrva1 from t = 0 to t = 10 sec. About how far did the body travel duting those 10 sec?

b. Sketch a graph of s as a function of t for 0 ,;; t ,;; 10 assum· ings(O) = O.

v ....

5

I

\

6

8

10

Time (sec) 10

10

3. Suppose that ~ak = -2 and ~bk = 25. Find the value of i-I

7. 8.

•

lim ~e,
liPlI~o k-I

+

lim

liPlI~o t=1

(cos ("--2k

II aXk, where P is a partition of [-'II", 0] ))

•

lim ~(sinek)(cos ek) .1xk , whereP is a partition of[O, 'lT/2]

liPlI~o k~1

t,

9. If 3f(x) dx = 12, f~,f(x) dx = 6, and fmd the values of the following.

a.J:

f(x) dx

l-,

'(f(x) + g(x)) 5

= 2,

d.i:(-'II"g(X))dx dx

101 g(x)dx =

a. J.'g(x) dx

b. !,'g(x) dx

c.l°f (x) dx

d. J.'

i-I

10

b. ~(bk - 3ak)

f~,g(x) dx

b. l'f(X) dx

10. If fo'f(x)dx = 'IT, fo'7g(x)dx = 7, and the values of the following.

V 4

6.

•

lim ~(2ek -1)-1/2.1xk,wherePisapartitionof[I,5]

liPlI~o k~1

e.

1\ 2

5.

c. J,-'g(x) dx

1/

o

In Exercises 5-8, express each limit as a defutire inregral. Then evaluare the integral to fmd the value of the limit. In each case, P is a partition of the given interval and the numbers Ck are chosen from the subinrervals of P.

e. J.'(g(x) - 3f(x)) dx

V2 f(x) dx

2, fmd

302

Chapter 5: Integration

Area Io Exercises 11-14, rmd the total area of the region between the graph of f and the x-axis. 11. f(x) ~ x 2

-

+ 3,

4x

12. f(x) ~ I - (x 2/4), 13. f(x) ~ 5 - 5x 2/',

0:5 x :5 3

33. Show that y

-2 '" x '" 3

y ~ l/x2,

16. y ~ x,

y ~ I/Vi, ~ I,

34. Show that y ~ problem

l'+

dl solves the initial value problen3

J;( I

+ 2v.;.;t) dl solves the initial value

d 2y -2 ~ Vsecxtanx; dx

~ 2

X

sin x aod

I dx 2 ~ 2 - x 2 ; y'(I) ~ 3, y(l) ~ I.

0:5 x :5 4

15. y ~ x,

~ x2 +

~

d'y

Find the areas of the regioos enclosed by the curves aod lines in Exercises 15-26.

+ Vy

Inlttal Value Problems

-1:5 x:5 8

14. f(x) ~ I - Vi,

17. Vi

32. Find the total area of the regioo between the curves y y = cosX' for 0 :5 x :s; 3w/2.

x ~ 2

x ~ 0, y ~ 0

y'(O) ~ 3, y(O) ~ O.

Express the solutions of the initial value probleD3S in Exercises 35 aod 36 in terms of integrals.

y

35. dxdy

~

sin x y () -x-, 5

36. :

~

V2 - sin2 x,

~-3

y(-I)

~2

Evaluating Indefinite Integrals Evaluate the integrals in Exercises 37-44. 37. /2(COSX)'I/2 sinx dx 39. /(28

18. x'

+ Vy

~ I,

x ~ 0, y ~ 0,

for

0:5 x '" I

y

x3

1

x

~

22. y2 ~ 4x

~

3

20. x ~ 4 - y2,

X~ 0

41. /

(I -t) (I + t)

43. /

VI sin (21'/2) dl

dl

~ 4x - 16

23. y ~ sinx, y ~ x,

0 '" x :5 'If/4

24. y ~ Isinxl, y ~ I,

45.1'<3X

'If)) d8

42. /

(I + ~~2 I

44. /

sec 8 IJln 8 VI

2

-

4x

+ 7) dx

47.

Jl

49.

{'.tit. Jl IVI

-'If/2:5 x :5 'If/2

25. y=2sinx, y=sin2x,

O::S;x::S;'11'

46.

-1

{24

y ~ 4x - 2

+ 4, y

(~ + 2 sec2 (28 -

-

dl

+ sec 8 d8

Evaluating Definite Integrals Evaluate the integrals in Exercises 45-70.

2y2, X ~ 0, y

21. y2 ~ 4x,

l))d8

40. /

+ Vy = I, O::s x ::s 1

o 19.

+ I + 2 cos (28 +

38. / (lJlnxr'/2 sec2x dx

1

51

48.

if dv

l

'(I

1

36 dx (18 + I)'

52

- 118 2

1

+

5) dx

+ VU)I/2 .r vu

du

d7

27. Find the area of the ''triangular'' regioo bounded 00 the left by x + y ~ 2,ontherightbyy ~ x 2,aod above byy ~ 2.

53. {' x·1/3(1 - x 2/3)'/2 dx

• ]0, V'(7 - 57)2 {'/2 54. Jo x'(i + 9x')"/2 dx

28. Find the area of the ''triangular'' regioo bounded 00 the left by y ~ Vi,ootherightbyy ~ 6 - x,aodbelowbyy ~ I.

55. ],,,,. 8in2 5r dr

56.

26. y ~ 8 cos x, y ~

see> x,

-'If/3:5 x :5 'If/3

29. Find the extren3e values of f(x) ~ x' - 3x 2 and rmd the area of the region enclosed by the graph off and the x-axis. 30. Find the area of the regioo cut frOO3 the rlrst quadrant by the curve x 1/2 + yl/2 ~ a1/2. 31. Find the total area of the region enclosed by the curve x ~ y2/3 aod the lines x ~ y and y ~ -I.

• ]0,

50.

{I (88'

Jo {27 Jl X-<13 dx

il/8

r/' see> 8 d8

57. Jo

l'w 61.1 59.

,w/.

58.

60.

secxtanxdx

62.

0

-/3

l f 1Jln2~d8 l

csc2 x dx

w/'

~ dx

cot'

],w/. cos2 (41 - f) dl

'w/'

w/'

csczcotzdz

Chapter 5 ./2 63.

/.

1

15 sin' 3x cos 3x dx

/.o

VI+3sin'x

./2 3 .

sm.xC05X

. /3

69.

/.o

dx

tan 8 d8 V2sec8

= nu:

83. Find dy/dx if Y = yom calculation.

+ b

over [-I, 1]

84.

72. Find the average value of

Y=

so. Y =

v'h over [0, 3]

b. y= ~aver[O,a] 73. Let I be a function that is differentiable on [a, b]. In Cbapter 2 we defmed the average rate of change of lover [a, b] to be

I(b) - I(a) b-a

KU

Find dy/dx if Y = in your calculation.

85. A new parking lot To meet the demand for parlcing, your town bas allocated the area shawn bere. As the town engineer, you bave been asked by the town council to find out if the lot can be built for $10,000. The cost to dear the land will be $0.10 a square foot, and the lot will cost $2.00 a square foot to pave. Can the job be done for $IO,OOO? Use a I~ sum estimate to see. (Answers may vary slightly, depending on the estimate used.) Oft

average value of f' on [a, b].

74. Is it true that the average value of an integrable function

51 ft

over an

49.5 ft

interval oflength 2 is half the function's integral over the interval? Give reasons for your answer.

Vertical spacing = 15 ft

Compute the average value of the temperature function I(x) = 37 sin

U:s

(x - 101») + 25

Speclfl. heat of. gas Specific heat Cv is the amount of beat required to raise the temperature of a given mass of gas with constant volume by 1"C, measured in units of cal/deg-mole (calories per degree gram molecule). The specific beat of oxygen depends on its temperature T and satisfies the formula Cv = 8.27

+ 10-' (26T -

Fmd the average value of C. for 20" perature at whicb it is attained.

:5

1.87T'). T :5 675· C and the tem-

54ft

64.4 ft

67.5 ft

for a 365-day year. This is one Wfrj to estimate the annual mean air temperature in Fairbanks, Alaska. The National Weather Service's official figure, a numerical average of the daily normal mean air temperatures for the year, is 25.7"F, which is slightly higher than the average value of the approximating function I(x).

D 76.

42ft

19nor
86. Skydivers A and B are in a helicopter hovering at 6400 ft. Skydiver A jumps and descends for 4 sec befure opening her parachute. The helicopter then climbs to 7000 ft and hovers there. Forty-fIVe seconds after A leaves the aircraft, B jumps and descends for 13 sec before opening his parachute. Both skydivers descend at 16 ft/sec with paracbutes open. Assume that the skydivers fu.lI freely (no effective air resistance) before their parachutes open. L

At what altitude does A's paracbute open?

b. At what altitude does B's parachute open?

DIfferentiating Integrals

•• Which skydiver lands fmlt?

In Exercises 77-80, fmddy/dx.

77. y =

v'1+7. Ex-

t,.% (1/(1 - I'» dt. Explain the main steps

Is this the case? Give reasons for your answer.

D 7S.

-,- I -dl t + 1

!.' "\I1+I2 dl. Explain the main steps in

and the instantaneous rate of change of I atx to be f'(x) . 1n this chapter we defmed the average va1ue of a function. For the new definition of average to be consistent with the old one, ..., should have I(b) - I(a) b_ a

l'

101 v'1+7 dx in terms of F and give a reason for your

press answer.

b. over [-k, k] L

6 - - .dl 3+1

82. Suppose that F(x) is an antiderivative of I(x) =

Average Value. 71. Find the average value of I(x) L

,

303

Theory and Examples 81. Is it true that every function y = I(x) that is differentiable on [a, b] is itself the derivative of some fuIK:tion on [a, b]? Give reasons for your answer.

~/2

67.

I

%

·/2

6S.

79. y=

5(sinx!'/2cosxdx

0

Practice Exercises

1% V2 + cos'ldl

78. Y

('''' = J, V2 + cos'ldl

304

Chapter 5: Integration

Chapter

~

Additional and Advanced Exercises

Theory and Examples

1.

If

L

1,'71 1,'/

(X) dx

(x) dx

b. If

~ 7,

does

1,'/

[v7Wdx =

V4 =

~ 17

(x) dx

= 4 and/(x)

9. Finding a curve Find the equation for the curve in the xy-plane that passes through the point (I, -I) if its slope at x is always

3x 2

+ 2.

10. Shoveling dirt You sling a shovelful of dirt up from the bottom of a hole with an initial velocity of 32 ft/ sec. The dirt must rise 17 ft above the release point to clear the edge of the hole. Is that enough speed to get the dirt out, or had you better duck?

'" 0, does

2?

Piecewise Continuous Functions

Give reasons for your answers. 2. suppose12/(x)dx

~

= 4,

= 2. hf'1(x)dx = 3,l'g(x)dx ~

Which, ifany, of the following statements are true?

(2 / (x) dx

L

k

=

b.1'(f(x)

-3

~

+ g(x)) ~

If' Jo I(t) sin a(x -

t) dt

solves the initial value problem

d'y dx 2

+ a'y

: =0

= I(x),

andy

sided limits exist and are finite at the endpoints of 1. All piecewise continuous functions are integrable. The points of discontinuity subdivide I into open and half-open subinterva1s on which 1 is continuous, and the limit criteria above guarantee that 1 has a cootinuous exteosion to the closure of each subinterval. To integrate a piecewise continuous function, we integrate the individual extensions and add the results. The integral of

= owhen x = O.

I-x

I(x)

(Hint: sin (ax - at) 4. Proportionality

=

=

x 2,

t

1 -1

Show that d'yldx 2 is proportional to y and rmd the constant of proportionality.

-1

~

x < 0

O~x<2 2~x~3

(Figure 5.31) over [-1,3] is 3/

VJkdt.

'

{ -I,

sin ax cos at - cos ax sin at.)

Suppose that x and yare related by the equation

x=

and

x~,

exist and are fInite at every interior point of I, and the appropriate one-

Show that

Y ~ 7i

lirn./(x)

9

c. I(x) :5 g(x) on the intervai-2 :5 x :5 5

3. Initial value problem

Although we are main\y interested in continuous functions, many functions in applications are piecewise continuous. A function I(x) is piecewise continuous on a closed interval I if 1 has only rmitely many discontinuities in I, the limits

(x)dx

=

1°(1 -

x)dx + f'x 2 dx + f'(-I)dx

-1

=

[x - ~l +

Jo

J2

[x;l: + [-xl:

S. Find 1(4) if ("

L

Jo

I(t) dt

[I(x)

=

x COS'lrx

b.

Jo

t 2 dt ~ x cos .".x.

y

6. Find 1(."./2) from the following informatioo.

i)

f

is positive and continuous.

ii) The area uoder the curve y

4

= I(x) from x = 0 to x = a is 3

7. The area of the region in the xy-plane eoclosed by the x-axis, the curvey = l(x),/(x) '" O,andthelinesx = I and x = biseqnai

to"\l'b'+l -

v'2 for all b > 1. Find I(x).

8. Prove that

-1

o -1

(Hint: Express the integral 00 the right-hand side as the differeoce of two integrals. Theo show that both sides of the equation have the same derivative with respect to x.)

3 y =-1

-

FIGURE 5.31 Piecewise continuous functions like this are integrated piece by piece.

Chapter 5 Additional and Advanced Exercises The Fundamental Theorem applies to piecewise continuous functions with the restriction that (d/dx)J: f(l) dl is expected to equal f(x) oo1y at values of x at which f is contiuuous. There is a similar restriction on Leibniz's Rule (see Exercise 27). Graph the functions in Exercises 11-16 and integrate them over their domains.

305

y

2/' 11. f(x) = { x_ : 4

13. g(t) =

{'0' x {I,.

14. h(z) =

~' { (7z _ 6)-1/3,

12. f(x) =

4,

~r-~~~-L--L--L--~~~x

o

0";/<1

smwt,

{

I - x 2,

,-II

Ii

--,-

Therefore, when n is large, S. will he close to 2/3 and we will have OSz<1

lszs2 -2";x<-1 -1";x<1

I, 15. f(x) =

12

Ii

lsxS2

2,

-lSr
{"

'~ , o

1

= Vt + v'2 +

Root sum

Vn = S.·n'/2 '" ~n'/2.

The following table shows how good the approximation can be. n

Root sum

(2/3)n'/2

Relative error

10 50 100 1000

22.468 239.04 671.46 21,097

21.082 235.70 666.67 21,082

1.386/22.468 '" 6% 1.4% 0.7%

0.07%

19. Evaluate

lim I' + 2' + 3' + ... +

n'

n6

11_00

by showing that the \intit is

2

18. Find the average value of the function graphed in the accompanying figure.

... +

l1x

5 dx

and evaluating the integral. 20. See Exercise 19. Evaluate + 2' + 3' + ... + n'). lim ~(1' 4

0---0

o

•

2

1

•

to approximate integrals. For example, let's estimate the sum of the square roots of the first n positive integers, Vt + v'2 + ... + Vn. The integral

~x'/21 = ~

is the lintit of the upper sums

S = II

n

21. Let f(x) he a contiuuous function. Express

3

Approximattng Finite Sums with Integrals In many applications of calculus, integrals are used to approximate finite sums-the reverse of the usual procedure of using finite sums

J.'VXdx =

11_00

as a delmite integral. 22. Use the result of Exercise 21 to evaluate •• lim

11_00

b. 1:.

~(2 + 4 + 6 + ... + 2n), 2 n

lim _1_ (1" + 2" + 3" + ... + n") 16

11_00

n

'

nw)

. I (. '11' • 2'11' . 3'11' . sm-+sm--+sm--+···+sm--. InnII_oon n n n n

What can be said about the following lintits?

fI.l+ ~.l+ ... + 'In ~.ln 'In n 'In n

Vt + v'2 + ... + Vn n'/2

d. lim _1_ (1" + 2" + 3" + ... + n") 11_00

n 17

e. lim _1_(1" + 2" + 3" + ... + n") 11_00

n 15

306

Chapter 5: Integration

23. a. Show that the area An of an n-sided regular polygon in a circle of radius r is

An =

nr2 .

27T

Tsmn·

b. Find the limit of An as n ~ 00 • Is this answer consistent with what you know about the area of a circle? 24. Let

can be evaluated directly, but the second cannot. We may fmd the derivative of either integral, however, by a formula called Leibniz's Rule.

Leibniz's Rule If J is continuous on [a, b] and if u(x) and v(x) are differentiable functions ofx whose values lie in [a, b], then

d dx

V (X)

dv du J(t) dt = J(v(x» dx - J(u(x» dx·

u(x)

To calculate lim,. ..... oo Sn, show that

and interpret Sn as an approximating sum of the integral

11

l

x 2 dx.

(Hint: Partition [0, 1] into n intervals of equal length and write out the approximating sum for inscribed rectangles.)

Figure 5.32 gives a geometric interpretation of Leibniz's Rule. It shows a carpet of variable width J(t) that is being rolled up at the left at the same time x as it is being unrolled at the right. (In this interpretation, time is x, not t.) At time x, the floor is covered from u(x) to vex). The rate du/dx at which the carpet is being rolled up need not be the same as the rate dv/dx at which the carpet is being laid down. At any given time x, the area covered by carpet is

Defining Functions Using the Fundamental Theorem

25. A function dermed by an integral The graph of a function J consists of a semicircle and two line segments as shown. Let g(x) = J(t) dt.

A(x) =

l

V(X)

J(t) dt.

u(x)

It

y

y

Uncovering

3

U(X)

i

A(x) =

a. Find g(1).

.(x)

b. Find g(3). c. Findg( -1). d. Find all values of x on the open interval ( - 3, 4) at which g has a relative maximum. e. Write an equation for the line tangent to the graph of gat x = -1. f. Find the x-coordinate of each point of inflection of the graph of g on the open interval (-3,4). g. Find the range of g.

i) y" = -sinx

+ 2 sin2x

dA dv du dx = J(v(x» dx - J(u(x» dx· At what rate is the covered area changing? At the instant x, A(x) is increasing by the width J(v(x» of the unrolling carpet times the rate dv/dx at which the carpet is being unrolled. That is, A(x) is being increased at the rate

7T.

J(u(x» : '

In applications, we sometimes encounter functions

[x2 (1 + t) dt

Jsmx

dv J(v(x» dx· At the same time, A is being decreased at the rate

ii) y = 1 andy' = -2 when x =

J(x) =

v(X)

FIGURE 5.32 Rolling and unrolling a carpet: a geometric interpretation ofLeibniz's Rule:

26. A differential equation Show that both of the following conditions are satisfied by y = sinx + Ix1/" cos 2t dt + 1:

Leibniz's Rille like

V(X)

f(t) dt

2Vi

and

g(x) =

1 Vi

sint 2 dt,

defmed by integrals that have variable upper limits of integration and variable lower limits of integration at the same time. The first integral

the width at the end that is being rolled up times the rate du/ dx. The net rate of change in A is

~

= J(v(x»: - J(u(x» : '

which is precisely Leibniz's Rule.

Chapter 5 Technology Application Projects To prove the rule, let F be an antiderivative off on [a, b]. Then

("V<)f(t) dt = 1.,,)

27. f(x) =

equation we want:

!

[F(v(x)) - F(u(x))

= F'(v(x)):Z = f(v(x)) dv dx

Chapter ~

1 1

I

tdt

l/x

Differentiatiog both sides of this equation with respect to x gives the

!!~)f(t) dt =

Use Leibniz's Rule to find the derivatives of the functions in Exercises 27-29. x

F(v(x)) - F(u(x)).

307

1

- F'(u(x))::

,",x

28. f(x) 29. g(y)

ChajnRulc

=

=

I - - 2 dt

coul-t

{2Vy sin t 2 dt

1Vy

30. Use Leibniz's Rule to fmd the value ofx that maximizes the value of the integral

du - f(u(x)) dx'

X +3

1 x

t(5-t)dt.

Technology Application Projects

Mathematica/Maple Modules: Usi1lg RieMtm1l Sums fD EslimauA-nIlS, Volum .., and Le1lgths o/Curves Visualize and approximate areas and volumes in Part I. Riema1l1l Slims, Definite 11I/egrals, and the FII1Idtmte1lfD1 Theorem o/Calcllills Parts I, II, and ill develop Riemann sums and defutite integrals. Part N contioues the development of the Riemann sum and defutite integral using the Fundamental Theorem to solve problems previously investigated. Rain Catchers, ElevatDrs, and Rockets Part I illustrates that the area under a curve is the same as the area of an appropriate rectangle for examples taken from the cbapter. You will compute the amount of water accumulatiog in basins of different sbapes as the basin is fllied and drained.

Motio1lAl01lg a Straight Line, Part II You will observe the sbape of a grapb through dramatic aninJated visualizations of the derivative relations among position, velocity, and

acceleration. Figures in the text can be animated using this software.

Be1lding 0/ Beams Study bent shapes of beams, determine their maximum deflections, concavity, and inflection points, and interpret the results in terms of a beam's compression and tension.

6 ApPLICATIONS OF DEFINITE INTEGRALS OVERVIEW In Chapter 5 we saw that a continuous function over a closed interval has a definite integral, which is the limit of any Riemann sum for the function. We proved that we could evaluate definite integrals using the Fundamental Theorem of Calculus. We also found that the area under a curve and the area between two curves could be computed as definite integrals. In this chapter we extend the applications of definite integrals to finding volumes, lengths of plane curves, and areas of surfaces of revolution. We also use integrals to solve physical problems involving the work done by a force, the fluid force against a planar wall, and the location of an obj ect's center of mass.

6.1

Volumes Using Cross-Sections In this section we define volumes of solids using the areas of their cross-sections. A cross-

y Cross-section S(x) with area A (x)

s

section of a solid S is the plane region formed by intersecting S with a plane (Figure 6.1). We present three different methods for obtaining the cross-sections appropriate to finding the volume of a particular solid: the method of slicing, the disk method, and the washer method. Suppose we want to find the volume of a solid S like the one in Figure 6.1. We begin by extending the definition of a cylinder from classical geometry to cylindrical solids with arbitrary bases (Figure 6.2). If the cylindrical solid has a known base area A and height h, then the volume of the cylindrical solid is Volume = area X height = A • h.

o

a

b

x

This equation forms the basis for defining the volumes of many solids that are not cylinders, like the one in Figure 6.1 . If the cross-section of the solid S at each point x in the interval [a, b] is a region S(x) ofareaA(x), and A is a continuous function of x , we can define and calculate the volume of the solid S as the definite integral of A (x). We now show how this integral is obtained by the method of slicing.

FIGURE 6.1 A cross-section Sex) of the solid S fonned by intersecting S with a plane Px perpendicular to the x-axis through the point x in the interval [a, b].

~= Plane region whose area we know

height

Cylindrical solid based on region Volume = base area x height = Ah

FIGURE 6.2 The volume of a cylindrical solid is always defined to be its base area times its height.

308

6.1 Volumes Using Cross-Sections

309

Slicing by Parallel Planes

y

We partition [a, b] into subintervals of width (length) fllk and slice the solid, as we would a loaf of bread, by planes perpendicular to the x-axis at the partition points a = Xo < Xl < ... < Xn = b. The planes P Xk ' perpendicular to the x-axis at the partition points, slice S into thin "slabs" (like thin slices of a loaf of bread). A typical slab is shown in Figure 6.3. We approximate the slab between the plane at Xk- l and the plane at Xk by a cylindrical solid with base area A(Xk) and height IlXk = Xk - Xk- l (Figure 6.4). The volume Vk of this cylindrical solid is A(Xk) • Ilxk, which is approximately the same volume as that of the slab:

S

fr' I I I

o

I I I

Volume of the kth slab x

FIGURE 6.3

A typical thin slab in the

~

Vk = A(Xk) fllk.

The volume V of the entire solid S is therefore approximated by the sum of these cylindrical volumes, n

solidS.

n

V ~ ~ Vk = ~A(Xk) IlXk. k=l

y Approximating cylinder based on S(xk) has height

Plane at Xk- l

k=l

This is a Riemann sum for the function A(x) on [a, b]. We expect the approximations from these sums to improve as the norm of the partition of [a, b] goes to zero. Taking a partition of [a, b] into n subintervals with Ilpll - 0 gives

~xk = xk - xk_ l

n~~ ~A(Xk) fllk =

/

l

bA (X)dx.

So we define the limiting definite integral of the Riemann sum to be the volume of the solid S.

o

,---

Xk - l

The cylinder's base is the region S(Xk) with area A(xk)

l

DEFINITION The volume of a solid of integrable cross-sectional area A(x) from x = a to x = b is the integral of A from a to b,

Plane atxk

~X

NUfTOSCALE

FIGURE 6.4 The solid thin slab in Figure 6.3 is shown enlarged here. It is approximated by the cylindrical solid with base S(Xk) having area A (Xk) and height Ilxk

This definition applies whenever A(x) is integrable, and in particular when it is continuous. To apply the definition to calculate the volume of a solid, take the following steps:

= Xk - Xk - l ·

Calculating the Volume of a Solid 1. Sketch the solid and a typical cross-section. 2. Find aformulafor A (x), the area ofa typical cross-section. 3. Find the limits of integration. 4. Integrate A (x) to find the volume.

EXAMPLE 1 A pyramid 3 m high has a square base that is 3 m on a side. The crosssection of the pyramid perpendicular to the altitude x m down from the vertex is a square x m on a side. Find the volume of the pyramid. Solution

1.

A sketch. We draw the pyramid with its altitude along the x-axis and its vertex at the origin and include a typical cross-section (Figure 6.5).

310

Chapter 6: Applications of Definite Integrals 2.

y

Aformulafor A(x). The cross-section at x is a square x meters on a side, so its area is A(x)

Typical cross-section

~

3.

The limits of integration. The squares lie on the planes from x = 0 to x = 3.

4.

Integrate to find the volume:

3 -e. _ x

= x 2.

V=

-- ~x(m)

13 o

A(x) dx =

13 0

3]3

x 2 dx = ~ 3

=

9 m 3•

0

•

EXAMPLE 2 A curved wedge is cut from a circular cylinder of radius 3 by two planes. One plane is perpendicular to the axis of the cylinder. The second plane crosses the first plane at a 45° angle at the center of the cylinder. Find the volume of the wedge. FIGURE 6.5

The cross-sections of the pyramid in Example I are squares.

2~

Solution We draw the wedge and sketch a typical cross-section perpendicular to the x-axis (Figure 6.6). The base of the wedge in the figure is the semi-circle with x ;:::: 0 that is cut from the circle x 2 + y2 = 9 by the 45° plane when it intersects the y-axis. For any x in the interval [0, 3], the y-values in this semi-circular base vary from

y = - ~ to y = ~. When we slice through the wedge by a plane perpendicular to the x-axis, we obtain a cross-section at x which is a rectangle of height x whose width extends across the semi-circular base. The area of this cross-section is A(x) = (height)(width) = (x)(2~)

=2x~. The rectangles run from x = 0 to x = 3, so we have

V = lbA(x) dx = =

12x~ 3

_l (9

_ x2)3/2]3

3

FIGURE 6.6 The wedge of Example 2, sliced perpendicular to the x-axis. The cross-sections are rectangles.

dx

0

Letu=9-x 2 , du = - 2x dx, integrate, and substitute back.

+ l(9)3/2

=

0

=

18.

3

•

EXAMPLE 3 Cavalieri's principle says that solids with equal altitudes and identical cross-sectional areas at each height have the same volume (Figure 6.7). This follows immediately from the definition of volume, because the cross-sectional area function A(x) and the interval [a, b] are the same for both solids.

b HISTORICAL BIOGRAPHY

__ - : : Same V hOlume ---_ ,r-

-----

Bonaventura Cavalieri (1598- 1647)

Same cross-section area at every level

FIGURE 6.7 Cavalieri's principle: These solids have the same volume, which can be illustrated with stacks of coins.

•

6.1 Volumes Using Cross-Sections

311

Solids of Revolution: The Disk Method

y

~Or-----~X~------~4~~~--X

The solid generated by rotating (or revolving) a plane region about an axis in its plane is called a solid of revolution. To find the volume of a solid like the one shown in Figure 6.8, we need only observe that the cross-sectional area A(x) is the area of a disk of radius R(x), the distance of the planar region's boundary from the axis of revolution. The area is then

A(x)

(a)

7T[R(x)f

= 7T(radius)2 =

So the definition of volume in this case gives y

y=Vx

Volume by Disks for Rotation About the x-axis

This method for calculating the volume of a solid of revolution is often called the disk method because a cross-section is a circular disk of radius R(x).

(b)

FIGURE 6.8 The region (a) and solid of revolution (b) in Example 4.

Yx,O :::; x

EXAMPLE 4 The region between the curve y = revolved about the x-axis to generate a solid. Find its volume.

:::; 4, and the x-axis is

Solution

We draw figures showing the region, a typical radius, and the generated solid (Figure 6.8). The volume is

V

=

=

l 14

= 7T

EXAMPLE 5

b

7T[R(x)]2 dx 7T[

Radius R(x) = Vx for rotation around x-axis

Yxy dx

f4 xdx

Jo

=

7Tx2]4 2

•

87T.

= 7T (4)2 =

0

2

The circle

is rotated about the x-axis to generate a sphere. Find its volume.

Solution

We imagine the sphere cut into thin slices by planes perpendicular to the x-axis (Figure 6.9). The cross-sectional area at a typical point x between -a and a is R(x) = ~ for

rotation around x-axis

Therefore, the volume is

V

=

a L_aA(x) dx

=

La - a7T(a 2 - x 2) dx

= 7T

[ 4 7Ta3 . a 23 x -];a- a = 3

•

The axis of revolution in the next example is not the x-axis, but the rule for calculating the volume is the same: Integrate 7T(radius? between appropriate limits.

EXAMPLE 6 by y =

Find the volume of the solid generated by revolving the region bounded

Yx and the lines y

= 1, x = 4 about the line

y = 1.

312

Chapter 6: Applications of Definite Integrals y

A(x) = 7T(a 2 -

x2)

/

x

FIGURE 6.9 The sphere generated by rotating the circle x 2 + y2 = a 2 about the x-axis. The radius is R(x) = y = ~ (Example 5).

SoLution We draw figures showing the region, a typical radius, and the generated solid (Figure 6.10). The volume is V

= =

14 14

1T[R(x)f dx 1T[ vi:

-

Radius R(x) = Vx - 1 for rotation around y = 1

Y

1 dx

Expand integrand.

=

2 32 / + X]4 1T [£.2- 2'-x 23 1

Integrate.

y R(x) =

Vx -

1

y

",,," I

/

'"

I I

~O+-----L---------xL-------4~X

(a)

FIGURE 6.10 The region (a) and solid of revolution (b) in Example 6.

(b)

•

To find the volume of a solid generated by revolving a region between the y-axis and a curve x = R(y), C :::; y :::; d, about the y-axis, we use the same method with x replaced by y. In this case, the circular cross-section is

and the definition of volume gives

313

6.1 Volumes Using Cross-Sections

Volume by Disks for Rotation About the y-axis v =

y

l

d

A(y) dy =

l

d

7T[R(y)f dy.

EXAMPLE 7 Find the volwne of the solid generated by revolving the region between the y-axis and the curve x = 2/y, 1 ::; y ::; 4, about the y-axis.

J.~

We draw figures showing the region, a typical radius, and the generated solid (Figure 6.11). The volwne is

Solution

4

v

y

R(y) =

2

0

=

14

7T[R(y)f dy

~

Radius R(y) =

~ for

rotation around y-axis

x

•

(a)

y

EXAMPLE 8 Find the volwne of the solid generated by revolving the region between the parabola x = y2 + 1 and the line x = 3 about the line x = 3.

4

We draw figures showing the region, a typical radius, and the generated solid (Figure 6.12). Note that the cross-sections are perpendicular to the line x = 3 and have y-coordinates from y = - V2 to y = V2. The volwne is Solution

(~,y)

y =

±

V2 when x =

3

Radius R(y) = 3 - (y2 + 1) for rotation around axis x = 3

x (b)

Expand integrand.

FIGURE 6.11 The region (a) and part of the solid of revolution (b) in Example 7. =

7T [4y -

y5]v2

4

"3 y3 + 5 -v2

Integrate.

647TV2 15 Y

R(y)

=3-

(y2

= 2_y2

+ 1)

y

~"2l (3, Y2)

Y2

=2_

y2

Y2

y

y

0

-Y2

R(y)

5

3 x

= y2 + 1

(3,-Y2) (a)

x

31---

0

5 x

-Y2 (b)

FIGURE 6.12 The region (a) and solid of revolution (b) in Example 8.

•

314

Chapter 6: Applications of Definite Integrals y

y

FIGURE 6.13 The cross-sections of the solid of revolution generated here are washers, not disks, so the integral A(x) dx leads to a slightly different fonnula.

lab

Solids of Revolution: The Washer Method If the region we revolve to generate a solid does not border on or cross the axis ofrevolution, the solid has a hole in it (Figure 6.13). The cross-sections perpendicular to the axis of revolution are washers (the purplish circular surface in Figure 6.13) instead of disks. The dimensions of a typical washer are y

Outer radius:

R(x)

Inner radius:

r(x)

The washer's area is

A(x) = 7T[R(x)f - 7T[r(x)f = 7T([R(xW - [r(xW). Consequently, the definition of volume in this case gives integration

Volume by Washers for Rotation About the x-axis

(a)

y

This method for calculating the volume of a solid of revolution is called the washer method because a thin slab of the solid resembles a circular washer of outer radius R(x) and inner radius rex).

EXAMPLE 9 The region bounded by the curve y = x 2 + 1 and the line y = -x is revolved about the x-axis to generate a solid. Find the volume of the solid.

+

3

SoLution We use the four steps for calculating the volume of a solid as discussed early in this section. Washer cross-section Outer radius: R(x) = -x + 3 Inner radius: r(x) = x 2 + 1

(b)

FIGURE 6.14 (a) The region in Example 9 spanned by a line segment perpendicular to the axis of revolution. (b) When the region is revolved about the x-axis, the line segment generates a washer.

1.

Draw the region and sketch a line segment across it perpendicular to the axis of revolution (the red segment in Figure 6. 14a).

2.

Find the outer and inner radii of the washer that would be swept out by the line segment if it were revolved about the x-axis along with the region. These radii are the distances of the ends of the line segment from the axis of revolution (Figure 6.14). Outer radius:

R(x)

=

Inner radius:

rex)

=

+3 x2 + 1 -x

6.1 Volumes Using Cross-Sections 3.

Find the limits of integration by finding the x-coordinates of the intersection points of the curve and line in Figure 6.14a.

x2 + 1

=

-x + 3

2

=

0

+ 2)(x - 1)

=

0

x2

(x

+X

-

x = -2,

4.

v=

R(y) =

x = 1

Limits of integration

Evaluate the volume integral.

y

4~

315

vY

l

b

7T([R(x)f - [r(x)f) dx

= l~7T«-x +

(2, 4)

3? -

Rotation around x-axis

Values from Steps 2 and 3

(x 2 + 1)2)dx

4

Simplify algebraically.

y

=

2xor

=

x= ~

2

y

=

x=vY (a)

= 2:y

y

R(y)

- 3x 2

x3

-

x 5 ]!

3 - 5

1177T

-2

5

•

x 2 or

----~~~------~2--~x

r(y)

7T [8x

To find the volume of a solid formed by revolving a region about the y-axis, we use the same procedure as in Example 9, but integrate with respect to y instead of x. In this situation the line segment sweeping out a typical washer is perpendicular to the y-axis (the axis of revolution), and the outer and inner radii of the washer are functions ofy.

= "vy EXAMPLE 10

The region bounded by the parabola y = x 2 and the line y = 2x in the first quadrant is revolved about the y-axis to generate a solid. Find the volume of the solid. First we sketch the region and draw a line segment across it perpendicular to the axis of revolution (they-axis). See Figure 6.15a. The radii of the washer swept out by the line segment are R(y) = yJ, r(y) = y/2 (Figure 6.15). The line and parabola intersect at y = 0 and y = 4, so the limits of integration are c = 0 and d = 4 . We integrate to find the volume:

Solution

Rotation around y-axis (b)

FIGURE 6.15 (a) The region being rotated about the y-axis, the washer radii, and limits of integration in Example 10. (b) The washer swept out by the line segment in part (a).

Substitute for radii and limits of integration.

= 7T

r

Jo

(y _ y2) dy 4

=

7T[y2 _ y3]4 2 12 0

=

~7T. 3

•

316

Chapter 6: AppLications of Definite Integrals

Exercises 6.1 Volumes by Slfdng Find the volumes of the solids in Exercises 1-10. 1. The solid lies between planes perpendiaular to the x-axis at x = 0 and x = 4. The cross-scctions perpendicular to tile axis on the in.tlnaJ. 0 :s x :s 4 8Ie sqLUIlCII whose ~ run from the parabolay = -v';;totheparabolay = Yx. 1. The solid lies between planes perpendi.cular to the x-axis at x - -1 and x - L The cross-sections perpendicular to the x-axis 8Ie circular disks whose di.ameteIs run from. the parabola y = x:1tothePlUUJolay = 2 - x'l.

,

8. The base of a solid is the region bmmded by the graphs ofY - v';; and Y - x12. The cross-sections perpendicular to the x-axis are L

isosceles triangles of height 6.

b. semi-circles with diameters running across the base of the solid 9. The solid lies between planes perpendicular to the y-axis at y = 0 and y = 2. The cross-sections perpendicular to the y-axis arc circular disks with diamebml running from the y-axis to the parabola x _ Vsy2. 10. The base of the solid is the diskx 2 + y:1 S L The cross-sections by planes perpendicular to the y-axis between y = - 1 and y = 1 arc isosceles right triangles with one leg in the disk.

,

o ---- ~----

x

3. The solid lies between planes perpendicular to the x-axis at x - - 1 and x - I. The cross-sections perpendicular to the x-axis between these planes arc squares whose bases run from the semicircle y = - ~ to the semicircle y = ~. 4. The solid lies benwen planes perpendicular to the x-axis at x = -I and x = 1. The cross-sections perpendicular to the x-axis between these planes ~s whose diagonals run from the semicircley = -VI - x 2 to the semicircley = ~.

o

11. Find the volume ofthc given tetrahedron. (Hint Consider slices perpendicular to one of tile labeled edges.)

5. The base of a solid is the :region between the ClU'Ve y = 2 ~ and the interval [0, 'Ir] on the x-axis. The cross-sections perpendicular to the x-axis are L

,

cquilateIal. triangles with bases running from. the x-axis to the curve as shown in the accompanying figure.

, / 11. Find the volume of the given pyramid, which has a square base of area 9 and height 5.

•

x

b. squares with bases running from the x-axis to the curve.

6. The solid lies between planes perpendi.cular to the x-axis at x - -'lr/3 andx - 'lr/3. The cross-sections perpendicular to the x-axis are circula:r disks with diameters running from the curve Y - tanxtothecurvey - secx. b. squares whose bases run from the curve Y - tan x to the L

curvey - secx.

7. The base of a solid is the region bounded by the graphs of Y - 3x, Y - 6, and x - O. The cross-sections perpendicular to the x-axis arc L rectangles of height 10. b. rectangles of perimeter 20.

13. A twisted lOUd A square of side length s lies in a plane perpendicularto alineL. One vertex of the square lies onL. As this square moves a distance II along L, the square turns one revolution about L to generate a carkscrew-li]re colmnn with square cross-sections. L

Find the volmne of the colmnn.

b. What will the volume be if the square turns twice instead of once? Give reasons foryom answer.

6.1 Volumes Using Cross-Sections 14. CJIYlIlisi'. principle A solid lies between planes perpendicular to the x-axis at x - 0 and x - 12. The cross-sections by planes perpendicular to the x-axis are circular disks whose diameters run from the line y - x/2 to the line y - x as shawn in the accompanying figure. Explain why the solid has the same volume as a right circular cone with base radius 3 and height 12.

317

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 27-32 about the y-axis. 27. Th.eregionenclosedbyx - y'Sy2, 28. Th.eregionenclosedbyx - y3/2,

0, y - -I, y - I

X -

X - 0, y - 2

29. Th.eregion enclosedbyx = v'2sin2y,

O:s y:S;

30. The region enclosed by x = Vcos('Ir}'/4),

'Ir/2, x

= 0

-2:s y:s; 0,

x=O 31. x = 2/(y 32. x -

+ 1), x = 0, y = 0, y = 3 V2Y/(yl + I), X - 0, y - 1

Volumes by the Washer Method Find the volumes of the solids generated by revolving the shaded regions in Exercises 33 and 34 about the indicated axes. 33. The x-axis

Volumes by the Disk Method In Exercises 15-18, rind the volume of the solid generated by revolving the shaded region about the given axis. 15. Aboutthex-ws 16. About the y-axis

,

,

18. About the x-axis

,

1f---

-7'

L _/

.% -

tan

1

sin.%COllI.%

23. y

o

~

1:

.

= V9 - xl, y = 0 22. y = x - x2 , Y = 0 = Vcosx, O:s x:s ",./2, y = 0, x = 0

24. y - secx, y - 0,

35. y - x, y - I, x - 0

36.y - 2Vx, y - 2, x - O 37.y=xl +l, y=x+3 38.y=4-xl , y=2-x

v'2,

40. y - secx, y - tanx,

Find the volumes of the solids gweratOO. by revolving the regions bounded by the lines and curves in Exercises 19-24 about the x-axis. 19. y = Xl, Y = 0, x = 2 20. y = x 3, Y = 0, x = 2

21. y

~----~~----~~ _'!!' 0 'II' . 2 1:

39. y - secx, y y -

(~y)

o

x - tany

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 35--40 about thex-axis.

-,I"'---------c!- • ~--~------~~ o 2 . o 3

,

,

~O~---------+I-- '

.%+2J=2

17. Aboutthey-ws

34. The y-axis

x - -'lr/4, x - 'Ir/4

-'lr/4:S x :s 'Ir/4 x - 0, x - I

In Exercises 41-44, rind the volume of the solid generated by revolving each region about the y-axis. 41. The region enclosed by the triangle with vertices (I, 0), (2, I), and (I, I) 42. The region enclosed by the triangle with vertices (0, 1), (I, 0), and (I, I) 43. The region in the rust quadrant bounded above by the parabola y - xl, below by thex-axis, and on the right by the line x - 2 44. The region in the rlISt quadrant bounded on the left by the circle xl + y2 _ 3, on the right by the line x and above by the

v'3,

line y -

v'3

In Exercises 25 and 26, rmd the volume of the solid generatJ;:d by revolving the region about the given line.

In Exercises 4S and 46, find the volume of the solid generated by revolving each region about the given axis.

25. The region in the first quadrant bounded above by the line y = below by the curve y = sec x tanx, and on the left by the y-axis, about the line y =

45. The region in the rust quadrant bounded above by the curve y - x 2, below by the x-axis, and on the right by the line x - I, about the line x - -1

26. The region in the rrrst quadrant bounded abave by the line y = 2, bclow by the curve y = 2sinx, 0 :s x :s ",./2, and on the left by the y-axis, about the line y = 2

46. The region in the second quadrant bounded above by the curve y - -x3,belowbythex-axis,andontheleftbythelinex - -I, about the line x - -2

v'2,

v'2

318

Chapter 6: AppLications of Definite Integrals

Volumes of Solids of Revolution 47. Find the volume of the solid generated by revolving the region bounded by y = Vx and the lines y = 2 and x = 0 about a. the x-axis.

b. the y-axis.

c. the line y = 2.

d. the line x = 4.

56. Designing a plumb bob Having been asked to design a brass plumb bob that will weigh in the neighborhood of 190 g, you decide to shape it like the solid of revolution shown here. Find the plumb bob's volume. If you specify a brass that weighs 8.5 g/cm3 , how much will the plumb bob weigh (to the nearest gram)? y(cm)

48. Find the volume of the solid generated by revolving the triangular region bounded by the lines y = 2x, y = 0, and x = 1 about a. the line x = 1.

y =.£ Y36 - x 2 12

b. the line x = 2.

49. Find the volume of the solid generated by revolving the region bounded by the parabola y = x 2 and the line y = 1 about a. the line y = 1.

c. the line y

=

b. the line y = 2.

x (cm)

-1.

50. By integration, find the volume of the solid generated by revolving the triangular region with vertices (0, 0), (b, 0), (0, h) about a. the x-axis.

b. the y-axis.

Theory and Applications 51. The volume of a torus The disk x 2 + y2 ::::; a 2 is revolved about the line x = b (b > a) to generate a solid shaped like a doughnut 2 - y2 dy = and called a torus. Find its volume. (Hint: J~a

57. Designing a wok You are designing a wok frying pan that will be shaped like a spherical bowl with handles. A bit of experimentation at home persuades you that you can get one that holds about 3 L if you make it 9 cm deep and give the sphere a radius of 16 cm. To be sure, you picture the wok as a solid of revolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? (1 L = 1000 cm3.)

Ya

y(cm)

'Tfa 2/2, since it is the area ofa semicircle of radius a.) /---- --...

52. Volume of a bowl A bowl has a shape that can be generated by revolving the graph of y = x 2/2 between y = 0 and y = 5 about they-axis.

/

I

I

x 2 + y2 = 16 2 = 256

' "

I

a. Find the volume of the bowl.

o

b. Related rates If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep?

x (cm)

__

53. Volume of a bowl

~~~ drep

a. A hemispherical bowl of radius a contains water to a depth h. Find the volume of water in the bowl. b. Related rates Water runs into a sunken concrete hemispherical bowl of radius 5 m at the rate of 0.2 m 3/ sec. How fast is the water level in the bowl rising when the water is 4mdeep? 54. Explain how you could estimate the volume of a solid of revolution by measuring the shadow cast on a table parallel to its axis of revolution by a light shining directly above it. 55. Volume of a hemisphere Derive the formula V = (2/3)'TfR 3 for the volume of a hemisphere of radius R by comparing its cross-sections with the cross-sections of a solid right circular cylinder of radius R and height R from which a solid right circular cone of base radius R and height R has been removed, as suggested by the accompanying figure. YR2_ h 2

h

~t

•

R

l' h

_ t_

R

58. Max-min The arch y = sinx, 0 ::::; x ::::; 'Tf, is revolved about the line y = c, 0 ::::; c ::::; 1, to generate the solid in the accompanying figure. a. Find the value of c that minimizes the volume of the solid. What is the minimum volume? b. What value of c in [0, 1] maximizes the volume of the solid?

D

c. Graph the solid's volume as a function of c, first for o : : ; c ::::; 1 and then on a larger domain. What happens to the volume of the solid as c moves away from [0, I]? Does this make sense physically? Give reasons for your answers. y

6.2 Volumes Using Cylindrical Shells 59. Consider the region R bounded by the graphs of y = f(x) > 0, x = a > 0, x = b > a, and y = 0 (see acoomanying figure). If the volume of the solid formed by revolving R about the x-axis is 4"., and the volume of the solid formed by revolving R about the line y = -I is 8"., fmd the area of R.

319

60. Consider the region R given in Exercise 59. If the volume of the solid formed by revolving R arouod the x-axis is 6"., and the volume of the solid formed byrevolvingR around the line y = -2 is 10"., fmd the area of R.

y

~ :

R

:

I I --;;-t-~-----:-~x

o

6.2

a

b

Volumes Using Cylindrical Shells

1:

In Section 6.1 we defmed the volume ofa solid as the defmite integral V = A(x) dx, where A(x) is an integrable cross-sectional area of the solid from x = a to x = b. The area A(x) was obtained by slicing through the solid with a plane perpendicular to the x-axis. However, this method of slicing is sometimes awkward to apply, as we will illustrate in our flISt example. To overcome this difficulty, we use the same integral definition for volume, but obtain the area by slicing through the solid in a different way.

Slicing with Cylinders Suppose we slice through the solid using circular cylinders of increasing radii, like cookie cutters. We slice straight down through the solid so that the axis of each cylinder is parallel to the y-axis. The vertical axis of each cylinder is the same line, but the radii of the cylinders increase with each slice. In this way the solid is sliced up into thin cylindrical shells of constant thickness that grow outward from their common axis, like circular tree rings. Umolling a cylindrical shell shows that its volume is approximately that of a rectangular slab with area A(x) and thickness ax. This slab interpretation allows us to apply the same integral defmition for volume as before. The following example provides some insight before we derive the general method. The region enclosed by the x-axis and the parabola y = j(x) = 3x - x 2 is revolved about the vertical line x = - 1 to generate a solid (Figure 6.16). Find the volume of the solid.

EXAMPLE 1

Solutton Using the washer method from Section 6.1 would be awkward here because we would need to express the x-values of the left and right sides of the parabola in Figure 6.l6a in terms ofy. (These x-values are the inner and outer radii for a typical washer, requiring us to solve y = 3x - x 2 for x, which leads to complicated formulas.) Instead of rotating a horizontal strip of thickness ,ly, we rotate a vertical strip of thickness ,lx. This rotation produces a cylindrical shell of height Yk above a point Xk within the base of the vertical strip and of thickness ,lx. An example ofa cylindrical shell is shown as the orange-shaded region in Figure 6.17. We can think of the cylindrical shell shown in the figure as approximating a slice of the solid obtained by cutting straight down through it, parallel to the axis of revolution, all the way around close to the inside hole. We then cut another cylindrical slice around the enlarged hole, then another, and so on, obtaining n cylinders. The radii of the cylinders gradually increase, and the heights of

320

Chapter 6: Applications of Definite Integrals

G t I-

\

__ 1. __ L __ 'L

-_-2~-_~ I--~---L--~~~-+x

I

I

1 I

1 I

r,

-- ~ --f01

-

_1_- _1 __

-1

Axis of revolution x =-1

Axis of revolution x = -1

-2

(b)

(a)

r

FIGURE 6.16 (a) The graph of the region in Example I, before revolution. (b) The solid formed when the region in part (a) is revolved about the axis of revolution x = - I.

~f 1

r

__

Yk

L_~_ J--~

T

° r~3

~"'"---l-I-'r x =-1

FIGURE 6.17 A cylindrical shell of height Yk obtained by rotating a vertical strip of thickness .:lxk about the line x = -I. The outer radius of the cylinder occurs at Xb where the height of the parabola is Yk = 3Xk - Xk2 (Example I).

x

the cylinders follow the contour of the parabola: shorter to taller, then back to shorter (Figure 6.16a). Each slice is sitting over a subinterval of the x-axis oflength (width) f:.xk. Its radius is approximately (1 + Xk), and its height is approximately 3Xk - x/. If we unroll the cylinder at Xk and flatten it out, it becomes (approximately) a rectangular slab with thickness f:.xk (Figure 6.18). The outer circumference of the kth cylinder is 21T' radius = 21T(1 + Xk), and this is the length of the rolled-out rectangular slab. Its volume is approximated by that of a rectangular solid, f:. Vk

= circumference

X height X thickness

= 21T( 1 + Xk) • (3Xk - Xk2) • f:.xk. Summing together the volumes f:. Vk of the individual cylindrical shells over the interval [0,3] gives the Riemann sum n

n

k=l

k=l

~f:.Vk = ~21T(Xk + 1)(3Xk - Xk2) f:.xk.

Outer circumference

L Radius ~

=

= 2'Fr • radius =

2 'Fr(1

+ Xk)

1 +xk

--- =--

I

-F

Llxk = thickness

FIGURE 6.18 Cutting and unrolling a cylindrical shell gives a nearly rectangular solid (Example 1).

6.2

Taking the limit as the thickness D..xk - 0 and n -

Volumes Using Cylindrical Shells

00

321

gives the volume integral

n

V = lim ~ 27T(Xk n~OO k=l

+ O(3Xk - Xk2) D..xk

= 1327T(X + O(3x - x2) dx = 1327T(3X2 + 3x - x3 - x2) dx

1 3

(2x 2 + 3x - x3) dx

=

27T

=

2 3 + -x2 3 27T [-x - -x 4

1]3

3

457T 2 .

240

•

We now generalize the procedure used in Example 1.

The Shell Method

I

The volume of a cylindrical shell of height h with inner radius r and outer radius R is

7TR 2h - 7Tr 2 h

=

27T (R ; r )
Suppose the region bounded by the graph of a nonnegative continuous function y = f(x) and the x-axis over the finite closed interval [a, b] lies to the right of the vertical line x = L (Figure 6.19a). We assume a ~ L, so the vertical line may touch the region, but not pass through it. We generate a solid S by rotating this region about the vertical line L. Let P be a partition of the interval [a, b] by the points a = Xo < Xl < ... < Xn = b, and let Ck be the midpoint of the kth subinterval [Xk- l, Xk]. We approximate the region in Figure 6.19a with rectangles based on this partition of [a, b]. A typical approximating rectangle has height f( Ck) and width D..xk = Xk - Xk- l. If this rectangle is rotated about the vertical line x = L, then a shell is swept out, as in Figure 6.19b. A formula from geometry tells us that the volume of the shell swept out by the rectangle is

D.. Vk = 27T X average shell radius X shell height X thickness

= 27T· (ck - L)· f(Ck) • D..xk.

Vertical axis of revolution Vertical axis of revolution

J~ Rectangle height = !(ck)

x=L

x (a)

(b)

FIGURE 6.19 When the region shown in (a) is revolved about the vertical line x = L, a solid is produced which can be sliced into cylindrical shells. A typical shell is shown in (b).

322

Chapter 6: Applications of Definite Integrals

We approximate the volume of the solid S by summing the volumes of the shells swept out by the n rectangles based on P:

The limit of this Riemann sum as each solid as a definite integral: V

= lim

Ln ~ V

k

=

n--->OO k=!

Ib

fllk -

0 and n -

00

gives the volume of the

27T(shell radius)(shell height) dx.

a

=

I

b

27T(X - L)f(x) dx.

We refer to the variable of integration, here x, as the thickness variable. We use the first integral, rather than the second containing a formula for the integrand, to emphasize the process of the shell method. This will allow for rotations about a horizontal line L as well.

Shell Formula for Revolution About a Vertical Line The volume of the solid generated by revolving the region between the x-axis and the graph of a continuous function y = f(x) 2: 0, L :5 a :5 x :5 b, about a vertical line x = L is

-l

V-

a

b

(shell ) ( shell ) dx 27T radius height .

EXAMPLE 2 The region bounded by the curve y = Vx, the x-axis, and the line x is revolved about the y-axis to generate a solid. Find the volume of the solid.

=

4

SoLution Sketch the region and draw a line segment across it parallel to the axis of revolution (Figure 6.20a). Label the segment's height (shell height) and distance from the axis of revolution (shell radius). (We drew the shell in Figure 6.20b, but you need not do that.)

y=

Vx (4,2)

y=

2

Vx

Shell height

o ~+-------+-~----~----~X

integration

x

Interval of integration (a)

(b)

FIGURE 6.20 (a) The region, shell dimensions, and interval of integration in Example 2. (b) The shell swept out by the vertical segment in part (a) with a width ~x.

6.2

Volumes Using Cylindrical Shells

323

The shell thickness variable is x, so the limits of integration for the shell formula are a = 0 and b = 4 (Figure 6.20). The volume is then

v= =

l sh~ll 14 Yx) 1 b

a

27T (

radIUs

) (

27T(X) (

s~ell )

heIght

dx

dx

]24 =~. 128 = 27T 4 x 3/ 2 dx = 27T [- 2 X 5/

o

5

0

5

•

So far, we have used vertical axes of revolution. For horizontal axes, we replace the x's withy's.

EXAMPLE 3 The region bounded by the curve y = Yx, the x-axis, and the line x = 4 is revolved about the x-axis to generate a solid. Find the volume of the solid by the shell method. Solution This is the solid whose volume was found by the disk method in Example 4 of Section 6.1 . Now we find its volume by the shell method. First, sketch the region and draw a line segment across it parallel to the axis of revolution (Figure 6.21a). Label the segment's length (shell height) and distance from the axis of revolution (shell radius). (We drew the shell in Figure 6.21b, but you need not do that.) In this case, the shell thickness variable is y, so the limits of integration for the shell formula method are a = 0 and b = 2 (along the y-axis in Figure 6.21). The volume of the solid is

v= =

l 1 1 b

a

sh~ll (s~ell

27T( ) ) d radIUs heIght Y

227T(Y)(4 - y2) dy

= 27T 2(4Y - y3) dy

y 2

y

o 0 .... OJ ''''c :>

~

B~

~ .5

12

y

4 _ y2 Shell height

[;:1<4.2)

o

y

y Shell radius

~ ~r-~-----------L-~x -0 4 W (a)

(b)

FIGURE 6.21 (a) The region, shell dimensions, and interval of integration in Example 3. (b) The shell swept out by the horizontal segment in part (a) with a width .ly .

•

324

Chapter 6: Applications of Definite Integrals

Summary of the Shell Method Regardless of the position of the axis of revolution (horizontal or vertical), the steps for implementing the shell method are these.

1. Draw the region and sketch a line segment across it parallel to the axis of revolution. Label the segment's height or length (shell height) and distance from the axis of revolution (shell redius). 2. Find the limits of integration for the thickness variable. 3. Integrate the product 21T (shell mdius) (shell height) with respect to the thickness variable (x or y) to find the volume.

The shell method gives the same answer as the washer method when both are used to calculate the volume of a region. We do not prove that result here, but it is illustmted in Exercises 37 and 38. (Exercise 60 in Section 7.1 outlines a proof.) Both volume formulas are actually special cases of a geneml volume formula we will look at when stodying double and triple integmls in Chapter 15. That geneml formula also allows for computing volumes of solids other than those swept out by regions of revolution.

Exercises 6.2 Revolution About the Axes

S. The y-axis

rn Exercises

l--{j, use the shell method to rmd the volumes of the solids generated by revolviog the shaded regioo about the indicated axis.

6. The y-axis y

y

Y=

9x

5

v9+9

o

3

2. y

~b 2 ~t---------~--~x

o

0

x

2

~f---'----4 x

0

Y

v2

2

........... -.......=::----1 ~-H

x=3-A

~f---------~~ -->x 3 ;J) 0

~

8.y~2x,

v3

x~r

x, y = -x/2, x ~ 2 y=x/2, x~1 9. y=x2 , y=2-x, x=O, forx~O 10. y = 2 - xl, Y = xl, X = 0 7. y

Y

Fv2

Revolution About the y-Axis Use the shell method to find the volumes of the solids generated by revolving the regions bouoded by the curves and lines in Exercises 7-12 about the y-axis.

4.

3.

0

2

v3

11. y ~ 2x 12. y ~

I, y ~ Yx, x ~ 0

3/{2Yx), y

=

0, x ~ I, x

=

4

6.2

13. Letj(x) = {(Sinx)fx,

0

1,

< x :5

25. y=x+2, y=x' L The line x = 2

'Tr

x = 0

a. Showthatxj(x) = sin x, 0:5

y

{

sinX

o<x

1,

x =

x'

S'11'

d. The lioe y = 4

26. y = x', y = 4 - 3x' L

The line x = I

c. The x-axis

In Exertises 27 and 28, use 1he shell metbod to fiod 1he volumes of 1he solids generated by revolving 1he shaded regions about 1he indicated axes.

27.

0

L

The x-axis

b. The line y = I

•• The line y = 8/5

o

= {o(tan, x)'/x,

0

1

< x :5 'Tr/4

x=o a. Showthatxg(x) = (tan x)', Os

X

b. The lioe y = -2/5

y

~~--------~~-+x

14. Letg(x)

325

b. Thelioex =-1

c. The x-axis

'Tr.

X S

b. Find the volwne of 1he solid generated by revolving the shaded regioo about the y-axis in 1he accompanying figure.

y=

Volumes Using Cylindrical Shells

x

= 12(y' -

y'J

:5 'Tr/4.

b. Find the volwne of the solid generated by revolving the

shaded region about 1he y-axis in the accompanying figure. y

tan' x

y

'1T

= { ---'O<x:Sx 4 x= 0

0,

~F---------~----------~x

28. a. The x-axis

b. The lioe y = 2

•• The line y = 5

d. The lioe y = -5/8

y ~I'---:o----+x

o

(2.2)

Revolution About the x-AxIs Use the shell me1hod to find the volwnes of1he solids generated by reo volving the regions bounded by 1he curves and lines in Exercises 15-22 about the x-axis.

15. x =

yY,

16. x = y2,

x = -y, y = 2 X = -y, y

=

17. x = 2y - y', x = 0

18. x = 2y - y', x = Y

19. y = lxi, y = I

20. y = x, y = 2x, y = 2

21. y=

Yx,

y=O, y=x-2

22.y=Yx, y=O, y=2-x Revolution About Horizontal and Vertical Lines In Exercises 23-26, use 1he shell me1hod to f'md 1he volwnes of the solids generated by revolving the regions bounded by 1he given curves

about the given lines. 23.y=3x, y=O, x=2 a. The y-axis •• The line x = -I e. Theliney = 7 24. y = x', y = 8, x = 0 a. The y-axis •• The line x = -2 e. Theliney = 8

~t-----~----~~x

2, Y ~ 0

o

Choosing the Washer Method or Shell Method For soroe regions, both the washer and shell me1hods work well for 1he solid generated by revolving the region about 1he coordinate axes, but this is not always the case. When a region is revolved about the y-axis, for example, and washers are used, we must integrate wi1h respect toy. It may not be possible, however, to express the integrand in terms ofy. In such a case, the shell method allows us to integrate with respect to x instead. Exercises 29 and 30 provide some insight

29. Compute 1he volwne of1he solid generated by revolving 1he regioo bounded by y = x and y = x' about eath coordinste axis using •• 1he shell me1hod.

b. The line x = 4

d. The x-axis f. The line y = -2

2

b. 1he washer method.

30. Compute the volwne of the solid generated by revolving 1he triangu1ar region bounded by the lines 2y = x + 4, y = x, and x = 0 about

•• 1he x-axis using 1he washer me1hod. b. The line x = 3

b. 1he y-axis using 1he shell method.

d. The x-axis f. Theliney=-I

•• 1he line x = 4 using 1he shell method. d. 1he line y = 8 using 1he washer me1hod.

326

Chapter 6: Applications of Definite Integrals

Io Exercises 31-36, find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so.

use to f"md the volume of the solid? How many integrals would be required in each case? Explain. y

31. The triangle with vertices (I, I), (I, 2), and (2, 2) about L

the x-axis

c. the line x

b. the y-axis ~

10/3

d. the line y

1 ~

I

32. The region bounded by y ~ Yx, y ~ 2, x ~ 0 about L

the x-axis

c. the line x

~~----~~--~~X

-2

0

b. the y-axis ~

4

d. the line y

~

2

33. The region in the f"ITst quadrant bounded by the curve x ~ y - y' and the y-axis about L

the x-axis

b. the line y

~

I

40. The region shown here is to be revolved about the y-axis to generate a solid. Which of the methods (disk, washer, shell) could you use to f"md the volume of the solid? How many integrals would be required in each case? Give reasons for your answers.

34. The region in the first quadrant bounded by x ~ y - y',x ~ I, and y ~ I about L

the x-axis

c. the line x

y

b. the y-axis ~

I

d. the line y

~

I

35. The region bounded by y ~ Yx and y ~ x 2/8 about L

the x-axis the y-axis

x2

o

and y ~ x about

b. the line x

~

I

37. The region in the f"ITSt quadrant that is bounded above by the curvey ~ I/x'/"ootheleftbythelinex ~ 1/16,andbelowby the line y

--;cf"""':....--+.-----> x

b. the y-axis

36. The region bounded by y ~ 2x L

= 1 is revolved about the x-axis to generate a solid.

-1

41. A bead is formed from a sphere of radius 5 by drilling through a diameter of the sphere with a dri11 bit of radius 3.

Find the volume of the solid by L

the washer method.

.. Find the volume of the bead. b. the shell method.

38. The region in the f"ITSt quadrant that is bounded above by the curve y ~ I/Yx, on the left by the line x ~ 1/4, and below by the line y = 1 is revolved about the y-axis to generate a solid. Find the volume of the solid by L

the wasber method.

b. the shell method.

Choosing Disks, Washers, or Shells 39. The region shown here is to be revolved about the x-axis to generate a solid. Which of the methods (disk, washer, shell) could you

6.3

(I, 1)

b. Find the volume of the removed portion of the sphere.

42. A Bundt cake, well known for having a ringed shape, is formed by revolving around the y-axis the region bouoded by the graph of y ~ sin (x 2 - I) and the x-axis over the interval I :5 x :5 vT+1T. Find the volume of the cake. 43. Derive the formula for the volume of a right circular cone of height h and radius r using an appropriate solid of revolution.

44. Derive the equation for the volume of a sphere of radius r using the shell method.

Arc Length We know what is meant by the length of a straight line segment, but without calculus, we have no precise dermition of the length of a general winding curve. If the curve is the graph of a continuous function dermed over an interval, then we can rmd the length of the curve using a procedure similar to that we used for derming the area between the curve and the x-axis. This procedure results in a division of the curve from point A to point B into II1lIl1Y pieces andjoining successive points of division by straight line segments. We then sum the lengths of all these line segments and derme the length of the curve to be the limiting value of this sum as the number of segments goes to inf'mity.

Length of a Curve y

~ f(x)

Suppose the curve whose length we want to find is the graph of the function y ~ f(x) from ~ a to x ~ b. In order to derive an integral formula for the length of the curve, we assume that fhas a continuous derivative at every point of [a, bl. Such a function is called smooth, and its greph is a smooth curve because it does not have any breaks, corners, or cusps.

x

6.3

y

327

Arc Length

ll.-l

--~-,--,-----,-----r#------~--~~--~~------~x

FIGURE 6.22 The length of the polygonal path PoPlP2... p. approximates the length of the curve y = f(x) from point A to pointB. y

I

AYk

I

I

L1

r ------AXl

P,

I

I I I

FIGURE 6.23 The arc P'-IP, of the curve y = f(x) is approximated by the straight line segment shown here, which has length L, = v'(ax.,)2 + (ay.,)2.

We partition the interval [a, b] into n subintervals with a = Xo < Xl < X2 < ... < X. = b. IfYk = j(Xk), then the corresponding point Pk(Xk, Yk) lies on the curve. Next we connect successive points Pk-l and Pk with straight line segments that, taken together, form a polygonal path whose length approximates the length of the curve (Figure 6.22). If Axk = Xk - Xk-l and ~Yk = Yk - Yk- t. then a representative line segment in the path bas length (see Figure 6.23)

so the length of the curve is approximated by the sum (1)

We expect the approximation to improve as the partition of [a, b] becomes rmer. Now, by the Mean Value Theorem, there is a point with Xk-l < Ck < X., such that

c'"

~Yk = f'(et)

Axk.

With this substitution for ~Y'" the sums in Equation (I) take the form

•

~ Lk

=

1=1

•

~ V(Axk)2 + (f'(Ck)~Xk)2 1=1

=

•

~ VI + [('(Ck»)' ~Xk'

(2)

1=1

Because VI + [('(x»)' is continuous on [a, b], the limit of the Riemann sum on the righthand side of Equation (2) exists as the norm of the partition goes to zero, giving

..

lim ~ Lk = lim ~ VI

n--i'OO 1=1

n_OO k=l

+

[('(et»)' Axk =

l

b

a

VI + [('(x»)' dx.

We derme the value of this limiting integral to be the length of the curve.

DEFINmON If j' is continuous on [a, b], then the length (arc length) of the curvey = j(X) from the point A = (a,j(a»tothepointB = (b,j(b» is the value of the integral

L

=

t

VI + [{'(x)l'dx

=

t ~I (:y +

dx.

(3)

328

Chapter 6: Applications of Definite Integrals

y

EXAMPLE 1

Find the length of the curve (Figure 6.24)

Y Solution

=

3/ 2 4v'Zx 3

I ,O";x";l.

-

We use Equation (3) with a = 0, b = I, and

y = 4v'Zx 3/ 2 3

I

_

x=1,y~O.89

dy = 4v'Z • lx'/2 = 2V2xl/2

FIGURE 6.24 The length of the curve is slightly larger than the length of the line segmeot joining points A and B (Example 1).

dx

(:Y

3

2

(2V2xl/2y =

=

&.

The length of the curve over x = 0 to x = I is

L= =

[)l+ (:Ydx= [~dx t· t(I +

&)3/2

J:

Eq. (3) with

a=O,b=l Letu = 1 + ~, integrate, and replace u by I + Ilx.

Ii '" 2.17.

=

Notice that the length of the curve is slightly larger than the length of the straight-lioe segment joining the points A = (0, -I) and B = (I, 4 v'Z/3 - I) on the curve (see Figure 6.24):

,112 + (1.089)2 '" 2.14

2.17>

EXAMPLE 2

Solution

•

Find the length of the graph of

f(x) y

Decimal approximations

=

x3

I

12 + x'

I ,,; x ,,; 4.

A graph of the function is shown in Figure 6.25. To use Equation (3), we find

f'(x)

=

x2

I

4 - x2

so

I + [f'(x)f I + (x24 _ l)2 I + (x'16 _ !2 + l) x x' x' I I (x2 I )2 =T6+z+ '= 4+ =

2

=

x

X

2

'

The length of the graph over [I, 4] is FIGURE 6.25 The curve in Example 2, where A = (1, 13/12) andB = (4,67/12).

L

=

=

l VI

+ [f'(x)]2dx

[~~ f

I

-

=

=

l (~

+ :2)dx

-

(~i - i) - C~ I) i~ 6. =

=

•

Dealing with Discontinuities in dy/dx At a point on a curve where dy/dx fails to exist, dx/dy may exist. In this case, we may be able to find the curve's length by expressing x as a function of y and applying the following analogue of Equation (3):

6.3

329

Arc Length

Formula for the Length of x = g(y), c ,;; y ,;; d Ifg'iscontinuouson[c,d],thelengthofthecurvex = g(y) from A = (g(c),c) to B = (g(d), d) is

L

EXAMPLE 3 Solution

y

=

t ~I

+

(~y dy =

t

(4)

Find the length of the curve y = (X/2)2/3 from x = 0 to x = 2.

The derivative

is not defined at x = 0, so we cannot find the curve's length with Equation (3). We therefore rewrite the equation to express x in terms of y:

y= y3/2

=

(~t Raise both sides to the power 3/2.

'! 2

x = 2y 3/2.

FIGURE 6.26 The graph of y = (X/2)2/3 from x = Otox = 2 is also the graph ofx = 2y 3/2 from y = 0 to y = I (Example 3).

VI + [g'(y)]2dy.

Solve forx.

From this we see that the curve whose length we want is also the graph of x = 2y 3/2 from y = 0 to y = I (Figure 6.26). The derivative

~ = 2(~}'/2 =

3y1/2

is continuous on [0, I]. We may therefore use Equation (4) to find the curve's length:

L = =

=

t ~I I 2 -'-(1 9 3

+

(~y dy = 1'v'i+9Ydy

+ 9y)3/2 ]'

0

{7 (IOv'iO - I) ,., 2.27.

Eq. (4) with c=O,d=l. Letu= 1 +9y, du/9 = dy, integrate, and substitute back.

•

The Differential Formula for Arc Length If y = f(x) and if I' is continoous on [a, b], then by the Fundamental Theorem of Calculus we can defme a new function

1 x

s(x) =

VI + [I'(t)f dt.

(5)

From Equation (3) and Figure 6.22, we see that this function s(x) is continoous and measures the length along the curve y = f(x) from the initial point Po(a, f(a» to the point Q(x, f(x» for each x E [a, b]. The function s is called the arc length function for y = f(x). From the Fundamental Theorem, the function s is differentiable on (a, b) and

:

=

VI + [1'(x)]2

=

~I +

(:r

330

Chapter 6: Applications of Definite Integrals

Then the differential of arc length is

y

(6) A useful way to remember Equation (6) is to write

ds = -4--------------~x

o

(aJ y

Vax' + dy',

(7)

which can be integrated between appropriate limits to give the total length of a curve. From this point of view, all the arc length fonnulas are simply dllferent expressions for the equation L = ds. Figure 6.27a gives the exact interpretation of ds corresponding to Equation (7). Figure 6.27b is not strictly accurate, but is to be thought of as a simplified approximation of Figure 6.27a. That is, ds RJ fl •.

J

EXAMPLE 4 A

=

Find the arc length function for the curve in Example 2 taking (1, 13/12) as the starting point (see Figure 6.25).

Solution

In the solution to Example 2, we found that

~--------------~X

1

o

+

+ ~,

[I'(x»)' = (x:

r

(bJ

Therefore the arc length function is given by FIGURE 6.27 Diagrams for remembering the equation tis = v'fix' + dy'.

.(x) =

[VI + 3 1

[I'(I»)'dl=

[(~+I~)dl

1]' =12-:X+12' 1

= [ 12 -I

x

I

3

11

To compute the arc length along the curve from A instance, we simply calculate 43

I

=

(1, 13/12) to B

=

(4,67/12), for

11

.(4) = 12 - 4 + 12 = 6.

•

TIris is the same result we obtained in Example 2.

Exercises 6.3 finding Lengths of Curves Find the lengths of the curves in Exercises 1-10. If you have a grapher, you may want to graph these curves to see what they look like. 1. y = (I/3)(x' + 2)3/2 from x = 0 tox = 3

y=x'/' from x=Otox=4 x = (y'/3) + 1/(4y) from y = I toy = 3 x = (y'/2/3) - yl/2 from y = I toy = 9 x = (y'/4) + 1/(8y') from y = I toy = 2 6. x = (y'/6) + 1/(2y) from y = 2 toy = 3 7. y = (3/4)x'/3 - (3/8)x'/3 + 5, I,;; x ,;; 8 8. y = (x'/3) + x' + x + 1/(4x + 4), 0,;; x ,;; 2

2. 3. 4. 5.

9. x = 10. y =

D

J.Y v'sec't -

I dt,

1\/:31'-=-! dt, -,

-1f/4,;; y '" 1f/4 -2';; x ,;; -I

Finding Integrals for Lengths of Curves In Exercises 11-18, do the following.

.. Set up an integral for the length of the curve. b. Graph the curve to see what it looks like.

c. Use your grapher's or computer's integral evaluator to find the curve's length numerically.

= x', -I '" x '" 2 Y = tanx, -",/3 '" x '" x = siny, O::s; y ::s; 'IT

24. Circumference of a circle Set up an integral to find the circumference of a circle of radius r centered at the origin. You will learn how to evaluate the integral in Section 8.3.

11. Y

12. 13.

0

14. x =~, -1/2 '" Y '" 1/2 15. y' + 2y = 2x + I from (-I, -I) to (7, 3)

sinx - x COS x,

16. y

=

17. Y

=

18. x

= J.Y V

I.'

tan I dl,

O:s; x :s;

25. If9x' = y(y - 3)', show that

, _ (y

ds-

+ I)' 4y

dy. ,

'11'

26. If 4x' - y' = 64, show that

0 '" x '" ",/6

sec' I - I dl,

331

Arc Length

6.3

ds' = 4, (5x' - 16) dx'.

-",/3 '" Y '" ",/4

y 27. Is t1rere a smooth (continuously differentiable) curve y

=

j(x)

whose length over the interval 0 '" x '" a is a1ways Yza? Give

reasons for your answer.

Theory and Example. 19••• Find a curve through the point (I, I) whose length integral (Equation 3) is

L =

[~1+ lxdx.

b. How many such curves are there? Give reasons for your answer.

20. •• Find a curve through the point (0, I) whose length integral (Equation 4) is

L=

28. Using tangent fins to derive the length formula for curves Assume that j is smooth on [a, h1and partition the interval [a, h1 in the usual way. In each subinterval [X.-I,x.tl, construct the langenlfin at the point (X.-b j(x._I)) , as shown in the accompanying figure . •• Show that the length of the kth tangent rm over the interval

[XH, x.tl equals V(&t.)'

+ (!'(XH) &t.)'.

b. Showthat

lim ±(lengthofkthtangentftn) =

[~I+ :.dy.

n-OO.t-l

l'

VI

a

+ (j'(x))'dx,

which is the lengthL of the curve y = j(x) from a to h.

b. How many such curves are there? Give reasons for your answer.

21. Find the length of the curve

y =

I.'

Vcos21dl

I

~ Tangent fin : with slope

(xk_l. !(xk_l»

from x = 0 to x = 1T/4. 22. The length of an astroid The graph of the equation x'i' + y'/3 = I is one of a family of curves called astroids (not "asteroids',) because of their starlike appeanmce (see the accompanying figure). Find the length of this particular astroid by rmding the length of half the first-quadrant portion, y = (I - X'/3)3/2, v'2/4 '" x '" I, and multiplying by 8.

-----f!""'--------:-Ilx-.----~, H'.-l) I I I

'.

29. Approximate the arc length of one-quarter of the unit circle (which is~) by computing the length of the polygonal approximation with n = 4 segments (see accompanying figure).

y y

1 x2i3

+ y'1J3 = 1

~~------~------~~ ,

-1

1

~~~~~~~~,

o

-1

23. Length of a line .egment Use the arc length formula (Equation 3) to rmd the length of the line segment y = 3 - 2x, 0 '" x '" 2. Check your answer by rmding the length of the segment as the hypotenuse of a right 1riangle.

0.25 0.5 0.75 1

30. Distance between two points Assume that the tv.<> points (Xl, Y1) and (x"Yz) lie on the graph of the slraight line y = nIX + h. Use the arc length formula (Equation 3) to rmd the distance between the two points. 31. Find the arc length function furthe graph ofj(x) = 2x 3/2using(O,O) as the stilting point What is the length ofthe curve from (0, 0) to (I, 2)?

332

Chapter 6: Applications of Definite Integrals

32. Find the arc length function for the curve in Exercise 8, using (0, 1/4) as the starting point. What is the length of the curve from (0, 1/4) to (1,59/24)?

c. Evaluate the length of the curve using an integral. Compare your approximations for n = 2, 4, 8 with the actual length given by the integral. How does the actual length compare with the approximations as n increases? Explain your answer.

COMPUTER EXPLORATIONS

33. f(x) =~,

In Exercises 33- 38, use a CAS to perform the following steps for the given graph of the function over the closed interval.

34. f(x)

a. Plot the curve together with the polygonal path approximations for n = 2, 4, 8 partition points over the interval. (See Figure 6.22.) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments.

= x l/3

+

35. f(x) = sin (1TX 2 ) , 36. f(x) =

x2

-I ~ x ~ I

x 2/ 3,

cos x,

0 ~ x ~ 2 0 ~ x ~

v2

0 ~ x ~ 1T

37. f(x) = x2 - I ' 4x + I

-l<x<1 2 -

38. f(x)

-I ~ x ~ I

=

x3

-

x 2,

Areas of Surfaces of RevoLution

6.4

When you jump rope, the rope sweeps out a surface in the space around you similar to what is called a surface of revolution. The surface surrounds a volume of revolution, and many applications require that we know the area of the surface rather than the volume it encloses. In this section we define areas of surfaces of revolution. More general surfaces are treated in Chapter 16.

Defining Surface Area

t!.x y

I""'t!.x-+! B

A

27TY

X

0

NOfroSCALE

(a)

(b)

FIGURE 6.28 (a) A cylindrical surface generated by rotating the horizontal line segment AB of length .:lx about the x-axis has area 21TY~X . (b) The cut and rolled-out cylindrical surface as a rectangle.

If you revolve a region in the plane that is bounded by the graph of a function over an interval, it sweeps out a solid of revolution, as we saw earlier in the chapter. However, if you revolve only the bounding curve itself, it does not sweep out any interior volume but rather a surface that surrounds the solid and forms part of its boundary. Just as we were interested in defining and finding the length of a curve in the last section, we are now interested in defining and finding the area of a surface generated by revolving a curve about an axis. Before considering general curves, we begin by rotating horizontal and slanted line segments about the x-axis. If we rotate the horizontal line segment AB having length Llx about the x-axis (Figure 6.28a), we generate a cylinder with surface area 27TyLlx. This area is the same as that of a rectangle with side lengths Llx and 27TY (Figure 6.28b). The length 27TY is the circumference of the circle of radius Y generated by rotating the point (x, y) on the line AB about the x-axis. Suppose the line segment AB has length L and is slanted rather than horizontal. Now when AB is rotated about the x-axis, it generates a frustum of a cone (Figure 6.29a). From classical geometry, the surface area of this frustum is 27TY* L, where y* = (Yl + Y2)/2 is the average height of the slanted segment AB above the x-axis. This surface area is the same as that of a rectangle with side lengths Land 27TY' (Figure 6.29b). Let's build on these geometric principles to define the area of a surface swept out by revolving more general curves about the x-axis. Suppose we want to find the area of the surface swept out by revolving the graph of a nonnegative continuous function Y = I(x), a ~ x ~ b, about the x-axis. We partition the closed interval [a, b] in the usual way and use the points in the partition to subdivide the graph into short arcs. Figure 6.30 shows a typical arc PQ and the band it sweeps out as part of the graph of f.

6.4 Areas of Surfaces of Revolution

333

L Y

I---'--+-L-_ x

o

NafTOSCALE

(a)

(b)

FIGURE 6.29 (a) The frustum of a cone generated by rotating the slanted line segment AB of length L about the x-axis has area

21TY· L. (b) The area of the rectangle for y.

Yl + Y2 = --2-' the average

height of AB above the x-axis.

x

FIGURE 6.30 The surface generated by revolving the graph of a nonnegative function y = f(x), a ~ x ~ b, about the x-axis. The surface is a union of bands like the one swept out by the arc PQ.

As the arc PQ revolves about the x-axis, the line segment joining P and Q sweeps out a frustum of a cone whose axis lies along the x-axis (Figure 6.31). The surface area of this frustum approximates the surface area of the band swept out by the arc PQ. The surface area of the frustum of the cone shown in Figure 6.31 is 21Ty*L, where y. is the average height of the line segment joining P and Q, and L is its length (just as before). Since I ~ 0, from Figure 6.32 we see that the average height of the line segment is

+ I(Xk»j2, and the slant length is L = V(~Xk)2 + (~Ykf Therefore, I(Xk- i) + I(Xk) .. I Frustum surface area = 21T· 2 . v (~xkf + (~Ykf

y. = (f(Xk- i)

=

1T(f(Xk- i)

+ J(Xk»V(~Xk)2 + (~Ykf

The area of the original surface, being the sum of the areas of the bands swept out by arcs like arc PQ, is approximated by the frustum area sum n

,L1T(f(Xk- i) k=i

+ I(Xk»V(~Xkf + (~Ykf.

(1)

We expect the approximation to improve as the partition of [a, b] becomes finer. Moreover, if the function I is differentiable, then by the Mean Value Theorem, there is a point (Ck, I( Ck» on the curve between P and Q where the tangent is parallel to the segment PQ (Figure 6.33). At this point, FIGURE 6.31 The line segment joining P and Q sweeps out a frustum of a cone.

~Yk

!'(Ck)

=

~Xk'

~Yk = !'(Ck) ~Xk. Segment length: p

L = Y(8xk)2

+ (8Yk)2

With this substitution for

~Yk ,

the sums in Equation (1) take the form

n

,L1T(f(Xk- i) k=i

+ I(Xk»V(~Xk)2 + (f'(ck) ~xkf

n

= ,L1T(f(Xk- i) + I(Xk»V1 + (f'(ck)f ~k .

(2) k=i These sums are not the Riemann sums of any function because the points Xk- i, Xb and Ck are not the same. However, it can be proved that as the norm of the partition of [a, b] goes to zero, the sums in Equation (2) converge to the integral FIGURE 6.32 Dimensions associated with the arc and line segment PQ.

l

b

21TI(x)V1

+ (f'(x»2 dx.

334

Chapter 6: AppLications of Definite Integrals

We therefore define this integral to be the area of the surface swept out by the graph of f from a to b.

DEFINITION

If the function f(x) 2:: 0 is continuously differentiable on [a, b], the area of the surface generated by revolving the graph of y = f(x) about the x-axis is (3)

Iff is smooth, the Mean Value Theorem guarantees the existence of a point Ck where the tangent is parallel to segmentPQ. FIGURE 6.33

The square root in Equation (3) is the same one that appears in the formula for the arc length differential of the generating curve in Equation (6) of Section 6.3.

EXAMPLE 1

y

1

:5 X :5

SoLution

Find the area of the surface generated by revolving the curve y 2, about the x-axis (Figure 6.34). We evaluate the formula b S = 27T

+

= 1,

y

l Y)1 (2Y

dx

=

2Vx,

Eq.(3)

with x

a

b = 2,

dy dx

= 2Vx,

1 Vx·

First, we perform some algebraic manipulation on the radical in the integrand to transform it into an expression that is easier to integrate. FIGURE 6.34 In Example 1 we calculate the area of this surface.

=

~1 + i = ~x ~ 1 = ~.

With these substitutions, we have

r

S = JI 27T·2Vx

~

~

]2

dx

2 = 47T· I(x + 1)3/2 I = 8;

{2

= 47TJI

~ dx

(3V3 - 2\12).

Revolution About the y-Axis For revolution about the y-axis, we interchange x and y in Equation (3).

Surface Area for Revolution About the y-Axis If x = g(y) 2:: 0 is continuously differentiable on [c, d], the area ofthe surface generated by revolving the graph of x = g(y) about the y-axis is

•

6.4 Areas of Surfaces of Revolution

335

°: :;

EXAMPLE 2 The line segment x = 1 - y, y :::; 1, is revolved about the y-axis to generate the cone in Figure 6.35. Find its lateral surface area (which excludes the base area).

y

A (0, 1)

Solution

Here we have a calculation we can check with a formula from geometry: base circumference

.

2

Lateral surface area =

X slant heIght =

~

;;:

7T V 2 .

To see how Equation (4) gives the same result, we take

dx dy

x = 1 - y,

d = 1,

c = 0,

-1,

x

\!I+(-I?=V2 FIGURE 6.35 Revolving line segmentAB about the y-axis generates a cone whose lateral surface area we can now calculate in two different ways (Example 2).

and calculate

s= =

l 27TX~1 Y 11 d

+ (:

27TV2

[Y - Y22

dy =

1

27TV2

=

27T(1 - y)V2 dy

(1 - t)

= 7TV2.

•

The results agree, as they should.

Exercises 6.4 Finding Integrals for Surface Area In Exercises 1- 8: a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.

D

b. Graph the curve to see what it looks like. If you can, graph the surface too.

D

c. Use your grapher's or computer's integral evaluator to find the surface's area numerically.

2. y = x 2,

O:s x :s 2;

x-axis

3. xy

l:s y :s 2;

y-axis

1,

4. x = siny,

O:s y :s

5. xl/2 + yl/2 = 3 6. y

+ 2vY =

x,

7T;

from

y-axis (4,I)to(l,4);

l:s y :s 2;

x-axis

Finding Surface Area 9. Find the lateral (side) surface area of the cone generated by revolving the line segment y = x/2, 0 :s x :s 4, about the x-axis. Check your answer with the geometry formula

t

X base circumference X slant height.

11. Find the surface area of the cone frustum generated by revolving the line segment y = (x/ 2) + 0 / 2), 1 :s x :s 3, about the x-axis. Check your result with the geometry formula

X base circumference X slant height.

7Th + r2)

X slant height.

12. Find the surface area of the cone frustum generated by revolving the line segment y = (x/2) + 0/2), 1 :s x :s 3, about the y-axis. Check your result with the geometry formula Frustum surface area =

y-axis

=

Lateral surface area =

t

Frustum surface area =

x loy tan t dt, O:s y :s 7T/3; y-axis 8. y = IX ~ dt, l:s x :s Vs; x-axis

7.

Lateral surface area =

O:s x :s 7T/4; x-axis

1. y = tanx, =

10. Find the lateral surface area of the cone generated by revolving the line segment y = x/2, 0 :s x :s 4, about the y-axis. Check your answer with the geometry formula

7Th + r2)

X slant height.

Find the areas of the surfaces generated by revolving the curves in Exercises 13- 23 about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. =

x 3/ 9,

O:s x :s 2;

14. y =

Yx,

3/4:S x :s 15/4;

15. y =

V2x -

13. y

x 2,

16. y =~,

x-axis x-axis

0.5:S x :s 1.5; l:s x :s 5;

x-axis

x-axis

17. x = y3/ 3, O:s y :s 1; y-axis 18. x = 0 / 3)y3/2 - yl/2, l:s y :s 3; y-axis

336

Chapter 6: Applications of Definite Integrals

19. x = 2~,

O:s y :s 15/4;

y (cm)

y-axis

,_ - -_,x 2 + y2 = 16 2 = 256 I

"'

/

I 1

' \ \

\ _ _

I

_---T-1

x (cm)

1

oi--- _ 1 1

20. x =~,

-

5/8:S y:S I;

x

28. Slicing bread Did you know that if you cut a spherical loaf of bread into slices of equal width, each slice will have the same amount of crust? To see why, suppose the semicircle y = ~ shown here is revolved about the x-axis to generate a sphere. Let AB be an arc of the semicircle that lies above an interval of length h on the x-axis. Show that the area swept out by AB does not depend on the location of the interval. (It does depend on the length of the interval.)

y-axis

y

y

y=v'r2-x 2

t1

-

(~, i)

5 8

A

x=~

0

x 21. y = (x 2/2)

+

0/2),

O:s x :s I; y-axis

22. y = 0/3)(x 2 + 2)3/2, O:s x :s V'2; y-axis (Hint: Express ds = dx 2 + dy2 in terms of dx, and evaluate the integral

S=

V J 2?TX ds with appropriate limits.)

23. x = (y4/4) ds =

S

=

+ 1/(8y 2), l:s y :s 2; x-axis (Hint: + dy2 in terms of dy, and evaluate the

V J 2?TY ds with appropriate limits.) dx 2

29. The shaded band shown here is cut from a sphere of radius R by parallel planes h units apart. Show that the surface area of the band is 2?TRh .

Express

~

integral

h

T

24. Write an integral for the area of the surface generated by revolving the curve y = cosx, -?T/2 :s x :s ?T/2, about the x-axis. In Section 8.4 we will see how to evaluate such integrals. 25. Testing the new definition Show that the surface area of a sphere of radius a is still 4?Ta 2 by using Equation (3) to find the area of the surface generated by revolving the curve y = Va 2 - x 2 , -a :s x :s a, about the x-axis. 26. Testing the new definition The lateral (side) surface area of a cone of height h and base radius r should be ?Tr~, the semi perimeter of the base times the slant height. Show that this is still the case by finding the area of the surface generated by revolving the line segment y = (r/h)x, O :s x :s h, about the x-axis.

D 27.

Enameling woks Your company decided to put out a deluxe version of a wok you designed. The plan is to coat it inside with white enamel and outside with blue enamel. Each enamel will be sprayed on 0.5 mm thick before baking. (See accompanying figure.) Your manufacturing department wants to know how much enamel to have on hand for a production run of 5000 woks. What do you tell them? (Neglect waste and unused material and give your answer in liters. Remember that I cm3 = I mL, so I L = 1000 cm3 .)

30. Here is a schematic drawing of the 90-ft dome used by the U.S. National Weather Service to house radar in Bozeman, Montana. a. How much outside surface is there to paint (not counting the bottom)?

D

b. Express the answer to the nearest square foot.

45 ft : Center

11/

t l 22.5 ftI

Radius

~ 45ft

.1 1!-=::::::::;;;:o~

6.5 Work and Fluid Forces

c. Show that the lateral surface area of the frustum of the cooe swept out by the tangent line segmeot as it revolves about the x-axisis2.".j(m.)"V'1 + (j'(m.»'I1x•.

31. An alternative derivation of the surface area formula Assume jis smooth 00 [a, b] andpartitioo [a, b] in the usual way. In the kth subinterval [x._" x..:J, cons1ruct the tangeot line to the curve at the ntidpointm. = (X'-1 + x.)/2, as in the accompanying figure.

d. Show that the area of the surface geoerated by revolving y = j(x) about the x-axis over [a, b] is

a. Show that

r,

= j(m.) - f'(m.)

t.;.

r,

aod

= j(m.)

337

+ f'(m.) 11;•.

±

lim

n_OO j;=1

(lateral surface area\ = of kth frustum )

l'

2.".j(x)"V'1

+ (f'(x»'dx.

a

b. Show that the leogth L. of the tangeot line segmeot in the kth

suhinterval is L. = "V'(I1x.)'

+ (f'(m.) 11x.)'.

32. The surface of an astroid Find the area of the surface geoerated by revolving about the x-axis the portioo of the astroid x'/3 + y'/3 = I abowo in the accompanying figure. (Hint: Revolve the first-quadrant portioo y = (I - x'/3)3/2,

0'"

X

s I, about the x-axis and double your result.) y

---<1•••---->1'

1<-1<

6.5

I

Work and FLuid Forces In everyday life, work means an activity that requires muscular or mental effort. In science, the term refers specifically to a force acting on a body (or object) and the body's subsequent displacement. This section shows how to calculate work. The applications run from compressing nrilroad car springs and emptying subterranean tanks to forcing electrons together and lifting satellites into orbit.

Work Done by a Constant Force When a body moves a distance d along a stnright line as a result of being acted on by a force of constant magnitude F in the direction of motion, we define the work W done by the force on the body with the formula

W=Fd

(Constant-force formula for work).

(1)

From Equation (1) we see that the unit OfWOlK in any system is the unit of force multiplied by the unit of distance. In S1 units (S1 stands for Systeme International, or International System), the unit offorce is a newton, the unit of distance is a meter, and the unit of work is a newton-meter (N • m). This combination appears so often it has a special name, the joule. In the British system, the unit of wotk is the foot-pound, a unit frequently used by engineers.

I

Joules The joule, abbreviated J and prooouoced ''jewel,'' is named after the English pbysicist James Prescott Joule (181S-1889). The defming equation is I joule = (I newtoo)(I meter). Insymbols,IJ= IN·m.

EXAMPLE 1 Suppose you jack up the side of a 2000-lb car 1.25 ft to change a tire. The jack applies a constant vertical force of about 1000 lb in lifting the side of the car (but because of the mechanical advantage of the jack, the force you apply to the jack itself is ouly about 30 lb). The total work performed by the jack on the car is 1000 X 1.25 = 1250 ft-lb. In S1 units, the jack has applied a force of 4448 N through a distance of 0.381 m to do 4448 X 0.381 '" 1695 J of work. _

338

Chapter 6: Applications of Definite Integrals

Work Done by a Variable Force Along a Line If the force you apply varies along the way, as it will if you are compressing a spring, the fonnuia W = Fd has to be replaced by an integral fonnula that takes the variation in F into account. Suppose that the force performing the work acts on an object moving along a straight line, which we take to be the x·axis. We assume that the magnitode of the force is a contino uous functionF of the object's positionx. We want to f'md the work done over the interval from x = a to x = b. We partition [a, b1in the usual way and choose an arbitrary point Cl in each subinterval [Xl-lo xii. If the subinterval is short enough, the continuous function F will not vary much from Xl-I to Xl. The amount of work done across the interval will be about F( Cl) times the distance axlo the same as it would be if F were constant and we could apply Equation (I). The total work done from a to b is therefore approximated by the Riemann sum Work

• L F(Cl) axl. 1- 1

RJ

We expect the approximation to improve as the nonn of the partition goes to zero, so we define the work done by the force from a to b to be the integral of F from a to b:

.

.~ ~ F(Cl)

axl =

l

b

a F(x) fix.

DEFINITION The work done by a variable force F(x) in the direction of motion along the x·axis from x = a to x = b is W

=

l

b

F(x) fix.

(2)

The units of the integral are joules ifF is in newtons and x is in meters, and foot·pounds if F is in pounds and x is in feet. So the work done by a force of F(x) = l/x 2 newtons in moving an object along the x·axis from x = I m to x = 10 m is

W=

1

10 I

I

x2

1]10

fix = - X

I

I

= - 10

+

I = 0.9 J.

Hooke's Law for Springs: F = Ioc F

Hooke's Law says that the force required to hold a stretched or compressed spring x units from its natoral (unstressed) length is proportional to x. In symbols, F= kx.

j 4 f-- -<'

Work done by F ~ 0 tox ~ 0.25

fromx

-o :-t"---:c':-:------~x

0.25

(ft)

(3)

The constant k, measured in force units per unit length, is a characteristic of the spring, called the fo",e constant (or spring constant) of the spring. Hooke's Law, Equation (3), gives good results as long as the force doesn't distort the metal in the spring. We assume that the forces in this section are too small to do that.

Amount compressed (b)

FIGURE 6.36 The force F needed to hold a spring under compression increases linearly as the spring is compressed (Example 2).

EXAMPLE 2 Find the work required to compress a spring from its natural length of I ft to a length of 0.75 ft if the force constant is k = 16lb/ft. Solution We picture the uncompressed spring laid out along the x·axis with its movable end at the origin and its f'IXed end at x = I ft (Fignre 6.36). This enables us to describe the

6.5 Work and Fluid Forces

339

force required to compress the spring from 0 to x with the formula F = 16x. To compress the spring from 0 to 0.25 ft, the force must increase from

F(O)

=

16·0

Olb

=

F(0.25)

to

=

16·0.25

=

4 lb.

The work done by F over this interval is

r.25

W

Jo

=

16x dx

Eq. (2) with

2]0.25 8x °

=

0.5 ft-Ib.

=

a = 0, b = 0.25, F(x) = 16x

•

EXAMPLE 3 A spring has a natural length of 1 m. A force of 24 N holds the spring stretched to a total length of 1.8 m. (a) Find the force constant k. (b) How much work will it take to stretch the spring 2 m beyond its natural length?

x=o

(e) How far will a 45-N force stretch the spring? SoLution

0.8 24N

(a) The force constant. We find the force constant from Equation (3). A force of 24 N maintains the spring at a position where it is stretched 0.8 m from its natural length, so

24 = k(0.8) k = 24/0.8

x(m)

FIGURE 6.37

A 24-N weight

stretches this spring 0.8 m beyond its unstressed length (Example 3).

Eq. (3) with F = 24,x = 0.8

30 N/m.

=

(b) The work to stretch the spring 2 m. We imagine the unstressed spring hanging along the x-axis with its free end at x = 0 (Figure 6.37). The force required to stretch the spring x m beyond its natural length is the force required to hold the free end of the spring x units from the origin. Hooke's Law with k = 30 says that this force is

F(x) = 30x. The work done by F on the spring from x = 0 m to x W

=

12

30x dx

=

15X 2]:

=

2 m is

=

60 J.

(e) How far will a 45-N force stretch the spring? We substitute F = 45 in the equation F = 30x to find 45 = 30x, or x = 1.5 m. A 45-N force will keep the spring stretched 1.5 m beyond its natural length.

•

The work integral is useful to calculate the work done in lifting objects whose weights vary with their elevation.

EXAMPLE 4 A 5-lb bucket is lifted from the ground into the air by pulling in 20 ft of rope at a constant speed (Figure 6.38). The rope weighs 0.08 lb/ft. How much work was spent lifting the bucket and rope? x

The bucket has constant weight, so the work done lifting it alone is weight X distance = 5·20 = 100 ft-Ib. The weight of the rope varies with the bucket's elevation, because less of it is freely hanging. When the bucket is x ft off the ground, the remaining proportion of the rope still being lifted weighs (0.08)· (20 - x) lb. So the work in lifting the rope is SoLution

20

1 2

Work on rope

o FIGURE 6.38

in Example 4.

= =

Lifting the bucket

[1.6x -

1 2

°(0.08)(20 - x) dx

0.04x2]~0

=

=

°(1.6 - 0.08x) dx

32 - 16

=

16 ft-Ib.

The total work for the bucket and rope combined is 100

+ 16 = 116 ft-Ib.

•

340

Chapter 6: Applications of Definite Integrals

Pumping Liquids from Containers How much work does it take to pump all or part of the liquid from a container? Engineers often need to know the answer in order to design or choose the right pump to transport water or some other liquid from one place to another. To find out how much work is required to pump the liquid, we imagine lifting the liquid out one thin horizontal slab at a time and applying the equation W = Fd to each slab. We then evaluate the integral this leads to as the slabs become thinner and more numerous. The integral we get each time depends on the weight of the liquid and the dimensions of the container, but the way we find the integral is always the same. The next example shows what to do. y = 2x or x = !y 2

EXAMPLE 5 The conical tank in Figure 6.39 is filled to within 2 ft of the top with olive oil weighing 57 Ibjft3. How much work does it take to pump the oil to the rim ofthe tank? Solution We imagine the oil divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0, 8]. The typical slab between the planes at y and y + a y has a volume of about av = 7T(radius)2(thickness) =

7T(~y

Y

~ y2 ay ft3.

ay =

The force F(y) required to lift this slab is equal to its weight,

F(y) FIGURE 6.39 Example 5.

The olive oil and tank in

=

57 av =

577T 2 TY ay lb.

Weight = (weight per unit

volume) X volume

The distance through which F(y) must act to lift this slab to the level of the rim of the cone is about (10 - y) ft, so the work done lifting the slab is about

aw =

577T 2 T(10 - y)y ayft-Ib.

Assuming there are n slabs associated with the partition of [0, 8], and that y = Yk denotes the plane associated with the kth slab ofthickness ayk. we can approximate the work done lifting all of the slabs with the Riemann sum

W::::;

±5~7T

(10 - Yk)Y/ aYdt-lb.

k=i

The work of pumping the oil to the rim is the limit of these sums as the norm of the partition goes to zero and the number of slabs tends to infinity:

W

=

. n 577T hm ~ - 4 (10 - Yk)Y/ aYk =

n~ook=i

1 8

0

= --..!!.. 57 4

577T - 4 (10 - y)y2 dy

1 8

(10y2 - y3) dy

0

= 577T [10y3 _ y4]8 ::::; 30561 ft-Ib 4

3

4

0

'

• .

Fluid Pressures and Forces

FIGURE 6.40 To withstand the increasing pressure, dams are built thicker as they go down.

Dams are built thicker at the bottom than at the top (Figure 6.40) because the pressure against them increases with depth. The pressure at any point on a dam depends only on how far below the surface the point is and not on how much the surface of the dam happens to be tilted at that point. The pressure, in pounds per square foot at a point h feet below the surface, is always 62.4h. The number 62.4 is the weight-density of freshwater in pounds per cubic foot. The pressure h feet below the surface of any fluid is the fluid's weightdensity times h.

6.5 Work and Ruid Forces

I Weight-deD.lity A fluid's weight-density w is i11l weight per unit volume. lYPical values (lb/ftl) are listed below.

-

lbe Preslure-Depth Equation In a fluid that is standing still, the pressure p at depth II is the fluid's weight-

density w times II:

p =wh.

42

(4)

84.

M"""",

Milk

64.5

In a container of fluid with a flat horizontal base, the total force exerted by the fluid against the base can be calculated by multiplying the area of the base by the pressure at the

100

MOWBe8

Olive oil Soa_ Freshwa1m

57 64

base. We can do this because total force equals force per unit area (pressure) times area. (See Figure 6.41.) If F,p, andA are the total force, pressure, and area, then

62.4

F - total force - force per unit area X area = pressure X area = pA = wltA.

1•

FIGURE 6.41

F=pA=wltA

These containers are filled

with water to the same dep1h and have the same base area. The total :force is therefore the same on the bottom of each container. The containers' shapes do not matter here.

, Surface of fluid

....

Submerged vertical.

y

•

--t::L(Y)

p = wh from Eq. (4)

Fluid Force on a Constant-Depth Surface

~

•

341

-1}., Strip

Strip lcng1h at lcvcJ. J

For example, the weight-density of freshwater is 62.4Ib/tP, 80 the fluid force at the bottom ofa 10 ft X 20 ftrectangular swimming pool 3 ft deep is F - whA - (62.41b/fI')(3 11)(10' 20 ft') = 37,440 lb.

For a flat plate submetged horizontaUy, like the bottom of the swimming pool just discussed, the downward force acting on its upper face due to liquid pressure is given by Equation (5). If tho plate is subm_d vertically, bowover, !bon tho pressure against it will be
!!J.F = (pressure along bottom edge) X (area)

- w • (strip depth). L(y) ay. Assume there are 11 strips associated with the partition of a ~ y ~ b and that Yk is the bottom edge of the kth strip having length L(Yk) and width !!J.Yk. The force against the entire plate is approximated by summing the forces against each strip, giving the Riemann sum

,

FIGURE 6.42 The force exerted by a fluid against roe side of a thin, flat horizontal strip ill about AF = pressure X area = w X (strip depth) X L(y)Ay.

(5)

F ~ '5'(w' (strip depth),. L(y,)) ay,. (:1

(6)

The sum in Equation (6) is a Riemann sum for a continuous :function on [a, b], and we expect the approximations to improve as the norm of the partition goes to zero. The force

against the plate is the limit of these sums:

,

lim '5'(w'(stripdepth)"L(YV)ay, ~

/J_OO

t=1

l' II

w·(strip depth)'L(y) dy.

342

Chapter 6: Applications of Definite Integrals

The Integral for Fluid Force Against a Vertical Flat Plate Suppose that a plate submerged vertically in fluid of weight-density w runs from y = a to y = b on the y-axis. Let L(y) be the length of the borizontal strip measured from left to right along the surface of the plate at levely. Then the force exerted by the fluid against one side of the plate is F =

y (ft)

y=xorX=Y / _P ,O ,O_l '_wf _ 8C _ e_at+ _ _ _ _+/.. ,Y~_5

Deptht.: 5-y

x=y

~

Y

11;1- ~ /

/

/ // y=3 (3,3)

JY(x, x)

~ (y, y)

/ 0

x (ft)

/

l

b W'

(strip depth). L(y) dy.

(7)

EXAMPLE 6 A flat isosceles right-triaogular plate with base 6 ft aod height 3 ft is submerged vertically, base up, 2 ft below the surface of a swimming pooL Find the force exerted by the water against one side of the plate. Solution We establish a coordinate system to work in by placing the origin at the plate's bottom vertex and running they-axis upward along the plate's axis of symmetry (Figure 6.43). The surface of the pool lies along the line y = 5 and the plate's top edge along the line y = 3. The plate's right-band edge lies along the line y = x, with the upper-right vertex at (3, 3). The length of a thin strip at level y is

L(y) = 2x = 2y. FIGURE 6.43 To fmd the force on one side of the submerged plate in Example 6, we can use a coordinate system like the one here.

The depth of the strip beneath the surface is (5 - y). The force exerted by the water against one side of the plate is therefore

F =

t

W'

u:~) 'L(y) dy

=

1362.4(5 - y)2ydy

=

124.81 (5y - y2) dy

=

5 124.8 [Zy2 -

Eq. (7)

3

y3]30 =

3

1684.8 lb.

•

Exercises 6.5 Springs 1. Spring constant It took 1800 J of work to s1retch a spring from its natora1length of2 m to a leogth of5 In. Find the spring's force constant. 2. Stretcbing. spring A spriog has a natora11ength of \0 in. Au 800·lb force stretches the spring to 14 in.

4. Stretcbing a spring If a force of 90 N s1retches a spring I m beyond its natora1 length, how much work does it take to stretch the spriug 5 m beyond its natora1leogth? 5. Subway car springs It takes a force of 21,714 Ib to compress a coil spriug assembly on a New York City Transit Authority subway car from its free height of8 in. to its fully compressed height of 5 in.

a. Find the force constant.

a. What is the assembly's force constant?

b. How much work is done in stretching the spring from lOin. toI2in.?

b. How much work does it take to compress the assembly the fIrSt half inch? the second half inch? Answer to the nearest in.·lb.

c. How far beyond its natora1length will a 1600-lb force stretch the spring? 3. Stretcbing. rubber band A force of 2 N will stretch a rubber band 2 em (0.02 m). Assuming that Hooke's Law applies, how far will a 4·N force stretch the rubber band? How much work does it take to stretch the rubber band this far?

6. Bathroom scale A b.1hroom scale is compressed 1/16 in. when a 150·lb person stands on it. Assmning that the scale behaves like a spriog that obeys Hooke's Law, how much does someone who compresses the scale 1/8 in. weigh? How much work is done compressing the scale 1/8 in.?

6.5

Work Done by a VariabLe Force 7. Lifting a rope A mountain climber is about to haul up a 50 m length of hanging rope. How much work will it take if the rope weighs 0.624 N/m? 8. Leaky sandbag A bag of sand originally weighing 144 Ib was lifted at a constant rate. As it rose, sand also leaked out at a constant rate. The sand was half gone by the time the bag had been lifted to 18 ft. How much work was done lifting the sand this far? (Neglect the weight of the bag and lifting equipment.) 9. Lifting an elevator cable An electric elevator with a motor at the top has a multi strand cable weighing 4.51b/ft. When the car is at the fIrst floor, 180 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top? 10. Force of attraction When a particle of mass m is at (x, 0), it is attracted toward the origin with a force whose magnitude is k/x 2 • If the particle starts from rest at x = b and is acted on by no other forces, find the work done on it by the time it reaches x = a, 0< a < b.

11. Leaky bucket Assume the bucket in Example 4 is leaking. It starts with 2 gal of water (16 Ib) and leaks at a constant rate. It finishes draining just as it reaches the top. How much work was spent lifting the water alone? (Hint: Do not include the rope and bucket, and find the proportion of water left at elevation x ft.) 12. (Continuation of Exercise 11.) The workers in Example 4 and Exercise 11 changed to a larger bucket that held 5 gal (40 Ib) of water, but the new bucket had an even larger leak so that it, too, was empty by the time it reached the top. Assuming that the water leaked out at a steady rate, how much work was done lifting the water alone? (Do not include the rope and bucket.)

Pumping Liquids from Containers 13. Pumping water The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.41b/tP . a. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full? b. If the water is pumped to ground level with a (5/11)horsepower (hp) motor (work output 250 ft-lb/sec), how long will it take to empty the full tank (to the nearest minute)? c. Show that the pump in part (b) will lower the water level lOft (halfway) during the first 25 min of pumping. d. The weight of water What are the answers to parts (a) and (b) in a location where water weighs 62.261b/ft3 ? 62.591b/f~? Ground

y

14. Emptying a cistern The rectangular cistern (storage tank for rainwater) shown has its top 10 ft below ground level. The cistern,

343

currently full, is to be emptied for inspection by pumping its contents to ground level. a. How much work will it take to empty the cistern? b. How long will it take a 1/2-hp pump, rated at 275 ft-lb/sec, to pump the tank dry? c. How long will it take the pump in part (b) to empty the tank halfway? (It will be less than half the time required to empty the tank completely.) d. The weight of water What are the answers to parts (a) through (c) in a location where water weighs 62 .261b/f~? 62.591b/tP?

y

15. Pumping oil How much work would it take to pump oil from the tank in Example 5 to the level of the top of the tank if the tank were completely full? 16. Pumping a half-full tank Suppose that, instead of being full, the tank in Example 5 is only half full. How much work does it take to pump the remaining oil to a level 4 ft above the top of the tank? 17. Emptying a tank A vertical right-circular cylindrical tank measures 30 ft high and 20 ft in diameter. It is full of kerosene weighing 51.2 Ib/ft3 . How much work does it take to pump the kerosene to the level of the top of the tank? 18. a. Pumping milk Suppose that the conical container in Example 5 contains milk (weighing 64.5 Ib/ft 3) instead of olive oil. How much work will it take to pump the contents to the rim? b. Pumping oil How much work will it take to pump the oil in Example 5 to a level 3 ft above the cone's rim? 19. The graph of y = x 2 on 0 :s x :s 2 is revolved about the y-axis to form a tank that is then filled with salt water from the Dead Sea (weighing approximately 73 Ibs/ft 3) . How much work does it take to pump all of the water to the top of the tank? 20. A right-circular cylindrical tank of height 10ft and radius 5 ft is lying horizontally and is full of diesel fuel weighing 53 Ibs/tP. How much work is required to pump all of the fuel to a point 15 ft above the top of the tank? 21. Emptying a water reservoir We model pumping from spherical containers the way we do from other containers, with the axis of integration along the vertical axis of the sphere. Use the figure here to find how much work it takes to empty a full hemispherical water reservoir of radius 5 m by pumping the water to a height of 4 m above the top of the reservoir. Water weighs 9800 N/m3. y

level

Work and Fluid Forces

344

Chapter 6: Applications of Definite Integrals

22. You are in charge of the evacuation and repair of the storage tank shown here. The tank is a hemisphere of radius 10 ft and is full of benzene weighing 561b/ft3 • A firm you contacted says it can empty the tank for 1/2¢ per foot-pound of work. Find the work required to empty the tank by pumping the benzene to an outlet 2 ft above the top of the tank. If you have $5000 budgeted for the job, can you afford to hire the firm?

and 3.5 in. across at the top (a standard size at Brigham's in Boston). The straw sticks up an inch above the top. About how much work does it take to suck up the milkshake through the straw (neglecting friction)? Answer in inch-ounces. y

(1.75,7) Y + 17.5 --14-

x Dimensions in inches

Work and Kinetic Energy 23. Kinetic energy If a variable force of magnitude F(x) moves a body of mass m along the x-axis from XI to X2, the body's velocity v can be written as dx/dt (where t represents time). Use Newton's second law of motion F = m(dv/dt) and the Chain Rule

30. Putting a satellite in orbit The strength of Earth's gravitational field varies with the distance r from Earth's center, and the magnitude of the gravitational force experienced by a satellite of mass m during and after launch is F(r) =

m~G. r

dv dt

dv dx dx dt

dv dx

-=--=v-

to show that the net work done by the force in moving the body from Xl to X2 is

Here, M = 5.975 X 10 24 kg is Earth's mass, G = 6.6720 X 10- 11 N • m 2 kg- 2 is the universal gravitational constant, and r is measured in meters. The work it takes to lift a 1000-kg satellite from Earth's surface to a circular orbit 35,780 km above Earth's center is therefore given by the integral

1

35,780,000

where VI and V2 are the body's velocities at XI and X2. In physics, the expression O/2)mv2 is called the kinetic energy of a body of mass m moving with velocity v. Therefore, the work done by the force equals the change in the body's kinetic energy, and we can find the work by calculating this change.

Work =

1000MG

6,370,000

r

2

.

dr Joules.

Evaluate the integral. The lower limit of integration is Earth's radius in meters at the launch site. (This calculation does not take into account energy spent lifting the launch vehicle or energy spent bringing the satellite to orbit velocity.)

In Exercises 24-28, use the result of Exercise 23 . 24. Tennis A 2-oz tennis ball was served at 160 ft/sec (about 109 mph). How much work was done on the ball to make it go this fast? (To find the ball's mass from its weight, express the weight in pounds and divide by 32 ft/sec 2 , the acceleration of gravity.)

Finding Fluid Forces 31. Triangular plate Calculate the fluid force on one side of the plate in Example 6 using the coordinate system shown here. y (ft)

25. Baseball How many foot-pounds of work does it take to throw a baseball 90 mph? A baseball weighs 5 OZ, or 0.3125 lb. 26. Golf A 1.6-oz golf ball is driven off the tee at a speed of280 ft/ sec (about 191 mph). How many foot-pounds of work are done on the ball getting it into the air? 27. On June 11,2004, in a tennis match between Andy Roddick and Paradom Srichaphan at the Stella Artois toumament in London, England, Roddick hit a serve measured at 153 mi/h. How much work was required by Andy to serve a 2-oz tennis ball at that speed?

/ /

/

/

_-..-________+-S_wf __a_ce_o_f~p_O_Ol~/-/~x(ft)

o

/ 5 /

Depth

Iyl ..----__-I-_ _.."L l~_-=2

-.J!'-----~~=t==:::;:~(X, y)

28. Softball How much work has to be performed on a 6.5-oz softball to pitch it 132 ft/sec (90 mph)? 29. Drinking a milkshake The truncated conical container shown here is full of strawberry milkshake that weighs 4/9 oz/in3. As you can see, the container is 7 in. deep, 2.5 in. across at the base,

32. Triangular plate Calculate the fluid force on one side of the plate in Example 6 using the coordinate system shown here.

6.5 Work and Fluid Forces

345

38. Semicircular plate A semicircular plate 2 ft in diameter sticks straight down into freshwater with the diameter along the surface. Find the force exerted by the water on one side of the plate.

y (ft)

Pool swfaa: at Y ~ 2

39. TIlted plate Calculate the fluid force 00 one side ofa 5 ft by 5 ft -- "3 :----::+----7..:-->. (ft) 3

square plate if the plate is at the bottom of a pool filled with water to a depth of 8 ft and •• lying flat on its 5 ft by 5 ft face.

b. resting vertically on a 5-ft edse.

33. Rectangular plate In a pool filled with water to a depth of 10ft, calculate the fluid force 00 ooe side of a 3 ft by 4 ft rectangular plate if the plate rests vertically at the bottom of the pool L

00

its 4·ft edge

b.

00

its 3·ft edge.

34. Semlcln:u111J' plate Calculate the fluid force 00 one side of a semicircular plate of radius 5 ft that rests vertically 00 its diame· ter at the bottom of a pool filled with water to a depth of 6 ft. y Surface of water

c. resting on a 5-ft edge and tilted at 45' to the bottom of the pool. 40. Tilted plate Calculate the fluid force on one side of a righttriangular plate with odses 3 ft, 4 ft, and 5 ft if the plate sits at the bottom of a pool filled with water to a depth of 6 ft on its 3-ft edge and tilted at 60' to the bottom of the pool. 41. The cubical meta1 tank sbown here has a parabolic gate held in place by bolts and desigoed to withstand a fluid force of 160 Ib without rupturing. The liquid you plan to store has a weightdensity of 50 Ib/tt' . L

6

What is the fluid force on the gate when the liquid is 2 ft deep?

b. What is the maximum height to which the container can be filled witheut exceeding the gate's desigo limitatioo?

5

y (ft) _ _L-_

_

- L_

_

-i_~.

(-I, 1)

(I, 1)

35. Triangular plate The isosceles triangular plate sbown here is submerged vertically I ft below the surface of a freshwater lake. L

----:-----''':0:+''''''---- -:--. (ft)

Find the fluid force against one face of the plate.

b. What would be the fluid force on one side of the plate if the water were seawater instead of freshwater? Surface level

&larged view of panlbolic gate

41. The end plates of the trough sbown here were desigoed to withstand a fluid force of 6667 lb. How many cubic fi:et of _er can the tank hold without exceeding this limitation? Round dawn to the Deatest cubic foot y (ft) (-4, 10) (4, 10)

36. Rotated triangular plate The plate in Exercise 35 is revolved 180' about line AB so that part of the plate sticks out of the lake, as shown here. What force does the water exert 00 one face of the plate now?

- --c' o r----> • (ft) End view of trough

view of trough

43. A vertical rectangular plate a units long by b units wide is sul>merged in a fluid ofweight-de:nsity w with its long edges parallel to the fluid's surface. Find the average value of the pressure along the vertical dimension of the plate. Explain your answer.

37. N... EagiaDd AqnarInm The viewiog portion of the rectaDguIar g1ass window in a typical fISh tank at the New England Aquarium in Boston is 63 in. wide and runs from 0.5 in. below the water's surface to 33.5 in. below the surface. Find the fluid foroe against this portion of the window. The \Wight-density of seawater is 64lb/tt'. (In case you """" wondering, the glass is 3/4 in. thick and the tank walls exteod 4 in. above the water to keep the fISh fromjumping out)

44. (Continuation ofExercise 43.) Show that the force exerted by the fluid on one side of the plaIA: is the average vslne of the pressure (found in Exercise 43) times the area of the plate. 45. Water pours into the tsnk sbawn bore at the rate of 4 tt'/min. The tank~ cross-scctioos are 4-ft-
346

Chapter 6: Applications of Definite Integrals 46. Watering trough The vertical eods of a watering trough are squares 3 ft OIl a side. a. Find the fluid force against the ends when the trough is full. b. How many inches do you have to lower the _or level in the

hole in the bottom at the rate of 5 It'/min. Will the movable end reach the hole before the tank overllows? Movable end

Water in

~'===d

trough to reduce the fluid force by 25%?

hole Side view

6.6

Moments and Centers of Mass Many structures and mechanical systems behave as if their masses were concentrated at a single point, called the center of mass (Figure 6.44). It is important to know how to locate this point, and doing so is basically a mathematical enterprise. For the moment, we deal with one- and two-dimensional objects. Three-dimensional objects are best done with the multiple integrals of Chapter 15.

Masses Along a Line We develop our mathematical model in stages. The first stage is to imagine masses m" m2, and m3 on a rigid x-axis supported by a fulcrum at the origin. Xl

• ml

..,• X,

X2

0

:.&< ':2

>X

Fulcrum at origin

The resulting system might balance, or it might not, depending on how large the masses are and how they are arranged along the x-axis. Each mass mk exerts a downward force mkg (the weight of mk) equal to the magnitude of the mass times the acceleration due to gravity. Each of these forces has a tendency to turn the axis about the origin, the way a child turns a seesaw. This turning effect, called a torque, is measured by multiplying the force mkg by the signed distance Xk from the point of application to the origin. Masses to the left of the origin exert negative (counterclockwise) torque. Masses to the right of the origin exert positive (clockwise) torque. The sum of the torques measures the tendency of a system to rotate about the origin. This sum is called the system torque. System torque = ml8X1

+

m28X2

+

m38X3

The system will balance if and only if its torque is zero. Ifwe factor out the g in Equation (I), we see that the system torque is g

---

.

(mlxl

+ m2X2 + m3x3).

a feature of the

a feature: of

environment

the system

(1)

6.6

Moments and Centers of Mass

347

Thus, the torque is the product of the gravitational acceleration g, which is a feature of the environment in which the system happens to reside, and the number (mlxl + m2X2 + m3x3), which is a feature of the system itself, a constant that stays the same no matter where the system is placed. The number (mlxl + m2X2 + m3x3) is called the moment of the system about the origin. It is the sum of the moments mlX!. m2X2, m3X3 of the individual masses.

Mo

=

Moment of system about origin = ~ mkXk

(We shift to sigma notation here to allow for sums with more terms.) We usually want to know where to place the fulcrum to make the system balance, that is, at what point x to place it to make the torques add to zero. Xl

x

X3

~

m3

X2

0

• ml

/\ • L__~~ m2

•

, X

Special location for balance

The torque of each mass about the fulcrum in this special location is distance) (downward) T orque 0 f mk ab out x- -- (Signed fi of mk fromx orce

When we write the equation that says that the sum of these torques is zero, we get an equation we can solve for x: ~ (Xk - x)mkg = 0

~ mkXk

_

x

Sum ofthe torques equals zero.

=

This last equation tells us to find the system's total mass:

~ mk .

Solved for x

x by dividing the system's moment about the origin by system moment about origin system mass

X=

(2)

The point x is called the system's center of mass.

Masses Distributed over a Plane Region FIGURE 6.44

A wrench gliding on

ice turning about its center of mass as the center glides in a verticalline. y

Suppose that we have a finite collection of masses located in the plane, with mass mk at the point (xt, Yk) (see Figure 6.45). The mass ofthe system is System mass: Each mass mk has a moment about each axis. Its moment about the x-axis is mkYk, and its moment about the y-axis is mkXk. The moments of the entire system about the two axes are Moment about x-axis:

Mx = ~ mkYk,

Moment about y-axis:

My = ~ mkXk.

The x-coordinate of the system's center of mass is defined to be Each mass mk has a moment about each axis. FIGURE 6.45

_

My M

x=-=

(3)

348

Chapter 6: Applications of Definite Integrals With this choice of X, as in the one-dimensional case, the system balances about the line = x (Figure 6.46). The y-coordinate of the system's center of mass is dermed to be

y

x

_ Mx y= M=

L

mkYk Lmk'

(4)

With this choice of y, the system balances about the line y = y as well. The torques exerted by the masses about the line y = y cancel out Thus, as far as balance is concerned, the system behaves as if all its mass were at the single point (x, y). We call this point the system's center of mass.

o

Thin, Flat Plates FIGURE 6.46 A two-dimensional array of masses balances on its center of mass.

y Strip of mass

.am

In many applications, we need to find the center of mass of a thin, flat plate: a disk of aluminum, say, or a triangular sheet of steel. In such cases, we assume the distribution of mass to be continuous, and the fonnulas we use to calculate x and y contain integrals instead of finite sums. The integrals arise in the following way. Imagine that the plate occupying a region in the .!J'-plane is cut into thin strips parallel to one of the axes (in Figure 6.47, the y-axis). The center of mass of a typical strip is (x, y). We treat the strip's mass dm as if it were concentrated at (x, y). The moment of the strip about the y-axis is then dm. The moment of the strip about the x-axis is y dm. Equations (3) and (4) then become

x

_ My Lxdm x=M=Ldm' -~----------~----~X

o

'i

FIGURE 6.47 A plate cut into thin sttips parallel to tire y-axis. The moment exerted by a typical sttip about each axis is the moment its mass .11m would exert if concentrated at the strip's center of mass

_

Mx

Y = -M- =

The sums are Riemann sums for integrals and approach these integrals as limiting values as the strips into which the plate is cut become narrower and narrower. We write these integrals symbolically as

x=

jxdm jdm

and

y=

jydm jdm'

(X',y).

Moments, Mass, aod Center of Mass of a Thin Plate Covering a Region in the xy-Plaoe

I Density A material's density is its mass per unit area For wires, rods, and narrow s1rips, we use mass per unit length.

Moment about the x-axis:

Mx= J ydm

Moment about the y-axis:

My=Jxdm

Mass:

M=

J

(5)

dm

Center of mass:

The differential dm is the mass of the strip. Assuming the density Il of the plate to be a continuous function, the mass differential dm equals ilie product Il dA (mass per unit area times area). Here dA represents the area of the strip. To evaluate the integrals in Equations (5), we picture the plate in the coordinate plane and sketch a strip of mass parallel to one of the coordinate axes. We then express the strip's mass dm and the coordinates (x, y) of the strip's center of mass in tenos ofx or y. Finally, we integrate y dm, x dm, and dm between limits of integration determined by the plate's location in the plane.

6.6 y(em)

2

Moments and Centers of Mass

349

EXAMPLE 1 The triangular plate shown in Figure 6.48 has a constant density of Il = 3 g/=2. Find

(1,2)

(8) the plate's moment My about the y-axis.

(b) the plate's mass M.

(c) the x-coordinate of the plate's center of mass (c.m.).

Solution

Method 1: Vertical Strips (Figure 6.49)

(8) The moment My: The typical vertical strip has the following relevant data.

FIGURE 6.48

center of mass (c.m.): length: width: area: mass: distance of C.m. from y-axis:

The plate in Example 1.

(1,2)

My = Strip c.m. is halfway.

J

xdm = 116x 2 dx = 2x 3 ]: = 2g·=.

(b) The plate's mass:

lx, y) ~ (x, x)

M =

->~+-I

dm =

_

Units in centimeters

My

x=M=

FIGURE 6.49 Modeling the plate in Example 1 with vertical strips.

6x dx = 3x 2 ]: = 3 g.

2g'cm 3g

2

=3='

By a similar computation, we could find Mx and y = Mx/M.

Method 2: Horizontal Strips

(Figure 6.50)

(8) The moment My: The y-coordinate of the center of mass of a typical horizontal strip is y (see the figore), so

y(em)

2

J 11

(c) The x-coordinate of the plate's center of mass:

~f----,,-~~--~x

o

x

The moment of the strip about the y-axis is xdm = x'6xdx = 6x 2 dx. The moment of the plate about the y-axis is therefore

y 2

(x,y) = (x,x) 2x dx dA = 2x dx dm = IldA = 3·2xdx = 6xdx = x

y=y. (1,2) Strip c.m. is halfway.

__) (Y+2 (X,Y --4- ,Y) ~

The x-coordinate is the x-coordinate of the point halfway across the triangle. This makes it the average ofy/2 (the strip's left-hand x-value) and 1 (the strip's right-hand x-value):

=

+ t

_

x =

(y/2) + 1 2

y 1 Y+2 = 4 + 2 = -4-'

dy

-o o+------!---~x (em)

FIGURE 6.50 Modeling the plate in Example 1 with horizootal strips.

We also have length:

1-

width:

dy

area: mass:

2=

2 - Y -2-

2-y

dA =--dy 2

2-y

dm = IldA = 3'--dy 2

_

distance of c.m. to y-axis:

Y

y+2

x = --4-'

350

Chapter 6: Applications of Definite Integrals

The moment of the sttip about the y-axis is

:tdm

=

y+2 -4-

0

3

2-y -2-dy

0

3 8(4 - y2)dy.

=

The moment of the plate about the y-axis is

My

=

J :tdm

f~(4 -

=

y2)dy

=

H4Y - y;]:

- y) dy

=

1

=

H~6) =

=

1(4 - 2)

2gocm.

(b) The plate's mass:

M

=

J

dm

=

( ' 1(2

102

[2 -

y2]2

2Y 20

(c) The x-coordinate of the plate's center of mass: _ My 2g ocm x= M = 3g

2

=

3 g.

2

=3 cm.

By a similar computation, we could fmd Mx and y.

•

If the distribution of mass in a thin, flat plate has an axis of symmetry, the center of mass will lie on this axis. If there are two axes of symmetry, the center of mass will lie at their intersection. These facts often help to simplify our work. y 4

EXAMPLE 2

Find the center of mass ofa thin plate covering the region bounded above

by the parabola y = 4 - x 2 and below by the x-axis (Figure 6.51). Assume the density of the plate at the point (x, y) is 8 = 2x 2 , which is twice the square of the distance from the point to the y-axis.

Solution The mass distribution is symmetric about the y-axis, so x = O. We model the distribution of mass with vertical strips since the density is given as a function of the varible x. The typical vertical strip (see Figure 6.51) has the following relevant data.

center of mass (c.m.): FIGURE 6051 Modeling tire plate in Example 2 with vertical sttips.

(:t, y)

=

(x, 4

~ x2)

length: 4 - x 2 width: ax area: ciA = (4 - x 2 ) ax mass: dm = 8 ciA = 8(4 - x 2 ) ax _ 4 - x2 distance from c.m. to x-axis: y=-2The moment of the strip about the x-axis is

y dm

=

4 - x2 8 - 2 - 08(4 - x 2) ax = 2(4 - x 2)2ax.

The moment of the plate about the x-axis is

Mx=JYdm= =

('06x 2

1-2

-

f2~(4_x2)'ax=

1-2

8>;4

+ x 6 ) ax

=

f\2(4-x2),ax

1-2

2C::58 1

6.6

Moments and Centers of Mass

351

Therefore,

_ Y

Mx 2048 15 8 M = 105 • 256 = 7'

=

The plate's center of mass is

• Plates Bounded by Two Curves Suppose a piate covers a region that lies between two corves y = g(x) and y = f(x), where f(x) 2: g(x) and a :5 x :5 b. The typical vertical s1rip (see Figure 6.52) has

y

"y

=

g(x)

I I I

I

I I I

~---L----~tc--~~x

o

a

dx

(X, y) = (x,! [[(x) + g(x)]) f(x) - g(x)

center of mass (c.m.): length: width: area: mass:

ax dA dm

= =

[[(x) - g(x)] ax Il dA = Il[[(x) - g(x)]

ax.

b

The moment of the plate about the y·axis is FIGURE 6.52 Modeling the plate bounded by _ curves with vertical s1rips. The s1rip

c.m. is halfway, so 'ji

=

t

[f(x)

My

J l xdm

=

+ g(x)].

b

=

xll[[(x) - g(x)]

ax,

and the moment about the x·axis is

Jy l l2 dm

Mx =

b

=

a

Il

b

=

t [[(x) + g(x)]'Il[[(x) - g(x)] ax

[f(x) - g2(x)]

ax.

These moments give the formulas

lib

X= M

a

Ilx [[(x) - g(x)]

ax

(6)

(7)

EXAMPLE 3 Find the center of mass for the thinpiate bounded by the curves g(x) = x/2 and f(x) = v;"O :5 x :5 I, (Figure 6.53) using Equations (6) and (7) with the density function Il(x) = x 2. Solutton We first compute the mass of the plate, where dm = Il[[(x) - g(x)] ax:

M

=

t 10

x2(v;, -

~) ax = t 2

10

(r'/2 _x2

3 )

ax

=

[2.x7/2 _ Ix,]! =~. 7 8 0 56

352

Chapter 6: Applications of Definite Integrals Then from Equations (6) and (7) we get X = 56

9

=

y

[x2.{Vx -~) (X7/2 _ ~4) dx

56 [ 9

= 56

dx

[2. x9/2 _

9 9

...Lx5]! 10

0

308

405'

and

~~------------~---->x

o

1

252 FIGURE 6.53

405'

The region in Example 3.

•

The center of mass is shown in Figure 6.53.

Centroids When the density function is constant, it cancels out of the numerator and denominator of the formulas for x and y. Thus, when the density is constant, the location of the center of mass is a feature of the geometry of the object and not of the material from which it is made. In such cases, engineers may call the center of mass the centroid of the shape, as in "Find the centroid of a triangle or a solid cone." To do so, just set {; equal to 1 and proceed to fmd x and y as before, by dividing moments by masses. y A typical small segment of wire has

tim

~

a tis ~ aad8.

a

-a (a)

EXAMPLE 4 Find the center of mass (centroid) of a thin wire of constant density {; shaped like a semicircle of radius a.

Va

length:

y

mass: distance ofc.m. tox·axis:

a c.m.

-a

ds = a dO dm = {; tis = {;a dO y = a sin O.

Mass per unit length times length

Hence,

(0. ~a)

0

x

2 2 (Figure 6.54). The We model the wire with the semicircle y = distribution of mass is symmetric about the y·axis, so x = O. To find y, we imagine the wire divided into short subarc segments. If (:t, y) is the center of mass of a subarc and 0 is the angle between the x·axis and the radial line joining the origin to (:t, y), then y = a sin 0 is a function of the angle 0 measured in radians (see Figure 6.54a). The length ds of the subarc containing (:t, y) subtends an angle of dO radians, so tis = a dO. Thus a typical subarc segment has these relevant data for calculating y:

Solution

a

(b)

FIGURE 6.54 The semicircular wire in Example 4.
_ fydm fo"a sinO'{;a dO y= = f dm fo" {;a dO

[;a2[ -cos 0]; {;a7f

2 = 7f

a.

The center of mass lies on the axis of symmetry at the point (0, 2a/7f), about two·thirds of the way up from the origin (Figure 6.54b). Notice how {; cancels in the equation for y, so we could have set {; = 1 everywhere and obtained the same value for y. • In Example 4 we found the center of mass of a thin wire lying along the graph of a differentiable function in the xy-plane. In Chapter 16 we willieam how to find the center of mass of wires lying along more general smooth curves in the plane (or in space).

6.6

Surface level of fluid

11 = centroid depth

Moments and Centers of Mass

353

Fluid Forces and Centroids Ifwe know the location of the centroid of a submerged flat vertical plate (Figure 6.55), we can take a shortcut to find the force against one side of the plate. From Equation (7) in Section 6.5,

l l

b

Plate centroid

F =

= W

FIGURE 6.55 The force against one side of the plate is w·1i . plate area.

w X (strip depth) X L(y) dy b

(strip depth) X L(y) dy

= w X (moment about surface level line of region occupied by plate) = w X (depth of plate's centroid) X (area of plate).

Fluid Forces and Centroids The force ofa fluid of weight-density w against one side ofa submerged flat vertical plate is the product ofw, the distance Ii from the plate's centroid to the fluid surface, and the plate's area: F=

whA.

(8)

EXAMPLE 5 A flat isosceles triangular plate with base 6 ft and height 3 ft is submerged vertically, base up with its vertex at the origin, so that the base is 2 ft below the surface of a swimming pool. (This is Example 6, Section 6.5.) Use Equation (8) to find the force exerted by the water against one side of the plate. Solutton

The centroid of the triangle (Figure 6.43) lies on the y-axis, one-third of the way from the base to the vertex, so Ii = 3 (where y = 2) since the pool's surface is y = 5. The triangle's area is

A = t(base)(height) = t(6)(3) = 9. Hence,

•

F = whA = (62.4)(3)(9) = 1684.8 lb. y

d

The Theorems of Pappus ----- :o -~ - --_

In the third century, an Alexandrian Greek named Pappus discovered two formulas that relate centroids to surfaces and solids of revolution. The formulas provide shortcuts to a number of otherwise lengthy calculations.

L(y) R

c

------~-~-----~ Centroid

-o~--------------'~~ X FIGURE 6.56 The region R is to be revolved (ooce) about the x-axis to generate a solid, A l700-year-old theorem says that the solid's volume can be calculated by multiplying the regioo's area by the distance traveled by its centroid during the revolution,

THEOREM 1

Pappus's Theorem for Volumes

If a plane region is revolved once about a line in the plane that does not cut through the region's interior, then the volume of the solid it generates is equal to the region's area times the distance traveled by the region's centroid during the revolution. If p is the distance from the axis of revolution to the centroid, then

v=

27TpA.

(9)

Proof We draw the axis of revolution as the x-axis with the region R in the first quadrant (Figure 6.56). We let L(y) denote the length of the cross-section of R perpendicular to the y-axis at y. We assume L(y) to be continuous.

354

Chapter 6: AppLications of Definite Integrals By the method of cylindrical shells, the volume of the solid generated by revolving the region about the x-axis:is

v

~

l' 2,,(she1l

rulius)(shell height) dy

~ 2,,1' y L(y) dy,

(10)

The y-coordinate of R's centroid is

_ l'

Distance from Will of revolution to centroid

Y

=

'ji dA

A

l'

y L(y) dy

y-

A

y, dA. - L(y)dy

so that

Substituting Ay for the last integral in Equation (10) gives V = 2'1Tj1A. With p equal to y, we have V = 27rpA. •

FIGURE 8.57 With Pappus's frrst theorem, we can fmd the volume of a torus without having to integrate (Example 6).

EXAM PLE 6 Find the volume of the torus (doughnut) generated by revolving a circular disk ofradius a about an axis in its plane at a distance b 2: a from its center (Figure 6.57). Solution We apply Pappus's Theorem for volumes. The centroid of a disk is located at its center, the area is A = 7ra 2, and p = b is the distance from the centroid to the axis of revolution (see Figure 6.57). Substituting these values into Equation (9), we fmd the volume of the torus to be

v-

The next example shows haw we can use Equation (9) in Pappus's Theorem to find one of the coordinates of the centroid of a plane region of knawn area A wbm we also know the volume V of the solid generated by revolving the region about the other coordinate axis. That is, if y is the coordinate we want to fmd, we revolve the region around the x-axis so that y - p is the distance from the centroid to the axis of revolution. The idea is that the rotation generates a solid of revolution whose volume V is an already knawn quantity. Then we can solve Equation (9) for p, which is the value of the centroid's coordinate y.

, •

,.

4 0 Centroid

--'----+-----':,{1i+. z -0

0

•

2,,(b)(,,.') - 2,,'00'.

ow

FIGURE 8.58 With Pappus's frrst theorem. we can locate the centroid of a semicircular region without having to integrate (Example 1).

EXAM PLE 7

Locate the centroid of a semiciroulsr region of radius a.

We consider the region ~ the semicirele y - Va 2 x 2 (Figure 6.58) and the x-axis and imagine revolving the region about the x-axis to generate a solid sphere. By symmetry, the x-coordinate of the centroid is x - O. With y - p in Equation (9), we have

Solution

_ V (4/3)".' 4 y - 27rA - 27r(1/2)1Ta2 - 37r a.

•

THEOREM 2 Pappus's Theorem for Surface Areas If an arc of a smooth plane curve is revolved once about a line in the plane that does not cut through the arc's interior, then the area of the surface generated by the arc equals the length L of the arc time, the _1m>e1ed by the ",,'s cen1roid during the revolution. If p is the distance from the axis of revolution to the centroid, then

s=

21TpL.

(11)

355

6.6 Moments and Centers of Mass

The proof we give assumes that we can model the axis of revolution as the x-axis and the arc as the graph of a continuously differentiable function of x.

y

Proof We draw the axis of revolution as the x-axis with the arc extending from x = a to x = b in the f"ITSt quadrant (Figure 6.59). The area of the surface generated by the arc is S=

o

l

x~b

lx~b

27T)1 tis = 2".

x- a

Y tis.

(12)

x- a

The y-coordinate of the arc's centroid is FIGURE 6.59

_ E:bytls E:bytls

Figure for proving

Pappus's Theorem for surface area. The arc lenglh differential tis is given by Equation (6) in Section 6.3.

y =

l

x- b

tis

L

=

L

= J ds is the arc's

length and

y ~ y.

x=a

Hence

E:bytls

=

YL.

Substituting yL for the last integral in Equation (12) gives S = 2"'YL. With p equal to y, = 2".pL. •

we have S

EXAMPLE 8

Use Pappus's area theorem tof"md the surface area of the torus in Example 6.

Solution From Figure 6.57, the surface of the torus is generated by revolving a circle of radius a about the z-axis, and b ;" a is the distance from the centroid to the axis ofrevolution. The arc length of the smooth curve generating this surface of revolution is the circumference of the circle, so L = 2".a. Substituting these values into Equation (II), we f"md the surface area of the torus to be

• Exercises 6.6 10. a. The region cut from the frrst quadrant by 1he circle x 2

Thin Plates with Constant Density In Exercises 1-12, find the cenrer of mass ofa thin plate of coostant deosity B covering the given region.

1. The region bounded by the parabola y = x 2 and the line y = 4 2. The region bounded by the parabola y = 25 - x 2 and the x·axis 3. The region bounded by the parabola y = x - x 2 and the line y = -x 4. The region enclosed by the parabolas y = x 2

3 and y = - 2x 2

+ y2

= 9

b. The region bounded by 1he x·axis and the semicircle

y=

vI9=7

Compare your answer in part (b) wi1h 1he answer in part (a). 11. The "ttiangular" region in 1he ftrst quadrant between the circle x 2 + y2 = 9 and 1he lines x = 3 and y = 3. (Hint: Use geometry to fmd the area.)

s.

12. The region bounded above by 1he curve y = I/x', below by the curve y = -I/x', and on 1he left and right by 1he lines x = I and x = a > 1. Also, f"md lima_ oo j.

7. The region bounded by the x·axis and the curve y = cos x,

Thin Plates with varying Density 13. Find the center of mass of a thin plate covering the region between the x·axis and 1he curve y = 2/X2, I '" x s 2, if the

-

The region bounded by the y-axis and the curve x = y - y', O"'y'" I 6. The region bounded by the parabola x = y2 - Y and the line y = x

-"./2 '"

x"''''/2

8. The region betweeo 1he curve y = sec' x, -"./4 s x s "./4 and the x-axis 9. The region bounded by the parabolas y = 2x 2

y=2x-x 2

-

4x and

plate's deosity at 1he point (x,y) is B(x) = x 2.

14. Find the cenrer of mass of a thin plate coveriog the region bounded below by 1he parabola y = x 2 and above by 1he line y = x if1he plate's deosity at the point (x,y) is B(x) = l2x.

356

Chapter 6: Applications of Definite Integrals

±4/Yx

15. The region bounded by the curves y = and the lines x = I and x = 4 is revolved about the y-axis to generate • solid. L

25. Variable density Suppose that the density of the wire in Example 4 is 8 = hin6 (kconstant). Find the center of mass.

16. Variable density Suppose that the density of the wire in Exam-

Find the volume of the solid.

b. Find the center of mass of. thin plate covering the region if the plate's density at the point (x, y) is 8(x) = I/x. c. Sketch the plate and show the center of mass in your sketch.

16. The region between the curve y = 2/x and the x-axis fium x to x = 4 is revolved about the x-axis to generate a solid

=I

a. Find the volume of the solid.

ple4is8 = I

Yx.

c. Sketch the plate aod show the center of mass in your sketch.

Centroids of Triangles 17. The centroid of a triangle lies at the intersection of the triangle's medians You may recall that the point inside a triaogie that lies ace-third of the way from esch side toward the opposite vertex is the point where the triaog\e's three medians interse<:t. Show that the centroid lies at the intersection of the medians by showing that it too lies one-tbitd of the way from each side toward the opposite vertex. To do so, take the following steps. I) Stand one side of the triangle on the x-axis as in part (b) of the accompanying figure. Express dm in terms of L and dy.

il) Use similar triangles to show thatL = (b/h)(h - y). Substitute this expression for L in your formula for dm. Ill) Showtbat y = h/3. Iv) Extend the argurneot to the other sides.

(kconstaot). Find the eeuterofmass.

Plates Bounded by Two Curves In Exercises 27-30, find the centroid of the thin plate bounded by the graphs of the given funetions. Use Equations (6) and (7) with 8 = I and M = area of the region covered by the plate. 17. g(x)

b. Find the center of mass of. thin plate covering the region if the plate's density at the point (x,y) is 8(x) =

+ kloos61

18. g(x) 19. g(x)

30. g(x)

x' and f(x) = x + 6 = x' (x + I), f(x) = 2, and x = 0 = x'(x - I) and f(x) = x' = 0, f(x) = 2 + sinx, x = 0, and

=

(Hint:

J

xsiux
x

= 2.".

+ C.)

Theory and Examples Verify the statemenls and formulas in Exercises 31 and 32.

31. The coordinates of the centroid of a differentiable plane curve are

_ J x tIs x

=

length'

_

fytIs

y = length·

y

y

, \ \

' //

\ \ \ \

h- y

dy h ~

..... \

-:.:

",/

,C"'~lI', \

T

y

.......... .....

I,

L4+ .. ,I

0

(aJ

b

32. Whatever the value of p > 0 in the equation y = x'/(4p), the y-coordinate of the centroid of the parabolic segment shown here is y = (3/5}o. y

(bJ

Use the result in Exercise 17 to fmd the centroids of the triangles whose vertices appear in Exercises 18-22. Assume a, b > O.

18. (-1,0), (I , 0), (0, 3) 19. (0,0), (I, 0), (0, I) 10. (0,0), (a, 0), (0, a) 21. (0,0), (a, 0), (0, b) 11. (0,0), (a , 0), (a/2 , b)

Thin Wires 23. Constant deoslty Find the moment about the x-axis of a wire of constant density that lies along the curve y = from x = 0 tox = 2.

Yx

14. Constant density Find the momeot about the x-axis of a wire of constant deosity that lies along the curve y

tox = I.

= x'

from x

=0

The Theorems of Pappus 33. The square region with vertices (0, 2), (2, 0), (4, 2), and (2, 4) is revolved about the x-axis to generate a solid. Fted the volume and surface area of the solid. 34. Use a theorem of Pappus to fmd the volume generated by revolving about the line x = 5 the triangular region bounded by the coordinate axe. and the line 2x + Y = 6 (see Exercise 17). 35. Find the volume of the torus generated by revolving the circle (x - 2)' + y' = I about the y-axis.

Chapter 6 Practice Exercises 36. Use the theorems of Pappus to fmd the lateral surface area and the volume of a right-circular cone. 37. Use Pappus's Theorem for sorface area and the fact that the surface area of a sphere of radius a is 4'1Ta 2 to fmd the centroid of the semicircle y ~ Va 2 - x 2 • 38. As found in Exercise 37, the centroid of the semicircle y ~ Va 2 - x 2 lies at the pcint (0, 2a/lr). Find the area of the sorface swept out by revolving the semicircle about the line y

~

a.

39. The area of the region R enclosed by the semiellipse y ~ (b/a)Va 2 - x 2 and the x-axis is (I/2)'lTab, and the volume of the ellipsoid generated by revolving R about the x-axis is (4/3)'lTab 2 . Find the centroid of R. Notice that the location is independent of a. 40. As found in Example 7, the centroid of the region enclosed by the x-axis and the semicircle y = Va 2 - x2 lies at the point (0, 4a/3'IT). Find the volume of the solid generated by revolving this region about the line y ~ -a.

357

42. As found in Exercise 37, the centroid of the semicircle y ~ Va 2 - x 2 lies at the pcint (0, 2a/'IT). Find the area of the sorface generated by revolving the semicircle about the line y = x-a. In Exercises 43 and 44, use a theorem of Pappus to fmd the centroid

of the given triangle. Use the fact that the volume of a cone of radius r and height h is V ~ ~ 'lTr2h. 43.

y (0, b)

~~--------~~x (0,0) (a, 0)

44.

y

(a, c)

41. The region of Exercise 40 is revolved about the line y = x - a to generate a solid. Find the volume of the solid.

Chapter

!

Questions to Guide Your Review

1. How de you define and calculate the volumes of solids by the method of slicing? Give an example. 2. How are the disk and waaber methods for calculating volumes derived from the method of slicing? Give examples of volume calculations by these methods. 3. Describe the method of cylindrical sbells. Give an example. 4. How do you fmd the length of the graph of a smooth function over a closed interva1? Give an example. What about functions that do not have continuous flISt derivatives? 5. How do you defme and calculate the area of the sorface swept out by revolving the graphofa smooth functiony ~ f(x), a :5 x:5 b, about the x-axis? Give an example.

Chapter

[j

6. How de you defme and calculate the work done by a variable force directed along a pcrtion of the x-axis? How do you calculate the work it takes to pomp a liquid from a tank? Give examples. 7. How de you calculate the force exerted by a liquid against a portion of a flat vertical wall? Give an example. 8. What is a center of mass? a centroid? 9. How do you locate the center of mass ofa thin flat plate ofmat<>rial? Give an example. 10. How de you locate the center of mass of a thin plate bounded by _ curves y ~ f(x) andy ~ g(x) over a :5 x :5 b?

Practice Exercises

Volumes Find the volumes of the solids in Exercises 1-16.

the solid perpendicular to the x-axis are equilateral triangles whose bases stretch from the line to the curve.

1. The solid lies between planes perpendicular to the x-axis at x ~ 0 and x ~ 1. The cross-sections perpendicu1ar to the x-axis between these planes are circular disks whose diameters run from the parabola y ~ x 2 to the parabola y ~ Yx.

3. The solid lies between planes perpendicular to the X-
2. The base of the solid is the region in the fust qoadrant between the line y ~ x and the parabola y ~ 2 Yx. The cross-sections of

4. The solid lies between pIanes perpendicular to the x-axis at x ~ 0 and x ~ 6. The cross-sections between the.. p1aoes

358

Chapter 6: Applications of Definite Integrals

are squares whose bases run from the x /2 + yl/2 = '

V6.

y

x-axis up to the curve

y ~ll~--~~-----r. ll~X

6

-2

2

Lengths of Curves Find the lengths of the curves in Exercises 17-20.

17. y

= x ' /2

- (1/3)x'/2, I '" y '" 8

18. x = y'f3,

5. The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections of the solid perpendicular to the x-axis between these planes are circular disks whose diaroeters run from the curve x' = 4y to the curve y' = 4x. 6. The base of the solid is the region bounded by the parabola y2 = 4x and the line x = 1 in the xy-plane. Each cross-section perpendicular to the x-axis is an equilatera\ triangle with one edge in the plane. (The triangles all lie on the sarae side of the plane.) 7. Find the volume of the solid generated by revolving the region bounded by the x-axis, the curve y = 3x4, and the lines x = 1 and x = -I about (a) the x-axis; (b) the y-axis; (c) the line x = I; (d) the liney = 3.

8. Find the volume of the solid generated by revolving the "triangular" region bounded by the curve y = 4/x' and the lines x = I and y = 1/2 about (a) the x-axis; (b) the y-axis; (c) the line x = 2; (d) the liney = 4. 9. Find the volume of the solid generated by revolving the region boonded 00 the left by the parabola x = y' + I and on the right by the Hoe x = 5 about (a) thex-axis; (b)they-axis; (c) the line x = 5. 10. Find the volume of the solid generated by revolving the region bouoded by the parabola y' = 4x and the Hoe y = x about (a) the x-axis; (b) they-axis; (c) the line x = 4; (d) the liney = 4. 11. Find the volume of the solid generated by revolving the "triangular" region bouoded by the x-axis, the line x = 'If/3, and the curve y = tan x in the fust quadrant about the x-axis. 12. Find the volume of the solid generated by revolving the region bounded by the curve y = sin x and the lines x = 0, x = '1T', and y = 2 about the line y = 2. 13. Find the volume of the solid generated by revolving the region be2x about (a) the x-axis; tween the x-axis and the curve y = (b) the Hoey = -I; (c) the line x = 2; (d) the line y = 2.

x' -

14. Find the volume of the solid generated by revolving about the x-axis the region bouoded by y = 2tanx,y = O,x = -'If/4,and x = 'If/4. (The region lies in the frrst and third quadrants and resembles a skewed bowtie.)

V3

15. Volum. of a .olid .ph.... hoi. A rouod hole of radius ft is bored through the center of a solid sphere of a radius 2 ft. Find the volume of material removed from the sphere.

_haD

16. Volume of a The prof!le of a fuothall resembles the ellipse shown here. Find the football's volU03e to the nearest cubic inch.

19. y

20. x

I '" x '" 4

= (5/12)x 6/' - (5/8)x 4/', I '" x = (y'/12) + (i/y), I '" y '" 2

'" 32

Areas of Surfaces of Revolution In Exercises 21-24, fmd the areas of the surfaces generated by revolving the curves about the given axes. 21. y

=~,

22. y

= x

23. x

=

24. x =

/3,

3

x-axis

0:5 x:5 1; x-axis

v'4y -

yy,

0 '" x '" 3;

y',

I '" y '" 2;

2 s y '" 6;

y-axis

y-axis

Work 25. Lif1ing equipmeot A rock climber is about to haul up 100 N (about 22.5 Ib) of equipment that has been hanging beneath her on 40 m of rupe that weighs 0.8 newtoo per meter. How much work will it take? (Hint: Solve for the rope and equipment separately, then add.) 26. Leaky tank truck You drove ao 800-gal tank truck of water from the base of Mt. Washington to the summit and discovered on arrival that the tank was only halffull. You started with a full tank, clmbed at a steady rate, and accomplished the 4750-ft .Ievation chaage in 50 min. Assuming that the water leaked out at a steady rate, how much work was spent in carrying water to the top? Do not couot the work done in getting yourself and the truck there. Water weighs 8 Ib/U.S. gal. 27. Stretching a spring If a force of20 Ib is required to hold a spring I ft beyond its unstressed length, how much work does it take to stretch the spring this far? An additional lOot? 28. Garage door spring A fore. of 200 N will stretch a garage door spring 0.8 m beyond its unstressed length. How far will a 300-N force stretch the spring? How much work does it take to stretch the spring this far from its unstressed length? 29. Pumping a ......-.oir A reservoir shaped like a right-circular cone, point down, 20 ft across the top and 8 ft deep, is full of water. How much work does it take to pump the water to a level 6 ft above the top? 30. Pumping a .....rvoir (Continuation of Exercise 29.) The reservoir is filled to a depth of 5 ft, aod the water is to be pumped to the same level as the top. How much work does it take?

31. Pumping a conical tauk A right-circular conical tank, point down, with top radius 5 ft aod height \0 ft is filled with a liquid whose weight-density is 60 Ib/tt' . How much work does it take to pump the liquid to a point 2 ft above the tank? If the pmnp is

Chapter 6 Additional and Advanced Exercises driven by a motor rated at 275 fl:-1b/sec (1/2 hp), how long will it take to empty the tank?

y

32. Pumping a cylindrical tank A storage tank is a right-circular cylinder 20 ft long and 8 ft in diameter with its axis horizonta1. If the tank is half full of olive oil weighing 57 lb/ft3 , find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank. Centers of Mass and Centroids 33. Find the centroid of a thin, flat plate covering the region enclosed by the parabolas y = 2x 2 and y = 3 - x 2 • 34. Find the centroid of a thin, flat plate covering the region enclosed by the x-axis, the lines x = 2 and x = -2, and the parabola y = x 2•

-L------~~----~~x

-4

FLuid Force 39. Trough of water The vertical triangular plate shown here is the end plate of a trough full of water (w = 62.4). What is the fluid force against the plate?

Chapter

4

40. Trough of maple syrup The vertical trapezoidal plate shown here is the end plate of a trough full of maple syrup weighing 75lb/ft3 . What is the force exerted by the syrup against the end plate of the trough when the syrup is 10 in. deep? y

y=x-2// ~----~~----~/

--~ _L 2----0+---/~~~-+x

36. Find the centroid of a thin, flat plate covering the region enclosed by the parabola y2 = x and the line x = 2y.

38. a. Find the center of mass of a thin plate of constant density covering the region between the curve y = 3/x 3/ 2 and the x-axis fromx = 1 tox = 9. b. Find the plate's center of mass if, instead of being constant, the density is 8(x) = x. (Use vertical strips.)

~~~~O

UNITS IN FEET

35. Find the centroid of a thin, flat plate covering the "triangular" region in the first quadrant bounded by the y-axis, the parabola y = x 2/4, and the line y = 4.

37. Find the center of mass of a thin, flat plate covering the region enclosed by the parabola y2 = x and the line x = 2y if the density function is 8(y) = 1 + y. (Use horizontal strips.)

359

/

/ / /

UNITS IN FEET

41. Force on a parabolic gate A flat vertical gate in the face of a dam is shaped like the parabolic region between the curve y = 4x 2 and the line y = 4, with measurements in feet. The top of the gate lies 5 ft below the surface of the water. Find the force exerted by the water against the gate (w = 62.4).

D 42.

You plan to store mercury (w = 849lb/fi3) in a vertical rectangular tank with a 1 ft square base side whose interior side wall can withstand a total fluid force of 40,000 lb. About how many cubic feet of mercury can you store in the tank at anyone time?

Additional and Advanced Exercises

VoLume and Length 1. A solid is generated by revolving about the x-axis the region bounded by the graph of the positive continuous function y = f(x), the x-axis, and the fixed line x = a and the variable line x = b, b > a. Its volume, for all b, is b 2 - ab. Find f(x). 2. A solid is generated by revolving about the x-axis the region bounded by the graph of the positive continuous function y = f(x), the x-axis, and the lines x = 0 and x = a. Its volume, for all a > 0, is a 2 + a. Find f(x). 3. Suppose that the increasing function f(x) is smooth for x ~ 0 and that f( 0) = a. Let s(x) denote the length of the graph of f from (0, a) to (x, f(x», x > O. Find f(x) if s(x) = Cx for some constant C. What are the allowable values for C? 4. a. ShowthatforO <

a:::;

'Tf'/2,

loa VI + cos 0 dO > Va 2

b. Generalize the result in part (a).

2

+ sin2 a.

5. Find the volume of the solid formed by revolving the region bounded by the graphs of y = x and y = x 2 about the line y = x. 6. Consider a right-circular cylinder of diameter 1. Form a wedge by making one slice parallel to the base of the cylinder completely through the cylinder, and another slice at an angle of 45° to the first slice and intersecting the first slice at the opposite edge of the cylinder (see accompanying diagram). Find the volume of the wedge.

360

Chapter 6: Applications of Definite Integrals

Surface Area

7. At points on the curve y = 2 Vx, line segments oflength h = y are drawn perpendicular to the xy-plane. (See accompanying figure.) Find the area of the surface formed by these perpendiculars from (0, 0) to (3, 2\13). y

10. Work and kinetic energy Suppose a 1.6-oz golf ball is placed on a vertical spring with force constant k = 2 lb/in. The spring is compressed 6 in. and released. About how high does the ball go (measured from the spring's rest position)?

Centers of Mass 11. Find the centroid of the region bounded below by the x-axis and above by the curve y = 1 - x n , n an even positive integer. What is the limiting position of the centroid as n ~ oo? 12. If you haul a telephone pole on a two-wheeled carriage behind a truck, you want the wheels to be 3 ft or so behind the pole's center of mass to provide an adequate "tongue" weight. The 40-ft wooden telephone poles used by Verizon have a 27-in. circumference at the top and a 43.S-in. circumference at the base. About how far from the top is the center of mass? 13. Suppose that a thin metal plate of area A and constant density 8 occupies a region R in the xy-plane, and let My be the plate's moment about the y-axis. Show that the plate's moment about the line x = b is

8. At points on a circle of radius a, line segments are drawn perpendicular to the plane of the circle, the perpendicular at each point P being of length ks, where s is the length of the arc of the circle measured counterclockwise from (a, 0) to P and k is a positive constant, as shown here. Find the area of the surface formed by the perpendiculars along the arc beginning at (a, 0) and extending once around the circle.

a. My - baA if the plate lies to the right of the line, and b. b8A - My if the plate lies to the left of the line. 14. Find the center of mass ofa thin plate covering the region bounded by the curve y2 = 4ax and the line x = a, a = positive constant, ifthe density at (x,y) is directly proportional to (a) x, (b) Iyl. 15. a. Find the centroid of the region in the first quadrant bounded by two concentric circles and the coordinate axes, if the circles have radii a and b, 0 < a < b, and their centers are at the origin. b. Find the limits of the coordinates of the centroid as a approaches b and discuss the meaning of the result. 16. A triangular comer is cut from a square 1 ft on a side. The area of the triangle removed is 36 in2 • If the centroid of the remaining region is 7 in. from one side of the original square, how far is it from the remaining sides?

y

Work 9. A particle of mass m starts from rest at time t = 0 and is moved along the x-axis with constant acceleration a from x = 0 to x = h against a variable force of magnitude F(t) = t 2• Find the work done.

Chapter •

Fluid Force 17. A triangular plate ABC is submerged in water with its plane vertical. The side AB, 4 ft long, is 6 ft below the surface of the water, while the vertex Cis 2 ft below the surface. Find the force exerted by the water on one side of the plate. 18. A vertical rectangular plate is submerged in a fluid with its top edge parallel to the fluid's surface. Show that the force exerted by the fluid on one side of the plate equals the average value of the pressure up and down the plate times the area of the plate.

TechnoLogy AppLication Projects

MathematicajMaple Modules: Using Riemann Sums to EstimateAreas, Volumes, and Lengths of Curves Visualize and approximate areas and volumes in Part I and Part II: Volumes of Revolution; and Part III: Lengths of Curves. Modeling a Bungee Cord Jump Collect data (or use data previously collected) to build and refine a model for the force exerted by a jumper's bungee cord. Use the work-energy theorem to compute the distance fallen for a given jumper and a given length of bungee cord.

7 TRANSCENDENTAL FUNCTIONS OVERVIEW Functions can be classified into two broad complementary groups called algebraic functions and transcendental functions (see Section 1.1). Except for the trigonometric functions, our studies so far have concentrated on the algebraic functions. In this chapter we investigate the calculus of important transcendental functions, including the logarithmic, exponential, inverse trigonometric, and hyperbolic functions. They occur frequently in many mathematical settings and scientific applications.

7.1

Inverse Functions and Their Derivatives A function that undoes, or inverts, the effect of a function f is called the inverse of f. Many common functions, though not all, are paired with an inverse. Important inverse functions often show up in applications. Inverse functions also playa key role in the development and properties of the logarithmic and exponential functions, studied in Section 7.3.

One-to-One Functions A function is a rule that assigns a value from its range to each element in its domain. Some functions assign the same range value to more than one element in the domain. The function f(x) = x 2 assigns the same value, 1, to both ofthe numbers -1 and + 1; the sines of 7T/3 and 27T/3 are both 0/2. Other functions assume each value in their range no more than once. The square roots and cubes of different numbers are always different. A function that has distinct values at distinct elements in its domain is called one-to-one. These functions take on anyone value in their range exactly once.

DEFINITION A function f(x) is one-to-one on a domain D if f(Xl) #- f(X2) whenever Xl #- X2 inDo Some functions are one-to-one on their entire natural domain. Other EXAMPLE 1 functions are not one-to-one on their entire domain, but by restricting the function to a smaller domain we can create a function that is one-to-one. The original and restricted functions are not the same functions, because they have different domains. However, the two functions have the same values on the smaller domain, so the original function is an extension ofthe restricted function from its smaller domain to the larger domain. (a) f(x) = Vx is one-to-one on any domain of nonnegative numbers because vx;: whenever Xl #- X2·

'V'.X;"

#-

(b) g(x) = sin x is not one-to-one on the interval [0,7T] because sin (7T/6) = sin(57T/6). In fact, for each element Xl in the subinterval [0, 7T/2) there is a corresponding element X2 in the subinterval (7T/2, 7T] satisfying sinxl = Sinx2, so distinct elements in

361

362

Chapter 7: Transcendental Functions y

the domain are assigned to the same value in the range. The sioe function is one-toone on [0, "/2], however, because it is an iocreasing function on [0, "/2] giviog distinct outputs for distinct ioputs. • The graph of a one-to-one function y = f(x) can iolersect a given horizontallioe at most once. If the function iolersects the lioe more than once, it assumes the same y-value for at least two different x-values and is therefore not one-to-one (Fignre 7.1).

(a) One-to-one: Graph meets each horizontal line at most once.

The Horizontal Line Test for One-to-One Functions A function y = fix) is one-to-one if and only if its graph iotersects each horizontallioe at most once.

y

Same y-value

0.5 --!--'~'---!-_ x

-1

I

\

5:;

0

x

y = sinx (b) Not one-to-one: Graph meets one or more horizontal lines more than once.

x'

FIGURE 7.1 (a)y = andy = Yxare one-to-one on their domains (-00,00) and [O,oo).(b)y =x2 andy = sin x are not one-la-one on their domains (-00, (0).

Inverse Functions Sioce each output of a one-to-one function comes from just one ioput, the effect of the function can be ioverted to send an output back to the ioput from which it came. Suppose that f is a one-to-one function on a domaio D with rnnge I is defmed by

DEFINITION

R. The inverse function

r

rl(h) = a if f(a) = h.

The dornaio of

r

l

is R and the rnnge of

r

l

is D.

r

r

The symbol l for the ioverse of f is read "f ioverse." The "-1" in l is not an exponent; rl(x) does not mean 1/fix). Notice that the domains and ranges of f and l are ioterchanged.

EXAMPLE 2

Suppose a one-to-one function y

= fix)

r

is given by a table of values

x fix)

r

A table for the values of x = l(y) can then be obtained by simply ioterchanging the values io the columns of the table for f:

y

•

rl(y)

r

If we apply f to send an ioputx to the outputf(x) and follow by applyiog l to fix) we get right back to x, just where we started Similarly, if we take some number y io the range of f, apply l to it, and then apply f to the resulting value rl(y) , we get back the

r

value y with which we began. Composiog a function and its ioverse has the same effect as doiog nothiog. U- l

U

0

0

f)(x) = x,

rl)(y) = y,

for all x io the domaio of f for all y io the domaio of f- I (or range of f)

Only a one-to-one function can have an ioverse. The reason is that if f(Xl) = y and f(X2) = y for two distinct ioputs Xl and X2, then there is no way to assign a value to rl(y) that satisfies both rIU(XI)) = XI and r IU(X2)) = X2.

7.1

Inverse Functions and Their Derivatives

363

A function that is increasing on an interval so it satisfies the inequality f(X2) > f(XI) when X2 > Xl, is one-to-one and has an inverse. Decreasing functions also have an inverse. Functions that are neither increasing nor decreasing may still be one-to-one and have an inverse, as with the function f(x) = l/x for X oF 0 and f(O) = 0, defined on ( - 00, (0) and passing the horizontal line test.

Finding Inverses The graphs of a function and its inverse are closely related. To read the value of a function from its graph, we start at a point X on the x-axis, go vertically to the graph, and then move horizontally to the y-axis to read the value of y. The inverse function can be read from the grapb by reversing this process. Start with a point y on the y-axis, go horizontally to the graph of y = f(x), and then move vertically to the x-axis to read the value of x = r'(y) (Figure 7.2). y

y

--~--~~----~-------.x

o

X DOMAINOF

o

f

X RANGE OF

1-1

r'

(b) The graph of is the graph off, but with x and y interchanged. To find the x that gave y. we start at y and go over to the curve and down to the x-axis. The domain off-1 is the range off, The range off-1 is the domain off.

(8) To find the value off at x, we start at X, go up to the curve, and then over to the y-axis.

y

y ~r'(x)

__1-__~O~/~__~~______~y

-------o~--~~------~x

/

/ / / / /

/ /

(c) To draw the graph ofr' in the

(d) Then we interchange the letters x andy. We now have a normal-looking graph ofj-l as a function of x.

more usual way. we reflect the

system. across the line y = x.

r'

FIGURE 7.2 Detennining tlie graph ofy ~ (x) from tlie graph ofy = j(x). The graph of is obtaioed by reflecting the graph of j about the line y ~ x.

r'

r

We want to set up the graph of I so that its input values lie along the x-axis, as is usually done for functions, rather than on the y-axis. To achieve this we interchange the x and y axes by reflecting across the 45° line y = x. After this reflection we have a new graph that represents The value of r'(x) can now be read from the graph in the usual way, by starting with a point x on the x-axis, going vertically to the graph, and then horizontally

r'.

364

Chapter 7: Transcendental Functions to the y-axis to get the value of r'(x). Figure 7.2 indicates the relationship between the The graphs are interchanged by reflection through the line y = x. graphs of I and Although this geometric treatment is not a proof, it does make reasonable the fact that the inverse of a one·to·one continuous function defmed on an interval is also continuous. The process of passing from I to 1-' can be summarized as a twc>-step procedure.

r' .

y

1.

Solve the equation y = I(x) for x. This gives a formula x = r'(y) where x is ex· pressed as a function ofy.

2.

Interchange x and y, obtaining a formula y = r'(x) where r' is expressed in the conventional format with x as the independent variable and y as the dependent variable.

y~2x-2

EXAMPLE 3

Find the inverse of y =

~ x + I, expressed as a function ofx.

Solution 1.

2. FIGURE 7.3 Graphingf(x) = (1/2)x + I and r'(x) = 2x - 2 together shows the graphs' symmetry with respect to the line y = x (Example 3).

Solve for x in terms ofy:

Interchange x and y:

y =

y =

2y

=

x

=

1 2x +

x

I

+2

2y - 2. 2x - 2.

The inverse of the function I(x) = (1/2)x + I is the function r'(x) = 2x - 2. (See Figure 7.3.) To check, we verify that both composites give the identity function:

r'(f(x»

=

2(~X + I) - 2 =

I(r'(x»

=

~(2x

- 2) + I

=

+2

x

- 2 = x

x - I + I

x.

=

•

y

EXAMPLE 4

Find the inverse of the function y = x 2 , X

;;"

0, expressed as a function

ofx.

Solution

We flISt solve for x in terms of y:

y = x2

Ixl ~~------------~X

FIGURE 7.4 The functions y = Vi xl, X ~ 0, are inverses of one another (Example 4).

because x

2:

0

We then interchange x andy, obtaining

,0

and y =

= x

y=

Yx.

The inverse of the functiony O,isthefunctiony = Yx (Figure 7.4). Notice that the function y = x 2 , X ;;" 0, with domain restricted to the nonnegative real numbers, is one·to·one (Figure 7.4) and has an inverse. On the other hand, the func· tion y = x 2, with no domain restrictions, is not one·to·one (Figure 7.lb) and therefore has = x 2,x;;"

•

no inverse.

Derivatives of Inverses of Differentiable Functions If we calculate the derivatives of I(x) = (1/2)x + 1 and

its inverse r'(x) = 2x - 2

from Example 3, we see that

~/(x)=~(~x+I)=~ ~r'(x)

=

~(2x

- 2)

=

2.

365

7.1 Inverse Functions and Their Derivatives

The derivatives are reciprocals of one another, so the slope of one line is the reciprocal of the slope of its inverse line. (See Figure 7.3.) This is not a special case. Reflecting any nonhorizontal or nonverticalline across the line y = x always inverts the line's slope. If the original line has slope m oF 0, the reflected line has slope I/m.

The slopes are reciprocal: (r') '(b) ~ _I_or (f-') '(b) ~ 1'(0)

IV

I '(b»

FIGURE 7.5 The graphs of inverse functions hsve reciprocal slopes at corresponding points.

r

The reciprocal relationship between the slopes of f and 1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = f(x) at the point (a,f(a»isj'(a) andj'(a) oF O,thentheslopeofy = (x) at the point (f(a), a) is the reciprocal 1/ j'(a) (Figure 7.5). Ifwe set b = f(a), then

r'

(r ' ),(b)

=

j'~a)

= j'(f \b))'

r'

If y = f(x) has a horizontal tangent line at (a, f(a» then the inverse function has a vertical tangent line at (f(a), a), and this wmite slope implies that is not differentiable at f(a). Theorem I gives the conditions under which is differentiable in its domain (which is the same as the range of f).

r'

r'

THEOREM I-The Derivative Rule for Inverses If f has an interval I as domain and j' (x) exists and is never zero on I, then 1 is differentiable at every point in its domain (the range of f). The value of(r y at a point b in the domain of ' isthereciprocalofthevalueofj' at the point a = r'(b):

r

r'

(f - 1)'(b) _

I - j'(f '(b))

(i)

or

dr' I ax

x- b -

_l_

dfl ax

x - r'(b)

Theorem 1 makes two assertions. The trrst of these has to do with the conditions under which f- 1 is differentiable; the second assertion is a formula for the derivative of

366

Chapter 7: Transcendental Functions

I -, when it exists. While we omit the proof of the first assertion, the second one is proved in the following way:

I(r'(x))

=

x

Inverse function relationship

fx I(r'(x))

=

I

Differentiating both sides

j'(r'(x))' ~ r'(x) dx d

dx

= 1

Chain Rule

1-'( ) _

I

Solving for the derivative

x - j'(j '(x»"

Thefunction/(x) = x 2 ,x;" oand its inverse r'(x) = Yxhavederiva· tivesj'(x) = 2xand(r'),(x) = 1/(2Yx). Let's verify that Theorem I gives the same fannula for the derivative of r'(x):

EXAMPLE 5

(r'),(x)

j'(j \x))

=

1'(x) ~ 2xwithxreplaced by r1(x)

I

--ccfL----!--~____!---"-----~ x

o

1

2

3

4

FIGURE 7.6 The derivative of r'(x) = Vi at the point (4, 2) is the reciprocal of the derivative of f(x) = x 2 at (2, 4) (Example 5).

Theorem I gives a derivative that agrees with the known derivative of the square root function. Let's examine Theorem 1 at a specific point. We pick x = 2 (the number a) and 1(2) = 4 (the value b). Theorem I says that the derivative of I at 2, j'(2) = 4, and the derivative of at 1(2), )'(4), are reciprocals. It states that

r'

(r'

I _ I (/ -')'(4) -- j'(j I '(4)) -_ j'(2) - 2x

I.-2-_4'I

•

See Figure 7.6.

y

6

y=x 3 -

(2, 6)

2 Slope 3.1;2 ~ 3(2)2 ~ 12

We will use the procedure illustrated in Example 5 to calculate fannulas for the derivatives of many inverse functions throughout this chapter. Equation (I) sometimes enables us to find specific values of dr'/dx without knowing a formula for r'.

Let I(x) = x 3 finding a fannula for r'(x).

EXAMPLE 6

Reciprocal slope:

lz

L _ J-- - - --<,ii(6,:22)

Solution

2. Find the value of dr'/dx at x = 6 = 1(2) without

-

We apply Theorem 1 to obtain the value of the derivative of

~-~-~-----~x

-dl

6

dx

I

= 3x 2

x=2

dr'l

dX FIGURE 7.7 The derivative of f(x) ~ x' - 2atx ~ 2 tells us the derivative of at x ~ 6 (Example 6).

r'

1

.=/(2) =

=

6:

12

x=2

dl I dx

See Figure 7.7.

I

r' at x =

1 12

Eq. (I)

.=2

•

7.1

367

Inverse Functions and Their Derivatives

Exercises 7.1 Identifying One-to-One Functions Graphically Which of the functions graphed in Exercises l--{j are one-to-one, and which are not?

1.

2.

y

Y = -3x3 X

y

W.

o

x

o 13.

y

y

14.

= intx

y

y =f(x) = sin x,

...

-¥sxs¥

-=> -=> -=>

y = f(x) = taox,

-¥<x<¥

1

-=> -=> -=>

15.

0

6.

y

y=

l x

\

=r-.. y

G, -

8. f(x) =

{2xx ++4,6,

x,

{ 1 - 2' x 9. f(x) = _x_ x + 2' 10. f(x) =

16.

r

2 2 -x ,

x,

y

6 f(x)

x
=6-

f(x)

lx,

={

X+I' 2

-2

OS x S 3

- IS xS O

+ "3x,

0< x< 3

- "--c+----£i..... x -I

3

..... x

--:-t----~

o

17.

7. f(x) =

x

"'2

y

In Exercises 7-10, determine from its graph if the function is

one-to-one.

x

"2

x

5.

= l-l,x>o

y

4.

x

y=f(x)

1

~~~~----~x

I

y

12.

y

y

y=x4_x2

3.

11.

3

B.

Graph the functionf(x) = ~,O '" x'" 1. What symmetry does the graph have?

v?

b. Show that f is its own inverse. (Remember that = x if 0.) 18. B. Graph the functionf(x) = I/x. What symmetry does the graph have?

x'"

b. Show that f is its own inverse.

x s-3 x> -3

Fonnulas for Inverse Functions Each of Exercises 19-24 gives a formula for a function y = f(x) and shows the grapha of f andr'. Find a formula for in each case.

r'

xsO

19. f(x) = x

x>O

2

+ I, x '" 0

y

y

x '" I x> I

Graphing Inverse Functions Each of Exercises 11-16 shows the graph of a function y = f(x). Copy the graph and draw in the line y = x. Then use symmetry with to your sketch. (It is reapect to the line y = x to add the graph of Identify the domain and not necessary to f'md a formula for rangeofr'.

r'.)

r'

20. f(x) =

x 2,

..... x

--:-t-~-----

X '"

0

368

Chapter 7: Transcendental Functions

22. f(x) ~ x 2 - 2:x:

21. f(x) ~ x' - I

+

I, x "" I

another.

Y

Y

40. a. Showthath(x) ~ x'/4andk(x) ~ (4x)'/3 are inverses of one

b. Graph h and k over an x-inrervallarge enough to show the graphs inrersecling at (2, 2) and (-2, -2). Be sure the pictore shows the required symmetry about the line y ~ X.

c. Find the slopes of the tangents to the graphs of h and k at (2,2) and (-2, -2). d. What lines are tangent to the curves at the origin? ~t-~~--------~x

o

41. Let f(x) ~ x' - 3x 2 - I, x "" 2. Find the value of dr'/dx at thepointx~ -I ~f(3). 42. Let f(x) ~ x 2 - 4x - 5,x the point x ~ 0 ~ f(5).

23. f(x) ~ (x

+ 1)2,

X "" -I 24. f(x) ~ x 2/3,

y

x "" 0

>

2. Find the value of dr'/dx at

43. Suppose that the differentiable function y ~ f(x) has an inverse and that the graph of f passes through the point (2, 4) and hss a slope ofl/3 there. Find the value of dr'/dx atx ~ 4.

y

44. Suppose that the differentiable function y ~ g(x) has an inverse and that the graph of g passes through the origin with slope 2. Find the slope of the graph of g-' at the origin

-1 -1

~~~----------~X

o

b. What can you conclude about the inverse of a function y ~ f(x) whose graph is a line through the origin with a

Derlvattves of Inverse Functions Each of Exercises 25--34 gives a formula for a function y ~ f(x). In each case, fmdr'(x) and identifY the domain and range of Ai; a check, showthatf(r'(x)) ~ r'(f(x)) ~ x.

rio

25. f(x) ~ x' 27. f(x) ~ x'

26. f(x) ~ x\

+

29. f(x) ~ Ilx 2, X

x "" 0

28. f(x) ~ (1/2)x - 712

I

>0

31. f(x) ~ x + 23

x-

30. f(x) ~ Ilx',

32. f(x) ~

x¢'o

'-Ix • r

vx - 3

33. f(x) ~ x 2 - 2:x:, x '" I (Hint: Complete the square.)

34. f(x) ~ (2:x:'

Inverses of Lines 45. a. Find the inverse of the function f(x) ~ m:x:, where m is a constant different from zero.

nonzero slope m? 46. Show that the graph of the inverse of f(x) ~ m:x: + h, where m and b are constants and m ¢' 0, is a line with slope 11m and y-intercept -him. 47. a. Find the inverse off(x) ~ x + l.Graphfanditsinverse together. Add the line y ~ x to your sketch, drawiog it with dashes or dots for contrast

b. Find the inverse of f(x) ~ x + h(b constant). How is the reiared to the graph of f? graph of

r'

c. What can you conclude about the inverses of functions whose graphs are lines parallel to the line y ~ x? 48. a. Find the inverse off(x) ~ -x + 1. Graph the line y ~ -x + I together with the line y ~ X. At whst angle do

+ 1)'/'

the lines intersect? In Exercises 35-38: L

Findr'(x).

b. Graph f and

b. Find the inverse of f(x) ~ -x + b (b constant). What angle does the line y ~ -x + h mske with the line y ~ x?

r' together.

c. Evaluaredfldxatx ~ aanddr'/dxatx ~ f(a) to show that at these points dr' I dx ~ I I (djI dx ) .

2:x: + 3, a ~ -I 36. f(x) ~ (1/5):x: + 7, a ~ -I 37. f(x) ~ 5 - 4x, a ~ 1/2 38. f(x) ~ 2:x: 2, X "" 0, a ~ 5

35. f(x) 39.

L

~

Show that f(x) ~ x' and g(x) ~

'Il'X

are inverses of one

c. What can you conclude about the inverses of functions whose graphs are lines perpendicular to the line y ~ x? Increasing and Decreasing Functions 49. Show that increasing functions and decreasing functions are aneto-one. That is, show that for any Xl and X2 in I, X2 =F Xl implies f(X2) ¢' f(x,).

another. Use the results of Exercise 49 to show that the functions in Exercises

b. Graph f and g over an x·inrervallarge enough to show the graphs in_ecting at (I, I) and ( -I, -I). Be sure the picture shows the required symmetry about the line y ~ X.

50-54 hsve inverses over their domains. Find a formula for dr'/dx using Theorem 1.

c. Find the slopes of the tangents to the graphs of f and g at (I, I) and (-I, -I) (four tangents in all).

52. f(x) ~ I - 8:x;'

d. What lines are tangent to the curves at the origin?

54. f(x) ~ x'/'

50. f(x) ~ (1/3)x

+ (516)

51. f(x) ~ 27x'

53. f(x) ~ (I - x)'

7.2

Natural Logarithms

369

Theory and Appl;cations

COMPUTER EXPLORATIONS

55. If f(x) is on.,.to-one, can anything be said about g(x) = - f(x)?

In Exercises 61-
Is it also one-to-one? Give reasons for your answer. 56. If f(x) is ooe-to-one and f(x) is never zero, can anything be said about h(x) = 1/f(x)? Is it also one-to-one? Give reasons for your answer. 57. Suppose that the range of g lies in the domain of f so that the composite fog is defmed. If f and g are one-ro.one, can anything be said about f 0 g1 Give reasons for your answer.

•• Plot the function y = j(x) together with its derivative over the giveo interval. Explain why you know that j is on<>-to-
58. If a composite fog is one-to-one, must g be one-to-one? Give reasons for your answer.

c. Find the equation for the tangeot line to j at the specified point

59. Assome that f and g are differentiable functions that are inverses of one another so that (g 0 f)(x) = x. Differentiate both sides of this equation with respect to x using the Chain Rule to express (g 0 f)'(x) as a product of derivatives of g and j. What do you fmd? (This is not a proof of Theorem 1 because we assume here is differentiable.) the theorem ~ conclusion that g =

d. Find the equation for the tsngent line to g at the point (f(Xo), xo) located symmetrically across the 45 0 line y = x (which is the graph of the ideotity function). Use Theorem I to find the slope of this tangent line.

r'

60. Equivalence of the wa.her and .hell methods for rmding volume Let j be differeotiable and increasing on the interval a '" x '" b, with a > 0, and suppose that j has a differentiable inverse, Revolve about the y-axis the regioo bounded by the graph of j and the lines x = a aod y = j(b) to geoerate a solid. Then the values of the integrals giveo by the washer and shell methods for the volume have ideotica1 values:

r' .

i

f

(') 1T«r'(y»' - a') t(y =

f(a)

l'

21TX(f(b) - j(x)) tIx.

To prove this equality, defme

i

e. Plot the functions j and g, the ideotity, the two tangoot lines, and the line segmoot joining the points (xo, j(xo» and (f(xo), xo). Discuss the symmetries you see across the main diagonal.

- 3x+2 62 ,y-2x-11'

-2< _ x<2 _,

4x

Xo =

3

Xo= 1/2

63. y = - , - - ,

-I '" x '" I, Xo = 1/2

x3 64. y = - , - - , x + I

-I '" x '" I, Xo = 1/2

65. Y

= x3

66. Y

=

f (')

'tT«r'(y))' - a') t(y

t '" x '" 4,

V3x - 2,

61. y =

x

a

W(t) =

(xo, j(xo)).

+

-

I

3x 2

1, 2 ~ x ~ 5, xo

-

2 - x - x 3,

-2 ~ x ~ 2, Xo

= 27 10 =

"23

f(a)

8(t) =

l'

2'tTx(f(t) - j(x)) tIx.

68. y Then show that the functions W aod 8 agree at a point of [a, b] and have ideotical derivatives 00 [a, b]. As you saw in Sectioo 4.7, Exercise 90, this will gusrantee W(I) = 8(1) for allIin [a, b]. In particular, W(b) = 8(b). (80urce: ''Disks and Shells Revisited," by Walter Carlip, American Mathematical Monthly, Vol. 98, No.2, Feb. 1991, pp. 154-156.)

7.2

-3

67. y = eX,

~

x

~

5, Xo = 1

= sinx,

In Exercises 69 and 70, repeat the steps above to solve for the functions y = j(x) and x = r'(y) defmed implicitly by the giveo equations over the interval.

69. y'/3 - I = (x

70. cosy

=

x l / S,

+ 2)3, -5 '" x '" 5, Xo = -3/2 0 ~ x ~ 1, Xo = 1/2

Natural Logarithms Historically, logarithms played important roles in arithmetic computations making possible the great seventeenth-century advances in offshore navigation and celestial mechanics. In this section we derme the natural logarithm as an integral using the Fundamental Theorem of Calculus. While this indirect approach may at first seem stnmge, it provides an elegant and rigorous way to obtain the key characteristics of logarithmic and exponential functions.

Definition of the Natural Logarithm Function The natural logarithm of any positive number x, written as In x, is defined as an integral.

370

Chapter 7: Transcendental Functions

DEFINITION

The natural logarithm is the function given by Inx =

il{"IT dl ,

x> O.

{l}

From the Fundamental Theorem of Calculus, In x is a continuous function. Geometrically, if x > 1, then In x is the area under the curve y = 1/1 from I = 1 to I = x (Figure 7.8). For 0 < x < 1, In x gives the negative of the area under the curve from x to 1. The function is not defined for x ,;; O. From the Zero Width Interval Rule for definite integrals, we also have In 1 =

il[11T dl =

y

HO <.< I, then In.

O.

~j X 1dt ~ -j \dt

gives the negative of this area.

If.> l,th.DIn.

~ jXjdt

gives this area.

1

1

y

=

Inx

FIGURE 7.8 The graph of y = Inx and its relation to the function y = I/x, x > O. The graph of the logarithm rises above the x-axis as x moves from I to the right, and it falls below the x-axis as x moves from 1 to the left.

Notice that we show the graph of y = l/x in Figure 7.8 but use y = 1/1 in the integral. Using x for everything would have us writing TABLE 7.1

TypicaL 2-pLace

lnx =

vaLues of Ln x

x

Inx

0 0.05 0.5 1 2 3 4 10

undefined

-3.00 -0.69 0 0.69 1.10 1.39 2.30

("I "dx,

il

with x having two meanings. Thus we change the variable of integration to I. By using rectangles to obtain tInite approximations of the area under the graph of y = 1/1 and over the interval between I = 1 and I = x, as in Section 5.1, we can approximate the values of the function In x. Several values are given in Table 7.1. There is an important number between x = 2 and x = 3 whose natorallogarithm equals 1. This number, which we now define, exists because In x is a continuous function and therefore satisfies the Intermediate Value Theorem on [2, 3].

DEFINITION

The number e is that number in the domain of the naturalloga-

rithm satisfying In(e) = 1.

7.2

371

Natural Logarithms

So the number e lies within the interval [2, 3] and satisfies ' I ]

1

tdt

I.

=

Interpreted geometrically, the number e corresponda to the point on the x-axis for which the area under the graph of y = I/t and above the interval [I, e] equals the area of the unit square. That is, the area of the region shaded blue in Figure 7.8 is I sq unit when x = e. In the next section, we will see that the number e can be calculated as a limit and has the numerical value e RJ 2.718281828459045 to 15 decimal places.

The Derivative of y

= In x

By the first part of the Fundamental Theorem of Calculus (Section 5.4),

d

dx lnx

=

d]'1

dx

1

tdt

=

x'

=

I

x·

For every positive value of x, we have

d dx Inx

I

and the Chain Rule extenda this formula for positive functions u(x):

iL lnu dx

=

iL lnu · du du dx

u> O.

EXAMPLE 1

(2)

We use Equation (2) to fmd derivatives.

dId I I (8) dx In 2x = 2x dx (2x) = 2x (2) = x' (b) Equation (2) with u = x 2

iL ln (x 2 dx

+ 3)

=

x

>0

_1_.iL(x 2 x 2 +3 dx

+ 3)

+ 3 gives =

_1_. 2x =~. x 2 +3 x 2 +3

•

Notice the remarkable occurrence in Example la. The function y = In 2x has the same derivative as the function y = Inx. This is true of y = In bx for any constant b, provided that bx > 0:

dI d -Inbx = -'-(bx) dx bxdx If x < 0 and b < 0, then bx b = -I we get

> d

=

I -(b) bx

I

=-.

0 and Equation (3) still applies. In particular, if x

dx In (-x)

=

1

X

for x

<

(3)

x

<

0 and

O.

Since Ix I = x when x > 0 and Ix I = -x when x < 0, we have the following important result, which says that In Ixl is an antiderivative of I/x, x oF O.

(4)

372

Chapter 7: Transcendental Functions

Properties of Logarithms HiSTORICAL BIOGRAPHY

John Napier (155{}-16l7)

Logarithms, invented by John Napier, were the single most important improvement in arithmetic calculation before the modern electronic computer. What made them 80 useful is that the properties oflogarithms reduce multiplication of positive numbers to addition of their logarithms, division of positive numbers to subtraction of their logarithms, and exponentiation of a number to multiplying its logarithm by the exponent.

THEOREM 2-Algebralc Properties of the Natural Logarithm For any numbers > 0 and x > 0, the natural logarithm satisfies the following rules:

b

1. Product Rule:

Inbx = Inb + Inx

2. Quotient Rule:

b Inx = Inb - Inx

3. Reciprocal Rule:

I Inx = -Inx

Rule 2 with b

4. Power Rule:

Inx T = rlnx

For r rational

=

1

For now we consider ouly rational exponents in Rule 4. In Section 7.3 we will see that the rule bolds for all real exponents as well.

EXAMPLE 2 (a) 1n4 + In sin x = In (4 sin x)

Product

x + I (b) 1n2x _ 3 = In(x + I) - 1n(2x - 3)

Quotient

(e) Int = -inS

Reciprocal

= -In23 = -3ln2

•

Power

We now give the proof of Theorem 2. The properties are proved by applying Corollary 2 of the Mean Value Theorem to each of them. Proof that In bx = In b + In x have the same derivative:

The argument starts by observing that In bx and In x

d bid dx In (bx) = bx = x = dx In x. According to Corollary 2 of the Mean Value Theorem, the functions must differ by a constant, which means that

Inbx = Inx + C for some constant C. Since this last equation holds for all positive values of x, it must hold for x = I. Hence, In(b'l) = Inl + C Inb=O+C

lnl~O

C = Inb.

By substituting we conclude that

Inbx = Inb + Inx.

•

7.2

Proof that Inx'

=

Natural Logarithms

3 73

rlnx (assuming r rational) We use the same-derivative argument

again. For all positive values of x,

l-"-

Eq. (2) with u

=

1 r-l -rx

General Power Rule for derivatives, r rational

=

r·k =

-"- In x T = T (x T ) dx x dx x'

=

x7

fx (rlnx).

Since In x' and r In x have the same derivative, Inx T = rlnx

+C

for some constant C. Taking x to be I identifies C as zero, and we're done. (Exercise 46 in Section 3.7 indicates a proof of the General Power Rule for derivatives when r is rational.) You are asked to prove Rule 2 in Exercise 86. Rule 3 is a special case of Rule 2, obtained by setting b = I and noting that In I = O. This covers all cases of Theorem 2. • y

We have not yet proved Rule 4 for r irrational; however, the rule does hold for all r, rational or irrational. We will show this in the next section after we def'me exponential functions and irrational exponents.

1 1 ~

-+______

______

~~

__

2

O~-------+1--------~2~~x


y

The Graph and Range of In x We displayed the graph of y = Inx in Figure 7.8. Let's verify its properties. The derivative d{lnx)/dx = I/x is positive for x > 0, so In x is an increasing function ofx. The second derivative, -1/x 2 , is negative, so the graph oflnx is concave down. We can estimate the value of In 2 by considering the area under the graph of y = I/x and above the interval [I, 2]. In Figure 7.9a, a rectangle of height 1/2 over the interval [I, 2] fits under the graph. Therefore the area under the graph, which is In 2, is greater than the area, 1/2, of the rectangle. So 1n2 > 1/2. Knowing this we have

y = lnx

1n2n = nln2 ~~~~-------------+x

>

n(!)

<

-n(!)

=

~

and Inr" = -nln2

=

-~.

It follows that (b)

lim Inx =

00

x~OO

FIGURE 7.9 (a) The rectangle of height y = 1/2 fits beoeath the graph of y = I/x for the interval I s x '" 2. (b) The graph of the natural logarithm.

and

lim Inx = x-o+

-00.

We defined In x for x > 0, so the domain of In x is the set of positive real numbers. The above discussion and the Intermediate Value Theorem show that its range is the entire real line giving the graph of y = In x shown in Figure 7.9b.

The Integral

J(1/u) du

Equation (4) leads to the following integral formula.

If u is a differentiable function that is never zero,

Jitdu

=

In

lui + C.

(5)

374

Chapter 7: Transcendental Functions

Equation (5) says that integrals of a certain form lead to logarithms. If u = f(x), then du = t'(x) and

ax

t'(x) f(x)

J

ax =

In If(x) I

+C

whenever f(x) is a differentiable function that is never zero. EXAMPLE 3

Here we recognize an integral of the form

Jti,:'.

1ti,:'=ln u ]-1 ax=1.~x2-5, l l Jo x - 5 .(0) ~ -5, =In I-II-In I-51 =Inl -In5 = -In5 f'-22x

-5

d.
_,

•

The Integrals of tan x, cot x, sec x, and esc x Equation (5) tells us how to integrate these trigonometric functions.

J

tan x ax =

J

sinx eosx ax =

•

-du

--..

-In lui

-In leosxl

+C

=

1n__ 1_ + C = In Iseexl 1cos x 1

+ C.

J

cot x ax =

=

Jeo~x ax J smx

= In lui

+C

=

(seex seex (seex

sec x

In lui

=

+

C

=In Isinxl

+ tanx) + tanx) ax -

J J J-:u =

=

esex

-In lui

= -In lesexl

+C

J

(esex (esex

+

+ eotx) ax + eotx)

C = -In lese

+ C.

+ tanx). 2

see x + seextanx seex + tanx ax u = secx + tanx du = (sec x tan x + se~x)dx

= In Iseex + tanxl + C + eotx).

Foresex, we multiply and divide by (esc x esex ax =

= sinx, = cosx dx

du

U

To integrate sec x, we multiply and divide by (sec x

J ax -J Jti,:' =

Reciprocal Rule

u

du

cou > 000 (-7r/2, 7r/2), = -sinxtix

~

du

=

For the cotangent,

+C

J

=

x+

J

ese2 x + esc x cot x esex + eotx

cot xl

+

C

ax

u = cscx + cotx flu = (-cscxcotx - csc 2 x)dx

Integrals of the tangent, cotangent, secant, and cosecant functions

J J

tanudu = In Iseeul

+C

eotudu = In Isinul + C

J J

seeudu = In Iseeu

+ tanul + C

eseudu = -Inleseu

+ eotul + C

7.2

375

Natural Logarithms

EXAMPLE 4 r~/6

10

r~/3

tan2xdx = =

10

d I r~/3 tanu.; = tanudu

Substitute u

110

I l~/3 Iln Isecul 0

=

I I(ln2 -Inl)

= 2x,

dx ~ du/2, u(O) ~ 0,

=

I Iln2

u('TT/6)

~

'TT/3

•

Logarithmic Differentiation The derivatives of positive functions given by formulas that involve products, quotients, and powers can often be found more quickly if we take the natural logarithm of both sides before differentiating. This enables us to use the laws oflogarithms to simplify the formulas before differentiating. The process, called logarithmic differentiation, is illustrated in the next example.

EXAMPLE 5

Finddy/dxif

y=

(x 2 + I)(x + 3)'/2 x - I

x> I.

Solution We take the natural logarithm of both sides and simplify the result with the properties of logarithms:

(x 2 + I)(x + 3)'/2

In y = In '-------'-'---;----'--x - I = In «x 2

+ I)(x + 3)'/2) - In (x - I)

Rule 2

=

In(x 2 + 1) + In (x + 3)'/2 -In(x - 1)

Rule!

=

In(x 2 + I) + tln(x + 3) -In(x - I).

Rule 4

We then take derivatives of both sides with respect to x, using Equation (2) on the left:

Idy I --= --.2x Ydx x 2 +1

I -I- - -I+ -. 2

x+3

x-I'

Next we solve for dy/dx:

dy (2x I I) dx=y x2 +1 +2x+6-x-1 . Finally, we substitute for y from the original equation: 2 2 dy = (x + I)(x + 3)1/ (~+ _1_ _ _ 1_) dx x-I \;2+1 2x+6 x - I '

•

A direct computation in Example 5, using the Quotient and Product Rules, would be much longer.

Exercises 7.2 Using the Algebraic Properties-Theorem 2 1. Express the following logaritluns in terms of In 2 and In 3. .. In 0.75

b. In (4/9)

c. In (1/2)

d. In V'9

e. In 3 V2

f. In

v'i3.5

2. Express the following logarithms in terms ofln 5 aod In 7.

•• In (1/125)

d. In 1225 f. (In 35

+

b. In 9.8

e. In 0.056 In (1/7))/(1n 25)

c. In 7 0

376

Chapter 7: Transcendental Functions

Use the properties of logarithms to simplify the expressions in Exercises 3 and 4. 3.

L

lnsinO - In

c.

~1n(4t4)

(S~O)

b. In (3x 2 - 9x) + In

47.

(iX)

lnsecO + In cos 0 b. 1n(8x + 4) - 2ln2 c. 31n~ -In(t + I)

Finding Derivatives In Exercises 5-36, find the derivative of y with respect to x, t, or 8, as appropriate.

6. y = In lx, k constant

S. y = In3x

7. y = In (t 3

= In (t 3/2) 10 = InX= In (20 + 2)

8. y

)

9.y=ln x

10. y

11. y = In (0 + I) 13. y = Inx 3

12. y

15. y = I(lnl)'

16. y

= IYmt

18. y

=

X4

19. y

14. y = (lnx)3 x4

-T6

17,y=4 lnx

Int

(x 2 lnx)4

20 =I+lnl • Y I

= -1-

Inx 21. y = I + Inx

xlnx 22. y = I + Inx

23. y = In (lnx)

24. y = In (In (lnx))

I

xvx + I

= I + Int 29 • y I - Inl

31. y = In (sec (In 0))

_ (x + I)')

28

.y

=llnl+x 2 I-x

= Yin vi

30. y

O) = In (Y";in----:-O-co-s-:c .y 1+2lnO

32

2

33. y-In

_~

vl-x

35. y = ['" In vi dl J../2

(x

= In

34. y

(x

+ I)' + 2)20

\Yr 36.y= ['lnldl

JVx

Evaluating Integrals Evaluate the integrals in Exercises 37-54. 2dx 0 3dx 37. -X 38. -1 3x - 2 -3

1

39.

J

secytany 2 + secy
1 cotldt r/ ~/2

J

2y
['" sint 41. Jo 2 - cosl dl

2 cot dO

.

J2Yx

52. Jo

3

~/2

53

12

0

54

dx

+ 2x

.

6 tan 3x dx

JYin

secxdx (sec x + tanx)

Logarithmic Differentiation In Exercises 55-
Y

57. y

= ~I ~

I

58.

59. y =

v'i+3 sinO

61. y

1(1

=

+

1)(1

+ 2)

y=~I(I~ I)

60. y = (tanO)v'29+1 62

• y

=

I(t

I 1)(1

+

0+5 63. y = OcosO

64.y= OsinO

65 =xW+! • y (x + 1)2/3

66. y

3

(x - 2) x 2 +1

+ 2)

~

68. y

(x

= =

+ 1)10

(2x 3

+ I)'

x(x + I)(x - 2) (x 2

+

1)(2x

+ 3)

Theory and Applications 69. Locate and identifY the absolute extreme values of

26. y = In(sccO + tanO) _~

1

~

67.y=

2S. y = O(sin (In 0) + cos (In 0))

27. y=ln

48.

~/4

-In2

L

2

3sec't 6 + 3tanl dl

50. 51.

4.

J

1

40. 42. 44

•

J

8rdr 4r2 - 5

/,o

~/3

l' 2

~

xlnx

1

16

46.

2

4' 0 sm dO 1-4cosO

dx

2x~

•• In (cos x) on [-1T/4,1T/3], b. cos (In x) on [1/2, 2].

70. a. Provethatf(x) = x -Inxisincressingforx b. Using part (a), show that Inx

> 1.

< xifx > 1.

71. Find the area between the curves y = Inx and y = In 2x from x=ltox=5.

72. Find the area between the curve y = tanx and the x-axis from x = -1T/4tox = 1T/3. 73. The region in the first quadrant bounded by the coordinate "'"'s, the line y = 3, and the curve x = 2/vy+! is revolved about

the y-axis to generate a solid. Find the volume of the solid. 74. The region between the curve y = ~ and the x-axis from x = 1T/6 to x = 1T/2 is revolved about the x-axis to generate a solid. Find the volume of the solid. 75. The region between the curve y = l/x2 and the x-axis from x = 1/2 to x = 2 is revolved about the y-axis to generate a solid. Find the volume of the solid.

76. In Section 6.2, Exercise 6, we revolved about the y-axis the region between the curve y = 9x/W+9 and the x-axis from x = 0 to x = 3 to generate a solid of volume 361T. What volume do you get if you revolve the region about the x-axis instead? (See Section 6.2, Exercise 6, for a graph.) 77. Find the lengths of the following curves.

a. y = (x 2/8) - Inx, 4 '" x '" 8 b. x = (y/4), - 21n (y/4), 4 '" y '" 12

7.3 Exponential Functions 78. Find a curve through the point (I, 0) whose length from x = I to x

[~I + :,dx.

L =

D 79.

D 85.

2is

=

I to x

=

=

a. Derive the linearization 10 (1

2. Give the coordinates to

b. Sketcb the region and sbow the centroid in your sketch.

+ lox is one-ro-one.

D 87• ••

84.

d'y

dx' =

I

I

+ x'

D 88.

y(1)

=

3

sec' x, y(O)

=

0

aod y'(O)

+ sinx) for a

= 2,

I

Does the graph of y = Yx - lox, x> 0, have an inflection point? Try to answer this question (a) by graphing, (b) by using calculus.

Having developed the theory of the function In x, we introduce its inverse, the exponential function exp x = eX. We study its properties and compute its derivative and integral. We prove the power rule for derivatives involving general real exponents. Finally, we introduce general exponential functions, aX, and generallogaritbmic functions, log.,x.

y y = In-ix

or 7

Graph y = sinxaod the curves y = In (a 4, 8, 20, and 50 together for 0 ,;; x ,;; 23.

Exponential Functions

7.3 8

=

O.

b. Why do the curves flatten as a increases? (Hint: Find an a-
Solve the initial value problems in Exercises 83 aod 84.

dx =

=

and 4 of Theorem 2, to prove the Quotient Rule property oflogarithms.

81. Use a derivative to sbow that f(x) = 10 (x' - I) is on.,.to-one.

dy

'" x at x

86. Use the same-derivative argument, as was done to prove Rules 1

h. Find the ceoter of mass if, instead of being constant, the densityfimctionisS(x) = 4/Yx. 82. Use a derivative to show that g(x) = v'x'

+ x)

c. Graph 10 (I + x) aodx togcther for 0 ,;; x';; O.5.Use different colors, if available. At what points does the approximation ofln (I + x) seem best? !.cast good? By reading coordinstes from the graphs, find as good ao upper bound for the error as your grapher will allow.

80. .. Find the ceoter of mass of a thin plate of constant deosity covering the region b _ the curve y = I/Yx aod the x-axis from x = ltox = 16.

83.

Tbelinelll'ization orin (1 + x) atx = 0 Instead of approximating In x near x = I, we approximate In (I + x) near x = O. We get a simpler formula this way. b. Estimate to five decimal places the error involved in replacing 10 (I + x) by x on the interval [0, 0.1].

a. Find the ceotroid of the region b _ the curve y = I/x and the x·axis from x two decimal places.

377

X =

lny

6

The Inverse of In x and the Number e The function In x, being an increasing function of x with domain (0, 00) and range (-00,00), has an inverse In-1x with domain (-00,00) and range (0, 00). The graph of In-I x is the graph of In x reflected across the line y = x. As you can see in Figore 7.10, lim In-I x = 00

lim In-I x

and

x_-oo

X~oo

= 0.

The function In-I X is usually denoted as exp x. We now show that exp x is an exponential function with base e. The number e was dermed tu satislY the equation In (e) = 1,80 e = exp (1). We can raise the number e to a rational power r in the usual algebraic way: e2

=

e·e,

e

-2 _

1 e

-2'

e l/2 =

Ye,

and 80 on. Since e is positive, e' is positive too, so we can take the logarithm of e'. When we do, we find that for r rational FIGURE 7.10 Thegraphsofy = lox andy = In-I x = expx. Thenumbereis 10-1 1 = exp (I).

Ine T

=

rlne

=

r-1

=

r.

Theorem 2, Rule 4

Then applying the function In-I to both sides of the equation In e' = r, we find that

e'=expr

forrratioual.

expisln-1.

(l)

We have not yet found a way to give an obvious meaning to eX for x irrational. But In-I X has meaning for any x, rational or irratioual. So Equation (1) provides a way to extend the

378

Chapter 7: Transcendental Functions definition of eX to irrational values of x. The function exp x is defined for all x, so we use it to assign a value to eX at every point.

DEFINITION For every real number x, we define the function to be eX = exp x.

natural

exponential

TypicaL vaLues of Ii'

x

eX (rounded)

-I

0.37 1 2.72 7.39 22026 2.6881 X 1043

0 1 2 10 100

For the first time we bave a precise meaning for an irrational exponent-we are nrising a specific number e to any real power x, rational or irrational. Since the functions In x and eX are inverses of one another, we bave the following relationships.

Inverse Equations for eX and In x

e lnx = x

In(eX)=x EXAMPLE 1

(all x

> 0)

(all x)

Solve the equation e 2x- 6 = 4 for x.

Solution

We take the natorallogarithrn of both sides of the equation and use the second inverse equation:

In (e 2x- 6 )

=

In 4

2x-6=ln4 2x = 6

Invorse relationship

+ 1n4

x = 3

+ Iln4 = 3 + 1n4 ' /2 2

x = 3

+ 1n2

•

The Derivative and Integral of eX According to Theorem I, the natural exponential function is differentiable because it is the inverse of a differentiable function whose derivative is never zero. We calculate its derivative using the inverse relationship and Chain Rule:

In (eX)

!L ax In (eX)

=

x

Invorse relationship

= 1

Differentiate both sides.

:X· :fx(eX)

=

I

Eq. (2), Section 72, with u

:fx eX

=

11'.

Solve for the derivative.

~

e"

That is, for y = eX, we find that dy/ ax = eX so the natural exponential function eX is its own derivative. Moreover, if I(x) = eX, then f' (0) = eO = I. This means that the natoral exponential function eX bas slope I as it crosses the y-axis at x = o. The Chain Rule extends the derivative result for the natural exponential function to a more general form involving a function u(x):

If u is any differentiable function of x, then

d u_ udu

axe-eax·

(2)

7.3 Exponential Functions

EXAMPLE 2

379

We fmd derivatives of the exponential using Equation (2).

(a) iL(5e X ) = 5 iL ex = 5e X dx dx (b) iLe-x = e-xiL(-x) = e-C«-I) = -e-C< dx dx

d· dx

. d eSlDX-(sinx) dx

(c)

_ e smx =

(d)

~(e'hX+I)

=

=

. eSlDX·cosx

e,hx+l. ~(v3x+l)

Eq. (2) with.

~

-x

Eq. (2) with.

~

sinx

Eq. (2) with. ~ ~

• Since eX is its own derivative, it is also its own antiderivative. So the integral equivalent of Equation (2) is the following.

The general antiderivative ofthe exponential function

J

eUdu

EXAMPLE 3 (a)

l

ln2

o

e 3x dx

=

lin' 0

1 eU'-du

3

• ~ 3>;,

eU + C

1 3d.

~

ax,

.(0)

~

0,

.(ln2) ~ 3ln2 ~ 1n2' ~ inS

rm' e du

=

1 3}o

=

leu]In' 3

=

U

0

=t(S-I)=!

(b)

Jo{"/2 esinxcosxdx = esinx ]"/2 0 = e ' - eO =

Antiderivative from Example 2c

•

e - 1

The derivative of eX exists and is everywhere positive, confmning that it is a continuous and increasing function as shown in Figure 7.10. Since the second derivative of eX is also eX and everywhere positive, the graph is concave up. Moreover, Figure 7.10 shows that the exponential function has the limits

limex=O x--i'-OO

and

lim eX =

00.

X~OO

From the flIst of these limits we see that the x-axis is a horizontal asymptote of the graph y = eX.

Laws of Exponents Even though eX is defmed in a seemingly roundabout way as In-I x, it obeys the familiar laws of exponents ftmn algebra. The following Theorem 3 shows us that these laws are consequences of the defmitions of In x and eX.

380

Chapter 7: Transcendental Functions

THEOREM 3

For all numbers x, the following laws:

X" and X2, the natoral exponential eX obeys 2. e -x

I

= eX

4. (e X' )'

=

e rx1 , ifr is mtional

Proof of Law 1 Let Yl = eX, and Y2 = eX,. Then

Xl = !nYl Xl

+ X2

X2 = !nY2

and

Inverse equations

+ !nY2

=

!nYl

=

!nY1Y2

Product Rule for logarithms

Exponentiate.

• Proof of Law 4

Let Y = (e")'. Then

!ny = !n (ex,)' =

r!n(e")

Power Rule for logarithms, mtional r

lne"

= rxl

=

uwithu

= Xl

Thus, exponentiating each side,

• Laws 2 and 3 follow from Law I. Like the Power Rule for logarithms, Law 4 holds for all real numbers r.

The General Exponential Function a" Since a = e lna for any positive number a, we can think of a Xas (e lna )' = eX Ina. We therefore use the function eX to derme the other exponential functions, which allow us to raise any positive number to an irrational exponent.

DEFINITION base a is

For any numbers a

> 0 and X,

the exponential function with

Theorem 3 is also valid for aX, the exponential function with base a. For example, aXle aX2

= eX1lna. eX2lna

Def'mition of a:%

= eXllna+X2lna

Law!

= e(J:l+xz)lna =

a X1 +X2 •

In particular, an. a-I = a n- l for any real number n.

Factor In a Def'mition of a:%

7.3 Exponential Functions

381

Proof of the Power Rule (General Version) The definition of the general exponential function enables us to make sense of raising any positive number to a real power n, rational or irrational. That is, we can define the power function y = x' for any exponent n. For any x > 0 and for any real number n,

DEFINmON

Because the logaritbm and exponential functions are inverses of each other, the definition gives In x' = n In x,

for all real numbers n.

That is, the Power Rule for the naturallogaritbm holds for all real exponents n, not just for rational exponents as previously stated in Theorem 2. The definition of the power function also enables us to establish the derivative Power Rule for any real power n, as stated in Section 3.3. General Power Rule for Derivatives For x > 0 and any real number n,

If x :5 0, then the formula holds whenever the derivative, x', and x·- 1 all exist.

Proof Differentiating x' with respect to x gives Definition of x", x > 0

Chain Rule for e", Eq. (2)

=

In short, whenever x

For x < 0, if y

x"·!! x

Definition and derivative of 1n x

> 0,

=

x n, y' , and X,,-l all exist, then

lnlyl

=

lnlxl'

n lnlx!-

=

Using implicit differentiation (which assumes the existence of the derivative y') and Equation (4) in Section 7.2, we have

Y' y

'!

X'

Solving for the derivative, y'

=

y

nX-

=

x'

nx

=

nx n- t .

382

Chapter 7: Transcendental Functions It can be shown directly from the defInition of the derivative that the derivative equals 0 wben x = 0 and n ;;,: I. This completes the proof of the general version of the Power Rule for all values of x. •

EXAMPLE 4

> O.

Differentiate /(x) = xx, x

Solution We cannot apply the power rule here because the exponent is the variable x mther than being a constant value n (mtional or irrational). However, from the def"mition of the general exponential function we note !bat /(x) = XX = exlnx , and differentiation gives /,(x) =

~ (e xInX ) dx

=

e xlnx ~ (x In x) dx

=

e Xlnx ( Inx

=

xX(lnx + I).

Eq. (2) with u

~

.In.

+ x.:})

•

.>0

The Number e Expressed as a Limit We have def"med the number e as the number for which In e = I, or equivalently, the value exp (I). We see that e is an important constant for the logarithmic and exponential functions, but what is its numerical value? The next theorem shows one way to calculate e as a limit. The number e can be calculated as the

THEOREM 4-The Number. as a Um;t

limit e = lim (I • -0

Proof If /(x)

=

+ x)l/x .

Inx, then /,(x) = I/x, so /,(1) = 1. But, by the def"mition of derivative,

/,(1) = lim /(1 h-O

=

lim In (I

x-o =

lim In (I

x-o

+ h) - /(1)

=

lim /(1

h

x-o

+ x)

- In I = lim

x-o

X

+ x)l/x

=

In [lim(1

x-o

+ x)

- /(1)

x

lin (I + x)

Inl~O

X

+ x)l/x]

In is continuous;

use Theorem lOin Chapter 2.

Because /,(1) = I, we have In [lim(1

.-0

+ X)I/x]

=

I

Therefore, exponentiating both sides we get lim (I + x)l/x .-0

=

e.

•

Approximating the limit in Theorem 4 by taking x very small gives approximations to e. Its value is e '" 2.718281828459045 to 15 decimal places as noted before.

The Derivative of aU To f"md tbis derivative, we start with the defIning equation aX = exlna . Then we have

383

7.3 Exponential Functions

We now see why eX is the exponential function preferred in calculus. If a = e, then In a = I and the derivative of aX simplifies to

With the Chain Rule, we get the following form for the derivative of the general exponential function.

If a > 0 and u is a differentiable function of x, then a" is a differentiable function ofx and

%x a"

= a"lna :,.

(3)

The integral equivalent of this last result gives the general antiderivative

y

=

lOx y

=

3x

y~ F ,....-----'--

-1

J

(4)

From Equation (3) with u = x, we see that the derivative of aX is positive if In a > 0, or a > I, and negative if In a < 0, or 0 < a < I. Thus, aX is an increasing function of x if a > I and a decreasing function of x if 0 < a < I. In each case, aX is one-to-one. The second derivative d2 d -(aX) = -(axIna) = (InaJ2a x dx 2 dx

1

FIGURE 7.11 Exponential functions decrease if 0 < a < 1 and increase if a> 1.Asx---+00,wehaveax---+Oif O 1.As x---+ -00, we have aX ---+ 00 if 0 < a < 1 and aX ---+ 0 if a > 1.

a" + C.

a"du = Ina

=2x

is positive for all x, so the graph of aX is concave up on every interval of the real line. Figure 7.11 displays the graphs of several exponential functions.

EXAMPLE 5 (a)

~3x

(b)

~rx

dx dx

We fmd derivatives and integrals using Equations (3) and (4).

= 3x ln3

=

rX(ln3)~(-x) dx

=

-rxln3

d· . d . (e) dx 30nx = 3,mx(ln3) dx (sinx) = 3,mx(ln3) cosx

(d) (e)

J J

2 x dx =

~x2 + C

2 .... cosxdx =

Eq.(3)wi1ha

~ 3,u ~-x

... ,u

~

2"du =

2 sinx

= In2

+

C

~"2

+

C

u

~

x

~

,inx

Eq.(4)wi1ha

J

3,u

Eq.(3)wi1ha

~ 2,u ~ x

~ sinx,du ~

COSXtU,

u replaced by sin x

andEq. (4)

•

Logarithms with Base Q If a is any positive number other than I, the function aX is one-to-one and has a nonzero derivative at every point. It therefore has a differentiable inverse. We call the inverse the logarithm ofx with base a and denote it by log.x.

384

Chapter 7: Transcendental Functions

DEFINITION

For any positive number a

'¢'

I,

log"x is the inverse function of aX.

~=-----;;of-,f-c:.~~~~-~x

The graph of y = log"x can be obtained by reflecting the graph of y = aX across the 45° line y = x (Figure 7.12). When a = e, we have logex = inverse of eX = Inx. (The function 10glO x is sometimes written simply as log x and is called the common logarithm of x.) Since log. x and aX are inverses of one another, composing them in either order gives the identity function.

FIGURE 7.12 Thegraphof2x andits inverse, log2 x.

Inverse Equations for aX and log.x

a log.,x = x

(x> 0)

log" (aX) = x

(all x)

The function log" x is actually just a nmnerical multiple of In x. To see this, we let Y = log"x and then take the natural logarithm of both sides of the equivalent equation

a'

=

x to obtain Y In a

= In x. Solving for y gives

Inx log"x = Ina'

TABLE 7.2

Rules for base a

Logarithms For any nmnbers x

> 0 and

y> 0, 1.

2.

3.

Product Rule: log" xy = log" x

+ log" y

The algebraic rules satisfied by log. x are the same as the ones for In x. These rules, given in Table 7.2, can be proved using Equation (5) by dividing the corresponding rules for the natural logarithm function by In a. For example, Inxy = Inx

+ Iny

Inxy Inx Ina = Ina

+

log" xy = log" x

Iny Ina

+

log" y.

Rule 1 for natura110garitbms ... . .. djvided by In a ...

. .. gives Rule 1 for base a logarithms.

Quotient Rule: log" ~ = log" x - log" Y

Derivatives and Integrals Involving log. x

Reciprocal Rule:

To f"md derivatives or integrals involving base a logarithms, we convert them to natural logarithms. If u is a positive differentiable function of x, then

I log" y = -log" Y 4.

(5)

Power Rule: log" x' = Y log" x

d d (lnu) I d I I du tU (Iog"u) = tU Ina = Ina tU (lnu) = lna'll tU'

d I I du -(Iog"u) = -.-tU In a u tU

7.3

Exponential Functions

385

EXAMPLE 6 d

(a) dx log1O(3x

+

lid

1) = In 10 • 3x

+

r°'6)X dx = ~2J In/ dx

(b)

=

~2

J

U

I dx (3x

u

3

+

1) = (In 1O)(3x

1)

~ :;

log,x

du

+

~ lnx,

du

~ ~tU

I u2 I (lnxj2 (lnxj2 =ln2T+ C =ln2-2-+ C = 2ln2 +C

•

Exercises 7.3 Solving Exponential Equations

35.

In Exercises 1-4, solve for I.

1. a.

e--()·3t

=

~

2. a. e.....011

b.e kt

27

1000

b. e

tt

I

=-

c.

e(lnO.2)t =

~

c.

e(Jn2)'

2

/0

~

0.4

t

37.

39. 41.

Finding Derivatives In Exercises 5--24, rmd the derivative of y with respect to x, I, or 8, as appropriate. 6. Y ~ e2>/3 5. y ~ 8. y = e(4v'i+x1) 7. y = e5 - 7:r:

e-"

10. y ~ (1

9.y=xe x -e x 11. Y ~ (x 2

-

2x + 2)e'

13. y ~ e'(sin 8 + cos 8) 15. Y ~ cos (e-6')

+ 2x)e-2> - fu + 2)e"

16. y ~

~ In (3te-l)

18. Y

~ In (2e-' sin I)

19. Y

~ In (----"'-) 1 + eO

20. y

~ inC :~)

21. y

= e(coIHlnt)

22. y ~ e""(InI 2 + I)

23. y

~ /,'.. sin e' dl

In Exercises 25--28, find 25. Iny

~

eY sinx

27. e2> ~ sin (x

+ 3y)

24. Y

~

l.

Inldl

26. In xy = e X +Y 28. !any

~

e'

+ Inx

Finding Integrals

31.

f l f

(e"

+ 5e~) fix

30.

8e«+I) fix

J-1n2

34.

f

38'] e;;

21 e-" dt

40.

f V/r f f e'l'

42.

7fix

44.

f f e~o(~+t) l f

dr

I'e(") dl e-II" ---;>fix

J.~j2 (1 + e""') csc2 8 d8

e sec frt sec 7ft tan wI dt

esc (."

+ I) cot (." + I) dl

2evcosevdv

48.

In ('11'/6)

49.

f f

fix

~/4

ln(wj2)

47.

44

/,v.;;,:

2 2 2xeX cos(eX )dx

0

e' I + e' dr

50.

f

fix I + e'

Initial Value Problems Solve the initial value problems in Exercises 51-54.

51. :;;

~ e'sin(e' -

52. :;;

~ e-lsec2 (."e-'),

d 2y

53. fix2 = 2e-',

d 2y

-2

dl

2), y(ln2)

0

y(ln4) = 2/."

y(O) = I

= I - e 2t ,

~

aod y'(O) ~ 0

y(l) = -I

aod y'(I) ~ 0

(2e' - 3e -2» fix

32. ('

1n2

33.

f

in,

e' fix

46.

54.

Evaluate the integrals in Exercises 2!1-50.

29.

45.

e

eV;

'b

e4Vi.

dy/ fix.

lln4

r /4 (1 + e""') see> 8 d8

lI'e-29 cos 58

17. y

36. 10

43. 10

12. y ~ (9x 2

14. y ~ In (38e-')

[m 16

{In' e,j2 fix

e~ fix

2e (2)-1) fix

Differentiation In Exercises 55-112, rmd the derivative of Y with respect to the given independent variable. 55.y~2'

56.y=3~

57. Y ~ 5V;

58. Y =

2(")

386

Chapter 7: Transcendental Functions Logarithmic Differentiation

59. y = x"

60. y = t l -

61. y = (cos8)v2

62. y = (ln8)"

63. y = 7sec 81n 7 65. y = 2sin3t

64. y = 3tan 8 In 3

67. y = IOg258

68. y = IOg3(l

66. y =

69. y = IOg4x + IOg4x2 71. y = x 3 10g lO x 73. y = log3 (

(~ ~

U

l n3)

e

In Exercises 111- 118, use logarithmic differentiation to find the derivative of y with respect to the given independent variable. 111. y = (x + I)X 112. y = x 2 + x2x

5 - cOS 21

+ 81n 3)

113. y = (VI)t

114. y = t Vi 116. y = x sinx 118. y = (lnx)lnx

70. y

=

IOg25 eX - IOg5vX

115. y = (sinx)X

72. y

=

log3 r· log9 r

117. y = sinxx

r-;----,--;-~

(3}~ 2 r 5

74. y = IOg5

Theory and Applications 119. Find the absolute maximum and minimum values of f(x) = eX - 2x on [0, I].

75. y = 8 sin (lOg7 8)

76

77. y

858 78. y = 2 - IOg5 8

10glO eX

=

= I

(sin8cos8) Og7 e 8 28

• y

79. y = 310g ,t

80. y = 3 logg (lOg2 t)

81. y = IOg2 (8t ln2 )

82. y = t IOg3 (e(Sin t)(ln 3))

120. Where does the periodic function f(x) = 2e sin (x/2) take on its extreme values and what are these values? y

~,

Integration Evaluate the integrals in Exercises 83- 92.

83.

J

SX dx

84.

J ~x 3

121. Let f(x) = xe- x.

a. Find all absolute extreme values for f.

3x dx

b. Find all inflection points for f. eX 122. Letf(x) = --2x' I +e a. Find all absolute extreme values for f.

(I3')

r,,/4

90.

Jo

92.

J

tan 1

b. Find all inflection points for f. sec2 t dt

where it is assumed.

x2x'

D 124.

---,dx I + 2x

Evaluate the integrals in Exercises 93- 106.

J 95.1 J i

3x v'3 dx

93.

94.

3

(0 + I)x v1dx

10glOx -x-dx

97.

103.

Jo

+ 2) + 2 dx 2 10glO (x + I) x + I dx

105.

x

r

-x-dx

126. Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = ex / 2 , below by the curve y = e - x/2 , and on the right by the line x = 2 In 2 .

21n 10 log 10 X x dx

127. Find a curve through the origin in the xy-plane whose length from x = 0 to x = I is

i i 1 1 J I

r210g2 (x

Jo

J

10

102.

log 10 (lOx) x dx

L =

11)1 +

:teXdx .

1/10

104.

3 2 IOg2 (x - I) x _ I dx

2

106.

dx x log 10 X

Graph f(x) = (x - 3)2e x and its first derivative together. Comment on the behavior of f in relation to the signs and values of f' . Identify significant points on the graphs with calculus, as necessary.

125. Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = e 2x , below by the curve y = eX, and on the right by the line x = In 3 .

410g2X e

100.

I

101.

x v2- 1dx

I

41n210g2X x dx

99.

J

96·iexOn2) - ldx

98.

123. Find the absolute maximum value of f(x) = x 2 1n (I/x) and say

128. Find the area of the surface generated by revolving the curve x = (e Y + e- Y )/2,0 ::s y ::s In 2, about the y-axis. eY + e-Y

dx x(logg x)2

X=---

2

Evaluate the integrals in Exercises 107- 110.

107. 109.

i i

lnx

I

I

l /X

I [dt,

x

I [dt,

x > 0

> I

108.

iex I I [dt

I 110. Ina

i

I

X

I [dt,

x> 0

x

387

7.4 Exponential Change and Separable Differential Equations In Exercises 129--132, find the length of each curve. 129. y = t(e X

+ e~)fromx

130. y = In (eX - I) -In(e'

137. Find the area of the region between the curve y = Zr/(l and the interval -2 ~ x ::s; 2 of the x-axis.

= Otox = I

+

I) from x = 1n2tox = 1n3

+ c.

graphing.

o 140. Could

Xln2 possibly be the same as 21n' for two functions and explain what you see.

b. Find the average value oflnx over [I, e].

134. Find the average value of f(x) = I/x on [I, 2]. 135. Thelinearizalion or.x at x = 0

a. Derive the linear approximation eX

~

1

+ x at x

= O.

x

eX

x :5 x :5 colors, if availahle. On what intervals does the approximation appear to overestimate eX? Underestimate eX?

136. The geometric, logarithmic, and arithmetic mean inequality

a. Show that the graph of e' is concave up over every interval of x-values. b. Show, by reference to the accompanying figure, that if 0< a < bthen

e(ln.+lnb)/2'(lnb -Ina) <

J.

lnb

e'dx < e

lnll

+e

lnb

'(lnb -Ina).

2

In.

x > O? Graph the

141. Thelinearizalion ofl'

o b. Estimate to five decimal places the magnitude of the error involved in replacing by I + on the interval [0, 0.2]. o c. Graph and I + together for - 2 2. Use different eX

2

= 2' bas three sclutions: x = 2, x = 4, and one other. Estimate the third solution as accurately as you can by

132. y = In (esc x) from x = 'IT/6 to x = 'IT/4 133. a. ShowthatJlnxdx =xlnx -x

138. Find the area of the region between the curve y = 2 1- x and the interval -1 :5 x :5 1 of the x-axis.

o 139. The equation x

131. y = In (cos x) from x = Otox = 'IT/4

+ x 2)

a. Find the linearization of f(x)

= 2' at x = O. Then round its coefficients to two decin2a1 places.

o b. Graph the linearization and function together for -3:5 3and-l 1. x:5

:5 x:5

142. Thelinearizalion oflog, x

a. Find the linearization of f(x) = log, x at x its coefficients to two decimal places.

=

3. Then round

o b. Graph the linearization and function together in the

window

O:5x:5 8and2:5x :54.

o 143. Which is bigger, 'IT' or e ? W

Calculators have taken scme of the mystery out of this once-challenging question. (Go ahead and check; you will see that it is a surprisingly close call.) You can answer the question without a calculator, though.

&.

Find an equation for the line through the origin tangent to the graph of y = In x.

E

B,

__~A~:Ina____~__~~__~~__~,x Ina+lnb

2

b. Give an argument hased on the grapha of y = In x and the tangent line to explain why In x < x/e for all positive x #' e.

c_ Show that In (x')

NorTOSCALB

c. Use the inequality in part (b) to conclude that

.rc

b-a a+b vab < Inb - Ina < --2--'

This inequality says that the geome1ric mean of two positive numbers is less than their logarithmic mean, which in turn is less than their arithmetic mean.

7.4

[-3,6] by [-3, 3]

d. Conclude that xt'

< x for all positive x #' e. < eX for all positive x =F e.

e. So which is bigger, ",e or e""?

o 144. A decimal representalion of.

Find e to as many decimal places as your calculator allows by sclving the equation Inx = I using Newton's method in Section 4.6.

Exponential Change and Separable Differential Equations Exponential functions increase or decrease very rapidly with changes in the independent variable. They describe growth or decay in many natural and industrial situations. The variety of models based on these functions partly accounts for their importance. We nOW investigate the basic proportionality assumption that leads to such exponential change.

388

Chapter 7: Transcendental Functions

Exponential Change In modeling many real-world situations, a quantity y increases or decreases at a rate proportional to its size at a given time I. Examples of such quantities include the amount of a decaying radioactive material, the size of a population, and the temperature difference between a hot object and its surrounding medium. Such quantities are said to undergo exponential change. If the amount present at time I = 0 is called Yo, then we can find y as a function of t by solving the following initial value problem:

dy dl

Differential equation:

y = Yo

Initial condition:

(la)

= ky

when

t

= O.

(lb)

Ify is positive and increasing, then It is positive, and we use Equation (Ia) to say that the rate of growth is proportional to what has already been accumulated. If y is positive and decreasing, then k is negative, and we use Equation (Ia) to say that the rate of decay is proportional to the amount still left. We see right away that the constant function y = 0 is a solution of Equation (Ia) if Yo = O. To find the nonzero solutions, we divide Equation (Ia) by y: Idy

y"O

y'dl=k

J~:dl=Jkdl In

Iyl Iyl Iyl

=

Integrate with respect to t;

kl + C

= ekt+c

J(1ju) du ~ In lui + C. Exponentiate.

= eC'e kt

= ±eCekt y = Ae kt .

y

If Iyl ~

T, tbeny

~

±T.

A is a shorter name for ±e c ,

By allowing A to take on the value 0 in addition to all possible values ±ec, we can include the solution y = 0 in the formula. We fmd the value of A for the initial value problem by solving for A when y = Yo and

1=0:

Yo

=

Ae k ' O = A.

The solution of the initial value problem is therefore y = yoe kt .

(2)

Quantities changing in this way are said to undergo exponential growth if It > 0, and exponential decay if k < O. The number k is called the rate constant of the change. The derivation of Equation (2) shows also that the only functions that are their own derivatives are constant multiples of the exponential function. Before presenting several examples of exponential change, let's consider the process we used to derive it.

Separable Differential Equations Exponential change is modeled by a differential equation of the form dy/dx = ky for some nonzero constant k. More generally, suppose we have a differential equation of the form

dy dx = f(x,y),

(3)

7.4 Exponential Change and Separable Differential Equations

389

where f is a function of both the independent and dependent variables. A solution of the equation is a differentiable function y = y(x) defined on an interval of x-values (perhaps infmite) such that d

dx y(x) = f(x,y(x»

on that interval. 1bat is, wheny(x) and its derivative y'(x) are substituted into the differential equation, the resulting equation is true for all x in the solution interval. The general solution is a solutiony(x) that contains all possible solutions and it always contains an arlJitrary constant. Equation (3) is separable if f can be expressed as a product of a function of x and a function of y. The differential equation then has the fonn dy dx = g{x)H(y).

g is a function of x; H is a function of y.

When we rewrite this equation in the form dy dx

g(x) h(y) ,

I H(y) ~ hey)

its differential fonn allows us to collect all y terms with dy and all x terms with dx: h(y) dy = g(x) dx.

Now we simply integrate both sides of this equation:

J

h(y) dy =

J

(4)

g(x) dx.

After completing the integrations we obtain the solution y defmed implicitly as a function

ofx. The justification that we can simply integrate both sides in Equation (4) is based on the Substitution Rule (Section 5.5):

J

h(y) dy =

=

=

EXAMPLE 1

J J J

h(y(x)) : dx g(x) h(y(x)) h(y(x)) dx

ely

g(x)


g(x) dx.

Solve the differential equation dy dx = (I + y)e", y

Solution Since I the variables.

+ y is never zero for y >

-I, we can solve the equation by separating

dy = (I dx

-

dy

=

> -\.

+ y)e"

(I + y)e" dx

~=e"dx 1+ Y 1+ y J ~=Je"dx In(i+y)=e"+C The last equation gives y as an implicit function of x.

Treat dy/dx as a quotient of diffi:rentials and multiply bolh sides by
+ y).

Integrate bolh sides. C represents the combined

constants of integration.

•

390

Chapter 7: Transcendental Functions

EXAMPLE 2 Solution

Solve the equation y(x

+

1):

=

X(y2 + I).

We change to differential form, separate the variables, and integrate:

y(x

+ I)dy

=

ydy y2+1

X(y2

+ I)dx

xdx x+1

J/+~2 J(I =

tln(1

x =F -1

x

~ I) dx

Dividexbyx + 1.

+ y2) =x -lnlx + II + C.

The last equation gives the solution y as an implicit function of x.

•

The initial value problem

dy dt

= ky,

y(O) = Yo

involves a separable differential equation, and the solution y = yoe kt expresses exponential change. We now present several examples of such change.

Unlimited Population Growth Strictly speaking, the nwnber of individuals in a population (of people, plants, animals, or bacteria, for example) is a discontinuous function of time because it takes on discrete values. However, when the nwnber of individuals becomes large enough, the population can be approximated by a continuous function. Differentiability of the approximating function is another reasonable hypothesis in many settings, allowing for the use of calculus to model and predict population sizes. If we assume that the proportion of reproducing individuals remains constant and asswne a constant fertility, then at any instant t the birth rate is proportional to the nwnber y(t) of individuals present. Let's asswne, too, that the death rate of the population is stable and proportioual to y(t). If, further, we neglect departures and arrivals, the growth rate dyjdt is the birth rate minus the death rate, which is the difference of the two proportioualities under our asswnptions. In other words, dy j dt = ky so that y = yoe kt , where Yo is the size of the population at time t = O. As with all kinds of growth, there may be limitations imposed by the surrounding environment, but we will not go into these here. The proportiouality dyj dt = ky models unlimited population growth. In the following example we asswne this population model to look at how the nwnber of individuals infected by a disease within a given population decreases as the disease is appropriately treated.

EXAMPLE 3 One model for the way diseases die out when properly treated asswnes that the rate dy j dt at which the nwnber of infected people changes is proportioual to the nwnber y. The nwnber of people cured is proportioual to the nwnber y that are infected with the disease. Suppose that in the course of any given year the nwnber of cases of a disease is reduced by 20%. If there are 10,000 cases today, how many years will it take to reduce the nwnber to 1000? We use the equation y = yoe kt • There are three things to find: the value of Yo, the value of k, and the time t when y = 1000. The value of Yo. We are free to count time beginning anywhere we want. If we count fromtoday,theny = 10,000 when t = O,soYo = 10,000. Our equation is now

Solution

(5)

7.4 Exponential Change and Separable Differential Equations

391

The value of It. When t = I year, the number of cases will be SO% of its present value, or SOOO. Hence,

SOoo = 1O,000ek(1)

ek

= O.S

In (e k ) = In O.S

k= 1n0.S

<

Eq. (5) wilh 1= 1 and y = 8000 Logs of both sides

O.

At any given time t, y = 10,000e(lno,')t.

(6)

The value oft that makes y = 1000. We set y equal to 1000 in Equation (6) and solve for t:

1000 = 1O,000e (In 0.')1 e(lno,')1 =

0.1

(lnO.S)t = 1n0.1

t

= :

~:~

Logs of both sides

'" 10.32 years.

It will take a little more than I 0 years to reduce the number of cases to 1000.

•

Radioactivity

I

For radon-222 gas, t is measured in days and k = 0.18. For radiwn-226, which used to be painted on watch dials to make them glow at night (a dangerous practice), t is measured in years and k = 4.3 X 10-4.

Some atoms are unstable and can spontaneously emit mass or radiation. This process is called radioactive decay, and an element whose atoms go spontaneously through this process is called radioactive. Sometimes when an atom emits some of its mass through this process of radioactivity, the remainder of the atom re-forms to make an atom of some new element. For example, radioactive carbon-14 decays into nitrogen; radium, through a number of intermediate radioactive steps, decays into lead. Experiments have shown that at any given time the rate at which a radioactive element decays (as measured by the number of nuclei that change per unit time) is approximately proportional to the number of radioactive nuclei present. Thus, the decay of a radioactive element is described by the equation dy/ dt = -Tty, It > O. It is conventional to use -k, with k > 0, to emphasize thaty is decreasing. If Yo is the number of radioactive nuclei present at time zero, the number still present at any later time t will be

It> O. The half-life of a radioactive element is the time required for half of the radioactive nuclei present in a sample to decay. It is an interesting fact that the half-life is a constant that does not depend on the number of radioactive nuclei initially present in the sample, but only on the radioactive substance. To see wby, let Yo be the number of radioactive nuclei initially present in the sample. Then the number ypresent at any later time twill be y = yoe-kt • We seek the value oft at which the number of radioactive nuclei present equals half the original number: yoe-kl = !YO 2

e-kl

=

!

2

-Itt = In! = -In2 2

Reciprocal Rule for logarithms

392

Chapter 7: Transcendental Functions

Half-life =

~2

(7)

For example, the half-life for radon-222 is half-life =

~l~

"" 3.9 days.

EXAMPLE 4 The decay of radioactive elements can sometimes be used to date events from the Earth's past. In a living organism, the ratio of radioactive carbon, carbon-14, to ordinary carbon stays fairly constant during the lifetime of the organism, being approximately equal to the ratio in the organism's atmosphere at the time. After the organism's death, however, no new carbon is ingested, and the proportion of carbon-14 in the organism's remains decreases as the catbon-14 decays. Scientists who do carbon-14 dating use a figure of 5700 years for its half-life. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed. Solution We use the decay equation y = yoe -11. There are two things to find: the value of k and the value oftwheny is 0.9yo (90% of the radioactive nuclei are still present). That is,findtwhenYoe-1I = 0.9Yo,ore-kt = 0.9. The value ofk. We use the half-life Equation (7):

ln2 k = half-life

ln2 5700

(about 1.2 X 10-4)

The value oft that makes e-1I = 0.9.

e-kt = 0.9 e -(1n2/5700)t = 0.9 ln2 - 5700 t = lnO.9

t

Logs of both sides

570~~ 0.9 "" 866 years

=

The sample is about 866 years old.

•

Heat Transfer: Newton's Law of Cooling Hot soup left in a tin cup cools to the temperatore of the surrounding air. A hot silver bar immersed in a large tub of water cools to the temperatore of the surrounding water. In situations like these, the rate at which an object's temperatore is changing at any given time is roughly proportioual to the difference between its temperatore and the temperature of the surrounding medium. This observation is called Newton s Law of Cooling, although it applies to warming as well. If H is the temperature of the object at time t and Hs is the constant surrounding temperatore, then the differential equation is dH at

=

-k(H - Hsl·

(8)

7.4 Exponential Change and Separable Differential Equations

393

If we substitute y for (H - Hs ), then dy d dH dt = dt (H - Hs) = dt

-

d dt (Hs)

dH _ 0 dt dH dt = -k(H- Hs) = -kyo =

Hs is a constant.

Eq. (8) H-Hs~Y

Now we know that the solution of dyjdt = -ky is Y = yoe-kt , where y(O) = Yo. Substituting (H - Hs) for y, this says that

H - Hs = (Ho - Hs)e-kt ,

(9)

where Ho is the temperature at t = O. This equation is the solution to Newton's Law of Cooling.

EXAMPLE 5 A hard-boiled egg at 98°C is put in a sink of 18°C water. After 5 min, the egg's temperature is 38°C. Assuming that the water has not warmed appreciably, how much longer will it take the egg to reach 20°C? Solution We find how long it would take the egg to cool from 98°C to 20°C and subtract the 5 min that have already elapsed. Using Equation (9) with Hs = 18 and Ho = 98, the egg's temperature t min after it is put in the sink is H = 18

+

(98 - 18)e-kt = 18

+

80e-kt •

To find k, we use the information that H = 38 when t = 5: 38 = 18

e

-5k

-5k k

+ 80e-5k

1

="4 =

1 = -ln4 In4

= tln4 = 0.2ln4

The egg's temperature at time tis H = 18 H= 20: 20 = 18

+

(about 0.28).

80e-(O.2ln4)t. Now rmd the time t when

+ 80e-(O.2ln4)t

80e-(O.2ln4)t = 2 e-(O.2ln4)t = ~ 40

1 -(0.2ln4)t = In 40 = -ln40 ln40 13' t = 0.2ln4 "" nun. The egg's temperature will reach 20°C about 13 min after it is put in the water to cool. Since it took 5 min to reach 38°C, it will take about 8 min more to reach 20°C. •

394

Chapter 7: Transcendental Functions

Exercises 7.4 Verifying Solutions In Exercises 1-4, show that each function y = f(x) is a solution of the accompanying differential equation.

+ 3y = a. y = e" c. y = e~

1. 2y'

e~

b. y = e"

+ Ce-(3/2)x

= y2

2. y'

I

I b. y = - - -

a. y =-X3. Y

+ e-{3/2~

I

ret

XlI Tdt, x 2y'

=

l' v'1+7 ,

4. y =

I

+ xy =

a. If t represents time in years and y represents tooth size, use

I

c. y=-x+C

x+3 eX

+ -2x- y I +x'

= I

c. What will be our descendants' tootlt size 20,000 years from now (as a percentage of our present tootlt size)?

Initial value Problems In Exercises 5--8, show that each function is a solution of the given initial value problem.

Differential

S.y'+y=

I

2

+ 4e 2r

+ Y = -sinx,

Suppose that tlte pressure at sea level is 1013 millibars (about 14.7 pounds per square inch) and that the pressure at an altitode of 20 km is 90 millibars.

y(-In2) = "-

y = e-'tan-' (2e')

a. Solve the initial value problen3

y(2) = 0

y = (x - 2)e-r'

cosx

y(f) = 0

y=-x-

y(e) = e

x y = lnx

(k a constant) h= 0

when

c. At what altitode does tlte pressure equal 900 millibars?

25. Fint-order chemical reactions In some chemical reactions, the rate at which the amount of a substance changes with time is

Separable Differential Equations Solve the differential equation in Exercises ~22.

• r- dy _ 9. 2vxy fix -I,

x,y > 0

dy = '-Y 11'fix e 13.

~=

15.

Vx ~ =

2

vY cos vY

eY+Y.;,

.,.-------,

17. fix - 2x V I - y-,

-I

dy

12.

~=

14.

~~ =

> 0 16.

x

proportional to the amount present For the chaoge of B-glucono lactooe into gluconic acid, for example,

dy _ 2. r 10. fix -x vy, y>O

dt

2

3x e -:Y

(&Ccx)

~

19. y2 ~ = 3x"y3 - 6>;2

~ = yer' + 2vY er'

I

dy

20. fix=xy+3x-2y-6

27. Working underwater The intensity L(x) oflightx feet beneatlt the surface of the ocean satisfies the differential equation

tiL = eX - Y

+ eX +

-0.6y

26. The inversion of sngar The processing of raw sugar has a step called ''inversion'' that changes the sugar's molecular s1ructw'e. Once the process has begun, the rate of chaoge of the amount of raw sugar is proportional to the ao30Unt of raw sugar remaining. If 1000 kg ofraw sugar reduces to 800 kg ofraw sugar during the 'IrSt 10 hours, how much raw sugar will remain after aoother 14 hours?

= e Y+';"


22. :

=

when t is measured in hours. If there are 100 grams of B-glucono lactooe present when t = 0, how many grams will be left after tlte first hour?

dy e 2r - y 18. dx = eX+Y

21. }

dp/dh = kp p = Po

b. What is the atmospheric pressure at h = 50 km?

x > I

dy-

Differential equation: Initial condition:

to express p in tenns of h. Determine tlte values of po aod k from tlte given altitude-pressure data.

x>O 8. x"y' = xy - y2,

24. Atmospheric pressure The earth's atmospheric pressme p is often modeled by asson3ing that the rate dp/dh at which p chaoges witlt the altitode h above sea level is proportional to p.

candidate

2

6. y' = e""" - 2xy 7. xy'

Solution

Initial condition

equation

theconditionthaty = 0.99yowhent = 1000 to find the value of k in the equation y = yoe it . Then use this value of k to answer the following questions.

b. In about how many years will humao teeth be 90% of their present size?

3

vT+7dt, y'

23. Human evolution continues The aoalysis of tooth shrinkage by C. Loring Brace and colleagues at the University of Michlgao's Musenm of Anthropology indicates that humao tooth size is continuing to decrease aod that tlte evolutionary process did not come to a halt some 30,000 years ago as many scientists contend In northern Emopeans, for example, tooth size reduction now has a rate of I % per 1000 years.

e-Y

+1

Applications and Examples

D The answers to most of the following exercises are in terms of logarithms aod exponentials. A calculator can be helpful, enabling you to express the aoswers in decimal form.

fix

=

-AL.

As a diver, you know from experience that diving to 18 ft in tlte Caribbean Sea cuts tlte intensity in half. You caonot work witltout artificial light when the intensity falls below on.,.tenth of the surface value. About how deep can you expect to work witltout artificiallight?

7.4 Exponential Change and Separable Differential Equations 28. Voltage in • discharging capacitor Suppose that electricity is draiuiug from a capacitor at a rate that is proportional to the voltage V across its tenninals and that, if t is measured in seconds,

Solve this equation for V, using Vo to denote the value of V when t = O. How long will it take the voltage to drop to 10% of its original value? 29. Chol.ra b.ct.ria Suppose that the bactetia in a colony can grow unchecked, by the law of expouential cbange. The colony starts with I bacterium aod doubles every ba1f-hour. How many bacteria will the colony contain at the end of24 hours? (Under fa-

vorable laboratory conditions, the number of cholera bacteria can double every 30 min. In ao infected petlion, many bacteris are destroyed, but this example helps explain why a person whn feels well in the morning may be dangerously ill by evening.) 30. Growth of bacteria A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are Ml,OOO. How many bacteria were present initially?

dis....

31. Th. incidence of. (Continuation of Example 3.) Suppose that in any given year the number of cases can be reduced by 25% instead of 20%.

a. How long will it take to reduce the number of cases to 1000? b. How long will it take to eradicate the disease, that is, reduce

the number of cases to less than I?

u.s.

32. Th. population The US. Census Bureau keeps a running clock totaling the US. popu1ation. On March 26, 2008, the tota1 was increasing at the rate of I petlion every 13 sec. The population figure for 2:31 P.M. EST on that day was 303,714,725.

a. Assuming exponential growth at a constaot rate, fmd the rate constaot for the population's growth (people per 365-day year).

b. At this rate, what will the US. population be at 2:31 P.M. EST on March 26, 2015? 33. Oil depl.tion Suppose the amount of oil pumped from one ofthe canyon wells in Whittier, California, decreases at the contiouous rate of 10% per year. When will the well's output fall to on.,.fifth

ofits present value? 34. Continuous price discounting To encourage buyers to place 100-unit ordetli, your firm's sales department applies a continuous discouot that makes the unit price a functionp(x) of the number of units x ordered. The discount decreases the price at the rate of $0.0 I per unit ordered. The price per unit for a lOO-unit order is p( 100) = $20.09.

a. Find p(x) by solving the following initial value problem: Differential equation: Initial condition:

dp I dx = -lOOP p( 100)

=

20.09.

b. Find the unit price p(1 0) for a 10-unit order and the uoit price P(90) for a 90-unit order. c. The sales department has asked you to find out if it is discouoting so much that the firm's revenue, r(x) = X' p(x) , will actually be less for a 100-unit order than, say, for a

395

90-unit order. Reassure them by showing that r has its maximum value atx = 100. d. Graph the reveoue function r(x) = xp(x) for 0 :5 x :5 200. 35. Plutonium-239 The balf-life of the plutooium isotope is 24,360 years. If 10 g of plutonium is released into the atmosphere by a nuclear accident, how many years will it take for 80% of the is()tope to decay? 36. Polonium-210 The balf-life of polonium is 139 days, but your sample will not be useful to you after 95% of the radioactive nuclei preseot on the day the sample arrives bas disintegrated. For about how many days after the sample arrives will you be able to

use the polonium? 37. Th. mean life of. radio.ctive nucleus Physicists using the radioactivity equation y = yoe -kt call the number I/k the mean life

of a mdioactive nucleus. The mean life of a radon nucleus is about 1/0.18 = 5.6 days. The meao life ofa carbon-14 nucleus is more than 8000 years. Show that 95% of the radioactive nuclei origiually preseot in a sample will disintegrate within three meao lifetimes, i.e., by time t = 3/k. Thus, the meonlife ofanucleus gives a quick way to eslimste how long the radioactivity of a sample will last. 38. Californiuno-252 What costs $27 million per gram and can be used to treat brain cancer, analyze coal for its sulfur content, and detect explosives in luggage? The answer is californium-252, a radioactive isotope so rare that only 8 g of it bave been made in the western world since its discovery by Gleno Seaborg in 1950. The ba1f-life of the isotope is 2.645 years--Iong enough for a us.,. ful service life and short enough to bave a high radioactivity per unit mass. One microgram of the isotope releases 170 million

neutrons per minute. a. What is the value of k in the decay equation for this isotope?

b. What is the isotope's meao life? (See Exercise 37.) c. How long will it take 95% of a sample's radioactive nuclei to disintegrate? 39. Cooting .oup Suppose that a cup of soup cooled from 90'C to 60°C after 10 min in a room whose temperature was 20D e. Use Newton's law of cooling to aoswer the following questions. a. How much longer would it take the soup to cool to 35'C?

b. Instead ofbeing left to staod in the room, the cop of90'C soup is put in a freezer whose temperature is -15°C. How long will it take the soup to cool from 90'C to 35'C? 40. A beam of unknown temperatore An aluminum beam was brought from the outside cold into a machine shop where the temperature was held at 65'F. After 10 min, the beam warmed to 35'F and after another 10 min it was 50'F. Use Newton's law of cooling to estimate the beam's initial temperature. 41. Surrounding medium of unknown temperatore A pan of warm water (46'C) was put in a refiigerator. Ten minutes later, the water's temperature was 39'C; 10 min after that, it was 33'C. Use Newton's law of cooling to estimate how cold the refiigerator was. 42. SiIv.r cooling in air The temperature of an ingot of silver is 60'C above room temperature right now. Twenty minutes ago, it

was 70D e above room temperature. How far above room temperature will the silver be

•• 15 min from now?

b. 2 hours from now?

c. When will the silver be lODe above room temperature?

396

Chapter 7: Transcendental Functions

43. The age of Crater Lake The charcoal from a tree killed in the volcanic eruption that formed Crater Lake in Oregon cootained 44.5% of the carbon-14 found in living matter. About how old is Crater Lake? 44. The sensitivity of carbon-14 dating to measurement To see the effect of a relatively sma1l error in the estimate of the amount of carbon-14 in a sample being dated, consider this hypothetical situation: L A fossilized bone found in centrallllinois in the year A.D. 2000 cootains 17% ofits original carbon-14 cootent. Estimate the year the animal died.

45. Carbon-14 The oldest known frozen humao mummy, discovered in the Schoalslal glacier of the Italian Alps in 1991 and called Olzi, was found wearing straw shoes and a leather coal with goat fur, and holdiog a copper ax and stone dagger. It was estimated that Otzi died 5000 years before he was discovered in the melting glacier. How much of the originsl carboo-14 remained in Otzi al the time of his discovery? 46. Art forgery A painting attributed to Vermeer (1632-1675), which should cootain no more than 96.2% of its originsl carboo14, contains 99.5% instead. Aboulhow old is the forgery?

b. Repeat part (a) assuming 18% instead of 17%.

c. Repeat part (a) assuming 16% instead of 17%.

7.5 HISTORICAL

Indeterminate Forms and L'Hopital's Rule

BIOGRAPHY

Guillaume Fran~ois Antoine de I'H6pital (\661-1704) Johano Bernoulli (1667-1748)

John (Joharm) Bernoulli discovered a rule using derivatives to calculate limits of fractions whose numerators and denominators both approach zero or + 00 . The rule is known today as I'HopitaI'. Rule, after Guillaume de I'HOpitai. He was a French nobleman who wrote the first introductory differential calculus text, where the rule 1mt appeared in print. Limits involving transcendental functions often require some use of the rule for their calculation.

Indeterminate Form 0/0 If we want to know how the function

F(x) = x - sinx x3 behaves near x = 0 (where it is undefined), we can examine the limit of F(x) as x --+ O. We carmol apply the Quotient Rule for limits (Theorem I of Chapter 2) because the limit of the denominator is O. Moreover, in this case, both the numerator and denominator approach 0, and % is unde1med. Such limits may or may not exist in general, but the limit does exist for the function F(x) under discussion by applying I'HOpital's Rule, as we will see in Example I d. If the continuous functions f(x) andg(x) are both zero at x = a, then lim f(x) x~a

g(x)

cannot be found by substitoting x = a. The substitotion produces 0/0, a meaningless expression, which we carmot evaluate. We use % as a notation for an expression known as an indeterminate form. Other meaningless expressions often occur, such as 00/00, 00 00, 0, 00 - 00, 0°, and 1 , which cannot be evaluated in a consistent way; these are called indeterminate forms as well. Sometimes, but not always, limits that lead to indetermiuate forms may be found by cancellation, rearrangement of terms, or other algebraic manipulations. This was our experience in Chapter 2. It took considerable analysis in Section 2.4 to find limx~o (sinx)/x. But we have had success with the limit

'( ) f a

=

Ii

m

x~a

f(x) - f(a) x a '

7.5

Indeterminate Forms and rH6pitars Rule

397

from which we calculate derivatives and which produces the indetenninant fann % when we substitute x = a. I:Hopitai's Rule enables us tu draw on our success with derivatives to evaluate limits that otherwise lead to indetenninate forms.

THEOREM 5- L'H6Jrital's Rule Suppose 1hatj(a) = g(a) = 0, thatj and g are differentiable on an open interval I containing a, and that g' (x) # 0 on !if x # a. Then lim j(x) g(x)

x~a

=

lim f'(x) x~a

g'(x) ,

assuming that the limit on the right side of this equation exists. We give a proof of Theorem 5 at the end of this section.

I

Caution

To apply I'H6pital's RnIe to Ilg, divide 1he derivative of I by 1he derivative of g. Do not fall into 1he trap of taking 1he detivative of IIg. The quotient to use is ng', not (f!g)'.

EXAMPLE 1 The following limits involve % indeterminate forms, so we apply I'HOpitai's Rule. In some cases, it must be applied repeatedly. (8) lim 3x - sinx = lim 3 - cosx = 3 - cosxl = Z x-o X x-o 1 1 x=o

1 (b) lim

x-o

(c) lim

yIJ+x - I

=

x

yIJ+x -

lim

x-o

zyIJ+x 1

I 2

1 - x/2

o

2

x-o

o

X

=

=

xr'/ x-o 2x . -(1/4)(1 + xr lim lim (I/Z)(I +

2

-

32 /

x-a

2

I/Z

Still ~; differentiate again.

I 8

=--

o

(d) lim x - sinx

x-o

Not~; limit is found

o

x3

=

lim 1 - cosx

x-o =

Still Q

o

3x 2

lim sinx

x-o 6x

Not~;limitisfound Here is a summary of the procedure we followed in Example 1. Using UHopital's Rule Tofmd

lim j(x) x-a g(x)

by I'HOpital's Rule, continue to differentiate j and g, so long as we still get the form % at x = a. But as soon as one or the other of these derivatives is different from zero at x = a we stop differentiating. I:HOpitai's Rule does not apply

when either the numemtor or denominator has a fmite nonzero limit.

•

398

Chapter 7: Transcendental Functions

EXAMPLE 2

Be careful to apply I'Hopital's Rule correctly:

o o

lim I - cosx

x-o =

X+X2

. sin x ;~ I + 2x

0

=

T = O.

Not ~; limit is found.

Up to now the calculation is correct, but if we continue to differentiate in an attempt to apply I'HOpital's Rule once more, we get

!

lim cosx =

x-o 2

2'

wbich is not the correct limit. I:HOpital's Rule can ouly be applied to limits that give indeterminate forms, and 0/ I is not an indeterminate form. • I:HOpital's Rule applies to one-sided limits as well.

EXAMPLE 3

In this example the one-sided limits are different.

o o

I Recall that thing.

00

Positive for x > 0

and + 00 mean the same

o

. sin x (b) Iu n -2-

o

x-o- x =

lim cosx x-o- 2x

Indeterminate Forms

=

-00

•

Negative for x < 0

ot:J / ot:J, ot:J •

0,

ot:J -

ot:J

Sometimes when we try to evaluate a limit as x --+ a by substitoting x = a we get an indeterminant form like 00/00, 00·0, or 00 - 00, instead of 0/0. We first consider the form

00/00. In more advanced treatments of calculus it is proved that I'HOpital's Rule applies to the indeterminate form 00/00 as well as to 0/0. Iff(x) --+ ±oo andg(x) --+ ±oo asx --+ a, then lim fix) = lim f'(x) x~.

g(x)

x~.

g'(x)

provided the limit on the right exists. In the notation x --+ a, a may be either finite or infinite. Moreover, x --+ a may be replaced by the one-sided limits x --+ a + or x --+ a -.

EXAMPLE 4 (a)

lim

x~"/2 I

Find the limits of these 00/00 forms: secx

(b) lim

+ tanx

x-OO

lnx

2Vx

• eX (c) lun 2'

x-oo x

Solution
secx

1

+ tanx

00 00

from the left

lim secxtanx x~("/2'- se~ x

lim sinx = I x~("/2)-

399

7.5 Indeterminate Forms and rH6pitars Rule

The right-hand limit is I also, with (-00 )/( -00) as the indeterminate form. Therefore, the two-sided limit is equal to I. I

Vx

• Next we turn our attention to the indetenninate forms 00 • 0 and 00 - 00. Sometimes these forms can be handled by using algebra to convert them to a % or 00/00 form. Here again we do not mean to suggest that 00 • 0 or 00 - 00 is a number. They are only notations for functional behaviors when considering limits. Here are examples of how we might work with these indeterminate forms.

EXAMPLE 5

Find the limits of these

00 • 0

forms:

Solution (8)

lim

x_OO

(x sin i)

=

(-hI sin h) =

lim

h--i>O+

lim sinh = I

QO'O;Leth

h

h--i>O+

00 • 0

=

lim x-O+

=

EXAMPLE 6

lim

x-o+

I/x -1/2x 3/ 2

(-2Vx)

I/x.

converted to

00/00

I'H8pital's Rule

•

= 0

Find the limit of this

00 -

00

lim ( _ I x-a sinx

Solution

~

form:

_1)

x·

Ifx--+ 0+, then sinx--+ 0+ and

I - -~oo I -.smx x

00

Similarly, if x --+ 0-, then sin x --+ 0- and

I - .I - - ---+ smx x

00 -

(-00) = -00

+

00.

Neither form reveals what happens in the limit. To find out, we flfst combine the fractions: I sin x

I

----

x

x - sinx x sin x

Common denominator is x sin x.

Then we apply I'HOpital's Rule to the result: lim

(_._1__ 1) =

x-a smx

x

lim x - .sinx

x-a x smx

= lim

I - cosx

x-a sinx + xcosx sin x = Q= 0 x_o2cosx-xsinx 2 .

= lim

o o Sti1I()

o

•

400

Chapter 7: Transcendental Functions

Indeterminate Powers Limits that lead to the indeterminate fonns 100, 0°, and 00° can sometimes be bandied by first taking the logarithm of the function. We use I'Hllpital's Rule to find the limit of the logarithm expression and then exponentiate the result to find the original function limit. This procedure is justified by the continuity of the exponential function and Theorem lOin Section 2.5, and it is fonnulated as follows. (The fonnula is also valid for one-sided limits.)

Iflimx~a

In j(x) = L, then lim j(x) = lim elnf(x) = eL. x-+a

x-a

Here a may be either finite or infinite.

EXAMPLE 7

Apply I'Hllpital's Rule to show thatlimx~o+ (I

+ x) I/x

= e.

The limit leads to the indeterminate fonn 100. We let j(x) = (I find limx~o+ In j(x). Since

Solution

In j(x) = In (I

+ x)I/X

+ x)I/X

and

i In (1 + x),

=

I'Hllpital's Rule now applies to give

+ x)

lim In j(x) = lim In (I

x-o+

x-o+

X

0 0

I I +x Iim - x-o+ 1

=+=1. Therefore, lim (I x-+o+

EXAMPLE 8

+ x)l/x =

lim j(x)

x-+o+

=

lim elnf~)

x-o+

= el =

e.

•

Find limx~oo xl/x.

Solution

The limit leads to the indeterminate fonn 00°. We let j(x) = xl/x and f"md lim_oo In j(x). Since

~x,

Inj(x) = Inxl/x =

I 'Hllpital's Rule gives lim In j(x) = lim Inx x_OO x

x_OO

=

=

Therefore lim xl/x x-+ 00

=

lim j(x) x-+ 00

=

00

I/x lim-

x-+oo

¥

=

lim elnf(x) x- 00

00

1

o. =

eO

= I.

•

Proof of L'Hiipital's Rule The proof of I'H6pital's Rule is based on Caucby's Mean Value Theorem, an extension of the Mean Value Theorem that involves two functions instead of one. We prove Caucby's Theorem first and then show how it leads to I'H6pital's Rule.

7.5

Indeterminate Forms and rH6pitars Rule

401

HisTORICAL BIOGRAPHY

Augustin-Louis Cauchy (178!1-1857)

THEOREM 6-Cauchy's Mean Value Theorem Suppose functions f and g are continuous on [a, bl and differentiable throughout (a, b) and also suppose g'(x) oF 0 throughout (a, b). Then there exists a number c in (a, b) at which

f'(c) g'(c)

f(b) - f(a) g(b) - g(a)"

Proof We apply the Mean Value Theorem of Section 4.2 twice. First we use it to show that g(a) oF g(b). For ifg(b) did equalg(a), then the Mean Value Theorem would give g'(c)

=

g(b) - g(a) b-a

=

0

for some c between a and b, which cannot happen because g'(x) oF 0 in (a, b). We next apply the Mean Value Theorem to the function

F(x)

=

f(b) - f(a) f(x) - f(a) - g(b) _ g(a) [g(x) - g(a)].

This function is continuous and differentiable where f and g are, and F(b) = F(a) = O. Therefore, there is a number c between a and b for which F' (c) = O. When expressed in terms of f and g, this equation becomes

F'(c)

=

f'(c) -

:~:~ =~~~ [g'(c)1 =

0

so that

f'(c) g'(c)

y j '(c) slope ~,«c) -

--.!

A (g(a).j(a)) ~--------------------~X

o

FIGURE 7.13 There is at least one point P on the curve C for which the slope of the tangent to the curve at P is the same as the slope of the secaot lioe joioing the poiots A(g(a), f(a)) aodB(g(b), f(b)).

f(b) - f(a) g(b) - g(a)"

•

Notice that the Mean Value Theorem in Section 4.2 is Theorem 6 with g(x) = x. Cauchy's Mean Value Theorem has a geometric interpretation for a general winding curve C in the plane joining the two points A = (g(a),/(a)) and B = (g(b), f(b)). In Chapter 11 you willleam how the curve C can be formulated so that there is at least one point P on the curve for which the tangent to the curve at P is parallel to the secant line joining the points A andB. The slope of that tangent line toms out to be the quotient f' /g' evaluated at the number c in the interval (a, b), as asserted in Theorem 6. Because the slope of the secant line joining A andB is

f(b) - f(a) g(b) - g(a)' the equation in Cauchy's Mean Value Theorem says that the slope of the tangent line equals the slope of the secant line. This geometric interpretation is shown in Figure 7.13. Notice from the figure that it is possible for more than one point on the curve C to have a tangent line that is parallel to the secant line joining A and B.

Proof of rHClpital's Rule We flfst establish the limit equation for the case x --> a +. The method needs almost no change to apply to x --> a -, and the combination of these two cases establishes the result. Suppose that x lies to the right of a. Then g'(x) oF 0, and we can apply Cauchy's Mean Value Theorem to the closed interval from a to x. This step produces a number c between a and x such that f'(c) g'(c)

f(x) - f(a) g(x) - g(a)"

402

Chapter 7: Transcendental Functions But f(a) = g(a) = 0, so

j'(c) g'(c)

f(x) g(x)"

As x approaches a, c approaches a because it always lies between a and x. Therefore, lim f(x) = lim j'(c) = lim j'(x)

.r-a+ g(x)

c-a+ g'{c)

.r-a+ g'(x) ,

which establishes I'HOpital's Rule for the case where x approaches a from above. The case where x approaches a from below is proved by applying Cauchy's Mean Value Theorem to the closed interval [x, aj, x < a. •

Exercises 7.5 Finding Limits in Two Ways In Exercises l--{j, use I'HOpital's Ru1e to evaluare the limit. Then evaluare the limil using a method studied in Chapter 2.

. 3x - I 30. lun~1 x-o~

31. lim In(x + I)

33. lim

3. lim 5x' - 3x x_OO 7x 2 + 1

%-0+

35. lim

5. lim I - cosx x-o x 2

8.

t2

1--3

-

+

15 t - 12

11. lim 5x' - 2x x_OO 7x 3 + 3

,

.

13. lim sm 1 1-0

15.

t

8x '

lim=~.

x-o cosX'

lim 20 - 'If · '~w/' cos (2'1f 0)

17

19

lim

I - sinO • 6-1T/2 1 + cos 26

,

10. lim

' 3"'" - I lun 27. ,~o 0

41

r

+ I))

(lnx)'

x~+ In (sinx) lim (_1___1_) x-I Inx

• x-I+

I' - I

x-a

t- 3

71'X

In (esc x) x~w/' (x - ('If/2))'

(3X + I__smX' .1_) X

42. lim (cscx - calx x-o+

e' -

<ex - I)' .

lim 0 - sinO cosO

49

• "~O

lanO - 0

r

SO. x~

sin3x-3x+x 2 sin x sin 2x

Indeterminate Powers and Products Find the limits in Exercise 51--{j6. 51. lim x 1/(1-x) x-I+

53. lim (lnx)l/x x~oo

lim

+ cosx)

t

1 sin 1

x-(n/2)-

"~O

40. lim x-o+

+ ' 45. lim _et_1_

lim

28. lim

%-0+

.

20.lim

1_01 -

,a>O

x-o x smx

r 30 + 'If . '~'!!;/' sin (6 + ('If/3))

26.

y

38. lim (lnx - In sin x)

48. lun

xl

x-? x_tInx - SIn

~-a

43. lim cos 0 - I 8_0efJ - 8-1 1_00

18

(-SInt

(x - ~2) sec x

•

36. lim

y~O

16. lim sinX' - x

24. lim

lim

39

14. lim sin 5/ 1-0 21

/(1 - cos I) 23. lim .

x-(7r/2)-

25 5

12. lim x - 8x' x_OO 12x2 + 5x

22.

25.

+

1-14t 3 -

21 lim x • x~o In (sec x) 1-0

x' -

lim

x--5 x

9. lim I' - 41

y

X~OO

log, x

34. lim In (eX - I) x-o+ Inx

+ 25 - 5

37. lim (In 2x - In(x

Use I'H6pital's rule 10 fmd the limits in Exercises 7-50. 7.lim x - 2 x_2x2 - 4

Y5y

r

• x!'.?o log, (x + 3)

In(x' + 2x) In x

y~O

Applying l'H6pital's Rule

32

lo~x

x_OO

1. lim x+2 x_-2x2 - 4

-

cost

(~2 - x) Ian x

(1/2)" - I 6

57. lim
+ 2x)I/(21u)

58. lim <ex x~o

x~oo

59.

lis x'

x~o

60.

lim

x-o+

+ x)l/>

(I + l)X X

7.5 Indeterminate Forms and rH6pitars Rule

61. lim %_00

63.

6S.

(Xx+- I2)'

62.

2

lim x 1nx

64.

x-o+

lim xtan(" x-o+ 2

X)

66.

lim x_OO

(X' + I )'IX

80. For what values of a and b is

x+2

X

lim (In x)' x-o+

"\I9x+1

D 81.

69.

lim (x-~)

X~OO

hy graphing f(x) val of x-values.

lim cotx • x-o+ escx

ZX - 3x 71. x~3x + 4x

x~oo

D 83.

i' 73• x~xex I· e

- v'x'

(W+i - Yx).

0/0 Form Estimate the value of lim 2x' - (3x

75. Which one is correct, and which one is wrong? Give reasons for

your answers.

+ x over a suitably large inter-

the new numerator. 82. Find lim

.

=x

b. Now confirm your estimate hy fmding the limit with I'H6pital's Rule. Ail the fust step, multiply f(x) hy the fraction (x + ~)/(x + v'x' + x) and simplifY

Yx

Ysinx

70

lim sec x %-(7f/2>- tanx

X

•• Estimate the value of

lim sin x olnx

%-0+

xl

oc - oc Form

X~O+

68. lim

x_OO ~

+ "- + sin bX) = O?

lim (tan 2x x-a x3

Theory and Applications I:H6pital's Rule does not help with the limits in Exercises 67-74. Try it-you just keep on eycling. Find the limits some other way. 67. lim

403

+

I)Yx + 2

1 hy graphing. Then confmn your estimate with I'H6pital's Rule. x-I

X

84. This exercise explores the difference between the limit

a. lim x-3 = lim 1...=1. x-3 x 2 - 3 x-3 2x 6 b. lim x-3 =Q=o %-3 x 2 - 3 6

and the limit

76. Which one is correct, and which one is wrong? Give reasons for

. (1+"I)X =e.

your answers.

lim

X~OO

2x - 2 a. lim x '-2x = l i m x_ox2 - sinx x-O 2x - cosx = lim x_o2 b.

2

+ sin x

•• Use I'H6pital's Rule to show that

lim x' - 2x = lim 2x - 2 x-O 2x cosx _ sinx

-2

.%_ox2

0 - 1 = 2

others wrong? Give reasons for your answers. a. lim xlnx = 0·(-00) = 0

x-o+ b. lim xlnx = 0'(-00) = -00 x-o+

lim Inx -00 c. x-'r!1l+ X In X = x~o+ (I/x) = """00 =

lim

2+0

77. Only one of these calculations is correct Which one? Why are the

d.

. (1+"I)X =e.

= _2_ = I

-

1

X~OO

D b. Graph

f(x)

=

(I +:S

lim (I

.... 00

a. f(x) = x,

g(x) = x',

(a,b) = (-2,0)

b. f(x) = x,

g(x)

(a, b) arbitrary

=

c. f(x) = x'/3 - 4x,

x',

g(x) = x',

(a, b) = (0,3)

79. Continuous OItension Find a value of c thst makes the function f(x) =

{

9X - 3sin3x, x # 0 5x'

c,

x= 0

continuous at x = O. Explain why your value of c works.

(I +~)'

85. Show thst

lim x In x = lim In/x X~O+ (I x)

78. Find all values of c thst satisfY the conclusion of Cauchy's Mean Value Theoreru for the given functions and interval.

=

together for x '" O. How does the behavior off compare with thst of g? Estimate the value oflim~oo f(x). c. Confmn your estimate oflimx->oo f(x) by calculating it with I'H6pital's Rule.

X~O+

= lim (I/x) = lim (-x) = 0 x-o+ (-1/x 2) x-o+

and g(x)

86. Given that x a. Xl/x

>

1")'

+k

=

e'.

0, fmd the maximum value, if any, of

b. xl/i'

c. xlix' (n a positive integer) d. Show that 1imx-oo xl/x· = 1 for every positive integer n. 87. Use limits to find horizontal asymptotes for each function.

•. y=xtan(~)

b. y

= { 0,

3x + e2x "'2x'--'+-'e~3x"

x # 0

e-11i'

88. Findf'(O) for f(x)

=

'

x

= O.

404

D 89.

Chapter 7: Transcendental Functions

D 90.

The continuous extension of(sinxf to [0, 'IT[

°

The function (sin x)... • (Continuation ofExercise 89.)

a. Graphf(x) ~ (sinxf on the interval :5 x:5 'IT. What value would you assign to f to make it continuous at x = O?

a. Graphf(x) ~ (sin x)"'" on the interva1-7 s x '" 7. How do you account for the gaps in the graph? How wide are the gaps?

b. VerifY your conclusion in part (a) by rmding lim,....o· f(x)

b. Now graph f on the interval :5 x '" 'IT. The function is not dermed at x ~ 'IT/2, hut the graph has no break at this point. What is going on? What value does the graph appear to give for f atx ~ 'IT/2? (Hint: Use I'H6pital~ Rule to rmd limf as x ..... ('IT/2)- and x ..... ('IT/2)+.)

°

with I'H6pital~ Rule. c. Returning to the graph, estimate the maximwn value off on [0, 'IT]. About where is max f taken on? d. Sharpen your estimate in part (c) by graphing f' in the same window to see where its graph crosses the x-axis. To simplifY your work, you might want to delete the esponential factor from the expression for f' and graph just the factor that has a zero.

Inverse Trigonometric Functions

7.6

Inverse trigonometric functions arise when we want to calculate angles from side measurements in triangles. They also provide useful antiderivatives and appear frequently in the solutions of differential equations. This section shows how these functions are derIDed, graphed, and evaluated, how their derivatives are computed, and why they appear as important antiderivatives.

y x

= siny

2) VDomain: 'IT'

.

Y = sin-Ix

-1';x,,1

Range: -wl2 :S y :S 'lT12

_~

c. Continuing with the graphs in part (b), rmd max f and min f as accurately as you can and estimate the values of x at which they are taken on.

1

x

2

FIGURE 7.14 The graph ofy ~ sin-1 x.

Defining the Inverses The six basic trigonometric functions are not one-ro-one (their values repeat periodically). However, we can restrict their domains to intervals on which they are one-ro-one. The sine function increases from -I at x = -1f/2 to + I at x = 1f/2. By restricting its domain to the interval [-1f/2, 1f/2], we make it on.,.to-one, so that it has an inverse sin- 1 x (Figure 7.14). Similar domain restrictions can be applied to all six trigonometric functions.

Domain restrictions that make the trigonometric functions one-to-one Y

Y

sin x

= sinx Domain: [-'IT/2, 'IT/2] Range: [-1,1]

y

= cosx Domain: [0, 'IT] Range: [-1, 1]

y

Y

2

i( ~ cotx Domain: (0, 'IT) Range: (-00, 00)

= secx Domain: [0, 'IT/2) U ('IT/2, 'IT] Range: (-00, -1] U [I, 00)

y

~ tao x Domain: (-'IT/2, 'IT/2) Range: (-00, 00)

y

sec x

-o;;t---,,;;;---;;,,~x

y

tan x

----;;,,;--c;cot---c,,;;-~x

-~

2

= cscx Domain: [-'IT/2, 0) U (0, 'IT/2] Range: (-00, -1] U [1, 00)

y

405

7.6 Inverse Trigonometric Functions

I

Since these restricted functions are now one-to-one, they have inverses, which we denote by

The "Arc" in Arcsine and Arccosine The accompanying figure gives a

geometric interpretation ofy = sin-I x andy = COS-IX for radian angles in the frrst quadrant. For a unit circle, the = r8 becomes s = (J, so cen1ral angles and the arcs they subtend have the same measure. If x = sin y, then, in addition to being the angle whose sine is.:t, y is also the length of arc on the unit circle that subtends an angle whose sine is x. So we call y ''the arc

equation s

whose sine is .:t,"

y = sin-Ix

or

y = cos- t x

or

y

= tan-I

x

or

y

= coC I X

or

sec- 1 x

or

Y = arccotx y = arcsecx

y = csc- I x

or

y

y =

Y = arcsinx Y = arccosx y = arctanx

= arccscx

These equations are read "y equals the arcsine of x" or "y equals arcsin x' and so on. Caution The - 1 in the expressions for the inverse means ''inverse.'' It does not mean reciprocal. For example, the reciprocal of sin x is (sinx)-I = l/sinx = cscx.

y

x2

+ y2 = 1

Arc whose sine is x

~-+--

Angle whose sine is x --~------~~---L~--t---~x

o

x

1

The graphs of the six inverse trigonometric functions are shown in Figure 7.15. We can obtain these graphs by reflecting the graphs of the restricted trigonometric functions through the line y = x, as in Section 7.1. We now take a closer look at these functions and their derivatives. Domain: -1 =S x =S 1 Range' _1!:s yS 'I!

Domain: -1:S X:S 1 Range: 0 :S Y :S '1r

y

y

.

2

Angle whose

2

Domain: -00 < x < Range:

OQ

-I
cosineisx

--+___

I'--+--~ x

----'---+-~--~x

-1

(b)

(a)

Domain: x:s -1 or x 2: 1 Range:

O=SY=S1T,Y=F

I

Domain: x:s -1 or x 2: 1 Range: -~=s;y=s;~,y*O

y

r -'I! " 2

Domain: _00 < x < 00 Range: O
y

~

-------

(c)

_______'1f'

y=sec-1x

-------

--~~--~~~~-c>x

-2

(d)

-I

(0)

1

2

(f)

FIGURE 7.15 Graphs of the six basic inverse trigonometric functions.

The Arcsine and Arccosine Functions We derme the arcsine and arccosine as functions whose values are angles (measored in redians) that belong to restricted domains of the sine and cosine functions.

DEFINmON y

y

= sin -1 X is the number in [-'IT/2, 'IT/2] for which sin y = cos -1 X is the number in [0, 'IT] for which cos Y = x.

=

x.

406

Chapter 7: Transcendental Functions

"

"

. y y=smx'-2:=Sx:S2: Domain: [-,,/2, ,,!2] Range: [-1, 1]

The graph of y = sin-1 x (Figure 7.16) is symmetric about the origin (it lies along the graph ofx = siny). The arcsine is therefore an odd function: sin- I( -x) = -sin-I x. (1) The graph of y = cos-I x (Figure 7.17) has no such symmetry.

(a)

EXAMPLE 1

y

Solution

x = siny

Y = sin-Ix '! )

2

(a) We see that

Domain: [-1, 1] Range: [-,,/2, ,,/2]

-(;01

x

because sin ('n/3) = "'1/3/2 and 7f/3 belongs to the range [-7f/2, 7f/2] of the arcsine function. See Figure 7.ISa

(b) Wehave

2

(b)

FIGURE 7.16 The graphs of
because cos(27f/3) = -1/2 and 27f/3 belongs to the range [0,7f] of the arccosine function. See Figure 7.ISb. • Using the same procedure illustrated in Example I, we can create the following table of common values for the arcsine and arccosine functions.

x = siny.

x y

Y= cosx,O:sxSTr Domain: [0, ,,]

Range:

1

[-1,1]

/

"'1/3/2

7f/3

7f/6

Yz/2

7f/4

7f/4

/

1/2

7f/6

7f/3

-1/2

-7f/6

27f/3

-Yz/2

-7f/4

37f/4

-"'1/3/2

-7f/3

57f/6

/

/

°

-1

(a)

y x = cosy

\

y

"2

°

cos-Ix Domain: [-1,1] Range: [0, ,,]

y

sin-1V3 2

y = '!!'

3

=

-+-L...,~-'----~x

--+~I-+--~x

-1

1

COS(~1T) = -~

(b)

(b)

FIGURE 7.17

FIGURE 7.18

(a)y

(Example 1).

The graphs of cosx, 0 ::;; x:5 71',and (hlits inverse,y = COS-Ix. The graph of cos-1 x, obtained by reflection across the line y = x, is a portion of the curve x = cosy. =

Values of the arcsine and arccosine functions

EXAMPLE 2 During an airplane flight from Chicago to St. Louis, the navigator determines that the plane is 12 mi off course, as shown in Figure 7.19. Find the angle a for a course parallel to the original correct course, the angle b, and the drift correction angle c = a + b.

7.6

Inverse Trigonometric Functions

407

From Figure 7.19 and elementary geometry, we see that 180 sin a = 12 and 62 sinb = 12, so

Solution

a = sin- 1

b = sin-

1

/io '"

~~

'"

0.067 radian '" 3.8°

0.195 radian '" 1l.2°

•

c=a+b"'15°.

Identities Involving Arcsine and Arccosine FIGURE 7.19 Diagramfordrift correction (Example 2), with distances rounded to the nearest mile (drawing not to scale).

All we can see from Figure 7.20, the arccosine of x satisfies the identity cos- 1 x + cos- 1( -x) = 1r,

(2)

or cos- 1 (-x) = TT - cos- 1 X. Also, we can see from the triangle in Figure 7.21 that for x sin-

1

X

+

cos-

1

X =

(3)

>

0,

1r/2.

(4)

Equation (4) holds for the other values ofx in [-I, I] as well, but we cannot conclude this from the triangle in Figure 7.21. It is, however, a consequence of Equations (I) and (3) (Exercise 113).

Inverses of tan x, cot x, sec x, and esc x The arctangent of x is an angle whose tangent is x. The arccotangent of x is an angle whose cotangent is x. The angles belong to the restricted domains of the tangent and cotangent functions. FIGURE 7.20 cos-1 x and cos- 1( -x) are

supplementary angles (so their sum is 'IT).

DEFINmON

y y

= tan-'xisthenumberin(-1r/2,1r/2)forwhichtany = x. = coC' x is the number in (0, 1r) for which coty = x.

We use open intervals to avoid values where the tangent and cotangent are undefined. The graph of y = tan-I X is symmetric about the origin because it is a branch of the graph x = tany that is symmetric about the origin (Figure 7.15c). Algebraically this means that tan-I (-x) = -tan- 1 x; FIGURE 7.21

sin-1 x and cos-1 x are

complementary angles (so their sum is 'IT/2).

the arctangent is an odd function. The gmph of y = coC 1 X has no such symmetry (Figure 7.15f). Notice from Figure 7.15c that the graph of the arctangent function has two horizontal asymptotes: one at y = TT/2 and the other at y = -1r/2.

EXAMPLE 3 tan-' x

x

The accompanying figures show two values of tan-I x. y tan-I_1_ = tan-I V3 = 1! y'3 3 6 ----,--~

v'3

'IT/3

I

'IT/4

v'3/3 -v'3/3

'IT/6 -1r/6

-I

-1r/4

-v'3

-1r/3

tan 1!

= _1_

6y'3

The angles come from the first and fourth quadmots because the range of tan-I x is (-1r/2, TT/2). •

408

Chapter 7: Transcendental Functions

Ixl" 1

Domain: Range:

OSY:S'1T.Y*~

The inverses of the restricted forms of sec x and esc x are chosen to be the functions graphed in Figures 7.15d and 7.15e.

y

3" ---------- ------------ 2'

~ "

---; ~-s~;-~--+---------,~-~-~-------- ~

Caution There is no general agreenIent about how to def'me sec-I x for negative values of x. We chose angles in the second quadrant between 1T/2 and 1T. This choice makes sec- I x = cos-I (l/x). It also makes sec- I x an increasing function on each interval of its domain. Sonre tables choose sec-I xto lie in [-1T, -1T/2) for x < 0 and some texts choose it to lie in [1T, 31T/2) (Figure 7.22). These choices simplify the fonnula for the derivative (our fonnula needs absolute value signs) but fail to satisfY the computational equation sec- I x = cos-I (l/x). From this, we can derive the identity

---------- ------------ -2 "

~

-IT

------------

by applying Equation (4).

,/

3" ------------ -2'

FIGURE 7.22 There are several logical choices for the left-hand branch of y = sec-1 x. With choice A, sec-I x = COS-I (I/x), a useful identity employed by many calculators. y

"

2

, "\

(5)

The Derivative of y = sin-1 u We know that the function x = siny is differentiable in the interval -1T/2 < y < 1T/2 and that its derivative, the cosine, is positive there. Theorem 1 therefore assures us that the inverse function y = sin-I x is differentiable throughout the interval - 1 < x < 1. We cannot expect it to be differentiable at x = 1 or x = - 1 because the tangents to the graph are vertical at these points (see Figure 7.23.) We find the derivative of y = sin-I x by applying Theorem 1 with j(x) = sin x and rl(x) = sin-I x:

= sin-Ix Domain: -1 :s x :s 1 Range: -'1T12 :s y :S 1Tfl

Y

(j-I),(X) =

f'(j~I(X))

Theorem 1

1 cos (sin

f'(u)

1 x)

=

casu

1

VI -

sin2 (sin 1 x)

1 FIGURE 7.23 TIre graph of y = sin-I X has vertical tangents at x = -1 and x = I.

If u is a differentiable function of x with Iu I

sin (sin-l x)

<

1, we apply the Chain Rule to get

d ( . -I ) _ 1 du ,.------,2 dx ' dx sm u - •vi - u

EXAMPLE 4

= X

lui <

1.

Using the Chain Rule, we calculate the derivative

•

7.6 Inverse Trigonometric Functions

409

The Derivative of y = tan-1 u We f'md the derivative of y = tan-1 X by applying Theorem I with f(x) = tan x and 1(x) = tan-1 x. Theorem I can be applied because the derivative of tan x is positive for

r

-TT/2 < x < TT/2: (r' ),(x)

Theorem 1

f'{f \x))

=

I

I

1'(") ~

+ tan2 (tan 1 x)

sec2 u

=

sec'" 1 + tan2 u

I I

+ x2

The derivative is defined for all real numbers. If u is a differentiable function of x, we get the Chain Rule form:

~ (tan- 1 u) = _I_du. dx I + u 2 dx

The Derivative of y = sec-1 U Since the derivative of sec x is positive for 0 < x < TTI2 and TT/2 < x < TT, Theorem I says that the inverse function y = sec-1 x is differentiable. Instead of applying the formula in Theorem I directly, we find the derivative of y = sec- 1 x, Ixl > I, using implicit differentiation and the Chain Rule as follows:

secy

=

x

Inverse function relationship

~(secy)

=

~x

Differentiate both sides.

dx

dx

dy

secytany dx = I

ClutinRule

dy

I

dx

secytany

Since Ixl > l,y lies in (0, ",/2) U (",/2, "') and secytany =F 0

To express the result in terms of x, we use the relationships secy = x

and

tany = ±Vsec2 y - I = ±~

to get

dy = ± I dx x~·

410

Chapter 7: Transcendental Functions

Can we do anything about the ± sign? A glance at Figure 7.24 shows that the slope of the graph y = sec-I X is always positive. Thus,

--------- 2"

+

______L-~+_~d-----~x -1

0

1

FIGURE 7.24 TIre slope of the curve y = sec-Ix is positive for both x <-1

and x > 1.

iLsec-I x

=

dx

I

x~

1

I

x~

if x > I ifx<-l.

With the absolute value symbol, we can write a single expression that eliminates the "±" ambiguity:

iL sec-I x

I

=

Ixl~

dx If u is a differentiable function of x with Iu I

.

> I, we have the fannula

lui>

EXAMPLE 5

I.

Using the Chain Rule and derivative of the arcsecant function, we tmd d sec-I ( 4) 5x = 5x 4

I I IV(5x4)2 -

dx

d ( 4) I dx 5x 5x 4 >1>O

4

•

Derivatives of the Other Three We could use the same techniques to find the derivatives of the other three inverse trigonometric fimctions---arccosine, arccotangent, and arccosecant---but there is an easier way, thanks to the following identities.

Inverse Function-Inverse Cofunction Identities cos-I x

=

TT/2 - sin- 1 x

coC I X = TT/2 - tan-I x csc- I x = TT/2 - sec-I x

We saw the tlISt of these identities in Equation (4). The others are derived in a similar way. It follows easily that the derivatives of the inverse cofunctions are the negatives of the

7.6 Inverse Trigonometric Functions

411

derivatives of the corresponding inverse functions. For example, the derivative of COS-I x is calculated as follows:

(1T2- sm. -I x )

d ( -I) d dx coS x =dx

Identity

1

Derivative of arcsine

\1'1=7

The derivatives of the inverse trigonometric functions are summarized in Table 7.3.

TABLE

7.3 Derivatives of the inverse trigonometric functions

d(sin- I u)

I

dx

1.

du

d(cos- I u)

2.

1

dx

I

dx dx

lui <

du

1

I

du

+ u 2 dx I

du

dx

lulVu2=! dx'

d(csc- I u)

1

dx

6.

I

-----

d(sec- I u)

5.

1

+ u 2 dx

I

d(coC I u)

4.

du

~dx'

d(tan- I u)

3.

lui <

~dx'

lui>

du

I

lui>

lulVu2=! dx'

I

Integration Formulas The derivative formulas in Table 7.3 yield three useful integration formulas in Table 7.4. The formulas are readily verified by differentiating the functions on the right-band sides.

TABLE

7.4 Integrals evaLuated with inverse trigonometric functions

The following formulas hold for any constant a oF O.

1.

2. 3.

Jv'a~u_ J~ J a2

u2 =

du

(*) + ~tan-I (*) +

C

(Valid for u

C

(Valid for all u)

= !sec-lllll

+

u 2 = sin-I

uv'u2 _ a 2

a

a

C

(Valid for

2

< a2)

lui> a > 0)

The derivative formulas in Table 7.3 have a = 1, but in most integrations a oF I, and the formulas in Table 7.4 are more useful.

412

Chapter 7: Transcendental Functions

EXAMPLE 6 V3/2


J

Vo/2

These examples illustrate how we use Table 7.4. dx

v'1=7

]V3/2

sin- t X

=

a

I, u

=

x in Table 7.4. Formula 1

=

Vo/2

sin-

=

1

('?) -

sin-

1

('?) a ~

=

~-

;

= ;;

V3, u ~ 'lx, and dul2 ~ fix

Table 7.4, Fonnula 1

tsin-1

=

(~)

(e) JVe:- JV::/~ J 6 =

= eX,du = eXdx, dx = du/ex = du/u,

a2

a~

! sec-1 I~ I + C

=

v6

Thbl. 7.4, Formula 3

~sec-l (~)

=

(a)

u

du uVu 2 - a2

=

EXAMPLE 7

+C

+C

•

Evaluate

J

dx V4x - x 2

J

4x 2

(b)

~x + 2

+

Solution (a) The expression V4x - x 2 does not match aoy of the fonnulas in Table 7.4, so we IlISt rewrite 4x - x 2 by completing the square: 4x - x 2 = -(x 2 - 4x) = -(x 2 - 4x + 4) + 4 = 4 - (x - 2)2. Then we substitute a = 2, u = x - 2, aod du = dx tu get

JV4:- JV4 - ~ JVa u x2

=

2)'

du

2 -

=

a

2

=

sin-1

(~)

=

sin-1

(x; 2) +

+C

+2

=

4(x 2 + x)

=

4

+2

2, u

=

x - 2, anddu

=

Table 7.4, Fonnula 1

C

(b) We complete the square on the binomial4x 2

4x 2 + 4x

=

=

(x + tY+ I

+ 4x: 4(x2 =

(2x

+X +

t) + t

+ 1)2 +

2 -

I.

ax

413

7.6 Inverse Trigonometric Functions Then,

J

~x + 2

4x + 2

= =

J

(lx

+

~)2 + 1

V

"2

=

~a

a = 1, u = 2x + I, anddu/2 ~ fix

2

(II) + C

!. !tan-1 2 a

Table 7.4, Formula 2

a

= l,u = 2x +

a

1

Exercises 7.6 Common Values Use reference triangles like those in Examples I and 3 to fmd the angles in Exercises 1-8.

1. a. tan- 1 1 2. a. tan- 1(

b. tan- 1( -v'3)

-I)

3.a.sm. -1 (-I) T

-t)"2 a. cos- (t)

4.a.sm . 5.

1

6. a. csc-1

V2

c. tan-

v'3

b. tan- 1

b. sinb. sin-

1

(~) (~)

(-v'3) (v'2 3) (v'2 3)

25. y ~ scc- 1 (2s

+

27. y ~

+ I), x> 0

28. y ~ csc-

29• Y -- sec-11t'

v'3

c. csc-1 2

40. y

~

c. scc-1( -2)

42. y

~ In(x2 + 4) -

1

(~)

b. coC 1 (v'3)

c. cot-1

(-I) v'3

10. sec

12. cot (Sin-

1

(-

35. y ~ csc- 1 (e')

Lim;ts Find the limits in Exercises 13-20. (If in doubt, look at the function's graph.) 13. lim sin- 1 x 14. lim cos-1 x x--l+

x-l-

COc1 }

lim csc-1 X

Rnd;ng Derivatives In Exercises 21-42, fmd the derivative ofy with respect to the appropriate variable. 22. y ~ cos-1 (I/x)

l,t

ooS-l (e-l)

COS-1 8

Sec- 1 8

tan- 1 X

-

xtan- 1

(f)

Evaluating Integrals

J J J

47

•

•

17

·

46

+ x2

dx xV2Sx 2

/. '

49

44

dx

•

2

-

4 tis

51

55

·

57

·

52

59

•

4y2 - I

y

3dr

(2x -

-

dx

- I)'

dx

4 tis

_~

v9-48 2

11 J J

~

Vi/3

54.

56

4(r -

+ (x

dx xVSx 2

• _24+3t 2

ely -v'r~===

JVI - I)' J J I)V(2x - I)' - 4 2

4x 2

dx 9+3x 2

2

~

-1

J J /o .

2 • /o8+2t .

1

dx

50.

'o~ 2

48 •

JVI 3Vi/4

Vi/2

20.

sin- 1

Evaluate the integrals in Exercises 43--j)6.

53.

X~OO

=

+ 38. Y = v.;z::-! tan-1 v.;z::-! + csc-1 x, X > I

39. y ~

15. lim tan- 1 x X~OO

30. y

36. y =

37. y ~ 8~

4S

'?))

sec- 1 S8

34. y = tan-1 (In x)

43 .~ dx

(COS-I t)

=

32. y = cot-1 vt=l

31. y ~ coC' Y; 33. y ~ In (tan- 1 x)

Evaluations Find the values in Exercises c}'-12.

21. y ~ cos- 1 (x 2 )

O
c. cos-1

b. csc-1 (-2)

26. y

f

-I) V2

(

8. a. coC ' (-I)

19. lim csc-1 x

1

. -1 c. sm

b. scc-

11. tan (Sin-+t))

cSC- 1 (x 2

I)

(-=l ) V2

b. cos-1

('?) )

24. y = sin- 1 (I - t)

. -1 - 2 c. sm

7. a. scc-1( -V2)

9. sin (cos-'

V2 t

(~)

1

1

c. tan-

1

23. y ~ sin-1

·

58

•

-2/3

dy yV9y2 - I 6dr

V4-(r+l)' dx I + (3x + 1)2

•

414 60 •

Chapter 7: Transcendental Functions

J

Integration Formulas

dx

(x + 3)V(x + 3)' - 25

61.1~/2 2 cos 0 dO

62.

63.

64.

~/2 I + (sinO)' r In v'3 e dx -

].

I +e

o

J · V1=7

65

2r

J.~/4 esc' x dx

~/6 I + (cotx)'

1'<" I( I 4+dlIn'l)

J

100. Jx'

I

J .

66

ydy

VerilY 1he integration fonnu1as in Exercises 99-102. tan-I x 1 tan-I x 99. ~dx ~ Inx - IIn(1 + x') - - x - + C

sec'ydy VI - tan'y

101.

COS-I

4

~ x4

5x dx

J

(sin-I x)' dx

102. J In (a' + x')dx

•

69 •

1

6dl

0

-I V3 - 21 - I'

71J •

73

dx

V-x' + 4x - 3

dy y' - 2y

{'

· JI

+5

Sdx x'-2x+2

75.JX+4 dx x' + 4 77. Jx' 79 ·

J

+ 2x -

x' + 9

68.J

72

104.

74

J

dy y'+6y+1O

r

· J,

dx (x + i)Vx' + 2x

78.

80

•

2dx x'-6x+ 10

1-2 dl 1'-61+10

J J

I' - 21' + 31 - 4 dl

I'

+

~ xln(a'

2x +

2~ sin-I X

+ x') - 2x +

+ C

2atan-l~ +

C

Solve 1he initial value problems in Exercises 103-106.

dy 103. ~ ~

76.J I dx

V2x - x'

70 6dl • 1/2 V3 + 41 - 41'

•

x4 dx

In;tial Value Problems

dx

11

+ -45J

VI - 25x'

~ x(sin-I x)' -

Evaluate 1he integrals in Exercises 67-1!0.

67J

5x

COS-I

I

dx

(x - 2)Vx' - 4x + 3

r.u.

dy

I VI-xl

y (0) ~ 0

.,.-------0'

I

~ ~

- , - - I, y(O)

""

x+1

dy I 105. ~ ~ ."------;' x r.u. xvx 2 - 1 106 dy

'dx

~

~

I

> I; y(2)

_ I_ _ 2 I+x' ~'

~

y(O)

~

'If

2

AppUcations and Theory 107. You are sitting in a classroom next to the wall looking at the blackboard at 1he froot of 1he room. TIre blackboard is 12 ft loog aod starts 3 ft from 1he wall you are sitting oext to.

a. Show 1hat your viewing angle is a = coC1 -.-£ - coCt !. IS 3 if you are x ft from 1he froot wall.

b. Find x so 1hat a is as large as possible.

T11 0

12'

~

1 3'

"-

I' L'Hop;tal's Rule

x-a

93. lim r~""

x

x

xtan-I~

. tan-I x 2 95. lun--x-O x sin- 1 X

92. lim x-l+

w-=-l sec 1 x

x

'I

erate a solid. Find 1he volume of 1he solid. y

94. lim 2 tan-I 3x'

7x 2

x-a 96. lim eX tan-I eX x_OO e2%

98.

Wall

108. The region between 1he curve y ~ sec-I x and the x-axis from x ~ I to x ~ 2 (shown here) is revolved about 1hey-axis to geo-

Find 1he limits in Exercises 91-9S. . -IS

91. lim 8m

You

a

+x

. -1 2

x x-o+(sin lx)2 lim

sm

3"

~t---~--~~x

o

2

7.6 Inverse Trigonometric Functions 109. The slant height of the cone shown here is 3 m. How large should the indicated angle be to maximize the cone's volume?

415

+ tan- I (l/x) is constant.

114. Show that the sum tan- I x 115. Use the identity

csc- Iu = ~ - sec- Iu 2 to derive the formula for the derivative of csc- I u in Table 7.3 from the formula for the derivative of sec- I u . 116. Derive the formula dy

dx

1

+ x2

for the derivative of y = tan- I x by differentiating both sides of the equivalent equation tan y = x.

110. Find the angle a .

117. Use the Derivative Rule, Theorem 1, to derive

d - I dx sec x

1

-lx-I-V---=x=2=-=I'

=

> 1.

Ixl

118. Use the identity 111. Here is an informal proof that tan- II Explain what is going on.

+ tan- I 2 + tan- I 3

coc l =

U

=

1T.

~ - tan- I u 2

to derive the formula for the derivative of coC I u in Table 7.3 from the formula for the derivative of tan- I u. 119. What is special about the functions

f( ) - . - I x-I x - sm x + l'

x;:: 0,

and

g(x) = 2 tan- I\IX?

Explain. 120. What is special about the functions

f(x)

=

sin- I

1

W+l

and

g(x)

=

tan- I}?

Explain. 121. Find the volume of the solid of revolution shown here. 112. Two derivations ofthe identity sec- 1 (-x) =

sec- 1 x a. (Geometric) Here is a pictorial proof that sec- I (-x) = 1T - sec- I x . See if you can tell what is going on. 7T -

y

:=17--

17

-V3 -3-

y

\/ ~~\--l / 1

__ -1 ______ _ 1 7T

y= _ _ 1_

~ V3

7------+--4--x

"2

1

_ _- L _ _- L _ _ _ _~--J---J-_+x

-x

-1

o

x

b. (Algebraic) Derive the identity sec- I (-x) = 1T - sec- I x by combining the following two equations from the text: cos- I x I sec- Ix = cos- (l/x)

cos- I (-x) =

1T -

122. Arc length Find the circumference of a circle of radius r using Eq. (3) in Section 6.3. Find the volumes of the solids in Exercises 123 and 124.

Eq. (3) Eq. (5)

113. The identity sin- 1 x + cos- 1 X = 7T /2 Figure 7.21 establishes the identity for 0 < x < 1. To establish it for the rest of [ - 1, 1], verify by direct calculation that it holds for x = 1, 0, and - 1 . Then, for values of x in (- 1, 0), let x = - a, a > 0, and apply Eqs. (1) and (3) to the sum sin- I( -a) + cos- I( -a).

123. The solid lies between planes perpendicular to the x-axis at x = - I and x = I . The cross-sections perpendicular to the x-axis are a. circles whose diameters stretch from the curve y =

-I/~tothecurvey

=

I/~ .

b. vertical s uares whose base edges run from the curve y = -1/ I + x 2 to the curvey = I/~ .

416

Chapter 7: Transcendental Functions

124. The solid lies between ~lanes perpendicular to the x-axis at x ~ - Yz/2 and x ~ Yz/2. The cross-sections perpendicular

to the x-axis are

D 125. Find the values of the following. c. coCl 2

D 126. Find the values of the following. •• sec- I( -3)

b. csc- I 1.7

C.

coCI (-2)

D 10 Exercises 127-129, fmd the domain and range of each composite function. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. 127••• Y ~ tan-I (tanx) b. y ~ tan (tan- I x) 128••• Y ~ sin-I (sin x)

7.7

D Use your graphing utility for Exercises 13()-134. sec.

b. squares whose diagonals stretch from the x-axis to the curve y ~ 2/~I-x'. b. csc- I (-1.5)

b. Y ~ cos (cos-I x)

130. Graph y ~ sec (sec-I x) ~ sec (cos-I(1/x)). Explain what you

a. circles whose diameters stretch from the x-axis to the curve y ~ 2/~I-x'.

•• sec-I 1.5

129... Y ~ cos-I (cos x)

131. Newton'. .erpentine Graph y ~ 4x/(x' + I), knowo as Newton's serpentine. Then graph y ~ 2 sin (2 tan-I x) in the same graphing window. What do you see? Explain. 132. Graph the rational function y ~ (2 - x')/x'. Then graph y ~ cos (2 sec-I x) in tire same graphing window. What do you see? Explain. 133. Graph f(x) ~ sin-I x together with its first two derivatives. Comment on the behavior of f and the shape of its graph in relation to the signs and values of f' and

r.

134. Graph f(x) ~ tan-I x together with its frrst two derivatives. Comment on the behavior of f and the shape of its graph in relation to the signs and values of f' and

r.

b. Y ~ sin (sin- I x)

Hyperbolic Functions The hyperbolic functions are formed by taking combinations of the two exponential functions eX and e -x. The hyperbolic functions simplify many mathematical expressions and occur frequently in mathematical applications. In this section we give a brief introduction to these functions, their graphs, and their derivatives.

Definitions and Identities The hyperbolic sine and hyperbolic cosine functions are defined by the equations

.

,ex~-c'e,--x_

sinhx ~ -

2

coshx ~

and

eX

+ e-x 2

.

We pronounce sinh x as "cinch x," rhyming with ''pinch x," and cosh x as "kosh x:' rhyming with "gosh x." From this basic pair, we defme the hyperbolic tangent, cotangent, secant, and cosecant functions. The defining equations and graphs of these functions are shown in Table 7.5. We will see that the hyperbolic functions bear many similarities to the trigonometric functions after which they are named. Hyperbolic functions satisfy the identities in Table 7.6. Except for differences in sign, these resemble identities we know for the trigonometric functions. The identities are proved directly from the defmitions, as we show here for the second one: 2 sinh x cosh x = 2 ( eX -2 e2:x -

e-X) (eX +2 e-X)

e-2:x

2 = sinh2x.

The other identities are obtained similarly, by substitoting in the defmitions of the hyperbolic functions and using algebra. Like many standard functions, hyperbolic functions and their inverses are easily evaluated with calculators, which often have special keys for that purpose.

7.7

HyperboLic Functions

417

The six basic hyperboLic functions

TABLE 7.5

y

Yy

=

coshx

eX 3

Y

= 2 2 Il ~ L "

Y

=

sinhx

-3-2-1

1 2 3

(a)

(b)

Hyperbolic sine: .

sinhx

(c)

Hyperbolic tangent:

Hyperbolic cosine:

eX - e-x

= "'---020"'--

coshx

+ e-x

eX

=

tanh x

2

= _sinh __ x = cosh x

.. e~x~-...'e'C~~ eX + e-x

Hyperbolic cotangent:

cosh x

cothx

y

Identities for hyperbolic functions

TABLE 7.6

cosh2 X - sinh2 X = I sinh 2x = 2 sinh x cosh x cosh 2x = cosh2 X + sinh2 X cosh2 X = cosh 2x 2

+

y=sechx (e)

(d)

Hyperbolic secant:

sechx = _1_ = 2 cosh x eX + e-x

Hyperbolic cosecant:

cschx = _._1_ = sinhx

Derivatives of hyperboLic functions

TABLE 7.7

:fx (sinh u) =

cosh u::

:fx (cosh u) =

sinh u::

JL (coth u) dx

2

eX - e-x

For any rea! number u, we know the point with coordinates (cos u, sin u) lies on the unit circle x 2 + y2 = 1. So the trigonometric functions are sometimes called the circular functions. Because of the first ideotity sinh2 U =

COSh2 U -

sech2

eX + e-x eX _ e-x

I

sinh2 X = cosh 2x - I 2 2 tanh X = I - sech2 X coth2 X = I + csch2 X

:fx (tanh u) =

= sinhx =

with u substituted for x in Table 7.6, the point having coordinates (cosh u, sinh u) lies on the right-hand branch of the hyperbola x 2 - y2 = 1. This is where the hyperbolic functions get their names (see Exercise 86).

Derivatives and Integrals of Hyperbolic Functions The six hyperbolic functions, being mtiona! combinations of the differentiable functions

u::

= -csch2 U

1,

du dx

d du dx (sech u) = -sech u tanh u dx

eX and e""", have derivatives at every point at which they are defined (Table 7.7). Again,

there are similarities with trigonometric functions. The derivative formulas are derived from the derivative of e U :

:fx (sinhu) = :fx (e

U -;

e U du/dx

e~)

+ e-u du/dx 2

dcsch () dx u = -csch u coth u •dx = coshu

du dx

Def'mition of sinh u

Derivative of e" Defmition of cosh u

418

Chapter 7: Transcendental Functions

IntegraL formulas for hyperboLic functions

TABLE 7.8

J J J J J J

coshu

+C

coshudu = sinhu

+C

sinhudu

=

sech'udu

= tanhu + C

csch'udu

= -cothu + C

sechutanhudu = -sechu

TIris gives the first derivative formula. From the definitioo, we can calculate the derivative of the hyperbolic cosecant functioo, as follows:

~(CSChU) = ~ (s~U)

Def'mition of csch u

coshu du sinh'u dx

Quotient Rule

I coshu du sinh u sinh u dx

Rearrange terms.

----

------

= -csch u coth u

+C

cschucothudu = -cschu + C

du dx

Def'mitions of csch u and coth u

The other formulas in Table 7.7 are obtained similarly. The derivative formulas lead to the integral formulas in Table 7.8.

EXAMPLE 1 (a)

:1 (tanh v'l+7)

=

(b) J

coth5x dx

t Jo

sinh5x

sinh'xdx = = =

(d)

u = sinh5x, du = 5 cosh5xrh

=JCOsh5Xdx=lJdU =

(c)

v'l+7. ~ (v'l+7) 1 sech' v'l+7 v'l+7

= sech'

tlo

lui

5

+C

{ ' cosh2x -

Jo

2Jo

{In' 4ex sinh x dx =

Jo

=

tlo I

sinh 5xl

=

+C

I dx

Table 7.6

2

1 {' (cosh2x

su:

u

2 -

~

_ 1) dx =

1 [Sinh2x 2

- xl ' 0

Evaluate with a calculator.

,., 0.40672

{In' 4ex eX - e .... dx = 2

Jo

[e 2x -

2

{In' (2e 2x

Jo

_

2) dx

2xl~' = (e'In' - 2102) - (I - 0)

= 4 - 2102 -

I,., 1.6137

•

Inverse Hyperbolic Functions The inverses of the six basic hyperbolic functioos are very useful in integratioo (see Chapter 8). Since d(sinhx)/dx = coshx > 0, the hyperbolic sine is an increasing function of x. We deoote its inverse by

y = sinh- 1 x. For every value of x in the interval - 00 < x < 00, the value of y = sinh-1 x is the number whose hyperbolic sine is x. The graphs of y = sinh x and y = sinh-1 x are shown in Figure 7.25a. The functioo y = coshx is not ooe-to-ooe because its graph in Table 7.5 does not pass the horizootal line test. The restricted function y = cosh x, x ;" 0, however, is oneto-ooe and therefore has an inverse, deooted by

y = cosh- 1 x.

7.7

419

Hyperbolic Functions

For every value of x ;", I, Y = cosh-I x is the number in the interval 0 ,,;; y < 00 whose hyperbolic cosine is x. The graphs of y = cosh x, x ;", 0, and y = cosh-I x are shown in Figure 7.25b. Y

y

y=siDhx y=x

Y = coshx, X2!=O y=x

y

/

/

/

// Y

=

sinh-I X

3

(x = sinhy)

/

o

=

/

/

/

/

/

y

/

(x ~ coshy,y" 0) x

/ /

/ /

cosh-1 X

/ / / /

y '" 0)

2

/

y

y=x

y = sech-1 x (x = seehy.

/

/

=

sechx

X2:0

~~+-~~~--> x

o

123 4 5 6 7 8

2

3

(c)

(b)

(a)

FIGURE 7.25 The graphs of the inverse hyperbolic sine, cosine, and secant ofx. Notice the symmetries about Ibe line y = x.

Like Y = coshx, the function y = sechx = I/coshx fails to be one-to-one, but its restriction to nonnegative values of x does have an inverse, denoted by y = sech- I x. For every value of x in the interval (0, II, y = sech- I x is the nonnegative number whose hyperbolic secant is x. The graphs of y = sech x, x ;", 0, and y = sech-I x are shown in Figure 7.25c. The hyperbolic tangent, cotangent, and cosecant are one-to-one on their domains and therefore have inverses, denoted by y = tanh-I x, y = coth- I x, y = csch- I X. These functions are graphed in Figure 7.26. y

y ,

,

x=cothy i y = coth-Ix :

x= ltanhy

Y = :tanh-1x

, , , ,

, , , ,

x

-I , , , , , , ,

(a)

TABLE 7.9

Identities for inverse hyperbolic functions

csch- I X = sinh-II x

FIGURE 7.26

1

y

il

x = cschy

y

,

0

(b)

x

=

esch-Ix

~

X

(c)

The graphs oflbe inverse byperbolic tangent, cotangent, and cosecant ofx.

Useful Identities We use the identities in Table 7.9 to calculate the values ofsech- I x, csch- I x, and coth- I x on calculators that give only cosh-I x, sinh-I x, and tanh-I x. These identities are direct consequences of the defmitions. For example, if 0 < x ,,;; I, then

420

Chapter 7: Transcendental Functions We also know that sech (sech- 1 x) = x, so because the hyperbolic secant is one-to-one on (0, I], we have

Derivatives of Inverse Hyperbolic Functions An important use of inverse hyperbolic functions lies in antiderivatives that reverse the derivative formulas in Table 7.10. TABLE 7.10

d(sinh- 1 u)

dx d(cosh- 1 u)

dx d(tanh- 1 u)

Derivatives of inverse hyperbolic functions I du ~dx I du w-=-ldx'

u

I du 1-u2dx '

lui <

1

1 du l-u 2dx '

lui>

1

----

dx d(coth- 1 u)

----

dx d(sech- 1 u)

dx d(csch- 1 u)

dx

>

I

1 du u~dx'

0< u < 1

du lul~dx'

u",O

1

The restrictions Iu I < 1 and Iu I > 1 on the derivative formulas for tanh-1 U and coth- 1 u come from the natural restrictions on the values of these functions. (See Figure 7.26a and b.) The distinction between lui < I and lui> I becomes important when we convert the derivative formulas into integral formulas. We illustrate how the derivatives of the inverse hyperbolic functions are found in Example 2, where we calculate d(cosh- 1 u)ldx. The other derivatives are obtained by similar calculations.

EXAMPLE 2

Show that if u is a differentiable function of x whose values are greater

than 1, then

d ( -1) _ I du ~dx' 2 dx cosh u - _vu - 1 Solution First we find the derivative of y = cosh-1 X for x > 1 by applying Theorem 1 with f(x) = coshx andr ' (x) = cosh- 1 x. Theorem I can be applied because the derivative of cosh x is positive for 0 < x.

(r' )'(x) = !'u I, (x)) 1 sinh (cosh

Theorem 1

1 x)

1 Ycosh2(cosh 1 x) - I

1

1'(u)

~

cosb2 U sinhu

=

sinh u -

sinh2 U

=

I,

v'cosh2 u - 1

421

Hyperbolic Functions

7.7 The Chain Rule gives the final result:

HisTORICAL BIOGRAPHY

Sonya Kova1evsky (!8S()-1891)

d ( sh- 1 ) _ dx co u -

du

1

•

~dx'

With appropriate substitutions, the derivative formulas in Table 7.10 lead tu the integration formulas in Table 7.11. Each of the formulas in Table 7.11 can be verified by differentiating the expression on the right-hand side. Integrals leading to inverse hyperbolic functions

TABLE 7.11

1.

1 1V2

du 2

Va + u

2.

3.

5.

a>O

du = sh-1 (1l)+c a ' u - a 2 co

1 1V 1 V2 2

-

a

2

u =

a

u2

'

< a2

u2 > a2

(*) +

u

du =_1 a sech-l(Il)+C a ' a2 - u 2

O
du = _1 CSCh-111l1 + C a a ' a +u

u"Oanda>O

u

EXAMPLE 3

Solution

u>a>O

r-'(')+c ~coth-l c,

du

a

4.

= sinh-1 (*) + C,

2

2

Evaluate

The indefinite integral is

1V3

2dx

+ 4x 2

=

1

du Va 2 + u 2

u

=Sinh-1(*)+C = sinh-

1

(0)

= lx,

du

= 2dx,

a

= Vi

Formula from Table 7.11

c.

+

Therefore, ('

Jo

2 dx

V3 + 4x2

= sinh-1

(~)]'

V3

= sinh-1

0

=

(~)

V3 sinh- 1 (~)

_ sinh-1 (0) - 0 '" 0.98665.

•

Exercises 7.7 Values and Identities Each of Exercises 1--4 gives a value of sinh x or coshx. Use the defi-

nitions aod the identity cosh' x - siob' x

=

I to fmd the values of the

remaining five hyperbolic functions.

1. siobx =

-*

2. siobx

=

!

3. coshx

=

17

15' x >

0

4. coshX'

=

13

5' x > 0

Rewrite the expressions in Exercises 5-10 in terms of exponentials and simplify the results as much as you can.

S.2coshQnx)

6. siob(21ox)

422

Chapter 7: Transcendental Functions

+ sinh 5x 9. (sinhx + coshx)4

8. cosh 3x - sinh 3x

7. cosh 5x

10. In (coshx

Evaluating Integrals Evaluate the integrals in Exercises 41-{j().

+ sinhx) + In (coshx -

siohx)

41.

11. Prove the identities

+ y) + y)

sinh (x cosh (x

~

=

+ coshxsinhy, coshx coshy + sinhx sinhy. sinhxcoshy

45.

Then use them to show thst L

sinh 2x

~

2 sinh x cosh x.

b. cosh 2x = cosh2 X

47.

+ sinh2 x.

12. Use the definitions of cosh x and sinh x to show thst

cosh2 X

-

43.

49.

51.

In Exercises 13-24, fmd the derivative of y with respect to the appropriate vatiable.

53.

14. y

15. y ~ 2Vttanh Vt

17. y

~

16. y

~ ~

k sinh (2x

+ 1) 55.

19. y ~ sechO(1 - InsechO)

20. y

~

cschO(1 - IncschO)

v - ktanh'v

22. y

~

In sinh

21. y = In cosh 23. y ~ (x'

+

6 cosh

tanh

- 1n3) dt

k) dt

sech Vt

sinhfdt

4 cosh (3x - In 2) dt

44.

46.

dt

sech'

J J J J J

42.

Vt dl

v - kcoth'v

{m' cothx dt

coth

48.

1m,

{-In' 2e' cosh 0 dO

J-1n4

1

~/4

' l 1:,

cosh (tan 0) sec' 0 dO

57.

cosh (In I)

I

dl

59.

cosh'

(~) dt

x) dt

csch (In I) coth (In I) dl

I

{In'

10

54.1

In

tanh 2x dt

' 4e-8 sinh 0 dO

56.1~/2 2 sinh (sin 0) cos 0 dO

58.1' flO

8 cosh Vi dt

Vi

I

I

~ dO

csch' (5 -

SO.

52.

~/'

I'tanht

18. y ~ In (coshz)

In (sinhz)

sinh2xdt

sinh2 X = 1.

Finding Derivatives

13. y ~ 6sinhf

J J (~ J ~ J (x J 7,

60.

4 sinh'

(~) dt

1) sech (lnx)

(Hinl: Before differentiating, express in tenns of exponentials and simplifY.)

24. y ~ (4x' - I) csch (In 2x)

Inverse Hyperbolic Functions and Integrals When hyperbolic function keys are not available on a calculator, it is still possible to evaluate the inverse hyperbolic functions by expressing them as logatithms, as shown here.

In Exercises 25-36, fmd the derivative of y with respect to the appropriate vatiable. 26. y ~ cosh-12Vx+l 25. y ~ sinh-l Vi 27. y ~ (1 - 0) tanh-I 0 29. y ~ (1 - I) coth- l Vt

cos- 1 X

=

33. y

~ csch-l (k)"

X

35. y = sinh-l (tanx) 36. y ~ cosh-l (sec x),

32. y ~ Inx

x

~

00

1

Ixl <

+ ~sech-lx

34. y ~ csch- 12'

0

<x <

-00

+ 20) tanh-I (0 + 1)

30. y ~ (1 - I') coth- l I

sech- 1 x

31. y

-

28. y ~ (0'

I

0<x:51 x#O

< x < '1f/2

Ixl >

Integration Fonnulas

1

VetifY the integration formulas in Exercises 37-40. 37.

38. 39. 40.

J J J J J L

sechx dt = tan-I (sinh x)

+C

b.

sechx dt = sin- l (tanh x)

+C

xsech-lxdt xcoth-lxdt tanh-lxdt

Use the formulas in the box here to express the ntanhers in Exercises 61-{i6 in tenns ofnaturallogatithms. 61. sinh- l (-5/12) 62. cosh- l (5/3)

~ ~ sech-lx - k~ + C ~ x' ~

I coth-lx

+~+C

~ X tanh-I x + kln(1

- x')

+C

63. tanh-I (-1/2)

64. coth- l (5/4)

65. sech-l (3/5) Evaluate the integrals in Exercises 67-74 in tenns of

a. inverse hyperbolic functions. b. natural logarithms.

7.7

67.

69

•

J.

'v3

v4 +

o

l'

68.

----""-

dx

1 --:-7'==='" l/S

73. J.~

xVI -

80. Volume

6 dx

J. VI + 9x' 12 70. J.1 ~ o 1- x

x'

5/41-x 2 3/13

71.

113

dx

.~

16>;'

cos x dx

•

l' 1

xVI

The region enclosed by the curve y = sech x, the

V3

81. An: length

Find the lengih of the graph of y = (1/2) cosh 2x

fromx = Otox = In

Vs.

82. Use the definitions of the hyperbolic funetions to fmd each of the following lintits.

1x~

74

(a) lim tanh

dx

lim tanhx

x~oo

+ (lnx)'

(c) lim sinhx

(d)

(e) lim sechx

(f) lim cotbx

x~oo

Appllcattons and Examples 75. Show that if a function f is defmed on an interval symme1ric ahout the origin (so that f is defmed at -x whenever it is defined at x), then

_ f(x) + f( -x) ( ) fx 2

+

f(x) - f( -x) 2

423

x-axis, and the lines x = ± In is revolved about the x-axis to generate a solid. Find the volume of the solid.

0

72·1'-,=dx===

o VI + sin'x

Hyperbolic Functions

(1)

+ f( -x))/2 is even and that (f(x) - f( -x))/2 is odd.

Then show that (f(x)

W+i)

76. Derive the fornrula sinh- 1 x = In (x + for all real x. Explain in your derivation why the plus sign is used with the square root instead of the minus sign. 77. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body's velocity I sec into the fall satisfies the differential equation

x~oo

lim sinhx

x_-oo x~oo

(g) lim coth x

x-o+

(i)

lim csch x

x_-oo

83. Hanging cable. ln3agine a cahle, like a telephone line or TV cable, strung from one support to another and hanging freely. The cahle's weight per unit lengih is a constant w and the horizontal tension at its lcmest point is a vector of length H. Ifwe choose a c0ordinate system for the plane of the cahle in which the x-axis is horizontal, the force of gravity is straight down, the positive y-axis points straight up, and the lowest point of the cahle lies at the point y = H/w on the y-asis (see accompanying fignre), then it can be shown that the cahle lies along the graph of the hypetbolic cosine

H

w

y = wcoshJjx.

dv , mdt=mg-kv, where k is a constant that depends on the body's aerodyuamic properties and the density of the air. CWe assume that the fall is short enough so that the variation in the air's density will not affect the outconre significantly.)

a. Showthat

satisfies the differential equation and the initial condition that v = o when I = O. b. Find the body's limiting velocity, Jim,....oo v. c. For a 160-lb skydiver (mg = 160), with time in seconds and distance in feet, a typical value for k is 0.005. What is the diver~ limiting velocity'l

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning "chain." a. LetP(x,y) denote an arbitrary point on the cable. The next accompanying fignre displays the tension at P as a vector of lengih (magnitude) T, as well as the tension H at the lowest pnintA. Show that the cable's slope atP is

dy . w tan = dx = sinhJjx.

78. Accelerations whose magnitndes IIl"O proportionlll to displacement Suppose that the position of a body moving along a coordinate line at time t is

y

y= l1cosh~x w H

a. s = acoskt + bsinkt. b. s = a coshkl + b sinhkl. Show in hoth cases that the acceleration d's/ dl' is proportional to s bot that in the frrst case it is directed toward the origin, whereas in the second case it is directed away from the origin. 79. Volume A region in the flIS! quadrant is hounded shove by the curve y = cosh x, below by the curve y = sinh x , and on the left and right by the y-axis and the line x = 2, respectively. Find the volume of the solid generated by revolving the region ahout the x-axis.

P(X,y)~.....

TcostfJ H

o

424

Chapter 7: Transcendental Functions

b. Using the result from part (a) aod the fact that the horizontal tension at P must equal H (the cable is not moving), show that

T = "y. Hence, the magnitude of the tension atP(x,y) is exactly equal to the weight of y units of cable. 84. (Continuation of Exercise 83.) TIre length of arc AP in the Exercise 83 figure is s = (1/a) sinh ax, where a = wlH. Show that the coordinates ofP may be expressed in terms of s as

x=

~Sinh-l as,

y =

a. Show that the areaA(u) of sector AOP is A(u) = tCOshuSinhu

~S2 + :2"

85. Area Show that the area of the region in the fIrSt quadrant enclosed by the curve y = (1/a) cosh ax, the coordinate axes, and the line x = b is the sarae as the area of a rectangle of height 1/a and length s, where s is the length of the curve from x = 0 to x = b. Draw a figure illustrating this result. 86. The hyperbolic in hyperbolic functions Just as x = cos u and y = sin u are identified with points (x, y) on the unit circle, the functions x = cosh u and y = sinh u are identified with points (x, y) on the right-hand branch of the unit hyperbola, X2 _

Another artalogy between hyperbolic and circular functions is that the variable u in the coordinates (cosh u, sinh u) for the points of the right-hand branch of the hyperbola x 2 - y2 = I is twice the area of the sector AOP pictured in the accompanying figure. To see why this is so, carry out the followiog steps.

-1°O""~ fix.

b. Differentiate both sides of the equation in part (a) with respect to u to show that A'(u) =

t.

c. Solve this last equation for A(u). What is the value of A(O)? What is the value of the constant of integration C in your solution? With C determined, what does your solution say about the relationship of u to A(u)? y

y 2=1. y

y x2

+

r =1

P(cos U, sin u)

"...--,-.

~

------t-~~------~x

Since cosh2 U - sinh2 U = I, the point (cosh u, sinh u) lies on the right-hand branch of the hyperbola x 2 - y2 = I for every value of u (Exercise 86).

7.8

One of the artalogies between hyperbolic and circular functions is revealed by these two diagrams (Exercise 86).

Relative Rates of Growth It is often important in mathematics, computer science, and engineering to compare the rates at which functions of x grow as x becomes large. Exponential functions are important in these comparisons because of their very fast growth, and logarithmic functions because of their very slow growth. In this section we introduce the liUle-oh and big-oh notation used to describe the results of these comparisons. We restrict our attention to functions whose values eventoa1ly become and remain positive as x ..... 00 .

y

160 140 120 100

Growth Rates of Functions

80 60 40 20

-d~~~~~~--x o 1 2 3 4 5 6 7 FIGURE 7.27 andx2 .

The graphs of eX, 2x,

You may have noticed that exponential functions like 2" and eX seem to grow more rapidly as x gets large than do polynomials and rational functions. These exponentials certaiuly grow more rapidly than x itself, and you can see 2" outgrowing x 2 as x increases in Figure 7.27. In fact, as x ..... 00, the functions 2X and eX grow faster than any power of x, even x',ooo,ooo (Exercise 19). In contrast, logarithmic functions like y = log2x and y = lnx grow more slowly as x ..... 00 than any positive power of x (Exercise 21). To get a feeling fur how rapidly the values of y = eX grow with increasing x, think of graphing the function on a large blackboard, with the axes scaled in centimeters. At x = I cm,

7.8

425

the graph is e ' '" 3 em above the x-axis. At x = 6 em, the graph is e 6 '" 403 em '" 4 m high (it is about to go through the ceiliog if it hasn't done so already). At x = 10 em, the graph is e 'O '" 22,026 em '" 220 m high, higher than most buildings. At x = 24 cm, the graph is more than halfway to the moon, and at x = 43 em from the origin, the graph is high enough to reach past the sun's closest stellar neighbor, the red dwarf star Proxima Centauri. By contrast, with axes scaled in centimeters, you have to go nearly 5 light-years out on the x-axis to find a point where the graph of y = lnx is even y = 43 em high. See Figure 7.28. These important comparisons of exponential, polynomial, and logarithmic functions can be made precise by defining what it means for a function I(x) to grow faster than another function g(x) as x --+ 00 .

y

y

Relative Rates of Growth

= eX

70 60 50 40 30 20

10 0

Y = lnx 10

20

30

40

50

60

FIGURE 7.28 Scale drawings of the graphs of e' and \0 x.

x

Rates of Growth as x --+

DEFINmON

QC)

Let I(x) andg(x) be positive for x sufficiently large.

1.

1 grows faster than g as x --+ 00 if lim I(x) =

00

x_oo g(x)

or, equivalently, if

.

g(x)

lim I( X ) = O.

x--+OO

2.

We also say that g grows slower than 1 as x --+ 00 . 1 and g grow at the same rate as x --+ 00 if lim I(x) = L x_oo

g(x)

where L is finite and positive.

According to these defmitions, y = 2x does not grow faster than y = x. The two functions grow at the same rate because lim

2x

x--+OO X

=1im2=2 x--+ OO

'

which is a finite, positive limit. The reason for this departure from more common usage is that we want "I grows faster than Ft' to mean that for large x-values g is negligible when compared with f.

EXAMPLE 1

Let's compare the growth rates of several common functions.

(a) eX grows faster than x 2 as x --+

e'

Iim -2 x

,.--+00

lim eX 2x

=

x-co

00/00 (b) 3' grows faster than

00

because

lim eX = 2

00.

x_OO

Using I'H6pital's Rule twice

00/00 2x

as x --+

00

because

. 3x

. (3)X -2 =

Inn 2x = lim

x--i>OO

x--i>OO

00.

426

Chapter 7: Transcendental Functions (e) x 2 grows faster than In x as x --->

00

because

2

lim x = lim k = lim 2>;2 = x---+OO Inx x--+OO l/x x--+OO (d) In x grows slower than x '/n as X --->

x_oo

j'H6pital's Rule

for any positive integer n because

00

lim Inx = lim x_oo xl/n

00

I/x (I/n)x(l/n)-'

j'H6pital's Rule

= lim --"-= O. x_ co x l / 1I

11

is constant.

(e) As Part (b) suggests, exponential functions with different bases never grow at the same rate as X---> 00. If a > b > 0, then a' grows faster than b'. Since (a/b) > I, lim

x_OO

ba: =

(-ba )' =

lim

x_OO

00.

>

(f) In contrast to exponential functions, logarithmic functions with different bases a and b > I always grow at the same rate as x ---> 00 :

I

. log" x . In x/In a Inb hm--=hm =10llbX x_OO Inx/lnb Ina'

x_OO

•

The limiting ratio is always finite and never zero.

x --->

If 1 grows at the same rate as g as x ---> 00, and g grows at the same rate as h as 00, then 1 grows at the same rate as h as x ---> 00. The reason is that lim _gl = L,

x_OO

and

together imply

If L, andL 2 are fmite and nonzero, then so is L,L2 • Show that ~ and (2Vx

EXAMPLE 2

-

1)2 grow at the same rate asx--->

00.

Solution We show that the functions grow at the same rate by showing that they both grow at the same rate as the function g(x) = x:

lim

x_OO

lim

~= lim~l+ x52 =1 ' x x_OO

(2Vx -

x-oo

1)2 = lim

x

x-oo

(2Vx Vi

1)2 = lim x-oo

(2 __1_)2 = Vi

4.

•

Order and Oh-Notation The "little-oh" and ''big-oh'' notation was invented by number theorists a hundred years ago and is now commonplace in mathematical analysis and computer science.

DEFINITION

A function

1

is of smaller order than g as x --->

lim g(/(X» = O. We indicate this by writing f

x-OO

x

00

= o(g) ("I is little-oh of g'').

if

7.8

Notice that saying f

=

o(g) as x --->

00

Relative Rates of Growth

427

is another way to say that I grows slower than gas

x~oo.

EXAMPLE 3 (a) lox

=

Here we use little-oh notation.

o(x) as x --->

(b) x 2 = 0(x 3

00

because

+ I) as x ---> 00 because

2

lim _x_= 0

x_ oo x 3

+ 1

•

DEFINmON Let I(x) and g(x) be positive for x sufficiently large. Then I is of at most the order of g as x --+ 00 if there is a positive integer M for which I(x) < M g(x) ,

for x sufficiently large. We indicate this by writing /

EXAMPLE 4

= O(g) ("I is big-oh of g").

Here we use big-oh notation.

(a) x

+ sinx

(b) eX

+ x2

=

=

O(x) as x ---> O( eX) as x --+

00

00

because because

x + sinx . x ,;; 2 for x sufficiently large. eX

+ x2 e

x

~lasx--+oo

X. ---> 0 as x ---> 00 . • e If you look at the defmitions again, you will see that I = o(g) implies 1= O(g) forfunctions that are positive for x sufficiently large. Also, if I and g grow at the same rete, then I = O(g) and g = OU) (Exercise ll). (c) x = O( eX) as x --->

00

because

Sequential vs. Binary Search Computer scientists often measure the efficiency of an algorithm by counting the number of steps a computer must take to execute the algorithm. There can be significant differences in how efficiently algorithms perform, even if they are designed to accomplish the saroe task. These differences are often described in big-oh notation. Here is an exarople. Websters International Dictionary lists about 26,000 words that begin with the letter a. One way to look up a word, or to learn if it is not there, is to read through the list one word at a time until you either fmd the word or determine that it is not there. This method, called sequential search, makes no particular use of the words' alphabetical armngement. You are sure to get an answer, but it might take 26,000 steps. Another way to find the word or to learn it is not there is to go straight to the middle of the list (give or take a few words). If you do not find the word, then go to the middle of the half that contains it and forget about the half that does not. (You know which half contains it because you know the list is ordered alphabetically.) This method, called a binary search, eliminates roughly 13,000 words in a single step. If you do not find the word on the second try, then jump to the middle of the half that contains it. Continue this way until you have either found the word or divided the list in half so many times there are no words left. How many times do you have to divide the list to find the word or learn that it is not there? At most 15, because (26,000/2 15 ) < 1. That certainly beats a possible 26,000 steps.

428

Chapter 7: Transcendental Functions For a list of length n, a sequential search algorithm takes on the order of n steps to find a word or determine that it is not in the list. A binary search, as the second algorithm is called, takes on the order of 10(U n steps. The reason is that if 2m - 1 < n ,;; 2m , then m - I < 10(U n ,;; m, and the number of bisections required to narrow the list to one word will be at most m = 10(U n the integer ceiling for 10(U n. Big-oh notation provides a compact way to say all this. The number of steps in a sequential search of an ordered list is O(n); the number of steps in a binary search is O(iog2 n). In OUI example, there is a big difference between the two (26,000 ys. 15), and the difference can only increase with n because n grows faster than log2 n as n --> 00.

r

1,

Exercises 7.8 Comparisons with the Exponential e' 1. Which of the following functions grow faster thao eX as x ..... 00 ? Which grow at the same rate as eX? Which grow slower?

3

b. x 3

c. Yx e. (3/2)'

d.4x

LX -

g. e

X

f.

/2

+ 8in2 x

ex/2

h. 10glOX

2. Which of the following functions grow faster thao eX as x ..... 00 ? Which grow at the same rate as eX? Which grow slower?

+ 30x +

L

lOx'

c.

vT+7

I

b. x Inx - x

6. Which of the following functions grow faster than In x as x .... oo? Which grow at the same rate as In x? Which grow slower?

a. logz (x 2 )

b. 10glO lOx

c. I/Yx e. x - 2Inx

d. l/x 2 f. e-x

g. In (In x)

b. In (2x

Ordering Functions by Growth Rates 7. Order the following functions from slowest growing to fastest growing as x ---+

d. (S/2)'

+ 5)

a.

00.

OX

c. (lnx)'

Comparisons with the Power i' 3. Which of the following functions grow faster than x 2 as x ..... 00 ? Which grow at the same rate as x 2 ? Which grow slower?

8. Order the following functions from slowest growing to fastest growing as x ---+ 00. a. 2% b. x 2 d.

c. (ln2)'

OX

Big-oh and Litt1e-oh; Order 9. True, or false? As x ---+

e. xInx g.

x 3e-x

X

f. 2 h.

&x 2 2

4. Which of the following functions grow faster than x as x ..... 00 ? Which grow at the same rate as x 2 ? Which grow slower? L

x2

+ Yx

b. IOx 2

d. 10glO (x 2 ) f. (1/10)'

g. (1.1)'

h. x 2

+ IOOx

a. x = o(x) c. x = O(x + 5) e. 0% = 0(02%)

b.x=o(x+S)

g. Inx = 0(1n 2x)

b.

10. True, or false? As x ---+

x! = o(}) c. } - ;tI = o(})

a.

e. eX

5. Which of the following functions grow faster thao In x as x ---+ oo? Which grow at the same rate as In x? Which grow slower? L

log,x

b. In 2x

c. InYx

d. Yx

e. x

f. Slnx

g.

l/x

h. eX

d. x = 0(2x) f. x

+

Inx = O(x)

W+5 =

O(x)

00,

3

2

Comparisons with the Logarithm In x

00,

+x

=

O(e X)

g. In (In x) = O(lnx)

d. 2

+ cosx

= 0(2)

f. x Inx = 0(x 2)

b. In (x) = 0(1n (x 2 + I))

11. Show that ifpositive functions f(x) and g(x) grow at the same rate as x"" 00, thenf= O(g) andg = O(f). 12. When is a polynomial f(x) of smaller order thao a polynomial g(x) as x ---+ oo? Give reasons for your answer. 13. When is a polynomial f(x) of at most the order of a polynomial g(x) as x ---+ oo? Give reasons for your answer.

Chapter 7 Questions to Guide Your Review 14. What do the conclusions we drew in Section 2.4 about the limits of rational functions tell us about the relative growth of polynomi· a1sasx~ oo?

429

Xl/l,ooo,ooo > Inx. You might start by observing that when x > 1 the equation Inx = X 1/1,000,000 is equivalent to the equation In (In x) ~ (lnx)/I,OOO,OOO.

o c. Even x takes a long time to overtake In x. Experiment with a calculator to fmd the value of x at which the graphs of x l/lO

o

Other Comparisons 15. Investigate

lim In (x + I) Inx

%_00

1/10

and

and Inx cross, or, equivalently, at which Inx ~ 10 In (lnx) . Bracket the crossing point between powers of 10 and then close in by successive halving.

. In (x + 999) lim In X .

.%_00

o d. (Continuation ofpart (c).) The value

ofx at which Inx ~ 10 In (lnx) is too far out for some graphers and root fmders to identify. Try it on the equipment available to you and see what happens. 22. The function In x grows slower than any polynomial Show that In x grows slower as x ---+ 00 than any nonconstant polynomial.

Then use I'HOpital's Rule to explain what you fmd. 16. (Continuation ofExercise 15.) Show that the value of

lim In (x + a) Inx

];_00

is the same no matter what value you assign to the constant a. What does this say about the relative rates at which the functions f(x) ~ In (x + a)andg(x) ~ Inxgrow? 17. Show that v'lOx + 1 and Vx+l grow at the same rate as x ---> 00 by showing that they both grow at the same rate as Vi as

Algorithms and Searches 23. •• Suppose you have three different algorithms for solving the same problem and each algorithm takes a number of steps that is of the order of one of the functions listed here:

n lo~ n, n 3/2,

X~OO.

18. Show that ~ and v'x' - x' grow at the same rate as x ---+ 00 by showing that they both grow at the same rate as x 2 as x ---+ 00.

19. Show that eX grows faster as x ---+ 00 than x" for any positive integer n, even x 1,000,000 . (Hint: What is the nth derivative of x"?) 20. The function e' outgrows any polynomial Show that e' grows faster as x ---> 00 than any polynomial

n(lo~ n? .

Which of the algorithms is the most efficient in the long ruo? Give reasons for your answer.

o b. Graph the functions in part (a) together to get a sense ofhow rapidly each one grows. 24. Repeat Exercise 23 for the functions n,

Vn log, n,

(log, n? .

o 25. Suppose you are looking for an item in an ordered list one million items long. How many steps might it take to fmd that item with a sequential search? A binary search?

21. a. Show that Inx grows slower as x ---> 00 than xii' for any positive integer n, even x 1/1,000,000 •

You are looking for an item in an ordered list 450,000 items long o b. Although the values ofxl/l,ooo,ooo eventually overtake the val- o 26. (the length of Webster. Third New International Dictionary).

ues of In x, you have to go way out on the x-axis before this happens. Find a value of x greater than I for which

Chapter

,

How many steps might it take to fmd the item with a sequential search? A binary search?

Questions to Guide Your Review

1. What functions have inverses? How do you know if two functions f and g are inverses of one another? Give examples of functions that are (are not) inverses of one another. 2. How are the domains, raoges, and graphs of functions and their inverses related? Give an example.

3. How can you sometimes express the inverse of a function of x as a function ofx? 4. Under what circumstances can you be sure that the inverse of a function f is differentiable? How are the derivatives of f and l related?

r

5. What is the natural logarithm function? What are its domain, raoge, and derivative? What arithmetic properties does it have? Comment on its graph. 6. What is logarithmic differentiation? Give an example.

7. What integrals lead to logarithms? Give examples. What are the integrals of tan x and cot x? 8. How is the exponential function e' defined? What are its domain, range, and derivative? What laws of exponents does it obey? Comment on its graph. 9. How are the functions a' and log.x defmed? Are there any r<>s1rictions on a? How is the graph oflog. x related to the graph of In x? What troth is there in the statement that there is really only one exponential function and one logarithmic function? 10. How do you solve separable first-order differential equations? 11. What is the law of exponential change? How can it be derived from an initial value problem? What are some of the applications of the law?

12. Describe I'HOpital's Rule. How do you know when to use the rule and when to stop? Give an example.

430

Chapter 7: Transcendental Functions

13. How can you sometimes handle limits that lead to indeterminate forms 00/00, 00' 0, and 00 - 001 Give examples.

14. How can you sometimes handle limits that lead to indeterminate forms rx>, 00, and 00 001 Give examples. 15. How are the inverse trigonometric functions deImed? How can you sometimes use right triangles to find values of these functions? Give examples. 16. What are the derivatives of the inverse trigonometric functions? How do the domains of the derivatives compare with the domains of the functions? 17. What integrals lead to inverse trigonometric functions? How do substitution and completing the square broaden the application of these integrals? 18. What are the six basic hyperbolic functions? Comment on their domains, ranges, and graphs. What are some of the identities relating them?

Chapter 'f)

1. y = LOe~/'

2. y

= VieW.

I. 3. Y -_I"" "4xe - 16 e.

4. y

=

=

In (sin2 0)

7. y = 10!!2 (x /2) 9. y = 8-1 11. y = 5x,·6

+ 2)"+2

15. y = sin-I~, 16. y = sin-I

x 221 e- x

6. y = In (see> 0)

2

13. y = (x

X~OO?

23. What roles do the functions e% and In x play in growth comparisons?

24. Describe big-oh and little-oh notation. Give examples. 25. Which is more efficient---
COe l

zsec- I z -

28. y

=

29. y

= (sinO)v6

31. j e% sin (e") fix

34.

=

Incos-Ix

22. y =

2Vx"=l sec-'Vi

23. y

=

24. y

=

csc-I (sec 0), 0 (1 + x 2 )e"'-' %

<

0

>

< '1f/2

2

I»)'

,

I

>2

=

~3X + 4 2x - 4

36. j csc2 x e"''' fix

Llt~4

38.

39.

f

40. (II' 2 cot '1fX fix 11/6

43.

Logarithmic Differentiation In Exercises 25--30, use 10gari1hmic differentiation to Imd the derivative ofy with respect to the approptiate vatiable.

_((I

+ 1) cot (eY + I)
tanffix

41. J.'_2_I_ dl o 12 - 25

I

26. y

eY csc (eY

32. j e' cos (3e' - 2) dl

37.

45.

_ 2(x + I) 25. y - .,----;:v cos 2x + 1)(1 27. y (I _ 2)(1 + 3)

j

35. j see> (x)e"'" fix

21 z

= (lnx)I/(In%)

30. y

Evaluating Integrals Evaluate the integrals in Exercises 31-78.

14. y = 2(lnx)"f2

I 17. y

w-=-l,

2u2"

W+1

33. j e% sec2 (e% - 7) fix

19. y = I tan-I I - tlnl

=

22. How do you compare the growth rates of positive functions as


0

(~), v>

20. y = (I + 12)

21. What integrals lead naturally to inverse hyperbolic functions?

8. Y = log, (It - 7) 10. y = 9" 12. y = y'l,,-Yz

18. y=zcos-Iz- ~

21. y

20. How are the inverse hyperbolic functions deImed? Comment on their domains, ranges, and graphs. How can you fmd values of sech-I X, csch- I x, and coth-I x using a calculator's keys for cosh- 1 X, sinh-1 x, and tanh-I x?

Practice Exercises

Finding Derivatives In Exercises 1-24, Imd the derivative ofy with respect to the approptiate vatiable.

5. y

19. What are the derivatives of the six basic hyperbolic functions? What are the corresponding integral formulas? What simi1arities do you see here with the six basic trigonometric functions?

47.

j j j }csc>

tan (In v) v dv

(lnx)-' -x-fix (I + Inr) dr

42.

[~fix

1

~/6

cos I

-",/2 1 - sm t

dl

44.j~ vlnv 46.

j

ln(x - 5) x 5 fix

48.

j

cos(I-lnv) v dv

49. jX3%' fix

SO. j 2"'" see> x fix

51. [ffix

52.

[2

;x fix

431

Chapter 7 Practice Exercises

53.

54.1 (~

l' (~+ ~)dx

55.1-

56.1

1

e --(>0+1) dx

/.ln5 e'(3e' + 1)-'/2 dr

' l 1

+ 7lnxt l/'

dx

(In (v + I»)' v+1 dv

1 1

58.

60

"

65.

-6- d6

1

'/4

-'/4

6dx

4x 2

66.

68

J

dy yV4y2 - I

74

dx V-2x-x 2 1

77 •

V

2

t-O+

4 - 25x2

-liS

~ 24dy

J I J

--=-r~==

yVy2 - 16

•

-2tVs

•

dl

(I + I)VI 2 + 21 - 8

78 •

dy :-::,~==

Iy IV5y2 - 3

dx V-x 2 +4x-1

76 11

+ 4v + 5

-1

J

3 dv 4v 2 + 4v + 4 dl

(31 + I)V912 + 61

Solving Equations In Exercises 751--84, solve for y. 79. 3" = 2"+1

80. 4" = 3"+2

81. ge'" = x 2

82. 3"

83. In (y - I)

106. lim e-l/Y lny

= x + Iny

y-o+

107. lim (eX + I x_OO

=

d. f(x)

=

e. f(x)

= csc-

= 1n5x

tan~

x-o x + smX'

· sinmx 90• Inn . x-o smnx

89. Inn - ( 2)

x-o tan x

l

x,

tan-I X

=

g(x) = I/x g(x)

eX

=

.. f(x)

= 3~,

b. f(x) = 1n2x,

g(x)

= 2~

g(x)

=

Inx 2

c. f(x) = lOx' + 2x 2, g(x) = eX d. f(x) = tan- l (1jx), g(x) = I/x

= sechx,

g(x) = l/x2 g(x) =

e~

.. ~2 + ~ = O(~) 2

b.

c. X = o(x + Inx) e. tan-I x = 0(1)

d. In (lnx) = o(lnx) f. coshx = O(e X )

x

x4

x

~ + ~ = O(~) x 2 x4 x4

~= O(~+~) x4 x2 x4

c. Inx = o(x + I) e. sec-l x = 0(1)

d. 1n2x = O(lnx) f. sinhx

rl

x~o

G'

g(x)

x,

= O(e

X

)

113. Thefimctioof(x) = eX + x,beingdifferentishleaodone-to-one, has a differentiable inverse (x). Find the value of /dx at the point f(ln 2).

x-n/2-

+ I - Vx2 - xl

,)

lim _x_ _ _ x_ 2-1 x 2 +1

'x_oo

....

Theory and Applications

lim sec 7x cos 3x

93. lim (esc x - cotx)

96

(1+ ?)X

112. Tme, or false? Give reasons for your answers.

2 8in X

X

x-o+

110. Does f grow faster, slower, or at the same rate as g as x ..... oo? Give reasons for your answers.

..

X~OO

108. lim

f. f(x) = sinh,

X

95. lim (VX2 +

)lnX

111. True, or false? Give reasons for your answers.

3 Inx

84. In (lOlny)

88. lim

91.

eX - 1

Comparing Growth Rates of Functions 109. Does f grow faster, slower, or at the same rate as g as x ..... oo? Give reasons for your answers. .. f(x) = 10l!2 x, g(x) = log, x I b. f(x) = x, g(x) = x + X g(x) = xe~ c. f(x) = x/100,

f. f(x)

' x2+3x-4 85. Inn I

•

+ 21)

I - In (1 2 t

e. f(x) = sin-l (l/x),

Apply;ng L:H6pUal's Rule Use I'HOpital's Rule to find the limits in Exercises 85-108. x-I

100. lim Z-.mx - I x-o eX-l

6dx

-V.tVs

72.

J 75.1- --,-~2~d~v_c: J -2

98. lim 3' - I 6-0 8

x

101. lim 5 - 5 cou x_oe x x 1 103. lim

dx

'8ln310g,6 d6 6

{'

70.

2/3 dy 71. /, :-::-r~= Yz/3IY IV9y2 - I •

1)1/2 d6

• Jv'33 + 12

• _24+31

73

1 1V 1/5

-:;:~~

'19 -

l"-Ix~

x-o

99. lim 2'"" - I x_oex-l

dw

/.1n9 e'(e' _

64'1

6712~2 69 •

e2w

97. lim lOX - I

62. ],4(1 + 1n1)llnldl

81084 6

63.

0

-In2

59.1' ~ (1 61.

-

3x

1

-2

57.

~) dx x2

8

drl

114. Find the inverse of the fimctioof(x) = I + (I/x),x # O.Then showthatr l (f(x» = f(r l (x» = x and that

drll dx

__1_ ft.) -

f'(x)"

432

Chapter 7: Transcendental Functions

In Exercises 115 and 116, fmd the absolure maximum and minimum values of each functioo 00 the given interval.

[i., fJ

115. y = xlo2x - x, 116. y = Hh:(2 - Inx),

(0,

127. y.y' = secy' sec' x

.'l = 2(lnx)/x and the

Find the area between the curve y x-axis from x = 1 to x = e.

117. Area

118. a. Show that the area hetweoo the curve y = I/x and the x-axis from x = 10 to x = 20 is the same as the area between the curve and the x-axis from x = 1 to x = 2. b. Show that the area between the curve y = I/x and the x-axis from lea to kh is the same as the area between the curve and the x-axis from x = a tox = b (0 < a < b, k > 0).

119. A particle is traveling upward and to the right along the curve y = Inx. Its x-coordinare is increasing at the rate (fix/dt) = Vi mlsec. At what rare is the y-coordinare changiog at the point (.',2)? 120. A girl is sliding dowo a slide shaped like the curve y = 9. -xl3 . Her y-coordinare is changiog at the rate dy/ dt = ( -1/4)~ Nsec. At approximarely what rate is her x-coordinate changiog when she reaches the bottom of the slide at x = 9 ft? (Take. 3 to be 20 and rouod your ans",", to the nearest sec.)

ftl

121. The rectangle shown here has one side

In Exercises 125-128 solve the differential equatioo. dy _, 3y(x + I)' 125. fix = v'Ycos' vy 126. y' = y I

00

the positive y-axis,

one side on the positive x-axis, and its upper right-hand vertex on the curve y = e -r. What dimensions give the rectangle its largest area, and what is that area? y

128. y cos' x dy

+ sin x fix

= 0

In Exercises 129-\32 solve the initial value problem.

.-'-Y-',

129. :

=

dy 130. fix

=-

131. xdy -

ylny

I

+x

"

y(O) = -2

y(0)

=•

(y + v'Y) fix =

132. y-' dyfix =

,

0, y(l) = I

,f--, y(O) = • +I

I

133. What is the age ofa sample of charcoal in which 90% of the carbon-14 origioally present has decayed? 134. Cooling a pie A deep-dish apple pie, whose inrerna\ temperature was 2200f when removed from the oven, was set out on a breezy 40°F porch to cool. Fifteen minures later, the pie's internal tcmperatore was 180°F. How long did it take the pie to cool from there to 70°F? 135. Locating a solar station You are under cootract to build a solar statioo at grouod level on tlte east-west line between the two buildings shown here. How far from the taller building should you place the statioo to maximize the number of hours it will be in the sun 00 a day when the sun passes directly overhead? Begio by observing that

8

= 'If

-

-1 X

cot

C1

60 - co

50 - x

~.

Then fmd the value of x that maximizes 8. ~~------~--~ x

o

60m 122. The rectangle shown here has one side

00

the positive y-axis,

one side on the positive x-axis, and its upper right-hand vertex on the curve y = (lnx)/x'. What dimeosions give the rectangle its largest area, and what is that area? 0.2 0.1

t ,_ ..

o

D

~

~I

II

'x

123. Graph the following functioos and use what you see to locate and estimare the extreme values, identify the coordinares of the inflectioo points, and identifY the intervals 00 which the grapha

are concave up and concave down. Then confrrm your estimates by working with the functions' derivatives.

a. y

=

(lnx)/Vi

D 124. Graph f(x)

b. y = .-..'

c. Y

=

(l

+ x).~

= x In x. Does the function appear to have an absolure minimum value? Cnnfmn your answer with calculus.

00 00 00 00

00 00

9

30m

~-+----~--~~~x

o

x

SOm

136. A rouod underwater transmissioo cable consists of a core of copper wires surrounded by noocooducting insulatioo. 1f x denores the ratio of the radius of the core to the thickness of the insulation, it is known that the speed of the transmission signal is given by the equation v = x'in (I/x). If the radius of the core is I em, what insulatioo thickness h will allow the greatest transmissioo speed? Insulation

Chapter 7 Additional and Advanced Exercises

Chapter

f)

Additional and Advanced Exercises

Limits

16. Let g be a function that is differentiable throughout an open interval containing the origin. Suppose g has the following properties:

Find the limits in Exercises l--{j.

[h dx b-l-jo ~

2. lim} {' tan-I I dl

1. lim

%_00

3. lim (cos Yx)l/x %-0+

4. lim (x

Jo

• I) g(x

x

+ ex)2/x

+n+2 +... +~) 2n

.~o

_1_

iii) lim g(h)

1n (e1/1l + e2/11 + ... + e(II-1)/1I + en/II)

volving the region about the x-axis. Find the following limits.

b. lim V(I)/A(t)

c. lim V(I)/A(I)

t-oo

1_00

1-0+

8. Varying a logarithm's bas.

o

a. Findlimlog.2asa .... 0+, 1-, 1+, and 00. b. Graph y ~ log. 2 as a function of a over the interval 0< a :5 4.

Theory and Examples 9. Find the areas between the curves y ~ 2(loll2x)/x and y ~ 2(log.,x)/x and the x-axis from x ~ I to x ~ e. What is the ratio of the larger area to the smaller?

0 10. Graph f(x) ~ tan-I x + tan-l(l/x) for -5

:5 x :5 5. Then use calculus to explain what you see. How would you expect f to behave beyond the interval [-5,5]? Give reasons for your answer.

11. For what x

>

0 does x(x') ~ (xx)'? Give reasons for your

answer. 0

12. Graphf(x) ~ (sinx)"nxover[O, 31T]. Explain what you see. 13. Findj'(2) if f(x)

~

e«X) andg(x)

~ (X_t_, dl.

l,

~

0

~ I

h

a. Showthatg(O)

7. l.e!A(t) be the area of the region in the 'ITst quadrant enclosed by the coordinate axes, the curve y = e -x, and the vertical line x ~ I, t > O. Let V(I) be the volume of the solid generated by rea. lim A(t)

g(x) + g(y) ~ I _ g(x)g(y) forallrealnurnbers x,y, and

+ y in the domain of g

h-O 1I_00

+ y)

il) lim g(h)

%_00

5. lim (_1_ 11_00 n+ 1

6. lim

433

1+ t

14. a. Find df/dx if

~

O.

b. Showthatg'(x) ~ I

+ [g(x)]'.

c. Find g(x) by solving the differential equation in part (b).

17. Center of mas. Find the center of mass ofa thin plate of constant density covering the region in the first and fourth quadnmts enclosed by the curves y ~ 1/(1 + x') and y ~ -1/(1 + x') and by the lines x ~ 0 and x ~ 1. 18. Solid of revolution The region between the curve y ~ 1/(2Yx) and the x-axis from x ~ 1/4 to x ~ 4 is revolved about the x-axis to generate a solid. •• Find the volume of the solid.

b. Find the centroid of the region. 19. Th. best branching angl•• for blood v ••••l. and pipe. When a smaller pipe branches offfrom a larger one in a flow systero, we may want it to run off at an angle that is best from SOD2O energysaving point of view. We might require, for instance, that energy loss due to friction be minimized along the sectionAOB showo in the accompanying figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe Ul'" stream from B, aod 0 is the point where the braochiog occurs. A law due to Poiseuille states that the loss of energy due to friction in nontorbulent flow is proportional to the length of the path aod inversely proportional to the fourth power of the radius. Thus, the loss along AO is (ledl)/R' and along OB is (Ied,)/r' , where /r; is a constant, dl is the length of AO, d, is the length of OB, R is the radius of the larger pipe, and r is the radius of the smaller pipe. The angle 8 is to be chosen to minimize the sum of these two losses:

f(x)~ ['2~ldt.

dl

d,

R

r

L~/r;4+/r;4·

b. Find f(O). c. What can you conclude about the graph off? Give reasons

for your answer. 15. Even-odd decompositions

a. Suppose that g is an even function of x and h is an odd function ofx. Show that if g(x) + h(x) ~ 0 for all x then g(x) ~ for aIIxandh(x) ~ for allx.

o

A

o

b. If f(x) ~ IE (x) + f 0 (x) is the sum of an even function h(x) and an odd function fo(x) , then show that lE(x) ~ f(x)

+ f( -x) 2

and

fo(x) ~ f(x) - f( -x)

2

c. What is the sigoificance of the result in part (b)?

In our model, we assume that AC Thus we have the relations

~

a and BC

d,sinO

~

b

~

b are fixed.

434

Chapter 7: Transcendental Functions

D 20.

so that d,=bcscO,

dl

=a

- d,cosO

=a

- beatO.

We can express the total loss L as a function of 0:

Urban gardening A vegetable gardeo 50 ft wide is to be growo between two buildiogs, which are 500 ft apart along an east-west line. If the buildiogs are 200 ft and 350 ft ta1l, where should the garden be placed in order to receive the maximum llUll1ber ofbours of sunlight exposure? (Hint: Determine the value of x in the accompa-

nying fIgure that maximizes sunlight exposure for the garden.)

L= k(a - bcotO + bCSCO). R4

L

74

00 00 00 00 00

Show that the critical value of 0 for which dL/ dO equals zero is

200 ft tall

b. If the ratio of the pipe radii isr/R = 5/6, estimate to the nearest degree the optimal branching angle given in part (a).

The mathematical analysis described here is also used to explain the angles at which arteries branch in an animal's body.

West

00 00 00 x

50

350 ft tall

East

8 TECHNIQUES OF INTEGRATION OVERVIEW The Fundamental Theorem tells us how to evaluate a definite integral once we have an antiderivative for the integrand function. Table 8.1 summarizes the forms of antiderivatives for many of the functions we have studied so far, and the substitution method helps us use the table to evaluate more complicated functions involving these basic ones. In this chapter we study a number of other important techniques for finding anti derivatives (or indefinite integrals) for many combinations of functions whose antiderivatives cannot be found using the methods presented before.

TABLE 8.1

1.

3.

4. 5. 6.

7. 8. 9. 10. 11.

Basic integration formulas

I

kdx = kx

+

I~ = In Ixl

I

eX dx = eX

lax

I I I I I I

dx =

+

C

+

12.

(n # -1)

13. 14.

C

C

15.

LIna + C

sinxdx = -cosx

cosxdx = sinx 2

(any number k)

sec xdx = tanx

(a>O,a#l)

+

+ +

csc2 xdx = -cotx

C

17.

C

18.

C

+

16.

19. C

20.

I I I I I I I v'ar:- x (~) I ~ ~tan-1 (~) I:!. I I v' tanxdx = In Isecxl

+

C

cotxdx = In Isinxl

+

C

secxdx

= In Isecx + tanxl + C

cscxdx

= -In Icscx + cotxl + C

sinh x dx

= coshx + C

coshxdx

= sinhx + C 2 =

a2

x

secxtanxdx = secx

+

cscxcotxdx = -cscx

C

+

x2 =

sin-

1

+

+

dx =1 a sec-1 a x 2 - a2

C

C

+

C

(a> 0) C

(x> a

>

0)

435

436

Chapter 8: Techniques of Integration

8.1

Integration by Parts Integration by parts is a technique for simplifying integrals of the fann

J f(x)g(x) dx. It is useful when f can be differentiated repeatedly and g can be integrated repeatedly without difficulty. The integrals

and

JxCOSXdx

are such integrals because f(x) = x or f(x) = x 2 can be differentiated repeatedly to become zero, and g(x) = cos x or g(x) = eX can be integrated repeatedly without difficulty. Integration by parts also applies to integrals like and

J Inxdx

J

eX cosxdx.

In the rlISt case,j(x) = In x is easy to differentiate and g(x) = I easily integrates to x. In the second case, each part of the integrand appears again after repeated differentiation or integration.

Product Rule in Integral Form If f and g are differentiable functions of x, the Product Rule says that

d dx [f(x)g(x)] = f'(x)g(x)

+ f(x)g'(x).

In terms of indermite integrals, this equation becomes

J

fx [f(x)g(x)] dx

= J

[f'(x)g(x)

+

f(x)g'(x)] dx

or

J

fx [f(x)g(x)] dx

= J

f'(x)g(x) dx

+ J f(x)g'(x) dx.

Rearranging the terms of this last equation, we get

J f(x)g'(x) dx = J

fx [f(x)g(x)] dx -

J f'(x)g(x) dx,

leading to the integration by parts formula

J f(x)g'(x) dx = f(x)g(x) - J f'(x)g(x) dx

{l}

Sometimes it is easier to remember the formula if we write it in differential form. Let u = f(x} and v = g(x}. Then du = f'(x) dx and dv = g'(x) dx. Using the Substitution Rule, the integration by parts fannula becomes

8.1

437

Integration by Parts

Integration by Part. Formula

j udv

=

uv - j vdu

(2)

J

J

This formula expresses one integral, u dv, in terms of a second integral, v du. With a proper choice of u and v, the second integral may be easier to evaluate than the first. In using the formula, various choices may be available for u and dv. The next examples illustrate the technique. To avoid mistakes, we always list our choices for u and dv, then we add to the list our calculated new terms du and v, and finally we apply the formula in Equation (2).

EXAMPLE 1

Find

j We use the formula j

Solution

u dv

xcosxd!:.

=

uv - j v du with

dv = c08xdx, v = sinx.

u =x, du = d!:,

Simplest antiderivative of cos x

Then

j xcosxd!:

=

x sin x - j sinxd!:

=

x sin x

+ cosx +

C.

•

There are four choices available for u and dv in Example 1:

1. Letu = 1 anddv = xcosxd!:. Letu = xcosx and dv = d!:.

2. 4.

3.

Letu = xanddv = cosxd!:. Letu = cosxanddv = xd!:.

Choice 2 was used in Example I. The other three choices lead to integrals we don't know how to integrate. For instance, Choice 3 leads to the integral

j{xcosx - x 2 sinx)d!:.

J

The goal of integration by parts is to go from an integral u dv that we don't see how to evaluate to an integral v du that we can evaluate. Generally, you choose dv first to be as much of the integrand, including d!:, as you can readily integrate; u is the leftover part. When finding v from dv, any antiderivative will work and we usually pick the simplest one; no aIbitrary constant of integration is needed in v because it would simply cancel out of the right-hand side of Equation (2).

J

EXAMPLE 2

Find

j Inxd!:. Solution

J udv = u du

Since

uv -

= lnx =

1

xd!:,

J In x d!: can J vduwith

be written as

Simplifies when differentiated

J In x • 1 d!:, dv

= d!:

v =x.

we use the formula Easy to integrate

Simplest antiderivative

438

Chapter 8: Techniques of Integration Then from Equation (2),

1

lnxdx

=

1

xlnx -

x·fdx

1

xlnx -

=

dx

=

xlnx - x + C.

•

Sometimes we have to use integration by parts more than once.

EXAMPLE 3

Evaluate

1

2

X

x e dx.

Solution

With u = x 2 , dv

= eX dx,

1

du = 2xdx, and v =

x 2e Xdx

x 2e x -

=

21

eX,

we have

xeXdx.

The new integral is less complicated than the original because the exponent on x is reduced by one. To evaluate the integral on the right, we integrate by parts again with u = x, dv = eX dx. Then du = tix, v = eX, and

1

xeXdx

=

xe X -

1

eXdx

=

xe X - eX +

c.

Using this last evaluation, we then obtain

1

x 2e Xdx

=

x 2e x -

= x 2e x -

21

xeXdx

2xe X

+

2e X

•

+ c.

J

The technique of Example 3 works for any integral x"e X dx in which n is a positive integer, because differentiating x" will eventually lead to zero and integrating eX is easy. Integrals like the one in the next example occur in electrical engineering. Their evaluation requires two integrations by parts, followed by solving for the unknown integral.

EXAMPLE 4

Evaluate

1

eXcosxdx.

Solution

Letu

=

eX and dv

1

=

cosxdx. Then du

eXcosxdx

=

eX sin x -

= eXdx,v

sinx,and

=

1

eXsinxdx.

The second integral is like the flISt except that it has sin x in place of cos x. To evaluate it, we use integration by parts with

dv Then

1

eX cosx dx

= sinxdx,

v =

du=exdx.

-C08X,

1 1

=

eX sin x - (-ex cosx -

=

eX sin x + eXcosx -

(-cosx)(e Xdx»)

eXcosxdx.

Integration by Parts

8.1

439

The unknown integral now appears on both sides of the equation. Adding the integral to both sides and adding the constant of integration give

2J

e'cosxdx

=

e'sinx + e'cosx + CI.

,.

Dividing by 2 and renaming the constant of integration give

J

e , cosx dx -- e

EXAMPLE 5

+' 2 e cosx +

SfiX

C.

•

Obtain a fonnula that expresses the integral

J

cos' xdx

in terms of an integral of a lower power of cos x. We may think of cos' x as cos·- I x • cos x. Then we let

Solution

u

=

cosn-

1X

dv

and

c08xdx,

=

so that

du = (n - I) cos·- 2 x(-sinx dx)

v = sinx.

and

Integration by parts then gives

J

cos'xdx = cos·-Ixsinx

+

(n

-1)J

sin2 xcos·- 2 xdx

J

= cos·- I x sin x

+ (n - 1)

= cos·-Ixsinx

+ (n

(1 - cos2 x) cos·- 2 X dx

-1)J

cos·- 2 xdx - (n -

1)J

cos·xdx.

Ifwe add (n

-1)J

cos'xdx

to both sides of this equation, we obtain

J

n

cos' xdx = cos·- I x sin x

+

J

(n - 1)

We then divide through by n, and the final result is

J

.-1' cos• x dx -- cos nx smx

IJ

+~ n

cos·- 2 xdx.

cosn-2 x dx .

•

The fonnula found in Example 5 is called a reduction formula because it replaces an integral containing some power of a function with an integral of the same fonn having the power reduced. When n is a positive integer, we may apply the fonnula repeatedly until the remaining integral is easy to evaluate. For example, the result in Example 5 tells us that

J

3

2

cos xdx =

cos xsinx 3 2

2J

+3

= tcos xsinx

cosxdx

+ tsinx +

c.

440

Chapter 8: Techniques of Integration

Evaluating Definite Integrals by Parts The integration by parts fonnula in Equation (I) can be combined with Part 2 of the Fundamental Theorem in order to evaluate definite integrals by parts. Assuming that both f' andg' are continuous over the interval [a, b], Part 2 of the Fundamental Theorem gives

Integration by Parts Formula for Definite Integrals

tf{x)g'{x) dx

=

f{x)g(x)

l: - tf'{x)g(x) dx

(3)

In applying Equation (3), we normally use the u and v notation from Equation (2) because it is easier to remember. Here is an example. y

EXAMPLE 6 Find the area of the region bounded by the curve y = xe-x and the x-axis fromx = Otox = 4. Solution

0.5

The region is shaded in Figure 8.1. Its area is

1'xe-X dx. -1

1

2

3

4

x

Letu

=

x,dv

e-xdx,v

=

-e-x,anddu = tix. Then,

1'xe-x dx

=

-xe -x l~

=

[-4e-4 - (O)] +

=

-4e-4 - e-x]~

=

-4e-4 - e-4 - (-eO)

=

FIGURE 8.1 The region in Example 6.

-14 (

-e -X) dx

14

e-X dx

= I -

5e-4 '" 0.91.

•

Tabular Integration

J

We have seen that integrals of the fonn f{x)g(x) dx, in which f can be differentiated repeatedly to become zero and g can be integrated repeatedly without difficulty, are natural candidates for integration by parts. However, if many repetitions are required, the calculations can be cumbersome; or, you choose substitutions for a repeated integration by parts that just ends up giving back the original integral you were trying to find. In situations like these, there is a way to organize the calculations that prevents these pitfalls and makes the work much easier. It is called tabular integration and is illustrated in the following examples.

EXAMPLE 7

Evaluate

J

2 x

x e dx.

Solution

With f{x) = x 2 and g{x) = eX, we list:

f(x) and its derivatives

g(x) and its integrals

Integration by Parts

8.1

441

We combine tbe products of tbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain

J

x'e' dx

x'e' - 2xe'

=

+ 2e' + C.

•

Compare tbis witb tbe result in Example 3.

EXAMPLE 8

Evaluate

J

x 3 sinxdx.

Witb f(x) = x 3 and g(x) = sin x, we list:

Solution

f(x) and its derivatives

g(x) and its integrals

x3

(+)

sin x

3x'

(-)

-cosx

(+) (-)

.... -sinx

6x

----------

6

----------

0

----------

~

cos x

~

sin x

Again we combine tbe products of tbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain

J

x 3 sinxdx

= -x 3 cosx

+ 3x'sinx + 6xcosx

- 6 sin x

+ C.

•

The Additional Exercises at tbe eod of!bis chapter show how tabular integration can be used wheo neitber function f nor g can be differentiated repeatedly to become zero.

Exercises 8.1 Integration by Parts Evaluate the integrals in Exercises 1-24 using integmtion by parts.

15. / x'exdx

(x' - 5x)e Xdx

1. / xsinIdx

2. /oeOS1rOdO

3. / t'eostdt

4. / x sinxdx

19. / x'exdx

6. /,e x31nx dx

21. /

5.

/,2

xlnxdx

7. /xeXdx 9.

/x e-

11. /

2

X

2

8. dx

tan-1ydy

13. / xsec2 xdx

17. /

10.

/xe

3x

12. /

23. / e lx cos 3x dx

dx

/(X 2- 2x + l)e sin-1 ydy

14. /4xsec 2 2xdx

e'sinOdO

2x

dx

16. / p

e" dp

4

18. /

(r' + r + l)e'dr

20. /

t'e 4t dt

22. /

e-Yeosydy

24. /

e-2x sin2xdx

Using Substitution Evaluate the integrals in Exercises 25-30 by using a substitutioo prior to integration by parts. 25. /

e v'hl9 tis

26.

J,lx~dx

442 27.

29.

Chapter 8: Techniques of Integration

J.

~/3

xtan2 xdx

0

j

28.

30.

sin (lnx) dx

j j

+ x 2)dx

In(x

d. What pattern do you see? What is the area b-.eo the curve

and the x-axis for 2 (~)

z(lnz)2 dz

n an arbitrary positive integer? Give reasons for your answer.

Evaluating Integrals Evaluate the integrals in Exercises 31-50. Some integrals do not require integration by parts.

31. jxsecX2 dx

32.

33. jXQnx)2dx

34.

45.

j ~: j j v?"+l j j e~ j Vx

47.

J.~/2 8 2 sin 28 d8

35. 37. 39. 41. 43.

49.

46.

48.

J.~/2 x 3 cos 2x dx

50.

J.

36.

x 3 e x'dx

38. dx

40.

sin 3x cos 2x dx

42.

sin eX fix

44.

dx

cos

tsec-Itdt

('

12;v3

(lnx)3

x

dx

x 2 sin x 3 dx sin 2x cos 4x dx

dx

e Vi dx

1;0.

0

~ 'IT.

55. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes and the curve y = cos x, 0 :5 x :s; 71'/2, about b. the line x = 1T/2. 56. Finding volume Find the volume of the solid generated by revolving the region bounded by the x-axis aod the curve y = xsinx,O:5 x:s; '11',about

a. the y-axis.

y=xsinx

5

o -5 52. Finding area Find the area of the region enclosed by the curve y = x cos x aod the x-axis (see the accompaoying figure) for

a. 1T/2 s x s 31T/2. 51T/2.

c. 51T/2 s x s 71T/2.

andx

=

e.

a. Find the area of the region. b. Find the volume of the solid formed by revolving this region

about the x-axis.

y

10

7T'.

(See Exercise 51 for a graph.)

nonnegative integer? Give reasons for your answer.

S

a. about the y-axis. b. about the line x = I.

57. Consider the region boonded by the graphs of y = In x, Y = 0,

d. What pattern do you see here? What is the area b-.eo the curve and the x-axis for mr s x s (n + 1)1T, n an arbitnlry

'" x

54. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes, the curve y = e -r, and the line x = 1

b. the line x =

1T'

b. 31T/2

53. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes, the curve y = eX. and the line x = In 2 about the line x = In2.

a. the y-axis. 2xsin-1 (x 2 )dx

:s x :s 21T. c. 21T :s x :s 31T.

b.

-10

-x-dx

51. Finding area Find the area of the region eoc1osed by the curve y = x sin x aod the x-axis (see the accompaoying figure) for ~

10

1_2 dx xQnx)

Theory and Examples

a. 0

y

jco~dx

j __ j j x' e" j j j -:x j Vx

dx

x3

«~) 2 'fr,

< '1T_X_

c. Find the volume of the solid formed by revolving this region about the line x = -2. d. Find the centroid of the region. 58. Consider the region bounded by the graphs of y aodx = 1.

= tan-I x, y = 0,

a. Find the area of the region. b. Find the volume of the solid formed by revolving this region about the y-axis. 59. Average value A retarding force, symbolized by the dashpot in the accompaoying fIgure, slows the motion of the weighted spring so that the mass's position at time t is

t '" O.

443

8.1 Integration by Parts Find the average value ofy aver the intervaJ. 0

:!;';

,

t

:!;';

2'11".

The idea is to take the most complicated part of the integral, in this case l(x), and simplify it fIrst For the integral ofln x, we get

r

J

lnxdx

~

J

, - lnx,

ye'dy

X _ 1I 7

dx - Il'rIy

+C x + C.

= ye' - II'

= xlnx -

o

,

For the integral of COlI-I X we get

f

-I

C08-l xdx - xoos-Ix

y

cosydy

=

coa-1x

= xoos-Ix - siny + C

Use the formula

60. Avenge value In a mass-spring-dashpot system l:ilcc the one in Exercise 59, the mass's position at time t is

y = ~-t(sint - cost),

t:2!:

Find the average value ofy aver the interval 0

O.

Reduction Formulas

61.

61.

f x~

cosxdt = x"sinx -

1

1

x· sinxQ = -x· cosx +

to cvaluate the integrals in Exercises 67-70. Express your answers in terms ob.

69.

f

70·/IO~Xdx

sec-I x ttt

r

1

1

x·- I COSXQ

If

(4)

fly) dy

Another way to integrate rl(x) (when course) is to use integration by parts with u rewrite the integral. of 1 as

X,,-I sinxQ

If

J

r'(x) dx - xr'(x) -

:s: t :s: 2'11".

In Exercises 61-64, use integration by parts to establish the reduction

-.....

J

rl(x)dt

=

xr1(x)

-I

r 1 is integrable, of r 1(x) and dv = dx to

x(!r1(x»)dt.

(5)

Exm:ises 71 and 72 compare the:results ofusing Equations (4) and (5). 64.

f

(lnx)"Q = x(lnx)· -

nf

71. Equations (4) and (5) give diffen:ul: form.ulas for the integral. of

(lnX),,-I Q

cos-1 x:

65. Show that

L

f

b.1 66. Use inqration by paI19 to obtain the formula

Integration by parts leads to a rule for integrating inverses that usually gives good results:

b.

J

~ yf(y) -


J J

~ xr'(x) -

fly) dy fly) dy

~

X COS-I X -

VI="? +

C

Eq.(!5)

tan-I X ttt =

X tan-I

x -lnsec(tan-1x) + C

ftan-IXQ=xtan-IX-ln\l'1+7"+C

Eq.(4) Eq.(5)

f'(y)dy

Intcgndion by pan. with u

Eq.(4)

Can both integrations be oorrcct? Explain.

, - rl(x), x - f(,)

yf'(y)dy

+ C

tan-I x:

.. I

~

sin(OOS-l x )

72. Equations (4) and (5) lead to different form.ulas for the integral of

Integratfng IlIY8I'Hs of functions

r'(x)dx

cos-1xttt =

-

Can both integrations be oorrcct? Explain.

1 dx. 2 2 VI="? J ~dt=lxVI="?+lJ

J

cos-Ixttt - x COS-I X

-"dv - !'(,)dy

Evaluate the integrals in Exercises 73 and 74 with (a) Eq. (4) and (b) Eq. (5). In each case, check your work by differentiating your answer with respect to x.

73.1

sinh-IxQ

74.1

tanh-Ixdx

444

Chapter 8: Techniques of Integration

8.2

Trigonometric Integrals Trigonometric integrals involve algebraic combinations of the six basic trigonometric functions. In principle, we can always express such integrals in terms of sines and cosines, but it is often simpler to work with other functions, as in the integral

J

sec?-xdx = tanx

+ C.

The general idea is to use identities to transform the integrals we have to f"md into integrals that are easier to work with.

Products of Powers of Sines and Cosines We begin with integrals of the form:

J

sin"' x cos' x dx,

where m and n are nonnegative integers (positive or zero). We can divide the appropriate substitotion into three cases according to m and n being odd or even.

Case 1 Ifm is odd, we write m as 2k + 1 and use the identity sin2 x = 1 - cos2 X to obtain sin"'x = sin2k+1X = (sin2 x)k sinx = (1 - cos2 x)ksinx.

(1)

Then we combine the single sin x with dx in the integral and set sinxdx equal to -d(cosx).

Case 2 If m i. even and n i. odd in

J

sin"' x cos' x dx, we write n as 2k + 1 2 2 and use the identity cos x = 1 - sin X to obtain cos'x = cos2k+ 1x = (COS2X)kcosx = (1- sin2X)kcosx.

We then combine the single cos x withdx and set cosx dx equal to d(sinx).

Case 3

If botb m and n are even in . 2

_ 1 - cos2x 2 '

smx-

J sin"' x cos' x dx, we substitote l+cos2x 2 cosx= 2

to reduce the integrand to one in lower powers of cos 2x.

Here are some examples illustrating each case.

EXAMPLE 1

Evaluate

J

3 2 sin xcos xdx.

(2)

445

8.2 Trigonometric Integrals Solution

This is an example of Case I.

f

f f f f

sin3 X COS2 X dx = =

=

=

EXAMPLE 2

sin2 X COS2 x sin x dx

m is odd.

(I - COS2X)cos2x(-d(cosx))

sinxdx~ -d(co. .)

2 2 U )(U )( -du)

(I 4

(U

-

f

Multiplytonns.

COSS xdx.

This is an example of Case 2, where m = 0 is even and n = 5 is odd.

cossxdx = =

=

=

EXAMPLE 3

Solution

du

~ co. .

Evaluate

f Solution

2 U )

U

f

cos'xcosxdx =

f f

f

(I - sin2xj2d(sinx)

cosxdx

(I - u 2 j2du (I - 2u 2

u

=

=

d(sinx)

sinx

+ u·) du

u - ~u3 + !u s + C 3 5 Evaluate

f f if if

= sinx -

~sin3x + !sinsx + C. 3 5

•

sin2xcos'xdx.

This is an example of Case 3.

f.

sm2xcos• x dx = =

=

=

+ COS2x)2dx 2

(I - COS2x)(I 2 (I - cos2x)(1

(I

i [x +

+

m and n both even

+ 2cos2x + cos2 2x)dx

cos2x - cos2 2x - cos3 2x)dx

tsin2x -

f(COS22X + COS3 2X)dx].

For the term involving cos2 2x, we use

f

2 cos 2xdx = t f (I = t

+

cos4x)dx

(x + i Sin4X).

Omitting the constant of

integration until the flnal result

446

Chapter 8: Techniques of Integration For the cos3 2x term, we have

J

cos3 2xdx = =

J V

u = sin2x,

(I - sin2 2x)cos2xdx 2

(I - u )du

=!

du

t

(Sin2x -

Sin32x ).

~

2cos2<",

Again

omitting C

Combining everything and simplifYing, we get

J

2

4

sin xcos xdx =

/6 (x - i Sin4x + t Sin32x) + C.

•

Eliminating Square Roots In the next example, we use the identity cos2 0 = (I

EXAMPLE 4

Evaluate

1"f4 VI + Solution

cos4xdx.

To eliminate the square root, we use the identity 20 1+ cos20 cos = 2

With 0 =

+ cos 20)j2 to eliminate a square root.

1+ cos20

or

= 2cos2 0.

2x, this becomes 1+ cos4x

= 2cos2 2x.

Therefore,

{"f4

10

VI

+

{"f4

cos4xdx = =

10

{"f4

10

2

V2cos 2xdx =

Vi {"f4 Icos2x1 dx = Vi {"f lo

= Vi

[S~2x

r

2

ViVcos 2xdx 4

lo

=

'?

[I - 0] =

cos2xdx

,?

cos2x 2: 0 on [0, "./4]

•

Integrals of Powers of tan x and sec x We know how to integrate the tangent and secant and their squares. To integrate higher powers, we use the identities tan2 x = sec' x - I and sec2 x = tan2 x + I, and integrate by parts when necessary to reduce the higher powers to lower powers.

EXAMPLE 5

Evaluate

J

4

tan xdx.

Solution

J

tan4 xdx = =

=

=

J J J J

tan2 x·tan2 xdx = 2

2

tan xsec xdx tan2 xsec2 xdx 2

2

tan xsec xdx -

J J J J

tan2 X' (sec2 x - I) dx

tan2 xdx (sec2 x - I)dx 2

sec xdx

+

J

dx.

447

8.2 Trigonometric Integrals In the first integral, we let

u

du = sec2 xdx

tanx,

=

and have

1

t

2

u du =

u3

+ Ct.

The remaining integrals are standard forms, so

1

tan4 xdx

EXAMPLE 6

3

= ttan x - tanx + x +

Evaluate

1

•

c.

3

sec xdx.

We integrate by parts using

Solution

u = secx,

dv

= sec2 x dx,

v = tanx,

du = secxtanxdx.

Then

1

se2 xdx = sec x tan x = sec x tan x

-

= sec x tan x +

1 1 1

(tanx}{secxtanxdx)

{sec2 x - I)secxdx secxdx -

Combining the two secant-cubed integrals gives

21

and

sec3 xdx = secxtanx +

tan2 x

=

sec2 x - 1

1 1

sec3 xdx.

secxdx

1

sec3 xdx = !secxtanx +!In Isecx + tan x I +

c.

•

Products of Sines and Cosines The integrals

1

sin mx sin nx dx,

1

sin mx cos nx dx,

and

1

cos mx cos nx dx

arise in many applications involving periodic functions. We can evaluate these integrals through integration by parts, but two such integrations are required in each case. It is simpler to use the identities sin mx sin nx = ! [cos {m - n)x - cos {m + n)x],

(3)

sin mx cos nx = ! [sin {m - n)x + sin {m + n)x] ,

(4)

cos mx cos nx =

2I [cos {m

- n)x + cos {m + n)x].

(5)

These identities come from the angle sum formulas for the sine and cosine functions (Section 1.3). They give functions whose antiderivatives are easily found.

448

Chapter 8: Techniques of Integration

EXAMPLE 7

Evaluate

J

sin 3x cos 5xdx.

Solution

From Equation (4) with m = 3 and n = 5, we get

J

sin3xcos5xdx

=

=

tJ tJ

=

[sin (-h) + sinRx]dx (sinRx - sinh)dx

_cosRx 16

+ cosh + C 4

•

.

Exercises 8.2 Powers of Sines and Co';nes Evaluate 1he integrals in Exercises 1-22. 1.

3. 5.

7. 9.

11.

13.

J J J J J J J

. x dx 2. f 0 3Sffi 3

cos3 x sin x dx

4.

sin3 xlix

sinS xlix

6.

cos' 4x dx

/.w/'

12.

cos2 x dx

14. f0P sin2 xdx

J J

16 8in2 xcos2 xdx

8 cos' 20 sin 20 dO

20.

0

VI -

cosx

l

w

3 cos' 3x dx

J

cos' 2x sin' 2x dx

33.

J J

7 cos' tdt

37.

8 cos' 2'ITx dx

/.W 8 sin' Y cos y th>

0

/.

OVI - cos20dO

32.

J~ (I

VI -

wp

44.

J Jsec' J J J J

4tan'xdx

46.

[/.

tan' xdx

4S.

sec' x tan x dx

34.

sec' x tan x dx

36.

sec' x tan2 x dx

38.

~/'

41.

22. f0P sin 20 cos' 20 dO 43.

Integrating Square Roots 45. cos2xdx

'w/.

J.

VI -

0

2

/.W VI -

/.

30.

J J J

39.1

2

24.

VI + sinxdx

0

sin2xdx

- cos2 t)'p dt

Powers of Tangents and Secants Evaluate 1he integrals in Exercises 33-50.

35.

Evaluate 1he integrals in Exercises 23-32.

/.2W ~I - 2cOS Xdx

2S.

cos x dx • sinx

J [P J J

2 sec' xdx

sec' 0 dO csc' 0 dO

W/'

23.

W/6

dx

WP

31.

sin'~dx

sin3 x cos3 x dx

IS.

29.

5,"/6

10.

17. /,wSSin'xdx

21.

sin' 2x cos 2x dx

cos3 xlix

8in2 x

(Hint: Multiply by ~: :: :::.)

J J

S. f

16.

J.

1r/2

w/3

cos2xdx

15. f0P sin'yth>

19.

27.

47.

49. [/3 cot' x dx w/'

40.

42.

sec x tan2 x dx x tan' x dx

sec4 x tan2 x dx

eX sec3 eX dx

3 sec' 3x dx 6

sec

X

dx

6tan'xdx

~/'

50.

J J

cot' 2x dx

8 cot' tdt

8.3 Products of Sines and Cosines Evaluate the integrals in Exercises 51-56.

65.

51.

J

52.

J

53.

J~ sin 11: sin 11: tb:

54.

Jor/2 sinx cosx dx

55.

J

sin 11: cos 2x tb:

cos 11: cos 4x tb:

56.

sin 2x cos 11: tb:

l

w/2

67.

59. 61.

•

J J J

sin' 8 cos 38 d8

58.

cos' 8 sin 28 d8

60.

sin 8 cos 8 cos 38 d8

62.

J

tan' x csc x tb:

66.

x sin' x tb:

68.

J J

cot x tb: cos2 X

x cos' x tb:

69. Arc lengtb Find the length of the curve

cos x cos 7x dx

~/2

y

3

sec x dx tan x

8.3

64.

J J J

cos' 28 sin 8 d8

In (secx),

0:5 x:5

",/4.

71. Volume Find the volume generated by revolving one arch of the curve y = sinx about the x-axis.

sin' 8 cos 28 d8

72. Area Find the area between the x-axis and the corve y

VI

sin 8 sin 28 sin 38 d8

J

~

70. Center of gravity Find the center of gravity of the region bounded by the x-axis, the curve y = sec x, and the lines x = -",/4,x ~ ",/4.

+ cos4x,O ~ x ~

~

'IT.

73. Centroid Find the centroid of the region bounded by the graphs ofy = x + cos x andy = o forO ~ x ~ 2w.

Assorted Integrations Use any method to evaluate the integrals in Exercises 63-{)8.

63

J J

449

Applications

Exercises 57-{)2 require the use of various trigonometric identities before you evaluate the integrals. 57.

Trigonometric Substitutions

.,

74. Volume Find the volume of the solid formed by revolving the region bounded by the graphs of y ~ sin x + sec x, Y ~ 0, x ~ 0, and x ~ ",/3 about the x-axis.

sm x tb: cos4 X

Trigonometric Substitutions Trigonometric substitutions occur when we replace the variable of integration by a trigonometric function. The most common substitutions are x ~ a tan IJ, x ~ a sin IJ, and x ~ a sec IJ. These substitutions are effective in transfurming integrals involving Va' + x', Va' - x', and x' - a'into integrals we can evaluate directly since they come from the reference right triangles in Figure 8.2.

V

Withx

~

atanlJ,

a' + x' Withx

~

~

a' + a'tan'lJ

~

a'(\ + tan'lJ)

~ a'sec'lJ.

~

a' - a' sin'lJ

=

a'(1 - sin'lJ)

~

asinlJ,

a' - x'

a' cos'lJ.

a x=atan8

x=asin8

x=asecO

Va' + x' ~ alsecol

Va' - x' ~ alcosol

Vx'- a'~ aitanOI

FIGURE 8.2 Reference triaog1es for the three basic substitotions identifying the sides labeled x and a for each substitotion.

450

Chapter 8: Techniques of Integration Withx = asecO,

x2 6

"2

-

a2

a 2 sec'0 - a 2

=

= a 2(sec2 0 -

1) = a 2 tan2 0.

We want any substitution we use in an integration to be reversible so that we can cbange back to the original variable afterward. For example, if x = a tan 0, we want to be able to set = tan-I (x/a) after the integration takes place. If x = a sin 0, we want to be able to set 0= sin-I (xja) when we're done, and similarly fur x = asecO. As we know from Section 7.6, the functions in these substitutions bave inverses only for selected values of 0 (Figure 8.3). For reversibility,

o

x

a

-2 "

x

=

(~)

with

-2 " < 0 < 2' "

sin-I ("') a

with

-2';; 0 ,;;

with

1:<0<;

atanO requires 0= tan-I

6 8

"2

= sin-l~

x=asinO requires

o=

x

a

-1

x

= asecO

requires

0= sec-I

(~)

71"

2' if ~ ~ 1,

a. 2<0';;" if "'<-I

-2 "

To simplify calculations with the substitution x = a sec 0, we will restrict its use to intograisinwhichx/a;" l.ThiswillplaceOin[O, ,,/2) and make tan 0 ;" O.Wewillthenbave V x 2 - a 2 = Va 2 tan2 0 = Ia tan 0 I = a tan 0, free of absolute values, provided a > O.

6

.-J "

71"

(J =

sec-1 ~

"2 x

-1

0 1

a

Procedure For a Trigonometric Substitution

ax, and specify the

FIGURE 8.3 The arctangeot, arcsine, and arcsecant of x/a, graphed as functions of

1. Write down the substitution for x, calculate the differential selected values of 0 for the substitution.

x/a.

2. Substitute the trigonometric expression and the calculated differential into the integrand, and then simplify the results algebraically. 3. Integrate the trigonometric integral, keeping in mind the restrictions on the angle 0 for reversibility. 4. Draw an appropriate reference triangle to reverse the substitution in the integration result and convert it back to the original variable x.

EXAMPLE 1

Evaluate

J~. Solution

We set

x = 2 tan 0,

ax =

+ x2

+ 4tan2 0

4

= 4

2 sec2 0 dO, = 4(1

_71"<0<" 2 2'

+ tan2 0)

= 4 sec2 O.

Trigonometric Substitutions

8.3

21'

Then

Jv'4+7 Jvi J J ln =ln17 +~I + ln =

2 se~ 0 dO 4 see2 0 =

=

sec 0 dO

=

IseeO

dx

2

FIGURE 8.4 Reference triangle for , = 2 tan 8 (Example I): tan8 = :t: 2

2

see 0 dO Isec 0 I sec(J

-

sec 8 -

v'4+7 2 .

>

ofor -~2 < (J < '!! 2

+ tan0l + C

From Fig. 8.4

C.

and

451

Notice how we expressed Isec 0 + tan 0 I in tenDs of x: We drew a reference triangle for _ the original substitution x = 2 tan 0 (Figore 8.4) and read the ratios from the triangle.

EXAMPLE 2

Solution

Evaluate

We set

_1r
dx = 3 cos 0 dO,

x = 3 sinO,

9 - x 2 = 9 - 9sin2 0 = 9(1 - sin2 0) = 9 eos2 O. Then

JV9=7 x2dx

=

=

J J J 9

sin2 0dO

= 9

FIGURE 8.5 Reference triangle for 3 sin 8 (Example 2):

, =

sin8

=

.! 3

and

co.8=

V9=7 3

EXAMPLE 3

2

9sin 0·3eosOdO 13 eosOI

I -

-

cosO> 0 for _'!! < 8 < '!! 2 2

~OS20 dO

=

~ (0

=

~(O

=

~ (Sin-It -

=

~Sin-It - ~V9=7 +

Sin 20) 2

+C

- sinO cosO) t·

+C

sin2(J

~) + C

Solution

We first rewrite the radical as

2 sin 8 cos 8

Fig. 8.5

-

C.

Evaluate

x

=

2

>5'

452

Chapter 8: Techniques of Integration to put the radicand in the form x'

- a'. We then substitute

2 x = Ssec 0,

is

x' - (~)' =

2 dx = Ssec 0 lanO dO,

sec' 0 -

~lan'O

=

25

25

~x' - (~)' ~ ItanOI ~lanO. =

J FIGURE 8.6 Ifx = (2/5)sec 6, 0< 6 < 'IT/2,then6 = sec- I (5.
=

=

tan 6 > o for 0< 0 < ",/2

=

With these substitutions, we have dx v'25x' - 4 =

<;

2~

~(sec'O - 1)

=

0< 0

J

dx 5v'x' - (4/25) =

J

(2/5) sec 0 lan 0 dO 5' (2/5) lanO

tJ tin + tin I~ + v'25~' - 41 + c. secOdO =

IsecO

+

tanOI

C Fig. 8.6

•

EXERCISES 8.3 Using Trigonometric Substitutions Evaluate the integrals in Exercises 1-2S.

1

·

1v'9+7

2

dx

.

31'~ • -,4+x'

1.'1' ~ 0

1 1 1Vy' 13.1 ~, 7. 9.

11.

8.

"\I25=7dt

dx , V4x' - 49 Y

x2 x 2

49

dy,

1

-

3dx

1.' 1.

21. 1

+ 9x'

4 dx • 0 S+2x' I 2dx /2Vi 6. 0 4x'

dx

5.

1VI

19 1 Sdw • w'V4 - w'

x>-7 2

y> 7 x>1

10 • 12. 14.

23.

VI _

1 1 1Vy' -

dy, Y

2dx

1

X3

>~

25

y'

_~, 2 - 1

vx

5

15·1~dx 9- x 2

1 · W+4

17

x'dx

18.

1 1

0

27.

x'~

'/2'

29 1 Sdx • (4x' + I)'

x'dx x2 - 1

x> 1 33 1 v'dv • (1 - v')'/2

x> 1

24.

26. 28.

1.' dx 0 (4 _ x')'/2 1

x'dx (x' _ 1

)'/2 '

x > 1

1(I-X')I/2 4 dx x

6dt (9t' + I)'

30 1

·

32 1 xdx • 25 + 4x' 34.

1 (1 - r')'/2 8 dr r

In Exercises 35-4S, use an appropriate substitution and then a 1rigonome1ric substitution to evaluate the integrals.

35 •

ln4

1. 0

[1/4

dx

22. 1 x Vx' - 4dx

4x' dx (1 _ x')'/2

1 (1 - x')'/2 6 dx x

31. 1

Assorted Integrattons

- -x', d x 4+x

1.V3/2

dx

V9

>5

Vse any method to evaluate the integrals in Exercises 15-34. Most will require 1rigonome1ric substitutions, but some can be evaluated by other methods. 16.

\00

+ 25x'

25.1 (x ,dx - 1)

v'1=9t2dt 5dx x V25x' - 9'

36

w~ W'dw

20. 1

37.

e

I dt

Ve u + 9 2 dt

JI/12 vi + 4tvl

36

• 38

lin

(4/')

In ('/4)

I dt

_-"e~~

(1

['

· JI

+ eU )'/2

dy

yVI

+ (\ny)'

453

8.4 Integration of RationaL Functions by PartiaL Fractions

.. J J · '\17=1 J · V1+7 . . J)4 ;x '7. JVx~
40.J~ l+x


·

J JVI

41

x
42 .~
43

x
44.

56. Consider 1he region bounded by the graphs of y - Vxtan-1 x and y - 0 for 0 :S X :s L Find 1he volume of the solid formed by revolving this region about the x-axis (see accompanying ftgure). y

(lnx)2. xlnx tb

-, _ V:xtan-1x


~fL-----4x

(Hint: Let x _ u 1.)

o

57. Evaluate

Initial Value Problems Solve the initial value problems in Exercises 49-52 for y as a fimction ofx.

dy ry-o 49. x dx - 'vx- 4, SO.

x~2.

y(2) = 0

.~dy­ vx- - 9 dx - 1, x> 3, y(5) = 1n3

51. (x 1

+ 4)

52. (xl

+ 1)2 :

:

= 3, y(2) = 0 -

W+I,

y(O) - 1

J x 3 ~_dx using

a. integration by p!lrts. b. a v-substitution.

e. a trigonometric substitution. 58. Path of. water dder Suppose that a boat is positioned at the mgin with a water skier tethered to the boat at the point (30. 0) on a rope 30 ft long. As the boat travels along the positiw y-axis. the skier is pulled behind fue boat along an unknown path y = f(x), as shown in the acccwnpanying figure. -\1'900 Xl L Showthatf'(x) = X (!lint: Assume that the skier is always pointed directly at the boat and the rope is on a line tangent to the path y = f(x).)

AppLications and Examples 53. Area Find the area of the :region in the first quadrant that is enclosed by the coordinate axes and the curve y = v'9 - x 2/3. 54. Area

I

b. Solve the equation in part (a) fur f(;&), using f(30) = y y - f{x) pGb. of mel"

Find the area enclosed by the ellipse

x2

o.

y2

2+2=1.

a b 55. ConsideIthcregi.on bmmdedbythe graphs ofy = sin-I :C,Y = 0, andx = 1/2.

a. Find the area of the region. b. Find the centroid of the regicm..

8.4

~~L-____~~~_ x

0];

(30,0)

Integration of Rational Functions by Partial Fractions This section shows how to express a rational function (a quotient of polynomials) as a sum of simpler fractions, called partial fractions, which are easily integrated. For instance, the ,..tiona! function (S.% - 3)/(x' - 2x - 3) can be rewri_ as --,-5~x"c-;-,3"--:o ~ _2_ Xl 2x 3 x+l

+ _3_

x-3'

You can verify this equation algebraically by placing the fractions on the right side over a common denominator (x + l)(x - 3). The skill acquired in writing rational functions as such a sum. is useful in other settings as well (for instance, when usmg certain transfonn methods to solve di:ffcrcntial equations). To integrate the rational function

454

Chapter 8: Techniques of Integration

(5x - 3)/(x 2

- 2x - 3) on the left side of our previous expression, we simply swn the integrals of the fractions on the right side:

J (x

5x - 3

dx = J_2_ dx x + I

+ I)(x - 3)

= 2ln

+ J_3_ dx x - 3

Ix + II +

3ln

Ix -

31 +

c.

The method for rewriting rational functions as a swn of simpler fractions is called the method of partial fractions. In the case of the preceding example, it consists of finding constants A and B such that

5x-3 =_A_+~ x 2 -2x-3 x+1 x-3·

(1)

(Pretend for a moment that we do not know that A = 2 and B = 3 will work.) We call the fractions A/(x + I) andB/(x - 3) partial fractions because their denominators are only part of the original denominator x 2 - 2x - 3. We call A and B undetermined coefficients until proper values for them have been found. To find A and B, we flfst clear Equation (I) of fractions and regroup in powers of x, obtaining

5x - 3 = A(x - 3) + B(x + I) = (A + B)x - 3A + B. This will be an identity in x if and only if the coefficients of like powers of x on the two sides are equal: A

+B

= 5,

-3A

+B

=

-3.

Solving these equations simultaneously gives A = 2 and B = 3.

General Description of the Method Success in writing a rational function f(x)/g(x) as a swn of partial fractions depends on two things:

• The degree of f(x) must be less than the degree of g(x). That is, the fraction must be proper. If it isn't, divide f(x) by g(x) and work with the remainder term. See Example 3 of this section. • We must how the factors of g(x). In theory, any polynomial with real coefficients can be written as a product of real linear factors and real quadratic factors. In practice, the factors may be bard to find. Here is how we fmd the partial fractions ofa proper fraction f(x)/g(x) when the factors ofg are known. A quadratic polynomial (or factor) is irreducible if it cannot be written as the product of two linear factors with real coefficients. That is, the polynomial has no real roots.

Method of Partial Fractions (f(x)/g(x) Proper) 1. Let x - r be a linear factor of g(x). Suppose that (x - r)m is the highest power ofx - r that dividesg(x). Then, to this factor, assign the swn of the m partial fractions:

Aj

(x - r)

+

A2 + ... + Am (x - r)2 (x - r)m·

Do this for each distinct linear factor of g(x).

continued

8.4 Integration of Rational Functions by Partial Fractions

455

2. Let x 2 + px + q be an irreducible quadratic factor of g(x) so that x 2 + px + q bas no real roots. Suppose that (x 2 + px + q)n is the highest power of this factor that divides g(x). Then, to this factor, assign the sum of the n partial fractions:

BIX

+

Cl

c--.~-~c-

~+~+~

+

B2X

+

C2

~+~+~

B.x + C. + ... + -;-c;c-"--~'-=

~+~+~.

Do this for each distinct quadratic factor of g(x). 3. Set the original fraction f(x)/ g(x) equal to the sum of all these partial fractions. Clear the resulting equation of fractions and arrange the terms in decreasing powers of x. 4. Equate the coefficients of corresponding powers of x and solve the resulting equations for the undetennined coefficients.

EXAMPLE 1

Use partial fractions to evaluate

J

x2+4x+1 (x - I)(x + I)(x + 3)

Solution

ax .

The partial fraction decomposition bas the fonn

2 x +4x+1 =_A_+~+~ (x - I)(x + I)(x + 3) x - I x + I x +3. To find the values of the undetermined coefficients A, B, and C, we clear fractions and get

x2

+ 4x + I

= A(x

+ I)(x + 3) + B(x - I)(x + 3) + C(x - I)(x + I) + 4x + 3) + B(x 2 + 2x - 3) + C(x 2 - I)

= A(x 2 = (A

+B +

C)x 2

+

(4A

+ 2B)x +

(3A - 3B - C).

The polynomials on both sides of the above equation are identical, so we equate coefficients of like powers of x, obtaining Coefficient of x 2: Coefficient of Xl: Coefficient ofxo:

A+ B+C=I 4A+2B =4 3A-3B-C=1

There are several ways of solving such a system oflinear equations for the unknowns A, B, and C, including elimination of variables or the use of a calculator or computer. Whatever method is used, the solution is A = 3/4, B = 1/2, and C = -1/4. Hence we bave

J

X2 + 4x + I (x - I)(x + I)(x

+

3)

ax -

=

J[l_l_ 4x - I

+ 1_1_ _ l _I _]ax 2x + I 4x +3

"In Ix - II + 21n Ix + II - "In Ix + 31 + 3

I

I

K,

where K is the arbitrary constant of integration (to avoid confusion with the undetennined coefficient we labeled as C). •

EXAMPLE 2

Use partial fractions to evaluate

J

6x+7

(x

+ 2)'

ax .

456

Chapter 8: Techniques of Integration Solution First we express the integrand as a sum of partial fractions with undetennined coefficients.

6x+ 7 (x + 2)'

=

_A_+ B x +2 (x + 2)'

+7

=

A(x

=

Ax

6x

+ 2) + B + (2A + B)

Multiply both sides by (x

+ 2)'.

Equating coefficients of corresponding powers of x gives

A=6

2A

and

+B

=

12

+B

7,

=

A

or

= 6

and

B = -5.

Therefore,

j

EXAMPLE 3

C! 2-

(~++2;2 dx =

j

=

6

=

6ln

j '! x

2)') dx

(x :

2 - 5

j

(x +

2r dx 2

Ix + 21 + 5(x + 2r t + C.

•

Use partial fractions to evaluate 3

2

-X-3 dx. j 2x x-4x-2x-3 2

Solution First we divide the denominator into the numerator to get a polynomial plus a proper fraction. 2x x 2 - 2x - 3)2x 3 - 4x 2 - X - 3 2x 3

-

4x 2

-

6x 5x - 3

Then we write the improper fraction as a polynomial plus a proper fraction.

2x 3 -4x 2 -x-3 x 2 -2x-3

~--.~=--~~

= 2x

5x-3 + --,-'~~---c2 x -2x-3

We found the partial fraction decomposition of the fraction on the right in the opening

example, so 3

2

2x -4x -X-3 dx =j2xdx+j 5x-3 dx x 2 -2x-3 j x 2 -2x-3 = j2xdx+ j x i l dx + =

EXAMPLE 4

x2

jX~3dx

+ 2ln Ix + 11 + 3ln Ix - 31 + c.

•

Use partial fractions to evaluate

j

-2x + 4 (x 2 + 1)(x _ 1)' dx.

Solution The denominator has an irreducible quadratic factor as well as a repeated linear factor, so we write

(x 2

-2x + 4 + 1)(x - 1)2

=

Ax

x2

+B +~ + D + 1 x-I (x - 1)'·

(2)

8.4 Integration of Rational Functions by Partial Fractions

457

Clearing the equation of fractions gives -Zl; + 4 = (Ax + B)(x - 1)2 + C(X - 1)(x2 + I) + D(x 2 + 1) = (A + C)x 3 + (-2A + B - C + D)x2 + (A - 2B + C)X + (B - C + D).

Equatiog coefficients oflike tenns gives Coefficients of x 3 :

O=A+C

°

2

-2A + B - C -2 = A - 2B + C 4=B-C+D

Coefficients of x : Coefficients of x I : Coefficients ofxo:

=

+D

We solve these equations simultaneously to find the values ofA,B, C, andD: -4 = -2A,

A = 2

Subtract fourth equation from second.

C = -A =-2

B = (A

From the flfSt equation

+ C + 2)/2

=

I

D=4-B+C=1.

From the third equation and C

~

-A

From the fourth equation

We substitute these values into Equation (2), obtaining -Zl; + 4 = Zl; + I _ _ 2_ + I (x 2 + I)(x - 1)2 x2 + I x - I (x - 1)2 . Finally, using the expansion above we can integrate:

!

-Zl;+4 dx_!(Zl;+I _ _ 2_+ I (x 2 + I)(x - 1)2 ~2 + I x - I (x - 1)2

)dx

_!C'~_+_I_ _ _2_+

-

=

EXAMPLE 5

~2 + I

x - I

I (x - 1)2

10 (x 2 + 1) + tan-I x - 210 Ix - II - x

~

)dx I + C. •

Use partial fractions to evaluate

! Solution

x2 + I

x(x 2

a:: 1)2 .

The fonn of the partial fraction decomposition is I =4+Bx+C+ Dx+E x(x 2 + 1)2 x x2 + I (x 2 + 1)2

Multiplying by x(x 2 + 1)2, we have I = A(x 2 + 1)2 + (Bx + C)x(x 2 + I) + (Dx + E)x = A(x4 + Zl;2 + I) + B(x 4 + x 2 ) + C(x 3 + x) + Dx 2 + Ex = (A + B)x4 + Cx 3 + (2A + B + D)x 2 + (C + E)x + A

If we equate coefficients, we get the system A+B=O,

C = 0,

2A

+B +D

=

0,

C+E=O,

A=l.

458

Chapter 8: Techniques of Integration Solving this system gives A = I,B = -I, C = O,D = -I, andE = O. Thus,

J

Iix

x(x 2

-

+

I)' -

=

J[1 ----=-'---Jx - J J x

+

x2 + I

Iix

+

(x 2

-x

+

xlix x2 + I -

I)'

] Iix

xlix (x 2 + I)' u =

du

+ I, 2xdx

Xl

=

I I =Inlxl-zlnlul +2u+ K

= In Ixl = In

t

ln (x 2

+ I) +

+

I

Ixl ~

2(x2

+ I)

2(X,'+ I)

+K

+ K.

HisTORICAL BIOGRAPHY

The Heaviside uCover-upu Method for Linear Factors

Oliver Heaviside

When the degree of the polynomial I(x) is less than the degree of g(x) and

•

(185()-1925)

g(x) = (x - r,)(x - r2)'" (x - rn) is a product of n distinct linear factors, each raised to the lITst power, there is a quick way to expand I(x)/g(x) by partial fractions.

EXAMPLE 6

Find A, B, and C in the partial fraction expansion

x + I =_A_+~+~ (x - I)(x - 2)(x - 3) x - I x - 2 x - 3. 2

Solution

(3)

If we multiply both sides of Equation (3) by (x - I) to get x2 + I B(x - 1) C(x - I) -;---"-::-cO---'----;:-c = A + + ----'--~ (x - 2)(x - 3) x - 2 x - 3

and set x = I, the resulting equation gives the value of A:

(I)' + I (1 - 2)(1 - 3)

= A

+ 0 + 0,

A=l. Thus, the value of A is the number we would have obtained if we had covered the factor (x - I) in the denominator of the original fraction

x2 + I (x - I)(x - 2)(x - 3) and evaluated the rest at x = I:

(1)' + I

A =

(:= x= -=I~ ) "1

[1 =

n

Cover

(~I--~2:-:)(-1-----:-:-3)

2 = I (-1)( -2)

(4)

8.4 Integration of Rational Functions by Partial Fractions

459

Similarly, we find tbe value of B in Equation (3) by covering tbe factor (x - 2) in Expression (4) and evaluating tbe rest at x = 2:

(2)' + I B = -(2---1-)~ I (~ x =- =2)= '1'---(2---3)

5 (1)(-1) = -5.

f

Covor

Finally, C is found by covering tbe (x - 3) in Expression (4) and evaluating tbe rest at

x

=

3: 10 (3)2 + I C = -:-(3---1-:-)(:-'3-'---2~ )1 =:'(x= _==C )"'11 = (2)( I) = 5. 3

•

f

Cover

Heaviside Method

1. Write the quotient with g(x)factored: f(x) g(x)

f(x) (x - r,)(x - r2)'" (x - rn) .

2. Cover the factors (x - ri) ofg(x) one at a time, each time replacing all tbe uncoveredx's by tbe number rio This gives a number Ai for each root ri:

A,

=

A2 =

f(r,) (r, - r2)'" (r, - rn) f(r2) (r2 - r,)(r2 - 1'3) ... (r2 - rn)

3. Write the partialjraction expansion off(x)/g(x) as f(x) g(x)

EXAMPLE 7

Use tbe Heaviside Metbod to evaluate

J

x3

Solution g(x) = x 3

x+4

+ 3x 2 - lOx

ax .

The degree of f(x) = x + 4 is less tban tbe degree oftbe cubic polynomial 3x 2 - lOx, and, witbg(x) factored,

+

x+4 x3

+ 3x 2 -

lOx

x+4 x(x - 2)(x + 5) .

460

Chapter 8: Techniques of Integration The roots ofg(x) are'l = 0,'2 = 2, and'3 = -5. We find Al

=

0 +4 = 4 = _~ 2)(0 + 5) (-2)(5) 5

GJ (0 f

Cover

A _ 2 -

2+4 2 1 (x - 2) (2 + 5)

6 (2)(7)

1

3 7

f

Cover

-5 + 4

A3 = -

-I (-5)( -7)

__-----"'--"--~===J

+ 5)

(-5)(-5 - 2) 1 (x

1

I 35 .

n

Cover

Therefore,

x +4 = _~ x(x - 2)(x + 5) 5x

+

I

3 7(x - 2)

35(x

+ 5)'

and

J

x+4

x(x _ 2)(x

+

2 5)dx = -slnlxl

3

I

+ 7 1n Ix - 21 - 35 In Ix + 51 + C.

•

Other Ways to Determine the Coefficients Another way to determine the constants that appear in partial fractions is to differentiate, as in the next example. Still another is to assign selected numerical values to x.

EXAMPLE 8

FindA,B, and Cin the equation

x - I (x

+

_A_

=

1)3

X

+

+

+

B

(x

I

+

C

(x

1)2

+

1)3

by clearing fractions, differentiating the result, and substitoting x = -I.

Solution

We fIrst clear fractions: x - I = A(x

+ I)' + B(x + I) + C.

Substituting x = - I shows C = -2. We then differentiate both sides with respect to x, obtaining

+

I = 2A(x

1)

+ B.

Substituting x = - I shows B = I. We differentiate again to get 0 = 2A, which shows A = O. Hence,

x - I (x

+

2

I

1)3

(x

+

1)3'

•

In some problems, assigning small values to x, such as x = 0, ±I, ±2, to get equations in A, B, and C provides a fast alternative to other methods.

EXAMPLE 9

FindA,B, and Cin the expression

x +I =_A_+~+~ (x - I)(x - 2)(x - 3) x - I x - 2 x - 3 2

by assigning numerical values to x.

8.4 Integration of Rational Functions by Partial Fractions Solution

461

Clear fractions to get x2

+I

3)

= A(x - 2)(x -

+ B(x

- I)(x - 3)

+ C(x

- I)(x - 2).

Then let x = 1,2,3 successively to f'mdA, B, and C:

(1)2 + 1 = A( -1)( -2) + B(O) + C(O)

x = 1:

x

2 = 2A A=1 (2)2 + 1 = A(O) 5 =-B

= 2:

+ B(I)( -1) + C(O)

B =-5 x

(3)2

= 3:

+

1 = A(O)

10 C

= =

+ B(O) + C(2)(l)

2C 5.

Cooclusioo: 2

x + 1 =_1_ _ _ 5_+_5_ (x - l)(x - 2)(x - 3) x-I x - 2 x - 3.

•

Exercises 8.4 Expanding Quotients into Partial Fractions

20 /

Expand the quotients in Exercises 1-8 by partial fractions.

1 5x-13 · (x - 3)(x - 2)

3.

x+4 , + I)

(x

5 z + I 'z'(z-I)

(' + 8 - 5t + 6

7. , (

2.

Sx - 7

, x -3x+2

4 2x + 2 'x'-2x+1

8.

(' + 9 t' + 9t'

11./ 13.

x+4 dx x'+Sx-6

/.8 ,

ydy

'Y -2y-3

15 / ·

dt ('+t'-2t

Irreducible Quadratic Factors

10./~ x + 2x

12./

2x+1 dx x' - 7x + 12

[' y

14.

1, '

dx . , (x + I)(x' + I)

23. /

10 Exercises 9-16, express the integrand as a sum of partial fractions and evaluate the integrals.

x'dx (x - I)(x' + 2x + I)

10 Exercises 21-32, express the integrand as a sum of partial fractions and evaluate the integrals.

21

6 z ·z3_ z 2_6z

Nonrepeated Linear Factors

9./~ I - x

.

y'+2Y +I (y' + I)' dy

25/ 23+2 tis . (8' + 1)(8 - I)'

+ 2 dx

27. /

x' - x

29.

x' -,--dx x - I

/

x' - I

+4

l'/2 y' + y dy x + 3 dx 2x' - 8x

16. /

Improper Fractions Repeated Linear Factors 10 Exercises 17-20, express the integrand as a sum of partial fractions and evaluate the integrals. 17

r

1 --o;-__x,-'... dx___

• Jo x2+2x+ 1

18 •

l' . .

~ x,-'__ dx___

-1

x2

-

2x + 1

10 Exercises 33-38, perfor03long division on the integrand, write the proper fraction as a sum. of partial fractions, and then evaluate the integral.

33./ 2x'-2x'+l dx x2 - x

34.

/

x' -,--dx x - I

462

Chapter 8: Techniques of Integration

35. /9X'-3X+ Idx x 3 - xl

37.

y' /

+ y2 - I , dY y +Y

56. The y-axis

16x' dx 4x -4x+1

36. /

2

_ 2 y - (x + 1)(2

y

2'

11--.

38 Y dY • / y' - y2 + y - I

x)

.....

Evaluating Integrals Evaluate the integrals in Exercises 39-50.

39 / •

etdt e" + 3e'

41. /

cosydY sin2 y + siny - 6

+2

40.

/

e

4t

+

2e

e2'

2t

e

-

+I

"""'O+---------""'"I--+ x

t

dt

(x - 2)'tan- 1 (2x) - 12x' - 3x (4x2 + I)(x _ 2)' dx

43. /

D 57.

Find, to two decimal places, the x-<>oordiuate of the centroid of the region in the first quadrant bounded the x-axis, the curve y = tan-Ix, and the line x = V3.

D 58.

Find the x-coordiuate of the centroid of this region to two decimal places.

sinO dO 0082 8 + cosB - 2

42 / •

by

(x + I)'tan- I (3x) + 9x' + x 44. /

+

(9x2

I)(x

+

y (3. 1.83)

dx

1)2

46./ I Vi (xl/3 - I)

dx

(5,0.98)

(Hint: Letx = u'.)

47.

(Hint: Let x

49./ I+ x(x'

+

x x

o

9

D 59.

I = u2.)

50/ I

dx

I)

