Thomas' Calculus, Single Variable (12th Edition)

THOMAS'

CALCULUS Twelfth Edition

Single Variable 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

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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 978-0-321-63742-0 I. CaicolUll-Textbooks. I. Hass, Joel. II. Thomas, George B. (George Brinton), 1914--2006. III. Thoroas, George B. (George Brinton), 1914--2006. Calcolus. Iv. Title V. Title: Calcolus. QA303.2.W452009b 515-dc22

2009023069

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ISBN-10: 0-321-63742-9 ISBN-13: 978-0-321-63742-0

CONTENTS Preface

1

I

ix

1

Functions 1.1 1.2 1.3 1.4

Functions and Their Graphs I Combining Functions; Shifting and Scaling Graphs Trigonometric Functions 22 Graphing with Calco1ators and Computers 30 QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

34

35 37

AoDffiONAL 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

AoDffiONAL 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 ofTrigonometric Functions 135 The Chain Rule 142

iii

iv

Contents

3.7 3.8 3.9

Implicit Differentiation 149 Related Rates 155 Linearization and Differentials

164

QuESTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES

ADDITIONAL AND ADvANCED EXERCISES

4

184

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

198

I Integration

246 5.1 5.2 5.3 5.4 5.5 5.6

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

6

180

Applications of Derivatives 4.1 4.2 4.3 4.4 4.5 4.6 4.7

5

175

176

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 VHopital's Rule 396 Inverse Trigonometric Functions 404 Hyperbolic Functions 416 Relative Rates of Growth 424 QuEsTIONS TO GUIDE YOUR REVIEW 429 PRACTICE EXERCISES 430 ADOffiONAL AND ADvANCED EXERCISES 433

Integration by Parts 436 Trigonometric Integrals 444 Trigonometric Substitutions 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 ADOffiONAL AND ADvANCED EXERCISES 491

453

496

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

Infinite Sequences and Series 10.1 10.2 10.3 10.4 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 '!bsts

532

563

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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 ofPlane 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 ofTaylor Series 589 The Binomial Series and Applications ofTaylor Series 596 QuESTIONS TO GUIDE YOUR REVIEW 605 PRACTICE EXERCISES 605 ADDITIONAL AND ADvANCED EXERCISES 607

Curves in Space and Their Tangents 707 Integrals ofVector Functions; Projectile Motion 715 Arc Length in Space 724 Curvatore andNorma1 Vectors ofa Curve 728 Tangential and Normal Components ofAcceleration 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 ofMass 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

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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 Al A.2 A3 AA A.5 A6 A7 A.8 A9

Real Numbers and the Real Line AP-I Mathematical Induction AP-6 Lines, Circles, and Parabolas AP-1O Proofs of Limit Theorems AP-18 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 1'H6pital'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.

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Preface

• Antiderivatives and Integration We maintain the organization of the eleventh edition in placing antiderivatives as the rmal 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 ofthe 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.com/thomas. • Series We retain the organizational structure and content ofthe 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 ofvectors 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 coonections of multivariable ideas with their single-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.

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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.

No matter what Y positive number £ is, the graph enters this band at x = and stays.

1

y=l £

I---~.,----------''--'-----

-=-----.---------"<,-------1-£

No matter what positive number £ is, the graph enters this band atx = and stays.

-1

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 MATHXL 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

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Preface calculus students, while mentioning deeper or subtler issues they would study in a more advanced COUIlle. 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 ofthese theorems on the completeness ofthe 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 RMEWS AND PROJECTS In addition to problems appearing after each section, each chapter culminates 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. These 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. WRmNG 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 COUIlle using the text, technology can be incorporated according to the taste of the instructor. Each section contains exercises 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 I-II), 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 Transcendentals 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 Transcelldelltals, ISBN 0-321-62718-01978-0-321-62718-6 In addition to including all of the answers present in the stodent editions, the Illstructor ~ Editiolls 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 Ulliversity Calculus texts are based on Thomas' Calculus and feature a streamlined presentation ofthe contents ofthe 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 Illstructor ~ Solutiolls Mallual 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 Studellt~ Solutio1lS Mallual 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 algebra and trigonometry skills are critical to mastering calculus, and Just-ill-TIme Algebra alld 1Hgollometry 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 algebra or trigonometry topics and pointing out potential problem spots. The easy-to-use table of contents has algebra and trigonometry topics arranged 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 Mallual by James Stapleton, North Carolina State University Mathematica Mallual by Marie Vanisko, Carroll College IT-Graphillg Calculator Mallual by Elaine McDonald-Newman, Sonoma State University These manuals cover Maple 13, Mathematica 7, and the TI-83 PluslTI-84 Plus and TI-89, respectively. Each manual provides detailed gnidance for integrating a specific software package or graphing calculator throughout the course, including syntax and commands.

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

WEB SITE www.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 instruction 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 ""ereises, 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"maI grades. You can also add offiine (paper-and-pencil) grades to the gradebook. o MathXL Exereise 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 MyMatbLab and MathXL.

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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. • Personalized Study Plan, generated when students complete a test or quiz, 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, auturnatically 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 auturnatically 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 Gll 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 ofthe 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.

PowerPointGll 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 through the Pearson Instructur 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 ofColorado 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 ofCentral 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

==

I(x)

("y equals I of x").

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

DEFINITION A function I from a set D to a set Y is a rule that assigns a unique (single) element I{x) E Y to each element XED. The set D of all possible input values is called the domain of the function. The set of all values of I(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

x""",,:-_.~

Input (domain)

f

""",,:~_.~f(x)

Output

(nmge)

FIGURE 1.1 A diagram showing a

function as a kind ofmachine.

~

~f(a) D

= domain set

f(x)

Y = set containing

the range

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

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 = 1fr 2 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 def'me 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 ofreal x-values for which the formula gives real y-values, the so-called natural 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 ofreal 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 I), the range is {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 realvalued. The domains and ranges ofmany real-valued functions of a real variable are intervals or combinations ofintervals. The intervals may be open, closed, or half open, and may be finite or infinite. The range of a function is not always 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 ofa function as a machine. For instance, the Y!;;key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the y!;; key. A function can also be pictured as an arrow diagram (Figure 1.2). Each arrow associates an element ofthe 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).

element in D.

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 = y!;;

y=~ y =

'\IT=7

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

[0, I]

Solution The formula y = x 2 gives a real 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;;,: 0. The formula y = I/x gives a real y-value for every x except x = 0. For consistency in the rules ofarithmetic, we cannot divide any number by zero. The range of y = I/x, the set ofreciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = I/(I/y). That is, for yolO the number x = I/y is the input assigned to the output valuey. The formula y = '\Ii; gives a real y-value only if x ;;,: 0. The range of y = '\Ii; 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 = '\f"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 '\f"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 - I to I. Outside this domain, I - x 2 is negative and its square root is not a real number. The values of I - x 2 vary from 0 to I on the given domain, and the square roots ofthese values do the same. The range of ~ is [0, I]. •

Graphs of Functions If j is a function with domain D, its graph consists of the pnints 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 ofpoints 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 1.4). y

y

---;01---'-;---'-;;--\--I-;---x

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

EXAMPLE 2

Graph the function y

=

FIGURE 1.4 If(x,y) lies on the graph of f, theo the value y = f(x) is the height of the graph above the point x (or below x if f(x) is uegative).

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). • How do we know that the graph of y = x 2 doesn't look like one ofthese curves? y

----'---'---~....-:---'---'---~x

-2

-I

FIGURE 1.5

Example 2.

0

2

Graph ofthe 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 numerieally, 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 tuning 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 plotred points gives a graph of the pressure function represeoted 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

-_C-f---~f---+----jl--x

---!---+--~_x

-I

1

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 grapb ofa function f(x) = ~. (c) The lower semicircle is the grapb ofa functiong(x) ~ -~.

y

y=x

Piecewise-Defined Functions __!____c:--!--c?I''----~__!____c:___x

-3 -2 -I

0

2

3

FIGURE 1.8 The absolute value function bas domain (- 00, 00) and range [0, 00). y

Sometimes a function is described by using different formulas on different parts of its domain. One example is the absolute value function

Ixl =

x;;"

°

x < 0,

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 < 0. Here are some other examples.

EXAMPLE 4

FIGURE 1.9 To grapb 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 real 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 :5 x :5 I, and y = I when x > I. The function, however, is just one jUnction whose • domain is the entire set ofreal numbers (Figure 1.9).

°

,

,'Y=x 3

,

,~

,':"""'"

2

~

y~LxJ

"

-_-!2~_~I-,~'-~-c2e--!3~--x

,

"

......-0

"

EXAMPLE 5 The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer 1100r function. It is denoted l x J. Figure 1.10 shows the graph. Observe that

l2.4J = 2, l2J = 2,

-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 e9uul 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

3

/ 'y=x

7

2

0---;-' ,

y

~

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.

rxl

/

---:----9----,If--:---7---;:-~x

,

-2 - I ,

I

2

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

3

e--;" -I

7/

-2

FIGURE 1.11 TIre graph ofthe leasl 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 (Appendix I). EXAMPLE 7 The function graphed in Figure 1.9 is decreasing on (-00,01 and increasing on [0, 1]. The function is neither increasing nor decreasing on the interval [1, 00) because ofthe 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

(aJ y

---~..,+;~--~x

(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 functioo) is symmetric ahout the origin.

The names even and odd come from powers of x. If Y is an even power of x, as in y = x 2 or Y = x4, it is an even function ofx because (-x)2 = x 2 and (-x)4 = x 4 . Ify is an odd power of x, as iny = X or y = x3, it is an odd function ofxbecause (-x)' = -x and (-x)' = -x'. The graph of an even function is symmetric abontthey-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 only 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 f(x) = x 2

f(x) = x 2

+I

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

+

I = x 2 + I 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: (-x)' = x 2 for all x; symmetry about y-axis.

+I

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

1.1 y

Functions and Their Graphs

7

y

--~"'-----~x


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

(b)

FIGURE 1.13
y =

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=2 y~2x

m=l

m=zI

y

2

y~~ 2

0

2

x


x

(bJ

FIGURE 1.14
DEFINmON Two varisbles y and x are proportional (to one another) if one is always a constant multiple ofthe other; that is, if y = kx for some nonzero constant k.

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 (-00,01 and increasing on [0, 00); the odd-powered functions are increasing over the entire real line (-00,00).

--;-'''''::'t-'--:---x

---:-~L--:-_x

-1

FIGURE 1.15

0 -1

-1

1

0

-1

Graphs of j(x) = x', n = 1,2,3,4,5, dermed 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). J

~,

)J

1

Domain: x =I: 0 Range: n' 0

1 ~~c--+~~~x

0

(a)

(b)

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

I I 3

1 Domain: x =# 0 Range: J> 0

~

x'forpart(a)a

~

-1

2

(c) a = 2'3'2,and 3 . The functions f(x) = x1/2 = Y; and g(x) = x ' / 3 = \Yx are the square root and cube 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 x2/3 = (x 1/3)2 .) Polynomials

A function p is a polynomial if

p(x)

=

a.x'

+ a._lX·- l + ... + a,x + ao

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

1.1

9

Functions and Their Graphs

y y

y

y

1 --oO+--I;----~x

---"""c;---~x

-----nc~:---~x

o

Domain: O:=s; x < 00 Range: 0 :=s; y < 00

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

FIGURE 1.17

1 Domain: O:=s; x < Range: 0 :S Y <

-----;;'l'---;--~x

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

00

00

t,t,~,andt.

leading coefficient a. oF 0 and n > 0, then n is called the degree ofthe polynomial. Linear functions with m oF 0 are polynomials ofdegree I. Polynomials ofdegree 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 shows the graphs of three polynomials. Techniques to graph polynomials are studied in Ouipter 4. x3 x2 1 Y~3-2-2x+3 y 4

y y

2

y

= 8x 4 -

14x 3 - 9x 2 + llx - 1

y ~ (x - 2)4(x

+ 1)3(x -

I)

16

2 ---c\-----j?--->,~--___l~-->x

-I

2

-4

FIGURE 1.18

(c)

(b)

(a)

Graphs of three polynomial functions.

Rational Functions A rational function is a quotient or ratio f(x) = p(x)/q(x), where p and q are polynomials. The domain of a rational function is the set of all real x for which q(x) oF O. The graphs of several rational functions are showo 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

NorroSCAlB

-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~

y

y ~ ~(x2 _ 1)213

W>x

----{-+----cf------>x

(a)

.«1 - x)'"

---+..--.:+---->x

(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

(a) f(x)

~

sin x

(b) f(x)

~

eou

FIGURE 1.21 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 I, 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

-I

-0.5

y

12

12

10

10

8

8

6

6

4

4

2

2

0 (a)

0.5

I

x

FIGURE 1.22 Graphs ofexpooential functions.

-I

-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~~----:;;O~~~:::y:::~~IOg,~X -1

y

=

x

lOglOx

-''--LL--_--:lc-----;OC-----;j---'L-''--x

FIGURE 1.23

Graphs offour logaritlunic

functions.

FIGURE 1.24 Graph ofa cateoary or hsnging cable. (The Latin word catena 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 derming the graph is discussed in Section 7.7.

Exercises 1.1 Functions

8. a.

Y

b.

y

10 Exercises !--{j, find the domain and range of each function.

+ x2 ~ v'Sx + 10

1. f(x) = !

2. f(x) = ! -

3. F(x)

4. g(x) ~

S. f(t)

~

4 - 3t -

6. G(t)

~

v;.

v'x 2 -

o

3x

2 -2-t - 16

10 Exercises 7 and 8, which of the graphs are graphs of functions ofx,

and which are not? Give reasons for your answers. 7.a.y b.y

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

o

--+---------.x o

Finding Formulas for Functions 9. Express the area and perimeter of an equilateral triangle as a function of the triangle's side lengthx. 10. Express the side length ofa square as a function of the length d of the square's diagonal. Then express the area as a function of the diagonal length.

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

o

--+---------.x o

11. Express the edge length ofa cube as a function of the cube's diagonal length d. Then express the surface area and volume of the cube as a function of the diagonal length.

12

Chapter 1: Functions

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

13. Consider the point (x, y) lying 00 the graph of the line 2x + 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 00 the graph of y = ~. Let L he the distsnce hetweeo the points (x,y) and (4, 0). WriteL as a functioo ofy. Functions and Graphs Find the domain and graph the functions in Exercises 15-20.

15. j(x) = 5 - 2x 17. g(x) =

16. j(x) = I - 2x - x' 18. g(x) = h

19. F(t) = tlltl

20. G(t) =

'\!Ix1

x~.

21. Find the domain ofy =

x' -

4 -

22. Find the range of y

~

2

lfItl

9

x' + -,--. x +4

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

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

b. y'=x'

a.lyl=x

24. Graph the following equatioos and explain why they are not graphs of functioos of x.

+ Iyl

a. Ixl

= I

b. Ix

b. rxl=O?

a.lxJ=O?

+ yl

j(x) =

{f:~:

x~o

x

< O.

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

Piecewise-Defined Functions Graph the functioos in Exercises 25-28.

25. j(x) = {x, 2-x, I-x 26. g(x) = { 2 _ x', 27. F(x) 28. G(x)

~ ~

37. y

= -x'

38.y=_I,

x:5 II x>

39. y

=-~

40.y=~

41. y

~

{4,-+x

', 2x ,

x

{l/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 O::sx:sl 1<x:s2

x < 0 0 :s x

43. y

Find a formula for each functioo graphed in Exercises 29-32. 29.a.

b.

y (I, I)

r--1

0

I I I

I I I

2

3

4

y

b.

Give reasons for your answer. 47. j(x) = 3 49. j(x) = x' + I 51. g(x) = x' + x 53. g(x)

3

5

42.y~h 44. y = -4"';; 46. y ~ (-x)'/3

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

X

30. a.

= x'/8

45. y ~ -x'j2

x

y

2

2

'\!Ix1

I

x

~

-3

SO. j(x) = x' + x 52. g(x) = x' + 3x' - I ~

-,--

54. g(x)

I t=1

56. h(t) = It'l

57. h(t) = 2t + I

58. h(t) = 21tl

55. h(t) = x

I x - I

48. j(x) = x- s

x

-,-x - I

+I

Theory and Examples 59. The variable s is proportiooa1 to t, and s = 25 wheo t = 75. Determine t wheo 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 K when v = 10 m/sec?

Functions and Their Graphs

13

66.a.y=5x y

61. The variables r and s are inversely proportional, and r = 6 when s = 4. Determine s when r = 10. 62. Boyle's Law Boyle's Law says that the volume V of a gas at constant temperature increases whenever the pressure P decreases, so that V and P are inversely proportional. If P = 14.71bs/in2 when V = 1000 in3, then what is VwhenP = 23.41bs/in2?

--===;:::::::::~--~--~x

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 comer and then folding up the sides as in the figure. Express the volume V of the box as a function ofx. I(

Tx

f---'

14

x

1

-Ix

D 67.

II

22 IX

XL

x

3_ xl

a. Graph the functions j(x)

= x/2 and g(x) = 1 + (4/x) together to identify the values of x for which x 4 2> 1 + x·

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

D 68.

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 of x. y

a. Graphthefunctionsj(x)

= 3/(x - l)andg(x) = 2/(x + 1) together to identify the values of x for which

_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.

-+__

---J.'-----_--'---_ _

-----.JL-_~~x

-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 = x 4

b. y = x 7

c. Y = x 10

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. 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. 1(--(----2mi----~.1 P x City

y

Power plant NOT TO SCALE

---~=-"'"o!-~""""=----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(j) n D(g», we define functions f + g, f - g, and fg by the formulas

(j

+ g)(x)

=

f(x)

+ g(x).

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

L)(x) = f(x) (g g(x)

(where g(x) oF 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) = cf(x).

EXAMPLE 1

The functions defined by the formulas

y;,

f(x) =

g(x) = ~

and

have domains D(j) = [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 ofthe two functions. We also write f' g for the product function fg.

Function

Formula

f+g f-g g-f fog

(j + g)(x) (j - g)(x) (g - f)(x) (j' g)(x) =

fig

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

glf

~(

f x

) =

Domain = = =

y;, + ~ y;, - ~ ~ - y;,

f(x)g(x) = Vr-:x(cI _-----ox)

g(x) = f(x)

~

"Y-f-=-i

lx

x

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

I] = D(j) 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(j) 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)

--;;t------;:---'---~'----~x

o

a

FIGURE 1.25

Graphical addition oftwo

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

functions.

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 io the domain of f.

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

g

x __

g

-g(x)_

f

FIGURE 1.27 Two functions can be composed at x whenever the value of one function at x lies in the domain ofthe other. The composite is denoted by

g(x)

FIGURE 1.28

fog.

f

_ _ f(g(x))

Arrow diagram for fog.

To evaluate the composite function g 0 1 (when defioed), we fmd I(x) first and then 0 1 is the set of numbers x io the domaio of 1 such that I(x) lies io the domain ofg. 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

Iff(x) = \/;;andg(x) = x

(h) (g

g)(x)

f)(x)

0

+ I, find

(C) (f

f)(x)

0

(d) (g

0

Solution Composite

g) (x) .

Domain

(a) (f

0

g)(x)

=

f(g(x))

=

v.gw = \Ix+l

(h) (g

0

f)(x)

=

g(f(x))

=

f(x)

(c) (f

0

f)(x)

=

f(f(x))

=

V1W = y:;.rx =

(d) (g

0

g)(x)

=

g(g(x))

=

g(x)

+

+

I = \/;; I = (x

+

+

[-1,00) I

[0, 00)

xl/'

[0, 00)

I)

+

I = x

+2

(-00,00)

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

0

g)(x)

=

(\/;;)2 =

x. However,

the domain of fog is [0, 00), not ( - 00, 00), since \/;; requires 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 Shifts y

y = f(x)

+k

Shifts the graph of f up k units if k Shifts it down Ikl units if k

>0

<0

Horizontal Shifts

y

=

f(x + h)

Shifts the graph of f left h units if h Shifts it right 1h I units if h

>0

<0

EXAMPLE 3 -+-~4+--~x

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).

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

+I

(h) Adding -2 to the right-hand side of the formula y = x 2 to get y = x 2 graph down 2 units (Figure 1.29).

-

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

-2

2 shifts the

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

1

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

(d) Adding

shifts the

1x - 21 -

I •

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

1.2 Combining Functions; Shifting and Scaling Graphs Add a positive

constant to x.

Add a negative 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 altift 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 Scallng and Reflecting Formulas For c

>

1, the graph is scaled:

y

=

cf(x)

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

y

=

cI f(x)

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

y

=

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) f( -x)

= -

y =

EXAMPLE 4

Reflects the graph of f across the x-axis. Reflects the graph of f across the y-axis. Here we scale and reflect the graph of y =

y/;;.

(a) Vertical: Multiplying the right-hand side of y = y/;; by 3 to get y = 3 y/;; 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 = 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 = ~ = 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.

v'3

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

h

y

y/;; across the x-axis, and • y

y 4

3 y~

2

1 ~~:::::::::~::::::"-YF

Yx

----"----=-F-~-_o____cL---'--~x

FIGURE 1.32 Vertically stretching and compressing the graph y = Vi hy a factor of3 (Example 4a).

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

0

2

3

-3

-2

-1

v;j3

~~~=;='x -1 0 1 2 3 4

-1

-_;;____c:-~_+___cc-_;;____c:-~x

4

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

(Example 4c).

18

Chapter 1: Functions Given the functioof(x) = x 4

EXAMPLE 5

-

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

20

-1

x

0 -10

=

16x4 + 32x 3 + 10 y

\

y

20

-2

10 x

1

0

-1

-10

-20

+ 2x3 - 5

y = _!x 4 2

x

0 -10

-20 (a)

(b)

(c)

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

c 2x 2 y

+ y2

= r 2.

(1)

y

y

r

-+---OO+-----f-~x

r

-r

r C

-r

(a) circle

FIGURE 1.36

-r

-r

(b) ellipse, 0 < c

<

1

Horizontal stretching or compression of a circle produces graphs ofellipses.

(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 gmph 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 (±r/c, 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 (±r/c, 0).1n both cases, the major axis is the longer line segment. Ifwe divide both sides of Equation (I) by r 2 , we obtain

y

b

--+---4>::----+-----> X -a

19

a

-b (2)

FIGURE 1.37 2 X

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

Graph of tire ellipse

y2

+ 2 = l.a > b,wherethemajor

a b axis is horizontal.

(x - h)2

+

-'----a"2-'--

(y - k)' b2 = 1.

(3)

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 Algobraic Combinations f·g· I. f(x) = x, 2. f(x) =

=

g(x) = v;=J v;+l, g(x) = v;=J

I, g(x)

=

I

9. f(x) = vx

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

- x

Let f(x) = x - 3, g(x) = \/;;, h(x) = x', and j(x) = lx. Express each of the functions in Exercises 11 and 12 as a composite involving one or more of f,g, h, andj.

11. a. y

c. y

+ \/;;

= \/;; -

3

= x l/ 4

12.a.y=lx-3 2

5. If f(x) = x + 5 andg(x) = x - 3, fmd the following. a. f(g(O» b. g(f(O»

c. f(g(x» e. f(f( -5))

d. g(f(x»

g. f(f(x»

b. g(g(x»

r.

b. y =

Z\/;;

d. y

4x

r.

e. y = V(x - 3)'

Composites of Functions

=

y = (lx - 6)'

b.y=x'j2

LY=~

d.y=x-6

e. y = Z~ r. y = W-=3 13. Copy and coroplore the following table.

g(g(Z»

(f

g)(x)

g(x)

f(x)

I/(x + I), fmd the following.

a. x - 7

\/;;

?

a. f(g(l/Z» c. f(g(x» e. f(f(Z»

b. g(f(I/Z»

b. x + Z

3x

?

d. g(f(x))

c.

~

~

r.

b. g(g(x»

x x - I

?

g. f(f(x»

x d. -I x-

e.

?

1+ 1 x

x

f.

x

?

x

6. If f(x)

=

x - I andg(x)

=

g(g(Z»

In Exercises 7-10, write a formula for fog

0

h.

+ I, g(x) = 3x, h(x) = 4 - x 8. f(x) = 3x + 4, g(x) = lx - I, h(x) = x 2

xI

x2 .~ g(x) = - 2 - - ' h(x) = V Z - x x +I

x+Z 10. f(x) = -3- ,

In Exercises 3 and 4, fmd the domains and ranges of f, g, fig, and glf. 3. f(x) = Z, g(x) = x 2 + I

4. f(x)

I

• ,---c-;

In Exercises I and Z, fmd tire domains and ranges of f, g, f + g, and

7. f(x) = x

?

I

0

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)

f(x)

a. -I-

Ixl

x - I

(f

0

g)(x)

?

b.

?

x - I x

c.

?

V;

d.

V;

?

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.

a. j(g( -I))

g(j(O» e. g(j( -2))

b.

d. g(g(2»

a. y

j(j( -I)) f. j(g(l))

c.

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

g(j(3» e. g(j(O»

a. j(g(O» d.

b.

j(j(2»

~ ~,

g(x)

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

~

e. g(g(-l» f.

j(g(1/2)) 0

f and

t

V;

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

x-

(f

0

d. Y = (x

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

17. f(x)

+2 + 3)' - 2

b. Y ~ (x - 2)2

1)2 - 4

+ 2)2 + 2 y

16. Evaluate each expression using the functions

f(x) = 2 - x,

~ (x -

e. y = (x

gXx) = x.

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

(I, -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)

ShITtfng Graphs 21. The accompanying figure shows the graph of y = -x 2 shi!red 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 grapb. Then sketch the original and sbifted graphs together, labeling each graph with its equation.

25. x 2 + y2 = 49 26. x 2 + y2 ~ 25 27. Y =

x'

28. Y ~ x

2

21

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, and sketch their graphs. y

Down 3, left 2 Up 3, left 4

Left I, down 1 Right I, down I

/'

29. Y ="I/x Left 0.81 30. y = -"I/x Right 3 31. Y

~

2x - 7 Up 7

32. Y

~

1 2(x

a. g( -I) c. g(l)

+

33. Y ~ Ilx

I)

+5

Down5,rightl

Up I, right 1

34. Y = IIx 2 Left 2, down 1

37. Y

~

36.y=~

Ix - 21

38. y ~

39.y=I+~ 41. y ~ (x

= '¥x -

47. Y

=x

1

42. Y ~ (x - 8)2/3

1)2/' 43.y=l-x 2/'

44.y+4=x 2/'

1- I

1 _ 2

46. y

= (x + 2)'/2 + 1

48. y

=x -

I

2

•

g. g(I - I)

h. -g(1 - 4)

57. Y = x 2

58. y = x

I

- (x - 1)2

-

I,

stretched vertically by a factor of3

-

I,

compressedhorizontallybyafactorof2

1 + --"

compressed vertically by a factor of2

60. y = I

1 + --"

stretched horizontally by a factor of3

x x

61. y ~ y';+J,

compressed horizontally by a factor of 4

62. Y ~ y';+J,

stretcbedverticallybyafactorof3

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

stretched horizontally by a factor of 2

50,y~x+2

64. Y ~

I 52.y=-2-1

65. y = I 66. y = I -

I 54. Y ~ (x + I)'

Graphing In Exercises 67-74, graph each function, not by plotting points, but by

X

I 53. Y ~ x2 + I

2

59. Y = I

63. y ~

I

51y-

d. I - g(l) f. g(1 - 2)

40.y=I-"I/x

+

45. Y

II - xl -

+3 + 2)

e. g( -I

Vertical and Horizontal Scaling 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.

Graph the functions in Exercises 35-54.

35.y=v;+4

b. -g(l)

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 ofthe following functions, and sketch their graphs. y

x', x',

compressed vertically by a factor of3

compressed horizontally by a factor of3 stretchedhorizontallybyafactorof2

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

70. Y = (I - x)'

1 71.y=2x- 1 73. Y =

2

0

•

y~~I-I +

2

2 72·y=x 2 + 1

-"\Yx

74. Y = (-2x)'/'

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

Ix2

-

"\IIx1.

II .

a. f(x) + 2

b. f(x) - 1

c. 2f(x)

d. -f(x)

e. f(x + 2)

f. f(x - I)

Ellipses Exercises 77-82 give equations ofellipses. Put each equation in standard form and sketch the ellipse. 77. 9x 2 + 25y 2 ~ 225 78. 1&2 + 7y 2 ~ 112

g. f( -x)

h. -f(x+ I) + I

79. 3x 2

+ (y

- 2)'

=3

80. (x

+

1)2

+ 2y 2

= 4

22

Chapter 1: Functions

+ +-k)'

81. 3(x - I)'

82.

+ 2(y + 2)'

+ ~)' +

~ 6

a. fg d. f' =

= 54

If

g. gof

83. Write an equation for the ellipse (x'/16) + (y'/9) = 1 shifted 4 units to the left and 3 units up. Slretch 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. Slretch the ellipse and ideotifY its center and major axis.

b. fig

e. glf

e. g' = gg

r.

h. fof

i. gog

fog

86. Can a function be both flNeo and odd? Give reasons for your answer.

D 87.

D 88.

Comlrin;ng Functions

\h

(Continuation of Example I.) Graph the functions f(x) ~ and g(x) ~ ~ together with their (a) sum, (b) product, (c) two differeoces, (d) two quotieots.

Let f(x) = x - 7 and g(x) = x'. Graph f and g together with

f

0

gandg

0

f.

85. Assume that f is an even function, g is an odd function, and both

f

andg are dermed on the eotire real line R. Which of the following (where dermed) are flNeo? odd?

Trigonometric Functions

1.3

TIris section reviews radian measure and the basic trigonometric functions.

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

C""1e of ,iJi"'"

<

s = rO

FIGURE 1.38 The radian measure of the ceotral aogleA'CB' is the number () ~ sir. For a unit circle ofradius r = 1,8 is the leogth of arc AB that ceotta! angle ACB cuts from the unit circle.

«() in radians).

(1)

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

radians = 1800

(2)

and

1 radian =

I~O ('" 57.3) degrees

or

1 degree =

I~O ('" 0.017) radians.

Table 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

o(radians)

-1T

-135

-31T 4

-90

-45

0

30

45

60

90

o

1T

1T

1T

6

4

3

120

135

150

1T

31T

51T

2

4

6

180

270

360

1.3 Trigonometric Functions

23

An angle in the xy-plane is said to be in standard position ifits 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 clockwise are assigned negative measures. y

y

Initial ray ----~,f___~---"--~x

--t~-~==--~-~x

FIGURE 1.39 Angles in standard 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

317 --+-jLL+-~x

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

----,1---,----> x

--+-h-+--~x

917

4 hypotenuse

opposite

adjacent sin {1

= opp

csc {1

= hyp

sec {1

=

hyp adj

cot {1

=

hyp d"

opp

COS{1=:~

opp

tan 9 ~ adj

adj

FIGURE 1.40 Nonzero radiao measures cao be positive or negative and cao go beyond 2".. 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 troe when angles are measured in degrees.

The Six Basic Trigonometric Functions

opp

FIGURE 1.41 Trigonometric ratios of an acute angle.

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 ofthe point P(x, y) where the angle's terminal ray intersects the circle (Figure 1.42).

y

sine:

sin () = )I r

cosecant:

r csc()=y

rx

secant:

sec () = ,.

~

cotangent:

cosine: cos 8 = tangent:

tan () =

x

cot () =

y

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

tan () = sin () cos ()

I cot () = tan ()

~() cos

I csc () = sin ()

sec() =

24

Chapter 1: Functions As you can see, tan II and sec II are not defined if x = cos II = O. This means they are not defined if II is ±".j2, ±3".j2, ... . Similarly, cot II and esc II are not defined for values ofll forwhichy = O,narnelyll = 0, ±1r, ±21r, .... The exact values of these trigonometric ratios for some angles can be read from the triangles in Figure 1.43. For instance,

~1

zj 2

'Tf -

'Tf -

4

. 1r

2

. 1r V3 smT=2

1

sm-=-

4v'2

1

".

1r - 1 cos-=

FIGURE 1.43 Radian angles and side lengths of two common triangles.

V3

cos 6 = 2

4v'2

tan ". = 1 tan 1r = _1_ tan ". = V3 4 6 V3 3 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 2".

cosT

=

I

tan

-2'

2; = -V3.

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

y

y S sin pas

A all pos x

T tanpos

C cospos

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

FIGURE 1.45 The triangle for calculating the sine and cosine of 21r/3 radiaos. The side lengths come from the geometry ofright triangles.

trigonometric functions are positive in each quadrant.

Using a simiiarmethod we determined the values ofsin II, cos II, and tan II shown in Table 1.3. TABLE 1.3

Degrees II (radians)

Values of sin II, cos II, and tan II for selected vaLues of II

-180 -".

sin II

0

colli

-I

tan II

0

-135 -3". 4

-v'2 2

-v'2 2

1

-90

-45

-1r

-".

2

4

-I

0

-v'2 2

v'2 2

-1

0

30

45

60

90

120

0

".

".

".

1r

21r

6

4

3

2

I 2

v'2

V3

2

2

I

I

V3

v'2

2

2

1 2

0

V3

1

V3

0

3

0

3

135 3". 4

150 5". 6

V3

v'2

2

2

I 2

-v'2

-V3

2

2

-1

-V3

1 2

-V3

3

180 270 360 3". 2". ". 2 0

-I

0

-I

0

I

0

0

1.3

25

Trigonometric Functions

Periodicity and Graphs of the Trigonometric Functions When an angle ofmeasure fJ and an angle of measure fJ + 2". are in standard position, their terminal rays coincide. The two angles therefore have the same trigonometric fimction values:

I

Period. ofTrigonometric Functions Period 'IT : tan (x + ".) ~ tan x cot(x + ".) = cotx Period 2'IT: sin(x + 2".) = sinx cos(x + 2".) ~ cosx "",,(x + 2".) = socx csc(x + 2".) = cscx

sin(fJ + 21T) = sin fJ, tan(fJ + 21T) = tan fJ, and so on. Similarly, cos(fJ - 21T) = cos fJ, sin(fJ - 2'IT) = sin fJ, 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 if there is a positive numberp 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 fJ. 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 - lanx

Domain: _00 < x < co

Even

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

cos ( -x) see ( -x)

= =

cosx

(a) y

secx

y = socx

n n

Domain-

Domain: -co < x < 00 Range: -1:s y:s 1 Period: 2'11" (b)

.X

y

3'lT -2'- 2.···

*+'1r +

Range:

-co

< y
Period:

'1T

(c) y

y - cotx

\v

----,t:-=-tn+-+-c":-±-~x

Odd

_3

sin( -x) tan(-x) csc( -x) cot (-x)

= = =

-sinx -tanx -cscx

= -COlx

0

Domain'• X "" -2'-2 + 'Ir + 31T •... Range: y:s -1 ory ~ 1 Period: 2'lT (d)

Domain: x "'" 0, ±1f, ±21f, ••• Range: y::5 -lory 2:: 1 Period: 27l'

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

(oj

FIGURE 1.46 Graphs ofthe 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 fJ that ray OP makes with the positive x-axis (Figure 1.42). Since xlr = cosfJ andylr = sinfJ, we have -t--~""","b~--t-~x

x = rcosfJ,

y

=

rsinfJ.

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

FIGURE 1.47 The referee.. triangle for a general angle (}.

(3)

26

Chapter 1: Functions TIris equation, true for all values of IJ, is the most frequently used identity in trigonometry. Dividing this identity in turn by cos2 1J and sin2 1J gives

I I

+ tan2 IJ + cot' IJ

= sec2 IJ = csc2 IJ

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

Addition 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 IJ for both A and B in the addition formulas gives

Double-Angle Formulas cos 21J = cos2 1J - sin2 1J sin 21J = 2 sin IJ cos IJ

(5)

Additional formulas come from combining the equations cos2 1J

+

sin2 1J = I,

cos2 1J - sin2 1J = cos 21J.

We add the two equations to get 2 cos2 1J = I + cos 21J and subtract the second from the first to get 2 sin2 IJ = I - cos 21J. TIris results in the following identities, which are useful in integral calculus.

HaIf-Angie Formulas cos2 1J = I

+

cos 21J 2

(6) (7)

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

(8)

TIris equation is called the law of cosines.

1.3

B(a cos 8, a sin 8)

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

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 ofthe distance between A and B is therefore

y

---~

Trigonometric Functions

+

(a sin 11)2

sin211)

+ b2 -

Zabcosll

I

......--:---?~;C-~X b A(b,O) a2

=

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

+

+ b2

-

Zabcosll.

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

7T/Z,

then cos II = 0

Transformations 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.

/Vcrtical

Vertical stIm:h or compression.; reflection about x-axis ifnCgative~

y = af(b(x

+ c)) + d

~

Horizontal stretch or CODIJlI"Ssion; / reflection abouty-axis if negative

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 ofthe various terms is revealing and given below. y

y = A sin

(2;<x - C») + D This axis is the liney=D.

o Two Special Inequalities 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 ioequalities, we picture fJ as a nonzero angle io standard position (Figure 1.49). The circle io the figure is a unit circle, solfJl equals the length ofthe circular arc AP. The length oflioe segmentAP is therefore less than IfJl. Triangle APQ is a right triangle with sides of length

y p

QP = IsiofJl,

From the Pythagorean theorem and the fact that AP

+--___co++~-+___c~X

A(l,O)

cos 6

AQ = I - cosfJ.

2

sio fJ

I-cosO

+

<

IfJ I, we get

(I - 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': sio2 fJ :5 FIGURE 1.49 From the geometry of this figure, drawn for 8 > 0, we get the inequality sin2 8 + (I - cos 8)2 :5 IP.

6'

and

(I - cos fJ)2 :5

6'.

By taking square roots, this is equivalent to sayiog that Isio fJl :5 IfJl

and

II - cosfJl :5 IfJl,

-lfJl :5 siofJ:5 IfJl

and

-lfJl :5 I - cosfJ :5 IfJl.

so

These inequalities will be useful io the next chapter.

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

4. If you roll a I-m-diameter wheel forward 30 em over level ground, through what aogle will the wheel tum? 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'fr/2

-2'fr/3

o

'fr/2

-'fr/6

'fr/4

S1f/6

sinO

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

xe[f,?T] 9ocosx=t, xe[-f,o] t, ['If, 7.

Sinx=~,

11. tanx =

fJ

-'fr/3

XE

3;]

8. tanx=2,

xe[o,f]

5 10ocosx=-1 3' xe[f,'T1'] 12. sinx =

-t,

XE

['If, 3;]

3'fr/4

sinO

cosO tan 8 cot 8 sec 0 esc 8 6. Copy aod complete the following table of function values. If the function is und.efmed 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

-1)

18. - cos 21TX

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"..

- 2

Graph the functions in Exercises 23-26 in the Is-plane (I-axis horizontal, s-axis vertical). What is the period of each function? What symmetries do the graphs have?

23. s

= cot 21

24. s

= -tao".1

25. s

= seC(~I)

26. s

= csc(f)

D 27.

D 28.

Graph y = taox and 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 and range of l sin x J?

= l sin x J together. What are the domain

30. Graph y = sinx and y and range of sin x 1?

= rsinx 1together. What are the domain

I-) + f)

= sinx

33. Sin(x

=

32. cos(x

~

+

34. Sin(x -

cosx

35. cos(A - B) = cosA cosB different derivation.)

36. sin(A - B)

1-) = -sinx f) -cosx

+ sinA sinB (Exercise 57 provides a

sinA cosB - cosA sinB

In Exercises 39-42, express the given quantity in terms of sin

cosx.

+ x)

40. sin(2". - x)

x)

42. cose;

+

x)

. 7". . (". ".) 43· Evaluatesffi12assm 4+3.

44.

Evaluate cos

45. Evaluate cos

If; ~.

~

0

tao(A

54. cos20

+ cosO

~

0

+ B)

taoA+taoB ~ I - taoA taoB'

Derive the formula.

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

as cos

x --+--=------'""""'---'-L--+--_+ o

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

38. What happeos if you take B = 2". in the addition formulas? Do the results agree with something you already koow?

41. Sin(3; -

53. sin20 - cosO

=

37. What happens if you take B = A in the trigonoroetric identity cos(A - B) = cosA cosB + sinA sinB? Does the result agree with something you already koow?

39. cos(".

52. sin2 0 = cos2 0

1

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

31. cos(x -

t

Theory and Examples 55. The tangent SDID formula The staodard formula for the taogent of the sum of two sogles is

a. Graph y = cos x and y = sec x together for - 3"./2 ,;; x ,;; 3"./2. Comment on the behavior of sec x in relation to the signs and values of cos x. b. Graphy = 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.

r

51. sin2 0 =

x and

0)

to obtain the addition formula for sin(A

+ B).

h. Derive the formula for cos (A + B) by suhstitoting - B for B in the formula for cos (A - B) froro Exercise 35. 59. A triangle has sides a = 2 and b = 3 and angle C = 60°. Find the length of side c.

60. A triangle has sides a

~

2 aod b

~

3 aod angle C

~

40°. Find

the length ofside c.

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

61. The law of sines

sinA

8inB

sin C

abe .

Use the accoropanying figures and the identity sin(". - 0) sin 0 , if required, to derive the law.

(-;f + 2:).

=

A

=

A

46. Evaluate sin ; ; .

Using the Douhle-Angle Formulas Find the function values in Exercises 47-50. 47• COS2 '! 8

48

'2'" 49• sm 12

" 2 3'1T 50• sm 8

2

• cos

5".

B L..--"'a--'-----' C

B £..----:a,.--J

~

62. A triangle has sides a = 2 aod b = 3 and aogle C = 60° (as in Exercise 59). Find the sine of angle 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'lDllerically small values ofx. In Sectioo 3.9, we will see why the approximatioo holds. The approximatioo error is less than I in 5000 if Ixl < 0.1. a. With your grapher in radian mode, graph y = sinx and y = x together in a viewiog window ahout 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

identify A, B, C, and D for the sine functioos in Exercises 65-<)8 and sketch their graphs. 65. y = 2 sin (x 67.

+ 1f)

- I

y=-%Sin(ft) +,\0

66. y

= kSin(1fX

68. y

= 21fsmT'

L

- 1f)

. 21ft

L

+k >0

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

Set the constants A = 3, C = D = o. a. Plot f(x) for the values B = I, 3, 21f, 51f over the interval -47f :5 x :5 47f. Descrihe 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.

69. TheperiodB

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 -47f :5 x :5 47f. Descrihe what happens to the graph of the general sine function as Cincreases 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 horlzootal shift? Confirm 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 -47f :5 x :5 47f. Descrihe 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. b. What happens to the graph for negative values ofA?

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 ofthe graph. We use the term square window when the units or scales on both axes are the same. Ths term does not mean that the display window itselfis square (usual1y it is rectangular), but instead it means that the x-unit is the same as the y-unit. When a graph is displayed in the default window, the x-unit may differ from the y-unit of scaling in order to ftt the graph in the window. The viewing window is set by specifying an interval [a, b] for the x-values and an interval [c, d] for the y-values. The machine selects equally spaced x-values in [a, b] and then plots the points (x, j(x)). A point is plotted if and

1.4 Graphing with Calcutators 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 2 + 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 interval ofx-values and the range of y-values for the window. The resulting graph is shown in Figure 1.50a. It appears that the window is cutting off the bottom part of the graph and that the interval ofx-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 2 + 28 in differeot viewing windows. Selecting a window that gives a clear picture of a graph is often a trial-and-error process (Example I).

(b) Now we see more features of the graph (Figure 1.50b), 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 1.50b 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 2 , 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 1.51b 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 1.5la and I.5lb, but the scaling on the x-axis has been compressed in Figure 1.5lb to make the window square. Figure I.5lc gives an • enlarged view of Figure 1.51b 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 ofthe perpendicular lines y ~ x andy ~ -x + 3v2,andthesemicircley ~ ~ appear distorred (a) in a nonsquare 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). 4

10

1o

-10

) -6

-4

-10

r

6

(b)


FIGURE 1.52 Grapbs of the function y ~ 2 ~ x' A verlicalline may appear without a careful choice of the viewing window (Example 3). Sometimes the graph ofa trigonometric function oscillates very rapidly. When a calcula-

tor or computer software plots the points of the graph and connects them, many of the maximwn and mininnun 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 1. 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 oscillaliog function (Example 4).

1.4 Graphing with Calculators and Computers

33

experiment with a smaller viewing window, say [-6,6] by [-I, I], 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 [-I, I] 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 significantly 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 ofthe cosine curve. _

Solution

1.02

0.8 (b)

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

io sin second lena, io sio

y = cosx

+

50x graphed io (a). The rerm cos x clearly dominates lhe

5Ox, which produces !he rapid oscillations along lhe

cosine curve. Bolh views are needed for a clear idea oflhe 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 ofthe 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 lhe graph ofy = x l/3 = ~inFigure 1.17, we see lhatthe left branch for 2

2

-2

-2

(a)

(b)

-3

FIGURE 1.55

The graph of y ~ x 1/3 is missing !he left branch in (a). In

(b) we graph lhe function f(x) ~ 1:1"lx I1/3 , obtaining bolh branches. (See

Example 6.)

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(l/'Jlnx. 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·lxl'/'.

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

Exercises 1.4 Choosing a Viewing Window

o In

19. f(x)

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.

~

2 x +2 x2 + I x - I

23. f(x) = 6x

[-2, 2] by [-5, 5]

[-5,5] by [-25, 15] [-3, 3] by [-10, 10]

[-S,S]by[-IO, 10]

8. f(x) ~ 4x' - x' 10. f(x) = x 2(6 - x')

12. y 14. y

Ix

16. y

x 17. y = x

-

+3 +2

Chapter

II

~ -2X - 9

x2 - 3 24. f(x) = - - 2

x-

= 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)

x 2 +2x=4+4y- y 2.

18. y

= x ' /'(x 2 - 8) = x 2/'(5 - x) ~ Ix 2 - xl =I

- x

I

+3

= 1.

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

[-10, 10] by [-10, 10]

13. y = 5x 2/' - 2x 15. y ~

+I

8

31. Graph the lower half of the circle dermed by the equation

[-2,6]by[-I,4]

11. y = 2x - 3x 2/' 2

1

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

Exercises 5-30, rmd an appropriate viewing window for the given functioo and use it to display its grsph. x2 x3 5. f(x) = x' - 4x' + IS 6. f(x) = T - 2 - 2x + I

9. f(x) = x~

22. f(x)

-

25. y

[-4,S]by[-IS,25]

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

o In

+ 10

+6

- 15x 4x 2 - lOx

[-20,20] by [-100,100]

Finding a Viewing Window

7. f(x) ~ x' - 5x'

2

x2

~ -2-

X

21. f(x) ~ -x2o"_--x--'_~6

1. f(x) ~ x' - 7x 2 + 6x

a. [-I, I]by[-I, I] b. e. [-10, 10] by [-10, 10] d. 2. f(x) ~ x' - 4x 2 - 4x + 16 a. [-I, I] by [-5,5] b. e. [-S, S] by [-10,20] d. 3. f(x) = S + 12x - x' a. [-I, I]by[-I, I] b. e. [-4,4] by [-20,20] d. 4. f(x) ~ v'S + 4x - x 2 a. [-2,2] by [-2,2] b. e. [-3, 7] by [0, 10] d.

20. f(x)

35. Graph the function f(x) ~ sin 2x

I + 1.

+ cos 3x.

36. Graph the function f(x) ~ sin' x.

o

Graphing in Dot Mode Another way to avoid incorrect conoections 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 37. y ~ x - 3

39. y ~

xlxJ

38. y

~ sin~

40. y

~ -2--

x3

-

X

-

1 I

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 grsph of a real-valued functioo of a real varisble? What is the vertical line test?

3. What is a piecewiSl>-dermed functioo? Give examples.

4. What are the important types of functions frequently encouotered 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 j and g are real-valued functions, how are the domains of j + g, j - g, jg, and j / g related to the domains of j and g? Give examples.

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 = j(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 = j(x) to compress or stretch the graph by a factor c > I? Reflect the graph across a coordinate axis? Give examples.

Chapter

21

13. Graph the six basic trigonometric functions. What symmetries do the graphs have? 14. What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions? 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 j(x) = A sin «(2'tT/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 ofa sphere as a function ofthe 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 tracked 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.

10 Exercises 19. y =

= x2

-

2x - I

8. y

= e-"

10 Exercises ~16, determine whether the function is even, odd, orneitheJ: 9. y = x 2 + I 10. y = x' - x 3 - x 11. y

=I

12. y

= sec x tsnx

13 .y

~ L±-L x 3 -2x

14. y

~x

15. y

= x + cosx

16. y

= xcosx

- cosx

- sinx

23. y

~ 2e~

25. y

~

27. y

= 10 (x

Ig

b.

18. If j(a - x) function.

13

~

j(a

c. I(sinx)

d. g(sec x)

+ x), show that g(x)

~

j(x

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

2

20. y = -2

x

2

22. y =

- 3

2sin(3x

24. y

+ 'tT)

- 3)

- I

+I

~

+~ +I

32 -,

tsn(2x - 'tT)

26. y ~ x 2/s 28. y = -I

+ "0/2 - x

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

c. Height above Earth's sealevel as a function ofa1mospheric pressure (assumed nonzero) d. Kinetic energy as a function ofa particle's velocity

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

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

= (x +

I)' d. R(x)~~

b. j(x)

Piecewise-Defined Functions 10 Exercises 31 and 32, fmd the (a) domain and (b) range.

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

a.

Ixl -

21. y = Vl6 -

10 Exercises 5--1l, determine whether the graph of the function is symmetric about the y-axis, the origin, or neither. 5. y = x'i' 6. y = x 2/'

7. y

1~28,

e.

31

_

{h,

Igl

+ a) is an even

-4 S x S 0 0<xs4

.y-~,

-x - 2 32.y=

x,

{

,

-x + 2,

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

36

Chapter 1: Functions

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

e. S1retch vertically by a 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).

-c<J---~--~x

o

4

Composition of Functions In Exercises 35 and 36, f"md

a. (jog)(-l).

b. (gof)(2).

c. (j ° flex).

d. (g

I x'

35. f(x) =

° g)(x).

I

g(x) = • ~ Yx

b. Y = f(4x)

c. Y = f(-3x)

d. Y = f(2x + I)

e. y = f(t) - 4

f. Y = -3f(x)

In Exercises 37 and 38, (a) write formulas for fog and g ° f and find the (b) domain and (c) range of each.

37. f(x) = 2 - x 2 , g(x) = v;+2 38. f(x) ~ V';;, g(x) ~ ~

51.

y~ -~I +~

53. y

~ ~2 +

39. f(x)

40.

f (x)

~ ~

{

+ I, x _ I,

X

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

-2:5 x < 0 0:5x:52

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

Ii(x) 41. x

Ixl

42. x 2

Ixl

54. Y

~

I-t (-5x)1/3

55. Y

= cos2x

56• y

. x = sm 2

57. y

~

58. Y

~ COST

sin1fX

1fx

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

59. Sketch the graph y = 2 cos 60. Sketch the graph y = I

sin

In Exercises 61--{j4, 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

Ix'i

44.x 2 +x

Ix

2

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

+ xl

14 - x

46.

xI

I Ixl

47.

V';;

vfxj

48. sinx

sinlxl

2

1

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~

y~

63. a. Express a in terms ofBand b.

43. x'

a. Up

±

Trigonometly In Exercises 55-58, sketch the graph ofthe given function. Whst is the pcriod of the function?

hex)

45.4-x2

52.

I

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

-x - 2 -I, , { x - 2,

+

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 1ransformation.

+2

g(x) = ~

36. f(x) = 2 - x,

a. y = f(x - 5)

65. Heigbt 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 10 m closer to the base of the pole than is B. If wire BTmakes an angle of 35° 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 Ian 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. Whst appears to be the period of litis function? right 3

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

d. Reflect about the x-axis

c. ConIrrm your f"mding in part (b) algebraically.

D 68.

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

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

Chapter 1

Chapter

= g I? 0

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 he 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 he said ahout g(O)? Give reasons for your answer. 5. Graph the equation Ix I + Iy I = I 6. Graph the equationy

+ Iyl

+ x. = x + 14

sin x

1

b. Uniqueness Show that there is only one W!I:j to write 1 as the sura of an even and an odd function. (Hint: One W!I:j is given in part (a). If also I(x) = E,(x) + 01(X) where E 1 is even and 01 is odd, show that E - El ~ 01 - O. Then use Exercise 11 to show that E = E1 and 0 = 0 1 .) 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

1 - cosx 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. a changes while b and c remain fixed?

Derivations and Proofs 7. Prove the following identities. 1 - cosx sin x

37

Additional and Advanced Exercises

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

a.

Additional and Advanced Exercises

2

tan

+ b)3 +

x

a. a changes while b and c remain fixed?

2

b. b changes (a and c flXCd, 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.)

cas

c. c changes (a and b flXCd, a # OJ? Geometry

15. An object's center ofmass moves at a constant velocity v along a straight line past the origin. The accompanying figure shows the coordinate system and the line ofmotion. The dots show positions

a

that are I sec apart. Why are the areas A j, A 2 , ... , As 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

a a

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 + c)/2isthe

10 Kilometers

15

16. a. 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). y

secoiperimeter of the triangle. 11. Show that if1 is hoth even and odd, then I(x) ~ 0 for every x in the domain of I. 12. a. Even-odd decompositions

Let

1 he

a function whose do-

main is symmetric about the origin, that is, -x belongs to the domain whenever x does. Show that 1 is the sura of an even function and an odd function:

I(x) = E(x)

+ O(x),

-o;i~-----~A"(a-,COO):>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 " IsmOcosO

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?

I sinO

0

< I < I cosO"

y (0, I)

-c-ILL8"'----_ _'c-+:=D--=-_~x A

Chapter

E

(1,0)

Technology Application Projects

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

MathematicafMaple Module: ModeIiIIg Change: Springs, Drivillg Safety, RJuJioactivity, 1Tees, 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 obj ect 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, Ga1i1eo 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 fall. It assumes negligible air resistance to slow the obj ect 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 Ga1i1eo'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 of time 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.com/thomas.

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


(b) during the I-sec interval between second I and second 2?

Solution

The average speed of the rock during a given time interval is the change in disdivided by the length of the time interval, al. (Increments like and al are tance, reviewed in Appendix 3.) Measuring distance in feet and time in seconds, we have the following calculations:

ay,

ay

ay al


(b) From sec I to sec 2:

=

ay M=

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

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 ofthe falling rock in Example I at! = I and 1= 2 sec.

Solution We can calculate the average speed ofthe rock over a time interval [10, 10 having length al = h, as

ay al

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).

TABLE 2.1

Average speeds over short time intervals [to, to + h] Average speed:

Length of time interval

h I

0.1 0.01 0.001 0.0001

ay al

=

16(10 + h)2 - 16102 h

Average speed over interval oflength h starting at to = 1

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

48 33.6 32.16 32.016 32.0016

80 65.6 64.16 64.016 64.0016

The average speed on intervals starting at 10 = I seems to approach a limiting value of 32 as the length ofthe 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.

Rates of Change and Tangents to Curves

2.1

Ifwe set to

=

41

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

I!.y I!.t

16(1

+ h)' -

16(1)'

16(1

+

2h

+ h2)

-

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 Q(x,,!(x,)) I I I

Given an arbitrary functioo 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, I!.y = j(X2) - j(XI), by the length box = X2 - XI = h of the interval over which the change occurs. (We use the symbol h for I!.x to simplifY the notatioo here and later on.)

I

Secant

I

I

',!!.Y

P(XI,!(XI)

I

--"~~-~-~--------j !!oX = h :

DEFINmDN boy

I!.x FIGURE 2.1 A secant to the graph = j(x). Its slope is I!.y/ At, the average rate of change of j over the interval [Xl, X,].

y

The average rate of change of y

=

j(x) with respect to Xover the

interval [Xl,X2] is j(X2) - j(xI) X2 Xl

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

h .. O.

Geometrically, the rate of change of j over [Xl, X2] is the slope of the line through the points P(x], j(XI» and Q(X2,f(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 XI 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 ifit 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 ofchange as the graph of a linear function. But what is meant by the slope ofa curve at a point P 00 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 ofthe tangent as the slope of the curve at P. So we need a precise meaning for the tangent at a point 00 a curve. For circles, tangency is straightforward A line L is tangent to a circle at a point P if L passes throughP perpendicu1ar to the radius at P (Figure 2.2). Such a line just touches the circle. But what does it mean to say that a lineL is tangent 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 ofthe 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 ofthe secant PQ and investigate what happens to the slope as Q approacbes P along the curve: ~y (2 Secant slope = =

+ h)2 - 22

~

h

h2 + 4h

+4- 4

= '"--'-='0-'--"--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

=

+ h)2 h

4

=

h + 4.

4

I I I

,Ay ~ (2

Ll

P(2,4) - -

+ h)' -

4

I Ax = h I +""'7'~l' --~--c----~x

------=.....

2+h

NOTTOSCALB

FIGURE 2.4 Finding 1he slope of1he parabola y

limit of secant slopes (Example 3).

= x 2 at 1he point P(2, 4) as 1he

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 = I and t = 2 are called instantaneous rates ofchange. 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 /ip 340 - 150 190 . Averagerateofchange:~ = 45 23 = 22 '" 8.6 flIes/day. p /

350 ~

V

300

1/ -!:J.-p-=r 190 / / /;.p V /;., - 8. ilie day /

" 250

'S

.ll

/ ' Q(45 340)

200

PI ,23,150)

.~ ISO Z

100 /

50

'/

~

&t =

22

./

o

10

20

30

40

50

Time (days)

FIGURE 2.5 Growth of a fruit fly population in a controlled experiment. The average rate of chaoge over 22 days is the slope /ip/ /it ofthe 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 changing 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 ofExample 4 growing

on day 23?

Solution To answer this question, we examine the average rates of change over increasingly short time interva1s 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

= ap/at

Q

Slope of PQ (flies / day)

(45,340)

340 - 150 '" 8 6 45 - 23 .

(40,330)

3~~ ::: ~~O

(35,310)

310 - 150 '" 13 3 35 - 23 .

~

h

/

Q(45,340)

~ '/ P; V

250

'a

i: z

350) // ;-;:-- / ff

300

P(23,

100 /

50

265 - 150 '" 164 30 - 23 .

(30,265) FIGURE 2.6

'" 10.6

B(35,

350

o

/

~~ /llJ

~

150) J rY

~

\

~

20

30

40

50

A(14,0) Time (days)

The positioos and slopes offour secants through the pointP 00 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 t-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

~;O_-I~

= 16.7 flies/day (approximately).

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

•

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 oflength 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 t approached the values t = I and t = 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 limit.

Exercises 2.1 Average Rate. of Change In Exercises I-j), fmd the average rate of chaoge ofthe functioo over the given interval or intervals. 1. f(x) ~ x 3

+

b.[-I,I]

2. g(x)

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

v'49+l;

6. P(O) = 03 - 40'

[0,2]

+ 50;

[1,2]

Slope of a Curve at a Point

I

a.[2,3] = x2

S. R(O) ~

b. [-2,0]

In Exercises 7-14, use the method in Exsmple 3 to find
7. y = x 2

-

3, p(2, I)

8. y = 5 - x 2 , b. ["./6, "./2]

9. y =

x2 -

10. y ~ x 2 b. [-".,,,,]

11. y ~ x 3,

-

P(1,4)

2x - 3, P(2, -3) 4x, p(1, -3) p(2, 8)

2.1

12. y = 2 13. y ~

x' -

x',

+ 4,

c. Usc your graph to estimate the rate at which the profita were changing in 2002.

P(2, 0)

D 18.

Instantaneous Rates of Change

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

s p

,

600

J;j-

D 19.

/ /

05101520 Elapsed time (sec)

ammging them in order in a table like the one in Figure 2.6. What are the appropriate units for these slopes?

= 20 sec.

16. The ""companying figure shows the plot of distance fallen versus time for an object that fell from the lunar landing module a distance 80 m to the surf... of the moon.

a. Estimate the slopes of the secanta PQ" PQ2,PQ" andPQ., ammging them in a table like the one in Figure 2.6. b. Aboot bow fast was the object going when it hit the surface? y

80 I

yI; forx

"" O.

d. Calco1ate the limit as h approaches zero of the average rate of change ofg(x) with reapect to x over the interval [I, I + h].

a. Estimate the slopes of secanta PQ" PQ2, PQ" and PQ.,

b. Then estimate the car's speed at time I

Letg(x) ~

a. Find the average rate of change ofg(x) with respect to x over the intervals [1,2], [I, 1.5] and [I, 1+ h]. b. MaIre a table of values of the average rate of change ofg with respect to x over the interval [I, I + h] for sorue values ofh approa<:hing zero, say h ~ 0.1,0.01,0.001,0.0001,0.00001, and 0.000001. c. What does your table indicate is the rate of change ofg(x) with respect to x at x = I?

Q, V 100

Maire a table of values for the function F(x) ~ (x + 2)/(x - 2) atthepointax = 1.2, x = l1/lO,x = 101/100, x = 1001/1000, x ~ 10001/10000, and x ~ I.

15. Speed of a car The ""oompanying figure shows the timetl>-distance graph for a sporta car accelerating from a standstill.

650

45

b. What is the average rate of increase of the profita between 2002 and 2004?

P(I, I)

12x, P(I, -11)

14. y = x' - 3x2

Rates of Change and Tangents to Curves

p

Q.

q'J f--

D 20.

Letf(l) ~ 1/1 for I '" O.

a. Find the average rate of change of f with respect to I over the intervals (i) from I = 2 to I = 3, and (ii) from I = 2 to I = T. b. MaIre a table of values of the average rate of change of f with respect to lover the interval [2, 11, for some values of T approaching 2, say T = 2.1,2.01,2.001,2.0001,2.00001, and 2.000001.

c. What does your table indicate is the rate of change off with respect to I at I = 2? d. Calco1ate the limit as T appro""hes 2 of the average rate of change of f with respect to lover the interval from 2 to T. Vou will have to do some algebra before you can substitote T = 2. 21. The ""companying graph shows the tota1 distance s traveled by a bicyclist after I hours.

Q /

o

....- ......

Q !/ V

'"

s

/'

10

5 Elapsed time (sec)

D 17.

./

.,......

The profita of a small company for e""h of the first five years of ita operation are given in the following table:

./ /

/ Vear

Profit in $1000.

1

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, and 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], and [2.5, 3.5]. b. Estimate the bicyclist's instantaneous speed at the times t =

I

~

2, and I

~

!,

3.

c. Estimate the bicyclist's maximum speed and 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, 3J, [0, 5J, and [7, IOJ. b. Estimate the instantaneous rate of gaaoline consomption at the times I ~ 1,1 ~ 4, and I ~ 8.

A

:; 16

r:

:(

,l!

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

.......

"-

I' I"-

4

o

1 2

3

4

.....

-

5 6 7

8

9 ill

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 informal 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 fmd ourselves interested in the function's behavior near a particular point xo, but not al 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 actoa1Iy 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 /iyjh for h closer and closer to zero. Here's a specific example where we explore numerically how a function behaves near a particu1ar point at which we carmot directly evaluate the function.

Lintits

V2,

y

2

---,~-C+---:----~x

o

EXAMPLE 1

How does the function

I(x)

y

=

2

x - 1 x-I

behave near x = I? 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 dermed (Example 1).

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 canceliog common factors: () Ix=

(x - I)(x + 1)

x-I

=x+I

for

x oF I.

The graph of I is the lioe 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 defmed, 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

2.2 The closer x gets to 1, the closer f(x) seems to get to 2

TABLE

=

47

(x' - 1)/(x - 1)

x' -1

Values of x below and above 1

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

0.9 l.l 0.99 1.01 0.999 1.001 0.999999 1.000001

1.9 2.1

x-

1.99 2.01 1.999 2.001 1.999999 2.000001

Let's generalize the idea illustrated in Example I. Suppose j(x) is defined on an open interval about xo, except possibly at Xo itself. If j(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, and write lim j(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 j(x) approaches the limit 2 as x approaches 1, and write lim f(x) = 2,

x-I

or

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

x-I

Essentially, the detmition 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 and suffICiently close are imprecise; their meaoing depeods on the context. (fo a machinist maoufacturing a piston, close may mean within a few thousandths ofan inch. To an astronoroer studying distant galaxies, close may mean within afew thousand light-years.) Nevertheless, the detmition is clear enough to eoable us to recogoize and 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 j has limit 2 as x --+ 1 even though f is not defined at x = 1. y

y

----.~-c+--:_x

----.r.--c+--:_x

(a) f(x)

~ ~~

t'

X2 -

y

1

- - , x,*l x-I

(b) g(x) ~ {

I,

---cr.---O+--~x

x

=

(c) h(x)

~

x

+1

1

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

48

Chapter 2: Limits and Continuity The functiong has limit 2 as x --+ 1 even though 2 # g(I). The function h is the onlyone of the three functions in Figure 2.8 whose limit as x --+ 1 equals its value at x = I. 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
lim f(x) = lim x = xo.

x-xo

y

X-Xo

(h) 1£ f is the constant function f(x) = k
°

------4-----'"--~x

xo

X--+Xo

For instances of each of these rules we have

limx=3

x~3

(b) Constant function

and

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

x--7

x-2

•

We prove these rules in Example 3 in Section 2.3.

FIGURE 2.g The functions in Example 3 have limits at all points xo.

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

1 X'

X~O

x*o

1,

1+-----'I II

I, I,

-------;;9------~x

----;:'Httt-t-~f----~x

°

{o. xso

0:,

y

~

I

I I,

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

(b) g(x)

(c)j(x)

°

FIGURE 2.10 None ofthese functions has a limit as x approaches (Example 4).

EXAMPLE 4

Discuss the behavior ofthe following functions as x --+ 0.

{O,I,

x
I

x#o

(h) g(x) =

x' { 0,


{o,

x:sO


. 1

x;;"

°

x=o

SIDX-, x>o

. 1

sm

x'

x>O

2.2 Limit of a Function and Lim;t 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, U(x) = O. For positive values of x arbitrarily close to zero, U(x) = I. 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 2.1Ob). (c) It oscillates too much to have a limit: f(x) has no limit as x --> 0 because the function's values oscillate between + I aod -I in every open interval containing O. The values • do not stay close to aoy one number as x --> 0 (Figure 2.IOc).

The Limit Laws When discussing limits, sometimes we use the notation 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 partico1ar discussion or example). Other times, such as in the statements ofthe 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 on the x-axis that may or may not belong to the domain of the function involved. To calculate limits of functions that are arithmetic combinations of functions having known limits, we cao use several easy rules.

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

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

x~c

1. Sum Rule:

lim g(x) = M,

then

x~c

lim(j(x)

+ g(x))

= L

+M

x~c

2. Difference Rule:

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

7. Root Rule:

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 "!' 0

lim[f(x))" =

L", n a positive integer

lim ~ =

"ifi =

x-c

x~c

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

>

L 1/", n a positive integer

0.)

In words, the Sum Ru1e 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 constant times a function is the constant times the limit of the function; the limit of a product is the product of the limits; the limit ofa quotient is the quotient ofthe limits (provided that the limit ofthe denominator is not 0); the limit ofa positive integer power (or root) ofa function is the integer power (or root) ofthe limit (provided that the root ofthe limit is a real number). It is reasonable that the properties in Theorem I 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'mition 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 Ru1e in Section 2.3, based on a precise def'mition oflimit. Ru1es 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....... c k = k and limx~c x = c (Example 3) and the properties oflimits to f"md the following limits. (a) lim(x' x~c

2 lim x '+X -1

+ 4x 2

-

3)

+ 4x 2

-

3) = lim x'

(b)

x~c

x2 + 5

(c) lim V4x 2 x--2

-

3

Solution (a) lim(x' x-c

+

lim 4x 2

=

c' + 40

lim(x' x~c

2

+x

-

x-c

X--i'C

3

-

2

lim 3

x-c

Sum and Difference Rules

Power and Multiple Rules

I)

-

Quotient Rule

lim(x 2 + 5)

x-e

lim x'

x-c

+

c' + c2 c2 (c)

lim V4x 2

x--2

-

-

lim x 2

lim I

-

x-c

x-c

I

Power or Product Rule

+5

3 =

Vx--2 lim (4x 2 -

=

Vx--2 lim 4x 2 -

=

V4(-2)2 - 3

=

Vi6=3

=

Vi3

Sum and Difference Rules

3)

Root Rule with Il

lim 3

Difference Rule

x--2

=2

Product and Multiple Rules

•

Two consequences ofTheorem I 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) = ancn

x-c

+ an_lc n- 1 + ... + ao.

THEOREM 3-L1mlts of Rattonal Functions IfP(x) and Q(x) are polynomials and Q(c) i' 0, then . P(x) x-c Q(x)

P(c) Q(c)"

lim - - = - -

2.2

EXAMPLE 6

.

x3

x~-l

Identifying Common Factors It can be shown that if Q(x) is a polynomial and Q(c) = 0, then (x - c) is a factor of Q(x). Thus, if

the numerator and denominator of a rational function of x are both zero atx = c,theybave(x - c)asa common factor.

+ 4x 2 - 3 x2 + 5

=

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

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 by substitution in the simplified fraction.

Evaluate lim x 2

+

X -

x2 -

x---+l

2

X

•

Solutton We cannot substitute x = I because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = 1. 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 i' I: (x - I)(x

(a)

+ 2)

x

x(x - I)

y y=x+2 x

-

Eliminating Zero Denominators Algebraically

EXAMPLE 7

---"_:l-2c----..ot-+--------> X

51

The following calculation illustrates Theorems 2 and 3: lim

I

Limit of a Function and Lim;t Laws

+2 x

if x i' 1.

'

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

3

lim x x--+l

-----=":l-c_---..ot-+-------->x

2

+x

x2

-

- 2 = lim x

+2

X

X

x--+l

= I

+2

= 3.

1

-

See Figure 2.11.

(b)

FIGURE 2.11 The graph of f(x) ~ (x' + x - 2)/(x' - x) in part (a) is the same as 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 as x ---. I (Example 7).

Using Calculators and Computers to Estimate Limits When we cannot use the Quotient Rwe in Theorero I because the limit ofthe denominator is zero, we can try using a calcwator or computer to guess the limit numerically as x gets closer and closer to c. We used this approach in Example I, but calcu1ators 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.

EXAMPLE 8

Estimate the value of lim

x-o

Yx' + 1,00 x

10.

Solutton 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

TABLE 2.3 Computer values of f(x) ~

x

v'x' + 100 ,

10

x

near x

=

0

!(x)

±I ±0.5 ±O.I ±0.01

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

±0.0005 ±O.OOOI ±O.OOOOI ±O.OOOOOI

0.080000) h? 0.000000 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 deoominator have no obvious common factors (as they did in Example 7). Sometimes, however, we can create a common factor algebraically.

EXAMPLE 9

Evaluate

" Yx2 + 100 - 10 Iun 2 •

x-o

X

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 x2

Yx 2 + 100 - 10 Yx 2 + 100 + 10 x2 • v'x 2 + 100 + 10 x 2 + 100 - 100 X

2

(Yx 2 + 100 + x

X

2

(Yx 2 +

10)

2

100 + 10)

I Yx 2 + 100 + 10'

Common

fuctor"

Cancel x2 for x

'¢

0

Therefore,

· Yx2 IlID x~o

+ 100 - 10 x2

=

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

Y02 + 100 + 10 =

I

20

=

Denominator not 0 at x = 0; substitute

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 deoominator 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 ofTheorem 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 g ~t------O--------->x

o

c

The following theorem enables us to calculate a variety oflimits. 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 (Figure 2.12). You will fmd a proof in Appendix 4.

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

FIGURE 2.12

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 Pinching Theorem. y~

1-

4x'

EXAMPLE 10

Given that 2

--'-------:--~--+----x

-I

I -

0

FIGURE 2.13 Aoy function u(x) whose graph lies io the region between y = I + (x'/2) andy = I - (x'/4) has

4x

:5 u(x) :5 I

Solution

* 0,

Since lim(1 - (x 2/4)) = I

and

x~O

lim(I

x~O

+ (x 2j2»

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

EXAM PLE 11

= I,

•

The Sandwich Theorem helps us establish several important limit rules:

(a) lim sin 0 = 0 ~------')IL----~8 -1r

for all x

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

limit I as x ---> 0 (Example 10).

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

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

lim sinO = O.

8~O

--_-!2c---c-I:-""~O¥-~Ic--2;\--8

(h) From Section 1.3,0:5 I - cosO :5101 for all 0 (see Figure 2.14b), and we have lims~o(1 - cosO) = 0 or

lim cos 0 = 1.

(b)

FIGURE 2.14 The Saodwich Theorem conflflDS 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~c 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

TIf(x) :5 g(x) for allxin some open interval containingc, except and the limits of f and g both exist as x approaches c,

= c itself,

then

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 2.14a 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 lhe function g(x) graphed here, fmd lhe following limits or explaia why lhey do not exist. 8.

lim g(x)

x-I

b. lim g(x) %-2

c. lim g(x) %-3

y

d. lim g(x) x-2.S

y

4. Which of lhe following statements about 1he function y graphed here are 1rue, aod which are false? 8.

lim f(x) docs not exist.

x~2

2. For lhc function f(/) graphed here, fmd 1he following limits or explaia why lhey do not exist. 8. lim f(/) b. lim f(/) c. lim f(/) d. lim f(/) 1--2

t--I

t-O

t---{}.S

s

= f(x)

b. lim f(x) x~2

=2

c. lim f(x) docs not exist. x~l

d. lim f(x) exists at every pointxo in (-I, I). x~

...

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

...

y s

~ j(r)

o ....-&-1

3. Which of lhe following slarements about lhe function y = f(x) graphed here are 1rue, aod which are false? 8.

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 (-I, I). x~

...

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

Extstence of Limits In Exercises 5 and 6, explaia why lhe limits do not exist.

5lim~ . x~o Ixl

61" I . x~ x - I

7. Suppose lhat a function f(x) is defined for all real values ofx except x = xo. Can anything he said about lhe existence of lim~... f(x)? Give reasons for your answer. 8. Suppose iliat a function f(x) is defmed for all x in [-I, I]. Cao any1hing he said about the existence of limx_o f(x)? Give reasons for your answer.

2.2 9. Iflimr-l f(x) ~ 5, must f be defmed at x ~ I 7 If it is, must f( I) = 57 Can we conclude anything about the values of f at x ~ I 7 Explain, 10. If f(l) ~ 5, must lim,~1 f(x) exist? If it does, then must limr-l f(x) = 57 Can we cooclude anything about limr-l f(x)7 Explain.

Limit of a Function and Limit Laws

Using Limit Rulos 51. Suppose limr-o f(x) = I and lim,~o g(x) = -5, Name the rules in Theorem I that are used to accomplish steps (a), (b), and (c) of the following calcu1atioo. lim 2f(x) - g(x) ~ ,~o (j(x)

+ 7)2/3

,~o

%--7 1-+6

,~o

14. lim

%-+-2

lx 2 + 4x + 8)

(x 3 -

x-o

x-o

z~o

y--+-3

21. lim

3

v'3h+1 + I

h~O

22. lim

v5h+4 -

4-0

2

limY5h(x)

27.lim

t2

I-I

+ t - 2 t2 - 1

x+3

+

32. lim

x-I

x-

33. lim

"4 u3 -

I

34. lim if - 8

1

16

%-9

37.lim

'\."x -

x - 9

x-o

3

36. lim

38. lim

x-I

39. lim

W+12 -

x-2

x -

41. lim 2 %-+-3

2

4

+3

c. lim (j(x)

,~,

3

%-4

x-c

+ 3g(x)) ~

d

48. lim (x 2

+ 3)

c. lim (g(x»2

-

1)(2 - cosx)

yI;+4 cos (x + 'IT) SO. lim Y 7 + seC> X x-o

~

- 3. Find

b. lim xf(x) %-4

r

d

%-4

•

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) a. lim (j(x) + g(x)) x-b

x-b

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

x-b

x-o

r f(x) • ,~ f(x) - g(x)

0 and lim.~4 g (x)

c. lim 4g(x)

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

5 2

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

54. Suppose lim.~d(x) a. lim (g(x)

x-I

5andlimr-,g(x) = -2. Find

x-c

,~o

x-o

Supposelim,~d(x) =

a. lim f(x)g(x)

Find the limits in Exercises

46. lim tan x

}&-+-'II"

53.

4-x '~45-~

45. lim sec x

x-o

V(5)(5)

42.lim

44. lim sin2 x

(c)

x-I

(1)(4 - 2)

2

x + 1

43. lim (2 sin x - I) ,~o

x-I

40. lim x+2 '~-2W+5 -3

Limits with trigonometric functions 43-50,

49. lim

,~1

1

W+8 -

x-I

y51im h(x)

X+1

x

(b)

( lim p(x»)( lim 4 - lim r(x»)

'\."x

%-+-1

-~ X

4 -

x ,~4 2 -

y';+J - 2

x-I

X

v-2 v 4 -

,~1

,~1

( lim p(x»)( lim (4 - r(x»))

16y 2

-

_1_ x - I

31. lim -'--I

35. lim

vlim 5h(x)

' x2 - 7x + 10 26. lun 2 %-2 x 2 28. lim t + 3t + 2 1-+-1 t 2 - t - 2

y-o 3y 4

! - I

(a)

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

,~1

5y 3 + 8y 2 30. lim --'-;c-----co

29. lim -lx - 4 %-+-2 x 3 + 2x 2

u-l

,~1

lim'VShW ,~1 p(x)(4 - r(x))

+ 4x + 3

%-+-3 Xl

x2+3x-1O lim 25. %--5 X + 5

7 4

h

24. lim

25

%-5 Xl -

(c)

lim 7»/3

x-o

52. Let limr-l h(x) = 5, lim'~IP(X) = I, and lim,~1 r(x) = 2. Name the rules in Theorem I that are used to accomplish steps (a), (b), and (c) of the following calcu1atioo.

Limits of quotients Find the limits in Exercises 23-42. 23.lim x - 5

+

(2)(1) - (-5) (I + 7j2/3

18.lim

%--1

x-a

(lim f(x)

y+2 y-2 y2 + 5y + 6 20. lim (Zz - 8)1/3

(b)

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

s-2/3

19. lim (5 - y)4/3

x-o

+ 7))2/3

( lim (J(x)

16. lim 3s(28 - I) 17. lim 3(lx - 1)2

(a)

+ 7j2/3

x-o %-2

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

g(x))

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

12. lim (-x 2 + 5x - 2)

+ 5)

l!'!'o (2f(x) lim (j(x)

Calculating Limits Find the limits in Exercises 11-22. 11. lim (lx

55

x-b

56. Suppose that limr--2 p(x) = 4,limr--2 r(x) = 0, and lim'~-2s(x) = -3. Find a. lim (P(x) + r(x) + s(x» %-+-2

b.

lim p(x)· r(x)' s(x)

%-+-2

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

+ 5r(x))/s(x)

56

Chapter 2: Limits and Continuity

Umits of Average Rates of Change 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, evaluste 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) =

v;,.

58. f(x) = x 2• X = -2 60. f(x) = I/x. x = -2

Using the Sandwich Theorem 63. If v'5 - 2x lim,-+o f(x).

2

'"

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

2

.

<

r xsinx? x~ 2 - 2cosx· Give reasons for your answer. Db. Graphy = I - (x 2/6), y = (x sinx)/(2 - 2 cos x). and y = I together for - 2 '" x '" 2. Comment on the behavior of the graphs as x .... O. 66. a. Suppose that the inequalities

2

x 2 < 1 - cosx < 1 24 x2 2

•

Give reasons for your answer. Db. Graph the equations y ~ (1/2) - (x 2/24), y = (I - cosx)/x 2 .andy = 1/2togetherfur-2 '" Comment on the behavior of the graphs as x .... O.

x'" 2.

Estfmatfng LImits

D You will fmd a graphing calculator useful fur Exercises 67-74. 67. Letf(x) = (x 2 - 9)/(x + 3). a. Make a table of the values off at the points x

2x - 3)/(x 2

-

4x

+ 3).

at if you evaluate h atx

= 3.1.3.01.3.001....

instead?

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

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

a. Make tables ofvalues ofF at values ofx that approach Xo = -2 from above and below. Then estimate lim,-+_' F(x).

lim 1 - cosx?

x2

-

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

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

a. Make a table of the values of h at x = 2.9.2.99.2.999. and so on. Then estimate lim,-+3 h(x). What estimate do you arrive

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

1_

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

'"

1- £

= -6.1. -6.01.

x-+ -6.

g(x) '" 2cosxforallx.fmdlim.~og(x). 65. a. It can be shown that the inequalities 64. If2 - x

2

V2.

c. Find lim,-+Vz g(x) algebraically. 69. Let G(x) = (x + 6)/(x 2 + 4x - 12). a. Make a table of the values of Gat x = -5.9. -5.99. -5.999, and so on. Then estimate lim.~.... G(x). What estimate do you arrive at ifyou evaluate G atx -6.001•... instead?

62. f(x) =~. x = 0

x= 7

b. Support your conclusion in part
b. Support your conclusion in part
a. Make a table of the values ofg at values of 0 that approach 00 = 0 from above and below. Then estimate limo~o g(O). b. Support your conclusion in part
= -3.1.

-3.01. -3.001. and so on as far as your calculator can go. Then estimate lim.~-3 f(x). What eslimate do you arrive at if youevaluatefatx ~ -2.9. -2.99, -2.999.... instead?

b. Support your conclusions in part
00

~

74. Let G(I)

O.

= (I

- cos 1)/1'.

a. Make tables ofvalues ofG at values oflthat approach 10 = 0 from above and below. Then estimate lim,....o G(I). b. Support your conclusion in part I. 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 Oefinition of a Limit 76. Supposethatg(x):5 f(x):5 h(x) for all x "!' 2 and suppose that

D 81.

lim g(x) = lim h(x) = -5.

%-2

%-2

Can we cooclude anything about lbe values of f, g, and h at x = 2? Could f(2) = O? Could limx_,j(x) = O? Give reasoos for your answers. .

77. Iflun

%-4

f(x) - 5 . 2 = I,f"md lim f(x). x x-4

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

~

I ,f"md

a. lim f(x)

b. lim

%-+-2

79. a. Iflim

%-+-2

f(x) - 5

%-2

.

b. If lim

X -

X -

2

= 3,f"md lim

%-2

f(x) - 5

;(-2

.

2

f~)

D

a. Graph g(x) ~ x sin (l/x) to estimate limx_Og(x), zooming in on the origin as necessary. b. ConIrrm your estimate in part (a) wilb a proof. 82. a. Graph h(x) = x 2 cos (l/x 3 ) to estimate limx_ o h(x), zooming in on the origin as necessary. b. ConIrrm your estimate in part (a) wilb a proof. COMPUTER EXPLORATIONS Graphical Estimates of Limits In Exercises 83-88, use a CAS to perform lbe following steps:

a. Plot lbe functioo near lbe point xo being approached. b. From your plot guess lbe value of lbe limit.

f(x).

.. = 4, fmd lun f(x). %-2

f(x)

83. lim x' - 16 %-2 x - 2

2.3

(x

%-+-1

2

-

+

x -9

87. lim I - :osx

88 lim

2x

b.

x~o

x

-

86.lim

x-o

a. lim f(x)

3

85.lim~-1

80. Iflun= I,f"md 2

x-o x

84. lim x

x-O

5x - 3 I?

2

%-3 ~ - 4

x

X SInX'

• x-o

3

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 "gets arbitrnrily close to" in the infonna1 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 enough" 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 = 2x - I near xo = 4. Intoitively it appears that y is close to 7 when x is close to 4, so limx~' (2x - I) = 7. However, how close to xo = 4 does x have to be so that y = 2x - 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 f----J'----1--=--c-~. Lower bound: y~5

71 = 1(2x -

I) -

71 <

2? To fmd the answer we

71 = 12x - 81·

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

12x - 81 <

2? To

12x - 81 <

FIGURE 2.15 Keeping x within I unit of

xo ~ 4 will keep Y within 2 units of Yo = 7 (Example I).

2 -2 < 2x - 8 < 2 6 < 2x < 10 3<x<5 -I < x - 4 < 1.

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

10

flO)

L L -

f(x) lies inhere

1

Definition of Limit

10 for all x

'* Xo

inhere

~

~~ x 0

-'0- 6

-'0

x

-'0+ 6

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+ IIO)?

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 E-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 E 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

DEFINmON 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------{c-iO-o>--+--> x

o

FIGURE 2.17 The relation of Band E in the dermition oflimit

X-'xo

if, for every number E 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 E, 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 The Precise Definition of a Limit y

L

y

y

I + j]j f-------f--

L

y

I + j]j f-----+L

L

o

----'+---'-/+--c'--+\-~X

o

Xo

Makelf(x) -LI <

E

Xo

Xo - al/IO Response:

The challenge:

~~

Ix -

--C:+-----::----> X

o

Xo + aI/tO

Xo

--C:+--7---::--(---> X o / Xo \ Xo - 81/100 Xo + 8 11100 Response:

New challenge:

Makelf(x) -LI <

Xo I < 81110 (a number)

y

L

I

+ 100 f - - - - - - k - -

L I L -100

L

----'::t------!-----> X

59

E

Ix - xol < 8 11100

~ I~

y

I

+ 1000

L

~f-----f---

I

+ 1000 ~f------,l---

L

L

o

---'c-f---''-:L-c'---> X

o

Xo New challenge: I f! = 1000

Xo

Response:

Ix-xol

y

<81/1000

y

y

L

+ E~f=e=-=__=-=-=-~-rF===

L-E

----'cf---~"----->X

o

----'c-t----';C"---> x

o

Xo

New challenge: I E = 100,000

--'-j---~l----~x

o

Xo

Response:

Ix - Xo I < EXAMPLE 2

Xo

New challenge: E= ...

81/100,000

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

x->I

Solution Set Xo = I ,f(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 tmd Il by working backward from the .-inequality:

y

1{5x - 3) - 21 = 15x - 51 < •

six -

II

<•

Ix - II < ./5. Thus, we can take Il = ./5 (Figure 2.18). If 0 < Ix - II < Il = ./5, then 1{5x - 3) - 21 = 15x - 51 = Six - II < 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 - II < Il imply 15x - 51 < E. 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
If fix) ~ 5x - 3, thea o < Ix - II < ./5 guaranrees that If(x) - 21 < • (Example 2). FIGURE 2.18

Prove the following results presented graphically in Section 2.2.

(b) lim k = k

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

o< + f!

implies

Ix - xol < •.

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

Xo + a f - - - - - - 7 t " - - - - Xo Xo - ~ f------c1l'-r-+---Xo -

Ix - xol < Il

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

y Xo


x-xo

f!

o<

Ik - kl < •.

implies

Ix - xol < Il

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

Finding Deltas Algebraically for Given Epsilons For the functioo fix) = x, wetmdthatO < Ix -xol < 8 will guaranree If(x) - Xo I < E wheaever 8 ,;; • (Example 3a). FIGURE 2.1g

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, tmd all> 0 that works for • = I. That is, find all> 0 such that for all x

k

y~k

+ E f-----,------'----

k-

k+---t-',..-,----I E

f----+----{----1f - - - -

! : I I I !

---+------'-,---L---'-----c---~x

o

FIGURE 2.20

1~-21 < 1.

0< Ix - 51 < Il

y

xo-8 Xo %0+8

For the functioof(x) we f"md that If(x) - k I < £for any positive 8 (Example 3b).

~

k,

Solution

We organize the search into two steps, as discussed below.

1. Solve the inequality I~ - 21 < I to find an interval containing Xo = 5 on which the inequality holds for all x "" xo.

1~-21<1 -I

<

~

-

2

<

1<~<3 l<x-I<9 2

<x<

10

I

61

2.3 The Precise Definition of a Limit 3

The inequality holds for all x in the open interval (2, 10), so it holds for all x this interval as well.

3

---c;",.....'....-.;_...... ' ..I~'----'---7.~>X 2

5

8

10

2. FIGURE 2.21

An open interval of

radius 3 about Xo = 5 will lie inside the open interval (2, 10).

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 (2, 10) is 3 (Figure 2.21). If we 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

I~ y

3

21 < I (Figure 2.22):

1~-21<1.

0< Ix - 51 < 3

f-------,------=~

* 5 in

>

0

If(x) - LI <

E

How to FlndAlgebralcaUy a 6 for a Given I, L, Xo, and.

•

The process of f"mding all> 0 such that for all x 0< Ix - xol < Il can be accomplished in two steps.

1. Solve the inequality If(x) - L I <

E to find an open interval (a, b) containing Xo on which the inequality holds for all x 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 If(x) - L I < • will hold for all x Xo in this Il-interval.

NOT TO SCALE

FIGURE 2.22

*

*

The function and intervals

in Example 4.

EXAMPLE 5

Prove that limx~2 f(x) = 4 if

{xI,

2

f(x) Solution

=

x*2

,

x

Our task is to show that given E > 0 there exists all> 0 such that for all x

If(x) - 41 <

0< Ix - 21 < Il y

4

1. Solve the inequality If(x) - 41 < which the inequality holds for all x

+. r - - - + - - - -

For x

= 2.

* Xo

= 2, we have

f(x)

=

to find an open interval containing Xo = 2 on

E

* xo.

x 2 , and the inequality to solve is Ix 2 Ix2

4

-

41

<

FIGURE 2.23 An interval containing x = 2 so that the function in Example 5 satisfies If(x) - 41 < E.

2.

E

41 < .:

x2 -

~<x<~. The inequality If(x) - 41 < ~) (Figure 2.23).

-

E

< 4 < E 2 4-.<x <4+. ~
(2,1)

E.

holds for all x

Assumes E < 4; sec below. An open interval about Xo that solves the inequality

~

2

* 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,

If(x) - 41 < e.

0< Ix - 21 < Il

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 iotervaI (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 precediog 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

Let e

+ g(x))

=

M,provethat

L + M.

> 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 + M)I Sioce limx~c f(x) =

Triangle Inequality: + (g(x) - M)I la + hi '" lal + Ihl :5 If(x) - L I + Ig(x) - MI· L, there exists a number Il, > 0 such that for all x

=

I(j(x) - L)

0< Ix - cl < Il,

If(x) - LI < e/2.

Similarly, sioce lim~c g(x) = M, there exists a number 112

o<

Ix -

cl <

> 0 such that for all x

Ig(x) - MI < e/2.

112

Letll = mio {1l,,1l2},thesmaIlerofll , 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 + M)I < ~ + ~ Thisshowsthatlim~c(j(x)+

g(x»

=

=

L + M.

cl <

Ill,

e.

•

Next we prove Theorem 5 of Section 2.2.

EXAMPLE 7 Given that lim~cf(x) = L and 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

>

63

The Precise Definition of a Limit

0 such that

- f(x» - (M - L)I < e

whenever

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.

>

9.

10.

Il.

M must be false. Therefore •

Exercises 2.3 Centering Intervals About a Point In Exercises l--{j, sketcb the interval (a, b) 00 the x-axis with the point Xo inside. Then find a value of Il > 0 such that for all x,O < Ix - xal < Il =* a < x < b.

1. a = I, 2. a

=

1,

b = 7,

b

3. a = -7/2,

4. a

~

-7/2,

=

7,

f(x) ~ y

Xo

=

y f(x) =

1

E

2VX+1

Xo ~ 3 L=4

L~I

Xo = 5

Xo

~

Yx

=!

4

= 0.2

E

2

b = -1/2, Xo = -3

b

~

-1/2, Xo

~

-3/2

5. a = 4/9, b = 4/7, Xo = 1/2 6. a = 2.7591, b = 3.2391, xa = 3

0

x

9

16 X

FInding Deltas Graphically In Exercises 7-14, use the graphs to [md all > 0 such that for an x 0< Ix-xol
NOfTOSCALE

11.

12.

y

f(x)

f(x) Xo

3

~

-2x+ 3

E

L = 7.5 € = 0.15

6 5.8

y

f(x)=4-x 2 Xo ~-I

~ x2

=2

L~3

L~4

Xo =-3 6.2

y

y

8.

7.

•

=1

0

I I I

-i--r-

7.35

o

--

I I I

-1--I

4

7.65 7.5 -::1----.f---7~----X

0.25

y=4-x 2

5

= -~x+ 3

~

/

v3

NOTTOSCALB NOf TO SCALE

2

"

v5

X

2.75

I I I I I I I I I I I I

v3 -T NUl' 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 f(x), a point XO, and a positive number E. Find L = lim f(x). Then find a number 8 > 0 such

y

f(x)~ ~ -'0

~

-I

f(x) ~ ~

-x

L~2 E

-'0 ~! 2

2.01

= 0.5

x~xo

that for all x

Ix -

0<

31. f(x) = 3 - 2x,

= 0.01

E

Xo = 3,

32. f(x) = -3x - 2, 2.5 I I

I I

I I I I I I I I I I I

-9

35. f(x) =

~ !2

0

X

0

36. f(x)

'--I

~

"V'1-=-5x, 4/x,

~

2x - 2,

17. f(x) ~

"\1;+1,

18. f(x) ~

V;,

-6,

L ~ 1/2,

L

I/x,

~

4,

Xo

~

xo ~ xo ~

L ~ 3,

20. f(x) ~ ~, ~

Xo

L ~ I,

19. f(x) ~ ~,

21. f(x)

5,

~

L

1/4,

Xo

~

23. f(x) ~ x 2 ,

L ~ 4,

Xo ~ -2,

25. f(x)

~ x2 -

26. f(x)

~

27. f(x)

~ »IX,

28. f(x)

~ »IX,

L

L

~

m

>

0,

m

>

0,

=

C

120/x,

+ b,

30. f(x) = »IX E ~ 0.05

+ b,

E

-I,

Xo

L ~ 11,

5,

29. f(x) ~ »IX Xo = 1/2,

~

m

~

~ 0.1

38. lim (3x - 7) x~3

1/4,

E

~ 0.1 E E

E ~

E ~

48.

x~o

Xo

~

2,

E ~

0.03

L

~

3m,

Xo

~

3,

E ~ C

>

~

(m/2)

+

x-I

x2 - 1 =2 x-I

xsin~ ~

if f(x)

~ {~;,

x
0

I

+ b,

b,

46.lim

I

2m,

L = m

=-2

0.1

~

> 0,

2 ,

y

L

L

{xI, xx'" -2

I

1~ f(x) ~ 0

49. lim

~

0,

E ~

~

x<1

~ I

E ~

24,

if f(x)

xO?::l

0.5

Xo ~ 4,

lim~ ~

x~l

4

~ I

0.05

E ~

-I,

.%--2

x '" I x =I

2

~ 0.1

E

~

42. lim f(x)

Xo = 1,

---r"+lAIH~f---x

0

2

x~o

if f(x) = {x , 2,

x~l

~

40. lim~=2

45. lim x -9=_6 x--3 x + 3

Xo

5,

>

~

0.4

E ~

2,

5

41. lim f(x) = I

43.

>0 m

~

• = 0.5

2

0.02

E

v'3,

Xo ~

I/x,

E ~

0,

4,

L ~ 3,

~

0.01

-2,

Xo ~ 23,

22. f(x) ~ x 2 ,

24. f(x)

E ~

Xo ~ 10,

L ~ 4, ~

Xo

x~.

Eacb of Exercises 15-30 gives a function f(x) and numbers L, XO, and E > O. In each case, f'md an open interval about Xo on which the inequality If(x) - L I < E holds. Then give a value for 8 > 0 sucb that forallxsatisfyingO < Ix - xol < 8 the inequality If(x) - LI < E holds.

16. f(x)

Xo = -3,

39. lim~=2

finding Deltas Algebraically

~

• = 0.05

Prove the limit statements in Exercises 37-50. x~4

L

Xo = -5,

1.99

2.01

I,

:5+ 5,

:

37. lim (9 - x)

+

• = 0.05

2

NOT TO SCALE

15. f(x) ~ x

• = 0.03

Xo = 2,

x-

34. f(x) = x

16 -25

-I

16

<

• = 0.02

Xo = -I,

x2 - 4 33. f(x) = - - 2 '

1.5

------1--+---I I I I I I I I I

1.99

2

--I--

If(x) - LI

xol < 8

L~2

E.

When Is a Number L Not the Lim;t of / (x) as x .... xo? Showing L is not a limit We can prove that lim_.. f(x) providing an E for all x,

>

65

The Precise Oefinition of a Limit

2.3

0 such that no possible 8

>

~

If(x) - LI <

0 < Ix -xol < 8

We accomplish this for our candidate 0 there exists a value of x such that

E

L hy

0 satisfies the condition E.

hy showing that for each

a>

0< Ix - xol < 8

If(x) - LI '"

and

E.

y

L

+ E f-,--------,,£----,----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 statement abont limits

h-O

+ c)

= L.

Show hy example that the fol-

oxo -a\xo xo 8

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<

Explain why the function in your example does not have the given value of L as a limit as x ~ xo. Show hy example that

Ix - xol < ~ and If(x) - LI '"

E

x< I 57. Let f(x) = {x, I x + , x>!. 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 <

o 55.

f-+---;c---+----

----;;-f--'--:;+~-___';__;;_----x

lowing statement is wrong.

54. Anotber wrong statement about limits the following starement is wrong.

E

f(x) +-+-:;;-'!'

E.

Explain why the function in your example does not have the given value of L as a limit as x ~ Xo.

2

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 = '/T(x/2)' and look for the interval in which you must hold x to make IA - 91:5 0.01. What interval do you fmd?

om

56. Manufactnriog 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, 1 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 1 is to be 5 ± 0.1 amp. In what interval does R have to lie for 1 to be within 0.1 amp of the value 10 ~ 5?

a. Let E = 1/2. Show that no possible 8 > 0 satisfies the following condition: For all x,

0 < Ix - 11 < 8

That is, for each 8 that

+

R

If(x) - 21 < 1/2.

> 0 show that there is a value of x such

0< Ix - II < 8

v

~

If(x) - 21 '" 1/2.

and

This will show that limx_ d(x)

b. Show that lim_1 f(x) ~ I. c. Show that1im_ 1 f(x) ~ 1.5.

~

2.

66

Chapter 2: Limits and Continuity

58. Leth(x)

~

{

X',

X

<

3, 2,

x

~

x>

y

2 2 2.

•

2

y

4

3

---1'---'------+--+ x

o

2

COMPUTER EXPLORATIONS -;;1..L-----:c-----+ x

o

In Exercises 61--{j6, you will further explore rmding deltas graphically. Use a CAS to perform the followiug steps:

a. Plot the functioo y = f(x) near the point Xo being approached. Show that a. lim h(x) #' 4

b. Guess the value of the limit L aod then evaluate the limit symbolically to see if you guessed correctly.

b. lim h(x) #' 3

c. Using the value. = 0.2, graph the haoding lines Yl = L - • aod Y2 = L + • together with the functioo f near Xo.

x~2 x~2

c. lim h(x) #' 2 x~2

59. For the functioo graphed here, explain why

d. From your graph in part (c), estimate a 8

0<

a. lim f(x) #' 4 x~3 b. lim f(x) #' 4.8 x~3 c. lim f(x) #' 3

Ix - xol < 8

~

>

0 such that for all x

If(x) - LI < •.

Thst your estimate by plotting f, Yl, aod y, over the interval 0< Ix - xol < 8. For your viewiog window use Xo - 28 '" x '" Xo + 28 aodL - 2. '" y '" L + 2•. If any functioo values lie outside the interval [L - " L + .J, your choice of 8 was too large. Try again with a smaller estimate.

x~3

y

e. Repeat parts (c) aod (d) successively for. = 0.1,0.05, and 0.001. 4.8 4

61. f(x) =

•

x· x

5x 3

3

62. f(x) = 2x' sin

81 3' Xo = 3

+ 9x' + ]X,'

Xo = 0

a

63. f(x) = 3X' Xo = 0 -ccI------o----~x

o

3

60. a. For the functioo graphed here, show that limx~-1 g(x) #' 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

64. f(x) =

x(1 - cosx) . , Xo = 0 x - smx

~-I 65. f(x) ~ x I ' Xo ~ I _ 3x' - (7x ( ) 66·fx-

+ l)~ + 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 dermed 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 One-Sided Limits y

J

•

F~

•

0

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)

x-e-

= M.

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)

x-o+

=

I

and

lim j(x)

x-o-

y

=

-1.

y

f(·)

L

o (a)

-C:+---~"'-----"-'------>.

o

•

c

lim f(.)

~c'"

M

f(·)

~

•

c

L

FIGURE 2.25 approaches c. (b) Left-hand limit as> approaches c. y

EXAMPLE 1 The domain of j(x) Figure 2.26. We have

=

V4=7 is [-2, 2]; its graph is the semicircle in and

~~---;;t---+--.

-2

0

FIGURE 2.26 lim

x--2+

2

lim

x-2-

v'4=7 =

v'4

-

.2 =

0 and

0 (Example I).

The function does not have a left-hand limit at x = -2 or a right-hand limit at x does not have ordinary two-sided limits at either - 2 or 2.

=

2. It •

One-sided limits have all the properties listed in Theorem I in Section 2.2. The righthand limit ofthe sum oftwo functions is the sum oftheir 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) .~c

= L

limj(x) .~c

=

L

and

lim j(x)

x-c+

= L.

68

Chapter 2: Limits and Continuity

EXAMPLE 2

y

For the function graphed in Figure 2.27, limx~o+ f(x) =

Atx = 0: 2

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-

Atx = 1:

•

-=+-~-----o-----o----'---x

o

2

3

f(x)

=

limx~l+ f(x) =

4

0 even though f(l) = 1, 1,

limx~l f(x) does not exist. The right- and left-hand limits are not

equal.

FIGURE 2.27 Graph of the function in Example 2.

limx~'- f(x) =

1, limx~2+ f(x) = 1, limx~2f(x) = 1 even though f(2) = 2. limx~r f(x) = limx~3+ f(x) = lim~3 f(x) = f(3) = 2. limx~4- f(x) = 1 even though f( 4) oF 1, limx~4+ f(x) and limx~4 f(x) do not exist. The function is not defined to the right of x = 4.

Atx = 2:

Atx = 3:

y

Atx = 4:

•

At every other point c in [0, 4], f(x) has limit f(c). L+<

f(Xl! f(x) lies

Precise Definitions of One-Sided Limits

inhere

L

The formal definition of the limit in Section 2.3 is readily modified for one-sided limits.

L-<

'*

for all x Xo in here , 8

DEFINmONS

-::+---o~--ii>---1)-_>X o "0 "0+8

We say that f(x) has right-hand limit L at xo, and write lim f(x) = L

x-xo+

(see Figure 2.28)

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-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 <

xo-Il<x<xo L+ <

f(x) f(x) lies inhere

L

E.

)

EXAMPLE 3

Prove that

L-<

lim

x-o+

'*

for all x Xo inhere ~

8

O.

Let E > 0 be given. Here Xo = 0 and L = 0, so we want to find all> 0 such that for all x

O<x<1l

"0 - 8

FIGURE 2.29 Intervals associated with the defmition ofleft-hand limit.

=

Solution

-=+---f-~x_---v-__ x

o

y!;

or

O<x<1l

y!;<

E.

2.4

One-Sided Limits

69

Squaring both sides ofthis last inequality gives

y f(x) ~

if

Vi

.f-------;;.....~----

0< x < Il.

If we choose Il = e 2 we have

y!;; < e,

O<x
f(x)

or

0<x<e2 FIGURE 2.30

lim

x-o+

Vi = 0 in Example 3.

IY!;;-Ol <e.

=>

According to the def"mition, this shows that limx~O+y!;; = 0 (Figure 2.30).

•

The functions examined so far have had some kind of limit at each point ofinterest. In general, that need not be the case.

EXAMPLE 4 Show that y = sin (I/x) has no limit as x approaches zero from either side (Figure 2.31). y

v FIGURE 2.31 Thefunctiony = sin (1/x) has neilher arighthand nor a left-hand limit as x approaches zero (Example 4). The graph here omits values very near lhe y-axis.

Solution As x approaches zero, its reciprocal, I/x, grows without bound and the values ofsin (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 0 (radians)

o

-_cl31T---="'--~'-----+---'l.~~"""'=-'3L1T--'O NOfTOSCAlB

FIGURE 2.32 The graph ofj(O) = (sinO)/e suggests1hatlherightand left-hand limits as e approaches 0 are holh 1.

70

Chapter 2: Limits and Continuity

y

THEOREM 7 (II in radians)

(1)

p

lane 1

Proof The plan is to show that the right-hand and left-hand limits are both I. 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