~~~~~--------~--~x

48./ ~dx +

/~dx

•

x' (x'

+ 4)

dx

by::.)

(Hint: Multiply

Initial Value Problems

3

Social diffusion Sociologists s0fi3ctimes use the phrase "social diffusion" to describe the way information spreads througb a population. The information might be a rumor, a cultural fad, or news about a tecbnica1 innovation. In a sufficiently large population, the nun3ber of people x who have the information is treated as a differentiable function of time t, and the rate of diffusion, dx/ dt, is asSun3ed to be proportional to the number of people who have the information times the number of people who do not. This leads to the equation

Solve the initial value problems in Exercises 51-54 for x as a function oft.

51. (t

2

-

3t + 2): = I

I):

52. (3t' + 4t 2 + 53. (12 54. (t

+ 2t):

+

2x

=

I ) : = x2

=

dx dt

(t > 2), x(3) = 0

2V3,

x(1) =

5

=

kx(N - x)

'

where N is the number of people in the population. Suppose t is in days, k ~ 1/250, and two people start a mor at time t ~ 0 in a population of N = 1000 people.

-1fV3/4

TU-

a. Find x as a function of t.

+ 2 (t,x > 0), x(1)

+ I (t > -I),

=

I

b. When will half the population have heard the rumor? (This is when the rumor will be spreading the fastest.)

D 60.

x(O) = 0

Applications and Examples In Exercises 55 and 56, fmd the volwne of the solid generated volving the shaded region about the indicated axis.

by re-

Seoond-order chemical reaction. Many chemical reactions are the result of the interaction of two molecules that uodergo a change to produce a new product. The rate of the reaction typically depends on the concentrations of the two kinds of moleoules. If a is the amount of substance A and b is the amount of substance B at time t ~ 0, and ifx is the an30unt ofproduct at time t, then the rate of formation of x may be given the differential equation

by

55. The x-axis y~

dx dt

3

V3x-x 2

Y (0.5. 2.68)

(2.5, 2.68)

2

k(a - x)(b - x),

or

I dx = k (a - x)(b - x) dt '

~~~----------~~x

o

=

0.5

2.5

where k is a constant for the reaction. Integrate both sides of this equation to obtain a relation between x aod t (a) if a ~ b, aod (h) if a #' b. Assume in each case that x = 0 when t = O.

8.5 Integral Tables and Computer Algebra Systems

8.5

463

Integral Tables and Computer Algebra Systems In this section we discuss how to use tables and computer algebra systems to evaluate integrals.

Integral Tables A Brief Table of Integrals is provided at the back of the book, after the index. (More extensive tables appear in compilations such as CRC Mathematical Tables, which contain thousands of integrals.) The integration fonnulas are stated in terms of constants a, b, c, m, n, and so on. These constants can usually assume any real value and need not be integers. Occasional limitations on their values are stated with the formulas. Formula 5 requires n oF - I , for example, and Formula 11 requires n oF -2. The formulas also assume that the constants do not take on values that require dividing by zero or taking even roots of negative numbers. For example, Formula 8 assumes that a oF 0, and Formulas 13a and 13b cannot be used unless b is positive.

EXAMPLE 1

Find

J

+ 5r l dx.

x(2x

We use Formula 23 at the back of the book (not 22, which requires n oF -I):

Solution

J

x(ax

+ b)-ldx

:2

=

~-

=

I - ~ In 12x + 51 + C.

In lax

+ bl + C.

With a = 2andb = 5,wehave

J

x(2x

EXAMPLE 2

+ 5r l dx

•

Find

Jxvb' Solution

We use Formula 29a:

J

dx xVax-b

=

~tan-l~ax Vb

- b b

+

C.

With a = 2 andb = 4, we have

JxV2x dx

EXAMPLE 3

=

4

~tan-l ~2x -

v'4

+

JV

n . -1 x n+ 1 . -1 a x sm axdx--+ Ism ax--+ n n I

J

C = tan-I ~x - 2 2

+ C.

Find

We begin by using Formula 106:

Solution

4

4

xn+l dx 1 _ a 2x 2

'

n oF -\.

•

464

Chapter 8: Techniques of Integration With" = 1 and a = 1, we have

/

1/

2 . -I x. -I dx _ x xsm x -2 sm Z

2 x dx

~.

Next we use Fonnula 49 to f"md the integral on the right:

/

x2 dx Va2-x2

=

2 a sin-I 2

(~)

_ lxVa 2 _ x 2 + C. 2

Witha = 1,

x2dx /

~

1. -I X - -1'''------'1 =-sm X V I - X -2 2 2

+ C.

The combined result is /

X

sin-I xdx = x; sin-I X

=

-

tx~ +

t (tsin-I x -

C)

x2 1). -I + 1 '''------'12 + C' . ( 2-4 sm x 4 XV1 -x-

•

Reduction Formulas The time required for repeated integrations by parts can sometimes be shortened by applying reduction fonnulas like

tannxdx

=

/ / (lnx)n dx sin"xcosmxdx=

_1_ tann-I x - /tan n- 2xdx " - 1

(1)

x(lnx)n - " / (lnx)n-I dx

(2)

=

m+1 1/ . n-I sm mX~O: X+~~n sin,,-2xcosmxdx

/

(" "" -m). (3)

By applying such a fonnula repeatedly, we can eventually express the original integral in terms of a JXlVI"f low enough to be evaluated directly. The next example illuatrates this procedure.

EXAMPLE 4

Find

/ tanSxdx. Solution

We apply Equation (1) with 11 = 5 to get

/ tan'xdx

=

itan4x - / tan3xdx.

We then apply Equation (1) again, with" = 3, to evaluate the remaining integral:

/ tan3xdx

2

= ttan x - /

tanxdx

2

= ttan x

+ In Icosxl +

C.

The combined result is

/ tan'xdx

=

itan4x - ttan2 x -In Icosxl + C'.

•

As their fonn suggests, reduction fonnulas are derived using integration by parts. (See Example 5 in Section 8.1.)

8.5 Integral Tables and Computer Algebra Systems

465

Integration with a CAS A powerful capability of compoter algebm systems is their ability to integrate symbolically. This is performed with the integrate command specified by the particular system (for example, int in Maple, Integrate in Mathematica).

EXAMPLE 5

Suppose that you want to evaluate the indef'mite integml of the function

f(x) = x 2Ya 2 + x 2. Using Maple, you first define or name the function:

> f:=

x A2 * sqrt (a A2

+ xA2);

Then you use the integmte command on f, identifYing the variable ofintegmtion:

> int(j,x); Maple retorns the answer

t

x(a 2 + x 2)3/2 - ta 2xYa 2 + x 2 -

ta 4Jn (x + Ya 2 + X2).

If you want to see if the answer can be simplified, enter

>

simplify(%);

Maple retorns

ta 2xYa 2 + x 2 +

t x3Ya2 + x 2 - ta 4Jn (x + Ya2 + X2).

If you want the def'mite integmlfor 0 :5 X :5 1T/2, you can use the format

>

int(j, x = 0.. Pi/2);

Maple will retorn the expression

~1T(4a2 + ,.2)(3/2) - ~a21TY4a2 + 1T2 + la 4 Jn(2) 64

32

-

ta 4Jn(1T +

8

Y4a 2 +,.2) +

1~a4Jn(a2).

You can also f'md the definite integral for a particular value of the constant a:

> a:= 1;

> int(j,x

=

0.. 1);

Maple retorns the numerical answer

iYz + tJn(Yz - I). EXAMPLE 6

Solution

•

Use a CAS to find

With Maple, we have the entry

>

int«sin A2)(x)

* (cosA3)(x),x);

with the immediate return -tsin(x) cos (X)4

+ 1~

cos(xj2sin(x)

+

2 15 sin(x).

•

Compoter algebm systems vary in how they process integmtions. We used Maple in Examples 5 and 6. Mathematica would have returned somewhat different results:

466

Chapter 8: Techniques of Integration

1.

In Example 5, given

In [1J:= Integrate [x'2' Sqrt [a'2

+ x'2],x]

Mathematica returns Out[lJ= Va 2 + x 2

(a~x + x;)

-

t

a4LOg[X

+ Va 2 + X2]

without having to simplify an intermediate result. The answer is close to Formula 22 in the integral tables. 2.

The Mathematica answer to the integral

In [2J:= Integrate [Sin [x]Az' Cos [x],3, x] in Example 6 is Sin [x] I. I . Out[2J= - - - -Sm[3x] - -Sm[5x] 8 48 80 differing from the Maple answer. Both answers are correct. Although a CAS is very powerful and can aid us in solving difficult problems, each CAS has its own limitations. There are even situations where a CAS may further complicate a problem (in the sense of producing an answer that is extremely difficult to use or interpret). Note, too, that neither Maple nor Mathematica retorns an arbitrary constant +C. On the other hand, a little mathematical thinking on your part may reduce the problem to one that is quite easy to handle. We provide an example in Exercise 67.

Nonelementary Integrals The development of computers and calculators that find antiderivatives by symbolic manipulation has led to a renewed interest in determining which antiderivatives can be expressed as finite combinations of elementary functions (the functions we have been stodying) and which cannot. Integrals of functions that do not have elementary antiderivatives are called nonelementary integrals. They require infinite series (Chapter 10) or numerical methods for their evaluation, which give only an approximation. Examples of nonelementary integrals include the error function (which measures the probability of random errors) 2 erf(x) = ...;:;;.

10f'

e-I'dt

and integrals such as

f

2

sinx d!;

and

that arise in engineering and physics. These and a number of others, such as

f

ex

-xdx,

f ~xd!;, f

in (in x) d!;,

f

sinx d!; x '

0< k < 1, look so easy they tempt us to try them just to see how they tom out. It can be proved, however, that there is no way to express these integrals as fmite combinations of elementary functions. The same applies to integrals that can be changed into these by substitution. The integrands all have antiderivatives, as a consequence of the Fundamental Theorem of Calculus, Part 1, because they are continuous. However, none of the antiderivatives are elementary. None of the integrals you are asked to evaluate in the present chapter fall into this category, but you may encounter nonelementary integrals in your other work.

8.5 Integral Tables and Computer Algebra Systems

467

Exercises 8.S Using Integral Tables Use the table of integrals at the back of the book to evaluate the integrals in Exercises 1-26.

1/ · xVx-=-3

2/ · xVx+4

dx

dx

xdx

4 / •

(2x

5./X~dx

6. /

x(7x + 5)'/2 dx

7·/7

8 / dx • x 2v'4x - 9

3 /

·

9. /

11 /

·

13.

\Ix=2

dx

/~dx

xv'4x -x2 dx

10.

dx xv7+?-

12 /

dx x~

·

/~ dx

14.

xdx + 3)'/2

/~dx

15. /

e 2t cos 31 dt

16. /

e-3t sin 4t dt

17. /

X COS-IX tb:

18. /

xtao-l xdx

19. / 21. /

X

2tao- l xdx

20.

sin 3x cos 2x dx

I

23. /

8 sin 4t sin

25. /

cos~cos£ dO

dt

2

38. /

40. /

x v'x 2 - 4x

+5

dx

Using Reduction Fonnulas Use reduction fonnulas to evaluate the integrals in Exercises 41-50. sin' 2x dx

42. /

8 cos' 27ft dt

43. / sin2 20 cos' 20 dO

44. /

2 sin2 t sec' t dt

41. /

45. /

4 tao' 2x dx

46.

47. /2sec'7fXdx 49. /

cscs x dx

24• /

. Sffi

26. /

cOS~COS70dO

t . t dt

3 S1n"6

51. /

e'sec' (e' - l)dt

/

29. / 31. / 33. /

34. / 36. /

(x 2 + 1)2 dx

28.

(x2

+ 3)' dx

sin- l Vidx Vi

~

0<

t

50. /

l6x'(ln x)2 dx

52. /

2 (r2 _ 1)'/2 r dr

56.

55. /, 1

<

7f/2

dt tao tv'4 - sin2 t

35 /

tao-I Vydy

37./

•

yv'3

dy

+ (lny)'

I dx v'x 2 + 2x + 5 (Hint. Complete the square.)

cs~ dO v'3/2 _ _dy~= /2 (I - y2)'

j,o j,o';v'3

dt

(12

+

1)7/2

Applications 57. Surface ....8 Find the area of the surface generated by revolv0 s x s about the x-axis. ing the curve y = 58. An: lengtb

v'2,

Find the length of the curve y = x 2 , 0 S

X

s

v3/2. 59. Centroid Find the centroid of the region cut from the first quadrant by the curve y = l/Vx+l and the line x = 3. 60. Moment about y-axis A thin plate of constaot density 6 = I occupies the region euclosed by the curve y = 36/(2x + 3) and the line x = 3 in the fust quadrant. Find the moment of the plate about the y-axis.

D 61.

dx

cottv'1 - sin2 tdt,

3 sec' 3x dx

54.

you can fmd in the table. Then evaluate the integral.

27.

tdt

48. /

53. [2W+1dx

W+2,

/X2+6x

/8 cot

Evaluate the integrals in Exercises 51-56 by making a substitution (possibly trigonometric) and then applying a reduction formula.

sin 2x cos 3x dx

22. /

v'5 - 4x - x 2 dx

x 2 v'2x - x 2 dx

/~~'xdx

Substitution and Integral Tables In Exercises 27-40, use a substitution to change the integral into one

X'+X+l

39. /

Use the integral table and a calculator to fmd to two decimal places the area of the surface generated by revolving the curve y = x 2 , -1 :s x :s 1, about the x-axis.

62. Volume The head of your fmn's accouoting department has asked you to fmd a formula she can use in a computer program to calculate the year.;:nd inventory of gasoline in the company's tanks. A typical tank is shsped like a right circular cylinder ofradius r and length L, mouoted horizontally, as shown in the accompanying figure. The data conre to the accounting office as depth measurements 1aken with a vertical measuring stick marl<ed in centimeters.

468

Chapter 8: Techniques of Integration

a. Show, in the notation of the figure, that the volume of gasoline that fills the tank to a depth d is

v = 2L

l:

J

d. What pattern do you see? Predict the formula for x 4 1n x dx and then see if you are correct by evaluating it with a CAS. e. What is the formula for xn In x dx, n 2: I? Check your answer using a CAS.

J

r d

+ Yr2 - y2 dy .

66. Evaluate the integrals

b. Evaluate the integral.

a.

y

J

lnxdx 2 X

b•

J

lnxd 3 X

c.

X

J

Inx x4

dx .

d. What pattern do you see? Predict the formula for - Measuring stick

r

(

and then see if you are correct by evaluating it with a CAS.

Z[d

I

= Depth of

't-;~~~~~-L-------"' )

e. What is the formula for

JI;:

gasoline

dx,

n

2:

2?

Check your answer using a CAS.

63. What is the largest value

67. a. Use a CAS to evaluate

1

,,/2

o

can have for any a and b? Give reasons for your answer.

. n sm x

sinn x

+ cosn X

d X

where n is an arbitrary positive integer. Does your CAS find the result?

64. What is the largest value

b. In succession, find the integral when n = 1,2,3,5, and 7. Comment on the complexity of the results. can have for any a and b? Give reasons for your answer.

65. Evaluate the integrals

J

xlnxdx

8.6

1

,,/2

COMPUTER EXPLORATIONS In Exercises 65 and 66, use a CAS to perform the integrations.

a.

c. Now substitute x = (7T/2) - u and add the new and old integrals. What is the value of

c.

J

x 3 In x dx.

o

sinn x . dx? smn x + cosn X

This exercise illustrates how a little mathematical ingenuity solves a problem not immediately amenable to solution by a CAS.

NumericaL Integration The antiderivatives of some functions, like sin (x 2), 1/ln x, and ~, have no elementary formulas. When we cannot find a workable anti derivative for a function f that we have to integrate, we can partition the interval of integration, replace f by a closely fitting polynomial on each subinterval, integrate the polynomials, and add the results to approximate the integral of f. This procedure is an example of numerical integration. In this section we study two such methods, the Trapezoidal Rule and Simpson's Rule. In our presentation we assume that f is positive, but the only requirement is for it to be continuous over the interval of integration [a, b].

Trapezoidal Approximations The Trapezoidal Rule for the value of a definite integral is based on approximating the region between a curve and the x-axis with trapezoids instead of rectangles, as in

8.6

Numerical Integration

469

Trapezoid area

~(Yl + Y2)6.X

'I

)111-1

Y2

)III

FIGURE 8.7 The 1'nIpezoidai Rule approximates short s1retches of the curve y ~ f(x) with line segments. To approximate the integral of f from a to b, we add the areas of the trapezoids made by joining the ends of the segments to the x-axis.

Figure 8.7. It is not necessary for the subdivision points Xo, X" X2, ... , Xn in the figure to be evenly spaced, but the resulting fonnula is simpler if they are. We therefore assume that the length of each subinterval is a.:

b-a n

~-­

The length ~x ~ (b - a)/n is called the step size or mesh size. The area of the trapezoid that lies above the ith subinterva1 is

a.: (

Y'-l 2+

Y')

a.: ~T(YI-l+YI)'

where Yi-l = f(X'-I) andy, = f(x,). This area is the length a.: of the trapezoid's horizontal "altitude" times the average of its two vertical "bases." (See Figure 8.7.) The area below the curve Y = f(x) and above the x-axis is then approximated by adding the areas of all the trapezoids:

T

=

I

I

Z(yo + ytla.: + Z(YI + 1

Y2)~x

+ ...

1

+ z(Yn-2 + Yn-l)a.: + z(Yn-l + Yn)a.: = a.:

=

(!YO + Yl + Y2 + ... + Yn-l + !Yn)

~x

T(YO + 2Yl + 2Y2 + ... + 2Yn-l + Yn),

where

Yo

=

f(a),

... ,

Yn-l

=

f(Xn-l),

Yn

=

f(b).

The Trapezoids! Rule says: Use T to estimate the integral of f from a to b. It is equivalent to the midpoint rule discussed in Section 5.1.

470

Chapter 8: Techniques of Integration

y

The Trapezoidal Rule

1: j(x) dx, use T ~ &0 + 2Yl + 2Y2 + ... + 2Yn-l + y.).

To approximate

=

The y's are the values of j at the partition points Xo = a,Xl = a

where

~ =

+

aX,X2 =

a

+ 2.6.x, ... ,Xn-l

=

a +

{n

Use the Trapezoidal Rule with n = 4 to estimate estimate with the exact value. 1

The trapezoidal approximation of the area under the graph of y = x 2 from x = I to x = 2 is a slight over-

estimate (Example I).

TABLE 8.2

y

1 5 4 6 4 7 4 2

1 25 16 36 16 49 16 4

=

b,

J,2 x 2 dx. Compare the

S 6 7 2 "4 "4 "4

FIGURE 8.8

x

=

(b - a)/n.

EXAMPLE 1

o

-l)ax,xn

x2

Solution Partition [1, 2] into four subintervals of equal length (Figure 8.8). Then evaluate y = x 2 at each partition point (Table 8.2). Using these y values, n = 4, and ~ = (2 - 1)/4 = 1/4 in the Trapezoidal Rule, we have

=! (1 + 2(i~) + 2 G~) + 2(in + 4) =

75 32

=

2.34375.

Since the parabola is concave up, the approximating segments lie above the curve, giving each trapezoid slightly more area than the corresponding strip under the curve. The exact value of the integral is

2x 2 dx=x3]2 =8- - -1= -7

1 1

3

1

3

3

3'

The T approximation overestimates the integral by about half a percent of its true value of 7/3. The percentage error is (2.34375 - 7/3)/(7/3) '" 0.00446, or 0.446%. • y

Simpson's Rule: Approximations Using Parabolas

FIGURE 8.9 Simpson's Rule approximates short stretches of the curve with parabolas.

Another rule for approximating the definite integral of a continuous function results from using parabolas instead of the straight line segments that produced trapezoids. As before, we partition the interval [a, b] into n subintervals of equal length h = ~ = (b - a)/n, but this time we require that n be an even number. On each consecutive pair of intervals we approximate the curve y = j(x) ;", 0 by a parabola, as shown in Figure 8.9. A typical parabola passes through three consecutive points (Xi-" Yi- d, (Xi, y;), and (Xi+l,Yi+l) on the curve. Let's calculate the shaded area beneath a parabola passing through three consecutive points. To simplify our calculations, we first take the case where Xo = -h, Xl = 0, and

8.6 y (0. y,) ...--r-..... (h. y,)

Y = Ax2

so the area under it from x y,

1:

y,

=

+ Bx + C,

=

his

2

+ Bx + C)dx

(Ax

2 Ax 3 =_+Bx +CX ]h 3 2 -h

----~--~--~----~x

0

-h to x

=

Ap

-h

471

X2 = h (Figure 8.10), where h = dx = (b - a)/n. The area under the parabola will be the same if we sbift the y-axis to the left or right. The parabola has an equation of the fonn

y =Ax2+Bx+ C

Yo

Numerical Integration

h

FIGURE 8.10 By integrating from -h to h. we fmd the shaded area to be

=

~h3 + 2Ch

=

~(Uh2 + 6C).

Since the curve passes through the three points (-h, Yo), (0, Yt), and (h,Y2), we also have Yo

=

Ah2 - Bh + C,

Yt

= C,

Y2

=

Ah2

+ Bh + C,

from wbich we obtain C = Yt.

Ah2 - Bh

=

Yo - Yt.

Ah2 + Bh = Y2 - Yt. Uh 2 = Yo + Y2 - 2Yt. Hence, expressing the area Ap in terms of the ordinates Yo, Yt , and Y2, we have

Now sbifting the parabola horizontally to its shaded position in Figure 8.9 does not change the area under it. Thus the area under the parabola through (xo,Yo), (xt.Yt) , and (X2' Y2) in Figure 8.9 is still

Similarly, the area under the parabola through the points (X2, Y2), (X3 ,Y3), and (X4,Y4) is

Computing the areas under all the parabolas and adding the results gives the approximation

= HISTORICAL BIOGRAPHY

Thomas Simpson (172()-1761)

3h (YO + 4Yt + 2Y2 + 4Y3 + 2Y4 + ... + 2Yn-2 + 4Yn-t + Yn)·

The result is known as Simpson's Rule. The function need not be positive, as in our derivation, but the number n of subintervals must be even to apply the rule because each parabolic arc uses two subintervals.

472

Chapter 8: Techniques ofIntegration

Simpson's Rule To approximate

1: f(x) dx, use

dx

S = T(YO

+ 4Yl + 2,Y2 + 4Y3 + ... + 2Yn-2 + 4Yn-l + Yn).

The y's are the values of f at the partition points

Xo = a,Xl = a

+

dx,X2

= a + 2dx, ""Xn-l = a + (n - I)dx,xn = b.

The number n is even, and dx = (b - a)/n. Note the pattern of the coefficients in the above rule: 1,4,2,4,2,4,2, ... ,4, I.

EXAMPLE 2

2

Use Simpson's Rule with n = 4 to approximate 10 5X 4 dx.

Solution Partition [0, 2] into four subintervals and evaluate Y = 5X 4 at the partition points (Table 8.3). Then apply Simpson's Rule with n = 4 and dx = 1/2:

TABLE 8.3

x

y

0 1 2 1 3 2 2

0 5 16 5 405 16 80

= 5>;4

S =

=

~ Vo + 4Yl + 2,Y2 + 4)13 + Y4)

i (0 + 4(;6) + 2(5) + 4(41°i) + 80)

= 32

I

12 ,

This estimate cliffers from the exact value (32) by ouly 1/12, a percentage error of less than three-tenths of one percent, and this was with just four subintervals. •

Error Analysis Whenever we use an approximation technique, the issue arises as to how accurate the approximation might be. The following theorem gives formulas for estimating the errors when using the Trapezoidal Rule and Simpson's Rule. The error is the difference between the apf(x) dx. proximation obtained by the rule and the actual value of the def"mite integral

1:

THEOREM l-Error Estimates In the Trapezoidal and Simpson's Rules If I" is continuous and M is any upper bound for the values of II" I on [a, b], then the error ET in the trapezoidal approximation of the integral of f from a to b for n steps satisfies the inequality <:

IET I -

M(b - a)3

12n

2

Trapezoidal Rule

If /4) is continuous and M is any upper bound for the values of 1/4 ) I on [a, b], then the error Es in the Simpson's Rule approximation of the integral off from a to b for n steps satisfies the inequality

IE I ,;; M(b s

- a)S 180n 4

Simpson's Rule

To see why Theorem 1 is true in the case of the Trapezoidal Rule, we begin with a result from advanced calculus, which says that if I" is continuous on the interval [a, b], then

l

a

b

b - a f(x) dx = T - - - . l"(c)(dx)2 12

8.6

Numerical Integration

473

for some number c between a and b. Thus, as Jlx approaches zero, the error defined by

ET

= -

b - a 'j"(c)(Jlx)' 12

approaches zero as the square of Jlx. The inequality

where max refers to the interval [a, b], gives an upper bound for the magnitode of the error. In practice, we usually cannot f'md the exact value ofmaxlf"(x) I and have to estimate an upper bound or ''worat case" value for it instead. If M is any upper bound for the values of If"(x) I on [a, b], so that Ij"(x)l :5 M on [a, b], then

Ifwe substitute (b - a)Jn for Jlx, we get

a)3

< M(b -

IET I -

l2n

2

To estimate the error in Simpson's rule, we start with a result from advanced calculus that says that if the fourth derivative /4) is continuous, then

l

b

j(x) dx

=

a

S _ b - a • f4)(c)(Jlx)4 180

for some point c between a and b. Thus, as Jlx approaches zero, the error,

approaches zero as the fourth power of Jlx. (This helps to explain wby Simpson's Rule is likely to give better results than the Trapezoidal Rule.) The inequality

where max refers to the interval [a, b], gives an upper bound for the magnitude of the error. Ai; with max Ij" I in the error formula for the Trapezoidal Rule, we usually cannot f'md the exact value of max If4 )(x) Iand have to replace it with an upper bound. If M is any upper bound for the values of If4) I on [a, b], then IEsl

:5

b - a

4

180 M(Jlx) .

Substituting (b - a)Jn for Jlx in this last expression gives

I :5 M(b

E

I EXAMPLE 3

s

- a)S 180n 4

Find an upper bound for the error in estimating Simpson's Rule with n = 4 (Example 2).

J025x 4 dx

using

Solution To estimate the error, we f'lISt f'md an upper bound M for the magnitude of the fourth derivative of j(x) = 5x 4 on the interval 0 :5 x :5 2. Since the fourth derivative has

474

Chapter 8: Techniques ofIntegration the constant value f4 l(x) = 120, we take M = 120. With b - a = 2 and n = 4, the error estimate for Simpson's Rille gives

E

I

I< s -

M(b - a)' = 120(2)' 180'44 180n4

1 12·

This estimate is consistent with the resillt of Example 2.

•

Theorem 1 can also be used to estimate the number of subintervals required when using the Trapezoidal or Simpson's Rilles if we specify a certain tolerance for the error.

EXAMPLE 4 Estimate the minimum number of subintervals needed to approximate the integral in Example 3 using Simpson's Rille with an error of magnitude less than 10-4. Solution Using the inequality in Theorem 1, if we choose the number of subintervals n to satisfy M(b - a)' 180n

-4

<

4

10 ,

then the error Es in Simpson's Rille satisfies IEsl < 10-4 as required. From the solution in Example 3, we have M = 120 and b - a = 2, so we want n to satisfy 120(2)'

---'-'0-4 180n

<-

1

10 4

or, equivalently,

n

4>64'10

4

3.

It follows that

n>

64)1/4 10 ( T

'"

21.5.

Since n must be even in Simpson's Rille, we estimate the minimum number of subintervals required for the error tolerance to be n = 22. •

EXAMPLE 5

As we saw in Chapter 7, the value of ln 2 can be calculated from the

integral ln2 =

[fax.

ft2

ax

Table 8.4 shows T and S values for approximations of (l/x) using various values of n. Notice how Simpson's Rille dramatically improves over the Trapezoidal Rille. TABLE 8.4 Trapezoidal Rule approximations (T,) and Simpson's Rule approximations (5,) of In 2 = (l/x) rJx

J:

IError I n 10 20 30 40 50 100

IError I

T.

less than ...

S.

less than ...

0.6937714032 0.6933033818 0.6932166154 0.6931862400 0.6931721793 0.6931534305

0.0006242227 0.0001562013 0.0000694349 0.0000390595 0.0000249988 0.0000062500

0.6931502307 0.6931473747 0.6931472190 0.6931471927 0.6931471856 0.6931471809

0.0000030502 0.0000001942 0.0000000385 0.0000000122 0.0000000050 0.0000000004

8.6

475

Numerical Integration

In particular, notice that when we double the value of n (thereby halving the value of h = ~x), the Terror is divided by 2 squared, whereas the S error is divided by 2 to the/our/h. This has a dramatic effect as ~ = (2 - 1)ln gets very small. The Simpson approximation for n = 50 rounds accurately to seven places and for n = 100 agrees to nine decimal places (billionths)! •

If f(x) is a polynomial of degree less than four, then its fourth derivative is zero, and

Es = - b - a t<4)(C)(~)4 = _ b - a (0)(~)4 = 0 180

180

.

Thus, there will be no error in the Simpson approximation of any integral of f. In other words, if f is a constant, a linear function, or a quadratic or cubic polynomial, Simpson's Rule will give the value of any integral of f exactly, whatever the number of subdivisions. Similarly, if f is a constant or a linear function, then its second derivative is zero, and 146 ft

Vertical spacing

ET =

~

20 ft

The dimensions of the in Example 6.

FIGURE 8.11 SWllDlp

b-a -----uf"(C)(~)

2

=

b-a

-----u-(O)(~x)

2

= O.

The Trapezoidsl Rule will therefore give the exact value of any integral of f. This is no surprise, for the trapezoids fit the graph perfectly. Although decreasing the step size ~ reduces the error in the Simpson and Trapezoidsl approximations in theory, it may fail to do so in practice. When ~ is very small, say ~ = 10-', computer or calculator round-off errors in the arithmetic required to evaluate S and T may accumulate to such an extent that the error formulas no longer describe what is going on. Shrinking ~ below a certain size can actually make things worse. Although this is not an issue in this book, you should consult a text on numerical analysis for alternative methods if you are having problems with round-off.

EXAMPLE 6 A town wants to drain and fill a small polluted swamp (Figure 8.11). The swamp averages 5 ft deep. About how many cubic yards of dirt will it take to fill the area after the swamp is drained? Solution To calculate the volume of the swamp, we estimate the surface area and multiply by 5. To estimate the area, we use Simpson's Rule with ~ = 20 ft and the y's equal to the distances measured across the swamp, as shown in Figure 8.11. ~

S=T~+~+~+~+~+~+~

= ;0 (146 + 488 + 152 + 216 + 80 + 120 + 13) = 8100 The volume is about(8100)(5) = 40,500 ft3 or 1500 yd3.

•

Exercises 8.6 Estimating Integrals The instractions for the integrals in Exercises 1-10 have two parts, one for the Trapezoidal Rule and one for Simpson's Rule. L Using the Trapezoidlll Rule

•. Estimate the integral with n = 4 steps and f'md an upper bound for IETI. b. Evaluate the integral directly and f'md

IETI .

c. Use the fannul. ( IETI I( trae value» X 100 to express as a percentage of the integral's true value.

IETI

II. Using Simpson'. Rule

.. Estimate the integral with n bound for IEsl.

= 4 steps and f'md an upper

b. Evaluate the integral directly and f'md

IEsl .

c. Use the formula (lEsI I(true value» X 100 to express as a percentage of the integral's trae value.

2.1'

(a - I) tb;

IEsl

476

Chapter 8: Techniques ofIntegration

3.1:

(x

2

+

4.1:

I) tb:

6.1:

5.1,2 (I' + I)dl 7.

2I

1

2 tis

8.

1 •

9.1,~ sinldl

(x 2

-

I) tb:

(I'

+

I)dl

J2t(s -I 1ftis

10.1,1 sin'11'tdt

Estimating the Number of Subintervals In Exercises 11-22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10-4 by (a) the Trapezoidal Rule and (b) Simpson's Rule. (The integrals in Exercises 11-18 are the integrals from Exereises 1-8.) 12·1' (Zl; - I)tb:

1l·1'xtb:

13.1:

(x

2

+

14.1:

I) tb:

15.1,2 (I' + I) dl 17.

1

16. 18.

2 tis

1 •

19.1,'~tb:

21.1,2

sin (x

(x

1:

2

(I'

-

I) tb:

+

I) dl

4

I

2

25.

+

1

2(.-1)2

1" 22.1: 20.

o

I) tb:

I

26.

tis

I • ~tb: yX + I cos (x

+ 1f) tb:

Position (ft) x

f

0

Depth (ft) h(x)

Depth (ft) h(x)

Zero to 30 mph 40 mph 50 mph 60 mph 70 mph 80 mph 90 mph 100 mph 110 mph 120 mph 130 mph

2.2 3.2 4.5 5.9 7.8 10.2 12.7 16.0 20.6 26.2 37.1

Yo = 1.5 ft, Yl = 1.6 ft, Y2 = 1.8 ft, y, = 1.9 ft, Y4 = 2.0 ft, Y5 = Y6 = 2.1 ft Horizontal spacing

Day

Oil consumption rate (liters/h)

Sun Mon The Wed Tbu Fri Sat Sun

0.019 0.020 0.021 0.023 0.025 0.028 0.031 0.035

Theory and Examples 27. Usable values of the sine-integral function jUnction,

0 5 10 15 20 25

6.0 8.2 9.1 9.9 10.5 11.0

30 35 40 45 50

11.5 11.9 12.3 12.7 13.0

24. Distance traveled The accompanying table shows time-1:Gspeed data for a sports car accelerating from rest to 130 mph. How far had the car traveled by the time it reached this speed? (Use trapezoids to estimate the area under the velocity curve, but be careful: The time intervals vary in length.)

= 1 ft

Oil consumption on Patbf"mder Island A diesel generator runs continuously, consuming oil at a gradually increasing rate until it must be teD3porarily shut down to have the filters replaced. Use the Trapezoidal Rule to estimate the amount of oil consumed by the generator daring that week.

30·h(x)tb:.

Position (ft) x

Time (sec)

W"mg design The design of a new airplane requires a gasoline tank of constant cross-sectional area in each wing. A scale drawing of a cross-section is shown here. The tank must hold 5000 Ib of gasoline, which has a density of 42 Ib/tt' . Estimate the length of the tank by SinJpson's Rule.

Estimates with Numerical Data 23. Volume of water in a swimming pool A rectangular swimming pool is 30 ft wide and 50 ft long. The acconJpanying table shows the depth h(x) of the water at 5-ft intervals from one end of the pool to the other. Estimate the volume of water in the pool using the Trapezoidal Rule with n = 10 applied to the integral

V=

Speed change

Si(x) =

1,% S~I dl,

The .ine-inlegral "Sineintegmlofx"

is one of the many functions in engineering whose fonnulas cannot be simplified. There is no elementary formula for the antiderivative of (sin 1)/1. The values ofSi(x), however, are readily estimated by numerical integration. Although the notation does not show it explicitly, the function being integrated is sinl

/(1) =

{

10#0 I' t = 0, I,

8.6 the continuous extension of (sin t)/t to the interval [0, x]. The function has derivatives of all orders at every point of its domain. Its graph is smooth, and you can expect good results from Simpson's Rule.

_

=

rsin Jo

The length of one arch of the curve y = sin x is given by L =

D 33.

. 0 -< x -< 20 m.

If the roofing is to be stamped from flat sheets by a process that does not stretch the material, how wide shou1d the original material be? To find out, use numerical integration to approximate the length of the sine curve to two decimal places.

[",/2.

-I

Your metal fabrication company is bidding for a contract to make sheets of corrugated iron roofing like the one shown here. The cross-sections of the corrugated sheets are to conform to the curve . 37r y -- sm 20 x,

a. Use the fact that If (4) I ,.;; 1 on [0, 7r/2] to give an upper bound for the error that will occur if Si ( ) =

+ cos2 xdx.

27r

7r

x

1'" VI

Estimate L by Simpson's Ru1e with n = 8. t dt

t

--1--'-

477

AppLications

D 32.

y

Si (x)

NumericaL Integration

Jo srr; t dt

is estimated by Simpson's Rule with n = 4.

Original sheet

y

Corrugated sheet

b. Estimate Si(7r/2) by Simpson's Rule withn = 4. c. Express the error bound you found in part (a) as a percentage of the value you found in part (b). 28. The error function

The error junction, 2 [X Y; Jo

erf(x) =

I

2

e- dt,

important in probability and in the theories of heat flow and signal transmission, must be evaluated numerically because there is 2 no elementary expression for the anti derivative of e -1 • a. Use Simpson's Ru1e with n = 10 to estimate erf(l). b. In [0, 1],

D 34.

Your engineering firm is bidding for the contract to construct the tumtel shown here. The tumtel is 300 ft long and 50 ft wide at the base. The cross-section is shaped like one arch of the curve y = 25 cos (7rx/50). Upon completion, the tumtel's inside surface (excluding the roadway) will be treated with a waterproof sealer that costs $1.75 per square foot to apply. How much will it cost to apply the sealer? (Hint: Use numerical integration to find the length of the cosine curve.)

Give an upper bound for the magnitude of the error of the estimate in part (a).

y = 25 cos (7rx/50)

y

J:

29. Prove that the sum T in the Trapezoidal Rule for f(x) dx is a Riemann sum for f continuous on [a, b]. (Hint: Use the Intermediate Value Theorem to show the existence of Ck in the subinterval [Xk-l. Xk] satisfying f(Ck) = (f(Xk-l) + f(Xk»/2.)

J:

30. Prove that the sum S in Simpson's Rule for f(x) dx is a Riemann sum for f continuous on [a, b]. (See Exercise 29.)

D 31.

Elliptic integrals

The length of the ellipse 2

£+L= 2 2 a

1

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.

b

turns out to be

[",/2

Length = 4a

Jo \II -

e 2 cos2 t dt,

wheree = Va 2 - b 2/a is the ellipse's eccentricity. The integral in this formu1a, called an elliptic integral, is nonelementary except whene = Oor1. a. Use the Trapezoidal Ru1e with n = 10 to estimate the length of the ellipse when a = 1 ande = 1/2. b. Use the fact that the absolute value of the second derivative of f(t) = e 2 cos2 t is less than 1 to find an upper bound for the error in the estimate you obtained in part (a).

VI -

x (ft) NOT TO SCALE

2

35. y = sin x, 36. y = x2/4,

0,.;; x ,.;; 7r 0,.;; x ,.;; 2

37. Use numerical integration to estimate the value of . -I

sm

_1°·6

0.6-

°

dx 2 vI - x

.~.

For reference, sin-I 0.6 = 0.64350 to five decimal places. 38. Use numerical integration to estimate the value of 7r = 4

t __ 1_2 dx. 1+x

Jo

478

Chapter 8: Techniques ofIntegration

8.7

I Improper Integrals Up to now, we have required definite integrals to have two properties. First, that the do· main of integration [a, b] be finite. Second, that the range of the integrand be rmite on this

y 0.2 0.1

o

2

3

4

6

x


domain. In practice, we may encounter problems that fail to meet one or both of these con· ditions. The integral for the area under the carve y = (lnx)/x 2 from x = 1 to x = 00 is an example for which the domain is inf"mite (Fignre 8.12a). The integral for the area under the curve of y = l/Yx between x = 0 and x = 1 is an example for which the range of the integrand is infmite (Fignre 8.12b). In either case, the integrals are said to be improper and are calculated as limits. We will see in Chapter 10 that improper integrals play an im· portant role when investigating the convergence of certain inf"mite series.

Infinite Limits of Integration y~_I­

v'.t

I

o

Consider the inf"mite region that lies under the curve y = e""",,/2 in the first quadrant (Figure 8.13a). You might think this region has infmite area, but we will see that the value is finite. We assign a value to the area in the following way. First find the areaA(b) of the portion of the region that is bounded on the right by x = b (Fignre 8.13b). A(b)

=

].b e""",,/2ax = _2e- 2J: = -2e-b/2 + 2 x/

(bJ

FIGURE 8.12 Are the areas under these inf"mite curves finite? We will see that the answer is yes for both curves.

Then fmd the limit of A(b) as b --> 00 lim A(b)

b-oo

=

lim (-2e-b/2 + 2)

b-"J>oo

=

2.

The value we assign to the area under the curve from 0 to 00 is 00

o

e""",,/2ax = lim].b e""",,/2ax = 2.

].

b_oo 0

y

DEnNmON Integrals with infmite limits of integration are improper integrals of Type I.

1. If !(x) is continuous on [a, 00), then

---+-----===--==-=>x

1

00

a


!(x) ax = lim

l

b-co a

b !(x) ax.

2. If !(x) is continuous on (-00, b], then

3. If !(x) is continuous on ( - 00, 00), then b (bJ

FIGURE 8.13 (a) The area in the rITS! quadran! under the curve y ~ e -xj2 (b) The area is an improper integral ofthe rrrs! type.

where c is any real number. In each case, if the limit is finite we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exist, the improper integral diverges.

8.7 Improper Integrals

479

It can be sbown that the choice of c in Part 3 of the definition is unimportant. We can evaluate or determine the convergence or divergence of J';, f(x) dx with any convenient choice. Any of the integrals in the above detmition can be interpreted as an area if f ;;,: 0 on the interval of integration. For instance, we interpreted the improper integral in Figure 8.13 as an area. In that case, the area has the tmite value 2. If f ;;,: 0 and the improper integral diverges, we say the area under the curve is infinite.

EXAMPLE 1 Is the area under the curve y = (lnx)lx 2 from x = I to x = so, what is its value?

y

00

finite? If

Solution We find the area under the curve from x = I to x = b and examine the limit as b --+ 00. If the limit is tmite, we take it to be the area under the curve (Figure 8.14). The area from I to b is Integration by parts with u = lnx, dv = dx/x 2, du ~ tb:/x, v ~ -1/x

FIGURE 8.14 The area under this curve

is an improper integral (Example I).

The limit of the area as b --+ 00 is

=_[limlnb]_O+1 b_oo b = - [lim

lib]

b-OO 1

+ I = 0 + I = 1.

l'H6pital's Rule

Thus, the improper integral converges and the area has tmite value I.

EXAMPLE 2

Evaluate dx

1

00

-oo1+x 2 '

Solution

According to the detmition (part 3), we can choose c = 0 and write

1

Lejeune Dirichlet (1805--1859)

dx

= -oo1+x 2

HISTORICAL BIOGRAPHY

00 - - -

1°

1°

00

- dx -2

_001+x

+

- dx - -2 l+x

'

Next we evaluate each improper integral on the right side of the equation above.

1 0

-00

-dx -1 + x2

-

=

. bm

1°

0_-00

-dxa 1 + x2

lim tan-I 0 _ - 00

x]oa

•

480

Chapter 8: Techniques of Integration

lo

OO

. -dx - - hm

o 1 + x2

-

lob -dx-

b_oo 0

lim tan-1

=

b-oo y

=

+ x2

1

x]b 0

lim (tan-1 b - tan- 1 0) =

b~oo

°

=

7f -

2

7f

2

Thus,

NOT TO SCALE

FIGURE 8.15 The area under this curve is fmite (Example 2).

Since I/(i + x 2 ) > 0, the improper integral can be interpreted as the (finite) area be• neath the curve and above the x-axis (Figure 8.15).

The Integral

l

""dx Ii X

1

The function y = l/x is the boundary between the convergent and divergent improper integrals with integrands of the form y = l/x p • Ai; the next example shows, the improper integral converges if p > 1 and diverges if p :5 1.

EXAMPLE 3

For what values ofp does the integral tegral does converge, what is its value?

Solution

Jt' dx/x

P

converge? When the in-

If p "" 1,

I

1

bdx -

xP

=

x -p+ 1

-p

+1

]b 1

= _1_ b-p+l -

1- p(

1 = -1- (1 - - - 1) ) 1 - P b P- 1

•

Thus,

=

b~oo [1 - p (b

I

lim _1_ _ 1__ 1 P- 1

= ) ]

p> 1

P - l' { co,

p

< 1

because lim _1_ = {O,

h-OO bP - 1

00,

p>1 p
Therefore, the integral converges to the value 1/(p - 1) if p p
>

1 and it diverges if

8.7 Improper Integrals

481

If p = 1, the integral also diverges:

=

=

j

b~oo 1

b

dx X

lim (lnb - In 1) =

b_co

•

00.

Integrands with Vertical Asymptotes

y

Another type of improper integral arises when the integrand has a vertical asymptote-an infmite discontinuity-at a limit of integration or at some point between the limits of integmtion. If the integmnd f is positive over the in1erval of integration, we can again interpret the improper integral as the area under the graph of f and above the x-axis between the limits of integration. Consider the region in the flISt quadmnt that lies under the curve y = 1/ Vx from x = 0 to x = 1 (Figure 8. 12b). First we fmd the area of the portion from a to 1 (Figure 8.16).

y~_l_

Yx

l' ~

~f-L----'---~x

o

lim

a

1

FIGURE 8.16 The area under this curve is an example of an improper integral of the second kind.

= 2Vx]: = 2 - 2Va.

Then we find the limit of this area as a --> 0+:

Therefore the area under the curve from 0 to 1 is finite and is defined to be

DEFINmON Integrals of functions that become infmite at a point within the interval of integration are improper integrals of Type II. 1. If fix) is continuous on (a, bl and discontinuous at a, then b fix) dx = lim+ib fix) dx.

i i a

C-Q

c

2. If fix) is continuous on [a, b) and discontinuous at b, then

a

b fix) dx = lim_ic fix) dx. c-b

a

3. If fix) is discontinuous at e, where a [a, c) U (e, bl, then [

fix) dx = [f{X) dx

+

<

t

e

< b,

and continuous on

fix) dx.

In each case, if the limit is finite we say the improper integral converges and that the limit is the value of the improper integral. If the limit does not exist, the integral diverges.

482

Chapter 8: Techniques of Integration

y

I 1 ..------;y=l-x

In Part 3 of the defmition, the integral on the left side of the equation converges if both in-

tegrals on the right side converge; otherwise it diverges .

EXAMPLE 4

Investigate the convergence of

1.'

I

~ x dx .

Solution The integrand f(x) = 1/(1 - x) is continuous on [0, I) but is discontinuous at x = I and becomes infinite as x --+ 1- (Figure 8.17). We evaluate the integral as

1

lim

-

o

b

b-l-

r_ - x

Jo

I

-I - dx = lim [-In II - xl]ob b-l-

1

=

The area beneath the curve aod above the x-axis for [0, 1) is not a real number (Example 4). FIGURE 8.17

lim [-In(1 - b) + 0]

b-I-

= 00.

•

The limit is infinite, so the integral diverges.

EXAMPLE 5

Evaluate

{' dx Jo (x - 1),/3' Solution The integrand bas a vertical asymptote at x = I and is continuous on [0, I) and (1,3] (Figure 8.18). Thus, by Part 3 of the definition above,

y

f'

_ 1 Y - (x - 1)213

dx

{'

Jo (x - 1)2/3

=

1 3

dx

dx

Jo (x - 1),/3 + 1 (x - 1)2/3'

Next, we evaluate each improper integral on the right-band side of this equation.

l'

(x _dxl )'/3

= =

o

_1_ b

FIGURE 8.18

=

c

Example 5 shows that the

area under the curve exists (so it is a real

1

b~_1b (x _dxl )2/3 lim 3(x - 1)1/3]b

b-l-

lim [3(b - 1)'/3 + 3]

b-l-

1

3 -:-~'-c-;;", dx -

1

. hm (x - 1),/3 - c~l+ =

number).

=

0

3

c

=

3

dx (x - 1),/3

lim 3(x - 1)1/3]3

c-l+

c

lim [3(3 - 1)'/3 - 3(c - 1)1/3]

c-l+

=

3V'2

We conclude that

f'

dx

Jo (x _ 1)2/3

=

3 + 3V'2.

•

Improper Integrals with a CAS Computer algebm systems can evaluate many convergent improper integrals. To evaluate the integral

rOO

J2

x +3 dx (x - l)(x 2+ I)

8.7 Improper Integrals

483

(which converges) using Maple, enter

> f:= (x + 3)/«x - 1). (x'2 +

I»;

Then use the integration command

>

int{f, x = 2 .. infinity);

Maple returns the answer

I -"2" + In (5) + arctan (2). To obtain a numerical result, use the evaluation command evalf and specify the number of digits as follows:

>

evalf(%, 6);

The symbol % instructs the computer to evaluate the last expression on the screen, in this case (-1/2)" + In (5) + arctan (2). Maple returns 1.14579. Using Mathematica, entering

In [1]:= Integrate [(x

+ 3)/«x

- l)(x'2

+ I)), {x, 2, Infinity}]

returns

Out [1]=

-or

"""2 + ArcTan [2] + Log [5].

To obtain a numerical result with six digits, use the command ''N[%, 6]"; it also yields

1.14579.

Tests for Convergence and Divergence When we cannot evaluate an improper integral directly, we try to determine whether it converges or diverges. If the integral diverges, that's the end of the story. If it converges, we can use numerical methods to approximate its value. The principal tests for convergence or divergence are the Direct Comparison Test and the Limit Comparison Test.

y

EXAMPLE 6 Solution

-

~t-----~----~~----~x

The graph of e~' lies below the graph of e ~ for x > I FIGURE 8.19

Does the integral

.K'" e --i' ax converge?

By definition,

We cannot evaluate this integral directly because it is nonelementary. But we can show that its limit as b --> 00 is f"mite. We know that fIb e --i' ax is an increasing function of b. Therefore either it becomes infinite as b --> 00 or it has a finite limit as b --> 00. It does not become infinite: For every value ofx 2: I, we have e--i' :5 e-x (Figure 8.19) so that

l

(Example 6).

b

e-" ax:5

l

b

e-x ax

=

-e-b + e-1 < e- 1

'"

0.36788.

Hence,

converges to some definite f"mite value. We do not know exactly what the value is except that it is something positive and less than 0.37. Here we are relying on the completeness • property of the real numbers, discussed in Appendix 6.

484

Chapter 8: Techniques of Integration The comparison of e --x' and e -x in Example 6 is a special case of the following test.

THEOREM 2-Dtred Comparison Test Let with 0 ,;; f(x) ,;; g(x) for all x ;" a. Then

HISTORICAL BIOGRAPHY

Karl Weierstrass (1815--1897)

1.

to

f(x) dx

converges if

2. [ " g(x) dx

f

and g be continuous on [a, 00)

converges.

[ " g(x) dx

diverges if

diverges.

[ ' f(x)dx

Proof The reasoning behind the argument establishing Theorem 2 is similar to that in Example 6. If 0 ,;; f(x) ,;; g(x) for x ;" a, then from Rule 7 in Theorem 2 of Section 5.3 we have

l

b

f(x) dx,;;

l

b

b

g(x) dx,

> a.

From this it can be argued, as in Example 6, that

to to

f(x) dx

converges if

[ " g(x)dx

Turning this around says that

EXAMPLE 7

g(x)dx

to

diverges if

f(x)dx

converges.

•

diverges.

These examples illustrate how we use Theorem 2.

2

(a) JOO sin x dx 1 x2

converges because

on

[1,00)

OOI

J

and

2dx

x

1

converges.

Example 3

diverges.

Example 3

OO

(b)

J Vx 1

1

Vx 2 -

I

2 -

dx

diverges because

0.1

>1 0.1 -

x

on

OO 1

[1,00)

J

and

1

If the positive functions f and g are con-

THEOREM 3-Llmlt Comparison Test tinuous on [a, 00), and if

lim f(x) = L g(x) ,

X~OO

then

1

xdx

00

O
1

00

f(x)dx

both converge or both diverge.

and

g(x)dx

•

485

8.7 Improper Integrals

We omit the more advanced proof of Theorem 3. Although the improper integrals of two functions from a to 00 may both converge, this does not mean that their integrals necessari1y have the same value, as the next example shows.

EXAMPLE 8

Show that

00 /.

I

1

y

ax + x2

converges by comparison with

f,oo (1/x 2 ) ax. Find and compare the two integral values.

The functions fix) ous on [1, 00). Also,

= l/x 2 and g(x) =

Solution

1/(1 + x 2) are positive and continu-

2 fix) . l/x = bm ---ccc'----;;:X~OO g{x) X~OO 1/(1 + x 2) .

lim -

=

o

2

FIGURE 8.20

_

lim X--i'OO

~------~----~----~~X

(~+ x2

. 1 + x-2 1u n x---+OO x2

1) = 0

+

1 = 1,

3

The functions in

a positive finite limit (Figore 8.20). Therefore,

/.00 ~ converges because /.00 ~ l+x x I

Example 8.

I

converges.

The integrals converge to different values, however: Example 3

and

/. I

00 -ax- - lun . /.b -ax1+x 2

- b - oo l

=

EXAMPLE 9 TABLE 8.5

b

/.'1 1

2 5

x e-

x

ax

10 100 1000 10000

0.5226637569 1.3912002736 2.0832053156 4.3857862516 6.6883713446 8.9909564376

100000

11.2935415306

1+x2

lim [tan-I b _ tan-II]

= 71' _ 71' = 71'.

b~oo

Investigate the convergence of,

/.

2

00 1 _ e-x x

4

4

•

ax.

The integrand suggests a comparison of f(x) = (1 - e-X)/x with g(x) = l/x. However, we cannot use the Direct Comparison Test because fix) ,,; g(x) and the integral of g(x) diverges. On the other hand, using the Limit Comparison Thst we fmd that

Solution

which is a positive fmite limit. Therefore,

/.00 1 -x e -x ax diverges because /.00 ~ di-

verges. Approximations to the improper integral are given in Table 8.5. Note that the values do not appear to approach any fIXed limiting value as b --> 00.

•

486

Chapter 8: Techniques of Integration

Types of Improper Integrals Discussed in This Section INFINITE LIMITS OF INTEGRATION: TYPE I

INTEGRAND BECOMES INFINITE: TYPE II

1. Upper limit

4. Upper endpoint

{I

10

dx = lim (x - 1)2/3 b- C

r (x -dx1)2/3

10

y

y

\

-

Y~r;: 1

~-l)'"

+ -'-------+x

o

2. Lower limit

5. Lower endpoint

1

1

. -dx - - lun 2 - a~ -00 --00 1 + x 0

0

a

j

-dx1 + x2

1

3

dx (X - 1),/3

r

=

e

dx

d~+1d (X - 1)2/3

Y Y

1

-

0

_ 1 Y- 1 +-<, x

x

6. Interior point

3. Both limits

e

10

dx (x - 1),/3 =

t

10

dx (x - 1),/3

+

j3 1

Y Y

1

Y~

1 (x -

1)'"

dx (X - 1)2/3

8.7

Improper Integrals

487

Exercises 8.7 Evaluating Improper Integrals Evaluate the integrals in Exercises 1-34 without using tables. 00 dx 00 dx 1 -2 -], 2 • 1 x1.001 • 0 x + 1

{I

3.

1

dx

Yx

Jo

{'

4·

1dx 2/3

6.

7.[~

8

5.

I

-1 X

00 ], 00

1

13 2xdx • ~ (x 2 + If

Jo

+ I d8 y'~ + 28 8

roo dx · Jo (I + x)Yx 19 roo dv 2 17

· Jo (I + v 21. J~ 8e'd8 23.

J~ e-i>1dx

rz

00

•

-1

1;3

dx 44· [ol -

{'~

48

10.1 ~

471°O~ 3 49],00

50

12.],00~ 2

51],00

· 10

rO. 999

•

dx · /.00 , Yx-I

x +1

1

--00

2

100

x2 t

+4 1

-

·0

+

I

Jo V4=7

18.100 1

20.

tis

I

x~ 1

J:"

30 •

dx

00

1

0

+e

52],00

W+1

55.1 2

],oo~6 dx

·2~

54],00 •

2

00

xdx

w-=-l

I +x ~inx dx 58. [I Inx dx

+;osx dx

56.1

~

dx

roo 16 tan-I x dx

10

•

53.[~dx 1 x2

14 xdx • ~ (x 2 + 4)3/2

16. {2 s

dv

·2~

+ x2

2xe -x' dx

4rdr

Jo v'1-=-7 ],' dt

2t~

57.

/.00 2 dt , t /2-1

2

3

60.

[00 In (In x) dx

61.

[ '~ [I

62.

[I

63.

f

64.

f

59.

1

dx

~dx

eX - x

1

dx ~W+1

1

~

e~_2;fdx

dx ex+--x e

Tbeory and Examples 65. Find the values ofp for which eacb integral converges.

],00

2 dx a. 1 1 x(lnx)'

1>.

..

2

dx x(lnx)'

.

66. r~ f(x)dxDHynotequJ. lim J!.f(x)dx Show that .~

2

32

v'fxT

d8 ~ + 58

X

46.1:-xlnlxldx

28. {'

dx

311'~

1

dx2 -o 1- x

1dx

26.],1 (-Inx) dx

Is~

33

dt. (Hint:t"'=sintfort"'= 0) t-smt

45. 1: In Ixldx

24.

29·12-~dx~= -1

43. [

22.],00 2e-8 sin 8 d8

· Jo V4=7

•

10

-, x

)(1 + tan Iv)

25.],1 xlnxdx 27

42. {I

2

2

9.1- ;dx -oox-l 2 11. -2--dv 2 v - v

15. {I

{~----r~d~t-:-­ Vi + sint

Jo

dx

~

Jo

I

41.

+6

dx ], ·o~

34

roo

· Jo

dx (x + 1)(x 2 + I)

diverges and booce that

00 2xdx

1

--00

Testing for Convergence In Exercises 35--64, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergeoce. If

x2

+

1

diverges. Thoo show that

more than one method applies, use whatever method you prefer. {~/2

35. 37.

Jo

r

Jo

{~/2

tan8d8

36.

sin8d8

38 ·

~

40.

Jo

cot8d8

1~/2

cos8d8

Exercises 67-70 are about the infmite regioo in the first quadrant between the curve y = e ---x and the x-axis.

67. Find the area of the region.

~/2 (1T - 28)1/3

68. Find the centroid of the region.

{' -Vi

69. Find the volwne of the solid gooerated by revolving the region about the y-axis.

Jo eYx

dx

488

Chapter 8: Techniques of Integration

70. Find the volume of the solid generated by revolving the region about the x-axis.

D

71. Find the area of the region that lies between the curves y = sec x andy = tan x from x = Otox = 'TT'/2.

b. Explore the convergence of

72. The region in Exercise 71 is revolved about the x-axis to generate a solid. a. Find the volume of the solid. b. Show that the inner and outer surfaces of the solid have infinite area. 73. Estimating the value of a convergent improper integral whose domain is infinite a. Show that

1

00

and hence that

e-3x dx =

te-9 <

00

r3 dx can be replaced by Joe -x

If it converges, what is its value? 76. Error function

b. Evaluate

f:

D

1

00

x2

1

00

o

• ,dt. V'TT'

77. Normal probability distribution

I ~ 2'TT'x\j I + -;tdx,

f( X )

2'TT'i~1 + :4dx > 2'TT'1

2e-t2

If it converges, what appears to be its value? You will see how to confirm your estimate in Section 15.4, Exercise 41.

which measures the surface area of the solid of revolution traced out by revolving the curve y = I/x, I ,.;; x, about the x-axis, diverges also. By comparing the two integrals, we see that, for every finite value b > I, b

2

2 -t V;dt,

a. Plot the error function for 0 ,.;; x ,.;; 25.

e- dx numerically.

1

r

Jo

b. Explore the convergence of

2

74. The infinite paint can or Gabriel's hom As Example 3 shows, the integral f,oo (dx/x) diverges. This means that the integral

1

The function erf(x) =

dx without introducing an error of magnitude greater than 0.000042.

D

dt.

called the error function, has important applications in probability and statistics.

2

2

1 s~t

0.000042,

hoo e -x dx < 0.000042. Explain why this

roo means that Joe -x

a. Plot the integrand (sin t)/t for t > O. Is the sine-integral function everywhere increasing or decreasing? Do you think Si (x) = 0 for x > O? Check your answers by graphing the function Si (x) for 0 ,.;; x ,.;; 25.

=

The function

- -I - e _1(=)2 2 u

u\,f2;

is called the normal probability density jUnction with mean f.L and standard deviation u. The number f.L tells where the distribution is centered, and u measures the "scatter" around the mean. From the theory of probability, it is known that

b

idx.

1:

y

f(x) dx

=

1.

In what follows, let f.L = 0 and u = I.

D

o

a. Draw the graph of j. Find the intervals on which f is increasing, the intervals on which f is decreasing, and any local extreme values and where they occur. b. Evaluate

However, the integral

forn = 1,2,and3. c. Give a convincing argument that

1:

for the volume of the solid converges.

f(x) dx = 1.

a. Calculate it. b. This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But ifwe fill the hom with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction. 75. Sine-integral function

1

e -x/2 for x

> I, and for b > I,

00

e-x/ 2 dx --+ 0

as

b --+ 00.)

78. Show that if f(x) is integrable on every interval of real numbers and a and b are real numbers with a < b, then

laoo f(x) dx both converge if and only if f~oo f(x) dx and hoo f(x) dx both converge.

x

Si(x) =

< f(x) <

a. f~oo f(x) dx and

The integral

1s~t

(Hint: Show that 0

dt,

called the sine-integral jUnction, has important applications in optics.

b.

oo

f~oo f(x) dx + fa f(x) dx

=

oo

f~oo f(x) dx + fb f(x) dx

when the integrals involved converge.

Chapter 8 Practice Exercises 00

COMPUTER EXPLORATIONS In Exercises 79--1!2, use a CAS to explore the integrals for various values ofp (include nonintegervalues). For what values ofp does the in-

79·1'XP InXdx

tegral converge? What is the value of the integral when it does converge? Plot the integraod for various values ofp.

81.

Chapter m

1

80.

xP lox dx

00

xP 10 x dx

82. /:, xPIo

Ixldx

Questions to Guide Your Review

1. What is the formula for integration by parts? Where does it come from? Why might you waot to use it? 2. When applying the formula for integration by parts, how do you choose the u aod dv? How cao you apply integration by parts to ao integral of the form f(x) dx? 3. Ifan integrand is a product of the fonn sin" x coslllx, where m and n are nonnegative integers, how do you evaluate the integral? Give a specific example of each case. 4. What substitotions are made to evaluate integrals of sin mx sin nx, sin mx cos nx, and cos mx cos nx? Give an example of each case.

J

5. What substitotions are soroetinres used to transform integrals in2 2 2 2 2 - x , - a into integrals + x 2, aod volving that can be evaluated directly? Give an example of each case.

Va

1

489

Va

Vx

6. What restrictions cao you place on the variables involved in the three basic trigonoroetric substitotions to ma1re sure the substitotions are reversible (have inverses)?

a. is a product of distinct linear factors? b. consists of a repeated linear factot?

c. contains an irreducible quadratic factot? What do you do if the degree of f is not less thao the degree of g? 9. How are integral tables typically used? What do you do if a particular integral you want to evaluate is not listed in the table? 10. What is a reduction formula? How are reduction formulas used? Give ao example. 11. You are collaborating to produce a short "how-to" manual for numerical integration, aod you are writing about the Trapezoidal Rule. (0) What would you say about the rule itself aod how to use it? How to achieve accuracy? (b) What would you say if you were writing about Simpson ~ Rule instead? 12. How would you coropare the relative merits of Simpson's Rule and the Trapezoidal Rule?

7. What is the goal of the method of partial fractions?

13. What is an improper integral ofType I? Type ll? How are the values of various types of improper integrals defmed? Give examples.

8. When the degree of a polyooroia1 f(x) is less thao the degree of a polyoomial g(x), how do you write f(x)/g(x) as a sum ofpartia1 fractions if g(x)

14. What tests are available for detemtining the convergeoce and divergence of improper integrals that carmot be evaluated directly? Give examples of their use.

!

Chapter

Practice Exercises

Integration by Parts Evaluate the integrals in Exercises 1-1l using integration by parts. 1. / 10 (x

+

2 2. / x loxdx

I)dx

3. / tao-I 3x dx

cos2 8 + cos 8 - 2 2 15. /3X + 4x + 4dx x3 + x

17. /

5. /

(x + I?e%dx

6. / x 2 sin(1 - x)dx

7. /

e%cos2xdx

8. /

e-2x sin3xdx

19 / · 21. /

Partial Fractions Evaluate the integrals in Exercises substitotion IlrSt 9 / .

xdx x2-3x+2

11 / dx • x(x + 1)2

~28.

10 / .

12./

sinOdO

13 / ·

v + 3 dv

2v 3 - 8v

dt 1'+412 +3 2 3 x +x dx 2 x +X - 2 3

2

14. /

cosOdO sin2 8 + sin8 - 6

4xdx x 3 + 4x (3v - 7)dv 18. / (v - l)(v - 2)(v - 3)

16. /

20. /

tdl t 4 _t 2 -2

22./ + l X3

x3

-

x

dx 2

3

It may be necessary to use a

23. /

x + 4x dx x 2 +4x+3

24. /

2x + x - 21x + 24 dx x2+2x-8

xdx x 2 +4x+3

25 /

dx x(3Vx+l)

26. /

dx x(1 +~)

x+1 dx x 2(x - I)

• 27. /

e'

~I

28. /

tis

W+1

490

Chapter 8: Techniques of Integration

Trigonometric Substitutions Evaluate 1he integrals in Exercises 29--32 (a) wi1hout using a trigon<>metric substitution, (b) using a trigonometric substitution. 29

•

J

30 •

ydy v'16-y'

31.J~ 4 - x2

32.

J Jv'4t'=1 xdx

J x_~,

tdt

35.J~ 9-x

J

x(9

39. 41.

43.

sin3 x cos4 .:tdx

38.

tan4 x sec2 x ax

40.

sin 58 cos 68d8

42.

v'1

(x - 101»)

+ 25

Find 1he average value of Cv fot 20' s T s 675'C and 1he temperature at which it is attained.

Evaluate 1he integrals in Exercises 37-44.

J J J J

U:s

Cv ~ 8.27 + 10-5 (26T - 1.87T').

Trigonometric Integrals 37.

~ 37 sin

50. Heat capacity of a gas Heat capacity Cv is 1he anlOunt ofbeat required to raise the temperature of a given mass of gas with constant volume by I 'C, measured in units of cal/deg-mol (calories pet degtee gtan3 molecu1ar weight). The heat capacity of osygen depends on its temperature T and satisfies 1he fotmula

~ x') 36.J~ 9-~

34.

9

!(x)

for • 365-day year. This is one way to estimate 1he armual mean air Ietoperature in Fairbanks, Alaska. The National Wea1her Service's official figure, a numerical average of the daily normal mean air temperatures fot 1he year, is 25.7'F, which is slightly higher 1han 1he avetage value of !(x).

v'4+x'

Evaluate 1he integrals in Exercises 33-36.

33.

49. Mean temperature Compute 1he average value of 1he temperature fune1ion

+ cos (t/2) dt

44.

J J J J

coss x sins x dx 3

tan

51. Fuel el'lkiency An automohile computet gives a digital readout of fuel consmnption in gallons pet hoor. During a trip, a passenger recotded 1he fuel consumption every 5 min for a full hoor of travel

x sec3 x ax

Time

e'v'tan' e'

+

I dt

Numerical Integration 45. According to 1he error-bound funnula for Simpson's Ru1e, howIIlllllY subintervals shou1d you use to be sure of estimating 1he va1ue of 1n3

Gal/h

Time

Gal/h

2.5 2.4 2.3 2.4 2.4 2.5 2.6

35 40 45 50 55 60

2.5 2.4 2.3 2.4 2.4 2.3

cos 38 cos 38 d8

~ [~dx

by Simpson's Ru1e wi1h an error of no more 1han 10-4 in absolute

value? (Remember iliat for Simpson's Ru1e, 1he numbet of subinterva1s has to be even.) 46. A brief calculation shows iliat if 0 '" x '" I, 1hen 1he second derivative of !(x) ~ v'1+7 lies between 0 and 8. Based on 1his, about how IIllIlIY subdivisions wou1d you need to estimate 1he integral of! from 0 to I wi1h an ertor no gtealet 1han 10-3 in aJ>. solute value using 1he Trapezoidal Ru1e?

0 5 10

15 20 25 30

a. Use 1he Trapezoidal Ru1e to approximate 1he total fuel consumption during 1he hoor. b. If1he automobile covered 60 n1i in 1he hour, what was its fuel efficiency (in n1iles per gallon) fot iliat portion of 1he trip? 52. A new parking lot To meet 1he demand fot parking, yoor town has allocated the area shown here. As the town engineer, you have been asked by 1he town council to f"md out if 1he lot can be built for Sl1,OOO. The cost to clear 1he land will be S0.10 • square foot, and 1he lot will cost $2.00 a square foot to pave. Use Simpson's Ru1e to f"md out if1hejob can be done fot SII,Ooo.

47. A direct calculation shows iliat

Oft

54ft

How close do you come to 1his value by using 1he Trapezoidal Ru1e wi1hn ~ 6?Simpson'sRu1ewi1hn ~ 6?Try1hemandf"mdout 48. You are planning to use Simpson's Ru1e to estimate 1he value of 1he integral

l'

!(x)dx

wi1h an ertor magnitude less 1han 10-'. You have deletmined 1hat Ij<')(x) I '" 3 1hroughout 1he interval of integration. How IIllIlIY subintetva1s shou1d you use to assure 1he required accoracy? (IWmembet iliat fot Simpson's Ru1e 1he number has to be even.)

51 ft

49.5 ft Vertical spacing

~

15 ft

54ft 64.4ft

67.5 ft 42ft Ignored

Chapter 8 Additional and Advanced Exercises Improper Integrals Evaluate the improper integrals in Exercises 53-JJ2. 53

dx

['

54.

· Jo V9 - x 2

55. 57.

[2

Jo

dy

[00

1

2du u -2u

2

85. dO

/

J-2 (0 + I)' 5

87.

58.

l

89.

60.

I~ xe" dx

62

1

003v-1 '2dv 4v - v

roo

· )-00

1 I~z 1 00

67.

e~

~

dz

66.

2dx + e~

f

z +I 2( 2 )dz z z +4

86. I x'eO<')dx

I I V9 I l

tdt

.

95

•

ooln y

- , dy

97.

I I

Y

· J~ x 2(1

90.

tan'tdt

92.

cotvdv

sin5tdt

96

dr

98.

vr

1+

dx

+ eX)

71

•

73. 75. 77. 79

•

I I I Io I

+ Vi

dx x(x 2 + I)'

9dv 81 - v'

cos (20

(I

81. I

+

sin 20 dO

I)dO

+ cos20)'

xdx

~

Chapter

l

721 · V-a I dx

2 - cosx + sin x dx sin2 x

I

(sin- 1 x)' dx

x2

102.

I (I

+x

I

+ x dx + x)' + v'f+Xdx

[/2 VI +

"v'1=7dx

108·lx'lnx:ln(lnX)dx 110.

I

112.

I~dx

x2

2

sin 0 dO cos2 (J 00 dx 76. /,2 (x - I )2

113... Show that Tr/2

x'dx

J.~/2 VI + cos4xdx

], o

J; f(x) dx = J; f(a -

[t +

x) dx.

.

sinX'

smX'

+ cosX' dx. 2

114. I

~/'

82·/~dV

(In x)""

b. Use part (a) to evaluate

78. I x 2 -a+1

116.

.

I

sinx

smX'

+ cosX'

dx

115. I

1

sin x2 dx

+ 8in x

I - cos x dx 1 + cosX'

Additional and Advanced Exercises

Evaluating Integrals Evaluate the integrals in Exercises I-JJ. 1.

4x' - 20x dx IOx 2 + 9

--,dx

106.

74.

80.

dv

104.

70. Ix' + 2dx 4 - x2

xdx

+ 4e'dO

100.

random order. I

Insmv

2

Assorted Integrations Evaluate the integrals in Exercises 69-116. The integrals are listed in

69. I

x dx

94.

+ (cos 5t)2

I

x'

I m:~' I I I e'V3 . I v?V=l I x' I I I VI

+ 3et + 2

e2t

xdx

V8 - a 2 -

88.

4t 2

-;'dt

roo

68

84 I

93. I ebrYxdx

+ 16

],00

00

dy y2-2y+2

1

4dx

x2

1

e'dt

91.

Which of the improper integrals in Exercises 63-j)8 converge and which diverge? 00 dO 64. e-cosudu 63. /.6 y'02+J 65.

•

lnxdx

Jo

56. ['

j3

(y - 1)2

J3

[0

83

491

2•

I

x(x

+

l)(x +dx2) .. ·(x

+ m) 4.

I

sin- 1 vY dy

1

In ~x) dx

492

Chapter 8: Techniques of Integration

5.! t-VI=7

.!~ • ;t" + 4

dt

20. Finding volume The infmite region bounded by the coordinate axes and the curve y - -In x in the flISt quadrant is revolved about the;t-axis to generate a solid. Find the volume of the solid.

Evaluate the limits in Exercises 7 and 8. 7. lim

1'"

x--->"" --.:

21. Centmid of a region Find the centroid of the region in the flISt quadrant that is bounded below by the »-axis, above by the curve y - In x, and on the right by the line;t - e.

8. lim ;til c~t dt l:--->o+ % t

sintdt

Evaluate the limits in Exercises 9 and 10 by identifying them with definite integrals and evaluating the integrals.

.-,

10oHm~

....... "" .1;-0

1

Vn 2 - t2

Applfcatlons 11. Finding are length Find the length of the curve Y

-1%

Vcos2tdt,

21. Centmid of a region Find the centroid of the region in the plane enclosed by the curves y - ±( I - ;t2tl/2 and the lines ;t - Oand;t - 1. 23. Length of. curve Find the length of the curve y - In;t from ;t - lto;t - e. 24. Finding mrfaee area Find the area of the surface generated by revolving the curve in Exercise 23 about the y-axis.

25. The.urfaee poenrted by an astroid The gnlph of the equation ;t2f3 + yl/3 _ I ill an tJ.ftroid (see accompanying ftgure). Find the area of the surface generated by revolving the curve about the x-Ws.

O:!;';;t:!;'; 'lr/4.

11. Finding are lengtb Find the length of the graph of the fimction y - In(1 _;t2), O:!;'; x:!;'; 1/2.

y

13. Finding volume The region in the ftrst ~t that is enclosed by the ;t-axis and the curve y = 1x"~ is revolved about the y-axis to generate a solid. Find the volume of the solid. 1... FIndIng volume The region in the first quadrant that is enclosed by the".ui>. the ""'" Y ~ '/(x~)o the .... ~ 1 and ;t = 4 is revolved about the »-axis to gwerate a solid. Find the volume of the solid.

""

x

15. FIndIng volume The region in the flIst quadrant enclosed by the coordinate axes, the curve y = er • and the line ;t = 1 is revolved about the y-axis to generate a solid. Find the volume of the solid. 16. Finding volume The region in the flISt quadrant that ill bounded above by the curve y - er - 1, below by the x-axis, and on the

-I

Z6. Lengthofaclll'ft Find the leug1h of the curve y= [%y'V,-ldt.

right by the line ;t - In 2 is revolved about the line ;t - In 2 to generate a solid. Find the volume of the solid.

27. For what value or values of a does

17. Finding volume Let R be the ''triangular"' region in the flIst quadrant that is bounded above by the line y = 1, below by the curve y = In;t, and on the left by the line ;t = 1. Find the

1-

volume of the solid generated by revolving R about L the ;t-axis. b. the line y - 1.

18. FInding volume (Continuation of Exercise 17.) Find the volume of the solid generated by revolving the region R about L the y-axis. b. the line x = 1. 19. FInding volume The region between the;t-axis and the curve Y

~

f(x)

~ {Oo

xlnx.

;t=0

0<x:s2

is revolved about the »-axis to generate the solid shown here. L Show that f is continuous at;t = O. b. Find the volume of the solid.

[G::

Isxsl6.

ix)dx

converge? Evaluate the corresponding integral(s). 28. For each ;t > O. let G(;t) for each x > O.

10

00

e --,.t dt. Prove that ;tG(;t) - 1

29. lnf"mite area IIDd (mite wlume What values of p have the folIawing property: The area of the region between the curve y - ;t---P, 1 :!;'; ;t < 00. and the ;t-axis is infmite but the volume of the solid generated by revolving theregi.on aboutthe;t-axis is finite. 30. Infbdte area and fbdte volume What values ofp have the :follawing property: The area of the region in the flISt quadrant enclosed by the curve y - ;t---p , the y-axis, the line x - I, and the interval [0, I] on the x-axis ill infinite but the volume of the solid genemted by revolving the region about one of the coon1ina1J; axes is finite.

The Gamma Function and Stlrlfng's Formula Euler's gamma fimction f(;t) ("gamma ofr'; f is a Greek capital g) uses an integral to extend the factorial function from the nonnegative integers to ot:heI" real. values. The formula is

_1

00

f(;t)

r1e-tdt,

;t> O.

For each positive;t. the number f{x) is the integral oftr-1e ~ with respect to t from. 0 to 00. Figure 8.21 shDws the graph of f near the origin. You will sec how to calculate f(l/2) if you do Additional Exercise 23 in Chapter 14.

Chapter 8 Additional and Advanced Exercises y

493

As you will see if you do Exercise 104 in Section 10.1, Equation (4) leads to the approximation

~"'~.

(5)

o b. Compare your calculator's value for n! with the value given by Stirling's approximation for n = 10, 20, 30, ... , as far as your calculator can go.

o c. A refmoment of Equation (2) gives

---r--±--+~+-~--~~~->X

o

2

3

-1

f(x) =

-2 -3

(~Y l#e1/(l2x)(1

or f(x) '"

FIGURE 8.21 Euler's gamma function f(x) is a continuous function ofx whose value at each positive integer n + 1 is n!. The deftning integral fonnula for f is valid only for x > 0, but we can extend f to negative noninteger values of x with the fonnula f(x) = (f(x + 1»/x, which is the subject of Exercise 31. 31. H II is a nonnegative integer, f(1I

+ 1)

(6) Compare the values given for 1O! by your calculator, Stirling's approximation, and Equation (6).

Tabular Integratton The technique of tabular integration also applies to integrals of the f(x )g(x) fix when neither function can be differentiated reform peatedly to become zero. For example, to evaluate

J

= II!

b. Then apply integration by parts to the integral for f(x show that f(x + I) = xf(x). This gives

+

f(4) = 3f(3) = 6

~andits derivatives

= nf(n) = n!

(1)

Use mathematical induction to verify Equation (1) for every nonnegative integer n.

32. Stirling'. formula

Scottish mathematician James Stirling (1692-1770) showed that

,~('iYfIrf(X)=

integral.

e'"'~

cos x

2e2x~inX (+) ~ -cosx

4e2¥

<E-

Stophere:Rowissameas [lI'St row except for multiplicative constants (4 on the left, -Ion tire right).

We stop differentiating and integrating as soon as we reach a row that

is the same as the fIrst row except for multiplicative constants. We in-

.(x) --+ 0 as x --+ oc.

(2)

f21r (ex)' \j--x

J

=

+(e2x siox) - (2e'"'(-cosx»

+ (Stirling'. formula).

(3)

a. Stirling'. approximation for II! Use Equation (3) and the fact that n! = nf(n) to show that

(~)" VZ;;;

terpret the table as saying

e2x cosx fix

Dropping .(x) leads to the approximation

n! '"

cos x and its

I,

so, for large x,

f(x) '"

e2;rcosxdt

we begin as before with a table listing successive derivatives of e2x and integrals of cos x:

f(3) = 2f(2) = 2

c.

J

I) to

f(2) = If(1) = I

+ 1)

(~Y l # e W2x),

which tells us that

a. Show that f(l) = 1.

[(n

+ .(x»

(Stirling'. approximation).

(4)

J

(4 e 2x)( -cos x) fix.

We take signed products from the diagonal arrows and a signed iot.,. gra1 for the last horizontal arrow. Transposing the integral on the righthand side over to the left-hand side now gives

5

J

e2xcos x dx

=

e2x sinx

+ 2e2xcosx

494

Chapter 8: Techniques of Integration

or

.

2z

(9)

smx=--2·

/

. " dx -- e " smX' +5 2" e cosX' e cosx

I

+ C'

+z

Finally, x ~ 2tan-1 z,so

after dividing by 5 and adding the constant of integration. Use tabullIr integration to evaluate the integrals in Exercises 33-40. 33. /

e2t"cos3xdt

34. /

e"'sin4xdx

35. /

sin 3x sinxdt

36. /

cos 5x sin 4x dx

37. /

e~sinbxdx

38. /

e~ cos bx dx

(10)

Examples _/ I dx~/I+.'~ • I+cosx 2 1+.'

=/dz=Z+C

40. / x2 1n (ax) dx

39. / In (ax) dx

= tan The Substitution z The substitotion

~

(f) + C

tan (x/2)

z = tan~ 2

(7)

I

b. /

2

+ sinx

dx = /

reduces the problem of integrating a rational expression in sin x and cos x to a problem of integrating a rational function of z. This in turn can be integrated by partial fractions. From the accornpaoying figure

2

I + z2 2dz + 2z + 2z2 1 + z2

= / .' +

~+ I = /

(z +

(I/~)2 + 3/4

=/,;~,; =

~tan-l (~)

+ C

= ~tan-12z + I + C yI3 yI3 2 _II + 2 tan (x/2) =yI3tan yI3 +C

Use the substitotions in Equations (7}-{1O) to evaluate the integrals in Exercises 41-48. Integrals like these arise in calculating the average aogular velocity of the output shaft of a uoiversal joint wben the input and output shafts are not aligned.

we can read the relation

tan'! 2

=

1

sin x

41. /

+ cosx·

To see the effect of the substitution, we calculate 43.

cosx

~ 2 cos

2

I cosX'

+

1-

(f) - I ~

sec2

~X/2) -

I - dx. smX'

W!2

I

W!2 45. /. 0

2 _1~_2_ _ 1 tan2 (x/2) I + z2

47. / .

z2

= --2' I +z

dx

. smX'

44.

dO 2+cosO

46

1

/.o

+

dl

smt - cost

(8)

~I~~ . ....dx'"'"c-_ _ + SInX' + cosX'

42. /

l

W !2

'TT/3

1

2w

• ",/2

48. /

/3

dx 1 - cosX' cos 0 dO sinO cos 0 + sinO

cosldl 1 - cost

Use the substilution z = tan (0/2) to evaluate the integrals in Exercises 49 and 50.

and

. . x x sin (x/2) 2 (x) smx ~ 2 sm Icosl ~ 2 cos (x/2) • cos I

~ 2tan"-.

I 2 sec (x/2) 2

2 tan (x/2) I + tan2 (x/2)

49. /

secOdO

50. /

cscOdO

Chapter 8 Technology Application Projects

Chapter

3

Technology Application Projects

Mathematica/Maple Modules: Riemllllll, Tropezoidol, IIIId SiIIIpsonApproxilllations Part I: Visualize the error involved in using Riemann sums to approximate the area under a curve.

l'lIrt II: Build a table of values and compute the relative magnitude of the error as a function of the step size 4.<. l'lIrt III: Investigate the effect of the detivative function on the error. l'lIrt. IV and V: Trapezoidal Rule approximations. l'lIrtVI: Simpson's Rule approximations. GtlIHes ojChanc", Explorillg tho Momo Carlo Probllbilistic T.chlliquojor NUIHerical Imqp'lltion

Graphically explore the Monte Carlo method for approximating dermite integrals. Computing Probabilities with Improper ImqpYlls

More explorations of the Monte Carlo method for approximating definite integrals.

495

9 FIRST-ORDER DIFFERENTIAL EQUATIONS OVERVIEW In Section 4.7 we introduced differential equations of the form dy/dx = f(x), where f is given and y is an unknown function of x. When f is continuous over some interval, we found the general solution y(x) by integration, y = f(x) dx. In Section 7.4 we solved separable differential equations. Such equations arise when investigating exponential growth or decay, for example. In this chapter we study some other types offirst-order differential equations. They involve only first derivatives of the unknown function.

J

9.1

Solutions, Slope Fields, and Euler's Method We begin this section by defining general differential equations involving first derivatives. We then look at slope fields, which give a geometric picture ofthe solutions to such equations. Many differential equations cannot be solved by obtaining an explicit formula for the solution. However, we can often find numerical approximations to solutions. We present one such method here, called Euler s method, upon which many other numerical methods are based.

General First-Order Differential Equations and Solutions A first-order differential equation is an equation

dy dx = f(x,y)

(1)

in which f(x, y) is a function of two variables defined on a region in the xy-plane. The equation is of first order because it involves only the first derivative dy/dx (and not higher-order derivatives). We point out that the equations

y' = f(x,y)

and

d dx Y = f(x,y)

are equivalent to Equation (I) and all three forms will be used interchangeably in the text. A solution of Equation (I) is a differentiable function y = y(x) defined on an interval I of x-values (perhaps infinite) such that d dxY(x) = f(x,y(x))

on that interval. That is, wheny(x) and its derivative y'(x) are substituted into Equation (1), the resulting equation is true for all x over the interval I. The general solution to a firstorder differential equation is a solution that contains all possible solutions. The general solution always contains an arbitrary constant, but having this property doesn't mean a solution is the general solution. That is, a solution may contain an arbitrary constant without being the general solution. Establishing that a solution is the general solution may

496

9.1 Solutions, Slope Fields, and Euler's Method

497

require deeper results from the theory of differential equations and is best studied in a more advanced course.

EXAMPLE 1

Show that every member of the family of functions

y

t+ 2

=

is a solution of the first-order differential equation

dy I dx = x(2 - y) on the interval (0, 00), where C is any constant.

Solution Differentiating y = C/x :

=

+ 2 gives

C! (i) + 0

= -

~.

We need to show that the differential equation is satisfied when we substitute intn it the expressions (C/x) + 2 for y, and -C/x 2 for dy/dx. That is, we need tn verifY that for all

XE(O,OO),

This last equation follows immediately by expanding the expression on the right-hand side:

1x [2 _(cx + 2)] 1(_ c) C x =

= _

X

Therefore, for every value of C, the function y = C/x equation.

x2·

+ 2 is a solution of the differential _

As was the case in finding antiderivatives, we often need a particular rather than the general solution to a first-order differential equation y' = f(x, y). The particular solution satisfying the initial condition y(xo) = Yo is the solution y = y(x) whose value is Yo when x = Xo. Thus the graph of the particular solution passes through the point (xo, Yo) in the ~-p1ane. A first...rder initial value problem is a differential equation y' = f(x, y) whose solution must satisfY an initial condition y(xo) = Yo.

EXAMPLE 2

Show that the function

Y=

(x + I) -

I -ex 3

is a solution tn the first-order initial value problem

dy dx=y-x,

Solution

The equation

dy dx=y-x is a first-order differential equation with f(x, y) = y - x. On the left side ofthe equation:

dy = ~ dxdx

(x +

1

_lex) 3

= 1 _

leX

3'

498

Chapter 9: First-Order Differential Equations y

On the right side of the equation: y

=

(x

+ 1) -

!e x

3

y - x = (x

te x

+ I) -

-

x = I - teX.

The function satisfies the initial condition because

1

I 2 I X y(O) = [(x+I)--e =1--=-. 3 x- o 3 3 Graph of the solution to the initial value problem in Example 2. FIGURE 9.1

•

The graph of the function is shown in Figure 9.1.

Slope Fields: Viewing Solution Curves Each time we specify an initial condition y(xo) = Yo for the solution of a differential equation y' = f(x, y), the solution curve (graph of the solution) is required to pass through the point (xo, Yo) and to have slope f(xo, Yo) there. We can picture these slopes graphically by drawing short line segments of slope f(x, y) at selected points (x, y) in the region of the .lJ'-plane that constitutes the domain of f. Each segment has the same slope as the solution curve through (x, y) and so is tangent to the curve there. The resulting pictore is called a slope field (or direction field) and gives a visualization of the general shape of the solution curves. Figure 9.2a shows a slope field, with a particular solution sketched into it in Figure 9.2b. We see how these line segments indicate the direction the solution curve!skes at each point it passes through.

(b)

FIGURE 9.2


curve through the point

=

y - x. (h) TIre particular solution

(0, ~)(Example 2).

Figure 9.3 shows three slope fields and we see how the solution curves behave by following the tangent line segments in these fields. Slope fields are useful because they display the overall behavior of the family of solution curves for a given differential equation. For instance, the slope field in Figure 9.3b reveals that every solution y(x) to the differential equation specified in the figure satisfies limx->±ooy(x) = O. We will see that knowing the overall behavior of the solution curves is often critical to understanding and predicting outcomes in a real-world system modeled by a differential equation.

499

9.1 Solutions, Slope Fields, and Euler's Method

(c) y'~ (l-x)y+ ~

y

y

~

L(x)

~

Yo + f(x .. yol(x - xo)

FIGURE 9.3 Slope fields (top row) and selected solution curves (bottom row). In computer renditions, slope segments are sometimes portrayed with arrows, as they are here. This is not to be taken as an indication that slopes have directioos, however, for they do not.

Yo

o FIGURE 9.4 The linearization L(x) of y ~ y(x) atx ~ Xo.

Constructing a slope field with pencil and paper can be quite tedious. All oUI examples were generated by a computer.

Euler's Method If we do not require or cannot immediately tmd an exact solution giving an explicit formula for an initial value problemy' = f(x,y),y(xo) = Yo, we can often use a computer to generate a table of approximate numerical values of y for values of x in an appropriate interval. Such a table is called a numerical solution of the problem, and the method by which we generate the table is called a numerical method.

Given a differential equation dyl ax = fix, y) and an initial condition y(xo) can approximate the solution y = y(x) by its linearization

L(x) The first Euler srep approximates y(x,) with y, ~ L(x,). FIGURE 9.5

y

Euler approximation (x2.

Yv

=

y(xo)

+ y'(xo)(x - xo)

,

,

True solution ClU'Ve y ~ y(x) : I I

I I

Three sreps in the Euler approximatioo to the solutioo of the initial value problem y' ~ f(x, y), y (xo) ~ Yo. FIGURE 9.6

As we take more steps, the errors involved

usually accumulate, but not in the exaggerated way shown here.

L(x)

=

Yo

Yo, we

+ f(xo,yo)(x - xo).

The function L(x) gives a good approximation to the solution y(x) in a short interval about Xo (Figore 9.4). The basis of Euler's method is to patch together a string oflinearizations to approximate the curve over a longer stretch. Here is how the method works. We know the point (xo, Yo) lies on the solution curve. Suppose that we specify a new value for the independent variable to be x, = Xo + ax. (Recall that ax = Ax in the definition of differentials.) If the increment ax is small, then

y, I

or

=

=

L(x,)

=

Yo + f(xo,Yo)

ax

is a good approximation to the exact solution value y = y(x,). So from the point (xo, Yo), which lies exactly on the solution curve, we have obtained the point (Xlo yd, which lies very close to the point (XloY(X,)) on the solution curve (Figore 9.5). Using the point (xloyd and the slope f(XloY,) of the solution curve through (XloY,), we take a second step. Setting X2 = + dx, we use the linearization of the solution curve through (Xlo Yt) to calculate

x,

Y2

=

y, + f(XloY,)

ax.

This gives the next approximation (X2, Y2) to values along the solution curve y = y(x) (Figure 9.6). Continuing in this fashion, we take a third step from the point (",Y2) with slope f(X2, Y2) to obtain the third approximation

Y3

=

Y2

+ f(X2, Yo) ax,

500

Chapter 9: First-Order Differential Equations and so 00. We are literally building an approximatioo to ooe of the solutioos by following the directioo of the slope field of the differential equatioo. The steps in Figure 9.6 are drawn large to illustrate the constroctioo process, so the would be small enough to make the red curve approximatioo looks crude. In practice, hug the blue one and give a good approximatioo throughout.

ax

EXAMPLE 3 Find the first three approximatioos Yl,Y2,Y3 using Euler's method for the initial value problem y(O) = I, y' = I + Y, starting at Xo = 0 with

ax =

0.1.

Solution We have the starting values Xo = 0 and Yo = I. Next we determine the = 0.1, values of x at which the Euler approximations will take place: Xl = Xo + X2 = Xo + 2 = 0.2, and X3 = Xo + 3 = 0.3. Then we find

ax

ax

ax

First:

= =

Second:

= 1.2

+ f(xl, Yl) ax Yl + (I + Yl) ax 1.2 + (i + 1.2)(0.1)

Y2 = Yl = =

Thjrd:

+ f(xo, Yo) ax Yo + (I + Yo) ax I + (I + 1)(0.1)

Yl = Yo

= 1.42

+ f(X2, Y2) ax Yo + (I + Y2) ax 1.42 + (i + 1.42)(0.1)

Y3 = Y2 = =

= 1.662

•

The step-by-step process used in Example 3 can be continued easily. Using equally spaced values for the independent variable in the table for the numerical solutioo, and generating n of them, set Xl = Xo

X2 = Xn

=

Xl

+ ax + ax

Xn-l

+ dx.

Then calculate the approximations to the solutioo,

Yl = Yo Yo = Yl

+ +

Yn = Yn-l

HISTORICAL BIOGRAPHY

Leonhard Euler (1703-1783)

ax f(xl, Yl) ax f(xo,yo)

+

f(xn-l,Yn-l)

ax.

The number of steps n can be as large as we like, but errors can accumulate if n is too large. Euler's method is easy to implement 00 a computer or calculator. A computer program generates a table of numerical solutions to an initial value problem, allowing us to input Xo and Yo, the number of steps n, and the step size It then calculates the approximate solution values Yl, Y2, ... ,Yn in iterative fashion, as just described Solving the separable equatioo in Example 3, we find that the exact solutioo to the initial value problem is Y = 2e x - I. We use this informatioo in Exarople 4.

ax.

9.1 Solutions, Slope Fields, and Euler's Method

EXAMPLE 4

501

Use Euler's method to solve

y' = I

+ y,

y(O)

=

I,

on the interval 0 :5 x :5 1, starting at Xo = 0 and taking

TABLE 9.1 Euler solution of y' = 1 step size dx = 0.1

y 4

3

2

y=2e x - l

+ y, y( 0)

= 1,

x

y (Euler)

y <eIact)

Error

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

I 1.2 1.42 1.662 1.9282 2.2210 2.5431 2.8974 3.2872 3.7159 4.1875

I 1.2103 1.4428 1.6997 1.9836 2.2974 2.6442 3.0275 3.4511 3.9192 4.4366

0 0.0103 0.0228 0.0377 0.0554 0.0764 0.1011 0.1301 0.1639 0.2033 0.2491

--~--------------~--~X

o

FIGURE 9.7 Thegraphofy = 2e' - I superimposed on a scatterplot of the Euler approximations shown in Tahle 9.1 (Example 4).

By the time we reach x = I (after 10 steps), the error is about 5.6% of the exact solution. A plot of the exact solution curve with the scatterplot of Euler solution points from Table 9.1 is shown in Figure 9.7.

(b) One way to try to reduce the error is to decrease the step size. Table 9.2 shows the results and their comparisons with the exact solutions when we decrease the step size to 0.05, doubling the number of steps to 20. As in Table 9.1, all computations are performed before rounding. This time when we reach x = I, the relative error is only about 2.9%. _ It might be tempting to reduce the step size even further in Exaruple 4 to obtain greater accuracy. Each additional calculation, however, not ouly requires additional computer time but more importantly adds to the buildup of round-off errors due to the approximste representations of numbers inside the computer. The analysis of error and the investigation of methods to reduce it when making numerical calculations are important but are appropriate for a more advanced course. There are numerical methods more accurate than Euler's method, usually presented in a further study of differential equations.

502

Chapter 9: First-Order Differential Equations

Euler solution of y' step size dx = 0.05

TABLE 9.2

x

=

1

+ y, y(O)

=

1,

y (eIllet)

Error

1

1

0

1.1

1.1025

0.0025

0.10

1.205

1.2103

0.0053

0.15

1.3153

1.3237

0.0084

0.20

1.4310

1.4428

0.0118

0.25

1.5526

1.5681

0.0155

0.30

1.6802

1.6997

0.0195

0.35

1.8142

1.8381

0.0239

0.40

1.9549

1.9836

0.0287

0.45

2.1027

2.1366

0.0340

0.50

2.2578

2.2974

0.0397

0.55

2.4207

2.4665

0.0458

0.60

2.5917

2.6442

0.0525

0.65

2.7713

2.8311

0.0598

0.70

2.9599

3.0275

0.0676

0.75

3.1579

3.2340

0.0761

0.80

3.3657

3.4511

0.0853

0.85

3.5840

3.6793

0.0953

0.90

3.8132

3.9192

0.1060

0.95

4.0539

4.1714

0.1175

1.00

4.3066

4.4366

0.1300

0 0.05

y(EuJer)

Exercises 9.1 Slope Fields In Exercises 1-4, match the differential equations with their slope fields, graphed here. y

T

x

1\\-/ /-\11 '.\ \'

-

- 4'"/ /4

/f,l.\(

\ \-/

'.\

\' \ \---/ /---\ \

~'\.7'i!

If~~/-1 \

lim Ilr 1111+1 (a)

~mmj

Tf111 ~ Ir \1 ~ 11+f L-!/ ~

,,""""""

:--...........

""''-''''"

(b)

///

..........................................................

nmm

/ / /2"/ / /4'"

.........................................................

~~~~~~~~ ~~~~~~~~ x

--,//1111 .......-,...-'/ I!!

Iqlq lf IHdldl qII HI Hdld ~

y

y

y

(mm ~ 111111 ,.

mm(( IIIIIIII

t

"'\ ......." -. . . .---/''/ ///// ! I

\\"-.,.......-// g \ \ \ " " .......-----v'" ........-----,./

\\\\" x

\\\ , '~

II!Hill!

/"/111

- m m -......--

I

/ / ); t

v / 21

!qm II \ ~~ \

........

"" .......-----,....'"/1/1 \ " . . . .---/''/ ! !

!IPI U

\\".......-/// \ \ \ ",.......-..?/ \\\\""........---/

III\\'.'<-

(d)

(e)

1.y'=x+y

2.y'=y+1

3.yl=_~

4. y' =y2_x 2

x

9.1 Solutions, Slope Fields, and Euler's Method In Exercises 5 and 6, copy the slope fields and sketch in some of the

solution curves.

s.

y'

503

19. Use the Euler method with dx ~ 0.5 to estimate y(5) if y' ~ y2/,Vx andy(l) ~ -1. What is the exact value ofy(5)? 20. Use the Euler method with dx ~ 1/3 to estimate y(2) if y' ~ xsinyandy(O) ~ 1. What is the exact value ofy(2)?

~(y+2)(y-2)

Y

21. Show that the solutioo of the initial value problem

lllllllt llllllll

~x+y,

y'

y(XO)~Yo

is y

x

iiiiii l~ 6. y'

~

+ (I + Xo + Yo) er

-I - x

",.

22. What integral equatioo is equivalent to the initial value problem y' ~ f(x), y(xo) ~ Yo?

iiiiiiii

COMPUTER EXPLORATIONS In Exercises 23-28, obtain a slope field and add to it graphs of the

solution curves passing through the given points.

+ I)(y - I)

y(y

~

23. y' Y

............-,,,.......-...-...,............. ...................'.,".......--..-......-....

y with b. (0,2)

a. (0, I)

111ll\1~ 11111111

-4""...-"...-"4 ...-"..........--

~

24. y'

~

a. (0, I) 25. y' x

~

y(x

b. (0,4)

c. (0,5)

+ y) with

a. (0, I)

:./...-"...-"Z ......................

c. (0, -I)

2(y - 4) with

b. (0, -2)

c. (0, 1/4)

d. (-1,-1)

b. (0,2)

c. (0, -I)

d. (0,0)

c. (0,3)

d. (I, -I)

26. y' ~ y2 with

mmttmimi

a. (0, I) 27. y' ~ (y - I)(x

a. (0, - I)

Integral Equations In Exercises 7-10, write an equivalent rust·order differential equatioo and initial conditioo for y.

7. y

~

8. y ~ 9. y 10. y

~

1 x

+

-I

(I - y(l))dl

1 2

t dl

_/,x

~~ with 2

x +4 a. (0,2)

b. (0, -6)

29. A logistic equation

(I

+ y(l))sinldl

O::s;x::S;4,

12. y'

~

x(1 - y),

13. y'

~

2xy

+ 2y,

y(l)

~

0,

y(O) ~ 3,

dx

~

(-2V3, -4)

y'

~ y(2 - y),

dx ~ 0.2

y(O) ~ 1/2;

O~y~3

-6 '" x '" 6,

-6 '" y '" 6

Exercises 31 and 32 have no explicit solution in tenDs of elementary functions. Use a CAS to explore graphically each of the differential equatioos.

31. y' ~ cos (2x - y), y(O) ~ 2; 0 s x '" 5, 0 '" y '" 5 32. A Gompertz equation y' ~ y(I/2 - lny), y(O) ~ 1/3; 0:5x:54, Q:5y:53

33. Use a CAS to rmd the solutioos of y'

+y

~

f(x) subjectto the

initial cooditiooy(O) ~ 0, if f(x) is

a. 2x

0.2

b. sin 2x

c. 3e x/2

d 2e -x/2 cos 2x.

Graph all four solutions over the interval -2 '" x '" 6 to coropare the results.

14. y' ~ y2(! + 2x), y( -I) ~ I, dx ~ 0.5 Is. y' ~ 2xe x', y(O) ~ 2, dx ~ 0.1

y' ~ yeX, y(O) ~ 2,

c.

30. y' ~ (sinx)(siny), y(O) ~ 2;

~ 1 + /,X y(l) dl

Using Euler's Method In Exercises 11-16, use Euler's method to calculate the rust three ap· proximatioos to the given initial value problem for the specified incre· ment size. Calculate the exact solutioo and investigate the accuracy of your approximations. Round your results to four decimal places. y 11. y' ~ I - x' y(2) ~ -I, dx ~ 0.5

D D 16.

b. (0, I)

In Exercises 29 and 30, obtain a slope field and graph the particular solutioo over the specified interval. Use your CAS DE solver to rmd the general solution of the differential equatioo.

XI

1

28. y'

+ 2) with

34. a. Use a CAS to plot the slope field of the differential equatioo

dx ~ 0.5

17. Use the Euler method with dx ~ 0.2 to estimate y(l) if y' andy(O) ~ 1. What is the exact value ofy(I)?

~

18. Use the Euler method with dx ~ 0.2 to estimate y(2) if y' and y(!) ~ 2. What is the exact value of y(2)?

y/x

~

y

,_3x2 +4x+2 y2(y-1) over the region -3 :5 x :5 3 and -3 :5 Y :5 3. b. Separate the variables and use a CAS integrator to rmd the general solution in implicit fonn.

504

Chapter 9: First-Order Differential Equations

c. Using a CAS implicit function grapher, plot solution curves for the arbitrary constsnt values C = -6, -4, -2, 0, 2, 4, 6.

d. Find and graph the solution that satisfies the specified initial condition over the interval [0, b].

d. Find and graph the solution that satisfies the initial condition y(O) = -1.

e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the x-intervai and plot the Euler approsimation superimposed on the graph produced in part (d).

In Exercises 35-38, use Euler~ method with the specified step size to estimate the value of the solution at the given point x .... Find the value of the exact solution at x" . 35. y'

=

2xe"',

36. y'

=

2y2(x - I), y(2)

37. y'

=

Yx/y,

38. y'

=

I

y(O)

+ y2,

=

2, dx =

y> 0, y(O) y(O)

=

0,

=

0.1, x·

-1/2, dx =

dx

I, =

dx

=

I

=

0.1, x·

=

3

=

0.1, x·

=

I

0.1, x·

=

f. Repeat part (e) for 8,16, and 32 subinterva1s. Plot these three Euler approximations superimposed on the graph from part (e).

g. Find the error (y(exact) - y(Euler)) at the specified point x = b for each of your four Euler approsimations. Discuss the improvemeot in the perceotage error.

I

Use a CAS to explore graphically each of the differential equations in Exercises 39-42. Perform the followiog steps to help with your explorations. •• Plot a slope field for the differential equation in the given

39. y' = x b = 1 40. y'

=

+ y,

y(O) = -7/10;

-x/y, y(O)

41. y' = y(2 - y),

=

2;

-4'" x '" 4,

-3 '" x '" 3,

y(O) = 1/2;

-4'" Y '" 4;

-3 s y s 3; b

=

2

0 s x s 4, 0 s Y s 3;

b=3

",,-wiodow. b. Find the general solution of the differential equation using your CAS DE solver.

42. y' = (sinx)(siny), y(O) = 2;

b

=

-6 s x s 6,

-6 s Y s 6;

37T/2

c. Graph the solutions for the values of the arbitrary constsnt C = -2, -I, 0, I, 2 superimposed on your slope field plot.

9.2

First-Order Linear Equations A first-order linear clifferential equation is one that can be written in the form

dy dx

+ P(x)y

=

(l)

Q(x),

where P and Q are continuous functions of x. Equation (1) is the linear equation's .tandard form. Since the exponential growth/decay equation dy/dx = ky (Section 7.4) can be put in the standard form

dy --ky=O dx '

we see it is a linear equation with P(x) = -k and Q(x) = O. Equation (I) is linear (iny) because y and its derivative dy/ dx occur only to the flISt power, they are not multiplied together, nor do they appear as the argument ofa function (SUCh as siny, eY, or

EXAMPLE 1

V dy/dx).

Put the following equation in standard form:

x

dY =x 2 +3y dx '

x> O.

Solution

Divide by x.

Standsrd funn withP(x) and Q(.)

=•

= -3/.

9.2

First-Order Linear Equations

Notice thatP(x) is -3/x, not +3/x. The standard form is y' + P(x)y nus sign is part of the formula for p(x).

=

505

Q(x) , so the mi•

Solving Linear Equations We solve the equation

dy dx + P(x)y

=

Q(x)

by multiplying both sides by a positive function v(x) that transforms the left-hand side into the derivative of the product v(x) •y. We will show how to find v in a moment, but fITst we want to show how, once found, it provides the solution we seek. Here is why multiplying by v(x) works:

dy dx + P(x)y

=

Q(x)

dy v(x) dx + P(x)v(x)y

=

v(x)Q(x)

MulDpIy by positi\Io v(x).

v(x)Q(x)

v(x) is chosen to make dy d vdx +Pvy~ dx(v· Y ).

d

dx (v(x)' y)

=

v(x)' y

=

y

=

Original equation is in standard fonn.

J

Integrate with respect

v(x)Q(x) dx

v(~)

tox.

J

v(x)Q(x) dx

(2)

Equation (2) expresses the solution of Equation (1) in tenDs of the functions v(x) and Q(x). We call v(x) an integrating factor for Equation (1) because its presence makes the equation integrable. Why doesn't the formula for P(x) appear in the solution as well? It does, but indirectly, in the construction of the positive function v(x). We have

d dx (vy)

dy v dx + Pvy

Condition imposed on v

dy dv dy vdx+Ydx=vdx+Pvy

Derivative Product Rule

dv Ydx=Pvy

dy The terms v dx cancel.

=

This last equation will hold if

dv dx

=

Pv

dv v

=

Pdx

Variables separated, v > 0

Integrate both sides.

lnv

=

J

Pdx

e lnv = ejPdx v = ejPdx

Since v > O. we do not need absolute value signs in In v. Exponentiate both sides to solve for v.

(3)

506

Chapter 9: First-Order Differential Equations Thus a formula for the general solution to Equation (I) is given by Equation (2), where v(x) is given by Equation (3). However, rather thau memorizing the formula, just remember how to find the integrating factor once you have the standard form so P(x) is correctly identified. Any antiderivative of P works for Equation (3).

To solve the linear equation y' + P(x)y = Q(x), multiply both sides by the integrating factor v(x) = ef P{x) dx and integrate both sides.

When you integrate the product on the left-hand side in this procedure, you always obtain the product v(x)y of the integrating factor and solution function y because of the way v is defined

EXAMPLE 2

Solve the equation

x HISTORICAL BIOGRAPHY

Solution

dY =x 2 +3y dx '

x>

o.

First we put the equation in standard form (Example I):

Adrien Marie Legendre

dy 3 dx - -xy

(1752-1833)

=

x,

so P(x) = -3/x is identified. The integrating factor is

v(x) =

efP«)dx = ef{-3/x)dx

Constant of integration is 0, so v is as simple as possible.

= e-3lnlxl

• >0

Next we multiply both sides of the standard form by v(x) and integrate:

L

(dY_ly) = ~.x x3

x3dxx

I dy 3 I -3- - - y = -2 x dx X4 x

~(Iy)=~2 dx \,;3 x

Left·handside is ::. (v· y).

Integrate both sides.

I I ,y = -x + c. x Solving this last equation for y gives the general solution:

y =x 3 (-} +

EXAMPLE 3

c)

=

-x 2 +

ex 3,

x>

Find the particular solution of

3xy' - y = lnx satisfying y( I) = - 2.

+ I,

x>

0,

o.

•

9.2

Solution

With x

>

First-Order Linear Equations

507

0, we write the equation in standard form:

, 1 Y - 3x Y

lnx

+1

3x

=

Then the integrating factor is given by

v = ef -ob/3. = e H / 3)ln. = x- I / 3 •

•

>0

Thus Left-hand side is try.

Integration by parts of the right-hand side gives x- I / 3y =

-x- I / 3(lnx

+ 1) +

J

x-4/3

dx

+ c.

Therefore

or, solving for y,

y When x

=

1 and y

=

=

-(lnx + 4) + ex 1/3 •

-2 this last equation becomes -2

=

-(0 + 4) + C,

so

c=

2.

Substitution into the equation for y gives the particular solution

y

= 2x 1/ 3 -

lnx - 4.

•

In solving the linear equation in Example 2, we integrated both sides of the equation after multiplying each side by the integrating factor. However, we can shorten the amount of work, as in Example 3, by remembering that the left-hand side always integrates into the product v(x) •Y of the integrating factor times the solution function. From Equation (2) this means that

v(x)y =

J

v(x)Q(x) dx.

(4)

We need only integrate the product of the integrating factor v(x) with Q(x) on the righthand side of Equation (1) and then equate the result with v(x)y to obtain the general solution. Nevertheless, to emphasize the role of v(x) in the solution process, we sometimes follow the complete procedure as illustrated in Example 2. Observe that if the function Q(x) is identically zero in the standard form given by Equation (1), the linear equation is separable and can be solved by the method of Section 7.4:

dy dx

+ P(x)y

= Q(x)

dy dx

+ P(x)y

=

dy Y

0

= -P(x) dx

Q(x) = 0 Separating the wriables

508

Chapter 9: First-Order Differential Equations

RL Circuits The diagram in Figure 9.8 represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can he closed to connect a constant electrical source of V volts. Ohm's Law, V = RI, has to be augmented for such a circuit. The correct equation accounting for hoth resistance and inductance is

v + ,---1 1---, 2> /

6

I

a

Switch

b

di L dt

L

R

+ R''

= V

(5)

,

where i is the current in amperes and t is the time in seconds. By solving this equation, we can predict how the current will flow after the switch is closed.

FIGURE 9.8 The RL circuit in Example 4.

EXAMPLE 4

The switch in the RL circuit in Figure 9.8 is closed at time t = O. How will the current flow as a function of time?

Solution Equation (5) is a flIst-order linear differential equation for i as a function of t. Its standard form is (6)

and the corresponding solution, given that i = 0 when t = 0, is (7)

0Ne leave the calculation of the solution for you to do in Exercise 28.) Since R and L are positive, -(RIL) is negative ande-(R/L)' ..... 0 as t ..... 00. Thus,

(1' -

lim i = lim 1_00

o

2~

L R

R

R

1_00

1'e-(R/L)') = l' R R

- 1'. 0 R

=

1'. R

At any given time, the current is theoretically less than VIR, but as time passes, the current approaches the steady-state value VIR. According to the equation

3~

R

FIGURE 9.9 The growth of the current in the RL circuit in Example 4. I is the current's steady-state value. The number t = L/R is the time coostant of the circuit The current gets to within 5% of its

L di dt

+ R' '

= V

,

steady-state value in 3 time constants

I = VIR is the current that will flow in the circuit if either L = 0 (no inductance) or dildt = 0 (steady current, i = constant) (Figure 9.9). Equation (7) expresses the solution of Equation (6) as the sum of two terms: a steady-state solution VIR and a transient solution -(VIR)e-(RjL)' that tends to zero as

(Exercise 27).

t~oo.

•

Exercises 9.2 First-Ordor Unoar Equations Solve the differentisl equations in Exercises 1-14.

l.x:+y=e x , x>O 3. xy'

2. eX :+2eX y= I

x

-'1f/2

fly I +2y=l- x, x>O dx

6. (1

+ x)y' + Y

=

Vx

7. 2y'

+ 2e 2x y

=

2x

9. xy' - y

8. e 2x y'

+ 3y = sm,x, x > 0

4. y' + (tanx)y = cos2 x,

5.x

<x<

'1f/2

fly cosx 10. x dx = -x- - 2y, x > 0

=

e x/2 =

+y 2xlnx

9.2

11. (I - 1)3:

+ 4(1 - 1)28 ~ 1 + I, 1 > I

~

12. (I + I) dsd + 2s 1

3(1 + I) + (

1

13. sin 6 ~;

+ (cos6)r ~

14. tan 6 ~;

+r ~

sin2 6,

I )"

+

I

1 > -I

Ld;+R'~O dt l ,

0 < 6 < 1r/2

which is Equation (5) with V

Solve the initial value problems in Exercises 15-20.

+ 2y

16. 1dy dl

~ 3,

c. Show that the value of the current when 1 ~ L/R is I/e. (The significaoce of this time is esplained in the next exercise.)

+ 2y -_ 3 I, 1 > 0, y (2)

~

I

6> 0, Y(1r/2) ~ I

dy _ 3 18. 6 d6 - 2y - 6 sec 6 tan 6,

6> 0,

dy

y(1r/3)

~

2

x'

19. (x

+ I) dx - 2(x 2 + x)y ~ x e+ I' x > -I, y(O) ~ 5

dy 20. dx

+ xy

~

O.

fall to half its original value?

y(O) ~ I

dy . 17. 6 d6 + Y ~ sm6,

~

a. Solve the equation to express i as a function of t. b. How loog after the switch is thrown will it take the current to

Solving In;tial Value Problems dy 15. dl

509

26. Current in an open RL circuit If the switch is thrown opeo after the current in an RL circuit has built up to its steady-state value I ~ VIR, the decaying current (see accompaoying figure) obeys the equatioo

0 < 6 < 1r/2

tan6,

First-Order Linear Equations

3!-

x, y(O) ~ -6

R

21. Solve the expooeotial growth/decay initial value probl= for y as • functioo of 1 by thinking of the differential equation as a rITStorder linear equation with P(x) ~ -kandQ(x) ~ 0:

dy dl

~

ky

(keonstant),

y(O)

~

Yo

22. Solve the following initial value probl= for u as a functioo of t.

~~ + ~u ~

0 (kaodm positive coostants), u(O)

~

Uo

27. TIme constants Engineers call the number L/R the time constanl of the RL circuit in Figure 9.9. The sigoificance of the time coostant is that the current will reach 95% of its final value within 3 time coostants of the time the switch is closed (Figure 9.9). Thus, the time coostant gives a built-in measure of how rapidly an individual circuit will reach equilibrium. •• Find the value of; in Equation (7) that correspoods to 1 ~ 3L/R and show that it is about 95% of the steady-state value I ~ V/ R.

b. Approximately what percentage of the steady-state current will be flowing in the circuit 2 time coostants after the switch is closed (i.e., wheo 1 ~ 2L/R)?

a. as a flIst-order linear equation. b. as a separable equatioo.

28. Derivation of Equation (7) in Example 4 a. Show that the solution of the equation

Theory and Examples 23. Is either of the following equatioos correct? Give reasoos for your

answers. b.

x/

tdx

~ xlnlxl + ex

24. Is either of the following equatioos correct? Give reasoos for your

answers. a.

co~x/ cosX' dx =

1/

b. cosX'

cosX' ax

=

is

; ~ l' + Ce -(R/Ll' tanx

+C

tanx

+

C

cosX'

25. Current in a closed RL circuit How maoy seeonds after the switch in an RL circuit is closed will it take the current ; to reach half of its steady-state value? Notice that the time depeods 00 R and L and not 00 how much voltage is applied.

R

b. Theo use the initial cooditioo ;(0) ~ 0 to determine the value of C. This will complete the derivation ofEquatioo (7). c. Showthati ~ VIR is a solution of Equation (6) and that i = Ce -{RILlt satisfies the equation

di dl

R.

+ I'

~

o.

510

Chapter 9: First-Order Differential Equations we have n ~ 2,

James Bernoulli

so thatu ~ yl-2 ~ y-l anddu/dx ~ -y-2dy/dx.Thendy/dx ~ -y 2 du/dx ~ -u-2 du/dx.

(1654-1705)

Substitution into the original equation gives

HiSTORICAL BIOGRAPHY

A Bernoulli differential equation is of the form

dy dx

Observe that, if n

+ P(x)y

~

Q(x)y".

or, equivalently,

0 or I, the Bernoulli equation is lioear. For other values of n, the substitution u = y I-n transforms ~

du dx

+ u = -e-x.

the Bernoulli equation into the linear equation

: + (I -

n)P(x)u

~ (1 -

This last equation is linear in the (unknown) dependent variable u.

n)Q(x).

Solve the Bernoulli equations in Exercises 2~32.

For example, in the equation dy - y dx

9.3

~ e~y2

29. y'

-y~

_y2

30. y' - Y ~xy2

31. xy'

+y

y-2

32.

~

x'y' + 2xy

~ y'

AppLications We now look at four applications of flfSt-order differential equations. The flfSt application analyzes an object moving along a straight line while subject to a fOrce opposing its motion. The second is a model of population growth. The third application considers a curve or curves intersecting each curve in a second family of curves orthogonally (that is, at right angles). The fmal application analyzes chemical concentrations entering and leaving a container. The various models involve separable or liuear flfSt-<mler equations.

Motion with Resistance Proportional to Velocity In SOIl3e cases it is reasonable to assume that the resistance encountered by a moving object,

such as a car coasting to a stop, is proportional to the object's velocity. The faster the object moves, the more its forward progress is resisted by the air through which it passes. Pictore the object as a mass m moving along a coordinate liue with position function s and velocity vat time t. From Newton's second law of motion, the resisting force opposing the motion is Force = rnass X acceleration

~

m ~~ .

If the resisting force is proportional to velocity, we have dv

m dt = -kv

or

dv k dt ~ - m v

(k

> 0).

This is a separable differential equation representing exponential change. The solution to the equation with initial condition v ~ Vo at t = 0 is (Section 7.4) (1)

What can we learn from Equation (I)? For one thing, we can see that if m is something large, like the rnass of a 20,000-ton ore boat in Lake Erie, it will take a long time for the velocity to approach zero (because t must be large in the exponent of the equation in order to make kt/m large enough for v to be small). We can learn even more ifwe integmte Equation (1) to find the position s as a function of time t. Suppose that a body is coasting to a stop and the only force acting on it is a resistance proportional to its speed. How far will it coast? To find out, we start with Equation (1) and solve the initial value problem ds _

dt - voe

-(klm)t

,

s(O)

~

O.

9.3

Applications

511

Integrating with respect to I gives v~m e-(klm)t

S = -

Substitoting S

~

0 when I

~

+ C.

0 gives

vom o ~--+ k

C

vom k

C~-

and

The body's position at time I is therefore

_ vom -("m)t ()st -Te "

vom _ vom (

+ T - T I-e

-(klm)t)

.

(2)

To find how far the body will coast, we fmd the limitofs(l) as 1--> 00. Since -(kim) < 0, we know that e -(klm)t --+ 0 as I --+ 00 , so that

lim S(I) ~ lim vom (I t_OO k

e-(klm),)

t_OO

~ v~m (I _ 0) ~ v~m .

Thus, Distance coasted

=

vom

T

(3)

The number vom/k is only an upper bound (albeit a useful one). It is true to life in one respect, at least: if m is large, the body will coast a long way.

l In the English system, where weight is measmed in pounds, mass is measured in slugs. Thus, Pounds

~

slugs X 32,

assuming the gravitational constant is 32 ft./sec'.

EXAMPLE 1

For a 192-lb ice skater, the k in Equation (I) is about 1/3 slug/sec and

m = 192/32 = 6 slugs. How long will it take the skater to coast from 11 ftjsec (7.5 mph)

to I ft/sec? How far will the skater coast before coming to a complete stop? Solution We answer the flISt question by solving Equation (I) for I: lIe-tl18 ~ e-t118 ~

Eq. (I) with k ~ 1/3, m = 6, Vo = 11, v = 1

I

1/11

-1/18

~

In (1/11)

~

I

~

18ln 11 '" 43 sec.

-In 11

We answer the second question with Equation (3):

. D.stance coasted

vom

=

T

=

198 ft.

~

11·6 1/3

•

Inaccuracy of the Exponential Population Growth Model In Section 7.4 we modeled popolation growth with the Law of Exponential Change: dP iii =

P(O)

!P,

~

Po

whereP is the population at time t, k > 0 is a constant growth mte, and Po is the size of the population at time I ~ O. In Section 7.4 we found the solution P ~ Poe kt to this model. To assess the model, notice that the exponential growth differential equation says that dPt

dl

= k

(4)

512

Chapter 9: First-Order Differential Equations

TABLE 9.3 WorLd popuLation (midyear)

f

7000

World population (1980-2008)

( 5000

~O~------~I~O---Av

I

30

)

t

FIGURE 9.10 Notice that the value of the solutionP ~ 4454e°.0 17t is 7169 when t ~ 28, which is nearly 7% more than the actual population in 2008.

Orthogooal trajectory

Year

Population (millions)

1980 1981 1982 1983 1984 1985 1986 1987 1988 1989

4454 4530 4610 4690 4770 4851 4933 5018 5105 5190

76/4454 80/4530 80/4610 80/4690 81/4770 82/4851 85/4933 87/5018 85/5105

'" '" '" '" '" '" '" '" '"

0.0171 0.0177 0.0174 0.0171 0.0170 0.0169 0.0172 0.0173 0.0167

Source: u.s. Bureau of the Census (Sept., 2007): WWW.ceDSUS .gov/ipc/www/idb.

is constant. This rate is called the relative growth rate. Now, Table 9.3 gives the world population at midyear for the years 1980 to 1989. Taking dt = I and tiP '" l1P, we see from the table that the relative growth rate in Equation (4) is approximately the constant 0.017. Thus, based on the tabled data with t = 0 representing 1980, t = I representing 1981, and so forth, the world population could be modeled by the initial value problem,

tiP Iii =

FIGURE 9.11 An nrthogonal trajectory intersects the family of curves at right angles, or nrthogonally.

l1P/P

0.017P,

P(O) = 4454.

The solution to this initial value problem gives the population function P = 4454eo.017t.1n year 2008 (so t = 28), the solution predicts the world population in midyear to be about 7169 million, or 7.2 billion (Figure 9.10), which is more than the actoa1 population of 6707 million from the U.S. Bureau of the Census. A more realistic model would consider environmental and other factors affectiog the growth rate, which has been steadily decliniog to about 0.012 since 1987. We consider one such model in Section 9.4.

Orthogonal Trajectories y

An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family at right angles, or orthogonally (Figure 9.11). For instance, each straight line through the origin is an orthogonal trajectory of the family of circles x2 + y2 = a 2 , centered at the origin (Figure 9.12). Such mutoa1ly orthogonal systems of curves are ofparticular importance in physical problems related to electrical potential, where the curves in one family correspond to strength of an electric field and those in the other family correspond to constant electric potential. They also occur in hydrodynamics and heat-flow problems.

EXAMPLE 2 Find the orthogonal trajectories of the family of curves xy a .. 0 is an arhitrary constant.

FIGURE 9.12 Every straight line through the origin is orthogonal to the family of circles centered at the origin.

= a, where

Solution The curves xy = a form a family of hyperbolas having the coordinate axes as asymptotes. First we find the slopes of each curve in this family, or their dy/ ax values. Differentiatiog xy = a implicitly gives

dy

xax+y=O

or

dy

Y

ax

X'

9.3

Applications

513

Thus the slope of the tangent line at any point (x, y) on one of the hyperbolas xy = a is y' = -y/x. On an orthogonal trajectory the slope of the tangent line at this same point must be the negative reciprocal, or x/yo Therefore, the orthogonal trajectories must satisfy the differential equation

dy dx

x y'

This differential equation is separable and we solve it as in Section 7.4:

ydy = xdx

Separate variables. Integrate both sides.

FIGURE 9.13 Each curve is orthogonal to every curve it meets in the other family

(Example 2).

(5)

where b = 2C is an arbitrary constant. The orthogonal trajectories are the family ofhyperbolas given by Equation (5) and sketched in Figure 9.13. •

Mixture Problems Suppose a chemical in a liquid solution (or dispersed in a gas) runs into a container holding the liquid (or the gas) with, possibly, a specified amount of the chemical dissolved as well. The nrixture is kept uniform by stirring and flows out of the container at a known mte. In this process, it is often important to know the concentration of the chemical in the container at any given time. The differential equation describing the process is based on the formula (mte at WhiCh) (mte at WhiCh) Rate of change of amount = chemical chemical in container arrives departs.

(6)

If y(t) is the amount of chemical in the container at time t and V(t) is the total volume of liquid in the container at time t, then the departure mte of the chemical at time t is y(t) Departuremte = V(t) • (outflow rate)

=

concentration in ) ( container at time t • (outflow rate) .

(7)

Accordingly, Equation (6) becomes

dy . . y(t) dt = (chenncal'samvalmte) - V(t) . (outflow rate).

(8)

If, say, y is measured in pounds, V in gallons, and t in minutes, the uuits in Equation (8) are pounds minutes

pounds pounds gallons minutes - gallons • minutes'

EXAMPLE 3 In an oil refinery, a storege tank contains 2000 gal of gasoline that initially has 100 lb of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 lb of additive per gallon is pumped into the tank at a rate of 40 gal/min.

514

Chapter 9: First-Order Differential Equations

The wt:ll-mixed solution is pumped out at a rate of 45 gal/min. How much of the additive is in the taJ:Jk 20 min after the pumping process begins (Figure 9.14)? 40 gaIImin containing 21b1pl

• FIGURE 9.14 The storage tank in Example 3 mixes input liquid with stored liquid to produce an output liquid.

Lety be the amolDlt (in pounds) of additive in the taJ:Jk at time I. We know that y = 100 when I = O. The number of gallons of gasoline and additive in solution in the taJ:Jk at any time I is

Solution

~ 2000 gal + (40: -

V(I)

~

45 : ) (tmin)

(2000 - 51) gal.

Thcrefon:,

y(1)

Rate out

=

- (200: ~

Eq.(7)

Y(I)' outflow rate

Ootflawrate ia45 gaJjmin and v - 2000 - St.

51) 45

45y Ib 2000 51 min .

Also, Hate in

~ (2~)(4O:) ~801b. mm

The diffcrcn:tial equation modeling the

mnture process is

dy 45y dt ~ 80 - 2000 51

Eq. (8)

in pmmds per minute. To solve this dift"erent:i.al equation, wt: rll"St write it in standard linear form: dy dt

45

51 Y ~ 80.

+ 2000

Thus, P(I) - 45/(2000 - 51) and Q(I) - 80. Th. integnUing factnr is v(l)

=

eJPdt

=

eJ~dt

= e-9In(2000-St)

~ (2000 -

5Ir'.

2000-5t> 0

9.3

Applications

515

Multiplying both sides of the standard equation by v(l) and integrating both sides gives 9 (2000 - 51r •

(2000 - 51r9

(t + 200~5_ Y)

t+

51

= 80(2000 - 51r

9

45(2000 - 51)-10 Y = 80(2000 - 51r9

1

:1 [(2000 - 51)-9y = 80(2000 - 51r9

(2000 - 51)-9y =

J

80(2000 - 51)-9 dl

-9 (2000 - 51r' (2000 - 51) Y = 80· (-8)(-5)

+ c.

The general solution is

y

= 2(2000 - 51)

+

C(2000 - 51)9.

Because y = 100 when I = 0, we can detennine the value of C: 100 = 2(2000 - 0)

C=

+

C(2000 - 0)9

3900 (2000)9·

The particular solution of the initial value problem is

The amount of additive 20 min after the pumping begins is y(20) = 2[2000 - 5(20)1 - ( 3900)9 [2000 - 5(20)1 9 .., 1342Ib. 2000

•

Exercises 9.3 Motion Along a Line 1. Coa.ting bicycle A 66·kg cyclist on a 7·kg bicycle starts coast· ing on level ground at 9 m/sec. The k in Equation (I) is about 3.9 kgfsec.

a. About how far will the cyclist coast before reaching a com· plete stop?

b. How long will it take the cyclist~ speed to drop to I m/sec? 2. Coa.ting battleship Suppose that an Iowa class battleship has rnass around 51,000 metric tons (51,000,000 kg) and a k value in

Equation (I) of about 59,000 kgfsec. Assume that the ship loses power when it is moving at a speed of9 mfsec. a. About how far will the ship coast before it is dead in the water?

b. About how long will it take the ship~ speed to drop to I mfsec? 3. The data in Table 9.4 were collected with a motion detector and a CBLTM by Valerie Sharritts, a mathematics teacher at St. Francis DeSales High School in Columbus, Ohio. The table shows the distance s (meters) coasted on in·line skates in t sec by her daughter Ashley when she was 10 years old. Find a model for Ashley's

516

Chapter 9: First-Order Differential Equations

position given by the data in Table 9.4 in the form of Equation (2). Her initial velocity was Vo ~ 2.75 mlsec, her mass m ~ 39.92 kg (she weighed 88Ib), and her total coasting distaoce was 4.91 m.

Orthogonal Trajectories In Exercises S-10, fmd the orthogonal trajectories of the family of curves. Sketch several members of each family. 6.y~cx2

5.y~mx

7. k:x;2

TABLE 9.4 Ashley Sharritts skating data

t (sec)

s(m)

t (sec)

s(m)

t (sec)

s(m)

0 0.16 0.32 0.48 0.64 0.80 0.96 1.12 1.28 1.44 1.60 1.76 1.92 2.08

0 0.31 0.57 0.80 1.05 1.28 1.50 1.72 1.93 2.09 2.30 2.53 2.73 2.89

2.24 2.40 2.56 2.72 2.88 3.04 3.20 3.36 3.52 3.68 3.84 4.00 4.16 4.32

3.05 3.22 3.38 3.52 3.67 3.82 3.96 4.08 4.18 4.31 4.41 4.52 4.63 4.69

4.48 4.64 4.80 4.96 5.12 5.28 5.44 5.60 5.76 5.92 6.08 6.24 6.40 6.56

4.77 4.82 4.84 4.86 4.88 4.89 4.90 4.90 4.91 4.90 4.91 4.90 4.91 4.91

+ y2

~

I

11. Show that the

8.2x2+y2~c2

10. y

9. y = ce --x curves 2x 2

+ 3y2

=

eb

~ S and y2 ~ x 3 are orthogonal.

12. Find the family of solutions of the given differential equation and the family of orthogonal trajectories. Sketch both families.

••

xdx+ydy~O

b.xdy-2ydx~0

Mixture Problems 13. Salt miItnre A tack initially contains \00 gal of brine in which SO Ib of salt are dissolved. A brine containing 2 Ib/gal of salt runs into the tack at the rate of S gal/min. The mixture is kept uniform by stirring and flows out of the tack at the rate of4 gal/min. •• At what rate (pounda per minute) does salt enter the tack at time I?

b. What is the volume of brine in the tank at time I? •• At what rate (pounda per minute) does salt leave the tack at

time t? d. Write down and solve the initial value problem describing ilie mixing process. e. Find the concentration of salt in ilie tank 25 min afler the

process starts. 4. Coasting to ••Iop Thhle 9.S shows the distaoce s (meters) coasU:d on in-line skates in terms of time t (seconds) by Kelly Schmilzer. Find a model for her position in the form of Equation (2). Her initial velocity was Vo ~ 0.80 mlsec, her mass m ~ 49.90 kg (110 Ib), and her total coasting distaoce was 1.32 m.

TABLE 9.5

•• At what time willilie tack be full?

b. At ilie time the tack is full, how many pounds of concentrate will it contain?

Kelly Schmitzer skating data

t (sec)

s(m)

t (sec)

s(m)

t (sec)

s(m)

0 0.1 0.3 0.5 0.7 0.9 1.1 1.3

0 0.07 0.22 0.36 0.49 0.60 0.71 0.81

1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9

0.89 0.97 1.05 1.11 1.17 1.22 1.25 1.28

3.1 3.3 3.5 3.7 3.9 4.1 4.3 4.5

1.30 1.31 1.32 1.32 1.32 1.32 1.32 1.32

9.4

14. MiItnre problem A 200-gal tack is half full of distilled water. At time t ~ 0, a solution containing O.Slb/gal of concentrate enters ilie tack at the rate of 5 gal/min, and the well-stirred mixture is wiilidrawn at ilie rate 00 gal/min.

15. Fertilizer mhture A tack contains 100 gal offresh water. A solution containing I Ib/gal of soluble \awn fertilizer ruos into the tack at ilie rate of 1 gal/min, and ilie mixture is pumped out ofilie tack at the rate 00 gal/min Find ilie maxinuun amount of fertilizer in ilie tack and the time required to reach the maxinuun. 16. Carbon monmide pollution An executive conference room of a corporation contains 4S00 It' of air initially free of carbon monoxide. Starting at time t ~ 0, cigarette smoke containing 4% carbon monoxide is blown into the racon at the rate of 0.3 It'/min. A ceiling fan keeps the air in the room well circulated and the air leaves ilie room at the same rate of 0.3 It'/min. Find the time when the concentration of carbon monoxide in the room reaches 0.01 %.

Graphical Solutions of Autonomous Equations In Chapter 4 we learned that the sign of the first derivative tells where the graph of a function is increasing and where it is decreasing. The sign of the second derivative tells the concavity of the graph. We can build on our knowledge of how derivatives determine the shape of a graph to solve differential equations graphically. We will see that the ability to

9.4 Graphical Solutions of Autonomous Equations

517

discern physical behavior from gmphs is a powerful tool in understanding real-world systems. The starting ideas for a gmphical solution are the notions of phase line and equilibrium value. We arrive at these notions by investigating, from a point of view quite different from that studied in Chapter 4, what happens when !be derivative of a differentiable function is zero.

Equilibrium Values and Phase Lines When we differentiate implicitly the equation

I

Sin (5y - 15)

= x

+ I,

we obtain

I( 5 )dY

S

5y - 15

dx = I.

Solving for y' = dy/dx we find y' = 5y - 15 = 5(y - 3). In this case the derivative y' is a function ofy only (the dependent variable) and is zero when y = 3. A differential equation for which dy/ dx is a function ofy only is called an autonomous differential equation. Let's investigate what happens when the derivative in an autonomous equation equals zero. We assume any derivatives are continuous.

DEnNmON If dy/dx = g(y) is an autonomous differential equation, then the values of y for which dy/ dx = 0 are called equilihrium values or rest points.

Thus, equilibrium values are those at which no change occurs in the dependent variable, so y is at rest. The emphasis is on the value of y where dy/ dx = 0, not the value of x, as we studied in Chapter 4. For example, the equilibrium values for the autonomous differential equation dy dx = (y

+

I)(y - 2)

arey = -I andy = 2. To construct a graphical solution to an autonomous differential equation, we fIrst make a phase line for the equation, a plot on the y-axis that shows the equation's equilibrium values along with the intervals where dy/dx and d'y/dx 2 are positive and negative. Then we know where the solutions are increasing and decreasing, and the concavity of the solution curves. These are the essential features we found in Section 4.4, so we can determine the shapes of the solution curves without having to find formulas for them.

EXAMPLE 1

Draw a phase line for the equation dy dx = (y

+ I)(y - 2)

and use it to sketch solutions to the equation.

Solution 1.

Draw a number line for y and mark the equilibrium values y = - I and y = 2, where dy/dx = O. ------1@)------------i@~--->, Y

-1

2

518

Chapter 9: First-Order Differential Equations 2.

IdentifY and label the intervals where y' > 0 and y' < 0, This step resembles what we did in Section 4.3, only now we are marking the y-axis instead of the x-axis, I I I I

y'>o

I I I I

y'
@

@

y'>o

,y

2

-1

We can encapsulate the information about the sign of y' on the phase line itself, Since y' > 0 on the interval to the left of y = -I, a solution of the differential equation with a y-value less than -I will increase from there toward y = -I, We display this information by drawing an arrow on the interval pointing to - I , -'--~~@~----~--'-------~@r-'--.I~ 'y

-1

2

Similarly, y' < 0 between y = -I and y = 2, so any solution with a value in this interval will decrease toward y = - L For y > 2, we have y' > 0, so a solution with a y-value greater than 2 will increase from there without bound, In short, solution curves below the horizontal line y = - I in the xy-plane rise toward y = - L Solution curves between the lines y = -I and y = 2 fall away from y = 2 toward Y = - L Solution curves above y = 2 rise away from y = 2 and keep going up,

3.

Calculate y' and mark the intervals where y' > 0 and y" entiate y' with respect to x, using implicit differentiation, y' = (y

+ I)(y - 2)

= y2 - Y -

2

<

0, To find y", we differ-

Formula for y' ...

y" = ~(y') = ~(y2 _ Y _ 2)

dx

y

=

y'>O y">O

dx

differentiated implicitly with respect to x

2Y.)1' - y'

= (2y -

I)y'

= (2y -

I)(y

+

I)(y - 2),

From this formula, we see that y" changes sign at y = - I, y = 1/2, and y = 2, We add the sign information to the phase line. y'>O y"
y'>O y"
y'O

I

-1

•

I I I

..

!

y'
y'>O y">O I

,y

2

2

FIGURE 9.15 Graphical solutions from Example I include the horizontal lines y = -I and y = 2 through the equilibrium values, No two solution curves can ever cross or touch each other.

4.

Sketch an assortment of solution curves in the xy-plane. The horizontal lines y = -I,y = 1/2, and y = 2 partition the plane into horizontal bands in which we know the signs of y' and y". In each band, this information tells us whether the solution curves rise or fall and how they bend asx increases (Fignre 9.15). The "equilibrium lines" y = - I and y = 2 are also solution curves. (The constant functions y = - I and y = 2 satisfY the differential equation.) Solution curves that cross the line y = 1/2 have an inflection point there. The concavity changes from concave down (above the line) to concave up (below the line). AB predicted in Step 2, solutions in the middle and lower bands approach the equilibrium value y = -I as x increases. Solutions in the upper band rise steadily away from the value y = 2. •

9.4 Graphical Solutions of Autonomous Equations

519

Stable and Unstable Equilibria Look at Figure 9.15 once more, in particular at the behavior of the solution curves near the equilibrium values. Once a solution curve has a value near y = -I, it tends steadily toward that value; y = - I is a stable equilibrium. The behavior near y = 2 is just the opposite: all solutions except the equilibrium solution y = 2 itself move away from it as x increases. We call y = 2 an unstable equilibrium. If the solution is at that value, it stays, but if it is offby any amount, no matter how small, it moves away. (Sometimes an equilibrium value is unstable because a solution moves away from it ouly on one side of the point.) Now that we know what to look for, we can already see this behavior on the initial phase line (the second diagnun in Step 2 of Example I). The arrows lead away from y = 2 and, once to the left of y = 2, toward Y = - I. We now present several applied examples for which we can sketch a family of solution curves to the differential equation models using the method in Example I.

Newton's Law of Cooling dH>O dt

: I

In Section 7.4 we solved analytically the differential equation

dH
--~~~~@~---'--~-+'H 15

FIGURE 9.16 First step in constructing the phase line for Newton's law of cooling. The temperature teods towards the equilibriwn (surrounding-mediwn) value in the long run.

I I

dH>O dt 2

d H <0 dt 2

I

:

: I

~-+-~@

dH at

= -k(H - H s),

k>O

modeling Newton's law of cooling. Here H is the temperature of an object at time t and Hs is the constant temperatore of the surrounding medium. Suppose that the surrounding medium (say a room in a house) has a constant Celsius temperatore of 15°C. We can then express the difference in temperatore as H(t) - 15. Assuming H is a differentiable function of time t, by Newton's law of cooling, there is a constant of proportionality k > 0 such that

dH = -k(H - IS) dt

dH <0 dt

-

(1)

2

d H >0 dt 2 •

)H

15

FIGURE 9.17 The complete phase line for Newton's law of cooling.

H

Initial

Temperature of SlUTOunding

medium

(millus kto give a negative derivative when H > IS). Since dH/dt = 0 at H = IS, the temperature 15°C is an equilibrium value. If H > IS, Equation (I) tells us that (H - IS) > 0 and dH/dt < O. If the object is hotter than the room, it will get cooler. Similarly, if H < IS, then (H - IS) < 0 and dH/ dt > O. An object cooler than the room will warm up. Thus, the behavior described by Equation (I) agrees with our intuition of how temperature should behave. These observations are captured in the initial phase line diagnun in Figure 9.16. The value H = IS is a stable equilibrium. We determine the concavity of the solution curves by differentiating both sides of Equation (I) with respect to t:

~ (a:) = ~(-k(H-

IS))

2

d H = _ledH dt 2 dt· Initial "'mperature

FIGURE 9.18 Temperature versus time. Regardless of initial temperature, the object's remperature H(t) teods toward ISoC, the temperature of the surrounding mediwn.

Since -k is negative, we see that d 2H/dt 2 is positive when dH/dt < 0 and negative when dH/dt> o. Figure 9.17 adds this information to the phase line. The completed phase line shows that if the temperatore of the object is above the equilibrium value of 15°C, the graph of H(t) will be decreasing and concave upward. If the temperature is below 15°C (the temperature of the surrounding medium), the graph of H(t) will be increasing and concave downward. We use this information to sketch typical solution curves (Figure 9.18).

520

Chapter 9: First-Order Differential Equations From the upper solution curve in Figure 9.18, we see that as the object cools down, the rate at which it cools slows down because 1iH/ dt approaches zero. This observation is implicit in Newton's law of cooling and contained in the differential equation, but the flattening of the graph as time advances gives an immediate visual representation of the phenomenon.

A Falling Body Encountering Resistance Newton observed that the rate of change in momentum encountered by a moving equal to the net force applied to it. In mathematical terms,

F

=

d

o~ect

is

(2)

dt (mv),

where F is the net force acting on the object, and m and v are the object's mass and velocity. Ifm varies with time, as it will if the o~ect is a rocket burning fuel, the right-hand side of Equation (2) expands to

using the Derivative Product Rule. In many situations, however, m is constant, dm/dt = 0, and Equation (2) takes the simpler form or

F=

'Ina,

(3)

s

known as Newton second law ofmotion (see Section 9.3). In free fall, the constant acceleration due to gravity is denoted by g and the one force acting downward on the falling body is

y

Fp FIGURE 9.19 An object falling under 1he influence of gravity wi1h a resistive force assumed to be proportional to 1he velocity.

=

mg,

the force due to gravity. If, however, we think of a real body falling through the air---1!ay, a penny from a great height or a parachutist from an even greater height-we know that at some point air resistance is a factor in the speed of the fall. A more realistic model of free fall would include air resistance, shown as a force F, in the schematic diagram in Figure 9.19. For low speeds well below the speed of sound, physical experiments have shown that F, is approximately proportional to the body's velocity. The net force on the falling body is therefore

F = Fp - F" giving

dv mdt=mg-ku dv k dt =g- m V '

(4)

We can use a phase line to analyze the velocity functions that solve this differential equation. The equilibrium point, obtained by setting the right-hand side of Equation (4) equal to

-

dv>O dt ____

I I I

dv v

---{@~~

mg

zero, is

mg v=T'

k

FIGURE 9.20 Initial pbase line for 1he falling body encountering resistance.

If the body is initially moving faster than this, dv/dt is negative and the body slows down. If the body is moving at a velocity below mg/k, then dv/ dt > 0 and the body speeds up. These observations are captured in the initial phase line diagram in Figure 9.20.

9.4 Graphical Solutions of Autonomous Equations

dv>O dt 2 d v dt 2

<0

I I I I I I I I

@

We determine the concavity of the solution curves by differentiating both sides of Equation (4) with respect to t:

dv <0

dt 2 d v dt 2

,

~:~ = ~ (g - ! v) =

>0

mg

521

'v

T FIGURE 9.21 The completed phase line for the falling body.

v

-

! ~~ .

We see that d 2v/dt 2 < 0 when v < mg/Ie and d 2v/dt 2 > 0 when v > mg/Ie. Figure 9.21 adds this infonnation to the phase line. Notice the similarity to the phase line for Newton's law of cooling (Figure 9.17). The solution curves are similar as well (Figure 9.22). Figure 9.22 shows two typical solution curves. Regardless of the initial velocity, we see the body's velocity tending toward the limiting value v = mg/k. This value, a stable equilibrium point, is called the body's terminal velocity. Skydivers can vary their terminal velocity from 95 mph to 180 mph by changing the amount of body area opposing the fall, which affects the value of k.

Logistic Population Growth In Section 9.3 we examined population growth using the model of exponential change. That is, if P represents the number of individuals and we neglect departures and arrivals, then velocity

dP

'JYpical velocity curves for a falling body encountering resistance. The value v = mg/k is the tenninal velocity.

dt

FIGURE 9.22

= kP,

(5)

where k > 0 is the birth rate minus the death rate per individual per unit time. Because the natural environment has only a limited number of resources to sustain life, it is reasonable to assume that only a maximum population M can be accommodated. As the population approaches this limiting population or carrying capacity, resources become less abundant and the growth rate Ie decreases. A simple relationship exhibiting this behavior is

k = r(M - P), where r > 0 is a constant. Notice that k decreases as P increases toward M and that Ie is negative if P is greater than M. Substituting r(M - P) for k in Equation (5) gives the differential equation

';J; I I I I

@ 0

dP >0 dt

I I I I

• , @

dP <0 dt

,p

M

FIGURE 9.23 The initial phase line for logistic growth (Equation 6).

= r(M - P)P = rMP - rp2.

(6)

The model given by Equation (6) is referred to as logistic growth. We can forecast the behavior of the population over time by analyzing the phase line for Equation (6). The equilibrium values are P = M and P = 0, and we can see that dP/ dt > 0 if 0 < P < M and dP/ dt < 0 if P > M. These observations are recorded on the phase line in Figure 9.23. We determine the concavity of the population curves by differentiating both sides of Equation (6) with respect to t:

(7)

If P = M/2, then d 2p/dt 2 = O. If P < M/2, then (M - 2P) and dP/dt are positive and d 2p/dt 2 > O. If M/2 < P < M, then (M - 2P) < 0, dP/dt > 0, and d 2p/dt 2 < O.

522

Chapter 9: First-Order Differential Equations

I I I

I

If P > M, then (M - 2P) and dP/dt are both negative and d 2p/dt 2 > O. We add this information to the phase line (Figure 9.24). The lines P = M/2 and P = M divide the llfst quadrant of the tP-plane into horizontal bands in which we know the signs of both dP/dt and d 2p/dt 2• In each band, we know how the solution curves rise and fall, and how they bend as time passes. The equilibrium lines P = 0 and P = M are both population curves. Population curves crossing the line P = M/2 have an inflection point there, giving them a sigmoid shape (curved in two directions like a letter S). Figure 9.25 displays typical population curves. Notice that each population curve approaches the limiting population M as t --+ 00.

dP >0 dt I

2

' 4X>o I dt 2

@l o

J'

2 : d P I

dt2

I,

<0 J

M

1:

FIGURE 9.24 The completed phase line for logistic growth (Equation 6).

P

FIGURE 9.25

Population curves for logistic growth.

Exercises 9.4 Phase Unes and Solution Curves 10 Exercises 1-8, L

IdentifY the equilibrium values. Which are stahle and which are unstable?

b. Cons1ruct a phase line. IdentifY the sigas of y' and yO. c. Sketch several solution curves. dy dy 2. dx 1. dx = (y + 2)(y - 3)

= y2

dy 3. dx

=

=

y' - Y

5.y'=Vy, y>O 7. y' = (y - I)(y - 2)(y - 3)

dy

4. dx

- 4

y' - 2y

6. y' = y - Vy, y > 8. y' = y3 _ y2

a

Models of Population Growth The autonomous diffi:rential equations in Exercises 9-12 represeot models for populatioo growth. For each exercise, use a phase line analysis to sketch solution curves for p(t), selecting different startiog values P(O). Which equilibria are stahle, aod which are uostable?

dP 9. dt = I - 2P dP 11. dt

=

2P(P - 3)

dP 10. dt

= P( I

and that the current populatioo Po is fairly close to the carrying capacity Mo. You ntight imagine a populatioo of fish living in a freshwater lake in a wilderuess area. Suddeoly a catastrophe such as the Mount St Helens volcauic eruption contaminates the lake and destroys a siguificant part of the food and oxygen 00 which the fish depend. The result is a new enviromnent with a carrying capacity Ml coosiderably less than Mo and, in fact, less than the current population Po. Startiug at some time before the catastrophe, slretch a ''before-and-after'' curve that shows how the fish populatioo responds to the change in environment 14. Controlling a population The fish and game depar1ment in a certain state is planning to issue huntiog permits to cootrol the deer population (one deer per permit). It is knowo that if the deer populatioo falls below a certain level m, the deer will become extioct. It is also knowo that if the deer population rises above the carrying capacity M, the population will decrease back to M throngh disease and maIoutrition.

a. Discuss the reasonableness of the following model for the growth rate of the deer population as a function of time:

dP dt

= rP(M -

P)(P - m),

- 2P)

12 dP = 3P(i _ P)(p • dt

_1) 2

13. Catastropbic change in logistic growth Suppose that a healthy population of some species is growing in a limited environment

where P is the population of the deer and r is a positive constant of proportionality. Ioclude a phase line.

b. Explain how this model differs from the logistic model dP/ dt = rP(M - Pl. Is it better or worse than the logistic model?

9.5

> Mforallt,thenlim,_ooP(t) What happens if P < m for all t?

c. ShowthatifP d.

~

M.

e. Discuss the solutions to the differential equation. What are the equilibrium points of the model? Explain the dependence of the steady-stste value of P on the initial values of P. About how many pcrruits should he issued?

Applications and Examples 15. Skydiving If a body of mass m falling from rest under the ac· tion of gravity encouoters an air resistsoce proportional to the square of velocity, then the body's velocity t seconds into the fall

satisfies the equation dv 2 m-=mg-kv dt '

Systems of Equations and Phase Planes

523

a. Discuss the reasonableness of the model. b. Construct a phase line identifying the signs of X' and X".

c. Sketch representative solution curves. d. Predict the value of X for which the information is spreading most rapidly. How many people eventually receive the infor-

mation? 19. Current in an RL-circuit The accoropanying diagram represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. From Section 9.2, we have

k>O

di

.

Ldt+RI~V,

where k is a constsot that depends on the body's aerodynamic properties and the density of the air. CWe assume that the fall is too short to be affected by changes in the air's density.)

where i is the current in amperes and t is the time in seconds.

v

a. Draw a phase line for the equation.

+ d~ -

b. Sketch a typical velocity curve.

~--b

c. For al60·lb skydiver (mg ~ 160) and with time in seconds and distsoce in feet, a typical value of k is 0.005. What is the diver's terminal velocity?

Switch

Vv

16. Resistance proportional to A body ofmassm is projected vertically downward with initial velocity Vo. Assume that the reo sisting force is proportional to the square root of the velocity and [mel the terruioal velocity from a graphical analysis. 17. Sailing A sailboat is running along a slraight course with the wind providing a constsot forward force of 50 lb. The only other force acting on the boat is resistance as the boat moves through the water. The resisting force is numerically equal to five times the boat's speed, and the initial velocity is 1 ft/sec. What is the maximum velocity in feet per second of the boat under this wind? 18. The spread of information Sociologists recognize a phenomenon called social diffUsion, which is the spreading of a piece of informa· tion, technological innovation, or cultoral fad among a population. The members of the population can he divided into tv.<> classes: those who have the information and those who do not In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the infor· mation times the number yet to receive it IfX denotes the nwnhcr of individuals who have the information in a population of N people, then a mathematical model for social diffusion is given by

L

Use a phase line analysis to sketch the solution curve assuming that the switch in the RL-circuit is closed at time t ~ O. What happens to the current as t ---> oo? This value is called the steady-state solution. 20. A pearl in shampoo Suppose that a pearl is sinking in a thick fluid, like shampoo, subject to a frictional force opposing its fall and proportional to its velocity. Suppose that there is also a resistive buoyant force exerted by the shampoo. According to Archimedes'principle, the buoyant force equala the weight of the fluid displaced by the pearl. Using m for the mass of the pearl and P for the mass of the shampoo displaced by the pearl as it desceods, coroplete the following steps. a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Figore 9.19. b. Using v(t) for the pearl's velocity as a func1ion of time t, write a differential equation modeling the velocity of the pearl as a falling body. c. Construct a phase line displaying the signs of v' and v".

dX

dt ~ kX(N - X),

d. Sketch typical solution curves.

where t represents time in days and k is a positive constant.

9.5

R

e. What is the terminal velocity of the pearl?

Systems of Equations and Phase Planes In some situations we are led to consider not one, but several first-order differential equations. Such a collection is called a system of differential equations. In this section we present an approach to understanding systems through a graphical procedure known as a phase-plane analysis. We present this analysis in the context of modeling the populations of trout and bass living in a common pond.

524

Chapter 9: First-Order Differential Equations

Phase Planes A general system of two first-order clifferential equations may tske the form

dx dt = F(x,y), dy dt = G(x,y). Such a system of equations is called autonomous because dx/ dt and dy/ dt do not depend on the independent variable time t, but only on the dependent variables x andy. A solution of such a system consists of a pair of functions x(t) and yet) that satisfies both of the differential equations simultaneously for every t over some time interval (finite or infinite). We cannot look at just one of these equations in isolation to find solutions x(t) or y(t) since each derivative depends on both x andy. To gain insight into the solutions, we look at both dependent variables together by plotting the points (x(t), yet»~ in the xy-plane starting at some specified point. Therefore the solution functions define a solution curve through the specified point, called a trajectnry of the system. The xy-plane itself, in which these trajectories reside, is referred to as the phase plane. Thus we consider both solutions together and study the behavior of all the solution trajectories in the phase plane. It can be proved that two trajectories can never cross or touch each other. (Solution trajectories are examples of parametric curves, which are studied in detail in Chapter 11.)

A Competitive-Hunter Model Imagine two species of fish, say trout and bass, competing for the same limited resources (such as food and oxygen) in a certain pond. We let x(t) represent the number of trout and yet) the number of bass living in the pond at time t. In reality x(t) and y(t) are always integer valued, but we will approximate them with real-valued differentiable functions. This allows us to apply the methods of clifferential equations. Several factors affect the rates of change of these populations. As time passes, each species breeds, so we assume its population increases proportionally to its size. Taken by itself, this would lead to exponential growth in each of the two populations. However, there is a countervailing effect from the fact that the two species are in competition. A large number of hass tends to cause a decrease in the number of trout, and vice-versa. Our model takes the size of this effect to be proportional to the frequency with which the two species interact, which in turn is proportional to xy, the product of the two populations. These considerations lead to the following model for the growth of the trout and hass in the pond:

dx dt = (a - by)x,

(la)

dy dt = (m - nx)y.

(lb)

Here x(t) represents the trout population, yet) the bass population, and a, b, m, n are positive constants. A solution of this system then consists of a pair of functions x(t) and yet) that gives the population of each fish species at time t. Each equation in (I) contains both of the unknown functions x and y, so we are unable to solve them individually. Instead, we will use a graphical analysis to study the solution trajectories of this competitive-hunter model. We now examine the nature of the phase plane in the trout-bass population model. We will be interested in the I st quadrant of the xy-plane, where x ;" 0 and y ;" 0, since populations cannot be negative. First, we determine where the bass and trout populations are both constant. Noting that the (x(t),y(t» values remain unchanged when dx/dt = 0 and dy/dt = 0, Equations (Ia and Ib) then become

(a - by)x = 0, (m - nx)y =

o.

This pair of simultaneous equations has two solutions: (x, y) = (0, 0) and (x, y) = (m/n, a/b). At these (x,y) values, called equilibrium or rest points, the two populations

9.5

525

Systems of Equations and Phase Planes

remain at constant values over all time. The point (0, 0) represeots a pond containing no members of either fish species; the point (m/n, a/b) corresponds to a pond with an unchanging number of each fish species. Next, we note that ify = alb, then Equation (Ia) implies dx/dt = 0, so the trout popillation x(t) is constant. Similarly, if x = min, theo Equation (Ib) implies dy/dt = 0, and the bass popu1ation y(t) is constant. This information is recorded in Figure 9.26. y Bass

y Bass

y Bass I

!!

<--~=o, dt

a

:::..

b

b

:(mn'ba)

I ----- ~ -----I

y Bass

I I ~--------~~x

~.r-----=-----==--~x

!!!

Trout

•

(b)

(a)

FIGURE 9.26 __~~x !!! Trout

Trout

(c)

Rest points in the competitive-hunter model giveo by Equations (Ia) and (I b).

L-----~

•

FIGURE 9.27 To the left of the line = the trajectories move upward, and to the rigbt they move

x min

downward.

y Bass

a b

_ L----------=Trou=--t~X

In setting up our competitive-hunter model, precise values of the constants a, b, m, n will not geoerally be known. Nonetheless, we can analyze the system of Equations (I) to learn the natore of its solution trajectories. We begin by determining the signs of dx/ dt and dy/ dt throughout the phase plane. Although x(t) represeots the number of trout and y(t) the number of bass at time t, we are thinking of the pair of values (x(t), y(t» as a point tracing out a trajectory curve in !be phase plane. Wheo dx/ dt is positive, x(t) is increasing and !be point is moving to the right in the phase plane. If dx/ dt is negative, the point is moving to !be left. Likewise, tlle point is moving upward where dy/ dt is positive and downward where dy/ dt is negative. We saw that dy/dt = 0 along the vertical line x = min. To the left of this line, dy/dt is positive since dy/dt = (m - nx)y and x < min. So the trajectories on this side ofthe line are directed upward. To the right of this line, dy/dt is negative and the trajectories point downward. The directions of the associated trajectories are indicated in Figure 9.27. Similarly, above the horizontal line y = alb, we have dx/dt < 0 and the trajectories head leftward; below this line they head rightward, as shown in Figure 9.28. Combining this information gives four distinct regions in the plane A, B, C, D, with their respective trajectory directions shown in Figure 9.29.

FIGURE 9.28 Above !be line y = alb the trajectories move to the left, and below it they move to the rigbt.

y Bass

8

A

B0

a b

'0 @ (0,0)

c

n

"'-

•

8 Trout

x

FIGURE 9.29 Composite graphical analysis of the trajectory directious in the four regious determined by = and y = alb.

x mi.

Next, we examine what happens near the two equilibrium points. The trajectories near (0, 0) point away from it, upward and to the right. The behavior near the equilibrium point (m/n, a/b) depeods on the region in which a trajectory begins. If it starts in region B, for instance, theo it will move downward and leftward towards the equilibrium point. Depeodiug on where the trajectory begins, it may move downward into region D, leftward into region A,

526

Chapter 9: First-Order Differential Equations

I;:.. (i

;..;

m

(0,0)

"

FIGURE 9.30 Motion along the trajectories near the rest points (0, 0) and(m/ll,alb). Y BRIls

Limltattons of the Phase-Plane Analysis Method

B~wm/ Ii

(~,t)

/ Troutwm

~

(0.0)

Tmut

or perhaps straight into the equilibrium point If it enters into regions A or D, then it will continue to move away from the rest point We say that both rest points are unstable, meaning (in this setting) there are trajectories near each point that head away from them. These features are indicated in Figure 9.30. It turns out that in each of the half-planes above and below the line y = alb, there is exactly one trajectory approaching the equilibrium point (m/n, a/b) (see Exercise 7). Above these two trajectories the bass population increases and below them it dem:ases. The two trajectories approaching the equilibrium point are suggested in Figure 9.3l. Our graphical anaIysis leads us to conclude that, under the ll&'WIllp1ions of the campetitive-buntermodel, it is unlikely that both species will reach equililnium levels. This is because it would be almost impossible for the fish populations to move exactly along one of the two approaching trajectories for all time. Furthennore, the initial populations point (.ro, YO) detennines which of the two species is likely to survive over time, and mutual coexistence of the species is highly improbable.

' %

FIGURE 9.31 Qualitative results of analyzing the competitive-hunter model. There an: exactly two trajectories approaching the point (mIn, alb).

Unlike the situation for the competitive-hunter model, it is not always possible to determine the behavior of trajectories near a rest point. For example, suppose we know that the trajectories near a rest point, chosen here to be the origin (0, 0), behave as in Figure 9.32. The infmmation provided by Figure 9.32 is not sufficient to distinguish between the three possible trajectories shown in Figure 9.33. Even if we could detennine that a trajectory near an equilibrium point resembles that of Figure 9.33c, we would still not know how the other trajectories behave. It could happen that a trajectory closer to the origin behaves like the motions displayed in Figure 9.33a or 9.33b. The spiraling trajectory in Figure 9.33b can never actually reach the rest point in a finite time period.

y

•

y

x

---+-{(~Lf-- x

y

--+-H++-~·x

FIGURE 9.32 Trajectory direction near the rest point (0, 0). (. )

y

(b)

(0)

FIGURE 9.33 Three posSIble trajectory motions: (a) periodic motion, (b) motion toward an asymptotically stable rest point, and (c) motion near an unstable rest point.

Another Type of Behavior The system

FIGURE 9.34 The 8OIOOon%2 + y2 - 1 is a limit cycle.

:~Y+X_x(x2+y'J.

(2.)

:=-x+y_Y(X 2 +y2)

(2b)

can be shown to have only one equiHbrium. point at (0, 0). Yet any trajectory starting on the unit circle traverses it clockwise because, whenx 2 + y2 = I, we have dy/dx = -x/y (sec Exercise 2). If a trajectory starts inside the unit circle, it spimls outward, asymptotically approaching the circle as t - 00. Ifa trajectory starts outside the unit circle, it spirals inward, again asymptotically approaching the circle as t _ 00. The circle x 2 + y2 = 1 is called a UmIt cycle of the system (Figure 9.34). In this system, the values ofx andy eventually become periodic.

9.5

Systems of Equations and Phase Planes

527

Exercises 9.5 1. List three important considerations that are ignored in the competitive-hunter model as presented in the text

2. For the system (2a) and (2b), show that aoy trajectory starting on the unit circle x 2 + y2 = 1 will traverse the unit circle in a periodic solution. First introduce polar coordinates and rewrite the system as dr/dt ~ r(1 - ,') and -dO/dt ~ -\. 3. Develop a model for the growth of trout and bass, assuntiog that in isolatioo trout demoostrate exponential decay [so that a < 0 in Equations (Ia) and (Ib)j and that the bass population grows logistically with a population limit M. Aualyze gnJphically the motion in the vicinity of the rest points in your model. Is coexisteru:e possible? 4. How might the competitive-hunter model be validated? Include a discussion of how the mons constants a, b, m, and n might be estimated How could state conservation authorities use the model to ensure the survival of both species? 5. Consider another competitive-hunter model dermed by

- ~) x- bxy' kl t ~ m(1 - :,)y - nxy, dx dt

~ a(1

c. Give an economic interpretation of the curves that determine the equilibrium points.

7. Two trajectories approa.h equilibrium Show that the two trajectories leading to (m/n, a/b) shown in Figure 9.31 are unique by carrying out the following steps. a. From system (Ia) and (Ib) apply the CbainRule to derive the following equation: (m - nx)y dy dx (a - ~»: b. Separate the mables, integrate, and exponentiate to obtain

=

yfJe-by

Kx"'e-nJC,

where K is a constant of integration.

•• Letf(y) ~ ya/ebyandg(x) ~ xM/e M • Show that f(y) has a ~ (a/eb)awheny ~ a/b as shown in unique maximum of Figure 9.35. Similarly, show that g(x) has a unique maximum M, ~ (m/en)mwhenx ~ m/n, also shown in Figure 9.35.

My

f(y)

where x andy represent trout and bass populations, respeetively.

a. What assumptions are implicitly being made about the growth of trout and bass in the absence of competition? b. Interpret the constants a, b, m, n, kh and k2 in terms oftbe physical problem. •• Perform a graphical analysis: i) Find the possible equilibrium levels. ii) Determine whether coexistence is possible. iii) Pick several typical starting points and sketch typical trajectories in the pbase plane. iv) Interpret the outcomes predicted by your graphical analysis in terms oftbe constants a, b, m, n, kb andk2 . Note: When you get to part (iii), you should realize that rIVe cases exist. You will need to analyze all five cases. 6. An economic model Consider the following economic model. Let P be the price of a single item 00 the market Let Q be the quantity of the item available on the market Both P and Q are functions of time. If one coosiders price and quantity as two interacting species, the following model might be proposed:

dP dt dQ

tit

~ ~

aP("-Q _p)' cQ(fP - Q),

where a, b, c, and f are positive coostants. Justify and discuss the

adequacy of the model. a. !fa ~ I,b ~ 20,OOO,c ~ I,andf ~ 30, find the equilibrium points of this system. !f possible, classify each equilibrium point with respect to its stability. If a point cannot be readily classified, give some explanation. b. Perform a graphical stability analysis to determine wbat will bappen to the levels of P and Q as time increases.

L-_____~ ~--------~y

b g(x)

M~

------

L-------~------------~x

'" n

FIGURE 9.35 Graphs of the functions f(y) ~ ya/e by andg(x) ~ xM/e M • d. Consider wbat bappens as (x, y) approaches (m/n, a/b). Take limits in part (b) as x .... m/n andy .... a/b to show that either lim

x-min

[(y--a)("-~)l ~ K x", eby

y-ajb

or My/M, ~ K. Thus aoy solution trajectory that approaches (m/n, a/b) must satisfy

.i'. ~ eby

(My) (x:). M;; e

e. Show that only one trajectory can approach (m/n, a/b) from below the line y ~ a/b. Pick Yo < a/b. From Fignre 9.35 you can see that f(yo) < which implies that

My,

528

Chapter 9: First-Order Differential Equations where a, h, c, d are positive constants. The values of these constants vary according to the specific situation being modeled. We can study the nature of the population changes without setting these constants to This in turn implies that

specific values.

9. What happens to 1he rabbit population if there are no foxes present?

xm

< Mx· e

----;a

10. What happens to the fox population if there are no rabbits present?

Figure 9.35 tells you that for g(x) there is a uoique value Xo < min satisfYing this last inequality. That is, for each y < alb there is a unique value ofx satisfYing the equation in part (d). Thus there can exist only one trajectory solution approaching (min, alb) from below, as shown in Figure 9.36. f. Use a sintilar argumeot to show that the solution trajectory leading to (min, alb) isuoique if yo > alb. y Bass

11. Show that (0, 0) and (cld, alb) are equilibrium points. Explain the meaning of each of these points. 12. Show, by differentiating, that the function C(t)

~

a lny(t) - by(t) - dx(t)

+ c lnx(t)

is constant when x(t) and y(t) are positive and satisfY the predatorprey equations. While x and y may change over time, C(t) does not Thus, C is a conserved quantity and its existence gives a conservation law. A trajectory that begins at a point (x,y) at time t ~ 0 gives a value of C that remains

a b Yo

-~ t Uniquexo

• m n

Trout

x

FIGURE 9.36 For any < alb only one solution trajectory leads to the rest point (min, alb). y

unchanged at future times. Each value of the constant C gives a trajectory for the autonomous system, and these trajectories close op, rather than spiraling inwards or outwards. The rabbit and fOx populations oscillate through repeated cycles along a fIXed trajectory. Figure 9.37 shows several trajectories for the predator-prey system.

Fox population

8. Show that the second-order differeutial equation y" ~ F(x, y, y') can be reduced to a system of two first-order difThreotial equations dy dx

= z,

liz dx ~ F(x, y, z).

o

c d

Rabbit population

Can something sintilar be done to the nth-order differeotial equationy(n) ~ F(x,y,y',y", ... ,y
Lotka-Volterra Equations for a Predator-Prey Model In 1925 Lotka and Volterra introduced the predator-prey equations, a

system of equations that models the populations of two species, one of which preys on the other. Let x(t) represeot the number of rabbits living in a region at time t, and y(t) the number of foxes in the same region. As time passes, the number of rabbits increases at a rate proportional to their population, and decreases at a rate proportional to the number of eocounters betweeo rabbits and foxes. The foxes, which compete for food, increase in number at a rate proportional to the

number of encouoters with rabbits but decrease at a rate proportional to the number of foxes. The number of encounters between rabbits and foxes is assumed to be proportional to the product of the two populations. These assumptions lead to the autonomous system dx dt dy dt

~ (a -

~

(-c

bylx

FIGURE 9.37

Some trajectories along which C is conserved.

13. Using a procedure similar to that in the text for the compctitiVl>hunter mode~ show that each trajectory is traversed in a counterclockwise direction as time t increases. Along each trajectory, both the rabbit and fox populations fluctuate between their maximum and minimum levels. The maximum and

utinimum levels for the rabbit population occur where the trajectory intersects the horizontal line y ~ alb. For the fox population, they occur where the trajectory intersects the vertical line x = c/d. When the rabbit population is at its maximum, the fox population is

below its maximum value. As the rabbit population declines from this point in time, we move counterclockwise around the trajectory, and the fox population grows until it reaches its maximum value.

At this point the rabbit population has declined to x

~

cld and is

no longer at its peak value. We see that the fox population reaches

its maximum value at a later time than the rabbits. The predator population lags behind that of the prey in achieving its maximum

+ dx)y

values. This lag effect is shown in Figure 9.38, which graphs both x(t) and y(t).

Chapter 9 Practice Exercises

529

14. At some time during a ttajectory cycle, a wolf invades the rabbit-fox territory, eats some rabbits, and then leaves. Does this mean that the fox population will from then on have a lower lIUlXimum value? Explain your answer.

>,Y

+------+ Lag time

FIGURE 9.38 The fox and rabbit populations oscillate periodically, with the lIUlXimwn fox population lagging the IIUlXimwn rabbit population.

Chapter ~

Questions to Guide Your Review

1. What is a frrst-order differential equation? When is a function a

solution of such an equation? 2. What is a general solution? A particular solution? 3. What is the slope field of a differential equation y' What can we learn from such fields?

= j(x, y)?

4. Describe Euler's method for solving the initial value problem y' = j(x, y), y(xo) = Yo numerically. Give an example. Comment on the method's accuracy. Why might you want to solve an initial

value problem numerically? 5. How do you solve linear ftrSt-order differential equations? 6. What is an orthogonal ttajectory of a family of curves? Deseribe how one is found for a given farnily of curves.

Chapter

ru

1. y' = xe"~

5. y'

+ xcos

2

ydx = 0

=~

11. .>:y'

=

+ 2y =

16.

10. What is an autonomous system of differential equations? What is a solution to such a system? What is a trajectory of the system?

xe xj2

I - x-I

Initial Value Problems

2. y' = .>:ye" 4.

a 2 dx -

tho

3vYcscxtho = 0

xe~-Y esc

y

8. y' = (y2 - \)x- I y' 10. 2: + Y = e-x sinX' 12. .>:y' - y

=

17. (x+l)dx+2y=x, dy

+ 2y

18. x dx

tho 19. dx

_ 2 - x

+ I,

+ h). = x 2,

x>-I, y(O) x

> 0,

+ 3y2) tho + Y dx = 0 (Hint: d(xy) = y dx + x tho) x tho + (3y - x-2 cosx) dx = 0, x> 0

20. x tho

21 . .>:y'

+ (y -

+ (x

22. y dx

cosx)dx = 0,

- 2)y

+ (3x

= h3e~,

- .>:y

y(l)

=

=

I

I

y(O) = -I

amx

13. (l + eX) tho + (ye x + e~) dx = 0 14. e~ tho + (e~y - 4x) dx = 0

15. (x

9. Why is the exponential model unrealistic for predicting long-term population grow1h? How does lbe logistic model correct for the deficiency in the exponential model for population grow1h? What is the logistic differential equation? What is the form of its solution? Desctibe the graph of the logistic solution.

In Exercises 17-22 solve lbe initial value problem.

6. y' =

7. x(x - I) tho - ydx = 0 9. 2y' - y

8. How do you construct lbe phase line for an autonomous differential equation? How does lbe phase line help you produce a graph which qualitatively depicts a solution to the differential equation?

Practice Exercises

In Exercises 1-16 solve the differential equation.

3. secxtho

7. What is an autonomous differential equation? What are its equilibtium values? How do they differ from ctitical points? What is a stable equilibtiwn value? Unstable?

+ 2) dy

y(~) y(l)

= 0,

=

= 0

0

y(2) = -I, y < 0

530

Chapter 9: First-Order Differential Equations

Euler's Method

Massm mgR2

In Exercises 23 and 24. use Euler's method to solve the initial value problem 00 the given interval starting at Xo with dx ~ 0.1.

0 23.

y' ~ y

+

cosx,

y(O) ~ 0;

0,;; x ,;; 2;

Xo

F~---

,2

~

~ 0

0 24. y' ~ (2 - y)(2x + 3), y(-3) ~ I; -3:sx:s -1; Xo =-3 In Exercises 25 and 26, use Euler's method with dx ~ 0.05 to estimate y(c) where y is the solution to the given initial value problem.

0 25. c ~ 3;

dy x-2y dx ~ X+T' y(O) ~ I

0 26. c ~ 4;

dy x 2 -2y+1 dx ~ x ,y(l) ~ I

In Exercises 27 and 28, use Euler's method to solve the initial value problem graphically, starting at Xo ~ 0 with L

0 27. dxdy

dx

~

0.1.

~

;

eX Y

+2'

b. dx

y(O)

~

~

-0.1.

2gR.

-

v'2iii

30. y' ~ I/x

29.y'=x

32. y' ~ I/y

xy

Autonomous Differential Equations and Phase Lines In Exercises 33 and 34: IdentifY the equilibrium values. Which are stahle and which are unstable?

b. Construct a phase line. IdentifY the signs of y' and y'. c. Sketch a representative selection of solution curves.

dy

dy

33'dx~y2-1

34.dx~y_y2

Applications 35. Escape velocity

+ VOl

The gravitationsl attraction F exerted by an

airless moon on a body of mass m at a distance s:from the moon's center is given by the equation F = -mg R 2s -2, where g is the acceleration of gravity at the moon's surface and R is the moon's radius (see accompanying figaro). The force F is negative because it acts in the direction of decreasing s.

Chapter

v'2iii,

The velocity Vo = is the moon's escape velocity. A body projected upward with this velocity or a greater one will escape from the moon's gravitational pull.

Slope Fields

L

2gR2 = -.-

Thus, the velocity remains positive as loog as Vo '"

In Esercises 2!1-32, sketch part of the equatioo's slope field. Then add to your sketch the solutioo curve that pssses through the point P( I, -I). Use Euler~ method with Xo ~ I and dx ~ 0.2 to estimate y(2). Round your answers to four decimal places. Find the exact value of y(2) for comparison. ~

v2

-2

dy x2 + Y 0 28. dx ~ - eY + x' y(O) ~ 0

31. y'

a. If the body is projected vertically upward from the moon's surface with an initial velocity Vo at time t = O. use Newton's second law, F ~ rna, to show that the body's velocity at position s is given by the equation

b. Show that if Vo ~

v'2iii, then 3vo

.~RI+2Rt (

)2/3 .

36. Coasting to a stop

Table 9.6 shows the distance. (meters) coasted 00 in-line skates in t sec by 10huathon Krueger. Find a model for his position in the form of Equatioo (2) ofSectioo 9.3. His initial velocity was Vo ~ 0.86 m/sec, his mass m ~ 30.84 kg (be weighed 68 Ib), and his total coasliog distance 0.97 m.

TABLE 9.6 Johnathon Krueger skating data

t (sec)

s(m)

t (sec)

s(m)

t (sec)

s(m)

0 0.13 0.27 0.40 0.53 0.67 0.80

0 0.08 0.19 0.28 0.36 0.45 0.53

0.93 1.06 1.20 1.33 1.46 1.60 1.73

0.61 0.68 0.74 0.79 0.83 0.87 0.90

1.86 2.00 2.13 2.26 2.39 2.53 2.66

0.93 0.94 0.95 0.96 0.96 0.97 0.97

Additional and Advanced Exercises

Theory and Applications 1. Transport through a cell membrane Under some conditions, the result of the movement ofa dissolved substance across a cell's membrane is described by the equation

dy A dt ~ kV(c - y). In this equation, y is the concentration of the substance inside the cell and dy/dtis the rate at whichy changes over time. The letter>

Chapter 9 Technology Application Projects k,A, V, and c stand for constants, kbeing the permeability coefficient (a property of the membrane), A the surface area of the membrane. V the cell's volume, and c the concentration of the

•• Vetify that y(x) problem

substance outside the cell. The equation says that the rate at which the cooceotratioo changes within the cell is proportional to the difference between it and the outside concentration.

b. For the integratiog factor v(x) = ef P«)"', show that

y'

In this equation, m is the mass of the system at time t, v is its velocity, and v + u is the velocity of the mass that is entering (or leaving) the aystern at the rate dm/ dt. Suppose that a rocket of initial mass mo starts from rest, but is driven upward by ftring some of its mass directly backward at the constant rate of dm/dt = -b units per second and at constant speed relative to the rocket u = -c. The ooly external force acting on the rocket is F = -mg due to gravity. Under these assumptions, show that the height of the rocket above the ground at the end of t secoods (t smal\ compared to molb) is

_ [ mo - bt In mo - bt]_ 1 2 y-ct+ b mo 2gt· 3••• Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Foodameotal Theorem of Calculus, Part I to show that aoy function y satisfYing the equation

v(x)y

=

J

v(x)Q(x) tb:

+C

for v(x) = ef p(.) '" is a solutioo to the tirst-<>rder Hoear equstion

dy tb: + P(x)y = Q(x).

J:'

b. If C = yov(xo) v(t)Q(t) dt, then show that any solutioo y in part (a) satisfies the initial conditioo y(xo) = Yo.

4. (Continuation ofExercise 3.) Assume the hypotheses of Exercise 3, and assume that Yl (x) and Y2(X) are both solutions to the firstorder Hoear equation satisfYing the initial condition y(xo) = Yo.

Chapter

y(xo) = O.

o.

c. Fnnn part (a), we have Yl (xo) - Y2(XO) = O. Since v(x) > 0 fora < x < b, usc part (b) to estshlish that Yl(X) - Y:z(x) = 0 on the interwl (a, b). ThusYl(X) = Y2(X) for all a < x < b.

tim

+ (v + u)Tt·

= F

+ P(x)y = 0,

Conclude that v(x )[Yl (x) - Y2(X)] = constant.

2. Height of • rocket If an exteroa\ force F acts upon a aystem whose mass varies with time, Newton's law of motion is

d(mv)

Yl (x) - Y2(X) satisfies the initial value

d tb: (V(X)[yl(X) - Y2(X)]) =

•• Solve the equation for y(t), using Yo to deoote y(O). b. Find the steady-state cooceotratioo, lim,...ooy(t).

----;It

=

531

Homogeneous Equations A rITst-order differeotiaI equation of the form

is called homogeneous. It can be transformed into an equation whose variables are separable by defIning the new variable v = y/x. Then, y = vxand dy dv tb:=v+xtb:. Substitution into the original differential equation and collecting terms with \ike vatiables then gives the separable equstioo tb:

dv

x + v _ F(v) =

O.

Afler solving this separable equation, the solution of the original equation is obtsined when we replace v by y/x .

Solve the branogeoeous equstioos in Exercises 5-10. First put the equation in the funn of a homogeneous equstioo. 5. (x 2 + y2)tb: + xydy = 0 6. x 2dy + (y2 - xy)tb: = 0 7. (xeYi'

+ y)tb: - xdy

= 0

8. (x +y)dy+ (x -y)tb: 9. y'

10.

Y

=i +

=

0

y-x cos - x -

(x sin ~ - ycos~) tb: + xcos ~ dy

Technology Application Projects

Mathematica/Maple Modules: Drllg Dosages: Are They Effective? Are They S"/e? Formulate and solve an initial value model for the absorption of a drug in the bloodstream. First-Ortkr Differenlifll Eqllfltions and Slope Fields Plot slope fields and solution curves for various initial conditions to selected rrrst-order differeotiaI equations.