(J cos 8 --;c1'-...l........;;=_..J.J~_--'l~~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 proofof Theorem 7. By detmition, TA/OA = taIl8, butOA = l,soTA = taIl8.

Area !J.OAP = t base X height = t (I )(sin II) = t sin II Area sector OAP =

1 r 211 = 1 (1)'11 = It 222

(2)

Area !J.OAT = tbase X height = t (l)(tan II) = ttanll.

I Equation (2) is where radian measure

Thus,

111 sm < 2 2I'll < 2I tan II .

comes in: The area of sector OAP is 8/2 only if 8 is measured in radians.

11ris last inequality goes the same way if we divide all three terms by the nwnber (1/2) sin II, which is positive since 0 < II < TT/2: I

II

I

< ~II sm < --II' cos

Taking reciprocals reverses the inequalities: I

sin II > -11>

cos II.

Since limtHo+ cos II = I (Example lib, Section 2.2), the Sandwich Theorem gives lim sin II = 1. 8

8--+0+

Recall that sin II and II are both odd jitnctions (Section 1.1). Therefore, /(11) = (sin 11)/11 is an even jitnction, with a graph symmetric about the y-axis (see Figure 2.32). 11ris 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 h-O

h

and

(b) lim sin 2x = ~.

x-a 5x

5

2.4 One-Sided Limits

71

Solution (a) Using the half-angle formula cos h = 1 - 2 sin2 (h/2), we calculate 2

lim cos h - 1 = lim _ 2 sin (h/2) h~O h h~O h · sin 0 . 0 = - 1un--sm O~O 0 =

LctO ~ hl2, Eq. (I) and Example H. in Section 2.2

-(1)(0) = 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: . sin 2x . (2/5)' sin 2x hm - - = hm -'--7~c-::-5x x~O (2/5)' 5x

x~O

= ~ lim sin2x

5 x-o

Now, Eq. (I) applies with

o ~ 2>.

2x

=~(I)=~ 5

•

5

' 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 = t~O

1 lim sin 1 • _1_. _1_ 3 t~O

31 =

1

1

3 (1)(1)(1)

cos 1 cos 21 =

1

Eq. (I) andExampleHb in Section 2.2

3'

•

Exercises 2.4 FInding Limits Graphically

y

1. Which of the following statements about the function y graphed here are true, aod which are false?

= f(x)

y

~

f(x)

•

y

-7-----;;1'--;----c2;----c;---> x 3

a.

lim f(x) = I

.%--1+

c. lim f(x) = I .%-0-

e. lim f(x) exists.

x-o

g. lim f(x) x-O

i. lim f(x) x-I

k.

.%-0-

r.

.%-0-

lim f(x)

x-o

h. lim f(x)

~

j. lim_ f(x)

0

a

..

d. lim f(x) = lim f(x)

=I

lim f(X) does not exist.

.%--1-

h. lim f(x) =

x-I

x-O+

2

L lim f(x) =

a

.%-2

x-2+

2. Which of the following statements about the function y graphed here are true, aod which are false?

c. lim f(x) = 2 e. lim f(x)

=I

g. lim f(x)

~

x-l+

x-o+

h. lim f(x) does not exist. x~2

d. lim f(x)

x~2

=0 =I ~

lim f(x) = I

x--I+

r.

x-r

=2

lim f(x) does not exist.

x~l

lim f(x)

x-o-

h. lim f(x) exists at every c in the opeo interval ( -I, I). x~,

L lim f(x) exists at every c in the opeo interval (I, 3). ~

f(x)

x~,

j.

lim f(x) =

.1'--1-

a

k. lim f(x) does 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)

= ~sin(I/x). y

1 y

a. 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,....,- f(x) and lim,....,+ f(x).

d. Does lim,...., f(x) exist? Ifso, what is it? Ifno~ why not?

4. Letf(X)=!:-X'

2'

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. 7. a. Graphf(x) ~ y

X 3, {

x =F 1 x = 1.

0,

b. Find lim,....,- f(x) and lim%~l+ f(x).

c. Does lim,...., f(x) exist? If so, what is it? Ifno~ why not?

{I2, - x

I

x ~ x-I. b. Find lim,....,+ f(x) and lim%~l- f(x). c. Does lim,...., f(x) exist? If so, what is it? If n~ why not?

8. a. Graphf(x) =

a. Find lim,....2+ f(x),

lim%~2- f(x)

Graph the functions in Exercises 9 and 10. Then answer these questions.

, and f(2).

a. What are the domain and range of f7 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. Find lim""'-l- f(x) and lim,....-l+ f(x).

c. At what points does only the left-hand limit exist?

d. Does lim""'-l f(x) exist? If so, what is it? Ifnot, why not?

5. Letf(x)

=

O'

{

d. At what points does only the right-hand limit exist?

x"'O

. I Sffi:x. x>

2 ,

"\,/'l-=7; 9. f(x) ~

O.

{

X,

y

10. f(x) ~

{

O"'x<1

I '" x < 2 x = 2 -l::S;x 1

I, 2,

°

I, 0,

Find;ng One-S;ded UmUs Algebra;cally

----,+-'-L+----c/-------> x

°

y~ I'

-1

I

o. . 1 {Sffij.

x'" °

Find the limits in Exercises 11-18. 11.

x>O

I

a. Does lim%~o+ f(x) exist? If so, what is it? If n~ why not? b. Does lim,...." f(x) exist? If so, what is it? Ifnot, why not? c. Does lim,....o f(x) exist? If so, what is it? If no~ why not?

lim x-+-0.5-

13

lim

~x + 2 x

1

x-I- x

15. lim Vh h-+O+

1

(_x) (2x ++ 5) + (_I + )(~)(~)

• x--2+ X

14. lim

+

x2

x

1

2

X

+ 4h + 5 h

7

-

V5

2.5

16

r

. h~1lJ

17. a.

V6 -

y'5h' + llh + 6

41. lim

o~o

h Ix + 21 . lim (x + 3 ) - + 2

%--2+

18. a. %-1+ lim

~(x - I)

Ix-

Ix + 21 + 3) x + 2

. b. ,.!,"2!,jx

X

~(x-I)

b. lim

II

,~l

Ix - II

Use tlIe graph oftlIe greatest integer functioo y = lx J, Figure 1.10 in Section 1.1, to help you tmd tlIe lintits in Exercises 19 and 20. 19. a. lim lOJ 0-3+ 8

b. lim lOJ

20. a. lim(t-ltJ)

b. limjt-ltJ)

o-r

1-4+

e-4

.-0

22. lim Sintkt

21. lim sin\/20

\/20

(kcoostant)

,....0

24. lim

25. lim tan 2x

26.

27. lim x csc 2x x-o cos5x

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 h-O sm

.h

h-o- sm 3h

x

~ ~t ,~o

,

33. lim sin (I - cos t) 1-0 1 - cost

+.SIDX

35. lim sinO

36. lim s~5x x-o sm4x

37. lim 0 cos 0 0-0 3x 39. lim tan 8x

38. lim sin 0 cot 20

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 tlIen know lim....... j(x)? Give reasons

for your answer. 44. !fyou know iliat lim......, j(x) exists, can you find its value by calculating limx-c+ j(x)? Give reasons for your answer. 45. Suppose iliat j is an odd functioo of x. Does knowing iliat lim,_o' j(x) = 3 tell youanytlIing about lim......o- j(x)? Give ",a46. Suppose iliat j is an even functioo of x. Does knowing iliat lim,_r j(x) = 7 tell you anytlIing about eitlIer lim,__r j(x) or

o~o

_sin=3Y,--,-co:::t-c5,yY cot4y

40. limy-+O

Formal Definitions of Dne-5lded Limits 47. Given. > O,tmd an interval I = (5,5 + S),S > O,suchiliatif x lies in I, tlIen ~ < •. What !intit is being verified and what is its value?

48. Given. > O,tmd an interval I = (4 - S, 4), S > 0, such iliat if x lies in I, tlIen ~ < •. What !intit is being verified and what is its value?

. sin 3y 23. hm-y-o 4y

8-0 sm28

0 cot 48

o~o sin' 0 cot' 20

limx_-2+ j(x)? Give reasons for your answer.

Find tlIe lintits in Exercises 21-42.

x-o

42. lim

73

sons for your answer.

(J

Using 11m sino fJ = 1

0-0

tan 0

0' cot 30

Continuity

Use tlIe dEtntitioos of right-hand and left-hand lintits to prove 1he 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,_40o' l x J and (b) lim,_40o- l x J ; tlIen use limit definitioos to verilY your tmdings. (0) Based 00 your cooclusions in parts (a) and (h), can you say anything about lim......400 l x J ? Give ",asoos for your answer.

< 00 x> .

52. One-sidedlintits Letj(x) = {:';in(I/X), x

vx,

Find (a) lim......o' j(x) and (h) lim,_o- j(x); tlIen use lintit definitions to verilY your tmdings. (0) Based on your cooclusions in parts
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 jwnp 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 I

p

Q

~'j' 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 !/

V

..... o

5

10

Elapsed time (sec)

FIGURE 2.34 Connecting plotted pointa by an unbroken curve from experimental

data QI , 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. Solution The function 1 is continuous at every point in its domain [0,4] except at x = 1, x = 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 ofthe function's domain. Points at which f is continuous:

y

lim I(x) = 1(0). x-o+

Atx = 3,

lim I(x) .-3

lim I(x) = I(c).

.-1

lim I(x) does not exist.

Atx = 1,

• 1

2

3

1(3).

Points at which f is not continuous:

1

0

=

.-c

AtO < c < 4,c "!' 1,2,

-fi!(X)

2

Atx = 0,

Atx = 2,

x

4

lim I(x) .-2 lim_/(x) .-4

Atx = 4, FIGURE 2.35 The function is continuous on [0, 4] except atx = I,x = 2, and x = 4 (Example I).

= =

1, but 1 "!' 1(2). 1, 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

DEFINmON 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=f(x)

:

I I

I I

I I

I

I

I

a

c

b

FIGURE 2.36 andc.

x-c

Endpoint: A function y )

X

Continuity at pointa 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

limj(x) = I(b),

x-b

respectively.

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.- c+ I(x) = l(c).1t is left-continuous (continuous from the left) at c if lim.- c- 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 j(x) = ~ is continuous at every point of its domain [-2,2] (Figure 2.37), including x = -2, where j is right-continuous, and x = 2, where j is left-continuous. _

~--;;t-----c~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),--

Continuity Test A function j(x) is continuous at an interior point x = e 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).

1.

j(e) exists

(e lies in the doroain of f).

2.

limx~,

(f has a limit as x --+ e).

3.

Y

Y~

2

LxJ

j(x) exists limx~, j(x) = j( e)

(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

3

_

We summarize continuity at a point in the form of a test.

--~R---~x

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 e of its doroain if and only if it is both right-continuous and left-continuous at e (Figure 2.36).

y

-2

Continuity

EXAMPLE 4 Thefunctiony = lxJ introduced in Section 1.1 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 l n J = n, the greatest integer function is right-continuous at every integer n (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 n - 1

< e < n, n an integer, then limlxJ = n - I = leJ.

x~,

-

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 j( 0) = I. The function in Figure 2.40c would be continuous if j(O) were I instead of 2. The discontinuities in Figure 2.4Qb and c are removable. Each function has a limit as x --+ 0, and we can remove the discontinuity by setting j(O) equal to this limit. The discontinuities in Figure 2.40d through f are more serious: lim_ o j(x) does not exist, and there is no way to improve the situation by changing j at O. The step function in Figure 2.40d has a jump discontinuity: The one-sided limits exist but have different values. The function j(x) = 1/x 2 in Figure 2.40. has an inrmite 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

(a)

(b)

x

(d)

y

J

Y~f(x)~l2

0

x

x x

(f)

(e)

FIGURE 2.40 The function in (a) is continuous at x are not

~

0; the functions in (b) through (f)

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 ofits domain. It has a point ofdiscontinuity at x = 0, however, because it is not defined there; that is, it is discontinuous on any interval containing x = O.

------;o+-----x

\"

FIGURE 2.41 The function y ~ l/x is continuous at every value of x except x = O. It has a point of discontinuity at x ~ 0 (Example 5).

(h) The identity function f(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 8-Propertfes of Continuous Functions If the functions f and g are continuous at x = c, then the following combinations are continuous at x = c. 1. Sums: 2. Differences:

3. Constant multiples: 4. Products: 5. Quotients:

f+g f-g k· f, for any number k

f'g fig,

6. Powers:

r,

7. Roots:

"ifj

providedg(c) oF 0

n a positive integer provided it is dermed 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, Thcorcm 1

x-c

+ g(c)

Continuity off. g at c

(j + g)(c).

+ g is continuous.

EXAMPLE 6 (a) Every polynomial P(x) = anx n + an_1Xn-1 lim P(x) = P(c) by Theorem 2, Section 2.2.

+ ... + ao

is continuous because

x~e

(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) o¢' 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». g

·f

Continuous at c g

f Continuous

•

e

FIGURE 2.42

ate

_-:;.O,-

Continuous atf(e)

f(e)

=..~-

g(f(e»

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 ofTheorem 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. (a) y

=

(0) y =

Yx 2 x -

x 2/3

2x - 5

-

+ x4

(b) Y = I

21

Ix2 _ 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 ofthe identity function squared; the denominator is an everywhere-positive polynomial. Therefore, the quotient is continuous. (0) The quotient (x - 2)/(x 2 - 2) is continuous for all x oF ± 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 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). •

y =

Theorem 9 is actually a consequence of a more general result which we now state and prove.

THEOREM to-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 number 8 , Ig(y) - g(b)1

<

whenever

E

Since limx~, fix) = b, there exists a 8

If(x) - bl < 8 ,

>

<

Iy - bl

<

Ix - cl

0

<

>

0 such

8,.

0 such that

whenever

0

<

8.

Ifwe let y = fix), we then have that Iy - bl

<

8,

whenever

0

<

Ix - cl

<

8,

which implies from the f"rrst statement that Ig(y) - g(b)1 = Ig(f(x» - g(b)1 < E whenever 0 < Ix - cl < 8. From the def"mition of limit, this proves that lim~,g(f(x» =

EXAMPLE 9

g(b). As an application ofTheorem 10, we have

lim COS(2x

X-'1T12

•

+ Sin(32'IT +

x))

= co/ lim 2x \x-7Tl'2 = cos ('IT

+

+ lim Sin(32'IT + X-'1T12

sin2'IT) = COS'IT = -1.

x)) •

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~, ~) _,!

t~, ~)

(~, ~) x

"2

0

2

(0, I)

f(x)

F(x)

(~, ~)

_,!

0

2

(a)

"2

x

(b)

FIGURE 2.44 The graph (a) of f(x) ~ (sinx)/x for -,,/2 :5 x :5 ,,/2 does nol 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. IfI(c) is not dermed, but limx~, 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 there the value of 5/4. Thus F is the continuous extension of f to x = 2, and

y 2

y~

x2

1 -1

+x-6

x2 _ 4

lim x x-2

I I

~

1

0

I

3

4

)

x

(a)

y y

5

= x+

3

4[--

-1

0

1

2

4

3

x

(b)

FIGURE 2.45

+X xl -

-

4

6 = lim f(x) = x-l

l. 4

The graph of f is shown in Figure 2.45. The continuous extension F has the same graph except with no hole at (2, 5/4). Effectively, F is the function f with its point of discontinu_ ity at x = 2 removed. Intermediate Value Theorem for Continuous Functions

x+ 2

2

2

Fuoctions that are continuous on intervals have properties that make them particularly useful in mathematics and its applications. One of these is the Intermediate ~lue Property. A function is said to have the Intermediate Value Property if whenever it takes on two values, it also takes on all the values in between.

(a) The graph of

f(x) and (b) the graph ofits

THEOREM 11-The Intermediate Yalue Theorem for Continuous Functions If f is a continuous function on a closed interval [a, bj, and if Yo is any value between f(a)andf(b),thenyo = f(c)forsomecin[a,bj.

continuous extension F(x) (Example 10).

y y

~

f(x)

f(b) - - - - - - - - - - - - - -

Yo

~----------

{(a)

"Ot--~a----c~---tb--x

y

3 2

-=-+-~I---'-----'-----L-~x

o

2

FIGURE 2.46

f(x) =

4

The functioo

{Z>: - 2, 3,

3

1:5 x

Theorem II says that continuous functions over finite closed intervals have the Intermediate Value Property. Geometrically, the Intermediate Value Theorem says that any horizontalline y = Yo crossing the y-axis between the numbers f(a) and f(b) will cross the curve y = f(x) at least once over the interval [a, b j. The proof of the Intermediate Value Theorem depends on the completeness property of the real number system (Appendix 6) and can be fouod in more advanced texts. The continuity of f on the interval is essential to Theorem II. If f is discontinuous at even one point ofthe interval, the theorem's conclusion may fail, as it does for the function graphed 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 that the graph of a function continuous on an interval cannot have any breaks over the interval. It will be connected-a single, uobroken curve. It will not have jumps like the graph of the greatest integer function (Figure 2.39), or separate branches like the graph of I/x (Figure 2.41). A Consequence for Root Finding We call a solution of the equation f(x) = 0 a root of the equation or zero of the function f. The Intermediate Value Theorem tells us that if f is continuous, then any interval on which f changes sign contains a zero of the function. In practical terms, when we see the graph of a continuous function cross the horizontal axis on a computer screen, we know it is not stepping across. There really is a point where the 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. - x-I. Since 1(1) = 1 - I - I = -1 < 0 and 1(2) = 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 1 and 2. Figure 2.47 shows the result of zooming in to locate the root near x = 1.32. _

Solution 3

2

-

Let I(x) = x 3

2- 1

=

5

1.6

-11::::=::~~

2

-2 (a)

(b)

0.003

1.330

1.3240

f--"----"----~,...,..,

1.3248

-0.003 (e)

FIGURE 2.47

(d)

Zooming in on a zero ofthe function f(.) = . ' - • - 1. The zero is near

• = 1.3247 (Example II).

EXAMPLE 12

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,

___-+__ o

"------'L-~.

c

2

FIGURE 2.48 The curves y ~ ~ and y = 4 - .2 have the same value at .~cwhere~~4-.2 (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 ofthe 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) = 2.24 and 1(2) = + 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. _

V5 '"

V9

82

Chapter 2: Limits and Continuity

Exercises 2.5 7. a. Is I dermed at x = 2? (Look at the dermition of I.)

Continuity from Graphs In Exercises 1-4, say whether the function graphed is continuous on [-I, 3]. If not, where does it fail to he continuous and why'!

L

~

b. Is I continuous at x

= 2?

8. At what values of x is I continuous? 9. What value should he assigned to 1(2) to make the extended function continuous at x = 2? 10. To what new value should 1(1) he changed to remove the discontinuity?

Applying the Continuity Test At which points do the functions in Exercises II and 12 fail to he continuous? At which points, if any, are the discontinuities removahle?

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?

1

I

13'Y~x_2-3x

15.y=

x+1 x 2 -4x+3

17. y ~

Ix - 11 + sinx

Exercises 5-10 refer to the function

x' I(x)

=

I,

2>:, I, -2>: + 4,

l

-lsx
19.y = cosX' x

2L y

~

esc 2>:

l<x<2 2<x<3

0,

graphed in the accompanying figure.

25.y=~

26. y =

27. y = (2): - 1)1/3

28. y = (2 - x)I/'

29. g(x)

=

2

3

30. I(x)

The graph for Exercises 5-10.

~,-

x¢'2,x¢' -2

4'

x~2

x

~-2

Umits Involving Trigonometric Functions Find the limits in Exercises 31-36. Are the functions continuous at the point being approached?

5. a. Does I( -1) exist? b. Does lim%~_I' I(x) exist?

= I(-I)?

d. Is I continuous at x = -I? 6. a. Does 1(1) exist? b. Does lim%~1 I(x) exist?

c. Does lim%~1 I(x) ~ I(I)? d. Is I continuous at x

l

,

4,

-1

c. Doeslim%~_I' I(x)

~

I

X'-X-6 x¢'3 x 3 ' { x = 3 5,

x' - 8

-+-~c((----L--1j.--O---->X

"'¢'3x -

~

I?

31.

}~sin(x -

sinx)

32.

~Sin(fcos(tant»)

33. lim sec (y sec' y - tan' Y - I) y~1

34.

%~ tan(~cos (Sinx 1/3»)

35. lim cos ( t-O

V 19 -

1f

3 sec 2t

)

36.

lim v'csc'x x-1T16

+ 5V3tanx

2.5

Continuous Extensions

Theory and Examples

37. Defmeg(3) in a way that extends g(x) = (x 2 continuous at x = 3.

-

9)/(x - 3) to be

= (1 2 + 31 -

38. Defmeh(2) in a way that extends h(l) to be continuous at 1 = 2.

10)/(1 - 2)

39. Defme f(l) in a way that extends f(s) ~ (s' - 1)/(s2 - I) to be continuous at s = 1. 40. Defme g(4) in a way that extends

g(x)

= (x 2 -

to be continuous at x

16)/(x 2

4.

=

53. Roots of a cubic has three solutions in the interval [-4,4].

55. Solving an equation If f(x) = x' - 8x + 10, show that there are values c for which f(c) equals (a) 'IT; (b) (c) 5,000,000.

~

2 -

I, x < 3

2ax.

x~3

{x

56. Explain why the following fIVe statements ask for the same infor-

mation. a. Find the roots of f(x)

continuous at every x? 42. For what value of b is X

g(x) =

{

b~2,

X '"

-2

~ {~~~ -

e. Solve the equation x~2

2a,

continuous at x

x
continuous at every x? 45. For what values of a and b is x:5-}

ax '-- b, { 3,

<x <

-I x

~

1

( ) =

f x

ax + 2b, x 2 + 3a - b, { 3x - 5,

x:50

TxT

so.

= IOI%~ -

f(x) ~ (I

0,

if x is rational if x is irrational

b. Is f right-continuous or left-continuous at any point? 60. If functions f(x) and g(x) are continuous for 0 ,;; x ,;; I, could f(x)/g(x) possibly be discontinuous at a point of [0, I]? Give rea-

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. 1f1he 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?

48. f(x)

{I,

is discontinuous at every point.

O<x:52 x>2

continuous at every x?

49. f(x) ~

2, and how you know the discontinuity is

a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function

I

continuous at every x?

sin x

=

59. A function discontinuous at every point

-Z

46. For what values of a and b is

W x- I

3x - I = O.

58. Nonremovable discontinuity Give an example of a function g(x) that is continuous for all values ofx except x = -1, where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not removable.

x 2 + b. x>o

=

x' -

removshle.

g(X)~{~~~'

47. f(x)

3x = 1.

57. Removable discontinuity Give an example of a function f(x) that is continuous for all values of x except x = 2, where it has a removable discontinuity. Explain how yoo know that f is dis-

x<2

continuous at every x?

o

-

x'

d. Find the x-coordinates of the points where the cubic curve y = x 3 - 3x crosses the line y = 1.

44. For what value of b is

=

- 3x - 1.

c. Find all the values of x for which x 3

43. For what values of a is

g(x)

= x'

b. Find the x-coordinates of the points where the curve y ~ crosses the line y = 3x + 1.

< -2

x

continuous at every x?

~

= x has at least one solution. Show that the equation x' - 15x + I = 0

52. Explain why the equation cos x

V3;

f(x)

f(x)

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 between x ~ 0 and x ~ I? lllustrate with a sketch.

54. A function value Show that the function F(x) ~ (x - a j2 • (x - b)2 + x takes on the value (a + b)/2 for same value ofx.

3x - 4)

-

41. For what value of a is

f(x)

83

(ontinuity

I

+ a)'/%

sons for your answer. 61. If the product function h(x) = f(x)'g(x) is continuous atx = 0, mustf(x) andg(x) be continuous atx ~ O? Give reasons for your

answer. 62. Discontinuous composite of continuous functions Give an example of functions f and g, both continuous 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. Stret
to prove !bat 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 !bat a function f is continuous on the closed interval [0, I] and that 0 '" f(x) '" I for every x in [0, I]. Show !bat there must exist a number c in [0, I] such !bat ftc) = c (c is called a f'lIed point of I). 66. The sign-preserving property of eontinuous functions Let f be defmed on ao interval (a, b) aod suppose !batf(c) 0 at some c where f is continuous. Show that there is an interval (c - 6,c + 6) about c where fhas the same sigo asf(c).

*

Solvtng Equations Graphically

D Use the Iotermediate Value Theorem in Exercises 6fJ-76 to prove !bat each equation has a solution. Thea use a graphing calculator or computer grapher to solve the equations. 69.x 3 -1>:-1=0 70.2>;3-2>;2_2>;+ I =0

71. x(x - 1)2 ~ I

67. Prove !bat f is continuous at c if aod only if lim ftc

h~O

+ h)

(one root)

72.x%=2

73. Y;+~~4 74. x 3

= f(c).

-

75. cosx

68. Use Exercise 67 together with the ideotities

ISx

+

=x

+ c) = sinhcosc + cosh sine, cos(h + c) = coshcosc - sinh sine

(three roots)

(one root). Make sure you are using radian mode.

76. 2 sin x = x

sin(h

I ~ 0

(three roots). Make sure you are using radiaumode.

Limits Involving Infinity; Asymptotes of Graphs

2.6

In this section we investigate the hehavior 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 lllbitrarily 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 (00) 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) = l/x is defined for all x 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, l/x again becomes small. We summarize these observations by saying that f(x) = l/x bas limit 0 as x --> 00 or x --> -00, or that 0 is a limit off(x) = I/x at i'!finity and negative i'!finity. Here are precise definitions.

*

4

3

y=l

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 nwnber E FIGURE 2.49

The graph of y = I/x

approaches 0 as x ~ 00 or x ~ -

00.

0, there exists a corresponding nwnber M such !bat 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,

the graph enters this band at x = and stays.

y=l E

lim k=k

and

x---+±OO

=

O.

(1)

lim

x---+-OO

i

=

o.

Solutton

<8> Let E > 0 be given. We must find a number M such that for all x

=-l

FIGURE 2.50 The geometry behind the argument in Example I.

i

Show that (b)

No matter what positive number E is, the graph enters this bandatx and stays.

lim

x~±OO

We prove the secood result and leave the tITSt to Exercises 87 and 88.

EXAMPLE 1

"--i--~---1-E

85

Intuitively, limx~oo f{x) = L if, as x moves increasingly far from the origin in the positive directioo, f{x) gets arbitIarily close to L. Similarly, lim.....-oo f{x) = L if, as x moves increasingly far from the origin in the negative direction, f{x) gets arbitIarily 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 tITSt 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

l

y=E f---'>__-"-'----

Limits Involving Infinity; Asymptotes of Graphs

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. (b> Let E > 0 be given. We must find a number N such that for all x

x
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. (8)

lim

x_OO

(5 + 1) x

=

lim

x_OO

5+ lim 1 x_OO X

=5+0=5 (b)

lim x_-OO

1TY' 3= x2 = =

lim x_-OO

lim

x---+-OO

Sum Rule

Known limits

1TY'3. 1. 1 X X 1TY'3.

lim

1. x---+-OO lim 1 X

x---+-OO X

1TY'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

-2

10

x

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.

EXAMPLE 3 These examples illustrate what happens when the degree of the numerator is less than or equal to the degree of the denominator.


5x 2 + 8x - 3 x~OO 3x 2 + 2 .

hm

8

llx + 2

Y~2x3_1

6

.

hm

5

x~OO

+ (8/x) - (3/x 2 ) 3 + (2/x2)

5+0-0 (b)

lim IIx

+

x~-OO 2);' y

=

5

Divide nmnerator sod

den . torhy , OInms •.

See Fig. 2.51.

3+ 0 3 2 3 2 = lim (lI/x ) + (2/x ) I x~-OO 2 - (I/x')

Divide numerator and

denominator by x'.

=0+0=0 2-0

Sec Fig. 2.52.

•

A case for which the degree of the numerator is greater than the degree of the denominator is illustrated in Example 8.

4

Horizontal Asymptotes ==----'-++--"----=-=~'_>x -4 -2 2 4 6 -2

If the distance between the graph of a function and some rIXed line approaches zero as a point on the graph moves increasingly far from the origin, we say that the graph approaches the line aaymptotically and that the line is an asymptote of the graph. Looking at f(x) = I/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 IxIincreases.

and on the left because lim

x~-OO

i

=

o.

We say that the x-axis is a horizontal asymptote of the graph of f(x) = I/x.

DEFINmON A line y = b is a horizontal asymptote of the graph ofa function y = f(x) if either lim f(x) = b

or

x~OO

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

87

Find the horizontal asymptotes of the graph of

x3 - 2 f(x) = Ixl3 + 1.

2

We calculate the limits as x --+ ± 00.

Solution

3 _

lim x ._00 Ixl 3

Forx 20 0:

Forx FIGURE 2.53 The graph of the fonctioo in Example 4 has two horizootal asymptotes.

<

2

+

1 - (2/x 3)

2

3 _

lim _x__ = lim ._00 x 3 + I ._00 I

I

;~ x3 - 2 __ lim x - 2 ~. --""-~--- = I ._-00 Ixl 3 + 1 ._-00 (-x)3 + 1 3

0:

+ (l/x 3)

=

1.

3 . 1 - (2/x ) lIm = -1. ._-00 -1 + (l/x 3)

The horizontal asymptotes are y = -1 and y = 1. The graph is displayed in Figure 2.53. Notice that the graph crosses the horizontal asymptote y = -1 for a positive value

-

~~ EXAMPLE 5

Find (8) lim sin (l/x)

and

x_co

(b)

lim xsin (l/x).

x--+±oo

Solution (8) We introduce the new variable t = l/x. From Example 1, we know that t--+O+ as x --+ 00 (see Figure 2.49). Therefore, y

I,

. 1

I'

.

1IIlSID:x= unsmt= );--+00 1-0+

0.

Likewise, we can investigate the behavior of y = f(l/x) as x --+ 0 by investigating y = f(t)ast--+±oo,wheret = I/x.

--;---+1'IlIlIt-f---+----. FIGURE 2.54 The line y = I is a horizootal asymptote of the fonctioo graphed here (Example 5b).

(b) We calculate the limits as x --+

00

and x --+

. 1 I'un--= sin t 1 lim xsm-= X 1-0+ t

and

x--+oo

-00:

· ,I lim sint xsm-= -=1 . Iun X 1-0- t

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

2

and lim.__ II/xl = 0, we have lim.__ (sinx)/x = 0 by the Sandwich Theorem. Hence, lim (2

X-±OO

A curve may cross ooe of its asymptotes infmitely o\teo (Esample 6).

FIGURE 2.55

+

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

x_ oo (x

EXAMPLE 7

Find lim

-

v'X

2

+ 16).

Solution Both of the tenns x and v 'x 2 + 16 approach infinity as x --+ 00, so what happens to the difference in the limit is unclear (we cannot subtract 00 from 00 because the symbol does not represent a real number). In this situation we can multiply the numerator and the denominator by the conjugate radical expression to obtain an equivalent algebraic result: lim

x_

OO

(x -

v'x 2

+ 16)

=

=

lim

x_

OO

(x -

v'x 2

2

2

. x hm

(x

-

x_oo x +

v'x 2

2

+ 16r + x

v'x

Yx

+

2

+ 16

+ 16

+ 16). -16 = hm --~~= + 16 x_oo x + v'x 2 + 16

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

+

v'x 2

+ 16

=

16 I·un ----rii==== x x_ oo x2 16

1+

1+

-+Xl

o

v'1+O

Xl

=

0

.

•

Oblique Asymptotes

If the degree ofthe numerator of a rational function is I 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

between curve and line goes to zero as x --+

in Figure 2.56.

00

~+I 2

5 4

x=2

2x -

3

4Jx2 -

x2

2

~--;;C+--;-*----;;C-7-~---~x

-1 0 -1

x2 - 3 2x - 4

Solution We are interested in the behavior as x--+ ±oo. We divide (2x - 4) into (x 2 - 3):

The vertical distance

6

=

2

3

4

x

-2 -3

-

3

2x 2x - 3 2x - 4 I

This tells us that

f(x)

=

~ == ~ (~ + I) + (~) . =

'---.--"

FIGURE 2.56 The graph ofthe function in Example 8 has eo oblique asymptote.

linear g(x)

remainder

As X --+ ± 00, the remainder, whose maguitode gives the vertical distance between the graphs of f and g, goes to zero, making the slanted line g(x) =

2x +

I

2.6 y You can get as high

as you want by taking x close enough to O. No matter how

B

\ ~gh B is, the graph higher.

res

~

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 conImn 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 ofthe 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_lt X

x

No matter how low -B is, the graph goes lower.

j8\

-B

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) x-o+

One-sided inf"mite limits: and lim 1 = -00. x_o-x

=

lim x-o+

i

= 00

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 andpositive 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

1 Y=x-t

lim f(x) x-o-

=

lim

x-o-

i

= -00.

Again, we are not saying that the limit exists and equais the nwnber - 00 . There is no real nwnber - 00 . We are describing the behavior of a function whose limit as x ---> 0- does not

1 ----:-----;;+---;1--2~----;;3---x

exist because its values become arbitrarily large and negative.

EXAMPLE 9

FIGURE 2.58 Near x = I, the function y = I/(x - I) behaves the way the function y = I/x behaves near x = o. Its grapb is the grapb of y = I/x sbifted I unit to the right (Example 9).

x-l+ x - I

· nI u

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 - I) ---> oo.Asx---> 1-, we have (x - I) --->0- and I/(x - 1)--->

y

-00.

} x 0

Find lim _1_

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) in Example 10 approaches inf"mity as x --+ O.

as

x ---> o.

Solution 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

lim f(x) x-o

=

lim \ x-Ox

= 00.

•

90

Chapter 2: Limits and Continuity The function y = l/x shows no consistent behavior as x --> o. We have l/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 lim._ o (l/x 2) = 00.

EXAMPLE 11

These examples illustrate that rational functions can behave in various

ways near zeros of the denominator.

.-2 .-2

.-2 .-2 +

(a) lim (x - 2)' = lim (x - 2)2 = lim x - 2 = 0 .-2 X 2 - 4 (x - 2)(x + 2) X + 2

(b) lim x - 2 = lim .-2 X 2 -

(c) y

4

x - 2 = lim _1_ = ! (x - 2)(x + 2) X 2 4

lim x - 3 = lim

.-2+ X 2 -

4

.-2+

(d) lim x - 3 = lim • -2- X 2 -

4

.-T

.-2

(e) lim x - 3 = lim • -2 X 2 -

Bf------/-----j----'\---

--;;-I;f----'c~--o:---C;\---~x

Xo

+8

.-2

(f) lim

4

x - 3 = _ 00 (x - 2)(x + 2)

The values are negative for x > 2,xnear2.

x - 3 = 00 (x - 2)(x + 2)

The values are positive forx < 2,xnear2.

x - 3 does not exist. (x - 2)(x + 2)

See pans (c) and (d).

2-x = lim -(x-2)=lim -I =-00 (x - 2)3 .-2 (x - 2)3 .-2 (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 requiring f(x) to lie arbitrarily close to a finite number L for all x sufficiently close to xo, the defInitions ofinfinite limits require f(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seen before. Figures 2.60 and 2.61 accompany these definitions.

DEFINmONS 1. We say that f(x) approaches infinity as x approaches xo, and write lim f(x) = 00, x-xo if for every positive real number B there exists a corresponding /l that for all x O
=)

>

0 such

f(x»B.

2. We say that f(x) approaches mious inf"mity as x approaches xo, and write lim f(x) = -00, X---+Xo

FIGURE 2.61

For Xo - 8

< x < XQ + 8,

the graph of f(x) lies below the line y = -B.

if for every negative real number - B there exists a corresponding /l that for all x f(x) < -B. o < Ix - xol < /l

>

0 such

The precise defmitions of one-sided infinite limits at Xo are similar and are stated in the exercises.

2.6

EXAMPLE 12 Solution

Prove that lim

x~o

Given B

>

Limits Involving Infinity; Asymptotes of Graphs

~ X

91

= 00

>

0, we want to find /l

0 such that 1 x

OB. Now,

~>B 2 x

if and only ifx 2

<~

or, equivalently,

Ixl< Thus, choosing /l =

I

vB'

1/vB (or any smaller positive number), we see that Ixl < /l implies ~ > ~;;,: B. x /l

Therefore, by definition,

• y

Vertical Asymptotes

Vertieal asympto'"

Horizontal

y=

l

Notice that the distance between a point on the graph of f(x) = I/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) = I/x is unbounded as x approaches 0 because and

~as;;ymp~tote~:::=-OO rt-=~==;=,>' Horizontal

lim 1 = x---+o-x

-00

;;;;;

asympto"'. y~O

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 ofboth branches ofthe hyperbola y ~ 1/•. FIGURE 2.&2

DEFINmON A line x y = f(x) if either

=

a is a vertical asymptote of the graph ofa function or

EXAMPLE 13

limj(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

The asymptotes are quickly revealed ifwe recast the rational function as a polynomial with a remainder, by dividing (x + 2) into (x + 3):

y

Vertical asymptote, x

=-2

6 5 4

I

x+3

y= x+2

~ 1

Horizontal

x + 2)" + 3 x+2 I

+ _1_

x+2

asymptote,

y- 1

-~-;--c";-':k-±--;-~--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 graph 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 The lines y = I and x = - 2 are asymptotes ofthe curve in Example 13.

EXAMPLE 14

Find the horizontal and vertical asymptotes ofthe graph of

I(x) y y~

Vertical

asymptote, x=-2

=

2

Horizontal asymptote, y

8 x2 _ 4'

Solution 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 graph is symmetric with respect to the y-axis.

__ 8_ x2 _ 4

Vertical asymptote, x

= -

=

0

(a) The behavior as x---> ±OO. Since limx~oo/(x) = 0, the line y = 0 is a horizontal asymptote of the graph to the right. By symmetry it is an asymptote to the left as well (Figure 2.64). Notice that the curve approaches the x-axis from ouly the negative side (or from below). Also, 1(0) = 2.

(b) The behavior as x---> ±2. Since

and the line x = 2 is a vertical asymptote both from the right and from the left. By symmetry, the line x = - 2 is also a vertical asymptote. FIGURE 2.64 Graph ofthe function in Example 14. Notice that the curve approaches the x-axis from only one side.

There are no other asymptotes because I has a finite limit at every other point.

EXAMPLE 15

Asymptotes do not have to he two-sided.

The curves

Y

=

secx

=

I

cosx

and

y

=

tanx

=

sin x cosx

both have vertical asymptotes at odd-integer multiples oftr12, where cos x y

y

=

y

secx

-+----'-+-~-+_--'--t-~x

3

•

=

0 (Figure 2.65).

y = tanx

-+~'----+-~Oo-+_~-t-~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

I f(x) '" 2>; - 4

y

For x numerically large, 2x

~ 4 is near O.

For x near 2, this term is very large.

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 tota1 value of f is insignificant. It behaves like 1/(2); - 4) when x is so close to 2 that 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.

If we want to know how

----;2!---~1 ~""'O'+--""-!----C:-2 ~x

-5 Let f(x) = 3x 4 - 2>;3 + 3x 2 - 5x + 6 and g(x) = 3x 4 • 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 (Figore 2.66a), but appear as virtually identical on a larger scale (Figore 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 --> ± 00 • We fmd that 2Oc O---;_loO:O"-""'O::+-~"OIOcO---;2f.oO~x

-_-C0

-100,000 (bJ

FIGURE 2.66 The graphs off and g are (a) distinct for Ixl small, and (b) nearly identical forlxllarge (Example 16).

lim f(x) = lim 3x 4 - 2>;3 + 3x 2 - 5x .~±OO g(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

Limits of Rational Funettons

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.

h. lim f(x)

i.

x~,

lim f(x)

%-0+

%-+-3

g. lim f(x) x~O

lim _ f(x)

e.

%--3+

x~oo

%-+-3

lim_ f(x)

In Exercises 13-22, f"md the limit of each rational function (a) as

x ---+

00

and (b) as x ---+

13. f(x)

=

+3 +7

14. f(x)

2x' + 7 = x-,~-"x"",~+'--'x'--+-7

15. f(x)

= -"'--±-l 2

16. f(x)

=~+7

17. h(x)

7x' =,x-3x+6:< ,

18. g(x) = -x~'---4'-x-+-1

2x Sx

x~o

lim f(x)

%_-00

y

x

+3

3

,-,/

19. g(x) = lOx'

1/

f

r

-6 - -4 - -i/o

,

s 6

3

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)

e.

c. lim_ f(x)

x-2+

x~4

x~,

x~,

r.

lim f(x)

x--3+

lim_ f(x)

%-+-3

g. lim f(x)

h. lim f(x)

i. lim_ f(x)

j. lim f(x)

k. lim f(x)

L

%-0+

%-+-3 x~o

21. h(x)

=

-2x' - 2x + 3 3x' + 3:<' - Sx

lim f(x)

X~-OO

23. x_oo,,~ lim ~ 25.

lim %-+--00

x

f

"-

1-

-6

-s -41-

1'-

-2 ~

{

29.

.....

4 5 6

2/3

II

3

26

x-

4. f(x)

3

= 7f

-

I

5. g(x)

= 2 + (I/x) -S

7. h(x) =

3 -

+ (7/x) (')

I/x

~

x'

. 2x'/3 - x 1/3 31. lun %_00 x 8/ S + 3x +

8. h(x)

4

+ ("1/2/x')

Find the limits in Exercises ~ 12. 9. lim sin2x x-+oo x U. lim 2-t+sint t---oo t + cos t

10.

lim 9-+-00

12

c03~

30. lim x x_oo

+7

y;

~ x +1

32.

• ,-+00

2r

+ smr

+ 7 - 5sinr

-

-I

x

+

x

-4

2 _X 3

~-Sx+3

lim

+ x 2/3

x_-oo

lim

~ x + 1

lim

4 - 3:<'

x_-oo~

38. lim :_

39. x_2-X lim ~2 -

40. x_3+Xlim ~3

41.

x-o- ~

X.!.~8+ x ~ 8

42.

4 (x - 7)'

45. a. lim x-o+

1)1/3

3

2 x -Sx x 3 +x-2

x_- oo 2x

34.

x-O+ ~

x~7

U

r

lim.

lim~

37. lim }_

43. lim

0

&x 2

Find the limits in Exercises 37-48.

3 - (2/x)

= -----'~c­

(x' + x-

Infinite Limits

I

6. g(x) = 8 _ (S/x')

lim

7

36.

3. f(x) =

4

+ x-I

~-~

lim

x_oo

2

=

'x_oo

X~-OO ~ +~

33. lim

In Exercises 3-8, f"md the limit of each function 00 and (b) as x ~ - 00. (You may wish to visualize your answer with a graphing calculator or computer.)

4

9x + x 2x +5x2 -x+6 -x 4 22. h(x) = --.--,--"----;;---c 4 x -7x'+7x'+9 20. h(x)

x_--oo

+ 7x

27. lim 2'\."x x_lXl 3x

/ 1\

I

24.

1 ----=-£ ' )' ( 2

y

3 1\ 2

x - 2

Limits as x ---+ 00 or x ---+ - OQ 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~o

X~OO

+ ~4 + 31 x

1/ 1/i 3

1 '"

/

-00.

2 /, 3x 1

X.!.~,- 2x ~ 10

44. lim

x~o

b. lim x-O-

2 /3 3x 1

-1 x'(x + I)

- 4

2.6

46. a. lim

x-O+X

47. lim

x-o

X

~/5

b.lim~ 15

Graphing Simple Rational Functions Graph the rational functions in Exercises 63--{j8. Include the graphs and equations ofthe asymptotes and don3inaot terms.

.%-0- X /

x-o x ~/,

;/5

48. lim

Find the limits in Exercises 4<}-52. 49. 51.

tanx

lim

SO.

x-('1f'/2r lim (I + esc 8)

lim

x-( --rr/2>+

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+ b.

c. x~ -2+

x~2-

d. x---+ -2-

I 63'Y~x_1

I 64'Y~x+1

I 65'Y=2x+4

66. y = - x-3

67. y

b. x-+ 1-

70. flO)

~) as d. x-+-l

0, lim f(x) x_±OO ~ -2 X-±OO -00,

%-1+

72. f(2)

-2, lim f(x) x~-oo

~

0, lim f(x) .%-0+

~

I, f( -I)

lim f(x)

x-a-

2

73.

a. x-+O+ c. x-+2-

d. x-+2

e. What, if anything, can be said about the limit as x --> O? x2 - 3x + 2 as 58. lim x 3 - 4x a. x-+2+

74. 75.

~

~

2, and

.%--1-

0, lim f(x)

~

Find the limits in Exercises 5'HJ2.

t~/3) as

0, lim f(x) .1'-0+

lim f(x) = 0, lim f(x) =

X_±lXl

%-2-

lim g(x) = 0, lim g(x) =

x_±OO

%-3-

~ 00,

I

00,

and lim f(x) = %-2+

-00,

00

and lim g(x) = x-3+

00

lim h(x) = -I, lim h(x) = I, lim h(x) = -I, and %-0-

.%_00

lim h(x) ~ I x~'" 76. lim k(x) ~ I, lim k(x) %-1-

~ 00,

and lim k(x) %-1+

~ -00

77. Suppose that f(x) and g(x) are polyoomials 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.

for your answer.

b. t-+ 0-

79. How many horizontal asymptotes can the graph of a given rational function have? Give reasons for your answer.

7) as b. t-+ 0-

t-+ 0+

~/3 + (x - 2 I)2/') as b. x-+O-

c.x-+l+

d.x---+l-

4/') I)

Finding Limits of Differences when x--> ±oo Find the limits in Exercises 80-86.

(v;+9 - v;+4) lim (Vx 2 + 25 - w-=!) lim (w-+3 + x) x_-oo lim (2x + Vrc 4X"2-+--c3x -_--c 2) x_-oo

80. lim

a. x-+O+

(x -

~

00,

78. Suppose that f(x) and g(x) are polyoomials in x. Can the graph of f(x)/g(x) have an asymptote if g(x) is never zero? Give reasons

t---+ 0+

I

and lim f(x)

lim f(x) =

%_-00

X-±OO

e. What, if anything, can be said about the limit as x --> O?

~/3 x

~ -00,

%_-00

c. x-+O-

(:/5 +

-I, and

In Exercises 73-76, fmd a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defmed in pieces if that will help.)

57 r x -3x+2 • un x 3 -2x 2 as

59. lim(2 -

~

.1'--1+ -00

%-1-

and lim f(x) =

];_00

1 + 4 as

c.x-+l+

62. lim(

~

2, f( -I)

lim f(x) =

-

Lx-+-2+

x

~

71. flO) = 0, lim f(x) = 0, lim f(x) = b. x-+O-

61. lim(

~

x~"-

c. x --> V'2

a.

~

lim f(x)

a. x-+O+

60. lim

2x 68·y=x+1

0, f(l) lim f(x) = I

69. flO)

c. x---+ -1+

a.

x+3

=x +2

x~oo

a. x-+ 1+

. x2 56. lun 2x

-3

Inventing Graphs and Functions In Exercises 6<}-72, slretch the graph of a function y = f(x) that satisfies the given conditions. No form:ulas are required--just label the coordinate axes and slretch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.)

54.lim+-as x - I

55. lim(x; -

95

Limits Involving Infinity; Asymptotes of Graphs

x~oo

81. 82.

as

83.

a. x ---+ 0+

b. x -+ 0-

c. x-+ 1+

d. x-+ 1-

x~oo

84. lim (V9x 2 x~oo

-

x - 3x)

96

Chapter 2: Limits and Continuity

(VX 2 + 3x - VX 2 - 2x) lim (Vx 2 + x - ~)

Use the formal defmitioos from Exercise 93 to prove the lintit statemeats in Exercises 94-98.

85. lim

x~oo

86.

94. lim

X~OO

x-o+

Using the Formal Deflnltlons Use the formal defmitioos of limits as x---> ±oo to establish the limits in Exercises 87 and 88. 87. Iff baa the coostant value f(x) = k, thea lim f(x) = k.

xl ~

95. lim

00

96. lim _1_ ~

-00

98. lim _1_ ~ %-1-1 - x 2

00

2

x_2-X -

1~

-00

x_o-x

97. lim _1_ ~

2

x_2+X -

00

X~OO

88. Iff baa the coostant value f(x) = k, thea lim f(x) = k. X~-OO

Use formal defmitions to prove the lintit statemeats in Exercises 8l}-92.

-I 89. lim -2 =

x-o

91. lim

x~3

90. lim

-00

-2 = (x - 3)'

x-

I -I1 =

00

x-o x

X

92. lim

-00

X~-,

(x

Oblique Asymptotes Grapb the rational functioos in Exercises 9l}-104. Include the grapbs and equations of the asymptotes. x2 x 2 +1 99. y ~ - - I 100. y ~ - - I

I

+ 5)2

=

101. y ~

x2

x-

x 2 -1

-4

-.t=T

102. y ~ 2x

+4

00

93. Here is the defmition oflnflnite rigbt-band limit.

x2

-1

103. y = - x -

104. y =

x 3 +1 --2X

Additional Graphing Exerdses

D Grapb We say that f(x) approacbes infmity as x approacbes Xo from the right, and write lim f(x) =

x-xo+

x

106. y

105. y = _ ,.-------, v4 - x 2

00,

if. for every positive real number B. there exists a corresponding nomber 8 > 0 sucb that for all x xo<x<xo+8

the curves in Exercises 105-108. Explain the relationship betweea the curve's formula and what you see.

f(x) > B.

107. y =

x 2/ 3

-I = _,.-------, 2

v4 - x

+ ~f3

108. y

x

= sin

(+-) x + I

D Grapb the functions in Exercises 109 and 110. Thea answer the following questioos.

a. How does the graph bebave as x ---> o+? Modify the defmition to cover the following cases.

a.

lim_f(x)

b. How does the graph behave as x ---> ± oo?

~ 00

c. How does the graph behave near x = I and x = -I?

X-Xo

lim f(x)

= -00

Give reasons for your answers.

c. limj(x)

= -00

109. y

b.

x-xo+ %-Xo

Chapter

r

~ &(x - ~

3(

X )2/3

110'Y~2 x - I

Questions to Guide Your Review

1. What is the average rate of change ofthe 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 calculatiog lintits? Give exam-

2. What lintit must be calculated to fmd the rate of cbaoge 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 does not exist? Give examples.

3. Give an informal or inmitive defmition of the lintit lim f(x) = L. x~X<>

Why is the definition "informal''? Give examples. 4. Does the existence and value of the limit of a function f(x) as x approaches Xo ever depead on what bappeos at x ~ xo? Explain and give examples. 5. What function behaviors might occur for which the limit may fail to exist? Give examples.

ples ofbow the theorems are used

8. What is the value oflim._o «(sinO)/O)? Does it matter whether 0 is measured in degrees or radians? Explain. 9. What exactly does lim~x<> f(x) ~ L mean? Give an example in whicb you fmd a 8 > 0 for a givea f, L, xo, and. > 0 in the precise defmition oflintit.

10. Give precise defmitions of the following statements. a. lim~2- f(x) = 5

b. lim~2' f(x) = 5

c. lim~2 f(x) =

d. lim~2 f(x) = -

00

00

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 function is continuous? 13. What does it mean for a function to be right-continuous at a point? Left-continuous? How are continuity and onl>-sided continuity related?

14. What does it mean for a function 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 grapbingandsolvingtheequationj(x) ~ O? 17. Under what circumstances can you extend a function j(x) to be continuous at a point x = c? Give an example. 18. What exact1y do limr _ oo j(x) = L and limr-~ j(x) = L mean? Give examples. 19. What are limr_± k (k a constant) and limr-± (I/x)? How do you extend these results to other functions? Give examples. 20. How do you rmd the limit of a rational function as x ---+ ± oo? Give examples. 21. What are horizontal and vertical asymptotes? Give examples.

Practice Exerdses

Limns and Continuity 1. Graph the function

In Exercises 5 and 6, rmd the value that limr-Og(x) must have if the given lintit statements hold.

I,

j(x) =

I

-~:

-x, I,

x:5

-I

5. lim (4 - xg(x)) = I

-1

<x<

x-O

0

X~O

~

= x'/3 = x-'/3

a. j(x) e. h(x)

O<x<1

x

6. lim (Xlim g(x)) = 2

x-o

x--4

7. On what intervals are the following functions continuous?

1.

b. g(x)

d. k(x)

= X'/4 = X-l/.

8. On what intervals are the following functions continuous?

Then discuss, in detail, limits, onl>-sided lintits, continuity, and one-sided continuity of j at x = -I, 0, and I. Are any of the discontinuities removable? Explain.

a. j(x)

= tanx

b. g(x)

= csex

e. h(x)

= xc~~

d. k(x)

= s~x

Finding limits

2. Repeat the instructions of Exercise I for

In Exercises 9-24, rmd the lintit or explain why it does not exist.

0, j(x) ~

I/x, 0, 1,

1

x:5 -1

0< Ixl < 1 x =I x>1.

9 lim x'-4x+4 • x 3 + 5x 2 - 14x b. asx----+2

a. asx----+O

x

10. lim ,

+x

2

4'

x+2x+x

3. Suppose that j(l) and g(1) are defined for all 1 and that lim,....,. j(l) ~ -7 and lim,.....g(l) ~ O. Find the lintit as 1---+10 of the following functions.

a. 3j(l)

b. (j(I)'f

e. j(I)' g(l)

j(l) d. g(l) _ 7

e. cos (g(1)) go j(l) + g(l)

f.

IMI

h. l/j(l)

4. Suppose the functions j(x) and g(x) are dermed for all x and that limr-oj(x) = 1/2 and limr_Og(x) = Find the lintits as x ---+ 0 of the following functions.

v'2.

a. -g(x)

b. g(x)' j(x)

e. j(x) + g(x)

d. I/j(x)

e. x

+ j(x)

97

f.

j(x)' cosx 1 x

b. asx----+-l

a. asx----+O

11. lim I x-I 1

, ,

v;.

12. lim '"----=-"-x-a x 4 - a 4

x

(x + hf - x' 13. lim h-O

h

1

1

(x + hf - x' r~O

----

15.lim2+x

x

x-O

2

1/3 - I 17. lim .. x~--r~' 1 . tan (2x) 19. x-O Ion tan ('TT'X )

v;. -

21.

x~ sin (~ + sin x)

23. lim 3 . x-O

ax

smx-x

h

14. lim 16. lim

(2

r~O

+ x)' x

- 8

'/3 - 16 x~----'''' 18. lim .. r~64

v;. -

8

20. lim cscx X-1r-

22. lim cos' (x - tanx) r~"

24. lim cos 2x - I x-o smx

98

Chapter 2: Limits and Continuity

In Exercises 25-28, fmd lhe limit of g(x) as x approaches lhe indicated value.

h. Solve lhe equatioo/(6) ~ 0 graphically with an error of magoitude at most 10-4.

25. lim (4g(x))I/3 = 2

c. It can he shown lhatlhe exaclvalue oflhe solutioo in part (h) is

26.

x-o+

.

27 lim

x~l

3x2+ I ~ g (X )

00

lim

x~y,

x

I

+ g(x)

28. lim 5-x x~-2

2

'V'iW

=2

[19 _I )1/3 _ ([19 )'/3 ("1/27 "1/27 + I .

~o

Evaluate this exact answer and COO3pare it wilh the value you found in part (h).

Continuous Extension 29. Can f(x) ~ x(x 2 - 1)/ Ix 2 - II he extended to he cootinuous at x = I or -I? Give reasoos for your answers. (Graph the functioo-you will fmd lhe graph interesting.)

Limits at Infinity Find lhe limits in Exercises 37-46.

o.

39.

D In Exercises 31-34, graph lhe functioo to see whether it appears to

41.

and Zoom to fmd a good candidate for lhe exteoded functioo's value at Q. If the function does not appear to have a continuous extension, can it he exteoded to he cootinuous from 1he right or left? !f so, what do you lhink lhe extended functioo's value should be?

1() x

x-I = x _ ~,

a = I

() 5cos6 32. g 6 = 46 2".'

lim x_- oo

have a continuous extension to the given point a. If it does, use Trace

31.

43

/

a = ". 2

D

Roots 35. l.et/(x)

+ 110 '/', a

= 0 34. k(x) = I

-:"zlxl'

+

42. lim

= x3 -

+ 3 x + 128

4

(!fyou have a grapher, try graphiog ~ x(cos (l/x) - I) near 1he origin to "see" lhe limit at infinity.)

I(x)

(J

+ sinx :- 2~ X

x

(!fyou have a grapher, try graphlog the function for -5 :5 x :5 5.)

lxJ

x_OO

3 7

I

2

x_co 12x 3

1

lim sinx

• x~oo

45. lim x

+ 8m x

23 / x_OO x 2j3

46. lim

x

+ X-I

+ cos2X

Horizontal and Vertical Asymptotes 47. Use limits to determine lhe equations for all vertical asymptotes.

b. Solve lhe equatioo I(x) = 0 graphically wilh an error of magoitude at most IO-s .

a.

2 =x +4 y x-3

c. It can he shown lhat lhe exact value of lhe solutioo in part (h) is

f3 (1- v69)' (12 + v69)'f3 18 + 2 18 .

xl

+X

2 b.f(x)=x -x-2 2 x -2x+1 -

6

c'Y~x2+2x_8

48. Use limits to determine the equations for all horizontsl asymptotes.

Evaluate this exact answer and compare it wilh lhe value you found in part (h).

D

7

2

X

2x' + +

x_co x - 7x + 1

lim ~

x_-oo

lim

x_-oo 5x 2

40. lim

3x

a= 0

hetweeo -I and 2.

a. y

l.et 1(6) ~ Ii' - 2IJ + 2. a. Use lhe Intermediate Value Theorem to show that 1 has a zero hetweeo -2 and O.

Chapter

38.

44. lim cos 6 - I

X - I. a. Use lhe Intermediate Value Theorem to show that 1 has a zero

D 36.

3 7

x2-4~+8

8_00

33. h(l) = (I

+ +

37. lim 2x x_oo 5x

30. Explain why lhe functioo I(x) = sin(l/x) has 00 cootinuous extension to x =

~

c. g(x)

2 I - x 2 x + I ~

'\1'7+4 x

h. I(x)

~ ~+4

d. y ~

"1/R+9 9x2+1

y';+4

Additional and Advanced Exercises b. Graph lhe function y = XX for 0 < X :5 I. Even lhough lhe function is not defmed for x :5 0, lhe graph will approach lhe y-axis from lhe right. Toward what y-value does it seem to be headed? ZoOO3 in to further support your idea.

I. Assigning a value to O' The rules of exponeots tell us lhat a' = I if a is any mnnber 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 conflictiog results. The fll'St rule would say O' ~ I, whercsslhe second would say O' = o. We are not dealing with a question of right or wroog 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 loog as we could persuade olhers to agree. What value would you like O' to have? Here is an example lhat might help you to decide. (See Exercise 2 below for anolher example.)

D

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

I(x) ~ (

xI )'/(InX)

as x increases? Here are two ways to f'md out.

a. Calculate rfor x = 0.1,0.01,0.001, and so 00 as far as your calculator can go. Record lhe values you get. What pattern do you see?

a. Evaluate 1 forx = 10, 100, 1000, and so 00 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 cm3) 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

stripes~~

~ 2.

'--=

about Imm wide

L = L O\jl -

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.

...-------.-...,. ,/

/

Liquid volume V = 361Th

(b)

r

x Exit rate y ft 3/min

Suppose that y = v;,/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 = 1 ft3/min? b. within 0.1 ft3/min of the rate Yo = 1 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 of laboratory 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 1-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

A l-L measuring cup (a), modeled as a right circular cylinder (b) of radius r = 6 cm

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,

= 1

Xo

9. h(x) =~,

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 1 and limx~xo f(x) = L2, then L 1 = L 2. 12. Prove the limit Constant Multiple Rule: lim kf(x) = k lim f(x) x~c

for any constant k.

x~c

13. One-sided limits find a. limx~o+ f(x 3 c. limx~o+ f(x

2

-

If limx~o+ f(x) = A and limx~o- f(x) = B,

x) x

4

)

b. limx~o- f(x 3

-

x)

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.

Iflimx~a

f(x) exists but limx~a g(x) does not exist, then + g (x)) does not exist.

limx~a(f(x)

b. Ifneither limx~a f(x) nor limx~a g(x) exists, then limx~a (f(x) + g(x)) does not exist. c. If f is continuous at x, then so is If I. d. Ifl f I is continuous at a, then so is f.

100

Chapter 2: Limits and Continuity

In Exercises IS and 16, use the formal defmition 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 continu.ous at only one point Let

()=

f x

{x, ifxis ratiooa1

0, if x is irratiooa1.

a. Show that f is continuous at x

= O.

b. Use the fact that everynonempty open interval of real num-

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 common factors greater than I. (We say that such a fraction is in lowest terms. For example, 6/4 written in lowest terms is 3/2.) Letf(x) be defmed for all x in the interval [0, I] by

= {I/n, ~x.~ .m/~ is a rational number in lowest terms 0,

a. Show that f is bounded onD if and only if there exists a numberBsuchthat If(x) I '" B for allxinD.

b. Suppose that f is bounded from above by N. Show that if lim_xof(x) = L, thenL '" N. c. Suppose that f is bounded from below by M. Show that if lim_xof(x) ~ L, thenL '" M. 24. Max {a, b} sod min {a, b} a. Show that the expression

bers contains both ratiooa1 and irratiooa1 numbers to show that f is not continuous at any nonzero value of x.

f(x)

exists, a lower bound for fonD and say that f is bounded from below by M. We say that f is bounded on D ifit is bounded from both above and below.

if x IS rrrational.

Forinstance,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, and so on.

a +b max {a, h} ~ - 2 -

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 (diametrically opposite) points on 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

21. Root. ofa 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: _ -I T+ ( a ) -

+~ a

'

r_(a)

~

-I-~ a .

Generalized Limits involving

b. What happens to r_(a) as a ---> 01 As a ---> -1+1

c. Support your conclusions by graphing r+(a) andr_(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, and 0.05.

+ 2 cos x

Ii:

8

The formula lime_o(sin 8)/8 = I can be genera1ized. If limx _, f(x) = 0 and f(x) is never zero in an open interval containing the point x = c, except possibly c itself, then lim sinf(x) = I f(x) .

x~,

Here are several examples.

.

a. lim. 8mx

,

=

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 lun • x-+-l X + 1 x-+-l (x 2 - X - 2) . (x' - x - 2) . (x + I)(x - 2) lim =1' lim =-3 x

+1

x--l

x

+1

. sin (I - '\!'x) . sin (I - '\!'x) I - '\!'x d. lun = lun = x-I x-I x-I 1_ ~ x-I

I. lim (I - '\!'x) (I + '\!'x) x~l (x - 1)(1 + '\!'x)

lim I-x x~l (x - 1)(1 +

'\!'x)

Find the limits in Exercises 25-30. sin (I - cosx) 25. lim x

a. What happens to r+(a) as a ---> 01 As a ---> -1+1

22. Root of so equation Show that the equation x has at least one solution.

2

b. Find a similar expression for min {a, h}, the smaller of a andb.

x--l

lim f(x)g(x). x~,

la - bl

equals a if a '" b and equals h if h '" a. In other words, max {a, hI gives the larger of the two numbers a andh.

a. Show that f is discontinuous 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 fmitely manyratiooa1 numbers rin [0, I] such thatf(r) '" E.)

+

= 0

23. Bounded function. A real-valued function f is bounded from above on a set D if there exists a number N such that f(x) '" N for an x in D. We call N, when it exists, an upper bound for f on D and say that f is bounded from above by N. In a similar manner, we say that f is bounded from below on D if there exists a number M such that f(x) '" M for an x in D. We call M, when it

x~o

27. lim

sin (sinx)

x

x~o

29. lim

x-2

28. lim

sin (x'

x

x~o

+ x)

sin ( '\!'x

sin (x' - 4) 2

30. lim

x

x-9

x

- 3) 9

Oblique Asymptote. Find all possible oblique asymptotes 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

101

Technology Application Projects

Mathematica,tMap1e Modules: 1IIke It to the Limit Part I Part II (Zero Raised to the Power Zero: What Does it Mean?) Part III (One-Sided Limits) Visualize and interpret the limit concept through graphical and numerical explorations.

Part IV (What a Difference a Power 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 (Rates of Growth) 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 function, 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.

Tangents and the Derivative at a Point

3.1

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). Ifthe 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. (xo, f(xo))

: _______ J

~-~',

h

: I

~-------''------......I.....--------~

o

FIGURE 3.1 The slope of the tangent r t p. r f(xo + h) - f(xo) Ine a IS h~ h

X

DEFINITIONS

The slope of the curve y

==

f{x) at the point P{xo, f{xo)) is the

number

. m==hm

h~O

f{xo

+ h) - f{xo) h

(provided the limit exists).

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

y=l slope is-~

I .'

(a) Find the slope of the curve y = I/x at any point x = a oF 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

I slope is-l

(a) Here fix) = I/x. The slope at (a, I/a) is

I

atx =-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).

-h

I·

h~ ha(a + h)

lim

1 a - (a + h)

h~O

h a(a + h)

lim

=

-I

h~O a(a + h)

I

a

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 4

=

I

+ hh - a =

a

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 = -I, the slope is -1/(-1)2 = -I (Figure 3.2).

y

slopels. 1

= lim h~O

(2. ~)

~'----t-~x sl . 1 opels-"4

FIGURE 3.3 The two tangent lines to y = I/x having slope -1/4 (Example I).

I a2

I 4'

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 oF 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 f 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 f 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 = fix) 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 1sec, 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? We let f(l) = 161 2. The average speed of the rock over the interval between 1 = 1 and 1 = 1 + h seconds, for h > 0, was found tu be

Solution

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

Tangent Lines with Spedfied Slopes

In Exercises 1-4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points PI

At what points do the graphs of the functions in Exercises 23 and 24 have horizontal tangeats?

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 tangeot to the curve y ~ I/(x - I). 26. Find an equatioo of the straight line having slope 1/4 that is tangeat to the curve y = ~.

1 I-

Rates of Change

r

27. Object dropped from a tower An o!!iec! is dropped from the top of a 100-m-high tower. Its height above ground after 1 sec is 100 - 4.91' m. How fast is it falling 2 sec after it is dropped? 28. Speed of a rocket At 1 sec after liftoff, the height of a rocket is 31' ft. How fast is the rocket climbing 10 sec after liftoff?

3.

4.

y

29. Circle's changing area What is the rate of chauge ofthe area of a circle (A ~ 'lTr') with respect to the radius whea the radius is r = 3?

y

30. Ball's changing volume What is the rate of change of the volume ofa ball (V ~ (4/3)'lTr') with respect to the radius whea the radius is r = 2?

3

31. Show that the line y = point (xo, nlXo + b).

nIX

+b

is its cwo tangeat line at any

32. Find the slope of the tangeat to the curve y ~ I/~ at the point where x = 4.

o

Testtng for Tangents 33. Does the graph of

In Exercises 5-10, fmd an equatioo for the tangeat to the curve at the givea point. Thea sketch the curve and tangeot together.

5. y ~ 4 - x', 7. y

= 2~,

9. y

= x',

(-1,3) (1,2)

(-2, -8)

6. y ~ (x - I)' 8. y 10. y

=

1"x

+

(I, I)

I,

34. Does the graph of

(-2,-t)

= :',

+ I,

(2,5)

x 13. g(x) = x _ 2' 15. h(l) ~ I',

(3,3)

(2,8)

17. f(x) =~, (4,2)

12. f(x) ~ x - 2>;',

8

14. g(x) = x"

=

5x 2 ,

X =

I 21. y = x _ I'

-1 x = 3

20. y

=

+ 31, (1,4) v;+l, (8,3)

1 - x 2,

x - I 22'Y=x+I'

x ~ 0 x =0

have a tangent at the origin? Give reasons for your answer.

Vertical Tangents (2,2)

In Exercises 19-22, find the slope of the curve at the point indicated. 19. y

g(x) ~ {x sin (1Ix), 0,

(1,-1)

16. h(l) ~ I' 18. f(x) =

x ~ 0 x= 0

have a tangent at the origin? Give reasons for your answer.

(-I, I)

In Exercises 11-18, fmd the slope of the function's graph at the givea point. Thea find an equatioo for the line tangeat to the graph there. 11. f(x) ~ x'

f(x) ~ {x' sin (1Ix), 0,

X =

2

x=O

We say that a cootinuous curve y ~ f(x) has a vertical tangent at the point where x = Xo if lim._o (/(xo + h) - f(xo))lh = 00 or -00. For example, y ~ xli' has a vertical tangeat at x ~ 0 (see accompanying figure): lim f(O h-O

+ h) - f(O) h = lim _1_ = 4-0 h2j3

00

.

106

Chapter 3: Differentiation y

36. Does lbe graph of

{O,1,

U(x) =

x<

0

x ~ 0 have a vertical tangent at 1he point (0, I)? Give reasons for your

answer.

-----cf----~x

D Graph 1he curves in Exercises 37-46. a. Where do 1he graphs appear to have vertical tangents? b. Conrrrrn your fmdings in part (a) wi1h limit calculations. But before you do, read 1he introduction to Exercises 35 and 36.

VERTICAL TANGENr AT ORIGIN

However, y = x'/' has no vertical tangent at x = 0 (see next figure):

. g(O + h) - g(O) 11m h-O h

~

. ..h'_/3~-----"O 1Imh-O h

37. 39. 41. 43.

y

= x'/'

y

45. y

= x'/' =

= lim _1_ 11-0

hlj3

= x'/' 40. y ~ x'/'

38. y

y ~ x /' ' y ~ 4x'/' -

a

42. y ~ x'/3 - 5x'/'

- (x - 1)'/'

{-vfxI, ~,

44. y

x'" 0 x>O

46. y

= x ' /3 + (x

- 1)'/3

= ~4 - xl

COMPUTER EXPLORATIONS

does not exist, because lbe limit is 00 from lbe right and left.

00

from lbe

Use a CAS to perform 1he following steps for lbe functions in Exercises 47-50:

a. Ploty = j(x) over1heinterval (xo - 1/2) '" x'" (xo

+ 3).

b. Holding Xo fixed, 1he difference quotient

-----=-f----~x

o

into your CAS workspace. e. Find lbe limit of q a, h ---. O.

NOVFRI1CAL TANGENT KI ORIGIN

d. Defme 1he secant lines y = j(xo) + q' (x - xo) for h = 3, 2, and I. Graph 1h= toge1her wi1h j and 1he tangent line over 1he interval in part (a).

35. Does lbe graph of

-I

j(x) =

{

xO

0:

have a vertical tangent at lbe origin? Give reasons for your answer.

3.2

= j(xo + h)

- j(Xo) h at Xo becomes a function ofthe step size h. Enter this function q(h)

47. j(x) = x' + a, Xo = 0 48. j(x) = x + 49. j(x) ~ x + sin (a), Xo ~ 7f/2 SO. j(x) = cosx + 4 sin (a), Xo = 7f

5 x' Xo

= I

The Derivative as a Function In 1he last section we dermed the derivative of y = f(x) at the point x = Xo to be the limit

H!SlORICAL ESSAY

j'(xo)

=

lim f(xo

+ h) - f(xo)

h~O

The Derivative

h

We now investigate the derivative as a function derived from f by considering the limit at each point x in the domain of f.

DEFINmON The derivative of the function f(x) with respect to the variable x is the function j' whose value at x is j'(x)

=

lim f(x h~O

provided the limit exists.

+

hl-

f(x),

107

3.2 The Oenvalive as a Function

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 dilTerentiable (has a derivative) at x. If I' exists at every point in the domain of f, we call f dilTerentiabIe. If we write z = x + h, then h = z - x and h approaches 0 ifand 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.

T

f(z) - f(x)

~"'--------i,~ ~h =

z-

X~:

I

I

.

.

--~z-x+h -~-

x

Alternative Formula for the Derivative

Derivative off at x is I'(x) ~ lim f(x + h) - f(x) IHO h ~

I'(x)

=

lim f(z) - f(x) .t-+X

FIGURE 3.4 quotient.

Z

lim f(z; - ~(x) . z~x

X

Two fonns for lhe difference

Calculating Derivatives from the Definition The process of calculating a derivative is called dilTerentiation. 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 Reeiproeal Funetion

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

L(l) ~ _l x"

~(l) =_l dx x x2 '

dx

X

x2'

0

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

=

(x + h) (x + h) _ I ,so

f(x + h) - f(x) h x+h

lim x

h~O

=

f(x + h)

+

h - I h

+ h) and

Defmition

x x - I

---

. I (x + h)(x - I) - x(x + h - I) l~ h . -'-----(c-x'--+----O-h--'--I~)-;-(x'-------=-I )c----'-

= lim h-O

1.

h (x

= lim h-O

(x

-h

SimplifY

+ h - I)(x - I)

+h

-I - I)(x - I)

-I (X-i)2'

Cancelli

'¢

0

•

108

Chapter 3: Differentiation

EXAMPLE 2

v'x for x > O. Find the tangent line to the curve y = v'x at x


(bl

I

4.

=

Solution

Derivative of the Square Root

Function

:fx y; =


I'(xl = lim f(z; - t(x) z-x

.1 r' x> 0 2vx

= lim

z-x

y

= lim

z-x

Vi - v'x (Vi - v'x)( Vi + v'x) I

_1_

Vi + v'x

2v'x .

(b) The slope ofthe curve at x = 4 is

1'(4) = _1_ =

2V4

FIGURE 3.5 The curve y = YX and its tangent at (4, 2). The tangent's slope is found by evaluating the derivative at x = 4 (Example 2).

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 +

•

1.

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 x=a =

d/l

I'(a) = dx

-

dx

= -

d f(x) I

dx

x=a

.

x=a

For instance, in Example 2

1'(4) =

~v'xl dx

= x=4

_I_I 2v'x

= x=4

_1_

2V4

=

1 4.

Graphing the Derivative We can often make a reasonable plot ofthe 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 ofthe function y = I(x) in Figure 3.6a.

Solution 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. • 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; 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. 2.

y

Slope 0 A

y ~f(X)/

.....

-10 "

Slo

",-1

B,\

'-

C

5-

/Slope- ~ = 2y-uni fs!x-ooit

: :I

Slope _1

4

- 8y-units

/

Slope 0

- 4 x-units I

10

5

0

I

15 I

I

(a)

x

I

I

I

Slope

4

y ~f'(xY

3

2 1

A'

D'

-1

;'

~~10 e'

-2

V~cal

coordinate

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 = I'(x) in (h) by plottiog slopes from lbe graph ofy = f(x) in (a). The vertical coordioste of B' is the slope atB aod so on. In (h) we see lbat lbe rate ofchange off is negative for x between A' aodD'; lberate 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 difJerentiable on an open interval (finite or inimite) if it has a derivative at each point of the interval. It is difJerentiable on a closed interval [a, b] if it is differentiable on the interior (a, b) and ifthe limits ----L---'-c-----o-"--o---L-~>X

a

a+h h>O

b+h h
FIGURE 3.7 Derivatives at endpoints are one-sided limits.

liro I(a h-O+

liro I(b h~O-

+ h) - I(a)

Right-hand derivative at a

h

+ h) - I(b) h

exist at the endpoints (Figure 3.7).

Left-hand derivative at b

110

Chapter 3: Differentiation Right-hand and left-hand derivatives may be deImed at any point of a function's domain. Because of Theorem 6, Section 2.4, a function bas 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 = mx right ofthe origin,

+b

is the slope m. Thus, to the

--------;~---_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 wbere the graph has a "comer" (Example 4).

d -(Ixl) dx

=

d

-(-x)

dx

=

d -(-I·x) dx

=-1

(Figure 3.8). There is no derivative at the origin because the one-sided derivatives differ there: Right-hand derivative of IxIat zero = lim

10 + hl-IOI h

h-O+

Ihl

Left-hand derivative of IxIat zero = lim

h

Ihl

= lim-l =-1. h---+O-

EXAMPLE 5

In Example 2 we found that for x

>

~

10 + hl-IOI

h--+O-

=

~

Ihl

lim -h

h-O+

hwhcnh

=

>0

Ihl

lim -h

h--+O-

-h whcnh < 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 ofthe secant lines joining the origin to the points (h, v'l.) on a graphofy = vX approach 00, the graph bas 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 bas a derivative at a point Xo if the slopes ofthe secant lines through P(XO, j(xo» and a nearby point Q on the graph approach a Imite limit as Q approaches P. Whenever the secants fail to take up a limiting 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 and - 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 rapidly 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-DlfferentiabllUy Implies Continuity

x

= C,

then / is continuous at x

=

If

f

has a derivative at

c.

Proof Given that t' (c) exists, we must show that limx _, f(x) = f( c), or equivalently, thatlimh_of(c + h) = f(c).Ifh oF O,then f(c

+ h)

= f(c)

+

(f(c

= f(c)

+

f(c

+ h)

+ h~

- f(c)) - f(c) 'h.

112

Chapter 3: Differentiation

Now take limits as h --+ O. By Theorem I of Section 2.2, lim f(c + h) = lim f(c) + lim f(c + h) - f(c) • lim h h-O

h-O

h-O

=

f(c) + I'(c)· 0

=

f(c) + 0

=

f(c).

h

h-O

•

Similar arguments with one-sided limits show that if f has a derivative from one side (right or left) at x = c then f 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 ofthe functions in Exercises I-j), Then Imd the values ofthe derivatives as specified.

In Exercises 17-18, differentiate the functions. Theo Imd an equation

of the tangeot line at the indicated point on the graph of the function.

S

I. j(x) = 4 - x 2 ; 1'(-3),1'(0),1'(1)

+

2. F(x) ~ (x - 1)2

3. g(l) =

1,; 1

4. k(z) = I 5. p(O) ~ 6. r(.)

I;

F'(-I),F'(0),F'(2)

g'( -I), g'(2), g'( v'3)

~/;

v30;

k'( -I), k'(l), k'( VZ)

p'(l),p'(3),p'(2/3)

= 'V'2,;+1;

r'(O), r'(I), r'(1/2)

In Exercises 7-12, Imd the indicated derivatives.

f' 7· dy , ' y~2x dx

dx 9. dl dp 11. dq

1 if '~21+1 if p=

I

yq:tl

8. dr dx

,'f r

~ .' -

10 dv • dt

if v

= t -

dz 12. dw

if z=

29 2

+3

I

t

I v'3w - 2

Using the Alternative Formula for Derivatives Use the formula I'(x) ~ lim j(z) - j(x)

Slopes and Tangent Lines

%-x

In Exercises 13-16, differentiate the functions and Imd the slope of

the tangeot line at the giveo value of the indepeodeot vatiable.

13. j(x) = x + 14. k(x) = 2

I

9

x' x

+ x' x

15.• = I' - 12,

x+3

= -3 = 2

t = -I

16. Y = -I- , x = -2

-x

Z

X

to Imd the derivative of the functions in Exercises 23-26. I

23. j(x) = x + 2 24. j(x) = x 2

-

]x

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 1 over the closed interval [- 2, 5]. i) The graph of1 is made of closed line segmoots joined ood to ood.

ii) The graph starts at the point (- 2, 3). iii) The derivative of 1 is the step function in the figore shown here. y' (b)

(0)

,

1

0

y'

0

-2

3

x

5

,

-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 annual percootage change y ~ 1(t) in the u.s. gross natioua1 product (GNP) for the years 1983-1988. Graph dy/dt (where defmed). 7% ,------..----,----,----,---,

6 -----",.r-<---~x

5

-----.
o

4

3 2

1 0'-

29.

30.

y

y

-'

1983 1984 1985 1986 1987 1988 34. Fruit m.. (Continuation of Example 4, Section 2.1.) Populations starting out in closed environmoots grow slowly at fust, whoo there are relatively few members, thoo more rapidly as the

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

number of reproducing individuals increases and resources are still ahundant, thoo slowly again as the population reaches the carrying capacity of the environmoot

31. a. The graph in the accompanying figore is made of line segmoots joined ood to ood. At which points of the interval [-4, 6] is f' not defined? Give reasons for your answer.

a. Use the graphical technique of Example 3 to graph the derivative of the fruit fly population. The graph of the population is reproduced here. p

y

,...

350

~

300

V

250

/

'a 200

~ 150

/

3 100

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 days do.. 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, hetweeo 6 A.M. and 6 P.M.

39.

40.

Y

Y

Y ~ J(x)

T

P(I, I)

80

G:'

.....

70

---7--7----x

"-

/ I-' ~

'"

----cn+--7----- X

o

3 9 A.M.

6 12 NOON

9 3 P.M.

12 6 P.M.

41. f(x) =

Tune (brs)

a. Estimate the rate of temperature change at the times i) 7 A.M. ii) 9 A.M. iii) 2 P.M. iv) 4 P.M. b. At what time does the temperature increase most rapidly? Decrease most rapidly? What is the rate for each of those times? c. Use the graphical technique ofEXlUIlple 3 to graph the derivative oftemperature 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.orglwiki/Jared_Fogle). A chart showing his possible dramatic weight loss is given in the accompanying figure.

42. g(x) =

{lx +- lxI, + 7, x2

{

X

xli',

x<

<0

0

DIfferentiability and Continuity on an Interval

a. differeotiable? b. continuous but not differeotiable? e. neither continuous nor differeotiable?

Give reasons for your answers. 44. Y

Y

D: -3:Sx:s2

425

2

~-c!c---!'~too---:-----;~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:Sxs3

'" .....

300 200

~

x

Each figure in Exercises 43-48 shows the graph of a function over a closed interval D. At what domain pointa does the function appear to be

Y ~J(x)

500

x~o

~ 0

2 3 X / ,

43.

w

i'n

y=x

O

In Exercises 41 and 42, determine if the piecewise defined function is differeotiable at the origin.

6 A.M.

~

y=l

ii)t~4

Y

Y Y ~J(x)

iii)t~l1

Y ~J(x) D: -2 S

D: -3:s x:s 3

b. Wheo does Jared lose weight most rapidly and what is this rate of weight loss?

x:s 3

e. Use the graphical technique ofEXlUIlple 3 to graph the derivative ofweight W.

One-Sided Derivatives Compute the right-hand and left-hand derivatives as limita to show that the functions in Exercises 37-40 are not differeotiable at the point P. 37.

Y

38.

Y

47.

Y ~ J(x)

Y~J(x)

Y~2x Y~ 2

48. Y

2

Y ~J(x) D: -1:s x:s 2

P(I.2)

P(O, 0)

-----c0;;t--~I--;2;-~x

-"-------;+----'-----!'---->x -I 0 2

Y Y ~J(x) D: -3s x s 3

3.3

D 60.

Theory and Examples In Exercises 4~52,

115

Graph y ~ 3x in a window that has - 2 '" x '" 2, 0 '" Y '" 3. 2

Then, on the same screen, graph

a. Find the derivative j'(x) of the given function y = f(x). 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 ofx-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 ahout this relationship in Section 4.3.)

49. Y

Differentiation Rules

= -x 2

SO. Y

51. y ~ x'/3

= -I/x

52. Y ~ x 4/4

53. Tangent to a parabola Does the parabola y = 2x 2 - 13x + 5 have a tangent wbose slope is -I ? If so, find an equation for the line and the point of tangency. Ifnot, wby 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, wbynot?

(x + h)' - x' Y =

h

for h = 2, 1,0.2. Then try h = -2, -I, -0.2. Explain wbat is goingon. 61. Derivative of y = Ixl Graph the derivative of f(x) = 14 Then graphy = Gxl- O)/(x - 0) = lxi/x. What can you conclude?

D 62.

Weierstrass's nowhere differentiable continu.ous function The som of the rltst 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 snm. Zoom in several times. How wiggly and bnmpy is this graph? SpecifY a viewing window in which the displayed portion of the graph is smooth.

COMPUTER EXPLORATIONS

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.

Use a CAS to perform the following steps for the functions in Exercises 63--{j8.

56. Derivative of multiples Does knowing that a function g(t) is differentiable at t = 7 tell you anything ahout the differentiability of the function 3g at t ~ 7? Give reasons for your answer.

b. Derme the difference quotient q at a general point x, with general step size h.

57. Limit of a quotient Suppose that functions g(t) and h(t) are dermed for all values of t and g(O) = h(O) = O. Can Itm,....o (g(t))/(h(t)) exist? If it does exist, must it equal zero?

d. Substi1ute the value x = Xo and plot the function y = f(x) together with its tangent line at that point.

Give reasons for your answers. 2

58. a. Let f(x) be a function satisfYing If(x) I '" x for -I '" x '" I. Show that f is differentiable atx ~ 0 and rmdj'(O).

X2 Sin}, X::F 0

f(x) = {

Graphy ~ 1/(2Vi) in a window that has 0'" the SaJ13e screeo, graph

v'XTI -

for h ~ 1,0.5,0.1. Then try h is going on.

3.3

e. Substitute various values for x larger and smaller than Xo into the formnla obtained in part (c). Do the nombers make sense with your piclure?

x'" 2. Theo, on

65. f(x) =

-2--'

66. f(x) ~

Vi

h ~

64. f(x) =

+ x 2 - x, Xo ~ I x 1/3 + x 2/3, Xo = I

63. f(x) ~ x'

0, x = 0 is differentiable atx = 0 and rmdj'(O).

y=

e. Take the limit as h .... O. What formula does this give?

f. Graph the formnla obtained in part (c). What does it mean wben its values are negative? 'hro? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer.

b. Sbowthat

D 59.

a. Plot y = f(x) to see that function~ global behavior.

-I, -0.5, -0.1. Explain what

4x

Xo = 2

x +I

x - I 3x

2

+

I

'

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 differentiate 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

Derivative of a Constant Function

c

If f has the constant value f(x) = c, then

df d dx = dx (c) = O. ~I-------"---"-_--'-c~-~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 ofa 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 rmd that

t'(x)

lim f(x + h) - f(x) h

=

h-O

= lim c h-O

- c h

= limO = h-O

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

H!SlORICAL 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,

t'(x)

= lim z_x

f(z) - f(x) zx

= lim(z·-l

= lim z' - x· z_x z x

+ z'-":,; + ... + zx·- 2 + x·- 1)

• terms

z~x

• The Power Rule is actoally 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.

3.3

Differentiation Rules

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,-l are defined.

EXAMPLE 1 (a)

Differentiate the following powers ofx.

x'

(d)

~4

(f) Yx2+w

(e) x-4/'

x

Solution (8) :fx(x') = 3x'-1 = 3x 2

(c) :fx(xv'i)

=

(e) ~(x-4/') = dx

(b) :fx(x 2/') = t x (2f3)-l = tx-I/' (d) :fx84) = :fx(x-4) = -4x-4-1 = -4x-s = -:s

v'2xv'i-l

_ix -(4/')-1 3

_i x - 7/'

=

3

(f) :fx(Yx2+w) = :fx(X 1+(W/2») =

(I +

;)x 1+(W/2)-1 =

t(2 +

1T)W

•

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 :fx(cu) = c:. y

In particular, if n is any real number, then

:fx (ex') = cnx,-t. Slope /

3

(I, 3)

~

3(:!x) -00; ~

6(1)

~

6

Proof d

.

cu(x

~_cu = hm

2

Slope

~

:!x

~

2(1)

~

2

+

h-O

UA.

h) - cu(x) h

Derivative def'mition with j(x) ~ ",,(x)

=

. u(x + h) - u(x) dun ---'-----;----'~ h~O h

Constant Multiple Limit Property

=

du c dx

u is differenti.able.

1(1, I)

•

1

EXAMPLE 2

1 1

1

----'~r='----__:_--____!--->x

o

1

2

FIGURE 3.10 Thegraphsofy ~ x'and y = 3x'. Tripling the y-coordinate triples the slope (Example 2).

(8) The derivative formula

~(3X2) dx

= 3·2:< = 6x

says that if we rescale the graph of y = x'bymultiplying eachy-coordinate by 3, then we multiply the slope at each point by 3 (Figure 3.10).

118

Chapter 3: Differentiation

(b) Negative ora function The derivative ofthe negative of a differentiable function u is the negative ofthe function's derivative. The Constant Multiple Rule with c = -I gives

•

l£(-u) = l£(-Iou) = -I.l£(u) = _du ~

I

Denoting Fnnctions by u and v The functions we are working with when we need a differentiation formula are likely to be denoted by letters like f aod g.

~

~~.

The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.

Derivative Sum Rule 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,

We do not want to use these same letters when statiog general differentiation rules, so we use letters like u aod v instead that are not likely to be already in use.