= 0

10 INFINITE SEQUENCES AND SERIES OVERVIEW Everyone knows how to add two numbers together, or even several. But how do you add infinitely many numbers together? In this chapter we answer this question, which is part of the theory of infinite sequences and series. An important application of this theory is a method for representing a known differentiable function f(x) as an infinite sum of powers of x, so it looks like a ''polynomial with infinitely many terms." Moreover, the method extends our knowledge of how to evaluate, differentiate, and integrate polynomials, so we can work with even more general functions than those encountered so far. These new functions are often solutions to important problems in science and engineering.

10.1 HrSlORICAL ESSAY

Sequences and Series

Sequences Sequences are fundamental to the study of infinite series and many applications of mathematics. We have already seen an example of a sequence when we studied Newton's Method in Section 4.6. There we produced a sequence of approximations Xn that became closer and closer to the root of a differentiable function. Now we will explore general sequences of numbers and the conditions under which they converge.

Representing Sequences A sequence is a list of numbers

in a given order. Each of aJ, a2, a3 and so on represents a number. These are the terms of the sequence. For example, the sequence 2,4,6,8, 10, 12, ... , 2n, ... has first term al = 2, second term a2 = 4, and nth term an = 2n. The integer n is called the index of an, and indicates where an occurs in the list. Order is important. The sequence 2, 4, 6, 8 ... is not the same as the sequence 4, 2, 6, 8 .... We can think of the sequence

as a function that sends 1 to al , 2 to a2, 3 to a3, and in general sends the positive integer n to the nth term an. More precisely, an infinite sequence of numbers is a function whose domain is the set of positive integers. The function associated with the sequence 2, 4, 6, 8, 10, 12, ... , 2n, ... sends 1 to al = 2, 2 to a2 = 4, and so on. The general behavior of this sequence is described by the formula an = 2n.

532

10.1 Sequences

533

We can equally well make the domain the integers larger than a given number no, and we allow sequences of!bis type also. For example, the sequence 12, 14, 16, 18,20,22 ... is described by the fonnula a" = 10 + 2n. It can also be described by the simpler fonnula bn = 2n, where the index n starts at 6 and increases. To allow such simpler fonnulas, we let the first index of the sequence be any integer. In the sequence above, {an} starts with a, while {b.} starts withb 6 • Sequences can be described by writing rules that specify their terms, such as • r an=vn,

bn = (_I)n+11 n'

dn = (_I)n+1,

n - I cn=-n-'

or by listing !enns:

{an}

=

{Vi, Vi, v'3, ... , Vn, ... }

{b n }

=

{1,-!,t,-i, .. ·,(-I r1 k, ... }

{cn}

=

1234 n-I } { 0'2'3'4'5'''''-n-''''

{dn }

= {I, -I, I, -I, I, -I, ... ,

(-I).+" ... }.

We also sometimes write

{a,,}

=

{Vn }:,.

Figure 10.1 shows two ways to represent sequences graphically. The fIrst marks the rrrs! few points from a" a2, a3,"" an,'" on the real axis. The second method shows the graph of the function defIning the sequence. The function is dermed only on integer inputs, and the graph consists of some points in the xy-plane located at (I, ad, (2, a2), ... , (n, a,,), ....

-+--~~~I--~I--~ 'n

2

3

4

5

FIGURE 10.1 Sequences can be represented as peints on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis all is its value.

Convergence and Divergence Sometimes the numbers in a sequence approach a single value as the index n increases. This happens in the sequence

534

Chapter 10: Infinite Sequences and Series whose terms approach 0 as n gets large, and in the sequence

{o,~, t, ~'~'"'' I

-

k, .. ·}

whose terms approach I. On the other hand, sequences like

{VI, Yz, V3, ... , V;;, ... } L-E

o

L L+E

•• ••• • (• • -I- )

~------------------- L+E

L

have tenDs that get larger than any number as n increases, and sequences like

------------(n,a,,)- ~ - . ---

L-E

{I, -I, I, -I, I, -I, ... , <-1),+1, ... } bounce back and forth between I and - I , never converging to a single value. The following definition captures the meaning of having a sequence converge to a limiting value. It says that if we go far enough out in the sequence, by taking the index n to be larger than some value N, the difference between a, and the limit of the sequence becomes less than any preselected number E > O.

DEFINmONS The sequence {a,} converges to the number L iffor every positive number E there corresponds an integer N such that for all n,

~~~~----~~~----~n

o

2 3

N

n

la, - LI <

n>N

In the representation ofa sequence as points in the plane, all ~ L if y = L is a horizontal asymptote of the sequence of points (in, a,)}. In this figure,

FIGURE 10.2

E.

Ifno such number L exists, we say that {an} diverges. If {an} converges to L, we write lim,~oo an = L, or simply a, --+ L, and call L the limit of the sequence (Figure 10.2).

all the a,/s after aN lie within E of L. The definition is very similar to the definition of the limit ofa function fix) asx tends to 00 (limx~oof<x) in Section 2.6). We will exploit this connection to calculate limits of HISTORICAL BIOGRAPHY

sequences.

Nicole Oresme (ca. l32()-1382)

EXAMPLE 1
lim

n--i'OO

Show that

1n =

0

(b) lim k = k

(any constant k)

n~OO

Solution
>

0 be given. We must show that there exists an integer N such that for all n,

n>N This implication will hold if (l/n) < E or n > l/E. If N is any integer greater than I/E, the implication will hold for all n > N. This proves that lim,~oo(I/n) = O.

(b) Let E

>

0 be given. We must show that there exists an integer N such that for alln,

n> N

=)

Ik - kl <

E.

Since k - k = 0, we can use any positive integer for N and the implication will hold. This proves that lim.~oo k = k for any constant k. •

EXAMPLE 2 Solution

Show that the sequence {I, -I, I, -1, I, -I, ... , (-1),+1, ... } diverges.

Suppose the sequence converges to some number L. By choosing E = 1/2 in the definition of the limit, all terms an of the sequence with index n larger than some N must lie within E = 1/2 of L. Since the number 1 appears repeatedly as every other term of the sequence, we must have that the number I lies within the distance E = 1/2 of L.

535

10.1 Sequences

M

.. . .. ~~------~--~. 0 123 N (a)

...

It follows that IL - II < 1/2, or equivalently, 1/2 < L < 3/2. Likewise, the number - I appears repeatedly in the sequence with arbitrarily high index. So we must also have that IL - (- I) I < 1/2, or equivalently, - 3/2 < L < - 1/2. But the number L cannot lie in both of the intervals (1/2, 3/2) and (-3/2, -1/2) because they bave no overlap. Therefore, no such limit L exists and so the sequence diverges. Note that the same wgument works for any positive number € smaller than I, not ~~

.

The sequence {vii} also diverges, but for a different reason. Ai; n increases, its terms become larger than any fixed number. We describe the behavior of this sequence by writing

lim n~OO

~~----~----~. o 123 . N

m

v;. =

00.

In writing infinity as the limit of a sequence, we are not saying that the differences between the terms an and 00 become small as n increases. Nor are we asserting that there is some number infinity that the sequence approaches. We are merely using a notation that captures the idea that an eventually gets and stays 1aIger than any fIxed number as n gets 1aIge (see Figure lO.3a). The terms of a sequence might also decrease to negative infinity, as in Figure 10.3b.

(b)

FIGURE 10.3
DEFINmON The sequence {an} diverges to infinity iffor every number M there is an integer N such that for all n 1aIger than N, an > M. If this condition holds we write or Similarly if for every number m there is an integer N such that for all n bave an < m, then we say {an} diverges to negative infinity and write

>

N we

or

A sequence may diverge without diverging to infinity or negative inrmity, as we saw in Example 2. The sequences {I, -2, 3, -4,5, -6,7, -8, ... } and {I, 0, 2, 0, 3, O, ... } are also examples of such divergence.

Calculating Limits of Sequences Since sequences are functions with domain restricted to the positive integers, it is not surprising that the theorems on limits of functions given in Chapter 2 bave versions for sequences.

THEOREM 1

Let {an} and ibn} be sequences of real numbers, and let A and B be real numbers. The following rules hold if Iim.~oo an = A and liml'l_oo b" = B.

1. Sum Rule:

lim.~oo(an

+ bn )

= A

2. Difference Rule:

lim.~oo(an

- bn )

= A - B

3. Constant Multiple Rule:

lim.~oo(k·bn) =

4. Product Rule:

lim.~oo(an· bn )

5. Quotient Rule:

an Iim.~oob n

+B

k·B (any number k) = A· B

A

= Jj

The proof is similar to that of Theorem I of Section 2.2 and is omitted.

536

Chapter 10: Infinite Sequences and Series

EXAMPLE 3 (a)

lim

,,---+00

By combining Theorem I with the limits of Example I, we have:

(_1)n

= -I' lim

11--+ 00

1n

= -1,0 = 0

ConstantMultipJeRuleandExampJela

n - I) = lim ( (b) lim ( -nI -I11 ) = lim I 11--+00 11--+00 11--+00

I = I - 0 = I andExampJela Difference Rule lim 11

11--+ 00

Product Rule

- 7 6 (4In 6 ) (d) lim 4 6 n=lim •

~OO

n

+3

.~OO I

+

-

7

(3In 6 )

0 - 7 =--=-7. I

Sum and Quotient Rules

+0

•

Be cautious in applying Theorem I. It does not say, for example, that each of the sequences {a.} and {b.} have limits if their sum {an + bn} has a limit. For instance, {an} = {1,2,3, ... } and {bn} = {-1,-2,-3, ... } both diverge, but their sum {an + bn} = {O, 0, O, ... } clearly converges to o. One consequence of Theorem I is that every nonzero multiple of a divergent sequence {an} diverges. For suppose, to the contrary, that {can} conveIges fur some number c oF o. Then, by taking k = II c in the Constant Multiple Rule in Theorem I, we see that the sequence

.... .

L

..

f· -. b~.~T-; •••~:::f:-i- /I

a;

..

{i·can}

=

{a.}

•• •• -

~~------------~n

o

FIGURE 10.4 TIre terms of sequence {b.} are sandwiched betweeo those of {a,,} and {c.}, forcing them to the same common limit L.

converges. Thus, {ca.} cannot converge uuless {a.} also converges. If {an} does not converge, then {can} does not converge. The next theorem is the sequence version of the Sandwich Theorem in Section 2.2. You are asked to prove the theorem in Exercise 109. (See Figure 10.4.)

THEOREM 2-The Sandwich Theorem for Sequences Let {an}, {bn} , and {cn} be sequences of real numbers. If a. :5 bn :5 Cnholds for all n beyond some index

N, and if lUnn_oo an

=

lilIln-+oo en

=

L, then limn_ oo b"

=

L also.

An immediate consequence of Theorem 2 is that, if IbnI :5 Cn and cn ..... 0, then -en ~ hn ~ en. We use this fact in the next example.

b" ---+ 0 because

EXAMPLE 4 (a) (b)

Since lin ..... 0, we know that

cosn~O

n

.1.n ..... 0 2

k. . . o

(c) (-l) n

because

_1<

because

0<: .1. <: - 2" -

because

_1n -<: (_l)n1n

n -

COS 11

n

<1. - n'

111'. <: 1 - n"

•

The application of Theorems I and 2 is broadened by a theorem stating that applying a continuous function to a convergent sequence produces a convergent sequence. We state the theorem, leaving the proof as an exercise (Exercise 110).

THEOREM 3-The Continuous Function Theorem for Sequences Let {an} be a sequence of real numbers. If a. ..... L and if / is a function that is continuous at L and dermed at all an. then /(an) ..... /(L).

537

10.1 Sequences

EXAMPLE 5

Show that Y(n

+

I)/n --+ I.

Solution We know that (n + I)/n --+ 1. Taking f(x) ~ gives V(n + I)/n --+ VI ~ I.

Vx

and L ~ I in Theorem 3 •

2"

EXAMPLE 6 The sequence {I/n} converges to o. By taking an ~ I/n, f(x) ~ and L = 0 in Theorem 3, we see that 2'/n ~ f(l/n) --+ f(L) ~ 20 ~ I. The sequence {2'/n} converges to I (Figure 10.5). •

Using L'Hopital's Rule o

•• I

I

3

2

•

>"

FIGURE 10.5 As n --+ 00, lin --+ 0 and 21/· --+ 2° (Example 6). The terms of {lin} are shown on the x-axis; the terms of {21/.} are shown as the y-values 00 the graph of f(x) ~ 2".

The next theorem formalizes the connection between Iim,,~oo an and limx~oo f(x). It enahles US to use I'Hllpital's Rule to f'md the limits of some sequences.

THEOREM 4 Suppose that f(x) is a function defined for all x ;;,: no and that {an} is a sequence of real numbers such that an ~ f(n) for n ;;,: no. Then ~

lim f(x)

lim an

L

x~OO

~

L.

n~OO

Proof Suppose that lim_oo f(x) her M such that for all x,

~

L. Then for each positive number E there is a numIf(x) - LI <

x>M

E.

Let N be an integer greater than M and greater than or equal to no. Then

n >N

EXAMPLE 7

=>

a.

~

f(n)

and

lao -

LI

~

If(n) - LI <

E.

•

Show that

Solution The function (lnx)/x is defined for all x ;;,: I and agrees with the given sequence at positive integers. Therefore, by Theorem 4, 1im,,~oo(1n n)/n will equal limx~oo(lnx)/x if the latter exists. A single application ofl'Hllpital's Rule shows that lim Inx x---+OO

~

X

Weconcludethatlim.~oo(lnn)/n ~

lim I/x 1

~

x---+ OO

Q 1

~

O.

•

O.

When we use I'Hllpital's Rule to f'md the limit of a sequence, we often treat n as a continuous real variable and dllferentiate directly with respect to n. This saves us from having to rewrite the formula for an as we did in Example 7.

EXAMPLE 8

Does the sequence whose nth term is an

~ (~)' n- 1

538

Chapter 10: Infinite Sequences and Series

Solution The limit leads to the indeterminate form 100. We can apply I 'Hllpital 's Rule if we first change the form to 00 • 0 by taking the natorallogarithm of an:

Ina. = In("" +- 1l)n Then, lim

11_00

In an =

lim

11_00

"In("n +- 11)

00·0 form

In(: ~ :)

()form

lim -"-:--;-----"'--

o

1/"

. -2/(,,2 - 1) = lim -'--'---c--.---'n~OO -1/,,2

rHilpital's Rule: diffi:rentiate numerator and denominator.

lim~=2. n2 - 1

=

11_00

Since

In an --> 2 and f(x) = eX is continuous, Theorem 4 tells us that an

= elnail ~

e2 .

•

The sequence {an} converges toe 2 •

Commonly Occurring Limits The next theorem gives some limits that arise frequently.

THEOREM 5 1.

lim 11- 00

The following six sequences converge to the limits listed below:

In" = 0 n

lim x ' /n = 1

3.

2. lim'%=1 n~OO

(x> 0)

n~OO

S.

lim 11_ 00

(1+'"-)"=e n

4. limxn=o n_OO

(lxl

<

1)

n X

(any x)

6. lim '"--- = 0 lI_

oon !

In Formulas (3) through (6), x remains rlXed as ,,-->

(any x)

00.

Proof The first limit was computed in Example 7. The next two can be proved by taking logarithms and applying Theorem 4 (Exercises 107 and 108). The remaining proofs are • given in Appendix 5.

EXAMPLE 9 (a)

In ~,,2) =

(b)

W

These are examples of the limits in Theorem 5.

2 ~"

--> 2 • 0 =

Formula 1

= ,,2/n = (,,1In),--> (1)2 = I

(c) "{i3;; = 3 1/n(,,1In) (d)

0

(-t)"--> 0

--> 1·1 =

1

Formula 2 Formula 3 with x

=

3 and Formula 2

Formula 4 with x

=

-! 2

10.1 Sequences

I Factorial Notation

(e)

The notation n! (Un factorial") means the product I . 2 . 3 ... n of the integers from I to n. Notice that (n + I)! = (n + 1). n!. Thus, 4! = 1'2'3'4 = 24 and 5! = 1·2·3·4'5 = 5'4! = 120. We derme O! to be I. Factorials grow eveo faster than exponeotiais, as the table suggests. The values in the table are rouoded.

It

e"

It!

1

3

5 10 20

148 22,026 4.9 X 108

1 120 3,628,800 2.4 X 10 '8

(n ~ 2)" (1 + --,.2r-~e-2 =

100'

(C) - - + 0

n!

Fonnula 5 with x

=

-2

Fonnula 6 with x

=

100

539

•

Recursive Definitions So far, we have calculated each a" directly from the value of n. But sequences are often dermed recursively by giving

1. The value(s) of the initial tenn or tenns, and 2. A rule, called a recursion Cormula, for calculating any later tenn from tenns that precede it.

EXAMPLE 10 (a) The statements a, = 1 and a" = a.-1 + 1 for n > 1 define the sequence 1,2,3, ... , n, ... of positive integers. With a, = 1, we have a2 = a, + 1 = 2, a3 = a2 + 1 = 3, and so on. (b) The statements a, = 1 and a. = n·a.-1 for n> 1 derme the sequence 1,2,6,24, ... , n!,... of factorials. With a, = 1, we have a2 = 2· a, = 2, a3 = 3'a2 = 6,,,. = 4'a3 = 24, and so on. (c) The statements a, = 1, a2 = 1, and a.+1 = a" + a.-1 for n > 2 define the sequence 1, 1,2,3,5, ... of Fibonacci numbers. With a, = 1 and a2 = 1, we have a3 = 1 + 1 = 2, a4 = 2 + 1 = 3, as = 3 + 2 = 5, and so on. (d) As we can see by applying Newton's method (see Exercise 133), the statements xo = Iandx.+1 = x. - [(sinx. - x.2 )J(cosx. - lx.)]forn > odefine a sequence that, when it converges, gives a solution to the equation sin x - x 2 = o. •

Bounded Monotonic Sequences Two concepts that play a key role in determining the convergence of a sequence are those of a bounded sequence and a monotonic sequence.

DEFINmONS A sequence {a.} is bounded Crom above if there exists a number M such that a. ,,; M for all n. The number M is an upper bound for {a.}. If M is an upper bound for {a.} but no number less than M is an upper bound for {a.}, then M is the least upper bound for {a,,}. A sequence {a,,} is bounded from below if there exists a number m such that a. ;" m for all n. The number m is a lower bound for {a,,}. If m is a lower bound for {a,,} but no number greater than m is a lower bound for {a.}, then m is the greatest lower bound for {a,,}. If {a.} is bounded from above and below, the {a.} is bounded. If {a.} is not bounded, then we say that {a.} is an unbounded sequence. EXAMPLE 11 (a) The sequence 1,2,3, ... , n, ... has no upper bound since it eventually surpasses every number M. However, it is bounded below by every real number less than or equal to I. The number m = 1 is the greatest lower bound of the sequence.

123

n

.

(b) Thesequence Z'3'4"'" n + 1"" lSbounded above by every real number greater than or equal to 1. The upper bound M = 1 is the least upper bound (Exercise 125). The sequence is also bounded below by every number less than or equal to its greatest lower bound

t,

which is •

540

Chapter 10: Infinite Sequences and Series

I

Convergent .equences are bounded

If a sequence {a,,} converges to the number L, then by detmition there is a number N such that Ia" - L I < 1 if n > N. That is,

L-l
a" M ~-------------

'.

forn>N.

If M is a number larger than L + 1 and all of the tmitely many numbers aj, a2, ... , aN, then for every index n we have an :5 M so that {an} is bounded from above. Similarly, ifm is a number smaller than L - 1 and all of the numbers aj, a2, ... , aN, then m is a lower bound of the sequence. Therefore, all convergent sequences are bounded. Although it is true that every convergent sequence is bounded, there are hounded sequences that fail to converge. One example is the bounded sequence {( _1)n+1) discussed in Example 2. The problem here is that some hounded sequences bounce around in the band determined by any lower bound m and any upperboundM (Figure 10.6). An important type of sequence that does not behave that way is one for which each term is at least as large, or at least as small, as its predecessor.

~~-------..----+'

o

m

123

f-"'---"---'.----~---

FIGURE 10.6 Some bounded sequences bounce around between their bounds and fail to converge to any limiting value.

DEFINITION A sequence {an} is nondecreasing if an :5 an+1 for all n. That is, at ~ a2 ~ a3 ~ .... The sequence is noninc::reasing if an ~ an+ 1 for all n. The sequence {an} is monotonic ifit is either nondecreasing or nonincreasing. EXAMPLE 12 (a) The sequence 1,2,3, ... , n, ... is nondecreasing.

123 n . d . (b) The sequence 2' 3' 4' ... , n + I"" IS non ecreasmg. 111

I

()Th c e sequence 1'2' 4' 8"'" 211""

... 18 nonmcreasmg.

(d) The constant sequence 3, 3, 3, ... ,3, ... is both nondecreasing and nonincreasing. (e) The sequence I, -I, I, -I, I, -I, ... is not monotonic. _ A nondecreasing sequence that is bounded from above always has a least upper bound. Likewise, a nonincreasing sequence bounded from below always has a greatest lower hound. These results are based on the completeness property of the real numbers, discussed in Appendix 6. We now prove that if L is the least upper bound of a nondecreasing sequence then the sequence converges to L, and that ifL is the greatest lower bound of a nonincreasing sequence then the sequence converges to L.

y y~M

M r-----------~--------y~L

... ..

L r-----------~. -~.~.~~

. ...

THEOREM 6-The Monotonic Sequence Theorem If a sequence {an} is both bounded and monotonic, then the sequence converges .

'

Proof Suppose {an) is nondecreasing, L is its least upper bound, and we plot the points (I, all, (2, a2), ... , (n, an), ... in the xy-plane. If M is an upper bound of the sequence, all ~----------------------->x

o

FIGURE 10.7 If the terms ofa

nondecreasing sequence have an upper bound M, they have a limit L :5 M.

these points will lie on or below the line y = M (Figure 10.7). The line y = L is the lowest such line. None of the points (n, an) lies above y = L, but some do lie above any lower line y = L - E, if E is a positive number. The sequence converges to L because (a) a,,:5 L for all values ofn, and

(b) given any E > 0, there exists at least one integer N for which aN > L -

E.

The fact that {a,,} is nondecreasing tells us further that for all n 2: N. Thus, all the numbers an beyond the Nth number lie within E of L. Ibis is precisely the condition for L to be the limit of the sequence {an} . The proof for nonincreasing sequences hounded from below is similar. _

10.1 Sequences

541

It is important to realize that Theorem 6 does not say that convergent sequences are monotonic. The sequence {( -l)'+1Jn} converges and is bounded, but it is not monotonic since it alternates between positive and negative values as it tends toward zero. What the theorem does say is that a nondecreasing sequence converges when it is bounded from above, but it diverges to infinity otherwise.

Exercises 10.1 Find;ng Terms of • Sequence Each of Exercises 1--6 gives a formula for the nth term an of a sequence {a,}. Find the values of a" a" a" aod /l4. 1. an

I - n

2.

= --,-

n

~ =

3. a, = 2n

I

4. a" = 2 6. a"

=

+ (-I)'

2' - I ----zn-

ten tenns of the sequence.

10. a,

Each positive integer

26. The sequence 0, 1, 1,2,2,3,3,4, ...

Each of Exercises 7-12 gives the fIrst tenn or_ ofa sequence along with a recursion formula for the remaining terms. Write out the first

7. a, 8. a, 9. a,

Alternating 1's and O's

25. The sequence 1,0, 1,0, 1, ...

I

I n.

(-1),+1

Cubes of positive integers divided by pOWOIS of 5

I 8 27 64 125 24. 25' 125' 625' 3125' 15,625''''

Convergence and Divergence Which of the sequences {a,} in Exercises 27-90 converge, aod which diverge? Find the limit of each convergent sequence. n + (-I)' 27. a" = 2 + (0.1)' 28. all = n I - 2n

29·a,,=I+2n

= I, a,+1 = a, + (1/2') = I, a,+1 = aJ(n + I) = 2, a,+1 = (-1),+1a,,/2 = -2, a,+1 = naJ(n + I) = a2 = 1, a n+2 = a n+l + an

repeated

30.a,=

2n

+. I1

1- 3 vn

31. a"

=. + ,

32

33

=n'-2n+1 n-I

34. all =

I 2 70 - 4n

36. a,

(-I)' (I -

.a"

I - 5n 4

n

8n

• all

=

n

+3

n2 +5n+6

n'

35. a"

=

Find;ng a Sequence's Formula In Exercises 13-26, find a formula for the nth term of the sequence.

37. a"

=

13. The sequence 1, -1, 1, -1, 1, .. .

I ~ with alternating signs

39. a"

=

( _1),+1 2n I

40. a,

=

k) (2 - i,)(3 + i,) (-t)"

14. The sequence -1, 1, -1, 1, -1, .. .

1's with alternating signs

41. a"

=

~n ~ I

42. a,

=

I (0.9)'

alternating signs

43. a"

=

Sin(f + k)

44. all

= n'11' cos (n'11')

Reciprocals of squares of the positive integers, with alternating signs Powers of2 divided by

45•

11.

at

12.

at =

2, a2

=

-1, all +2

15. The sequence I, -4,9, -16,25, ...

16. The sequence I, -

I I

;V9' -

I I 16' 25''''

1 2 2 ' 2 ' 24

17.

+ (-I)'

I

= an+l/all

9' IT' 15' 18' 21' ...

Squares of the positive integers, with

multiples of3

Integers differing by 2 divided by prndncts of consecutive integers

(n;;, 1)(1 - k) 38. a,

=

_ sinn

an -

n

3' 48. all =3 n

n 47. a" = 2' 49.

=

a,

=

In(n+l)

Vn

Inn

50. a, = 1n2n (0.03)'/'

19. The sequence 0, 3, 8, 15,24, ...

Squares of the positive

51. a"

=

8 ' /'

52. a,

20. The sequence -3, - 2, -I, 0, I, ...

integers diminished by I httegers, beginning with -3

53. a"

=

(I + 'f,)"

54.a,=(I-k)"

21. The sequence I, 5, 9, 13, 17, ...

Every other odd positive

integer

22. The sequence 2,6, 10, 14, 18, ...

Every other even positive integer httegers differing by 3 divided by factorials

55. a" = "()"iOn 57. a"

=

(Ii3)'/' Inn

59. a" = n /'

'

=

56. a, =

%2

58. a,

(n

=

+ 4),/(,+4)

60. a, = Inn - In(n + I)

542

Chapter 10: Infinite Sequences and Series

61.a,=~ 63. a,

=

62. a,

(-4)' n.

=

n! 10 611

= 211 .3"

67. all

=

(krOn')

=1n(1 + k)'

69. all

= (~)' 3n - I

71. all

= (~r 2n + I '

66.

all

n!

70.a'=(n~S

:s

=(I (10/11)" 74. a, =(9/10)' + (11112)' 76. a, =sinh (In n) 78. n(l - cask) 80. a, =(3' + 50)1/0 72. a,

all =

82• a,=

75. an

= tanhn

77. an

= 2n

79. a,

=y;, sin ~

81. an

= tan-I n

83. a,

I =e)' 3 + "IIi"

85. an

= --n-

=(Iny;,n)'

I 3 7 17 a a + 2b l' 2' S' 12'···· b' a + b , .... x>O

3",6" 2-no n!

=

1-1n y;,tan

n2 • 1 _ 1 smli

(Inn)""

• ~

90. an

I = /,' pdx,

Recursively Deftned Sequences

limit cz,,+1 =

1

72

+ all

a ll +1

93. al = -4,

a,+1 =

94. al = 0,

a,,+1 =

VS + Za, V S + Za,

95. al = 5, 96. al = 3,

a,,+1 =

vSa;;

a,,+1

12 -

=

v;;.

I

I

2

+

What is that limit? (Hint: Use part (a) to show that r,2 - 2 = ±(i/y,)2 and that Yo is not less than n.)

Vi, VI

+

f(x,)

= Xn

-

Xn? -

a. X'o = I,

X n +l = Xn -

b. X'o = I,

X n +l = X n -

c. X'o = 1,

Xn+l = XII -

d. a,

2+1

Vi, VI

b. The fractions Tn = xJYlI approach a limit as n increases.

!'(x )· n

2

~ =

1

Xn

2: + Xn

tanxn - 1

sec2 XII 1

c. a,

97.2,2+ , 2 + - -,2+ I , ... 2 1 2+-2 2+--

98.

-I,

102... Suppose that f(x) is differentiable for all x in [0, I] and that f(O) = O. Define sequence {a,} by the rule a, = nf(i/n). Show that lim,_ooao = 1'(0). Use the result in part (a) to fmd the limits of the following sequences {a,}.

all + 6 = all + 2

92. at = -1,

or

respectively.

p>1

1 X

In Exercises 91-9S, assume that each sequence converges and fmd its

I

(a + 2b)2 - 2(a + b)' = +1

Do the sequences converge? If so, to what value? In each case, begin by identifying the functioo f that generates the sequence.

1

2,

2b 2 = -I or + I , then

-

101. Newton'. method The following sequences come from the recursioo formula for Newtoo's method,

vn 2 - 1 - vn 2 +n

=

generally, that if a 2

Xn+l

1/,'1xdx 89. all = 11

91. at

Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let XII and y, he, respectively, the numerator and the denominator of the nth fractioo r, = xJy,. •• VerilY that X1 2 - 2Y12 = -I,xi - 2yi = +1 and, more

87.a1J=n-~

= • ,,------;

+,·,+xlI •

Write out enough early terms of the sequence to deduce a general formula for X, that holds for n '" 2. 100. A sequence of rational numhers is described as follows:

73. all

84.a,=~

88. a,

X II +1 =Xl +X2

65. an

all = - - , -

86. a,

Theory and Examples 99. The first tenn of a sequence is Xl = 1. Each succeeding term is the sum of all those that come hefore it:

n! (Hint: Compare with I/n.) n

64.

68. a,

= V'32n + 1

VI + Vi,

VI + VI + VI + Vi, ...

=

nln(1

= n(e 1/'

-

I)

+~)

103. Pythagorean triple. A triple of positive integers a, b, and c is called a Pythagorean triple if a 2 + b 2 = c 2. Let a he an odd positive integer and let

b=l~2J

and

c=r~21

be, respectively, the integer floor and ceiling for a 2/2.

543

10.1 Sequences

l~J

2' - I --y-

117. a"

~

119. a"

~ «-I)' + l)(n ~

120. The fl!St tenD of a sequence is XI ~ cos (I). The next terms are X2 = XI or cos (2), whichever is larger; and X3 = X2 or cos (3), whichever is larger (farther to the right). In general, X,+l ~

6

121. a"

a

a. Show that a 2 + b 2 ~ c 2 • (Hint: Let a ~ 2n band c in tenDs ofn.)

+

I and express

b. By direct calculation, or by appealing to the accompanying figure, fmd

104. The 11th root of II!

a. Showthat1imn~oo(2mr)1/(2n) ~ I aodhence, using Stirling's approximatioo (Chapter 8, Additiooal Exercise 32a), that

Test the approximatioo in part (a) for n ~ 40,50, 60, ... , as far as your calculator will allow.

105... Assuming that lim.~oo(l/n') ~ 0 if c is any positive coo· stant, show that

b. Prove that limn~oo(l/n') ~ 0 if c is aoy positive coostant (Hint: IfE ~ 0.001 andc ~ 0.04, how large shouldNbeto ensure that Ii/n' - 01 < E ifn > N?)

106. The zipper theorem Prove the "zipper theorem" for sequences: If {an} and {b,} both cooverge to L, then the sequence

converges to L. 108. Prove that lim.~ooXl/' ~ I, (x

> 0).

110. Prove Theorero 3.

is bounded.

3n + I n+T

113. an

=-, n. 2"3"

(2n + 3)!

112. a" ~ (n

+

2

2an

-

3

126. Uniqueness of least upper bounds Show that if MI and M2 are least upper bouods for the sequence {an}, then Ml ~ M2. That is, a sequence cannot have two different least upper bounds.

128. Prove that if {an} is a convergent sequence, then to every positive number E there corresponds an integer N such that for all m

andn,

m > N and n > N

~

I .... -

a" 1

<

E.

130. Limits and subsequences If the terms of one sequence appear in another sequence in their given order, we call the first sequence a subsequence of the second. Prove that if two subsequences of a sequence {a,,} have different limits LI L2, then {an} diverges.

*

verge? Give reasons for your answers. I

116.a,,~n-1i

SI>-

133. Sequence. generated by N_on'. method Newton's method, applied to a differentiable functioo f(x), begins with a starting value Xo and coostructs from it a sequence of numbers {x,} that under favorable circumstances converges to a zero of f. The

recursion formula for the sequence is Xn+l

I

Which of the sequences in Exercises 115-124 converge, and which di· 115.a,~I-1i

a n +l =

I)!

114. a" ~ 2 - Ii - 2'

I

1,

al =

125. The sequence {II/(II + I)} bas a lea.t upper bound of 1 Show that if M is a number less thao I, then the terms of {n/(n + I)} eventually exceed M. That is, if M < I there is ao inreger N such that n/(n + I) > M whenever n > N. Since n/(n + 1) < I for every n, this proves that I is a least upper bouodfor {n/(n + I)}.

132. Prove that a sequence {a,} converges to 0 ifaod only if the quence of absolure values {I an I} cooverges to O.

In Exercises 111-114, detennine if the sequence is monotooic aod if it ~

124.

131. For a sequence {a,,} the tenDs of even index are denored by au and the terms of odd index by au+ I. Prove that if au ---+ L and a2k+1 -+ L, then a" -+ L.

107. Prove that lim.~oo 'C",; ~ 1.

111. a,

4"+ 1 + 3" 4'

n+ I 122. an = - n -

129. Uniqueness of limit. Prove that limits of sequences are uoique. That is, show that if LI aod L, are numbers such that On -+ LI and all -+ L2, then Ll = L,..

if c is any positive constant.

109. Prove Theorero 2.

~ I +~

123. a" ~

+ I)}.

max {x.. cos (n

127. Is it true that a sequence {a,} of positive numbers must cooverge if it is bouoded from above? Give reasons for your answer.

vi,;! ~ ~ for large values of n.

D b.

I)

=X

f(xn) ll -

!'(x

)'

lI

.. Show that the recursioo formula for f(x) ~ x 2 caobewrittelIaSxn+l ~ (xn + a/x,)/2.

D b.

-

a, a

> 0,

Starting with Xo ~ I aod a ~ 3, calculate successive tenDs of the sequence uotil the display begins to repeat. What number is being apprnsimared? Explain.

544

Chapter 10: Infinite Sequences and Series

D 134. Arecursi.. defioi1ionof"./2

Ifyou start with Xl = I anddefme the subsequent terms of {x,} by the rule x, = X,_I + COSX,_I, you geoerate a sequeoce that converges rapidly to 1f/2. (a) Try it. (b) Use the accompaoying figure to explain why the conver-

a. Calculate and theo plot the frrst 25 terms of the sequeoce. Does the sequeoce appear to be bouoded from above or be-

gence is so rapid.

b. If the sequeoce converges, find an integer N such that Ia" - L I :5 0.01 for n '" N. How far in the sequeoce do you have to get for the terms to lie within 0.0001 of L?

low? Does it appear to converge or diverge? If it does cooverge, what is the limit L?

y

135. a, =

COS~_l

o

I

138.

an+l = an

at = =

1,

(I + 0~5)"

= an + 5"

QII+I

+ (-2)11 140. a"

sinn

sinn

COMPUTER EXPLORATIONS Use a CAS to perform the followiog steps for the sequeoces in Exercises 135-146.

10.2

136. a" =

137. at = I,

139. an

~¥-~~--J------>x

%

=

n sin

k

Inn

141. a, = --n-

142. a" = "...

143. a, = (0.9999)'

144. a" = (123456)1/,

145. an

=

n 41

8' I

146. a" = 19"

n.

Infinite Series An infinite series is the sum of an infmite sequence of numbers

al + a2 + a, + ... + a, + ... The goal of this section is to understand the meaning of such an infinite sum and to develop methods to calculate it. Since there are infmitely many terms to add in an infinite series, we cannot just keep adding to see what comes out. Instead we look at the result of summing the first n terms of the sequence and stopping. The sum of the first n terms Sn =

al

+

a2

+ a3 + ... + all

is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n geta larger, we expect the partial sums to get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit, as discussed in Section 10.1. For example, to assign meaning to an expression like

1+

1

1

1

2 + 4 + "8 +

1

16 + ...

we add the terms one at a time from the beginning and look for a pattern in how these partial sums grow.

Partial sum

Value

Suggestive expression for partial sum

First:

Sl =

1

1

2 - 1

Second:

s2=1+ 21

Third:

1 1 s,=I+-+2 4

3 2 7 4

1 2-2 1 2-4

nth:

1 1 1 s, =1+-+-+···+-2 4 2,-1

2' - 1 2n - 1

2 -n1-1 2 -

10.2

Infinite Series

545

Indeed there is a pattern. The partial sums form a sequence whose nth tenn is SrI

=

1 2 - 2,,-1.

This sequence of partial sums converges to 2 because 1im,,~00(I/2·-1) = O. We say "the sum of the infinite series 1

+ !2 + !4 + ... + _1_ + ... 2.- 1

is 2."

Is the sum of any fmite numberoftenns in this series equal to 2?No. Can we actually add an infinite number oftenns one by one? No. But we can still define their sum by defming it to be the limit of the sequence of partial sums as n --> 00, in this case 2 (Figure 10.8). Our knowledge of sequences and limits enables us to break away from the confines of finite sums.

1

114

~

o

112

FIGURE 10.8 As the lengths 1, ,/2, '/.,

1/8

2

'I" ... are added one by one, the sum

approaches 2.

HISTORICAL BIOGRAPHY

Blaise Pascal (1623-1662)

DEFINmONS

Given a sequence of numbers {a.}, an expression of the form

al+a2+ a3+ .. ·+ a.+ .. · is an inf"mite series. The number a. is the 11th term of the series. The sequence {s.} defmed by

is the sequence of partial sums of the series, the number s. being the 11th partial sum. If the sequence of partial sums converges to a limit L, we say that the series converges and that its sum is L. In this case, we also write 00 al + az + ... + a. + ... = ~ a. = L . •- 1

If the sequence of partial sums of the series does not converge, we say that the series diverges.

When we begin to study a given series al + az + ... + an + .. " we might not know whether it converges or diverges. In either case, it is convenient to use sigma notation to write the series as or

A useful shor1hand when smnmation from 1 to 00 is

understood

546

Chapter 10: Infinite Sequences and Series

Geometric Series Geometric series are series of the form 00

a + aT + ar2 + ... + ar n- t + ...

=

L ar n- 1 11=1

in which a and r are fixed real numbers and a oF ~:o ar". The ratio r can be positive, as in

o. The

I I

(1)"-1 + ...

1-t+~-···+

(-tf' + ....

1+-+-+···+ 2

4

2

series can also be written as

r~I/2,a~1

'

or negative, as in r~ -1/3,a~

1

If r = I, the nth partial sum of the geometric series is

s"

= a + a(l) + a(I)2 + ... + a(I)"-1 = na,

and the series diverges because lim"~oo s" = ± 00, depending on the sign of a. If r = -I, the series diverges because the nth partial sums alternate between a and O. If Ir I oF 1, we can determine the convergence or divergence of the series in the following way:

rSn

=

+ aT + ar2 + ... + arl'l-l aT + ar2 + ... + arn- 1 + ar1!

Sn - rSn

=

a - ar"

Sn =

a

s.(1 - r) = a(1 - r") 8n =

Multiply s. by r. Subtract rSn from Sn. Most of the terms on the right cancel.

Factor.

a(1 - r") 1- r '

(r oF 1).

We can solve for 311 if r =F 1.

If Irl < I, then rn --> 0 as n --> 00 (as in Section 10.1) and s" --> a/(I - r). Iflrl then Ir"I--> 00 and the series diverges.

Iflrl < 1, the geometric series a to a/(I - r):

+ ar + ar2 + ... + ar"-l + ...

> I,

converges

00

L ar n - 1 =

11=1

_a_, 1- r

Irl

< I.

If Ir I ;;,: I, the series diverges.

We have determined when a geometric series converges or diverges, and to what value. Often we can determine that a series converges without knowing the value to which it converges, as we will see in the next several sections. The formula a/ (I - r) for the sum of a geometric series applies only when the summation index begins with n = I in the expression ~:"l ar"-l (or with the index n = 0 if we write the series as ~:o ar").

EXAMPLE 1

The geometric series with a

I

I

I

= 1/9 and r = 1/3 is

I (I )"-'

9 + 27 + Sf + ... = ~ 9 3 EXAMPLE 2

00

1/9 I - (1/3)

The series 00

(-I)n5 4"

5

5

5

4

16

64

~--=5--+---+···

n~O

I 6·

•

10.2

is a geometric series with a

a

547

= 5 and r = -1/4. It converges to 5

a

+ (l/4)

1- r

ar - -

Infinite Series

=

•

4.

EXAMPLE 3 You drop a ball from a meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than 1. Find the total distance the ball travels up and down (Figure 10.9).

--_( -_ _ _

ar2 __ _

Solution

The total distance is

ar 3 - - S

2ar = a + 2ar + 2ar 2 + 2ar 3 + ... = a + - = all+r -l-r -r This sum is 2ar/ (I - r) .

If a = 6 m and r = 2/3, for instance, the distance is (a)

1

S

EXAMPLE 4

~ o o

Solution

o o a

+ (2/3)

(5/3)

= 6 1 _ (2/3) = 6 1/3

•

= 30 m.

Express the repeating decimal 5.232323 ... as the ratio of two integers.

From the definition of a decimal number, we get a geometric series

5.232323 ... = 5

+ 12030 + ~ + ~ + .. .

o

(l00)

(l00)

1

( 1 100

23 (

Q

= 5 + 100 1 + 100 + o

)2 + ... )

a = 1, r = 1/ 100

1/ (1 - 0.01)


•



.,.,

-..

.,., ., .,A ., ., .,

.'"

23 ( 1 )

23

= 5 + 100 0.99

518

= 5 + 99 = 99

•

Unfortunately, formulas like the one for the sum of a convergent geometric series are rare and we usually have to settle for an estimate of a series' sum (more about this later). The next example, however, is another case in which we can find the sum exactly. 00

(b)

FIGURE 10.9 (a) Example 3 shows how to use a geometric series to calculate the total vertical distance traveled by a bouncing ball if the height of each rebound is reduced by the factor r. (b) A stroboscopic photo of a bouncing ball.

EXAMPLE 5

Find the sum of the "telescoping" series

L ( ~ n=i n n

1)'

Solution We look for a pattern in the sequence of partial sums that might lead to a formula for Sk. The key observation is the partial fraction decomposition

1 n(n

+ 1)

----

n

n

+

l'

so

1 L--,---''----:n=i n(n + 1) k

k

,&i

(1

1)

n-n+l

and

Removing parentheses and canceling adjacent terms of opposite sign collapses the sum to Sk =

We now see that Sk ~ 1 as k ~

00.

1- k

1

+

1.

The series converges, and its sum is 1: 00

L 1 n=i n(n + 1)

= 1

.

•

548

Chapter 10: Infinite Sequences and Series

The nth-Term Test for a Divergent Series One reason that a series may fail to converge is that its terms don't become small.

EXAMPLE 6

The series

00 L,,+I=I+l+1+ ... +,,+I+ ...

•-1"

1

2

3

"

diverges because the partial sums eventually outgrow every preassigned number. Each term is greater than I, so the sum of" terms is greater than ". • Notice that lim.~oo a. must equal zero if the series ~::'~1 a. converges. To see why, let S represent the series' sum and s. = a, + a2 + ... + a. the nth partial sum. When n is large, both s. and S.-1 are close to S, so their difference, a., is close to zero. More formally,

an

= 3 11

-

~

3 11 -1

S- S

=

O.

Difference Rule for sequences

This establishes the following theorem.

I Caution

Theorem 7 does not say that l:;1 an converges if tJn ---+ O. It is possible for a series to diverge when an ---+ 0 .

00

If ~ an converges, then an ----i> O.

THEOREM 7

11=1

Theorem 7 leads to a test for detecting the kind of divergence that occurred in Example 6.

The nth-Term Test for Divergence

00

L a. diverges if lim 11=1

an fails to exist or is different from zero.

lI--i'OO

EXAMPLE 7

The following are all examples of divergent series.

00

(a)

L ,,2 diverges because ,,2 --> 00. n=1

(b)

00,,+1. ,,+1 L -,,diverges because - , , - --> I. n=1

(c)

L (- 1).+1 diverges because lim.~oo( - 1).+1 does not exist.

00

.-1 00

-n

(d) ~ 2n

+

EXAMPLE 8

.

.

-n

5 diverges because lim.~oo 2n

+5

1

= -

2 oF o.

•

The series

IIIIII II I 1+-+-+-+-+-+-+···+-+-+···+-+··· 224444 2'2' 2' ~

2 terms

4 terms

2" terms

diverges because the terms can be grouped into infinitely many clusters each of which adds to I, so the partial sums increase without bound. However, the terms of the series form a sequence that converges to O. Example 1 of Section 10.3 shows that the harmonic series also behaves in this manner. _

10.2 Infinite Series

549

Combining Series Whenever we have two convergent series, we can add them term by term, subtract them term by term, or multiply them by constants to make new convergent series.

If ~an = A and ~bn = B are convergent series, then

THEOREM 8

~(an + bn) = ~a. + ~bn = A + B

1. Sum Rule:

2. Difference Rule: 3. Constant Multiple Rule:

(any number k).

Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section lO.1. To prove the Sum Rule for series, let ~=~+_+

...

+~

~=b,+~+···+~.

Then the partial sums of ~(an + bn) are Sn = (a, + b,) + (a2 + b2) + ... + (an + bn) = (a, + ... + an) + (b, + ... + bn) =

An

+ Bn·

Sinee An --+ A and Bn --+ B, we have Sn --+ A + B by the Sum Rule for sequences. The proof of the Difference Rule is similar. To prove the Constant Multiple Rule for series, observe that the partial sums of ~kan form the sequence Sn = ka, + ka2 + ... + kan = kia, + a2 + ... + an) = kAn.

•

which converges to kA by the Constant Multiple Rule for sequences. All corollaries of Theorem 8, we have the following results. We omit the proofs.

1. Every nonzero constant multiple of a divergent series diverges. 2. If ~an converges and ~bn diverges, then ~(an + bn) and ~(a. diverge.

- bn) both

Remember that ~(an + bn) can converge when ~a. and ~bn both diverge. = I + I + I + "'and~bn = (-I) + (-I) + (-1) +"'diverge, whereas ~(an + bn) = 0 + 0 + 0 + ... converges to O. Caution

Forexample,~an

EXAMPLE 9 00

(8)

L n- l

Find the sums of the following series.

3n-' _ 1 6n ,

00

=

L n- l 00

=

L

11=1

(

1 1 ) 2n-' - 6n-'

1

00

L

I

2n-' - 11=1 6n-' I 1

1 - (1/2)

=2- Q =± 5

5

1 - (1/6)

Difference Rule Geometric series with a ~ 1 andr ~ 1/2,1/6

550

Chapter 10: Infinite Sequences and Series

(b)

Cons_ Multiple Rule

=

4C - ~1/2»)

Geometric series with a

I, r

=

=

1/2

•

=8

Adding or Deleting Terms We can add a finite number of terms to a series or delete a rmite number of terms without altering the series' convergence or divergence, although in the case of convergence this will usually change the sum. If ~:;"~1 a, converges, then ~:;;'ka. converges for any k> land 00

Lan n=1

00

+

= at

a2

+ ... +

Conversely, if ~:;"-ka. converges for any It

>

ak-t

+

Lan" 1I=k

I, then ~:::'1 a, converges. Thus,

and

HISTORICAL BIOGRAPHY

Richard Dedekind (1831-1916)

Reindexing As long as we preserve the order of its terms, we can reindex any series without altering its convergence. To raise the starting value of the index h units, replace the 11 in the fonnula fora, by 11 - h: 00

00

L an = 11-L1+11: an-h = 11 - 1

at

+

a2

+ a3 + ....

To lower the starting value of the index h units, replace the 11 in the fonnula for a. by 11 00

+ h:

00

:L an = 11=1-11: :L an+h = 11=1

at

+

a2

+

Q3

+ ....

We saw this reindexing in starting a geometric series with the index 11 = 0 instead of the index 11 = 1, but we can use any other starting index value as well. We usually give preference to indexings that lead to simple expressions.

EXAMPLE 10

We can write the geometric series

1

00

~ 2,-1

1

=

I

1

+ 2 + 4 + ...

as 00

I

~2n'

00

1

~ 2,-5' 11=5

00

or even

1

n~2n+4'

The partial sums remain the same no matter what indexing we choose.

•

10.2 Infinite Series

551

Exercises 10.2 Finding nth Partial Sums

Using the nth-Tenn Test

Iu Exercises l--{j, fmd a formula for the nth partial sum of each series and use it to fmd the series' sum if the series converges.

Iu Exercises 27-34, use the nth-Term Test for divergence to show that

1. 2+~+~+l+ ... +~+ ... 3 9 27 3,-1 2 ~ + _9_ + _9_ + ... + _9_ + ... • 100 100' 100' 100' 3.1_

1 + 1 _ 1 +"'+(_1),-1_1_+ ... 2

4

8

2,-1

4.1- 2 +4 - 8 + ... + (-1),-12,-1 + ...

the series is divergent, or state that the test is inconclusive. 00 n ~ n(n + I) 27. ~ n + 10 28. ,-I (n + 2)(n + 3) 00 00 I 30. ~+29. ~~4 11""0 n 11- 1 n + 3 00 , 00 I 31. ~ cos" 32. ~+-+ 11"'0 e n

,-I

34. ~ cosmT

I I I I 5. 2. 3 + 3· 4 + 4· 5 + ... + (n + I)(n + 2) + ... 5 5 5 5 6. 1'2 + 2'3 + 3'4 + ... + n(n + I) + ...

Series with Geometric Terms Iu Exercises 7-14, write out the fITst few tenDs of each series to show how the series starts. Then fmd the sum of the series. 00 (-I)'

7. ~-4'

, -0

00 7

(5 + I) (I + -----sn (-I)') ~

m (~y +

+

(~)' + (~)' + .. .

1& (~2y + (~2)' + (~2)' + (~2)' + (~2)6 + ... Repeating Decimals Express each of the numbers in Exercises 19-26 as the ratio of two integers. 19. 0.23 = 0.23 23 23 ... 20. 0.234 = 0.234 234 234 ... 21. 0.7 = 0.7777 .. . 22. O.d = O.dddd ... , where d is a digit

= 0.06666 .. .

24. 1.414

(I"-~I I) n

00 37. ~ (In v;;-+!

,-I

(t) + (t)' + (t)' + (t)' + (t)' + ...

23. 0.06

11- 1

00 40. ~

211

16. 1+ (-3) + (-3)2 + (-3)' + (-3)' + .. . 17.

00 35. ~

-

36.~(3

3) (n + I)'

'-1~'

In Vn)

311

In Exercises 15--18, determine if the geometric series converges or diverges. If a series converges. fmd its sum.

15. I +

Iu Exercises 35-40, fmd a formula for the nth partial sum of the series and use it to detennine if the series converges or diverges. If a series converges, ftnd its sum.

11=1

00 11. ~ 211 00

Telescoping Series

38. ~ (tan (n) - tan (n - I))

9. ,-I ~4'

13.

11=0

1.414414414 ...

=

25. 1.24123

=

26. 3.142857

1.24123 123 123 .. .

= 3.142857 142857 .. .

(Vn+4 - Vn+3)

Find the sum of each series in Exercises 41-48. 41

~.

4

. ::1 (4n - 3)(4n + I)

43

~ 40n • ,~1 (2n - 1)'(2n + I)'

45.

00(1 I) ~ Vn - v;;-+!

47

~.

'.;:1

(

42

~

. ::1 (2n

6

- 1)(2n + I)

44.~2n+1 ,~1 n'(n

46.

00(1 ~ 21/,

+ I)' I) - 21/(,+ 1)

I _ I ) In(n+2) In(n+l)

48. ~(tan-l (n) - tao-I (n + I))

,-I

Convergence or Divergence Which series in Exercises 49--{j8 converge, and which diverge? Give reasons for your answers. If a series converges, f"md its sum.

00(1)' 49. ~ • r.:: 11=0 v2 51.

~ (-1)'+1

11=1

00 53. ~ cosmT 11=0

3, 2

00 SO. ~(V2r n=O

52. ~ ( -1)'+1n n=l

00 54. ~ cos:'11' n=O 5

552

Chapter 10: Infinite Sequences and Series

00

,-I

59.

61.

~~

11=-1

83. Show by example that ~(aJb,) may diverge even though and ~b, converge and no b, equals O.

56. ~ In 3'

00

57.

I

00

e-'" ,-0

55. ~

00

58.

10"

00

2"_ 1

00

1

84. Find convergent geometric series A ~ ~a, and B ~ ~b, that illustrate the fact that ~a"b. may converge without being equal toAB.

I

~"' Ixl> 11=0 X

I

I)'

~3"

00 ( 60.~ I-Ii

~ I~O'

00

•

00

21'.

85. Show by example that ~(a,/b,) may converge to something other than A/Beven when A ~ ~a., B ~ ~b. #' 0, and no b, equals O.

62. ~"1Ft n!

2"+ 3" 63. ~11=1 4" 00

64.

86. If ~a, converges and a, > 0 for all n, can anything be said about ~(l/a,,)? Give reasons for your answer.

+ 4"

~311+411

87. What happens if you add a fmite number of terms to a divergent series or delete a ftnite number of tenns :from a divergent series? Give reasons for your answer.

65.~1n(n :1) 11=-1

Geometric Series with a Variable x In each of the geometric series in Exercises 69-72, write out the fl!St few terms of the series to fmd a and r, and fmd the sum of the series. Then express the inequality Ir I < I in terms of x and fmd the values of x for which the inequality holds and the series converges.

69. ~(-I)'x'

70. ~(-I)'x'"

.-0

(-I)' (

~-2- 3 + 00

88. If ~a" converges and ~b. diverges, can anything be said about their term-by-term sum ~(a, + b,)? Give reasons for your answer. 89. Make up a geometric series ~ar"-l that converges to the number 5if a. a

=2

b. a ~ 13/2.

90. Find the value of b for which

1 + eb + e2JJ + e3b + ... = 9. 91. For what values of r dues the infinite series

,-0

72.

~a"

I

sinx

)' converge? Find the sum of the series when it converges.

10 Exercises 73-78, find the values of x for which the given geometric senes converges. Also, fmd the sum of the series (as a function ofx) for those values of x. 00

00

73. ~2'x'

74. ~(-!)y'"

,-0

,-0

00

92. Show that the error (L - so) obtained by replacing a convergent geometric series with one of its partial sums s" is ar"/( 1 - r). 93. The accompanying figure shows the first five of a sequence of squares. The outermost square has an area of 4 m2 • Each of the other squares is obtained by joining the midpoints of the sides of the squares before it Find the sum of the areas of all the squares.

I)"(x - 3)' 00 ( - 76. ~

75. ~(-I)'(x + I)' 11=0

11=0

00

00

.-

2

78. ~(lnx)'

77. ~ sin"x

Theory and Examples 79. The series in Exercise 5 can also be written as

~

I

,~I (n + I)(n + 2)

d ~ an ,~_I (n

Write it as a sum beginning with (a) n (0) n ~ 5.

~

I

+ 3)(n + 4) . -2, (b) n

~

94. Helga von Kocb's snowflake CIIl"W Helga von Koch's snowflake is • curve of infmite length that encloses a region of fmite area. To see why thls is so, suppose the curve is generated by starting with an equilateral triangle whose sides have length 1.

0,

80. The series in Exercise 6 can also be written as 00

5

~ n(n + I)

00

and

~ (n +

5 !)(n

Write it as a sum beginning with (a) n (0) n ~ 20.

+ 2) . ~

-I, (b) n

~

3,

81. Make up an infinite series of nonzero terms whose sum. is L I b . -3 c. O. 82. (Continuation o/Exercise 81.) Can you make an write series of nonzero terms that converges to any number you want? Explain.

.. Find the length L, of the nth curve C, and show that limn. . . oo L" = 00. b. Find the area An of the region enclosed by C, and show that lim,~ooA. ~ (8/5)A , .

LOO O C2

c,

10.3 The Integral Test

10.3

553

The Integral Test Given a series, we want to know whether it converges or not. In this section and the next two, we study series with nonnegative terms. Such a series cooverges if its sequence of partial sums is bounded. If we establish that a given series does converge, we generally do not have a formula available for its sum, so we investigate methods to approximate the sum instead.

Nondecreasing Partial Sums Suppose that ~~,
Since the partial sums form a noodecreasing sequence, the Moootonic Sequence Theorem (Theorem 6, Sectioo 10.1) gives the following result.

Corollary of Theorem 6 A series ~;:"- 1 an of nonnegative terms cooverges if and ouly if its partial sums are bounded from above.

EXAMPLE 1

The series

i1= 1 +1+1+ ... +1+ ...

n~1

n

2

n

3

is called the harmonic series. The harmonic series is divergent, but this doesn't follow from the nth-Term Test. The nth term l/n does go to zero, but the series still diverges. The reasoo it diverges is because there is no upper bound for its partial sums. To see why, group the terms of the series in the following way:

The sum of the first two terms is 1.5. The sum of the next two terms is 1/3 + 1/4, which is greater than 1/4 + 1/4 = 1/2. The sum of the next four terms is 1/5 + 1/6 + 1/7 + 1/8, which is greater than 1/8 + 1/8 + 1/8 + 1/8 = 1/2. The sum of the next eight terms is 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16, which is greater than 8/16 = 1/2. The sum of the next 16 terms is greater than 16/32 = 1/2, and so 00. In general, the sum of 2n terms ending with 1/2n+ 1 is greater than 2n/2n+ 1 = 1/2. The sequence of partial sums is not bounded from above: If n = 21, the partial sum Sn is _ greater than k /2. The harmonic series diverges.

The Integral Test We now introduce the Integral Test with a series that is related to the harmooic series, but whose nth term is l/n 2 instead of l/n.

EXAMPLE 2

Does the following series converge?

i~= 1 +1+1+~+ ... +~+ ... 2 2

n- l

n

4

9

16

n

554

Chapter 10: Infinite Sequences and Series

We determine the convergence of ~:::'I(I/n2) by comparing it with hoo(l/x2) dx. To carry out the comparison, we think of the terms of the series as values of the function f(x) = l/x2 and interpret these values as the areas of rectangles under the curve y = l/x 2 • As Figure 10.10 shows,

y

Solution (1,f(I)) Graph off(x) ~ ~ I

s

(2,f(2))

l'

,

1

2

3

4

...

-.!.. + -.!.. + -.!..2 + .. . + -.!..2

=

f(1)

12

+ f(2) + f(3) + ... + f(n)

11

n - 1 n ...

< I + FIGURE 10.10 The sum of the areas of the rectangles under the graph of !(x) ~ I/x' is less thao the area under the graph

(Example 2).

I Caution The series and integral need not have the same value in the convergent case. As we noted in Example 2, ~:;;'I (I/n') = ...'/6 while 1,00(I/x') dx = 1.

n

3

22

Rectangle areas sum to less than area under graph.

< f(1) + (" J,dx

~-1~~~~~~~~ x

o

=

x

roo J,dx

11

< I + I

=

x 2.

As in Section 8.7, Example 3, ll°O(I/x 2 ) d.Jc ~ 1.

Thus the partial sums of ~~I (l/n 2) are bounded from above (by 2) and the series con· _ verges. The sum of the series is known to be ""/6 '" 1.64493.

THEOREM 9-The Integral Test Let {a,} be a sequence of positive terms. Suppose that a" = f(n), where f is a continuous, positive, decreasing function of x for all x 2: N (N a positive integer). Then the series ~~Na, and the integral f(x) dx both converge or both diverge .

f:

Proof We establish the test for the case N = I. The proof for general N is similar. We start with the assumption that f is a decreasing function with f(n) = a" for every n. This leads us to observe that the rectaogles in Figure lO.11a, which have areas a" a2, ... , a", collectively enclose more area than that under the curve y = f(x) from x = I tox = n + 1. That is,

y

y

al

2

0

~

f(x)

1'+1

a, n

3

a" n

+I

f(x) dx ,,; al

+ a2 + ... + a,.

x

(a)

In Figure 1O.llb the rectangles have been faced to the left instead ofto the right. Ifwe mo· mentarily disregard the first rectaogle of area ai, we see that

y

a2 al

2

3

____ x

Subject to the conditions of the Integral Test, the series ~:::'Ia. and 00 the integral (x) dx both converge or both diverge.

11

al

n-l n (b)

FIGURE 10.11

l'

f(x)dx.

If we include ai, we have

., a,

~~~~ _ _L -_ _~~a~'L-

o

+ a3 + ... + a,";

+ a2 + ... + a,,";

al

+

l'

f(x)dx.

Combining these results gives

{"+I

11

f(x) dx ,,; al

+ a2 + ... + a"

,,; al

(" + 11 f(x) dx.

These inequalities hold for each n, and continue to hold as n --> 00 • If f1°O f(x) dx is fmite, the right·hand inequality shows that ~a" is fmite. If oo h f(x) dx is infmite, the left·hand inequality shows that ~a, is infmite. Hence the series _ and the integral are both finite or both infinite.

10.3 The Integral Test

EXAMPLE 3

555

Show that the p-series

~~= I~+~+~+"'+~+'" (p a real constant) converges if p

Solution

If p

>

>

I, and diverges if p

I.

:5

I, then j(x) = l/xP is a positive decreasing function ofx. Since

1

00

1

-I dx = xP

1

-p+1 ]b

00

lim

x -p dx =

1

[

~x,"----~

-P + 1

b_OO

= _1_ lim ( _ I I - P b~oo b P - 1

1

I) bP-l~ooasb--+oo

= _1_(0 _ 1) = _1_

I-p

I ~ 1 I The p-series 1I~1 nP converges if p

>

1, diverges if p :5 1.

p-I'

because p - 1 > O.

the series converges by the Integral Test. We emphasize that the sum of the p-series is not I/(p - I). The series converges, but we don't know the value it converges to. Ifp < I,thenl-p > oand

roo ~dx = x

11

P

_1_ lim (bl-p - I) = 1 - Pb_OO

00.

The series diverges by the Integral Test. If p = I, we have the (divergent) hannonic series I I I 1+-+-+···+-+··· 2 3 n We have convergence for p

>

I but divergence for all other values ofp.

•

The p-series with p = I is the harmonic series (Example I). The p-Series Test shows that the hannonic series is just barely divergent; if we increase p to 1.000000001, for in-

stance, the series converges! The slowness with which the partial sums of the hannonic series approach infinity is impressive. For instance, it takes more than 178 million terms of the hannonic series to move the partial sums beyond 20. (See also Exercise 43b.)

2+ I)) is not ap-series, but it converges by the

The series ~::'~I (1/(n Integral Test. The function j(x) = 1/(x 2 x ~ 1,and

EXAMPLE 4

+

I) is positive, continuous, and decreasing for

{"" _1_ dx = lim [arctan x2 + I b~oo

11

=

x]b 1

lim [arctan b - arctan I]

b~oo

1T

1T

1T

2

4

4'

---

Again we emphasize that 1T/4 is not the sum of the series. The series converges, but we do not know the value of its sum. •

Error Estimation If a series ~a. is shown to be convergent by the Integral Test, we may want to estimate the size of the remainder R. between the total sum S of the series and its nth partial sum s•. That is, we wish to estimate Rn =

S-

9"

= an+l +

tln+2

+

an+3

+

556

Chapter 10: Infinite Sequences and Series To get a lower bound for the remainder, we compare the sum of the areas of the rectangles with the area under the curve y = f(x) for x ;;,: n (see Figure 1O.lla). We see that

1

00

R. = a.+1

+

a.+2

+

a,,+3

+ ... ;;,:

f(x) fix .

• +1

Similarly, from Figure 10.llb, we fmd an upper bound with

1

00

R.

= a,,+1

+

a.+2

+

a,,+3

+ ... ,,;;

f(x) fix.

These comparisons prove the following result giving bounds on the size of the remainder.

Bounds for the Remainder in the Integral Test Suppose {at} is a sequence of positive terms with at = f(!), wbere f is a continuous positive decreasing function of x for all x ;;,: n, and that l:a" converges to S. Then the remainder R. = S - s. satisfies the inequalities

1

00

1

00

f(x) fix ,,;; R. ,,;;

11+1

f(x) fix.

(1)

11

Ifwe add the partial sum s. to each side of the inequalities in (I), we get

1

00

s.

+

1

00

f(x) fix ,,;; S ,,;; s.

+

n+l

f(x) fix

(2)

n

since s. + R. = s. The inequalities in (2) are useful for estimating the error in approximating the sum of a convergent series. The error can be no larger than the length of the interval containing S, as given by (2).

EXAMPLE 5 n = 10.

Estimate the sum of the series l:(I/n 2 ) using the inequalities in (2) and

Solution We have that

Using this result with the inequalities in (2), we get SIO

Taking S10 = I equalities give

I<S< + IT - -

SIO

+1 10·

+ (1/4) + (1/9) + (1/16) + ... + (1/100)

'" 1.54977, these last in-

1.64068 ,,;; S,,;; 1.64997. If we approximate the sum S by the midpoint of this interval, we find that 00

I

~ 2: '" 1.6453.

n=1 n

The error in this approximation is less !ban half the length of the interval, so the error is • less !ban 0.005.

10.3 The Integral Test

557

Exercises 10.3 Applying the Integral Test Use the Integral Test to determine if the series in Exercises 1-10 con· verge or diverge. Be sure to check that the conditions of the Integral Test are satisfied 00 I 00 I

1. ~2 11 - 1 n

2. ~0:2 11"'1 n

I 3. ~-211=-1 n + 4

4. i-In +4

00

41

~ 00 • /I _ I

43.

B.

.-2

.-1 00

00

7.~+1Ft n + 4

In

(n 2)

10.

i

2

1I"'2n

D b.

n - 4 -2n+l

Give reasons for your answers. (When you check an answer, remem-

00

00

11. ~_I 1Ft 10"

00

i.

~ 00

29.

19.

11=1

Inn

00

21.

2'

~3'

22.

00

24'~2~1 11- 1 n

-2 ----=t=l n I

Vn(Vn +

27. I)

~

(lin)

.~, (lnn)Yln2 n - I

00 I 33. "5"-nsinJi

::1 00

+

II- I

(1 I( v;;-+l)) diverge.

Use the accompanying graph to show that the partia1 sum Sso =

~;'!,1

(I/( Vn+J)) satisfies

2'

25. ~----=t=l

• -1

00

n

1 Vx+l 51

I

dt

Conclude that 11.5

I (ln3)'

30.

~

32.

i n(1 + 1n n) .-1 I

< Sso < {SO. ~

Jo

2

<

12.3 .

2

1

f(x)= .~

Y ntan" 00

sso

1

vx + 1

::1

36'~-1+ • II - I e

<

dt.

vx+ I

y

I

00

34.

positive numbers? Give reasons for your answers. 46. (Continuation ofExercise 45.) Is there a "largesf' convergent series of positive numbers? Explain. B.

00

:x dt = I + Inn.

45. Is it true that if~:l On is a divergent series of positive numbers, then there is also a divergent series ~~ I hn of positive numbers with bIt < On for every n? Is there a "smallest" divergent series of

47. ~:~ 1

.-

'1

1

44. Are there any values of x for which ~:::'I(i/(nx)) converges? Give reasons for your answer.

00

&; 4' 5'+ 3

1

There is absolutely no empirical evideoce for the divergeoce of the hsrmooic seties even though we know it diverges. The partial sums just grow too slowly. To see what we mean, suppose you had started with SI = I the day the ooiverse was formed, 13 billion years ago, and added a new term every second. About how large would the partial sum s. he today, assuming a 365-day year?

11=2

00 ( I)" 28.~1+"

Inn

•

~_e_ 11 - 1 1 e ln

i ~n

I I I :xdtSI+z+"'+"

1

iVn 11=2

I

~ (ln2)'

31. £J

35.

.~1 nVn

00 -8 18. ~n

00

.-1

16.~~

Vn

00

26.

00

15. ~_3 .~1

~Vn ~

13'~n~1

11=1

00

11 - 0

00

12. ~e~

14. ~_5_ 1Ft n + 1

23.

1

'" I +

her that there may he more than one way to determine the seties' con· vergence or divergeoce.)

00

1

·+1

Determining Convergence or Divergence

20.

Draw illustrations lilre those in Figures 10.7 and 10.8 to

In(n+I)=

Which of the series in Exercises 11-40 cooverge, and which diverge?

-

0 0(1 42. ~ -11- 3 n - l n+l

+4

11=2

2

00

n

8. ~-n-

9. ~-"-11 - 1 e"/3

~

+2

n

I

00

2a)

_a_ _ _ I_)

6. ~--2 n(1n n)

5. ~ e-2•

17.

(

show that the partial sums of the hsrmooic seties satisfY the inequalities

11",1

00

00

Theory and Examples For what values of a, if any, do the seties in Exercises 41 and 42 converge?

I ••• I I I -+~-LJ-L-L-~-LJ-L-->x

o

2 3 4 5 ···48495051

00

38. 00

39. ~ sechn 11=1

~+11"'1 n + 1

b. What should n he in order that the partial sum

00

40. ~ sech2 n 11=1

s. = ~~~1

(

I/( v'i+J)) satisfY s. >

1000?

558

Chapter 10: Infinite Sequences and Series

48. ~:= 1 (l/n') converges a. Use the accompanying graph to determine the error if S30 = ~~~1 (I/n') is used to estimate the value Of~:::'l (I/n'). y

f(x)

29

x

31

I

00

I I 1+-+···+2 n

= 1,

30

I

00

c. ~--, d. ~--, ,-2 nln(n ) ,-2 n(lnn) 57. Euler's constant Graphs like those in Figure 10.11 suggest that as n increases there is little change in the difference between the sum

and the integral 32

... 33

x

Inn

["I dx x .

= Jl

To explore this idea, carry out the followiog steps. b. Find n so that the partial sum s, = ~~-1 (I/i') estimates the value Of~~l (l/n') with an error of at most 0.000001.

a. Bytakingf(x) In(n

49. Estimatethevalueof~:=l (I/n') to within 0.01 of its exact value. 50. Estimate the value of ~~2 (l/(n 2 act value.

+ 4)) to within 0.1

of its ex-

51. How many terms of the convergent series ~:-1 (l/nl.l) should be used to estimate its value with error at most 0.00001?

54. Use the Cauchy coodensatioo test from Exercise 53 to show that 00

I

a. ~ - - diverges; n:z2nlnn b.

~.~ cooverges if p

::1.

> I and diverges if p

'" I.

I

1,00 x(:x)P cooverges if and ooIy if p

an

1+

=

"2I + ... + IiI - Inn

is bounded from below and from above. h. Showthat

1'+11x

I I <, n+

dx = In (n

+ 1) - Inn,

and use this result to show that the sequence {a,} in part (a) is decreasing. Since. decreasing sequence that is bounded from below converges, the numbers a, deImed in part (a) converge: I

I

+ 2 + ... + " - Inn --> 1·

The number 1, whose value is 0.5772 ... ,is calledEuler~

(p a positive coostant)

>

I

Thus. the sequence

I

a. Show that the improper integral

I

0< In (n + I) - Inn '" I + 2 + ... + " - Inn '" 1.

55. Logarithmic p-series constant.

58. Use the Integral Test to show that the series

1.

b. What implicatioos does the fact in part (a) have for the convergence of the series

f-I-? ,-2 n(1n n)P . Give reasons for your answer. 56. (Continuation ofExercise 55.) Use the result in Exercise 55 to determine which of the following series converge and which diverge. Support your answer in each case. 00 1 00 1 a. ~n(lnn) b. ~n(lnn)1.01

10.4

I

+ I) '" I + 2+ "'+,,'" I + Inn

or

52. How many terms of the convergent series ~~,(I/n(lnn)') should be used to estimate its value with error at most 0.01? 53. The Cauchy condeusation test The Cauchy coodensatioo test says: l.ct {a,} be • nooincreasing sequence (a, '" a,+1 for all n) of positive terms that converges to O. Then ~an converges if and ooIy if ~2'a2" converges. For example, ~(l/n) diverges because ~2" (1/2') = ~I diverges. Show why the test works.

l/xintheproofofThe0ren39,showthat

=

converges.

59. a. For the series ~(I/n'), use the inequalities in Equatioo (2) with n = 10 to Imd an interval cootaining the sum S. b. Ai; in Example 5, use the midpoint of the interval found in part (a) to approximate the sum of the series. What is the maximum error for your approximation? 60. Repeat Exercise 59 using the series ~(I/n').

Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence of many more series by comparing their tenns to those of a series whose convergence is known.

10.4 Comparison Tests

THEOREM lO-The Comparison Test Let ~an. ~cn. and nonnegative terms. Suppose that for some integer N

~d"

559

be series with

n >N.

for all
y

Proof In Part (a), the partial sums of ~a" are bounded above by 00

M = at

+ a2 + ... + aN +

:L Cn • n=N+l

They therefore form a nondecreasing sequence with a limit L ,;; M. That is, if ~c" converges, then so does ~a". Figure 10.12 depicts this result, where each term of each series is interpreted as the area ofa rectangle Gust like we did for the integral test in Figure 10.11). In Part (b), the partial sums of ~a" are not bounded from above. If they were, the partial sums for ~d" would be bounded by 00

FIGURE 10.12 If the total area ~c" of the taller en rectangles is rmite, then

so is the total area ~all of the shorter a" rectangles.

M'

= dl +

d2

+ ... +

dN

+

~ a" n=N+l

•

and ~d" would have to converge instead of diverge.

EXAMPLE 1

We apply Theorem 10 to several series.


~ 5n

5 - I

diverges because its nth term

5

5n - I

=_1_>1 n-

I

n

S

is greater than the nth term of the divergent harmonic series. HisTORICAL BIOGRAPHY

Albert of Saxony (ca. \316-1390)

(b) The series

~l= I +l+l+l+ ... .-0 n! l! 2! 3! converges because its terms are all positive and less than or equal to the corresponding terms of

I

I

I

2

2

2

00

1+~n=I+I+-+2+"" "~O

The geometric series on the left converges and we have 00

I

I

+ ~ 2"

I =

I

+

I _ (1/2) = 3.

The fact that 3 is an upper bound for the partial sums of~;::'o (lin!) does not mean that the series converges to 3. As we will see in Section 10.9, the series converges to e.
5+~+1+1+ I + I + I + ... + I + ... 37 2+V1 4+V2 8+V3 2"+Vn converges. To see this, we ignore the first three terms and compare the remaining terms with those of the convergent geometric series ~:;'~0(i/2"). The term I/{ 2" + Vn) of

560

Chapter 10: Infinite Sequences and Series the truncated sequence is less than the corresponding term 1/2' of the geometric series. We see that tenD by term we have the comparison

1+

1

2+V1

+

1

4+0

+

1

8+v3

+ ... :51+1+1+1+ ....

2

4

8

So the truncated series and the original series converge by an application of the Comparison Test. _

The Limit Comparison Test We now introduce a comparison test that is particularly useful for series in which rational function of n.

THEOREM ll-Llmlt Comparison Test all n

2:

Suppose that a,

>

0 and b,

>

a. is a

0 for

N (N an integer).

a

1. If lim b, = c n

11_00

>

a

0, then

2. If lim b, = 0 and 11- 00

~a, and ~b, hoth converge or hoth diverge.

~b, converges, then ~a, converges.

II

a

3. If lim b, = 11_00

00

11

and ~b, diverges, then

~a. diverges.

Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 55a and b. Since c/2 > 0, there exisla an integer N such that for all n

-

Limit defInition with E = c/2,L = c,and a" replaced by a"fb,

n> N ~ I~ cl < I. Thus, for n

>

N,

If ~b, converges, then ~(3c/2)b, converges and ~a. converges by the Direct Comparison Test. If ~b, diverges, then ~(c/2)b, diverges and ~a. diverges by the Direct Comparison Test. _

EXAMPLE 2

Which of the following series converge, and which diverge?

(a) 1 + ~ + l

4

9

+ ~ + ... 16 25

(b) 1 + 1 + 1 +

1

3

7

~ + ... = 15

00

=

00

~ 2n + I = ~

,~l

(n + 1)'

14

2n + I n2 + 2n + 1

5: _1 2' - 1

,~l

(c) 1 + 2m 2 + 1 + 3m 3 + 1 + 4m 4 + ... 9

,~l

21

=

5: 1 + +n 5 n m

,~2

n2

10.4 Comparison Tests

561

We apply the Limit Comparison Test to each series. (8) Let a" = (2n + I )/(n 2 + 2n + I). For large n, we expect a" to behave like 2n/n2 = 2/n since the leading terms dominate for large n, so we let b. = I/n. Since 00 00 I ~b. = ~ n diverges

Solution

n- l

n- l

and lim a" = lim 2n2 + n = 2, bn 11--+00 n2 + 2n + 1

11--+00

~a. diverges by Part 1 of the Limit Comparison Test. We could just as well have taken b. = 2/n, but I/n is simpler.

(b) Let a. = 1/(2' - I). For large n, we expect a. to behave like 1/2", so we let b. = 1/2'. Since 00 00 1 ~bn = ~ 21l converges

and lim a. = lim ~ b ll 11_00 211 1

11_00

=

lim

.~OO

1 1 - (1/2')

= 1, ~a.

converges by Part I of the Limit Comparison Test. (0) Leta" = (1 + nlnn)/(n 2 + 5). For large n, we expecta. to behave like (nlnn)/n 2 = (lnn)/n,whichisgreaterthanl/nforn;;" 3,soweletb. = l/n.Since 00 00 I ~b. = ~ n diverges ,,=2

11=2

and

= 00, ~a.

diverges by Part 3 of the Limit Comparison Test.

EXAMPLE 3

00

Does ~

11=1

Inn 3/2

n

-

converge?

Because In n grows more slowly than n' for any positive constant c (Section 10.1, Exercise 105), we can compare the series to a convergent p-series. To get the p-series, we see that Inn nl /4 1 Solution

-3 2< =32 n

n

n'/4 3 for n sufficiently large. Then taking a. = (In n)/n /2 and b. = I/n'/" we have /

/

lim a. = lim Inn btl 11--+00 n 1/ 4

11_00

=

=

lim

l/n

c-~----;o,-;

.~OO (1/4)n 3/4

lim ;/4 II-con

=

Since ~b. = ~(I/n'/4)isap-serieswithp of the Limit Comparison Test.

l'H6pital's Rule

O.

> I,itconverges,so ~a.convergesbyPart2 _

562

Chapter 10: Infinite Sequences and Series

Exercises 10.4 Comparison Test

00

10 Exercises 1--1!, use the Comparison Test to determine if each series

37.~.

converges or diverges. I

00

1. ~ ~,.--o-::-::11=-1 n + 30 3.

~

I

'~'Vn-I 00

,

7.

~~n/4

+4

n

11=1

4.~~+2

41. ~-2' 11 - 1 n

1Ft

n - n

42.

~In(n: I)

I

00

8.~Vn+1

(Hint: First show that (lin!) '" (1/n(n - I» for n '" 2.) 00 I 00 • I ~ (n - I)! 45. ~ sm n 46. ~!ann 44. 11=-1 £- (n + 2)1. 00 tan-I n 00 -1 00 th 47. "5"_ -,-.,48 ")'_ sec n 49. ~_ co 2 n

,~'W+3

,-,

~ 00

50.

-

tanhn

~~n, + I

• 11=1

n +2 (Hint: Limit Comparison with ~:::, (1/Vn)) ~ n(n+1) 00 2' 11. ~ (n' + I)(n _ I) 12. ~ 3 + 4'

51.

n

1

~

nl.3 I

00

~-2-

53~

,-,

• ~

n

11=1

(Hint: Limit Comparisoo with ~::"~, (lin'» 10.

2"+311

43. ~., ~ n.

n-2 n2 + 3

n3

1

00 I 6. ~-3' 11=1 n

series converges or diverges.

~

+

3'

40·~3'+4'

oo2n _n

10 Exercises ~ 16, use the Limit Comparison Test to detennioe if each

11=-1

00

+ 3n 5n

n

3"-1

::1

~ ~ + I .-.L

39.

Umit Comparison Test

9.

(Xl

38. Y

+1

~ n-I 2. £ - - . - /I -I n + 2 11=2

5.~co:n 11=1 n /2

,I

~3n

00

~ .,r

~~

52.

lI=lnvn

I

+ 2 + 3 + ... + n

S4~ •

11=1

n

dr

11=1 n

1 + 22

I

+ 32 + ... + n 2

11=1

S' 13 ~.• .-=1 Vn4n 00 I 15. ~·-In ~ n 00

14.

~ 00

Theory and Examples 55. Prove (a) Part 2 and (h) Part 3 of the Limit Comparison Thst. S6. If ~:l all is a convergent series of nonnegative numbers, can anything be said about ~';,(aJn)? Explain.

(2nSn ++ 43)'

57. Suppose that a" > 0 and b, > 0 for n '" N (N an integer). If 1in3.~oo(a./b,) ~ 00 and ~a, converges, can anything be said

shout ~ b, ? Give reasons for yoor answer.

58. Prove that if ~all is a convergent series of nonnegative tenns,

(Hint: Limit Comparisoo with ~::"~, (lIn»

then ~a,2converges.

16. ~ In (I + 1,) n

59. Suppose that an

11=-1

(Hint: Limit Comparisoo with ~::"~, (lIn'»

20.

~

I

,~,

2Vn +

~

I +

-%

~osn

18. 21.

n

11=1

~

3 ,~, n + Vn

19.

~~

22.

n=t

23~

~ (3n:

S

26.

~~

29~.

31~.

32.

:.:1

n

I

":.:1 1 + 1nn

34.

~ 11=-1

n

,Vn

+

,~,

2

~n 1Ft

+ I n 2Vn

27.

j,1n(~n)

30.

~.----,p:-

00

• ~

35. 1

~ Sin:n

Sn'-3n , _, n'(n - 2)(n' + S)

(Inn)' 28. ~. - , 00

= 00.

,~OO

Prove that ~an diverges.

60. Suppose that a" > 0 and 1in3 n'a" ~ O. Prove that ~a, converges. 61. Show that ~::"~2 «In n)qlnP) converges for p>1.

-00


00

and

(Hint: Lin3it Comparison with ~:;;'2 lin' for I < r < p.) 62. (Continuation of Exercise 61.) Show that ~';2 «lnnY'/nP ) divergesfor-oo < q < 00 and 0

(Hint: Lin3it Comparison with an appropriate p-series.)

24.~

IOn + I • ,_, n(n + I)(n + 2)

25.

3n - 1

0 and lim an n~OO

Determining Convergence or Divergence Which of the series in Exercises 17-S4 converge, and which diverge? Use any method, and give reasoos for yoor answers. 17.

>

00

VnI Inn

~. In (n++1 1)

33.

~I

36.

~ n- l

n

-2,n n

(Inn)'

,t:( n

10 Exercises 63--68, use the results of Exercises 61 and 62 to deter-

mine if each series converges or diverges. 00 (In n)' 63. ~-4n

, -2 00

~. ~ 2

65. ~

~n+

67

~n /I _ I

n -1

2'

n22n

11=2

~

(In n)'OOO n

1.00'

I

. , -2 nl.l(lnn)'

00 (In n)'/S 66. ~----0:99

11=-2

n

10.5 The Ratio and Root Tests 70• •• Use Theorem 8 to show that

COMPUTER EXPLORATIONS 69. It is not yet known whether the series

s~f 1 +f(~. - 1 n(n + 1) .-1 n2 converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps.

where S ~ ~:I (l/n

L

1

2 3 11- 1 n sin n

2),the sum ofa convergentp-series. 1

00

S~I+L

• ~I

.

s.

What happens when you try to fmd the limit of as k --> 00 ? Does your CAS fmd a closed form answer for this limit? b. Plot the fITst 100 points (I<; s.) for the sequence of partial

2

n (n

+ 1)

•

c. Explain why taking the fITst M terms in the series in part (b) gives a better approximation to S than taking the fITSt M terms in the otigina1 series ~::'- 1 (I/n 2 ). d. The exact value of Sis known to be ".2/6. Which of the sums

sums. Do they appear to converge? What would you estimate the limit to be?

1000000 1

L ., .~1 n

c. Next plot the fITst 200 points (I<; s.). Discuss the behavior in your own words.

1 ) n(n + I)

b. From Example 5, Section 10.2, show that

a. Derme the sequence of partial sums s.~ •

563

1000

or

1+

1

L 'C2~'--'C .~1 n (n + 1)

gives a better approximation to S?

d. Plot the fITst 400 points (I<; s.). What happens when k ~ 355? Calculate the number 355/113. Explain from your calculation what happened at k ~ 355. For what values of k

would you guess this behavior might occur again?

10.5

The Ratio and Root Tests The Ratio Test measures the mte of growth (or decline) of a series by examining the mtio a.+J/a•. For a geometric series ~ar', this mte is a constant ({ar·+ 1)j{ar·) = r), and the series converges if and only if its mtio is less than 1 in absolute value. The Ratio Test is a powerful rule extending that result.

THEOREM 12-The Ratto Test suppose that

Let ~a. be a series with positive terms and

. a.+l_ 1un n~~ - p. -n

11_00

Then (8) the series converges if p < 1, (h) the series diverges if p finite, (c) the test is inconclusive if p = 1.

>

1 or p is in-

Proof (8) P < 1. Let r be a number between p and 1. Then the number E = r - p is positive. Since

a.+J/a. must lie within E of p when 11 is large enough, say for alln ;" N.1n particular, al'l+l a;;
=

r,

when n ;" N.

564

Chapter 10: Infinite Sequences and Series

1bat is,

< < <

aN+l aN+2 aN+3

raN, raN+ 1 raN+2

< <

2 r aN, r3 aN,

These inequalities show that the tenns of our series, after the Mh tenn, approach zero more rapidly than the terms in a geometric series with ratio r < I. More precisely,

consider the series

~cn,

where en

= an

2

for n = 1,2, ... , N and CN+l for all n, and

= raN, CN+2 =

r aN, ... ,CN+m = r"'aN, ... . Now Cln :5 en 00

~c. =

+ a2 + ... + aN-I + aN + raN + r 2aN + ...

al

n=l

+ a2 + ... + aN-I + aN(I + r + r2 + ...). I + r + r2 + ... converges because Ir I < I,

= al

The geometric series verges. Since an :5 en, ~an also converges. (b) 1

<

P

:S

so ~c. con-

oc. From some index M on,

> 1

an+l

a.

and

The terms of the serles do not approach zero as n becomes infinite, and the serles diverges by the nth-Term Test. (e) p = 1. The two serles 00

I

~" .-1

and

show that some other test for convergence must be used when p = I. ~ I For ~ ,,:

a.+1

~ I

a.+1

For ~

I/(n + 1) n I/n = n + I

a.

I/(n + 1)2 = l/n2

n2: a.

--> 1.

(_n_)2 --> + n

I

2= I

1.

In both cases, p = I, yet the first serles diverges, whereas the second converges.

_

The Ratio Test is often effective when the tenns of a series contain factorials of expressions involving n or expressions raised to a power involving n.

EXAMPLE 1 (a)

Investigate the convergence of the following serles.

~ 2'3~ 5

Solution

(2n)! n!n!

00

(b)

~

00

~

(c)

4' , ,

(2:)!'

We apply the Ratio Test to each series.

(a) For the serles ~~o (2' 1

a.+1

(2.+ (2'

a.

+

+ 5)/3.+ + 5)/3'

1

5)/3', =

1. 3

2.+ 1 + 5 = 1. (2 2' + 5 3 I

+ 5'2--) -->1. l + 5' 2-3 I

=

l 3.

The series converges because p = 2/3 is less than 1. This does not mean that 2/3 is the sum of the series. In fact, 00 2' + 5 00 (2)' ~ -.- = ~ 3 .-0 3

.~O

5

+ .~ n _03 00

=

I I - (2/3)

5

+ ------"'--c-oI - (1/3)

21 2'

10.5 The Ratio and Root Tests

565

(2n)! (2n + 2)! (b) If a, = -n.n. ,-, ' then a,+1 = ( n + 1)'( . n + I)'. and n!n!(2n + 2)(2n + 1)(2n)! (n + I)!(n + 1)!(2n)!

a,+1 a,

(2n + 2)(2n + I) 4n + 2 = ------>4 (n + I)(n + I) n + I .

=

The series diverges because p = 4 is greater than I. (c) If a, = 4'n!n!/(2n)!, then

4'+I(n + I)!(n + I)! (2n)! (2n + 2)(2n + 1)(2n)!· 4'n!n!

a.+l

a. =

4(n + I)(n + I) 2(n + I) (2n + 2)(2n + I) = 2n + I ..... I.

Because the limit is p = I, we cannot decide from the Ratio Test whether the series converges. When we notice that a.+1/a, = (2n + 2)/(2n + I), we conclude that a.+l is always greater than a. because (2n + 2)/(2n + 1) is always greater than I. Therefore, all terms are greater than or equal to a, = 2, and the nth term does not ap_ proach zero as n ..... 00. The series diverges.

The Root Test The convergence tests we have so far for ~a, work best when the fonnula for a, is relatively simple. However, consider the series with the terms

n/2', a, = { 1/2',

nodd

n even.

To investigate convergence we write out several terms of the series:

fa=l+l+~+l+~+l+2+ ... 3 5 6

, - 1'

2'

22

113

2

24

2

15

2

27

17

= 2 + 4 + 8 + 16 + 32 + 64 + 128 + ... Clearly, this is not a geometric series. The nth term approaches zero as n ..... 00, so the nth-Tenn Test does not tell us if the series diverges. The Integral Test does not look promising. The Ratio Test produces

a,+1 =

a.

12~'

n+ I 2 '

n odd n even.

All n ..... 00, the ratio is alternately small and large and has no limit. However, we will see that the following test establishes that the series converges.

THEOREM 13-The Root Test and suppose that

Let

~a,

be a series with

a. ;",

Then (8) the series converges if p < I, (b) the series diverges if p finite, (0) the test is inconclusive if p = I.

0 for n ;", N,

>

lor P is in-

566

Chapter 10: Infinite Sequences and Series

Proof (8) P < 1. Choose an e > 0 so small that p + e < I. Since %;; ---> p, the terms %;; eventually get closer than e to p. In other words, there exists an index M 2: N such that

%;;
whenn2:M.

Then it is also true that

< (p + e)'

a"

forn2:M.

Now, ~:M (p + e)', a geometric series with ratio (p + e) comparison, ~:M a" converges, from which it follows that

""

~an

<

I, converges. By

"" + ... + aM-l + ~an

= at

n-l

n- M

converges.

< P :S "". For all indices beyond some integer M, we have %;; > I, so that a. > I for n > M. The terms of the series do not converge to zero. The series di-

(b) 1

verges by the nth-Term Test. (c) p = 1. The series ~:1 (I/n) and ~:;"~1 (l/n2) show that the test is not conclusive when p = 1. The first series diverges and the second converges, bot in both cases %;; ---> I. _

. the senes · · WI t h terms an = CODSl'der agam

EXAMPLE 2

{n/2" /.

I 2,

nodd

n even.

Does ~a. converge?

Solution We apply the Root Test, finding that

%: = {"'I;/n/2, a.

nodd n even.

1/2,

Therefore,

t ,; %;; ,; fn· Since "'I;/n ---> 1 (Section 10.1, Theorem 5), we have Iim"~",, %;; = 1/2 by the Sandwich _ Theorem. The limit is less than I, so the series converges by the Root Test.

EXAMPLE 3 ""

(a)

2

~ ~.

(b)

.-1

Solution

Which of the followiog series converge, and which diverge?

~ 2: n

"" (

~

(c)

n=-1

1

1

+n

).

We apply the Root Thst to each series.

"" n 2

(a) ~ 2' converges because _1

"" 2' .

(b) ~ --, diverges because ,,=1 n

""(I)'

(c) ~ -+I ,,=1 n

1f.n W ("'I;/ny 12 2' = _.r.:;. = --2- ---> 2 < I. 2

v~

~2' --, n

2

r)3

= ( •• Vn

converges because

2

---> --,

1

> I.

~(1)' 1In ---> 0 < I. -+I = -+ n

-

10.5 The Ratio and Root Tests

567

Exercises 10.5 Using the Ratio Test In Exercises 1-8, use the Ratio Test to determine if each series con-

verges or diverges. 00 2' 1. /I - I n. ~ (n - I)! 3. ~ 2 ,~I (n + I) S. 7.

2.

~

in~2 00

4.

4

6.

~

37.

3

11",1

~:II 00

+ 2)!

8

2"+1 n3',-1

f. 3Inn i (2n + 3)n5'In (n + I) 11

2

+

Using the Root Test In Exercises 9-16, use the Root Thst to detennine if each series con· verges or diverges. 00 7 00 4' 10. (3 )" 9. 11 - 1 (2n + 5 )' 11"'1 n

L

L

3n - 5

11 - 1

13 ~ • ,;';:1 (3 15.

i

,-I

+

(I -

8 (1In))2n

14.

i i

(In(e sin'

2

+

k) r'

(~)

k)'"

(Hint: lim (I + xln)' ,~OO

= e")

I

00

16.

12.

"5'.

~n

1+,

Determining Convergence or Divergence In Exercises 17-44, use any method to detennine if tire series con·

verges or diverges. Give reasons for your answer. 00 V2 00 17.

"5' n ,

:.:1

00

00

19. Ln!e-' 11=1

21.

L InO' ,-I 00

23.

~

2

+ (-I)' 1.25'

00 ( 25'LI-

,-I

27.

i

~ Inri,

22.

i (n; 2)' ,-I

3)'

n

(-2)' , 24. ,-I 3 00

L-

OO(

I)'

26'LI-11"'1

3n

(Inn)' L-,n 00

In;

28.

11 - 1 n

(I I)

II - I

(I I )'

29. ~ li- n 2

30. ~ Ii - n 2

~. ~n

~_ n~n

00

31. 33.

32.

,;":1 ~ ~

/I - I

(n

00

+ I)(n + 2) , n.

,;":1

+ I)!

3'n.,

00

,

~n 00

L L

40

L n ,~2 (In n )(,/2)

42.

3' L--'--;; n2

•

00

11- 1

(n!? (2n)'.

00

43.

~

n2'(n

11 _

38. "5' n;

I)!

,,=-2

L

,,"" I

~

44.~ ,-I

(2n + 3)(2' + 3) 3'+2

Recunively Def"med Terms Which of the series ~;l an defmed by the formulas in Exercises 45-54 converge, and which diverge? Give reasons for your answers. all +l =

al =

1,

a,,+1 =

a, =

I

47.

3'

""+I =

48.

al

49.

al =

2,

SO.

a, =

5,

51.

a,

= 3,

1 + sinn n

all

1 + tan-I n

46.

n

a"

3n - I

2n + 5 "" n

an+1 = n

+ 1 a"

n

+ Inn

= I,

I

2'

=

a,,+1 =

a, t, ""+1 54. a, t, ""+I 53.

,

20.

I.

00

,

~ (2n:

52. al

18. Yn 2e-

~ 2

oo

36.

(Inn )' n 00 'In 41 n. n . ,_I n(n + 2)!

39.

• ,~I

3211

11. i (4n + 3)'

+ 3)! 3!n!3"

(n

00

11=2

n2 (n n!

~ 00

L, 00

00

35.

n

+ 10

an

~

=

=

=

= (",,),+1

Convergence or Divergence Which of the series in Exercises 55--62 converge, and which diverge? Give reasons for your answers. 00 2' , , 00 (3n)! 55 L~ 56. ~ n!(n + I)!(n + 2)! • ,~I (2n)! 00 (n!)" 57. II- I n ')2

L( 00

59. Y

,

n( ')

:.:1 2 00

IJ

1.3 .... '(2n - 1) 4"2IJn !

61.

~

62.

~ [2 -4- ...• (2n)](3' + I)

00 00

1-3- ... '(2n -I)

568

Chapter 10: Infinite Sequences and Series

Theory and Examples 63. Neither the Ratio Test nor the Root Thst helps with p-series. Try them an

n/2',

65. Let a, = { 1/2', Does

I ~11=1 nP 00

~an

if n is a prime number otherwise.

converge? Give reasons for your answer.

66. Show that ~:_ I 2(.i')/n! diverges. Recall from the Laws ofExponents that 2('~ = (2')'.

and show that both tests fail to provide information shout convergence. 64. Show that neither the Ratio Test nor the Root Test provides information about the convergence of

(p constant).

10.6

Alternating Series, Absolute and Conditional Convergence A series in which the terms are alternately positive and negative is an alternating series. Here are three examples:

1

1

1

1

2 + 3 - 4 + 5 - ... +

I -

(_1),+1

n

+ ...

(1)

1 1 1 (-1)'4 -2 + 1 - 2 + 4 - 8 + ... + ----z'" + ... 1- 2

(2)

+ 3 - 4 + 5 - 6 + ... + (-I)·+l n + ...

(3)

We see from these examples that the nth term of an alternating series is of the fonn

an

= (_1).+l u•

or

a. = (-I)·u.

where u. = Ia.1 is a positive number. Series (1), called the alternating harmonic series, converges, as we will see in a moment Series (2), a geometric series with ratio r = -1/2, converges to -2/[1 + (1/2)] = -4/3. Series (3) diverges because the nth term does not approach zero. We prove the convergence of the alternating barmonic series by applying the Alternating Series Test. The Test is for convergence of an alternating series and cannot be used to conclude that such a series diverges.

THEOREM 14-The Alternating Series Test (Leibniz's Test)

The series

00

~(_1)'+IU. = UI - U2

+ U3

-

U4

+ ...

n=1

converges if all three of the following conditions are satisfied: 1. The u. 's are all positive. 2. The positive un's are (eventuslly) nonincreasing: u. ;;" U.+I for all n ;;" N, for some integer N.

3. u...... O. Proof Assume N = I. If n is an even integer, say n = 2m, then the sum of the {lISt n tenns is S2m =

(UI - U2)

+

(U3 -

= UI - (U2 - U3) -

U4) (U4 -

+ ... +

(U2m-1 -

us) - ... -

U2m)

(U2m-2 -

U2m-I) -

U2m.

10.6 Alternating Series, Absolute and Conditional Convergence

569

The first equality shows that S2m is the sum of m nonnegative terms since each tenn in parentheses is positive or zero. Hence S2m+2 ;;,: S2m, and the sequence {S2m} is nondecreasing. The second equality shows that S2m ,;; U1. Since {S2m} is nondecreasing and bounded from above, it has a limit, say (4)

If n is an odd integer, say n = 2m S2m+l

=

82m

+

I, then the sum of the first n terms is

+ U2m+l. Sinceun~O,

and,asm----i>00,

+

S2m+l = 82m

U2m+l --+ L

+0=

Iim,,~oo Sn ~

Combining the results of Equations (4) and (5) gives Exercise 131).

EXAMPLE 1

(S)

L.

L (Section 10.1, _

The alternating harmonic series 00

L(-l)n+11 =

n~l

I

n

_1 + 1_1 + ... 2

3

4

clearly satisfies the three requirements of Theorem 14 with N verges.

~

I; it therefore con-

-

Rather !ban directly verifying the defInition Un ;;,: Un+1o a second way to show that the sequence {un} is nonincreasing is to define a differentiable function j(x) satisfying j{n) = Un. That is, the values of j match the values of the sequence at every positive integer n. If j' (x) ,;; 0 for all x greater than or equal to some positive integer N, then j(x) is nonincreasing for x ;;,: N. It follows that j(n) ;;,: j(" + I), or Un ;;,: Un+ 10 for" ;;,: N.

EXAMPLE 2 Consider the sequence where Un ~ IOn/(,,2 + 16). Detme IOx/(x 2 + 16). Then from the Derivative Quotient Rule, j '() X

j(x) ~

2

~

1O{16 - x ) <: 0 (x 2 + 16)' -

whenever x

~

4.

It follows that Un ;;,: Un+1 for " ;;,: 4. That is, the sequence {un} is nonincreasing for _

n;;':~

+", "2

+", ~-U4-

o

x

"

FIGURE 10.13 The partial sums of an alternating series that satisfies the bypotheses of Theorem 14 for N ~ 1 straddle the limit from the beginning.

"

A graphical interpretation of the partial sums (Figure 10.13) shows how an alternating series converges to its limit L when the three conditions of Theorem 14 are satisfied with N ~ I. Starting from the origin of the x-axis, we lay off the positive distance Sl ~ U1. To tmd the point corresponding to S2 ~ U1 - U2, we back up a distance equal to U2. Since U2 ,;; U1, we do not back up any farther than the origin. We continue in this seesaw fashion, backing up or going forward as the signs in the series demand. But for n ;;,: N, each forward or backward step is shorter than (or at most the same size as) the preceding step because Un + 1 ~ Un. And since the nth term approaches zero as n increases, the size of step we take forward or backward gets smaller and smaller. We oscillate across the limit L, and the amplitude of oscillation approaches zero. The limit L lies between any two successive sums s. and s.+ 1 and bence differs from Sn by an amount less !ban U.+ 1. Because

IL -

Snl

<

U n +1

forn

~

N,

we can make useful estimates of the sums of convergent alternating series.

570

Chapter 10: Infinite Sequences and Series

THEOREM lS-Tbe Alternating Series Estimation Tbeoll!m

If the alternating series ~:1 (-I)o+l uo satisfies the three conditions of Theorem 14, then for n ~N, So = U1 -

U2

+ ... + (- 1)0+ 1uo

approximates the sum L of the series with an error whose absolute value is less than Uo +1, the absolute value of the first unused term. Furthermore, the sum L lies between any two successive partial sums 8" and 8 11 + 1, and the remainder, L - So, bas the same sign as the first unused term.

We leave the verification of the sign of the remamder for Exercise 61.

EXAMPLE 3

We try Theorem 15 on a series whose sum we know:

i(-I),l = I - 1 + 1_ 1 + l _ l + l __1_ i + _1__ ... 2' 2 4 8 16 32 64 128 i 256 .

,~O

The theorem says that if we truncate the series after the eighth term, we throw away a total that is positive and less than 1/256. The sum of the flISt eight terms is s. = 0.6640625 and the sum of the first nine terms is S9 = 0.66796875. The sum of the geometric series is I 1-(-1/2)

I 3/2

2 3'

andwenotethatO.6640625 < (2/3) < 0.66796875. The difference, (2/3) - 0.6640625 = 0.0026041666 ... ,is positive and is less than(I/256) = 0.00390625. •

Absolute and Conditional Convergence We can apply the tests for convergence studied before to the series of absolute values of a series with both positive and negative terms.

DEFINITION A series ~a" converges absolutely (is absolutely convergent) if the corresponding series of absolute values, ~I a,l, converges. The geometric series in Example 3 converges absolutely because the corresponding series of absolute values

il=I+1+1+1+ ...

,~ ~

2

4

8

converges. The alternating harmonic series does not converge absolutely because the corresponding series of absolute values is the (divergent) harmonic series.

DEFINITION A series that converges but does not converge absolutely converges conditionally. The alternating harmonic series converges conditionally. Absolute convergence is important for two reasons. First, we bave good tests for convergence of series of positive terms. Second, if a series converges absolutely, then it converges, as we now prove.

10.6 Alternating Series, Absolute and Conditional Convergence

571

00

If ~ Ia,1 converges, then

THEOREM 16-The Absolute Convergence Test 00

L a. converges. 11=1 Proof For each n, -I a,1 ,;; a, ,;; Ia. I,

so

If ~::'~,la,1 converges, then ~::,~,2Ia,1 converges and, by the Direct Comparison Test, the nonnegative series ~:;;'da, + la,l) converges. The equality a, = (a, + la,l) -la,1 now lets us express ~:;;'I a. as the difference of two convergent series: 00

00

00

00

La, = L(a, + 1a.1-la,l) = L(a, + la,l)

n=1

n=1

L la,l· n=1

-

n=1

-

Therefore, l::1 an converges.

Caution We can rephrase Theorem 16 to say that every absolutely convergent series converges. However, the converse statement is false: Many convergent series do not converge absolutely (such as the alternating harmonic series in Example 1).

EXAMPLE 4

This example gives two series that converge absolutely.

00

(8) For

L (-1 )'+ 1 ~ = n

,-I

1-

1 + 1 - ~ + "', the corresponding series of absolute 4

9

16

values is the convergent series

il=I+1+1+~+ ... n2 4 9 16 .

,~I

The origiual series converges because it converges absolutely.

~sinn sin1+sin2 sin3 . b 0 th pOSItive . . and nega-= - + - + . . . which contains (b) For ~ 2 1'1=- 1 n 1 4 9 ' tive tenDs, the corresponding series of absolute values is

il l

,-I

sin2n

=

n

Isinll

+

Isin21

1

+ "',

4

which converges by comparison with ~::'~I (1/n 2 ) because Isinnl ,;; 1 for every n. The origiual series converges absolutely; therefore it converges. _

EXAMPLE 5 Ifp is a positive constant, the sequence {l/nP } is a decreasing sequence with limit zero. Therefore the alternatingp-series 00

~

(_1),-1 1 1 1 nP =1-2P+Y-4P+"',

converges. If p > 1, the series converges absolutely. If 0 tionally. Conditional convergence: Absolute convergence:

p>O

< P ,;; 1, the series converges condi-

1 __1_+_1___1_+ ...

v'2

v'3

v'4

1 __1_+_1___1_+ ... 23(2

33/2

4 3/ 2

-

572

Chapter 10: Infinite Sequences and Series

Rearranging Series We can always rearrange the terms of a finite sum. The same result is true for an inimite series that is absolutely convergent (see Exercise 68 for an outline of the proof).

THEOREM 17-The Reammgement Theorem for Absolutely Convergent Series If ~:::, a. converges absolutely, and bj, b2, ... , b., ... is any arrangement of the sequence {a.} , then ~b. converges absolutely and 00

00

Lb. = n=1 La.·

11=1

If we rearrange the terms of a conditionally convergent series, we get different results. In fact, it can be proved that for any real number r, a given conditionally convergent series can be rearranged so its sum is equal to r. (We omit the proof of this fact.) Here's an example of summing the terms of a conditionally convergent series with different orderings, with each ordering giving a different value for the sum.

EXAMPLE 6 We know that the alternating harmonic series ~:::, (-I).+I/n converges to some number L. Moreover, by Theorem 15, L lies between the successive partial sums 82 = 1/2 and 83 = 5/6, so L '¢' O. If we multiply the series by 2 we obtain

2L=2~

=2(1_1+1_1+1_1+1_1+1_~+~_ ...) .-1(_1).+1 n 2 3 4 5 6 7 8 9 10 11 212121212 =2-1 +3-2+5-3+7-4+9-5+0- .. ·.

Now we change the order of this last sum by grouping each pair of tenns with the same odd denominator, but leaving the negative terms with the even denominators as they are placed (so the denominators are the positive integers in their natural order). This rearrangement gives

(2 - 1) -

! + (t -t) - t + (t -t) - i + (% - t) - i + ... (1 -! + t - t + t - i + t - i + ~ - 1~ + 1\ _ .. -) 00

L 11=1

(-1).+1 n

= L.

So by rearranging the terms of the conditionally convergent series ~:::,2( _I).+I/ n, the series becomes ~:;O~I ( -1)'+ '/ n, which is the alternating harmonic series itself. If the two • series are the same, it would imply that 2L = L, which is clearly false since L '¢' O. Example 6 shows that we cannot rearrange the terms of a conditionally convergent series and expect the new series to be the same as the original one. When we are using a conditionally convergent series, the terms must be added together in the order they are given to obtain a correct result. On the other hand, Theorem 17 guarantees that the terms of an absolutely convergent series can be summed in any order without affecting the result.

Summary of Tests We have developed a variety of tests to determine convergence or divergence for an infinite series of constants. There are other tests we have not presented which are sometimes given in more advanced courses. Here is a summary of the tests we have considered.

10.6 Alternating Series, Absolute and Conditional Convergence

573

1. The 11th-Term Test: Unless a, --> 0, the series diverges. 2_ Geometric series: ~ar' converges if Ir I < I; otherwise it diverges. 3. p-series: ~ I/n" converges if p > I; otherwise it diverges.

4. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test or the Limit Comparison Test. ~la,1 converge? If yes, so does ~a, since absolute convergence implies convergence.

5. Series with some negative terms: Does 6. Alternating series:

~a. converges if the series satisfies the conditions of the

Alternating Series Test.

Exercises 10.6 Determining Convergence or Divergence

00

In Exercises 1-14, detennine if the alternating series converges or

27. L( -1)'n'(2/3)'

diverges. Some of the series do not satisfY the conditions of the Alternating Series Test.

29.

,~I

::1

4. L ( - I ) ' - )' ( 11=2 In n 00 '5 6. L(-I),+I

;-±n + 4

11=1

10"

(Xl

8.

,&1 (-I)' (n + I)!

10. i(-l)n+I_I-

Inn

11"'2

12. i(-l)nln(1

14.

~(-I)'+I

+

k)

3v;;-:t:l" Vn+1

~

00

33.

00

37.

00

19.

L (- 1),+1

1Ft 00

21.

++ n

1

I

(-I)' ,~I I + Vn

n; n

5

+n +n

00 (-2),+1 24·L +S' 11"'1 n

25. i(_I),+11

~n

26. L( - I)n+I("'\Ylo)

23. i(-I)'+l 3 11 - 1

11=-1

n

00

,-I

11-

(_1),-1

~,c'-='---1 n + 2n + 1

00

36. ,~_. CO~n1T ~

(-I)'(n + I)' (2n)n

00

38.

~

(-I),+I(n!)' (2n)!

00 (nl)'3' 40. ~(-I)' (2n.+ I)!

00

00

11=1

11=1

41. L(-I)'(v;;-:t:l" - Vn)42. L(-I)'(~ - n) 00

43. L(-I)'(Vn

+ Vn - Vn)

11=1

44.

L _r 11""1

(-I)' _ r-:--c

00

45. L(-I)' sechn

,-I

vn+ vn+l

00

46. L(-I)'cschn

,-I I

I

I

I

I

I

"4 - 6 + "8 - 10 + IT - 14 + ...

48. I +

1111111 2S + 36 - 49 - 64 + ...

"4 - "9 - 16 +

•

22. L(-I)n sm,n 11=1

~

L

2

11=1 00

~(-I)' n + 3

I

00

20. L( - I)n+l

00

34.

00 ,(2n)! 39. ~(-I) 2"n!n

47.

18. i

~

35. ~_ cos mr ~ nVn

n-I

17. i(-I),_I,~I Vn

32. Y(-st'

(-100)' I n.

00

Which of the series in Exercises 15-48 converge absolutely, which converge, and which diverge? Give reasons for your answers. 00 00 (0 I)' 15. L(_I),+I(O.I)' 16. L(-I)'+I---j,

n

~ I

00

Absolute and Conditional Convergence

~-1

L /I _ I

Inn

n - Inn

00

31. Y(-I)'

4

00

3. i(-I)'+I-.!.". 1Ft n3

11- 1

00

2. i(-I)'+I+ 11=1 n /2

Vn

30. i(-I)'

1

n

nlnn

11=2

i( -I)' ~-I+ n

II- I

1. i(_I),+I_I-

28. i(_I),+I_I-

,-I

Errnr Estimation In Exercises 4g...S2, estimate the magoitude of the error involved in using the sum of the first four tenns to approximate the sum of the entire series.

574

Chapter 10: Infinite Sequences and Series is the same as the sum of the ftrSt n terms of the series

N. you will see in Section 10.7, the sum is In(l.Ol).

52. I

~t~

i.(

I

-I)'t',

0< t

<

I

~(-I)' 11=1

55.

_2_1n +3

~(_I)'+I

54.

1Ft

~(-I)'+1 (n + ~Y;;)'

+n + 1

~(-I)' In(ln(! + 2))

56.

65. Show that if~::;' 1 a" and ~::;, 1 b, brrth converge shsolutely, then so do the following.

.. ~(a. + b,)

b. ~(a, - b,)

,-I

,-I

c. ~ ka" (k any number)

As you will see in Section 10.9, the sum is cos l,thecosine ofl radian.

1

57. ~(-1)'(2 )' 11"'0 n .

b. Use Theorem 17 to find the sum of the series in part (a).

66. Show ~ example that ~~=1 a"b ll may diverge even if ~~=1 all and ~'_I b, brrth converge. 67. In the alternating harmonic series, suppose the goal is to arraoge the terms to get a new series that converges to -1/2. Start the new arraogement with the fifllt negative term, which is -1/2. Whenever you have a sum that is less than or equal to -1/2, start introducing positive terms, taken in order, until the new total is greater thao -1/2. Then add negative terms until the total is less than or equal to -1/2 again. Continue this process until your partial sums have been above the target at least three times aod fInish at or below it. If 9" is the sum of the fIrst n terms of your new series, plot the points (n, so) to illustrate how the sums are behaving.

The!intit L of ao alternating series that satisfies the conditions of Theorem 14 lies between the values of any two consecutive par-

68. OutHne of the proof of the Rearrangement Theorem (Theorem17)

~(-I)'J, n.

As you will see in Section 10.9, the sum is e-1 .

11- 0

Theory and Examples 59. L The series

1_1 +

l _ l + l _ l + ... + l _ l + ... 32942783'2'

does not meet one of the conditions ofTheorern 14. Which

one?

D 60.

I

64. Show that if ~::;'I a, converges absolutely, then

tude less than 5 X 10....

58.

I

63. Show that if~::;'1 a, diverges, then ~:;O_ lla,1 diverges.

D Approximate the sums in Exercises 57 and 58 with an error of magni00

I

Do these series converge? What is the sum of the fIrSt 2n + I tenns of the fIrst series? If the series converge, what is their swn?

In Exercises 53-56, determine how many terms should be used to estimate the sum of the entire series with an error ofless than 0.001.

53.

I

1·2 + 2·3 + 3·4 + 4'5 + 5·6 + ....

tial sums. This suggests using the average Sli

+2

3 11 +1 _

-

911

a. Let e be a positive real number, let L = l:~- 1 an. and let

+ 1 (-1),+2

an+1

2

D

Sk

=

~!""1 an· Show that for some index Nl and for some

index N2

O?::

Nt,

to estimate L. Compute

E

ISN, - LI < 2' I

I

s20+2'2[ as an approximation to the sum of the alternating harmonic series. The exact sum is In2 ~ 0.69314718 .... 61. The sign of the remainder of an alternating series that satisfies the conditions of Theorem 14 Prove the assertion in Theorem 15 that whenever ao alternating series satisfying the conditions of Theorem 14 is approximated with one ofits partial sums, then the remainder (sum of the unused terms) has the sarne sign as the first unused term. (Hint: Group the remainder's terms in consecutive pairs.)

62. Show that the sum of the flfst 2n terms of the series 111111111

1-2+2-J+J-~+~-J+J-~+'"

Since all the terms ah a2 •••• , aN: appear somewhere in the sequence {b,}, there is an index N, '" N2 such that if n ~ N3 • then (:~k- l bk ) - SNI is at most a sum of terms elm withm O?:: NI. Therefore,ifn O?:: N3,

:5

~la.1 k - Nt

+ ISN, - LI <

E.

b. The argument in part (a) shows that if~:;O_ 1 a, converges absolutely then ~~ I bn converges and ~~ 1 bn = ~: 1 an· Now show that because ~::;'lla,1 converges, ~::;'llb,1 converges to ~:;O~II a" I.

10.7

575

Power Series

Power Series

10.7

Now that we can test many infinite series of numbers for convergence, we can study sums that look like "infinite polynomials;' We call these sums power series because they are de[med as infmite series of powers of some variable, in our case x. Like polynomials, power series can be added, subtracted, multiplied, differentiated, and integrated to give new power series.

Power Series and Convergence We begin with the formal definition, which specifies the notation and terms used for power series.

A power series about x

DEFINmONS

= 0 is a series of the form

00

~Cnxn

=

Co

+ CIX + C2X2 + ... + cnx" + .. ".

(1)

11=0

A power series about x = a is a series of the form 00

~c.(x - a)' = Co

+ Ct(x - a) + C2(X - af + ... + c.(x - a)' + ...

(2)

11=0

in which the center a and the coefficients Co, CI, cz, ... , en, . .. are constants.

Equation (I) is the special case obtained by taking a ~ 0 in Equation (2). We will see that a power series defines a function I(x) on a certain interval where it converges. Moreover, this function will be shown to be continuous and differentiable over the interior of that interval.

EXAMPLE 1

Taking all the coefficients to be I in Equation (I) gives the geometric

power series 00

~x'~ I +x+x 2 + .. ·+x·+ .. ·. 11=0

This is the geometric series with [lISt term I and ratio x. It converges to 1/(1 - x) for I. We express this fact by writing

Ix I <

1

--~I+x+x

I-x

2

+"'+x• + ...

'

-I<x
(3)

•

I

Reciprocal Power Series

1

00

--~ ~x', 1- x /FO

Ixl<

1

Up to now, we have used Equation (3) as a formula for the sum of the series on the right. We now change the focus: We think of the partial sums of the series on the right as polynomials p.(x) that approximate the function on the left. For values of x near zero, we need take only a few terms of the series to get a good approximation. As we move toward x ~ I, or -I, we must take more terms. Figure 10.14 shows the graphs of I(x) ~ 1/(1 - x) and the approximating polynomials y. ~ Pix) for n = 0, 1,2, and 8. The function I(x) ~ 1/(1 - x) is not continuous on intervals containing x ~ I, where it has a vertical asymptote. The approximations do not apply when x 20 I.

576

Chapter 10: Infinite Sequences and Series

1

Y=l-x

5 4

Y2=1+x+xZ

3

~Y'~I+X

2

Yo -I

=

1

x

0

FIGURE 10.14 The grnphs of f(x) ~ 1/(1 - x) in Example I and four of its polynomial approximations.

EXAMPLE 2

The power series

(_i)"

1 (x

- 2) + 1 (x - 2)' + ... + (x - 2)" 2 4 2

I -

+ ...

(4)

= 2, Co = I, c, = -1/2, C2 = 1/4, ... , c" = (-1/2)". This

matches Equation (2) with a

is a geometric series with flIst term I and ratio r = - x ; 2 . The series converges for

IX;21

<

lorO

<x < 4. The sum is I I - r

I 1+ x - 2 2

(x - 2)' 4 - ...

+ ( -2

2

x'

so y

x2 =

I -

(x - 2) 2

+

I)"

(x - 2)"

+ ... ,

0< x < 4.

Series (4) generates useful polynomial approximations of f(x) = 2/x for values of x near 2:

Y2~3-¥+~

2

Yo ~ 1 o

2

Po(x) = I

Y~ ~

(2, I)

?Y' ~ 2 _ ~

P,(x) = I -

- 2) = 2 - "2

I( x - 2 ) P2() x = I - 2

3

FIGURE 10.15 The graphs of f(x) ~ 2/x and its [lISt three polynomial approximations (Example 2).

12 (x

+ 4I(x

3x - 2 )2 = 3 - 2

2

x + 4'

•

and so on (Figure 10.15).

The following example illustrates how we test a power series for convergence by using the Ratio Test to see where it converges and diverges.

EXAMPLE 3 00

(al ~(-I) "~,

For what values of x do the following power series converge? n-l

xn

x2

x3

2

3

,,=x--+--'"

10.7

00

(b) L(-I)

.-1

(c)

x" L = • ~O n! 00

x 2n - t 2n - 1

n-l

X3

xS

3

5

Power Series

577

--=x--+--'" X2

X3

2!

3!

+ x + - + - + ...

1

00

(d) Ln!x' = 1

+ x + 2!x2 + 3!x 3 + " .

•- 0

Solution Apply the Ratio Test to the series ~I u.l, where u. is the nth tenn of the power series in question. (Recall that the Ratio Test applies to series with nonnegative terms.)

The series converges absolutely for Ix I < 1. It diverges if Ix I > 1 because the nth tenn does not converge to zero. At x = 1, we get the alternating harmonic series 1 - 1/2 + 1/3 - 1/4 + "', which converges. At x = -1, we get -1 - 1/2 1/3 - 1/4 - "', the negative of the harmonic series; it diverges. Series (a) converges for - 1 < x ,;; 1 and diverges elsewhere. I

o

-1

x2n+1

(b)

I2n + 1

.2n -

11

2(.

x 2n - 1

+ 1) - 1 ~ 2. + 1

The series converges absolutely for x 2 < 1. It diverges for x 2 > 1 because the nth tenn does not converge to zero. At x = 1 the series becomes 1 - 1/3 + 1/5 - 1/7 + "', which converges by the Alternating Series Theorem. It also converges at x = -1 because it is again an alternating series that satisfies the conditions for convergence. The value at x = - 1 is the negative of the value at x = 1. Series (b) converges for - 1 ,;; x ,;; 1 and diverges elsewhere.

(c)

IU~:11

'x

o

-1

X·+I

n!

I(n + 1)!'x'

I=

n

Ixl

+

1--+0foreveryx.

n! (.+1)!

1·2·3···n 1·2·3 .. ·.·(.+1)

The series converges absolutely for all x.

o (d)

IU~:ll

(n

I

+

1)!x'+ nIx'

l

l=

(n

+

1)lxl--+oounlessx = O.

The series diverges for all values of x except x = O.

•

o

, x

•

The previous example illustrated how a power series might converge. The next result shows that if a power series converges at more than one value, then it converges over an entire interval of values. The interval might be finite or inf'mite and contain one, both, or none of its endpoints. We will see that each endpoint of a finite interval must be tested independently for convergence or divergence.

578

Chapter 10: Infinite Sequences and Series

THEOREM 18-The Convergence Theorem for Power Series If the power series co La,x' ~ ao + a,x + a2x2 + ... converges at x ~ c # 0, then it converges ,-0 absolutely for all x withlxl < lei. If the series diverges atx = d, then it diverges for all x with Ixl > Idl.

Proof The proof uses the Comparison Test, with the given series compared to a converging geometric series. Suppose the series ~:.o a.c' converges. Then lim,~co a.c' = 0 by the nth-Term Test. Hence, there is an integer N such that Ia.c'l < 1 for all n > N, so that forn>N. Now take anyx such thatlxl

<

(5)

Icl, so thatlxl/lc I < 1. Multiplying both sides of Equation (5)

by Ix I' gives la,llxl'

<

lxi'

~

forn

> N.

r

_______ - _____

-Idl

-R

.

-lei

•

0

lei

____ -

R

~

x

Idl

FIGURE 10.16 Convergence of ~a"x' at x = c implies absolute convergence on the interval -I c I < x < IcI; divergence at x ~ d implies divergence forlxl > Idl. The corollary to Theorem 18 asserts the

existence of a radius of convergence R '" O.

Since Ixlc I < 1, it follows that the geometric series ~~o Ixl c converges. By the Comparison Test (Theorem 10), the series ~~o la,llx'l converges, so the original power series ~~o a,x' converges absolutely for -I c I < x < Ie I as claimed by the theorem. (See Figure 10.16.) Now suppose that the series ~::'~o a,x' diverges at x ~ d. If x is a number with Ixl > Idl and the series converges at x, then the first half of the theorem shows that the series also converges at d, contrary to our assumption. So the series diverges for all x with Ixl > Idl· • To simplify the notation, Theorem 18 deals with the convergence of series of the form - a)' we can replace x - a by x' and apply the results to the series ~a.(x')'. ~a,x'. For series of the form ~a,(x

The Radius of Convergence of a Power Series The theorem we have just proved and the examples we have studied lead to the conclusion that a power series ~c.(x - a)' behaves in one of three possible ways. It might converge only at x ~ a, or converge everywhere, or converge on some interval of radius R centered at x ~ a. We prove this as a Corollary to Theorem 18.

COROLLARY TO THEOREM 18 The convergence of the series ~c,(x - a)' is described by one of the following three cases: 1. There is a positive number R such that the series diverges for x with Ix - a I > R but converges absolutely for x with Ix - a I < R. The series may or may not converge at either of the endpoints x ~ a - R and x ~ a + R. 2. The series converges absolutely for every x (R ~ 00). 3. The series converges at x ~ a and diverges elsewhere (R = 0).

10.7

Power Series

579

Proof We first consider the case where a = 0, so that we have a power series ~:o c"x' centered at O. If the series converges everywhere we are in Case 2. If it converges only at x = 0 then we are in Case 3. Otherwise there is a nonzero number d such that ~:;"~o c.d· diverges. Let S be the set of values of x for which ~:o c"x' converges. The set S does not include any x with Ixl > since Theorem 18 implies the series diverges at all such values. So the set S is bounded. By the Completeness Property of the Real Numbers (Appendix 7) S has a least upper bound R. (This is the smallest number with the property that all elements of S are less than or equal to R.) Since we are not in Case 3, the series converges at some number b oF 0 and, by Theorem 18, also on the open interval (-Ibl, Ibl). ThereforeR > O. Iflxl < R then there is a number c in S with Ixl < c < R, since otherwise R would not be the least upper bound for S. The series converges at c since c E S, so by Theorem 18 the series converges absolutely at x. Now suppose Ixl > R. If the series converges at x, then Theorem 18 implies it converges absolutely on the open interval (-lxi, Ixl), so that S contains this interval. Since R is an upper bound for S, it follows that Ix I :5 R, which is a contradiction. So if Ix I > R then the series diverges. This proves the theorem for power series centered at a = O. For a power series centered at an arbitrary point x = a, set x' = x - a and repeat the argument above replacing x with x' . Since x' = 0 when x = a, convergence of the series ~:olc'(x') I' on a radiusR open interval centered at x' = 0 corresponds to convergence oftheseries~:olc.(x - a)I' onaradiusR open interval centeredatx = a. _

14

R is called the radius of convergence of the power series, and the interval of radius R centered at x = a is called the interval of convergence. The interval of convergence may be open, closed, or half-open, depending on the particular series. At points x with Ix - a I < R, the series converges absolutely. If the series converges for all values of x, we say its radius of convergence is infinite. If it converges only at x = a, we say its radius of convergence is zero.

How to Test a Power Series for Convergence

1. Use the Ratio Test (or Root Test) to find the interval where the series converges absolutely. Ordinarily, this is an open interval Ix - al

I

Test each endpoint ofthe (rmite)

2.

interval of convergence.


or

a - R

< x < a + R.

Ifthe interval of absolute convergence is finite,

test for convergence or divergence at each endpoint, as in Examples 3a and b. Use a Comparison Test, the Integral Test, or the Alternating Series Test.

3.

Ifthe interval ofabsolute convergence is a - R < x < a + R, the series diverges for Ix - a I > R (it does not even converge conditiona1ly) because the nth term does not approach zero for those values of x.

Operations on Power Series On the intersection of their intervals of convergence, two power series can be added and subtracted term by term just like series of constants (Theorem 8). They can be multiplied just as we multiply polynomials, but we often limit the computation of the product to the 1rrst few terms, which are the most important. The following result gives a formula for the coefficients in the product, but we omit the proof.

580

Chapter 10: Infinite Sequences and Series

THEOREM

A(x)

=

19-The Series Multiplication Theorem for Power Series

~::'~o a.x· and B(x) = ~:;;,o b.x· converge absolutely for Ixl

c. = aob.

If

< R, and

•

= ~akb.-.,

+ a,b.-, + a2b.-2 + ... + an-,b, + anbo

k~O

then ~::'~o c.x· converges absolutely to A(x)B(x) for Ixl

< R:

Finding the general coefficient c. in the product of two power series can be very tedious and the term may be unwieldy. The following computation provides an illustration of a product where we find the first few terms by multiplying the terms of the second series by each term of the fITst series:

00 ) • (00 .+1 ) ~(-l)'_x( ~x' ~o ~o n + I Multiply second series ...

(x -

f + X; - .. -) +

(x2 -

by!

~ + X; - .. -) +

t

(x3 -

by.

+

X; - .. -) + ...

by.' and gather the IllS! four powers.

We can also substitute a function j(x) for x in a convergent power series.

If ~:;;'o a.x· converges absolutely for Ix I < R, then ~:o an (j(x))" converges absolutely for any continuous function j on Ij(x) I < R.

THEOREM

20

Since I/(i - x) = ~:;;'ox' converges absolutely for Ixl < I, it follows from Theorem 20 that 1/(1 - 4x 2 ) = ~::'_o(4x2)'convergesabsolutelyforI4x21 < 1 orlxl < 1/2. A theorem from advanced calculus says that a power series can be differentiated term by term at each interior point of its interval of convergence. THEOREM 21-The Term-by-Term Differentiation Theorem

radius of convergence R

>

If ~cn(x - a)' has

0, it defines a function

00

j(x)

= ~cn(x

- a)'

on the interval

a - R

<

x

<

a

+ R.

11=0

This function j has derivatives of all orders inside the interval, and we obtain the derivatives by differentiating the original series term by term:

00

j'(x)

= ~ncn(x 11=1

r(x)

= ~n(n

- a)'-"

00

- l)c.(x - a)n-2,

n=2

and so on. Each of these derived series converges at every point of the interval

a-R<x
10.7 EXAMPLE 4

Power Series

581

Find series for f'(x) and f"(x) if

+ x + x 2 + x 3 + x4 + ... + xn + ...

f(x) = _1_ = I I - x 00

=

Lx n,

<x<

-I

I

11=0

Solution

We differentiate the power series on the right term by term:

f'(x) =

I (I -

+ 21: + 3x 2 + 4x 3 + ... + ""n-l + ...

I

= X)2

<x<

-I

f"(x) =

2 (I - x)3

=

2

I;

+ 6x + 121:2 + ... + n(n - l)x n - 2 + ...

00

=

Ln(n - l)x n - 2 ,

<x<

-I

•

I

n=2

Caution Term-by-term differentiation might not work for other kinds of series. For example, the trigonometric series

i

sin (~!x) n

1'1 =- 1

converges for allx. But if we differentiate term by term we get the series

i

n!cos (n!x)

n2

n=1

'

which diverges for all x. This is not a power series since it is not a sum of positive integer powers ofx. It is also true that a power series can be integrated term by term throughout its interval of convergence. This result is proved in a more advanced course.

THEOREM 22-The Term-by-Term Integration Theorem

Suppose that

00

f(x) = LCn(x - a)n n- O

converges fora - R

< x < a + R (R > 0). Then 00

~CIl converges for a - R

J EXAMPLE 5

n+ 1

< x < a + R and f(x)

fora - R

(x _ a)n+l

ax =

00

~Cn

(x - a)n+l n + I

+C

< x < a + R. Identify the function 00

f(x)=~

(-I)n X2n+l 2n+1

x3 =x-

3

x' +j-"',

-I,,;;x";;l.

582

Chapter 10: Infinite Sequences and Series Solution

We differentiate the original series term by term and get f'(X)=I-X 2 +X 4 _X 6 + ... , -I<x
This is a geometric series with trrst term I and ratio -

2 X ,

so

We can now integrate f'(x) = 1/{1 + x 2 ) to get

j

f'(X)

j~ +

ax =

x2

I

= tan-I

x + C.

The series for f(x) is zero when x = 0, so C = O. Hence

1"-4 = tan-II = n-=o2n+l ~ (-I)'

f(x) = x -

x3

xS

x7

3 + 5 - 7 + ... =

_

-I<x
tan I x,

(6)

It can be shown that the series also converges to tan-I X at the endpoints x = ± I, but we omit the proof. _

Notice that the original series in Example 5 converges at both endpoints of the original interval of convergence, but Theorem 22 can guarantee the convergence of the differentiated series only inside the interval.

EXAMPLE 6

The series I 2 -I 3 + ... --=1-1+1 I +1

converges on the open interval -I

1n{1+x)=

l'

< 1<

I. Therefore,

12+ 13- - + 14 ' " ]' - -I d l = I - 01+1 234 0

=x-

x2

x3

Theorem 22

X4

Z + 3 - 4 +'"

or

I

00

(_1),-1 ~ n 00

I ln2

=

In (I

,-I

+ x)

=

~

(_1),-1 x' n '

-I<x
n=1

It can also be shown that the series converges at x = I to the number In 2, but that was not guaranteed by the theorem. _

Exercises 10.7 Intervals of Convergence

00

In Exercises 1-36, (a) fmd the series' radius and interval of coovergence. For what values ofx does the series converge (b) absolutely, (0) conditionally'/ 00

2. ~(x 11=0

00

3. ~(-I)'(4x

,-0 00

.:=to

(x - 2)'

10"

+

00

I)"

4. ~

,-I

+ 5)'

(-I)"x' 11.~ ,

12.

(3x - 2)"

00

6. ~(2x)'

00

10.

00

13.

n.

~ 4n~2lI 00

~

00

14. "

15 "5'. ----;c~x==

. .:=toW+]

~

11"'1 (X,l

16.

+ 2)'

n (x - I)"

Vn

f 311~n 11"'0

,-I

(-I)'(x

,-I

1 nVn3n

11 - 0

n

00

8. ~

,

9. ~_x_ 11 -

11=0

5."5'

00

00

1. ~x'

,

7. ~--"'---11 - 0 n + 2

n.

(x - I)' 3"

n 3

(_l)"x n+l

Y~_~,­

:=6 vn+3

10.7

00

17.

~ 00

+ 3)'

n(x

00

,

18 ~

.

5' ~JI

nx &0 4'(n 2 + I)

2 )' 47.~ ~ 00

48.

(

11=0

3

Power Series

583

2 )' ~ £-=-! 00

(

11=0

2

00

20. ~ %(lx

19. ~-3'

,-I

, -0

+ 5)'

Theory and Examples 49. For what values ofx does the series

00

21.

~

,-I 00

22. "5'.

~

23.

(2 + (-I)')-(x + 1),-1

I - 3) + 4(x I - 3)2 + ... + ( -2I)'(x - 3)" + ... 1- 2(x

(-I)' 32o(x - 2)' 3 n

~ (I + k)" x'

00

24. ,&1(lnn)x'

00

00

26. ~n!(x - 4)'

25. Lnllx"

,-I 00

27. ~

,-0 (_

11=-1

1),+I(x + 2)' 2' n

00

29.

,

~_X_

,-2n(lnn)'

SO. If you integrate the series in Exercise 49 term by term, what new series do you get? For what values of x does the new series converge. and what is another name for its sum? 51. The series

. x 3 x S x' x9 smx = x - 3! + 5! - 7! + 9! -

11=0

Get the information you need about ~ I/(o(ln 0)') ftmn Section 10.3,

xli

TiT +

...

converges to sin x for all x.

Exercise 55.

a. Find the flISt six terms of a series for cos x. For what values of x should the series converge?

~_x_

Get the infonnation you need about ~ 1/(0 In n) ftmn Section 10.3,

b. By replacing x by lx in the series for sin x, fmd a series that

n=2nlnn

Exercise 54.

00

30.

00

28. ~(-2)'(n + I)(x - I)'

converge? What is its sum? What series do you get if you differentiate the given series term by term? For what values of x does the new series converge? What is its sum?

,

00

32.

~

+

1),+1 2n+2

(3x

converges to sin lx for all x. c. Using the result in part (a) and series multiplicatioo, calculate the first six tenns of a series for 2 sin x cos x. Compare your answer with the answer in part (b). 52. The series x2

x3

x4

XS

e' = I + x + 2! + 3! + 4! + 5! +

converges to e~ for all x . •• Find a series for (d/dx)e'. Do you get the series for e'? Explain your answer. b. Find a series for eX dx. Do you get the series for eX? Explain your answer.

35. 00

36. ~

,-I

(v;;-+i - Vn)(x -

J

3)'

1:.

In Exercises 37-40, find the series' radius of convergence.

~ n! II 37. ,-I ~ 3'6'9 ... 3nx 2.4'6 ... (2n) 38. ,-I 2·5·8 .. ·(3n - I) 00 (n!)' 39. ,&12'(2n)! x'

L

00

(

53. The series

)2 x"

tanx

00

(x - 1)20 4'

43.

~

45.

i (Vx2 - I)'

+

lx'

15 +

17x'

6lx'

315 + 2835 + ...

c. Check your result in part (b) by squsring the series given for sec x in Exercise 54. 54. The series

00

, -0

x'

3

b. Find the frrst five terms of the series for sec2 x. For what values of x should this series converge?

In Exercises 41-48, use Theorem 20 to fmd the series' interval of coovergence and. within this interval, the sum of the series as a function ofx. 41. ~ 3'x'

+

•• Find the fll'St five terms of the series for In 1sec x I. For what values of x should the series converge?

(Hint: Apply the Root Test)

,, - 0

=x

converges to tan x for -'If/2 < x < 'If/2.

40'~(n~Jx'

00

Replace x by -x in the series for eX to fmd a series that converges to e -x for all x. Then multiply the series for eX and e-x to find the fITst six terms ofa series for e-x· eX.

42. ~ (e' - 4)'

,-0 00

44.

~

(x

+ 1)20 9'

00

46. ~(lnx)'

_

secx-I+

x2

2

5 4 61. 277 8 + 24 x + 720 x + 8064 x + ...

converges to sec x for -'If/2 < x < 'If/2. •• Find the fll'St five terms of a power series for the functioo In 1sec x + tan x I. For what values of x should the series converge?

584

Chapter 10: Infinite Sequences and Series b. Show that if ~~=o all xll = 0 for all x in an open interval (-c, c), then a" = 0 for every n.

b. Find the ftrst four terms of a series for sec x tan x. For what

values ofx should the series converge? c. Check your result in part (b) by multiplying the series for

56. The sum of the series ~::"-o (11'/2') To fmd the sum of this series, express 1/(1 - x) as a geometric series, differeotiate both sides of the resulting equatioo with respect to x, multiply both sides of the result by x, differeotiate again, multiply by x agsin, and set x equal to 1/2. What do you get?

secx by the series given for tan x in Exercise 53. 55. Uniqueness of convergent power series L

Shaw that if two power series ~~=o anx" and ~:o bnx"are convergent and equal for all values ofx in an open interval (-c, c), thea a. = b. for every n. (Hint: Let I(x) = ~::"~oa.x· = ~:o b.x·. Differeotiate term by term to show that a. and b. both eqoa1 I(·)(O)/(n!).)

10.8

Taylor and Maclaurin Series This section shows how functions that are infmitely differentiable generate power series called Taylor series. In many cases, these series can provide useful polynomial approximations of the generating functions. Because they are used routinely by mathematicians and scientists, Taylor series are considered one of the most important topics of this chapter.

Series Representations We know from Theorem 21 that within its interval of convergence the sum of a power series is a continuous function with derivatives of all orders. But what about the other way around? If a function I(x) has derivatives of all orders on an interval I, can it be expressed as a power series on I? And if it can, what will its coefficients be? We can answer the last question readily if we assume that I(x) is the sum of a power series 00

I(x) = ~a.(x - a)' =

ao

+ al(x -

a)

+ a,(x

+ ... + a.(x

- a)'

- a)'

+ ...

with a positive radius of convergence. By repeated term-by-term differentiation within the interval of convergence I, we obtain I'(x) = al

+ 2a,(x

/,,(x) = I' 2a,

- a)

+ 3a3(X

+ 2' 3a3(X

I"'(x) = 1·2· 3a3

- a)

+ 2·3· 4",,(x

- a)'

+ ... + na,,(x

+ 3' 4",,(x -

a)'

+ ... ,

+ ... ,

+ 3·4· 5as(x

- a)

- a)·-l

- a)'

+ ... ,

with the nth derivative, for all n, being

"'l(x)

=

lI!a.

+ a sum of terms with (x - a) as a factor.

Since these equations all hold at x

=

a, we have

/,,(a) = I· 2a"

I"'(a) = 1'2'3a3,

and, in general,

"'l(a)

=

n!a•.

These formulas reveal a pattern in the coefficients of any power series ~::"-o an(x - a)' that converges to the values of Ion I (''represents I on 1"). If there is such a series (still an open question), then there is only one such series, and its nth coefficient is

"'l(a)

an = -----"y-.

585

10.8 Taylor and Maclaurin Series

If I has a series representation, then the series must be

I(x)

=

I(a)

/,,(a)

+ f'(a){x - a) + ~ (x - a)' I"l(a)

+ ... + --,-(x - a)" + ....

(1)

n.

But if we start with an arbitrary function I that is infinitely differentiable on an interval I centered at x = a and use it tn generate the series in Equation (1), will the series then converge tn I(x) at each x in the interior of I? The answer is maybe-for some functions it will but for other functions it will not, as we will see.

Taylor and Maclaurin Series The series on the right-hand side of Equation (1) is the most important and useful series

we will study in this chapter.

HIsTORICAL BIOGRAPIllES

Brook Taylor (1685-1731) Colin Maclaurin (1693-1746)

DEFINmONS Let I be a function with derivatives of all orders throughout some interval containing a as an interior point. Then the Taylor series generated by/atx = ais 00

Ikl(a)

/,,(a)

~ ~(x -

a)' = I(a) + f'(a)(x - a) + ~(x - a)' I("l(a)

+ ... + - - I-(x - a)" + .... n.

The Maclaurin series generated by I is 00 l(kl(O) ~ ~xk = 1(0)

1"(0)

l"l(O)

+ f'(O)x + ~X2 + ... + ----;!Ix" +"',

the Taylor series generated by I at x =

o.

The Maclaurin series generated by I is often just called the Taylor series of I.

EXAMPLE 1 Find the Taylor series generated by I(x) = l/x at a = 2. Where, if anywhere, does the series converge to l/x? Solutton

We need tu find 1(2), 1'(2), /"(2), .... Taking derivatives we get

so that

1( 2)

= Z-I =

-21, 1'(2)

= -

1"(2)

212 '

~ =

_ 2 3

=

l"l(2) n!

1 23 '

(-1)" 2",+1 .

The Taylor series is

1(2)

/,,(2)

l"l(2)

2!

n!

+ f'(2){x - 2) + - - ( x - 2)2 + ... + - - ( x - 2)" + ... =

1 _ (x - 2) + (x - 2)' _ ... + (-1)" (x - 2)" + ... 2

22

23

2"+1

586

Chapter 10: Infinite Sequences and Series This is a geometric series with first tenn 1/2 and ratio r = -(x - 2)/2. It converges absolutely for Ix - 21 < 2 and its sum is 1/2 I

+ (x -

I (x - 2)

+

2

2)/2

I

x·

In this example the Taylor series generated by I(x) = I/x at a = 2 converges to I/x for Ix-21 <2orO<x<4. •

Taylor Polynomials The linearization of a differentiable function I at a point a is the polynomial of degree one given by Pt(x) = I(a)

+

f'(a)(x - a).

In Section 3.9 we used this linearization to approximate I(x) at values ofx near a. If I bas derivatives of higher order at a, then it bas higher-order polynomial approximations as well, one for each available derivative. These polynomials are called the Taylor polynomials of j.

DEFINITION Let I be a function with derivatives of order k for k = I, 2, ... , N in some interval containing a as an interior point. Then for any integer n from 0 through N, the Taylor polynomial of order n generated by I at x = a is the polynomial Pn(x) = I(a)

+

f'(a)(x - a)

y

/"(a) + ~ (x +

- a)'

f-k)(a) -L-, -(x -

a)k

+ ... + ... +

A.

f-n)(a)

- - I-(x

n.

- a)n.

3.0

We speak of a Taylor polynomial of order n rather than degree n because f- n)( a) may be zero. The tlISt two Taylor polynomials of I(x) = cos x at x = 0, for example, are Po(x) = I and Pt(x) = I. The first-order Taylor polynomial bas degree zero, not one. Just as the linearization of I at x = a provides the best linear approximation of I in the neighborhood of a, the higher-order Taylor polynomials provide the "besf' polynomial approximations of their respective degrees. (See Exercise 40.)

EXAMPLE 2

Find the Taylor series and the Taylor polynomials generated by I(x) = eX

atx = O. Since f-n)(x) = eX and f-n)(O) = I for every n = 0, 1,2, ... , the Taylor series generated by I at x = 0 (see Figure 10.17) is

Solution --~~--~--~~--~--->x

-0.5

0

0.5

1.0

The graph of f(x) and its Taylor polynomials FIGURE 10.17

Pl(X) = I P2(X) = I P,(x) = I

1(0)

= e'

+x + x + (x 2/2!) + x + (x 2/2!) + (x'/3!).

Notice the very close agreement near the center x = 0 (Example 2).

+

f'(0)x

/"(0)

f-n)(O)

2!

n!

+ - - x 2 + ... + - - xn + ... x2 2

x" n!

= I +x+-+···+-+··· 00

=

k

L~,· k- O •

This is also the Maclaurin series for eX. In the next section we will see that the series converges to eX at every x.

10.8 Taylor and Maclaurin Series

587

The Taylor polynomial of order n at x = 0 is

x2

Xli

EXAMPLE 3

•

+".+,. n.

P.(x)=I+x+2

Find the Taylor series and Taylor polynomials generated by I(x) = cosx

atx = O.

Solution

The cosine and its derivatives are

I(x) =

cos x,

f"(x) =

-cos x,

=

-sin x,

p(x) =

sin x,

I'(x)

1o.)(x) = (-I)n cosx, At x = 0, the cosines are I and the sines are 0, so

10.+1)(0) =

10.)(0) = (-I)n,

o.

The Taylor series generated by 1 at 0 is

1(0)

+ 1'(04 + l'i!O) x 2 + f"~~0) x3 + ... + I:~o) xn + ... = I

+ O'x

00

=

~

(_

x2

- 2!

x4

x2n

+ 0'x 3 + 4! + ... + (-I)n (2n)! + ...

I)kx2k

(2k)!

.

This is also the Maclaurin series for cos x. Notice that only even powers of x occur in the Taylor series generated by the cosine function, which is consistent with the fact that it is an even function. In Section 10.9, we will see that the series converges to cos x at every x. Because 10.+1)(0) = 0, the Taylor polynomials oforders 2n and 2n + I are identical: x2

Po.(x) = Po.+1(X) = I - 2!

+

x4

4! - ...

x2n

+ (-I)n (2n)!

.

Figure 10.18 shows how well these polynomials approximate I(x) = cosx near x = O. Only the right-hand portions of the graphs are given because the graphs are symmetric • about the y-axis. y

2

~--~~~--~--~~~--~~~~--~~X

o

-I

-2

FIGURE 10.18

The polynomials n

P2n(x) = ~

(-I)<x2k (2k)!

converge to cos x as n ~ 00. We can deduce the behavior of cos x arbitrarily far away solely from knowing the values of the cosine and its derivatives at x = 0 (Example 3).

588

Chapter 10: Infinite Sequences and Series

It can be shown (though not easily) that

EXAMPLE 4 y

j(x) =

o {e~l/x',

x = 0 x oF 0

(Figure 10.19) has derivatives of all orders at x = 0 and that I-n)(o) means that the Taylor series generated by j at x = 0 is -----'---~+...L-'---------'----x -2

-1

0

= 0 for all n. This

1"(0) I-n)(o) j(O) + 1'(04 + ~ x 2 + ... + ----;;! xn + ...

2

= 0 + O·x + 0·x 2 + ... + O·x n + ... =0+0+···+0+···.

The graph of the continuous extension of y = e-1/;;- is so flat at the origin that all of its derivatives there are zero (Example 4). Therefore its Taylor series is not the function itself. FIGURE 10.19

The series converges for every x (its sum is 0) but converges to j(x) only at x = O. That is, the Taylor series generated by j(x) in this example is not equal to the function j(x) itself. •

Two questions still remain.

1.

For what values of x can we normally expect a Taylor series to converge to its generating function?

2.

How accurately do a function's Taylor polynomials approximate the function on a given interval?

The answers are provided by a theorem of Taylor in the next section.

Exercises 10.8 Finding Taylor Polynomials

25. f(x) = x' + x 2 + I,

In Exercises 1-10, find the Taylorpolynomia1s of orders 0,1,2, and 3 generated by f at a.

26. f(x)

1. f(x) = e"', a = 0 3. f(x) = Inx, a = 1

4. f(x) = In (I + x),

5. f(x) = I/x,

6. f(x) = I/(x

a= 2

7. f(x) = sinx, a = '1T/4 9. f(x) = Yx, a = 4

a= 0

2. f(x) = sinx,

8. f(x)

=

+

tanx,

2),

a

=

10. f(x) =~,

a= 0 a = 0

'1T/4 a= 0

finding Taylor Series at x = 0 (Maclaurin Series) Find the Maclaurin series for the functions in Exercises 11-22. 11. e-x 12. xe~ 13. I

1

16. sin ~

17. 7 cos (-x)

18. 5 cos '1TX

19. cosh x 21. x' -

=

eX

z.3 -

+ e--;t" 2

5x

+4

.

20. sinhx =

22. x

x2

+

=

a = I 1/(1 - x)3, a

=

-

2,

a

=

-I

0

29. f(x) = e', a = 2 30. f(x) = 1', a = I

31. f(x) = cos (z. + ('1T/2», a = '1T/4 32. f(x) = Vx+1, a = 0

Theory and Examples 37. Use the Thylor senes generated by e' at x = a to show that

eX-e-x 2

e'

I

Finding Taylor and Maclaurin Series In Exercises 23-32, fmd the Taylor senes generated by f at x 23. f(x) = x 3 - Z. + 4, a = 2 24. f(x) = z.3 + x 2 + 3x - 8, a = 1

=

l/x 2,

a = -2

+ z.3 + x 2

verges absolutely. 33. f(x) = cosx - (2/(1 - x» 34. f(x) = (I - x + x 2 ) e' 35. f(x) = (sin x) In (1 + x) 36. f(x) = x sin2 x

I - x

15. sin 3x

3x' - x'

In Exercises 33-36, fmd the first three nonzero terms of the Maclaurin series for each function and the values of x for which the series con-

14. 2 + x

+x

27. f(x) 28. f(x)

=

=

a.

= eO

[I + (x _ a) + (x ;1 a)' + ... ].

38. (Continuation of Exercise 37.) Find the Thylor senes generated by e' at x = 1. Compare your answer with the fonnu1a in Exercise 37. 39. Let f(x) have derivatives through order n at x = a. Show that the Taylor polynomial of order n aod its fIrSt n derivatives have the same values that f and its first n derivatives have at x = a.

589

10.9 Convergence of Taylor Series 40. Apprmrlmation properties of Taylor polynomials Suppose that j(x) is differentiable on an ioterval centered at x ~ a and that g(x) ~ bo + bl(X - a) + ... + bo(x - a)' is • polynomial of degree n with constant coefficients ho, ... , b". Let E(x) = j(x) - g(x). Show that ifwe impose on g the conditions

i) E(a) = 0

The approximation error is zcroatx

ii) lim ( E(x) ). x-a X -

a

0,

=

Thus, the Taylor polynomial p.(x) is the only polynomial of degree less than or equal to n whose error is both zero at x = a and negligible when compared with (x - a)'. Quadrati. Approsimation. The Taylor polynomial of order 2 geoerated by a twice-differeotiable function j(x) at x ~ a is called the quadratic approximation of j at x ~ a. In Exercises 41-46, fmd the (a) lioearization (Taylor polynomial of order I) and (h) quadratic approximation of j at x ~ O.

a.

The error is negli.gible when compared to (x - a)".

then

f"(a) g(x) ~ j(a) + f'(a)(x - a) + 2i(x - a)' +

~

e.mx

41. j(x) ~ In (coox)

42. j(x)

43. j(x) ~ 1/\.f1=7 45. j(x) ~ siox

44. j(x) ~ coshx

46. j(x)

~

tanx

j(')(a) +--,-(x-a)o. n.

10.9

l~c_ on_v_ e~ ~ en_c_e_o_ fl_a~Y_ LO_ r_ se_n_ 'e_ s ______________________-----In the last section we asked when a Taylor series for a function can be expected to converge to that (generatiog) function. We answer the question in this section with the following theorem.

/0)

If j and its first n derivatives j', !", ... , are continuous on the closed interval between a and b, and is differentiable on the open interval between a and b, then there exists a number c between a and b such that

THEOREM 23-Taylor's Theoll!m

f(b)

=

/0)

!"(a)

f(a) + j'(a)(b - a) + ~ (b - a)2 + ...

+

AO)(a)

_l'_ _ (b _

n!

a

)"

+

Ao+l)(C) (b _ )"+1 + 1)1 a .

l'

(n

Taylor's Theorem is a generalization of the Mean Value Theorem (Exercise 45). There is a proof of Taylor's Theorem at the end of this section. When we apply Taylor's Theorem, we usually want to hold a fixed and treat b as an independent variable. Taylor's formula is easier to use in circumstances like these if we change b to x. Here is a version of the theorem with this change.

Taylor's Formula If f has derivatives of all orders in an open interval I contaioing a, then for each positive integer n and for each x in I,

f(x) ~ f(a)

!"(a) (x - a)2 + ... + j'(a)(x - a) + ~ /o)(a)

0

+ --, -(x - a) + Ro(x) , n.

(1)

where

R.{x)

=

/o+I)(C) (n + I)! (x - a)o+1

for some C between a and x.

(2)

590

Chapter 10: Infinite Sequences and Series When we state Taylor's theorem this way, it says that for each x

f(x)

=

E

I,

P.{x) + R.(x).

1.+

1 The function R.(x) is detennined by the value of the (n + I)st derivative ) at a point c that dependa on both a and x, and that lies somewhere between them. For any value of n we want, the equation gives both a polynomial approximation of f of that order and a formula for the error involved in using that approximation over the interval I. Equation (I) is called Taylor's formula. The function R.(x) is called the remainder of order n or the error term for the approximation of f by p.(x) over I.

If R.(x) --+ 0 as n --+ 00 for all x E I, we say that the Taylor series generated by f at x = a converges to f on I, and we write 00

f(x)

=

Ik)(a)

L -It'-(x k- O •

a)k.

Often we can estimate R. without knowiog the value of c, as the followiog example illustrates.

EXAMPLE 1 Show that the Taylor series generated by f(x) f(x) for every real value of x. Solution

= eX at x = 0 converges to

The function has derivatives of all orders throughout the interval I = = eX and a = 0 give

(-00,00). Equations (I) and (2) withf(x)

x2 xn e X = I +x+-+···+-+R(x) 2! n! n

Polynomial from Section 10.8, Example 2

and

_ eC 11+1 R.(x) - (n + 1)1 x

for some c between 0 and x.

Since eX is an increasing function of x, e' lies between eO = I and eX. When x is negative, soisc,ande' < I. When x is zero, eX = landR.(x) = O.Whenxispositive,soisc,and e' < eX. Thus, for R.(x) given as above,

Ixl·+ 1

IR.(x) I ,;; (n + 1)1

whenx

~

0,

and

x n+ 1

IR.(x) I < eX (n + 1)1

when x

>

O.

Finally, because X ·+

lim

.~OO (n

1

+ 1)1

= 0

for every x,

Section 10.1, Theorem 5

lim R.{x) = 0, and the series converges to eX for every x. Thus, .~OO

(3)

I The Number. as • Serie. 00

• We can use the result of Example I with x = I to write

1

e= ~/FO n!

e

=

I

+ I + -2!I + ... + -n!I + Rn (I),

10.9 Convergence of Taylor Series

591

where for some c between 0 and I, R.(I)

C

= e (n

11)! < (n 11)!'

Estimating the Remainder It is often possible to estimate R.{x) as we did in Example I. This method of estimation is

so convenient that we state it as a theorem for future reference.

THEOREM 24-The Remainder Estimation Theorem

If there is a positive constant M such that 1.1.+ 1)(1) I :5 M for all 1 between x and a, inclusive, then the remainder term R.(x) in Taylor's Theorem satisfies the inequality

Ix - al·+ 1

IR.(x) I :5 M (n

+ 1)1

If this inequality holds for every n and the other conditions of Taylor's Theorem are satisfied by f, then the series converges to fix).

The next two examples use Theorem 24 to show that the Taylor series generated by the sine and cosine functions do in fact converge to the functions themselves.

EXAMPLE 2 Solution

Show that the Taylor series for sin x atx

=

0 converges for allx.

The function and its derivatives are fix) =

f"(x) =

sin x, - sin x,

cos x,

f'(x) =

f'"(x)

=

- cos x,

so

The series has only odd-powered terms and, for n . x3 smx = x - 3!

xS

+ 5! - ... +

= 2k

+ I, Taylor's Theorem gives

(_I)kx 2k+1 (2k + I)!

+ R2k+1(X).

All the derivatives of sin x have absolute values less than or equal to I, so we can apply the Remainder Estimation Theorem with M = I to obtain

IR2k+l(X) I :5

Ixl 2k+2

I· (2k + 2)!'

From Theorem 5, Rule 6, we have (lxl 2k+2/(2k + 2)!) ---> 0 as k---> 00, whatever the value of x, so R2k+ 1(x) ---> 0 and the Maclaurin series for sin x converges to sin x for every x. Thus, 00 (_I)k 2k+1 3 5 7 x . '" xxx smx = ~ (2k + I)! = x - 3! + 5! - 7!

+ ...

(4)

•

592

Chapter 10: Infinite Sequences and Series

EXAMPLE 3

Show that the Taylor series for cos x at x = 0 converges to cos x for every

value ofx. Solution We add the remainder term to the Taylor polynomial for cos x (Section 10.8, Example 3) to obtain Taylor's formula for cos x with n = 2k: x2

x 2k

x4

cosx = I - 2! + 4! - ... + (-I)k (2k)! + R2I.{x). Because the derivatives of the cosine have absolute value less thao or equal to I, the Remainder Estimation Theorem with M = 1 gives I R 2k(x) I

For every value of x, R2k(x) --+ 0 as k--+ value of x. Thus, 00

cosx = ~

,;;

Ixl2k+ 1

I· (2k + I)! .

00. Therefore, the

series conveIgeS to cos x for every

I)kx2k x 2 x' x. (2k)! = 1 - 2! + 4! - 6! + ...

(-

(5)

• Using Taylor Series Since every Taylor series is a power series, the operations of adding, subtracting, and multiplying Taylor series are all valid on the intersection of their intervals of convergence.

EXAMPLE 4 Using known series, f"md the first few terms of the Taylor series for the given function using power series operations. 1 (8) 3(2x + x cos x)

(b) eX cos x

Solution 1 21(X2X4 x2k) (8) 3(2x + x cos x) = 3 x + 3 x 1- 2! + 4! - ... + (-I)k (2k)! + ...

2

3

4

x + -x+ ' " x ( l + x +2!- + 3! 4!

= 1

x3

x4

3

6

+x----+ .. ·

)

-

•

By Theorem 20, we can use the Taylor series of the function f to find the Taylor series of f(u(x» where u(x) is any continuous function. The Taylor series resulting from this substitution will converge for all x such that u(x) lies within the interval of convergence of the Taylor

10.9 Convergence of Taylor Series

593

series of f. For instance, we can find the Taylor series for cos 2x by substituting 2x for x in the Taylor series for cos x: (-1)1(2x)21 (2k)! =

00

cos2x =

=

~

22X 2

24x4

(2x)'

(2x)4

(2x).

Eq. (5) with 2xfor x

I - 2! + 4! - 6 ! + ... 2.x.

1 -2!- +4!- - -6!+ .. ·

For what values ofx can we replace sinxby x - (x 3/3!) with an error of magnitude no greater than 3 X 1O-4?

EXAMPLE 5

Solution Here we can take advantage of the fact that the Taylor series for sinx is an alternating series for every nonzero value of x. According to the Alternating Series Estimation Theorero (Section 10.6), the error in truncating

. = x - -x 3 ! smx 3! I I

5

+ -x - ... 5!

after (x 3/3!) is no greater than

I~; I

lxi' 120 .

Therefore the error will be less than or equal to 3 X 10-4 if

lxi' < 120

3 X 10-4

or

Ixl <

~360 X 10 4 '" 0.514.

Rmmded down, to be safe

The Alternating Series Estimation Theorem tells us something that the Remainder Estimation Theorem does not: namely, that the estimate x - (x 3/3!) for sin x is an underestimate when x is positive, because then x'/120 is positive. Figore 10.20 shows the graph of sin x, along with the gmphs of a number of its approximating Taylor polynomials. The gmph of P3(X) = x - (x 3/31) is almost indistin• guishable from the sine curve when 0 ,;; x ,;; I. y 2 1 ~~-L--~~~~~-.f-~~~~~~~~x

o

-1

-2

FIGURE 10.20 The polyoomials

• (_1)<,;2>+1

P2.+1(X)

=

~ (2k + I)!

converge to sin x as n -+ 00. Notice how closely P3(X) approximates the sine curve for x '" 1 (Example 5).

594

Chapter 10: Infinite Sequences and Series

A Proof of Taylo(s Theorem We prove Taylor's theorem assuming a The Taylor polynomial

Pn(x)

=

< b. The proof for a > b is nearly the same.

f"(a) j(n)(a) f(a) + f'(a)(x - a) + ~(x - a)' + ... + ----;;r-(x - a)n

and its flISt n derivatives match the function f and its first n derivatives at x = a. We do not disturb that matching if we add another term of the form K(x - a )n+ 1 , where K is any constant, because such a term and its first n derivatives are all equal to zero at x = a. The new function

cf>.{x)

Pn(x) + K(x - a)n+l

=

and its frrst n derivatives still agree with f and its first n derivatives at x = a. We now choose the particular value of K that makes the curve y = cf>.{x) agree with the original curve y = f(x) at x = b. In symbols,

f(b)

=

Pn(b) + K(b - ar , '

or

K

=

--:f(--,-b)_---cP~n(b,_'_) (b - a)n+l

(7)

With K defined by Equation (7), the function

F(x)

=

f(x) - cf>.{x)

measures the difference between the original function f and the approximating function cf>n for each x in [a, b]. We now use Rolle's Theorem (Section 4.2). First, because F(a) = F(b) = 0 and both FandF' are continuous on [a, b], we know that

F'(Cl)

=

0

for some Cl in (a, b).

Next,becauseF'(a) = F'(Cl) = OandbothF'andF" are continuous on [a, c,],weknow that

F"(C2)

=

0

for some C2 in (a,

Cl).

Rolle's Theorem, applied successively to F", F"', ... , p(n-l) implies the existence of

C3

such that F"'(C3) = 0,

in (a, C2)

such thatF(4)(C4) = 0,

C4 in (a, C3) Cn

in (a, cn -,)

Finally, because F(n) is continuous on [a, cn] and differentiable on (a, cn), and = F(n)(cn) = 0, Rolle's Theorem implies that there is a number Cn+l in (a, cn) such that

F(n)(a)

(8)

Ifwe differentiate F(x) = f(x) - P.{x) - K(x - a)n+l a total ofn

F(n+l)(X)

=

+

I times, we get

j
(9)

Equations (8) and (9) together give

K

=

j
for some numberc =

Cn+l

in (a, b).

(10)

10.9

595

Convergence of Taylor Series

Equations (7) and (10) give

f(b) = P.(b)

+

I'n+I)(c) (n + I)! (b - a)n+l.

•

This concludes the proof.

Exercises 10.9 Find;ng Taylor Series Use substitution (as in Example 4) to find the Taylor series at x = 0 of the functions in Exercises 1-10. 1. e-ll 3. 5 sin (-x) 2• e -x/2

5. cos 5x 7. !n(1

+x

2

2

8. tan-I (3x')

)

6. cos (x2/3 /Yz)

9 I . 1+

lx 3

I 10. -2-

-x

38. If coox is replaced by I - (x 2/2) and Ixl < 0.5, what estimate can be made of the error? Does I - (x 2/2) tend to be too large, or

too small? Give reasons for your answer. 39. How close is the approximation sinx = x when Ixl which of these values ofx is x < sinx?

<

IO-'? For

40. The estimate ~ = I + (x/2) is used when x is small. Estimate the error when Ix I < 0,01. 41. The approximation eX = I + x + (x 2/2) is used when x is small. Use the Remainder Estimation Theorem to estimate the error whenlxl < 0.1.

42. (Continuation of Exercise 41.) When x < 0, the series for eX Use power series operations to fmd the Taylor series at x = 0 for the functions in Exercises 11-28.

.

14. smx - x

+ x' 3!

2

2

17. cos x (Hint: cos x 2 8in x

21.

I (1 - xj2

15.

16. x 2 cos (x 2 )

XCOS'11'X

+ x 2)

I

+ cosx

= (I + cos2:<)/2.) 19'1_2:<

24. sinX" cosx 27. t!n (1

13.

x2

18.

2

x T -

12. x 2 sinx

20. xln (1

+ 2:<)

22. ( I-x)'

28. !n(1

26. cosX' - sinx

+ x) -!n(1 -

x)

Find the fIrst four nonzero terms in the Maclaurin series for the functions in Exercises 29-34.

29.

e~

sin x

30.

!n(1 +x) I - x

2

Theory and Examples 43. Use the identity sin2 x = (1 - cos 2:<)/2 to obtain the Maclaurin series for sin2 x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin 2x.

44. (Continuation of Exercise 43.) Use the identity cos2 x cos 2x + sin2 X to obtain a power series for cos2 X.

2

25.ex+I~x

is an alternating series. Use the Alternating Series Estimation The0reD3 to estimate the error that results from replacing eX by I + x + (x 2/2) when -0.1 < x < O. Compare your estimate with the one you obtained in Exercise 41.

31. (tan-I X)2

45. Taylor's Theorem and the Mean Value Theorem Explain how the Mean Value Theorem (Section 4.2, Theorem 4) is a special case of Taylor's Theorem.

46. Linearizations at inflection points Show that if the graph of a twic.,.differentiable function I(x) has an inflection point at x = a, then the linearization of f at x = a is also the quadratic approximation of 1 at x = a. This explains why tangent lines fit so well at inflection points. 47. The (second) .e.ond derivative test

34. sin (tan- 1 x)

32. cos x' sinx

I(x) Error Estimates 35. Estimate the error if P,(x) valueofsinxatx = 0.1.

=

x - (x' /6) is used to estimate the

36. EstimatetheerrorifP,(x) = I + x + (x 2/2) + (x'/6) is used to estimate the value of eX at x = 1/2.

+ (x'/24)

37. For approximately what values of x can you replace sin x by x - (x'/6) with an error of magnitude no greater than 5 X 1O-4?

Give reasons for your answer.

=

=

I(a)

Use the equation

/"(C2)

+ /,(a)(x - a) + -2-(x -

aJ>

to establish the followiug test. Let 1 have continuous fust and second derivatives and suppose that/,(a) = O. Then

•. 1 bas a local maximum at a if/"'"

0 throughout an interval

whose interior contains a; b.

1 bas a local mininrum at a if /" whose interior contains a.

'" 0 throughout an interval

596

Chapter 10: Infinite Sequences and Series

48. A cubic appl'flIimation Use Taylor's formula with a = 0 and n = 3 to fmd the standard cubic approximation of f(x) = 1/( I - x) at x = O. ffive an upper bound for the magnitode of the error in the approximation whenlxl :5 0.1. 49.

L

Use Taylor's formula withn = 2 to find the quadratic approximationoff(x) = (I + x)k atx = (kaconstant).

o

b. If k = 3, for approximately what values oh in the interval [0, I] will the error in the quadratic approximation be less than 1/100?

50. Improving appl'flIimations of 'If L

D b.

Try it with a calculator.

51. The Taylor .eries generated by f(x) = ~::'~o a.x· is ~:=o allX' A function defined by a power series ~:o QnX" with a radius of convergence R > 0 has a Taylor series that converges to the function at every point of (-R, R). Show this by showing that the Taylor selies generated by f(x) = ~::'-o a.x· is the series ~:o a"x" itself. An immediate consequence of this is that series like •

2

6

8

5!

7!

x4 -+ x- - -x + ...

3!

approximation with an error less than 10-2 ? b. What is the maximum error we could expect ifwe replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions
53-58. Step 1: Plot the function over the specified interval.

Let P be an approximation of 'IT accurate to n decimals. Show that P + sin P gives an approximation correct to 3n decimals. (Hint: LetP = 'If + x.)

xsmx=x

and cubic approximations. In these exercises \\e explore the errors associated with these approximations. We seek answers to two questions: L For what values of x can the function be replaced by each

and

Step 2: Find the Taylor polynomials PI (x), Pz(x), and P,(x) at

x = O. Step 3: Calculate the (n + I)st derivative /,·+1)(c) associated with the remainder tenn for each Taylor polynomial. Plot the derivative as a function of c over the specified interval and estimate its maximum absolute value, M. Step 4: Calculate the remainder R.(x) for each polynontial. Using the estimate M from Step 3 in place of1·+1)(c) , plot R.(x) over the specified interval. Then estimate the values of x that answer question (a). Step 5: Compare ynur estimated error with the actual error E.(x) = If(x) - p.(x) I by plotting E.(x) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5.

obtained by multiplying Taylor selies by powers of x, as well as selies obtained by integration and differentiation of convergent power series, are themselves the Taylor selies generated by the functions they represent. 52. Taylor serie. for even functions and odd functions
Iff is even, then al = a, = a, = ... = 0, i.e., the Taylor series for f at x = 0 contains ouly even powers of x.

b. Iff is odd, then ao = az = Cl4 = ... = 0, i.e., the Taylor series for f at x = 0 contains ouly odd powers of x.

1 53. f(x) = .~' vl+x 54. f(x) = (1 55. f(x) =

+ x)'/2,

Ixl s

3

4

-t:5

x

S

2

+-, +

Ixl s 2 x I (cosx)(sin2:x:), Ixl:5 2

56. f(x)

=

57. f(x)

= e~cos2:x:,

Ixl:5 I

58. f(x)

=

e'/3 sin 2:x:,

Ixl:5 2

COMPUTER EXPLORATIONS Taylor's formula with n = I and a = 0 gives the linearization of a function at x = O. With n = 2 and n = 3 we obtain the standard quadratic

l().l()~T_ h_ e _B_in_o_m _i_a_ L_ se_n_ ·e_s_a_n_d_A~p~p_"_c_ ati _·o _n_s_o_f_l_a~ yL_o_ r_ S_ en _·e _s_______________ In this section we introduce the binomial series for estimating powers and roots of binomial expressions (I + x)m. We also show how series can be used to evaluate nonelementary integrals and limits that lead to indeterminate fonns, and we provide a derivation of the Taylor series for tan-I x. This section concludes with a reference table of frequently used series.

10.10 The Binomial Series and Applications of Taylor Series

597

The Binomial Series for Powers and Roots The Taylor series generated by f(x) = (I

I

+ mx +

m(m - I) 2 2! x

+ x)", when m is constant, is

+

m(m - I)(m - 2) 3 3! x

+

m(m - I)(m - 2)'" (m - k k!

+ ... +

I)

k

x + ...

This series, called the binomial series, converges absolutely for series, we first list the function and its derivatives:

f(x)

=

(I + x)"

f'(x)

=

m(1

f"(x)

=

m(m - 1)(1

f"'(x)

=

m(m - I)(m - 2)(1

P)(x)

=

m(m - I)(m - 2) .. · (m - It

Ix I <

(1)

I. To derive the

+ x),,-t + X),,-2

+ x)m-3

+ 1)(1 + x),,-k.

We then evaluate these at x = 0 and substitute intu the Taylor series fonnula to obtain Series (I). If m is an integer greater than or equal to zero, the series stops after (m + I) terms because the coefficients from It = m + I on are zero. If m is not a positive integer or zero, the series is inf'mite and converges for Ix I < I. To see why, let Uk be the term involving xk. Then apply the Ratio Thst for absolute convergence to see that UH11Im-kl Ilk = It + I x

I

ask---

---Ixl

00.

Our derivation of the binomial series shows ouly that it is generated by (I + x)" and converges for Ix I < I. The derivation does not show that the series converges to (I + x)m . It does, but we omit the proof. (See Exercise 64.)

The Binomial Series

For-I

<x< I,

where we define

(7)

=

m,

m) = m(m - I) (2 2!'

and

m) = m(m - I)(m - 2) .. ·(m - k (k It!

EXAMPLE 1

+

I)

for It ;" 3.

rfm = -I,

(-I) I

= -I

'

(-I)2

=

-1(-2) 2!

=

I,

598

Chapter 10: Infinite Sequences and Series

and

-I) _ -1(-2)(-3)"'(-1 - k + I) __ k (k!) _ _ k k! - ( I) k! - ( 1). ( k With these coefficient values and with x replaced by -x, the binomial series formula gives the familiar geometric series 00

(I +xr! = I + ~(_I)kxk= I -x +X 2 -X 3 + ... + (-Ilxk+ ....

•

k-!

EXAMPLE 2 We know from Section 3.9, Example I, that v'1+x RJ I + (x/2) for Ix Ismall. With m = 1/2, the binomial series gives quadratic and higher-order approximations as well, along with error estimates that come from the Alternating Series Estimation Theorem:

(I + x)!/2 = I +

f + (t)~-t) x2 + (t)(-~)(-~) x

3

(t)(-t)(-~)(-%) x4 + ...

+

4! 2

3

4

x x =1+ -- +x - - -5x+ ' " 2 8 16 128

Substitution for x gives still other approximations. For example, for Ix 2 1 small

~I_IRJI_~ 2x

x

__I8x2

for

I} Ismall, that is, Ix I large .

•

Sometimes we can use the binomial series to f"md the sum of a given power series in terms of a known function. For example, x6

2 X

3! +

-

x lO

x!4

51 - 7f + ... =

(x 2)3

2

(x 2)'

(x2)7

3! + 51 - 7! +

(x ) -

.

2

... = smx .

Additional examples are provided in Exercises 59-62.

Evaluating Nonelementary Integrals Taylor series can be used to express nonelementary integrals in terms of series. Integrals like sin x 2 dx arise in the study of the diffraction oflight.

J

Express J sinx 2 dx as a power series.

EXAMPLE 3

From the series for sin x we substitute x 2 for x to obtain

Solution

. 2 2 x6 smx = x - 3!

+ -x

lO

5!

- -X14 + -x 7!

I8

9!

- ...

Therefore,

. J

2

-

smx dx - C +

x3

x7

XU

XIS

x lO

3 - 7-3! + 11-5! - 15-7! + 19-9! - ....

•

10.10 The Binomial Series and Applications of Taylor Series

EXAMPLE 4 Solution

Estimate

599

fol sinx 2 dx with an error ofless than 0.001.

From the indetmite integral in Example 3,

I I 10t.smx2 dx=:3-7'3!

I

I

I

+ 1l'5! -15'7! + 19'9! - ....

The series alternates, and we find by experiment that I

II • 5! "" 0.00076 is the first term to be numerically less than 0.001. The sum of the preceding two terms gives

11

t - 12 "" 0.310.

2

sinx dx ""

With two more terms we could estinlate

11

sinx 2 dx "" 0.310268

with an error ofless than 10--< . With ouly one term beyond that we have

10t.smx 2 dx "" :3I -

I 42

I

I

I

+ 1320 - 75600 + 6894720 "" 0.310268303,

with an error of about 1.08 X 10-9 • To guarantee this accuracy with the error formula for the Trapezoidal Rule would require using about 8000 subintervals. •

Arctangents In Section 10.7, Example 5, we found a series for tan-I x by differentiating to get

~tan-I x

= _1_ = I -

dx

I

+ x2

x2

+ X4

-

x6

+ ...

and then integrating to get 3

5

7

3

5

7

x x x tan-1 x=x--+---+'" However, we did not prove the term-by-term integration theorem on which this conclusion depended. We now derive the series again by integrating both sides of the finite formula

+ t4

_1_ = I - t 2

l+t

2

-

(

+ ... + (-I )nt o. + -

t6

I)n+l t 2n+2

l+t 2

(2)

in which the last term comes from adding the remaining terms as a geometric series with tITSt term a = (_l)n+l t o.+2 and ratio r = -t 2 . Integrating both sides of Equation (2) fromt = Otot = x gives 3 tan-1 x = x - -x 3

S

7

2n 1

x + + -x5 - -x7 + ... + (-I) 2n+1 + Rn (x) ,

where

-l

X

R.{x) -

D

11

(_l)n+l t o.+2

I

+t

2

dt.

The denominator of the integrand is greater than or equal to I; hence

IR.{xli

:5

fixi o. 2 Ix1o. +3 t + dt = 2n + 3 .

10

600

Chapter 10: Infinite Sequences and Series If Ix I :5 1, the right side of this inequality approaches zero as n --> lim"~""R"(x) = 0 iflxl :5 I and

(-I)"x 2n + 1 tanx- 11=0 2+1' n -I

_

""

L

Ixl

3

5

7

Therefore

1.

:5

x 3 x S x7 tan-1 x=x--+---+···

00.

(3)

'

We take this route instead of finding the Taylor series directly because the formulas for the higher-order derivatives of tan-I x are unmanageable. When we put x = 1 in Equation (3), we get Leibniz'. formula: ~

-=

4

1 1 1 1 (-If 1--+---+--···+--+···

3

5

7

2n+l

9

Because this series converges very slowly, it is not used in approximating ~ to many decimal places. The series for tan-I X converges most rapidly when x is near zero. For that reason, people who use the series for tan-I X to compute ~ use various trigonometric identities. For example, if

a = tan-II 2

{3 = tan-II 3'

and

then

tan(a + {3)

=

tana+tan{3 1 _ tanatan{3

I

I

2+3 =

1 _1

=

1

~

=

6

tan 4

and ;

=

a

+ {3

=

tan-It

+ tan-It·

Now Equation (3) may be used with x = 1/2 to evaluate tan-I (1/2) and with x = 1/3 to give tan-I (1/3). The sum of these results, multiplied by 4, gives ~.

Evaluating Indeterminate Forms We can sometimes evaluate indeterminate forms by expressing the functions involved as Taylor series.

EXAMPLE 5

Evaluate

· nInx -Iu

x---+lx-l"

Solution We represent In x as a Taylor series in powers of x-I. This can be accomplished by calculating the Taylor series generated by In x at x = 1 directly or by replacing x by x - I in the series for In (1 + x) in Section 10.7, Example 6. Either way, we obtain Inx = (x - 1) -

t

(x - 1)2

+ ... ,

from which we rmd that

lim

x---+l

EXAMPLE 6

~ = x-I

lim

x-l

(1 _I(x - 1) + ...) 2

Evaluate lim sinx - tanx

x-o

x3

=

1.

•

10,10 The Binomial Series and Applications of Taylor Series Solution

601

The Taylor series for sin x and tan x, to terms in x 5, are

. = x - -x 3 smx 3!

S

+ -x - ... 5!

tanx

'

=

x3 3

2>;5

x+- +-

15

+ .. ,

Hence,

sinx-tanx=-x; _x; _"'=x3(_!_X; _ .. -) and

I 2'

•

If we apply series to calculate li1llx~o((I/sinx) - (lIx)) , we not only tmd the limit successfully but also discover an approximation formula for esc x,

EXAMPLE 7

(~ smx

Find lim

x---+O

1), x

Solution I sin x

I

----

x

x- (x - ~ + ~ - "-)

x - sinx x sin x

x' (x - ~; + ~; - .. -)

Therefore, 2

---+ .. ,) I x

lim _1__ 1 lim x 3! 5! x~JSinX x ) x~o ( I _ ~ + .. , =

=

0

'

From the quotient on the right, we can see that if Ix I is small, then

-I- - -I sin x x

~

I x·-

3!

=x -

6

or

I

cscx '" X

x

+ 6'

•

Euler's Identity As you may recall, a complex number is a number of the form a

real numbers and i = use the relations

yCJ, Ifwe substitute x

=

+ hi, where a and h are

iO (0 real) in the Taylor series for eX and

602

Chapter 10: Infinite Sequences and Series

and so on, to simplify the result, we obtain

e

'0 .21fl .303 .404 .505 .606 '6= I +'--+-'- + - '- + - '- + - '- + - '- + ... 1! 2! 3! 4! 5! 6!

(I - ~

+ :; -

:~ + .. )

+;(0 - ~~ + ~; - .. )

= cosO + isinO.

This does not prove that e'6 = cos 0 + i sin 0 because we have not yet defined what it means to mise e to an imaginary power. Rather, it says how to define e '6 to be consistent with other things we know.

DEFINITION ForanyrealnumberO,e'6 = cosO + isinO.

(4)

Equation (4), called Euler's identity, enables us to define e a+bl to be ea. e bl for any complex number a + hi. One consequence of the identity is the equation e ifT = -1. When written in the form e'" + I = 0, this equation combines five of the most important constants in mathematics.

TABLE 10.1 Frequently used Taylor series I I-x

--= 1

+x+x 2 +···+x"+···=

00

~xn

~

,

Ixl < I

I 00 - - = I - x + x 2 - ... + (_x)n + ... = ~(_l)nxn, 1+x n=O Xl

x" n!

00

x"

eX = I +x+-+ .. ·+-+ .. ·= " ' 2!

~n!'

Ixl <

. x3 x5 x 2n + l smx = x - 3! + 5! - ... + (-I)n (2n + I)! + ... = x2 X4 x2n 00 cos x = 1 -2-, +-4' -"'+(-I)n(2 )' + ... = ~ .. n. 11=0

Ixl < I

00

00

(_I)nx 2n+l

~ (2n + I)! ' I )nx2n (2 )' '

Ixl <

00

(-

n.

Ixl <

00

x2 x3 xn 00 (_1)n-l x n In(I+x)=x-- +-3 - ... +(_1)n-ln+"·=~ n ' 2 n=1

-1<x:51

3 5 2n+l 00 (_I)n 2n+l X tan-I x = x - "'-- + "'-- - ... + (- I)n x + ... = ~ -'--..--'--cc---.3 5 2n+1 n-O 2n+I'

Ixl :5 I

Exercises 10.10 Binomial Series Find the fIrSt four terms of the binomial series for the functions in Exercises 1-10.

1. (1

+ x)!/2

2. (1

+ x)!/3

3. (1 - xt!/2

4. (1 - lx)!/2

7. (1

+ x 3t!/2

8. (1

+ x 2 )-!/3

10.10 The Binomial Series and Applications of Taylor Series

x 10•• ,r.-;-

+ X1)'/2

9. ( I

39. lim

x-a

vl+x

Find the binomial series for the functions in Exercises 11-14.

+ x)4

11. (1

13. (I - 2>;)'

14. (I -

2

40. lim In (1

x-a

I)'

I

I

I

(t)' + (t)' + (t)' + (t)" + ...

Approximattons and Nonelementary Integrals

42.

error of magnitude less than 10-'. (The answer section gives the integrals' values rounded to five decimal places.) e -x - 1 2 15. sinx 16.

32 - - +3-4 - - - 3'+ · · · 4 3 . 124 '2! 44 '4! 4"6!

0.2

17.

/.0.' ~ax ° + I

D Use

/.0.2 ° --x-ax

ax

18.

/.O·2S~ax

x4

series to approximate the values of the integrals in Exercises 0.1 .

20.

/. 0.1

21.

°

"\I'l+7 ax

° e--r ax /.

22. / . ' I - ~osx

°

/.

x

23. Estimate ~e error ifcos 12 is approximated by I cos 12 dl. integral

fo

VI is approximated by I in the integral fo' cos VI dl.

24. Estimate the error if cos

2

46.

ax

49.

-1' + ~! t + ~! 4

8

in the

2

-

,

-

~!

10-' . 25. F(x) =

2'22 3"3!

xln(1

28. F(x)

=

°

I

+

/.

I)

2'

2'

31. lim

x

Xl -

3,>:2 21

4

' Y - tao-I Y 33. lun ,

y-o

dl,

(a) [0,0.5]

(b) [0, I]

35. lim x 2(e- 1/ x'

-

x~oo

1n(1 + x ) 37. lim """"'"''-----"x-a 1 cosX' 2

sinB - B + (11'/6) 32. 9-0 lim "' tr '

34. lun

y-o

y

I)

3'x'

+ 41 - 61 +'" X S + x 7 - x 9 + XII - •••

51. -I +2>;-3x 2 +4x'-5x 4 + ...

Theory and Examples 53. Replace x by -x in the Taylor series for In (I + x) to ob1ain a series for In (I - x). Then subtract this from the Thylor series for In (I + x) to showthatforlxl < I,

54. How many lem3s of the Taylor series for In (I + x) should you

2

t

34x4

+ ...

add to be sure of calculatiog In (1.1) with ao error of magnitude

less than 10-8? Give reasons for your answer.

I - cos 1 - (1 2/2)

1-0

2'

2 3 22.t-4 23x 5 2 4x 6 50. x -2>; +21-"""31+41-'"

(b) [0, I]

Indeterminate Forms

x-a

3"7!

In I + _ x =2 ( x+ x' + x' + . . .) . I x 3 5

Use series to evaluate the limits in Exercises 29--40. eX - (I + x) 29. lim

3"5!

"3 - 3"3 + 3"5 - 3"7 x 3 + X4 + x 5 + x 6 + ...

=

I

_1_ _ _1_ + ... 3'2' 4'24

1r - -r- + -1f'5- - - 1T'7 - + ...

. 3

48.1-

10 Exercises 25--28, fmd a polynomial that will approximate F(x) throughout the given interval with ao error of magnitude less than

/.X sin 12 dl, [0, I] 26. F(x) /,xI2e -t'dl, [0, I] 27. F(x) = /.X tao-I dl, (a) [0,0.5]

. 2

47.

0.1

s~x dx

0

1. __1_ +

44

4S

19--22 with an error of magnitude less than 10-8 .

19.

+ x') sinx 2

41. I + I + 2! + 3! + 4! + ...

D 10 Exercises 15--18, use series to estimate the integrals' values with an

/.°

X'

Using Table 10.1 10 Exercises 41-52, use Table 10.1 to fmd the sum of each series.

+ x 2)'

12. (1

. 3

sm x 1 cos 2x

603

_tao_-_'~y_-_sin ____ y

, y cosy

36. lim (x + I) sin _+1 I x_OO x

55. According to the Alternatiog Series Estimation The0ren3, how many lem3s of the Thylor series for tao-I I ""uld you have to add to be sure of fmding 'If/4 with ao error of magnitude less than

10-31 Give reasons for your answer. 56. Show that the Thylor series for !(x) Ixl> I.

D 57.

= tao-I x

diverges for

Estimating Pi About how many tenns of the Taylor series for tao-I x v.<>uld you have to use to evaluate each term on the righthand side of the equation

_ 48

'If -

-1

tao

I + 18

32

-1

tao

I 20 -1 I 57 tao 239

with ao error of magnitude less than IO.... ? 10 contrast, the convergence of ~:~I(I/n2) to ~/6 is so slow that even 50 terms will not yield two-place accoracy.

604

Chapter 10: Infinite Sequences and Series

58. Integrate the rlrst three nonzero terms of the Taylor series for tan t from 0 to x to obtain the fIrst three nonzero terms of the Taylor series for In sec x.

59.

L

by integrating the series

Use the binomial series and the fact that

in the first case from x to tox. to generate the fIrst four nonzero terms of the Taylor series

for sin-1 x. What is the radius of convergence? b. Series for cos-1 x Use your result in part (a) to rmd the rrrst five nonzero terms of the Taylor series for 008- 1 x. 60.

L

Seri.. for sinh-1 x Taylor series for

Find the rrrst four nonzero terms of the

sinb-1 x

=

r

dt

Jo \IT+7

+ x)'

from the series for

+ x).

62. Use the Taylor series for 1/(1 - x') to obtain a series for 22/(1 - x')'.

D 63.

00

Euler's Identity 67. Use Equation (4) to write the followiog powers of e in the form a + hi. a. e-l'll' 68. Use Equation (4) to show that

69. Establish the equations in Exercise 68 by combining the formal Taylor series for e ifJ and e-ifJ • 70. Show that

a. coshiO = cosO,

estimation error. 61. Obtain the Taylor series for 1/(1

and in the second case from -

.

D b. Use the rrrst three terms of the series in part (a) to estimate sinb-1 0.25. Give an upper bound for the magnitude of the

-1/(1

00

Estimating Pi The English mathernaticiao Wallis discovered the fom2Ula ~

2,4,4,6,6,8, .. .

4

3,3,5,5,7,7, ... .

b. sinhi8

= isin8.

71. By multiplying the Taylor series for eX and sin x, rmd the terms through x 5 of the Taylor series for eX sin x. This series is the imaginary part of the series for eXoeix

=

e(l+i~.

Use this fact to check your anawer. For what values of x should the series for eX sin x converge? 72. Wheo a aod b are real, we derme e(a+ib).< with the equation e(a+ib)x

=

etIXoeibx

=

etIX(cosbx

+ isinbx).

Find ~ to two decin2al places with this formula. Differentiate the right-hand side of this equation to show that

64. Use the followiog steps to prove Equation (I). L

:fx

Differentiate the series f(x)

=

I

+

~ (:)x'

e(a+ib).<

Thus the familiar rule (d/dx)e a

= !tea holds for k complex as

well as real.

to show that

f '(x) = mf(x) 1 + x'

-I

73. Use the dermitioo of and 62 •

< x < I.

a.

b. Defineg(x) = (I +x)-mf(x)aodshowthatg'(x) = O.

c. From part (b), show that f(x) = (I

e" to show that for any real nwnbers 6, 61,

ei6tei~ = ei(6t+6i),

toshowthatforlxl

J

+ x)m.

Integrate the binomial series for (I - x')-I/2

< I,

b.

e-i6 =

1/e i6 .

74. 1\vo complex numbers a + ib aod c + id are equal ifaod only if a = c aod b = d. Use this fact to evaluate etlXcosbxdx

65. Serie. for sio-1 x

. _

= (a + ib )e(a+ib).<.

and

J

etIXsinbxdx

from

001__'~3_'5~.~ .._._.,::(2=n~---c-=I"-) X2n+l 2·4·6· .. · • (2n) 2n + I'

.-1

smlx=x+~-

66. Serie. for tan-I x for Ixl

>

tan-Ix = ~ _1 + _1___1_ + ... 2X3x3sxS' tan-Ix

where C

1 Derive the series

= -~ _1 + _1___1_ + ...

2X3x3sxS'

x>

I

x <-I

,

=

Cl

+ iC2 is a complex constant of integration.

Chapter 10

Chapter

a!J

Practice Exercises

605

Questions to Guide Your Review

1. What is an infinite sequence? What does it mean for such a sequence to converge? To diverge? (live examples.

18. What is an alternating series? What theorem is available for determining the convergence of such a series?

2. What is a monotonic sequence? Under what circumstances does

19. How cao you estimate the error involved in approximating the

soch a sequence have a limit? (live examples.

3. What theorems are available for calculating limits of sequences? (live examples.

sum ofao alternating series with one of the series' partial sums? What is the reasooing behind the estimate?

20. What is absolute convergeoce? Conditional convergence? How

4. What theorem sometimes enables us to use I'H6pital's Ru1e to calcu1ate the limit of a sequence? (live ao example.

S. What are the six commonly occurting limits in Theorem 5 that

arise frequently when you work with sequences and series? 6. What is an write series? What does it mean for such a series to cooverge? To diverge? (live examples.

7. What is a geometric series? When does such a series converge? Diverge? When it does converge, what is its sum? Give examples. 8. Besides geometric series, what other convergent and divergent series do you know?

9. What is the nth-Thnn Thst for Divergeoce? What is the idea be-

are the two related?

21. What do you know about rearranging the terms of ao absolutely convergeot series? Of a conditionally convergeot series?

22. What is a power series? How do you test a power series for convergeoce? What are the possible outcomes?

23. What are the basic facts about a. sums, differences, and products of power series? b. substitution of a function for x in a power series? •• term-by-term differeotiation of power series?

d. term-by-term integration of power series? (live examples.

hind the test?

10. What cao be said about term-by-term soms aod differeoces of coovergeot series? About coostsot mu1tiples of coovergeot and di-

24. What is the Taylor series generated by a function f(x) at a point x ~ a? What informatioo do you need about f to coostruet the

series? Give an example.

vergent series?

11. What happeos if you add a fmite number of terms to a coovergent series? A divergeot series? What happeos if you delete a finite

number of terms from a convergent series? A divergent series?

25. What is a Maclaurin series? 26. Does a Taylor series always cooverge to its geoerating function? Explain.

12. How do you reindea a series? Why might you want to do this?

27. What are Thylor polynomials? Of what use are they?

13. Under what circumstsoces will ao infmite series of nonnegative terms converge? Diverge? Why study series of nonnegative terms?

28. What is Taylor's formu1a? What does it say about the errors involved in using Taylor polyoornia1s to approximate functioos? In particu1ar, what does Taylor's formo1a say about the error in a linearization? A quadratic approximation?

14. What is the Integral Test? What is the reasooing behind it? (live an example of its use. 15. Wheo do p-series cooverge? Diverge? How do you know? (live examples of convergeot and divergeot p-series.

16. What are the Direct Comparisoo Test aod the Limit Coroparison Test? What is the reasooing behind these tests? (live examples of their use.

17. What are the Ratio and Rnot Tests? Do they always give you the information you need to determine convergence or divergence?

29. What is the hinomial series? On what interval does it converge? How is it used?

30. How can you sometimes use power series to estimate the values of nooelernentary definite integrals?

31. What are the Taylor series for 1/(1 - x), 1/(1 + x), eX, sinx, cosx,ln (I + x), and tso-I x? How do you estimate the errors involved in replacing these series with their partial soms?

(live examples.

Chapter

ail

Practice Exerdses

Determining Convergence of Sequence. Which of the sequences whose nth terms appear in Exercises 1-18 cooverge, aod which diverge? Find the limit of each convergent sequence.

( -I)1. a,,=l+-nI - 2-

3.an=~

2.a.~

4. a.

I - (-I)-

~ I

Vn

+ (0.9)-

5. a.

~

. n1r sm

7. a.

~-n-

2

In (n 2 )

_ n + Inn 9.a,,n 11. a.

~

5)-

n ( -n-

6. an

=

8. all

=

sin n71' 1n(2n+I)

10. all = 12.a_~

n

In (2n'

n

+

1)

(1+,.1)--

606

Chapter 10: Infinite Sequences and Series

13. all =

./¥ 'In

15. a. = n(2 1/.

17.

-

I)

(n + I)!

all =

Maclaurin Series Each of the series in Exercises 51-56 is the value of the Taylor series at x = 0 of a function f(x) at a particular point What function and what point? What is the swn of the series?

3)1/,

14. all

= (n

16. a.

= 'ZY2n +

I

(-4)' n.

51 1_1 + -.L - ... + (-I)'-.L + ... · 4 16 4'

18. an = - , -

I

n.

52 ~_~+~_ ... +(_I)'-I~+ ... • 3 18 81 n3'

Convergent Series Find the sums of the series in Exercises 151--24.

~

19 •

~(2n

21

f

20~.

-2 • ,;':1 n(n + I)

I - 3)(2n -I)

22

9

.•-1 (3n - 1)(3n + 2)

f

-8 •• _, (4n - 3)(4n

00

+ I)

.3

00

,,4 ,.2n ,.2 54. 1- - - + - - - ... + (-1)'--- + ... 9'2! 81'4! 32o (2n)! (lnZj' (lnZ)' 55. I + 1n2 + ----z! + ... + ----n! + ...

56. _1_ _ _ 1_+_1_ _ ... \1'3 9\1'3 45\1'3

24. L(-I) 4' •- 1

+ (_1).-1

Detennining Convergence of Series Which of the series in Exercises 25-40 converge absolutely, which converge conditionally, and which diverge? Give reasons for your answers. 00

25.

28.

I

L.,n

11=-1 V

~._I :.:12n 3

00

26.

f

35.fn~1 11=1 00

37.

39.

n. (-3)'

L-,11=-1 n.

f v'n(n + II)(n + 2) .-1

00

Ln 00

29. '5'.

In: 11=1 n 00 (-I)' 33. '5' .~ ~nvn2+ 1 31.

-5

27.

,-I

L ., Vn 00

Inn

00

~ In (Inn)

00 (-I)'3n2 34. '5' '---c,='--''''-~ n + 1 2 + I) 00 (-I)'(n 36. ~2'-=-C'1Ft 2n + n - 1

L

38.

40.

2"3" n

L-. 00

/I - I

f I .-2n~

,&1

00

(-I).-I(3x - I)' n2

45.

L X. 11=-1 n

00

00

47.

~

•

(n + I)x 2o - 1 3'

00

49. ,&1(cSChn)x'

=

0 for the functions in Exercises 57-64.

I 57· 1 _ 2x

58 _1_ 'I+x'

59. sin 1TX

60

61. cos (x'/')

62. cos

63. e(=/2)

64. e-x'

. 2x

• 8m

T

x' v'5

Taylor Series In Exercises 65-68, find the flISt four nonzero terms of the Taylor series generated by f at x = a. 65. f(x) = ~ at x = -I 66. f(x) = 1/(1 - x)

67. f(x)

=

I/(x

68. f(x) = I/x

at x = 2

+ I) at at x

=

x = 3 a

>0

Nonelementary Integrals 651--72 with an error of magnitude less than 10-8 • (The answer section gives the integrals' values rounded to 10 decimal places.)

Power Series

,&1

Find Taylor series at x

Use series to approximate the values of the integrals in Exercises

In Exercises 41-50, (a) fmd the series' radius and interval of conver· gence. Then identify the values of x for which the series converges (b) absolutely and (0) conditionally. 00 (x + 4)' 00 (x _ 1)20-2 41.,&1 n3' 42. (2n - I)! 43.

I + ... (In - 1)(\1'3)20-1

I

30. '5' - - 2 ,;':1 n (Inn)

::lin (n + I) 32.

(-I)'

11=-1

(-I)'

~+1

'ITs

w3

53 , , - - + - - ... +(-1)' + ... · 3! 5! (2n + I)!

~. (n + 1)(2x + I)'

44':=6 00

(Zn + 1)2' X

•

46·L.Vn ,

11 - 1 00

48.~

(-I)'(x _ 1)2.+1 Zn+1

69.

0

/.

1/2e-x' fix 1f2

71.

/o .

tan-I x -x-fix

70.

/.10 x sin (x') fix 1/64 tan-I x

72.

fix

Vx

Using Series to Find Limits In Exercises 73-78: a. Use power series to evaluate the limit

D b.

Then use a grapher to support your calculation. . e 9 - e- 9 - 26 73. lim 7 sin x 74. J~ 0 sinO .%"-0 e'h - 1

00

50. ,&1 (coth n)x'

•,

/o .

75 lim ( • 1-0

I _ -.L) 2 - 2eost t 2

76. lim

h-O

(sinh)/h - cosh h

2

Chapter 10 Additional and Advanced Exercises 1 - cos, z z~o In (I - z) + sinz

77 lim

.

,

78. lim ------y'---~­ y-o cosy - coshy

b. Show that the function defined by the series satisfies a differential equation of the form

d'y

tJx' ~

Theory and Examples 79. Use a series representation of sin 3x to find values of rand s for which

x'Y + b

and Imd the values of the constants a andb. 86••• Find the Maclaurin series for the function x

b. Does the series converge at x

D 80.

Compare the accuracies of the approximations sinx '" x and sinx '" 6>:/(6 + x') by comparlng the graphs off(x) ~ sinx - x andg(x) ~ sinx - (6):/(6

+ x'». Describe what you Imd.

81. Find the radius of convergence of the series

~

'/(1 + x).

I? Explain.

87. If ~:1 all and ~~=1 b" are convergent series of nonnegative numbers, can anything be said about ~~I Glib,,? Give reasons

for your answer. 88. If~~=l all and ~~=1 bll are divergent series of nonnegative numbers, can anything be said about ~: 1 anbll ? Give reasons for

your answer.

2.5.8 .... '(3n - I) ~ 2·4·6· .. · • (2n) x' . ,-I 00

82. Find the radius of convergence of the series

3.5.7 .... '(2n + 1) ~ 4.9.14 ..... (5n _ I) (x -

607

00

I)'.

83. Find a closed-form formula for the nth partial sum of the series ~::,~,In (I - (l/n'» and use it to determine the convergence or divergence of the series.

89. Prove that the sequence {x,} and the series ~:1 (Xk+l

-

x.)

both converge or both diverge.

90. Prove that ~::;'1 (aJ(1 + a,» converges if a" > 0 for all nand ~~=1 a" converges. 91. Suppose that at, al, a3, . .. , all are positive numbers satisfying the following conditions:

i)

Dl ~ a2 ~ a3 ~ ••• ;

ii) the series a2

+ "4 + as + al6 + ...

diverges.

Show that the series

84. Evaluate ~:2(1/(k' - I» by Imding the limits as n --> 00 of the series' nth partial sum. 85. a. Find the interval of convergence of the series

y

~ 1+ +

Chapter

ix' + I~Ox6 + ...

I

1·4·7· .... (3n - 2) (3n)! x"

ill

diverges.

92. Use the result in Exercise 91 to show that

+ ...

+ ~_I11_ 2

Additional and Advanced Exercises

Determining Convergence of Series

Choosing Centers for Taylor Series

Which of the series ~: 1 all dermed by the fonnulas in Exercises 1-4 converge, and which diverge? Give reasons for your answers. 1 00 1 00 (tan- n)'

Taylor's fonnula

1.

~ (3n

3.

~(-I)'tanhn

_ 2),+(1/2)

2.

~

4.

~ log.~n!)

11 - 1

11=-2

n'

a,+1 ~ (n

n(n + 1) + 2)(n +

(Hint: Write out several terms, see which factors cancel, and then genera1ize.)

n

= a2 = 7,

a,+1 ~ (n _ I)(n

7. al

= a2 = 1,

all+i -

8. a,

~

+ I) a"

ifn ~ 2

_ _1_ if >2 1 + all n -

1/3' ifn is odd, a,

~

n/3'

/,n)(a)

ifn is even

r(a) 2 ! (x

- a)' +

/,'+1)(c)

+ --, -(x - a)' + ( I)' (x - a)'+1 n. n+ .

n

3) a,

6. al

f(x) ~ f(a) + f'(a)(x - a) +

+1

Which of the series ~::;'1 a" deImed by the formulas in Exercises 5--1! converge, and which diverge? Give reasons for your answers.

S. al ~ I,

nlnn

diverges.

expresses the value of f at x in terms of the values of f and its derivatives at x = a. In numerical computations, we therefore need a to be a point where we know the values of f and its derivatives. We also need a to be close enough to the values of f we are interested in to make (x - a )'+ 1 so small we can neglect the remainder.

In Exercises <),,14, what Taylor series would you choose to represent the function near the given value of x? (There may be more than one good answer.) Write out the Ill'St four nonzero terms of the series you choose. 9. cosx

11. eX

near x

=1

near x = 0.4

13. cosx

near x = 69

10. sinx near x = 6.3

12. In x

near

14. tan-I x

x

~

1.3

near x = 2

608

Chapter 10: Infinite Sequences and Series

Theory and Examples 15. Let a and b be constants with 0 < a < b. Does the sequence {(an + b n)l/n} converge? Ifit does converge, what is the limit?

Raabe:r Test: If~~=l un is a series of positive constants and there exist constants C, K, and N such that

-""--

16. Find the sum of the infmite series I

",,+1

+~+~+ -"-- + ~+~+-"--+ ~ 10

10'

10'

10'

10'

10'

10'

+~+-"--+ ... 10'

10'

00

In+1

11=011

~

n~1

27. If~:lanconverges,andifCln =F 1 andan >

n

n~OO

(

1-

ofor all n,

a. Show that ~: 1 a,? converges.

nx

(n + 1)(2); + I)n

b. Does ~:;"~1 an/(I - an) converge? Explain.

28. (Continuation of Exercise 27.) If ~::;'1 a" converges, and if I> a" > Oforalln,abowthat~:lln(1 - an)eonverges.

Does the value of

.

n2 •

(2n - I)' Ul ~ I, Un+l ~ (2n)(2n + I) Un·

converges absolutely.

hm

j(n)

n

Apply Raabe's Test to determine whether the series converges.

00

L

+ f: +

where Ij(n)l < K for n '" N, then ~::;'1 Un converges ifC > I and diverges if C '" I. Show that the results of Raabe's Test agree with what you know about the series ~:;"- 1 (I/n') and ~:;"- 1 (I/n).

I - - , dx. l+x

18. Find all values of x for which

D 19.

I

26. (Continuation ofExercise 25. ) Suppose that the terms of ~:;"- 1 Un are defmed recursively by the formulas

17. Evaluate

~

~

(Hint: First show that Iln(1 - an) I '" a"f(1 - a,,).)

eos(a/n»)n a constant, n •

appear to depend on the value of a? If so, how?

I

b. Does the value of . ( cos (a/n»)n hm 1bn ' 11_00

a and b constant, b

Prove Nicole Oresme's Theorem that

29. Nicole Oresme's Theorem

¢'

0,

+ 1. 2

2

+ 1 . 3 + ... + --"-+ ... ~ n 1 4

2

-

4.

(Hint: Differentiate both sides of the equatioo 1/(1 - x) 1 + ~:_ IX".)

~

30. a. Show that

appear to depend on the value of b? If so, how? c. Use calculus to confirm your fmdings in parts (a) and (b). 20. Show that if ~:;"~1 a" converges, then

00

~I n-

n(n

+ I) Xn

2>;'

~(x - I )'

for Ix I > I by differentiating the identity 00

LXn+

,

1

IFI

converges.

=_x_ 1 - x

twice, multiplying the result by x, and then replacing x by I/x.

21. Find a value for the eonstant b that will make the radius of con-

vergence of the power series

b. Use part (a) to fmd the real solmioo greater than I of the equation

x~fn(n;l). 11=1

equal to 5.

x

31. Quality control

22. How do yoo know that the functioos sinx, lnx, and e' are not polyoomia\s? Give reasons for your answer.

23. Find the value of a for which the limit lim sin (ax) - sinx - x

x-a

x3

is fmite and evaluate the limit. 24. Find values of a and b for which . cos(ax)-b hm , 2x

x-o

~-l.

25. Raahe'. (or Gau••'.) Te.t The fo\lowiog test, which we state without proof, is an extension of the Ratio Test.

a. Differentiate the series _1_= 1 I-x

+X+X2+ ... +X"+'"

toobtainaseriesforl/(l-x)'.

b. In ooe throw of two dice, the probability of gettiog a roll of7 is p ~ 1/6. If you throw the dice repeatedly, the probability that a 7 will appear for the fll'St time at the nth throw is qn-lp , where q ~ I - p ~ 5/6. The expected number of throws until a 7 frrst appears is ~:=1 nq"-Ip . Find the sum of this series. c. As an engineer applying statistical control to an industrial operation, yoo inspect items taken at randem from the assembly line. You classifY each sampled item as either "good" or

Chapter 10 Technology Application Projects ''bad.'' If the probability of an item's beiog good is p and of an item's beiog bad is q ~ I - p, the probability that the fust bad item found is the nth one inspected is p,,-lq. The average number iospected up to and iocluding the fust bad itero found is ~:lnpll-lq.Evaluatethis sum, assuming 0 < P < 1. 32. EIpected value Suppose that a random variable X may assome the values 1, 2, 3, ... , with probabilities Pt,P2,P3, ... , where Pi is the probability that X equals k (k ~ I, 2, 3, ... ). Suppose also that Pk '" 0 and that ~:'Pk ~ 1. The OIpected value of X, denoted by E(X), is the number ~:,kp1' provided the series converges. In each of the following cases, show that ~~'Pk ~ I and fmdE(X) ifit exists. (Hint: See Exercise 31.)

C. P1~

D 33.

c. Ifk ~ 0.01 h- 1 and to ~ IOh, fmd the smallestn such that

R. > (1/2)R. (Source: Prescribing Safe and Efficnve Dosage, B. Horelick and S. Kooot, COMAP, Inc., Lexington, MA.)

34. Time b_een drug do... (ContinUfJtion of Exercise 33.) If a drug is known to be ineffective below a concentration CL and harmful above some higher concentration CH , one needs to fmd values of Co and to that will produce a concentration that is sare (not above CH ) but effective (not below CL ). See the accompanyiog figure. We therefore want to fmd values for Co and to for which

R

51--1

~

CL

and

Co

+R

b.Pk~1

a.Pk=2--j;

6

I

I

"§ :0 C H

.S

Safe and effective dosage The concentration io the blood resultiog from a siogle dose of a drug nonnaily decreases with time as the drug is eliotioated from the body. Doses may therefore need to be repeated periodically to keep the conccn1ration from dreppiog below some particular level. One model for the effect of repeated doses gives the residual concentration just before the (n + I)stdoseas

Coe-kto

+ Coe-2kto + ... + Coe-nkto ,

where Co ~ the change io conceotratioo achievable by a single dose (mgjmL), k ~ the elimination constant (h-1) , and to ~ time between doses (h). See the accompanying figure.

~

CH.

C

I

k(k+ 1) ~k- k+ I

RII =

609

.g•

~ ~

Highest safe !evel

t

Co

CL

.j,

I

I

Lowest effective level

~ to~ 0

Time

Thus Co ~ CH - CL . Wheo !bese values are substituted io !be equatioo for R ohtsioed io part (a) of Exercise 33, the resultiog equatioo simplifies to

C

To reach an effective level rapidly, one might administer a "Ioadiog" dose that would produce a conccn1ration of CH mg/mL. This could be followed every to hours by a dose that raises the conceotratioo by Co ~ CH - CL mg/mL. a. VerifY the preceding equation for to.

Time (h)

a. Write R" in closed form as a single fraction, and fmd R =

litnn_oo R".

b. CaicuiateR, and RIO for Co ~ I mg/mL, k ~ 0.1 h- 1 , and to ~ 10 h. How good an estimate of R is RIO?

Chapter

2J!)

b. If k ~ 0.05 h- 1 and the highest safe conccn1ration is e times the lowest effective concen1ration, fmd the leogth of time b _ doses that will assure safe and effective concentrations. c. GivenCH~ 2mg/mL,CL ~ O.5mg/mL,andk~ 0.02h- 1 , determine a scheroe for administeriog the drug. d. Suppose that k ~ 0.2 h-1 and that the sma11est effective coocen1ratioo is 0.03 mgjmL. A siogle dose that produces a coocen1ratioo of 0.1 mg/mL is administered. About how long will the drug reroaio effective?

Technology Application Projects

Mathematica/Maple Modules: Bouncing BaH The model predicts the height of a bounciog ball, and the time until it stops bounciog. ~Ior Po/ynomiDIApproximll/ion.

ofII Function

A graphical animation shows the convergence of the Taylor polyoomials to functioos baviog derivatives of all orders over an ioterval io their domaios.

11 PARAMETRIC EQUATIONS AND POLAR COORDINATES OVERVIEW In this chapter we study new ways to define curves in the plane. Instead of thinking of a curve as the graph of a function or equation, we consider a more general way of thinking of a curve as the path of a moving particle whose position is changing over time. Then each of the x- and y-coordinates of the particle's position becomes a function of a third variable t. We can also change the way in which points in the plane themselves are described by using polar coordinates rather than the rectangular or Cartesian system. Both of these new tools are useful for describing motion, like that of planets and satellites, or projectiles moving in the plane or space. In addition, we review the geometric definitions and standard equations of parabolas, ellipses, and hyperbolas. These curves are called conic sections, or conics, and model the paths traveled by projectiles, planets, or any other object moving under the sole influence of a gravitational or electromagnetic force.

11.1

Parametrizations of PLane Curves In previous chapters, we have studied curves as the graphs of functions or equations involving the two variables x and y. We are now going to introduce another way to describe a curve by expressing both coordinates as functions of a third variable t.

Parametric Equations Figure 11.1 shows the path of a moving particle in the xy-plane. Notice that the path fails the vertical line test, so it cannot be described as the graph of a function of the variable x. However, we can sometimes describe the path by a pair of equations, x = f(t) and y = g(t), where f and g are continuous functions. When studying motion, t usually denotes time. Equations like these describe more general curves than those like y = f(x) and provide not only the graph of the path traced out but also the location of the particle (x, y) = (f(t), g(t» at any time t.

DEFINITION

Ifx andy are given as functions

x = f(t),

FIGURE 11.1 The curve or path traced by a particle moving in the xy-plane is not always the graph of a function or single equation.

610

y = g(t)

over an interval loft-values, then the set of points (x, y) = (f(t), g(t» defined by these equations is a parametric curve. The equations are parametric equations for the curve.

The variable t is a parameter for the curve, and its domain I is the parameter interval. If I is a closed interval, a :5 t :5 b, the point (f(a), g(a») is the initial point ofthe curve and (f(b), g(b) is the terminal point. When we give parametric equations and a parameter

11.1

611

Parametrizations ofplan. Curves

interval for a curve, we say that we have parametrized the curve. The equations and interval together constitute a parametrization of the curve. A given curve can be represented by different sets of parametric equations. (See Exercises 19 and 20.)

EXAMPLE 1

Sketch the curve defined by the parametric equations

y = I

+ I,

-00 < 1<00.

Solution We make a brief table of values (fable 11.1), plot the points (x, y), and draw a smooth curve through them (Figure 11.2). Each value of I gives a point (x, y) on the curve, such as I = I giving the point (I, 2) recorded in Table ILL Ifwe think of the curve as the path of a moving particle, then the particle moves along the curve in the direction of the arrows shown in Figure 11.2. Although the time intervals in the table are equal, the consecutive points plotted along the curve are not at equal arc length distances. The reason for this is that the particle slows down at it gets nearer to the y-axis along the lower branch of the curve as I increases, and then speeds up after reaching the y-axis at (0, I) and moving along the upper branch. Since the interval of values for I is all real numbers, there is no initial point and no terminal point for the curve. _ y

TABLE 11.1 Values of x~ t'andy~ t+ 1 for

t

,3 -

~

(9.4)

selected values of t. t

x

y

-3 -2 -I 0 I 2 3

9 4 I 0 I 4 9

-2 -I 0 I 2 3 4

t

=0 t~

::r---_

(9. -2)

t

FIGURE 11.2

x

(4. -1)

~-3

The curve given by the and y = t

parametric equations x =

,2

+1

(Example I).

EXAMPLE 2 Identify geometrically the curve in Example I (Figure 11.2) by eliminating the parameter I and obtaining an algebraic equation in x and y. Solution We solve the equation y = I + I for the parameter I and substitote the result into the parametric equation for x. This procedure gives I ~ Y - I and x ~ I' ~ (y - I)' = y' - 2y

y

+ I.

The equation x = y' - 2y + I represents a parabola, as displayed in Figure 11.2. It is sometimes quite difficult, or even impossible, to eliminate the parameter from a pair of parametric equations, as we did here. _

EXAMPLE 3 (a) x t

=

31T

2

FIGURE 11.3 The equations x ~ cos t and y = sin t describe motion on the circle + y' ~ 1. The arrow shows the direction of increasing t (Example 3).

x'

~

cos I,

(b)x~acosl,

Graph the parametric curves

y

=

Y

=

sint, asint,

o ,,;; I ,,;; 2".. o ,,;; I ,,;; 2"..

Solution (a) Since

x' + y' ~ cos'l + sin' I ~

I, the parametric curve lies along the unit circle x' + y' = I. As I increases from 0 to 2"., the point (x,y) = (cos I, sin I) starts at (1,0) and traces the entire circle once counterclockwise (Figure 11.3).

612

Chapter 11: Parametric Equations and Polar Coordinates (b) For x = a cos I, y = a sinl, 0 :5 1:5 2"., ""have x' + y' = a'cos' I + a' sin'l = a'. The parametrization describes a motion iliat begins at lhe point (a, 0) and 1Iaverses lhe circle x' + y' = a' once counterclockwise, returning to (a, 0) at I = 2".. The graph is a circle centered at 1he origin wilh radius r = a and coordinate points (a cos I, a sin I). •

EXAMPLE 4 The position P(x, y) of a particle moving in the xy-plaoe is given by the equations and parameter interval x

=

Vi,

y

=

t,

I '"

o.

Identify the path traced by the particle and describe the motion.

------~~~------~x

o

Starts at t =0

FIGURE 11.4 The equations x = Vi and y = t and lhe interval t '" 0 describe lhe palh of a particle iliat traces 1he right· hand half oflhe parabola y =

x'

(Example 4).

y

Solution We try to identify the path by eliminating I between the equations x = Vi and y = I. With any luck, this will produce a recogoizable algebraic relation between x and y. We find that

y = I =

(Vi), = x'.

Thus, the particle's position coordinates satisfy the equation y = x', so the particle moves along the parabola y = x'. It would be a mistake, however, to conclude that the particle's path is the entire parabola y = x'; it is only haIfthe parabola. The particle's x·coordinate is never negative. The particle starts at (0, 0) when I = 0 and rises into the frrst quadrant as I increases (Figure 11.4). The parameter interval is [0, 00) and there is no tenninal point. • The graph of any function y = /(x) can always be given a natural parametrization x = I and y = /(1). The domain of the parameter in this case is the same as the domain of the function/.

\ <-2.4) 1=:::2 \

EXAMPLE 5

x

--------~L---------+x

FIGURE 11.5 The palh defmed by x = t,y = t 2, -00 < t < 00 is the entire parabola y = (Example 5).

x'

A parametrization of the graph of the function /(x) = x' is given by =

t,

y = /(1) = I',

-00 < I < 00.

When I ;;" 0, this parametrization gives the same path in the JY·plane as we had in Exam· pie 4. However, since the parameter I here can now also be negative, we obtain the left· hand part of the parabola as well; that is, we have the entire parabolic curve. For this parametrization, there is no starting point and no terminaI point (Figure 11.5). • Notice that a parametrization also specifies when (the value of the parameter) a parti· cle moving along the curve is localed at a specific point along the curve. In Example 4, the point (2, 4) is reached when I = 4; in Example 5, it is reached "earlier" when I = 2. You can see the implications of this aspect of parametrizations when considering the possibility of two objects coming into collision: they have to be at the exact same location point P(x,y) for some (possibly different) values of their respective parameters. We will say more about this aspect of parametrizations when we stody motion in Chapter 13.

EXAMPLE 6

Find a parametrization for the line through the point (a, b) having slope m.

Solution A Cartesian equation of the line is y - b = m(x - a). If we set the parameter I = x - a, we find that x = a + I andy - b = mi. That is, x = a + t,

y=b+ml,

-00 < 1<00

parametrizes the line. This parametrization differs from the one "" would obtain by 1he tech· nique used in Example 5 when I = x . However, both parametrizations give the same line .

•

11.1 Parametrizations of Plane (urves

TABLE 11.2 Values of x = t + (l/t) and y = t - (l/t) for selected

EXAMPLE 7

Sketch and identifY the path traced by the point P(x, y) if x

values of t. I

1/1

x

y

0.1 0.2 0.4 1.0 2.0 5.0 10.0

10.0 5.0 2.5 1.0 0.5 0.2 0.1

10.1 5.2 2.9 2.0 2.5 5.2 10.1

-9.9 -4.8 -2.1 0.0 1.5

613

=

t

I

+ t'

Y

= I -

I

I> O.

t'

Solution We make a brief table of values in Table 11.2, plot the points, and draw a smooth curve through them, as we did in Example I. Next we eliminate the parameter I from the equations. The procedure is more complicated than in Example 2. Taking the difference between x and y as given by the parametric equations, we fmd that x - y =

(I + +) - (I - +) t. =

If we add the two parametric equations, we get

4.8 9.9

x

+y

=

(I + +) + (I - +)

=

21.

We can then eliminate the parameter I by multiplying these last equations together: y

10

(x - y)(x

+ y)

(t)<21)

=

=

4,

or, multiplying together the terms on the left-hand side, we obtain a standard equation for a hyperbola (reviewed in Section 11.6):

5

x2

-

y2 =

4.

(I)

Thus the coordinates of all the points P(x, y) described by the parametric equations satisfY Equation (I). However, Equation (I) does not require that the x-coordinate be positive. So there are points (x, y) on the hyperbola that do not satisfY the parametric equation -5 x = I + (1/1), I > 0, for which x is always positive. That is, the parametric equations do not yield any points on the left branch of the hyperbola given by Equation (I), points where the x-coordinate would be negative. For small positive values of I, the path lies in the -10 fourth quadrant and rises into the first quadrant as I increases, crossing the x-axis when t = 0.1 I = I (see Figure 11.6). The parameter domain is (0, 00) and there is no starting point and _ FIGURE 11.6 The curve for x = t + (l/t), no terminal point for the path. y = t - (I/t), t > 0 in Example 7. (The Examples 4,5, and 6 illustrate that a given curve, or portion ofi!, can be represented by part shown is for 0.1 s t '" 10.) different parametrizations. In the case of Example 7, we can also represent the right-hand branch of the hyperbola by the parametrization ~~~----~------~--~x

x=

\1'4+7,

y

=

t,

-00


00,

which is obtained by solving Equation (I) for x ;" 0 and letting y be the parameter. Still another parametrization for the right-hand branch of the hyperbola given by Equation (I) is x = 2 sec t,

y = 2 tan I,

1r

1r

-2 < 1<2'

This parametrization follows from the trigonometric identity sec2 I - tan2 I x2

HISTORICAL BIOGRAPHY

Christian Huygeos (1629-1695)

- y2 =

4 sec2 I - 4 tan2 I

=

4(sec2 I - tan2 I)

=

=

I, so

4.

As I runs between -1r/2 and TT/2, x = sec I remains positive and y = tan I runs between - 00 and 00, so P traverses the hyperbola's right-hand branch. It comes in along the

branch's lower half as 1--> 0- , reaches (2, 0) at I = 0, and moves out into the first quadrant as I increases steadily toward TT/2. This is the same hyperbola branch for which a portion is shown in Figure 11.6.

614

Chapter 11: Parametric Equations and Polar Coordinates

Cycloids

c:::::_ )

The prchlem with a pendulum clock whose bob swings in a circular arc is that the frequency of the swing depends on the amplitude of the swing. The wider the swing, the longer it ta1res the bob to return to center (its lowest position). This does not happen if the bob can be made to swing in a cycloid. In 1673, Christian Huygens designed a pendulum clock whose bob would swing in a cycloid, a curve we define in Example 8. He hung the bob from a fine wire constrained by guards that caused it to draw up as it swung away from center (Figure 11.7).

.:/

... .. Cycloid

•

FIGURE 11.7 In Huygens' pendulum clock, the bob swings in a cycloid, so the frequency is independent of the amplitude. y

P(x,y)

=

(at

EXAMPLE 8 A wheel of radius a rolls along a horizontal straight line. Find parametric equations for the path traced by a point P on the wheel's circumference. The path is called a cycloid.

+ acosO,a + a sin 0)

---------~

Solution We take the line to be the x-axis, mark a point P on the wheel, start the wheel with P at the origin, and roll the wheel to the right. As parameter, we use the angle t thrcugh which the wheel turns, measured in radians. Figure 11.8 shows the wheel a short while later when its base lies al units from the origin. The wheel's center C lies at (ai, a) and the coordinates of P are

x = at + a cos 6,

FIGURE 11.8 The position ofP(x,y) on the rolling wheel at angle t (Example 8).

y = a

To express 0 in terms of I, we observe that I + 0 y

o=

=

+ asinO.

3fT/2 in the figure, so that

3; - t.

This makes

FIGURE 11.9 The cycloid curve x = a(t - sint),y = all - cost), for t ~ O.

cosO

=

cos

(3; - I)

=

-sinl,

'0 = sm. (3fT T - I) =

sm

-cosl.

The equations we seek are x = at - asint,

y=a-acost.

These are usually written with the a factored out: x = a(1 - sinl),

B(a7T,2a)

y

y = a(1 - cos I).

Figure 11.9 shows the 1mt arch of the cycloid and part of the next.

FIGURE 11.10 To study motion along an

upside-down cycloid uoder the influence of gravity, we tum Figure 11.9 upside down. This points the y-axis in the direction of the gravitational force and makes the downward y-coordinates positive. The equations and parameter

interval for the cycloid are still x y

= a(t - sint), = a(J- cost),

t'" O.

The arrow shows the direction of increasing t.

(2)

•

Brachistochrones and Tautochrones If we turn Figure 11.9 upside down, Equations (2) still apply and the resulting curve (Figure 11.10) has two interesting pbysical properties. The 1mt relales to the origin 0 and the point B at the boltom of the first arch. Among all smooth curves joining these points, the cycloid is the curve along which a frictionless bead, subject only to the force of gravity, will slide from 0 to B the fastest. This makes the cycloid a braehistoehrone (''brah-kiss-toe-krone''), or shortest-time curve for these points. The second prcperty is that even if you start the bead partway down the curve toward B, it will still take the bead the same amount of time to reach B. This makes the cycloid a tautoehrone ("Iaw-toekrone"), orsarne-time curve for 0 andB.

Parametrizations of Plane Curves

11.1

615

Are there any other brachistochrones joining 0 and B, or is the cycloid the ooly one? We can fonnulate this as a mathematical question in the following way. At the start, the kinetic energy of the bead is zero, since its velocity is zero. The work done by gravity in moving the bead from (0, 0) to any other point (x, y) in the plane is mgy, and this must equal the change in kinetic energy. That is,

Thus, the velocity of the bead when it reaches (x, y) has to be v =

vzg;;.

That is, tis is the arc length differential along the bead's path.

or ds VI + (dy/dx)2 dx dl = - - = --~"=-"-----

vzg;;

vzg;;

The time Tf it takes the bead to slide along a particular path y = j(x) from 0 to B(a'IT, 2a) is _lx~aw

Tf-

I

+ (dy/dx)2 2 gy

x- o

(3)

dx.

What curves y = j(x) , if any, minimize the value of this integral? At flISt sight, we might goess that the straight line joining 0 and B would give the shortest time, but perhaps not There might be some advantage in having the bead fall vertically at first to build up its velocity faster. With a higher velocity, the bead coold travel a longer path and still reach B flISt. Indeed, this is the right idea. The solution, from a branch of mathematics known as the calculus o/varialions, is that the original cycloid from 0 to B is the one and only brachistochrone for 0 and B. While the solution of the brachistochrone problem is beyond our present reach, we can still show why the cycloid is a tautochrone. In the next section we show that the derivative dy/dx is simply the derivative dy/dl divided by the derivative dx/dl. Making the derivative calculations and substituting into Equation (3) (we oruit the details of the calculations here) gives Tcycloid =

=

l

I

x~aw

+ (dy/dx) 2 gy

x- O

dx

'=" a (2 - 2 cos I)

1

1=0

2ga( I - cos I)

dl

From Equations (2), tIx/d, ~ a(l - cost),

dy/d, ~ a sin', and Y = a(1 - cost)

Thus, the amount of time it takes the ftictiooless bead to slide down the cycloid to B after it is released from rest at 0 is 'IT v;;jg . Suppose that instead of starting the bead at 0 we start it at some lower point on the cycloid, a point (xo, YO) corresponding to the parameter value 10 > O. The bead's velocity at any later point (x, y) on the cycloid is v =

V2g(y -

YO) =

V'2g-a-(c-c-oS-I-o---Co-s--c I).

y ~ a(l -

cost)

616

Chapter 11: Parametric Equations and Polar Coordinates Accordingly, the time required for the bead to slide from (xo, YO) down to B is

T=

l

w

a'(2 - 2 cos I) ~lw dl= 2ga(cos lo - cost) g ~

~

_~r - "{gi"

w

I

2 sin'(t/2) dl '1(2 cos' (10/2) - I) - (2 cos' (t/2) - I)

fc; rw

=

V~il'

fc;l' V~

sin (1/2) dl Vcos'(to/2) - cos'(t/2)

1=1,

u -2 du c

-2du

=w

=

1 - cos I dl cosio - cosl

Ve' -

u'

= cos (t/2) = sin (t/2) dt = cos (to/2)

. _lU]'=W -_~[ g -sm C 1=10

~

fc; [ . -1

cos (t/2) ]W cos (to/2) I,

=

"V g

=

~ (-sin-1 0 + sin-1 1) = '1T~.

-sm

y

This is precisely the time it takes the bead to slide to B from O. It takes the bead the same amount of time to reach B no matter where it starts. Beads starting simultaneously from 0, A, and C in Figure 11.11, for instance, will all reachB at the same time. This is the reason that Huygens' pendulum clock is independent of the amplitode of the swing.

FIGURE 11.11 Beads released simultaneously on the upside·down eycloid at 0, A, and C will reach B at the same time.

Exercises 11.1 Finding Cartesian from Parametric Equations

15. x = sec'l - I,

Exercises 1-18 give parametric equations and parameter intervals for the motion of a particle in the xy-plane. Identify the particle's path by fmding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direetion of motion.

17. x

1. x 2. x

= 31, Y = -Vi,

= 9t 2 ,

-00

y = I,

I '"


4.x=3-3t, y=2t, 5. x

= cos2t,

y

= 4 cos t,

y

=

8. x = 4sint, y = 5 COS I,

9. x

to.

x

=

sint, Y

=

1 + sint, y

=

11. x = t 2 , Y = t 6

cos2t, =

14. x

00

os I s '1t

O:s t:s 21T

0:5

t:5

-

2'7T

-f:5 It:5

\Il=7, = '\I't+l, y = Vi,

-00


t:5w 00

-I <1< I -I

s

I '"

I

°

s

-cosht, Y

=

-'tt/2 < 1< 'tt/2

sinht,

18. x = 2 sinh I, Y = 2 cosh I,

-00


00

-00 < 1<00

a. once clockwise.

b. once counterclockwise.

c. twice clockwise. d. twice counterclockwise. (There are many ways to do these, so your answers may not be

Os I s '1t

cost- 2, 0:5

2t4, I 1-2 12. x=t=1' y= 1+ I' 13. x = I, Y =


sin('1t - I),

y = 28int,

=

19. Find parametric equations and a parameter interval for the motion ofaparticle that starts at (a, 0) and traces the circle x' + y' = a'

0:51:51

= sin21,

6. x = cos('1t - I),

7. x

-00

16. x = -seCI, Y = tanl,

-'1t/2 < I < 'tt/2

finding Parametric Equations

00

°

3. x = 21 - 5, Y = 41 - 7,

y = tan I,

°

the same as the ones in the back of the book.) 20. Find parametric equations and a parameter interval for the motion of a particle that starts at (a, 0) and traces the ellipse (x'/a') + (y'/b') = I

a. once clockwise. c. twice clockwise.

b. once counterclockwise.

d. twice counterclockwise.

(As in Exercise 19, there are many correct answers.) In Exercises 21-26, fmd a parametrization for the curve. 21. the line segment with endpoints (-I, -3) and (4, I) 22. the line segment with endpoints (-1,3) and (3, -2)

11.1 23. the lower half of the parabola x - I = y2 24. the left half of the parabola y

=

x2

+ 2x

25. the ray (half line) with initial point (2, 3) that passes through the point (-I, -I)

Parametrizations ofplane Curves

617

34. Find aparametrizatioo for the circle x' + y' = 1 startiog at (1,0) aod moving counterclockwise to the temtinal point (0, I), using the aogle 8 in the accoropaoying figure as the parameter. y

26. the ray (half line) with initial point (- I, 2) that passes through the point (0, 0)

(0, 1)

27. Find parame1ric equations aod a parameter interval for the motioo of a particle startiog at the point (2, 0) aod tracing the top half of

the circle x 2 + y2 = 4 four times. 28. Find parame1ric equations aod a parameter interval for the motioo of a particle that moves along the graph of y = x' in the following way: beginning at (0, 0) it moves to (3, 9), and then travels back and forth from (3, 9) to (-3, 9) infmitelymaoy times. 29. Find parametric equations for the semicircle x2

+ y2

using as parameter the slope t at (x,y).

=

a 2,

y

> 0,

= dy/ fix of the taogent to the curve

30. Find parame1ric equatioos for the circle x2

+ y2 =

a2,

using as parameter the arc length s measured counterclockwise from the point (a, 0) to the point (x, y). 31. Find a parametrizatioo for the line segroent joining points (0, 2) aod (4,0) using the aogle 8 in the accompanying figure as the parameter. y

(1,0) x

-2

35. The witch of Maria Agnesi The bell-shaped witch of Maria Agoesi cao be constructed in the following way. Stsr! wi1h a circle of radius 1, centered at 1he point (0,1), as showo in the accompanying figure. Choose a pointA on the line y = 2 aod connect it to the origin with a line segment Call the point where the segment crosses 1he circle B. Let P be the point where 1he vertical line through A crosses the horizontal line through B. The witch is 1he curve traced by P as A moves along 1he line y = 2. Find pararne1ric equations aod a panuneter interval for the witch by expressing the coordinates ofP in terms of t, the radian measure of the angle that segment D A makes with 1he positive x-axis. The following equalities (which you may asswne) will help.

b.y=2-ABsint

.. x =AQ c. AR·DA = (AQ)2

y

Q

A

~~~--------~~--+x

o

4

32. Find a parametrization for the curve y = Vx with temtinal point (0, 0) using the aogle 8 in the accompanying figure as the parameter.

----------~~~-L---------+x

o

y

y=Vx

/~

/ ~~/ ~8~____________-+x

o

33. Find a parametrizatioo for the circle (x - 2)' + y2 = I startiog at (I, 0) aod moving clockwise once around the circle, using the central aogle 8 in the accompaoying figure as the parameter. y

o

36. Hypocycloid When a circle rolls 00 1he inside of a fixed circle, any point P on the circmnference of 1he rolling circle describes a hypocycloid. Let 1he fixed circle be x' + y2 = a', let the radius of the rolling circle be b, and let the initial position of the tracing point P be A (a, 0). Find parame1ric equations for the hypocycloid, using as the parameter the angle 8 from the positive x-axis to the line joiniug 1he circles' centers. In particular, if b = a/4, as in 1he accompaoying figure, show that 1he hypocycloid is 1he astroid

x = a cos3 0, y = a 8in3 O. y

618

Chapter 11: Parametric Equations and Polar Coordinates

37. As the point N moves along the line y = a in the accompanying figure, P moves in such a way that OP = MN. Find parametric equations for the coordinates of P as functions of the angle I that the line ON makes with the positive y-";s.

+ 3,

y

=

12 - I,

t - sin t, y

=

1 - cos t,

21

=

=

-2:5 I :5 2 over

•• Osls2'1f C. '1T

y

x

43. Parabola

44. Cycloid x

::s t ::s 31T.

45. Deltoid

A(O, a)

x = 2cosl

+ cos21,

y = 2sinl - sin21;

0 SIS 2'1f

What happens if you replace 2 with -2 in the equations for x and

y? Graph the new equatioos and fmd out 46. A nic. curve x = 3cosl

--------~~o ~~----------~x

+ cos 31,

y = 3 sin I - sin31;

0 SIS 2'1f

What happens if you replace 3 with - 3 in the equations for x and 38. Trochoid. A wheel of radius a rolls along a horizontal straight line without slipping. Find parametric equations for the curve traced out by a pointP 00 a spoke of the wheel b units from its ceoter. As parameter, use the aogle 8 through which the wheel turns. TIre curve is called a trochoid, which is a cycloid v.lren b = a. Distance U';ng Parametric Equations 39. Find the point on the parabola x = I,y = 12, -00 < I < 00, closest to the point (2, 1/2). (Hinl: Minimize the square of the distance as a function of I.)

40. Find the point on the ellipse x = 2 cos t,Y = sin t, 0 ::s; t ::s; 2w closest to the point (3/4, 0). (Hinl: Minimize the square of the distance as a function of I.)

D GRAPHER EXPLORATIONS If you have a parametric equatioo grspher, grsph the equatioos over the given intervals in Exercises 41-48.

41. Ellipse x = 4 cos t, Y a. 0 ~ t ~ 2'17"

= 2 sin t,

47••• Epicycloid

x = 9cosl - cos 91, Y = 9sinl - sin91;

0 SIS 2'1f

b. Hypocycloid

x = 8east + 200841, y = 8sint - 2sin4t;

O::s t::s 2w

c. Hypotrocboid

x = cost

+

500831,

48••• x = 6cosl

o :s; t

c. x = 6cosl =

= 6cost -

+ 5 cos 31,

o

d. x

y

58in3t;

O::s t::s 2w

y = 6sinl - 5 sin 31;

:s; 2w

b. x = 6 cos 21 ::s t ::s 1T

o ::s t

over

+

5 cos 61,

+ 5 cos 31,

y = 6 sin 21 - 5 sin 61;

y = 6 sin 21 - 5sin3t;

::s 2w

6 cos 2t

+ 5cos6t,

y = 6sin4t - 5sin6t;

OStS1T

c. -'1f/2 :5 I :5 '1f/2. 42. Hyperbola brancb x = sec I (enter as I/cos (I)), y ter as sin (I)/COS (I», over •• -1.5:5 1:5 1.5

y? Graph the new equatioos and fmd out

= tan I (en-

b. -0.5 :5 I :5 0.5

c. -0.1 :5 I :5 0.1.

11.2

Calculus with Parametric Curves In this section we apply calculus to parametric curves. Specifically, we fmd slopes, lengths, and areas associated with parametrized curves.

Tangents and Areas A parametrized curve x = /(1) and y = g(l) is differentiable at 1 if / and g are differentiable at I. At a point on a differentiahle parametrized curve where y is also a differentiable function of x, the derivatives dy/dl, dx/dt, and dy/dx are related by the Chain Rule:

dy dy dx dt=dx'dt'

619

11.2 Calculus with Parametric Curves Ifdx/dloF 0, we may divide both sides oflhis equation by dx/dl to solve fordy/dx.

Parametric Formula for dy / dx

If all three derivatives exist and dx/ dl oF 0, dy dx

dy/dl dx/dl·

(1)

If parametric equations define y as a twice-differentiable function of x, we can apply Equation (1) to the function dy/dx = y' to calculate d 2y/dx 2 as a function of I: d'y d dy'/dl dx 2 = dx (y') = dx/dl .

Eq. (\) with y' in place ofy

Parametric Formula for d'y/dx'

If the equations x = /(1), y = g(1) define y as a twice-differentiable function of

x, then at any point wbere dx/dl oF 0 andy' = dy/dx,

y

d 2y dx 2

2

EXAMPLE 1

dy'/dl dx/dl·

(2)

Find the tangent to the curve

--~--~+---~--~X

o

x = sect,

_7T
y = tan I,

2

2'

at the point (v2, 1), where 1= 7T/4 (Figure 11.12). FIGURE 11.12

The curve in Example I

is the right-hand branch of the hyperbola xl -

y2

=

Solution

The slope of the curve at I is

1.

dy dx

sec'1 sec I tan I

dy/dl dx/dl

sec I tan r

Eq. (\)

Setting I equal to 7T/4 gives dy I sec (7T/4) dx t~"/4 = tan (7T/4) =

'?

=

v2.

The tangent line is y - I = y = y =

EXAMPLE 2

v2 (x - v2) v2x - 2 + 1 v2x - I.

Findd'y/dx 2 asafunctionoflifx = I - 12,y = I - 13.

Solution

1.

Express y' = dy/ dx in terms of I. dy y' = dx

dy/dl dx/dl

1 - 31 2 1 - 21

•

620

Chapter 11: Parametric Equations and Polar Coordinates Differentiate y' with respect to t.

2.

I Finding d'y/dx' in Term. of t

1. Expressy' = dy/dx in terms oft. 2. Finddy'/dt. 3. Divirk dy'/dtby dx/dt.

2 2 (I 3t ) = 2- 6t + 6t 2t (I - 2t)2 dt I

=.4.

dy' dt

3.

Derivative Quotient Rule

Divide dy' / dt by dx/ dt.

+ 6t 2)/( I

(2 - 6t

2-6t+6t 2 (I - 2t)3

- 2tj2

I - 2t

EXAMPLE 3

y

•

Find the area enclosed by the astroid (Figure 11.13)

x

=

cos3 t,

y

=

o ,,;; t ,,;; 2'1f.

sin3 t,

Solution By symmetry, the enclosed area is 4 times the area beneath the curve in the first quadrant wbere 0 ,,;; t ,,;; 'If/2. We can apply the def"mite integral formula for area studied in Chapter 5, using substitution tu express the curve and differential dx in terms of the parameter t. So,

~~----~+-----~~~X

-1

A

-1

Figure 11.13

Eq. (2)

The astroid in Example 3.

= 411

= 4

ydx

("/2

10

sin3 t'3 cos2 t sin tdt

= 121"/2

y

=

"210

=

~ [1"/2 (I

=

+

2t

3 {"/2 "210 (I

=

c

e-~os ye ~os

=

3 {"/2

Substitution for yand ax

2t) dt

sin4 t

- 2 cos 2t + cos2 2t)(1 + cos 2t) dt

(I - cos 2t - cos2 2t + cos3 2t) dt - cos 21) dt - 1"/2 cos2 21 dt

~ [(I - t sin 2t) -

t (I + i

sin 2t)

+

=

(1 - 2cos 21)'

Expand square term.

Multiply tenns.

+ 1"/2 cos3 2t dl]

t

(Sin2t - tSin321)r

M(; -0- 0- 0) - t (; + 0- 0- 0) + t(O - 0- 0+ 0)] 3'1f

Section 8.2, Example 3 Evaluate.

•

8'

Length of a Parametrically Defined Curve P,

-co+---------------------~x

FIGURE 11.14 The smoo1h curve C defmed parametrically by the equations x = f(l) aod y = g(t), a :5 t :5 b. The length of the curve from A to B is approximated by the sum of the leagths of the polygonal path

(s1raight line segmeols) starting alA thea to P, , aod so on, ending at B

=

Po,

= p •.

Let C be a curve given parametrically by the equations

x

=

/(t)

and

y

=

g(t),

a";;

I";; b.

We assume the functions / and g are continuously differentiable (meaning they have continuous fIrst derivatives) on the interval [a, b]. We also assume that the derivatives /' (I) and g' (I) are not simultaneously zero, which prevents the curve C from having any comers or cusps. Such a curve is called a smooth curve. We subdivide the path (or arc) AB into n pieces at points A = Po, P lo P2 , ••• , Pn = B (Figure 11.14). These points correspond to a partition of the interval [a, b] by a = to < t1 < t2 < ... < tn = b,

11.2 Calculus with Parametric Curves

621

where Pk = (j(lk),g(lk». Join successive points of this subdivision by straight line segments (Figure 11.14). A representative line segment bas length

y

P, ~ (t(t,), g(,,»

Lk = Y(dxk)'

I

: AYk

= Y[j(lk) -

I

_________ -.-l

Ax,

P'-1

~ (J("-I),

g("-I»

~--------------------~X

+

(dYk)2

/(lk_I)]2 + [g(lk) - g(lk-l)l'

(see Figure 11.15). If Mk is small, the length Lk is approximately the length of arc Pk-IPk. By the Mean Value Theorem there are numbers I; and I;' in [10-- h Iii such that

o

dxk = /(Ik) - /(Ik-I) = 1'(1;)

FIGURE 11.15 The arc P.-1P. is approximated by tire straight line segment shown bere, which baa length

L, =

V(ax.), + (dy,)' ,

M..

dYk = g(lk) - g(lk-l) = g'(I;') dlk'

Assuming the path from A to B is traversed exactly once as I increases from I = a to I = b, with no doubling back or retracing, an approximation to the (yet to be defined) "length" of the curve AB is the sum of all the lengths Lk:

Although this last sum on the right is not exactly a Riemann sum (because I' and g' are evaluated at different points), it can be shown that its limit, as the norm of the partition tends to zero and the number of segments n ---> 00, is the definite integral

.

lim ~ Y[j' (1;)1'

11111~O

t;1

+ [g'(I;')l' dlk

=

l

b

Y[j'(I)l'

a

+ [g'(I)l'dl.

Therefore, it is reasonable to define the length of the curve from A to B as this integral.

DEFINITION If a curve C is derfied parametrically by x = /(1) and Y = g(1), a :5 1:5 b, where I' and g' are continuous and not simultaneously zero on [a, b], and C is traversed exactly once as I increases from I = a to I = b, then the length of C is the definite integral b L = Y[j'(1)]2 + [g'(1)]2 dl.

l

A smooth curve C does not double back or reverse the direction of motion over the time interval [a, b] since (j,)2 + (g'), > 0 throughout the interval. At a point where a curve does start to double back on itself, either the curve fails to be differentiable or both derivatives must simultaneously equal zero. We will examine this phenomenon in Chapter 13, where we study tangent vectors to curves. If x = /(1) and Y = g(I), then using the Leibniz notation we have the following result for arc length: L = [

(:)'

+

(~)' dl.

(3)

What if there are two different parametrizations for a curve C whose length we want to find; does it matter which one we use? The answer is no, as long as the parametrization we choose meets the conditions stated in the definition of the length of C (see Exercise 41 for an example).

622

Chapter 11: Parametric Equations and Polar Coordinates

EXAMPLE 4

Using the definition, find the length of the circle of radius r defined para-

metrically by

x Solution ence is

=

rcost

y

and

o ,,;; t

rsint,

=

,,;; 21T.

As t varies from 0 to 21T, the circle is traversed exactly once, so the circumfer2"

1

L=

dx)2 + (dy)2 dt dt. ( dt

o

We find

dx dt

=

dy dt = rcost

. -rsmt,

and

So

L =

EXAMPLE 5

{2"

10 W

r[t l~" =

dt =

•

21Tr.

Find the length of the astroid (Figure 11.13)

x = cos3 t,

y

o ,,;; t

3 t,

= 8in

,,;; 21T.

Solution

Because of the curve's symmetry with respect to the coordinate axes, its length is four times the length of the first-quadrant portion. We have

x

=

(:Y

cos 3 t,

= [3 cos

dy ( dt dx)2 ( dt

+

)2 =

y 2

3 = 8in

t

t( -sin t)f = 9 cos' t sin2 t

[3sin2 t(cost)12 = 9sin'tcos2 t

2 2 2 (dy)2 dt = Y9cos tsin t(cos t

+ sin2 t) 1

=

Y9 cos2 t sin2 t

= 31costsintl =

cos tsin t ~ 0 for t S 'Tr/2

3costsint.

QS

Therefore, Length of first-quadrant portion =

1"/2

3 cos t sin t dt

3 ("/2

=

2. 10

=

-~Cos2tr = l

The length of the astroid is four times this: 4(3/2) = 6.

sin 2t dt

cos tsin t = (1/2) sin 2,

•

11.2 Calculus with Parametric Curves

623

HISTORICAL BIOGRAPHY

Length of a Curve y = !(x)

Gregory SI. Vincent

The length funnula in Section 6.3 is a special case of Equation (3). Given a continuously differentiable function y = I(x), a :5 x :5 b, we can assign x = I as a parameter. The graph of the function I is then the curve C defmed parametrically by

(1584-1667)

x=1

a

y = 1(/),

and

~

t

~

b,

a special case of what we considered before. Then,

ax = dl

I

and

dy

dy/dl

:

= /'(/).

From Equation (I), we have

ax = ax/dl =

/'(1),

giving

(:)' + (:)' =

I

+ [('(/)]2

=1+[{'(x)12.

t~x

Substitution into Equation (3) gives the arc length formula for the graph of y = I(x), consistent with Equation (3) in Section 6.3.

The Arc Length Differential Consistent with our discussion in Section 6.3, we can defme the arc length function for a parametrically defmed curve x = 1(1) and y = g(1), a :5 I :5 b, by

S(/) =

l'

V[{'(z»), + [g'(z»), dz.

Then, by the Fundamental Theorem of Calculus,

lis dz

=

V[{'(/»)' + [g'(/»)'

=

~(ax)2 + dz

(dy)2 dz'

The differential of arc length is

ds

~ ( : ) ' + (:)' dl.

=

(4)

Equation (4) is often abbreviated to

lis

=

Vax 2 + dy2.

Just as in Section 6.3, we can integrate the differential lis between appropriate limits to find the total length of a curve. Here's an example where we use the arc length formula to fmd the centroid of an arc.

EXAMPLE 6

Find the centroid of the first-quadrant arc of the astroid in Example 5.

Solution We take the curve's density to be [; = I and calculate the curve's mass and moments about the coordinate axes as we did in Section 6.6.

624

Chapter 11: Parametric Equations and Polar Coordinates

y

The distribution of mass is symmetric about tbe line y = x, so X = ment oftbe curve (Figure 11.16) has mass

y. A typical seg-

B(O,I) ,

"

(x,j)

xX, ds l

,-? I

,

.

= (cos3 t, sin3 t)

dm= l'ds= )(:)' + (t)'dl=3COSISinldl.

From Example 5

',

c,m. ,

,

""

The curve's mass is

~~------~~~x

o

A(I,O)

M= 10{"/2 dm = 10{"/2 3 cos I sin I dl = 2'3

FIGURE 11.16 The centroid (c.m.) of the astroid arc in Example 6.

Again from Example 5

The curve's moment about the x-axis is Mx =

("/2 ydm = 10 sin J

3

= 3

,,/2

1 o

1 '3coslsinldl

sin4 I cos I

dl = 3. sm.551]"/2 = -.53 0

It follows tbat

__ Mx _ 3/5 _ I y - M - 3/2 - 5'

•

The centroid is tbe point (2/5,2/5).

Areas of Surfaces of Revolution In Section 6.4 we found integral formulas for tbe area of a surface when a curve is revolved about a coordinate axis. Specifically, we found tbat tbe surface area is S = !21fY tis for revolution about tbe x-axis, and S = !21fx tis for revolution about tbe y-axis. If tbe curve is parametrized by tbe equations x = 1(1) and y = g(1), a ,;; I ,;; b, where I and g are continuously differentiable and 0 on [a, b], tben tbe arc lengtb differential tis is given by Equation (4). This observation leads to tbe following formulas for area of surfaces of revolution for smootb parametrized curves. Area of Surface of Revolution for Parametrized Curves

If a smootb curve x = 1(1),y = g(I), a ,;; I';; b, is traversed exactly once as I increases from a to b, tben tbe areas of tbe surfaces generated by revolving tbe curve about tbe coordinate axes are as follows.

1. Revolution about the x-axis (y

;0:

0):

""c----o-;'"

S=

[21TY (:)' + (t),dl

2. Revolution about the y-axis (x

;0:

(5)

0): ~~c--~~

S=

[21TX (:)' + (t),dl

(6)

As witb lengtb, we can calculate surface area from any convenient parametrization tbat meets tbe stated criteria.

11.2 Calculus with Parametric Curves Circle

625

EXAMPLE 7 The standard parametrization of the circle of radius 1 centered at the point (0, 1) in the xy-plane is

y

x = cost,

y = 1

o~

+ sint,

t

~

21T.

Use this parametrization to find the area of the surface swept out by revolving the circle about the x-axis (Figure 11.17). Solution

We evaluate the formula Eq. (5) fOT revolution about the x-axis; y= 1 + sint;;;: 0

dx)2 (dy)2 ( dt + dt dt

x

{27r

= 21TJo (1 + sin t) dt FIGURE 11.17 In Example 7 we calculate the area of the surface of revolution swept out by this parametrized curve.

•

Exerdses 11.2 Tangents to Parametrized Curves

18. x sin t

In Exercises 1-14, find an equation for the line tangent to the curve at the point defmed by the given value of t. Also, fmd the value of d 2y / dx 2 at this point. 1. x = 2 cos t, Y = 2 sin t, t = Tr/4 2. x = sin 2Trt, y = cos 2Trt, t = -1/6 3. x = 4 sin t, y = 2 cos t, t = Tr/4

19. x = t 3

v3 cos t,

4. x = cos t, Y = 5. x = t,

y = tant,

'\I't+l,

9. x = 2t 2 + 3, 10. x = l/t,

y = t

t = 2Tr/3

1

v'3t, ,

Area 21. Find the area under one arch of the cycloid x = aCt - sin t),

t

14. x = t + e t , y = 1 - e t ,

x = a cos t,

17. x

+ 2x 3/ 2 =

t2

t = 2 t = 0

+ 3)3/2/3, y 8 cos t + 8t sin t

Yr, t = 4 y'\l't+l + 2tV;; =

yet - 1) =

+ t,

0

~

t

~

2Tr.

x 3 over [0, 1] using the following

Find the lengths of the curves in Exercises 25-30. 25. x = cos t, Y = t + sin t, 0 ~ t ~ Tr 26. x = t\ y = 3t 2/2, 0 ~ t ~ v'3 27. x = t 2/2, Y = (2t + 1)3/2/3, 0 ~ t ~ 4

t = Tr/2

Implicitly Defined Parametrizations

Vi,

=

Lengths of Curves

t = Tr/3

Assuming that the equations in Exercises 15-20 defme x and y implicitly as differentiable functions x = jet), y = g(t), find the slope of the curve x = jet), y = g(t) at the given value of t. 15. x 3 + 2t2 = 9, 2y 3 - 3t 2 = 4, t = 2

Y5 -

y = b sin t,

24. Find the area under y parametrizations.

28. x = (2t

16. x =

+ e-t •

t = 1

+ sint,

y =~,

y = a(l - cos t).

23. Find the area enclosed by the ellipse

t = -1

+ In t,

t = 0

x = t - t 2, Y = 1

t= 3

y = 1 - cos t,

12. x = cost, Y = 1

t+1'

4

t sin t - 2t = y, t = Tr = 2x + t 2, t = 1

+ 2t 3

20. t = In (x - t), Y = tet,

t = Tr/6

y =

Y = -2

11. x = t - sin t,

13. x =

Y

22. Find the area enclosed by the y-axis and the curve

t = 1/4 6. x = sec2 t - 1, y = tan t, t = -Tr/4 8. x = -

= t,

Yr,

Y =

7. x = sect,

+ 2x + t,

4,

t = 0

29. x = y = 8 sin t - 8t cos t, o ~ t ~ Tr/2

= t

+

t 2/2,

0 ~ t ~ 3

30. x = In (sec t + tan t) - sin t y = cos t, 0 ~ t ~ Tr/3

Surface Area Find the areas of the surfaces generated by revolving the curves in Exercises 31-34 about the indicated axes. 31. x = cos t, Y = 2 + sin t, 0 ~ t ~ 2Tr; x-axis

626

Chapter 11: Parametric Equations and Polar Coordinates

32. x = (2/3)1'/2, y = 2V1,

0 S I '"

\13;

y

y-axis

2

33. x = I + V2, y = (1 /2) + V2I, - V2 '" I'" V2; y-axis 34. x = In (sec I + tan I) - sin I, Y = cos I, 0 '" I '" 1f/3; x-axis

3

35. A cone frustum

The line segment joining the points (0, I) and (2, 2) is revolved about the x-axis to generate a frustum of a cone. Find the surface area of the frustum using the parametrization x = 21,y = 1 + 1,0'" 1 S 1. Checkyourresuitwiththegeometry fonnula: Area = 1f(r, + ,.,)(slant height).

36. A cone The line segment joining the origin to the point (h, r) is revolved about the x-axis to generate a cone of height h and base radius r. Find the cone's surface area with the parametric equations x = hI, Y = rt, 0 :5 t :s; 1. Check your result with the

44. The curve with parametric equations

x=l, y= I-cost,

geome1ry fonnula: Area = 1fr(slant height).

0:5t:52w

Centroids

is called a sinusoid and is shown in the accompanying figure. Find the point (x, y) where the slope of the tangent line is

37. Find the coordinates of the cen1roid of the curve

.. largest

x

= cost + t sin t,

b. smallest

Y = sint - teost, 0:5 t :511'/2.

y

38. Find the coordinates of the cen1roid of the curve

x = et cos t, y = et sin t, 0:5 t :s

2

1T'.

39. Find the coordinates of the cen1roid of the curve x = cos t, Y

D 40.

= t + sin t,

0:5 t

:5 'Tl'.

Most cen1roid calculations for curves are done with a calculator or computer that has an integral evaluation program. As a case in point, fmel, to the nearest hundredth, the coordinates of the cen1roid of the curve

x = I', Y = 312/2,

0

SIS

--~~------------~~--+x

o

D

0.

Theory and Examples 41. Length is independent of parametrization To illustrate the fact that the ournbers we get for length do not depend on the way we parametrize our curves (except for the mild res1rictions preveoling

2~

The curves in Exercises 45 and 46 are called Bawdilch curves or Lissajous figures. In each case, find the point in the interior of the fust quadrant where the tangent to the curve is horizontal, aod find the equations of the tvro tangents at the origin.

45.

46. y

y

x = sint y = sin2t

x = sin2t y=sin3t

doubling hack mentioned earlier), calculate the length of the sentiwith these tvro different parametrizations: circle y =

'\1'1=7

•• x b. x

= cos2t, = sin 1f1,

= sin21, Y = cos 1f1, y

0 S IS 1f/2 -1/2 '" I S 1/2

42. •• Show that the Cartesisn formula 47. Cycloid for the length of the curve x = g(y), c S y '" d (Section 6.3, Equation 4), is a special case of the parametric length formula

.. Find the length of one arch of the cycloid x = a(1 - sin I),

y = a(1 - cos I).

b. Find the area of the surface generated by revolving one arch of the cycloid in part (a) about the x-axis for a = 1. Use this result to fmd the length of each curve.

b. x

= y'/2,

c. x

2

=

t

0

S

Y

S

4/3

48. Volume Find the volume swept out by revolving the region bounded by the x-axis and one arch of the cycloid

x=t-sint, y=l-cost

y f3,

0'" y '" I about the x-axis.

43. The curve with parametric equations x = (I

+

2 sin 8) cos 8,

y = (I

+ 2 sin 8) sin 8

is called a limafon and is shown in the accompanying figure. Find the points (x,y) and the slopes of the tangent lines at these points for

•• 8 = O.

b. 8 = 1f/2.

c. 8 = 4'1f/3.

COMPUTER EXPLORATIONS In Exercises 49-52, use a CAS to perfonn the following steps for the given curve over the closed interval.

.. Plot the curve together with the polygonal path approximations for n = 2, 4, 8 partition points over the interval. (See Figure 11.14.)

11.3

_1

b. Find the corresponding approximation to the length of the

curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for n ~ 2, 4, 8 with the actual length given by the integral. How does the actual length compare with the approximations as n increases? Explain your answer.

3

I ,

49.x- 3 1,

Y~21,

SO. x ~ 21 3

161'

-

Polar Coordinates

627

0"'1"'1

+ 251 + 5,

y ~ I'

+ 1-

OSls6 51. x = t - cost, Y = 1 + sint, -1r::S IS 52. x = etcost, y = elsint, O::s t::s 'T1'

3, 1T'

Polar Coordinates

11.3

In this section we study polar coordinates and their relation to Cartesian coordinates. You will see that polar coordinates are very useful for calcolating many multiple integrals studied in Chapter 15.

~""

Origin (pole)

J

8

)X

Initial ray

FIGURE 11.18 To derme polar coordinates for the plane, we start with an origin, called the pole, and an initial ray.

Definition of Polar Coordinates To derme polar coordinates, we first fix an origin 0 (called the pole) and an initial ray from 0 (Figure 11.18). Then each point P can be located by assigning to it a polar coordinate pair (r, II) in which r gives the directed distance from 0 to P and II gives the directed angle from the initial ray tu ray OP.

Polar Coordinates

P(r, II) Directed distance from 0 to P

o ~---,-L----------~x

Initial ray 9~O

FIGURE 11.19 Polar coordinates are not unique.

FIGURE 11.20 Polar coordinates can have negative r-values.

"

Directed angle from initial ray to OP

As in trigonometry, II is positive when measured counterclockwise and negative when measured clockwise. The angle associated with a given point is not unique. While a point in the plane has just one pair of Cartesian coordinates, it has infinitely many pairs of polar coordinates. For instance, the point 2 units from the origin along the ray II ~ 1r/6 has polar coordinates r ~ 2, II ~ 1r/6. It also has coordinates r = 2, II = -111r/6 (Figure 11.19). In some situations we allow r to be negative. That is wby we use directed distance in derming P(r, II). The point P(2, 71r/6) can be reached by turning 71r/6 radians counterclockwise from the initial ray and going forward 2 units (Figure 11.20). It can also be reached by turning 1r/6 radians counterclockwise from the initial ray and going backward 2 units. So the point also has polar coordinates r = -2, II = 1r/6.

EXAMPLE 1 ~ o~---'----"----;;----;o;->x

)

Find all the polar coordinates of the point P(2, 1r/6).

Solution We sketch the initial ray of the coordinate system, draw the ray from the origin that makes an angle of 1r/6 radians with the initial ray, and mark the point (2, TT/6) (Figure 11.21). We then find the angles for the other coordinate pairs of P in which r = 2 andr = -2. For r = 2, the complete list of angles is 7T

6'

7T

7T

7T

6 ± 21r, 6 ± 41r, 6 ± 61r, ....

628

Chapter 11: Parametric Equations and Polar Coordinates

(2. *) ~ (_2._5;) ~ (-2. 7;)

For r = -2, the angles are

_ 5'1f ± 4'Il" 6 '

5'1f -T± 6'1f, ....

The corresponding coordinate pairs of P are

(2, ~ + 2n'lf).

n

=

0, ±l, ±2, ...

n

=

and

FIGURE 11.21 The point P(2. 'IT/6) has inImitely many polar coordinate pairs (Example 1).

5'1f + ( -2'-6

2n'lf) '

0, ±l, ±2, ....

When n = 0, the formulas give (2,'If/6) and (-2, -5'1f/6). When n = I, they give (2, 13'1f/6) and (-2, 7'1f/6) , and so on. _

Polar Equations and Graphs If we hold r fixed at a constant value r = a oF 0, the point P(r, 6) will lie Ia Iunits from the origin O. As 0 varies over any interval oflength 2'1f, P then traces a circle of radius Ia I centered at 0 (Figure 11.22). If we hold 6 fixed at a constant value 6 = 60 and let r vary between - 00 and 00, the point P(r, 0) traces the line through 0 that makes an angle of measure 00 with the initial ray. FIGURE 11.22 The polar equation for a circle is r = a.

Equation

Graph

r=a 6 = 60

Circle of radius Ia Icentered at 0 Line through 0 making an angle 60 with the initial ray

y

(a)

1::sr:S2.0:S(J:S~

EXAMPLE 2 ----c+---1'-'-~x

o

1

2


(h) 6 = 'If/6,6 = 7'1f/6, and 6 = -5'1f/6 are equations for the line in Figure 1l.21. y

(b)

~

~

'8 =~

4'

--~'c--=-~x

_

Equations of the form r = a and 6 = 60 can be combined to derme regions, segments, and rays.

EXAMPLE 3

Graph the sets of points whose polar coordinates satisfy the following

conditions.
0';;6';;;

and

~ ~

(e)

2" 3

Y

(h) -3';;r,;;2 < 6 < 5'1f (0) 2'1f 3 - 6

and

6

=

'If 4

(no restriction on r)

----:"'\..:---~x

Solution

The graphs are shown in Figure 11.23.

-

Relating Polar and Cartesian Coordinates FIGURE 11.23 The graphs of typical inequalities in r and 8 (Example 3).

When we use both polar and Cartesian coordinates in a plane, we place the two origins together and take the initial polar ray as the positive x-axis. The ray 6 = 'If/2, r > 0,

11.3 y

Polar Coordinates

629

becomes the positive y-axis (Figure 11.24). The two coordinate systems are then related by the following equations.

Ray9=1! 2 p(x, y) ~ P(r, 8)

Equations Relating Polar and Cartesian Coordinates r

Common origin

y

x

(J 8=O,r~O ~------~~~~~~~~-+ x

o

x

y

rcosO,

=

=

tan II = )'

rsinO,

x

Initial ray

FIGURE 11.24 The usua1 way to relate polar and Cartesian coordinates.

The first two of these equations uniquely determine the Cartesian coordinates x and y given the polar coordioates r and II. On the other hand, if x and y are given, the third equation gives two possible choices for r (a positive and a negative value). For each (x,y) oF (0,0), there is a unique liE [0, 2".) satisfying the first two equations, each then giving a polar coordinate representation of the Cartesian point (x, y). The other polar coordinate representations for the point can be determined from these two, as in Example I.

EXAMPLE 4 Here are some equivalent equations expressed in terms of both polar coordinates and Cartesian coordioates. Polar equation

Cartesian equivalent

rcosli = 2

x=2

r cos II sin Ii = 4 r2 cos2 Ii - r2 sin2 Ii = I

xy = 4 x 2 - y2 = I y2-3x 2 -4x-1 =0

2

r = I

+ 2rcosli X4

r=l-cosli

+ y4 + 2x"y2 + 2x3 + 2xy2 _

y2 = 0

Some curves are more simply expressed with polar coordinates; others are not. Find a polar equation for the circle x 2

EXAMPLE 5 Solution

•

+ (y - 3)' = 9 (Figure 11.25).

We apply the equations relating polar and Cartesian coordinates:

+ (y - 3)2 + y2 - 6y + 9 x 2 + y2 - 6y x2

x2

= 9 = 9

Expand(y - 3)'.

= 0

Cancellation

r2 - 6rsinli = 0 r=O

FIGURE 11.25 The circle in Example 5.

or

r-6sinli=0 r = 6 sin Ii

Includes both possibilities

• EXAMPLE 6

Replace the following polar equations by equivalent Cartesian equations and identifY their graphs.
(b) r2 = 4r cos Ii 4 (c) r = 2eDse _ sinO Solution

We use the substitutions r cos II = x, r sin Ii = y, r2 = x 2


rcosli = -4

x =-4 V erticalline through x = -4 on the x-axis

+ y2.

630

Chapter 11: Parametric Equations and Polar Coordinates

(b) r2 = 4r cos 0 The Cartesian equation:

r2 = 4r cos 0 x 2 + y2 = 4x x 2 -4x+y2=0 x 2 -4x+4+y2=4 (x - 2)'

The graph: (c) r

=

+ y2

=

Completing Ihe squaro

4

Circle, mdius 2, center (h, k) = (2, 0)

4 2cosO - sinO

The Cartesian equation:

r(2 cos 0 - sin 0) = 4

2rcosO - rsinO = 4

2x-y=4 y=2x-4 The graph:

Line, slope m = 2, y-intercept b = -4

•

Exercises 11.3 Polar Coonltnates 1. Which polar coordinate pairs label the same point?

•. (3,0)

b. (-3,0)

c. (2,2w/3)

d. (2,7w/3)

e. (-3, w)

f. (2, w/3)

g. (-3, 2w)

h. (-2, -w/3)

2. Which polar coordinate pairs label the same point?

•• (-2, w/3) d. (r,9

+ w)

g. (-r, 9

+ w)

b. (2, -w/3) •• (-r, 9)

c. (r,9) f. (2, -2w/3)

h. (-2, 2w/3)

3. Plot the following points (given in polar coordioates). Then rmd all the polar coordinates of each point.

•. (2, w/2)

b. (2,0)

c. (-2, w/2)

d. (-2,0)

4. Plot the following points (given in polar coordioates). Then rmd all the polar coordinates of each point.

•. (3, w/4)

b. (-3, w/4)

d. (-3, -w/4)

Polar to Cartesian Coordinates 5. Find the Cartesian coordinates of the points in Exercise I. 6. Find the Cartesian coordioates of the following points (given in polar coordinates).

•. (-3,5"./6)

f. (5, tan-I (4/3»

g. (-1,7".)

h.

(2\13, 2".13 )

Cartesian to Polar Coordinates 7. Find the polar coordinates, 0 :5 9 < 2w aod r '" 0, of the following points given in Cartesian coordinates.

•. (I, I)

b. (-3,0)

c. (\13, -1)

d. (-3,4)

8. Find the polar coordinates, -w :5 9 < w and r '" 0, of the following points given in Cartesian coordinates. •• (-2, -2)

b. (0,3)

c. (-\13,1)

d. (5, -12)

9. Find the polar coordinates, 0 :5 9 < 2w aod r :5 0, of the following points given in Cartesian coordinates.

•• (3,3)

c. (-I,

\13)

b.(-I,O)

d. (4, -3)

10. Find the polar coordinates, -w :5 9 < 2w and r :5 0, of the following points given in Cartesian coordinates. •. (-2,0)

b. (I, 0)

c. (0, -3)

d.

(~,k)

Graphing in Polar Coordinates Graph the sets of points whose polar coordinates satislY the equations and inequalities in Exercises 11-26.

•. ("\12, w/4)

b. (1,0)

11.r=2

c. (0, w/2)

d. (- v'2, "./4)

13. r '" I

12.0:5r:52 14.I';;r:52

631

11.4 Graphing in Polar Coordinates 15. 0 :5 0 :5 '11'/6,

r "" 0

16. 0 ~ 2'11'/3,

r:5-2

43.

r2 72

+ 2r 2 cos8sin8 = -4r cos 8

=

1

44. cos' 0 ~ sin' 0

17.0 = '11'/3,

-Is r s 3

18. 0

~

11'11'/4, r"'-I

45.

19. 0 = '11'/2,

r "" 0

20. 0

~

'11'/2,

47. r

~

8 sinO

48. r = 3 cosO

49. r

~

2cosO + 2 sinO

50. r = 2 cosO - sinO

21. 0 :5 0 :5 '11',

22. 0 s 0

r = I

23. '11'/4 s 0 :5 3'11'/4,

S

r:5 0 '11',

r

~

-I

46. r' = -6rsinO

51.rSin(0+~)=2

0 s r s I

52. rsin (2; -

0)

= 5

24. -'11'/4 s 0 '" '11'/4,

-I s r s I

25. -'11'/2 :5 0 :5 '11'/2,

1:5 r :5 2

Cartesian to Polar Equations

Irl :5

Replace the Cartesian equations in Exercises 53-66 with equivalent polar equations.

26. 0 :5 0 :5 '11'/2,

1:5

2

Polar to Cartesian Equations Replace the polar equations in Exercises 27-52 with equivaleot Cartesian equations. Theo describe or ideotifY the graph.

28. rsinO

27. rcosO = 2

29. rsinO

~

0

32. r

+ rsinO

~

I

~

=

56.x-y=3

57. x' + y' = 4

58. x' - y' ~ I

x2

63.

rcos8

35.r'=1

36. r' ~ 4rsin 0

5 37.r=·0 0 sm - 2 cos

38. r'sin20 = 2

39. r = cotOcscO

40. r

41. r = esc (J eTcos 9

42. rsinO = Inr

11.4

55.x=y

y2

9+4=

60.xy=2

I

61. y' = 4x

-3 sec 0

34. 78in8

S4.y=1

59.

-I

30. rcosO = 0

31. r=4cscO

33. rcosO

=

53.x~7

62. x'

x' + (y -

65. (x - 3)'

2)' = 4

+ (y +

I)' = 4

+ xy + y'

= I

64. (x-5)'+y'=25 66. (x

+ 2)' + (y

- 5)'

=

16

67. Find all polar coordinates of the origin.

68. Verticlll and horizontlllline.

~

•• Show that every vertical1ine in the xy-plane has a polar equa-

4tanOsecO

+ In cos 0

tionoftheformr = asec8. II- Find the analogous polar equation for horizontal lines in the

xy-plane.

Graphing in PoLar Coordinates It is often helpful to have the graph of an equation in polar coordinates. This section de-

scribes techniques for grapbing these equations using symmetries and tangents to the graph.

Symmetry Figore 11.26 illustrates the standard polar coordinate tests for symmetry. The following summary says how the symmetric points are related.

Symmetry Tests for Polar Graphs 1. Symmetry about the x-axis: If the point (r, 0) lies on the graph, then the point (r, -0) or (-r, '11' - 0) lies on the graph (Figore 11.26a). 2. Symmetry about the y-axis: If the point (r, 0) lies on the graph, then the point (r, '11' - 0) or (-r, -0) lies on the graph (Figore 11.26b). 3. Symmetry about the origin: If the point (r, 0) lies on the graph, then the point (-r, 0) or (r, 0 + '11') lies on the graph (Figore 11.26c).

632

Chapter 11: Parametric Equations and Polar Coordinates

Slope

y

The slope of a polar curve r = 1(0) in the xy-plane is still given by dy/dx, which is not r' = dl/dO. To see why, think of the graph of I as the graph of the parametric equations

(r,8)

----~r---4-~X

o

x

I I I

(r, -6)

or(-r,'1r- 9)

r cos 0

=

1(0) cos 0,

y = rsinO = 1(0) sinO.

If lis a differentiable function of 0, then so are x andy and, when dx/dO oF 0, we can calculate dy/dx from the parametric formula

(a) About the x-axis (r,"- 8) or (-r, -8)

=

Y (r,8)

dy

dy/dO

Section 11.2, Eq. (1)

dx

dx/dO

witht=(J

10 U(O)· sin 0)

o

d dO U(O)· cos 0)

df.

dijsmO

(b) About the y-axis

+ 1(0) cosO

df

y

.

Product Rule for derivatives

dO cosO - I(O)smO (r,6)

Therefore we see that dy/dx is not the same as dl/dO.

o (-r, 8) or (r, B + ,,)

Slope of the Curve r = 1(0)

(c) About the otigin

FIGURE 11.26

dy I /,(0) sin 0 + 1(0) cos 0 dx (r,B) = /,(0) cosO - 1(0) sinO ,

Three tests for

symmetry in polar coordinates.

provided dx/dO oF Oat (r, 0).

If the curve r = 1(0) passes through the origin at 0 = 00 , then 1(00 ) = 0, and the slope equation gives

I

/,(00) sin 00 dy dx (0, B,) = /' (00) cos 00 = tan 00' If the graph of r = 1(0) passes through the origin at the value 0 = 00, the slope of the curve there is tan 00. The reason we say "slope at (0, (0)" and not just "slope at the origin" is that a polar curve may pass through the origin (or any point) more than once, with different slopes at different O-values. This is not the case in our first example, however.

EXAMPLE 1 Solution

Graph the curve r = 1 - cos O.

The curve is symmetric about the x-axis because

(r, 0) on the graph => r = 1 - cos 0 => r = 1 - cos ( -0) => (r, -0) on the graph.

cos 6

~

cos (-6)

11,4

(J

o

0 1 2

3

"'2 2"

3

3

'2

"

2 (a)

Graph the curve r2 = 4 cos o.

EXAMPLE 2

(", 2) -+---~--,,¥--~

x

Solution The equation r2 = 4 cos 0 requires cos 0 20 0, so we get the entire graph by running 0 from -1T/2 to 1T/2. The curve is symmetric about the x-axis because

(7, 0) on the graph

(b)

r = 1-

cos '!. 8 _~

633

As 0 increases from 0 to 1T, cos 0 decreases from I to -I, and , = I - cos 0 increases from a minimum value of 0 to a maximum value of 2. As 0 continues on from 1T to 21T, cos 0 increases from -I back to I and 7 decreases from 2 back to O. The curve starts to repeat when 0 = 21T because the cosine has period 21T . The curve leaves the origin with slope tan (0) = 0 and returns to the origin with slope tan (21T) = O. We make a table of values from 0 = 0 to 0 = 1T, plot the points, draw a smooth curve through them with a horizontal tangent at the origin, and reflect the curve across the x-axis to complete the graph (Figure 11.27). The curve is called a cardioid because of its heart shape. _

r=l-cos(J

'!

Graphing in Polar Coordinates

=) 7

2

= 4 cos 0

=)

r2 = 4 cos (-0)

=)

(r, -0) on the graph.

__~

ce. 8

= cos (-0)

The curve is also symmetric about the origin because (r, 0) on the graph

=)

r2 = 4 cos 0

=)

(-r)2 = 4cosO

=)

(-r, 0) on the graph.

('" 2) -+----c2.-----.""~--.x

(c)

FIGURE 11.27 The steps in graphing the cardioid, = I - cos 6 (Example I), The

Together, these two symmetries imply symmetry about the y-axis. The curve passes through the origin when 0 = -1T/2 and 0 = 1T/2. It has a vertical tangent both times because tan 0 is infinite. For each value of 0 in the interval between -1T/2 and 1T/2, the formula r2 = 4 cos 0 gives two values of r: r = ±2Ycos O.

arrow shows the direction of increasing 8.

We make a short table of values, plot the corresponding points, and use information about symmetry and tangents to guide us in connecting the points with a smooth curve (Figure 11.28).

y

I

8

0 +'!

-6

+'!

cos 8 r= ±2Vcos6 j 1

V3 2 1

-4

Yz

+'!

!

+'!

0

-3

-2

"=400.8

0:2

2

-

±1.9

=

±1.7

-

±1.4

0

/

Loop for, = -2Vcos 8,

\ Loop for, = 2Voo.8,

(a)

(b)

FIGURE 11.28 The graph of ,2 = 4 cos 6. The arrows show the directioo of increasing 6. The values of, in the table are rounded (Example 2).

_

634

Chapter 11: Parametric Equations and Polar Coordinates

A Technique for Graphing

(a) ,2

One way to graph a polar equation, = /(6) is to make a tahle of (7, 6)-values, plot the corresponding points, and connect them in order of increasing 6. This can work well if enough points have been plotted to reveal all the loops and dimples in the graph. Another method of graphing that is usually quicker and more reliable is to

l

1.

flISt graph r = /(6) in the Cartesian r6-plane,

2.

then use the Cartesian graph as a "table" and guide to sketch the polar coordinate graph.

No square roots of negative numbers

This method is better than simple point plotting because the firat Cartesian graph, even when hastily drawn, shows at a glance where r is positive, negative, and nonexistent, as well as where ,is increasing and decreasing. Here's an example.

(b) ,

Y~+P_from

~---,-lj------j---'-+'_

8

squareroots

EXAMPLE 3

Graph the lemniscate curve ,2 = sin26.

,~ -v'sin28

(c)

Solution Here we begin by plotting r2 (not r) as a function of 6 in the Cartesian r 26-plane. See Figure 11.29a. We pass from there to the graph of r = ± Vsin 26 in the rIJ-plane (Figure 11.29b), and then draw the polar graph (Figure 11.29c). The graph in Figure 11.29b "covers" the final polar graph in Figure 11.29c twice. We could have managed with either loop alone, with the two upper halves, or with the two lower halves. The double covering does no harm, however, and we actually learn a little more about the behavior of the function this way. •

y

USING TECH NOLOGY FIGURE 11.29 To plot 7 = /(8) in the Cartesiao r8-plane in (b), we 'ITst plot ,2 = sin 28 in the ,28-plane in
Graphing Polar Curves Parametrically

For complicated polar curves we may need to use a graphing calculator or computer to graph the curve. If the device does not plot polar graphs directly, we can convert r = /(6) into parametric form using the equations

x = rcos6 = /(6) cos 6,

y = rsin6 = /(6) sin6.

Then we use the device to draw a parametrized curve in the Cartesisn xy-plane. It may be necessary to use the parameter t rather than 6 for the graphing device.

Exercises 11.4 Symmetries and Polar Graphs IdentifY the symmetries of the corves in Exercises 1-12. Then sketch

Graph the lemniscates in Exercises \3-16. What symmetries do tlrese

the curves. 1. , = I + cos8

13. ,2 = 4 cos 28 15. 72 = -sin 28

2. ,=2-2c088

3. , = I - sin8

4. , = I + sin8

S. , = 2 + sin8

6.,=1+2sin8

7. , = sin (8/2)

8. , = cos (8/2)

9.

11.

r2 = r2

cosO

= -sin 8

10. ,2 = sin8 12.

r2

= -cos 8

curves have?

14.72 = 4 sin 28 16. 72 = -cos 28

Slopes of Polar Curves Find tire slopes of the corves in Exercises 17-20 at the given points. Sketch the corves along with their taogeots at these points. 17. Cardioid ,= -I + cos8; 8 = ±'1r/2 18. Cardioid r

=

-1

+ sin 8;

8

=

0, '1T'

19. Four-l....d ros. 7 = sin28; 8 = ±'1r/4, ±3'1r/4 20. FOllJ'oleaved rose 7 = cos 28; 8 = 0, ±'1r/2, '1r

11.5 Graphing Uma~ons Graph the lima90DS in Exercises 21-24. L~on ("lee-ma-sahn") is Old French for "snail." You will understand the name when you graph the limal'O!ls in Exercise 21. Equations for lima~ons have the form

r = a ± b cos (} or r = a ± b sin (}. There are four basic shapes. Lima~ons

21.

b.r=-I+sin6

D 31. D 32.

23. Dimpled lim~on. 3

b. r =

r=I+2sin~ .

D 33. D 34.

Graphing Polar Reg;ons and Curves 25. Sketch the region defmed by the inequalities -I :5 r :5 2 and -1T/2 '" 6 :5 1T/2. 26. Sketch the region defined by the inequalities 0 '" r '" 2 sec 6 and -1T/4 :5 6 :5 1T/4.

Ro.e.

Graph the rosesr = cosm6form = 1/3,2,3, and 7.

Spiral. Polar coordinates are just the thing for defming spirals. Graph the following spirals. a.7=(}

b. r = -6

c. A logarithmic spiral: r = eO/lo d. A hyperbolic spiral: r = 8/6

e. An equilaterallryperbola: r = ± IO/Vo (Use different colors for the two branches.)

In Exercises 27 aod 28, sketch the region defmed by the inequality.

11.5

Graph the equation r = I - 2 sin 36.

The nephroid of Freeth Graph the nephroid of Freeth:

2

b. r = -2 + sin6

27. 0:5 r :5 2 - 2 cos 6

A rose within • ro.e

:3_ sin6

24. Owl lim~on.

•. r=2+cos8

Which of the following has the same graph as r = cos 26?

.. r = -sin (26 + 1T/2) b. r = -cos (6/2) Coofmn your answer with algebra.

22. Cardioids

•. 7="2+oos 8

Which of the following has the same graph as r = I - cos 6?

r = -1 - cosO b.r=l+cos6 Coofmn your answer with algebra.

b.r=~+sin6

a.r=1-cos8

635

L

D 30.

with an inner loop

a.r=~+cos6

D 29.

Areas and Lengths in Polar Coordinates

28. 0 ~,2 ~ cos(J

Areas and Lengths in PoLar Coordinates This section shows how to calculate areas of plane regions and lengths of curves in polar coordinates. The defining ideas are the same as before, but the formulas are different in polar versus Cartesian coordinates.

y

Area in the Plane

ors

/"", = 1(6)

The region in Figure 11.30 is bounded by the rays 0 = " and 0 = f3 and the curve r = f(O). We approximate the region with n nonoverlapping fan-shaped circular sectors based on a partition P of angle The typical sector has radius = f(O.) and central angle of radian measure !J.O•. Its area is !J.OJ21T times the area of a circle of

r.

ros.

radius 71, or 9=a

t

o

t

A. = rl fl.(}. = (J(O.)), fl.(}•.

~~~--~~-------------+,

The area of region

ors is approximately

FIGURE 11.30 To derive a formula for the area of region ars, we approximate the region with fan-shaped circular sectors.

If f is continuous, we expect the approximations to improve as the norm of the partition P goes to zero, where the norm of P is the largest value of !J.O•. We are then led to the following formula defining the region's area:

636

Chapter 11: Parametric Equations and Polar Coordinates

y

A

• 1 2 lim ~ -2 (I( Ok») t..(Jk

=

IIplI~o~1

tt

=

(1(0»)' dO.

Area of the Fan-Shaped Region Between the Origin and the Curve , = /(0), a ,,; 0 ,,; fJ FIGURE 11.31 The area differential dA for the curve r = flO).

A

=

l

P t,2dO.

This is the integral of the area differential (Figore 11.31)

t

y

dA =

--~---------+~x

4

r2 dO

=

t

(1(0»)' dO.

EXAMPLE 1 Find the area of the region in the plane enclosed by the cardioid r = 2(1 + cosO). Solution We graph the cardioid (Figure 11.32) and determine that the radius OP sweeps out the region exactly once as 0 runs from 0 to 21T. The area is therefore

FIGURE 11.32

The cardioid in Example 1.

1

8- 2"

1 2r 2 dO =

8-0

y

1 1 1 1

2 "

I 2'4(1 + cos 0)2 dO

0

2

=

"

2( I + 2 cos 0 + cos2 0) dO

2

=

"(2 + 4cosO + 2 1 +

~OS20) dO

2

= ~~------------~X

o

FIGURE 11.33 The area of the shaded region is calculated by subtracting the area of the region between '1 and the origin from the area of the region betweeo '2 and the origin. y

Tl=1-cos8

Upper limit (J=",/2

"(3 + 4 cos 0 + cos 20) dO

. 20

= [30+4sinO+s~

]2"0 =61T-0=61T.

•

To find the area of a region like the one in Figure 11.33, which lies between two polar curves rl = rl(O) and r2 = r2(0) from 0 = a to 0 = f3, we subtract the integral of (1/2h2 dO from the integral of (1/2)r,' dO. This leads to the following formula.

Area of the Region 0 ,,; '1(0)";,,,; ,,(O),a"; 0,,; A =

l

P

tr,'dO -

l

P

trNO

=

l

P

fJ

th2 - r?) dO

(1)

-r-----r----~~--~~x

EXAMPLE 2 Find the area of the region that lies inside the circle r = 1 and outside the cardioid r = I - cos O.

FIGURE 11.34 The region aud limits of integration in Example 2.

Solution We sketch the region to determine its boundaries and find the limits of integration (Figore 11.34). The outer curve is r2 = 1, the inner curve is rl = I - cos 0, and 0 runs from -1T/2 to 1T/2. The area, from Equation (I), is

11.5 Areas and Lengths in Polar Coordinates

637

Symmetry

,,/2 =

0

/.

(I - (I - 2 cos II

+

cos2 (1))dll

Square rl.

= /."/\2 cos II - cos2 (1) dll = /'''12 (2 cos II _ I

= [ 2sinll-

6' sm 26]"/2

Z-

4

0

+

~os 26) dll

•

=2-;.

The fact that we can represeot a point in different ways in polar coordinates requires extra care in deciding when a point lies on the graph of a polar equation and in determining the points in which polar graphs intersect. CWe needed intersection points in Example 2.) In Cartesian coordinates, we can always f'md the points where two curves cross by solving their equations simultaneously. In polar coordinates, the story is different. Simultaneous so· lution may reveal some intersection points without revealing others, so it is sometimes dif· ficult to find all points of intersection of two polar curves. One way to identify all the points of intersection is to graph the equations.

Length of a Polar Curve We can obtain a polar coordinate fannula for the length ofa curve r = /(11), a ,,; II ,,; by parametrizing the curve as

x = rcos6 = /(6) cos 6,

y = rsin6 = /(6) sin 6,

a";

6 ,,; f3.

f3, (2)

The parametric length fannula, Equation (3) from Section 11.2, then gives the length as

L =

J.{P

(dx)2 (dy )2 d6 + d6 dll.

This equation becomes

when Equations (2) are substituted for x and Y (Exercise 29).

Length of a Polar Curve If r = /(6) has a continuous first derivative for a ,,; 6 ,,; f3 and if the point P(r,6) traces the curve r = /(6) exactly once as 6 runs from a to f3, then the length of the curve is

y

(3) 8 ----+---~~--~~--~x

EXAMPLE 3

FIGURE 11.35 Calculating the length of a cardioid (Example 3).

Find the length of the cardioid r = I - cos II.

Solution We sketch the cardioid to determine the limits of integration (Figure 11.35). The point P(r, 6) traces the curve once, counterclockwise as II runs from 0 to 21T, so these are the values we take for a and f3 .

638

Chapter 11: Parametric Equations and Polar Coordinates With r = 1 - cosO,

dr

. 0

dO=sm,

we have r2 +

(~;)' = (1

- cos 0)' + (sin 0)'

= 1 - 2cosO + cos2 0 + sin2 0 = 2 - 2cosO and

L=

J. a

p

r2+

(dd; )2 dO= Jof2W V2-2cosOdO 1-

COS 8

~ 2 sin' (8/2)

sin(8/2) '" 0

for 0 s 8 s 2..

= [ -4cos 0 l2W =4+4=8. 20

•

EXERCISES 11.5 Finding Polar Al'I!as

4. Inside the cardioid r = a( I

Find the areas of the regions in Exercises 1-8.

1. Bounded by the spiral r

~

8 for 0

:5

8

____

a> 0

cos 8),

~

S. Inside ooeleafofthe four-Ieavedroser

:5 ..

6. Inside ooe leaf of the 1Irree-Ieaved rose r

Y T=8

+

cos28

~

cos 38

y

I

..

( , )

2. Bounded by the circle r = 2 sin 8 for ../4

:5

8

:5 ../2

y

7. Inside ooe loop of the lemniscate r2 ~ 4 sin 28 8. Inside the six-leaved rose

r' ~ 2 sin 38

Find the areas of the regions in Exercises 9-16.

9. Shared by the circles r 10. Shared by the circles r 11. Shared by the circle r

~

~

~

12. Shared by the cardioids r 3. Inside the ovallima~oo r = 4

+ 2 cos 8

2 cos 8 and r I and r

~

~

2 sin 8

2 sin 8

2 and the cardioid r ~

2(1

~

+ cos 8) and r

2(1 - cos 8) ~

2(1 - cos 8)

13. Inside the lemniscare r' = 6 cos 28 and outside the circle r ~

V:l

11.6

Conic Sections

639

14. Inside the circle r ~ 3a cos 0 and outside the cardioid r ~ a(1 + cosO),a > 0

27. Thecurver ~ cos'(0/3),

15. Inside the circle r = -2 cos (J and outside the circle r = 1

29. The length of the curve r = /(9), a '" 9 '" fJ Assuming that the necessary derivatives are continuous, show how the substitutions x ~ /(0) cosO, y ~ /(0) sinO

16. Inside the circle r

6 above the line r

~

17. Inside the circle r r=sec8

~

18. Inside the circle r r=3csc8

~

3 csc 0

4 cos 0 and to the right of the vertical line ~

4 sin 0 and below the horizontal line

os 0 s '1r/4 28. The curve r ~ v'1 + sin 20, 0 s 0 s '1rYz

(Equations 2 in the text) transform

L~t (O:OY+(:YdO

19. •• Find the area of the shaded region in the accompanying figure. y

r=tan9

-¥ < 8 < ¥

into L --1~ a r2

~----~~~----~~X

-1

0

1

b. It looks as if the graphofr ~ tan 0, -'1r/2 < 0 < '1r/2, could be asymptotic to the lines x ~ I and x ~ -I. Is it? Give reasons for your answer.

20. The area of the region that lies inside the cardioid curve r = cos (} + 1 and outside the circle r = cos (} is not

1].27

20 [(cosO + I)' - cos2 01dO ~

'1r.

21. The spiral r ~

02,

22. The spiral r ~

.'(V2,

23. The cardioid r

~

0 s 0 '"

L

r= a

b. r=acos8

I

Theory and Examples 31. Average valne If/is contiouous, the average value of the polar coordinate r over the curve r ~ /(8), a s 8 s fJ, with respect to 0 is given by the formula

0 s 0

25. Theparabolicsegrneotr ~ 6/(1 26. Theparabolicsegrneotr

11.6

~

roy ~ {3 -I a

L

S '1r

1~ a /(8) d8.

The cardioid r ~ a( I - cos 8)

b. The circle r = a

+ cos 0

24. The curve r ~ a sin2 (0/2),

c. r=asinO

Use this formula to find the average value of r with respect to 8 over the following corves (a > 0).

Vs

os 0

(dr)2 dO dO.

30. Circumference. of circle. As usual, when faced with a new formula, it is a good idea to try it on familiar objects to be sure it gives results consisteot with past experience. Use the length for· mula in Equation (3) to calculate the circon3fereoces of the fol· lowing circles (a > 0).

Why not? What is the area? Give reasons for your answers. Finding Lengths of Polar Curves Find the leogths of the curves in Exercises 21-28.

+

S '1r,

+ cosO),

c. The circle r

a> 0 0'" 0 '" '1r/2

2/(1 - cosO), '1r/2 '" 0 s

'1r

~ a

cos 8,

-'1r/2 s 8 s '1r/2

32. r = /(9) .s. r = 2/(9) Can anything be said about the relative lengths of the curves r ~ /(8), 8 '" (3, and r ~ 2/(8), a ~ (J ~ f3? Give reasons for your answer.

a'"

Conic Sections In this section we detme and review parabolas, ellipses, and hyperbolas geometrically and

derive their standard Cartesian equations. These curves are called conic sections or conics b.,. cause they are formed by cutting a double cone with a plane (Figure 11.36). This geometry method was the only WH:f they could be described by Greek mathematicians who did not have our tools of Cartesian or polar coordinates. In the next section we express the conics in polar coordinates.

Parabolas DEFINITIONS A set that consists of all the points in a plane equidistant from a given tlXed point and a given fixed line in the plane is a parabola. The fixed point is the focus of the parabola. The fixed line is the directrix.

640

Chapter 11: Parametric Equations and PoLar Coordinates

Circle: plane perpendicular to cone axis

Ellipse: plane oblique to cone axis

Hyperbola: plane cuts both halves of cone

Parabola: plane parallel to side of cone (a)

Point: plane through cone vertex only

Pair of intersecting lines

Single line: plane tangent to cone (b)

FIGURE 11.36 The standard conic sections (a) are the curves in which a plane cuts a double cone. Hyperbolas come in two parts, called branches. The point and lines obtained by passing the plane through the cone's vertex (b) are degenerate conic sections.

If the focus F lies on the directrix L, the parabola is the line through F perpendicular to y

L. We consider this to be a degenerate case and assume henceforth that F does not lie on L. A parabola has its simplest equation when its focus and directrix straddle one of the coordinate axes. For example, suppose that the focus lies at the point F(O, p) on the positive y-axis and that the directrix is the line y = -p (Figure 11.37). In the notation of the figure, a point P(x, y) lies on the parabola if and only if PF = PQ. From the distance formula,

The vertex lies ~"""~~-+------+x halfway between directrix and focus. P L Directrix: y = -P Q(x, -p)

PF

=

PQ

=

Y(x - of + (y - pf = Yx 2 + (y - p)2 Y(x - xf + (y - (-p)f = Y(y + p)2.

When we equate these expressions, square, and simplify, we get FIGURE 11.37 The standard form of the parabola x 2 = 4py, p > o.

x2

y = 4p

or

x2

=

4py.

Standard form

(1)

These equations reveal the parabola's symmetry about the y-axis. We call the y-axis the axis of the parabola (short for "axis of symmetry"). The point where a parabola crosses its axis is the vertex. The vertex of the parabola x 2 = 4py lies at the origin (Figure 11.37). The positive number p is the parabola's focal length.

11.6

y

~

Conic Sections

641

If the parabola opens downward, with its focus at (0, -p) and its directrix the line p, then Equations (I) become 2

x y=--

and

4p

x 2 ~ -4py.

By interchanging the variables x and y, we obtain similar equations for parabolas opening to the right or to the left (Figure 11.38). y

y

Directtix

Directtix

x=-p

x=p

Vertex

Vertex

(b)

(a)

FIGURE 11.38 (a) The parabola y2 ~ 4px. (b) The parabolay 2 ~ -4px.

EXAMPLE 1 Solution

Find the focus and directrix of the parabolay2 ~ lOx.

We rmd the value ofp in the standard equation y2 ~ 4px:

4p = 10,

so

Then we find the focus and directrix for this value ofp: Vertex

Vertex

Focus: Directrix: FIGURE 11.39 Points on the focal axis of an ellipse.

(p, 0)

x

~ (~, 0 )

~-p

or

x

~

5

-2'

•

Ellipses

y b

FIGURE 11.40 The ellipse defined by the equation PFl + PF2 ~ 2a is the grapb of the equation (x 2/a 2) + (y2/b 2) ~ I, where b 2 = a 2 - c 2•

DEFINmONS An ellipse is the set of points in a plane whose distances from two fixed points in the plane have a constant sum. The two fixed points are the foci of the ellipse. The line through the foci of an ellipse is the ellipse's focal ans. The point on the axis halfway between the foci is the center. The points where the focal axis and ellipse cross are the ellipse's vertices (Figure 11.39).

If the foci are Fl( -c, 0) andF2(c, 0) (Figure 11.40), and PFI + PF2 is denoted by 2a, then the coordinates of a point P on the ellipse satisfY the equation

Y(x + c)2 + y2 + Y(x - c)2 + y2 ~ 2a.

642

Chapter 11: Parametric Equations and Polar Coordinates To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaming radical, and square agam, obtaining

x2

-2 + a

y2 2

a - c

2=

1.

(2)

Since PFI + PF2 is greater than the length FIF2 (by the triangle inequality for triangle PFIF2), the number 2a is greater than 2c. Accordingly, a > c and the number a2 - c 2 in Equation (2) is positive. The algebraic steps leading to Equation (2) can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < c < a also satisfies the equation PFI + PF2 = 2a. A point therefore lies on the ellipse if and only if its coordinates satisfy Equation (2). If

Va 2 - c2 ,

b = then a

2

-

c

2

=

(3)

2

b and Equation (2) takes the form x 2 y2

,+,=1.

(4) b Equation (4) reveals that this ellipse is symmetric with respect to the origin and both coordinate axes. It lies inside the rectangle bounded by the lines x = ±a and y = ±b. It crosses the axes at the points (±a, 0) and (0, ±b). The tangents at these points are perpendicular to the axes because

a

dy

b 2x

ax

a 2y

Obtained from Eq. (4) by implicit differentiation

is zero if x = 0 and inf'Inite if Y = O. The major axis of the ellipse in Equation (4) is the line segment oflength 2a joining the points (±a, 0). The minor axis is the line segment of length 2b joining the points (0, ±b). The number a itself is the semimajor axis, the number b the semiminor axis. The number c, found from Equation (3) as

c

=

Va 2 -

b2 ,

is the center-to-Cocus distance of the ellipse. If a = b, the ellipse is a circle. 2 y2 Y 1 I6+9~ (0,3)

x

EXAMPLE 2

The ellipse

Vertex

(5)

(-4,0)

(Figure HAl) has

a=

Sernimajor axis:

v'i6 = 4,

Center-to-focus distance: (0, -3)

FIGURE 11.41 An ellipse with its major axis horizontal (Example 2).

Foci:

(±c, 0)

Vertices: Center:

=

c =

Serniminor axis:

b

=

v'9 = 3

v'l6=9 = 0

(± 0,0)

(±a,O)

= (±4,0)

(0,0).

•

Ifwe interchange x andy in Equation (5), we have the equation

x2

y2

9+16=1.

(6)

The major axis of this ellipse is now vertical instead of horizontal, with the foci and vertices on the y-axis. There is no confusion in analyzing Equations (5) and (6). Ifwe f'md the intercepts on the coordinate axes, we will know which way the major axis runs because it is the longer of the two axes.

11.6

Conic Sections

643

Standard-Form Equations for Ellipses Centered at the Origin 2

Foci on the x-axis:

X, a

2

+ y2

=

b

I

(a

> b)

Center-to-focus distance:

2

X, b

Va 2 -

b2

=

Va 2 -

b2

(±a, 0) 2

+ y2 a

=

I

(a

> b)

Center-to-focus distance: Foci:

=

(±e, 0)

Foci:

Vertices:

Foci on they-axis:

e

e

(0, ±e) (0, ±a)

Vertices:

In each case, a is the semimajor axis and b is the semiminor axis.

Hyperbolas DEFINmONS A hyperbola is the set of points in a plane whose distances from two fixed points in the pIane have a constant difference. The two f'lxed points are the foci of the hyperbola. The line through the foci of a hyperbola is the focal axis. The point on the axis halfway between the foci is the hyperbola's center. The points where the focal axis and hyperbola cross are the vertices (Figure 11.42).

FIGURE 11.42 Poiots 00 the focal axis of a hyperbola.

If the foci are F,( -e, 0) and F2(c, 0) (Figure 11.43) and the constant difference is 2a, then a point (x, y) lies on the hyperbola if and only if (7)

y

x=-a

x=a

To simplifY this equation, we move the second radical to the right-band side, square, isolate the remaining radical, and square again, obtainiog (8)

FIGURE 11.43 Hyperbolas have _ branches. For poiots 00 the right-hand branch of the hyperbola shown here, PF, - PF2 = 2a. For poiots on the lefthand branch, PF, - PF, = 2a. We then leth = Ve' - a'.

So far, this looks just like the equation for an ellipse. But now a 2 - e 2 is negative because 2a, being the difference of two sides of triangle PF,F2 , is less than 2e, the third side. The algebraic steps leading to Equation (8) can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < a < e also satisf'les Equation (7). A point therefore lies on the hyperbola if and only if its coordinates satisfy Equation (8). Ifwe let b denote the positive square root of e 2 - a 2 ,

b = then a 2

-

Ve 2 -

a2 ,

(9)

e 2 = -b 2 and Equation (8) takes the more compact form

x2 y2 a2 - b 2 = I.

(10)

644

Chapter 11: Parametric Equations and Polar Coordinates The differences between Equation (10) and the equation for an ellipse (Equation 4) are the minus sign and the new relation FromEq. (9)

Like the ellipse, the hyperbola is symmetric with respect to the origin and coordinate axes. It crosses the x-axis at the points (±a, 0). The tangents at these points are vertical because

b 2x

dy dx

Obtained from Eq. (10) by implicit differentiation

a'y

is iufinite when y = O. The hyperbola has no y-intercepts; in fact, no part of the curve lies between the lines x = -a and x = a. The lines

y =

±%x

are the two asymptotes of the hyperbola deimed by Equation (10). The fastest way to find the equations of the asymptotes is to replace the I in Equation (10) by 0 and solve the new equation for y:

x2

y2

2

2

---=

a

b

I --->

x2

ofor I

hyperbola

EXAMPLE 3

y y

b 0 ---> y = ±a x .

y2

;;I - b2 =

asymptotes

The equation

~ v'5.

x2

2

y2

4-5=

(11)

I

is Equation (10) with a 2 = 4 and b 2 = 5 (Figure 11.44). We have Center-to-focus distance: Foci: Center:

b2

Vertices:

= V4+5 =

3

(±a,O) = (±2, 0)

(0,0)

x2

TIre hyperbola and its asymptotes in Example 3.

= Va 2 +

(±c,O) = (±3,0),

Asymptotes: FIGURE 11.44

c

y2

4 - 5

= 0

or y =

Y5

•

±T x .

If we interchange x and y in Equation (II), the foci and vertices of the resulting hyperbola will lie along the y-axis. We still find the asymptotes in the same way as before, but now their equations will be y = ±2x/Y5.

Standard-Form Equations for Hyperbolas Centered at the Origin Foci on the x-axis:

x2 y2 -2 - -2 = I a b

Center-to-focus distance: Foci:

(±c,O)

Vertices:

(±a,O)

c =

Va 2 + b 2

x2

2

b2

a

Center-to-focus distance: Foci:

I

c =

Va 2 + b 2

(0, ±c)

Vertices:

(0, ±a)

x2 y2 b - 2 - - 2 = 0 or y = ±-ax Asymptotes: a b Notice the difference in the asymptote equations (b/a in the imt, a/b in the second). Asymptotes:

y2

---=

Foci on the y-axis:

y2

x2

- 2 - - 2 = 0 a b

or y =

±"x b

11.6 Conic Sections

We shift conics using the principles reviewed in Section 1.2, replacing x by x ybyy + k. Show that the equation x 2 - 4y2 perbola. Find its center, asymptotes, and foci.

EXAMPLE 4

Solution follows:

+ 2x +

645 + h and

8y - 7 = 0 represents a by-

We reduce the equation to standard form by completing the square in x and y as (x 2 (x 2

+

2x

+ I) -

+ 2x)

- 4(y2 - 2y) = 7

4(y2 - 2y (X

+ 1)2 4

+ 1)

=

7 + I - 4

- (y - 1)2 = I.

This is the standard form Equation (lO) of a hyperbola with x replaced by x + 1 and y replaced by y - I. The hyperbola is shifted one unit to the left and one unit upward, and it has center x + 1 = oandy - I = O,orx = -1 andy = I. Moreover,

so the asymptotes are the two lines

x2 + -1 - (y - 1) = 0 The shifted foci have coordinates ( -1 ±

and

x ~ 1 + (y _ 1)

=

O.

v'5, 1).

•

Exercises 11.6 Identity;ng Graphs Match the parabolas in Exercises 1-4 with the following equations:

x' = 2y, x' = -6y, y'

= 8x,

y'

Match each conic section in Exercises 5-8 with one of these equations:

= -4x.

Then f'md each parabola's focus and direc1rix.

1.

Y

y

2.

Then fmd the conic section's foci and vertices. If the conic section is a hyperbola, f'md its asymptotes as well. S. Y 6. Y

~f---->x

3.

Y

4.

Y -+-+--+~X

- -....~--->x

646

Chapter 11: Parametric Equations and Polar Coordinates

Shifting Conic Sections

8.

7.

You may wish to review Section 1.2 before solving Exercises 31}-56.

y

y

39. The parabola y2 = 8x is shifted down 2 units aod right I unit to generatetheparabola(y + 2)2 = 8(x - I) . •• Find the new parabola's vertex, focus, aod direc1rix. b. Plot the new vertex, focus, and directrix, aod sketch in the parabola.

40. The parabola x 2 = -4y is shifted left I uoit aod up 3 units to generate the parabola (x + I)" = -4(y - 3) . •• Find the new parabola's vertex, focus, aod direc1rix.

Parabolas Exercises 1}-16 give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Ioclude the focus aod direc1rix in your sketch. 9. y2 = 121: 12. y2 = -21: 15. x

10. x 2 = 6y

11. x 2 = -8y

= 4x 2

14. Y = -8x 2

13. Y

= -3y2

b. Plot the new vertex, focus, and directrix, aod sketch in the parabola. 41. The ellipse (x 2/16) + (y2/9) = 1 is shifted 4 units to the right aod 3 units up to generate the ellipse (x - 4)2 16

16. x = 2y2

+

(y - 3)2 9

= 1.

a. Find the foci, vertices, and center of the new ellipse.

EIUpses Exercises 17-24 give equations for ellipses. Put each equation in staodard form. Then sketch the ellipse. Ioclude the foci in your sketch. 17. lfu;2 + 25y2 = 400 19. 21:2 +y2=2

18. 7x 2 + 16y2 = 112 20. 21: 2 +y2=4

21. 3x2

= 6

22. 9x 2

= 54

24.

23. fu;2

+ 2y2 + 9y2

b. Plot the new foci, vertices, aod center, aod sketch in the new ellipse. 42. The ellipse (x 2/9) + (y2/25) = I is shifted 3 units to 1he left aod 2 units down to generate 1he ellipse

(x + 3)2

+ IOy2 = 90 169x 2 + 25y2 = 4225

9

+

(y

+

2)2 = 1.

25

a. Find the foci, vertices, and center of the new ellipse. b. Plot the new foci, vertices, and center, and sketch in the new

Exercises 25 aod 26 give information about the foci and vertices of ellipses centered at the origin of the xy-plane. 10 each case, f"md the ellipse's staodard-form equation from the given information.

ellipse. 43. The hyperbola (x 2/16) - (y2/9) right to generate the hyperbola

=I

is shifted 2 units to the

25. Foci: (± 0,0) Vertices: (±2, 0) 26. Foci: (0, ±4) Vertices: (0, ±5)

27. x 2 - y2 = I

28. 9x 2 - 16y2 = 144

29.y2_ X2=8

30.y2_X2=4

-

y2

16

-"9 =

1.

a. Find the center, foci, vertices, and asymptotes of the new

Hyperbolas Exercises 27-34 give equations for hyperbolas. Put each equation in staodard form aod f"md the hyperbola's asymptotes. Then sketch the hyperbola. Ioclude the asymptotes aod foci in your sketch.

31. 8x2

(x - 2)2

2y2

=

hyperbola.

b. Plot the new center, foci, vertices, aod asymptotes, aod sketch in the hyperbola. 44. The hyperbola (y2/4) - (x 2/5) = 1 is shifted 2 units down to generate the hyperbola

32.y2_3x 2 =3

16

(y

4

34. 64x 2 - 36y2 = 2304

33. 8y2 - 21: 2 = 16

+ 2)'

x2

"5

=

1.

a. Find the center, foci, vertices, and asymptotes of the new Exercises 35-38 give information about the fuci, vertices, aod asymptotes of hyperbolas centered at 1he origin of 1he xy-plane. 10 each case, f"md 1he hyperbola~ staodard-furm equation from the information giveo. 35. Foci:

36. Foci:

(0, ±0)

Asymptotes: 37. Vertices:

y = ±x

(±3, 0)

Asymptotes:

y

=

(±2,0)

Asymptotes: 38. Vertices:

±~ x 3

y = ±_I- x

v3

(0, ±2)

Asymptotes:

y

=

hyperbola.

b. Plot the new center, foci, vertices, aod asymptotes, aod sketch in the hyperbola.

Exercises 45-48 give equations for parabolas aod tell how many units up or down and to the right or left each parabola is to be shifted. Find ao equation for the new parabola, aod f"md 1he new vertex, focus, and directrix. 45. y2

±l2 x

= 4x,

47. x 2 = 8y,

= -121:,

left 2, down 3

46. y2

right I, down 7

48. x 2 = 6y,

right 4, up 3

left 3, down 2

11.6 Exercises 4~52 give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an

equation for the new ellipse, and fmd the new foci, vertices, and center. x2 49. (;

y2

+9

2

50.

x T + y2

51.

3 +T

52.

16 + 25

x2

y2

Xl

y2

~

left 2, down I

I,

~ I,

right 3, up4

~

right2,up3

I,

~ I,

x2

y2

16 - 9

~

left 4, down 5

I,

left 2, down I

55. y2 - x 2 ~ I,

left I, down I

y2

56.

3 -

x 2 ~ I,

647

70. Suspension bridge cable. hang in parabolas The suspension bridge cable shown in the accorupanying figure supports a uniform load of w pounds per horizontal foot It can be shown that if H is the horizontal tension of the cable at the origin, then the curve of the cable satisfies the equation

Show that the cable hangs in a parabola by solving this differential equation subject to the initial condition that y ~ 0 when x ~ O.

Exercises 53-56 give equations for hyperbclas and tell how many units up or down and to the right or left each hyperbcla is to be shifted. Find an equation for the new hyperbola, and fmd the new center, foci, vertices, and asymptotes. x 2 y2 53. 4 - 5 ~ I, right 2, up2 54.

Conic Sections

l-~ 71. The width of a parabola .t the fucu. Show that the number 4p is the width of the parabola x 2 ~ 4py (p > 0) at the focus by showing that the line y ~ p cuts the parabola at points that are 4p units apart

72. The asymptote. of (x'Ia') - (yl Iii') = 1 Show that the vertical distsnce between the line y ~ (bla}x and the upper half of the right-hand branch y ~ (bla)Vx 2 - a 2 of the hyperbola (x 2/a 2) - (y2/b 2) ~ I approaches 0 by showing that

right I, up 3

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises 57--68. 57. x 2 + 4x + y2 ~ 12 58. 2x2 + 2y2 - 28x

+

12y

+

Sintilar results hold for the remaining portions of the hyperbola andthelinesy ~ ±(b/a}x.

114 ~ 0

59.X2+2x+4y-3~0

60. y2 - 4y - 8x - 12 ~ 0

61. x 2 + 5y2

62. 9x 2 + 6y2

+ 4x

~ I

+

36y ~ 0

63. x 2 + 2y2 - 2x - 4y ~ -I 64. 4x 2 + y2 65.

+

8x - 2y ~-I

x2_y2_2x+4y~4

67. 2x2

-

y2

+ 6y

~ 3

66. x2_y2+4x-6y~6

68. y2 - 4x 2

+

16x ~ 24

Theory and Examples 69. lflines are drawn parallel to the coordinate axes through a point P on the parabola y2 ~ Irx, /r; > 0, the parabola partitions the rectsngular region bounded by these lines and the coordinate axes into two arualler regions, A and B. a. If the two arualler regions are revolved about the y-axis, show that they generate solids whose volumes have the ratio 4: 1. b. What is the ratio of the volumes generated by revolving the regions about the x-axis? y p

B ~~----------~~X

o

73. Area Find the dimensions of the rectsng1e of largest area that can be insctibed in the ellipse x 2 + 4y2 ~ 4 with its sides parallel to the coordinate axes. What is the area of the rectangle? 74. Volume Find the volUD2e of the solid generated by revolving the region enclosed by the ellipse 9x' + 4yl ~ 36 about the
W+1,

V2,

80. Surf... are. The curve y ~ 0 '" x '" which is part of the upper branch of the byperbola y2 - x 2 ~ 1, is revolved about the x-axis to generate a surface. Find the area of the surface.

648

Chapter 11: Parametric Equations and Polar Coordinates

81. The reflective property of parabolas The accompanying figure shows a typical pointP(xo,yo) on 1he parabolay2 = 4px. The line L is tangent to 1he parabola at P. The parabola's focus lies at F(p, 0). The ray L' ex1ending from P to 1he right is pam1lel to 1he x-axis. We show 1hat light from F to P will be reflected out along L' by showing 1hat f3 equals a. Establish this equality by taking 1he following steps.

This reflective property of parabolas is used in applications like car headlights, radio telescopes, and satellite TV dishes. y L

L'

•• Show1hattanf3 = 2p/yo. b. Show 1hat tan 4> = Yo/(x, -

pl·

c. Use 1he identity tan 4> - tan f3 tana = 1 + tan4>tanf3 to show 1hat tan a = 2p/yo.

Since a and f3 are bo1h acute, tan f3 = tan a implies f3 = a.

11.7

Conics in PoLar Coordinates Polar coorilinates are especially important in astronomy and astronautical engineering because satellites, moons, planets, and comets all move approximately along ellipses, parabolas, and hyperbolas that can be described with a single relatively simple polar coordinate equation. We develop that equation here after first introducing the idea of a conic section's eccentricity. The eccentricity reveals the conic section's type (circle, ellipse, parabola, or hyperbola) and the degree to which it is "squashed" or flattened.

Eccentricity Although the center-to-focus distance c does not appear in the equation

x2 2 a

y2

+ 2= b

1,

(a> b)

Va

a

2 2 for an ellipse, we can still determine c from the equation c = - b • If we fix and vary c over the interval 0 :5 c :5 a, the resulting ellipses will vary in shape. They are circles if c = 0 (so that a = b) and flatten as c increases. If c = a, the foci and vertices overlap and the ellipse degenerates into a line segment. Thus we are led to consider the ratio 2 e = We use this ratio for hyperbolas as well, ouly in this case equals + b2

cia.

instead of

c

Va

2

-

2

Va

b , and derme these ratios with the somewhat familiar term eccentricity.

DEFINmON The eccentricity of the ellipse (x 2/a 2)

+ (y2/b 2)

c e=-= a

c e=-= a The eccentricity of a parabola is e = I.

= I

(a > b) is

11.7 Conics in Polar Coordinates y Directrix 1 x

Whereas a parabola has one focus and one directrix, each ellipse has !we foci and !we ±a/e from the center. The parabola has the property that

Directrix 2

=-~

X=

b

directrices. These are the lines perpendicular to the major axis at distaoces

~

PF

D2

-b c=ae a

directrix 2 to focus F2.

=-~

Y

Direc1rix 2 X=

I'PD

(1)

for any point P on it, where F is the focus and D is the point nearest P on the directrix. For an ellipse, it can be sbown that the equations that replace Equation (I) are (2)

FIGURE 11.45 The foci and directrices of the ellipse (x 2/a 2) + (y2/b 2) ~ 1. Directrix I corresponds to focus F1 and

x

~

a

Ii

DiIeclrix 1

649

Here, e is the eccentricity, P is any point on the ellipse, FJ and Fz are the foci, and DJ and Dz are the points on the directrices nearest P (Figure 11.45). In both Equations (2) the directrix and focus must correspond; that is, if we use the distaoce from P to F 1 , we must also use the distaoce from P to the directrix at the same end of the ellipse. The directrix x ~ -ale corresponds to F1{ -c, 0), and the directrix x ~ a/e corresponds to Fz{c, 0). Ai; with the ellipse, it can be shown that the lines x ~ ±a/e act as directrices for the hyperbola and that

~

e

and

PFz = e·PDz.

(3)

Here P is any point on the hyperbola, F1 and Fz are the foci, and D1 and Dz are the points nearest P on the directrices (Figure 11.46). In both the ellipse and the hyperbola, the eccentricity is the ratio of the distaoce between the foci to the distaoce between the vertices (because cia = 2c/2a).

E

FIGURE 11.46 The foci and directrices of the hyperbola (x 2/a 2) - (y2/b 2) ~ 1. No matter where Plies 00 the hyperbola,

PF1

~

e·PD 1 andPF2

~

e·PD2 •

tri·ty distaoce between foci ccen C1 - distaoce between vertices

In an ellipse, the foci are closer together than the vertices and the ratio is less than 1. In a hyperbola, the foci are farther apart than the vertices and the ratio is greater than 1. The "focus-directrix" equation PF ~ e' PD unites the parabola, ellipse, and hyperbola in the following way. Suppose that the distance PF of a point P from a f"",d point F (the focus) is a constaot multiple of its distaoce from a fixed line (the directrix). That is, suppose

PF~

e'PD,

(4)

where e is the constant of proportionality. Then the path traced by P is

(al aparabola if e

~

I,

(bl an ellipse of eccentricity e if e

<

(c) a hJPerbola of eccentricity e if e

I, and

> 1.

There are no coordinates in Equation (4), and when we try to translate it into coordinate form, it translates in different ways depending on the size of e. At least, that is what happens in Cartesian coordinates. However, as we will see, in polar coordinates the equation PF = e' PD translates into a single equation regardless of the value of e. Given the focus and corresponding directrix of a hyperbola centered at the origin and with foci on the x-axis, we can use the dimensions sbown in Figure 11.46 to fmd e. Knowing e, we can derive a Cartesian equation for the hyperbola from the equation PF ~ e' PD, as in the next example. We can find equations for ellipses centered at the origin and with foci on the x-axis in a similar way, using the dimensions sbown in Figure 11.45.

650

Chapter 11: Parametric Equations and Polar Coordinates

EXAMPLE 1 Find a Cartesian equation for the hyperbola centered at the origin that bas a focus at (3, 0) and the line x = I as the corresponding directrix.

y

x=l

Solution We first use the dimensions shown in Figure 11.46 to find the hyperbola's eccentricity. The focus is

(c,O) = (3,0), ----+-~~~

__

--~x

c = 3.

so

The directrix is the line

x

=

ea =

a

so

1,

=

e.

When combined with the equation e = cia that defines eccentricity, these results give FIGURE 11.47 The hyperbola and directrix in Example 1.

e

c

a

=

=

3

e'

e2

so

3 and e =

=

v'3.

Knowing e, we can now derive the equation we want from the equation PF = e' PD. In the notation of Figure 11.47, we bave

PF = e'PD

V(x - 3)2 + (y - 0)2 x2

-

6x + 9 + 2x 2 -

x2

v'3lx - II

=

y2 =

Eq. (4)

3(x 2

-

e~

v'3

2x + 1)

y 2=6 y2

•

3-6=1. Polar Equations Directrix

D

To find polar equations for ellipses, parabolas, and hyperbolas, we place one focus at the origin and the corresponding directrix to the right of the origin along the vertical line x = k (Figure 11.48). In polar coordinates, this makes

PF= r --~~~---r----~ X

k

and

PD = k-FB = k- rcos6. x=1c

The conic's focus-directrix equation PF = e' PD then becomes

r = elk - rcos6), FIGURE 11.48 If a conic section is put in the position with its focus placed at the origin and a direc1rix perpendicular to the initial ray and right of the origin, we can fmd its polar equation from the conic's focus-directrix equation.

which can be solved for r to obtain the following expression.

Polar Equation for a Conic with Eccentricity e ke

r=l+ecosO' where x = k

>

(5)

0 is the vertical directrix.

EXAMPLE 2 Here are polar equations for three conics. The eccentricity values identifying the conic are the same for both polar and Cartesian coordinates. I

e

= 2:

ellipse

e

=

parabola

e

= 2:

I:

hyperbola

k 2 + cos6 k r= I + cos6 2k r= 1+2cos6 r=

•

11.7 Conics in Polar Coordinates

651

You may see variations of Equation (5), depencling on the location of the directrix. If the directrix is the line x = -k to the left of the origin (the origin is still a focus), we replace Equation (5) with

ke r=l-ecosO" The denominator now has a ( - ) instead of a ( + ). If the directrix is either of the lines y = k or y = -k, the equations have sines in them instead of cosines, as shown in Figure 11.49. r =

ke

r =

1+ecos8

ke

l-ecos8

----""-1+ - x

(b) r =

ke 1+esin9

Ice

r =

71'\

y Direclrixy

l-esin9

\

=k

Focus at origin

117too~

~y=-k

(c)

(d)

FIGURE 11.49 Equations for conic sections wilh eccentricity e > 0 but different locations of1he directrix. The graphs here show a parabola, so e = 1.

EXAMPLE 3 = 2.

Find an equation for the hyperbola with eccentricity 3/2 and directrix

x

Solution

We use Equation (5) withk = 2ande = 3/2: 2(3/2) r = cl -c+-(O-::3-i:/2°c)'--co-s--:C O

EXAMPLE 4

r = 10

Solution

•

Find the directrix of the parabola

Directrix

x=k

6 r = "2-+~3c-c-o-s~O .

or

+

25 10cosO"

We divide the numerator and denominator by 10 to put the equation in standard

polar form: r = 1

5/2 cosO'

+

This is the equation

ke a e

FIGURE 11.50 In an ellipse wilh

semimajor axis a, the focus-directrix distance is k = (a/e) - ea, so Ire = a(1 - e').

r=1+ecos8 with k = 5/2 and e = 1. The equation of the directrix is x = 5/2.

•

From the ellipse diagram in Figure 11.50, we see that k is related to the eccentricity e and the semimajor axis a by the equation

k

=

ea -

ea.

652

Chapter 11: Parametric Equations and Polar Coordinates From this, we find that ke = a(1 - e 2 ). Replacing ke in Equation (5) by a(1 - e 2 ) gives the staodard polar equation for an ellipse.

Polar Equation for the Ellipse with Eccentricity e and Semimajor Axis a a(1 - e 2 ) r = 1 + e cos 6

(6)

Notice that when e = 0, Equation (6) becomes r = a, which represents a circle.

Lines y

Suppose the perpendicular from the origin to line L meets L at the point Po(ro, (0), with ro 2: 0 (Figure 11.51). Then, if P(r, 0) is any other point on L, the points P, Po, and 0 are the vertices of a right triangle, from which we can read the relation ro = rcos(O - (0).

The Standard Polar Equation for Lines lf the point Po(ro, ( 0 ) is the foot of the perpendicular from the origin to the line L, and ro 2: 0, then an equation for L is We can obtain a polar equation for line L by reading 1he relation ro ~ r cos (9 - ( 0 ) from 1he right triaogle

r cos (0 - (0) = ro.

(7)

FIGURE 11.51

For example, if 00 = 1T/3 and ro = 2, we find that

OPoP.

rcos(0-;)=2 r(cosocos;

+

!rcosO

sinO Sin;) = 2

+

~ rsinO =

2,

or

x

+

V3y =

4.

Circles y

To find a polar equation for the circle ofredius a centered at Po(ro, (0), we let P(r, 0) be a point on the circle and apply the Law of Cosines to triangle opoP (Figure 11.52). This gives

a2 = rl + r2

- 2rorcos(0 - (0)'

If the circle passes through the origin, then ro = a and this equation simplifies to

FIGURE 11.52 We can get a polar equation for this circle by applying 1he Law of Cosines to triaogle OPoP.

a 2 = a 2 + r2 - 2arcos (0 r2 = 2ar cos (0 - (0) r = 2a cos (0 - ( 0 ),

(0)

If the circle's center lies on the positive x-axis, 00 = 0 and we get the further simplification r = 2a cos O.

(8)

11.7 Conics in Polar Coordinates

653

If the center lies on the positive y-axis, 8 = 7r/2,cos(8 - 7r/2) = sin8, and the equation r = 2a cos (8 - (0) becomes

r=2asin8.

(9)

Equations for circles through the origin centered on the negative x- and y-axes can be obtained by replacing r with - r in the above equations.

EXAMPLE 5 Here are several polar equations given by Equations (8) and (9) for circles through the origin and baving centers that lie on the x- or y-axis. Center (polar coordinates)

Radius 3 2

(3,0) (2,7r/2) (-1/2,0) (-1,7r/2)

1/2 I

Polar equation r=6cos8 r = 4sin8 r = -cos8 r=-2sin8

•

Exercises 11.7 =3

Ellipses and Eccentridty

21. &;' - 2y' = 16

22. y' - 3x'

10 Exercises 1-8, fmd the eccentricity of the ellipse. Then fmd and graph the ellipse's foci aod directrices.

23. 8y' - 2x'

24. 64x' - 36y'

1. 16>:' + 25y' = 400

2. 7x' + 16y' = 112

3.2x'+y'=2

4.2x'+y'=4

5. 3x'

+ 2y'

= 6

7. 6>:' + 9y' = 54

6. 9x'

+ IQy'

= 90

8. 169x' + 25y' = 4225

Exercises 9-12 give the foci or vertices and the eccentricities of ellipses centered at the origin of the xy-plane. 10 each case, find the ellipse's standard-form equation in Cartesian coordinates.

9. Foci:

(0, ±3) Eccentricity: 0.5

11. Vertices: (0, ±70) Eccentricity: 0.1

10. Foci: (±8,0) Eccentricity: 0.2

12. Vertices: (±IO,O) Eccentricity: 0.24

( v5, 0)

Directrix:

x

= ..:is

15. Focus: (-4,0) Directrix: x = -16

14. Focus:

Vz,

Vz

Hyperbolas and Eccentridty 10 Exercises 17-24, fmd the eccentricity of the hyperbola. Then fmd

=I

19. y' - x'

=

8

18. 9x' - 16y' = 144

20. y' -

x' =

Exercises 25--28 give the eccentricities and the vertices or foci of hyperbolas centered at the origio of the xy-plaoe. 10 each case, fmd the hyperbola's standard-form equatinn in Cartesian coordinates. 25. Eccentricity: 3 26. Eccentricity: 2 Vertices: (0, ±i) Vertices: (±2,0) 27. Eccentricity: 3 Foci: (±3,0)

4

28. Eccentricity: 1.25 Foci: (0, ±5)

Eccentridties and Directrices Exercises 29-36 give the eccentricities of conic sections with one focus at the origin a100g with the directrix corresponding to that focus. Find a polar equation for each conic section. 31. e

=

5, y

33. e

=

1/2, x

=

30. e = I, y = 2

-6 =

I

35. e = 1/5, y = -10

32. e=2, 34. e

=

x=4

1/4, x

=

-2

36. e = 1/3, y = 6

I 37. r= I + cos8

38.r=2+cos8

25 39. r = 10 - 5cosB

4 40. r = 2 - 200sB

400

and graph the hyperbola's foci and directrices.

17. x' - y'

2304

Sketch the parabolas and ellipses in Exercises 37-44. Ioclude the directrix that corresponds to the focus at the origin. Label the vertices with appropriate polar coordinates. Label the centers of the ellipses as well.

6 x = 13

16. Focus: ( 0) Directrix: x = -2

=

Parabolas and Ellipses

(4,0)

Directrix:

16

29. e = I, x = 2

Exercises 13-16 give foci and correspnnding directrices of ellipses centered at the origin of the xy-plane. In each case, use the dimensions in Figure 11.45 to fmd the eccentricity of the ellipse. Then fmd the ellipse's standard-form equation in Cartesian coordinates. 13. Focus:

=

6

12

41. r = 16

+ 8sinB

42. r= 3

+ 3sinB

43.r=2

8 2sinB

44. r = 2

4 sinB

654

Chapter 11: Parametric Equations and Polar Coordinates

+ 2 sin B)

+ 2 ces B)

Lines

73. r = 1/(1

Sketch the lines in Exercises 45-48 and rmd Cartesian equations for them.

75. Perihelion and aphelion A planet travels about its SOD in an ellipse whose semimajor axis has length a. (See accompanying figure.)

45. rcos

(B - f) =

47. r cos (B - 2:)

V2 =3

(B + 3:) = I 48. rces (B + f) = 2

V2x + V2y = 6

1/(1

b. Use the data in the table in Exercise 76 to rmd how close each planet in our solar system comes to the sun and how far away each planet gets from the sun. Perihelion

Aphelion

50. V3x-y= 1 52.x=-4

51. y = -5

=

a. Show that r = a( I - e) wheo the planet is closest to the SOD and that r = a(l + e) wheo the planet is far1hest from the SOD.

46. rces

Find a polar equation in the form r ces (B - Bo) = ro for each of the lines in Exercises 4!1-52.

49.

74. r

(farthest

(closest

from sun)

to sun)

Circles Sketch the circles in Exercises 53-56. Give polar coordinates for their cooters and idootiJY their radii.

53. r = 4cosB 55. r = -2cosB

54. r = 6sinB 56. r = -8 sinB

Find polar equations for the circles in Exercises 57-64. Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations.

57. (x - 6)'

+ y2

76. Planetary orbits Use the data in the table below and Equation (6) to rmd polsr equations for the orbits of the planets.

+ 2)2 + y2 = 4 2 60. x + (y + 7)' = 49 58. (x

= 36

59. x 2 + (y - 5)2 = 25 61. x 2 +Zx+y2=0

62. x 2 - 16x

63. x 2 +y2+ y =0

64. x 2 + y2

+ y2 = 0 4 - :fy = 0

Examples of Polar Equations

D Graph the lines and conic sections in Exercises 65-74. 67. r = 4sinB 69. r = 8/(4 + cos B)

66. r = 4 sec (B + 'TT/6) 68. r = -2C08B 70. r = 8/(4 + sinB)

71. r = 1/(1 - sin B)

72. r = 1/(1

65. r

=

3 sec (B - 'TT/3)

Chapter

rm

+

Planet

Semimajor axis (astronomical units)

Eccentricity

Mercury

0.3871

0.2056

Venus

0.7233

0.0068

Ear1h

1.000

0.0167

Mars

1.524

0.0934

Jupiter

5.203

0.0484

Saturn

cos B)

9.539

0.0543

Uranus

19.18

0.0460

Neptone

30.06

0.0082

Questions to Guide Your Review

1. What is a parametrization of a curve in the xy-plane? Does a function y = f(x) always have a parametrization? Are parametrizations of a curve unique? Give examples.

2. Give some typical parametrizations for lines, circles, parabolas, ellipses, and hyperbolas. How might the parametrized curve differ from the graph ofits Cartesian equation? 3. What is a cycloid? What are typical parametric equations for cycloids? What physical pruper1ies account for the importance of cycloids? 4. What is the formula for the slope dy/dx of a parametrized curve x = f(t), y = get)? When does the formula apply? Whoo can you expect to be able to rmd d'y/dx 2 as well? Give examples. 5. How can you sometimes rmd the area bounded by a parametrized

curve and one of the coordinate axes? 6. How do you rmd the leogth of a smooth parametrized curve x = f(t), y = get), a :5 I :5 b? What does smoothness have to do with leogth? What else do you need to know about the parametrization in order to rmd the curve~ length? Give examples.

7. What is the arc length function for a smooth parametrized curve? What is its arc leogth differential? 8. Under what conditions can you find the area of the surface generated by revolving a curve x = f(I), Y = g(1), a :5 I :5 b, about

the x-axis? the y-axis? Give examples. 9. How do you rmd the centroid of a smooth parametrized curve x = f(t), y = g(t), a :5 I :5 b? Give an example. 10. What are polar coordinates? What equations relate polar coordinates to Cartesian coordinates? Why might you want to change from one ceordinate system to the other? 11. What cousequooce does the lack of uniqueoess of polsr coordinates have for graphing? Give an example. 12. How do you graph equations in polar coordinates? Include in your discussiou syounetry, slope, behavior at the origin, aud the use of Cartesisn graphs. Give examples. 13. How do you rmd the area of a region 0 '" r,(B) :5 r :5 r2(B), a :5 B :5 {j, in the polar coordinate plane? Give examples.

Chapter 11 14. Under what conditions can you fmd the length of a curve r = f( 6), Ol '" 6 s ,B, in the polar coordinate plane?
15. What is a parabola? What are the Cartesian equations for parabolas whose vertices lie at the origin and whose foci lie on the coordinate axes? How can you fmd the focus and directrix of such a parabola from its equation? 16. What is an ellipse? What are the Cartesian equations for ellipses centered at the origin with foci on one of the coordinate axes?

How can you fmd the foci, vertices, and directrices of such an ellipse from its equation?

Chapter

[[1

655

17. What is a hyperbola? What are the Cartesian equations for hyperbolas centered at the origin with foci on one of the coordinate axes? How can you f'md the foci, vertices, and directrices of such an ellipse from its equation? 18. What is the eccentricity of a conic section? How can you classify conic sections by eccentricity? How are an ellipse's shape and eccentricity related?

19. Explain the equation PF = e' PD. 20. What are the siandard equations for lines and conic sections in

polar coordinates? Give examples.

Practice Exercises

Identifying Parametric Equations in the Plane Exercises I-J) give parametric equations and parameter intervals for the motion of a particle in the ",,-plane. Identify the particle's path by finding a Cartesian equation for it Graph the Cartesian equation and indicate the direction of motion and the portion traced by the particle.

1. x = 1/2, Y = I

Practice Exercises

+

I;

-00

2.x=Vt, y=I-Vt;


00

3. x = (1/2) ian I, Y = (1/2) sec I; -'1T/2 4. x = -2 cos t, Y = 2 sin t; 0 ~ t ~ '1T S. X = -cost, Y = cos2 t; O:s; t:5 'T1' 6. x = 4 cos t, y = 9sint; 0:5 t:5 2'7T

+ (I/y), I s y '" 2 17. x = 5eosl - cos 51, y = 5 sin I - sin51, 18. x = t 3 - 6t 2, Y = t 3 + 6t 2, 0:5 t :5 I

16. x = (y'/12)

19. x = 3 cos 6, y = 3 sin 6,

1"'0

< I < '1T/2

I s x s 32

15. y = (5/12)x6/' - (5/8)x4/',

os I s '1T/2

Os 6 s 3;

20. Find the length of the enclosed loop x = 12,y = (1'/3) - I shown here. The loop starts at I = - V3 and ends at I = V3.

finding Parametric Equations and Tangent Lines 7. Find parametric equations and a parameter interval for the motion of a particle in the ",,-plane that traces the ellipse 16x2 + 9y2 = 144 once counterclockwise. (There are many ways to do this.) 8. Find parametric equations and a parameter interval for the motion of a particle that starts at the point (-2, 0) in the ",,-plane and traces the circle x2 + y2 = 4 three times clockwise. (There are many ways to do this.)

In Exercises 9 and 10, fmd an equation for the line in the xy-plane that is iangent to the corve at the point corresponding to the given value of I. Also, fmd the value of d'y/dx 2 at this point.

9. x = (1/2) ian I, Y = (1/2) sec I; I = '1T/3 10. x = I + 1/12, y = I - 3/1; I = 2 11. Eliminate the parameter to express the corve in the form y = f(x). a. x = 4t 2, Y = t 3 - I b. x = cos t, Y = tan t 12. Find parametric equations for the given curve. a. Line through (I, -2) with slope 3 b. (x - I)' + (y + 2)' = 9 c.y=4x 2 -x d. 9x'

+ 4y'

=

y2/3,

1:5 Y :5 8

21. x

=

t 2/2,

22. x = I'

+

y

=

2t,

0:5 t :5

1/(21), y = 4vt,

V5;

x-axis

1/0", I '" I; y-axis

!'olaf to Cartesian Equations

for each line.

Lengths of Curves Find the lengths of the curves in Exercises 13-19.

14. x

Find the areas of the surfaces generated by revolving the corves in Exercises 21 and 22 about the indicated axes.

Sketch the lines in Exercises 23-28. Also, fmd a Cartesian equation

= 36

13. y = xl/2 - (1/3)x'/2,

Surface Areas

I

s xs 4

23. rcos (6

+

f)

= 2V3

24. rcos

(6 - 3:) V; =

-\12 sec 6

25.r=2see6

26. r =

27. r

28. r = (3V3) cse6

=

-(3/2) csc 6

656

Chapter 11: Parametric Equations and Polar Coordinates

Find Cartesian equations for the circles in Exercises 29--32. S1retch each circle in the coordinate plaoe and label it with both its Cartesian aod polar equations.

29.

r~

-4sinO

31. r ~ 2Vz cos 0

Area in Polar Coordinates Find the areas of the regions in the polar coordinate plaoe described in Exercises 47-50.

30. r ~ 3\13 sinO

47. Euc10sed by the I~on r = 2 - cos 0

32. r

4S. Enclosed by one leaf of the three-leaved rose r

~

-6cosO

Cartestan to Polar Equations Find polar equations for the circles in Exercises 33--36. Sketch each circle in the coordinate plane aod label it with both its Cartesiao aod polar equations.

~

sin 30

49. Inside the "figure eight" r = I + cos 20 and outside the circle r = I 50. Inside the cardioid r ~ 2(1 + sin 0) aod outside the circle

r=2sin8

33. X2+y2+5y~0

34.X 2 +y2_2y=0

Length In Polar Coordinates

x2+y2-3x~0

36.x 2 +y2+4x=0

Find the lengths of the curves given by the polar coordinate equations in Exercises 51-54.

35.

Graphs In Polar Coordinates Sketch the regions defmed by the polar coordinate inequalities in Exercises 37 and 38.

37. 0 :s; r

:5

6 cos 8

53. r ~ 8 sin' (0/3),

3S. -4sin0:5 r:5 0

Match each graph in Exercises 39--46 with the appropriate equation (a)-(I). There are more equations thao graphs, so some equations will not be matched.

6

a. r = cos28

b. rcosO = I

•• r=I-200s0

d. r = sin20

e. r = 8

&.r=l+cosO

h. r = I - sinO

j.

k. r = -sinO

f. ,2 = cos28 i. r = ~__2"----" I cosO Lr=2cosO+1

,2 =

sin28

39. Four-leaved rose

40. Spiral y

y

51. r~ -I + cosO 52. r = 2 sinO + 2 cosO,

v'1

54. r =

0:5 0:5 'If/2

0:5 0 :5 'If/4

+ cos 20,

-'If/2 :5 0 :5 'If/2

Graphing Conic Sections Sketch the parabolas in Exercises 55-58. Include the focus and directtix in each sketch.

55. x 2 ~ -4y

56. x 2 ~ 2y

57. y2

5S. y2

=

3x

=

-(S/3)x

Find the eccentricities of the ellipses and hyperbolas in Exercises 59--62. Sketch each conic section. Include the foci, vertices, and asymptotes (as appropriate) in your sketch. 59. 1,,2

+ 7y2

60. x 2 + 2y2 = 4

= 112

61.3x 2 - y 2=3

62. 5y2 - 4x 2 = 20

Exercises 63-68 give equations for conic sections aod tell how maoy units up or down and to the right or left each curve is to be shifted Find

an equation for the new conic section, and fmd the new foci, vertices, 41.

Lin!a~on

centers, and asymptotes, as appropriate. If the curve is a parabola, fmd the new directtix as well.

42. Lemoiscate

y

63. x 2 ~ -12y, 64. y2 = lOx,

-I1f-4-- -+ --> x

x2

65.

"9 2

44. Cardioid

43. Circle y

left 1/2, down I

y2

+ 25 ~ I,

left 3, down 5

y2

66.

1~9

67.

If - "2 =

y2

y

right 2, up 3

+ 144 = I,

right 5, up 12

x2

I,

x2 y2 6S. 36 - 64 ~ I,

right 2, up 2Vz left 10, down 3

Identifying Conic Sections 45. Parabola

Complete the squares to identify the cunic sections in Exercises 69--76. Find their foci, vertices, centers, and asymptotes (as appropriate). If the curve is a parabola, fmd its directtix as well.

46. Lemoiscate y

69. x 2 -4x-4y2=0 -~-i-=-

__ x

70.4x 2 - y 2+4y=8 72. x 2 - 2x + 8y = -17

71. y2 - 2y + I" = -49 73. 9x 2 + 16y2 + 54x - 64y = -I 74. 25x 2 + 9y2 - lOOx + 54y ~ 44 75. X2+y2_2x-2y=0

76. X2 +y2+4x+2y= I

657

Chapter 11 Additional and Advanced Exercises Conics in Polar Coordinates Sketch the conic sections whose polar coordinate equations are given in Exercises 77-80. Give polar coordinates for the vertices and, in the case of ellipses, for the centers as well.

2

8

77. r = I + cos 0 6

78. r = 2 + cos 0 12 80.r=3+sinO

79.r=I-2cosO

86. TIre ''triangular'' regioo in the fIrSt quadrant bounded by the x-axis, the line x = 4, and the byperbola 9x 2 - 4y2 = 36 is revolved about the x-axis to generate a solid. Find the volume of the solid. 87. Show that the equatioos x = r cos 0, y = rsin 0 1ransfonn the polar equatioo k

r = ~1-+c-"',-co-s~O

Exercises 81-84 give the eccentricities ofeanic sections with one focus at the origin of the polar coordinate plane, along with the directrix for that focus. Find a polar equation for each conic section. 81. , = 2,

83. , = 1/2,

rcosO = 2 rsinO = 2

82. ,= I,

into the Cartesian equation

(! - ,2)x 2 + y2 + 2kex - k 2 = O.

rcosO = -4

84. ,= 1/3,

rsinO = -6

Theory and Examples

88. Archimedes spirals The graph of an equatioo of the fonn r = a8, where a is a nonzero constant, is called an Archimedes spiral. Is there anything special about the widths betwoen the soc-

85. Find the volume of the solid generated by revolving the region enclosed by the ellipse 9x 2 + 4y2 = 36 aboot (a) the x-axis, (b) the y-axis.

Chapter

mO!

cessive turns of such a spiral?

Additional and Advanced Exercises

Finding Conic Sections

10. Show that the line

1. Find ao equatioo for the parabola with focus (4, 0) and directrix x = 3. S1retch the parabola together with its vertex, focus, and di-

b 2xxi

2. Find the vertex, focus, and directrix of the parabola -

6x - l2y + 9 = O.

3. Find an equation for the curve traced by the point p(x, y) if the distsnce fromP to the vertex of the parabola x 2 = 4y is twice the distsnce from P to the focus. IdentifY the curve. 4. A line segment oflength a + b runs from the x-axis to the y-axis. The point P on the segment lies a units from one end and b units from the other end. Show that P traces an ellipse as the ends of

the segment slide along the axes.

5. The vertices of ao ellipse of eccentricity 0.5 lie at the points (0, ±2). Where do the foci lie? 6. Find an oquation for the ellipse of eccentricity 2/3 that has the line x = 2 as a directrix aod the point (4, 0) as the corresponding focus.

7. One focus of a hyperbola lies at the point (0, -7) and the corr<>sponding directrix is the line y = -I. Find an equation for the byperbola if its eccentricity is (a) 2, (b) 5. 8. Find an equatioo for the byperbola with foci (0, -2) and (0, 2) that passes through the point (12, 7). 9. Show that the line

is tsngent to the ellipse b 2x 2 + a2y2 - a 2b 2 = 0 at the point (XI, YI) on the ellipse.

a7.Y1 - a 2b 2

=0

is tsngent to the hyperbola b'x 2 - a2y2 - a 2b 2 (XI, YI) on the byperbola.

rectrix.

x2

-

= 0 at the point

Equations and Inequalities What points in the xy-plane satisfY the equations aod inequalities in Exercises 11-16? Draw a ftgul'e for each exercise. 11. (x 2 - y2 - l)(x 2 + y2 - 25)(x 2 + 4y2 - 4) = 0 12. (x

13.

+ y)(x2 + y2

(x 2/9)

- I) = 0

+ (y2/16) :5 I

14. (x 2/9) - (y2/16) 15. (9x 2 16. (9x 2

+ 4y2 + 4y2

:5 I

- 36)(4x 2 - 36)(4x 2

+ +

9y2 - 16) :5 0 9y2 - 16)

>0

Polar Coordinates 17.

L

Find an equation in polar coordinates for the curve

x = e 2 tcost. y = e2t sint;

-00


00.

= 2". = 2 sin' (0/3), 0 :5 0

b. Find the length of the curve from t = 0 to t 18. Find the length of the curve r the polar coordinate plane.

s 3", in

Exercises 19-22 give the eccentricities of conic sections with one focus at the origin of the polar coon1inate plane, along with the directrix for that focus. Find a polar equation for each conic section. 19. e = 2,

20. e

rcos6 = 2

21. e = 1/2, rsinO

=

2

22.

=

1,

,= 1/3,

rcos8 = -4

rsinO

=

-6

658

Chapter 11: Parametric Equations and Polar Coordinates

Theory and Examples 23. Epicycloid. When a circle rolls externa1ly along the circumference of a second, ftxed circle, any point P on the circumference of the rolling circle describes an epicyC/oid, as shown here. Let the IlXed circle have its center at the origin 0 and have radius a. y

Since I/J = t/> - 0 from (I), tanI/J

= tan(t/>

tant/> - tanO

- 0)

= I + tant/>tanO'

Furthermore,

p

dy

dy/dO

tant/> = is the slope of the curve at P. Also,

Y tanO=x'

--~~----~~~X

o

A(a, 0)

Hence Let the radius of the rolling circle be b aod let the initial position of the tracing point P be A(a, 0). Find parametric equations for the epicycloid, using as the parameter the aogle 0 from the positive x-axis to the line through the circles' centers.

tanI/J =

24. Find the centroid of the region enclosed by the x-axis and the cy-

I

cloid arch x

~

a(t - sint), y

~

a(I - cost);

~

0

+ I/J,

(4)

+ x
The numerator in the last expression in Equation (4) is found from Equations (2) and (3) to be

0:5 t:5 2"..

The Angle Between the Radius Veetor and the Tangent Line to a Polar Coordinate Curve In Cartesian coordinates, when we want to discuss the direction of a curve at a point, we use the angle cfJ measmed counterclockwise from the positive x-axis to the tangent line. In polar coordinates, it is more convenient to calculate the angle I/J from the radius vector to the tangent line (see the accoropanying figore). The aogle t/> can then be calculated from the relation

t/>

dy/dO y
----

(1)

Similarly, the denominator is

When we substitute these into Equation (4), we obtain

which comes from applying the Exterior Angle Theorem to the triangle in the accompanying figore.

r

(5)

tanI/J = dr/dO'

y

This is the equation we use for fmdiog I/J as a function of O. 25. Show, by reference to a figure, that the angle f3 between the tangents to two curves at a point of intersection may be found from

the formula (6)

o

When will the two curves intersect at right angles?

Suppose the equation of the curve is given in the form r where flO) is a differentiable function of O. Then x

= rcos8

and y = 7sin8

= f(O), (2)

are differentiable functions of 0 with
+ cos.adr dO'

dy a ' a dr dO ~ rcosu + smu dO'

(3)

26. Find the value oftanI/J for the curver ~ sin4 (0/4). 27. Find the angle between the radius vector to the curve r and its tangent when 0 = "./6.

D 28.

~

2a sin 30

a. Graph the hyperbolic spiral rO = I. What appears to happen to I/J as the spiral winds in around the origin?

b. Confmn yoar fmdiog in part (a) analytically. 29. The circles r = v'3 cos 0 and r = sin 0 intersect at the point (0/2, "./3). Show that their tangents are perpendicular there. 30. Find the angle at which the cardioid r ~ a(I - cos 0) crosses the ray 0 = "./2.

Chapter 11 Technology Application Projects

Chapter [il

Technology Application Projects

Mathematica/Maple Module: Radar 'Ii'ackillg oftJ Movillg Object Part I: Convert from polar to Carlesian coordinates.

I'tmlmetric tJIId PoItJr EqUtJliollS with tJ Fi1P're SktJter Part I: Visualize positioo, velocity, and acceleratioo to analyze motioo defmed by parametric equations. Part II: Find and analyze the equatioos of motion for a figure skater 1racing a polar plot.

659

12 VECTORS AND THE GEOMETRY OF SPACE OVERVIEW To apply calculus in many real-world situations and in higher mathematics, we need a mathematical description of three-dimensional space. In this chapter we introduce three-dimensional coordinate systems and vectors. Building on what we already know about coordinates in the xy-plane, we establish coordinates in space by adding a third axis that measures distance above and below the xy-plane. Vectors are used to study the analytic geometry of space, where they give simple ways to describe lines, planes, surfaces, and curves in space. We use these geometric ideas later in the book to study motion in space and the calculus of functions of several variables, with their many important applications in science, engineering, economics, and higher mathematics.

Three-DimensionaL Coordinate Systems

12.1

z

z = constant

I

(O,y,z)

(x, 0, z)

01___

P(x,y,z)

- - - __ (O, y,O)

------'y

~ y = constant x

x = constant

(x, y, 0)

FIGURE 12.1 The Cartesian coordinate system is right-handed.

660

To locate a point in space, we use three mutually perpendicular coordinate axes, arranged as in Figure 12.1. The axes shown there make a right-handed coordinate frame. When you hold your right hand so that the fingers curl from the positive x-axis toward the positive y-axis, your thumb points along the positive z-axis. So when you look down on the xy-plane from the positive direction of the z-axis, positive angles in the plane are measured counterclockwise from the positive x-axis and around the positive z-axis. (In a left-handed coordinate frame, the z-axis would point downward in Figure 12.1 and angles in the plane would be positive when measured clockwise from the positive x-axis. Right-handed and left-handed coordinate frames are not equivalent.) The Cartesian coordinates (x, y, z) of a point P in space are the values at which the planes through P perpendicular to the axes cut the axes. Cartesian coordinates for space are also called rectangular coordinates because the axes that define them meet at right angles. Points on the x-axis have y- and z-coordinates equal to zero. That is, they have coordinates of the form (x, 0,0). Similarly, points on the y-axis have coordinates of the form (O,y, 0), and points on the z-axis have coordinates of the form (0, 0, z). The planes determined by the coordinates axes are the xy-plane, whose standard equation is z = 0; the yz-plane, whose standard equation is x = 0; and the xz-plane, whose standard equation is y = O. They meet at the origin (0, 0, 0) (Figure 12.2). The origin is also identified by simply 0 or sometimes the letter O. The three coordinate planes x = 0, y = 0, and z = 0 divide space into eight cells called octants. The octant in which the point coordinates are all positive is called the first octant; there is no convention for numbering the other seven octants. The points in a plane perpendicular to the x-axis all have the same x-coordinate, this being the number at which that plane cuts the x-axis. The y- and z-coordinates can be any numbers. Similarly, the points in a plane perpendicular to the y-axis have a common y-coordinate and the points in a plane perpendicular to the z-axis have a common z-coordinate. To write equations for these planes, we name the common coordinate's value. The plane x = 2 is the plane perpendicular to the x-axis at x = 2. The plane y = 3 is the plane perpendicular to the y-axis

12.1 Three-Dimensional Coordinate Systems

661

z xz-plane: y = 0

\

//

/ /

___ yz-plane: x

=

0

/

xy-plane: z = 0 - - - -- ____ /~gin

~

(0,3,0)

:\ : (0, 0, 0)

Z y

y

I

x

Line x = 2, y = 3

FIGURE 12.2 The planes x = O, y = O,andz = space into eight octants.

°

divide

FIGURE 12.3 The planes x = 2, y = 3, and z = 5 determine three lines through the point (2,3,5).

at y = 3. The plane z = 5 is the plane perpendicular to the z-axis at z = 5. Figure 12.3 shows the planes x = 2, y = 3, and z = 5, together with their intersection point (2, 3, 5). The planes x = 2 and y = 3 in Figure 12.3 intersect in a line parallel to the z-axis. This line is described by the pair of equations x = 2, y = 3. A point (x, y, z) lies on the line if and only if x = 2 and y = 3. Similarly, the line of intersection of the planes y = 3 and z = 5 is described by the equation pair y = 3, z = 5 . This line runs parallel to the x-axis. The line of intersection of the planes x = 2 and z = 5, parallel to the y-axis, is described by the equation pair x = 2, z = 5 . In the following examples, we match coordinate equations and inequalities with the sets of points they define in space.

EXAMPLE 1 (a) z ~

We interpret these equations and inequalities geometrically.

°

The half-space consisting of the points on and above the xy-plane.

(b) x = -3

(c) z = 0, x

z The circle

x 2 + y2

= 4,

z= 3

/

~

0, y

~

(d) x ~ O,y ~ 0, z ~ (e) -1 ~ y ~ 1 (f) y = - 2, z = 2

EXAMPLE 2

°

°

The plane perpendicular to the x-axis at x = - 3 . This plane lies parallel to the yz-plane and 3 units behind it. The second quadrant of the xy-plane. The first octant. The slab between the planes y = -1 and y = 1 (planes included). The line in which the planes y = - 2 and z = 2 intersect. Alternatively, the line through the point (0, -2, 2) parallel to the x-axis. _

What points P(x, y, z) satisfy the equations and

x

FIGURE 12.4 The circle x 2 the plane z

=

3 (Example 2).

+ y2

=

4 in

z = 3?

Solution The points lie in the horizontal plane z = 3 and, in this plane, make up the circle x 2 + y2 = 4 .We call this set of points "the circle x 2 + y2 = 4 in the plane z = 3" or, _ more simply, "the circlex 2 + y2 = 4, z = 3" (Figure 12.4).

663

12.1 Three-Dimensional Coordinate Systems

quadratic as a squared linear expression. Then, from the equation in standard form, read off the center and radius. For the sphere here, we have

(X2

+ 3x +

GY)

x 2 + y2 + Z2 + 3x - 4z + I = 0 (X 2 + 3x) + y2 + (Z2 - 4z) = -I

(-;4 y) -I + (~y + (-;4Y

+ y2 + (Z2 - 4z +

=

21 (x+ 23)2 +y2+(z-2)2=-1+ 49 +4=4' From this standard form, we read that Xo = -3/2, Yo = 0, Zo = 2, and a = v21/2. The centeris (-3/2,0,2). The radius isv21/2. • EXAMPLE 5 Here are some geometric interpretations of inequalities and equations involving spheres. (a) x 2 + y2 + z2 < 4 The interior of the sphere x 2 + y2 + z2 = 4. (h) x 2 + y2 + z2 :5 4 The solid ball bounded by the sphere x 2 + y2 + z2 = 4. Alternatively, the sphere x 2 + y2 + z2 = 4 together with its interior. 2 (e) x + y2 + z2 > 4 The exterior of the sphere x 2 + y2 + z2 = 4. 2 (d) x + y2 + z2 = 4, z :5 0 The lower hemisphere cut from the sphere x 2 +

y2

+ z2

= 4 by the xy-plane (the plane

z = 0) .

•

Just as polar coordinates give another way to locate points in the xy-plane (Section 11.3), alternative coordinate systems, different from the Cartesian coordinate system developed here, exist for three-dimensional space. We examine two of these coordinate systems in Section 15.7.

Exercises 12.1 Geometric Interpretations of Equations

Geometric Interpretations of Inequalities and Equations

In Exercises 1-16, give a geometric description of the set of points in

In Exercises 17-24, describe the sets of points in space whose coordinates satisfY the given inequalities or combinatioos of equatioos aod inequalities.

space whose coordinates satisfY the given pairs of equations. 1. x = 2, Y = 3 2. x = -1, z = 0

= 0, 2 x + y2

3. y

z = 0

S.

=

2

4,

z = 0

+z 2 =4, y=O 9. x 2 + y2 + z2 = I, x = 0

7.

X

11. 12. 13. 14. 15.

16.

+ y2 = 4, z = 8. y2 + z2 = I, x =

6. xl

+ y2 + z2 = 25, Y = -4 x 2 + y2 + (z + 3)' = 25, z = 0 x 2 + (y - 1)2 + z2 = 4, Y = 0 x 2 + y2 = 4, z = Y x 2 + y2 + z2 = 4, Y = x Y = x 2, z = 0 z = y2, X = 1

10. x 2

17. a. x

4. x = I, Y = 0 -2 0

~

0, y

0,

~

Z

18.•. 0 '" x '" 1 c. 0 ~ x ~ 1, 0

~

y

= 0

~

b. x

1,

~

+ y2 + z2

:s 1

20. a. x 2

+ y2 :s 1,

Z

+ y2:5

norestrictiononz

a.

C. x 2

1,

= 0

~

b. 0 '" x '" I, 0 ~z~ 1

x2

19.

0, y

b.

Xl

b. x 2

0,

Z

= 0

0 '" Y s 1

+ y2 + z2 >

1

+ y2 :s

=3

1,

Z

21. a. 1 :sx 2 +y2+z2 :S4 b. x 2

+ y2 + z2 :s

22. a. x = y,

23. a. y ~ x 2,

z = 0 Z

24. a. z = 1 - y, b. z =

y3,

X

~ 0

I, z ~ 0 b. x

=

y,

b. x

S

y2,

no restriction onx

=2

no restriction on z

0

S Z

s 2

662

Chapter 12: Vectors and the Geometry of Space

Distance and Spheres in Space

Z

The formula for the distance between two points in the xy-plane extends to points in space.

The Distance Between P1(XhYh Zl) and P 2(X2,Y2, Z2) is

+

IPI P21 = V(X2 - XI)2

x

FIGURE 12.5 We find the distance between PI and P 2 by applying the Pythagorean theorem to the right triangles PIAB and PIBP2.

(Y2 - YI)2

+

(Z2 - ZI)2

Proof We construct a rectangular box with faces parallel to the coordinate planes and the points PI and P2 at opposite comers of the box (Figure 12.5). If A(X2, YI, zd and B(X2, Y2, Zl) are the vertices of the box indicated in the figure, then the three box edges P I A, AB, and BP2 have lengths

Because triangles PIBP2 and PIAB are both right-angled, two applications of the Pythagorean theorem give IPIP212 = IPIBI 2 + IBP21 2

and

(see Figure 12.5). So IPIP212 = IPIBI 2 + IBP21 2 IPIA 12

+

IABI2

+

Substitute IP]BI 2 = IP]AI 2 + IABI2 .

IBP212

IX2 - xl1 2 + IY2 - YI1 2 + IZ2 - zI1 2 =

(X2 - XI)2

+

(Y2 - Ylf

+

(Z2 - zlf

Therefore

• EXAMPLE 3

The distance between P I (2, 1,5) and P2( -2,3,0) is IPIP21 = V(-2 - 2f

Z

Po(Xo, Yo, zo)

\

P(x,Y,z)

al

I

=

V16

=

V45

+4+ I"::j

+

(3 - 1)2

+

(0 - 5)2

25

•

6.708.

We can use the distance formula to write equations for spheres in space (Figure 12.6). A point P(x, y, z) lies on the sphere of radius a centered at Po(xo,Yo, zo) precisely when IPoPI = a or

I

-- I f"

,,/ 1--

The Standard Equation for the Sphere of Radius a and Center (xo,Yo, zo) (x - xof

+ (y -

YO)2

+

(z - zO)2

= a2

Y x

FIGURE 12.6 The sphere of radius a centered at the point (xo,Yo, zo).

EXAMPLE 4

Find the center and radius of the sphere x2

+ y2 + z2 +

3x - 4z

+ 1 = O.

SoLution We find the center and radius of a sphere the way we find the center and radius of a circle: Complete the squares on the X-, Y-, and z-terms as necessary and write each

664

Chapter 12: Vectors and the Geometry of Space

In Exercises 25-34, describe the given set with a single equatioo or

with a pair of equations. 25. The plaoe perpendicular to the L

x-axis at (3, 0, 0)

b. y-axis

c. z-axis

27. The plaoe througb the point (3, -I, I) parallel to the xy-plane

b. yz-plane

c. xz-plane

28. The circle of radius 2 centered at (0, 0, 0) and lying in the L

xy-plane

b. yz-plane

c. xz-plane

29. The circle of radius 2 centered at (0, 2, 0) and lying in the L

xy-plane

b. yz-plane

parallel to the xy-plane

b. yz-plane

Sphel'l!s Find the centers and radii of the spheres in Exercises 47-50.

+2)' + +(z - 2)' = 8 48. I)' +&+tY +(z +3)' 25 49. (x - Vz)' +{y - Vz)2 +(z +Vz)' 50. x+ &+tY +(z-tY ~ 47. (x

y2

(x -

=

2

b. y-axis

2

=

Find equations for the spheres whose centers aud radii are given in Exercises 51-54.

c. xz-plaoe

Center

Radius

31. The line through the point (I, 3, -I) parallel to the

LX-axis

=

c. planey = 2

30. The circle of radius I centered at (-3,4, I) and lying in a plane L

P2(2, -2, -2)

46. P I (5, 3, -2), P2(0, 0, 0)

26. The plaoe througb the point (3, -1,2) perpendicular to the

L

P 2(2, 3, 4)

45. PI(O, 0, 0), b. y-axis at (0, -I, 0)

c. z-axis at (0, 0, -2) LX-axis

44. P I (3, 4, 5),

c. z-axis

32. The set of points in space equidistant from the origin and the point (0, 2, 0)

53.

33. The circle in whicb the plane through the point (I, 1,3) perpendicular to 1Ire z-axis meets the sphere of radius 5 centered at the origin 34. The set of points in space that lie 2 units from 1Ire point (0, 0, I) and, at 1Ire saroe time, 2 units from the point (0, 0, -I) Inequalities to Describe Sets of Points Write inequalities to describe the sets in Exercises 35-40.

35. The slab bounded by the planes z = 0 and z = I (planes

Vi4

51. (1,2,3) 52. (0, -1,5)

2

(-I,t,-t)

4 9

54. (0, -7,0)

7

Find the centers and radii of the spheres in Exercises 55-58. 55. x 2 y2 z2 4x - 4z = 0

+++ 56. ++ + 57. + + +++ 9 58. 3x + + + 9 x2

lx

y2

2 2

z2 -

8z = 0

6y

2y2

2z2

X

3y2

3z2

2y -

Y

z =

2z =

included) 36. The solid cube in the frrst octant bounded by 1Ire coordioate plaoes and 1Ire planes x = 2, y = 2, and z = 2

37. The half-space consisting of the points 00 and below the xy-plane 38. The upper hemisphere of the sphere of radius I centered at the origin

39. The (a) interior and (b) exterior of the sphere of radius I centered at the point (I, I, I) 40. The closed regioo bounded by the spheres of radius I and radius 2 centered at the origin. (Closed means the spheres are to be included. Had we wanted the spheres left out, we would have asked for the open regioo bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out)

Distance In Exercises 41-46, fmd the distsnce between points PI and P2. 41. Pl(l, I, I),

P 2(3, 3, 0)

42. Pl( -I, 1,5), P2(2, 5, 0) 43. PI(I, 4, 5),

P2(4, -2,7)

Theory and Examples 59. Find a formula for the distance from the point P(x, y, z) to the

a. x-axis

c. z-axis

b. y-axis

60. Find a formula for the distance from the point P(x, y, z) to the

a. xy-plane

c. xz-plane

b. yz-plane

61. Find the perimeter of the triangle with vertices A( -1,2, I), B(I, -I, 3), and C(3, 4, 5). 62. Show that the point p(3, I, 2) is equidistant from the points A(2, -1,3) andB(4, 3, I). 63. Find an equation for the set of all points equidistant from the planesy = 3 andy = -\. 64. Find an equation for the set of all points equidistant from the point (0, 0, 2) and the xy-plane. 65. Find the point on the sphere x 2 nearest •• thexy-plane.

+(y - + + 3)'

(z

5)' = 4

b. thepoint(O, 7, -5).

66. Find the point equidistant from the points (0, 0, 0), (0,4, 0), (3, 0, 0), and (2, 2, -3).

12.2

12.2

Vectors

665

I~ ~_ ct_ or_ s __________________________________ Some of the things we measure are determined simply by their magnitudes. To record mass, length, or time, for example, we need only write down a number and name an appropriate unit of measure. We need more information to describe a force, displacement, or velocity. To describe a force, we need to record the direction in which it acts as well as how large it is. To describe a body's displacement, we have to say in what direction it moved as well as how far. To describe a body's velocity, we have to know where the body is headed as well as how fast it is going. In this section we show how to represent thiogs that have both magnitude and direction in the plane or in space. Terminal

FIGURE 12.7

The directed line segment

AB is called a vector.

Component Form A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment (Fignre 12.7). The arrow points in the direction of the action and its length gives the magnitude of the action in terms of a suitably chosen uuit. For example, a force vector points in the direction in which the force acts and its length is a measure of the force's strength; a velocity vector points in the direction of motion and its length is the speed of the moving object. Figure 12.8 displays the velocity vector v at a specific location for a particle moving along a path in the plane or in space. (This application of vectors is studied in Chapter 13.) y

y B _______ D

A

C p ----~~~-----~x

o

FIGURE 12.9 The four arrows io the plane (directed lioe segments) shown here have the same length and direction. They therefore represeot the same vector, aodwewriteAB = cD = OP = EF.

~~~~----~y

o

F

E

o

~--------~X

x

(a) two dimensions

(b) three dllnensions

FIGURE 12.8 The velocity vector of a particle moving aloog a path
DEFINmONS The vector represented by the directed line segmen.!."AB has initial point A and terminal point B and its length is denoted by lAB I. Two vectors are equal if they have the same length aod direction.

FIGURE 12.10 A vector PQ io standard position has its initial point ~the origin. The directed lioe segments PQ and v are parallel and have the same length.

The arrows we use when we draw vectors are understood to represent the same vector if they have the same length, are parallel, and point in the same direction (Fignre 12.9) regardless of the initial point. In textbooks, vectors are usually written in lowercase, boldface letters, for example u, v, and w. Sometimes we use uppercase boldface letters, such as F, to denote a force vector. In handwritten form, it is customary to draw small arrows above the letters, for example ii, ii, lV, andF. We need a way to represent vectors algebraically so that we can be more precise shout the direction of a vector. Let v = There is one directed line segment equal to whose initial point is the origin (Fignre 12.10). It is the representative ofv in standard position and is the vector we nonnally use to represent v. We can specify v by writing the

PQ.

PQ

666

Chapter 12: Vectors and the Geometry of Space

coordinates of its terminal point (v" V2, v,) when v is in standard position. Ifv is a vector in the plane its terminal point (V" V2) has two coordinates.

DEFINITION If v is a two-dimensional vector in the plane equal to the vector with initial point at the origin and terminal point (V" V2), then the component form of v is v = (V" V2).

Ifv is a three-dimensional vector equal to the vector with initial point at the origin and terminal point (V" V2, v,), then the componentform ofv is v = (V" V2, V3).

So a two-dimensional vector is an ordered pair v = (v" V2) of real numbers, and a three-dimensional vector is an ordered triple v = (V" V2, V3) of real numbers. The numV2, and V3 are the components ofv. bers If v = (V" V2, V3) is represented by the directed line segment PQ, where the initial point is P(x"y" Z,) and the terminal point is Q(X2,Y2, Z2), then Xl + VI = X2, Yl + V2 = Y2, and Zl + V3 = Z2 (see Figure 12.10). Thus, VI = X2 - X" V2 = Y2 - Yl, and V3 = Z2 - Zl are the components of PQ. In summary, given the points P(x" y" Zl) and Q(X2,Y2, Z2), the standard position vector v = (v" V2, v,) equal to PQ is

v"

v = (X2 - X"Y2 - y"Z2 - Zl).

If v is two-dimensional with P(X"YI) and Q(X2,Y2) as points in the plane, then v = (X2 - X" Y2 - Yl). There is no third component for planar vectors. With this understanding, we will develop the algebra of three-dimensional vectors and simply drop the third component when the vector is tw(}-dimensional (a planar vector). Two vectors are equal if and only if their standard position vectors are identical. Thus (u" U2, U3) and (v" V2, V3) are equal if and only iful = v" U2 = V2, and U3 = V3. The magnitude or length of the vector PQ is the length of any of its equivalent directed line segment representations. In particular, if v = (X2 - X" Y2 - y" Z2 - Zl) is the standard position vector for PQ, then the distance formula gives the magnitude or length of v, denoted by the symbollvl orllvll.

The magnitude or length of the vector v = Ivl = Yv,'

+ vl + vl

= Y(X2 -

PQ is the nonnegative number X,)2

+

(Y2 - Yll'

+

(Z2 - zll'

(see Figure 12.10).

The only vector with length 0 is the zero vector 0 = (0, 0) or 0 = (0, 0, 0). This vector is also the only vector with no specific direction.

EXAMPLE 1 Find the (a) component form and (b) length of the vector with initial point P( - 3, 4, 1) and terminal point Q( -5, 2, 2). Solution

(a) The standard position vector v representing PQ has components VI

=

X2 - Xl = -5 - (-3) = -2,

V2 = Y2 - Yl = 2 - 4 = -2,

12.2

Vectors

667

and V3 = Z2 -

Z1 =

2 - I = I.

~

The component form of PQ is v = (-2, -2, I). ~

y

(b) The length or magnitude of v = PQ is

t

I

Ivl

I

I I I

I

FIGURE 12.11 The force pulling the cart forward is represented by the vector F

=

Y(_2)2

+ (-2)2 + (1)2

=

V9 =

•

3.

EXAMPLE 2 A small cart is being pulled along a smooth horizontal floor with a 20-lb force F making a 45° angle to the floor (Figure 12.11). What is the effective force moving the cart forward? Solution

The effective force is the horizontal component ofF = (a, b), given by

a

= IFI cos 45° = (20)(,?) "" 14.14 lb.

whose horizontal component is the effective force (Example 2).

•

Notice that F is a two-dimensional vector.

Vector Algebra Operations Two principal operations involving vectors are vector addition and scalar multiplication. A scalar is simply a real number, and is called such when we want to draw attention to its differences from vectors. Scalars can be positive, negative, or zero and are used to "scale" a vector by multiplication.

DEFINmONS

Let u = (u"

U2, U3)

and v = (v"

V2, V3)

be vectors with k a

scalar. Addition: Scalar multiplication:

+ v = (U1 + v" U2 ku = (ku" ku2, ku3)

u

+ V2, U3 + V3)

We add vectors by addiog the correspondiog components of the vectors. We multiply a vector by a scalar by multiplying each component by the scalar. The definitions apply to planar vectors except there are ouly two components, (u" U2) and (Vb V2). The defioition of vector addition is illustrated geometrically for planar vectors io Figure 12.12a, where the initial poiot of one vector is placed at the termioal poiot of the other. Another ioterpretation is showo io Figure 12.12b (called the parallelogram law of y

y

__ ..... .- ..... ....---I -

I

/

I

I

I I

I

~~--~----~--------~x

(0)

FIGURE 12.12 vector addition.

o (b)

(a) Geomelric inlefpretation of the vector swn. (b) The parallelogram law of

668

Chapter 12: Vectors and the Geometry of Space

~

addition), where the smn, called the resultant vector, is the diagonal of the parallelogram.

//

/ /u~

In physics, forces add vectorially as do velocities, accelerations, and so on. So the force

acting on a particle subject to two gravitational forces, for example, is obtained hy adding the two force vectors. Figure 12.13 displays a geometric interpretation of the product ku of the scalar k and vector u. If k > 0, then ku has the same direction as u; if k < 0, then the direction of ku is opposite to that of u. Comparing the lengths of u and ku, we see that

lkul

FIGURE 12.13 Scalar multiples ofu.

= V(ku,)2 + (ku2)2 + (ku 3)2 = Vk2(u,2 + ui + u,2) =

WVU,2 + ui + u/

=

Ikllul.

The length of ku is the absolute value of the scalar k times the length of u. The vector ( -I)u = -u bas the same length as u but points in the opposite direction. The difference u - v of two vectors is defmed hy u-v=u+(-v). Ifu = (u" U2, U3) and v = (v" V2, V3), then u - v = (u, - v" U2 - V2, U3 - V3).

(a)

Note that (u - v) + v = u, so adding the vector (u - v) to v gives u (Figure 12.14a). Figure 12.14b shows the difference u - vas the smn u + (-v).

EXAMPLE 3

v

(a) 2u

+ 3v

Let u = (-1,3,1) and v = (4,7,0). Find the components of (h) u - v

(e)

I!ul.

Solution (a) 2u + 3v = 2(-1,3, 1) + 3(4,7,0) = (-2,6,2) + (12,21,0) = (10,27,2)

FIGURE 12.14 (a)Thevector u - v, when added to v, gives u. (b)u - v

= u + (-v).

(h) u - v= (-1,3,1) - (4,7,0) = (-1 - 4,3 -7,1 - 0) = (-5,-4,1) (e)

I!ul I(-!,~,!)I ~(-!y + (~y + (!y !V1l. =

=

•

Vector operations have many of the properties of ordinary arithmetic.

Properties orVeetor Operations Let u, v, w be vectors and a, b be scalars. 1.u+v=v+u 3. u + 0 = u 5. Ou = 0 7. a(bu) = (ab)u 9. (a + b)u = au + bu

2. 4. 6. 8.

(u + v) + w = u + (v + w) u + (-u) = 0 lu = u a(u + v) = au + av

These properties are readily verified using the definitions of vector addition and multiplication by a scalar. For instance, to establish Property 1, we have u

+v

= = = = =

(u," U2, U3) + (v," V2, V3) (u, + v" U2 + V2, U3 + V3) (v, + U" V2 + U2, V3 + U3) (v," V2, V3) + (u," U2, U3) v + u.

12.2 Vectors

669

When three or more space vectors lie io the same plane, we say they are coplanar vectors. For example, the vectors u, v, and u + v are always coplanar.

Unit Vectors A vector v of length 1 is called a unit vector. The standard unit vectors are i = (1,0,0),

j = (0,1,0),

and

k = (0,0,1).

Aoy vector v = (Vh V2, V3) can be written as a linear combination of the standard unit vectors as follows: v = (Vh V2, V3) = (VI> 0, 0) = vl(l, 0, 0) = Vii

+ v2j

+ (0, V2, 0) + (0,0, V3)

+ V2(0, 1,0) + V3(0,0, 1) + V3k.

We call the scalar (or number) VI the i-component of the vector v, V2 the j-component, and V3 the k-
x

FIGURE 12.15 The vector from PI to P2 + (Y2 - YI)j +

is P-;/>2 ~ (X2 - xl)i (Z2 - ZI)k.

(Figure 12.15). Wheneverv oF O,itslength Ivl is not zero and

That is, v/lvl is a unit vector io the direction of v, called tbe direction of the nonzero vectorv.

EXAMPLE 4 P2(3, 2, 0). Solution

Fiod a unit vector u io the direction of the vector from P,(l, 0, 1) to

We divide P-;P2 by its length:

P-;P2 IP-;P21

= (3 -

l)i

= Y(2j2

+

+

(2 - O)j

(2)2

+

+

(0 - I)k = 2i

(-1)2 = Y4

+4+

+ 2j 1=

~

- k

v'9 =

3

•

The unit vector u is the direction of P,P2.

EXAMPLE 5 If v = 3i - 4j is a velocity vector, express v as a product of its speed times a unit vector io the direction of motion. Solution

Speed is the magnitude (length) of v: Ivl = Y(3j2

HIsTORICAL BIOGRAPHY

Hermann Grassmann (180!l-1877)

+

(-4j2 =

V9+16 =

The unit vector vI Iv I has the same direction as v: v

3i-4j 3.4. 5 = 5' - 5 J ·

5.

670

Chapter 12: Vectors and the Geometry of Space

So

4i

v = 3i -

=

/ Length

5Gi - ti).

•

Direction of motion

(speed)

In summary, we can express any nonzero vector v in terms of its two importsnt features, length and direction, by writing v = Iv II; I .

IfvoF 0, then 1. 1;1 is a unit vector in the direction of v;

2. the equation v = Ivll;1 expresses v as its length times its direction.

EXAMPLE 6 A force of 6 newtons is applied in the direction of the vector v = 2i + 2i - k. Express the force F as a product of its magnitude and direction.

Solution

The force vector has magnitude 6 and direction I; I ' so

F = 6_v_ = 6 Ivl

21_._+-c2c:--i_-_k 2i + 2i - k = 6_ 2 2 Y2 +2 +(-1)2 3

-t

=6(t i +t i

•

k ).

Midpoint of a Line Segment Vectors are often useful in geometry. For example, the coordinates of the midpoint of a line segment are found by aversging.

The midpoint M of the line segment joining points PI(X"y"ZI) and P2(X2, Y2, Z2) is the point Xl (

+ X2 2

Yl

+ Y2 Zl + Z2)

'

2

'

2

.

To see why, observe (Figure 12.16) that

o FIGURE 12.16 The coordinates of tire midpoint are the averages of the coordinates ofPI andP2.

Xl

EXAMPLE 7

+ X2 • 2

1

+

Yl

+ Y2 • 2

I

Zl + Z2

+ -2- k .

The midpoint of the segment joining P I (3, -2,0) andP2(7, 4, 4) is

(

4 4) = (5"1 2).

0 + 3 + 7 -2 + 2' 2 '2

•

12.2 Vectors

671

Applications An important application of vectors occurs in navigation.

EXAMPLE 8 Ajet airliner, flying due east at 500 mph in still air, encounters a 70-mph tailwind blowing in the clirection 60° north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?

N

500

u

NOfroSCAIE

FIGURE 12.17 Vectors representing tire velocities of the airplane u and tailwind v in Example 8.

Solution lfu = the velocity of the airplane alone and v = the velocity of the tailwind, then lui = 500 and Ivl = 70 (Figure 12.17). The velocity of the airplane with respect to the ground is given by the magnitode and clirection of the resultant vector u + v. Ifwe let the positive x-axis represent east and the positive y-axis represent north, then the component forms of u and v are u = (500,0)

v = (70 cos 60°, 70 sin 60°) = (35,35\1'3).

and

Therefore, u lu

+v

=

(535,35\1'3)

+ vi

=

\15352

+

3~~

'" 6.50

= 535i

+ 35\1'3 j

(35V'3)2 '" 538.4

and (J =

tan-I

Figure

12.17

The new ground speed of the airplane is about 538.4 mph, and its new clirection is about _ 6.5° north of east. Another important application occurs in physics and engineering when several forces are acting on a single object.

40'

55'

EXAMPLE 9 A 75-N weight is suspended by two wires, as shown in Figure 12.18a. Find the forcea F 1 and F 2 acting in both wirea. Solution

The force vectors F 1 and F 2 have magnitudes IF 11 and IF 21 and components that are measured in Newtons. The resultant force is the sum Fl + F2 and must be equal in magnitude and acting in the opposite (or upward) clirection tu the weight vectur w (see Figure 12.18b). It follows from the figure that

(a)

Fl = (-IFI Icos 55°, IFdsin55°) Since Fl

/ /

+ F2

and

F2 = (IF2Icos40°, IF2Isin400).

= (0,75), the resultant vectur leads to the system of equations

/ /

-IFl Icos 55° IFdsin55°

+ IF21cos40° + IF21sin400

= 0 = 75.

Solving for IF21 in the 1mt equation and substituting the result into the second equation, we get

w

~

(0. -75)

(b)

FIGURE 12.18 in Example 9.

IF21 =

IFdcos55° cos 40°

. ° IF llsm55

and

+

IFdcos55°. ° _ cos 40° sm40 - 75.

It follows that

The suspeuded weight

IFll = sin 55°

+

75 cos 55° tan 40° '" 57.67N,

672

Chapter 12: Vectors and the Geometry of Space and

IF2 I =

75 cos 55° sin 55° cos 4()0 + cos 55° sin 40° 75 cos 55° sin(550 + 400) '" 43.18 N.

•

The force vectors are then F, = (-33.08,47.24) and F2 = (33.08,27.76).

Exercises 12.2 Vectors In the Plane

Geometric Representations

In Exercises 1--1!, letu = (3, -2) and v = (-2,5). Find the (a) compooent form and (b) magnitude (length) of the vector.

In Exercises 23 and 24, copy vectors U, v, and ... head to tail as needed to sketch the indicated vector.

2. -2v

1. 3u

3. u

+v

4. u - v 6. -2u

5.2u-3v 7.

3

4

5u + 5v

8.

+ 5v

5

12

-TI u + TIv

In Exercises !l-16, f"md the compooent form of the vector. L

9. The vector PQ, where P = (1,3) and Q = (2, -I)

oP

10. The vector where 0 is the origin and P is the midpoint of segmentRS, whereR = (2, -I) andS = (-4,3)

U

+V

b.u+v+w

c. u - v

d. u - w

24.

11. The vector from the point A = (2, 3) to the origin

cD,

where A=(I,-I),B=(2,0), 12. The sum of AB and C = (-1,3),andD = (-2,2) 13. The unit vector that makes an angle 8 = 27T/3 with the positive

u

x-axis 14. The unit vector that makes an angle 8 = -37T/4 with the positive

x-axis 15. The unit vector obtained by rotating the vector (0, I) 120" coun-

terclockwise about the origin 16. The unit vector obtained by rotating the vector (I, 0) 135" coonterc10ckwise about the origin L

Vectors In Space In Exercises 17-22, express each vector in the form v = l'2j + v3k.

18. P-;P2 if P, is the point (I, 2, 0) and P2 is the point (- 3, 0, 5) 20.

b.u-v+w

V

c. 2u - v Vii

17. P-;P2 if P, is the point (5, 7, -I) and P2 is the point (2, 9, -2) 19.

U -

AB ifA is the point (-7, -8, I) andB is the point (-10,8, I) AB ifA is the point (1,0,3) andB is the point( -1,4,5)

d.u+v+w

+ Length and DIrection In Exercises 25-30, express each vector as a prodoct of its length and direction. 25.2i+j-2k

26.9i-2j+6k

27.5k

28.~i+!k

21. 5u - vifu = (I, I, -I) and v = (2,0,3) 22. -2u+ 3vifu = (-1,0,2) and v = (1,1,1)

29. _I_i - _I_j __I_k

v'6

v'6

v'6

i

30.

j

k

v'3 + v'3 + v'3

12.2 31. Find the vectors whose lengths and directions are given. Try to do the calculations without writiog. Length

30-

Vectors

673

45-

Direction

a. 2 b.

c.

V3

-k

1

1j + i

"2

5

5

k

46. Consider a 50-N weight suspended by two wires as shown in the accompanying figure. If the magoi1ude of vector F I is 35 N, fmd angle a and the magni1ude ofvectorF,.

.6. i _.2. j +1 k

d. 7

777

32. Find the vectors whose lengths and directions are given. Try to do the calculations without writiog. Length

Direction

a. 7

-j

b.

v2 13

_li - i

k

5

5

3.

4.

12 k

c'IT

TI 1 - TIl - TI

d. a> 0

_I_i

v2

+ _I_j

V3

__I_k

V6

47. Consider a woN weight suspended by two wires as shown in the accompanying figure. If the magoi1ude of vector F, is 100 N, fmd w aod the magoi1ude of vector Fl.

33. Find a vector ofmagoi1ude 7 in the directioo ofv = 12i - 5k. 34. Find a vector of magoi1ude 3 in the directioo opposite to the directionofv = (1/2)i - (1/2)j - (1/2)k.

40'

35'

Direction and Midplrints In Exercises 35--38, find

a. the direction of p-;P, aod b. the midpoint ofline segment PI P,.

35. PI( -I, I, 5)

P,(2, 5, 0)

36. PI(I, 4, 5)

P,(4, -2,7)

37. P I (3, 4, 5)

P,(2, 3, 4)

38. PI(O, 0, 0)

P,(2, -2, -2)

39. If AB 40. IfAB

48. Consider a 25-N weight suspended by two wires as shown in the accompaoying figure. If the magoi1udes of vectors F I and F, are both 75 N, then angles a and fJ are equal. Find a. a

fl

= i + 4j - 2k aodB is the point (5, 1,3), fmdA. = -7i + 3j + 8kaodAisthepoint(-2,-3,6),fmdB.

Theory and Applications

41. Linear combination Let u = 2i + j. v = i + j, and w = i - j. Find scalars a and b such that u = av + bw. 42. Linear combination Let u = i - 2j, v = 2i + 3j, aod w = i + j. Write U = UI + U2. where UI is parallel to v and U2 is parallel to w. (See Exercise 41.) 43. Velocity An airplane is flying in the directioo 25' west of north at 800 kmfh. Find the compooent form of the velocity of the air-

plane, assuming that the positive x-axis represents due east and the positive y-axis represents due north. 44. (Continuation of Example 8.) Wbat speed and directioo should the jetliner in Example 8 have in order for the resultant vector to be 500 mph due east? 45. Consider a lOO-N weight suspended by two wires as shown in the accompanying figure. Find the magoi1udes and compooents of the force vectors FI aodF,.

49. Location A bird flies from its nest 5 km in the directioo 60' north of east, where it stops to rest 00 a tree. It then flies 10 km in the directioo due southeast aod lands atop a telephone pole. Place an xy-coordinate system so that the origin is the bird's nest, the

x-axis points east, and the y-axis points north. a. At what point is the tree located? b. At what point is the telephone pole?

50. Use similar triangles to fmd the coordinates of the point Q that divides the segment from PI(XIoYI, ZI) to P,(x"y" z,) into two leogths whose ratio is p/q = r. 51. Medians of a triangle Suppose that A, B, and C are the corner points of the thin 1riangn1ar plate of constant density shown here. a. Find the vector from Cto the midPointM ofsideAB. b. Find the vector from C to the point that lies lwD-thirds of the way from C to M 00 the median CM.

674

Chapter 12: Vectors and the Geometry of Space

c. Find the coordinates of the point in which the medians of I!.AJJC intersect According to Exercise 17, Section 6.6, this

53. Let ABCD be a general, not necessarily planar, quadrilateral in space. Show that the two segments joining the midpoints of oppo-

point is the p1ate~ center of mass.

site sides of ABCD bisect each other. (Hint: Show that the segments have the same midpoint) 54. Vectors are drawn from the center of a regular n-sided polygon in the plane to the vertices of the polygon. Show that the sum of the vectors is zero. (Hint: What happens to the sum if you rotate the polygon about its center?)

C(I, I, 3)

· c.m.

55. Suppose that A, B, and C are vertices of a triangle and that a, b, and c are, ~ectively, the midpoints of the opposite sides. Show thatAa + Bb + Co ~ o.

~y

56. Unit veeton io the plane Show that a unit vector in the plane can be expressed as u ~ (cosO)i + (sinO)j,obtainedbyrotating

B(l, 3, 0) M

i through an angle 0 in the couoterclockwise direction. Esplain why this form gives every unit vector in the plane.

A(4, 2, 0)

52. Find the vector from the origin to the point of intersection of the medians of the triangle whose vertices are

A(I,-1,2),

12.3

B(2,1,3),

and

C(-1,2,-1).

The Dot Product If a force F is applied to a particle moving along a path, we often need to know the magnitode of the force in the direction of motion. Ifv is parallel to the tangent line to the path at

v

Length ~

IF I cos Ii

FIGURE 12.19 The magnitude of the force F in the direction of vector v is the length IFI cos 0 oftheprojectioo ofF ontov.

the point where F is applied, then we want the magnitode ofF in the direction ofv. Figure 12.19 shows that the scalar quantity we seek is the length IFI cosO, where 0 is the angle between the two vectors F and v. In this section we show how to calcoIate easily the angle between two vectors directly from their components. A key part of the calcoIation is an expression called the dot product. Dot products are also called inner or scalar products because the product results in a scalar, not a vector. After investigating the dot product, we apply it to finding the projection of one vector onto another (as displayed in Figure 12.19) and to finding the work done by a constant force acting through a displacement.

Angle Between Vectors When two nonzero vectors u and v are placed so their initial points coincide, they form an angle 0 of measure 0 ,,; 0 ,,; 1f (Figure 12.20). If the vectors do not lie along the same line, the angle 0 is measured in the plane containing both of them. If they do lie along the same line, the angle between them is 0 if they point in the same direction and 1f if they point in opposite directions. The angle 0 is the angle between u and v. Theorem I gives a fonnula to detennine this angle.

v

THEOREM i-Angle Between Two Vectors vectors u =

(Ulo U2, U3)

The angle 0 between two nonzero and v = (Vlo V2, V3) is given by

_

o-

-I (utVt +

cos

U2 V2

lullvl

+ U3V3)

.

FIGURE 12.20 The angle between u and v.

Before proving Theorem 1, we focus attention on the expression UI VI + U2 V2 + U3 V3 in the calcoIation for O. This expression is the sum of the products of the corresponding components for the vectors u and v.

12.3

The Dotproduct

675

DEFINmON The dot product u' v ("u dot v") of vectors u = (Ul, U2, U3) and v = (Vl, V2, V3) is

EXAMPLE 1 (8) (1,-2,-1)'(-6,2,-3)

+ (-2)(2) + (-1)(-3) -6 - 4 + 3 = -7

(1)(-6)

= =

(b)

(!i +

+

3j

k)

'(4i -

j

+ 2k)

=

(!)t4) +

+

(3)(-1)

(1)(2) =

I

•

The dot prodoct of a pair of two-dimensional vectors is dermed in a similar fashion:

(Ul, U2)' (Vl, V2) = U1 V1

+ U2V2.

We will see throughoot the remainder of the book that the dot product is a key tool for many important geometric and physical calculations in space (and the plane), not just for rmding the angle between two vectors. Proof of Theorem 1 Applying the law of cosines (Equation (8), Section 1.3) to the triangle in Figure 12.21, we find that

Iwl2 = lul 2 + Ivl2 - 21ullvl cosO 21ullvl cosO = lul 2 + Ivl 2 - Iw1 2 .

Law of cosines

Becausew = u - v,thecomponentformofwis(U1 - Vl,U2 - V2,U3 - V3).SO lul 2 = (VU12 Ivl 2 = (Vv,'

FIGURE 12.21 The parallelognun law of addition of vectors gives w = u - v.

=

+ ul + ul)' = u,' + ul + uJ' + vl + vl)' = v,' + vl + vi (V(U1 - V1)2 + (U2 - V2)' + (U3 - V3)2)' (U1 - v,)2 + (U2 - vd + (U3 - V3)2

=

u? -

Iwl 2 =

2UtVt

+

Vt

+ ui -

2

2U2V2

+ vi + ul -

2U3V3

+ vl

and Therefore, 21ullvl cosO = lul 2 + Ivl2 - Iwl2 = 2(U1V1 lullvl cos 0 = U1 V1 UtVt

cosO = Since 0

:5 ()

<

1T,

+ U2V2 + U3V3)

+ U2V2 + U3V3 +

U2'V2

+

U3V3

lullvl

we have V U2 V2 + U3 V3) 0 -- cos-1 (U1 1 +lullvl .

In the notation of the dot prodoct, the angle between two vectors u and v is

o=

1 cos- C:I'I:I)'

•

676

Chapter 12: Vectors and the Geometry of Space

EXAMPLE 2 Solution

Find the angle between u = i - 2j - 2k and v = 6i

+ 3j + 2k.

We use the fonnula above:

+

u'v = (1)(6)

+ (-2)(2) =

(-2)(3)

6 - 6 - 4 = -4

lui = VO)' + (-2), + (-2), = v9 = 3 Ivl = V(6)' () =

+

(3)'

+

(2)2 =

1

cos- C:I'I:I) = cos-

1

v49 = 7

(3)~») '" 1.76radians.

-

The angle fonnula applies to two·dimensional vectors as well. y B(3,5)

EXAMPLE 3 Find the angle () in the triangle ABC detennined by the vertices = (0, O),B = (3,5), and C = (5,2) (Figure 12.22).

A

~

~

Solution The angle () is the angle between the vectors CA and CB. The component fonns of these two vectors are

eA

=

(-5, -2)

and

cE

=

(-2,3).

First we calculate the dot product and magnitudes of these two vectors. FIGURE 12.22 The 1riangle in

eA'cE =

Example 3.

leAl

=

(-5)(-2) + (-2)(3)

=

4

V(-5)' + (-2), = v29

IcEl = V(-2),

+ (3)'

=

v'13

Then applying the angle formula, we have

'" 78.10

or

1.36 radians.

-

Perpendicular (Orthogonal) Vectors Two nonzero vectors u and v are perpendicular or orthogonal if the angle between them is For such vectors, we have u' v = 0 because cos (1f/2) = O. The converse is also true. If u and v are nonzero vectors with u' v = lullvl cos () = 0, then cos () = 0 and () = cos- 1 0 = 1f/2.

1f/2.

DEFINITION

Vectors u and v are orthogonal (or perpendicular) if and only

ifu'v = O.

EXAMPLE 4

To determine if two vectors are orthogonal, calculate their dot product.

+ (-2)(6) = O. (b) u = 3i - 2j + k and v = 2j + 4k are orthogonal because u' v = (3)(0) + (-2)(2) + (1)(4) = O. u = (3, -2) and v = (4,6) are orthogonal because u' v = (3)(4)

12.3

The Dotproduct

677

(c) 0 is orthogonal to every vector u since

o· u

= (0,0, O) • (UI. U2, U3) = (O)(Ul) =

+

(0)(U2)

+ (0)(U3)

o.

•

Dot Product Properties and Vector Projections The dot product obeys many of the laws that hold for ordinary products of real numbers (scalars).

Properties of the Dot Product

Ifu, v, and w are any vectors and c is a scalar, then 1. U'V = V'U 2. (eu)'v = u'(ev) = e(u'v) 3. u'(v

+ w)

= U'V

+ U'w

4.

U'U

= lul 2

5.0·u=0.

Carl Friedrich Gauss

Proofs of Properties 1 and 3 The properties are easy to prove using the detmition. For instance, here are the proofs of Properties 1 and 3.

(1777-1855)

1.

U'V

3.

U'

HIsTORICAL BIOGRAPHY

(v

=

UlVl

+ w)

+ U2V2 + U3V3

=

= (UIo U2, U3)' (VI = Ul(Vl

VlUl

+

+ V2U2 + V3U3

WI. V2

=

V'U

+ W2, V3 + W3)

+ WI) + U2(V2 + W2) + U3(V3 + W3)

Q

/1.,

P

•

=u·v+u·w

S

R

We now return to the problem of projecting one vector onto another, posed in the opening to this section. The ~ctor projection of u = PQ onto a nonzero vector v = (Figure 12.23) is the vector PR determined by dropping a perpendicular from Q to the line PS. The notation for this vector is

PS

• R

FIGURE 12.23 u onto v.

P

S

proj. u

The vector projection of

/

("the vector projection ofu onto v").

If u represents a force, then projv u represents the effective force in the direction of v (Figure 12.24). If the angle Ii between u and v is acute, projv u has length Iu I cos Ii and direction Iv I (Figure 12.25). If Ii is obtuse, cos Ii < 0 and proj. u has length -I u I cos Ii and direction Ivl. In both cases,

vi

-vi

proj.u = (lui cos Ii) v

1:1

proj,.u

FIGURE 12.24 Ifwe pull on the box with force u, the effective force moving the box forward in the direction v is the projection ofo ontov.

u.v) v ( Ivl Ivl u.v) ( Ivl2 V.

lui cos 8 ~

lullvl cos 8 Ivl

u',

Ivl

678

Chapter 12: Vectors and the Geometry of Space

v

v

Length ~ lui cos 9

Length ~ -lui cos 9

(a)

(b)

FIGURE 12025 Tbe\ength ofprojyu is (a) lui cos II if cos II '" (b)-lui cos II if cos II < O.

oand

The number Iu I cos Ii is called the scalar component of u in the direction of v (or of u onto v). To summarize,

The vector projection of u onto v is the vector

.

prOJyU =

(uov) Ivl2 v.

(1)

The scalar component of u in the direction of v is the scalar lui cos Ii =

u·v lVT =

v

(2)

UOT;!"

Note that both the vector projection of u onto v and the scalar component of u onto v depend only on the direction of the vector v and not its length (because we dot u with v / Iv I , which is the direction of v).

EXAMPLE 5 Find the vector projection of u = 6i and the scalar component ofu in the direction ofv. Solution

+ 3j +

2k onto v = i - 2j - 2k

We find projy u from Equation (1):

.

uov

6-6-4. . proJy u = v 0v v = 1 + 4 + 4 (. - 2J - 2k) =

-1(i - 2j - 2k) = -1i

+ ~j + ~k

9999'

We f"md the scalar component of u in the direction of v from Equation (2): lui cos Ii = UOI;I = (6i =

2 - 2 -

+

3j

+ 2k)0

(t

i - t j - tk)

4 34 = -3'

•

Equations (1) and (2) also apply to two-dimensional vectors. We demonstrate this in the next example.

EXAMPLE 6 Find the vector projection of a force F = 5i the scalar component ofF in the direction ofv.

+ 2j

onto v = i - 3j and

12.3 The Dotproduct

Solution

679

The vector projection is proj. F =

(~;I~)v

=

5 - 6 (i - 3j) 1+9

=

-TO'

1 .

+

=

-...!.. (i 10

- 3j)

3 . lO J .

The scalar component ofF in the direction ofv is IFI cosO

F·v

5 - 6

=-1-1 =.~ v vi + 9

1

•

---

v'W.

A routine calculation (see Exercise 29) verifies that the vector u - proj. u is orthogonal to the projection vector proj. u (which has the same direction as v). So the equation . u = pro).u

+

. (u - pro).u) = (u.v) Ivl2 V

+

( u - (u.v)) Ivl2 V

Parallel to v

Orthogonal to v

expresses u as a sum of orthogonal vectors.

Work

1----1 I

: I I I

I

p

I

D

In Chapter 6, we calculated the work done by a constant force of magnitude F in moving an object through a distance d as W = Fd. That formula holds ouly if the force is directed along the line of motion. If a force F moving an object through a displacement D = has some other direction, the work is performed by the component of F in the direction of D. If 0 is the angle between F and D (Figure 12.26), then

PQ

Q

IFI cos 6

component OfF)(1 gth f ) Work = (scalar en 0 D in the direction of D

FIGURE 12.26 The wotk done by a constant force F during a displacement D is ( IF I cos 8) ID I, which is the dot product

(lFI cosO)IDI F·D.

F·D.

DEFINmON

The work done by a constant force F acting through a displace-

mentD = PQis

W= F·D.

EXAMPLE 7 IflFI = 40N(newtons), IDI F in acting from P to Q is

= 3m,andO = 60°, the work done by

Work = F·D

De:fmition

= IFIIDI cosO = (40)(3) cos 60° = (120)(1/2) = 60 J (joules).

Given values

•

We encounter more challenging work problems in Chapter 16 when we learn to f"md the work done by a variable force along a path in space.

680

Chapter 12: Vectors and the Geometry of Space

Exercises 12.3 a. Showthat

Dot Product and Projections In Exercises 1-8, fmd L

V'

cos a =

u, lvi, lui

a Fr'

cos,B =

and cos2 a

c. the scalar component ofn in the direction ofv

the direction cosines of v.

2. v = (3/5)i + (4/5)k,

u = 5i + 12j

3. v = lOi + 11j - 2k,

u = 3j + 4k

direction cosines ofv.

16. Water main construction A water main is to be constructed with a 20% grade in the nurth direction and a 10% grade in the east direction. Determine the angle 8 required in the water main for the turn from nurth to east.

4. v = 2i + 10j - 11k, u = 2i + 2j + k 5. v = 5j - 3k, 6.v=-i+j,

c

Fr'

b. Unit vectors are built from direction cosine. Show that if v = ai + bj + ck is a unit vector, then a, b, and c are the

d. the vector projy u. u = -2i + 4j - Vsk

cos 1 =

+ cos2 fj + cos2 'Y = 1. These cosines are called

b. the cosine of the angle between v and u

1. v = 2i - 4j + Vsk,

b

Fr'

u = i + j + k u=v2i+V3j+2k

+ j, u

7. v

=

5i

& v

=

(:n, ~),

2i

=

+ u

Wj

=

(:n,- ~)

Angle Between Vectors

D Find the angles between the vectors in Exercises !l-12 to the nearest huodredth of a radian. 9. u=2i+j,

Theory and Examples 17. Sums IUld dilJerence. In the accompanying figore, it looks as ifvl + V2 and VI - V2 are orthogonal. Is this mere coincidence, or are there circumstances uoder which we may expect the som of two vectors to be urthogona1 to their difference? Give reasons for

v=i+2j-k

10. u = 2i - 2j + k, 11. u = V3i - 7j,

v = 3i + 4k v = V3i + j - 2k

12. u=i+v2j-v2k,

your answer.

v=-i+j+k

13. TrilUlgle Find the measures of the angles of the mangle whose vertices are A = (-1,0), B = (2, I), and C = (I, -2). 14. Rectangle Find the measures of the angles between the diagonals of the rectangle whose vertices are A = (I, O),B = (0,3), C = (3,4), andD = (4, I). 15. Direction angles and direction cosines The direction angles a, {3, and "y of a vector v = ai + bj + ck are dermed as follows:

a is the angle between v and the positive x-axis (0 :s; a :s; '11')

,B is the angle between v and the positive y-axis (0 '" ,B '" 'IT) 1 is the angle between v and the positive z-axis (0 '" 1 '" 'IT). 18. Orthogons1ity on a circle Suppose that AB is the diameter of a circle with center 0 and that C is a p-oint on one of the two arcs joiningA andB. Show that and CB are urthogona1.

cA

A ~---:U:-~ O'-----;D:-+l B

y

19. Diagonals of a rhombus Show that the diagonals of a rhombus (parallelogram with sides of equa1lengtil) are perpendicular.

12.3

20. Perpendicular diagonals Show that squares are the only rec· tangles with perpendicular diagonals. 21. When parallelograms ore rectangles Prove that a parallelo· gram is a rectangle if and only if its diagonals are equal in length. (This fact is often exploited by carpenters.) 22. Diagonal ofporallelogram Show that the indicated diagonal of the parallelogram determined by vectors u aod v bisects the aogle betweenuandviflul ~ Ivl.

The Dot Product

681

Equations for Unes in the Plane 31. Line perpendicular to a vector Show that v ~ ai + bj is perpendicular to the line ax + by ~ c by establishing that the slope

of the vector v is the negative reciprocal of the slope of the given line. 32. Line porallel to a vector Show that the vector v ~ ai + bj is parallel to the line bx - ay ~ c by establishing that the slope of the line segment representing v is the sarne as the slope of the given line. In Exercises 33-36, use the result of Exercise 31 to fmd an equation for the line 1hrough P perpendicular to v. Then sketch the line. Include v in your sketch as a vector starting at the origin.

23. Projectile motion A gun with muzzle velocity of 1200 ft/sec is frred at an angle of 8' above the horizontal. Find the horizontal and vertical components of the velocity. 24. tnctined plane Suppose that a box is being towed up ao inclined plane as shown in the figure. Find the force w needed to ma1re the component of the force parallel to the inclined plane equal to 2.S lb.

33.P(2,1),

v~i+2j

35. P( -2, -7),

~

v

-2i + j

34.P(-1,2),

v~-2i-j

36. P(ll, 10),

v

~

2i - 3j

In Exercises 37-40, use the result of Exercise 32 to fmd ao equation for the line through P parallel to v. Then sketch the line. Include v in your sketch as a vector starting at the origin.

37. P(-2, I), 39. P(I,2),

v v

~

~

38. P(O, -2),

i - j

40. P(I, 3),

-i - 2j

v v

~

~

2i + 3j

3i - 2j

Work 41. Work along a tine Find the work done by a force F ~ Si (magnitnde S N) in moving ao object along the line from the otigin to the point (I, I) (distance in meters). 25. a. Cauchy-Schwartz inequality Since u' v ~ Iu II v I cos 9, show that the inequality Iu • v I '" Iu II v I holds for any vectors u and v. b. Under what circumstances, if any, does Iu • v I equal Iu II v I ?

Give reasons for your answer. 26. Copy the axes and vector shown here. Then shade in the points (x,y) for which (xi + yj) • v '" O. JustifY your answer.

42. Locomotive The Union Pacific's Big Boy locomotive could pull 6000-ton 1l'ains with a tractive effort (Pull) of 602,148 N (13S,375 Ib). At this level of effort, about how much work did Big Bay do on the (approximately straight) 60S-kin journey from

San Francisco to Los Angeles? 43. lnelined plane How much work does it take to slide a crate 20 m along a loading dock by pulling on it with a 200 N force at an aogle of 30' from the horizontal? 44. Sailboat The wind passing over a boat's sail exerted a 1000-lb magnitude force F as shown here. How much work did the wind perform in moving the boat forward I mi? Answer in foot-pounds.

y

60'

27. Orthogonal unit vectors If Ul and tors and v = aUI + bU2,f'mdv·ut.

U2

magnitude f"",e

are orthogonal unit vec-

28. CanceDation in dot products In real-number multiplication, if U'VI = U\l2 and u =F 0, we can cancel the u and conclude that ~ V2. Does the same rule hold for the dot product? That is, if U'VI = U'V2andu =F O,canyouconcludethatvl = v2?Give

v,

Angles Between Lines In the Plane The acute angIe between intenecting lines that do not cross at right aogles is the same as the angle determined by vectors normal to the lines or by the vectors para1lel to the lines.

reasons for your answer. 29. Using the definitioo of the projection ofu onto v, show by direct calculation that (u - proj. u) • projy u ~ O. 30. A force F ~ 2i + j - 3k is applied to a spacecraft with velocity vector v ~ 3i - j. Express F as a sum of a vectur para1lel to v and a vector orthogonal to v.

F

L2 L2

682

Chapter 12: Vectors and the Geometry of Space

Use this fact and the results of Exercise 31 or 32 to rmd the acute angles betweeo the lines in Exercises 45-50.

45.

3x+y~5,

46. y~

2x-y~4

V3x-I,

12.4

y~ -V3x+2

47. 48. 49.

V3x - y ~ -2, x - V3y ~ 1 x + V3y ~ 1, (1 - V3)x + (1 3x-4y~3,

50. 12x + 5y

~

1,

+ V3)y ~ 8

x-y~7

2x - 2y

~

3

~I_Th_e_ C _rn_s_s _Pr_o_du_c_t _________________________________ In studying lines in the plane, when we needed to describe how a line was tilting, we used the notions of slope and angle of inclination. In space, we want a way to describe how a plane is tilting. We accomplish this by multiplying two vectors in the plane together to get a third vector perpendicular to the plane. The direction of this third vector tells us the ''inclination'' of the plane. The product we use to multiply the vectors together is the vector or cross product, the second of the two vector multiplication methods. We study the cross product in this section.

The Cross Product of Two Vectors in Space We start with two nonzero vectors u and v in space. Ifu and v are not parallel, they determine a plane. We select a unit vector n perpendicular to the plane by the right-hand rule. TIris means that we choose n to be the unit (normal) vector that points the way your right thumb points when your f'mgers curl1hrough the angle 0 from u to v (Figure 12.27). Then the cross product u X v ("u cross v") is the vector defined as follows.

DEFINITION u X v ~ (lullvl sinO) n FIGURE 12.27

The construction of

u X v. Unlike the dot product, the cross product is a vector. For this reason it's also called the vector product of u and v, and applies only to vectors in space. The vector u X v is orthogonal to both u and v because it is a scalar multiple of n. There is a straightforward way to calculate the cross product of two vectors from their components. The method does not require that we know the angle between them (as suggested by the definition), but we postpone that calculation momentarily so we can focus first on the properties of the cross product. Since the sines of 0 and 1f are both zero, it makes sense to define the cross product of two parallel nonzero vectors to be O. If one or both of u and v are zero, we also define u X v to be zero. TIris way, the cross product of two vectors u and v is zero ifand only ifu and v are parallel or one or both of them are zero. Parallel Vectors Nonzero vectors u and v are parallel if and only if u X v

~

O.

The cross product obeys the following laws. Properties of the Cross Product If u, v, and w are any vectors and r, s are scalars, then

+ w) + w) X u

+u Xw +w Xu

1. (ru) X (sv) ~ (rs)(u X v)

2. u X (v

~

3. v X u ~ -(u X v)

4. (v

~ v X u

S.OXu~O

6. u X (v X w) ~ (u'w)v - (u'v)w

u X v

12.4

-D~ ~

W

Xu

FIGURE 12.28

The construction of

v XU.

The (ross Product

683

To visualize Property 3, for example, notice that when the fingers of your right hand curl through the angle 0 from v to u, your thumb points the opposite way; the unit vector we choose in forming v X u is the negative of the one we choose in forming u X v (Figure 12.28). Property I can be verified by applying the definition of cross product to both sides of the equation and comparing the results. Property 2 is proved in Appenrux 8. Property 4 follows by multiplying both sides of the equation in Property 2 by -I and reversing the order of the products using Property 3. Property 5 is a detmition. As a rule, cross product multiplication is not associative so (u X v) X w does not generally equal u X (v X w). (See Additioual Exercise 17.) When we apply the definition to calculate the pairwise cross products of i, j, and k, we find (Figure 12.29)

i Xj =

z k~ixj~-(jxi)

-0

X i) = k

j X k = -(k X j) = i

/

k X i = -(i X k) = j

Diagram for recalling these products

and

i Xi = j X j = k X k =

o.

Iu X V I Is the Area of a Parallelogram Because n is a unit vector, the magnitode of u X v is FIGURE 12.29 The pairwise cross products ofi, j, and k

lu X

Area

= base' height ~ ~

lui· Ivllsin 01 luxvl

\

u

FIGURE 12.30 The parallelogram detennined by u and v.

vi

lullvl sinO.

This is the area of the parallelogram determined by u and v (Figure 12.30), Iu I being the base of the parallelogram and Iv II sin () I the height.

Determi nant Form ula for u X v Our next objective is to calculate u X v from the components of u and v relative to a Cartesian coordinate system. Suppose that u = uji

+

u2j

+

and

U3k

v = vji

+

v2j

+

V3k.

Then the distributive laws and the rules for multiplying i, j, and k tell us that u X v = (uji

+

u2j

= Uj vji X i

+

U3k) X (vji

+ Uj V2i

+ u2vd X i + + u3vjk X i +

X j

v2j

+ Uj V3i

u2v2j X

+

V3k)

X k

+ U2V3 j X k j + U3V3k X k

j

U3V2k X

= (U2V3 - U3V2)i -

+

U3Vj)j

(UjV3 -

+

(UjV2 -

u2 vj)k.

The component terms in the last line are hard to remember, but they are the same as the terms in the expansion of the symbolic determinant

j

k

684

Chapter 12: Vectors and the Geometry of Space

I Determinant.

So we restate the calculation in this easy-to-remember form.

2 X 2 and 3 X 3 determinants are evaluated as follows:

I: !I

= ad - be

Calculating the Cross Product as a Determinant Ifu = uli + u2j + U3k and v = vii + v2j + V3k, then

EXAMPLE

I_~ ~

I

=

u X v = (2)(3) - (1)( -4)

i

j

k

Ul

U2

U3

VI

V2

V3

=6+4=10

EXAMPLE 1

Find u X v and v X u ifu = 2i + j + k and v = -4i + 3j + k.

Solution

_a,lbl b'l + a,lbl b'l Ct

Ct

C3

uXv=

C2

EXAMPLE

-5

3

I

2 -4

I 3

I I

=(-5)I~

=

-5(1 - 3) - 3(2

~I

+ 4)

PQ = PR =

R(-I, 1,2)

, O. // ;(.-

":/ :..

,,,"

FIGURE 12.31 The vector PQ X PR is perpendicular to the plane of triangle PQR (Example 2). The area of 1riang1e PQR is

•

2i + 6j - 10k

Solution The vector PQ X PR is perpendicular to the plane because it is perpendicular to both vectors. In terms of components, (2 - l)i + (1 + I)j + (-I - O)k = i + 2j - k (-I - l)i + (1 + I)j + (2 - O)k = -2i + 2j + 2k

.............. PQ X PR =

z

=

EXAMPLE 2 Find a vector perpendicular to the plane of P(i, -I, 0), Q(2, I, - I), and R( -1,1,2) (Figure 12.31).

(For more information, see the Web site at www.aw.comlthomas.)

ha\fof IPQ X PRI (Example 3).

1 = I 3

v X u = -(u X v)

+ 1(6 + 4) = 10 - 18 + 10 = 2

------

II

k

= -2i - 6j + 10k

:1

- (3)1_~ :1 + (I)I-~

2 -4

.

1J 3

1 -2

j 2 2

-~2

12

=

2

-11'2' _I -21 •

=6i+6k.

EXAMPLE 3 Find the area of the triangle with vertices P(I, -1,0), Q(2, 1, -I), and R( -I, 1,2) (Figure 12.31). Solution

The area of the parallelogram determined by P, Q, andR is

IPQ X PRI

= 16i + 6kl

Values from Example 2

= V(6)' + (6)'

v'2-36 =

=

6v'2.

•

The triangle's area is half of this, or 3v'2.

EXAMPLE 4 Find a unit vector perpendicular to the plane of P(I, -1,0), Q(2, I, -1), andR(-I,I,2). Solution Since PQ X PR is perpendicular to the plane, its direction n is a unit vector perpendicular to the plane. Taking values from Examples 2 and 3, we have ~

n=

~

PQ X PR

IPQ X PRI

=

6i + 6k 6v'2

=

I. I v'2' + v'2 k.

•

12.4

The Cross Product

685

For ease in calculating the cross product using detenninants, we usually write vectors in the fonn v = VI i + v2j + V3k rather than as ordered triples v = (VI, V2, V3).

Torque When we turn a bolt by applying a force F to a wrench (Figure 12.32), we produce a torque that causes the bolt to rotate. The torque vector points in the direction of the axis of the bolt according to the right-hand rule (so the rotation is counterclockwise when viewed from the tip of the vector). The magnitude of the torque depends on how far out on the wrench the force is applied and on how much of the force is perpendicular to the wrench at the point of application. The number we use to measure the torque's magnitude is the product of the length of the lever ann r and the scalar component of F perpendicular to r. In the notation of Figure 12.32, Component of F perpendicular to r. Its length is IFI sin

ii:----

or Ir X F I. If we let n be a unit vector along the axis of the bolt in the direction of the torque, then a complete description of the torque vector is r X F, or Torque vector

FIGURE 12.32 The torque vector describes the tendency of the force F to drive the bolt forward.

(I r II FI sin 8) n.

Recall that we defined u X v to be 0 when u and v are parallel. This is consistent with the torque interpretation as well. Ifthe force F in Figure 12.32 is parallel to the wrench, meaning that we are trying to turn the bolt by pushing or pulling along the line of the wrench's handle, the torque produced is zero.

EXAMPLE 5

The magnitude of the torque generated by force F at the pivot point P in

p'-------.--~ Q

FIGURE 12.33

=

Figure 12.33 is

3 ft bar

20lb magnitude force

Ir II FI sin 8,

Magnitude of torque vector =

IPQ

X

FI

F

The magnitude of the

torque exerted by F at P is about 56.4 ft-lb (Example 5). The bar rotates counterclockwise around P.

=

IPQIIFI sin 70°

~

(3)(20)(0.94)

~

56.4 ft-Ib.

In this example the torque vector is pointing out of the page toward you.

•

Triple Scalar or Box Product The product (u X v)· w is called the triple scalar product ofu, v, and w (in that order). As you can see from the fonnula

I(u

X v)· w I =

Iu

X v II w II cos 81,

the absolute value of this product is the volume of the parallelepiped (parallelogram-sided box) determined by u, v, and w (Figure 12.34). The number Iu X vi is the area of the base uXv

Area of base --=Iuxvl u

Volume = area of base . height = lu X vi IwlIcos III = I(u X v)· wi

FIGURE 12.34 The number I(u Xv)· w I is the volume of a parallelepiped.

686

Chapter 12: Vectors and the Geometry of Space

parallelogram. The number Iw II cos () I is the parallelepiped's height. Because of this geometry, (u X v), w is also called the hox product ofu, v, and w. By treating the planes of v and w and of w and u as the base planes of the parallelepiped detennined by u, v, and w, we see that (u X v)· w = (v X w)· u = (w X u)· v. Since the dot product is commutative, we also have 1

The dot and cross may be interchaoged in a 1riple sca1ar product without altering its value.

(u X v)·w = u'(v X w). The triple scalar product can be evaluated as a determinant: (u X v)·w = [I::

::Ii

Ul

U2

-I~: ~:Ij + I~: ~:Ik J-w U3

Vl

V2

V3

Wl

W2

W3

Calculating the Triple Scalar Product as a Dcterminant

(u

EXAMPLE 6 v = -2i Solution

X

v)· w

=

Ul

U2

U3

VI

V2

V3

Find the volume of the box (parallelepiped) determined by u = i + 2j - k, = 7j - 4k.

+ 3k,andw

Using the rule for calculating determinants, we rmd

I 2 (uXv)·w= - 2 0 o 7

-1 3 =-23. -4

The volume is I(u X v)· wi = 23 units cubed.

•

Exercises 12.4 Cross Product Calculations

7. u = -8i - 2j - 4k,

In Exercises 1-8, rmd the length and direction (when dermed) of

uXvandvXu. 1. u=2i-2j -k, v=i-k 2. u = 2i + 3j, v = -i + j 3. u=2i-2j+4k,

v=-i+j-2k

4.u=i+j-k, v=O S. u

=

2i,

v

6. u

=

i X j,

=

v

-3j =

j Xk

8. u = ti -

t j + k,

v = 2i + 2j + k v = i + j + 2k

In Exercises 9-14, s1retch the coordinate axes and then include the vectors u, v, and u X vas vectors starting at the origin.

9. u = i,

v= j

11. u = i - k,

v= j + k

13. u = i + j, v = i - j

10. u = i - k,

v= j

12. u = 2i - j, v = i + 2j 14. u = j + 2k,

v= i

12.4

Triangles in Space In Exercises 15--18,

e. (-u) X v

andR.

b. Find a unit vector perpendicular to plane PQR.

16. P(I, I, I),

Q(2,0, -I), R(O, 2, I)

Q(2, 1,3), R(3, -I, I)

17. P(2, -2, I),

Q(3, -1,2), R(3, -I, I)

18. P( -2,2,0),

Q(O, I, -I), R( -1,2, -2)

687

-(u X v)

d. (en)'v = u'(cv) = c(n'v)

a. Find the area of the triangle detennined by the points P, Q,

15. P(1, -1,2),

=

The (ross Product

(aoynumbere)

e. c(n X v) = (en) X v = u X (cv)

(aoynumbere)

f. U'U = lul 2 g. (uXn)·n=O h. (uXv)'u=v'(uXv)

29. Given nonzero vectors u, v, and w, use dot product and cross product notatioo, as appropriate, to describe the followiog. a. The vector projection of u onto v

b. A vector orthognoal to u and v

Triple Scalar Products In Exercises

1!l-22, verifY that (u X v)· w = (v X w)· u = (w X u)· v and fmd the volume of the parallelepiped (box) determined by n, v, and w. u

v

w

2j

2k

20.i-j+k

2i+j-2k

-i + 2j - k

21.2i+j

2i-j+k

i + 2k

22.i+j-2k

-i - k

2i+4j-2k

19. 2i

e. A vector orthogonal to u X v and u X w f. A vector oflength In I io the direction ofv

30. Compute (i X j) X j aod i X (j X j). What cao you cnoclude about the associativity of the cross product? 31. Let u, v, and w be vectors. Which of the followiog make sense,

and which do not? Give reasons for your answers. a. (uXv)'w b. u X (v'w)

Theory and Examples 23. ParaI1el and perpendieular vecton I.et n = 5i - j + k, v = j - 5k,w = -15i + 3j - 3k. Whieh vectors, ifaoy, are (a) perpendicular? (b) Parallel? Give reasons for your aoswers. 24. Parallel and perpendieular vectors I.ct u = i + 2j - k, v = -i + j + k, w = i + k, r = -('/T/2)i - '/Tj + ('/T/2)k. Which vectors, if aoy, are
reasons for your answers.

e. u X (v X w) d. u'(v'w) 32. Cross products of three veetors Show that except in degenerate cases, (u X v) X w lies in the plane of u and v, whereas n X (v X w) lies in the plane ofv and w. What are the degener-

ate cases? 33. Cancellation in cross products If u X v = u X w and u =F 0, then does v = w? Give reasons for your answer.

In Exercises 25 and 26, fmd the magoitude of the torque exerted by F on the bolt atP if IPQI = 8io. aod IFI = 301b. Answer in foot-

pounds.

25.

c. A vector orthogonal to u X v and w

d. The volume of the parallelepiped detennined by u, v, and w

34. Double eaneellalion If n # 0 aod if u X v = u X w aod u· v = U· lV, then does v = w? Give reasons for your answer.

Area of a Parallelogram Find the areas of the parallelograms whose vertices are given in Exer-

26.

cises 35--40.

~

p • ~ p

27. Which of the following are always true, and which are not always true? Give reasons for your answers.

a. Iu I

= y;;-:-;;-

35. A(1,O),

B(O, I),

C( -1,0),

36. A(O, 0),

B(7,3),

C(9, 8),

37. A( -1,2),

B(2,0),

38. A( -6,0),

B(l, -4),

39. A(O, 0, 0),

B(3, 2, 4),

40. A(1, 0, -I),

h. u'n = lui

D(O, -I)

D(2,5)

C(7, I),

D(4,3)

C(3, I),

D( -4,5)

C(5, 1,4),

B(l, 7, 2),

D(2, -1,0)

C(2,4, -I),

D(O, 3, 2)

e.uXO=OXn=O

Area of a Triangle Find the areas of the triangles whose vertices are given in Exercises

d. u X (-n)

41-47.

=

0

e.uXv=vXu

41. A(O, 0),

f. u X (v + w) = u X v + u X w

42. A( -I, -I),

B( -2,3),

C(3, I)

B(3,3),

C(2, I)

g. (uXv)'v=O

43. A( -5,3),

B(I, -2),

h. (uXv)'w=u'(vXw)

44. A( -6,0),

B(IO, -5),

C( -2,4) C(O,O,-I)

C(6, -2)

45. A(I,O,O),

B(0,2,0),

true? Give reasons for your answers.

46. A(O, 0, 0),

B( -I, I, -I),

a.u·v=v·n

47. A(I, -I, I),

28. Which of the following are always true, and whieh are not always b.uXv=-(vXu)

B(O, I, I),

C(3, 0, 3) C(I,O, -I)

688

Chapter 12: Vectors and the Geometry of Space

48. Find the volume of a parallelepiped if four of its eight vertices are A(O, 0, 0), B(l, 2, 0), C(O, -3,2), and D(3, -4, 5).

50. Triangle area Find a coacise formula for the area of a triangle in the xy-plane with vertices (a" a2), (b" IJ,.), and (q, e2).

49. TrilUlgie area Find a formula for the area of the triangle in the xy-plane with vertices at (0,0), (ai, a2), and (b" b2). Explain

your work.

12.5

Lines and Planes in Space This section shows how to use scalar and vector products to write equations for lines, line segments, and planes in space. We will use these representations throughout the rest of the book.

Lines and Line Segments in Space

o y

In the plane, a line is detennined by a point and a number giving the slope of the line. In space a line is detennined by a point and a veclor giving the direction of the line. Suppose that L is a line in space passing through a point Po(xo, Yo, zo) paralle.!10 a vector v = Vii + v2j + V3k. Then !:...is the set of all points P(x, y, z) for which PoP is parallel to v (Figure 12.35). Thus, PoP = tv for some scalar parameter I. The value of I depends on the location of the point P along the line, and the domain of I is (- 00, 00). The expanded form of the equation P-;P = tv is

(x - xo)i FIGURE 12.35 A point P lies on L through Po parallel to v if and ooly if P-;P is a scalar multiple ofv.

+ (y

- yoH

+

(z - zo)k = t(vli

+ V2j + V3k),

which can be rewritten as

xi

+ yj + zk

= xoi

+ yoj + zok + I(vli + v2j + V3k).

(1)

!fr(l) is the position vector ofa pointP(x,y, z) on the line and ro is the position vector of the point Po(XO,YO, zo), then Equation (1) gives the following vector form for the equation of a line in space.

Vector Equation for a Line

A vector equation for the line L through Po(Xo, Yo, xo) parallel to v is r(l) = ro

+ tv,

-00 < 1< 00,

(2)

where r is the position vector of a point P(x, y, z) on L and ro is the position vector of Po(xo, Yo, zo).

Equating the corresponding components of the two sides of Equation (1) gives three scalar equations involving the parameter I:

X=Xo+tvt,

y=YO+IV2,

These equations give us the standard parametrization of the line for the parameter interval -00<1<00.

Parametric Equations for a Une The standard parametrization of the line through Po(xo, Yo, xo) paraDel to v = Vii + V2j + v3kis

x

=

Xo

+ lVI, y

=

Yo

+ IV2, Z = Zo + IV3, -

00 < I < 00

(3)

12.5

Lines and Planes in Space

689

EXAMPLE 1 Find parametric equations for the line through (-2,0,4) parallel to + 4j - 2k (Figure 12.36).

v = 2i V O(-2, 0, 4) 4 _

2

o

With Po(XO,YO,zo) equal to (-2,0,4) and vii 2i + 4j - 2k, Equations (3) become

Solution

"\;1=0 NO, 4, 2) G

t

x = -2

+ 21,

Y

=

+ v2j + V3k

•

z = 4 - 21.

41,

=1

EXAMPLE 2

2

equal to

Find parametric equations for the line through P( -3,2, -3) and

Q(1, -1,4).

4

Solution

v=2i+4j-2k

The vector

Pg FIGURE 12.36 Selected points and parameter values on the line in Example 1. The arrows show the direction of increasing t.

= =

(I - (-3))i + (-I - 2)j + (4 - (-3»k 4i - 3j + 7k

is parallel to the line, and Equations (3) with (xo, Yo, zo) = (-3,2, -3) give

x = -3

+ 41,

y = 2 - 31,

+ 71.

z = -3

We could have chosen Q( I, -I, 4) as the ''base point" and written

x = ]

+ 41,

y = -] - 31,

z = 4

+ 71.

These equations serve as well as the first; they simply place you at a different point on the line for a given value of I. • Notice that parametrizations are not unique. Not ouly can the ''base poinf' change, but so can the parameter. The equations x = -3 + 41 3 ,y = 2 - 31 3,andz = -3 + 71 3 also parametrize the line in Example 2. To parametrize a line segment joining two points, we first parametrize the line through the points. We then find the I-values for the endpoints and restrict I to lie in the closed interval bounded by these values. The line equations together with this added restriction parametrize the segment.

EXAMPLE 3 Parametrize the line segment joining the points P( -3,2, -3) and Q(1, -1,4) (Figure 12.37).

Q(I,-1,4)

Solution We begin with equations for the line through P and Q, taking them, in this case, from Example 2:

x = -3

"'t~O

P(-3, 2, -3)

+ 41,

z

y=2-31,

=

-3

+ 71.

We observe that the point

Y

FIGURE 12.37 Example 3 derives a parametrization of line segmentPQ. The arrow shows the direction of increasing t.

(x,y,z) = (-3

+ 41,2 - 31, -3 + 71)

on the line passes through P( -3,2, -3) at I = 0 and Q(I, -1,4) at I = \. We add the restriction 0 :5 I :5 ] to parametrize the segment:

x = -3

+ 41,

z

y=2-31,

=

-3 + 71,

0:51:5\.

•

The vector fonn (Equation (2» for a line in space is more revealing if we think of a line as the path of a particle starting at position Po(xo, Yo, zo) and moving in the direction ofvec!or v. Rewriting Equation (2), we have r(l) = ro = ro

/

Initial position

+

Iv

+ Ilvl I~I'

! \ "'-

Time

Speed

(4)

Direction

690

Chapter 12: Vectors and the Geometry of Space

In other words, the position of the particle at time I is its initial position plus its distance moved (speed X time) in the direction vi Ivl of its straight-line motion.

EXAMPLE 4 A helicopter is to fly directly from a helipad at the origin in the direction of the point (I, I, I) at a speed of 60 ft/sec. What is the position of the helicopter after 10 sec? Solution We place the origin at the starting position (helipad) of the helicopter. Then the unit vector

u = _I_i + _I_j + _I_k v3 v3 v3 gives the flight direction of the helicopter. From Equation (4), the position of the helicopter at any time I is r(l) = ro + I(speed)u = 0 + 1(60)(_I_ i + _I_j + _I_ k ) v3 v3 v3 = 20v3I(i + j + k). When I = 10 sec, r(lO) = 200v3 (i + j + k) = (200v3, 200v3, 200v3). After 10 sec of flight from the origin toward (I, I, I), the helicopter is located at the point (200v3, 200v3, 200v3) in space. It has traveled a distance of (60 ft/secHIO sec) = _ 600 ft, which is the length of the vector r(IO).

The Distance from a Point to a Line in Space

s

To f'md the distance from a point S to a line that passes thmugJI a point P parallel to a vecPS in the direction of a vector normal to the line (Figure 12.38). In the n.Q1ation of the figure, the absolute value of the ~ IPS X vi scalar component is IPSI sinO, which is Ivl

tor v, we find the absolute value of the scalar component of

FIGURE 12.38 The distance from S to the line through P parallel to v is IPS I sin 8, where 8 is the angle between

Distance from a Point S to a Line Through P Parallel to v

d=

IPS X vi Ivl

(5)

PSandv.

EXAMPLE 5

Find the distance from the point S(I, 1,5) to the line

L:

x = 1

+ I,

y = 3 - I,

z = 21.

Solution We see from the equations for L that L passes thmugh P(I, 3, 0) parallel to v=i-j+2k.With

PS =

(I - I)i + (I - 3)j + (5 - O)k = -2j + 5k

12.5

6g1

Lines and Planes in Space

omd

J

I PS X v = 0 1

-2 -1

k 5 = i 2

+ 5j + 2)[,

Equation (5) gives

d

~ IPS X vi ~ Ivl

v'1 + 25 + 4 v'1 + 1 + 4

~ v':iO ~ Vs

v'6

•

.

An Equation for a Plane in Space

1-

.....,M

I

FIGURE 12.39 The standard equation for a plane in space is defined in terms of. vector normal to the plane: A point Plies in the plane tbrough Po normal to a if and only ifa' P-;P - o.

A plane in space is determined by knowUlg a point on the plane and its "tilf' or orientation. This "tilf' is defmcd by specifying a vector that is perpendicular or normal to the plane. Suppose that plane M passes through a point Po(xo,Yo, zo) and is normal to the n~ vector a = AI + Bj + CIL Then M is the set of all p~ts p(x, y, z) for which PoP is orthogonal to a (Figure 12.39). Thus, the dot product n' PoP = o. This equation is equivalent to (AI

+ BJ + Ck)'[(x -

xo)1

+ (y

- Yo)J

+ (z - zo)kj

~

0

or

A(x - Xo)

+ B(y -

Yo)

+ C(z -

Equation for a Plane The plane throughPo(XO,)'O, zo) normsl to n = Ai

~

o.

+ Bj + Ckhas

n'PoP = 0 A(x - Xo) + B(y - Yo) + C(z - zo) Ax + By + Cz = D, where D - Axo + l1.Yo + Czo

Vector eqnation: Componeat equation: Component equation IimpIified:

EXAMPLE 6

Zo)

~

0

Find an equation for the plane through P o(- 3, 0, 7) perpendicular to

n=5i+2j-k. Solution

The component equation is 5(x - (-3))

+ 2(y -

0)

+ (-I)(z

- 7)

~

O.

Simplifying, we obtain 5x

+

15

+ 2y - z + 7 5x

+ 2y -

=

0

z = -22.

•

Notice in Example 6 how the components of a = 51 + 2J - k. became the coefficients of x, y, and z in the equation 5x + 2y - z = -22. The vector a = AI + BJ + Ck is normal to the plane Ax + By + Cz = D.

692

Chapter 12: Vectors and the Geometry of Space

EXAMPLE 7

Find an equation for the plane throughA(O, 0,1), B(2, 0, 0), and qo, 3, 0).

SoLution We find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equation for the plane. The cross product

j

---"

AB X AC =

2

°

°

k -1 = 3i

3

-1

+

2j

+

6k

is normal to the plane. We substitute the components of this vector and the coordinates of A(O, 0, 1) into the component form of the equation to obtain

3(x - 0)

°

+ 2(y - 0) + 6(z - 1) = 3x + 2y + 6z = 6.

-

Lines of Intersection Just as lines are parallel if and only if they have the same direction, two planes are parallel if and only if their normals are parallel, or 01 = k02 for some scalar k. Two planes that are not parallel intersect in a line.

EXAMPLE 8 Find a vector parallel to the line of intersection of the planes 3x - 6y - 2z = 15 and 2x + y - 2z = 5. SoLution The line of intersection of two planes is perpendicular to both planes' normal vectors 01 and 02 (Figure 12.40) and therefore parallel to 01 X 02. Turning this around, 01 X 02 is a vector parallel to the planes' line of intersection. In our case,

01

X

02 =

3

2 Any nonzero scalar multiple of 01 X

FIGURE 12.40 How the line of intersection of two planes is related to the planes' normal vectors (Example 8).

j -6 1 02

k -2 = 14i -2

+

2j

+

15k.

-

will do as well.

EXAMPLE 9 Find parametric equations for the line m which the planes 3x - 6y - 2z = 15 and 2x + y - 2z = 5 intersect. SoLution We fmd a vector parallel to the line and a point on the line and use Equations (3). Example 8 identifies v = 14i + 2j + 15k as a vector parallel to the line. To find a point on the line, we can take any point common to the two planes. Substituting z = in the plane equations and solving for x andy simultaneously identifies one of these points as (3, -1,0). The line is

°

x = 3

°

+

14t,

y = -1

+ 2t,

z = 1St.

The choice z = is arbitrary and we could have chosen z = 1 or z = -1 just as well. Or we could have let x = and solved for y and z. The different choices would simply give _ different parametrizations of the same line.

°

Sometimes we want to know where a line and a plane intersect. For example, if we are looking at a flat plate and a line segment passes through it, we may be interested in knowing what portion of the line segment is hidden from our view by the plate. This application is used in computer graphics (Exercise 74).

12.5

EXAMPLE 10

Lines and Planes in Space

693

Find the point where the line

8 x=3+ 21,

y = -21,

z = 1

+1

intersects the plane 3x + 2y + 6z = 6. Solution

The point

(~ +

21, -21, 1 + I)

lies in the plane if its coordinates satisfy the equation of the plane, that is, if

3(~ + 21) +

2(-21) + 6(1 + I) = 6

8+61-41+6+61=6 81 = -8 1 = -1. The point of intersection is

(~- 2,2,1

(x,y,z)lt=-1 =

- 1) = (f,2,O).

•

The Distance from a Point to a Plane If P is a point on a plane with normal n, then the distance from any point S to the plane is the length of the vector projection of pS onto n. That is, the distance fromS to the plane is

d =

Ips· 1:11

(6)

where n = Ai + Bj + Ck is normal to the plane.

EXAMPLE 11

Find the distance fromS(I, 1, 3) to the plane 3x + 2y + 6z = 6.

Solution We find a point P in the plane and calculate the length of the vector projection of PS onto a vector n norma! to the plane (Figure 12.41). The coefficients in the equation 3x+2y+6z=6give n=3i+2j+6k. z

I

S(I,

3x + 2y + 6:z: = 6

0=31+2j+61<

~~~ ____ _

(0,0, 1) /

,/

0)1/

/

~

Distance from S to the plane

/

/

/ /

/

'(2,0,0)

y

x

FIGURE 12.41 The distance from Sio the plane is the length of the vector projection of liS onto n (Example II).

694

Chapter 12: Vectors and the Geometry of Space

The points on the plane easiest to find from the plane's equation are the intercepts. If we take P to be the y-intercept (0, 3, 0), then

n

=

(1 - O)i

=

i - 2j

+

(1 - 3)j

+ (3 - O)k

+ 3k,

101 =V(3)2 + (2? + (6)2 =v'49 =7. The distance from S to the plane is

d=ln·I:11 1

(i - 2j

length ofprojn

+ 3k)'

11_ i + lil 7

7

=

7

(t + t i

j

*

+ k)

Ps

1

11 7 .

•

Angles Between Planes The angle between two intersecting planes is defined to be the acute angle between their normal vectors (Figure 12.42). Find the angle between the planes 3x - 6y - 2z EXAMPLE 12 2x + y - 2z = 5. Solution

= 15 and

The vectors 01

= 3i - 6j - 2k,

02 =

2i

+j

- 2k

are normals to the planes. The angle between them is FIGURE 12.42 The angle between two planes is obtained from the angle between their normals.

COS-IC:I\I:~I) (2 1)

() =

4

=

cos- I

~

1.38 radians.

•

About 79 deg

Exercises 12.5 Lines and Line Segments

9. The line through (0, -7,0) perpendicular to the plane x + 2y + 2z = 13

Find parametric equations for the lines in Exercises 1- 12.

1. The line through the point P(3, -4, - 1) parallel to the vector i +j +k

10. The line through (2, 3, 0) perpendicular to the vectors u = i 2j + 3k and v = 3i + 4j + 5k

2. The line through P(1, 2, - 1) and Q( -I, 0, I)

11. The x-axis

+

12. The z-axis

3. The line through P( -2,0,3) and Q(3, 5, -2) 4. The line through P(I, 2, 0) and Q(1, I, - 1) 5. The line through the origin parallel to the vector 2j

+

k

6. The line through the point (3, -2, 1) parallel to the line x = I + 2/, y = 2 - I, z = 31 7. The line through (1, 1, 1) parallel to the z-axis 8. The line through 3x + 7y - 5z = 21

(2,4,5)

perpendicular

to

the

plane

Find parametrizations for the line segments joining the points in Exercises 13- 20. Draw coordinate axes and sketch each segment, indicating the direction of increasing 1 for your parametrization.

13. (0, 0, 0),

(1, 1,3/2)

14. (0, 0, 0),

(1,0,0)

15. (1, 0, 0),

(1, 1,0)

16. (1, 1,0),

(1, I, 1)

17. (0,1,1), (0,-1,1)

18. (0, 2, 0),

(3,0,0)

19. (2, 0, 2),

20. (1,0, - 1),

(0,3,0)

(0,2,0)

12.5

Planes Find equations for the planes in Exercises 21-26.

Angles Find the aogles between the planes in Exercises 47 and 48.

21. The plane through Po( 0, 2, - I) normal to n = 3i - 2j - k

47. x + Y = I, 2x + y - 2z = 2 48. 5x + y-z = 10, x -2y + 3z=-1

22. The plane through (I, -1,3) parallel to the plane

D Use a calculator to fmd the acute angles between the planes in Exer-

3x+y+z=7

cises 49-52 to the nearest hundredth of a radian.

23. The plane through (I, I, -I), (2, 0, 2), and (0, -2, I)

49.2x+2y+2z=3,

24. The plane through (2, 4, 5), (I, 5, 7), and (-1,6,8)

SO. x

25. The plane through Po(2, 4, 5) perpendicular to the line

x

=

5

+ t,

y

=

+ 31,

1

695

Lines and Planes in Space

z

=

+y +Z

= I,

52. 4y

26. The plane through A(I, -2, I) perpendicular to the vector from the origin to A

+

3z

°

z = (thexy-plane) x + 2y + z = 2

51. 2x + 2y - z = 3,

4t

2x-2y-z=5

= -12,

3x+ 2y+ 6z

=6

Intersecting Lines and Planes

27. Find the point of intersection of the lines x = 21 + I, y = 31 + 2, z = 4t + 3, and x = • + 2, y = 2s + 4, z = -48 - I, and then fmd the plane detennined by these lines.

In Exercises 53-56, fmd the point in which the line meets the plane.

28. Find the point of intersectino of the lines x = I, Y = -I + 2, z = 1 + I, and x = 2s + 2, y = • + 3, Z = 5. + 6, and then fmd the plane determined by these lines.

55.x=I+21, y=I+5t, z=31; x+y+z=2 56. x = -1 + 31, y = -2, z = 5t; 2x - 3z = 7

In Exercises 29 and 30, fmd the plane detennined by the intersecting

lines. 29. Ll:x

=

-1

+ t, y

=

L2: x

=

1 - 4s, Y

=

30. LI: x = I,

+ t, 1 + ls, 2

Y = 3 - 31,

L2: x = 1 + s, Y = 4 +

Z

S,

1 - t;

Z =

z

=

-00

2 - 2s;

= -2 - I;

<s< < 1 < 00

-00

< s<

y=31,

54. x = 2, Y = 3

+

z=I+I;

21, z = -2 - 21;

2x-y+3z=6 6x + 3y - 4z = -12

Find parametrizations for the lines in which the planes in Exercises 57-{;0 intersect.

57. x + Y + z = I, x + Y = 2 58.3x-6y-2z=3, 2x+y-2z=2

00

-00

-00

z = -1 + s;

< t<

53.x=I-I,

00

59. x - 2y

+ 4z = 2,

60. 5x-2y= 11,

00

31. Find a plane throughPo(2, I, -I) and perpendicular to the line of inletsection of the planes 2x + y - Z = 3, x + 2y + Z = 2. 32. Find a plane through the points PI(I, 2, 3), P2(3, 2, I) and petpendicular to the plane 4x - y + 2z = 7.

x

+ y - 2z = 5

4y-5z= -17

Given two lines in space, either they are parallel, or they inletseet, or they are skew (imagine, for example, the flight paths of two planes in the sky). Exercises 61 and 62 each give three lines. In each exercise, determine whether the lines, tsken two at a time, are parallel, intersect, or are skew. If they intersect, find the point of intersectino. 61. LI: x = 3

+ 21,

y = -I

+ 41,

z = 2 - I;

-00

Distances

L2:x= I + 4.,y= I + 2s,z= -3 +48;

-00

In Exercises 33-38, find the distance from the point to the line.

L3: x

33. (0,0,12);

x = 4t,

34. (0,0,0);

x = 5

35. (2, 1,3);

x = 2

Y = -21,

+ 31, y + 21, y

36. (2,1,-1);

x=21,

37.(3,-1,4);

x=4-1,

38. (-1,4,3);

x = 10

= 5 = I

Z

y= I +2t,

+

41,

z = -3 - 5t

y = -3,

z=21

40. (0,0,0), 41. (0, I, I), 42. (2,2,3), 43. (0, -1,0), 44. (1,0, -I),

z = 4t

T, Z =

-2 + 2r;

<. < 00 < r < 00 < 1 < 00

-00

T,

Z

= 8 + 3r;

-00

< r<

00

64. Use the coroponent form to generate an equation for the plane through P I (4, I, 5) normal to nl = i - 2j + k. Then generate another equatioo for the same plane using the point P2(3, -2, 0) and the normal vector.2 = + 2 Vzj - \12k.

+ 2y + 2z = 13 3x + 2y + 6z = 6 4y + 3z = -12 2x + y + 2z = 4 2x + y + 2z = 4 -4x + y + z = 4 x

45. Find the distance from the plane x x + 2y + 6z = 10.

2+

Theory and Examples 63. Use Equations (3) to generate a pararoe1rizatioo of the line through P(2, -4,7) parallel to VI = 2i - j + 3k. Then generate another pararoe1rizatioo of the line using the point P2( -2, -2, I) and the vector V2 = -i + (I/2)j - (3/2)k.

z=-5+31

In Exercises 39-44, find the distance from the point to the plane.

39. (2, -3,4),

=

L3: x = 5 + 2r, y = 1 -

z = 3

y=3+2t,

3 + 2r, y

00

62. LI: x = I + 21, y = -I - I, z = 31; -00 L2:x=2-s, y=3s, z=l+s; -oo<s
= 21

+ 4t, + 6t,

=

< 1<

Yzi

+ 21, y = -I - I, z = 31 meets the coordinste planes. Describe the reasoning behind your answer.

65. Find the points in which the line x = I

+ 2y + 6z

= I to the plane

46. Fiod the distance from the line x = 2 + I, y = I z = -(1/2) - (I/2)ttotheplanex + 2y + 6z = 10.

+

I,

66. Find equations for the line in the plane z = 3 thst mskes an angle of'IT/6rad with i and an aogle of'IT/3 rad withj. Describe the reasooing behind your answer. 67. Is the line x = I - 2t, Y = 2 + 5t, z = -31 parallel to the plane 2x + y - z = 8? Give reasons for your answer.

696

Chapter 12: Vectors and the Geometry of Space

68. How can you tell when two planes Alx + Bly + Clz = DI and A,x + B,y + C'z = D, are parallel? Perpeodicular? Give reasons for your answer.

b. Test the formulas obtained for y and z in part (a) by investigatiog their behavior at XI = and XI = Xo and by seeing what happeos as xo ..... 00. What do you find?

°

69. Find two differeot planes whose intersectioo is the line x = 1 + t,y = 2 - t, Z = 3 + 2t. Write equations for each plane in the form Ax + By + Cz = D.

z

70. Find a plane through the origin that is perpendicular to the plane M: 2x + 3y + z = 12 in a right angle. How do you know that your plane is perpeodicular to M?

P(O,y, z)

71. The graph of (x/a) + (y/b) + (z/c) = 1 is a plane for any nonzero numbers a, h, and c. Which planes have an equation of this form?

:

y

72. Suppose LI and L, are disjoint (noointersecting) nooparallellines. Is it possible for a nonzero vector to be perpeodicu1ar to both LI and L,.? Give reasons for your answer. 73. Perspective in computer graphics In computer graphics and perspective drawing, we need to represent oi!jects seen by the eye in space as images on a two-dimensional plane. Suppose that the eye is at E(xo, 0, 0) as sbown bere and that we want to represent a point PI(XI.YI. Zl) as a point on the yz-plane. We do this by projecling PI onto the plane with a ray from E. The point PI will be portrayed as the point p(0, y, z). The problem for us as graphics designers is to f'mdy andz givenE andP I . L

Write a vector equation that holds between BP and BPI' Use the equation to expressy andz in terms OfXO,Xl,YI. andzi.

12.6

x

74. Hidden lines in computer graphics Here is another typical problem in computer graphics. Your eye is at (4, 0, 0). You are looking at a triangular plate whose vertices are at (1,0, 1), (I, 1,0), and ( -2, 2, 2). The line segment from (1, 0, 0) to (0, 2, 2) passes through the plate. What portioo of the line segmeot is hiddeo from your view by the plate? (This is an exercise in f'mding intersections of lines and planes.)

Cylinders and Quadric Surfaces Up to now, we have studied two special types of surfaces: spheres and planes. In this section, we extend our inventory to include a variety of cylinders and quadric surfaces. Quadric surfaces are surfaces defmed by second-degree equations in x, y, and z. Spheres are quadric surfaces, but there are others of equal interest which will be needed in Chapters 14-16.

Cylinders z

Generating curve (in lbe yz-plane)

A cylinder is a surface that is generated by moving a straight line along a given planar curve while holding the line parallel to a given fixed line. The curve is called a generating curve for the cylinder (Figure 12.43). In solid geometry, where cylinder means circular cylinder, the generating curves are circles, but now we allow generating curves of any kind. The cylinder in our first example is generated by a parabola

EXAMPLE 1 Find an equation for the cylinder made by the lines parallel to the z-axis that pass through the parabola y = x 2 , Z = 0 (Figure 12.44).

lines through

generating curve parallel to x-axis

FIGURE 12.43 A cylinder and generatiog curve.

The point Po(xo, xo2, 0) lies on the parabola y = x' in the xy-piane. Then, for Solution any value of z, the point Q(xo, x02, z) lies on the cylinder because it lies on the line x = xo,y = xo' through Po parallel to the z-axis. Conversely, any point Q(xo,xo2,z) whose y-coordinate is the square of its x-coordinate lies on the cylinder because it lies on the line x = Xo, Y = xo' through Po parallel to the z-axis (Figure 12.44). Regardless of the value of z, therefore, the points on the surface are the points whose coordinates satisfY the equation y = x 2 • This makes y = x 2 an equation for the cylinder. Because of this, we call the cylinder ''the cylinder y = x 2 ." •

12.6

,

Cylinder.; and Quadric Surfaces

6g7

As Example 1 suggests, any curve f(x,y) = c in the xy-p1ane defines a cylinder paraIle! to the z-axis whose equation is also f(x,y) = c. For instance, the equation x 2 + y2 = 1 def'mcs the circular cylinder made by the lines parallel to the z-axis that pass through the circle x 2 + y2 = 1 in the xy-plane. In a similar way. any curve g(x, z) = c in the xz-plane defines a cylinder parallel to they-axis wh08C space equation is also g(x,z) = c. Any curve h(y,z) = c def'mes a cylinder parallel to the x-axis whose space equation is also hey, z) = c. The axis of a cylinder need not be parallel to a coordinate axis, h~.

Quadric Surfaces A quadric I.rface is the graph in space of a second-degree equation in x, y, and z. We fucus on the special equation ..(xl

FIGURE 12.44 Everypointofthe cylinder in Example 1 has coordinates of the form (ZO. xOl, z). We call it "the cylinder y _ xl."

+Byl

+ Cz1 + Dz - E,

where A, B, C, D, and E are oonstmJ.ts. The bosic quadric smfaces are ellipsoids, paraboloids, elliptical cones, and hyperboloid.. Spheres are special cases of ellipsoids. We present a few examples illustrating how to sketch a quadric surface, and then give

II. S1DD-

mmy table of graphs of the basic types.

EXAMPLE 2

The ellipsoid

(Figure 12.45) cuts the coonlinatc axes at (± a, O. 0). (0. ± b, 0), and (0, 0, ± c). It lics within the rectangular box defmcd by the inequalities Ix I :S a, Iy I :S h. and Iz I :s c. The surface is symmetric with respect to each of the coordinate planes because each variable in the defmmg equation is squared.

Elliptical cross-section in the pIane z - lo

,

,

/ c

\

,,

/' ,,, , : f .. ~

,

, ~

I

--- I --- ~

.

. r1 " , " >t-:.......;..--;,,:',,1"".......... I -- -::. I ~ _

1

r

, ,,

y

, I ..'

''f

FIGURE 12.45 The ellipsoid

in Example 2 has elliptical Cf'OfIS-sections in each of the three coordinate planes.

The curves in which the three coordinate planes cut the surface are ellipses. For example,

when

z

=

o.

698

Chapter 12: Vectors and the Geometry of Space

= zo, IZo I < c, is the ellipse

The curve cut from the surface by the plane z

x2 ----::-----=----:2,2

a (1 - (zoic) )

+

y2 2

2

b (1 - (zoic) )

= 1.

If any two of the semiaxes a, b, and c are equal, the surface is an ellipsoid of revolu_ tion. If all three are equal, the surface is a sphere.

EXAMPLE 3

The hyperbolic paraboloid

y2

x2

2

a2

---

b

c>

c'

0

has symmetry with respect to the planes x = 0 and y = 0 (Figure 12.46). The crosssections in these planes are

x

=

0:

y = 0:

the parabola z = c2 y2 . b

(1)

the parabola z = - c2 x 2 •

(2)

a

In the plane x = 0, the parabola opens upward from the origin. The parabola in the plane y = 0 opens downward. If we cut the surface by a plane z = Zo > 0, the cross-section is a hyperbola,

y2 b2

x2

Zo

a2

C'

---

with its focal axis parallel to the y-axis and its vertices on the parabola in Equation (1). If Zo is negative, the focal axis is parallel to the x-axis and the vertices lie on the parabola in Equation (2).

2

2

Part of the hyperbola y 2 - X 2 = 1 b a

z

/ "\-----,.. y /

/' The parabola z = - ~ x2

a

22

Part of the hyperbola \

/ ' in the plane z = -c

a

~y

- y2 = 1 b

in the xz·plane

FIGURE 12.46 The hyperbolic paraboloid (y2/b 2) - (x 2/a 2) = z/c, c > o. The cross-sections in planes perpendicular to the z-axis above and below the xy-plane are hyperbolas. The cross-sections in planes perpendicular to the other axes are parabolas.

Near the origin, the surface is shaped like a saddle or mountain pass. To a person traveling along the surface in the yz-plane the origin looks like a minimum. To a person traveling the xz-plane the origin looks like a maximum. Such a point is called a saddle point of a surface. We will say more about saddle points in Section 14.7. _ Table 12.1 shows graphs of the six basic types of quadric surfaces. Each surface shown is symmetric with respect to the z-axis, but other coordinate axes can serve as well (with appropriate changes to the equation).

12.6

699

Cylinder.; and Quadric Surfaces

TABLE 12.1 Graphs of Quadric Surfaces

,

, Tho PEIobolu _ ~l:3

ill Ibo ",,"pImo

*~~+,, -T'---, ,, ,,

" ......

, 'Ibe p,nbolu _

'~/

~:I

____

;"'1110 1f'1Ilmo

,

,

• ELLIPSOID

ELLIPTICAL PARABOLOID

•

,/

)f--r-"

HYPERBOLOID OF ONE SHEET

ELLIPTICAL CONE

,

. .,.

'Ibe e1Hpee -

+- _1

.'f - cV'i

,

,

. HYPERBOLOID OF 1WO SHEETS

HYPERBOLIC PARABOLOID

% C'

.>0

700

Chapter 12: Vector.; and the Geometry of Space

Exercises 12.6 Matchtng Equations with Surfaces In Exercises 1-12, match the equation with the surface it defines.

k.

,

L

Also, identify each surface by type (paraboloid, ellipsoid, etc.) The swfaces arc labeled (a)-{I).

1. x:1 +y:1 + 4z2 = 10 3. 9y2 +z:1 = 16 5. x _ y 2_ z 2 7.x2 +2z 2 _ S

2.z:1+4y 2_4x:1=4 4.y:1+ z 2=X 2 6.x _ _ 2_ 2 z y 8.z 2 +x:1- y 2 _ 1 10. z = _4.1: 2 _ y2

9. X=z2_ y :1 11. x:1 + 4z:1 = y2

12. 9x:1

+ 4y:1 + 2z2

Drawtng

Sketch the surfaces in Exercises 13-44.

= 36

,

b.

L

,

CYUNDERS 13. X2 +y2=4 15. x2+~ _ 16

14.z=y2-1 16. 4x 2 + y2 _ 36

ELUPSOIDS 17. 9;t2 + y2 + z2 = 9 19. 4x 2 + 9y2 + ~ _ 36

18. 4x:1 20. 9;t2

+ 4y:1 + z:1 = 16 + 4y2 + 36z2 _ 36

PARABOLOIDS AND CONES

,

,.

,

d.

:n.z=S-:x2 -y:1

21. z=x:1+4y:1 23. x=4-4y:1-~ 25. X:1+ y 2=z:1

24.y=I-:x 2 -z2 26. 4x:1 + ~ = 9y2

HYPERBOLOIDS 28.y:1+ z 2_ x 2=1 30. (y'/4) - (.'/4) -

27.x:1+ y 2- z :1=1 29. z2_ X:1_ y 2= 1

y

z'

~

1

HYPERBOUC PARABOLOIDS

••

,

f•

31. y2-:x:1 _ z

,

ASSORTED

y

..

,

h•

x2

38.x2+~ _ y

40. 16y2 + 9z:1 _ 4x 2 42.y:1_ x 2_ z 2=1 44.x 2 +y:1=z

+z2

_ 1 41. z = _(X2 + y2) 43.4y 2+ z 2_4x:1=4

39.

y

34. 4x:1 + 4y:1 = z2 36. 1&2 + 4y2 _ 1

33. z= 1 +y2_X:1 35. Y _ _ (x:1 + z2) 37. x 2 +y2_z2 _ 4

Theory and Examples 45. L Express the area.4. of the cross-section cut from. the ellipsoid

, ,

x:1+L+~=1 4 9

L

,

J.

,

by the plane z - c as a function of c. (The area of an ellipse with semiaxesa andbis7rab.) b. Use slices perpendicular to the z-axis to fmd the volume of

the ellipsoid in part (a). c. Now find the volume of the ellipsoid

x2

y2

z:1

-+-+- 1. a:1 b2 c2 y

Does your formula give the volume of a sphere of radius a if a - b - c?

Chapter 12

701

h. Express your answer in part (a) in terms of h and the areas Ao and Ah of the regions cut by the hyperboloid from the planes z = 0 andz = h.

46. The barrel shown here is shaped like an ellipsoid with equal pieces cut from the ends by planes perpendicular to the z-axis. The crosssections perpendicular to the z-axis are circular. The barrel is 2h units high. its midsection radius is R, and its end radii are both r. Find a formula for the barrel's volume. Then check two things. First, suppose the sides of the barrel are straightened to turn the barrel into a cylinder of radius R and height 2h. Does your formula give the cylinder's volume? Second, suppose r = 0 and h = R so the barrel is a sphere. Does your formula give the sphere's volume?

z

Questions to Guide Your Review

c. Show that the volume in part (a) is also given by the formula V=

"6h (Ao + 4Am + Ah ),

where Am is the area of the region cut by the hyperboloid from the plane z = h/2.

Viewing Surfaces

D Plot the surfaces in Exercises 49-52 over the indicated domains.

If

you can, rotate the surface into different viewing positions.

-2 ~ x ~ 2,

49. z = y2,

-2 ~ x ~ 2,

50. z = 1 - y2, y

51. z = x 2 52. z = x 2

47. Show that the volume ofthe segment cut from the paraboloid x2 y2 z -2+ - = a b2 c

by the plane z = h equals half the segment's base times its altitude. 48. a. Find the volume of the solid bounded by the hyperboloid x2 y2 z2 -2+ -2- - = 1 a b c2

and the planes z = 0 and z = h, h

Chapter

>

~

c. -2 d. -2

~

x

~

x

-2 ~ y ~ 2

x ~ 3,

-3 ~ y ~ 3

~

3,

-3

~

y

~

3

~x~

1, 2,

-2

~

y

~

3

~

-2

~

y

~

2

~

2,

-1

~y~

1

x

COMPUTER EXPLORATIONS Use a CAS to plot the surfaces in Exercises 53-58. Identify the type of quadric surface from your graph. x2

53.

y2

9 + 36

55. 5x 2

o.

+ y2, -3 ~ + 2y2 over

a. -3 h. -1

-0.5 ~ y ~ 2

=

9 -

1 - 25

z2 - 3y2

x2

57.

z2 =

y2

1

=

54.

56. z2

16 + 2

x2

z2

9 - 9 y2

16

58. y -

=

1-

=

1-

x2

y2

16

9 +z

V4 -

z2 = 0

Questions to Guide Your Review

1. When do directed line segments in the plane represent the same vector? 2. How are vectors added and subtracted geometrically? Algebraically? 3. How do you find a vector's magnitude and direction? 4. If a vector is multiplied by a positive scalar, how is the result related to the original vector? What if the scalar is zero? Negative? 5. Define the dot product (scalar product) of two vectors. Which algebraic laws are satisfied by dot products? Give examples. When is the dot product of two vectors equal to zero? 6. What geometric interpretation does the dot product have? Give examples. 7. What is the vector projection of a vector u onto a vector v? Give an example of a useful application of a vector projection. 8. Define the cross product (vector product) of two vectors. Which algebraic laws are satisfied by cross products, and which are not? Give examples. When is the cross product of two vectors equal to zero? 9. What geometric or physical interpretations do cross products have? Give examples.

10. What is the determinant formula for calculating the cross product of two vectors relative to the Cartesian i, j, k-coordinate system? Use it in an example. 11. How do you find equations for lines, line segments, and planes in space? Give examples. Can you express a line in space by a single equation? A plane? 12. How do you find the distance from a point to a line in space? From a point to a plane? Give examples. 13. What are box products? What significance do they have? How are they evaluated? Give an example. 14. How do you find equations for spheres in space? Give examples. 15. How do you find the intersection of two lines in space? A line and a plane? Two planes? Give examples. 16. What is a cylinder? Give examples of equations that define cylinders in Cartesian coordinates. 17. What are quadric surfaces? Give examples of different kinds of ellipsoids, paraboloids, cones, and hyperboloids (equations and sketches).

702

Chapter 12: Vectors and the Geometry of Space

Chapter

Practice Exercises

Vedor Calculations in Two Dimensions In Exercises 1-4, let u = (-3,4) and v = (2, -5). Find (a) the componeot form of the vector aod (b) its magnitude.

In Exercises 25 and 26, fmd (a) the area of the parallelogram determined by vectors u and v and (b) the volume of the parallelepiped determined by the vectors u, v, and w.

1.3u-4v

2.u+v

2S.u=i+j-k,

3. -2u

4. 5v

26.u=i+j,

In Exercises 5-8, fmd the component form of the vector.

5. The veetor obtained by rotating (0, I) througb an angle of21f/3

radians 6. The unit vector that makes an angle of 1f/6 radian with the positivex-axis 7. The vector 2 units loog in the directioo 4i - j

v=2i+j+k,

v=j,

w=i+j+k

Lines, Planes, and Distances 27. Suppose that n is DOnnai to a plane and that v is parallel to the plane. Describe how you would fmd a veetor n that is both perpendicular to v and parallel to the plane. 28. Find a veetor in the plane parallel to the line ax + by = c. In Exercises 29 and 30, fmd the distance from the point to the line.

8. The veetor 5 units loog in the directioo opposite to the directioo of(3/5)i + (4/5)j

29. (2,2,0); x

=

-t, Y

=

t,

Z =

Express the vectors in Exercises 9-12 in terms of their lengtha aod directioos.

12. Velocity vector v = (etcost - etsint)i + (etsint + etcost)j wheol = In 2.

Vedor Calculations in Three Dimensions Express the veetors in Exercises 13 and 14 in terms of their leogtha and directions.

+ 6k

14. i

Z

= I

31. Parametrize the line that passes througb the point (I, 2, 3) parallel to the vector v = - 3i + 7k. 32. Parametrize the line segment joining the points P(I, 2, 0) and

10. -i - j

11. Velocityveetorv = (-2sinl)i + (2cosI)jwheol = 1f/2.

13. 2i - 3j

+t

-1

30. (0,4, I); x = 2 + I, y = 2 + I,

9. v2i + v2j

w=-i-2j+3k

+ 2j - k

15. Find a vector 2 units long in the direction ofv = 4i - j + 4k. 16. Find a vector 5 units loog in the direction opposite to the directioo ofv = (3/5) i + (4/5)k.

Q(I,3, -I). In Esercises 33 and 34, fmd the distance from the point to the plane.

33. (6,0, -6),

x - y = 4

34. (3,0, 10),

2x

+

3y

+Z

= 2

35. Find an equation for the plane that passes through the point (3, -2, 1) normal to the veetor n = 2i + j + k. 36. Find an equatioo for the plane that passes through the point (-1,6,0) perpendicular to the line x = -I + I,y = 6 - 21, Z = 31. In Exercises 37 and 38, fmd an equatioo for the plane througb points

P,Q,andR. In Esercises 17 and 18, fmd lvi, lui, V'u, U'v, v X u, u X v,

37. P(i, -1,2),

Iv X u I, the angle between v and u, the scalar compooent of u in the direction afv, and the vector projection ofo onto v.

38. P(i, 0, 0),

17. v = i + j

18. v = i + j + 2k

u=2i+j-2k

u=-i-k

In Exercises 19 and 20, fmd proj. u.

19. v=2i+j-k u=i+j-5k

Q(2, 1,3),

R( -1,2, - 1)

Q(O, 1,0), R(O, 0, I)

39. Find the points in which the line x = I + 21, y = -I - I, z = 3t meets the three coordinate planes. 40. Find the point in which the line througb the origin perpendicular to the plane 2x - Y - Z = 4 meets the plane 3x - 5y + 2z = 6. 41. Find the acute angle between the planes x = 7 and x + y = -3.

v2z

20. u=i-2j

v=i+j+k

+

42. Find the acute angle betweeo the planes x + y = I and

y+z= 1. In Exercises 21 and 22, draw coordinate axes and then sketch u, v, and

u X v as vectors at the origin. 21. u=i.,

v=i+j

22. u=i-j,

v=i+j

23. !flvl = 2, Iwl = 3,andthe angie between v andw is 1f/3,fmd Iv - 2wl· 24. For what value or values of a will the vectors u = 2i + 4j - 5k and v = -4i - 8j + ak be parallel?

43. Find pararne1ric equations for the line in which the planes x + 2y + Z = I and x - y + 2z = -8 intersect. 44. Show that the line in which the planes

x+2y-2z=5

and

5x-2y-z=0

intersect is parallel to the line

x = -3 + 21, y = 31, z = I + 41.

Chapter 12 45. The planes 3x L

+ 6z = I and 2x + 2y - z = 3 intcncct in a Iinc.

Shaw that the planes are orthogonal.

L (21 - 3J + 3k)' b. x - 3-t,

b. Find equatiOlli for the line of intersection.

046. Find an equation for the plane that pasSCll through the point (1,2, 3)paraIleltou - 2i + 3j + kandv - i - j + 2k. 47. Is v = 2i - 4j + k related in any !pCcial way to the plane 2x + Y = 5? Give reasons for your answer. 48. The equation.' P-;P = 0 represents ~plane through Po normal to D. What set docs the inequality D' PoP> 0 tcpICscn.t?

e. (x

«x + 2)1 + (y -

y - -Ilt,

+ 2) + l1(y

Practice Exercises

703

I)J + zk) - 0

z - 2-3t

- 1) - 3z

d. (2i - 3j + 3k) X «x + 2)i + (y - I)j + zk) = 0 e. (2i - j + 3k) X (-3i + k)·«x + 2)i + (y - I)j + zk) ~O

6l. The puallelogram !hawn here has vertices at A(2, -1,4), B(I, 0, -I), C(I, 2, 3), andD. Find

,

49. Find the distance from the point P(I, 4, 0) to the plane through A(O. o. 0). B(2. o. -1). "'" C(2. -1.0).

D),

50. Find the distance from the point (2, 2, 3) to the plane 2x + 3y + 5z - O. 51. Find a vector parallel to the plane 2x - y - z - 4 and orthogonaltot+J+k. A(2,-l,4)

51. Find a unit vector orthogonal to A in the plane of B and C if A - 2i - j + k,B - i+ 2j + k,andC - i + j - 2k.

C(l.2, 3)

53. Find a vector of magnitude 2 parallel to the line of intcncction of theplane8x + 2y +z - I = and x -y + 2z+ 7 = O.

o

54. Find the point in which the line through the origin perpendicular totheplane2x - y - z = 4mcctsthep1mc3x - 5y +22' = 6.

y B(l, 0. -1)

55. Find the point in which the line through p(3, 2, I) normal ttl the plane 2x - y + 2z = -2 meets the plane. 56. What angle doc! the line of intcncction of the planca 2x + Y - z = 0 and x + Y + 2z = 0 make with the positive

x-axis?

L the coordinates of D, b. the cosine of the interior angle atB,

e. the vector projection of BA. onto iiC,

57. The line

d. the area of the parallelogram, L:

x=3+2t,

y=2t,

z=t

in1eDcctll the plane x + 3y - z = -4 in a point P. Find the coonIin.atcs of P and find equations for the line in the plane through P perpendicular ttl L.

58. Shaw that for every real number k the plane

x - 2y+ z + 3 + k(2x - y- z + I) = 0

contains the line of intmscction of the planes

c. an equatioo for the plane of the parallelogram. f. the areas of the orthogonal projections of the panillelogram

on the three coord.i.na.tJ; planes. 63. Distmte betweea lineI Find the distance bctwccn the line L, through the points .4(1,0,-1) and B(-I,I,O) and the line L2 through the points C(3, I, -I) andD(4, 5, -2). The distance is ttl be meuured along the line perpendicular In the two lines. FltBt fmd • vector D perpendicular to both lines. Thm project AC onto D. 64. (Continuatio1l o/Exerci.!e 63.) Find the distance between the line through A(4, 0, 2) and B(2, 4, 1) and the line through C(t, 3, 2)

""'D(2. 2. 4). x - 2y + z + 3 = 0 and 2x - y - z + I = O. Quadric SumCt!!s 59. Find an equation for the plane through A(-2,O, -3) and B(I, -2, I) that lies parallel to the line through

Identify and sketch the surfaces in Exercises 65-76.

C( -2. -13/'. 26/') ""'D(16/'. -13/'. 0). 60. Is the line x = 1 + 2t, y = -2 + 31, z = -5t related in any way to the plane -4x - 6y + 10z - 91 Give reasoos for your

67.4x 2 +4y 2+ Z l=4 69. z = _(x 2 + y2)

""""'.

61. Which of the following are equations for the plane through the points P(I, I, -I), Q(3, 0, 2), andR(-2, I, O)?

65. x 2 + y2 + z2 = 4

71. X2 +y2=z2 73. X2 +y2_z2=4 75. yl_x2_zl= 1

66. x 2 + (y - 1)2 + z2 = I 68. 3fu:l + 91 2 + 4z2 = 36

70. Y = _(Xl

+ zZ)

71. Xl+zZ=yl

74. 4yl+zZ-ob:l =4 76. z2 _x 2 _ y2 = I

704

Chapter 12: Vectors and the Geometry of Space

Chapter

Additional and Advanced Exercises

1. Submarine hunting Two surface ships on maneuvers are trying to determine a submarine's course and speed to prepare for an aircraft intercept. As shown here, ship A is located at (4, 0, 0), whereas ship B is located at (0, 5, 0). All coordinates are given in thousands of feet. Ship A locates the submarine in the direction of the vector 2i + 3j - (I/3)k, and ship B locates it in the direction of the vector 18i - 6j - k. Four minutes ago, the submarine was located at (2, -1, -1/3). The aircraft is due in 20 min. Assuming that the submarine moves in a straight line at a constant speed, to what position should the surface ships direct the aircraft?

clockwise when we look toward the origin fromA. Find the velocity v of the point of the body that is at the position B(l, 3, 2).

z

IA~(:: l,~l,~l!-)

_---I.

B

I

:~: : ----.J!___ v: / I

z

---

x

(l, 3, 2)

I

/

I

1/

I~

I /

-----_

I

-------1//

~y

//

ShipB

Ship A

(4,0,0)

(0,5,0)

y

x

5. Consider the weight suspended by two wires in each diagram. Find the magnitudes and components of vectors F I and F 2, and angles a and {3.

a.

-\ Submarine

Nor TO SCALE

2. A helicopter rescue Two helicopters, HI and H 2, are traveling together. At time t = 0, they separate and follow different straight-line paths given by

HI:

x = 6 + 40t,

Y = -3 + lOt,

z = -3 + 2t

H2:

x = 6 + 11 Ot,

Y = -3 + 4t,

z = -3 + t.

Time t is measured in hours and all coordinates are measured in miles. Due to system malfunctions, H2 stops its flight at (446, 13, 1) and, in a negligible amount of time, lands at (446, 13,0). Two hours later, HI is advised of this fact and heads toward H2 at 150 mph. How long will it take HI to reach H2? 3. Torque The operator's manual for the Toro® 21 in. lawnmower says "tighten the spark plug to 15 ft-lb (20.4 N· m)." If you are installing the plug with a lO.5-in. socket wrench that places the center of your hand 9 in. from the axis of the spark plug, about how hard should you pull? Answer in pounds.

b.

(Hint: This triangle is a right triangle.)

6. Consider a weight of w N suspended by two wires in the diagram, where T I and T 2 are force vectors directed along the wires.

a. Find the vectors T I and T 2 and show that their magnitudes are w cos (3 ITII = sin (a + (3) 4. Rotating body The line through the ongm and the point A( 1, 1, 1) is the axis of rotation of a right body rotating with a constant angular speed of3/2 rad/sec. The rotation appears to be

and T

-

121-

wcosa sin (a+{3)

705

Chapter 12 Additional and Advanced Exercises b. For a fIxed {j determine the value of a which minimizes the

magnitude IT ,I. c. For a fIxed a determine the value of (j which minimizes the magnitude IT21· 7. Determinants and planes

X

X2 -

X

X3 -

x

Y' - Y Y2 - Y y, - Y

Zt -

z

Z2 -

Z

Z3 -

z

d

b. What set of points in space is described by the equation

Y

Z

Xl

Yl

Zl

X2

Y2

Z2

X3

Y3

Z3

8. Determinant. and line.

lax, + by,

-

cl

~ ~'--;=.c===.<~

Va 2 + b 2

•

I

I

~ '-
Ax----"'--;+;=B:c:Y~,=+=:c=c=z,~-"D-'-1 VA2+B2+C 2

~O

is an equation for the plane through the three noncollinear points Pt(xt, Yh Zt), P2(X2,)I2, Z2), and P3(X3,Y3, Z3).

x

d

12. a. Use vectors to ahow that the distance from P,(XhYh z,) to the plane Ax + By + Cz ~ Dis

a. Showthat XI -

11. Use vectors to show that the distance from P,(XhY') to the line ax+by=cis

13. a. Show that the distance between the parallel planes Ax + By + Cz ~ D, and Ax + By + Cz ~ D, is d

~

07

.

b. Find an equation for the sphere that is tangent to the planes x + Y + z ~ 3 and x + Y + z ~ 9 if the planes 2x - Y ~ 0 and 3x - z ~ 0 pass through the center of the sphere.

~

ID, - D21 IAi + Bj + Ckl .

b. Find the distance between the planes 2x + 3y - z 2x + 3y - z ~ 12.

I

~

6 and

c. Find an equation for the plane parallel to the plane 2x - Y + 2z ~ -4 if the point (3, 2, -I) is equidistant from the two planes.

Show that the lines

d. Write equations for the planes that lie parallel to and 5 units away from the plane x - 2y + z ~ 3.

and

14. Prove that four points A, B, C, and D are coplanar (lie in a common plane) if and only if AD . (AB X BC) ~ O. intersect or are parallel if and only if

15. The projection of a veetor on a plane Let P be a plane in space and let v be a vector. The vector projection ofv onto the plane P, projp v, can be defmed informally as follows. Suppose the suo is shining so that its rays are norma1 to the plane P. Then projp v is the "shadow" ofv onto P. If P is the plane x + 2y + 6z ~ 6 and v ~ i + j + k,fmdprojpv.

b1-d1

at

Cl

a2

C2

b2

a3

C3

b3 -d3

d2

-

O.

=

9. Consider a regular tetrahedron of side length 2.

a. Use vectors to fmd the angle 0 formed by the base of the tetrahedron and any one of its other edges.

16. The accompanying figure shows nonzero vectors v, W, and z, with z orthogonal to the line L, and v aod w ma1cing equal angles {j with L. Assuming Iv I ~ Iwi, fmd w in tenns ofv and z.

D

2

p

2

C

1

B

b. Use vectors to fmd the angle 0 formed by any two adjacent faces of the tetrahedron. This angle is commonly referred to as a dihedral angle.

10. In the fIgure here, D is the midpoint of side AB oftriang1e ABC, and E is one·third of the way between C and B. Use vectors to prove that F is the midpoint of line segment CD. C

~

A

D

B

17. Triple vector products The triple vector products (u X v) X w and u X (v X w) are usoally not equal, although the formulas for evaluating them from components are similar: w

~

(u'w)v - (v·w)u.

u X (v X w)

~

(u'w)v - (u·v)w.

(u

X

v)

X

VerilY each formula for the following vectors by evaluating its two sides and coroparing the results. u v w

a. 2i b. i-j+k c. 2i+ j d. i+j-2k

Zj Zi+j-Zk

Zk -i+Zj-k

2i-j+k -i - k

i + 2k 2i+4j-2k

706

Chapter 12: Vectors and the Geometry of Space

18. Cro.s and dot products

Show that if u, v, w, and r are any

21. Use vectors to prove that

vectors, then

(a' + b')(c' + d') '" (ac + bd)'

LUX~X~+vX~X~+wX~X~-O

for any four nurohers a, b, c, and d. (Hint: Let u - ai + bj and + dj.)

b. u X v - (u'v X i)i + (u'v X j)j + (u'v X k)k c.(uXv).(wxr)_lu,W

u·r

19. Cro•• and dot products

V'WI. v"r

v - ci

22. Dot multiplication is positive def"utite

~

for every vector u and that U' u - 0 if and only ifu - O.

Prove or disprove the formula

u X (u X (u X v))·w - -Iul'u'v X w.

Show that dot multipli-

cation of vectors is positive definite; that is, show that u . u 23. Show that Iu + v I '" Iu I + Iv I for any vectors u and v.

24. Show that w - Iv Iu + Iu Iv bisects the angle between u and v. 20. By forming the cross product of two appropriate vectors, derive the trigonometric identity

25. Showthat Ivlu + lulvand Ivlu - lulvare orthogonal.

sin(A - B) - sinAcosB - cosAsinB.

Chapter

Technology Application Projects

Mathematica/Maple Module: Using ~ctors to Represent Lilies and Find Distances Parts I and II: Learn the advantages of interpretiog lines as vectors. Part ill: Use vectors to f"md the distance from a point to a line. Putting a Sc~,,~ in Thn~ DiMensions onto Il 'IWo-DiIIIensional Caln'llS Use the concept of planes in space to obtain a two-dimensional image.

Getting Started in Plotting in 3D Part I: Use the vector def"mition of lines and planes to generste graphs and equations, and to compare different forms for the equations of a single line. Part II: Plot functions that are def"med implicitly.

0

13 VECTOR-VALUED FUNCTIONS AND MOTION IN SPACE OVERVIEW Now that we have learned about vectors and the geometry of space, we can combine these ideas with our earlier study of functions. In this chapter we introduce the calculus of vector-valued functions. The domains of these functions are real numbers, as before, but their ranges are vectors, not scalars. We use this calculus to describe the paths and motions of objects moving in a plane or in space, and we will see that the velocities and accelerations of these objects along their paths are vectors. We will also introduce new quantities that describe how an object's path can turn and twist in space.

13.1

Curves in Space and Their Tangents When a particle moves through space during a time interval I, we think of the particle's coordinates as functions defined on I: x

z

= I(t),

y = get),

z

= h(t),

tEl.

(1)

The points (x,y, z) = (f(t), get), h(t», tEl, make up the curve in space that we call the particle's path. The equations and interval in Equation (1) parametrize the curve. A curve in space can also be represented in vector form. The vector ret) =

x F1GU ~ 13.1 The position vector r = OP of a particle moving through space is a function of time.

oF =

I(t)i

+ g(t)j + h(t)k

(2)

from the origin to the particle's position P(f(t), g(t), h(t)) at time t is the particle's position vector (Figure 13.1). The functions I, g, and h are the component functions (components) of the position vector. We think of the particle's path as the curve traced by r during the time interval I. Figure 13.2 displays several space curves generated by a computer graphing program. It would not be easy to plot these curves by hand. z

z

z

x

x

y

x

y r(t) = (sin3t)(cost)i + (sin3t)(sint)j + tk

r(t) = (cost)i

+ (sint)j + (sin2t)k

r(t) = (4

(4

+ sin20t)(cost)i + + sin20t)(sint)j +

(cos20t)k (a)

FIGURE 13.2

(b)

(c)

Space curves are defined by the position vectors ret).

707

708

Chapter 13: Vector-Valued Functions and Motion in Space Equation (2) defines r as a vector function of the real variable t on the interval 1. More generally, a vector-valued function or vector function on a domain set D is a rule that assigns a vector in space to each element in D. For now, the domains will be intervals of real numbers resulting in a space curve. Later, in Chapter 16, the domains will be regions in the plane. Vector functions will then represent surfaces in space. Vector functions on a domain in the plane or space also give rise to "vector fields," which are important to the study of the flow of a fluid, gravitational fields, and electromagnetic phenomena. We investigate vector fields and their applications in Chapter 16. Real-valued functions are called scalar functions to distinguish them from vector functions. The components of r in Equation (2) are scalar functions of t. The domain of a vector-valued function is the common domain of its components.

z

EXAMPLE 1

ret)

)

~ t = 2-rr

I I

........ t .. ,

\

r -..

/ "-{'

= (cos t)i + (sin t)j + tk

is defined for all real values of t. The curve traced by r winds around the circular cylinder x 2 + y2 = 1 (Figure 13.3). The curve lies on the cylinder because the i- and j-components ofr, being the x- andy-coordinates ofthe tip ofr, satisfy the cylinder's equation:

,,

"

I

0)..

The vector function ret)

,

------:------

(

SoLution

= (cos t)i + (sin t)j + t k.

= 7r

I I I I I I /~

Graph the vector function

•

x 2 + y2

t = :1!. 2

'

The curve rises as the k-component z = t increases. Each time t increases by 21T, the curve completes one turn around the cylinder. The curve is called a helix (from an old Greek word for "spiral"). The equations

p

-;::1

(1,0,0)

= (cos t)2 + (sin t)2 = 1.

y

x

x

= cost,

y = sint,

z = t

parametrize the helix, the interval -00 < t < 00 being understood. Figure 13.4 shows more helices. Note how constant multiples of the parameter t can change the number of turns per unit of time. _

FIGURE 13.3 The upper half of the helix ret) = (cos t)i + (sin t)j + tk (Example 1).

z

z

x

x

z

x

y ret) = (cos t)i

FIGURE 13.4

+ (sin t)j + tk

y ret) = (cos t)i

+ (sin t)j + O.3tk

ret)

=

(cos 5t)i

+ (sin 5t)j + tk

Helices spiral upward around a cylinder, like coiled springs.

Limits and Continuity The way we define limits of vector-valued functions is similar to the way we define limits of real-valued functions.

13.1 Curves in Space and Their Tangents

709

DEFINmON Let r(t) = I(t)i + g(t)j + h(t)k be a vector function with domain D, and L a vector. We say that r has limit Last approaches to and write

lim r(t)

= L

1-/0

if, for every number E for all tED Ir(t) -

IfL = Lji

>

0, there exists a corresponding number II

LI <

whenever

E

<

0

It - tol

<

+ L2j + L3k, then it can be shown that lim,....tor(t)

>

0 such that

ll.

= L precisely when

and We omit the proof. The equation lim r(t) =

1-/0

(lim I(tj\ + (lim g(t)'j + (lim h(t)\ I-I{)

')

I-I{))

(3)

1-10)

provides a practical way to calculate limits of vector functions.

EXAMPLE 2

Ifr(t) = (cost)i

+

(sint)j

lim r(t) = (lim cos t)i

t---1>'1T/4

t---1>'1T/4

+

+

tk, then

(lim sin t)j t-w/4

+

lim t)k ( l-w/4

=V2·+V2·+1T k J 2'

2

•

4'

We derme continuity for vector functions the same functions.

WH:j

we derme continuity for scalar

DEFINmON A vector function r{t) is continuous at a point t = to in its domain if lim,....~ r(t) = r(to). The function is continuous if it is continuous at every point in its domain. From Equation (3), we see that r(t) is continuous at t = to if and only if each component function is continuous there (Exercise 31).

EXAMPLE 3 (a) All the space curves shown in Figures 13.2 and 13.4 are continuous because their component functions are continuous at every value of t in ( - 00, 00).

(b) The function g(t) = (cost)i

+

(sint)j

+

ltJk

is discontinuous at every integer, where the greatest integer function discontinuous.

l t J is •

Derivatives and Motion Suppose that r(t) = I(t)i + g(t)j + h(t)k is the position vector of a particle moving along a curve in space and that I, g, and h are differentiable functions of t. Then the difference between the particle's positions at time t and time t + !J.t is ~r =

r(t

+ !J.t) - r(t)

710

Chapter 13: Vector-Valued Functions and Motion in Space

(Figure 13.5a). In terms of components, ar = r(t =

= ~----------------~y

(a)

+

M) - r(t)

+ at)i + g(t + M)j + h(t + M)k] - [/(t)i + g{t)j + h(t)k] [j(t + at) - I(t)]i + [g(t + M) - g{t)]j + [j(t

[h(t

+

M) - h(t)]k.

As at approaches zero. three things seem to happen simultaneously. First. Q approaches P along the curve. Second, the secant line PQ seems to approach a limiting position tangent to the curve atP. Third, the quotient arl M (Figure 13.5b) approaches the limit

lim arat [lim I(t + at)at -

I(t)]i

=

.1.,-0

[lim g{t + at)at - g(t)]J.

+

.l.t-O

+

[lim h(t

+ at) - h(t)]k

.1.,-0 =

.l.t-O

M

g [d/]i + [d ].J + [dh]k dt' dt dt

We are therefore led to the following definition. ~----------------~y

(b) x

FIGURE 13.5 As M .... O. thepointQ approaches the point P along the curve C. In the limit, the vector PQ/ at becomes the tangent vector r'(t).

DEFINmON The vector function r(t) = I(t)i + g{t)j + h(t)k has a derivative (is differentiable) at t if I, g, and h have derivatives at t. The derivative is the vector function '(t) r

= dr =

l' r(t dt,J!!!o

+ at)

- r(t)

at

dl. dt I

+ dg. + dt J

dh k dt'

A vector function r is differentiable if it is differentiable at every point of its domain. The curve traced by r is smootb if dr/dt is continuous and never 0, that is, if I, g, and h have continuous first derivatives that are not simultaneously O. The geometric significance of the definition of derivative is shown in ~ure 13.5. The points P and Q have position vectors r(t) and r(t + at), and the vector PQ is represented by r(t + at) - r(t). For at > 0, the scalar multiple (1/at)(r(t + at) - r(t» points in the same direction as the vector As at ..... 0, this vector approaches a vector that is tangent to the curve atP (Figure l3.5b). The vector r'(t), when different from 0, is defined to be the vector tangent to the curve at P. The tangent line to the curve at a point U(to), g{to), h(to» is defined to be the line through the point parallel to r'(to). We require dr/dt "" 0 for a smooth curve to make sure the curve has a continuously turoing tangent at each point. On a smooth curve, there are no sharp corners or cusps. A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smooth (Figure 13.6). Look once again at Figure 13.5. We drew the figure for at positive, so ar points forward, in the direction of the motion. The vector arl at, having the same direction as ar, points forward too. Had M been negative, ar would have pointed backward, against the direction of motion. The quotient arI at, however, being a negative scalar multiple of ar, would once again have pointed forward. No matter how ar points, arl at points forward and we expect the vector dr/dt = lim&--+o arl M, when different from O. to do the same. This means that the derivative drI dt, which is the rate of change of position with respect to time, always points in the direction of motion. For a smooth curve, drI dt is never zero; the particle does not stop or reverse direction.

PQ.

FIGURE 13.6 A piecewise smooth curve made up offive smooth curves connected end to end in • continuous fashion. The curve here is not smooth at the points joining the five smooth curves.

=

13.1 Curves in Space and Their Tangents

711

DEFINmONS If r is the position vector of a particle moving along a smooth curve in space, then dr V(I) = dl is the particle's velocity vector, tangent to the curve. At any time I, the direction of v is the direction of motion, the magnitude of v is the particle's speed, and the derivative 8 = dv/dl, when it exists, is the particle's acceleration vector. In

summary, dr v = dl'

1. Velocity is the derivative of position:

2. Speed is the magnitude of velocity:

Speed = Ivl.

3. Acceleration is the derivative of velocity: 4. The unit vectorv/lvl is the direction of motion at time I.

EXAMPLE 4 Find the velocity, speed, and acceleration of a particle whose motion in space is given by the position vector r( I) = 2 cos I i + 2 sin I j + 5 cos2 I k. Sketch the velocity vector V(77T/4). Solution

The velocity and acceleration vectors at time I are V(I) = r'(I) = -2sinli + 2coslj - 10 cos I sinlk

z

= -2 sin I i + 2 cos I j - 5 sin 21 k, 8(1) = r"(I) = -2cosli - 2sinlj - 10cos21k, and the speed is

y t

=

7," 4

IV(I) I = Y(-2sinl)2 + (2 cos 1)2 + (-5 sin 21)2 = Y4 + 25si02 21. When 1= 77T/4, we have

x

FIGURE 13.7 The curve and ilie velocity vector when t = 7'IT/4foriliemotion given in Example 4.

V(7:)=Vzi+ Vz i+ 5 k, A sketch of the curve of motion, and the velocity vector when 1= 77T/4, can be seen in Figure 13.7. • We can express the velocity of a moving particle as the product of its speed and direction: Velocity =

IVIC~I) =

(speed)(direction).

Differentiation Rules Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions.

712

Chapter 13: Vector-Valued Functions and Motion in Space

Differentiation Rules for Vector Functions Let u and v be clifferentiable vector functions of t, C a constant vector, c any scalar, and / any differentiable scalar function. d

1. Constant Function Rule:

dtC

2. Scalar Multiple Rules:

:t [cu(t)]

=

0

cu'(t)

=

:t [f(t)u(t)]

I When you use the Cross Product Rule,

remember to preserve the order of the factors. Ifu comes fIrst on the left side of the equation, it must also come first on the right or the signs will be wrong.

!,(t)u(t) + /(t)u'(t)

=

3. Sum Rule:

:t [u(t) + v(t)]

4. Difference Rule:

dt [u(t) - v(t)]

S. Dot Product Rule:

:t [u(t)· v(t)]

6. Cross Product Rule:

dt [u(t) X v(t)]

7. Chain Rule:

:t [u(f(t))]

d

=

d

=

=

u'(t) + v'(t)

=

u'(t) - v'(t)

u'(t)· v(t) + u(t)· v'(t) =

u'(t) X v(t) + u(t) X v'(t)

!'(t)u'(f(t))

We will prove the product rules and Chain Rule but leave the rules for constants, scalar multiples, sums, and differences as exercises. Proof of the Dot Product Rule

Suppose that

u = ul(t)i

+ U2(t)j + u3(t)k

and Then d

dt(u'v)

=

d

dt(U1Vl + U2 V2 + U3 V3)

= U1VI

+ ulv2 +

U3V3

+ UtVl + U2V2 + U3V3.

u' ·v

u·v'

•

Proof of the Cross Product Rule We model the proof after the proof of the Product Rule for scalar functions. According to the detmition of derivative,

E. (

)_ .

dtUXV-~

u(t + h) X v(t + h) - u(t) X v(t) h

To cbange this fraction into an equivalent one that contains the difference quotients for the derivatives ofu and v, we subtract and add u(t) X v(t + h) in the numerator. Then

:t

(u X v) = 1~~u(~t_+~h~)_X~v(~t_+~h~)_-~u(~~~X~v~(t_+~h)~+~u~(t~)_x_v~(t~+~h~)_-~u~(t~)_x_v~(~t)

~o

.

= lun

h~O

=

h [U(t

+ h)h - u(t) Xvt+h ( ) () +ut

X

v(t

+ h)h - V(t)]

I~ u(t + h) - u(t) X I~ v(t + h) + I~ u(t) X I~ v(t + h) - v(t) h-O

h

h-O

h-O

h-O

h

13.1

713

Curves in Space and Their Tangents

The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist (Exercise 32). As h approaches zero, vet + h) approaches v(t) because v, being differentiable at t, is continuous at t (Exercise 33). The two fractions approach the values of du/dt and dv/dt at t. In short,

d (

tit

IAs an a1~braic convenience, we

sometimes write the product of a scalar c and a vector v as vc instead. of cv. Thill permi11l us, for instance, to write the Chain Rule in a familiar form.:

~

)

du Xv oX dv =tit + dt·

•

Proof of the Chain Rule Suppose that o(s) = a(s)i + b(s)j + c(s)k is a differentiable vector function ofs and thats = J(t) is a differentiable scalar function oft. Then a, b, and c are differentiable functions of t, and the Chain Rule for differentiable real-valued functions gives

~[u(')l ~

du da ds (ji - didt'

whore,

oXv

:1 +
da da db da de da - dadti+dadtJ+dadt k

f(t).

~:(~I+:i+~k) ~

da du dtda

~

/'(t)u'(f(t)).

I

•

- let)

Vector Functions of Constant Length When we tmck a particle moving on a sphere centered at the origin (Figure 13.8), the position vector has a constant length equal to the radius of the sphere. The velocity vector dr/tit, tangent to the path of motion, is tangent to the sphere and hence perpendicular to r. This is always the case for a differentiable vector :function of constant length: The vector and its Imt derivative are orthogonal. By direct calculation.

r(t)·r(t) = c 2

r'(t)· ret)

Ir(t)l - cisconstant.

~[r(t)'r(t)l ~

0

Dtil"=mWoboll"'d
+ r(t)· r'(t)

0

Rulc!iwithr{t)

~

=

aCt)

=

vet)

2r'(t)· ret) ~ O.

The vectors r'(t) andr(t) are orthogonal because their dot product is o. In summary, If r is a differentiable vector function of t of constant length, then FIGURE 13.8 Ifa particle moves on a sphere in such a way that its position r is a differentiable function of time, then r·(drjdt) = o.

dr

(4)

r·dt=O.

We will use this observation repeatedly in Section 13.4. The converse is also true (see _ .. 21).

Exercises 13.1 Motion fn the Plane In Exercises 1--4, r(t) is the position of a particle in the xy-plane at time t. Find an equation in x and y whose graph is the path of the particle. Then fmd the particle's velocity and acceleration vectors at the given value oft.

1. r(t) - (t

+ l)i + (rl - l)j,

t - 1

.. ret) - _t_i + II' t - -1/2 t+1 t' 3. r(t) - e'l

+ je 2tJ,

4. r(t) - (cos2t)i

t -

ln3

+ (3sin2t)J,

t - 0

714

Chapter 13: Vector-Valued Functions and Motion in Space

Exercises 5--8 give the position vectors of particles moving along various curves in the xy-plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve.

x' + y' ~ 1

5. Motion on the circle r(t)

~

(sint)i + (cost)j;

~ ~

'IT/4and'IT/2

i) Does the particle have constant speed? If so, what is its con-

(4 cos!} + (4Sin!)j;

7. Motion on the cycloid x r(t)

~

x' + y' ~ 16

6. Motion on the circle r(t)

t

~

~ 'lTand3'IT/2

t

t - sin t, Y

(t - sin t)i + (1 - cos t)j;

8. Motion on the parabola y ~

t

1 - cos t

iiI) Does the particle move clockwise or counterclockwise

~

'IT and 3'IT/2

iv) Does the particle begin at the point (I, OJ?

t ~ -1,0, and I

Motion ;n Space In Exercises !l-14, r(t) is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Then fmd the particle's speed and direction of motion at the given value oft. Write the particle's velocity at that time as the product of its speed and direction. 9. r(t) ~ (t + l)i + (t' - I)j + 2tk, ,

10. r(t)

t ~ I

3

~ (I + t)i + ~j +}-k, t ~ I

11. r(t) ~ (2 cos t)i + (3 sin t)j + 4tk,

12. r(t)

~

(sect)i + (tant)j + jtk,

13. r(t) ~ (2 In (t + I))i + t'j +

T,

~

'IT/6

t ~ I

k,

14. r(t) ~ (e -1)i + (2 cos 3t)j + (2 sin 3t)k,

t ~ 0

In Exercises 15-18, r(t) is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time

t

~

around the circle?

a. r(t)

~

(cost)i + (sint)j,

b. r(t)

~

cos (2t)i + sin (2t)j,

c. r(t)

~

cos (t - 'IT/2)i + sin(t - 'IT/2)j,

d. r(t)

~

(cost)i - (sint)j,

O.

~

('?

t)i+

('?

vi'+! k

Tangents to Curves As mentioned in the text, the tangent line to a smooth curve r(t) ~ f(t)i + g(t)j + h(t)k at t ~ to is the line that passes through the point (f(to), g(to), h(to)) parallel to v(to) , the curve's velocity vector at to. In Exercises 19-22, fmd paramettic equations for the line that is tangent to the given curve at the given parameter value t ~ to. 19. r(t) ~ (sin t)i + (t' - cos t)j + e'k, to ~ 0 20. r(t) ~ t'i + (2t - l)j + t 3k, to ~ 2

~

Inti +:

~ ~j

+ tlntk,

to

22. r(t) ~ (cost)i + (sint)j + (sin2t)k,

~

t'" 0

24. Motion along a eircle Show that the vector-valued function r(t)

~

(2i + 2j + k)

I

to ~

f

k)

describes the motion of a particle moving in the circle of radius I centered at the point (2,2, I) and lying in the plane x + y - 22 ~ 2.

25. Motion along a parabola A particle moves along the top of the parabola y' ~ 2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point (2, 2). 26. Motion along a .ycloid A particle moves in the xy-plane in such a way that its position at time t is

D a.

18. r(t) ~ ~ (I + t)'/2i + ~ (I - t)3/2j + 1 tk 9 9 3

21. r(t)

t'" 0

e. r(t) ~ cos (t')i + sin (t )j, t '" 0

r(t)

t - 16t')j

17. r(t) ~ (In (t' + I))i + (tan- 1 t)j +

t '" 0

2

15. r(t) ~ (3t + l)i + y'3tj + t'k

16. r(t)

t'" 0

I. I.) + . (I . + I . + I + cost ( y'2' - y'21 smt y'3' y'31 y'3

t ~ 'IT/2

t

stant speed? h") Is the particle's acceleration vector always orthogonal to its velocity vector?

~

x' + 1

r(t) ~ ti + (t' + I)j;

Theory and Examples 23. Motion along a eircle Each of the following equations in parts (aHe) describes the motion of a particle having the same path, lUIJ1lely the unit circle x 2 + y' ~ I. Although the path of each particle in parts (aHe) is the same, the behavior, or "dynamics:' of each particle is different For each particle, answer the following questions.

~

(t - sin t)i + (I - cos t)j.

Graph r(t). The resulting curve is a cycloid.

b. Find the maximum and minimum values ofl v Iand I_I. (Hint:

Find the extreme values ofl v I' and Ia I' flISt and take square roots later.)

27. Let r be a differentiable vector function of t. Show that if r' (dr/dt) ~ 0 for all t, thenlrl is constant. 28. Derivatives of triple ••alar product.

a. Show that ifu, v, and w are differentiable vector functions of t, then d du dv dw dt(u'v X w) ~ lit' v X w + U· dt X w + u·v X lit. b. Showthat

E. (r. dr dt

X d'r) dt dt'

~

3 r' (dr X d r) dt dt 3 '

(Hint: Differentiate on the left and look for vectors whose products are zero.)

13.2 Integrals of Vector Functions; Projectile Motion 29. Prove the two Scalar Multiple Rules for vector functions. 30. Prove the Sum and Difference Rules for vector functions. 31. Component Test for Continuity at a Point Show that the vector function r defined by r(l) = /(I)i + g(l)i + h(l)k is contiouous att = 10 ifaod only if /,g, aod h are contiouous atto.

ll. Limits of cross products of vector functions Suppose that r,(I) = J,(I)i + h(lli + h(I)k, r2(1) = g,(I)i + g2(1)i + g,(I)k, lim,.....r'(I) = A,aodlim,.....r2(1) = B. Use the deterotinant formula for cross products aod the Limit Product Rule for scalar functions to show that

35. r(l)

(sint - lcost)i + (cost &IT, 10 = 3",/2

=

o '" I

34. Constant Function Rule Prove that if u is the vector function with the constaot value C, then do/dl = O. COMPUTER EXPLORATIONS Use a CAS to perform the followiog steps in Exercises 35-38. L

Plot the space curve traced out by the position vector r.

b. Find the components of the velocity vector dr/dl.

c. Evaluate dr/dl at the given point to and deterotine the equation of the taogent line to the curve at r(lo).

+ I sin tli + 12k,

'"

+ e'i + e-lk, -2 '" I '" 3, 10 = 1 (sin21)i + (10(1 + t))j + Ik, 0'" I'" 41r,

36. r(l) = V2li

37. r(l) = 10 = ",/4

38. r(l) = (10 (12 + 2))i + (t..-' 31)i -3 ::;; t:5 5, to = 3

+

v't'+l k,

10 Exercises 39 and 40, you will explore graphically the behavior of the helix r(l)

33. Differentiable vector functions are continuous Show that if r(l) = /(I)i + g(t)i + h(l)k is differentiable at I = 10, then it is continuous at to as well.

715

= (cosal)i +

(sinal)i

+ blk

as you change the values of the constaots a aod b. Use a CAS to perform the steps in each exercise. 39. Set b = 1. Plot the helix r(l) together with the taogent line to the curve at I = 3",/2 for a = 1, 2, 4, and 6 over the interval o '" I '" 41r. Describe in your owo words what happens to the graph of the helix and the position of the tangent line as a increases through these positive values. 40. Set a = 1. Plot the helix r(l) together with the taogent line to the curve at I = 3",/2 for b = 1/4, 1/2,2, aod 4 over the interval o '" I '" 4",. Describe in your owo words what happens to the graph of the helix aod the position of the taogent line as b increases through these positive values.

d. Plot the taogen! line together with the curve over the given interval.

13.2

Integrals of Vector Functions: Projectile Motion In this section we investigate integrals of vector functions and their application to motion along a path in space or in the plane.

Integrals of Vector Functions A differentiable vector function R(t) is an antiderivative of a vector function r(t) on an interval I if dR/ dt = r at each point of!. If R is an antiderivative of r on I, it can be shown, working one component at a time, that every antiderivative of r on I has the form R + C for some constant vector C (Exercise 41). The set of all antiderivatives of r on I is the indeUnite integral of ron!.

DEFINmON The indef"mite integral of r with respect to t is the set of all antiderivatives ofr, denoted by r(t) dt. IfR is any antiderivative ofr, then

J

J

r(t) dt

=

R(t) + C.

The usual arithmetic rules for indefinite integrals apply.

716

Chapter 13: Vector-Valued Functions and Motion in Space

EXAMPLE 1

To integrate a vector function, we integrate each of its components.

J

((cosl)i

+j

(J

- 2lk)dl =

COSldt}

(J dl)i - (J

+

- (1 2

+ C,)i + (I + C2)j (sin I)i + Ij - 12k + C

= (sin I =

+

2ldl)k

(1)

C 3 )k

(2)

c ~ CII + C,l - C,k

As in the integration of scalar functions, we recommend that you skip the steps in Equations (I) and (2) and go directly to the fmal form. Find an antiderivative for each component and add a conslanl veclor at the end. _ Defmite integrals of vector functions are best defmed in tenns of components. The definition is consistent with how we compute limits and derivatives of vector functions.

DEFINITION Ifthe components ofr(l) = !(I)i + g(l)j + h(l)k are integrable over [a, b], then so is r, and the dermite integral ofr from a to b is

t EXAMPLE 2

r(l)dl =

(t !(I)dl) + (tg(l)dt)i + (th(l)dl)k.

As in Example I, we integrate each component.

J.w((COSI)i

+j

- 2lk)dl = =

(J.w

COSldt}

[sinll~ i + [Il~ j

= [0 - O]i

+ [1T -

(J.w dl)i -

+ -

(J.w2Idl)k

[12l~ k

OJi - [1T 2 - 02]k

1T 2k

= 1Tj -

The Fundamental Theorem of Calculus for continuous vector functions says that

t

r(l)

dl

-

= R(I) l: = R(b) - R(a)

where R is any antiderivative ofr, so that R'(I) = r(l) (Exercise 42).

EXAMPLE 3 Suppose we do not know the path of a hang glider, but only its acceleration vectora(l) = -(3 cos I)i - (3 sin I)j + 2k. We also know that initially (at time I = 0) the glider departed from the point (3, 0, 0) with velocity v(O) = 3j. Find the glider's position as a function of I. Solution

Our goal is to find r(l) knowing

d2~

dl

= -(3 cos I)i - (3 sin I)j

The differential equation:

a =

The initial conditions:

v(O) = 3j

and

r(O) = 3i

+

(3 cos I)j

+ 21k +

C,.

We use v(O) = 3j to find C,: 3j = -(3 sin O)i 3j = 3j

c,

=

o.

+

C,

+

(3 cos O)j

+

(O)k

+

2k

+ OJ + Ok.

Integrating both sides of the differential equation with respect to I gives V(I) = -(3 sin I)i

+

C,

13.2 Integrals of Vector Functions; Projectile Motion

717

The glider's velocity as a function of time is

~~ z

= V(I) = -(3 sin I)i

+

(3 cos I)j

+ 21k.

Integrating both sides of this last differential equation gives

r(t) = (3 cos t)i

+ (3 sin t)j + t 2k + C 2.

We then use the initial condition r( 0) = 3i to f"md C2: 3i = (3 cos O)i 3i = 3i x

- ._--"-"y

FIGURE 13.9 The path of the hang glider in Example 3. Although the path spirals around the z-axis, it is not a helix.

y

+ (O)j

+ (3 sin O)j + + (O)k + C2

(02)k

+ C2

C 2 = O. The glider's position as a function of t is

r(t) = (3 cos t)i

+

(3 sin t)j

+ t 2 k.

This is the path of the glider shown in Figure 13.9. Although the path resembles that of a helix due to its spiraling natore around the z-axis, it is not a helix because of the way it is rising. (We say more about this in Section 13.5.) Note: It turned out in this example that both of the constant vectors of integration, C, and C 2, are O. Exercises 15 and 16 give examples for which the constant vectors of integration are not O. •

The Vector and Parametric Equations for Ideal Projectile Motion

----~L-~~----~x

r=Oat/lvolcosai timet = 0 .~

-gJ


y

A classic example of integrating vector functions is the derivation of the equations for the motion of a projectile. In physics, projectile motion describes how an object f"rred at some angle from an initial position, and acted upon by only the force of gravity, moves in a vertical coordinate plane. In the classic example, we ignore the effects of any frictional drag on the object, which may vary with its speed and altitode, and also the fact that the force of gravity changes slightly with the projectile's changing height. In addition, we ignore the long-distance effects of the Earth torning beneath the projectile, such as in a rocket launch or the f"lring of a projectile from a cannon. Ignoring these effects gives us a reasonable approximation of the motion in most cases. To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projectile during its flight is the constant force of gravity, which always points straight down. We assume that the projectile is launched from the origin at time t = 0 into the f"ITS! quadrant with an initial velocity Vo (Figure 13.10). Ifvo makes an angle a with the horizontal, then Vo = (lvolcosa)i

+

(lvolsinali.

If we use the simpler notation Vo for the initial speed Ivol, then Vo = (vo cos a)i "'~------------------~ x

+

(vosina)j.

(3)

The projectile's initial position is

i<-----R------>i'l

ro = Oi

+ OJ

=

o.

(4)

Horizontal range (b)

FIGURE 13.10 (a) Position, velocity, acceleratioo, and lauocb angle at t ~ o. (b) Positioo, velocity, and acceleratioo at a later time t.

Newton's second law of motion says that the force acting on the projectile is equal to the projectile's mass m times its acceleration, or m(d2 r/dt 2 ) ifr is the projectile's position vector and I is time. If the force is solely the gravitatioual force -mgj, then

d 2r . m= -mgJ 2 dt

and

d'r

.

dt 2 = -gJ

718

Chapter 13: Vector-Valued Functions and Motion in Space where g is the acceleration due to gravity. We find r as a function of I by solving the following initial value problem.

d'r

Differential equation:

dl 2 =

r = ro

Initial conditions:

.

-gJ

dr

and

dl = Vo

when I = 0

The first integration gives

~;

-(gt)j

=

+ Vo·

A second integration gives r = _!gt2j

+ vol + roo

Substituting the values ofvo and ro from Equations (3) and (4) gives

+ (vo cos a)li + (vosina)lj + O.

r = -tgl1

vat Collecting terms, we have Ideal Projectile Motion Equation r = (vocosa)li

+

(voSina)1 - !gt2)i.

(5)

Equation (5) is the veclor equalion for ideal projectile motion. The angie a is the projectile's launch angle (firing angie, angle of elevation), and Vo, as we said before, is the projectile's initial speed. The components ofr give the parametric equations x=(vocosa)1

y=(vosina)I-!gt2,

and

(6)

where x is the distance downnmge and y is the height of the projectile at time I ;;,: O.

EXAMPLE 4 A projectile is fired from the origin over horizontal ground at an initial speed of 500 m/ sec and a launch angle of 60°. Where will the projectile be 10 sec later? Solution We use Equation (5) withvo = 500,a = 60°,g = 9.8, and I = 10 to find the projectile's components 10 sec after firing. r = (vocosa)li

+

= (500) (t)(iO)i '" 2500i

+

(voSina)1 - !gt2)i

+

(500)('{3)1O - (!)<9.8)(100))i

3840j

Ten seconds after f"rring, the projectile is about 3840 m above ground and 2500 m downrange from the origin. • Ideal projectiles move along parabolas, as we now deduce from Equations (6). Ifwe substitute I = x/( Vo cos a) from the first equation into the second, we obtain the Cartesiancoordinate equation

y=-(

g 2

2

2vo cos a

TIris equation has the form y = ax 2

)x 2 +(tana)x.

+ bx, so its graph is a parabola.

13.2 Integrals of Vector Functions; Projectile Motion

719

A ]ITOjectile reaches its highest point when its vertical velocity component is zero. When fired over horizontal ground, the projectile lands when its vertical component equals zero in Equation (5), and the range R is the distance from the origin to the point of impact. We summarize the results here, which you are asked to verify in Exercise 27.

Height, Flight Time, and Range for Ideal Projectile Motion For ideal ]ITOjectile motion when an object is launched from the origin over a horizontal surface with initial speed Vo and launch angle a:

y

Maximum height:

YIllIIX

=

(vosina)' 2g

t

=

---"~-­

Flight time: Range:

R

2vo sin a

g

vl .

= gsm2a.

~oi-------------------~~X

FIGURE 13.11

The path ofa projectile

rITed from (xo, Yo) with ao initial velocity Vo at an angle of a degrees with the horizootal.

Ifwe rlre our ideal projectile from the point (xo,Yo) instead of the origin (Figure 13.11), the position vector for the path of motion is r = (xo

+ (vocosa)t)i +

&0 + (vosina)t -

tgt2)j,

(7)

as you are asked to show in Exercise 29.

Projectile Motion with Wind Gusts The next example shows how to account for another force actiug on a ]ITOjectile, due to a gust of wind. We also assume that the path of the baseball in Example 5 lies in a vertical plane.

EXAMPLE 5 A baseball is hit when it is 3 fi above the ground. It leaves the bat with initial speed of 152 fi/sec, making an angle of 20° with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of -S.Si (fi/sec) to the ball's initial velocity (S.S fi/sec = 6 mph). (a) Find a vector equstion (position vector) for the path of the baseball.

(b) How high does the baseball go, and when does it reach maximum height? (c) Assuming that the ball is not caught, find its range and flight time.

Solution (a) Using Equation (3) and accountiug for the gust of wind, the initial velocity of the baseball is Vo = (vocosa)i

+

= (152cos200)i

(vosina)j - S.Si

+

(152sin200)j - (S.S)i

= (152 cos 20° - S.S)i

The initial position is ro = Oi

+

~~

+

(152sin20°)j.

3j. Integration of d'rldt 2 = -gj gives = -(gt)j

+

Vo.

720

Chapter 13: Vector-Valued Functions and Motion in Space A second integration gives

I 2.J r = -2,gt

+ vol + roo

Substituting the values of Vo and ro intu the last equation gives the position vectur of the baseball. r =

I 2.J + vot + ro -Zgt

= -1612j

+ (152 cos 20°

= (152 cos 20° - 8.8)ti

- 8.8)li

+

(3

+

+ (152 sin200)lj + 3j

(152sin200)t - 16t 2)j.

(b) The baseball reaches its highest point when the vertical component of velocity is zero, or dy . dl = 152sm20° - 32t = O.

Solving for I we find 20° '" I 62 t = 152 sin 32 . sec. Substituting this time intu the vertical component for r gives the maximum height ynw< = 3

+ (152 sin 20°)(1.62)

- 16(1.62)2

'" 45.2 ft. That is, the maximum height of the baseball is about 45.2 ft, reached about 1.6 sec after leaving the bat. (c) To find when the baseball lands, we set the vertical component for r equal to 0 and solve for t:

3

(152sin200)t - 16t 2 = 0

+

3

+

(51.99)t - 16t 2 = O.

The solution values are about I = 3.3 sec and 1= -0.06 sec. Substituting the positive time into the horizontal component for r, we fmd the range

R = (152 cos 20° - 8.8)(3.3) '" 442 ft. Thus, the horizontal range is about 442 ft, and the flight time is about 3.3 sec.

_

In Exercises 37 and 38, we consider projectile motion when there is air resistance slowing down the flight.

Exercises 13.2 Integrattng Vedor-Valued Functtons Evaluate 1he integrals in Exercises 1-10.

1./,\t'1

+ 7j + (t + I)k] dt

2. [[(6 -

6t)i

+ 3Vtj +

(~)k] dt

~/4

3.

1 l3

[(sin t)i

+ (i + cos t)j + (sec't)k] dt

~/4

r

4. Jo 5.

[(secttant)i + (tant)j + (2sintcost)k]dt

[[~i + 5 ~ ti + itk]dt

13.2

Integrals of Vector Functions; Projectile Motion

721

6. J . 1 [ b i +

20. Finding muzzle speed Find the muzzle speed of a gun whose maximum range is 24.5 kIn.

7. J.'rlet'i + e-' j + k] dl

21. Flight time and height A projectile is fued with an initial speed of 500 mfsec at an angle of elevation of 45'.

V3,k]dl 01-1'1+1

a. Wheo and how far away will the projectile stti1re? 1n3

8.

1 1

b. How high overhead will the projectile he wheo it is 5 kIn downrange?

[Ie'i + e'j + lnlk]dl

9.

J.~/2[COSli -

10.

J.~[secti +

c. What is the greatest height reached by the projectile? sin21j + sin'lk]dl

22. Throwing a baseball A baseball is thrown from the stands 32 ft above the fIeld at an angle of 30' up from the horizontal. Wheo and how far away will the ball strike the grouod if its initial speed is 32 ftl sec?

tan'lj - Isinlk]dl

Initial Value Problems Solve the initial value problems in Exercises 11-16 for r as a vector function of I.

11. Differeotial equstion: Initial condition: 12. Differeotial equstion: Initial condition:

dr

dl

=

. • k -11 - IJ - I

r(O) = i

+ 2j +

~~ = (1801)i +

3k (1801 - 161')j

r(O) = lOOj

23. Firing golf balls A spring gun at grouod level fues a golf ball at an angle of 45'. The balllands 10 m away.

a. What was the ball's initial speed? b. For the saroe initial speed, fmd the two fIring angles that ma1re the range 6 m. 24. Beaming eleetrnns An electron in a TV tuhe is beanred horizontally at a speed of 5 X 10' mlsec toward the face of the tube 40 em away. About how far will the electron drop before it hits?

dr = 1(1 + 1)1/'. + -,. + _I_k dl2 leJ I+ 1

25. Equal-range fIring angles What two angles of elevation will enable a projectile to reach a target 16 kIn downrange on the sanre level as the gun if the projectile's initial speed is 400 mfsec?

Initial condition:

r(O) = k

26. Range and height versus speed

14. Differeotial equstion:

~~ = (13

Initial condition:

r(O) = i

13. Differeotial equstion:

15. Differeotial equstion: Initial conditions:

Initial conditions:

+j

dl' r(O) = lOOk and

I = 8i +

8j

t""O

d'r = -(i

dl'

+ j + k)

r(O) = lOi + IOj + 10k and dr dt

It=O --

a. Show that doubling a projectile's initial speed at a giveo lauoch angle multiplies its range by 4. b. By about what perceotage should you increase the initial speed to double the height and range?

d'r = -32k

dr dt 16. Differeotial equstion:

+ 41)i + Ij + 21'k

0

27. Verify the results given in the text (following Example 4) for the maximum height, flight time, and range for ideal projectile motion. 28. Colliding marbles The accoropanying figure shows an experimoot with two marbles. Marble A was launched toward marble B with launch angle a and initial speed Vo. At the saroe instant, marble B was released to fall from rest at R tan a units directly above a spot R units dowurange from A. The marbles were found to collide regardless of the value of Vo. Was this mere coincidence, or must this happen? Give reasons for your answer.

Motton Along a Straight Line 17. At time I = 0, a particle is located at the point (I, 2, 3). It travels in a straight line to the point (4, 1,4), has speed 2 at (I, 2, 3) and constant acceleration 3i - j + k. Find an equation for the position vector r(1) of the particle at time I. 18. A particle traveling in a straight line is located at the point (I, -1,2) and has speed 2 at time I = O. The particle moves toward the point (3, 0, 3) with constant acceleration 2i + j + k. Find its position vector r(l) at time I.

Projectile Motion Projectile flights in the following exercises are to be treated as ideal unless stated otherwise. All launch angles are assumed to be measured from the horizontal. All projectiles are assumed to be launched from the origin over a horizontal s'Wface unless stated otherwise. 19. Travel time A projectile is fIred at a speed of 840 mlsec at an angle of60'. How long will it take to get 21 kIn downrange?

29. Firing from ("""Yo)

x

=

Derive the equations

Xo

+ (vo cos a)t,

y = Yo + (vosina)1 -

kg,'

(see Equation (7) in the text) by solving the following initial value problem for a vector r in the plane.

722

Chapter 13: Vector-Valued Functions and Motion in Space d 2r dt 2

Differential equation:

-=

Initial conditions:

reO)

.

-gJ

= xoi

+ yoj

/

/

" / "

--

..- ..-"-

Pin

T

/ / /

:~ (0)

/

= (vo

cos a)i

45 ft

+ (vo sin a)j

30. Where trajectories crest For a projectile fired from the ground at launch angle a with initial speed vo, consider a as a variable and Vo as a fixed constant. For each a, 0 < a < 7T/2, we obtain a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse

1

Tee 1(

)1

369 ft NOT TO SCALE

33. Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6-ft-high net. It leaves the point of impact with an initial velocity of 35 ft/sec at an angle of 27° and slips by the opposing team untouched. a. Find a vector equation for the path of the volleyball.

where x

~

O.

b. How high does the volleyball go, and when does it reach maximum height? y

c. Find its range and flight time. d. When is the volleyball 7 ft above the ground? How far (ground distance) is the volleyball from where it will land?

Ellipse

,

,," /"

......

-----

I

/

I

\

/ Parabolic trajectory

" ............

--------'-'~, ~ -~~~~~--------------~~~-+x

o

31. Launching downhill

An ideal projectile is launched straight down an inclined plane as shown in the accompanying figure. a. Show that the greatest downhill range is achieved when the initial velocity vector bisects angle AOR.

b. If the projectile were fired uphill instead of down, what launch angle would maximize its range? Give reasons for your answer.

e. Suppose that the net is raised to 8 ft. Does this change things? Explain. 34. Shot put In Moscow in 1987, Natalya Lisouskaya set a women's world record by putting an 8 Ib 13 oz shot 73 ft lOin. Assuming that she launched the shot at a 40° angle to the horizontal from 6.5 ft above the ground, what was the shot's initial speed? 35. Model train The accompanying multiflash photograph shows a model train engine moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fired into the air by a spring in the engine's smokestack. The marble, which continued to move with the same forward speed as the engine, rejoined the engine 1 sec after it was fired. Measure the angle the marble's path made with the horizontal and use the information to fmd how high the marble went and how fast the engine was moving.

A

:u,au

32. Elevated green A golf ball is hit with an initial speed of 116 ft/ sec at an angle of elevation of 45° from the tee to a green that is elevated 45 ft above the tee as shown in the diagram. Assuming that the pin, 369 ft downrange, does not get in the way, where will the ball land in relation to the pin?

lLElSJ

-

36. Hitting a baseball under a wind gust A baseball is hit when it is 2.5 ft above the ground. It leaves the bat with an initial velocity of 145 ft/sec at a launch angle of 23°. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of -14i (ft/sec) to the ball's initial velocity. A 15-fthigh fence lies 300 ft from home plate in the direction of the flight. a. Find a vector equation for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height?

13.2 Integrals of Vector Functions; Proje
Theory and Examples 39. Establish the following properties of integrable ve
The C01l3tanJ Scalar Multiple Rule:

•• Has the batter hit a home run? Explain.

l'kr(t) dt

Projectile Motion wtth Llne.r Drag

The main force affeeting the motion of a projeetile, other than gravity, is air resistance. This slowing down force is drag force, and it acts in a dire
723

=

l'

k

(aoy sealar k)

r(t) dI

The Rulefor Negatives,

1'(

-r(t» dt =

-1'

r(t) dt,

is obtained by taking k = -I . b. The Sum and Difference Rules:

y

l'(r'(t) ± r2(t» dl = l'r'(t) dt ± l'r2(t) dt Velocity

"- The Co1l3tanJ Veclor Multiple Rules:

l'

C· r(t) dt

C

=

,1'

r(t) dt

(any constaot vector C)

and

l'

37. Linear drag Derive the eqnations

Vo • x = T (1 - e- ')cosa Va

g

-k.

-k

Y = - (1- e ')(sma) +-(1 - kt- e ') k k2 by solving the following initial value problem for a vector r in the

plaoe. DiiTercntial eqnation:

-d2r. = -gJ 2

Initial conditions:

r(0)

dl

=

- tv

= -gj

t t=O

= Vo = (vocos a)1

(any constaot vector C)

b. If u and r are both differentiable on [D, b], show that ur is differentiable on [a, b] aod that

d

)

dr

Iii (ur = u dt + r

du dt '

41. Antiderivati... of vector function.

+ ("
The drag coefficient k is a positive constant representing re~ sistance due to air density, Vo and a are the projectile's initial speed and laWlch angle, and g is the acceleration of gravity. 38. Hitting a baseball with line.. drag Consider the baseball problem in Example 5 when there is linear drag (see Exercise 37). Assume a drag coefficieotk = 0.12, hut no gust of wind. L

X l ' r(t) dl

•• Show that ur is continuous on [a, b] if u aod r are continuous on [a,b].

L

ddrl

=C

40. Product. of ""alar and vector function. Suppose that the scalar fun
dr - kdt

0

C X r(/) dl

From Exercise 37, find a ve
b. How high does the baseball go, and wbeo does it reach

max.imum height? e. Find the range aod flight time of the baseball d. When is the baseball 30 It high? How far (ground distaoce) is thc baseball from home plate at that height? •• A IO-It-high outfield fenee is 340 It from horne plate in the dire
Use Corollary 2 of the Meao Value Theorem for scalar fun<.. tions to show that if two vector fun
b. Use the result in part (a) to show that ifR(t) is aoy aotiderivative ofr(t) on I, then aoy other aotiderivative ofr onI equals R(t) + C for some constaot vector C.

42. The Fundamental Theorem of ClIlculu.

The Fundamental Theorem of Calculus for scalar fun
d1'

dl • r(T) dr = r(1)

at every point t of (a, b). Then use the conclusion in part (b) of Exercise 41 to show that if R is any aotiderivative of r on [a, b] thea

l'r(l) dl = R(b) - R(a).

724

Chapter 13: Vector-Valued Functions and Motion in Space

43. Bitting. baseball with linear drag under a wind gust Consider again the baseball problero in Example 5. This time assume a drag coefficient of 0.08 and an instantaneous gust of wind that adds a component of -17.6i (ft/seo) to the initial velocity at the instant the baseball is hit. L

Find a vector equation for the path of the baseball.

b. How high does the baseball go, and when does it reach maximum height? c. Find the range and flight time of the baseball.

e. A 20-ft-high outfield fence is 380 ft from home piate in the direction of the flight of the baseball. Has the batter hit a home run? If"yes." what change in the horizontal component of the ball's initial velocity would bave 1reptthe ball in the park? If'-"o:' what change would bave allowed it to be a homerun? 44. Heigbt versus time Show that a projectile attains three-quarters of its maximum height in balf the time it lakes to reach the maximum height

d. When is the baseball 35 fthigh? How far (ground distance) is the baseball from home piate at that height?

13.3 Basepoint

~

~1

r~

Arc Length in Space In this and the next two sections, we study the mathematical features of a curve's shape that describe the sharpness of its turning and its twisting.

/t ..,

"

Arc Length Along a Space Curve

4

s

FIGURE 13.12 Smooth corvos can be scaled like number lines, the coordioate of each point being its directed distance along the curve from a preselected base point.

One of the features of smooth space and plane curves is that they have a measurable length. This enables us to locate points along these curves by giving their directed distance s along the curve from some base point, the way we locate points on coordinate axes by giving their directed distance from the origin (Figure 13.12). This is what we did for plane curves in Section 11.2. To measure distance along a smooth curve in space, we add a z-term to the formula we use for curves in the plane.

DEFINITION The length of a smooth curve r(l) = X(I)i + Y(I)j + z(l)k, a ,;; I ,;; b, that is traced exactly once as I increases from I = a to I = b, is (1)

Just as for plane curves, we can calculate the length of a curve in space from any convenient parametrization that meets the stated conditions. We omit the proof. The square root in Equation (1) is lvi, the length of a velocity vector dr/dl. This enables us to write the formula for length a shorter way.

Arc Length FormuJa (2)

EXAMPLE 1 Agliderissoaringupwardalongthehelixr(l) = (cosl)i How long is the glider's path from I = 0 to I = 2fT?

+ (sint)j + tk.

726

Chapter 13: Vector-Valued Functions and Motion in Space Unlike Example 2, the arc length parametrization is generally difficult to find ana1ytically for a curve already given in tenns of some other parameter I. Fortunately, however, we rarely need an exact fonnula for s(t) or its inverse t(s).

HISTORICAL BIOGRAPHY

Josiah Willard Gibbs (1839--1903)

Speed on a Smooth Curve Since the derivatives beneath the radical in Equation (3) are continuous (the curve is smooth), the Fundamental Theorem of Calculus tells us that s is a differentiable function of I with derivative (4)

Equation (4) says that the speed with which a particle moves along its path is the magnitude of v, consistent with what we know. Although the base point P(/o) plays a role in defining s in Equation (3), it plays no role in Equation (4). The rate at which a moving particle covers distance along its path is independent of how far away it is from the base point. Notice that ds/dt > 0 since, by definition, Ivl is never zero for a smooth curve. We see once again that s is an increasing function of I.

Unit Tangent Vector We already know the velocity vector v = dr/dt is tangent to the curve r(/) and that the

vector v

T=N is therefore a unit vector tangent to the (smooth) curve, called the unit tangent vector (Figure 13.15). The unit tangent vector T is a differentiable function of I whenever v is a differentiable function of I. As we will see in Section 13.5, T is one of three unit vectors in a traveling reference frame that is used to describe the motion of objects traveling in three dimensions. EXAMPLE 3 FIGURE 13.15

Find the unit tangent vector of the curve

We rmd the unit tangent

vector T by dividing v by Iv I.

r(t)

= (3 cos t)i + (3 sin t)j + 12k

representing the path of the glider in Example 3, Section 13.2. Solution

In that example, we found v =

~~

= -(3 sin I)i + (3 cos I)j + 2/k

and Ivl =

y

Y9 +

4/2.

Thus,

T=~=Ivl (1,0) x

3sint

Y9 +

4/2

i+

3cost

Y9 +

j+

4/2

For the coonterclockwise motion

r(t)

= (cos I)i + (sin t)j

around the unit circle, we see that FIGURE 13.16

Counterclockwise motioo around the unit circle.

v = (-sint)i + (costH is already a unit vector, so T = v (Figure 13.16).

21

Y9 +

k. 4/2

•

13.3

z

The path segment during this time corresponds to one full turn of the helix (Figure 13.13). The length of this portion of the curve is

~

L =

1....... .t =7T I .....,

2 7T

"

: :

=

\

This is stands.

I

(

,,-

725

Solution

1

t=

Arc Length in Space

_---t----_

° I

r

.....

t_

1T

a

{27T

b

Ivldt

127T

= Jo

Y(-sint)2

+

(cost)2

+

(l)2dt

V2 dt = 21TV2 units oflength .

V2 times the circumference of the circle in the xy-plane over which the helix _

p-2"

/

(1,0,0)

l

y

If we choose a base point P( to) on a smooth curve C parametrized by t, each value of t determines a point P(t) = (x(t), y(t), z(t» on C and a "directed distance"

x

s(t) = ltlv(r)ldr, to

FIGURE 13.13

The helix in Example I, r(t) = (cos t)i + (sin t)j + tk.

z

measured along C from the base point (Figure 13.14). This is the arc length function we defined in Section 11.2 for plane curves that have no z-component. If t > to, s(t) is the distance along the curve from P(to) to P(t). If t < to, s(t) is the negative of the distance. Each value of s determines a point on C and this parametrizes C with respect to s. We call s an arc length parameter for the curve. The parameter's value increases in the direction of increasing t. We will see that the arc length parameter is particularly effective for investigating the turning and twisting nature of a space curve.

Arc Length Parameter with Base Point P(to)

s(t) = ltY[x1(r)]2

1___..:r_--->7 P(t)

+

[y'(r)]2

+

[zl(r)]2dr = l tlv(r)ldr

~

(3)

~

Base

7-~------POint

P(to)

x

FIGURE 13.14 The directed distance along the curve from P(to) to any point P(t) is

We use the Greek letter r ("tau") as the variable of integration in Equation (3) because the letter t is already in use as the upper limit. If a curve r(t) is already given in terms of some parameter t and s(t) is the arc length function given by Equation (3), then we may be able to solve for t as a function of s: t = t(s). Then the curve can be reparametrized in terms of s by substituting for t: r = r( t(s». The new parametrization identifies a point on the curve with its directed distance along the curve from the base point.

EXAMPLE 2 This is an example for which we can actually find the arc length parametrization of a curve. If to = 0, the arc length parameter along the helix r(t) = (cos t)i

+

(sin t)j

+ tk

from to to t is

s(t) = ltlv(r)ldr

Eq. (3)

to

= =

1

tV2dr

Value from Example I

V2t.

s/V2.

Solving this equation for t gives t = Substituting into the position vector r gives the following arc length parametrization for the helix:

r(t(s» = (cos

~} + (sin ~} + ~ k.

_

13.3 Arc Length in Space

727

The velocity vector is the change in the position vector r with respect to time t, but how does the position vector change with respect to arc length? More precisely, what is the derivative dr/ds? Since ds/dt > 0 for the curves we are considering, s is one-to-one and has an inverse that gives t as a differentiable function of s (Section 7.1). The derivative of

the inverse is

dt ds

1

1

ds/dt

Ivl'

This makes r a differentiable function of s whose derivative can be calculated with the Chain Rule to be

dr ds

=

dr dt dt ds

=

vTVl1 =

v

[Vj

=

(5)

T.

This equation says that dr/ds is the unit tangent vector in the direction of the velocity vector v (Figure 13.15).

Exercises 13.3 Finding Tangent Vectors and Lengths In Exercises 1-8, find the curve's unit tangent vector. Also, f'md the

length of the indicated portion of the curve. 1. r(t) = (2 cos t)i + (2 sin t)j + Vstk, 2. r(t) = 3. r(t) = 4. r(t) = 5. r(t) = 6. r(t) =

0 s t s 'IT

(6 sin 2t)i + (6 cos 2t)j + 5tk, 0 '" t '" 'IT ti + (2/3)t'/2k, 0 s tS 8 (2 + t)i - (t + I)j + tk, 0 S t S 3 (eos't)j + (sin't)k, 61'i - 2t'j - 3t'I<,

7. r(t) = (t cos t)i

0 S tS 'IT/2 I S t S 2

+ (t sin t)j + (2V2/3 )t'I'k, 0 S t S 'IT

(I sin t + cos t)i + (t cos t - sin t)j, 9. Find the point on the curve 8. r(t)

=

r(t)

= (5 sin t)i +

Yz '" t '" 2

(5 cos t)j + 12tk

at a distance 267T units along the curve from the point (0, 5, 0) in the direction of increasing arc length. 10. Find the point on the curve r(l)

=

(12 sin I)i - (12 cos t)j

+ 5tk

at a distance 13'IT units alnng the curve from the point (0, -12,0) in the direction opposite to the direction of increasing arc length. Arc Length Parameter In Exercises 11-14, find the arc length parameter along the curve from the point where t = 0 by evaluating the integral s =

!.'IV(T)ldT

from Equation (3). Then f'md the length of the indicated portion of the curve. 11. r(t) = (4 cos t)i + (4 sin I)j + 3tl<, 0 S t S 'IT/2

Theory and Examples 15. Arc length Find the length of the curve

r(t) = (Yzt)i + (YzI)j + (I - I')k from (0, 0, I) to (Yz, Yz, 0) .

16. Length of heIiI The length 2'IT Yz of the tum of the helix in Example I is also the length of the diagonal of a square 2'IT units on a side. Show how to obtain this square by cutting away and flatteoing a portion of the cylinder around which the helix winds. 17. Ellipse •• Show that the curve r(t) = (eost)i + (sint)j + (I - eost)k, o '" I '" 2'IT, is an ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find equations for the cylinder and plane. b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors at t = 0, 'IT/2, 'IT, and 3'IT/2.

c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for I = 0, 'IT/2, 'IT, and 3'IT/2 to your sketch. d. Write an integral for the length of the ellipse. Do not 1ry to evaluate the integral; it is nonelementary. D e. Numerical integrator Estimate the length of the ellipse to two decimal places. 18. Length is independent of parametrization To illustrate that the length of a smooth space curve does not depend on the parame1rization you use to compute it, calculate the length of one tum of the helix in Example I with the following parame1rizations. •• r(t) = (eos41)i + (sin4t)j + 41k,

12. r(t) = (cost + tsint)i + (sint - tcost)j, 'IT/2 '" t S 'IT 13. r(t) = (eteosl)i + (etsint)j + e'k, -In.4 S t S 0

b. r(t) = [cos (1/2)]i

14. r(t) = (I + 2t)i + (1 + 3t)j + (6 - 6t)k,

c. r(t) = (cos t)i - (sin t)j - tk,

-IS t S 0

Os I S 'IT/2

+ [sin (t/2)]j + (1/2)k, 0 S I S 4'IT -2'IT S tS 0

728

Chapter 13: Vector-Valued Functions and Motion in Space

19. The involute of a circle If a string wonod around a f"lxed circle is uowound while held taut in the plaoe of the circle, its end P traces ao involute of the circle. In the accompanying f"lgure, the circle in question is the circle x 2 + y2 = 1 and the tracing point starts at (I, 0). The UOv.
cost

=

+ I sin t, y

=

sint - teost,

t>

0

20. (Continuation ofExercise 19.) Find the unit taogent vector to the involute of the circle at the point p(x, y). Show that if u is a unit vector, then the arc length parameter a100g the line r(t) ~ Po + tu from the point Po(Xo, Yo, zo) where t ~ 0, is t itself.

21. Distance along a line

22. Use Siropsoo's Rule with n ~ 10 to approximate the length of arc ofr(t) ~ ti + t2 j + t 3k from the origin to the point (2, 4, 8).

of the point P(x, y) for the involute.

13.4

Curvature and Normal Vectors of a Curve In this section we study how a curve turns or bends. We look frrst at curves in the coordinate plane, and then at curves in space.

y

Curvature of a Plane Curve As a particle moves along a smooth curve in the plane, T ~ dr/ ds turns as the curve bends. Since T is a unit vector, its length remains constant and only its clirection changes as the particle moves along the curve. The rate at which T turns per unit of length along the curve is called the curvature (Figure l3.17). The traditional symbol for the curvature function is the Greek letter K (''kappa'').

s

-t Po

T

~~------------------~X

o

FIGURE 13.17 As P moves aloog the curve in the direction of increasing arc length, the unit tangent vector tums. The value ofldT/dsl atP is called the curvature of the curve at P.

DEFINITION

IfT is the unit vectur of a smooth curve, the curvature function

of the curve is

Ifl dT/ds Iis large, T turns sharply as the particle passes throughP, and the curvature at P is large. If IdT/ ds I is close to zero, T turns more slowly and the curvature at P is smaller.

729

13.4 Curvature and Normal Vectors of a Curve

If a smooth curve r(t) is already given in terms of some parameter t other than the arc length parameter s, we can calculate the curvature as K =

=

I~I = I~~ ~I

CbainRule

Ids~dtll~~1

I IdTI =TVlTt·

Formula for Calculating Curvature If r(t) is a smooth curve, then the curvature is K =

I~II~~I,

(1)

where T = vJlvl is the unit tangent vector.

Testing the definition, we see in Examples I and 2 below that the curvature is constant for straight lines and circles.

EXAMPLE 1 A straight line is parametrized by r(t) = C + tv for constant vectors C and v. Thus, r'(t) = v, and the unit tangent vector T = vJlvl is a constant vector that always points in the same direction and has derivative 0 (Figure 13.18). It follows that, for any value of the parameter t, the curvature of the straight line is K

EXAMPLE 2

I~II~~I = I~IIOI = o.

=

•

Here we find the curvature ofa circle. We begin with the parametrization

r(t) = (a cos t)i + (a sin t)j Along a straight line, T always points in the same direction. TIre curvature, IdT/ tis I, is zero (Example I). FIGURE 13.18

of a circle of radius a. Then, v =

~~

=

-(a sin t)i + (a cos t)j

Ivl = Y(-asint)'

+ (a cos t)2

=

W

= lal = a.

Since a > 0,

lal ~

a.

From this we find T =

~~

I~~I

I~I

= -(sint)i

+

(cost)j

= -(cost)i - (sint)j 2

= Ycos t

+ sin2 t

=

1.

Hence, for any value of the parameter t, the curvature of the circle is K

=

I~II~~I = ~(i) = ~ = ra~us·

•

Although the formula for calculating K in Equation (I) is also valid for space curves, in the next section we find a computational formula that is usually more convenient to apply.

730

Chapter 13: Vector-Valued Functions and Motion in Space Among the vectors orthogonal to the unit tangent vector T is one of particular significance because it points in the direction in which the curve is turning. Since T has constant length (namely, 1), the derivative dT/ds is orthogonal to T (Equation 4, Section 13.1). Therefore, if we divide dT/ tis by its length K, we obtain a unit vector N orthogonal to T (Figure 13.19).

DEFINITION At a point where a smooth curve in the plane is FIGURE 13.19 The vector dT/d>, normal to the curve, always points in the direction in which T is turning. The unit normal vector N is the direction of dT/d>.

K

oF 0, the principal nnit normal vector for

The vector dT/ tis points in the direction in which T turns as the curve bends. Therefore, if we face in the direction of increasing arc length, the vector dT/tis points toward the right if T turns clockwise and toward the left if T turns counterclockwise. In other words, the principal normal vector N will point toward the concave side of the curve (Figure 13.19). If a smooth curve r(t) is already given in terms of some parameter t other than the arc length parameter s, we can use the Chain Rule to calculate N directly:

dT/tis N = IdT/tisl

(dT/dt)(dt/tis) IdT/dtl Idt/tis I dT/dt ---

d.

1


IdT/dtl'

>

0 cancels.

This formula enables us to find N without having to 1md K and s first

Formula for Calculating N If r(t) is a smooth curve, then the principal unit normal is

dT/dt

(2)

N = IdT/dtl' where T = v / Iv Iis the unit tangent vector.

EXAMPLE 3

Find T and N for the circular motion

ret) Solution

= (cos 2t)i

+

(sin 2t)j.

We 1mt find T: v = -(2 sin 2t)i Ivl = V4sin2 2t

+

(2 cos 2t)j

+ 4cos2 2t

= 2

T =

I~I

~~

= -(2 cos2t)i - (2 sin2t)j

= -(sin2t)i

+

(cos2t)j.

From this we find

I~~I

2 = V4cos 2t

+ 4sin2 2t

= 2

13.4 Curvature and Normal Vectors of a Curve

731

and

dT/dt N = IdT/dtl = -(cos 2t)i - (sin 2t)j.

Eq. (2)

Notice that T' N = 0, verifYing that N is orthogonal to T. Notice too, that for the circular motion here, N points from r(t) towards the circle's center at the origin. •

Circle of Curvature for Plane Curves The circle of curvatore or osculating circle at a point P on a plane curve where K oF 0 is the circle in the plane of the curve that

1. is tangent to the curve at P (has the same tangent line the curve has) 2. has the same curvature the curve has at P 3. lies toward the concave or inner side of the curve (as in Figure 13.20). The radius of curvatore of the curve at P is the radius of the circle of curvatore, which, according to Example 2, is Radius of curvature = p = FIGURE 13.20 The osculating circle at P(x, y) lies toward the inner side of the curve.

k.

To find p, we find K and take the reciprocal. The center of curvatore of the curve at P is the center of the circle of curvature.

EXAMPLE 4

Find and graph the osculating circle of the parabola y = x 2 at the origin.

Solution We parametrize the parabola using the parameter t = x (Section 11.1, Example 5)

r(t) = ti

+ t 2j.

First we find the curvature of the parabola at the origin, using Equation (I): v = dr = i dt

+ 2tj

Ivl = VI + 4t 2 so that

From this we find

At the origin, t = 0, so the curvatore is

K(O)

=

IvtO)II~~ (0)1 + 2jl

=

-1-IOi

=

(1)\l0 0 2'+:-2="2 = 2.

VI

Eq. (\)

732

Chapter 13: Vector-Valued Functions and Motion in Space y

Therefore, the radius of curvature is 1/K = 1/2. At the origin we have t = 0 and T = i, so N = j. Thus the center of the circle is (0, 1/2). The equation of the osculating circle is therefore

----------~~~--~--~ x

FIGURE 13.21 The osculating circle for the parabola y = x 2 at the origin (Example 4).

You can see from Figure 13.21 that the osculating circle is a better approximation to the parabola at the origin than is the tangent line approximation y = O. •

Curvature and Normal Vectors for Space Curves If a smooth curve in space is specified by the position vector r(t) as a function of some parameter t, and if s is the arc length parameter of the curve, then the unit tangent vector T is dr/ ds = v/ Iv I. The curvature in space is then defined to be

IdTI

IdTI

1 K=ds=~dt

(3)

just as for plane curves. The vector dT/ds is orthogonal to T, and we define the principal unit normal to be 1 dT dT/dt N = Kds = IdT/dtl '

z

EXAMPLE 5

Find the curvature for the helix (Figure 13.22)

r(t) = (a cos t)i

I

_/~127Tb /

~ 7T t = 27T

I I I I I I

Solution

-..: ....

....

+ btk,

a, b

2::

0,

We calculate T from the velocity vector v:

T

2

=

+

(a cos t)j

+ bk

1 [-(a sin t)i a 2 + b2

FIGURE 13.22 The helix

K

=

*,I~~I

=

V

+ btk,

drawn with a and b positive and t (Example 5).

~

+

(a cos t)j

+ bk].

Then using Equation (3),

x

(a sin t)j

_IvI = V v

y

+

(a sin t)j

Iv I = Va 2 sin2 t + a 2 cos2 t + b 2 = Va 2 + b 2

\

t= ~

ret) = (a cos t)i

+

v = -(a sin t)i

,

(4)

0

1

a2

=

-_

a

2a

+

+ b2

b

IV

1

a2

+

b2

[-(a cos t)i - (a sin t)j] I

21-(cost)i - (sint)jl

2 a 2 Y( cos t )2 a +b

+ (.sm t )2 -_ 2 a 2 . a +b

From this equation, we see that increasing b for a fixed a decreases the curvature. Decreasing a for a fixed b eventually decreases the curvature as well. If b = 0, the helix reduces to a circle of radius a and its curvature reduces to 1/a, as it should. If a = 0, the helix becomes the z-axis, and its curvature reduces to 0, again as it should. •

13.4 Curvature and Normal Vectors of a Curve

EXAMPLE 6 Solution

733

Find N for the helix in Example 5 and describe how the vector is pointing.

We have

dT dt

V

[(a cos t)i

I

+ (a sin t)i]

a' + b' dT I = I Va' cos' t + a' sin' t = _r=;ja====.i I dt Va' + b' Va' + b' N =

dT/dt IdT/dt I

Eq. (4)

Vrca'..----+-b"' a = -(cost)i -

Example 5

V

I

a' + b'

[(a cos t)i + (a sin t)j]

(sintH.

•

Thus, N is parallel to the xy-plane and always points toward the z-axis.

Exercises 13.4 Pla.e Curves Find T, N, and" for the plane curves in Exercises 1-4. =

3. r(t) =

+

< t<

'1f/2

+ tj, -'1f/2 < t < (2t + 3)i + (5 - t')j

'1f/2

1. r(t) = ti

2. r(t)

(lncost)j,

-'1f/2

(In sec t)i

4. r(t) = (cost + tsint)i + (sint - tcost)j,

t> 0

5. A formula for the curvature of the graph of. function in the xy-plane = I(x) in the xy-pIane automatically has the parametrization x = x, y = I(x) , and the vector formula r(x) = xi + I(x)j. Use this fonnula to show that if1 is a twice-
a. The graph y

,,(x) =

I/"(x) I

[I + (/'(x»']'/2

.

b. Use the fonnula for" in part (a) to fmd the curvature of y = In (cos x), -'If/2 < x < '1f/2. Compare your answer

with the answer in Exercise 1. c. Show that the curvature is zero at a point of inflection. 6. A formula for the curvature of. parametrized plane curve

a. Show that the curvature of a smooth curve r(t) = I(t)i + g(t)j defmed by twice-differentiable functions x = I(t) and y = g(t) is given by the formula

yxl ,,= (x'liY+-y')'/2' The dots in the formula denote differentiation with respect to t, one derivative for each dot. Apply the formula to fmd the curvatores of the following curves. b. r(l) = Ii

+ (In sin t)j,

c. r(t) = [tan-I (sinht)]i

0


'1f

+ (lncosht)j.

7. Normals to plane curves •• Showthatn(t) = -g'(t)i + f'(t)j and -n(t) = g'(t)if'(t)j are both normal to the curve r(t) = I(t)i + g(l)j at the point (f(t), g(t)). To obtain N for a particular pIane curve, we can choose the one of n or -n from part (a) that points toward the concave side of the curve, and make it into a mtit vector. (See Figure 13.19.) Apply this method to fmd N for the following curves. b. r(t) = ti

c. r(t) =

+ eUj

v'4=f2 i + tj,

-2 '" t s 2

8. (Continuation ofExercise 7.) •• Use the method of Exercise 7 to fmd N for the curve r(t) Ii + (I/3)t' j when t < 0; when t > o.

=

b. Calculate N for t#'O directly from T using Equation (4) for the curve in part (a). Does N exist at t = O? Graph the curve and explain what is happening to N as t passes from negative to positive values.

Space Curves Find T, N, andK for the space curves in Exercises 9-16. 9. r(t) = (3 sin I)i 10. r(t)

+ (3 cos t)j + 4tk

= (cost + t sin t)i + (sin I - ICOSI)j + 3k + (e'sint)j + 2k = (6sin2t)i + (6cos2t)j + Sik = (t'/3)i + (t'/2)j, I> 0 = (COS'I)i + (sin't)j, 0 < t < '1f/2 = Ii + (a cosh (t/a))j, a > 0 = (cosht)i - (sinht)j + tk

11. r(t) = (e'cost)i

12. r(t) 13. r(t) 14. r(t) 15. r(t)

16. r(t)

More on Curvature 17. Show that the parabola y = ax', a #' 0, has its largest curvature at its vertex and has no minimmn curvature. (Note: Since the curvature ofa curve remains the same if the curve is translated orrotated, this result is true for any parabola.)

734

Chapter 13: Vector-Valued Functions and Motion in Space

18. Show that the ellipse x = a cos I, y = b sin I, a > b > 0, has its largest cmvature on its major axis and its smallest curvature on its minor axis. (As in Exercise 17, the same is true for any ellipse.) 19. Muimizing the curvature of. helli In Example 5, we found the curvature of the helix r(l) = (a cos I)i + (a sin I)j + blk (a, b '" 0) to be " = a/(a' + b'). What is the largest value" can have for a given value of b? Give reasons for your answer. 20. Total curvature We find the tutal curvature of the portion of a smooth curve that runs from s = So to s = 91 > So by integrating K from So to Sl • If the curve has some other parameter, say t, then the total curvature is

where to and t, correspond to So and L

s, .

Find the tota! curvatures of

The portion of the helix r(l) = (3 cos I)i 0:5 t:5 41T.

h. The parabola y = x 2 ,

-00

<X <

+ (3 sin I)j + Ik,

Plot the plane curve given in parametric or function form over the specified interval to see what it looks like.

b. Calculate the curvature K of the curve at the givoo value 10 using the appropriate formula from Exercise 5 or 6. Use the parametrization x = I andy = f(l) if the curve is giveo as a function y = f(x).

•• Find the unitnorma1 vectorN atto. Notice that the sigus of the componeots ofN depood on whether the unit tangeot vector T is turuing clockwise or counterclockwise at 1= 10. (See Exercise 7.) d. If C = ai + bi is the vector from the origio to the cooter (a, b) of the osculatiog circle, fmd the cooter C from the vector equation

C = r(lo)

22. Find an equation for the circle of curvature of the curve r(l) = (2101)i - [I + (I/I)]j, e-',;; I';; e', at the point (0, -2), where 1= 1.

D The formula

+ _(I) N(lo). K

10

The point P(xo, Yo) on the curve is giveo by the position vector r(lo) . e. Plot implicitly the equation (x - a)' + (y - b)' = I/K'ofthe osculatiug circle. Thoo plot the curve and osculatiug circle together. You may need to experimoot with the size of the viewing window, but be sure it is square.

+ (5 sin I)j, 0,;; I ,;; 2'1f, 10 = '1f/4 1)i + (sin3 1)j, 0,;; I ,;; 2'1f, 10 = '1f/4

27. r(l) = (3 cos I)i ,,(x)

=

28. r(l) =

1f"(x)1

[I + (f'(x»']3/2'

derived in Exercise 5, expresses the curvature K(X) of a twicedifferentiable plane curve y = f(x) as a function of x. Find the curvature function of esch of the curves in Exercises 23-26. Then graph f(x) together with K(X) over the given interval. You will fmd some surprises. =

L

00.

21. Find an equation for the circle of curvature of the curve r(l) = Ii + (sinl)j at the point ('1f/2, I). (The curve parametrizes the graph ofy = sin x in thexy-plane.)

23. y

COMPUTER EXPLORATIONS In Exercises 27-34 you will use a CAS to explore the osculatiug circle at a pointP on a plane curve where" oF O. Use a CAS to perform the following steps:

x 2,

-2:S; x

25. Y = sinx,

13.5

:5

2

O::s x:S 21T

24. Y

=

x 4/4,

26. Y = eX,

-2:5 x

:5

2

-1::s x:s 2

(COS

3

29. r(l) = l'i

+ (1 3

30. r(l) = (1 3

-

-

31)j,

21' - I)i

+

-4';; I ,;; 4,

hi, +

I I' (2 - 2 cos I)j,

31. r(l) = (21 - sin I)i + 10 = 3'1f/2 32. r(l) = (e-tcosl)i + (e-'sinl)j,

33. Y

=

34. y

=

10 = 3/5 -2';; I ,;; 5,

10 = I

0,;; I ,;; 3'1f,

0,;; I';; 6'1f,

10 = '1f/4

x, -2 ~ x ~ 5, Xo = 1 x(i - x)'/5, -I';; x ,;; 2, Xo = 1/2 x2 -

Tangential and Normal Components of Acceleration

z

If you are traveling along a space curve, the Cartesian i, j, and k coordinate system for representing the vectors describing your motion is not truly relevant to you. What is meaningful instead are the vectors representative of your forward direction (the unit tangent vector T), the direction in which your path is turning (the unit normal vector N), and the tendency of your motion to ''!wisf' out of the plane created by the.. vectors in the direction perpendicular to this plane (defined by the unit binormal vector B = T X N). Expressing the acceleration vector along the curve as a linear combination of this TNB frame of mutually orthogonal unit vectors traveling with the motion (Figure 13.23) is particularly revealing of the nature of the path and motion along it.

Po FIGURE 13.23 The TNB frame of mutually orthogonal unit vectors traveling along a curve in space.

The TNB Frame The binormal vector of a curve in space is B = T X N, a unit vector orthogonal to both T and N (Figure 13.24). Together T, N, and B derme a moving right-handed vector frame that plays a significant role in calculating the paths of particles moving through space. It is called the Frene! ("fre-nay") frame (after Jean-Frederic Frenet, 1816-1900), or the TNB frame.

13.5

Tangential and Normal Components of Acceleration

735

Tangential and Normal Components of Acceleration B

When an ohject is accelerated by gravity, brakes, or a comhination of rocket motors, we usually want to know how much of the acceleration acts in the direction of motion, in the tangential direction T. We can calculate this using the Chain Rule to rewrite v as

p

_ dr _ dr ds _ T ds dt'

v - dt - ds dt -

Then we differentiate hoth ends of this string of equalities to get

~ ( ds) _ d 22T s + ds dT a-dt-dt T dt -dt dtdt _ dv _

FIGURE 13.24 The veclon! T, N, and B (in that order) make a right-handed frame of mutually orthogooal unit vectors in space.

=

DEFINmON

2 d 2s T + ds (dT ds) dt dt ds dt

=

2 d 2s T + ds (KN ds) dt dt dt

If the acceleration vector is written as (1)

then and

(2)

are the tangential and normal scalar components of acceleration.

FIGURE 13.25 The tangential and normal components of acceleration. The acceleratioo a always lies in the plane ofT and N, orthogonal to B.

FIGURE 13.26 The tangential and normal components of the acceleration

Notice that the hinormal vector B does not appear in Equation (1). No matter how the path of the moving object we are watching may appear to twist and tum in space, the acceleration a always lies in the plane of T and N orthogonal to B. The equation also tells us exactly how much of the acceleration takes place tangent to the motion (d 2s/dt2) and how much takes place normal to the motion [K(ds/dt)'] (Figure l3.25). What information can we discover from Equations (2)? By definition, acceleration a is the rate of change of velocity v, and in general, both the length and direction ofv change as an o~ect moves along its path. The tangential component of acceleration aT measures the rate of change of the length ofv (that is, the change in the speed). The nOTmai component of acceleration aN measures the rate of change of the direction of v. Notice that the normal scalar component of the acceleration is the CUTVatore times the square of the speed. This explains wby you have to hold on when your car makes a sharp (large K), high-speed (large Ivl) tum. If you double the speed of your car, you will experience four times the normal component of acceleration for the same CUTVatore. If an object moves in a circle at a constant speed, d 2s/dt 2 is zero and all the acceleration points along N toward the circle's center. If the object is speeding up or slowing down, a has a nonzero tangential component (Figure l3.26). To calculate aN, we usually use the formula aN = Vlal 2 - aT', which comes from solving the equation Ia 12 = a' a = aT' + aN' for aN. With this formula, we can find UN without having to calculate K first.

Formula for Calculating the Normal Component of Acceleration

of an object that is speeding up as it moves

counterclockwise around a circle of radius p.

aN =

Vlal 2

-

aT'

(3)

736

Chapter 13: Vector-Valued Functions and Motion in Space

EXAMPLE 1

Without finding T and N, write the acceleration of the motion

r(t)

+ tsint)i + (sint -

= (cost

t >0

tcost)i,

in the form a = aTT + lINN. (The path of the motion is the involute of the circle in Figure 13.27. See also Section 13.3, Exercise 19.)

Solution

We use the rlISt of Equations (2) to rmd aT: v =

~;

=

(-sint + sint + tcost)i + (cost - cost + tsint)i

= (tcost)i

+ (tsint)j

Ivl =

Vt 2 cos2 t + t 2 sin2 t

aT =

dt

d

d Ivl = dt(t)

=

=

W

It I = t

=

t> 0 Eq. (2)

1.

Knowing aT, we use Equation (3) to find aN:

FIGURE 13.27 The tangential and normal components of the acceleration of

a = (cost - tsint)i

laI 2 =t2 +1 aN = Vlal 2

themotionr(l) = (cos I + I sin I)i + (sin I - I cos I)j, for I > O. If a string

After some algebra

~~~

wound around a ftxed circle is unwound while held taut in the plaoe of the circle, its endP traces an involute of the circle (Example I).

+ (sint + tcost)i

=

V(t2

al

-

+ 1) - (I)

=

W

t.

=

We then use Equation (I) to rmd a: a = aTT

+ lINN

=

(I)T + (t)N = T + tN.

•

Torsion How does dB/lis behave in relation to T, N, and B? From the rule for differentiating a cross product, we have

dB = d(T X N) = dT X N

lis

lis

+ T X dN

lis

ds'

Since N is the direction of dT/Iis, (dT/ds) X N = 0 and

dB = 0

lis

+T

X dN = T X dN.

ds

ds

From this we see that dB/lis is orthogonal to T since a cross product is orthogonal to its factors. Since dB/lis is also orthogonal to B (the latter bas constant length), it follows that dB/lis is orthogonal to the plane of B and T. In other words, dB/ds is parallel to N, so dB/lis is a scalar multiple ofN.1n symbols,

dB

ds

= -TN.

The negative sign in this equation is traditional. The scalar 'I' is called the torsion along the curve. Notice that

dB

ds • N

= -TN' N = -'1'(1) =

We use this equation for our next definition.

-'I'.

13.5

I

Rectifying

plane

Binormal

DEFINmON

Let B

=

Tangential and Normal Components of Acceleration

737

T X N. The torsion function of a smooth curve is

Normal plane

T=

Principal normal

< A

Unit tangent

FIGURE 13.28 The names of the three planes determined by T. N, and B.

dB -dS·N.

(4)

Unlike the curvature K, which is never negative, the torsion T may be positive, negative, or zero. The three planes determined by T, N, and B are named and shown in Figure 13.28. The curvature K = IdT / ds Ican be thought of as the rete at which the normal plane turns as the point P moves along its path. Similarly, the torsion T = -(dB/ds)' N is the rete at which the osculating plane turns about T as P moves along the curve. Torsion measures how the curve twists. Look at Figure 13.29. If P is a !min climbing up a curved track, the rete at which the headlight turns from side to side per unit distance is the curvature of the tmck. The rete at which the engine tends to twist out of the plane formed by T and N is the torsion. In a more advanced course it can be shown that a space curve is a helix if and only if it has constant nonzero curvature and constant nonzero torsion.

The tmsion

I ~ ~ l(dT/d,)I· "

/

The curvature at P

s increases

~-

p

I

8=0

FIGURE 13.29 Every moving body travels with a TNB frame that characterizes the geometty of its path of motion.

Computational Formulas The most widely used formula for torsion, derived in more advanced texts, is

x x T=

y )i

x' y

Iv X al

z z 'z' 2

(ifv X a .. 0).

(5)

The dots in Equation (5) denote differentiation with respect to t, one derivative for each dot. Thus, x ("x dot") means dx/dt, x ("x double dof') means d 2x/dt 2 , and x ("x triple dot'') means d 3x/dt 3 • Similarly, y = dy/dt, and so on. There is also an easy-to-use formula for curvature, as given in the following summary table (see Exercise 21).

738

Chapter 13: Vector-Valued Functions and Motion in Space

Computation Formulas for Curves in Space v

[Vj

Unit tangeot vector:

T

Principal unit normal vector:

dT/dt N = IdT/dtl

Bioormal vector:

B=TXN

=

Curvatore:

Torsion:

T =

Tangeotial and normal scalar componeots of acceleration:

a

dB

tis

d aT=dt

aN =

Y

z

x

ji "ji

z

x ·Z· 2 Iv x al

--'N =

= aTT

i:

+ aNN

lvl

Klvl2 = Vlal 2-

al

Exercises 13.5 Finding Tangential and Normal Components In Exercises 1 and 2, write a in the form • finding T and N. 2. r(l) = (I

+ 31)i + (I -

4. r(l) = S. r(l) = 6. r(l) =

+

14. r(l) =

(sin I)j

+

II<,

I = 0 0

1= '1f/4 I= 0

In Exercises !l-16 ofSectioo 13.4, you found T, N, aod K. Now, in the following Exercises !l-16, rmd B aod T for these space curves.

+

9. r(l) = (3 sin I)i 10. r(l) = (cos I

+

(3 cos I)j

Isinl)i

11. r(l) = (e' cos I)i

+

+ 41k

(sin I - Icosl)j

+ (e' sin I)j +

2k

Ik

17. The speedometer 00 your car reads a steady 35 mph. Could you be accelerating? Explain.

finding the TNB Frame

(sinl)j - I<,

+

at the given

In Exercises 7 aod 8, rmd r, T, N, aod B at the giveo value of I. Theo find equatioos for the osculating, nonnal, and rectifying plaoes at that value oft.

+ +

'1f/2

a > 0

Physical Applications

+ 21j + 12k, I = I (I cos I)i + (I sin I)j + 12k, I = 0 12; + (I + (I/3)1')j + (I - (I/3)1')k, (e' cos I)i + (e' sin I)j + Y2e'l<, I =

7. r(l) = (cosl)i

+ (a cosh (I/a»j,

16. r(l) = (cosh I)i - (sinh I)j

+ tlNN

l)i

8. r(l) = (cos I)i

13. r(l) = (1'/3)i 15. r(l) = Ii

2)j - 31k

In Exercises 3--6, write • in the form a = aTT value oflwithout rmding T aod N.

3. r(l) = (I

+ lINN without

+ (a sin I)j + blk

1. r(l) = (a cos I)i

+ (6 cos 21)j + 51k + (12/Z)j, I> 0 (cos'l)i + (sin'l)j, 0 < I <

12. r(l) = (6 sin 21)i = aTT

+ 3k

18. Can anything be said shout the acceleration of a particle that is

moving at a constant speed? Give reasons for your answer. 19. Can anything be said about the speed of a particle whose acceleration is always orthogonal to its velocity? Give reasons for your

answer. 20. An object of mass m travels aloog the parabola y = x 2 with • coostaot speed of \0 mtits/sec. What is the force 00 the object due to its acceleratioo at (0, O)? at (Z I 12, Z)? Write your answers in terms ofi aod j. (Remember Newton's law, F = ma.)

Theory and Examples 21. Vectnr furmula for curvature For a smooth curve, use Equatioo (I) to derive the curvature formula X K =

Iv al lvi' .

13.6 Velocity and Acceleration in Polar Coordinates 22. Show that a moving particle will move in a straight line if the normal component of its acceleration is zero.

v'k = 0 for all tin [a, b]. Show that h(t) = 0 for all tin [a, b]. (Hint: Start with a = d'r/dt' and apply the initial conditions in reverse order.). 28. A formula that calculate. T from B and v If we start with the defmition T = -(dB/tis)· N and apply the Chain RnIe to rewrite dB/ds as

23. A sometime shortcut to curvature If you already know I"" I and lvi, then the formula"" = Klvl' gives a convenient way to fmd the curvature. Use it to fmd the curvature and radius of curvature of the curve

r(t)

=

(cost + tsint)i + (sint - tcost)j,

(Take"" andlvl from Example I.)

= (:<0

dBdt dt ds

dB tis

t> O.

dB I dt lvi'

we arrive at the formula

24. Show that K and T are both zero for the line

r(t)

739

+ At)i + (YO + Htli + (zo + Ct)k

25. What can be said about the torsion of a smooth plane curve r(t) = j(t)i + g(t)j? Give reasons for your answer. 26. The torsion of a helix

The advantage of this formula over Equation (5) is that it is easier to derive and state. The disadvantage is that it can take a lot of work to evaluate without a computer. Use the new formula to fmd the torsion of the helix in Exercise 26.

Show that the torsion of the helix

r(t) = (a cos t)i + (a sin t)j + btk, a, b '" 0 is T = b/(a' + b'). What is the largest value given value of a? Give reasons for your answer.

T

COMPUTER EXPLORATIONS

can have for a

Rounding the answers to four decimal places, use a CAS to find v, ., speed, T, N, B, K, T, and the tangential and normal components of acceleration for the curves in Exercises 29-32 at the given values of t.

27. Differentiable curves with zero torsion lie in planes That a sufficiently differentiable curve with zero torsion lies in a plane is a special case of the fact that a particle whose velocity remains perpendicular to a fixed vector C moves in a plane perpendicular to C. This, in turn, can be viewed as the following result. Suppose r(t) = j(t)i + g(t)j + h(t)k is twice differentiable for all t in an interval [a, b], that r = 0 when t = a, and that

13.6

29. r(t) = (t cos t)i + (t sin t)j + tk, t = v'3 30. r(t) = (e'cost)i + (e'sint)j + e'k, t = 102 31. r(t) = (t - sint)i + (I - cost)j + v=tk, t = -3". 32. r(t) = (3t - t')i

+ (3t')j + (3t + t')k, t

= I

Velodty and Acceleration in Polar Coordinates In this section we derive equations for velocity and acceleration in polar coordinates. These equations are useful for calculating the paths of planets and satellites in space, and we use them to examine Kepler's three laws of planetary motion.

y

Motion in Polar and Cylindrical Coordinates When a particle at P(r, IJ) moves along a curve in the polar coordinate plane, we express its position, velocity, and acceleration in terms of the moving unit vectors u, = (cos lJ)i

+

Ue

(sin lJ)j,

= -(sinlJ)i

+

(coslJ)j,

(1)

~

shown in Figure 13.30. The vector u, points along the position vector OP, so r = ru,. The vector Ue, orthogonal to u" points in the direction of increasing IJ. We fmd from Equations (1) that

FIGURE 13.30 The length ofr is the positive polar coordinate r of the point P. Thus, Un which is rllrl, is also r/r. Equations (I) express u, and lItJ in terms ofiandj.

~~ = ~~e

-(sin lJ)i

+

(cos lJ)j = Ue

= -(cos lJ)i - (sinlJ)j = -u,.

When we differentiate u, and lie with respect to t to fmd how they change with time, the Chain Rule gives .

u,

=

du,.

di IJ

.

= lJue,

. Ue

=

due·

.

dlJ IJ = -lJu,.

(2)

740

Chapter 13: Vector-Valued Functions and Motion in Space

Hence, we can express the velocity vector in terms of u, and U9 as

Y

v

~~~~~------------~X

o

=

v

I

v=

+ rU. . +rlJ· Do· r = rUr

(i'u, + ro,) + (riJu. + rOll9 + riJ~).

When Equations (2) are used to evaluate 0, and ~ and the components are separated, the equation for acceleration in terms of u, and 119 becomes

= ;0,. + rOu".

s = (i' - riJ2)u,

+ (rii + 2riJ)U9.

To extend these equations of motion to space, we add zk to the right-hand side of the equation r = ru,. Then, in these cylindrical coordinates, we have

z

r = TDr

= rUr

rUr

See Figure 13.31. As in the previous section, we use Newton's dot notation for time derivatives to keep the formulas as simple as we can: 0, means du,/ dt, iJ means dfJ/ dt, and soon. The acceleration is s =

FIGURE 13.31 In polar coordinates, the velocity vector is

. d( ) .

r = dt

+ zk ru, + riJll9 + zk

r = rUr k

+ zk

u,

"

v =

(3)

s = (i' - rif)u, + (rO + 2riJ)u9 + zk. The vectors u" 119, and kmake a right-handed frame (Figure 13.32) in which urXuo=k,

FIGURE 13.32 Positioo vector and basic

Planets Move in Planes

unit vectors in cylindrical coordinates. Noticethatlrl #' rifz #' O.

Newton's law of gravitation says that if r is the radius vector from the center of a sun of mass M to the center of a planet of mass m, then the force F of the gravitational attraction between the planet and sun is F=_GmM~

Irl2 Irl (Figure 13.33). The mnnber G is the universal gravitational constant. Ifwe measure mass in kilograms, force in newtons, and distance in meters, G is about 6.6726 X 10-11 Nm2 kg-2. Combining the gravitation law with Newton's second law, F = mi', for the force acting on the planet gives ..

mr

..r

Cd) F=_GmM~

Irl' Irl

GmM r -Irf jrj'

GM = -- - rIrl21rl"

The planet is accelerated toward the sun's center of mass at all times. Since i' is a scalar multiple ofr, we have rXi'=O. From this last equation, t dd~X~=txt+rXi'=rxi'=~ o

FIGURE 13.33 The force of gravity is

~

directed aloog the line joining the centers of mass.

=

It follows that

rXt=C for some constant vector C.

(4)

13.6 Velocity and Acceleration in Polar Coordinates

741

Equation (4) tells us that r and t always lie in a plane perpendicular to C. Hence, the planet moves in a fixed plane through the center of its sun (Figure 13.34). We next see how Kepler's laws describe the motion in a precise way.

c= rxr

Kepler's First Law (Ellipse Law) Kepler's first law says that a planet's path is an ellipse with the sun at one focus. The eccentricity of the ellipse is FIGURE 13.34 A planet that obeys Newton's laws of gravitation and motion travels in the plane through the sun's center ofmassperpendiculllrtoC = r X t.

rovo2

and the polar equation (see Section 11.7, Equation (5» is

+ e)ro

(1 r

HISTORICAL BIOGRAPHY

Johaones Kepler

(5)

e= GM-I

= 1

+

(6)

ecosO"

Here Vo is the speed when the planet is positioned at its minimum distance ro from the sun. We omit the lengthy proof. The sun's mass Mis 1.99 X 1030 kg.

(1571-1630)

Kepler's Second Law (Equal Area Law) Planet

Kepler's second law says that the radius vector from the sun to a planet (the vector r in our model) sweeps out equal areas in equal times (Figure 13.35). To derive the law, we use Equation (3) to evaluate the cross product C = r X t from Equation (4):

-

C=rXt=rXv

X (hI, + rOu.)

= ru,

FIGURE 13.35 The line joining a planet to its sun sweeps over equal areas in equal

= rT(u, Xu,)

times.

=

+ r(rO)(u,

Eq. (3),

z~ 0

X De)

o

k

r(rO)k.

(7)

Setting t equal to zero shows that

C = [r(rO)lt~o k = rovok. Substitoting this value for C in Equation (7) gives 2'

rovok = r Ok,

r 2'()

or

=

rovo.

This is where the area comes in. The area differential in polar coordinates is

dA

=

2 dO !r 2

(Section 11.5). Accordingly, dA/ dt has the constant value

dA

dt

=

1

2'

2r 0

=

1

2 rovo.

(8)

So dA/ dt is constant, giving Kepler's second law.

Kepler's Third Law (Time-Distance Law) The time T it takes a planet to go around its sun once is the planet's orbital period. Kepler's third law says that T and the orbit's semimajor axis a are related by the equation

T2 a3

4,.' GM'

Since the right-hand side of this equation is constant within a given solar system, the ratio of T2 to a 3 is the same for every planet in the system.

742

Chapter 13: Vector-Valued Functions and Motion in Space Here is a partial derivation of Kepler's third law. The area enclosed by the planet's elliptical orbit is calculated as follows:

loT dA

Area =

Eq. (8)

I

2 Trovo.

=

If b is the semiminor axis, the area of the ellipse is 7Tab, so

T=

Planet

27fab TOVO

=

27fa 2

• ~

rovo vI - e-.

For any ellipse, b~.~

(9)

It remains only to express a and e in terms of ro, vo, G, and M. Equation (5) does this for e. For a, we observe that setting 0 equal to 7f in Equation (6) gives Sun Tmax

FIGURE 13.36 The length of1he major axis of the ellipse is 2a = TO + Tmax_

= TO

1+ e 1 - e"

Hence, from Figore 13.36, 2a = ro

+ r"""

2ro

= I _ e =

2roGM 2· 2GM - rovo

(10)

Squaring both sides of Equation (9) and substituting the results of Equations (5) and (lO) produces Kepler's third law (Exercise 9).

Exercises 13.6 In Exercises 1-5, f'md the velocity and acceleration vectors in terms of "'and .... d8 1.r~a(l-cos8) and dl~3

2. r = a sin 28 and : 3. r=e"

4. r = a(1

and

and 8 = I - e-'

and 8

~

21

6. Type of orbit For wbat values of Vo in Equatioo (5) is 1he orbit in Equatioo (6) a circle? An ellipse? A parabola? A hyperbola? 7. Circular orbits Show 1hat a plaoet in a circular orbit moves with a constant speed (Hint: This is a consequence of one of Kepler~ laws.)

Chapter

meots and limits for 1he validity of 1he equatioo

= 21

:~2

+ sin I)

S. r = 2 cos 41

8. Suppose 1hat r is 1he positioo vector of a particle moving aloog a plane curve aod dA/ dl is 1he rate at which 1he vector sweeps out area. Without introducing coordinates, and assuming the necessary derivatives exist, give a geometric argument based on incre-

9. Kepler'. third law Complete 1he derivatioo of Kepler's 1hird law (1he part following Equatioo (10)). 10. Find 1he length of 1he major axis of Ear1h's orbit using Kepler's 1hird law aod 1he fact ilia! Ear1h's orbital period is 365.256 days.

Questions to Guide Your Review

1. State 1he rules for differentiating and integratiog vector functioos. Give examples. 2. How do you defme aod calculate 1he velocity, speed, direction of motion, aod acceleratioo of a body moving along a sufficieotiy differentiable space curve? Give an example.

3. What is special about the derivatives of vector functions of constsnt length? Give an example.

4. Wbat are 1he vector and parametric equatioos for ideal projectile motioo? How do you fmd a projectile~ maximmn height, flight time, and raoge? Give examples.

Chapter 13 5. How do you derme and calculate the length of a segment of a smooth space curve? Give an example. What mathematical assumptions are involved in the def'mition?

6. How do you measure distance along a smooth curve in space from a preselected base point? Give an example. 7. What is a differentiable curve's unit tangent vector? Give an example. S. Derme curvature, circle of curvature (osculating circle), center of curvature, and radius of curvature for twice·differentiable curves in the plane. Give examples. What curves have zero curvature? Constant curvature?

Practice Exercises

743

10. How do you define N and K for curves in space? How are these quantities related? Give examples. 11. What is a curve's binonnal vector? Give an example. How is this vector related to the curve's torsion? Give an example.

12. What formulas are available for writing a moving body's acceleration as a sum of its tangential and normal components? Give an example. Why ntight one want to write the acceleration this way? What if the body moves at a constant speed? At a constant speed around a circle?

13. State Kepler's laws.

9. What is a plane curve's principal normal vector? When is it deImed? Which way does it point? Give an example.

Chapter

:g)

Practice Exercises

Motton in the Plane In Exercises 1 and 2, graph the curves and sketch their velocity and acceleration vectors at the given values of t. Then write • in the form a ~ aTT + "NN without finding T andN, and rmd the value OfK at the given values of I.

b. Find v and. at I your sketch.

~

0, 1,2, and 3 and add these vectors to

c. At any given time, what is the forward speed of the topnlOst point of the wheel? Of C? y

1. r(l) ~ (4cosl)i + (V2sinl)j,

I~ Oand'Il/4 2. r(l) ~ {V3 sec I)i + (V3tanl)j, t ~ 0 3. The position of a particle in the plane at time tis r ~

I

,

+

I

,

v'J+t2' v'J+t21.

c 7rt

r

Find the particle's highest speed.

1

4, Suppose r(l) ~ (e' cos l)i + (e' sin I)j. Show that the angle between r and a never changes. What is the angle? 5. Finding cnrvatnre At point P, the velocity and acceleration of a particle moving in the plane are v ~ 3i + 4j and • ~ Si + ISj. Find the curvature of the particle's path atP. 6. Find the point on the curve y = eX where the curvature is greatest 7. A particle moves around the unit circle in the xy-plane. Its position at time I is r ~ xi + yj, where x and y are differentiable functions of I. Find dy/dl ifv' i ~ y. Is the motion clockwise or

counterclockwise?

Projectile Motion 11, Shot put A shot leaves the thrower's hand 6.S ft above the ground at a 4S' angle at 44 ft/sec. Where is it 3 sec later?

12, Javelin A javelin leaves the 1hrow\:r's hand 7 ft above the grouod at a 4S' angle at 80 ft/sec. How high does it go? 13. A golf ball is hit with an initial speed Vo at an angle a to the horizontal from a point that lies at the foot of a straight-sided hill that

8, You send a message through a pneumatic tube that follows the

x'

(distance in meters). At the point (3, 3), v' i ~ 4 curve 9y ~ and.'i ~ -2. Find the values ofv' j and.'j at (3, 3).

is inclined at an angle to the horizontal, where

0<



1T

< a < 2'

9, Characterizing circular motion A particle moves in the plane so that its velocity and position vectors are always orthogonal. Show that the particle moves in a circle centered at the origin. 10, Speed along. cycloid A circular wheel with radius I ft and center C rolls to the right along the x-axis at a half-torn per second. (See the accompanying figore.) At time I seconds, the position vector of the point P on the wheel's circumference is r ~ (1TI - sin1TI)i

+ (I

- COS1TI)j.

a. Sketch the curve traced by P during the interval 0 '" I '" 3.

Show that the ball lands at a distance

2vo'cosa . ( 2

goos

sma-q"

)

measured up the face of the hill. Hence, show that the greatest range that can be achieved for a given Vo occurs when a = (<1>/2) + (1T/4), Le., when the initial velocity vector bisects the angle between the vertical and the hill.

744

D 14.

Chapter 13: Vector-Valued Functions and Motion in Space

Javelin In Potsdam in 1988, Petra Felke of (then) Baat Germany set a women's world record by throwing ajavelin 262 ft 5 in. L

Assuming that Felke launched the javelin at a 40° angle to the horizonta16.5 ft above the ground, what was the javelin's initial speed?

30. Radius of curvature Show that the radius of curvature of a twice-differentiable plane curve r(t) = !(t)i + g(t)j is given by the formula p =

.;:2+»2 Vii2

+ y2 _ .2'

where

• =

d

-'1'.;:2 + »2 dt .

b. How high did the javelin go? 31. An alternative dermition of curvature in the plane An alternative definition gives the curvature of a sufficiently differentiable plane curve to be Id/dsl, where is the angle between T and i (Figure 13.37a). Figure 13.37b shows the distance s measured col.IDterclockwise around the circle x 2 + y2 = a 2 from the point (a, 0) to a pointP, along with the angle at P. Calculate the circle's curvature using the alternative definition. (Hint: = 8 + '1f/2.)

Motion In Space Find the lengths of the curves in Exercises 15 and 16. 15. r(t) = (2 cos t)i + (2 sin t)j + t 2k, 0 '" t '" '1f/4 16. r(t) = (3 cos t)i + (3 sin t)j + 2t 3/2k, 0 '" t '" 3 In Exercises 17-20, fmd T, N, B, 1(, andT at the given value oft.

17. r(t) =

~ (1

+ t)3/2i +

18. r(t) = (e'sin2t)i

~ (1 -

t)3/2j +

t

tk,

+ (e'cos2t)j + 2e'k,

19. r(t) = Ii + te2'j,

Y

t = 0

t = 0

t = 1n2

20. r(t) = (3 cosh 2t)i + (3 sinh 2t)j + 6tk,

fi

~--------------~x

o

(a)

t = In 2 y

In Exercises 21 and 22, write a in the form a = aTT + IlNN at t = 0 without fmding T and N. 21. r(t) = (2 + 3t + 3t 2)i + (4t + 4t 2)j - (6 cos t)k

22. r(t) = (2 + tli + (t + 2t 2)j + (1 + t 2)k 23. Find T, N, B, K, and T as functions of t if

r(t)

= (sin t)i +

(v'2 cos t)j + (sin t)k. (b)

24. At what times in the interval 0 '" t '" '1f are the velocity and acceleration vectors of the motion r(t) = i + (5 cos t)j +(3 sin t)k orthogonal?

25. The position of a particle moving in space at time t r(t) = 2i + (4Sint) + (3 -

~

0 is

ir)k.

Find the first time r is orthogonal to the vector i - j. 26. Find equations for the osculating, normal, and rectifYing planes of the curve r(t) = ti + t 2j + t 3k at the point (I, I, I). 27. Find parametric equations for the line that is tangent to the curve r(t) = e'i + (sint)j + 1n(1 - t)katt = O. 28. Find parametric equations for the line tangent to the helix r(t) (v'2cost)i + (v'2sint)j + tkatthepointwheret

=

= '1f/4.

FIGURE 13.37 Figures for Exercise 31. 32. The vi.... from Skylsb 4 What percentage of Earth's surface area could the astronauts see when Skylab 4 was at its apogee height, 437 Ian above the surface? To find out, model the visible surface as the surface generated by revolving the circular arc GT, shown here, about the y-axis. Then carry out these steps: 1. Use sintilar triangles in the figure to show that yo/6380 6380/(6380 + 437). Solve for Yo. 2. To four significant digits, calculate the visible area as

VA =

f802'1fX~1 +

(:Y

dy.

3. Express the result as a percentage of Earth's surface area. y

Theory and Examples 29. Synchronous curve.

S (Skylab)

By eliminating a from the ideal projectile

equations

437 {G

~ V (6=3=80"')2'_-y'2 C

x

T

x = (vocosa)t,

y = (vosina)t - tgt2, 6380

show that x 2 + (y + gt 2/2)' = vo2t 2. This shows that projectiles launched simultaneously from the origin at the same initial speed will, at any given instant, all lie on the circle of radius vot centered at (0, -gt2/2), regardless of their launch angle. These circles are the synchronous curves of the launching.

--~----~-----+~x

o

=

746

Chapter 13: Vector-Valued Functions and Motion in Space b. Showthat dU r

(jif

k

~ 110

and

dU(J

(jif

~

-u,.

c. Assuming that the necessary derivatives with respect to t exist, express v = i' and a = r in terms ofuy• Us. k, r, and jJ. 10. Conservation of IUlgular momentom Let r(l) denote the position in space of a moving object at time t. Suppose the force acting on the object at time t is

F(I) 9 r

(r. (J, 0)

L Show that u,., 110, and k, in this order, form a right-handed frame afunit vectors.

Chapter

~ -Ir(~) I' r(l) ,

where c is a constant. In physics the angular momentum of an object at time 1 is defmed to be L(I) ~ r(l) X mV(I) , where m is the mass of the object and v(l) is the velocity. Prove that angular momentum is a conserved quantity; i.e., prove that L(I) is a constant vector, independent of time. Remember Newton's law F ~ rna. (This is a calculus problem, not a physics problem.)

Technology Application Projects

Mathematica/Maple Module: RadIJr Trilcking ofa Moving Object Visualize positi~ velocity, and acceleration vectors to analyze motion.

PtmJlltetric allll Polar Equation. with a Figun Skat., Visualize positi~ velocity, and acceleration vectors to analyze motion.

Moving in Thr~~ DiMensions Compute distance traveled, speed, curvature, and torsion for motioo along a space curve. Visualize and compute the tangential, normal, and

binormal vectors associated with motion along a space curve.

Chapter 13 Additional and Advanced Exercises

Chapter

00

Additional and Advanced Exercises

Appllcattons 1. A frictionless particle P, starting from rest at time t point (a, 0, 0), slides down the helix r(8)

= (a cos 8)i + (a sin 8)j + b8k

(a, b

= 0 at the

Motton In Polar and Cylindrical Coordinates 3. Deduce from the orbit equation r=

> 0)

under the influence of gravity, as in the accompanying figure. The 8 in this equation is the cylindrical coordinate 8 and the helix is the curve r = a, Z = b8, (} O?:: 0, in cylindrical coordinates. We asswne 8 to be a differentiable function of t for the motion. The law of conservation of energy tells us that the particle's speed after it has fallen straight down a distance z is v'2g., where g is the constant acceleration of gravity.

a. Find the angu!arvelocity dOjdt when 0 = 2".. b. Express the particle's 0- and z-coordinates as functions of t.

c. Express the tangential and normal components of the velocity drjdt and acceleration d'rjdt' as functions of t. Does the acceleration have any nonzero component in the direction of the binormal vector B?

x

r - _ . /p

The helix //r=a.z=b8

(I + .)r. 1 + ecos8

that a planet is closest to its sun when 8 at that time.

D 4.

= 0 and show that r = r.

A Kepler equation The problem oflocating a planet in its orbit at a given time and date eventually leada to solving "Kepler" equations of the form

I(x) = x - I - ksinx = O.

a. Show that this particular equation has a solution between x = 0 and x = 2. b. With your computer or calculator in radian mode, use Newton's method to fmd the solution to as many places as you can. 5. In Section 13.6, we found the velocity of a particle moving in the plane to be v

y

745

= i i + yj

=

rUr + rlJutJ.

•• Express x and j> in terms of r and riJ by evaluating the dot produets v· i and v· j.

r

x

b. Express and r iI in terms of and j> by evaluating the dot products v • Ur and v· lie.

6. Express the curvatore of a twice-differentiable curve r = 1(0) in the polar coordinate plane in terms of 1 and its derivatives. 7. A slender rod through the origin of the polar coordinate plane r0tates (in the plane) about the origin at the rate of 3 rad/min. A beetle starting from the point (2, 0) crawls along the rod toward the origin at the rate of I in/min.

Positive z-axis .-------- points down.

1 z

2. Suppose the curve in Exercise I is replaced by the conical helix r = aO, z = bO shown in the accompanying figure.

a. Express the angular velocity dOjdt as a function ofO. h. Express the distance the particle travels along the helix as a function of o.

•• Find the beetle's acceleration and velocity in polar form when it is halfway to (I in. froru) the origin.

D b.

To the nearest tenth of an inch, what will be the length of the path the beetle has traveled by the time it reaches the origin? 8. Arc length in cylindrical coordinate. a. Show that when you express ds' = dx' + ely' + dz'in terms of cylindrical coordinates, you get ds' = dr' + r'do' + dz'. b. Interpret this result geometrically in terms of the edges and a

diagonal of a box. Sketch the box. c. Use the result in part (a) to fmd the length of the curve r = e 6,z = e 6,O::S O::s 8InS. Conical helix T

1

= a8,z =

be

Conez= ~r

Positive z-axis points down.

z

9. Unit vecton for position and motion in cylindrical coordinates When the position of a particle moving in space is given in cylindrical coordinates, the unit vectors we use to describe its position and motion are

u, = (cosO)i + (sinO)j,

... = -(sinOli +

(cosO)j,

and k (see accompanying figure). The particle's position vector is then r = rUr + zk, where r is the positive polar distance coordinate of the particle's position.

14 PARTIAL DERIVATIVES OVERVIEW Many functions depend on more than one independent variable. For instance, the volume of a right circular cylinder is a function V = 7Tr 2h of its radius and its height, so it is a function V(r, h) of two variables rand h. In this chapter we extend the basic ideas of single variable calculus to functions of several variables. Their derivatives are more varied and interesting because of the different ways the variables can interact. The applications of these derivatives are also more varied than for single-variable calculus, and in the next chapter we will see that the same is true for integrals involving several variables.

14.1

Functions of SeveraL VariabLes In this section we define functions of more than one independent variable and discuss

ways to graph them. Real-valued functions of several independent real variables are defined similarly to functions in the single-variable case. Points in the domain are ordered pairs (triples, quadruples, n-tuples) of real numbers, and values in the range are real numbers as we have worked with all along.

DEFINITIONS Suppose D is a set ofn-tuples of real numbers (x], X2, " " xn). A real-valued function f on D is a rule that assigns a unique (single) real number w =

f(Xl , X2, " . ,

xn)

to each element in D. The set D is the function's domain. The set of w-values taken on by f is the function's range. The symbol w is the dependent variable of f, and f is said to be a function of the n independent variables Xl to X n . We also call the x/ s the function's input variables and call w the function's output variable.

If f is a function of two independent variables, we usually call the independent variables X and y and the dependent variable z, and we picture the domain of f as a region in the xy-plane (Figure 14.1). Iff is a function of three independent variables, we call the independent variables x, y, and z and the dependent variable w, and we picture the domain as a region in space. In applications, we tend to use letters that remind us of what the variables stand for. To say that the volume of a right circular cylinder is a function of its radius and height, we might write V = f(r , h). To be more specific, we might replace the notation f(r, h) by the formula that calculates the value of V from the values of rand h, and write V = 7Tr 2h. In either case, rand h would be the independent variables and V the dependent variable of the function.

747

748

Chapter 14: Partial Derivatives

f f(a, b) --~~~~--~~~Z

o

FIGURE 14.1

f(x,y)

An arrow diagram for the functionz = j(x, y).

As usual, we evaluate functions defined by fannulas by substituting the values of the independent variables in the fannula and calculating the corresponding value of the 2 + y2 + z2 at the point dependent variable. For example, the value of f(x,y,z) = (3,0,4) is

Yx

f(3,0,4) =

Y(W + (0)2 + (4)2 = V25 =

5.

Domains and Ranges In defining a function of more than one variable, we follow the usual practice of excluding x 2,y cannot inputs that lead to complex numbers or division by zero. If f(x,y) = 2 be less than x . If f(x,y) = I/(xy),xy cannot be zero. The domain of a function is assumed to be the largest set for which the defining rule generates real numbers, unless the domain is otherwise specified explicitly. The range consists of the set of output values for the dependent variable.

Yy -

EXAMPLE 1 (a) These are functions of two variables. Note the restrictions that may apply to their domains in order to obtain a real value for the dependent variable z. Function z =

Yy -

x2

I z=xy z

=

sinxy

Domain

Range

y ~ x2

[0, 00)

xy '" 0

(-00,0) U (0, 00)

Entire plane

[-1,1]

(b) These are functions of three variables with restrictions on some of their domains.

Function w=

Yx 2 + y2 + z2

w=

~---'o---~

W

I

x2

+ y2 + z2

= xyInz

Domain

Range

Entire space

[0, 00)

(x,y,z) '" (0,0,0)

(0, 00)

Half-space z

>

0

(-00,00)

•

Functions of Two Variables Regions in the plane can have interior points and boundary points just like intervals on the real line. Closed intervals [a, b] include their boundary points, open intervals (a, b) don't include their boundary points, and intervals such as [a, b) are neither open nor closed.

14,1 Functions of Several Variables

749

DEFINmONS A point (xo, YO) in a region (set) R in Ibe .!J'-plane is an interior point of R if it is Ibe center of a disk of positive radius !bat lies entirely in R (Figure 14.2). A point (xo, YO) is a boundary point of R if every disk centered at (xo, YO) contains points !bat lie outside of R as well as points !bat lie in R. (The boundary point itself need not belong to R.) The interior points of a region, as a set, make up Ibe interior of Ibe region. The region's boundary points make up its boundary. A region is open ifit consists entirely of interior points. A region is closed if it contains all its boundary points (Figure 14.3).

(a) Interior point

y

y

y

, 1

R

«.' yol

o

/

o

(b) Boundary point

FIGURE 14,2 Interior points and boundary points of a plane regioo R, An interior point is necessarily a point of R. A

Ix

((x,y) 2 + y2< 1) Open unit disk. Every point an interior point

{(x,Y)lx2+y2~ 1)

Boundary of unit disk. (The unit circle.)

Ix

((x,y) 2 + y2,,;; 1) Closed unit disk. Contains all boundary points.

boundary point of R need not belong to R. FIGURE 14.3

Interior points and boundary points of 1he unit disk in 1he plane.

As wilb a half-open interval of real numbers [a, b), some regions in Ibe plane are neiIber open nor closed. If you start wilb Ibe open disk in Figure 14.3 and add to it some of but not all its boundary points, Ibe resulting set is neilber open nor closed. The boundary points !bat are Ibere keep Ibe set from being open. The absence of Ibe remaining boundary points keeps Ibe set from being closed.

DEFINmONS A region in Ibe plane is bounded if it lies inside a disk offIxed radius. A region is unbounded if it is not bounded.

y Interior points, wherey - x 2 > 0

/

Examples of bounded sets in Ibe plane include line segments, triangles, interiors of triangles, rectangles, circles, and disks. Examples of unbounded sets in Ibe plane include lines, coordinate axes, Ibe graphs of functions defined on infinite intervals, quadrants, half-planes, and Ibe plane itself.

EXAMPLE 2

Describe Ibe domain oflbe function f(x,y) =

Yy -

x 2•

Since f is defmed only where y - x 2 ;" 0, Ibe domain is Ibe closed, unbounded region shown in Figure 14.4. The parabola y = x 2 is Ibe boundary oflbe domain. The points above Ibe parabola make up Ibe domain's interior. •

Solutton

------~~~~--~----~x

-1

0

FIGURE 14,4 The domain of f(x, y) in Example 2 consists of1he shaded regioo and its bounding parabola.

Graphs, Level Curves, and Contours of Functions of Two Variables There are two standard ways to picture Ibe values of a function fix, y). One is to draw and label curves in Ibe domain on which f has a constant value. The olber is to sketch Ibe surface z = fix, y) in space.

750

Chapter 14: Partial Derivatives

DEFINITIONS The set of points in the plane where a function f(x, y) has a constant value f(x, y) = c is called a level curve of f. The set of all points (X, y, f(x, y) in space, for (X, y) in the domain of f, is called the graph of f. The graph of f is also called the surface z f(x,y) .

=

The surface z =f(x,y) = 100 - x 2 _ y2 is the graph off.

f(x, y) = 75

X'

X

f(x,y) = 51 (a typical level curve in the function's domain)

y

EXAMPLE 3 Graph f(x , y) = 100 - x 2 - y2 and plot the level curves f(x,y) f(x, y) = 51, and f(x, y) = 75 in the domain of f in the plane.

SoLution The domain of f is the entire xy-plane, and the range of f is the set of real numbers less than or equal to 100. The graph is the paraboloid z = 100 - x 2 - y2, the positive portion of which is shown in Figure 14.5. The level curve f(x, y) = 0 is the set of points in the xy-plane at which f(x,y)

The contour curve f(x , y) = 100 - x 2 - y2 = 75 is the circle x 2 + y2 = 25 in the plane z = 75.

f1~"

\

= 75

~

-P~

z = 100 - x 2 _ y2

><~

=

100 - x 2 - y2 = 0,

or

x 2 + y2

=

100,

which is the circle of radius 10 centered at the origin. Similarly, the level curves f(x,y) = 51 andf(x,y) = 75 (Figure 14.5) are the circles

x

FIGURE 14.5 The graph and selected level curves ofthe function f(x, y) in Example 3.

= 0,

f(x,y)

=

100 - x 2 - y2 = 51,

or

f(x,y)

=

100 - x 2 - y2 = 75,

or

x 2 + y2

=

49

The level curve f(x, y) = 100 consists of the origin alone. (It is still a level curve.) If x 2 + y2 > 100, then the values of f(x, y) are negative. For example, the circle x 2 + y2 = 144, which is the circle centered at the origin with radius 12, gives the constant value f(x, y) = -44 and is a level curve of f. • The curve in space in which the plane z = c cuts a surface z = f(x, y) is made up of the points that represent the function value f(x, y) = c. It is called the contour curve f(x, y) = c to distinguish it from the level curve f(x, y) = c in the domain of f. Figure 14.6 shows the contour curve f(x,y) = 75 on the surface z = 100 - x 2 - y2 defined by the function f(x, y) = 100 - X2 - Y 2. The contour curve lies directly above the circle x 2 + y2 = 25, which is the level curve f(x,y) = 75 in the function's domain. Not everyone makes this distinction, however, and you may wish to call both kinds of curves by a single name and rely on context to convey which one you have in mind. On most maps, for example, the curves that represent constant elevation (height above sea level) are called contours, not level curves (Figure 14.7).

Functions of Three Variables

/

x The level curvef(x, y) = 100 - x 2 - y2 = 75 is the circle x 2 + y2 = 25 in the xy-plane.

FIGURE 14.6 A plane z = c parallel to the xy-plane intersecting a surface z = f(x, y) produces a contour curve.

In the plane, the points where a function of two independent variables has a constant value f(x,y) = c make a curve in the function's domain. In space, the points where a function of three independent variables has a constant value f(x, y, z) = c make a surface in the function's domain. DEFINITION The set of points (x, y, z) in space where a function of three independent variables has a constant value f(x, y, z) = c is called a level surface of f.

Since the graphs of functions of three variables consist of points (x,y, z, f(x,y, z)) lying in a four-dimensional space, we cannot sketch them effectively in our three-dimensional frame of reference. We can see how the function behaves, however, by looking at its threedimensional level surfaces. EXAMPLE 4

Describe the level surfaces of the function

f(x,y, z) = Yx 2

+ y2 + z2 .

14.1

Functions of Several Variables

751

FIGURE 14.7 Contours on Mt. Washington in New Hampshire. (Reproduced by permission from the Appalachian Mountain Club.) FIGURE 14.8 The level surfaces of f(x,y,z) = Yx 2 + y2 + z2 are concentric spheres (Example 4).

(a) Interior point

(b) Boundary point

FIGURE 14.9 Interior points and boundary points of a region in space. As with regions in the plane, a boundary point need not belong to the space region R.

Solution The value of f is the distance from the origin to the point (x, y, z). Each level surface x 2 + Y 2 + Z 2 = c, c > 0, is a sphere of radius c centered at the origin. Figure 14.8 2 + y2 + z2 = 0 shows a cutaway view of three of these spheres. The level surface consists of the origin alone. We are not graphing the function here; we are looking at level surfaces in the function's domain. The level surfaces show how the function's values change as we move through its domain. If we remain on a sphere of radius c centered at the origin, the function maintains a constant value, namely c. If we move from a point on one sphere to a point on another, the function's value changes. It increases if we move away from the origin and decreases if we move toward the origin. The way the values change depends on the direction we take. The dependence of change on direction is important. We return to it in Section 14.5. •

Y

Yx

The definitions of interior, boundary, open, closed, bounded, and unbounded for regions in space are similar to those for regions in the plane. To accommodate the extra dimension, we use solid balls of positive radius instead of disks.

DEFINITIONS A point (xo, Yo, zo) in a region R in space is an interior point of R if it is the center of a solid ball that lies entirely in R (Figure 14.9a). A point (xo, Yo, zo) is a boundary point of R if every solid ball centered at (xo, Yo , zo) contains points that lie outside of R as well as points that lie inside R (Figure 14.9b). The interior of R is the set of interior points of R. The boundary of R is the set of boundary points of R. A region is open if it consists entirely of interior points. A region is closed if it contains its entire boundary.

Examples of open sets in space include the interior of a sphere, the open half-space z > 0, the first octant (where x, y, and z are all positive), and space itself. Examples of closed sets in space include lines, planes, and the closed half-space z 2: O. A solid sphere

752

Chapter 14: PartiaL Derivatives with part of its boundary removed or a solid cube with a missing face, edge, or comer point is neither open nor closed. Functions of more than three independent variables are also important. For example, the temperature on a surface in space may depend not only on the location of the point p(x, y, z) on the swface but also on the time 1 when it is visited, 80 we would write T = /(x, y, z, t).

computer Graphing Three-dimensional graphing programs for computers and calculators make it possible to graph functions of two variables with only a few keystrokes. We can often get information more quickly from a graph than from a formula. w

EXAMPLE 5 The temperature w beneath the Earth's surface is a function of the depth x beneath the swface and the time t of the year. If we measure x in feet and t as the number of days elapsed from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function w = 008(1.7 X 10-21 - O.2x)e-o·~.