d du dv ~(u+v)=~+~.

For example, if y = x 4 + 12x, then y is the sum of u(x) = x 4 and v(x) = 12x. We thenbave

dy d ~ = ~(X4)

d

d ~(I2x) = 4x 3 + 12.

We apply the def"mition of the derivative to f(x) = u(x) + v(x):

Proof ~_

+

() [u x

+v

( ) . [u(x + h) + v(x + h)] - [u(x) + v(x)] x ] = lim h h--i'O

UA,

= lim [u(x + h) - u(x) + v(x + h) - v(x)] h~O

h h . u(x + h) - u(x) . v(x + h) - v(x) = lim h +1Jm h h--i'O

h~O

du

dv

UA,

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:

l£(u _ v) = l£[u + (-I)v] = du + (-I) dv = du _ dv ~

~

~

~

~

~.

The Sum Rule also extends to finite sums of more than two functions. If Ul, "2, ... , Un

are differentiable at x, then so is "1 + d

~ (Ul + U2 + ... + un) =

dUl

Iii

U2

+

+ ... + Un, and

dU2

Iii

+ ... +

dUn

Iii'

For instance, to see that the rule holds for three functions we compute

d

d

d

~ (Ul + U2 + u,) = ~ ((Ul + U2) + u,) = ~ (Ul + U2) +

du,

Iii

=

dUl

Iii

+

dU2

Iii

+

du,

Iii'

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 3 +

= ;"x

3

= 3x 2 +

+ ;"

2 (jcx ) - ;" (5x)

i. 2x 3

+ ;" (I)

- 5 + 0 = 3x 2 +

llx 3

jc x 2 -

5x + 1.

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

=

x4

-

2x 2 + 2 have any horizontal tangents? If so,

The horizontal tangents, if any, occur where the slope dy/dx is zero. We have

dy

Now solve the equation dx

=

0 for x: 4x 3 - 4x = 0

(-I, 1)

-1

(I, 1)

4x(x 2

I) = 0

-

x=O,I,-1.

0

FIGURE 3.11 The curve in Example 4 and its horizontal tangents.

The curve y = x 4 - 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, I). 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 oftheir derivatives, the derivative of the product of two functions is not the product oftheir derivatives. For instance,

d

d

dx (x), dx (x) = I, I = 1.

while

The derivative of a product oftwo 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. Inprime notation, (uv)' = uv' + vu'. In fimction notation,

:fx [f(x)g(x)] = EXAMPLE 5

Find the derivative ofy

+ g(x)j'(x).

f(x)g'(x) = (x 2

+

1)(x 3

+ 3).

Solution

(a) From the Product Rille with u

:fx [(x 2 + l)(x 3 + 3)] =

=

x2

(x 2 4

+ I and v

+

1)(3x 2 ) 2

=

+ 4

x3

(x 3

+ 3, we f'md + 3)(2x)

=3x +3x +2x +fix = 5x 4

+ 3x 2 + fix.

d

-(uv) dx

~


udx

+ vdudx

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

(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 5 + x 3 + 3x 2 + 3

+ h)

= 5x

:

f---------, u(x+h)..1.v /+------'--'---'----. ---

4

+

3x 2

+ 6x.

&v

u(x)v(x)

o

Iv(x) 4u

Proof of the Derivative Product Rule

it (uv)

/

4u \ u(x) u(x + h)

Then the change in the product uv is the difference in areas ofthe larger and smaller "squares," which is the sum ofthe

upper and right-hand reddish-shaded rectangles. That is, .i(uv)

~ ~

u(x u(x

.i(uv) -h-

it( ) _ ' dx

uv -11m

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

= u(x + h)T + v(x)T'

The limit as h .... 0+ gives the Product

h) - u(x)v(x)

h

h-O

To change 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(x)4u. 4V

+ h)v(x +

lim u(x

=

dx

Divisioo by h gives

Rille.

•

This is in agreement with our flISt calculation.

v(x)

h-O

x+

h) I'

.

=

- v(x)

h

+

( ) u(x +vx

h) - v(x)

h

()'

+ v x ·1=

h

h-O

+ h)

- U(X)] u(x

+

h) - u(x)

h

h-O

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 ofthe quotient oftwo 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

it dx

('!)

du =

dv

vfiX - ufiX

if'

V

In function notation,

it [f(X)]

= g(x)t'(x) - f(x)g'(x)

dx g(x)

EXAMPLE 6

Solution

Find the derivative of y

g2(x)'

t2 - I t + I

= -3--'

We apply the Quotient Rule with u dy dt

=

t2

-

I and v

(t 3 + I)' 2t - (t 2 - I). 3t 2 (t 3 + 1)2

+ 2t - 3t 4 + 3t 2 (t 3 + 1)2 -t 4 + 3t 2 + 2t (t 3 + I)'

=

t 3 + I:

~

(Il)

dt v

~ v(du/dt) - u(du/dt)

if

2t 4

•

3.3

Differentiation Rules

121

Proof of the Derivative Quotient Rule u(x + h)

('!)

~

lim v(x + h)

=


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 ~

('!)

=


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

h-O

( ) vx

u(x

h

( ) v(x -ux

+ h)

+ h)

- v(x)

h

----'"----~~~~--'"-----

v(x

+ 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

-

2x)

x'

expand the numerator and divide by x': y=

(x - 1)(x 2

-

2x)

x'

Then use the Sum and Power Rules: dy = -x-2


3( -2)x-3 + 2( -3)x-4

-

6 = -1 -+ -6 - -4 2 3 x

x

x

-

°

Second- and Higher-Order Derivatives If y = j(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 f". So f" = (f')' . The function f" is called the second derivative of j because it is the derivative ofthe rlISt derivative. It is written in several ways: d'y

d (dY )
f"(x) =
=

dy'
=

y"

=

n 2(f)(x)

=

n/ j(x).

The symbol n 2 means the operation of differentiation is performed twice. Ify = x 6 , theny' = 6x' and we have y" =

dy' ax =

d


=

3Ox'.

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 prime" 'y double prime"

y' y" d'y

y(n)

"d squared y dx squared"

dx'

ym

d"y

"d to the n of y by dx to the n"

dx"

"D to the n"

D"

dny

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"

y(n)

d

= dxy(n-l) = dx n = Dny

EXAMPLE 8

The rlISt four derivatives of y = x' - 3x'

+ 2 are

First derivative:

y' = 3x' - 6x

Second derivative:

y" = 6x - 6

Third derivative:

y'" = 6

Fourth derivative:

y(4) =

O.

The function has derivatives of all orders, the fifth and later derivatives all being zero.

_

Exercises 3.3 Derivative Calculations

25.v=l+x;4v'S

In Exercises 1-12, Imd the IITSt and second derivatives.

1. y = -x' + 3 3. s =

5t 3 -

2. Y = x' + x + 8

3t 5

4.

W

= 3z'

- 7z 3 + 21z2

4x 3 5·y=T- x

x3 x2 x 6·y=T+T+4

7. w = 3z-2

8.

-}

8

2s

x4 3 29. y=T-Ix'-x

=4

31. Y

- 2s - x-,

8

8'

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 differeotiate.

13. y = (3 - x')(x' - x + I) 14. Y

= (2x + 3)(5x'

- 4x)

15. Y = (x' + I) (x + 5 +}) 16. y = (I + X')(X'/4 - x-') Find the derivatives of the functions in Exercises 17-28.

2x + 5 17. Y = 3x - 2

18.

x2 - 4 19. g(x) = x + 0.5

20. /(/) =

21. v

= (I

23. /(8)

- /)(1

= v'S -

v'S +

22. w

I

24 =5x+1 • u 2v'S

= (2x

28. Y

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

= ~---c~-~

x5 30. y= 120

- I)(x' + 3x - 5) 32. Y

= (4x' + 3x)(2

- x)

Find the IITSt and second derivatives of the functions in Exercises 33-40.

=/'+5/-1

33• Y =,,'+7 x

34. 8 , /

(8 - 1)(8'

35. r

=

37. w

=

,

+ 8 + I)

eq';/z)<33)(q4-- I) 8

z)

+ (l2q

~

36.u=

(x' + x)(x' - x + I) 4 x

38. w = (z + I)(z - I)(z' + I)

q' + 3

40. P = (q _ I)' + (q + I)'

41. Suppose u and v are functions ofxthat are differeotiable atx and that

/' - I

/' + / -

= (x

39. P =

4 - 3x z=-,-3x + x

+ /'t l I

(x' - I)(x' + x + I)

= -2/- 1 + ~

12 4 I 12.r=---+4

11. r=_I_-~

38'

10. Y

I

=

Find the derivatives of all orders of the functions in Exercises 29- 32.

/'

9. y = 6>:' - lOx - 5x-'

27. Y

26.r=2(.vo+VO)

u(O)

2

- 7t'(x

+ 5)

= 5,

u'(O)

= -3,

v(O)

= -I,

v'(O)

=0

= 2.

Find the values of the following derivatives at x = O. d

a. dx (uv)

d d. dx (7v - 2u)

3.3

passing through the point (3, 8).

u(i) = 2, u'(I) = 0, v(l) = 5, v'(I) = -I.

d

~

123

52. Find all points (x,y) on the graph of f(x) ~ x' with tangent lines

42. Suppose u and v are differentiable functioos of x and that

Find the values of the following derivatives at x

Oifferentiation Rules

f(x) ~ x'

I.

d d. dx (7v - 2u)

a. dx (uv)

Slopes and rangents 43. a. Normal to aeurve Find an equation for the line perpendicular to the tangent to the curve y = x' - 4x + I at the point (2, I). b. Smallest slope What 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 equatioos for the horizontal tangents to the curve y = x' - 3x - 2. Also !md equations for the lines that are perpendicular to these tangents at the points of tangency.

b. Smallest slope What 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" serpentine (graphed here) at the origin and the point (I, 2). y

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 tangent line together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point's coordiuates. Dc. ConImn your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).

54. a. Find an equation for the line that is tangent to the curve y = x 3 - 6x 2 + Sx at the origin. D b. Graph the curve and tangent together. The tangent intersects the curve at another point. Use Zoom. and Trace to estimate the point~ coordiuates. Dc. ConImn your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).

Theory and Examples 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 converting 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.

g(x)

x

47. Quadratic tangent to identity function The curve y = ax' + bx + c passes through the point (I, 2) and is tangent to the line y = x at the origin. Find a, b, and c. 48. Quadratics 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. 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.

~

ax { x" - 3x,

if x < 0 if x '" 0

58. Find the values of a and b that make the following function differentiable for all x-values.

ax + b, () = { bx'-3, fx

x> x:5

-I -1

59. The general polynomial of degree n has the fonn

P(x) where a.

¢'

=

anx n + an_tXn-1

+ ... + a2x 2 + alX + ao

O. Find P' (x).

60. The body's reaction to medicine The reaction of the body to a dose ofmedicine 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 ofmedicine 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 dR/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? What does this say about the Derivative Constant Multiple Rule?

64. Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is, d ( -m) _ -m-l dx x - -mx

where m is a positive integer.

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 P =

62. The Reciprocal Rule

dx

'

in which a, b, n, and R are constants. Find dP/ dV. (See accompanying figure.)

a. The Reciprocal Rule says that at any point where the function v(x) is differentiable and different from zero,

!£

2

nRT _ an V - nb V2

(1) __ -.L v

-

dv v 2dx ·

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 gives the formula d dv -(uv) = u dx dx

The Derivative Product Rule

+

du vdx

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 A(q) =

ofx.

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 1U2 u3 U4 offour differentiable functions of x?

c. What is the formula for the derivative of a product Ul U2 U3 ... 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 dA/dq and d 2A/dq 2.

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 (f(x + h) - f(x))/h as the average rate of change in f over the interval from x to x + h, we can interpret its limit as h ~ 0 as the rate at which f is changing at the point x.

3.4

DEFINmON

The Derivative as a Rate of Change

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 ofchange.

EXAMPLE 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 m 2/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

11s--->!'1

~

/(t).

The displacement ofthe object over the time interval from tto t

+ At

•

=

tis

) S

s+l!.s~f(t+l!J)

FIGURE 3.12 The positions ofa body moving along a coordinate line at time t and shortly later at time t + b.t. Here the coordinate line is borizontal.

=

+ tit (Figure 3.12) is

/(t + tit) - /(t) ,

and the average velocity of the object over that time interval is V av

=

displacement tis /(t travel time = tit =

+ tit) - /(t) tit

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 + tit as tit 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 + tit) - /(t) ti 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 ofmotion. When the object is moving forward (s increasing), the velocity is positive; when the object is moving backward (s 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

I ~------_)I

o

s increasing:

s decreasing:

positive slope so

negative slope so moving downward

moving upward

FIGURE 3.13 For motioo s ~ 1(1) along a straight line (the vertical axis), v = dsldt is positive when s increases and negative whens decreases. The blue curves represent positioo along the lioe over time; they do not portray the path of motion, which lies along the s-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 speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.

DEFINmON

Speed is the absolute value ofvelocity. Speed = Iv(I)1 =

I~I

EXAMPLE 2 Figure 3.14 shows the graph of the velocity v = 1'(1) ofa particle moving along a horizontal line (as opposed to showing a position function s = 1(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

I

I

I FORWARD I I AGAIN I

I I

(v> 0)

I I

Speeds~Steady~ Slows

up

:(v =

consO:

Velocity v

I

I

down

J

~ 1'(')

I

i

:

I

I

I

I

I

I

I

I

r

Stands

,

(v

I

I I

till

,S

I I

2

: (v> 0)

~ Speeds ~ I up I

I I

o

I

5

3,

~

I

I

~I

I 0) ,

6

, , , , , , , ,

I ,

l

I

Speeds

4- Slows ~ down

I

MOVES BACKWARD

, I

IUp

I

,

I

, I

I

:

(v

< 0)

,

:

FIGURE 3.14 The velocity graph ofa particle moving aloog a horizootalline, discussed in Example 2. HIsroRICAL BIOGRAPHY

Bernard Balzano (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 1is s = /(1), then the body's acceleration at time 1is 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 significant. The value of g in the equation s = (1/2)gI2 depends on the units used to measure 1and s. With 1in secoods (the usua1 unit), the value ofg determined by measurement at sea level is approximately 32 ft./sec 2 (feet per second squared) in Eoglish units, and g = 9.8 m/sec2

128

Chapter 3: Differentiation

r (seconds)

=0 t=1 t

•.. ' ,

'-'

(meters per second squared) in metric units. (These gravitationai constants depend on the distance from Earth's center of mass, and are slightly lower on top ofMt. Everest, for example.) The jerk associated with the constant acceleration of gravity (g = 32 ft/sec 2) is zero:

s (meters)

0 5 10

t

~

2

15

..

' , '-'

j =

30

~

3

'-'

O.

EXAMPLE 3 Figure 3.15 shows the free fail of a beavy bail bearing released from rest at time I = Osee.

35 t

=

An object does not exhibit jerkiness during free fail.

25

'..,

~ (g)

20

40


45

(b) What is its velocity, speed, and acceleration when I = 2?

Solution

FIGURE 3.15 A ball bearing falling from rest (Example 3).


s

v(l) = .'(1) =

~(4.91 2 ) =

9.81.

At I = 2, the velocity is

v(2) = 19.6 m/sec in the downward
a(l) = v'(I) = ."(1) = 9.8 m/sec2. Att = 2, the acceleration is 9.8 m/sec2.

•

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 ~ 1601 - 161'


5

-160

10

v~cfs~160-32t til

(b)

FIGURE 3.16 (a) The rock in Example 4. (b) The graphs of s aod v as functions of time;. is largest when v = ds/dl = O. The graph of. is not the path ofthe rock: It is a plot ofheight versus time. The slope ofthe plot is the rock's velocity, graphed here as a straight line.


Solution
tis d 2 = dl = 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 f"mt f"md 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 2 a -_dv_d( dt - dt 160 - 32t)_ - -32 ft/ 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 ofthis 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 . c(x = hm dx h~O ~I-----'--------L~x

o

x x+h Production (tons/week)

FIGURE 3.17 Weekly steel production: c(x) is the cost ofproducing x tons per week. TIre cost of producing an additional h lons is c(x + h) - c(x). y

+ h)

h

- 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

c(x

ax

+ I) -

c(x)

I

which is approximated by the value of dcl dx at x. This approximation is acceptable if the slope ofthe graph ofc does not change quickly near x. Then the difference quotient will be close to its limit Ik/dx, which is the rise in the tangent line if = I (Figure 3.18). The approximation works best for large values of x. Economists often represent a total cost function by a cubic polynomial

ax

c(x) = ax 3

+

(3x 2

+ 'Yx + {J

where {J represents fIXed costs such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs such as the costs of raw materials, laxes, 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

--;ol---'--------"~---~x

o

x

x+ 1

FIGURE 3.18 Tlremarginalcostdcldds approximately the extra cost ac of producing ax = I more unit.

dollars to produce x radiators when 8 to 30 radiators are produced and that

r(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) =

:Ix (x 3 -

6x 2

15x) = 3x 2

+

c'(IO) = 3(100) - 12(10)

+ 15

-

12x

+

15

= 195.

The additional cost will be about $195. The marginal revenue is

r'(x) =

:Ix (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 language of marginal rates, consider marginal tax rates. If your marginal income tax rate 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 ofincrease 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 rate will increase.

Sensitivity to Change When a amall change in x produces a large change in the value of a function j(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 plants, provided the first scientific explanation ofbybridization. His careful records showed that if p (a number between 0 and I) is the frequency of the gene for smooth skin in peas (dominant) and (I - p) is the frequency of the gene for wrinkled skin in peas, then the proportion of smooth-skinned peas in the next generation will be y = 2p(1 - p)

+ p2

= 2p _ p2.

The graph 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 graph 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 I.

dyldp 2 y

-o::T-------:--p (a)

-o::+---~-~p

(b)

FIGURE 3.19 (a) The graph ofy = 2p - p2, describing the proportion of smooth-skinoed peas in the next generation. (b) The graph ofdyjdp (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 generations 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 Coordfnattl Un. Excroises 1--6 give the positions 9 = jet) ofa bodynwving ona coordinate Iinc, with s in meters and t in seoondlI.

L Find the body's displaoement and average velocity for the given tinu: interval. b. Find the body's speed and acoelemtion at the endpoint! ofthe interval.

c. When, if ever, duriog the in1:aval does the body change diIectiDn? 1."-12 -31+2,OSIS2 1. " - 6t

-,z,

5.s=-25_~ lStS5 t2 t' t

+ 5'

13. Free flIll from the Tower of Ph. Had Galileo dropped a cannonball from the Tawer ofPisa, 179 ft abave the ground, the ball's height abave the ground 1 s« into the fall would have been s - 179 - 16t2.. L

0 SIS 6

3.9=-13 +312 -31,OSIS3 4. 9 = (1 4/4) - t 3 + 12., O:S t :S 3

25 .. , - - -

11. Speeding boDet A 45-caliber bullet shot s1raight up from the sutface of the moon would reach a height ofs - 8321 - 2.61 2 ft after t sec. On Earth, in the absence of air, its height would be s - 8321 - 1612 ft after t scc. Haw long will the bullet be aloft in each case? Haw high will the bullet go?

-4:StSO

7. Particle motion At time I, the position ofa body moving along the,,·axis iu - 13 - 61 2 + 91m.

L Find the body's acceleration each time the velocity is Zt.l'O. b. Find the body's speed each tinu: the acceleration is zero. Co Find the total distance traveled by the body from t = 0 to t = 2. 8. Particle motion At time t ~ 0, the velocity of a body moving along the horizontal. ,,-axis is v = t2. - 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 tabn the ball to hit the ground?

c.

What~have

been the ballll velocity at the D1OOll'lllt ofitq]act?

14. GJiliIeo'J free-fall formula Galileo developed a formula for a body's velocity during free fall by rolling balls from rest down incteasingly steep inclined planks and looking for a limiting formula that would predict a ball's behavior when the p1aDk was vertical and the ball fell :freely; see part (a) of the accompanying fJ.gUre. He found that, for any given angle ofthe plank, the ball's velocity t sec into motion was a constant multiple of I. That is, the velocity was given by a formula ofthe form v - kt. The value of the constant k depended on the inclination ofthe plank. In modern notation--part (b) ofthe fi.gmc----with distance in meters and time in seconds, what Galileo determined by experiment was that, for any given angle 9, the ball's velocity 1 sec into the roll was

v - 9.8(sin9)tmfsec.

b. When is the body moving forward? Backward?

c. When is the body's velocity increasing? Decreasing?

Fre..-Fall Applications ,~ , ,

9. Free fa1J. on Man and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (3 in meters, t in seoondlI) are 9 = 1.86t2. on Mars and 9 = n.44t2. on Jupiter. Haw long does it tab a rock falling from rest to reach a velocity of 27.8 mfsec (about 100 kmjh) on each planct? 10. Lunar proJectDe 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 9 = 24t - 0.8t 2 minI sec. L Find the rock's velocity and acceleration at time t. (The acceleration in this case is the acceleration of gravity on the moon.) b. Haw long does it take the rock to reach its highest poinfl

-

c. Hawhighdoestherockgo? d. Haw long does it take the rock to reach half its maximum

e. Haw long is the rock aloft? 11. FindiDKg (III .1IIUIII1Iirlen planet Explorers on a small airless planet used a spring gun to laundJ a ball bearing vertically upward from the surliu:e at a launch velocity of 15 mfs«. Because the acceleration of gmvity at the plaDet's surlBce was & miser?, the explorers expected the ball bearing to reach a beigb1 of s =- 1St - (l/2)g.,zmts«1aIer.Theballbearingreacheditsmaximum heigh120 sec after being lanoclJril What was the value ofg, ?

,

if (.) L

(b)

What is the equation for the ball's velocity during:free fall?

b. Building on your work in part (a), what constant acceleration does a freely fiilling body experlencenear the surface of Earth?

Undem:andfng Motton from Graphs 15. The accompanying fIgUre shows the velocity v = dr/dt = 1(1) (m/sec) ofa body moving along a coord.i.natl; line.

v (m1scc) v -fit)

L When does the body reverse direction?

b. When (approximately) is the body nwving at a constant speed?

3.4 The Derivative as a Rate of Change c. Graph the body's speed for 0

:S

t

:S

10.

18. The accompanying figure shows the velocity v cle moving on a horizontal coordinate line.

d. Graph the acceleration, where defined.

16. A particle P moves on the number line shown in part (a) ofthe accompanying figure. Part (b) shows the position of P as a function of time t. _ _-----'I_ _ ........

P

............ ~-~) S

=

133

f(t) of a parti-

v

(em)

o (a)

s (em)

a. When does the particle move forward? Move backward? Speed up? Slow down? t (sec)

----=-1L------'----'---~L----'-----':------':-~

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.

Ii·

I.

a. How fast was the rocket climbing when the engine stopped? b. For how many seconds did the engine burn?

•

200 150 ~

u

II)

100

~

~

.~

..s II)

>-

50 0 -50 -100 0

2

4 6 8 10 Time after launch (sec)

! ! ! ! '!

-I

12

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)?

J I 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 l60-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 trud< The accompanying graph shows the positioo s of a truck traveling 00 a highway. The truck starts at t ~ 0 aod returns 15 h later at t = 15.

Economics 23. Marginal co.t Suppo.e that the dollar co.t of prodocing x washing machines is c(x) ~ 2000 + 10(1,; - 0.lx 2 •

a. Use the technique desctihed in Sectioo 3.2, Example 3, to

a. Find the average cost per machine of producing the rlrst 100 washing machines.

graphthetruck~velocityv ~


b. Find the marginal cost when 100 washing machines are prodoced.

b. Supposethats ~ 15t 2 - t'.Graph
500

/ '\ / \ I

100

/ o

/

e. Show that the marginal cost when 100 washing machines are produced is approximately the eo.t of producing one more washing machine after the rrrst 100 have been made, by calculating the latter co.t 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 Elapsed time, t (hr)

b. Use the function r'(x) to estimate the increase in revenue that will result from increasing prodoction from 100 machines a week to 101 machines a week.

15

21. The graphs in the accoropaoying figure show the positioo 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. Bacteriom population When a bactericide 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 hegan to decline. The size of the population at time t (hours) was b ~ 10' + lO't - I O't 2 • Find the growth rates at

a. t = 0 hours. b. t = 5 bours.

c. t = 10 hours. 26. Draining a tank The number of gallons of water in a taok t mioutes atler the taok bas started to drain is Q(t) ~ 200(30 - tf. 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 10 min?

D 27. 22. The graphs in the accompanying figure show the positioo 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/h) at which the taok is draining at time t. b. When i. the fluid level in the taok falling faste.t? Slowest? What are the values of dy/dt at the.. times? e. Graphy and dy/dt together and discuss the hebavior ofy in relation to the .igns and values of dy/dt. 28. Inflating a balloon The volume V = (4/3)".r' of a spherical balloon changes with the radius.

a. At wbat rate (fl'/ft) doc. the volume change with respect to the radius when r = 2 ft? b. By approximately how much does the volume increase when the radius change. from 2 to 2.2 ft?

3.5

135

velocity fimction v(t) ~ d>/dt ~ I'(t) and the acceleration function a(t) = d 2s/dt2 = I"(t). 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 1ravels along a runway before takeoff is given by D ~ (l0/9)t 2 , where D is measured io meters from the starting point and t is measured io seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km/h. How long will it take to become airbome, and what distance will it lravel 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 polot shot lava 1900 ft straight ioto the air (a Hawaiian record). What was the lava's exit velocity io feet per second? In utiles per hour? (Hint: If Vo is the exit velocity of a patticle of lava, its height t sec later will be s ~ vot - 16t2 ft. Begin by fmdingthe time at wbich d>/dt ~ O. Neglect air resistance.)

e. When does it change direction? d. Wheo does it speed up and slow dowo? e. When is it moviog fastest (highest speed)? Slowest? f. Wheo is it farthest from the axis origin?

16t 2 , 0 '" t '" 12.5 (a heavy object fIred straight up from Earth's surface at 200 ft/ sec)

31. s

= 200t -

32.s=t2 -3t+2,0"'t"'5 33.s=t'-6t2 +7t,0"'t"'4

Analyzing Motion Using Graphs

D Exercises 31-34 give the position fimction s =

/(t) of an o~ect moviog along the s-axis as a fimction of time t. Graph / together with the

3.5

Oerivatives ofTrigonometric Functions

34. s

=4

- 7t

+ 6t 2

-

t',

0 '" t '" 4

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 Sa and Theorem 7 in Section 2.4 with the angle sum identity for the sine function: sin (x

+ h)

= sinxcosh

+

cosxsinh.

Jf f(x) = sinx, then

f'(x)

= lim

f(x + h) - f(x)

+ h)

+ cos x sin h) -

. coshh I· (smx· = h~ I'

=smx·h~

cos h - I

h

1)

Derivative def'mition

sinx = lim sin x (cosh - 1)

h

h-+O

- sinx

h

h-+O

= lim (sinxcosh

,

= lim sin (x

h

h-+O

h-+O

+ cosxsinh

h

+ h-O lim ( cosx-Sinh) h-

+ cosx·h~-h-=smx· I' sin h , 0 + cosx· 1

limit 0

limit 1

The derivative of the sine function is the cosine function:

:x

(sinx) = cosx.

=C08X.

Example Sa and Theo=n 7, Section 2.4

136

Chapter 3: Differentiation

EXAMPLE 1

We find derivatives of the sine function involving differences, products,

and quotients. (a) y = x 2

dy

sinx:

-

-

dx

(b) y = x 2 sinx:

(c) Y =

:

-(smx) dx

= 2x -

cosx.

Difference Rule

:fx (sin x) + 2x sin x

=

x2

=

x 2 cos x + 2x sinx.

x'~(sinx) dx

dy

sin x --x-:

d,

= 2x -

Product Rule

sinx'l

dx

Quotient Rule

xcosx - sinx

•

x2

Derivative of the Cosine Function With the help ofthe angle sum formula for the cosine function, cos (x + h) = cosxcosh - sin x sinh,

we can compute the limit ofthe difference quotient: ~( )-1' cos(x+h)-cosx cosx - un h dx h~O =

= =

Deri:wtive definition

, (cosx cos h - sin x sin h) - cosx lun '--------~---'------

h~O

h

Cosine angle sum identity

, cosx(cosh - I) - sinxsinh lun =~='-"-~~'----===-:c h-'O h lim cosx' ~co~s~h~-~I - lim sin x • _sin_h h h~O h

h~O

. I'l I Dsinh =cosx· lim cosh - I -smx· .-h-O

y'

h

=

cosx·O - sin x 1

=

-sinx.

h-O

h

Example Sa and Theorem 7, Section 2.4

0

-----,/L---c?!..---'--F----. The derivative ofthe cosine function is the negative of the sine function: FIGURE 3.20 The curve y' = -sin x as the graph ofthe slopes ofthe tangents to the curve y = cos x.

:fx (cos x) =

-sinx.

3.5 Derivatives ofTrigonometric Functions

137

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.
+ cosx: dy

d dx = dx (5x)

d

+ dx (cos x)

Sum Rule

= 5 - sinx. (b) y = sin x cosx:

dy

.

d

dx = SfiX dx (cos x) = sinx( -sin x) =

<e) y =

cos2 X

-

+

d

.

cosx dx (SfiX)

Product Rule

+ cosx(cosx)

8in2 x.

cosx . 1 - sinx'

dy

(1 - Sinx)1x(cosx) - cosx1x(l - sinx)

dx

(1 - sinx)2

Quotient Rule

(1 - sinx)( -sinx) - cosx(O - cosx) (1 - sinxf

1 - sinx

(1 - sinxf 1 1 - sinx'

•

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 .imple 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

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).

.=5cosl.

Rest

What are its velocity and acceleration at time I?

Solution We have Position: • = 5 cos I 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 The 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 ofthe motion is 5. The period ofthe motion is 21T, the period of the cosine function.

2.

The velocity v = -5sint attains 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, that 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 ofmotion.

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 ofgravity 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 ofthe 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 ofthe 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 formnlas for the cofunctions.

The derivatives ofthe other trigonometric functions:

:Ix (tanx) =

sec2 x

:Ix (cotx) =

-csc2 X

:Ix (sec x) =

secxtanx

:Ix (csex) =

-cscx cot x

To show a typical calculation, we fmd the derivative ofthe tangent function. The other derivations are left to Exercise 60.

3.5 Derivatives ofTrigonometric Functions

EXAMPLE 5 Solution

139

Find d(!JJnx)jax.

We use the Derivative Quotient Rule to calculate the derivative:

~

~

_ (Sinx) _ cosx:Ix (sinx) - sin x 2 ax (!JJnx) - ax cosx COS x

:Ix (cos x)

Quotient RDIe

cos x cos x - sinx(-sinx)

cos2 x 2 008 X

+

2 8m 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 rnle for secant function

d y' = ax (secx!JJnx)

d

= sec x ax (!JJnx) = secx(sec2 x)

d + !JJnx ax (sec x)

+ !JJnx(secx!JJnx)

Derivative Product Rule

Derivative roles

•

2

3

= sec x + secx tan x

The differentiability of the trigonometric functions throughout their domains gives another proof oftheir 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 x~o

Y2

Y2

+ secx = + sec 0 cos(or - !JJnx) cos(or - !JJn 0)

V2+l

cos(or - 0)

=

V3 = -V3 -I

Exercises 3.5 Derivatives In Exercises 1-18, fmddy/dx.

1. Y

= -lOx + 3 eosx

3. y = x 2 cosx

5. Y

= esex

- 4v'i + 7

6. y = x 2 cotx -

~ X

2. Y

= f + 5 sinx

4. Y

= ~secx + 3

7. f(x) ~ sin x tanx 8. g(x) 9. y = (seex + tanx)(seex - tanx) 10. Y ~ (sinx + cosx) sec x

~

csex catx

•

140

Chapter 3: Differentiation

cotx 11. Y = I + cotx

12.

4 I 13. y = cosx + tanx

14. y

cosX'

y=~~­

1

+ sinx

cosX'

= -x- +

x cosx

In Exercises 45 and 46, fmd an equation for the tangent to the the horizontal tangeot to the curve at Q. curve at P and

(b>

+ 2x cosX' - 2 sinx

15. Y = x 2 sinx

44. Find all points on the curve y = colx, 0 < x < 1t, where the tangent line is parallel to the line y ~ -x. Sketch the curve and tangent(s) together, labeling each with its equation.

2

16. y = x cosX' - 2x sin x - 2 cosX' 17. j(x) =x'sinxcosx 18. g(x) = (2 -x)tan2 x

45. y

46. y

In Exercises 19-22, fmdds/dt.

19.s=tant-t

20. s = t 2

21. s = I + csct

sin t 22.s=1 - cos t

1 - csct

-

sect + I

In Exercises 23-26, fmd dr/dO.

23. r ~ 4 - 0 2 sinO 25. r = secOcscO

24. r ~ osinO + cosO 26. r = (I + sec 0) sinO

In Exercises 27-32, fmddp/dq. I

27. p = 5 + cotq 29. p = 31. p =

sinq

28. p = (I + cscq) cosq

+ cosq

cosq

qsinq -2--

q - I

tanq

30. p = I + tanq

32. p =

3q + tanq qsecq

Trigonometric Limits Find the limits in Exercises 47-54. 47. lim sin .%-2

33. Find y" if

a. y

= cscX'.

b. y = secx.

x--,"/6

34. Find y(4) = d 4y/dx 4 if

a. y

= -2 sinx.

b. y = geosx.

Tangent Lines In Exercises 35-38, graph the curves over the given intervals, together with their tangents at the given values of x. Label each curve and tangent with its equation. 35. y = sin x,

x

36. y x 37. y x 38. y

= -'TT,

-371/2:5 x :s; 2w

0, 3w/2

-1t/2 < x < 1t/2 ~ -1t/3, 0, 1t/3 = secx, -1t/2 < x < 1t/2 = -1t/3,1t/4 = I + cosx, -31t/2';; x ,;; 21t = tanx,

X ~

-1t/3,31t/2

lim

48.

49.

(1- 1) X

2

V~I-+-co-s-(~1t-C-sc-x~)

sinO - 1

~2

lim

9-nj4

51. lim sec [cos x + 1t tan

x-o

· sm . ( 52• IlDl

x-o

53. lim tan 1-0

(-4 1t )

1t + tanx ) tanx - 2 sec x

(I _ t t) sin

54. lim cos ( 1t8.)

sm"

9-+0

Theory and Examples The equations in Exercises 55 and 56 give the position s = j(t) of a body moving on a coordinate line (s in meters, t in seconds). Find the body's velocity, speed, acceleration, andjerk at time t = 1t/4 sec.

55. s

=2

- 2sint

56. s

= sint + cost

57. Is there a value of c that will make

tangents in the interval 0 :5 x :5 2w? If so, where? If not, why not? Visualize your fmdings by graphing the functions with a grapher.

j(x) ~

S:23X, { c,

x =F 0

x = 0

continuous at x = O? Give reasons for your ans'WeI'. 58. Is there a value of b that will make

42.y~x+2cosx

43. Find all points on the curve y = tanx, -1t/2 < x < 1t/2, where the tangent line is parallel to the line y ~ 2x. Sketch the curve and tangeot(s) together, labeling each with its equation.

~I

8 - "4

- I]

seCX'

D Do the graphs of the functions in Exercises 39-42 have any horizontal 39. y = x + sinx 40. y = 2x + sinx 41. y = x - cotx

lim tan 0

50.

8 - 6

8-1T/6

g(x) continuous at x your answers.

~

~

{x

+ b, x <

cosx,

x

~

O? Differeotiable at x

0 0 ~

O? Give reasons for

3.5

59. Find d 999/ dl;999 (COS x).

141

See the accoropanying figure.

60. Derive the formula for the derivative with respect to x of a. secx.

Oerivatives ofTrigonometric Functions

b. cscx.

y

c. cotx.

Slope

61. A weight is attached to a spring aod reaches its equilibrium position (x ~ 0). It is then set in motion resulting in a displacement of

c

~

f'(x)

~

Slope ~ f(x

+ h) - f(x) h

x= IOcost,

where x is measmed in centimeters and t is measmed in seconds. See the accompanying figure.

Slope ~ f(x

+ h) - f(x -

h)

2h

o Equilibrium

a. To see how rapidly the centered difference quotient for j(x) = sinx converges to f'(x) = cosx, graphy = cosx together with

position atx=O

sin(x

a. Find the I =

spring~

displacement when I

~ 0, I ~

'Tr/3, aod

3'Tr/4.

b. Find the spring~ velocity when I = 0, 1= 'Tr/3, aod I = 3'Tr/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 I = 0, I = 'Tr/2, aod I =

'Tr.

b. Find the particle~ velocity when I = 0, I = 'Tr/2, aod I =

'Tr.

o 63. Graph y = cos x for y

+ h) -

sinx

h

~

over the interval [-'Tr, 2'Tr] forh = 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'Tr. On the same screen, graph

-'Tr ,;; X ,;;

sin(x

h) - siu(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, aod 0.1. Then, in a new window, tty -I, -0.5, aod -0.3. What happens ash"" O+? Ash .... O-?

over the interval [-'Tr, 2'Tr] 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?

064.

Graph y graph

= -sinx

for

-'Tr ,;; X ,;;

2'Tr. On the same screen, j(x

cos(x

+ h)

y=

j(x - h)

2h

- cosx

= I,

may have a lintit as h .... 0 when j has no derivative at x. As a case in point, takej(x) ~ Ixlaodcalcu1ate

What phenomenon is being illustrated here?

o 65. Centered differenee qnotient. j(x

+ h) -

(Continuation

h

0.5, 0.3, aod 0.1. Then, in a new window, tty h = -I, -0.5, aod -0.3. What happens ash"" O+? Ash"" O-? for h

66. A caution about centered difference quotients ofExercise 65.) The quotient

+ h)

- 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, aod (2) it usually gives a better approximation of f' (x) for a given value of h thao the difference quotient j(x

+ h) h

lim 10

The cenlered difference quotienl

- j(x)

h-O

+ hl-

10 - hi

2h

As you will see, the lintit exists even thoughj(x) ~ Ixlhas no deo. Moral: Before using a centered difference quotient, be sure the derivative exists.

rivative at x =

= taox aodits derivative together on (-'Tr/2, 'Tr/2). Does the graph of the taogent function appear to have a smallest slope? A largest slope?

067. Slope. on the graph of the tangent function 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 runction Graph y = cot x and its derivative together for D < x < 'If. Does the graph of the cotsogent function appear to have a smallest slope? A !lugest slope? Is the slope ever positive? Give reasons for your answers.

b. With your grapher still in degree mode, estimate

lim cosh - I h .

h-O

Exploring (sin kxl/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 asia? What would you espect the graphs of y ~ (sin5x)/x and y ~ (sin(-3x))/x to do as x---'D? 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 derivatioo using degree-mode lintits. What formula do you obtain for the derivative? d. Work through the derivatioo 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.

e. The disadvantages of the degree-mode formulas become apparcot as you start taking derivatives of higher order. Try it What are the second and third degree-mode derivatives of sin x and cosx?

a. With your graphing calculator or coroputer graphcr in degree

mode, graph f(h) = sinh h and estimate lilllj,~o 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 fog of two functions y = flu) = sin u and u = g(x) = x 2 - 4 that we koow how to differentiate. The answer, given by the Chain Rule, says that the derivative is the product of the derivatives of f 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'

dy/dx

= 3/2 = (1/2)(3) =

(dy/du)(du/dx).

1 2'

and

du = 3

dx

.

3, we see in this case that

dy

c: y turns B: u turns A: x toms

FIGURE 3.23 Whee 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 du

dy du

dx = du' dx'

If we think of the derivative as a rate of change, our intuition allows us to see that this relationship is reasonable. If y = f(u) changes half as fast as u and u = g(x) cbanges three times as fast as x, then we expect y to change 3/2 times as fast as x. This effect is much like that of a multiple gear train (Figure 3.23). Let's look at another example.

EXAMPLE 1

The function

y

=

(3x 2 + 1)'

3.6 The Chain Rule is the composite of y = f(u) = u 2 and u = g(x) = 3x 2 see that

dy du du dx

-.- = =

+

143

1. Calculating derivatives, we

2u'6x 2(3x 2

= 36x

3

+ 1)'6x + 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 ofthe composite functionf(g(x» atx is the derivative of f atg(x) times the derivative ofgat 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'----

Rate of change at x is g'(x), --4'---u

~

g(x)

Rate of change at g(x) is f'(g(x)).

y

~

f(u)

~

f(g(x)

FIGURE 3.24 Rates of change multiply: The derivative of fog alx is the derivative off alg(x) times the derivative ofg alx.

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) = f'(g(x))' g'(x).

lnLeibniz'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 ax ---> 0:

dy dx

lim ay 4x~O

ax

lim ay, au

ax

4x~O au

lim ay, lim au

=

4x~O

4x~O

au

ax (Note that..:1u --+ 0 as ~- 0 since g is continuous.)

lim ay, lim au ..:1u_o.6.u .:1%-0 ax dy du =du'dx· =

The problem with this argument is that it could be true that au = 0 even when ax o¢' 0, so the cancellation of au 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 t ;;,: 0 is given by x(t) = cos(t 2 + I). Find the velocity ofthe object as a function oft. Solution We know that the velocity is dx/dt. In this instance, x is a composite function: x = cos(u) andu = t 2 + I. Webave

dx du du dt

= -sin(u)

x ~ cos(u)

2t.

=

By the Chain Rule,

dx dx du dt=du'dt tb:

= -sin(u)' 2t = -sin(t 2 + 1)' 2t

du evaluated at u

•

= -2tsin(t 2 + I).

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 = f(g(x)), then

dy dx

=

f'(g(x))' g'(x).

In words, differentiate the "outside" function f 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

:Ix sin (x 2 + x) = inside

cos (x 2

+ x), (2x +

inside left alone

1).

derivative of 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)

~(tan (5

- sin 2t))

Derivati:~ of tan u with u=5-sin2t

= sec2 (5 -

sin2t)· :t(5 - sin2t)

Derivative of 5 - sin u with u = 2t

2 = sec (5 -

sin2t)·

=

(0 - cos2t· ~(2t))

= sec2 (5 -

sin2t)'(-cos2t)'2 -2(cos2t) se~(5 - sin2t).

=

•

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-I

du dx'

The Power Chain Rule simplifies computing the derivative of a power of

an expression. (a) ~(5x3 - x 4)7 = 7(5x 3 - x4)6~(5x3 - x 4) dx dx = 7(5x 3 - x 4)6(5' 3x 2 - 4x 3) =

(b)

~ (_I-) = dx 3x-2

Power Chain Rule with u

= 5;t3

- X 4 ,1I

=7

7(5x 3 - x 4)6(15x 2 - 4x 3) 4<3x dx

2r

l

-1(3x - 2r2~(3x - 2) dx 2 (3) = -1(3x 3 (3x - 2)2 =

2r

Power Chain Rule with

u=3x-2,n=-1

In part (b) we could also find the derivative with the Derivative Quotient Rule.

d (c) dxd (sin'x) = 5 sin4 x' dx sinx

=5sin4 xcosx

PowerChainRuicwithu

~

sinx,_

~ 5,

bccKusc sinll x means (sinX)", 11 :# -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 =

W, we can derive the following formula:

I Derivative of the

~(lxl) = ~W

Absolu.te Valu.e Function

Powor Chain Rule with

_1_. !£(x 2) dx 1 = 21xI'2x =

x,
=

EXAMPLE 7

2W

x TXT'

u=;t2,n= 1/2,x:#O

•

x,< O.

Showthattheslopeofeverylinetangenttothecurvey = 1/(1 - 2x)3is

positive.

Solution

We fmd the derivative:

dy = !£(I - 2xr3 dx dx =

-3(1 - 2xr4 • ~ (I - 2x)

=

-3(1 - 2xr4 '(-2)

Power Chain Rule with u

~

~-3

(1 - 2>:),.

6

(I - 2x)4 At any point (x, y) on the curve, x ,< 1/2 and the slope ofthe tangent line is

dy dx

6

(1 - 2x)4'

•

the quotient oftwo positive numbers.

EXAMPLE 8 The formulas for the derivatives ofboth 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 ofthe 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 ofmdian measure in computations. y

I

Y = sinx

ISO

FIGURE 3.25 Sin (x') oscillates ooly 71"/180 times as often as sin x oscillates. Its maximum slope is 7I"j180 atx = o(Example 8).

147

3.6 The Chai" Rule

Exercises 3.6 Derivative Calculations In Exercises 1-8, given y = f(u) and u = g(x), rmd dy/dx =

49. y = sin (cos (21 - 5))

f'(g(x))g'(x) . 1. y = 6u - 9,

u = (1/2»:4

+1

3. y

=

sinu,

u

=

3x

S. y

=

cosu,

u

=

sinX'

7. y=tanu,

u= IOx-5

2. y = 2u',

u = 8x - I

4. y

=

cosu,

u

-x/3

6. y

=

sinu,

u =x - cosX'

8. y

= -sec u. u = x 2 + 7x

=

In Exercises 9-18, write the function in the form y = f(u) and u = g(x). Then rmd dy/dx as a function ofx.

9. y

= (2x +

11.y= 13. y~ 15. y

~

I)'

(I-t)-7 (

= (4

10. y

12.y=

~2 +x-x1)4

sec(tanx)

- 3x)'

()-10 I-I +6

~ cot(". - ~)

21. s

20. q = V'2r - r 2

~ 3~ sin 31 + 5~ cos 51

23. r = (cscO + cotO)-1 25. y = x2 sin4 x

+ xcos-2 x

. (3".1) 22.s~smT

il (3x - 2)7 + (4 - ~2

27. y

=

28. y

= (5 -

29. y ~ (4x

2xt' +

(3".1) +cosT

24. r = 6(secO - tanO)'f2 26. Y

r

= }sin-.sx - j-cos3 x

31. h(x) = x tan (2VX) + 7

. 0

~u;,os 0

)2

30. y ~ (2x - 5)-I(x 2 - 5x)' 32. k(x) =

37. r ~ sin (0 2 ) cos(20) 39. q = sin

(vf+--J)

36

• g

38. r

x2seC(~)

(I) = (I + sin 3/)-1 3 21

~ secW tan (i)

40. q =

cot(S~I)

41. y = sin2 (".1 - 2)

42. y = sec'".1

43. y = (I + cos 21)-4

44. y = (i + cot(I/2)t2 46. y = (1-'/4 sin 1)4/3

47. y

~

,12- -)' (I - 41

Vi)

V~31-+-y'--;r=2~+~V~I==-==I

Second Derivatives

(I +~)'

60. y = (I - VX)-1

I 61. y = 9cot(3x - I)

63. y ~ x(2x

(t)

62. y = 9tan

+ 1)4

64. y ~ x 2 (x' - I)'

Finding Derivative Values 0

g)' at the given value ofx.

65. flu) = u' + I,

u = g(x) = VX, x = I

I 66.f(u)=I-ij,

I u=g(x)=l_x' x=-I

67. flu) = cot 7~,

u = g(x) = 5VX, x = 1

1 68. flu) = u + -2-' cos u

-f"-, u +1 ~:

U

= g(x) = ".x,

X

u = g(x) = IOx2 +

r

X

= 1/4

+ I, x = 0

u = g(x) = :2 - I, x = -I

71. Assumethatf'(3) = -1,g'(2) = 5,g(2) = 3, andy = f(g(x)). Whatisy'atx = 2? 72. Ifr = sin (J(t)),f(O) = "./3,andf'(0) = 4,thenwhatisdr/dl att = O?

73. Suppose that functions f and g aod their derivatives with respect to x have the following values at x = 2 and x = 3.

x

f(x)

g(x)

f'(x)

g'(x)

2 3

8 3

2

1/3 2".

-3

-4

5

Find the derivatives with respect to x of the following combinations at the given value of x.

In Exercises 41-58, finddy/dl.

45. y = (I tan 1)10

58. y =

70. flu) = (:

tan3x 34. g(x) ~ (x + 7)4 35. flO) = ( I

57. y = 31 (212 - 5)4

69. flu) =

H~ + I)'

+ 3)4(x + It'

54. y=4sin(v'1 + 56. y = cos4 (sec' 31)

In Exercises 65-70, rmd the value of(J

Find the derivatives of the functions in Exercises 19-40.

v'3=t

52. y = i(1 + cos2 (71))'

53. y = v'1 + cos (12) 55. y = tan2 (sin' I)

59. y=

18. y = 5cos-4 x

19. P =

(li))'

(t))

Find y" in Exercises 59-64.

14. y ~ v'3x 2 - 4x 16. y

4 51. y = (I + tan

50. y = cos (5 sin

31 -

4)-'

48.y~ ( 51+2

a. 2f(x), x = 2 c. f(x)' g(x),

x

b. f(x)

=3

e. f(g(x)),

x = 2

g. l/g 2(x),

X

=3

+ g(x),

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

r. v'iW,

x

x

=3

=2

x=2

h. v'f(x) + g2(x),

X

= 2

148

Chapter 3: Differentiation

74. Suppose that lhe functioos f and g and their dctivatives wilh respect to x have lhe following values at x ~ D and x ~ I.

83. Rnnning machinery too fast Suppose that a piston is moving straight up and down and that its position at time t sec is

s

x

f(x)

D I

I 3

1'(x)

g'(x)

I

5

-4

-1/3

1/3 -8/3

g(x)

Find lhe derivatives wilh respect to x of lhe following combina-

tions at the given value of x. a. 5f(x) - g(x), x ~ I f(x) c. g(x) + I' x = I

b. f(x)g'(x),

e. g(J(x)), x = 0

f. (x 11

g. f(x

+ g(x)),

d. f(g(x)),

with A and b positive. The value of A is the amplitude of the motion, and b is lhe frequeocy (number oftimes lhe piston moves up and down each second). What efree! does doubliog lhe frequeocy have on the piston's velocity, acceleration, and jerk? (Once you fmd out, you will know wby some machinery breaks wheo you roo it too fast.) 84. Temperatures in Fairbanks, Alaska The graph in lhe accompanying figure shows the average Fahrenheit temperature in Fairbanks, Alaaks, doting a typical 365-day year. The equation that approximates lhe temperature on day x is

x ~ 0 x = 0

+ f(x)t 2 , x

= I

y = 37 sin

x = 0

75. Find d>/dtwheo 8 ~ 3'1f/2 ifs ~ cos8andd8/dt ~ 5. 76. Find eIy/dt wheox ~ I ify ~ x 2

+ 7x

- 5andd>/dt ~ 1/3.

What happens if you can write a function as a composite in different

b. y

= x by using lhe Chain Rule withy as a compos-

= (u/5) + 7 = I + (I/u)

u and u and

= 5x - 35 = I/(x - I).

78. Find eIy/dx if y = x'/2 by using the Chain Rule wilh y as a composite of

a. y = u' b. y

= 'VU

and and

101)] + 25

and is graphed in the accompanying figure.

a. On what day is the temperature increasing the fastest? ing when it is increasing at its fastest? y

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.

a. y

[;t7s (x -

b. About how many degrees per day is the temperature increas-

Theory and Examples

77. Find dy/dx ify ite of

= A cos (2'1fbt) ,

u = vX

u =x'.

79. Findthetangeottoy = «x - I)/(x 80. Find the tangeot to y =

y'x 2

-

X

+

1))2 atx = O.

+ 7 at x

= 2.

81. a. Find lhe tangeot to lhe 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

a. Find equations for the tangeots to the curves y = sin 2x and y ~ -sin(x/2) at lhe origin. Is lhere anything special about how the tangeots are related? Give reasons for your answer. b. Can anything be said about lhe tangeots to lhe curves y ~ sinmxandy ~ -sin(x/m) at the origin (m a constant '¢' O)? Give reasons for your answer. c. For a giveo m, what are the largest values the slopes of lhe 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, 2'1f1, lhe 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.

60 H-+-+-+--bo>""i-.,..d-+-+-+-+-+-H .... fi:'

i

I....'

c... 40 HH-+~'-+-+-+-+' "~-+--+--+-+-+-I

~

--~,-----~ - - - - 1'\ ----~--I>x

20 - - - V0

-til

-20

,,~
V+

85. Particle motion ordinate line is s =

position of a particle moving along a co1 4t. with s in meters and t in seconds. Find lhe particle~ velocity and acceleration at t = 6 sec.

86. Constant acceleration Suppose that the velocity of a fa1liog body is v = kVS m/sec (k a constant) at the instant the body has falleo s m from its starliog point. Show that the body's accelera-

tion is constant. 87. Falling meteorite The velocity of a heavy meteorite eotering Earth's atmosphere is inversely proportional to VS wheo it is s km from Earth's ceoter. Show that the meteorite's acceleration is inversely proportional to S2 • 88. Particle acceleration A particle moves along lhe x-axis wilh velocity dx/dt ~ f(x). Show that lhe particle's acceleration is

f(x)1'(x). 89. Temperature and the period of a pendnIum For oscillations of small amplitude (short swings), we may safely model the relationship betweeo the period T and the leogth L of a simple peodulum with the equation

T

~ 2'1f~,

where g is the constant acceleration of gravity at the peodulum's

location. If we measure g in centimeters per second squared, we measure L in ceotimeters and T in seconds. If lhe peodulum is

Implicit Differentiation

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

a. Graph dg/dt (where defmed) over [-"., ".].

b. Findd//dt. c. Graph d// dt. Where does the approximation of dg/dt by d//dt seem to be best? Least good? Approximations by 1rigonome1ric polynomials are important in the theories of heat and oscillation, but we must not expect too much of them, as we see in the next exercise.

constant, tiL du = kL. 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 composites

(j

0

g)(x) =

lxi' =

are both differentiable at x

o

and

(g

0

/)(x) =

Ix'i =

91. The derivative of sin 2x Graph the function y = 2 cos 2x for -2 '" x '" 3.5. Then, on the same screen, graph

+ 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. 92. The derivative of cos (x') Graph y = -2x sin (x') -2 '" x '" 3. Then, on the same screen, graph cos ((x

y=

+ h)') -

~

g(t)

'~f(t)

/"

o

-"

= 0 even though g itself is not differ-

sin 2(x

, /

x'

entiable at x = O. Does Ibis contradict the Chain Rule? Explain.

y=

o

x'

,

Ixl. Then the

Suppose that /(x) = x' and g(x) =

149

"

94. (Continuation ofExercise 93.) In Exercise 93, the 1rigonome1ric polynomial /(t) that approximated the sawtooth function g(t) on [-"., ".j 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 approximating 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.

cos (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.

,

COMPUTER EXPLORATIONS

I

Trigonometric Polynomials

-"

93. As the accompanying figure shows, the 1rigonometric "polynomial"

k(t)

~h(t)

1

-2"

~

"2

0

"

-1

s = /(t) = 0.78540 - 0.63662 cos 21 - 0.07074cos6t - 0.02546 cos lOt - 0.01299 cos 14t

a. Graph dk/dt (where defmed) over [-"., ".j.

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 fann 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 technique.

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 usua1 way. Then we differentiate the equations inJplicitly, 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 implicitly as a differentiable function of x so that dyjax exists.

EXAMPLE 1 FIGURE 3.26 The curve + y' - 9xy ~ Dis not the graph of

x'

anyone function of x. The curve can,

however, be divided into separate arcs that are the graphs offunctions ofx. This particular curve, called afolium, dates to Descartes in 1638.

Find dyjax if y 2 = x.

The equation y2 = x defmes two differentiable functions of x that we can actoally find, namely y, = YX and Y2 = -YX (Figure 3.27). We know how to calculate the derivative of each of these for x > 0:

Solution

dYl ax

I 2YX

dY2 ax

and

1 2YX .

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 dyjax, 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

= -t

2Y2

x

The Chain Rule give. ::.

ax

-Yx

dy ax

t

=--

2Yx

FIGURE 3.27 The equationy2 - x = 0, or y2 = X as it is usually written, dermes two differentiable functions ofx on the interval x > O. Example I shows how to find the derivatives of these functions without solving the equation y2 = x for y.

(y 2 ) ~

d 2 dy dx [f(x)] ~ 2f(x)f'(x) ~ 2y dx'

2y dy = 1

Q(x, -vX)

Y2~

/

Slope

=

I 2y'

This one fannula gives the derivatives we calculated for both explicit solutions y, = YX andY2 = -YX:

dYl ax

EXAMPLE 2

I 2Yl

I 2YX

and

d)i2 ax

I

2( -YX)

I 2YX .

•

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 Y2, 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 2V25=9

3 4'

3.7 Implicit Differentiation y

151

We can solve this problem more easily by differentiating the given equation of the cITcle implicitly with respect to x:

~(X2) + ~(y2) dx

dx

=

dx

dy 2x+ 2y dx=O

--_5*---~Ot----f;5;--~X

dy dx The slope at (3, -4) is--!I = Y (3, ....) FIGURE 3.28 The circle combines the graphs of two functions. The graph of Y2 is the lower semicircle and passes through (3, -4).

~(25)

x y'

-~ = 1. -4 4

Notice that unlike the slope formula for dy,)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 ofother implicitly defmed functions, we proceed as in Examples I and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides ofthe defming equation.

Implldt Differentiation 1. Differentiate both sides ofthe 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.

EXAMPLE 3

y

Find dy/dx if y2 = x 2

+ sinxy

(Figure 3.29).

T~i'+sinxy

Solution

We differentiate the equation implicitly.

y2 :. (y2)

FIGURE 3.29 The graph of y 2 = x 2 + sin>;)' in Example 3.

2Y: -

(cosxy)

=

x 2 + sinxy

= :.

(x 2) + :. (sinXY)

2y dy dx = 2x

+

d (xy ) (cosxy) dx

2y: = 2x

+

(cosxy)&

(x :)

=

2x + (cosxy)y

dy (2y - xcosxy) dx = 2x dy dx

+ x :)

Differentiate both sides with respect to x ...

. .. treatingY as a function of and using the Chain Rule.

x

Treatxy as a product

Collect terms with dy/dx.

+ ycosxy

2x + Y cosxy 2y - x cosxy

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 also be used to find higher derivatives.

EXAMPLE 4

Find d'y/dx 2 if 2x' - 3y 2 = 8.

Solution To start, we differentiate both sides ofthe equation with respect to x in order to findy' = dy/dx.

JI (2x' dx

JI (8)

- 3y 2) =

dx

1leatyasa functianofx.

6y'y' = 0

6>;2 -

2

y'

x y,

=

whenyoF 0

Solvefory'.

We now apply the Quotient Rille to fmd y" . y" =

Finally, we substitute y'

=

fx (~)

y ;,x2 ' =

= 2xy

~ - ;~.y'

x 2/y to express y" in terms ofx andy. 2

2

_ 2x x (x- ) x' _ 2x y" ----Y

Curve of lens smface

y2

Y

Y

•

wheny oF 0

y"

Lenses, Tangents, and Normal Lines In the law that describes how light changes direction as it enters a lens, the important an-

gles 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 to 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 perpendicu1ar to the tangent ofthe profile curve at the point of entry. FIGURE 3.30 TIre profile ofa leos, showing the beoding (refraction) ofa ray oflight as it passes through the leos

surface.

y

EXAMPLE 5 Show that the point (2, 4) lies on the curve x' + y' - 9xy find the tangent and normal tu the curve there (Figure 3.31).

=

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 firat use implicit differentiation to fmd a fonnu1a for dy/dx: x'+y'-9xy=0

3x

2

JI (x') + JI (y') dx

dx

-

JI (9xy)

=

JI (0)

dy + 3y 2 dx

- 9 (d x dx

Y

+ Y dx) dx

=

0

dy (3y 2 - 9x) dx

FIGURE 3.31 Example 5 shows how to fmd equations for the taogeot and normal to the folium ofDescartes at (2, 4).

dx

+ 3x 2

-

dx

9y = 0

3(y2 - 3x) dy = 9y - 3x 2 dx dy 3y-x 2 dx

y2 - 3x'

Differentiate both sides with respect to x. TreatX)' as a product andy

as 8 function of x.

Solve for dy/d<.

153

3.7 Implicit Differentiation We then evaluate the derivative at (x, y) = (2,4):

dyl dx

3(4)-22 8 4 = 42 - 3(2) = 10 = 5'

3Y-X21 (2,4) = y 2 -

3x

(2,4)

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 perpendico1ar 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 - 27x 3 -

4-

x ; -

)~6 - 27X3) J.

Using in3plicit differentiation in Example 5 was much sin3pler than calco1ating 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

Find dr/dO in Exercises 15-18.

Use implicit differentiatioo to fmd dy/ dx in Exercises 1-14. 1. x'y + xy' ~ 6 2. x 3 + y3 ~ ISx)' 3. 2xy + y' ~ x + Y 4. x 3 - xy + y3 ~ I

+ 7)'

S. x'(x - y)' ~ x' - y'

6. (3xy

,_x-I 7·y-x+1

3_ 2x - y 8.x- x + 3y

9. x = tany 11. x

+ tan (xy)

13. ysin

G) ~

~ 6y

I - xy

12. x'

+ siny

14. xcos(2x

= x 3y'

+ 3y)

+ r 1/2

17. sin (rO)

=

2v8 =

= I

16. r -

t

18. cosr + cotO

10'/3 2

+ 1 03/. 3

= TO

Second Derivatives

10. xy = cot(xy) = 0

15. 01/2

~ ysinx

In Exercises 19-26, use implicit differentiatioo to fmd dy/ dx and then d'y/dx'. 20. x'/3 + y'/3 = I 19.x'+y'=1 22. y'-2x~ 1-2y 21. y' ~x' + 2x 23. 2yY ~ x - y

24.xy+y'~1

154

Chapter 3: Differentiation

25. Ifx' 26. Ifxy

+ y' ~ 16, fmd the value ofd"y/dl;2 at tire point (2, 2). + y2 = I,findthe value ofd"y/dl;2 at the point (0, -I).

43. The devll's curve (Gabriel Cramer, 1750) Find the slopes of tire devil's curve y' - 4y 2 ~ x' - 9x 2 at tire fuur indicated points.

In Exercises 27 aod 28, fmd the slope of the curve at the given points.

+ x 2 ~ y' (x 2 + y2j2 = (x

27. y2

2x

28.

-

at

yj2 at (1,0) aod (I, -I)

Slopes, Tangents, and Normals In Exercises 29-38, verifY that tire given point is on the curve aod fmd the lines that are (a) taogent aod (b) normal to the curve at the given point.

29. x 2 + xy - y2 ~ I, 30. x 2 31.

+ y2

~ 25,

xV ~ 9,

--f--*----tc-->

X

(2,3)

(3, -4)

(-1,3)

32. y2 - 2x - 4y - I = 0, 2

35. 2xy + 'If siny = 2'1f,

44. The folium of Deseartes

(-2, I)

+ 3xy + 2y 2 + 17y - 6 = 0, 2 34. x - V3xy + 2y 2 = 5, ( V3, 2 ) 33. 6x

y

(-2, I) aod(-2,-1)

(-1,0)

(I, 'If/2)

36. xsin2y = ycos2x, ('If/4,'If/2) 37. y ~ 2 sin ('IfX - y), (1,0) 38. x 2 cos2 y - siny ~ 0,

(See Figure 3.26.)

a. Find the slope of the folium ofDescartes x'

+ y'

- 9xy ~ 0

at the points (4, 2) aod (2, 4). b. At what point other thao the origin does the folium have a horizontal tsngent?

c. Find the coordinates of the point A in Figure 3.26, where the folium has a vertical tsngent.

(0,'If)

39. Parallel tangents Find the two points where the curve x 2 + xy + y2 = 7 crosses the x-axis, aod show that the taogents to the curve at tlrese points are parallel. What is the common slope of these taogents?

Theory and Examples 45. Intersecting normal The line that is normal to tire curve x 2 + 2xy - 3y 2 = 0 at (I, I) intersects the curve at what other point?

40. Normals parallel to a line Find the normals to the curve xy + 2x - y ~ 0 that are parallel to the line 2x + y ~ O.

46. Power rule for rational exponents 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,

41. The eight curve Find the slopes of the curve y' ~ y2 - x 2 at the two points shown here.

~ dl;Xpl.

y

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

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 accompaoying diagram, then a must be greater thao 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 cissoid ofDioeles (from about 200 B.c.) Find equations for the tsngent and normal to tire cissoid of Diocles y2(2 - x) ~ x' at (1,1). y

= EqX "'1.)-1 .

3.8

Related Rates

155

x' D 10 Exercises 51 and 52, fmd both dy/dx (treatingy as a differeotiable

48. Is there anything special about the tangents to the curves y' ~ and 2;x' + 3y 2 = 5 at the points (I, ± I)? Give reasoos for your

answer. y

function of x) and dx/dy (treating x as a differeotiable function ofy). How do dy/dx and dx/dy seern to be related? Explain the relationship geometrically in terms of the graphs. 51. xy'

+ x'y

52. x'

= 6

+ y2

= sin2 y

COMPUTER EXPLORATIONS Use a CAS to perform the following steps in Exercises 53-{)(). a. Plot the equation with the implicit plotter ofa CAS. Check to see that the given point P satisfies the equation.

-f----=+
b. Using implicit differeotiation, find a formula for the derivative dy/dx and evaluate it at the given point P .

49. Verify that the following pairs ofcurves meet orthogouaIly. a. x 2 + y2 = 4, x 2 = 3y 2 b. x = I - y',

x =

13 y '

x'

SO. The grapb of y2 = is called a semi.obi.a! parabola and is shown in the accoropanying figure. Determine the constant b so that the line y = x + b meets this graph orthogouaIly.

-l

•• Use the slope found in part (b) to fmd an equation for the tangent line to the curve at P. Then plot the implicit curve and tangent line together on a single graph. 53.

x' -

+ y'

~ 7,

P(2, I)

54. x' + y'x + yx2 + y4 55. y

2

57. x

59.

60.

= 4,

P(I, I)

_2+x

+ Y - I _ x' P(O, I)

+ cos xy

~ x 2,

+ tan (~)

= 2,

56. y'

P(l,O)

P(I,

';l)

0)

~ I, p(;f, 2y' + (xy)l/' ~ x' + 2, P(l, I) xvT+2,Y + y = x2, P(I,O)

58. xy'

3.8

xy

+ tan (x + y)

Related Rates In this section we look at problems tbat 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 f'md an equation relating the rates of change of V and r,

So if we know the radius r of the balloon and the rate dV/tit at which the volume is increasing at a given instant of time, then we can solve this last equation for dr/tit 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 illustmted in the following example.

EXAMPLE 1 Water nms 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 risiDg when the water is 6 ft deep? Solution

Figure 3.32 shows a partially tilled 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 are the dimensions of the tank. We are asked for dy/dt when FIGURE 3.32 The geometry of the conical tank and the rate at which water :rills the tank determine how fast the water level rigell (Example 1).

y

~

6 ft

dJ, ~ 9 ft'/min.

and

The water forms a cone with volume

This equation involves x as well as V and y. Because 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 _

di Finally, use y

=

6 and dV/tit

=

1T 3 2dy _ 1T 2dy 12' ~ tit -"4Y dt'

9 to solve for dy/dt.

9~~(6)': dy

dt =

1 1T Rl

0.32

At the moment in question, the water level is rising at about 0.32 ft/min.

•

Related Rates

3.8

157

Related Rate. Problem Strategy 1. Draw a picture and name the variable" and constants. 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 whatyou 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 Imd 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 angle is w/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 - tr/4

,

[" -, d'

wben8 - '1d4

FIGURE 3.33 The rate of change ofthe balloon's height is :related to the rate of change of the angle the range finder makes with the ground (Example 2).

Solution We answer the question in six steps. 1. Draw a picture and name the variables and constants (Figure 3.33). The variables in the picture are 6 = 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 6 and y are differentiable functions oft. The one constant in the picture is the distance from the range finder to the liftoffpoint (500 ft). There is no need to give it a special symbol. 2.

Write down the additional numerical information.

-::

~

0.14md/min

8

wben

~

:

3.

Write down what we are tofind. We want dy/dt when 8 - w/4.

4.

Write an equation that relates the variables y and 8.

y 500 - tan8 5.

or

y - 500 tan 8

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

~

500 (sec' 8) d8 dt

6. Evaluate with 8 = v/4andd8/dt = 0.14tofinddy/dt. :

~ 500(Vi)2(0.14) ~

140

At the moment in question, the balloon is rising at the rate of 140 ft/min.

•

EXAMPLE 3 A police cnUser, approaching a right-angled intersection from the north, is chasing a speeding car that has twned the corner and is now moving straight east. When the cruiser is 0.6 mi north ofthe 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 ofmeasurement, what is the speed ofthe 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 highway and the positive y-axis as the southbound highway (Figure 3.34). We let I represent time and set

x

y dy dt

= position of car at time I

y = position of cruiser at time I

~-60 -nl---~X:lD-"_x o dx_? x tit - .

FIGURE 3.34 The speed of the caris related to the speed of the police cruiser and the rate ofchange of the distance between them (Example 3).

s = distance between car and cruiser at time I. We assume that x, y, and s are differentiable functions of I. We want to f'md dx/dl when

x

= 0.8mi,

y

dy

= 0.6mi,

tis dl = 20 mph.

dl = -60mph,

Note that dy/dl 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 2sdl=2xdl+2Ydl

~=}(x:+yt) I

=

Yx2

+ y2

(x

dx dl

+ y dY )

dl'

Finally, we use x = 0.8, y = 0.6, dy/dl = -60, tis/dl = 20, and solve for dx/dl. 20 = Y

I (0.8: (0.8)' + (0.6)'

dx 20Y(0.8)2 ~=

+

(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 ofradius 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 along 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 I 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 orr/2 radians in 1/2 min, or orr rad/min. In other words, dO/dl = -orr, with I being measured in minutes. The negative sign appears because 0 is decreasing over time. Setting x(l) to be the distance at time I from the point Q to the origin, we want to f'md dx/dlwhen

x

= 20

ft

and

~

= -orr 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

dx dO dt = 10 sec 0 tan 0 dt = -1O'IT sec 0 tan O. Note that dx/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. •

A

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 ofsight ofa 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 approaching 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?

_"------"--------'-~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 dx/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 offect for our distance units, the last equation translates to

x

=

12,000 5280 cotO.

Differentiation with respect to t gives

dx 1200 _2 dO dt = - 528 cs.- 0 dt' When 0 = 'IT/6, sin2 0 = 1/4, so esc' 0 = 4. Converting dO/dt = 2/3 deg/sec to radians per hour, we find

dO 2('IT) dt = 3 180 (3600) radJhr.

1 br

~

3600 sec, 1
~

1T/180 rad

Substitotion into the equation for dx/dt then gives

dx '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? We let OM be the horizontal line of length x ft from a point 0 directly below the pulley to the worker's handM at any instant of time (Figure 3.37). Let h be the height of the weight Wabove 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): SoLution

o

(a)

+z + x2

20 - h 20 2

j

If we solve for z = 25 we have

z

20ft

+

45

=

Total length of rope is 45 ft.

= z2

Angle at 0 is a right angle.

h in the first equation, and substitute into the second equation,

20 2

h j.

+ x2 =

(25

+ hf.

(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).

+ 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)2

=

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.

•

Exerdses 3.8 1. Area Suppose that the radius r and area A = 7Tr 2 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 = 47Tr 2 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. Assume that 2x

+ 3y

= 12 anddy/dt = -2. Finddx/dt.

8. Ifx 2y 3 = 4/27 anddy/dt = 1/2, then what is dx/dt when x = 2? 9. If L = Yx 2 + y2, dx/dt = -1, and dy/dt = 3, find dL/dt when x = 5 andy = 12. 10. If r + s2 + v 3 = 12, dr/dt = 4, and ds/dt = -3, find dv/dt whenr = 3 ands = 1. 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

5. Ify = x 2 anddx/dt = 3, then what is dy/dtwhenx = -I?

a. surface area change?

6. Ifx =y3 - yanddy/dt = 5,thenwhatisdx/dtwheny = 2? 7. Ifx 2 + y2 = 25anddx/dt = -2,thenwhatisdy/dtwhenx = 3 andy = -4?

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? h. 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 ofa 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? h. 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?

Related Rates

161

a. How is dA/dt related to d6/dt if a andb are constant? h. How is dA/dtrelated to d6/dt and da/dt ifouly b is constant? c. How is dA/dt related to d6/dt, da/dt, anddb/dt ifnone of a, b, and 6 are 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. Changing dimensions in a rectangle The length I of a rectangle is decreasing at the rate of2 cro/sec while the width w is increasing at the rate of 2 cro/ sec. When I = 12 cro and w = 5 cro, fmd the rates of change of (a) the area, (b) the perimeter, and (c) the lengths ofthe diagonals ofthe rectangle. Which ofthese quantities

are decreasing. and which are increasing?

15. Changing voltage The voltsge V (volts), curreot 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 denote time in seconds.

+I~­ ,1--,

~---ll

R

a. What is the value of dV/dt? h. 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 = 12 volts and I ~ 2 amp. Is R increasing, or decreasing? 16. Eleetrical power The power P (watts) of an electric circuit is related to the circuit's resistance R (ohms) and curreotI (amperes) by the equation P ~ RI'.

a. How are dP/dt, dR/dt, and dI/dt related ifnone ofP, R, and I

22. Changing dimension. in a rectangular box Suppose that the edge lengths x, y, and z of a closed rectangular box are changing at the following rates:

dx dt = 1m/sec,

dy dt = -2 m/sec,

dz dt = 1m/sec.

Find the rates at which the box's (a) volume, (b) surface area, and

V

x' + y' + z' are changing at the (c) diagonal length s = instant when x = 4,y = 3, andz = 2. 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 hase is 12 ft ftum the house, the hase is moving at the rate of5 ft/sec. a. How fast is the top ofthe ladder sliding down the wall then? h. At what rate is the area of the triangle formed by the lAdder, wall, and ground changing then? c. At what rate is the angle 6 between the ladder and the ground changing then? y

y(t) •

are constant? t 3-ft ladder

h. How is dR/dt related to dI/dt if P is constant? 17. Distance

s ~

Let x and y be differentiahle functions of t and let

Vx' + y' he the distance hetween the points (x, 0) and (0, y)

in the xy-plane.

6

----'--------'---='~x

a. How is tis/dtreJAted to dx/dt ify is constant? h. How is tis/dtreJAted 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, and z are lengths ofthe edges of a rectangular hox, the common length of the hox's diagonals is s ~

Vx'

+ y' +z'. a. Assuming that x, y, and z are differentiahle functions of t, how is tis/dtrelated to dx/dt, dy/dt, anddz/dfl h. How is tis/dtreJAted 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 and b enclosing an angle ofmeasure 8 is A = tabsin6.

o

x(t)

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 angles. PlAne A is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nautical mile is 2000 yd). PlAne B is approaching the intersection at 481 knots. At what rate is the distance between the planes changing 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 of3oo ft, the wind carrying the kite horizontally away ftum her at a rate of25 ft/sec. How fast must she let out the string when the kite is 500 ft away ftum her? 26. Boring a cylinder The mechanics at Lincoln Automotive are rehoring a 6-in.-deep cylinder to fit a new piston. The machine

they are using increases the cylinder's mdius one thousandth ofan 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 10m3jmin 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 3jmin from a shallow concrete conical reservoir (vertex down) ofbase 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 ftjsec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ftj 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 3jmin from a reservoir shaped like a hemispherical bowl of radius 13 m, shown here in profile. Answer the following questions, given that the volume ofwater in a hemispherical bowl ofradius R is V = ('77"j3)y2(3R - y) when the water isy 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 in3jmin. 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 100'77" ft3jmin. 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 ftjsec. a. How fast is the boat approaching the dock when 10ft of rope are out? b. At what rate is the angle (J changing at this instant (see the figure)? Ring at edge of do;;ck;qd-__ I I

e

I

6':

r-

I .....,r----------I

35. Cardiac output In the late l860s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wiirzberg, Germany, developed one ofthe 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 Ljmin. At rest it is likely to be a bit under 6 Ljmin. If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 Ljmin. Your cardiac output can be calculated with the formula Q

y=

15'

3.8

where Q is the number of milliliters of CO2 you exhale in a minute and D is the difference between the C02 concentration (mlfL) in the blood pumped to the lungs and the Co, concentration in 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/rnm,

Related Rates

163

40. A building's shadow On a morning of a day when the suo will pass directly overhead, the shadow of an 80-ft building on level ground is 60 ft long. At the moment in question, the angle 6 the sun makes with the ground is increasing at the rate of 0.27" /min. At wbat rate is the shadow decreasing? (Remember to use radians. Express your answer in inches 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. Kenneth Herd, M.D., Quillan College ofMedicine, East Tennessee State University.) Suppose that when Q = 233 and D = 41, we also know that D is decreasing at the rate of 2 units a minute but that Q remains uncbanged. What is bappening to the cardiac output? 36. Moving along a parabola A particle moves along the parabola y = xl in the Ill'St quadrant in such a way that its x-coordinate (measured in meters) increases at a steady 10 m/sec. How fast is the angle of inclination 6 of the line joining the particle to the origin changing when x = 3 m? 37. Motion in the plane The coordinstes of a particle in the metric xy-plane are differentiable functions of time t with dx/ dt = -Im/secanddy/dt = -5m/sec. How fast is the particle's distance from the otigin cbanging as it passes thtough the point (5, 12)? 38. Videotaping a moving ear You are videotaping a race from a stand 132 ft from the track, following a car that is moving at 180 mi/h (264 ft/sec), as shown in the accompanying figure. How fast will your camera angle 6 be changing when the ear is right in front of you? A balf second later? Camera 9

A spherical iron ball 8 in. in diameter is coated with a layer ofice ofunifonn thickness. If the ice melts at

41. A meltiog 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 decreasing? 42. ffighway patrol A highway patrol plane flies 3 mi above a level, straight road at a steady 120 mi/h. The pilot sees an oncoming car and with radar determines that at the instant the lin.,.ofsight distance from plane to car is 5 mi, the line-
I

1

a. At what rate is the player's distance from third base cbanging when the player is 30 ft from fust base?

I

1 I

1

b. At what rates are angles 6, and 62 (see the figure) cbanging at that time?

1132' I I I

e. The player slides into second base at the rate of 15 ft/sec.At what rates are angles 6, and 62 cbanging as the player touches base?

1 1 I

d Car 39. A moving shadow A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft away from the light. (See accompanying figure.) How fast is the shadow of the ball moving along the ground 1/2 sec later? (Assume the ball falls a distance s = 16t2 ft in t see J

Second base V

90'

9~Player 30'

Light

, Ball at time t

~

0

1/2 sec later

Home

50-ft pole

Shadow "='>----~--------<_·----"~-x

o

First

< base

30

x(t)

NOf TO SCALE

44. Ships Two ships are stearning straight away from a point 0 along routes that make a 120" angle. 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 moving 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 dyJdx 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 briefinterva1 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

=

0~-=="====='===cV2

o

2x-l at(l,l).

Tangent and curve very close near (1, 1).

1.2

1.003

0.8~=~=='==='==VI.2

0.997 S;="=~="==='==='c=0 1.003 0.997

0.8

Tangent and curve very close throughout entire x-interval shown.

y

Tangent and curve closer still. Computer screen cannot distinguish tangent from curve on this x-interval.

FIGURE 3.38 TItemore we magnify the grapb ofa function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent

In general, the tangent to y = I(x) at a point x = a, where 1 is differentiable (Figure 3.39), passes through the point (a, I(a)), so its point-slope equation is y = I(a) --;cOt----aL-------~x

FIGURE 3.39 TIre tangent to the curve y = f(x) atx = a is the line L(x) ~ f(a) + I'(a)(x - a).

+ f'(a)(x - a).

Thus, this tangent line is the graph of the linear function

L(x) = I(a)

+ f'(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).

165

Linearization and Differentials

3.9

If f is differentiable at x = a, lben lbe approximating function

DEFINmDNS

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

\I'lh atx =

Find lbe linearization of f(x) =

0 (Figure 3.40).

y

=1+ ~

y

~

l.l

y= 1 + ~ 2

7

2

y~Vf+X

1.0

---!-----:ct----=---:----::----:------> x -1 0 2 3 4

0.9

FIGURE 3.40 The graph of y = y'l+-.;: and its linearizatioos atx ~ 0 and x ~ 3. Figure 3.41 showse

FIGURE 3.41 Magnified view of the window in Figure 3.40.

L -_ _-"--_ _---"-_ _----'

-0.1

0

0.1

0.2

magoified view of the small window about loathe y-axis.

Solution

Since

f'(x)

=

to +

xr' /2 ,

webavef(O) = landf'(O) = 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 \I'lh '" I + (x/2) from Example I is for some values of x near O. All we move away from zero, we lose accuracy. For example, for x = 2, lbe linearization gives 2 as !be approximation for which is not even accurate to one decimal place.

V3,

True value

Approximation

v'i2 '" I + O:}

Vl.Os '"

I

v'1.005 '" I

+

Og5

+ O.l~()5

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 !bat wbatever 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 yI]h for x close to 0 and can tolerate the small amount of error involved, 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 yI]h '" I + (x/2) will probably be too crude to be useful near x = 3. There, we need the linearization at x = 3.

EXAMPLE 2 Solution

yI]h at x

Find the linearization of I(x) =

= 3.

We evaluate the equation defining L(x) at a = 3. With

1(3) = 2,

/,(3)

=

1 (I + x)-1/21 2

= x-3

1, 4

we have

( )= 2 Lx

I

+ 4(x -

5

x

•

4 + 4'

3) =

At x = 3.2, the linearization in Example 2 gives

yI]h

=

VI + 3.2 '"

which differs from the true value linearization in Example I gives

yI]h

=

%+ 342 =

+ 0.800

1.250

= 2.050,

v'4.2 '" 2.04939 by less than one one-thousandth. The

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 I(x) = cosx at x = ",/2 (Figure 3.42).

Solution Since 1(",/2) = cos(",/2) = 0, /,(x) = -sinx, and/,(",/2) = -sin(",/2) = - I , we find the linearization at a = ",/2 to be

L(x) = I(a) + /,(a)(x - a) --;;-r--~c-----,--x

FIGURE 3.42 The graph of f(x) = cosx and its linearizatioo at x ~ ",/2. Near x ~ ",/2, cos x '" -x + (",/2) (Example 3).

+ (-I)

=

0

=

'" -x + 2'

(x - ~) •

An important linear approximation for roots and powers is (I

+ x)k

'" I

+ kx

(xnearO;anynumberk)

(Exercise 13). This approximation, good for values of x sufficiently close to zero, has broad application. For example, wben x is smail,

yI]h '"

I

+

tx

-1_1- = (I - X)-1 '" I

-x

k ~ 1/2

+ (-I)(-x)

'\0/1 + 5x 4 = (I + 5x 4)1/3 '" I + 1(5x 4) 3 I

v'l=7

=

= I

=

k

I + ~x4 3

(I - x 2r1/2 '" I + (_1)(_x 2) 2

+x

=

~

-I; replace x by -x.

k ~ 1/3; replace x by 5:c'.

I + l x2 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. Contrllry to its appearance, 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 = j'(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 ofthe function I, then these values determine the numerical value of dy.

EXAMPLE 4 (a) Find dy ify = x S

+ 37x.

(b) Find the value of dy when x = I and ax = 0.2. Solution

(8) dy = (5x' + 37) ax (b) Substitoting x = I and ax = 0.2 in the expression for dy, we bave

dy = (5 -I'

+

•

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

+ fix)

- f(a)