FIGURE 14.10 Thisgrapbshowsthe seasonal variation of the temperatuIe below ground as a fraction of surface

tmnpenrtmc (Exampl' 5).

(The temperature at 0 ft is scaled to vary from +1 to -1, so that the variation atx feet can be interpreted as a fraction of the variation at the surface.) Figure 14.10 shows a graph of the function. At a depth of 15 ft, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 25 ft, there is almost no variation during the year. The graph also shows that the temperature 15 ft below the surface is about half a year out of phase with the surface temperature. When the temperature is lowt:st on the surface (late January, say), it is at its highest 15 ft below. Fifteen feet below the ground, the seasons are reversed. _ Figure 14.11 shows computer-generated graphs of a number of functions of two variables together with their level curves.

y

x

(a) z - sin x

FIGURE 14.11

+ 2 sin]

(b) l - (4x2

+ ]2.}o!-r-r

Computer-generated graphs and level curves of typical functions of two variables.

14.1 Functions of Several Variables

753

Exercises 14.1 Domain, Range, and Level Curves

21. f(x,y)

~

23. f(x,y)

~ \I'

In Exercises 1-4, find the specific function values.

+ xy'

1. f(x,y) ~ x'

b. f(-I, I)

a. f(O,O) c. f(2,3) 2. f(x,y) ~ sin (xy)

d. f(-3,-2)

16 -

25. f(x,y) ~ In (x' 27. f(x,y)

f( -3, ;;) d. f( -f, -7)

22. f(x,y) ~ y/x'

xy I

x' _ y'

+ y')

~ sin-1 (y -

x)

+ y' -

24. f(x,y)

~ \1'9 - x' - y'

26. f(x,y) ~ .""",'+Y') 28. f(x,y)

~ tan-I (~)

b.

29. f(x,y) ~ In (x'

a. f(3, -1,2)

b.

Exercises 31-36 show level curves for the functions graphed in (aHf) on the following page. Match each set of curves with the appropriate function.

c. f(O,-t,O)

d. f(2,2, 100)

a. f(2,');) c.

f(,."t)

3. f(x,y,z)

~

1) 30. f(x,y) ~ In (9 - x' - y')

Matching Surfaces with Level Curves

x-y -,--, y +z

f(l,t, -t)

31.

32. y

4. f(x,y,z) ~ \1'49 - x' - y' -

a. f(O,O,O)

z'

I(~t

b. f(2,-3,6)

c. f(-1,2,3)

d.

( 4 5 6)

f \1'2' \1'2' \1'2

~¥~'

In Exercises 5-12, fmd and sketch the domain for each function.

5. f(x,y) ~ \l'y - x - 2 6. f(x,y) ~ In (x'

+ y'

(x - I)(y

- 4)

+ 2)

7. f(x,y) ~ (y _ x)(y _ x') sin (xy)

8. f(x,y)

~,

9. f(x,y)

~ cos- 1 (y

x

+y

10. f(x,y) ~ In(xy

,

33.

34. y

- 25

- x')

+x - y -

1)

11. f(x,y) ~ \I'(x' - 4)(y' - 9)

I

12. f(x,y) ~ In (4 _ x' _ y') In Exercises 13-16, fmd and sketch the level curves f(x, y) ~ c on the same set of coordinate axes for the given values of c. We refer to these level curves as a contour map.

13. f(x,y)

~

x

+y

- I, c

~

-3, -2, -I, 0, 1,2,3

14. f(x,y) ~ x' + y', c ~ 0, 1,4,9, 16,25 15. f(x,y)

~

xy,

c

~

-9, -4, -I, 0, 1,4,9

16. f(x,y) ~ \1'25 - x' - y',

c ~ 0, 1,2,3,4

In Exercises 17-30, (a) fmd the function's domain, (b) fmd the function's range, (oj describe the function's level curves, (dJ fmd the boondary of the function's don3ain, (eJ determine if the don3ain is an open reo gion, a closed region, or neither, and (f) decide if the don3ain is boonded or unbounded

17. f(x,y)

~

y - x

19. f(x,y) ~ 4x'

+ 9y'

18. f(x,y) ~ ~ 20. f(x,y) ~ x' - y'

35.

36. y

1111

IIlIJi'

754

Chapter 14: PartiaL Derivatives

L

L

b.

,.

Functions of Two variables Display the values of the :functions in Exercises 37--48 in two ways: (a) by sketching the surfacez = f(",y) and (b) by drawing an assortment oflevel curves in the function's dmnain. Label each level curve with its function value.

Vx

37. f(x,y) = yl

38. f(x,y) =

39. f(x,y) _ xl + yl 41. f(x,y) _ xl - y

40. f(x,y) _ V"x,cc+c-y" 42. f(x,y) _ 4 - "l _ yl

43. f(x,y) _ 4"l ... f(x.y) - 1

47. f(",y) =

+ yl

44. f(x,y) - 6 - 2x - 3y

-Iyl

... f(x.y) - 1

yGx ,i'+'---y,OC+-04

48. f(x,y) =

-Ixl -Iyl

y,,2 + yl

4

FInding Level Curves d.

In Exercises 49-52, fmd an equation for and sketch the graph of the level curve of the function f(X, y) that passes through the given point.

,

4•• !(x.y) = 16 - x' - y'.

(20.0)

50. !(x.y) -

Vx'=l.

(1.0)

51. f(",y) =

y" + yl

3,

52. f(x,y) - x

(3,-1)

2y -"

+ Y + I' (-1, I)

y

Sketching Level Surfaces In Exercises 53-60, sketch a typical level surface for the :functioo..

••

+ yl + z2 54. f(x,y,z) _ 10(,,2 + yl + z2) 56. f(x,y,z) = z S5. f(x,y,z) =" + z 58. f(x,y,z) _ y2 + z2 57. f(x,y,z) _,,2 + yl 53. f(x,y,z) _ ,,2

59. f(x,y,z) - z - x2 _ y2 60. f(x,y,z) = (x 2/25) + (y2/16)

+ (z2/9)

FInding Level Surfaces In Exercises 61-64, fmd an equation for the level surface of the function through the given point.

61. f(",y,z) = ~ - loz, 62. f(x,y,z) _

1o(x 2

(3, -1, I)

+ y+z2), (-1,2,1)

14.2 63. g(x,y,z)

= Yx 2 + y2 + Z2, (I, -I, v'2)

64. g(x,y, z)

= ,_

Limits and Continuity in Higher Dimensions

71. f(x, y) = sin (x + 2 cos y), - 2". '" X '" 2"., -2". "'y '" 2"., P(".,,,.) 72. f(x,y) = e(z'Ly) sin (x 2 + y2), 0'" X '" 2".,

x-y+z + y-z ,(1,0, -2)

.LA.

-2". '" Y ",,,.,

In Exercises 65--{j8, rmd and sketch 1he domain of f. Then find an equation for 1he level curve or surface of1he function passing through 1he given point.

65. f(x,y)

=

it. ~Y. 00

66.g(x,y,z)=~

(x

1 1~ dO

vl-OZ

Y

68. g(x,y,z) =

%

I

+t

(0, I)

+ {' ~, (0, I, \13) Jo 4 - OZ

COMPUTER EXPLORATIONS Use a CAS to perfonn 1he following steps for each of 1he functioos in Exercises 6!1--72. L

Plot 1he surface over the given rectangle.

c. Plot 1he level curve of f through 1he given point.

69. f(x, y) = x sin ~

+ Y sin 2>:,

0 '" x '" 5".,

0 '" Y '" 5".,

14.2

- (cosy)Yx 2

0 '" x '" 5".,

+ z2

=

2

some parameter interval I, you can sometimes describe surfaces in

u,

0:5 u:S; 2,

78.x=ucosv, y=usinv, z=v, o :5 v :s; 2'1T

0::;; u:s; 2,

=

u cos v, y :s 21T

o :5 v =

=

u sin v, z

(2 + cosu) cos v, y

o :s; u :s; 2'17",

70. f(x,y) = (sinx)(cosy)e Vbl"/', o '" Y '" 5"., P(41T, 4".)

I

space wi1h a1riple of equations x = f(u, v),y = g(u, v),z = h(u, v) defined on some parameter rectangle a ::;; u :s; h, c :s; v :s; d. Many computer algebra systeros permit you to plot such surfaces in parametric mode. (parametrized surfaces are discussed in detail in Sectioo 16.5.) Use a CAS to plot 1he surfaces in Exercises 77-80. Also plot sevetallevel curves in 1he xy-plaoe.

79. x

P(3"., 3".)

(~)

74. x 2 + z2 = I

I

=

Parametrized Surfaces Just as you describe curves in 1he plaoe parametrically wi1h a pair of equations x = f(t),y = g(t) dermed on

77. x

b. Plot several level cutves in 1he rectangle.

+ z2)

+ y2 - 3z 2 =

76. sin

(ln4,ln9,2)

_ r.--:;,'

%

75. x

+ y)' I"

y

67. f(x, y) =

(1,2)

P("., -".)

Use a CAS to plot 1he implicitly dermed level surfaces in Exercises 73-76. 73. 4ln (x 2 + y2

n.z

11=0

755

O:s;

V

=

=

(2 + cosu) sin v, z

=

sinu,

:s; 2'17"

80. x = 2cosucosv, y = 2 cos u sin v, z o :5 u :s; 2'1T, 0:5 v :s; '11'

=

28inU,

Limits and Continuity in Higher Dimensions This section treats limits and continuity for multivariable functions. These ideas are analogous to limits and continuity for single-variable functions, but including more independent variables leads to additional complexity and important differences requiring some new ideas.

Limits for Functions of Two Variables If the values of f(x, y) lie arbitrnrily close to a f"lxed real number L for all points (x, y) suff"lciently close to a point (xo, yo), we say that f approaches the limit L as (x, y) approaches (xo, yo). This is similar to the informal definition for the limit of a function of a single variable. Notice, however, that if (xo, Yo) lies in the interior of j's domain, (x, y) can approach (xo, yo) from any direction. For the limit to exist, the same limiting value must be obtained whatever direction of approach is taken. We illustrate this issue in several examples following the detmition.

756

Chapter 14: Partial Derivatives

DEFINITION We say that a function /(x, y) approaches the limit L as (x, y) approaches (xo, yo), and write lim

(x, y)-(x. Yo)

/(x,y) = L

if, for every number E > 0, there exists a corresponding number 6 for all (x, y) in the domain of /,

<

I/(x,y) - LI

o<

whenever

E

>

0 such that

V(x - xo)' + (y - Yo)' < 6.

The detmition oflimit says that the distance between /(x, y) and L becomes arbitrarily small whenever the distance from (x, y) to (xo, Yo) is made sufficiently small (but not 0). The definition applies to interior points (xo, Yo) as well as boundary points of the domain of /, although a boundary point need not lie within the domain. The points (x, y) that approach (xo, Yo) are always taken to be in the domain of f. See Figure 14.12.

y

f

--'----I • }-------> z o L-ELL+E

FIGURE 14.12 In the limit defInition, 8 is the radius ofa disk centered at (xo, yo). For all points (x, y) within this disk, the function values f(x,y) lie inside the corresponding interval (L - E, L + E).

As for functions of a single variable, it can be shown that lim

X=Xo

lim

y=Yo

lim

k=k

(x, y)--+(xo •.vo)

(x, y) - (xo, Yo) (x, y) - (xo, Yo)

(any number k).

For example, in the first limit statement above, /(x, y) = x and L = Xo. Using the definition of limit, suppose that E > 0 is chosen. Ifwe let 6 equal this E, we see that

o<

V(x -

XO)2

+ (y - YO)2 < 6

= E

implies

V(x - xo)' <

Ix - xol < I/(x,y) - xol <

E

(x - xo)' s (x - xo)'

E

W~lal

E

x

~

+ (y -

Yo)'

!(x,y)

That is, I/(x,y) -

xol <

E

whenever

o<

V(x - xo)' + (y - Yo)' < 6.

14.2

Limits and Continuity in Higher Dimensions

757

So a Il has heen found satisfying the requirement of the def"mition, and f(x,y) = lim x = Xo. lim (x.y)~(Xo,y,)

(x,y)~(x"y,)

All with single-variable functions, the limit of the sum of two functions is the sum of their limits (when they both exist), with similar results for the limits of the differences, constant multiples, products, quotients, powers, and roots.

THEOREM l-Properties of Limits of Functions of Two variables

The fol-

lowing rules hold if L, M, and k are rea! numbers and lim

(x, y)~(x. Yo)

f(x,y) = L

1. Sum Rule:

lim

and

(x,y)~(Xo,y,)

4. Product Rule:

lim

(f(x,y) - g(x,y)) = L - M

lim

kf(x,y) = kL

lim

(any number k)

(f(x,y)' g(x,y)) = L· M

lim

f(x,y)

(x,y)~(x"y,) g(x,y)

L

M# 0

= M'

[f(x,y)]n = Ln, n a positive integer

lim

6. Power Rule:

= L

(x,y)-(x"y,) (x, y)-(x. y,)

5. Quotient Rule:

+M

(f(x,y)

(x,y)~(x"y,)

3. Constant Multiple Rule:

+ g(x,y))

lim

(x,y)~(x"y,)

2. Difference Rule:

g(x,y) = M.

(x,y)-(x"y,)

lim

7. Root Rule:

-.;:/f(x,y) =

-.;:Ii

= Llfn,

(x, y)-(x"y,)

n a positive integer, and if n is even, we assume that L

> O.

While we won't prove Theorem I here, we give an informal discussion ofwby it's true. If (x, y) is sufficiently close to (XO,yo), then f(x, y) is close to Land g(x, y) is close to M (from the informa! interpretation of limits). It is then reasonable that f(x, y) + g(x, y) is closetoL + M;f(x,y) - g(x,y) is close to L - M; kf(x,y) is close to kL; f(x,y)g(x,y) is close toLM; and f(x,y)/g(x,y) is close to L/Mif M # O. When we apply Theorem I to polynomials and rational functions, we obtsin the useful result that the limits of these functions as (x, y) ---> (xo, Yo) can be ca!culated by evaluating the functions at (xo, yo). The only requirement is that the rational functions be def"med at (xo, yo).

EXAMPLE 1 In this example, we can combiae the three simple results following the limit def"mition with the results in Theorem I to calculate the limits. We simply substitule the x and y values of the point being approached into the functional expression to f"md the limiting value. ( ) a

(h)

x - xy + 3

lim

(x,y)~(O,l) x2y + 5xy - y3 lim

(x,y)~(3,

EXAMPLE 2

-4)

Vx 2

+ y2

=

0 - (0)(1) + 3 = -3 (0)2(1) + 5(0)(1) - (1)3

V(3)2

+ (-4)2

=

v'25 =

-

xy

Find

x2 lim

(x,y)-(O,O)

.Vr .r· x - VY

5

•

758

Chapter 14: Partial Derivatives

Vx - vY

Solution Since the denominator approaches 0 as (x,y) --+ (0, 0), we cannot use the Q1!otient Rule from Theorem 1. If we multiply numerator and denominator by + Vy, however, we produce an equivalent fraction whose limit we can find:

Vx

(x 2 - .!JI)( Vx + vY) (x, y)~(O,O) Vx - vY (x,y)~(O,O) (Vx - vY)(Vx + vY) x2

lim

-

n,

lim

-J

x(x lim

Y)(Vx + vY) x

(x,y)-(O,O)

=

lim

(x,y)~(O,O)

Algebra

y

Cancel the nonzero fa
x(Vx + vY)

Known limit values

We can cancel the factor (x - y) because the path y = x (along which x - y = 0) is not in the domain of the function

• · d 'lim Fm

EXAMPLE 3

x

(x, y)~(O,O)

4.!J12· ·· 2 I f It exJsts. +Y

2

Solution We first observe that along the line x = 0, the function always has value 0 when y '" O. Likewise, along the line y = 0, the function has value 0 provided x '" O. So if the limit does exist as (x, y) approaches (0, 0), the value of the limit must be O. To see if this is true, we apply the dermition of limit. Let E > 0 be given, but arbitrary. We want to rmd all> 0 such that 4xy2

1x2 + y2

-

01 <

E

o<

whenever

Yx 2 + y2 < Il

or

o<

whenever Since y2 ,;; x 2

Yx 2

+ y2 <

Il.

+ y2 we have that

41xly2 2 2';; 41xl X +Y

=

So if we choose Il = e/4 and let 0

• r? ".-------. 4vx 2 ,;; 4Yx 2 + y2.

< Yx 2 + y2 <

Il, we get

It follows from the definition that 4 2 = O. 2 (x,y)-(O,O) x + y2 lim

.!JI

•

14.2 Limits and Continuity in Higher Dimensions

EXAMPLE 4

Jff(x,y)=~,

do..

lim

(x,y) ...... (O,. 0)

759

f(x,y)exist1

Solution The domain off does not include the y-axis, so we do not consider any points (x,y) where x - 0 in the approach toward the origin (0, 0). Along the x-axis, the value of the function is I(x, 0) - 0 for all x"'" O. So if the limit does exist as (x,y) ~ (0, 0), the value of the limit must be L - O. On the other hand, along the line y - x, the value of the function is. I(x, x) - x/x - 1 for all x .;. O. That is, the function f approaches the value 1 along the line y - x. This means that for every disk of ramus B centered at (0, 0), the disk. will contain points (x, 0) on the x-axis where the value of the function is 0, and also points (x, x) along the line y - x where the value of the function is 1. So no matter how small we choose B as the radius of the disk in Figme 14.12, there will be points within the disk. for which the function values differ by 1. Therefore, the limit cannot exist because we can take E to be any nwnber less than 1 in the limit definition and deny that L - 0 or I, or any other real nwnber. The limit does not exist because we have different limiting values along different paths approaclring the point (0, 0), •

Continuity & with functions of a single variable, continuity is dermed in terms of limits.

DEFINmON 1. 2,

3,

f

A function f(x,y) is COntinUOUI at the point ~o,Y.) if

is defined at (X'o,Yo), lim f(x, y) exists,

(x,y)-
lim

(x,y)-(ro.)'o)

f(x,y) - f(xo,,,),

A function is continuoul if it is continuous at every point of its domain. (.)

AB with the definition of limit, the def'mition of continuity applies at boundary points well as interior points of the domain off. The only requirement is that each point (x,y) near (xo, YO) be in the domain off. A consequence of Theorem 1 is that algebraic combinatioos of continuous functions are continuous at every point at which all the functions involved an: dermed. This means that sums, differences, constant multiples, products, quotients, and powers of continuous functions are continuous where defined In particular, polynomials and rational functions of two variables are continuous at every point at which they are defmed.

y

88

EXAMPLE 5

Show that

(b)

FIGURE14.13

I(;t,Y) -

(a) The graph of

,2xy " :x + y { 0,

(x,y)

~ (0,0)

(x,y) = (0,0),

The fimction is continuous at every point except the origin. (b) The values off are different constants along each line

Y -

1rU, X '"

0 (Example 5).

f(x,y)

=

{xl;y,' 0,

(x,y) " (0,0) (x,y)

=

(0,0)

is continuous at every point except the origin (Figure 14.13).

Solution The function! is continuous at any point (x,y) #- (0,0) because its values lD'e then given by II. rational function ofX' and y and the limiting value is obtained by substituting the values ofX' andy into the functional expression.

760

Chapter 14: Partial Derivatives

At (0, 0), the value of f is defined, but f, we claim, has no limit as (x, y) -- (0, 0). The reason is that different paths of approach to the origin can lead to different results, as we now see. For every value of m, the function f has a constant value on the "punctured" line y = mx, x i=- 0, because

f(x,y) I

=

y=mx

2 2.xy 2

X

+Y

2m

I y=mx

Therefore, f has this number as its limit as (X, y) approaches (0, 0) along the line: lim (x,y)->(O,O)

f(x,y) =

lim (x , y)->(O,O)

[f(X,Y) I

y=mx

] = 1 +2mm 2 .

aJongy=mx

This limit changes with each value of the slope m. There is therefore no single number we may call the limit of f as (x, y) approaches the origin. The limit fails to exist, and the function is not continuous. _ Examples 4 and 5 illustrate an important point about limits of functions of two or more variables. For a limit to exist at a point, the limit must be the same along every approach path. This result is analogous to the single-variable case where both the left- and right-sided limits had to have the same value. For functions of two or more variables, if we ever find paths with different limits, we know the function has no limit at the point they approach. z

1\vo-Path Test for Nonexistence of a Limit Ifa function f(X,y) has different limits along two different paths in the domain of f as (x, y) approaches (xo, Yo), then lim(x, y)->(xO,Yo) f(x, y) does not exist.

EXAMPLE 6

Show that the function

f(x,y) = y

x

2x 2y X

4

+

Y

2

(Figure 14.14) has no limit as (x,y) approaches (0, 0). SoLution The limit cannot be found by direct substitution, which gives the indeterminate form 0/0. We examine the values of f along curves that end at (0, 0). Along the curve y = kx 2 , X i=- 0, the function has the constant value

(a)

y

o

Therefore,

o

~_....:..O~X

lim

(x, y)->(O,O)

aJongy=kx2

f(x,y) =

lim

(x,y)->(O ,O)

[f(x,y) I

y=kx2

] =~. 1 + k2

This limit varies with the path of approach. If (x, y) approaches (0, 0) along the parabola y = x 2, for instance, k = 1 and the limit is 1. If (x, y) approaches (0, 0) along the x-axis, k = 0 and the limit is o. By the two-path test, f has no limit as (x, y) approaches (0, 0). _ (b)

FIGURE 14.14 (a) The graph of f(x,y) = 2x 2y/(x 4 + y2). (b) Along each path y = kx 2 the value of f is constant, but varies with k (Example 6).

It can be shown that the function in Example 6 has limit 0 along every path y = mx (Exercise 53). We conclude that

Having the same limit along all straight lines approaching (xo, Yo) does not imply a limit exists at (xo, yo).

14.2

Limits and Continuity in Higher Dimensions

761

Whenever it is correctly dermed, the composite of continuous functions is also continuous. The only requirement is that each function be continuous where it is applied. The proof, omitted here, is similar to that for functions of a single variable (Theorem 9 in Section 2.5).

Continuity of Composites If ! is continuous at (xo, YO) and g is a single-variable function continuous at !(XO,yo), then the composite function h = g , f dermed by h(x,y) = g(f(x,y)) is continuous at (xo, yo).

For example, the composite functions

xy

e x-y ,

In (1

cos--, 2

+

X

1

+ X'y2)

are continuous at every point (x, y).

Functions of More Than Two Variables The definitions of limit and continuity for functions of two variables and the conclusions about limits and continuity for sums, products, quotients, powers, and composites all extend to functions of three or more variables. Functions like

In (x

+ Y + z)

ysinz x-I

and

are continuous throughout their domains, and limits like

e x +z

lim P-(I,O,-I) z2

+

cos

1

v':ry

(_1)2

+

2'

cosO

where P denotes the point (x, y, z), may be found by direct substitution.

Extreme Values of Continuous Functions on Closed, Bounded Sets The Extreme Value Theorem (Theorem I, Section 4.1) states that a function of a single variable that is continuous throughout a closed, bounded interval [a, b] takes on an absolute maximum value and an absolute minimum value at least once in [a, b]. The same holds true of a function z = !(x, y) that is continuous on a closed, bounded set R in the plane (like a line segment, a disk, or a filled-in triangle). The function takes on an absolute maximum value at some point in R and an absolute minimum value at some pointinR. Similar results hold for functions of three or more variables. A continuous function w = !(x, y, z), for example, must take on absolute maximum and minimum values on any closed, bounded set (solid bailor cube, spherical shell, rectangular solid) on which it is dermed. We will learn how to rmd these extreme values in Section 14.7.

Exercises 14.2 Limits Mth Two variables Find the limits in Exercises 1-12. 3x 2 - y2 + 5 2, lim ....£.. lim 1 • (x.y)~(O,O) x 2 + y2 + 2 (x,y)~(O,4)

VY

3.

lim

(x,y)~(3,4)

v'x 2

+ y2

-

I

4,

lim

(x,y)~(2, -3)

(1. + Y1.)2 x

+ y3 +y + I

x2

s.

lim.

(x,y)~(O,"/4)

sec x tany

6,

lim

(x,y)~(O,O)

cos ~--'--~ X

762 7. 9.

11.

Chapter 14: Partial Derivatives lim

(x,y)-(O,1n2)

e X- Y

8.

lim

10.

(x.y)~(o.O)

xsiny -(x,y)-(l, '1f'/6) xl + 1 lim

12.

lim

(;;,,)_(1,1)

In 11 + x'y'l

lim

(x.y)~(I/'7."')

cos~

cosy + 1 .

lim

x' -

(x.YI~h.l)

x

2xy + y' y

lim

14.

(x.y)~(I.I)

x""y

15. 16.

17.

xy-y-2x+2

(x,y)-(l,I) x¢l

x-I

lim

20.

x

y' y

~

37. a. h(x,y, z)

~ xy sin}

x+y-4

lim

x

(x.y)~('.3)

lim

(.%,,)-(2,0) 2x-y:;t:4

27. 28. 30.

y' I

z'

+ y' + z2

-

9

x'

42. f(x,y) ~ - , - - , x +Y

lim

22.

(x.y)~(o.O)

lim

lim

z

I - cos (xy)

xy

(.%,,)-(0,0)

+ y3

--+(x.y)~(1.-I) x Y

Umits with Three Variables Find the limits in Exercises 25-30. 25.

I b. h(x,y, z) ~ Ixyl + Izl

z

x3

23.

X

41. f(x,y) ~

x:;t:y+l

21.

2

+z - 1

No Umit at a Point By considering different paths of approach, show that the functions in Exercises 41-48 have no limit as (x,y) --> (0, 0).

1

y

I

2 X

Y4 - x' 4 - Yx'

Vx - "\I'Y+l

lim

b. h(x,y, z) ~

b. h(x, y, z) ~ -:---=-r~~C==~==:'

2VY 19.

Vx+Y - 2

(x.y)~(2.2) x+y:;t:4

b. f(x,y,z) ~ eX+Ycosz

Inxyz

I 38. a. h(x,y,z) ~ Iyl + Izl

Vx - VY

(x.y)~(o.O)

+ y2 - I

36. a. f(x,y, z)

40. a. h(x,y,z) ~

xy

+ 2Vx -

x - y

lim

b. f(x,y, z) ~ Yx 2

39. a. h(x,y, z) ~ In (z - x' - y' - 1) I b. h(x,y,z) ~ • ~ z - vx 2 + y2

y + 4 + 4x 2 - 4x

(x.y)~(2 ..... ) x2y y:;t:-4,xt:-i'-

x¢y

18.

x' -

x;b.y

lim

continuous? 35. a. f(x,y, z) ~ x' + y' - 22'

(;;,y)-(,"'/2,O) y - smx

Umits of Quotients Find the limits in Exercises 13-24 by rewriting the fractions lIrst. 13.

Continuity in Space At what points (x, y, z) in space are the functions in Exercises 35-40

lim

P-(1,3,4)

(~+ yl + })

lim

P-(1J",'IT,O)

2xy

+ yz

(sin'x + cos'y + sec'z)

lim

P~(-I/'.w/2,2)

lim

P-(2,-3.6)

tan-I xyz

29.

34. a. g(x,y)

x +Y x-y 45. g(x,y) ~ x + y

lim.

ze -2y cos 2x

InYx' + y' + z'

~ ~

x2

-

y

46. g(x,y) ~ x - y x2y

48. h(x,y) ~ - , - - , x

+Y

Theory and Examples In Exercises 49 and 50, show that the limits do not exist. xy'-I xy+1 49. lim 50. lim (x,y)-(l,l) y - 1 (x,Y)-(1,-I)x2 _ y2

I, y '" x' I, yS 0 { 0, otherwise. Find each of the following limits, or explain that the limit does not exist 51. l.etf(x,y) ~

~

xy

44. f(x,y) ~ Ixyl

P-('IT,O,3)

sin (x + y)

x+y 32. a. f(x,y) ~ x - y

33. a. g(x,y)

X4 _ y2 ~, ,

xl + y 47. h(x,y) ~ - y -

Continuity In the Plane At what points (x,y) in the plane are the functions in Exercises 31-34

continuous? 31. a. f(x,y)

43. f(x,y)

sin

x2

b. f(x,y)

;y

xl -

b. f(x,y) ~ In (x' + y')

y -,-x + I

x + y b. g(x,y) ~ 2 + cou

+ y2 3x

~

+

a. b.

I b. g(x,y) ~ - , - x - y 2

c.

lim

(x.y)~(O.I)

f(x,y)

(x.YI~2,3/(x, y) lim

(x.y)-{O,O)

f(x, y)

14.2

52. Letf(x,y)

=,x,

0 0

X ===

X2,

{

x

<

Limits and Continuity in Higher Dimensions

763

59. (Continuation ofExample 5.)

.

a. Reread Example 5. Then substitute m = tan 8 into the formula

Find the following limits. ••

1in3

(x, y)-(3, -2)

f(x,y)

b.

(x,y)~2, l/(X,y)

c.

(x, y~o, O/(X, y)

and sin3plify the result to show how the value of f varies with the line's angle of inclination.

53. Show that the function in Example 6 has limit 0 along every straight line approaching (0, 0). 54. If f(Xo, Yo) = 3, what can you say about 1in3

(x,y)-(xo,Yo)

f(x,y)

if f is continuous at (xo, Yo)? If f is not continuous at (xo, Yo)?

Give reasons for your answers. The Sandwich Theorem for functions of two variables states that if g(x, y) '" f(x, y) s h(x, y) for all (x, y) ¢' (xo, Yo) in a disk centered at (xo, Yo) and if g and h have the san3e fInite limit L as (x,y) ..... (XO,yo), then 1in3

(x,y)~(xo,"')

f(x,y) = L.

Use this result to support your answers to the questions in Exercises 55-58,

b. Use the formula you obtained in part (a) to show that the limit of f as (x, y) ..... (0, 0) along the line y = mx varies from -I to I depending on the angle of approach.

60. Continuous extension Derme f(O, 0) in a way that extends x2 _ y2 f(x,y) = xy '+ ' x y to be continuous at the origin. Changing to Polar Coordinate. If you canoot mske any headway with linl(x,y)~(o,o)f(x,y) in rectangular coordinates, try changing to polar coordinates. Substitute x = r cos 6, y = r sin 8, and investigate the limit of the resulting expression as r ..... 0, In other words, try to decide whether there exists a number L satisfying the following critl>rion: Given < > 0, there exists a 6 > 0 such that for all r and 8, Irl < 6

lim

(x,y)-(O,O)

tan-I xy

1--3-<~<1

If(r,8) - LI < <.

(1)

If such an L exists, then

55. Does knowing that

x'y'

~

f(x,y)

=

lim f(r cos 8, r sin 8)

r-O

=

L.

For instance,

tell you anything about tan-I xy --? (x,y)~(O,O) xy

lim

To verify the last of the.. equalities, we need to show that Equation (I) is satisfled with f(r, 8) = r cos' 8 and L = O. That is, we need to show that given any < > 0, there exists a 6 > 0 such thatfor all r and 8,

Give reasons for your answer. 56. Does knowing that

Irl<6

x 2y2

21xyl-

6

< 4 - 4 cos ~ < 21xyl

~

Ircos'8-01«·

Since Ircos'81 = Irllcos'81

tell you anything about

lim

4 - 4cos~xyl

(x,y)~(O,O)

Ixyl

?

s Irl'l

= Irl,

the in3plication holds for all r and 8 ifwe take 6 = <. In contrast,

Give reasons for your answer. 57. Does knowing that Isin (I/x) I '" I tell you anything about

, Iun

(x,y)-(O,O)

. I? ysm-. x

Give reasons for your answer. 58. Does knowing that Icos (I /y)l

lim

(x,y)~(O,O)

Give reasons for your answer.

'" I tell you anything about xcos 1 ? y

takes on all values from 0 to I regardless of how small Ir I is, so that linl(x,y)~(O,O)X'/(x' + y') does not exist. In each of these instances, the existence or nonexistence of the limit as r ..... 0 is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) 8 = constant and yet fail to exist in the broader sense. Example 5 illustrates this point In polar coordinates, f(x,y) = (2x'y)/(x' + y') becomes . f(rcos8,rsm8)

rcos8 sin 28 =,rcos8+sm8 4

. ,

764

Chapter 14: Partial Derivatives

for r # O. If we hold 6 constant and let r --+ 0, the limit is O. On the

path y

=

x 2 , however, we have r sin (}

.

f(rcos6,rsm6) =

= ,2

cos2 (J and

rcos8sin28 2 • ( 2)2 r cos (} + Teos (} 2rcos2 (} sin (} 2r2 cos4 (}

v'x 2 + y2 < 8

r sinO ,2 cos2 (J

69. f(x,y) = x 2 + y2,

I.

70.

In Exercises 61-
61. f(x,y) = xl2 - xy22

62. f(x,y) = cos (x: - y:)

63. f(x,y) =

64. f(x,y) =

65. f(x,y)

x+y y2 2

x+y

2

x+y 2x

2

x+x+y

2

= tan-I (I~I + I~I) x +y x2 _

66. f(x,y) =

2 X

2

= In (

3x2 - x"y2 2 X

+Y

+

3y2 )

2

3x"y

68. f(x,y) =

2 X

+Y

73.

If(x,y) - f(O, 0)1 < •.

0.Ql

f(x,y) = y/(x 2 + I), E = 0.05 f(x,y) = (x + y)/(x 2 + 1), • = 0.Ql f(x,y) = (x + y)/(2 + cos x), • = 0.02 xy2 f(x,y) = - 2 - - 2 andf(O,O) = 0, • = 0.04 x +Y x3

74. f(x,y) =

+ y4

-2--2

x

+Y

andf(O,O) = 0,

If(x,y, z)

2

75. f(x,y, z) = x 2 + y2

+ z2, •

= 0.Ql5

76. f(x,y, z) = xyz,

- f(O, 0, 0) I <

E.

• = 0.008 x+y+z

77. f(x,y, z)

=

78. f(x,y,z)

=

79. Show that (XO,YO, zo).

2

2

2

x+y+z+l

'

E =

+ tan2y + tan2 z, f(x,y,z) = x + y - z is tao2 x

80. Showthatf(x,y,z) = x 2 + y2

14.3

• = 0.02

~

2 y"x "--:-+-y"2'C+z"2 < 8

In Exercises 67 and 68, defme f(O, 0) in a way that extends f to be continuous at the origin.

67. f(x,y)

71. 72.

E =

~

Each of Exercises 75-78 gives a function f(x, y, z) and a positive nwnher E. In each exercise, show that there exists a 8 > 0 such that for all (x,y, z),

y2

+y

Using the Limit Definition Each of Exercises 69-74 gives a function f(x,y) and a positive number •. In each exercise, show 1hat there exists a 8 > 0 such that for all (x, y),

0.Ql5 E =

0.03

contiouous at every point

+ Z2 is contiouous at the origio.

Partial Derivatives The calculus of several variables is similar to single-variable calculus applied to several variables one at a time. When we hold all but one of the independent variables of a function constant and differentiate with respect to that one variable, we get a "partial" derivative. This section shows how partial derivatives are defined and interpreted geometrically, and how to calculate them by applying the rules for differentiatiog functions of a single variable. The idea of differentiability for functions of several variables requires more than the existence of the partial derivatives, but we will see that differentiable functions of several variables behave in the same way as differentiable single-variable functions.

Partial Derivatives of a Function of Two Variables If (xo, YO) is a point in the domain of a function f(x, y), the vertical plane Y = Yo will cut the surface z = f(x, y) in the curve z = f(x, YO) (Figure 14.15). This curve is the graph of the function z = f(x, YO) in the plane y = Yo. The horizontal coordinate in this plane is x; the vertical coordinate is z. The y-value is held constant at Yo, so Y is not a variable. We defme the partial derivative of f with respect to x at the point (xo, YO) as the ordinary derivative of f(x, YO) with respect to x at the point x = Xc. To distinguish partial derivatives from ordinary derivatives we use the symbol arather than the d previously used. In the definition, h represents a real number, positive or negative.

14.3

Partial Derivatives

765

Vertieal axis in ----the plane y

=

Yo

P(Xo. Yo./(x .. yOJ) z~/(x.Y)

The curve z = f(x, Yo) in the plane y = Yo

o Yo (xo. Yo)

Y

(Xo + h. Yo)

Horizontal axis in the plane y = Yo

FIGURE 14.15 The intersection of the plane y ~ Yo with the surface z ~ f(x, y), viewed from above the fIrSt quadrant ofthexy-plane.

DEFINmON

The partial derivative of f(x, y) with respeet to x at the point

(XO,yo) is

of

ox

=

1

(xo.Yo)

lim f(xo

+ h,yo) - f(xo,yo) h

h-O

'

provided the limit exists.

An equivalent expression for the partial derivative is

~ f(x, Yo) x~x; 1

The slope of the curve z = I(x, Yo) at the point P(xo, Yo, I(xo, YO» in the plane y = Yo is the value of the partial derivative of I with respect to x at (xo, yo). (In Figure 14.15 this slope is negative.) The tangent line to the curve at P is the line in the plane y = Yo that passes through P with this slope. The partial derivative aI/ax at (xo, YO) gives the rate of change of I with respect to x wheny is held fixed at the value Yo. We use severa! notations for the partial derivative:

az

1

ax <X.y,)'

and

The definition of the partial derivative of I(x, y) with respect to y at a point (xo,Yo) is similar to the definition of the partial derivative of I with respect to x. We hold x fixed at the value Xo and take the ordinary derivative of I(xo, y) with respect to y at Yo .

DEFINmON

The partial derivative of I(x, y) with respect to y at the point

(XO,Yo) is

all a Y

=

<x. y,)

A.. ( )1 dy / Xo,y y-,.

provided the limit exists.

=

l'

un

h~O

I(xo,yo

+ h)

h

- I(xo,yo)

'

766

Chapter 14: Partial Derivatives The slope of the curve z = I(xo,y) at the point P(xo,Yo, I(xo,yo)) in the vertical plane x = Xo (Figure 14.16) is the partial derivative of I with respect to y at (xo,Yo). The tangent line to the curve at P is the line in the plane x = Xo that passes through P with this slope. The partial derivative gives the rate of change of I with respect to y at (xo, Yo) when x is held fixed at the value Xo. The partial derivative with respect to y is denoted the same way as the partial derivative with respect to x:

Vertieal axis in the plane

z

1

'~"

P(xo, Yo,/(xo, Yo»

al ay (XO,yo),

z ~/(x,y)

Yo

Xo x

(Xo, Yo) (Xo,Yo+ k)

The curve:z:

= f(xo. Y) in the plane x=xo

~/ /

Y

al ay'

Iy(xo,yo),

Notice that we now have two tangent lines associated with the surface z = I(x, y) at the point P(xo,yo, I(xo, Yo)) (Figure 14.17). Is the plane they detennine tangent to the surface at P? We will see that it is for the differentiable functions defined at the end of this section, and we wilileam how to f'md the tangent plane in Section 14.6. Firat we have to learn more about partial derivatives themselves.

~

z

Horizontal axis in the plane x = Xo

r

This tangent line P(xo, y.'/(xo, Yo» has slope I,(xo, Yo).

FIGURE 14.16 The intersection of the plane x = Xo with the surface z = /(x, y), viewed from above the frrst quadrant of the xy-plane.

This tangent line has slope fx(xQ. Yo).

The curve z = f(xo. Y) in the plane x = Xo ~~"\

The ClU'Ve z = f(x, Yo) in the plane y = Yo

z~

y

=

Yo

(xo,Yo)

/(x,y)

x = Xo

FIGURE 14.17 Figores 14.15 aod 14.16 combined. The tangent lines at the point (XO,YO, /(XO,Yo» detennioe a plaoe that, in this picture at least, appears to be tangent to the surface.

Calculations The definitions of ai/ax and al/ay give us two different ways of differentiating I at a point: with respect to x in the usual way while treating y as a constant and with respect to y in the usual way while treating x as a constant. As the following examples show, the values of these partial derivatives are usually different at a given point (xo, yo).

EXAMPLE 1

Find the values of ai/ax andal/ay at the point (4, -5) if

I(x,y) = x 2

Solution

+ 3xy + y

- I.

To f'md ai/ax, we treaty as a constant and differentiate with respect to x:

al a ax = ax (x 2

+ 3xy + y

- 1) = 2x

The value of ai/ax at (4, -5) is 2(4)

+ 3(-5)

+ 3·1· Y + 0 =

-7.

0

=

2x

+ 3y.

14.3

Partial Derivatives

767

To find 011oy, we treat x as a constaot and differentiate with respect to y:

01

ay

0

= oy (x

2

+ 3lY + y

- I) = 0

The value ofolloy at(4, -5) is 3(4)

EXAMPLE 2 Solution

+

+ 3 ·x·1 +

I - 0 = 3x

+

I.

•

I = 13.

Fiodolloyasafunctionif/(x,y) = ysiolY.

We treat x as a constaot and I as a product of y and sio lY:

01 oy

= oy Y smlY = Y ay smlY

o.

+ (.smlY ) oy 0 ( ) Y

= (y cos lY) ~ (xy)

+ sio lY

= lY cos lY

0 (

.

)

+

•

sio lY.

Fiod Ix and Iy as functions if

EXAMPLE 3

2y

I(x,y) = y

Solution

+ cosx'

We treat I as a quotient. With y held constao!, we get

f

= x

~J

2y

) = (y

o

0

+ cos x) ax (2y) - 2y ax (y + (y + cosxJ2

ox

\Y + cosx

(y

+ cosx)(O) - 2y(-siox) (y + cosxJ2

cosx)

2ysiox

(y

+ cosxJ2'

With x held constao!, we get

I = ~( y

2y

) = (y

o

0

+ cosx)ay(2y) - 2y ay (y + cosx) (y + cosxJ2

oy

\Y + cosx

(y

+ cosx)(2) - 2y(l) (y + cosx)2

2cosx (y + cosx)2'

•

Implicit differentiation works for partial derivatives the way it works for ordinary derivatives, as the next example illustrates.

EXAMPLE 4

Fiod ozlax if the equation

yz - lnz = x

+y

dermes z as a function of the two iodependent variables x and y and the partial derivative

exists. Solution We differentiate both sides of the equation with respect to x, holding y constaot and treating z as a differentiable function of x: o 0 ox -(yz) - -lnz = ox ox ox

oy +ox

oz_!oz=I+O Yox Zox

With Y constant,

a

az

ax (yz) ~ y ax .

&-})~~=I oz ox

z yz-I'

•

768

Chapter 14: Partial Derivatives

EXAMPLE 5 The plane x = 1 intersects the paraboloid z = x 2 + y2 in a parabola. Find the slope of the tangent to the parabola at (1, 2, 5) (Figure 14.18). z

Solution

t

The slope is the value of the partial derivative iJz/iJy at (1, 2): iJz iJy

Surface

V

Z=X2+ y2

= iL(x 2 + y2)1

1

iJy

{1,2}

= 2yl {1,2}

= 2(2) = 4. (I,2)

As a check, we can treat the parabola as the graph of the single-variable function z = (1)2 + y2 = 1 + y2 in the plane x = 1 and ask for the slope at y = 2. The slope, calculated now as an ordinary derivative, is

-dz 1

dy y=2

y

= -d (1 + y 21 ) dy

y=2

= 2y 1

y=2

= 4.

•

x

FIGURE 14.18 The tangent to the curve of intersection of the plane x = I and surface z = x 2 + Y 2 at the point (I, 2, 5) (Example 5).

Functions of More Than Two Variables The definitions of the partial derivatives of functions of more than two independent variables are like the definitions for functions of two variables. They are ordinary derivatives with respect to one variable, taken while the other independent variables are held constant.

EXAMPLE 6

If x, y, and z are independent variables and j(x,y,z)

=

xsin(y

+

3z),

then

~~ =

:z

[x sin (y

= x cos (y +

R,

+

3z)] = x

:z

sin (y

+

3z)

iJ 3z) iJz (y + 3z) = 3x cos (y + 3z).

•

EXAMPLE 7 Ifresistors of R I , R2, and R3 ohms are connected in parallel to make an R-ohm resistor, the value of R can be found from the equation

R2

'-.

l

R3 Jy

(Figure 14.19). Find the value of iJR/iJR 2 when RI = 30, R2 = 45, and R3 = 90 ohms.

) 1-

FIGURE 14.19 Resistors arranged this way are said to be connected in parallel (Example 7). Each resistor lets a portion of the current through. Their equivalent resistance R is calculated with the formula

Solution To find iJR/iJR 2, we treat RI and R3 as constants and, using implicit differentiation, differentiate both sides of the equation with respect to R2 :

1 iJR -- = 0 - - 1+ 0 R2 iJR2

When RI = 30, R2 = 45, and R3 = 90,

Ri

14.3 soR =

Partial Derivatives

769

15 and aR = aR2

(15)2 (1)2 19' 45 3 =

=

Thus at the given values, a sma11 change in the resistance R2 leads to a change in R about 1/9th as large. _

Partial Derivatives and Continuity A function I(x, y) can have partial derivatives with respect to hoth x and y at a point without the function being continuous there. This is different from functions of a single variahle, where the existence of a derivative implies continuity. If the partial derivatives of I(x, y) exist and are continuous throughout a disk centered at (xo,Yo), however, then I is continuous at (xo,Yo), as we see at the end of this section.

z~ {o,xy#=o I, xy

=

0

EXAMPLE 8

Let

z

{~:

I(x,y) =

.xy ,. 0 .xy = 0

(Figure 14.20). (8) Find the limit of I as (x, y) approaches (0, 0) along the line y = x. (h) Prove that I is not continuous at the origin.

(c) Show that hoth partial derivatives aI/ax and af/ay exist at the origin.

Solution FIGURE 14.20 The graph of

f(

x,y

)=

{a,I,

xy,. xy =

°°

(8) Since I(x,y) is constantly zero along the line y = x (except at the origin), we have

lim

(.,y)~(O,o)

consists of the lines L, and L, and the four open quadrants of the xy·plane. The function has partial derivatives at the origin but is not continuous there (Example 8).

I(x,y)

I

y-.

lim

(.,y)~(O,o)

0=

o.

(h) Since 1(0,0) = I, the limit in part (a) proves that I is not continuous at (0, 0). (c) To find aI/ax at (0, 0), we holdy rlXed at y = O. Then I(x,y) = I for all x, and the graph of I is the line Ll in Figure 14.20. The slope of this line at any x is aI/ax = o. In particular, aI/ax = 0 at (0, 0). Similarly, af/ay is the slope of line L2 at any y, so al/ay = 0 at (0, 0). _

Example 8 notwithstanding, it is still true in higher dimensions that differentiability at a point implies continuity. What Example 8 suggests is that we need a stronger requirement for differentiability in higher dimensions than the mere existence of the partial derivatives. We derme differentiability for functions of two variables (which is slightly more complicated than for single-variable functions) at the end of this section and then revisit the connection to continuity.

Second-Order Partial Derivatives When we differentiate a function I(x, y) twice, we produce its second-order derivatives. These derivatives are usually denoted by

a21

2 ax

or

a21

lxx,

ay2

and

or

Iyy,

a21

a)"lx or Ixy·

770

Chapter 14: Partial Derivatives The defining equations are

:y, :x (:~). =

and so on. Notice the order in which the mixed partial derivatives are taken:

a2f

Differentiate first with respect to y, then with respect to x.

axay

f", HiSTORICAL BIOGRAPHY

Means the same thing.

= (jy)x

EXAMPLE 9

Iff(x,y)

=

x cosy + ye~f"mdthesecond-orderderivatives

a2f

Pierre-Simon Laplace (1749--1827)

ayax' Solution

af ax

a2f

and

axay'

The first step is to calculate both first partial derivatives.

=

a ax (x cosy + ye X)

=

cosy + ye X

af ay

=

a ay(xcos y + ye X)

=

-x siny + eX

Now we find both partial derivatives of each first partial:

2 af ayax

f ay ax

=

iL (a )

:~ =

:x (:~)

= -siny

=

ye

2 af axay

+ eX

=

f

iL (a ) ax ay

= -siny

:~ = ~ (~) =

x •

-x

+ eX

cosy.

•

The Mixed Derivative Theorem You may have noticed that the ''mixed'' second-order partial derivatives

alf ayax

and

a2f axay

in Example 9 are equal. This is not a coincidence. They must be equal whenever f, fx. f y , fxy, andf", are continuous, as stated in the following theorem.

THEOREM 2-The Mixed Derivative Theorem If f(x,y) and its partial derivatives f x. f y , f xy, and f", are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then fxy(a, b)

HiSTORICAL BIOGRAPHY

Alexis C\airaut (1713-1765)

=

f",(a, b).

Theorem 2 is also known as Clairaut's Theorem, named after the French mathematician Alexis Clairsut who discovered it. A proof is given in Appendix 9. Theorem 2 says that to calculate a mixed second-order derivative, we may differentiate in either order, provided the continuity conditions are satisfied. This ability to proceed in different order sometimes simplifies our calculations.

EXAMPLE 10

Find a2w/axay if

eY w=XY+-2--' y + 1

14.3

Partial Derivatives

771

Solution The symbol a'w/ axiJy tells us to differentiate first with respect to y and then with respect to x. However, if we interchange the order of differentiation and differentiate f'lISt with respect to x we get the answer more quickly. In two steps,

aw ax

=

and

y

a'w ayax

=

1.

Ifwe differentiate f'lISt with respect toy, we obtaina'w/axay = I as well. We can differentiate in either order because the conditions of Theorem 2 hold for w at all points (xo, yo) . •

Partial Derivatives of Still Higher Order Although we will deal mostly with first- and second-order partial derivatives, because these appear the most frequently in applications, there is no theoretical limit to how many times we can differentiate a function as long as the derivatives involved exist. Thus, we get third- and fourth-order derivatives denoted by symbols like

a3f

axiJy 2

=

f yyx,

a'f ax2ay2

=

fyyxx,

and so on. As with second-order derivatives, the order of differentiation is immaterial as long as all the derivatives through the order in question are continuous. EXAMPLE 11

Findf_ if f(x,y,z)

=

1 - 2xy2z + x~.

Solution We f'lISt differentiate with respect to the variable y, then x, then y again, and f'ma11y with respect to z: fy = -4xyz + x 2 fyx = -4yz + 2x fyry = -4z

f_=

•

-4

Differentiability The starting point for differentiability is not the difference quotient we saw in studying singlevariable functions, but rather the idea of increment. Recall from our work with functions of a single variable in Section 3.9 that ify = f(x) is differentiable atx = Xo, then the change in the value off that results from changing x from Xo to Xo + dx is given by an equation of the form

ay

=

f'(xo)ax + Edx

in which E ---> 0 as dx ---> O. For functions of two variables, the analogous property becomes the def'mition of differentiability. The Increment Theorem (proved in Appendix 9) tells us when to expect the property to hold. THEOREM 3-The Increment Theorem for Functions of Two Variables Suppose that the f'rrst partial derivatives of f(x, y) are defined throughout an open region R containing the point (xo, Yo) and that fx and fy are continuous at (xo, yo). Then the change az = f(xo + dx,yO + ay) - f(xo,yo)

in the value of f that results from moving from (xo, Yo) to another point (xo + dx, Yo + ay) in R satisfies an equation of the form

az

=

fx(xO,yo)dx + f,(xo,Yo)ay +

in which each of EJ, E2 ---> 0 as both dx, ay ---> o.

ElaX

+ E2ay

772

Chapter 14: Partial Derivatives You can see where the epsilons come from in the proof given in Appendix 9. Similar results hold for functions of more than two independent variables.

DEFINITION A function z = I(x,y) is differentiable at (xo,Yo) if Ix(xo,yo) and I,(xo,yo) exist and f1z satisfies an equation of the form f1z = fx(xo,yo)flx

+ Iy(xo, Yo) fly + E1/lx + E2fly

in which each of Et. E2 --> 0 as both /lx, fly --> O. We call I differentiable if it is differentiable at every point in its domain, and say that its graph is a smooth surface.

Because of this definition, an immediate corollary of Theorem 3 is that a function is differentiable at (xo, yo) if its first partial derivatives are continuous there.

COROLLARY OF THEOREM 3 If the partial derivatives Ix and/y ofa function I(x, y) are continuous throughout an open region R, then I is differentiable at every point of R.

flz

If z = I(x, y) is differentiable, then the definition of differentiability assures that I(xo + /lx,Yo + fly) - I(xo, Yo) approaches 0 as /lx and fly approach O. This tells

=

us that a function of two variables is continuous at every point where it is differentiable.

THEOREM 4-Dlfferentlabillty Implies Continuity If a function I(x, y) is differentiable at (xo,Yo), then I is continuous at (xo,Yo). As we can see from Corollary 3 and Theorem 4, a function/(x,y) must be continuous at a point (xo,Yo) if Ix and Iy are continuous throughout an open region containing (xo,Yo). Remember, however, that it is still possible for a function of two variables to be discontinuous at a point where its first partial derivatives exist, as we saw in Example 8. Existence alone of the partial derivatives at that point is not enough, but continnity of the partial derivatives guarantees differentiability.

Exercises 14.3 Calculating First-Order Partial Derivatives In Exercises 1-22, fmd ai/ax andal/ay.

1. 3. 4. 5. 7. 9. 11. 13.

I(x,y) ~ 2x2 - 3y - 4 2. I(x,y) ~ x 2 - xy I(x,y) ~ (x 2 - I)(y + 2) I(x,y) ~ 5xy - 7x 2 - y2 + 3x - 6y + 2 I(x,y) ~ (xy - I)' I(x,y) ~ v'x 2 + y2

+ y2

6. I(x,y) ~ (2x - 3y)'

8. I(x,y) ~ (x'

+ (y/2»,/' x/(x 2 + y2)

I(x,y) ~ I/(x + y) 10. I(x,y) ~ I(x,y) ~ (x + y)/(xy - 1) 12. I(x,y) ~ tan-I (y/x) I(x,y) ~ e(>+,+1) 14. I(x,y) ~ e~ sin (x + y)

+ y)

15. I(x,y)

~

17. I(x,y)

~ sin2 (x

In (x

- 3y)

16. I(x,y)

~

e"lny

18. I(x,y) ~ cos (3x - y2) 2

19. I(x,y)

~

21. I(x,y)

~

20. I(x,y) ~ log, x

x'

l'

(g cootinuous for all t)

g(t) dt

22. I(x,y) ~ ~(xy)'

.-0

(Ixyl <

I)

In Exercises 23-34, find Ix. I" and I,. 23. I(x,y,z) ~ I + xy2 - 2z2 24. I(x,y,z) ~ xy

+ yz + xz

25. I(x,y, z) ~ x - v'y2 + z2 26. I(x,y, z) ~ (x 2 + y2 + z2)-I/2 27. I(x,y,z) ~ sin-I (xyz) 29. I(x,y, z)

~

In (x

28. I(x,y,z) ~ sec-1 (x

+ 2y + 3z)

+ yz)

14.3 31. /(x,y,z) = e-(>'+y'+r)

30. /(x,y,z) = yZIn(xy) 32. /(x,y,z)

33. /(x,y, z) = tanh (x + 2y + 3z) 34. /(x,y, z)

In Exercises 35-40, fmd the partial derivative of the function with reo spect to each variable. 36. g(u, v) = v 2e(lu/v) 35. /(t, a) = cos (2".t - a)

37. h(p, t/>, 6) = p sin t/> cos 6 39. Work done by the heart

Using the Partial Derivative Definition

57. /(x,y)

sinh (xy - z2)

=

- x

_

+ 2x -

58. /(x,y) - 4

59. /(x,y)

(Section 3.9, Exercise 49)

=

60. /(x,y)

=

2

a/

A(c, h, k,m, q)

kin

=

q + em + T

Find all the second-order partia1 derivatives of the functions in Exercises 41-50.

+ Y + xy

x"y

=

44. h(x,y)

= xe Y

46. s(x,y)

= taJ.-1 (y/x)

+y + I

48. w = ye r - y

45. r(x,y)

In (x

+ y)

49. w = x sin (x"y)

x-y

= -2-X +y

+ 3y) xy2 + x"y' + x'y4

w,..

52. w = e'

fIrst? Try to answer without writing anything down.

a. /(x,y)

=

b. /(x,y)

= I/x =y +

c. /(x,y) e.

xsiny

+ eY

f. /(x,y)

+ y2

'

ay

and

a/ ay

a/ iJy

at(I,2) at (-2, I)

at (-2,3)

(x,y) oF (0,0) (x,y) = (0,0),

0,

a/

a/ ax

and

iJy

at

(0,0)

62. Let /(x,y) = x 2 + y'. Find the slope of the line tsngent to this surface at the point (-I, I) and lying in the a. plane x = -I b. plane y = I.

63. Three \'lUiabl.. Let w = /(x, y, z) be a function of three independent varisbles and write the formal defInition of the partial derivative a//az at (xo,Yo, zo). Use this defirtition to fmdaf/az at (I,2,3)for/(x,y,z) =x"yz2. 64. Three \'lUiabl.. Let w = /(x, y, z) be a function of three independent varisbles and write the formal defmition of the partial derivative a/lay at (XO,YO, zo). Use this defmition to fmda//ay at (-1,0,3) for /(x,y, z) = _2xy2 + yz2

defmes z as a function of the two independent variables x and y and the partial derivative exists.

66. Find the vaIue ofiJx/az at the point (I, -I, -3) if the equation

xz + ylnx - x 2 + 4 = 0 defmes x as a function of the two independent varisbles y and z and the partial derivative exists. Exercises 67 and 68 are about the triangle shown here.

(x/y)

+ x"y + 4y' - In (y2 + I) 2 /(x,y) = x + 5xy + sinx + 7e' =

+ y4)

ax

a/

xy+z'x-2yz=0

+ xlny + ylnx w = 54. w = x siny + y sinx + xy Which order of differeotiation will calculate f", faster: x first or y

51. w = 1n(2x

d. /(x,y)

and

a/

2

I,

and

Differentiating Implldtly 65. Find the vaIue ofaz/iJx at the point (I, I, I) if the equation

In Exercises 51-54, vetifY that w", =

55.

=

47. w=x 2 tan(xy)

Mixed Partial Derivatives

53.

= sinxy

+ cosy + ysinx

43. g(x,y)

50. w

42. /(x,y)

ax

{

ax

3y - xy,

+ 3y -

x2

a/

- 3x"y,

61. Let /(x, y) = 2x + 3y - 4. Find the slope of the line tsngent to this sorface at the point (2, -I) and lying in the a. plane x = 2 b. plane y = -I.

hq

Calculating Second-Order Partial Derivatives

41. /(x,y) = x

V2x

sin (x'

VIlv zg

(Section 4.5, Exercise 51)

+y

=I

38. g(r, 6, z) = r(1 - cos 6) - z

W(P, V, 6, v, g) = PV+ 40. Wilson lot size formula

773

In Exercises 57-{j(), use the limit defmition of partial detivative to compute the partial derivatives of the functions at the specifled points.

e-"'"

=

Partial Derivatives

y

= xlnxy

56. The fIfth-order partial derivative a'//iJx 2ay' is zero for each of the followiog functioos. To show this as quickly as possible, which variable would you differentiate with respect to first: x or y? Try to answer without writing anything down. a. /(x,y) = y'x 4e' + 2 b. /(x,y) = y2 + y(sinx - x 4) c. /(x,y) = x 2 + 5xy + sinx + 7e' d. /(x,y) = xe Y'/2

B~ C

b

A

67. Express A iroplicitly as a function of a, b, and c and calculate aA/aa and aA/ab. 68. Express a iroplicitly as a function of A, b, and B aod calculate aa/aA and aa/aB. 69. Two dependent variable. Express v, in terms of u and y if the equations x = v In u and y = u In v define u and v as functions of the independent variables x and y, and if v, exists. (Hint: Differeotiate both equations with respect to x and solve for v, by elintinating u,.)

774

Chapter 14: Partial Derivatives

70. Two dependent variable. Find ax/au and ay/au if the equations u = x 2 - y2 and v = x 2 - y defme x and y as functions of the indepeodeot variables u and v, and the partial derivatives exist (See the hint in Exercise 69.) Then let s = x 2 + y2 and fmd

as/au.

73. I(x,y, z) = x 2 + y2 - 2z2 74. I(x,y, z) = 2z 3 - 3(x 2 + y2)z 75. I(x, y) =

y3, 71. Let I(x,y) = { 2 -y,

y '" 0 0 y< .

Find I~ Iy. I"" and Iyx, and state the domain for each partial derivative. 72. Let I(x,y) =

Show that each function in Exercises 73-80 satisfies a Laplace equation.

Yx { X.2 '

x~0 X < O.

Find I ~ I yo I "Y' and I yx, and state the domain for each partial derivative. Theory and Examples

The three-dimensional Laplace equation

e -2y cos

2x

76. I(x,y) = lnv'x 2 + y2 77. I(x,y)

=

3x

+ 2y

78. I(x,y) = tan-I

- 4

Y

79. I(x,y, z) = (x 2 + y2 + z2)-I/2 80. !(x,y, z)

=

e 3r+4y cos 5z

The Wave Equation If we stand on an ocean shore and take a snapshot of the waves, the picture shows a regular pattern of peaks and valleys in an instant of time. We see periodic vertical motion in space, with respect to distance. If we stand in the water, we can feel the rise and fall of the water as the waves go by. We see periodic vertical motion in time. In pbysics, this beautiful symmetry is expressed by the

one-dimensional wave equation

a'w= e 2a'w 2 2 is satisfied by steady-state temperature distributions T = I(x, y, z) in space, by gravitational potentials, and by electrostatic potentials. The

two-dimensional Laplace equation

al

ax '

where w is the wave height, x is the distance variable, t is the time variable, and e is the velocity with which the waves are propagated.

a21 a21 ax ay

2+-2=0,

w

obtained by dropping the a2 f!az 2 term from the previous equation, describes potentials and steady-state temperature distributions in a plane (see the accompanying figure). The plane (a) maybe treated as a thin slice of the solid (b) perpeodicular to the z-axis.


In our example, x is the distance across the ocean's surface, but in other applications, x ntight be the distance along a vibrating string, distance through air (sound waves), or distance through space (light waves). The number c varies with the medium and type of wave. Show that the functions in Exercises 81-87 are all solutions of

the wave equation. 81. w = sin (x + el) 82. w = cos (2x + 2el) 83. w = sin (x + el) + cos (2x + 2el) 84. w = In (2x + 2el) 85. w = tan (2x - 2el) 86. w = 5 cos (3x + 3el) + e x+a 87, w = I(u), where I is • differeotiable function of u, and u = (b)

a(x

+ ct), where a is a constant

88. Does a function I(x, y) with continuous fIrst partial derivatives throughout an open region R have to be continuous on R? Give

reasons for your answer.

Boundary temperatures controlled

89. If a function I(x, y) has continuous second partial derivatives throughout an opeo region R, must the first-order partial derivatives off be continuous on R? Give reasons for your answer.

14.4

90. The heat equation An important partial differential equatioo that descrihes the distributioo of heat in a regioo at time t can be represented by the one-dimensional heat equation

Show that u(x, t) ~ sin (ax)' e-~t satisfies the heat equatioo for coostants a and (3.

91. Let/(x,y)

~

14.4

775

Show that liO, 0) and 1,(0,0) exist, but I is not differentiable at (0,0). (Hint: Use Theorem 4 and show that I is not cootinuous at (0,0).)

O, x' < Y < 92. Let/(x,y) ~ { l o t h . , erwtse.

a'i ax' .

al at

The Chain Rule

2>;'

Show that liO, 0) and 1,(0,0) exist, but I is not differentiable at (0,0).

xy'

x' + y4' { 0,

(X,y): (0,0) (x,y) - (0,0).

I~ Th_e_C_ ha_in_R_u_le_________________________________ The Chain Rule for functions of a single variable studied in Section 3.6 says that when w = f(x) is a differentiable function of x and x = g( I) is a differentiable function of I, w is a differentiable function of I and dw/ dl can be calculated by the formula dw dl

dwdx dxdt'

For functions of two or more variables the Chain Rule has several forms. The form depends on how many variables are involved, but once this is taken into account, it works like the Chain Rule in Section 3.6.

Functions of Two Variables The Chain Rule formula for a differentiable function w = f(x,y) when x = X(I) and y = y(t) are both differentiable functions of I is given in the following theorem.

I

Each

Of:~, Z.Jx indicates the partial

derivative of I with respect to x.

THEOREM 5-Chain Rule for Functions of Two Independent Yariables If w = f(x,y) is differentiable and if x = x(I),y = Y(I) are differentiable functions of I, then the composite w = f(x(t), y(I» is a differentiable function of t and dw dt

= fx(x(t),y(t))· x'(t)

+ fy(x(I),y(t))· y'(t),

or dw

af dx

af dy

= axdt - - +aydt' -dt

Proof The proof consists of showing that if x and y are differentiable at t = 10, then w is differentiable at 10 and

where Po = (x(to),y(to». The subscripts indicate where each of the derivatives is to be evaluated.

776

Chapter 14: Partial Derivatives Let ~x, ~y, and ~w be the increments that result from changing t from to to to Since f is differentiable (see the definition in Section 14.3),

b

=

+

~t.

to

(~; ~x + (~1, ~y + El~ + E2~y,

where E), E2 --> 0 as ~x, ~y --> O. To find dw/dt, we divide this equation through by M and let ~t approach zero. The division gives

~w

Tt Letting

~t

=

(aw) ~ ax Po ~t

+

(aw) ~y ay Po ~t

~x ~y + El ~t + E2 ~t .

approach zero gives

Often we write aw/ax for the partial derivative af/ax, so we can rewrite the Chain Rule in Theorem 5 in the fonn

dw = awd!: dt axdt

I To remember the Chain Rille picture the diagram below. To fmddw/dt, start atw and read down each route to t, multiplying derivatives along the way. Then add the products. CbalnRuIe Dependent

w ~f(x,y)

variable

aw ax

aw ay

x

y

dx dt

dw dt

=

Intermediate variables

dy

dt t awdx + awdy ax dt ay dt

Independent

+

awdy aydt'

However, the meaning of the dependent variable w is different on each side of the preceding equation. On the left-hand side, it refers to the composite function w = f(x(t),y(I» as a function of the single variable t. On the right-hand side, it refers to the function w = f(x,y) as a function of the two variables x andy. Moreover, the single derivatives dw/dl, d!:/dl, and dy/dl are being evaluated at a point 10, whereas the partial derivatives aw/ax and aw/ay are being evaluated at the point (xo, Yo), with Xo = x(to) and Yo = y(lo). With that understanding, we will use both of these forms interchangeably throughout the text whenever no confusion will arise. The branch diagram in the margin provides a convenient way to remember the Chain Rule. The ''true'' independent variable in the composite function is I, whereas x and y are inlermediale variables (controlled by I) and w is the dependent variable. A more precise notation for the Chain Rule shows where the various derivatives in Theorem 5 are evaluated:

variable

dw af d!: dI(to) = ax (xo, yo) • dl(lo)

EXAMPLE 1

+

af dy ay(Xo,Yo)' dt(to).

Use the Chain Rule to fmd the derivative of w = xy

with respect to I along the path x = cosl,y = sinl. What is the derivative's value at = 7f/2?

t

Solution

We apply the Chain Rule to fmd dw/ dt as follows:

dw = awd!: dt ax dt =

+

awdy iJy dt

a(xy) d a(xy) d . ax . dl (cos I) + ay . dl (sm I)

+ (x)(cost) (sinl)(-sinl) + (COSI)(COSI)

= (y)(-sint) =

= -sin2 t = cos2t.

+ cos2 t

14.4 The Chain Rule

777

In this example, we can check the result with a more direct calculation. As a fimction of I, . = 2sm I. 2I, w = xy = coslsml so

dw dt

=

(I. ) 1

d 2 sm21 dl

= 2' 2 cos 21 = cos2t.

In either case, at the given value of I, = (dw) dl t~~/2

cos (2'

TT)

= COSTT =

2

•

-1.

Functions of Three Variables You can probably predict the Chain Rule for functions of three variables, as it only involves adding the expected third term to the two-variable formula.

THEOREM 6-Chain Rule for Functions of Thl'l!e Independent variables If w = f(x,y, z) is differentiable and x, y, and z are differentiable functions of I, then w is a differentiable function of I and

dw =

awdx ax dt

dt

I

Here we have three routes from w to t instead of two, but rmding dw/dt is still the same. Read down each route, multiplying derivatives aloog the way; then add.

Dependent

variable iJw iJx

iJw

EXAMPLE 2

ay x

z

y

dx dt

dy dt

+ z,

x = cost,

y

=

sint,

z = I.

In this example the values ofw(l) are changing along the path ofa helix (Section 13.1) as I cbanges. What is the derivative's value at t = O? Using the Chain Rule for three independent variables, we have

dw =

Intennediare

dl

variables

awdx ax dl

+

awdy ay dl

= (y)(-sinl) =

Independent

=

so

(r;;; to

+ awdz az dl

+ (x)(COSI) + (1)(1) (sint)(-sint) + (cost)(cost) + 1 -sin2 1 + cos2 1 + I = 1 + cos 21,

dz

dt

t variable dw=iJwdx +iJwdy +iJwdz dt ax dt iJy dt iJz dt

az dt'

Find dw/dl if w = xy

Solutton

iJw

ay dt

The proof is identical with the proof of Theorem 5 except that there are now three intermediate variables instead of two. The branch diagram we use for remembering the new equation is similar as well, with three routes from w to t.

ChainRuI. w ~ J(x. y. z)

+ awdy + awdz

= 1

+

cos (0) = 2.

Substitute for the intermediate wriables.

•

For a physical interpretation of cbange along a curve thiok of an object whose position is cbanging with time I. If w = T(x, y, z) is the temperatore at each point (x, y, z) along a curve C with parametric equations x = x(I), Y = y(I), and z = z(I), then the composite function w = T(x(t),y(t),z(t» represents the temperatore relative to t along the curve. The derivative dw/ dl is then the instantaneous rate of change of temperatore due to the motion along the curve, as calculated in Theorem 6.

Functions Defined on Surfaces If we are interested in the temperature w = f(x, y, z) at points (x, y, z) on the earth's surface, we might prefer to thiok of x, y, and z as functions of the variables r and s that give

778

Chapter 14: Partial Derivatives

the points' longitudes and latitudes. If x = g(r, s), y = h(r, s), and z = k(r, s), we could then express the temperature as a function of r and s with the composite function w = /(g(r, s), h(r, s), k(r, s».

Under the conditions stated below, w has partial derivatives with respect to both r and s that can be calculated in the following way.

THEOREM 7-Chain Rule for Two Independent variables and Three Intennediate variables Supposetbatw = /(x,y,z),x = g(r,s),y = h(r,s),andz = k(r,s). If all four functions are differentiable, then w has partial derivatives with respect to rands, given by the formulas

+ away + awaz

aw = awax ar ax ar aw = awax as ax as

+

ay ar away ayas

+

az ar awaz azas'

The first of these equations can be derived from the Chain Rule in Theorem 6 by holding s fixed and treating r as t. The second can be derived in the same way, holding r fixed and treating s as t. The branch diagrams for both equations are shown in Figure 14.21. w

Dependent

variable

aw ax

-----} :0 0----0: ,

Intermediate

,

'--W-____

variables

g

Independent

h

k

aw ax

aw az

aw ay

x

J

y

ax ar

w ~ fv:, y, z)

~f(x,y,z)

x

z

ay ar

y

ay as

ax as

az ar

aw az

aw ay

z

k

as

r, s

variables w

~

f(g(r, ,), h (r, ,), k (r,

aw=awax+iJw~+iJw~

or

iJx iJT

iJy ar

as

iJz or

ax as

(b)

(a)

FIGURE 14.21

s ow=owax +iJwiJy +iJwaz

r

,»

oy as

iJz as

(c)

Composite function and branch diagrams for Theorem 7.

EXAMPLE 3

Expressaw/ar andaw/as in tenns ofr ands if w =

Solution

X

+ 2y + z2,

y =

r2

+ Ins,

z = 2r.

Using the fonnulas in Theorem 7, we fmd aw = awax ar ax ar

+ away + awaz ay ar

az ar

=

(l)(i) + (2)(2r) + (2z)(2)

=

1 S

+ 4r + (4r)(2)

aw = awax as ax as =

=

1 S

+ 12r

Substitute for intermediate variablez.

+ away + awaz ay as

az as

(1)(-'"-) + (2)(!) + (2z)(O) 82 S

=

~S -

'"32

•

14.4 The Chain Rule

If f is a function of two variables instead of tbree, each equation in Theorem 7 becomes correspondingly one term shorter.

CbainRui. w ~f(x,y)

iJw

779

iJw

ax

ay

Ifw ~ f(x,y),x ~ g(r,s), andy = h(r,s), then y

+ away

iIw = iIw ax

ar

ax

ar

ax ar

aw ~ ilwax as ax as

and

ayar

+ away iJy as .

iJT

r

aw=iJwax+iJway

aT

ax ar

oy ar

FIGURE 14.22 tire equation

aw

~

ar

Branch diagram for

Figure 14.22 shows the branch diagram for the first of these equations. The diagram for the second equation is similar; just replace r with s.

EXAMPLE 4

Express iIw/ar and aw/as in terms ofr ands if

awax + away ax ar ay ar'

y = r + s.

x = r - s, Solution

The preceding discussion gives the following. iIw = aw ax

ar

+

iIw ay

iIw ~ iIw ax

ayar

as

ax ar ~

(2x)(I) + (2y)(I)

=

2(r - s)

= 4r

ay as

~

+ (2y)(l) -2(r - s) + 2(r + s)

~

4s

~

+ 2(r + s)

+ away

ax as (2x)(-I)

Substitute for the intermediate

variables.

If f is a function of x alone, our equations are even simpler.

•

Ifw ~ f(x) and x ~ g(r,s),then

aw ar

CbainRui.

and

aw as

dwax ax as'

In this case, we use the ordinary (single-variable) derivative, dw/ ax. The branch diagram is shown in Figure 14.23.

w ~f(x)

dw

Implicit Differentiation Revisited

dx

The two-variable Chain Rule in Theorem 5 leads to a formula that takes some of the algebra out of implicit differentiation. Suppose that

x

ax ar

,

r iJw

dw iJx

or

dx or

iJw

dw iJx

a,

dwax ax ar

dx

1.

The function F(x, y) is differentiable and

2.

The equation F(x, y) ~ 0 defines y implicitly as a differentiable function of x, say y ~ h(x).

Since w ~ F(x, y) ~ 0, the derivative dw/ ax must be zero. Computing the derivative ftmn the Chain Rule (branch diagram in Figure 14.24), we find

as

FIGURE 14.23 Branch diagram for differentiating f as a composite function of r and s with one intennediate variable.

dw

o ~ ax

~ Fx

ax

ax + Fy dy ax dy

=

Fx'l

+ Fy ' ax'

Theorem 5 with t

ami! ~ F

~x

780

Chapter 14: Partial Derivatives w

~

If Fy = aw/ay .. 0, we can solve this equation for dy/dx to get

F(x,y)

iJw = F. x

dy dx

F. = iJw

ax

ay

Y

Fx F' y

x

We state this result formally. dy

dx

~ h'(x)

THEOREM 8-A Formula for Implldt Differentiation Suppose that F(x, y) is differentiable and that the equation F(x, y) = 0 defines y as a differentiable function of x. Then at any point where Fy .. 0,

x

dW=F.l+F. dy dx.t 'dx

FIGURE 14.24 Branch diagram for differentiating w ~ F(x, y) with respect tox. Settingdw/dx ~ oleads to a simple computational fonnula for implicit differentiation (Theorem 8).

dy dx

EXAMPLE 5 Solution

Fx F' y

(1)

Use Theorem 8 to find dy/dx if y2 - x 2 - sinxy = O.

Take F(x,y) = y2 - x 2 - sinxy. Then

dy dx

Fx Fy

-2x-ycosxy 2y - x cosxy

2x + ycosxy 2y - xcosxy' This calculation is significantly shorter than a single-variable calculation using implicit differentiation. _ The result in Theorem 8 is easily extended to three variables. Suppose that the equation F(x, y, z) = 0 defines the variable z implicitly as a function z = I(x, y). Then for all (x,y) in the domain of I, we have F(x,y,f(x,y» = o. Assuming thatF and I are differentiable functions, we can use the Chain Rule to differentiate the equation F(x,y, z) = 0 with respect to the independent variable x:

0= aFax ax ax = Fx 'l

+

aFay ay ax

+ aFaz az ax

+ Fy'O +

az Fz ' ax'

y is constant when differentiating with respect to x.

so

Fx

az

+ Fz ax

=

o.

A similar calculation for differentiating with respect to the independent variable y gives

Fy

az

+ Fz ay

=

o.

Whenever Fz .. 0, we can solve these last two equations for the partial derivatives of z = I(x, y) to obtain

az ax

and

az ay

(2)

14.4

781

The Chain Rule

An important result from advanced calculus, called the Implicit Function Theorem, states the conditions for which our results in Equations (2) are valid If the partial derivatives F., Fy , and Fz are continuous throughout an open region R in space containing the point (xo, Yo, zo), and if for some constant c, F(Xo, Yo, zo) = c and Fixo, Yo, zo) '" 0, then the equation F(x, y, z) = c defines z implicitly as a differentiable function of x and y near (xo, Yo, zo), and the partial derivatives of z are given by Equations (2).

EXAMPLE 6 Solution

Find

~;

~;

and

+ z2 + ye + zcosy XZ

= O.

+ z2 + ye + zcosy. Then

LetF(x,y,z) = x 3

Fy

at (0, 0, 0) if x 3 XZ

exz

=

-

zsiny,

and

Fz = 2z

+

xyeXZ

+ cosy.

Since F(O, 0, 0) = 0, FiO, 0, 0) = 1 '" 0, and all rlISt partial derivatives are continuous, the Implicit Function Theorem says that F(x,y, z) = 0 dermes z as a differentiable function ofx andy near the point (0, 0, 0). From Equations (2), e XZ

2z

+ xye + cosy xz

and

2z

-

z

siny

+ xye + cosy' xz

At (0, 0, 0) we rmd and

oz oy

- = - -1 =-1

1

.

•

Functions of Many Variables We have seen several different forms of the Chain Rule in this section, but each one is just a special case of one general formula. When solving particular problems, it may help to draw the appropriate branch diagram by placing the dependent variable on top, the intermediate variables in the middle, and the selected independent variable at the bottom. To rmd the derivative of the dependent variable with respect to the selected independent variable, star! at the dependent variable and read down each route of the branch diagram to the independent variable, calculating and multiplying the derivatives along each route. Then add the products found for the different routes. In general, suppose that w = f(x, y, ... , v) is a differentiable function of the variables x,y, ... , v (a rmite set) and the x,y, ... , v are differentiable functions of p, q, ... , t (another rmite set). Then w is a differentiable function of the variables p through t, and the partial derivatives of w with respect to these variables are given by equations of the form

oW

op

=

oW oX + owoy + ... + owov

oxop

oyop

avop'

The other equations are obtained by replacing p by q, ... , t, one at a time. One way to remember this equation is to think of the right-hand side as the dot product of two vectors with components and Derivatives ofw with respect to the

intermediate variables

Derivatives of the intermediate variables with respect to the selected independent variable

782

Chapter 14: Partial Derivatives

Exercises 14.4 Chain Rule: One Independent Variable In Exercises l--{j, (a) express dw/dt as a function of t, both by using the Chain Rule and by expressing w in terms of t and differentiating directly with respect to t. Then (b) evaluate dw/ dt at the given value of t.

1. w = x 2 + y2, X = cos t, Y = sin t; t = '11' 2. w=x 2 +y2, x=cost+sint, y=cost-sint;

i

3. w =

+~. x = cos2 t. Y =

2 8in t,

Z

=

=

sint, z

= tan-II,

=

z

4Vt;

= e';

=

x

z - sinxy,

=

t, Y

=

Z = e t - 1;

In t,

t =

1

Chain Rule: Two and Three Independent Variables In Exercises 7 and 8, (a) expressaz/au andaz/av as functions ofu and v both by using the Chain Rule and by expressing z directly in terms of u and v before differentiatiog. Then (b) evaluate az/au and az/av at the given point (u, v).

7. z = 4e'lny, x = In(ucosv), y (u, v) = (2, '1f/4) 8. z = tan-I (xly), x (u, v) = (1.3, 1f/6)

=

u cos v, y

= usinv; =

u sin v;

In Exercises 9 and 10, (a) express aw/au and aw/av as functions of u and v both by using the Chain Rule and by expressing w directly in terms of u and v before diffurentiatiog. Then (b) evaluate aw/au and aw/av at the given point (u, v). 9. w = xy + yz + xz, (u, v) = (1/2, I)

10. w = In(x 2 + y2 z = ue v ; (u, v)

x = u

+ v,

y

=u

- v, z = uv; y = ue V cos u,

In Exercises II and 12, (a) express au/ax, au/ay, and au/az as functions of x, y, and z both by using the Chain Rule and by expressing u directly in terms of x, y, and z before differentiating. Then (b) evaluate au/ax, au/iJy, and au/az at the given point (x,y, z). p - q 11. u = q _ T' P = X + Y + z, q = x - y + z,

r

=

x

+y

- z;

(x,y,z)

=

(v'3,2, I)

12. u = e qr sin-1p, p = sinx, q = z 2 1ny, (x,y,z) = (1f/4, 1/2, -1/2)

r = lIz;

Using a Branch Diagram In Exercises 13-24, draw a braoch diagram and write a Chain Rule fannula for each derivative.

dz 13. dl forz

= f(x,y),

x

= g(t),

y

dz 14. dl for z 15.

= f(u, v, w), u = g(t), v = h(t), aw aw forw = h(x,y,z), x = f(u,v), au and

av z=k{u,v)

aw aw 18. ax and iJy forw

u

az

az

= g(u, v),

= f(x,y),

ay 20. ar for y = f(u),

y

= k(x,y)

Y

= h(l, s)

= g(p,q),

y

= h(p,q),

= g(r),

y

= h(s)

u x

h(s, I)

=

v = k(p, q)

aw aw 23. ar and iii for w aw 24. as for w

v

u = g(r, s)

= f(x,y,z,v),

z = j(p, q),

= g(l, s),

x

aw aw 21. as and at for w = g(u), aw 22. ap forw

y=k(u,v)

= h(x,y),

= f(x,y),

= g(x,y),

x

x

= h(r, s, I),

y

= k(r, s, I)

Implidt Differentiation Assuming that the equations in Exercises 25-Z8 derme y as a differentiable function ofx, use Theorem 8 to f"md the value of dy/tix at the given point.

25. x' - 2y2

+ X)'

=

(I, 1) 0, (-I, I)

0,

26. X)' + y2 - 3x - 3 = 27. x 2 + X)' + y2 - 7 = 0, Y

+ sinX)' + y

Find the values of

- 1n2

=

0,

(0, InZ)

az/ax and az/ay at the points in Exercises Zg..3Z.

29.

z' - X)' + yz + y' -

30.

xI + yI + ZI

31. sin (x

(1,2)

- I = 0,

+ y) +

sin(y

2 = 0,

(I, I, I)

(2,3,6)

+ z) +

+ z) = 0,

sin (x

32. xe Y + ye' + 2lnx - 2 - 3ln2 = 0,

34. Find aw/av x = v 2ju,y = 35. Find X

=

U -

U

when u = -I, v + v,z = cosu.

=2

aw/av when u = 0, v = 0 2v + 1, y = 2u + v - 2.

u=

0, v

=

1

37. Find az/au and az/av when Stan-Ix and x=e"+lnv.

= g(u,v),

38. Find az/au and az/av wheo u and q = "V'iJ+3 tan-1 u.

(1f,1f,1f)

(I, In 2, 1n3)

Finding Partial Derivatives at Specified Points 33. Find aw/ar wheo r = I,s = -I if w x = r - s,y = cos(r + s),z = sin(r + s).

= k{t)

w

= h(x,y),

s

k{x,y)

36. Find az/au when x = u 2 + v 2 ,y = uv.

= h(l)

= g(x,y),

r

x=h(u,v),

28. xe

+ z2), X = uevsinu, = (-2,0)

= f(r, s, I),

aw () 17. aw au andavforw =gx,y,

=3

t

I = I

6. w

t =

19. al and as for z

In(x 2 + y2 + z2), X = cost, Y 1=3 5. w = 2ye' -Inz, x = In (1 2 + I), y

4. w

= 1ft;

1=0

aw aw 16. ax and iJy forw

= (x + Y + z'/,

if if

w w

= X)' + Inz,

= x 2 + (y/x),

if z = sinxy

+ x siny,

u = In 2, v = I

if z =

= I,

v

=

-Z if z

= In q

14.4

Theory and Examples 39. Assume thatw = f(s'

a

+ t') andf'(x)

_(,S)af

= .'. Find

_

a

a7 and a;'

af

_x'2-

aw

aw as'

783

47. Extreme value. on a helh Suppose that the partial derivatives of a function I(x, y, z) at points on the helix x = cos t, Y = sin t,

z=tare

40. Assume that w - f Is 't ' ax (x,y) - xy, and ay (x,y) F'md ai and

The Chain Rule

= sint,

/y

/x=cost,

f,

= t'

+ t - 2.

At what pcints on the curve, if any, can f take on extreme values?

41. Changing voltage in a circuit TIre voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery out At the same time, the resistance R is increasing as the resistor heats up. Use the equatioo

we""

dV dt

=

aVdl aI dt

+ avdR aR dt

to fmd how the curreot is chaoging at the instaot when R =

48. A space curve Let w = x'.'y cos 3z. Find the value of dw/dt at the point (l,ln2,O) 00 the curve x = cost,y = In(t+ 2),

z

t.

=

49. Temperature on a circle Let T = f(x, y) be the temperature at the point (x, y) on the circle x = cos t, Y = sin t, 0 :5 t :5 2'7T and suppose that

600 ohms, 1= 0.04 amp, dR/dt = 0.5 ohm/sec, and dV/dt = -om volt/sec.

aT -=8x- 4:y ax '

aT ay=8y-4x.

a. Find where the maximum and minimum temperatures on the circle occur by examining tire derivatives dT/ dt and d 2T/ dt 2• b. Suppose that T = 4x' - 4xy + 4y'. Find tire maximum and minimum values of T on the circle.

I

50. Temperature on an ellipse Let T = g (x, y) be tire temperature at the point (x, y) 00 the ellipse

R

42. Changing dimension. in a bOI The lengths a, b, and c of the edges of a rectaogular box are chaoging with time. At the instaot in questioo, a = 1m, b = 2 m, c = 3 m, tla/dt = db/dt = I mfsec, and dc/dt = -3 mfsec. At what rates are tire box's volume V and surface area S changing at that instao!? Are the box's interior diagonals increasing in length or decreasing? 43. If f(u, v, w) is differentiable and u = x - y, v = y - z, and w = z - x, show that

af af af ax + iIy + az =

o.

44. Polar coordinat.. Suppose that we substitute polar coordinates x = r cos 8 and y = r sin 8 in a differentiable function

x

= 2V2 cos t,

o~ t

y = V2 sint,

~

2w,

aod suppcse that

aT

ax

=

aT ay

y,

=

x.

a. Locate the maximum and minimum temperatures on the ellipse by examining dT/ dt and d'T/ dt'. b. Suppose that T = xy - 2. Find the maximum and minimum values of T on the ellipse. Differentiating Integrals

Under mild cootinuity restrictions, it is

true that if

w = f(x,y).

a. Showthat

~;

F(x) = = f,cosB

+ fysinB

and

then F'(x) = }: =

-f, sin B + fy cos B.

/.b

g,(t,x) dt. Using this fact and the Chain Rule, we

F(x)

c. Showthat =

(~;)' + r\ (~;)'.

45. Laplaee equation. Show that if w = f(u, v) satisfies the Laplace equatioo f .. + fvv = 0 aod if u = (x' - y')/2 aod v = xy, then w satisfies the Laplace equation W;g- + Wyy = O. 46. Laplaee equations Let w = f(u) + g(v), where u = x + iy, v = x - iy, and i = \1'=1. Show that w satisfies tire Laplace equation W;a- + w". = 0 if all the necessary functions are differentiable.

g(t,x) dt,

can fmd the derivative of

b. Solve the equations in part (a) to express f, and fy in terms of aw/ar and aw/aB.

(/.)' + (fr)'

/.b

=./.'"')

g(t, x) dt

by letting

G(u, x)

= /." g(t,

x) dt,

where u = f(x). Find the derivatives of the functions in Exercises 51 and 52. 51. F(x)

=

I. 0

,'VI'+7

dt

784

Chapter 14: Partial Derivatives

14.5

Directional Derivatives and Gradient Vectors If you look at the map (Figure 14.25) showing contours along the Hudson River in New York, you will notice that the tributary streams flow perpendicular to the contours. The streams are following paths of steepest descent so the waters reach the Hudson as quickly as possible. Therefore, the fastest instantaneous rate of change in a stream's elevation above sea level has a particular direction. In this section, you will see why this direction, called the "downhill" direction, is perpendicular to the contours.

FIGURE 14.25 Contours along the Hudson River in New York show streams, which follow paths of steepest descent, running perpendicular to the contours.

Directional Derivatives in the Plane We know from Section 14.4 that if I(x,y) is differentiable, then the rate at which I changes with respect to t along a differentiable curve x = get), y = h(t) is

Y Line x =xo + sU!,Y = Yo + sU2

~

dl al dx al dy -= ---+ ---dt

R

--~--~~--------------~x

o

FIGURE 14.26 The rate of change of fin the direction of u at a point Po is the rate at which f changes along this line at Po .

ax dt

ay dt .

At any point Po(xo, Yo) = Po(g(to), h(to)), this equation gives the rate of change of I with respect to increasing t and therefore depends, among other things, on the direction of motion along the curve. If the curve is a straight line and t is the arc length parameter along the line measured from Po in the direction of a given unit vector u, then dl I dt is the rate of change of I with respect to distance in its domain in the direction of u. By varying u, we find the rates at which I changes with respect to distance as we move through Po in different directions. We now define this idea more precisely. Suppose that the function I(x, y) is defined throughout a region R in the xy-plane, that Po(xo, Yo) is a point in R, and that U = Ul i + u2i is a unit vector. Then the equations x = Xo

+ SUh

y = Yo

+ SU2

parametrize the line through Po parallel to u. If the parameter S measures arc length from Po in the direction ofu, we find the rate of change of I at Po in the direction ofu by calculating dl Ids at Po (Figure 14.26).

785

14.5 Directional Derivatives and Gradient Vectors

DEFINmON

The derivative of I at Po(xo,Yo) in the direction of the unit

vector u

+ u2j is the number

= uli

dl) ( tis uRo

I(xo + su"Yo + SU2) - I(xo,yo)

= lim 8-0

S

(1) '

provided the limit exists.

The directional derivative defmed by Equation (1) is also denoted by "The derivative of/at Po in the direction of a"

The partial derivatives Ix(xo, YO) and /y(xo, YO) are the directional derivatives of I at Po io the i andj directions. This observation can be seen by comparing Equation (1) to the definitions of the two partial derivatives given io Section 14.3.

EXAMPLE 1

Usiog the definition, fmd the derivative of

I(x,y)

x 2 + xy

=

at Po(l, 2) io the direction of the unit vector u = (l/Yz)i Solution

+

(l/Yz)j.

Applying the defioition io Equation (1), we obtaio

dl) ( ds a.Po

= lim I(xo 8-0

+ su"yo + SU2)

I(xo,yo)

-

1(1 + S._I_, 2 + s._I_) = lim 8-0

= lim 8-0

Eq. (\)

s

Yz

- 1(1,2)

Yz S

(

S 1+_ Yz

)2+(1+_3)(2+_3)-(12+ 1'2) Yz

Yz

S

~+32 = lim Yz ,~o 3

The rate of change of I(x,y) = x 2

= lim ,~o

(_5_ + 3) _5_. =

Yz

Yz

+ xy atPo(l, 2) io the direction u is 5/V2.

•

Interpretation of the Directional Derivative The equation z = I(x, y) represents a surface S io space. If Zo = I(xo, yo), then the poiot P(xo, Yo, zo) lies on S. The vertical plane that passes through P and Po(xo, YO) parallel to u FIGURE 14.27 The slope of curve Cat Po is lim slope (PQ); this is the Q~P

directional dctivative

(~)"Po = (D.f)po·

iotersects S io a curve C (Figure 14.27). The rate of change of I io the direction ofu is the slope of the tangent to C at P io the right-handed system formed by the vectors u and k. When u = i, the directional derivative at Po is ai/ax evaluated at (XO, YO). When u = j, the directional derivative at Po is al/ay evaluated at (xo, yo). The directional derivative generalizes the two partial derivatives. We can now ask for the rate of change of I io any direction u, not just the directions i and j.

786

Chapter 14: Partial Derivatives For a physical interpretation of the directional derivative, suppose that T = I(x, y) is the temperature at each point (x, y) over a region in the plane. Then I(xo, YO) is the temperature at the point Po(xo, YO) and (D./)po is the instantaneous rate of change of the temperature at Po stepping offin the direction u.

Calculation and Gradients We now develop an efficient formula to calculate the directional derivative for a differentiable function f. We begin with the line

x =

Xo

+

(2)

Y=YO+SU2,

SUl,

through Po(xo, Yo), parametrized with the arc length parameter s increasing in the direction of the unit vectur u = uli + u2j.Thenhythe Chain Rule we f"md Chain Rule for differentiable f From Eqs. (2), dx/ds ~ ", and dy/ds ~ ",

(3) Gradient of / at Po

Direction u

Equation (3) says that the derivative of a differentiable function I in the direction of u at Po is the dot product ofu with the special vectur called the gradient of I at Po.

DEFINITION

The gradient vector (gradient) of I(x, y) at a point Po(xo, YO)

is the vectur

VI = al i + al. ax

ayJ

obtained hy evaluatiog the partial derivatives of I at Po.

The notation VI is read "grad f' as well as "gradient of f" and "del I:' The symbol V hy itself is read "del." Another notation for the gradient is grad f.

THEOREM 9-The Directional Derivative Is a Dot Product differentiable in an open region containing Po(xo, Yo), then

(:to

the dot product of the gradient

=

(Vj)po'u,

(4)

VI at Po and u.

EXAMPLE 2

Find the derivative of I(x,y) = xe Y direction ofv = 3i - 4j. Solution

If I(x,y) is

+ cos (xy) at the point (2,0) in the

The direction ofv is the unit vector obtained by dividing v by its length: u =

v

v

3.

4.

[Vj = 5 = 5' - 5 J·

14.5

Directional Derivatives and Gradient Vectors

787

The partial derivatives of I are everywhere cootinuous and at (2, 0) are given by /x(2,0) = (e Y - y sin (XY»)(2.0) = eO - 0 = 1

2

jj,(2,0) = (xe Y - x sin (XY))(2.0) = 2e o - 2· 0 = 2. 1

The gradient of I at (2, 0) is V/I(2.0) = fx(2,0)i + I y (2,0)j = i + 2j

lor--'~~~~~~~~r x

(Figure 14.28). The derivative of I at (2, 0) in the directioo ofv is therefore

(DuI) 1(2.0) = V/I(2.0)·U

Eq.(4)

-1

=

FIGURE 14.28 Picture Vf as a vector in the domain of f. The figure shows a number oflevel curves off. The rate at whicb f changes at (2, 0) in the directioo U = (3/5)i - (4/5)i is Vf'u = -\

(i + 21')' (Ii - 15 1,) 5

=

I5 - ~5 =

-1

.

•

Evaluating the dot product in the fonnul.

D.I =

V/'u = IV/llulcosO = IVII cos 0,

where 0 is the angle between the vectors u and VI, reveals the following properties.

(Example 2). Properties of the Directional Derivative Dul = VI' U = IVf Icos 0

1. The functioo I increases most rapidly when cos 0 = I or when 0 = 0 and U is the direction of Vj. That is, at each point P in its domain, I increases most rapidly in the directioo of the gradient vector VI atP. The derivative in this directioo is

Dul

= I VII cos (0) = IV/I·

2. Similarly, I decreases most rapidly in the directioo of - Vj. The derivative in this directioo is Dul = IVI Icos ('IT) = -I VI

I·

3. Any directioo U orthogonal to a gradient V/oF 0 is a direction of zero change in I because 0 then equals 'IT/2 and

Dul

o.

= IVII cos ('IT/2) = IV/I'O =

As we discuss later, these properties hold in three dimensions as well as two.

EXAMPLE 3

Find the directions in which I(x,y) = (x 2/2) + (y2/2)

(a) increases most rapidly at the point (I, 1). (h) decreases most rapidly at (1, 1). (c) What are the directioos ofzero change in I at (1, I)? Solution

(a) The functioo increases most rapidly in the directioo of VI at (1, 1). The gradient there is (Vf)(l.l) =

(xi

+ yj)(l.l)

=

i

+ j.

Its directioo is i+j i+j u = Ii + jl = \/(1)2 + (1)2 =

1.1.

y'2'

+

y'21'

(h) The functioo decreases most rapidly in the directioo of - VI at (1, 1), which is -u = __I_i __1_j.

y'2

y'2

788

Chapter 14: Partial Derivatives (0) The directions ofzero change at (I, I) are the directions orthogonal to Vj: n = - I- ,.

Vi

+ -I J.

and

Vi

I.

I.

-n = - , - - J .

Vi

Vi

•

See Figure 14.29. I

k 1

I

-Vf

"'~ (I, I): Y

Most rapid ~ decreaseinf

x

Gradients and Tangents to Level Curves

I

/-- ' - : ~ : I '/

I

Zero change

inf

/

Mostrapid/ increase in!

If a differentiable function I(x, y) has a constant value c along a smooth curve r = g(t)i + h(t)j(making the curve a level curve off), then/(g(t), h(t» = c. Differentiating both sides of this equation with respect to t leads to the equations

!

d

Vf~I+J

FIGURE 14,29 The direction in which f(x,y) increases most rapidly at (I, I) is the direction of VII (1,1) = i + j.Itcorresponds to the direction of steepest ascent on the surface at (I, I, I) (Example 3).

The level curve f(x, y)

~

d/(g(t), h(t))

+

a/. ( ax'

d

dt (c)

=

al dg al dh axdt+aydt=O al .) • (dg . + dh .) = 0 • ayJ dt' dt J

Chain Rule

(5)

dr dt Equation (5) says that VI is normal to the tangent vector dr/dt, so it is normal to the curve.

Vf

f(xo, Yo)

At every point (xo, YO) in the domain of a differentiable function I(x, y), the gradient of I is normal to the level curve through (XO,Yo) (Figure 14.30).

FIGURE 14.30 The gradient ofa differentiable function of two variables at a point is always normal to the function's level curve through that point

Equation (5) validates our observation that streams flow perpendicular to the contours in topographical maps (see Figure 14.25). Since the downflowing stream will reach its destination in the fastest way, it must flow in the direction of the negative gradient vectors from Property 2 for the directional derivative. Equation (5) tells us these directions are perpendicular to the level curves. TIris observation also enables us to find equations for tangent lines to level curves. They are the lines normal to the gradients. The line through a point Po(xo, YO) normal to a vector N = Ai + Bj has the equation

A(x - xo)

+ B(y - YO)

=

0

(Exercise 39). If N is the gradient (V/li-.Yol = fx(xo,yo)i the tangent line given by

1.(xo,Yo)(x - xo)

y

Vf(-2, I)

~

-I

+ 2J

x-2y=-4

EXAMPLE 4

+ /y(XO,yo)j, the

+ /y(XO,yo)(Y - YO)

=

O.

Find an equation for the tangent to the ellipse

4x' + y'

= 2

(Figure 14.31) at the point ( -2, I).

Solution

The ellipse is a level curve of the function

I(x,y) FIGURE 14.31 We can fmd the tangent to the ellipse (x'/4) + y' = 2 by treating the ellipse as a level curve of the function f(x,y) = (x'/4) + y' (Example 4).

=

x' + y'. 4

The gradient of I at ( -2, 1) is

V/I(_"I) =

(~i + 2yj ) (-',I) =

-i

+ 2j.

equation is

(6)

14.5 Directional Derivatives and Gradient Vectors

789

The tangent is the line

(-I)(x

+ 2) +

(2)(y - I) = 0

x - 2y

Eq. (6)

•

-4.

=

If we know the gradients of two functions I and g, we automatically know the gradients of their sum, difference, constant multiples, product, and quotient. You are asked to establish the following rules in Exercise 40. Notice that these rules bave the same form as the corresponding rules for derivatives of single-variable functions.

Algebra Rules for Gradients

+ g)

1. Sum Rule: 2. Difference Rule:

V(j

3. Constant Multiple Rule:

V(kj) = kVI

4. Product Rule:

V(jg) = IVg

v(L)

5. Quotient Rule:

EXAMPLE 5

= VI

+

Vg

V(j - g) = VI - Vg

=

(anynumberk)

+ gVI

gVI - IVg

g

g2

We illustrate two of the rules with

I(x,y) = x - y

g(x,y) = 3y

VI = i - j

Vg = 3j.

We have

1.

V(j - g) = V(x - 4y) = i - 4j = VI - Vg

2.

V(jg) = V(3xy - 3y2) = 3yi

+ (3x -

+ 3yj + (3x j) + (3x - 3y)j j) + (x - y)3j =

=

3y(i - j)

=

3y(i -

=

3y(i -

Rule 2

6y)j

6y)j

gVI

+ IVg

Rule4 •

Functions of Three Variables For a differentiable function I(x, y, z) and a unit vector u =

U1 i

+ u2j + U3k in space, we

have

and Duf = V/·u =

al ax

-U1

al + -U3. al + -U2 ay az

The directional derivative can once again be written in the form

Dal

= V/·u = IV/liul cosO = IV/I cosO,

so the properties listed earlier for functions of two variables extend to three variables. At any given point, I increases most rapidly in the direction of VI and decreases most rapidly in the direction of - Vf. In any direction orthogonal to VI, the derivative is zero.

790

Chapter 14: Partial Derivatives

EXAMPLE 6
-

xy2 -

Z

at Po(1, 1,0) in the direction of

(b) In what directions does I change most mpidly at Po, and what are the mtes of change in these directions?

Solution
Ivl = Y(2)' + (-3)2 + (6)2 = v'49 = 7 v 2. 3.+6 k u=fVj=,p-,p 7'

The partial derivatives of I at Po are

Ix = (3x 2 -

y2)(l,l,O)

= 2,

Iy = -2xy1 (1 ,1,0) = -2,

Iz =

-11(1,1,0) = -\.

The gmdient of I at Po is

VII 0,1,0)

= 2i - 2j - k.

The derivative of I at Po in the direction ofv is therefore (DaI)(l,l,O) =

V/b,l,O)'U

=

7 + 7 -7

4

6

=

(2i - 2j - k)' (ti -

=

7'

6

~j + ~k)

4

(b) The function increases most mpidly in the direction of VI = 2i - 2j - k and decreases most mpidly in the direction of - Vf. The mtes of change in the directions are, respectively,

IV/I

= Y(2)' + (-2), + (-1)2 = V9 = 3

and

-I VII

= -3,

•

Exercises 14.5 Calculattng Gradients

finding Directional Derlvattves

In Exercises 1- 6, find the gradient of the function at the given point. Then sketch the gradient together with the level curve that passes through the point.

In Exercises 11-18, fmd the derivative of the function at p. in the di-

1. f(x,y) = y - x,

(2, I)

3. g(x,y) = xy', (2, -I) 5. f(x,y)

= V2x + 3y,

6. f(x,y) = tan- 1

'7,

2, f(x,y) = In (x' + y'), (1,1) ,

4. g(x,y) = x 2

y'

Z' (v'2,I)

rection of u. 11. f(x,y) = 2xy - 3y', P.(5,5), u = 4i + 3j 12. f(x,y)

=

2x'

+ y',

u = 3i - 4j

x-y 13. g(x,y) = xy + 2' p.(I, -I), u = 12i + 5j

(-1,2) (4, -2)

p.( -I, I),

14. h(x,y) = tan-I i,y/x) u=3i-2j

+ v'3 sin- l (xy/2),

15. f(x,y, z) = xy + yz + n:, p.(I, -1,2), In Exercises 7-10, fmd Vf at the given point.

7. f(x,y,z) = x' + y' - 2z' + zlnx, (I, I, I) 8. f(x,y,z) = 2z' - 3(x' + y')z + tan- 1Xl, (1,1,1) 9. f(x,y,z) = (x' + y' + z')-l/2 + In (xyz), (-1,2, -2) 10. f(x,y,z) = e'+'cosz

+ (y + I) sin-Ix, (O,O,1T/6)

p.(I, I),

u = 3i + 6j - 2k

16. f(x,y,z) = x' + 2y' - 3z', p.(I, I, I), u = i + j + k 17. g(x,y,z) = 3e'cosyz, P.(O, 0, 0), u = 2i + j - 2k 18. h(x,y, z) = cosxy + eYO + Inn:, P.(i,O, 1/2), u=i+2j+2k

791

14.6 Tangent Planes and Differentials In Exercises 19-24, fmd the directions in which the functions increase and decrease most rapidly at Po. Then rmd the derivatives of the functions in these directions.

+ >;)' + y', Pot -I, I) x'y + e'" siny, Po(!,O)

19. I(x,y) = x'

20. 21. 22. 23. 24.

I(x,y) = I(x,y, z) = g(x,y,z) = I(x,y, z) = h(x,y, z) =

(x/y) - yz, Po(4, I, I) xe Y + z', P o(1,ln2,1/2) In>;)' + Inyz + Inxz, Po(!, I, I) In (x' + y' - I) + y + 6z, Po(1, 1,0)

Tangent LInes to Level Curves

31. Zero directional derivative In what directioo is the derivative of I(x, y) = >;)' + y' at P(3, 2) equal to zero? 32. Zero directional derivative In what directions is the derivative of I(x,y) = (x' - y')/(x' + y') atp(l, I) equal to zero? 33. Is there a direction n in which the rate of change of I(x, y) = 3>;)' + 4y' at P(I, 2) equals 14? Give reasoos for your

x' -

answer. 34. Changing temperatnre along a circle Is there a direction u in which the rate of change of the temperatore functioo T(x, y, z) = 2>;Y - yz (temperature in degrees Celsius, distance in feet) at P(I, -I, I) is -3°C/It? Give reasons for your answer.

In Exercises 25-28, sketch the curve I(x, y) = c together with VI and the tangent line at the given point Then write an equation for the tangent line.

35. The derivative of I(x,y) at Po(I,2) in the direction ofi + j is 2V2 and in the direction of -2j is -3 . What is the derivative of I in the directioo of -i - 2j? Give reasons for yoor answer.

x' + y' = 4, (v'2, v'2) 26.x'-y=l, (v'2,I)

36. The derivative of I(x, y, z) at a point P is greatest in the direction of v = i + j - k. In this direction, the value of the derivative is

25.

27. >;)' = -4,

2v'3.

(2, -2)

28.x'->;),+y'=7,

a. What is Vf at P? Give reasons for your answer.

(-1,2)

b. What is the derivative of I at P in the direction ofi

Theory and Examples 29. Let I(x,y) = x' - >;)' + y' - y. Find the directions u and the values of D. 1(1, -I) for which

b.

c. Du/(1, -I) e. Du/(1, -I)

d. D./(!,-I)

30. Let I(x,y) =

D./(-H)

= =

0 -3

~~ ~ ~~. Find the directions u

=

4

derivative of a differentiable function I(x,y, z) at a point Po in the direction of a unit vector u related to the scalar component of (V/)p, in the direction ofn? Give reasoos foryoor answer.

+~) is largest

b. D.

c. Du/(

+~) = 0

d.

38. Directional derivative. and partial derivative. Assuming that the necessary derivatives of I(x, y, z) are dermed, how are DI/, DJ/, and D.I related to I., I y , and I.? Give reasons for your

answer.

and the values of

39. Line. in the xy-plane Show that A(x - xo) + B(y - Yo) = 0 is an equation for the line in the xy-plane through the point (xo,Yo) norma1 to the vectorN = Ai + Bj.

for which

a. Du/(

e.

37. Directional derivatives and scalar components How is the

Do I(!, -I) is smallest

a. Du 1(1, -I) is largest

1(+ ~)

is smallest

D./(-t,~) =-2

40. The algebra rule. for gradient. gradients

Du/(+~) = I

14.6

+ j?

VI

= al i + al. + al k, ax

ayJ

az

Given a constant k and the

ag ag ag Vg=-i+-j+-k, ax ay az

establish the algebra rules for gradients.

Tangent Planes and Differentials In this section we derme the tangent plane at a point on a smooth surface in space. Then we show how to calculate an equation of the tangent plane from the partial derivatives of

the function defining the surface. This idea is similar to the dermition of the tangent line at a point on a curve in the coordinate plane for single-variable functions (Section 3.1). We then study the total differential and linearization of functions of several variables.

Tangent Planes and Normal Lines If r = g(t)i + h(t)j + k(t)k is a smooth curve on the level surface f(x, y, z) = c of a differentiable function f, then f(g(t), h(t), k(t)) = c. Differentiating both sides of this

792

Chapter 14: Partial Derivatives

equation with respect to t leads to

d dtf(g(t), h(t), k(t))

=

d dt (c)

+ af dh + af dk

=

0

af dg ax dt

FIGURE 14.32 The gradient VI is orthogonal to the velocity vector of every smooth curve in the surface through Po. The velocity vectors at Po therefore lie in a common plane, which we call the tangent plane at Po.

f ( a~ i

+ af. + af ~J

'

~

ay dt

az

Chain Rule

dt

k)' (d~g i + dh. + dk k) O. ~J ~ =

~

(1)

' --~

VI

dr/dt

At every point along the curve, Vf is orthogonal to the curve's velocity vector. Now let us restrict our attention to the curves that pass through Po (Figure 14.32). All the velocity vectors at Po are orthogonal to Vf at Po, so the curves' tangent lines all lie in the plane through Po normal to Vf. We now define this plane.

DEFINITIONS The tangent plane at the point Po(xo, Yo, zo) on the level surface f(x, y, z) = c of a differentiable function f is the plane through Po normal to Vflpo' The normal line of the surface at Po is the line through Po parallel to Vf

IPo'

From Section 12.5, the tangent plane and normal line have the following equations:

Tangent Plane to f(x,y, z)

fx(Po)(x - xo)

=

cat Po(xo,Yo, zo)

+ fy(Po)(y - Yo) + fz(Po)(z - zo) =

0

(2)

Normal Line to f(x,y, z) = cat Po(xo,Yo, zo)

x z

Po(l, 2, 4)

1

:/

EXAMPLE 1

The surface

xZ+r+z-

9 =O

I

=

Xo

+ fiPo)t,

z = Zo

+ fAPo)t

(3)

Find the tangent plane and normal line of the surface

f(x,y, z)

=

x 2 + y2

+z

- 9 = 0

A circular paraboloid

at the point Po(1, 2, 4).

I

Nonnalline

SoLution The surface is shown in Figure 14.33. The tangent plane is the plane through Po perpendicular to the gradient of f at Po . The gradient is

Vflpo = (2xi

+

2yj

+

k)(1 ,2,4)

= 2i + 4j + k.

The tangent plane is therefore the plane x

/

FIGURE 14.33 The tangent plane and normal line to this surface at Po (Example I).

2(x - 1)

+ 4(y

- 2)

+

(z - 4) = 0,

or

2x

+ 4y + z =

14.

The line normal to the surface at Po is

x = 1

+

2t,

y = 2

+ 4t,

z = 4

+

t.

•

To find an equation for the plane tangent to a smooth surface z = f(x, y) at a point = f(xo, Yo), we first observe that the equation z = f(x, y) is

Po(xo, Yo, zo) where Zo

793

14.6 Tangent Planes and Differentials

equivalent to f(x,y) - z = O. The surface z = f(x,y) is therefore the zero level surface of the function F(x, y, z) = f(x, y) - z. The partial derivatives of Fare

o

Ix -

F. = ox (f(x,y) - z) =

0 =

Ix

o

Fy = oy (f(x,y) - z) = /y - 0 = /y

o

Fz = oz(f(x,y) - z) = 0 - 1 = -1.

The formula F'(Po)(x - xo)

+ Fy(Po)(y - Yo) + F.{Po)(z - zo)

=

0

for the plane tangent to the level surface at Po therefore reduces to f.(xo,yo)(x - xo)

+ fy(xo,yo)(Y - Yo) - (z - zo)

=

O.

Plane Tangent to a Surface z = f(x,y) at (Xo,Yo, fVco,yo» The plane tangent to the surface z = f(x, y) of a differentiable function f at the poiot Po(xo,Yo, zo) = (xo,Yo, f(xo,yo)) is f.(xo,yo)(x - xo)

EXAMPLE 2

+

fY(xo,yo)(Y - Yo) - (z - zo) = O.

(4)

Fiod the plane tangent to the surface z = xcosy - ye' at (0,0,0).

Solution We calculate the partial derivatives of f(x, y) = x cos y - ye' and use Equation (4): f.(O,O) = (cosy - ye')(o.o) = 1 - 0·1 = 1

/y(O,O)

=

(-x sioy - e')(o.o)

0- 1

=

-1.

=

The tangent plane is therefore 1'(x - 0) - 1'(Y - 0) - (z - 0) = 0,

or x - y - z =

EXAMPLE 3

Eq.(4)

o.

•

The surfaces f(x,y,z) = x 2

+ y2

- 2 = 0

A cylinder

and g(x,y,z) = x

+z -

4 = 0

ApJane

meet in an ellipse E (Figure 14.34). Fiod parametric equations for the lioe tangent to E at the poiot Po(I, 1,3). Solution The tangent lioe is orthogonal to both Vf and Vg at Po, and therefore parallel tov = Vf X Vg. The components ofv and the coordinates of Po give us equations for the lioe. We bave Vfl(l.l~) = (2xi

Vgl (l.l~)

FIGURE 14.34 This cylinder and plane intersect in an ellipse E (Example 3).

=

(i

+ 2yj)(1.1.3)

+ k)(1.1.3)

v = (2i

=

i

=

2i

+ 2j

+k

+ 2j) X (i + k)

=

2 1

j 2 0

k 0 = 2i - 2j - 2k. 1

794

Chapter 14: Partial Derivatives The tangent line is

y=I-2t,

x=I+2t,

•

z=3-2t.

Estimating Change in a Specific Direction The directional derivative plays the role of an ordinary derivative when we want to estimate how much the value of a function / changes if we move a small distance tis from a point Po to another point nearby. If / were a function of a single variable, we would have

d/

=

f'(Po) tis.

Ordinary derivative x increment

For a function of two or more variables, we use the formula

d/

=

(V/lpo"u) tis,

Directional derivative x increment

where u is the direction of the motion away from Po.

Estimating the Change in / in a Direction u To estimate the change in the value of a differentiable function / when we move a small distance tis from a point Po in a particular direction u, use the formula

d/

=

(V/lpo" u)

tis

Directional Distance derivative increment

EXAMPLE 4

Estimate how much the value of

/(x,y,z) = ysinx + 2yz will change if the point P(x,y, z) moves 0.1 unit from Po(O, 1,0) straight toward Pt(2, 2, -2).

Solution We first fmd the derivative of / at Po in the direction of the vector P-;;Pt = 2i + i - 2k. The direction of this vector is

The gradient of / at Po is

V/I(O.1.0)

=

((ycosx)i

+

(sinx

+ 2zli + 2yk)(O.1.0)

= i

+ 2k.

Therefore,

V/lpo·u = (i

+

2k)· (t + ti - tk) t -t -to i

=

=

The change d/ in / that results from moving tis = 0.1 unit away from Po in the direction of u is approximately

d/

=

(V/lpo"u)(tis) =

(-t)tO.I) '"

-0.067unit.

•

How to Linearize a Function of Two Variables Functions of two variables can be complicated, and we sometimes need to approximate them with simpler ones that give the accuracy required for specific applications without being so difficult to work with. We do this in a way that is similar to the way we fmd linear replacements for functions of a single variable (Section 3.9).

14,6 Tangent Planes and Differentials

A point _ near (xQ. Yo)

(x, y)

Suppose the function we wish to approximate is z = I(x, y) near a point (xo, Yo) at which we know the values of I, Ix> and Iy and at which I is differentiable. Ifwe move from (xo, Yo) to any nearby point (x, y) by increments Ax = x - Xo and ~y = y - Yo (see Figure 14.35), then the defmition of differentiability from Section 14.3 gives the change

I(x,y) - I(xo,yo) A point where f is differentiable

795

=

fx(xo,Yo)Ax

+ ly(xo,Yo)~y + EtAx +

E2~y,

where E1, E2 --> 0 as Ax, ~y --> O. If the increments Ax and ~y are small, the products EJAx and E2~y will eventoally be smaller still and we have the approximation

/

I(x,y) "" I(xo,yo)

+ fx(xo,Yo)(x - xo) + /y(xo,Yo)(y - yo). L(x,y)

If! is differentiable at (xo, yo), then the value of! at aoy point (x, y) nearby is approximately !(xo,Yo) + fx(xo,Yo)/!x + !y(xO,yo)4y. FIGURE 14.35

In other words, as long as as the linear function L.

DEFINmONS

I

~x

and ~y are small, I will have approximately the same value

The linearization of a function I(x, y) at a point (xo, Yo) where

is differentiable is the function

L(x,y)

I(xo,yo) + fx(xo,Yo)(x - xo) + /y(xo,Yo)(y - yo).

=

(5)

The approximation

I(x,y) "" L(x,y) is the standard linear approximation of I at (xo, yo).

From Equation (4), we fmd that the plane z = L(x,y) is tangent to the surface z = I(x, y) at the point (xo, yo). Thus, the linearization of a function of two variables is a tangent-plane approximation in the same way that the linearization of a function of a single variable is a tangent-line approximation. (See Exercise 63.)

EXAMPLE 5

Find the linearization of

I(x,y)

=

x 2 - xy

+ ty2 + 3

at the point (3, 2).

Solutton

We first evaluate 1./.. and Iy at the point (xo, Yo) = (3,2):

1(3,2)

(x2 - xy

=

fx(3,2)

= :

/y(3,2)

=

x

+ 1 y2 +

(x2 - xy

aa (x2 - xy Y

2

3)

+ -21 y2 +

= (3,2)

3)(3,2)

+ -21 y2 + 3)

(3,2)

8 =

(2x - y)(3,2)

=

(-x

+ Y)(3,2)

=

4

=

-I,

giving

L(x, y)

=

I(xo, Yo) + fx(xo, Yo)(x - xo) + /y(xo, Yo)(Y - Yo)

=

8

+ (4)(x - 3) + (-I)(y - 2)

The linearization of I at (3, 2) is L(x,y) = 4x - y - 2.

=

4x - y - 2.

•

When approximating a differentiable function I(x, y) by its linearization L(x, y) at (xo, Yo), an important question is how accurate the approximation might be.

796

Chapter 14: Partial Derivatives

y

If we can f'md a common upper bound M for II", I, IIyy I, and II", I on a rectangle R centered at (xo,Yo) (Figure 14.36), then we can bound the error E throughout R by using a simple formula (derived in Section 14.9). The error is def'med by E(x,y) =

I(x,y) - L(x,y).

R ~--------------~X

o

FIGURE 14.36 The rectangular region R: Ix -.tol '" h, Iy - yol '" kinthe xy·plane.

The Error in the Standard Linear Approximation If I has continuous first and second partial derivatives throughout an open set containing a rectangle R centered at (xo, YO) and if M is any upper bcund for the values of 1/",1, I/yyl, and 1/",1 on R, then the error E(x, y) incurred in replacing I(x, y) on R by its linearization

L(x, y)

=

I(xo, YO) + fx(xo, yo)(x - xo) + /y(xo, YO)(y - YO)

satisfies the inequality

IE(x,y) I'" t M(lx - xol + Iy - yoll'·

To makeIE(x,y) I small for a givenM, we just make Ix - xol andly - yol small.

EXAMPLE 6 Find an upper bound for the error in the approximation I(x, y) '" L(x, y) in Example 5 over the rectangle R:

Ix - 31 '" 0.1,

Iy - 21'" 0.1.

Express the upper bound as a percentage of 1(3, 2), the value of I at the center of the rectangle. Solution

We use the inequality

IE(x,y) I'" t M(lx - xol + Iy - yoll'· To f'md a suitable value for M, we calculate I ",,j"', and I yy, f'mding, after a routine differentiation, that all three derivatives are constant, with values 1/",1 = 121 = 2,

I/yyl = III = I.

1/",1 = I-II = I,

The largest of these is 2, so we may safely takeMto be 2. With (xo, Yo) = (3,2), we then know that, throughout R,

IE(x,y) I '" t(2)(lx - 31

+ Iy

- 21)2 = (Ix - 31

+ Iy

- 21)'.

Finally, since Ix - 31 '" 0.1 andly - 21 '" 0.1 onR, we have

IE(x,y) I '" (0.1 + 0.1)2

= 0.04.

As a percentage of 1(3,2) = 8, the error is no greater than

0.~4

X 100 = 0.5%.

•

Differentials Recall from Section 3.9 that for a function of a single variable, y = I(x), we defined the change in I as x changes from a to a +
t:.1 = I(a

+
and the differential of I as

dl

=

f'(a)
14.6 Tangent Planes and Differentials

797

We now consider the differential of a function of two variables. Suppose a differentiable function f(x, y) and its partial derivatives exist at a point (XO,yo). If we move to a nearby point (xo + at,yO + Jly), the change in f is Jlf = f(xo

+ Jlx,yo + Jly) - f(xo,yo).

A straightforward calculation from the definition of L(x, y), using the notation x - Xo = Jlx and y - Yo = Jly, shows that the corresponding change in L is

+ at,yO + Jly) - L(xo,yo) fx(xO,yo)at + /y(XO,yo)Jly.

JlL = L(xo =

The differentials ax and dy are independent variables, so they can be assigned any values. Often we take ax = Jlx = x - Xo, and dy = Jly = y - Yo. We then have the following def'mition of the differential or total differential of f. DEFINmON If we move from (XO,Yo) to a point (xo + ax,yO + dy) nearby, the resulting change df

=

fx(xo , YO) ax + /y(XO,yo) dy

in the linearization of f is called the total differential of f.

EXAMPLE 7 Suppose that a cylindrical can is designed to have a redius of 1 in. and a height of 5 in., but that the radius and height are off by the amounts dr = +0.03 and dh = -0.1. Estimate the resulting absolute change in the volume of the can. 7Tr 2h, we use

Solution

To estimate the absolute change in V

With V,

JlV'" dV = V,(ro, ho) dr + Vh(ro, ho) dh. 27Trh and Vh = 7Tr 2, we get

=

dV

=

27Troho dr + 7Tr02 dh

=

0.37T - 0.17T

=

=

=

27T(1)(5)(0.03) + 7T(1)2( -0.1)

0.27T '" 0.63 in3

-

EXAMPLE 8 Your company manufactures right circular cylindrical molasses stomge tanks that are 25 ft high with a redius of 5 ft. How sensitive are the tanks' volumes to small variations in height and redius? Solution With V = 7Tr 2h, the total differential gives the approximation for the change in volume as

dV = V,(5, 25) dr + Vh(5, 25) dh 2 = (27Trh)(,,2') dr + (7Tr )(,,2') dh = 2507T dr + 257T dh.

r= 5

/'

r=

(aJ

25

(bJ

FIGURE 14.37 The volume of cylinder
Thus, a I-unit change in r will change V by about 2507T units. A I-unit change in h will change Vby about 257T units. The tank's volume is 10 times more sensitive to a small change in r than it is to a small change of equal size in h. As a quality control engineer concerned with being sure the tanks have the correct volume, you would want to pay special attention to their radii. In contrast, if the values of r and h are reversed to make r = 25 and h = 5, then the total differential in Vbecomes dV = (27Trh)(2',5) dr + (7Tr 2)(2',5) dh = 2507T dr + 6257T dh. Now the volume is more sensitive to changes in h than to changes in r (Figure 14.37). The general rule is that functions are most sensitive to small changes in the variables that generate the largest partial derivatives. _

798

Chapter 14: Partial Derivatives

EXAMPLE 9 The volume V = 7Tr'-h of a right circular cylinder is to be calculated from measured values ofr and h. Suppose that r is measured with an error of no more than 2% and h with an error of no more than 0.5%. Estimate the resulting possible percentage error in the calculation of V. Solution

We are told that

I~XIOOI';2

1~h X 1001 ,; 0.5.

and

Since

dV V

27Trhdr + 7Tr 2 dh 7Tr 2h

=

=

2dr r

+ dh h'

we have

Id;1

=

12~ + ~hl

,;

12~1 + I~I

,; 2(0.02) + 0.005

=

0.045.

•

We estimate the error in the volume calculation to be at most 4.5%.

Functions of More Than Two Variables Analogous results hold for differentiable functions of more than two variables.

1.

The linearization of I(x,y, z) at a point Po(xo, Yo, zo) is L(x,y,z) = I(Po)

2.

+

fx(Po)(x - xo)

- YO)

+ iz(Po)(z

- zo).

Suppose that R is a closed rectangular solid centered at Po and lying in an open region on which the second partial derivatives of I are continuous. Suppose also that 1/",1, I/)'}'I, Ilrl, 1/",,1, Il",l, andl/yzl are all less than or equal to Mthroughout R. Then the error E(x,y, z) = I(x,y,z) - L(x,y, z) in the approximation of Iby L is bounded throughout R by the inequality lEI,; !M(lx - xol

3.

+ ly(Po)(y

+ Iy - yol +

Iz - zoll'.

If the second partial derivatives of I are continuous and if x, y, and z change from xo,Yo, andzo by small amounts ax, dy, anddz, the total differential

ax + I,(Po) dy + IAPo) dz

dl = fx(Po)

gives a good approximation of the resulting change in I.

EXAMPLE 10

Find the linearization L(x,y, z) of I(x,y,z) = x 2

at the point (xo,Yo, zo) = (2, by L on the rectangle

I

R:

Solution

Ix -

-

xy

+ 3 sinz

1,0). Find an upper bound for the error incurred in replacing

21 ,; 0.01,

Iy -

11 ,; 0.02,

Izl';

0.01.

Routine calculations give

1(2, 1, 0)

= 2,

1.(2, 1, 0)

= 3,

/y(2, 1,0)

=

-2,

1.(2, 1, 0)

= 3.

799

14.6 Tangent Planes and Differentials

Thus,

L(x,y, z) = 2

+ 3(x

- 2)

+ (-2)(y

- 1)

+ 3(z -

0) = 3x - 2y

+ 3z

- 2.

Since

In =

2,

Iyy =

In

0,

Ixy

= -3 sinz,

=

-I,

Ixz =

0,

Iyz =

0,

andl-3 sinzl:5 3 sin 0.01 '" .03, we may take M = 2 as a bound on the second partials. Hence, the error incurred by replacing I by L on R satisfies lEI :5

t

(2)(0.01

+ 0.02 + 0.01)'

= 0.0016.

•

Exercises 14.6 Tangent Planes and Normal Lines to Surfaces In Exercises 1--1l, find equations for the

Estimating Change 19. By about how much will

/(x,y,z) = InYx' + y' + z'

(a) tangent plane and (b) nonnalline at the pnint Po on the given sorface.

+ y' + z' 2. x' + y' - z' 1. x'

3. 2z -

x' =

0,

change if the point P(x, y, z) moves from Po(3, 4,12) a distance of ds = 0.1 unit in the direction of3i + 6j - 2k?

Po(l, I, 1)

= 3,

20. By about how much will

Po(3, 5, -4)

= 18,

f(x,y,z) = eXcosyz

Po(2, 0, 2)

4. x' + 2xy - y' + z' = 7, Po(l, -I, 3) 5.

COS?TX -

x'y + e

= 0,

6. x' - xy - y' - z

change as the point P(x, y, z) moves from the origin a distance of = 0.1 unit in the direction of 2i + 2j - 2k?

ds

+ yz = 4, Po(O, 1,2)

D

21. By about how much will

Po(l, I, -I)

+ Y + z = I, Po(O, 1,0) 8. x' + y' - 2xy - x + 3y - z = -4, Po(2, -3, 18)

7. x

In Exercises c}'-12, fmd an equation for the plane that is tangent to the given surface at the given point. 9. z = In (x'

(1,0,0) 10. z = e-("+Y~,

+ y'),

11. z =~,

(1,2, 1)

12. z

=

4x'

+ y',

(0,0, I)

(I, I, 5)

Tangent Lines to Space Curves In Exercises 13-18, fmd parametric equations for the line tangent to the curve of intersection of the surfaces at the given point. 13. Surfaces: x Point:

+ y' + 2z =

Point: Point:

x' + 2y' + 3z' =

x' + 2y + 2z = 4, + y' + z

=I

y = I

+ 3x'y' + y' + 4xy + y2 + z2 = 11

z' = 0,

(I, 1,3)

+ y2

=

4,

(Yz, Yz, 4 )

x2

+ y2

-

h(x,y,z) = cos (=y)

+ xz'

change if the point P(x, y, z) moves from Po( -I, -I, - 1) a distance of ds = 0.1 unit toward the origin? 23. Temperatore change along a circle Suppose that the Celsius temperature at the point (x,y) in thexy-plane is T(x,y) = x sin2y and that distance in the xy-plane is measured in meters. A particle is moving clockwise around the circle of radius 1 m centered at the origin at the constant rate of2 m/sec.

b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?

(1/2, I, 1/2)

18. Surfaces: x 2 Point:

= 2,

22. By about how much will

p( 1/2, \/3/2 'fI

6

(I, I, 1/2)

17. Surfaces: x' x2 Point:

y

change if the pnint P(x, y, z) moves from Po(2, -1,0) a distance ofds = 0.2 unit toward the pointP,(O, 1,2)?

•• How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

(I, I, I)

16. Surfaces: x Point:

x = I

(I, I, I)

14. Surfaces: xyz = I, 15. Surfaces:

4,

g(x,y,z) = x + xcosz - ysinz + y

Z =

0

24. Changing temperature along a space curve The Celsius temperature in a region in space is given by T(x,y, z) = 2x' - xyz. A particle is moving in this region and its position at time t is given by x = 212, Y = 31, Z = -12, where time is measured in seconds and distance in meters. a. How fast is the temperature experienced by the particle changing in degrees Celsius per meter when the particle is at thepointP(8,6, -4)? b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?

800

Chapter 14: Partial Derivatives

Finding Unearizations

34. f(x,y) = (1/2).<2

In Exercises 25--30, f"md the linearization L(x, y) of the funetion at

each point. 25. f(x,y)

26. f(x,y) 27. f(x,y)

28. f(x, y) 29. f(x,y) 30. f(x,y)

= x 2 + y2 + I at = (x + Y + 2)2 at = 3x - 4y + 5 at = x'y4 at = .'cosyat = .2y-, at

a. (0,0),

b. (I, I)

•• (0,0),

b. (1,2)

•• (0,0),

b. (I, I)

a. (I, I),

b. (0,0)

a. (0,0),

b. (0,1f/2)

a. (0,0),

b. (1,2)

n=

35.74

R:

R:

partial wind chill chart is given.

I

5

0

-5 -11

-5 -10 -16 -22

7

15

9

3

-4 -10 -16 -22 -28

15 19 13 v (mph) 20 17 11

6

0

-7 -13

4

-2

25 16

9

3

-4

-11

-17 -24 -31

-37

30 15

8

I

-5

-12 -19 -26 -33

-39

35 14

7

0

-7

-14 -21

L

-19

-26 -32

-9 -15 -22

-29 -35

-27

-34 -41

Use the table to f"md W(20, 25), W(30, -10), and W(15, 15).

b. Use the formula to f"md W(IO, -40), W(50, -40), and W(60, 30).

c. Find the linearizationL(v,

n of1he funetion W(v, n at1he

point (25, 5).

d. Use L( v, values.

n in part (c) to estimate the fullowing wind chill

i) W(24,6)

n

n

32. Find the linearization L( v, of 1he function W( v, in Exercise 31 at 1he point (50, -20). Use it to estimate the following wind chill values. b. W(53, -19)

c. W(60, -30)

Bounding the Error in Unear Approximations In Exercises 33-38, find the linearization L(x, y) of 1he function f(x, y) at Po. Then f"md an upper bound for the magnitude lEI of the error in the approximation f(x, y) '" L(x, y) over the rectang1e R. 33. f(x,y) = x 2 - 3xy + 5 at Po(2, I),

R:

Ix - 21 s 0.1,

R:

= eX cosy at

Ixl s 0.1,

Iy - II s 0.1

ly-2Is0.1 Po(O, 0),

Iyl s 0.1

(Use eX s 1.11 aodI cosYI s I in estimating E.) 38. f(x,y) = Inx R:

+ Inyat

Ix - II '" 0.2,

Po(\, I),

Iy - II '" 0.2

Unearizations for Three Variables Find 1he linearizations L(x, y, z) of the funetions in Exercises 39-44 at 1he given points.

39. f(x,y, z) = xy

+ yz + xz at

=

b. (1,0,0) x2

c. (0,0,0)

+ y2 + z2 at

a. (I, I, I)

b. (0, 1,0)

41. f(x,y, z) = \lx 2

c. (1,0,0)

+ y2 + z2 at

a. (1,0,0)

c. (1,2,2)

b. (I, I, 0)

42. f(x,y, z) = (sinxy)/z at

a. (1f/2, I, I)

b. (2,0, I)

43. f(x,y, z) = e' + cos (y + z) at a. (0,0,0)

b. (0,1',0)

44. f(x,y, z) = tan-1 (xyz) at

a. (1,0,0)

c. (I, I, I)

b. (I, I, 0)

In Exercises 45-48, find the linearization L(x, y, z) of 1he function f(x, y, z) at Po. Then f"md ao upper bouod for the magnitude of the error E in the approximation f(x, y, z) '" L(x,y, z) over the region R.

45. f(x,y, z) = xz - 3yz R:

Ix - II s 0.01, 46. f(x,y, z) = x 2 + xy R:

Ix - II s 0.01,

+2

at Po(\, 1,2),

Iy - II s 0.01,

+ yz + (1/4)z2 Iy - II s 0.01,

Iz - 21 s 0.02 at P o(\, 1,2), Iz - 21 s 0.08

47. f(x,y, z) = xy + 2yz - 3xz at P o(\, 1,0),

il) W(27,2)

iii) W(5, -10) (Explain why this value is much different from 1he value fouod in 1he table.)

•• W(49, -22)

Iyl s 0.2

Ix-llsO.I,

40. f(x,y, z)

19 13

10 21

Ixl s 0.2,

a. (I, I, I) T("F)

5 25

Ix - 21 '" 0.1, Iy - 21'" 0.1 + y + xcosyat Po(O,O),

37. f(x,y)

+ 0.6215 T - 35.75 v O•16

10

Po(2,2),

(Useleosyl s I andlsinyl s I inestimatiogE.)

where T is air temperature in OF and v is wind speed in mph. A

15

+ 4 at

36. f(x,y) = xy2 + Y cos (x - I) at P o(\,2),

+ 0.4275 T· v°.16,

30 25 20

3x - 3y

35. f(x,y) = I

31. Wmd chill factor Wmd chill, a measure of the apparent temperature felt on exposed skin, is a funetion of air temperature and wind speed. The precise formula, updated by the National Weather Service in 200 I and based on modern heat transfer theory, a human face model, and skin tissue resistance, is

W = W(v,

R:

+ xy + (I/4)y2 +

R:

Ix - II s 0.01,

48. f(x,y, z) =

R:

Iy - II s 0.01,

v'2 cos x sin (y + z)

Ixl s 0.01,

Iyl s 0.01,

Izl s 0.01

at Po(O, 0, 1f/4),

Iz - 1f/41 s 0.01

Estimating Error; Sensitivity to Change 49. Estimatiog muimum error Suppose that T is to be fouod from 1he formula T = x (e Y + e -Y), where x aod y are fouod to be 2 and In 2 with maximmn possible errors of Idx I = 0.1 and I
14.6 Tangent Planes and Differentials 51. Consider a closed rectangular box with a square base as shown in the accompanying figure. If x is measured with error at most 2% and y is measured with error at most 3%, use a differential to estimate the corresponding percentage error in computing the box's a. surface area b. volume.

801

55. You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer. 56. a. Around the point (1, 0), is f(x, y) = X 2(y + 1) more sensitive to changes in x or to changes in y? Give reasons for your answer. b. What ratio of dx to dy will make df equal zero at (1, O)?

y

57. Error carryover in coordinate changes y P(3 ± 0.01,4 ± 0.01)

52. Consider a closed container in the shape of a cylinder of radius 10 cm and height 15 cm with a hemisphere on each end, as shown in the accompanying figure. 15 cm JlOcm

- -l<----''-----.l...--__+x

o

3

I

I I

The container is coated with a layer of ice 1/2 cm thick. Use a differential to estimate the total volume of ice. (Hint: Assume r is radius with dr = 1/2 and h is height with dh = 0.) 53. Maximum percentage error If r = 5.0 cm and h = 12.0 cm to the nearest millimeter, what should we expect the maximum percentage error in calculating V = 7Tr 2h to be? 54. Variation in electrical resistance The resistance R produced by wiring resistors of RJ and R2 ohms in parallel (see accompanying figure) can be calculated from the formula

a. If x = 3 ± 0.01 and y = 4 ± 0.01, as shown here, with approximately what accuracy can you calculate the polar coordinates rand () of the point P(x, y) from the formulas r2 = x 2 + y2 and () = tan- J (y/x)? Express your estimates as percentage changes of the values that r and () have at the point (XO,yo) = (3,4). b. At the point (xo, YO) = (3,4), are the values of rand () more sensitive to changes in x or to changes in y? Give reasons for your answer. 58. Designing a soda can A standard 12-fl-oz can of soda is essentially a cylinder ofradius r = 1 in. and height h = 5 in. a. At these dimensions, how sensitive is the can's volume to a small change in radius versus a small change in height?

a. Show that

b. Could you design a soda can that appears to hold more soda but in fact holds the same 12 fl oz? What might its dimensions be? (There is more than one correct answer.) 59. Valueofa2 X 2 determinant Iflalismuchgreaterthanlbl,lcl, and 1d I, to which of a, b, c, and d is the value of the determinant

b. You have designed a two-resistor circuit like the one shown to have resistances of RJ = 100 ohms and R2 = 400 ohms, but there is always some variation in manufacturing and the resistors received by your firm will probably not have these exact values. Will the value of R be more sensitive to variation in R J or to variation in R2? Give reasons for your answer.

c. In another circuit like the one shown you plan to change R J from 20 to 20.1 ohms and R2 from 25 to 24.9 ohms. By about what percentage will this change R?

f(a, b, c, d) =

I: !I

most sensitive? Give reasons for your answer. 60. Estimating maximum error Suppose that u = xe Y + Y sin z and that x, y, and z can be measured with maximum possible errors of ±0.2, ±0.6, and ±7T/180, respectively. Estimate the maximum possible error in calculating u from the measured values x = 2, Y = In 3, z = 7T/2. 61. The Wilson lot size formula The Wilson lot size formula in economics says that the most economical quantity Q of goods (radios, shoes, brooms, whatever) for a store to order is given by the formula Q = Y2KM/h, where K is the cost of placing the order, M is the number of items sold per week, and h is the weekly holding cost for each item (cost of space, utilities, security, and so on). To which of the variables K, M, and h is Q most sensitive near the point (Ko, Mo, ho) = (2, 20, 0.05)? Give reasons for your answer.

802

Chapter 14: Partial Derivatives

62. Surveying a triangular field The area of a triangle is (I/2)ab sin C, where a and b are the lengths of two sides of the triangle and C is the measure of the included angle. In surveying a triangular plot, you have measured a, b, and C to be 150 ft, 200 ft, and 60°, respectively. By about how much could your area calculation be in error if your values of a and b are off by half a foot each and your measurement of C is off by 2°? See the accompanying figure. Remember to use radians.

64. Change along the involute of a circle Find the derivative of f(x, y) = X 2 + Y 2 in the direction of the unit tangent vector of the curve r(t) = (cost

+ t sin t)i + (sint

t> O.

65. Change along a helix Find the derivative of f(x,y, z) = x 2 + y2 + z2 in the direction of the unit tangent vector of the helix r(t) = (cos t)i

a = 150± 1ft 2

+

(sin t)j

+

tk

at the points where t = -1T/4, 0, and 1T/4. The function f gives the square of the distance from a point P(x, y, z) on the helix to the origin. The derivatives calculated here give the rates at which the square of the distance is changing with respect to t as P moves through the points where t = -1T/4, 0, and 1T/4.

Theory and ExampLes 63. The linearization of f(x, y) is a tangent-plane approximation Show that the tangent plane at the point Po(xo, Yo, f(xo, Yo» on the surface z = f(x, y) defined by a differentiable function f is the plane

fAxo, yo)(x - xo)

- tcost)j,

+ fY(xo, yo)(y - Yo) - (z - f(xo, Yo»

66. Normal curves A smooth curve is normal to a surface f(x, y, z) = c at a point of intersection if the curve's velocity vector is a nonzero scalar multiple of Vf at the point. Show that the curve r(t) =

Vti + Vtj -

i(t

+ 3)k

= 0

is normal to the surface x 2 + y2 -

or

Thus, the tangent plane at Po is the graph of the linearization of f at Po (see accompanying figure).

Z

= 3 when t = 1.

67. Tangent curves A smooth curve is tangent to the surface at a point of intersection if its velocity vector is orthogonal to VI there. Show that the curve r(t) =

Vti + Vtj + (2t -

is tangent to the surface x 2 + y2 -

Z

l)k

= I when t = I.

z =f(x, y)

, I I

(!'

Z=

L(x, y)

I

_ _-r-_ _ I- I _ Y I

I I I I I I I

x

14.7

Extreme VaLues and SaddLe Points

HISTORICAL BIOGRAPHY

Simeon-Denis Poisson (1781- 1840)

Continuous functions of two variables assume extreme values on closed, bounded domains (see Figures 14.38 and 14.39). We see in this section that we can narrow the search for these extreme values by examining the functions' first partial derivatives. A function of two variables can assume extreme values only at domain boundary points or at interior domain points where both first partial derivatives are zero or where one or both of the first partial derivatives fail to exist. However, the vanishing of derivatives at an interior point (a, b) does not always signal the presence of an extreme value. The surface that is the graph of the function might be shaped like a saddle right above (a, b) and cross its tangent plane there.

14.7

Extreme Values and Saddle Points

803

Derivative Tests for Local Extreme Values To find the local extreme values of a function of a single variable, we look for points where the graph has a horizontal tangent line. At such points, we then look for local maxima, local minima, and points of inflection. For a function I(x, y) of two variables, we look for points where the surface z = I(x, y) has a horizontal tangent plane. At such points, we then look for local maxima, local minima, and saddle points. We begin by defining maxima and minima. FIGURE 14.38

z

The function

(cosx)(cosy)e-Vx2+y2

=

has a maximum value of I and a minimum value of about -0.067 on the square regionlxl ~ 31T/2,lyl ~ 31T/2.

DEFINITIONS (a, b). Then

Let I(x, y) be defined on a region R containing the point

1. I(a, b) is a local maximum value of I if I(a, b) ~ I(x,y) for all domain points (x,y) in an open disk centered at (a, b).

2. I(a, b) is a local minimum value of I if I(a, b) :5 I(x, y) for all domain

points (x,y) in an open disk centered at (a, b).

z Local maxima correspond to mountain peaks on the surface z = I(x, y) and local minima correspond to valley bottoms (Figure 14.40). At such points the tangent planes, when they exist, are horizontal. Local extrema are also called relative extrema. As with functions of a single variable, the key to identifying the local extrema is a first derivative test.

Local maxima (no greater value off nearby)

/ .

FIGURE 14.39

z

=

The ''roof surface"

±(llxl-lyll-lxl-lyl)

has a maximum value of 0 and a minimum value of -a on the square region Ix I ~ a,

Iyl ~

(no smaller value off nearby)

a. FIGURE 14.40 A local maximum occurs at a mountain peak and a local minimum occurs at a valley low point.

z of

I

ox

=

~ z =f(x,y)

. - --.£

o a

(x) =}(x, b) I b

--___

:~

-.. ,

I ,/

/ ; x

of oy

THEOREM to-First Derivative Test for Local Extreme Values =

0

If I(x, y) has a local maximum or minimum value at an interior point (a, b) of its domain and if the first partial derivatives exist there, then IxCa, b) = 0 and Iy(a, b) = o.

h(y) = f(a, y)

------.. y

(a, b, 0)

FIGURE 14.41 Ifa local maximum of! occurs at x = a, y = b, then the fIrst partial derivatives !x(a, b) and !y(a, b) are both zero.

Proof If I has a local extremum at (a, b), then the function g(x) = I(x, b) has a local extremum at x = a (Figure 14.41). Therefore, g'(a) = 0 (Chapter 4, Theorem 2). Now g'(a) = Ix(a, b), so IxCa, b) = O. A similar argument with the function h(y) = I(a,y) shows thatly(a, b) = o. • Ifwe substitute the values IxCa, b) = 0 and Iy(a, b) = 0 into the equation

IxCa, b)(x - a) + Iy(a, b)(y - b) - (z - I(a, b» = 0

14.7 Extreme Values and Saddle Points

805

z

THEOREM ll-Secnnd Derivative rest for Local Extreme Values Suppose that I(x, y) and its first and second partial derivatives are continuous throughout a disk centered at (a, b) and thatlx(a, b) = /y(a, b) = O. Then i) I has a local maximum at (a, b) if I", ii) I has a local minimum at (a, b) if I", iii) I has a saddle point at (a, b) iflx,dyy

< Oand/",lyy - Ix'? > oat (a, b). > Oandlxxlyy - Ix,' > Oat(a,b). - Ix'? < oat (a, b).

iv) the test is inconclusive at (a, b) if 1",lyy - Ix'? = 0 at (a, b). In this case, we must fmd some other way to determine the behavior of I at (a, b).

(a)

The expression Ixxlyy - Ix'? is called the discriminant or Hessian of I. It is sometimes easier to remember it in determinant form,

Ixxlyy

-!xl =

I~: ~:I·

Theorem 11 says that if the discriminant is positive at the point (a, b), then the surface curves the same way in all directions: downward if Ixx < 0, giving rise to a local maximum, and upward if Ixx > 0, giving a local minimum. On the other hand, if the discriminant is negative at (a, b), then the surface curves up in some directions and down in others, so we have a saddle point. (b)

(a) The origin is a saddle point of the function j(x, y) = y2 - x 2. There are no local extreme values (Example 2). (b) Level curves for the function j io Example 2. FIGURE 14.44

EXAMPLE 3

Find the local extreme values of the function

I(x,y) = xy - x 2

y2 - 2x - 2y

-

+ 4.

Solution The function is defined and differentiable for all x and y and its domain has no boundary points. The function therefore has extreme values only at the points where Ix and Iy are simultaneously zero. This leads to Ix = y - 2x - 2 = 0,

Iy = x - 2y - 2 = 0,

or

x =y

=

-2.

Therefore, the point ( -2, -2) is the only point where I may take on an extreme value. To see if it does so, we calculate

Ixx = -2,

Iyy = -2,

Ixy = I.

The discriminant of I at (a, b) = (-2, -2) is

Ixxlyy - Ixy2 = (-2)(-2) - (1)2 = 4 - 1 = 3. The combination

Ixx tells us that

I

I( -2, -2)

8.

=

EXAMPLE 4 Solution

<0

and

has a local maximum at (-2, - 2). The value of

I

at this point is

• Find the local extreme values of I(x,y) = 3y2 - 2y 3

-

3x 2 + 6xy.

Since I is differentiable everywhere, it can assume extreme values only where

Ix = 6y - 6x = 0

and

Iy = 6y - 6y2

+

6x = O.

804

Chapter 14: PartiaL Derivatives

,

fur the tangent plane to the surface z = I(x,y) at (a, b), the equation reduces to O·(x - a)

+ O·(y - b) - z + f(a, b) z

y

~

~

0

f(a, b).

Thus, Theorem 10 says that the surface does indeed have a horizontal tangent plane at a local extremum, provided there is a tangent plane there.

DEFlNmON An interior point of the domain of a function I(x,y) where both I" and I y are zero or where one or both of I J: and Iy do not exist is a critical point

off. Theorem 10 says that the only points where a ftmction I(x,y) can assume extreme values are critical points and boundary points. As with differentiable ftmctions of a single variable, not every critical point gives rise to a local extremum. A differentiable :function of a single variable might have a point of inflection. A differentiable function of two variables might have a saddle point.

FIGURE 14.42 origin.

Saddle points at the

,

EXAMPLE 1

j,.,-: 11

DEFlNmON A differentiable ftmction I(x, y) has a .utDe point at a critical point (a, b) if in every open disk centered at (a, b) there are domain points (x,y) when: f(x,y) > f(a, b) and domain points (x.y) when: f(x,y) < f(a,b). The corresponding point (a, b, I(a, b» on the surface z = I(x,y) is called a saddle point of the surface (Figure 14.42).

2

3

+ 9.

Solution The domain of I is the entire plane (so there are no boundary points) and the partial derivatives I" = 2x and Iy = 2y - 4 exist everywhere. Therefore, local extreme values can occur only where

IJ:

4·'

x 2 FIGURE 14.43 Thegraphofthefimction f(x,y) - x 2 + y2 - 4y + 9ma plUUx:lloid which has a local minimwn value of 5 at the point (0, 2) (Example 1).

Find the local extreme values of I(x,y) = x 2 + y2 - 4y

=

2x

=

0

and

Iy

=

2y - 4

=

o.

The only poooibility is the point (0, 2). when: the value of f is 5. Sinc. f(x,y) ~ x 2 + (y - 2)2 + 5 is never less than 5, we see that the critical point (0, 2) gives a local minimum (Figure 14.43). _

EXAMPLE 2

Find the local extreme values (if any) of I(x, y) =

y2 -

x 2•

Solution The domain of I is the entire plane (so there are no boundary points) and the partial derivatives I" = -2x and Iy = 2y exist everywhere. Therefore, local extrema can occur only at the origin (0, 0) where I" = 0 and Iy = O. Along the positive x-axis, however, I has the value I(x, 0) = -x2 < 0; along the positive y-axis, I has the value I(O,y) = y2 > O. Therefore, every open disk in the xy-plane centered at (0, 0) contains points where the function is positive and points where it is negative. The :function has a saddle point at the origin and no local extreme values (Figure 14.44a). FigLD"e 14.44b displays the level curves (they are hyperbolas) of I, and shows the ftmction decreasing and increasing in an alternating fashion among the four groupings of hyperbolas. _ That I" = Iy = 0 at an interior point (a, b) of R does not guarantee I has a local ex1reme value there. If I and its Imt and second partial derivatives are continuous on R, however, we maybe able to learn more from the following theorem, proved in Section 14.9.

806

Chapter 14: Partial Derivatives

z

From the first of these equations we find x equation then gives

6x - 6x 2 + 6x = 0

= y, and substitution for y into the second 6x (2 - x) = O.

or

The two critical points are therefore (0, 0) and (2, 2). To classify the critical points, we calculate the second derivatives: y

Ixx

Iyy

-6,

=

=

6 - 12y,

Ixy

=

6.

The discriminant is given by

Ixxlyy - Ixy2 FIGURE 14.45 The surface z = 3y2 - 2y 3 - 3x 2 + 6xy has a saddle point at the origin and a local maximum at the point (2, 2) (Example 4).

=

(-36

+ 72y) - 36

=

72(y - 1).

At the critical point (0, 0) we see that the value of the discriminant is the negative number -72, so the function has a saddle point at the origin. At the critical point (2, 2) we see that the discriminant has the positive value 72. Combining this result with the negative value of the second partial I xx = -6, Theorem 11 says that the critical point (2, 2) gives a local maximum value of 1(2,2) = 12 - 16 - 12 + 24 = 8. A graph of the surface is shown in Figure 14.45. •

Absolute Maxima and Minima on Closed Bounded Regions We organize the search for the absolute extrema of a continuous function I(x, y) on a closed and bounded region R into three steps.

1.

List the interior points of R where I may have local maxima and minima and evaluate at these points. These are the critical points of f.

I 2.

List the boundary points ofR where I has local maxima and minima and evaluate I at these points. We show how to do this shortly.

3.

Look through the lists for the maximum and minimum values of f. These will be the absolute maximum and minimum values of I on R. Since absolute maxima and minima are also local maxima and minima, the absolute maximum and minimum values of I appear somewhere in the lists made in Steps 1 and 2.

EXAMPLE 5

y

Find the absolute maximum and minimum values of

I(x, y) = 2

B(0,9)

+

2x

+ 2y - x 2 - y2

on the triangular region in the fIrst quadrant bounded by the lines x = 0, y = 0, y = 9 - x .

y=9-x

SoLution Since I is differentiable, the only places where I can assume these values are points inside the triangle (Figure 14.46) where Ix = Iy = 0 and points on the boundary.

x=O

(a) Interior points. For these we have

(1, 1)

•

---O~~--y-=-On-----~----~x

A(9,0)

FIGURE 14.46 This triangular region is the domain of the function in Example 5.

Ix

=

2 - 2x

=

0,

h

=

2 - 2y

=

0,

yielding the single point (x, y) = (1, 1). The value of I there is

1(1,1)

=

4.

(b) Boundary points. We take the triangle one side at a time: i) On the segment OA, y

= O. The function

I(x,y) = l(x,O) = 2

+

2x - x 2

14.7 Extreme Values and Saddle Points

807

may now be regarded as a function of x dermed on the closed interval 0 :5 x :5 9. Its extreme values (we know from Cbapter 4) may occur at the endpoints

x=O

where

1(0,0) = 2

x=9

where

1(9,0) = 2

+

18 - 81 = -61

and at the interior points where f' (x, 0) = 2 - 2x = O. The only interior point where f'(x,O) = 0 is x = I, where

l(x,O) = 1(1,0) = 3.

ii) On the segment OB, x = 0 and I(x,y) = I(O,y) = 2

+ 2y

- y2.

We know from the symmetry of I in x and y and from the analysis we just carried out that the candidates on this segment are

1(0,0) = 2,

1(0,9) = -61,

1(0,1) = 3.

iii) We have already accounted for the values of I at the endpoints of AB, so we need only look at the interior points of AB. With Y = 9 - x, we have

I(x,y) = 2

+ 2x + 2(9

- x) - x 2 - (9 -

xf =

-61

+

18x - 2x 2.

Setting f'(x, 9 - x) = 18 - 4x = 0 gives

x

9

18

= 4=2'

At this value of x,

y=9- 29 =29

and

I(x,y)

=

I

99) =-2' (2'2 41

Summary We list all the candidates: 4, 2, -61,3, -(41/2). The maximum is 4, which I assumes at (I, I). The minimum is-61, which I assumes at (0, 9) and (9, 0). • Solving extreme value problems with algebraic constraints on the variables usually requires the method of Lagrange multipliers introduced in the next section. But sometimes we can solve such problems directly, as in the next example.

EXAMPLE 6 A delivery company accepts only rectangular boxes the sum of whose length and girth (perimeter of a cross-section) does not exceed 108 in. Find the dimensions of an acceptable box of largest volume. Let x, y, and z represent the length, width, and height of the rectangular box, respectively. Then the girth is 2y + 2z. We want to maximize the volume V = xyz of the box (Figure 14.47) satisfying x + 2y + 2z = 108 (the largest box accepted by the delivery company). Thus, we can write the volume of the box as a function of two variables:

Solution Girth = distance

V=~and

V(y,z) = (108 - 2y - 2z)yz

x~108-2y-2z

= 108yz - 2y 2z - 2yz2.

Setting the first partial derivatives equal to zero, x

FIGURE 14.47

y

The box in Example 6.

V,(y,z) Vz(y,z)

= IOSz = 108y -

4yz - 2z2 2y2 - 4yz

= (108 = (108 -

=0 4z)y = 0,

4y - 2z)z 2y -

808

Chapter 14: Partial Derivatives gives the critical points (0, 0), (0, 54), (54, 0), and (18, 18). The volume is zero at (0, 0), (0,54), (54, 0), which are not maximum values. At the point (18, 18), we apply the Second Derivative Test (Theorem II):

Vyy

=

v,. =

-4z,

108 - 4y - 4z.

Then

Vyy V,. - V,.2

= 16yz -

16(27 - Y -

z}'.

Thus,

Vyy(18, 18) = -4(18)

<

0

and

[Vyy V,. - V,.2](18.18)

= 16(18)(18) - 16(-9)2

>

0

imply that (18,18) gives a maximum volume. The dimensions of the package are x = 108 - 2(18) - 2(18) = 36 in.,y = 18 in., andz = 18 in. The maximum volume is V = (36)(18)(18) = 11,664 in3, or 6.75 W. • Despite the power of Theorem II, we urge you to remember its limitations. It does not apply to boundary points ofa function's domain, where it is possible for a function to have extreme values along with nonzero derivatives. Also, it does not apply to points where either Ix or Iy fails to exist.

Summary of Max-Min Tests The extreme values of I(x, y) can occur only at i) boundary points of the domain of I ii) critical points (interior points where Ix = Iy = 0 or points where Ix or Iy fails to exist). If the first- and second-order partial derivatives of I are continuous throughout a disk centered at a point (a, b) and Ix(a, b) = Iy(a, b) = 0, the natore of I(a, b) can be tested with the Second Derivative Test: i) ii) iii)

iv)

I., < 0 and/.,lyy - Ix/ > 0 at (a, b) => local maximum I., > 0 and/.,lyy - Ix/ > 0 at (a, b) => local minimum 1.,lyy - Ix/ < 0 at (a, b) => saddle point 1.,1yy - Ix/ = 0 at (a, b) => test is inconclusive

Exercises 14.7 finding Local Extrema Find all the local maxima, local minima, and saddle points of the functions in Exercises 1-30.

1. I(x,y) 2. I(x,y) 3. I(x,y) 4. I(x,y)

+ :x:y + y2 + 3x - 3y + 4 = 2xy - 5x2 - 2y2 + 4x + 4y 2 = x + :x:y + 3x + 2y + 5 = 5:x:y - 7x 2 + 3x - 6y + 2 =

x2

4

I(x,y) = 2xy - x 2 - 2y2 + 3x + 4 6. I(x,y) = x 2 - 4:x:y + y2 + 6y + 2

s.

7. I(x,y) = 2x 2 + 3:x:y + 4y2 - 5x + 2y 8. I(x,y) = x 2 - 2xy + 2y2 - 2x + 2y + I 9. I(x,y) = x 2 - y2 - 2x + 4y + 6 10. I(x,y) = x 2 + 2xy

Exbeme Values and Saddle Points

14.7

11. f(x,y) = "';56.1:2

15. f(x,y) 16. f(x,y) 17. f(x,y) 18. f(x,y)

16.1:

31

+I

39. Find two munbcrs a and b with a s b !Nch that

- 8x

[b(6 -x

V'x2 + y2

11. f(x,y) _ I 13. f(x,y) - x 3 14. f(x,y) _ x 3

8y2

-

y3 - 2xy

+6

+ 3xy + y3 2x 3 + 3y2 + 6xy 3 = x + y3 + 3.1: 2 - 3y2 - 8 = x 3 + 3xy2 - 15x + y3 - 15y - 2x 3 + 2y 3 - 9x 2 + 3y2 - 12y

l

2 X

1

1

2

- 1

23. f(x,y) = ysinx 25. f(x,y) 27. f(x,y) _ 29. f(x,y) = 38. f(x,y) =

22. f(x,y) = X

1 + xy + Y

11:x.y) = x2

24. f(x, y) = e2.: cosy

f(x.y) ~ xy in the open flISt quadnurt (x OIl. a miIrimum thcI'C:.

finding Absolute Extrema 31. f(x,y) = 2x 2 - 4x + y2 - 4y + I on the closed triangular plate bounded by the lines x = 0, y = 2, Y = 2x in the fll'St quodnrot - xy + y2 + Ion the closed triangularplatcin the fmt quadnntbounded by the lines x = O,y = 4,y = x 33. f(x,y) - x 2 + y2 on the cl.o&cd triangular plate bounded by the Iin.ca x = 0, y = 0, y + 2x = 2 in the fmt quadIant

31. D(x,y) = x2

on

+2

the

dr

rectangular

+ 2y2

- x.

42. Find the critical point of

In Exerciscs 31-38, fmd the absolute maxima and minima of the :functions on the given d.omain9.

34. 11:x, y) - x + xy + y2 - 6x o :!;'; x :!;'; 5, -3 :!;'; y :!;'; 3 35. 11:x, y) - x 2 + xy + y2 - 6x o :!;'; x :!;'; 5, -3 :!;'; y :!;'; 0

X 2 )1/3

Find the tcmpcraturcs at the hottest and coldest points on the plate.

er+r-.v 26. f(x.y) - " - y<' 28. f(x,y) _ r(x 2 _ y2) e"7(xl + y2) 2lnx + Iny - 4x - Y In (x + y) + x 2 - Y

2

(24 - 2x -

41. TemperatDrel A flat circular plate 1m the shape of the region x 2 + yl :!;'; I. The plaIe, including the bouodaIy where x 2 + yl _ 1,isheated80thatthetaopemtureatthepoin1:(x,y)is

+ y4 + 4zy +Y

b

has its largest value.

19. f(x,y) _ 4xy - x4 _ y4

21. f(x,y) =

-x2)dr

has its largest value. 40. Find two IIUIDbers a and b with a :!;'; b such that

= 6.1: 2 -

20. f(x,y) _ x4

809

plate

on the rectangular plate

36. f(x,y) - 48xy - 32x3 - 24yl on the rectangular plate O:!;';x:!;'; 1,0:!;';y:!;'; I 37. f(x,y) - (4x - x 2) cosy on the rectangular plate I:!;'; x :!;'; 3, -'fr/4:!;';y:!;'; 'fr/4 (see accompanying figure).

+ "" - lnx,> > O,y > 0) and shaw that f

takes

Theory and Examples 43. Find the muima, minima, and saddle points of f(x, y), if any, given
f" = 2x - 4y

b. f" = 2x - 2

and fy = 2y - 4x and fy = 2y - 4

c. f" - 9%2 - 9 and fy - 2y + 4 Describe your reasoning in each case.

..... Thediscriminantfnf»- - A/iszero at the origin for each of the following functiOlUl, so the Second Derivative Test fails there. Determine whether the function Iwl a maximum, a minimum, or neither at the origin by jmagining what the surface z - f{x, y) looks like. Describe your reasoning in each case. L

f{x,y) - x1y2

c. f{x,y) - xy2 e. f{x,y) - xy3

b. f{x.y) - 1 - x1yl

II. f{x.y) _ xy2

f. f{x.y) - x"y4

45. Shaw that (0, 0) is a critical point of f(x, y) = x 2 + kty + y2 no matter what valuc the constant k has. (Hint: Consider two cases:

k - Oandk.;. 0.) 46. Fox what values of the constant k docs the Second Derivative Test gu.anurtcc that f(x,y) = x2 + kty + y2 will have a saddle point at (0, O)? A local minimum at (0, O)? For what values of k is the Second Derivative Teat inconclusive? Give :reasons for your

......,..

47. Iff,,(a,b) - fy{a,b) - 0. must f have a local maximwn or minimmn value at (a, b)? Give rcuons for your amiwet'.

x

38. f(x,y) = 4x - 8xy + 2y + 1 on the triangularplatcbmmdcdby the linen = 0, y = 0, x + y = 1 in the fmt quadnurt

48. Can you. conclude anything about f(a, b) iff and its fmrt and second partial derivatives are continuous throughout a disk centered at the critical point (a, b) and fxJ..a, b) andfyy(a, b) differ in sign? Give rea&Orul for your anawer. 49. Among all the points on the graph of z - 10 - x 2 - y2 that lie

above the plane x the plane.

+ 2y + 3z

-

0.

find the point farthest from

810

Chapter 14: Partial Derivatives

50. Find the point on the graph of z ~ x 2 + y2 planex + 2y -z~ O. 51. Find the point on the plane 3x origin.

+ 2y + z

~

+

10 nearest the

63. Function: I(x, y) ~ xy Curves:

6 that is nearest the

52. Find the minimum distance from the point (2, -I, I) to the plane

i)Thelinex~2t,

y~t+1

~

2t, y

~

t

iii) The line segment x

~

2t, y

~

t + I,

x+y-z~2.

53. Find three numbers whose sum is 9 and whose sum of squares is a minimum. 54. Find three positive numbers whose sum is 3 and whose product is a maximum. 55. Find

the

maximum

of s

value

~

xy + yz + xz where

x+y+z~6.

point (-6,4,0). 57. Find the dimensions of the rectangular box of maximum volume that can be inscribed inside the sphere x 2 + y2 + z2 ~ 4. 58. Among all closed rectangolar boxes of volume 27 em3, what is the smallest surface area? 59. You are to construct an open rectangular box from 12 It' of material. What dimensions will result in a box of maximum volume? 60. Consider the function I(x, y) ~ x 2 + y2 over the square 0 s x ::s 1 and 0 s y ::s 1. L

+ 2xy

- x - y

+

-Is t s 0

Os t s I

a. I(x,y) ~ x 2 + y2 b. g(x,y) ~ 1/(x 2 + y2) Curves: i)Thelinex~t,

y~2-2t

~

t, Y

~

2 - 2t,

0 '" t s 1

65. Least squores and regression lines When we try to fit a line y ~ mx + b to a set of numerical data points (XhYI), (X2,)'2), ... , (X.,y.) (Figore 14.48), we usually choose the line

that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means fmding the values of m and b that minimize the value of the function

w ~ (mxl + b - YIl' + ... + (mx. + b - y.)2 .

(1)

Show that the values of m and b that do this are

I

(Lx,) (LY.) - nLxkY.

Show that I has an absolute minimum along the line segment 2x + 2y ~ I in this square. What is the absolute minimum value?

(LX'), -nLxl b

E:rtreme Value. on Panunetrized Curves To fmd the extreme values of a function I(x, y) on a curve x ~ x(t), y ~ y(t), we treat I as a function of the single variable t and use the Chain Rule to fmd where dl/dt is zero. As in any other single-variable case, the extreme values of I are then found among the values at the critical points (points where dl/dt is zero or fails to exist), and

b. endpoints of the parameter domain.

(2)

m~

b. Find the absolute maximum value of I over the square.

L

I,

64. Functions:

ii) The line segment x

56. Find the minimum distance from the cone z ~ \lx 2 + y2 to the

+

ii)The line segment x

~

*

(LYk - mLx,),

(3)

with all sums rwming from /r; ~ I to /r; ~ n. Many scientific calculators have these formulas built in, ensbling you to fmd m and b with only a few keystrokes after you have entered the data The line y ~ mx + b derermined by these values of m and b

is called the least squares line, regression line, or trend line for the data under study. Finding • least squares line lets you

1. summarize data with a simple expression, Find the absolute maximum and minimorn values of the following functions on the given curves.

2. predict values ofy for other, experimentally untried values of x, 3. handle data analytically.

61. Functions: L

I(x,y)

~

c. h(x,y) ~

+y 2x 2 + y2

x

b. g(x,y) ~ xy y

Curves: i) The semicircle x 2 + y2 ~ 4,

y '" 0 4, x ~ 0, y"'O Use the parametric equations x = 2 cos t,y = 2 sin t.

ii) The quarter circle x 2

+ y2

=

62. Functions:

2x + 3y c. h(x,y) ~ x 2 + 3y2

L

I(x,y)

~

b. g(x,y) ~ xy

Curves: i) The semiellipse (x 2/9)

+

ii) The quarter ellipse (x 2/9)

(y2/4) ~ I, y '" 0

+

Use the parametric equations x

(y2/4) ~ I, x '" 0, y '" 0 3 cos t,y = 2 sin t.

=

-1------------> x

o

FIGURE 14.48 To fit a line to noncollinear points, we choose the line that minimizes the sum of the squares of the deviations.

14.8 Lagrange Multipliers

In

Exercises 6tH;8, use Equations (2) and (3) to rmd the least squares line for each set of data points. Then use the linear equation you obtain to predict the value ofy that would correspond tox = 4. 66. (-2,0), (0,2), (2,3) 67. (-1,2), (0, I), (3, -4) 68. (0, 0), (1,2), (2, 3)

In Exercises 69-74, you will explore functions to identifY their local

extrema. Use a CAS to perform the followiog steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle.

c. Calculate the function's Itrst partial derivatives and use the CAS equation solver to rmd the critical points. How do the critical points relate to the level corves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer.

14.8

d. Calculate the function's second partial derivatives and rmd the discriminant lui", - 1,0/. e. Using the max-min tests, classifY the critical points found in part (c). Are your rmdings consistent with your discussion in part (c)? 69. I(x,y) = x 2 + y' - 3~, -5';; x ,;; 5, -5';; Y ,;; 5 70. I(x,y)

COMPUTER EXPLORATIONS

811

=

x' - 3~2 + y2, -2';; x ,;; 2, -2';; Y ,;; 2 x' + y2 - ax 2 - 6y + 16, -3';; x';; 3,

71. I(x,y) = -6 ,;; Y ,;; 6

72. I(x,y) = 2>;4 + y' - 2>;2 - 2y2 + 3, -3/2'" x'" 3/2, - 3/2 ,;; Y ,;; 3/2 73. I(x,y) = 5x' + lax' - 3Ox' + 3Oxy2 - 12Ox', -4 :s x :s 3, -2:s y :s 2 (x,y) # (0,0) 74. I(x,y) = {~:In(x2 + y2), (x,y) = (0,0)' -2 :s x :5 2, -2:s y :5 2

Ir-La~g~~_n= ge__ M_ UL_ ti~ Pl_ ie_~________________________________

HIsTORICAL BIOGRAPHY

Joseph Louis Lagrange (1736-1813)

Sometimes we need to find the extreme values of a function whose domain is constrained to lie within some particular subset of the plane---a disk, for example, a closed triangular region, or along a curve. In this section, we explore a powerful method for rmding extreme values of constrained functions: the method of Lagrange multipliers.

Constrained Maxima and Minima We first consider a problem where a constrained minimum can be found by eliminating a variable.

EXAMPLE 1

FindthepointP(x,y,z)ontheplane2x

+ y -z -

5 =

othat is closest

to the origin.

Solution

The problem asks us to rmd the minimum value of the function

IoFl = Y(x - 0)2 + (y = Yx2 + y2 + z2

- 0)2

+

(z - 0)2

subject to the constraint that

2x+y-z-5=0.

oF I has a minimum value wherever the function

Since I

f(x,y, z) = x 2

+ y2 + z2

has a minimum value, we may solve the problem by finding the minimum value of f(x, y, z) subject to the constraint 2x + y - z - 5 = 0 (thus avoiding square roots). Ifwe regard x and y as the independent variables in this equation and write z as z=2x+y-5, our problem reduces to one of finding the points (x,y) at which the function

h(x, y) = f(x, y, 2x

+y

- 5) = x 2

+ y2 +

(2x

+y

- 5)'

812

Chapter 14: Partial Derivatives has its minimum value or values. Since the domain of h is the entire xy-plane, the First Derivative Test of Section 14.7 tells us that any minima that h might have must occur at points where hy = 2y + 2(2x + y - 5) = O.

hx = 2x + 2(2x + y - 5)(2) = 0, This leads to

lOx + 4y = 20,

4x

+ 4y

=

10,

and the solution

x

5

=

3'

We may apply a geometric argument together with the Second Derivative Thst to show that these values minimize h. The z·coordinate of the corresponding point on the plane

z=2x+y-5is 5 6' Therefore, the point we seek is Closest point:

•

The distance fromPto the origin is 5jV6 '" 2.04.

Attempts to solve a constrained maximum or minimum problem by substitotion, as we might call the method of Example 1, do not always go smoothly. This is one of the rea· sons for learning the new method of this section.

z

EXAMPLE 2

Find the points on the hyperbolic cylinder x 2 - z2 - I = 0 that are clos·

est to the origin. Solution 1 The cylinder is sbowo in Figure 14.49. We seek the points on the cylinder closest to the origin. These are the points whose coordinates minimize the value of the function

/(x,y,z) = x 2 + y2

+

z2

Square of the distance

subject to the constraint that x 2 - z2 - I = O. Ifwe regard x andy as independent vari· abIes in the constraint equation, then

and the values of /(x,y, z) = x 2

+ y2 + z2 on the cylinder are given by the function

h(x, y) = x 2 + y2 + (x 2 - I) = 2x 2 + y2 - 1. To f'md the points on the cylinder whose coordinates minimize /, we look for the points in the :ry·plane whose coordinates minimize h. The only extreme value of h occurs where FIGURE 14.49

x' - z' -

The hyperbolic cylinder

I = 0 in Example 2.

hx =4x=0

and

hy =2y=0,

that is, at the point (0, 0). But there are no points on the cylinder where both x and y are zero. What went wrong? What happened was that the First Derivative Test found (as it sbonld have) the point in the domain of h where h has a minimum value. We, on the other hand, want the points on the cylinder where h has a minimum value. Although the domain of h is the entire

14.8 Lagrange Multipliers The hyperbolic cylinder x 2 On this part,

x= v?+l

- Z2 =

1

On this part,

z x=-v?+l

xy-plane, the domain from which we can select the first two coordinates of the points (x, y, z) on the cylinder is restricted to the "shadow" ofthe cylinder on the xy-plane; it does not include the band between the lines x = -1 and x = 1 (Figure 14.50). We can avoid this problem if we treat y and z as independent variables (instead of x and y) and express x in terms of y and z as

+

x 2 = z2

k(y, z)

1.

+ y2 + z2 becomes

With this substitution, I(x,y, z) = x 2

FIGURE 14.50 The region in the ;ry-plane from which the first two coordinates of the points (x,y, z) on the hyperbolic cylinder x 2 - z2 = 1 are selected excludes the band -1 < x < 1 in the xy-plane (Example 2).

813

= (z2 + 1) + y2 + z2 = 1 + y2 + 2z2

and we look for the points where k takes on its smallest value. The domain of k in the yzplane now matches the domain from which we select the y- and z-coordinates of the points (x, y, z) on the cylinder. Hence, the points that minimize k in the plane will have corresponding points on the cylinder. The smallest values of k occur where

ky = 2y = 0 or where y

kz

and

=

4z = 0,

= z = O. This leads to x2

= z2 + 1 = 1,

x = ±1.

The corresponding points on the cylinder are (± 1,0,0). We can see from the inequality

k(y, z)

= 1 + y2 + 2z2

~ 1

that the points (± 1, 0, 0) give a minimum value for k. We can also see that the minimum distance from the origin to a point on the cylinder is 1 unit.

\

Solution 2 Another way to find the points on the cylinder closest to the origin is to imagine a small sphere centered at the origin expanding like a soap bubble until it just touches the cylinder (Figure 14.51). At each point of contact, the cylinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and cylinder are represented as the level surfaces obtained by setting and

x

/

FIGURE 14.51 A sphere expanding like a soap bubble centered at the origin until it just touches the hyperbolic cylinder x 2 - z2 - 1 = 0 (Example 2).

g(x,y,z)

= x 2 - z2 - 1

equal to 0, then the gradients VI and Vg will be parallel where the surfaces touch. At any point of contact, we should therefore be able to find a scalar A ("lambda") such that

VI = AVg, or

2xi

+

2yj

+

2zk = A(2xi - 2zk).

Thus, the coordinates x, y, and z of any point of tangency will have to satisfy the three scalar equations

2x

= 2Ax,

2y

= 0,

2z

= -2Az.

For what values of A will a point (x, y, z) whose coordinates satisfy these scalar equations also lie on the surface x 2 - z2 - 1 = O? To answer this question, we use our knowledge that no point on the surface has a zero x-coordinate to conclude that x O. Hence, 2x = 2Ax only if

*'

2 = 2A,

or

A=1.

For A = 1, the equation 2z = -2Az becomes 2z = -2z. If this equation is to be satisfied as well, z must be zero. Since y = 0 also (from the equation 2y = 0), we conclude that the points we seek all have coordinates of the form

(x, 0, 0).

814

Chapter 14: Partial Derivatives What points on the surface x 2 points (x, 0, 0) for which

x2

-

(0)'

z2 =

-

=

I have coordinates oflhis fonn? The answer is the

x2

I,

=

I,

or

x

=

± I.

The points on the cylinder closest to the origin are the points (± 1,0,0).

•

The Method of Lagrange Multipliers In Solution 2 of Example 2, we used the method of Lagrange multipliers. The method says that the extreme values of a function I(x, y, z) whose variables are subject to a constraint g(x,y,z) = 0 are to be found on the surface g = 0 among the points where VI = AVg for some scalar A (called a Lagrange multiplier). To explore the method further and see why it works, we first make the following observation, which we state as a theorem.

THEOREM 12-The Orthogonal Gradient Theorem

Suppose that I(x, y, z) is differentiable in a region whose interior contains a smooth curve C:

r(t)

=

g(t)i

+ h(t)i + k(t)k.

If Po is a point on C where I has a local maximum or minimum relative to its values on C, then VI is orthogonal to C at Po.

Proof We show that VI is orthogonal to the curve's velocity vector at Po. The values of I on C are given by the composite I(g(t), h(t), k(t», whose derivative with respect to tis

dl dt At any point Po where curve, dl/dt = 0, so

I

=

al dg ax dt

al dh

al dk

+ ay dt + az dt

= VI' v.

has a local maximum or minimum relative to its values on the

V/'Y

=

o.

•

By dropping the z-tenns in Theorem 12, we obtain a similar result for functions of two variables.

COROLLARY OF THEOREM 12 At the points on a smooth curve r(t) = g(t)i + h(t)j where a differentiable function I(x, y) takes on its local maxima and minima relative to its values on the curve, VI' Y = 0, where Y = dr/dt.

Theorem 12 is the key to the method of Lagrange multipliers. Suppose that I(x, y, z) and g(x, y, z) are differentiable and that Po is a point on the surface g(x, y, z) = 0 where I has a local maximum or minimum value relative to its other values on the surface. We assume also that Vg '" 0 at points on the surface g(x,y,z) = o. Then I takes on a local maximum or minimum at Po relative to its values on every differentiable curve through Po on the surface g(x,y, z) = o. Therefore, Viis orthogonal to the velocity vector of every such differentiable curve through Po. So is Vg, moreover (because Vg is orthogonal to the level surface g = 0, as we saw in Section 14.5). Therefure, at Po, VI is some scalar multiple A ofVg.

14.8

815

Lagrange Multipliers

The Method of Lagrange Multipliers Suppose that I(x, y, z) and g(x, y, z) are clifferentiable and Vg '" 0 when g(x,y,z) = O. To find the local maximum and minimum values of I subject to the constraint g(x, y, z) = 0 (if these exist), f"md the values of x, y, z, and A that simultaneously satisfY the equations

VI

AVg

=

g(x,y, z)

and

o.

=

(1)

For functions of two independent variables, the condition is similar, but without the variable z.

Some care must be used in applying this method. An extreme value may not actually exist (Exercise 41). y

EXAMPLE 3 X2

Y2

8+

Find the greatest and smallest values that the function

y2

I(x,y)

2~1

=

xy

takes on the ellipse (Figure 14.52) x

0

2Y2 We want to find the extreme values of I(x, y)

Solution FIGURE 14.52 Example 3 shows how to f"md the largest and smallest values of the product"" on this ellipse.

g(x,y)

=

x'

y'

"8 + "2 - I

xy subject to the constraint

=

=

o.

To do so, we first find the values ofx,y, and A for which

VI

=

AVg

g(x,y)

and

=

O.

The gradient equation in Equations (1) gives yi

+ xj

=

Ay,

=

~ xi + Ayj,

from which we find y

xy=-2

xy=2

x

so that y

-~~i~;~~~~~~

=

0 or A = ±2. We now consider these two cases.

Case 1: If y = 0, then x = y = O. But (0, 0) is not on the ellipse. Hence, y '" O. Case2: If y '" 0, then A = ±2 and x = ±2y. Substituting this in the equation g(x, y) = 0 gives

x

(±2y)' -8xy=2/

FIGURE 14.53

eonstraiot g(x, y)

and

y'

+ "2

=

I,

4y' + 4y'

=

8

and

y

=

±l.

The function I(x,y) = xy therefore takes on its extreme values on the ellipse at the four points (±2, 1), (±2, -I). The extreme values are xy = 2 andxy = -2.

xy=-2

When subjected to the = x'/8

+ y'/2 - I

=

0, The Geometry of the Solution The level curves of the function I(x, y) = xy are the hyperbolas xy = c (Figure 14.53). The farther the hyperbolas lie from the origin, the larger the absolule value of I. We want to f"md the es1reme values of I(x, y), given that the point (x, y) also lies on the ellipse x' + 4y' = 8. Which hyperbolas intersecting the ellipse lie farthest from the origin? The hyperlJolas that just graze the ellipse, the ones that are tangent to it, are

the functioo f(x, y) = "" takes 00 extreme values at the four points (±2, ± I ~ These are the points 00 the ellipse when Vf (red) is a scalar multiple ofVg (blue) (Example 3).

816

Chapter 14: Partial Derivatives farthest. At these points, any vector nonna! to the hyperbola is nonna! to the ellipse, so VI = yi + xj is a multiple (A ~ ±2) ofVg ~ (x/4)i + yj. At the point (2, I), for example, VI

~

i

+

Vg ="2' 1. + J, .

2j,

aod

VI = 2Vg.

aod

VI

At the point ( - 2, I), VI = i - 2j,

EXAMPLE 4 3x

Vg =

-ti +

j,

~

-2Vg.

•

Find the maximum aod minimum values of the function I(x,y) = ~ 1.

+ 4y on the circle x 2 + y2

Solution

We model this as a Lagraoge multiplier pmblem with

g(x,y) ~ x 2 + y2 - I

+ 4y,

I(x,y) ~ 3x

aod look for the values ofx,y, aod A that satisfy the equations VI ~ AVg:

g(x,y) ~

+ 4j = 2xAi + 2yAj 0: x 2 + y2 - I = O. 3i

The gradient equation in Equations (I) implies that A "" 0 aod gives

x

3 = 2A'

Y

~

2

X'

These equations tell us, among other things, that x aod y have the same sign. With these values for x aod y, the equation g(x, y) ~ 0 gives

(iAY + (iY -

I

~ 0,

so

9

+ 16

= 4A2,

4A2 = 25,

5 A ~ ±"2'

aod

Thus,

y

3

3

x~2A=±S'

2

4

Y ~ X = ±S'

aod/(x,y) = 3x + 4yhas extreme values at (x,y) ~ ±(3/5,4/5). By calculating the value of 3x + 4y at the points ±(3/5, 4/5), we see that its maximum aod minimum values on the circle x 2 + y2 ~ I are 3x+4y=5

aod 3x

+ 4y =-5

FIGURE 14.54 The functionj(x,y) ~ 3x + 4y takes 00 its largest value 00 the unitcircleg(x,y) ~ x 2 + y2 - I ~ Oat the point (3/5, 4/5) and its smallest value at the point (-3/5, -4/5) (Example 4). At each of these points, Vj is a scalar multiple ofVg. The figure shows the gradients at the Il!St point but not the second.

The Geometry of the Solution The level curves of I(x, y) ~ 3x + 4y are the lines 3x + 4y ~ c (Figure 14.54). The farther the lines lie from the origin, the larger the absolute value of f. We want to find the extreme values of I(x,y) given that the point (x,y) also lies on the circle x 2 + y2 ~ I. Which lines intersecting the circle lie farthest from the origin? The lines taogent to the circle are farthest. At the points oftaogency, aoy vector nonna! to the line is normal to the circle, so the gradient VI ~ 3i + 4j is a multiple (A ~ ±5/2) of the gradient Vg ~ 2xi + 2yj. At the point (3/5, 4/5), for example, VI = 3i

+

4j,

Vg

~

6. + 8. S' SJ,

aod

5

VI = "2Vg.

•

14.8 Lagrange Multipliers

817

Lagrange Multipliers with Two Constraints Many problems require us to find the extreme values of a differentiable function I(x, y, z) whose variables are subject to two constraints. If the constraints are and

c

\

and gl and g2 are differentiable, with Vg 1 not parallel to Vg2, we find the constrained local maxima and minima of I by introducing two Lagrange multipliers A and JL (mu, pronounced "mew"). That is, we locate the points P(;x, y , z) where I takes on its constrained extreme values by finding the values of x, y , z, A, and JL that simultaneously satisfy the equations

gl = 0

FIGURE 14.55 The vectors Vg 1 and Vg2 lie in a plane perpendicular to the curve C because Vg 1 is normal to the surface gl = 0 and Vg2 is normal to the surface

g2 = O.

(2)

°

°

Equations (2) have a nice geometric interpretation. The surfaces gl = and g2 = (usually) intersect in a smooth curve, say C (Figure 14.55). Along this curve we seek the points where I has local maximum and minimum values relative to its other values on the curve. These are the points where VI is normal to C, as we saw in Theorem 12. But Vg 1 and Vg2 are also normal to C at these points because C lies in the surfaces gl = and g2 = 0. Therefore, VI lies in the plane determined by Vg 1 and Vg2, which means that VI = AVgl + JL Vg2 for some A and JL . Since the points we seek also lie in both surfaces, their coordinates must satisfy the equations gl(X, y, z) = and g2(X,y, z) = 0, which are the remaining requirements in Equations (2).

°

°

EXAMPLE 5 The plane x + y + z = 1 cuts the cylinder x 2 + y2 = 1 in an ellipse (Figure 14.56). Find the points on the ellipse that lie closest to and farthest from the origin. Solution

We find the extreme values of

I(x,y, z)

=

x 2 + y2

+ z2

(the square of the distance from (x, y , z) to the origin) subject to the constraints

gl(X, y , Z)

=

x 2 + y2 - 1

g2(X, y, z)

=

x

=

°

(3)

+ Y + z - 1 = 0.

(4)

The gradient equation in Equations (2) then gives

= AVg 1 + JLVg2 2xi + 2yj + 2zk = A(2xi + 2yj) + JLO + j + k) 2xi + 2yj + 2zk = (2Ax + JL)i + (2Ay + JL)j + JLk VI

or 2x = 2Ax FIGURE 14.56 On the ellipse where the plane and cylinder meet, we find the points closest to and farthest from the origin. (Example 5).

+ JL,

2y

=

2Ay

+ JL,

2z = JL .

(5)

The scalar equations in Equations (5) yield

+ 2z~(1 - A)x = z, 2Ay + 2z~(1 - A)y = z.

2x = 2Ax

2y

=

(6)

°

Equations (6) are satisfied simultaneously if either A = 1 and z = or A*-1 and x = y = z/(1 - A). If z = 0, then solving Equations (3) and (4) simultaneously to find the corresponding points on the ellipse gives the two points (1, 0, 0) and (0, 1,0). This makes sense when you look at Figure 14.56.

818

Chapter 14: Partial Derivatives If x = y, then Equations (3) and (4) give X2 +X 2

x+x+z- 1=0

-1=0 2>;2 = I

z=I-2>;

x=±v'Z 2

z= I 'f

v'Z.

The corresponding points on the ellipse are PI

=

('?, '?,

1-

v'Z)

and

Here we need to be careful, however. Although PI and P2 both give local maxima of f on the ellipse, P2 is farther from the origin than PI. The points on the ellipse closest to the origin are (I, 0, 0) and (0, 1,0). The point on the ellipse farthest from the origin is P2 • •

Exercises 14.8 Two Independent Variables with One Constraint 1. Extrema on an ellipse Find the points 00 the ellipse x 2 + 2y2 = I where f(x, y) = xy has its extreme values. 2. Extrema on a circle Find the extreme values of f(x, y) = xy subject to the constraint g(x, y) = x 2 + y2 - 10 = 0. 3. Maximum on a line Find the maximum value of f(x, y) = 49 - x 2 - y2 00 the line x + 3y = 10.

4. Extrema on a line Find the local extreme values of f(x, y) 00 the line x + y = 3.

= x7

S. Constrained minimum nearest the origin.

Find the points on the curve xy2 = 54

6. Constrained minimum

Find the points

00

the curve x7 = 2

nearest the origin. 7. Use the method of Lagrange multipliers to find L

Minimum on a hyperbola The minimum value of x subject to the coostraints xy = 16, x > 0, Y > 0

+ y,

b. Muimum on a line The maximum value ofxy, subject to the constraint x + y = 16.

Comment on the geometry of each solution. 8. EItrema on a curve Find the points on the curve x + xy + y2 = I in the xy-plane that are nearest to aod farthest from the origio. 2

9. Minimum .urface area with ("ned volume Find the dimensioos of the closed right circular cylindrical cao of smallest surface area whose volume is 16'7T cm.3• 10. Cylinder in a .phere Find the radius aod height of the open right cireolar cylinder oflargest surface area that cao be inscribed in a sphere of radius a. What is the largest surface area? 11. Rectaogle of greate.t area in an ellipse Use the method of Lagrange multipliers to f"md the dimensions of the rectsng1e of greatest area that can be inscribed in the ellipse x 2/16 + y2/9 = I with sides parallel to the courdinate ases.

ll. Rectaogle of longest perimeter in an ellipse Find the dimensioos of the rectangle oflargest perimeter that cao be inscribed in

the ellipse x 2/a 2 + y2/b 2 = I with sides parallel to the coordinate ases. What is the largest perimeter? 13. Extrema on a circle Find the maximum aod minimum values of x 2 + y2 subject to the coostraint x 2 - 2x + y2 - 4y = O. 14. Extrema on a circle Find the maximum and minimum values of 3x - y + 6 sui!ject to the constraint x 2 + y2 = 4. 15. Ant on a metal plate TIre temperature at a point (x, y) on a metal plate is T(x, y) = 4x 2 - 4xy + y2. An aot on the plate wa1ks arouod the circle of radius 5 centered at the origio. What are the highest aod lowest temperatures encoontered by the aot? 16. Cheape.t storage taok Yoor f"mn has been asked to design a storage tauk for liquid petroleum gas. The customer's specifications call for a cylindrical tank with hemispherical ends, aod the tank is to hold 8000 m' of gas. The customer also wants to use the smallest arooont of materisl possible in building the tank. What radius and height do you recommend for the cylindrical portion of the tank?

Thllle Independent Variables with One Cons!ra;nt 17. Minimnm distance to a point Find the point x + 2y + 3z = 13 closest to the point (I, I, I).

00

the plane

18. Maximnm distaoce to a point Find the point on the sphere x 2 + y2 + z2 = 4 farthest from the point (I, -I, I). 19. Minimnm di.laoce to the origin

from the surface x 2

Find the minimum distance

z2 = 1 to the origin.

- y2 -

20. Minimnm distance to the origin z = xy + 1 nearest the origin.

Find the point on the surface

21. Minimum di.laoce to the origin Find the points on the surface z2 = xy + 4 closest to the origin. 22. Minimnm distaoce to the origin face xyz = I closest to the origio.

Find the point(s) on the sur-

23. Extrema on a sphere Find the maximoro aod minimum values of

f(x,y,z) on the sphere

x2

=

+ y2 + z2

x - 2y = 30.

+ 5z

819

14.8 Lagrange Multipliers 24. Extrema on a sphere Find the points on the sphere x 2 + y2 + z2 ~ 25 where I(x,y, z) ~ x + 2y + 3z has its

maximum and minimum values.

34. Minimize the function I(x,y,z) ~ x 2 + y2 + z2 subject to the constraints x + 2y + 3z ~ 6 and x + 3y + 9z ~ 9.

25. Minjmjzing a sum of squares Find three real numbers whose sum. is 9 and the sum of whose squares is as small as possible.

35. Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes y + 2z ~ 12 and x + y ~ 6.

26. Maximizing a product Find the largest product the positive numbersx,y,andz can have if x + y + z2 = 16.

36. Muimum value on line of intersection Find the maximum

27. Rectangular bOJ[ of largest volume in a sphere Find the di· mensions of the closed rectangular box with maximum volume that can be inseribed in the unit sphere.

37. EItrema on a curve of intersection Find the extreme values of

28.

])OJ[ with vertex on a

plane Find the volume of the largest closed rectangular box in the fIrSt oclaot having three faces in the coordinareplanesandavertexontheplanex/a + y/b + z/c ~ I,where a> O,b > o,andc > O.

29. Hottest point on a space probe A space probe in the shape of the ellipsoid 4x 2 + y2 + 4z2 ~ 16 enters Earth's atmosphere and its surface begins to heat After I hour, the temperature at the point (x,y, z) on the probe's surface is

T(x,y,z) ~

ax 2 + 4yz -

16z + 600.

Find the hotrest point on the probe's surface.

30. Extreme temperatores on a sphere Suppose that the Celsius temperature at the point (x, y, z) on the sphere x 2 + y2 + z2 ~ I is T ~ 40Oxyz2 . Locare the highest and lowest temperatures on the sphere. 31. Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts x and y of two capital goods G1 and G2 is sometimes measured by a function U(x, y). For example, Gl and (h might be two chemicals a pharmaceutical coropany needs to have on hand and U(x, y) the gain from manufactoring a product whose synthesis requires different amounts of the chemicals depending on the process used. If Gl costs a dollars per kilogram, (h costs b dollars per kilogram, and the total amount allocared for the purchase of G 1 and G2 together is c dollars, then the company's managers want to maximize U(x, y) given that ax + by ~ c. Thus, they need to solve a typical Lagrange multiplier problem. Suppose that

U(x,y) ~ xy and that the equation ax

+

by

2x + Y

~

~

+ 2x

c simplifies to

30.

Find the maximum value of U and the corresponding values of x and y subject to this latter constraint

32. Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you waot to place it where the magnetic field of the plauet is weakest. The plauet is spherical, with a radius of 6 units. Based on a coordinare system whose origin is at the center of the plauet, the streugth of the magnetic field is given by M(x, y, z) ~ 6x - y2 + xz + 60. Where should you locare the radio relescope?

Extreme value. Subject to Two Constraints

33. Maximize the function I(x,y, z) ~ x 2 + 2y - z2 subject to the constraints 2x - y ~ 0 aod y + z ~ O.

value that I(x, y, z) ~ x 2 + 2y - z2 can have on the line of intersection of the planes 2x - Y ~ 0 and y + z ~ O.

I(x, y, z) ~ x'yz

+

1 on the intersection of the plane z ~ I with

the sphere x 2 + y2 + z2 = 10. 38. a. Muimum on line of intersection Find the maximum value of w ~ xyz on the line of intersection of the two planes x + y + z ~ 40 and x + y - z ~ O. b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of w.

39. EItrema on a circle ofintenection Find the extreme values of the function/(x,y, z) ~ xy + z2 on the circle in which the plane y - x = 0 intersects the sphere x 2 + y2 + z2 = 4.

40. Minimum distance to the origin Find the point closest to the origin on the curve of intersection of the plane 2y the cone z2 = 4x 2 + 4y2.

+ 4z

~

5 and

Theory and Examples 41. The condition VI = AVgisnotsnllicient Although VI ~ AVg is a necessary condition for the occurrence of an extreme value of I(x, y) subject to the conditions g(x, y) ~ 0 and Vg #' 0, it does

not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to fmd a maximum value of I(x, y) ~ x + y subject to the constraint that xy ~ 16. The method will identifY the two points (4, 4) and (-4, -4) as candidares for the location of extreme values. Yet the sum (x + y) has no maximUD2 value on the hyperbola xy ~ 16. The farther you go from the origin on this hyperbola in the first quadrant, the larger the sUD2/(x,y) ~ x + ybecomes.

42. A lea.t squa...s plane The plaue z ~ Ax "fitted" to the following points (x" Yb z,):

(0,0,0),

(0, I, I),

+ By + C

is to be

(1,0, -I).

(I, I, I),

Find the values of A, B, and C that minimize

f-1• (Ax, + By, + C -

z,)"

the sum of the squares of the deviations. 43. a. Muimum on a sphere Show that the maximum value of a 2b 2c 2 on a sphere of radius r centered at the origin of a Cartesian abc-coordinare systero is (r 2/3)3 b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers a, b, and c,

a (abc ) 1/3 < -

+b +c. 3

'

that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean. 44. Sum of products Let at. a2, ... , an be n positive numbers. Find the maximum of l:?=1 a,x, subject to the constraint l:1=1 x? = 1.

820

Chapter 14: Partial Derivatives

COMPUTER EXPLORATIONS

46. Minimize J(x, y, z) = :x:yz subject to the constraints x 2 + y2 - 1 = 0 and x - z = O. 47. Maximize J(x, y, z) = x 2 + y2 + z2 sui!ject to the constraints 2y + 4z - 5 = 0 and4x 2 + 4y2 - z2 = O.

In Exercises 45--50, use a CAS to perform the following steps implementing tire method of Lagrange multipliers for fmding constrained extrema: L

Formthefunctionh = J - AlgI - A2g2,whereJ is the function to optimize subject to tire constraints gl = 0 and g2 = O.

48. Minimize J(x,y,z) = x 2 + y2 + z2 subject to the constraints x 2 - :x:y + y2 _ z2 - I = 0 andx 2 + y2 - I = O. 49. Minimize J(x,y,z, w) = x 2 + y2 + z2 +w 2 subject to the

h. Determine all the f!rst partial derivatives of h, including the partials with respect to Al andA2, and set them equal to O.

2:x:-y+z-w-I=O

constraints

and

x+y-z+

w - I = O.

c. Solve the system of equations found in part (b) for all the unknowns, including Al and A2.

50. Determine the distance from the line y = x + I to the parabola y2 = x. (Hint: Let (x, y) be a point on tire line and (w, z) a point on the parabola. You want to minimize (x - w)' + (y - z)'.)

eI. Evaluate J at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the

exercise. 45. Minimize J(x,y,z) = :x:y + yz sui!ject to x 2 + y2 - 2 = 0 andx 2 + z2 - 2 = O.

14.9

rl

the

constraints

_~~y~Lo_(_s_F_o_ rm_u_La__ ro_r_Tw _ o__ Va_n_'a_b_Le_s______________-----------In this section we use Taylor's formula to derive the Second Derivative Test for local extreme values (Section 14.7) and the error formula for linearizations of functions of two independent variables (Section 14.6). The use of Taylor's formula in these derivations leads to an extension of the formula that provides polynomial approximations of all orders for functions of two independent variables.

Derivation of the Second Derivative Test t

~

1 S(a

+ h,b + k)

Parametrized

Let I(x, y) have continuous partial derivatives in an open region R containing a point P( a, b) where Ix = Iy = 0 (Figure 14.57). Let h and k be increments small enough to put the

point S(a + h, b segment PS as

+ k) and the line segment joining it to P inside R. We parametrize the x=a+th,

If F(t) = I(a

O:5t:51.

y=b+tk,

+ th, b + tk), the Chain Rule gives

,

t= 0

F (t)

P(a, b)

=

ax

Ix dt

dy

+ Iy dt

=

hlx

+ k/y.

Part of open region R

FIGURE 14.57 We begin the derivation of the Second Derivative Test atp(a, b) by parametrizing a typical line segment from P to a point S nearby.

Since Ix and Iy are differentiable (they have continuous partial derivatives), F' is a differentiable function of t and

F"

=

aF' ax + aF'dy axdt aydt

= h 21xx

=

~(hf + kf)'h + ~(hf + kf)'k ax x y ay x y f", ~ fyx

+ 2hklxy + elyy .

Since F and F' are continuous on [0, I] and F' is differentiable on (0, 1), we can apply Taylor's formula with n = 2 and a = 0 to obtain

F(I)

=

F(O) + F'(O)(1 - 0) + F"(c)

F(I)

=

F(O) + F'(O)

t

+ F"(c)

(1

2

0)' (1)

14.9 Taylor's Formula for Two Variables

821

for some c between 0 and 1. Writing Equation (1) in terms of I gives I(a

+ h, b + k)

= I(a, b)

+ hlx(a, b) +

kly(a, b)

+ ~ (h 2lxx + 2hkl"1 + k 2lyy) I

.

(2)

(a+ch, b+cl)

Since Ix(a, b) = I,(a, b) = 0, this reduces to I(a

I(a

+ h,b + k)

- I(a,b) =

~(h2Ixx + 2hkl"1 + k 2lyy) I(a+ch, b+ck).

(3)

The presence of an extremum of I at (a, b) is determined by the sigo of k) - I(a, b). By Equation (3), !bis is the same as the sigo of

+ h, b +

Q(c) = (h 2lxx

+ 2hkl"1 + k2Iyy)l(a+do.b+ck)'

Now, if Q(O) oF 0, the sigo of Q(c) will be the same as the sigo of Q(O) for sufficiently small values of h and k. We can predict the sigo of

Q(O)

2

= h Ixx(a, b)

+ 2hkl"1(a, b) + k 2Iyy(a, b)

(4)

from the signs of Ixx and fulyy - 1"12 at (a, b). Multiply both sides of Equation (4) by Ixx and rearrange the right-band side to get (5)

From Equation (5) we see that

1. If Ixx < 0 and Ixxlyy - 1"12 > 0 at (a, b), then Q(O) < 0 for all sufficiently small nonzero values of h and Ie, and I has a local maximum value at (a, b). 2. If Ixx > 0 and Ixxlyy - 1"12 > 0 at (a, b), then Q(O) > 0 for all sufficiently small nonzero values of h and Ie, and I has a local minimum value at (a, b). 3. If Ixxlyy - 1"12 < 0 at (a, b), there are combinations of arbitrarily small nonzero values of h and k for which Q( 0) > 0, and other values for which Q( 0) < O. Arbitrarily close to the point Po(a, b, I(a, b)) on the surface z = I(x,y) there are points above Po and points below Po, so I has a saddle point at (a, b). 4. If I xxlyy - 1"12 = 0, another test is needed. The possibility that Q(O) equals zero prevents us from drawing conclusions about the sigo of Q(c).

The Error Formula for Linear Approximations We want to show that the difference E(x, y), between the values of a function I(x, y), and its linearization L(x, y) at (xo, Yo) satisfies the inequality IE(x,yli:5

~M(lx -

xol

+ Iy

- yoll'·

The function I is assumed to have continuous second partial derivatives throughout an open set containing a closed rectangular region R centered at (xo, yo). The number M is an upperbouod forl/xxl, I/yyl, and 1/"11 onR. The inequality we want comes from Equation (2). We substitute Xo and Yo for a and b, and x - Xo and y - Yo for h and k, respectively, and rearrange the result as I(x,y) = l(xo,Yo)

+ fx{xO,yo)(x

- xo)

+ Iy(xo,yo)(y

- Yo)

linearization L(x, y) 1

+ -2 (x

- Xol'lxx

+ 2(x

- xo)(y - yo)I"1

+ (y

error E(x, y)

- Yol'lyy)l(Xo + c (X-Xo.Yo ) + c (Y-Yo » .

822

Chapter 14: Partial Derivatives

TIris equation reveals that

IEI:5 !Ox - xo1 2 1/",1 + 21x - xolly - yoll/.,,1 + Iy -

YOI21/yy!).

Hence, if Mis an upper bound for the values ofl I", I, 1/.,,1, andl/yyl onR,

IEI:5 !Ox -

xol 2M

!M
=

+ 21x -

xolly - yolM

+ Iy -

Yo I2M)

xol + Iy - yol)'.

Taylor's Formula for Functions of Two Variables The formulas derived earlier for F' and F" can be obtained by applying to I(x, y) the operators

(h~ax + k~) ay

and

(

h~ + k~ ax

2

ay )

2

=

2

2

h2~2 + 2hk_a- + k2~. ax

ax ay

ay2

These are the first two instances of a more general formula, F (")()-~()-(~ t - dt" F1 t - h ax

~)"() + kay I X,y ,

(6)

which says that applying d"/dt" to F(t) gives the same result as applying the operator

+ k~)" (h~ ax ay to I(x, y) after expanding it by the Binomial Theorem.

If partial derivatives of I through order 11 + I are continuous throughout a rectangular region centered at (a, b), we may extend the Taylor formula for F(t) to

F(t) and take t

=

=

F"(O)

+ F'(O)t + ----z! t 2 +

F(O)

F(")(O)

+ ----,;y- t(") + remainder,

I to obtain F( I)

=

F( 0)

F"(O)

F(")(O)

+ F' (0) + ----z! + ... + ----,;y- + remainder.

When we replace the first,. derivatives on the right of this last series by their equivalent expressions from Equation (6) evaluated at t = 0 and add the appropriate remainder term, we arrive at the following formula.

Taylor'. Formula for I(x,y) at the Point (a, b) Suppose I(x, y) and its partial derivatives through order,. + I are continuous throughout an open rectangular region R centered at a point (a, b). Then, throughout R, I(a

+ h,b + k)

= I(a, b)

i,

+ (hi. + kly ) I(a. b) +

2 (h /",

+ 2hkl." + k2Iyy)l(a.b)

+ 3\. (h 3/= + 3h2klxxy + 3hk21xyy + k3/yyy)l(a,b) + ... + ~ (h aa + kaa 11. ct Y +

I

(,. + 1)'

(

h~ + k~ ) ax

ay

"+1

I

1

(a+ch,b+ck)'

)"/1

(a, b)

(7)

823

14.9 Taylor's Formula for Two Variables

The IlISt n derivative terms are evaluated at (a, b). The last term is evaluated at some point (a + ch, b + ck) on the line segment joining (a, b) and (a + h, b + k). If (a, b) = (0, 0) and we treat h and k as independent variables (denoting them now by x and y), then Equation (7) assumes the following simpler form.

Taylor'. Formula for I(x,y) at the Origin

I(x,y) = 1(0,0)

+ +

+ xix + yly +

i!

(x 3/=

(n

+ I)!

I

i!

(x 2lxx

+ 2xy1", + y2lyy)

+ 3x'ylxxy + 3xy21xyy + y3lyyy)

I +n!

+

(a x - + ya - )nI ax ay

(a x-+ y a - )n+lI I ax

ay

(8)

(ox.",)

The first n derivative terms are evaluated at (0, 0). The last term is evaluated at a point on the line segment joining the origin and (x, y). Taylor's formula provides polynomial approximations of two-variable functions. The IlISt n derivative terms give the polynomial; the last term gives the approximation error. The first three terros of Taylor's formula give the function's linearization. To improve on the linearization, we add higher-power terms.

EXAMPLE 1 Find a quadratic approximation to I(x,y) = sinxsiny near the origin. How accurate is the approximation if Ixl ,,;; 0.1 and Iyl ,,;; O.I? Solution

We take n = 2 in Equation (8):

Calculating the values of the partial derivatives,

1(0,0)

= sinxsinyl(o.o) =

0,

1",,(0,0)

= -sinxsinyl(o.o) =

0,

1,(0,0)

= cosxsinyl(o.o) =

0,

1",(0,0)

= cosxcosyl(o.o) =

1,(0,0)

= sinxcosyl(o.o) =

0,

lyy(O,O) = -sinxsinyl(o.o) = 0,

I,

we have the result sinxsiny '" 0

+ 0 + 0 + !(x 2(0) + 2xy(I) + y2(0)),

or

sin x siny '" xy.

The error in the approximation is

The third derivatives never exceed I in absolute value because they are products of sines and cosines. Also, Ixl ,,;; 0.1 andlyl ,,;; 0.1. Hence

IE(x,y)1 ,,;; i«0.1)3

+ 3(0.1)3 + 3(0.1)3 + (0.1)3)

=

~(0.1)3 ,,;; 0.00134

(rounded up). The error will not exceed 0.00134 iflxl ,,;; 0.1 andlyl ,,;; 0.1.

•

824

Chapter 14: Partial Derivatives

Exercises 14.9 Finding Quadratic and Cubic Approximations In Exercises 1-10, use Taylor's formula for f(x,y) atthe origin to fmd quadratic and cubic approximatioos of f near the origin.

1. f(x,y) = xeY

2. f(x,y) = ,'cosy

3. f(x,y) =ysinx

4. f(x,y) = sinxcosy

5. f(x,y) 7. f(x,y)

= ,rln(1 + y) = sin (x 2 + y2)

+ y + I) 2 cos (x + y2)

6. f(x,y) = In (2x

8. f(x,y) =

I 9. f(x, y) = ~I----''--­

-x-y

I 10. f(x,y) = I -x-y

+ xy

11. Use Taylor's formula to fmd a quadratic approximation of f(x, y) = cos x cos y at the origin. Estimate the error in the approximationiflxl'" 0.1 andlyl '" 0.1. 12. Use Taylor's formula to fmd a quadratic approximation of siny at the origin. Estimate the error in the approximation iflxl '" 0.1 andlyl'" 0.1.

,r

14.1 0 I-p_a_rti_'a_L_D_e_ri_v_ati_'v_e_S_Wl_'_th_ C_on_s_t_ra_in_e_d_V_a_"_'a_b_Le_s_ _ _ _ _ _ _ _ __ In finding partial derivatives of functions like w = f(x, y), we have assumed x andy to be independent. In many applications, however, this is not the case. For example, the internal energy U ofa gas may be expressed as a function U = f(P, Y, 1') ofpressureP, volume Y, and temperature T. If the individual molecules of the gas do not interact, however, P, Y, and T obey (and are constrained by) the ideal gas law

PY= nRT

(n and R constant),

and fail to be independent. In this section we learn how to find partial derivatives in situations like this, which occur in economics, engineering, and physics.·

Decide Which Variables Are Dependent and Which Are Independent If the variables in a function w = f(x,y, z) are constrained by a relation like the one imposed on x, y, and z by the equation z = x 2 + y2, the geometric meanings and the numerical values of the partial derivatives of f will depend on which variables are chosen to be dependent and which are chosen to be independent. To see how this choice can affect the outcome, we consider the calculation of iJw/ ax when w = x 2 + y2 + z2 and z = x 2 + y2.

EXAMPLE 1

Find iJw/ax if w =

x2 + y2 + z2 and z

=

x2 + y~

Solution

We are given two equations in the four unknowns x, y, z, and w. Like many such systems, this one can be solved for two of the unknowns (the dependent variables) in terms of the others (the independent variables). In being asked for iJw/ax, we are told that w is to be a dependent variable and x an independent variable. The possible choices for the other variables corne down to

Dependent w,z w,y

Independent x,y x,z

In either case, we can express w explicitly in terms of the selected independent variables. We do this by using the second equation z = x 2 + Y 2 to eliminate the remaining dependent variable in the flISt equation. *This section is based on notes written for MIT by Arthur P. Mattuck.

14.10 Partial Derivatives with Constrained Variables

825

In the f"lISt case, the remaining dependent variable is z. We eliminate it from the first equation by replacing it by x' + y' . The resulting expression for w is

+ y' + z' = x' + y' + (x' + y')' x' + y' + x4 + 2x'y' + y4

w = x' =

and

~;

= 2x

+ 4x 3 + 4xy'.

(l)

This is the formula for aw/ ax when x and y are the independent variables. In the second case, where the independent variables are x and z and the remaining dependent variable is y, we eliminate the dependent variable y in the expression for w by replacing y' in the second equation by z - x' . This gives

,

w = x'

+ y' + z'

=

x'

+ (z - x') + z'

=

z

+ z'

and aw = 0 ax .

FIGURE 14.58

If P is constrained to

lie on the paraboloid z = x 2 + y~ the value of the partial derivative of w = x 2 + y2 + z2 with respect to x at P depends on the direction of motion (Example I). (I) Ail x changes, with y = 0, P moves up or down the surface on the parabola z = xl in the xz-plane with itw/iJx ~ 2x + 4x'. (2)Ailxchanges, with z = 1, P moves on the circle

x' + y' ~

This is the formula for aw/ ax when x and z are the independent variables. The formulas for aw/ax in Equations (I) and (2) are genuinely different. We cannot change either formula into the other by using the relation z = x' + y'. There is not just one aw/ax, there are two, and we see that the original instruction to f"md aw/ax was incomplete. Which aw/ ax? we ask. The geometric interpretations of Equations (I) and (2) help to explain why the equa+ y' + measures the square of the distance from the tions differ. The function w = point (x, y, z) to the origin. The condition z = x' + y' says that the point (x, y, z) lies on the paraboloid of revolution shown in Figure 14.58. What does it mean to calculate aw/ax at a point P(x, y, z) that can move only on this surface? What is the value of aw/ax when the coordinates of P are, say, (I, 0, I)? If we take x and y to be independent, then we f"md aw/ ax by holding y fixed (at y = 0 in this case) and letting x vary. Hence, P moves along the parabola z = x' in the xz-plane. As P moves on this parabola, w, which is the square of the distance from P to the origin, changes. We calculate aw/ax in this case (our first solution above) to be

x'

x

I,z ~ I,andaw/ax ~ O.

(2)

z'

aw ax = 2x

+ 4x 3 + 4xy'.

At the point P(I, 0, I), the value of this derivative is

~;

= 2

+4 +

0 = 6.

If we take x and z to be independent, then we f"md aw/ax by holding z fixed while x varies. Since the z-coordinate of P is I, varying x moves P along a circle in the plane z = I. As P moves along this circle, its distance from the origin remains constant, and w, being the square of this distance, does not change. That is,

aw = 0 ax '

as we found in our second solution.

•

How to Find rJw/rJx When the Variables in w = I(x, y, z) Are Constrained by Another Equation As we saw in Example 1, a typical routine for finding aw/ ax when the variables in the function w = /(x,y, z) are related by another equation has three steps. These steps apply to finding aw/ay and aw/az as well.

826

Chapter 14: Partial Derivatives

1. Decide which variables are to be dependent and which are to be independent. (In practice, the decision is based on the physical or theoretical context of our work. In the exercises at the end of this section, we say which variables are which.)

2. Eliminate the other dependent variable(s) in the expression for w. 3. Differentiate as usual.

Ifwe cannot carry out Step 2 after deciding which variables are dependent, we differentiate the equations as they are and try to solve for awjax afterward. The next example shows how this is done.

EXAMPLE 2

Findawjax at the point (x,y, z) = (2, -1, I) if w = x2

+ y2 + z2,

Z3 - xy

+ yz + y3

= I,

and x and y are the independent variables.

It is not convenient to eliminate z in the expression for w. We therefore differentiate both equations implicitly with respect to x, treating x and y as independent variables and w and z as dependent variables. This gives

Solution

aW=2x+2z az ax ax

(3)

and

2aZ 3z - - y ax

az + yax -+0

=

o.

(4)

These equations may now be combined to express awjax in terms of x, y, and z. We solve Equation (4) for azjax to get

az ax

y y+3z2

and substitute into Equation (3) to get

aw=2x+ ax

2yz y+3z2 '

The value of this derivative at (x,y, z) = (2, -I, I) is

aw) = 2(2) + 2(-1)(1) = 4 + -2 = 3. ( ax (2,-1.1) -I + 3(1)2 2

•

HISTORICAL BIOGRAPHY

Notation

Sonya Kovalevsky (185{)-1891)

To show what variables are assumed to be independent in calculating a derivative, we can use the following notation:

(~;)y

awjaxwithx andy independent

(:~t

af jay with y, x and t independent

14.10 Partial Derivatives with Constrained Variables

EXAMPLE 3 Solution

Find (iIw/ax)y,zifw =

x2 + y

827

- z + sin I and x + y = I.

With x, y, z independent, we have I = x

(~;)y,z =

+ y,

w

=

x 2 + y - z + sin (x + y)

2x + 0 - 0 + cos (x + y)!(x + y)

•

2x + cos (x + y).

=

Arrow Diagrams In solving problems like the one in Example 3, it often helps to start with an arrow diagram tbat shows how the variables and functions are related. If w = x2

+y

- z

+ sinl

x+y=1

and

and we are asked to findilw/ax whenx,y, andz are independent, the appropriate diagram is one like !bis:

(5)

Independent variables

Intermediate variables

Dependent variable

To avoid confusion between the independent and intermediate variables with the same symbolic names in the diagram, it is helpful to rename the intermediate variables (so they are seen as jUnctions of the independent variables). Thus, let u = x, v = y , and s = z denote the renamed intermediate variables. With !bis notation, the arrow diagram becomes

(;)

m

--->

Independent

variables

w

--->

Intennediate variables and

(6)

Dependent

variable

relations u=x v=y s=z

t

= x +y

The diagram shows the independent variables on the left, the intermediate variables and their relation to the independent variables in the middle, and the dependent variable on the right. The function w now becomes w = u2

+

V -

s + sint,

where

u =x,

v =y,

s = z,

and

I =

x

+ y.

828

Chapter 14: Partial Derivatives To find aw/iJx, we apply the four-variable fonn of the Chain Rule to w, guided by the arrow diagram in Equation (6):

=

(2u)(1) + (1)(0) + (-1)(0) + (cost)(I)

=2u+cost =

2x + cos (x + y).

Substituting the original independent = x and t = x + y

variables u

Exercises 14.10 finding Partial Derivatives with Constrained Variables In Exercises 1-3, begin hy drawing a diagram that shows the rela-

tions among the variables. 1. Ifw = x 2 + y2 + z2 andz

x2

=

each give aw/ax, depending on which variables are chosen to be dependent and which variables are chosen to be independent IdentifY the independent variables in each case.

+ y2, fmd Theory and Examples

2. Ifw

=

x2 + y - z + sintandx + y

t.

=

9. Estsblish the fact, widely used in hydrodynamics, that if j(x,y, z) = 0, then

t,find

(:t (:t

••

c.

(aa~)x"

f.

(~~)Y,' .

3, Let U = j(P, V, 1') be the internal energy of a gas that obeys the ideal gas law PV = nRT (n and R constant). Find

(~~)v

L

t.

(~~)v'

t.

(:)Y

(Hint: Express all the derivatives in terms of the formal partial derivatives ajlax, aj/iJy, and aj/az.) 10. Ifz = x

+ j(u), where u = xy, show that ilz

iJz

x ax - y ay

=

x.

4, Find

(~;)Y

L

11. Suppose that the equation g(x, y, z) = 0 determines z as a differentiable function of the independent variables x and y aod that g, O. Show that

*

at the point (x,y,z) = (O,I,1T)if

ysinz + zsinx

and

=

az) ag/iJy ( iJy , = - ag/az .

O.

5. Find 12. Suppose thatj(x,y,z, w) = 0 andg(x,y,z, w) = 0 determinez and w as differentiable functions of the independent variables x and y, and suppose that

at the point (w,x,y, z) = (4,2, I, -I) if

w

= x'y' + yz -

z'

aj ag aj ag -----*0

and

6. Find (au/ay), at the point (u, v) = y = uv.

iJz iJw

(\12,1), if x =

u'

+ v' and

w=x 2

_

(:)Y =

(~:);

8. Suppose that

aj ag aj ag axiJw awax aj ag aj ag ----aziJw iJwiJz -----

nates. Find

(~; ),

.

Show that

7. Suppose that x2 + y2 = r2 and x = r cos 0, as in polar coordi-

and

aw iJz

and

y 2+4z+t

and

x+2z+t=25.

iJw -=2x-1 ax

ajag az ay aj ag

ajag ay az aj ag

iJz ilw

ilwaz

-----

Show that the equatioos and

aW=2x_2 ax

(:), =

-----

Chapter 14 Practice Exercises

Chapter

00

Questions to Guide Your Review

1. What is a real-valued function of two independent variables? Three independent variables? Give examples. 2. What does it mean for sets in the plane or in space to be open? Closed? Give examples. Give examples of sets that are neither open nor closed. 3. How can you display the values ofa function f(x,y) of two independent variables graphically? How do you do the same for a function f(x,y, z) of three independent variables? 4. What does it mean for a function f(x, y) to have limit L as (x, y) ... (XQ, yo)? What are the basic properties of limits of functions of two independent variables? 5. When is a function of two (three) independent variables continuous at a point in its domain? Give examples of functions that are

continuous at some points but not others. 6. What can be said about algebraic combinations and composites of

continuous functions? 7. Explain the two-path test for nonexistence of limits. and a/lay of a function dermed? How are they interpreted and calculated?

8. How are the partial derivatives

/(x,y)

a/lax

9. How does the relation between rITst partial derivatives and continuity of functions of two independent variables differ from the relation between rITst derivatives and continuity for real-valued functions of a single independent variable? Give an example.

10. What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives of second and higher orders? Give examples. 11. What does it mean for a function /(x, y) to be differentiable? What does the Increment Theorem say about differentiability?

12. Haw can you sometimes decide from examjning Ix and /, that a function /(x, y) is differentiable? What is the relation between the differentiability of/and the continuity of / at a point? 13. What is the general Chain Rule? What form does it take for functions of two independent variables? Three independent variables? Functions derfied on surfaces? How do you diagram these differ-

14. What is the derivative ofa function /(x,y) at a point Po in the direction of a unit vector u? What rate does it describe? What geometric interpretation does it have? Give examples. 15. What is the gradient vector of a differentiable function /(x, y)? How is it related to the function's directional derivatives? State the analogous results for functions of three independent variables. 16. How do you rmd the tangent line at a point on a level curve of a differentiable function /(x, y)? How do you rmd the tangent plane and normal line at a point on a level surface of a differentiable function f(x, y, z)? Give examples. 17. How can you use directional derivatives to estimate change? 18. How do you linearize a function /(x, y) of two independent variables at a point (xo, yo)? Why right you want to do this? How do you linearize a function of three independent variables? 19. What can you say about the accuraey of linear approximations of functions of two (three) independent variables? 20. If(x,y) moves from (XQ,Yo) to a point (xo +
row your search for these values? Give examples. 23. How do you rmd the extrema of a continuous function /(x, y) on a closed borrnded region of the xy-plane? Give an example. 24. Describe the method of Lagrange multipliers and give examples. 25. How does Taylor's formula for a function f(x, y) generate polynomial approximations and error estimates? 26. Ifw = /(x,y,z), where the variables x, y, and z are cons1rained by an equation g(x, y, z) = 0, what is the meaning of the notation (ow/ax)y? How can an arrow diagram help you calculate this par-

tial derivative with constrained variables? Give examples.

ent forms? Give examples. What pattern enables one to remember all the different forms?

Chapter

ai

Practice Exercises

Domain, Range, and Level Curves In Exercises 1-4, rmd the domain and range of the given function and identifY its level curves. Sketch a typical level curve. 1. /(x,y) = 9x 2 + y2 2. f(x,y) = e X+Y 3. g(x,y) = I/xy

829

4. g(x,y)

=

7. h(x,y,z) = 8. k(x,y,z) =

~

In Exercises 5-1!, rmd the domain and range of the given function and identifY its level surfaces. Sketch a typical level surface. 5. /(x,y, z) = x 2 + y2 - z 6. g(x,y, z) = x 2 + 4y2

+ 9z2

2 X

I

2

+Y + z

2

I 2 2 2 x+y+z+1

Evaluating Um;ts Find the lirrtits in Exercises 9-14. 9. lim e" cosx 10. (x,y)~(~,ln2)

lim

(X,Y)~(O,O) X

+Y + cosy

2

830

11. 13.

Chapter 14: Partial Derivatives

~ 2

lim

(xoY)-(1,I) x

lim

P-(1,-l,e)

12.

y2

_

(x~)~(I.I)

14.

Inlx + y + zl

X 3y 3 -

lim

xy

lim

P-(I,-l,-l)

1

34. Show that if w = I(s) is any differentiable funetioo of s and if s=y+5x,then

1

tan-I (x + Y + z)

By considering different paths of approach, show that the limits in Exercises 15 and 16 do not exist

Implidt Differentiation Assuming that the equatioos in Exercises 35 and 36 defme y as a differentiable functioo of x, fmd the value of dy/dx at point P.

17. Continuous extension Let I(x,y) = (x 2 - y2)/(X 2 + y2) for (x, y) #' (0, 0). Is it possible to defme 1(0, 0) in a way that makes 1 cootinuous at the origin? Why?

Let

18. Continuous extension

(x,y)

=

37. I(x,y) = cosxcosy, P o(,,-/4, ,,-/4), v = 3i 38. I(x,y) = x 2e-2y, Po(i,O), v = i + j

+ rsinO

(x 2 + y2) + tan-I

1

21. I(R" R2, R,) = RI

1

v = 2i

1

r(l) = (cos 31)i + (sin 31)j + 31k

att

(the ideal gas law)

1/T

= 2rZ"'ifW

+Y

26. g(x,y) = eX + y sinx

43. Directional derivative. with given value. At the point (1, 2), the funetioo I(x, y) has a derivative of 2 in the direetioo toward (2, 2) and a derivative of -2 in the direetioo toward (1, 1). a. Find Ix(i, 2) and/y(I, 2).

44. Which of the following statements are true if I(x, y) is differentiable at (xo, yo) ? Give reasons for yoor answetli.

27. I(x,y) = x + xy - 5x' + In(x + 1)

= y2

,,-/3.

b. Find the derivative of1 at (1, 2) in the direction toward the point (4, 6).

x

2

28. I(x,y)

=

42. Mnimnm directiolllli derivative What is the largest value that the direetiooa1 derivative of I(x,y, z) = xyz can have at the point (1, 1, I)?

Find the secood-order partia1 derivatives of the functioos in Exercises

y

Po(O, 0, 0),

41. Derivative in velocity direction Find the derivative of I(x,y, z) = xyz in the direction of the velocity vector of the helix

25--28. =

+ 4j

v=i+j+k

~

Second-Order Partials

25. g(x,y)

increases and

+ 3j + 6k

+ R2 + R,

~T

f

+ 3y + 6z), P o( -1, -1, 1),

40. I(x,y, z) = x 2 + 3xy - z2 + 2y + z + 4,

22. h(x,y,z) = sin (2,,-x + y - 3z) 23. P(n, R, T, V) =

P(O,I)

P(O, In 2)

vector v.

39. I(x,y, z) = In (2x

19. g(r,O) = rcosO

0,

direetioo. Also, find the derivative of 1 at Po in the direction of the

(0,0).

Partial Derivatives

24. l(r,Z,T,w)

+' - 2 =

decreases most rapidly at Po and find the derivative of f in each

Ixl + Iyl #' 0

Ixl + Iyl ' { 0,

In Exercises II}-24, fmd the partial derivative of the funetioo with respect to each variable.

tin

X

Directional Derivatives

Is 1 continuous at the origin? Why?

20. I(x,y) =

+e

36. 2xy

In Exercises 37-40, fmd the directions in which

Sin(X- y ) I(x,y) =

35. 1 - x - y2 - sinxy = 0,

- 3xy

a. If u is a unit vector, the derivative of 1 at (xo, Yo) in the direc-

+ cosy + 7<'

tioo ofu is (fx(xo, Yo)i + I,(xo,yo)j)' u. b. The derivative of 1 at (Xo, Yo) in the direetioo of u is a vector.

Chain Rule Calculations 29. Find dw/dl at 1 = 0 if w = sin (xy + ,,-), x = e 1n(1+ 1).

t

,

c. The direetiooa1 derivative of1 at (xo, yo) has its greatest value in the directioo of VI.

and y =

d. At (xo, Yo) , vector VIis normal to the curve I(x, y) = I(xo, yo).

30. Find dw/dl at 1 = 1 if w = xe' + y sinz - cosz, x = 2vt, y = 1 - 1 + Inl,andz = ,,-1. 31. Findaw/arandaw/aswhenr = ,,-ands = Oifw = sin(2x - y),

x

=

32. Find

r

+ sins,y = rs.

aw/au

and

Grad;ents, Tangent Planes, and Normal Unes

aw/av

u= v =

when

0

if

w=

1n'\l'1+7 - tan-Ix and x = 2e"cosv. 33. Find the value of the derivative of/(x,y,z) = xy + yz + xzwith

respecttotonthecurvex

=

cost,y

=

sint,z

=

cos2tatt

=

1.

In Exercises 45 and 46, sketch the surface I(x,y, z) = c together with VI at the given points.

+ y + z2 = 0; (0, -1, ±I), (0,0,0) y2 + z2 = 4; (2, ±2, 0), (2, 0, ±2)

45. x 2 46.

Chapter 14

831

Practice Exercises

In Exercises 47 and 48, rmd an equation for the plane tangent to the level surface f(x,y, z) ~ c at the point Po. Also, find parametric equations for the line that is nonnal to the surface at Po. 47. x 2 - y - 5z ~ 0, P o(2, -I, I)

62. Muimum error in estimating the area of an ellipse If a ~ 10 em and b ~ 16 em to the nearest millimeter, what should you expect the maximum percentage error to be in the calculated areaA ~ '1Tab of the ellipsex2/a 2 + y2/b 2 ~ I?

48. x 2 + y2

63. Error in estimating a product Let y = uv and z where u and v are positive independent variables.

~ 4,

+Z

Po(l, 1,2)

In Exercises 49 aod 50, rmd an equation for the plane tangent to the surface z ~ f(x, y) at the given point. 49. z ~ In (x 2 50. z ~ 1/(x

2

+ y2), + y2),

(0, 1,0)

+ 2z

~ 4,

(1/2, I, 1/2)

Linearizations In Exercises 55 and 56, rmd the linearization L(x, y) of the function f(x, y) at the point Po. Then rmd an upper bound for the magnitude of the error E in the approximation f(x, y) '" L(x, y) over the rectangle R. 55. f(x,y) ~ sinxcosy,

Ix - *1

S

0.1,

56. f(x,y) ~ xy - 3y2

R:

Ix - II

S

0.1,

PO('1T/4, '1T/4)

Iy - *1

+ 2,

Iy -

S

0.1

Po(l, 1)

II

S

0.2

Find the linearizations of the functions in Exercises 57 and 58 at the given points. 57. f(x,y, z) ~ xy + 2yz - 3xz at (1,0,0) and (I, 1,0) 58. f(x,y,z) ~ v2cosxsin(y

cardiac output body surface area .

~ ~-~-"-----

The body surface area B of a person with weight w and height h is approximated by the formula B

= 71.84wo.42Sho.72S,

y ~ I

(I, I, 1/2) Point: 54. Surfaces: x + y2 + Z ~ 2, y ~ I

R:

64. Cardiac index To ma1re different people comparable in studies of cardiac output, researchers divide the measured cardiac oulput by the body surface area to rmd the cardiac index C:

C

Tangent Unes to Curves In Exercises 53 and 54, rmd parametric equations for the line that is tangent to the curve of intersection of the surfaces at the given point

Point:

a. Ifu is measured with an error of2% and v with an error of3%, about what is the percentage error in the calculated value ofy?

P o('1T, 1)

53. Surfaces: x 2 + 2y

+ v,

b. Show that the percentage error in the calculated value of z is less than the percentage error in the value of y.

(I, I, 1/2)

In Exercises 51 and 52, rmd equations for the lines that are tangent aod normal to the level curve f(x,y) ~ c at the point Po. Then sketch the lines and level curve together with Vf at Po. 51. y - sinx ~ I,

= u

+ z)at(O, 0, '1T/4) and ('1T/4, '1T/4, 0)

which gives B in square centimeters when w is measured in kilograms and h in centimeters. You are about to calculate the cardiac index of a person 180 em tall, weighing 70 kg, with cardiac output of 7 L/min. Which will have a greater effect on the calculation, a l·kg error in measuring the weight or a l·em error in measuring the height? Local Extrema Test the functions in Exercises 65-70 for local maxima and minima and saddle points. Find each function's value at these points. 65. f(x,y) ~ x 2 - xy + y2 + 2x + 2y - 4 66. f(x, y) ~ 5x 2 + 4xy - 2y2 + 4x - 4y 67. f(x,y) ~ 2x' 68. f(x,y) ~ x' 69. f(x,y) ~ x' 70. f(x, y) ~ x'

60. Sensitivity to change Is f(x,y) ~ x 2 - xy + y2 - 3 more sensitive to changes in x or to changes in y when it is near the point (I, 2)? How do you know? 61. Change in an etectrical circuit Suppose that the current I (am· peres) in an electrical circuit is related to the voltage V(volts) and the resistance R (ohms) by the equation I ~ VIR. If the voltage drops from 24 to 23 volts and the resistance drops from 100 to 80 ohms, will I increase or decrease? By about how much? Is the change in I more sensitive to change in the voltage or to change in the resistance? How do you know?

3xy

+ 2y'

Absolute Extrema In Exercises 71-78, rmd the absolute maximum and minimum values of f on the region R. 71. f(x,y) ~ x 2

Estimates and Sensitivity to Change 59. Measuring the volume of a pipeline You plan to calculate the volume inside a stretch of pipeline that is about 36 in. in diameter and I mile long. With which measurement should you be more careful, the length or the diameter? Why?

+

+ y' - 3xy + IS + y' + 3x2 - 3y2 - 8x 2 + 3y2 - 6y

+ xy + y2

- 3x

+

3y

R: The triangular region cut from the rITSt quadrant by the line x+y~4

72. f(x,y) ~ x 2 - y2 - 2x

+ 4y + 1

R: The rectangular region in the rll'St quadrant bounded by the coordinate axes and the lines x = 4 and y = 2 73. f(x,y) ~ y2 - xy - 3y

+ 2x

R: The square region enclosed by the lines x 74. f(x,y) ~ 2x + 2y - x 2 - y2

~

±2 and y

~

±2

R: The square region bounded by the coordinste axes and the lines x ~ 2, y ~ 2 in the rITSt quadrant 75. f(x,y) ~ x 2 - y2 - 2x + 4y R: The triangular region bounded below by the x-axis, above by the line y ~ x + 2, and on the right by the line x ~ 2

832

Chapter 14: Partial Derivatives

76. f(x,y) ~ 4X)' - x4 - y4

+

16

R: The triangu1llr region bounded below by 1he line y ~ -2, above by 1he line y ~ x, and on 1he right by 1he line x ~ 2 77. f(x,y) ~ x 3 + y3 + 3x2 - 3y2

R: The square region enclosed by 1he lines x 78. f(x,y) ~ x 3 + 3xy + y3 + I

~

R: The square region enclosed by 1he lines x

~

± I and y ± I and y

~

~

93. If a and b are constants, w = u 3 + tanh U + cos U, and U = ax + by, show iliat

±I ±I

Lagrange MultipUers 79. Extrema on a circle Find 1he extreme values of f(x, y) ~ x 3 + y2 on the circle x 2 + y2 = 1.

94. Using the Chain Rule Ifw ~ In (x 2 + y2 + 2z), x ~ r + s, Y = r - s, and z = 2rs, fmd Wr and w" by the Chain Rule. Then check your answer another way. 95. Angle between vectors The equations e" cos v - x = 0 and e" sin v - y ~ def"me u and v as differentiable functions of x and y. Show 1hat 1he angle between 1he vectors

°

80. Extrema on a circle Find 1he extreme values of f(x, y) ~ X)' on the circle x 2 + y2 = 1. 81. Extrema in a disk Find 1he extreme values of f(x, y) ~ x 2 + 3y2 + 2y on 1he urnt disk x 2 + y2 '" 1. 82. Extrema in a disk Find 1he extreme values of f(x, y) ~ x 2 + y2 - 3x - xy on the diskx 2 + y2 ~ 9. 83. Extrema on a sphere Find 1he extreme values of f(x,y, z) ~ x - y + zontheunitspherex 2 + y2 + z2 = 1.

OUi+OU j iix ay

is constant. 96. Polar coordinates and second derivatives Introducing polar coordinates x ~ r cos 8 and y ~ r sin 8 changes f(x, y) to g(r,8). Find 1he value of a2g/ali' at 1he point (r, 8) ~ (2, ",/2), given iliat af ax

84. Minimum distance to origin Find 1he points on 1he surface x 2 - zy = 4 closest to the origin.

85. Minimjzjng cost of. bOI A closed rectangular box is to have volume V em3• The cost of the material used in the box is a cents/em2 for top and bottom, b ceots/em2 for front and back, and c cents/cm2 for the remaining sides. What dimensions minimize the total cost of materials?

and

af

a2f iix 2

a2f ay2

-~-~-~-~l

ay

at iliat point. 97. Normal line parallel to a plaoe

Find 1he points on 1he surface

(y + z)2 + (z - x)'

86. Leastvolume Find 1he plane x/a + y/b + z/e ~ liliatpasses 1hrough 1he point (2, I, 2) and cuts off 1he least volume from 1he f"lrSt octant.

98. Tangent plane parallel to xy-plane face

87. EItrema on curve of intersecting surfaces Find the extreme

X)'

valuesoff(x,y,z) ~ x(y + z)on1hecurveofintersectionof1he right circular cylinder x 2 + y2 ~ I and 1he hyperbolic cylinder :x:z~1.

88. Minimum distance to origin on curve of intersecting plane and cone Find the point closest to the origin on the curve of intersection of the plane x + y + z = 1 and the cone z2 = 2:x;2

+

2y2.

Partial Derivatives wtth Constrained Variables In Exercises 89 and 90, begio by drawing a diagram iliat shows 1he

relations among the variables. 89. Ifw ~ x 2.". andz ~ x 2 - y2 f"md

where 1he nonnalline is parallel to 1he yz-plane.

+ yz + z:x:

L

(~~)p

~ (~~)T'

Theory and Examples 2 + y2, and 8 ~ tan-I (y/x). Find 91. Let w ~ f(r, 8), r ~ aw/iix andaw/ay and express your answers in terms ofr and 8.

Vx

92. Let z ~ f(u, v), u ~ ax + by, and v ~ ax - by. Express Zx and Zy in terms of /u,/v, and the constants a andb.

Find 1he points on 1he sur-

- x - z2 ~

°

where 1he tangent plane is parallel to 1he xy-plane.

99. When gradient is parallel to position vector Suppose that Vf(x,y, z) is always parallel to 1he position vector xi + yj + zk. Show iliat f(O, 0, a) ~ f(O, 0, -a) for aoy a. 100. One-sided directional derivative in all directions, but no gradient The one-sided directional derivative off at P (xo, Yo, zo) in the direction u = uti + u2j + U3k is the number

lim f(xo

+ SUhYO + SU2,ZO + SU3) - f(xo,Yo,zo)

3-0+

S

•

Show 1hat 1he one-sided directional derivative of f(x,y, z) ~

90. Let U ~ f(P, V, T) be 1he internal energy of a gas iliat obeys 1he ideal gas law PV ~ nRT (n and R constant). Find

~ 16

Vx2 + y2 + z2

at 1he origio equals I in aoy direction but iliat f has no gradient vector at the origin. 101. Normal line through origin Show that 1he line normal to 1he surfaceX)' + z ~ 2 at1he point (I, I, I) passes 1hrough 1he origin. 102. Tangent plane aod normal line a. Sketch 1he surface x 2 - y2

+ z2

~ 4.

~ Find a vector nonnal to 1he surface at (2,

-3,3). Add 1he

vector to your sketch. c. Find equations for 1he tangent plane and normal line at

(2, -3,3).

Chapter 14 Additional and Advanced Exercises

Chapter

833

Additional and Advanced Exercises

Partial Derivatives

6. Surface in polar coordinates

1. Function with saddle at the origin If you did Exercise 60 in Section 14.2, you know that the function

f(x,y)

{XY:: :

=

sin6r

f(r,

~:,

{

(x,y)

= (0,0)

r#O

6r'

(J) =

(x,y) # (0,0)

0,

Let

r

I,

0,

=

where r and (J are polar coordinates. Find

(see the accompanying figure) is continuous at (0, 0). Find fxy(O, 0) andfyx(O, 0).

a. lim f(r, (J) r--->O

b. fr(O,O)

c. fe(r, (J),

r # 0.

z

y

Gradients and Tangents 2. Finding a function from second partials Find a function w = f(x,y) whose first partial derivatives are aw/ax = I + eX cosy and aw/ay = 2y - eX siny and whose value at the point (In 2, 0) is In 2. 3. A proof of Leibniz's Rule Leibniz's Rule says that if f is continuous on [a, b] and if u(x) and v(x) are differentiable functions ofx whose values lie in [a, b], then V

d l (X) dv du dx f(t) dt = f(v(x» dx - f(u(x» dx . u(x)

Let r = xi

7. Properties of position vectors r = Irl.

+ yj + zk

and let

a. Show that Vr = r/r. b. Show that V(rn) = nr n - 2 r. c. Find a function whose gradient equals r. d. Show that r • dr = r dr. e. Show that V(A· r) = A for any constant vector A.

8. Gradient orthogonal to tangent Suppose that a differentiable function f(x, y) has the constant value c along the differentiable curve x = g(t),y = h(t);thatis,

Prove the rule by setting

g(u, v)

=

l

f(g(t), h(t»

v

u

f(t) dt,

=

u(x),

v

=

v(x)

and calculating dg/ dx with the Chain Rule. 4. Finding a function with constrained second partials Suppose that f is a twice-differentiable function of r, that r = x 2 + y2 + z2 , and that

+ fyy + fzz

=

0.

c

for all values of t. Differentiate both sides of this equation with respect to t to show that Vf is orthogonal to the curve's tangent vector at every point on the curve.

9. Curve tangent to a surface

V

fn

=

r(t) = (In t)i

Show that the curve

+ (t In t)j +

tk

is tangent to the surface

Show that for some constants a and b,

f(r)

=

XZ2 -

ra + b.

yz

+ cos xy

= I

at (0,0, I). 5. Homogeneous functions A function f(x, y) is homogeneous of degree n (n a nonnegative integer) if f(tx, ty) = tnf(x,y) for all t, x, and y. For such a function (sufficiently differentiable), prove that

af a. x ax

af

+ y ay

= nf(x,y)

10. Curve tangent to a surface

Show that the curve

is tangent to the surface

x3 at (0, -I, I).

+ y3 + z3

-

xyz =

°

834

Chapter 14: Partial Derivatives

Extreme Values 11. Extrema on a .urfa.. Show 1hat the only possible maxima aod minima of z on the surface z = x 3 + y3 - 9xy + 27 occur at (0, 0) and (3, 3). Show that neither a maximum nor a minimum occurs at (0, 0). Determine whether z has a maximum or a minimum at (3, 3).

12. Muimum in closed rIrst quadrant Find the maximum value of f(x, y) =

fuye -(2%+3y)

in the closed first quadrant (includes the

nonnegative axes). 13. Minimum volume cut from f"IrSt octant Find the minimum volume for a regioo bounded by the plaoes x aod a plaoe taogeot to the ellipsoid 2

2

= 0, y = 0, Z = 0

a. Amoog all the possible directions taogentia1 to the surface S

14. Minimnm distance from a line to a parabola in """plaoe By minimizing the functioo f(x,y, u, v) = (x - uJ' + (y - vJ' subject to the coostraints y = x + I and u = v 2, fmd the minimum distance in the xy-plane from the line y = x + I to the parabola y2 = x.

Theory and Examples 15. Boundedne.. of ra.t partials implies continuity Prove the following theorero: If f(x,y) is defmed in an opeo regionR of the xy-plaoe and if fx andfy are bounded ooR, theo f(x,y) is cootinuous on R. (The assumptioo ofboundedoess is esseotial.) 16. Suppese 1hat r(l) = g(l)i + h(l)j + k(l)k is a smooth corve in the domain of a differeotiable function f(x, y, z). Describe the relation betweendf/dl, Vf, and v = dr/dl. What can be said aboot Vf aod v at interior points of the corve where f has extreme values relative to its other values on the curve? Give reasons for your aoower.

17. Finding function. from partial derivatives Suppese 1hat f aod g are functions of x and y such that and

af ax

ax

=

0,

f(1,2)

=

g(I, 2) = 5

at the point (0, 0, 10), which directioo will make the rate of change oftemperatore at (0, 0,10) a maximum?

b. Which directioo taogeotial to S at the point (I, I, 8) will make the rate of chaoge oftemperatore a maximum? 22. DriIIing anotber borebole On a flat surface oflaod, geologists drilled a borehole straight dowo and hit a mineral deposit at 1000 ft. They drilled a secood borehole 100 ft to the north of the fIrSt aod hit the mineral deposit at 950 ft. A third borehole 100 ft east of the fIrSt borehole struck the mineral deposit at 1025 ft. The geologists have reasoos to believe 1hat the mineral deposit is in the shape of a dome, aod for the sake of econoroy, they would like to fmd where the deposit is closest to the sorface. Assuming the surface to be the xy-plaoe, in what direction from the first borehole would you suggest the geologists drill their fourth borehole? Tbe one-
ag ay'

This equation is called the one-dimensional heat equation. The value of the positive coostaot c 2 is determined by the material from which the wire is made.

aod suppose that

af

angle of incidence. Assuming no loss of speed, what is the velocity of the particle after the ricochet? SimplilY your aoswer. face 1hat is the graph of f(x, y) = 10 - x 2 - y2 . Suppose 1hat the temperatore in space at each point (x, y, z) is 11:x,y,z) = x'y + y2z + 4x + 14y + z.

2

at a point in the fIrst octant

ag ax

20. Velocity after a ricocbet A particle traveling in a straight line with coostaot velocity i + j - 5k passes through the point (0, 0, 30) aod hits the surface z = 2x 2 + 3y2. The particle ricochets off the sorface, the angle of reflection being equal to the

21. Directional deri. .tive. tangent to a .urface Let S be the sur-

£+L+~=I a 2 b 2 c2

af ay

19. Path of a beat-seeking particle A heat-seeking particle has the property 1hat at any point (x,y) in the plane it moves in the direction of maximum temperatore increase. If the teroperatore at (x,y) is 11:x,y) = -e-2Ycosx,findaoequatiooy = f(x) for the path ofa heat-seeking particle at the point ('1f/4, 0).

aod

f(O,O)

=

4.

Find f(x,y) andg(x,y). 18. Rate of cbange oftbe rate ofcbange We know 1hat if f(x,y) is a function of two variables and ifu = ai + bj is a unit vector, then Duf(x,y) = fX<x,y)a + fy(x,y)b is the rate ofchaoge off(x,y) at (x,y) in the directioo ofu. Give a similar formula for the rate of chaoge afthe rale afchonge aff(x,y) at (x,y) in the direction u.

23. Find all solutions of the ooe-dimeosionsl heat equatioo of the form w = ert sin '7TX, where r is a constant. 24. Find all solutions of the ooe-dimensional heat equation 1hat have the form w = erl sin kx and satislY the conditions 1hat w(O, I) = 0 aod w(L, I) = O. What happeos to these solutions as t --. oo?

Chapter 14 Technology Application Projects

Chapter

00

835

Technology Application Projects

Mathematica/Maple Module: Plotting Surfaces

Efficiently generate plots of surfaces, contours, and level curves. Exploring the Mathematics Behind Slcateboarding: Analysis ofthe Directional Derivative The path of a skateboarder is introduced, Itrst on a level plaoe, then on a ramp, aod fmally on a paraboloid. Compute, plot, aod aoalyze the directional derivative in terms of the skateboarder. Looking for Patterns IlndApplying the Method ofLellSt SqulU'eS to Real Datil Fit a line to a set of numerical data points by choosing the line that minimizes the sum of the squares of the vertical distaoces from the points to the line. Logrange Goes SIcatebOllrding: HIIHI High Does He Go? Revisit and analyze the skateboarders' adventures for maximum and minimum heights from both a graphical and analytic perspective using Lagraoge multipliers.

15 MULTIPLE INTEGRALS OVERVIEW In this chapter we consider the integral of a function of two variables f(x, y) over a region in the plane and the integral of a function of three variables f(x, y, z) over a region in space. These multiple integrals are defined to be the limit of approximating Riemann sums, much like the single-variable integrals presented in Chapter 5. We illustrate several applications of multiple integrals, including calculations of volumes, areas in the plane, moments, and centers of mass.

DoubLe and Iterated IntegraLs over RectangLes

15.1

In Chapter 5 we defined the definite integral of a continuous function f(x) over an interval [a, b] as a limit of Riemann sums. In this section we extend this idea to define the double integral of a continuous function of two variables f(x, y) over a bounded rectangle R in the plane. In both cases the integrals are limits of approximating Riemann sums. The Riemann sums for the integral of a single-variable function f(x) are obtained by partitioning a finite interval into thin subintervals, multiplying the width of each subinterval by the value of f at a point Ck inside that subinterval, and then adding together all the products. A similar method of partitioning, multiplying, and summing is used to construct double integrals.

Double Integrals We begin our investigation of double integrals by considering the simplest type of planar region, a rectangle. We consider a function f(x, y) defined on a rectangular region R, R:

d

R Ak IlYk

c:5

Y

:5

d.

We subdivide R into small rectangles using a network of lines parallel to the x- and y-axes (Figure 15.1). The lines divide R into n rectangular pieces, where the number of such pieces n gets large as the width and height of each piece gets small. These rectangles form a partition of R. A small rectangular piece of width ilx and height ily has area LlA = ilxily. If we number the small pieces partitioning R in some order, then their areas are given by numbers LlAb LlA 2 , • •• , LlAn' where ilAk is the area of the kth small rectangle. To form a Riemann sum over R, we choose a point (Xh Yk) in the kth small rectangle, multiply the value of f at that point by the area ilAk, and add together the products:

Y

I I e-

a:5 x :5 b,

HXk'Yk)

Ilxk c ~--L-------------------~X

o

a

FIGURE 15.1 Rectangular grid partitioning the region R into small rectangles of area AAk = Axk AYk.

836

b

n

Sn =

:L f(Xh Yk) ilAk. k=l

Depending on how we pick (Xh Yk) in the kth small rectangle, we may get different values for Sn.

15.1

Double and Iterated Integrals over Rectangles

837

We are interested in what happens to these Riemann sums as the widths and heights of all the small rectangles in the partition of R approach zero. The norm of a partition P, written Ilpll, is the largest width or height of any rectangle in the partition. If Ilpll = 0.1 then all the rectangles in the partition of R have width at most 0.1 and height at most 0.1. Sometimes the Riemann sums converge as the norm of P goes to zero, written Ilpll - o. The resulting limit is then written as n

L j(Xk,Yk) flAk. IIPII->O k= i lim

As Ilpll - 0 and the rectangles get narrow and short, their number n increases, so we can also write this limit as n

L j(Xk, Yk) flAk, n->OO k= i lim

with the understanding that Ilpll - 0, and hence flAk - 0, as n - 00. There are many choices involved in a limit of this kind. The collection of small rectangles is determined by the grid of vertical and horizontal lines that determine a rectangular partition of R. In each of the resulting small rectangles there is a choice of an arbitrary point (Xk, Yk) at which j is evaluated. These choices together determine a single Riemann sum. To form a limit, we repeat the whole process again and again, choosing partitions whose rectangle widths and heights both go to zero and whose number goes to infinity. When a limit of the sums Sn exists, giving the same limiting value no matter what choices are made, then the function j is said to be integrable and the limit is called the double integral of j over R, written as

11

j(x,y) dA

11

or

j(x,y) dxdy.

R

R

It can be shown that if j(x, y) is a continuous function throughout R, then j is integrable, as

in the single-variable case discussed in Chapter 5. Many discontinuous functions are also integrable, including functions that are discontinuous only on a finite number of points or smooth curves. We leave the proof of these facts to a more advanced text.

Double Integrals as Volumes

z

y

x

FIGURE 15.2 Approximating solids with rectangular boxes leads us to define the volumes of more general solids as double integrals. The volume of the solid shown here is the double integral of I(x, y) over the base region R.

When j(x, y) is a positive function over a rectangular region R in the xy-plane, we may interpret the double integral of j over R as the volume of the 3-dimensional solid region over the xy-plane bounded below by R and above by the surface z = j(x, y) (Figure 15.2). Each term j(Xk, Yk) flAk in the sum Sn = L j(Xk, Yk) flAk is the volume of a vertical rectangular box that approximates the volume of the portion of the solid that stands directly above the base flAk. The sum Sn thus approximates what we want to call the total volume of the solid. We define this volume to be Volume

=

lim Sn

n~OO

=

JJRff j(x,y) dA,

where flAk- 0 as n - 00. As you might expect, this more general method of calculating volume agrees with the methods in Chapter 6, but we do not prove this here. Figure 15.3 shows Riemann sum approximations to the volume becoming more accurate as the number n of boxes Increases.

838

Chapter 15: Multiple Integrals

(a)n _ 16

(e) n _ 256

(b)n _ 64

FIGURE 15.3 As n increases, the Riemann 9UDl approximations approach the total volume of the solid shown in Figure 15.2.

,

Fubini's Theorem for calculating Double Integrals Suppose that we wish to calculate the volume 1UI.der the plane z - 4 - x - y over the recs Y ::s; I in the xy-plane. Ifwe apply the method ofslicing from Section 6.1, with slices perpendicular to the x-axis (Figure 15.4), then the volume is

4

tangular region R: 0 ::s; x ::s; 2,0

1,-0'~' A(x)

dx,

(1)

where A(x) is the cross-sectional area at x. For each value of x, we may calculate A(x) as the integral A(x)

1

2 '

} A(x) -

l' ' \

~

,·, 1

(4 - x - y) dy,

(2)

y·O

'

J,'.' (4-x-y)dy , ·0

FIGURE 15.4 To obtain the crosssectional areaA(x), we holdx flXCd and integrate with respect toy.

which is the area under the curve z = 4 - x - Y in the plane of the cross-section at x. In calculatingA(x), x is held fIxed and the integration takes place with respect to y. Combining Equations (1) and (2), we see that the volume of the entire solid is Volume

~

1,·, 1,·, [

A(x) dx

~

x- O

=

x- O

=

1'·' (l

• ' (4

- x - y)dy ) dx

y'r·' 1'·' (7

x- o

4y-xy-2

[~x - x22]: =

Y

5.

1- 0

dx=

y- O

x- O

)

- - x dx 2 (3)

If we just wanted to write a formula for the volume, without carrying out any of the integrations, we could write

The expression on the right, called an iterated or repeated integral, says that the volume is obtained by integrating 4 - x - y with respect to y from y = 0 to Y = I, holding x f"txed, and then integrating the resulting expression in x with respect to x from x = 0 to x = 2. The limits of integration 0 and 1 are associated withy, so they are placed on the integral closest to dy. The other limits of integration, 0 and 2, are associated with the variable x, so they are placed on the outside integral symbol that is paired with dx.

15.1

,

Double and Iterated Integrals over Rectangles

839

What would have happened if we had calculated the volume by slicing with planes perpendicular to the y-axis (Figure 15.5)? As a function of y, the typical cross-sectional

4

'""" is

A(y) ~

1,·,

[' ]'.'

(4 - x - y) d>; ~ 4x - "- - xy 2

",- 0

~ 6 - 2y,

(4)

x- O

The volume of the entire solid is therefore

Volum, ~

y

,

,

\ 1.'.' ,·0

A(y) _

A(y) dy ~

1',-0·'

(6 - 2y) dy ~ [6y - y']: ~ 5,

in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing

(4-x-)')d.t

FIGURE 15.5 To obtaW. the cross-sectional. area A(y), we hold y fDWd and integrate with respect to x.

1,-0,·,

Volume =

11102.(4 - x -

y) duly.

The expression on the right says we can find the volume by intcgIating 4 - x - y with respect tox from x - 0 tax - 2 as in Equation (4) and integrating the result with respect to y from y - 0 to Y - 1. In this. iterated integral, the order of integration is fll'S1: x and then y, the reverse of the order in Equation (3). What do these two volume calculations with iterated integrals have to do with the double integral

jj(4 - x - y) dA R H:rsWRICAL BIOGRAPHY

Guido Fubini (1879-1943)

over the rectangle R: 0 ~ x ~ 2, 0 ~ y ~ I? The answer is that both iterated integrals give the value of the double integral. This is what we would reasonably expect, since the double integml measures the volume of the smne region as the two iterated integrals. A theorem published in 1907 by Guido Fubini says that the double integral of any continuous functioo aver a rectangle can be calculated as an itemted integral in either order of integration. (Fubini proved his theorem in greater generality, but this is what it says in our setting.)

THEOREM l-Fubini's Theorem
jj /(x,y) dA ~ 1'1'/(x,y) d>;dy ~ 1'1' /(x,y) dy

d>;,

R

Fubini's Theorem says that double integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time. Fubini's Theorem also says that we may calculate the double integral by integrating in either order, a genuine convenience. When we calculate a volume by slicing, we may use either planes perpendicular to the x-axis or planes perpendicular to the y-axis.

EXAMPLE 1

Calculate .O'R/(x,y)

f(x,y) ~ 100 -

6x'y

dA for and

R:

O::Sx::S2,

-1::Sy::sl.

840

Chapter 15: Multiple Integrals

,

.: -100 -6.1:7

Solution

Figw-e 15.6 displays the volume beneath the surface. By Fubini's Theorem,

~ 1\200 -, - 16y) dy ~

[201ly - 8y']:,

~ 400.

Reversing the order of integration gives the same answer: x

R

,

1 ,

{' {\100 -

Jo J-l

6x'y)dydx

~ ('[lOlly - 3'V~::, dx

Jo

~ 1,\(100 -

FIGURE 15.6 The double integral f(x, y) cIA. gives the volume under this surface aver the n:ctangular region R (Example 1).

DR

3x') - (-100 - 3x')] dx

-£2

200 dx _ 400.

EXAM PLE 2

Find the volume of the region bounded above by the ellipitical paraboloid

z - 10 + x2 + 3y 2 andbelowbytherectangleR:0::S; x S 1,0

Sy::s;

2.

Solution The surface and volume a:re shown in Figure 15.7. The volume is given by the double integral

R

ff + -l' [lOy

FIGURE 15.7 The double inWgral f(x, y) gives the volume under this smfaoc aver the n:ctangular region R

DR

•

(10

V=

dA

x 2 + 3y2)dA =

11 1

2

(10

+ x 2 + 3y2) dydx

R

+ x'y + y'~:: dx

(Example 2).

=

Jor (20 + 2x

2

+ 8)dx = [2Ox + ~X3 + 8x]1 3

•

= 86.

3

0

Exercises 15.1 ~~tingH8~In~~

11.

In Exercises 1-12, evaluate the iterated integral.

1.

J.'110f'

3.1:1: la la 3

5.

2. ].' {' (x - y)dydr.

2xydydr.

(x

+y +

l

(4 -

l)dxdy

y~dydx

Y 7. ].0 '].' 0 1 +xydxdy

9. ].0 "'11[b' e2>:+y dy dx

01-1 x' -+ Y') 4. ]. '].' (1 - dr.dy o 0

12·1 la" lfr

y sin;t dx dy

(sin;t

+ cosy)dxdy

Evaluating Double Integra~ over Rectangles In Exercises 13-20, evaluate the double integral. aver t1u: given regionR.

2

13.

f' (x'y 01-2

6. ].'

1-,'],."'

2xy) dy dr.

jj(6y2 -2x)cIA.,

R:

O:!;';;t:!;'; I,

0

:!;';y:!;'; 2

R

14.

jj (v:)cIA., R

15.

R:

jjxycosYdA, R: R

0 S;t S 4,

1 Sy S 2

Y

-IS;tSI,

OSys",.

15.2 Double Integrals over General Regions

16.

JJ JJ JJ

y sin (x

+ y) dA,

R:

-1T':S

x :s 0,

O:s y ::s

1T'

R

17.

.'-Y dA,

0 '" x '" In 2,

R:

0 '" Y '" In 2

R

18.

xyexi' dA,

0 '" x '" 2, 0 '" Y '" I

R:

R

19.

ff

,xy3

}} x + I

dA,

0 '" x '" I,

R:

0 '" Y '" 2

R

20.

JJ X'y: +

I dA,

R:

0 '" x '" I,

0 '" Y '" I

R

In Exercises 21 and 22, integrate f over the given region.

21. Square f(x,y) = l/(xy) over the square I '" x'" 2, I'" Y '" 2 22. Rectangle f(x,y) = y cosxy over the rectangle 0 '" x '" 'fr,

O"'y"'l

841

Volume Beneath a Surface 23. Find the volume of the region bounded above by the paraboloid z = x' + y' and below by the square R: -I '" x '" I, -I "'y"'i. 24. Find the volume of the region bounded above by the ellipiticai paraboloid z = 16 - x' - y' and below by the square R: 0 :s x :s 2,0 :s y :s 2. 25. Find the volume of the region bounded above by the plane z = 2 - x - y and below by the square R: 0 '" x '" 1,

o '" Y '" i. 26. Find the volume of the region bounded above by the plane z = y/2 and below by the rectangle R: 0 '" x '" 4,0 '" Y '" 2. 27. Find the volume of the region bounded above by the surfaco z = 2 sin x cos y and below by the rectangle R: 0 '" x '" 'fr/2, o '" Y '" 'fr/4. 28. Find the volume of the region bounded above by the surfaco z = 4 - y' and below by the rectangle R: 0 '" x '" 1, o '" Y '" 2.

z= f(x,y)

Double Integrals over General Regions

15.2

In this section we detme and evaluate double integrals over bounded regions in the plane which are more general than rectangles. These double integrals are also evaluated as iterated integrals, with the main practical problem being that of determining the limits of integration. Since the region of integration may have boundaries other than line segments parallel to the coordinate axes, the limits of integration often involve variables, not just constants.

Double Integrals over Bounded, Nonrectangular Regions

"


R

~Yl

I .

.u,

r-zx.t'Yk

/

FIGURE 15.8 A rectangular grid partitioning a bounded nonrectangular region into rectangular cells.

To define the double integral of a function f(x, y) over a bounded, nonrectangular region R, such as the one in Figure 15.8, we again begin by covering R with a grid of small rectangular cells whose union contains all points of R. This time, however, we cannot exactly fIll R with a tmite number of rectangles lying inside R, since its boundary is curved, and some of the small rectangles in the grid lie partly outside R. A partition of R is formed by taking the rectangles that lie completely inside it, not using any that are either partly or completely outside. For commonly arising regions, more and more of R is included as the norm of a partition (the largest width or height of any rectangle used) approaches zero. Once we have a partition of R, we number the rectangles in some order from I to n and let aA. be the area of the kth rectangle. We then choose a point (x., Y') in the kth rectangle and form the Riemann sum

As the norm of the partition forming S. goes to zero, Ilpll --> 0, the width and height of each enclosed rectangle goes to zero and their number goes to intmity. If f(x, y) is a continuous function, then these Riemann sums converge to a limiting value, not dependent on any of the choices we made. This limit is called the double integral of f(x,y) over R:

842

Chapter 15: Multiple Integrals

The nature of the boundary of R introduces issues not found in integrals over an interval. When R has a curved boundary, the n rectangles of a partition lie inside R but do not cover all of R. In order for a partition to approximate R well, the parts of R covered by small rectangles lying partly outside R must become negligible as the norm of the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made from polygons, circles, ellipses, and from continuous graphs over an interval, joined end to end. A curve with a "fractal" type of shape would be problematic, but such curves arise rarely in most applications. A careful discussion of which type of regions R can be used for computing double integrals is left to a more advanced text.

Volumes If I(x, y) is positive and continuous over R, we define the volume of the solid region betweenR and the surface z = I(x,y) to be JJRI(x,y) dA, as before (Figure 15.9). If R is a region like the one shown in the xy-plane in Figure 15.10, bounded "above" and ''below'' by the curves y = g2(X) and y = gl (x) and on the sides by the lines x = a, x = b, we may again calculate the volume by the method of slicing. We first calculate the cross-sectional area A(x) =

l

y =g2(X)

I(x,y) dy

y=gl(X)

and then integrate A(x) from x

V=

l

=

a to x = b to get the volume as an iterated integral:

bA(x)dx= lblg2(X) I(x,y)dydx.

a

a

(1)

gl(X)

z z =f(x,y)

o

y

x

Volume

=

lim !fiXk, Yk) aAk =

If f(x, y) dA R

FIGURE 15.9 We define the volumes of solids with curved bases as a limit of approximating rectangular boxes.

FIGURE 15.10 The area of the vertical slice shown here is A(x). To calculate the volume of the solid, we integrate this area from x = a to x = b:

l

a

b

A(x)dx =

lblg2(X) a

g,(x)

f(x,y)dydx.

15.2 Double Integrals over General Regions

843

Similarly, if R is a region like the one shown in Figure 15.11, bounded by the curves x = h2(y) and x = hl(y) and the lines y = c and y = d, then the volume calculated by slicing is given by the iterated integral

z z =f(x,y)

x

Il d

Volume =

y

hb )

(2)

I(x,y) dxdy.

h,(y)

c

That the iterated integrals in Equations (1) and (2) both give the volume that we defined to be the double integral of lover R is a consequence of the following stronger form of Fubini 's Theorem.

FIGURE 15.11 shown here is

I

c

The volume of the solid

dA(y)dy = Idlh2(v) j(x,y)dxdy. h,(v) c

TH EOREM 2-Fubini's Theorem (Stronger Form)

Let I(x, y) be continuous on a

regionR .

1. If R is defined by a on [a , b] , then

:5

x

:5

b, gl (x)

Y

:5

:5

r iirrI(x, y) dA = laig,(x)

For a given solid, Theorem 2 says we can calculate the volume as in Figure 15.10, or in the way shown here. Both calculations have the same result.

g2

b

g2(X) , with gl and g2 continuous

I(x, y) dy dx.

(x)

R

2. If R is defined by c on [c, d], then

:5

Y

:5

d, hl(y)

iJif I(x,y) dA = I

x

:5

:5

h2(y), with hI and h2 continuous

dlh2(y)

I(x,y) dxdy. hICy)

c

R

EXAMPLE 1 Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in the plane z = I(x, y) = 3 - x - y.

Solution See Figure 15.12. For any x between 0 and l,y may vary from y = 0 to y = x (Figure 15.12b). Hence, V=

=

11 1

x

o

(3 - x - y) dy dx

11 [

=

0

1

1 (

o

3y - xy - -y2]Y=X dx 2 y =O

0

3

3x - ~2) dx = 2

[3~ 2

-

2

:!.-3 ]X=I 2

= 1.

x=o

When the order of integration is reversed (Figure 15.12c), the integral for the volume is

V =

= =

11 [ 1111 11 (3 - ~ 11 (% - ~ 2) [% o

(3 - x - y) dx dy

3x - ~2 - xy ]X=I dy

=

0

y

y - 3y

4y

+

Y

x=y

+ Y22 + y2) dy

dy =

The two integrals are equal, as they should be.

y - 2y2

+ ~3 I:~ =

1.

•

844

Chapter 15: Multiple Integrals z y

(3,0,0)

FIGURE 15.12 (a) Prism with a triangular base in the xy-plane. The volume of this prism is defined as a double integral over R. To evaluate it as an iterated integral, we may integrate first with respect to y and then with respect to x, or the other way around (Example I). (b) Integration limits of

l

x=11Y=x f(x, y) dy dx. x=o y=o

If we integrate first with respect to y, we integrate along a vertical line through R and then integrate from left to right to include all the vertical lines in R. (c) Integration limits of

l

Y=11X=1 f(x,y) dx dy. y=o X=Y

If we integrate first with respect to x, we integrate along a horizontal line through R and then integrate from bottom to top to include all the horizontal lines in R.

Although Fubini's Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen.

EXAMPLE 2

Calculate

11 si~x

dA,

R

where R is the triangle in the xy-plane bounded by the x-axis, the line y = x, and the line x = 1.

15.2 y x

Double Integrals over General Regions

845

Solution The region of integration is shown in Figure 15.13. Ifwe integrate IlISt with respect to y and then with respect to x, we Imd

=1

l' (lS~XdY)dx l' &s~x IJdx =

= -cos (I) ~~------~--~X

o

+

I

RJ

=

1'SinXdx

0.46.

If we reverse the order of integration and attempt to calculate

J.

o

FIGURE 15.13 The region of integration in Example 2.

111. y

s~x dx dy,

we run into a problem because J«sinx)/x) dx cannot be expressed in terms of elementary functions (there is no simple antiderivative). There is no general rule for predicting which order of integration will be the good one in circumstances like these. If the order you IlISt choose doesn't work, try the other. Some_ times neither order will work, and then we need to use numerical approximations.

y

Finding Limits of Integration We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works. Using Vertical Cross-sections When faced with evaluating lfR f(x,y)dA, integrating IlISt with respect to y and then with respect to x, do the following three steps:

(a)

y

/

Leaves at y~~ Enters at y=l-x

1.

Sketch. Sketch the region of integration and label the bounding curves (Figure 15.14a).

2.

Find the y-limits of integration. Imagine a vertical line L cutting through R in the direction of increasing y. Mark the y-values where L enters and leaves. These are the y-limits of integration and are usually functions of x (instead of constants) (Figure 15.14b).

o

3. Find the x-limits of integration. Choose x-limits that include all the vertical lines x

throughR. The integral shown here (see Figure 15.14c) is

ff

(b)

f(x,y)dA

=

E~IL~~ f(x,y)dydx.

R

y

Leaves at

/y~~ Enters at

y=l-x

Using Horizontal Cross-sections To evaluate the same double integral as an iterated integral with the order of integration reversed, use horizontal lines instead of vertical lines in Steps 2 and 3 (see Figure 15.15). The integral is

JJrr f(x,y)dA R

=

J.lkr~ f(x,y)dxdy. 0

y

~t---~--~~--~x

l

x

/

Smallest x

Largest x

isx=O

isx=l

Largesty Y isy = 1 ~1

Enters at

x=l-y

(c)

FIGURE 15.14 Finding the limits of integration when integrating first with respect to y and then with respect to x.

Y --r---~~~--~

Smallest y isy ~ 0

"

~ Leaves at x

~

v'l=Y'

--;;ct-------~~--~x

o

FIGURE 15.15 Finding lhe limits of integration when integrating first with respect to x and then with respect to y.

846

Chapter 15: Multiple Integrals

EXAMPLE 3

Sketch the region of integration for the integral

1212x(4x + 2)dydx and write an equivalent integral with the order of integration reversed.

Solution The region of integration is given by the inequalities x 2 :5 y :5 2x and o :5 x :5 2. It is therefore the region bounded by the curves y = x 2 and y = 2x between x = 0 and x = 2 (Figure 15.16a). To find limits for integrating in the reverse order, we imagine a horizontal line passing from left to right through the region. It enters at x = y/2 and leaves at x = To include all such lines, we lety run from y = 0 to Y = 4 (Figure 15.16b). The integral is

...;y.

y

r

rv,.(4x + 2) dx dy. 101Y/2

•

The common value of these integrals is 8.

Properties of Double Integrals ~~----~----~X

Like single integrals, double integrals of continuous functions have algebraic properties that are useful in computations and applications.

If f(x, y) and g(x, y) are continuous on the bounded region R, then the following properties hold

FIGURE 15.16 Region of integration for Example 3.

1. constantMUltiPle:ff cf(x,y)dA = cff f(x,y)dA R

(any number c)

R

2. Sum and Difference: ffU(X,Y) ± g(x,y)) dA

ff f(x,y) dA ± ff g(x,y) dA

=

R

R

R

3. Domination: y

(a)

if

ff f(x,y) dA ;" 0

f(x,y) ;" 0 onR

R

(b)

ff f(x,y)dA;" ff g(x,y)dA R

o

II f(x, R

y) dA

II

II

~ f(x, y) dA + f(x, y) dA RI

f(x,y) ;" g(x,y) onR

R

4. Additivity: ff f(x,y) dA --~--*-----~---------->x

if

R

=

ff f(x,y) dA + ff f(x,y) dA Rl

•

if R is the union of two nonoverlapping regions R, and R2

R2

FIGURE 15.17 The Additivity Property for rectangular regioos holds for regions bounded by continuous curves.

Property 4 assumes that the region of integration R is decomposed into nonoverlap· ping regions R, and R2 with boundaries consisting of a finite number of line segments or smooth curves. Figure 15.17 illustrates an example of this property.

15.2 Double Integrals over General Regions

847

The idea behind these properties is that integrals behave like swns. If the function f(x, y) is replaced by its constant multiple cf(x, y), then a Riemann swn for f n

Sn =

z

L f(xk, Yk) flAk k=!

is replaced by a Riemann swn for cf n

L

k=!

c

=

L

k=!

f(xk, Yk) flAk

",, \

\... I I

(a)

y

y

=

4x- 2

2

~r-----~~------L-~x

(b)

FIGURE 15.18 (a) The solid "wedgelike" region whose volume is found in Example 4. (b) The region of integration R showing the order dx dy.

Find the volwne of the wedgelike solid that lies beneath the surface z = 16 - x 2 - y2 and above the region R bounded by the curve Y = 2Vx, the line Y = 4x - 2, and the x-axis. Solution Figure 15.18a shows the surface and the "wedgelike" solid whose volwne we want to calculate. Figure 15.18b shows the region of integration in thexy-plane. If we integrate in the order dy ax (first with respect to y and then with respect to x), two integrations will be required because y varies from y = 0 to y = 2vX for 0 :::; x :::; 0.5, and then varies from y = 4x - 2 to Y = 2vX for 0.5 :::; x :::; 1. So we choose to integrate in the order ax dy, which requires only one double integral whose limits of integration are indicated in Figure 15 .18b. The volwne is then calculated as the iterated integral:

11

(16 - x 2 - y2) dA

= =

1 1

2 1(Y+2)!4

o

2 [

o

(16 - x 2 - y2)dxdy

y./4 x3 ]X=(Y+2)!4 16x - - - xy2 dx 3 x=y./4

t [

= Jo

4(y

_ [191Y

+

2) -

63y2

- ---z4 + ~ -

(y + 2)3 3 . 64 145y 3 ~

-

(y

+ 2)y2 - 4y2

4

49y4 768

+

y5 20

+

y6 3' 64

+ o

-

Sketching Regions of Integration In Exercises 1- 8, sketch the described regions of integration. 1. 0 :::; x :::; 3, 0:::; y ~ 2x

x-I ~ y ~ x 2

7. 0 ~ y ~ 1,

0 ~ x ~ sin- I Y

8. 0 ~ y ~ 8,

i

3. -2 ~ y ~ 2, y2 ~ X ~ 4 ~

y

~

1, y

~

x

~

2y

y

~ x ~ Y 1/3

y4]

+ 4 dy

7 ]2 _ 20803

Y 1344

Exercises 15.2

4. 0

cSn .

EXAMPLE 4

R

2. -1 ~ x ~ 2,

=

Taking limits as n ~ 00 shows that c limn-><Xl Sn = C ffR f dA and limn-><Xl cSn = fJR C f dA are equal. It follows that the constant multiple property carries over from swns to double integrals. Y The other properties are also easy to verify for Riemann sums, and carryover to double integrals for the same reason. While this discussion gives the idea, an actual proof that these properties hold requires a more careful analysis of how Riemann swns converge.

.J......., _

x

n

c f(xk, Yk) flAk

1680 ~ 12.4.

•

848

Chapter 15: Multiple Integrals

Finding Limits of Integration In Exercises 1}-18, write an iterated integral for lfR dA over the described region R using ( a) vertical cross-sections, (b) horizontal cross-

sections.

29.1:1"

2 dp dv

(the pv-plane)

{'{~

8t dt dx

30. Jo Jo

(the st-plane)

10.

9.

31.1~/' {E"3 cos t du dt

y

(the tu-plane)

-1r/3io

32.

1,o

'/2~4'2' 4 2 ~dvdu 1 v

(theuv-plane)

Reversing the Order of Integration --~~------~--~X

In Exercises 33-46, sketch the region of integration and write an equivalent double integral with the order of integration reversed.

33. 12.

11.

y

35.

11 1, o

4

'2.<

dy dx

34.

2

1,o11VY

dx dy

36.

y

Jl

dy dx

x=2

= Vi, y = 0, and x = 9 Boundedbyy = tanx,x = O,aody = I Bounded by y = e~,y = I, and x = In 3 Bounded by Y = O,X = O,y = I, andy = Inx Bounded by y = 3 - 2>:,y = x, aodx =

15. 16. 17.

°

18. Bounded by y = x 2 andy = x

+2

r

20. Jo Jo 21.

1, 1vG' -VI=7 Jl Jo

45.

1,'1"

y dy dx

Jtr'r' Jo e +Y dx dy X

(' p'

25. Quadrilateral I(x, y) = x/y over the region in the flIst quadrant bounded by the lines y = x, y = 2>:, x = I, and x = 2

26. Triaogle I(x, y) = x2 + y2 over the ttiaogular region with vertices (0, 0), (I, 0), aod (0, I)

27. Triaogle I(u, v) = v - Vu over the ttiangular region cut from the flISt quadraot of the uv-plane by the line u + v = I 28. Curved region I(s, t) = e' In t over the region in the flISt quadraot of the st-plane that lies above the curve s = In t from

t=llot=2 Each of Exercises 21}-32 gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral.

3ydxdy

42.

ydxdy

1v'k'

2

44.

1,o .v'k' ~/6L1/2

1,o

6x dydx

xy 2 dydx

sinr

{V3 {,m,-"

+ y) dx dy

46. Jo

Jo

v;Jdxdy

In Exercises 47-56, sketch the region of integration, reverse the order of integration, aod evaluate the integral.

r Jxrsiny --y dy dx

48.

47. Jo

1,111 2y'j;31y'j;3 51. 1, ,/2

x 2e"" dx dy

o

50.

1,2122y2 sinxy dy dx 2j,4'X' xe ~

1,o

4 _

0

52. {' {' Jo

ex' dx dy

JWi

Y

dy dx

.r dy dx

53. {1/16 {1/2 cos (I&n-x') dx dy

Jo

23. Jo Jo 3y'e"" dx dy In Exercises 25-28, integrate 1 over the giveo region.

dy dx

dx dy

{' r·r

xy dy dx (x

J"

40. Jo Jo

{' {Inx

43.

49.

{.;nx

1-x

16x dy dx

Jo

o

finding Regions of Integration and Double Integrals In Exercises 11}-24, sketch the region of integration and evaluate the integral.

0

dx dy

I,I.il'X'

1

41.

13. Bounded by y 14.

y-2

38. Jo

{'/2 {"4x' 39. Jo

0

{1n2 {'

{' {" 37. Jo

1,2:£0

54.

),1/4

l,'h o

2

-dydx -+I

~ y4

55. Square region lfR (y - 2>:2) dA where R is the region boundedbythesquarelxl + Iyl = I 56. Triaogular region

thelinesy

= ;t,Y =

lfRXY dA where R is the region hounded by +y = 2

a,andx

Volume Beneath a Surface z

=

/(x, y)

57. Find the volume of the region bounded above by the paraboloid z = x2 + y2 and below by the ttiangle enclosed by the lines y = x,x = 0, and x + y = 2 in thexy-plane. 58. Find the volume of the solid that is hounded above by the cylinder z = x 2 aod below by the region enclosed by the parabola y = 2 - x 2 and the line y = x in the xy-plane.

15.2 Double Integrals over General Regions 59. Find the volume of the solid whose base is the region in the xyplane that is bounded by the parabola y ~ 4 - x 2 and the line y = 3x, while the top of the solid is bounded by the plane z = x + 4. 60. Find the volume of the solid in the fIrSt octant bounded by the coordinate planes, the cylinder x 2 + y2 = 4, aod the plaoe z + y = 3. 61. Find the volume of the solid in the flIst octant bounded by the coordinate planes, the plaoe x = 3, aod the parabolic cylinder z ~ 4 - y2. 62. Find the volume of the solid cot froro the fIrSt octant by the surfacez = 4 - x 2 - y.

63. Find the volume of the wedge cut from the fIrSt octaot by the cylinder z ~ 12 - 3y2 and the plaoe x + y ~ 2. 64. Find the volume of the solid cot froro the square colunm Ixl + Iyl'" I by the planes z = Oaod3x + z = 3. 65. Find the volume of the solid that is bounded 00 the froot aod back by the plaoes x ~ 2 aod x = I, 00 the sides by the cylioders y = ± I/x, aod above aod below by the plaoes z = x + I aod z ~ O. 66. Find the volume of the solid bounded 00 the froot aod back by the planes x = ±'1T/3, on the sides by the cylinders y = ±secx, above by the cylinder z ~ I + y2, aod below by the xy-plaoe. In Exercises 67 and 68, sketch the region of integratioo aod the solid whose volume is giveo by the double integral. 67.

/.J-

/'

':£'116- v'25 /.o y Y

68.

(I - tx -

2r

-v'16-

'

l

Integrals over Unbounded Regions Improper double integrals cao often be coroputed similarly to improper integrals of ooe vatiable. The flIst iteratioo of the following improper integrals is cooducted just as if they were proper integrals. One thea evaluates ao improper integral of a single vatiable by taking appropriate limits, as in Sectioo 8.7. Evaluate the improper integrals in Exercises 6')..72 as iterated integrals. 69.

1

00'1. '

I

70.

74. f(x,y) = x + 2y over the regioo R inside the circle (x - 2)' + (y - 3)' = I using the partition x = 1,3/2,2,5/2, 3 aody ~ 2, 5/2, 3, 7/2, 4 with (Xk,Yk) the center (centroid) in the kth suhrectangle (provided the suhrectaogle lies within R)

Theory and Examples

v'

75. Cin:ular sector Integrate f(x, y) ~ 4 - x 2 over the smaller 2 sector cut from the disk x + y2 S 4 by the rays 8 = '1T/6 aod 8 = '1T/2. 76. Unbounded region Integrate f(x,y) ~ 1/[(x 2 - x)(y - 1)2/3]

aver the infmite rectangle 2 ::s x <

V

= /"/'

-1

~ ~ (x 2

Y

(x 2 + y2) tb;,ry

+

1

2 2 Y /. - (x 2

+ y2) tb; dy.

78. Converting to a double integral Evaluate the integral /.2(tao-1'I/'X

tao-I x) tb;.

_

(Hint: Write the integraod as an integral.)

79. M,xjmjzing a double integral What region R in the xy-plane maximizes the value of

//(4 -

x 2 - 2y2) dA?

R

Give reasons for your answer. 80. Minimizing. double integral

What regioo R in the xy-plane

minimizes the value of Jj(x 2

+ y2

- 9) dA?

R

(2y

+

+

I),rytb;

I 1)(y2

+ I) tb; dy

/.oo/.oo xe-{r+2Y)tb;,ry

Approximating Integrals with Finite Sums In Exercises 73 aod 74, approximate the double integral of f(x,y) over the region R partitiooed by the given vertical lines x = a aod horizontallioes y ~ c. In each suhrectaogle, use (Xto y,) as indicated for your

approximation.

0 ::s y ::s 2.

Sketch the base region R aod express the cylioder's volume as a single iterated integral with the order of integratioo reversed. Thea evaluate the integral to fmd the volume.

y

-1/~ 100'1'/~ 00

00.

77. Noncircular cylinder A solid right (noncirco1ar) cy1ioder has its base R in the xy-plaoe aod is bounded above by the paraboloid z = x 2 + y2. The cylioder's volume is

I -,-,ry tb;

e1 i X

71.1 1 72.

73. f(x,y) = x + y over the regionR bounded above by the semicircle y = ~ and below by the x-axis, using the partitioo x = -I, -1/2,0, 1/4, 1/2, I aody = 0, 1/2, I with (Xk,Yk) the lower left corner in the kth suhreCW!gle (provided the suhrectangle lies within R)

ty),rytb;

x 2 - y2 tb; ,ry

849

Give reasons for your answer. 81. Is it possible to evaluate the integral of a continuous functioo f(x, y) over a rectaogular regioo in the xy-plane aod get different

answers depending on the order of integration? Give reasons for your answer. 82 How would yrru evaluate the double integral of a continuous function f(x,y) over the regionR in thexy-plaoe enclosed by the triangle with vertices (0, I), (2, 0), aod (I, 2)? Give reasoos for your

answer. 83. Unbounded region

1'-ooJ-ooroo e~'-Y' (00

Prove that tb;,ry

~

lim

1'1' e~'-Y'

b_oo -b -b

tb;,ry

850

Chapter 15: Multiple Integrals

84. Improper double integnol

Use a CAS double-integral evaluator to fmd the integrals in Exercises 8~94. Theo reverse the order of integratioo and evaluate, again wilh a CAS.

Evaluate the improper integral

].1].' 2 o

0

x dydx (y - 1)2/' .

COMPUTER EXPLORATIONS Use a CAS double-integral evaluator to estimate the values of the integrals in Exercises 85-1!8.

85. 87.

{' (' I xydydx

86.

Jl Jl

].1].1 tan-I xy dy dx

15.3

88.

].1].1

1

].1o J2Y{'ex'dx dy

90. ] . ' [ xcos(y2)dydx

91.

].o2~4v2Y (x"y - xy2) dx dy

92.

].0

94.

1

y'

e-(x'+y') dy dx

2

93.

1 ] . \I'i=:;>

-1

89.

3\IJ -

x 2 - y2dydx

x

1:1.

' _+1 dydx lOX Y

1

2].4-Y' 0

eX)! dx dy

2~s y'

I v'x2

+ y2

dxdy

0

Area by Double Integration In this section we show how to use double integrals to calculate the areas of bounded regions in the plane, and to find the average value of a function of two variables.

Areas of Bounded Regions in the Plane If we take !(x, y) = I in the defmition of the double integral over a region R in the preceding section, the Riemarm sums reduce to

S. =

• • L !(XhY.) aA. = L aA•. i- I i- I

(1)

11ris is simply the swn of the areas of the small rectangles in the partition of R, and approximates what we would like to call the area of R. As the norm of a partition of R approaches zero, the height and width of all rectangles in the partition approach zero, and the coverage of R becomes increasingly complete (Figure 15.8). We define the area of R to be the limit lim . aA. = 1IJ1I~o~

DEFINITION

iJif dA.

(2)

R

The area of a closed, bounded plane region R is

As with the other defmitions in this chapter, the definition here applies to a greater variety of regions than does the earlier single-variable definition of area, but it agrees with the earlier defmition on regions to which they both apply. To evaluate the integral in the definition of area, we integrate the constant function !(x, y) = I over R.

EXAMPLE 1 quadrant.

Find the area of the region R bounded by Y = x and y = x 2 in the frrst

15.3

851

Area by Double Integration

Solution

y

We sketch the region (Figure IS.19), noting where the two curves intersect at the origin and (1, 1), and calcolate the area as

A = =

!oJ dyax 10f' (x -

x

2 )

=

[~]:ax

ax = [; 2-

3]'0

~

=

1 6'

ax,

FIGURE 15.19

The region in Example I.

Notice that the single-variable integral Jo' (x - x 2 ) obtained from evaluating the inside iterated integral, is the integral for the area between these two curves using the method of Section S.6. • Find the area of the regionR enclosed by the parabolay = x 2 and the line

EXAMPLE 2 y = x + 2. Solution

Ifwe divide culate the area as

y

Rinto the regions R, and R2 shown in Figure IS.20a, we may cal-

On the other hand, reversing the order of integration (Figure IS.20b) gives A =

----=~~~x o
This second result, which requires only one integral, is simpler and is the only one we would bother to write down in practice. The area is A =

y

12-1]..r'+2 dy ax.

2[ ],+2 ax 12 1 Y

-1

(x + 2 - x 2 )

=

x'

-1

ax =

•

Average Value

------'~""'--I------->

o

x

(b)

FIGURE 15.20 Calculating this area ta1res (a) two double integrals if the rust integration is with respect to x, but (b) only one if the ftrSt integration is with respect to y (Example 2).

The average value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length of the interval. For an integrable function of two variables defmed on a bounded region in the plane, the average value is the integral over the region divided by the area of the region. This can be visualized by thinking of the function as giving the height at one instant of some water sloshing around in a tank whose vertical walls lie over the boundary of the region. The average height of the water in the tank can be found by letting the water settle down to a constant height. The height is then equal to the volume of water in the tank divided by the area of R. We are led to define the average value of an integrable function f over a region R as follows:

10fRJJ IdA.

Averagevalueof/overR = area

(3)

R

If I is the temperatore of a thin piate covering R, then the double integral of lover R divided by the area of R is the plate's average temperature. If I(x, y) is the distance from the point (x, y) to a flXed point P, then the average value of lover R is the average distance of points inR fromP.

852

Chapter 15: Multiple Integrals

EXAMPLE 3 R: 0 :5 x :5

Find the average value of 'IT,

0

:5

f(x,y)

= xcosxy

over the rectsngle

f

= sinxy + C

Y :5 1.

The value of the integral of f over R is

Solution

1"1' o

xcosxy dydx =

0

1"[

sinxy

0

]Y-l dx

xcosxydy

y=O

= l"(SinX - 0)

I

dx = -cosx

= I

+

I = 2.

•

The area of R is 7f. The average value of f over R is 2/7f.

Exercises 15.3 Area by Double Integrals In Exercises 1-12, sketch 1he region bounded by 1he given lines and curves. Then express the region's area as an iterated double integral and evaluate 1he integral.

1. The coordinate axes and 1he line x

+y

= 2

2. The lines x = O,y = 2x,andy = 4

3. The parabola x

_y2 and 1he line y

=

x

=

+2

4. The parabola x = y - y2 and 1he line y = -x 5. The curve y = eX and 1he lines y = 0, x = 0, and x = In 2 6. Thecurvesy = lox andy = 2loxand1helinex = e,in1hefrrst quadrant

Find;ng Average Values 19. Find 1he average value of f(x, y) = sin (x

a. the rectangle 0 ::s x :s

1r.

b. 1he rectangle 0 ,;; x ,;; 7f,

O:s y :s

+ y) over 1r.

0,;; y ,;; 7f/2.

20. Which do you think will be larger, 1he average value of f(x,y) = xy over 1he square 0 '" x '" 1,0 '" Y '" I, or 1he average value of f over 1he quarter circle x 2 + y2 ,;; I in 1he first quadrant? Calculate 1hern to fmd out 21. Find 1he average height of 1he paraboloid z = x 2 + y2 over 1he

square 0 :s x :s 2, 0 :s y :s 2. 22. Find 1he average value of f(x,y) = I/(xy) over 1he square 102'" x '" 2102,102'" Y '" 2102.

7. The parabolas x = y2 and x = 2y _ y2 Theory and Examples 8. The parabolas x = y2 - I and x = 2y2 - 2 9. The linesy = x,y = x/3, andy = 2 10. The lines y = I - x and y = 2 and 1he curve y = eX 11. Thelinesy = 2x,y = x/2,andy = 3 - x 12. The lines y = x - 2 and y = -x and 1he curve y =

Vx

Identlfytng the Region of Integration The integrals and sums of integrals in Exercises 13-18 give 1he areas of regions in 1he xy-plane. Sketch each region, label each bounding curve with its equati~ and give the coordinates of the points where 1he curves intersect Then find 1he area of1he region.

11 1"1'1. 1°1 '6

II

o

2Y

K

dxtty

ylj3

0

o

0%

16.

ttydx

sinx X

17.

-1

18.

ttydx

+

Ox 2 -4

12:£'-x ttydx -x/2 0

-2;(

12:£0

-x Y+2

00

15.

131X(2-X) ttydx

dydx

+

14:£Vx 00

ttydx

- y. 1~

dx tty

23. Baeterium population If f(x,y) = (IO,OOOe Y )/(\ + IxV2) represents 1he "population density" of a certain bactetiun3 on 1he xy-plane, where x and y are measured in centimeters, fmd the tota! population of bactetia wi1hin 1he rectangle -5 '" x '" 5 and -2 '" Y '" O. 24. Regional population If f(x, y) = 100 (y + 1) represents 1he population density of a planar region on Earth, where x and y are measured in miles, find the number of people in the region bouoded by 1he curves x = y2 aod x = 2y _ y2. 25. Average temperature in Teus According to 1he Texas Almanac, Texas has 254 counties and a National Weather Service station in each county. Assume that at time to, each of the 254 wea1her stations recorded 1he local teroperature. Find a formula that would give a reasonable approximation of the average temperature in Texas at time to. Your answer should involve information that you would expect to be readily available in 1he Texas Almanac. 26. If Y = f(x) is a nonnegative contiouous function over 1he closed interval a '" x '" b, show that 1he double integral defInition of area for 1he closed plane region bounded by the graph of f, 1he vertical lines x = a and x = b, and the x-axis agrees with the defmition for area beuea1h 1he curve in Section 5.3.

853

15.4 Double Integrals in Polar Form

15.4

~I_D_ou_b_Le__ In_~~g~~_~_l_ 'n_P_o_ ~_ r_ ~_r_ m___________________________ Integrals are sometimes easier to evaluate if we change to polar coordinates. This section shows how to accomplish the change and how to evaluate integrals over regions whose boundaries are given by polar equations.

Integrals in Polar Coordinates When we dermed the double integral of a function over a region R in the xy-plane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. These were the natoralshapes to use because their sides have either constant x-values or constant y-values. In polar coordinates, the natural shape is a ''polar rectangle" whose sides have constant r- and O-values. Suppose that a function I(r, 0) is dermed over a region R that is bounded by the rays = a and 0 = {3 and by the continuous curves r = gM) and r = g2(0). Suppose also that 0 :5 gl(O) :5 g2(O) :5 a for every value of 0 between a and {3. Then R lies in a fanshaped region Q dermed by the inequalities 0 :5 r :5 a and a :5 0 :5 {3. See Figure 15.21.

o

r=a

o~

,,--"---------------''''''''-----------~o ~

o

0

FIGURE 15.21 TheregionR:g,(8) '" r s g2(8),a s 8 s ,B,is contained in the fanshaped region Q: 0 :5 r S a, a s 8 s ,B. The partition of Q by circular arcs aod rays induces a partition of R.

We cover Q by a grid of circular arcs and rays. The arcs are cut from circles centered at the origin, with radii !J.r, 2!J.r, ... , m!J.r, where !J.r = aim. The rays are given by () =

Q,

0= a + M,

0= a + 2M,

... ,

o=

a + m'!J.O

= {3,

where !J.O = ({3 - a)lm'. The arcs and rays partition Q into small patches called ''polar rectangles." We number the polar rectangles that lie inside R (the order does not matter), calling their areas !J.A" !J.A 2, ... , !J.A •. We let (rk, Ok) be any point in the polar rectangle whose area is !J.Ak. We then form the sum

If I is continuous throughout R, this sum will approach a limit as we refine the grid to make !J.r and !J.O go to zero. The limit is called the double integral of lover R. In symbols, lim S. =

n~OO

JJRrr I(r, 0) dA.

854

Chapter 15: Multiple Integrals To evaluate this limit, we first have to write the sum S. in a way that expresses aA. in terms of!!J.r and!!J.lI. For convenience we choose r. to be the average of the radii of the inner and outer arcs bounding the kth polar rectangle aA•. The radius of the inner arc bounding aA. is then r. - (!!J.r/2) (Figure 15.22). The radius of the outer arc is r. + (!!J.r/2). The area of a wedge-shaped sector of a circle having radius r and angle II is

A = Large sector

o FIGURE 15.22 aA k =

The observation lbat

1 11 • r2 2

'

as can be seen by multiplying 7Tr 2 , the area of the circle, by 1I/27T, the fraction of the circle's area contained in the wedge. So the areas of the circular sectors subtended by these arcs at the origin are

Car::e:!m) -(sn:::!or)

leads to lbe formula aA, = r, !!J.r !!J.O.

)2 M

Inner mdius:

!!J. "21 ( r.---f

Outer radius:

"21 ( r. + T!!J.r)2 !!J.II.

Therefore,

aA.

= area oflarge sector - area of small sector =

y

~II [ (r. + ~r)' -

(r. -

~)'] = ~O (2r. !!J.r) =

r. ar M.

Combining this result with the sum defming S. gives

•

S. = ~ j(r., lI.h ar M.

~~----------->x

As n --> integral

00

and the values of ar and ao approach zero, these sums converge to the double

o

lim S. =

(a)

n_OO

rr

JJ R

y

Leaves at r

~

2 r sin 8 : : ~ r =

j(r, II) r dr dll.

A version of Fubini's Theorem says that the limit approached by these sums can be evaluated by repeated single integmtions with respect to r and II as

2

=

L

R

V2 t-.....,,\t-~

Y2 esc 9

Enters atr

=

V2 esc 8

8

~~~--------->x

o

Largest 6 is

I'

L

2

/

// Y

=x

V2 I--+-....:/~ / / / /

/

R

1

,~plr~"'(.)

8=a

j(r,lI) r dr dll. r=g,(8)

Finding Limits of Integration The procedure for finding limits of integration in rectangular coordinates also works for polar coordinates. To evaluate lfR j(r, II) dA over a region R in polar coordinates, integmting fITst with respect to r and then with respect to II, take the following steps.

(b) y /

iJif

j(r,lI) dA =

~ Smallest 6 is ~.

1. Sketch. Sketch the region and label the bounding curves (Figure 15.23a). 2. Find the r-limits of integration. Imagine a my L from the origin cutting through R in the direction of increasing r. Mark the r-values where L enters and leaves R. These are the r-limits of integmtion. They usually depend on the angle II that L makes with the positive x-axis (Figure 15.23b).

/

~~-L--------->x

o

(c)

FIGURE 15.23 Finding !be limits of integration in polar coordinates.

3.

Find the II-limits of integration. Find the smallest and largest II-values that bound R. These are the II-limits of integmtion (Figure 15.23c). The polar iterated integral is

iJif R

j(r, II) dA =

1

·~"/21r~2

9='1fj4

r=V2csc6

j(r, II) r dr dll.

15.4 Double Integrals in Polar Form

855

EXAMPLE 1 Find the limits of integration for integrating fir, 0) over the region R that lies inside the cardioid r = I + cos 0 and outside the circle r = I.

y

Solutton

O=-I

L

Enters at r=1

Leaves at

1. We flfst sketch the region and label the bounding curves (Figure 15.24). 2. Next we fmd the r-limits of integration. A typical ray from the origin enters R where r = I and leaves where r = I + cos O. 3. Finally we find the O-limits ofintegration. The rays from the origin that intersectR run from 0 = -7f/2 to 0 = 7f/2. The integral is

r=I+0089

1

-fTj2

FIGURE 15.24 Finding the limits of

integratioo in polar coordinates for the regioo in Example 1.

~/211+'0"

•

fir, 0) r dr dO.

1

If fir, 0) is the constant function whose value is I, then the integral of f over R is the areaofR.

Area in Polar Coordinates

I Area Differential in Polar Coordinate.

The area of a closed and bounded region R in the polar coordinate plane is

dA = rdrd6

A = IfrdrdO. R

This fonnula for area is consistent with all earlier fonnulas, although we do not prove this fact. y

Leaves at r=

V4 co, 28

" I. ~------~~------+---~X

Enters at r= 0

r 2 =400829

To integrate over the shaded regioo, we run r from 0 to y'4 cos 26 and 6 from 0 to ",/4 (Exarople 2). FIGURE 15.25

EXAMPLE 2

Find the area enclosed by the lemniscate r2

=

4 cos 20.

Solutton We graph the lemniscate to determine the limits of integration (Figure 15.25) and see from the symmetry of the region that the tota1 area is 4 times the first-quadrant portion.

J.

A=4

~/4:£V4"'26

J.~/4

00

=

J.

4

0

[r2]'=V4'0'2'

rdrdO=4"2 0

~/4

2 cos 20 dO

]~/4

=

4 sin 20

0

dO

r=O

=

4.

•

Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integralifR fix, y) dx dy into a polar integral has two steps. First substitute x = r cos 0 and y = r sin 0, and replace dx dy by r dr dO in the Cartesian integral. Then supply polar limits of integration for the boundary of R. The Cartesian integral then becomes

II R

fix, y) dx dy =

II

fir cos 0, r sin 0) r dr dO,

G

where G denotes the same region of integration now described in polar coordinates. This is like the substitution method in Chapter 5 except that there are now two variables to substitute for instead of one. Notice that the area differential dx dy is not replaced by dr dO but by r dr dO. A more general discussion of changes of variables (substitutions) in multiple integrals is given in Section 15.8.

856

Chapter 15: Multiple Integrals

EXAMPLE 3

Evaluate

II

e

x2 y2

+ dy dx,

R

where R is the semicircular region bounded by the x-axis and the curve y = ~ (Figure 15.26). y

y=~

SoLution In Cartesian coordinates, the integral in question is a nonelementary integral and there is no direct way to integrate e x2 +y2 with respect to either x or y. Yet this integral and others like it are important in mathematics--in statistics, for example--and we need to find a way to evaluate it. Polar coordinates save the day. Substituting x = r cos e, y = r sin e and replacing dy dx by r dr de enables us to evaluate the integral as

0=0 --~~------~~------~

__ x

= Jot'l"2 (e - 1) de

FIGURE 15.26 The semicircular region in Example 3 is the region

o :s r :s

1,

o :s ()

:s

7T.

= ;

(e - 1).

The r in the r dr de was just what we needed to integrate e r2 • Without it, we would have been unable to find an antiderivative for the first (innermost) iterated integral. _

EXAMPLE 4

SoLution

Evaluate the integral

Integration with respect to y gives

t (x2~+

Jo

(1 - X2) 3/2) 3 dx,

an integral difficult to evaluate without tables. Things go better if we change the original integral to polar coordinates. The region of integration in Cartesian coordinates is given by the inequalites 0 ::; y ::; ~ and o ::; x ::; 1, which correspond to the interior of the unit quarter circle x 2 + y2 = 1 in the first quadrant. (See Figure 15.26, first quadrant.) Substituting the polar coordinates x = r cos e, y = r sin e, 0 ::; e ::; 7T/2 and 0 ::; r ::; 1, and replacing dx dy by r dr de in the double integral, we get

z

_1

-

o

7T /

2

_1

[r4 ]r=1 4 de -

r=O

7T /

2

!4 de -_ 1T8·

0

Why is the polar coordinate transformation so effective here? One reason is that x 2 simplifies to r2. Another is that the limits of integration become constants.

EXAMPLE 5 z x

FIGURE 15.27 Example 5.

_

Find the volume of the solid region bounded above by the paraboloid

= 9 - x 2 - y2 and below by the unit circle in the xy-plane.

The region of integration R is the unit circle x 2 + y2 = 1, which is described in polar coordinates by r = 1,0 ::; e ::; 21T. The solid region is shown in Figure 15.27. The volume is given by the double integral

SoLution The solid region in

+ y2

15.4 Double Integrals in Polar Form

=

10f'''10{' (9r -

r

3

)

857

dr dO

• EXAMPLE 6 Using polar integration, find the area of the regionR in the xy-plane eoclosed by the circle x 2 + y2 + 4, above the line y = 1, and below the line y = 0x. Solutton

A sketch of the region R is showo in Figure 15.28. First we note that the line y = 0x has slope 0 = tan 6, so 6 = 7T/3. Next we observe that the line y = 1 intersects the circle x 2 + y2 = 4 wheo x 2 + 1 = 4, or x = 0. Moreover, the radial line from the origin through the point (0,1) has slope 1/0 = tan 0, giving its angle of inclination as 6 = 7T/6. This information is showo in Figure 15.28. Now, for the regionR, as 6 varies from 7T/6 to 7T/3, the polar coordinate r varies from the horizontal line y = 1 to the circle x 2 + y2 = 4. Substituting r sin 0 for Y in the equation for the horizontal line, we have r sin 6 = 1, or r = esc 6, which is the polar equation of the line. The polar equation for the circle is r = 2. So in polar coordinates, for 7T/6 :5 6 :5 7T/3, r varies from r = esc 0 to r = 2. It follows that the iterated integral for the area theo gives

y

~~--~~------~~X

o

2

FIGURE 15.28 Example 6.

1 2

/3

{{ dA = {" Jrr/6

JJR

The region R in

- {"/3 - Jrr/6

rdrd6

esc (J

[1 r2l'~2 2

d6

r- csc9

,,/3 1

csc20] dO

=

1

=

21 [46 + cot 6 ],,/ "/6

=

1 (47T +

,,/6

2 [4 -

3

_1_) _1(47T + 0)

23026

= 7T -

Exercises 15.4 Regions in Polar Coordinates

4.

3.

In Exercises 1-1l, describe the given region in polar coordinates.

L

y

1 y --~--~~----~->x

-1

--4-----+-----,--> x 9

--------+-~--4-->x

o

4

0

1

0

3'

•

858

Chapter 15: Multiple Integrals Area in Polar Coordinates

6.

5. y

27. Find the area of the region cut from the fIrst quadrant by the curve r = 2(2 - sin 26)112.

y

28. Cardioid overlapping a clrcle Find the area of the region that lies inside the cardioid r = 1 + cos 0 and outside the circle r = 1.

2t--- - - - - ,

29. One leaf of a rose r = 12 cos 36.

30. SnaR shen Find the area of the region eoclosed by the positive x-axis aod spiral r = 46/3,0 '" 6 '" 21T. The region looks like a anall shell.

2

7. The region enclosed by the circle x 2

8.

+ y2 = 21:. Theregionenclosedbythesemicirclex 2 + y2 =

2y,y '" O.

In Exercises 9-22, change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.

1 11,~

11. Jo Jo

I

2

-~ 1

+

~

-Vd'-x'

22

YX

+ y2

/,v22/,'v4=Y'

Find the awrage height of

Ya 2 - x2 -

y2 above the disk

34. Average height ofa cone Find the average height of the (single) dxdy

dydx

conez = Yx 2

+ y2 above the diskx 2 + y2

'" a 2 inthexy-plane.

35. Average distance from interior of disk to center Find the average distance from a point P(x, y) in the disk x 2 + y2 '" a 2 to the origin.

2

2 X + y2f

+

dydx

°

1 1Vi-v'I=Yl =Y'

f(r, 6) r dr d6.

R

33. Awrage height of a hemisphere

1n2:£V(ln2)'-" • ~ • vx'+,' dx dy

1,°

JJ

36. Average distance squared from a point in a disk to a point in

-~ (I

1

20.

1

16.

dydx

18.111~ 19.

12.

Are~(R)

a1~ dydx

the hemispherical surface z = x 2 + y2 ~ a 2 in the xy-plane.

17.1°1° -1

+ y2) dx dy

+ y2) dx dy

Average values In polar coordinates, the average value of a function over a region R (Section 15.3) is giveo by

0

x dx dy

V3r Jl

-1

(x 2

0

(x

13. 1,61,' 1

1,11,Vi=Y'

0

{'{VH

15.

10.

dy dx

-1

31. Cardioid in the r"st quadrant Find the area of the region cut from the frrst quadrant by the cardioid r = 1 + sin 6. 32. Overlapping cardioids Find the area of the region conunon to the interiors of the cardioids r = I + cos 6 and r = I - cos 6.

Evaluating Polar Integrals

9.

Find the area enclosed by one leaf of the rose

its houndary Find the average value of the square of the distance from the point p(x, y) in the disk x 2 + y2 '" I to the boundary pointA(I, 0).

Theory and Examples In (x 2

+ y2 + I) dx dy

-1

37. Converting to a polar integral Integrate f(x,y) = [In (x 2 + y2)]fYx 2 + y2 OWl the region 1 '" x 2 + y2 S e.

21.1,11~ (x + 2y)dydx

38. Converting to a polar integral Integrate f(x, y) = [In (x 2 + y2)]/(x 2 + y2) 0'"' the region I '" x 2 + y2 S .2.

22. 121,~

39. Volume of noncircular right cylinder The region that lies inside the cardioid r = 1 + cos 8 and outside the circle r = I is the base of a solid right cylinder. The top of the cylioder lies in the plane z = x. Find the cylinder's volume.

1

0

1 (X 2

+ y2)2

dydx

In Exercises 23-26, sketch the region of integration and convert each polar integral or sum of integrals to a Cartesian integral or sum of in· tegrals. Do not evaluate the integrals. 12 {' 23. Jo Jo r3 sin 6 cos 6 dr d6

r

24.

r121='

r2 cos 6 dr d6

i1r/6 I /4 {''''''' 25. Jo Jo

r

40. Volume of noncircular right cylinder The region enclosed by the leoutiscate r2 = 2 cos 26 is the base of a solid tight cylinder whose top is bounded by the sphere z = ~. Find the cylinder's volume. 41. Converting to polar integrals

a. The usual way to evaluate the improper integral 00 I = e --;f1 dx is frrst to calculate its square:

10

r' sin2 6 dr d6 Evaluate the last integral using polar coordinates and solve the resulting equation for 1.

15.5 Triple Integrals in Rectangular Coordinates b. Evaluate lim erf(x) =

x-HX)

lim

!O0

X

2 - /' • e/ dt.

X-H)()

46. Area Suppose that the area of a region in the polar coordinate plane is

1

3Tr/412sin6

A =

V 11'

Tr/4

42. Converting to a polar integral

Evaluate the integral

43. Existence Integrate the function j(x, y) = 1/(1 - x 2 - y2) over the disk x 2 + y2 ~ 3/4. Does the integral of j(x, y) over

rdrdO. esc 6

Sketch the region and find its area. COMPUTER EXPLORATIONS In Exercises 47- 50, use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise.

a. Plot the Cartesian region of integration in the xy-plane.

the disk x 2 + y2 ~ I exist? Give reasons for your answer.

b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for r andO.

44. Area formula in polar coordinates Use the double integral in polar coordinates to derive the formula

1

859

.6 1

2

c. Using the results in part (b), plot the polar region of integration in the rO-plane.

for the area of the fan-shaped region between the origin and polar curve r = j(O), a ~ 0 ~ {3.

d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility.

A =

a

- r 2 dO

45. Average distance to a given point inside a disk Let Po be a point inside a circle of radius a and let h denote the distance from Po to the center of the circle. Let d denote the distance from an arbitrary point P to Po. Find the average value of d 2 over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and Po on the x-axis.)

15.5

47.

49.

1111 1l o

x

1

o

Y

- 2 - - 2 dy

+Y

x

Y 3 /

-y/3

dx

Y

Yx 2 + y2

48.

l1 1

o

x 2 x / ---dydx

0

x 2 + y2

dxdy

Triple Integrals in Rectangular Coordinates Just as double integrals allow us to deal with more general situations than could be handled by single integrals, triple integrals enable us to solve still more general problems. We use triple integrals to calculate the volumes of three-dimensional shapes and the average value of a function over a three-dimensional region. Triple integrals also arise in the study of vector fields and fluid flow in three dimensions, as we will see in Chapter 16.

Triple Integrals If F(x, y, z) is a function defined on a closed, bounded region D in space, such as the region occupied by a solid ball or a lump of clay, then the integral of F over D may be defined in the following way. We partition a rectangular boxlike region containing D into rectangular cells by planes parallel to the coordinate axes (Figure 15.29). We number the cells that lie completely inside D from 1 to n in some order, the kth cell having dimensions ~Xk by ~Yk by ~Zk and volume ~ Vk = ~k~Yk~Zk. We choose a point (Xk, Yk, Zk) in each cell and form the sum n

Sn = ~ F(xk, Yk , Zk) ~ Vk.

(1)

k=l

FIGURE 15.29 Partitioning a solid with rectangular cells of volume ~ Vk •

We are interested in what happens as D is partitioned by smaller and smaller cells, so that ~k, ~Yk, ~Zk and the norm of the partition Ilpll, the largest value among ~k, ~Yk, ~Zk, all approach zero. When a single limiting value is attained, no matter how the partitions and points (Xk, Yk, Zk) are chosen, we say that F is integrable over D. As before, it can be

860

Chapter 15: Multiple Integrals shown that when F is continuous and the bounding surface of D is fonned from tmitely many smooth surfaces joined together along tmitely many smooth curves, then F is integrable. As Ilpll --> 0 and the number of cells n goes to 00, the sums S. approach a limit. We call this limit the triple integral of F over D and write

•

~S. =

iiiF(x,y,Z)dV

lim S. =

or

IWII~O

D

rrr F(x,y,z)

111 D

dxdy dz .

The regions D over which continuous functions are integrable are those having ''reasonably smooth" boundaries.

Volume of a Region in Space If Fis the constant function whose value is 1, then the sums in Equation (1) reduce to

As dxk, t.Yk, and t.Zk approach zero, the cells t. V. become smaller and more numerous and ml up more and more of D. We therefore define the volume of D to be the triple integral

DEFINITION

The volume of a closed, bounded region D in space is

V= iff dV. D

TIris detmition is in agreement with our previous detmitions of volume, although we omit the verification of this fact. As we see in a moment, this integral enables us to calculate the volumes of solids enclosed by curved surfaces.

Finding Limits of Integration in the Order dz dy dx We evaluate a triple integral by applying a three-dimensional version of Fubini's Theorem (Section 15.2) to evaluate it by three repeated single integrations. As with double integrals, there is a geometric procedure for finding the limits of integration for these single integrals. To evaluate

iff F(x,y,z)dV D

over a region D, integrate first with respect to z, then with respect to y, and finally with respect to x. (You might choose a different order of integration, but the procedure is similar, as we illustrate in Example 2.)

1.

Sketch. Sketch the region D along with its "shadow" R (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces of D and the upper and lower bounding curves of R.

15.5 Triple Integrals in Rectangular Coordinates

861

z z = fz(x, y)

D

y

x

2.

Find the z-limits a/integration . Draw a line M passing through a typical point (x,y) in R parallel to the z-axis. As z increases, M enters D at z = f 1(x, y) and leaves at z = h(x,y). These are the z-limits of integration. z

D

y

a b

--- ---

x

(x,y)

3.

Find the y-limits a/integration. Draw a line L through (x,y) parallel to the y-axis. As y increases, L enters Rat y = gl(X) and leaves at y = g2(X). These are the y-limits of integration. z

D

Enters at y = g,(x) a

y

x b

x

Leaves at y = g2(x)

862

Chapter 15: Multiple Integrals

4.

Find the x-limits of integration. Choose x-limits that include all lines through R parallel to the y-axis (x = a and x = b in the preceding figure). These are the x-limits of

integration. The integral is

l

x=bly=g2(X)lz= h(x,y)

x=a

F(x, y, z) dz dy dx. y=gl(X)

z= !J(x, y)

Follow similar procedures if you change the order of integration. The "shadow" of region D lies in the plane of the last two variables with respect to which the iterated integration takes place. The preceding procedure applies whenever a solid region D is bounded above and below by a surface, and when the "shadow" region R is bounded by a lower and upper curve. It does not apply to regions with complicated holes through them, although sometimes such regions can be subdivided into simpler regions for which the procedure does apply.

EXAMPLE 1

Find the volume of the region D enclosed by the surfaces z = x 2

+

3y2

and z = 8 - x 2 - y2. SoLution

The volume is

v=

111

dzdydx,

D

the integral of F(x, y, z) = lover D. To find the limits of integration for evaluating the integral, we first sketch the region. The surfaces (Figure 15.30) intersect on the elliptical cylinder x 2 + 3y2 = 8 - x 2 - y2 or x 2 + 2y2 = 4, z > O. The boundary of the region R, the projection of D onto the xy-plane, is an ellipse with the same equation: x 2 + 2y2 = 4. The ''upper'' boundary of R is the curve y = V(4 - x 2)/2. The lower boundary is the curve y = -V(4 - x 2 )/2. Now we find the z-limits of integration. The line M passing through a typical point (x,y) in R parallel to the z-axis enters D at z = x 2 + 3y2 and leaves at z = 8 - x 2 - y2. M

The curve of intersection

'-(-2, 0,4)

I I

Enters at

z =x 2 +

3y2

I

I

J-----

-- I I , I '.,,1

Enters at~-----=_ y = -V(4 - x 2 )/2

x

R

Leavesat

~

L

y

y = V(4 - x 2 )/2

FIGURE 15.30 The volume of the region enclosed by two paraboloids, calculated in Example 1.

15.5 Triple Integrals in Rectangular Coordinates

863

Next we find the y-limits of integration. The line L through (x, y) parallel to the y-axis enters Rat y = - Y(4 - x 2 )/2 and leaves at y = Y(4 - x 2 )/2. Finally we find the x-limits of integration. As L sweeps across R, the value of x varies from x = -2 at (-2,0,0) to x = 2 at (2, 0, 0). The volume of Dis

v=

JJJ

dzdydx

D

1 11 1

21V(4 - X2)/ 218 - X2 - y2

=

- 2 - V(4 - x 2)/ 2 x2 +3y2

V (4 - X2)/ 2

2

=

dzdydx

(8 - 2x2 - 4y2) dy dx

- 2 - V(4 - x 2)/ 2 2 [

=

-2

=

4 ]y=V(4 - X2)/ 2 (8 - 2x 2)y - _y3 dx 3 y= - V(4 - x 2)/ 2

81TY2.

•

After integration with the substitution x = 2 sin u

In the next example, we project D onto the xz-plane instead of the xy-plane, to show how to use a different order of integration. z __________ (0, 1,1)

EXAMPLE 2 Set up the limits of integration for evaluating the triple integral of a function F(x, y, z) over the tetrahedron D with vertices (0, 0, 0), (1 , 1,0), (0, 1, 0), and (0, 1, 1). Use the order of integration dy dz dx. Solution

We sketch D along with its "shadow" R in the xz-plane (Figure 15.31). The upper (right-hand) bounding surface of D lies in the plane y = 1. The lower (left-hand) bounding surface lies in the plane y = x + z. The upper boundary of R is the line z = 1 - x. The lower boundary is the line z = 0. First we find the y-limits of integration. The line through a typical point (x, z) in R parallel to the y-axis enters D at y = x + z and leaves at y = 1. Next we find the z-limits of integration. The line L through (x, z) parallel to the z-axis enters R at z = and leaves at z = 1 - x. Finally we find the x-limits of integration. As L sweeps across R, the value of x varies from x = to x = 1. The integral is

°

(1,1,0)

°

ll i

x

Finding the limits of integration for evaluating the triple integral of a function defined over the tetrahedron D (Examples 2 and 3).

l

o

FIGURE 15.31

l x -

0

l

•

F(x,y, z) dy dz dx.

x+ z

EXAM PLE 3 Integrate F(x, y , z) = 1 over the tetrahedron D in Example 2 in the order dz dy dx, and then integrate in the order dy dz dx. Solution

First we find the z-limits of integration. A line M parallel to the z-axis through a typical point (x, y) in the xy-plane "shadow" enters the tetrahedron at z = and exits through the upper plane where z = y - x (Figure 15.32). Next we find the y-limits of integration. On the xy-plane, where z = 0, the sloped side of the tetrahedron crosses the plane along the line y = x . A line L through (x, y) parallel to the y-axis enters the shadow in the xy-plane at y = x and exits at y = 1 (Figure 15.32).

°

864

Chapter 15: Multiple Integrals

z

Finally we find the x-limits of integration. As the line L parallel to the y-axis in the previous step sweeps out the shadow, the value of x varies from x = 0 to x = 1 at the point (1, 1,0) (see Figure 15.32). The integral is

ri

Jo

x

(y-x

l

Jo

F(x,y,z) dzdydx.

For example, if F(x,y, z) = 1, we would find the volume of the tetrahedron to be

v=

1

lillY- Xdzdydx l

= l i\y-x)dY dx x

FIGURE 15.32 The tetrahedron in Example 3 showing how the limits of integration are found for the order dz dy dx.

1

6' We get the same result by integrating with the order dy dz dx. From Example 2,

V

=

l i o

l l X l dydzdx l 0

r r-

x+z

x

=

Jo Jo

=

1 --(1 - x)3

6

(1 - x - z) dz dx

]1

•

0

Average Value of a Function in Space The average value of a function F over a region D in space is defined by the formula Average value of F over D =

vol~e of D

Jff

F dV.

(2)

D

For example, if F(x,y, z) = Yx 2 + y2 + z2, then the average value of F over D is the average distance of points inD from the origin. If F(x,y, z) is the temperature at (x,y, z) on a solid that occupies a region D in space, then the average value of F over D is the average temperature of the solid.

865

15.5 Triple Integrals in Rectangular Coordinates

EXAMPLE 4 Find the average value of F(x,y, z) = xyz throughout the cubical region D bounded by the coordinate planes and the planes x = 2, Y = 2, and z = 2 in the tIrst octant. Solution We sketch the cube with enough detail to show the limits of integration (Figure 15.33). We then use Equation (2) to calculate the average value of F over the cube. The volume of the region Dis (2)(2)(2) = 8. The value of the integral of F over the cube is

D

121212

xyzdxdydz =

000

FIGURE 15.33 in Example 4.

The region of integration

1212 [2 12 12

~ yz ]X~2 dydz

00

=

[]Y~2 y2z dz=

o

= 12122yzdydz

x- O

y=O

00

[]2

4zdz= 2z2

0

=8.

0

With these values, Equation (2) gives / Average value of = lYz over the cube vo ume

rrr

111 cube

(-81 )(8)

xyz dV =

=

I.

In evaluating the integral, we chose the order dx dy dz, but any of the other five possible orders would have done as well. •

Properties of Triple Integrals Triple integrals have the same algebraic properties as double and single integrals. Simply replace the double integrals in the four properties given in Section 15.2, page 846, with triple integrals.

Exercises 15.5 Triple Integrals In Different Iteration Orders 1. Evaluate 1he integral in Example 2 taking F(x,y, z) = 1 to Imd 1he volmne of1he tetrahedron in 1he order dz dx dy. 2. Volume of rectangular .olid Write six different iterated triple

Evaluating Triple Iterated Integrals Evaluate 1he integrals in Exercises 7-20.

7. 1, '1, '1,

'<x 2

+ y2 + z2) dz ely dx

integrals for 1he volume of 1he rectangular solid in 1he fIrst octant bounded by 1he coordinate planes aod 1he planes x = I, Y = 2, and z = 3. Evaluate one of1he integrals.

9.

3. Volume of tetrahedron Write six different iterated triple inre. grals for 1he volmne of 1he tetrahedron cut from 1he IlrSt octant by 1he plane 6x + 3y + 2z = 6. Evaluate one of 1he integrals.

4. Volume of solid Write six different iterated triple integrals for 1he volmne of 1he region in 1he first octant enclosed by 1he cylinder x 2 + z2 = 4 and 1he plane y = 3. Evaluate one of 1he integrals.

10.

1,'1,'-"'1.'-"'-'

12.1'Jo{'Jo

{2(x

13.

{i' {<'

I xyzdxelydz

ysinzdxelydz 1,~/'1.'1' 00-2

+ Y + z)elydxdz

1,o'1.~1~ dzelydx 0

14.

{2 {V47' {'-+, dz dx ely

Jo l-v'Rlo

0

16. 1,'1,hJ,4-x'-'

different triple iterated integrals for 1he volmne of D. Evaluate one of1he integrals.

Write triple iterated integrals in 1he order dz dx ely and dz ely dx iliat give 1he volmne ofD. Do not evaluate ei1her integral.

11.

,J, J,

-I

5. Volume enclosed by paraboloid. Let D be 1he region bounded by 1he paraboloids z = 8 - x 2 - y2 and z = x 2 + y2. Write six

6. Volume inside paraboloid beneath a plane Let D be 1he region bounded by 1he paraboloid z = x 2 + y2 and 1he plane z = 2y.

dzdydx

000

I

1

17.1,~1,~1,~ cos(u + v + w)dudvdw 18.

{'IV.J,{'

Jo

1

(In 1)2 se'lnr-I-dldrds

(uvw-space)

(rsl-Space)

x dz ely dx

866

19. 20.

Chapter 15: Multiple Integrals

1.,/' [>-·1" r 0 10

--or;>

171'-1~

T

!

dx dt dv

24. The regicm. in the flrst octant bounded by the coordinate planes andthep1anesx + z - l,y + 2z - 2

(tux-space)

1 tip dq fir

,

(pqr-space)

Finding Equivalent Iterated Integrals 21. Here is the region of integration of the integral

1'1, '1.'-' r

-1

,

dzdydx.

0

25. The region in the flrst octant bounded by the coontinate planes, the plane y + z = 2, and the cylinder x = 4 _ y2

,

Top:y+z=1 1

/-

;....---"> (-1, 1,0)

,

1

x

,,

(I, I, 0)

Rewrite the integral as an equivalent iterated integm in the order L dydzdx b. dydxdz c, dxdydz d. dxdzdy e, dzdxdy. 22. Here is the regicm. of integraticm. of the integral

1. ,1-1f'1.T o

26. The wedge cut from the cylinder x 2

z - -yandz - 0

x

1

z _ ,'1.

27. The tetrahedron in the first octantboundedbythe coordinate planes and the plane passing through (1, 0, 0), (0, 2, 0), and (0, 0, 3)

,

o , (1,-1,0)

- 1 by the planes

,

,

(0,-1, I)

+ y2

,

dzdydx.

0

(1,-1, I)

.

1 x

Rewrite the integral. as an equivalent iterated integral. in the order dydzdx b. dydxdz c. dxdydz d. dxdzdy e. dzdxdy.

(0,2,0)

,

L

Finding Volumes UsIng TrIple Integrals Find the VOiUlllCS ofb regions in Exercises 23-36. 23. The region between the cylinder z = y'- and the xy-plane that is bounded by the planes x = O,x = l,y = -1,y = 1

(1,0,0) ,/

x 28. The region in the fmt octant bounded by the coordinate planes, b plane Y = 1 - x, and the surface z = cos ('/I"xj2), Osxsl

,

,

, x

15.5 Triple Integrals in Rectangular Coordinates 29. The region common to the interiors of the cylinders x 2 + y2 = 1 and x 2 + z2 = 1, one-eighth of which is shown in the accompanying figure

33. The region between the planes x z = 4 in the first octant

+ y + 2z

34. The finite region bounded by the planes z y = 8, andz = 0

=

2 and 2x

867 + 2y +

= x, x + z = 8, z = y,

35. The region cut from the solid elliptical cylinder x 2 + 4y2 ::::; 4 by the xy-plane and the plane z = x + 2

xZ+r=l

36. The region bounded in back by the plane x = 0, on the front and sides by the parabolic cylinder x = 1 - y2, on the top by the paraboloid z = x 2 + y2, and on the bottom by the xy-plane

Average Values In Exercises 37-40, find the average value of F(x, y, z) over the given region. 37. F(x,y, z) = x 2 + 9 over the cube in the first octant bounded by

the coordinate planes and the planes x

x

30. The region in the first octant bounded by the coordinate planes and the surface z = 4 - x 2 - Y

= 2, y = 2, and z = 2

38. F(x, y, z) = x + Y - z over the rectangular solid in the first octant bounded by the coordinate planes and the planes x = l,y = 1, andz = 2 39. F(x,y,z) = x 2 + y2 + z2 over the cube in the first octant bounded by the coordinate planes and the planes x = l,y = 1, andz = 1 40. F(x, y, z) = xyz over the cube in the first octant bounded by the coordinate planes and the planes x = 2, y = 2, and z = 2

z

Changing the Order of Integration Evaluate the integrals in Exercises 41-44 by changing the order of integration in an appropriate way.

10 12 1o r r1112xzezy2 Jo Jo 4

41.

0

y

42.

x

1

4 cos (x • r 2 Yz

2y

1hI x2l

2 )

dxdydz

dy dx dz

x2

31. The region in the first octant bounded by the coordinate planes, the plane x + y = 4, and the cylinder y2 + 4z2 = 16

z

43. 44.

1

o

{2

~

ln3

'Tfe

2x

0

r- r si~2z

Jo Jo Jo

4

sin 7TJ'2

Y

2

dxdydz

dy dz dx

z

Theory and Examples 45. Finding an upper limit of an iterated integral y

114-a-x214-X2-y

1 o

x

32. The region cut from the cylinder x 2 + y2 = 4 by the plane z = 0 and the plane x + z = 3

z

0

Solve for a:

_

~ dzdydx - 15'

a

46. Ellipsoid For what value of c is the volume of the ellipsoid x 2 + (y/2)2 + (z/c)2 = 1 equal to 8'Tf? 47. Minimizing a triple integral mizes the value of the integral JJJ(4x 2 + 4y2

What domain D in space mini-

+ z2

- 4) dV?

D

Give reasons for your answer. 48. Maximizing a triple integral mizes the value of the integral

What domain D in space maxi-

JJJ(1 - x 2 - y2 - z2) dV? y

x

D

Give reasons for your answer.

868

Chapter 15: Multiple Integrals

COMPUTER EXPLORATIONS In Exercises 49-52, use a CAS integration utility to evaluate the triple integral of the given :function aver the specified solid region.

49. F(x,y,z) = xly2z over the solid cylinder bounded by x 2 + y2 _ 1 and the planesz - Oandz - 1 50. F(x, y, z) = Ixyz Iover the solid bounded below by the paraboloid z = x 2 + y2 and above by the plane z = I

15.6

2 : 2 3/2 over the solid bounded below by (x +y +z) the cone z = v'x 2 + y2 and above by the plane z = 1

51. F(x,y, z) =

52. F(x,y, z) = x" + yl + i2 over the solid sphere x2 + yl + zl :S 1

I~M~ om =e=n~u~a=n=d~c~ en=t~ e~ ~of~M ~a=s~ s __________________________ This section shows how to calculate the masses and moments of two- and threedimensional objects in Cartesian coordinates. Section 15.7 gives the calculations for cylindrical and spherical coordinates. The defmitions and ideas are similar to the sDlgle-variable case we studied in Section 6.6, but now we can consider more realistic situations.

,

Masses and First Moments If B(x, y, z) is the density (mass per unit volwne) ofan object occupying a region D in space, the integral of Ii over D gives the mall of the object. Th see why, imagine partitioning the object into n mass elements like the one in Figure 15.34. The object's mass is the limit

,

y

FIGURE 15.34 To defme an object's mass, we fll8t imagine it to be partiti.oru:d

The first moment of a solid region D about a coordinate plane is defined as the triple integral """ D of the
into a fmite number of mass elements

<1m.

MY'

~

JJ1

x8(x,y,z) dV.

D

,

The center ofma33 is found from the fll'Sl: moments. For instance, the x-coordinate of the center of mass is x - Myz/M. For a two-dimensional object, such as a thin, flat plate, we calculate fll'St moments about the coordinate axes by simply dropping the z-coordinate. So the first moment about the y-axis is the double integral over the region R fanning the plate of the distance from the axis multiplied by the density, or M,

~

11

>8(x, y) dA.

R

Table 15.1 summarizes the formulas.

, FIGURE 15.35 Finding the center of mass of a solid (Example 1).

EXAM PLE 1 Find the center of mass of a solid of constant density B bmmded below by the disk R: Xl + y2 :s 4 in the plane z = 0 and above by the paraboloid z = 4 - x 2 - y2 (Figure 15.35).

15.6

TABLE

Moments and Centers of Mass

869

15.1 Mass and first moment formulas

THREE-DIMENSIONAL SOLID

a~a(x,y,z)isthedensityat(x,y,z).

Mass:M= JJJ8dV D

First moments about the coordinate planes:

My<

=

JJJ x8dV,

Mxz

=

D

JJJ y8dV,

Mxy

=

D

JJJ z8dV D

Center of mass:

_

_ Mxy z=-

Mxz

y=M'

M

TWO-DIMENSIONAL PLATE Mass:

M = JJ8 ciA

a

~ a(x,y) is the density at (x,y).

R

First moments:

Center of mass:

Solution

By symmetry x = y = 0. To f"md Z, we first calculate

Mxy

=

~~~

z=o

R

=

z-4-x'-1

z8dzdydx

=

~if [z2 ]z- 4-x'-1

2

R

~JJ(4 -

z=o

8dydx

x 2 - y2)2dydx

R

2(4-r2)2rdrdlJ 8/02"1 2

=-

0

=

r 210

~

Polar coordinates simplify the integration.

0

2 " [_1(4 _ 6

r2)3]'~2 dO ,~O

= IM!2" dO = 321T8. 3 3

10

A similar calculation gives the mass

M

=

rrr-x'--18

1110

dz dy dx

=

81T8.

R

Thereforez = (Mxy/M) = 4/3 and the center of mass is (x, y, z) = (0,0,4/3).

•

When the density of a solid object or plate is constant (as in Example 1), the center of mass is called the centroid of the object. To f"md a centroid, we set 8 equal to I and proceed to find x, y, and z as before, by dividing first moments by masses. These calculations are also valid for two-dimensional objects.

EXAMPLE 2 Find the centroid of the region in the first quadrant that is bounded above by the line y = x and below by the parabola y = x 2•

870

Chapter 15: Multiple Integrals

y

Solution We sketch the region and include enough detail to detennine the limits of integration (Figure 15.36). We then set 8 equal to 1 and evaluate the appropriate fonnulas from Table 15.1: M =

111xx 2

1 dy dx =

o

11 []y=x 11 y

dx =

y=~

0

(x - x 2 ) dx

=

0

1 6

FIGURE 15.36 The centroid of this region is found in Example 2.

1 12·

From these values of M, Mx , and My, we find

_ x

My =

M =

1/12 1 1/6 = 2"

__ Mx _ 1/15 _ 1 1/6 - 5·

and

y - M -

•

The centroid is the point (1/2,2/5).

Moments of Inertia An object's first moments (Table 15.1) tell us about balance and about the torque the object experiences about different axes in a gravitational field. If the object is a rotating shaft, however, we are more likely to be interested in how much energy is stored in the shaft or about how much energy is generated by a shaft rotating at a particular angular velocity. This is where the second moment or moment of inertia comes in. Think of partitioning the shaft into small blocks of mass .:lmk and let rk denote the distance from the kth block's center of mass to the axis of rotation (Figure 15.3 7). If the shaft rotates at a constant angular velocity of w = dO/dt radians per second, the block's center of mass will trace its orbit at a linear speed of

Vk

=

d dt (rk O)

dO rk dt

=

=

rkw.

The block's kinetic energy will be approximately FIGURE 15.37

To find an integral for the amount of energy stored in a rotating shaft, we first imagine the shaft to be partitioned into small blocks. Each block has its own kinetic energy. We add the contributions of the individual blocks to find the kinetic energy of the shaft.

1A 2 2"/j.mkVk

=

2"1 /j.mk rkw)2 A

(

=

1 2rk 2 /j.mk A 2"W ·

The kinetic energy of the shaft will be approximately ~ 1 2 2 £.. 2"W rk

A

/j.mk·

The integral approached by these sums as the shaft is partitioned into smaller and smaller blocks gives the shaft's kinetic energy: KEshaft =

J

1 22 2"W r dm

=

12/2 2"W r dm.

(1)

The factor

1=

J

r 2 dm

is the moment of inertia of the shaft about its axis of rotation, and we see from Equation (1) that the shaft's kinetic energy is _ 1 2 KEshaft -

2" Iw

.

15.6

z

x

FIGURE 15.38 Distances from dVto the coordinate planes and axes.

Moments and Centers of Mass

871

The moment of inertia of a shaft resembles in some ways the inertial mass of a locomotive. To start a locomotive with mass m moving at a linear velocity v, we need to provide a kinetic energy ofKE = (l/2)mv 2 • To stop the locomotive we have to remove this amount of energy. To start a shaft with moment of inertia I rotating at an angular velocity w, we need to provide a kinetic energy of KE = (l/2)Iw2 • To stop the shaft we have to take this amount of energy back out. The shaft's moment of inertia is analogous to the locomotive's mass. What makes the locomotive hard to start or stop is its mass. What makes the shaft hard to start or stop is its moment of inertia. The moment of inertia depends not only on the mass of the shaft but also on its distribution. Mass that is farther away from the axis of rotation contributes more to the moment of inertia. We now derive a formula for the moment of inertia for a solid in space. If r(x, y, z) is the distance from the point (x,y, z) in D to a line L, then the moment of inertia of the mass !J..mk = 8(xhYk,zk)!J..Vk about the line L (as in Figure 15.37) is approximately !J..h = r2(xh Yk, Zk) !J..mk. The moment of inertia about L of the entire object is

If L is the x-axis, then r2 = y2

Ix

+ z2 (Figure 15.38) and =

JJJ(y2 + z2) 8(x, y, z) dV. D

Similarly, if L is the y-axis or z-axis we have and Table 15.2 summarizes the formulas for these moments of inertia (second moments because they invoke the squares of the distances). It shows the definition of the polar moment about the origin as well.

EXAMPLE 3

Find Ix, Iy , Iz for the rectangular solid of constant density 8 shown in

Figure 15.39. SoLution

The formula for Ix gives

e

x

A

r '~Y a

block

b

FIGURE 15.39 Finding Ix, Iy, and Iz for the block shown here. The origin lies at the center of the block (Example 3).

Ix

=

lc/21b/21a/2(y2 + z2) 8 dx dy dz. -c/2 -b/2 -a/2

We can avoid some of the work of integration by observing that (y2 + z2)8 is an even function of x, y, and z since 8 is constant. The rectangular solid consists of eight symmetric pieces, one in each octant. We can evaluate the integral on one of these pieces and then multiply by 8 to get the total value.

r/

r/

Jor/ Jo Jor / (y2 + z2) 8 dx dy dz = 4a8Jor/ Jo 2

Ix = 8

2 a2

2

2 (y2 + z2) dy dz

M = abel>

872

Chapter 15: Multiple Integrals

TABLE 15.2

Moments of inertia (second moments) formulas

THREE-DIMENSIONAL SOLID About the x-axis:

Ix =

fJJ(y2 + Z2) /l dV

About the y-axis:

Iy =

fJJ(X 2+ z2) /l dV

h

fJJr 2/l dV

6

~

6(z,y,z)

About the z-axis:

About a line L:

=

rex, y, z) = distance from. the point (x, y, z) to line L

TWO-DIMENSIONAL PLATE About the x-axis:

Ix =

About the y-axis:

Iy =

About a line L:

h

ff ff ff r2(x,y) 2

6

y /ldA

~

6(z,y)

2

x /l dA

=

rex, y) = distance from (x,y)toL

/l dA

10 = ff(x 2 + y2)/ldA = Ix + Iy

About the origin (polar moment):

Similarly, and

Iz

_ M (2 a + b2) . -T2

•

EXAMPLE 4 A thin plate covers the triangular region bouoded by the x-axis and the lines x = I and y = 2x in the first quadrant. The plate's density at the point (x, y) is /l(x, y) = 6x + 6y + 6. Find the plate's moments of inertia about the coordinate axes and the origin. y 2

(1,2)

Solution We sketch the plate and put in enough detail to determine the limits of integration for the integrals we bave to evaluate (Figure 15.40). The moment of inertia about the

x-axis is

nf----+--~x

=

J.' [

2xy3

3 y' + 2y 3 lY~2z dx +_

o

FIGURE 15.40 The triangular region covered by the plate in Example 4,

=

[s.x S + 4x'l~

y~O

2

=

12.

=

J.' 0

(4Ox'

+

16x 3) dx

15.6

Moments and Centers of Mass

873

Similarly, the moment of inertia about the y-axis is

Beam A

Iy =

Axis

39 10t 10(2x x 2S(x,y) dy dx = s·

Notice that we integrate y2 times density in calculating Ix and x 2 times density to find IySince we know Ix and Iy, we do not need to evaluate an integral to tmd 10; we can use the equation 10 = Ix + Iy from Table 15.2 instead: BeamB

10 = 12

+ 39 5

=

60

+ 39 5

99 5'

•

Axis

The moment of inertia also plays a role in detennining how much a horizontal metal beam will bend under a load. The stiffness of the beam is a constant times I, the moment of inertia of a typical cross-section of the beam about the beam's longitudinal axis. The greater the value of I, the stiffer the beam and the less it will bend under a given load. That FIGURE 15.41 The greater the polar is why we use I-beams instead of beams whose cross-sections are square. The flanges at moment of inertia of the cross-section of a the top and bottom of the beam hold most of the beam's mass away from the longitudinal beam about the beam's longitudinal axis, the axis to increase the value of I (Figure 15.41). stiffer the beam. Beams A aod B have the same cross-sectional area, but A is stiffer.

Exercises 15.6 Plates of Constant Density 1. Finding a center of mass Find the center of mass of a thin plate of density B = 3 bounded by the lines x = 0, y = x, and the parabola y = 2 - x 2 in the tITst quadrant 2. Finding moments of inertia Find the moments of inertia about the coordinate axes of a thin rectangular plate of constant density B bounded by the lines x = 3 and y = 3 in the rITst quadraot. 3. Finding a centroid Fiod the centroid of the region in the first quadrant bounded by the x·axis, the parabola y2 = 2x, and the lioex+y=4. 4. Finding a centroid Find the centroid of the triangular region cut from the rITSt quadrant by the line x + y = 3. 5. Finding a centroid Fiod the centroid of the region cut from the rITSt quadrant by the circle x 2 + y2 = a 2. 6. Finding a centroid Find the centroid of the region b&ween the x-axis and the archy = sin x, 0:5 x :s; 71'. 7. Finding moments of ioertia Find the moment of inertia about the x·axis of a thio plate of deosity B = I bounded by the circle x 2 + y2 = 4. Then use your result to rmd 1, and 10 for the plate. 8. Finding a moment of inertia Find the moment of inertia with respect to the y·axis of a thin sheet of constant density B = I bouoded by the curve y = (sin2 x)/x 2 and the interval 'TT' :s; x :s; 2w of the x-axis.

9. The centroid of an infinite region Find the centroid of the inrmite region io the second quadrant enclosed by the coordinate axes and the curve y = e~. (Use improper integrals in the massmoment formulas.)

10. The fin! moment of an ioImite plate Find the rITSt moment about the y-axis of a thin plate of density B(x, y) = I covering the 2 infmite region under the curve y = e -x j2 in the first quadrant.

Plates with Varying Density 11. Finding a moment of inertia Find the moment of inertia about the x·axis of a thio plate bounded by the parabola x = y - y2 and the line x + y = if B(x,y) = x + y.

ma..

o

12. Finding Find the mass of a thin plate occupying the smaller region cut from the ellipse x 2 + 4y2 = 12 by the parabola x = 4y2 if B(x, y) = 5x. 13. Finding a center of mass Find the center of mass of a thin triangular plate bounded by the y-axis aod the lines y = x and y = 2 -xifB(x,y) = 6>:+ 3y+ 3. 14. Finding a center of mass and moment of inertia Find the center of mass and moment of inertia about the x-axis of a thin plate bounded by the curves x = y2 and x = 2y - y2 if the density at the point (x. y) is B(x,y) = y + 1. 15. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin rectangular plate cut from the rITSt quadrant by the lioes x = 6 and y = I if B(x,y) = x + Y + 1. 16. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin plate bounded by the line y = I and the parabola y = x 2 if the density is B(x,y) = y + 1. 17. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin plate bounded by the x·axis, the lines x = ± I, and the parabola y = x 2 if B(x,y) = 7y + 1.

874

Chapter 15: Multiple Integrals

18. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the x-axis of a thin rectangular plate bounded by the linesx = 0, x = 20, Y = -I, and y = I if B(x,y) = I + (x/20).

19. Center of mass, moments of inertia Find the center of mass, the moment of inertia about the coordinate axes, and the polar moment of inertia of a thin triangular plate bounded by the linesy = x,y = -x, andy = I ifB(x,y) = Y + I. 20. Center of mass, moments of inertia B(x,y) = 3x 2 + I.

Repeat Exercise 19 for

l---..

y

Solids with Constant Den';ty 21. Moments of inertia Find the moments of inertia of the rectangular solid shown bere with respect to its edges by calculating Ix, [Y' and /z·

25. a. Center of mass Find the center of mass of a solid of constant density bounded below by the paraboloid z = x 2 + y2 and above by the plane z = 4. b. Find the plane z = c that divides the solid into "'" parts of equal

volume. This plane does not pass through the center of mass. 26. Moments A solid cube, 2 units on a side, is bounded by the

c I I

-- b

)-/

/ /

planes x

= ±I,z = ±I,y = 3,

and y

= 5.

Find the center of

mass and the moments of inertia about the coordinate axes.

Y

27. Moment of inertia about aline A wedge like the one in Exercise 22 bas a = 4, b = 6, and c = 3. Make a quick sketch to

a

22. Moments of inertia

The coordinate axes in the figure run through the centroid of a solid wedge parallel to the labeled

edges. Find Ix, Iy, and Iz if a

=

b

=

6 and c

=

4.

check for yourself that the square of the distance from a typical point (x, y, z) of the wedge to the line L: z = O,y = 6 is r2 = (y - 6)2 + z2 Then calculate the moment of inertia of the wedge about L. 28. Moment of inertia about aline A wedge like the one in Exercise 22 bas a = 4, b = 6, and c = 3. Make a quick sketch to check for yourself that the square of the distance from a typical is point (x, y, z) of the wedge to the line L: x = 4, Y = r2 = (x - 4)2 + y2. Then calculate the moment of inertia of the wedge about L.

°

Centroid 81(0,0,0)

c

~

3

a

23. Center of mas. and moment. of inertia A solid ''trough'' of constant density is bounded below by the surface z = 4y2, above by the plane z = 4, and on the ends by the planes x = I and x = -I. Find the center of mass and the moments of inertia with

Solids with varying Den';ty In Exercises 29 and 30, fmd

a. the mass of the solid. b. the center of mass. 29. A solid region in the flIst octant is bounded by the coordinate planes and the plane x + y + z = 2. The density of the solid is B(x,y,z) = 2x.

°

respect to the three axes. 24. Center of ma.. A solid of constant density is bounded below by the plane z = 0, on the sides by the elliptical cylinder x 2 + 4y2 = 4, and above by the plane z = 2 - x (see the accompanying figure).

a. Find>, andy. b. Evaluate the integral Mxy =

1'J -

,,- /

(1j2)~J.2-x

-(1/2)~ 0

zdzdydx

using integral tables to carry out the fmal integration with respect to x. Then divide Mxy by M to verify that z = 5/4.

°

30. A solid in the flIst octant is boWlded by the planes y = and z = andbythesurfacesz = 4 - x 2 andx = y2 (seethe accompanying figure). Its density function is B(x, y, z) = ir:xy, /r; a constant

///

2 /

,/

//

'

X=y2

~

} (2, Yz,O)

Y

15.7 Triple Integrals in Cylindrical and Spherical Coordinates In Exercises 31 and 32, fmel

31.

32.

33.

34.

a. the mass of the solid. b. the center of mass. c. the moments of inertia about the coordinate axes. A solid cube in the fIrSt octant is bounded by the coordinate planes and by the planes x = I,y = I,andz = l.Thedensityof the cube is 8(x,y,z) = x + Y + z + l. A wedge like the one in Exercise 22 has dimensions a = 2, b = 6, and c = 3. The density is 8(x,y,z) = x + l. Notice that if the density is constant, the center of mass will be (0, 0, 0). Mass Find the mass of the solid bounded by the planes x + z = I, x - z = -I, y = 0 and the surface y = The density of the solid is 8(x,y,z) = 2y + 5. Mass Find the mass of the solid region bounded by the parabolic surfaces z = 16 - 2x 2 - 2y2 and z = 2x 2 + 2y2 if the density of the solid is 8(x,y,z) = v'x 2 + y2.

875

~1

L

I

.........v=xi+yJ

- ',

(x,y, z)

-

~ ~~/ tcJn. /

I I I

(h, 0, 0)

Yz.

.y D

I

b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line L c.m . along the z-axis and the line L perpeodicular to the xy-plaoe at the point (h, 0, 0). Let D be the region of space occupied by the body. Then, in

the notation of the fIgure, Theory and Examples The PIII'IIlle1 ADs Theorem Let L,m. be a line through the center of mass of a body of mass m aud let L be a parallel line h units away from L c.m • The Parallel Axis Theorem says that the moments of inertia lc.m. aud h of the body about L,m. aud L satisfY the equation

(2) As in the two-dimensional case, the theorem. gives a quick way to calculate one moment when the other moment and the mass are

known. 35. Proof of the PIII'IIi1e1 ADs Theorem a. Show that the fIrst moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the yz-plane. What does the formula x = M,,/M then tell you?)

15.7

=

ifflv -hil

2

dm.

D

Expand the integrand in this integral and complete the proof. 36. The moment of inertia about a dismeter of a solid sphere of conslant density and radius a is (2/5)ma 2, where m is the mass of the sphere. Find the moment of inertia about a line tangent to the sphere. 37. The moment of inertia of the solid in Exercise 21 about the z-axis is I, = abc(a 2 + b 2 )/3.

•• Use Equation (2) to fmd the moment of inertia of the solid about 1he line parallel to 1he z-axis through the solid's center of mass. b. Use Equation (2) and 1he result in part (a) to fmd the moment of inertia of 1he solid about the line x = 0, y = 2b. 38. If a = b = 6 and c = 4, 1he moment of inertia of the solid wedge in Exercise 22 about the x-axis is I, = 208. Find 1he moment of inertia of the wedge about 1he line y = 4,z = -4/3 (1he edge of the wedge's narrow end).

Triple Integrals in Cylindrical and Spherical Coordinates When a calculation in physics, engineering, or geometry involves a cylinder, cone, or sphere, we can often simplify our work by using cylindrical or spherical coordinates, which are introduced in this section. The procedure for transfonning to these coordinates and evaluating the resulting triple integrals is similar to the transformation to polar coordinates in the plane studied in Section 15.4.

z P(r, 8, z)

o

h

z

Integration in Cylindrical Coordinates We obtain cylindrical coordinates for space by combining polar coordinates in the xy-plane with the usual z-axis. This assigns to every point in space one or more coordinate triples of the form (r, fJ, z), as shown in Figure 15.42.

DEFINmON

Cylindrical coordinates represent a point P in space by ordered

triples (r, fJ, z) in which The cylindrical coordinates of a point in space are r, 6, andz. FIGURE 15.42

1. r and fJ are polar coordinates for the vertical projection of P on the xy-plane 2. z is the rectangular vertical coordinate.

876

Chapter 15: Multiple Integrals

The values of x, y, r, and () in rectangular and cylindrical coordinates are related by the usual equations. Equations Relating Rectangular (x,y, z) and Cylindrical (r, 8, z) Coordinates x

= rcos(),

y

= rsin(),

z = z,

tan() = y/x

z 0=00 ,

randzvary-

zoo

Z=

rand 0 vary

I

In cylindrical coordinates, the equation r = a describes not just a circle in the xy-plane but an entire cylinder about the z-axis (Figure 15.43). The z-axis is given by r = o. The equation () = ()o describes the plane that contains the z-axis and makes an angle ()o with the positive x-axis. And, just as in rectangular coordinates, the equation z = Zo describes a plane perpendicular to the z-axis. Cylindrical coordinates are good for describing cylinders whose axes run along the z-axis and planes that either contain the z-axis or lie perpendicular to the z-axis. Surfaces like these have equations of constant coordinate value:

r=4 y

() =

7T

Plane containing the z-axis

3 z = 2.

r = a,

oand z vary

FIGURE 15.43 Constant-coordinate equations in cylindrical coordinates yield cylinders and planes.

Cylinder. radius 4. axis the z-axis

Plane perpendicular to the z-axis

When computing triple integrals over a region D in cylindrical coordinates, we partition the region into n small cylindrical wedges, rather than into rectangular boxes. In the kth cylindrical wedge, r, () and z change by tJ..rk, tJ..()k, and tJ..Zk, and the largest of these numbers among all the cylindrical wedges is called the norm of the partition. We define the triple integral as a limit of Riemann sums using these wedges. The volume of such a cylindrical wedge tJ.. Vk is obtained by taking the area !J..A k of its base in the rO-plane and multiplying by the height tJ..z (Figure 15.44). For a point (rk, ()k, Zk) in the center of the kth wedge, we calculated in polar coordinates that !J..A k = rk tJ..rk tJ..()k. So tJ.. Vk = tJ..Zk rk tJ..rk tJ..()k and a Riemann sum for lover D has the form n

Sn =

I Volume Differential in Cylindrical Coordinates

dV

=

dzrdrdO

'L I(rk, ()k, Zk) tJ..Zk rk tJ..rk tJ..()k. k=!

The triple integral of a function lover D is obtained by taking a limit of such Riemann sums with partitions whose norms approach zero:

n~~Sn =

Iff D

IdV=

1IIIdzrdrd(). D

Triple integrals in cylindrical coordinates are then evaluated as iterated integrals, as in the following example.

z r !:J.r!:J.O

rM

L

T!:J.z

~~ ~!:JL . __ ~----J ~ - - -

r------1~ !:J.r

EXAMPLE 1 Find the limits of integration in cylindrical coordinates for integrating a function I(r, (), z) over the region D bounded below by the plane z = 0, laterally by the circular cylinder x 2 + (y - 1)2 = 1, and above by the paraboloid z = x 2 + y2. SoLution The base of D is also the region's projection R on the xy-plane. The boundary of R is the circle x 2 + (y - 1f = 1. Its polar coordinate equation is

+ (y - 1)2 = 1 2 x + y2 - 2y + 1 = 1 r2 - 2r sin () = 0 r = 2 sin(). x2

FIGURE 15.44 In cylindrical coordinates the volume of the wedge is approximated by the product a v = az r ar ao.

15.7 Triple Integrals in Cylindrical and Spherical Coordinates

Top Cartesian: z = .x2 Cylindrical: z = ,2

877

The region is sketched in Figure 15.45. We find the limits of integration, starting with the z-limits. A line M through a typical point (r, 0) in R parallel to the z-axis enters D at z = 0 and leaves at z = x 2 + y2 = r2. Next we find the r-limits of integration. A ray L through (r, 0) from the origin enters R at r = 0 and leaves at r = 2 sin O. Finally we find the O-limits of integration. As L sweeps across R, the angle 0 it makes with the positive x-axis runs from 0 = 0 to 0 = 7T. The integral is

+I

iii

fer, 0, z) dV = l1r12sinolr2 fer, 0, z) dz r dr dO.

•

D i-=----~y

L

x

Example 1 illustrates a good procedure for finding limits of integration in cylindrical coordinates. The procedure is summarized as follows.

How to Integrate in Cylindrical Coordinates To evaluate

iff

FIGURE 15.45 Finding the limits of integration for evaluating an integral in cylindrical coordinates (Example l).

fer, 0, z) dV

D

over a region D in space in cylindrical coordinates, integrating first with respect to z, then with respect to r, and finally with respect to 0, take the following steps.

1.

Sketch . Sketch the region D along with its projection R on the xy-plane. Label the surfaces and curves that bound D and R. z

,. - -...... /

z = g2(r, 0)

y x r = h2(O)

2.

Find the z-limits of integration. Draw a line M through a typical point (r, 0) of R parallel to the z-axis. As z increases, M enters D at z = g\(r, 0) and leaves at z = g2(r, 0). These are the z-limits of integration. z z = gir, 0)

y x

878

Chapter 15: Multiple Integrals

3.

Find the r-limits of integration. Draw a ray L through (r, 0) from the origin. The ray enters R at r = hI (0) and leaves at r = h2( 0) . These are the r-limits of integration. z

4.

Find the O-limits of integration. As L sweeps across R, the angle 0 it makes with the positivex-axis runs from 0 = a to 0 = (3. These are the O-limits of integration. The integral is

hrJ!

f(r,O,z)dV=

l

O=f31r=h2(O)lz=g2(r, o)

f(r,O,z)dzrdrdO .

O=a

D

r=h,(O)

z=gk, O)

EXAMPLE 2 Find the centroid (8 = 1) of the solid enclosed by the cylinder x 2 + y2 = 4, bounded above by the paraboloid z = x 2 + y2, and bounded below by the xy-plane. 4

Centroid

R

x

FIGURE 15.46 Example 2 shows how to find the centroid of this solid.

SoLution We sketch the solid, bounded above by the paraboloid z = r2 and below by the plane z = 0 (Figure 15.46). Its base R is the disk 0 ::; r ::; 2 in the xy-plane. The solid's centroid (x, y, :z) lies on its axis of symmetry, here the z-axis. This makes x = y = O. To find :z , we divide the first moment Mxy by the mass M. To find the limits of integration for the mass and moment integrals, we continue with the four basic steps. We completed our initial sketch. The remaining steps give the limits of integration. The z-limits. A line M through a typical point (r, 0) in the base parallel to the z-axis enters the solid at z = 0 and leaves at z = r2. The r-limits. A ray L through (r,O) from the origin enters R at r = 0 and leaves at r = 2. The O-limits. As L sweeps over the base like a clock hand, the angle 0 it makes with the positive x-axis runs from 0 = 0 to 0 = 27T. The value of Mxy is

Mxy

r7T {2 r

-1 1

{27T {2 [ 2]r2

2

Jo Jo Jo

=

z dz r dr dO

_1

27T 2r 5

-

o

0

2 dr dO -

=

Jo Jo

27T [~]2 12 dO

z2

0

_1

-

r dr dO

27T l§.

000

3 dO -_ 327T 3 .

The value of Mis

M

r7T {2 r

=

Jo Jo Jo

=

Jo Jo

{27T {2

{27T {2 [

2

dz r dr dO

r3 dr dO =

=

Jo Jo

z

]r2 0

{27T [ 4]2 ~ 0 dO =

Jo

r dr dO

r7T

Jo

4 dO = 87T.

879

15.7 Triple Integrals in Cylindrical and Spherical Coordinates Therefore,

z P(p,

_ Mxy 321T 1 4 z = M = -3- 81T = "3 '

cP, 0)

•

and the centroid is (0, 0, 4/3). Notice that the centroid lies outside the solid.

z=pcoscp

7----y /

x FIGURE 15.47 The spherical coordinates 1>, and (J and their relation to x, y, z, and r.

Spherical Coordinates and Integration Spherical coordinates locate points in space with two angles and one distance, as shown in Figure 15.47. The first coordinate, p = 1oFl, is the point's distance from t~ origin. Unlike r, the variable p is never negative. The second coordinate, 4>, is the angle OP makes with the positive z-axis. It is required to lie in the interval [0, 1T]. The third coordinate is the angle 8 as measured in cylindrical coordinates.

p,

DEFINITION Spherical coordinates represent a point P in space by ordered triples (p, 4>, 8) in which 1. p is the distance from P to the origin. 2. 4> is the angle oF makes with the positive z-axis (0 ::; 4> ::; 1T). 3. 8 is the angle from cylindrical coordinates (0 ::; 8 ::; 21T).

z

cp

=

CPo,

p and 0 vary

~

On maps of the Earth, 8 is related to the meridian of a point on the Earth and 4> to its latitude, while p is related to elevation above the Earth's surface. The equation p = a describes the sphere of radius a centered at the origin (Figure 15.48). The equation 4> = 4>0 describes a single cone whose vertex lies at the origin and whose axis lies along the z-axis. (We broaden our interpretation to include the xy-plane as the cone 4> = 1T/2.) If 4>0 is greater than 1T/2, the cone 4> = 4>0 opens downward. The equation 8 = 80 describes the half-plane that contains the z-axis and makes an angle 80 with the positive x-axis.

Equations Relating Spherical Coordinates to Cartesian and Cylindrical Coordinates

= p sin 4>, z = p cos 4>, r

FIGURE 15.48 Constant-coordinate equations in spherical coordinates yield spheres, single cones, and half-planes.

EXAMPLE 3 Solution

x = rcos8 = psin4>cos8, y = rsin8 = psin4>sin8,

Find a spherical coordinate equation for the sphere x 2 + y2

(1)

+ (z

- 1)2

We use Equations (1) to substitute for x , y , and z:

x 2 + y2 + (z - 1)2

=

1

+ p2 sin2 4> sin2 8 + (p cos 4> - 1)2 = 1 p2 sin2 4>( cos2 8 + sin2 8) + p2 cos2 4> - 2p cos 4> + 1 = 1 p2 sin2 4> cos2 8

p2

=

Eqs. (I)

2pcos 4>

p = 2 cos 4> .

p> 0

=

1.

880

Chapter 15: Multiple Integrals z x2

+ y2 + (z

2

- 1)2 = 1 p=2cose/>

The angle cP varies from 0 at the north pole of the sphere to 1T/2 at the south pole; the angle (J does not appear in the expression for p, reflecting the symmetry about the z-axis (see Figure 15.49). •

EXAMPLE 4

Find a spherical coordinate equation for the cone z = v'x 2 + y2.

SoLution 1 Use geometry. The cone is symmetric with respect to the z-axis and cuts the first quadrant of the yz-plane along the line z = y. The angle between the cone and the positive z-axis is therefore 1T/4 radians. The cone consists of the points whose spherical coordinates have cP equal to 1T/4, so its equation is cP = 1T/4. (See Figure 15.50.)

p

Use algebra. Ifwe use Equations (1) to substitute for x,y, andz we obtain the

SoLution 2

same result:

z FIGURE 15.49

The sphere in Example 3.

=

P cos cP =

v'x2

+ y2

v'p2 sin2 cP

P cos cP = P sin cP

z

Example 3 p

>

0, sine/> ;;, 0

cos cP = sin cP

e/>=JI 4

cP

•

1T

=

4·

Spherical coordinates are useful for describing spheres centered at the origin, half-planes hinged along the z-axis, and cones whose vertices lie at the origin and whose axes lie along the z-axis. Surfaces like these have equations of constant coordinate value:

y

x

p=4

Sphere, radius 4, center at origin

cp=1T 3

Cone opening up from the origin, making an angle of 71/3 radians with the positive z-axis

(J = FIGURE 15.50

1T 3.

The cone in Example 4.

I Volume Differential in Spherical Coordinates

dV = p2 sin e/> dp de/> dO

z p sin e/>

Half-plane, hinged along the z-axis, making an angle of 71/3 radians with the positive x-axis

When computing triple integrals over a region D in spherical coordinates, we partition the region into n spherical wedges. The size of the kth spherical wedge, which contains a point (Ph CPh (h), is given by the changes liPh li(Jh and liCPk in p, (J, and cpo Such a spherical wedge has one edge a circular arc of length Pk liCPh another edge a circular arc of length Pk sin CPk li(Jh and thickness lipk. The spherical wedge closely approximates a cube of these dimensions when liPh li(Jh and liCPk are all small (Figure 15.51). It can be shown that the volume of this spherical wedge Ii Vk is Ii Vk = pl sin CPk liPk liCPk li(Jk for (Ph CPh (h) a point chosen inside the wedge. The corresponding Riemann sum for a function f(p, cP, (J) is n

Sn

= ~ f(Ph CPh (Jk) Pk2 sin CPk liPk liCPk li(Jk. k=l

As the norm of a partition approaches zero, and the spherical wedges get smaller, the Riemann sums have a limit when f is continuous:

n~ Sn =

111

f(p,

cP, (J) dV =

D

111

f(p,

cP, (J) p2 sin cP dp dcp d(J.

D

In spherical coordinates, we have dV = p2 sin cP dp dcp d(J. FIGURE 15.51

dV

= =

In spherical coordinates

dp· p de/> • p sin e/> dO p2 sin e/> dp de/> dO.

To evaluate integrals in spherical coordinates, we usually integrate first with respect to p. The procedure for finding the limits of integration is as follows. We restrict our attention to integrating over domains that are solids of revolution about the z-axis (or portions thereof) and for which the limits for (J and cP are constant.

15.7 Triple Integrals in Cylindrical and Spherical Coordinates

881

How to Integrate in Spherical Coordinates To evaluate

111

f(p,

cP, 0) dV

D

over a region D in space in spherical coordinates, integrating first with respect to p, then with respect to cp, and finally with respect to 0, take the following steps.

1.

Sketch . Sketch the region D along with its projection R on the ~-plane. Label the surfaces that bound D. z

y x

2.

Find the p-limits of integration. Draw a ray M from the origin through D making an angle cp with the positive z-axis. Also draw the projection of M on the xy-plane (call the projection L). The ray L makes an angle 0 with the positive x-axis. As p increases, M enters D at p = gl(CP, 0) and leaves at p = g2(CP, 0). These are the p-limits of integration. z

(J =

0.-----

y

x

3.

Find the cp-limits of integration. For any given 0, the angle cP that M makes with the z-axis runs from cP = CPmin to cP = CPmax. These are the cP -limits of integration.

882

Chapter 15: Multiple Integrals

Find the a-limits of integration. The ray L sweeps over R as a runs from a to (3. These are the a-limits of integration. The integral is

4.

hrJ!

f(p, cp, a) dV

lI= a

D

p

R

FIGURE 15.52 Example 5.

4>=4>mm

II)

f(p, cp, a) p2 sin cp dp dcp dO.

p=g,(4), II)

EXAMPLE 5

z

x

1

1l=f314>=4>maxlP=gz(q"

=

L

Y

The ice cream cone in

::5

Find the volume of the "ice cream cone" D cut from the solid sphere I by the cone cp = 1T/3.

SoLution The volume is V = fffvp2 sin cp dp dcp dO, the integral of f(p, cp, a) = I over D . To find the limits of integration for evaluating the integral, we begin by sketching D and its projection R on the xy-plane (Figure 15.52). The p-limits of integration . We draw a ray M from the origin through D making an angle cp with the positive z-axis. We also draw L, the projection of M on the xy-plane, along with the angle athat L makes with the positive x-axis. Ray M enters D at p = 0 and leaves at p = 1. The cp-limits of integration. The cone cp = 1T/3 makes an angle of 1T/3 with the positive z-axis. For any given a, the angle cp can run from cp = 0 to cp = 1T/3. The a-limits of integration. The ray L sweeps over R as a runs from 0 to 21T. The volume is

r

7T

{7T/3 [p3]1

{27T {7T/31 '3 sin cp dcp dO

=

Jo Jo

=

/ 1 + -1) dO = --coscp ]7T 3dO = 127T (-10 27T [1 3 63 0 0

3

0

sin cp dcp dO =

Jo Jo

1 -(21T) = 1T .

6

3

•

EXAMPLE 6 A solid of constant density 8 = 1 occupies the region D in Example 5. Find the solid's moment of inertia about the z-axis. SoLution

In rectangular coordinates, the moment is

In spherical coordinates, x 2 + y2 = (p sin cp cos 0)2 Hence,

+ (p sin cp sin a?

=

p2 sin2 cpo

For the region in Example 5, this becomes

=

1r sJo

7T

(

1

-2 + 1 +

SJo{27T 245 dO = 241 (21T)

1 1) 1 24 - '3 dO =

=

1T

12'

•

15.7 Triple Integrals in Cylindrical and Spherical Coordinates

883

Coordinate Conversion Formulas C'YuNDRICAL TO

SPHERICAL TO

SPHERICAL TO

REcTANGULAR

RECTANGULAR

CYLINDRICAL

X = rcosO y = rsinO z=z

= psincpcosO

r=psincp

Y = psincpsinO

z=pcoscp

z=pcoscp

0=0

X

Corresponding fonnulas for dV in triple integrals:

dV= dxdydz = dzrdrdO = ; sin cp dp dcp dO In the next section we offer a more general procedure for detennining dV in cylindrical and spherical coordinates. The results, of course, will he the same.

Exercises 15.7 Evaluating Integrals in Cylindrical Coordinates Evaluate the cylindricsl coordinate integrals in Exercises l--{j.

1.

J.

'~J.1J~ dzrdrd6

2.

J.'~ {'/2~ {3+24,.' dz r dr d6

4.

OOr

3. 5.

Jo Jo J.'~J.'Jl~3 dzrdrd6

6.

J.'~

00,.1/3

o

1 {11 o Jo -1/2

/2(r'sin'6

J.'~J.'f,v'i8::;> dzrdrd6 J.~ {'/~ {'~ z dz r dr d6 0

Jo l-v'h'-

Changtng the Order of Integration In Cylindrical Coordinates

9. 10.

J.'~ {' {'/3 r'drdzd6 o

Jo Jo

0

0

J.

1J.V,Jo{'~ r-2

I 1J.'wJo{1+~"4rdrd6dz -I 0

(r' cos' 6 + z')r d6 dr dz

J.'l~/"~ o

8.

(rsin6

•• dzdrd6

b. drdzd6

Finding Iterated Integrals in Cylindrical Coordinates 13. Give the limits of integratioo for evaluatiog the integral

JJJ

f(r, 6, z) dz r dr d6

+ I)rd6dzdr

14. Convert the integrsl

[J.vq[(x' +

a. dz drd6

y') dz dx dy

to an equivalent integrsl in cylindricsl coordinates and evaluate

the result In Exercises 15--20, set up the iterated integral for evaluatiog fffnf(r, 6, z) dz r dr d6 over the given regionD.

15. D is the right circular cylinder whose base is the circle r = 2 sin 6 in the xy.plane and whose top lies in the plane z = 4 - y.

z

0

11. Let D be the region bounded below by the plane z = 0, above by + y' + z' = 4, aod on the sides by the cylinder the sphere + y' = 1. Set up the triple integrals in cylindricsl coordi· nates that give the volume of D using the following orders of in· tegration.

x'

c. d6dzdr

as an iterated integral over the region that is bounded below by the plaoe z = 0, on the side by the cytioder r = cos 6, aod on top by the paraboloid z = 3r'.

+ z')dzrdrd6

The integrals we have seen so far suggest that there are preferred or· ders of integration for cylindricsl coordinates, but ot1rer orders usually work well and are occasionally easier to evaluate. Evaluate the integrsls in Exercises 7-10.

7.

integrals in cylindricsl coordinates that give the volwne of D using the following orders of integration.

z=4-y

x'

b. dr dz d6

c. d6 dz dr

12. Let D be the regioo bounded below by the cone z = y'x' + y' and above by the paraboloid z = 2 y'. Set up the triple

x' -

/

884

Chapter 15: Multiple Integrals

16. D is the right circular cylinder whose base is the circle r = 3 cos 0 and whose top lies in the plane z = 5 - x .

z

z z=5-x

y

"r=3cos8

x

x

y=x

17. D is the solid right cylinder whose base is the region in the xy-plane that lies inside the cardioid r = I + cos 0 and outside the circle r = I and whose top lies in the plane z = 4.

z

x

Evaluating Integrals in Spherical Coordinates Evaluate the spherical coordinate integrals in Exercises 21- 26.

rrr {7r {2sin'"

21.

Jo Jo Jo

24.

Jo Jo Jo

25.

1211'111'/312

r=1+cos8

18. D is the solid right cylinder whose base is the region between the circles r = cos 0 and r = 2 cos 0 and whose top lies in the plane z = 3 - y.

r 7r/2 {7r e5p3sin 1> dp d1> dO o

z

p2 sin 1> dp d1> dO

3

3p2 sin 1> dp d1> dO

sec'"

0

{27r {7r/4 {sec'"

26.

Jo Jo Jo

(p cos 1» p2 sin 1> dp d1> dO

Changing the Order of Integration in Spherical Coordinates The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals in Exercises 27- 30. 27.

121-11' 111'11'/4/2 0

p3 sin 21> d1> dO dp

o

19. D is the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = I and whose top lies in the plane z = 2 - y .

z

1 11'/312 esc "'1211'p2 sin 1> dO dp d1> 11'/6 esc'" 0

29.1 11 1 7r

z=2-y

2

28.

ix I

7r /\2P sin 3 1> d1> dO dp

30. {7r/2 {7r/212 5p4 sin31> dp dO d1> J7r/6 J- 7r/2 esc'" 31. Let D be the region in Exercise 11. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.

a. dp d1> dO x

20. D is the prism whose base is the triangle in the xy-plane bounded by the y-axis and the lines y = x and y = I and whose top lies in the plane z = 2 - x.

b. d1> dp dO

v'

32. Let D be the region bounded below by the cone z = x 2 + y2 and above by the plane z = 1. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.

a. dp d1> dO

b. d1> dp dO

885

15.7 Triple Integrals in Cylindrical and Spherical Coordinates Finding Iterated Integrals in Spherical Coordinates In Exercises 33- 38, (a) find the spherical coordinate limits for the in-

volume of D as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find V.

tegral that calculates the volume of the given solid and then (b) evaluate the integral.

41. Let D be the smaller cap cut from a solid ball of radius 2 units by a plane I unit from the center of the sphere. Express the volume of D as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals.

33. The solid between the sphere p = cos 4> and the hemisphere p = 2,z ~

°

z p

= coscf> _

42. Express the moment of inertia Iz of the solid hemisphere x 2 + y2 + z2 ::s I, z ~ 0, as an iterated integral in (a) cylindrical and (b) spherical coordinates. Then (c) find I z •

_ _

1-

Volumes Find the volumes of the solids in Exercises 43-48. 43.

x

y

34. The solid bounded below by the hemisphere p = I, Z above by the cardioid of revolution p = I + cos 4>

44.

z

z ~

0, and

z p

=

1

x x

45.

46.

z

x

35. The solid enclosed by the cardioid of revolution p = I - cos 4> 36. The upper portion cut from the solid in Exercise 35 by the xy-plane

r= 3cos(J"-..:

37. The solid bounded below by the sphere p = 2 cos 4> and above by 2 + y2 the cone Z =

y

x

' r = -3 cos (J

x

Yx

47.

y

48.

z

z

y

x

x

38. The solid bounded below by the xy-plane, on the sides by the sphere p = 2, and above by the cone 4> = 1T/3

x

y

r

z

=

cos (J

49. Sphere and cones Find the volume of the portion of the solid sphere p::S a that lies between the cones 4> = 1T/3 and 4> = 21T/3 . p=2 x

y

50. Sphere and half-planes Find the volume of the region cut from the solid sphere p ::s a by the half-planes () = and () = 1T/6 in the first octant.

°

51. Sphere and plane Find the volume of the smaller region cut from the solid sphere p ::s 2 by the plane z = 1.

Finding Triple Integrals 39. Set up triple integrals for the volume of the sphere p = 2 in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. 40. Let D be the region in the first octant that is bounded below by the cone 4> = 1T/4 and above by the sphere p = 3. Express the

52. Cone and planes Find the volume of the solid enclosed by the x 2 + y2 between the planes z = I and z = 2. cone z =

Y

53. Cylinder and paraboloid Find the volume of the region bounded below by the plane z = 0, laterally by the cylinder x 2 + y2 = I, and above by the paraboloid z = x 2 + y2.

886

Chapter 15: Multiple Integrals

54. Cylinder and p .....boloids Find 1he volwne of the regioo bouoded below by the paraboloid z ~ + y', laterally by the cylinder x' + y' = l,andaboveby1l1eparaboloidz =x' + y' + 1.

73. Moment of inertia of solid cone Find the moment of inertia of a right circular cooe of base radius I and height I about an axis 1hrough 1I1e vertex parallel to 1I1e base. (Take 8 = 1.)

55. Cylinder and cones Find the volume of the solid cut from the thick-walled cylinder 1:5 + y' :5 2 by the cooes z ~

x'

74. Moment of inertia of solid sphere Find the moment of inertia ofa solid sphere of radius a about a diameter. (Take 8 = 1.)

56. Sphere and cylinder Find 1he volume of the regioo that lies inside the sp\rere + y' + z, ~ 2 and outside the cylinder

75. Moment of inertia of solid cone Find the moment of inertia of a right circular cone of base radius a and height h about its asis. (Hint: Place the cone with its vertex at the origin and its axis aloog 1I1e z-axis.)

x'

±Vx' + y'.

x'

x'+y'=1. 57. Cylinder and planes Find the volume of the regioo enclosed by the cylinder + y' ~ 4 and the planes z ~ 0 and y + z ~ 4.

x'

58. Cylinder and planes Find the volume of the regioo enclosed by the cylinder + y' = 4 and the planes z = 0 and x + y + z = 4.

x'

76. Variable density A solid is bounded on 1I1e top by 1I1e paraboloidz = r',oothebottombytheplanez = O,andonthesidesby 1I1e cylinder r = 1. Find 1I1e center of mass and 1I1e moment of inertia about the z-axis if the density is

a. 8(r, 0, z) = z

b. 8(r, 0, z) = r.

59. Region trapped by paraboloids Find the volume of the regioo bounded above by 1he paraboloid z = 5 y' and below by the paraboloid z = 4x' + 4y'.

77. Variable density A solid is bounded below by the cone z = + y' and above by the plane z = I. Find 1I1e center of mass and the moment of inertia about the z-axis if the density is

60. Paraboloid and cylinder Find the volume of the regioo bounded above by the paraboloid z = 9 y', below by the xy-plane, and lying outside the cylinder x' + y' ~ 1.

78. Variable density A solid ball is bounded by the sphere p Find the moment of inertia about the z-axis if the density is

x' -

x' -

61. Cylinder and sphere Find 1he volume of the region cut from the solid cylinder + y' :5 I by the sphere + y' + z, = 4.

x'

x'

62. Sphere and paraboloid Find the volume of the region bounded above by the sphere + y' + z, ~ 2 and below by the paraboloidz = x 2 + y2.

x'

Average Values 63. Find the average value of the functioo I(r, 0, z) = r over the regioo bounded by the cylinder r ~ I between the planes z ~ -I andz = 1. 64. Find the average value of the function I(r, 0, z) = r over the solid ball bounded by the sphere + z, = 1. (This is 1he sphere + y' + = 1.)

x'

r'

z'

65. Find the average value of the function I(p,
~

p cos

Masses, Moments, and Centroids 67. Center of mas. A solid of constant density is bounded below by the plane z = 0, above by the cone z = r, r O?:: 0, and on the sides by the cylinder r = 1. Find the center of mass. 68. Centroid Find the centroid of thV"gioo in the fl1'St octant that is bounded above by 1I1e cooe z = + y', below by 1I1e plane z = 0, and 00 the sides by 1I1e cylinder + y' = 4 and the planes x ~ 0 and y ~ O.

x'

69. Centroid

x'

Find 1I1e centroid of1l1e solid in Exercise 38.

70. Centroid Find the centroid of 1I1e solid bounded above by 1I1e spherep ~ a and below by the cooe

Vx'

a. 8(r, 0, z) = z

b. 8(r, 0, z) = z'-

a. 8(p,
~

a.

b. 8(p,
79. Centroid of solid .emiellipsoid Show that the centroid of the solid semiellipsoid ofrevolutioo (r'/a') + (z'/h') :5 I,z '" 0, lies on 1I1e z-axis 1hree-eigh1l1s of1l1e way from the base to 1I1e top. The special case h = a gives a solid hemisphere. Thus, the centroid of a solid hemisphere lies on 1I1e axis of symmetry 1hreeeighths of the way from the base to the top. 80. Centroid of solid cone Show that 1I1e centroid of a solid right circular cooe is one-four1h of 1I1e way from 1I1e base to 1I1e vertex. (In general, the centroid of a solid cone or pyramid is one-four1h of1l1e way from 1I1e centroid of1l1e base to 1I1e vertex.) 81. Density of .enter of a planet A planet is in 1I1e shape of a sphere of radius R and total mass M with spherically symmetric

density distribution that increases linearly as one approaches its center. What is the density at the center of this planet if the density at its edge (surface) is tsken to be zero? 82. Mas. of planet's atmosphere A spherical p1anet of radius R bas an atmosphere whose density is p. = l-I<Je ~', where h is the altitude above 1I1e surface of1he planet, I-I<J is 1I1e density at sea level, and c is a positive constant Find the mass of1he p1anet's atmosp\rere.

Theory and Examples 83. Vertical planes in cylindrical coordinate.

a. Show that planes perpendicular to 1I1e x-axis bave equations of the form r

~

a sec 0 in cylindrical coordinates.

b. Show that planes perpendicular to the y-axis bave equations of1l1e form r = b csc O. 84. (Continuation of Exercise 83.) Find an equation of 1I1e form r = 1(0) in cylindrical coordinates for 1he plane ax + by = c, c #' O.

71. Centroid Find 1I1e centroid of 1he region that is bounded above by 1I1e surface z = on the sides by the cylinder r = 4, and below by the xy-plane.

85. Symmetry What symmetry will you fmd in a surface that bas an equatioo of the form r = I(z) in cylindrical coordinates? Give

Find the centroid of the region cut from the solid ball r' + z':5 I by the balf-planes 0 = -'If/3,r '" O,andO = 'If/3, r O?:: O.

86. Symmetry What symmetry will you fmd in a surface that bas an equation of the form p = I(
Yr,

72. Centroid

reasons for your answer.

reasons for your answer.

15.8 Substitutions in Multiple Integrals

15.8

887

I~S_ Ub_S_ ti_ tu_ti_ ·o_ns_l_ ·n_M_u_~~·p~L_ e _In_t~ eg~~_L_ S ------------------------

o

_ (u. v)

G

The goal oftbis section is to introduce you to the ideas involved in coordinate transformations. You will see how to evaluate multiple integrals by substitution in order to replace complicated integrals by ones that are easier to evaluate. Substitutions accomplish this by simplifying the integrand, the limits of integration, or both. A thorough discussion ofmultivanable transformations and substitutions, and the Jacobian, is best left to a more advanced course following a study oflinear algebra.

Substitutions in Double Integrals ----~+-----------~u

o

Cartesian uv-plane

y

x

~ g(u. 0)

y

~

1

h(u, v)

The polar coordinate substitution of Section 15.4 is a special case ofa more general substitution method for double integrals, a method that pictures changes in variables as transformations of regions. Suppose that a region G in the uv-plane is transformed one-to-one into the region R in the xy-plane by equations of the form

x = g(u, v),

y = h(u, v),

as suggested in Figure 15.53. We call R the image of G under the transformation, and G the preimage of R. Any function fix, y) defined on R can be thought of as a function f(g(u, v), h(u, v» dermed on G as well. How is the integral of f(x,y) over R related to the integral of f(g(u, v), h(u, v» over G? The answer is: If g, h, and f have continuous partial derivatives and J(u, v) (to be discussed in a moment) is zero only at isolated points, if at all, then Cartesian xy-p1ane

The equatioos x = g(u,v) andy = h(u,v) allow us to change an integral over a regioo R in the xy-plane into an integral over a regioo G in the uv-plane by using Equatioo (1). FIGURE 15.53

ff

f(x,y)dxdy =

ff

(1)

f(g(u,v),h(u, v»IJ(u, v)ldu dv.

G

R

The factor J(u, v), whose absolute value appears in Equation (I), is the Jacobian of the coordinate transformation, named after German mathematician Carl Jacobi. It measures how much the transformation is expanding or contracting the area around a point in G as G is transformed into R.

DEFINmON The Jacobian determinant or Jacobian of the coordinate transformationx = g(u, v),y = h(u, v) is

J(u, v) =

ax au

ax av

ay au

ay av

ax ay auav

ay ax au av'

-----

(2)

The Jacobian can also be denoted by

a(x,y) J(u, v) = a(u, v) HIsTORICAL BIOGRAPHY

Carl Gustav Jacob Jacobi (1804-1851)

to help us remember how the determinant in Equation (2) is constrocted from the partial derivatives ofx andy. The derivation of Equation (I) is intricate and properly belongs to a course in advanced calculus. We do not give the derivation here.

EXAMPLE 1 Find the Jacobian for the polar coordinate transformation x = r cos 0, y = r sin 6, and use Equation (I) to write the Cartesian integral ffR fix, y) dx dy as a polar integral.

888

Chapter 15: Multiple Integrals

o "f--~,------,

2

G

Solution Figure 15.54 shows how the equations x = r cos 6, y = r sin 6 transfonn the rectangle G: 0 :5 r :5 1,0 :5 0 :5 .,/2, into the quarter circle R bounded by x 2 + y2 = 1 in the first quadrant of the ~-plane. For polar coordinates, we have r and 0 in place of u and v. With x = r cos 6 and y = r sin 0, the Jacobian is

ax J{r,O) =

Cartesian rO-plane

or

oX 00

oy or

oy 00

sin 0

11

f(x,y)dxdy =

R

0=

~

1

+

sin2 0) = r.

Since we assume r ;;,: 0 when integrating in polar coordinates, IJ(r, 0) I = Irl = r, so that Equation (1) gives

I x=rcos8 ~ y=rsin9

y

-rSin°1 = r(cos2 0 r cos 0

= Icoso

11

f(r cos 6, rsinO) rdrdO.

(3)

G

This is the same formula we derived independently using a geometric argument for polar area in Section 15.4. Notice that the integral on the right-hand side of Equation (3) is not the integral of f(r cos 0, r sin 0) over a region in the polar coordinate plane. It is the integral of the product of f(r cos 0, r sin 0) and r over a region G in the Cartesian rO-plane. _ Here is an example of a substitution in which the image of a rectangle under the coordinate transfonnation is a trapezoid. Transformations like this one are called linear trans-

formations. Cartesian xy-plane

EXAMPLE 2

Evaluate

FIGURE 15.54 The equations x =

1o

r cos 8, y = rsin 8 transform G into R.

4:LX~(Y/2)+1

x-y/2

2x - y --dxdy 2

by applying the transformation

2x - Y u=--2-'

y 2

v=-

(4)

and integrating over an appropriate region in the uv-plane.

Solution

We sketch the region R of integration in the xy-plane and identify its boundaries (Figure 15.55).

y

v 2

u=o

v=2

4

u

G

=1

~t--~-L--~.

o

V -

0

1

x=u+v y = 2v )

o \

1

y~O

FIGURE 15.55 The equatioos x = u + v and y = 2v transfonn G into R. Reversing the transformatioo by the equatioos u = {2x - y)/2 and v = y/2 transforms R into G (Example 2).

15.8 Substitutions in Multiple Integrals

889

To apply Equation (I), we need to fmd the corresponding uv-region G and the Jacobian of the transformation. To find them, we first solve Equations (4) for x andy in terms of u and v. From those equations it is easy to see that

x

u + v,

=

y

~

2v.

(5)

We then find the boundaries of G by substituting these expressions into the equations for the boundaries of R (Figure 15.55).

.>y-equations for the boundary of R

= =

x y/2 x (y/2) y=O y=4

+

1

Corresponding uv-equations for the boundary of G

Simplified uv-equations

+v =2v/2 =v u + =(2v/2) + 1=v + 1 2v =0 2v =4

u=O u v=O v=2

u v

=1

The Jacobian of the transformation (again from Equations (5» is

ax au J(u, v) = ay au

ax av ay av

a au(u + v)

a av(u + v)

a au (2v)

a av (2v)

=

I~ ~I

=

2.

We now have everything we need to apply Equation (I):

:L

4

1 o

v

X

J

- (Y/2 +I

2x - y -2-dxdy

1 1"-1 2

V

=

-

x- yf2

ul.J(u,v)ldudv

v- O

u=O

=f[(U)(2) du dv =f

v=u

I

[u

2

]:

dv

=f dv =2.

--c1{--+--~U

• EXAMPLE 3

v=-2u

Evaluate

{' {'-x

-2

10 10

~ (y

- 2x)' dy dx.

Solution

We sketch the region R of integration in the xy-plane and identify its boundaries (Figure 15.56). The integrand suggests the transformation u = x + y and v = y - 2x. Routine algebra produces x and y as functions of u and v:

y

x

=

u

v

3 - 3'

y

2u

=

v

3 + 3'

(6)

From Equations (6), we can find the boundaries of the uv-region G (Figure 15.56).

x=o R -o ~--' \--~~X

.>y-equations for the boundary of R

Corresponding uv-equations for the boundary of G

Simplified uv-equations

y~O

FIGURE 15.56 The equations x ~ (u/3) - (v/3) andy ~ (2u/3) + (v/3) transform G into R. Reversing the transformation by the equations u ~ x + y and v ~ y - 2x transforms R into G (Example 3).

x+y=1 x=O y=O

(1-¥)+(2;+¥)=1 '!_"=O 3 3 2u+,,=0 3 3

u

=I

v=u

v

=-2u

890

Chapter 15: Multiple Integrals The Jacobian of the transfonnation in Equations (6) is

J(u, v) =

ax au

ax

I

I

av

3

3

ay au

ay av

2 3

3

I 3.

I

Applying Equation (I), we evaluate the integral:

=

1'1" o

-2u

(I)

1'£' [I

u'(2v 2 3

dv du = 3

0

u'(2 -v3 3

]V-"

du

v - -2u

• In the next example we illustrate a nonlinear transfonnation of coordinates resulting from simplifying the fonn of the integrand Like the polar coordinatea' transfonnation, nonlinear transfonnations can map a straight line boundary of a region into a curved boundary (or vice versa with the inverse transfonnation). In general, nonlinear transfunnations are more complex to ana1yze than linear ooes, and a complete treatment is left to a more advanced course. y

EXAMPLE 4

Evaluate the integral

2 1----\~-"'--_;,f___:

1, l1/r VX~eVxYaxdy. 2

y

1

Solution

The square root terms in the integt'lll!d suggest that we might simplify the inteand v = vYf;:. Squaring these equatioos, we readily gratioo by substituting u = have u 2 = xy and v 2 = y/x, which imply that u 2if = y2 and u 2/if = x 2. So we obtain the transfonnation (in the same ordering of the variables as discussed before)

~~--~----~--~.x

o

1

2

'\IxY

u x=v

FIGURE 15.57 The region of integration R in Example 4.

y = uv.

and

Let's firat see what happens to the integrand itself under this transfonnatioo. The Jacobian of the transfonnation is

u=1

J-----

J(u, v) =

xy=l

/

v=l

1

2

ax av

ay au

ay av

I

-u

v

u

v if

2u v'

If G is the region of integration in the uv-plane, then by Equation (I) the transfonned

I

y=x ,

double integral under the substitution is

I

0

ax au

U

FIGURE 15.58 The boundaries of the region G correspond to those of region R in Figure 15.57. Notice as we move counterclockwise around the region R, we also move counterclockwise around the region G. The inverse transformation equations u ~ v'i,Y, v ~ vY7i prodoce the regioo G from the region R.

11 ~ R

e VxY ax dy =

11

ve"

~ du dv =

G

11

2ue" du dv.

G

The transfonned integrand function is easier to integrate than the original one, so we proceed to detennine the limits of integratioo for the transfonned integral. The regioo of integratioo R of the original integral in the xy-plane is shown in Figore and v = vYf;:, we see that the image of 15.57. From the substitution equations u = the left-hand boundary xy = I for R is the vertical line segment u = 1,2 ;;,: v ;;,: I, in G (see Figure 15.58). Likewise, the right-hand boundary y = x of R maps to the horizontal line segment v = I, I :5 U :5 2, in G. Finally, the horizontal top boundary Y = 2 of R

'\IxY

15.8 Substitutions in Multiple Integrals

891

maps to uv = 2, 1 ::; v ::; 2, in G. As we move counterclockwise around the boundary of the region R, we also move counterclockwise around the boundary of G, as shown in Figure 15.58. Knowing the region of integration G in the uv-plane, we can now write equivalent iterated integrals:

1P ~ 2

I

ll /y \jx

e vXY dx dy

=

11 2

I

2U /

2ue U dv duo

Note the order of integration.

I

We now evaluate the transformed integral on the right-hand side,

11 2

2U /

u

2ue dv du

=

212 ]~:~/u 212 212 (2 -

=

2[(2 - u)e U

=

2(e 2

= =

vue U

(2e U

du

ue U ) du

-

u)e U du

(e

-

+

+ et:~

Integrate by parts.

e» = 2e(e - 2).

•

Substitutions in Triple Integrals The cylindrical and spherical coordinate substitutions in Section 15.7 are special cases of a substitution method that pictures changes of variables in triple integrals as transformations of three-dimensional regions. The method is like the method for double integrals except that now we work in three dimensions instead of two. Suppose that a region G in uvw-space is transformed one-to-one into the region D in xyz-space by differentiable equations of the form

x

=

g(u, v, w),

y

= h(u, v, w),

z = k(u, v, w),

as suggested in Figure 15.59. Then any function F(x, y, z) defined on D can be thought of as a function F(g(u, v, w), h(u, v, w), k(u, v, w»

= H(u, v, w)

defined on G. If g, h, and k have continuous first partial derivatives, then the integral of F(x, y, z) over D is related to the integral of H(u , v , w) over G by the equation

fff

F(x, y, z) dx dy dz =

D

fff

H(u, v, w) IJ(u , v, w) Idu dv dw.

G

z

w x = g(u, v, w) y = h(u, v, w) z = k(u, v, w) )

u

Cartesian uvw-space

x

Cartesian xyz-space

FIGURE 15.59 The equations x = g(u, v, w),y = h(u, v, w), and z = k(u, v, w) allow us to change an integral over a region D in Cartesian xyz-space into an integral over a region G in Cartesian uyw-space using Equation (7).

(7)

892

Chapter 15: Multiple Integrals

z

Cube with sides parallel to the coordinate axes

The factor J(u, v, w), whose absolute value appears in this equation, is the Jacobian

determinant

J(u, v, w) =

()

r

iJx iJu

iJx iJv

iJx iJw

iJy iJu

iJy iJv

iJy iJw

iJz iJu

iJz iJv

iJz iJw

iJ(x,y,z) iJ(u, v, w)·

Cartesian r()z-space

x = rcos ()

= I yz=z

rsin ()

This determinant measures how much the volume near a point in G is being expanded or contracted by the transformation from (u, v, w) to (x, y, z) coordinates. As in the twodimensional case, the derivation of the change-of-variable formula in Equation (7) is omitted. For cylindrical coordinates, r, 8 , and z take the place of u, v, and w. The transformation from Cartesian r8z-space to Cartesian ~z-space is given by the equations

z

x

= rcos8,

y

=

z=z

rsin 8,

(Figure 15.60). The Jacobian of the transformation is

J(r, 8, z) = x

Cartesian xyz-space

FIGURE 15.60 The equations x = r cos 8, y = r sin 8, and z = z transform the cube G into a cylindrical wedgeD.

iJx iJr

iJx iJ8

iJx iJz

iJy iJr

iJy iJ8

iJy iJz

cos 8 sin 8 0

iJz iJr

iJz iJ8

= r cos2 8

-r sin 8

0

r cos 8

0

0

iJz iJz

+

r sin2 8 = r.

The corresponding version of Equation (7) is

fff

F(x,y,z)dxdydz

fff

=

H(r,8,z)lrldrd8dz.

G

D

We can drop the absolute value signs whenever r ;:::: O. For spherical coordinates, p,
y = p sin
z

=

pcos
(Figure 15.61). The Jacobian of the transformation (see Exercise 19) is

J(p,
=

iJx iJp

iJx iJ
iJx iJ8

iJy iJp

iJy iJ
iJy iJ8

iJz iJp

iJz iJ
iJz iJ8

= p2 sin
The corresponding version of Equation (7) is

fff D

F(x,y, z) dx dy dz

=

fff G

H(p,
15.8 Substitutions in Multiple Integrals

I)

893

Cube with sides parallel to the coordinate axes x=psinc/>cosl)

y=psinc/>sinl) z=pcosc/> )

I)

p

Cartesian pc/>I)-space

x

Cartesian xyz-space

FIGURE 15.61 The equations x = p sin cos 8, y = p sin sin 8, and z = p cos transfonn the cube G into the spherical wedge D.

We can drop the absolute value signs because sin is never negative for 0 ::::; ::::; 7T. Note that this is the same result we obtained in Section 15.7. Here is an example of another substitution. Although we could evaluate the integral in this example directly, we have chosen it to illustrate the substitution method in a simple (and fairly intuitive) setting.

w

EXAMPLE 5

Evaluate

1o3141X=

y

l

u

(y/ 2l+ 1

(2x y --+ -z) dxdydz 2

x=y/ 2

0

3

by applying the transformation

x=u+ y Y = 2y

u =

z = 3w

(2x - y)/2,

v = y/2,

w

= z/3

(8)

and integrating over an appropriate region in uvw-space.

z

Solution We sketch the region D of integration in xyz-space and identify its boundaries (Figure 15.62). In this case, the bounding surfaces are planes. To apply Equation (7), we need to find the corresponding uvw-region G and the Jacobian of the transformation. To find them, we first solve Equations (8) for x, y, and z in terms ofu, v , and w. Routine algebra gives

I

x

~

x = u

~ /

Y

Front plane:

+ v,

y = 2v,

z = 3w.

(9)

We then find the boundaries of G by substituting these expressions into the equations for the boundaries of D:

x = ~ + I, ory = 2x - 2

FIGURE 15.62 The equations x = u + v, y = 2v, and z = 3w transfonn G into D . Reversing the transfonnation by the equations u = (2x - y)/ 2, v = y/ 2, and w = z/3 transfonns D into G (Example 5).

xyz-equations for the boundary of D

Corresponding uvw-equations for the boundary of G

+

Simplified uvw-equations

y=4

v = (2v/2) + 1 = v 2v = 0 2v = 4

z=O

3w = 0

u=O u= 1 v=O v=2 w=O

z = 3

3w = 3

w=1

x =y/2 x = (y/2)

y=O

u

+

1

u

+

v = 2v/2 = v

+

894

Chapter 15: Multiple Integrals The Jacobian of the transfonnation, again from Equations (9), is

J(u, v, w)

=

ax au ay au az au

ax av ay av az av

ax aw ay aw az aw

I 0 0

I 2 0

0 0 = 6. 3

We now bave everything we need to apply Equation (7):

1 1

4 :£X- (Y/2)+ 1

3

00x=y/2

(2x---y + -z) dxdydz 23

= l'121\u + w)lJ(u,v,w)ldudvdw = [lJ(u+w)(6)dudvdw=61'1'[~2 +uwl>vdw =

61'1' (! + w) dvdw = 6[ [¥ + vw

=

6[w + w2l~ = 6(2) = 12.

l:

dw = 6[(1 + 2w) dw

•

Exercises 15.8 Jacoblans and Transfonned Regions In the Plane 1. a. Solve the system u = x - y,

v=2x+y

for x and y in terms of u and v. Then rmd the value of the Jacobian a(x, y)/a(u, v). b. Find the image under the transformatioo u = x - y, v = 2x + Y of the triangolarregioo with vertices (0, 0), (I, I), and (I, -2) in the xy-plaoe. Sketcb the traosformed regioo in the uv-plane. 2. a. Solve the system

u= x

+ 2y,

v=x-y

for x and y in terms of u and v. Then rmd the value of the Jacobiaoa(x,y)/a(u, v). b. Find the image uoder the traosformatioo u ~ x + 2y, v ~ x - y of the triaogu1ar region in the xy-plaoe bouoded by the lines y ~ 0, y ~ x , aod x + 2y = 2. Sketch the traosformed region in the uv-plane. 3. a. Solve the system

u = 3x

by the x-axis, the y-axis, and the line x transformed region in the uv-p1ane.

+y

= I. Sketch the

4. .. Solve the system

u

v = -x + y

2x - 3y,

~

for x aod y in terms of u aod v. Then rmd the value of the Jacobiao a(x,y)/a(u, v). b. Find the image uoder the traosformation u = 2x - 3y, v ~ -x + y of the parallelogram R in the xy-plaoe with bouodsries x = -3, x = O,y = x, aod y ~ x + I. Sketch the traosformed region in the uv-plaoe.

Substitutions In Double Integrals 5. Evaluate the integral 4:£X=(Y/2)+1 2x - y /.o ;;- y/2

--dxdy 2

from Example I directly by integration with respect to x and y to confmn that its value is 2. 6. Use the traosformation in Exercise I to evaluate the integral

+ 2y,

v = x

+ 4y

for x aody in terms ofu and v. Then rmd the value of the Jacobiaoa(x,y)/a(u, v). b. Find the image uoder the traosformatioo u ~ 3x + 2y, v ~ x + 4y of the triaogo1arregion in thexy-plane bounded

JJ(2x 2 - xy -

y2) dx dy

R

for the regioo R in the rIrSt qoadrant bounded by the lines y = -2x + 4,y = -2x + 7,y = x - 2, aody = x + I.

15.8 Substitutions in Multiple Integrals 7. Use the transfonnation in Exercise 3 to evaluate the integral

16. Use the transformation x = u' - v',y = 2uvto evaluate the integral

t

//(3X' + 14xy + 8y') dx dy

8. Use the transformation and para1IeiogramR in Exercise 4 to evaluate the integral

(Hint Show that the image of the triangular region G with vertices (0, 0), (I, 0), (I, I) in the uv-plane is the region ofintegration R in the xy-plaue defmed by the limits of integration.)

Finding Jacobians a. x = ucosv, y = usinv

R

9. Let R be the region in the frrst quadrant of the xy-plane bounded by the byperbolas xy = I, xy = 9 and the lines y = x, y = 4x. Use the transformation x = u/v,y = uv with u > 0 and v > 0 to rewrite

// (Ix +

V;;)dxdy

R

as an integral over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G. 10. a. Find the Jacobian of the transfonnation x = u, y = uv and sketch the region G: 1 ::s; u ::s; 2, 1 ::s; uv ::s; 2, in the uv-plane.

b. Then use Equation (I) to transform the integral

b. x = usinv, Y = ucosv. 18. Find the Jacobiana(x,y, z)/a(u, v, w) of the transfonnation

a. x = u cos v, y = u sin v, z = w b. x = 2u - I, Y = 3v - 4, z = (1/2)(w - 4). 19. Evaluate the appropriate determinant to show that the Jacobiao of the transformation from Cartesian p. 20. Substitutions in single integrals How can substi1utions in single defInite integrals be viewed as transfonnations of regions? What is the Jacobian in such a case? lllustrate with an example.

Substitutions In Triple Integrals 21. Evaluate the integral in Example 5 by integrating with respect to x,y,andz.

[[~dYdx into an integral over G, and evaluate both integrals. 11. Polar moment of inertia of an elliptical plate A thin plate of constant density covers the region bounded by the ellipse x'/a' + y'/b' = I, a > 0, b > 0, in the xy-plane. Find the frrst moment of the plate about the origin. (Hint: Use the transformationx = arcos8,y = brsin8.) 12. The area of an ellipse The area '/Tab of the ellipse x'/a' + y'/b' = 1 can be found by integrating the function f(x, y) = I over the region bounded by the ellipse in the xy-plane. Evaluating the integral directly requires a trigonometric substi1ution. Au easier way to evaluate the integral is to use the transformation x = au, Y = bv and evaluate the transformed integral over the disk G: + s 1 in the uv-plaue. Find the area this way.

22. Volume of an ellipsoid

= au, y = 00, and z = cwo Then find the volume of an appropriate region in uvw-space.)

23. Evaluate

over the solid ellipsoid

x2

=

(x 2

24. Let D be the region in xyz-space defmed by the inequalities

1

15. Use the transformation x = u/v,y = uv to evaluate the integral som

1

+ y2) dx dy + 1414~ (x 2 + y2) dx dy. 2

y/4

= cw. Then integrate over an

v to evaluate the

by lust writing it as an integral over a region G in the uv-plane.

Y

abc

= au, y = bv, and z appropriate region in uvw-space.)

' [,,+4)/2 y'(2x _ y)e(2r-y )' dx dy ]. o }yf2

11l/y

z2

(Hint: Let x

by frrst writing it as an integral over a region G in the uv-plane.

+ (i/2)v,y

y2

2+2+2~1.

13. Use the transformation in Exercise 2 to evaluate the integral

].'/3l'-2y(x + 2y)e,,-r) dx dy

Find the volume of the ellipsoid

(Hint: Let x

u' v'

14. Use the transformation x = u integral

Yx' + y'dydx.

17. Find the Jacobiana(x,y)/a(u, v) of the transformation

//2(X - y)dxdy.

2

{2VFx

10 10

R

for the region R in the fIrst quadrant bounded by the lines y = -(3/2)x + I, y = -(3/2)x + 3, y = -(I/4)x, and y = -(1/4)x + I.

895

~

x

~

2,

0

~

xy

~

2,

0

~

z

~

1.

Evaluate

ff/(x'y

+ 3xyz) dx dy dz

D

by applying the transfonnation

u

= x,

v

= xy,

w

= 3z

and integrating over an appropriate region G in uvw-space.

896

Chapter 15: Multiple Integrals

25. Centroid of a solid semiellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base toward the top, show, by traosforming the appropriate integrals, that the center of mass of a solid semiellipsoid (x 2/a 2) + (y2/b 2) + (z2/e2) '" I, Z '" 0, lies on the z-axis three-eighths of the way from the base toward the top. (You cao do this without evaluating aoy of the integrals.)

Chapter

iI!"3

26. Cylindrical shell. In Section 6.2, we learned how to rmd the volume of a solid of revolution using the shell method; namely, if the region between the curve y = f(x) aod the x-axis from a to b (0 < a < b) is revolved about the y-axis, the volume of the 2'1fxf(x) dx. Prove that rmding volumes by resulting solid is using triple integrals gives the same result. (Hint: Use cylindrical coordinates with the roles ofy andz cbanged.)

1:

Questions to Guide Your Review

1. Derme the double integral of a function of two vatiables over a

7. How are double and triple integrals in rectangular coordinates

2. How are double integrals evaluated as iterated integrals? Does the

used to calculate volumes, average values, masses, moments, and centers of mass? Give examples.

order of integration matter? How are the limits of integration de-

8. How are triple integrals dermed in cylindrical and sphetical coor-

bounded region in the coordinate plane.

termined? Give examples.

3. How are double integrals used to calculate areas and average values. Give examples.

dinates? Why might one prefer working in one of these coordinate systems to worldng in rectangular coordinates? 9. How are triple integrals in cylindrical and sphetical coordinates evaluated? How are the limits of integration fouod? Give examples.

4. How can you change a double integral in rectangular coordinates into a double integral in polar coordinates? Why might it be worthwhile to do so? Give an example.

10. How are substitutions in double integrals pictured as transformations of two-dimensional regions? Give a sample calculation.

S. Derme the triple integral of a function f(x, y, z) over a bounded region in space.

11. How are substitutions in triple integrals pictured as transformations ofthree-dimensiona\ regions? Give a sample calculation.

6. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example.

Chapter

ilM

Practice Exercises

Evaluating Double Iterated Integrals

Areas and Volumes Using Double Integrals

In Exercises 1-4, sketch the region of integration and evaluate the double integral.

13. Area between line and parabola Find the area of the region enclosed by the line y = 2x + 4 and the parabola y = 4 - x 2 in the xy-plane.

['0 {1/Y

1.

3.

J1 Jo ye" dx dy f2:Lv9=4t' ' tdsdt /, o

-V9-4t1

2./,1/," 4.

eY/r dydx

1J2/,ovY

vY xydxdy

In Exercises 5-8, sketch the region of integration and write an equivalent integral with the order of integration reversed. Then evaluate both integrals.

4:Lr,-4)f2

s. /, o -V4=Y 7.

' /,o

dxdy

6.

-V9-4,,'

Jo iX2

{2 {,-r'

f2

:Lv9=4Y' ydxdy

[' {\.;;. dy dx

8.

Jo Jo

2xdydx

Evaluate the integrals in Exercises ~12.

rz h~

9. {'

8

11. /, o

4 cos (x 2 ) dx dy

dydx ~ y4 + 1 2

h

10. 12.

{2 (1 er' dx dy

h~

/,1h 1

2'If sm . 'IfX 2 dx dy o"o/Y x2

14. Area bonnded by lines and parabola Find the area of the ''triangular" region in the xy-plane that is bounded on the right by the parabola y = x 2 , on the left by the line x + y = 2, and above by the line y = 4. 15. Volume of the region under a p ...aboloid Find the volume under the psraholoid Z = x 2 + y2 above the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xy-plane.

16. Volume of the region under p ...abolic cylinder Find the volume under the parabolic cylinder z = x 2 above the region enclosed by the parabola y = 6 - x 2 and the line y = x in the xy-plane.

Average Values Find the average value of f(x, y) and 18.

= xy over the regions in Exercises 17

17. The square bounded by the lines x

=

I, Y

=

I in the first

quadrant 18. The quarter circle x 2 + y2 '" I in the fIrst quadrant

Chapter 15

Polar Coordinates Evaluate the integrals in Exercises 19 and 20 by changing to polar coordinates.

19.

2dydx 1 11~ _~ (I + x2 + y2)2

In (x 2 + y2

+

31. Cylindrical to rectangular coordinates

21. Integrating over lemniscate Integrate the function f(x, y) = I/O + x 2 + y2)2 over the region enclosed by one loop of the lemniscate (x 2 + y2)2 - (x 2 - y2) = O. 22. Integrate f(x,y) = I/O + x 2 + y2)2 over and

The triangle with vertices (0, 0), (1, 0),

(I, v3).

b. First quadrant

The first quadrant of the xy-plane.

cos (x

l 710ln 21ln4 25.1 1'1 in

24.

26.

In 5

In6

0

1

x

j I

r ~ 0

r

to (a) rectangular coordinates with the order of integration dz dx dy and (b) spherical coordinates. Then (c) evaluate one of the integrals. 32. Rectangular to cylindrical coordinates (a) Convert to cylindrical coordinates. Then (b) evaluate the new integral.

33. Rectangular to spherical coordinates (a) Convert to spherical coordinates. Then (b) evaluate the new integral.

+ Y + z) dx dy dz

r dzdydx 1 11~ -~ JVx'+y'

e(x+y+z) dz dy dx

- I

x y + (2x - y - z) dz dy dx

0

io(2"ior./2!~3 dz r dr dO,

-~

elxlz2Y 3dydzdx I

Convert

21xy2 dz dy dx 1o 11~1(X'+Y') - (x'+y')

Evaluating Triple Iterated Integrals Evaluate the integrals in Exercises 23- 26.

23.1"1"1"

29. Average value Find the average value of f(x,y,z) = 30xz ~ over the rectangular solid in the first octant bounded by the coordinate planes and the planes x = I, y = 3, z = I.

Cylindrical and Spherical Coordinates

1) dx dy

- I

a. Triangular region

897

30. Average value Find the average value of p over the solid sphere p ::s a (spherical coordinates).

- I

20.111v'-v'JJ=Y2 =Y2

Practice Exercises

Z

Volumes and Average Values Using Triple Integrals 27. Volume Find the volume of the wedge-shaped region enclosed on the side by the cylinder x = -cosy, -7r/2 ::s y ::s 7r/2, on the top by the plane z = - 2x, and below by the xy-plane.

z

z=-2x

34. Rectangular, cylindrical, and spherical coordinates Write an iterated triple integral for the integral of f(x, y, z) = 6 + 4yover the region in the first octant bounded by the cone z = v'x 2 + y2 , the cylinder x 2 + y2 = I, and the coordinate planes in (a) rectangular coordinates, (b) cylindrical coordinates, and (c) spherical coordinates. Then (d) find the integral of f by evaluating one of the triple integrals. 35. Cylindrical to rectangular coordinates Set up an integral in rectangular coordinates equivalent to the integral

{"/2

r.f3 {~ r 3(sin 0 cos

io il il

O)z2 dz dr dO.

Arrange the order of integration to be z first, then y, then x. 36. Rectangular to cylindrical coordinates

1

2iva- x'iv4- x' - y'

o

2 7T

'2

x

y

28. Volume Find the volume of the solid that is bounded above by the cylinder z = 4 - x 2, on the sides by the cylinder x 2 + y2 = 4, and below by the xy-plane.

z

0

The volume of a solid is

dzdydx.

- V4 - x'- y'

a. Describe the solid by giving equations for the surfaces that form its boundary. b. Convert the integral to cylindrical coordinates but do not evaluate the integral. 37. Spherical versus cylindrical coordinates Triple integrals involving spherical shapes do not always require spherical coordinates for convenient evaluation. Some calculations may be accomplished more easily with cylindrical coordinates. As a case in point, find the volume of the region bounded above by the sphere x 2 + y2 + z2 = 8 and below by the plane z = 2 by using (a) cylindrical coordinates and (b) spherical coordinates.

Masses and Moments 38. Finding /z in spherical coordinates Find the moment of inertia about the z-axis of a solid of constant density 8 = I that is bounded above by the sphere p = 2 and below by the cone cf> = 7r/3 (spherical coordinates).

898

Chapter 15: Multiple Integrals

39. Moment of inertia of. "thick" sphere Find the moment of inertia of a solid of constant density 8 bounded by two concentric spheres of radii a and b (a < b) about a diameter. 40. Moment of inertia of an apple Find the moment of inertia about the z-axis of a solid of density 8 ~ I enclosed by the spherical coordinate surface p ~ 1 - cos"'. The solid is the red curve rotated about the z-axis in the accompanying figure.

z

bounded by the line y ~ x and the parabola y ~ x 2 in the xy-plane if the density is 8(x, y) ~ x + 1 . 47. Plate witb variable density Find the mass and first moments about the coordinate axes of a thin square plate bounded by the linesx ~ ± I, Y ~ ± I in the xy-plane if the density is 8(x, y) ~ x 2 + y2 + 1/3. 48. Triangles with same inertial moment Find the moment of inertia about the x-axis of a thin triangular plate of constant density 8 whose base lies along the interval [0, b1on the x-axis and whose vertex lies on the line y = h above the x-axis. As you will see, it

does not matter where on the line this vertex lies. All such triangles have the same moment of inertia about the x-axis. 49. Centroid Find the centroid of the region in the polar coordinate plane defined by the inequalities 0 :5 r :5 3, -7T/3 :5 9 :5 7T/3. 50. Centroid Find the centroid of the region in the first quadrant bounded by the rays 9 ~ 0 and 9 ~ 7T/2 and the circles r ~ I andr ~ 3.

51. a. Centroid Find the centroid of the region in the polar coordinate plane that lies inside the cardioid r = 1 + cos (} and outside the circle r 41. Centroid Find the centroid of the "triangular" region bounded by the lines x ~ 2, Y ~ 2 and the byperbola xy ~ 2 in the xy-plane. 42. Centroid Find the centroid of the region between the parabola x + y2 - 2y ~ 0 and the line x + 2y ~ 0 in the xy-plane. 43. Polar moment Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 ~ 3 bounded by the y-axis and the lines y ~ 2x and y ~ 4 in the xy-plane. 44. Polar moment Find the polar moment of inertia about the center of a thin rectangular sheet of constant density 8 ~ 1 bounded by the lines

a. x

~

1>. x

~

±2, Y ±a, y

~

~

± I in the xy-plane ±b in the xy-plane.

(Hint: Find Ix. Then use the formula for Ix to fmd Iy and add the two to find 10).

~

1.

b. Sketch the region and show the centroid in your sketch.

52. a. Centroid Find the centroid of the plane region defined by the polar coordinate inequalities 0 ~ r ~ a, -a ~ (J ~ a (0 < a::;; '7T).Howdoesthecentroidmoveasa~'7T-? b. Sketch the region fora ~ 57T/6 and show the centroid in your sketch.

Substitutions S3. Show that if u

~

x - y and v

~

y, then

54. What relationship must hold between the constants a, b, and c to mske

45. Inertial moment Find the moment of inertia about the x-axis of a thin plate of constant density 8 covering the triangle with vertices (0, 0), (3, 0), and (3, 2) in thexy-plane. 46. Plate with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin plate

Chapter

iI5

(Hint: l.et s ~ ax + ,By and t ~ oyx ac - b 2. Thenax2 + 2bxy + cy2 =

+ 8y, where (a8 - ,Boy)' 2 2 8 + t .)

~

Additional and Advanced Exerdses

Volumes 1. Sand pile: double and triple integrals The base of a sand pile covers the region in the xy-plane that is bounded by the parabola x 2 + y ~ 6 and the line y ~ x . The height of the sand above the point (x, y) is x 2• Express the volume of sand as (a) a double integral, (b) a triple integral. Then (0) fmd the volume.

3. Solid eylindrioal region between two planes Find the volume of the portion of the solideylinderx 2 + y2:5 I that lies between the planes z ~ o and x + y + z ~ 2.

2. Water in a hemispherioal bowl A hemispherical bowl of radius 5 em is filled with water to within 3 em of the top. Find the volume of water in the bowl.

5. Two paraboloid. Find the volume of the region bounded above by the paraboloid z ~ 3 - x 2 - y2 and below by the paraboloid z ~ 2x 2 + 2y2.

4. Sphere and paraboloid Find the volume of the region bounded above by the sphere x 2 loidz = x 2 + y2.

+ y2 + z2

~ 2 and below by the parabo-

Chapter 15 Additional and Advanced Exercises 6. Spherical coordinates Find the volume of the region enclosed by the spherical coordinate surface p = 2 sin cP (see accompanying figure).

z p=2sincp

Similarly, it can be shown that

1l"

X

V

em(x-t) f(t) dt du dv = lx (x - t)2 em(x-t) f(t) dt. 2 l0 0 0 0

14. Transforming a double integral to obtain constant limits Sometimes a multiple integral with variable limits can be changed into one with constant limits. By changing the order of integration, show that

11

y

X

f(x) ( l g(x - y)f(y) dY ) dx =

7. Hole in sphere A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is

=

11 (1

1

f(y)

g(x - y)f(x) dx) dy

Ire Jo

g( Ix - yl )f(x)f(y) dx dy.

2Jo

Masses and Moments

[2" [v'3 [~

Jo Jo Jl

V= 2

899

15. Minimizing polar inertia A thin plate of constant density is to occupy the triangular region in the first quadrant of the xy-plane having vertices (0, 0), (a, 0), and (a, l/a). What value of a will minimize the plate's polar moment of inertia about the origin?

r dr dz dO.

a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral.

8. Sphere and cylinder Find the volume of material cut from the solid sphere r2 + z2 ::::; 9 by the cylinder r = 3 sin o. 9. Two paraboloids Find the volume of the region enclosed by the surfaces z = x 2 + y2 and z = (x 2 + y2 + 1)/2.

16. Polar inertia of triangular plate Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 = 3 bounded by the y-axis and the lines y = 2x and y = 4 in the xy-plane.

10. Cylinder and surface z = xy Find the volume of the region in the first octant that lies between the cylinders r = 1 and r = 2 and that is bounded below by the xy-plane and above by the surfacez = xy.

17. Mass and polar inertia of a counterweight The counterweight of a flywheel of constant density 1 has the form of the smaller segment cut from a circle of radius a by a chord at a distance b from the center (b < a). Find the mass of the counterweight and its polar moment of inertia about the center of the wheel.

Changing the Order of Integration

18. Centroid of boomerang Find the centroid of the boomerangshaped region between the parabolas y2 = -4(x - 1) and y2 = -2(x - 2) in thexy-plane.

11. Evaluate the integral

1

00

e-ax -

o

dx. Theory and Examples

(Hint: Use the relation

e -ax - e -bx x

e-bx

x

=

l

19. Evaluate b

e-XYdy

a

to form a double integral and evaluate the integral by changing the order of integration.) 12. a. Polar coordinates that

ra Jo

sin {3

Show, by changing to polar coordinates,

[Y';-I- In (x 2 + y2) dx dy = a 2{3 (Ina -

Jy cot{3

where a

where a and b are positive numbers and

t),

20. Show that

> 0 and 0 < (3 < 'TT" /2.

b. Rewrite the Cartesian integral with the order of integration reversed.

over the rectangle Xo ::::; x ::::; xl. Yo ::::; Y ::::; yl. is

13. Reducing a double to a single integral By changing the order of integration, show that the following double integral can be reduced to a single integral: l

x l

u

em(x-t) f(t) dt du = l

x (x - t)em(x-t) f(t) dt.

21. Suppose that f(x,y) can be written as a product f(x, y) = F(x )G(y) of a function of x and a function of y. Then

900

Chapter 15: Multiple Integrals

the integral off over the rectangle R: a ~ x ~ b, c be evaluated as a product as well, by the formula

ff

dx)

!(x,y) dA = ( [ F(x)

(t

~

y

G(y) dy ).

~

d can

a. If you have not yet done Exercise 41 in Section 15.4, do it now to show that

(1)

1=

Joroo e f dy = Y; ;rr.

R

The argument is that

ii

!(x,y) dA

=

R

= =

t ([ dx) t (G(Y{ t ([ dx )G(Y) F(x)G(y)

b. Substitute y = Vi in Equation (2) to show that r( 1/2) = 2I = Y-ii. dy

F(x)dx) dy

F(x)

dy

= ( [ F(X)dx)t G(y)dy. L

(i)

(ii) (iii)

(iv)

Give reasons for steps (i) through (iv).

vertical.

When it applies, Equation (I) can be a time-saver. Vse it to evaluate the following integrals. b.

{1n2 {~/2 eX cosy dy dx

Jo Jo

c.

24. Total electriclll chorge over circul... plate The electrical charge distribution on a circular plate of radius R meters is O"(r, 0) = kr(1 - sinO) coulomb/m2 (k a constant). Integrate 0" over the plate to fmd the total charge Q. 25. A parabolic rain gauge A bowl is in the shape of the graph of z = x 2 + y2 from z = 0 to z = 10 in. You plan to calibrate the bowl to make it into a rain gauge. What height in the bowl would correspond to 1 in. of rain? 3 in. of rain? 26. Water in ...temte dish A parabolic satellite dish is 2 m wide and 1/2 m deep. Its axis of symmetry is tilted 30 degrees from the

{2 {I x dx dy

11 i-I Y 2

22. Let Du! denote the derivative of !(x,y) = (x 2 + y2)/2 in the direction of the unit vector u = U1 i + U2j. Finding average value Find the average value of Du! over the triangular region cut from the fIrSt quadrant by the line x+y=l. b. Average value and centroid Show in general that the average value of Du! over a region in the lJ'-plane is the value of D.! at the cen1roid of the region. 23. The value ofr(I/2) The gamma function, L

a. Set up, but do not evaluate, a triple integral in rectangular coordinates that gives the amount ofwator the satellite dish will hold. (Hint: Put your coordinate system so that the satellite dish is in "standsrd position" and the plane of the water level is slanted) (Caution: The limits of integration are not "nice.") b. What would be the anta1lest tilt of the satellite dish so that it holds no water? 27. An infinite half-cylinder Let D be the interior of the infmite right circular half-cylinder of radius I with its single-end face suspended I unit above the coigin and its axis the ray from (0, 0, I) to 00. Vse cylindrical coordinates to evaluate

iff

z(r2

+ z 2t'/2 dV.

D

rex) = 1,00 1"-1 e-t dt,

I: dx

extends the factorial function from the noonegative integers to other real values. Of particular interest in the theery of differential equations is the number

r(1)

=

2

Chapter

n~

roo t(I/2)-1 e-' dt = roo e-t dt.

Jo

Jo Vt

(2)

28. Hypervolume We have learned that I is the length of the interval [a, b] on the number line (one-dimensional space), ffR I dA is the area ofregionR in thexy-plane (two.
Technology Application Projects

Mathematica/Maple Module:

rue Your Chane..: 1ly the MOllte Carlo Techllique for Numerical Integratioll ill Three Dimemioll' V se the Monte Carlo technique to integrate numerically in three dimensions. Means alld MomeIrls and Exploring New Plotting Techlliques, Part II

V se the method of moments in a form that makes use of geometric symmetry as well as multiple integration.

16 INTEGRATION IN VECTOR FIELDS OVERVIEW In this chapter we extend the theory of integration to curves and surfaces in space. The resulting theory ofline and surface integrals gives powerful mathematical tools for science and engineering. Line integrals are used to find the work done by a force in moving an object along a path, and to find the mass of a curved wire with variable density. Surface integrals are used to find the rate of flow of a fluid across a surface. We present the fundamental theorems of vector integral calculus, and discuss their mathematical consequences and physical applications. In the final analysis, the key theorems are shown as generalized interpretations of the Fundamental Theorem of Calculus.

16.1

Line Integrals

z

To calculate the total mass of a wire lying along a curve in space, or to find the work done by a variable force acting along such a curve, we need a more general notion of integral than was defined in Chapter 5. We need to integrate over a curve C rather than over an interval [a, b]. These more general integrals are called line integrals (although path integrals might be more descriptive). We make our definitions for space curves, with curves in the .:ry-plane being the special case with z-coordinate identically zero. Suppose that f(x, y, z) is a real-valued function we wish to integrate over the curve C lying within the domain off and parametrized by ret) = g(t)i + h(t)j + k(t)k, a ::5 t ::5 b. The values of f along the curve are given by the composite function f(g(t), h(t), k(t)). We are going to integrate this composite with respect to arc length from t = a to t = b. To begin, we first partition the curve C into a finite number n of subarcs (Figure 16.1). The typical subarc has length /lsk. In each subarc we choose a point (Xk, Yh Zk) and form the sum n

Sn

=

,Lf(XhYh Zk) /lSk, k=!

x

FIGURE 16.1 The curve r(t) partitioned into small arcs from t = a to t = h. The length of a typical subarc is !:J.Sk.

which is similar to a Riemann sum. Depending on how we partition the curve C and pick (Xh Yh Zk) in the kth subarc, we may get different values for Sn. If f is continuous and the functions g, h, and k have continuous first derivatives, then these sums approach a limit as n increases and the lengths /lsk approach zero. This limit gives the following definition, similar to that for a single integral. In the definition, we assume that the partition satisfies /lsk -- 0 as n -- 00. If f is defined on a curve C given parametrically by ret) + h(t)j + k(t)k, a ::5 t ::5 b, then the line integral of f over Cis

DEFINITION g(t)i

=

(1)

provided this limit exists.

901

902

Chapter 16: Integration in Vector Fields If the curve C is smooth for a :5 t:5 b (so v = dr/dt is continuous and never 0) and the function f is continuous on C, then the limit in Equation (I) can be shown to exist. We can then apply the Fundamental Theorem of Calculus to differentiate the arc length equation,

l'

s(t) =

Eq. (3) of Section 13.3

IV(T) I dT,

with to

=

a

to expreSS tis in Equation (I) as tis = IV(t) I dt and evaluate the integral of lover C as

11

(X,y,Z)dS =

l

b

l(g(t),h(t),k(t»lv(t)1 dt.

(2)

Notice that the integral on the right side of Equation (2) is just an ordinary (single) definite integral, as defined in Chapter 5, where we are integrating with respect to the parameter t. The formula evaluates the line integral on the left side correctly no matter what parametrization is used, as long as the parametrization is smooth. Note that the parameter t defines a direction along the path. The starting point on C is the position r(a) and movement along the path is in the direction of increasing t (see Figure 16.1).

How to Evaluate a Line Integral To integrate a continuous function I(x, y, z) over a curve C: 1. Find a smooth parametrization of C,

r(t) = g(t)i

+ h(t)j + k(t)k,

a

:5

t

:5

b.

2. Evaluate the integral as

1

I(x,y, z) tis =

t

=

l

b

I(g(t), h(t), k(t))lv(t) I dt.

If I has the constant value I, then the integral of I over C gives the length of C from = b in Figure 16.1.

a to t

EXAMPLE 1 Integrate I(x,y,z) = x - 3y2 origin to the point (I, I, I) (Figure 16.2).

+z

over the line segment C joining the

(I. 1. 1) I I I I I

I I I

\ \ \ \ \

I I I

~

(1. 1.0)

FIGURE 16.2

Example 1.

We choose the simplest parametrization we can think of:

I ~Y

--At'--

X

Solution

The integration path in

r(t) = ti

+ ti + tk,

The components have continuous first derivatives and Iv(t)1 =

VI

2

+

12

+

12 =

Ii + j + kl

=

v'3 is never 0, so the parametrization is smooth. The integral of I

over Cis

1

I(x,y, z) ds = / . ' I(t, t, t)( v'3) dt

=

=

l'

Eq. (2)

(t - 3t 2 + t)v'3 dt

v'3.Jo{' (2t -

3t 2) dt =

v'3 [t 2 - t 3]01 = o.

•

904

Chapter 16: Integration in Vector Fields (See Section 13.3.) The spring's or wire's mass, center of mass, and moments are then calculated with the formulas in Table 16.1, with the integrations in terms of the parameter I over the interval [a, b]. For example, the formula for mass becomes

M =

t

lJ(x(I),y(t), z(t))

~ ( : ) ' + (~)' + (:)' dt.

These formulas also apply to thin rods, and their derivations are similar to those in Section 6.6. Notice how alike the formulas are to those in Tables 15.1 and 15.2 for double and triple integrals. The double integrals for planar regions, and the triple integrals for solids, become line integrals for coil springa, wires, and thin rods.

TABLE 16.1 Mass and moment formulas for coil springs, wires, and thin rods lying along a smooth curve Cin space Mass:

lJ

= 8(x, y, z) is the density at (x, y, z)

First moments about the coordinate planes: Myz =

1

x lJ ds,

Mxy =

1

zlJds

Coordinates of the center of mass:

x=

Myz/M,

y

=

Mxz/M,

Moments of inertia about axes and other lines:

Ix

1 lr

(y2

=

/r=

+ z2)lJ ds,

2 lJdS

r(x,y, z)

= distance from the point (x,y, z) to line L

Notice that the element of mass dm is equal to lJ ds in the table rather than lJ dV as in Table 15.1, and that the integrals are taken over the curve C. EXAMPLE 3 A slender metal arch, denser at the bottom than top, lies along the semicircle y2 + z2 = I, z ;" 0, in theyz-plane (Figure 16.4). Find the center of the arch's mass if the density at the point (x, y, z) on the arch is lJ(x, y, z) = 2 - z.

z

C.In.

-1

Solution We know that x = 0 and y = 0 because the arch lies in the yz-plane with its mass distributed symmetrically about the z-axis. To find z , we parametrize the circle as

r(l)

FIGURE 16.4 Example 3 shows how to find 1he center of mass of a circular arch of variable deosity.

For this parametri7ation,

so ds

=

Ivldt =

dt.

= (cos lli

+

(sin I)k,

o :5 I

:5 'fr.

16.1

Line Integrals

903

Additivity Line integrals have the useful property that if a piecewise smooth curve C is made by joining a finite number of smooth curves C" C2 , ••• , Cn end to end (Section 13.1), then the integral of a function over C is the sum of the integrals over the curves that make it up:

{Ids= {Ids+ {Ids+ ... + (Ids.

~

k (1, 1, 1)

EXAMPLE 2 line segments

(0,0,0)

".I. -~-

~

(3)

~

Figure 16.3 shows another path from the origin to (I, I, 1), the union of x - 3y2 + z over U C2.

C1 and C2. Integrate I(x,y, z) =

C1

Solution We choose the simplest parametrizations for C 1 and C2 we can find, calculating the lengths of the velocity vectors as we go along:

+ Ij, i +j +

C 1:

r(l) = Ii

C2:

r(l) =

0:5 I :5 I; Ik,

Ivl

= VI 2

Ivl

0:5 1:5 1;

+

Y2

12 =

= V02

+

02

+

12 = 1.

With these parametrizations we fmd that FIGURE 16.3 Example 2.

The path of integration in

{

I(x,y,z)ds

=

}C,UC2

{

lei

I(x,y,z)ds + ( I(x,y,z)ds

Eq. (3)

iC2

=

11 I{I, I, 0)Y2 dl + 11 I{I, 1,1)(1) dl

=

11 (I -

=

Y2 [12 _ /3]1 2

31 2

+

0

O)Y2dl + 11 (I +

Eq. (2)

- 3 + 1)(I)dl

[12 _ 21]1 = _ Y2 _ .3.. 2

0

2

2

•

Notice three things about the integrations in Examples I and 2. First, as soon as the components of the appropriate curve were substituted into the formula for I, the integration became a standard integration with respect to I. Second, the integral of lover Cl U C2 was obtained by integrating I over each section of the path and adding the results. Third, the integrals of lover and U 2 had different values.

C C1 C

The value of the line integral along a path joining two points can change if you change the path between them.

We investigate this third observation in Section 16.3.

Mass and Moment Calculations We treat coil springs and wires as masses distributed along smooth curves in space. The distribution is described by a continuous density function Il(x, y, z) representing mass per unit length. When a curve C is parametrized by r(l) = X(I)i + Y(lli + Z(I)k, a :5 1 :5 b, thenx,y, andz are functions of the parameter I, the density is the function /l(X(I), y(1), z(I)), and the arc length differential is given by

ds=~(:Y + (drY + (~ydl.

16.1

Un. Integrals

905

The formulas in Table 16.1 then give

M= 18ds= 1(2-z)cis= !."(2-Sint)dt=21T-2 Mxy

=

1 1 z8 cis

=

z(2 - z) cis

= !."(2Sint - sin2 t) dt _ Z =

Mxy M

=

=

8 - 1T I -2-'21T 2

=

!."(Sint)(2 - sint) dt

8;

1T

8 41T

1T

4'" 0.57.

z

With to the nearest hundredth, the center of mass is (0, 0, 0.57).

•

Line Integrals in the Plane There is an interesting geometric interpretatioo for line integrals in the plane. If C is a smooth curve in the ~-plane parametrized by r(t) = x(t)i + y(t)j, a ,;; t ,;; b, we generate a cylindrical surface by moving a straight line aloog C orthogonal to the plane, holding the line parallel to the z-axis, as in Sectioo 12.6. If z = I(x,y) is a nonnegative cootinuous functioo over a region in the plane cootaining the curve C, then the graph of I is a surface that lies above the plane. The cylinder cuts through this surface, forming a curve on it that lies above the curve C and follows its winding nature. The part of the cylindrical surface that lies beneath the surface curve and above the ~-plane is like a "winding wall" or "fence" standing on the curve C and orthogonal to the plane. At any point (x, y) along the curve, the height of the wallis I(x,y). We show the wall in Figure 16.5, where the "top" of the wall is the curve lying on the surface z = I(x, y). (We do not display the surface formed by the graph of I in the figure, ouly the curve 00 it that is cut out by the cylinder.) From the definitioo

height!(x. y)

y

x

t

=b

Ie /

FIGURE 16.5 The line integral tis gives the area of the portion of the cylindrical surface or ''wall'' beneath z = /(x,y) '"

o. where !lsk~ 0 as n ~ shown in the figure.

00,

we see that the line integral

lei ds is the area of the wall

c.

d.

Exercises 16.1 Graphs of Vector Equations

Match the vector equations in Exercises \-8 with the graphs (a)-(h) given here. b.

L

z

z

z

2

-1

1

y x

x

y

y

906

Chapter 16: Integration in Vector Fields

e.

z

f.

z

z

2 (0,0, I)

r""

1 ~~~

I--

x

(0,0,0) ~

-I

(I, I, I) (0,0,0)

~y

~

y x

y

-2

(1.1,-1)

(b)

(a)

b.

g.

The paths of integration for Exercises 15 and 16.

z

z

16. integrate f(x, y, z) ~ x + vY - z' over the path from (0, 0, 0) to (I, I, I) (see accompanying figure) given by r(l) ~ tk,

C1: y y

1. r(t)

~

+ (1

Ii

2. r(t) ~ i

3. r(t) ~ (2 cos t)i

4. r(t) ~ Ii,

+ (2 sin t)j,

0 '" t '" 2 0 '" t '" I

+ 21k, - I s I '" I (2 cos I)i + (2 sin I)k, 0 s I s

".

Ie (x

t,y

~

(1 - t),z

0 '" t '" I

~

O,from(O, 1,0)to(1,0,0).

Ie

10. Evaluate (x - y + z - 2) tis where C is the straight-line segmentx ~ t,y ~ (1 - t),z ~ l,from(O, I, l)to(I,O, I).

Ie

(XY + Y + z) tis along the curve r(t) ~ 21i 11. Evaluate Ii + (2 - 21)k, 0 '" I '" I.

+

Ie \Ix' + y' tis along the curve r(t) ~ (4 cos I)i + + 3tk, -2". '" I '" 2".. 13. Find the line integral of f(x,y, z) ~ x + Y + z over the straight12. Evaluate (4 sin l)j

line segment from (I, 2, 3) to (0, -I, I). 14. Find the line integral of f(x,y, z) ~ V3/(x' the curve r(t) ~ Ii + ti + tk, I '" t '" 00.

+ y' + z')

over

15. integrate f(x, y, z) ~ x + vY - z' over the path from (0, 0, 0) to (I, I, I) (see accompanying figure) given by Cl:

r(l) ~ Ii

C,:

r(l) ~ i + j + tk,

+ t'j,

~

o~ t

(a cos I)j + (a sin I)k,

~

I, Y

~

~

2'17".

1/2, from (0, 0) to (4, 2).

0 '" t '" I 0 '" t '" I

Ie

\Ix + 2y tis, where Cis a. the straight-line segment x ~ t, Y ~ 4t, from (0, 0) to (I, 4). b. C1 U C,; C1 is the line segment from (0, 0) to (I, 0) and C, is the line segment from (I, 0) to (I, 2). 21. Find the line integral of f(x,y) ~ ye'" along the curve r(t) ~ 4ti - 3tj, -I '" t '" 2. 22. Find the line integral of f(x, y) ~ x - y + 3 along the curve r(t) ~ (cos I)i + (sin I)j, 0 '" I '" 2"..

20. Evaluate

Evaluating Line Integrals over Space Curves 9. Evaluate + y) tis where C is the straight-line segment ~

0 '" t '" I

h. the parabolic curve x ~ I,y ~ I', from (0,0) to (2, 4).

7. r(t) ~ (I' - I)j

x

r(t) ~ Ii + j + k,

Line Integrals over Plane Curves 19. Evaluate leX tis, where Cis a. the straight-line segment x

0 '" t '" 2".

-I '" t '" I

6. r(t) ~ tj + (2 - 2t)k,

~

r(l) ~ ti + k,

C,:

r(t)

-I '" t '" I

5. r(t) ~ Ii + ti + tk,

8. r(l)

C,:

17. integratef(x,y,z) ~ (x + Y + z)/(x' + y' + z') over the path r(t) ~ Ii + ti + tk, 0 < a '" t '" h. 18. integratef(x,y,z) ~ -\Ix' + z'overthecircle

- t)j,

+ j + tk,

0 '" I '" I

23.

Evaluatel:~3 tis, eY

2

where C is the curve x = t , Y = t

3

,

for

1 :s; t :s; 2. 24. Find the line integral of f(x,y) ~ 0/x along the curve r(t) ~ I'i + I'j, 1/2 '" t '" I. 25. Evaluate figure.

Ie (x + 0) tis where C is given in the accompanying y

907

16.2 Vector Fields and Line Integrals: Work. Circulation. and Flux

26. Evaluate

1

2

eX

12

+y + I

tis where C is given in the accompanying

figure.

yz-plaoe. Find the moments of inertis of the rod about the three

coordinate axes. 39. Two springs of constant density A spring of constant density 8 lies along the helix

y

o :s t :s 2w.

r(t) = (cos t)i + (sin t)j + tk, (0, I)' +--E--~ (I, I)

B.

~~~--~--~~~X

(0,0)

(1,0)

In Exercises 27-30, integrate f over tire given curve. 27. f(x,y) = x 3/y, C: y = x 2/2, 0 '" x'" 2 28. f(x,y) = (x + y2)/vT+7, C: y = x 2/2 from (I, 1/2) to (0,0) 29. f(x,y) = x + y, C: x 2 + y2 = 4 in tire first quadrant from (2, 0) to (0, 2)

30. f(x,y) = x 2 - y, C: x 2 + y2 = 4 in the first quadrant from (0, 2) to (v'2, v'2) 31. Find the area of one side of the "winding wall" standing orthogo· nally on the curve y = x 2• 0 :s x :s 2. and beneath the curve on the surface f(x, y) = x +

Findlz •

b. Suppose that you have another spring of constant density 8 that is twice as long as the spring in part (a) and lies along the helix for 0 '" t '" 4,.. Do you expect Ix for the longer spring to be tire same as that for the shorter one, or should it be different? Check your prediction by calcularing Ix for the longer spring. 40. Wire of constant density along the curve

r(l) = (tcost)i

Masses and Moments 33. Mass of a wire Find the mass of a wire that lies along the curve r(t) = (12 - I)j + 2tk, 0 '" t '" I, if tire density is 8 = (3/2)t.

34. Center of ma.. of a cuned wire A wire of density 8(x,y, z) = ISv',Y+2 lies along the curve r(t) = (t 2 - I)j + 2tk, -I '" t '" 1. Find its center of mass. Then sketch the curve and center of mass together.

V2ti

42. Center of mass and moments of inertia for wire with variable density Find the center of mass and the moruents of inertia about the coordinate axes of a thin wire lying along the curve

r(t) = ti + if the density is 8

t 312j +

= I/(t +

~k,

Os t'" 2,

I).

a. Find tis = Iv(t) I dtforthepathr(t) = g(t)i

+ h(t)j +

k(1)k. b. Express the integraod f(g(t), h(t), k(1)) Iv(t) I as a function of the parameter t.

43. f(x,y,z) =

Je f tis using Equation (2) in the text VI

+ 30. 2 +

VI

+ x 3 + Sy3; r(t)

lOy;

r(t) = Ii + t 2j + 3t 2k,

0:51:52

44. f(x,y,z) =

37. Moment of inertia of wire hoop A circular wire hoop of constant density 8 lies along the circlex 2 + y2 = a 2 in thexy-plane. Find the hoop's moment of inertia about the z-axis.

45. f(x,y,z)

16.2

2r

COMPUTER EXPLORATIONS In Exercises 43-46, use a CAS to perfOr03 tire following steps to evaluate the liue integrals.

36. Center of mass of wire with variable density Find the center of mass of a thin wire 1ying along the curve r(l) = Ii + 2tj + (2/3)t 312k, 0 '" t S 2, if the density is 8 = 3'\15+t.

38. Inertia of a slender rod A slender rod of constant density lies along the line segment r(t) = Ij + (2 - 2t)k, 0 '" I '" I, in the

0'" t '" 1.

41. The arch in Eilimple 3 Find Ix for the arch in Example 3.

c. Evaluate

35. Ma•• of wire with variable density Find the mass of a thin wire lying along the curve r(t) = + Yztj + (4 - t 2)k, o s tS 1, if the density is (a) 8 = 3t and (b) 8 = 1.

+ (tsint)j + (2Yz/3)t 312k,

= I lies

Find z and I,.

v'Y.

32. Find the area of one side of the "wall" standing orthogonally on the curve 2x + 3y = 6, 0 '" x '" 6, and beneath the curve on the surface f(x, y) = 4 + 3x + 2y.

A wire of constant density 8

= ti

+

tt

2

j

+ Vtk,

0:51:52

o :5 t

:5

46. f(x,y,z) t'12k,

xVy - 3z 2 ; r(l) 2w =

=

(I

+ ~zlf3

)'/4;

=

(cos2t)i

+ (sin21)j + Stk,

r(l) = (cos2t)i

+ (sin21)j +

0 s I '" 2,.

Vector Fields and Line Integrals: Work, Circulation, and Flux Gravitational and electric forces have both a direction and a magnitude. They are represented by a vector at each point in their domain, producing a vector field. In this section we show how to compute the work done in moving an object through such a field by using a line integral involving the vector field. We also discuss velocity fields, such as the vector

908

Chapter 16: Integration in Vector Fields field representing the velocity of a flowing fluid in its domain. A line integral can be used to find the rate at which the fluid flows along or across a curve within the domain.

Vector Fields

FIGURE 16.6 Velocity vectors of a flow around an airfoil in a wind tunnel.

Suppose a region in the plane or in space is occupied by a moving fluid, such as air or water. The fluid is made up of a large number of particles, and at any instant of time, a particle has a velocity v. At different points of the region at a given (same) time, these velocities can vary. We can think of a velocity vector being attached to each point of the fluid representing the velocity of a particle at that point. Such a fluid flow is an example of a vector field. Figure 16.6 shows a velocity vector field obtained from air flowing around an airfoil in a wind tunnel. Figure 16.7 shows a vector field of velocity vectors along the streamlines of water moving through a contracting channel. Vector fields are also associated with forces such as gravitational attraction (Figure 16.8), and to magnetic fields, electric fields, and also purely mathematical fields. Generally, a vector field is a function that assigns a vector to each point in its domain. A vector field on a three-dimensional domain in space might have a formula like F(x,y,z)

FIGURE 16.7 Streamlines in a contracting channel. The water speeds up as the channel narrows and the velocity vectors increase in length.

= M(x,y,z)i + N(x,y,z)j + P(x,y,z)k.

The field is continuous if the component functions M, N, and P are continuous; it is differentiable if each of the component functions is differentiable. The formula for a field of two-dimensional vectors could look like F(x,y)

= M(x,y)i + N(x,y)j .

We encountered another type of vector field in Chapter 13. The tangent vectors T and normal vectors N for a curve in space both form vector fields along the curve. Along a curve ret) they might have a component formula similar to the velocity field expression vet)

= f(t)i + g(t)j + h(t)k.

If we attach the gradient vector Vf of a scalar function f(x, y, z) to each point of a level surface of the function, we obtain a three-dimensional field on the surface. If we attach the velocity vector to each point of a flowing fluid, we have a three-dimensional field defined on a region in space. These and other fields are illustrated in Figures 16.9-16.15. To sketch the fields, we picked a representative selection of domain points and drew the

FIGURE 16.8 Vectors in a gravitational field point toward the center of mass that gives the source of the field.

FIGURE 16.9 A surface, like a mesh net or parachute, in a vector field representing water or wind flow velocity vectors. The arrows show the direction and their lengths indicate speed.

16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux

909

y

1i / \ \ 1 t i

-! I

f(x, y, z) = c \

FIGURE 16.10 The field of gradient vectors Vj on a surface j(x, y, z) = c.

FIGURE 16.11 The radial field F = xi + yj of position vectors of points in the plane. Notice the convention that an arrow is drawn with its tail, not its head, at the point where F is evaluated.

FIGURE 16.12 unit vectors

F

=

A "spin" field of rotating

(-yi

+ xj)/(x 2 + y2)1/2

in the plane. The field is not defined at the origin.

vectors attached to them. The arrows are drawn with their tails, not their heads, attached to the points where the vector functions are evaluated.

Gradient Fields The gradient vector of a differentiable scalar-valued function at a point gives the direction of greatest increase of the function. An important type of vector field is formed by all the

x

FIGURE 16.13 The flow of fluid in a long cylindrical pipe. The vectors v = (a 2 - r2)k inside the cylinder that have their bases in the xy-plane have their tips on the paraboloid z = a 2 - r2. y

(\

I

...,0:+------ x

WIND SPEED, FIGURE 16.14 The velocity vectors v(t) of a projectile's motion make a vector field along the trajectory.

o

2

4

6

8

Mis

10 12 14 16+

FIGURE 16.15 NASA's Seasat used radar to take 350,000 wind measurements over the world's oceans. The arrows show wind direction; their length and the color contouring indicate speed. Notice the heavy storm south of Greenland.

910

Chapter 16: Integration in Vector Fields gradient vectors of the function (see Section 14.5). We define the gradient field of a differentiable function f(x, y, z) to be the field of gradient vectora

Vf =

of.

of.

of

ax' + ayJ + az k .

At each point (x, y, z), the gradient field gives a vector pointing in the direction of greatest increase of f, with magnitude being the value of the directional derivative in that direction. The gradient field is not always a force field or a velocity field.

EXAMPLE 1

Suppose that the temperature T at each point (x, y, z) in a region of space

is given by

T = 100 - x 2 - y2 - z2, and that F(x, y, z) is defined to be the gradient of T. Find the vector field F. Solution The gradient field F is the field F = VT = -2xi - 2yj - 2zk. At each point in space, the vector field F gives the direction for which the increase in temperature is greatest. _

Line Integrals of Vector Fields In Section 16.1 we defmed the line integral of a scalar function f(x,y, z) over a path e. We turn our attention now to the idea of a line integral of a vectur field F along the curve

e.

Aasume that the vectur field F = M(x,y, z)i + N(x,y, zli + P(x,y, z)k has continuous components, and that the curve C has a smooth parametrization r(t) = g(t)i + h(tli + k(1)k, a ,,; I ,,; h. Aa discussed in Section 16.1, the parametrization r(l) defines a direction (or orientation) along C which we call the forward direction. At each point along the path C, the tangent vector T = dr/ ds = v/ Iv Iis a unit vector tangent to the path and pointing in this forward direction. (The vector v = dr/ dl is the velocity vectur tangent to C at the point, as discussed in Sections 13.1 and 13.3.) Intuitively, the line integral of the vector field is the line integral of the scalar tangential component of F along e. This tangential component is given by the dot product

dr ds'

F'T = F ' -

so we have the following formal definition, where 16.1.

f

= F· T in Equation (I) of Section

DEFINITION Let F be a vectur field with continuous components defined along a smooth curve C parametrized by r(t), a ,,; t ,,; h. Then the line integral of F along Cis

We evaluate line integrals of vectur fields in a way similar to how we evaluate line integrals of scalar functions (Section 16.1).

16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux

Evaluating the Line Integral of F

c: r(t) = g(t)i + h(t)j + k(t)k

= Mi +

Nj

911

+ Pk along

1. Express the vector field F in terms of the parametrized curve Cas F(r(t)) by substituting the components x = g(t),y = h(t), z = k(t) ofr into the scalar components M(x,y,z), N(x,y,z), P(x,y, z) ofF. 2. Find the derivative (velocity) vector dr/dt. 3. Evaluate the line integral with respect to the parameter t, a :5 t :5 b, to obtain

EXAMPLE 2

Evaluate feF • dr, where F(x,y, z) = zi + ~ j - y 2k along the curve C

given by r(t) = t'i + tj + Vt k, 0 :5 t :5 1.

Solution

We have F(r(t)) = Vti + t 3j - t 2k

and dr=2ti+j+_l_k. dt 2Vt Thus,

1 l' l' ( F • dr =

F(r(t))· : 2t 3/2

=

=

+ t3

dt -

tt 3/ 2 ) dt

[G)(~tS/2) + tt4]: = ~~.

•

Line Integrals With Respect to the xyz Coordinates It is sometimes useful to write a line integral ofa scalar function with respect to one of the coordinates, such as fe M dx. This integral is not the same as the arc length line integral feM tis we defined in Section 16.1. To define the new integral for the scalar function M(x,y,z), we specify a vector field F = M(x,y,z)i over the curve C parametrized by r(t) = g(t)i + h(tli + k(t)k, a :5 t :5 b. With this notation we have x = g(t) and dx = g'(t) dt. Then,

dr F·dr = F· dtdt = M(x,y,z)g'(t)dt = M(x,y,z)dx. So we define the line integral of M over C with respect to the coordinate x as

1

M(x,y,z) dx =

1

F·dr,

where

F = M(x,y,z)i.

In the same way, by defming F = N(x,y,z)j, or F = P(x,y, z)k, we obtain the integrals feN dy and feP dz. Expressing everything in terms of the parameter t, we have the following fonnulas for these integrals:

912

Chapter 16: Integration in Vector Fields

1 1 1

M(x,y, z) ax =

N(x,y,z) dy =

P(x,y, z) dz =

l l l

b M(g(t), h(t), k(t)) g'(t) dt

(1)

N(g(t), h(t), k(t» h'(t) dt

(2)

P(g(t), h(t), k(t)) k'(t) dt

(3)

b

b

It often happens that these line integrals occur in combinatioo, aod we abbreviate the nota-

tion by writing

1

M(x,y,z) ax

EXAMPLE 3 ret) = (cos t)i

+

1

N(x,y,z) dy

+

1

Evaluate the line integral

+ (sin t)j + tk, 0

:5

t

P(x,y,z) dz =

1

Max

+ N dy + Pdz.

fe -yax + z dy + 2x dz, where C is the helix

:5 21T.

Solution We express everything in terms of the parameter t, so x = cos t, Y = sin t, z = t, aod ax = - sin t dt, dy = cos t dt, dz = dt. Then,

1 1 2

l-yax

+ zdy + 2x dz

=

=

"

2 "

[(-sintl( -sint)

[Z cos t

= [ZSint =

[0

=

1T.

+ t cos t + 2 cos t] dt

+ t cos t + sin2 t] dt

+ (tsint + cost) +

(t - s~2t)

r

+ (0 + I) + (1T - 0)] - [0 + (0 + 1) + (0 - 0)]

•

Work Done by a Force over a Curve in Space Suppose that the vector field F = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k represents a force throughout a regioo in space (it might be the force of gravity or ao electromagnetic force of some kind) aod that ret) = g(t)i

FIGURE 16.16 The work done along the subarc shown here is approximately F.. T. dah where F. = F(XhY., z.) and T. = T(xh Yh z.).

+ h(t)j + k(t)k,

a :5 t :5 b,

is a smooth curve in the regioo. The formula for the work dooe by the force in moving ao object along the curve is motivated by the same kind ofreasooing we used in Chapter 6 to b derive the formula W = fa F(x) ax for the work dooe by a cootinuous force of magnitude F(x) directed aloog ao interval of the x-axis. For a curve C in space, we define the work dooe by a cootinuous force field F to move ao object along C from a point A to aoother point B as follows. We divide C into n subarcs P'-lP, with lengths fl.s h starting at A aod ending at B. We choose aoy point (Xk, Yh Zk) in the subarc Pk--lPk aod let T(xh Yk, Zk) be the unit tangent vector at the chosen point. The work Wk dooe to move the object aloog the subarc Pk-lPk is approximated by the tangential compooent of the force F(Xk' Yk, Zk) times the arclength fl.sk approximating the distance the object moves along the subarc (see Figure 16.16).

913

16.2 Vector Fields and Line Integrals: Work. Circulation. and Flux

The tota1 work done in moving the object from point A to point B is then approximated by summing the work done along each of the subarcs, so n

W

RJ

n

L w. k-Ll F(x., Yh z.) • T(xh Yh Zk) fls•. i- I RJ

For any subdivision of C into n subarcs, and for any choice of the points (x., Y., each subarc, as n --> 00 and fls. --> 0, these sums approach the line integral

Zk)

within

L F·TtL.. This is just the line integral of F along C, which is defined to be the total work done.

DEFINmON Let Cbe a smooth curve parametrized by r(t), a :5 t:5 b, and F be a continuous force field over a region containing C. Then the work done in moving an object from the point A = r(a) to the pointB = r(b) along Cis

T F

W = L F' T ds =

A t= a

FIGURE 16.17 The wotk done by a force F is the line integral of the scalar

component F . T over the smooth curve fromA toB.

t

F(r(t))·

~~ dt.

(4)

The sigo of the number we calculate with this integral depends on the direction in which the curve is traversed. If we reverse the direction of motion, then we reverse the direction ofT in Figore 16.17 and change the sigo ofF' T and its integral. Using the notations we have presented, we can express the work integral in a variety of ways, depending upon what seems most suitable or convenient for a particular discussion. Table 16.2 shows five ways we can write the work integral in Equation (4).

Different ways to write the work integral for F = /ofi + Nj + Pk over the curve C: r(t) = g(t)i + h(t)j + k(t)k, a :5 t:5 b

TABLE 16.2

The definition

W= LF'TtL. =

=

~

(I, I, 1)

Vector differential form

F·dr

rF.drdt dt

Parametric vector evaluation

Ja

r (Mdx dt

+N

dy dt

1.

=

LMdx+NdY+Pdz

dt

Parametric scalar evaluation Scalar differential form

Find the work done by the force field F = (y - x 2)i + (z _ y2)j + (x - z2)k along the curve r(t) = Ii + t 2j + t 3k, 0 :5 t :5 I, from (0, 0, 0) to (I, I, I) (Figore 16.18).

EXAMPLE 4

Solution

First we evaluate F on the curve r(t): F = (y - x 2)i + (z - y2)j =

FIGURE 16.18 The curve in Example 4.

+ pdz)dt

=

z

(0,0,0)

L

(t 2

-

o

t 2 )i

+ (t 3

-

t 4 )j

+

(x - z2)k

+ (t -

t 6 )k.

Substitute x = t, y=t 2 ,z=t3•

914

Chapter 16: Integration in Vector Fields Then we find drJdt,

~~

~ (Ii +

=

t 2j + t 3k)

=

i + 2tj + 3t 2k.

Finally, we find F . drJdt and integrate from t = 0 to t = 1: F. dr = [(t 3 =

(t 3

t 4)j + (t - t 6 )k]' (i + 2tj + 3t 2k)

-

dt

+ (t - t 6 )(3t 2 )

t4)(2t)

-

2t4 - 2t'

=

+ 3t 3

-

3t 8

so, Work =

l'

(2t 4 - 2t 5 + 3t 3

-

3t 8) dt

• EXAMPLE 5 Find the work done by the force field F = xi + yj + zk in moving an object along the curve C parametrized by r(t) = cos ('ITt) i + t 2j + sin ('ITt) k, 0 :5 t :5 I. Solution

We begin by writing F along C as a function of t, F(r(t)) = cos ('ITt) i

+ t 2j + sin ('ITt) k.

Next we compute drJ dt,

~~

=

-'IT

sin ('ITt) i

+ 2tj + 'IT cos ('ITt) k.

We then calculate the dot product, F(r(t».: =

-'IT

sin ('ITt) cos ('ITt)

+ 2t 3 + 'IT sin ('ITt) cos('lTt)

= 2t 3•

The work done is the line integral

l

a

b

F(r(t». dr dt = {' 2t 3 dt = t4]' =

dt

10

2 0

1.

•

2

Flow Integrals and Circulation for Velocity Fields Suppose that F represents the velocity field of a fluid flowing through a region in space (a tidal basin or the turbine cbamber of a hydroelectric generator, for example). Under these circumstances, the integral ofF' T along a curve in the region gives the fluid's flow along the curve.

DEFINmONS Ifr(t) parametrizes a smooth curve C in the domain of a continuous velocity field F, the flow along the curve from A = rea) to B = reb) is F1ow= IF'Tds.

(5)

The integral in this case is called a flow integral. If the curve starts and ends at the same point, so that A = B, the flow is called the circulation around the curve.

16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux

915

The direction we travel along C matters. If we reverse the direction, then T is replaced by - T and the sign of the integral changes. We evaluate flow integrals the same way we evaluate work integrals.

EXAMPLE 6 r(l) = (cosl)i Solution

A fluid's velocity field is F = xi (sinl)j + Ik,O:5 1:5 7r/2.

+ zj + yk. Find the flow along the helix

+

We evaluate F on the curve,

F = xi

+ zj + yk

= (cos I)i

+

Ij

+

(sin I)k

Substitute x = cos t, Z = t, Y = sin t.

and then f"md dr/ dl:

+ (cosl)j +

= (-sinl)i

:

k.

Then we integrate F' (dr/dl) from I = 0 to I = ; : F' :

= (cosl)(-sinl) = -sinlcosl

+ (I)(COSI) + (sinl)(I)

+ Icost + sinl

so, Flow =

y

- '- \ / / - - ~, //. /~~--+---~~~X ~/ /~...---c

I

l~~b F' ~~ dl =

=[CO~21 + EXAMPLE 7 r(l) = (cos I)i

+

1"12

ISinir

(-sin I cos I

+

I cos I

+

sin I) dl

=(0 + ;) - (t + 0) =; - t.

Find the circulation of the field F = (x - y)i (sin lli, 0 :5 I :5 27r (Figure 16.19).

+ xj

-

around the circle

I j \ " ~

Solution

Onthecircle,F = (x - y)i

FIGURE 16.19 ThevectorfieldFand

+ xj

= (cos I - sinl)i

:

= (-sinl)i

~~

= -sin I cos I

+

+

(cosl)j,and

(cosl)j.

curve r(l) in Example 7.

Then F·

+ sin2 t + cos2 1 1

gives

Jor " F· drdl dl = Jor " (I 2

Circulation =

= [I -

2

s~2lr =

sin I cos I) dl

27r.

As Figure 16.19 suggests, a fluid with this velocity field is circulating counlerclockwise around the circle. _ Not simple, not closed

Not simple,

closed

FIGURE 16.20 Distinguishing curves that are simple or closed. Closed curves are also called loops.

Flux Across a Simple Plane Curve A curve in the xy-plane is simple if it does not cross itself (Figure 16.20). When a curve starts and ends at the same point, it is a closed curve or loop. To find the rete at which a fluid is entering or leaving a region enclosed by a smooth simple closed curve C in the xy-plane,

916

Chapter 16: Integration in Vector Fields

we calculate the line integral over C of F' n, the scalar component of the fluid's velocity field in the direction of the curve's outward-pointing nonnal vectnr. The value of this integral is thef/ux ofF across C. Flux is Latin forf/ow, but many flux calculations involve no motion at all. If F were an electric field or a magnetic field, for instance, the integral of F • n would still be called the flux of the field across C.

DEFINITION If C is a smooth simple closed curve in the domain of a continuous vectnr field F ~ M(x, y)i + N(x, y)j in the plane, and if n is the outwardpointing unit normal vector on C, the flux of F across Cis Flux ofF across C

z For clockwise motion, k X T points outward.

y

~

1

F· n

tis.

(6)

Notice the difference between flux and circulation. The flux of F across C is the line integral with respect to arc length ofF' n, the scalar component ofF in the direction of the outward nonnal. The circulation of F around C is the line integral with respect to arc length of F • T, the scalar component of F in the direction of the unit tangent vectnr. Flux is the integral of the normal component of F; circulation is the integral of the tangential component ofF. To evaluste the integral for flux in Equstion (6), we begin with a smooth parametrization

c x = g(I),

z For counterclockwise motion, T X k points outward.

y x

c

a

:5

t

:5

b,

that traces the curve C exactly once as I increases from a to h. We can f'md the outward unit normal vectnr n by crossing the curve's unit tangent vector T with the vector k. But which order do we choose, T X k or k X T? Which one points outward? It depends on which way C is traversed as I increases. If the motion is clockwise, k X T points outward; if the motion is counterclockwise, T X k points outward (Figure 16.21). The usual choice is n ~ T X k, the choice that assumes counterclockwise motion. Thus, although the value of the integral in Equation (6) does not depend on which way C is traversed, the fonnulas we are about to derive for computing n and evaluating the integral assume counterclockwise motion. In terms of components, n

FIGURE 16.21 To fmd an outward unit normal vector for a smooth simple

curve C in the xy-plane that is traversed counterclockwise as t increases, we take n = T X k. For clockwise motion, we taken ~ k X T.

y = h(I),

ifF = M(x,y)i

~

T X k = (cb; i

ds

+ dy ds

j)

X k = dy i _ cb;

tis

tis

j.

+ N(x,yJi, then F'B

=

dy cb; M(x,y) tis - N(x,y) ds'

Hence,

IF'BdS~ l(M~-N:)tIs~

fMdY-Ncb;.

c We put a directed circle 0 on the last integral as a reminder that the integmtion around the closed curve C is to be in the counterclockwise direction. To evaluate this integral, we express M, dy, N, and cb; in terms of the parameter I and integrate from I ~ a to I ~ h. We do not need to know B or tis explicitly to find the flux.

917

16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux

Calculating Flux Across a Smooth Closed Plane Curve (Flux ofF = Mi

+ Nj across C)

f

=

M dy - N ax

(7)

C

The integral can be evaluated from any smooth parametrizatioo x = g(t),y = h(t), a :5 t :5 b, that traces C counterclockwise exactly ooce.

Find the flux ofF = (x - y)i + xi across the circle x 2 + y2 = I in the xy-plaoe. (The vector field and curve were showo previously in Figure 16.19.)

EXAMPLE 8

The parametrizatioo r(t) = (cost)i + (sint)j,O:5 t:5 2'IT, traces the circle counterclockwise exactly ooce. We can therefore use this parametrizatioo in Equatioo (7). With M = x - y = cost - sint, dy = d(sin t) = cos t dt

Solution

ax

N=x=cost,

d(cost)

=

-sintdt,

=

we find Flux = =

1M r Jo

w

1

2w

dy - N ax =

cos2 t dt =

(cos2 t - sintcost

+

cos t sin t) dt

2w r I + 2cos 2t dt [12 + sin42t]2W Jo =

Eq.(7)

= 'IT.

0

The flux of F across the circle is 'IT. Since the answer is positive, the net flow across the • curve is outward A net inward flow would have given a negative flux.

Exercises 16.2 Vector Fields

.. The slraight-line path Cl: ret)

= Ii + tj + tk,

Find the gradient fields of the functions in Exercises 1-4.

1>. The curved path C,: ret) = ti

+ t'j + t 4k,

1. f(x,y,z)

=

(x' + y' + z')-l/2

2. f(x,y,z) = InVx' + y' + z' 3. g(x,y, z) = e' - In (x' + y') 4. g(x,y, z) = xy + yz + xz 5. Give a fonnula F = M(x, y)i + N(x, y)j for the vector field in the plane that has the property that F points toward the origin with magnitude inversely proportional to the square of the distance from (x, y) to the origin. (The field is not defined at (0, 0).) 6. Give a fonnula F = M(x, y)i + N(x, y li for the vector field in the plane that has the properties that F = 0 at (0, 0) and that at any other point (a, b), F is tangent to the circle x' + y' = + b' and points in the clockwise direction with magnitode IFI = Va' + b'.

a'

c. The path C, U C4 consisting of the line segment from (0, 0, 0) to (I, 1,0) followed by the segment from (I, 1,0) to (I, I, I)

7. F = 3yi + 2xj + 4zk 9. F = Vzi - 2xj + Vyk

8. F = [1/(x' + l)]j 10. F = xyi + yzj + X2k

11. F = (3x' - 3x)i + 3zj + k 12. F

= (y + z)i +

(z

+ xli + (x + y)k z

(0,

o. 0)

~

Line Integrals nf Vector Fields In Exercises 7-12, fmd the line integrals of F from (0, 0, 0) to (I, I, I) over each of the following paths in the accompanying figore.

0 s t '" I 0 '" t '" I

(I, 1.0)

918

Chapter 16: Integration in Vector Fields

Une Integrals with Respect to x, y, and z

Une Integrals in the Plane

In Exercises 13-16, fmd the line integrals along the given path C.

23. Evaluate Ie xy d< + (x + y) dy along the curve y ~ x' from (-I,l)to(2,4).

13.1 (x -

y)d<,w1rereC:x

~ I,y ~ 21 +

l,forO:5 1:5 3

14.1'J;dy, where C: x ~ I,y ~ I', for I '" 1"'2 15.

1

(x' + y') dy, where C is given in the accompanying figore.

24. Evaluate Ie (x - y) d< + (x + y) dy counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, I). 25. Evaluate Ie F • T ds for the vector field F ~ x'i - yj along the curve x ~ y' from (4, 2) to (I, -I). 26. Evaluate Ie F • dr for the vector field F ~ yi - xj counter+ y' ~ I from (1, 0) to (0, I). clockwise along the unit circle

x'

y

Work, Circulation, and Flux in the Plane (3,3)

c

1Vx+Y

~

xyi + (y - x)j

28. Work Find the work done by the gradient of f(x, y) ~ (x + y)' counterclockwise around the circle + y' ~ 4 from (2, 0) to itself.

x'

~~t---~~~--~x (0,0) (3.0)

16.

27. Work Find the work done by the force F over the straight line from (I, I) to (2, 3).

29. Circulation and flu Fl

d<, where C is given in the accompanying figore.

= xi + yj

a. The circle r(l)

c

and

F2

= -yi + xj

around and across each of the following curves.

b.

y

Find the circulation and flux of the fields

~

(cos t)i + (sin I)j,

0 '" I :5 21f

The ellipse r(l) ~ (cos I)i + (4 sin I)j,

0 '" I '" 21f

30. Flu acro•• a circle Find the flux of the fields Fl

~

2xi - 3yj

F,

and

~

+ (x - y)j

2xi

across the circle

o '" I '" 21f.

r(l) ~ (a cos I)i + (a sin I)j, ~~---~x

(0,0)

17. Along the curve r(l) ~ Ii - j + I'k, 0 :5 I :5 I, evaluate each of the following integrals.

a.l(X +

y -z)d<

b.l(x+ y c.

-Z)dy

(x + y - z)dz

b.l

xzdy

c·l

xyzdz

Work In Exercises 1~22, fmd the work done by F over the curve in the

direction of increasing t. 19. F

~

xyi

+ Y.i -

xi

33. F

~

-yi +xj

+ y'j

34. F ~ -y'i

+ x'j

b. The line segment from (I, 0) to (-1,0)

c. The line segment from (I, 0) to (0, -I) followed by the line segment from (0, -I) to (-1,0) 36. Flu across a triangle Find the flux of the field F in Exercise 35 outward across the triangle with vertices (I, 0), (0, I), (-1,0). 37. Find the flow of the velocity field F ~ y'i + 2xyj along each of the following paths from (0, 0) to (2, 4).

a.

b.

y

yzk

r(l) ~ Ii + I'j + Ik, 0 '" I '" I 20. F ~ 2yi + 3xj + (x + y)k r(t) ~ (cost)i + (sinl)j + (1/6)k, 21.

32. F ~ x'i

+ yj

~

a. The upper half of the circle x' + y' ~ I

18. Alongthecurver(l) ~ (cosl)i + (sinl)j - (cosl)k,O '" I '" 1f, evaluate each of the following integrals. xzd<

line segment r,(I) ~ Ii, -a '" I :5 a. 31. F

35. Flow integral. Find the flow of the velocity field F ~ (x + y)i - (x' + y')jalongeachofthefollowingpathsfrom(I,O) to (-1,0) in thexy-plane.

1

a.l

In Exercises 31-34, fmd the circulation and flux of the fieldF around and across the closed semicircular path that consists of the semicircular arch rl(l) ~ (a cos I)i + (a sin I)j, 0 '" I '" 1f, followed by the

y

(2,4)

(2, 4)

0:5 t '" 21f

F~zi+xj+yk

r(t) ~ (sin t)i + (cos I)j + tk,

0 '" I '" 21f

22. F ~ 6zi + y'j + 12xk r(l) ~ (sin I)i + (cos I)j + (1/6)k,

0 '" I '" 21f

cc-~----L-~ X

~CC+"------L-~x

(0, 0)

(0, 0)

2

2

c. Use any path from (0, 0) to (2, 4) different from parts (a) and (b).

16,2 Vector Fields and Une Integrals: Work, Circulation, and Flux 38. Find the circulation of the field F = yi + (x + 2y)j around each of the following closed paths.

a.

y

(-1,1)

(1, 1)

(-I, -1)

b.

919

43. Unit vectors pointing toward the origin Find a field F = M(x, y)i + N(x, y)j in the xy-plane with the property that at each point (x,y) #' (0,0), F is a unit vector pointing -.rd the origin. (The field is undefmed at (0, 0).)

44. Two "central" fields Find a field F = M(x, y)i + N(x, y)j in the '9'-plane with the property that at each point (x,y) #' (0,0), F points toward the origin and IF I is (a) the distance from (x, y) to the origin, (h) inversely proportional to the distance from (x, y) to the origin. (The field is undefined at (0, 0).) 45. Work and area Suppose that 1(1) is differentiable and positive for a :5 1 :5 b. Let C be the path r(l) = ti + 1(I)j, a :5 1 :5 b, and F = yi. Is there any relation betweeo the value of the work integral

(1, -1)

y

and the area of the region bounded by the l-axis, the graph of I, and the lines t = a and I = b? Give reasons for your answer. 46. Work done by a radial force with constant magoitnde A particle moves along the smooth curve y = I(x) from (a, I(a» to (b, I(b». The force moving the particle has constant magoitude k and always points away from the origin. Show that the work done by the force is

--+-----~----_r~x

c. Use any closed path differeot from parts (a) aod (h).

Vector fields In the Plane Draw the spin field

Flow Integrals in Space In Exercises 47-50, F is the velocity field ofa fluid flowing through a

39. Spin field

F=-

Y

v'x2

+ y2

i+

x

v'x2

+ y2

j

region in space. Find the flow along the giveo curve in the direction of increasing t.

47. F (sec Figure 16.12) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y2 = 4.

40. Radial field Draw the radial field F=xi+yj (sec Figure 16.11) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y2 = 1.

41. A field oftangent vectors a. Find a field G = P(x, y)i + Q(x, y)j in the '9'-piane with the property that at any point (a, b) #' (0, 0), G is a vector of magoitude v'a 2 + b 2 tsngent to the circle x 2 + y2 = a 2 + b 2 and pointing in the counterclockwise direction.

=

r(l)

-4xyi + 8yj + 2k

= ti + 12j + k,

0:5 I :5 2

48. F = x2; + yzj + y 2k r(t) = 31j + 41k, 0:5 1 :5 I 49. F = (x - z)i + xk r(l) = (cos I)i + (sin I)k,

0:5 I :5

'IT

SO. F=-yi+xj+2k r(l) = (-2 cos t)i + (2 sin I)j + 2tk,

0 :5 I :5

51. Circulation Find the circulation of F = 2xi + 2zj + 2yk around the closed path consisting of the following three curves traversed in the direction of increasing t.

C,:

r(l) = (cost)i + (sinl)j + tk,

C2:

r(t)

C,:

r(t)

= j + ('IT/2)(1 = ti + (I - I)j,

t)k,

0:5 1:5

0:5

0:5

t :5 I

(0,1,

¥)

(The field is undefmed at (0, 0).) b. How is Grelated to the spin fieldF in Figure 16.12?

z

42. A field of tangent vectors

a. Find a field G = P(x, y)i + Q(x, y)j in the '9'-piane with the property that at any point (a, b) #' (0,0), G is a unit vector tangeot to the circle x 2 + y2 = a 2 + b 2 and pointing in the clockwise direction. h. How is Grelated to the spin fieldF in Figure 16.12?

2'lT

C2 (1, 0'70)_:::::..._---"'~ (0, 1,0)

C,

y

t:5 I

'IT/2

920

Chapter 16: Integration in Vector Fields

52. Zero circulation Let C be the ellipse in which the plane 2x + 3y - z = 0 meets the cylinder x 2 + y2 = 12. Show, without evaluating either line integral directly, that the circulation of the field F = xi + yj + zk around C io either direction is zero. 53. Flow along a curve The field F = xyi + yj - yzk is the velocity field of a flow in space. Find the flow from (0, 0, 0) to (I, I, I) along the curve ofiotersection of the cylioder y = x 2 aod the plane z = x. (Hint: Use t = x as the parameter.)

56. F

g(t)i

+ h(t)j + k(t)k.

=

=

xy6 i t

~

1

F • dr.

+ 3x(xy 5 + 2)j; ret)

=

(2 cos t)i

=

(cost)i

+

(sin t)j,

27T

- 3 -2i I+x

+ -2-2j; r(t) I+y

+

(sint)j,

O~t~7T

-----...'

58. F = 2xyi - y2j I ~ t~ 4

54. Flow of a gradient field Fiod the flow of the field F

=

57. F = (y + yzcosxyz)i + (x 2 + xzcosxyz)j + (z + xy cosxyz)k; r(t) = (2 cos t)i + (3 sin t)j o ~ t ~ 27T

x =

V(xy 2z 3):

a. Once around the curve C io Exercise 52, clockwise as viewed from above b. Along the line segment from (1, I, I) to (2, I, -1).

16.3

c. Evaluate

o~

~ '~" 1 --T\----~r:~------ ~y 1-

a. Fiod drfor the path ret)

b. Evaluate the force F along the path.

55. F

z

L-

COMPUTER EXPLORATIONS

In Exercises 55--60, use a CAS to perform the following steps for findiog the work done by force F over the given path:

+ zexk; r(t)

59. F = (2y + siox)i + (z2 r(t) = (sint)i + (cost)j

= -ti

+ k,

+ Vtj + 3tk,

+ O/3)cosy)j + x 4 k; + (sin2t)k, -7T/2 ~ t

60. F = (x 2y)i + tx3j + xyk; r(t) = (cost)i (2 sin2 t - 1)k, 0 ~ t ~ 27T

~

7T/2

+ (siot)j +

Path Independence, Conservative FieLds, and PotentiaL Functions A gravitational field G is a vector field that represents the effect of gravity at a point in space due to the presence of a massive object. The gravitational force on a body of mass m placed in the field is given by F = mG. Similarly, an electric field E is a vector field in space that represents the effect of electric forces on a charged particle placed within it. The force on a body of charge q placed in the field is given by F = qE. In gravitational and electric fields, the amount of work it takes to move a mass or charge from one point to another depends on the initial and final positions of the object-not on which path is taken between these positions. In this section we study vector fields with this property and the calculation of work integrals associated with them.

Path Independence If A and B are two points in an open region D in space, the line integral of F along C from A to B for a field F defined on D usually depends on the path C taken, as we saw in Section 16.1 . For some special fields, however, the integral's value is the same for all paths from A toB.

DEFINITIONS Let F be a vector field defined on an open region D in space, and suppose that for any two points A and B in D the line integral F . dr along a path C from A to B in D is the same over all paths from A to B. Then the integral F . dr is path independent in D and the field F is conservative on D.

Ie

Ie

The word conservative comes from physics, where it refers to fields in which the principle of conservation of energy holds. When a line integral is independent of the path C from

16.3 Path Independence, Conservative Fields, and Potential Functions

point A to point B, we sometimes represent the integral by the symbol usual line integral symbol

I:

921

rether than the

Ie' This substitution helps us remember the path-independence

property. Under differentiability conditions nonnally met in practice, we will show that a field F is conservative if and only if it is the gradient field ofa scalar function I-that is, if and only if F = VI for some j. The function I then has a special name.

DEFINmON If F is a vector field defined on D and F = VI for some scalar function I on D, then I is called a potential function for F.

A gravitatioual potential is a scalar function whose gmdient field is a gravitatioual field, an electric potential is a scalar function whose gradient field is an electric field, and so on. As we will see, once we have found a potential function I for a field F, we can evaluate all the line integrals in the domain of F over any path between A and B by

1B

F·dr =

1B

V/'dr = I(B) - I(A).

(1)

If you think of VI for functions of several variables as being something like the derivative j' for functions of a single variable, then you see that Equation (1) is the vector calculus analogue of the Fundamental Theorem of Calculus fannula

l

b

j'(x) dx = I(b) - I(a).

Conservative fields have other remarkable properties. For example, saying that F is conservative on D is equivalent to saying that the integral ofF around every closed path in D is zero. Certain conditions on the curves, fields, and domains must be satisfied for Equation (1) to be valid. We discuss these conditions next.

Assumptions on Curves, Vector Fields, and Domains In order for the computations and results we derive below to be valid, we must assume certain properties for the curves, surfaces, domains, and vector fields we consider. We give these assumptions in the statements of theorems, and they also apply to the examples and exercises unless otherwise stated. The curves we consider are piecewise smooth. Such curves are made up of fmitely many smooth pieces connected end to end, as discussed in Section 13.1. We will treat vector fields F whose components have continuous first partial derivatives. The domains D we consider are open regions in space, so every point in D is the center of an open ball that lies entirely in D (see Section 13.1). We also assume D to be connected. For an open region, this means that any two points in D can be joined by a smooth curve that lies in the region. Finally, we assume D is simply connected, which means that every loop in D can be contracted to a point in D without ever leaving D. The plane with a disk removed is a two-dimensioual region that is 1101 simply connected; a loop in the plane that goes around the disk cannot be contracted to a point without going into the ''hole'' left by the removed disk (see Figure l6.22c). Similarly, if we remove a line from space, the remaining region D is 1101 simply connected. A curve encircling the line cannot be shrunk to a point while remaining inside D.

922

Chapter 16: Integration in Vector Fields

y

Connectivity and simple connectivity are not the same, and neither property implies the other. Think of connected regions as being in "one piece" and simply connected regions as not having any "loop-cawhing holes." All of space itself is both connected and simply connected. Figure 16.22 illustrates some of these properties. Caution Some of the results in this chapter can fail to hold if applied to situations where the conditions wtl've imposed do not hold In particular, the component test for conservative fields, given later in this section, is not valid on domains that are not simply connected (see

Simply connecred ---r--~~~====--- ,

ExampleS).

(.)

Line Integrals in Conservative Fields Gradient f1Clds F are obtamed by different.:iatiDg a scalar function I. A theorem. analogous to the Fundamental. Theorem of Calculus gives a way to evaluate the line integmls of gradient fields.

THEOREM I-fund.mental Theon!!m of Line Integnlls Let C be a smooth curve joining the point A to the point B in the plane or in space and parametrized by r(t). Let I be a differentiable function with a continuous gradient vector F = VI on a domain D containing C. Then

1

(h)

F·dr - /(B) - /(A).

y

o.

Like the Fundamental Theorem, Theorem 1 gives a way to evaluate line integmls without having to take limits of Riemann sums or fmding the line integral by the procedure used in Section 16.2. Before proving Theorem 1, we give an example.

Not simply connected

---r------------------,

EXAMPLE 1

Suppose the force field F = VI is the gradient of the function

(0)

I(x,y,z) = -

2 X

o

2

2·

Find the work done by F in moving an object along a smooth curve C joining (1, 0, 0) to (0, 0, 2) that does not pass through the origffi.

/ /

Solution An application ofTheore:m 1 shows that the work done by F along any smooth curve C joining the two points and not passing through the origin is

---7l"---,====-

,

1

+Y +z

Not simply connected

y

(d)

FIGURE 16.22 Four connected regions. In (a) and (b), the regions arc simply connectJ::d. In (c) and (d), the regions arc not simply connected becllllSe the curves Ct and C2 cannot be contracted to a poiIrt inside the regions containing them.

I

F ' dr - /(0,0,2) - /(1,0,0) -

-i -

(-1) -

t·

•

The gravitational force due to a planet, and the electric force associated with a chaxged particle, can both be modeled by the field F given in Example 1 up to a constant that depends on the units of measurement.

Proof of Theorem 1 Suppose that A and B are two points in region D and that C: r(t) = g(t)i + h(t)j + k(t)k, a ~ t ~ b, is a smooth curve in D joining A to B.

16.3

923

Path Independence, Conservative Fields, and Potential Functions

We use the abbreviated form r(t) = xi + yj + zk for the parametrization of the curve. Along the curve, I is a differentiable function of t and Chain Rule in Section 14.4 withx ~ g(t), y ~ hIt), z ~ k(t)

dl al dx al dy al dz dt=axdt+aydt+azdt

=V/·(dxi+dYj+dzk)=V/·dr=F.dr ~

~

~

~~.

Therefore,

1 e

F'dr=

l

j

bdl '=b F·..!.dt= d -dt '=a dt a dt

= I(g(t), h(t), k(t))

J:

=

Because F

«0)

~

=

Vf

A, rIb)

~

B

I(B) - I(A).

•

So we see from Theorem 1 that the line integral of a gradient field F = VI is straightforward to compute once we know the function f. Many important vector fields arising in applications are indeed gradient fields. The next result, which follows from Theorem 1, shows that any conservative field is of this type. THEOREM 2-Conservative Fields are Gradient Fields Let F = Mi + Nj + Pk be a vector field whose components are continuous throughout an open connected region D in space. Then F is conservative if and only if F is a gradient field VI for a differentiable function f.

Theorem 2 says that F = VI if and only iffor any two points A andB in the regionD, the value ofline integral IcF' dr is independent of the path CjoiningA toB inD.

Proof of Theorem 2 If F is a gradient field, then F = VI for a differentiable function I, and Theorem I shows that F' dr = I(B) - I(A). The value of the line integral does not depend on C, but only on its endpoints A and B. So the line integral is path independent and F satisfies the definition of a conservative field. On the other hand, suppose that F is a conservative vector field. We want to f"md a function I on D satisfYing VI = F. First, pick a point A in D and set I(A) = O. For any other point BinD define I(B) to equal F • dr, where C is any smooth path in D from A to B. The value of I(B) does not depend on the choice of C, since F is conservative. To show that VI = Fweneedto demonstrate thatallax = M,allay = N,andallaz = P. Suppose that B has coordinates (x, y, z). By def"mition, the value of the function I at F • dr, where Co is any path from A to Bo. We a nearby point Bo located at (XO,y, z) is take a path C = Co U L from A to B formed by first traveling along Co to arrive at Bo and then traveling along the line segment L from Bo to B (Figure 16.23). When Bo is close to B, the segment L lies in D and, since the value I(B) is independent of the path from A toB,

Ie

Ie

Xo --------1--) X

________

I

/

j./

x

FIGURE 16.23 The function fIx, y, z) in the proof ofTheorem 2 is computed by a line integral e• F • dr ~ f(Bo) fromA to BOo plus a line integra1h. F· dr along a line segment L parallel to tire x-axis and joining Bo to B located at (x,y, z). The value of f atA is f(A) ~ O.

f

Ie,

I(x,y,z) =

1.

F·dr +

1

F·dr.

Differentiating, we have

:xl(x,y,z) = :x

(1.

F·dr +

1

F'dr).

924

Chapter 16: Integration in Vector Fields Only the last tenn on the right depends on x, so

a

a1

ax f(x,y, z) = ax

L

F' dr.

Now parametrize L as r(t) = Ii + yj + zk, Xo :5 t :5 x. Then dr/dt = i, F· dr/dt = M, and F· dr = M(t, y, z) dt. Substitution gives

JL

J:'

a

a

axf(x,y,z) = ax

x

Ix,

M(t,y,z)dt = M(x,y,z)

by the Fundamental Theorem of Calculus. The partial derivatives af/ay = N and af/az = P follow similarly, showingthatF = Vf. •

EXAMPLE 2

Find the work done by the conservative field

F

= yzi

+ xzj + xyk

=

Vf,

f(x, y, z) = xyz,

where

along any smooth curve C joining the pointA( -1,3,9) tu B(I, 6, -4). With f(x,y, z) = xyz, we have

Solution

1 1B F·dr = =

F

Vf'dr

~

VI and pa1h

independeoce

f(B) - f(A)

Theorem 1

= XYZ I(l,6,-4) - XYZI(-1~,9) =

(1)(6)(-4) - (-1)(3)(9)

=

-24

+ 27

=

•

3.

A very useful property of line integrals in conservative fields comes into play when the for integration path of integration is a closed curve, or loop. We often use the notation C around a closed path (discussed with more detail in the next section).

f

THEOREM 3-Loop Property of Conservative Fields are equivalent. 1.

f

The following statements

F' dr = 0 around every loop (that is, closed curve

q

in D.

C

2. The field F is conservative on D.

Proof that Part 1 => Part 2 We want tu show that for any two points A and Bin D, the integral of F • dr has the same value over any two paths C 1 and C2 from A to B. We reverse the direction on C2 to make a path -C2 fromB toA (Figure 16.24). Together, C, and -C2 make a closed loop C, and by assumption,

FIGURE 16.24 Ifwe have two p.ths from A to B, one of them cao be reversed to make.loop.

Thus, the integrals over C , and C2 give the same value. Note that the definition of F' dr shows that changing the direction along a curve reverses the sign of the line integral.

16.3 B

B

925

Path Independence, Conservative Fields, and Potential Functions

Proof that Part 2 => Part 1 We want to show that the integral of F • dr is zero over any closed loop C. We pick two points A and B on C and use them to break C into two pieces: Cl from A to B followed by C2 from B back to A (Figure 16.25). Then

f

F ' dr =

c

r

lei

F.dr

+

r

icz

F.dr =

r F.dr - JA-r F·dr

JA.

=

O.

•

The following diagram summarizes the results of Theorems 2 and 3. A

A

Theorem 2

If A andB lie on a loop, we cao reverse part of the loop to lIIlIke two patha from A to B. FIGURE16.25

F = VIon D

=

Theorem 3

= fF'dr

F conservative onD

=

0

c

over any loop in D

Two questions arise:

1.

How do we know whether a given vector field F is conservative?

2.

If F is in fact conservative, how do we f"md a potential function I (so that F = V/)?

Finding Potentials for Conservative Fields The test for a vector field being conservative involves the equivalence of certain partial derivatives of the field components.

Component Test for Conservative Fields Let F = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k be a field on a connected and simply connected domain whose component functions have continuous first partial derivatives. Then, F is conservative if and only if

oP oy

ON

oM

OP

oz'

iJz

ax'

oN oX

and

oM oy'

(2)

Proof that Equations (2) hold if F is conservative There is a potential function I such that

F = Mi

+ Nj + Pk

01

= oX

i

+

01

oyj

+

01

oz k.

Hence, 2

oP _ l.. (01) _ 0 1 oy - oy OZ - oyoz

all

Mixed Derivative Theorem, Section 14.3

ozoy

=:z (:;)

=

~~.

The others in Equations (2) are proved similarly.

•

The second half of the proof, that Equations (2) imply that F is conservative, is a consequence of Stokes' Theorem, taken up in Section 16.7, and requires our assumption that the domain of F be simply connected.

926

Chapter 16: Integration in Vector Fields Once we know that F is conservative, we usually want to find a potential function for F. This requires solving the equation Vj = F or

oj i ox

+ oj. + oj k oyJ

OZ

Mi

=

+ N· + Pk J

for j. We accomplish this by integrating the three equations

oj oX =M,

oj oy =N,

oj oz =P,

as illustrated in the next example.

EXAMPLE 3 Show that F = (eX cosy + yz)i + (xz - eX siny)j servative over its natural domain and find a potential function for it.

+ (xy + z)k is con-

Solution The natoral domain of F is all of space, which is connected and simply connected. We apply the test in Equations (2) to M = eXcosy

+ yz,

N

=

xz - eXsiny,

P=xy+z

and calculate

OP oN oy =x= oz'

oM

iiZ

OP

oN = -exsiny oX

= y = ox'

+z

= oM.

oy

The partial derivatives are continuous, so these equalities tell us that F is conservative, so there is a function j with Vj = F (Theorem 2). We fmd j by integrating the equations

oj = eXcosy oX

-

+ yz

'

oj oy = xz - eX siny,

oj oz = xy

+ z.

(3)

We integrate the first equation with respect to x, holding y and z fIXed, to get

j(x,y, z) = eX cosy

+ xyz + g(y, z).

We write the constant of integration as a function of y and z because its value may depend ony andz, though not onx. We then calculate ojjoy from this equation and match it with the expression for ojjoy in Equations (3). This gives

-eX siny + xz +

:~ =

xz - eX siny,

so ogjoy = O. Therefore, g is a function of z alone, and

j(x,y, z) = eX cosy

+ xyz + h(z).

We now calculate ojjoz from this equation and match it to the formula forojjoz in Equations (3). This gives

dh xy+ dz =xy+z,

or

so Z2

h(z) =

2 + c.

dh dz

=

z,

16.3

Path Independence, Conservative Fields, and Potential Functions

927

Hence,

f(x,y,z) = e'cosy

Z2

+ ~z + 2 + c.

-

We have infinitely many potential functions ofF, one for each value of C.

EXAMPLE 4

+ (cosz)k is not conservative.

Show that F = (h - 3)i - zj

We apply the component test in Equations (2) and find immediately that

Solution

oP 0 oy = oy (cosz) = 0,

oN = -(-z) 0 = -\. oz oz

-

-

The two are unequal, so F is not conservative. No further testing is required.

EXAMPLE 5

Show that the vectnr field F =

2 -y 2 i x +y

+

2 X 2j x +y

+ Ok

satisfies the equations in the Component Test, but is not conservative over its natural domain. Explain why this is possible. Solution We have M = -y/(x 2 Component Test, we find

oP=O=oN oy oz'

+ y2), N

=

x/(x 2 + y2), and P = O. If we apply the

oP=O=oM ax

and

iJz'

oM oy

y2 - x 2 (x 2 + y2)2

oN oX .

So it may appear that the field F passes the Component Test. However, the test assumes that the domain of F is simply connected, which is not the case. Since x 2 + y2 cannot equal zero, the natoral domain is the complement of the z-axis and contains loops that cannot be contracted In a point. One such loop is the unit circle C in the ~-plane. The circle is parametrized by r(l) = (COSI)i + (sinl)j,O :5 I :5 2,.. This loop wraps around the z-axis and cannot be contracted to a point while staying within the complement of the z-axis. To show that F is not conservative, we compute the line integral F • dr around the loop C. First we write the field in terms of the parameter I: C

f

F =

x

2

- y2.1 +Y

+ x

2

x. 2J =

+Y

Next we f'md dr/dl = (-sinl)i

f C

F·dr =

-sint. __ 2 1 + cos-t

. 2 sm I

+

+

. 2 sm I

cost . ( . ). _2 J = -sml 1 + co.-t

+ (cosl).l·

(COSI)j, and then calculate the line integral as

f F'~~ 1 dl

=

2 2 " (sin t

+ cos2 t) dt

=

2,..

C

Since the line integral ofF around the loop C is not zero, the field F is not conservative, by Theorem 3. _ Example 5 shows that the Component Test does not apply when the domain of the field is not simply connected. However, if we change the domain in the example so that it is restricted to the ball of radius 1 centered at the point (2, 2, 2), orin any similar ball-shaped region which does not contain a piece of the z-axis, then this new domain D is simply connected Now the partial derivative Equations (2), as well as all the assumptions of the Component Test, are satiafied In this new situation, the field F in Example 5 is conservative on D.

928

Chapter 16: Integration in Vector Fields Just as we must be careful with a function when determining if it satisfies a property throughout its domain (like continuity or the intennediate value property), so must we also be careful with a vector field in determining the properties it mayor may not have over its assigned domain.

Exact Differential Forms It is often conveuient to express work and ciroulation integrals in the differential form

lMdx+Ndy+Pdz discussed in Section 16.2. Such line integrals are relatively easy to evaluate if M dx + N dy + P dz is the total differential of a function f and C is any path joining the two points from A to B. For then

{

( af ax dx

Jc M dx + N dy + P dz = Jc =

1B

+

af ay dy

Vf'dr

= f(B) - f(A).

+

af az dz

Vfisconservative. Theorem 1

Thus,

1B

df= f(B) - f(A),

just as with differentiable functions of a single variable.

DEFINmONS Any expression M(x,y,z)dx + N(x,y,z)dy + P(x,y,z)dz is a differential form. A differential form is exaet on a domain D in space if af

af

af

Mdx+Ndy+Pdz=~dx+~dy+~dz=~

for some scalar function f throughout D.

Notice that if M dx + N dy + P dz = df on D, then F = Mi + Nj + Pk is the gra. dient field of f on D. Conversely, if F = V/, then the form M dx + N dy + P dz is exact. The test for the form's being exact is therefore the same as the test for F being conservative.

Component Test for Exactness of M dx + N dy + p dz The differential form M dx + N dy + P dz is exact on a connected and simply connected domain if and only if

ap ay

aN az'

aM

ap

iJz

ax'

and

This is equivalent to saying that the field F = Mi

aN ax

aM ay'

+ Nj + Pk is conservative.

16.3

EXAMPLE 6

Path Independence, (onservative Fields, and Potential Functions

929

+ x dy + 4 dz is exact and evaluate the integral

Show that y dx

<2.3,-')

1

ydx + xdy + 4dz

(1,1,1)

over any path from (I, I, I) to (2, 3, - I).

Solution

We let M = y, N = x, P = 4 and apply the Test for Exactness:

aN=I=aM ax ay'

+ x dy + 4 dz is exact, so

These equalities tell us that y dx

y dx + x dy + 4 dz

dl

=

for some function I, and the integral's value is 1(2, 3, -I) - 1(1, I, I). We f'md 1 up to a constant by integrating the equations

al ax = y,

al ay = x,

al az = 4.

(4)

From the first equation we get

I(x,y,z)

xy + g(y,z).

=

The second equation tells us that

al ag ay=x+ay=x,

ag

or

iJy =

o.

Hence, g is a function of z alone, and

I(x,y, z)

=

xy

+ h(z).

or

h(z)

The third of Equations (4) tells us that

al az

=

0

+ dh

= 4 dz'

=

4z +

c.

Therefore,

I(x,y,z)

=

xy

+ 4z +

C.

The value of the line integral is independent of the path taken from (I, I, I) to (2, 3, -I), and equals

1(2,3, - I) - I(I, I, I)

=

2 + C - (5 + C)

-3.

=

•

Exercises 16.3 Testing for Conservative fields Which fields in Exercises 1--6 are conservative, and which are not? 1. F =yzi +xzj +xyk 2. F = (y sinz)i + (x sinz)j + (xy cosz)k 3. F = yi + (x + z)j - yk 4.F=-yi+xj S. F = (z + y)i + zj + (y + x)k 6. F = (eX cosy)i - (eX siny)j + zk

finding Potential Functions In Esercises 7-12, rmd a poteotial function f for the fieldF.

7. F=2xi+3yj+4zk

8. F = (y + z)i + (x + z)j + (x + y)k

9. F

=

eY+"(i

+ xi + 2xk)

10. F = (y sinz)i + (x sinz)j + (xy cosz)k 11. F = (lnx + sec2 (x + y))i +

~)j + y~k +z

( sec'(x + y) + Y +z 12.F=

Y i+( x + I + x 2 y2 1+ x 2 y2

VI _z y2 z 2 )i+ (

VI -Y y2 z 2 +.!)k z

930

Chapter 16: Integration in Vector Fields

Exact Diffell!ntial Fonns In Exercises 13-17, show that 1he differential fonns in 1he integrals are exact Then evaluate 1he integrals. (2,"-<)

13.

1

2xtb;

L

+ 2ydy + 2zdz

10

15.

1

2xy tb;

+ (x'

16.

1

(0.0.0)

~

z

~

1

°'"

°

yztb;+xzdy+:rydz

z

- z') dy - 2yz dz

(0.0.0) (,·,·1)

= 0, 0

+ (sin I)j + (1/2".)k, I '" 2". c, The x-axis from (I, 0, 0) to (0, 0, 0) followed by 1he parabola z ~ x',y ~ from (0, 0, 0) to (I, 0, I)

(1,1,') (1~")

The line segment x = l,y

b. The helix r(l) ~ (cos I)i

(0,0,0) (',',0)

14.

29. Work along different paths Find 1he work done by F ~ (x' + y)i + (y' + x)j + ze'k over 1he following pa1hs from (1,0,0) to (I, 0, I).

4

2x tb; - y' dy - - - d z I + Z,

(0.1.1)

17.

sinycosxdx + cosy sin x dy + dz

10

.........,,(0:',,::-O'c::0,-)_ - . " y

(1,0,0)

finding Potential Functions to Evaluate Line Integrals AI1hough they are not defmed on all of space R', 1he fields associated wi1h Exercises 18-22 are simply connected and 1he Component Test can be used to show they are conservative. Find a potential function for each field and evaluate 1he integrals as in Example 6. 18.

1

(1,7/2,')

2 cosy tb;

(I

)

+ y -

2xsiny dy

I

+ zdz

30, Work along different paths Find 1he work done by F ~ e"'i + (xze'" + zcosy)j + (:rye'" + siny)k over 1he following pa1hs from (1,0, I) to (I, "./2, 0).

a. The line segrototx

~

I,y ~ ".1/2,z ~ I - 1,0 '" I '" I

(0.'.1)

19.

1o

(1~")

z

,

3x'tb;

+~

dy

+ 2z lny dz

(1.1.1)

20.

('.1.1)

10

(1.'.1)

21.

Io 1

('~~)

(2xlny - yz)tb;

I

(

,

)

+ ~ - xz dy -:rydz

(I

x) Y -tb; + z - - dy - -dz Y y2 z2 ('~~) 2x tb; + 2y dy + 2z dz (-I, -1, -1) x 2 + y2 + z2

"2 ~~-----/ ~~y

(1,1,1)

22.

Applications and Examples 23. Revisiting Example 6

Evaluate 1he integral

('·',-1)

10

ytb;

x

r;~o)

1

b. The line segrotot from (I, 0, I) to 1he origin followed by 1he line segrotot from 1he origin to (I, "./2, 0)

z

+ xdy + 4dz

(I,I,I)

from Example 6 by fmding paramemc equations for 1he line segment from (I, I, I) to (2,3, -I) and evaluating 1he line integral ofF ~ yi + xi + 4k along 1he segroent Since F is conservative, 1he integral is independent of1he pa1h. 24. Evaluate

1

x' tb;

Independence of path Show that 1he values of 1he integrals in Exercises 25 and 26 do not deptod on 1he pa1h taken from A toB.

26.

l

A

z'tb;

r;~o)

+ yz dy + (y'/2) dz

along 1he line segroent C joining (0, 0, 0) to (0, 3, 4),

25.1"

"2 ~~~----/ ~~y

°

z

+ 2ydy + 2xzdz

1

+ ydy + zdz --:-r~==:'=~ \/x' + y' + z,

"Xtb;

e;/)i,

c. The line segrotot from (I, 0, I) to (I, 0, 0), followed by 1he x-axis from (I, 0, 0) to 1he origin, followed by 1heparabola y ~ ".x'/2,z ~ from1here to (1,,,./2,0)

In Exercises 27 and 28, fmd a pottotial function for F.

27. F

~ :;" i +

28. F

~ (e'lny)i + (~ + Sinz)j + (ycosz)k

{(x,y): y > OJ

16.4 Green's Theorem in the Plane 31. Evaluating a work integral two ways Let F ~ V(x'y2) and let C be the path in the xy·plane from (-I, I) to (I, I) that consists of the line segment from (-I, 1) to (0, 0) followed by the line segment from (0, 0) to (I, I). Evaluate leF'dr in two ways.

34. Gradient of a line integral vative vector field aod

32. Integral along different path. Evaluate the line integral Ie 2x cos y dx - x 2 sin y dy along the following paths C in the xy-plane.

a. The parabola y ~ (x - 1)2 from (1,0) to (0, I) b. The line segment from (-I,,,.) to (I, 0) c. The x-axis from (-I, 0) to (I, 0) d. The astroid r(l) ~ (eos'l)i + (sin' I)j, 0 ,;; I';; 2"., counterclockwise from (I, 0) back to (I, 0) y

Vj is a conser-

~

L

F·dr.

(0.0.0)

ShowthatVg

b. Usej(x,y) ~ x'y2 as a potential function forF.

~

«3.)

g(x,y,z)

a. Find parametrizations for the segments that make up C and evaluate the integral.

Suppose that F

931

~

F.

35. Path of lea.t work You have beeo asked to find tlte patlt along which a force field F will perform tlte least work in moving a particle betweeo two locations. A quick calculation on your part shows F to be conservative. How should you respond? Give rea-

sons for your answer. 36. A revealing experiment By experiment, you fmd that a force field F performs only half as much work in moving an object along patlt C1 from A to B as it does in moving tlte object along path C2 from A to B. What can you conclude about F? Give rea-

sons for your answer. 37. Work by a constant force Show that tlte work done by a constant force field F ~ ai + bj + ck in moving a particle along anypatltfromA toB is W ~ F·AB.

(0. 1)

38. Gravitational field (-1,0)

(1,0)

~-E~--~--~~~x

a. Find a potential function for the gravitational field F

~

-GmM

_~:nc.'_+-,y~j_+~zk~ (x2 + y2 + z2)'/2

(0, -1)

(G, m, aodM are eoostants). 33, a. Exact differential form How are tlte constants a, b, and c related if the following differeotial form is exact? (ay2 + 2czx) dx + y(bx + cz) dy + (ay2 + ex 2) dz For what values of b and c will F ~ (y2 + 2=)i + y(bx + cz)j + (y2 + ex 2 )k

b. Gradient field

b. Let P1 aod P2 be points at distance 81 and 82 from the origin. Show that tlte work done by tlte gravitational field in part (a) in moving a particle from PI to Pz is

GmM(1. _1.). 82

Sl

be a gradient field?

16.4

Green's Theorem in the Plane If F is a conservative field, then we know F = Vf for a differentiable function f, and we can calculate the line integral of F over any path C joining point A to B as fcF'dr = f(B) - f(A). In this section we derive a method for computing a work or flux integral over a closed curve C in the plane when the field F is not conservative. This method, known as Green's Theorem, allows us to convert the line integral into a double integral over the region enclosed by C. The discussion is given in terms of velocity fields of fluid flows (a fluid is a liquid or a gas) because they are easy to visualize. However, Green's Theorem applies to any vector field, independent of any particular interpretation of the field, provided the assumptions of the theorem are satisfied. We introduce two new ideas for Green's Theorem: divergence and circulation density around an axis perpendicular to the plane.

Divergence Suppose that F(x, y) = M(x, y)i + N(x, y)i is the velocity field of a fluid flowing in the plane and that the first partial derivatives of M and N are continuous at each point of a region R. Let (x, y) be a point in R and let A be a small rectangle with one corner at (x, y) that, along with its interior, lies entirely in R. The sides of the rectangle, parallel to the coordinate axes, have lengths of dx and
932

Chapter 16: Integration in Vector Fields

(X,y

+ Oy)

t

(x

+ Dx,Y + Oy)

F'j>O ... F ·(-1) < 0 A

F'(-j)
...

(x

+ Ox,y)

k-I,-----ox------->l,1 FIGURE 16.26 The rate at which the fluid leaves the rectangular region A across the bottom edge in the direction of the outwardnormai-j is approximately F(x, y). (-j) At, which is negative for the vector field F shown here. To approximate the flow rate at the point (x, y), we calculate the (approximate) flow rates across each edge in the directions of the red arrows, sum these rates, and then divide the sum by the area ofA. Thking the limit as At ..... 0 and ay ..... 0 gives the flow rate per unit area.

change sign tbroughout a small region containing the rectangle A. The rate at which fluid leaves the rectangle across the bottom edge is approximately (Fignre 16.26)

F(x,Y)'(-j) ax

=

-N(x,y)ax.

TIris is the scalar component of the velocity at (x,y) in the direction of the outward nonna! times the length of the segment. If the velocity is in meters per second, for example, the flow rate will be in meters per second times meters or square meters per second. The rates at which the fluid crosses the other three sides in the directions of their outward normals can be estimated in a siruilar way. The flow rates may be positive or negative depending on the signs of the components of F. We approximate the net flow rate across the rectangolar boundary of A by summing the flow rates across the four edges as defined by the following dot products.

Fluid Flow Rat.s:

F(x, y + ay)· j ax = N(x, y + ay)ax F(x,Y)'(-j) ax = -N(x,y)ax F(x + ax,y)'i ay = M(x + ax,y)ay F(x,y)' (-i) ay = -M(x,y)ay.

Top: Bottom: Right: Left:

Summing opposite pairs gives Top and bottom:

(N(x,y

Right and left:

(M(x

+

(~~ ay)ax

+ ay) - N(x,y»ax '" ax,y) - M(x,y))ay '"

(~~ ax )ay .

Adding these last two equations gives the net effect of the flow rates, or the

ax + ON) oy axay.

Flux across rectangle boundary '" ( oM

We now divide by axay to estimate the total flux per unit area or flux density for the rectangle: Flux across rectangle boundary '" rectangle area

(OM + ON). oX

oy

16.4 Green's Theorem in the Plane Source: div F (xo. Yo) > 0 A gas expanding at the point (%". Yo).

~r/

/1",>

<--

-----

SInk: div F ('0. Yo)

933

Finally, we let ~ and Jly approach zero to define the flux density ofF at the point (x,y). In mathematics, we call the flux density the divergence of F. The symbol for it is div F, pronounced "divergence ofF' or "div F."

DEFINmON

The divergence (flux density) of a vector field F = Mi

+ Nj at

the point (x, y) is div F

<0

=aM+aN ax ay'

(1)

A 888 compressing at the point ('0. Yo).

~l/ 7r~

FIGURE 16.27 If a gas is expanding at a point (:to. Yo), the lines of flow have positive divergence; if the gas is

compressing, the divergence is negative.

A gas is compressible, unlike a liquid, and the divergence of its velocity field measures to what extent it is expanding or compressing at each point. Intuitively, if a gas is expanding at the point (xo, Yo), the lines of flow would diverge there (hence the name) and, since the gas would be flowing out of a small rectangle about (xo, Yo), the divergence of F at (xo, Yo) would be positive. If the gas were compressing instead of expanding, the divergence would be negative (Figure 16.27).

EXAMPLE 1 The following vector fields represent the velocity of a gas flowing in the xy-plane. Find the divergence of each vector field and interpret its physical meaning. Figure 16.28 displays the vector fields.

FIGURE 16.28

Velocity fields ofa gas flowing in the plaoe (Example I).

(8) Uniform expansion or compression: F(x. y) = cxi

(b) Uniform rotation: F(x,y) = -cyi

+ cxj

(c) Shearingflow: F(x.y) = yi

(d) Whirlpool effect: F(x. y) =

2 -y 2 i + 2 X 2j x +y x +y

+ cyj

934

Chapter 16: Integration in Vector Fields Solution (a) div F = :x (ex) sion; if c

+ ~ (cy)

the gas is undergoing unifonn expan-

< 0, it is undergoing unifonn compression.

(b) div F = :x (-cy)

(c) div F =

> 0,

= 2c: If c

:x (y)

+ ~ (ex)

= 0: The gas is neither expanding nor compressing.

= 0: The gas is neither expanding nor compressing.

( -y ) + ~( x ) = 2xy 2xy = (d) divF = ~ oX x 2 + y2 oy x2 + y2 (x2 + y2)2 (x2 + y2)2 the divergence is zero at all points in the domain of the velocity field.

o.

. Again,

• Cases (b), (c), and (d) of Figure 16.28 are plausible models for the two-dimensional flow of a liquid. In fluid dynamics, when the velocity field of a flowing liquid always bas divergence equal to zero, as in those cases, the liquid is said to be incompressihle.

Spin Around an Axis: The k-Component of Curl The second idea we need for Green's Theorem bas to do with measuring how a floating paddle wheel, with axis perpendicular to the plane, spins at a point in a fluid flowing in a plane region. This idea gives some sense of how the fluid is circulating around axes located at different points and perpendicular to the region. Physicists sometimes refer to this as the circulation density of a vector field F at a point. To obtain it, we retorn to the velocity field

F(x,y)

=

M(x,y)i + N(x,y)j

and consider the rectangle A in Figure 16.29 (where we assume both components ofF are positive). (x.y

+ Oy)

(x

+ ox.y + Oy)

~

F·(-I)
I

F·(-j)
f

A

F(x.y)

(x,y)

I,

(x

+ ox,y)

Ox

,I

FIGURE 16.29 The rate at which a fluid flows along the bottcm edge of a rectangular regioo A in the direction i is approximately F(x, y) • i ax, which is positive for the vector field F shown here. To approximate the rate of circulation at the point (x. y), we calcn1ate the (approximate) flow rates along each edge in the directions of the red arrows, sum these rates, and then divide the sum by the area of A. Taking the limit as 0 aod 0 gives the rate of the circulation per unit area.

ax ....

ay ....

The circulation rate of F around the boundary of A is the sum of flow rates along the sides in the tangential direction. For the bottom edge, the flow rate is approximately

F(x,y)·iAx

=

M(x,y)Ax.

935

16.4 Green's Theorem in the Plane Vertical axis

kt

cb

This is the scalar component of the velocity F(x, y) in the tangent direction i times the length of the segment. The flow rates may be positive or negative depending on the components ofF. We approximate the net circulation rate around the rectangular boundary of A by summing the flow rates along the four edges as dermed by the following dot products. Top: Bottom: Right: Left:

----

CurlF (xo,Yo)' k > 0 Counterclockwise circulation

We sum opposite pairs to get

Top and bottom:

Vertical axis

kt

cb

+----

F(x,y + Jly)' (-i) Jlx = -M(x,y + Jly)Jlx F(x,y)·i Jlx = M(x,y)Jlx F(x + Jlx,y). j Jly = N(x + Jlx,y)Jly F(x,y)' (-j) Jly = -N(x,y)Jly.

Right and left:

-(M(x,y (N(x

+ Jly) - M(x,y»Jlx '" -

+ Jlx,y) - N(x,y))Jly '"

(~~ JlY)Jlx

(~~ Jlx )Jly.

Adding these last two equations gives the net circulation relative to the counterclockwise orientation, and dividing by JlxJly gives an estimate of the circulation density for the rectangle:

CurlF (xo,Yo)' k < 0 Clockwise circulation

Circulation around rectangle aN aM rectangle area '" ax - ay'

FIGURE 16.30 In the flow of an incompressible fluid over a plane region, the k-component of the curl measures the rate of the fluid's rotation at a point. The k-component of the curl is positive at points where the rotation is counterclockwise and negative where the rotation is clockwise.

We let Jlx and Jly approach zero to define the circulation density of F at the point (x,y). If we see a counterclockwise rotation looking downward onto the xy-plane from the tip of the unit k vector, then the circulation density is positive (Figure 16.30). The value of the circulation density is the k-component of a more general circulation vector field we derme in Section 16.7, called the curl of the vector field F. For Green's Theorem, we need only this k-component.

DEFINmON The circulation density of a vector field F = Mi point (x, y) is the scalar expression aN ax

aM ay'

+ Nj at the (2)

---

This expression is also called the k-component of the curl, denoted by (curl F) • k.

If water is moving about a region in the :ry-plane in a thin layer, then the k-component of the curl at a point (xo, YO) gives a way to measure how fast and in what direction a small paddle wheel spins if it is put into the water at (xo, YO) with its axis perpendicular to the plane, parallel to k (Figure 16.30).

EXAMPLE 2 Find the circulation density, and interpret what it means, for each vector field in Example 1. Solution (8) Uniform expansion: (curl F)' k = at very small scales.

:x (ey) -

~ (ex)

= O. The gas is not circulating

936

Chapter 16: Integration in Vector Fields y

(b) Rotation: (curl F)' k =

:x

(ex) -

~ (-cy)

indicates rotation at every point. If c the rotation is clockwise.

(0) Shear: (curl F)' k = -

:o ~

--- -

~ (y)

>

= 2c. The constant circulation denaity

0, the rotation is counterclockwise; if c

<

0,

= -I. The circulation density is constant and negative,

so a paddle wheel floating in water undergoing such a shearing flow spins clockwise. The rate of rotation is the same at each point. The average effect of the fluid flow is to push fluid clockwise around each of the small circles shown in Figure 16.31.

(d) Whirlpool: FIGURE 16.31 A shearing flow pushes the fluid clockwise around each point (Example 2c).

(curlF)'k=~( oX

2 x )_~( -y )= y2_x2 _ y2_x =0 2 x + y2 oy x2 + y2 (x2 + y2)2 (x2 + y2)2 .

The circulation density is 0 at every point away from the origin (where the vector field is undefined and the whirlpool effect is taking place), and the gas is not circulating at any point for which the vector field is defined _

Two Forms for Green's Theorem In one form, Green's Theorem says that under suitable conditions the outward flux of a vector field across a simple closed curve in the plane equals the double integral of the divergence of the field over the region enclosed by the curve. Recall the formulas for flux in Equations (3) and (4) in Section 16.2 and that a curve is simple if it does not cross itseIf.

THEOREM 4-Green's Theorem (Flux-DI""rgence or Normal Form) Let C be a piecewise smooth, simple closed curve enclosing a region R in the plane. Let F = Mi + Nj be a vector field with M and N baving continuous first partial derivatives in an open region containing R. Then the outward flux of F across C equals the double integral of div F over the region R enclosed by C.

f

F • nds =

e

fMdY-Ndx= JJ(oo~+~~)dxdy e

Divergence integral

O-.rdflux

We introduced the notation

(3)

R

fe in Section 16.3 for integration around a closed curve.

We elaborate further on the notation here. A sirople closed curve C can be traversed in two possible directions. The curve is traversed counterclockwise, and said to be positively oriented, if the region it encloses is always to the left of an object as it moves along the path. Otherwise it is traversed clockwise and negatively oriented. The line integral of a vector field F along C reverses sign if we cbange the orientation. We use the notation

f

F(x,Y)'dr

e

for the line integral when the sirople closed curve C is traversed counterclockwise, with its positive orientation. A second form of Green's Theorem says that the counterclockwise circulation of a vector field around a sirople closed curve is the double integral of the k-<:omponent of the curl of the field over the region enclosed by the curve. Recall the defming Equation (2) for circulation in Section 16.2.

937

16.4 Green's Theorem in the Plane

THEOREM 5-Green's Theorem (Circulation-Curl or Tangential Form) Let C be a piecewise smooth, simple closed curve enclosing a region R in the plane. Let F = Mi + Nj be a vector field with M and N having continuous first partial derivatives in an open region containing R. Then the counterclockwise circulation of F around C equals the double integral of (curl F) • k over R.

f

F ' Tds

f

=

C

Mdx + Ndy

=

ff(~~-~~)dxdY

(4)

R

C

Curl integral

Counterclockwise circulation

The two forms of Green's Theorem are equivalent. Applying Equation (3) to the field + Mj gives

G, = Ni - Mj gives Equation (4), and applying Equation (4) to G, = -Ni Equation (3).

Both forms of Green's Theorem can be viewed as two-dimensional generalizations of the Net Change Theorem in Section 5.4. The outward flux of F across C, defmed by the line integral on the left-hand side of Equation (3), is the integral of its rate of change (flux density) over the region R enclosed by C, which is the double integral on the right-hand side of Equation (3). Likewise, the counterclockwise circulation ofF around C, defmed by the line integral on the left-hand side of Equation (4), is the integral of its rate of change (circulation density) over the region R enclosed by C, which is the double integral on the right-hand side of Equation (4).

EXAMPLE 3

Verify both forms of Green's Theorem for the vector field

F(x,y) = (x - y)i

+ xj

and the region R bounded by the unit circle C:

Solution

ret)

= (cost)i

+

(sint)j,

o ,,; t ,,; 21T.

Evaluating F(r(t» and differentiating components, we have M= cost - sin I,

dx

=

N = cost,

dy

= d(sin t) = cos t dt,

aM = I ax '

d( cos t)

aM =-1 ay ,

= -sin t dt,

aN = 0

ay

.

The two sides of Equation (3) are

f

Mdy - Ndx

=

C

'~2" 1=0

1

2"

JJ (~~ + ~~)

dxdy

(cost - sin t)(costdt) - (cost)(-sintdt)

cos2 tdt

=

0 /.

=

f f (I + O)dxdy

R

= 1T

R

=

f f dx dy R

= area inside the unit circle =

1T.

938

Chapter 16: Integration in Vector Fields y

The two sides of Equation (4) are

f Mdx + Ndy c

=

=

JJ (~~ - ~~)

dxdy

=

R

FIGURE 16.32 The vector field in Example 3 has a counterclockwise circulatioo of2". around the unit circle.

1~~2w (cost -

12w JJ

sintH-sintdt) + (cos tHcos tdt)

(-sintcost + 1) dt

(1 - (-1» dxdy

=

R

2".

=

2JJ dxdy

2"..

=

R

•

Figure 16.32 displays the vector field and circulation around C.

Using Green's Theorem to Evaluate Line Integrals If we construct a closed curve C by piecing together a number of different curves end to end, the process of evaluating a line integral over C can be lengthy because there are so many different integrals to evaluate. If C bounds a region R to which Green's Theorem ap· plies, however, we can use Green's Theorem to change the line integral around C into one double integral over R.

EXAMPLE 4

Evaluate the line integral

f xydy - y 2 dx, c where C is the square cut from the first quadrant by the lines x = 1 and y = 1.

Solution

We can use either form of Green's Theorem to change the line integral into a double integral over the square. 1.

With the Normal Form Equation (3): Taking M = xy, N = y2, and C and R as the square's boundary and interior gives

fXYdy - y2 dx= JJ(y + 2y)dxdy= l'1'3Ydxdy C

R

=

l ' []'~1 l' 3xy

dy

=

3ydy

=

0 .- 0 0

2.

With the Tangential Form Equation (4): Taking M

=

3]' 3

_y2

=-.

20

_y2 and N

=

2

xy gives the same

result:

f_ y 2 dx + xydy C

=

JJ(y - (-2y))dxdy

=~.

•

R

EXAMPLE 5 Calculate the outward flux of the vector field F(x,y) = xi the square bounded by the lines x = ± 1 and y = ± I.

+ y2j

across

16.4 Green's Theorem in the Plane

939

Solution

Calculating the flux with a line integral would take four integrations, one for each side of the square. With Green's Theorem, we can change the line integral to one double integral. WithM = x, N = y2, Cthe square, andR the square's interior, we have Flux = f

F·nd. = f

Mdy - Ndx

C

C

= ff (aM + aN) dxd JJ ax ay Y

Green's Theorem

R

LL

=

(i

+ 2y)dxdy

[1 (2 + 4y) dy =

=

L[x

[2Y + 2y2]1

~-1

y

=

+ 2xy [~t dy

•

= 4.

-1

Proof of Green's Theorem for Special Regions Let C be a smooth simple closed curve in the ;ry-plane with the property that lines parallel to the axes cut it at no more than two points. Let R be the region enclosed by C and suppose

I I I I

I

:C'

that M, N, and their fIrst partial derivatives are continuous at every point of some open region containing C and R. We want to prove the ciroulation-curl fonn of Green's Theorem,

I

1 1 ' Y - J1X I I I

I I

f Mdx + Ndy

~r-L-------L-----r----'x

o

a

11 (~~ -~~)dxdY.

_ '() I P,(x.f,Vc))

x

b

=

C

(5)

R

Figure 16.33 shows Cmade up of two directed parts:

FIGURE 16.33 The boundary curve Cis made up ofC" the graph ofy ~ ''(x), and C2. the graph of y ~ J,(x).

Ct : Y = /t(x),

C2 : y = j,(x),

a:5 x :5 b,

b

~

x

~

a.

For any x between a and b, we can integrate aM/ay with respect to y from y = /t(x) to = h(x) and obtain

Y

h (X) aM

1

h")

Tdy

=

M(x,y)

r=

Y

!z(x)

=

M(x,h(x» - M(x,/t(x)).

y= h(x)

We can then integrate this with respect to x from a to b:

l

b

flo") aaM dy dx

a ih(x)

~

y

Ci: x ~ 8,(Y) ---- \ --~ --__

y

----

----------

[M(x, h(x)) - M(x, /t(x))] dx

a

= d

l -l b

=

a

M(x, h(x» dx -

l

b

M(x, /t(x)) dx

=-fMdx-f Mdx

ic'l

Q2(8,(y).y)

Q,(8,(Y). y)

lei

=-fMdx. C

c

Therefore

~r-----------------~x

o

FIGURE 16.34 The boundary curve Cis made up ofCl. the graphofx ~ g,(y), and C;, the graph of x ~ g2(Y).

f Mdx C

=

11 (-~~)dxdy.

(6)

R

Equation (6) is half the result we need for Equation (5). We derive the other halfby integrating

aN/ax trrst with respect to x and then with respect to y, as suggested by Figure 16.34.

940

Chapter 16: Integration in Vector Fields

y

TIris shows the curve C of Figure 16.33 decomposed into the two directed parts C\: x = gt(Y), d ;;,: Y ;;,: c and Co: x = g2(Y), c ,,; Y ,,; d. The result oflhis double integration is

e

f

~

N dy =

e

II ~~

(7)

dx dy.

R

Summing Equations (6) and (7) gives Equation (5). TIris concludes the proof.

o

•

Green's Theorem also holds for more general regions, such as those shown in Figures 16.35 and 16.36, but we will not prove this result here. Notice that the region in Figure 16.36 is not simply connected. The curves C t and C, on its boundary are oriented so that the region R is always on the left-hand side as the curves are traversed in the directions shown. With this convention, Green's Theorem is valid for regions that are not simply connected. While we stated the theorem in the xy-plane, Green's Theorem applies to any region R contained in a plane bounded by a curve C in space. We will see how to express the double integral over R for this more general form of Green's Theorem in Section 16.7.

(a)

y b

a

y

~O~--~~~b--~X a (b)

FIGURE 16.35 Other regions to which Green's Theorem applies.

-t--+-~r-';--;' I->X

FIGURE 16.36 Green's Theorem may be applied to the aonular region R by sununing the line integrals along the houodaries Ct

and C, in the directions shown.

Exercises 16.4 Verifying Green's Theorem 10 Exercises 1-4, verifY the conclusion of Green's Theorem by evaluating both sides of Equations (3) and (4) for the fieldF = Mi + Nj. Take the domains of integration in each case to be the disk R: x 2 + y2 ,,; a 2 and its boundiog circle C: r = (a cos t)i + (a sin t)j, t '" 2,.. l.F=-yi+xj 2.F=yi

9. F

=

(xy

+ y2ji + (x

- y)j

10. F

=

+ 3yji + (2x

(x

- y)j

y

y

°'"

3. F=2xi-3yj

~~----~----~~X

4. F = -x'yi + xy2j

-2

2

-I

Clrculatton and Flux 10 Exercises 5-14, use Green's Theorem to fmd the counterclockwise circolation and outward flux for the field F and curve C. S. F = (x - y)i + (y - x)j C: The square bounded by x = O,X = I,y = O,y = I

11. F =X 3y 2 i

+ kX4yj

12. F = ~i+ (tan-ty)j I+y

y

y

6. F = (x 2 + 4y)i + (x + y2)j C: The square bounded by x = O,X = I,y = O,y = I

(2,2)

7. F = (y2 - x 2)i + (x2 + y2)j C: The triangle houoded by y = 0, x = 3 , and y = x 8. F = (x + y)i - (x 2 + y2)j C: The triangle houoded by y = 0, x = I , and y = x

-I

~~~--~-------+X

(0,0)

-I

16.4 Green's Theorem in the Plane 13. F

~

(x

+ eX siny)i + (x + eX cosy)j

C: The right·hand loop of the lemniscate r2 ~ cos 28 14. F

Green'. Theorem Area Formula

~ (tan-l~} + In(x 2 + y2)j

AreaofR

tf

~

15. Find the counterclockwise circulation and outward flux of the field F ~ xyi + y2j around and over the boundary of the region enclosed by the curves y ~ x 2 and y ~ x in the frrst quadrant

The reason is that by Equation (3), run backward,

R

~

17. Find the outward flux of the field

~

f

(y2 dx

(y - 2)2 ~ 4

(3ydx

0, x

~

+y

~

I,y

~

0

(6y

(2x

7T,

0 :s y :s sinx

+ x)dx + (y + 2x)dy

C: The circle (x - 2)2

f

.. f f

j(x) dx

I - cost

+ g(y) dy

ky dx

+ hx dy (k and h constants).

c 30. Integral dependent only on .....

Show that the value of

+ (x2y + 2x) dy

xy2 dx

around any square depends ouly on the area of the square and not on its location in the plane. 31. What is special about the integral

f

4x 3y dx

+ x' dy?

Give reasons for your answer.

f

+ 2xdy)

c

24.

~

32. What is special about the integral

C: The boundary of 0 :s x :s

f

I - sinl, Y

29. Let C be the boundary of a region on which Green's Theorem holds. Use Green~ Theorem to calculate

c

c

23.

~

28. One arch of the cycloid x

0 s I :5 2".

c

+ x 2 dy)

C: The mangle bounded by x

f

0:5 I :5 2".

f

4y)j

C

22.

+ (b sin I)j, + (sin3 1)j,

b.

Using Green's Theorem Apply Green~ Theorem to evaluate the integrals in Exercises 21-24.

21.

27. The astroid r(l) ~ (COS3 1)i

(a cos I)i

c

C: The boundary of the ''triangular'' region in the frrst quadrant enclosed by the x-axis, the line x = 1, and the curve y = x 3

C: The circle (x

0:5 1 :5 2".

~

26. The ellipse r(l) ~ (a cos I)i

once counterclockwise around the given curve. 19. F ~ lxy':i + 4x2y2j

+ (2x - 2)2 +

+ (a sin l)j,

25. The circle r(l)

+ cos8),a > O.

Work In Exercises 19 and 20, fmd the vrork done by F in moving a particle

(4x - 2y)i

txdy - tydx.

Use the Green's Theorem area formula given above to fmd the areas of the regions enclosed by the curves in Exercises 25-28.

18. Find the counterclockwise circulation of F ~ (y + eX In y)i + (eX/y)j around the boundary of the region that is bounded above by the curve y ~ 3 - x 2 and below by the curve y ~ x' + 1.

~

f

+ t)dydx

R

c

(3xy - ~)i + (eX + tan-I y)j I +y

aerossthe cardioidr ~ a(1

II (t

~ JJ dydx ~

AreaofR

16. Find the counterclockwise circulation and the outward flux of the field F ~ (-siny)i + (xcosy)j around and over the square cut from the frrst quadrant by the lines x ~ "./2 and y ~ "./2.

F

xdy - ydx

c

C: The boundary of the region defmed by the polar coordinate inequalities 1 :s r :s 2, 0 ::s 8 ::s 11'

20. F

941

+ y2) dx +

+ (y -

(lxy

_y3dy+x 3 dx?

c Give reasons for your answer. 33. Are. as a Une integral Show that if R is a region in the plane bounded by • piecewise smooth, simple closed curve C, then

3)' ~ 4

+ 3y) dy

c C: Any simple closed curve in the plane for which Green's Theorem holds Calculating A .... with Green" Theo...m If a simple closed curve C in the plane and the region R it encloses satisfy the hypotheses of Green's Theorem, the area of R is given by

AreaofR

~

f

xdy

~

c

-

f

c

ydx.

34. Definite integral .s • Une integral Suppose that a nonnegative functiony ~ j(x) has a continuous frrstderivative on [a, b]. Let C be the boundary of the region in the xy-plane that is bounded below by the x-axis, above by the graph of j, and on the sides by the lines x ~ a and x ~ b. Show that

/.b

j(x)dx

~

-

f

c

ydx.

942

Chapter 16: Integration in Vector Fields

35. Area and the centroid Let A be the area and xthe x-coordinate of the centroid of a region R that is bounded by a piecewise

a. Let f(x,y) ~ In (x 2 + y2) and let Cbe the circle x 2 + y2 = a 2. Evaluate the flux integral

smooth, simple closed curve C in the ~-p1me. Show that

f

Vf·nds.

c 36. Moment of inertia Let Iy be the moment of ioertia about the y-axis of the region ioExercise 35. Show that

37. Green's Theorem and Laplace's equation Assuming that all the necessary derivatives exist and are continuous, show that if f(x,y) satisfies the Laplace equation

a2f a2f '+-2=0. ax ay then

b. Let Kbe an arbitrary smooth, simple closed curve in the plane that does not pass through (0, 0). Use Green's Theorem to show that

hss two possible values, depending on whether (0, 0) lies inside K or outside K.

40. Bendi::J:80n's criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid's individual particles. The vectors F ~ M(x, y)i + N(x, y)j of the flow's velocity field are the tsngent vectors of the streamlines. Show that if the flow takes place over a simply connected region R (no holes or missiog points) and that if M, + Ny ¢ 0 throughout R, then none of the streamlines in R is closed In other words, no particle of fluid ever has a closed trajectory in R. The criterion M, + Ny ¢ 0

is called Bendi:J:son's criterion for the nonexistence of closed trajectories. 41. Establish Equation (1) to rmish the proof of the special case of for all closed curves C to which Green's Theorem applies. (The converse is also true: If the line integral is always zero, !ben f satisfies the Laplace equation.)

38. Mnjrnjzjng work Among all smooth, simple closed curves in the plane, oriented counterclockwise, find the one along which the work done by

Green's Theorem. 42. Curl component of conservative fields

Can anytbiog be said

about the curl component of a conservative two-dimensional vector field? Give reasons for your answer.

COMPUTER EXPLORATIONS In Exercises 43-46, use a CAS and Green's Theorem to rmd the counterclockwise circulation of the field F around the simple closed curve C. Perform the followiog CAS steps.

a. Plot C in the xy-p1me. is greatest. (Hint: Where is (curl F) • k positive?) 39. Regions with many holes Green's Theorem holds for a region R with any rmite number of holes as long as the bounding curves

are smooth, simple, and closed and we integrate over each component of the boundary io the direction that keeps R on our immediate left as we go along (see accompanying figure).

b. Determine the integrand (aN/ax) - (aM/ay) for the curl form of Green's Theorem. c. Determioe the (double integral) limits of integration from your plot in part (a) and evaluate the curl integral for the circulation. 43. F ~ (2x - y)i + (x + 3y)j, C: The ellipse x 2 + 4y2 ~ 4 44. F ~ (2x' - y')i

45. F ~ x-'e"i

+ (x' + y')j,

2 C: The ellipse ~

I

+ (e"lnx + 2x)j,

C: The boundary of the region dermed by y ~ 1 andy ~ 2 (above) 46. F ~ xeYi

y2

+"9 ~

+ x' (below)

+ (4x 2 Iny)j,

C: The triangle with vertices (0, 0), (2, 0), and (0, 4)

16.5

16.5

I

Surfaces and Area

g43

Surfaces and Area We have dermed curves in the plane in three different ways:

11

II -

constant

__+,::-::;_ --"11 _ constant (II, v)

f(x)

Explicitform:

y

Implicitfonn:

F(x,yl

Paramc1ric vector fonn:

r(tl

=

~

0

~ f(t~

+ g(tll,

a" t "

h.

We have analogous def'mitions of surfaces in space:

R

;;!---+---. o , Curve 11 _ CODStInt

Explicitfonn:

z ~ f(x, y)

Implicitform:

F(x,y,z) -

o.

There is also a parametric form for surfaces that gives the position of a point on the surface as a vector function of two variables. We discuss this new form in this section and apply the form to obtain the area of a surface as a double integral. Double integral formulas fur areas of surfaces given in implicit and explicit forms IU'e then obtained as special cases

of the more general parametric formula.

Curve" _ constaD1 %

r(M_ 11) _ f(M, 11)1 + g(lI, v)J + 11(11,1/)11:, Position vector to surface point

~y FIGURE 16.37 A parametrized swfacc S ~sscd IS a vector function of two variables dcfmed on a region R.

Parametrizations of Surfaces Suppose

r(a,v) - f(a,v)1

+ g(a,v)l + h(a,v)k

(1)

is a continuous vector function that is dermed on a region R in the uv-plane and one-toone on the interior of R (Figure 16.37). We call the range of r the .urface S defined or traced by r. Equation (1) together with the domain R constitute a parametrization of the surface. The variables u and v are the parameten, and R is the parameter domain. To s:implify our discussion, we take R to be a rectangle def"med by inequalities of the form a SUS b, c S v S d. The requirement that r be one-to-one on the interior of R eDSW'eS that S does not cross itself. Notice that Equation (1) is the vector equivalent of three parametric equations:

x = /(u, v), EXAMPLE 1 Cone:

y = g(u,v),

z = h(u, v).

Find a parametrization of the cone

J;

OSzS 1.

.t - W+"J2

Solution Here, cylindrical coordinates provide a parametrization. A ~t (x,y, z) on the cone (Figure 16.38) has x = rcos6,y = rsinl1, and z = Yx 2 + y2 = r, with S r S 1 and 0 S 8 S 2'IT. Thking u = r and v = 6 in Equation (1) gives the parametrization

o

.

"

r(r, 6)

(,"os 6)1 + (r sin 6)1 + rk,

o :S r :S

1,

O:s 11

S

2'IT.

The parametrization is one-to-one on the interior of the domain R, though not on the boundary tip of its cone where r = O. • y

EXAMPLE 2 FIGURE 16.38 The cone in Example I can be parametrized using cylindrical coordinates.

~

Find a parametrization of the sphere x 2

+ y2 + z2

= a 2.

Solution Spherical coordinates provide what we need. A typical point (x, y, z) on the sphere (Figure 16.39) has x=asin~cos8, y=asinq,sin8, and z=acosq"

944

Chapter 16: Integration in Vector Fields

o ~ t/J ~ 7T, Q ~ 0 ~ 27T. Taking U = t/J and v

= 0 in Equation (1) gives the parame-

1rization r(, 9) - (a sin co, 9)i

+ (asin.p,in9)J + (ao ..
Again, the parametrization is one-to-one on the interior of the domain R, though not on its

boundary "poles" where t/J - 0 or t/J EXAMPLE 3

7T.

•

Find a parametrization of the cylinder x2

FIGURE 18.39 The sphere in Example 2 can be parametrized using spherical coordinates.

-

+ (y -

Q ~ z ~ 5.

3)2 = 9,

Solution In cylindrical coordinates, a point (x, y, z) has x - rcosO,y - rsinO, and z - z. For points on the cylinder x 2 + (y - 3)2 - 9 (Figure 16.40), the equation is the same as the polar equation for the cylinder's base in the xy-p1ane: x2

+ (y2

- 6y

+ 9) - 9 ,,2+yl=rl,

r2 - 6rsinB = Q z

Cylindcr: x Z + (y - 3)Z = 9

or

oc r=6ainO

,

y=rllinO

r = 6 sinO,

Q:sO:S7T.

A typical point on the cylinder therefore has

x - rcos6 - 6sinBcos6 - 3 sin 26 Y = rsinB = 6sin2 B z = z.

TaldDg u - B and v - z in Equation (1) gives the one-to-one parametrization

r(B,z) - (3sin28)I+(6sinZB)j+zk, FIGURE 18.40 The cylinder in Example

3 can be parametrized using cylindrical coordinates.

O:SB:S7T,

Q:Sz:S5.

•

Surface Area Our goal is to fwd a double integral for calculating the area of a curved surface S based on the parametrization r(u, v) = f(u, v)i

+ g(u, v)j + h(u, v)k,

a

~

u

~

b,

c ~ v ~ d.

We need S to be smooth for the construction we are about to carry out The deItnition of smoothness involves the partial derivatives of r with respect to II and v:

or

of

og

iIr

a!

ag

ah

au - au l + au j + auk

r ll

-

r"

= (Iv =

au 1+ ov j

+

ah (Iv

k.

DEFINmON Aparametrizedswfacer(u,v) ~ !(u,v); + g(u,v)i H(u,v)k is smooth if r II and r" are continuous and r II X r" is never zero on the interior of

the parameter domain. The condition that r" X r" is never the zero vector in the defInition of smoothness means that the two vectors rv and rv are nonzero and never lie along the same line, so they always determine a plane tangent to the surface. We relax this condition on the boundary of the domain, but this does not affect the area computations.

16.5

g45

Surfaces and Area

Now consider a small rectangle J:1A. w inR with sidcs on the lincs u - Uo," - Uo + .6.u, v - Vo, and v - 110 + &v (Figure 16.41). Each side of .:1A.rw maps to a curve on the surface S, and together these four curves bound a "curved patch element" dul/V' In the notation of the figure, the side v - Vo maps to curve Ct. the side u - Uo maps to C2. and their common vertex (uo. vol maps to Po.

,

,.

Parametrization

v d

Va + Au

"", 0

---- ~

----R

x

•

"

f

No+A"

b

• y

'.

FIGURE 1&.42 A magnified view ofa surface patch clement 40'....

FIGURE 16.41 A rectangular area element M..., in the uv-plane maps onto a curved patch elemtnt !J.u"" on S.

Figure 16.42 shows an enlmged view of .6.0'111/' The partial derivative vector r,,(1IO. vol is tangent to Cl at Po. Likewise, r v(lIO. vol is tangeDtto C2I1.tPO. The cross product ru X r" is normal to the surface at Po. (Here is where we begin to use the assmnption that S is smooth. We want to be sure that r 1/ X r 1/ '# 0.) We next approximate the surface patch element 4uvv by the pa.ral.lelogmm on the tangent plane whose sides are determined by the vectors .6.ur" and bovr" (Figure 16.43). The area oftbis plll'Bllelogmm is (2)

A partition of the region R in the llV-plune by rectangu1a:r regions .:UIIII induces a portition of the surface S into surface patch elements 40'wv' We defme the area of each s1D'face patch

i"", A,'C' x

y

FIGURE 16.43

The area of the

parallelogram dctcmrincd by the vccton Aur. and .&vr"is dcImcd to be the area of

element 40'wv to be the pamllelogram area in Equation (2) and sum these areas together to obtain an approximation of the surface area of S:

~ Ir, X

,

r.1

4.4v.

(3)

As 4u and 4v approach zero independently, the number of area elements 1'1 tends to 00 and the continuity of ru and r ll guarantees that the sum in Equation (3) approaches the double integral !cd lab Iru X rill du area of the surface S.

dv. This double integral over the regionR derIDeS the

the surface patch element Au.....

DEFlNmON

«•. v)

~

The area of the smooth surface

f( •• v)i

+ g(•• v)i + h(•• v)"

is A

~ jJ1r, X r.1 dA ~ R

a~u~b.

1'1' Ir,

X

r.1

d.dv.

c~v~d

(4)

946

Chapter 16: Integration in Vector Fields

We can abbreviate the integral in Equation (4) by writing du for Ir. X rv I du dv. The surface area differential du is analogous to the arc length differential tis in Section 13.3.

Surface Area Differential for a Parametrized Surface

JJdU s

du = Ir. X rvl dudv

Differential formula for surface area

Surface area differential

EXAMPLE 4

(5)

Find the surface area of the cone in Example 1 (Figure 16.38).

In Example 1, we found the parametrization

Solution

r(r, 0)

= (r cos O)i

+

(r sin O)j

+ rk,

0 ,,;; r ,,;; 1,

0,,;; 0 ,,;; 2"..

To apply Equation (4), we first fmd r, X r.:

r, X r9

=

i cos 8

j sinO r cos 0

-rsin /I =

A =

=

0

-(rcosO)i - (rsinO)j

12"1' Ir, X r.1 drdO

+ (rcos2 /1 + rsin2 0)k. ,

~

Eq.(4)withu

2 12"1' VZr drdO = 1 "'(j dO =

EXAMPLE 5 Solution

k 1

,

T,V

,

~8

'(j (2".)

=

".VZ units squared.

Find the surface area ofa sphere of radius a.

We use the parametrization from Example 2: r(cf>, 0)

=

(asincf>cosO)i + (asincf>sinO)j o ,,;; cf> ";;,,., 0,,;; 0 ,,;; 2"..

+ (a cos cf»k,

For r., X re, we get

r", X r.

i

j

k

=

a cos cf> cos 0 -a sin cf> sin 0

a cos cf> sin 0 a sin cf> cos 0

-a sin cf> 0

=

(a sin cf>cosO)i

2 2

+

2

+ (a 2sincf>coscf»k.

(a 2 sin cf> sinO)j

Thus, Irq, X r.1

2 + a sin cf>sin20 + a sin2cf>cos2cf> sin cf> + a sin2cf> cos2cf> = y'a sin2cf> (sin2cf> + cos2cf»

=

y'a 4 sin4 cf>cos /1

=

y'a 4

=

a 2 y'sin2 cf>

4

4

4

=

a 2 sin cf>,

4

4

4

•

16.5 Surfaces and Area

since sin cf>

;:::: 0 for 0 :::; cf> :::; A= =

947

1T. Therefore, the area of the sphere is

121rl1r a 2sin cf> dcf> de 121r [-a 2cos cf>

I

de

=

121r 2a 2de =

41Ta

2

units squared.

•

This agrees with the well-known formula for the surface area of a sphere.

z

EXAMPLE 6 Let S be the "football" surface formed by rotating the curve x = cos z, y = 0, -1T/2 :::; z :::; 1T/2 around the z-axis (see Figure 16.44). Find a parametrization for S and compute its surface area.

A

~

/ () -

x

r = cos z is the radius of a circle

J. at height z ________ (x, y, z)

y

Solution Example 2 suggests finding a parametrization of S based on its rotation around the z-axis. If we rotate a point (x, 0, z) on the curve x = cos z, y = 0 about the z-axis, we obtain a circle at height z above the xy-plane that is centered on the z-axis and has radius r = cos z (see Figure 16.44). The point sweeps out the circle through an angle of rotation e,o :::; e :::; 21T. We let (x, y, z) be an arbitrary point on this circle, and define the parameters u = z and v = e. Then we have x = r cos e = cos u cos v, y = r sin e = cos u sin v, and z = u giving a parametrization for S as r (u, v )

. v J. + u, k = cos u cos V I.+ cos u sm

_1T<'U<.1T 2 - 2'

0 :::; v:::; 21T.

Next we use Equation (5) to find the surface area of S. Differentiation of the parametrization gives FIGURE 16.44 The "football" surface in Example 6 obtained by rotating the curve x = cos z about the z-axis.

ru

=

-sin u cos v i-sin u sin v j

+

k

and rv

= -cos u sin v i + cos u cos v j

Computing the cross product we have

ru X rv

=

=

j -sinusinv cosucosv

-sinucosv -cosusinv

k 1 0

-cos u cos v i-cos u sin v j - (sin u cos u cos2 V

+

cos u sin u sin2 v)k.

Taking the magnitude of the cross product gives

Ir u

X rv

I=

V cos2 U (cos2 V

+

sin2 v)

+

sin2 U cos2 U

= Vcos2 U (1 + sin2 u) = cosu VI + sin2 u.

cos U

~

0 for

From Equation (4) the surface area is given by the integral

A =

1 21rl1r/2cosu VI o -1r/2

+ sin2 ududv.

7T -2:S

u :s

27T

948

Chapter 16: Integration in Vector Fields

To evaluate the integml, we substitute w = sin u and dw = cos u du, - 1 ~ w ~ 1. Since the surface S is symmetric across the xy-plane, we need only integrate with respect to w from 0 to 1, and multiply the result by 2. In 1IUIllIllII.l"), we have A - 2

Jof" Jo[I Vl+w

1

dwdv

=2£lW[;VI +w2+t1n(w+ VI +w )]:av 2

~

f

~

2..[Vi + In(l + Vi)].

Integral Thblo Formula 35

2[tVi +tln(l + Vi)]dV.

•

Implldt Surfaces Surfaces are often presented as level sets of a function, described by an equation such as F(x, y, z)

Surface F~,:y, .t) _ c

=

c,

for some constant c. Such a level surface does not come with an explicit parametrization, and is called an implicitly defined surface. Implicit surfaces arise, for example, as equipotential surfaces in electric or gravitational fields. Figure 16.45 shows a piece of such a surface. It may be difficult to fmd Cllplicit formulas for the functions /, g, and h that describe the surface :in the form r{u, v) = /(u, v)1 + g(u, v)J + h(u, v)k. We now show how to compute the surface area differential for implicit surfaces. Figure 16.45 shows a piece of an implicit surface S that lies above its "shadow" region R:in the plane beneath it. The surface is defined by the equation F(x,y, z) = c and p is a unit vector normal to the plane region R. We assume that the surface is smooth (F is differentiable and VF is nonzero and continuous on S) and that VF' P =F 0, so the surface never folds back over itsel£ Assume that the normal. vector P is the unit vector k, so the region R in Figure 16.45 lies in the xy-plane. By assumption, we then have VF' P = VF' k = Fz =I- 0 on S. An advanced calculus thoorem called the Implicit Function Theorem implies that S is then the graph of a differentiable function z = h(x, y), although the function h(;K, y) is not Cllplicitly known. Define the parameters u and v by u = x and v = y. Then z = h(u, v) and

au

Tho ."",.;'" oroiodion or "IIhadow" of Son.

r(u, v) = vi

coordinate plane

FIGURE 18.45 At. we soon see, the area of a surface S in space can be calculated by evaluating a relatM double integral over the vertical projection or "shadow" of S on a coordinate plane. The unit vector p is normal to the plane.

+ vj + h(u, v)k

(6J

gives a parame1rization of the surface S. We use Equation (4) to f"md the area of S. Calculating the partial derivatives of r, we f"md ru - i

ah + auk

ru - j

ah + avk.

Applying the Chain Rule for implicit differentiation (see Equation (2) in Section 14.4) to F(x,y, z) = c, where x = u,y = v, andz = h(u, v), we obtain the partial derivatives

ah

au

F, ~

F,

and

ah

au

F, ~

F;

Substitution of these derivatives :into the derivatives of r gives

""d

16.5

Surfaces and Area

949

From a routine calculation of the cross product we find

1

+

=

F. (F;rl

Fyj

=

VF = VF F" VF'k

+

F"k)

VF - VF·p·

po,

Therefore, the surface area differential is given by IVFI

du ~ Ir. X r.1 dudv ~ IVF.pldxdy.

u - xaruiv - y

We obtain similar calculations if instead the vector p = j is normal to the xz-plane when Fy =# 0 on S, or if P = i is normal to the yz-plane when Fx =I- 0 on S. Combining these results with Equation (4) then gives the following general formula.

Formula for the SurfaeeArea of an ImpUclt Surface The area ofthc surface F(x,y,z) = c over a closcdand bounded p1anercgionR is

Surface area ~

rr IVF.pl IVFI dA.

JJ

(7)

R

wherep - i,J, ork is normal toR and VF'p "'" O.

Thus, the area is the double integral over R of the magnitude of VF divided by the magnitude of the scalar component of VF normal to R. We reached Equation (7) under the assumption that VF' P =I- 0 throughoutR and that VF is continuous. Whenever the integral exists, however, we derme its value to be the area of the portion of the surface F(x,y,z) = c that lies over R. (Recall that the projection is asswned to be one-to-one.)

, EXAMPLE 7 x2

+ y2

-

Z

Find the area of the surface cut from the bottom of the pamboloid = Obytheplanez = 4.

Solution We sketch the surface S and the region R below it in the xy-p1anc (Figure 16.46). The surface S is part of the level surface F(x,y, z) - x 2 + y2 - Z - 0, and R is the disk x 2 + yl :s; 4 in the xy-plane. To get a unit vector normal to the plane of R, we can take p - k. At any point (x,y, z) on the surface, we have F(x,y,z) = Xl + y2 - z VF - 2xI+2yJ-k

IVFI ~ "11(2<)' FIGURE 16.46 The area of this parabolic surface is calculated in Example 7.

IVF·pl

+ (2y)' + (

=

V4x 2 + 4yl

~

IVF·kl

~

+

1-11

1 ~

I.

1)'

950

Chapter 16: Integration in Vector Fields In the region R, dA = dx dy. Therefore,

ff

Surface area =

IVFI IVF _p I dA

11 R

11

Eq. (7)

V4x 2 + 4y2 + I dxdy

"+;"4 = =

10f2"10!' V4r2 + Irdrd6 f2"

[l

Polar coordinates

(4r2 + 1)3/2]2 d6

10 12 0 !''' 1~(173/2_I)d6= ~(17V17-I). = 10

•

Example 7 illustrates how to find the surface area for a function z = I(x, y) over a region R in the xy-plane. Actually, the surface area differential can be obtained in two ways, and we show this in the next example.

EXAMPLE 8 Derive the surface area differential du of the surface z = I(x, y) over a region R in the ~-plane (a) parametrically using Equation (5), and (b) implicitly, as in Equation (7). Solution (a) We parametrize the surface by taking x = u, y = v, and z = I(x, y) over R. This gives the parametrization r(u, v) = ui + vj + I(u, v)k Computing the partial derivatives gives r.

r.

X

rv

= -

+ i + I. k, r v

I.i - I,J + k.

1

o

= j j

k

0 1

f. f.

+ Iv k and

Thenlr. X rvldudv = V/. 2 + Iv 2 + I dudv. Substilutingforuandvthengives the surface area differential

du

=

VI.' + 1/ + I dxdy.

(b) We define the implicit function F(x, y, z) = I(x, y) - z. Since (x, y) belongs to the regionR, theunitnonnal to the plane ofR is p = k. Then VF = Ix i + Iyj - k so that

IVF-pl = I-II = I,IVFI = V/x 2 + face area differential is again given by

du

=

1/ + l,andIVFI/IVF-pl = IVF~ Thesur-

VI.' + 1/ + I dxdy.

•

The surface area differential derived in Example 8 gives the following fonnula for calculating the surface area of the graph of a function deIfied explicitly as z = I(x, y).

Formula for the Surface Area oCa Graph z = f (x, y) For a graph z = I(x, y) over a region R in the ~-plane, the surface area fonnula is

A

=

11 R

VI.' +

f/ + Idxdy.

(8)

951

16.5 Surfaces and Area

Exercises 16.5 Finding Parametrizations In Exercises 1-16, f"md a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.) 1. The paraboloid z = x 2

+ y2, Z

2. Theparaboloidz = 9 - x 2

-

~ 4

y2,z ~ 0

3. Cone frustum The first-octant portion of the cone z v'x' + y'/2 between the planes z = 0 and z = 3 4. Cone frustum The portion of the cone z = between the planes z = 2 and z = 4 5. Spherical cap Ve cap cut from the sphere bytbeconez= x 2 +y2

=

2v'x' + y'

x' + y' + z' =

9

x'

6. Spherical cap The portion of the sphere + y' + z' = 4 in the fIrst octant between the xy-plane and the conez = v'x' + y' 7. Spherical band

The portion of the sphere

between the planes z

x' + y' + z' =

3

= v'3/2 and z = - v'3/2

8. Spherical cap The upper portion cut from the sphere + y' + z' = 8 by the plane z = -2

x'

9. Parabolic cylinder between plane. The surface cut from the parabolic cylinder z = 4 - y' by the planes x = 0, x = 2, and z=O 10. Parabolic cylinder between plane. The surface cut from the parabolic cylinder y = by the planes z = 0, z = 3, and y = 2

x'

11. Circular cylinder band The portion ofthe cylinder y' + z' = 9 between the planes x = 0 and x = 3

x'

+ z' = 4 12. Circular cylinder baud The portion of the cylinder above the xy-plane between the planes y = - 2 and y = 2 13. Tilted plane inside cylinder z= I L

Inside the cylinder

x' +

y'

The portion of the plane x + y =

+

The portion of the plane

x-y+2z=2 Inside the cylinder

x' +

z' = 3

I>- Inside the cylinder y' + z' = 2 15. Circular cylinder baud The portion of the cylinder (x - 2)' + z' = 4 between the planes y = 0 and y = 3 16. Circular cylinder band The portion of the cylinder y' (z - 5)' = 25 between the planes x = 0 and x = 10

20. Cone frustum The portion of the cone z = between the planes z = I and z = 4/3

v'x'

21. Circular cylinder band The portion of the cylinder between the planes z = I and z = 4

x' + y' =

22. Circular cylinder band The portion of the cylinder 10 between the planes y = -I and y = I by the cone z =

v'x'

+

+

y'/3 I

x' + z' =

The csp cut from the parsholoid z = 2 -

23. Parabolic cap

x' - y'

y'

24. Parabolic band The portion of the paraboloid z = between the planes z = I and z = 4

x' + y'

25. Sawed-off sphere The lower portion cut from the sphere x' + y' + z' = 2 by the cone z = v'x' + y' 26. Spheric.. baud The portion of the sphere between the planes z = -I and z = v'3

x' + y' +

z' = 4

Planes Tangent to Parametrized Surfaces The tangent plaoe at a poiot Po(f(Uo, vol, g(uo, vol, h(uo, vo)) on a parametrized surface r(u, v) = f(u, v)i + g(u, v)j + h(u, v)k is the plane througbPo normal to the vectorr.(uo, vol X rv(uo, vol, the cross product of the tangent vectors r.(uo, vol aod rv(uo, vol at Po. In Exercises 27-30, f"md an equation fur Ibe plane tangent to Ibe surface at Po. Theo f"md a Cartesiao equation fur the surface aod sketch the surface aod tangent plane together. 27. Cone The cone r(r,8) = (r cos 8)i + (rsin 8)j + rk, r '" 0, 8 '" 21T at Ibe point Pot 2) corresponding to

o '"

(r,8)

v'2, v'2,

= (2,1T/4)

+

(1T/6,1T/4) 29. Circular cylinder The circular cylinder r(8, z) = (J sin 28)i + (6sin'8)j +zk,O '" 8 '" 1T, at the point Po(3v'3/2,9/2,0) corresponding to (8, z) = (1T/3, 0) (See Example 3.) 30. Parabolic cylinder The parsholic cylinder surface r(x, y) = xi + yj - x 2k, -00 < X < 00, -00 < Y < 00, at the point Po(1,2,-I)correspondingto(x,y) = (1,2)

More Parametrizatlons of Surfaces 31.

Surface Area of Parametrized Surfaces In Exercises 17-26, use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the sarne values, however.) 17. TIlted plane inside cylinder The portion of the plane y + 2z = 2 inside the cylinder + y' = I

x'

v'x' + y'

v'2, v'2,

I>- Inside the cylinder y' + z' = 9

L

x'

19. Cone frustum The portion of the cone z = 2 between the planes z = 2 and z = 6

28. Hemisphere The hemisphere surface r(, 8) = (4 sin cos 8)i + (4sinsin8)j + (4 cos '" 1T/2,0 '" 8 '" 21T, at the poiot Pot 2v'3) corresponding to (,8) =

9

14. TIlted plane inside cylinder

18. Plane inside cylinder The portion of the plane z = -x inside + y' = 4 the cylinder

L

A torus ofrevolution (doughnut) is obtained byrotatiog a circle e in the xz-plane about the z-axis in space. (See the accompanying fIgure.) If e has radius r > 0 and center (R, 0, 0), show that a parametrization of the torus is

r(u,v)

where 0 fIgure.

S

= «R + rcosu)cosv)i + «R + rcosu)sinv)j

+ (rsinu)k,

u :s 2'1T and 0 :s v :s 21T are the angles in the

952

Chapter 16: Integration in Vector Fields

b. Show that the surface area of the torus is A: = ~Rr.

34. Hyperboloid of one meet L

c c.~-----+--~-Lt- ·

Find a parametrization for the hyperboloid of one sheet :x2 + y2 - z2 _ I in terms of the angle II associated with the circle:x 2 + y2 _ ,2 and the hyperbolic parameter u associated with the hypeIbolic function,2 - z2 - 1. (Hint. cosh 2 v - sinh 2 v - 1.)

b. Generalize the result in part (a) to the hyperboloid (xl/a 2) + (y2/b 2) - (z2/ c 2) _ 1.

35. (Contbwation ofExercise 34.) Find a Cartesian equation for the plane tangent to the hyperboloid:x 2 + y2 - z'2 - 2S at the point (::to,Yo, 0), where:xo2 + Y02 - 25. 36. Hyperboloid of two lbeet. Find a parametrization of 1he hyperboloid of two sheets (z2/c2) - (:x 2/a 2) - (y2/b 2) - 1.

32. Parametrization of a mrfaee of revolution Suppose that the parametrized curve C: (f(v), g(u» is revolved about the :x-axis, whereg(v) > ofora :!;'; u:!;'; b.

a. Showthat r(v, v) - I(v)i

+ (g(V)C08V)j + (g(v)sinv)k

is a parametrization of tlu: :resulting surface of revolution, where 0 :5: v :5: 2'1T is 1he angle from the xy-plane to the point r(u, v) on the sur6we. (See 1he accompanying figure.) Notice that I(v):measures distance aJongthe axis of revolution and g(u) measures distancefrom 1he axis ofrevolution.

,

abave the triangle bounded by the lines :x - V3,y - 0, and y = x in tlu: xy-plane. 41. Find the area of the surface:x2 - 2y - 2z - 0 that lies above the triangle bounded by the lines:x - 2, y - 0, and y - 3:x- in the xyplane. 41. Find the areaoftlu: cap cut from the spbere:x2 + y2 + z2 _ 2by the canez _ Y:x2 + y2. 43. Find 1he area of the ellipse cut from. the plane z - ex (c a constant) by 1he cylinder:x2 + y2 - 1.

(/(11), 8(N), 0)

r(II. v)

\

"-f(u)

Surface Ani for Implklt and ExpUdt Forms 37. Find the area of the surface cut from. the paraboloidx2 + y2 - Z = Obytlu: planez = 2. 38. Find the area of the band cut from. the paraboloid x 2 + y2 - Z = Obytlu: planes z = 2andz = 6. 39. Find the area of tlu: region cut from the plane:x + 2y + 2z - 5 by the cylinder whose walluI'C x = y 2 andx = 2 - y~ 40. Find the area of the portion of the surface :x2 - 2z - 0 that lies

/ c

....-:;

b. Find a parametrization for the surface obtained by revolving the curve:x - y2,y ~ 0, about the:x-axis.

33. a. Parametriation of an eDiplOid Recall the parametrization :x - a C06 lI,y - bsinll,O:!;'; lI:S 2'1r frrthe ellipse (:x 2/a 2) + (y2/b 2) _ I (Section 3.9, Example S). Using the angles 9 and tfJ in spherical coordinates, show that

+ (bsinllc06tfJ)j + (csintfJ)k isapamnetrization oftheeUipsoid (xl/a 2) + (y2/b 2) + r(9,tfJ) - (acos8costfJ)i

(z2/ C2) = 1. b. Write an integral for the surface area of the ellipsoid, but do not evaluate the integral.

44. Find the area of the upper portion of the cylinder x 2 + z2 = 1 that lies between theplaneu - ±1/2 andy - ±1/2. 45. Find 1he area of 1he portion of 1he paraboloid :x _ 4 - y2 - z2 that lies above the ring 1 :5: y2 + z2 :s 4 in the yz-plane.

46. Findtheareaof1hesurfacecutfrom.1heparaboloiib2 + y + z2 _ 2 by the plane y - O. 47. Find the area of the surface:x 2 - 2ln:x + z - 0 abave tlu: square R: I :!;';:X :s 2, 0 :!;'; y :!;'; I, in the xy-plane. 48. Find the area of the surface 2:x3/2 + 2y3/2 - 3z - 0 above the square R: 0 :!;'; x :!;'; I, 0 :s y :!;'; I, in the xy-plane.

Wy -

Find the area of the surfaces in Exercises 49-54. 49. The surface cut from the bottom. of the paraboloid z _ :x2 + y2 by tlu: plane z - 3 SO. The surface cut from. the ''nose'' of the paraboloid :x - 1 y2 _ z'2 by the yz-plane 51. The portion of the cane z - V:x 2 + y2 that lies over the region between the circle:x 2 + y2 _ 1 and the ellipse 9:%-2 + 4y2 - 36 in the xy-plane. (Hint: Vie formulas from. geometry to fmd the area of the region.) 51. The triangle cut from tlu: plane 2:x + 6y + 3z - 6 by the bounding planes of the flfSt octant. Calculate the area three ways, using di:ffurent explicit forms. 53. The surface in the fust octant cut from the cylinder y - (2/3)z3/2 by tlu: planeu - I and y - 16/3

16.6 Surface Integrals

953

,

st. The portion of the plane Y + z = 4 thatli.es above the region aut from the fll8t quadrant of the xz-plane by the parabola

x=4-z 2

55. Use the parametrization

~';;01--~ J

r(x, z) = xi

+ f(x, z)j + zk

and Equation (5) to derive a furmula fur du associated with the

explicit form y = I(x, z). 56. Let S be the surface obtained by rotating the smooth cmrve y = 1'<'), a :s x :s h, about the x-axis, when: I(x) 2: O. L

Shaw that the vector function

r(x,9) =xi

+ f'<')cos8j + f(x)sin8k

b. Use Equation (4) to shaw that the surface area of this surface of revolution is given by

is II. parametrization of S, when: 8 is the angle of rotation around the x-axis (see the accompanying figure).

16.6

rl

A

~ /.' 2,
_ Su_ria _c_e_l_ nt~ eg~~_~________________________________ To compute quantities such as the flow ofliquid across a curved membrane or the upward furoe on a falling parachute, we need to integrate a function over a curved surface in space. This concept of a surface integral is an extension of the idea of aline integml for integrating over a curve.

Surface Integrals

Avril

1);

Ao:t -

----- (X,l;,y,l;,q)

Suppose that we have an electrical charge distributed over a surface S, and that the function G(x,y, z) gives the charge density (charge per unit area) at each point on S. Then we can calculate the total charge on S as an integral in the following way. Assume, as in Section 16.5, that the surface S is defined. parametrically on a region R in the uv-p1ane, r(u,v) = f(u,v)i

+ g(u,v)j + h(u,v)k,

(u,v)eR.

Au...

FIGURE 16.47 The area of the patch 4Uk is the area of the tangent paral.lelogram determined by the vectors 4u r .. and 4v r". The point (XhYb Z}) lies on the !UIfaoe patch, beneath the paral.lelogram shawn 1u:rc.

In Figure 16.47, we see how a subdivision of R (considered as a rectangle for simplicity) divides the surface S into corresponding curved surface elements, or patches, of area

tJ.u""

Rl

Ir.. X rvl

dudv.

As we did for the subdivisions when defming double integrals in Section 15.2, we number the surface element patches in some order with their areas given by dul. tJ.uz, ••• , tJ.ull• To form a Riemann sum over S, we choose a point (X},y,.. z.t) in the kth patch, multiply the value of the function G at that point by the area tJ.U,b and add together the products:

•

~

G(Xob Job

z.t) tJ.u,t.

Depending on how we pick (Xob Job z.t) in the kth patch, we may get different values for this Riemann sum. Then we take the limit as the number of surface patches increases, their areas shrink to zero, and both 4u --+ 0 and tJ.v --+ O. This limit, whenever it exists independent of all choices made, delmes the .urface integral of G over the Ianace 80s (!)

954

Chapter 16: Integration in Vector Fields Notice the analogy with the definition of the double integral (Section 15.2) and with the line integral (Section 16.1). If S is a piecewise smooth surface, and G is continuous over S, then the surface integral defined by Equation (I) can be shown to exist. The formula for evaluating the surface integral depends on the manner in which S is described, parametrically, implicitly or explicitly, as discussed in Section 16.5.

Formulas for a Surface Integral

1. For a smooth surface S defined parametrically as r(u, v) = I(u, v)i + g(u, v)j + h(u, v)k, (u, v) eR, and a continuous function G(x, y, z) defined on S, the surface integral of Gover S is given by the double integral over R,

JJ G(x,y,Z)dCT

=

JJ G(/(u, v),g(u, v),h(u, v)) Iru X rvl dudv.

S

(2)

R

2. For a surface S given implicitly by F(x, y, z) = c, where F is a continuously differentiable function, with S lying above its closed and hounded shadow region R in the coordinate plane beneath it, the surface integral of the continuous function G over S is given by the double integral over R,

JJ

G(x,y,Z)dCT

S

=

JJ G(x,y'Z)I~;~~1

(3)

dA,

R

where p is a unit vector normal to R and VF' p

'¢'

O.

3. For a surface S given explicitly as the graph of Z = I(x, y), where I is a continuously differentiable function over a region R in the xy-plane, the surface integral of the continuous function G over S is given by the double integral over R,

JJ G(x, y,z) dCT S

=

JJ G(x, y,f(x, y))

V/. 2 + I; + I dxdy.

(4)

R

The surface integral in Equation (I) takes on different meanings in different applications. If G has the constant vaIue I, the integral gives the area of S. If G gives the mass density of a thin shell of material modeled by S, the integral gives the mass of the shell. If G gives the charge density of a thin shell, then the integral gives the total charge.

EXAMPLE 1

Integrate G(x, y, z) = x 2 over the cone Z =

Vx 2 + y2, 0 :5 Z

:5

I.

Solution Using Equation (2) and the calculations from Example 4 in Section 16.5, we have Irr X rei = Vir and

JJX 2 dCT= /.2w/"(r 2 cos2 6)(Vir)drd6 S

x~rcos9

955

16.6 Surface Integrals

Surface integrals behave like other double integrals, the integral of the sum of two functions being the sum of their integrals and so on. The domain Additivity Property takes the form

11

Gda =

11

Gda

+

s]

S

11

Gda

11

+ ... +

Gda.

Sn

S2

When S is partitioned by smooth curves into a finite number of smooth patches with nonoverlapping interiors (i.e., if S is piecewise smooth), then the integral over S is the sum of the integrals over the patches. Thus, the integral of a function over the surface of a cube is the sum of the integrals over the faces of the cube. We integrate over a turtle shell of welded plates by integrating over one plate at a time and adding the results.

EXAMPLE 2 Integrate G(x, y, z) = xyz over the surface of the cube cut from the first octant by the planes x = 1, y = 1, and z = 1 (Figure 16.48).

z

J :

I I I I I

1-_

/ 0

x

SoLution We integrate xyz over each of the six sides and add the results. Since xyz = 0 on the sides that lie in the coordinate planes, the integral over the surface of the cube reduces to

~

--_

///

/ /f

SMd

/

11 --- ~ i idec

Y

SideB

FIGURE 16.48

11

xyz da =

xyz da

11

+

Side A

Cube surface

xyz da

+

SideB

11

xyz da.

SideC

Side A is the surface f(x, y , z) = z = lover the square region Rxy: 0 1, in the xy-plane. For this surface and region,

:5 X :5

o :5 Y :5

p = k,

Vf

=

k,

The cube in Example 2.

da =

IVfl

=

1,

IVf'pl

IVfl IVf ' pl dA

=

1 Tdxdy = dxdy

=

Ik'kl

=

1,

1

xyz = xy(1) = xy and

Symmetry tells us that the integrals ofxyz over sides Band C are also 1/4. Hence,

ff

}}

xyz da =

1

1

1

"4 + "4 + "4

=

3

•

"4 '

Cube surface

EXAMPLE 3 Integrate G (x, y , z) = VI - x 2 - y2 over the "football" surface S formed by rotating the curve x = cos z, y = 0 , -1T/2 :5 z :5 1T/2, around the z-axis. SoLution The surface is displayed in Figure 16.44, and in Example 6 of Section 16.5 we found the parametrization X

=

cos u cos v,

y = cos u sin v,

z = u,

1T

1T

-"2:5 U :5"2

and

0:5

v

:5 21T,

where v represents the angle of rotation from the xz-plane about the z-axis. Substituting this parametrization into the expression for G gives

The surface area differential for the parametrization was found to be (Example 6, Section 16.5) da = cos u

VI

+ sin2 u du dv.

956

Chapter 16: Integration in Vector FieLds

These calculations give the surface integral

fJif

\11 -

x2

-

y2

du

=

1

21T 11T/2

o

S

-1T/2

Isin u Icos u \11 + sin2 ududv

{21T {1T/2 Jo sin u cos u

= 2 Jo

_1

-

w=1+sin2 u

00

21T

12 ~

o

1

I d d Vw W v

2 32 = 21T 3 w / 0

\11 + sin2 u du dv dw = 2 sin u c;s u du Whenu=Ow=l. ' When u = 'TT/2, w = 2.

]2 = 4; (2V2 - 1).

•

1

Orientation

FIGURE 16049 Smooth closed surfaces in space are orientable. The outward unit normal vector defines the positive direction at each point. c

d

a

b

Start

We call a smooth surface S orientable or two-sided if it is possible to define a field n of unit normal vectors on S that varies continuously with position. Any patch or subportion of an orientable surface is orientable. Spheres and other smooth closed surfaces in space (smooth surfaces that enclose solids) are orientable. By convention, we choose n on a closed surface to point outward. Once n has been chosen, we say that we have oriented the surface, and we call the surface together with its normal field an oriented surface. The vector n at any point is called the positive direction at that point (Figure 16.49). The Mobius band in Figure 16.50 is not orientable. No matter where you start to construct a continuous unit normal field (shown as the shaft of a thumbtack in the figure), moving the vector continuously around the surface in the manner shown will return it to the starting point with a direction opposite to the one it had when it started out. The vector at that point cannot point both ways and yet it must if the field is to be continuous. We conclude that no such field exists.

Surface Integral for Flux Suppose that F is a continuous vector field defined over an oriented surface S and that n is the chosen unit normal field on the surface. We call the integral of F n over S the flux of F across S in the positive direction. Thus, the flux is the integral over S of the scalar component of F in the direction of n. 0

FIGURE 16050 To make a Mobius band, take a rectangular strip of paper abed, give the end be a single twist, and paste the ends of the strip together to match a with e and b with d. The Mobius band is a nonorientable or one-sided surface.

DEFINITION The flux of a three-dimensional vector field F across an oriented surface S in the direction of n is Flux =

11

Fondu.

(5)

s

The definition is analogous to the flux of a two-dimensional field F across a plane curve C. In the plane (Section 16.2), the flux is

1

Fonds,

the integral of the scalar component ofF normal to the curve. If F is the velocity field of a three-dimensional fluid flow, the flux of F across S is the net rate at which fluid is crossing S in the chosen positive direction. We discuss such flows in more detail in Section 16.7.

16.6 Surface Integrals

EXAMPLE 4

z

y

=

x

2

,

0

:5 X

957

Find the flux of F = yzi + xj - z 2 k through the parabolic cylinder :5 1,0 :5 z :5 4, in the direction n indicated in Figure 16.51.

SoLution On the surface we have x = x, y = x 2 , and z = z, so we automatically have the parametrization rex, z) = xi + x 2j + zk,O :5 x :5 1,0 :5 z :5 4. The cross product of tangent vectors is k

rx X rz = 1

j 2x

o

0

1

I

n"-- ~-. 1

__~ /'

,1

, y

j.

The unit normal vectors pointing outward from the surface as indicated in Figure 16.51 are n =

x

FIGURE 16051 Finding the flux through the surface of a parabolic cylinder (Example 4).

o = 2xi -

rx X rz Irx X rzl

2xi - j

= -----;====;::=====

\/4x 2 + 1·

On the surface, y = x 2 , so the vector field there is

Thus, F n 0

= \/4X; + 1 ((x 2z)(2x) + (x)( -1) + (-z2)(O)) 2x 3z - x

\/4x 2 + 1 . The flux of F outward through the surface is

1 ]X=1x=o dz z - -x 2 2 1o 10 (2x z - x) dxdz = 14 [1-x 4 1

=

3

4

0

=

41- (z -

1

o 2

1

2

0

1) dz

1]4 (z - 1)2

= -

4

0

1

= 4(9) - 4(1) = 2. If S is part of a level surface g(x, y, z) fields

•

= c, then n may be taken to be one of the two Vg

(6)

n = ±IVgl'

depending on which one gives the preferred direction. The corresponding flux is Flux =

11

Fondu

s

=

rr (F0 ±Vg) IVgopl IVgl dA

11

1Vg1

Eqs. (6) and (3)

R

=

rr FOIVgopl ±Vg dA.

11 R

(7)

958

Chapter 16: Integration in Vector FieLds EXAMPLE 5 Find the flux of F = yzj + z 2k outward through the surface S cut from the cylinder y2 + z2 = 1, z 2':: 0, by the planes x = and x = 1.

I;

Rx> t

(1, -1, 0)

°

SoLution The outward normal field on S (Figure 16.52) may be calculated from the gradient of g(x,y, z) = y2 + z2 to be

s

1 ----- y

(1, 1,0)

Vg +IVgl

n =

With P

=

=

2yj

+ 2zk

~ I.

2vl

•

= Yl + zk.

k, we also have

x

FIGURE 16.52 Calculating the flux ofa vector field outward through the surface S. The area of the shadow region Rry is 2 (Example 5).

2yj + 2zk Y4y2 + 4z2

du =

IVgl IVg.kl

2

We can drop the absolute value bars because z The value of F • n on the surface is F • n = (yzj =

y2z

=

z.

12z1

dA =

2'::

1 dA = zdA.

°on S.

+ z 2k) • (yj + zk) + z3 = z(y2 + z2)

The surface projects onto the shadow region Rxy, which is the rectangle in the xy-plane shown in Figure 16.52. Therefore, the flux ofF outward through S is

• Moments and Masses of Thin Shells Thin shells of material like bowls, metal drums, and domes are modeled with surfaces. Their moments and masses are calculated with the formulas in Table 16.3. The derivations are similar to those in Section 6.6. The formulas are like those for line integrals in Table 16.1, Section 16.1. TABLE 16.3 Mass and moment formulas for very thin shells

Mass:

M =

11

8 du

8 = 8(x,y,z) = density at (x,y,z) as mass per unit area

s

First moments about the coordinate planes:

Myz

=

11

Mxz

x8du,

s

=

11

y 8 du,

s

Coordinates of center of mass:

Moments of inertia about coordinate axes:

r(x,y, z)

=

distance from point (x,y, z) to line L

16.6 Surface Integrals

EXAMPLE 6

959

Find the center ofmass of a thin hemispherical shell of radius a and con-

""'" density 6.

Solution We model the shell with the hemisphere

s

•

•

f(x,y, z) = x" y

x

FIGURE 16.53 Thecentl:rofmassofa thin hemispherical shell of constant density lies on the axis of symmetry haI:fway from the base to the ~ (Example 6).

+ y2 + z2 =

a 2,

z"o

(Figure 16.53). The symmetry of the surface about the z-axis tells us thatj = mains only to findi from the fonnulaz = Mxy/M• The mass of the shell is

JJs

M -

JJs

6 du - 6

O. Itre-

du - (6)(are. of S) - 2"0'6.

To evaluate the integral for MJiY' we take p

=

k and calculate

IVII

~

12xi + 2yj + 2m1

IVI'pl

~

IVI'kl ~ 12z1 ~ 2z IVII 0 IVI'pl dA ~ zdA.

da ~

Y=

~ 2V,'

+ y' + z,

~ 20

Then

Mxy -

JJ

Jf z~dA JJ

z6 du -

- 8a

I)

S

R

2 3 dA. - &(7Ta ) - &iTa

R

z--M --21Ta 8 --

_

Mxy

'1f'a 3l) 2

a

2'

•

The shell's C"",", ofDUlllS is the paID! (0, 0, 0/2).

EXAMPLE 7 Find the center of mass of a thin shell of density 8 cone z = Yx 2 + y2 by the planes z = 1 andz = 2 (Figure 16.54).

=

Solution The symmetry of the surface about the z-axis tells us that Z = Mxy/M. Working as in Example 4 of Section 16.5, we have

x= y =

r(r,6) = (rc088)i

+ (rsine)j + rk.

1

~

r

:S

2,

O:s

and

FIGURE 16.54 The cone frustmn formed 2 + y 2 iscutby whentheconez the planes z - 1 !IIld z - 2 (Example 7).

Therefon:,

Yx:

- VZJ,'"[lnr];d6 - VZJ,'"1n2d6 ~ 2..VZ1n2,

1/z2 cut from the

O. We :fmd

(J :S 217',

960

Chapter 16: Integration in Vector Fields

Mxy =

ff

Bzdu = 12'1r12 :2rV2rdrd6

S

=

Vi12'1r12 drdf)

1,"

~ v'? _

M,.

z~-=

M

dO

~ 2..v'?,

2..v'? v'? In 2

2..

1

=-

1n2'

•

The shell's center ofmass is the point (0, 0, I/ln2).

Exercises 16.6 Surface Integrals In Exercises 1--8, integrare the given function over the given surface.

1. Parabolie cy6nder G(x,y,z) = x, over the parabolic cylinder y = Xl, 0 :S X :S 2,0 :S z :S 3

2. Circalar cyIID.der

y2 +z2 =

G(x,y,z) = z, am: the cylindricalsuxfacc

I, 1)

4,z~ O,I:Sx:S4

3. Spbere G(x,y,z) =x 2,ovcrthe1lllitsphcR:x2 + yl +z2 = 1 4. Hemisphere G(x,y,z) = z2, aver the hcmisphcIe x 2 + y2 + z2=a 2,z2:0 5. Portion of plaDe F(x,y, z) = z, am: the portion of the plane x + y + z = 4 that lies .baw: the square 0 s x s I, o s y s I, in the xy-plane 6. Coae F(x,y, z) = z - x, am: the cone z = Yx 2 + y2, OSzsl

(0,0, y

yS - 4z, over the parabolic

7. Parabolie dome H(r,y,z) - x 2 domez - 1 - x 2 - y2,z 2: 0

8. SpberiuJ. eap H(x,y,z) - yz, over the part of the sphere x2 + y2 + 1'2 _ 4 that lies above the cone z _ Yx2 + y2 9. Integrate G(x,y,z) - x + Y + z over the surface of the cube cut from the fust octant by the planes x - a, y - a, Z - a. 10. Intqp-ate G(r,y, z) = y + I' aver the surl'ace of the wedge in the frrst octant bomded by the coordinate planes and the planes x - 2andy+z - 1. 11. Intqp-ate G(x,y,z) = X)'% over the surface of the rectangular solid cut from the frrst octant by the planes x - a, y - b, and z - c. 12. Intqp-ate G(x,y,z) = X)'% over the surface of the rectangular solid bounded by the planes x - ±a, y - ±b, and z - ±c.

(I, I, 0)

16. Integrate G(x,y,z) = x am: the !IUIfaoe given by

z=X2+y for Osxsl, -lSysl. 17. IntegrateG(x,y,z) - xyz am: the triangular surface with vertices (1, 0, 0), (0, 2, 0), om! (0, 1, 1).

I, 1)

13. Intqp-ate G(x,y,z) = x + Y + z aver the portion of the plane b + 2y + z = 2 that lies in the fmt octant

14. Intqp-atc G(x,y, z) = xVyl + 4 aver the surface cut from the parabolic cylinder y2 + 4z = 16 by the planes x = 0, x = I, andz = O. 15. Intqp-atc G(x,y,z) = z - x over the portion of the graph of z = x + y2 above the triangle in the xy-plane having vertices (0. o. 0). (1. 1.0). om! (0. 1. 0). (s.. " " _ figure.)

18. Integrate G(x,y,z) - x - y - z aver the portion of the plllllC x + Y = 1 in the fmt octant between z = 0 and z = 1 (lICe the accompanying f1jUIC).

16.6

Surface Integrals

961

+k + zyj + z 2k xi + yj + zk xi + yj + zk V2 2 x + y2 + z

33. F(x,y,z) = yi - xj

z

34. F(x,y,z) = zxi 35. F(x,y,z)

=

(1,0, 1)

36. F(x,y,z) = - ~"""-- y

x

Finding FLux Across a Surface In Exercises 19-28, use a parametrization to find the flux ffsF· n du across the surface in the given direction. 19. Parabolic cylinder F = z 2i + xj - 3zk outward (normal away from the x-axis) through the surface cut from the parabolic cylinder z = 4 - y2 by the planes x = 0, x = 1 , and z = 0 20. Parabolic cylinder F = x 2j - xzk outward (normal away from the yz-plane) through the surface cut from the parabolic cylinder y = x 2 , -1 :::; x :::; 1 , by the planes z = 0 and z = 2 21. Sphere F = zk across the portion of the sphere x 2 + y2 + z2 = a 2 in the first octant in the direction away from the origin zk across the sphere x 2

22. Sphere F = xi + yj + in the direction away from the origin

24. Cylinder F = xi + yj + zk outward through the portion of the cylinder x 2 + y2 = 1 cut by the planes z = 0 and z = a 25. Cone F = xyi - zk outward (normal away from the z-axis) through the cone z = Yx 2 + y2, 0 :::; Z :::; 1 26. Cone F = y 2i + xzj - k outward (normal away from the z-axis) through the cone z = 2Yx 2 + y2, 0 :::; Z :::; 2 27. Cone frustum F = -xi - yj + z 2k outward (normal away from the z-axis) through the portion of the cone z = Yx 2 + y2 between the planes z = 1 and z = 2 28. Paraboloid F = 4xi + 4yj + 2k outward (normal away from the z-axis) through the surface cut from the bottom of the paraboloid z = x 2 + y2 by the plane z = 1 In Exercises 29 and 30, find the flux of the field F across the portion of the given surface in the specified direction. =

-i

+ 2j + 3k

S: rectangular surface z = 0, direction k 30. F(x, y, z) = yx 2i - 2j + xzk S: rectangular surface y = 0, direction - j

0:::; x :::; 2,

0:::; y :::; 3,

=

zk

32. F(x, y, z)

=

-yi

+ xj

39. Let Sbe the portion of the cylinder y = eX in the first octant that projects parallel to the x-axis onto the rectangle Ryz: 1 :::; y :::; 2, o :::; z :::; 1 in the yz-plane (see the accompanying figure). Let n be the unit vector normal to S that points away from the yz-plane. Find the flux of the field F(x, y, z) = -2i + 2yj + zk across S in the direction of n.

z 1 --_

1 --x S

~y / ' "

40. Let S be the portion of the cylinder y = In x in the first octant whose projection parallel to the y-axis onto the xz-plane is the rectangle Rxz: 1 :::; x :::; e, 0 :::; z :::; 1. Let n be the unit vector normal to S that points away from the xz-plane. Find the flux of F = 2yj + zk through S in the direction ofn. 41. Find the outward flux of the field F = 2xyi + 2yzj + 2xzk across the surface of the cube cut from the first octant by the planes x = a,y = a,z = a. 42. Find the outward flux of the field F = xzi + yzj + k across the surface of the upper cap cut from the solid sphere x 2 + y2 + z2 :::; 25 by the plane z = 3. Moments and Masses 43. Centroid Find the centroid of the portion of the sphere x 2 + y2 + z2 = a 2 that lies in the first octant. 44. Centroid Find the centroid of the surface cut from the cylinder y2 + z2 = 9, z ~ 0, by the planes x = 0 and x = 3 (resembles the surface in Example 5).

-1:::; x :::; 2,

2:::; z :::; 7,

In Exercises 31-36, find the flux of the field F across the portion of the sphere x 2 + y2 + z2 = a 2 in the first octant in the direction away from the origin.

31. F(x,y,z)

38. Find the flux of the field F(x,y,z) = 4xi + 4yj + 2k outward (away from the z-axis) through the surface cut from the bottom of the paraboloid z = x 2 + y2 by the plane z = 1.

+ y2 + z2 = a 2

23. Plane F = 2xyi + 2yzj + 2xzk upward across the portion of the plane x + y + z = 2a that lies above the square o :::; x :::; a, 0 :::; y :::; a , in the xy-plane

29. F(x,y, z)

37. Find the flux of the field F(x,y, z) = z2j + xj - 3zk outward through the surface cut from the parabolic cylinder z = 4 - y2 by the planes x = O,X = 1, andz = O.

45. Thin shell of constant density Find the center of mass and the moment of inertia about the z-axis of a thin shell of constant density {) cut from the cone x 2 + y2 - z2 = 0 by the planes z = 1 andz = 2. 46. Conical surface of constant density Find the moment of inertia about the z-axis of a thin shell of constant density {) cut from the cone 4x 2 + 4y2 - z2 = 0, Z ~ 0, by the circular cylinder x 2 + y2 = 2x (see the accompanying figure).

962

Chapter 16: Integration in Vector FieLds 47. Spherical shells

z

a. Find the moment of inertia about a diameter of a thin spherical shell of radius a and constant density 8. (Work with a hemispherical shell and double the result.) b. Use the Parallel Axis Theorem (Exercises 15.6) and the result in part (a) to find the moment of inertia about a line tangent to the shell. 48. Conical Surface Find the centroid of the lateral surface of a solid cone of base radius a and height h (cone surface minus the base). y

x

16.7

or r=2cosO

Stokes' Theorem

-/CurlF «" ffJ FIGURE 16.55 The circulation vector at a point (x, y, z) in a plane in a tbreedimensional fluid flow. Notice its right-hand relation to the rotating particles in the fluid.

As we saw in Section 16.4, the circulation density or curl component of a two-dimensional field F = Mi + Nj at a point (x, y) is described by the scalar quantity (aN/ax - aM/ay). In three dimensions, circulation is described with a vector. Suppose that F is the velocity field of a fluid flowing in space. Particles near the point (x,y, z) in the fluid tend to rotate around an axis through (x,y, z) that is parallel to a certain vector we are about to define. This vector points in the direction for which the rotation is counterclockwise when viewed looking down onto the plane of the circulation from the tip of the arrow representing the vector. This is the direction your right-hand thumb points when your fingers curl around the axis of rotation in the way consistent with the rotating motion of the particles in the fluid (see Figure 16.55). The length ofthe vector measures the rate of rotation. The vector is called the curl vector and for the vector field F = Mi + Nj + Pk it is defined to be curl F =

(ap + (aM _ ap)j + (aN _ aM)k. ay _ aN)i az az ax ax ay

(1)

This information is a consequence of Stokes' Theorem, the generalization to space ofthe circulation-curl form of Green's Theorem and the subject of this section. Notice that (curl F)· k = (aN/ax - aM/ay) is consistent with our definition in Section 16.4 when F = M(x,y)i + N(x,y)j. The formula for curl F in Equation (1) is often written using the symbolic operator

V

.a +.a =lax Jay

+k

a az·

(The symbol V is pronounced "de1.") The curl of F is V X F : j

k

a VXF= ax

a a ay az M N P

_ aN)i + (aM _ ap). + (aN _ aM)k ( ap ay az az ax J ax ay =

curl F.

(2)

16.7

Stokes'Theorem

curl F = V X F

EXAMPLE 1 SoLution

Find the curl ofF = (x 2 - z)i

963

(3)

+ xeZj + xyk.

We use Equation (3) and the determinant form, so curlF = V X F j

k

a

a

a

ax 2 x - z

ay

az

xe z

xy

- ~(xeZ))i - (~(xy) - ~(x2 (~(xy) ay az ax az + (~(xeZ) - ~(x2 - Z))k ax ay

z))j

= (x - xeZ)i - (y + l)j + (e Z - O)k =

x(I - e)i - (y

+

I)j

+

•

e~

As we will see, the operator V has a number of other applications. For instance, when applied to a scalar function f(x, y, z), it gives the gradient of f:

Vf =

af. ax.

af.

af

+ ayl + azk.

It is sometimes read as "del f' as well as "grad f."

Stokes' Theorem

s - ~-

x) G

...... - - - - - -

n

'/

FIGURE 16.56 The orientation of the bounding curve C gives it a right-handed relation to the normal field n. If the thumb of a right hand points along n, the fingers curl in the direction of C.

Stokes' Theorem generalizes Green's Theorem to three dimensions. The circulation-curl form of Green's Theorem relates the counterclockwise circulation of a vector field around a simple closed curve C in the xy-plane to a double integral over the plane region R enclosed by C. Stokes' Theorem relates the circulation of a vector field around the boundary C of an oriented surface S in space (Figure 16.56) to a surface integral over the surface S. We require that the surface be piecewise smooth, which means that it is a finite union of smooth surfaces joining along smooth curves.

THEOREM 6-Stokes'Theorem

Let S be a piecewise smooth oriented surface having a piecewise smooth boundary curve C. Let F = Mi + Nj + Pk be a vector field whose components have continuous first partial derivatives on an open region containing S. Then the circulation of F around C in the direction counterclockwise with respect to the surface's unit normal vector n equals the integral of V X F· n over S.

f

F·dr =

c

Counterclockwise circulation

11

V X F·nda-

s

Curl integral

(4)

964

Chapter 16: Integration in Vector Fields Notice from Equation (4) that if two different oriented surfaces SI and S2 have the same boundary C, their curl integrals are equal:

Both curl integrals equal the counterclockwise circulation integral on the left side ofEquation (4) as long as the unit normal vectors 01 and 02 correctly orient the surfaces. If C is a curve in the xy-plane, oriented counterclockwise, and R is the region in the xy-plane bounded by C, then du = dx dy and

(V

x

F)' 0 = (V

x

F)' k = ( -aN - -aM) . ax ay

Under these conditions, Stokes' equation becomes

f

F'dr

=

11 (~~ -~~)

dxdy,

R which is the circulation-curl form of the equation in Green's Theorem. Conversely, by reversing these steps we can rewrite the circulation-curl form of Green's Theorem for two-dimensional fields in del notation as C

f

Circulation

F·dr =

11

V

(5)

X F·kdA.

R

C

See Figure 16.57. Evaluate Equation (4) for the hemisphere S: x 2 + y2 its bounding circle C: x 2 + y2 = 9, z = 0, and the field F = yi - xj.

EXAMPLE 2

+ z2 =

9, z

2:

0,

Solution

FIGURE 16.57 Comparison of Green's Theorem and Stokes' Theorem.

The hemisphere looks much like the surface in Figure 16.56 with the bounding circle C in the xy-plane (see Figure 16.58). We calculate the counterclockwise circulation around C (as viewed from above) using the parametrization reO) = (3 cos O)i + (3 sin O)j, :5 0 :5 21T:

°

dr = (-3 sin 0 dO)i

+ (3 cos 0 dO)j

F = yi - xj = (3 sin O)i - (3 cos O)j

f

F'dr = -9 sin2 0 dO - 9 cos2 0 dO = -9 dO F • dr = 127T-9dO=-181T.

C

For the curl integral ofF, we have

V X F

= (ap _ aN)i + (aM _ ap). + (aN _ aM)k ay

=

y 0=

x

du FIGURE 16.58 A hemisphere and a disk, each with boundary C (Examples 2 and 3).

V

az

+ (0 + yj + zk Yx 2 + y2 + z2

(0 - O)i xi

az

O)j =

3

= zdA

X F'odu = _ 2zl dA = -2dA

3 z

ax J

ax

+ (-1 - l)k + yj + zk

=

xi

ay

-2k Outer unit nonnal

3 Section 16.6, Example 6, with a = 3

16.7 Stokes'Theorem and

II

II

V X F'nder =

s

965

-2dA = -18,..

.r+y2.s9

The circulation around the circle equals the integral of the curl over the hemisphere, as it should. _ The surface integral in Stokes' Theorem can be computed using any surface having boundary curve C, provided the surface is properly oriented and lies within the domain of the field F. The next example illustrates this fact for the circulation around the curve C in Example 2.

EXAMPLE 3 Calculate the circulation around the bounding circle C in Example 2 using the disk of radius 3 centered at the origin in the xy-plane as the surface S (instead of the hemisphere). See Figure 16.58. Solution Ai; in Example 2, V X F = - 2k. For the surface being the described disk in the xy-plane, we have the normal vector n = k so that V X F'nder = -2k'kdA = -2dA and

II

II

V X F'nder =

s

-2dA = -18,.,

.r+y2.s9

a simpler calculation than before. z

_

Find the circulation of the field F = (x 2 - y)i + 4zj + x 2k around the curve C in which the plane z = 2 meets the cone z = Vx 2 + y2, counterclockwise as viewed from above (Figure 16.59).

EXAMPLE 4

C:X 2 +y2=4. z=2

/

Solution Stokes' Theorem enables us to fmd the circulation by integrating over the surface of the cone. Traversing C in the counterclockwise direction viewed from above corresponds to tsking the inner normal n to the cone, the normal with a positive k-component. We parametrize the cone as r(r, 0) = (r cos O)i + (r sin O)j + rk,

s: r(t) ~ (r cos 6)1 + (r sin 6)J + rk x

FIGURE 16.59 in Example 4.

o :5 r :5 2,

0:5 0 :5 2,..

We then have

"-. y

n=

The curve C and cone S

=

Vz der

=

r, X r.

-(rcosO)i - (rsinO)j + rk

Ir, X r.1

rVz

Section 16.5, Example 4

(-(COSO)i - (sinO)j + k)

rVz drdO

Section 16.5, Example 4

VXF=-4i-2xj+k

Example 1

= -4i - 2rcosOj + k.

x = rcos(J

Accordingly, V X F'n =

=

Vz Vz

(4 cosO + 2r cos 0 sinO + I) (4 cosO + rsin20 + I)

966

Chapter 16: Integration in Vector Fields and the circulation is

f

F·dr =

e

II

V X F'ndu

Stokes'Theorem,Eq.(4)

S =

/.2"/.2 Vz (4

cosO

+ rsin20 + l)(rV2drdO)

= 41T.

•

EXAMPLE 5 The cone used in Example 4 is not the easiest surface to use for calculating the circulation around the bounding circle Clying in the plane z = 3.lfinstead we use the flat disk of radius 3 centered on the z-axis and lying in the plane z = 3, then the normal vector to the surface S is n = k. Just as in the computation for Example 4, we still bave V X F = -4i - 2xj + k. However, now we get V X F· n = 1, so that

II S

V X F· n du =

II

The shadow is the disk of radius 2 in the xy-plane.

1 dA = 41T.

x2+y2S4

TIris result agrees with the circulation value found in Example 4.

Paddle Wheel Interpretation of V

•

F

X

Suppose that F is the velocity field of a fluid moving in a region R in space containing the closed curve C. Then

f

F·dr

e

is the circulation of the fluid around C. By Stokes' Theorem, the circulation is equal to the flux of V X F through any suitably oriented surface S with boundary C:

f

F·dr =

II

V X F·ndu.

s

e

Suppose we fIX a point Q in the region R and a direction u at Q. Take C to be a circle of radius p, with center at Q, whose plane is normal to u. If V X F is continuous at Q, the average value of the u-component of V X F over the circular disk S bounded by C approaches the u-component of V X F at Q as the radius p --> 0: (V X F'u)g = lim

-l,JJrr V X F·udu.

p-o trp

s

If we apply Stokes' Theorem and replace the surface integral by a line integral over C, we get CurlF _ _

(V X F'u)g = lim

l..2f F·dr.

p-O'lrp

(6)

e

The left-hand side of Equation (6) has its maximum value when u is the direction of V X F. When p is small, the liruit on the right-hand side of Equation (6) is approximately -l-fF'dr 7f'p2

FIGURE 16.60 The paddle wheel interpretation of curl F.

e

'

which is the circulation around C divided by the area of the disk (circulation density). Suppose that a small paddle wheel of radius p is introduced into the fluid at Q, with its axle directed along u (Figure 16.60). The circulation of the fluid around C affects the rate

16.7 Stokes'Theorem

967

of spin of the paddle wheel. The wheel spins fastest when the circulation integral is maximized; therefore it spins fastest when the axle of the paddle wheel points in the direction of V X F.

EXAMPLE 6 A fluid of constant density rotates around the z-axis with velocity F = w( -yi + xj), where w is a positive constant called the angular velocity ofthe rotation (Figure 16.61). Find V X F and relate it to the circulation density. SoLution

+

With F = -wyi

wxj, we find the curl

P(x,y, z)

V X F = (ap _ aN)i ay az =

(0 - O)i

+

+

(aM _ ap). az ax J

+ (w -

(0 - O)j

+

(aN _ aM)k ax ay

(-w»k = 2wk.

By Stokes' Theorem, the circulation of F around a circle C of radius p bounding a disk S in a plane normal to V X F , say the xy-plane, is F = w(-yl + xj)

x

/

P(x,y, O)

f

F·dr =

e

y

11

V X F·odO" =

s

11 s

Thus solving this last equation for 2w, we have

FIGURE 16.61 A steady rotational flow parallel to the xy-plane, with constant angular velocity w in the positive (counterclockwise) direction (Example 6).

2wk·kdxdy = (2w)( 1Tp2).

(V X F)· k = 2w =

-1-f 1Tp2

F· dr,

e

•

consistent with Equation (6) whenu = k.

EXAMPLE 7 Use Stokes' Theorem to evaluate Je F· dr, ifF = xzi + xyj + 3xzk and C is the boundary of the portion of the plane 2x + Y + z = 2 in the first octant, traversed counterclockwise as viewed from above (Figure 16.62). z

SoLution

y

The plane is the level surface I(x, y, z) = 2 of the function I(x, y, z) = 2x

+

+ z. The unit normal vector o =

(0,0,2)

VI IVI I =

(2i 12i

+ j + k) + j + kI =

1 (. 21

V6

•

+J +

)

k

is consistent with the counterclockwise motion around C. To apply Stokes' Theorem, we find

j

:x

ay

:z

xz

xy

3xz

V X F = (x - 3(2 - 2x - y»j

+ yk

=

(7x

6 + y)

=

~ (7X + 4y -

curl F = V X F =

x y

FIGURE 16.62 Example 7.

k

a

=

(x - 3z)j

+ yk.

On the plane, z equals 2 - 2x - y, so

The planar surface in

+

3y - 6)j

+ yk

and

V X F· 0

=

~ (7X + 3y -

The surface area element is dO" =

IV/I IVI. kl

V6

dA = -1- dx dy.

6).

968

Chapter 16: Integration in Vector Fields The circulation :is

f

ff

F·dr =

C

V X Fonda

Stnkcs''Ibcarem, Eq. (4)

S

11

1 2 2x -

=

~ (7X + 4y -

6)\16 dydx

f' {'-'" (7x+4y-6)dydx~-1.