~~~l~,n j : When dx is a small change in x, I the corresponding change in : the linearization is precisely dy.

Tangent line

I

-j------'------"----------_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

= j'(a) ax.

+ fix)

a] - I(a) L(a)

168

Chapter 3: Differentiation 11lat is, the change in the linearization of I is precisely the value of the differential dy when x = a and dx = Ax. Therefore, dy 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

= f'(x) dx

in place of dy = f'(x) dx, calling dl the differential of I. For instance, ifI(x) = 3x' - 6, then

dl

=

d(3x' - 6)

= 6x dx.

Every differentiation formula like

+ u)

d(u

dx

du du =dx+dx

or

d(sin u) du dx = cosu dx

or

d(sinu) = cosu du.

has a corresponding differential form like

d(u

+ u)

= du

+ 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 + I)'

•

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)

'" I(a)

+ dy

when dx = ~x. Thus the approximation ~y '" dy can be used to estimate I(a when I(a) is known and dx is small.

= 0.1

+ 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 differeotial dA gives the estimate A(a

+ dr) = 1I"a' + dA

(Example 6).

Solution

Since A = 1I"r', the estimated increase is

dA = A'(a) dr = 211"a dr = 2".(10)(0.1) = 211" m'. Thus, since A(r

+

~r) '" A(r)

A(IO

+ dA, we have

+ 0.1) '" A(IO) + 2". = ".(10)' + 2".

= 102"..

3.9 Linearization and Differentials

169

The area ofa circle of radius 10.1 m is approximately 102'IT m' . The true area is

A(IO.I) = 'IT(IO.l)' = 102.01'IT m'

.

•

The error in our estimate is 0.01'IT m' , which is the difference aA - 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

=

I'(a) Ax.

How well does dl approximate Ill? We measure the approximation error by subtracting dI from Ill: Approximation error = III - dI = =

III - l'(a)Ax I(a + Ax) - I(a) - l'(a)llx /if

(!(a +

Il~

- I(a) _ I'(a»). Ax

Call this part E. = eo

Ilx.

As Ax --> 0, the difference quotient

I(a

+ Ilx) - I(a) Ax

approaches I'(a) (remember the definition of I'(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.

l'(a)llx

+ e Ax

true

estimated

error

change

change

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 = II

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.

=

I'(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

=

-dA

-

+ 0.01 'IT) m2

so the approximation error is aA - dA = E/1r = 0.01'IT and O.OI'ITjO.1 = O.I'IT m.

error E

=

0.01 'lTj /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 1 is differentiable at g(xo) , then the composite is differentiable at Xo and the derivative ofthe composite satisfies the equation

: Ix~xo =

I'(g(xo))' g'(xo).

Let /1x be an increment in x and let /1u and /1y be the corresponding increments in u andy. Applying Equation (I) we have

/1u where El

--->

=

+ El /1x

=

(g'(xo)

l'(uo)/1u + E2/1u

=

(I'(uo) + E2)/1u,

g'(xo)/1x

+ E,J/1x,

0 as /1x ---> o. Similarly,

/1y

=

where E2 ---> 0 as /1u ---> O. Notice also that /1u ---> 0 as /1x ---> O. Combining the equations for /1u and /1y gives

/1y

=

(j'(uo) + E2)(g'(Xo) + El)/1x,

so

Since El and E2 go to zero as /1x goes to zero, three ofthe four terms on the right vanish in the limit, leaving

dyl

-

ax

= X=Xo

lim -/1y = I'(uo)g'(xo) = I'(g(xo))' g'(xo).

J1x~O 4x

•

Sensitivity to Change The equation df = I'(x) dx tells how sensitive the output of 1 is to a cbange in input at different values of x. The larger the value of I' at x, the greater the effect of a given cbange dx. As we move from a to a nearby point a + dx, we can describe the change in 1 in three ways:

Absolute change Relative change Percentage change

True

Estimated

/11 = I(a + dx) - I(a) /11 I(a)

dl = dl

/11 I(a) X 100

I'(a)dx

I(a)

dl

I(a) X 100

3.9 Linearization and Differentials

171

You want to calculate the depth of a well from the equation s = 16t 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 = 32tdt depends on how big t is. If t = 2 sec, the change caused by dt = 0.1 is about

ds = 32(2)(0.1) = 6.4 ft. 1bree seconds later at t = 5 sec, the change caused by the same dt 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 = /cr', says that the volume V of fluid flowing through a small pipe or tube in a unit of time at a rIXed 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 uoblock 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 = 4/cr 3 dr.

The relative change in V is 3

dV = 4/cr dr = 4dr V /cr' 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,

d dv F = dt (mv) = m dt = 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 ofa body increases with velocity. In Einstein's corrected formula, mass bas the value

mo

m = -:-Vrl=_~if~/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, if/ 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)m oif 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 L1nearlzattons

Approxlmatton Error

In Exercises 1-5, find the linearizationL(x) of f(x) atx = a.

In Exercises 29-34, each function f(x) changes value when x changes from xo to Xo + dx. Find

1. f(x) = x' - 2x

2. f(x) ~

+ 3, a

W+9, 1

3. f(x)=x+x'

= 2

a ~ -4

a. the change 4f

~

a

tanx,

(b) cosx

+ 2x, Xo

xo

8. f(x) = X-I, 9. f(x) =

Find the lineariza-

lif ~ f(xo + dx)

(c) tanx

10. f(x)

~

1

+ x, Xo

3,

f(xo)

--c-I----L-----'--~------>x

Xo

xo+dx

~

+ 2x, Xo = I, 2x + 4x - 3, Xo

~ l' Xo

30. f(x) = = -I, dx = 0.1 31. f(X) = x' - x, Xo = 1, dx = 0.1

8.1

32. f(X) = x" Xo = 1, dx = 0.1 33. f(X) = x-I, Xo = 0.5, dx = 0.1

= 1.3

+ x)'

at x ~ 0 is

14. Use the linear approximation (I + xl' '" 1 + Ax to find an approximation for the functionf(x) for values ofxnear zero.

a. f(x) = (1 - x)'

~ +x (4 + 3X)'/3

c. f(x) = •

dx = 0.1

2

13. Show that the linearization of f(x) ~ (I L(x) = 1 + Ax.

d. f(x) =

v'2+7 "

f. f(x) =

34. f(x) ~ x' - 2x

+ 3, xo

~ 2,

dx ~ 0.1

Dlfferenttal Estimates of Change In Exercises 35-40, write a differential formula that estimates the

given change in volume or s'Wface area. 35. The change in the volume V = (4/3)'lTr' of a sphere when the radius changes from ro to ro + dr

h. f(x) = 1 :. x

V1

e. f(x) =

-

fix

29. f(x) = x 2

Xo = -0.9

11. f(x) =~, Xo = 8.5

12. f(x) = x

L

Tangent

o

~ 0.1

df.,~ f'(xo) fix

(xo,f(xo)

= 0.9

+ 4x -

2x 2

- f(xo);

y

~ 'IT

L1nearlzatton for Approxlmatton 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. 7. f(x) ~ x 2

+ dx)

c. the approximation error 14f - dfl.

6. Common Hnear appnIlimations at x = 0 tions of the following functions at x = O.

(a) sinx

f(xo

h. the value ofthe estimate df = 1'(xo) dx; and

a= 1

4. f(x) =~, a = -8 5. f(x)

~

1(1 -

,,,

2

~ x)'

36. The change in the volume V = lengths change from xo to Xo + dx

x' of a

cube when the edge

37. The change in the surface area S ~ 6x 2 of a cube when the edge lengths change from xo to Xo + dx

+ xl' '"

38. The change in the lateral surface area S = 'lTrv'r 2 + h 2 of a right circular cone when the radius changes from ro to ro + dr and the height does not change

16. Find the linearization of f(x) ~ y';+J + sinx atx ~ O. How is it related to the individual linearizations of y';+J and sin x atx = O?

39. The change in the volume V = 'lTr 2 h of a right circular cylinder when the radius changes from ro to ro + dr and the height does notchange

Derivatives in Differenttal Form

40. The change in the lateral surface area S = 2'lTrh of a right circular cylinder when the height changes from ho to ho + dh and the

15. Faster than a calculator Use the approximation (I 1 + Ax to estimate the following.

a. (1.0002)"

h. '\7"1.009

In Exercises 17-28, find dy.

17. y ~

x' -

3v'i

2x l+x 21. 2y'/2 + -9' - x = 0

19.

Y~--2

18. y ~ 20.y~

24. y

25. y

26. y

= cos (x ) = sec(x 2 -

27. y

= 3 csc (1

28. y

=2

- 2v'i)

Applicattons

2v'i

( ./C) 3 I + vX 22. -9'2 - 4x'/2 - y

= sin (5v'i) = 4 tan (x'/3)

23. y

radius does not change

xv'l=7

41. The radius of a circle is increased from 2.00 to 2.02 m.

=0

a. Estimate the resulting change in area.

h. 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 ofmaterial in a cylindrical shell with length 30 in., radius 6 in., and shell thickness 0.5 in. 0.5 in. """'--/

6in.L.....

~30in.-'>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 ofyour investigation, you decide to compare the effect on W of a given change dg on the moon, where g = 5.2 ft/sec2 , with the effect the same change dg would have on Earth, where g = 32 ft/sec 2 • Use the simplified equation above to find the ratio of dWmoon to dWEarth· 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 Ii.T, 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 = 7Th 3 . The volume is to be calculated with an error of no more than I % 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 em 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+

b. If g increases, will T increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 1DO-em pendulum is moved from a location where g = 980 cm/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 ofthelinearizationL(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

. E(x) 2.hm x - a =0 x->a

The error is negligible when compared withx - a.

The linearization, L(x): y = f(a) + f'(a)(x - a)

\

Some other linear approximation, g(x): y = m(x - a) + c y =f(x)

-------'-------~)

+ gb (a, b constant).

I

x

54. Quadratic approximations a. Let Q(x) = bo + bl(x - a) + Mx - a? be a quadratic approximation to f(x) at x = a with the properties:

i) Q(a) = f(a) ii) Q'(a) = f'(a) iii) Q"(a) = j"(a).

W = a

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.

vat? 2g ,

where W is the work per unit time, P is the average blood pressure, V is the volume ofblood pumped out during the unit of time, a ("delta") is the weight density of the blood, v is the average velocity ofthe exiting blood, and g is the acceleration of gravity. When P, V, a, and v remain constant, W becomes a function ofg, and the equation takes the simplified form

=

Determine the coefficients bo, bl' and b2.

Chapter 3 Questions to Guide Your Review b. Find the quadratic approximation to j(x) x =

D c.

o.

~

a. Plot the function j over 1.

1/(1 - x) at

b. Fiod the linearization L of the function at the point o.

Grapb j(x)

= 1/( I

- x) and its quadratic approximation at

c. Plot j and L together on a single graph.

x ~ O. Then zoom in on the two graphs at the point (0, I). Comment on what you see.

D d.

Find the quadratic approximation to g(x) = I/x at x = I. Graph g and its quadratic approximation together. Comment on what you see.

D e.

Find the quadratic approximation to h(x) = ~ at x ~ O. Graph h and its quadratic approximation together. Comment on what you see.

f. What are the linearizations of j, g, and h at the respective points in parts (b), (d), and (e)?

COMPUTER EXPLORATIONS In Exercises 55-58, use a CAS to estimate the magnitude of the error

in using the lineatization in place of the function over a specified interval 1. Perform the following steps:

Chapter

2. What role does the derivative play in defming slopes, tangents, and rates of change? 3. How can you sOl13etimes 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 geOI13Oltica11y when a function typically does not have a derivative at a point 7. How is a function's differentiability at a point related to its continuity there, ifat all? 8. What rules do you know for calculating derivatives? Give sOl13e examples. 9. Explain how the three formulas

tL (x') ~ dx

d

(UI

e. From your graph in part (d), estimate as large a 8 > 0 as you can, satisfying

Ix -

Ij(x) - L(x)1 <

01 < 8

E

for E ~ 0.5, 0.1, and 0.01. Then check graphically to see if your 8-estimate holds true. SS. j(x) = x' + x 2 - z,;, [-1,2], 0 = I 56. j(x)

=

4: -+11' 2

H, I].

57. j(x) = x 2f3(x - 2),

58. j(x) =

VX -

sinx,

!

0

=

[-2,3],

0

= 2

[0,2".],

0

= 2

+ U2 + ... + Un)

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?

15. Give examples of still other applications of derivatives. 16. What do the limits Ii.m,,~o«sinh)/h) and J.im,,~o«cosh - 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 ofthese functions? 18. At what points are the six basic trigonometric functions continuous? How do you know?

19. What is the rule for calculating the derivative of a cOl13posite of two differentiable :functions? How is such a derivative evaluated? Give examples. 20. If U is a differentiable function of x, how do you fmd (d/dx)(u') if n is an integer? If n is a real number? Give examples.

ox,-I

21. What is implicit differentiation? When do you need it? Give examples. 22. How do related rates problems arise? Give examples.

du d b. dx (cu) ~ c dx

c. dx

d. Plot the ahsolute error Ij(x) - L(x) lover I and fmd its maximum. value.

Questions to Guide Your Review

1. What is the derivative of a function j? How is its domain related to the doruain of j? Give examples.

s.

175

=

dul

dU2

dun

dX + dX + ... + dX

enable us to differentiate any polynOl13ial. 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 instantaneous rates of change? Give an example.

23. Ontline a strategy for solving related rates problems. illustrate with an example. 24. What is the linearizationL(x) ofa functionj(x) at a point x ~ o? What is required of j at 0 for the linearization to exist? How are linearizations used? Give examples.

25. If x moves from 0 to a nearby value 0 + dx, how do you estimate the corresponding change in the value of a differentiable function j(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' - O.l25x 2 + 0.25x 2. Y = 3 - 0.7x' + 0.3x' ,

3. Y ~ x - 3(x

+

5. Y = (x 7. y = (6' 9. • = I

2

_,.r.; 4. y - x + V 7x -

2

+".-)

1)2(X 2 + 2x)

+ sec 6 +

1f + I

6. y = (2x - 5)(4 - xt

I)'

Vi + Vi

I

8. y = (-I _

'

cs~ 6 - ~)'

I

10. • =

Vi -

53. Find d'y/dx 2 by implicit differentiation:

+ y'

.. x'

= I

~

b. y2 = I -

x 2 - y2 ~ I

54. .. By differentiating dy/dx = x/yo b. Then show that d'y/dx 2

implicitly,

show that

= -I/y'.

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 = 1.

I

11. Y ~ 2tan2 x - sec'x

2 12. Y ~ - 2- - - . 5in X smx

x

f(x)

g(x)

j'(x)

g'(x)

o

14.'~COt'(t)

I 5

1/2

13•• ~ cos' (I - 2t)

I 3

-3

I

1/2

-4

15•• ~ (sec t

+ tan t)'

I

16•• ~ csc'(I - t

17. r ~ v'26sin6

+ 3t2 )

18. r ~ 26v'cos 6

v28

20. r

= sin (6 + v'8+1)

21• Y --"2x I 2 csc 2

22. Y

= 2'VX sin 'VX

23. y = x- 1f2 sec(2x)' 25. Y ~ 5 cotx2

24. Y

= 'VXcsc(x +

19. r = sin

x

27. Y = x 2 sin2 (2x 2)

26. Y ~ x 2 cot 5x 28. Y = x -2 sin2 (x')

29•• =

30••

31. y = 33. y =

C: S2 (;":J

~

+ I)~

(5x 2

y

34. Y

sin 6 )2 cos6 - I

37. Y = (2x 39. Y

32.

~X2x~ x

35. r = (

3

+ sin2x)'f2

=

45.

+ 4xy

ViY ~

- 3y'I' ~ 2x

I

= 4xv'x + =

38. y

= 20(3x -

+

xV ~

X= 0

b. f(X)g2(X),

x ~ I

d. f(g(x)),

e. g(f(x)), x = 0 g. f(x

f. (x

+

x ~ 0 x = I

f(x))'f2,

+ g(x)), x = 0

56. Suppose that the function f(x) and its frrst derivative have the following values at x = 0 and x = 1.

x

j'(x)

f(x)

o

9

-2

I

-3

1/5

'VX

Find the flTSt derivatives of the following comhinations at the given value of x.

(I + Sin6)2 I - cos6

40. Y ~ (3

48. y2 =

x 47.y2 -x+1

x = I

2'VX )2 2'VX + I

36. r

46.

f(x) c. g(x) + I'

-I 15(15t - I)'

=(

44. 5x'I'

.. 6f(x) - g(x),

I)'

.. 'VXf(x),

4)'/'(3x - 4t'/'

c. f('VX),

+ cos' 3x)-I/3

Impl;dt Diffel'l!ntlation 10 Exercises 41-48, fmd dy/dx by implicit differentiation. 41. xy + 2x + 3y = I 42. x 2 + xy + y2 - 5x = 2 43. x'

Find the flTSt derivatives of the following comhinations at the given value of x.

e. 2

x

+f(x) cosx'

ViW,

x ~ 0

x ~ I

b.

=I

d. f(1 - 5 tan x),

_ x - 0

. f. IOsm

(=) T

2

57. Find the value ofdy/dt att = Oify = 3sin2xandx = t 2

+ 1f. + 5t and t =

58. Find the value of ds/du at u = 2 if • = t 2 (u 2 + 2U)'/3.

59. Find the value of dw/ds at • ~ 0 if w ~ sin (vr r = 8sin(. + 1f/6).

I

60. Find the value of dr/dt at t = 0 if r = (6'

-

2) and

+ 7)'/3 and

6't+6~1.

+_ Xx

+ Y = 2cosx,fmdthevalueofd'y/dx 2 atthepoint(O,I). Ifx ' /3 + y'/' = 4,findd'y/dx 2 atthepoint(8,8).

61. Ify' 62.

10 Exercises 49 and 50, fmddp/dq. 49. p' + 4pq - 3q 2 = 2 50. q = (5p 2

_

x - I

f (x),

IOy·I' ~ 15

~II

x = 0

+ 2pt'f2

Apply;ng tile Derivative Definftfon 10 Exercises 63 and 64, fmd the derivative using the defmition.

10 Exercises 51 and 52, fmddr/d•.

51. rcos2s

+ sin2 s

=

1r

52. 2rs - r -

S

+ 82 =

-3

63. f(t) = 2t

I

+I

64. g(x)

= 2x 2 +

I

65. a. Graph the function

f(x) h. Is f continuous at x

=

X2, { -X 2 ,

~

07

-1:5x
c. Is f differentiable at x = 07

66. a. Graph the function

= {x,

h. Is f continuous at x

tanx,

~

c. Is f differentiable at x

-I :5x
07

Give reasons for your answers.

h. Is f continuous at x

86.

I7 ~

(3,2)

x? = 2>: + 4, (6,2) + v'iY ~ 6, (4, I) x 3/2 + 2y 3/2 ~ 17, (1,4)

85. x

2 - x,

87. Find the slope of the curve x 3y and (I, -I).

17

Give reasons for your answers. 68. For what value or values of the constant m, if any, is

f(x) = {Sin2>:, x:5 0 mx,

a. continuous at x

+ y2 = 2, (I, I)

84. (y -

= {x,

c. Is f differentiable at x

+ I)

80. Normal to a eire\e Show that the normal line at any point of the circle x 2 + y2 = a 2 passes through the origin.

83. xy + 2>: - 5y = 2,

~

he taogeot

78. Slope of tangent Show that the tangeot to the curve y = x 3 at any point (0, 0 3) meets the curve again at a point where the slope is four times the slope at (0, 0 3).

82. x 3

67. a. Graph the function

f(x)

+ C is to

77. Tangent parahola The parabola y ~ x 2 to the line y = x. Find C.

In Exercises 81-86, fmd equations for the lines that are taogent and nonna! to the curve at the given point. 81. x 2 + 2y 2 = 9, (1,2)

07 ~

177

79. Tangent curve For what value of cis the curve y = c/(x taogent to the line through the points (0, 3) and (5, -2)7

Give reasons for your answers.

f(x)

Chapter 3 Practice Exercises

3

+ y2

~ X

+y

at the points (1, I)

88. The graph shown suggests that the curve y = sin (x - sinx) ntight have horizontal tangents at the x-axis. Does it7 Give reasons for your answer.

x>O

= O?

y 1

h. differentiable at x = 07

y

= sin (x -

sin x)

Give reasons for your answers. ~J..."o"'-----+---~:+-L----+--~~~x

Slopes, Tangents, and Nonnals 69. Tangeots with specified slope Are there any points on the curve y = (x/2) + 1/(2): - 4) where the slope is -3/27 If so, fmd them.

-2w

2w

-1

70. Tangeots with specified slope Are there any points on the curve y = x - 1/(2):) where the slope is 37 If so, fmd them. 71. Horizontal tangents Find the points on the curve y ~ 2>:3 - 3x 2 - 12>: + 20 where the tangeot is parallel to the x-axis. 72. Tangeot intercepts Find the x- and y-intercepts of the line that is tangeot to the curve y = x 3 at the point (-2, -8).

73. Tangents perpendicular or parallel to lines Find the points on the curve y ~ 2>:3 - 3x2 - 12>: + 20 where the tangent is

a. perpendicular to the line y h. parallel to the line y

~

= V2 -

Analyzing Graphs Each of the figures in Exercises 89 and 90 shows two graphs, the graph of a function y = f(x) together with the graph of its derivative f' (x). Which graph is which7 How do you know? 89.

y

90.

y

I - (x/24). 12>:.

74. Intersecting tangents Show that the taogents to the curve y ~ ('IT sin xlix at x ~ 'IT and x ~ -'IT intersect at right angles. 75. Normals parallel to a line Find the points on the curve y = tan x, -'IT/2 < x < 'IT/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 taogent and normal to the curve y ~ I + cos x at the point ('IT/2, I). Sketch the curve, taogent, and normal together, labeling each with its equation.

_+-_---1-__-'-----> X o

-1

@

-2

2

178

Chapter 3: Differentiation

91. Use the following information to graph the function y

~

f(x) for

-1~x~6.

Trigonometric Um;ts Find the limits in Exercises 95-102.

i) The graph of f is made ofline segroents joined end to end.

95. lim

iii) The derivative of f, where dermed, agrees with the step function shown here. y

~f'(x)

L -+-2 y

x-a 2

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?

-I[

( 0, 1000

I

o

I

50

I - 2 cot' 0

-7cotO - 8 102. lim I - ;os 0

x sinx 2cosx

,~o

100

103. g(x) =

tan (tan x) tan x

104. f(x) =

tan (tan x) . (.

sm

SUlX

)

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 = 2.".r2 + 2.".rh.

a. 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 and dh/dt ifneither rnor h is

constant? d. How is dr/dtrelated to dh/dt ifSis constant? 106. Right circular cone The lateral surface area S of a right circular cone is related to the base radius r and height h by the equationS = .".rVr 2 + h 2 •

a. 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 ....a The radius of a circle is changing at the rate of -2/.". rnfsec. At what rate is the circle's area changingwhenr ~ 10m? 108. Cube's changing edges The volume of a cube is increasing at the rate of 1200 em' /min at the instant its edges are 20 em long. At what rate are the lengths ofthe edges changing at that instant?

-----

,

,"iii

I

Show how to extend the functions in Exercises 103 and 104 to be con-

93. a. What is the value of the derivative of the rabbit population when the number of rabbits is largest? Smallest?

I i.I

+

tinuous at the origin.

Exercises 93 and 94 are about the accompanying graphs. The graphs in part (a) show the numbers ofrabbits 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.

I

4 tan2 0 + tan 0 '~(rrf2)tan2 0 + 5

. I1m

101. lim

92. Repeat Exercise 91, supposing that the graph starts at (-I, D) instead of(-1,2).

-,

99.

'~O+5cot'0

-1~ 1 2 3 ~

x-a

98. lim sin (sin 0) 6-0 8

97. lim sin r 1'-0 tan 2r

100. lim

~ c~-=;7--;'0~) x -1

96. lim 3x - atan 7x

s:' x

x-a 2x - x

ii) The graph starts at the point (-I, 2).

150

200

Time (days) (s)

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

I

I

I

R = R, + R2 50

o

1---- - - - -

,~q.~-~ I

I I I I

I I I :

RZIR

I

I

-50

I

I

:

I 1

50

100

150

Time (days)

Derivative of the rabbit population (b)

I

-

~

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 andR2 = 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 = R 2 + X 2 • 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-p1ane 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 11 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 ft;3 /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 = -TT/4

b. sec x atx = -TT/4.

Graph the curves and linearizations together. 118. We can obtain a useful linear approximation of the function j(x) = 1/(1 + tan x) at x = 0 by combining the approximations _1_ 1+x

~ 1- x

and

tanx

~x

to get 1 Exit rate: 5 ft 3/min

1

+ tanx

~ 1 - x.

Show that this result is the standard linear approximation of 1/(1 + tanx)atx = O. 119. Findthelinearizationofj(x) = ~

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 = rO to find how fast (radians per second) the spool is turning when the layer ofradius 1.2 ft is being unwound.

115. Moving searchlight beam The figure shows a boat 1 km offshore, sweeping the shore with a searchlight. The light turns at a constantrate,dO/dt = -0.6rad/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?

120. Find the linearization of j(x) = 2/(1 atx = O.

+ sinx - 0.5 atx = O. - 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 h o to h o + dh and the radius does not change.

V= !'lTr 2h 3 S = 'lTrVr 2 + h 2 (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 ofno more than 2%?

180

Chapter 3: Differentiation

b. Suppose that the edge is measured with the accuracy required in part (a). About bow 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 measurement.

124. Finding height To rmd the height of a lamppost (see accompaoying figure), you staod a 6 ft pole 2D ft froro the lamp and measure the length a of its shadow, rmding it to be 15 fl, 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 spbere. Estimate the percentage errors in the calculated values of

h

a. the radius. b. the surface area. c. the volume.

Chapter

Jil

Additional and Advanced Exercises

1. An equation like sin' 8 + cos' 8 = 1 is called an identity because it holds for all values of 8. An equation like sin 8 = D.5 is not an identity because it holds only for selected values of 8, not all. 1f 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.

= 2 sinO cosO b. cos 28 = cos2 0 - 8in2 8 2. Ifthe identity sin (x + a) = a. 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. a. Find values for the constaots a, b, and c that will make

I(x) = cosx aod g(x) = a + bx + ex' satisfy the conditions

1(0) = g(O), /,(0) = g'(O),

and

/"(0) = g"(O).

b. Find values for b and c that will make

I(x) = sin(x + a)

g(x) = b sinx + c cos x

and

satisfy the conditions

1(0)

~

g(O)

and

/,(0)

~

c. For the determined. values of a, b, and c, what happens for the third and fourth derivatives of 1 and g in each ofparts (a) aod (b)? 4. Solutions to differential equations

y"

+y

~

O.

equation y"

+ 4y = O?

Geueralize this result. S. An osculating circle Find the values of h, Ie, and a that make the circle (x - h)' + (y - k)' = a' taogent to the parabola y = + 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 like this one that are taogent to a curve and have the same secood derivative as the curve at the point oftangency are called osculating circles (from the Latin osculari, meaning ''to kiss''). We encouoter them again in Chapter 13.

x'

6. Marginal revenue A bus will hold 6D 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 terms. For example, let u = I(t)

g'(O).

a. Shawthaty = sinx,y = cosx, andy = acosx (a and b constaots) all satisfy the equation

b. How would you modify the functions in part (a) to satisfy the

+ bsinx

be the number ofpeople in the lshor force at time t in a given industry. CWe 1reat this function as though it were differentishle even though it is an integer-valued step function.) Let v = g(t) be the average production per person in the labor force at time t. The total production is then y = uv. Ifthe lshor 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 ofthe balloon with cables tangent to the surface ofthe balloon, as shown. Two ofthe 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

=

lOcos(t

2:

181

0 of a particle

+ 7r/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 ----;.j I*-Width NOT TO 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 + 18t + 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 1 (2 2m v - VQ 2) = 2k xQ - x 2) , where k, VQ, and XQ are constants. Show that whenever v =f. 0,

dv = -kx mdt . 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 y= { mx + b, x

< 7r 2:

7r

is a. continuous at x = 7r. b. differentiable at x = 7r.

16. Does the function 1 - cosx x=f.O x'

j(x) = Mike McCarthy ofLondon jumped from the Tower ofPisa 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.)

{

0,

have a derivative at x = O? Explain.

x=O

182

Chapter 3: Differentiation 26. Leibniz's rule for hlghe,,"order derivatives of products Leibniz's rule for higher-order derivatives of products of differentiable functions says that

17. a. For what values of a and b will

f(x)

~

ax { ax; - bx

+

x < 2 3, x '" 2

d'(uv)

a. be differentiable for all values of x?

=

du dv

d'v

+ 2 dx dx + u dx"

+ 3 d'u dv + 3 dud'v + ud'v.

dx' dx dx dx' do-lu dv c - - IJ= - vIJ+ n - - - + ' " • dx dx dx"-l dx

18. a. For what values of a and b will ()

d'u dx' v

b. d'(uv) = d'u v dx' dx' dO(uv) dOu

b. Discuss the geometry of the resulting graph of f.

g x

----.J;2 =

{ax + b, x '" -I ax3 + x + 2b, x>-1

n(n - I) .. · (n - k

+

dx'

I) dO-'u d'v 1 dx1

+ ~-~-k';!---~dxo dOv +"'+U dxA '

be differentiable for all values of x? b. Discuss the geometry of the resulting graph ofg.

19. Odd differentiable functions Is there anything special about the derivative ofan odd differentiable function of x? Give reasons 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 mathe03Stical induction, using

20. Even differentiable functions Is there anything special about the derivative of an eveo differentiable function of x? Give reasons for your answer.

m) (k

21. Suppose that the functions f and g are defmed throughout an open interval containing the point Xo, that f is differentiable at XO, 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 IxlLl' differentiable at x ~ O. 22. (Continuation of ExercLl'e 21.) Use the result of Exercise 21 to show that the followiog functions are differentiable at x = O.

a. Ixl sin x

b. x'/' sinx

c."%(1 - cosx)

d. h(x) = {x' sin (I/x), x '¢' 0 0, x = 0

23. Is the derivative of h(x) ~ {x' sin (I/x), 0,

x '¢' 0 x= 0

continuousatx ~ O? How about the derivative ofk(x) Give reasons for your answers.

~

xh(x)?

24. Suppose that a function f satisfies the followiog conditions for all real values ofx andy: i) f(x + y) ~ f(x)·f(y). il) f(x) = I

+ xg(x) , where 1in3r-+Og(x)

= I.

Show that the derivative f'(x) exists at every value ofx and that f'(x) = f(x). 25. The generalized product rule

Use mathematical induction

to prove that if Y = "1"2'" Un is a finite product of differentiable functions, then y is differentiable on their common domain and dy dx

dUI

=

du,

duo

liX U2 "'un + UldX"'U + ... + u t u2 "'un- l liX' II

+

(m)

\k + I

m! = k!(m - k)!

m!

+ (k + I)!(m - k - 1)1'

27. The period of a clock pendulum The period T of a clock pendulum (time for one full swing and back) is giveo by the formula T' ~ ~L/g, where Tis measured in seconds, g ~ 32.2 ft/sec', and L, the length of the peodulum, is measured in feet. Find approximately a. the length of a clock peodulum 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. If we call its edge length s, its volume is V = 3 3 and its surface area is 63 2 . We assume that V and 3 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 constant. (It probably depends on many things, such as the relative homidity of the surrouoding air, the air temperatore, and the incidence or absence of sunlight, to name only a few.) Assume a particular set of conditions in which the cobe lost 1/4 of its volume during the first hour, and that the volume is Vowhen 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. Derivatives, Slopes, Tangent Lines, and Making Movies Part. I-ill. You will visualize thc derivative at a point, the linearization of a function, and thc 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 Movie.). Parts N and V of the module can he used to animate 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. ModonAlong II Straight Line: Position .... Velocity .... Acceleration Ohserve 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)

~

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

----I---.....---~x

FIGURE 4.1 Absolute extrema for the sine and cosine functions on [ -'IT/2, 'IT /2] . These values can depend on the domain of a function.

184

~

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

185

Extreme Values of Functions

EXAMPLE 1 The absolute ex1rema 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.

•

----"<,L-----:c- x

----'--<,L-----:c- x

----"......-----:c- x 2

(a) abs min only

FIGURE 4.2

HISWRICAL BIOGRAPHY

Daoiel Bernoulli (170
o.

(b) y = x 2

x ----"¥-----:c2

(b) aba max and min

2

(c) abs max only

I

2

(d) no max or min

Graphs for Example 1.

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 Extn!me Yallle Theoll!m Iff is continuous on a closed interval [a, h], then f attains both an absolute maximum 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)

-"------'-----\---+-+---+_x a

a Maximum and minimum at endpoints

(Xl. m) Maximum. and minimum. at interior points

-=---'-------"-~>X

a 2 b Maximum at interior point, minimum at endpoint

FIGURE 4.3

:m

-"-----~------"-~>X

a

Xl

b

Minimum at interior point, maximum at endpoint

Some possibilities for a continuous function's maximwn and

minimwn on a closed interval [a, b].

y

No largest value

"

need not hold Example 1 shows that an absolute extreme value may not exist if the interval fails to be both closed and finite. Figure 4.4 shows that the continuity requirement cannot 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,

Q::=:;x
0,

x= 1

is continuous at every point of[O, 1] except x ~ I, yet its graph over [0, I] does not have a highest point.

Figure 4.5 shows a graph with five points where a function has extreme values on its domain [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 I(c) a maximum locally. The function attains its absolute maximum at d. We now derme what we mean by local extrema.

DEFINmONS A function 1 has a local mBIimum value at a point c within its domain D if I(x) :5 I( c) for all XED lying in some open interval containing c. A function 1 has a local minimum value at a point c within its domain D if I(x) ;;,: I( c) for all XED lying in some open interval containing c.

lf the domain of 1 is the closed interval [a, bJ, then 1 has a local maximum at the endpoint x = a, if I(x) :5 I(a) for all x in some half-open interval [a, a + Ii), Ii > O. Likewise, 1 has a local maximum at an interior point x = c if I(x) :5 I(c) for all x in some open interval (c - Ii, c + Ii), Ii > 0, and a local maximum at the endpoint x = b if I(x) :5 I(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 1 has local maxima at c and d and local minima 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 function I(x) = sin (l/x) on the interval (0, lJ. CWe graphed this function in Figure 2.40.)

4.1 Extreme Values of Functions

187

Absolute maximum. No greater value off anywhere. AlBo a local maximum.. Local minimum No smaller value : off nearby. Absolute minimum. No smaller value of f anywhere. AlBo a

Local minimum. I No smaller value of :fnearby. : I

I

local minimum. : ~

_ _ _ _ _ _ _ _L I

I

:

L I-----";'-~':----_>x

a

FIGURE 4.5 ~ x ~ b.

: :

e

d

b

How to identifY types ofmaxima and minima for a functioo willi domain

a

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 ofits domain, and if I' is defined at c, then

THEOREM 2-The First Derivative Theorem for Local Extreme Values I

I I I I

I'{c) = O.

,

Secant slopes

2:

0

Secant slopes =s 0 (never po~itive)

(never ,negative) I I I I

I I I I

I

!

)

e

X

FIGURE 4.6 A curve with a local maximum. value. The slope at c,

simultaneously the limit ofnoopositive

Proof To prove that I' (c) is zero at a local extremum, we show first that I' (c) cannot be positive and second that I' (c) cannot be negative. The only number that is neither positive nor negative is zero, so that is what I' (c) must be. To begin, suppose that f has a local maximum value at x = c (Figure 4.6) so that f{x) - ftc) ,;; 0 for all values of x near enough to c. Since c is an interior point of !'s domain, I' (c) is derfied by the tw<>-sided limit

r

numbers and nonnegative numbers, is zero.

f{x) - ftc)

x~

xc'

This means that the right-hand and left-hand limits both exist at x When we examine these limits separately, we find that <: 0 f '{ C ) = I"lITI f{x)x- cftc) ' x-c+

=

c and equal I'{c).

Because (x - c) > 0 and f(x) '" f(e)

(1)

Similarly, f '{ C ) = I"llD_ f{x)X - Cftc) x-e

:>

0

-.

Because (:::t - c) < 0 and f(x) " f{e)

(2)

Together, Equations (I) and (2) imply I'{c) = O. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use f{x) 2: ftc}, which reverses the inequalities in Equations (1) and (2). •

188

Chapter 4: Applications of Derivatives y

----'--

+-"=------'---~x

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 I can possibly have an extreme value (local or global) are

1. 2. 3.

interior points where f' = 0, interior points where f' is undefined, endpoints of the domain of I.

The following definition helps us to summarize. (al

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 I 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 I at all critical points and endpoints. 2. Take the largest and smallest of these values.

EXAMPLE 2 [-2,1].

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 = 1: Critical point value:

1(0) = 0

Endpoint values:

1(-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

7

Find the absolute maximum and minimum values of g(l) = 81 - 14 on

EXAMPLE 3 [-2,1].

y (1,7)

189

Extreme Values of Functions

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- t4

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 /(x) interval [-2,3]. FIGURE 4.8 The extreme values of = 81 - 14 on [-2, I] (Example 3).

g(l)

We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The 1rrst derivative

j'(x) = ~x-I/3 = _2_ 3 3\Yx

y =x 2l3 , -2sxs 3

Absolute maximum; also a local maximum.

has no zeros but is undefined at the interior point x = O. The values of / at this one critical point and at the endpoints are

-o--"---~<----'---o--c'--~x

-2

-I

x 2/' on the

Solutton

y

Local

=

0 ~ 1 2 3 Absolute minimum; also a local minimum

Critical point value:

/(0)

Endpoint values:

/( -2)

/(3)

FIGURE 4.9 The extreme values of f(x) ~ x 2/3 on [-2, 3] occuratx ~ oand x ~ 3 (Example 4).

=

0 =

=

(-2)2/'

(3j2/'

=

=

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--{j, 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+-a~-~c---;b-~x

-;;-I---'----"------;-~x

o

a

c

b

o

a

c

b

190 5.

Chapter 4: Applications of Derivatives 6.

y

In Exercises 15-20, sketch tire graph of each function and determine

y

whether the function has any absolute extreme values on ils domain. Explain how your answer is consistent with Theorem 1.

lxi, -I <x<2 6 = -,--, -I < x < I

15. f(x) = 16. Y

-ccI---'----"-------!--+ x

o

a

c

-ccI---'----'-------:-~x

o

b

a

b

c

x

17. g(x) ~

+2

-x, o~x< 1 I I 2 {x -. ~ x::=:;

In Exercises 7-10, fmd tire absolute extreme values and where they occur.

7.

8.

y

18. h(x) =

y

19. Y

1 x

-1

9.

0

-2

-1

10.

y

2

x

y

0

= 3 sinx,

20. f(x) = {x

0 < x < 2"

+

I,

-I '" x

<

0

< x< 0 - "

cos x,

Z

Absolute Extrema on finite Closed Intervals

_=----j+----'_x

-3

2

-1

2

In Exercises 11-14, match tire table with a graph.

12. - - - - - -

11.

of each function on the given interval. Then graph the function. IdentifY the points on the graph where the absolute extrema occor, and inelude their coordinates. 2

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

x

/,(x)

x

/,(x)

a b c

0 0 5

a b c

0 0

-5

21. f(x) = 3x - 5,

-2'" x '" 3

22. f(x) = -x - 4,

-4 '" x '" I

14.

x

/'(x)

x

/,(x)

a b c

does not exist 0

a b c

does not exist does not exist -1.7

-2

-I '"

24. f(x) = 4 - x',

-3 '" x '" I

I 25. F(x) ~ - " x I 26. F(x) = -x'

-2'" x '" -I

27. h(x) ~"\Yx,

-I '" x '" 8

0.5'" x '" 2

-I '" x '" I

29. g(x) =~, -2 '" x '" I 30. g(x) = -~,

. 8 3 1. f(8) --sm,

-vs '" x'" 0

- '! 2<- :8 5 5" 6

32. f(8) = tao 8, 33. g(x) = cscx, 34. g(x) = secx,

(a)

x'" 2

23. f(x) = x' - I,

28. h(x) = -3x'/3, 13.

<

-1:5 x

x ~, 0"'x"'4

In Exercises 21-36, find the absolute maximum and minimum values

5

o

I, {

3 " '"

8 '"

4 "

l' '" x '" 2; -1' '" x '" ~

35. f(t) = 2 -

Itl,

-I'" t '" 3

36. f(t) = It -

51,

4'" t '" 7

(b)

In Exercises 37-40, fmd the function's absolute maximum and minimum values and say where they are assomed. 37. f(x) = x 4/ 3, -I '" x '" 8 abc (c)

(d)

38. f(x) = x'/3, -I '" x '" 8 39. g(8) = 83/', -32'" 8 '" I 40. h(8) = 38'/3, -27 '" 8 '" 8

Extreme Values of Functions

4.1

finding Critical Points In Exercises 41-48, determine all critical points for each functioo.

41. y = x' - 6x + 7 43. f(x) = x(4 - x)' 45. Y = x' +

x2

47. Y ~ x' - 32~

42. f(x) = 6x' - x'

44. g(x) = (x - I)'(x - 3)'

Va -

V(x) = x(1O - a)(16 - a), x'

a. Find the extreme values of V.

In Exercises 4!1-58, f"md the extreme values (ahsolute and local) of the functioo and where they occur. 50.y~x'-a+4

b. Interpret any values found in part (a) in terms of the volume of the box.

f(x)

53. y

= W-=!

54. y

= x - 4~

55. Y

=~

56. y

= V3 + a - x'

58

=

.y

61. y=x~

_ {4 - a, x '" 1 x + 1, x > 1

63

60. y

75. Maximum height of a vertically moviog body body moving vertically is given by

64

.y

~

66.y~

_ {3 - x, •Y -

3

+a -

1

s=-2gt2+vot+so,

The height of a

g>O,

with s in m_s and t in secoods. Find the body's maximum height

- 4)

62. y=x'~

• Y -

65

= x'/3(x'

+ bx' + ex + d.

b. How many local extreme values can f have?

x+1 x'+a+2

Local Extrema and Critical Points

= x'/'(x + 2)

= ax'

a. Show that f can have 0, 1, or 2 critical points. Give examples and graphs to support your argument.

In Exercises 59-{j6, f"md the critical points, domain eodpoints, aod local extreroe values (absolute and local) for each functioo.

59. y

Coosider the cubic functioo

74. Cnbic functions

51.y=x'+x'-&;+552.y=x'(x-5)'

57 =_x_ .y x'+1

0< x < 5,

models the volume of a box.

finding Extreme Values

49.y~a'-&;+9

72. We know how to f"md the extreroe values of a continuous functioo f(x) by investigating its values at critical points and endpoints. But what if there are no critical points or endpoints? What happeos then? Do such functioos really exist? Give reasoos for your answers. 73. The functioo

x'

46. f(x) = x - 2 48. g(x) ~

191

x < 0 x', x '" 0

-x' - a + 4 x'" 1 { -x'+6x-4,, x>1

D Graph the functioos in Exercises 77-80. Then f"md the extreme values of the functioo 00 the interval and say where they occur.

_IX' _Ix + li, x'" 1 4 2 4 { x'-6x'+&;, x>1

In Exercises 67 and 68, give reasoos for your answers.

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

76. Peak alternating current Suppose that at any given time t (in seconds) the current i (in amperes) in an alternating current circuit is i = 2 cos t + 2 sin t. What is the peak current for this circuit (largest magnitude)?

77. f(x) = Ix - 21 + Ix+ 31,

-5 "'x '" 5

78. g(x) = Ix - II - Ix - 51, 79. h(x) = Ix + 21 - Ix - 31,

-00

80. k(x) = Ix + 11 + Ix - 31,

-00

-2 '" x '" 7

<x < <x<

00 00

COMPUTER EXPLORATIONS

= 2.

c. Does the result in part (h) cootradict the Extreme Va10e Theorem? d. Repeatparts(a)and(h)forf(x) ~ (x - a)'/3,replacing2bya.

68. Le!f(x) ~ lx' - 9xl.

In Exercises 81-86, you will use a CAS to help find the absolute extrema of the given functioo over the specified closed interval. Perform the following steps. a. Plot the functioo over the interval to see its general behavior there.

h. Does f' (3) exist?

b. Find the interior points where f' ~ O. (In some exercises, you may have to use the numerical equation solver to approximate a solutioo.) You may want to plot f' as well.

c. Does f' (- 3) exist?

c. Find the interior points where f' does not exist.

d. Determine all extreroa of f.

d. Evaluate the functioo at all points found in parts (h) and (c) and at the endpoints of the interval.

a. Does f' (0) exist?

Theory and Examples 69. A minimum with no derivative The functioo f(x) = Ixl has an ahsolute 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. Even functions If an eveo 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.

e. Find the function's absolute extreme values on the interval and identifY where they occur.

81. f(x) ~ x· -

&;' +

4x + 2,

[-20/25,64/25]

82. f(x) ~ -x· + 4x' - 4x + 1, [-3/4,3] 83. f(x) = x'/3(3 - x), [-2,2] 84. f(x) = 2 + a - ]x'/3, [-1, 10/3]

85. f(x) = ~ + 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 tn the graph is horizontal and the derivative is zero (Figure 4.10). We now state and prove this result. ~I-P-------'--~L\-____>x

o

(a)

THEOREM 3-Rolle's Theorem Suppose that y = f(x) is continuous at every point of the closed interval [a, b] and differentiable at every point of its interior (a, b). If f(a) = f(b), then there is at least one number c in (a, b) at which I'(c) = o.

y

I'(c,)

~

0

-o+P--"-------L----'--~---->x

(b)

FIGURE 4.10 Rolle~ Theorem says that a differentiable curve bas at least one horizontal tangent between any two points where it crosses a horizontal line. It may bavejust ooe (a), or it may bave more (b).

HISTORICAL BIOGRAPHY

Michel Rolle (1652-1719)

Proof Being continuous, f assumes absolute maximum and minimum values on [a, b] by Theorem 1. These can occur only 1.

at interior points where I' is zero,

2.

at interior points where I' does not exist,

3.

at the endpoints of the function's domain, in this case a and b.

By hypothesis, f has a derivative at every interior point. That rules out possibility (2), leaving us with interior points where I' = 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 I'(c) = 0 by Theorem 2 in Section 4.1, and we have found a point for Rolle's Theorem. If both the absolute maximum and the absolute minimum occur at the endpoints, then because f(a) = f(b) it must be the case that f is a constant function with f(x) = f( a) = f(b) for every x E [a, b]. Therefore I' (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 Xv b (b) Discontinuous at an interior point of [a, b]

J\:

-+---'--'-----'--:--''----~x

a

Xv

b

(c) Continuous on [a, b] but no! differentiable at an interior point

FIGURE 4.11 There may be no horizontal tangent ifthe hypotheses ofRolle's Theorem do not hold. Rolle's Theorem may be combined with the Intermediate Value Theorem tn show when there is ouly one real solution of an equation f(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

----c---;--;;i------;-~x

f(x) = x 3

+

3x

+ I.

Since f( -1) = - 3 and f(O) = I, the Intermediate Value Theorem tells us that the graph of f crosses the x-axis somewhere 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&weeo -I and 0 (Example I).

I'(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 I' 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

~/

Slope!"(c)

B

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

I

Slope f(b) - f(a) I b-a I I I

--c-f-c/-"-----'-------l'----> x o c b

Geometrically, the Mean Value Theorem says that somewhere b&weeo a and b the curve has at least ooe tangeot parallel to chord AB.

f(bl

=~(a)

= I'(c).

(1)

FIGURE 4.13

Proof We picture the graph of f and draw a line through the points A(a, f(a» and B(b,

feb»~.

(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 graphs of f and g at x is HIsTORICAL BIOGRAPHY

Joseph-Louis Lagrange (1736-1813)

hex) = f(x) - g(x) = f(x) - f(a) -

feb) - f(a) b (x - a).

-a

(3)

Figure 4.15 shows the graphs of f, g, and h together. The function h satisfies the bypotheses 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, heal = h(b) = 0 because the graphs of f and g both pass through A and B. Therefore h'(c) = 0 at some point c E (a, b). This is the point we want for Equation (1).

194

Chapter 4: Applications of Derivatives

B(b,f(b))

B

I I I I I I I I

a

y

t

y ~ Y1=7,-I" ,,5 I I

'x

b

FIGURE 4.14 The graph of j and the chordAB over the interval [a, b].

h(x)

~

f(x) - g(x)

/

:............... i ............

a

x

b

'x

FIGURE 4.15 The chordAB is the graph of the functiong(x). The function h(x) = j(x) - g (x) gives the vertical distance between the graphs of j and g at x.

-f---C:+--~~x

-I

0

I

FIGURE 4.16 The function j(X) = ~ satisfies the hypotheses (and conclusion) of the Mean Value Theorem on [-I, I] even though j is not differentiable at-land!.

To verify Equation (1), 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 ofEq. (3) ...

h'(e)

=

I'(e) _ I(b) - I(a) b a

... withx=c

y B(2,4)

4

;J

3

y

0= I'(e) _ I(b) - I(a) b-a

= x2

I'(e)

=

I(bl-

~(a),

h'(c)

~

0

Rcarnmged

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 = 1 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

240

H i3 160

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.

&J~

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

The Mean Value Theorem

195

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 j'(x) = 0 at each point x of an open interval (a, b), then j(x) = C for all x E (a, b), where C is a constant.

Proof We want to show that j has a constant vaIue 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 j(x,) = j(X2). Now j satisfies the hypotheses of the Mean Value Theorem on [Xl, X2]: It is differentiable at every point of [Xl. X2] and hence continuous at every point as well. Therefore,

at some point c between Xl and X2. Since j' = 0 throughout (a, b), this equation implies successively that

j(X2) - j(XI) X2 Xl = 0,

j(X2) - j(XI) = 0,

and

j(XI) = j(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 vaIues on the interval have a constant difference.

y

COROLLARY 2 Ifj'(x) = g'(x) at each point X in an open interval (a, b), then there exists a constant C such that j(x) = g(x) + C for all X E (a, b). That is, j - g is a constant function on (a, b).

Proof At each point X E (a, b) the derivative ofthe difference function h = j - g is h'(x) = j'(x) - g'(x) = O. Thus, h(x) = C on (a, b) by Corollary I. That is, j(x) - g(x) = C on (a, b), so j(x) = g(x) + C. • ----\c-\---"-+-"'--f+----->X

-2

FIGURE 4.19 From a geometric point of view, Corollary 2 of the Meao Value Theorem says that the graphs offunctions with identical derivatives on an interval cao differ only by a vertical shift there. The graphs ofthe functions with derivative Z. are the parabolas y ~ x' + C, shown here for selected values of C.

Corollaries 1 and 2 are also true 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 j(x) = x 2 on (-00, 00) is z., any other function with derivative 2x on ( - 00, 00) must have the fonnula x 2 + C for some vaIue of C (Figure 4.19).

EXAMPLE 4 Find the function j(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 j and g have the same derivative. Corollary 2 then says that j(x) = -cosx + C for some

196

Chapter 4: Applications of Derivatives constant C. Since the graph of j passes through the point (0, 2), the value of C is determined from the condition that j( 0) = 2:

j(O)

= -cos

The function isj(x) = -cosx

+

(0)

+C

=

2,

C = 3.

so

•

3.

Finding Velodty 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(/) 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(/) is some function whose derivative is 9.8. We also know that the derivative of g(/) = 9.81 is 9.8. By Corollary 2,

+

v(/) = 9.81

C

for some constant C. Since the body falls from rest, v(O) = O. Thus 9.8(0)

+

C = 0,

and

C = O.

The velocity function must be v(/) = 9.81. What about the position function 8(/)? We know that 8(/) is some function whose derivative is 9.8/. We also know that the derivative of j(/) = 4.9/2 is 9.8/. By Corollary 2,

8(1) = 4.9/ 2 + C for some constant C. Since 8(0) = 0, 4.9(0)'

+

C = 0,

and

C = O.

The position function is 8(/) = 4.9/ 2 until the body hits the ground The ability to find functions from their rates 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 Checldng the Mean Yalue Theoll!m

Find the value or values of c that satisfY the equation f(b) - f(a) ~ f'(c) b-a

Which of the functions in Exercises 7-12 satisfY the hypotheses of the Mean Value Theorem on the giveo interval, aod which do not? Give reasons for your answers. 7. f(x) = x 2/3 , [-1,8] [0, I]

in the conclusion of the Mean Value Theorem for the functions and in-

8. f(x) = x'I',

tervals in Exercises l--{j. 1. f(x) ~ x 2 + Zl: - I,

9. f(x) = v'x(1 - x),

2. f(x) = x 2/3 ,

3. f(x)

4. f(x) =

[t,2]

v;=J,

S.f(x)~x3-x2,

6. g(x)

~

X3 {

x2:

SinX

[0, I]

=x+~,

[1,3] [-1,2]

-2::=:; x::=:; 0 < x '" 2

°

[0, I]

[0, I] 10. f(x) =

{

-'T1' ~x

x'

0,

0

x=o

x2 - x 11. f(x) ~ { Zl:2 _ ~x - 3,

12. f(x) = {Zl: - 3,2 6x - x

<

-

7,

-2::=:; x

-1 <

°'<" 2

~

x::=:;

x '" 2 x ::;; 3

-} 0

4.2

197

The Mean Value Theorem

28. Suppose that f(O) ~ 5 and that f'(x) ~ 2 for all x. Must f(x) ~

13. The function

<

f(x) = {x, 0:5 X 0, x = 1

2x + 5 for all x? Give reasons for your answer.

I

29. Suppose thatf'(x) = 2x for allx. Findf(2) if

~

a. f(O) = 0

b. f(1) = 0

e. f(-2) = 3.

0 and x ~ I and differentiable on (0, I), but its derivative on (0, 1) is never zero. How can this be? Doesn't Rolle's Theorem say the derivative has to be zero somewbere 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.

is zero at x

14. For what values of a, m, and b does the function

=

f(x)

3' -x 2 + 3x { mx+b,

x = 0 0 < x

+ a,

< I

1~x~2

satisfY the bypotheses of the Mean Value Theorem on the interval [0,2]?

Give reasons for your answer.

= x2

y' = x 3

31. a. y' = x

b. y'

32. a. y' = 2x

b.y'=2x-1

c.y'=3x 2 +2x-1

,_ I 33 . a. y - - 2 x

I b . y ' = I - -2 x

c.y'=s+---\x

C.

I

.1 r 2vx

b. y'

= _1_

c'Y'=4x-~

Roots (Zeros)

35. a. y' = sin21

b. y'

= cos.!2

c. y' = 8in2t

15. a. Plot the zeros of eacb polynomial on a line together with the zeros of its first derivative.

36. a. y' = sec' 0

b. y'

= v8

e. y' =

34. a. y' =

~

+ cost

v8 -

sec2 0

i)y~x2-4

+ 8x + 15 3 x - 3x 2 + 4 =

ii) y = x 2 iii) y =

iv) y = x

3

-

(x + I)(x - 2)2 2 33x + 21& = x(x - 9)(x - 24)

b. Use Rolle's Theorem to prove that between every two zeros of

In Exercises 37-40, fmd the function with the given dctivative wbose grapb passes through the point P.

37. f'(x)

= 2x -

38. g'(x)

~ 1, + 2x,

39. 7'(0)

~8-

x" + QII_tX"- 1 + ... + QtX + ao there lies a zero of nx"- 1

+ (n

- 1)all_tx"-2

+ ... + at.

16. Suppose thatf" is continuous on [a, b] and that f has three zeros in the interval. Sbow thatf" has at least one zero in (a, b). Gener-

alize this result. 17. Sbow that if f" > 0 throughout an interval [a, b], then f' bas at most one zero in [a, b]. What iff" < 0 throughout [a, b] instead? 18. Sbow that a cubic polynontial can have at most three real zeros. Sbow that the functions in Exercises 19-26 have exactly one zero in the given interval.

20. f(x)

~

21. g(l) =

x3

41. v ~ 9.81

+ 5,

42. v = 321 - 2,

44. v

= IjTcos W'

23. 7(0) = 0

+

sin2

(~)

24.7(0) = 20 - oos2 0 I 25. 7(0) = sec 0 - 03

(0,00) 3.1,

(-I, I)

sin 'frl, 2

8,

+ v2,

+ 5,

tanO - cotO - 0,

(-00,00) (-00,00)

(0, 'fr/2) (0, 'fr/2)

Find;ng Functions from Derivatives 27. Suppose that f( -I) ~ 3 and that f'(x) ~ 0 for all x. Must f(x) = 3 for all x? Give reasons for your answer.

,(0) 21

~

0

,

s('7T""")

=

1

Exercises 45-48 give the acceleration a = d 2,/dI 2 , initial velocity, and initial position of a body moving on a coordinate line. Find the body's position at time I. 45. a = 32,

-

,(0) ~ 10 ,(0.5) = 4

(-00,0)

~ I + v'1+t -

~

F;nd;ng Position from velocity or Acceleration

+ ~ + 7,

4,

P(O, 0)

Exercises 41-44 give the velocity v ~ ds/dl and initial position ofa body moving along a coordinate line. Find the body's position at time t.

~

vt + v'1+t -

(-;r, 0)

40. 7'(1) = secltanl - I,

43. v

x

p(-I, I)

csc'O,

[-2, -I]

22. g(l) = I

26. 7(0)

x

I,

+ 3x +

19. f(x) ~ x'

I, P(O, 0)

v(O) = 20, v(O) ~ -3,

,(0) = 5

46. a

~

9.8,

47. a

~

-4 sin 21,

v(O) ~ 2,

,(0)

9 31 48. a = "" cos 'fr'

v(O) = 0,

,(0) = -I

,(0)

~

0 ~

-3

Applkations 49. Temperature change 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 somewbere along the way the 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 bours she had covered 159 mi 00 a toll road with speed limit 65 mph. The trucker was cited for speeding. Why? 51. Classical accounts tell us that a 17
60. Parallel tangents Assume that I and g are differentiable 00 [a, b] aod that/(a) ~ g(a) aod/(b) ~ g(b). Show that there is at least one point between a and b where the tangents to the graphs of I and g are parallel or the same line. lllustrate with a sketch. 61. Suppose that l'(x):5 I for I :5 x :5 4. Show that 1(4)1(1) :5 3. 62. Suppose that 0 < I'(x) < 1/2 for all x-values. Show that 1(-1) < 1(1) < 2 + 1(-1).

52. A marathoner rao the 26.2-mi New York City Marathoo in 2.2 hours. Show that at least twice the roarathooer was running at exactly II mph, assuming the initial aod fmal speeds are zero.

63. Show that Icoax - II :5 Ix I for all x-values. (Hint: Coosider I(t) = cos t 00 [0, x].)

53. Show that at some instaot during a 2-hour automobile trip the car's speedometer reading will equal the average speed for the trip.

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 ofgravity

65. If the graphs of two differentishle functions I(x) andg(x) start at the same point in the plane aod 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 Ister? Theory and Examples 55. The geometric mean of a and b The ~metric mean of two positive numbers a and b is the number Vab. Show that the value of c in the cooclusioo of the Meao Value Theorem for I(x) ~ I/x 00 an interval ofpositive numbers [a, b] is c ~ y;;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 cooclusioo of the Meao Value Theorem for I(x) ~ x 2 00 any interval [a,b] 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 I'(x) :5 I for all x-values. 67. Assumethatl is differentiable ona :5 x:5 bandthat/(b) Show that f' is negative at some point between a and b.

68. Let/be a functioo defmed 00 ao interval [a, b]. What conditioos could ynu plsce on I to guarantee that

. I' < I(b) - I(a) < I' mm b-a _max,

Graph the functioo

I(x) ~ sin x sin (x

+ 2)

- sin2 (x

+

I).

What does the graph do? Why does the functioo behave this way?

Give reasons for your answers. 58. RoUe's Theorem

a. Construe! a polynomial I(x) that has zeros atx

~

-2, -I, 0,

I, and 2. b. Graph I and its detivative I' togetber. How is what you see related to Rolle~ Theorem?

e. Do g(x) ~ sin x aod its derivative g' illustrate the same phenomenoo as I and I'? 59. Unique solution Assume that I is cootinuous 00 [a, b] and differentiable on (a, b). Also assume that I(a) aod I(b) have opposite sigus and that I' ~ 0 between a aod b. Show that I(x) = 0 exact1y ooce between a aod b.

4.3

< I(a).

where min I' aod max I' refer to the minimum and maximum values of I' on [a, b]? Give reasons for your answers.

D 69.

Use the inequalities in Exercise 68 to estimate 1(0.1) if I'(x) = 1/(1 + x'cosx) forO:5 x:5 0.1 aod/(O) ~ I.

D 70.

Use the inequalities in Exercise 68 to estimate 1(0.1) if I'(x) = 1/(1 - x') forO:5 x:5 0.1 aod/(O) = 2.

I be differentishle at every value of x and suppose that 1(1) = I, that I' < 0 on (-00, I), and that I' > 0 on (1,00).

71. Let

a. Show that I(x) '" I for all x. b. Mustl'(l) = O?Explain.

72. Let I(x) = px2 + qx + r be a quadratic function defined on a closed interval [a, b]. Show that there is exactly ooe 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 differentiable 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 3 (a, h). If I'(x)

Suppose that

I

is continuous on [a, h] and differentiable on

> 0 at each point x E (a, h), then I

is increasing on [a, h].

Ifl'(x) < 0 at each point x E (a, h), then/is decreasing on [a, h].

Proof Let XI andx2 be any two points in [a, h] with XI < X2. The Mean Value Theorem applied to I on [Xh X2] says that

I(X2) - l(xI)

=

l'(e)(x2 - XI)

for some e between XI and X2. The sign of the right-hand side of this equation is the same as the sign of I'(e) because X2 - X, is positive. Therefore, I(X2) > l(xI) if I' is positive • on (a, h) and/(x2) < I(x,) 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 e in (a, h). Ifl'(e) > 0, thenl'(x) > 0 for all x in (a, h) so/isincreasing on [a, h] by Corollary 3; if I'(e) < 0, then I is decreasing on [a, h]. The next example illustrates how we use this procedore. y y~x'-I2x - 5

Find the critical points of I(x) = x 3 - 12x - 5 and identify the intervals on which I is increasing and on which I is decreasing.

EXAMPLE 1

20

Solution

The function I is everywhere continuous and differentiable. The f"mt derivative

I'(x)

FIGURE 4.20 The function fix) ~ - 12x - 5 is monotonic on three separate intervals (Example 1).

=

3x 2 - 12

=

3(x + 2)(x - 2)

=

3(x2 - 4)

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 gmph of I is given in Figure 4.20.

x3

Interval I' evaluated Sign ofl' 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 1. 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, t' < 0 immediately to the left and t' > 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, t' > 0 immediately to the left and t' < 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 t' (x) changes. Absolute max

f' undefined 1

1

:

No extremum f'~ 0

1

1

11'<0 I

!'>o: 1 I

1

Local min :

:

: 1'= Absolute min _ _ _ _ _--'-

--'-

---L

a

FIGURE 4.21

:1'<0 1

1

I I --'-


c,

0:

I

:

:

c,

b

Local min

I I I I I I "-I_ _~I_ ____'I

_>"

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.

FIn! Derivative Test for Local. Extrema Suppose that c is a critical point ofa 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 t' changes from negative to positive at c, then I has a local minimum at c; 2. if t' changes from positive to negative at c, then I has a local maximum at c; 3. if t' does not change sign at c (that is, t' 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 (1). Since the sign of t' changes from negative

< c < b, t' < 0 on (a, c), and 0 on (c, b). IfxE (a, c), then/(c) < I(x) because t' < 0 implies that I is decreasing on [a, c]. If x E (c, b), then I( c) < I(x) because t' > 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 minito positive at c, there are numbers a and b such that a

t' >

mum ate. Parts (2) and (3) are proved similarly.

•

4.3

EXAMPLE 2

201

Monotonic Functions and the First Derivative Test

Find the critical points of I(x) = x l /3(x - 4) = x'/3 - 4x l / 3.

Identify the intervals on which I is increasing and decreasing. Find the function's local and absolute extreme values. Solution The function I is continuous at all x since it is the product of two continuous functions, x 1/3 and (x - 4). The first derivative f'(x) =

~ (X 4/3 - 4xl/3) ~

_i

-3x

=

i xl/3 _ i 3

3

2/

x- 3

-2/3( _ ) _ 4(x - 1) x 1 2/3 3x

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 I 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 of I between and at the critical points, as summarized in the following table.

y 4

Interval Sign off' Behavior of f

FIGURE 4.22 The functionf(x) ~ x 1/3 (x - 4) decreases when x < I and increases when x > I (Example 2).

x
O<x<1

decreasing

decreasing

x> +

1

increasing

Corollary 3 to the Mean Value Theorem tells us that I decreases on (- 00, 0], decreases on [0,1], and increases on [1, 00). The First Derivative Test for Local Extrema tells us that I does not have an extreme value at x = 0 (f' does not change sign) and that I has a local minimum at x = 1 (f' changes from negative to positive). The value of the local minimum is 1(1) = 11/ 3(1 - 4) = -3. This is also an absolute minimum since I 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 I has a vertical tangent at the origin.

•

Exercises 4.3 Analyzing Functions from Derivatfves Answer the following questions about the functions whose derivatives are given in Exercises 1-14:

a. What are the critical points of fI

1.

3. S.

6.

(x - 2)(x 8. f'(x) = (x + I)(x

+ 4)

9. f'(x) = I - :', x

b. On what intervals is f increasing or decreasing?

11. f'(x)

= x-1/3(x + 2)

e. At what points, if any, does f assume local maximum and minimum values? f'(x) = x(x - I) 2. f'(x) = (x - I)(x + 2) f'(x) ~ (x - I)'(x + 2) 4. f'(x) ~ (x - l)'(x + 2)' f'(x) ~ (x - I)(x + 2)(x - 3) f'(x) = (x - 7)(x + I)(x + 5)

13. f'(x)

~

14. f'(x)

~

7. f'(x)

=

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

* -2

3)' x

* -1,3

*0

10. f'(x) = 3 12. f'(x)

~,

= x- 1/2(x

x

*0

- 3)

(sinx - 1)(2 cos x + I), D :5 x :5 2,,(sinx + cosx)(sinx - cos x), 0:5 x :5 2,,-

Identffylng 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

y = f(x)

\

,

-3

-2

-1

1

1

2

'\

1 1 2 3 1 1

1 2

1\

-f'iC. -1

1

T

2

/

-2 -1

1/1

1 2-

21. h(x)

=

I

I

T 2

-x'

31

-

+ 2x

+3

20. g(l) = -31

2

22. h(x)

= 3r' +

27. j(x) = x 4

-

=

+ 91 + 5

6'

28. g(x) = x 4

-

4x'

30. K(I) = 151' - IS

31. j(x) = x - 6~ 33. g(x)=x~

32. g(x) = 4~ - x

=

57. j(x) =

1

29. H(I) = ll4 - I' 2

35. j(x)

36. j(x)

x-

=

:5 1f 0:5

+

taox,

-1f

T

X

:5 21f

0:5 x :5 21f 1f <x<2

I - 2sin I , 0:5 x:5 21f

58. j(x) = -2 cos x - cos2 X, 59. j(x) = csc2 X 2

60. j(x) = sec

-

X -

2 cot x,

-'T1'::=:; X ~ 'T1'

O<X<1T

-1f 1f 2 ian x, T<x<2

Theory and Examples

+ 4x 2

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 functioo has.

+3

x'+ I

62. h(6)

6

2, 5 sin ~,

61. h(6) = 3cos 2

34. g(x) = X2~

x2 - 3 - - 2 ' x oF 2

X

V3 cos x + sinx,

55. j(x) =

56. j(x) = -2x

26. h(r) = (r + 7)'

&:2 + 16

0:5

2x' - I&:

24. j(6) = 66 -

16r

Graph the functioo and its derivative together. Coonnent 00 the behavior of j in relatioo to the signs and values of 1'.

1 3

1

-2 2

D b.

1

1

23. j( 6) = 36' - 46' 25. j(r)

1 2

-3 -2 1-1

V 2/3

19. g(l) = -1

1\ /'

11

T

\

2- /

Y =fVc)

I

Y - fl;<)

\

I

1

1

and say where they occur.

54. j(x) = sinx - cosx,

Y

y

< x :5 I

a. Find the local extrema of each function 00 the given interval,

53. j(x) = sin 2x,

18.