•

~ JoJo

Let the surface S be the ellipitica1 paraboloid z = x 2 + 4y21ying beneath the plane z = 1 (Figure 16.63). We define the orientation of Sby taking the innernonnal vector n to the surface, which is the normal having a positive k-component. Find the flux of the curl V X F across S in the direction n for the vector field F = yI - xzj + xz2JL

EXAM PLE 8

Solution We use Stokes'Theorem to calculate the curl integral by finding the equivalent counterclockwise circulation of F around the cmve of intersection C of the paraboloid z - x 2 + 4y2 and the plane z - I, as shown in Figure 16.63. Note that the orientation of S is consistent with tntversiDg C in a counterclockwise direction around the z-axis. The cmve Cis the ellipse Xl + 4yl - 1 in the plane z - 1. We can parametrize the ellipse by x - cos t,y - sin t, z - 1 for 0 S t S 2v, so C is given by

!

y

FIGURE 1&.&3 The portion of the ellipiticai paraboloid in Example 8, showing its curve of intersection C with the p1.am: z = I and its inner normal orientation by n.

r(')

~

(cost)1

t

To compute the circulation integral iF dr, we evaluate F along C and fmd the velocity vectDr dr/ dt: 0

F(r('» -

t

(sin ')1 - (..st)J

-:;;

~

-(sin ')1

and

•

+

t

+ (cost)k

(cos Ill·

Thco,

f

f2.

F·dr - Jo

c D

c

B

o :S t :S 21T.

+ (sin 'll + k,

dr F(r(,»· dt dt

f2'(1-"lsin t - "lco;t 1)

=

Jo

=

-t 12w

2

dt

dJ = -1T.

(. )

Therefore the flux of the curl across S in the direction n for the field F is

ff

V X F'nd"

~ -v.

s

•

Proof of Stokes' Theorem for Polyhedral Surfaces Let S be • polyItednU surlioce consmting of • ,mire mnobee of plane regions '" fuceo. (See Figure 16.64 for examples.) We apply Green's Theorem to each separate face of S. There

are two types of faces: (b)

FIGURE 18.84 (a)Partofapolyhedral surface. (b) Other polyhedral. surfaces.

1. 2.

Those that are surrounded on all sides by other faces. Those that have one or more edges that are not adjacent to other faces.

969

16.7 Stokes'Theorem

The boundary Jl of S consists of those edges of the type 2 faces that are not adjacent to other faces. In Figure 16.64a, the triangles EAB, BCE, and CDE represent a part of S, with ABCD part of the boundary Jl. We apply a generalized tangential form of Green's Theorem to the three triangles ofFigore 16.64a in tum and add the results to get

In the generalized form, the line integral of F around the curve enclosing the plane region R normal to n equals the double integral of (curl F) • n over R. The three line integrals on the left-hand side of Equation (7) combine into a single line integral taken around the periphery ABCDE because the integrals along interior segments cancel in pairs. For example, the integral along segment BE in triangle ABE is opposite in sign to the integral along the same segment in triangle EBC. The same holds for segment CEo Hence, Equation (7) reduces to

f

F·dr

=

AllCDE

II

VX

F·nda-.

AllCDE

When we apply the generalized form of Green's Theorem to all the faces and add the results, we get

f

F ' dr =

IlvxF.nda-. s

~

This is Stokes' Theorem for the polyhedral surface S in Figore 16.64a. More general polyhedral surfaces are shown in Figore 16.64b and the proof can be extended to them. General smooth surfaces can be obtained as limits of polyhedral surfaces.

Stokes' Theorem for Surfaces with Holes Stokes' Theorem holds for an oriented surface S that has one or more holes (Figore 16.65). The surface integral over S of the normal component of V X F equals the sum of the line integrals around all the boundary curves of the tangential component of F, where the curves are to be traced in the direction induced by the orientation of S. For such surfaces the theorem is unchanged, but C is considered as a union of simple closed curves.

An Important Identity The following identity arises frequently in mathematics and the physical sciences.

curl grad I = 0

FIGURE 16.65 Stokes' Theorem also holds for oriented surfaces with holes.

or

V X Vf= 0

(8)

This identity holds for any function I(x, y, z) whose second partial derivatives are continuous. The proof goes like this:

i

a V X VI = ax al ax

j

k

a ay al ay

a az al az

= (fry - Iyz)i - (fzx -

Ixzli + (fp;

- Iry)k.

If the second partial derivatives are continuous, the mixed second derivatives in parentheses are equal (Theorem 2, Section 14.3) and the vector is zero.

970

Chapter 16: Integration in Vector Fields

Conservative Fields and Stokes' Theorem In Section 16.3, we found that a field F being conaervative in an open region D in space is equivalent to the integral of F around every closed loop in D being zero. This, in turn, is equivalent in simply connected open regions to saying that V X F = 0 (which gives a test for determining if F is conservative for such regions).

(a)

THEOREM 7-Curl F = 0 Related to the Closed-Loop Property

If V X F = 0 at every point of a simply connected open region D in space, then on any piecewisesmooth closed path C in D,

f

•

Sketch of a Proof Theorem 7 can be proved in two steps. The first step is for simple closed curves (loops that do not cross themselves), like the one in Figure 16.66a. A theorem from topology, a branch of advanced mathematics, states that every smooth simple closed curve C in a simply connected open region D is the boundary of a smooth two-sided surface S that also lies in D. Hence, by Stokes' Theorem,

f

(b)


F·dr = O.

c

FIGURE 16.66

F·dr =

c

11

V X F'ndCT

=

O.

s

The second step is for curves that cross themselves, like the one in Figure 16.66b. The idea is to break these into simple loops spanned by orientable surfaces, apply Stokes' Theorem one loop at a time, and add the results. _ The following diagram summarizes the results for conservative fields defined on connected, simply connected open regions. Theorem 2, Section 16.3

F conservative on D Theorem 3,

Section 16.3

<=>

II "I

.n-


fcFodr=O over any closed pathinD

F =

fonD Vector identity (Eq. 8) (continuous second partial derivatives)

<==

XF

=

0 throughout D

Theorem 7

Domain's simple connectivity and

Stokes' Theorem

Exercises 16.7 Using Stokes' Theorem to Find Line Integrals In Exercises l-{i, use the surface integral in Stokes' Theorem to calculate the circulation of the field F around the curve C in the indicated direction.

1. F=x'i+2xj +z'k C: The ellipse 4x' + y' = 4 in the xy-plane, counterclockwise when viewed from above

2. F = 2yi + 3xj -z'k C: The circle x 2 + y2 = 9 in the xy-plane, counterclockwise when viewed from above 3, F=yi+xzj +x'k C: The boundary of the triangle cut from the plane x + y + z by the first octant, counterclockwise when viewed from above

=1

16.7 4. F ~ (y' + z')i + (x' + z')j + (x' + y')k

Stokes' Theorem for Parametrized Surfaces

C: The boundary of the triangle cut from the plane x + y + z ~ 1 by the IITSt octant, counterclockwise when viewed from above 5. F ~ (y' + z')i + (x' + y')j + (x' + y')k

In Exercises 13-18, use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field F across the surface S in the direction of the outward unit normal n.

2zi + 3xj + 5yk r(r, 8) ~ (r cos 8)i + (rsin8)j + (4 - r')1<, S r :s 2, O:s 8 :s 21T 14. F ~ (y - z}i + (z - x)j + (x + z)k

C: The sqUllIe bounded by the lines x ~ ± 1 and y ~ ± 1 in the xy-plane, counterclockwise when viewed from above

13. F S:

6. F~x'y'i+j+zk C: The inlersectioo of the cylinder + y' ~ 4 and the hemisphere x 2 + y2 + z2 = 16, z ~ 0, cmmterclockwise when viewed

S:

x'

from above

~ 36,

z

~

0,

and let F ~ yi + x'j + (x' + y4)'/2 sin e v.;,;; k. Find the value of

II

8. Let n be the ooler unit normal (nonnal away from the origin) of the parabolic shell 4x'+y+z'~4,

y '" 0,

and let

~

(-z

+ 2

Find the value of

~ x} +

II

1

(tan- y)j +

(x

+ 4

~ z)k.

Vx F·ndu.

s

x'

9. Let S be the cylinder + y' ~ a',O '" z '" h, together with its top, x 2 + y2 ::;; a 2 , z = h. Let F = -yi + xj + x 2k. Use Stokes'Theorero to Imd the flux of V X F outward through S. 10. Evaluate

JJs

V X (yi)'ndu,

where S is the hemisphere x 2 11. Flux of curl F

+ y2 + z2

=

1, z

O?::

O.

Show that

JJs

r(r, 8) ~ (r cos 8)i + (r sin 8)j + (9 - r')1<, O:s 8 :s 21T

o :s r :s 3,

r(, 8) ~

('\13 cos
(Hint: One parametrizatioo of the ellipse at the hase of the shell is x ~ 3 cost,y ~ 2 sint, Os t '" 2,,-.)

F

o

S:

VX F·ndu.

S

s:

~

15. F ~ x'yi + 2y'zj + 3zk S: r(r, 8) ~ (r cos 8)i + (r sin 8)j + rl<, o :s r :s I, O:s 8 :s 21T 16. F ~ (x - y)i + (y - z)j + (z - x)k S: r(r, 8) ~ (r cos 8)i + (rsin8)j + (5 - r)1o, o :s r :s 5, O:s 8 :s 21T 17. F ~ 3yi + (5 - 2x)j + (z' - 2)k

Flux of the Curl 7. Let n be the ooler unit normal of the elliptical shell

s: 4x' + 9y' + 36z'

971

Stokes'Theorem

('\13 sin cos 8)i + ('\13 sinsin8)j 0 '" '" ,,-/2,

+

0 '" 8 s 2,,-

18. F ~ y'i + z'j + xk S: r(,8) ~ (2 sin cos 8)i + (2 sin sin8)j + (2 cos s '1r/2, 0 s 8 S 2'1r Theory and Examples 19. Zero circulation Use the identity V X VI ~ 0 (Equation (8) in the text) and Stokes' Theorero to show that the circu1ations of the following fields around the boundary of any smooth orientable surface in space are zero. •• F~2rl+2yj+2zk b. F ~ V(xy'z') c. F~VX(xi+yj+zk) d. F ~ VI 20. Zero circulation Let I(x, y, z) ~ (x' + y' + z')-I/2. Show that the clockwise circulation of the field F ~ VI around the circle x 2 + y2 = a2 in the xy-plane is zero •• by taking r ~ (a cos t)i + (a sin t)j, 0 s t S 2'1r, and integratiug F • dr over the circle. b. by applying Stokes' Theorero. 21. Let C be a simple closed smooth curve in the plane 2x + 2y + z ~ 2, oriented as shown here. Show that

f

2ydx

+ 3zdy - xdz

c

V X F'ndu

has the same value for all oriented surfaces S that span C and that

induce the same positive direction on C. 12. Let F be a differentiable vector field deImed on a regioo cootaining a smooth closed oriented surface S and its interior. Let n be the unit normal vector field on S. Suppose that S is the union of two surfaces SI and S, joined along a smooth simple closed curve e. Can anything be said about

JJs

V X F'ndu?

Give reasons for your answer.

depends only on the area of the region enclosed by C and not on the positioo or shape of C. 22. Show that ifF ~ xi + yj + z1o, then V X F ~ O.

972

Chapter 16: Integration in Vector Fields

23. Find a vector field with twice-differentiable compooents whose curl is xi + yj + zk or prove that no such field exists. 24. Does Stokes'Theorem say anything special about circulatioo in •

26. Zero cur~ yet field not conservative

F =

field whose curl is zero? Give reasons for your answer. 25. Let R be a regioo in the xy-plane that is bouoded by a piecewise smooth simple closed curve C and suppose that the moments of inertia of R about the x- and y-axes are koowo to be I, and lyo

Show that the curl of

Y - 2-2 I. + - 2x - -· 2+ J k Z x+y

x+y

is zero hut that

Evaluate the integral

f

4

V(r )·nd>,

is not zero if C is the circle x 2 + y2 = I in the xy-plane. (Theerem 7 does not apply here because the domain of F is not simply coooected. The field F is not defmed along the z-axis so there is no way to contract eto a point without leaving the domain ofF.)

C

wherer

=

Yx 2 + y2. in terms of I;; 8I1dly-

16.8

The Divergence Theorem and a Unified Theory The divergence form of Green's Theorem in the plane states that the net outward flux of a vector field across a simple closed curve can be calculated by integrating the divergence of the field over the region enclosed by the curve. The corresponding theorem in three dimensions, called the Divergence Theorem, states that the net outward flux of a vector field across a closed surface in space can be calculated by integrating the divergence of the field over the region enclosed by the surface. In this section we prove the Divergence Theorem and show how it simplifies the calculation of flux. We also derive Gauss's law for flux in an electric field and the continuity equation of hydrodynamics. Finally, we unify the chapter's vector integral theorems into a single fundamental theorem.

Divergence in Three Dimensions The divergence ofa vector fieldF = M(x,y,z)i function divF = V'F = aM

ax

+ N(x,y,z)j + P(x,y,z)k is the scalar + aN + ap. ay

az

(1)

The symbol ~'div F" is read as "divergence ofF" or "div F." The notation V· F is read "del dot F." Div F bas the same physical interpretation in three dimensions that it does in two. If F is the velocity field of a flowing gas, the value of div F at a point (x, y, z) is the rate at which the gas is compressing or expanding at (x, y, z). The divergence is the flux per unit volume or flux density at the point.

EXAMPLE 1 The following vector fields represent the velocity of a gas flowing in space. Find the divergence of each vector field and interpret its physical meaning. Figure 16.67 displays the vector fields. (a) Expansion: F(x, y, z) = xi

+ yj + zk

(b) Compression: F(x,y,z) = -xi - yj - zk

16.8 The Divergence Theorem and a Unified Theory

(c) Rotation about z-axis: F(x,y,z) = -yi

973

+ xj

(d) Shearing along horizontal planes: F(x, y, z) = zj

z

z

.'

...-+++

...-+++ Y

----.::::-----"--

(d)

(c)

-

FIGURE 16.67 Velocity fields ofa gas flowing in space (Example I).

Solutton (8) divF = ;x(x)

+ ~(y) + ;z(z)

= 3: The gas is undergoingunifonn expansion at all

points.

(b) divF = !(-x)

+ ~(-y) + ;z(-z)

=

-3: The gas is undergoing unifonn com-

pression at all points. (c) div F = ;x (-y)

+ ~ (x)

= 0: The gas is neither expanding nor compressing at any

point. (d) div F =

~ (z)

= 0: Again, the divergence is zero at all points in the domain of the ve-

locity field, so the gas is neither expanding nor compressing at any point.

•

Divergence Theorem The Divergence Theorem says that under suitable conditions, the outward flux of a vector field across a closed surface equals the triple integral of the divergence of the field over the region enclosed by the surface.

974

Chapter 16: Integration in Vector Fields

THEOREM 8-Divergence Theorem

Let F be a vector field whose components have continuous first partial derivatives, and let S be a piecewise smooth oriented closed surface. The flux of F across S in the direction of the surface's outward unit normal field n equals the integral of V • F over the region D enclosed by the surface: (2)

/ / F'ndu = ///V'FdV. D

S Outward flux

EXAMPLE 2 F = xi

Divergence integral

Evaluate both sides of Equation (2) for the expanding vector field = a 2 (Figure 16.68).

+ yj + zk over the sphere x 2 + y2 + z2

SoLution The outer unit normal to S, calculated from the gradient of l(x,y, z) = x 2 + y2 + z2 - a 2, is xi

+ yj + zk a

Hence,

Therefore, FIGURE 16.68 A unifonnly expanding vector field and a sphere (Example 2).

/ / F' n du = / / a du = a/ / du = a(47Ta 2) = 47Ta 3. S

S

S

The divergence of F is

v.F

a (x) + -a (y) + -a (z) ay az

= -

ax

=

3

'

so

• EXAMPLE 3 Find the flux ofF = xyi + yzj + xzk outward through the surface of the cube cut from the first octant by the planes x = 1, y = 1, and z = 1. SoLution Instead of calculating the flux as a sum of six separate integrals, one for each face of the cube, we can calculate the flux by integrating the divergence

a

V' F = ax (xy)

+

a

ay (yz)

+

a

az (xz) = y

+z+x

over the cube's interior: The Divergence Theorem

Flux = / / F'ndu = / / / V'FdV Cube surface

{I {I {I =

Jo Jo Jo (x

Cube interior

3

+ Y + z)dxdydz ="2.

Routine integration

•

16.8 The Divergence Theorem and a Unified Theory

975

Proof of the Divergence Theorem for Special Regions To prove the Divergence Theorem, we take the components of F to have continuous flI'St partial derivatives. We flI'St assume that D is a convex region with no holes or bubbles, such as a solid ball, cube, or ellipsoid, and that 8 is a piecewise smooth surface. In addition, we assume that any line perpendicular to the xy-plane at an interior point of the region Rzy that is the projection ofD on the xy-plane intersects the surface 8 in exactly two points, producing surfaces

(x,y) inRzy

8 t : z = It(x,y), z

8'l:

FIGURE 16.69 We prove the Divergence Theorem for the kind of three-dimensional region shown here.

h(x,y),

(x,y) inRzy,

with It :S h. We make similar assumptions about the projection of D onto the other coordmate planes. See Figure 16.69. The components of the unit normal vector 0 = ntl + n21 + n3k are the cosines of the angles a, /3, and 'Y that D makes with i, j, and k (Figure 16.70). This is true because all the vectors involved are unit vectors. We have nt

,

0 1 = 101111 cosu

=

0

=

cosu

., ~ n'i

~

Inllilc08,B

~

co.,B

o'k

=

1011klcos'Y

=

cos'Y.

n3 =

"

=

Thus, D

= (cos ali

+ (008/3)j + (cos 'Y)k

and

+ Ncos/3 + Pcos'Y.

F'O = Mcosa

In component form, the Divergence Theorem states that FIGURE 16.70 The components ofn arc p, and 'Y that it

the cosines of the angles a, makes with I. J, and k.

ff

(Mcosa

+ Ncos/3 + P cos 'Y) dO"

s

=

.

Jff (a: + a:: + ~) ~dydz. D

.,;.--~--"

divF

F'n

We prove the theorem by proving the three following equalities:

,

JJ

M co. a du

~ JJP:: dx dy dz

S

D

JJ Nco.,B du ~ JJJa::

dxdydz

(4)

~ JJJa;: dx dy dz

(5)

D

S

du

JJ

n

(3)

P cos 1 du

D

S

Proof of Equation (5) W. provo Equatioo (5) by coovertiog 1110 surface integml 00 the left to a double integral over the projection Rzy of D on the xy-plane (Figure 16.71). The surface 8 consists of an upper part s,. whose equation is z = h(x,y) and a lower part 81 whose equation is z = ft(x,y). Ons,., the outer normal 0 has a positive k-component and FIGURE 16.71 TheregionDenclosedby the surfaces 81 and /h projects vertically onto Rry in the xy-plane.

cos'Y du =

~dy

because

~

dxdy cos 'Y'

976

Chapter 16: Integration in Vector Fields

See Figure 16.72. On S1, the outer normal D has a negative k-component and cos-ydu - -dxdy. Thcrefon:, Here 'Y u. acute, so du _ dx tty/cos 'Y.

ff Pcos,du ~ ffpce"dU + ff Pce"du 8

S:!.

81

~ ff P(x,y,j,(x,y»dxdy - ff p(x,y,j,(x,y»dxdy

...

... ~ ff [p(x,y,j,(x,y» ...


so du - -dx dykOlJ 1.

- p(x,y,h(x,y))] dxdy

~ ff [£:) a:: dz1dx dy ~ JJJ a:: dz dx dy. R.

FIGURE 1&.72 Ancnlal&cdvi.ewofthc area patches in Figure 16.71. The relations du = ±dx dyjCO! "I come from Eq. (7) in Section 16.S.

D

This proves Equation (5). The proofs for Equations (3) and (4) follow the same pattern; or just permute x,y, z; M, N, P; a, p, 7, :in order, and get those results :from Equation (5). This proves the Divergence Theorem for these special regions.

•

Divergence Theorem for Other Regions

,

y

The Divergence Theorem can be extended to regions that can be partitioned into a lmite number of simple regions of the type just diSCWIsed and to regions that can be defined as limits of simpler regions in certain ways. For an cump1e of one step :in such a splitting process, suppose that D is the region between two concentric spheres and that F has continuously d:ifferentiable components throughout D and on the bounding surfaces. Split D by an equatorial plane and apply the Divergence Theorem to each half separately. The bottom half, Dt. is shown in Figure 16.73. The surface SI that bounds DI consists ofan outer hemisphere, a plane washer-shaped base, and an inner hemisphere. The Divergence Theorem says that (6)

FIGURE 18.73 Thelowerhalfofthe solid region between two concentric spheres.

The unit normal 01 that po:ints outward from Dl points away from the origin along the outer surface, equals k along the flat base, and points toward the origin along the inner surface. Next apply the Divergence Theorem to~, and its surface S2 (Figure 16.74):

,

ff F·.,du, ~ fff

)

s,

y

ff F·.du ~ fff V'FdY, S

......

",

(7)

& we follow 02 over s,., pointing outward from Dz , we sec that D2 equals - k along the washer-shaped base in the xy-p1ane, points away :from the origin on the outer sphere, and points toward the origin on the inner sphere. When we add Equations (6) and (7), the integrals over the flat base cancel because of the opposite signs of 01 and 02. We thus arrive at the result

"

FIGURE 1&.74 Theupperhalfofthe solid region between two concentric

V·FaY,.

D,

D

with D the region between the spheres, S the boundary of D consisting of t\W spheres, and D the unit normal to S directed outward from D.

16.8 The Divergence Theorem and a Unified Theory

EXAMPLE 4

Find the net outward flux of the field xi F

=

+ yj + zk 3

P

'

across the boundary of the region D: 0 < a 2

Solution

FIGURE 16.75 Two concentric spheres in an expanding vector field.

977

:::;

x2

+ y2 + z2

:::;

b 2 (Figure 16.75).

The flux can be calculated by integrating V . F over D. We have

and

a (xp- 3)

-aM -_ -

ax

ax

2 -_ p - 3 - 3xp- 4 -ap -_ - 1 - -3x

ax

p3

p5 .

Similarly,

aN

and

ay

ap

az

Hence,

and

111

V'FdV= O.

V' F = divF

D

So the integral of V . F over D is zero and the net outward flux across the boundary of D is zero. There is more to learn from this example, though. The flux leaving D across the inner sphere Sa is the negative of the flux leaving D across the outer sphere Sb (because the sum of these fluxes is zero). Hence, the flux of F across Sa in the direction away from the origin equals the flux of F across Sb in the direction away from the origin. Thus, the flux of F across a sphere centered at the origin is independent of the radius of the sphere. What is this flux? To find it, we evaluate the flux integral directly. The outward unit normal on the sphere of radius a is xi

+ yj + zk a

Hence, on the sphere,

and

The outward flux of F across any sphere centered at the origin is 41T.

•

978

Chapter 16: Integration in Vector FieLds

Gauss's Law: One of the Four Great Laws of Electromagnetic Theory There is still more to be learned from Example 4. In electromagnetic theory, the electric field created by a point charge q located at the origin is

E(x,y, z}

=

1 q 4m:o ~

(r) TrT

r

q

q

4m:o ~ = 4m:o

=

xi+yj+zk p3 '

where EO is a physical constant, r is the position vector of the point (x, y, z), and p = Ir I = x 2 + y2 + z2. In the notation of Example 4,

V

q E =-4-F.

7TEO

The calculations in Example 4 show that the outward flux ofE across any sphere centered at the origin is q/ EO, but this result is not confined to spheres. The outward flux of E across any closed surface S that encloses the origin (and to which the Divergence Theorem applies) is also q/EO. To see why, we have only to imagine a large sphere Sa centered at the origin and enclosing the surface S (see Figure 16.76). Since

V·E

=

V·-q-F 47TEo

=

-q-V·F 47TEo

=

°

when p > 0, the integral of V· E over the region D between S and Sa is zero. Hence, by the Divergence Theorem,

11

E·ndu

=

0,

Boundary ofD

FIGURE 16.76 A sphere Sa surrounding another surface S. The tops of the surfaces are removed for visualization.

and the flux of E across S in the direction away from the origin must be the same as the flux of E across Sa in the direction away from the origin, which is q/ EO. This statement, called Gauss s Law, also applies to charge distributions that are more general than the one assumed here, as you will see in nearly any physics text.

11

Gauss's Law:

E· n du

=

:0

s

Continuity Equation of Hydrodynamics Let D be a region in space bounded by a closed oriented surface S. Ifv(x, y, z) is the velocity field of a fluid flowing smoothly through D, 8 = 8(t, x, y, z) is the fluid's density at (x, y, z) at time t, and F = 8v, then the continuity equation of hydrodynamics states that

V·F

T

a8

+ at

=

0.

If the functions involved have continuous first partial derivatives, the equation evolves naturally from the Divergence Theorem, as we now see. First, the integral

11

F·ndu

s

s FIGURE 16.77 The fluid that flows upward through the patch !iu in a short time !it fills a "cylinder" whose volume is approximately base X height =

v· n!iu !it.

is the rate at which mass leaves D across S (leaves because n is the outer normal). To see why, consider a patch of area au on the surface (Figure 16.77). In a short time interval at, the volume a v of fluid that flows across the patch is approximately equal to the volume of a cylinder with base area au and height (vat) . n, where v is a velocity vector rooted at a point of the patch:

a v ~ v· n au at.

16.8 The Divergence Theorem and a Unified Theory

979

The mass of this volume of fluid is about Ilm "" {iv' n Ilrr Ill, so the rate at which mass is flowing out of D across the patch is about

This leads to the approximation

as an estimate of the average rate at which mass flows across S. Finally, letting Ilrr --+ 0 and III --+ 0 gives the instantaneous rate at which mass leaves D across S as

~7

=

11 11

{iv' n drr,

s

which for oUI particular flow is

~7 =

F·ndrr.

s

Now let B be a solid sphere centered at a point Q in the flow. The average value of V'FoverBis

vOl~e of B

lff

V· F dV.

B

It is a consequence of the continuity of the divergence that V· F actually takes on this value at some point Pin B. Thus,

11

1 of B ( V'F )p = volume

F'ndrr

iJifJ V'FdV= volume ofB S

B

rate at which mass leaves B across its surface S volume ofB

(8)

The last term of the equation describes decrease in mass per unit volume. Now let the radius of B approach zero while the center Q stays f'1xed. The left side of Equation (8) converges to (V, F)Q, the right side to (-a{ijal)Q' The equality of these two limits is the continuity equation

The continuity equation "explains" V· F: The divergence of F at a point is the rate at which the density of the fluid is decreasing there. The Divergence Theorem

11 S

F'ndrr =

111

V'FdV

D

now says that the net decrease in density of the fluid in region D is accounted for by the mass transported across the surface S. So, the theorem is a statement about conservation of mass (Exercise 31).

980

Chapter 16: Integration in Vector Fields

Unifying the Integral Theorems If we think of a two-dimensional field F = M(x, y)i + N(x, y)i as a three-dimensional field whose k-component is zero, then V' F = (aM/ax) + (aN/ay) and the normal form of Green's Theorem can be written as

f

F ' nds =

11(aa~+~~)dxdy=

11

V ' FdA .

e R R

Similarly, V X F· k = (aN/ax) - (aM/ay), can be written as

f

F·dr =

80

11 (~~ -~~)

the tangential form of Green's Theorem

dxdy =

11

V X F·kdA.

e R R

With the equations of Green's Theorem now in del notation, we can see their relationships to the equations in Stokes' Theorem and the Divergence Theorem.

Green's Theorem and Its GeneraIization to Three Dimensions

f

Normal form of Green's Theorem:

11

Divergence Theorem:

f f

F· n dU" =

F· dr =

n

a

=I

b

FIGURE 16.78 The outward unit normals atthe boundary of [a, b] in on<>-

dimensional space.

V X F· k dA

R

F· dr =

C

n =-1

V' F dV

D

C

Stokes' Theorem:

V' F dA

R

S

Tangential form of Green'. Theorem:

11 111 11 11

F· n ds =

C

V X F· n dU"

S

Notice how Stokes' Theorem generalizes the tangential (curl) form of Green's Theorem from a flat surface in the plane to a surface in three-dimensional space. In each case, the integral of the normal component of curl F over the interior of the surface equals the circulation of F around the boundary. Likewise, the Divergence Theorem generalizes the normal (flux) form of Green's Theorem from a two-dimensional region in the plane to a three-dimensional region in space. In each case, the integral of V • F over the interior of the region eqnals the total flux of the field across the boundary. There is still more to be learned here. All these results can be thought of as forms of a single .fUndamental theorem. Think back to the Fundamental Theorem of Calculus in Section 5.4. It says that if f(x) is differentiable on (a, b) and continuous on [a, bl, then

l

a

bdf dx dx

=

f(b) - f(a).

If we let F = f(x)i throughout [a, bl, then (df/dx) = V' F. Ifwe define the unit vector field n normal to the boundary of [a, b] to be i at band -i at a (Figure 16.78), then

+ f(a)i' (-i) + F(a)'n

f(b) - f(a) = f(b)i' (i) = F(b)'n

= total outward flux of F across the boundary of [a,

bI.

981

16.8 The Divergence Theorem and a Unified Theory The Fundamental Theorem now says that

F(b)·n + F(a)'n

J

=

V·Fdx.

[.,bl

The Fundamental Theorem of Calculus, the normal form of Green's Theorem, and the Divergence Theorem all say that the integral of the differential operator V' operating on a field F over a region equals the sum of the normal field components over the boundary of the region. (Here we are interpreting the line integral in Green's Theorem and the surface integral in the Divergence Theorem as "sums" over the boundary.) Stokes' Theorem and the tangential form of Green's Theorem say that, when things are properly oriented, the integral of the normal component of the curl operating on a field equals the sum of the tangential field components on the boundary of the surface. The beauty of these interpretations is the observance of a single unifYing principle, which we might state as follows.

A Unifying Fundamental Theorem The integral of a differential operator acting on a field over a region equals the sum of the field components appropriate to the operator over the boundary of the region.

Exercises 16.8 Calculating Divergence

9. Portion ofsphere

In Exercises 1-4, find the divergence of the field.

1. The spin field in Figure 16.12 2. The radial field in Figure 16.11

+ 3xzk

+ 2xy)i + (2y + x 2z)j + 4x"y'k The regioo cut from the frrst octant by the cylioder x 2 + y2 = F = (6x 2

10. Cylindrical can

3. The gravitational field in Figure 16.8 and Exercise 38a in Section 16.3

D:

4 and the plane z = 3

4. The velocity field in Figure 16.13

Calculating Flux U';ng the Divergence Theorem In Exercises 5-16, use the Divergence Theorem to ftnd the outward flux ofF across the boundary of the regiooD. 5. Cube

F = x'i - 2xyj

D: The region cut from the frrst octant by the sphere x 2 + y2 + z2 = 4

F = (y - xli

+ (z

- y)j

+ (y -

12. Sphere

x)k

D: The cube bouoded by the planes x = ± I, Y = ± I, and z = ±I

F = x':i

b. Cube D: The cube bounded by the planes x Y = ±l, andz = ±I

= ± I,

D: The region cut from the solid cylinder x 2 + y2 '" 4 by the planes z = 0 and z = I

F = yi + xyj - zk D: The regioo inside the solid cylinder x 2 + y2 '" 4 between the plane z = 0 and the paraboloid z = x 2 + y2

=

4

+ y'j + z'k + y2 + z2 ::s a 2

F =

D: The region 1 :s

a. Cube D: The cube cut from the first octant by the planes x = I,y = I, andz = I

+z

D: The solid sphere Xl

13. Thick sphere

6. F=x'i+y2j+z 2k

c. Cylindrical can

11. Wedge F = 2xzi - xyj - z 2k D: The wedge cut from the frrst octant by the plane y andtheellipticalcylinder4x2 + y2 = 16

14. Thick spbere

Yx 2 + y2 + z2 (xi + yj + zk) x2

+ y2 + z2 :s

2

= (xi + yj + zk)/Yx 2 + y2 + z2

F

D: The region 1 :s x 2

15. Thick spbere F (5z' + e' cosz)k

+ y2 + z2 :s

= (5x' +

12xy2)i

4

+ (y' + e'sinz)j +

D: The solid region between the spheres x 2 + y2 x 2 + y2 +z2 = 2

+ z2 =

I and

7. Cylinder and paraboloid

8. Sphere F = x'i + xzj + 3zk D: The solid sphere x 2 + y2 + z2 :s 4

16. Thickcylinder

F = In(x 2

+ y2)i

-

(~tan-I~)j +

zYx 2 + y 2 k D: The thick-walled cylioder I '" x 2 + y2 '" 2,

-I '" z '" 2

982

Chapter 16: Integration in Vector Fields

Properties of (url and Divergence 17. div (curl G) is zero a. Show that if the necessary partial derivatives of the components of the field G = Mi + Nj + Pk are continuous, then V·V X G = O. b. What, if anything, can you conclude about the flux of the field V X G across a closed surface? Give reasons for your answer.

18. Let F] and F 2 be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.

+ bF2) = aV·F] + bV·F2 (aF] + bF2) = aV X F] + bV X

b. Let n be the outward unit normal vector field on S. Show that it is not possible for F to be orthogonal to n at every point of S. 24. Maximum flux Among all rectangular solids defined by the inequalities 0 :s x :s a, 0 :s y :s b, 0 :s z :s I, find the one for which the total flux of F = (-x 2 - 4xy)i - 6yzj + 12zk outward through the six sides is greatest. What is the greatest flux? 25. Volume of a solid region Let F = xi + yj + zk and suppose that the surface S and region D satisfy the hypotheses of the Divergence Theorem. Show that the volume of D is given by the formula

a. V·(aF] b. V X

23. a. Show that the outward flux of the position vector field F = xi + yj + zk through a smooth closed surface S is three times the volume of the region enclosed by the surface.

F2

c. V· (F] X F 2) = F 2 • V X F] - F] • V X F2

19. Let F be a differentiable vector field and let g(x, y, z) be a differ-

VolumeofD =

entiable scalar function. Verify the following identities. a. V·(gF) = gV·F

+

+

Vg X F

20. If F = Mi + Nj + Pk is a differentiable vector field, we define the notation F • V to mean a a a M ax + N ay + P az .

26. Outward flux of a constant field Show that the outward flux of a constant vector field F = C across any closed surface to which the Divergence Theorem applies is zero. 27. Harmonic functions A function j(x, y, z) is said to be harmonic in a region D in space if it satisfies the Laplace equation

a2j V2j = V· Vj = ax 2

For differentiable vector fields F] and F 2, verify the following identities. a. V X (F] X F 2) = (F2 • V)F] - (F] • V)F2 (V· F])F2 b. V(F]· F 2) = (F] • V)F2 F2 X (V X F])

+

(F2· V)F]

+ F]

+ (V· F 2)F] X (V X F 2)

F·ndlT.

s

Vg·F

b. V X (gF) = gV X F

tff

-

+

Theory and Examples

a2j

+-

ay2

a2j az 2

+-

=

0

throughout D. a. Suppose that j is harmonic throughout a bounded region D enclosed by a smooth surface S and that n is the chosen unit normal vector on S. Show that the integral over S of Vj . n, the derivative of j in the direction ofn, is zero.

b. Show that if j is harmonic on D, then

21. Let F be a field whose components have continuous first partial derivatives throughout a portion of space containing a region D bounded by a smooth closed surface S. If IF I :s I, can any bound be placed on the size of

JJJ

V·F dV?

D

f f jVj·ndlT

=

fff IVjI2dV. D

S

28. Outward flux of a gradient field Let S be the surface of the portion of the solid sphere x 2 + y2 + z2 :s a 2 that lies in the firstoctantandletj(x,y,z) = Inv'x 2 + y2 + z2. Calculate

Give reasons for your answer. 22. The base of the closed cubelike surface shown here is the unit square in the xy-plane. The four sides lie in the planes x = 0, x = I, y = 0, and y = 1. The top is an arbitrary smooth surface whose identity is unknown. Let F = xi - 2yj + (z + 3)k and suppose the outward flux ofF through Side A is I and through Side B is - 3. Can you conclude anything about the outward flux through the top? Give reasons for your answer.

f f Vj·ndlT. S

(V j . n is the derivative of j in the direction of outward normal n .) 29. Green's first formula Suppose that j and g are scalar functions with continuous first- and second-order partial derivatives throughout a region D that is bounded by a closed piecewise smooth surface S. Show that

z

ffjvg.ndlT= fff(JV 2g+Vj.Vg)dV.

1

S

(9)

D

Equation (9) is Green's first formula. (Hint: Apply the Divergence Theorem to the field F = j Vg.) 30. Green's second formula (Continuation of Exercise 29.) Interchange j and g in Equation (9) to obtain a similar formula. Then subtract this formula from Equation (9) to show that

f f (J Vg - gVj)· n dlT =

/

x

Side A

•

(1,1,0)

SideB

S

JJJ

(J V 2g - gV 2j) dV.

D

This equation is Green's second formula.

(10)

Chapter 16 31. Conservation of mass Let v(l, x, y, z) be a continuously differentiable vector field over the region Din space and let p(l, x,y, z) be a continuously differentiable scalar function. The variable 1 represents the time domain. The Law of Conservation of Mass asserts that

!JJJp(l,x,y,Z)dV= -JJ pv·ndCT, D

S

Practice Exercises

983

(In the frrst term V' pv, the variable 1 is held fixed, and in the second term ap/al , it is asswned that the point (x, y, z) in Dis held fixed.) 32. Tbe be.t diffusion equation Let 1'(1, x, y, z) be a function with continuous second derivatives giving the temperature at time t at the point (x, y, z) ofa solid occupying a regionD in space. Ifthe solid's heat capacity and mass density are denoted by the constants c and p, respectively, the quantity cpT is called the solid's

where S is the surface enclosing D.

heat energy per unit volume.

a. Give a physical interpretation of the conservation of mass law ifv is a velocity flow field and p represents the deusity of the fluid at point (x,y, z) at time I.

b. Let -kVT denote the energy flux veetor. (Here the constant k is called the conductivity.) Assuming the Law of Conserva-

a. Explain why - VT points in the direction of heat flow.

tion of Mass with -kVT = v and cpT derive the diffusion (heat) equation

b. Use the Divergence Theorem and Leibniz's Rule,

!JJJp(l,x,y,Z)dV= JJJ?,dV, D

aT = Kv-T, ..." at

D

where K = k/(cp) > 0 is the diffosivity constant. (Notice that if 1'(1, x) represents the temperature at time 1 at position x in a uuifunn conducting rod with perfectly insulated sides, then V'T = a2T/ax2 and the diffusion equation reduces to the oru>dimensional heat equation in Chspter 14's Additional Exercises.)

to show that the Law of Conservation of Mass is equivalent to the continuity equation, ap

V'pv

Chapter

t3

= p in Exercise 31,

+ al = o.

Questions to Guide Your Review

1. What are line integrals? How are they evaluated? Give examples. 2. How can you use line integrals to f'md the centers of mass of springs? Explain.

3. What is a vector field? A gradient field? Give examples. 4. How do you calculate the work done by a force in moving a particle along a curve? Give an example. S. What are flow, circulation, and flux? 6. What is special about path independent fields? 7. How can you tell when a field is conservative? 8. What is a potential function? Show by example how to fmd a poteotial function for a conservative field. 9. What is a diffurential form? What does it mean for such a form to be exact? How do you test for exactness? Give examples. 10. What is the divergence of a vector field? How can you interpret it? 11. What is the corl of a vector field? How can you interpret it?

13. How do you calculate the area of a parame1rized surface in space? Of an implicitly defmed surface F(x, y, z) = O? Of the surface which is the graph of z = f(x, y)? Give examples. 14. How do you integrate a function over a parametrized surface in space? Of surfaces that are defmed implicitly or in explicit form? What can you calculate with surface integrals? Give examples. 15. What is an otiented surface? How do you calculate the flux of a three-dimensional vector field across an otiented surface? Give an example. 16. What is Stokes' Theorem? How can you interpret it? 17. Smmnarize the chapter's results on conservative fields. 18. What is the Divergence Theorem? How can you interpret it? 19. How does the Divergence Theorem generalize Green's Theorem? 20. How does Stokes' Theorem generalize Green's Theorem? 21. How can Green's Theorem, Stokes'Theorem, and the Divergence Theorem be thought of as forms of a single fundamental theorem?

12. What is Green's Theorem? How can you interpret it?

Chapter

m

Practice Exercises

Evaluating Lin. Integrals 1. The accompanying figore shows two polygonal paths in space joining the otigin to the point (I, I, I). Integrate f(x,y, z) = 2x 3y2 - 2z + 3 over each path.

(0,0,0)

O'O'O)

(I, I, 1)

,

,

,------------(I, 1,0) ,

\,

,,

\1

Path 1

y x

z

~ /

(1,1,1)

~y (1, 1,0)

Path 2

984

Chapter 16: Integration in Vector Fields

2. The accompanying figure shows three polygonal paths joining the origin to the point (1, 1, 1). Integrate f(x,y,z) = x 2 + Y - z over each path.

z

b. Find the area of the portion of the cylinder that lies inside the hemisphere. (Hint: Project onto the xz-plane. Or evaluate the integral Jh ds, where h is the altitude of the cylinder and ds is the element of arc length on the circle x 2 + y2 = 2x in the xy-plane.)

z

(0,0,0)

(0,0,0)

(1, 1, 1)

c;--------..

(1, 1, 1)

~Y

Y

(1, 1,0)

x

14. a. Hemisphere cut by cylinder Find the area of the surface cut from the hemisphere x 2 + y2 + z2 = 4, z ~ 0, by the cylinder x 2 + y2 = 2x.

(1, 1,0)

x

z (0,1,1)

y x

x

3. Integrate f(x,y, z)

=

~ over the circle

r(t) = (a cos t)j

4. Integrate f(x, y, z) = r(t) = (cost

+

(a sin t)k,

o :s

Cylinder r = 2 cos 0

- tcost)j,

O:s t:s

vi

Evaluate the integrals in Exercises 5 and 6.

1

(4,- 3,0)

5.

(- 1,1,1)

6.

{(lO,3,3)

+ dy + dz Yx + Y + z

16. Parabolic cylinder cut by planes Integrate

dx

yz

a. g(x,y,z) = -~r /~=

7. Integrate F = -(ysinz)i + (xsinz)j + (xycosz)k around the circle cut from the sphere x 2 + y2 + z2 = 5 by the plane z = - 1, clockwise as viewed from above. 8. Integrate F = 3x 2yi + (x 3 + l)j + 9z 2k around the circle cut from the sphere x 2 + y2 + z2 = 9 by the plane x = 2. Evaluate the integrals in Exercises 9 and 10. 9.1sxsinydx - Sycosxdy

C is the square cut from the first quadrant by the lines x andy = 7T/2.

= 7T/2

1

b. g(x,y,z)

=

v'4Y~ +

1

over the surface cut from the parabolic cylinder y2 - z the planes x = 0, x = 3 , and z = O.

=

1 by

Integrate g(x, y, z) = = 25 that lies in the first octant between the planes x = 0 and x = 1 and above the plane z = 3.

17. Circular x 4y(y2

cylinder

cut by planes

+ z2) over the portion of the cylinder y2 + z2

18. Area of Wyoming The state of Wyoming is bounded by the meridians 111 °3 / and 104°3/ west longitude and by the circles 41 ° and 45° north latitude. Assuming that Earth is a sphere of radius R = 3959 mi, find the area of Wyoming.

Parametrized Surfaces

+ x 2 dy

C is the circle x 2

+

v4y2

~dZ

}(l,I,I)

10.1 y2 dx

15. Area of a triangle Find the area of the triangle in which the plane (x/a) + (y/b) + (z/c) = 1 (a,b,c > 0) intersects the first octant. Check your answer with an appropriate vector calculation.

-----;===dx - jjdY -

y

t :s 27T.

v'x 2 + y2 over the involute curve

+ t sin t)i + (sint

~

+ y2

= 4.

Finding and EvaLuating Surface Integrals 11. Area of an elliptical region Find the area of the elliptical region cut from the plane x + y + z = 1 by the cylinder x 2 + y2 = 1. 12. Area of a parabolic cap Find the area of the cap cut from the paraboloid y2 + z2 = 3x by the plane x = 1. 13. Area of a spherical cap Find the area of the cap cut from the top of the sphere x 2 + y2 + z2 = 1 by the plane z = V2/2.

Find parametrizations for the surfaces in Exercises 19- 24. (There are many ways to do these, so your answers may not be the same as those in the back of the book.) 19. Spherical band The portion of the sphere x 2 between the planes z = - 3 and z = 3 \13

+ y2 + z2

20. Parabolic cap The portion of the paraboloid z above the plane z = - 2

= -(x 2

21. Cone

The cone z

=

1

+

v'x 2

+ y2, z

=

36

+ y2)/2

:s 3

22. Plane above square The portion of the plane 4x + 2y + 4z = 12 that lies above the square 0 :s x :s 2,0 :s y :s 2 in the first quadrant

Chapter 16 23. Portion of p .....boloid The portion of the paraboloid y = 2(x 2 + z2), y :5 2, that lies above the xy·plane

24. Portionofbemisphere Theportionofthebemispberex 2 + y2 z2 = 10, y '" 0, in the Ilrst octant

+

25. Surf.ce area Find the area of the surface r(u, v) = (u + v)i + (u - v)j + vk,

o :s; u :5

1, 0

:s;

38. Flow along different path. V(x 2ze')

Practice Exercises

985

Find the flow of the field F =

a. once around the ellipse C in which the plane x + y + z = I intersects the cylinder x 2 + z2 = 25, clockwise as viewed from the positive y-axis. b. along the curved boundsry of the helicoid in Exercise 27 from (1,0,0) to (I, 0, 2,,).

v :s; 1.

26. Surf.ce integral Integrate /(x,y,z) = xy - z2 over the sur· face in Exercise 25. 27. Area of a helicoid Find the surface area of the helicoid r(r, 6) = (r cos 6)i + (rsin 6)j + 6k, in the accompanying figure.

0:5 6

S

2", 0 '" r '" I,

In Exercises 39 and 40, use the surface integral in Stokes' Theorem to Imd the circulation of the field F around the curve C in the indicated direction. 39. Cireulation around an ellipse F = y>;' - yj + 3z2k C: The ellipse in which the plane 2x + 6y - 3z = 6 meets the cylinder x 2 + y2 = I, counterclockwise as viewed from above

40. Cireulation around a circle F = (x 2 + y)i + (x + y)j + (4y2 - z)k C: The circle in which the plane z = -y meets the sphere + y2 + z2 = 4, counterclockwise as viewed from above

x2

Masses and Moments 41. Wire with different doosities Find the mass of a thin wire lying along the curve r(l) = v2ti + v2tj + (4 - 12)k, o :5 I :5 I, if the deosity at I is (a) B = 31 and (b) B = 1. y

x

28. Surf.ce integral Evaluate the integral ffs v'x 2 + y2 + I du, wbere S is the helicoid in Exercise 27.

42. WI1'O with _ble density Find the center ofmass of a thin wire lying along the curve r(l) = Ii + 21j + (2/3)I'/2k, 0 :5 1:5 2, if the density at I is B = 3'\I'5+t. 43. Wire with _ble density

Find the ceoter of mass and the momeots of inertia aboot the coordioate axes of a thin wire lyiog along the curve

Conservative Fields Which of the fields in Exercises 29-32 are conservative, and which are not?

r(l) = Ii + 2Vz 1'/2j + f k, 3 2

0:5t:52,

ifthedensityatlisB = 1/(1+ 1).

29. F =xi + yj +zk 30. F = (xi + yj + zk)/(x 2 + y2 + z2)'/2

31. F = xe'i + ye%j + zeXJi. 32. F = (i + zj + yk)/(x + yz)

v'

mass.

Find potential functions for the fields in Exercises 33 and 34. 33. F = 2i + (2y + z)j + (y + i)k 34. F = (zcosxz)i + e'j + (xcosxz)k Work and C;rcuiation In Exercises 35 and 36, Imd the work done by each field along the paths from (0, 0, 0) to (1,1,1) in Exercise 1. 35.F=2xyi+j+x 2k 36.F=2xyi+x2j+k 37. Finding work in two way. Find the work done by xi + yj F = C-;;----C,,= (x2

44. Center of m ••• of an arch A slender metal arch lies along the a 2 - x 2 in the xy-plane. The density at the semicircle y = point (x, y) on the arch is B(x,y) = 2a - y. Find the center of

+ y2)'/2

over the plane curve r(l) = (e' cos I)i + (e' sin I)j from the point (1,0) to the point (e 2w , 0) in two ways: a. By using the parame1rization of the curve to evaluate the work integral. b. By evaluating a potential function for F.

45. Wire with constant density A wire of constant density B = 1 lies along the curve r(l) = (e' cos I)i + (e' sin I)j + e'k, 0:5 I :5 In 2. Find ii and I •. 46. Helical wire with constant density Find the mass and center of mass of a wire of constant density B that lies along the helix

r(l) = (2 sin I)i + (2 cos I)j + 31k, 0 :5 I :5 2". 47. Inertia and cooter of mas. of a shell Find I. and the center of mass of a thin shell of density B(x, y, z) = z cut from the upper portion of the spbere x 2 + y2 + z2 = 25 by the plane z = 3. Find the moment of inertia about the z-axis of the surface of the cube cut from the Ilrst octant by the planes x = I,y= I,andz= I if the density is B = 1.

48. Moment of inertia of a cube

Flux Across a Plane Curve or Surface Use Green's Theorem to Imd the counterclockwise circulation and ootward flux for the fields and curves in Exercises 49 and 50. 49. Square F = (2xy + xli + (xy - y)j C: The squsre bounded by x = O,X = I,y = O,y = 1

986

Chapter 16: Integration in Vector Fields

50. TrilUlgle

F ~ (y - 6x')i

+ (x + y')j

C: The triangle made by the lines y

~

55. Spherical cap

0, y

~

x , and x

~

f

~

0

c

for any closed curve C to which Greeo~ Theorem applies.

b. Let n he the outward unit nonnal vector to a closed curve to which Greeo's Theorem applies. Show that it is not possible for F ~ xi + yj to he orthogonal to n at every point of C.

In Exercises 53-56, fmd tire outward flux of F across tire bouodary ofD. F

~

+

2xyi

2yzj

+ 2x2k

D: The cuhe cut from the first octant by tire p1anes x z =1

54. Spberical cap

F

~

xri

~

I,y

~

I,

+ yzj + k

D: The entire surface of the upper cap cut from the solid sphere + y' + z':5 25 by the plane z ~ 3

x'

Chapter

+ zk

57. Hemisphere, oylinder, and plane Let S he the surface that is bounded on tire leftby the hemisphere x' + y' + z' ~ a',y:5 0, in the middle by the oylinder x' + z' ~ a', 0 s y s a, and on the right by the plaue y ~ a. Find the flux of F ~ yi + zj + xk outward across S. 58. Cylinder aod plane. Find the outward flux of the fteld F ~ 3X2'i + yj - z'k across the surface of the solid in the first octant that is bouoded by the oylinder + 4y' ~ 16 and the planesy ~ 22,x ~ 0, andz ~ O.

x'

59. Cylindrical can Use the Divergence Theorem to fmd the flux ofF ~ xy'i + x'yj + yk outward through the surface of tire region enclosed by the oylinder + y' ~ 1 and the planes z ~ 1 audz ~ -1.

x'

60. Hemisphere Find the flux ofF ~ (3z + I)kupwardacrossthe hemisphere x 2 + y2 + z2 = a 2• z ~ 0 with the Divergence Theorem aod (b) by evaluating the flux integral directly.

Additional and Advanced Exercises

Finding Areas with Green's Theorem Use the Green~ Theorem area formula in Exercises 16.4 to fmd the areas of the regions enclosed by the curves in Exercises 1-4.

1. The lima<;oo x

3yj

planes

52. a. Outward flux and area Show that tire outward flux of the position vector field F = xi + yj across any closed curve to which Greeo ~ Theorem applies is twice the area of the region enclosed by the curve.

53. Cube

~ -2xj -

D: The upper region cut from the solid sphere x' + y' + z' :5 2 by the paraboloid z ~ x' + y' 56. Cone and oylinder F ~ (6x + y)i - (x + z)j + 4yzk D: The region in the ftrst octant bouoded by the cone z ~ v'x' + y', tire oylinder x' + y' ~ I, and the coordinate

I

51. Zero line integral Show that cosy lnxsinyt(y - -x-tix

F

~

OSt:521T

2 cos 1 - cos 21, Y

~

3. The eight curve x

~

(1/2)sin21,y

~

sinl,O:5 1:5 'IJ"(oocloop)

y

2 sin 1 - sin 21,

y

--t---~-~-~x

o

4. The teardrop x 2. The deltoid x OSt:521T

~

2 cos 1 + cos 21, y y

~

= 2a cos t

-

y

2 sin 1 - sin 21,

-4-~--~~~ x 3

a sin 21, y = b sin t, 0 :s t :s 2'17"

b

Chapter 16 Additional and Advanced Exercises Theory and Applications S••• Give an example of a vector field F (x, y, z) that has value 0 at only ooe point and such that curl F is nonzero everywhere. Be sure to identify the point and compute the curl.

b. Give an example ofa vector fieldF (x,y, z) that has value 0 on precisely one line and such that curl F is nonzero everywhere. Be sure to identify the line and compute the curl.

Force

6. Find all points (a, b, c) 00 the sphere x 2 + y2 + z2 ~ R2 where the vector field F ~ yz'i + xz2j + 2xyzk is nannal to the surface and F(a, b, c) # O. 7. Find the mass of a spherical shell of radius R such that at each point (x, y, z) 00 the surface the mass density ~(x,y, z) is its distance to some fixed point (a, b, c) of the surface. 8. Find the mass of a helicoid

0:5 r:5 1,0:5 2v'x 2

~

(r cos 8)i

(J:5

+ (r sin 8)j + Ok,

2w, if the density function is 8(x,y,z)

13. Archimedes' principle If an object such as a ball is placed in a liquid, it will either sink to the bottom, float, or sink a certain distance and remain suspended in the liquid. Suppose a fluid has coostant weight density w and that the fluid's surface coincides with the plane z ~ 4. A spherical ball remains suspended in the fluid and occopies the region x 2 + y2 + (z - 2)' :5 1. •• Show that the surface integral giving the magnitude of the tota1 force on the ball due to the fluid's pressure is

c. Give an example ofa vector fieldF (x,y, z) that has value 0 on a surface and such that curl F is nonzero everywhere. Be sure to identify the surface and compute the curl.

r(r, 8)

=

+ y2. See Practice Exercise 27 for a figure.

9. Amoog all rectangular regioos 0 :5 x :5 a, 0 :5 Y :5 b, fmd the one for which the total outward flux of F ~ (x 2 + 4xy)i - 6yj across the four sides is least. What is the least flux? 10. Find an equatioo for the plane through the origin such that the circulatioo of the flow field F ~ zi + xj + yk around the circle of intersection of the plane with the sphere x 2 + y2 + z2 = 4 is a maxinnun. 11. A string lies along the circle x 2 + y2 ~ 4 from (2, 0) to (0, 2) in

~

lim ±w(4 - z.) dCT. n_OO k- l

•• Partition the string into a fmite number of subarcs to show that the work done by gravity to move the string straight down to the x-axis is given by

~

lim

±

n_ OO k- l

g x.t)'.2 ds•

~

lc{ g xy2 ds,

where g is the gravitational constant.

Buoyantforce

x-axis. 12. A thin sheet lies along the portion of the plane x + y + z ~ I in the fl!St octant. The density of the sheet is ~ (x,y, z) ~ xy. •• Partition the sheet into a fmite number of sUbpieces to show that the work done by gravity to move the sheet straight down to the xy-plane is given by Work

±

~ n_OO lim g x.t)'.z. dCT. ~ {{ g xyz dCT, }=1 JJ S

where g is the gravitational constant.

b. Find the tota1 work done by evaluating the surface integral in part (a). c. Show that the total work done equals the work required to move the sheet's center of mass (x,y, z) straight down to the xy-plane.

w(4 JJ({ S

z) dCT.

~

JJ

w(z - 4)k' 0 dCT,

s where n is the outer uuit nonnalat (x, y, z). This illustrates Archimedes' principle that the magnitude of the buoyant force on a submerged solid equals the weight of the displaced fluid. •• Use the Divergence Theorem to fmd the magnitude of the bunyant force in part (b).

14. Fluid force on a curved surface A cone in the shape of the stn'face z ~ v'x 2 + y2, 0 S z :5 2 is filled with a liquid of constant weight density w. Assuming the xy-plane is "groood level," show that the tota1 force on the portion of the cone from z z ~ 2 due to liquid pressure is the surface integral

F

~

JJ

~

I to

w(2 - z)dCT.

s Evaluate the integral. 15. Faraday's Law If E(t,x,y,z) and B(t,x, y, z) represent the electric and magnetic fields at point (x,y, z) at time I, a hasic principle of electromagnetic theory says that V X E ~ -aB/al. In this expression V X E is computed with I held fixed and aB/al is calculated with (x, y, z) fIXed. Use Stokes' Theorem to derive Faraday~ Law,

b. Find the tota1 work done by evaluating the line integral in part (a). c. Show that the total work done equals the work required to move the string's center of mass (x, y) straight down to the

~

b. Since the ball is not moving, it is being held up by the bunyant force of the liquid. Show that the magnitude of the buoyant force on the sphere is

the fl!St quadrant. The density of the string is p (x,y) ~ xy.

Work

987

fE'dr

~ -f,

c

JJs

B'odCT,

where C represents a wire loup through which current flows counterclockwise with respect to the surface's unit nonnal n, giving rise to the voltage

fE'dr

c around C. The surface integral on the right side of the equation is called the magnetic flux, and S is any oriented surface with boundaryC.

16. Let F~_GmMr

Irl' be the gravitatioual force field defined for r # O. Use Gauss's Law in Section 16.8 to show that there is no continuously differentiable vector field H satisfying F ~ V X H.

988

Chapter 16: Integration in Vector Fields surface. Also, the magnitude du = Idul is the element of surface area (by Equatioo 5 in Section 16.5). Derive the identity

17. If f(x, y, z) and g(x, y, z) are continuously differentiable scalar functions defmed over the oriented surface S with boundary curve C, prove that

jJ(Vf

X

Vg)·ndu

s

=

f

du = (EG - F2)1/2 du dv where

fVg·dr.

c

18. Suppose that V· F, = V· F2 and V X F, = V X F2 over a region D enclosed by the oriented surface S with outward unit normal n aod that F, • n = F2 • nonS. Prove that F, = F2 throughootD. 19. Prove or disprove that if V • F = 0 aod V X F = 0, then F = O.

E

Ir,12,

F

=

r,·r.,

and

G

=

Ir.12.

21. Sbow that the voluroe V of a region D in space enclosed by the oriented surface S with ootward normal n satisfies the identity

20. Let S be ao oriented surface parametrized by r(u, v). Defme the notation du = r" du X rv dv so thatdu is a vector normal to the

Chapter

=

v=tjJr.ndU,

s where r is the position vector of the point (x,y, z) inDo

Technology Application Projects

Mathematica/Maple Module: Work in Conse1'VllliPe and N01U!onse1'Vlllive Force Fields Explore integration over vector fields aod experiment with conservative aod nonconservative force functions along different paths in the field.

How Can You VrslUl/ize Onen's Theorem7 Explore integration over vector fields aod use parame1rizations to compute line integrals. Both forms of Green's Theorem are explored.

Vrsulllizing flnd Interpreting the Divergence Theorem VerilY the Divergence Theorem by formulating and evaluating certain divergence aod surface integrals.

SECOND-ORDER DIFFERENTIAL EQUATIONS OVERVIEW In this chapter we extend oUI study of differential equations to those of second order. Second-order differential equations arise in many applications in the sciences and engineering. For instance, they can be applied to the study of vibrating springs and electric circuits. You will learn how to solve such differential equations by several methods in this chapter.

Second-Order Linear Equations An equation of the form P(x)y"(x)

+ Q(x)y'(x) + R(x)y(x)

= G(x),

(1)

which is linear in y and its derivatives, is called a second-order linear differential equation. We assume that the functions P, Q, R, and G are continuous throughout some open interval 1. If G(x) is identically zero on I, the equation is said to be homogeneous; otherwise it is called nonhomogeneous. Therefore, the form of a second-order linear homogeneous differential equation is

P(x)y"

+

Q(x)y'

+ R(x)y

=

O.

(2)

We also assume thatP(x) is never zero for any x EI. Two fundamental results are important to solving Equation (2). The first of these says that if we know two solutions Yl and Y2 of the linear homogeneous equation, then any linear combination y = ClYl + C2Y, is also a solution for any constants Cl and C2.

THEOREM i-The Superposition Prindple

If Yl(X) and y,(x) are two solutions Cl and C2, the

to the linear homogeneous equation (2), then for any constants

function

y(x) = cm(x)

+

C2Y2(X)

is also a solution to Equation (2).

17-1

17 -2

Chapter 17: Second-Order Differential Equations Proof Substituting Y into Equation (2), we have

P(x)y"

+ Q(x)y' + R(x)y = P(X)(CIYI + C2Y2)" + Q(X)(CIYI + C2Y2)' + R(X)(CIYI + C2Y2) = P(X)(CIY!" + C2Y2") + Q(X)(CIYI' + C2Y2') + R(X)(CIYI + C2YV = CI(P(X)YI" + Q(X)YI' + R(X)YI) + C2(P(X)Y2" + Q(X)Y2' + R(X)Y2) =

=

CI(O)

Therefore, Y = CIYI

0, YI is a solution

=

0, Y2 is a solution

+ C2(0) = O. + C2Y2 is a solution of Equation (2).

•

Theorem I immediately establishes the following facta concerning solutions tu the linear homogeneous equation.

1.

A sum of two solutions YI

+ Y2

tu Equation (2) is also a solution. (Choose CI =

C2 = \.) 2. 3.

A constant multiple kYI of any solution YI to Equation (2) is also a solution. (Choose CI = k and C2 = 0.) The trivial solution y(x) = 0 is always a solution tu the linear homogeneous equation. (Choose CI = C2 = 0.)

The second fundamental result about solutions to the linear homogeneous equation concerns its general solution or solution containing all solutions. This result says that there are two solutions YI and Y2 such that any solution is some linear combination of them for suitable values of the constants CI and C2. However, not just any pair of solutions will do. The solutions must be linearly independent, which means that neither YI nor Y2 is a constant multiple of the other. For example, the functions f(x) = eX and g(x) = xe x are linearly independent, whereas f(x) = X2 and g(x) = 7X2 are not (so they are linearly dependent). These results on linear independence and the following theorem are proved in more advanced courses.

THEOREM 2

If P, Q, and R are continuous over the open interval I and P(x) is never zero on I, then the linear homogeneous equation (2) has two linearly independent solutions YI and Y2 on 1. Moreover, if YI and Y2 are any two linearly independent solutions of Equation (2), then the general solution is given by y(x) = CIYI(X) + C2Y2(X), where CI and C2 are arbitrary constants.

We now turn our attention tu rmding two linearly independent solutions tu the special case of Equation (2), where P, Q, andR are constant functions.

Constant-Coefficient Homogeneous Equations Suppose we wish to solve the second-order homogeneous differential equation

ay"+by'+cy=O,

(3)

17-3

17.1 Second-Order Linear Equations

where a, b, and c are constants. To solve Equation (3), we seek a function which when multiplied by a constant and added to a constant times its first derivative plus a constant times its second derivative sums identically to zero. One function that behaves this W
ar2e n: + bre1"X + ce rx

=

o.

Since the exponential function is never zero, we can divide this last equation through by e"'. Thus, y = e'" is a solution to Equation (3) if and ouly if r is a solution to the algebraic equation

ar2 + br + c

=

O.

(4)

Equation (4) is called the auxiliary equation (or charaeteristic equation) of the differential equation ay" + by' + cy = O. The auxiliary equation is a quadratic equation with roots

-b

+ Vb 2 - 4ac 2a

and

'2 =

There are three cases to consider which depend on the value of the discriminant b 2

-

4ac.

Case 1: b2 - 4ac > O. In this case the auxiliary equation has two real and unequal roots rl and r2. Then YI = e'" and Y2 = e"" are two linearly independent solutions to Equation (3) because e"" is not a constant multiple of e'" (see Exercise 61). From Theorem 2 we conclude the following result.

THEOREM 3 equation ar2

If rl and r2 are two real and unequal roots to the auxiliary = 0, then

+ br + c

is the general solution to ay"

EXAMPLE 1

+ by' + cy

= O.

Find the general solution of the differential equation

y" - y' - 6y

=

O.

Solution Substitution of y = e'" into the differential equation yields the auxiliary equation which factors as (r - 3)(r

+ 2)

=

O.

The roots are r1 = 3 and r2 = - 2. Thus, the general solution is y = cle 3x

+ C2e- 2x•

•

17-4

Chapter 17: Second-Order Differential Equations

Case 2: b2 - 4ac = O. In this case rl = r2 = -bI2a. To simplify the notation, let r = -bI2a. Then we have one solution YI = en< with 2ar + b = O. Since multiplication of e'" by a constant fails to produce a second linearly independent solution, suppose we try multiplying by afonction instead. The simplest such function would be u(x) = x, so let's see if Y2 = xe'" is also a solution. Substituting Y2 into the differential equation gives

aY2" + by,' + CY2

=

a(2re'" + r 2xe"') + b(e'" + rxe"') + cxe'"

= (2ar + b)e'" + (ar2 + br + c)xe'" = O(e"') + (O)xe'"

=

O.

The first term is zero because r = -bI2a; the second term is zero because r solves the auxiliary equation. The functions YI = e'" and Y2 = xe'" are linearly independent (see Exercise 62). From Theorem 2 we conclude the following result.

THEOREM 4

ar2

+ br + c

=

If r is the only (repeated) real root to the auxiliary equation 0, then

is the general solution to ay"

EXAMPLE 2

+ by' + cy

=

o.

Find the general solution to

y" + 4y' + 4y

=

o.

Salution The auxiliary equation is

r2+4r+4=0, which factors into

(7 + 2)2

=

o.

Thus, r = -2 is a double root. Therefore, the general solution is

Y

=

cle-2x +

C2 xe - 2x•

•

2

Case 3: b - 4ac < O. In this case the auxiliary equation has two complex roots rl = a + i{3 and r2 = a - i{3, where a and {3 are real numbers and i 2 = -1. (These real numbers are a = -bl2a and {3 = Y4ac - b 2/2a.) These two complex roots then give rise to two linearly independent solutions

Yl = e(a+ip)< = eax(cos {3x

+ i sin {3x)

and Y2 = e(a-ip)< = eax(cos {3x - i sin {3x).

(The expressions involving the sine and cosine terms follow from Euler's identity in Section 9.9.) However, the solutions YI and Y2 are complex valued rather than real valued Nevertheless, because of the superposition principle (Theorem 1), we can obtain from them the two real-valued solutions

Y3

=

I +1 ZYI ZY2

=

Q e ax cos,...x

and

_I I _ax." Y4 - 2i YI - 2i Y2 - e sm ,...x.

The functions Y3 and Y4 are linearly independent (see Exercise 63). From Theorem 2 we conclude the following result.

17-5

17.1 Second-Order Linear Equations

If rl = a + if3 and r2 = a - if3 are two complex roots to the auxiliary equation ar2 + br + c = 0, then

THEOREM 5

y = e""(cI cos f3x

+ by' + cy

is the general solution to ay"

EXAMPLE 3

+ =

C2Sin f3x)

o.

Find the general solution to the differential equation

y" - 4y' + 5y

=

O.

Solution The auxiliary equation is

r2-4r+5=0. The roots are the complex parr r = (4 ± V16 - 20)/2 or Thus, a = 2 and f3 = 1 give the general solution

rl =

2 + i and r2

=

2 - i.

• Initial Value and Boundary Value Problems To determine a unique solution to a {lISt-order linear differential equation, it was sufficient to specify the value of the solution at a single point. Since the general solution to a secondorder equation contains two arbitrary constants, it is necessary to specify two conditions. One way of doing this is to specify the value of the solution function and the value of its derivative at a single point: y(xo) = Yo and y'(xo) = YI. These conditions are called initial conditions. The following result is proved in more advanced texts and guarantees the existence of a unique solution for both homogeneous and nonhomogeneous second-order linear initial value problems.

If P, Q, R, and G are continuous throughout an open interval f, then there exists one and only one function y(x) satisfying both the differential equation

THEOREM 6

P(x)y"(x)

+ Q(x)y'(x) + R(x)y(x)

=

G(x)

on the interval f, and the initial conditions

y(xo)

=

Yo

and

y'(xo)

=

Yl

at the specified point Xo E f.

It is important to realize that any real values can be assigned to Yo and YI and Theorem 6 applies. Here is an example of an initial value problem for a bomogeneous equation.

17-6

Chapter 17: Second-Order Differential Equations

EXAMPLE 4

Find the particular solution to the initial value problem

y"-2y'+y=0,

y(O)

= 1,

y'(0)

= -I.

Solution The auxiliary equation is

+ I

r2 - 2r

y

= (r -

1)2 = O.

The repeated real root is r = I, giving the general solution

Y

= CteX

+

C2 xex .

Then,

y' = cle x

-4

+ C2(X +

I)e x •

From the initial conditions we have -6 -8

FIGURE 17"1

for Example 4.

and Thus, CI = I and C2 = -2. The unique solution satisfying the initial conditions is

y

Particular solution curve

=

eX - be x .

•

The solution curve is shown in Figure 17.1.

Another approach to determine the values of the two arbitrary constants in the general solution to a second-order differential equation is to specify the values of the solution function at two different points in the interval!. That is, we solve the differential equation subject to the boundary values and where Xl and X2 both belong to I. Here again the values for YI and Y2 can be any real numbers. The differential equation together with specified boundary values is called a boundary value problem. Unlike the result stated in Theorem 6, boundary value problems do not always possess a solution or more than one solution may exist (see Exercise 65). These problems are studied in more advanced texts, but here is an example for which there is a unique solution.

EXAMPLE 5

Solve the boundary value problem

y"

+ 4y

=

0,

y(O)

= 0,

y(;;)

= I.

Solution The auxiliary equation is r2 + 4

= 0, which has the complex roots r = ±2i. The general solution to the differential equation is

y = CI cos 2x + C2 sin 2x. The boundary conditions are satisfied if y(0)=cI"I+C2"0=0

It follows that CI = 0 and C2 = 2. The solution to the boundary value problem is

y

= 2 sin2x.

•

17-7

17.1 Second-Order Linear Equations

EXERCISES 17.1 In Exercises 1-30, fmd the general solution of the given equation.

a 4y ~ a

1. y' - y' - 12y 3. y'

+ 3y'

-

~

2. 3y" - y'

5.y·-4y~0

~

41. y" - 2y' - 3y 43. 4y"

6.y"-64y~0

45. 4y'

~

0

12y

11.y·+9y~0

12. y"

+ 4y' + 5y

13.y·+25y~0

14. y"

+Y = 0

+

19. y'

+ 5y + 2y' + 4y + 4y' + 9y

21. y"

=

a a ~ a

0

49. 4y'

42.6y·-y'-y~0

+

lOy'

+y

+ 4y' + 5y

~

~

44. 9y'

53. 9y"

a

+

24y'

+ 9y ~ + 16y ~

55. 6y" - 5y' - 4y

~

48. 6y'

0

0

51. 16y" - 24y'

0

+

12y'

+ 4y

+ \3y' - 5y

a a

52.

+ 16y' + 52y

18. y" -2y'

+3y~0

In Exercises 56-60, solve the initial value problem.

20. 4y" - 4y'

+ \3y ~ 0 + 16y ~ 0

56. y" - 2y'

d'y

dy

d'y

dy

d'y

23. tb;2+4tb;+4y~0 25. tb;2+6tb;+9y~0

d'y

dy

d'y

dy

dy

dy

26. 4tb;2 - 12tb;

58.

d'y

27.4tb;2+4tb;+y~0

59.

+ 9y

~ 0

dy

28.4tb;2-4tb;+y~0

d'y

29. 9tb;2+6tb;+y~0

dy

30. 9 tb;2 - 12 tb;

+ 4y

~

0

In Exercises 31-40, fmd the uoique solution of the second-order

initial value problero. 31. y" + 6y' + 5y ~ 0, y(O) ~ 0, y'(O) ~ 3 32. y" + 16y ~ 0, y(0) ~ 2, y'(O) ~ -2 33. y" + 12y ~ 0, y(0) ~ 0, y'(O) ~ I 34. 12y' + 5y' - 2y ~ 0, y(O) ~ I, y'(O) ~ -I 35. y" + 8y ~ 0, y(0) ~ -I, y'(O) ~ 2 36. y"

+ 4y' + 4y

~

+ 4y ~ 38. 4y" - 4y' + y ~ 37. y" - 4y'

d'y

39.4 tb;2

d'y

+

dy 12tb;

dy 40.9 tb;2 - 12tb;

+

0, y(O) 0, y(O) 0, y(O)

~ ~ ~

0, y'(O) I, y'(O) 4, y'(O)

~

I

~

a

~

0, y(0)

~

0, y'(O)

~

60.

~

0, y(0)

~

2

61. Prove that the two solution fimctions in Theorem 3 are linearly independent. 62. Prove that the two solution fimctions in Theorem 4 are linearly independent. 63. Prove that the two solution fimctions in Theorem 5 are linearly independent. 64. Prove that if Yl and Y2 are linearly independent solutions to the homogeneous equation (2), then the fimctions Y3 ~ Yl + Y2 and y, ~ Yl - Y2 are also linearly independent solutions. 65. •• Show that there is no solution to the boundary value problem ~

0, y(O)

~

0, y(1T)

~

I.

b. Show that there are infmitely many solutions to the bouodary

value problero y" + 4y

~

0, y(O)

~

0, y(1T)

~

~

O.

66. Show that if a, b, and c are positive constants, then all solutions of the homogeneous differential equation ay"+by'+cy=O

+ 4y

a

+ 2y' + y ~ 0, y(0) ~ I, y'(O) ~ I 4y" - 4y' + y ~ 0, y(O) ~ -I, y'(O) ~ 2 3y" + y' - 14y ~ 0, y(O) ~ 2, y'(O) ~ -I 4y" + 4y' + 5y ~ 0, y('IJ') ~ I, Y'(1T) ~ a

y" + 4y

~ 4 dy 9y ~ 0, y(0) ~ 2, tb;(0) ~ I

dy -I, tb;(0)

~

57. y"

24. tb;2-6tb;+9y~0

d'y

+ 2y

0

a

16. y"

~

+ 8y'

~

6y'-5y'-6y~0

54. 4y'

~

22. y"

0

50.y·+4y'+6y~0

17. y'

16y

~

46.y·+2y'+2y~0

15. y' - 2y'

0

~

~

~

0

~

+ 4y' + y ~ a + 20y ~ 0

47. 25y"

+

10. 3y' - 2Oy'

0

In Exercises 41-55, fmd the general solution.

4.y"-9y=O

8. 9y" - y

7.2y·-y'-3y~0

9. 8y" - lOy' - 3y

0

=

I

approach zero as x ~

00.

17-8

Chapter 17: Second-Order Differential Equations

Nonh

Linear In this section we study two methods for solving second-order linear noohomogeneous differential equations with constant coefficients. These are the methods of undetermined coefficients and variation ofparameters. We begin by considering the form of the general solution.

Form of the General Solution Suppose we wish to solve the noohomogeneous equation

ay"

+ by' + ey

(1)

= G(x),

where a, b, and c are constants and G is continuous over some open interval l. Let Ye = ClYl + C2Y2 be the general solution to the associated complementary equation

ay"

+ by' + cy

=

o.

(2)

(We learned how to find Ye in Section 17.1.) Now suppose we could somehow come up with a particular function YP that solves the noohomogeneous equation (I). Then the sum

Y = Ye

+ YP

(3)

also solves the noohomogeneous equation (I) because

a(ye

+ Yp)" + b(ye + Yp)' + c(ye + Yp) = (aye" + bye' + eye) + (ayp' + byp' + eyp) y, solves Eq. (2) and YP solves Eq. (\) = 0 + G(x) = G(x).

Moreover, if Y = y(x) is the general solution to the noohomogeneous equation (I), it must have the form of Equation (3). The reason for this last statement follows from the observation that for any function YP satisfying Equation (I), we have

a(y - Yp)"

+ b(y = (ay'

+ c(y + by' + ey)

- Yp)'

= G(x) - G(x) =

- Yp) - (ayp"

+ byp' + eyp)

o.

Thus, Ye = Y - YP is the general solution to the homogeneous equation (2). We have established the following result.

THEOREM 7

The general solution y = y(x) to the noohomogeneous differential equation (I) has the form

y = Ye

+ YP'

where the complementary solution Ye is the general solution to the associated homogeneous equation (2) and Yp is any particular solution to the noohomogeneous equation (I).

17.2

Nonhomogeneous Linear Equations

17-9

The Method of Undetermined Coeffidents This method for finding a particular solution YP to the nonhomogeneous equation (1) applies to special cases for which G(x) is a sum of terms of various polynomials p(x) multiplying an exponential with possibly sine or cosine factors. That is, G(x) is a sum of terms of the following forms: P2(x)e UX cos {3x,

P3(x)eax sin {3x.

For instance, 1 - x, e2x , xe x, cos x, and Sex - sin 2x represent functions in this category. (Essentially these are functions solving homogeneous linear clifferential equations with constant coefficients, but the equations may be of order higher than two.) We now present several examples illustrating the method.

EXAMPLE 1

Solve the nonhomogeneous equation y" - 2y' - 3y = 1 - x 2.

Solution The auxiliary equation for the complementary equation y" - 2y' - 3y = 0 is r2 - 2r - 3 = (r

+ l)(r - 3)

=

O.

It has the roots r = - 1 and r = 3 giving the complementary solution

Now G(x) = 1 - x 2 is a polynomial of degree 2. It would be reasonable to assume that a particular solution to the given nonhomogeneous equation is also a polynomial of degree 2 because if y is a polynomial of degree 2, then y" - 2y' - 3y is also a polynomial of degree 2. So we seek a particular solution of the form yp=Ax2+Bx+C.

We need to determine the unknown coefficients A, B, and C. When we substitote the polynomial YP and its derivatives into the given nonhomogeneous equation, we obtain 2A - 2(2Ax + B) - 3(Ax 2 + Bx + C) = 1 - x 2 or, collecting terms with like powers of x, -3Ax 2 + (-4A - 3B)x

+ (2A -

2B - 3C) = 1 - x 2.

This last equation holds for all values of x if its two sides are identical polynomials of degree 2. Thus, we equate corresponding powers of x to get -3A = -1,

-4A - 3B = 0,

and

2A - 2B - 3C = 1.

These equations imply in turn that A = 1/3, B = -4/9, and C = 5/27. Substituting these values into the quadratic expression for our particular solution gives

By Theorem 7, the general solution to the nonhomogeneous equation is

_

y-Yc

+ yp-cle _ -. +

C2e

3.

+ :31 x 2 -94 x + 27' 5

•

17 -1 0

Chapter 17: Second-Order Differential Equations

EXAMPLE 2

Find a particular solution of y" - y' = 2 sin x.

Solution If we try to find a particular solution of the fann yp=Asinx and substitute the derivatives of yp in the given equation, we f"md that A must satisfy the equation

-A sin x

+ A cos x

= 2 sin x

for all values of x. Since this requires A to equal both -2 and 0 at the same time, we conclude that the nonhomogeneous differential equation has no solution of the form A sin x. It turns out that the required fann is the sum

yp = A sin x

+ B cosx.

The result of substitoting the derivatives of this new trial solution into the differential equation is

-A sin x - B cos x - (A cos x - B sin x) = 2 sin x or (B - A) sin x - (A

+ B) cos x

= 2 sinx.

This last equation must be an identity. Equating the coefficients for like terms on each side then gives

B-A=2

A

and

+B

=

O.

Simultaneous solution of these two equations gives A = - 1 and B = 1. Our particular solution is

YP

EXAMPLE 3

=

•

cosx - sinx.

Find a particular solution of y' - 3y'

+ 2y

= 5e x •

Solution If we substitute YP = Ae

x

and its derivatives in the differential equation, we f"md that

Aex

-

3Aex

+

2Ae x = 5e x

or

0= 5e x • However, the exponential function is never zero. The trouble can be traced to the fact that y = eX is already a solution of the related homogeneous equation

y" - 3y'

+ 2y

=

O.

The auxiliary equation is

r2 - 3r

+2

= (r -

1)(r - 2) = 0,

which has r = 1 as a root. So we would expect Ae x to become zero when substitoted into the left-hand side of the differential equation. The appropriate way to modify the trial solution in this case is to multiply Aex by x. Thus, our new trial solution is x YP = Axe .

17.2

Nonhomogeneous Linear Equations

17-11

The result of substituting the derivatives of this new candidate into the differential equation is (Axe X + ZAe") - 3(Axe X + Ae")

+ ZAxe x

= 5e x

or Thus, A

=

-5 gives our sought-after particular solution YP

=

•

-5xe x.

Find a particular solution of y" - 6y' + 9y

EXAMPLE 4

= e 3x •

Solutton The auxiliary equation for the complementary equation

+9

r2 - 6r

= (r - 3)2 = 0

as a repeated root. The appropriate cboice for yp in this case is neither Ae 3x nor Axe because the complementary solution contains both of those terms already. Thus, we choose a term containing the next higher power of x as a factor. When we substitute

has r

= 3

3x

YP =

Ax2e 3x

and its derivatives in the given differential equation, we get (9Ax 2e 3X

- 6(3Ax 2e 3x

+ lZAxe 3x + ZAe~

+ ZAxe 3") + 9Ax 2e 3x

= e 3x

or Thus, A

=

1/2, and the particular solution is YP =

•

!x 2e"':

When we wish to fmd a particular solution of Equation (1) and the function G(x) is the sum of two or more terms, we choose a tria1 function for each term in G(x) and add them.

EXAMPLE 5

Find the general solution to y" - y'

= 5e x -

sin 2x.

Solutton We fITSt check the auxiliary equation r2 - r = O.

Its roots are r = I and r mogeneous equation is

=

O. Therefore, the complementary solution to the associated hoYc

= cle

x

+

C2·

We now seek a particular solution Yp. That is, we seek a function that will produce 5e x - sin 2x when substituted into the left-hand side of the given differential equation. One part of YP is to produce 5e', the other -sin 2x. Since any function of the form CteX is a solution of the associated homogeneous equation, we choose our trial solution YP to be the sum Yp = Axe'

+ B cos 2x + C sin 2x,

x

including xe where we might otherwise have included only eX. When the derivatives of YP are substituted into the differential equation, the resulting equation is (Axe'

+ ZAe'

- 4B cos 2x - 4C sin 2x) - (Axe'

+ Ae'

- 2B sin 2x

+ 2C cos 2x)

= 5e' -

sin 2x

17 -12 Chapter 17: Second-Order Differential Equations or

Ae' - (4B + 2C) cos 2x + (2B - 4C) sin 2x = 5e' - sin 2x. TIris equation will hold if

A = 5,

4B

+ 2C

=

2B-4C= -I,

0,

or A = 5, B = -1/10, and C = 1/5. Our particular solution is

I .2x I 2x+ Ssm. Yp - 5 xe' - TOcOS The general solution to the dllferential equation is y =Yc + Yp =

cle'

+

C2

+ 5xe' -

l~cos2x

+ tsin2x.

•

You may find the following table helpful in solving the problems at the end of this section.

TABLE 17.1

The method of undetermined coefficients for seLected equations of the form oy"

IT G(x) has a term that is a constant multiple of ...

e'"

+ by' + cy

=

And if

G(x). Then include this expression in the trial function for Y,.

r is not a root of the auxiliary equation

Ae'"

r is a single root of the auxiliary equation

Axe'"

r is a double root of the auxiliary equation sin kx, cos kx

ki is not a root of the auxiliary equation

Bcoskx + Csinkx

px 2 +qx+m

ois not a root of the

Dx2+Ex+F

auxiliary equation

ois a single root of the

Dx 3 + Ex 2 + Fx

auxiliary equation

ois a double root of the

Dx'

+ Ex 3 + Fx 2

auxiliary equation

The Method of Variation of Parameters TIris is a general method for finding a particular solution of the nonhomogeneous equation (I) once the general solution of the associated homogeneous equation is known. The method consists of replacing the constants Cl and C2 in the complementary solution by functions VI = VI(X) and V2 = V2(X) and requiring (in a way to be explained) that the

17.2

Nonhomogeneous Linear Equations

17-13

resulting expression satisfy the nonhomogeneous equation (I). There are two functions to be detennined, and requiring that Equation (I) be satisfied is only one condition. As a second condition, we also require that (4)

Then we have

Y = VIYI + V2Y2,

y' = y" =

+ V2Y2', VIY(" + V2Y2" + VIYl'

Vl'Yl'

+ V2'Y2'.

lfwe substitote these expressions into the left-hand side of Equation (I), we obtain

vI(aYI"

+ bYI' +

CYI)

+ v2(aY2" + by2' + eY2) + a(vI'YI' + ViY2')

=

G(x).

The first two parenthetical terms are zero since YI and Y2 are solutions of the associated homogeneous equation (2). So the nonhomogeneous equation (I) is satisfied if, in addition to Equation (4), we require that

+ v2'yi)

a(VI'YI'

G(x).

=

(5)

Equations (4) and (5) can be solved together as a pair Vt'Yt

+

V2'Y2

= 0,

for the unknown functions VI' and V2'. The usual procedure for solving this simple system is to use the method ofdeterminants (also known as Cramer ~ Rule), which will be demonstrated in the examples to follow. Once the derivative functions VI' and V2' are known, the two functions VI = VI(X) and V2 = V2(X) can be found by integration. Here is a summary of the method.

Variation of Parameters Procedure To use the method of variation of parameters to find a particular solution to the nonhomogeneous equation

ay"

+ by' + cy

=

G(x),

we can work directly with Equations (4) and (5). It is not necessary to rederive them. The steps are as follows. 1.

Solve the associated homogeneous equation

ay"+by'+cy=O 2.

to tmd the functions YI and Y2. Solve the equations

+

Vt'Yt

Vt'Yt'

3. 4.

V2'Y2 =

+ 112'y2'

=

0, G(x)

-a-

simultaneously for the derivative functions VI' and V2'. Integrate VI' and V2' to find the functions VI = VI(X) and V2 = V2(X). Write down the particular solution to nonhomogeneous equation (I) as

YP =

VJ)'1

+ V2Y2·

17 -14 Chapter 17: Second-Order Differential Equations EXAMPLE 6

Find the general solution to the equation

y" + Y

= tanx.

Solution The solution of the homogeneous equation

y" + Y

=

0

= Cl COS X

+

Cz

is given by

Yc

sinx.

Since Yl(X) = cos x and Y2(X) = sin x, the conditions to be satisfied in Equations (4) and (5) are Vt'

-Vt'

+

V2'

sinx = 0,

sinx +

V2'

cosx = tanx.

cos x

a

= 1

Solution of this system gives

Vl' =

0 sin x I tan x cosx I -c"--co-,,-x--S-in--"XC1

-smx

-tan x sin x cos2 x + 8in2 x

cosx·

cos X 1

Likewise,

I c~sx V2' =

-smx

I-smx c~sx

~xl sinx cos X

=

I

sinx.

After integmting VI' and V2', we have Vl(X) =

j

= =

and

V2(X) =

-Sin2 x

cos x dx

j

(sec x - cos x) dx

-In Isecx

j

+ tanxl + sinx,

sin x dx = -cos x.

Note that we have omitted the constants of integration in detennining Vl and V2. They would merely be absorbed into the arbitrary constants in the complementary solution. Substitutiog Vl and V2 into the expression for YP in Step 4 gives

+ tanxl + sin x] cos x + (-cos x) sin x (-cos x) In lsecx + tanxl.

YP = [-In Isecx =

The general solution is

Y =

Cl

cos x

+ C2 sin x

- (cos x) In Isecx

+ tanxl.

•

17.2

EXAMPLE 7

Nonhomogeneous Linear Equations

17-15

Solve the nonhomogeneous equation

y" + y' - 2y

=

xe x.

Solution The auxiliary equation is r2

+r

- 2 = (r

+ 2)(r

- I) = 0

giving the complementary solution

Yc

c}e- 2x

=

+ cze x .

The conditions to be satisfied in Equations (4) and (5) are Vt'e- 2x + v2'e x = 0,

+ v2'e x =

-2vt'e-2x

xe x.

a

= 1

Solving the above system for VI' and vi gives

Ix~x

Vt' =

I-2ee

eXI eX

2x

2x

eXI eX

_ - xe 2x _ - 3e-x -

I

-3 xe

'-'

.

Likewise,

Integrating to obtain the parameter functions, we have VI(X) =

J-t

=

xe 3x dx

_t(X~3X

- Je;x dx )

_ l( 1 3xe )'-' -27 , and

Therefore,

The general solution to the differential equation is

Y=

lx cle-

+

C2ex -

~xex +

kx

2 X e ,

where the term (1/27)e Xin YP has been absorbed into the term C2ex in the complementary solution. •

17 -16 Chapter 17: Second-Order Differential Equations

EXERCISES 17.2 Solve the equations in Exercises 1-16 by the method of undetermined coefficients. L y" - 3y' - lOy = -3 2_ y" - 3y' - lOy = 2x - 3 4. y" + 2y' + Y

3. y" - y' = sinx 5. y"

+Y

7. y" - y' - 2y = 20 cos x

+x

= e2x + Y = a+ 3ex + 2y' + Y = 6 sin 2x

6. y" + Y

= cos3x

9. y" - y = eX

= x2

2

8. y"

10. y"

11. y" - y' - 6y = e-x - 7 cosX' 12. y"

+ 3y' + 2y

d'y

= e-x

+ e-2.t" -

dy,

+ 5 dx

13_ dx'

x

d'y

d'y dy]x 15_ dx' - 3 dx = e - 12x

d'y

dy _

16_ dx'

,

+ 7 dx - 42x + 5x + I

Solve the equations in Exercises 17-28 by variation of parameters. 17. y"

+ y'

18. y"

+Y

19. y"

+ Y = sinx + 2y' + Y =

21. y"

= x =

20. y"

23. y" - y = eX 25. y"

+ 4y' + Sy

d'y 27. dx'

+Y

d'y 28. dx 2

-

=

= secx,

dy dx

=

+ 2y' + Y

= eX

22.y"-y=x 24. y" - y = sinx

e-x

10

26. y" - y' 1f

=

2x

eXcosx, x> 0

31. y" + Y = 2 cosx 32. y"

+ y'

+ Bxe~

+ sinx, Yp = A cosX' + B sinx + sinx, YP = Ax cosX' + Bx sinx

- 2y = xe x , YP = Ax 2e x

+ Bxe x

coefficients. dy

33 - - = eX • dx' dx

+ e-r

d'y dy 35. dx,-4 -5y =ex +4

dx

=

cscx,

39. y" - 8y' =

o<x<w

43. y"

+

45. y"

+Y

42. y"

2y' = x 2 =

46. y" - 3y'

e~

-

44. y"

secxtanx,

+

+ 4y = sinx + 4y' + 5y = x + 2 + 9y = 9x - cosX'

40. y"

e!b:

41. y" - y' = x 3

_'!! 2

< x < '!! 2

2y = eX - e'h

The method of undetermined coefficients can sometimes be used to solve flISt-order ordinary differential equations. Use the method to solve the equations in Exercises 47-50.

47. y' - 3y = eX

48. y'

+ 4y = x

49. y' - 3y = 5e'"

SO. y'

+Y

d'y 51. dx'

+ Y = sec x, -2 < x < 2;

d'y 52. dx'

+ Y = e2x;

,1f

y(O)

= 0,

1f

= sinx

y(O)

= y'(O) = I

2

=5

y'(O)

d'y

53. y"

x' - x, y(0) = 0, + y' = x, YP = T

54. y"

+ Y = x, YP = 2 sin x + x,

55.

In Exercises 33-36, solve the given differential equations (a) by variation of parameters and (h) by the method of undetermined d'y

cotx, O<X<'7T

isfYing the equation and the given initial conditions.

In each of Exercises 29-32, the given differential equation has a particular solution YP of the form given. Determine the coefficients in Yp.

30. y" - y' = cosX'

=

In Exercises 53-58, verify that the given function is a particular solution to the specified nonhomogeneous equation. Find the general solution and evaluate its arbitrary constants to f'md the unique solution sat-

1f

-2 < x < 2

Then solve the differential equation. 29. y" - 5y' = xeil:, YP = Ax 2e 5.l:

38. y"

+Y +Y

Solve the differential equations in Exercises 51 and 52 sui!ject to the given initial conditions.

_'!! < x < '!! 2 2

tanx,

37. y"

dy

14_ dx' - dx = -8x + 3

= 15x

Solve the differential equations in Exercises 37-46. Some of the equations can be solved by the method of undetermined coefficients, but others cannot.

dy

34. - 4dx' dx

+

4. = 2e 2x '

d'y dy 9x 36. dx' - 9dx = ge

ty" + y' + Y

57. y" - 2y' 58. y" - 2y'

YP

+Y +Y

= xexJnx,

= 0,

=0

y'(O)

=0

4el{cosx - sinx),

=

YP = 2e x cos x, y(0) 56. y" - y' - 2y

y(O)

y'(O)

=I

= 0,

y' (0)

- 2x, YP

=I

=x

- I, y(O)

= 0,

y'(O)

=

2e x, YP = x 2e x, y(O) = 1, y'(O) = 0

=

x -leX, x

y(1)

=I

> 0, = e, y'(I) = a

In Exercises 59 and 60, two linearly independent solutions YI and Y2 are given to the associated homogeneous equation of the variable-

coefficient nonhomogeneous equation. Use the method of variation of parameters to fmd a particular solution to the nonhomogeneous equation. Assmne x > 0 in each exercise. 59. x7" + 2xy' - 2y = x 2, Yl = x- 2 , Y2 = x 60.

x7" + xy'

- Y=

X,

YI = x-I, Y2 = x

17.3

I ,,,

Applications

17-17

AppLications In this section we apply second-order differential equations to the study of vibrating springs and electric circuits.

Vibrations A spring has its upper end fastened to a rigid support, as shown in Figure 17.2. An object of mass m is suspended from the spring and stretches it a length s when the spring comes to rest in an equilibrium position. According to Hooke's Law (Section 6.5), the tension force in the spring is ks, where k is the spring constant. The force due to gravity pulling down on the spring is mg, and equilibrium requires that massm

ks = mg.

at equilibrium

y

FIGURE 17.2

Massm stretches a spring by length s to the equilibrium position at y = O.

(1)

Suppose that the object is pulled down an additionsl amount Yo beyond the equilibrium position and then released We want to study the object's motion, that is, the vertical position of its center of mass at any future time. Let y, with positive direction downward, denote the displacement position of the object away from the equilibrium position y = 0 at any time t after the motion has started. Then the forces acting on the object are (see Figure 17.3)

Fp

=

mg,

F, = k(s

F,=ll

the propulsion force due to gravity,

+ y),

the restoring force of the spring's tension,

dy , dt

a frictional force assumed proportional to velocity.

The frictional force tends to retard the motion of the object. The resultant of these forces is F = Fp - F. - F" and by Newton's second law F = ma, we must then have

a"ly m dt 2

=

dy mg - ks - ky - Il dt'

By Equation (1), mg - ks = 0, so this last equation becomes

dly m dt 2 y

Yo

start

position y

FIGURE 17.3 Thepropulsion force ("";ght) Fp pulls iIle mass downward, but !he spring restoring force F, aod fric1iooa1 force F, pullille mass upward. The motion starts at Y = Yo with iIle mass vihratiog up aod down.

+

dy Il dt

+ ky

=

(2)

0,

subject to the initial conditions y(O) = Yo and y' (0) = O. (Here we use the prime notation to denote differentiation with respect to time t.) You might expect that the motion predicted by Equation (2) will be oscillatory about the equilibrium position y = 0 and eventually dsmp to zero because of the retarding frictional force. This is indeed the case, and we will show how the constants m, Il, and k determine the nature of the dsmping. You will also see that if there is no friction (so Il = 0), then the object will simply oscillate indefinitely.

Simple Harmonic Motion Suppose first that there is no retarding frictional force. Then Il = 0 and there is no dsmping. lfwe substitote w = Vkj;. to simplify our calculations, then the second-order equation (2) becomes with

y(O) = Yo

and

y'(O) = O.

17 -18 Chapter 17: Second-Order Differential Equations The auxiliary equation is r2

+ if

0,

=

having the imaginary roots r = ±wi. The general solution to the differential equation in (2) is

Y =

Cl

cos rut

+

sin wt.

C2

(3)

To fit the initial conditions, we compute

y'

-Cl Cd sin

=

rut +

C2W cos

wt

and then substitute the conditions. This yields c, = Yo and C2 =

o. The particular solution

Y=Yocoswl

(4)

describes the motion of the object. Equation (4) represents simple harmonic motion of amplitude Yo and period T = 27T/W. The general solution given by Equation (3) can be combined into a single term by using the trigonometric identity sin(WI

+ cf»

+ sin wi cos cf>.

wi sin cf>

= cos

To apply the identity, we take (see Figure 17.4)

c,

=

C sin cf>

and

C2

=

C cos cf>,

where

C= Vc,2 FIGURE 17_4 = Ceosc/>.

C2

c, =

C sin c/> and

+

co'

_

and

-1 Cl

cf> - tan

C2·

Then the general solution in Equation (3) can be written in the alternative form

Y = C sin (WI

+ cf».

(5)

Here C and cf> may he taken as two new arbitrary constants, replacing the two constants c, and C2. Equation (5) represents simple harmonic motion of amplitude C and period T = 27T/W. The angle wi + cf> is called the phase angie, and cf> may be interpreted as its initial value. A graph of the simple harmonic motion represented by Equation (5) is given in Figure 17.5.

y

Period

~ T=~ ~

C Csin."

I I

I I

o -C y=Csin(w,+.,,)

FIGURE 17.5 Simple hannonie motion of amplitude C and period T wi1h initial phase angle c/> (Equation 5).

17.3 Applications

17 -19

Damped Motion Assume now that there is friction in the spring system, so [; "" O. If we substitute w = ~ and 2b = [;/m, then the differential equation (2) is

y"

+ 2by' + .Jy

=

O.

(6)

The auxiliary equation is ,2+2b,+w2 =O,

with roots , = -b ± yCb"2-_-w'''. Three cases now present themselves, depending upon the relative sizes of b and w. Case 1: b = Cd. The double root of the auxiliary equation is real and equals, = w. The general solution to Equation (6) is

y = (Cl

+ c2t)e-"".

This situation of motion is called critical damping and is not oscillatory. Figure 17.6a shows an example of this kind of damped motion. Case 2:

b>

-b + is given by

'1 =

The roots of the auxiliary equation are real and unequal, given by 2 Vb - w' and '2 = -b - Vb 2 - w'. The general solution to Equation (6) Cd.

y = Cl e (-b+v'b'-w'p

+ C2e (-b-v'b'-w'Jt.

Here again the motion is not oscillatory and both '1 and '2 are negative. Thus y approaches zero as time goes on. This motion is referred to as overdamping (see Figure l7.6b). Case 3: b , =

<

Cd.

-b ± iVw2

The roots to the auxiliary equation are complex and given by b 2• The general solution to Equation (6) is given by

-

y = e- bl(cl cosVw' - b 2 t

+ C2SinVw' - b 2 t).

This situation, called underdamping, represents damved oscillatory motion. It is analogous to simple harmonic motion of period T = 21T/V w' - b 2except that the amplitude is not constant but damped by the factor e -bl. Therefore, the motion tends to zero as t increases, so the vibrations tend to die out as time goes on. Notice that the period T = 21T/Vw2 - b 2 is larger than the period To = 21T/W in the friction-free system. Moreover, the larger the value of b = [;/2m in the exponential damping factor, the more quickly the vihrations tend to become unnoticeable. A curve illustrating underdamped motion is shown in Figure l7.6c. y

y

y

t

o

y = e-t sin(St + 1114)

y=(l +t)e-t (a) Critical damping

(b) Overdamping

(c) Underdamping

FIGURE 17.6 Three examples of damped vibra1xny motion for a spring system with friction, so B oF O.

17-20

Chapter 17: Second-Order Differential Equations An external force F(t) can also be added to lbe spring system modeled by Equation (2). The forcing function may represent an external disturbance on lbe system. For instance, if lbe equation models an automobile suspension system, lbe forcing function might represent periodic bumps or polboles in lbe road affecting lbe performance of lbe suspension system; or it might represent lbe effects of winds when modeling lbe vertical motion of a suspension bridge. Inclusion of a forcing function results in lbe second-order nonhomogeneous equation

d 2y m dt 2

dy

+ lJ dt + ky

= F(t).

(7)

We leave lbe study of such spring systems tu a more advanced course.

Electric Circuits The basic quantity in electricity is lbe charge q (analogous to lbe idea of mass). In an electric field we use lbe flow of charge, or current I = dql dt, as we might use velocity in a gravitational field. There are many similarities between motion in a gravitational field and lbe flow of electrons (lbe carriers of charge) in an electric field. Consider lbe electric circuit shown in Figure 17.7. It consists of four components: voltage source, resistor, inductor, and capacitor. Think of electrical flow as being like a fluid flow, where lbe voltage source is lbe pump and lbe resistor, inductor, and capacitor tend to block lbe flow. A battery or generator is an example of a source, producing a voltage that causes lbe current to flow through lbe circuit when lbe switch is closed. An electric light bulb or appliance would provide resistance. The inductance is due to a magnetic field that opposes any change in lbe current as it flows through a coil. The capacitance is normally created by two metal plates that alternate charges and lbus reverse lbe current flow. The following symbols specify lbe quantities relevant to lbe circuit: q: charge at a cross section ofa conductor measured in coulombs (abbreviated c);

I: current or rate of change of charge dq/dt (flow of electrons) at a cross section of a conductor measured in amperes (abbreviated A);

E: electric (potential) source measured in volt. (abbreviated V); V: difference in potential between two points along lbe conductor measured in volts (V). R, Resistor

Voltage E soun:e

L.lnd~

C, Capacitor

FIGURE 17.7 An elec1ric circuit. Ohm observed lbat lbe current I flowing through a resistor, caused by a potential difference across it, is (approximately) proportional to lbe potential difference (voltage drop). He named his constant of proportionality IIR and called R lbe resistance. So Ohm ~ law is

1 I=JiV,

17.3

Applications

17 -21

Similarly, it is known from physics that the voltage drops across an inductor and a capacitor are and

q C'

where L is the inductance and C is the capacitance (with q the charge on the capacitor). The German physicist Gustav R. Kirchhoff (1824-1887) formulated the law that the sum of the voltage drops in a closed circuit is equal to the supplied voltage E(t). Symbolically, this says that

RI

+L

dI dt

q

+C

= E(t).

Since I = dqldt, Kirchhoff's law becomes

d 2q dq I L dt 2 +R dt +Cq=E(t).

(8)

The second-order differential equation (8), which models an electric circuit, has exactly the same form as Equation (7) modeling vibratory motion. Both models can be solved using the methods developed in Section 17.2.

Summary The following chart summarizes our analogies for the physics of motion of an object in a spring system versus the flow of charged particles in an electrical circuit.

Linear Second-Order Constant-Coefficient Models Mecbllllical System

my" y: y/:

y":

m: 8: k: F(t):

+ 8y' + ky

Electrical System

= F(t)

displacement velocity acceleration mass damping constant spring constant forcing function

Lq" q: q/:

q": L: R: I/C: E(t):

+ Rq' + lq C

= E(t)

charge current change in current inductance

resistance where C is the capacitance voltage source

EXERCISES 17.3 1. A 16·lb weight is attached to the lower eod of a coil spring sus· peoded from the ceiling aod having a spring coostaot of I Ib/ft. The resistance in the spring-mass system is numerically equal to the instaotaoeous velocity. At t = 0 the weight is set in motion from a position 2 ft below its eqnilibrium position by giving it a downward velocity of2 ft/sec. Write ao initial value problem that models the giveo situation.

2. An g·lb weight stretches a spring 4 ft. The spring-mass system resides in a medium offering a resistance to the motion that is numerically equal to 1.5 times the instantaoeous velocity. If the weight is released at a position 2 ft above its equilibrium position with a downward velocity of3 ft/sec. write ao initial value problem modeling the giveo situation.

17-22

Chapter 17: Second-Order Differential Equations

3. A 2()'lb weight is bung on an IS-in. spring and stretches it 6 in. The weight is pulled dawn 5 in. and 5 lb are added to the weight !f the weight is now released with a downward ",Iocity of". in./sec, write an initial value problem modeling the vertical disp1acement 4. A I ()'Ib weight is suspended by a spring that is stretched 2 in. by the weight. Assume a resistance whose magnitude is 20/ Vg Ib times the instantaneous velocity v in feet per second. !fthe weight is pulled down 3 in. below its equilibrium position and released, formulate an initial value problem modeling the behavior of the spring-mass system. S. Ao (opea) electrical circuit coosists of an inductor, a resistor, and a capacitor. There is an initial cbarge of 2 coulombs on the capac-

itor. At the instant the circuit is closed. a current of 3 amperes is present and a voltage of E(t) = 20 cos 1 is applied. In this circuit the voltage drop across the resistor is 4 times the instantaneous cbange in the charge, the voltage drop acroas the capacitor is 10 times the charge, and the voltage drop across the inductor is 2 times the instantaneous change in the current Write an initial value problem to model the circuit 6. Ao inductor of 2 henrys is connected in series with a resistor of 12 obms, a capacitor of 1/ 16 farad, and a 300 volt battery. Initially, the charge on the capacitor is zero and the current is zero. Formulate an initial value problem modeling this electrical

circuit. Mechanical units in the British and metric systems may be helpful in doing the following problems. Unit

Brlli.h System

MKSSystem

Distance

Feet (ft) Slugs Seconds (sec) Pounds (lb) 32 ftjsec"

Meters (m) Kilugrams (kg) Seconds (sec)

Msss

Time Force g(earth)

Newtoos(N) 9.81 m/sec"

7. A 16-lb weight is attached to the lower end of a coil spring suspeoded from the ceiling and having a spriog coostant of Ilb/ ft. The resistance in the spring-maas system is numerically equal to the instantaneous velocity. At t = a the weight is set in motion from a position 2 ft below its equilibrium position by giving it a downward ",Iocity of 2 ft/sec. At the end of 'IT sec, determine whether the mass is above or below the equilibrium position and by what distance. 8. Ao 8-lb weight stretches a spring 4 ft. The spring-msss system resides in a medium offering a resistance to the motion equal to 1.5 times the instantaneous velocity. If the weight is released at a position 2 It shove its equilibrium position with a downward velocity of 3 ft/ sec, fmd its position relative to the equilibrium position 2 sec later. 9. A 2()'lb weight is hung on an 18-in. spring stretching it 6 in. The weight is pulled down 5 in. and 5 lb are added to the weight. (fthe weight is now released with a downward velocity of". in./sec, frod the position of mass relative to the equilibrium in terms of". and valid for any time 1 '" O.

10. A mass of 1 slug is attached to a spring whose constant is 25/4 Ib/ ft. Initially the mass is released I ft above the equilibrium position with a downward ",Iocity of 3 ftjsec, and the subsequent motion takes place in a medium that offers a dsmping force numerically equal to 3 times the instantaneous velocity. Ao extema1 force /(1) is driving the system, but assume that initially /(1) ~ O. Formulate and solve an initial value problem that models the given system. Ioterpret your results.

11. A IO-lb weight is suspended by a spring that is stretched 2 in. by the weight. Assume a resistance whose magnitude is 40/Vg Ib times the instantaneous velocity in feet per second. If the weight is pulled down 3 in. below its equilibrium position and released, fmd the time required to reach the equilibrium position for the fll'St time.

12- A ...,ight stretches a spring 6 in. It is set in motion at a point 2 in. below its equilibrium position with a downward ",Iocity of2 in./sec.

a. When docs the weight retum to its starting position? b. When docs it reach its highest point1

c. Show that the maximum velocity is 2V2g + 1 in./sec. 13. A weight of 10 Ib stretches a spring lOin. The weight is drawn down 2 in. below its equilibriwn position and given an initial velocity of 4 in./sec. Ao identical spring has a different weight attached to it. This second weight is drawn down from its equilibrium position a distance equal to the amplitude of the frrst motion and then given an initial velocity of 2 ft/sec. If the amplitude of the second motion is twice that of the frrst, what weight is attached to the second spriog?

14. A weight stretches one spring 3 in. and a second weight stretches another spring 9 in. If both weights are simultaneously pulled down I in. below their respective equilibrium positions and then released, fmd the frrst time after 1 = 0 whco their velocities are equal. IS. A weight of 16 Ib stretches a spriog 4 ft. The weight is pulled dawn 5 ft below the equilibrium position and then released. What initial velocity ". given to the weight would have the effeet of doubling the amplitude of the vibration?

16. A mass weighing 8 Ib stretches a spring 3 in. The spring-mass system resides in a medium with a dampins constant of 2 Ib-sec/ft. If

the mass is released from its equilibrium position with a velocity of 4 in./sec in the downward direction, flud the time required for the mass to return to its equilibrium position for the frrst time. 17. A weight suspended from a spring execute. damped vibrations with a period of2 sec. !fthe damping factor decreases by 90% in 10 sec, fmd the acceleration of the weight when it is 3 in. below its eqoilibrium position and is moving upward with a speed of2 ft/sec. 18. A 1()'lb weight stretches a spring 2 ft. If the weight is pulled down 6 in. below its equilibrium position and released, fmd the highest point reached by the weight. Assume the spring-maas system residea in a medium offering a resistance of IO/ Vg lb times the instantaneous ",Iocity in feet per second.

17.4 Euler Equations 19. An LRC circuit is set up wi1h an inductance of liS henry, a resist· ance of! ohm, aod a capacitance of516 farad. Assuming 1he initial charge is 2 coulombs and 1he initial current is 4 amperes, fmd 1he solutioo fimction descnbing 1he charge 00 1he capacitor at any time. What is 1he charge 00 1he capacitor after a loog period of time? 20. An (opoo) electrical circuit coosists of an inductor, a resistor, aod a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is preseot but no external voltage is being applied. In 1his circuit 1he voltage drops at 1hree points are numerically related as follows: across the capacitor. 10 times the charge; across the resistor, 4 times the instantaneous change in the charge; and across the inductor, 2 times 1he instantaneous chaoge in 1he current. Find 1he charge on the capacitor as a function of time. 21. A 16-lb weight stretches a spring 4 It This spring-mass system is in a medium wi1h a damping constant of 4.5Ib-sec/ft, and ao external force giveo by /(t) = 4 + e- 2t (in pouods) is being applied. What is 1he solution fimction describing 1he position of 1he mass at any time if 1he mass is released from 2 ft below 1he equilibrium position wi1h ao initial velocity of 4 ft/ sec downward? 22. A IO-kg mass is a_hed to a spring hsving a spring coostant of 140 N/m. The mass is started in mntioo from 1he equilibrium position wi1h ao initial velocity of 1 mf sec in 1he upward direction and wi1h an applied extemal force giveo by /(t) = 5 sin t (in newtons). The mass is in a viscous medium. with a coefficient of resistance eqoa1 to 90 N-sec/m. Formulate ao initial value problem iliat models 1he given system; solve 1he model aod interpret 1he results. 23. A 2-kg mass is a_hed to 1he lower end of a coil spring suspended ftom 1he ceiling. The mass comes to rest in its equilibrium

17-23

positioo 1hereby stretcbing 1he spring 1.96 m. The mass is in a viscous medium. that offers a resistance in newtons numerically equal to 4 times the instantaneous velocity measmed in meters per secood. The mass is 1hoo pulled down 2 m below its equilibrium positioo and released wi1h a downward velocity of3 mfsec. At this same instant an external force givoo by /(t) = 20 cos t (in newtons) is applied to 1he system. At 1he ood of 1r sec detennine if1he mass is above or below its equilibrium position aod by how much. 24. An 8-lb weight stretches a spring 4 ft. The spring-mass system r<>sides in a medium offering a resistance to the motion equal to 1.5 times the instantaneous velocity, and an extemal force given by /(t) = 6 + e -, (in pounds) is being applied. If 1he weight is released at a position 2 ft above its equilibrium positioo wi1h downward velocity of 3 ft/sec, fmd its position relative to 1he equilibrium after 2 sec hsve elapsed. 25. Suppose L = 10 henrys, R = 10 ohms, C = 1/500 farads, E = 100 volts, q(O) = 10 coulombs, and q'(O) = i(O) = O. Formulate aod solve an initial value problem iliat models 1he given LRC circuit. Interpret your results. 26. A series circuit consisting of an inductor, a resistor, and a capacitor is open. There is an initial charge of 2 coulombs on 1he capacitor, and 3 amperes of current is present in the circuit at the instant 1he circuit is closed. A voltage giveo by E(t) = 20 cos t is applied. In this circuit 1he voltage drops are numerically eqoa1 to 1he following: across the resistor to 4 times the instantaneous change in the charge, across the capacitor to 10 times the charge, and across the inductor to 2 times the instantaneous change in the current. Find 1he chsrge 00 1he capacitor as a fimctioo of time. Determine 1he chsrge 00 1he capacitor and 1he current at time t = 10.

uations In Section 17.1 we introduced the aecond-order linear homogeneous differential equation

P(x)y"(x)

+ Q(x)y'(x) + R(x)y(x)

= 0

and showed how to solve this equation when the coefficients P, Q, and R are constants. If the coefficients are not constant, we cannot generally solve this differential equation in tenns of elementary functions we have studied in calculus. In this section you will learn how to solve the equation when the coefficients have the special forms

P(x)

= ax 2,

Q(x)

= bx,

and

R(x)

= c,

where a, b, and c are constants. These special types of equations are called Euler equations, in honor of Leonhard Euler who studied them and showed how to solve them. Such equations arise in the study of mechanical vibrations.

The General Solution of Euler Equations Consider the Euler equation

ax"y"

+ bxy' + cy

=

0, x

> O.

(l)

17-24

Chapter 17: Second-Order Differential Equations To solve Equation (I), we rlISt make the change of variables

z = Inx

y(x) = Y(z).

and

We next use the chain rule to find the derivatives y'(x) and y"(x):

y' (x) =

fx Y(z)

=

~Y(z):

= Y' (z){

and

-~2 Y'(z) + ! y"(z)dz

y"(x) = i£y'(x) = i£y'(z)! x = dx dx

x

x

dx

=

-~2 Y'(z) + ~2 Y"(z). x x

Substituting these two derivatives into the left-hand side of Equation (I), we rmd

ax"y"

+ bxy' +

:2

cy = ax 2 ( = aY"(z)

Y'(z)

+ (b

:2

+

+ bX({ Y'(Z)) +

Y"(Z))

- a)Y'(z)

cY(z)

+ cY(z).

Therefore, the substitutions give us the second-order linear clifferential equation with constant coefficients

aY"(z)

+ (b

- a)Y'(z)

+ cY(z)

=

o.

(2)

We can solve Equation (2) using the method of Section 17.1. That is, we find the roots to the associated auxiliary equation

ar2

+ (b

- a)r

+C

= 0

(3)

to find the general solution for Y(z). After rmding Y(z), we can detennine y(x) from the substitutionz = Inx.

EXAMPLE 1

Find the general solution of the equation x 2y"

+

2xy' - 2y =

o.

Solution This is an Euler equation with a = I, b = 2, and C = -2. The auxiliary equation (3) for Y(z) is r2

+ (2

+ 2)

- I)r - 2 = (r - I)(r

=

0,

with roots r = -2 and r = 1. The solution for Y(z) is given by

Y(z) = Ct e -2z

+ C2ez.

Substituting z = In x gives the general solution for y(x):

y(x) = cte-2lnx

EXAMPLE 2

+ c2e lnx

= CtX- 2

Solve the Euler equation x"y" - 5xy'

+

+ C2X

•

9y = O.

Solution Since a = I, b = -5, and C = 9, the auxiliary equation (3) for Y(z) is r2

+ (-5

- I)r

+9

= (r -

3)2 = O.

The auxiliary equation has the double root r = 3 giving

Y(z) = cte 3z

+

c2ze3~

Substituting z = In x into this expression gives the general solution

y(x) = Cte3lnx

+ c2lnxe3lnx

= CtX3

+ C2x3lnx

•

17.4 Euler Equations

EXAMPLE 3 Find the particular solution to x"y" - 3xy' initial conditions y(1) = 0 andy'(I) = 1. Solution Here a = I, b = -3, and gives

C

+ 68y

17-25

= 0 that satisfies the

= 68 substitoted into the auxiliary equation (3)

r 2 -4r+68=0. The roots are r = 2

+

8i and r = 2 - 8i giving the solution

Y(z) = e2z(cl cos 8z

+ C2 sin 8z).

Substitoting z = In x into this expression gives

y(x) =

e

21nx

( Cl

cos (8 In x)

+ C2 sin (8 In x»).

From the initial condition y(l) = 0, we see that Cl = 0 and

y(x) = C2x2 sin (8 In x).

y

To fit the second initial condition, we need the derivative

10

y'(x) = c2(8x cos (8 In x) 5

Since y'(I) = I, we immediately obtain C2 = 1/8. Therefore, the particular solution satisfying both initial conditions is

o

2

10

y(x) =

2

-5

+ 2xsin(8lnx»).

t

x 2 sin (8 In x).

y= "ssin(8lnx)

Since -I :5 sin (8 In x) :5 I, the solution satisfies -10

x2

-8 :5 y(x) FIGURE 17.8 Example 3.

:5

x2

8'

Graph of the solution to

A graph of the solution is shown in Figure 17.8.

EXERCISES 17.4 In Exercises 1-24, rmd the general solution to the given Euler equation. Assume x > 0 throughout

1. x"y" + 2xy' - 2y

=

0

=

0

3.x"y"-6y=0 5. xV - Sxy'

+ 8y

=

0

4.x"y"+xy'-y=0

7. 3xV + 4xy' = 0 9. x"y" - xy' + Y = 0 11. x"y" - xy' + Sy = 0

+ 3xy' + lOy = 0 15. 4xV + 8xy' + Sy = 0 17. x"y" + 3xy' + Y = 0 19. xV + xy' = 0 13. x"y"

2. x"y" + xy' - 4y

+ 7xy' + 2y = 0 8. xV + 6xy' + 4y = 0 10. x"y" - xy' + 2y = 0 12. x"y" + 7xy' + l3y = 0 14. x"y" - Sxy' + lOy = 0 16. 4x"y" - 4xy' + Sy = 0 18. x"y" - 3xy' + 9y = 0 20. 4xV + Y = 0 6. 2xV

21. 9x"y' + ISxy' 22. 16xV - 8xy'

+Y = 0 + 9y = 0 23. 16xV + S6xy' + 2Sy = 0 24. 4x"y' - 16xy' + 2Sy = 0 In Exercises 25-30, solve the given initial value problem.

25. xV 26. 6x"y' 27. xV

+ 3xy' - 3y = 0, y(l) = I, y'(I) = -I + 7xy' - 2y = 0, y(1) = 0, y'(I) = I - xy' + Y = 0, y(l) = I, y'(I) = I

28. xV + 7xy' + 9y = 0, y(l) = I, y'(I) = 0 29. xV - xy' + 2y = 0, y(1) = -I, y'(I) = I 30. xV + 3xy' + Sy = 0, y(l) = I, y'(I) = 0

•

17 -26

Chapter 17: Second-Order Differential Equations

Power-Series SoLutions In this section we exteod our stody of second-order linear homogeoeous equations with variable coefficieots. With the Euler equations in Section 17.4, the power of the variable x in the nonconstant coefficieot had to match the order of the derivative with which it was paired: x 2 with y', Xl with y', and x O(=I) with y. Here we drop that requiremeot so we ean solve more geoeral equations.

Method of Solution The poweMeries method for solving a second-order homogeneous differential equation consists of finding the coefficieots of a power series 00

y(x) = ~ c.x' = Co

.-0

+ ClX + C2x2 + ...

(1)

which solves the equation. To apply the method we substitote the series and its derivatives into the differential equation to determine the coefficieots Co, Ct, C2, .... The technique for finding the coefficieots is similar to that used in the method of uodetermined coefficieots preseoted in Section 17.2. In our first example we demonstrate the method in the setting of a simple equation whose general solution we already know. This is to help you become more comfortable with solutions expressed in series form.

EXAMPLE 1

Solve the equation y'

+Y

= 0

by the power-series method.

Solution We assume the series solution takes the form of

and calculate the derivatives 00

y'

=

~

00

ncll x n -

1

y"

and

n(n - 1)c.x·- 2.

= ~

11=1

11=2

Substitotion of these forms into the second-order equation gives us 00

00

11 - 2

11 - 0

~ n(n - l)c.x·- 2 + ~ c.x· = O. Next, we equate the coefficients of each power of x to zero as summarized in the followiog table.

Powerofx

Coefficient Equation

1

xo

2(I)C2

+ Co

=

0

or

C2 = -ZCo

xl

3(2)C3

+ Cl

=

0

or

C3 = -3-2 Cl

x2

4(3)C4

+ C2

=

0

or

C4 = -4'3 C2

x3

5(4)cs

+ C3

=

0

or

Cs = -5'4 C3

x4

6(5)C6

+ C4

=

0

or

C6 = -6'5 C4

+ C.-2

=

0

or

en =

X,,-2

n(n - l)c.

1 1 1 1

1 n(n - 1) C.-2

17.5 Power-Series Solutions

17-27

From the table we notice that the coefficients with even indices (n = 2k, k = 1,2,3, ... ) are related to each other and the coefficients with odd indices (n = 2k + 1) are also interrelated. We treat each group in turn. Even indices: Here n = 2k, so the power is x 2k--2. From tlie last line of the table, we have 2k(2k - l)c2k

+ C2k-2

=

0

or

C2k =

I 2k(2k _ I) C2k-2·

From this recursive relation we find

c2k = [ 2k(2;- 1)][ (2k - 2;(2k - 3)]-" [-

=

4(~)][-!]co

(-1)1 (2k)! co·

Odd indices: Here n = 2k line of the table yields (2k

+ 1, so the power is X2k-I. Substituting this into the last

+ 1)(2k)c2k+ I + C2k-1

=

0

or

c2k+ I =

(2k

I

+ I )(2k) C2k-I·

Thus,

C2k+1 = [ (2k+ \)(2k)] [ (2k -

=

1)~2k -

2J . ht4)] ht2)]cl

(_1)1 (2k + I)! CI·

Writing the power series by grouping its even and odd powers together and substituting for the coefficients yields

00

=

~

<-0

00

C2kx2k

+ ~ C2k+l X2H1 1- 0

00 (-1)1 '" (2k)! X 2k -_ CotEl,

00

(_1)1

+ CI tEl, '" (2k + I)! x 2.<+1.

From Table 9.1 in Section 9.10, we see that the trrst series on the right-hand side of the last equation represents the cosine function and the second series represents the sine. Thus, the general solution to y" + Y = 0 is y = Co cos x

+ Cl sinx.

•

17-28 Chapter 17: Second-Order Differential Equations EXAMPLE 2

+ ~' + Y

Find the general solution to y"

= O.

Solution We assume the series solution form 00

y = ~CnXll n=O

and calculate the derivatives 00

y'

L ncn x

=

.-1

ll

00

t

-

y"

and

=

Ln(n - l)c.x·- 2•

.-2

Substitution of these forms into the second-order equation yields 00

00

~ n(n - 1)cnx"- 2

+

n=2

L

ncnX

n+

11=1

00

L

n=

CnX

o.

11=0

We equate the coefficients of each power of x to zero as summarized in the following table. Powerofx

Coefficient Equation

+ Co 3(2)C3 + Cl + CI 4(3)C4 + 2C2 + C2 5(4)cs + 3C3 + C3 6(5)C6 + 4C4 + C4

=

+ 2)(n + l)c.+2 + (n + l)c.

2(I)c2

x'

(n

= = = = =

or or or

C3

or

Cs

=

0 0 0 0 0

or

C6

=

0

or

Cn+2 =

= = =

C2

C4

-!co -let -lc2

-lC3 -~C4 -

I

n + 2 en

From the table notice that the coefficients with even indices are interrelated and the coefficients with odd indices are also interrelated. Even indices: Here n = 2k - 2, so the power is x2k-2. From the last line in the table,

we have C2k

= -

I 2k C2k-2·

From this recurrence relation we obtain

=

(2)(4)(6) ... (2k) Co·

Odd indices: Here n we have

=

2k - 1, so the power is X2k-I. From the last line in the table,

C2k+1 = -

1 2k + 1 C2k-I·

From this recurrence relation we obtain C2k+l =

=

(-2k

~ 1)(-2k ~

(-I)' (3)(5)'" (2k

+ 1) CI.

J . (-t) (

-t)Cl

17-29

17.5 Power-Series Solutions

Writing the power series by grouping its even and odd powers and substituting for the coefficients yields 00

Y

=

00

L C2kx2k + L C21-+ 1X2k+ 1 1=0

1=0

(_I)k

00

~

- 0&'0 (2)(4) .. ·(2k) -

EXAMPLE 3

C

X2k

+C

(_I)k

00

~

2k+ l

1&'0 (3)(5)"'(2k + I) X

•

.

Find the general solution to

(I -

X

2

Ixl <

)y" - &y' - 4y = 0,

1.

Solution Notice that the leading coefficient is zero when x = ± 1. Thus, we assume the solution interval I: -I < x < I. Substitution of the series form

and its derivatives gives us

(I - X2)

00

L

n(n - I )C.X·- 2 - 6

11=2 00

L

n=2

L

00

nc.x' - 4

n=1

00

n(n - I )C.X·- 2 -

00

L

n=2

L

c.x· = 0,

n=O

00

n(n - I )C.X' - 6

L 00

nc.x' - 4

n=1

L

c.x· = O.

n=O

Next, we equate the coefficients of each power of x to zero as summarized in the following table.

Powerofx

Coefficient Equation

xO Xl

x2 x3 x'

(n

_4

- 4co=0

or

Cz - leO

- 6(I)Cl - 4Cl = 0 3(2)C3 4(3)C4 - 2(I)c2 - 6(2)C2 - 4C2 = 0 5(4)c, - 3(2)C3 - 6(3)C3 - 4C3 = 0

or

C3 = iCl

or

C4 -

2(I)c2

+ 2)(n + I)C.+2 - [n(n - I) + 6n + 4]c. (n + 2)(n + l)c.+2 - (n + 4)(n + I)c.

or

_6

4C2

_7

Cs -

SC3

= 0 = 0

or

n+4

Cn+2

= n + 2 en

Again we notice that the coefficients with even indices are interrelated and those with odd indices are interrelated Even indices: Here n = 2k - 2, so the power is x2k. From the right-hand column and last line of the table, we get

=

(2\~ 2)(2/~ 2)(;::= ~)- .. ~(~)CO

= (k

+

I)co.

17 -30

Chapter 17: Second-Order Differential Equations

Odd indices: Here n

=

2k - I, so the power is x 2k+l. The right-hand column and last

line of the table gives us

=

=

(2k + 3)(~)(~) 2k+1 2k-1 2,\;-3

... l(i) 53

Cl

2k + 3 --3- C1 •

The general solution is 00

Y

=

L cnx" .-0

=

L C2kx2k + L C2k+ 1x 2k+ 1

00

00

k~O

k~O

00

= CO~

(k

+

00

~O

EXAMPLE 4

2k+3

•

+ Cl~ -3-X2k+1.

l)x2k

~O

Find the general solution to y" - 2>y'

+Y

=

o.

Solution Assuming that

substitution into the differential equation gives us 00

00

~ n(n - l)c.x·- 2

-

n=2

00

2 ~ nc.x· n=1

+

~ c.x· =

o.

11=0

We next determine the coefficients, listiog them in the following table.

Powerofx

Coefficient Equation

2(I)c2 3(2)C3 -

2Cl

Co =

0

or

C2 = -Zco

+

Cl =

0

or

c3

3 c4=4_3 c2

4(3)C4 - 4c2

+ C2

=

0

or

5(4)c, - 6C3

+ C3

=

0

or or

x·

(n

+ 2)(n + l)c.+2 -

I

+

(2n - I)c. = 0

I

=3_2

c1

5 c'=5_4 c3 7 c6=6_5 c4 2n - I

or

C.+2 = (n

+ 2)(n + I) c.

17.5 Power-Series Solutions

17-31

From the recursive relation

2n - I

+ 2)(n +

e.+2 = (n

I) e.,

we write out the first few terms of each series for the general solution: Y =

eO(1 - lx 2

2

-

lx' 41

n61 x• -

... )

• EXERCISES 17.5 In Exercises 1-18, use power series to f'md 1he general solution of1he

differential equation.

1. y" 2. y"

+ 2y' = 0 + 2y' + Y =

0

+ 2y = 0 2xy' + 2y = 0

11. (x 2

-

-

+ 2xy' x"y

l)y" - 6y

- 2y = 0

0

=

0

=

15. y" -2xy' +3y=O

6.y"-.>;Y'+y=O

7. (I + x)y" - Y = 0 8. (1 - x 2)y" - 4.>;Y' + 6y

l)y"

+ y'

+ 2)y' + 2y = 0 2 13. (x - l)y" + 4'>;y' + 2y = 0 14. y" - 2xy' + 4y = 0

4. y" - 3y'

xV -

-

10. y"

12. .>;y" - (x

3.y"+4y=O

5.

9. (x 2

16. (I - x 2)y" - .>;y'

=0

=

0

+ 6y

=

17. y" - xy' + 3y 18.

xV -

4.>;Y'

+ 4y 0

= 0

ApPENDICES Real Numbers and the Real Line

A.1

This section reviews real numbers, inequalities, intervals, and absolute values.

Real Numbers Much of calculus is based on properties of the real number system. Real numbers are numbers that can be expressed as decimals, such as

- 4"3 =

"31 =

v2 =

-0.75000 . .. 0.33333 . . . 1.4142 .. .

The dots ... in each case indicate that the sequence of decimal digits goes on forever. Every conceivable decimal expansion represents a real number, although some numbers have two representations. For instance, the infinite decimals .999 . .. and 1.000 ... represent the same real number 1. A similar statement holds for any number with an infinite tail of9's. The real numbers can be represented geometrically as points on a number line called the real line.

Rules for inequalities I

If a, b, and e are real numbers, then: l.a
2. a < b

~

a - e

< b- e 3. a < b and e > 0 ~ ae < be 4. a < b and e < 0 ~ be < ae Special case: a < b ~ -b < -a 5. a> 0

~

7i> 0

1

6. If a and b are both positive or both

. then a < b negatIve,

~

b1 < 7i1

-2

I

o

1

3

2

I I

I

37T

4

)

The symbollR denotes either the real number system or, equivalently, the real line. The properties of the real number system fall into three categories: algebraic properties, order properties, and completeness. The algebraic properties say that the real numbers can be added, subtracted, multiplied, and divided (except by 0) to produce more real numbers under the usual rules of arithmetic. You can never divide by O. The order properties of real numbers are given in Appendix 6. The useful rwes at the left can be derived from them, where the symbol ~ means "implies." Notice the rules for multiplying an inequality by a number. Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign. For example, 2 < 5 but -2 > -5 and 1/2 > 1/5. The completeness property of the real number system is deeper and harder to defme precisely. However, the property is essential to the idea ofa limit (Chapter 2). Roughly speaking, it says that there are enough real numbers to "complete" the real number line, in the sense that there are no "holes" or "gaps" in it. Many theorems of calcwus would fail if the real number system were not complete. The topic is best saved for a more advanced course, but Appendix 6 hints about what is involved and how the real numbers are constructed.

AP-l

AP-2

Appendices

We distinguish three special subsets of real numbers.

1. 2.

The natural numbers, namely 1,2,3,4, ...

3.

The rational numbers, namely tbe numbers tbat can be expressed in tbe form of a fraction min, where m and n are integers and n oF o. Examples are

The integers, namely 0, ± 1, ±2, ±3, ...

1 3'

4 9

-4 9

4 -9'

200 13'

57 = 57 1

and

The rational numbers are precisely tbe real numbers witb decimal expansions tbat are eitber

(a) terminating (ending in an infinite string of zeros), for example,

43

=

0.75000 ... = 0.75

or

(b) eventually repeating (ending witb a block of digits tbat repeats over and over), for example 23

IT

=

2.090909... = 2.09

The bar indicates the block of repeating digits.

A terminating decimal expansion is a special type of repeating decimal, since tbe ending zeros repeat. The set of rational numbers has all tbe algebraic and order properties of tbe real numbers but lacks tbe completeness property. For example, tbere is no rational number whose should be. square is 2; tbere is a ''hole'' in tbe rational line where Real numbers tbat are not rational are called irrational numbers. They are characterized by having nonterminating and nomepeating decimal expansions. Examples are 'fr, and 10g!O 3. Since every decimal expansion represents a real number, it should be clear tbat tbere are infmitely many irrational numbers. Botb rational and irrational numbers are found arbitrarily close to any point on tbe realline. Set notation is very useful for specifying a particular subset of real numbers. A set is a collection of objects, and tbese objects are tbe elements oftbe set. If S is a set, tbe notation a E S means tbat a is an element of S, and a I'! S means tbat a is not an element of S. If S and T are sets, tben S U T is tbeir union and consists of all elements belonging eitber to S or T (or to botb S and T). The intersection S n T consists of all elements belonging to botb S and T. The empty set 0 is tbe set tbat contains no elements. For example, tbe intersection of tbe rational numbers and tbe irrational numbers is tbe empty set. Some sets can be described by listing tbeir elements in braces. For instance, tbe set A consisting of tbe natural numbers (or positive integers) less than 6 can be expressed as

v2

v2, ¥S,

A = {1,2,3,4,5}. The entire set of integers is written as {O, ± 1, ±2, ±3, ... } . Anotber way to describe a set is to enclose in braces a rule tbat generates all tbe elements of tbe set. For instance, tbe set

A = {xix is an integer and 0 is tbe set of positive integers less tban 6.

<x<

6}

Appendix 1 Real Numbers and the Real Line

AP-3

Intervals A subset of the real line is called an interval if it contains at least two numbers and contains all the real numbers lying between any two of its elements. For example, the set of all real numbers x such that x > 6 is an interval, as is the set of all x such that - 2 :5 X :5 5. The set of all nonzero real numbers is not an interval; since 0 is absent, the set fails to contain every real number between -I and I (for example). Geometrically, intervals correspond to rays and line segments on the real line, along with the real line itself. Intervals of numbers corresponding to line segments are finite intervals; intervals corresponding to rays and the real line are infinite intervals. A finite interval is said to be closed if it contains both of its endpoints, half-open if it contains one endpoint but not the other, and open if it contains neither endpoint. The endpoints are also called boundary points; they make up the interval's boundary. The remaining points of the interval are interior points and together comprise the interval's interior. Infinite intervals are closed if they contain a finite endpoint, and open otherwise. The entire real line n;l is an infinite interval that is both open and closed Table A.I summarizes the various types of intervals.

TABLE A.l Types of intervals Notation

Set description

Type

(a, b)

{xla < x < b}

Open

[a, b]

{xla

:5

x

Closed

[a, b)

{xla

:5

x < b}

(a, b]

{xla < x

(a, 00)

{xix> a}

Open

: a

[a, 00)

{xix;;,: a}

Closed

a

(-00, b)

{xlx
Open

(-oo,b]

{xlx:5b}

Closed

(-00,00)

n;l (set of all real numbers)

Both open and closed

:5

:5

b}

Pie!ure

Half-open

b}

Half-open

--

a

b

a

b

a

b

a

b

• • • •

-

b

b

-

Solving Inequalities The process of rmding the interval or intervals of numbers that satisfy an inequality in x is called solving the inequality.

EXAMPLE 1 (a) 2x - I

Solve the following inequalities and show their solution sets on the real line.

<x +3

x

(b) - 3

< 2x + 1

(c) _6_;;,: 5 x - I

AP-4

Appendices

Solution

! ! ____~~----~--~)X

I

0

4

2x-I<x+3

(a)

(a)

<x+ x < 4

2x : 3

-7

!

] X

0

Add I to both sides.

Subtract x from both sides.

The solution set is the open interval (-00,4) (Figure A.la).

(b) --~--~~____~--~)X

-~ 3

(b)

11

0

4

5

+

I

+ 0< 7x + 3 < 7x

3

Multiply both sides by 3.

3

Add x to both sides.

-x

(c)

FIGURE A.1 Solution sets for the inequalities in Example I.

-

<

2x

<

6x

Subtract 3 from both sides.

_17 < x

Divide by 7.

The solution set is the open interval ( - 3/7, 00) (Figure Al b). 2: 5 can hold only if x > I, because otherwise 6/(x - I) is undermed or negative. Therefore, (x - I) is positive and the inequality will be preserved if we multiply both sides by (x - I), and we have

(c) The inequality 6/(x - I)

__6_> 5

x-I-

62:5x-5 2: 5x

Multiply both sides by (x - I).

11

Add 5 to both sides.

11> x T.

OrX:s~l.

The solution set is the half-open interval (1,11/5] (Figure Alc).

Absolute Value

-

The absolute value of a number x, denoted by Ix I, is dermed by the fonnula Ixl = {

EXAMPLE 2

131 = 3,

101 = 0,

x, -x,

x2:0 x < O.

I-51 = -(-5) = 5,

Hall = lal

-

Geometrically, the absolute value of x is the distance from x to 0 on the real number line. Since distances are always positive or 0, we see that Ixl 2: 0 for every real number x, andlxl = oif and ouly if x = O.Also, Ix - yl = the distance between x andy ~ 1-51~5 I

-5

) ,

131 ---->

I

I

o

3

on the real line (Figure A2). Since the symbol Va always denotes the nonnegative square root of a, an alternate definition of Ix Iis

+---14 - II ~ II - 41 ~ 3 ~

I

I

4

FIGURE A.2 Absolute values give distances between points on the number line.

Ixl=W. It is important to remember that W = Ia I. Do not write W = a unless you already know that a 2: O. The absolute value has the following properties. (You are asked to prove these properties in the exercises.)

Appendix 1 Real Numbers and the Real Line

AP-5

Absolute Value Properties 1.1-al = lal

A number and its additive inverse or negative have the same absolute value.

2. labl = lallbl

The absolute value of a product is the product of the absolute values.

lal bal =jbj l 4. la + bl ,;; lal

The absolute value of a quotient is the quotient of the absolute values.

3.

~ a --~)¥I,--- a ~ ~~~.~----~I------~~

-a

x 0 ~ Ixl --->l

a

FIGUREA.3 Ixl < a means x lies between -a and a.

+ Ibl

Note that I-al oF -Ial. For example, 1-31 = 3, wbereas -131 = -3. Ifa and b differ in sign, then la + bl is less than lal + Ibl. In all other cases, la + bl equals Ia I + Ib I. Absolute value bars in expressions like 1-3 + 51 work like parentheses: We do the arithmetic inside before taking the absolute value. EXAMPLE 3 1-3 13

AbsoLute vaLues and intervals Ixl = a

~

6. Ixl < a

~

7.lxl>a

=

8. Ix I ,;; a 9. Ix I ;;" a

=

~

x = ±a -a < x < a x>aorx<-a -a';; x ,;; a x;;" a or x ,;; -a

+ 51 = + 51 =

1-3 - 51

If a is any positive number, then

s.

The triangle inequality. The absolute value of the sum of two numbers is less than or equal to the sum of their absolute values.

121

= 2 < 1-31 + 151 = 8

181 = 131

+ 151

•

= HI = 8 = HI + I-51

The inequality Ix I < a says that the distance from x to 0 is less than the positive number a. This means that x must lie between -a and a, as we can see from Figure A.3. The statements in the table are all consequences of the def"mition of absolute value and are often belpful when solving equations or inequalities involving absolute values. The symbol ~ is often used by mathematicians to denote the "if and only if" logical relationship. It also means ''implies and is implied by." EXAMPLE 4

Solution

Solve the equation 12x - 31 = 7.

By Property 5, 2x - 3 = ± 7, so there are two possibilities: 2x - 3 = 7 2x = 10

2x = -4

x = 5

x =-2

The solutions of 12x - 31 EXAMPLE 5

Solution

2x - 3 = -7

Equivalent equations without absolute values

Solve as usual.

•

= 7 are x = 5 andx = -2.

Solve the inequality 15 - fl < 1.

We have <1=-1<5-

=-6 < =3>

1

~3

2

x <1

-x2 <-4

f> 2 1

<x
Property 6

Subtract 5.

Multiply by -

~.

Take reciprocals.

AP-6

Appendices

Notice how the various rules for inequalities were used here. Multiplying by a negative number reverses the inequality. So does taking reciprocals in an inequality in which both sides are positive. The original inequa1ityholds if and ouly if(I/3) < x < (1/2). The solution set is the open interval (1/3, 1/2). •

Exercises A.l 1. Express 1/9 as a repeating decimal, using a bar to indicate the repeating digits. What are the decimal representations of2/9? 3/9? 8/9? 9/9?

Solve the inequalities in Exercises 18-21. Express the solution sets as intervals or unions of intervals and show them on the real line. Use the result W = Ia Ias appropriate.

2. If2 < x < 6, which of the following statements about x are necessarily true, and which are not necessarily true?

18. x 2 < 2

19. 4 < x 2 < 9

20. (x - I)' < 4

21. x 2

LO<x<4

~O<x-2<4

e. l
d·

e. 1<¥<3

f. Ix - 41 < 2

g. -6 < -x < 2

h. -6 < -x < -2

I

I

I

6 <:'<2

S·2x-

>4

24. A proof of the triangle inequolity Give the reason justifYing each of the nwnhered steps in the following proof of the triangle inequality. la + bl 2 = (a + b)'

2 "'7x+

7

4

6. S(x - 2)

6

I

< 3(x

2

'" a + 21allbl + b

8·121+51=4

= lal + 21allbl + Ibl

- 6)

13.I~-11

11·11-11"'3

'"

16. II - xl> I

A.2

I

14.1 3 -}I < t 17.lr~

(2) 2

la + bl '" lal + Ibl 9. 18 - 391 =

(3)

= (Ial + Ibl)'

~

Solve the inequalities in Exercises 10--17, expressing the solution sets as intervals or unions of intervals. Also, show each solution set on the real line. 10. Ixl < 2

2

2

Solve the equations in Exercises 7-9. 7. Iyl = 3

(1)

=a 2 +2ab+b 2

4.5x-3"'7-3x

I

x< 0

23. Solve the equation Ix - II = I-x.

In Exercises 3--6, solve the inequalities and show the solution sets on the rea1line. 3. -2x

-

22. Do not fall into the trap of thinking I-al = a. For what real numbers a is this equation true? For what real numbers is it false?

12. 13y - 71 < 4 15. 1291'" 4

(4)

25. Prove that Iab I = Ia II b I for any nwnbers a and b. 26. Ifl xl '" 3 and x > -1/2, what can you say abootx? 27. Graph the inequalitylxl + Iyl '" 1. 28. For any number a, pruve that I-a I = Ia I. 29. Let a he any positive nwnber. Prove that Ix I > a if and only if x>aorx<-a. 30. a. Ifb is any nonzero real nwnber, prove that I I/bl = I/Ibl. ~ Prove that

II '" I

bal

l

fbi

= lal

for any numbers a and b '" O.

Mathematical Induction Many formulas, like

1

+ 2 + ... + n

n(n + 1)

= ='-c2~='-'

can be shown to hold for every positive integer n by applying an axiom called the mathematical induction principle. A proof that uses this axiom is called a proofby mathematical induction or a proofby induction. The steps in proving a fonnula by induction are the following:

1.

Check that the fonnula holds for n = I.

2.

Prove that if the fonnula holds for any positive integer n = k, then it also holds for the next integer, n = k + 1.

AP-7

Appendix 2 Mathematical Induction

The induction axiom says that once these steps are completed, the formula holds for all positive integers n. By Step I it holds for n = I. By Step 2 it holds for n = 2, and therefore by Step 2 also for n = 3, and by Step 2 again for n = 4, and so on. If the flfst domino falls, and the kth domino always knocks over the (k + I)st when it falls, all the dominoes fall. From another point of view, suppose we have a sequence of statements S" S2, ... , So, ... , one for each positive integer. Suppose we can show that assuming any one of the statements to be true implies that the next statement in line is true. Suppose that we can also show that SI is true. Then we may conclude that the statements are true from SI on.

EXAMPLE 1

Use mathematical induction to prove that for every positive integer n,

I Solution 1.

n(n

+

1)

+ 2 + ... + n = -'-~2~---

We accomplish the proof by carrying out the two steps above.

The formula holds for n = I because I =+_1___) I = --,I(__ 2

2.

If the formula holds for n = k, does it also hold for n = k + I? The answer is yes, as we now show. If I

+ 2 + ... + k = _k(,--k~+----,-I) 2

'

then I

+ 2 + ... + k + (k +

I)

= k(k + 1) + (k + 2

(k

+

I)(k 2

+ 2)

I)

= k' + k + 2k + 2

(k

2

+ I)«k + I) +

I)

2

The last expression in this string of equalities is the expression n(n n = (k + I).

+

1)/2 for

The mathematical induction principle now guarantees the original formula for all positive integers n. _ In Example 4 of Section 5.2 we gave another proof for the formula giving the sum of the first n integers. However, proof by mathematical induction is more general. It can be used to fmd the sums of the squares and cubes of the first n integers (Exercises 9 and 10). Here is another example.

EXAMPLE 2

Show by mathematical induction that for all positive integers n, I I I -+-+ ... +-= 21 22 2'

Solution 1.

I 1--

2'·

We accomplish the proofby carrying out the two steps of mathematical induction.

The formula holds for n = I because I

-=

21

I 1-2 1.

AP-8

Appendices

2.

If

then

~+~+ ... +~+_I_= I_~+_I_= I_~+_I_ 2' 22 2k 2k+l 2k 2k+l 2k .2 2k+l =

2

+

I - 2k+l

Thus, the original formula holds for n = (k

+

I

I

2k+l = I - 2k+l·

I) whenever it holds for n = k.

With these steps verified, the mathematical induction principle now guarantees the • formula for every positive integer n.

Other Starting Integers Instead of starting at n = I some induction arguments start at another integer. The steps for such an argument are as follows.

1.

2.

Check that the formula holds for n = n, (the first appropriate integer). Prove that if the formula holds for any integer n = k n = (k

+ I).

20

n"

then it also holds for

Once these steps are completed, the mathematical induction principle guarantees the formula for all n 20 n,.

EXAMPLE 3 Solution

Show that n!

>

3" if n is large enough.

How large is large enough? We experiment: n n! 3"

I I 3

2 2 9

3

4

6

24 81

27

5 120 243

7 5040 2187

6

720 729

> 3" for n 20 7. To be sure, we apply mathematical induction. We take 7 in Step I and complete Step 2. Suppose k! > 3k for some k 20 7. Then

It looks as if n!

n, =

(k + I)! = (k + I)(k!)

Thus, for k

20

>

(k + 1)3k

>

7.3 k

>

3k+l.

7,

k! > 3k implies

(k + I)! > 3k+l.

The mathematical induction principle now guarantees n!

20

3" for all n

Proof of the Derivative Sum Rule for Sums of Finitely Many Functions We prove the statement

20

7.

•

AP-9

Appendix 2 Mathematical Induction

by mathematical induction. The statement is true for n = 2, as was proved in Section 3.3. This is Step 1 of the induction proof. Step 2 is to show that if the statement is true for any positive integer n = k, where k ;", no = 2, then it is also true for n = k + 1. So suppose that d

dx (U1

+ U2 + ... + Uk)

=

dU1 dU2 ax + ax +

(1)

Then d dx

(U1

+ U2 + ... + uk + UH 1) ~

Call this function v.

Call the function delmed by this sum u.

+ Uk) + dU1

dU2

dUk

=dx -++ dx

d

Sum Rule for ax (u + v)

dUk+1

dx

dUk+1

+dx- + dx

Eq. (1)

With these steps verified, the mathematical induction principle now guarantees the Sum Rille for every integer n ;", 2.

Exercises A.2 1. Assuming that the triangle inequality Ia for aoy two numbers a and b, show that IX1

+ X2 + ... + x,1

:5 Ixd

+ b I :5

Ia I + Ib I holds

6. Showthatn! 7. Show that 2'

+ IX21 + ... + lx, I

8. Show that 2" '" 1/8 for n '" -3. 9. Sum. of .quare.

for aoy n numbers.

positive integers is

2. Show that if r #' I, then

for every positive integer n.

d

dv

3. Use the Product Rnie, dx (uv) = u dx = I to show that

:! (x')

du

+ v dx' and the fact that

= nx,-l for every positive inte-

gern.

4. Suppose that a function fIx) has the property that f(xlxz) = f(xl) + f(x2) for aoytwo positive nurohers Xl andxz. Show that

Show that the sum of the squares ofilie fmt n

+ t}n +

1 - r"+l 1 +r+r2+"'+r"=~--"-­ I r

:! (x)

> n'ifnislargeenough. > n Z ifn is large enuugh.

+

I)

3

10. Sum. of cub.. Show that ilie sum of the cubes of the I!rst n positive integers is (n(n + 1)/2),. 11. Rules for Imite .um. Show that ilie following Imite sum rules hold for every positive integer n. (See Section 5.2.)

a.

" "" L(a,+ b,) = La,+ Lb, k=i

k=i

k=l

f(xlxz" ·x,) = f(xl) + f(xz) + ... + fIx,)

for the product ofany n positive numbers Xl. X2 •••• , xn.

(aoy number c)

S. Showthat

2 2 2 I -+-+···+-=1-1 2

3

3

for all positive integers n.

3"

(if

a, has ilie constant value c)

3"

12. Show that Ix"1 = Ix I" for every positive integer n and every real numberX'.

AP-l0

Appendices

Lines, Circles, and Parabolas

A.3

TIris section reviews coordinates, lines, distance, circles, and parabolas in the plane. The notion of increment is also discussed. y

b

-------o P(a,b) I I I I I I I I I

Positive y-axis 3

~ 2 Origin

Negative x-axis

:

~--~"~~~~~~~~~x -3 -2 -1 0 1 \2 a 3 -1

Positive x-axis

Negative y-axis 2

~ -3

FIGURE A.4 Cartesian coordinates in the plane are based on two perpendicular axes intersecting at the origin.

HISTORICAL BIOGRAPHY

Rene Descartes (1596-1650)

y

(1,3) 3

Second quadrant (-, +)

•

•

First 2

quadrant (+, +)

•

1

(-2, 1)

-2

(0,0) -1

(2, 1)

~

(1,0)

0

2

x

(-2,-1)

• Third

-1

quadrant (-, -)

-2

•

Fourth quadrant (+, -)

(1, -2)

FIGURE A.S Points labeled in the xy-coordinate or Cartesiao plane, The points on the axes all have coordinate pairs but are usually labeled with single real numbers, (so (I, 0) on the x-axis is labeled as I). Notice the coordinate sign patterns of the quadrants.

Cartesian Coordinates in the Plane In Appendix I we identified the points on the line with real numbers by assigning them coordinates. Points in the plane can be identified with ordered pairs of real numbers. To begin, we draw two perpendicular coordinate lines that intersect at the O-point of each line. These lines are called coordinate axes in the plane. On the horizontal x-axis, numbers are denoted by x and increase to the right. On the vertical y-axis, numbers are denoted by y and increase upward (Figure AA). Thus ''upward'' and "to the right" are posi-

tive directions, whereas "downward" and ''to the left" are considered as negative. The origin 0, also labeled 0, of the coordinate system is the point in the plane where x and y are both zero. IfP is any point in the plane, it can be located by exactly one ordered pair of real numbers in the following way. Draw lines through P perpendicular to the two coordinate axes. These lines intersect the axes at points with coordinates a and b (Figure AA). The ordered pair (a, b) is assigned to the pointP and is called its coordinate pair. The first number a is the ;»-coordinate (or abscissa) of P; the second number b is the y-eoordinate (or ordinate) of P. The x-coordinate of every point on the y-axis is O. The y-coordinate of every point on the x-axis is O. The origin is the point (0, 0). Starting with an ordered pair (a, b), we can reverse the process and arrive at a correspondingpointPin the plane. Often we identifY Pwith the ordered pair and write P(a, b). We sometimes also refer to "the point (a, b)" and it will be clear from the context when (a, b) refers to a point in the plane and not to an open interval on the real line. Several points labeled by their coordinates are shown in Figure A.5. TIris coordinate system is called the rectangular coordinate system or Cartesian coordinate system (after the sixteenth-century French mathematician Rene Descartes). The coordinate axes of this coordinate or Cartesian plane divide the plane into four regions called quadrants, numbered counterclockwise as shown in Figure A.5. The graph of an equation or inequality in the variables x and y is the set of all points P(x, y) in the plane whose coordinates satisfY the equation or inequality. When we plot data in the coordinate plane or graph formulas whose variables have different units of measure, we do not need to use the same scale on the two axes. If we plot time vs. thruat for a rocket motor, for example, there is no reason to place the mark that shows I sec on the time axis the same distance from the origin as the mark that shows I Ib on the throst axis. Usually when we graph functions whose variables do not represent pbysical measurements and when we draw figures in the coordinate plane to study their geometry and trigonometry, we try to make the scales on the axes identical. A vertical unit of distance then looks the same as a horizontal unit. As on a surveyor's map or a scale drawing, line segments that are supposed to have the same length will look as if they do and angles that are supposed to be congruent will look congruent. Computer displays and calculator displays are another matter. The vertical and horizontal scales on machine-generated graphs usually differ, and there are corresponding distortions in distances, slopes, and angles. Circles may look like ellipses, rectangles may look like squares, right angles may appear to be acute or obtuse, and so on. We discuss these displays and distortions in greater detail in Section 1.4.

Increments and Straight Lines When a particle moves from one point in the plane to another, the net changes in its coordinates are called increments. They are calculated by subtracting the coordinates of the

Appendix 3 Lines. Circles, and Parabolas

starting point from the coordinates of the ending point. If x changes from x, to X2, the increment in x is

y C(5,6)

6

5

I

B(2, 5)

4

EXAMPLE 1 In going from the point A( 4, -3) to the point B(2, 5) the increments in the x- and y-coordinates are

';';~j

3

~x =

2 1

AP-ll

~y=5-(-3)=8.

2 - 4 = -2,

From C(5, 6) to D(5, 1) the coordinate increments are

D(5, I)

~O+-~I--~~~~--5~-->X

~=5-5=0,

~Y =

1- 6

=

-5.

-I

See Figure A6.

-2

Given two points PI(Xt.YI) and P2(X2,Y2) in the plane, we call the increments ~ = X2 - XI and ~Y = Y2 - YI the run and the rise, respectively, between PI and P2. Two such points always determine a unique straight line (usually called simply a line) passing throogh them both. We call the line PIP2. Any nonverticalline in the plane has the property that the mtio

-3

~~_ A(4,-3)

(2,-3)

I1x =-2

FIGURE A.6 Coordinate increments may be positive, negative, or zero (Example I).

•

~Y

rise

Y2 - YI

m=run=~=X2

y

XI

has the same value for every choice of the two points PI(Xt.YI) and P2(X2,n) on the line (Figure A 7). This is because the mtios of corresponding sides for similar triangles are equal.

!!.y

(rise)

!!.y'

PI (xl' Yl)

<JI-----:--~

!!.x

Q(x"y,)

DEFINITION

The constant ratio rise

m = run =

~Y ~

Y2 - YI

= X2

XI

(run) -----------~ Q'

is the slope of the nonverticalline PIP2.

!!.x' ----~~--------------~x

o

FIGURE A.7 Triangles P,QP, and Pl'Q'P2' are similar, so the ratio of their sides has the same value for any two

points on the line. Ibis common value is the line's slope.

The slope tells us the direction (uphill, downhill) and steepness of a line. A line with positive slope rises uphill to the right; one with negative slope falls downhill to the right (Figure A8). The greater the absolute value of the slope, the more mpid the rise or fall. The slope of a vertical line is undefined. Since the run ~ is zero for a vertical line, we cannot form the slope ratio m. The direction and steepness of a line can also be measured with an angle. The angle of inclination of a line that crosses the x-axis is the smallest counterclockwise angle from the x-axis to the line (Figure A9). The inclination of a horizontal line is 0°. The inclination of a vertical line is 90°. If c/> (the Greek letter phi) is the inclination of a line, then o :5 c/> < 180°. The relationship between the slope m of a nonverticalline and the line's angle of inclination c/> is shown in Figure A.10:

m

= tanc/>.

Straight lines have relatively simple equations. All points on the vertical line through the point a on the x-axis have x-coordinates equal to a. Thus, X = a is an equation for the vertical line. Similarly, Y = b is an equation for the horizon/alline meeting the y-axis at b. (See Figure All.) We can write an equation for a nonvertical straight line L if we know its slope m and the coordinates of one point PI(Xt. YI) on it. If Pix, y) is any other point on L, then we can

AP-12

Appendices

use the two points PI and P to compute the slope,

y

Y - YI

m = x -

Xl

so that

Y - YI

=

mix - Xl),

Y =YI + mix -Xl).

or

The equation

Y

= YI

+ mix -

XI)

is the point-slope equation of the line that passes through the point (Xj,YI) and has slope m. FIGURE A.S The slope of LI is dy 6 - (-2) 8 m=dX= 3 0 3'

y

That is, y increases 8 units every time

x increases 3 units. The slope of L2 is dy 2- 5 -3 m = dx = 4 - 0 = T' That is, y decreases 3 units every time

~.s

x increases 4 units. this

--~~--~~x I

X

I

. . . __ ~//Jnotthis

--~----------~x

Inotthis

FIGURE A.9 Angles of inclination are measured counterclockwise from the x-axis.

EXAMPLE 2

FIGURE A.I0 The slope ofanonver1ical line is the tangent of its aog1e of inclination.

Write an equation for the line through the point (2, 3) with slope - 3/2.

Solution We substitute XI = 2,YI = 3, and m = -3/2 into the point-slope equation and obtain

When X = 0, Y = 6

EXAMPLE 3

y

6

Along this line, x=2

Solution

3 -Zx + 6.

=

80

the line intersects the y-axis at Y = 6.

or

Y

=

•

Write an equation for the line through (-2, -I) and (3,4).

The line's slope is

-I - 4

5 4

3

3 - z(x - 2),

Y

-5

m = -2 _ 3 = -5 = I. Along this line, y~3

+ f-- +-'-- - -

We can use this slope with either of the two given points in the point-slope equation:

(2,3)

2

With (x"yU = (-2, -1)

Y = -I

+ I'(x - (-2))

y=-I+x+2 FIGURE A.II The standard equations for the vertical and horizontal lines through (2,3) are x = 2 andy = 3.

Y = 4

+ I'(x - 3)

y=4+x-3

y=x+l -..............Sam.....u1t

Either way, Y = x

With (x"yU = (3,4)

+

/ y=x+l

1 is an equation for the line (Figure A.12).

•

Appendix 3 Lines, Circles, and Parabolass

AP-13

The y-coordinate of the point where a nonverticalline intersects the y-axis is called the y-intercept of the line. Similarly, the x-intercept of a nonhorizootalline is the x-coordinate of the point where it crosses the x-axis (Figure A. 13). A line with slope m and y-intercept b passes through the point (0, b), so it has equation

y = b

+ m(x

- 0),

or, more simply,

y

=

mx + b.

The equation

y=mx+b FIGUREA.12 The line in Example 3.

is called the slope-intercept equation of the line with slope mandy-intercept b.

Lines with equations of the form y = mx have y-intercept 0 and so pass through the origin. Equations oflines are called linear equations. The equation

(A and B not both 0)

Ax+By=C

is called the generailinear equation in x and y because its graph always represents a line and every line has an equation in this form (including lines with undefined slope).

Parallel and Perpendicular Lines

FIGURE A.13 LineLhasx-intercept a and y-intercept h.

Lines that are parallel have equal angles of inclination, so they have the same slope (if they are not vertical). Conversely, lines with equal slopes have equal angles of inclination and so are parallel. If two nonverticallines Ll and L2 are perpendicular, their slopes ml and m2 satisfY ml m2 = -I, so each slope is the negative reciprocal of the other: mt =

Y

I -m2'

To see this, notice by inspecting similar triangles in Figure A.14 that ml = a/h, and m2 = -h/a. Hence, mlm2 = (a/h)( -h/a) = -\.

Distance and Circles in the Plane The distance between points in the plane is calculated with a formula that comes from the Pythagorean theorem (Figure A.15).

FIGURE A.14 4ADCissimi1arto !!J.CDB. Hence 1 is also the upper angle in !!J.CDB. From the sides of !!J.CDB, we read tan 1 ~ a/h.

Y

Y2

This distance is

d~

Vrel'-2---'1~12C-+-cI-Y2---Y~112

~

V ('2-'1)2+ (Y2-Yl)2

\ Yl

o

'2

FIGURE A.15 To calculate the distance between P(XloYl) and Q(X2,Y2) , apply the Pythagorean theorem to triaogle PCQ.

AP-14

Appendices

Distance Formula for Points in the Plane The distance between P(Xl ,Yl) and Q(X2,Y2) is

+ (~y)2

d = Y(Jlx)2

=

Y(X2 - Xl)'

+ (Yo - Yl)'.

EXAMPLE 4

y

(al The distance between P( -1,2) and Q(3, 4) is

Y(3 - (-1)' + (4 - 2)'

a

=

Y(4)' + (2)'

v20 = v'4=5 =

=

2v'5.

(bl The distance from the origin to P(x, y) is

C(h, k)

• (x - h)'

+ (y -

k)' ~ a'

~--~--~~--~----~x

o

By def'mition, a circle of radius a is the set of all points P(x, y) whose distance from some center C(h, k) equals a (FigureA.16). From the distance formula, P lies on the circle if and only if

FIGURE A.16 A circle of radius a in the xy-plane, with center at (h, k).

Y(X - h)2 + (y - k)2

a,

=

so

(X - h)2 + (y - k)2

=

a 2.

{ll

Equation (1) is the standard equation ofa circle with center (h, k) and radius a. The circle of radius a = 1 and centered at the origin is the unit circle with equation

X2+y2=1.

EXAMPLE 5 (al The standard equation for the circle ofmdius 2 centered at (3, 4) is

(X - 3)2 + (y - 4)2

=

22

=

=

3

4.

(bl The circle

(X - 1)'

+ (y + 5)'

hash = 1,k= -5, and a = v'3.The center is the point (h,k) = (1,-5) and the • radius is a = v'3.

If an equation for a circle is not in standard form, we can find the circle's center and radius by first converting the equation to standard fonn. The algehmic technique for doing so is completing the square.

EXAMPLE 6

Find the center and mdius of the circle

X2

+ y2 + 4x

- 6y - 3 = O.

Appendix 3 Lines, Circles, and Parabolas

AP-15

Solution We convert the equation to standard form by completing the squares in x and y: x 2 + y2 + 4x - 6y - 3 = 0 Start with the given equation. (x 2 + 4x)

+ (y2 - 6y)

=

Gather terms. Move the con_ t to the right-hand side.

3

(X2+4X+ (~Y) + &2_ 6 Y+ (~6Y) =

3+ (~y + (~6y (x 2 + 4x

(x

+ 4) + (y2 - 6y + 9)

+ 2)' +

=

3

Add the square of half the coefficient of x to each side of the equation. Do the same for y. The parenthetical expressions on the left-hand side are now perfect

squares.

+4+9 Write each quadratic as a squared

(y - 3)2 = 16

linear expression.

•

The center is ( - 2, 3) and the radius is a = 4. The points (x, y) satisfYing the inequality Interior: (x - h)2

+ (y

_ k)2

< .2

~--------~----------~x

o

h

(x - h)2

make up the interior region of the circle with center (h, k) and redius a (Figure A.17). The circle's exterior consists of the points (x,y) satisfying

FIGURE A_17 The interior and exterior of the crrcle (x - h)' + (y - k)' = a 2.

y y =x2 (2,4)

+ (y - k)' < a 2

(x - h)2

+ (y - k)' > a 2.

Parabolas The geometric defmition and properties of general parabolas are reviewed in Section 11.6. Here we look at parabolas arising as the gmphs of equations of the furm y = ax 2 + bx + c. EXAMPLE 7

Considertheequationy = x 2 • Some points whose coordinates satisfy this

equation are (0, 0), (I, I),

(~, ~). (-I, I), (2,4), and (-2,4). These points (and all oth-

ers satisfying the equation) make up a smooth curve called a parabola (Figure A.IB).

•

--~--~~~~--~~X

-2

-1

FIGURE A.18 (Example 7).

0

2

The gmph of an equation of the form

y

Theparabolay = ,2

= ax 2

is a parabola whose axis (axis of symmetry) is the y-axis. The parabola's vertex (point where the parabola and axis cross) lies at the origin. The parabola opens upward if a > 0 and downward if a < O. The larger the value ofla I, the narrower the parabola (Figure A. 19). Generally, the gmph of y = ax 2 + bx + c is a shifted and scaled version of the parabola y = x 2 • We discuss shifting and scaling of gmphs in more detail in Section 1.2.

TheGraphofy=ax2 +bx+c,

a*'O

The graph of the equation y = ax 2 + bx + c, a oF 0, is a parabola. The parabola opens upward if a > 0 and downward if a < O. The axis is the line (2)

The vertex of the parabola is the point where the axis and parabola intersect. Its x-coordinate is x = -b/2a; itsy-coordinate is found by substitoting x = -b/2a in the parabola's equation.

AP-16

Appendices Notice that if a = 0, 1hen we have y = bx + c, which is an equation for a line. The axis, given by Equation (2), can be found by completing 1he square.

EXAMPLE 8

Graph the equation y = -

I a =--

-

+ 4.

X

+

+ c we see that

bx

b = -1,

2'

c = 4.

< 0, the parabola opens downward. From Equation (2) the axis is the vertical b

x=--=

2a

FIGURE A.19 Besides determining 1he direction in which 1he parabola y = ax 2 opens, the number a is a scaling factor. The parabola widens as a approacbes zero aod narrows as Ia Ibecomes large.

2

Comparing the equation with y = ax 2

Solution

Since a line

tx

( -1) 2(-1/2)

=

-1.

Whenx = -1, we have

y = -t(_1)2 - (-I)

+ 4 =~.

The vertex is (-1,9/2). The x-intercepts are where y = 0:

_lx 2 2

-

x+4

=

0

x 2 +2x-8=O (x - 2)(x + 4) = 0 x = 2, x =-4

-k-;;--;;---;--;;i---;-~t---~x

,\2

We plot some points, sketch the axis, and use the direction of opening to complete the graph in Figure A.20. •

i

Intercepts at x=-4andx=2

FIGURE A.20 The parabola in Example 8.

Exercises A.3 Distance#, Slopes, and Lines In Exercises 1 aod 2, a particle moves from A to B in 1he coordinate plaoe. Find 1he incremeots dx and .1y in 1he particle's coordinates. Also rmd 1he distance from A toB.

1. A( -3,2), B( -1, -2)

+ y2 =

I

S. A( -1,2), B( -2, -I)

6. A(2,3), B( -1,3)

2. A( -3.2, -2), B( -8.1, -2)

Describe 1he graphs of 1he equations in Exercises 3 and 4.

3. x 2

Plot 1he points in Exercises 5 and 6 and find 1he slope (if any) of 1he line 1hey detennine. Also rmd 1he coramon slope (if any) of1he lines perpeodicular to line AB.

4. x 2

+ y2

'"

3

In Exercises 7 aod 8, find ao equation for
7. (-1,4/3)

8.

(0, -VZ)

Appendix 3 Lines, Circles, and Parabolas In Exercises 9--15, write an equation for each line described.

Theory and Examples In Exercises 37-40, graph the two equations and fmd the points at which the graphs intersect 37. y = 2x, x 2 + y2 = I 38. y - x = I, y = x 2

9. Passes through ( -I, 1) with slope -I

10. Passes through (3, 4) and (-2,5) 11. Has slope -5/4 and y-intercept 6 12. Passes through (-12, -9) and hss slope 0

-x 2, Y 40. x 2 + y2 = I,

39. y

13. Has y-intercept 4 and x-intercept -I 14. Passes through (5, - 1) and is parallel to the line 2x

AP-17

+ 5y =

IS

15. Passes through (4, 10) and is perpendicular to the line 6x - 3y = 5.

=

=

2x2 - I (x - 1)2

+ y2

=

I

41. Insulation By measuriog slopes in the figure, estimate the temperature change in degrees per inch for (a) the gypsum wallboard; (b) the fiberglass insulation; (0) the wood sheathing. 80'

In Exercises 16 and 17, fmd the line's x- andy-intercepts and use this information to graph the line.

16.

]x

+ 4y

= 12

17.

70'

V2x - v3y = v'6

18. Is there anything special about the relationship hetween the lines Ax + By = C, and Bx - Ay = C2 (A "" O,B "" OJ? Give reasons for your answer. 19. A particle starts at A( -2,3) and its coordinates change by incrementsdx = 5,dy = -6. Find its new position. 20. The coordiuates of a particle change by dx = 5 and dy = 6 as it moves fromA(x,y) to B(3, -3). Find x andy.

60'

E 50' ~ 40' R

~

30' 20'

Circles In Exercises 21-23, fmd an equation for the circle with the given center C(h, k) and radius a. Then skeroh the circle in the '>;y-ptane. Include the circle's center in your skeroh. Also, label the circle's x- and y-intercepts, if any, with their coordinate pairs.

21. C(0,2),

a= 2

23. C( -v3, -2), a

22. C( -1,5), a = =

v'iO

10' 0'

o

1

2

2 3 4 5 Distance lbrough wall (incbes)

6

7

The ten2perature changes in the wall in Exercises 41 and 42.

Graph the circles whose equations are given in Exercises 24-26. Label each circle's center and intercepts (if any) with their coordinate pairs. 24. x 2 + y2 + 4x - 4y + 4 = 0

25. x 2 + y2 - 3y - 4

=

0

26. x 2 + y2 - 4x

+ 4y

=

0

Parabolas Graph the parabolas in Exercises 27-30. Label the vertex, axis, and intercepts in each case. 27.y=x2 -2x-3 28.y=-x 2 +4x

29. y

=

-x 2

-

6x - 5

30. y

=

I 2x2

+X +4

Inequalities Describe the regions defmed by the inequalities and pairs of inequalities in Exercises 31-34. 32. (x - 1)2 + y2 :5 4 31. x 2 + y2 > 7

33. x 2 + y2 34. x 2 + y2

42. Insulation According to the figure in Exercise 41, which of the materials is the best insulator? The poorest? Explain. 43. Pressure under water The pressure p experienced by a diver under water is related to the diver's depth d by an equation of the form p = kd + I (k a constant). At the surface, the pressure is I atmosphere. The pressure at 100 meters is about 10.94 atmospheres. Find the pressure at 50 meters. 44. Reflected light A ray of light comes in along the line x + Y = I from the second quadrant and reflects off the x-axis (see the accompanying figure). The angle of incidence is equal to the angle of reflection. Write an equation for the line along which the d<>parting light travels. y

x+y=l

> I, x 2 + y2 < 4 + 6y < 0, y>-3

35. Write an inequality that describes the points that lie inside the circle with center ( - 2, I) and radius

v'6.

36. Write a pair of inequalities that describe the points that lie inside or on the circle with center (0, 0) and radius and on or to the right of the vertical line through (I, 0).

V2,

--~------~L-----------~x

o

The path of the light ray in Exercise 44. Angles of incidence and reflection are measured from the perpendicular.

AP-18

Appendices

45. Fahrenheit vo. Celsius

47. By ealeulatiog the lengths ofits sides, show that the triangle with vertices at the points A(I, 2), B(5, 5), and C(4, -2) is isosceles but not equilateral.

In the FC·plane, sketch the grnph of the

equation C =

~(F - 32)

48. Show that the triangle with verticesA(O, 0), B( I, is equilatera1.

9

linking Fahrenheit and Celsius temperatures. On the same graph

49. Show that the poiots A(2, -I), B(I, 3), and C( - 3, 2) are vertices of a square, and fmd the fourth vertex.

sketch the line C = F. Is there a temperature at which a Celsius thermometer gives the same numerical reading as a Fahrenheit

50. Three differeot parallelograms have vertices at (-I, I), (2, 0), and (2, 3). Sketch them and fmd the coordinates of the fourth ver-

thermometer? If so, fmd it. 46. The Mt. Washington Cog Ranway Civil eogioeers calculate the slope of roadhed as the ratio of the distance it rises or falls to the distance it ruos horizootally. They eall1his ratio the grade of the roadbed, usually wtitteo as a percentage. Aloog the coast, commercial rallroad grades are usually less than 2%. In the mountaios, they may go as high as 4%. Highway grades are usually less than 5%. The steepest part of the Mt Washingtoo Cog Railway in New Hampshire has an exceptiooa1 37.1 % grade. Aloog this part of the track, the seats in the froot of the car are 14 ft above those in the rear. About how far apart are the front and rear rows of seats?

A.4

v'3), and C(2, 0)

tex of each.

51. For what value of /r; is the line 2x + ~ = 3 perpeodicular to the lioe 4x + y = I? For what value of k are the lioes parallel? 52. Midpoint of a line segment nates Xl (

Show that the point with coordi-

+ X2 2

+ .Y2)

Yl

'

2

is the midpoint of the line segment joining P(Xl. Yl) to Q(X2, Y2).

Proofs of Limit Theorems This appendix proves Theorem I, Parts 2-5, and Theorem 4 from Section 2.2.

THEOREM l-Llmlt Laws

If L, M, c, and k are real numbers and

lim f(x) = L

and

x--"'c

lim g(x) = M,

then

x--"'c

1. Sum Rule:

lim (f(x) x-c

+ g(x»

= L

2. Difference Rule:

lim (f(x) x-c

- g(x»

= L - M

3. Constant Multiple Rule:

lim (k' x-c

4. Product Rule:

lim (f(x) x-c

f(x))

=

-g(x»

+M

k' L = L· M

M .. 0

5. Quotient Rule: 6. Power Rule:

lim [j(x)]' = L', n a positive integer x-c

7. Root Rule:

lim x-c

ifiW = "'?'L =

(If n is even, we assume that lim f(x) = L x-c

>

L 11" n a positive integer

0.)

We proved the Sum Rule in Section 2.3 and the Power and Root Rules are proved in more advanced texts. We obtain the Difference Rule by replacing g(x) by -g(x) and M by -Min the Sum Rule. The Constant Multiple Rule is the special case g(x) = k of the Product Rule. This leaves only the Product and Quotient Rules.

Procrf crf the Limit Product Rule We show that for any E > 0 there exists a lJ that for all x in the intersection D of the domains of f and g,

0< Ix - cl <

lJ

=>

If(x)g(x) - LMI <

E.

>

0 such

Appendix 4 Proofs of Limit Theorems Suppose then that

E

AP-19

is a positive number, and write f(x) and g(x) as

f(x)

L + (f(x) - L),

=

g(x)

=

M + (g(x) - M).

Multiply these expressions together and subtract LM:

f(x)· g(x) - LM

= =

=

(L + (f(x) - L»)(M + (g(x) - M)) - LM LM + L(g(x) - M) + M(f(x) - L) + (f(x) - L)(g(x) - M) - LM L(g(x) - M) + M(f(x) - L) + (f(x) - L)(g(x) - M).

(l)

Since f and g have limits L and M as x --> c, there exist positive numbers Ill, 1l2, 1l3, and Il. such that for allx in D

o< o< o<

Ix Ix Ix 0< Ix

-

cl cl cl cl

< < < <

III 112 113 Il.

=> => => =>

If(x) - LI < yI;f3 Ig(x) - MI < yI;f3 If(x) - LI < E/(3(1 + IMI» Ig(x) - MI < E/(3(i + ILl».

(2)

lfwe take Il to be the smallest numbers III through Il., the inequalities on the right-hand side of the Implications (2) will hold simultaneously for 0 < Ix - cl < Il. Therefore, for all x inD, 0 < Ix - cl < Il implies

::~~tqualityaPPlied

If(x)· g(x) - LMI

+ IMllf(x) - LI + If(x) - Lllg(x) - MI ,;; (I + ILI)lg(x) - MI + (i + IMI)lf(x) - LI + If(x) - Lllg(x) - MI ,;; ILIIg(x) - MI

Values from (2)

•

This completes the proof of the Limit Product Rule.

Proof of the Limit Quotient Rule We show that lim..... ,(I/g(x))

=

l/M. We can then

conclude that

lim f(x) g(x)

.~,

= lim

.~,

(f(X)

._1_) g(x)

= lim

.~,

f(x) • lim

.~,

_1_ g(x)

=

L.l M

=

£

M

by the Limit Product Rule. Let E > 0 be given. To show that lim.~,(l/g(x)) = I/M, we need to show that there exists all> 0 such that for all x

0< Ix - cl < Il

=>

Ig(~) -

11

<

E.

Since IMI > 0, there exists a positive number III such that for all x

0< Ix - cl < III

=>

M

Ig(x) - MI < 2.

(3)

For any numbers A andB it can be shown that IAI-IBI,;; IA - BI and IBI-IAI,;; IA - BI, from which it follows that IIAI-IBII ,;; IA - BI. With A = g(x) and B = M, this becomes Ilg(x) I - IMII ,;; Ig(x) - MI,

AP-20

Appendices

which can he combined with the inequality on the right in Implication (3) to get, in turn,

IMI IIg(x)I-IMII
IMI

- 2 < 19(x)I-IMI < 2

I~I

< Ig(x) I <

31~1

IMI < 21g(x) I < 31MI _1_<~<_3_

Ig(xJl Therefore, 0

IMI

(4)

Ig(x)l"

< Ix - c I < III implies that I I I Ig(x)-M=

1M - g(x) I Mg(x)

I

I

I

"'IMI"lg(x)I"IM-g(x)1

2

(5)

Inequality (4)

< IMI"IMI"IM - g(x)l·

Since (1/2) IMI2E > 0, there exists a number 112 > 0 such that for all x (6)

Ifwe take Il to be the smaller of III and 1l2' the conclusions in (5) and (6) both hold for all x such that 0 < Ix - c I < Il. Combining these conclusions gives

•

This concludes the proof of the Limit Quotient Rule.

THEOREM 4-The Sandwich Theorem Suppose that g(x) ", f(x) ", h(x) for all x in some open interval I containing c, except possibly at x = c itself. Suppose also that lim.~c g(x) = lim_ c h(x) = L. Then lim.~c f(x) = L.

Procrf for Right-Hand Limits Suppose lim.~c+ g(x) = lim.~c+ h(x) = L. Then for any E > 0 there exists all> 0 such that for all x the interval c < x < c + Il is contained in I and the inequality implies L -

E

<

g(x)

<

L

+

and

E

L -

E

<

h(x)

<

L

+

E.

These inequalities combine with the inequality g(x) ", f(x) ", h(x) to give

L - E < g(x) ", f(x) ", h(x) < L L -

E

-

E

< <

f(x)

<

L

f(x) - L

+ <

Therefore, for all x, the inequality c < x < c

+ E,

E, E.

+ Il

implies If(x) - L I <

E.

Appendix 5 Commonly Occurring Limits

AP-21

Proof for Left-Hand Limits Suppose limx~c- g(x) = lim~c- h(x) = L. Then for any e > 0 there exists all> 0 such that for all x the interval c - Il < x < c is contained in I and the inequality implies L -

E

<

g(x)

<

+

L

L -

and

E

We conclude as before that for all x, c - Il

<x<

E

< h(x) < L +

E.

c implies If(x) - L I <

E.

Proof for Two-Sided Limits Iflimx~cg(x) = limx~ch(x) = L,theng(x) and h(x) both approachL as x --> c+ and as x--> c-; so lim~c+ f(x) = Land lim..... c- f(x) = L. Hence limx~c f(x) exists and equals L. •

Exercises A.4 5. Limits of rational function. Use Theorem I and the result of Exercise 4 to show that if I(x) and g(x) are polynomial functions andg(c) #' 0, then

1. Suppose that functions /1(x), h(x), and f,(x) have limits L" L" and L" respectively, as x ..... c. Show that their sum has limit Ll + L, + L,. Use mathematical induction (Appendix 2) to generalize this result to the sum of any fmite number of functions. 2. Use mathematical induction and the Limit Product Rule in Theorem I to show that if functions /!(x), h(x), ... , l.(x) have limits Lb L 2, ••• ,L" as x---+ c, then

.

I(x)

I(c) g(c)"

lun-~­

,~c g(x)

6. Composite. of continuons function. Figure A.21 gives the diagram for a proof that the composite of two cootinuous functions is continuous. Recons1ruct the proof from the diagram. The stat.,. ment to be proved is this: Iff is continuous at x = c and g is continuous at I(c), then g 0 1 is continuous at c. Assume that c is an interior point of the don3ain of 1 and that I(c) is an interior point of the dumain of g. This will make the !in3its involved two-sided. (The arguments for the cases that involve on.,.sided limits are similar.)

3. Use the fact that lirnx-cx = c and the result of Exercise 2 to show that limx_ c x" = en for any integer n > 1. 4. Limit. of polynomial. Use the fact that !in3,~,(k) ~ k for any number k together with the results of Exercises I and 3 to show that !in3,~,/(x) ~ I(c) for any polynomial function

go!

FIGURE A.21

The diagraro for a proof that the composite of two continuous functions

is continuous.

A.S

Commonly Occurring Limits This appendix verifies limits (4)-(6) in Theorem 5 of Section 10.1. Limit 4: If Ixl

<

..

We need to show that to each E > 0 there corresponds e for all n greater than N. Since e lfn --> 1, while Ixl < 1,

1, lim x" = 0 "~

an integer N so large that Ixnl < there exists an integer N for which El/N

>

Ix!- In other words, (1)

AP- 22

Appendices

This is the integer we seek because, iflxl

<

Ix"1

<

IxNI

Combining (I) and (2) produces Ix" I <

E

I, then for all n

for all n

> >

N.

(2)

N, concluding the proof.

•

._00 (1 + ~)' = e" Let

Limit 5: For any number x, lim

Then

as we can see by the following application of I'H6pital's Rule, in which we differentiate with respect to n:

lim n 1n(1

+ ,x:)

,,---+00

=

n

lim 1n(1 ,,--+00

=

+ x/n)

lin

. C+\/n)- (- :2) . x

Iw

-lIn 2

11_00

lim

=

n--+OO

1

+ xln

=x

.

Apply Theorem 3, Section 9.1, with f(x) = eX to conclude that

• n

Limit 6: For any number x, lim iL, = 0 n_1XI n.

Since

Ixl" x" Ixr n! - n! - n! '

--<-<-

all we need to show is that Ixl"/n! --> O. We can then apply the Sandwich Theorem for Sequences (Section 10.1, Theorem 2) to conclude that x"/n! --> O. The first step in showing that Ixr/n! --> 0 is to choose an integer M > lxi, so that (Ixl/M) < I. By Limit 4, just proved, we then have (Ixl/M)" --> O. We then restrict our attention to values of n > M. For these values of n, we can write Ixr n!

Ixr

1·2· ... 'M'(M+ I)'(M+ 2)' ... 'n (. - M) factors

Ixl" = Ixl"MM = MM (I~)" - MlM" M MlM" Ml \.M .

<

Thus,

(Ixl)"

< ~ < MM O -n.' - M. l M '

Now, the constant MM/Ml does not change as n increases. Thus the Sandwich Theorem tells us thatlxl"/n! --> 0 because (Ix I/M)" --> O. •

Appendix 6 Theory of the Real Numbers

A.6

AP-23

Theory of the Real Numbers A rigorous development of calculus is based on properties of!be real numbers. Many results about functions, derivatives, and integrals would be false if stated for functions dermed ouly on the rational numbers. In this appendix we briefly examine some basic concepts of the theory of the reals that bint at wbat might be learned in a deeper, more theoretical study of calculus. Three types of properties make the real numbers wbat they are. These are the algebraic, order, and completeness properties. The algebraic properties involve addition and multiplication, subtraction and division. They apply to rational or complex numbers as well as to the reals. The structure of numbers is built around a set with addition and multiplication operations. The following properties are required of addition and multiplication. Al

A2 A3 A4

+ (b + c) = (a + b) + c for all a, b, c. + b = b + a for all a, b. There is a number called "0" such that a + 0 = a for all a. For each number a, there is a b such that a + b = O. a a

M2

a(bc) = (ab)cforalla,b,c. ab = ba for all a, b.

Ml M3

There is a number called "1" such that a' 1 = a for all a.

M4

For each nonzero a, there is a b such that ab = 1.

D

a(b

+

c) = ab

+ bc for all a, b, c.

Al and Ml are associative laws, A2 and M2 are commutativity laws, A3 and M3 are identity laws, and D is the distributive law. Sets that bave these algebraic properties are examples of fields, and are studied in depth in the area of theoretical mathematics called abstract algebra. The order properties allow us to compare the size of any two numbers. The order properties are 01

For any a and b, either a :5 b or b :5 a or both.

02

lfa:5 lfa:5 lfa:5 lfa:5

03 04

OS

bandb:5 athena = b. bandb:5 c then a :5 c.

bthena

+

c:5 b

+

c.

bandO:5 c then ac :5 bc.

03 is the transitivity law, and 04 and 05 relate ordering to addition and multiplication. We can order the reals, the integers, and the rational numbers, but we cannot order the complex numbers. There is no reasonable way to decide whether a number like i = yCl is bigger or smaller than zero. A field in which the size of any two elements can be compared as above is called an ordered field. Both the rational numbers and the real numbers are ordered fields, and there are many others. We can think of real numbers geometrically, lining them up as points on a line. The completeness property says that !be real numbers correspond to all points on the line, with no ~'holes" or "gaps." The rationals, in contrast, omit points such as and 1T' , and the integers even leave out fractions like 1/2. The reals, baving the completeness property, omit no points. What exactly do we mean by this vague idea of missing holes? To answer this we must give a more precise description of completeness. A number M is an upper bound for a set of numbers if all numbers in the set are smaller than or equal to M. M is a least upper bound if it is the smallest upper bound. For example, M = 2 is an upper bound for the

Vi

AP-24 Appendices negative nmnbers. So is M = 1, showing that 2 is not a least upper bound. The least upper bound for the set of negative nmnbers is M = O. We define a complete ordered field to be one in which every nonempty set bounded above has a least upper bound. If we work with just the mtional nmnbers, the set of nmnbers less than v'2 is bounded, but it does not have a mtionalleast upper bound, since any mtional upper bound M can be replaced by a slightly smaller mtional nmnber that is still larger than So the mtionals are not complete. In the real nmnbers, a set that is bounded above always has a least upper bound. The reals are a complete ordered field. The completeness property is at the heart of many results in calculus. One example occurs when searching for a maxin3mn value for a function on a closed interval [a, bI, as in Section 4.1. The function y = , - x' has a maximmn value on [0, II at the point x satis· fYing I - 3x 2 = 0, or x = Vf13. If we limited our considemtion to functions defined only on mtional nmnbers, we would have to conclude that the function has no maximmn, since Vf13 is irrational (Figure A.22). The Extreme Value Theorem (Section 4.1), which in3plies that continuous functions on closed intervals [a, bI have a maxm.mn value, is not true for functions dermed only on the mtionals. The Intermediate Value Theorem in3plies that a continuous function 1 on an interval [a,blwith/(a) < Oand/(b) > omust be zero somewhere in [a, bl.The function values cannot jmnp from negative to positive without there being some point x in [a, bI where I(x) = O. The Intermediate Value Theorem also relies on the completeness of the real nmnbers and is false for continuous functions defined only on the mtionals. The function I(x) = 3x 2 - 1 has 1(0) = -I and 1(1) = 2, but if we consider 1 only on the mtional an nmnbers, it never equals zero. The only value ofx for which I(x) = 0 is x = irrational nmnber. We have captored the desired properties of the reals by saying that the real nmnbers are a complete ordered field. But we're not quite finished. Greek mathematicians in the school of Pytbagoms tried to in3pose another property on the nmnbers of the real line, the condition that all nmnbers are mtios of integers. They learned that their effort was doomed when they discovered irrational nmnbers such as \12. How do we know that our efforts to specny the real nmnbers are not also flawed, for some unseen reason? The artist Escher drew optical illusions of spiral staircases that went up and up until they rejoined them· selves at the bottom. An engineer trying to build such a staircase would find that no strue· tore realized the plans the architect had dmwn. Could it be that our design for the reals contains some subtle contradiction, and that no construction of such a nmnber system can be made? We resolve this issue by giving a specific description of the real nmnbers and verify· ing that the algebraic, order, and completeness properties are satisfied in this model. This is called a construction of the reals, and just as stairs can be built with wood, stone, or steel, there are several approaches to constructing the reals. One construction treats the reals as all the inf'mite decimals,

Vz.

y 0.5

~~~~~~~~~~~~~x

0.1

0.3

0.5

I

0.7

0.9 1

vV3 FIGURE A.22 The maximum value of = , - ,'on [0, I] occurs at 1he irrational number x = V1j3. y

V1!3,

a.dt d2d,d4 • •• In this approach a real nmnber is an integer a followed by a sequence of decimal digits dlo d2, d" ... , each between 0 and 9. This sequence may stop, or repeat in a periodic pattern, or keep going forever with no pattern. In this form, 2.00, 0.3333333 . .. and 3.1415926535898 ... represent three familiar real nmnbers. The real meaning of the dots " ... " following these digits requires development of the theory of sequences and series, as in Chapter 10. Each real nmnber is constructed as the limit of a sequence ofmtional nmnbers given by its rmite decin3al approxin3ations. An inf'mite decimal is then the same

as a series dl d2 a+-+-+ .. · 10 100 This decirnal construction of the real nmnbers is not entirely straightforward. It's easy enough to check that it gives nmnbers that satisfy the completeness and order properties,

Appendix 7 Complex Numbers

AP-25

but verifying the algebraic properties is rather involved. Even adding or multiplying two numbers requires an inf"mite number of operations. Making sense of division requires a careful argument involving limits of rational approximations to infinite decimals. A different approach was taken by Richard Dedekind (1831-1916), a German mathematician, who gave the first rigorous construction of the real numbers in 1872. Given any real number x, we can divide the rational numbers into two sets: those less !ban or equal to x and those greater. Dedekind cleverly reversed tIris reasoning and defined a real number to be a division of the rational numbers into two such sets. This seems like a strange approach, but such indirect methods of constructing new structures from old are common in theoretical mathematics. These and other approaches can be used to construct a system of numbers having the desired algebraic, order, and completeness properties. A final issue that arises is whether all the constructions give the same tiring. Is it possible that different constructions result in different number systems satisfying all the required properties? If yes, which of these is the real numbers? Fortunately, the answer turns out to be no. The reals are the only number system satisfying the algebraic, order, and completeness properties. Confusion about the nature of the numbers and about limits caused considerable controversy in the early development of calculus. Calculus pioneers such as Newton, Leibniz, and their successors, when looking at what happens to the difference quotient ~y ~

f(x +

~x)

- f(x)

~

as each of ~y and ~ approach zero, talked about the resulting derivative being a quotient of two inf"mitely small quantities. These "inf"mitesimals;' written dx and dy, were thought to be some new kind of number, smaller !ban any fixed number but not zero. Similarly, a def"mite integral was thought of as a sum of an infinite number of inf"mitesimals

f(x)-dx as x varied over a closed interval. While the approximating difference quotients ~y/ ~x were understood much as today, it was the quotient of infinitesimal quantities, rather than a limit, that was thought to encapsulate the meaning of the derivative. This way of thinking led to logical difficulties, as attempted def"mitions and manipulations of infinitesimals ran into contradictions and inconsistencies. The more concrete and computable difference quotients did not cause such trouble, but they were thought of merely as useful calculation tools. Difference quotients were used to work out the numerical value of the derivative and to derive general formulas for calculation, but were not considered to be at the heart of the question of what the derivative actually was. Today we realize that the logical problems associated with infinitesimals can be avoided by defining the derivative to be the limit of its approximating difference quotients. The ambiguities of the old approach are no longer present, and in the standard theory of calculus, infinitesimals are neither needed nor used.

A.7

Complex Numbers Complex numbers are expressions of the form a + ib, where a and b are real numbers and i is a symbol for v'=I. Unfortunately, the words ''real'' and ''imaginary'' have connotations that somehow place v'=I in a less favorable position in our minds !ban v'2. As a matter of fact, a good deal of imagination, in the sense of inventiveness, has been required to construct the real number system, which forms the basis of the calculus (see Appendix A.6). In this appendix we review the various stages of this invention. The further invention of a complex number system is then presented.

AP- 26

Appendices

The Development of the Real Numbers The earliest stage of number development was the recognition of the counting numbers 1,2,3, ... , which we now call the natural numbers or the positive integers. Certain simple arithmetical operations coo be performed with these numbers without getting out· side the system. That is, the system of positive integers is closed under the operations of addition ood multiplication. By this we meoo that if m ood n are any positive integers, then

m+n

=

P

ood

mn

= q

(1)

are also positive integers. Given the two positive integers on the left side of either equation in (1), we coo find the corresponding positive integer on the right side. More thoo this, we cao sometimes specify the positive integers m ood P ood tmd a positive integer n such that m + n = p. For instance, 3 + n = 7 coo be solved when the ouly numbers we know are the positive integers. But the equation 7 + n = 3 cannot be solved unless the number system is enlarged. The number zero ood the negative integers were invented to solve equations like 7 + n = 3. In a civilization that recognizes all the integers

... , -3, -2, -1, 0,1,2,3, ... ,

(2)

educated person coo always find the missing integer that solves the equation = P when given the other two integers in the equation. Suppose our educated people also know how to multiply any two of the integers in the list (2). If, in Equations (1), they are given m ood q, they discover that sometimes they coo find n ood sometimes they cannot. Using their imagination, they may be inspired to invent still more numbers ood introduce fractions, which are just ordered pairs min of integers m ood n. The number zero has special properties that may bother them for a while, but they ultimately discover that it is haody to have all ratios of inte· gers min, excluding only those having zero in the denominator. This system, called the set of rational numbers, is now rich enough for them to perform the rational operations of arithmetic: 00

m

+n

1. (a) addition (b) subtraction

1

FIGURE A.23 With a straightedge and compass, it is possible to construct a segment of irrationalleogth.

2. (a) multiplication (b) division

on ooy two numbers in the system, except that they cannot divide by zero because it is meaoingless. The geometry of the unit square (Figure A.23) ood the Pythagoreoo theorem showed that they could construct a geometric lioe segment that, in terms of some basic unit of length, bas length equal to Thus they could solve the equation

Vz.

x2

= 2

by a geometric constroction. But then they discovered that the lioe segment representing is 00 incommensurable quaotity. This meoos that cannot be expressed as the ratio of two integer multiples of some unit oflength. That is, our educated people could not tmd a rational number solution of the equation x 2 = 2. There is no rational number whose square is 2. To see why, suppose that there were such a rational number. Then we could find integers p ood q with no common factor other tbao I, ood such that

Vz

v'2

(3)

Since p ood q are integers, p must be even; otherwise its product with itself would be odd. In ayrobols, p = 2PI, where PI is 00 integer. This leads to 2pl2 = q2 which says q must be even, say q = 2qj, where ql is 00 integer. This makes 2 a factor of both p ood q, contrary to our choice ofp ood q as integers with no common factor other tbao 1. Hence there is no rational number whose square is 2.

AP-28

Appendices

and are familiar with the further fact that when the discriminant, b 2 - 4ac, is negative, the solutions in Equation (9) do not belong to any of the systems discussed above. In fact, the very simple quadratic equation

x2 + 1

=

0

is impossible to solve if the only number systems that can be used are the three invented systems mentioned so far. Thus we come to the fourth invented system, the set of all complex numbers a + ib. We could dispense entirely with the symbol i and use the ordered pair notation (a, b). Since, under algebraic operations, the numbers a and b are treated somewhat differently, it is essential to keep the order straight. We therefore might say that the complex number system consists of the set of all ordered pairs of real numbers (a, b), together with the rules by which they are to be equated, added, multiplied, and so on, listed below. We will use both the (a, b) notation and the notation a + ib in the discussion that follows. We call a the real part and b the imaginary part of the complex number (a, b). We make the following definitions. Equality a+ib=c+id if and only if a = candb = d.

Two complex numbers (a, b) and (c, d) are equal if and only if a = candb = d.

Addition (a + ib) =

+ (c + id) (a + c) + i(b + d)

The sum of the two complex numbers (a, b) and (c, d) is the complex number (a + c, b + d).

Multiplication (a + ib)(c + id) = (ac - bd) + i(ad c(a

+ ib)

=

ac

The product of two complex numbers (a, b) and (c, d) is the complex number (ac - bd, ad

+ bc)

+ i(bc)

+ be).

The product of a real number c and the complex number (a, b) is the complex number (ac, be).

The set of all complex numbers (a, b) in which the second number b is zero has all the properties of the set of real numbers a. For example, addition and multiplication of (a, 0) and (c, 0) give

(a,O)

+ (e,O)

=

(a

+ c,O),

(a, 0)' (e, 0) = (ae, 0), which are numbers of the same type with imaginary part equal to zero. Also, if we multiply a "real number" (a, 0) and the complex number (c, d), we get (a, 0)' (c, d) = (ac, ad) = a(c, d). In particular, the complex number (0, 0) plays the role of zero in the complex number

system, and the complex number (1, 0) plays the role of unity or one. The number pair (0, 1), which has real part equal to zero and imaginary part equal to one, has the property that its square,

(0,1)(0,1) = (-1,0), has real part equal to minus one and imaginary part equal to zero. Therefore, in the system of complex numbers (a, b) there is a number x = (0,1) whose square can be added to unity = (1,0) to produce zero = (O,O),thatis,

(0, 1)2

+ (1,0)

=

(0,0).

Appendix 7 Complex Numbers

AP-27

Although our educated people could not f"md a rational solution of the equation x 2 = 2, they could get a sequence of rational numbers

I I'

7 5'

41 29'

239 169'

... ,

(4)

whose squares form a sequence

I I'

49 25'

57,121 28,561'

1681 841 '

... ,

(5)

that converges to 2 as its limit. This time their imagination suggested that they needed the concept of a limit of a sequence of rational numbers. If we accept the fact that an increasing sequence that is bounded from above always approaches a limit (Theorem 6, Section 10.1) and observe that the sequence in (4) has these properties, then we want it to have a limit L. This would also mean, from (5), that L 2 = 2, and hence L is not one of our rational numbers. If to the rational numbers we further add the limits of all bounded increasing sequences of rational numbers, we arrive at the system of all ''real'' numbers. The word real is placed in quotes because there is nothing that is either "more real" or "less real" about this system than there is about any other mathematical system.

The Complex Numbers lmagioation was called upon at many stsges during the development of the real number system. In fact, the art of invention was needed at least three times in constructing the systems we have discussed so far:

1.

The first invented system: the set of all integers as constructed from the counting numbers.

2.

The second invented system: the set of rational numbers min as constructed from the integers.

3.

The third invented system: the set of all real numbers x as constructed from the rational numbers.

These invented systems form a hierarchy in which each system contsins the previous system. Each system is also richer than its predecessor in that it permits additional opera· tions to be performed without going outside the system:

1.

In the system of all integers, we can solve all equations of the form

x

+a

= 0,

(6)

where a can be any integer. 2.

In the system of all rational numbers, we can solve all equations of the form

ax + b

=

0,

(7)

provided a and b are rational numbers and a .. O. 3.

In the system of all real numbers, we can solve all of Equations (6) and (7) and, in ad· dition, all quadratic equations

ax 2

+ bx + c

= 0

having

a" 0

and

b2

-

4ac ;" O.

(8)

You are probably familiar with the formula that gives the solutions of Equation (8), namely,

x=

-b ± Vb 2 2a

-

4ac

(9)

Appendix 7 Complex Numbers

The equation

x

2

+I

=

AP-29

°

therefore bas a solution x = (0, I) in this new number system. You are probably more familiar with the a + ib notation than you are with the notation (a, b). And since the laws of algebra for the ordered pairs enable us to write

(a, b) = (a,O)

+ (0, b)

= a(l,

0)

+ b(O, I),

while (I, 0) behaves like unity and (0, I) behaves like a square root of minus one, we need not hesitate to write a + ib in place of (a, b). The i associated with b is like a 1Iacer element that tags the imaginary part of a + ib. We can pass at will from the rea1m of ordered pairs (a, b) to the realm of expressions a + ib, and conversely. But there is nothing less "real" about the symbol (0, 1) = i than there is about the symbol (1, 0) = I, once we have learned the laws of algebra in the complex number system of ordered pairs (a, b). To reduce any rational combination of complex numbers to a single complex number, we apply the laws of elementary algebra, replacing i 2 wherever it appears by -I. Of course, we cannot divide by the complex number (0, 0) = + iO. But if a + ib oF 0, then we may carry out a division as follows:

°

+ id a + ib e

(e (a

+ iri)(a + ib)(a

- ib) - ib)

(ae

+

bd)

+

i(ad - be)

a2 + b 2

The result is a complex number x + iy with

ae

x

= a2

+ bd + b2 '

y

=

ad - be a2 + b 2 '

anda 2 + b 2 oF O,sincea + ib = (a, b) oF (0,0). The number a - ib that is used as multiplier to clear the i from the denominator is called the complex conjugate of a + ib. It is customary to use (read "z bar") to denote the complex conjugate of z; thus

z

z

=

a + ib,

z=

a - ib.

Multiplying the numerator and denominator of the fraction (e + iri)/(a + ib) by the complex conjugate of the denominator will alwsys replace the denominator by a real number.

EXAMPLE 1

We give some illustrations of the arithmetic operations with complex

numbers. (8) (2 + 3i) + (6 - 2i) = (2 + 6) + (3 - 2)i = 8 + i

(b) (2 + 3i) - (6 - 2i) = (2 - 6) + (3 - (-2»i = -4 + 5i (c) (2 + 3i)(6 - 2i) = (2)(6) + (2)( -2i) + (3i)(6) + (3i)( -2i) = 12 - 4i + 18i - 6i 2 = 12 + 14i + 6 = 18 + 14i

(d) 2 + 3i = 2 + 3i 6 + 2i 6-2i 6-2i6+2i 12 + 4i + 18i + 6i2 36 + 12i - 12i - 4i2 _6+22i_l+!l. 40 -20 20'

Argand Diagrams There are two geometric representations of the complex number z = x + iy: L

as the point P(x, y) in the xy-plane

2_

as the vector

oF from the origin to P.

•

AP-30

Appendices

y

In each representation, the x-axis is called the real axis and the y-axis is the imaginary axis. Both representations are Argand diagrams for x + iy (Figure A.24). In tenns of the polar coordinates ofx andy, we have

x

rcosB,

=

y

rsin8,

=

and

+ isinll). (10) We define the absolute value of a coroplex number x + iy to be the length r of a vector z = x

FIGURE A.24 This Argand diagram represents z = x + iy both as a point P(x, y) and as a vector liP .

+ iy

= r(cosll

oP from the origin to P(x, y). We denote the absolute value by vertical bars; thus, Ix

+

iyl = Yx 2

+ y2.

If we always choose the polar coordinates r and II so that r is nonnegative, then

r

= Ix

+

iyl.

The polar angle II is called the argument of z and is written II = arg z. Of course, any integer multiple of 2'IT may be added to II to produce another appropriate angle. The following equation gives a useful formula connecting a complex number z, its conjugate z, and its absolute value Izl, namely,

z·z =

Iz12.

Eulefs Formula The identity ei(J =

cos 8 + isin8,

called Euler'. formula, enables us to rewrite Equation (10) as

This formula, in tum, leads to the following rules for calculating products, quotients, powers, and roots of coroplex numbers. It also leads to AIgand diagrams for ,/9. Since cos II + i sin II is what we get from Equation (10) by taking r = I, we can say that ,i9 is represented by a unit vector that makes an angle II with the positive x-axis, as shown in Figure A.25.

Yei6=cos8+isin6

Yeil':l=cos6+isin8 \

-t----~o~-L---f~x

(a)

(cos 6, sin 6)

-t----~o~~~-t->x

(b)

FIGURE A.25 AIgand diagrams for eil = cos 6 vector and (b) as a poio!.

+ i sio 6 (a) as a

Products To multiply two complex numbers, we multiply their absolute values and add their angles. Let

(11)

Appendix 7 Complex Numbers

AP-31

so that

y

Then

and hence IZIZ,I = rlr, = Izd'lz,1

+ 9,

arg (ZIZ,) = 91 FIGURE A.26 When ZI and z, are multiplied,lzlz,1 = rl'r,and arg (ZIZ,) = 81 + 8,.

= arg ZI

+ arg z,.

(12)

Thus, the product of two complex numbers is represented by a vector whose length is the product oftbe lengths oftbe two factors and whose argument is tbe sum of their arguments (Figure A.26). In particular, from Equation (12) a vector may be rotated counterclockwise through an angle 9 by multiplying it by e i'. Multiplication by i rotates 90°, by -I rotates 180°, by -i rotates 270°, and so on.

v'3 -

EXAMPLE 2 Let ZI = I + i, z, = i. We plot these complex numbers in an Argand diagram (Figure A.27) from which we read off the polar representations

y I I

Z2 =

:v3-1

2e -i'lf/6 .

I

~~--L-r-~~'-----~-r~xThen

1+v3

ZIZ, = 2v'Zexpe: - i n =

2v'ZexpG~)

-I

= 2v'Z

FIGURE A.27

To multiply two complex

numbers, multiply tbeir absolute values and add tbeir arguments.

(COS;; + i sin ;;) '" 2.73 + 0.73i. •

The notation exp(A) stands fore A •

Quotients Suppose,., '" 0 in Equation (11). Then

Hence

7,

ZII = 71 = Iz,l IZII z, I

and

arg (ZI) z, = 91 - 9, = argzi - argz,.

That is, we divide lengths and subtract angles for the quotient of complex numbers.

EXAMPLE 3

Letzi = 1

1+i v'3 _i

_ -

+ iandz,

=

v'3 -

i,asinExample2.Then

v'Ze'w/4 _ v'Z ''''/1' 0707 ( 57T + . . 57T) 2e 'w/6 2 e "'. cos 12 I sm 12

'" 0.183

+ 0.683i.

•

AP-32

Appendices

Powers If n is a positive integer, we may apply the product fonnulas in Equation (12) to fmd zll = z·z· ... ·Z.

With z

=

rej(J, we

11

factors

obtain • summands

(13) The length r = Iz Iis raised to the nth power and the angle 0 = arg z is multiplied by n. If we take r = I in Equation (13), we obtain De Moivre's Theorem.

De Mcrivre's Theorem (cosO

+ isinO)n

= cosnO

+ isinnO.

(14)

If we expand the left side of De Moivre's equation above by the Binomial Theorem and reduce it to the form a + ib, we obtain formulas for cos nO and sin n6 as polynomials of degree n in cos 6 and sin 6.

EXAMPLE 4

Ifn = 3 in Equation (14), we have (cosO

+

isin(J)3 = cos 36

+ isin30.

The left side of this equation expands to cos3 6

+

3i cos2 6 sin 6 - 3 cos 6 sin2 6 - i sin3 6.

The real part of this must equal cos 36 and the imaginary part must equal sin 36. Therefore, cos 36 = cos3 6 - 3 cos 6 sin2 6, sin 36 = 3 cos2 6 sin 6 - sin3 6.

•

Roots If z = rei' is a complex number different from zero and n is a positive integer, then there are precisely n different complex numbers Wo, WI, ... , Wn-l, that are nth roots ofz. To see why, let W = peia be an nth root of z = rei', so that wI! =

z

or

Then

p =

"0

is the real, positive nth root of r. For the argument, although we cannot say that na and 6 must be equal, we can say that they may differ only by an integer multiple of2'lT. That is,

na

=

6 + 217r,

k

=

0, ±I, ±2, ....

Therefore,

a

6

2'lT

=,,+ kn.

Appendix 7 Complex Numbers

AP-33

Hence, all the nth roots of z = re iO are given by

y

It = 0, ±1, ±2, ....

----j---~r-=~-_;;_t_~x

There might appear to be infmitely many different answers corresponding to the infinitely many possible values of k, but k = n + m gives the same answer as k = m in Equation (15). Thus, we need only take n consecutive values for k to obtain all the different nth roots of z. For convenience, we take

k W2

FIGURE A.28 z = re iIJ ,

The three cube roots of

= 0, 1,2, ... ,n -

1.

All the nth roots of re iO lie on a circle centered at the origin and having radius equal to the real, positive nth root of r. One of them has argument a = (I/n. The others are uniformly spaced around the circle, each being separated from its neighbors by an angle equal to 27T/n. Figure A.28 illustrates the placement of the three cube roots, wo, WI> W2, of the complex number z = re llJ •

EXAMPLE 5

y

(15)

Find the four fourth roots of - 16.

Solution As our first step, we plot the number -16 in an Argand diagram (Figure A.29) and determine its polar representation re iO • Here, z = -16, r = + 16, and (I = 7T. One of the fourth roots of 16e i " is 2e i"/4. We obtain others by successive additions of 27T/4 = 7T/2 to the argument of this first one. Hence, '¢'16

.

expz1T -

2

.(7T 37T 57T 77T)

expz

4'4'4'4 '

and the four roots are ~---- +----~-~--r~X

-16

Wo = 2 [cos

FIGURE A.29 of-16.

The four fourth roots

~ + isin ~ 1= Vz(1 + i)

Wj

= 2 [cos

3; + isin 3;] = Vz(-1 + i)

W2

= 2 [cos

5; + isin 5; 1= Vz(-1 -

W3

= 2 [cos

7; + isin 7; 1= Vz(1 -

i)

i).

•

The Fundamental Theorem of Algebra

v=1

One might say that the invention of is all well and good and leads to a number system that is richer than the real number system alone; but where will this process end? Are we also going to invent still more systems so as to obtain ~, ~,and so on? But it turns out this is not necessary. These numbers are already expressible in terms of the complex number system a + ib. In fact, the Fundamental Theorem of Algebra says that with the introduction of the complex numbers we now have enough numbers to factor every polynomial into a product of linear factors and so enough numbers to solve every possible polynomial equation.

AP-34

Appendices

The Fundamental Theorem of Algebra Every polynomial equation of the form anzll

+ Cln_lZIl - 1 + ... +

alZ

+ ao

=

0,

in which the coefficients 110, aj, ... , an are any complex numbers, whose degree n is greater than or equal to one, and whose leading coefficient a. is not zero, has exactly n roots in the complex number system, provided each multiple root of multiplicity m is counted as m roots.

A proof of this theorem can be found in almost any text on the theory of functions of a complex variable.

Exercises A.7 Operations with Complex Numbers 1. How computers multiply compleJ[ numbers Find (a, b)· (e, d)

+ be). (2,3)'(4,-2) c. (-I, -2)·(2, 1)

= (ae - bd,ad

b. (2,-1)'(-2,3)

L

15. cos 48

2. Solve the following equations for the real numbers, x andy. L

(3

b.

(II-I + ~)' + _1_. I + x+lY

- 2(x - iy) = x

+ iy

=

+ 4 = O. 22. Find the six solutions of the equation z6 + 2z3 + 2 = O. 23. Find all solutions of the equatioo x' + 4x' + 16 = O. 24. Solve the equatioo x' + I = O.

b. (-z)

c. -z d. l/z 4. Show that the distance b&ween the two points ZI and z, in an Argand diagram is IZI - z,l. In Exercises 5-10, graph the points z

=

x

+ iy that satisfy the given

conditions. 5.

L

Izl = 2

11 = 2 Iz + 11 = Iz - 11 Iz + 11'" Izl

<

2

c. Izl > 2

8.

9.

.. Extend the results of Exercise 26 to show that f(Z) = f(z) if II

\1'=3)'

13. I + fV'3 1- fV3

f(z)

=

a,.zn

+ a,._tz,.-l + ... + atZ + ao

is a polynomial with real coefficients ao•...• a,..

Express the complex numbers in Exercises 11-14 in the form re i6 , with r '" 0 and -'IT < 8 '" 'IT. Draw an Argand diagram for each calculation.

+

25. Complex numben and vecton in the plane Show with an Argand diagram that the law for adding complex numbers is the same as the para1lelogram law for adding vectors. 26. Complex arithmetic with conjugate. Show that the coojugate of the sum (product, or quotient) of two complex numbers, ZI and z" is the same as the sum (product, or quotient) of their conjugates. complex-conjugate pairs

11 = I Iz + il = Iz -

7. Iz +

11. (1

Theory and Examples

27. Complex roots of polynomials with real coefficients come in b. Izl

6. Iz -

10.

19. Find the three cube mots of -8i.

21. Find the four solutioos of the equation z· - 2z'

i

Graph;ng and Geometry 3. How may the followIDg complex numbers be obtained from z = x + iy geometrically? Sketch.

z

18. Find the two square roots of i. 20. Find the six sixth roots of 64.

c. (3 - 2i)(x + iy) = 2(x - 2iy) + 2i - I

L

16. sin 48

17. Find the three cube roots of I.

(This is how complcx numbers are multiplied by computers.)

+ 4iJ>

Powers and Roots Use De Moivre's Theorem to express the trigonometric functions in Exercises 15 and 16 in tenDs of cos 8 and sin 8.

12L±l . I - i 14. (2

+ 3i)(1

- 2i)

= O,wheref(z) is a polynomial with real coefficients as in part (a), show that the conjugate z is also a root of the equation. (Hint: Let f(z) = u + iv = 0; then both u and v are zero. Use the fact that f(z) = f(z) = u - iv.) 28. Absolutevalueofaconjngate Showthatlzl = Izi. 29. When z = i If z and z are equal, what can you say about the location of the point z in the complex plane? b. Ifzis a root of the equatioof(z)

Appendix 8 The Distributive Law for Vector Cross Products 30. Real and imaginary parts Let Re(z) denote the real part of z and Im(z) the imaginary part. Show that the following relations hold for any complex nwnbers z, Zl, and Z2. a. z

+Z=

b. z -

z=

A.8

c. IRe(z)I :s Izl + z21 2 = Iztl 2 + IZ212

d. IZI e. IZI

2Re(z)

+ z21 :s

Iztl

AP-35

+ 2Re(zlz2)

+ IZ21

2iIm(z)

The Distributive Law for Vector Cross Products In this appendix we prove the Distributive Law u X (v

+ w)

=

u X v

+u

X w,

which is Property 2 in Section 12.4.

Proof To derive the Distributive Law, we construct u X v a new way. We draw u and v from the common point 0 and construct a plane M perpendicular to u at 0 (Figure A.30). We then project v orthogonally onto M, yielding a vector v' with length Iv Isin 8 . We rotate v' 90° about u in the positive sense to produce a vector v" . Finally, we multiply v" by the length of u. The resulting vector Iu Iv" is equal to u X v since v" has the same direction as u X v by its construction (Figure A.30) and lullv"l = lullv'l = lullvlsin8 = lu X vi.

M

FIGURE A.30

As explained in the text, u X v = Iu Iv" .

Now each of these three operations, namely,

1.

projection onto M

2.

rotation about u through 90°

3.

multiplication by the scalar Iu I

when applied to a triangle whose plane is not parallel to u, will produce another triangle. If we start with the triangle whose sides are v, w, and v + w (Figure A.31) and apply these three steps, we successively obtain the following:

1.

2.

+ w)' satisfying the vector equation v' + w' = (v + w)' A triangle whose sides are v", w" , and (v + w)" satisfying the vector equation v" + w" = (v + w)" A triangle whose sides are v', w', and (v

AP-36

Appendices

(the double prime on each vector has the same meaning as in Figure A.30)

M

FIGURE A.31 The vectors, v, W, v + w, and their projections ooto a plane perpendicular to u.

3. A triangle whose sides are lulv",lulw", andlul(v + w)" satisfying the vectorequation lulv"

+

lulw" = lul(v

+

w)".

Substituting lulv" = u X v, lulw' = u X w, and lul(v from our discussion above into this last equation gives u X v

+u

X w = u X (v

+ w)"

= u X (v

+ w)

+ w),

•

which is the law we wanted to establish.

A.9

The Mixed Derivative Theorem and the Increment Theorem This appendix derives the Mixed Derivative Theorem (Theorem 2, Section 14.3) and the Increment Theorem for Functions of Two Variables (Theorem 3, Section 14.3). Euler rlISt published the Mixed Derivative Theorem in 1734, in a series of papers he wrote on hydrodynamics.

THEOREM 2-Tbe Mixed Derivative Theorem If I(x, y) and its partial derivatives Ix. Iy, I"", and Iyx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then

I",,(a, b)

=

Iyx(a, b).

Proof The equality of 1",,( a, b) and Iyx( a, b) can be established hy four applications of the Mean Value Theorem (Theorem 4, Section 4.2). By hypothesis, the point (a, b) lies in the interior of a rectangle R in the xy-plane on which I, Ix. Iy, I"", and Iyx are all defined. We let h and k be the numbers such that the point (a + h, b + k) also lies in R, and we consider the difference ~ =

F(a + h) - F(a),

(1)

I(x, b + k) - I(x, b).

(2)

where

F(x)

=

Appendix 9 The Mixed Derivative Theorem and the Increment Theorem

AP-37

We apply the Mean Value Theorem to F, which is continuous because it is differentiable, Then Equation (1) becomes

a where

=

hF'(c,),

(3)

c, lies between a and a + h, From Equation (2), F'(x)

fx(x,b + k) - fx(x,b),

=

so Equation (3) becomes

a

=

h[j.(cl, b + k) - fx(cl, b)l,

(4)

Now we apply the Mean Value Theorem to the functiong(y) = f.(c"y) and have

g(b

+ k) - g(b)

kg'(d,),

=

or

fx(cl, b + k) - fx(c!, b) for some d, between b and b

=

kfry(c!, d,)

+ k, By substituting this into Equation (4), we get

a

=

hkfry(cl, d,)

(5)

for some point (c), d,) in the rectangle R' whose vertices are the four points (a, b), (a + h, b), (a + h,b + k), and (a, b + k). (See FigureA.32.) By substituting from Equation (2) into Equation (1), we may also write

y

R

•

(a, b)

a=

R'

k

= =

h

f(a + h,b + k) - f(a + h,b) - f(a,b + k) + f(a,b) [j(a + h,b + k) - f(a,b + k)l - [f(a + h,b) - f(a,b)l cf>(b + k) - cf>(b),

(6)

where -o~------------------->x

FIGURE A.32 The key to proving fry(a, b) = f",(a, b) is that no matter how small R' is, fry and f", tske on equal values somewhere inside R' (although not necessarily at the same point).

cf>(y)

=

f(a + h,y) - f(a,y).

(7)

The Mean Value Theorem applied to Equation (6) now gives

a for some d2 between b and b

=

k¢'(d2 )

(8)

+ k. By Equation (7),

cf>'(y)

=

Na + h,y) - fy(a,y).

(9)

Substitoting from Equation (9) into Equation (8) gives

a=

k[jy(a + h, d2) - fy(a, d2)].

Finally, we apply the Mean Value Theorem to the expression in brackets and get (10)

for some C2 between a and a + h. Together, Equations (5) and (10) show that

fry(c), d,)

=

f",(c2, d2),

(11)

where (c), d,) and (C2, d2) both lie in the rectangle R' (Figure A.32). Equation (11) is not quite the result we want, since it says only that fry has the same value at (c), d,) that f yx has at (C2, d2 ). The numbers h and k in our discussion, however, may be made as small as we wish. The hypothesis that fry and fyx are both continuous at (a, b) means that fry(c),d,) = fry(a,b) + €,andfyx(c2,d2) = fyx(a,b) + €2,whereeachof€),€2-->Oas both h, k--> O. Hence, if we let hand k--+ 0, we have fry(a, b) = fyx(a, b). • The eqnality of f ry( a, b) and f yx( a, b) can be proved with hypotheses weaker than the ones we assumed. For example, it is enough for f, fx> and fy to exist in R and for fry to be continuous at (a, b). Then fyx will exist at (a, b) and eqnal fry at that point.

AP-38

Appendices

THEOREM 3-The Increment Theorem for Functions of Two variables Suppose that the first partial derivatives of f(x, y) are defined throughout an open region R cootaining the point (xo, YO) and that fx and fy are cootiouous at (xo, yo). Then the change


f(xo + At,yO + ay) - f(xo,yo)

=

in the value of f that results from moving from (xo, YO) to another point (xo + At,yO + ay) inR satisfies an equatioo of the fonn


=

in which each of Et. C(""

fx(xO,yo)ax + fy(xo,yo)ay + EJAt + E2ay E2 -->

0 as both At, ay --> o.

+ '\'x, Yo + ,\.y)

,~~. B(xo

/

1

+ ax, Yo)

Proof We work within a rectangle T centered at A(xo, YO) and lying within R, and we assume that At and ay are already so small that the line segment joining A to B(xo + At, YO) and the line segment joining B to C(xo + At, Yo + ay) lie in the interior of T (Figure A.33). We may think of
=

f(xo + At, Yo) - f(xo, Yo)

is the change in the value of f from A to B and T

aZ2 The rectangular region Tin the proof of the Increment Theorem. The figure is drawn for At and ay positive, but either increment might be zero or negative. FIGURE A.33

=

f(xo + At,yO + ay) - f(xo + At,yo)

is the change in the value of f from B to C (Figure A.34). On the closed interval ofx-values joining Xo to X. + At, the functioo F(x) = f(x,yo) is a differentiable (and hence contiouous) functioo of x, with derivative

F' (x)

=

fx(x, YO) .

By the Mean Value Theorem (Theorem 4, Sectioo 4.2), there is an x-value c between Xo and Xo + At at which

F(xo + ax) - F(xo)

=

F'(c) At

or

f(xo + ax,yo) - f(xo,yo)

fx(c,yo) ax

=

or


=

fx(c,yo) At.

(12)

Similarly, G(y) = f(x. + At,y) is a differentiable (and hence cootiouous) functioo ofy on the closed y-interval joining Yo and Yo + ay, with derivative

G'(y)

=

fix. + At,y).

Hence, there is ay-value d betweeo Yo and Yo

G(yO

+ ay at which

+ ay) - G(yo)

=

G'(d) ay

or

f(xo + ax,yO + ay) - f(x. + At,y)

=

fixo + ax, d) ay

or

(13)

AP-39

Appendix 9 The Mixed Derivative Theorem and the Increment Theorem

,

1_ - --;.s

r.

:=f<Xoy)-Q _:f--- r

Po

'},---' I I I

0

:,I

-----------ii+: Q1~4r ~I

I I

I

: P' :

I A(Zo. Yo)

,

:, Yo

I I

,: ~ .

4<:

I

I I

4:1

'

----,:-----:

--------..

,

B:

I

~

'10+4'1

'1

~ C("+ .... y'+.y)

FIGURE "-34 Part ofthc surface z = !(x,y) ncar Po(xo, Yo. !(xo,Yo)). The points Po, P' • and pw have the same height Zo - !(;to. Yo) abave the .xy-plane. The change in z is 4z - P'S. The change

4z\ = !(;to + Ax. Yo) - !(xo. Yo). shawnasPWQ - P'Q'. is ClIIlSed by cbangingx from.;toto;to holdingyequal to Yo. Then, with x beld cqual toX(l + .:lx,

+

Axwbile

+ .... Yo + "y) - f(", + .... Yo) is the change in z caused by changing Yo from Yo + b.y, wbicb. is represented by "" - f(",

Q'S. The total changeinzis the sum. of 4z1 and 4zz. Now, lUI both dx and b.y--+O, we know that c-XO and d-,Yo. Therefore, since h and h are continuous Ilt (xo,,Yo) , the quantities
<,

1.(c.Yo) - 1,(XQ,Yo), ~ J,(XO + 4>, d) - l,(x"Yo)

(14)

both approach zero as both dx and !l.y- O. Finally,

Ilz

Ilzl + Ilzz ~ j,(c,Yo)4>

=

=

+ I,(x, + 4>, d)"y [{,(XQ,Yo) + <1]4> + [{,(x"Yo) + <,]"y IJ...xo.yo)dx + ly(xo,yo)tJ.y + El!l.x + E2!l.y,

FromEqa. (12) and (13) From Eq. (14)

where both El and E2 --. 0 as both!l.x and!l.y - 0, which is what we set out to prove.

•

Analogous results hold for functions of any finite number of independent variables. Suppose that the first partial derivatives ofw - I(x,y, z) are deImed throughout an open region containing the point (xo, Yo. zo) and that I,;, I", and Is are continuous at (xo, Yo. zo). Then

tJ.w

I(xo + tJ.x,,Yo + !l.y,zO + Ilz) - I(xo.yo.zo) = Ixdx + I,,!l.y + Izllz + Eldx + E 2 tJ.y + E31lz, =

(15)

AP-40

Appendices

where £1, E2, £3 ~ 0 as .6.x, .6.y, and t:.z ~ O. The partial derivatives fx> fy, fz in Equation (15) are to be evaluated at the point (xo, Yo, zo) . Equation (15) can be proved by treating ~w as the sum of!hree increments, ~Wl =

~W2 = ~W3 =

+ f(xo + f(xo + f(xo

~,Yo,zo)

~,Yo ~,Yo

+ +

- f(xo,Yo,zo) ~y,zo) - f(xo + ~,Yo,zo) ~y,zo + ~z) - f(xo + ~x,Yo

(16)

+

~y,zo),

(17) (18)

and applying the Mean Value Theorem to each of these separately. Two coordinates remain ~ W2, ~W3. In Equaconstant and only one varies in each of these partial increments ~ tion (17), for example, only Y varies, since x is held equal to Xo + ~ and z is held equal to zoo Since f(xo + ~,Y, zo) is a continuous function ofy with a derivative fy, it is subject to the Mean Value Theorem, and we have

w"

~W2 = fy(xo

for some Yl between Yo and Yo

+

~y.

+

~x,y"zo) ~y

ANSWERS TO ODD-NuMBERED EXERCISES ~

29. (a) f(x)

CHAPTER 1 Section 1.1, pp. 11-13 1. D:(-oo,oo), R: [1, 00) 3. D:[-2,00), R:[O,oo) 5. D:(-00,3)U(3,00), R:(-oo,O)U(O,oo) 7. (_) Not a function ofx because some values ofx have two values ofy (b) A function ofx because fur every x 1Irere is only one possible y

_ \13

2

_

2

-/Ox +

_

9. A - TX' p - ] x 11. x 13. L = v'2Ox

15. (-00,00)

d

\13'

2

A ~ 2d,

(b) f(x) ~

~

31. (a) f(x)

d' V~ 3\13

0, I,,; x < 2 2,2::s;x<3 / 0,3::s;x:54 -x -I s x < 0 I,' 0<x s I { -~x+~, l<x<3 -2:sxsO -lx + 2, 0< x s I

!x.

(b) f(x) ~

25

{ -I,

l<xs3 33. (_) 0 ,,; x < I (b) -I < x ,,; 0 35. Yes 37. Symmetric about the origin 39. Symmetric about the origin

17. (-00,00)

,

{x' 2 0 ,,; x ,,; I -x +, 1 < x ~ 2 2,0";X<1

,

,

4

3

N) - -JJXf

-.5-4-3-2-~1

I 2 3 4.5

x

-2 -2

Inc. -00

<x<

Oaud

O<x
,

19. (-oo,O)U(O,oo) 2

4321

Dec.

-00

<x <

00

41. Symmetric about the y-axis

F(t) . i...

,

1.1

43. Symmetric about the origin

,

1234

~~

4

21. (-00,-5)U(-5,-3]U[3,5)U(5,00) 23. (_) For each positive value of (b) For each value of x#'O, x, there are two values ofy. there are two values ofy.

,

,

Dec.

-00

inc. 0 :5x

< x ::s 0; < 00

Inc.

-00

<x<

00

45. No symmetry

,

---;---*---;-~,

"ok-~~~~' -1

-2

-,

25.

,

27.

~)=l:x:

OS-tSI 2-x, l<xS2

~f-----7-----~2--~'

,

... ..,

Dec.O:5x
A-l

A-2

Chapter 1: Answers to Odd-Numbered Exercises

Section 1.2, pp. 19-22 1. Df : -00 < x < 00, Dg : x

I

~

1, Rj


: -00

33.y-l=x_1 00,

y

R.: y '" 0, D f +. = D f .• = DIP Rf +.: y '" I, Rf .• : y '" 0 3. Df: -00 < x < 00, D.: -00 < x < 00, Rf:Y = 2, R.:y'" I, Dff.: -00 < x < 00, Rff.: 0< y '" 2, D.ff: -00 < x < 00, R.fj: y '" 112 5. (a) 2 (b) 22 (0) x 2 + 2 (d) x 2 + lOx + 22 (e) 5

(f) -2

+ 10

(g) x

7. 13 - 3x

9.

11. (a) f(g(x»

2 --:-----nt---~.

+6

~~~ ! : (b) j(g(x»

(e) g(h(f(x») 13.

(b) x' - /ix 2

y

(0) g(g(x»

(d) j(j(x»

y

,

(f) h(j(f(x)))

g(x)

(f

f(x)

0

g)(x)

2

~~----~~----~~~

(a) x - 7 (b) x + 2 (0) x2

3x + 6

I

17. (a) f(g(x» =

y

y

x

x

x

(d)

a

(e) -I

~~ + I,g(f(x»

=

~

(0) -2

~Ol-:-,--C:2------:c,~.

x

1+1 x

(b) 2

x

(f)

a

45.

47.

y

y

)' _ ~_1

+I

(b) Df•• = (-00,-1] U (0, oo),D•• f = (-1,00) (0) Rj•• = [0, I)U(I, oo),R• • j = (0,00) 2x 19. g(x) = x - I

y

y

-1 -1

)'+1 _ (.1:+1)

-2

49.

51.

y

,

,

- 2-

, . ]1;3

1

y

,•

"--

-2

v'x + 0.81

31. y

=

53.

t+l J~ .,

2x

10

-----,hb---->.

-1 0

1

2

,

y

y

~~t-~---~. -0.81 4

1 y - <X_1)2

2

, . {+2

•

=

\y-¢,~

2

-_~'---C+---C:--="'2....-0::-,~.

21. (a) y = -(x + 7)2 (b) y = -(x - 4)2 23. (a) Position 4 (b) Position I (0) Positioo 2 (d) Position 3 25. (x + 2)' + (y + 3)2 = 49 27. y + I = (x + I)'

29. y

]=1+,-/'%-:

w-=s

x x - I

~

15. (a) I

~

3x

\.-'x-=s

(d) __ x_ x - I I (e) x - I

(f)

Vx

t;:l)~

1 2

•

55. (8) D: [0,2], R: [2, 3]

(b) D: [0,2], R:[-I,O]

y

y

, - f(x)-l

0

0

2

, •

•

•

A-3

Chapter 1: Answers to Odd-Numbered Exercises

,

(0) D: [0,2],

R: [0,2]

(d) D: [0,2], R:[-I,O]

,

75.

,

, •~

77.

4

+

~ _1

, - -f(x)

2

0

0

,

2

x

x

(e) D: [-2,0], R:[O,I]

,

-4

-.

-1

-1

(f) D: [1,3], R:[O,I]

,

,

79.

,

81.

2 2

--_~2-_~'--~~~-+~X

y - /(u2)

y=;N-l)

x

-2

(g) D: [-2,0], R:[O,I]

,

x

0

-4

(h) D: [-I, I], R:[O,I]

,

2

83.

2 , · -f(x+l)+l

y=f(-Jl)

(x

+ 4)'

+

16

63. y

~ ~4 - ~

67.

Center: (-4,3)

,

10

,

..--~'

61.y~~

~I-

,

I

(H4Y + (,_3)2 _ 1 8

3

65. y

3)'

The major axis is the line segment between (-8,3) and (0,3).

10

~ 3>;2 -

9

----_~,--~O~--~--~x

-----_~2~--~--~~x

57. y

(y -

(x-I? <1+2)2 --2- + - - , - _ 1

27x'

69.

, 85. (8) Odd (f) Even

2

(b) Odd (g)Even

(0) Odd

(h) Even

(d) Even (i) Odd

(e) Even

Section 1.3, pp. 28-30 -_~'-_~2~_I~-+'~2~'-+4~'~x

1. (8) 8"m

, 4 3 2

,

sin 8

a

,

cosB

-I

tsno

a

colO

UND

sec 0

-I

esc 8

UND

73. 4

Y=~-l

1

~4-_~3~2~~~1_-'~~-+'-4~X

2

3. 8.4 in.

9

-"

5. 0 71.

(h) 55" m

-2,,/3

0

V3

0

2 I 2

V3 I

V3 -2 2

V3

7. cosx

~

9. sinx

~ - VB tsnx ~ -VB 3 '

11. sinx

I 2 = --,COSX =--

-4/5, tsnx

Vs

~

-3/4

Vs

,,/2

3,,/4 I

Yz 0

I

Yz

0

UND

-I

UND

0

-I

UND

-Yz Yz

UND

A-4

Chapter 1: Answers to Odd·Numbered Exercises

13. Period".

15. Period2

67. A

,

y

---t----~~--_.r.---~,

-I~j--\

17. Period 6

~

2 -"..,B

~

4,C

O,D

~

1

~"..

__+--I__+--L~.,

19. Period 2". y

y

Section 1.4, p. 34

, __sin-¥-

1. d 3. d 5. [-3, 5] by [-15,40]

7. [-3,6] by [-250, 50]

'0

23. Period"./2, symmetric about the origin

10

y , , , ,

2

1 ..

~2t i

,

\

\

~'~0}-.L-~~~----~7~.~' -4 "4 4 4

• I,

25. Period 4, symmetric about

\,

0

t

\

q \

--.*2-.~,--t-+-~-t.M--~' ·10

,

,

, , , ,

, , , , ,

,-

I

I

11. [-2, 6] by [-5,4]

9. [-3, 3] by [-6, 6]

i

,

f(Jl)=X~

,

29. D: (-00, 00), R:y~-I,O,I

they-axis

y , , ,

I='~¥

13. [-2,8] by [-5,10]

I,

, , , , ,

, , , , ,

, , , , ,

I

I

I

I

39. -cos x

41. -cosx

2+ Vz 47. 4 53.

'1T '11'

49.

51T 31T

6'2'6'2

65. A

~

2,B

,

~

43.

v'3

24

...

V6+Vz Vz+V6 45. 4 4 ". 2". 4,,- 5". 51- 3'3'3'3

~

-".,D

--.4~.*2-+~~~~~~~' -2 ·4 4 ,

17. [-10, 10] by [-10, 10]

~

59. v7 '" 2.65

2"., C

,

·1

i(\2!-1 i(\2! , , , , ,

15. [-3, 3] by [0, 10]

,

, , ,

~

63. a

~

1.464

19. [-4,4] by [0, 3]

,

y

,.,

-I 8 6

y . 2sin(';+1t)-1

]=

10 8 6-4

2 4 6 8

'D

-'1:+2

•

2'

f(x)=

~+2 "+1

2D 246810

,

, 1.0 0' -4 -3 -2 -I

1

2,

•

,

Chapter 1: Answers to Odd-Numbered Exercises

21. [-10, 10] by [ -6, 6]

23. [-6, 10] by [-6,6]

y

, ,• 8

10 ,

, , ,

37.

39. y

y

, ~) - ~

!(x) _ 6.1: 2 15%+6 4x 2 -10%

,

r-x-6

-, -, 4 6 8 10

\

x

10

-, -4

x

2 -1

-, -,

4

8

3

\

S 6

4

• •

1;5'1;5] by

7

1 x-3

y-

/ /

'4321

1234':%

-I

-,

27. [-100"., 100.".] by

[-1.25, 1.25]

[-1.25, 1.25]

,

y

1.0

Practice Exercises, pp. 35-36 , - ~(~)

-300

-0.' -1.0

29. [-

, ,

x

-4

25. [-

,

. .

4

A-5

~, ~]by

31.

[-0.25, 0.25]

1. A = .".r' c= 2.".r A = C' 3. x = tanO,y= tan'O , '41T 5. Origin 7. Neither 9. Even 11. Even 13. Odd 15. Neither 17. (8) Even (b) Odd (0) Odd (d) Even (0) Even 19. (8) Domain: allreals (b) Range: [-2, 00) 21. (a) Domain: [-4,4] (b) Range: [0,4] 23. (8) Domain: all reals (b) Range: (- 3, 00) 25. (a) Domain: allreals (b) Range: [-3, I] (b) Range: all reals 27. (8) Domain: (3, 00) 29. (a) Increasing (b) Neither (0) Decreasing (d) Increasing 31. (a) Domain: [-4,4] (b) Range: [0,2] 33. f(x) =

{I-X, 2-x,

35. (a) 1 (d)

a :5x<

1

1:5x:52

~ v'2.5 "1/5

(b)_I_=

(o)x,x¢O

1

v'1/v'XTI + 2 37. (a) (j 0 g)(x) = -x, x '" -2, (g 0 f)(x) = v'4-=-? (b) Domain(j 0 g): [-2, 00), domain (g 0 f): [-2,2] (0) Range (j 0 g): (-00,2], range (g 0 f): [0,2] 33.

35. ., f~)

39.

,

,

=-tan It

, 4

,

• ,x -,

-,

-.

-20

41. Replace the portion for x < 0 with mirror image of the portion for x > 0 to make the new graph synnne1ric with respect to the y-axis.

A-6

Chapter 2: Answers to Odd-Numbered Exercises

43_ Reflects the portion for y < 0 across the x-axis 45_ Reflects the portion for y < 0 across the x-axis 47_ Adds the mirror image of the portion for x > 0 to make the new graph symmetric with respect to the y-axis 49_ (_) y = g (x - 3) (0) y

= g(-x)

+

t

(d) y

(b) y = g (x

= -g(x)

,

' --VI +~

, -,

~ -4 -3 -2-1

-1

55_ Period 'IT

Y=~+1

1

,

3 •

+h "\1'1+1; ("\1'1+1; - 1)/h

Y _ 2col(x_~)

I

-, b

I

b

63_ (-) a = tanB 65_ '" 16.98 m

= V3

(b) a = 2V3/3 a (b) c = sinA

67. (b)

c

= 4V3/3

1. Yes.Forinstancc:f(x) = I/xandg(x) = I/x,orf(x) = 2x andg(x) = x/2, or f(x) = eX andg(x) = lnx. 3. Iff(x) is odd, theng(x) = f(x) - 2 is not odd. Norisg(x) evcn, unless f(x) = a for all x. If f is even, theo g(x) = f(x) - 2 is

also even. ,

I)/h

(b) g(x) =

Vx

1.1

1.01

1.001

1.0001

1.04880

1.004987

1.0004998

1.0000499

0.4880

0.4987

0.4998

0.499

1.00001

1.000001

1.000005

1.0000005

0.5

0.5 (0) 0.5

("\1'1+1; -

(d) 0.5

21. (8) IS mph, 3.3 mph, 10 mph (0) 20 mph when t = 3.5 br

(b) 10 mph, 0 mph, 4 mph

4'IT

Additional and Advanced Exerdses, pp. 37-38

5.

(b) '" 50 m/sec or 180 km/h 17. (8)

(b) '" $56,000/year (0) '" $42,000/year 19. (_) 0.414213,0.449489,

=

are m/sec.

,

,

61_ (_) a

5. I

x

,

y _ cos2l'

3~

(b) y = 4x - 7 (b) Y = 2x - 7 (b) y = 12x - 16 (b) Y = -9x - 2 15. Your estimates may not completely agree with these. (8) PQ, PQ, PQ, PQ. The appropriate units 43 46 49 50

57_ Period2

,

59_

x

(b) -

7. (_) 4 9. (8) 2 11. (_) 12 13. (8) -9

,

53_

(b) I

!

3. (_) -

= 5g(x)

(f) y = g(5x) 51_

Section 2.1, pp. 44-46 1. (8) 19

+ ~) - 2

(0) y

CHAPTER 2

Section 2.2, pp. 54-57 1. (8) Does not exist As x approaches I from the right, g(x) approaches o. As x approaches I from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x --+ 1. (b) I (0) a (d) 1/2 3. (8) True (b) True (0) False (d) False (0) False (f) True (g) True 5. Asx approaches a from the left, x/lxl approaches -1. Asx approaches a from the right, x/lxl approaches 1. There is no single number L that the function values all get arbitrarily close to as x---+ O. 7. Nothing can be said. 9. No; no; no 11. -9 13.-8 15. 5/8 17. 27 19. 16 21. 3/2 23. 1/10 25.-7 27. 3/2 29. -1/2 31. -I 33. 4/3 35. 1/6 37. 4 39. 1/2 41.3/2 43. -I 45. I 47. 1/3 49. ~

Chapter 2: Answers to Odd-Numbered Exercises

51. (s) (c) 53. (8) 55. (s) 57. 2 65. (8) 67. (s)

QuotientRule (b) DitferenceandPowerRules Sum and Constant Multiple Rules -10 (b) -20 (c) -I (d) 5/7 4 (b) -21 (c) -12 (d) -7/3 59. 3 61. 1/(2V7) 63. The limit is 1. f(x) = (x 2 - 9)/(x + 3)

Vs

Vi?),

25. (v'i5,

2 27. (2 _ 0.03 m' 29.

Vi? -

8 =

+

0.03)

•

m'

(k-,i",i,+t),

A-7

4 ... 0.12 0.03

=

m

U

8=,i,

31.L=-3,

8=0.01

33. L

= 4, 8 = 0.05 35. L = 4, 8 = 0.75 55. [3.384, 3.387]. To be safe, the left endpoint was rounded up aod the right endpoint rounded down. 59. The limit does not exist as x approaclres 3.

Section 2.4, pp. 71-73 lim f(x)

(c)

69. (s) G(x)

=

+ 6)/(x 2 + 4x

(x

1. (a) (f) (1<) 3. (a)

-6

=

x--3

- 12)

lim G(x) = -1/8 = -0.125

(c)

(x 2

=

-

(c) False (b) False

(b) True (g) False (I) False

(b) No, lim f(x) x-2+

(c) 3,3 5. (a) No

(d) Yes, 3 (b) Yes, 0

7. (s)

y

(d) True (i) False

(0) True

(j) False

lim f(x) x-z-

¢

(c) No (b) I, I

(c) Yes, I

9. (a) D: Os x'" 2,R: 0 < y s I andy = 2 (b) (0,1) U (I, 2) (c) x = 2 (d) x = 0

x--6

71. (s) f(x)

True True True 2, I

I)/(Ixl- 1)

y

2

•

~'

y-

lim f(x)

(c)

x--i

73. (s) g(O)

l

0<<<1

1,

IS.t<2

2,

x ,. 2

2

=

o

= (sin 0)/0

2

11. v3 13. I 15. 2/Vs 17. (s) I (b)-I 19. (s) I (b) 2/3 21. I 23. 3/4 25. 2 27. 1/2 29. 2 31. 0 33. I 35. 1/2 37. a 39. 3/8 47. 8 = .2, lim v;=s = 0 x-s+ 51. (a) 400 (b) 399 (c) The limit does not exist

41. 3 lim g(O)

.~o

75. c 77. 7

=I

= O,I,-I;thelimitisOatc = O,andlatc = 1,-1. 79. (s) 5

(b) 5

Section 2.3, pp. 63-66 1. 8

=

2

-+,(___-'----+) ~" , 7

3. 8 = 1/2

)

5. 8 = 1/18

( 419

7. 15. 19. 23.

).t

-112

-712 -3

) ). 112

4n

Vs -

8 = 0.1 9. 8 = 7/16 11. 8 = 2 17. (-0.19,0.21), (3.99,4.01), 8 = 0.01 (3,15), 8 = 5 21. (10/3,5), 8 = 2/3 (W), 8 = 2 ... 0.12

v'4.5, -

v'4.5 -

13. 8 = 0.36 8 = 0.19

Section 2.5, pp. 82-84 1. 3. 7. 13. 17. 21. 23. 25. 31. 35. 39. 45. 71.

No; discontinuous at x = 2; not defined at x = 2 Continnons 5. (s) Yes (b) Yes (c) Yes (d) Yes (s) No (b) No 9. a 11. I, nonremovablo; 0, removable All x except x = 2 15. All x except x = 3,x = I Allx 19. All x except x = 0 All x except x = mr/2, n any integer All x except mr/2, n an odd integer All x '" -3/2 27. Allx 29. AIlx 0; continuousatx = 71' 33. 1; continuous aty = 1 V2/2; continuous at t = 0 37. g(3) = 6 f(l) = 3/2 41. a = 4/3 43. a = -2,3 a = 5/2, b = -1/2 69. x ... 1.8794, -1.5321, -0.3473 x'" 1.7549 73. x ... 3.5156 75. x ... 0.7391

A-8

Chapter 2: Answers to Odd-Numbered Exercises

Section 2.6, pp. 94-96 1. (.) 0 (b) -2 (0) 2 (d) Does not exist (0)-1 (f) 00 (g) Does not exist (b) I (i) 0 3. (a) -3 (b) -3 5. (.) 1/2 (b) 1/2 7. (a) -5/3 (b) -5/3 9. 0 11. -I 13. (.) 2/5 (b) 2/5 15. (.) 0 (b) 0 17. (.) 7 (b) 7 19. (.) 0 (b) 0 21. (.) -2/3 (b) -2/3 23. 2 25. 00 27. a 29. I 31. 00 33. I 35. 1/2 37. 00 39. -00 41.-00 43. 00 45. (a) 00 (b) -00 47. 00 49. 00 51.-00 53. (.) 00 (b) -00 (0) -00 (d) 00 55. (.) - 00 (b) 00 (0) 0 (d) 3/2 57. (a) -00 (b) 1/4 (0) 1/4 (d) 1/4 (0) It will be -00. 59. (.) - 00 (b) 00 61. (.) 00 (b) 00 (0) 00 (d) 00

63.

85. 5/2 81. 0 83. -3/4 93. (.) For every positive real nomber B there exists a corresponding nomber S > 0 such that for all x

xo - S < x < Xo => f(x) > B. (b) For every negative real number -B there exists a corresponding nomber S > a such that for all x Xo

Xo - S

<x<

=> f(x) < -B.

Xo

101. y , , , , ,

6

,--',-,

,

+ S => f(x) < -B.

99.

4

10

Xo

(0) For every negative real number -B there exists a corresponding nomber S > 0 such that for all x

65.

,

<x<

,

I I I

2

I( / I

, _ .1:+ 1//

~

~

x2

I

y= x_1=x+l+ x-I

I

-~~~:~~~~~~~~~_~:+'~i'~2~'-74-'~" 103.

105.

-10

,

y ,

:x--2 67.

,

69. Here is one possibility.

2

, . X,,/

,

y=-';~

/////// y=:r?-;l

,

----~+-~~+-----~,

-2

x=2

107.

71. Here is one possibility.

,

y

73. Here is one possibility.

,

-2

-,

109. At 00: 00, at -00: 0

Practice Exerdses, pp. 97-98 1. Atx

75. Here is one possibility.

,

lI(x) . -.:!.... ::t~O

1'1

= -I:

79. At most one

lim f(x) = I, so

%--1+

lim f(x) = I = f(-I);contiuoousatx = -\.

Atx

= 0:

'1-----o

lim f(x) =

.%--}-

Atx

=

I:

.%--1

lim f(x) = lim f(x) = 0, so lim f(x) =

x-O-

*

x-O

.%-0+

o.

However, f(O) 0, so f is discontinuous at x = O. The discontinuity cau be removed by redefming f(O) to he O. lim f(x) = -I and lim f(x) = I, so lim f(x) %-1-

x-l+

x-I

does not exist. The function is discontinuous at x = I, and the discontinuity is not removable.

A-9

Chapter 3: Answers to Odd-Numbered Exercises

,

9. Y

-

y=j(x)

(b) 49

(g) -7

(h)

-7

11.

2

13. 2x

(0) I

(I) 7 (-2,-8)

(h) [0,00)

9. (8) Does not exist I

(d) I

5.4

7. (8) (-00,+00) (d) (0,00)

17. 2/3

21. I

19. 2/'IT

23. 4

25. 2 27. 0 29. No in both cases, because 100 j(x) does not x~' exist, and 100 j(x) does not exist. x--i

31. Yes, f does have a continuous extension, to a = 1 with j(l) ~ 4/3.

37. 2/5

47. (s) x

~

39. 0

3 (b) x

~

41. -00

1 (c) x

~

-,

(C) (-00,0) and (0, 00)

(b) 0 I

15. - 4

33. No

43. 0 45. 1

11. m

~

13. m 15. m

~

~

4,y - 5 ~ 4(x - 2) -2,y - 3 ~ -2(x - 3) 12, Y - 8 ~ 12(1 - 2)

I I 17. m~4,y-2~4(x-4) ~ -10 21. m ~ -1/4 23. (-2,-5) Y ~ -(x + 1),y ~ -(x - 3) 27. 19.6 m/sec 6'1T 33. Yes 35. Yes 37. (8) Nowhere (8) At x ~ 0 41. (8) Nowhere 43. (8) At x (8) Atx ~ 0

19. m

25. 29. 39. 45.

3. 0; the left-hand 100it was needed because the function is undefmed for v > c. 5. 65 < 1 < 75; within 5°F (h) A (c) A (d) A 13. (8) B 21. (8) 100 r+(a) ~ 0.5, lim r+(a) ~ I

11. 2(q

Q_-l+

27. 1

31. Y

29. 4

~

~

~

x, y

~

-x

(h)

~

,,

1. P,:

m,

~

I,P,:

m,

~

32

,

f' on (-4, 6)

5

3. P,:

y~2x+5

m,

~

5/2, P,:

~

x

+

7. Y

,

m,

~

-0.7

-1/2

-33

I

,

35. (8) i) 1.5 of/hr ii) 2.9 of/hr iii) O°F/hr iv) -3.7°F/hr (h) 7.3 ~/hrat 12 P.M., -11 °F/hrat6p.M. (c)

Slopo

,

('F/hr) 6

3

-3

x

0

,

3

•

x

0 -3

-.-, -12

'-&-

0.'

~

86-420~:X:

Section 3.1, pp. 105-106

15. 31' - 21,5

x

"

CHAPTER 3

5.

13. 1- 9,,0

9 I • (21 + I)'

-4 y-4~-1(x-6) 19. 6 (x - 2)V;=Z' 2 -1 -I 21. 1/8 23. ( )' 25. (x - I )' 27. b 29. d x+2 31. (a) x ~ 0,1,4 33. y'

1

33. Y

2x

-~

+ 1) q + I

7. 6>;'

17.

a--l+

(h) 100 r_(a) does not exist, 100 r_(a)

25. 0

I

2 I 2 3. -1',2'-4'- 3V3

5 __3_ _ _ 3_ 1 _3_ . 2y'36' 2V3' 2' 2v2

Additional and Advanced Exercises, pp. 98-100

a-O

~

Section 3.2, pp. 112-115

-4

1. -2x,6,0,-2

a-O

,

------~,~+L----~x

(c) 0

I

12x + 16 , . I2¥+16

----_~,--~---L--~x

3. (8) -21

~

'(!mo)

84

c-:J

0-0

87 88

x

A-tO

Chapter 3: Answers to Odd-Numbered Exercises

37 S'

h!S+

•

IDee

r

flO + h) - f(O) ~ I h

while lim flO

+~ -

f(O)

+ h)

- flO)

41. (a) 13

43. (a) y

~ 0,

(0) Y

(b) -7

~

(0) 7/25

i

~ -~ +

(b) m

8x - 15,y

~

~

(d) 20

-4 at (0, I)

8x + 17

.~o

1'(0)

~

lim

flO

h

.~o

ferentiable at x

~

39. Since lim f(1 +

45.y~4x,y~2

does Dot exist and!(x) is Dot dif-

O.

~-

f(l)

~ 2 while

. f(1 + h) - f(1) lim_ h

I, . f(1 -2' ! (I) ~ lim

h-O

43. 45. 47.

51. (0, 0), (4, 2) 57. a

53. (a) y

~

2x + 2

(0) (2, 6)

-3

~

59. P'(x) ~ na"x·- 1 + (n - I)a.-lx'-' + ... + 2a,x + al

'~IJ'

41.

49. (2,4)

55. 50

47.a~l,b~l,c~0

+ h) - f(l)

h-O

h

does not exist and !(x) is Dot differentiable at x ~ 1. Since !(x) is not continuous at x ~ 0, !(x) is not differentiable atx = O. (0) None (a) -3 '" x '" 2 (b) None (0) x ~ 0 (b) None (a) -3 "'x < 0,0 <x s 3 (b) x ~ 0 (0) None (a) -I '" x < 0, 0 < x '" 2

61. The Product Ru1e is then the Constant Mu1tiple Ru1e, so the latter is a special case of the Product Ru1e. 63. (a) :" (uvw) (b)

~

+ uv'w + u'vw

uvw'

~ (UIU2U3U4)

= UlU2U3",,'

+ UtU2U3'1l4 + Ut U2'U31l4 +

Ul'U2 U3 U4

(c)

~ (Ul ••• Un) = UIU2'"

Un -lU,,'

+ UIU2'"

U,,-2U,,-I'U1l

+

... + UI'U2"'U

Il

Section 3.3, pp. 122-124

~

1 dy • tbc

3. dsd I

_,_ d'y

"", tbc'

~

~ -2 301 - 601'

dy_, d'y_ 5. tbc - 4x - I, tbc' - 8x 7 dw~_~+~d'w~~_~ • dz z3 z2' dz2 z4 z3

~ 12x -

9. :

11 dr ~ -2 • tis 38 3

10 + IOx-',

+~ 28 2 '

Z, ~

12 - 3Ox--4

d'r ~ ~ _ ~ 83 ds 2 34

13. y' ~ -5x' + 12x' - 2x - 3 , I 15. y' ~ 3x + lOx + 2 - 17. y'

~

x'

'(x)

19.

~ x' + x + 4

dv 21 • dl

(x + 0.5)'

g

23. I'(s)

27'~ • Y

~

nRT

. dV-

(V-nb)'

V'

Section 3.4, pp. 132-135

~

lSI' - lSI" d's, dl

+ 2an'

65 dP _ _

I

1'-21-1 (I + I')'

~ '--:c~~~

25. v'

Vs('Vs + I)'

-19 ---'"''---0 (3); - 2)'

~

1. (a) -2 m, -I m/sec (b) 3 m/sec, 1m/sec; 2 m/sec', 2 m/sec' (0) Changes direction at I ~ 3/2 sec 3. (a) -9m, -3 m/sec (b) 3 m/sec, 12 m/sec; 6 m/sec', -12 mjsec' (0) No change in direction 5. (a) -20 m, -5 m/sec (b) 45 m/sec, (1/5) m/sec; 140 m/sec', (4/25) m/sec' (0) No change in direction 7. (a) a(l) ~ -6 m/sec', a(3) ~ 6 mjsec' (b) v(2) ~ 3 m/sec (0) 6 m 9. Mars: '" 7.5 sec, Jupiter: '" 1.2 sec 11. g, ~ 0.75 m/sec' 13. (a) v ~ -321,lvl ~ 321ft/sec, a ~ -32ft/sec' (b) I '" 3.3 sec (0) v '" -107.0 ft/scc 15. (a) I ~ 2, I ~ 7 (b) 3 '" I '" 6 (d)

(0) Ivl (m/!ICC)

"

S,...

-J, + 2x-'/2 x

,, 4

-4x'-3>;'+1 (x' - 1)'(x' + x + I)'

~

y(')

=

0

yen) =

33. y' 35 dr 'd8 37. ':;;

=

a for n '" 5

3x 2

+ 4x

- S,y" Oforn O?:: 4

2x - 7x-2 ,y"

=

6x

=

2

+

+ 4,ym

=

6,

14x-3

'dB'

-z-' - I, ~';

dpi

39. dq ~ 7;q

I

~

,

4

6

,

10

~ 1.-' d'PII zq--4 - 5q-<

+ 7;q-' + q-', dq' ~ 7; -

t(sec)

0 -1

-,-,

1 2 3 4 S

7 8 9 10 ~

~

-4

17. (a) 190 ft/sec (b) 2 sec (0) 8 sec, 0 ft/sec (d) 10.8 sec, 90 ft/sec (0) 2.8 sec (f) Greatest acceleration happeos 2 sec after lauoch (g) Constant acceleration between 2 and 10.8 sec, -32 ft/sec' 19. (a)

~ 38--4 d'r ~ -12e-'

~

d,

1

29. y' ~ 2x' - 3x - I,y' ~ 6>;' - 3,ym ~ 12x,y(4) ~ 12, 31. y'

dv a=-

~ sec, 280 em/sec

(b) 560 em/sec, 980 cm/sec'

(0) 29.75 flashes/sec 21. C ~ position, A ~ velocity, B ~ acceleration 23. (a) $1l0/machine (b) $80 (0) $79.90 25. (a) b'(O) ~ 104 hacteria/b (b) b'(5) ~ 0 bacteriajb (0) b'(IO) ~ -104 bacteria/b

Chapter 3: Answers to Odd-Numbered Exercises dy I 27. (a) dl ~ IT

-

1

(0) The

dt is 0 m/h when ~ 12 and the smallest value of dt is -I m/h when I ~ O.

(b) The largest value of

(e)

I

o~ect cbanges direction at I ~

A-ll

6 ± 3V15 sec.

. speeds up on (6-V15 (d) The o~ect 3 , 2 ) U (6+V15 3 ,4

1

and slows down on [0, 6 - 3V15) U ( 2, 6 + 3V15) .

Y

~ a and I ~ 4 when it is . 7 ' / dl movmg umts sec an s owest at t = 6±V15 3 sec.

(0) The o~ect is moving fastest at I

(f) When I

~ 6 + 3V15 the object is at position s '" -6.303

units and fartlrest from the origin.

Section 3.S, pp. 139-142 29. I

~

25 sec

1. -10 - 3 sin x

D ~ 6250 m

9

5. -cscxcotx -

31.

11.

-csc' x (1 + ootx)'

17. 3x 2 sinxcosx

3. 2xcosx -x 2 sinx

~

7. sinxsec2 x + sinx

13. 4 tan x sec x - csc2x +X

3

2 COS X

9.0

15. x 2 cosx

-x 3 sin2 x

19. sec'l - I

21. -2csclcotl 23. -0 (OoosO + 2 sinO) (I - CSCI)' 25. sec 0 cscO (tan 0 - cot 0) ~ sec2 0 - csc2 0

(a) v ~ 0 when I ~ 6.25 sec. (b) v > 0 when 0 :5 t < 6.25 => tire object moves up; v < 0 when 6.25 < I :5 12.5 => the object moves down. (e) The object changes direction at t ~ 6.25 sec. (d) The o~ect speeds up on (6.25, 12.5] and slows down on [0,6.25). (e) The object is moving fastest at the endpoints I ~ 0 and I ~ 12.5 when it is traveling 200 ft./sec. It's moving slowest at t ~ 6.25 when the speed is O. (f) When I ~ 6.25 the object is 8 ~ 625 m from the origin and farthest away. 33.

(a) v

~

0 when I

(b)v
6+

;'15

~

0 when 0 ::;; t

<

6 - V15 3 or

< t :5 4 => the object moves right.

Y-'

., - -::1:-'11"

37.

39. Yes, at x

6± V15 3 sec

6-V15 6+V15 . 3 <1< 3 =>theobJect

>

27. sec' q 29. sec' q q3cos q - q2 sinq - qcosq - sinq 31. 2 2 (q - I) 33. (a) 2 csc' x - cscx (b) 2 sec' x - secx 35. Y

~ 'If

41. No

A-12

Chapter 3: Answers to Odd·Numbered Exercises

41. 2'1fsin ('1f1 - 2)eos('1f1 - 2)

43.

45. 101 10 tan' I seO' 1+ 101' tan lO I

8 sin (21) , (I + cos 21) -31' (1 2 + 4) (I' _ 41)'

47.

49. -2 cos (cos (21 - 5» (sin (21 - 5»

53.

57. 3(21 2 - 5)'(181 2 - 5)

45. (a) y ~ -x + '1f/2 + 2 (b) y ~ 4 - V3 47. a 49. 0/2 51- -I 53. 0 55. - Yz m/see, Yz m/see, Yz rnjseO', Yz rnjsee' 57. c ~ 9 59. sinx 61. (a) i) 10 em ii) 5 em iii) -5V2 '" -7.1 em (b) i) 0 em/sec ii) -50 '" -8.7 crnfsec iii) -5V2 '" -7.1 crnfsec

Section 3.6, pp. 147-149 1. Iz,,' 3. 3 cos (3x + I) 7. 10seO'(1(1< - 5) 9. Witb u

~ (z" +

I),y

5. -sin (sinx) eou

~ u': : ~

iu: ~

5u'· 2 ~

~

( I - (x/ 7» ,y

~

u

-7

dy : fix

~

dy du du fix

dy dydu 13. Witb u ~ «x 2/8) + x - (I/x»,y ~ u': fix ~ du fix ~

(~ +

Section 3.7, pp. 153-155

I + :2)

~ 4(~2 +

X -

})'(~ +

11.

13.

I + :2)

(seeutanu)(seO'x) ~ see(tanx)tan(tanx)see2x

u I

2~

21.

~

u':: ~ iu: ~

#(cos 31 -

sin 51)

3u 2 eosx 23

21.y'

~

esc 6 • cotO + esc 8

25. 2xsin4 x + 4x 2 sin3 xcosx + cos-2 x + 2xcos-3 xsinx (4x + 3)'(4x + 7) 27. (3x - 2)' ( I I 29. (x + I )' xl 4-z,,2

)2

33. xsecxtanx + seCX' 2V7 + neex

35.

2sin6 2 (I + cos 6)

39. (2(::

37. -2 sin (6') sin 26 + 26 cos (26) cos (6')

1~'j2) cos (~)

x

25. -2 29. (a) y 31. (a) y 33. (a)

+I

y2 - (x + 1)2 y3

"

vY I , y' ~ -~--vY + I 2(vY + I)' 27. (-2, I): m ~ -I, (-2, -I):m 7 I 4 29 ~4x ~

-

2

3x + 6

(b) y

6

6

+ '1f

~

'1f -2x

37. (a) y

~

2'1fx - 2'1f

~

~

I

-'[x+7

I

8

(b)Y~-3x+3

y~'[x+'[

35. (a) y

(b) y ~

39. Points: (- Y?, 0) and 41.

YSin(~) _ eos~) + xy

-r

~-y-,y

23. y' ~

9. cos2 y

8

17.

dy du 15. Witb u ~ tanx,y ~ secu: fix ~ du fix ~

7 I 'y(x+1)2 _y2

-cos2 (xy) - y x

dy

17. Witb ~ sinx,y 3 sin2X (eou)

1- 2y

2

X + 2xy -z,,' + J>;2y - xy2 + x 5. x'y-x'+y

-7U·,·(-t) ~ (I-tf 4u'·

(f) Yz/24 (g) 5/32 (b) -5/(3Vt7) 75. 5 (b) I 79. y ~ I - 4x 77. (a) I 81. (a) y ~ '1fX + 2 - '1f (b) '1f/2 83. It multiplies tbe velocity, aeceleratioo, and jerk by 2, 4, and 8, respectively. 2 4 2 85. v(6) ~ Sm/see, a(6) ~ - 125 m/sec

1.

~

+~)

59. ;, (I + })(I

61. 2esO'(J>; - l)eot(J>; - I) 63. 16(z" + 1)2(5x + I) 65. 5/2 67. -'1f/4 69. a 71.-5 73. (a) 2/3 (b) 2'1f + 5 (c) 15 - 8'1f (d) 37/6 (.)-1

_2xy _ y2

10(z" + I)' 11. Witb u

55. 6 tan (sin' I) sec2 (sin' I) sin2 I cos I

I sin (12) v'1 + cos (12)

-67 x -67 2

2

(b) Y ~ 'if x - 'if

'1f

+2

x I (b) Y ~ - 2'1f + 2'1f

(Y?, 0), Slope:

-2

m~-Iat (~,~). m~V3at (~,t)

Chapter 3: Answers to Odd-Numbered Exercises

43. (-3,2):m

(3, -2): m 45. (3, -I) dy y' 51. dx

= _2;; (-3, -2): m = 2;; (3,2): m = 2;;

17. (3x

27

= - If

+ 2xy

= x2 + 3xy2'

=

13. (a)

dx

x2

+ 3xy2

dx

dy

y' + 2xy'

dy

21. I dy/dx

1T72 ~;

(b)

dV 2dh (0) dt = 1T7 dt

15. (a) I volt/sec

dR

(0) dt =

+


7. -3/2 (b) -135m'/min

21Th7:

dr 21Th7 dt

I (b) - 3 amp/sec

dI)

(d) 3/2 ohms/sec, R is ioereasing. ds x dx 17. (a) dt = v'x 2 + y2 dt

(b) ds = dt

dx ydy (0) - = - - dt x dt

I d8 dA 19. (a) dt = 2:ab cos 8 dt

29. (a)

2~~ m/min

(0) :

=

(b) -59.5 ft'/sec

11.19 em/min

= -

(b)

(0) -I rad/sec

~=

14.92 cm/min

- 2Vx) cot(l - 2Vx» dx

1. 5x 4 - 0.25x + 0.25 3. 3x(x - 2) 5. 2(x + 1)(2x 2 + 4x + I) 7. 3(11'- + sec 8 + 1)2(28 + sec 8 tao 8)

I

11.2sec'xtaox

. 2v't( I + v't)2

13. 8 cos' (I - 2t) sio (I - 2t) 15. 5(sec t)(sec t

+ tao t)' 19. cos V20 • v28

=

17.

8 cos 8 + sio 8 • r.c::-;-: v28sio8

21.

xcsc(~) + csc(~)cot(~)

23. tx1/2sec(2x)'[ 16tao(2x)' - x-2 ] 25. -IOxcsc'(x2 ) 27. 8x'sio(2x2) cos (2x 2 ) + 2xsio2 (2x 2) -(t + I) I - x -I 29. 8t' 31. (x + I)' 33. ( )1/2 2x 2 I + 1 x

(b) 7 = v'26y - y2 m

28~1T m/min

31. 1ft/min, 401T ft'/min 33. 11 ft/sec 35. Jncreasiog at 466/1681 L/min2 37. -5 m/sec 39. -1500 ft,Isec

41

~ (csc (l

(a) Al (b) A (c) .01 (a) .231 (b).2 (0) .031 (a) -1/3 (b) -2/5 (0) 1/15 35. dV = 417702 d7 tiS = 12xo dx 39. dV = 21T70h d7 (a) 0.081T m2 (b) 2% 43. dV'" 565.5 io' I 45. (a) 2% (b) 4% 47. 3% 49. 3%

(0) -14/13 em/sec, deereasiog

~~

2vx

29. 31. 33. 37. 41.

9

dAI d81.da (b) dt = 2:ab cos 8 dt + 2: bsm8 dt dA I ~ I . • I . • (0) dt = 2:abcos8 dt + 2: bsm8 dt + 2: asm8 dt 21. (a) 14 em2/sec, iocreasiog (b) 0 em/sec, coostaot

27. (a)

.5 r cos (5Vx) dx

23.

dx

Practice Exercises, pp. 176-180

x dx + Y dy v'x2 + y2 dt v'x2 + y2 dt

23. (a) -12 ft/sec 25. 20 ft/sec

+x

51. The ratio equals 37.87, so a change io the acceleration of gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth.

I (dV V dt - I dt

I

• r 3 vy

25. (4x2)sec'(X;)dx 27.

3. 10 5. -6 11. (a) -180m2/min


2 - 2x 2 19. ( 2\2 dx I +x I

.3 r) dx

-

2vx

d7 21T7 dt

9. 31/13

2

(b) 1.003

I-y

Section 3.8, pp. 160-163 dA 1. dt

15. (a) 1.01

35.

5. / . 10. 2/ . • 721Tm. tnlD,Tm mm

-2 sio8 2 (cos8-1)

37.

3~

-3x 2

y+2 41. - x + 3

43.

43. (a) -32/v'i3 '" -8.875 ft/sec (0) d8J/dt = 1/6 rad/sec, d821dt = -1/6 rad/sec

d7 51. ds = (27 - l)(tan2s)

7. 2x

3. L(x) = 2 I 4 11. 12 x + 3

lOx - 13

9. -5

+ 1)2

d'y

Section 3.9, pp. 173-175

=

I

5. L(x)

=x

- 1T

13. 1(0) = 1. Also, I'(x) = k (l + x)k-l, so 1'(0) = k. This means the linearization at x = 0 is L(x) = I + kx.

53. (a) dx 2 =

4y

dp 49. dq

47.

2y (x

-

39. -9[

+2

45.

4x _ 4y 1/'

(b) d8J/dt = 8/65 rad/sec, d821dt = -8/65 rad/sec

1. L(x)

A-13

6q-4p + 4q

d'y (b)

L2

_2xy2 - I =

4 ,

""'

57. 0

59.

v'3

(c) 5/12 61.

-xY

3p2

-2xy' - 2x4 y'

55. (a) 7 (b) -2 (g) 3/4

5x + cos2x (5x 2 + sio2x)5/2

-t

X

(d) 1/4

63.

Y

(e) 12

-2 (2t

+ I)

2

(f) 9/2

1

A-14

Chapter 3: Answers to Odd-Numbered Exercises

65_
67_
,

117.
(b) Yes

1r - 2

+ -2-

(0) No

!(x).I ~,-IS.x

(0) Yes

(H) G,-t) and

71- (-I, 27) and (2, 0)

(b) (0,20),(1,7)

73_
,

-1

, - -ix-j-l 119. L(x) ~ 1.5x

123.
79_ 4

I

81- Taogent: y ~ -'Ix

9

+ 4' normal:

y ~ 4x - 2

83_ Tangent: y

~ 2x -

4, normal: y

~

-tx i

85_ Taogent: y

~ - ~x + 6, normal:

y

~ ~x

87_ (I, I): m

89_ B

~

I -2:; (I, -I):

graph of t, A

~

,

91-

~

121. tIS ~

+ 0.5

(b) 8%

1rrho

dh

\lr' + ho'

(0) 12%

Additional and Advanced Exercises, pp. 180-182 1.
+

1

- 15

mnotdefined

graph off'

+

cos 6 (2 cos 6); 2 cos 26 ~ -2 sin' 6 + 2 cos' 6; cos 26 ~ cos2 (} - 8in2 (J (b) cos26 ~ cos'6 - sin'6; -2 sin 26 ~ 2 cos 6 (-sin 6) - 2 sin 6 (cos 6); sin26 ~ cos 8 sin 8 + sin 8 cos 8; sin 28 = 2 sinO cos 0 I 3.
5. h

9 2:' a

~

5Vs -2-

7.
93_
tIS

(0) dt ~ (41rr

dh

dr

11.
+ 21rh) dt + 21rr dt

15.
dr r dh
~ ~h 5

109_ 0.02 ohm/sec

(b) -

111. 22 mfsec

125 ft/min 14411"

~ kmJsec or 600 m/sec

o

(b)~ rpm

~

17.
b

-'if

3

4' b

(b) 12.31 sec, 393.85 ft

(b) m ~

9

4

~

-I,b

~

1r

19. todd=> f' is eveu

23. h' is defmed but not continuous at x = 0; k' is def'med and continuous at x = O. 27.
A-1S

Chapter 4: Answers to Odd-Numbered Exercises 29. Absolute maximum: 2;

CHAPTER 4

31. Absolute maximum: I; absolute minimum: -1

absolute minimum: 0

,

Section 4.1, pp. 189-191

,

(1/'12, 1) Abl IIWt

1. Absolute minimum at x = Cz; absolute maximum. at x = b 3. Absolute maximum at x = c; no absolute minimum 5. Absolute minimum at x = a; absolute maximum at x = c

7. No absolute minimum; no absolute maximum 9. Absolute maximum at (0,5) 11. (c) 13. (d) 15. Absolute minimum at x = 0; no absolute maximum

--~--~-+--~----->%

17. Absolute maximum at x = 2; no absolute

minimum

(-2, 0) -1 Ab. min

-I

0

(-11'/2,-1) Absmin

-1

33. Absolute maximum: 2tV?'; absolute minimum: 1

,

Y - A'(.x)

,

Abllmax

1.0 0.8

,

AbIIIDIIX

(013. 2I"'l (2<13. 2I"'l ~

12 0.6

35. Absolute maximum: 2; absolute minimum: -I (0,2) Aba

y=esc.% (7(/2, 1) '11"/3 :S x:!li: 211"/3 Ab. min

OA 02

19. Absolute maximum at x x ~ 3'1T/2

~

-cO+------m~'~-.~n~2~.13~%

'1T/2; absolute minimum at

,

37. Increasing on (0,8), decreasing on (-1,0); absolute maximum: 16 at x ~ 8; absolute minimum: 0 at x ~ a 39. Increasing on ( - 32, I); absolute maximum: I at 8 ~ I; absolute minimum: -8 at 8 ~ -32 41. x

~

3

43. x= l,x=4 45. x ~ I 47. x ~ 0 and x ~ 4 21. Absolute maximum: - 3 ; absolute minimum: -19/3

,

-2 -I 0

23. Absolute maximum: 3; absolute minimum: -I

,

1

2

,

(2, 3) AbII

%

-, (-2,-19/3) Abo min

-2:Sx:S3

x~O

-7

~: t__l;':~:<2 -, (O.s,-4)

Absmin

61.

27. Absolute maximum: 2; absolute minimum: -1

,

-c+--L~----------->%

~

(~, - i~ )

I.

Critical point Derivative Extremum or endpoint 4 x =-5

min

-4

53. Minimum value is 0 at x ~ -I and x 55. There is a local minimum at (0, I).

, . jx-s

,

2.

51. Local maximum at (-2, 17); local minimum at

59.

(3,-3)

25. Absolute maximum: -0.25; absolute minimum: -4

~

"7• lYli1A.I.UJ.um ,,--~ value'IS "2I at x = I ' . val' I at x ;, ; tl1lDllD.Utn De 18 -"2

-I

-2

49. Minimum value is I at x

2

y='l,fx

-I. '<:.: %: ,<: ,'_---- (8,2) Aba ~

-

--1~~1~2~,-74-'~6~7~'~% -I

(-1,-1) Absmin

~

-2

~-0 ~

0

x~2

63.

Critical point or endpoint

x

~

Local max

Undermed Local min

Critical point or endpoint

x x x

a

I

= - I.

VlIlue 12 101/3 '" I 034 25 . 0

Derivative

Extremum

VlIlue

Undermed 0 0 Undermed

Local max Minimum Maximum Local min

0 -2 2 0

Derivative

Extremum

VlIlue

Undermed

Minimum

2

A-16 65_

Chapter 4: Answers to Odd-Numbered Exercises

Critical point or endpoint x x x

~

~ ~

-I I 3

Derivative

EItremum

Volu. 5 I 5

a

Maximum

UndeImed

Local min

a

Maximum

51.

67_ (a) No (b) The derivative is deIfied and nonzero for x '" 2. Also, f(2) ~ Oandf(x) > a for all x '" 2. (0) No, because (-00,00) is not a closed interval. (d) The answers are the same as parts (a) and (b) with 2 replaced by a. 69. Yes 71. g assumes a local maximum at -c. 73. (a) Maximum value is 144 atx ~ 2. (b) The largest volume of the box is 144 cubic uoits, and it occurs when x = 2. v0

2

55.

8.5 'C/sec ~ T'(lo), the rate at which the temperature was changing at I = 10 as measured by the rising mercury on the thennometer. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. The conclusion of the Mean Value Theorem yields I I b - a = _~~c2(a = b~c = Viib. b - a c2 ab f(x) must be zero at least once between a and b by the Iutermediate Value Theorem. Now suppose that f(x) is zero twice between a and b. Then, by the Mean Value Theorem, I' (x) would have to be zero at least once between the two zeros of f(x), but this can't be true since we are given that I'(x) '" 0 on this interval. Therefore, f(x) is zero once and only once between a and b. 1.09999 '" f(O.1) '" 1.1

-b) a_

59.

69.

+ 80

75. 2g

77. Maximum value is 11 at x = 5; minimum value is 5 on the interval [-3, 2]; local maximum at (-5, 9). 79. Maximum value is 5 on the interval [3, (0); minimum value is -5 on the interval (-00, -2].

Section 4.2, pp. 196-198 1. 1/2

3. I

5. t(1

+

v'7) '" 1.22,t(1

v'7) '" -0.549

-

7. Does not; f is not differentiable at the interior domain point x ~ O. 9. Does 11. Does not; f is not differentiable at x ~ -1. 15. (a) i) • • •

o

-2

2

-,• • •

ii)

-4

-3

•1 •0

iii)

•

iv)

•4

0

29. (a) 4

27. Yes x2

31. (.)

T +C

33. (a)

xI + C

(b) 3

•

(0) -tcoS21

x

I

+X+

+C

•

24

18

.. ..

(0) 3 x4

(b)~ + C (b) x

-

•

9

3

35. (a) -tcoS21

37. f(x) = x 2

•2

.. ..

(0)

4 +

(b) 2 sin

I+C

xI + C

39. r(6) = 86 43. 8 =

+ cot6

- 21T - I

I - cos (111) 1T

47.8= sin(21) - 3

49. If T(I) is the temperature of the thermometer at time I, then T(O) = -19'CandT(14) = 100 'C. From the MeanValue Theorem, there exists a 0

<

10

and

e:,

4:}decreaSing oo

<

14 such that

T(l4) - T(O)

14 _ 0

x =

(O,T). (T'

2:).

(~,21T)

(c) Local maximum. at x

+ 2Sinf + C

41. 8 = 4.912 + 51 + 10 45. 8 = 1612 + 201 + 5

1. (.) 0, I (b) Iucreasing on (-00,0) and (1,00); decreasing on (0, I) (0) Local maximum at x ~ 0; local minimum at x = I 3. (.) -2, I (b) Iucreasing on (-2, 1) and (1,00); decreasing on (-00, -2) (0) No local maximum; local minimum atx =-2 5. (a) -2, I, 3 (b) Iucreasing on (-2, I) and (3,00); decreasing on (-00, -2) and (1,3) (0) Localmaximumatx = I; local minima atx = -2,3 7. (a) 0,-1 (b) Iucreasing on (-00, -2) and (1,00); decreasing on (-2,0) and (0, I) (0) Local minimum atx = I 9. (a) -2,2 (b) Iucreasing on (-00, -2) and (2,00); decreasing on (-2,0) and (0, 2) (0) Localmaximumatx ~ -2; local minimum atx = 2 11. (.) -2,0 (b) Iucreasing on (-00, -2) and (0,00); decreasing 00 (-2,0) (0) Localmaximumatx ~ -2; local minimum atx = a 71' 2'I'T 4'I'T 13. (a) :f'~'~ (b) Iucreasingou

C

(0) 5x -

C

Section 4.3, pp. 201-203

21T

~andx

=

41
= 21T

15. (.) Iucreasing on (-2,0) and (2, 4); decreasing on (-4, -2) and (0, 2) (b) Absolute maximum at (-4, 2); local maximum at (0, I) and (4, -I); absolute minimum at (2, -3); local minimum at (-2, 0) 17. (a) Iucreasing on (-4, -I), (1/2, 2), and (2, 4); decreasing on (-1,1/2)

A-17

Chapter 4: Answers to Odd-Numbered Exercises (b) Absolute maximum at (4, 3); local maximum at (-I, 2) and (2, I); no absolute minimum; local minimum at (-4, -I) and (1/2, -I) 19. (a) Increasing on (-00, -1.5); decreasing on (-1.5, 00) (b) Local maximum: 5.25 atl = -1.5; absolute maximum: 5.25 atl = -1.5 21. (a) Decreasing on (-00,0); increasing on (0, 4/3); decreasing on (4/3,00) (b) Local minimum atx = o (0, 0); local maximum at

23. (a) (b)

25. (a) (b) 27. (a) (b)

29. (a) (b)

31. (a) (b)

x = 4/3 (4/3, 32/27); no absolute extrema Decreasing on (-00,0); increasing on (0, 1/2); decreasing on (1/2,00) Local minimum at 6 = 0 (0, 0); local maximum at 6 = 1/2 (1/2, 1/4); no absolute extrema Increasing on (-00,00); never decreasing No local extrema; no absolute extrema Increasing on (-2,0) and (2,00); decreasing on (-00, -2) and (0, 2) Local maximum: 16 at x = 0; local minimum: 0 at x = ±2; no absolute maximum; absolute minimum: 0 at x = ±2 Increasing on (-00, -I); decreasing on (-1,0); increasing on (0, I); decreasing on (I, 00) Local maximum: 0.5 atx = ±I; local minimum: 0 atx = 0; absolute maximum: 1/2 at x = ± I; no absolute minimum Increasing on (10, 00); decreasing on (I, 10) Local maximum: I atx = I; local minimum: -8 atx = 10; absolute minimum: -8 at x = 10

33. (a) Decreasing on (-2V2, -2); increasing on (-2,2); (b)

35. (a)

(b)

37. (a)

decreasing on (2, 2V2) Localminims:g(-2) = -4,g(2V2) = 0; local maxima: g(-2V2) = 0,g(2) = 4; absolute maximum: 4atx = 2; absolute minimum.: -4 at x = -2 Increasing on (-00, I); decreasing when I < x < 2, decreasing when 2 < x < 3; discontinuous at x = 2; increasing on (3,00) Local minimum atx = 3 (3,6); local maximum at x = I (1,2); no absolute extrema Increasing on (-2,0) and (0,00); decreasing on (-00, -2)

(b) Local minimum: -6V'2atx = -2; no absolute maximum; absolute minimum: -6V'2 atx = -2

39. (a) Increasing on (-00, -2/V7) and (2/v7, 00); decreasing on (-2/v7, 0) and (0, 2/v7) (b) Local maximum: 24V'2/77/ 6 '" 3.12atx = -2/v7;local minimum: -24V'2/77/6 '" -3.12 atx = 2/v7; no

absolute extrema 41. (a) (b) 43. (a) (b) 45. (a) (b) 47. (a) (b) 49. (a)

Local maximum: I at x = I; local minimum: 0 at x = 2 Absolute maximum: I at x = I; no absolute minimum Local maximom: I at x = I; local minimom: 0 at x = 2 No absolute maximum; absolute minimum: 0 atx = 2 Local maxima: -9 at t = -3 and 16 at I = 2; local minimum: -16 at I = -2 Absolute maximum: 16 at I = 2; no absolute minimum Local minimum: 0 at x = 0 No absolute maximum; absolute minimum: 0 atx = 0 Local maximum: 5 at x = 0; local minimum: 0 at x = -5 and x = 5

(b) Absolute maximum: 5 at x = 0; absolute minimum.: 0 at x = -5 and x 51. (a) Local maximum: 2 at x = 0.;

5

=

local minimum:.j3 at x = 2 - V:J 4 3 - 6 (b) No absolute maximum; an absolute minimum at x=2-V:J 53. (a) Local maximum: I atx = ",/4;

local maximum: 0 at x = '7T; local minimum: 0 at x = 0; local minimum: -I atx = 3",/4 55. Local maximum: 2 at x = ",/6; local maximum: V:J at x = 2",; local minimum: -2 atx = 7",/6; local minimum: V:J at x = 0 57. (a) Local minimum: (",/3) - V:J atx = 2",/3; local maximum: 0 at x = 0; local maximum: '7T at x = 2'17" 59. (a) Local minimum: a at x = ",/4 61. Local maximom: 3 at 6 = 0; local minimum: - 3 at 6 = 2", 63.

,16 ,=,(;<)

I

/ \

l¥ V~' ' -~

,I

,-f<» ,

0'(;<)

I

b' ~,(;<)

,I

\,

01010101

(a)

(b)

65. (a)

(b)

,

o

(e)

,

o

2

(d)

2

69. a = -2, b = 4

Section 4.4, pp. 211-213 1. Local maximum: 3/2atx = -I; local minimum: -3 atx = 2; point of inflection at (1/2, -3/4); rising on (-00, -I) and (2,00); falling on (-1,2); concave up on (1/2,00); concave down on (-00, 1/2) 3. Local maximum: 3/4 at x = 0; local minimum: a at x = ± I; points of inflection at (-V:J,

3f)

and (V:J,

3f}

rising on (-1,0) and (1,00); falling on (-00, -I) and (0, I); concave up on (-00, -V:J) and (V:J, 00); concave down on (-V:J, V:J) 5. Localmaxima:

-i'" + ~

atx = -2",/3,f

x = ",/3; local minims: - f -

~ atx =

+

~ at

-",/3, 2; -

~

at x = 2",/3; points of inflection at( -",/2, -",/2), (0, 0), and (",/2, ",/2), rising on (-",/3, ",/3); falling on (-2",/3, -",/3)

A-1S

Chapter 4: Answers to Odd-Numbered Exercises

,

and ('1f/3, 2'1f/3), concave up 00 (-'1f/ 2, 0) and ('1f/2, 2'1f/3); coocave down on (-2'1f/3, -'1f/2)and (0, '1f/2) 7. Local maxima: 1 at x ~ -'1f/2 and x ~ '1f/2, a at x ~ - 2'1f and x ~ 211'; local minima: -I atx ~ -3'1f/2andx ~ 3'1f/2, oat x ~ 0; points ofinflectioo at (-'1f, 0) and ('1f, 0); rising on (-3'1f/2, -'1f/2), (0, '1f/2), and (3'1f/2, 2'1f); falling on ( -2'1f, - 3'1f/2), (-'1f/2, 0), and ('1f/2, 3'1f/2); coocave up 00 ( -2'1f, -'1f) and ('1f, 2'1f); concave down 00 (-'1f, 0) and (0, '1f)

,

11.

Absmax (m'4,112)

'1O',M~~~~~~~7.-~.~~'

"" ....

(3m'4,-112) Ab!min

-1

,

,

,

31.

,---'{T.i

2

2

~~~~~~L-~~-L~,

-4-3-2-

\1234 I -2

-2

(2,-1)

-2

Absmin

,

,

,

,

4

,

(2, S) Lac max

2

InfI

(1,-1)

"" ....

2

(1,1)

-~~-_1L-O~~~~~--~4->'

-c~L--_~,--~IO~,~~~'--~~~~'

c...

-1

,"",min

(0,-3)

"" ....

,

,

,

,

17.

(0,0) !nil

(0. 4) Abs max

4

""(-2".0)

--~~----+-----"--~x

--~~~~-7~2'~~,,~,o,~~x

(4,0)

(-4. 0) AlII min

Ab! min

""min

23. ,

,

41.

y_

---;0;f-'==~.--------;2~.~X

25.

, S

Infl /

51rl3,S-fJ"'lT13-1) Loc min

(3'1112. 3..J3'IT12)

(./2. ~"/21 --'0~~~--.L-~~~2.L-->, (0,-2) Ab!min

x-2

4

V

: (3. 6) Loc min

:(1. 2) Loc max

-.-. -s

10

•

~-3

•

43.

,

Ab._ {2"'. "'I (2,2)

r -......."'"

Chapter 4: Answers to Odd-Numbered Exercises

,

45.

,

47.

, . 'f.;l

A-19

~ -~ (x + 1)-'/3

65. y"

--~4--~3--~2~-~'~~~'~2~3~4~' c.., Abo min

-2

(-1,0)

49. y"

=

2

(1,0)

_min

,

Abomin

1 -2>;

51. y"

~

3(x - 3)(x - 1)

xO

-2 69. y" = { 2,'

""-

,

71.

""min ------~-.f_-I------>,

x=-1

53. y" ~ 3(x - 2)(x

\

+ 2)

""-

~ InflX_2/

~~'-2

~

x- -2..J3

73.

x=2..J3

55. y" ~ 4(4 - x)(5x 2

-

,

16x + 8)

""x '" 8/5

75.

77.

"~,

1 ese2 6 59• Y • -- -2 2' 0<6<2".

57. y" = 2 sec2 x tan x

8='It

. - -....... Absmu

),=2

61. y' ~ 2tan6see2 6,

IL

3

: I,

--2

---~

,=,""U-I

4

~-I

", ,

--:-~-------

,

1

\:,=x-l

-f < 6 < f 79.

,

,

81.

,___,_

y _ _ x2 _2 ;1;2_1

1

" ' - xl-I:: ,,

,2-1

,

. ~~ ... '"

, , ,, ,

,

, . -1

63. y" = -sin t. 0 ::s t ::s 21T x - -1

, _ 1

, . 1

A-20

Chapter 4: Answers to Odd·Numbered Exercises

,

83.

115. The zeros of y' = 0 and y" = 0 are extrema and points of inflection, respectively. Inflection at x ~ local maximum atx = -2; local minimum atx = O.

,

x - -l

V'2;

y _ ;'-x+ 1 ,-1

,

87.

,

89.

Section 4.5, pp. 219-225

x=-1

x =-2

1. 16 in., 4 in. by4in. 3. (a) (x, I - x) (b) A(x) ~ 2x(1 - x) (0)

,, , :.1:=1

2

square units, I by

14 X 35 X 1·m 2450.m3 7. 80,000 m',' 400 m by 200 m ' 3 3 3 "27 9. (a) The optimum dimensions of the tank are 10 ft on the base edges and 5 ft deep. (b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel walls would likely be determined by other considerations such as s1ructural requirements.

y=Blif+4)

11.9XI8in.

Point p

y' -

y"

+ +

a

Q R

a

s

-

T

,

95.

+

7

-

4

-3

~

X - 3.3944487 Y - 1309.9547

(0) Maximum volume'" 1309.95 in3 when x '" 3.39 in. (d) V'(x) ~ 24x' + 864, so the critical point is at

336x

x ~ 7 which confirms the result in part (c). (e) x ~ 2 in. or x ~ 5 in. 19. '" 2418.40 em3 21. (a) h ~ 24,w ~ 18 (b) v (24. 10368)

V'i3,

AbsIlWt 10000

sooo

109. -1,2

111. a

(b) Domain: (0, 9)

-------O~-+'--4~~6--~'

(d) Positive: 5 '" t '" 7, 13 '" t '" 15; negative: 0 ~ t ~ 5, 7 ~ t ~ 13 99. '" 60 thousand units 101. Localminimumatx ~ 2; inflection points atx ~ I and x ~ 5/3 ~

-

1600 (6.7)

97. (a) Towards origin: 0 '" t < 2 and 6 '" t '" 10; away from origin: 2 '" t '" 6 and 10 '" t '" 15 (b) t ~ 2, t ~ 6, t ~ 10 (0) t ~ 5, t ~ 7, t ~ 13

105. b

15.h:r~8:"

13.f

17. (a) V(x) ~ 2x(24 - 2x)(18 - 2x) v

---------oor-~----~,

93.

t

5

,

91.

t

I, b

~

3, c

113. The zeros of y'

~

~

6000

9 0 aod y'

~

a are extrema and points of

inflection, respectively. Inflection at x x = 0, local minimum at x

,

=

3, local maximum at

4000 2000

o~~,-f.'0~1'~~~~~~~~~~'

= 4.

23. Ifr is the radius of the hemisphere, h the height of the cylinder,

and V the volume, then r 25. (b) x ~

27. Radius

851

~

3V)'/3 and h ~ (3'iTV) '/3 . (8"

(0) L '" 11 in.

~ v2 m, height ~

I m, volume

~

2:; m 3

A-21

Chapter 4: Answers to Odd-Numbered Exercises

29. I

31. 9+

33.

3

2X

2

9~

m, triangle,

3"

b~

9+

35. (a) 16

m, circle

3"

(b)-I

37. (a) v(O) = 96 ft/sec (b) 256 ft att = 3 sec (0) Velocity when s = 0 is v(7) = -128 ft/sec. 41. (a) 6 X 6v3 in. 39. '" 46.87 ft 43. (a) 4v3 X 4V6 in. 45. (a) 10" '" 31.42 em! sec; when I = 0.5 sec, 1.5 sec, 2.5 sec, 3.5 sec; s = 0, acceleration is O. (b) 10 em from rest position; speed is 0 47. (a) s = «(12 - 121)2 + 64( 2)1/2 (b) -12 knots, 8 knots (0) No (e) This limit is the square root of the sums of the squares of the individual speeds. a ka 2 c 49. x=2,v=T 51. 2 +50

4V13.

53. (a)

l':'

(b)

l':'

57.4 X 4 X 3 ft, $288

59.

C M="2

Section 4.7, pp. 236-239

,

1. (a) x 2

x' - x2 + X 3

(b) "3

(0) -

3. (a) x-,

(b)

_1 x -'

-:xI

(b)

-:x

5. (a)

7. (a) W

3 5

3 (0) x-1/3

(b) xl/3

11. (a) cos(=) 13. (a) tanx

(b) The minimwn distance is from the point (3/2,0) to the point (I, I) on the graph of y = Vx, and this occurs at the value x = I, where D(x), the distance squared, has its minimum value.

(b) -3 cos:<

17.

2 x

T

+

X

(t)

(b) 2 tan

15. (a) -cscx

(b)

t csc (5x) 3

+ C

23.

X x -:x1 - "3 - 3 +

27.

~x'/2

31. x 2 +

+

~x4/'

~+

(0)

-~tan (~)

(0) 2 csc ( ' ; )

t2

C

o

39. 3cotx + C

43. 4 sec x - 2tanx + C

f + s~ 41 + C

29. 4y2 -

zVt - ~ +

33.

21.

x4

s.x 2

T - ""2 +

7x + C

3 25. 2x2/' + C

C

+ C

37. -21 cos 3 + C

47.

I (0) -,.cos(=) + cos (3x)

19. t +"4 + C

3

D(x)=x2-lt+~

i

(0) 2W + 2Vx

(b) Vx

9. (a) x 2/3

(0) -ix-' +x2 + 3x (0) 2x +

65. (a) y =-1

" di .v5 67()Th •• e nnrumum stance 18 2 .

y,D(x)

17. ±1.3065629648764, ±0.5411961001462 19. x '" 0.45 21. 0.8192 23. The root is 1.17951. 25. (a) For Xo = - 2 or Xo = -0.8, X, ..... -I as i gets large. (b) For Xo = -0.5 or Xo = 0.25, Xi .... a as i gets large. (0) For Xo = 0.8 or Xo = 2, X, .... I as i gets large. (d) For Xo = - Y2I/7 or Xo = Y2I/7, Newton's method does not converge. The values of x, alternstc between - Y2I17 and Y2I/7 as i increases. 27. Answers will vary with machine speed.

~y'/4 C

+ C 35. -2 sin I + C

I 41. -2cscO + C

I

45. -2cos2x + cotx + C

49. tan 0 + C

51. -cotx -

X

+ C

53. -cos 0 + 0 + C 61. (a) Wrong:

Section 4.6, pp. 228-230 5 13

-3' 21

51 5763

-31' 4945

1.

X2

=

7.

XI>

and all later approximations will equalxo. ,

9.

3.

X2

=

5.

X2

2387 = 2000

x· ,-1-.& -

,x
11. The points of intersection of y = x' and y = 3x + I or y = x 3 - 3x and y = 1 have the same x-values as the roots of part (i) or the solutions of part (iv). 13. 1.165561185 15. (a) Two (b) 0.35003501505249 aod -1.0261731615301

.

xsmx

:ix (x; sinx + C) = ~ sinx + ~2

COSX

=

x2

+ TeosX'

(b) Wrong:

:ix (-x cos:< + C) = -cou + x sinx

(0) Right::ix (-x cos:< + sinx + C) = -cos:< + x sin x +

cosX' = x sin x . A. (2x + I)' ) _ 3(2x + 1)'(2) _ 63. (a) Wrong. dx 3 + C 3 Z(2x + I)' (b) Wrong:

:ix «2x + I)' + C) = 3(2x + 1)'(2) =

6(2x + I)' (0) Right:

:ix «2x +

I)' + C) = 6(2x + 1)2

A-22

Chapter 4: Answers to Odd-Numbered Exercises (a) Local maximum at x = 4; local minimum at x = -4; infIectioo point at x = 0 (b) x=4

Right

67_ (b)

69_ y=x'-7x+ 10

I x' I 71_ Y=-X+T-2

73_ y

=

75_ s

9xl/' + 4

77_ r = cos("O) - I

t + sin t + 4 I I 79_ v = 2 sect + 2 =

83_r=~+2t-2

81_ y=-x'+x'+4x+1 85_ y = x' - 4x' 87_ y 89_ 93. 95_ 97.

=

+

x - -4

(a) Local maximum at x = 0; local minima at x = -I and x = 2; infIectioo points at x = (1 ± v7)/3

5

-sint+ cost + t ' - I

y=2x'/'-50

(b)

y = -sinx - cosx - 2 (a) (i) 33.2 units, (Ii) 33.2 units, (Iii) 33.2 units

t= 88/k,k= 16 99. (a) v = IOt'/' - 6t l /2

(b) True

x=-1

37. (a) Localmaximumatx = -v'2;localminimumatx = v'2; infIectioo points at x = ± I aod 0

(b) s = 4t'/' - 4t'/2

(b)

Practice Exerdses, pp. 240-242 1. No 3. No minimum; absolute maximum: 1(1) 5. Yes,exceptatx = points: x = I aodll/3 7. No 11. (b) One 13. (b) 0.8555996772

=

a

""-

16; critical

19. GlObaiminimumvalueortatx = 2

(b) t = 3,9 21. (a) t = 0,6,12 (d) a < t < 6, 12 < t < 14

23.

""-

91_y=x-x 4/'+t

< t < 12

(c) 6

43. 25.

45. y

y- x+, -J~·L ,

y

x-3

y. x2;l :

x-3:

-H {

:

, ----- i ---

3

,

:

------ I

-~~~~+-~~~~_x

--=~_"'I.Jc-"-:-+:-"',--_x

-, 27.

-4 -3 -2

-v ,'-I

//A~~

y

47.

49. Y

Y

,,

4

,~/

"

2

x=-V3 :,

:x=~ :

-,

-,

y

51. (a) 0,36

(b) 18,18

53. 54 sqUlll"O units

55. Height = 2, radius = v'2

Vs hundred'" 276 tires, 2(5 - Vs) hundred'" 553 tires

57. x = 5 y =

59. Dimensions: base is 6 in. by 12 in., height maximum volume = 144 in3

),=1

----T----

1

31.

~-4

i'= x2-3

,

, ,

~~!:-~ -"-~~~~I~'~'~x

y

:

4

,'~ - ~ :'

29.

1 2 3 4

=

2 in.;

Chapter 5: Answers to Odd-Numbered Exercises

5 x, = 2.195823345 63. "4 + Ix2 - 7x + C 65. 21'/2 - t + C 67. - ~ 5 + C 69. (82 + 1)'/2 + C X4

61.

r

71.

t

75. -

5•

· SID'11' -

. 8m

'" + sm . '"

2

JzC'C

73. 10 tan :0 + C

\128 + C

79. y = x - } - I

77.

kX - .in I + C

81. r = 41'/2 + 41'/2 - 81

=

V32-

2

7• All 0 fthem

9. b

6 4 '

11.

~k

13.

.t-l

(1 + x 4)'/4 + C

3

A-23

17. 19. 21. 29. 31. 33.

~~ i'

15.

k- I

(s) -15 (b) I (s) 55 (b) 385

-56

~ (_1)1+11.

(s)

(b) y

y

,

(2,3)

1. The function is constant on the interval. 3. The extreme points will not be at the end of an open interval. 5. (s) A local minimwn atx = -I; points of inflection at x = 0 and x = 2 (b) A local maximum at x = 0 and local minima

(0) 16

(0) I (d) -11 (0) 3025

23. -73 25. 240 27. 3376 (b) 3500 (0) 2620 (b) en (0) (n 2 - n)/2

(s) 21 (s) 4n

Additional and Advanced Exercises, pp. 243-245

k

k- l

~) _ ;J;2_1.

OSl::S2

O:Sx:S2

Left·....

2

(~')

--!(x) . Xl-I, :

2

finfl' . atx=- I and x= 2 ;pomtso ectionatx= I ± 3v'7 9. No 11. a = I, b = 0, e = I 15. Dril1 the hole at y = h/2. 17. r =

19. (s)

2(H~ R) forH > 2R,r = e-b --ze

(d) e

e+b

(b) -2-

13. Yes RifHS 2R 2

2

(0) b -2be +e +4ae

4e

(0)

y

,

+b+1 2

21. ma = 1 -

I

7i' ml

23. (s) k = -38.72 25. Yce.,y = x

=

(2.3)

-

f~) _ x2_1,

O:Sx:!i:2

I 7j

2

(b) 25 ft

27• Vo = 2\12 3 b'/4

o '" c: "2 -:

"3

"4

,

CHAPTER 5

-1

Section 5.1, pp. 253-255 1. 3. 5. 9. 13.

(s) 0.125 (b) 0.21875 (0) 0.625 (d) 0.46875 (s) 1.066667 (b) 1.283333 (0) 2.666667 (d) 2.083333

0.3125,0.328125 7. 1.5,1.574603 (s) 87 in. (b) 87 in. 11. (s) 3490 ft (b) 3840 ft (s) 74.65 ft/sec (b) 45.28 ft/sec (0) 146.59 ft 31 15. 16 17. I 19. (s) Upper = 758 gal, lower = 543 gal (b) Upper = 2363 gal, lower = 1693 gal (0) '" 31.4h, '" 32.4 h 21. (s) 2 (0) 8

35. (s)

(b) y

f~) - sin::t, -'IT

l y

f(;<) - .mx.

Sx S 1'1"

-'II":SxSw

Left·Jmnd

Right-hand

1~

• (0)

(b) 2\12 '" 2.828

Sin(~:) '" 3.061

(d) Each arca is less than the area of the circle, "'. As n increases, the polygon area approaches 'T1' •

Section 5.2, pp. 261-262 6(1) 6(2) 1'T+1+2+1=7 3. cos(1)", + co.(2)", + cos(3)", + cos(4)", = 0

37. 1.2

39.

t -in - 6~2'

.5.+6n+1 43 . 6 6n 2 '

5 6

45.

1

~ I

41. 12 + 27;": 9, I

I

I + Ii + 2n2 ' I

12

A-24

Chapter 5: Answers to Odd-Numbered Exercises

Section 5.3, pp. 270-274 1. ]"X'dx

7.1

{'

3.1:(x' - 3x) dx

5.

I

J,

I-x dx

0

secxdx

--rr/4

9. 11. 13. 17. 21. 27. 35. 45. 51.

(a) 0 (b) -8 (c) -12 (d) 10 (e) -2 (I) 16 (a) 5 (b) 5V3 (c) -5 (d)-5 (a) 4 (b) -4 15. Area ~ 21 square units Area = 9,,/2 square units 19. Area = 2.5 square units Area ~ 3 square units 23. b'/4 25. b' (a) 2" (b)" 29. 1/2 31. 3".'/2 33. 7/3 1/24 37. 3a'/2 39. b/3 41. -14 43.-2 -7/4 47. 7 49. 0 Using n subintervals ofleogth!u ~ bin aod right-<mdpoint

a'

values: Area

65. (a) True. Since 1 is cootinuous, g is differeotiable by Part I of the Fundameotal Theorem of Calculus. (b) True: g is cootinuous because it is differeotiable. (c) True, sineeg'(I) ~ 1(1) ~ O. (d) False, sinceg"(I) ~ 1'(1) > O. (e) True, sinee g'(I) ~ 0 aodg"(I) ~ 1'(1) > O. (I) False: g" (x) = I' (x) > 0, so g" never cbanges sign. (g) True, since g' (I) ~ 1(1) ~ 0 aod g' (x) ~ I(x) is an increasing func1ioo ofx (because I'(x) > 0).

Section 5.5, pp. 290-291 1.

+C

4)'

13.

= bin and right-<mdpoint

t

11. -6(1 - r')'/2 + C

i

(x'/2 - I) -

15. (a) -

77. J.'/(X)dx '" J.'Odx

~0

1/4

69. b 4/4 - a 4/4

63. c(b - a) 65. b 3/3 - a 3/3 67. 9 71. a ~ 0 aod b ~ I maximize the integral. 73. Upperbound ~ I, lower bound ~ 1/2 75. For example, ].' sin (x') dx '" ].' dx

~

~I

79. Upper bound

13. -,,/4

15. I -

21. -3/4

23.

29. (cosVi)

43. 1/2

f

v'2 -

C~)

35. -tx-'/2sinx 45."

49. d,sincey'

~

7. I

5. 8

fg + I 31. 41'

~ 1/2

11.

33.

a

19. -8/3

25. -I

39. I

37. 0 47.

9. 2V3

17. 2 -4 v'2

27. 16

V1+7

25. tSin'

~

41.28/3

+C

+

C

. (I) +

33. -sm t - I I (I 16

I (x - 1)12 43 . IT

47.

~ (x'

I 31. 2 cos (21 + I) + C sin' (1/8) 2 + C

+C

I + rr(x - I) 11 + C

+ 1)'/2 -

t (x'

+ 1)3/' + C

6

(b)

+ tan3 x + C 6

3

2+tanx

6

2

-I , ,+C 4(x - 4)

49.

+ tan3 x

+C

+C

-3

59. s

59. a. 1'(0) = 70'F, 1'(16) = 76'F 1'(25) = 85'F b. av(T) ~ 75'F 61. 2x - 2 63. -]x + 5

(~: - I)' + C

1)'/2 + C 39. 32(2 - :x

57. s = 41 - 2 sin (21 +

2 55. 3 bh

I 23. 3tan(3x + 2) + C

53. isin Y3(2r - I)' + 6 + C ~

+ C

-~(I - 8')'/4 + C

35. -

C

4I (csc' 28)

I 4 2 45. -g(1 - x)' + 7(1 - x)' - 3(1 - x)' + C

51. b,sineey' = secxaody(O) = ]'°seCldl+ 4 = 4

r 53. y = J, sec 1dl + 3

19.

27.

44 + C + I)

2( I - x3)'/2 41. 27 3

(c) -

J.~ I ~ tdl - 3

(t)

51. (a) - 2

VJ:"

I :xaody(,,)

(b) -

2 ('/2 ) 29. -3cOS x + I + C

Section 5.4, pp. 282-284 3. -10/3

+ C

21. (-2/(1 + Vi» + C

37.

1. 6

sin (2x'/2 - 2) + C

4I (cot' 28)

17. -t(3 - 2x)'/2 55. av(f) ~ 0 57. av(f) ~ -2 59. av(f) ~ I 61. (a) av(g) ~ -1/2 (b) av(g) ~ I (c) av(g)

-t(x' + 5)-3 + C

3.

5. 110 (3x' + 4x)' + C 9. t sec 21 + C

~ ].']X' dx ~ b'

53. Using n subintervals ofleogth!u values:

i (2x +

~ sin (21- f)

~)

55. s

~

t(31' - 1)4 - 5

+ 9

+ 1001+ I

61.6m

57. $9.00

Section 5.6, pp. 297-300 1. (a) 14/3 (b) 2/3 5. (a) 15/16 (b) 0 11. (a) 1/6 (b) 1/2

3. (a) 1/2 (b) -1/2 7. (a) 0 (b) 1/8 9. (a) 4 13. (a) 0 (b) 0 15. 2V3

(b) 0

A-25

Chapter 6: Answers to Odd-Numbered Exercises

19. 3'/2 - I 21. 3 23. 'IT/3 25. 16/3 29. 'IT/2 31. 128/15 33. 4/3 37. 38/3 39. 49/6 41. 32/3 43. 48/5 47. 8 49. 5/3 (There are three intersection points.) 53. 243/8 55. 125/6 57.2 59. 104/15 4 4 63.4 65. 3- 17 67. 'IT/2 61. 56/15

17. 3/4 27. 2'/2 35. 5/6 45. 8/3 51. 18

y

,

_~'-~_I~-+-~-~'~X

69. 2 71. 1/2 73. I 75. (8) (±Vc, c) (b) c ~ 4'/3 77. 11/3 79. 3/4

17. 1/2

Practice Exercises, pp. 301-303 1. (_) About 680 ft

15. 13/3

25. (a) 0 (0) Y

(b) h (feet) 600

23. (b) 'lTr'

(b) -I (0) -'IT (d) x ~ I 2x + 2 - 'IT (I) X ~ -I,x ~ 2 sin4y siny 29 - - - -

(g) [-2'IT,0]

~

27. 2/x

700

21. /.If(x) dx

19. 1/6

. Vy

soo

2Vy

400

300 200

100 ----;;0t-""-.;----7-;---;-~ t

3. (8) -1/2 5.

l'(2x -

9. (_) 4

11. 8/3

(b) 31

~2

1)-1/2 dx

(b) 2

(0) 13

(d) -2'IT

V2 1 23. ~ 32 +-2--

17. 1/6

29. Min: -4, max: 0, area: 27/4

35. Y~ J,X(S~I)dl - 3 39.

t cos (21'/2) + C

53. 27V3/160

61. -I 63.2 65. -2 71. (8) b (b) b 75. 25'F

83.

41.

77. V~2-+-c-o~s"x

-v'\+?

~3 + i

57. V3

7. f(x) ~ ~ x2 + 1

-6

79. - - . 3 +x

9. Y ~ x'

+ 2x - 4

I 2 13'2- 17 y

_~8--_';-~0~~~X

-.

y=-4

5. (a) 2V3 13. (8) s'h

19. 3;'IT

51. 8

49. (a)

16'1T

15

53. (a) V ~

7'IT

6

(b)

117'IT 37. -5-

56'1T

(0)

29. 2'IT

64'IT

15

31. 3'IT

39. 'IT ('IT - 2) (0) ~

51. V

8'IT

T

4'IT 41. T (d)

, , 2a b'1T

I (b) 120'IT m/sec

'lTh'(3a - h) 3

~ 3308 em'

15. 2:;

32'IT (b) -5-

47. (8) 8'IT

(b) 36

23. 'IT

27. 2'IT

15

7. (8) 60

(b) s'h

21. 36'1T

2'IT 35. T

45.

(b) 8

59. 4 - ~ + a

81. Yes

\o/l2

(b)

y

17. 4 - 'IT

57. V

5. (a) 1/4

11. 36/5

11. 10

69.V2-1

67. I

Additional and Advanced Exerdses, pp. 304-307 (b) No

9. 8'IT

43. 8'IT

59. 6V3 - 2'IT

85. Cost '" $10,899 using a lower sum estimate

1. (_) Yes

6 3. 13

--' - 2'IT 33. .,.

+ C 49. I

47.2

1. 16

1 25. 'IT(f + 2V2 - 13 )

37. -4(oosx)l/2 + C

45. 16

55. 'IT/2

21. 9/8

31. 6/5

0' + 6 + sin (26 + 1) + C

43. -

(e) 8/5

19. 18

8~-7

27.

25.4

CHAPTER 6 Section 6.1, pp. 316-319

(d) 0

J~COSIdx ~ 2

7.

(0) -2

15. I

13. 62

(sec)

Section 6.2, pp. 324-326 1. 6'1T

11.

i~

17. 8:;

3. 2'IT

5•

14'1T 3

7. 8'IT

13. (b) 4'IT

15.

19. 4:;

I~

21.

I~; (3V2 +

(d) 241r

(b) 32'IT (e) 60'1T

(0) 28'IT (I) 48'IT

2~'IT

(b) 27'IT

(0) 72'IT

23. (8) 16'1T 25. (8)

27. (8) 6;

2 (b)

7

5'IT 9• 6

5 (0) 2'IT

5)

(d) 108'IT

5

(d) 2'IT

29. (8) About the x-axis: V

~ ~~; about the y-axis: V ~ ~

(b) About the x-axis: V

~ ~~; about the y-axis: V ~ ~

2241r

15

A-26 31_ (a)

Chapter 6: Answers to Odd-Numbered Exercises

5;

(b) 4". 3

(d) 2".

(c) 2".

3

5. (a) 2'lf1'(3 - x ' /2j2VI

4". 15

(b) 7". 30 24". (b) 48". 35_ (a) -55 9". 37- (a) 16 (b) 9". 16 39_ Disk: 2 integrals; washer: 2 integrals; shell: I integral

33_ (a)

41_ (a) 25:".

(b)

+ (I -

3x l/2j2dx

(0) S'" 63.37

(b) ,

,

~". .~--;----;;---"'.~.,

Section 6.3, pp. 330-332 3• 53 6

5 123 • 32

11. (a)

1>11

+ 4x2 dx

13. (a)

/,W,Ii + cos2y t(y

1• 12

15. (a) 17. (a)

1:

V I

[wi'

Jo

+

(y

secxdx

+

7 99 • 8

9 2 •

(b)

~

±X

(0) '" 3.82

0.8

(0) '" 9.29

.I: -!oJ Uintdt

(0) '" 0.55

31. {7 (\03/2 - I)

o '--coO'".l""0~2""0~3""0'",""o~,""0.'".""0.~7... ,

9. 4".Ys

11. 3'lfYs

17. 'If(Vs - 1)/9 23. 253"./20

Jo[wi' (tau) VI

+ sec' xdx

1. 5. 7. 15.

,

25. 31. 35. 39.

(b)

~ VI

27. Order 226.2 liters of each color.

3. 4 em, 0.08 J (b) 905 in.-Ib, 2714 in.-Ib 780 J 9. 72,900 ft-Ib 13. 160 ft-Ib (a) 1,497,600 ft-Ib (b) I hr, 40 min (d) At 62.26Ib/ft': a) 1,494,240 ft-Ib b) I hr, 40 min At 62.59Ib/ft': a) 1,502,160 ft-Ib b) I hr, 40.1 min 400 N/m

+ y-4 t(y

19. 23;611" ft-Ib

85.1 ft-Ib 27. 1684.81b 33. 1164.8 Ib 37. (a) 12,4801b

41. (a) 93.33 Ib

3. (a) 2".1

15. 2'lf

21. (2'lf/3)(2Vz - I)

(a) 7238 Ib/in.

17. 37,306 ft-Ib

2

13. 98'lf/81

19. 35".Ys/3

Section 6.5, pp. 342-346

Section 6.4, pp. 335-337

(b)

,

0.' 1)2 t(y

+ C where C is any real number

1. (a) 2'lf

(0) S'" 2.08

(0) '" 6.13

19. (a) y ~ Vi from (I, I) to (4, 2) (b) Only one. We know the derivative of the function and the value of the function at one value ofx. 21. I 27. Yes, f(x)

/,w/3 (J,' tan t dt) sec y dy

7. (a) 2'lf

21. 7,238,229.47 ft-Ib

98.35 ft-Ib 29. 91.32 in.-oz (a) 6364.8Ib (b) 5990.4lb 1309 Ib (b) 8580lb (0) 9722.3 Ib

(b) 3 ft

wb 43 . 2

45. No. The tank will overflow because the movable end will have moved only 3! ft by the time the tank is full.

(0) S'" 5.02

, 2

Section 6.6, pp. 355-357 1. x ~ O,y ~ 12/5 3. x ~ I,y 5. 9.

13. 7-~~~~~~~' 0.6 0.7 0.8 0.9

15.

~

-3/5

x ~ 16/105,y ~ 8/15 7. x ~ O,y ~ 'If/8 x ~ IS ~ -2/5 11. x ~ Y ~ _2_ 4-". x ~ 3/2,y ~ 1/2 (a) 2~ (b) x ~ 2,y ~ 0

Chapter 7: Answers to Odd-Numbered Exercises

,

(0)

11. D: (0, I]

R: [0,

15. D: [0,6]

R: [0,3]

,

4

A-27

13. D: [-I, I] R: [-,../2,,../2]

00)

y. ---±."

,

0

(2,0)

r

4

y. --±."

-4

35.

= Y = 1/3 21. j = a/3,y = b/3 23. 0- mr x= ,Y =4 j = 1/2,y = 4 29. j = 6/5,y = 8/7 V = 32,.., S = 32v'2,.. 4,..' 37. j = O,y = ~ 39• x- = 0,Y =

41.

v'2,..a 3(4 + 3,..)/6

19. 25 . 27.

33.

j

43.

=

j

~:o

3.,..'

7. (8) 2,.. 9. (a) 8,..

6

4b

3'17"

~,y = ~

~----~----~6~r

19. r'(x)

23. r'(x)

5. 7;;

25. r'(x)

(b),.. (0) 12,../5 (d) 26?r/5 (b) 1088,../15 (0) 512,../15

27. r'(x) range:

11. ,..(3\13 - ,..)/3

13. (a) 16,../15 15. 21. 27. 31. 35. 39.

2~'" It'

(b) 8,../5

0 17. 13

(0) 8,../3

29. r'(x)

(d) 32,../5

range:

33. r'(x) range:

Additional and Advanced Exemses, pp. 359-360

5 -,..'300

11.

j

7. 28/3

<

-00

= -v'C2=1 x + a, where C '"

Y

<

00

~; domain: x>

=

I -

=

-00

0; range: y

+ 13 ; domain: -00 <

x-00 < Y <

35. (8) r'(x) (b)

3. j(x)

= -v;=J 21. r'(x) = ~ = Vx - I = ~; domain: -00 < x < 00; range: -00 < y < = 5-v;=J; domain: -00 < x < 00;

31. r'<x) = 2x

19. 2:5

28,..v'2/3 23. 4,.. 25. 4640 J 10 ft-Ib, 30 ft-Ib 29. 418,208.81 ft-Ib 22,500,.. ft-Ib, 257 sec 33. j = O,y = 8/5 j = 3/2,y = 12/5 37. j = 9/5,y = 11/10 332.81b 41. 2196.481b

~ 1. j(x) = ,,-----..---,..-

,

,

136/6

Practice Exerdses, pp. 357-359 1.

17. (8) Symmetric aboul the liney=x

OO,y

>

0

x < oo,x # I;

=F 2

Vx+l; domain: -I

:5

x

<

00;


I -~

37. (8) r'(x)

=

(b)

,

,

I

9 4h\l3mh • 3

= O,y = 2n ~ I' (0, 1/2)

15. (8) j = Y = 4(a' (b) (2a/,..,2a/,..) 17. "'2329.6Ib

+ ab + b')/(3,..(a + b)) (0) 2,1/2

39. (b)

CHAPTER 7 Section 7.1, pp. 367-369 1. One-lo-one 3. Not one-ro-one 7. Nolone-lo-one 9. One-ro-one

5. One-lo-one

,

(0) -4, -1/4

-~ + ~

00

A-28

Chapter 7: Answers to Odd-Numbered Exercises

(0) Slope off at (I, I) : 3; slope of g at (I, I): 1/3; slope off at

(-I, -I): 3; slope ofg at (-I, -I): 1/3 (d) Y = 0 is tangent to Y = x' at x = 0; x = 0 is tangent to

y=~atx=O. 41. 1/9

65 xW+!

. (x + 1)2/3

43. 3

45. (a) rl(x) =

0+5[1 I ] 63·0cosO 0+5-i1+ tanO

,k x l

(b) The graphofr is 1he line through 1he origin wi1h slope I/m. 47. (a) rl(x) = x - I

[1 + _x_ x + I 3(x + I) (1 + ~) 2

2

X

]

67 l..jx(x - 2) _1__ "3 x 2 +1 x x-2 x 2 +1 69. (a)Max = Oatx = O,min = -In2atx = 1f/3 (b)Max = I atx = I, min = cos (In 2) atx = 1/2 and x = 2

71. In 16 73. 41fln4 75. 1f1n 16 77. (a) 6 + 1n2 (b) 8 + 1n9 79. (a) x '" 1.44,'y '" 0.36 (b) ,

(b) rl(x) = x-h. The graph of r l is a line parallel to 1he graph of f. The graphs of f aod rille 00 opposite sides of 1he line y = x aod are equidistant from 1hat line. (0) Their graphs will be parallel to ooe aoother aod lie 00 opposite sides of1he line y = x equidistant from 1hat line.

85. (b) 0.00469

83. Y = x + Inlxl + 2

51. increasing, therefore ooe-1:<>-ooe; drl/dx = ~x-2/3 53. Decreasing, 1herefore one-to-one; drl/dx = -tx-2/'

Section 7.3, pp. 385-387 1n2

1. (a) 1= -lOln3

Section 7.2, pp. 375-377 1. (a) In3 - 2ln2 (d) tln3

3. (a) In 5 5. I/x

(e) 1n3 + tln2 (b) In (x - 3)

7. 2/1

9. -I/x

27.

3x + 2 2x(x + I)

23. xlnx

43. (ln2j' 45. ~4 49. 1n2 51. 1n27

57. (t))1

19. I -2lnl

25. 2 cos (In 0) 0

VL

41. In 3

13. 2e' cos 0

47. 1n16 + 3 tan II + C 53. 1n(1 + Yx) + C

x+

2x + I 2v'x(x + 1)

111) = 2Vt(/+ 1)'/2

59. v'8+3(sin 0) (2(0

~ 3) + cot 0 )

2 61. 1(1 + 1)(1 + 2)[+ + 111 + 112] = 31 + 61 + 2

15. 20e -0' sin (e-o')

19. 1/(1 + e') 21. e""(1 - I sin I) yeY cos x 2e 2x - cos (x + 3y) 23. (sinx)jx 25. I - ye.rv smx • 27. ( ) 3 cos x + 3y 29.

t

e:Ix - 5e ~ + C

37. 2e";; + C 45.

31. I

39. -e-l' + C

~e~~' + C

47. I

51. Y = I - cos(e' - 2)

57. (ln5 )5";;

2Yx

59.

33. 8e(x+l) + C

35. 2

41. -ell' + C

43. e

49. In (1 + e') + C 53. y = 2(e-' + x) - I 55.2' 1n2

1fx(~-I)

61. -VlcosO(v2-')sinO

63. 7'" '(In 7)'(seo otan 0) 65. (3cos31)(2.m'')ln2 1 3 x2 2 67. Oln2 69. xln4 71. In 10 + 3x IOglOX

-2

I)(~ + _I-I) =

~ I (+ -

tan (In 0)

31.

1 35. 2xlnlxl- xln !x;;o2

39. Inly2 - 251 + C

1 55. (-2)v'X(X +

13. 3/x

11. 0 + I

2 29. 1(1 _ In I)'

10x 33. 2 + (I ) x +1 21-x 37. In(t)

(f) t(3ln3 -In2)

I

I

x(l + Inx)'

(0) -In2

(0) In (12) I

17. x'lnx

15. 2(ln1) + (lnl? 21.

3. 4(lnx)2

(b) 2(ln2 -In3)

1n.4

(b) 1 = - - () o 1=1n.2 k 5. -5e-'" 7. -7e(5-',) 9. xe'

73. (x + I)(x _ I)

77. 1n\0

79. +(Iog,3)3 Iog"

5' 83· ln5 +C 91. 32760

I

75. sin (log, 0) + In 7 cos (log, 0)

97. In I10 (lnxl') -2+ C 103. In 10

81. +

I I as' 2ln2 87' ln2 3x(v'3+l) 93'.;;0 +C v3 + I 99. 2(1n 2)'

105. (In 10) Inllnxl + C

107. In (In x), x > I

109. -Inx

6 89' ln7

101. 3 ~2

Chapter 7: Answers to Odd-Numbered Exercises

111. (x + I)'

(X ~

I + In (x +

1))

115. (sinx)'(lnsinx + xcotx) ~

119. Maximum: I at x

evt)' (~I + k)

113.

121. (a) Absmax: }atx

~

~

~

(b) Y

I

~~

0, Y

~

89. (a) We should assigo the value I to I(x) continuous at x = o.

117. cosxx.x-'(1 + Inx)

0, minimum: 2 - 2 10 2 at x

~

87. (a) Y

83. -I

10 2

A-29

(sin x)' to make it

y

I

(b) (2, :,)

123. Abs max ofl/(2e) assumed at x ~ 125. 2 127. Y ~ e X/ ' - I

1jVe

e' 2~

I 131. In (\12 + I) d I 133. (a) tb;(xlnx -x + C) ~x·x + Inx - I + 0 ~ Inx

129.

I (b) e - I

(e) The maximum value of I(x) is close to I near the point x '" 1.55 (see the graph in part (a».

135. (b) Ierror I '" 0.02140 (e) L(x) ~ x + I never overestimates eX.

Section 7.6, pp. 413-416 1. 3. 5. 7.

(a) (a) (a) (a)

'1f/4 (b) -'1f/3 -'1f/6 (b) '1f/4 '1f/3 (b) 3'1f/4 3'1f/4 (b) '1f/6

9. 1/\12 19.0 137. 21n 5

139. x '" -0.76666

141. (a) L(x)

~

25.

I + (102).< '" 0.69x + I

29.

1

~C

y'/' - x'/2

17. y

~ sin (x' +

21. 41o(Vy 23. 25. 31. 33. 39. 41.

+ 2) ~

15. e-Y + 2eV. ~ C 19. tlnlY' - 21

C) ex'

13. '1f/2

15. '1f/2

17. '1f/2

~ 23.~ I-x' 1-21'

21. I

.~

Ils + IIv8' + 8 -I

31.

v!=7

-2s

27.

(x' + I)Vx' + 2s'

-I 2Vt (1 + I)

33

I • (tan 1x)(1 + x')

35.

11. eY - eX ~ C

13. -x + 2tanvY ~ C

11. -1/Y3

-e' -I 37 -ls" • ".------; • .,.-------, 39. 0 le'IV(e')' - I ve" - I v I - 8' 41. sin- 1x 43. sin- 1 ~3 + C 45. _1_ tan-I _x_ + C

Section 7.4, pp. 394-396 9.

(e) '1f/6 (e) -'1f/3 (e) '1f/6 (e) 2'1f/3

47.

~ x' + C

~sec-ll~1

53. -7f/12

+C

(a) -0.00001 (b) 10,536 years (e) 82% 54.88 g 27. 59.8 ft 29. 2.8147498 X 10 14 (a) 8 years (b) 32.02 years

55.

v'i7

+ C

49.2'1f/3

51. '1f/16

~sin-12(r - I) + C

57. ,?tan-1(xvi)+c

15.28 years 35. 56,562 years (a) 17.5 min (b) 13.26 min -3°C 43. Aboot 6658 years 45. 54.44%

v'i7

59.tsec-112X;II+c

61. 7f

63. 7f/12

65. k sin-1y' + C

69. '1f

71. ktan-1

f;

I) + C

67. sin-1(x - 2) + C

73. 27f

75.11n(x'+4)+2tan-l~+C 2

Section 7.5, pp. 402-404 1. 1/4 3. 5/7 5. 1/2 7. 1/4 9. -23/7 15. -16 17. -2 19. 1/4 21. 2 23. 3 27. In 3 39.

-00

51. I/e 63. 0

I

31. In 2

29. In 2 41. -1/2 53. I 65. +1

75. (b) is correct

43. -I

55. I/e 67. 3

33. I

69. I

77. (d) is correct.

71. 0

0 77. x + 10 (x' + 9) - 13 tan-I 79. sec-1 Ix +

37. 1n2

47. 0 59. I

13. 0

25.-1

35. 1/2

45. I

57. e 1/2

11. 5/7

2

83. t(sin- ' xl' + C

49.2 61.

73. 00 27

79.c~TO

e' 81. (b)

11 +

89. 1tan-1 -I

T

97. I

(tan-~

103. y

C

j

+ C

81. e""-'x + C 85. Inltan-1yl + C

Vi) + C

~ sin-1x 105.

91. 5 y

87. Y3 - I

93. 2

~ sec-1x +

95. I 2;,x

>I

A-30

Chapter 7: Answers to Odd-Numbered Exercises

107_ (b) y = 3Ys

109_ 8 = COS-I

(~) "" 54.7'

3. sinhx

121. w2/2 123. (a) w2/2 (b) 2" 125. (a) 0.84107 (b) -0.72973 (0) 0.46365 127. (a) Domain: all real numbers except those having the fonn

f + k1r where !is an integer Range:

= 8/15, tanh x = 8/17, cothx = 17/8, sechx = 15/17,

cschx = 15/8 I 5. x + x

7.

0 s.

r. tanhVt 15. sech v t + Vt 2-

25.

,

I 2Vx(1 + x)

35.

<x<

00;

17. cothz

Range:

-00


00

lsecxl

21. tanh'v

23. 2

27. I +1 8 - tanh-I 8

33.

co~2x +

41.

1n2

31. -sech-I x

29 _1_ - coth-IVt • 2Vt

-00

3

19. (In sech 8)(sech 8 tanh 8)

"
(b) Domain:

13. 2cosh x

9. 0""

C

(f -ln3) + 45. 7 I + 0~/11 + 47. tanh (x - t) + 49. -2 sechVt + 51. % 43. 12 sinh

In 0'/1

C

C

53. ;2 + 1n2

61.

In (2/3)

55.

00;

(b) Domain: -I '" x s I; Range: -I s y s I

,

,

2

0-

-In 3 63. -2-

1

57. 3/4

65.

In 3

0;)

In

59.

~

+ 1nV2

(b) In (Y3 + 2)

69. (a) coth-I(2) - coth-I(5/4)

71. (a) -sech-I

C

C

0 -

67. (a) sinh- I(V3) 129. (a) Domain: -00 < x < Range: 0 ~y ~ '11'

~I + (t)"

+ sech-I

(b)

(t) In(t)

(~)

)'=001(008-1 x)

(b)

- (12/13)2) (I + VI - (4/5)2) -In ( I + VI(12/13) + In (4/5) -InG) + In (2) In (4/3) =

=

-2

n

73. (a) 0 131. The grnphs are identical.

,

133.

,

77. (b)

(b) 0 (0) 80Ys "" 178.89 ft/sec

79. 2"

81.

6

5

Section 7.8, pp. 428-429

,--'(~)'

In

Section 7.7, pp. 421-424 1. coshx

= 5/4, tanh x =

-3/5, cothx

sechx = 4/5, cschx = -4/3

1. (a) Slower (b) Slower (0) Slower (d) Faster (0) Slower (f) Slower (g) Same (b) Slower 3. (a) Same (b) Faster (0) Same (d) Same (0) Slower (f) Faster (g) Slower (b) Same 5. (a) Same (b) Same (0) Same (d) Faster (0) Faster (f) Same (g) Slower (b) Faster 7. d, a, c, b 9. (a) False (b) False (0) True (d) True (0) True (f) True (g) False (b) True 13. Wheo the degree of f is less than or equal to the degree of g. 15. I, I 21. (b) (ol7
= -5/3,

A-31

Chapter 8: Answers to Odd-Numbered Exercises

23. (a) TIre algorithm that takes 0 (n log, n) steps (b)

117. I 119. I/e m/sec 121. 1/v2 units long by I/Ve units bigh, A = I/v'];, '" 0.43 units2 123. (a) Absolute maximum of 2/e at x = e 2; inflection point (e'/3, (S/3)e-4/3); concave up on (e'I', 00); concave duwn on (0, e'/3)

, 2500 2000

,

"00 1000

,oo~~~~~;,:_~~~~ 20

40

60

80

0.'

100

-0.5

-I

25. It could take one million for a sequential search; at most 20 steps for a binary search.

-1.5

(b) Absolutemaximumofl atx = 0; inflection points (±I/v2, I/Ve); concave up on (-00, -1/v2) U (i/v2, 00); con-

Practice Exerdses, pp. 430-432 1.

-2e~/'

3.

xo""

sin (J

13. (x + 2),,+2(1n (x + 2) + I) -I

~2

15.

I - u

19. tan-I (I) + _1_ - ~ 1+1221

~ + sec-Iz z2 -

,

11. 18x 2.•

9. -S-1(ln S)

21.

cave down on (-I/v2,I/v2)

2 7. (ln2)x

5. 2 sin 02COS 0 = 2cotO

-_"'"-_..,,2"--!I---'----!1~2:--,'!---+x

23.-1

1

(0) Absolute maximum of I at x = 0; inflection point (I, 2/e); concave up on (I, 00); concave duwn on ( - 00, I)

+ I) ~-2-2x 25. 2(x •~ + tan 2x ] vcos2x + I 2

1)]' [I I t+T

[(I

+ 1)(1 27. 5 (I _ 2)(1 + 3) 29.

~(SinO)V.(In~ + ocot 0)

45. -t(lnx)-2 + C

2~3 (3"')

55. e -

I

51.3ln7

57. 1/6

59. 9/14

69. sec-' 12yl + C

79. Y

=

1n2 1n(3/2)

75. ",/2

81. Y

=

-2

-4

37.

-~

-.

47. -cot(i + Inr) + C

+ C

73. sin-I (x + I) + C

y=(l+J:)e"""

----c!-~::p-~--x 3 2 1 2 3

7

43. -[lnlcos(lnv)ll + C

61. H(ln4)' - (In 2)'] ort(ln2)'

"'/'1/3

I]

31. -cos eX + C

35. e tonx + C

41. 1n(9/25)

39. inS

67.

I

+ I- I - I- 2 - I+ 3

33. tan (eX - 7) + C

49.

,

53.15/16 + 1n2

125. Y= (tan-I (x ~ C))'

9~2

65. '"

77.

Inx - 1n3

t sec- I (I~ I) + C 83. Y

=

I -I--x - e

85. 5 87. 0 89. I 91. 3/7 93. 0 95. I 97. In 10 99. In 2 101. 5 103.-00 105. I 107. I 109. (a) Same rate (b) Sarae rate (0) Faster (d) Faster (0) Saraerate (f) Samerate 111. (a) True (b) False (0) False (d) True (e) True (f) True 113. 1/3 115. Absoluternaximum = 0 at x = e/2, ahsolutemininrum = -0.5 atx = 0.5

x+ C)

131. y = 4x - 4Vx + I 135. 20(5 - v17) m

Additional and Advanced Exercises, pp. 433-434 1. ",/2

71. ",/12

= sin-I (2 tan

129. y = -2 + In (2 - e~)

133. IS,935 years 63.

127. y2

I

3. I/Ve I

5. In 2

7. (a) I

9. 1n2' 2ln2' 2: I

11. x = 2

x = 11.:,.4, ji = a

19. (b) 61'

17.

(b) ",/2

13.2/17

CHAPTER 8 Section 8.1, pp. 441-443 1. -2x cos (x/2) + 4 sin (x/2) + C 3 3. 12 sin I + 21cosl - 2sinl + C 5. 1n4 - 4

7. XO X - eX + C 9. -(x 2 + 2x - 2) e~ + C

(0) '"

A-32 11. 13. 15. 17. 19.

Chapter 8: Answers to Odd·Numbered Exercises 1

ytan- I (y) - 1n\l'1+7 + C xtanx + In Icosxl + C (x' - 3x 2 + 6x - 6)e' + C (x 2 - 7x + 7)e' + C (x' - 5x 4 + 2Ox' - 60.2 + 120x - 120)e' + C

41. ttan8 + t,ec28tan8 + C

39. 2\13 + In (2 + \13)

43.4/3 45. 2tan2 x - 21n(1 + tan2 x) + C 1 1 47. 4tan4 x - :2tan2 x + In lsecxl + C 4.r. 1 1 49. 3 -Inv3 51. -TOcos5x - :2co,x + C

21. t(-e'cos8 + e'sin8) + C e'" 23. T3(3sin3x + 2cos3x) + C

1sm ' 7x 55• "21. smx + 14

25. t ( V3,;+9 e y,;:;:o - e y,;:;:o) + C

53."

+C

-l...sin58 + C . 16 ,in 38 _lsin8 4 20

57

21- ,,\13 _ In (2) _ ,,2 18

3

,

37. 3tan x + C

59.

29. t[-xcos(lnx) + x sin (lnx)] + C

-~CO"8+C

1 1 61. 4 cos 8 - 20 cos 58 + C

1 31. :21n Isecx2 + tanx 21 + C

63. ,ecx - In Icsex + cOlxl + C 65. cosx + ,ecx + C 1 2 1. 1 69. In (1 + V2) 67. 4x -4xsm2x-gco,2x+C

1 1 1 33. -x 2 (lnx)2 - -x 2 lnx + -x 2 + C 2 2 4 1 1 35. -x lnx - x + C 37- 4e 1 '" + C

71. ,,2/2

13 x 2 (x 2 +

41.

-~sin3xsin2x - ~COS3xcos2x +

1)'/2 -

43. -cos e' + C

Section 8.3, pp. 452-453 1. Inl~ + xl + C 3. ",/4

~ (x 2 + 1)'/2 + C

39.

15

C

9.

51. (a)" (b) 3" (c) 5" (d) (2n + 1)" 53. 2,,(1 - 1n2) 55. (a) "(,, - 2) (b) 2" 57. (a) 1 (d)

59.

(b) (e - 2",. =

j

t(e

+ I),)'

2

2~ (1 - e-2~)

t(e -

2

+ 9)

19.

69.

sec-Ix -Inlx +

Vx2=11

+ C

71. Yes

27.

X

sinh-Ix - (1 +

X 2 )1/2

+ C

1. tsin2x + C 7. -CDSX'

3. -tcos4 x + C

1

5. 3COS3 X -

+ tcos3x - tcos5x + C

11. tSin4x - iSin'x + C

cosx +

9. sinx - tsin3x

29

±

(1)'/2 _!Il _ ~ (1)'/2

. 5 2

35

7

2

+c

41.

Vx2=1 +

45. 4 sin-I

49. Y

31.

C

V2

=

11

ViO)

29. 2tan- l 2x +

+ C

4x + C (4x 2 + 1)

v )' + C 33. 31( ~

37. ,,/6

39. sec-Ilxl + C

43. tlnlv'1+7 + x 21 + C

V; + Yx~ + C

47. t'in-I Yx -

13. tx + tSin2x + C

15. 16/35 17. 3" 19. -4 sin x cos3 x + 2 cosX' sin x + 2x + C 21. -cos4 28 + C 23. 4 25. 2 27 Q- ~3 • \j:2

c

13.~ + C

Vx2=1

-~ (~)' + C

35. 1n9 - In (1 +

+ C

21. 10 tan-I ~ + C 3 x +C

25.

12 1 1n Ix 2 31. :2 x +:2

Section 8.2, pp. 448-449

+C

17. t (x 2 + 4)'/2 - 4-vY"+4 + C

-2~ +C

23. 4\13 - 4:

+ cos (sin-1 x) + C

73. (a) x sinh-I x - cosh (Sinh-I x) + C (b)

1

,ec-I(~) ]

15. -~+C

2)

67. x sin-1 x

e flX dx

=

=

=

l' (e

49 -

49

61. u = x', dv = cosx fix

63. u X

x", dv

(c)

tlnl~ + v'4X~ -

11. {v'y 27-

5. ",/6

t~ + C

7. 2; ,in-'(t) +

45. 2Yx sinYx + 2 cos Yx + C

49. 5" -9 3\13

47. ,,2;; 4

73 - _ 4" - _ 8,,2 + 3 12" • x - 3 ,y -

tYx~(1

z[ ~ - (I) ] sec-I

- 2x) + C

A-33

Chapter 8: Answers to Odd-Numbered Exercises

Section 8.S, pp. 467-468

I~ ('IT + 6\13 - 12)

55. (a) (b)

x=

57. (a)

-t

3\13 - 'IT , Y = .,:> + 12\13'IT - 72 4(". + 6\13 - 12) 12(". + 6\13 - 12) 2 x 2 (1 - x 2)3/2 - 15 (I - x 2)'/2 + C

(b) _1 (1 - x 2)3/2 3

15 (I

(0)

- x 2)'/2 -

+ 1 (1 - x 2)'/2 5

13 (1

- x 2)3/2

+C

+C

1.

-h (tan-I~X

; 3) + C

. ,------;:: (2(X 3- 2) +4 )

+C

3. vx-2 5.

(2x - 3)3/2(X 5

7.

-~ x

9.

+ I)

+C

21~-31 - -3 In • ~ v9-4x+3

(x + 2)(2x -

6

6)~

+C

(x - 2) +C

. -I +4sm -2-

Section 8.4, pp. 461-462 1 _2_+_3_ 3 _1_ + 3 . x-3 x-2 •x + I (x + I)' -2 -I 2 7. I +~+ -12 S. z+~+ z-l 1-3 1-2

11. - 0lnlV7 + f7+71 + C 13. v'4 - x 2

2lnl2 +

-

11. tlnl(x + 6)2(X - 1)'1 + C

15. 13(2 cos 31 + 3 sin3t) + C 17.

13. (lnI5)/2

~ COS-Ix + tSin-lx - tx~ + 3

-~lnlll +

ilnll + 21 +

tin II -

II + C

17. 3ln2 - 2

CO: 5x -

co;x + C

I

25. -(8 - If2 + (8 - I)-I + tan-I 8 + C

2 Ix -

3 1n

29. t In 31.

I

+ + 6 1n Ix 2 + x + II - •v r.:3 tan-I (2x\13

11

+ 28 + 2

2

33. x + Inlx

~

41.

y2

2

) ZI In (x 2 + I +

29.

(x -

2

~ + 7lnlx - II + C I

1

41.

39. In(e't e

I) C

+ + +2

+C

47.

- II + C '\I'X+l + I

x; 2 v'5 _ 4x _

x2

sin4 2x cos 2x

+

~

+ C

sin-I

47. 2v'1+x + In I '\I'X+l

51.

(x ; 2) + C

2 8in2 2x cos 2x _ 4 cos 2x

+

C

15

10

(sec ".x) (tan ".x) I 'IT + "iT Inl sec 'lTX + tan".x I + C -cs~xcotx

4

ZI [sec (e'

Ix.x~ II + C

3cscxcotx

8

3

glnlcscx + cotxl + C

- I) tan (e' - I) +

Inlsec (e' - I) + tan(e' - 1)11 + C

51. x = Inll - 21 - Inll - II + 1n2

53. x

55. 3". In 25

1000e

57. 1.10

v'x 2 + 2x + 51

15 2 3 3 sin 28 cos 26 + sin 26 + C 43. 10 15 2 45. tan 2x - 2 Inlscc2x1 + C

49. 45. InIVX-11 +C VX + I

V"!------:sin~2'1 I + C

35. In Ilny + v'3 + (lny)' I + C

39.

(tan- I 2x)2 6 43. 4 -3Inlx-21+x_2+C

49. tin

t) sin-IVx + ~v'x - x + C sml

II + C

2

x I-I 2(1 + x 2 ) + ztan x + C

33. v'1 - sin2 I _ In II +

-Inlyl + ZIn(1 + y2) + C Y

~sin(78/12) + C

27.

37. In Ix + I +

I ISin sin siny + 3

1+ C

31. sin-IVx - v'x - x 2 + C

+ In (0'- + 28 + 2) - tan-I (8 + I) + C

35. 9x + 2lnlxl +

37.

I) + C

I~ ~ : I + ~ tan-I x + C

0'- -I

23. 8 [Sin (;1/2) _ sin (:t/2)

0

25. 6 sin (8/12) +

+

I

2

21. ('IT + 2 In 2)/8

23. tan-I Y - -2-1- + C y

C

19. ~ tan-Ix - ~ + 61n(1 + x 2 ) + C 21. -

x l 19. 1ln l + lx +C 4 x- I 2(x 2 - I)

27.

+ C

e2t

I

9. Z[lnll + xl-Inll - xl] + C

15.

~I

59. (a) x

=

=

4l •

499 + e '

61 1+2 - I (b) 1.55 days

53. V2 + 1n(V2 + I) 57. 2'IT\13 + 'lTV21n 59.

x=

4/3,

55. "./3

("\12 +

Y = 1nV2

\13)

61. 7.62

63. "./8

67. 'IT/4

A-34

Chapter 8: Answers to Odd-Numbered Exercises

Section 8.6, pp. 475-477

7. 2eX~in2x + eXc~s2x + C

1. I: (8) 1.5, a (b) 1.5, a (0) 0% II: (8) 1.5, a (b) 1.5, a (0) 0% 3. I: (8) 2.75,0.08 (b) 2.67,0.08 (0) 0.0312 '" 3% II: (8) 2.67,0 (b) 2.67,0 (0) 0% 5. I: (8) 6.25,0.5 (b) 6,0.25 (0) 0.0417 '" 4% II: (8) 6,0 (b) 6, a (0) 0% 7. I: (8) 0.509, 0.03125 (b) 0.5,0.009 (0) 0.018 '" 2% II: (8) 0.5, 0.002604 (b) 0.5, 0.0004 (0) 0% 9. I: (a) 1.8961,0.161 (b) 2,0.1039 (0) 0.052 '" 5% II: (8) 2.0045, 0.0066 (b) 2,0.00454 (0) 0.2% 11. (8) I (b) 2 13. (8) 116 (b) 2 15. (8) 283 (b) 2 17. (8) 71 (b) 10 19. (a) 76 (b) 12 21. (8) 82 (b) 8 23. 15,990 ft' 25. '" 10.63 ft 27. (a) "'0.00021 (b) "'1.37079 (0) "'0,015% 31. (8) "'5.870 (b) IETI '" 0.0032 33. 21.07 in. 35. 14.4

3. 2

5. 6

7. '1f/2

1')

9. In 3

11. In 4

13.

I

13. _lin looso - II + C 300s0+2 15. 4lnlxl- kln(x2 + I) + 4tan-'x + C 17.

a

V3

25. 35. 43. 51. 59. 65. 67. 75.

-1/4 27. -rr/2 29. -rr/3 31. 6 33. In 2 Diverges 37. Converges 39. Converges 41. Converges Diverges 45. Converges 47. Cooverges 49. Diverges Converges 53. Converges 55. Diverges 57. Converges Diverges 61. Converges 63. Cooverges (a) Converges wben p < I (b) Converges when p > I I 69. 2'1f 71. In 2 73. (b) "'0.88621

19. 1n(1 +

21. -I

y

1.8

I..

08

1.4 1.2

,

Si{.t)=

/,,'

.

O.

IilD.tdt

'

0.8

y _ !lint

02

0.' 0.' 0.2 0

,

o.

,

10

(b) -rr/2 77. (8)

I'

20

os

,

,

0 -0.2

I(v - 2)'(v

v'

x2

+ 2)1

4 2 + 3 1n Ix + 21 + 31n Ix - II + C

21.

T

23.

x2 9 T - zln Ix

3

+ 31 + zln Ix + II + C

25. llnl Vx+l - II + C 3 Vx+l+1

"

27. Inll - e-'I + C

31. -kin 14 - x 21 + C 35.1ln1X + 31 + C 6 x - 3 ,

7

37. _ oo~ x

+ co;

41. eo;o -

eo~~IO + C

x

69.

2x;/2 - x + 2Vx -

Diverges

.

39.

~

x

+C

43. 4YI - cos (t/2) + C

47. T = -rr, S = -rr 49. 25'F (b) "'24.83 mi/gal 57. In 3 59. 2 61. -rr/6 65. Diverges 67. Converges

(8) "'2.42 gal -rr/2 55. 6

At least 16

,

+C

45. 51. 53. 63.

71 Inl 20

+C

~ tan-' --h + C

23. I

(8)

y

I

16 1n

19. ktan-' t -

15.

17. -rr

+ C

11. In Ixl - In Ix + II + x + I + C

Section 8.7, pp. 487-489 1. '1f/2

11

9. 2lnlx - 21 - Inlx -

2 In ( Vx + I) + C

Vx 1_ 1 ( X )2 + W+1 2 W+1

C

73. -2 eotx - Inlescx + eotxl + escx + C

vi

3 75• .Lin 1 + + ltan-'1J + C 12 3 - v 6 3 Y

77.

osin (20 2

81. 2(

85.

(b) "'0.683, "'0.954, "'0.997

+

co. (20 + I) 4 + C

(~)3 _ 2~) + C

i ln Izl -1z - i [k

1 87. --4 v'9=4t' + C

Practice Exerdses, pp. 489-491 1. (x + 1)(1n (x + I» - (x + I) + C

3. xtan-' (3x) - iln(1 + 9x 2 ) +

+ I)

c

5. (x + I)'e' - 2(x + I)e' + 2e' + C

93.

tx3/2 + C

95.

-t

ZI x 2 -

I

Zln (x

2

I

4 sec' 0

89. In(e't + I) + C e +2 tan-' cos (5t) + C

+ I) + C

+ C

83. tan-'(y - I) + C

ln (Z2 + 4) + k tan-'

97. 2Vr - 21n (1 + Vr) + C

99.

79.

ml

+ C

91. 1/4

A-35

Chapter 9: Answers to Odd-Numbered Exercises 2 101. 31n Ix

+

II

I 2 + 61n Ix

+

- x

_I_ tan- (2x - I) + C

7. y'

+

II

103.

v'3

*

+ Yx)7/2

(I

IYx +

105. 21n

107. Inx -In II

109.

kx'u + C ~

113. (b)

111.

115. x -

5.

x 2 sin-1 x

2

+

x~ - sin-Ix

4

H (I - V1=/') In

sin-l I)

7. 0 9. In (4) - I 11. I 17.
~

2

I, Yl ~ 2.0, Y2 ~ 2.0202, Y3 ~ 2.0618

+

23.

m.

;~~m ~~ ;~~i1

:;.:;.:;.:; ::;,;.:~~~ :;-::::::~ ::;,::::;,:

,;;.;:,:

25.

27.

,

+C

+

C 15. 2",

13. 32",/35

16)

+ 27

(e- 4+-I' -e2- -2) 2

21.

29·k
e'" sin3x 33. 13(3 e~

a

2

+b

I

+C

2(asinbx-bcosbx)+C

2 ( I ) l-tanx2

+

47. -II n Itan(I/2) Yz tan (1/2)

43. I

C

+ +

I + tan (8/2) I 49. In I _ tan (8/2)

I

S

+ 2 cos 3x) + C

35. cosxsin3x -8 3 sin x cos 3x

41.

-(I + y)sinx;y(O)

vT+7 -In(~ +~) - Yz + 1n(1 + Yz)

25. I;'"

37.

~

I - ~, Yl ~ -0.25, Y2 ~ 0.3, Y3 ~ 0.75

15. y(exact) ~ or'

(Yz tan x ) + C

tan-I

Additional and Advanced Exerdses, pp. 492-494 1. x(sin-l x)' + 2(sin- 1 x)~ - 2x + C 3.

9. y'

19. y '" -0.2272, exact value is 1/( I - 2Vs) '" -0.2880.

kin II - ~I + C

Jz

-I

17. y '" 2.48832, exact value is e.

+C

Inxl

~

~

13. y(exact) ~ 3e*+2>, Yl ~ 4.2, Y2 ~ 6.216, y, ~ 9.697

- ~ (I + Yx)'/2 + ~ (I + Yx)3/2 + C

~I + C

+

x - y;y(I)

11. y(exact)

l

v'3

~

I I

+

+C

Yzi Yz

45.

+C

35. Euler's method gives y '" 3.45835; the exact solution is y ~ I + , '" 3.71828. 37. y '" 1.5000; exact value is 1.5275.

Section 9.2, pp. 508-510 39.xln(ax)-x+C

1.

eX

+C

y~-x-'

cosx' x 3• y -- C - 3 x

x>O

v393'"

7. y

9. y

~

11••

~

I'

/

C

(/ - I)' (/ - I)' 13. r ~ (csc 8)(lnlsec 81 + C), 0 < 8 < ",/2 _31_ I '" 15. y -22 e 2 ' 1 7. y -_-Ocos8 + 28 21.y~yoekt

23. (b) is correct, but
Section 9.1, pp. 502-504

27.
,

+

3(/ - I)'

CHAPTER 9 3. (a)

= ~xeX/2 + Ce x/2

x(lnx)2 + ex

2 e x1 19. y~6er -x+1

1. (d) 5.

>0

~ ~ - ~e-' ~ ~(I

I 29. y ~ I + Ce~

25. I

~ ~ In 2 sec

- e-') '"

31. y' ~ I

Section 9.3, pp. 515-516 1.
+ Cx-'

0.95~amp

(b) 86%

A-36

Chapter 9: Answers to Odd-Numbered Exercises 3. y' = y' - y = (y + I)y(y - I) (a) y = -I and y = I are unstable equilibria and y = 0 is a stable equilibrium. (b) y' = (3y2 - l)y'

5_X 2 +y2=C y

y

3( y

=

1)( y

+

,'<0

-1.'

+

1/v'3}( y

,>.

,'<0

1/v'3)(y -

-

I)

,>. y ",

-1

:,'>0

,'<0

,,'>0

,

1

1

-:r, 9_ y = ±V2x

+C

:r,

y

(0) 1., 1'<0.,'<0

1'>0.,,'>0

1'<0.,'<0

-1.5

13- (a) IOlb/min

(b) (100 + t)gai (0) 4(10:+ t)lb/min dy 4y (d) dt = 10 - 100 + t' y(O) = 50,

ISO

y = 2(100 + t) - (

S. y' = yy,y > 0 (a) There are no equilibrium values. (b) y' =

,'>0

)4

t

•

1+ 100 . y (25) 188.6 (0) Coocen1ratioo = am!. brine in tank = 125

IS. y(27.8) '" 14.8Ib, t

'"

12

'"

2

,'>0

(0)

Y

1.5lb/gai

27.8 min

Section 9.4, pp. 522-523 1. y' = (y + 2)(y - 3) (a) y = -2 is a stable equilibrium value and y = 3 is an unstable equilibrium.

(b)

y'

,

-4

=

2(y

+ 2)&

,>.

-

t)
,>.

)"<0

~,

l ,'<0

, ,

,

,">0

,

•

2

,"<0

..,

!, , , ,

-2

2

4

7. y' = (y - I)(y - 2)(y - 3) (a) y = I and y = 3 are unstable equilibria and y = 2 is a stable equilibrium. (b) y' = (3y2 - 12y + ll)(y - I)(y - 2)(y - 3) =

4) "

)">0

(

3(y - I) y -

6 - 3v'3) (y -

>,'<0

(0)

y

4

,

,>0,,">0

6

•

,'<0

, ,

>,'>0

1~

, ,

:,'>0:

,

2) (y

,'<0

~"'1.42

2~

>,'<0

t ,,'>0

,

, ,

, ,

3~

:,'<0:

~"'2.S8

,>. l ,'>0

,

4

Y

Chapter 9: Answers to Odd-Numbered Exercises

. d2

,

(0)

(k)

A-37

(k) (g - /iiV k 2)

ConCavIty: dt v2 = -2 /iiv
b

(b)

~>o .2"

~
)

r(

d2v

•

v

""

->0

"" <0

f'ii Veq "' ...JT ,

_~,----L---~--~2--~3~'

9.

'!

= I - 2P bas a stable equilibrium at P =

2

d P

dt 2

=

dP

-2 dt

t;

(0)

v.."..""

(T6O

= ~ o:oos = 178.9 It/sec = 122 mph

-2(1 - 2P).

=

17. F

p

= Fp

- F,;ma

= 50

dv - 51vl; dt

maximum velocity occurs when ~~

1.,

I = /ii(50

- 51vl). The

0 or v

10 It/sec.

=

=

19. Phase line: 0

dt

p'>o,p' <0

0

dii "" <0

1.25

1.5

1.75

ieq -

11. dP = 2P(P - 3) bas a stable equilibrium at P = 0 and an undt 2 stableequi1ibriumatP = 3; = 2(2P - 3) dP = dt t 4P(2P - 3)(P - 3).

d;'

P'>O

:

O~5)

"

1"<0

]>"<0

I dt

.~

f

J

"" > 0

If the switch is closed at t = 0, then i(O) = 0, and the graph of the solution looks like this:

d

:

(I l!S 2,"<02~5

o~

di -<0

•

J

1">0

~3~5)4P P'>O

P">O

As

t~ 00, i(t) ---+ isteadysta.te

=

~.

P 4

F:::::---===:::='~P'>O'P' >0 2

0.4

0.5

-1

0.6

0.7 P'>O,P' <0

Section 9.5, pp. 527-529 1. Seasonal variations, nonconfonnity of the environments, effects of other interactions, unexpected disasters, etc. 3. This model assumes that the number of interactions is proportional to the product of>; and y: fix

dt = (a -

-2

13. Before tire catastrophe, the population exhibits logistic growth and P(t) increases toward M o, the stable equilibrium. After tire catastrophe, the population declines logistically and p(t) decreases toward M the new stable equilibrium. " p

p

, , ,

--~

----t----

,

dv _ k 2 15. dt - g - /iiv, Equilibrium' dv .~

----

~,

'--

g,k,m > Oandv(t) '" 0

=g

-

~ v 2 = 0 ~ v = rmg m ~~

dy =

bylx,

a

< 0,

m(1 _l')y - nxy y(m - my - nx) =

dt MM' Rest points are (0, 0), unstable, and (0, M), stable. 5. (8) Logistic growth occurs in the absence of the competitor, and involves a simple intemction between the species: growth dontinales the competition when either population is sma11, so it is difficult to drive either species to extinction. (b) a: per capita growth rate for trout m: per capita growth rate for bass b: intensity of competition to the trout n: intensity of competition to the bass k,: enviroomental carrying capacity for the trout k2: enviroomental carrying capacity for tire bass

~: growth versus competition or net growth of trout

W: relative survival of bass

A-38

Chapter 10: Answers to Odd-Numbered Exercises dx

(0) dt ~ 0

when

ely dt ~ 0 when

x~o

or y

y~O

or y

a

33. (a) y = -I is stable and y = I is unstable.

a bk,x,

~

b-

~

k2n k2 - /ilx.

d"'y ely (b) dx 2 = 2y dx = 2y(y2 - I) )' _,_1

By picking alb > k2 and min > k" we insure that an equilibrium pcint exists inside the fIrst quadrant

Y

I

~>o dx

•

: d7>O : d2y
: IIr

Practice Exerdses, pp. 529-530 1. y

~ -In(c - ~(x -

2)5/2

+C

3. !any = -x sin x - cosx

x-I 7. y = Cx-

-1(X - 2)'/2) 5. (y

,

(0)

+ I)." =

-Inlxl

+

: tfi2 , _0

2

C

x2

4" /2 + c.'/2

9. y =

x2-2x+C e-r + C 15. '9' + y' = C 13. y = 1 + eX 2x 2 2x'+1>;2+6 19. y = 40..,') 17. y = 6(x + I)' 21. y ~ .~(3x' - 1>;2) 11. y=

t(I -

23.

-2

Additional and Advanced Exerdses, pp. 530-531 x

y

x

y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0 0.1000 0.2095 0.3285 0.4568 0.5946 0.7418 0.8986 1.0649 1.2411 1.4273

1.1

1.6241 1.8319 2.0513 2.2832 2.5285 2.7884 3.0643 3.3579 3.6709 4.0057

1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

c + (YO - c).-k(AjY)1 (b) Steady-state solution: yoo = c 5. x 2(x 2 + 2y2) = C

1. (a) y =

7. In Ixl + ."/' = C 9. In Ixl -In lsec(y/x - I)

+ tan(y/x - 1)1

CHAPTER 10 Section 10.1, pp. 541-544 1. a, = 0, a2 = -1/4, a, = -2/9, flo, = -3/16 3. a, = l,a2 = -1/3,a, = 1/5,,,,- = -1/7 5. a, = 1/2, a2 = 1/2, a, = 1/2, flo, = 1/2 3 7 15 31 63 127 255 511 1023 7. 1'1' 4' 8' 16' 32' 64' 128' 256' 512

25. y(3) '" 0.8981 27.

(.)~

IIIIIIII 9. 2, I, -1'-4' 8'16'- 32' - 64' 128' 256

11. I, 1,2,3,5,8, 13,21,34,55 [ 0.2, 4.5] by [ 25,O.S]

(b) Note that we choose a small interval ofx-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x :5 -I. (This occurs because the analytic solution is y = -2 + 1n(2 - .~), which has an asymptote atx = -In 2 '" -0.69. Obviously, the Euler approximations are misleading for x :5 -0.7.)

~ .

~

..

[-1,0.2] by [-la, 2]

( ) 29. yexact 31. y(exact)

123

.1

= IX -I;y(2) '" 0.4; exact value IS I' = -.(>'-')/2;y(2) '" -3.4192; exact value is

-.'/2 '" -4.4817.

= C

15. a. = (- I)·+l(n)2, n '" I

19. an =

n2 -

1, n ~ 1

_ 3n + 2 > 23.an n! ,n_l 27. 33. 39. 45. 51. 57.

13. a.

= (-I)'+',n 211 -

'" I

1

17. a. = 3(n + 2)'n '" I

21. an = 4n - 3, n ~ 1 I + (_1)'+' 25.an = 2 ,n~l

29. Converges, -I 31. Converges, -5 Diverges 37. Converges, 1/2 41. Converges, 43. Converges, I 47. Converges, 0 49. Converges, 0 53. Converges,. 7 55. Converges, I 59. Diverges 61. Converges, 4 63. Converges, 0 65. Diverges 67. Converges, e-1 69. Converges, .2/' 71. Converges, x (x > 0) 73. Converges, 0 75. Converges, I 77. Converges, 1/2 79. Converges, I 81. Converges, 'If/2 83. Converges, 0 85. Converges, 0 87. Converges, 1/2 89. Converges, 0 Converges, 2 Diverges 35. Converges, 0 Converges, 0 Converges, I Converges, I

91. 8

93. 4

95. 5

v'2

97. I

+

v'2

99. x.

= 2.- 2

A-39

Chapter 10: Answers to Odd-Numbered Exercises

101. (a) f(x) = x' - 2,1.414213562 "" v'2 (b) f(x) = tan (x) - 1,0.7853981635 "" '1T/4 (0) f(x) = eX, diverges 103. (b) 1 111. Nondecreasing, bounded 113. Not nondecreasing, bounded 115. Converges, nondecreasing sequence theorem 117. Converges, nondecreasing sequence theorem 119. Diverges, defmition of divergence 121. Converges 123. Converges 133. (b) v'3

17. Converges; geometric series, r 19. Diverges; Integral Test 21. Converges; geometric series, r 23. Diverges; Integral Test

2(1 - (1/3)') 1-(1/3),3

7

IJ_con

Converges; Integral Thst Converges; Integral Test Converges; Integral Thst (a) y

~2 >

1

33. Diverges; ntb-Thnm Thst 37. Converges; Integral Test 41. a = 1

1114 7. 1 - 4 + 16 - 64 + ..., 5

G t) +

35. s. = 1 - n

s. f =

~

(n ~

45. 1

37. s, =

2}

21. 7/9

In Vn+1; diverges (b) "" 41.55 45. True 47. (b) n '" 251,415

converges,-~ ~2

47. -

49. Converges, 2 +

v'2

55. Converges, - ,e'- e - 1 57. Converges, 2/9 59. Converges, 3/2 61. Diverges 63. Converges, 4 65. Diverges 67. Converges, '11' ~ e 51. Converges, 1

,

+

1; converges, 1

oos-1

43. 5

53. Diverges

49.

s.

=

• I ~.,

,,- 1 n

""

1.195

51. 10 60

59. (a) 1.20166 s S s 1.20253

(b) S"" 1.2021, error < 0.0005

Section 10.4, pp. 562-563

=

1. Converges; compare witb ~ (I/n')

=

3. Diverges; compare witb ~

1, r = -x; converges to 1/(1 + x) forlxl < 1 3, r = (x - 1)/2; converges to 6/(3 - x) for x in (-1,3) 1 1 1 73.lxl<2'1_2x 75. -2<x<0'2+x

77. x "" (2k + I) 00

,~, (n

f, k an integer; 1 1 (b)

~ (n

1 + 2)(n + 3)

(I/Yn)

5. Converges; compare witb ~ (l/n 3/2)

. 7. Converges; compare WItb

1sinx 00

+ 4)(n + 5) 00 1 (0) ~ (n - 3)(n - 2)

79. (a)

2+' I "" 0

1 - (-1/2)'

15. Converges, 5/3 17. Converges, 1/7 19. 23/99 23. 1/15 25. 41333/33300 27. Diverges 29. Inconclusive 31. Diverges 33. Diverges

69. a 71. a

25. Diverges; lim

29. Diverges; geometric series, r =

31. 35. 39. 43.

(t + ~) + (i + i7) + ..., 2i 13. (1 + 1) + (L 1) + (1 + ~) + (L _1_) + ... 17 2 5 4 25 8 125 ' 6

41. 1

2/3 < I

7 7 7 16 + 64 + ..., 3

4+

11. (5+ 1) +

39.

=

< 1

3.s.= 1_(_1/2),2/3

111 5. s. = 2 - n + 2' 2 9.

i

27. Diverges; 1im._00 (Yn/In n) "" 0

Section 10.2, pp. 551-552 1. s.=

=

9. Converges

13.

~

~n+4n .ro 1 - 4 - - = V 5 ~ ..", n +0

11. Diverges; limit comparison witb

Diverges; limit comparison witb ~(I/Yn)

17. Diverges; limit comparison witb ~ ( I/Yn)

89. (a) r = 3/5

19. Converges; compare witb ~(1/2') 21. Diverges; ntb-Term Test 23. Converges; compare witb ~ (l/n')

Section 10.3, pp. 557-558

25. Converges; (3n: I)' < (3:)'

(b) r = -3/10 1 + 2r , 91. Irl < 1 , - - , 93. 8m 1- r

5. Converges

1. Converges

3. Converges

9. Converges

11. Converges; geometric series, r = 110 < 1

n_oon +n 1

13. Diverges; lim

= 1 ""

0

7. Diverges

15. Diverges; p-series, p < 1

27. 29. 31. 33.

n

=

(t)"

Diverges; direct comparison witb ~(I/n) Diverges; limit comparison witb ~(I/n) Diverges; limit comparison witb ~(I/n) Converges; compare witb ~(I/n3/2)

f~

~

(l/n)

15. Diverges

A-40

Chapter 10: Answers to Odd·Numbered Exercises

35. Converges;

n~'

,;;

i,

37. Converges; -3'~'"-+-1

< -3,_1__,

39. Converges; comparison with ~ (l/5n')

~ n(n ~

n'

<

n'

45. Converges absolutely; sech n 0'

Section 10.5, pp. 567-568 Converges 3. Diverges 5. Converges 7. Converges 11. Diverges 13. Converges 15. Converges Converges Converges; Ratio Test 19. Diverges; Ratio Test Converges; Ratio Thst Converges; compare with ~(3/(1.25)')

~)' = 0-3 # a

Converges; compare with ~(l/n') Diverges; compare with ~(I/(2n)) Diverges; compare with ~(I/n) 33. Converges; Ratio Test Converges; Ratio Test 37. Converges; Ratio Test Converges; Root Thst 41. Converges; compare with ~(I/n') Converges; Ratio Test 45. Converges; Ratio Test Diverges; Ratio Test 49. Converges; Ratio Test

I )('/,0 =:3 ... I (

51. Converges; Ratio Test

53. Diverges; a,

55. Converges; Ratio Thst 59. Converges; Root Test

57. Diverges; Root Test 61. Converges; Ratio Test

65. Yes

Section 10.6, pp. 573-574 1. 3. 5. 7. 9. 11. 13. 15.

Converges by Theorem 16 Converges; Alternating Series Thst Converges; Alternating Series Test Diverges; all + 0 Diverges; an + 0 Converges; Alternating Series Test Converges by Theorem 16 Converges absolutely. Series of absolute values is a convergent geometric series.

17. Converges conditionally;

l/Vn'" 0but ~:, ~ diverges.

19. Converges absolutely; compare with ~:::',(l/n'). 21. Converges conditionally; I/(n + 3)'" 0 but diverges (compare with ~:~,(I/n».

· 3 23• DIverges; 5

+ + nn ~ I

31. Diverges; a. -1+ 0

I = I(-nl!~+11 = n!/2

diverges (compare with ~(l/Vn)). 43. Diverges, a, ... 1/2 # 0

-,- =

27. 29. 31. 35. 39. 43. 47.

> I/n

37. Converges absolutely by Root Thst 39. Diverges; a, ... oc 41. Converges conditionally; Vn+1 - Vn = I/(Vn + Vn+1) ... 0, but series of absolute values

1r/2

--u

20'

,~OO

0 but (I + n)/n'

(convergent p·series)

53. Converges; limit comparison with ~(I/n') 63. Converges 65. Converges 67. Converges

25. Diverges; lim (I -

k) ...

35. Converges absolutely since c:v;;-

49. Converges; compare with ~(l/n') 51. Diverges; limit comparison with ~(I/n)

1. 9. 17. 21. 23.

+

I

I)

or limit comparison with ~ (I/n') 45. Diverges; limit comparison with ~(I/n)

tan-' n 47. Converges; -,-,-

C,

27. Converges absolutely; Ratio Test 29. Converges absolutely by Integral Test 33. Converges absolutely by Ratio Test

41. Diverges; comparison with ~ (I/n)

43. Converges; comparison with

25. Converges conditionally;

~:, n ~

3

2

n. a term

from

0

=

2

1J

+

2e n

~ = -,.--

e e e +1 . . a convergent geometric senes.

<

~ (-I) 2(n ~ I) converges by Alter· nating Series Thst; ~ 2(n ~ I) diverges by limit comparison

47. Converges conditionally;

with ~ (I/n). 49. IError I < 0.2 51. IError I < 2 X 10-11 53. n '" 31 55. n '" 4 57. 0.54030 59. (a) a, '" a,+, (b) -1/2

Section 10.7, pp. 582-584 1. (a) I, -I < x < I (b) -I < x < I (0) none 3. (a) 1/4, -1/2 < x < 0 (b) -1/2 < x < 0 (0) none 5. (a) 10, -8 < x < 12 (b) -8 < x < 12 (0) none (b) -I < x < I (0) none 7. (a) I, -I < x < I 9. (a) 3, - 3 ,;; x ,;; 3 (b) -3 ,;; x ,;; 3 (0) none 11. (a) oc,forallx (b) forallx (0) none (b) -1/2 < x < 1/2 13. (a) 1/2, -1/2 ,;; x < 1/2 (0) -1/2 (0) x = -I 15. (a) I, -I s x < I (b) -I < x < I 17. (a) 5, -8 < x < 2 (b) -8 < x < 2 (0) none 19. (a) 3, - 3 < x < 3 (b) -3 < x < 3 (0) none 21. (a) I, -2 < x < 0 (b) -2 < x < 0 (0) none 23. (a) I, -I < x < I (b) -I < x < I (0) none 25. (a) O,X = 0 (b) x = 0 (0) none 27. (a) 2, -4 < x ,;; 0 (b) -4 < x < 0 (0) x = 0 29. (a) I, - I s x'" I (b) -I'" x s I (0) none 31. (a) 1/4, I '" x '" 3/2 (b) I '" x '" 3/2 (0) none 33. (a) oc,forallx (b) forallx (0) none (b) -I < x < I (0)-1 35. (a) I, -I '" x < I 37. 3 39. 8 41. -1/3 < x < 1/3, 1/(1 - 3x) 43. -I < x < 3, 4/(3 + 2x - x') 45. 0 < x < 16, 2/(4 - Yx) 47. 49. I

v'2 < x < v'2,

3/(2 - x')

< x < 5, 2/(x - I), I < x < 5, -2/(x - I)' x2 x4 x6 x S x lO I - 2! + 4! - 6! + 8! - IO! + ... ; converges for all x

51. (a) cosx

=

(b) Same answer as part (c) 23x 3 2sx s 2'x' 29x 9 (0) 2x - 3! + 51 - 7! + 9!

x - IT! + Zll ll

...

Chapter 10: Answers to Odd-Numbered Exercises x2

X4

6

X

3lx 10

17x 8

'11'

00

'11'

IT + 45 + 2520 + 14175' -2 < x < 2

53.
2

(b) I + x +

2x4

T

17x'

4s

+

62x' + 315 + "',

'JT

3.

~

-2 < x < 2

(-

5. ~

P,(x)

O,P,(x)

~x

4 IX'

- I,P2(X)

~

(x - I) - t(x - I)',

~ (x - I) _l(x - I)' + l(x - I)' 2 I

3

I I 5. Po(x) ~ "2'P,(x) ~"2 - 4"(x - 2),

t-

P,(x)

~ l_l(X - 2) + l(X - 2)2 - ..!..(X - 2)'

7. Po(x) P2(X) P,(x)

~

t(X - 2) + t(X - 2)2,

4

,?,Pdx )

~

'? + '? f),

2

2

~ V2 + V2 (x 2

2

_ "'-) _

4

_ "'-) _ 4

~

2,P,(x) ~ 2

P2(X) ~ 2 + P,(x)

11

00

~

2+

4"I (x

t

~ (x

21.

23. 25.

x4

n!

2!

3!

4!

00

31.

~(_I)'+l
00

r

I, I 4 35. x -"2x +"6x

+ "',-1

41. L(x) ~ 0, Q(x) ~ -x 2/2 45. L(x) ~ x, Q(x) ~ x

()

2

",(x - 2)" 11=0 n.

52,;2 2.

·

x 2 - ~x4

3

5'x'

3.

xU

xS

49.

45

105

I

4

-gX + ... I

104'4! < 4.2 X 10-6 Ixl < (0.06)'/' < 0.56968 IError I < (10-')'/6 < 1.67 X 10-10, -10-' < x < 0 IError I < (3°")(0.1)'/6 < 1.87 X 10-4 k(k - I) 2
35. IError I 37. 39. 41.

+ ...

+ 23 x 6 _ ~x8 + ...

I 2 33.I+x+"2x :5

Section 10.10, pp. 602-604 x x2

<x< I

+ - , - - , + ...

- -x'3 + -5 - -xiS7+ ...

(1, +

1. I +"2

43. L(x) ~ I, Q(x) ~ I

(-I)'~x,"+l

~

x4ll-1

x3

31

Section 10.9, pp. 595-596 00 (-5x)' 1. Y--,- ~ I - 5x :=6 n.

2' 2n!

29. x+x2+T - 30

5 2 - "',-1 <x< I 33. -1-2x-"2x 2

~

11"'0

+ ...

29. ~

00

(-i)')x' ~ 2 + :lx2 - .s.x' + 25 x4 - ... n. 2 6 24 00 (_I)"-lx 2n+1 x3 xS x' 27 ~ ~---+-- ... · ,~, 3n 3 6 9

25.

'x'"+,

11=0

::1

,-,

2 1)'3'"+ 00 (-I )"x ' 00 '" 19. ~ (;,,)! ~ (2n + I)! 17. 7~ (2n)! x4 - 2x' - 5x + 4 8 + lO(x - 2) + 6(x - 2)' + (x - 2)' 21 - 36(x + 2) + 25(x + 2)' - 8(x + 2)' + (x + 2)4

27. ~(-I)'(n + I)(x - I)'

'JT'X7

(-I)'(2x)'"

- 4)2 + 5:2 (x - 4)'

x3

13. ~(-I)"x' ~ I - x + x 2 - x' 15.

xlO

21. ~nx'-' ~ I + 2x + )x2 + 4x' + ...

- 4),

x2

(_

x8

6! + 8! - IO! + ...

11=0

00

00

x6

+ ...

+ -4' ~ (2 )' . - -6' . + ... ~ ,~n.

00

(_X)"

11=0

xS

4'

00

~--~I-x+---+--'"

• 11=0

x4

3'x 9 -

19. x2~(2x)' ~ x 2 + 2x' + 4x4 + ...

I - 4) - 64 (x - 4)2,

(x - 4) -

'JT4X'

x3

...

-2' + -3' + -4' + ... ...

23. ~ (-1)'+1-- ~x, ,~, 2n - I

I + 4"(x

x8

00

",-)2 4' V2 (x _ ",-)2 4 4 V2 (x _ 4

- ~ (x -f)' 9. Po(x)

x6

(2x)' (2x)4 (2x)' (2x)' 1- 2'2! + 2'4! - 2'6! + 2'8! - ...

(x -

~ V2 + V2 (x

2

00

17. I + Y.

16

8

x4

00 X"+1

7r'x'

~

625x'

+ 41 - ...

••

11. ~ - , = x + x + 11=0 n. 00 (-I)"x211 x4 13. ~ (2n)! ~ 4! 15. x - -2' .

P2(X)

2

3! 5! 7! 1)'(5x 2)," 25x4 (2n)' ~ I - 21

7. ~ (-I)'+In~x2- - + - - - + ,~, 2 3 4 2 00 3 3+ 3 x6 9. ~ (-1)" (3)' - x 311 = 1 - -x ,-0 4 4 42

1. Po(x) ~ I, P,(x) ~ I + 2x, P 2(x) ~ I + 2x + 2x 2,

~

~~

11"'0· 00 x2n

Section 10.8, pp. 588-589

3. Po(x)

5( -1),+Ix,"+1 (2n + I)!

00

5x' 5x' 5x 7 ~-5x+---+-+ ...

'JT

00

P,(x) ~ I + 2x + 2x 2 +

5(-1)'( -x),"+1 (2n + I)!

A-41

+ x 2/2

X3

-"8 + 16 3x 2

x3

23.

x3

3x 6

5x 9

7.1- 2 + 8 - 16 1+..!.. __1_+_12x 8x 2 16>;' (I + X)4 ~ I + 4x + 6>;2 + 4x' + X4 (I - 2x)' ~ I - 6>; + 12x2 - 8x' 0.00267 17. 0.10000 19. 0.09994 21. 0.10000 1 x3 x' xU 13'6! '" 0.00011 25. T - 7'3! + 11'5!

5.I-x+T-2 9 · 11. 13. 15.

13 5 3. 1+"2 x + gX 2 + 16 x 3 + ...

A-42

Chapter 11: Answers to Odd-Numbered Exercises

27_ (a)

x2

x4

T -

(b)

13. Wi1h a = 22'1f, cosx = I -

12

x2

x4

T -

3-4

+

29. 1/2

31. -1/24

39. 3/2

41. e

49.

I

x'

(b)

x8

+ ... + (-I)" 31.32

5-6 - 7-8

35. -I

37. 2

45. V3/2

47. I

33. 1/3

43. cos ~

I - 6! (x - 22".)'

x 32

~x

23. a

x3

,-I

2 - x - 6 - 40 - ill

61.1-2x+3x 2 -4x'+··· 67. (a) -I

71 x + •

(b)

(1/v2)(1 + ,)

(o)-j

,-0

~

( _ 1),,.z.+IX2n+l

(2n

(x

00

+ I)!

+

1) 65. 2 - - 2 I' • .

61.

3(x

+ I?

+, 2 ·2!

~

(-I)'x,o.j3 (2n)!

00

63.

(n ;;,

85. (a) 00

(0) I/q

(b) 6

~

Section 11.1, pp. 616-618 1.

3.

,

5.

7.

, 2

n!

+ I)' + 2, ·3! + ...

I): the seties converges to In (b) a = I, b = 0

"'-.L . , 2

t=~ -2

t=O

-,

2

,

-2

(t) . 9.

,

+

Y.f

(x - '1f/3)'

3. Diverges; n1h-Term Test 7. Diverges; nth-Term Test

V3 I -2- (x - '1f/3) - 4 (x - '1f/3?

+ ... 2

11. Wi1h a = 0, e' = I

3

+ x + x2! + x3! + ...

0

t='IT

-,

75. 1/12

87. It converges.

1. Converges; Comparison Test 5. Converges; Comparison Test

2I -

,

~2+-i _ l

Additional and Advanced Exerdses, pp. 607-609

9. Wi1h a = '1f/3, cosx =

,

------~o~~--~,

«'1fx)/2)'

l_l(x 2

83. In

27. (b) Yes

9(x

3) +l(x - 3? _l(x - 3)' 4 4 4' 4' 69. 0.4849171431 71. 0.4872223583 73. 7/2 77. -2 79. r = -3, s = 9/2 81. 2/3 67.

-7/6

±sI

21. b =

CHAPTER 11

Converges to I 3. Converges to -I 5. Diverges Converges to 0 9. Converges to I 11. Converges to e-s Converges to 3 15. Converges to In 2 17. Diverges 1/6 21. 3/2 23. e/(e - I) 25. Diverges Converges conditiona11y 29. Converges conditionally Converges absolutely 33. Converges absolutely Converges absolutely 37. Converges absolutely Converges absolutely (0) x = -7 (a) 3, -7 '" x < -I (b) -7 < x < -I (0) None (a) 1/3,0 '" x '" 2/3 (b) 0 '" x '" 2/3 (a) 00, for all x (b) For all x (0) None (a) V3, -V3 < x < V3 (b) -V3 < x < V3 (0) None 49. (a) e, -e < x < e (b) -e < x < e (0) Empty set 00 I I 4 X 57. ~2'x' 51. 1 +x'4'5 53. sinx,'11',O 55. e ,ln2,2 1. 7. 13. 19. 27. 31. 35. 39. 41. 43. 45. 47.

59.

17. "./2

~xs + ...' forallx 30

+3

1x 3 -

Practice Exercises, pp. 605-607

00

+ 1! (x - 22'1f)'

33. (a) R, = Coe-"'(I - e"""")/(I - e-Ito ), R = Co(e -"')/(1 - e -Ito) = Co/(e lto - 1) (b) R, = I/e '" 0.368, RIO = R(I - e- ZO ) '" R(0.9999546) '" 0.58195; R '" 0.58198; 0 < (R - Rzo)/R < 0.0001 (0) 7

5x 7

3x 5

=

31. (a) ~nx'-1

-I 55. 500 terms 57. 4 terms (I + X)2 . x3 3x 5 5x 7 6 + 40 + ill' radius of convergence = I

'IT

2,L

(x - 22'1f?

00

•

+

x2

=

t

+ ...

15. Converges, limit = b

51

+ x2

59. (a) x

x6

--~f---+---~~~,

11.

,

"

4

,

Chapter 11: Answers to Odd-Numbered Exercises

13.

15. y

3

-- ,

19. I

y=-b -?

t=O

29.

"

\

\

-,

\

21. 3a'1T

23. ab1T

31. 8~

521T 33. -3-

,

'1r

37. (j,y)

r

•

d"y

fix' = 108

13. Y = 9x - I,

y

=

39. (j,y) =

(~

-

(t,1T -1)

43. (a) x = I, Y = 0,

.. •

, I

,

I

,,/

45. r

\ \

\

\

\

,, ,,

(~, I).

y = -a sin I, 0 s I S 21T y = asint, O:s; t:S; 2'I'T y = -asint, O:s; t:S; 4'1T y = a sin I, Os IS 41T x = -I + 51, y = -3 + 41, x = I' + I, Y = I, I s 0

35. x=2cott, y=2sin2 t, asin'ltanl, y

=

a sin'l,

Os I

d"y

fix' =

1T

r

n an integer (d) (2,(2n + 1)1T) and (-2,2n1T),n an integer

5. (a) (3,0) O:s; t :s; 2w

(0)(3,0)

< 1T/Z

39. (I, I)

7. (a)

(b) (-3,0) (f)

(V2, f) 1~1T)

d"y

11.

I 2

j'3+z,

(I, \13) (-1,\13)

(d) (b)

1

1)

(b) (-1,0)

(d) (-5, 1T - tan-I

Y

13.

~) Y .~

+-C+---t-~r

Z=-4

(g)(-3,0)

(d) (5'1T - tan-

(a)

(0) (-Z, 5;)

(-I, \13)

(b) (3, 1T)

9. (-3V2, 5:)

V2

(0)

(1,\13)

+ 4' fix' = -2 7. Y = lx - \1'3, fix' = -3\1'3

11.y=\I'3x-

__

(~O)

(b) (Z,2n1T)and(-Z,(2n + 1)1T),n an integer

-4

d"y

d"y y=x-4, fix'

-,

(a) (2'f + 2n1T) and (-z,f + (2n + 1)1T ),naninteger

O
fix

9.

3.

(-~~)

(0) (2,

5. Y = x

dy 2\1'3 - I fix = \1'3 - 2

(b)~

Os I s I

Section 11.2, pp. 625-627 d'y 1. Y = -x + ZV2, - , = -V2

I

dy = 0 fix

(0) (2, 3; + 2n1T) and (-2, 3; + (2n + 1)1T ).

33. Possible answer: x = 2 - cos t, Y = sin t,

-2 x + 2V2,

(b) x = 0, y = 3,

(-~O)

Os 8 < 1T/2 anilx = O,y = 2 if8 = 1T/2

3. y =

dy I fix = 2

(b) 1T

~~+~

x = 2 - 31, Y = 3 - 41, I '" 0 x = 2 cos I, Y = 2 Isin II, 0 s I s 41T -at a x = _~' y = _ ~' vl+I' vl+I'

I

41. (a) 1T

Y = lxall = 0, y = -lxall = 1T

1. a,e;b,g;c,h;d,f

4 4 tan 8 31. Possible answer: x = I + 2 tan 8' Y = I + 2 tan 8'

=

_" 35. 31T v 5

Section 11.3, pp. 630-631

-oo
37. x

27. 12

\1'3 - I 3 - \1'3 Z ,y = 2 '

47. (a) 8a

j

-1

19. (a) x = a cos I, (b) X = a cost, (c) X = a cost, (d) x = a cos I, 21. Possible answer: 23. Possible answer: 25. Possible answer: 27. Possible answer: . 29. PossIble answer:

(0) x =

,

Y

25. 4

17.-6

!'!,!'! - 2)

-3

17.

3

-16

15.

A-43

1

--+-+---+~~.r

A-44

Chapter 11: Answers to Odd-Numbered Exercises

,

,

17_

7. x-axis, y-axis, origin

5_ y-axis

6=¥

y

-lsrs3

r - !li.n(812)

-~o1"""'-----+ ' -I

,

19_

2

,

21_

8-

i

,. 1

,

0

,

9. x-axis, y-axis, origin

~S6S~

,

,

0

,

25_

11. y-axis, x-axis, origin

,

0""6""11'

,~O

23_

_~It---~~E---~~,

_'It-----r----.b>'

,

-I

13. x-axis, y-axis, ongm 15. Origin 17_ The slope at(-I, '1f/2) is -I, at(-I, -'1f/2)is 1.

,

27_ 31_ 33. 35_ 37_ 39_ 41_ 43_ 45_ 47_ 49_

x ~ 2, vertical line through (2, 0) 29_ y ~ 0, the x-axis y ~ 4, horizootalline through (0, 4) x + y = 1,line,m = -l,b = 1 x 2 + y2 ~ I, circle, C(O, 0), radius I y - 2x ~ S,line,m ~ 2,b ~ S y2 ~ x, parabola, vertex (0, 0), opens right y ~ eX, graph ofnatura1 expooential function x + y ~ ±I,twostraightlinesofslope-I,y-interceptsb ~ ±I (x + 2)2 + y2 ~ 4, circle, C(-2, 0), radius 2 x 2 + (y - 4)2 ~ 16, circle, C(O, 4), radius 4 (x - 1)2 + (y - 1)2 ~ 2, circle, C(I, I), radius v'2

51_ 57_ 61_ 65_

v3y +x ~ 4 53_ rcosO ~ 7 55_ 0 ~ '1f/4 r ~ 20rr ~ -2 59_ 4r2 cos2 0 + 9r 2 sin2 0 ~ 36 rsin2 0 ~ 4cosO 63_ r ~ 4 sinO r2 ~ 6rcosO - 2rsinO - 6 67_ (O,O),wbere Oisany angle

--~-----~2~'

19_ The slope at (I, '1f/4) is -I, at (-I, -'1f/4) is I, at (-I, 3'1f/4) is l,at(l, -3'1f/4)is -1. y

---~~~~---->,

Section 11.4, pp. 634-635 1. x-axis

,

3_ y-axis

, 21_ (a)

,

(b)

r=1+cos6

*----*~,

*"-I;-----+~~' 2

-I

-2

,

A-45

Chapter 11: Answers to Odd-Numbered Exercises

23. (a)

(b)

y

-,, l

13.

y

15.

y

y

1 2

1 2

x=~

12

x

1' - ~-sin8

x

-, 25.

1/4

0

5

x

~ 12

0

2

x

--h

dinIctrix Y

27.

y

,

17.

19.

,

Osrs2-2cos6

"

4

~----------~o~~x -5 -2 -4

29. (a)

21.

,

23.

,

Section 11.5, pp. 638-639 I'", I . (;

13.

3V3 - '"

3 19. (a) 2: 25.

'" 5. 8

3. 18", 15.

- 4'"

f + '?

21. 19/3

3( v'2 + In( I + v'2))

'" 9. 2

7. 2

17. 8:

-

I

11. 5", - 8

V3

+

23. 8 '" 27. 8

+ 83

x2 y2 25. 4+T~ I

27. Asymptotes: y

,

Section 11.6, pp. 645-648 1. y2 ~ s,;,

3.

5.

",2

~ -6y,

Xl

y2

4 - 9 ~

F(2, 0), directrix: '" ~ -2

7.

2

T +y ~

9.

±X

29. Asymptotes: y

,

~

±X

F(O, -3/2), directrix: y ~ 3/2

I,

asymptotes: y ",2

~

F(n/ i3, 0),

V(±2,0),

~ ±~ '"

I, F(±I, 0),

,

------~~~,-------->x

,,

v(nl2,o)

,

11.

x _ _3

),=2

2

----_,~~+-~~-->x

,,

2

-~rl-2---->x

31. Asymptotes: y

,

~

±2x

33. Asymptotes: y

,

,,

,,

,,

,

~ ±X/2

A-46

Chapter 11: Answers to Odd·Numbered Exercises

57. C(-2, 0), a = 4 59. V(-I,I), F(-I,O) . . (x+2)2 2 __ 61. Ellipse. 5 + Y - I, C( 2, 0), F(O, 0) and 39. (a) Vertex: (I, -2); focus: (3, -2); directrix: x = -I (b) ,

V(v'5 -

F(-4, 0),

2,O)andV(-v'5 - 2,0)

. (x - 1)2 63. Ellipse: 2 2

F(O, I),

~O++'I~2~'~--------~' -2

2

+ (y - I) = I, C(I, I), F(2, V(v'2 + 1,1) and V( -v'2 + 1,1)

65. Hyperbola: (x - 1)2 - (y - 2)2 = I,

"'- _F(3,_2)

I) and

C(I,2),

F( I + v'2, 2) andF(1

- v'2,2), V(2, 2) and asymptotes: (y - 2) = ±(x - I) (y - 3)' x2 67. Hyperbola: 6 - T = I, C(O, 3), F(O, 6) and

V(1,-2)

·4

V(O, 2);

41. (a) Foci: (4 ± (b) ,

0, 3); vertices: (8, 3) and (0, 3); center: (4,3)

F(O, 0),

+ 3)andV(O,-v'6 + 3); v'2x + 3 or y = - v'2x + 3

V(O, v'6

asymptotes: y =

69. (b) I: I 73. Length = 2 v'2, width = v'2, area = 4 75. 241T 77. x = O,y = O:y = -2x;x = O,y = 2:y = 2x + 2; x = 4,y = O:y = 2x - 8

79. .>;=0,

y=

j!

Section 11.7, pp. 653-654 43. (a) Center: (2, 0); foci: (7, 0) and (-3, 0); vertices: (6,0) and

(-2,0); asymptotes: y = (b)

±~ (x

1. e

=

3

S'

. directrices are x =

- 2)

3. e

F(±3,O);

25

±T'

=

I "1/2;

F(O, ± I);

directrices arey = ±2.

,

,

-s .........-/2

5. e 45. (y + 3)2 = 4(x + 2), V(-2, -3), F(-I, -3), directrix: x = -3 47. (x - 1)2 = 8(y + 7), V(1, -7), F(I, -5), directrix: y = -9 (x + 2)' (y + 1)2 49. 6 + 9 = I, F( -2, ± v3 - I), V(-2, ±3 - I),

51.

53.

(x - 2)2 3

4

I y'3; F(O, ±I);

directrices arey

,

= ±3.

7. e

=

~;

F(±v3, 0);

directrices are x = ±3v3.

,

C(-2, -I)

(y - 3)2

+ 2 V(±v3 + 2,3),

(x - 2)'

=

-

= I, F(3, 3) andF(I, 3), C(2,3)

(y - 2)2 5 = I,

.-16 .-13

C(2,2),

F(5, 2) and F( -1,2),

v'5

V(4, 2) and V(O, 2); asymptotes: (y - 2) = ± T (x - 2)

55. (y + 1)2 - (x + 1)2 = I, C(-I,-I), F(-I, v'2 -I) andF(-I,-v'2 - I), V(-I,O)andV(-I,-2); asymptotes (y + I) = ±(x + I)

A-47

Chapter 11: Answers to Odd-Numbered Exercises

17. e =

v'2;

F(±v'2,O);

~,__ . Ull~ctrices

are x =

19. e =

1

± v'2 .

v'2;

F(O, ±4);

directrices are y

=

±2.

y

y

y

f-~ _ I --------~

6

, -./8-

-- 2

,

__ ____

,_

--~--~~~--~--~->x ~ :~~/~J 2 4 ---2 --./8,

.

-4 F,

49. rcos(o 21. e

= Vs;

~,__

F( n/iii, 0);

.

UllOctriceS are x

23. e =

2

Vs;

F(O,

±vio);

53.

.;f)

=

3

51. rcos(o

+

f)

55.

y

5

= y

directrices arey = ±2 V'TIr

= ± V'TIr

r=-2005 8

, .

------~~~--_f--->x

Radin!

--::J~~~~: ~~~,_~___

•

.

=1

0/2 ---~----

RadinI=2

--~--~~~·~·--~--_7->x

-4

c->~---.----+-------_>x

->

4

57. r

=

12 cosO

59. r

=

108inO

y

•

y

r= IOIinB

J-+(y-Sl - 2S

(;<-61.:/ - 36 r=12C0118

x' 25.y'-g=1 31. r =

2 29. r= 1 + cosO

30 1 - 5 sinO

33. r

37.

1

=

2

+ cosO

35. r

=

------~~~~------>x

10 5 - sinO

39. y

y

61. r = -2 cos 0

x- I

63. r

= -sin (J y

y

,= --'-1+0018 ;+~+t)2 -

(.;+li+l=l r _ _20016

t

r _ -sinO

--~--.---t--------->x

--------__~._------~.x

65.

67.

43.

41. y

y

y

y 4

r=4sinO

------~~~~------>x

A-48

Chapter 11: Answers to Odd-Numbered Exercises

71-

,

,

11. (a) y

~

±lxl'/2

(b)

-8- - 1

y~ ±~

~~--_~,--~-~l>. ---~-1~~---~'

,--'-

,

,

-1

x=2

4+WI!I9

, ---~--~~-~.

27. y

~ -~

29. x 2 + (y + 2)2 ~ 4

,

,

_ 1 1 2sin8

r -

2

75_ (b) - - - - - - - - - - - - - -

Planet

Perihelion

Aphelion

Mercury

0.3075 AU 0.7184 AU 0.9833 AU 1.3817 AU 4.9512 AU 9.0210 AU 18.2977 AU 29.8135 AU

0.4667 AU 0.7282 AU 1.0167 AU 1.6663 AU 5.4548 AU 10.0570 AU 20.0623 AU 30.3065 AU

Venus Eartb Mars

Jupiter Saturn Uranus Neptune

•

,

r _ -4!in8

:s?-+(y+2r- - 4

•

2

>'=-1 31.

(x - v'2)' + y2 =

,

33. r = -5 sin 8

2

,

r _ _.5sinS

Practice Exercises, pp.6SS-6S7 1.

T=

3.

,

Moo! 8

;.+~+tY=-¥

,

x

41-4..'=1

1 1=0

2

----~o+----~·

35.r=3cos8

37.

,

,

5. \

\

7. x

= 3 cos t,

y'3 9. y ~2x

I

Y = 4 sin t, 1 1

+4'4

---~r----~6~'

", ,,

), _ .1:

\ \

,

0

~ t ~

2'1T

39. d

41. 1

51. 8

53.

'IT -

43. k

3

45. i

47.

29 'IT

49.2

'IT

+4

Chapter 11: Answers to Odd-Numbered Exercises

55. Focus is (0, -I), ~

directrix is y

57. Focus is

1.

(~, 0).

directrix is x

,

,

4 81.r~I+2cose

-l

~

85. (a) 241T

,_1

2 83.r~2+sine

(b) 161T

Additional and Advanced Exercises, pp. 657-658

,

61. e

~

2; the asymptotes

arey ~ ±V3x.

,

3.

3x2+3y2-8y+4~O

7. (a)

(y - 1)2 x2 16 - 48 ~ I

5. F(O,±I)

&+~y

(b)

(i~)

,

--r---~----+~c-~'

11. x2+4,r-4 - 0

,

13.

,2->,-1 - .

\

-4

63. (x - 2)2 ~ -12(y - 3), V(2, 3), F(2, 0), directrix is y ~ 6.

65.

(x + 3)2

9

+

(y + 5)2

25

~

1,C(-3,-5),F(-3,-I)and

F(-3, -9), V(-3, -10) and V(-3, 0).

67.

(y -

2Yz)' 8

A-49

(x - 2)2 2 ~ I, c( 2, 2Yz),

F( 2, 2 Yz ± Vio), v( 2, 4 Yz) and V(2, 0), the asymptotes

,

15.

arey ~ 2x - 4 + 2Yzandy ~ -2x + 4 + 2Yz.

9X2+4y2-36=O

69. Hyperbola: C(2, 0), V(O, 0) and V(4, 0), the foci are

F( 2 ± Vs, 0) and the asymptotes are y

~ ±x

""

; 2.

71. Parabola: V(-3,1),F(-7,I)aodthedirectrixisx

~

1.

73. Ellipse: C(-3,2),F(-3 ± 0,2), V(I,2)aodV(-7,2) 75. Circle: C(I, I) and radius ~ v'2 77. V(I,O) 79. V(2, 'II') aod V(6,1T)

,

,

17. (a) r ~ e 2B

,=_2_ l+cosO

----~~~----~, (1,0)

,- --'-1-20019

(b)

4 19.r~I+2cose

Vs (e"" 2

I)

2 21.r~2+sine

23. x ~ (a + b)cose - bCOS(a; be), y~ (a + b)sine - bSin(a; be) 27.

1T

2

x2

---~

(~)

,

I

A-50

Chapter 12: Answers to Odd-Numbered Exercises 23. The vector v is horizontal and 1 in. long. The vectors u and w

:~ in. loog.

CHAPTER 12

are

Section 12.1, pp. 663-664

horizootal. All vectors must be drawn ta scale.

1. The line through the paint (2, 3, 0) parallel to the z-axis 3. The x-axis 5. The circle x 2 + y2 = 4 in the xy-plane 2 7. The circle x + z2 = 4 in the xz-plane 9. The circle y2 + z2 = I in the yz-plane 11. The circle x 2 + y2 = 16 in the xy-plane 13. The ellipse formed by the intersectian afthe cylinder x 2 + y2 = 4 and the plane z = y 15. The parabala y = x 2 in the xy-plane 17. (a) The first quadrant afthexy-plane (b) The fourth quadrant afthe xy-plane 19. (a) The ball afradius I centered at the arigin (b) All points more than I unit from the origin 21. (a) The ball afradius 2 centered at the arigin with the interior af the ball af radius I ceotered at the origin removed (b) The solid upper hemisphere af radius I ceotered at the origin 23. (a) The regioo on or inside the parabala y = x 2 in the xy-plane and all points above this region (b) The regioo 00 or to the left afthe parabala x = y2 in the xyplane and all paints above it that are 2 units or less away from the xy-plane 25. (a) x = 3 (b) Y = -I (0) z = -2 27. (a) z = I (b) x = 3 (0) Y = -I 29. (a) x 2 + (y - 2)' = 4, z = 0 (b) (y - 2)' + z2 = 4, x = a (0) x 2 + z2 = 4, Y = 2 31. (a) y = 3, z = -I (b) x = I, z = -I (0) x = I, Y = 3 2 33. x + y2 +z2 = 25,z= 3 35. a I 37. a

(a)~

"'z '"

z'"

< I

39. (a) (x - 1)2 + (y - I)' + (z - I)'

>

(b) (x - 1)2 + (y - 1)2 + (z - 1)2

&-t)'

53. (x + I)' +

55. C(-2,0,2),a

= VB

59. (a) v'y2 + z2

61. v'i7 + V:l3 65. (a) (0,3, -3)

t)' ~~ c(-i,-i,-i).a = 5~

+ (z + 57.

=

(b) v'x 2 + z2

(0)

3(ti+

29.

~(~i - ~j - ~k)

31. (a) 2i

+6

(12, -19)

9. (I, -4)

.=

(b) V 505

11. (-2, -3)

15. \ -

Yj,-t)

21. 3i

+ 5j - 8k

3. (a) (1,3) 7. (a)

13. \

~

•• __

27.5(k)

(b) -\13k

(0) l.j +.2.k 10 5

(d) 6i - 2j + 3k

7

33. 13 (12i - 5k) 35. (a) _3_ i + _4_ j - _I_k 5V2 5V2 V2 37. (a) -

(b) (1/2,3,5/2)

~i - ~j - ~k

39. A(4, -3, 5)

3

2' b

41. a =

=

G,H)

(b)

I

2

43. "'(-338.095,725.046) 1()() cos 45° 45. IFd = sin 75' '" 73.205 N, =

I~;~~~O'

'" 89.658 N,

FI = (-IFd cas 30', IFd sin 30') '" (-63.397,36.603), F2 = (IF2 Icas45',IF2 Isin45') '" (63.397,63.397) 100 sin 75' 47. W = cas 40' '" 126.093 N, IFd =

ws:;;~'

'" 106.933 N

(t,

5r)

(b)

/ I 14) \5'"5

2

'

2 (b) i + j - 2k

(0) (2,2, I)

v'iO (b)

-t, Yj)

17. -3i + 2j - k

=

5 + 1OV2 5\13 - 1OV2)

51. (a) ti + t j - 3k

Section 12.2, pp. 672-674

5. (a)

-w

(b) (5 cas 60' + IOcas315',5sin60' + IOsin315')

63. Y = I (b) (0, 5, -5)

(b) 3v'i3

tk)

tj -

49. (a) (5 cas 60', 5 sin 60') =

(0) v'x 2 + y2

w

~~ (d)

(

1. (a) (9, -6)

8+1'+"

25.

IF21

51. (x - 1)2 + (y - 2)2 + (z - 3)' = 14

(b)

u

I

41. 3 43. 7 45. 2\13 47. C( -2,0,2), a = 2V2 49. C( V2, V2, - V2), a = V2

w is vertical and u makes a 45' angle with the

19. -3i + 16j

vm 5

Section 12.3, pp. 680-682 1. (a) -25,5,5

(b) -I

3. (a) 25, IS, 5

(b)

.=.1::

5. (a) 2, v34, v3 (d) 1\ (5j -

3k)

(0) -5

t

(0)

t

(b)

2 \l3V34

(d)

(d) -2i + 4j - Vsk

~ (lOi + (0)

2 V34

llj - 2k)

Chapter 12: Answers to Odd-Numbered Exercises

7. (8) 10 (0) 10

+ vTI, V26, V21 +

vTI

V26

9. 0.75 rad

(b) 10

9.

+ vTI

(d)

11.

,

V546 vTI 10 +26 (5i + j)

___ j+k

11. 1.77 rad

13. AngleatA = cos-1 B = cos- 1

G) '"

C = cos- 1

(~)

A-51

-=-:1'---~-----+,

(~)

'"

63.435degrees,aogleat

53.130 degrees, aogle at '"

13.

63.435 degrees.

23. Horizootal compooeot '" 1188 fi/sec, vertical componeot: '" 167 fi/sec 25. (8) Since Icos 81 '" I,wehavelu·vl = lullvllcos81 s lullvl (I) = lullvl· (b) We have equality precisely wheu Icos 81 = I or wheu ooe or both ofu and v are O. In the case of nonzero vectors, we have equality wheu 8 = 0 or 'IT , that is, whoo the vectors are parallel. 27. a 35. -2x + Y = -3 33.x+2y=4

,

,

17. (a)

-21+ J i+2j

15. (8) 2v6

--~~~--~o~--*~--~, 2

v'2 2

(b) ± _1_ (2i v6

+ j + k)

(b) ± _1_ (i - j)

v'2

19. 8 21. 7 23. (8) None (b) u aod w 25. 1OV3 fi-Ib 27. (8) True (b) Not always true (0) True (d) True (e) Not always true (f) True (g) True (b) True o·v 29. (a) projyu= v.vv (b) ±uXv (oj ±(uXv)Xw

-2.;+y - -3

(d) 1(0 X v)·wl

~O4-~--------~4~~'

37. x+y=-I

39. 2x - y = 0

,

,

P(l,2)

1'(-2, 1)

------f-----~.,

--~2--~~--~~--~, -I -

x+y=-1

43. 3464 J

(I) lu l :

l 1 (b) No (0) Yes (d) No 31. (8) Yes 33. No, v need not equal w. For example, i + j '" -i + j, but i X (i + j) = i X i + i X j = 0 + k = k and i X (-i + j) = -i X i + i X j = 0 + k = k. • ,.;;:;; 11 25 35. 2 37. 13 39. v 129 41. T 43. T 45.

t

47.

vp

49. If A = ali + a2j aod B = hli + h2j, thoo

2J

i -I

41. 5 J

(e) (u X v) X (u X w)

j

k

I-J

45.

'IT

"4

47.

'IT

"6

aod the triangle's area is

49. 0.14

tlA BI ±t I:: ::1· X

Section 12.4, pp. 686-688 1. 10 x vi = 3,directionisti +

tj

7.IOXVI=6Vs,directinois~idirection is - _1_ i + ....2.... k Vs Vs

The applicable sign is ( + ) if the acute angle from A to B runs couoterclockwise in the xy-plane, and ( - ) if it runs clockwise.

+ tk;lv x ul = 3,

directioois-.2. i _Ij -.2.k 3 3 3 3. lu x vi = 0, no direclino; Iv X ul = 0, no direction 5. 10 X vi = 6, direction is -k; Iv X 01 = 6, directioo is k .vsk;lvxol=6Vs,

=

Section 12.5, pp. 694-696 1. 3. 5. 7.

x=3+1, y=-4+1, z=-I+I x = -2 + 51, y = 51, z = 3 - 51 x=O, y=21, z=1 x = I, y = I, z = I + I 9. x = I, Y = -7 + 21, z = 21 11. x = t, Y = 0, z = 0

A-52 13•

Chapter 12: Answers to Odd·Numbered Exercises

X=I~y=l.

O S I S

Z

= 2' I,

15. x

I

=

I. Y

z=O.

(I,l, l)

I I I

17.

= I + I•

-I SISO

-7iL-----r----+,

,

/

/ / /

I I I

~c_=_>----4il, 1, 0)

'-"-"*"""- :-~/~-~" I I

,

(1.0.0)

/

_____ J/

23.

21.

/

, 17. x

= 0,

z = I.

19. x = 2 - 2t. Y = 21. z=2-21. OSlsl

y = I - 21.

0

,

S

I '" I

<0.-1, 1)

<0.1, 1)

, ,

29. Y + z = 3

9~

47. 1r/4

27.

25.

,

21. 3x - 2y - z = -3 25. x + 3y + 4% = 34

37.

1

23. 7x - 5y - 4z = 6 27. (1,2,3), -2Ox + 12y + z = 7

31- x - y + z =

a

33.

2V30

35.

a 1

39. 3

41- 19/5

49. 1.38 nul

43. 5/3

51. 0.82 rad

53.

45. 9/V41

G, -to t)

55. (I. 1,0) 57. x = I - t. Y = I + t. %= -I 59. x = 4. Y = 3 + 6t, z = I + 31 61. LI intersects L2; L2 is pamllel to L3; LI and L3 are skew. 63. :x = 2 + 21, y = -4 - I, Z = 7 + 31; x = -2 - t, Y = -2 + (1 / 2)1, z = I - (3/2)t

,

,

29.

31.

65. (o,-t.-t).H.O.-3),(I.-1.0) 69. Many possible answotli. One possibility: x + y = 3 and 2y + % = 7. 71. (x/a) + lJi/b) + (z/c) = 1 describes all planes except those through the origin or paraIlel to a coordinate axis.

Section 12.6, pp. 700-701 1. (d), ellipsoid 7. (b). cylinder 13. '

"'1'_'

r

33.

35.

3. (a), cylinder 5. (I), hyperbolic paraboloid 11. (b). cone 9. (k). hyperbolic paraboloid 15.

,

,

Chapter 12: Answers to Odd-Numbered Exercises

37.

39.

A-53

23. 2v7 25. (a) v'i4 (b) I 29. v7s/3 31. x = I - 3t, Y = 2, z = 3 + 7t 33. \12 35.2x+y+z=5 37. -9x + y + 7z = 4 39.

(0,-t,-~),(-1,0,-3),(I,-I,O)

41. 'IT/3

y

41.

43. 45. 47. 49.

43.

53.

~ (5i -

55.

ei,296,_~)

y

45. (a)

2'IT(9 - e 2 ) 9

x = -5 + 5t, y = 3 - t, Z = -3t (b) x = -12t, Y = 19/12 + 15t, Z = 1/6 + 6t Yes; v is parallel to the plaoe. 3 51. -3j + 3k j - 3k)

57. (I, -2, -1); x = 1 - 5t, Y = -2 + 3t, Z = -I + 4t 59.2x+7y+2z+IO=O 61. (a) No (b) No (e) No (d) No (e) Yes 63. l1/Ylo7 65. (e) 4'ITabc

(b) S'IT

3

2

Practice Exerdses, pp. 702-703 1. (a) (-17,32)

3. (a) (6, -S) 5. (7.

'?'-

Vi3i3

(b) (b) 10

t) [assuming counterclockwise1

(~,- In)

" I. 2directionlS 9. Length =, \12'

+

I. \121.

11. v ('IT/2) = 2(-i)

· . 2.1 13. Length =, 7 directionlS 7 15. _S_i - _2_j

v'33

v'33

3. + '1' 6k -.p

+ _S_k

v'33

17. Ivl = \I2,lul = 3,v'u =

U'V

= 3,v X

U

= -2i + 2j - k,

uX v= 2i - 2j + k,lvx ul= 3,8 = cos- I lulcos8 = 19.

~(2i

~,projvU

=

~(i

(Jz)

=-';f,

+ j)

+j - k)

21.uxv=k

I x (i+J)=k

---'f---........- - y

Additional and Advanced Exerdses, pp. 704-706 1. (26,23,-1/3)

3.IFI=20Ib

5. (a) IFd = SO Ib, IF21 = 60 Ib, FI = (-4S, 64), I+J

F2 = (4S, 36), a =

tan-I~, f3

= tan-I

t

A-54

Chapter 13: Answers to Odd-Numbered Exercises

2400 13 ""

(b) IFd =

184.615Ib, IF21 =

/ -12,000 28,800) FI = \ 169 '169

""

1000 13 ""

76.923Ib,

13. v =

9. (a) 0 = tan-I V2 "" 54.74'

(-71.006,170.414),

(e) 2x - Y

13. (b) _6_

v'i4

v(l) =

(b) 0 = tan-I 2V2 "" 70.53'

+ 2x =

S

15. ",/2

" 32. 23 • 13 k I ,. 41' + 411 - 41

17. ",/2

= -1,

Z

=1+

t

21. x = t, Y

= "3I t,

Z

= t

23. (a) (i): It has constant speed I (ii): Yes (iii): Counterclockwise (iv): Yes (b) (i): It has constant speed 2 (ii): Yes

(iii): Counterclockwise (iv): Yes (e) (i): It has coostant speed I (ii): Yes (iii): Counterclockwise (iv): It starts at (0, -I) instead of (I, 0) (d) (i): It has coostant speed I (ii): Yes (iii): Clockwise (iv): Yes (e) (i): It has variable speed (ii): No (iii): Counterclockwise (iv): Yes

17. (a) 0,0 (b) -IOi - 2j + 6k, -9i - 2j + 7k (e) -4i - 6j + 21<, i - 2j - 4k (d) -IOi - 10k, -12i - 4j - Sk 19. The fonnula is always true.

25. v = 2Vsi + Vsj

CHAPTER 13 Section 13.1, pp. 713-715 1. y=x 2 -2x,

3. y =

V6(~i + ~j + ~k)

19. x = t, Y

(d) x - 2y + z = 3 + 5V6 and x - 2y + z = 3 - 5V6

2tj + tk; a = (

-2 2)i + 2j + k; (t + I) ,f0 · I. 2. Ik speed: v 6dire 0; ction: V6' + V6 1 + V6 ;

/ 12,000 5000) \169' 169 "" (71.006, 29.586)

F2 =

(~I)i + t +

~X2,

V

Section 13.2, pp. 720-724

v=i+2j, = 3i + 4j,

a=2j 1. (1/4)i + 7j + (3/2)k

a = 3i + Sj

" '" v =V2. V2. 2,. - V2 2 ·., ,. t =4: T'-T I , a = -V2 2 2 1 t = ",/2: v = -j, a = -i

3. ('" + ;V2} + 2k

5. (ln4)i + (ln4)j + (ln2)k 7 e-1.+e-1.+ k . 2' e 1 9• ,· - 1'+"'k 4

y

11. r(t) =

(-t + I} + (-t + 2)j + (-t + 3)k

13. r(t) = «t + 1)3/2 - I)i + (-e- t + I)j + (in(t + I) + I)k 15. r(t) = Sti + Stj + (-16t 2 + lOO)k

17. r(t) = (tt2+ .Jut + I}- (tt2+ Jut-2} 7. t

= 'IT: v

= 2i,

• = -j;

t =

3:: v = i - j, a

~)(3i

=-i

y

2

r ,. (I-!in 1)1 + (1- oos I)J

o

•

19. 21. 23. 25.

31. 33.

9. v = i + 2tj + 2k; a = 2j; speed: 3; direction: t i + t j +

tk;v(l) = 3(t i + tj + tk) 11. v = (-2 sin t)i + (3 cos t)j + 4k; a = (-2 cos t)i - (3 sint)j; speed: 2Vs; direction: ( -I/Vs)i + (2/Vs)k; v(",/2) = 2Vs[( -l/Vs)i + (2/Vs)k]

35.

+(lt2 + _2_t + 3)k = (It 2 + 2 v'i1 2 v'i1 - j + k) +(i + 2j + 3k) 50 sec (a) 72.2 sec; 25,510 m (b) 4020 m (e) 637S m (a) Vo "" 9.9 m/sec (b) a "" IS.4' or 71.6' 39.3' or 50.7' (b) Vo would bisect LAOR. (a) (Assuming that "x" is zero at the point of impact) r(l) = (x(l))i + (Y(I))j, where x(t) = (35 cos 27')t and y(t) = 4 + (35 sin 27')t - 16t 2 • (b) At t "" 0.497 sec, it reaches its maximum height of about 7.945 ft. (e) Range"" 37.45 ft; flight time "" 1.201 sec (d) At t "" 0.254 and t "" 0.740 sec, when it is"" 29.554 and "" 14.396 ft from where it will land (e) Yes. It changes things because the ball won't clear the net. 4.00 ft, 7.S0 ft/ sec

A-55

Chapter 13: Answers to Odd-Numbered Exercises ~

43. (aJ r(t)

(x(t))i

+ (y(t))i; where

9. T ~ 3c~sti - 3S~li + ~k, N ~ (-sint)i - (cost)i,

x(t)

~ (0.~8)(1- e-o· ")(152 cos 20° -

y(t)

~3+

0

17.6) and

K ~

(152)(1 - e-o·08')(sin200) 0.08

11. T

~

(cos I.vzSin I} +

N

~

(-costv2- Sint). + (-Sintv2+ cost).I,

+(~)(I 2

(b) (0)

(dJ (eJ

- 0.081 - e-O•O") 0.08 At I '" 1.527 sec it reaches a maximwn height of about 41.893 feet. Range'" 351.734 It; flight time '" 3.181 sec At I '" 0.877 and 2.190 sec, when it is about 106.028 and 251.530 It from home plate No

Section 13.3, pp. 727-728 1. T 3.

T~

I

Vl+t

i+

5. T

~

-cos Ii + sin Ik, ~

7 T

=

(cost -

•

_ I.

v 12 + I

K ~

~

5t, L

3V1

_ I.

v3, L ~ -4-

1. a

5". T 15.

v2 +

In(1 +

v2)

17. (aJ Cylinder isx 2 + y2 ~ I, plane is x + z ~ 1. (bJ and (cJ

(tanh ~)i. ~ (-tanh~} + (sech~)i'

~

(sech

J. 0

2i ~ lsech a a

~

VI+sin2 ldl

(cost)i - (sint)i,

3 T ~

I

.

v'1+t2 i

I

v'1+t2 '

K

i -

~

N

~

~

laiN

~

(eJ L '" 7.64

(-sinl)i - (cosl)i,

liN ~

-I

v'1+t2 '

v'1+t2

I

2( v'1+t2)'

-2e 21 VI + 4e"

(oj N

i+

(x - 'I)' +

I

VI + 4e 4t

~ -t(v'4=t'i + ti)

3. a(1)

N(~") ~ ~

y2

~

I

25. K(X) ~ Isinxl/(I + cos2x)'/2

~

4 2v'S IT + -3-N

v2 · _ 2 1

k,T(1!) 4

,?i - '?i,B(f)

-I; nonual plane: -x + y

t,p

5. (bJ coox 7. (bJ N

21.

~

5. a(O)

~

2N

+ v2 1· ~ _ v2 2 i 2'

~ k;oscu1atingplane:

0; rectifYing plane:

v2

9• B ~ (icosI)i - (isinl)i - .3.k,T ~-~ 5 5 5 25 11. B ~ k, T ~ 0 13. B ~ -k, T ~ 0 15. B ~ k, T 17. Yes. If the car is moving on a curved path (K oF 0), then tIN ~ Klvl2 oF 0 and a oF O.

Section 13.4, pp. 733-734 1. T

~)i+

7 r(1!) ~ v2 i + 42 •

z

, (dJL~

v 12 + I

----cc-'----c= t(t 2 + 1)'/2

x+y~

2~

v 12 + I

I

Section 13.5, pp. 738-739

l

13. S(I) ~ v3e' -

Ii

T~ _~i+ _~i, N~ _~

23. K(X) ~ 2/(1 + 4x2)'/2

+ (v2t /2)k,"'>2 + ". 1+1 ~

13.

K

(Sint1+1 + teast). I

ISint).1 +

11. s(t)

I

19. 1/(2b)

1+1

9. (0, 5, 24'IT)

1

e'v2

N

VI k,52 Vl+t 3

(cost~Sint),

K~--

15. T

~ (-~Sint} + (~COSI)i + ~k,3'"

3 25

i

i-

K

= cost

~

0

23. K ~ ~I 29. Components of v: -1.8701,0.7089,1.0000 Components of a: -1.6960, - 2.0307 , 0 Speed: 2.2361; Components ofT: -0.8364,0.3170,0.4472 Components ofN: -0.4143, -0.8998, -0.1369 Components ofB: 0.3590, -0.2998, 0.8839; Curvature: 0.5060 Torsion: 0.2813; Tangential component of acceleration: 0.7746 Nonual cOO3ponent of acceleration: 2.5298 31. Components ofv: 2.0000, 0, -0.1629 Components of a: 0, -1.0000, -0.0086; Speed: 2.0066 Components ofT: 0.9967, 0, -0.0812 Components ofN: -0.0007, -1.0000, -0.0086 Components ofB: -0.0812, 0.0086, 0.9967; Curvature: 0.2484 Torsion: 0.0411; Tangential component of acceleration: 0.0007 Nom3al component of acceleration: 1.0000

A-56

Chapter 14: Answers to Odd-Numbered Exercises

Section 13.6, p. 742

(0) V(I) =

1. v ~ (3a sin O)u, + 3a(1 - cos 0)0. a ~ 9a(2 cos 0 - I)u, + (18a sin 0)0. 3. v = 2ae a6u r + 2ea6u6 a ~ 4e a8(a 2 - I)u, + 8ae a8u. S. v ~ (-8sin41)U, + (4 cos 41)0. a ~ (-40 cos 41)u, - (32 sin 41)0,

d2r dl 2

-=

x2

y2

2

)2

bg ( bgt T+a 2 N Va2 + b2 a + b2

There is no component in the direction ofB. S. (a) :

= ;-cosO - r8sinO,

." (b) d dt =rXCOSu

Practice Exerdses, pp. 743-744 1. 16 +

gbl T; Va 2 + b 2

d:

= ;-sinO

,,"dO + YSInu,r dt

+ r8cosO

"" " = -xStnu +. ycosu

7. (a) a(1) ~ -9u, - 6u" v(1) = -0,. + 30, (b) 6.5 in. 9. (0) v = ;-0,. + r8u. + £k, • = (r - riP)u, + (rii + 2;-6)0, + iik

~ I y

2

.Jjt--=~ 1

o

-4

«0) •

.(0)

x

CHAPTER 14

-1

Section 14.1, pp. 753-755 At I

~

0: aT

At t

=

4: aT

1r

~

~

0, "" 7

3' tlN

=

4, K

=

= 2;

4\12 -3-, K

4\12 ------rr-

=

3. Ivl~ = I S. K ~ 1/5 7. dy/dl ~ -x; clockwise 11. Shot put is on the ground, about 66 ft 3 in. from the stopboard.

,

y

f~1+ ~ + In(f + ~I + ~)

IS. Length

=

17. T(O) =

~i - ~i + tk;

B(O)

1. (a) 0 (b) 0 (0) 58 (d) 33 3. (a) 4/5 (b) 8/5 (0) 3 (d) 0 5. Domain: all points (x,y) on 7. Domain: all points (x,y) or above line y = x + 2 not lying on the graph of y=xory=x 3

3\12 19. T(102) = _ b i vl7

///,'

,,/ .,',-r'

N(O) = .vzi + .vzi;

~ _ _I_i + -I-i + _4-k; 3\12

+ _ ~i; N(102)

~--~~~~'~--~ x

K ~ \12. 7 3'

3\12

= - _ ~i

vl7

vl7

=

1 6

+ _ bi; vl7

,

.vz cos) - (sin I)i + ( .vz cost)k;

N = (- .vzSinl} - (cosl)j - (.vzSinl)k; B = _I_i - _1-k;K = _1_;7 = 0 \12 \12 \12 2S. 1r/3 27. x = I + I, y ~ I, Z = -I

31. K ~ I/a

Additional and Advanced Exerdses, pp. 745-746 1. (a) dJdV

I

to=211'

=

1rgb 2~ a+b 2 2

gbl2 (b) 0 - 2(a 2 + b 2)'

z

11. Dnmain: all points (x, y) for whicb (x - 2)(x + 2)(y - 3)(y + 3) '" 0 3

----~~--~--~ 2--~X

J=-3

-3

gb 212

= ---'~~ 2 2 2(a

+

b )

~

9. Domain: all points (x,y) satisfyingx 2 -I,,;;y ,,;;x 2 + I

= lOT + 6N

= (

/

/

______I ______ /1C.~.-I)

v~ '

17W

23. T

"

I ' /"/'

~~----~------~x

B(102) = k;K = _8_;7 = 0 21. a(O)

I/J

,-;c ",iii. I)

A-57

Chapter 14: Answers to Odd-Numbered Exercises 13.

15.

39. (a)

(b)

y

y

x+y-l=c ~-~,~~~~~~~~.% 03

,

-,

1

o

-1

-,

y

-3

17. (a) (0) (e) 19. (a) (0)

(d) (f) 21. (a) (0)

23.

25.

27.

29.

31. 37.

(d) (f) (a) (0) (d) (e) (a) (0) (d) (e) (a) (b) (0) (d) (e) (a) (b) (0) (d) (e) (f) (a)

All points in the xy-plane (b) Allreals TIre lines y - x ~ c (d) No boundary points Both opeo and closed (f) Unbounded (b) z '" a All points in the xy-plane Farj(x,y) ~ 0, !he origin; for j(x, y) 0, ellipses with the center (0,0), andmajar andminar axes along the x- and y-axes respectively No boundary points (e) Both opeo and closed Unbounded All points in the xy-plane (b) All reals Far j(x,y) ~ 0, !he x- andy-axes; far j(x,y) 0, hyperbolas with the x- and y-axes as asymptotes No boundary points (e) Both opeo aod closed Unbounded All (x,y) satisfying x' + y' < 16 (b) z '" 1/4 Circles centered at the origin with radii r < 4 Boundary is the circle + y' ~ 16 Opeo (f) Bounded (x,y) (0,0) (b) All reals The circles with center (0, 0) and radii r > 0 Boundary is the single point (0, 0) Opeo (f) Unboonded All (x, y) satisfying -I :5 Y - x :5 I -",/2 :5 z :5 ",/2 Straight lines of the fonn y - x ~ c where -I :5 c :5 I Boundary is two straight lines y ~ I + x and y ~ -1 + x Closed (f) Unboonded Domain: all points (x, y) ootside !he circle x' + y' ~ 1 Range: all reals Circles centered at the origin with radii r > I + y' ~ 1 Boundary: Opeo (f) Unboonded 33. (a) 35. (d) (b) y

41. (a)

(b) z=-3

,. -" :1:". -1

*

:;:=0

,. 1 ,=, ,. 3

-1

-,

*

-3

43. (a)

(b)

y z _ 16

4

x'

*

%

45. (a)

(b) t=

y

1-lyl

,

(0.,0,1)

~

,. 1 0

x'

,·0

, 47. (a)

,

:1:". -1

,-0

z--1

(b)

,

, ------~------~J

-4

%

A-58

Chapter 14: Answers to Odd·Numbered Exercises 51.x+y2~4

49. x2+y2~ 10

,

,

>W

.>W

2

>W x

x

4

·2

->W 53.

,

55.

37. 39. 41. 43. 45. 47. 49. 51. 55. 59. 63. 69. 75.

(a) All (x,y,z) with z "" 0 (b) All (x,y,z) withx 2 + z2 "" I (a) All points (x, y) satisfYing z > x 2 + y2 + I Coosider paths aloog y = x, x > 0, and aloog y = x, x < O. Coosider the paths y ~ kx 2 , k a coostaot. Coosider the paths y ~ mx, m a constaot, m "" -I. Coosiderthepathsy = kx 2 ,tacoostaot,k "" O. Coosider the pathsx = I and y ~ x. (a) I (b) 0 (0) Docs not exist 57. The lintit is O. The limit is 1. (a) l(x,y)IY~~ = sin 20 where tao 0 = m 61. 0 Docs not exist 65. 7f/2 67. 1(0,0) = In 3 8 = 0.1 71. 8 ~ 0.005 73. 8 ~ 0.04 8 = Y 0.Q15 77. 8 = 0.005

Section 14.3, pp. 772-775 al _ al _ 1. ax -4x,ay - -3

y

/

~,y,iJ

al _ 3. ax - z,,(,»

al al 5. ax = 2y(xy - I), ay = z,,(xy - I)

- ,;+% - 1

7 al = • ax

x

al

y

13 al 'ax y

y

1

1

1

al ay 19.

1 ~

z"

level curve: sin-1 y - sin-1 X = '!! 2

Section 14.2, pp. 761-764 I. 5/2 IS. -I

al 'ay

3. 2\16 5. I 7. 1/2 9. I 11. 1/4 13. 0 17. 2 19. 1/4 21. I 23. 3 25. 19/12 (b) All (x, y) except (0, 0) 27. 2 29. 3 31. (a) All (x, y) 33. (a) All (x,y) except where x ~ 0 or y ~ 0 (b) All (x,y) 35. (a) All (x,y,z) (b) All (x,y,z) except the interior of the cylinder x 2 + y2 = I

~

al

-I (x + y)' -x 2 - I ~ ----''-----'0 (xy - I)'

= .x+y+1

15 al 'ax

= _1_

al x+y'ay

= _1_ x+y

-6 sin (x - 3y) cos (x - 3y) al

ax = yxY-l, ay = xYlnx

21.

al

al

ax = -g(x), ay = g(,»)

23. Ix 25. Ix

= y2, Iy = 2xy, I, = -4z = 1,ly = _y(,»2 + z 2t '/2, I, = -z(,»2 + z2)-I/2

27. Ix

=

29/= •

x

31 • / x

level curve: y

= .x+y+l

Yx2 + y2

al 17. ax = 2 sin (x - 3y) cos (x - 3y),

61. ~ - Inz ~ 2 63. x 2 + y2 + z2 ~ 4 65. Domain: all points (x,y) 67. Domain: all points (x,y) satisfYing -I :5 x :5 I and satisfYing Ixl < Iyl -1:5y:51 y-x

y

Yx2 + y2'ay

-I al 9 al = • ax (x + y)" ay al _y2 - I al 11 - ~ • ax (xy - 1)2'ay

y

al _ 2 -x - I

+ 2)'ay

yz

,Iy V I - x'y2z 2

=

xz

,I, V I - x'y2z 2

VI

= _2xe-W+r + , IY= :_Zue-{x2+r+zl) / l' , z1

39.

=

- x'y2z 2

I f= 2 f= 3 x+2y+3z' Y x+2y+3z' % x+2y+3z = _2ze-{x2+y2+z1)

)

Z

33. Ix = sech2(x + 2y + 3z),jy = 2 sech2(x + 2y I, = 3 sech2 (x + 2y + 3z) al al 35. iii = -27f sin (27ft - a), aa = sin (27ft - a) ah 37• iJp

xy

= -:r~~~

. ./. oah Sffi'l'cos 'iJq,

=

./. oah P cos." cos 'iJO

=

''/''0 -psm."sIn

Wp(P, V, 8, v, g) ~ V, Wy(P, V, 8, v, g) ~ P

Wa(P,V,6,v,g )

+ 3z),

2 + 8v 2g'

Vv2( )V8v =z.g, Wv P,V,6,v,g =T'

V8v 2 Wg(P, V, 8, v, g) = - 2g2

Chapter 14: Answers to Odd-Numbered Exercises

aj 41. ax

aj

a2j ay

a'j ax

a2j axay

a'j ayax

~

5. (a)

= I + y, iJy = I + x, 2 = 0, -2 = 0,

7. (a) :: = 4 cos vln (u sin v)

--=-=1

ag_

43. ax - 2xy

a2g -

ag_

+ Y cosx, ay -

211 - ysmx

iJx2:r

a2g

a'g ayax 45.

47.

a2g -

.

=

=

ar

axay

ay2

,

X

-1

ayax

axiJy

(x + y)'

= x'y sec2 (xy) =

a'w

axay

=

+

a'w

=

ay2

49. :

a'wa'w ayax = axay

a~ ax

-I a'r (x + y)" iJy2

-I

(x

+ y)"

au

'av

2

-y (z - y)'

au ziJu 11. (a) ax = 0, iJy = (z _ y)" az

= 0 au = 1 au = -2 'ay 'iJz azdx + azely

(b) au

tan (xy),

~;

13 dz = axdl • dl

= x' sec2 (xy),

a,

+ a'y' sec' (xy) tan (xy) + 2 tan (xy)

ay

.,

'"


= x' cos (x'y),

aydl

a,

,

,


15. aw

= aw ax + aw iJy + aw az au ax au iJy au az au' aw = awax + away + awaz av ax av ay av az av

2 2 .. • . 2.. 3x cos(Xy) - aysm(xy)

6xy cos (x'y) - 4x'y2 sin (x'y)

=

= -2V2 (In 2 - 2)

ax

+ a'y cos (x'y), ~; =

smv

au

a' sec2 (xy) tan (xy)

= sin (x'y)

+ 4u~OS2v

2 = 2u + 4uv, iJw av = -2v + 2u aw = 3 aw = _.3.


= 4xy sec' (xy)

+ 2), :~

4 cos v,

9. (a) au

a'y sec2 (xy) tan (xy) + ]x2 sec' (xy)

a~ ax

a

= -4u sin vln (u sin v)

= '1r + 1

iJw

'

(b)

x+y'ax2

a2r

a'w

-cosy

=

+

uV

+ smx,

smy

-

- Ia'r ---

x+y'ay

ayax

~z

•.

(b) :: = V2 (ln2

cPr ~;

2

a + cosx

=

-lar --

ax

(b)~ (I)

= 41 tan-I 1+ 1,

•

•

a'w = -x' sin (x'y) ay2

aw

51. ax =

53.

3

a + 3y' ay

~; = y2 +

a'w

a'w

-6

a + 3y' ayax

axiJy

(a + 3y)'

aw

2

2xy' + 3x'y',

~; = 2xy + 3x'y2 + 4x'y',

a'w = a'w = 2y + 6xy2 + la2.., ay ax ax iJy y 55. (a) x flISt (b) Y flISt (0) x flISt (d) x flISt (e) y flISt (f) Y flISt 57. J.(1,2) = -13, j,(I, 2) = -2 59. j,(-2, 3) = 1/2,j,(-2,3) = 3/4 61. (a) 3 63 12 •

65. -2

67 aA • aa

=

,

_a_ aA be sinA' ab

=

" 17. aw = awax + awiJy aw = awax + away au ax au iJy au' av ax av ay avo

a.

(b) 2

,

ecosA - b be sinA

j",,(x,y) = j",(x,y) = 0 forallpoints(x,y)

dw

3. (a) lii

= I,

(b)

~ (3)

,

. a,

"a"

"a,

a, ay

x

Section 14.4, pp. 782-783 dw (b) lii('1r)

y

y

a_ _ "-

19 az = azax + azay az = azax + azay • al ax al ay aI' as ax as ayas

89. Yes

0,

"

.

.." ". •

a.

"

71. j,(x,y) = 0 forallpoints(x,y), 3y2, y '" 0 j, (x,y) = { -2y, y < 0'

=

" a,

lnv 69. v, = (lnu)(ln v) - I

dw 1. (a) lii

•

y

ax =

0

= 1

"a,

di

iJy

di

x

..
a, ay y

iJy

"

A-59

A-60

Chapter 14: Answers to Odd-Numbered Exercises

dwau aw du as' al

aw

dwau du al

---

21_ as

•

7. Vf = 3i + 2j - 4k

• dw

dw

I I ,I " ,I " dii

dii

"

"

•

19. u = -

1.+I·(Di) I . - y'21, I. y'21 y'21' .., p. = ''''2 V,; -u = y'21

az

~ -I

az

'ay

~ -I

37.

::~2,:~~1

39.

aa~

,

~ a, - o

""

33. 12

35.-7

+ j + k), (D-J)p, = -2Y3

,

ay

,

VI - 2...J2i + 2>/2j

= 21 ."+P,

29. (a) u =

~;

= 3s 2 ."+P

41. -0.00005 amps/sec 47. (cos I, sin I, I) and (cos( -2), sin( -2), -2)

Yj, Yj) and (Yj, -Yj} minimum at ( Yj, Yj) and (- Yj, -Yj)

49. (a) Maximum at (-

(b) Max 51. 2xv'x'

~

6,min

+ x' +

~

2

r' k d l Jo 2 +x t4

Section 14.5, pp. 790-791 1.

~ (i

a-

,

j + k), (DJ)p. = 2Y3;

d

27. -4/5

25. 4/3

. ax

-y'2

~ (i +

-u = -

•

Tr

~ a, - o

" 31

a_

, a,

17. 2

.I",i_ .5",j_ .1",k,(DJ)p. = 3Y3; 3v3 3v3 3v3 -u = _ _ I_i + _5_ j + _I_k, (D-J)p = -3Y3 3Y3 3Y3 3Y3 •

ay

,

15. 3

21. u=

a-

Tr

;!j - ;! k

13. 21/13

23. u =

a_

;~ i+

11. -4

(D-uf)p. =

a"

a"

9. Vf = -

, VJ- -I+J

,

~i

~j,Duf(l, -I)

-

(b) u =

-~i+ ~j,D.f(l, -I)

(c) u =

li + i

(d) u

~

j u = _li - i j 5' 5 5 . 24.7. -I, u = 25 1 - 25 I

. 7.+24. () e u=-l,u=25 1 251 31. u ~ _7_i - _2_ j , -u ~ _ _ 7_i

v'53

v'53

v'53

+ _2_j

33. No,themaximumrateofchangeisv'i8s 35. -7/Vs

v'53

<

14.

3

Section 14.6, pp. 799-802

,

1. (a) x + y + z ~ 3 (b) x = I + 21,y = I + 21,z = 1+21 3. (a) 2x - z - 2 ~ 0 (b) x = 2 - 41,y = O,z ~2 + 21 5. (a) 2x + 2y + z - 4 = 0 (b) x ~ 21,y ~ I + 21,z ~ 2 + 1 7. (a)x+y+z-I=O (b)x=l,y=I+I,z=1 9. 2x-z-2~0 11.x-y+2z-1~0 13. x= I,y= I +21,z= 1-21 I 15. x= 1-21,y= l,z=2+21

17. x

~

I + 901,y

~

I - 901, z

19. df = 11 ,830 '" 0.0008 23. (a) ----+--~-~, 2

=-5

5

9

(-1,2)

= 5

~ sin Y3 -

(b) Y3 sin v'3 25. (a) L(x,y) = I

27. (a) L(x,y)

=

~

3

21. dg = 0

kces v'3 '" 0.935'C/ft

-

cos Y3 '" 1.87·C/sec (b) L(x,y) = 2x + 2y - I ]x - 4y + 5 (b) L(x,y) = ]x - 4y

+5

Chapter 14: Answers to Odd-Numbered Exercises

29. (a) L(x,y) = I

+x

(b) L(x,y) = -y

+

T

31. (a) W(20, 25) = 11°F, W(30, -10) = -39°F, W(15, 15) = OaF (b) W(IO, -40) "" -65.5°F, W(50, -40) "" -88°F, W(60,30) "" 10.2°F (0) L(v, 1) "" -0.36 (v - 25) + 1.337(T - 5) - 17.4088 (d) i) L(24,6) "" -15.7°F ii) L(27,2) "" -22.1°F iii) L(5, -10) "" -30.2°F 33. L(x,y) = 7 + x - 6y; 0.06 35. L(x,y) = x + y + I; 0.08 37. L(x,y) = I + x; 0.0222 39. (a) L(x,y,z) = 2x + 2y + 2z - 3 (b) L(x,y,z) = y + z (0) L(x,y,z) = 0 I I 41. (a) L(x,y,z) = x (b) L(x,y,z) = v2 x + v2 y (0) L(x,y,z) =

43. (a) L(x,y,z)

(b) L(x,y,z)

=2 +X =x - y

21r + + 21r +

- z+

(0) L(x,y,z) = x - y - z

I I

L(x, y, z) = 2x - 6y - 2z + 6, 0.0024 L(x,y,z) = x + y - z - 1,0.00135 Maximum error (estimate) s0.31 in magnitude (a) ±5% (b) ±7% "" ±4.83% Pay more attention to the smaller of the two dimensions. It will generate the larger partial derivative. 57. (a) 0.3% 59. f is most sensitive to a change in d. 61. Q is most sensitive to a change in h. 45. 47. 49. 51. 53. 55.

11'

1T'

11'

1T'

65. At- 4 ,- 2v2;atO,0;at 4 '2v2

Section 14.7, pp. 808-811 1. f(-3,3) = -5,localminimum

5.

f(3,~) =

3. f(-2,1),saddlepoint

1;,localmaximum

7. f(2,-I) = -6,localminimum 11.

f

9. f(I,2),saddlepoint

6 (\6, 0) = _17 , local maximum

13. f(O, 0), saddle point; f (

-~, ~)

=

(4, -2) 37. Absolute maximum: 4 at (2, 0); absolute minimum:

3~ at

(3,-f). (3,f). (I,-f).aud (I'f) 39. a

=

-3, b

41. Hottest is 2

2

=

r

at (

-t, '?)

and (

-t, -'?}

coldest is

-f (t, 0). at

43. (a) f(O, 0), saddle point (b) f(l, 2), local minimum (0) f(l, -2), local minimum; f( -I, -2), saddle point 49.

(E, 3;:)

51.

(t,H)

53.3,3,3

55. 12

59. 2 ft X 2 ft X I ft

61. (a) On the semicircle, max f = 2v2atl = 1r/4,minf =-2 atl = 1r.Oothequartercircie,maxf = 2v2atl = 1r/4, minf = 2atl = 0,1r/2. (b) On the semicircle, maxg = 2 att = 1r/4, ming = -2 at 1= 31r/4. 00 the quarter circle, maxg = 2atl = 1r/4, ming = a at 1= 0,1r/2. (0) On the semicircle, max h = 8 at I = 0, 1r; min h = 4 at I = 1r/2. 00 the quarter circle, max h = 8 at I = 0, minh = 4atl = 1r/2. 63. i) minf = -1/2atl = -1/2; no max ii) maxf = Oat 1= -1,O;minf = -1/2atl = -1/2 iiI') maxf = 4at 1= I;minf = Oatl = 0 9 71 20 67. y = -IT x + IT' yl.-. =-IT

Section 14.8, pp. 818-820 1. (±

Jz,t).( Jz,-t) ±

3.39

5. (3, ±3v2)

7. (a) 8 (b) 64 9. r = 2 em, h = 4 em 11. Length = 4v2, width = 3v2 13. f(O,O) = 0 is minimum,f(2, 4) = 20 is maximum. 15. Lowest = 0°, highest = 125°

17.

G, V 2,

19. I

21. (0, 0, 2), (0, 0, -2)

23. f(l, -2, 5) = 30 is maximum, f( -1,2, -5) = -30 is

I;~ , local maximum

15. f(O, 0) = 0, local minimum; f(l, -I), saddle point 17. f(O, ±Vs), saddle points; f(-2, -I) = 30, local maximum; f(2, I) = - 30, local minimum 19. f(O, 0), saddle point; f(1, I) = 2, f(-I, -I) = 2, local maxima 21. f(O, 0) = -I, local maximum 23. f(n1r, 0), saddle points, for every integer n 25. f(2, 0) = e -4, local minimum 27. f(O, 0) = 0, local minimum; f(O, 2), saddle point 29.

35. Absolute maximum: II at (0, -3); absolute minimum: -10 at

444 57. \13 X \13 X \13

22 x + JY + 3 z 31

A-61

f(t, I) = 1nen - 3, local maximum

31. Absolute maximum: I at (0,0); absolute minimum: -5 at (1,2) 33. Absolute maximum: 4 at (0, 2); absolute minimum: 0 at (0, 0)

minimwn.

25. 3, 3, 3

~ by ~ by ~ uoits

27.

31. V(8, 14) = $128

29. (±4/3, -4/3, -4/3) 33. f(2/3, 4/3, -4/3) =

37. Maximum is I

4

3

35. (2,4,4)

+ 6\13 at (± \16, \13, I), minimum is

1- 6\13 at (±\16, -\13, I). 39. Maximum is 4 at (0, 0, ±2), minimumis2 at (±v2, ±v2, 0).

Section 14.9, p. 824 1. Quadratic: x

t

+ xy; cubic: x + xy + xy'

3. Quadratic: xy; cubic: xy

A-62

Chapter 14: Answers to Odd-Numbered Exercises

t

5_ Quadratic: y +

(2xy - y2);

t (2xy - y2) + i (3x'y - 3xy2 + 2y 3) 7_ Quadratic: t (2x 2 + 2y2) = x 2 + y2; cubic: x 2 + y2 cubic: y +

9_ Quadratic: I + (x + y) + (x + y)2; cubic: I + (x + y) + (x + y)' + (x + y)3

,

. 11 1 2 ;E( ) 11. Quadratic: Z"X 2 - Z"y x,y '" 0.00134 9. -2 11. 1/2 13. I 15. Let y = kx 2 , k #' I 17. No; 1im(x3)_(0.0) I(x,y) does not exist. ag ag 19. or = cosB + sin 8, a8 = -r8m8 + reas8

Section 14.10, p. 828 1. (8) 0

(b) I

+ 2z

(0) I

V) 3. (a) au ap + au( aT nR

5. (a) 5 7.

+ 2z

(b) au (nR)

ap

V

+ au aT

al 21. aR l

I al R12' aR2

-

I al - R," aR3

I

R,'

23 ap = RT ap = nT ap = nR ap = _nRT • an v'aR v'aT v'av V2 a2g a'g 2x a2g a2g I 25. - = 0 - = - - = - = - ax2 'ay2 y3'vyax eJxcJy y2

(b) 5

(:).=ooS6

a21 2 - 2x2 a21 a'i a21 27 - 2 = -3Ox + 2 - = 0 -- = -- = I • ax (x + I)" ay2 'ay ax ax ay

Practice Exercises, pp. 829-832 1. Domain: all points in the xy-plane; range: z '" O. Level curves are ellipses with major axis along the y-axis and minor axis along the x-axis.

,

29.

~to =-1

31.

~wl =2,~wl =2-,,r (r,I) - ('II',O) S (7,') - ('11',0)

33.

dd't i t- I =

-(sin I + cos2)(sinl) + (cos I + cos2)(cosl) - 2(sin I + cos 1)(sin2)

t=9

35.

3. Domain: all (x,y) such that x #' oandy #' O;range:z #' O. Level curves are hyperbolas with the x- aod y-axes as asymptotes.

,

dyl

fix

=

-I

(%,,) - (0,1)

37. Increases most rspidly in the direction u

= -

decreasesmostrapidlyinthedirection-u

Vz

Vz

=

'? '? ,?i + '?j; i -

j;

7

Du! = Z;D_u! = --2-;D•.f = -TO whereul = 39. Increases most rspidly in the direction u

=

~i +

decreases most rapidly in the direction -u = 5. Domain: all points in xyz-space; range: all real numbers. Level surfaces are paraboloids of revolution with the z-axis as axis. t--+--1f(Z-Y, t:) _ l:l + 1-: __1 m

.t=x1+I+ 1

, 7. Domain: all (x,y,z) such that (x,y, z) #' (0,0,0); nmge: positive real numbers. Level surfaces are spheres with center (0, 0, 0) and radius r > O.

_.2. i

D.I = 7; D..,.I = -7; D• .! = 7 where Ul = 41. '1f/Vz 43. (8) lil,2) = 1,(1,2) = 2 , 45.

(b) 14/5

vfl(O,_l,I) - J+211:

, vfl(o,_l,_lj - J-2k

t

j

-

v TVi

+ ~ k;

.3. j

-

.6: k;

777

v TVi

Chapter 15: Answers to Odd-Numbered Exercises

47. Tangent: 4x - y - 5z = 4; normal line: x=2+4t,y= -l-t,z= 1-51 49. 2y - z - 2 = 0 51. Tangent: x + y = 'IT + I; normal line: y = x Y

17. f(x,y) =

~ + 4,g(x,y)

=

A-63

I +~

19. y = 2lnlsinxl + ln2 'IT

+ I

, - -.;+'11"+ I

b

21. (8) _ (2i + 7j) v53 1 2 23. w = e -C -rr t sin 1TX

(b)

~ (98i -

127j + 58k)

29,097

CHAPTER 15 53. x= 1-21,y= I,z= 1/2+21 55. Answers will depend 00 the upper bound used for

57. 59. 61.

63. 67. 69.

71. 73. 75. 77. 79. 81.

Section 15.1, pp. 840-841

Ifni, Ifryl, If", I· WithM = v2/2, lEI'" 0.0142. With M = I, lEI'" 0.02. L(x,y,z) = y - 3z,L(x,y,z) = x + y - z - I Be more careful with the diameter. dI = 0.038, % change in 1= 15.83%, more sensitive to voltage change (8) 5% 65. Local minimum of -8 at( -2, -2) Saddle point at (0, O),f(O, 0) = 0; local maximum of 1/4 at (-1/2, -1/2) Saddle point at (0, 0), f(O, 0) = 0; local minimum of -4 at <0,2); local maximum of 4 at (-2,0); aaddle point at (-2, 2), f(-2,2) = 0 Absolute maximum: 28 at (0, 4); absolute minimum: -9/4 at <3/2,0) Absolute maximum: 18 at (2, -2); absolute minimum: -17/4 at (-2,1/2) Absolute maximum: 8 at (-2, 0); absolute minimum: -I at (I, 0) Absolute maximum: 4 at(l, 0); absolute minimum: -4 at (0, -I) Absolute maximum: I at (0, ±I) and <1,0); absolute minimum: -I at(-I,0) Maximum: 5 at (0, I); minimum: -1/3 at (0, -1/3)

83. Maximum:

v'3 at ( ~, -

~,~} minimum: - v'3 at

(-~,~,-~) 85. Width =

(c::t,

depth = (b::t,beight =

(a:t

minimum+t (- 0 ' 0 ' - v2) and ( 0 ' - 0 ' v2)

(0) (I

Section 15.2, pp. 847-850 1.

(b) x'eY'& -

~)

fi

_ sinOaw aw r iJ6' ay

= . oaw + sm aT

cosoaw r

iJ6

2

-2

5.

7.

,

13.

V~abc =

----------~~--

ly -iII' I I I

~------------~~x

<8>

(b) 11. (8)

O:5x:5 2,x 3 :5y:5 8

a '" y a '" x

'" 8, a '" x '" yl/3 '" 3, x' '" y '" 3x

(b) 0 '" y '" 9, ~ '" x '"

13. (8) a '" x (b) a '" y 15. (8) 0 '" x I (b) "3 '" y 17. (8) 0 "'x

19.

(b)

a '" y

~ T

+ 2

,

-4~----------~~x

VY

'" 9, a '" y '" Vx '" 3,y' '" x '" 9 '" In 3, e~ '" y '" I '" I, -Iny '" x '" In 3

'"

I,X "'y '" 3 - 2x

3-y '" I, a '" x '" y U I '" y '" 3, a '" x '" - 2 21. 8ln8 - 16 + e

,

In,

('11",'11")

Additional and Advanced Exercises, pp. 833-834

r; = t(x' + y' + z')

-----------

, 1

I I I I

•

7. (0)

-----------

--+------------r.4~X

97. (I, -I ± 4,1),larealnumber

1. f",(O,O) = -I, f,.(O, 0) = I

,

-----

+ x'y)e"

= Oaw 91 • aw ax cos aT

3.

,

9.

87. Maximum:t at (0'0,v2)and(- 0 ' - 0'-v2}

89. (8) (2y + x'z)eY'

1. 24 3. I So 16 7. 2ln2 - I 9. (3/2) (5 - e) 11. 3/2 13. 14 15. 0 17. 1/2 19. 2ln2 21. (ln2)' 23. 8/3 25. I 27. v2

(In In 8, In 8)

I _ Iny

0

•

x 0

In In ,

A-64

Chapter 15: Answers to Odd-Numbered Exercises

23_ e - 2

51.2

y

f - - - - - y(1. 1)

y

(1,1) .1: _ ,2

>

0

---co,¥-- ---i---» o

25. ~ln2 2 29. 8

27. -1/10 53. 1/(80".)

31. 2".

•

u

-,

v _ _p

u_p

-,

/,2

p

y

(-11'/3,2)

(O.5.0.~

(~-'l)

4/.(4-Y)/2 dxdy

,•

-,•

o

>

0.'

-'1:-), _ 1

-x-]= 1 -1

y (1,1)

4 -,=4-11: 2

57. 4/3 67.

y=x

y-'"

(1,2)

61. 16

II I I I

/

(l,t)

z=l-

tx-t,

,L-----

16>; dx dy

y

+ In2)

•

/

'

,

/

~------------~~,~>

/

/ /

9

/

3

(1,1)

---::ot---;----»

-01-o +~ -----»

41.1' [~3Ydydx

43. /.'

-1)0

y

__

65. 2(1

['[(~ 39. Jo Jo

Jl 1my

~~

63. 20

2

37. [' [' dx dy

,+ ,-e"

59. 625/12

y

-.ot-+----------»

~+-

0

,

L'

xy dx dy

69.

1 71.".'

77.

/. '1'->

(x'

o

x

73. -

;2

+ y')dydx

75.

20r

=4 -

3

y

y

____

-L~>

~----_f.--------~-»

45.

>

-1

35. lL>dydx

0

y

-1

x+y - l

-x+y - I

(-2,-2)

33.

,

55. -2/3

y

[" [' (x+y)dydx

it Jinx

y

47.2 y (11",'11")

'IT

y- >

o ~+.,------------~-»

.'

•

79. Risthe set of points (x,y) such that x' + 2y' < 4. 81. No, by Fubini's Theorem, the two orders of integration must give the same result. 85. 0.603 87. 0.233

Chapter 15: Answers to Odd-Numbered Exercises

Section 15.3, p. 852 1.

1,'1,'-' dydx = I,'I,'-Y dxdy =

13. 12

2 or

3

1'1-Y' dxdy=Jt, 2

'-2y-2

2

15.

,

-12

(12,6)

,2=3%

y=~

, _ linx

12

x ~

0

NOf TO SCALE

•

y - x+2

x- -y'J(-4,-2)

-,

17.

1

(11'/4, -v2J2)

2

0

---~4----~----~0~'

,

,

y= COlI X

,

(-1,1)

,

V2 -

A-65

3 2

x

,

--oJ o ------~,--~X

{'"'r 5. Jo Jo dy dx

=

7.

1

'1,y-r dxdy-3 _1 I, o

,

' \ (2,-1)

r

(1, I)

---cojL-----~----+x

--;Cot--:"lnco,--------+x

------~~~~,--~x

y - -~

19. (a) a (b) 4/~ 21. 8/3 23. 40,000(1 - .-')In (7/2) '" 43,329

Section 15.4, pp. 857-859 1.

9.I,'1'Y 1 dxdy = 4

I

11.

'1" 1'Y

1,o x/2 1 dydx + ' 1,o y/2 Idxdy+

,

, _2

,

~r~

9

3.

*" ~ (} ~ 3:. 0 ~ r ~ esc 8

;;/2

x

1

)'/2

1

~

'"

2' a '" r

13. 36

23.

= 3

~

'" 2 cos 0

v'3

15. 2 -

19. (2 In 2 - I) (~/2)

/,'l'-x dydx 2 /,'l'-Y Idxdy=2 1

-2 '" 0

11. 2~

I I

,

2w, 0

~

7.

y·ll:

y_x

~

5. 0 '" 0 '"

{' r 1 dydx + i2f' )x/3(' 1 dydx = 4 ,

(J

~, 1 '" r '" 2 v'3 sec 0; 671'
or

Jo Jx/3 Y

f~

2 (I

21.

9.

2

17. (I - In 2) ~

+

3

V2)

Y

or

3

-+------------'---+r

1,o'1,~ xydydx

,

ll,vR xy dx dy

or

0

25.

,

,

2

x=2 y=x

------------

~~--~--~,~--~,~, , · 3-x or x - 3-)'

~L-----------~,~x

1,'1,' y'(x' + y')dydx 27. 2(~ - I)

29. 12~

or

1,'1'y,(x, + y')dxdy

31. (3~/8)

+1

2a 33'"3

2a

35'"3

A-66

Chapter 15: Answers to Odd·Numbered Exercises

37- 2".(2 -

Ve)

43. ".In 4, no

39.!

t

45.

y;

+ 5:

(b) 1

41. (a) -2-

(a 2 + 2h2)

2) _ (2 2) _ (2 2)

Section 15.5, pp. 865-868 1. 1/6

23.

3. ].1].2-",].,-",-,y/2 lizdythe, ].2].I-Y/2 ].'-"'-'Y/2 lizthedy, 1].'-"'].2-"'-1213 ]. 000

].'].1-'13].2-"'-1213

dy liz the,

00

/2].1-1/2-1'/3 thelizdy,

dy the liz,

0

2].3-31 ]. o

0

1

].o

0

29. (a) 4/3 31. (a) 5/2

37. (a) 1,=

the dy liz.

8 2 5. 1 1v'h'1 -"-Y'llizthedy, -2 -~ r2+yl 121v'4=r18-Xl-y2 -2 _~

I the liz dy

81VS=;1V8-'-Y\ 1 -va=; 4

the dy liz

218-"1~ _vs_z_x11 dy dz dx

1 -vs=; -V8-,-,' 181VS=;1~ 4

12141v' + -2 -v'00'' + ].41Yi1v'0' 12 r1~ + _~1 +

1.

I the liz dy,

].41Yil~

I the dy liz,

-Yi _~I

0

9. 6

15. 7/6 21. (a)

5(2 - yIJ) 4

11.

11].1-"11-. -I 0

dylizthe

VY the dy liz ].o1].1-'1-VY VY Y (e) ]. 11 ].I- lizthedy 0

23. 2/3

13. 18

(b)

13.

15.

].11y'i=; r1-. dy the liz

17.

= 4/5,y = Z = 2/5 = y = z = 8/15 (0)

j j

(d) ].

1].I-Y1VY

o

0

-VY

2 2 abc(a + b ) 12 ,R,.m

=

27. I

12

I)

3.

I~'"

5. "'(6Yz - 8)

(b)

[Jv'J

(0)

1.']'~].2~ r d8 liz dr

rdrlizd8

....].'" ....].4-"in. ]. ~/211+"" 'j,4 1 1

~/2].

-w/2

0

[w[,].VR

29. 16/3

0

0

/(r,8,z)lizrdrd8

/(r, 8, z) liz r dr d8

I

0

21. thelizdy

31. 8". -

32 T

Section 15.6, pp. 873-875

= 5/14,y = 38/35 3. x = 64/35,y = 5/7 = Y = 4a/(3".) 7. I, = Iy = 4?T,I0 = 8". = -I,y = 1/4 11. I, = 64/105

33.

rdrlizd8

/(r, 8, z) liz r dr d8

0

~].2

0

+

33. 2 35. 4?T 37. 31/3 39. I 41. 2 sin 4 43.4 45. a = 3 or a = 13/3 47. The domain is the set of all points (x,y, z) such that 4x 2 + 4y2 + z2 ~ 4.

1. x 5. x 9. x

= Iy = I. = 11/6

~a2 + b 2

0

25. 20/3

I,

].2~].1].~ r liz dr d8

-w/2

o -~Jr

Xl

(0)

o -v'Y

11. (a)

I ". 19':2 - If

17. 0

= 2Yz

9. "./3

dy the liz.

The value of all six integrals is 16,,-. 7. I

=

4?T(Yz 3

dy dz dx,

-2izl

I dytheliz

(b) (b)

(b) c

Section 15.7, pp. 883-886

-Vi-~

0

-2 4

I liz the dy,

Xl+yl

y'

-VS-z_yl

'" 75.28,

2 2 _ abc(a + 7b ) _ ~a2 + 7b 2 (b) h 3 ,RL 3

The value of all six integrals is I.

1-2 4 -V8-.-y'

= Y = 0, z = 12/5,1, = 7904/105

25. (a) j = y = 0, z = 8/3 27. h = 1386

0

218-Y'1~

j

Iy = 4832/63 '" 76.70,1, = 256/45 '" 5.69

33. 3

0

'].2-2z/3].1- /2-1'/3

13. x = 3/8,y = 17/16 15. j = 11/3,y = 14/27,1y = 432 17. j = O,y = 13/31,1y = 7/5 19. j = O,y = 7/10; I, = 9/IO,Iy = 3/10,10 = 6/5 (b2 M M 21 • I x-- M 3 +c ,1'-3 a +c ,1%-3 a +b

].o2~].~/212cos,p p2 sinq,dpdq,d8 = 3~'" 0

".2

23. "./3

A-67

Chapter 15: Answer.; to Odd-Numbered Exercises I I I 3. (a) x = 5(20 - v),Y = 1O(3v - u); 10 (b) Triangular region with boundaries 3v = u, v

2u, and

=

3u+v=1O

7. 64/5 11.

/,'/,'(u + v) 2:: dudv =

9.

'lTab(a' + b') 4

1

uc~svl

-usmv

cosv

1

-V3

43. 8",/3

41r( 8 - 3'\1:3) 3

1. ge - 9

dzdydz

3", - 4 47. -18

49.

3

57. 16'IT

63. 2/3

65. 3/4

= 1 = 0, Z = 3/8 X = 1 = 0, Z = 5/6

67. X

69. (x,1, z)

71.

73. I,

77. (a) (x,1,') = (0, 0, ~ (b)

(x,1,')

=

(d) 51r/3

),1,

=

2~a'

51. 5",/3

= U

ucos2 v

=

-u

3. 9/2

,

-3

(1.1)

= (0,0,3/8)

= 'IT/4

10 (1110, 10)

59. 51r/2

75.

a:~1r

~

--:cf--o----~,

o

5.

1 NOT TO SCALE

1°1',-..' ~+4

-2

dydz= -4 3

7.

1

31(112)~

-3

° ,

9 ydydz =-2

(O,O,~).I, = ~

81. 3M 1rR' 85. The surface's equationr = f(z) tells us that the point (r, 6, z) = (f(z), 6, z) will lie on the surface for all 6. In particular, (f(z), 6 + 'IT, z) lies on the surface whenever (f(z), 6, z) lies on the surface, so the surface is symmetric with respect to the z-

axis.

-,

--+.--+--->, 9. sin 4 21. 'IT

~

t

13. 4/3

15. 4/3

23. 0

25. 8/35

27. 'IT/2

11. In

2

1 1VG'jV'-"-"

o

31. (a)

!W

-V2

-~

V..,2+ y 2

3 dz dz dy

'-6+.

Section 15.8, pp. 894-896 v __ v - 2u.1 1• (a) x = u + 3 ,Y 3' 3 (b) Triangular region with boundaries u u+v=3

+

/, v31~/'V'-''-'' ° 1

= 0, v = 0, and

37. (a)

8'IT( 4\.12 3

1

- 5)

3

'

17. 1/4 19. 'IT 2(31 - 3'12) 29. 3

Vi

(fW.'.<) I I I I I I

12~

1

45. 9/4 55.

= -u sin2 V -

2

Practice Exercises, pp. 896-898

4(2\/2 - 1)'IT

53. 1r/2

61.

-~

2

15.

a 2b 2c 2 23. -6-

21. 12

/,V.-''-''

3)

+ .' '" 0.4687

ucosv

(b) I sin v

v31~

1

-USinvl = ucos v + usin v

17. (a) Icosv smv

(c)

1(

'3

13.

8 + 5; 1n2

~~dzdydz

A-68

Chapter 16: Answers to Odd·Numbered Exercises

Section 16.2, pp. 917-920 1. Vf = - (xi + yj + zk)(x 2 + y2 + z2)-3/2

817~(b' - a')

15

39. I, = 41. x

= ji = 2 -\n4

43. 10

= 104

= 2~

45. 1%

3. Vg

49.x=3~,ji=0

47. M=4,M%=O,My =0

5. F

- - 1517 + 32 - - 0 51 . () a x - 61T + 48 ,y (b)

7. 9. 11. 13. 19. 29.

Y

G2 ~ y2 } - G2 ! y2) + e'k

= -

=

(x2

fa

ky /2i,anyk> 0 (x2 + y2)' (c) 9/2 (0) 0

/2i -

+ y2)'

(a) 9/2 (b) 13/3 (a) 1/3 (b) -1/5 (a) 2 (b) 3/2 (0) 1/2 -15/2 15. 36 17. (a) -5/6

(b) 0 (0) -7/12 1/2 21. -17 23. 69/4 25. -39/2 27. 25/6 (a) Cire, = 0, eire, = 217, flux, = 217, flux2 = 0 (b) Cire1 = 0, cirC2 = 8'I'T, flUXI = 8w, flUX2 = 0 31. Circ = 0, flux = a 217 33. Cire = a 217, flux = 0

1"

35. (a) 39.

.,

(0) I

(b) 0

37. (a) 32

(b) 32

(0) 32

Y

Additional and Advanced Exerdses, pp. 898-900

11 %' 2

I. (a) _,

6

%

-

11-%'].%' 2

2

(b)

x dy fix

_,

6

0

%

dz dy ~ --

(0) 125/4

3. 217

7. (a) Hole radius = I, sphere radius = 2

5. 317/2

(b) 4V317

9. 17/4

11.

In(£)

15.

I/~

41. (a) G 43. F

17. Mass 10 19.

= a 2 oos-'(£)

= ~4 cos-'(£)

';b (eo'h' -

25. h =

I)

- bv'a 2 - b 2,

~ Va2 _

_

21. (b) I

=

(t) '-?J

Section 16.1, pp. 90S-907

11.~

9. Vi

(£)

17. V3ln 21.

!~ (e16

13. 3\1 14

31.

i

(17'/2 - I)

27.

\O~ -

= 217Vi~

(b) Vi (b) I,

+

51. 0

53.

"2I

+2+2z +C

15. -16 =

5. Not couservative

2

+C

V(x2; I)

+ tau (x + y) + tln(y2 + z2) + C

17. 1

19.9ln2

29. (a) I

21.0

(b) 1

23.-3

(0) I

f(B) - f(A) = a(xb - xa)

+ b(yb

- ya)

+ c(zb

- za) =

Section 16.4, pp. 940-942

2

29. 8

In (I + Vi)

= 417Vi~

+ y2F

F·lk.

33. 2Vi - I

35. (a) 4Vi - 2 39. (a) I,

+ 9)

23. 17 (40'/2 - 13 3/2)

25. i(5 3/2 - 7Vi - 1)

v'x 2

31. (a) 2 (b) 2 33. (a) c = b = 2a (b) c = b = 2 35. It does not matter what path you use. The work will be the same on any path because the field is conservative. 37. The force F is couservative because all partial derivatives of M, N, andP are zero. f(x,y, z) = ax + by + cz + C; A = (xa,ya, za) andB = (xb,yb, zb). Therefore, JF' dr =

(b) .1..(173/2 - I) 12

19. (a) 4Vs

- e64 )

7. Graph (f)

15. i(5Vs

3y2

11. f(x,y,z) = xlnx - x

CHAPTER 16 5. Graph (d)

=

3. Not couservative 2

9. f(x, y, z) = xeY+:z.

27. F

1

47.48

7. f(x,y,z)=x

27. 217[t -

3. Graph (g)

xi+yj v'x2 + y2

1. Conservative

13. 49

1. Graph (c)

(b) G

-yi

Section 16.3, pp. 929-931

b2 _ b: (a2 _ b2)'/2

(c) 0

V20 in., h = V60 in.

+ xi

=

37. I,

41. 1%

=

= 217

217~a3 - 2

1. 5. 9. 13. 17. 29.

Flux = 0, eire = 2'7Ta 2 3. Flux = -71'a 2 , eire = 0 Flux = 2, eire = 0 7. Flux = -9, eire = 9 Flux = -11/60, eire = -7/60 11. Flux = 64/9, eire = 0 Flux = 1/2, eire = 1/2 15. Flux = 1/5, eire = -1/12 0 19. 2/33 21. 0 23. -1617 25. l7a 2 27. 317/8 (a) 0 if C is traversed counterclockwise (b) (h - k)(area oftheregiou) 39. (a) 0

A-69

Chapter 16: Answers to Odd-Numbered Exercises

Section 16.5, pp. 951-953 1. r(r, 0) ~ (r cas O)i

+ (r sin O)j + r'I<,O

:5 r :5 2,

O::S;(J::S;21T

3. r(r, 0) ~ (r cas O)i + (r sin O)j + (r/2)1<, 0 :5 r :5 6, o :5 0 '" ",/2 5. r(r, 0) ~ (r cas O)i + (r sin O)j + v'9=? I<, 0:5 r:5 3Vi/2,0:5 0 '" 2,,,;AIsa:r(4),0) ~ (3 sin 4> casO)i + (3sin4>sinO)j + (3 cas 4»1<, 0 :5 4>:5 ",/4, o :5 0 '" 2", 7. r(4),O) ~ (V3 sin 4> casO)i + (V3sin4>sinO)j +

~ 1'~1~ sin' 4> cas' 0 d4> dO ~ ~

3. II x' du

s

~ 1'1' (4 -

5. IlzdU (falX

~

U,Y

~

17.

[IV;

~

rdrdO

19.1'~1' TV's drdO ~

7. II x'v'S - 4zdu

21.1'~14 I dudv ~ 61T

8",vS

23. Jo Jo uv'4u' + I du dv

['~J.~ 2 sin 4> d4> dO ~

Jo

~ 1'1'~ u'cas'v·v'4u' + I·

s

+ Idvdu~ [' ['~ u'(4u' +

uv'4u'

~~

11.

9. 9a'

1 15. 30 (Vi 19. -32

abe 4 (ab +

+ 6v'6)

,

21. "';

27. -73,,/6

ae

+ be)

I) cos' v dvdu

~

11" 12

13. 2

17. v'6/30 23. 13a 4/6

,

29. 18

31. "'; 39. -4

25. 2",/3

,

33. "': 41. 3a 4

"'r

['~ ['

25.

~ 3V3

v)

(V3 cos 4»1<, ,,/3 :5 4> :5 2,,/3,0 :5 0 :5 2",

9. r(x,y) ~ xi + yj + (4 - y')k, 0 :5 X :5 2, -2 '" Y '" 2 11. r(u, v) ~ ui + (3 cas v)j + (3 sin v)k, 0 :5 u :5 3, o s v ::s 21T 13. (a) r(r, 0) ~ (r cas O)i + (r sin O)j + (1 - rcosO - r sin 0)1<, 0:5 r :5 3,0'" 0 :5 2" (b) r(u, v) ~ (I - ucasv - usinv)i + (ucasv)j + (usinv)k, 0:5 u:5 3,0:5 v:5 2w 15. r(u, v) ~ (4cas'v)i + uj + (4 cas v sinv)k,0 :5 u:5 3, -(,,/2):5 v:5 (",/2); anatherway: r(u, v) ~ (2 + 2cosv)i + uj + (2 sin v)k, 0 '" u '" 3,0 '" v '" 2",

u - v)V3dvdu

s

(4

~

1. 47r 3. -S/6 5. 0 7. -61T 9. 2",a' 15. -,,/4 17. -15" 25. 161, + 16I,

(SvS - I) 6 '"

13. 12",

Section 16.8, pp. 981-983

+ 2Vi)"

1. 0 3. 0 5. -16 7. -8", 9. 3" 11. -40/3 13. 4S", 15. 12"'(4 Vi - I) 21. The integral ~ value never exceeds the sorface area af S.

,"/4

27.

Section 16.7, pp. 970-972

29.

Practice Exercises, pp. 983-986 1. Path I: 2V3; path 2: I + 3Vi 9.0

7. 8"sin(l)

,

33. (b) A

15. abe

~ 1'~1~[a'b'Sin'4>cas'4> + b'e'cas4 4> cos' 0 +

a'e' cos4 4> sin' 0]'/2 d4> dO 35. xox

+ YoY ~ 25

37. 13,,/3

39. 4

",W+!

45. ~ (17v'i7 - SvS)

49.

~(13v'i3 -

I)

51.5",Vi

41. 6v'6 - 2Vi 47. 3

+ 21n 2

53. t(5vS - I)

s

f' ['

17. 50 b c ~ (6sin4>casO)i + (6sin4>sinO)j

~

~)

+ (6cas4»k,

2;,0 :5 0 :5 2"

(r cas O)i + (rsin O)j + (1 + r)k, 0 '" r :5 2,

o :5 (} :5 2?'T

~ Jo Jo uv'4u' + I dudv ~

Osv:5w

39. 0

41. (a) 4Vi - 2 (I 16 2) I

'15'3;;;

• X,Y,z =

17v'i7 - I 4

1

25. v'6 27. "'[ Vi + In (I + Vi) 29. Conservative 31. Natconservative 33. /(x,y,z) ~ y' + yz + 2x + z 35. Path I: 2; path 2: 8/3 37. (a) 1- e-'w (b) I - e-'~

43 (- - -)

Section 16.6, pp. 960-962

[[

13. 2,,(1 -

+ .l2 + .l2

2

~ :5 4> '" 21. r(r, 0)

5. 0

23. r(u, v) ~ (u cas v)i + 2u'j + (u sin v)k, 0 '" u :5 I,

43.

1. JJ xdu

"al.l

2 19. r(4),O)

11."V3

3. 4a'

_ 3 45. z ~ 2' I,

~

7V3 -3-

(b) Vi =

232 I 45' Y

47. (x,y, z)

~

+ 1n(1 + Vi) =

64 I 15'

Z

=

956

(0,0,49/12), I,

~

640",

A-70

Appendices: Answers to Odd·Numbered Exercises

49. Flux: 3/2; cire: -1/2 57. 0

53. 3

7. (.) x

55. 2: (7 - 8v'2)

11. y

59.".

~

-I

~

5 -4x

(b) y

+6

~

13. y

4/3 ~

9. y

+4

4x

~

-x

15. y

~

-2x + 12

17. x-intercept ~ v'3,y.intereept ~ - v'2

Additional and Advanced Exerdses, pp. 986-988

y

I. 6". 3. 2/3 5. (.) F(x,y,z) ~ zi + xj + yk (b) F(x,y,z) ~ zi + yk (0) F(x,y, z) ~ zi 16'1TR' 7. 3 9. a ~ 2, b ~ 1. The minimum flux is -4. -2

6

11. (b) 13 g

~

(0) Work

4 13. (0) } '1TW

(1

gxy

ds)or ~

19. False ifF

~

19. (3, -3) 21. x 2 + (y - 2)2 ~ 4

gl xy ds ~ g

yi

2

6 13

23.

(x + v'3)2 + (y + 2)' ~ 4

H'.o)

+ xj

y

---CC-7-~+~" -4

APPENDICES Appendix 1, p. AP-6 -

-

-

25. x 2

1. 0.1,0.2, 0.3, 0.8, 0.9 or I 3. x < -2

+ (y

- 3/2)' ~ 25/4

27.

2

7. 3, -3

(3,0)

2

9. 7/6,25/6

C(O.3/2)

(2.0)

-2

"

4

15. (-00,-2]U[2,00) -2

~~~-+-L~~~ ,

13. 0 :5 z :5 10

11. -2:5 t :5 4

10

17. (-00,-3]U[I,00)

"

2

)

o

-,

1

,

134

-I (0.-1)

-2

f + (y- 31Zl = 25/4

,, 29.

19. (-3,-2)U(2,3) 21. (0,1) 23. (-00 ,I] 27. The graph of Ixl + IYI :5 I is the interior and boundary of the "diamond-shaped" region.

y y _ _ f_fi.%_S

V(-3,4)

4

y

(0.-5)

-1

0, centered at the origin 33. The washer between the circleox 2 + y2 ~ I aodx 2 + y2 ~ 4 (points with elistaoce from the origin between 1 and 2) 35. (x + 2)' + (y - I)' < 6

31. Exterior points ofa circle of radius

Appendix 3, pp. AP-16-AP-18 1. 2, -4; 2

5.

Vs

3. Unit cirele

I

37.

m~ ~-} y

(Vs' Vs). (- Vs,- Vs)

39. (-

~,-t). (~,-t)

41. (.) '" -2.5 degrees/inch (0) '" -8.3 degrees/inch -----Y'--'---+~.,

(b) '" -16.1 degrees/inch 43. 5.97 alln

;!;

A-71

Appendices: Answers to Odd-Numbered Exercises

45. Yes: C = F = -40' c

3. (8) By reflecting z across the real axis (b) By reflecting z across the imaginary axis (0) By reflecting z in the real axis and then multiplying the length of the vector by 1/ Izl' 5. (8) Points 00 the circle + y' = 4 (b) Points inside the circle + y' = 4 (0) Points ootside the circle + y' = 4 7. Points 00 a circle of radius I, center (-1,0) 9. Points 00 the line y = -x 11. 4e',,;/3 13. le'w./3

x'

51. k

=

-8, k

=

1/2

Appendix 7, pp. AP-34-AP-35 1. (8) (14,8)

(b) (-1,8)

(0) (0, -5)

x' x'

I,-k ± ~ V;± ,?i,- V;± ,?i

15. cos'O - 6 cos' 0 sin'O

+ sin'O

19.2i,-v3-i,v3-i 23. I ± v3i, -I ± v3i

21.

17.

i

ANSWERS TO ODD-NuMBERED EXERCISES CHAPTER 17

+ c2e 4x 3. y = cle-4x + clex cte-lx + c2e 1x 7. Y = cte---.l" + C2e 3x/2 3xj2 cle---x/4 + c2e 11. Y = Ct cos 3x + C2 sin 3x c, cos 5x + c, sin 5x 15. y = .X(c, cos 2x + c, sin 2x)

1. Y

= cte-3x

5. y

=

9. y = 13. y =

y = e-2t(ct cosX' + c2sinx) + 2 27. y = Acosx + Bsinx + xsinx + cos x In (cos x)

25.

Section 17.1, p. 17-7

17. y = .-x(c, cos v'3x

_

29. y-c,+c" 31• Y =

Cl COS X

33. Y = ct

+ c,sin v'3x)

.5.t"

1 2.5.t" 1 5.l' +lOx • -25xe .

I - 'ix cosx

+ C2 smx

.

+ x smx

I

+ c2e x + 'ie-x + xe x

19. y = • -2x( c, cos Vsx

35• Y = cle 5.l'

21. Y =

37. y =

Ix + C2e -x - ge -"54 c, cos x + c, sin x - (sinx)[ln (cscx +

39. Y =

Cl

25. y = 29. y = 33. y =

+ c, sin Vsx) Ct + C2X 23. Y = cle-2f + c2xe-ll: cle-3x + c2xe-3x 27. y = cte-x/2 + C2xe-x/2 cte---xj3 + c2xe---xj3 31. y = +

-te-S% te-X

.1,-Sin2v'3x 2v3

35. y = -cos 2 v'2x +

Vz

37. y = (I - 2x)e2x

39. y = 2(1

sin 2

v'2x + 2x).-"/2

+ c2e 3x 43. y = cte-x/2 + c2xe-x/2 45. y = ct cos Vsx + C2 sin VSx 47. Y = cle -xIs + C2xe --xIs 49. y = .-x/2(c, cosx + c, sinx) 51. y = C,."/4 + c,xe"/4 53. y = cte-4xj3 + c2xe-4xj3 55. y = cte---xj2 + c2e 4xj3 57. y = (I + 2x)e-x 59• Y = 11 13'-7x/' + !! 13' 2x 41. y = cte---.l"

Section 17.2, p. 17-16 1. y =

c,." + c,.-2x +

I .

Ct

+ C2e + Z-cosx - Zsmx

5• Y =

Ct

cosx

-

I 3x gcas

7. Y = c1e:h + C2e-x - 6 cosx - 2 sin x 9. Y = Clex + c2e---.l" - x 2 - 2 + txex " 11• Y - cle 13. y = Ci

I_x 7. + C2e -2x -"4e + 49 50cOSX + 50smx

+ c,.-" + x' + ~' -

Ct

+ C2e3x +

17. Y = Ct

+ c2e-x +

15. Y =

19. y = ct cosx

21. y = (c,

+ C2

2,x2

43. y

+ c,.-2x -

45. y = c, cos x = Cl cosx 47. y = celt" 49. y = celt"

+

51. Y = 2cosx

53. y =

x 4/4 - x 3

t.x+

-

3x 2 - 6x

x'/6 - x'/4

+ x/4

+ c, sin x + (x - tan x) cosx - sin x In (cos x) + C2' sinx + x cosx - (sinx) In (cos x) ~ex 5xe 3x

+ sinx - 1 + sinxIn(secx + tanx)

_.-x + I + tx' -x

55. Y = 2(e X - e~ cosx - 3e -x sinx 57. y = (I - x + x')ex

-ix'

59. YP =

3• Y =

. + C2SmX

+ c2e x -

3 10

I

x

+ C2ellx + kxe llx

41. Y = ct = Ci

cot x)]

Ex

+ ~x + txe3X

tx2 - x sinX' - !x cosx

+ C2X).~ + tx"-x

Section 17.3, pp. 17-21 to 17-23 1. my" + y' + Y = 0, y(O) = 2, y'(O) = 2 25"

5,

Vo

+ 40y = 0, y(0) = 12' y (0) = 12 5. 2q" + 4q' + 10q = 20 cos I, q(O) = 2, q'(O) = 3. 32 Y

3

7. 0.0864 It (above equilibrium) Vo

.

= 0.2917 cos (7.15521) + 85.8623 sm (7.15521) (in feet), or y = 3.5 cos (7.15521) +

9. y(l)

0.~;98 sin (7.15521) (in inches). 11. 0.308 sec

13. 8.3341b 15. 24.4949 It/sec 17. -1.56 It/sec' (acceleratioo upward) 19. q(l) = -8. -3t + 10. -2t, lim q(l) = 0

21. y(l)

= I + 2.-t -

te-

2t _

,--

~-8t

23. Y("IT) = -2 m (above equilibrium)

25. q()l -- 5I + (49VT99 . VT99 995 sm 2 I + 49 5cOS VT99) 2 I .-'/2

A-l

A-2

Chapter 17: Answers to Odd-Numbered Exercises

Section 17.4, p. 17-25

3. Y = co(1 - 21;' + ... ) + C1 (x - tx' + ... ) _

Ct

3. Y - 2 X

2

+

3 C2 X

4

7. Y = CtX- 1j3 + 5. Y = CIX + C2X 9. Y = x(Cj + c,lnx) 11. Y = X[C1 cos (Z In x) + c,sin(Zlnx)] 13. Y

=

15. Y =

=

C2

,k(C1

+ c,lnx) x

1

2S' Y =21;'+I

19. Y

27.y=x

29. Y = x[ -cos (In x)

+ Z sin (In x)]

Section 17.5, p. 17-31 1. Y = Co

+ C1(X - x 2 +

=co-

Ct -2%

2e

~x3 -

11. Y = co(1 - 3x'

= C1 + c,lnx

23. Y = x· 5I'(c1

... )

+ t x'

- ix'

+ ... ) + C1 (x +

ix' + ... )

9. Y = CO(I - x' + 15Zx' - ... ) + C1X

~[C1 cos (In x) + c,sin(lnx)] + c,lnx)

+ clsin2x + C2X 2

cocos2x CtX

7. Y = cO(1

~[Cj cos (3 In x) + c, sin (3 In x)]

1 17. Y = i(C1

21. Y

=

S. y =

+ c,lnx)

13. Y = CO(I

+ Ct

+ ... ) + C1(X

- x')

+ x' + tx' + ... )

(x + x + ~xs + ... ) 3

15. Y = CO(I -

~x' +

17. Y = CO(I -

~x' + ix' +

... ) + C1 (x

- tx'

+ ... )

... ) + C1 (x

- tx')

INDEX

a,logarithms with base, 38J-384 Abscissa, AP-l 0 Absolute change, 170 Absolute cODveIgence, 570 Absolute ConveIgencc Test, 571 Absolute extrema, fmding, 188-189 graphs of, 183 Absolute (global) maximum, 184, 187, 189, 806-808 Absolute (global) minimum, 184, 187,189, 806-808 Absolute value, AP-4-AP-6, AP-30 properties of, AP-5 Absolute wlue function, as piccewise-defined function. 5 Acceleration, 127, 129 derivative of Oork), 127 as derivative of velocity, 127, 128-129 free fall and, 128 normal component of, 734-738 in polar coordinates, 739-742 in space, 711 tangential component of, 734-738 velocity and position from, 196 Addition, of functions, 14--15 ofveckrrs,667-668 Addition fannulas, trigonometric, 26 Additivity, 903 fur definite integrals, 266 double integrals and, 846 Albert of Saxony, 559 Algebra, Fundamental Theorem of, AP-3J-AP-34 Algebra operations, vector, 667-669 Algebra rules for f'mite stUnS, 257-258 fur gradients, 789 for the naturallogaritbm, 372 Algebra systems, computer. See Computer algebra systems (CAS) Algebraic functions, 10 Alternating series, 568 harmonic, 568-569 Alternating Series Estimation Theorem, 570,593 Alternating Series Test, 568 ~eoonvention.23

Angle of elevation, 718

Angle of inclination, AP-11 Angles, 22-23 betweeu planes, 694 diIOction, 680 in standard position, 23 between vectors, 674-676 Angular velocity of rotation, 967 Antiderivative linearity rules, 232 Antiderivatives,230-235 defutition of, 230 and indefutite integrals, 235-236 motion and, 233-234 of vector function, 715 Antidifferentistion, 230 Applied optimizstion, 214-219

of area of rectangle, 216 examples from economics, 218-219 examples from mathematics and physics, 216-219 using least material, 215-216 volume of box, 214 Approximations, differential, error in, 169-170 by differential~ 168 error analysis of, 472--475 lioear, error formula for, 796, 821--ll22 Newton's Method for roots, 226 for roots and powers, 166--167 by Simpson~ Rule, 472-475 standard linear, 165, 795 tangent line, 795 by Taylor polynomials, 586 trapezoidal, 469-470 by Trapezoidal Rule, 474, 475 using parabolas, 470-472 AlbitraIy constan~ 231 AJc1ength,326-330 along curve in space, 724-726 differential fomutia for, 329-330 ofa function, 327, 726 and line integrals, 901-902 AJc length differential, 62J-624 AJc length formula, 329-330, 621, 724 AJc length panuneter, 725 Arccosine function, 405-407 identities involving, 405-407 AJcsecan~ 450

Arcsine function, 405-407 identities involving, 407 Arctangeu~ 599-600 Area, 246-248 of bounded regions in plane, 850--851 cross-sectional, 308, 309 uoder curve or graph, 247 betM:en curves, 294-296 substitution and, 291-297 ss defutite integral, 246 defutition of, 268 by double integration, 850--ll52 enclosed by astroid, 619 estimation of, lower sum and, 247-248 upper sum and, 246--247 f'mite approximations for, 248, 249 under graph of nonnegative function, 253 by Green~ The=, 940 infutite, 479 optimizing, of rectangles, 216-217 ofpamlle1ogram. 945 vector cross product as, 683 in polar coordinates, 855 ofmnoothsurface, 945-946 surfaces and, 334, 624-625, 943-953 of surfaces of revolution, 332-335, 624-625 toUU,247,280-282 Area differential, 636 Argand diagrams, AP-29-AP-30 Argumen~ AP-30 Arrow disgram fur. function, 2, 827-828 Associative laws, AP-23 Astroid, 620 length of, 622 Asymptotes, 9 fmding,209 harizonad,84,86-88,92 of hyperbolas, 643-644 oblique or slant, 88-89 vertical, 84, 91--92 def'mition of, 91 f'mding equation for, 91-92 integraods with, 481-482 ti', def'mition of, 380 derivative of, 382-383 Average rates of change, 41, 43

1-1

1-2

Index

Average speed, 39-41 over short time intervals, 40 Average value, 851-852 of continuous functions, 269-270

nonnegative, 252-253 of multivariable functions, 851-852, 864-865 Average velocity, 125 Axis(es), coordinate, AP-1O moments of inertia about, 904, 958 of ellipse, 641

slicing and rotation about, volumes by, 30&-315 spin around, 934-936

Base, a, logarithms with, 383-384 of cylinder, 308 of exponential fooctioo, 380 Bernoolli, Danie~ 185 Bernoolli, Iohann, 145 Binary search, 427-428 Binmnia\ scties, 596-598 Binomial vector, 738 Bolzano, Bernan!, 127 Boundary points, 806-807, 808, AP-3 for regions in plane, 749 for regions in space, 751

Bounded intervals, 6 Bounded regions, 749 absolute maxima and minima on, 806-808 areas of, in plane, 850-852 noorocUmgular, 841-1!42

Bounded sequences, 539-540 Box produc~ 684-686

Center of curvature, for plane curves, 731 Center of mass, 347 centroid, 352 coordinstes of, 34&-349, 904, 958 moments and, 346-355, 86S-873 of solid, 869 of thin flat plate, 34&-351 of thin shell, 959-960 of wire or spring, 904

Centroids, 352, 869-870 fluid forces and, 353

Pappoa's Theorems and, 353-354 Chain Rule, 142-146, 156, 157, 168, 378, 383, 408,409,410,619,730,739,775-781, 786,820,828,948 derivative of composite function, 142-143

for functions of three variables. 77S-779 for functions of two variables, 775-777 for implicit differentiation, 648 intuitive ''proof'' of, 143--144 "ootside-inside" rule and, 144 w;th powers offooction, 145-146 proofof,I70 repeated use of, 145 Sobstitotion Rule and, 285-286 for two independent variables and three

intermediate variables, 778-779 for vector functions, 712, 713 Change, of base in a logarithm, 384 estimating, in special direction, 794 esponential,387-388

rates of, 39-44,103-104,124-125

Brachistochrones, 614-616

sensitivity to, 131, 170--172 Charge, electrical, 978

Branch diagram(s), for multivariable Chain

Circle of curvature, for plane curves,

fides, 776,777,778, 779,780 Cable, hanging, II Calculators, to estimate !intits, 51-53 grapblng w;th, 30-34 Carbon-14 decay, 392 Cardioid, in polar coordinates, area enclosed by, 636 graphing of, 636 length of, 637--038 Carrying capacity, 521 Cartesian coordinate systems, 660--663

Cartesian coordinates, conversion to/from polar coonlUuUes,855-857 in plane, AP-IO related to cylindrical and spherical coordi-

nates, 879 related to cylindrical coordinates, 876 related to polar coordinate~ 623--030

three-dimensional. See'Ibrce-dimensional coordinate systems triple integraia in, 859-865

Cartesian integrals, changing into polar integrals, 855-857

CAS. See Computer algebra systems CAST rule, 24 Catenary, 11,423 Caucby, Augnstio-Loui~ 401

Cauchy's Mean Value Theorem, 400-401 Cavalieri, Bonaventura, 310 Cavalieri's principle, 310

731-732 Circle ofmdius,AP-14 Circles, length of, 622 osculating, 731 in plane, AP-13-AP-15 polar equation for, 652--053 standard equation for, AP-14, 18 Circulation, flux versus, 916 Circulation density, 934-936

Circulation for velocity fields, 914--915 Clairau~

Alexis, 770

Claimut's Theorem, 770 Closed curve, 915 Closed regioo, 749, 751

Coefficients, binomial, 597 determination for partia\ fractions, 460-461 of polynomial, &-9 ofpower series, 575 undetermined, 454

Combining functions, 14--22 Combining series, 549--550

Common functions, 7-11 Common logarithm fooctioo, 384 Commutativity laws, AP-23

Comparison tests, for convergence of improper integraia, 484-485

for convergence of series, 558-561 Competitive-hunter model, 524-526 Complete ordered field, AP-23-AP-24 Completeness property of real numbers, AP-23 Completing the sqnare,AP-I4-AP-15

Complex conjugate, AP-29 Complex numbers, AP-25-AP-34 imaginary part of, AP-28 real part of, AP-28

Component equation, for plane, 691 Component form of vectors, 665-667 Component functions, 7f17, 908 Component (scalar) ofu in direction of v, 678 Component test, for conservative fields, 925, 927

for exact differential form, 928 Composite fooctiona, 15-16

continuity of, 77, 761 def'mition of, 15 derivative of, 142-143 !intit of, 78 Compressing a graph, 16 Compression of a gss, uniform, 934 Compotational formulas, for torsioo, 737 Compoter algebra systems (CAS), in evalnation of improper integraia, 482-483 integral tables and, 463-464 integrate command, 465 integration w;th, 465-466 Compoter graphing, 30-34 offunctions of two variables, 752 Compoters, to estimate limits, 51-53 Concave down graph, 203 Concave up graph, 203 Concavity, 203-206

second derivative test for, 203 Conditional convorgence, 570-571 Cones, elliptical, 697, 699

parametrization of, 943 surface area of, 946 Conics, occentricity of, 648-650 in polar coordinates, 639, 648-653 polar equations of, 650--652 Connected regioo, 921

Connectedness, 80 Conservative fields, component test for, 925,927

fmding potentials for, 925-928 as gradient fields, 923 line integrals in, 927 loop property of, 924

potentials for, 925-928 and Stokes' theorem, 970 arbitrary, 231

ConsUm~

nonzero, 232 rate, 388

spring,338 eonaUmt-depth sorface, fluid force on, 341 ConsUmt force, 337, 338, 339 work done by, 679

Constant Function Rule, 712 Constant functions, def'mition of, 7 derivative of, 116 limit of, 48

Constant Multiple Rules, for antiderivatives, 232,235 furdctivatives,117-118

and divergent series, 549 for f'mite sums, 257 for gradients, 789 fur integrals, 266, 846

Index fur limits, 49 for limits of functions of two variables, 757 for limits of sequences, 536 for series, 549 Cons1rainod maximum, 811--1! 14

Constrained minimum, 811-814 Constrained variables, 824-828 Construction ofreals, AP-24-AP-25

Continuity, 39-100 See also Discontinuity of composites, 761 differentiability and, 111, 772

ofa function at a point, 74-76 at interior point, 74 on an interval, 76 of inverse functions, 364 atleft endpoin~ 74 furmultivariable functions, 75~761 partial derivatives and, 769 of vector functions, 708-709 Continuity equation of hydrodynamics, 97&--979

Continuity Test, 75 Continuous extension, 79-80 Continuous function theorem for sequences, 536

Continuous functions, 76-77 absolute extroma of, 18&--189,761 average wlne of, 269--270, 851, 864 nonnegative, 252-253 defutition of, 76, 709 differontiability and, 111-112

extreme values of, on closed bounded sets, 185, 761 integrability of, 264

Intermediate Value Theorem for, 80-81, 217. 274-275 limits of, 73 at point, 759

properties of, 76 Continuous vector field, 908 Contour curve, 750 Convergence, 264 absolute, 570 conditional,570-571 of improper integrals, 478, 480-481 interval of, 579 ofpower series, 578-579 mdiua of, 57&--579

of sequence, 533-535 of series, 554 geometric, 546 power, 575-579 of Taylor Seties, 58~595 tests fur, 483-485, 572-573 Convergence of Riemann sums, 264 Convergence Theorem for Power Series, 578

Coordinate axes, AP-l 0 moments of inertia about, 904, 958 Coordiuate conversion formulas, 883 Coordiuate ftame, left-banded, 660 right-banded, 660 Coordiuate pair, AP-IO

Coordinate planes, 660 f'rrst moments about, 904, 958 Coordinate systems, three-dimensional. See Three-dimensional coordinate systems

Coordinates, of center of mass, 348-349, 904, 948 polar, integrals in, 853--1!54 xyz, line integrals and, 911-912 Coplanar vectors, 669 Corner, 111 Cosecant, 23 Cosecant function. integral of, 374 inverse of, 407 Cosine(s), 23 integrals of products of, 447-448 integrals of products of powers of, 444-446 law of, 26-27, 675

Cosine function, derivative of, 135-137 graph of, 10 integral of, 235 inverse of, 405 Costs, f"1XCd, 130

variable, 130 Cotangent function, 23 integral of, 374 Cooraot, RichaId, 116 Critical poin~ 188, 189,208,804,808 Cross product, with determinants, 683--685 proof of distnbutive law for, AP-35-AP-36 properties of, 682--083 right-band rule for, 682 of two vectors in space, 682--683 Cross Product Rule for derivatives of vector funetiona, 712-713

Cross-sections, 308-315 horizontal, limits of integration and, 845--1!46 vertical, lintits of integration and, 845 Cube, integral over sorface of, 955 Cube root function, 8 Cubic functions, 9 Curl, k-component of, 934-936 Curl vector, 962-963

Curvature, calculation of, 729, 738 center of, 731 of plane curves, 728-733

radius of, 731 in space, 732 Curved patch element, 945 Curves, area between, 294-296 substitution and, 291-297 area under, 479

assumptions for vector integral calculus, 921-922 closed, 915 contour, 750 genemting for cylinder surface, 696 initial point of, 610 level,78&--789 negatively oriented, 936

panuneDic,610-611 panune1rically defined, leog1h of, 620-622 panune1rized, 611, 624-625 piecewise smooth, 710 plane, curvature of, 72&--733 flux across, 915-917 lengths of, 327, 329, 621--024 parametrizations of, 610-616, 707 plates bounded by two, 351-352

points of inflection of, 204--206

1-3

polar, graphing of, 624 leog1h of, 637--038

positively oriented, 936 secant to, 41 sigmoid shape, 522 ....hing, 203-210

slope of, defmition of, 41-43, 102 fmding,42, 103,632 smooth, 3-4, 326-327, 620-621 curvature of, 728--729 leog1h of, 724 speed on, 726

torsion of, 737 in """e, 707-713 an: leog1h along, 724-726 binonnals to, 734 fonnulas fur, 738 normals to, 730 parnmetric equations for, 707 vector equations for. See Vector functions tangent line to, 102 tangcllts to, 3~, 726, 78&--789

terminal point of, 610 work done by force over, 912-914 y ~ fIx), leog1h of, 327 Cusp, 111 Cycloids,614 Cylinder(s),696--097

base of, 308 pambolic, flux through, 957 parametrization of, 944 slicing with,

31~321

volume of, 308 Cylindrical coordiuates, defutition of, 875 integration of, 877--1!79

motion in, 739-740 parametrization by, 944 to rectangular coordinates, 876, 883 from sphetical coordiuates, 883 1riple integrals in, 875-879

volume differential in, 876 Cylindrical shells, 319

volumes using, 319-324 Cylindrical solid, volume of, 308 Cylindrical surface, 332

De Moivre's Theorem, AP-32 Decay, exponential, 388 Decay rate, radioactive, 391 Dectoasing function, 6, 19~200 Dedekiud, Richard, 295, 550, AP-25 Defutite integrals, 262-270 applications of, 308--360 defutition of, 246, 262-263, 285 evaluation of, by parts, 440 existence of, 262-264 Mean Value Tbrorem fur, 274-275 notation for, 263 properties of, 265-267 substitution in, 291-293 of symmetric funetiona, 293-294 of vector function, 716 Defutite integration by parts, 440 by substitution, 292 Degree, of polynomial, 9 DeI(V),786,966-968

1-4

Index

Density, 348 circulation, 934-936

flux,933 Dependent variable of function, I, 747, 824-1!25 Delivati"" product rule, 119-120, 520 Derivative quotient rule, 120--121, 145,569 Derivative rule for inverses, 365 Derivative sumrule,1l8-119,AP-8

Derivative tests, for local extreme values, 187-188,635-638

Derivatives, of absolute value function, 146 alternate formula for, 107 applicationso~184-245

calculation from def'mition, 107 of composite function, 142-143,712 of constant function, 116 eonstantmultiplerule for, 117-118

of cosine function, 136-137 Cross Product Rule, 712 def'mition of, 106 directional. See Directional derivatives Dot Product Rule, 7\2 inoconomics, 129-131 of exponential functions, 378-379

as function, 102, 106-112 general Power Rule for, 117,381 graphical behavior of functions from, 210 graphing of, 10&-109 bigher-order, 121-122, 152 ofbyperbolic functions, 417-418 ofintegmi, 276, 280

of inverse differentiable functions, 364-366 of inverse hyperbolic functions, 420-421 of inverse trigonometric functions, 411 involving log. x, 384-385, AP-27 left-handed, 109-110 Leibniz's Rule, 306

notations for, 108 nth, 122

Difference Rules, 115-122, 235 for combining series, 549 for derivatives, 115-116 for gradieo~ 789 for integmis, 266 for limits, 49 for limits of functions with two variables, 757 for limits of sequences, 536, 548 for vector functions, 712 Diff=ntiability, 109-111,764,769, 771-772 Differentiable functions, 107,618,764,772 constant multiple rule of, 117 contiouous, 111-112,620 derivatives of inverses of, 364-366 grapbof,203 on interval, 109-110 rules for, 115-121, 143, 712 Taylor's formula for, 586 Differential approximation, error in, 169--170 Differential equations, autonomous, 517 f"",t-order,496-531 applications of, 510-515 solutions, 496-498 initial value problems and, 23:l-234 separable, 38&-390 system(s) of, 523-526 trajectory of, 524 Diff=ntial forms, 92&-929 Differential formula, short form of arc length, 329-330 Diff=ntials, 164, 167-168, 796-798 definition of, 167 estimating with, 16&-169 surface area. for parametrized surface, 946 total, 797, 798

Differentiation, 102-183 of function, 107 implicit, 149-153, 779-781

one-sided, 109-110

and integration, as inverse processes, 280

partial. See Partial derivatives

logarithmic, 375

atpoin~

102-104, 110-111 of power series, 580 as rate of change, 124-131 of reciprocal function, 107 right-handed, 109-110

sccond-order, 121-122 of sino function, 135-136 of square root function, 108 symbols for, 122 of~tvecUx,

730

third, 122

of trigonometric functions, 135-139 ofvoctor function, 709-711 as velocity, 125,711 ofy = lnx, 371 ofy = sec- 1 u, 409-410 ofy = sin-1 u,408 ofy = tan-I u, 409

term-by-term for power series, 580 of vector functions, rules for, 711-714 Differentiation rules, 115-122 Direct Comparison Tes~ 483, 484 Directed line segments, 665 Direction cosines, 680 Direction, estimating clnmge in, 794 along a path, 610-611, 901-902 of vectors, 668 Directional derivatives, 784-790 calculation of, 786-788 defined, 785 as dot product, 786 estimating change with, 794 and gradients, 786 interpretstion of, 785-786 in plane, 784-785 properties of, 787

jump, 75 oscillating, 75 point of, 74 removable, 75

Discriminant (Hessian) of function, 805 Disk method, 311-312 Displacemen~ 125 def"mition of, 251-252, 279 versus distance traveled, 254 Display window, 30-33 Distance, in plane, AP-13-AP-15 and spheres in space, 662--663

in three-dimensional Cartesian coordinates, point to line, 690 point to plane, 691--692, 693--U94 point to point, 662 Distance formula, 662, AP-14 Distance traveled, 24&-250 versus displacement, 251-252, 279 estimating with fmite sums, 249--250 total,251-252 Distributive Law, AP-23 proof of, AP-35-AP-36 for vector cross products, 682 Divergence, of improper integrals, 478, 480-481 11th-term test for, 548 of sequence, 533-535 to infmity, 535 to negative inf'mity, 535 of series, 545, 548 tes~for,483-485,572-573

of vector field, 931-934, 972 Div~ceThe~,974

for other regions, 976-977 for special regions, 975--1)76 Divergent sequence, 534 Divergent series, 545, 548 Domain, connected, 921 of function, 1-3, 747,748 natural,2 of vector field, 908, 921 connected, 921 simply connected, 921 Dominant terms, 93

Domination, double integrals and, 846 Domination Rule for def'mite integmis, 266 Dot prodoc~ 674-<>79 def'mition of, 675

directional derivative as, 786 of orthogonal vectors, 517 properties of, 677--U79

Dot Product Rule for vector functions, 712 Double-angle formulas, trigonometric, 26 Double integrals, Fubini's theorem for calculating, 83&-840

over bounded nonroctangular regions, 841-842

Descartes, Rene, AP-1O

Directrix (directrices), ofellipse, 649

over rectangles, 836-840

calculating the cross 686 Jacobian, 887, 889, 890, 892 Difference quotient, 103, 107 forms for, 107 limit of, 104

ofbyperbola, 649 of parabola, 649, 651 Diricbl~ Lejeune, 479

in polar form, 853--l!57 properties of, 846-847 substitutions in, 887--891 as volumes, 837, 838

~t(s),

produc~

Discontinuity, III See also Continuity in dy/dx, 32&-329 infmite,75

Double integration, area by, 850-852 Dummy variable in integmis, 264

Index e, deImition of number, 370 .. limi~ 382 natura1 exponential and, 377-378

as series, 590-591 Eccenlricity, 648-{;53

of ellipse, 648--649 ofhyperbola, 648 ofparabola, 648

polar equation for conic with, 650 Economics, derivatives in, 129-131 examples of applied optimization from, 218-219 Einstein, Albert, 172

Electric field, 920 Electromagnetic theory (Gauss' Law), 978 Elements of set, AP-2 ffilipse Law (Kepler's First Law), 741

Ellipses, center of, 19,641 center-to-focus distance of, 642 deImition of, 488 eccentricity of, 648--649 focal axis of, 641 graphs of, 18-19

major axis of, 19,642 minor axis of, 19,642 polar equatiooa of, 651--{j52

standard equation of, 19 vertices of, 641 ffilipsoids, 697--{j98, 699

ofrevolution, 698 ffiliptical cooes, 697, 699 ffiliptical paraboloids, 699 Empty~AP-2

Endpoint extreme values, 188 Endpoint values of function, 74, 188, 189 Energy, convmsion of mass to, 172 Equal Aloa Law (Kepler's Secood Law), 741 Equations, autonomous, graphical solutions of, 516-522

differential. See Differential equations for ellipses, 643, 652

Error estimation for integral test, 555-556 Error fonnula, for linear approximations, 169,

Fligbt time, 719 Flow integrals, 914-915

796,821-822 Error term in Taylor's formula, 590 Eoler, Leoobard, 500, AP-36 Eoler's formula, AP-30 Euler's ideotity, 601--{j02

Fluid flow rates, 932 Fluid forces, and centroids, 353 on constant-depth surface, 341 integral for, against vertical flat plate, 342 work and, 337-342, 913 Fluid pressures and forces, 340-342 Flux, across plane curve, 915-917 across rectangle boundary, 932-933

Euler's method for differential equations, 49<}-502

Evaluation Theorem (Fundamental Theorem, Part 2), 277-278 Even functions, 6-7 e:f, derivative and integral of, 378-379 Exact differeotisl forms, 928-929

Expansion, unifann, of a gas, 934 Exponentisl cbange (growth or decay), 387-388 Exponentisl functiooa, 377-385

deImition of, 377 derivatives of, 378-379 description of, 10 general antiderivative of, 379 general a:f, 380 natura1, 377 Exponentisl populatioo growth model, 511-512 Exponents, Lawa of, 37<}-380 Extrema, globs! (absolute), 184-186, 188 locsl (relative), 186, 187, 200-201, 206, 803

Extreme Value Theorem, 185-186,761, AP-24 Extreme values, constrained, by Lagrange multipliers, 814

at endpoints, 188 of functions, 184-189,803-804 local (relative), dctivative tests, 187,200,206

for several variables, 803, 805 for single variable functions, 186-187

629

relating rectangular and cylindrical coordi-

nates, 876 relating spherical coordinates to Cartesian and cyliodticsl coordinates, 879 Equilibria, stable and unstable, 519 Equilibtium values, 517, 524-525 Error analysis for numerical integration, 472-475

for linear approximation, 798 in standard linear approximation, 169,796

coostao~

337, 338, 339

field of, 912 fluld pressures and, 340-342

variable along line, 338 work done by, over curve in space, 912-914 Ihroogb displacemeo~ 679 Free fall, Gallileo's law fur, 39, 127-128 Freo~ Jean-Frederic, 734 Freoet frame, 734

computational fannulas, 737 torsion in, 736-737 Fubini, Guido, 839 Fubini's llieorem for double integrals, 838-840, 843-845,854,860

Functions, 1-38 absolute value, 5

addition of, 14-15 algebraic, 10, 361

combiniog of, 14-22 commoo, 7-11

636

Fidds,~tive,920,922-928,970

pressure-depth, 341

Foci. 639--641 Forces, addition of, 667--668

Falling body, encountering resistance, 520-521 Fan-shaped region in polar coordinates, area of,

ideal projectile motion and, 718 inverse, 384 linear,AP-13 parametric. See Parametric equations for plane in space, 691--{j92

relating polar and Cartesian coordinates,

dermitioo of, 916, 956 surface integral for, 956-958 Flux density (divergence), of vector field, 933,972

Factorial notation, 539

Fermat, Pierre de, 42 Fermat's principle in optics, 217 Fibonacci numbers, 539

polar for circles, 652 polar for lioea, 652

calculatioo of, 936, 957--958

versus circulation, 916

arcsine and arccosine, 405-407 arrow diagram of, 2, 748

Eoler's ideotity, 601-602 fucua-directrix, 649 for hyperbolaa, 644

point-slope, AP-12

1-5

electric, 920 gradieo~ 923 gravitatiooal, 920

number, AP-23 ordeced, AP-23

vector, 910 Finite (bounded) intervala, 6, AP-3 Finite soms, algebra rules for, 257-260 estimating willi, 246-253 limits of, 258-259

and sigma notation, 256-258 Firing angle, 718 First Derivative Tes~ 198-201,207,803, 812-813

First derivative theorem for local extreme values, 187-188

First moments, 346--348, 868--869 about coordinate axes, 348, 869

about coordinate planes, 869, 958 Flat plate, center of mass of, 348-351, 869 vertical, fluid force and, 342

component, 707 composite. See Composite functions coostao~ 7, 48, 116,252 cootinuity of, 74, 709, 759

continuous. See Continuous functions continuous at endpoint, 74 continuous at point, 79, 709, 759 continuous extension of, 79 continuous on interval, 76 continuously differentiable, 326, 334, 620-621

cosine,23,136-137 critical point of, 188,804 coberoo~8

cubic, 8-9 decreasing, 6 dermed by formulas, 14 deImed on surfaces, 777-779 deImition of, 1 dependent variable of, 1,747 derivative as, 106-112 derivative of, 102, 103, 107, 110-111,710 differentiable. See Differentiable functions discontinuity of, 74-75, 759

domain of, 1-3, 14,747,748 even. 6-7

1-6

Index

Functions (continued) exponential. See Exponential functions

square root, 8 sytD[Dcbic,6-7,293-294,631

extreme values of, 184-189, 803--l!08, 814,817

of three variables, 750-752, 777, 789-790 total area under graph of, 281

general sine, AP-7 gradient of, 786 graphs of, 3-4, 750 sealing of, 16-18 greatest integer,S growth rates of, 424-426 Hessian offunction of two variables, 805 hyperbolic. See Hyperbolic functions

total cost, 130

identity, 7, 48, 969 implicitlydef"med, 150-151,779 increasing, 6 independent wrisble of, I, 747 input variable of, 1,747 integer ceiling, 5 integer floor, 5 integrable, 264, 716,837,859 inverse. See Inverse functions least integer, 5

left-continuous, 74 limit of, 46-54, 756 linear, 7 linearization of, 164-165,794-796

logarithmic. See Logarithmic functions machine diagram of, 2 of many variables, 781 msrginal cos~ 129-130

ttanocendental,II,361-428

trigonometric. See Trigonometric functions of two variables, 748-749, 752, 771 Cbalo Rule(s) for, 775-777

Increment Theorem of, 771 limits for, 755-759 linearization of, 794-796 partial detivatives of, 747-752, 764-766 Thylor's formula for, 820-823 unit step, 49

value of, 2 vector. See Vector functions velocity, 196,251-252,711 vertical line test for, 4-5

Fundamental Theorem of Algebra, AP-3J-AP-34 Funtlsmental The= of Calculus, 274-282, 329,435,623,921,937,981

for line integrals, 902 Part I (derivative ofintegra1), 275-276 proof of, 276-277 Part 2 (Evaluation Theorem), 277-278

Net Change Theorem, 279 proof of, 277-278

Fundamental Theorem of Line Integrals, 922

maximum and minimum values of, 184--186, 188,200,206,806-807 rnODoknllc, 198-201

of more than two variables, 761, 768-769, 79&-799

multiplication of, 14 natural exponential, deImition of, 378 natwoUlogaritlmn,369-371

nondifferentiable, 110 nonintegrable, 264 nonnegati~, area under graph of, 268-269 continuous, 252

numerical representation of, 4 odd, 6

one-to-one, 361-362 output variable of, 1,747 piecewise-<:ontinuous, 264, 304 piecewise-defined, 5 piecewise-smooth, 921 polynomial. See Polynomial functions position,S positive, area under graph of, 253 potential, 921

power, 7-8 quadratic, 9 range of, 1-3,747,748 rational. See Rational functions ",ai-valued, 2, 708, 747

:representation as poWI:lI' series, 584 rlght-continuous, 74 scalar,708

scatterplot of, 4 of several variables, 747-752 shift fonnulas for, 16 sine, 23, 135-136 in space, average value of, 864-865

Galileo Galilei, 39 free-fall fonnula, 39, 127-128

law of, 39 Gauss, Carl Ftiedrich, 258, 677 Gauss's Law, 978 General linear equation, AP-13 General Power Rule for derivatives, 381 General sine function, 27

General solution of differential equation, 233,389

Geometric series, 546-548 convergence of, 546 Gibbs, Josiah Williud, 726 Global (absolu"') maxUmun, 184, 806-808 Global (absolu"') minimum, 184,806-808 Gradient Theorem, Orthogonal, for constrained

extrerna,814 Gradient vector fields, 909-910

conservative fields as, 923 GradientvecUns,784-787 algebra rules for, 789 curlof,969

definition of, 786 to level curves, 788--789 Graphical solutions, of autonomous equations, 516-522 Graphing, willi calculators and computers, 30-34

consequence for, 80 Graphing windows, 30-33 Graphs, asymptotes of, 84-93 of common functions, 7-11 connectedness and, 80 of derivatives, 108--109 of equation, AP-IO

offunctions,3-4,14-22 offunctions with several variables, 747-752

offunctions with three variables, 750-751 offunctions with two variables, 749-750 ofpanuncbicequations, 634 in polar coordinates, 628, 631--634 synunetry tests for, 631 tocbnique for, 634 of polar curves, 634 of sequence, 533 surface area of, 950 sytD[Detric about origin, 6, 631 sytD[Detric about x-axis, 6, 631 sytD[Detric about y-axis, 6, 631 trigonometric, transformations of, 27 of trigonometric functions, 25, 32-34 ofy ~ f(x), s1rategyfor, 19&-200 Grnssmann, Hermann, 669 Gravitation, Newton's Law of, 740 Gravitstiooal coostan~ 740 Gravitstiooal field, 920 vectors in, 908 Greatest integer function, 5 as piecewise-defmed function. 5 Green's Theorem, area by, 940 circulation curl or tangential fonn, 937, 939, 964,980 comparison with Divergence Theorem, 972, 980 comparison willi Stokes'Theorem, 963, 980 divergence or normal form of, 936, 972, 980 to evaluate line in"'graIs, 93&-939 funns for, 936-938 generalization in three dimensions, 980 and llie Net Change Theorem, 937 in plane, 931-940 proof of, for special regions, 939--940 Growth, exponential, 388 logistic, 521, 522 Growth .."'s, 388, 424-428. relative, 511-512 Half-angle formulas, trigonometric, 26 Half-life, 391-392 Halley, Edmund, 200

Harmonic series, 533, 553 alternating,56&-569 Heat equation, 775 Heat 1ransfer, 392-393

Heaviside, Oliver, 458 Heaviside "cover-up" method, for linear factors, 45&-460

Height, maximum in projectile motion, 719 Helix,708

Hessian of function, 805 Higher-orderderivatives, 121-122,152,771

Hooke's law of springs, 338-339 Horizontal asympto"'s, 84, 86-88, 92, 209

Horizontal scaling and reflecting formulas, 17

Horizontal shift of function. 16 Horizontal strips, 349-350

Huygens, Christian, 613, 614 Hydrodynamics, continuity equation of, 97&-979

Index Hyperl>olas, 643-M5 branches of, 640 center of, 643

def"mition of, 643 directrices, 649 eccentricity of, 648, 649

equation of, in Cartesian coordinates, 649-{j50

focal axis of, 643 foci of, 643 polar equation of, 650

standard-fonn equations for, 644--645 YMices of, 643 Hyperl>olic functions, 416-421

def'mition of, 416-417 derivatives of, 417-418, 420-421 graphs of, 419 identities for, 416-417, 419-420 integrals of, 417-418 """",",418-419,420-421 six basic, 417 Hyperl>olic paraboloid, 698, 699 Hyperl>oloids, 697, 699 i-component of vector, 669 Ideal gas law, 824 Identity function, 7, 48, 969 Image, 887 Implicitdiff=ntiation, 149-153,779-781 furmuia for, 780

Implicit Function Theorem, 781, 948 Implicit surfaces, 948-950 Improper integrals, 478-486 approximations to, 485 of Type I, 478, 486 ofJYpe II, 481, 486 Increasing function, 6, 199--200 Increment Theorem for Functions of Two Variable., 771,AP-38-AP-40 Increments, AP-IO-AP-13 Indef'mite integrals, 235-236 See also Antiderivatives defmition of, 235, 284, 285 evaluation with substitution rule, 284--290 Indepeodentvariableoffunction, 1,747, 824-825 Indeterminate furms oflimils, 396-402, 600-@1 Indeterminate powers, 400 Index of sequence, 532 Index of summation, 256, 550 Induction, mathematical, AP-6--AP-9 Inequalities, rules for, AP-l solving of, AP-~AP-4 Inertia, moments of, 870-873 Infinite discontinuities, 75 Iofinite (unboonded) interwia, 6, AP-3 Iofinite limits, 89-90 def"mition of, precise, 90-93 of integration, 478-480 Infinite sequence, 532-533. See also Sequences Infinite series. 544-550 Iofinitesimals, AP-19 Infinity, divergence of sequence to, 535 limits at, 84-93 of rational functions, 86

Inflection, point of, 188, 204-206, 209 Initial poin~ of curve, 6 \0 of vector, 665

Initial ray in polar coordinates, 627 Initial speed in projectile motion, 718 Initial value problems, 233, 390, 497 Inner products_ See Dot product Input variable of function, 1,747 Instantaneous rates of change, 43--44 derivative as, 124-125 Instantaneous speed, 39-41 Instantaneous velocity, 125-126 Integer ceiling function (Least integer function), 5 Integer floor function (Greatest integer function), 5 Integers, AP-26 positive, ~rule for, 116-117 starting, AP-8 Integrable functions, 264-265, 837, 859 Integral furm, product rule in, 436-439 Integral sigo, 235 Integral tables, 46~64, T-I-T-6 Integral test, 55~556 error estimation, 555-556 remainder in, 555-556 Integral1heorcms for vector fields, 980-981 Integrals, approximation of, by Riemann sum, 261 by lower swns, 248 by midpoint rule, 248, 469 by Simpson's Rule, 470-472 by Trapezoidal Rule, 469-470 by upper sums, 248 Brief Table of, 435 def'mite. See Defmite integrals double. See Double integrals evaluated with inverse trigonometric functiona, 411-412 f(l/u) du, 373-374 for fluid foree against vortical flat plate, 342 of hyperbolic functions, 417-418 improper, 478-486 approximations to, 485 of Type I, 478, 486 of Type II, 481, 486 indef'Inite, 235-236, 284-285 involving lo&zx, 384-385 iterated, 838 line. See Line integrals logaritlun def'med as, 370 multiple, 836-1!37 substitution in. 887-894 nonelemeotary, 466, 598-599 polar, changing Cartesian integrals into, 855--1!57 in polar coordinates, 853-854 of powers of tan x and soc x, 446-447 of rate, 278-280 repeated, 838 of sin2 x and cos2 x, 289-290

surface, 953-960, 965 table of, 463-464, T-I-T-6 of tan x, cotx, sec x and cscx, 374-375 trigonometric, 444-448 triple. See Triple integrals

1-7

of vector fields, 910 of vector functions, 715-717 work, 337-338, 727-728, 912-914 Integraods, 235 with vortical asymptote~ 481-482 Integrate connnand (CAS), 465 Integrating factor, 505-507 Integration, 246-307 basic fannulas, 435 with CAS, 465-466 in cylindrical coordinates, 875-879 and diffi:rentiation, relationship between, 280 formulas, 411-413 limits of. See Limits, of integration numerical,468-475 by parts, 436-441 by parts formula, 436-437 of rational functions by partial fractions, 45~61

with respect to y, area between CID'VeS, 296-297 in spbctical coordinates, 881--1!83

tabular, 440-441 techniques of, 435-495 term-by-term forpower selie., 581 by trigonometric substitution, 449-452 variable of, 235, 264 in vector fields, 901-981 of vector function, 716-717 Intetior poin~ 806, AP-3 continuity at, 74 for regions in plane, 749 for regions in space, 751 Intermediate Value Property, 80 Intermediate Value Theorem, AP-24 for continuous functions, 80-81, 217, 274-275 Intermediate variable, 776-777 Intersection,lines of, 692-693 Intersection of sets, AP-2 Interval of convergence, 579 Intervals, AP-3 differentiable on, 109-110 parameter, 6HJ-{i11 types of, AP-3 Inverse equations, 378, 384 Inverse function-inverse cofunction identities, 410-411 Inverse functions, 11,361,362-363 derivative rule for, 365 and derivatives, 364-366 fmding, 36~364 bypeIi>olic,418-419,420-421 one-to-one, 361 trigonometric. See Inverse trigonometric functions Inverse trigonometric functions, 404-413 cofunctionidentities, 410-411 def'mition of, 404-405 derivatives of, 411 integrals evaluated with, 411-412 Inverses, def'Ining of, 404-405 integration and differentiation operations, 280 oflnx and number e, 377-378 of tan x, cob, sec x, and csc x, 407-408

1-8

Index

Inational numbers, AP-2 as exponents, 377-378

Limit circle, 526

Jacobi, Carl Gustav Iacob, 887

Limit Comparison Ths~ 483. 484. 560-561 Limit Laws. 46-54. 757 llieorem, 49 Limit Power Rule, 49 Limit Product Rule, proof of, AP-18--AP-19 Limit Quotient Rule. proof of. AP-I<}-AP-20

Jacobian determinan~ 887. 889. 890. 892 Jerk, 96. 97

Limit Root Rule, 49 Limit theorems, proofs of, AP-18--AP-21

Joule, James Prescott, 337

Limiting population. 521 Limits.3<}-IOO commouly occurring. 538-539. AP-21-AP-22 of continuous functions, 78

Irreducible quadratic polynomial. 454-455 ltera"'d in"'gral. 838

j-<:omponent ofvcctor, 669

Joules. 337

Jwnp discontinuity, 75 k-component of curl. 934-936

k-<:omponent of vector, 669 Kepler. Joh8llllJ:S. 741 Kepler's First Law (Ellipse Law). 741 Kepler's Second Law (Equal Area Law). 741

Kepler's Third Law (Time-Distance Law), 741-742

Kinetic energy, 172 KDvalevsky. Sonya, 421. 826 kth subinterval of partition. 260. 263 Lagrange. Joseph-Louis. 193. 811 Lagrange multipliers. 807. 811-ll18 method of. 814-816 with two constraints, 817-818

Laplace, Pierre-Simon, 770 Laplace equstion, 774 Launch angle. 718 Law of cooling. Newton's. 392-393 Law of cosines, 26-27, 675 Law of Exponential Chauge. 511

Law of refraction. 218 Laws of exponents. 37<}-380 Laws oflogarithms, 384 Least integer function, 5 Least upper bound, 539. AP-2:l-AP-24 Left-continuous functions, 74 Left-hand derivatives. 10<}-11 0 Left-hand limits. 67

deflnition of, infannal, 66--67 precise, 57--63, 68 proofof.AP-21 Left-handed coordina'" fumte. 660 Legendre, Adrien Marie, 506 Leibniz. Gottfried, AP-25 Leibniz's formula, 600 Leibniz's notation, 108. 144. 164. 167.263.621 Leibniz's Rule. 306

Leibniz's Theorem. 568 Length. along curve in space. 724-725

constant, vector functions of, 713 of curves, 326-328, 621-624

of parametrically def'med curve, 620--622 of polar coordinate curve, 637--638 of vector (magoitude). 665. 667-668 Lenses.lightenu.ing.152 Level curves, of functions of two variables, 750

Level surface, of functions oftbree variables, 750 l'H6pital. Guillaume de. 396 l'H6pital's Rule. 396. 397-399. 400-402 Iuuling limits of sequences by. 537-538 proof of. 400-402

definition of, informal, 46-47 precise, 57--63 proving llieorems willi. 62--03 testing of. 51Hi0 del..... Imding algebraically. 60-62 of difference quotient, 104 estimation of, calculators and computers for, 51-53 • (llie number) as. 382 off"mite sums, 258--259 offunction values, 46--54

for functions of two variables, 755-759 indeterminate forms of. 396-399 inImi"'. 8<}-90

precise definitions of, 90--91 at inImity, 84-93 of integration, for cylindrical coordinates, 877-879 for defini'" in"'grals. 292-293 rmding of for multiple integrals, 845-846, 854-855.860-862.877-878.881-882 wmite. 478-480. 486 for polar coordina"'s. 854-855 for rectangular coordinates, 859, 860-864 for spherical coordinates. 881-ll82 left-hand See Left-hand limits nonexistence of, two-path test for functions of two variables, 760

one-sided. See One-sided limits of polynomials. 50

power rule for. 49 of rational functions, 50 ofRiemmnn sums, 262-264 right-haod. See Right-hand limits

root rule for, 49 Saodwich Theorem. 53-54

of sequences, 534, 536 of (sin 0)/9. 6<}-71 two-sided, 66 of vector-valued functions. 708-709 Line in"'grals. 901-905 additivity and, 903 definition of, 901 evaluation of. 902. 911 by Green's Theorem. 938-939 fundamental theorem of, 922 interpretation of, 905

mass and moment calculations and, 903-905 in plane, 905 vector fields and, 907-917 xyz coordinates and, 911-912

Line segments, directed, 665 midpoint of, finding with vectors, 670 in space, 688--694 Linear approximations, error formula for, 796, 821-822 standaId, 79>-796

Linear equations, AP-13 Ill"St-order differential. 504-508 standaId fonn. 504-505

solving of, 505-507 Linear factors, Heaviside "cover-up" method for. 458-460

Linear functions, 7 Linear transformations, 888-889 Linearization, 164-166,795 offunctions of two variables, 794-796, 798 Lines, ofinter&ection. for planes, 692--693 masses along, 346-347 motion along, 125-129 nonnal, 152 tangent planes and, 791-793 para11el. AP-13 parame1ric equations for. 681Hi89

perpendicular, AP-13 and planes, in space, 688--694 polar equation for, 652 secant, 41 straight line equations, 688, AP-IO--AP-13 tangen~ 43-44. 102 vector equations for. 688-{)90 vertical, shell formula for rovolction ~ 322 work done by variable force along, 338

Liquids, incompressible, 934 pumping from containers, 340 Inx, and change of base, 384 derivative of, 371 graph and muge of. 373 int
inverse equation for, 378, AP-24 inverse of, and number e, 377-378, AP-23 properties of. 372 Local extrema. 186.200-201 Ill"St derivative test for. 188. 200-201

flI'St derivative theorem for, 187 second derivative test for, 206 Local extreme values, defmed, 186, 803 derivative tests for, 187, 803--804, 805 Local (relative) tnaJ
Logaridunic functions, 369--375 common. 384 description of, 11 natunU. 369-371. 372 range of, 373 Logarithms. algebraic properties of, 372 willi base •• 38:l-384

inverse properties of, 384 natunU. 36<}-375 Logistic population growth, 521-522 Loop. 915 Lower bound, 267 Lower sums. 247-248

Machine diagnun of function. 2 Maclaurin, Colin, 585

Index Maclaurin series, 585-586, 587

Newton"ecood law of, 171-172, 5\0, 520

Magnitude (length) of vector, 665, 667--668 Marginalcos~ 129-130,218 Marginal profi~ 218 Marginal revenue, 218 Marginals, 129 Mass. See also Center of mass along line, 346-347 conversion to enCIgy, 172 distributed over plane region, 347-348 formulas for, 348, 351, 869, 904 by line integral, 904 measurement of, 511

in polar and cylindrical coordinates, 739-740

with resistance proportional to velocity, 510-511

simple harmonic, 137-138 vector functioos and, 707, 709-711 Multiple integrals. See Double integrals; triple integrals Multiplication, of functions, 14 of power series, 580 ~,ofvec~,667-668

Multiplier (Lagrange), 807, 811-818

and moment calculations, line integrals and, 903-905 multiple integrals and, 869, 872 swn of moments of, 348 of thin shells, 958-960 value of, 172 of wire or thin rod, 903--904 Mathematical induction, AP-6-AP-9 Max-Min Inoquality Rule for dellDi'" in"'grals, 266,267,274 Max-Min Tests, 200, 206, 803, 805, 808 Maximum, absolute (global), 184, 806-808 coostrained, 811-814 local (relative), 186, 187, 803, 808 Mean value. See Average value Mean Value Theorems, 192-196, 327, 401, 621, AP-36-AP-40 Caucby~, 401 corollary I, 195 corollary 2, 195-196,372 corollary 3, 199-200,201 fur definite integrals, 274-275 for derivatives, 277 interpretation of, 194,274 mathematical consequences of, 195-196 Meode~ Gregor Johann, 131 Mesh size, 469 Midpoint of line segment in space, f"mding with vectors, 670 Midpoint rule, 247, 248 Mioimum, absolute (global), 184,806-808 coostrained, 811-814 local (relative), 187,803,808 Mixed Derivative Theorem, 770, AP-36 Mixtore problems, 513-515 Mobius band, 956 Moments, and centers of mass, 346-355, 868-873,904,958 f",,~ 868-870, 958 ofim:rtia, 870-873, 958

and mass calculations, line integrals and, 903-905

of solids and plates, 872 of system about origin. 347 of thin shells, 958-960

of wires or thin rods, 903-904 Monotonic function, 198--201 Moooinalc Sequeoce Theorem, 540-541, 553 Monotonic sequences, 539-541 Motion, along curve in space, 709-710, 734 along line, 125-129 antiderivatives and, 233-234 directioo of, 711

Napier, John, 372 Natoral domain offunctioc, 2 Natoral expooeotial functioc, 378 derivative of, 378 graph of, 377 power series for, 586, 590

Naturallogaritbm function, algebraic properties of, 372 def"lDitioo of, 369-371 power series for, 582, 600 Natorallogatithms,369-375 Natural numbers, AP-2

Negative rule, for antiderivatives, 232 Net Change Theorem, 279 and Green's Theorem, 937 Newton, Sir Issac, AP-25 Newton-Raphsoo me1hod, 225-228 law of cooling, 392-393, 519-520 Newton~ law ofgravitatioc, 740 Newton's method, 225-228 applying, 226-227 convergence of approximations, 228 procedure for, 225-226 Newton's second law of motion, 171-172,510, 520 Noodecreasing partial sums, 545 Nondecreasing sequences, 540 Newton~

Nondift"erentiable function, 110 Nooelemectary integrals, 466, 598-599

Nonintegrable functions, 264 Norm ofpartitioo, 261, 837, 876

Normal component ofacceleration, 734-738 Normal line, 152,792 Normal plane, 737 Normal vector, 732-733 Notatioos, for derivative, 108, 765-766 nth partial sum, 544-545 nth-term test for divergence, 548 Numerical integratioo, 468-475 Numerical method, 499

Numerical representation of functions, 4 Numerical solution, 499

One-tcHme functions, def"mition of, 361 horizontal line test for, 362

OPec region, 749, 751 Optics, Fermat's principle in, 217 Snell~ Law of,

218

Optimization, applied See Applied optimization Orbital period, 741

Order, comparison of functions, 426-427 Order of Integration Rule, 266 Ordered field, AP-H-AP-24

Oresme, Nicole, 534 Oriectable sorface, 956 Origio, of coor
Outside-Inside interpretation of chain rule, 144-145

.p-limits ofintegrstioo, fmding of, 881-882 p-series, 555 Paddle wheel, 966--968 Pappus (Greek, 3rd ceotury), 353

Pappus's theorem for surface area. 354-355 Pappus's theorem of volumes, 353-354 Parabola(s), 611, AP-I5-AP-16 approximatioos by, 470-472 axis of, 640, AP-15 deflDitioo of, 639 directrix of, 639, 641, 651 eccentricity of, 648, 649 focal length of, 640 focus of, 639, 641 parametrization of, 611--612 senticubical, 155 slope of, 42 vertex of, 640, AP-15 Paraboloids, 697 elliptical, 699 hyperbolic, 698, 699 volume of region enclosed by, 862-863 Parallel lines, AP-13, 692 Parallel planes, 692 slicing by, 309-310 Parallel vectors, cross product of, 682 Parallelogram, area of, 683 law of additioo, 667-068, 675 Paonn_ domain, 610, 943 Parameter interval, 610--611 Param~,610,943

Parametric curve, 611, 612 Oblique (slaot) asymptote, 88-89 Octaots, 660 Odd functions, 6-7 Oh-notatioc, 426-427 One-sided. derivatives, 109-110 One-sided limits, 66-71. See also Left-hand limits; Right-band limits

def"mition of, informal, 66 precise, 68

1-9

arc length of, 620-622, 724-725 calculus with, 618-625 differentiable, 618 graphing of, 634 Parametric equations, 610--613 of circle, 611, 622

for curves in space, 707 of cycloid, 614 of hyperbola, 613, 619

1-10

Index

Parametric equations (continued) oflines, 689--690 for projectile motion, 717-719

Parametric formulas for derivatives, 619 Parametrization, of cone, 943 of curves, 610--616, 707 of cylinder, 944 ofline, 681Hi89 of sphere, 943-944 and surface area, 944-948 of surfaces, 943--948 Partial derivatives, 747-835 calculations of, 766-768

with constrained variables, 824-828 and continuity, 769

continuous, identity for function with, 969 defmitions of, 765 equivalent notations for, 765 of function of several variables, 747-752 of function of two variables, 764-766 bigher-order, 771 socond-order, 769-770 Partial fractions, def'mition of, 454 integration of rational functions by, 453-461 method of, 454-458 Partial sums, nondecreasing, 553 nth of series, 544--545 sequence of, 545 Particular solution, of differential equation, 233 Partitiona, 836 kth subinterval of, 260 norm of, 261, 837 for Riemann sums, 259--261 parts, integration by, 436-441

Pascal, Blaise, 545 Path independence, 920-921 Path integrals. See Line integrals Path of particle, 707 Peak voltage, 289 Pendulum clock, 614 Percentage change, 170 Periodicity, of trigonometric functions, 25 Perpendicular line., AP-\3 Perpendicular (orthogonal) vectors, 676-677 Phase lines, 517 Phaso-plane anslysis, 523 limitations of, 526 Phase planes, 524 Physics, examples of applied optimization from, 216-218 Piecewise-continuous functions, 264, 304-305 Piecewise-defmed functions,S Piecewise-smooth curves, 710, 921 Piecewise-smooth surface, 954, 963 Pinching Theorem. See Sandwich Theorem Plane areas for polar coordinates, 635--637 Plane curves, circle of curvature for, 731-732 lengths of, 326-329 pararnelrizations of, 610-616 Plane regions, interior point, 749 DlIlBSes distributed over, 347-348 Plane tangent to surface, 792, 793 Planes, angles betwoen, 694 Cartesian coordinates in, AP-I0 directional derivatives in, 784-785 distance and circles in, AP-13--AP-15

equation for, 691 Green's Theorem in, 931-940 horizontal tangent to surface, 803 line integrals in, 905 lines of intersection for, 692-693 motion of planets in, 740-741 nonnal,737 osculsting, 737 pars1lel, 692 =tifying, 737 in space, 688--694 Planetary motion, Kepler's First Law (Ellipse Law) of, 741 Kepler's Second Law (Equal Aroa Law) of, 741 Kepler's Third Law (Time-Distance Law) of, 741-742 as planar, 740-741 Plate(s), thin fla~ ofmsss of, 34S-35 I two-dimensional, 869, 872 Point-slope equation, AP-12 Points, boundary, 751 of discontinuity, defmition of, 74 ofirrflection, 188,204-206 interior, 751 in three-dimensional Cartesian coordinate system, distance to plane, 693--694 ~seuille,Iean, 171 ~sson, Simeon-Denis, 802 Polar coordinate pair, 627 Polar coordinates, 627-630 ares in, 855 area of polar region, 636 conics in, 639, 648--653 definition of, 627 graphing in, 628, 631-634 symmetry "'sts for, 631 initial ray of, 627 integrals in, 853--1!54 length of polar curve, 637 motion in, 739-740 pole in, 627 related to Cartesian coordinates, 628--630 slope of polar curve, 632-633 velocity and acceleration in, 739-742 Polar equations, of circles, 652--653 of conic sections, 650--652 graphing of, 628 of lines, 652 Polyhedral surfaces, 968-969 Polynomial functions, defmition of, 8 Polynomials, coefficients of, 8--9 degree of, 9 derivative of, 11 S-119 limits of, 50 quadratic irreducible, 454-455 Taylor, 586-588, 593, 594 Population growth,logistic, 521-522 unlimited, 390-391 Population growth model, exponential, 511-512 Position, of particle in space over time, 707 Position function, acceleration and, 196 Position vector, 665 Positive integers, AP-26 power rule for, 116-117 Potential function, 921 Potentials, for conservative fields, 925-928

con""

Power Chain Rule, 145-146 Power functions, 7-8 Power Rule, for derivatives, 115-116 for limits, 49 for limits of functions of two variables, 757 natnmllogarithms,372 for positive integers, 116-117 proof of, 381 Power series, 575-582 convergence of, 575-578 radius of, 578 testing of, 578 multiplication of, 579, 580 reciprocal,575-576 term-by-term differentiation of, 580 term-by-term integration of, 581 Powers, AP-32 binomial series for, 597-598 indeterminate, 400 of sines and cosines, products of, 444-446 Preimsge, 887 Pressure-depth equation, 341 Pressures, fluid, and forces, 340-342 Principal unit normal vector, 732, 738 Product Rule, for derivatives, 119-120 fur gradien~ 789 in in"'gra1 form, in~tion by parts, 436-439 for limits, 49 of functions with two variables, 757 proof of, 120 for natural logarithms, 372 for power series, 580 for sequences, 535 Products, of comples numbers, AP-30-AP-31 of powers of sines and cosines, 444--446 and quotients, derivatives of, 119-120 of sines and cosines, 447-448 Projectile motion, vector and parametric equations for, 717-719 with wind gusts, 719-720 Projection, of vectors, 677--679 Proportionality relationahip, 7 Pumping liquids from containers, 340 Pyramid, volume of, 309 Pythagorean theorem., 25, 27, 28, AP-13, AP-26 Quadnmts, of coordina'" sys_, AP-I 0 Quadratic approximations, 586 Quadratic polynomial, irreducible, 454-455 Quadric surfaces, 697--699 Quotient Rule, 51 for derivatives, 120-121, 145 for gradien~ 789 for limits, 49 of functions with two variables, 757 proof of, AP-I9-AP-20 for natural logarithms, 372 for sequences, 535 Quotients, AP-31 products and, derivative. of, 119-120

p-limits of integration, :fmding of, 881 r-limits of integration, 854, 878 Radian measure, 146

Index Radians,22-23,24 Radioactive decay, 391 Radioactive elements, half-life of, 391-392 Radioa
Radius, circle of, AP-14

ofconver,gence, 579 ofpower series, 578-579 of curvature, for plane curves, 731-732 Radius units, 22 ~,of~ction,I-3,747, 748 in projectile motion, 719, 720 Rate constant, exponential change, 388

Remainder, estimating of, in Taylor's Theorem, 590,591-592 in integral test, 555-556

of order n, def'mition for Taylor's fonnula, 590

Remainder Estimation Theorem, 591, 593 Removable discontinuities, 75 Representation of function, power series, 584-585

Resistance, falling body encountering, 520--521 proportional to velocity, motion with, 510-511

~s,ofchange,39-44

Rest mass, 172

average,41 instantaneous, derivative as, 43-44 Ratio, in geometric series, 546

Rest points, 517, 524-525

Ratio Tes~ 563-565, 576-577, 579, 597

Revolution, about y-axis, 334-335 areas of surfaces of, 624--625 ellipsoid of, 698 solids of, disk method, 311-313

Rational ~ctions, def'mition of, 9 domain of, 9 integration of, by partial fractions, 453-461 limits of, 50 at inImity, 86 Ratiooal numbers, AP-2, AP-26

Real numbers, construction of reals and, AP-24-AP-25 development of, AP-26-AP-27

properties of, algebraic,AP-l, AP-23 completeness, AP-l, AP-23 order,AP-l,AP-23 andrealline,AP-1 1heory of, AP-23-AP-25 Real-valued func1ioos, 2, 747, 546

Reals, construction of, AP-24-AP-25 Rearrangement theorem. for absolutely ~entseries,572

Rcciproea1 Rule for natorallogari1hms, 372

Rectangles, approximating area of, 247-248 def'tning Riemann sums, 260--261 double integrals over, 836-840

optimizing area of, inside circle, 216-217 Rectangular coordinates. See Cartesian coordinates Reetifying plane, 737

Recursion fonnula, 539 Recursive def'mitions, 539 Reduction fannula, 439, 464 Reflection of graph, 16-18 Regiooa, bounded 749 closed, 749, 751 counecl
general, double integrals over, 841-847 open, 749, 751, 970

plane, interior point, 749 masses distributed over, 347 simply connected, 921 solid, volume of, 842-845 in space, interior point, 751 volume of, 860 special, divergence theorem for, 975-976 Green's Theorem for, 939--940 oobounded, 749

Reindexing inf'mite series, 550 ReJal
Relative change, 170 Relati.. growth ral<, 511-512 Relative (local) extrema, 186, 187,803

unstable, 526 Resultant wolar, 667--068

washer method, 314-315 Riemano, Georg Friedrich Bemluud, 259

Riemann sums, 259-261, 278, 280, 309, 320, 322,333,340,341,348,876,880,901, 953

convergence of, 264 limits of, 262-264

Right-<:ontinuous functions, 74 Right-hand limits, 67

def'tnition of, 68 proof of, AP-20 Right-handed coordinal< frame, 660 Right-handed derivatives, 1O<}'-110

Rise,AP-11 RL circuits, 508 Rolle, Michel, 192 Rolle's Theorem, 192-193 Root Imding, 80 Root rule, for limits, 49 for limits of functions of two variables, 757 RootTes~ 565-566, 579 Roots, AP-32-AP-33 binomial series for, 597-598 rmding by Newton's Method, 225-228 and Intermedial< Value The=. 80-81 FUrtatioo,311-314 uniform, 934 Run,AP-11 Saddle points, 698, 804, 805, 806, 808

Sandwich Theorem, 53, 69, 70, 87 proof of, AP-20

for sequences, 536 Sealar func1ioos, 708

Scalar Multiple Rules for vector ~ctions, 712 Scalar multiplication ofvectOIs, 667--668 Scalar products. See Dot prodocts Scalars, def'mition of, 667 Scaling, of func1ioo graph, 16-18 Scatterplo~ 4

Sec x, integrals of powers of, 446--447 inverse of, 407-408 Secant, trigonometric function, 23 Secant function, inverse of, 418-419 Secant lines, 41 Seeant slope, 42

1-11

Second derivative test, 203, 206, 808 derivation of, two-variable function, 820--821 Second moments, 870--872, 958 Second-order differential equation topics cuvered online applications ofsecond-order equations auxiliary equation boundary-value problems complementary equation damped vibrations critical damping averdamping underdamping Euler equation Euler:r method existence of second-order solutions form of second-order solutions forced vibrations general solution for linear equations homogeneous equations linear combination linearly independent solutions linearity method ofundetermined coejficients 1Wnhomogeneous equations power series solutions U1Iiqueness ofsecond-order solutions second-order differential equations second-order initial value problems second-order linear equations second-order series solutions simple harmonic motion solution ofconstant-coejficient second-order linear equations superposition principle theorem on general solution form variation ofparameters Second-order partial derivatives, 769-770 Separable differential equatiooa, 38S-389

Sequences, 532-541 boooded, 53<}'-540 calculatioo of, 535-538 convergence of, 533-535 divergence of, 533-535 to inImity, 535 to negative inf'mity, 535

index of, 532 inImite, 532-533 limits of, 534, 536 by Continuous Function Theorem, 536 by I'H8pital's Rule, 537-538 by Sandwich Theorem, 536

monotonic, 540 of partial smns, 545 nondecreas~,540

reCUISively def'mcd, 539 Sequential search, 427-428

Series, absolutely convergent, 570 adding or delc1iog terms, 550 albrrnating,56S-570 harmonic, 56S-569 binomial,596-598 omnbining,549-550 cooditiooally coovergen~ 570

convergence of, comparison tests for, 55S-560

1-12

Index

Series (continued) convetgcn~

545, 570 divergen~ 545, 548 goome1ric,546-548 harmonic, 533, 553 infinite, 544-550 integral tes~ 554 error estimation. 556 Maclaurin, 585-586

p-,555 partial sum of, 544-545 power, 575-582 :rearrangement of, S72 ",indexing, 550 representations, of functions of power, 584-585 sum of, 544-545 Thylor, 585-586, 589, 592-593

tests, for absolute convergence, 571 alternating, 568 comparison, 559

convergence, 558-560 integral, 553-555

limit comparison, 560--561 ratio, 563-565 roo~ 565-566 summary of, 572-573 Set,AP-2

Shearing flow, 934 Shell fonnula for revolution, 322 Shell method, 321-324 Shells, thin, masses and moments of, 958-960

three-dimensional, masses and moments, 869,872

volume of, calculation of, 309 by disk method, 311-313 by donble integra1a, 837, 838, 842-845 by method of slicing, 30&-315 by triple integrala, 859-860 by waaher method, 314-315 Solids of revolution, by disk method, 311-313 by washer method, 314-315

Solution, of differential equation, 233, 389 general, 496-498 particular, 497-498

Subnrn£tion,ofveckxs,668 Sum Rule, for antiderivatives, 235 for combining series, 549 for dellnite integrals, 266 dorivative,II8-119

of geometric series, 546 fur gradients, 789

for limits, 49, 62 of functions of two variables, 757 of sequences, 534 for vector functions, 712 Sums, and difference, of double integrals, 846

Solution curves, 498 Space, curve in, wmk done by force over,

:f"tnite,572 estimation with, 246--253

912-914 Speed, 126-127, 158

limits of, 25&-259 lower, 247-248

along smooth curve, 726 average, 39-41 over short time intervals, 40 instantaneous, 39-41 of particle in space, 711 Spheres, concentric, in vector field, 976-977 parametrization of, 943-944 in space, distance and, 662--663 standard equation for, 662 surface area of, 946-947 Spherical coordinates, definition of, 879 triple integrals in, 879-883 Spin around axis, 934-936

partial, sequence of, 545

Spring constao~ 338 Springs, Hooke's law for, 33&-339

Riemann, 259-261

upper, 247, 248 Surface area, defining of, 332-334 differential for panunetrized snrface, 946

of explicit surface, 954 of graph, 950 of implicit surface, 949--950, 954 parametrization of, 944-948 for sph..., 947 Surface integrals, 953-960, 965

computation of, 955-956 fur flux, 956-958 formulas for, 954 Surfaces, aod area, 334, 624-625, 943-953

Shift formulas for functions, 16

mass of, 903-904

constant depth, fluid force on planar, 341

Shifting, of furu:tion graph, 16 Short differential formula, arc lensfh, 329-330 SIonits, 337

work to stretch, 339

furu:tiona defmed on, 777-779 with holes, 969 implici~ 94&-950

Sigma notation, 256--258 Simple harmonic motion, 137-138 Simply coonected region, 921

Simpson, Thomas, 471 Simpson's Rule, approximations by, 47()-472 error analysis and, 472-475 Sine(s), 23, 24 integrals of products of, 447-448 integrals of products of powers of, 444-446

Sine function, derivative of, 135-136 graph of, 10 integral of, 444 inverse of, 405

Sinusoid formula, 27 Slant (oblique) asymptote, 88-89 Slicing, with cylinders, 319-321 by parallel planes, 309-310 volume by, 30&-314

Slope, of curve, 41-43, 103--104 of nonverticalline, AP-l1 of parametrized curves, 612--613 of polar coon:l.inate curve, 632--633 Slope fields, 498, 499 Slugs, 511 Smooth curves, 3-4, 326--327, 710

Smooth surface, 944-945,948 Snell van Royen, Willebrord, 217

Snell's Law, 218 Solids, Cavalieri's principle of, 310 cross-section of, 308

Square root function, 8 derivative of, 108 Square roots, elimination of, in integrals, 446 Squeeze Theorem. See Sandwich Theorem Sl Vincen~ Gregory, 623 Staodsrd lioear approximation, 165,795 Staodsrd onit voctors, 669 Steady-state value, 508 Step size, 469 Stokes' Theorem, 925, 962-970, 980, 981

comparison with Green's Theorem, 962, 963, 964

conservative fields and, 970 for polylrednsl surfaces, 96&-969 snrface integral in, 965

for surfaces with holes, 969 Stretching a graph, 17 Substibltion, 285-289

implicitly deImcd, 948 level, 750

orientable, 956 parametrization of, 943-948

piecewise smooth, 954, 963 plane tangent to, 792-793 quadric, 697--{)99

smooth, 944--945, 948 Iw<>-sidod, 956 of two-variable functions, 750 Symmetric functions, 6 definite integrals of, 293-294 graphs of, 6-7 Symmetry tests, for graphs in polar coordinates, 631

System. torque, systems of masses, 346

and area between curves, 291-297 in double integrals, 887-891 rectangular to polar coordinates, 854

in multiple integrals, 887-894 trigonometric, 449-452 in triple integrals, 891-894 roctaognlar to cylindrical coordinates, 876 rectaognlar to spherical coordinates, 880

Substibltion fonnula for definite integrals, 291-293

Substibltion Rule, 286-287, 292 evaluation of indefinite integrals with, 284-290

8-limits of integration, Imdiog of, 854-855, 878 Thble of integrals, 463-464, T-I-T-6

Tabular integration, 440--441 Tan x and cot x, derivatives of, 138 integra1a of, 374-375

inverses of, 407

Tan x and sec x, derivatives of, 138 integrala of, 374 powers, 446-447 Thngent(s), to curves, 19, 39-44 of curves in space, 707-713 to graph of function, 102-103

Index to level curves, 788-789 andnonnals,152-153 at poin~ 102-104 slope of, 41 values of, 24 vertical, 111 Tangent line approximation, 164,795 Tangent lines, 43-44 to curve, 102 Tangent plane approximation, 795 Tangontphmes,horizontU,803-804 and normal lines, 791-793 to a parametric surface, 945 Tangent vector, 710 Tangential component of acceleration, 734-738 Tautochrones, 614-615 Taylor, Brook, 585 Taylor polynomials, 586-588, 593, 594 Taylor series, 585-586 applying of, 592-593, 59!H;02 convergence of, 589--595 frequently used, 602 as representations of functions, 585 Taylor's Formula, 589, 590 for functions of two variables, 820-823 Taylor's Theorem, 589 proof of, 594-595 Tenn of a sequence, 532 Tenn of a series, 545 Tenn-hy-term differentiatioo, 580 Tenn-hy-term integratioo, 581 Tenninal point, of curve, 610 of vector, 665 Tenninal velocity, 521

Theorem, Absolute Convergence Test, 571 Algebraic Properties ofNatum1 Lcgarilhm, 372 Alternating Series Estimation, 570, 593 angle between two vectors, 674 Chain Rule, 142-146 for functions of three variables, 777 for functions of two variables, 775 for two independent variables and three intermediate variables, 778 Comparisoo Ths~ 559 conservative fields are gradient fields, 923 Continuous function for sequences, 536 Convergence, for Power Series, 578 corl F ~ 0 rotated to loop property, 970 De Moivre's, AP-32 Derivative Rule for Inverses, 365 Differentiability implies cootinuity, III, 772 Diroct Comparison Tes~ 560

Divergence, 974 Evaluation, 277-278 Exactness of differential forms, 928 Extrome Value, 185-186, AP-24

First derivative test for local extreme values, 187,803

Formula for implicit Differentiatioo, 780 Fubini's, 839, 843--845 FundwmentU,ofCalcruus, part 1,275-277 part 2, 277-278 FundwmentU, of Algebra, AP-33-AP-34 of Line Integntla, 922

Green's, 937 Implicit Function, 781, 948 Increment, for Functions of Two Variables, 771, AP-38-AP-40 Integntl Ths~ 554 Intennediate Value, 80, 81, 192-193, AP-24 Leihoiz's,568 l'H6pitU's Rule, 397, 400, 537-538 limi~ proofs of, AP-I8-AP-21 Limit Comparison Test, 560 loop property of conservative fields, 924 MeanValue, 192-196, 327, 401, 621 corollary I, 195 corollary 2, 195 for def'mite integrals, 274-275 Mixed Detivative, 770, AP-36 Monotonic Sequence, 540--541, 553 Multiplication ofpower series, 580 Net Change, 279 Nondecreasing Sequence, 540 number e as limit, 382 Orthogooal gradien~ 814 of Pappus, 353-354 Properties of continuous functions, 76

Properties of limits of functions of two variables, 757 Ratio Test, 563 Rearrangement, for Absolutely Coovergent

1-13

Trapezoidal Rule, 468-470 error analysis and, 472-475 Triangle inequslity, AP-5 'nigonometric functions, 10 angles, 22-23 derivatives of, 135-139 graphs of, 25, 32-34 transformations of, 27 integntls of, 374, 444-448 inverse, 404-413 periodicity of, 25 six basic, 23-24 'nigonometric identities, 25 'nigonometric substitutions, 449-452

'niple integrals, in cylindrical coordinates, 875-879 properties of, 859-860, 865 in roctsngu1ar coordinates, 859-865 in spherical coordinates, 879--883 substitutions in, 891-894 Triple scalar product (box product), 685--086 Tboing fork dsta, 4 nw.<Jimensiooal vec-. cornpooent form of, 666 Two-path test for nonexistence of limit, 760 Two-sided lintils, 66 proofof,AP-21 1\vo-sided surface, 956

Series, 572

Remainder Estimatioo, 591, 593

Unbounded intervals, 6

Rolle's, 192-193 RootTes~ 565-566 Sandwich, 53, 69, 70, 87, 536, AP-20

Unbounded region, 749

Second derivative test for local extreme values, 206, 805

Unbounded sequence, 539

Undetermined coefficients, 454 Unified tlreory, 980 Union of set, AP-2

Stokes',963

Unit binormal vector, 734

Substitution in def'mite integrals, 292 Substitution Rule, 286

Unit circle, AP-14 Unit normal vector, 730 Unit step functions, limits and, 49

Taylor's, 589, 594-595

Term-by-Term Differentiation, 580 Tenn-hy-Tenn Integratioo, 581 Thickness variable, 322-323

Thin shells, moments and masses of, 958-960 Three-dimensional coordinate systems, 66Q-{j63 Cartesian, 660

coordinate planes, 660 cylindricsl, 877-879 right- and left-handed, 660 spherical, 881-882 Thrco-dimensional solid, 869, 872

Three-dimensional vectors, component funn of, 666

Time-Distance Law (Kepler's Third Law), 741-742 TNB frame, 734 Torque, 346-347, 685 Torsion, 736-737, 738 Torus, 951 TotU difThrential, 797, 798 TranacendentU functiooa, 11,361-428 Transformations, Jacobian of, 889, 890 linear, 888-889 of trigonometric graphs, 27 Transitivity law for real numbers, AP-23 Trapezoid, area of, 269

Unit tangent vector, 726-727, 738

Unit vectors, definition of, 669 writing vectors in terms of, 669-670 Universal gravitational constant, 740 Upper hound, 267, AP-23 Upper sums, 247, 248 Value(s), absolute, AP-4-AP-6, AP-30 average,252 OXUonle, 184-188,802-808 of function, 2-3, 864-865 of improper integntl, 478, 481 local maximum, 186, 803 local minimum, 186,803 Variable force, along curve, 912-913 along line, 338 Variable of integration, 235, 264

Variables, constrained, partial derivatives with, 824-828 dependen~

I, 824-825 functioos ofsevera1, 747-752, 761, 768-769, 781 independent, I, 775, 777, 824-825 inpu~ 747 intennediate, 776-777 outpu~ 747 proportiooal, 7

1-14

Index

Variables (continued) thickness, 322-323 three, functions of, 750-752, 78~790 Clutin Rule for, 777 two, functions of, 748-749, 752, 771 Chain Rule for, 775-777 independen~ and three intermediate, 778 limits for, 755-759 linearization of, 794-796, 798 partial derivatives of, 764-766 Thylor's formula for, 820-823 Vector equations, for curves in space, 707 for lines, 688, 68~90 of plane, 691 for projectile motion, 717-719 Vector fields, 908-909 ~tive.920.924-925

continuous, 908 curl of, 962

differentiable, 908 divergence of, 933--934

equality of, 665 and geome1ry in space, 660-699 gmdien~ 786 in gravitational field, 908 i-component of, 669 initial point of, 665 j-component of, 669 k-component of, 669 length (magnitude) of, 665, 667--{j68 in navigation, 671 nonna!, of curve, 732-733

notation for, 665 paml1el, 682 perpendicular (orthosOllJl!), 676-677 in physics and engineering, 671--{j72 position, standard, 665--{j66 priocipal unit normal, 730-731, 738 projection of, 677-fJ79 resultan~

667--{j68

scalar multiplication of, 667--668 standard position, for a point, 665-666

eleelric, 920 flux density of, 933 gradieo~ 9~lO,

direction of, 669 dot product, def'mition of, 674

922--923

gravitstiOllJl!, 920 integration in, 901--981 line integrals of, def'mition, 910 and line integrals, 907--917 potential function for, 921 Vector functions, 707-713 antiderivatives of, 715 of constant length, 713 continuity of, 708-709 defmite integral of, 716-717 derivatives of, deImition of, 710 differentiable, 71 0 differentiation rules for, 711-714 indeImite integral of, 715-716 integrals of, 715-720 limits of, 708-709 Vector product. See Cross product Vector-valued functions. See Vector functions Vectors, 665-672 aocelemtion, 711, 734 addition of, 667--668, 675 algebra operations with, 667--669 angle between, 674-676 applications of, 671---672

bdnornuil,ofcurve, 734 component fonn of, 665-667 coplanar, 669

cross product, as area of parallelogram. 683 in component fonn, 683--685 definition of, 682 as determinan~ 683--{j85 right-band rule for, 682 of two vectors in space, 682--683 corl, 962--963 defmition of, 665

standard unit, 669

subtraction (differeru:e) of, 668 tangen~ ofcorve, 710 terminal point of, 665 three-dimensional, 666 torque, 685

triple scalar product of, 685-686 two-dimensional, 666, 676 uni~ definition of, 66~70 derivative in direction of, 785 writing vectors in terms of, 669--670 unit binoonal, 734 unit nonnal, 732 unit tangen~ 726-727 velocity, 665, 711 zero vector, 666 Velocity, 129

along space curve, 711 angular, of rotation, 967 average, 125 free fall and, 128 instantaneous, 125-126 in polar coordinates, 739--742 resistance proportional to, motion with, 510-511 terminal,521

Velocity fields, circulation for, 914--915 flow integral, 914-915 Velocity function, 251-252, 711 acceleration and, 196, 711 Vertical asymptote~ 84, 91-92. See also Asymptotes

definition of, 91 Vertical flat plate, integral for fluid force on, 342

Vertical line tes~ 4-5 Vertical scaling and reflecting formulas, 17 Vertical shift of function, 16

Vertical slrip, 319, 349 Vertical tangents, 111 Viewing windows, 30-33 Voltsge, peak, 289 Volume, of cylinder, 308 differential, in cylindrical coordinates, 876

in spherical coordinates, 880 by disks for rotation about axis, 311, 313

double integrals as, 837, 838 by iterated integrals, 842-1143

Pappus's Theorom for, 353-354 of region in space, 860 by slicing, 308-315 of solid region, 842-845

of solid with known cross-section. 309 triple integrals as, 860 using cross-sections, 308--315 using cylindrical shells, 319--324 by washers for rotation about axis, 314

Washerme1hod, 314-315, 324 Wave equation, 774 Weierstrass, Karl, 484 Weight-density, of fluid, 340-341 Whirlpool efl'e~ 934 Wmdows, graphing, 30-33 Work, by constant force, 337-342 and fluid forces, 337-342 by force over curve in space, 912-914 by force 1hroogh displanement, 679 and kinetic energy, 344 in pumping liquids, 340 by variable force along curve, 913 by variable force along line, 338 x-coordinate, AP-I0

x-intercept, AP-13 x-limits of integration, 862, 864 >y-plane, 524,660 solution curves in, 518 xz-plane, 660 y, integration with respect to, 296-297 Y ~ !(x), graphing of, 206-208 length of, 326-328, 623 y = InX', derivative of, 371 y = sec-I u, derivative of, 409-410 y = sin-1 u, derivative of, 408 y = tan-I u, derivative of, 409 y-axis, revolution about, 334-335 y-coordinate, AP-I0 y-intercept, AP-13 y-limits of integration, 861, 863 )"-plane, 660

z-limits of integration, 861, 862, 863, 877 Zero denomina_, algebraical eliminatioo of, 51 Zero vector, 666 Zero Width Interval Rule, 266

A BRIEF TABLE OF INTEGRALS Basic Forms 1.

j

kdx

+ C (anynumberk)

kx

=

j~ + j dx ~xa + > 7. j + j + j + j eotxdx + j + j a2 dx+ x2 ltan-l~ + a a j Va dx+ x sinb-l~ + 3.

=

5.

aX

In Ixl

C (a

=

In Isinxl

13.

=

15.

eosbxdx = sinbx

17.

2

2

10.

C

12. 14.

C

C

16.

C

=

19.

8.

C

esexeolxdx = -esex

11.

6.

0, a "" 1)

C

ese2 xdx = -eolx

=

j

xn+l

x"dx=n+I+C (n "" -I)

4.jeX dx=e X +C

C

eosxdx = sinx

9.

2.

18.

C (a> 0)

20.

j sinxdx j j j tanxdx j sinbxdx

-eosx

=

see2 xdx = tanx

+C

+C

seextanxdx = seex In Iseexl

=

j

• / 2dx va

x2

-

j V dx j Vx dx a x

x2 2

-

eosbx

=

-

a2 2

+C +C

+C

=. sm-1 ~ a + = =

C

~ see-1 I~ I + C eosh- 1 ~

+

C (x

> a > 0)

Forms Involving ax + b 21. 22.

j (ax + b)" dx

(ax + b)"+1 a(n + 1) + C, n "" -1

=

1

j

" _ (ax + b)"+1 [ax + b b x(ax + b) dx a2 n + 2 - n + 1 + C,

23. j(ax + bt 1dx = 25.

27.

j x(ax + W 2dx j

=

~In lax + bl

+C

1

:2 [ In Iax + b I + ax : b + C

" 2 (Vax + b (Vax + b) dx = a n+2

(2

+ C, n "" -2

n "" -1,-2

24.

jx(ax+b)-ldx=~- :21nIax+bl

26.

j x(ax~ b) tin lax: bl + C

28.

+C

=

j~ dx=2~+bj xVaxdx + b T-l

T-2

A BriefTable of Integrals

29. (a) / 30.

ax

x~

Vb

2

X

/

34. /

a

2

ax +x

0

2

- Vb I + C

Vax+b+Vb

/~ ax=- ~ x

Forms Involving 32.

_I_In I~

=

+'!/ 2

ax

x~

(b) /

+ C

_ atan I -1 x a+ C

ax

Va + x

2

sinh-l~ + C =

=

35. /Va2+x2ax=IVa2+x2+

31 /

·

ax x 2Vax + b

33 /

ax (a 2 + x 2)2

•

In (x + Va 2 +

x2 )

Va 2 + x 2 _ a In Ia + V: + x21 + C

38. /

2 Va: 2+ x ax

=

In (x + Va 2 + x 2) _ Va /

39./

2 2 2 x ax=_a 1n (x+Va 2 +x2)+xVa +x2+c Va2 + x2 2 2

42.

/

44. /

0

2

2

41. /

43/

Ix

·

sin-l~ + C

46. /x 2Va 2 - x 2 ax =

~sin-l~ -

ax

(a 2 _ x 2)2 2

2

x2

Forms Involving il 52. / 53. /

V. 2ax x

-

a2

=

Vx 2 - a 2 ax

Va 2 - x 2 2 +C ax 0

2

In Ix + Vx 2 - a 2 + C 1

=

,+

2

Va x +C ax

IVx2 - a2 -

=

x

2a2(a2 _ x2) 2

+_I_Inlx+al +C 4a3

2+

x- a

~2 sin-l~ + C

kXVa 2 - x 2 (a 2 - 2x2) + C

2

=

= _

45. / Va - x ax = IVa - x

50. /

.

+C

x + _I-tan-l~ + C 2a 2(a 2 + x 2) 2a3 a

2

x 2Va + x 2

2

ax x 2Va 2 -

=

ax2

47. /Va -x ax=Va2_x2_alnla+V:2-x21+c 48. /Va: -x

51 /

ax

xVax + b

il

-

I In x + a 2 ax _ x 2 = 2a - al a +C

2 x

2b

2 x +C

-!In la + V:2 + x21 + C

=

Va:X- x 2 =

-"-/

C

2

=

Forms Involving

~ bx

+

ag' In (x + Va 2 + X2) + C

2 2 Va x + x ax

xVa + x 2

= -

b

+C

37. /

ax2

~tan-l~ax Vb b

~1n(x+Va2+x2)+C

~(a2 + 2x 2)Va 2 + x 2 -

36. /x 2Va 2 + x 2 ax =

40. /

=

+ il

2 -

2

ax

x~

~21n Ix +

Vx 2 - a 2 + C 1

2

ax=-sin-l~-

2 2 Va -x +C x

ax = _! In Ia + V:2 - x21 + C xVa 2 - x 2

A BriefTable ofIntegrals

54.

(Yx 2 - a 2)" dx =

J

55J dx • (Yx 2 -a 2)"

x(Yx2 - a 2)n n+ I -

(v' 2 =xx-a

2t

(2-n)a 2

nn~

J

2

I

(Yx 2 - a2)"-2 dx,

n'¢' -I

n

_n-3J dx (n-2)a 2 (Yx 2 -a 2)"

56. Jx(YX 2 - a 2)" dx

(Yx2 - a 2 )"+2 = n+2

57. J x 2Yx 2 - a 2 dx

i(2x 2 - a 2)Yx 2 - a 2 - as'1n Ix + Yx 2 - a 21 + C

58.

2

J

Yx2x - a dx = v'x 2

2 59. J YX: 2- a dx 60. 61.

=

J J "Ii

Yx2 _ a2

x

-a2

dx

x2

dx

-

-

a sec-I Ixl a +C

1- YX 2x-

2 a + C

=

In Ix + Yx 2 _ a 2

=

2 a 1n Ix + Yx 2 - a 21 + "-Yx 2 - a 2 + C 2 2

2

x

a2

-

+ C, n'¢'-2

62.

aI sec-llxl a

Trigonometric Forms 63. JSinaxdx

65. J sin2axdx 67.

68•

J J

sin" ax ax

J

(b)

J

(c)

72. 74.

J

. J J J

64.

C

=

~ - s~~ax + C

=

smn-I : cos ax

.

· ax + n dx -- cosn axI sm cosaX' na

69. (a)

70.

-~cosax +

=

IJ "--=---IJ

+

n ;

n

2

-

x

C

sin2ax

cosn-2 ax dx

a2

4

b2

r

2 2 a '¢' b

cos2ax 4a + C

~In Isin axl

2

cosaxdx-2+~+C

sin(a - b)x sin(a + b)x 2(a _ b) + 2(a + b) + C,

C?sax dx = smax

~sinax +

cos ax dx =

2 2 a ,¢,b

sm ax cos ax dx = -

71.

+ C

73.

sinax I cosaxdx = -aln Icosaxl + C

J

J J

.. sin(a - b)x sin(a + b)x smaxsmbxdx= 2(a-b) - 2(a+b) +C, cosaxcosbxdx =

2

dx = Yx - a + C 2 x 2Yx 2 -a 2 ax

sin,,-2 ax dx

cos(a + b)x cos(a - b)x . smaxcosbxdx = - 2(a + b) - 2(a _ b) + C,

. n-I ax cosm+1 ax sm am ( +n) 76.

66.

J

J J

+ -n + - I m

sin",+l ax COS",-l ax m- 1 sinn ax cosmax dx = aim + n) + m+ n

n

.

,,-2

SID

J J

On _sinn+lax sm axcosaxdx - (n + I)a + C, PI·

-

n'¢'-I

cosn+1ax

cos axsmaxdx - - (n + I)a + C,

m dx axcos ax , n '¢' -m

sinn ax cosm-2 ax dx,

m '¢' -n

(reduces sinn ax) (reduces cosmax)

n'¢'-I

T-3

T-4

A BriefTable of Integrals

77. J b + :inax

=

78. J

=

79.

81.

dx. b + csmax

J J

1 + dx sinax =

dx

=

dx b + ccosax

=

83. J 1+ %sax

89. J tanaxdx

~In Isecaxl

X,-I cos ax dx

~In Isinaxl

x + C

92. J cot'axdx

=

94. J cot" ax dx

= -

:'~ '0 - J ~In Isec ax +

n" 1

tan,-2 ax dx,

96. Jcscaxdx

tanaxl + C

J ~tanax 2J " J cs~~:~c~ax if , ----.uzJ sec,,-2 ax tan ax n a{n _ 1) + n _ 1

100. J csc'axdx

sec ax tan ax dx -

sec"ax

,-2

sec

+ : ::::

=

x: sin ax -

~sinax + ~J

=

X,-I sin ax dx

+ C

-~cotax -

x + C

:(:~ '0 - J

-~In Icscax +

=

C

-~cotax +

Cot"-2 ax dx, n" 1

cotaxl + C

C

ax dx, n" 1

n" 1

csc"-2 ax dx,

n"O

+ C,

=

98. J csc2 axdx

+ C

sec ax dx -

:2 cos ax +

=

90. J cotaxdx

=

101.

86. J xcosaxdx

-~cota; + C

=

+ C

93. J tan' ax dx

99.

%sax

=

~tanax -

sec2 axdx =

b2
88. J x' cos ax dx

=

97.

~J

_x: cos ax +

91. J tan2 ax dx

=

84. J 1 C

ltan(7T + ax) + C a 4 2

=

b2 > c 2

InIC+bcosax+Vc2-b2sinaxl+c b + c cos ax '

C

b2 < c2

ttx.smax

1

~tan a; +

+ C '

1

tan-I

=

95. Jsecaxdx

[~bb+c - ctanax] 2

J

2

aVb2-c2

87. J x' sinaxdx =

80.

+ C

a\Sinax-~cosax+

85. JXSinaxdx=

2 2 b > c

+ C,

v'

aVc2 _ b2

=

T)]

2 c2 In Ic + bsinax + - b cosaxl + C, aVc2 _ b2 b + csmax

-1

-a1 tan (7T 4 - 2ax)

b+ccosax

82.J

aVb~2_ c 2 tan-I [~~ ~ ~ tan(; -

102.

, J

- ----.uz-

csc ax cot ax dx - -

csc"ax

+ C,

n"O

Inverse Trigonometric Forms 103. J sin-Iaxdx

=

X sin-I ax +

~Vl

105. J tan-I ax dx

=

Xtan-I ax -

2~ In (1

J J J

11

•

-1

-

x ll + 1

• -1

- a 2x2 + C + a2x 2 ) + C a

J J J

106.

x sm axdx--+1 sm ax--+1 n n

107.

X,+I a x'cos-Iaxdx = n + 1 cos-lax + n + 1

108. x' tan-I ax dx

=

x"+l

104. J cos-Iaxdx = X COS-I ax -

a

n + 1 tan-I ax - n + 1

x"+ 1 dx

,I

'

vI - a 2x 2 X,+I dx VI _ a2x 2 '

x"+l dx 1 + a 2x 2'

n"-1 n"-1

n"-1

~Vl

- a2x 2 + C

A BriefTable ofIntegrals

Exponential and Logarithmic Forms 109.

111.

j eax dx j

=

xe ax dx

~eax + c

=

eax

-;;0 (ax

110.

-

114. je ax sinbx dx =

/ax 2(acosbx+bsinbx)+C

116.jlnaxdx=Xlnax-x+c

a +b

n+1

x '(lnax)mdx

b > 0, b oF 1

b > O,b oF 1

117. jX"(lnax)mdx = x"+l(lnax)m - ----"'----jX"(lnax)m-' dx

j

Ibax

a In b + C,

2eax 2 (a sinbx - b cos bx) + C a +b

l1s.jeaxcosbxdx=

118.

j

=

- I) + C

~i:,: a~bjX"-lbaxdx,

113. jX"baxdx =

bax dx

n+1

(lnax)m+l m + 1 + C,

=

m oF -I

n oF-I

'

119.

dx xlnax

j

=

lnllnaxl + C

Forms Involving Y2ax - xl, a> 0 120. 121.

j Y2axdx - x sin-1 (x ~ a) + C j Y2ax - x 2 dx x ; a y2ax - x 2 + 'fsin-1 (x ~ a) + C 2

=

=

"

2 122. j ( Y2ax - x ) dx = 123

dx

•j

124. 125.

(Y2ax - x2)"

=

na 2 j ( (x - a)( Y2ax - X2)" 2)"-2 n+ 1 + n+ 1 Y2ax - x dx

(x - a)(Y2ax - x2)2-" n - 3 j + ----""-~2 2 (n - 2)a (n - 2)a

dx

(Y2ax _ x2)" 2

j xY2ax _ x 2 dx (x + a)(2x - ~a)Y2ax - x 2 + ~ sin-1 (x ~ a) + C j Y2~ - x 2 dx Y2ax _ x 2 + a sin-1 (x ~ a) + C j Y2a;2 - dx -2 ~2a ; sin-1 (9) + C j x Y2axdx -x j Y2axxdx- a sin-l (x ~ a) - Y2ax - x 2 + C =

=

2

126. 127.

x

x -

=

x2

=

128.

~coshax + C

130.

= 2

_1~2a a

- x +C x

Hyperbolic Forms 129. jSinhaxdx = 131. 133.

j sinh2axdx j sinh" ax dx

=

S~2ax -

=

sinh"-l ~ cosh ax - n -;; 1

I+C

j cosh ax dx j cosh2ax dx n °

132.

j sinh"-2 ax dx,

~ sinh ax + C

=

oF

=

s~;ax + I + C

T-5

T-6

A BriefTable of Integrals

134. J cosh" ax dx

cosh"-l:: sinh ax

=

J cosh"-2 ax dx,

+ n -;;

1

135.Jx sinh ax dx ~ cosh ax - :2 sinh ax + C 137.Jx" sinh ax dx x; cosh ax - ~Jx"-l cosh ax dx 139.Jtanhaxdx ~ln(coshax) + C 141.J tanh2 ax dx x - ~ tanh ax + C =

=

~~-;;;:

+

144.] coth" ax dx

=

(~th~-;;;:

+

145'] sechaxdx

J J

tanh"-2 ax dx,

n i' 1

coth"-2 ax dx,

n i' 1

=

sech"-2 ax tanh ax (n _ l)a

=

=

2eax

[ ebx e-bx ] a + b - a _ b

2eax

[ ebx a +b

+

bx ea _ b

~ sinh ax -

:2 cosh ax

x" cosh ax dx = x; sinh ax -

~ln Isinh axl =

x -

X

=

w/2

1

157. o

f(n) = (n - 1)1,

1W

/2

sin" x dx =

0

cos" x dx =

n

x"-l sinh ax dx

+

C

-~cothax + C

2 1

4

T-

T-

n i' 0

+

152.

J

csch" ax coth ax dx = -

csch" ax

na

C,

1+ C,

>

°

156.1""e-ax ' dx

1.3.5 ..... (n - 1) 7r •2·4·6· .... n 2' 2 • 4 • 6 ••••• (n - 1) 3.5.7 •...• n '

1

J

~

~ coth ax + C

Some Definite Integrals

155.1""x"-le- dx

C

+C

~ln Itanh~1

cschaxdx =

+

J sech"-2 ax, dx 1 n n - 2J h"-2 dx 41 ax, n n- 1 esc

+ nn -_

= -

. 153.Je ax sinhbxdx 154.Jeaxcoshbxdx

=

csch2 axdx =

ax coth ax (n1 )a 50J esch" ax dx CSCh"-2 151.J sech" ax tanh ax dx sec~= ax + C,

1 .

J

146.

~ tanh ax + C

147.J sech2 ax dx 149.J sech" ax dx --

138.

J 148. J

~sin-l(tanhax) + C

=

Jx cosh ax dx =

=

=

136.

140. J cothaxdx 142. J coth2 ax dx

=

143'] tanh" ax dx

n i' 0

=

t.J%,

if n is an even integer ;,,2 ifn is an odd integer;,,3

a

>0

+

C,

ni'O

CREDITS Page ii, photo, Forest Edge, Hokuto, Hokkaido, Japan 2004 © Michael Kenna; Page 1, photo, Getty Images; Page 39, photo, Getty Images; Page 102, photo, Getty Images; Page 133, Section 3.4, photo for Exercise 19, PSSC Physics, 2nd ed., DC Heath & Co. with Education Development Center, Inc.; Page 178, Chapter 3 Practice Exercises, graphs for Exercise 94, NCPMF "Differentiation" by W. U. Walton et al., Project CALC, Education Development Center, Inc.; Page 181, Chapter 3 Additional and Advanced Exercises, photo for Exercise 9, APIWide World Photos; Page 184, photo, Getty Images; Page 246, photo, Getty Images; Page 308, photo, Getty Images; Page 347, Figure 6.44, PSSC Physics, 2nd ed., DC Heath & Co. with Education Development Center, Inc.; Page 361, photo, Getty Images; Page 435, photo, Getty Images; Page 496, photo, Getty Images; Page 532, photo, Getty Images; Page 547, Figure 10.9b, PSSC Physics, 2nd ed., DC Heath & Co. with Edocation Development Center, Inc.; Page 610, photo, Getty Images; Page 660, photo, Getty Images; Page 707, photo, Getty Images; Page 722, Section 13.2, photo for Exercise 35, PSSC Physics, 2nd ed, DC Heath & Co. with Education Development Center, Inc.; Page 747, photo, Getty Images; Page 836, photo, Getty Images; Page 901, photo, Getty Images; Page 908, Figures 16.6 and 16.7, NCFMF Book ofFilm Notes, 1974, MIT Press with Education Development Center, Inc.; Page 909, Figure 16.15, InterNetwork Media, Inc., and NASAlJPL; Page AP-l, photo, Getty Images.

C-1

6653_AWLThomas_EPp01-07 7/29/09 7:30 PM Page 3

BASIC ALGEBRA FORMULAS Arithmetic Operations asb + cd = ab + ac,

ac a#c = b d bd

a c ad + bc + = , b d bd

a>b a d = #c b c>d

Laws of Signs a a -a = - = b b -b

-s - ad = a, Zero Division by zero is not defined.

0 If a Z 0: a = 0,

a 0 = 1,

0a = 0

For any number a: a # 0 = 0 # a = 0 Laws of Exponents a ma n = a m + n,

sabdm = a mb m,

sa m dn = a mn,

n

a m>n = 2a m =

n aBm A2

If a Z 0, am = a m - n, an

a 0 = 1,

a -m =

1 . am

The Binomial Theorem For any positive integer n, sa + bdn = a n + na n - 1b + +

nsn - 1d n - 2 2 a b 1#2

nsn - 1dsn - 2d n - 3 3 a b + Á + nab n - 1 + b n . 1#2#3

For instance, sa + bd2 = a 2 + 2ab + b 2,

sa - bd2 = a 2 - 2ab + b 2

sa + bd3 = a 3 + 3a 2b + 3ab 2 + b 3,

sa - bd3 = a 3 - 3a 2b + 3ab 2 - b 3.

Factoring the Difference of Like Integer Powers, n>1 a n - b n = sa - bdsa n - 1 + a n - 2b + a n - 3b 2 + Á + ab n - 2 + b n - 1 d For instance, a 2 - b 2 = sa - bdsa + bd, a 3 - b 3 = sa - bdsa 2 + ab + b 2 d, a 4 - b 4 = sa - bdsa 3 + a 2b + ab 2 + b 3 d. Completing the Square If a Z 0, ax 2 + bx + c = au 2 + C The Quadratic Formula

au = x + sb>2ad, C = c -

If a Z 0 and ax 2 + bx + c = 0, then x =

-b ; 2b 2 - 4ac . 2a

b2 b 4a

6653_AWLThomas_EPp01-07 7/29/09 7:30 PM Page 4

GEOMETRY FORMULAS A = area, B = area of base, C = circumference, S = lateral area or surface area, V = volume Triangle

Similar Triangles c'

c h

a'

Pythagorean Theorem a

c

b

b' b

b

a' b' c' abc

A  1 bh 2

Parallelogram

Trapezoid

a a2  b2  c2

Circle a

h h

A  ␲r 2, C  2␲r

r

b b A  bh

A  1 (a  b)h 2

Any Cylinder or Prism with Parallel Bases

Right Circular Cylinder r

h

h h

V  Bh

B

B V  ␲r2h S  2␲rh  Area of side

Any Cone or Pyramid

Right Circular Cone

h h

Sphere

h s r

V  1 Bh B

3

B

V  1 ␲r2h 3 S  ␲rs  Area of side

V  43 ␲r3, S  4␲r2

6653_AWLThomas_EPp01-07 7/29/09 7:30 PM Page 5

LIMITS General Laws

Specific Formulas

If L, M, c, and k are real numbers and

If Psxd = an x n + an - 1 x n - 1 + Á + a0 , then

lim ƒsxd = L

and

x:c

lim gsxd = M,

x:c

then

lim Psxd = Pscd = an c n + an - 1 c n - 1 + Á + a0 .

x:c

lim sƒsxd + gsxdd = L + M

Sum Rule:

x:c

lim sƒsxd - gsxdd = L - M

Difference Rule:

x:c

If P(x) and Q(x) are polynomials and Qscd Z 0, then

lim sƒsxd # gsxdd = L # M

Product Rule:

Psxd Pscd = . Qscd x:c Qsxd lim

x:c

Constant Multiple Rule: Quotient Rule:

lim sk # ƒsxdd = k # L

x:c

lim

x:c

ƒsxd L = , M gsxd

M Z 0 If ƒ(x) is continuous at x = c, then

The Sandwich Theorem

lim ƒsxd = ƒscd.

x:c

If gsxd … ƒsxd … hsxd in an open interval containing c, except possibly at x = c, and if lim gsxd = lim hsxd = L,

x:c

x:c

lim

x:0

then limx:c ƒsxd = L.

sin x x = 1

and

lim

x:0

1 - cos x = 0 x

Inequalities

L’Hôpital’s Rule

If ƒsxd … gsxd in an open interval containing c, except possibly at x = c, and both limits exist, then

If ƒsad = gsad = 0, both ƒ¿ and g¿ exist in an open interval I containing a, and g¿sxd Z 0 on I if x Z a, then

lim ƒsxd … lim gsxd.

x:c

x:c

lim

x:a

Continuity

ƒsxd ƒ¿sxd = lim , gsxd x:a g¿sxd

assuming the limit on the right side exists.

If g is continuous at L and limx:c ƒsxd = L, then lim g(ƒsxdd = gsLd.

x:c

6653_AWLThomas_EPp01-07 7/29/09 7:30 PM Page 6

DIFFERENTIATION RULES General Formulas

Inverse Trigonometric Functions

Assume u and y are differentiable functions of x. d Constant: scd = 0 dx dy d du Sum: + su + yd = dx dx dx dy d du Difference: su - yd = dx dx dx d du Constant Multiple: scud = c dx dx d dy du Product: + y suyd = u dx dx dx

d 1 ssin-1 xd = dx 21 - x 2

d 1 scos-1 xd = dx 21 - x 2

d 1 stan-1 xd = dx 1 + x2

d 1 ssec-1 xd = dx ƒ x ƒ 2x 2 - 1

d 1 scot-1 xd = dx 1 + x2

d 1 scsc-1 xd = dx ƒ x ƒ 2x 2 - 1

du dy y - u d u dx dx a b = dx y y2 d n x = nx n - 1 dx d sƒsgsxdd = ƒ¿sgsxdd # g¿sxd dx

Quotient: Power: Chain Rule:

Hyperbolic Functions d ssinh xd = cosh x dx d stanh xd = sech2 x dx d scoth xd = - csch2 x dx

d scosh xd = sinh x dx d ssech xd = - sech x tanh x dx d scsch xd = - csch x coth x dx

Inverse Hyperbolic Functions Trigonometric Functions d ssin xd = cos x dx d stan xd = sec2 x dx d scot xd = - csc2 x dx

d scos xd = - sin x dx d ssec xd = sec x tan x dx d scsc xd = - csc x cot x dx

d 1 ssinh-1 xd = dx 21 + x 2

d 1 scosh-1 xd = 2 dx 2x - 1

d 1 stanh-1 xd = dx 1 - x2

d 1 ssech-1 xd = dx x21 - x 2

d 1 scoth-1 xd = dx 1 - x2

d 1 scsch-1 xd = dx ƒ x ƒ 21 + x 2

Exponential and Logarithmic Functions

Parametric Equations

d x e = ex dx d x a = a x ln a dx

If x = ƒstd and y = gstd are differentiable, then

d 1 ln x = x dx d 1 sloga xd = dx x ln a

y¿ =

dy>dt dy = dx dx>dt

and

d 2y dx

2

=

dy¿>dt . dx>dt

6653_AWLThomas_EPp01-07 7/29/09 7:30 PM Page 7

INTEGRATION RULES General Formulas a

Zero:

ƒsxd dx = 0

La a

Order of Integration:

b

ƒsxd dx = -

Lb b

Constant Multiples:

ƒsxd dx

La

b

kƒsxd dx = k

La

La

b

b

-ƒsxd dx = -

La

La

b

sƒsxd ; gsxdd dx =

La b

Additivity:

sk = - 1d

ƒsxd dx

b

Sums and Differences:

sAny number kd

ƒsxd dx

La

b

ƒsxd dx ;

c

ƒsxd dx +

La

gsxd dx

c

ƒsxd dx =

ƒsxd dx La Lb La Max-Min Inequality: If max ƒ and min ƒ are the maximum and minimum values of ƒ on [a, b], then b

min ƒ # sb - ad …

La

ƒsxd dx … max ƒ # sb - ad.

b

Domination:

ƒsxd Ú gsxd

on

[a, b]

implies

La

b

ƒsxd dx Ú

La

gsxd dx

b

ƒsxd Ú 0

on

[a, b]

implies

La

ƒsxd dx Ú 0

The Fundamental Theorem of Calculus x

Part 1 If ƒ is continuous on [a, b], then Fsxd = 1a ƒstd dt is continuous on [a, b] and differentiable on (a, b) and its derivative is ƒ(x); x

F¿(x) =

d ƒstd dt = ƒsxd. dx La

Part 2 If ƒ is continuous at every point of [a, b] and F is any antiderivative of ƒ on [a, b], then b

La

ƒsxd dx = Fsbd - Fsad.

Substitution in Definite Integrals b

La

Integration by Parts gsbd

ƒsgsxdd # g¿sxd dx =

Lgsad

ƒsxd g¿sxd dx = ƒsxdgsxd D a -

b

ƒsud du

La

b

b

La

ƒ¿sxdgsxd dx

6653_AWLThomas_Last3Bkpgs 7/29/09 7:31 PM Page 2

tan A + tan B 1 - tan A tan B tan A - tan B tan sA - Bd = 1 + tan A tan B

Trigonometry Formulas

tan sA + Bd = y

1. Definitions and Fundamental Identities y 1 sin u = r = Sine: csc u Cosine:

x 1 cos u = r = sec u

Tangent:

y 1 tan u = x = cot u

P(x, y) r 0

y

␪ x

x

sin aA +

p b = cos A, 2

sec2 u = 1 + tan2 u,

sin 2u = 2 sin u cos u, 1 + cos 2u , 2

csc2 u = 1 + cot2 u

cos 2u = cos2 u - sin2 u sin2 u =

1 - cos 2u 2

sin A cos B =

1 1 sin sA - Bd + sin sA + Bd 2 2

sin A + sin B = 2 sin

1 1 sA + Bd cos sA - Bd 2 2

sin A - sin B = 2 cos

1 1 sA + Bd sin sA - Bd 2 2 1 1 sA + Bd cos sA - Bd 2 2

cos A - cos B = - 2 sin

cos sA - Bd = cos A cos B + sin A sin B

1 1 sA + Bd sin sA - Bd 2 2

y

y y  sin x

y  cos x

Trigonometric Functions Radian Measure

Degrees

1

兹2 45

C ir

兹2

1

␲

3␲ 2␲ 2

x

–␲ – ␲ 2

␲ 2

0

␲

3␲ 2␲ 2

x

Domain: (–, ) Range: [–1, 1]

1

␲ 4

90

r

␲ 2

y

1

y

y  tan x

it cir cl

y  sec x

e

Un

1

θ

␲ 2

0

y  sinx Domain: (–, ) Range: [–1, 1]

␲ 4

45 s

–␲ – ␲ 2

Radians

p b = - sin A 2

cos aA +

1 1 cos sA - Bd + cos sA + Bd 2 2

cos A + cos B = 2 cos

cos sA + Bd = cos A cos B - sin A sin B

p b = sin A 2

1 1 cos sA - Bd - cos sA + Bd 2 2

sin sA + Bd = sin A cos B + cos A sin B sin sA - Bd = sin A cos B - cos A sin B

cos aA -

cos A cos B = cos s -ud = cos u

sin2 u + cos2 u = 1,

cos2 u =

p b = - cos A, 2

sin A sin B =

2. Identities sin s -ud = - sin u,

sin aA -

s cle of ra diu

1

r ␲ 6

30 兹3

2

60

u s s or u = r , r = 1 = u 180° = p radians .

90 1

– 3␲ –␲ – ␲ 2 2 兹3

2 ␲ 3

␲ 2

0 ␲ ␲ 3␲ 2 2

x

Domain: All real numbers except odd integer multiples of ␲/2 Range: (–, )

– 3␲ –␲ – ␲ 0 2 2

␲ ␲ 3␲ 2 2

x

Domain: All real numbers except odd integer multiples of ␲/2 Range: (–, –1] h [1, )

1 y

y

y  csc x

The angles of two common triangles, in degrees and radians. 1 –␲ – ␲ 0 2

y  cot x

1 ␲ 2

␲ 3␲ 2␲ 2

Domain: x  0, ␲, 2␲, . . . Range: (–, –1] h [1, )

x

–␲ – ␲ 0 2

␲ 2

␲ 3␲ 2␲ 2

Domain: x  0, ␲, 2␲, . . . Range: (–, )

x

6653_AWLThomas_Last3Bkpgs 7/29/09 7:31 PM Page 3

SERIES Tests for Convergence of Infinite Series 1. The nth-Term Test: Unless an : 0, the series diverges. 2. Geometric series: g ar n converges if ƒ r ƒ 6 1; otherwise it diverges. 3. p-series: g 1>n p converges if p 7 1; otherwise it diverges. 4. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test or the Limit Comparison Test.

5. Series with some negative terms: Does g ƒ an ƒ converge? If yes, so does gan since absolute convergence implies convergence. 6. Alternating series: gan converges if the series satisfies the conditions of the Alternating Series Test.

Taylor Series q

1 = 1 + x + x 2 + Á + x n + Á = a x n, 1 - x n=0

ƒxƒ 6 1

q

1 = 1 - x + x 2 - Á + s -xdn + Á = a s - 1dnx n, 1 + x n=0

ƒxƒ 6 1

q

ex = 1 + x + sin x = x cos x = 1 -

xn x2 xn + Á + + Á = a , 2! n! n = 0 n!

ƒxƒ 6 q

q s - 1dnx 2n + 1 x5 x3 x 2n + 1 + - Á + s -1dn + Á = a , 3! 5! s2n + 1d! n = 0 s2n + 1d! q s - 1dnx 2n x4 x2 x 2n + - Á + s -1dn + Á = a , 2! 4! s2nd! s2nd! n=0

ln s1 + xd = x -

ƒxƒ 6 q

ƒxƒ 6 q

q s - 1dn - 1x n x2 x3 xn + - Á + s -1dn - 1 n + Á = a , n 2 3 n=1

-1 6 x … 1

x 2n + 1 x5 1 + x x3 x 2n + 1 + Á + = 2 tanh-1 x = 2 ax + + + Á b = 2a , 5 1 - x 3 2n + 1 n = 0 2n + 1 q

ln

tan-1 x = x -

q s - 1dnx 2n + 1 x5 x3 x 2n + 1 - Á + s -1dn + + Á = a , 5 3 2n + 1 2n + 1 n=0

ƒxƒ 6 1

ƒxƒ … 1

Binomial Series s1 + xdm = 1 + mx +

msm - 1dsm - 2dx 3 msm - 1dx 2 msm - 1dsm - 2d Á sm - k + 1dx k + Á + + Á + 2! 3! k!

q m = 1 + a a b x k, k=1 k

ƒ x ƒ 6 1,

where m a b = m, 1

msm - 1d m a b = , 2! 2

msm - 1d Á sm - k + 1d m a b = k! k

for k Ú 3.

6653_AWLThomas_Last3Bkpgs 7/29/09 7:31 PM Page 4

VECTOR OPERATOR FORMULAS (CARTESIAN FORM) Formulas for Grad, Div, Curl, and the Laplacian Cartesian (x, y, z) i, j, and k are unit vectors in the directions of increasing x, y, and z. M, N, and P are the scalar components of F(x, y, z) in these directions.

The Fundamental Theorem of Line Integrals 1.

Let F = Mi + Nj + Pk be a vector field whose components are continuous throughout an open connected region D in space. Then there exists a differentiable function ƒ such that 0ƒ 0ƒ 0ƒ i + j + k F = §ƒ = 0x 0y 0z B

2.

if and only if for all points A and B in D the value of 1A F # dr is independent of the path joining A to B in D. If the integral is independent of the path from A to B, its value is

0ƒ 0ƒ 0ƒ §ƒ = i + j + k 0x 0y 0z

Gradient

B

LA Divergence

§#F =

F # dr = ƒsBd - ƒsAd.

0N 0P 0M + + 0x 0y 0z Green’s Theorem and Its Generalization to Three Dimensions

Curl

Laplacian

i

j

k

0 § * F = 4 0x

0 0y

0 4 0z

M

N

P

§2ƒ =

0 2ƒ 0x

2

0 2ƒ + 0y

2

Normal form of Green’s Theorem:

C

Divergence Theorem:

0 2ƒ + 0z

2

F # n ds = § # F dA F 6 6

Tangential form of Green’s Theorem:

su *

= sv *

u * sv * wd =

su # wdv -

= sw *

ud # v

Stokes’ Theorem:

9

§ # F dV

D

F # dr = § * F # k dA F 6

C

wd # u

F # n ds =

S

Vector Triple Products vd # w

R

R

F # dr = § * F # n ds F 6

C

S

su # vdw

Vector Identities In the identities here, ƒ and g are differentiable scalar functions, F, F1 , and F2 are differentiable vector fields, and a and b are real constants. § * s§ƒd = 0 §sƒgd = ƒ§g + g§ƒ § # sgFd = g§ # F + §g # F § * sgFd = g§ * F + §g * F § # saF1 + bF2 d = a§ # F1 + b§ # F2 § * saF1 + bF2 d = a§ * F1 + b§ * F2 §sF1 # F2 d = sF1 # §dF2 + sF2 # §dF1 + F1 * s§ * F2 d + F2 * s§ * F1 d

§ # sF1 * F2 d = F2 # § * F1 - F1 # § * F2 § * sF1 * F2 d = sF2 # §dF1 - sF1 # §dF2 + s§ # F2 dF1 - s § # F1 dF2 § * s§ * Fd = §s § # Fd - s § # §dF = §s§ # Fd - §2F 1 s§ * Fd * F = sF # §dF - §sF # Fd 2

---

Thomas' Calculus (12th Edition)

Read more

Thomas' Calculus (12th Edition)

Read more

Thomas' Calculus (12th Edition)

Read more

Thomas' Calculus, Multivariable (12th Edition)

Read more

Thomas' Calculus, Multivariable (12th Edition)

Read more

Thomas' Calculus, Single Variable (12th Edition)

Read more

Thomas' Calculus 13TH Edition

Read more

Thomas' Calculus, 11th Edition

Read more

Thomas' Calculus, 11th Edition

Read more

Thomas' Calculus, 11th Edition

Read more

Thomas' calculus

Read more

Thomas' calculus

Read more

Thomas Calculus

Read more

Solution Manual Thomas' Calculus, 11th Edition

Read more

Thomas' Calculus 13th

Read more

Thomas' Calculus. SOLUTION MANUAL

Read more

Thomas Calculus 13th [Solutions].pdf

Read more

International Economics, 12th Edition

Read more

Human Biology, 12th Edition

Read more

Managerial Accounting, 12th edition

Read more

International Economics, 12th Edition

Read more

Social Psychology 12th Edition

Read more

Understanding Nutrition (12th edition)

Read more

Understanding Nutrition (12th edition)

Read more

Managerial Accounting, 12th edition

Read more

Human Biology, 12th Edition

Read more

Human Biology, 12th Edition

Read more

International Economics, 12th Edition

Read more

Social Psychology 12th Edition

Read more

Managerial Accounting, 12th edition

Read more

Recommend Documents

Thomas' Calculus (12th Edition)

---

Thomas' Calculus (12th Edition)

THOMAS' CALCULUS Twelfth Edition Based on the original work by George B. Thomas, Jr. Massachusetts Institute of Techn...

---

Thomas' Calculus (12th Edition)

---

Thomas' Calculus, Multivariable (12th Edition)

THOMAS' CALCULUS Twelfth Edition Multivariable Based on the original work by George B. Thomas, Jr. Massachusetts Ins...

---

Thomas' Calculus, Multivariable (12th Edition)

THOMAS' CALCULUS Twelfth Edition Multivariable Based on the original work by George B. Thomas, Jr. Massachusetts Ins...

---

Thomas' Calculus, Single Variable (12th Edition)

THOMAS' CALCULUS Twelfth Edition Single Variable Based on the original work by George B. Thomas, Jr. Massachusetts I...

---

Thomas' Calculus 13TH Edition

classic calculus text book by Thomas

---

Thomas' Calculus, 11th Edition

4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 1 Chapter 1 PRELIMINARIES OVERVIEW This chapter reviews the basic ...

---

Thomas' Calculus, 11th Edition

4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 1 Chapter 1 PRELIMINARIES OVERVIEW This chapter reviews the basic ...

---

Thomas' Calculus, 11th Edition

4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 1 Chapter 1 PRELIMINARIES OVERVIEW This chapter reviews the basic...

---

×

Report "Thomas' Calculus (12th Edition)"

Your name

Email

Reason -Select Reason- Pornographic Defamatory Illegal/Unlawful Spam Other Terms Of Service Violation File a copyright complaint

Description

Close Send

×

Sign In

Email

Password

Remember me Forgot password?

Sign In

---

Login with Facebook

Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close