17.

-3

1

1 2 3

2

Y

1\.1

/

4

..I

-2

In Exercises 53-60:

YFf x)

1

x2 -2' 4-x

52. g(x) =

Y

=

0:5 6:5 21f, 0:5 6 :5 1f,

at6 =

oand 6 =

aU = 0 and 6 = 1f

= j(x)

63. Sketch the graph of a differentiable functioo y the point (I, I) if 1'(1) = 0 and

3x2

37. j(x) = x'/'(x + 8)

38. g(x) = x 2/'(x + 5)

a. f'(x) > Oforx < landf'(x) < Oforx> I;

39. h(x) = x /'(x 2 '

40. k(x) = x 2/'(x 2

b. f'(x)

4)

-

-

4)

a. IdentifY the functioo's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute?

Support your fmdings with a graphing calculator or compurer grapher.

41. j(x)

=

2x - x 2 ,

-00

42. j(x) = (x

+ I)',

43. g(x) = x 2

-

44. g(x) = -x 2 45. j(l) = 121 -

48. k(x)

=

x' T x'

:5 2

<

:5 0

-00

2x 2

+ 4x,

I

<

2

x 2,

=

a. a local minimum at (I, I) and 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) and (3, 3);

<

> 2,andg'(x)---+-1+asx---+2+; 2, g' < 0 for x < 2, g'(x) ---> -00 as x---> 2-,

g' < Ofcrx

b. g(2) = g' > o far x > 2, andg'(x)---+

00

asx---+2+.

66. Sketch the graph of a cootinuous functioo y = h(x) such that

<

-00

X

0 for x oF I.

64. Sketch the graph of a differentiable functioo y = j(x) that has

-1

00

00

<X

-5:5 x :5 5

Yx 2 - 2x - 3, 3:5 x-2 51. g(x) = - 2 - - ' 0:5 X < I x - I

50. j(x)

<

< I :5 3 X

<

a. g(2) = 2,0 < g' < I for x < 2, g'(x) ---> 1- as x ---> T,

00

00

0:5

d. f'(x)

d. local minima at (I, I) and (3, 3).

<

X

+ 3x + 3x + I,

49. j(x) = Y25 -

I;

65. Sketch the graph ofa cootinuous functiooy = g(x) such that

-4:5 x

-3:5

I',

X

1:5

6x - 9,

46. j(l) = I' - 31 2, 47. h(x) =

<X

-00

+ 4,

4x -

< Oforx < landf'(x) > Oforx>

through

c. f'(x) > 0 for x oF I;

In Exercises 41-52:

D c.

21f

<

:5 0

a. h(O)

= 0, -2 :5 h(x) :5 2 for all x, h'(x) ---> and h'(x) ---> 00 as x ---> 0+;

00

as x ---> 0-,

b. h(O) = 0, -2 :5 h(x) :5 0 for all x, h'(x) --->

00

as x ---> 0-,

and h'(x) ---+ -00 as x ---+ 0+. 00

67. Discuss the extreme-value behavior of the functioo j(x) = x sin (I/x), x oF 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). 70. Determine the values of constants a, b, c, and d so that f(x) = ax 3 + bx2 + ex + d has a local maximum at the point (0,0) and a local minimum at the point (I, -I).

absolute minimum? An absolute maximum? (See Exercise 49 in

Sectioo 2.3.) 68. Find the intervals on which the functioo f(x) a

-=F

203

~ ax 2 + bx + c, 0, is increasing and decreasing. Describe the reasoning be-

hind your answer.

Concavity and Curve Sketching

4.4 y

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 aboot 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 features of functions. Identifying and knowing the locations of these features 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 coocave down on (- 00, 0) and concave up 00 (0,00) (Example la).

As you can see in Figure 4.23, the curve y = x 3 rises as x increases, but the portions de-

fmed 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 ofthe 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 (a)

The graph of a differentiable function y

= j(x)

is

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 1.

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/" > 0 on I, and decreases if/" < o.

The Second Derivative Test for Concavity Let y = j(x) be twice-differentiable on an interval 1. 1. If/" 2. If/"

> 0 on I, the graph of j over I is concave up. < 0 on I, the graph of j over I is concave down.

If y = j(x) is twice-differentiable, we will use the notations /" and y" interchangeably when denoting the second derivative.

204

Chapter 4: Applications of Derivatives

EXAMPLE 1

y

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 derivative y" = 2 is always positive. _ (8) The curve

EXAMPLE 2

y

=

Detennine the concavity of y = 3

+ sin x on [0, 21T].

-+-+~:+<"'--!---!-~x

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 (0, 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

Points of Inflection

2

(Example Ib).

y

y=3+sinx

I I I

:~

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 tangent line of slope -I at the point (1T, 3). This point is called a point ofinflection of the curve. Notice from Figure 4.25 that the graph crosses its tangent 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.

DEFINmON A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection.

---;;cot..,;:-'-~-'---:;; A",IC--~~~~2"'''-x

-1 - -------- . y" = -sinx

Using lbe 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 I"(c) = O. This follows immediately from the Intermediate Value Theorem whenever I" 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 I"(c) = 0, provided the second derivative exists (although a more advanced agrument is required in this noncontinuous case). Since a tangent line must exist at the point of inflection, either the first derivative f' (c) exists (is finite) or a vertical tangent exists at the point. At a vertical tangent neither the flISt nor second derivative exists. In summary, we conclude the following result.

At a point of inflection (c, f(c», either I"(c) = 0 or I"(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. The graph of f(x) = x S/ 3 has a horizontal tangent at the origin because f'(x) = (5/3)x2/3 = 0 when x = O. However, the second derivative

EXAMPLE 3

I"(x) = FIGURE 4.26 The graph of f(x) = x'/3 has a horizontal taogent at lbe origin where the concavity changes, although j" does not exist at x = 0 (Example 3).

~G·X2/3)

=

10 x - 1/ 3 9

failstoexistatx = O. Nevertheless, I"(x) < ofor x < Oandl"(x) > ofor x > 0, so the second derivative changes sign at x = 0 and there is a point ofinflection at the origin. The graph is shown in Figure 4.26. _

Concavity and Curve Sketching

4.4 y

205

Here is an example showing that an inflection point need not occur even though both derivatives exist and j" = 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. •

x'

FIGURE 4.27 The graph of y = has 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 l/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

------"+----~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

s(l)

=

21 3

+ 221 -

141 2

-

5,

12: O.

Find the velocity and acceleration, and describe the motion of the particle. Solutton

The velocity is

V(I)

=

=

61 2

V'(I)

=

S'(I)

-

281 + 22 = 2(1 - 1)(31 - 11),

and the acceleration is

a(l)

=

S"(I)

=

121 - 28 = 4(31 - 7).

When the function S(I) is increasing, the particle is moving to the right; when S(I) is decreasing, the particle is moving to the left. Notice that the first derivative (v = s') is zero at the critical points 1 = I and 1 = 11/3.

0<1<1

Interval Sign ofv

I < 1 < 11/3

11/3 < 1

increasing

decreasing

increasing

right

left

right

=.'

Behnvior of.

+

Particle motion

+

The particle is moving to the right in the time intervals [0, I) and (11/3, 00), and moving to the left in (I, 11/3). It is momentarily stationary (at rest) att = I and 1 = 11/3. The acceleration a(l) = '"(1) = 4(31 - 7) is zero when 1 = 7/3. Interval

0< 1 < 7/3

Sign of a = ." Graph of.

concave down

<1 +

7/3

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 f' 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 Exbema on an open interval that contains x = c.

Suppose I" is continuous

1. Iff'(c) = Oandj"(c) < O,thenfhasalocalmaximumatx = c. 2. Iff'(c) = Oandl"(c) > O,thenfhasalocalminimumatx = c. 3. If f' (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.!" < 0 ~localmax

f'~O.!,,>O ~

local min

Proof Part (1). If I"(c) < 0, then I"(x) < 0 on some open interval I containing the point c, since I" is continuous. Therefore, f' is decreasing on 1. Since f' (c) = 0, the sign of f' 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 frrst and second derivatives are zero at x = O. Yet the function y = x' 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 I" only al c ilself and not in an interval about c. This makes the test easy to apply. That's the good news. The bad news is that the test is inconclusive if I" = 0 or if j" does not exist at x = c. When this happens, use the First Derivative Test for local extreme values. Together f' 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 toming or bending as defined by its concavity. We use this information to sketch a graph ofthe function that captures its key features.

EXAMPLE 7

Sketch a graph ofthe function

using the following steps. (8) 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 ofthe 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

The function f is continuous since j'(x) = 4x' - 12x 2 exists. The domain of f is (- 00, 00), and the domain of j' is also (- 00, 00). Thus, the critical points of f occur only at the zeros of j' . Since Solutton

j'(x)

4x' - 12x 2

=

the flISt derivative is zero at x = 0 and x vals where f is increasing or decreasing.

Interval

=

=

4x 2(x - 3)

3. We use these critical points to define inter-

x
0<x<3

3<x

decreasing

decreasing

increasing

+

Signofl'

Behavior off

(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) j"(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 ofthe curve is shown in the accompanying figure.

15 (0, 10) ,

decr -----;c---;;f-----;---\~-__::_-t:_-~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 exIreme 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 graph 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

I+x

2 •

Solution

1.

The domain of f is (- 00, 00) and there are no symmetries about either axis or the origin (Section 1.1).

2. Find j' and I" . x) = (x + 1)2 ( f 1+ x 2

'(x)

=

f

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

.x:-intcrcept at.l" = -I, y-inl=ept (y ~ 1) at x = 0

+ 1)2.2x

2(1 - x 2 )

(I + x 2 )'

I"(x)

=

Critical pointa: x=-l,x=l

(I + x 2),. 2( -2x) - 2(1 - x 2)[2(1 + x 2). 2x] (I + x2)4 4x(x 2

(I +

-

3)

x 2 )3

After some algebnl

3. Behavior at critical points. The critical points occur only at x = ± I where j' (x) = 0 (Step 2) since j' exists everywhere over the domain of f. At x = -I, I" (- I) = I > 0 yielding a relative minimum 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 j' (x) < 0, and the curve is decreasing. On the interval (-I, I), j' (x) > 0 and the curve is increasing; it is decreasing on (I, 00) where j'(x) < 0 again.

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 I" is zero when x = - V3, 0, and V3. The second derivative changes sign at each of these points: negative on ( - 00, - V3) , positive on ( - V3, 0), negative on (0, V3), and positive again on ( V3, 00 ). Thus each point is a point of inflection. The curve is concave down on the interval ( - 00, - V3), concave up on ( - V3, 0), concave down on (0, V3), and concave up again on ( V3, 00 ) . 6. Asymptotes. Expanding the numerator of I(x) and then dividing both numerator and denominator by x' gives

+ I)' + x'

(x) = (x I

1

I

+

Expanding numerator

(2/x) (I/x')

Point of inflection wherex ~ 0 2 (1,2)/ Y

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).

---~~_"II'-i----;----_x

Point of inflection wherex ~ -0

FIGURE 4.30

The graph of y

=

(x + I)' , l+x

+ (I/x') +I

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 to y = I asx --+ 00. •

EXAMPLE 9

Sketch the graph of I(x) =

x' + 4 ----zx-.

Solution

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 ofthe function, but rITst rewrite it in order to simplify our computations:

I(x)

=

I'(x)

=

x' ~ 4

t-:'

=

t+~

=

x'z:,

4

Function simplified for differm.tiation

Combine fractions to solve easily /,(x)

=

o.

Exists throughout the entire domain off

3.

The critical points occur at x = ±2 where I'(x) = O. Since 1"(-2) < 0 and 1"(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

There are no points of inflection because I"(x) < 0 whenever x < 0, f"(x) > 0 whenever x > 0, and f" 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-

6.

y=! 2

From the rewritten formula for f(x), we see that

_L-_L-~¥=-----L_--'--~x

-4

-2

2

~ FIGURE 4.31

lim (~+~) x-o+ 2 x

4

The graph ofy =

=

+00

and

lim (~+~) x-o2 x

00

-00' 00,

so the y-axis is a vertical asymptote. Also, as x -+ or as x -+ the graph of f(x) approaches the line y = x/2. Thus y = x/2 is an oblique asymptote.

2

x +4 ----zx-

=

7.

The graph of f is sketched in Figure 4.31.

•

(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 first 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 determine 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 following figure summarizes how the derivative and second derivative affect the shape of a graph.

y

rJd" 7

> 0 ~ rises ftom.

Differentiable ~ smooth, connected; graph may rise and fall

y'

,/

(

y" > 0

or

"

~ 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

or

y" < 0

left to righ~ may be wavy

\ :/

~ concave down throughout; no waves; graph may rise or fall

f\

and y"
y'~O

~

y' < 0 ~ falls from.

y" changes sign at an inflection point

V

0 and y">O at a point; graph has local mjnjmum

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 1-8. IdentifY the intervals on which the functions are concave up and concave down.

1.

2.

16. y = I - (x + I)' 17. y = x 4 - a 2 = x 2(x 2 - 2) 18. y = -x 4 + 6x 2 - 4 = x 2(6 - x 2) - 4 19. y ~ 4x' - x 4 ~ x'(4 - x) 20. y ~ x 4 + a' ~ x'(x + 2) 21. y ~ x' - 5x 4 ~ x 4(x - 5) 22. y

= x(~ -

5)'

= x + sinx, O::s x ::s 21T 24. y = x - sinx, O::s x ::s 2w

23. Y

-I--lt----+--.x

25. y

= V3x -

2 cos x,

4 26. y = 3x - tanx,

3.

1T

2 <x<2

27. Y = sin x cosx. O:s x s 11' 28. y = cosx + V3 sin x, 0,;; x ,;; 2,,-

4.

y ~ ~(x2 _ 1f"

0,;; x ,;; 2,,-

-11'

30. y = x 2/'

29. y = xii'

'vJV" , 3 y

3

x W+1

33. y =

a -

35. y

6. y=tanx-4x-1!:<x<'!! , 2 2

5. y=x+sin2x_21Tsx:s;2'1T

31. y ~

Y

3x2/'

~ x 2/3(t -

sin Ixl. -21T:S x :S 21T Y

.~ 3.. y=2coax- v2x, -1T:sx:ST y --'---~\---+~x

-..

0

34. y = 5x 2/' -

x)

36. y

42. y =

x-

= - 2ax- x +4 = Ix2 - II • r.-c

{h , x

v IxI ~ •VX r,

'\Yx'+l

x< 0 x~0

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 graphing procedure on page 208 to sketch the general shape of the graph off.

51. y' =

11. y = x' - 3x + 3 13. y ~ -a' + 6x 2 - 3 15. y = (x - 2)' + 1

~

44. y=_5_ x4 + 5 46. y = Ix 2 - 2xl

49.y'=2+x-x 2

9.y~x2_4x+3

5)

48.y~~

NOI'TOSCALE

Graphing Equations Use the steps of the graphing procedure 00 page 208 to graph the equatioos in Exercises 9-48. Include the coordinates of any local and ahsolute extreme points and inflectioo points.

~ x 2/3(x -

a

40. y = x 2 +

45. y

8.

I

39. y = VI6 - x 2 x2 - 3 41. y=--2

47. y ~ =

a +

38. y ~ (2 - x 2)'/2

3

7. y

\I'l=7

37. y ~ x~

43. y -;/------;;1-----;;24 .. x

32. y ~

x(x -

53. y' = x(x 2 55. y' = 57. y'

-

(ax -

= sec2 x.

10.y~6-a-x2

58. y' = tanx,

12. y = x(6 - a)2 14. y ~ I - 9x - 6x 2 - x'

59. y' =

cot~,

50. y' =x 2 -x-6

52. y' = x 2 (2 - x)

3)'

54. y' = (x - 1)2(a + 3) 56. y' = (x 2 - a)(x - 5)'

12)

5x2)(4 - x)2 _,!!

<x < '!! 2

_,!!

<

2

2

0

61. y' = tan2 0 - I,

x

< '!! 2

< 0 < 2,,- 60.

-1' <

0

<

y' =

l'

csc'~,

0

< 0 < 2,,-

212

Chapter 4: Applications of Derivatives

62. y' ~ I - cor-O, 0 < 0 < 63. y' = cost, O:s t:s 21T 64. y' = sint, O:s t:s 21T 65. y' 67. y'

'If

91. y

= (x + n-'/' = x-'/'(x - I)

69. y' =

21xl

-2x

= { z,;,'

o ,_ {-Xl,

7. Y -

89. y

66. y' = (x - 2t l /3 68. y' = x-4/5(x + I)

92. y

90. y

x - I 8

= -,-x +4 4x = -,-x +4

x - I = --i'-----'~ x'(x - 2)

(Agnesi's witch) (Newtoo's serpentine)

x:s 0

Theory and Examples

x>O

93. The accompanying figure shows a portioo of the graph of a twicedifferentiable functioo y = f(x). At esch of the five labeled points, classifY y' and y" as positive, negative, or zero.

x:s; 0

x2,

x = -,--

x> 0

y

Sketching y from Graphs of y' and y" Each of Exercises 71-74 shows the graphs of the first and second derivatives of a functioo y ~ f(x). Copy the picture and add to it a sketch of the approximate graph of f, given that the graph passes through the point P. 71.

72.

s

~f(X)

~ R

p

T

Q

Y

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

o

94. Sketch a smooth coooected curve y x

Y ~r(x)

73.

Y

f(-2) = 8,

1'(2) = 1'( -2) = 0,

flO) = 4,

I'(x) < 0 for

f(2) = 0,

r(x) < 0

I'(x)

P

= f(x) with

>

0

for

Ixl > 2,

r(x)

>

0

Ixl <

for x

2,

< 0,

for x> O.

95. Sketch the graph of a twice-differentiable function y = f(x) with the following properties. !.abel coordinates where possible.

x 74.

Y

y

Derivatives

x<2

y'

2<x<4

Graphing Rational Functions Graph the rational functions in Exercises 75-92.

75 .y

~z,;'+x-I x'-I

77 ~x4+1 • y x2

78.y

I 79. y=-,-x - I x2 - 2

81.Y=-x'_1

x'

83·y=x+1 85

• y

=x'-x+1

87.y=

x

76.y~

I

x'-3x'+3x-1 x'+x-2

80.

x'-49 x' + 5x - 14

~x'+4

x - I x2 - 4 82. y = -,-x - 2 x2 - 4 84. y= - - x + I

86• y =_x'-x+1 x I 88• y =x'+x-2 ,

x-x

y"

y'

>

0,

y"

4

y' > 0, y"

4<x<6 6 x>6

7

y' y' y'

>

0

>

> 0,

<

=

<

<

y" 0, y" 0, y"

0

=0 <

0 0 0

96. Sketch the graph of a twice-differentiable function y = f(x) that passes through the points (-2,2), (-I, I), (0,0), (I, I), and (2, 2) and whose f'lrst two derivatives have the following sign

patterns.

z,;

x' y = -,--

0,

4

-ccI'------,r--'\---"----r~X

p

<

y' = 0, y" > 0

2

y':

y":

+

+ 2

0

2

+ -I

Motion Along a line The graphs in Exercises 97 and 98 show the position s = f(l) of an object moving up and down on a coordinate line. (a) When is the object moving 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

97.

213

> 0, sketch a curve y ~ f(x) that has f(l) ~ 0 and j'(x) = I/x. Can anything be said about the concavity of such a curve? Give reasons for your answer.

103. For x

s

I

104. Can anything be said about the graph ofa functiooy ~ f(x) that has a continuous second derivative that is never zero? Give rea-

].

sons for your answer.

is 0

5

10

15

Ttme (sec)

105. If b, c, and d are constants, 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

98.

Concavity and Curve Sketching

.. 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 rea-

sons for your answers.

5l<

is

0

10

5

15

Tnne(sec)

107. Quadratic curws What can you say about the inflectioo points of a quadratic curve y = ax 2 + bx + c, a =F O? Give

reasons for your answer. 108. Cubic curws

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?

What can you say about the inflectioo points of -=F O? Give reasons

a cubic curve y = ax 3 + bx2 + ex + d, a for your answer.

109. Suppose that lbe secood derivative of the functioo y = f(x) is

= (x +

y"

I)(x - 2).

For what x-values does the graph of f have an inflectioo point?

c

110. Suppose that the secoodderivative of the functiooy = f(x) is

+ 3). For what x-values does the graph of f have an inflectioo point? y" ~ x 2(x - 2)'(x

'-'-cLclc~~L-L~c"->X

20 40 60 80100120 Thousands of ooits produced

100. The aeeompanying graph shows the moothly reveoue of the Widget Corporatioo for thc last 12 yeas>. Doring approximately what time intervals was the marginal revenue increasing? Decreasing?

111. Find the values of coostants a, b, and c so that the graph of y = ax' + bx2 + ex has a local maximum at x = 3, local minimumatx = -I, and inflectioo point at (I, II). 112. Find the values of coostants a, b, and c so that the graph of y = (x 2 + a)/(bx + c) has a local minimum at x = 3 and a 1<>cal maximum at (-I, -2).

COMPUTER EXPLORATIONS

y

10 Exercises 113-116, fmd the inflectioo points (if any) 00 the graph of the function and the coordinates of the points on the graph where the functioo has a local maximum or local minimum value. Thou graph the function in a region large eoough to show all these points simultan<>ously. Add to your picture the graphs of the functioo's fll'St and secood derivatives. How are the values at which these graphs intersect the x-axis related to the graph of the function? 10 what other ways are the grapha ofthe derivatives related to the graph ofthe functioo? 101. Suppose the derivative of the functioo y = f(x) is

y'

= (x

- I)'(x - 2).

At what points, if any, does the graph of f have a local minimum, local maximum, or point of inflectioo? (Hint: Draw the sign pattern for y' .) 102. Suppose the derivative of the functioo y ~ f(x) is

y'

= (x

- 1)2(X - 2)(x - 4).

At what points, if any, does the graph of f have a local mini-

mum, local maximum., or point ofinflection?

113. y ~ x' - 5x' - 240 115. Y

= ~x' +

116. Y

x4 x 3 =4 - 3 -

16x 2

-

4x 2

114. Y ~ x' - 12x 2

25

+ 12x +

20

117. Graph f(x) = 2x' - 4x 2 + I and its frrst two derivatives l<>gether. Commeot on the behavior off in relatioo to the signs and values of j' and f" . 118. Graph f(x) = x cos x and its second derivative together for o,;; x ,;; 2".. Commeot 00 the behavior of the graph of fin r<>latioo to the signs and values of f" .

214

Chapter 4: AppLications of Derivatives

Applied Optimization

4.5

What are the dimensions of a rectangle with lured 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. Dnnv a picture. Label lIllY part that may be important to the problem.. 3. h1troduce 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 o/the unknown. 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 corners 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 ofthis variable:

tin What size comers maximize the box's volmne (Example I)?

, y - x(12 -

2x'f,

O:Sx:S6

o

"""/ , 6

FIGURE 4.33 The volume of the box in Figure 4.32 graphed as a fimction of x.

Y=IIlw

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 miDimum. value of 0 at x = 0 and x = 6 and a maximum near x = 2. To learn more, we examine the first derivative of V with respect to x: :

~

144 - 9&

+ 120' ~

12(12 - 8x

+ x') ~

12(2 - x)(6 - x).

Of the two zeros, x - 2 and x - 6, only x - 21ies in the interior of the function's domain

and :makes the critical-point list. The values of Yat this one critical point and two endpoints are Critical-point value: Endpoint values:

Y(2) - 128 Y(O) ~ 0,

Y(6)

~

O.

The:JDaXimum 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? SoLution Volume ofcan: 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 = 27Tr 2

+ 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

=

27fr2

Tall and thin

+

2000, r r

/> 0

o Short and wide

FIGURE 4.35

The graph of A = 27fr 2

+ 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 27Tr 2 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 rrrst derivative is zero.

dA dr

=

4'ITr _ 2000 r2

o=

4'ITr _ 2000

SeldA/dr ~

r2

4'1Tr 3 = 2000 r =

o.

Multiply by r'.

~5~0

'" 5.42

Solve forr.

What happens at r = "¥500/'IT? The second derivative 2

d A = 4'IT dr 2

+

4000

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.

=

7T

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, Y/4 -

X

2)

2 ---cL-L-----cciL----"--c~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 -

Area:2x~.

x 2,

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~ = ~

-2x 2 + 2(4 - x 2)

0

= 0

2

8-4x =0 x 2 = 20rx

=

±v'2.

4.5 Applied Optimization

V2

V2,

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:

2V2v14=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

Medium 1 -oo+---x----""''k---+--~x

Medium 2

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 .!)I-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

d-x B

Time = distance rate

FIGURE 4.37 Alightrayrefracted (deflected from its palh) 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

= C1

=

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: I = 11

+ 12

=

Va 2 + x 2 C,

+

Vb 2 + (d - x)' C2

This equation expresses I as a differentiable function of x whose domain is [0, d]. We want to find the absolute minimum value of I on this closed interval. We find the derivative

dl dx dtIdx negative

I

~-

o

-

dtIdx zero

-

-

-

~x

Xo

d-x

+ x2

and observe that it is continuous. In terms of the angles 61 and 62 in Figure 4.37,

dtldx positive

~ ++++ +++( •

x c,Va2

dl dx

sin 6,

sin 62

-----

Cl

C2·

d

FIGURE 4.38 The siga pattero of dt/dx in Example 4.

The function I 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 ofx (Exercise 62). At this unique point we then have sin II,

c,

sin 112 C2

•

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 ofthe 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 value 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) = r'(x)

--;cl""'-L-------:c---L----c-------->x

o

Items produced

FIGURE 4.39 The graph of a typical cost function starts concave dowo BOd laler turns concave 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 occurring where c' (x) = r' (x). Farther to the right, cost exceeds revenue (perllaps because of a combination ofrising labor and material costs BOd market saturation) and production levels become unprofitable again.

4.5 Applied Optimization

219

Supposethatr(x) = 9xandc(x) = x 3 - 6x 2 + 15x, where x represents millions ofMP3 players produced. Is there a production level that maximizes profit? If so, what is it?

EXAMPLE 5

y

Solution

Notice that r'(x) = 9 and c'(x) = 3x2 -

3x 2 -

IZl; + IS.

IZl; + IS = 9

Selc'(x) ~ T'(X).

2

3x -1Zl;+6=0 The two solutions of the quadratic equation are

vTi

= 2 -

V2 '" 0.586

vTi

= 2 +

V2 '" 3.414.

12 -6 : Maximum : for profit

12 +6

I I I

Local maximum for loss "---'---;=----J---'----;cc---~ 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"(xl since r"(x) is everywhere zero. Thus, p"(x) = 6(2 - x), which is negative atx = 2 + v'2

V2.

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. •

Exercises 4.5 Mathematical Applications Whenever you are maximizing or minimizing a function ofa single variable, we wge 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 uoits long. &.

Express the y-<:oordinate of P in terms of x. (Hint: Write an equation for the line AB.)

b. Express the area of the rectangle in terms of x. c. What is the largest area the rectangle can have, and what are its dimensions? y

---!''-----'-------cc+-----'------'~X

-1

4. A rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12 - x 2 • What is the largest area the rectangle can have, and what are its dimensions? 5. You are plauoing to make an open rectangular box froro an 8-in.by-IS-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume?

6. You are planning to close off a comer of the fIrst quadrant with a line segment 20 uoits long running froro (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 he bouoded 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. Tbe sbortest fenee A 216 m' 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. What dimensions for the outer rectangle will require the sma11est total length of feoce? How much fence will be needed? 9. Designing a tank Your iron works bas contracted to desigo and build a 500 ft', square-based, open-top, rectangular steel holding tank for a paper coropany. 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?

I+-x~

b. Briefly describe how you took weight into account.

~

....:l

-< U

10. Catching rainwater A 1125 ft 3 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.

tI:l

~ ~

z

I+-x+j

T x

II

1-

10"

Base

1

Lid

T x l

T x l I+-x~

I+-x+j

If

JI

15"

a. If the total cost is c

a. Write a formula V(x) for the volume of the box.

= 5(x 2 + 4xy) + 10xy,

b. Find the domain of V for the problem situation and graph V over this domain.

what values ofx andy 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 comers 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 in3 . 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 (J. What value of (J will maximize the triangle's area? (Hint: A = (1/2)ab sin (J.) 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

+ 2'7T'rh

rather than the A = 2'7T'r 2 + 2'7T'rh 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 10-in. side as shown in the figure. Two equal rectangles are removed from the other comers so that the tabs can be folded to form a rectangular box with lid.

I 1

x

x

24"

I 1 24"

x

I(

)1

36"

x

~18"~

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 = -'7T' to x = '7T'. 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 ofthe 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 (J can be varied. What value of (J will maximize the trough's volume?

Square end

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-11-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

A

p

B

w 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 halfas much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions ofthe window that will admit the most light. Neglect the thickness ofthe 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 of x and y 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. Conltruet1ng eonel A right triangle whose hypotenuse is v'3 m long is revolved about one of ita legs to genmate a right circular cone. Find the radiull, height, and vulume of the cone of grearest volume that can be made this way.

39. Shortest beam The 8-ft wall shawn hme stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side ofthe building from the ground outside the wall

h Building

, 28. Find the point on the line ~

+

i

= 1 that is closest to the origin

29. Find a positive mnnber for which the sum of it and ita reciprocal is the smallest (least) posllible. 30. Find a postitive mnnber fur which the sum of its reciprocal aDd four times ita square is the smallest possible. 31. A wire b m long is cut into two pieces. One piece is bent into an equilateml triangle and the other is bent into a circle. If the sum of the areas enclosed by each part is a minimum, what il the length of each part?

1~27'~1 40. Motion OD a One The positions of two particles on the s-axis are Sl - sintandS'}. - sin.(t + 'If/3), with 31 and 32 in meters and t in seconds. L At what time(s) in the interval. 0 ~ t ~ 2'1f do the particles m<et1 b. What is the fartbest apart that the particles ever get?

c. When in the interval 0

~ t ~ 2'1f is the distance between the particles changing the fastest?

32. .Answl:lr Exercise 31 if one piece is bent into a square and the

41. The intensity of illumination at any point from a light source is proportional tD the square of the reciprocal of the distance 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?

other into a circle.

33. Determine the dimensions of the rectangle of ialBest area that can be :inscribed in the right triangle shawn in the accompanying figure.

• 3

34. Determine the dimensions of the rectangle of largest area that can be inscnbed in a semicircle of radius 3. (See accompanying ftgure.)

35. What value ofa IDIlkcs f(x) = x 2 L

•

R =

+ (a/x) have

b. a point of inflection at x-I ?

L

+ a;t2 + bx have

a local maximum at::c - -1 and a local minimmn at x - 3?

b. a local minimum. at::c = 4 and a point of inflection at::c = I?

PhyricalAppUcations 37. Vert1em motion The height above ground of an object moving vertically is given by S -

-161 2

vJ.

-gsm2a.

wlli:re g is the downward acceleration due tD gravity. Find the angle a for which the range R is the qest possible.

a local minimum. at x = 2?

36. What values of a and b make f(x) - x 3

41. ProJectDe motiOD The runge R of a projectile ftIed from the origin aver horizontal ground is the distance from the origin to the point of impact. If the projectile is flIed with an initial velocity vo at an angle a with the horizonlal, then in Chapter 13 we find that

+ 96t + 112,

o 43. Strength of. beam

The strength S of a rectangular wooden beam is proportional to ita width times the square of its depth. (See the accompanying figun:.) L

Find the dimensions of the strongest beam that can be cut from a 12-in..-diameter cylindrical log.

b. GraphSas a function of the beam's width w, assuming the proportionality constant to be k - 1. Reconcile what }'OIl see with your aDllweI in pert (a). c. On the same screen, graph S as a function ofthe beam's depth

d, again taking k - 1. Compare the graphs with one another and with your answer inpart (a). What would be the effi:ct of changing to some other value of k? Try it.

with s in feet and t in seconds. Find L

the object's velocity when t - 0

b. its maximmn height and when it occurs

c. its velocity when" - o. 38. Quickest route lane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Whete should she land her boat to reach the village in the least amount of time?

12"

d

•

4.5 Applied Optimization

o 44. Stltrneu of. beam

The stiffiless S of a rectangular beam is proportional to its width times the cube ofits 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 ofthe beam's width w, assuming the proportionality constant to be k - I. Reconcile what you see with your lDlSwel in part (a).

e. On the same screen, graph S as a fimction of the beam's depth d, again taking k - I. Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it. 45. FrtctioDlell urt A small frictionless cart, att:aclled to the wall by a spring. is pulled lO em. from its rest position and released at time t - 0 to roll back and :forlh for 4 sec. Its position at time tis 9 = lOoos'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 ofthe acceleration then? b. Where is the cart when the:m.agni.tude 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 -I :s t ::s 3, using different colors ifposSIble. 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 horizontal. 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 its relati.on to the ships' individual. speeds? 48. Fermat.. prlDdple iD optks Light from. a source A is reflected by a plane mirror to a receiver at point B, as shown in the accompanying rJgUl'C. Shaw that for the light to obey Fermat's principle, the angle of incidence must equal the angle of reflection, both measured from. the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purely goometric argument, which you may prefer.) Nomml

Angle of

Angle of

incidence

rdlection

"

B

0,

"

46. 'IWo masses hanging side by side from springs have positions 91 = 2 sin t and 92 = sin 2t, respectively. L At what times in the interval 0 < t dD the masses pass each other? (Hint: sin 2t = 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: casU = 2co:flt - 1.)

, 47. DistaJu:e between two ddpi At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (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 aU day. L Start counting time with t = 0 at noon and Cll:press the distance 9 betwoon the ships as a fimction oft. b. How rapidly was the distance between the ships changing at noon? One hour later?

49. Tin peat When metallic tin is kept below 13.2°C, it slowly becomes brittle and crumbles to a gray powder. T'm 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 Contagi0U5, 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 controls the rate of reaction without undergoing any permanent change in itself. An autoc4tQlytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at Hrst utile amount of catalyst present is sma1l 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 proporti.onal. both to the amount of the original substance present and to the amount ofproduct. That is, v may be considered to be a fimction of x alone, and v = kx(a - x) = Irax - kx 2, wh...

x - the amount ofproduct

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 maximum value of v? SO. Airplane landing path An airplane is flying at altitude H when it begins its descent to an aiIport IUIlWltY that is at horizoDtaJ. ground distance L from the airplane, as shown in the fJgUre. Assume that the

224

Chapter 4: Applications of Derivatives

landing path of the airplane is the graph of a cubic polynl>mial fimctioo y ~ ax' + fu;2 + ex + d, where y (- L) ~ H and

y(O) = O. a. What is dy/dx atx = O? b. What is dy/dx atx = -L? c. Use the values forth'/dx atx ~ oand x y(O) = Oandy(-L) = Htoshowthat

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 margiuaI cost 55. Showthatifr(x) = 6xandc(x) =

x' - 6x

2

+

15xareyourrev-

enue and cost functions, then the best you can do is break. even ~

-Ltogetherwith

(have revenue equal cost). 56. Production level Suppose that c(x) = x' - 20>:2 + 20,000>: is the cost of maoufacturing x iterus. Find a productioo level that

will minimize the average cost ofmaking x items. 57. You are to construct an open rectangular box with a square base y

Landing path

H = Cruising altitude

Airport

I<-I,-------L---------->I Business and Economics 51. It costs you c dollars each to maoufacture and distribute hackpacks. If the backpacks sell at x dollars each, the number sold is given by

n = x ~ c

and a volume of48 ft'. Ifmaterial for the bottom costs $6/ft' and materisl for the sides costs $4/ft', what dimensions will resolt in

+ b(100

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? 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: $200 per persoo if 50 people (the minimum number to book the tour) go on the tour.

a. When we cough, the trachea (windpipe) contracts to increase the velocity ofthe 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 persoo is reduced by $2.

velocity and whether it really cootraets that much when we cough.

Under reasonable assumptions about the elasticity ofthe

It costs $6000 (a fixed cost) plus $32 per person 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 frictioo, 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

hq

length of the tracbea. Show that v is greatest when r = (2/3)ro; that is, when the trachea is about 33% coottaeted. The remarkable fact is that X-ray photographs conf'mn that the trachea cootracts 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 he), k is the cost of

placing an order (the same, no matter how often you order), c is

(a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to fmd the quantity that will minimize A(q). What is it? (The formula you get for the answer is called the Wilson lot sizefimnula.)

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?

< r - roo

where ro is the rest radius ofthe trachea in centimeters and c is a positive constant whose value depends in part on the

q + cm + T'

the cost of ooe item (a constant), m is the number of items sold each week (a constant), and h is the weekly holding cost per item

70 < 2: -

v = c(ro - r)r 2 em/sec,

D

h. Take ro to he 0.5 and c to he I and graph v over the interval 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

Show that if a, b, c, and d

are positive integers, then (a 2

+

1)(b 2 + 1)(c 2

abed

+

1)(d2

+

I) > 16.

4.6

62. The derivative dt/dx in Example 4

b. Is f ever negative? Explain.

f(x)~~

+x

mum value on the interval 0 < x <

o b. Graph

is an increasing function of x.

the function and compare what you see with your an-

66. a. The function y = tanx + 3 cotx has an absolute minimum value on the ioterval 0 < x < "./2. Find it.

d-x

o b. Graph the function and compare what you see with your

is a decreasing function of x.

answer in part (a).

c. Showthat

67. a. How close does the curve y = ~ come to the point (3/2, OJ? (Hint: If you minimize the square of the distance, you can avoid square roots.)

d-x

x

c,Va'+x' c,Vb' + (d -

x)'

o b. Graph the distance function D(x) and y

= ~ together and reconcile what you see with your answer 10 part (a).

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

G,o)

68. a. How close does the semicircle y = ~ come to the point v'3)?

)X

b

(1,

64. You have been asked to determine whether the function f(x) 3 + 4 cos x + cos 2x is ever negative.

4.6

has an absolute maxiFind it.

'11'.

swer in part (a).

g(x) = ---r~~~~ Vb' + (d - x)'

dx

= cot x - v'2 esc x

65. a. The function y

b. Showthat

dt

225

a. Explain why you need to consider values ofx only 10 the 10terval [0, 2".].

a. Showthat a

Newton's Method

~

o b. Graph the distance function andy

= VI6 - x'togetherand reconcile what you see with your answer in part (a).

Newton's Method

s

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 Newlon's method for estimating a solution of an equation f(x) = 0 is to produee a sequence of approximations that approach the solution. We pick the first number Xo ofthe 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

x,

the point where the tangent meets the x-axis (Figure 4.41). The mnnber Xl is usually a better approximation to the solution than is xo. The point X2 where the tangent to the curve at (Xl. j(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., j(x.)) is

y

y = j(x.)

+ /'(x.)(x - x.).

We can fmd where it crosses the x-axis by setting y = 0 (Figure 4.42): (X2.!(X2» Root

sought

o=

\

'\

x, Second

"0 First

j(x.) - /'(x.)

=

x-x.

x

=

j(x.) x. - /'(x.)

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.

j(x.) + /'(x.)(x - x.)

If f'(x.) " 0

TIris value of x is the next approximation x.+,. Here is a sunnnary of Newton's method.

Newton's Method y Point:

1. Guess a flISt approximation to a solution of the equation j(x) = O. A graph of y = j(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 - !(x.) ~ f'(x.)(x - x.) Tangent line (graph of

X.+1 = X. -

j(x.) /'(x.) ,

if /'(x.) "" O.

(1)

linearization offatxn)

Applying Newton's Method

-;;o+-----:;;<1?--,,;-\-->--------> x !Vc.)

Xn+1 = xlJ - f'(x

) ll

FIGURE 4.42 The geometry of the successive steps of Newton's method. From Xn we go up to the curve and follow the tangent line down to fmdx.+1.

Applications of Newton's method generally involve many numerical 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 positiverootoftheequationj(x) = x 2 - 2 = O.

EXAMPLE 1

Find the positive root of the equation

j(x) Solution

Withj(x) = x 2

-

=

x2

-

2 = O.

2and/,(x) = 2x,Equation (I) becomes

x. I =x--+n 2 Xn

x. I =2+ x.'

4,6 y

Newton's Method

227

The equation X n +l =

x.

I

2 + Xn

enables us to go from each approximation to the next with just a few keystrokes. With the starting value Xo = I, we get the results in the first column of the following table. (To five decimal places, = 1.41421.)

V2

Error Xo =

I

-0.41421

X, = 1.5

FIGURE 4.43 The graph of f(x) = x 3 - x - I crosses the x-axis once; this is the root we want to rmd (Example 2).

X2

= 1.41667

X3 =

Number of correct digits

1.41422

0.08579 0.00246 0.00001

I I 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 I 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

V2

(1.5,0.875)

Find the x-coordinate of the point where the curve y = x 3 - x crosses the horizontal line y = I.

EXAMPLE 2

Root sought

~.

Xl)

---e---~~-H-----_x

The curve crosses the line when x 3 - x = I or x 3 - X - I = O. When does f(x) = x - X - I equal zero? Since 1(1) = -I and 1(2) = 5, we know by the Intermediate Value Theorem there is a root in the interval (I, 2) (Figure 4.43). We apply Newton's method to 1 with the starting value Xo = I. 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(l)showsthat/(x.) = O. We have found a solution of/(x) = 0 toninedecimals. _ Solutton

3

1.3478

FIGURE 4.44 The rust three x-values in Table 4.1 (four decimal places).

TABLE 4.1 The resuLt of appLying Newton's method to f(x) = with Xo = 1

y

25

20

15

10 B,(2.12,6.35)

5

-1/v3

n

x.

I(x.)

0 I 2 3 4 5

I

-I

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

f'(x.)

x' -

x - 1

X.+1

= X.

-

I(x.) f'(x.)

1.5 1.3478 26087 1.3252 00399 1.324718174 1.324717957 1.324717957

Root sought

1/v3 \ 1.6 2.12

FIGURE 4.45 Any starting value Xo to the right ofx ~ 1/\13 will lead to the root

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 (I) repeatedly as before, with I(x) = x 3 - X - I and i'(x) = 3x 2 - I, 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 bawd, the values x. get arlJitrarily close to the desired root r. (Ibis notion is similar to the idea ofthe limit ofa functiong(t) as t approaches inImity, as defmed in Section 2.6.) In practice, Newton's method usually gives convergence with impressive speed, but this is not guaranteed. One way to test convergence is to begin bY graphing 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 cOlWerge. You go from Xo to X, and back to .:to, never getting any closer to r.

{-~' ~,

x


x~r,

the graph 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:SO 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:SO 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 found

\

----:It----4p-------jIf"--=el'''-----~x

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 Xo ~ -I for the left-hand solution and with Xo = I for the solution on the right. Then, in each case, f'm dx2_ 2. Use Newton's method to estimate the one real solution of x' + 3x + I = O.Startwithxo = Oandthentmdx2.

3. Use Newton's method to estimate the two zeros of the function j(x) ~ x· + x - 3. Start with Xo ~ -I for the left-hand 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 j(x) = 2x - x 2 + I. Start with Xo = 0 for the left-hand zero and wi1l1 Xo = 2 for the zero on 1I1e right. Then, in each case, [md X2. S. Use Newton's method to [md the positive fourth root of2 by solving the equation x· - 2 = O. Start with Xo = I and [md X2.

6. Use Newton's method to [md 1I1e negative fourth root of2 by solving the equation x· - 2 ~ O. Start with Xo ~ -I and tmd X2. 7. Guessing a root Suppose that your [lISt guess is lucky, in the sense that Xo is a root of j(x) = O. Assmning that f' (xo) is de-

[med and not 0, what happens to XI and later approximations? 8. Estimating pi You plan to estimate 'ff/2 to five decimal places by using Newton's method to solve 1I1e equation cosx = O. Does it matter what your starting value is? Give reasons for your answer.

Theory and Examples 9. Oscillatioo Show that ifh > 0, applying Newton's method to

f(x) =

'\Ix,

x '" 0

{ ~,

x
leads to x, = -h ifxo = h and to x, picture that shows what is going on.

= h ifxo = -h.

Draw a

4.6 10. Approximations that get wone and wone Apply Newton's method to j(x) = x'/3wilhxo = I and calculate x, , X2, x" andx,. Find a fonnula for Ix. I . What happens to Ix. I as n .... oo? Draw a picture thet shows what is going 00.

11. Explain why Ihe following four statements ask for Ihe same infor-

mation: i) Find Ihe roots of j(x)

= x'

- 3x - 1.

ii) Find Ihe x-coordinates of Ihe intersections oflhe corve y = x'wilhtheliney = 3x + 1.

iiI") Find the x-coordinates of the points where the corve y ~ x' - 3x crosses the hotizonta1line y ~ I. iv) Find the values ofx where the derivative ofg(x) = (1/4)x' - (3/2)x 2 - X + 5 equals zero.

Newton's Method

229

\h

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 Intermediste Value Theorem from Section 2.5 to show that j(x) ~ x' + 2x - 4 has a root between x ~ I and x ~ 2. Then fmd the root to five decimal places. 24. Faetoring a quartic

Find the approsimate values of rl through

in the factorization 8x' - 14x' - 9x 2 + llx - I y

y~

= 8(x -

r,

r,)(x - r2)(x - r,)(x - r,).

Ilx" -14x' - 9x'+

llx-l

~f-------1F---'.c-'----t--~x

12. Locating a planet To calculate a planet's space coordinates, we have to solve equations like x = I + 0.5 sinx. Graphing the function j(x) = x - I - 0.5 sinx suggests thet the function has a root near x ~ 1.5. Use one application of Newton's method to improve this estimate. That is, start with Xo = 1.5 and fmd Xl . (The value of the root is 1.49870 to five decimal places.) Remem-

ber to use radians.

o 13. Intersecting corve.

The curve y = tao x crosses Ihe line y = 2x between x = 0 and x = 'IT/2. Use Newion's method to fmd where.

014. Real .olutioos or a qnartic Use Newton's method to fmd the two real solutions ofthe equstion x' - 2x' - x 2 - 2x + 2 = O. 15. a. How many solutions does the equstion sin 3x = 0.99 - x 2

o

have? b. Use Newton's method to fmd them.

16. Intersection of curves a. Does cos 3x ever equal x? Give reasons for your answer. b. Use Newton's method to fmd where. 17. Findthefourrealzerosofthefunctionj(x) = 2x' - 4x 2 + 1.

o 18. Estimating pi

Estimate 'IT to as many decimal places as your calcolator will display hy using Newton's method to solve the equstiontaox = Owithxo = 3.

19. Intersection orcnrve.

At what value(s) ofx does cosx = 2x?

025. Converging to different zero. Use Newion's method to fmd the zeros of j(x) = 4x' - 4x 2 using the given starting values. a. Xo = -2andxo = -0.8,lyingin(-00,-v2/2) b. Xo = -0.5 and Xo = 0.25, lying in ( - v21/7, v21/7)

c. Xo = 0.8 andxo = 2, lying in (v2/2, 00) d. Xo = - v21/7 and Xo = v21/7 26. Tbe .onobuoy problem In submarine location problems, it is often necessary to fmd a submarine's closest point of approach (CPA) to a sonobuoy (soood detector) in the water. Supposethet the submariue travels on the parabolic palh y = x 2 and thet the buoy is located at the point (2, -1/2). a. Show that the value of x thet minimizes the distaoce between the submarine and the buoy is a solution of the equation x ~ 1/(x 2 + I). b. Solve the equstion x = 1/(x 2 + I) with Newion's method. y

20. Intersection ofcnrve. 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.

----~t_"-____;--'-c;--c:---->x

o

1', 2

SOOobu~"(2, -!)

o 27. Corve. that are nearly flat at tbe root

Some corves are so flat that, in practice, Newton's method stops too far from the root to give a useful estimate. Try Newion's method on j(x) = (x - 1)40 with a starting value of Xo = 2 to see how close your machine comes to the root x = 1. See the accompanying graph.

230

Chapter 4: Applications of Derivatives

28. The accompanying figure shows a circle ofradius r with a chord ofleogth 2 sod so arc S ofleogth 3. Use Newton's method to solve for r sod 8 (radians) to four decimal places. Assume 0 < 6 < TT.

y

y ~ (x - 1)40

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 ofchange). For instaoce, 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 DEFINmON

A function F is an antiderivative of

f on an interval I if

F'(x) = 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 ofg, and so forth.

EXAMPLE 1 Find an antiderivative for each of the following functions. (8J f(x) = 2x (bJ g(x) = cosx
<8J F(x)

= x2

(bJ G(x)

=

sinx


= x2

+ 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 dnes 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 interva1I, then the most general antiderivative off on I is

F(x) + C where C is an arbitrary 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

Find an antiderivative of f(x) = 31;2 that satisfies F(l) = -I.

Since the derivative of x' is 3x 2 , the general antiderivative

Solution C~2 C~l

F(x)

=

C~O C~-l

..--------C=-2

x' + C

gives all the antiderivatives of f(x). The condition F(I) = -I determines a specific value for C. Substituting x = I into F(x) = x' + C gives

---t!-=+-''--ff-----> •

F(I) SinceF(I) = -I, solving I

+

=

(I)'

The curves y ~ . ' + C fill the coordinate piane without overlapping. In Example 2, we identify the curve y = x 3 - 2 as the one that passes through the given point (I, -I).

=

1

+

C.

C = -I for C gives C = -2. So

F(x)

FIGURE 4.48

+C

=

x' - 2

is the antiderivative satisfying F(I) = -I. Notice that this assignment for C selects the particular curve from the family of curves y = + C that passes through the point (I, -I) in the plane (Fignre 4.48). •

x'

By working backward from assorted differentiation rules, we can derive formulas and rules for antiderivatives. In each case there is an arbitrary constant C in the general expression representing all antiderivatives of a given function. Table 4.2 gives antiderivative formulas for a number ofimportant 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 /ex)/k + Cis sec2 kx, whatever the value of the constants Cork '" 0, and this establishes Formula 4 for the most general antiderivative of sec2 /ex.

EXAMPLE 3 (a)

f(x)

=

xS

Find the general antiderivative of each of the following functions. (b) g(x) =

1 y;,

(c) 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+1

+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)

(b) g(x)

=

6x +

X

H(x)

Fonnula 1

C

withn=5

=x-l/2 ,so l/2

G(x) = 1/2 (0)

+C

In each case, we can use one of the formulas listed in Table 4.2.

Solution (a)

+C

-cos 2x 2

=

(d) f(x) =

+C

sin (x/2) 1/2

=

•

r

2Vx

Fonnula 1 with. ~ -1/2

+C

Formula 2

+C

+

withk~2

. x C = 2 sm 2

+

Formula 3 with k ~ 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) =

V; +

sin 2x.

Solution We have that fix) = 3g(x) + h(x) for the functions g and h in Example 3. Since G(x) = 2Y!;; 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· 2-V:; = 6-V:; is an antiderivativeof3g(x) = 3/-V:;. 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

= 6-V:; -

is the genem1 antiderivative fannula for j(x), where C is an arbitrary constant.

•

Initial Value Problems and Differential Equations Antiderivatives play sevem1 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 j(x) is the same problem as finding a functiony(x) that satisfies the equation

dy dx = j(x). This is called a differential equation, since it is an equation involving an unknown 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 j(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 j(x) gives the general solution y = F(x) + C of the differential equation dy/dx = j(x). The general solution gives all the solutions ofthe 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 particular 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) = -1.

Antlderivatives 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 5(1) 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 dl

= -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(0) - 12

Initial condition:

BIlloon initially riliq;

This is our mathematical model for the package's motion. We solve the initial value problem to obtain the velocity ofthe package.

1. Solve the differential equation: The general formula for an antideri:vative of -32 is

I

~=-32

v - -32t

+ C.

Having found the general solution of the differential equation, we use the initial condition to find the particular solution that solves our problem.

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

Initialcondi.tianv{O)

12

=

The solution of the initial value problem is

v = -321

+ 12.

Since velocity is the derivative ofheight, and the height ofthe package is 80 ft at time

t = 0 when it is dropped, we now have a second initial value problem.. Differential equation:

= -321

:

+ 12

Seh> =

rbjdtinw

pIC\'iouI cquatioo.

s(0) - 80

Initial condition:

We solve this initial value problem to fmd the height as a function of t.

1.

Solve the differential equation: Finding the gmeral antiderivative of - 32t s - -16t

2.

2

+

12t

+ 12 gives

+ C.

Evaluate C: 80 - -16(0)' C

~

+ 12(0) + C

Initial cmditiou(O) - 80

80.

The package's height above ground at time t is

s

=

-16t 2

+

12t

+ 80.

Use the solution: To fmd how long it takes the package to reach the ground, we set s equal to 0 and solve for t: -16t 2

+

-4t2

12t+80-0

+3t+20=0

-3 ± v329 t8 t ~ -1.89, t ~ 2.64. The package hits the ground about 2.64 sec after it is dropped from the balloon. (The negative root has no physical meaning.) •

235

4.7 Antiderivatives

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

Solutton If we recognize that (x 3/3) - x 2 we can evaluate the integral as

+ 5)dx.

+ 5x is

an antiderivative of x 2

-

2x

+

5,

antiderivati~

J

(x 2

-

2x

x3

+ 5)dx = 3 -

x2

+

5x

+ £.

lUbitrary 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:

J (x

2

-

2x

+ 5) dx

=

2

J x dx - J 2x dx

+ J 5 dx

2 = JX dx-ZJXdx+5Jldx =

(;3 +

=

x 3 + C, - x 2 - ZC2 + 5x + 5C3.

3

C,) _ Z (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 5C3 into a single arbitrary constant C = C, - 2C, + 5C3, the formula simplifies to

x3 --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

3

=

x 3 - x' + 5x + C.

Find the simplest antiderivative you can for each part and add the arbitrary constant of integration at the end _

Exercises 4.7 finding Antlderlvatlves

finding Indefinite Integrals

10 Exercises 1-16, f"md an antiderivative for each function. Do as many as you ean mentally. Check your answers by differentiation.

10 Exercises 17-54, find 1he most general antiderivative or indeflnite integral. Check your answers by differentiation.

1.a.2x 2. a. 6x

e.x'-2x+1 e. x 7 - 6x + 8

b.

3. a. -3x-4

b. x-4

b.x' x7

e. x-4

-3

+ 2x + 3

b. ~ +x 2

C.

I 5. a. 2

b.~ x'

5 c. 2- 2 x

2 6. a. - x3

b. _1_

I x e. y!; + _1_

4. a. 2x-3

X

2

3

b. 1x-'/3

e. -tx-4/3

10. a. 12x -112

b. _lx-3j2

e. -~x-5j2

11. a. -w sin '11'X

b. 3 sinx

c. sin 71'X

12. a.

'T1'

cos '7TX

e. ~+_I-

2

'1fX

'T1'X

c. COST

2

+

e. -see' 3x 2

14. a. ese' x

b. _lese' 3x

e. 1-8ese'2x

15. a. cscxcotx

b. -esc 5x cot 5x

c. -w esc 1TX cot '7TX 2 2

16. a. secxtanx

b. 4 see 3x tan 3x

c. sccTtanT

=

-

5x

+ 7)dx

27. 29.

1(8Y-)I')dy

x- 1/3 dx (Y!;

+

33.

1

tV;;t;

20.

~) dx

V; dt

'1fX

37.

17Sin~

dO

39·/(-3eSC'x)dx 41.

1

ese °2eot 0 dO

1(~

+ 4t 3 ) dt

22.1(1 - x' 1(t - :3 +

3x')dx

24.

26.

35. 1(-2 eost) dl

13. a. see' x

2

3

t) dt

18·/(5-6x)dx

Ix -'I'

2x) dx

dx

I(Yf + ~)dx 30. I(t- )I')dy 32. IX-3(x+ I)dx 28.

34.

wooSX'

2 e' -x b. -se 3 3 2

1(2x

+

31. 12x(l - x -3) dx 3 sin 3x

-

1 1 1

25.

~

2

+ I)dx

23·/8,-x'-t)dx

y!;

3

b. 1rcos

(3t'

c. x - 3

9. a. 3x 2 -1/3

8.a. 4~ x

19. 21.

3

~ y!;

(x

-x-3 + x-I

2x 3 b. _1_ 2Y!; b. _1_ 3~

7. a.

17.1

36.

1 ~3V; 4

1(-5

dt

sin I) dt

38. 13 eos 50 dO 40.

42.

1(_~'X)dx

1~seeOtanOdO

4.7

I

43. 1(4seextmx - 2see'x)dx 44.

45.

47. 49.

I I l(i I I

(sin 2x - esc' x) dx I

cos 2x - 3 sin3x) dx

48·/1-~s6tdt

+ tm'B) dB

50./(2

+ tm' B =

sec' B)

eot?-x)dx

+ cot?- x = esc' x)

eosB(tmB

+ sec B) dB

+ tm'B) dB

l(i I

52.

eot?-xdx

(Hint: I

53.

1(2

46.

+ ~oS4t dt

(Hint: I

51.

k (esc' x - esex eotx) dx

54.

55. 56.

57.

58.

59. 60.

I I I I I I

'

(7x - 2) dx

=

+ 5)-'dx

(3x

(7x - 2)4 28

(3x

~

+

(x ~

I) dx = -3 cot

I

(x

+ I)' dx

(x

+ I)' dx

= (2x

+ I)' + C

c. 16(2x

+ I)' dx

= (2x

+ I)' + C

64. Right, or wrong? Say which for each formula aod give a briefrea-

son for each answer.

a.

I~dx=

b.

l~dx=~+C

c.

I~dx=t(~)'+c

+C

(x ~

+C

I)

3)' +c

66. Right, or wrong? Give a briefreason why.

_ ')-c-c-s_in_(_x') _ _ sin (x') _x_co_s_(_x__ dxx , x

+c

Initial Yalue Problems 67. Which of the following graphs shows the solution of the initial value problero

dy dx=2x, y=4whenx=l?

+C y

y

+I +C

x

I

+I +C

= x

4

61. Right, or wrong? Say which for each formula aod give a briefreason for each answer.

a. IXSinXdx =

~2 sinx + C

(1,4)

3

3 2

2 1 -1 0

b. IXSinXdx=-XCOSX+C

c. IXSinXdx

_-",15",(x,--+---,:"3)_'_ dx~ (x -+(X-2)4 x-2

I

I

= - x

v'x'+x+ C

65. Right, or wrong? Give a briefreason why.

I

sec'(5x - I)dx = ttm(5x - I) esc'

+ I)' dx

C

+ 5)-1 3

b. 13(2x

eseB. dB cscO - 8m8

Checking Antiderivative Formulas VerifY the formulas in Exercises 55--j)O by differentiation.

237

Antiderivatives

=

-xcosx

x

1

(a)

+ sinx + C

62. Right, or wrong? Say which for each formula and give a briefrea-

-1 0

x

x

(b)

(c)

Give reasons for your answer. 68. Which of the following graphs shows the solution of the initial value problero

son for each answer.

a. b. c.

I I I

tmBsee'BdB

~

tmBsee'BdB

~ ktm'B + C

""':J'B

tmBsee'BdB = ksec'B

+C

63. Right, or wrong? Say which for each formula aod give a briefreason for each answer.

a.

I

(2x

+ I)' dx

=

(2x

+ I)' 3

dy dx = -x, Y = I whenx = -I?

+C

+C

y

,~. o

(a)

y

x

x

(b)

Give reasons for your answer.

Hy y

o

(c)

x

238

Chapter 4: Applications of Derivatives

Solve the initial value problems in Exercises 69-1l8. dy 69. dx = 2x - 7,

Solution (Integral) Curves Exercises 91-94 show solution curves of differential equations. 10 each exercise, fmd an equation for the curve through the labeled point

y(2) = 0

dy 70. dx = 10 - x, y(O) = -I dy 1 71. dx = x 2

+ X,

dy

72. dx = 9x 2

~

73• dy dx

I • /'

dy dx

~

77.

+ 5, y(-I)

+ COSI,

cos I

~; ~

+ sinl,

dr

78. dO ~ cos,,-o,

d'y

82.

-2

dx

d'y dx 2

83. ddl r2 2

84 d s 'd1 2

87.

~

0

I

v(O) = I

y'(O)

y'(O)

= 2,

~

4, y(O)

y(O)

~

tis dl

r(l)

=I

~

tis I

~

s(4)

~4

8'

= 0;

dlt~4

3 '

Applications 95. Finding displacement from an antiderivative of velocity

a. Suppose that the velocity of a body moving along the s-axis is

= I•

31.

I

=0

-drl dt 1=1

85. dx' ~ 6; d'O

~

2 = -I';

d'y

86. dl'

s(,,-)

0"(0)

= -2,

0'(0)

-sin t + cos t; ym(o) = 7, y"(O) = y'(O)

=

88. y(4) = -cosx + 8 sin2x; ym(o) ~ 0, y"(O) ~ y'(O)

~

-z'I0(0). = v2

r.;

y(O)

I, y(O)

~

=0 3

~

9.81 - 3.

90. Uniqneness of solutions If differentiable functions y = F(x) and y = G(x) both solve the initial value problem y(xo) = Yo.

on an interval I, must F(x) = G(x) for every x in

~

I to I

~

3 given

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 oftime t. Is it true that once you know an antiderivative of the velocity function tis/dlyou 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-plane that passes through the point (9, 4) and whose slope at each point is 3 \1';;.

dy dx = f(x),

v

i1) Find the body's displacement from I thats = -2 when 1= O.

y(4) =

= -I,

~

i) Find the body's displacement aver the time interval from I ~ I tol ~ 3giventhats ~ 5 when I ~ O.

y"(O) ~ -8, y'(O) ~ 0, y(O) ~ 5

for your answer.

94. y

r(O) ~ 1

2 - 6x;

= 0;

fly = sinx _ cos x dx y

-5

r(O)

dv I 79. dl = Zsecltanl,

-2 ~

~

= 0

s(O) ~ 4

-,,-sin,,-O,

81.

93.

92.

y(4) ~ 0

2vx

75. dl ~ I

y

0; y(2) = I

3x -2/3 , y(-I)

74. dx ~

76. :

4x

-

>

X

d2~ 1_ ~x1J3 dx 3

91.

n Give reasons

96. Liftoff from Earth A rocket lifts offthe surface of Earth with a constant acceleration of 20 rnjsec'. How fast will the rocket be going I min later? 97. Stopping a car in time You are driving along a highway at a steady 60 mph (88 ft/sec) when you see an accident ahead and slan3 on the brakes. What constant deceleration is required to stop your car in 242 ft? To fmd out, carry out the following steps.

239

Chapter 4 Questions to Guide Your Review 1. Solve the initial value problem Differential equation:

d 2s 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

S=21

tis ~ 88ands ~ Owheot ~ 0 dt .

Initial conditions:

~

Differential equation:

O. (The answer will

3. Find the value of k that makes s = 242 for the value of t you found in Step 2. 98. Stopping a motorcycle 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?

Chapter

tis ~ Oands ~ 4wheot ~ 0 dt

VO and s

=

So when 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.

o 100. The hammer aod the feather

Initial conditions:

=

(2)

a. the velocity v = tis/dt in terms oft

d 2s - 2 ~ -5.2 ft/sec 2 dt

tis dt

102. Free fall near the surface of a plaoet For free fall near the sorface of a planet where 1he acceleration due to gravity has a constant msgnitude ofg leng1!l-units/sec', Equation (I) in Exercise 10I takes the form

b. the position s in terms of t.

Differential equation:

d 2s dt 2 = a

Initial conditions:

99. Motion along a coordinate line A particle moves on a coordinate line with acceleration a ~ d 2s/dt 2 ~ 15vt - (3/vt) , subject to the conditions that tis/dt = 4 and s = 0 wheo t = I. Find

WheoApollo 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) acceleration, he dropped them from about 4 ft above the ground. The television footage of the eveot shows the hammer and 1he feather falling more slowly than on Ear1h, where, in a vacuum, 1hey would have taken 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 following initial value problem for s as a function of t. Then fmd 1he value of t that makes s equal to O.

(1)

+vot+so.

where Vo and So are the body's velocity and position at time t = O. Derive this equation by solving the initial value problem

Measuring time and distance from when the brakes are applied

2. Find the value of t that makes tis/dt involve k.)

2

COMPlITER EXPLORATIONS Use a CAS to solve the initial problems in Exercises 103-106. Plot the solution curves.

103. y' = cos2 x + sinx, y(".) = I 104·y'=t+ x, y(I)=-1 105. y' = 106. y"

I

V4 - x 2

~ ~ +~,

,y(O) = 2 y(l)

~

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. S. What are the hypotheses and conclusion of the Mean Value Theorem? What pbysical interpretations might 1he theorem have?

6. State 1he Mean Value Theorem's three corollaries. 7. How can you sometimes identify a function f(x) by knowing 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 shape 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 would take to graph a polynomial functioo. Illustrate with an example.

19. What is an indefinite integral? How do yeu evaluate ooe? What general formulas do yeu know for fmding indefmite integrals?

14. What is a cusp? Give examples.

20. How can you sometimes solve a differential equation of the form

15. List the steps you would take to graph a ratiooal functioo. 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

dy/dx

Practice Exercises

= cscx + 2 cotx have

b. Now factor I'(x) and show that j has a local maxinnun atx = 1.70998 andlocalminirnaatx = ±v3 '" ±1.73205.

"\7'5 '"

c. Zoom in on the graph to fmd a viewing window that shows the and x = v3. presence of the extreme values at x =

"\7'5

any local maximum values?

Give reasons for your answer. 3. Does j(x) = (7 + x)(11 - 3X)I/3 have an absolute minimum

The moral here is that without calculus the existence of two of the three extreme values would probably have gone unnoticed. On any normal graph of the function, the values would lie close enough together to fall within the dimensions of a single pixel on the screen. (Source: Uses o/Technology in the Mathematics Curriculum, hy Benny Evans and Jerry Johuson, Oklaboma State University, published in 1990 under Natiooal Science Foundation Grant USE-8950044.)

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 any of these local maximum values also be local minimum values of f7 Give reasons for your answer.

6. a. Give an example of a differentiable function 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 function y ~ I/x does not take on either a maximum or a minimum on the interval 0 < x < 1 even though the function is continuous on this interval. Does this contradict the Extreme Value Theorem for continuous functions? Why?

8. What are the maximum and minimum values of the function y ~ Ix I on the imerval -I :5 x < I? Notice that the interval is not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why?

D

j(x)?

22. Ifyeu 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.

Extreme Values 1. Does j(x) = x 3 + 2>: + tan x have any local maximum or minimum. values? Give reasons for your answer. 2. Does g(x)

~

21. What is an initial value problem? How do yeu solve one? Give an example.

9. A graph that is large enough to show a function's global behavior may fail to reveal important local features. The graph of j(x) = (x'/8) - (x 6/2) - x' + 5x 3 is a case in point.

a. Graph j over the interval -2.5

:5

x

:5 2.5. Where does the

graph appear to have local extreme values or points of inflection?

D 10.

(Continuation ofExercise 9.) a. 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 1.2599. and a local minimum value at x ~

--rz '" c. Zoom in to fmd a viewing window that shows the presence of the extreme values at x = ~ and x = --rz.

The Mea. Value Theo",m 11. a. Show that g(l) = sin2 1 - 31 decreases on every interval in its domain.

b. How many solutions does the equation sin2 t - 3t = 5 have? Give reasons for your answer. 12. a. Show that y

~

tan 8 increases on every interval in its domain.

b. If the conclusion in part (a) is really correct, how do you explain the fact that tan". = 0 is less than tan ("./4) = I? 13. a. Show that the equation lution on [0, I].

D b.

x' + 2>:2 -

2 ~ 0 has exactly one so-

Find the solution to as many decimal places as yeu can.

14. a. Show that j(x) = x/(x domain. b. Show that j(x) = x 3 mum values.

+

I) increases on every interval in its

+ 2>:

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-ft 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 ft' , the volume that would cover I acre to the depth of I ft. 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) backward?

21.

s

16. The formula F(x) ~ 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. 17. Show that

Graphs and Graphing Graph the curves in Exercises 23-32.

even though

I

x

x+I"-x+I' Doesn't this contradict Corollary 2 of the Mean Value Theorem?

23. Y = x' - (x'/6) 24.y=x'-3x'+3 25. y= -x'+6x'-9x+3

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?

26. Y ~ (l/8)(x'

+ 3x' - 9x -

27. Y ~ x'(8 - x) 28. y 29. Y

= x'(2x' - 9) = x - 3x'/3 = x l /3(x - 4)

Analyzing Graphs In Exercises 19 and 20, use the graph to answer the questions.

30. y

19. IdentifY any global extreme values of f and the values of x at which they occur.

31.y~x~ 32. Y

y y

~

f(x)

(1, 1)

"-....H)

-"O+----~x

20. Estimate the intervals on which the function y = f(x) is

27)

= xv'4=7

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 mininuon, or inflection point? (b) Sketch the general abape of the graph.

x'

33. y' ~ 16 -

34. y' =x 2 -x-6

35. y'

~

6x(x

+ I)(x -

37. y'

=

x4

2x 2

-

2)

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

39. y = x'/3 + (x - 1)1/3 41. Y = x l /3

1)1/3

40. Y = x'/3

+ (x -

1)'/3

42. Y = x'/3 - (x - 1)1/3

Sketch the grapha of the rational functions in Exercises 43-50.

x

(2, 3)

+ (x -

+

I

43,y~x-3

45• y

~x'+1 x

2x

44,y~x+5

46 ~x'-x+1 • y x

3

x 4 -1

47. Y =

x +2 ----zx-

48. Y = ~

~

x2 - 4 -,-x-3

50. Y

49. Y

~

x2 -,-x-4

-2 OptImization 51. The sum of two noonegative numbers is 36. Find the numbers if Each of the grapha 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 oftheir square roots is to be as large as possible.

b. the sum oftheir 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. 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 ft3, to be welded from quarter-inch plate, and to weigh no more than necessary. What dimensions do you recommend? 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 rand h will give the smaller cone the largest possible volume?

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) = x 4 - 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.

r

63.

1 2'

65.

1

67

•

J J J

3

(x + 5x - 7) dx

~) dt

(3Vt + dr

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

73.

75. 40 - lOx

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 + 7r/4).

77. 78.

60. The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the comer

J ~dB 2

+ x 4)-1/4 dx

72.

2

sec :0 ds

74.

esc \l2B cot \l2B dB

76.

sin2

~ dx

2

I dx

cos

3

+ t) dt

-

6 dr

(r -

J J J

(Hint: sin2 B = 1 -

7

+

0

(2 - x)3/5 dx

csc

2

sec

7TS

ds

~ tan ~ dB

~os 2B)

Solve the initial value problems in Exercises 79-82. dy

79. -d =

x

o 59. Open-top box

An open-top rectangular box is constructed from a 10-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.

70.

68

(8t

InitiaL VaLue ProbLems

a. What is the farthest apart the particles ever get? b. When do the particles collide?

J J J J J

x 3(1

•

66.

+ 5)2

(r

J ~ JC!;; -:4) J \12)3

64.

69. J30W+! dO 71.

y=

(8,6)

6

=

80. : 2

81.

d r

-2

dt

82.

d 3r -3 dt

x2

+

1

- - 2- ,

x

y(l) = -1

(x + } Y.

y(1) = 1

" r 3 = 15vt + "r; V t

= -cos t;

r'(l) = 8,

r"(O) = r'(O) = 0,

r(l) = 0

r(O) = -1

dt

Chapter 4 Additional and Advanced Exercises

Additional and Advanced Exercises

Chapter

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

11. Finding a function Use the following information to find the values of a, b, and e in the formula I(x) = (x + a)/ (bx 2

+

ex

+

2).

i) The values of a, b, and e are either 0 or 1.

ii) The graph of I passes through the point ( - 1, 0) . iii) The line y = 1 is an asymptote of the graph of I.

12. Horizontal tangent will the curve y = x 3 tal tangent?

For what value or values of the constant k 3x - 4 have exactly one horizon-

+ kx 2 +

13. Largest inscribed triangle

I

to identify the points where values.

has local maximum and minimum

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 AB C is largest when the triangle is isosceles? How do you know?

14. Proving the second derivative test

5. Local extrema a. Suppose that the first derivative of y = I(x) is y' = 6(x

The Second Derivative Test for Local Maxima and Minima (Section 4.4) says:

a.

+ 1)(x - 2)2.

At what points, if any, does the graph of I have a local maximum, local minimum, or point of inflection?

b. Suppose that the first derivative of y = I(x) is y' = 6x(x

+ 1)(x -

2).

At what points, if any, does the graph of I have a local maximum, local minimum, or point of inflection? 6. If I' (x) ~ 2 for all x, what is the most the values of I can increase on [0, 6]? Give reasons for your answer. Suppose that I is continuous on [a, b] and that e is an interior point of the interval. Show that if I'(x) ~ 0 on [a, e) and I'(x) ~ 0 on (e, b], then I(x) is never less than I(e) on [a, b].

7. Bounding a function

I has a local maximum value at x

= e if I' (e) = 0 and

I"(e) < 0 b.

I has a local minimum value at x

= e if I' (e) = 0 and

I"(e) > O. To prove statement (a), let e = (1/2) II" (e) I.Then use the fact that .

I"(e) = hm

+ h) -

I'(e

h~O

I'(e

+ h)

h

> 0,

I '(e + h

S

.

= hm - - - h~O

h

to conclude that for some S

o < Ihl <

I'(e)

h) <1"(e)+e
Thus, I'(e + h) is positive for -S < h < 0 and negative for o < h < S. Prove statement (b) in a similar way.

15. Hole in a water tank You want to bore a hole in the side of the

8. An inequality a. Show that -1/2 ~ x/(1 + x

2

)

~ 1/2 for every value ofx.

b. Suppose that I is a function whose derivative is I' (x) = x/( 1 + x 2 ). Use the result in part (a) to show that

I!(b) - !(a)1 ~ ~Ib - al for any a and b. 9. The derivative of I(x) = x 2 is zero at x = 0, but I 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 of x.

Let h = I g be the product of

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--

In each case, if the answer is yes, give a proof If the answer is no, give a counterexample.

h

/Exitvelocity

a. If I and g are positive, with local maxima at x = a, and if I' and g' change sign at a, does h have a local maximum at a? b. If the graphs of I and g have inflection points at x = a, does the graph of h have an inflection point at a?

y

=

Y64(h - y)

y

..------

.........

Ground F.----....,....----------.+x

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 > 0 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 ofthe reciprocal of a number a without ever dividing by a if youapplyNewton'smethodtothefunctionj(x) = (1/x) - a. For example, if a = 3, the function involved is j(x) = (1/x) - 3. a. Graphy = (1/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 j(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" = O. 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 j(x), g(x) be two continuously differentiable functions satisfying the relationships j'(x) = g(x) and j"(x) = - j(x). Let h(x) = f2(x) + g2(x). Ifh(O) = 5, findh(IO). 25. Can there be a curve satisfying the following conditions? d 2y/dx 2 is everywhere equal to zero and, when x = 0, y = 0 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 + (1/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 = O.

Vi - (l/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' + 2hx + c with 0 > O. By coosidering the minimum, prove that f(x) '" 0 for all real x if, and ooly if, b 2 - ac :s; O. 30. Schwarz's inequality

and deduce Schwarz's inequality: (01 h,

+ o,h, + ... + a"h.)' '" (0" + 0,' + ... + a,,')(h,' + h,' + ... + h.').

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.

a. In Exercise 29, let

Chapter

245

Technology Application Projects

MathematicatMap1e Modules: ModonAlong 0 Straight Line: Position .... Velocity .... Acceleration You will observe the shape of a grapb through dramatic animated visualizatioos of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated

Newton's Method: Estimate 'fr to How Many Places? 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.

Area and Estimating with Finite Sums

5.1

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.

y

0.5 R

~---_.L...--_-_....:L....-_~X

a

0.5

FIGURE 5.1 The area of the region R cannot be found by a simple formula.

246

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 !,

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

-oI---~=----'!--~x

o

o

0.5

0.25

0.5

(a)

0.75

(b)

FIGURE 5.2
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 ,., I·

I

15

I

3

I

7

I

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 i1lSide the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1/4 as before, hut the rectangles are y

y

1

1

0.5

0.5

c+----;o~--f;:----;o!:_:_-+~x

o

0.25

0.5

(a)

0.75

(U)

c;i--'---o~--L--;;''o--'-;;-!c:-"-~x

o

0.25 0.5 0.75 1 0.125 0.375 0.625 0.875 (b)

FIGURE 5.3
248

Chapter 5: Integration shorter and lie entirely beneath the graph of j. The function f(x) = I - x 2 is decreasing on [0, I], 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 minimum value of f(x) for a point x in each base subinterval gives a lower sum approximation to the area, A ,.,

y

15

I

3

I

7

I

I

16 . 4 + 4 . 4 + 16 . 4 + o· 4 =

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

(aJ

<

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 In-_ _

In each of our computed sums, the interval [a, b] over which the function f is defined was subdivided into n subintervals of equal width (also called length) dx; = (b - a)/n, and f was evaluated at a point in each subinterval: Cj in the first subinterval, C2 in the second subinterval, and so on. The finite sums then all take the form f(cj) ~x

(b)

FIGURE 5.4 (a) A lower sum using 16 rectangles of equal width dx = 1/16. (b) An upper sum using 16 rectangles.

+

f(c2) ~x

+

f(C3) ~x

+ ... + f(cn)

~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 because the rectangles taken together contain R. The midpoint rule for 16 rectangles gives a total area approximation of 0.6669921875, but it is not immediately clear whether this estimate 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) ofa car moving down a highway, without changingdirection, and want to know how far it traveled between times 1 = a and 1 = b.lfwe already 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--» t

12

13

(sec)

b

with the subintervals all of equal length /1t. Pick a number t1 in the first interval. If /1t is so small that the velocity barely changes over a short time interval of duration /1t, then the distance traveled in the first time interval is about v(t,) /1t. If t2 is a number in the second interval, the distance traveled in the second time interval is about V(t2) /1t. The sum ofthe distances traveled over all the time intervals is

D '" v(t,) /1t

+ V(t2)

/1t

+ ... +

v(tn ) /1t,

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 I, with / evaluated at left endpoints giving an upper sum: tl

12

•

•

•

1

2

o

~4t>l

13 I)

3

t

250

Chapter 5: Integration With / evaluated at t = 0, I, and 2, we have

D '" /(It) at

+ /(t2)

+ /(t3) at + [160 - 9.8(1)](1) +

at

= [160 - 9.8(0)](1)

[160 - 9.8(2)](1)

= 450.6.

(b) Three subintervals oflength I, with / evaluated at right endpoints giving a lower sum:

o

11

t2

•

•

13

.) t

1

2

3

~<1t~1 With/ evaluated at t = 1,2, and 3, we have

D '" /(tt) at

+ /(t2)

+ /(t3) at + [160 - 9.8(2)](1) +

at

= [160 - 9.8(1)](1)

[160 - 9.8(3)](1)

= 421.2.



With six subintervals oflength 1/2, we get

These estimates give an upper sum using 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 ofthe true value. Error magnitude =

1true

value - calculated value 1

= 1435.9 - 435.671 = 0.23.

0.23 - 0. 050/
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

5.1

Area and Estimating with Finite Sums

251

Displacement Ve..us Distance r ....led If an object with position function set) 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 displacement, a( b) - s(a), the difference between its initial and (mal positions. To see why using the velocity ftmction in the summation process gives an estimate to the displacement, partition the time interval [0, b] into small enough equal subintervals ti.t so that the object's velocity does not change very much from time /101 to 'i. Then v(t.0 gives a good approximation of the velocity throughout the interval. Accordingly, the change in the object's position coordinate during the time interval is about

•

_(,,)
1_(',)1


The total distance traveled is approximately the sum

1_(',)1

II=OL~>--_-


+ 1_('2)1<1., + ... + 1_('.)1
We revisit these ideas in Section 5.4.

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

t

v(~

t

v(~

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

160 144 128 112 96 80 64 48 32

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 ofthe rock at any time dwing its motion to be v(t) = 160 - 321 ft/sec. The rock was 256 ft above the ground 2 sec after the explosion, continued upwards to reach a maximum height of400 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 cbange 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 ground is being estimated from the velocity function. On the other band, 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 ofthe 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 ofthe velocity function (that is, the speed ofthe TOck). As an illustration ofour discussion, we subdivide the interval [0, 8] into sixteen subintervals oflength at = 1/2 and take the right endpoint of each subinterval in OUT calculations. Table 5.3 shows the values ofthe 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)'-21 - 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 + 112 + 96 + 80 + 64 + 48 + 32 + 16 + 0 + 16 + 32 + 48 + 64 + 80 + 96)' Error magnitude

=

544 - 528

=

t

=

528

16

If we take more and more subintervals of [0, 8] in our calculations, the estimates to 256 ft and 544 ft improve, approaching 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 f 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

en-y=c

0

a

(a)

b

x

o

a

b

(b)

x

FIGURE 5.6 (a) The average value of f(x) = c on [a, b] is the area ofthe rectangle divided by b - a. (b) The average value ofg(x) on [a, b] is the area beneath its graph divided by b - a.

y

f(x)

~

sin x

What ifwe 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 ofthe 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 ofthe 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 the area under f(x) ~ sinxbetween

oand'1T' to compute the average value of sin x over [0, 1f1, using eight rectangles (Example 4).

Estimate the average value of the function f(x)

=

sin x on the interval

'IT].

Solution Looking at the graph ofsin 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 ofthe 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

5.1 Area and Estimating with Finite Sums

253

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 ofthe 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

R:$

. 1T + sm. 1T + sm. 31T + sm. 1T + sm. 1T + sm. 51T + sm. 31T + s. m 71T)- ' 1T ( sm848228488

'" (.38

+ .71 +

.92

+ 1 + 1 + .92 + .71 + .38)' ~

=

(6.02)' ~ '" 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 ofa 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 1 as if it were constant over each particular subinterval. Then we multiply the width of each subinterval by the value of 1 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 l(ck) is the value of 1 at the chosen point Ck in the kth subinterval, this process gives a fmite sum of the form

+ ... + l(c.) ~x. The choices for the Ck could maximize or minimize the value of 1 in the kth subinterval, or I(c,)

~

+ I(c,)

~x

+ I(c,)

~

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 ofthinner width.

Exercises 5.1 = I andx = 5.

Area

3. f(x) = I/x bctweco x

In Exercises 1-4, use ftnite approximations to estimate the area under the graph ofthe function using

4. f(x) = 4 - x'bctwecox = -2 and x = 2.

a. b. c. d.

a lower sum with two rectangles ofequal width. a lower sum with four rectaogles of equal width. an upper sum with two rectangles ofequal width. an upper sum with four rectangles ofequal width.

= 0 and x x'betwecox = oand x

1. f(x) = x'betwecox

= I.

2. f(x) =

= I.

Using rectaogies whose height is givco by the value of the function at the midpoint of the rectaogie's base (the midpoint rule), estimate the area under the graphs of the followiog functions, using 'ITSt two and thea four rectangles. 5. f(x) ~ x'bctwecox ~ oand x ~ I. 6. f(x) = x'bctwecox = oand x = I. 7. f(x) ~ I/x bctweco x ~ I andx ~ 5. 8. f(x) ~ 4 - x' betwecox ~ -2 and x ~ 2.

254

Chapter 5: Integration

Distance 9. Distance traveled The accompanying table shows the velocity of a model train engine moving along a track for 10 sec. Estimate the distance traveled by the engine using 10 subinterva1s oflength I with

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).

a. left-endpoint values. b. right-endpoint values. TIme (sec)

Velocity (in·/see)

TIme (see)

Velocity (in·/sec)

0 I 2 3 4 5

0 12 22 10 5 13

6 7 8 9 10

11 6 2 6 0

Time (h)

Velocity (mi/h)

TIme (h)

Velocity (mi/h)

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

mi/br

160

10. Distance traveled upstream You are sitting on the bank of a tidal river watching the incoroing tide carry a bottle 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 suhioterva1s oflength 5 with

140 120 100

a. left-endpoint values.

80

b. right-endpoint values. TIme (min)

Velocity (m/sec)

0 5 10 IS 20 25 30

I 1.2 1.7 2.0 1.8 1.6 1.4

TIme

(min)

Velocity (m/sec)

35 40 45 50 55 60

1.2 1.0 1.8 1.5 1.2 0

a. left-endpoint values.

20 0

0 10 20 30 40 50 60

0 44 IS 35 30 44 35

TIme (sec) 70 80 90 100 110 120

IS 22 35 44 30 35

I

0.002 0.004 0.006 0.008

om

hOlUll

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 resistance An object is dropped straight down from a helicopter. 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:

a Velocity (converted to fI Isee) (30 mi/h = 44 fl/see)

/

40

~

b. right-endpoint values.

TIme (sec)

/

60

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

Velocity (converted to ft/see) (30 mi/h = 44 fl/sec)

~

I

2

3

4

5

19.41

11.77

7.14

4.33

2.63

0 32.00

a. Find an upper estimate for the speed when I

= 5.

b. Find a lower estimate for the speed when t = 5. c. Find an upper estimate for the distance fallen when t = 3. 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' 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 interva1 by partitioning the interva1 into four subinterva1s of equa1length and eva1uatiog 1 at the subinterva1 midpoints.

15. I(x) = x' on [0,2] 17./(1) = (1/2)

+

16. I(x) = I/x

sin2 "'1

on

on

[1,9]

255

Area and Estimating with Finite Sums

Measuremeots are takeo at the eod of each month detennioing the rate at which pollutants are released into the a1mosphere, recorded as follows.

Month

Jan

Feb

Mar

Apr

May

Jon

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

= !2 + sin2 '1Tt

(tons/day)

Month

o

1

Pollutant

2

release rate 18. 1(1) = I - (cos

~t)'

(tons/day) on

[0,4] a. Assurniog a 30-day month and that new scrubbers allow only 0.05 ton/day to be released, give ao upper estimate of the tota! tonnage ofpollutants released by the end of June. What is a lower estimate?

y

1

b. In the best case, approximately wheo 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 and compute the area of the polygon for the followiog values of n:

234

a. 4 (square)

Examples of Estimations 19. Water pollution Oil is leaking out of a tanker 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)

Leakage (gal/h)

o

I

2

3

4

50

70

97

136

190

5

6

7

8

Leakage (gal/h)

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 ofone of the n congruent triangles formed by drawiog radii to the vertices ofthe 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), and (c) with the area of the circle.

00.

c. Repeat the computations in parts (a) and (b) for a circle of radius r.

COMPUTER EXPLORATIONS 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. e. The tanker contioues to leak 720 gai/h after the first 8 hours. If the tanker 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 poDntion A power plant 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 eventoa1ly they must be replaced when the amount of pollution released exceeds government standsrds.

In Exercises 23-26, use a CAS to perform the following steps. a. Plot the functions over the giveo interval.

b. Subdivide the interva1 into n = 100,200, and 1000 subintervals of equa1leogth aod evaluate the function at the midpnint of each subinterva1. c. Compute the average value of the function values generated in part (b). d. Solve the equation/(x) ~ (average value) for x using the average value calculated in part (c) for the n = 1000 partitioning. 24. I(x) = sin2 X on [0, ".] 23. I(x) = sin x on [0, ".]

25. I(x)

~ x sin

t [;f,,,.] on

26. I(x)

~ x sin2t on

[-;r,,,.]

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 instauce). 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), stauds 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 kth k = 1 ak lS8 .Lormwa..L0l e term.

"

The index k starts at k

= 1.

Thus we can write 11

12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 11 2 = ~k2, 1=1

and 100

1(1) + 1(2) + 1(3) + ... + 1(100) = ~/(i). j-l

The lower limit of summation does not have to be I; it can be any integer.

EXAMPLE 1 A sum in sigma notation

The Slim written alit, one term for each value of k

The vallie of the sum

~k k-l 3

1+2+3+4+5

15

(- 1)1(1) + (-1j2(2) + (-1)3(3)

-I + 2 - 3 = -2

I 2 TTI+2+1 42 52 --+-4- I 5- I

l+I=l 236

,

~(-I)kk

2

~_k_

k~1

,

k+ I k2

~k- I

k~4

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, hut the tenns generated remain the same. It is often simplest to start with k = 0 or k = I, hut we can start with any integer. 4

Starting with k = 0:

1

+3+

5

+7+9

= ~(2k k-O

Starting with k = 1:

1

+3+

5

+7+9

= ~(2k

+ 1)

S

- 1)

k=1 6

= ~(2k - 3) k-2 1 9 = ~ (2k + 7) 1=-3

Starting with k = 2:

1

+3+

5

+7+9

Starting with k = -3:

1

+3+

5

+7+

•

When we have a sum such as

we can rearrange its terms, 3

~(k

+ e)

=

(1

+ 12) + (2 + 22) + (3 + 32)

=

(1

+ 2 + 3) +

k-l

3

= ~.k

.r:1

+

(12

+ 22 + 32)

Regroup terms.

3

~.k2.

.r:1

This illustrates a general rule for finite sums:

•

••

~(ak + bk) = ~ak

i-I

1-1

+ ~bk 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

•

(8) ~(3k -

(Any number cJ

•

~_c=n·c ~

(c is any constant value.)

We demonstrate the use of the algebra rules.

e) =

••

3~k - ~e

Difference Rule and Constant Multiple Rule Constant Multiple Rule

258

Chapter 5: Integration 3 3 3

(0) ~(k

+ 4)

=

~k + ~4

=

(I

= 6

(d) HISTORICAL BIOGRAPHY

Carl Friedrich Gauss (1777-1855)

L• -nI =

1=1

Sum Rule

+ 2 + 3) + (3·4) + 12 = 18

Constant Value Rule

I n·- = I

Constant Value Rule (lin is constant)

n

•

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 ofthe 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

±k

= n(n

Solution

+ I). 2

k-l

The formula tells us that the sum of the first 4 integers is

(4~5)

=

10.

Addition verifies this prediction: I

+2+3+4

=

10.

To prove the formula in general, we write out the terms in the sum twice, once forward and once bsckward.

+ +

+ n + + I + If we add the two terms in the first column we get I + n = n + I. 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 + I. When we add the n columns together we get n terms, each equai to n + I, for a tota1 ofn(n + I). Since this is twice the desired quantity, the sum of the 1mt n integers is (n)(n + 1)/2. • I

n

2 (n - I)

+ +

3 (n - 2)

Formulas for the sums ofthe squares and cubes ofthe 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 ofthe subintervals go to zero and their number grows to infinity.

EXAMPLE 5 Find the limiting value oflower sum approximations to the area ofthe region R below the graph of y = I - x 2 and above the interval [0, I] on the x-axis using equai-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 equai width Ax = (I - Olin, and then we see what happens as n --+ 00. We start by subdividing [0, I] into n equai 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 - I)/n, kin] is I(k/n) = 1 - (k/n)2, giving the sum

We write this in sigma notation and simplify,

Difference Rule

Constant Value and Constant Multiple Rules Sum ofthe 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~1 I(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 ofthe 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 equai 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 - 1 points {X" X2, X3, ••• , X._I} between a and h and satisfying

a<

Xl

<

X2

< ... <

X.-I

< h.

260

Chapter 5: Integration

y

To make the notation consistent, we denote a by Xo and b by x., so that

a

=

Xo

<

XI

< X2 < ... < X.-l < x.

=

b.

The set is called a partition of [a, b]. The partition P divides [a, b] into n closed subinterva1s

[xo,xd, [XhX2],"" [X.-hX.]. FIGURE 5.8 A typical continuous function y = f(x) over a closed interval [a, b].

The first of these subinterva1s is [xo, xd, the second is [Xl, X2], and the kth subinterval of Pis [Xk- h xd, for k an integer between I and n.

-----1__
_--+------1I.:ilrth~su:iiibin:iiiterv~a1iooll_---+_-_____1f_ • Xo = a

Xl

Xk_l

Xi;

XII_l

XII =

b

The width of the first subinterva1 [xo, xd is denoted ,hI, the width of the second [XI,X2] is denoted ~X2, and the width of the kth subinterva1 is ~k = Xk - Xk-l. If all n subintervals have equal width, then the conunon width ~is equal to (b - a)/n.

~ <1.,----->1

I

I

Xk_l

Xl

In each subinterval we select some point. The point chosen in the kth subinterval [Xk-l, Xk] is called Ck. Then on each subinterval we stand a vertical rectangle that stretches from the x-axis to touch the curve at (Ck, f(ck»' 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(ck) = 0 (Figure 5.9). On each subinterval we form the product f(ck)' ~Xk. This product is positive, negative, or zero, depending on the sign of f( q). When f( q) > 0, the product f( q) . ~k is the area of a rectangle with height f(ck) and width ~k. When f(ck) < 0, the product f(Ck) • ~k is a negative number, the negative of the area of a rectangle of width ~k that drops from the x-axis to the negative number f( q) .

y

lrth rectangle

c.

(C2,

ftc,»)

FIGURE 5.9 The rectangles approximate the region between the grapb 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.

5.2 Sigma Notation and Limits of Finite Sums

261

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, h]. There are many such sums, depending on the partitioo 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)jn to partition [a, b], and then choose the point Ck to be the right-hand endpoint of each subinterval when forming the Riensann sum (as we did in Example 5). This choice leads to the Riemann sum formula y

"........... :-., y~fl

,

oa

S. = I'

b

lJ

x

.... 1..

~f(a + kb; 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)jn, 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 ofa 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)

to,

EXAMPLE 6 The set P = 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

!<-
--+-- --+-- ---+-- I

--;.-n';.----~----_;_----____.c7----____,;_->x

o

o

U

M

15

2

b x

a

The lengths of the subintervals are Ax! = 0.2, Ax, = 0.4, Ax, = 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)

The curve of Figure 5.9 with rectangles from rmer partitioos of [a, b]. Finer partitioos create collections of rectangles with thinner bases that FIGURE 5.10

approximate the regioo betweeo the grapb off and the x-axis with increasing

accuracy.

Any Riensann 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 ofrectangles that approximate this region with increasing accuracy, as suggested by Figore 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. ,

6k

l'~k+1

'-I, 3. ~ coskw '-I,

5. ~(_1)k+1 Sinf

2

7. Which of the following express I + 2 + 4 + 8 + 16 + 32 in sigma notation?

±

. '-I,

k-I k

4. ~ Sink1T

'-I,

6. ~(-I)'coskw

8. Which of the following express I - 2 sigma notation?

,

a. ~(-2)k-l k=1

, b. ~(-l)'2' k=O

+ 4 - 8 + 16 - 32 in

, c. ~(-1)k+12k+' k=-2

262

Chapter 5: Integration 5

9. Which formula is not equivalent to the other two? 4 (-I)k-! ~--

a.

b.

~ k- I

2 (-I)' ~-

1 (-I)' c.~-

~k+ I

'~1 k + 2

10. Which formula is not equivalent to the other two?

,

4

'-I

+ I)'

11. I + 2 + 3 + 4 + 5 + 6

12. I + 4 + 9 + 16

I I I I 13. 2 + 4 + 8 + 16

14. 2 + 4 + 6 + 8 + 10

I

2

3

16'-:5+:5-:5+:5-:5

1-1

k-l

•

• b,

b. ~b'1 6

1=1

•

c. ~(a, + b,) 1=1

" e. ~(b, - 2a,)

d. ~(a, - b,) k=l

~

" 0 and ~b, ~

1=1

I. Fiod the values of

"

c.

'-I

'-I " d. ~(b, '-I

I)

I)

Evaluate the sums io Exercises 19-32. 10

19. a. ~k k-I 13

20. a. ~k k-I

7

21. ~(-2k) ...1

•

23. ~(3 - k 2 ) ...1

5.3

264

500

30. a. ~k

b. ~k2

c. ~k(k - 1)

" 31. a. ~4

b. ~c

i=1 17

k=3 71

k=18

k=3

b"

"

" c.~(k-I)

'-I " ~(k+2n) b. '-I ~*

'-I

" k

c. ~2 k-l

n

RIemann Sums Partition the ioterval ioto four suhiotervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum ~~-d(c,) .1.<.. given that c, is the (a) left-hand endpoiot, (b) righthand endpoint, (c) midpoint of the kth suhinterval. (Make a separate sketch for each set ofrectang1es.)

33. I(x) ~ x 2 ~

-

siox,

I,

[0,2] [-orr, orr]

34. I(x) ~ -x2 , 36. I(x)

~

[0, I]

siox + I,

[-orr, orr]

37. Fiod the norm of the partition P

~

{O, 1.2, 1.5,2.3,2.6, 3}.

38. Fiod the norm of the partition P

~

{-2, -1.6, -0.5, 0, 0.8, I}.

1=1

b. ~250b,

" + ~(a,

7 k' -~k=1 4

c. ~1O

35. I(x)

b'1

" 18. Suppose that ~a,

=1

In Exercises 33-36, graph each function I(x) over the given interval.

" " 17. Suppose that ~a, = -5 and ~b, = 6. Fiod the values of a. ~3a,

~k

b. ~7

32. a.

5

CY

28.

29. a. ~3

...1

4

i=1

C)' ~k

+

b'1

Values of Finite Sums

•

5 k' ~ 225

k--3

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.

I I I I 15.1- 2 + 3 - 4 +:5

27.

,.

~k2

C.

k--l

k=1

26. ~k(2k + 1)

7

-1

b. ~ (k

a. ~(k - 1)2

7

25. ~k(3k + 5)

10

10

b. ~k2

c. ~k'

k=1 13

k=1 13

b. ~k2

c. ~k'

'-I

k-I

5 orrk 22. ~T5

'-I • 24. ~(k2 '-I

Limits of RIemann Sums For the functions in Exercises 39-46, fmd a formula for the Riemann sum obtained hy dividing the ioterval [a, b] ioto n equal suhintervals 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

+ I over the interval [0, 3].

42. I(x) ~ 3x overtheioterval[O,I]. 2

43. I(x) = x + x 2 0ver the interval [0, I]. 44. I(x) = 3x + 2x 2 over the ioterval [0, I].

5)

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] using n subintervals of equal width (or length), (b - a)ln. 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 integral of a function over a closed interval [a, b].

Definition of the Definite Integral The definition of the definite 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 the number J provided that the nann ofits partition is sufficiently small (so that all of its subintervals have thin enough widtha). We introduce the symbol E as a small positive number that specifies how close to J the Riemann sum must be, and the symbolll as a second smaIl positive number that specifies how small the 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 the defmite integral of f over [a, hI and that J is the limit of the Riemann sums ~~=tf(ck) Axk ifthe following condition is satisfied: Given any number E > 0 there is a corresponding number Il > 0 such that for every partition P = {xo, x" ... , x.} of [a, hI with Ilpll < Il and any choice of Ck in [Xk-" xii, we have

The defmition involves a limiting process in which the nann of the partition goes to zero. In the cases where the subintervals all have equal width ~x = (h - a)/n, we can fann

each Riemann sum as

•

• (h ; a ),

S. = ~ f(ck) ~Xk = ~ f(ck)

ax, = ax = (b - all, for all k

where Ck is chosen in the subinterval Axk. If the limit of these Riemann sums as n ---> 00 exists and is equal to J, thenJ is the defmite integral of f over [a, hI, so J = lim

n --+ 00

~f(ck)(h; a)

=

t:1

lim

n--+ 00

~f(ck) Ax.

{:1

ax = (b - all.

Leibniz introduced a notation for the defmite integral that captures its construction as a limit of Riemann sums. He envisioned the fmite sums ~~=1 f(Ck) Axk becoming an infinite sum of function values f(x) multiplied by "infinitesimal" subinterval widths dx. The sum symbol ~ is replaced in the limit by the integral symbol f, whose origin is in the letter "S." The function values f(ck) are replaced by a continuous selection of function values f(x). The subinterval widths Axk become the differential dx. It is as ifwe are summing all products of the fann f(x) •dx as x goes from a to h. While this notation captures the process ofconstructing an integral, it is Riemann's definition that gives a precise meaning to the definite integral. The symbol for the number J in the defmition of the definite integral is

[f(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 with respect to x." The component parts in the integral symbol also have names: The function is the integrand. Upper limit of integration/

-""f;(x) ,h~r-"-... . ~ Lower limit of mtegration

y

Integral

, W h e n you :lind the value of the integral, you have

off from a to b ~ evaluated the integral.

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) dx 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)

1I111~O

...,

~k =

J =

l bl

(X) dx.

a

When each partition has n equal subintervals, each of width ~ = (b - a)Jn, we will also write n

lim ~/(ck) ~ = J =

11_ 00

k=1

lb a

I(x) dx.

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

[1(X)dx.

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 ofiwnp discontinuities on [a, b]. (The latter are called piecewise-continuous fUnctions, 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.

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) dx exists and I is integrable over [a, b].

THEOREM I-Integrability of Continuous Functions

t

The idea behind Theorem I 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- 1, Xk], resulting in an upper swn. Likewise, we can choose CO to give the minimwn value of Ion [Xk-1, 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 sufficieotly 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

f(x)

{I, 0,

=

ifxisrational 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._ I> xtl then the correspondiog Riemann sum is

c.

•

•

U = ~f(C.) At. = ~(I) ~x. = I,

sioce each subioterval [Xk-- I> x.] contaios a rational number where f(c.) = I. Note that the lengths of the iotervals io the partition sum to I, ~f~l~X' = I. So each such Riemann sum equals I, and a limit of Riemann sums using these choices equals I. On the other hand, if we pick to be the poiot giving the minimum value for f on [x._1> xtl, then the Riemann sum is

c.

•

L = ~ f(c.)

At.

•

~(O) At.

=

=

0,

sioce each subioterval [x.-I>xtl 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 process of taking antiderivatives.

Properties of Definite Integrals

t

In defming f(x) dx as a limit of sums ~f~ 1/(c.) At., we moved from left to right across the ioterval [a, b]. What would happen ifwe iostead move right to left, starting with Xo = b and endiog at x. = a? Each At. io the Riemann sum would change its sign, with x. - X'-l now negative iostead ofpositive. 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

= -

Ibn

[f(X)dx.

Although we have only 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 iotegrals, given as rules that they satisfy, iocludiog the two just discussed. These rules become very useful io the process of computing iotegrals. 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 ofpositive functions, but the rules apply to general iotegrable functions.

THEOREM 2 Whenf andg are iotegrable over the ioterva1 [a, b], the definite iotegral satisfies the rules io Table 5.4.

266

Chapter 5: Integration

TABLE 5.4 RuLes satisfied by definite integrals

1.

bf OrderolIntegration: l"f(x) dx = - l (X) dx

l

2. Zero Width Interval:

A Definition

a

A Def'mition when/(o)oxists

f(x) dx - 0

"-

bf l\f(x)dx = kl (x)dx

Anycons_k

4. Sum and Difference:

bf l bU(x) ± g(x)) dx = l (x) dx ± l

b g(x) dx

5. Additivity:

l

6.

If f bas maximum value max f and minimum value minf on [a, b], then

3.

Constant Multiple:

Max-Min Inequality:

bf

(x) dx

cf

+l

(x) dx = l

cf (x) dx

bf minf·(b - a) " , l (x)dx '" maxf·(b - a). 7.

y

Domination:

y

y

~

f(x)

20

g(x) on [a, b] => l

f(x)

20

Oon[a,b] => l

bf

(x) dx

bf (x)dx

20

bg

20

l

(x) dx

0

(Special Case)

y

2/(x)

y

~ I(x)

y

~

+ g(x)

g(x)

f------1y ~ I(x) --001-----'-----> x

~---"--------!_x 0 b

-Oo+Jo'----------!-b- x

(a) Zero Width Interval:

(b) Constant Multiple: (k

2)

(c) Sum: (areas add)

f(x) rb:

l'(/(x) + g(x))rb:

o

o

0

[f(x)rb:

~

l ' kf(x) rb:

0

y

1

~k

~

b

y

~ l'f(x)rb: +

y

maxi min! y

-oO+---"o-----cb:-----c-'------>x

(d) Additivityfor definite integrals:

[f(X)rb: + [f(x)rb:

~ [f(x)rb:

--001--'--------+---> x

o

b

0

00

b

(f) Domination:

mmf'(b - 0)

f(x)

f(x)rb:

'" maxf'(b - a) FIGURE 5.11 Geometric interpretations of Rules 2-7 in Table 5.4.

g(x)

--oo1-'--------!-~x

(e) Max-Min Inequality:

"'1'

~

=>

20

l'

g(x) on [a, b]

f(x) rb:

20

l'

g(x) rb:

1

bg

(x)rb:

267

5.3 The Definite Integral

While Rules I and 2 are def"mitions, Rules 3 to 7 ofTable 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 ofthe points Ck, min/'(b - a) = mini'

• L aXk k=1

~.6.Xl = b -

a

Constant Multiple Rule

minf S

f(c,)

f(c,) s maxf

Constant Multiple Rule

maxi' (b - a).

=

In short, all Riemann sums for I on [a, b] satisfy the inequality

•

axk ,,;

min I' (b - a) ,,; ~/(ck)

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/ (X) dx

= 5,

-2,

=

f\(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)

2

Rule 1

+ 3[h(X) dx

+ 3(7)

=

Rules 3 and 4

31

j4, /

=

5

+ (-2)

Show that the value of

JoL\/I

+

cosxdx is less than or equal to

=

+

f'/(x) dx

1-,

(x) dx

=

3

•

Rule 5

vi

Solution 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. The maximum value of

VI

J:

+

1'VI

cosx on [0, I] is

+

cosxdx,,;

v'l+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 subinterval----1l11 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 ofthe 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.

DEFINmON If Y = f(x) is nonnegative and integrable over a closed interval [a, b], then the area under the curve y f(x) over [a, bl is the 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 interval

[O,b],b>O.

Solution

The region ofinterest is a triangle (Figure 5.12). We compute the area in two ways.

(a) To compute the def'mite integral as the limit of Riemann sums, we calculate limi''U~o ~~, f(ck) dxk for partitions whose norms go to zero. Theorem 1 tells us that --o-I'--------c!-~x

o

b

FIGURE 5.12 The region in Example 4 is a triangle.

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 subintervals 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,

*'

2:, 3:, ...

,n:}andck

= ,,:. So

fCc,)

~

c,

Constant Multiple Rule

=

b2 n2

n(n •

+ 1) 2

Sum ofFinit 11 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~


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 fannula 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. •

J:

Example 4 can be generalized to integmte f(x) = x over any closed interval [a,b], O < a < b.

y

RuleS

Rule I --f---+~~_x

ExampI.4

In conclusion, we have the following rule for integmting fix) = x: (b) 2 bxdx = b 2 1a

y

-t-------;,f.;---;--~x

_

a2 2'

a
(1)

This computation gives the area ofa tmpezoid (Figure 5.13a). Equation (1) remains 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 tmpezoid dropping down to the line y = x below the x-axis (Figure 5.13b). When a < and b > 0, Equation (1) is still valid and the defmite integral gives the difference between two areas, the area under the graph and above [0, b] minus the area below [a, 0] and over the gmph (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).

°


gives the negative of the area ofthis

trapezoidal regioo. (0) The defmite integral in Equatioo (1) gives the area ofthe blue ttiaogu\ar regioo added to the negative ofthe area ofthe gold ttiaogu\ar regioo.

1 1\2 b

C

dx = c(b - a),

a

c any constant

3- a3 3'

dx = b 3

a
(2)

(3)

Average Value of a Continuous Function Revisited In Section 5.1 we introduced informally the average value of a nonnegative continuous function f over an interval [a, b], leading us to defme this average as the area under the gmph of y = f(x) divided by b - a. In integral notation we write this as

Average

=

b

~ a 1bf(x)dx.

We can use this formula to give a precise definition of the average 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 average of n numbers is their sum divided by n. A continuous function f on [a, b] may have infinitely many values, but we can still sample them in an orderly way.

270

Chapter 5: Integration We divide [a, h] into n subintervals of equal width ~ = (h - a)/n and evaluate I at a point Ck in each (Figure 5.14). The average ofthe n sampled values is

y

I(c,)

+ 1(c2) + ... + I(c.) _ !

~ ( ) -n~/Ck

n

~l

=

~

•

h - a L/(ck) k-l

I

=

FIGURE 5.14 A sample of values ofa function on an interval [a, b].

b-a

1

fix

fix~-.-,soii~ b-a

•

-_h L l(ck) ~ a

Constant Multiple Rule

k=1

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»J:I(x) dx. Both points of view lead us to the following definition.

DEFINmON If I is integrable on [a, h], then its average value on [a, h], also called its mean, is y 2 f(x)~V4-x2 -y~

av(f) = h _I a

"-2 EXAMPLE 5

l

a

b

I(x) dx.

Find the average value of I(x) = V4=7 on [-2,2].

-_2!;--_1:--j---:---C2~X

FIGURE 5.15 The average value of f(x) = ~on[-2,2]is"'/2 (Example 5).

Solution We recognize I(x) = V4=7 as a function whose graph is the upper semicircle ofradius 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

Because I is nonnegative, the area is also the value of the integral of I from - 2 to 2,

1Y4 2

x 2 dx

= 2"..

-2

Therefore, the average value of I is

av(f) = 2 _1(_2)l:V4=7dx = t(2".) = ; . 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 a. Integrals Express the limits in Exercises 1-8 as defmite integrals.

n

3.

lim Y(ck2 - 3Ck) lixh whereP is a partition off-7,5]

IW~~o~

n

1.

lim ~cllix" where P is a partition of [0,2]

IWII~o k~l

4.

n

2.

lim 'Y2c,i' Iix" where P is a partition of[-1,0]

IWII~ot;1

5.

±(1 ) ±

lim 111'1- 0 1-1 lim

IW~~O""

Iix" where P is a partition of [1,4]

k

-1_1_lixh whereP is a partition of [2, 3] - Ck

5.3

6. 7. 8.

•

lim Y

1~1~0~

v'4 - c,' ax" where P is a partition of[O, I]

•

lim ~(secck) ax" whereP is a partition of[ -·,,/4,0]

I~I~OFI

Using Known Areas to Find Integrals 10 Exercises 15-22, graph the integrands aod use areas to evaluate the integrals.

lim ~(tanck) ax" where P is a partition of [0, ,,-/4]

I~I~OFI

UsIng the DefinIte Integral Rules 9. Suppose that f aod g are integrable aod that

~ -4,

1'f(X) dx

1'f(X) dx

~ 6,

1'g(x) dx

~ 8.

E(I

3) dx

16.

17.1:~dx

18.

19. DX1dx

20.

15.

•

21.

+

1:(2 -

Ixl) dx

11/2'/2

22.

(-2x+4)dx

1~v'16 - x'dx

1:(1 1: + '\!'l=7) Ixl) dx

(I

Use the rules in Table 5,4 to rmd

Use areas to evaluate the integrals in Exercises 23-28.

a. j,'g(x) dx

b. j,'g(x) dx

23.],' Idx,

e.1'3f (X)dx

d. j,'f(X)dx

25.1'2sds, O
l'

e.

27. f(x) = [f(x) - g(x)] dx

f.1'[4f(X) - g(x)] dx

10. Suppose that f aod h are integrable aod that

l'

~ -I,

f(x) dx

/,'f(x) dx

~ 5,

/,'h(x) dx

~ 4.

l'

a.

-2f(x) dx

e. 1'f(x) dx

+ h(x)] dx

d. j,'f(X) dx

11. Suppose that J,'f(x) dx

l'

a. 1'f(U) du

b.

e. j,'f(l)dl

d.1'[-f(X)]dx

12. Suppose that

t, g(l) dl

=

Y3f(z) dz

v'2. Find

a. ],-'g(l) dl

b. l:g(u) du

e·lo[

d.lo ~~

-g(x)] dx

-3

-3

13. Suppose thatf is integrable aod that

a.

f(z) dz

l:

v2

dr

Jo' f(z) dz = 3 aod

= 7. Find

f (z) dz

b. /.' f(l) dl

14. Suppose that h is integrable aod that J~l h(r) dr = 0 aod

l, h(r) dr ~ 6. Find a.1'h(r)dr

J,

'\!'4=7 + '\!'l=7

xdx

35.

Jf'/2o

rdr

Jo

I'dl

l

v'3a xdx

0

on a. [-2,2], b. [0,2] on a. [-I,D], b. [-I, I]

30.

j,'" xdx

J.

34.

Jo["" 8' ds

0.5

,r, 33.], 36.

x'dx

Jor/20' dO

w

OdO

37·1"xdx

>Yb 39.],

,w

31.

x'dx

40.

Jo[" x'dx

Use the rules in Table 5,4 aod Equations (1)-(3) to evaluate the integrals in Exercises 41-50.

41. ]" 7 dx

42.],'sxdx

43.],'(21 - 3) dl

44.

Jo (I - v'2) dl

45. j,' (I + I) dz

46.

],0(2z - 3) dz

47·1'3U'dU

48.

f' 24u' du J'/2

49.

],\3x' + x - 5) dx

[0

50.1°(3X'

+x -

5) dx

FInding Area by Definite Integrals 10 Exercises 51-54, use a definite integral to rmd the area of the region between the giveo curve aod the x-axis on the interval [0, b].

51. Y = b. _],'h(U)dU

>

26.1'3Idl, O
M.

32.

f. j,'[h(X) - f(x)] dx

= 5. Find

24.],'4Xdx, b

0

dx

Evaluattng DefinIte Integrals Use the results of Equations (I) aod (3) to evaluate the integrals in Exercises 29-40.

38..

10

>

28. f(x) = 3x

29.

b. /,' [f(x)

e. /,'[2f(x) - 3h(x)] dx

4

b

['/2

Use the rules in Table 5,4 to rmd

271

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 56. f(x) =

x' -T

on

[0,0]

10' sin (x') ax cannot possibly be 2.

76. Show that the value of fo' and 3.

~ ax lies between 2 v'2

77. Integrals of nonnegative functions

57. f(x) = -3x' - I

[0,3]

on

75. Show that the value of

on

[0, I]

58. f(x) ~ 3x' - 3 on [0, I]

'" 2.8

Use the Max-Min inequal-

ity 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) ~ t' - t on 61. g(x) =

62. h(x) =

Ixl -Ixl

[-2, I] on a. [-1,1], h. [I, 3], and c. [-1,3]

I

78. Integrals of nonpositi"" functions then

Show that if f is integrable

on a. [-1,0], h. [O,I],and c. [-1,1]

f(x) '" 0 on

=*

[a, b]

J.bf(X) ax '" O.

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.

(3x' - 2x + I) ax

1:

68.1:

10'

80. The inequality sec x '" I + (x'/2) holds on (-"./2, ",/2). Use it to find a lower bound for the value of fo' sec x

(x - x') ax

81. Ifav(f) rea11y 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

+

ax.

x'ax J.b av(j) ax

69. J.b x' ax,

a< b

ax.

I) ax

64. /,' (2x

a< b

79. Use the inequality sin x '" x, which holds for x '" 0, to fmd an upper bound for the value of sin x

~ J.bf(x) ax?

70. /,' (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(j + g) = av(j) + av(g) h. av(kf)

= k av(j)

c. av(j) '" av(g)

(Hint: Where is the integrand positive?) 72. What values of a and b miuimize the value of

(any nornber k) if f(x) '" g(x)

on

[a, b].

Do these rules ever hold? Give reasons for your answers. 83. Upper and lower sums for increasing functions

a. Suppose the graph of a continuous function f(x) rises steadily as x moves from left to right across an interval [a, b]. Let P be a partition of [a, b] into n subintervals oflength ~ (b - alin. Show by referring to the accompanying figure that the difference between the upper and lower soros for f on this partition can be represented graphically as the area of a rectangleR whose dimensiona are [f(b) - f(a)] by (Hint: The difference U - L is the sum of areas of rectangles whose diagonals QOQb Q,Q" ... , Q._,Q.lie along the

ax

73. Use the Max-Min Inequality to fmd upper and lower bounds for the value of

' I /,o 1 + x 2

--ax•

74. (Continuation of Exercise 73.) Use the Max-Min Inequality to fmd upper and lower bounds for 0,' - -I a2 x /o , 1 x

+

and

ax.

curve. There is no overlapping when these rectangles are shifted horizontally onto R.)

/,' - -Ia x . 2 0.5 1 + x

ax.

h. Suppose that instead ofbeing equal, the lengths of the subintervals ofthe partition of [a, b] vary in size. Show that

Add these to arrive at an improved estimate of

' - -I a x /,o 1 + x 2 •

U- L '" If(b) - j(a)lax.....

where

ax,... is the norm of P, and hence that lirn~~o = O.

(U - L)

5.3

The Definite Integral

273

y

y

1

f (b) - f(a)

Rj --->I~I<-

-

y

84. Upper and lower sums for decreasing functions (Continuation ofExercise 83.) a. 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

[77

subintervals of equalleogth. Find an expression for U - L that is analogous to the one you found for U - L in Exercise 83a.

b. Suppose that instead ofbeing equal, the lengths ax,ofthe

/

V'

7

I'---

7

a

0

7

Xk

Xk + 1

b

Xl

Xk+l

b

subintervals ofP vary in size. Show that the inequality U-L'" If(b) - f(a) I ax,.,.

x

y

of Exercise 83b still holds and hence that liml"U~o (U - L) ~ O.

85. Use the fonuula sinh

+ sin2h + sin3h + ... +

sinmh

cos (h/2) - cos ((m

+ (l/2))h)

2 sin (h/2) to fmdthe area uoder the curvey = sin x from x = Otox = '1r/2 in two steps:

a. Partition the interval [0, '1r/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 contiuuous and nonnegative over [a, b], as in the accompanying figure. By inserting points XloX2,··· , X,t-lo X,t, "', Xn-l

ax, x, -

as shown, divide [a, b] into n subintervals ofleng!bs = a, dx2 = X2 - Xl, . .. , dxn = b - Xn-I, which need not be equal.

a. !fm, ~ min {f(x) for x in the kth subinterval}, explain the connection between the lower sum L

=

mt

axt + m2 dx2 + ... + mIl 4xn

and the shaded regions in the first part of the figure. b. !fM, = 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.

Ic:::o;p;;p__9 I,

b - a

,1

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 contiuuous. Use this and the figure for Exercise 86 to show that iff is contiuuous and E > 0 is given, it is possible to make U - L '" E' (b - a) by making the largest of the sufficiently small.

ax,'s

88. If you average 30 mijh on a 150-mi trip and then return over the same 150 mi at the rate of 50 mifh, what is your average speed for the trip? Give reasons for your answer. COMPUTER EXPLORATIONS !fyour CAS can draw rectangles associated with Riemann sums, use it to drsw rectangles associated with Riemann sums that converge to the integrals in Exercises 8~94. Use n ~ 4, 10,20, and 50 subintervals of equallengtb in each case. {' I 89. Jo (I - x) dx ~ 2

274

Chapter 5: Integration

90.j,'<X' + I)

dl; =

91.

J~ cos x dl; =

92.

Jo

93.

[iX I

1

b. Partition the interval into n = 100, 200, and 1000 subintervals of equallengtb, and evaluate the function at the midpoint of each subinterval.

0

c. Compute the average value of the function values generated in part (b).

[~/4

d. Solve the equationf(x) ~ (average value) for x using the average value calculated in part (c) for the n = 1000 partitioning.

sec' x dl; = I dl;

94.1' ~dl;

~

95. f(x) = sinx

I

(The integra1'svalue is about 0.693.)

10 Exercises 95-98, use a CAS to perform the following steps:

a. Plot the functions over the given interval.

5.4

I

y

f(e), average heighl --.1--'----~-----O~x

a

b

----->I

FIGURE 5.16 The value f(e) in the Mean Value Theorem. is, in a sense, the average (or mean) height off on [a, b]. When f '" 0, the area of the rectangle is the area under the graph of f from a to b, f(e)(b - a) =

98.f(x)=xsin'~

[~, 'IT]

on

on

[~,'IT]

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 than by taking limits of Riemann sums as we did iu Section 5.3. Leibniz and Newton exploited this relationship and started mathematical 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.

Sir Isaac Newton (1642-1727)

c ~b- a

~

[0, 'IT]

on

The Fundamental Theorem of Calculus

HISTORICAL BIOGRAPHY

o

97. f(x) = x sin

[0, 'IT]

on

96. f(x) ~ sin' x

l'

f(x) dl;.

Mean Value Theorem for Definite Integrals In the previous section we defiued the average value of a continuous function over a I(x) dx divided by the length or width closed iuterva1 [a, hI as the def'mite iutegral 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, h]. 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.

J:

THEOREM 3-The Mean Value Theorem for Definite Integrals ous on [a, hI, then at some poiut c iu [a, hI,

l

If 1 is continu-

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

Avm~evw~ln ~~~-------~notassumed

"t---t----;;------x o 2 FIGURE 5.17

A discontinuous function

275

Since 1 is continuous, the Intermediate Value Theorem for Continuous Functions (Section 2.5) says that 1 must assume every value between min 1 and max f. It must therefore assume the value (l/(b - a))J:/(x) dx at some pointe in [a, b]. • The continuity of 1 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 asswne its average value.

b/(x) dx

=

0,

then I(x) = 0 at least once in [a, b]. Solution

The average value of 1 on [a, b] is

1 b

avU) = b _I a a I(x) dx = b _I a . 0 = O. By the Mean Value Theorem, 1 assumes this value at some point e

E

[a, b].

•

Fundamental Theorem, Part 1 area

y

~

If 1(/) is an integrable function over a finite interval I, then the integral from any f'1xed number a E I to another number x E I defines a new function F whose value at x is

F(x)

F(x) = [ / ( / ) dl.

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 > Q.

(1)

For example, if 1 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 ofintegration of an integral, but F is just like any other real-valued function ofa real variable. For each value ofthe input x, there is a well-defmed numerical output, in this case the defmite integral of1 from a to x. Equation (I) gives a way to defme new functions (as we will see in Section 7.2), but its importance now is the connection it makes between integrals and derivatives. If1 is any continuous function, then the Fundamental Theorem asserts that F is a differentiable function of x whose derivative is 1 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 1 ;;,: 0 on [a, b], then the computation of F'(x) from the defmition of the derivative means taking the limit as h ---> 0 ofthe difference quotient F(x b In Equation (I), F(x) is the area to the left of x. Also, F(x + h) is the area to the left of

FIGURE 5.19

+ h. The difference quotient [F(x + h) - F(x)l/h is then

x

approximately equal to f(x), the height ofthe rectangle shown here.

+ h) - F(x) h

For h > 0, the numerator is obtained by subtracting two areas, so it is the area under the graph of1 from x to x + h (Figure 5.19). If h is small, this area is approximately equal to the area ofthe rectangle 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 ofthis 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 1 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:

di

F'(x) = ax

X

(2)

I(t) dt = I(x).

a

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 Theorero to tmd dyJax if

X

(t 3

+ I)dt

(b) y = 153tSintdt

(c) Y =

1'"

costdt

We calculate the derivatives with respect to the independent variable x.

i

X

(a) -dy = -d (t 3 + I)dt = x 3 + I ax ax a 5 dy = ax d1x 3tsintdt = ax d ( f ' 3tsintdt ) (b) ax

-15

= -

d (X ax15 3tsintdt

= -

3x sin X

Eq.(2)withf(r) ~ r'

+

1

Table 5.4, Rule 1

Eq. (2) with f(t) ~ 31Sin t

(C) The upper limit of integration is not x but

x 2• This

makes y a composite of the two

functions,

y =

1"

and

costdt

We must therefore apply the Chain Rule when finding dyJax.

dy dy du ax=du'ax =

(~1" costdt), du 1

= casu'

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 f(x) for each x in (a, b). Thus, Fi,,-(x_+~h)'----_Fi-"(x-'-)

F'(x) = lim _ h~O

=

h

l~d [[+hf(t) dt l

= lim -h h--+O

l

[f(t) dt]

x h

+ f(t) dt

Table 5.4. Rule 5

x

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 f in the interval between x and x + h. That is, for some number c in this interval,

I1

h

x

X

+h f(t) dt = ftc).

(4)

All h --+ 0, x + h approaches x, forcing c to approach x also (because c is trapped between x and x + h). Since f is continuous atx, f(c) approaches f(x): lim ftc) = f(x).

(5)

h~O

In conclusion, we have l

F'(x) = lim -h h-O

l

x h

+ f(t) dt

x

= lim ftc)

Eq. (4)

= f(x).

Eq. (5)

h~O

Ifx = a or b, then the limit ofEquation (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 ofthe 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.

If f is continuous at every point in [a, b] and F is any antiderivative of f on [a, b], then

THEOREM 4 (Continued)-The Fundamental Theorem of Calculus, Part 2

l

bf

(x) dx

=

F(b) - F(a).

Proof Part I of the Fundamental Theorem tells us that an antiderivative of f exists, namely G(x) = [f(t) dt.

Thus, if F is allY antiderivative of f, 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.bI(t) dt - J."/(t) dt

=

J.bl(t) dt - 0

=

J.bI(t) dt.

•

The Evaluation Theorem is important because it says that to calculate the definite integral of lover an interval [a, b] we need do only two things: 1.

Find an antiderivative F of I, and

2.

Calculate the number F(b) - F(a), which is equal to

1: I(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 1 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.
loU cosx dx =

sin x

I

d .

dx sm;t

= COS;t

d dxsecx

=

= sin". - sinO = 0 - 0 = 0

(b)

r

sec x tan x dx = 8Ocx]O

J-w/4

= 8OcO- sec ( - ; ) = 1-


l' (~Vi - x~)

sec x tan x

-wj4

dx =

[X 3/2+ i [(4)3/2

=

Vi

I

+ ~] - [(1)3/2 +

[8 + I] - [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 I, 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 nel 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 ofits 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 '(1) moves along a coordinate line, its velocity is v(l) = .'(1). Theorem 5 says that

j

t'V(I) dl = .(12) - .(11),

t,

so the integral of velocity is the displacement over the time interval I, :5 I :5 12. On the other hand, the integral of the speed IV(I) I is the total distance 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] equais its initial value F(a) plus its net change over the interval. So ifv(l) represents the velocity function of an object moving along a coordinate line, this means that the object's final position .(12) over a time interval II :5 I :5 12 is its initial position .(11) 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 I during its motion was given as v(l) = 160 - 321 ft/sec. (a) Find the displacement of the rock during the time period 0 :5 I :5 8.

(b) Find the total distance traveled during this time period. Solution (a) From Example 4b, the displacement is the integral

10rV(I) dl = 10r(160 -

]8

321) dl = [1601 - 161 2

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. I.

280

Chapter 5: Integration

(h) As we noted in Table 5.3, the velocity function v(l) 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

1'lv(I)ldl = [lv(I)ldl + l'IV(I)ldl = 1'(160 - 321) = [1601 -

dl - 1'(160 - 321) dl

1612]~ - [1601 - 161 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 1 :5 8 is (i) the maximum height of400 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

f(l) dl

=

f(x),

which says that ifyou frrst integrate the function f and then differentiate the result, you get the function f back again. Likewise, replacing b by x and x by 1in Equation (6) gives

i

X

F'(I) dl

=

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 f 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 = f(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 f( Ck) ,lxk that give the area of a rectangle when f(ck) is positive. When f(ck) is negative, then the product f( Ck)
EXAMPLE 6 g(x) = 4 -

Figure 5.20 shows the graph of f(x) = x 2 - 4 and its mirror image across the x-axis. For each function, compute

x 2 reflected


Solution

f2
=

[x33 -

4x ]2-2

=

(83 - ) (8 -3 + ) 8

-

8

=

32 -3'

5.4 The Fundamental Theorem of Calculus

281

and

1

2g(x) dx

=

3 [4X _ X ]2 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. Al_ though the def"mite 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 def"mite integra! 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 showsthegraphofthefunctionf(x) = sin x between x = and x = 2".. Compute

°

(a) the def"mite integral of fix) over [0, 2".].

(b) the area between the graph of fix) and the x-axis over [0, 2".]. -----J'-------\-~~-_lc_-~x

Solution

The definite integral for fix) = sinx is given by

].2" sinxdx = -cosx):" = -[cos2". FIGURE 5.21 The tota1 area between = sinx and the x-axis for o ~ x ::=:; 21T is the sum of the absolute values of two integrals (Example 7). y

cosO] = -[I - I] = 0.

The def"mite integral is zero because the portions of the graph above and below the x-axis make canceling contributions. The area between the graph of fix) 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 f"md 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 absolute values ofthe integrals.

EXAMPLE 8 x2

Find the area of the region between the x-axis and the graph of 2x, -I :5 x :5 2.

fix) = x 3

-

Solution

First f"md the zeros of f. Since

-

fix) = x 3

-

x2

-

2x = x(x 2

-

X -

2) = xix

+

I)(x - 2),

282

Chapter 5: Integratian

Area =

y

&

the zeros are x = 0, -I, 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

---c-f------:'t-------j~x

-1 (x

3

f\x 3

10

2

- 2x) dx

-

x

_

x 2 _ 2x) dx

[I I ] 5

2 = [x44- ;3]0 - x -1 = 0 - 4 + 3 - I = 12

= [x

4

4

3

x

_

3

X 2 ]2

-

~-

= [4 -

0

3

4] - 0

=-~ 3

The total enclosed area is obtained by adding the absolute values of the calculated integrals. FIGURE 5.22

curve y = x 3 (Example 8).

-

The region between the xl - 2x and the x-axis

t2 + 1- ~ 1

Total enclosed area =

=

i~

•

Exercises 5.4 Evaluating Integrals

Derivatives of Integrals Find the derivatives in Exercises 29-32

Evaluate the integrals in Exercises 1-28. 1.

a. by evaluatiog the integral and differentiatiog the re,ult.

D2x + S)dx

2.1:(S-I)dx

3. 1,2 x(x - 3) dx

5.

[(3x - ~)dx

7. l,'(x 2 +

Vi) dx

9. 1,7/32 sec' x dx

11.

'7/4 /.7/4 0

13.

esc BcotB dB

1+

/ '1rj2 .

~as

21

dl

[7/4 2 15. Jo tan x dx

b. by differentiatiog the integral directly.

4.1: (x' - 2x

+ 3) dx

d [Vi 29. dxJo cas 1 dl

d ["n 30. dxJ, 31' dl

6. Dx' - 2x

+ 3)dx

d [" 31. dlJo VUdu

d [tm. 2 32. dBJo sec ydy

8./,32x ..../Sdx 10. 1,7(i + easx) dx 12. 14.

1,7/' 0 7

1

4 sec u tanu du

/'1 - eas21 2 dl

--rrj3

r/6 16. Jo (seex + tan x)' dx

L: (4 4

17. f/'Sin2xdx

19./,-1(r 21.

+ 1)2<Jr

Jo' (7

"-- - -I ) du

23. /, 1

2

uS

0 S2 +

'lis

,dx

25.

/.7 s~

27.

1

) x l dx

18.

sec' 1 +

20.1~(t + 1)(1 22. 24.

1-1 _,

/,'

2

~)dl

+ 4)dl

yS - , 2y dy y

(x'/3 + 1)(2 - x 2/') 1/' dx

S I X

2x dx 7/2 2 smx 4

Finddy/dx in Exercises 33-40.

r/' (easx + secx)' dx rl 28. Jo 2 (easx + lcosxl)dx 26. Jo

33. y

~ 1,''\Il+7 dl

35. y

=

37. y

=

l '-,--

38. y

=

(1,' (I'

39. y

=

I, .n" .,.------.,'

34. y ~

[0 sin (12) dl

1v'i

12

-1 1

+4

dl -

36. y

[" = x J,

x>

0

sin (I') dl

j,' - ,+- - dl 2 1

,

1

4

+ 1)10 dl)' dl

7f

Ixl < -2

v l - t2

o

J,['I,dl,

40.y= [0 ~ Jtau 1 + t Area Ia Exercises 41-44, fmd the total area between the region and the

x-axis. 41. y = -x 2 42. y

= 3x 2 -

-

lx,

3,

43. y=x'-3x 2 44. y

= x ' /' - x,

-3:5 x

-2:S; x

+2x,

:5

:5

2

2

0"'x"'2

-I '" x '" 8

5.4 The Fundamental Theorem of Calculus

Find 1he areas of1he shaded regions in Exercises 45-48. 45.

283

56. Show that if k is a positive constant, then the area between the x-axis and one arch of1he curve y = sin kx is 2/k.

y

57. Co.tfrom marginal eo.t The marginal cost ofprintiog a poster when x posters have been printed is x = 1T ~f-------""-L~x

2v'i

58. Revenue from marginal revenue Suppose that a company's marginal revenue from 1he manufaeture and sale of eggbeaters is

y

~ .. 6

47.

I

dx

dollars. Find e(100) - e(l), 1he eost ofprintiog posters 2-100.

o

46.

de

:

48.

+

2 - 2/(x

1)2,

where r is measured in thousands of dollars and x in thousands of units. How much money should 1he company expect from a produetion run of x = 3 1housand eggbeater>? To fmd out, integrate 1he marginal revenue from x = 0 to x = 3. 59. The temperature T(OF) ofaroom at time I minutes is given by

6

y

=

y

2

T = 85 - 3'\!25=t for 0 '"

25.

a. Find 1he room's temperature when 1= 0, 1= 16, and 1= 25.

= sec 8 tan 8

y

I '"

b. Find 1he room's average temperature for 0 '" I '" 25. --;;;;f-----Jl~--:oor:----> 6

60. The height H (It) of a palm tree after growing for I years is given

4

by H

-V2

=

1

'1

-dl - 3

d. Y =

-I

I

=

x'

for 0 '" I '" 8.

Y('IT)

=

b. Find 1he tree's average height for 0 '" I '" 8. f(l) dl = x 2 - 2x + I. Find f(x).

61. Suppose 1hat

J:

50. y'

-3

=

f(x) = 2 -

' I

w

/.

=

4

["

g(x) = 3 + JI

atx Express 1he solutions of1he initial value problems in Exercises 53 and 54 in terms ofintegrals. dy 53. dx

=

seex, y(2)

=

54. :

=

'\1'1+7,

y(l)

9

I

I + dl

64. Find 1he linearization of

I 52. y' = x' y(l) = -3

51. y' = sec x, y(O) = 4

['+1

J2

atx = I.

t dl - 3

seex, y(-I)

Jt

62. Findf(4)if f(l)dl = xeoS'ITX. 63. Find 1he linearization of

b. Y = /.' secldl + 4

1 I

e. y = 1'seeldl+4 dy 49. dx

51 1/3

a. Find 1he tree's height when I = 0, I = 4, and I = 8.

Initial Value Problems Each of 1he following functions solves one of 1he initial value problems in Exercises 4l}-52. Whieb function solves which problem? Give briefreasons for your answers.

a. y

'\It+1 +

=

3

sec (I - I)dl

= -1.

65. Suppose that f has a positive derivative for all values of x and that f( I) = O. Which of 1he following statements must be true of 1he function g(x) = /.' f(l) dl?

=

-2 Give reasons for your answers.

Theory and Examples 55. Archimede.' area formula for parabolie arebe. Archimedes (287-212 B.C.), inventor, ntilitary engineer, pbysicist, and 1he greatest mathematician ofclassical times in the Western world, discovered that the area under a parabolic arch is two-thirds the base times 1he height. Sketch 1he parabolic arch y = h - (4hlb 2 )x 2, -b/2 '" x '" b/2, assunting that h and b are positive. Then use calculus to fmd 1he area of the region enclosed between 1he arch and the x-axis.

a. g is a differentiable function of x. b. g is a continuous function of x.

e. The graph ofg has a horizontal taogent at x d. g has a local maximum at x

=

I.

e. g has a local minimum at x

=

1.

f. The graph ofg has an inflection point at x

=

=

I.

I.

g. The graph of dgl dx erosses 1he x-axis at x = I.

284

Chapter 5: Integration

66. Another proof of The Evaluation Theorem

= Xo < Xl

< X2 ••• < X n = b be any partition of [a, b], and letFbe any antiderivative of j. Show that

a. Let a

, F(b) - F(a)

= "5'. t:1

[F(x,) - F(X'-l)]'

67. j(x) ~ x' - 4x'

e. From part (b) and the defmition of the defmite integral, show that

F(b) - F(a) COMPUTER EXPLORATIONS

~ /.' j(x) dx.

J:

In Exercises 67-70, letF(x) = j(t) dt for the specified function j and interval [a, b]. Use a CAS to perform the following steps and

answer the questions posed. &.

69. j(x) = sin2xcos~,

b. SolvetheequationF'(x) = O. What can you see to be true about the graphs of j and F at points where F' (x) ~ O? Is your observation borne out by Part I of the Fundamental Theorem coupled with information provided by the ftrst derivative? Explain your answer.

[O,~]

[0,21f]

[0,21f]

In Exercises 71-74, letF(x) =

J:V<) j(t) dt for the specifteda, u, and

f. Use a CAS to perform the following steps and answer the questions posed. &.

Find the domain ofF.

b. Calculate F'(x) and determine its zeros. For what points in its domain is F increasing? Decreasing?

c. Calculate F"(x) and determine its zero. IdentilY the local extrema and the points of inflection ofF. d. Using the information from parts (a)-(c), draw a rough handsketch ofy ~ F(x) over its domain. Then graphF(x) on your CAS to support yoor sketch.

u(x) = x',

j(x) = ~

72. a = 0, u(x) = x',

j(x) = ~

71. a = I,

Plot the functions j and F together over [a, b].

[0,4]

68. j(x) = 2x' - 17x' + 46x' - 43x + 12,

70. j(x) = XCOS1fX, b. Apply the Mean Value Theorem to each term to show that F(x,) - F(XH) = j(c,)(x, - XH) for some c, in the interval (XH,X,). Then show thatF(b) - F(a) is a Riemano sum for j on [a, b].

+ 3x,

73. a = 0,

u(x) = I - x,

74. a = 0,

u(x) = I - x',

j(x) = x' - 2x - 3 j(x) = x' - 2x - 3

In Exercises 75 and 76, assume that j 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 {*J 75. Calculate dx}. j(t) dt and check your answer using a CAS.

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<) j(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 f with respect to x as the set of all antiderivatives of f, symbolized by

J

f(x)dx.

Since any two antiderivatives of f differ by a constant, the indefinite integral means that for any antiderivative F of f,

J

f(x) dx = F(x)

+

J notation

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 f, 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 f(x) dx is a number. An indef"mite integral f(x) dx is afUnction plus an arbitrary constant C. So far, we have only been able to find antiderivatives of functions that are clearly recognizable as derivatives. In tbis section we begin to develop more general techniques for f"mding antiderivatives.

J:

J

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 dx n + I - u dx· From another point of view, tbis same eqoation says that u·+ 1/(n + I) is one of the antiderivatives of the function u·(du/dx). Therefore,

I

u n+ 1

du

u· dx dx

=

n + 1 + C.

(1)

The integral in Eqoation (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 ofthe founders ofcalculus, 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'

(x' + x)5(3x 2 + 1) dx.

+ x. Then du

so that by substitution we have

I

I

(x' + x)5(3x 2 + 1) dx

=

:dx

=

I

=

(3x 2 + I)dx,

uS du

Letu

"

="--+C (x'

+ x)" 6

EXAMPLE 2 Solution

Find

I~

x 3 + x,d"

=

(:h 2 + 1)dx.

Integrate with respect to u.

6

=

=

+C

dx.

The integral does not fit the formula

IU'dU,

Substitute.:{3 + x fOT 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~.~ du

u

= tju

12 /

dU

Letu

I u 3/ 2

=

2 3/2 + C

=

1(2x 3

=

2x + I,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 ofthe following general rule.

If u = g(x) is a differentiable function whose range is an interval I, and f is continuous on I, then

THEOREM 6-The Substitution Rule

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 :fxF(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. 3.

287

Integrate with respect to u. Replace u by g(x) in the result.

EXAMPLE 3 Solution

Find

f

seel (51 + 1)' 5 dl.

We substitute u = 51 + I and du = 5 dl. Then,

f

EXAMPLE 4

Find

f

Letu~5t+l.du~5dt.

sec2 udu

sec2 (51+ 1)'5dl= f

= tan u + C

d du tanu ~

= tan (51 + I) + C

Substitute 5t + 1 for u.

see'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, f cos (70 + 3) dO = t f

cos (70 + 3)'7dO

Place factor 1/7 in ftoot ofintegra1.

78 + 3,du

=tfcoSUdU

Let"

=tsinu+C

Integrate.

= t sin (70 + 3) + C

Substitute 78 + 3 for u.

=

=

7d6.

There is another approach to iliis problem. Wiili u = 70 + 3 and du = 7 dO as before, we solve for dO to obtain dO = (1/7) duo Then ilie integral becomes

f

cos (70 + 3) dO = f

Letu~78+3.du~7d8.andd8~(1/7)du

cosu·t du

= t sin u + C

Iotegrate.

= t sin (70 + 3) + C

Substitute 78

+ 3foru.

We can verify this solution by differentiating and checking that we obtain ilie original • function cos (70 + 3).

EXAMPLE 5 Sometimes we observe that a power of x appears in ilie integrand that is one less 1han ilie power of x appearing in ilie argument of a function we want to integrate. This observation immediately suggests we try a substitution for ilie higher power of X. This situation occurs in the following integration.

f

x 2 sin (x') dx =

f

= f

sin (x') •x 2 dx

Sinu·t du

= x 3 ,du = 3x 2 dx, (1/3) du ~ . ' dx.

Letu

=t!SinUdU =

3I (-cosu)

+ C

= -tcos(x') + C

Iotegrate.

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 ofthis situation.

EXAMPLE 6

Evaluate

Jx~ ax.

Solution Our previous integration in Example 2 suggests the substitution u = 2x with du = 2 ax. Then,

+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 + 1 to obtsin x = (u - 1)/2, and find that

x~ax

t(u - 1)·tV;;du.

=

The integration now becomes

Jx~ ax t J tJ =

(u -

l)V;; du

=

(u 3/ 2

-

=

t J(u -

l)u 1/2 du

u ' /2 ) du

Mu1tiplytcnns.

=

l(2. u S/2 - 2. u 3/2) + 453

=

~ (2x + l)s/2 - 1 (2x + 1)3/2 + C 10

6

Substitute.

C

Integrate.

Replace u by 2l; +

1.

•

The success of the substitution method depends on finding a substitution that changes an integral we cannot evaluate directly into one that we can. Ifthe 1rrst substitution fails, try to simplify the integrand further with additional substitutions (see Exercises 51 and 52).

EXAMPLE 7

Evaluate

J

2zdz

.l~' 2

vz + 1

Solution We can use the substitution method of integration as an exploratory tool: Substitute for the most troublesome part ofthe integrand and see how things work out. For the integral here, we might try u = z2 + 1 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

+ 1. LetU=Z2+ I, du = 2zdz.

In the form

Ju

ll

du

Integrate.

Replace u by Z2 + 1.

289

5.5 Indefinite Integrals and the Substitution Method SoLution 2: Substitute u

V'z 2 +

=

I instead.

J

2zdz = V'z 2 + 1

J

2 3u du u

=

3JUdU

=

u2 3' 2

+C

=

1(z2

+

2

1)'/3

Lctu=~. .. 3 = Z2

+ I,

3u 2 du = 2zdz.

Integrate.

+

C

Replaceuby(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

~os lx dx

I -

sin2 x

tJ(l - coslx)

=

Ix _Isinlx + C

= '! _

J

I + sin4lx + C

(b) JCOS2Xdx =

I

2

dx

2

+ ~OSlx dx

1 - cos2x

2

=

2

=

=

2

sinlx 4

+

C 2

cosx=

J + cos2t 2

•

EXAMPLE 9 We can model the voltage in the electrical wiring of a typical home with the sine function

v

v=

v = Vmax sin 1201Tt

Vmax sin 1201TI,

which expresses the voltage V in volts as a function oftime 1 in seconds. The function runs through 60 cycles each second (its frequency is 60 hertz, or 60 Hz). The positive constant Vmax (''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 Vav

FIGURE 5.23 The graph of tire voltage V over a full cycle. Its average value over a half-cycle is 2Vmu!".. Its average value over a full cycle is zero (Example 9).

1.

=

1 12 I . / • (1/120) _ O. Vrnax sm 1201TI dl

=

]1/12. 1 120Vrnax [- 1201T cos 1201TI •

=

Vrnax ---.r [-cos 1T + cos 0] 2Vrnax 1T .

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 ofthe average value ofthe square of the voltage, namely 2

Vnns = V(V )av'

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

2 (V ).v

I

= (1/60) _ 0

1.

0

1 60 /

2 (Vmax)2 s in 1201Ttdt

(V,

)' 7',

=

(Exercise 64, part c), the rms voltage is "

=

'nna

~(Vmax)2

=

2

Vmax

v'z'

The values given for household currents and voltages are always rms values. Thus, "115 volts ac" means that the rms voltage is 115. The peak voltage, obtained from the last equation, is

Vmax

=

v'z Vnna = v'z ·115

'" 163 volts,

•

which is considerably higher.

Exercises 5.5 Evaluating Indefinite Integrals Evaluate the indeImite integrals in Exercises 1-16 by using the given substitutions to reduce the integrals to standard form.

+ 4)' dx, u

1. /2(2x

2. /

7~ dx,

3. /2x(X'

+ 5)-4dx, u

~ x' + 5

4 3 (X. : I)' dx, u = x'

5. /

(3x

~

sin 3x dx,

9. /

sec2ttan2tdt,

10. / 11.

/

u

~ 3x

7. /

u

9r'dr vI - r 3

12. /12(Y'

/VJ=Z.;ds

18.

/~ds 58 + 4

19.

/ofIJ -

20.

/3 yv'7 - 3y'
25.

u

~ 2x'

27.

30.

~ I- cosf

u

+ 2y)
~ x 3f2 -

13. /

v'X sin' (x 3f2

14. /

:,cOS'(~)dx, u=-~

- 1) dx,

u

= y'

/

.'x 26. / tan7X 2 sec2 dx

28. Jr' (7 - ~~)' dr

dr

(V-

secztanz dz

33. /

~COS(+-I)dt

34. /

~ cos(Vt + 3) dt

35. /

~ sin~cos~dO

36. /

I

37. / t 3(1 39. / b. Using u = esc 20

(V -

esc - 2 1T) - cot - 2 1T) - dv 32. /

/

+ 4y 2 + I

15. / esc' 20 cot 20 dO a. Using u = cot 20

24. / tan' x sec' x dx

. 'x x dx sm 3 cos 3

Jr' (~; -I)'

+ 4) dz

sin(2t+ I) dt cos' (2t + I)

u ~ I - r3 1)'(y3

+ 2) dx

22. /cos (3z

29. / x 112 sin (x 3f2 + I) dx

~ 2t

31.

+ 4y 2 +

/

If'dO

v'Xx(l I+ v'X' dx x)

23. / sec' (3x

8. / x sin (2x') dx,

(I - cosf)' sinfdt, .~,

~ 3x' + 4x

u~I+~

dx,

b. Usingu= ~

17.

21. /

I

+ 2)(3x' + 4x)' dx, u

(1 + ~)1/3 /

+

+8

a. Usingu = 5x

Evaluate the integrals in Exercises 17-50.

u = 7x - I

4. /

6.

+4

= 2x

16./~ 5x + 8

41.

+ t')3 dt

:,~2-~dx

/ "----;Il dx ~

38. /

40.

~

cos\18

\18 sin' \18

~x ;, I dx

/1.~x2 3 x

42.

dO

/~x

X 3 '

- I dx x2

- I

dx

5.6 Substitution and Area Between Curves

43. 45. 47. 49.

! ! ! x'W+1 ! ~

x(x - 1)'· dt

44.

(x + 1)'(1 - x)' dt

46.

(x 2

dt

48.

4)' dt

SO.

!X~dt

! ! 3x'W+! ! ~

(x + 5)(x - 5)'/3 dt

(x

dt

4)' dt

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplilY 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.

51.

!

2

2

18 tan X sec x dt (2 + tan' x)2 a. u = tan x , followed byv = u', then byw = 2

b. u ~ tan'x,followedbyv ~ 2

+v

58.

~; =

59.

~;~ ~

(~- 0)'

2

300s

-4 sin (21 -

!

+U

VI + sin (x - I) sin (x - I) cos (x - I)dt

b.

!

57.

~ = 8sin2 (I + ~).

5.6

-2u du

!

-u 2 +

C2

u = cosx = -cos2 x

sin2xdx

+ C2

2sin:rcosx=sin2x

a. Show by evaiuatiog the integral in the expression I (1/60) -

I)', .(1) = 3

~ 4x (x 2 + 8r'/',

=

+ Ct

Can all three integrations be correct'l Give reasons for your answer. 64. (Continuation afExample 9.)

Solve the initial value problems in Exercises 55-60.

56. :

= 8in2 X

+ Ct

c. !2 sin x COS X ax = !

In;tlal Value Problems -

2 sin x cos x dx

u= sinx

= u2

=

sin YO dO VOcos'v1i

= 121 (31

~0

~ _ 00~2x + C,.

!

~

100, .(0)

a. !2SinXCOSXdx = !2UdU

Evaluate the integrals in Exercises 53 and 54. (2r - I)cos V3(2r - I)' + 6 53. dr V3(2r - I)' + 6

55.

~

63. It looks as if we can integrate 2 sin x cosx with respect to x in three different ways:

2

2

.'(0)

f),

Theory and Examples 61. The velocity of a particle moving back and fotth on a line is v ~ tis/dl ~ 6 sin 21 m/sec for all I. If. ~ 0 when I ~ 0, fmd the value of. when I = 'IT/2 sec. 62. The acceleration of a patticle moving back and fotth on a line is a ~ d 2./dI 2 ~ ~ cos 'lT1 m/sec' for all I. If • ~ 0 and v ~ 8 m/sec when I = 0, fmd. when I = I sec.

a. u = x - I, followed by v = sinu,thenbyw = I + if b. u = sin (x - I), followed by v = I + u 2 C. U = I + sin2 (x - I)

54.!

f

d 2y 60. dt 2 = 4 sec2 2xtan 2x, y'(O) = 4, y(O) = -I

c.u=2+tan3 x

52.

r(O) =

291

y(O)

['/60

oj.

V~sinI20'ITldl

that the average value of V = is zero.

V~sin

120 'lT1 over a full cycle

b. The circoit that rons your electric stove is rated 240 volts rms.

What is the peak value of the allowable voltage?

~0

c. Showthat

['/60

J.

.(0) = 8

2

(V.".)'sin 120'ITldl=

(V,

)'

;;; .

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-Substftutlon In Definite Integrals If g' is continuous on the interval [a, b] and 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 Itt F(g(x)) ~

tf(g(x)). g'(x) dx = F(g(x)) [ :

~

=

F(g(b)) - F(g(a))

=

F(u)

]

.~g(b) .~g(a)

9(b)

=

1

F(g(x))g'(x) f(g(x))g'(x)

Fundamental

f(u) duo

Theorem, Part 2

g(al

•

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+1 dx.

We have two choices.

Method 1: Transform the integral and evaluate the transformed integral with the transformed limits given in Theorem 7. Let" = x 3 + l,du = 3,x 2 dt.

[3x

2

W+1dx

=

[VUdU

=

Z 3/ 2 _u 3

When x ~ -I,. ~ (-I)' + I ~ O. Whenx ~ I,. ~ (I)' + I ~ 2.

]2

Evaluate the new deImite integral.

0

= Hz3/2 -

0

3/2] = HzV2] =

4'(2

Method 2: Transform the integral as an indefinite integral, integrate, change back to x, and use the original x-limits.

J W+1 dx JVU 3x

2

=

=

'2. u 3/2 + C

=

'2. (x 3 + 3

=

3

+

= H «(1)3

= HZ3/2 -

1)3/2

+C

1)3/2L

+

=

x 3 + l,du

=

3x 2 dx.

Integrate with respect to u.

3

[3x W+1 dx t(x 2

Letu

du

Replace. by

Usc the integral just found, with limits of intcgmtion for x.

1)3/2 - «-1)3

0

x' + l.

+

3/ 2] = HZV2] =

1)3/2]

4'(2

•

5.6 Substitution and Area Between Curves

293

Which method is better----evaluating the transformed definite integral with transfanned limits using Theorem 7, or transfanning the integral, integrating, 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 of transfanning the limits of integration. Let u ~ cot 0, du ~

("/2 cotlicsc'lidli

=

lOU' (-du)

J"/4

1

=

WhenO

~

When 0

~

-cae' 0 dO,

-du ~ esc' 0 dB. ",/4,u ~ cot (",/4) ",/2, u ~ cot (",/2)

~

I.

~

O.

-lOUdU

• Definite Integrals of Symmetric Functions

y

The Substitotion Fannula in Theorem 7 simplifies the calculation of definite integrals of even and odd functions (Section 1.1) over a synuuetric interval [-a, a] (Figure 5.24).

THEOREM 8 -a

o

a

(a)

Let f be continuous on the aymmetric interval [ -a, a].

Iff is even, then 1>(X) dx

(a)

y

--!--+--"f-----\c----r.ao-~x

(b) Iff is odd, then

FIGURE 5.24

(a)f"""", J~f(x)
(b)fodd, J~af(x)
f(x) dx

z!oaf(X) dx.

= O.

Proof of Part (a)

i (b)

1:

=

f(x) dx

=

=

1:

f(x) dx

+ [f(X) dx

-!o-af(x) dx + [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 Integration Rule Letu = -x,du = -dx. Whcnx = O,u = O. "Whcnx = -0, U = Q.

lis even, so

I( -u)

~

I(u).

The proof of part (b) is entirely similar and you are asked to give it in Exercise 86.

•

294

Chapter 5: Integration

The assertions ofTheorem 8 remain true when f is an integrable function (rather than baving the stronger property of being continuous).

EXAMPLE 3

Evaluate

1:

(x 4

4x 2 + 6) dx.

-

Solution Since f(x) = x 4 - 4x 2 ric interval [- 2, 2] , so

+

6 satisfies f( -x) = f(x) , it is even on the symmet-

y

-/ b

x

•

Lower curve y ~ g(x)

FIGURE 5.25 The region betweoo the curves y = f(x) and y = g(x) and the lines x = a and x = b.

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 bave a sbape whose area we could f'md with geometry, but if f and g are arbitrary continuous functions, we usually bave to find the area with an integral. To see wbat the integral should be, we first approximate the region with n vertical rectangles based on a partition P = {xo, XI, ••• , x.} of[a, b] (Figure 5.26). The area of the kth rectangle (Figure 5.27) is

y y

a=

~o

~

x,-"L .".

Il1\2

/(x)

7'''''

It XII_I

T-

....1>- /_- x. !.-y

~k = height X width =

g(x)

[/(01) - g(01)] 4>:k.

We then approximate the area ofthe region by adding the areas ofthe n rectangles: FIGURE 5.26 We approximate the region with rectangles perpeodicular to the x-axis.

A '"

L• ~k = L• [((O1) -

k~l

g(01n 4>:k.

Riemann sum

~l

1:

(c•• /(c.)

T

A

/(c.) - g(c.)

I

a

b

H

n

As Ilpll ...... 0, the sums on the right approach the limit [/(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,

y

(c•• g(c.)

ax.

FIGURE 5.27 The area ,uk of the kth rectangle is the product of its height, f(q) - g(c.) , and its width, 4>:•.

X

=

.

lim L[/(ck) - g(Ck)] 4>:k =

11P1I~o k~l

l

b

[((x) - g(x)] dx.

a

DEFINmON If f and g are continuous with f(x) ;" g(x) throughout [a, b], then the area of the region between the curves y = f(x) and y = g(x) from a to b is the integral of (f - g) from a to b:

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 integration if they are not given. You may need to find where the curves intersect to

5.6 Suhstitution and Area Between Curves y

295

detennine the limits of integration, and this may involve solving the equation f(x) = g(x) for vslues of x. Then you can integrate the function f - g for the area between the intersections.

EXAMPLE 4

Find the area of the region enclosed by the parabola y = 2 - x 2 andthe

liney = -x. Solution 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. 2 - x 2 = -x

Equate f(x) and g(x).

2

x -x-2=O

FIGURE 5.28 The region in Example 4 with a typical approximating rectangle.

(x

+ I)(x - 2)

Rewrite.

0 x=2.

x=-l,

=

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

[f(x) - g(x)]

-1

HIsTORICAL BIOGRAPHY

=

Richard Dedekind (1831-1916)

EXAMPLE 5

Area

2

~ {(Yx - x + 2)dx J,'

' l

Area ~ 0 Yxdx

(x,f(x» I

\

+x

- x 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 fonnula changes and apply the fonnula for the area between curves to each subregion.

y =

y

(2

ax = 1~[(2 -

(x,f(x»

y~Yx \ (4,2) B

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 fonnulas 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 f(x) andg(x).

x = (x - 2)2 = x 2

y~x-2

-

4x

+4

x 2 -5x+4=O (x - I)(x - 4) = 0

e+-~---oc--!:-------!:-->x o / y~O 2 4

x

(x, g(x»

FIGURE 5.29 When the fonnu1a for a bounding curve changes, the area integral changes to become the sum ofintegrals to match, one integral for each ofthe shaded regions shown here for Example 5.

= 1,

Square both sides. Rewrite. Factor.

x

=

4.

Solve.

Only the vslue x = 4 satisfies the equation vX = x - 2. The vslue x = I is an extraneous root introduced by squaring. The right-hand limit is b = 4. ForO For2

:5 X :5 :5

x

:5

2: 4:

f(x) - g(x) = f(x) - g(x) =

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 Vi dx Tota1 area = ]. 0

+ {'(Vi - x + 2) dx

J2

.... orB

area of A

=

['1. X3/2]2 3

+

0

['1. X3/2 _ 3

2 x 2

+ 2x]4 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 Ifa 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)~===i

use the fannula

A = y

2

(4,2) (g(y), y)

l

d

[f(y) - g(y)] dy.

In this equation f always denotes the right-hand curve and g the left-hand curve, so f(y) - g(y) is nonnegative.

x=y+2

EXAMPLE 6

-;;f--::---:!:----:--x o y~O 2 4 FIGURE 5.30

It talres two

integrations to fmd the area ofthis region ifwe integrate with respect to x. It talres only one lfwe integrate with respect to y (Example 6).

Find the area ofthe region in Example 5 by integrating with respect toy.

Solution We first sketch the region and a typical horizontal rectangle based on a partition 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 ofintegration is y = O. We find the upper limit by solving x = y + 2 and x = y2 sinJultaneously for y: y + 2 = y2 y2 _ Y _ 2 = 0 (y

+

I)(y - 2) = 0

y=-l,

y=2

Equa1o/(y) ~ y

+ 2 and

g(y) ~ y'.

Rcwri1o. Factor.

Solve.

The upper limit of integration is b = 2. (The value y = - I gives a point of intersection below the x-axis.)

5.6

297

Substitution and Area Between Curves

The area of the region is

A

=

l

d

[f(y) - g{y)] dy

= ].2[y + Z _ y2] dy =

].\z +

=

[Zy + y; - ~3J:

y - y2] dy

=4+!-~= ~O.

•

This is the result of Example 5, found with less work.

Exercises 5.6 17.

Jor/ oos-3 26sin26d6 Jor 5(5 -

6

Evaluating Definite Integrals

Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises 1-24.

1. ..

t v':Y+l

dy

1~ v':Y+l dy Lr~dr

21. J.' (4y - y2 + 4y 3 + 1)-2/3 (12y 2 - 2y + 4) dy

I>.

3...

Jor/

1>.1

tan x sec' x dl;

-fr/4

4. a. J.'" 3 cos2 x sin x dx

b.

1

4 cos 1)'/4 sin I dl 20.

sin 21)3/2 cos 21 dl

22. J.\y3 + 6y 2 - 12y + 9t'/2 (y2 + 4y - 4) dy

0

x x

Jo{~/4 (I -

I>.

J.'r~dr tan sec' dl;

13~/2 ~ cot' (6) 6 sec' (6) 6 d6

19.

2...

4

18.

3~

3 cos2 x sinx dx

23. J.Y/,;iv'8 cos2 (63/2) d6

24.1~'/21-2 sin2 (I + +) dl

2~

5

J.'r

6

J.

3

Vi

7.

8... {'

Jo

9

4

{v'3

1>.1

-Vi

I>.

5r 2)' dr

+r lOy,;

(I + v 3/2)2

... Jo

0

I(1 2 + 1)'/3 dl

,,1'( -1

I>. 1>3(1 + 14)3 dl

+ 14)3 dl

(1

4x

W+1

dv

I>.

dl;

I>. I>.

11

I>.

12 13. a. 14. ..

E/2(2 {2'"

Jo

1°

5r

J.o (4 + r 2 )'

y

dr y = (1 - cau) sinx

j4

lOy,; dv , (I + ,,'/2)2 v'3

I

4x

0

dl;

-_C02t---+..c------;c-~·

x

2 + tanf)sec fd' I>.

cos z dz \14 + 3 sinz

b.

I>.

dw

15. J.' v't'+2t (514 + 2) dl

16

1 (I 1-~/2~/2(Z 1-~ \14 J.

o

"

dl;

cos31)sin3ldl

I) I + tan z sec'zdl

~

cosz dz + 3 sinz ~/2 . smw dw o (3+2cosw)'

j4

~-"--------'---->.

3

1-,\10+9 ~/6

sinw

26. y

~/3

cos31)sin3ldl

~/2 (3 + 2 cos w)'

25.

-v'3W+1

10... {' k d l ; 4 +9

Jo x 6 Jor/ (I -

'

1(12 + 1)'/3dl

Area Find the total areas of the sbaded regions in Exercises 25-40.

dy

. , 2yY(I + yY)'

28.

27. y

y = ~(COSX)(sin(1T + 1Tsinx»

y

298

Chapter 5: Integration

29.

36.

30.

Y

y

y

y=!sec 2 t 2

y

1

~

1

2

"2

0

"

o

x

37.

38. Y

sf

(-3,5)

-4

y=x 2 -4 y

31. (-2,8)

(2,8)

8

39.

40. Y

NOTTOSCAIE

32.

Y

y=~

3

(3, 1)

~lf-~oi<""--'L-~3--~x

o 33.

Find the areas ofthe regions enclosed by the lines and curves in Exercises 41-50.

41. Y = x 2

x2

42. Y = 2x -

34.

35. Y

Y

and y = 2

2

-

and y = -3

43.y=x' and y=8>: 44.y=x2 -2x and y=x 45. y=x2 and y= -x 2 +4x 46. y~7-2x2 and y~x2+4 47. y ~ x' - 4x 2 + 4 and y = x 2 48. Y

=

xv'a 2

49. Y

=

v'R

x 2,

a

>

0,

and

5y

=

x

-

and y = 0

+6

(How many intersection points

are there?)

50. y

= 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 ~ y2

and x ~ y

+2

y

53. y2 - 4x = 4

and

4x - y = 16

54. x - y2 = 0

and x

+ 2y 2 = 3

55. x = y2 - Y

and x = 2y 2 - 2y - 6

56. x - y2j3 ~ 0 57. x = y2 - I

and x

+ y4

~ 2

and x = lyl'V'l=Y'

58. x = y' - y2

and x = 2y

-oi-""'-:,--!---+-_x

o

Find the areas ofthe regions enclosed by the curves in Exercises 59-{j2. 59.4x 2 +y=4 and x 4 _y= I 60. x'-y~O 61. x 62. x

299

Substitution and Area Between Curves

3x2_y~4

and

+ 4y 2 ~ 4 + y2 ~ 3

and x and 4x

+ y4 + y2

~ I, ~

for

x "" 0

0

2

79. The figrrre here shows triangle AOC inscribed in the region cut from the parabola y ~ x 2 by the line y ~ a 2 • Find the limit of the ratio of the area of the triangle to the area of the parabolic region as a approaches zero. y

Find the areas of the regions enclosed by the lines and curves in Exercises 63-70.

63. y

= 2 sin x and

64. y ~ 8 cosx

y

= sin2x, o::=:;

x::=:;

'IT

and y ~ sec' x,

65. y = cos ('1Tx/2)

-'1T/3:5 x :5 '1T/3 and y = I - x 2

66. y = sin ('1Tx/2)

and y = x

67. y ~ sec' x, 2

68. x = tan y

y ~ tan2 x,

X

~ -'1T/4, 2

and x = -tan y,

69. x = 3 siny Vcosy 70. y ~ sec' ('1Tx/3)

-'1T/4:5 y:5 '1T/4

and x = 0,

and y ~ xli',

and x ~ '1T/4

0:5 Y :5 '1T/2 -1:5 x :5 I

71. Find the area ofthe 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 - ylj3 ~ o and x - yl/5 ~ o.

80. Suppose the area of the region between 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 = f(x) and y = 2f(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 frrst quadrant bounded by the liney

= x,the line x = 2,thecurvey = l/x2 ,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. 75. The region bounded below by the parabola y ~ x 2 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.

a. Sketch the region and draw a line y

~ 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.

82. True, sometimes true, or never true? The area of the region between the graphs of the continuous functions y = f(x) and y = g(x)andtheverticallinesx = 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.)

l\f(x) - g(x)] dx.

Give reasons for your answer.

2

76. Find the area ofthe region between the curve y = 3 - x and the line y ~ -I by integrating with respect to (8) x, (b) y. 77. Find the area of the region in the fJrst quadrant bounded on the left by the y-axis, below by the line y = x/4, above left by the curvey ~ I + v'X,andaboverightbythecurvey ~ 2/v'X. 78. Find the area of the region in the fJrst quadrant bounded on the left by they-axis, below by the curve x = 2yY, above left by the curve x ~ (y - 1)2, and above right by the line x ~ 3 - y.

Theory and Examples 83. Suppose thatF(x) is an antiderivative of f(x) Express {' sin2x dx x

11 in terms ofF.

= (sinx)/x,

x

> O.

300

Chapter 5: Integration

84. Show that iff is continuous, then

l

1f (X) dx =

l

y

1f (1 - x) dx.

85. Suppose that

Find 1>(X)dx

if(a) f is odd, (h) f is even. 86. a. Show that if f is odd on [-a, a], then

89. Use a substitution to verify Equation (I). 90. For each of the following functions, graph f(x) over [a, b] and f(x + c) over [a - c, b - c] to convince yourself that Equation (I) is reasonable.

L:f(X)dx = O. 1>. Test the result in part (a) with f(x)

87.

~

sinxanda

~

'fr/2.

a. f(x) ~ x 2 , a ~ 0, b ~ I,

Iff is a continuous function, fmd the value ofthe integral ('

1=

Jo

by making the substitution u integral to 1.

~

%

a - x and adding the resulting

COMPUTER EXPLORATIONS Io Exercises 91-94, you will find the area hetween curves in the plane when you cannot fmd their points of intersection using siruple algebra. Use a CAS to perform the following steps:

a. Plot the curves together to see what they look like and how many points ofintersection they have. 1>. Use the numerical equation solver in your CAS to fmd all the points ofintersection. c. Iotegrate If(x) - g(x) I over consecutive pairs ofintersection values. d. Sum together the integrals found in part (c).

""1t dt = 1'1t dt . 1

The Shift Property for Definite Integral. A basic property of definite integrals is their invariance under translation, as expressed by the equation [f(x) dx =

t~'f(x + c) dx.

(1)

The equation holds wbeoever f is integrable and defmed for the necessary values ofx. For example, in the accompanying figure show that -1(x 1 -2

+ 2)' dx

=

(l x 'dx

Jo

because the areas of the shaded regions are congruent.

Chapter

v;=4,

c. f(x) =

88. By using a substitution, prove that for all positive numbers x andy,

1

a = 0, b = 'fr, C = 'fr/2 a = 4, b = 8, c = 5

1>. f(x) = sinx,

f(x) dx f(x) + f(a - x)

c ~ I

3

2

91. f(x) =

x x T - 2 -

92. f(x) ~

2x· -

3x'

2x

+

1 + 3'

10,

g(x) = x - I

g(x) ~ 8 - 12x

93. f(x) ~ x

+ sin (2x),

94. f(x) ~

cos x, g(x) ~ x' - x

x2

g(x) ~ x'

Questions to Guide Your Review

I. How can you sometimes estimate quantities like distance traveled, area, and average value with 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'lDll?

4. What is the norm of a partition of a closed interval? 5. What is the defmite integral ofa function f over a closed interval [a, b]? When can you he sure it exists?

6. What is the 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 with defntite integrals (Table 5.4). Give examples. 9. What is the Fundamental Theorem of Calculus? Why is it so iruportant? Illustrate each part of the theorem with an example. 10. What is the Net Change Theorem? 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" ofeach 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.

13. How is inregration by substitution relared to the Chain Rule?

Chapter

16. How do you defme and calculare the area of the region between the graphs of two continuous functions? Give an example.

Practice Exerdses

Finite Sums and Estimates 1. The accompanying figure shows the graph ofthe velocity (fr/sec) ofa model rocket for the first 8 sec after launch. The rocket accelerared straight up for the fIrSt 2 sec and then coasred to reach its maximum height at t = 8 sec.

-

'0

20

20

20

a. ~3al

b. ~ (al

k-l

0

50 0

2

Definite Integrals 6

4

8

Time after launch (sec)

I

!;- -..J

I

+-H'

1"\

I

+ 1-+ 2

+ f-\ ++

4

6

8

10

Time (sec) 10

10

3. Suppose that ~al = -2 and ~bl = 25. Find the value of k-I

•

lim ~(2el-I)-If2.1>;1,wherePisapartitionof[I,5]

1V'1I~0 1~1

•

6. lim ~e.<e.' -1)1/3.1>;.,wherePisapartitionof[I,3] 1V'1I~0 1-1 7. 8.

+

lim

1V'1I~0 t:1

111 (cos ("--2 aXl, where P is a partition of [-"IT, 0]

))

•

lim ~(sinel)(cos el) .1>;1, whereP is a partition of[O, "lT12]

1V'1I~0 1~1

t,

9. If 3f(x) dx = 12, J~,f(x) dx = 6, and J~,g(x) dx = 2, fmd the values of the following.

a.J:

\ 1\

1/1

5.

f(x) dx

e. 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.

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 doting the time inrerval from t ~ 0 to t ~ 10 sec. About how far did the body travel doting those 10 sec? b. Sketch a graph of s as a function of t for 0 ,;; t ,;; 10 assumings(O) ~ O. 5

+ b1 )

d. ~(al- 2)

••

.~

>

k-l

'0

~100

OJ

10 - - b d. ~ 1 i-I 2

I)

4. Suppose that ~al = 0 and ~bl = 7. Find the values of

200 150 1 ill

(5 )

10

e. ~(al + b1 1-1

e.

l

b.l'f(X)dx d.1:(-"lTg (X))dx

'(f(x) + g(x)) 5 dx

-,

10. If Jo'f(x)dx = "IT, fo'7g(x)dx = 7, and the values of the following.

101g(x)dx =

a. J.'g(x) dx

b. j,'g(x) dx

e.l°f (x) dx

d. J.'

i-I

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 ofthe region between the graph of f and the x-axis. 11. f(x) ~ x 2 - 4x + 3, 0:5 x :5 3

32. Find the total area of the regioo between the curves y = sin x and y = cosx for 0 ::5 x :s; 3w/2.

12. f(x) = I - (x 2/4),

33. Show that y

-2:5 x :5 3

13. f(x) ~ 5 - 5x 2/',

dx 2

0:5 x :5 4

Find the areas of the regioos enclosed by the curves and lines in Exercises 15-26. 15. y

= x,

y

16. y ~ x,

17.

Vi +

=

l/x2,

I/Vi,

y ~

vY = I,

x

x ~

= 0,

l'+

dl solves the initial value problen3

I - x 2 ; y'(I)

J;( I

=0

= 3,

y(l)

= I.

+ 2~) dl solves the initial value

d 2y -2 ~ Vsecxtanx; dx

2 y

=2

34. Show that y = problen3

=2

X

~ x2 +

d'y

-1:5 x:5 8

Vi,

14. f(x) ~ I -

Initial Value Problems

y'(O) ~ 3, y(O) ~ O.

Express the solutions of the initial value probleD3S in Exercises 35 and 36 in terms of integrals.

y

sin x y () 35. dy dx = -x-, 5 =-3 = V2 - sin2 x,

36. :

y(-I) = 2

Evaluating Indefinite Integrals Evaluate the integrals in Exercises 37-44. 37. /2(COSX)-1/2 sinx dx

--:cr-----=--:-~x

o

+ I + 2 cos (28 +

39. /(28 18. x'

+

vY = I,

x

= 0,

y

= 0,

for

0:5 x :5 I

y

1

x3

= 2y 2,

X

= 0,

+ 4,

y

=3

20. x

=4

- y2,

X

=0

y = 4x - 16

23. y = sinx, y = x,

(~ + 2 sec2 (28 -

41. /

(I -

43. /

Vtsin(2I'/2) dl

t) + t) (I

dl

'If)) d8

+

~~2 -

42. /

(I

I dl

44. /

sec 8 tJln 8 VI

+ sec 8 d8

Evaluating Definite Integrals Evaluate the integrals in Exercises 45-70.

21. y2 ~ 4x, y ~ 4x - 2 22. y2 = 4x

l))d8

40. /

+ vY = I, 0 =s x =s 1

o 19. x

38. / (tJlnxt'/2 sec2x dx

0:5 x :5 'If/4

24. y = Isinxl, y = I, -'If/2:5 x :5 'If/2 25. y=2sinx, y=sin2x, O::5x::5'11' 26. y = 8 cosx, y = see> x,

-'If/3 :5 x :5 'If/3

2 45. 1>3X - 4x

{2

47. 49. 51 •

J1

+ 7) dx

48.

{'~

1.

50.

IVt

1

36 dx

(2) + I)'

0

27. Find the area of the ''triangular'' regioo hounded 00 the left by x + y = 2, on the right by y = x 2,and above byy = 2.

53. {' x-1/3(1 - x 2/3)'/2 dx

28. Find the area of the ''triangular'' regioo bounded 00 the left by y ~ Vi, 00 the right by y ~ 6 - x,andbelowbyy ~ I. 29. Find the extren3e values of f(x) ~ x' - 3x 2and rmd the area of

55.

the region enclosed by the graph off and the x-axis. 30. Find the area ofthe regioo cut frOO3 the rlrst quadrant by the curve x 1/2 + y1/2 ~ a1/2. 31. Find the total area of the region enclosed by the curve x = y2/3 and the linesx = y and y = -1.

il/8

57. 59.

J.1J'

J. 1~

8in2 5r

61.1

'~

dr

•

54. 56.

sec> 8 d8

58.

(88' - 12>2

+

J1 X

-01/3

5) dx

'(I l 1. 1. 1

cof ~dx

60.

secxtanxdx

62.

dx

+ VU)l/2 •r vu

du

dr

V'(7 - 5r)2 1/2 0 x'(1 + 9x')-'/2 dx 0

J.~/' cos2 (41 -

1

'~/'

1

'~/'

csc2 x

f)

ax

~/'

0

-/3

1

1

52

~/'

0

1.

{27

4

if dv

J1

46.

f tJln2~d8 ~/'

csczcotzdz

dl

Chapter 5 Practice Exercises

63.

r~/2

Jo

79. Y =

S(sinx)'/2cosxdx

' l %

1 1o 69. 1o

~/2

1S sin' 3x cos 3x dx

65. 67.

3 .

smxcosx dx

VI+3sin2 x

~/3

tan 8 dB V2""'-8

press

71. rmd the average value of I(x)

= nu; + b

y

answer. 83. Find dy/dx if Y = 84. Find dy/dx if Y = in your caleulation.

72. rmd the average value of

v'1+7. Ex-

fal v'1+7 dx in terms of F md give a reason for your

your caleulation.

over[-l,l]

b. over [-k, k] L

ICC&"

I

-,-- dt t +1

82. Suppose that F(x) is m antiderlvative of I(x) =

Average Valu.. L

, 1

80. Y =

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 function on [a, b]? Give reaSODS for your answer.

~/2 ~/2

- -6, d l 3 +t

303

t

V1+I'dl. Explain the main steps in

t..% (1/(1 - 12»

dt. Explain the main steps

85. A new parking lot To meet Ibe demand for parking, your town has allocated the area shown here. As the town engineer, you have been asked by the town council to find out if lbe lot can be buih

= "\I'h aver [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 ofcbange of lover [a, b] to bc

I(b) - I(a) b-a

for $10,000. The cost to elearlbe land will be $0.10 a square foot, and lbe lot will cost $2.00 a square foot to pave. Can lbe job be done for $1O,000? Use a lower sum estimate to see. (Answers may vary slightly, depending on lbe estimate used.)

md thc instantaneous rate ofcbange of I at.< to be I'(x). In this chapter we defmed the average va1ue ofa function. For the new definition ofaverage to be consistent wilb !be old one, we sbould have

Oft

I(b) - I(a) b_ a = average value of I' on [a, b]. 51 ft

Is this lbe ease? Give reasons for your answer. 74. Is it true that !be average value of an integrable function over an interval oflengtb 2 is half the function's integral aver tbe interval? Give reasons for your answer.

D 75.

49.5 ft Vertical spacing = IS ft

Compute !be average value of!be temperature function

I(x) = 37 sin

U:s

64.4ft

(x - 101») + 2S

67.5ft

for a 36s-day year. This is one way to estimate tbe annual meBD air temperature in Fairbanks, Alaska. The National Weatber Service's official figure, a numerical average of lbe daily normal mean air temperatures for the year, is 2S.7"F, whicb is slightly higher 1han the average value of!be approximating function/(x).

D 76.

Specific beat of. gas Specific heat C. is the amount ofbeat required to raise the temperature of a given mass of gas with COIlstant volume by I"C, measured in units ofea1/deg-mole (calories per degree gr.un molecule). The specific beat of oxygen depends on its temperature T md satisfies !be formula

C.

= 8.27

+ 10-' (UT -

J.87T2 ).

rmd the average value of C. for 20" :5 T :5 67S·C and the ternperatun: at which it is attained.

54ft

42ft

86. Skydivers A md B are in a helicopter havering at 6400 It Skydiver A jumps and descends for 4 sec befure opening her parachute. The helicopter then climbs to 7000 It md hovers there. Forty-fIVe seconds after A leaves lbe aircraft, B jumps and descends for l3 sec before opening his parachute. Both skydivers descend at 16 f1/sec with parachutes open. Assume that the skydivers f.ill freely (no effective air ",sistance) before their parachutes open. •• At what altitude does A's parachute open? b. At what altitude does B's parachute open?

DHf~adnglnt~b

c. Which skydiver lands fJ",t?

In Exercises 77-80, find dy/dx.

77. y =

1% V2 + cos'ldt

78. Y =

J2r'" V2 + cos'tdt

304

Chapter 5: Integration

Chapter

Additional and Advanced Exercises 9. Finding a curve Find the equation for the curve in the xy-plane if its slope at x is always that passes through the point (I,

Theory and Examples

1. a. If 1'7f(X) dx

b. If 1 'f (x) dx

~ 7,

= 4 andf(x)

[v7Wdx =

~ 17

does 1 'f (x) dx

V4 =

3:<2

-I)

+ 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 27

Piecewise Continuous Functions

Give reasons for your answers. 2. suppose12f(x)dx

~

= 4,

= 2. hf'f(x)dx = 3,l'g(x)dx ~

Which, ifany, of the following statements are true?

a. ],2 f (x) dx

,

~

b.1'(J(x)

-3

~

+ g(x)) ~

Show that

y ~ 7i1[' Jo f(l) sin a(x -

I) dl

solves the initial value problem

d'y dx 2

+ a'y

: =0

= f(x),

(Hint: sin (ax - at) 4. Proportionality

=

andy

= owhen x = O.

t

Show that d'yldx 2 is proportional to y and f'md the constant of proportionality.

x~c+

I-x x 2, '

=

-1

~

x < 0

O~x<2

{ -I,

2~x~3

(Figure 5.31) over [-1,3] is

~

1°(1 -1

x)dx + {'x 2 dx + f'(-I)dx

Jo

12

~ [x - ~l + [x;l: + [-xl: ~&+t-I~I:'

S. Find f(4) if

r" a. Jo f(l) dl

f(x)

3f(x)dx 1 -1

VJkdl.

1in2 f(x)

and

exist and are f'mite at every interior point of I, and the appropriate onesided 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 f is continuous, and the limit eriteria above guarantee that f has a continuous extension to the closure of each subinterval. To integrate a piecewise continuous function, we integrate the individual extensions and add the results. The integral of

sin ax cos at - cos ax sin at.)

Suppose that x and yare related by the equation

x=

limJ(x)

x~c

9

c. f(x) :5 g(x) on the interval-2 :5 x :5 5

3. Initial value problem

Although we are main1y interested in continuous functions, many functions in applications are piecewise continuous. A function f(x) is piecewise continuous OD a closed interval I if f has only f'mitely many discontinuities in I, the limits

~ x co. .rx

b.

[I(x) 2

Jo

1 dl ~ x cos '1tX.

y

6. Find f('1t12) from the following information.

i)

f

is positive and continuous.

ii) The area under the curve y

4

= f(x) from x = 0 to x = a is 3 2

7. The area of the region in the xy-plane enclosed by the x-axis, the curvey = f(x),f(x) '" O,andthelinesx = I and x = bisequal to

y=l-x

'\Ib'+l - v'2 for all b > I. Find f(x).

8. Prove that

1'(1'f(')d')du = 1'f(u)(x - u)du. (Hinl: Express the integral on the right-hand side as the difference of two integrals. Then show that both sides of the equation have the same derivative with respect to x.)

-1

0 -1

2 y

-

3

~-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(t) dt is expected to equal f(x) only 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. -4',

{

12. f(x) =

{~'

O~x~3

-4:5x

x-4,

13. g(t) = { '. smwt,

{ (7z _ 6)-1/3, ~' I, {

2, I' -

{

0 --;ot----:----c:--"---L--"---'-:--:-~x

o

-}:5r
7 2,

1,

17. Find the average value of the function graphed in the accompanying figure.

'~, o

I

IYt

Root sum

= v'1 + v'2 +

... +

y;, = 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

l1xs

dx

and evaluating the integral. lim ~(I' + 2' + 3' + ... + n'). n4

11_00

"-__~I~---<.~-~>x

r

2

J.'VXdx =

21. Let f(x) he a contiuuous function. Express

3

For example, let's estimate the sum of the square rools ofthe first n positive integers, v'1 + v'2 + ... + y;,. The integral

~x'/21 = ~

is the lintit of the upper sums 11

-.-

20. See Exercise 19. Evaluate

Approxfmattng Finite Sums w1t11 Integrals In many applications of calculus, integrals are used to approximate finite sums--the reverse of the usual procedure afusing finite sums to approximate integrals.

S =

.-11

Ii

2

18. Find the average value of the function graphed in the accompanying figure.

o

12

Ii

Therefore, when n is large, S. will he close to 2/3 and we will have

O:5z< I I :5 z :5 2 -2:5x<-1 -I :5 x < I 1:5x:52

I - x 2,

T

16. h(7) =

<

Q:5x:::S;3

0 :5 t < I 1 :s; t :s; 2

t

15. f(x) ~

Y

-8 ~ x < 0

X l/3

11. f(x) =

14. h(z) =

305

as a def'mite integral. 22. Use the result ofExercise 21 to evaluate a. lim ~(2 + 4 + 6 + ... + 2n), 11_00 n 2 b. 1:.

lim _1_ (I" + 2" + 3" + ... + n") n 16 '

n_OO

.

I (.

'T1'

•

2w

. 3w

What can be said about the following lintits?

fI.l+ ~.l+ ... + 'In ~.ln 'In n 'In n

v'1 + v'2 + ... + y;, n'/2

.

n71')

Inn- sm-+sm-+sm-+···+sm-. lI_oon n n n n

d. lim _1_ (I" + 2" + 3" + ... + n") 11_00 n 17 e. lim _1_ (I" + 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 =

nr 2

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 f 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 f(t) dt = f(v(x» dx - f(u(x» dx·

u(x)

To calculate lim"..... oo Sn, show that

and interpret Sn as an approximating sum of the integral

1

l

1

2

x 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 f(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 f consists of a semicircle and two line segments as shown. Let g(x) = f(t) dt.

A(x) =

l

V(X)

f(t) dt.

u(x)

It

y

y Uncovering

3

u(x)

i

A(x) =

a. Find g(l).

u(x)

b. Find g(3). c. Findg( -1). d. Find all values ofx on the open interval (- 3, 4) at which g has a relative maximum. e. Write an equation for the line tangent to the graph ofgat x = -1. f. Find the x-coordinate of each point of inflection of the graph ofg on the open interval (-3,4). g. Find the range ofg.

i) y" = -sinx

+ 2 sin2x

dA dv du dx = f(v(x» dx - f(u(x» dx· At what rate is the covered area changing? At the instant x, A(x) is increasing by the width f(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.

f(u(x» : '

In applications, we sometimes encounter functions

[x2 (1 + t) dt

Jsmx

dv f(v(x» dx· At the same time, A is being decreased at the rate

il) y = 1 andy' = -2 when x =

f(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 + fx71" cos 2t dt + 1:

Leibniz's Rule like

V(X)

f(t) dt

the width at the end that is being rolled up times the rate du/dx. The net rate of change in A is and

defmed by integrals that have variable upper limits of integration and variable lower limits of integration at the same time. The first integral

~

= f(v(x»: - f(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

(""")f(l) dl = F(v(x)) - F(u(x)).

Use Leibniz's Rule to find the derivatives of the functions in Exercises 27-29.

27. f(x) =

I tdl

1/%

Differeotiatlog both sides of this equation with respect to x gives the

d{"W f(l) dl =
- F'(u(x))::

Chain Rule

.ox

28. f(x)

~

29. g(y)

= {2Vy sin 12 dl

equation we want:

Chapter

1 %

J.",,)

307

( _1_ dl }cosxl-t 2

JVy

30. Use Leibniz's Rule to fmd the value ofx that maximizes the value of the integral

du - f(u(x))
%+3

1 %

1(5-I)dl.

Technology Application Projects

MathematicafMaple Modules: Using Riellumn Sums fD EstimareAretlS, Volumes, and Lengths ofCurves Visualize and approximate areas and volumes in Part I. Riemann Sums, Definite 11IIegrals, and the Fundame1lfD1 Theorem ofCakulus Parts I, II, and ill develop Riemann sums and defutite integrals. Part N contloues the development of the Riemann sum and defutite integral using the Fundamental Theorem to solve problen3S 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 chapter. You will compute the amount of water accumulatlog in basins of differeot shapes as the basin is f1lled and drained.

Motion Along a Straight Line, Part II You will observe the shape of a graph through dramatic animated visualizations of the derivative relations among position, velocity, and acceleration. Figures in the text can be animated using this software. Bending ofBeams Study bent shapes ofbeams, determine their maximum. deflections, concavity, and inflection points, and interpret the results in terms ofa 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 object's center of mass.

6.1

Volumes Using Cross-Sections

y Cross-section S(x) ·with area A(x)

s

In this section we define volumes of solids using the areas of their cross-sections. A crosssection 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

J

I

a

b

x

==

area X height

==

A • h.

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 Sex) 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 S(x) of the solid S formed by intersecting S with a plane Px perpendicular to the x-axis through the point x in the interval [a, b]. I

Plane region whose area we know

h = 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

Volumes Using Cross-Sections

6.1

309

Slicing by Parallel Planes

y

We partition [a, b] into subintervals of width (length) dXk 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 PXk ' 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 dXk == Xk - Xk-l (Figure 6.4). The volume Vk of this cylindrical solid is A(Xk) • dXk, which is approximately the same volume as that of the slab:

o

Volume of the kth slab ~ Vk == A(Xk) dXk. x

FIGURE 6.3 solid S.

A typical thin slab in the

The volume V of the entire solid S is therefore approximated by the sum of these cylindrical volumes, n

V~

y Approximating cylinder based on S(xk) has height

Plane at Xk-l

n

L Vk == k=l LA(Xk) dXk. 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

AXk = xk - xk-l

/

lim ±A(Xk) Axk = lbA(X)dxo n~OO

k=l

a

So we define the limiting definite integral of the Riemann sum to be the volume of the solid S.

o

I----~j

DEFINITION

Planeatxk

l~~x The cylinder's base

from x

==

a to x

The volume of a solid of integrable cross-sectional area A(x) ==

b is the integral of A from a to b,

is the region S(xk) with area A(xk) NOT TO SCALE

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 aXk

=

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 a formula for A(x), the area ofa typical cross-section. 3. Find the limits ofintegration. 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

A formula for A(x). The cross-section atx is a square x meters on a side, so its area is A(x) == x 2 •

Typical cross-section

3.

~

4.

x

o ..

The limits ofintegration. The squares lie on the planes from x == 0 to x == 3. Integrate to find the volume:

3

x

V==

x

--~x(m) 3

FIGURE 6.5 The cross-sections of the pyramid in Example 1 are squares.

1o

3A(x)dx== 13x

2

3]3 ==9m

dx==£

3

0

0

3

.

•

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. 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

2~

A(x) == (height)(width) == (x)(2~) y

==2x~. x

The rectangles run from x == 0 to x == 3, so we have

V= lbA(X)dx = -3

(x,-~)

132x~dx

== -

FIGURE 6.6 The wedge of Example 2, sliced perpendicular to the x-axis. The cross-sections are rectangles.

== 0

~ (9 3

- x 2)3/2]3 0

Let u = 9 - x 2 , du = - 2x dx, integrate, and substitute back.

+ ~ (9)3/2 3

== 18.

•

EXAM PLE 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. ____ Same volume

b

---

HISTORICAL BIOGRAPHY

Bonaventura Cavalieri (1598-1647)

-

\

I

I

I I

a~

__

Same cross-section area at every level

FIGURE 6.7 Cavalieris principle: These solids have the same volume, which can be illustrated with stacks of coins.

•

Volumes Using Cross-Sections

6.1

311

Solids of Revolution: The Disk Method

y

R(x) = ~ (T\\

-+----

o

x

---------I.--v~x 4

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

Qj)

A(x)

(a)

==

1T(radius)2

==

1T[R(x)]2.

So the definition of volume in this case gives y

y=~

Disk

___~-R~/-(-x) =

~

><

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.

EXAMPLE 4 The region between the curve y == -v.;,O :::; x :::; 4, and the x-axis is revolved about the x-axis to generate a solid. Find its volume. We draw figures showing the region, a typical radius, and the generated solid (Figure 6.8). The volume is

SoLution

l 1 b

V =

1T[R(x)f dx

4

=

1T[

-v.;y dx

RadiusR(x) = v.; for rotation around x-axis

• EXAMPLE 5

The circle

is rotated about the x-axis to generate a sphere. Find its volume. 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

SoLution

R(x) =

~for

rotation around x-axis

Therefore, the volume is V =

La a L_aA(x) dx = -a

1T(a

2

-

x

2

)

dx =

[3]a 4 x - ; -a = 3

1T a

2

1Ta3 .

•

The axis of revolution in the next example is not the x-axis, but the rule for calculating the volume is the same: Integrate 1T(radius)2 between appropriate limits.

EXAMPLE 6 by y

==

Find the volume of the solid generated by revolving the region bounded -v.; 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 - x 2 )

x 2 +y2=a2

/ ~-1-.

-a

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).

We draw figures showing the region, a typical radius, and the generated solid (Figure 6.10). The volume is

SoLution

1 1 '-h - 1Y 1 2'-h 1] 4

V=

1T[R(x)]2 dx

4

=

1T[

v.; -

1 Radius R(x) = for rotation around y = 1

dx

4

= 1T ==

+

[x -

2

7T £ - 2· -2 X 3/2 23

[

Expand integrand.

dx

+ X ]4 1

77T 6·

Integrate.

y R(x) = ~ - 1

y

y=l

I

0

~

IQ)) I I

",'" I /'" I / I

x

4

fl-

0

y

=

1

x x

FIGURE 6.10 The region (a) and solid of revolution (b) in Example 6.

•

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

A(y) and the definition of volume gives

==

7T[radius]2

==

7T[R(y)]2,

6.1

313

Volumes Using Cross-Sections

Volume by Disks for Rotation About the y-axis v =

y

ld

A(y) dy =

ld

1T[R(y)]2 dy.

EXAMPLE 7 Find the volume 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 volume is

SoLution

4

y

R(y) =

. 2 RadIus R(y) = Y for

~

rotation around y-axis

-+--------L----+x

o

2

•

(a)

y

EXAMPLE 8 the parabola x

==

Find the volume of the solid generated by revolving the region between y2 + 1 and the line x == 3 about the line x == 3.

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 == - \!2 to y == \!2. The volume is SoLution

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. Integrate.

y

R(y) = 3 - (y2 =

2 - y2

+ 1) ~~

\12

(3,

Y

R(y) = 2 - y2

\12)

y

y

-+----+-------+--------L..--+x

o

5

3

-\12

(3,

(a)

-\12)

-+----+--------'L...--------L..--+x

o

-\12 (b)

FIGURE 6.12 The region (a) and solid of revolution (b) in Example 8.

•

314

Chapter 6: Applications of Definite Integrals y

y

(x, R(x))

y

o x

Washer

FIGURE 6.13 The cross-sections of the solid of revolution generated here are washers, not disks, so the integral Jab A(x) dx leads to a slightly different formula.

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

Consequently, the definition of volume in this case gives

Volume by Washers for Rotation About the x-axis

(a)

v=

l

b

A(x) dx

=

l

b

7T([R(x)f - [r(x)f) dx.

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). The region bounded by the curve y == x 2 + 1 and the line y is revolved about the x-axis to generate a solid. Find the volume of the solid.

EXAMPLE 9

x

==

-x

+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)

==

-x

Inner radius:

r (x)

==

x2

+3 +1

Volumes Using Cross-Sections

6.1

3.

Find the limits of integration by finding the x-coordinates of the intersection points of the curve and line in Figure 6.14a.

+ 1 == -x + 3

x2 x (x

4.

l

v=

R(y) =

2

+X

+ 2)(x

2

==

0

- 1)

==

0

x

==

-2,

-

x

==

1

Limits of integration

Evaluate the volume integra1.

y

$

315

vy =

(2,4) 4

b

7T([R(x)]2 - [r(x)]2) dx

1~7T« -x + 3? -

Rotation around x-axis

Values from Steps 2 and 3

2

(x + I?) dx

~

Simplify algebraically.

0

',p C\S l-;

OJ)

,E

Y

.S ~

= 7T [8x - 3x

0

ca

2

-

x3

x 5 ]1

3 - 5

1171T

-2

•

5

~

~

~

0

x

2 (a)

r(y) =

2:Y

Y

R(y) =

...

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 of y.

i

vY

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.

EXAMPLE 10

SoLution First we sketch the region and draw a line segment across it perpendicular to the axis of revolution (the y-axis). See Figure 6.15a. The radii of the washer swept out by the line segment are R(y) == 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:

0,

o 2

x

(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).

v=

l

d

7T([R(y)f - [r(y)f) dy

Rotation around y-axis

Substitute for radii and limits of integration.

= 7T

r

Jo

(y _ y2) dy 4

==

1T[y2 _ y3]4 2 12 0

=

~7T. 3

•

316

Chapter 6: AppLications of Definite Integrals

Exercises 6.1 Volumes by Slfdng Find the voI1IJlli:s of the solids in Exercises 1-10. 1. The solid lies between planes perpendi.aular to the x-axis at x = 0 and x = 4. The cross-sections perpendicular to tile axis on the in.rervaJ. 0 :s x :s 4 are squares whose ~ run from the parabolay = -v';;totheparabolay = Yx. 2. The solid lies between planes perpendi.cular to the x-axis at x - -I and x - L The cross-sections perpendicular to the x-axis are circular disks whose di.ameteIs run from. the parabola y = x:l to the PlUUJola Y = 2 - x'l.

,

8. The base ofa solid is the region bmmded by the graphs of y - v';; and y - xll. The cross-sections perpendicular to the x-axis are

isosceles triangles ofheight 6. b. semi-circles with diameIers running lICl'OSS the base of the solid 9. The solid lies between planes perpendicuIar to the y-axis at y = 0 and y = 2. The cross-sections perpendicular to the y-axis are circular disks with diamebmi running from the y-axis to the parabola L

x _ y'Sy'l.

10. The base of the solid is the disk-x'l + y:l S L The cross-sections by planes perpendicuIar to the y-axis between y = - I and y = I are isosceles right triangles with one leg in the disk.

,

o x

3. The solid lies between planes petpendicuIar to the x-axis at x - - I and x - I. The cross-sections perpendicuIar to the x-axis between these planes are squares whose bases run from the semicircle y = - V'l="? to the semicircle y = V'l="? 4. The solid lies bctRen planes JlCIPCIldicular to the x-axis at x = -I and x = I. The cross-sections perpendicular to the x-axis between these planes ~s whose diagonals run from the

o

11. Find the volume of the given tetrahedron. (Hint Consider sliCCll perpendicular to one of tile labeled edges.)

semicircley = -VI - x'l to the semicircley = ~.

5. The base ofa 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 ClU'Ve 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 - -v/3 andx - v/3. The cross-sections perpendicular to the x-msare L

circular disks with diameters running from the curve y - tanxtothecurvey - secx.

b. squares whose bases run from the curve y - tan x to the

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 are L rectangles of height 10. b. rectangles of perimeter 20.

13. A twisted lOUd A square of side length & lies in a plane perpendicularto alineL. One vertex of the square lies onL. As this square moves a distance 11 along L, the square turns one revolution about L to generate a carkscrew-lilre 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 for yom lIllSWef.

317

6.1 Volumes Using Cross-Sections 14. CJmI1iIri'. 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 di8ks 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.

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.

X - 0, y - -I, y - I

27. Theregionenclosedbyx - y'Sy2,

x - 0, y - 2

28. Theregionenclosedbyx - y3/2,

29. The region enclogedbyx = V2sin2y,

O:s y:S;

30. The region enclosed by x = Vcos('Ir}'/4),

x=O 31. x = 2/(y + 1), x 32. x -

= 0,

V2Y/(y2 + I),

= 0,

y

X -

0,

'Ir/2, x

= 0

-2:s y:s; 0,

=3

y

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.

o

33. The x-axis

12

34. The y-axis

' J

J

y-VCOBJ:

Volumes by tile Disk Method In Exercises 15-18, f'1nd the volume of the solid generated by revolving the shaded region about the given axis. 16. About the y-axis 15. About thex-ws

~

4

x - tany 0

_,!!"

~

,

1:

2

1

0

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 35-40 about thex-axis.

.%+2J=2

35. y - x, y - I, x - 0

-;;t--'------"~,

o

2

36. y - 2v',;:, y - 2, :x - 0 37.y=x2 +1, y=x+3 38.y=4-x2, y=2-x

18. About the x-axis

17. Aboutthey-axis

39. y - secx, y -

J

If------7" L_/

;p.

tan

(~y)

o

~

,

1:

= V9 - x 2, y = 0 22. y = x - x2 , Y = 0 = Vcosx, O:s x:s ",./2, y = 0, x = 0

24. y - secx, y - 0,

-'1r/4:S x :s '1r/4

In Exercises 41-44, f'1nd the volume of the solid generated by revolving each region about the y-axis.

Find the volumes of the solids gweratJxl. by revolving the regions bounded by the lines and curves in Exercises 19-24 about the x-axis. 19. y = x 2, Y = 0, x = 2 20. y = x 3, Y = 0, x = 2

21. y 23. y

v'2,

40. y - secx, y - tanx, :x - 0, x-I

1

o

,

x - -'1r/4, x - '1r/4

41. The region enclosed by the triangle with vertices (I, 0), (2, I), and (I, 1) 42. The region enclosed by the triangle with vertices (0, 1), (I, 0), and (I, 1) 43. The region in the f'lI'St quadrant bounded above by the parabola y - x 2, below by thex-axis, and on the right by the line x - 2 44. The region in the f'tISt quadrant bounded on the left by the circle x 2 + y2 _ 3, on the right by the line x and above by the

\1'3,

line y -

v'3

In Exercises 25 and 26, f'md the volume of the solid generatJ;:d by revolving the region about the given line.

In Exercises 45 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 f'lI'St 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 f'lI'St quadnnt bounded abave by the line y = 2, below 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,andonthe1eftbythelinex - -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 and the lines y = 2 and x = 0 about bounded by y =

-v:;

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. Ifyou specify a brass that weighs 8.5 g/cm3 , how much will the plumb bob weigh (to the nearest gram)? y(em)

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 (em)

-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 ofrevolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? (l L = 1000 cm3.)

Va

y(em)

7fa 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 1

x (em)

~L!- ~ drep

53. Volume of a bowl 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. 3

55. Volume of a hemisphere Derive the formula V = (2/3)7fR for the volume of a hemisphere of radius R by comparing its cross-sections with the cross-sections of a solid right circular cylinder ofradius 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.

h

•

R

x 2 + y2 = 16 2 = 256

I

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?

~t

...

I

a. Find the volume of the bowl.

YR 2 _

"

h2 h

_t_

58. Max-min The arch y = sinx, 0 ::::; x ::::; 7f, 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 ofthe 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 around 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

~ i I

R

: 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 ofthe 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 = f(x) = 3x - x 2 is revolved about the vertical line x = - 1 to generate a solid (Figure 6.16). Find the volume ofthe 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 6.16a 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 ofheight 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 tbe 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 y

-2

-1

0

3

2

-1 Axis of revolution x =-1

Axis of revolution x =-1

-2

(b)

(a)

FIGURE 6.16 (a) The graph of the region in Example 1, before revolution. (b) The solid formed when the region in part (a) is revolved about the axis of revolution x = - 1.

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 of length (width) dXk. Its radius is approximately (1 + Xk), and its height is approximately 3Xk - Xk 2 • If we unroll the cylinder at Xk and flatten it out, it becomes (approximately) a rectangular slab with thickness dXk (Figure 6.18). The outer circumference of the kth cylinder is 27T· radius == 27T(1 + Xk), and this is the length of the rolled-out rectangular slab. Its volume is approximated by that of a rectangular solid,

d Vk FIGURE 6.17 A cylindrical shell of height Yk obtained by rotating a vertical strip of thickness aXk about the line x = -1. The outer radius of the cylinder occurs at Xk, where the height of the parabola is Yk = 3Xk - Xk2 (Example 1).

==

circumference X height X thickness

==

27T( 1

+ Xk) • (3Xk - Xk2) • dXk.

Summing together the volumes d Vk of the individual cylindrical shells over the interval [0, 3] gives the Riemann sum n

n

LdVk

==

k=l

L27T(Xk

+ 1)(3xk - Xk2) dXk.

k=l

rL

AX k

1

11

(3xk - xk 2 )

jJ

Outer circumference = 21T . radius = 21T(1 Radius = 1 + x k

~

-, #-

1.-'

+ xk)

AXk =

thickness

_

FIGURE 6.18 Cutting and unrolling a cylindrical shell gives a nearly rectangular solid (Example 1).

6.2

Volumes Using Cylindrical Shells

Taking the limit as the thickness dXk ~ 0 and n ~

00

321

gives the volume integral

n

V

==

lim

n~OO

=

1 1 1

27T(3x 2 + 3x - x 3 - x 2 ) dx 3

=

==

Xk2 ) dXk

327T (x + I)(3x - x 2 ) dx 3

=

L 21T(Xk + 1)( 3Xk k=l

27T

(2x 2

2 3 21T [-x 3

+ 3x - x 3) dx

1]3

3 2 - -x 4 + -x 24

451T 2 .

0

•

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 2h = 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 dXk == 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

==

21T X average shell radius X shell height X thickness

==

21T· (Ck - L) • f(Ck) • dXk.

Vertical axis of revolution

~"2J

Vertical axis of revolution

J~

Y =f(x)

% Rectangle height = !(ck)

--+---L.--~"""'---b..L...----+x

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: n

V ~ LdVk. k=l

The limit of this Riemann sum as each dXk ~ 0 and n ~ solid as a definite integral: V

==

lim n~OO

LdV n

k

k=l

==

lb l

00

gives the volume of the

21T{shell radius){shell height) dx.

a

b

=

21T(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) ~ 0, L :::; a :::; x :::; b, about a vertical line x == L is

-l

V -

a

b ( shell ) ( shell 21T. . ) dx.

radIus

heIght

EXAMPLE 2 The region bounded by the curve y == \/;, 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=~

y

~~ 2

Shell radius

/

(4,2)

y=~

2 Shell height

o 0

4

x

x

-4

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

ax.

Volumes Using Cylindrical Shells

6.2

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 == {b 27T ( sh~ll ) ( s~ell

Ja

1 1

radIus

4

=

== 27T

21T(X) (

heIght

) dx

~) dx

4 x 3/ 2 dx == 27T [- 2 ] 4 ==~. 128 X 5/2 050 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 == ~,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. 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 SoLution

y

Shell height

2 Y

y

/

~4 -y 2 y=~

(4,2)

4 - y2 Shell height 0 ~

o Cd

~

0

',p

r

(4,2)

~ ~ y

~~ ~

~

~

• ...-1

x

y Shell radius

4

0

(a)

.f?:~x ~

(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 ay.

•

324

Chapter 6: Applications of Definite Integrals

Summary of the SheU 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 regioll alld sketch a lille segmellt across it parallel to the axis ofrevolution. Label the segment's height or length (shell height) and distance from

the axis ofrevolution (shell redius). 2. Filld the limits ofilltegratioll for the thickness variable. 3. Illtegrate 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 general volume formula we will look at when stodying double and triple integrals in Chapter 15. That general formula also allows for computing volumes of solids other than those swept out by regions ofrevolution.

Exercises 6.2 Revolution About the Axes

S. The y-axis

In Exercises l--{j, use the shell method to rmd the volumes of the

solids generated by revolving the shaded region about the indicated

6. The y-axis y

y

Y=

axis.

9x

5

v9+9

o

3

2. y

~b 2

--:c+------;'c---x

o

x

2

0

3.

0

x

Revolution About the y-Axis Use the shell method to find the volumes ofthe solids generated by revolving the regions bounded by the curves and lines in Exercises 7-12 about the y-axis.

4. y

V2

2

0

7. y = x, y = -x/2, x=2 8.y=2x, Y = x/2, x = I 9. y = x 2 , y=2-x, x = 0, forx 10. y = 2 - x 2 , y = x 2 , x~O

y

FV2

v3

x~r

2

~-H

y~v3

x=3-i 0

v3

~-+x

3 ;))

~

0

11. y = 2x - I, Y = -rx, x=O 12. Y ~ 3/{2-rx), y

=

0, x

=

1,

x~4

6.2

0 < x :5 x = 0

13. Letf(x) = {(Sinx)fx, 1,

a. Showthatxf(x) = sin x, 0:5

25. y=x+2, y=x' a. The line x = 2

'Tr

X

:5

y sinX O<X:$'11' x' { 1, x = 0

d. The line y = 4

26. y = x 4 , Y = 4 - 3x' a. The line x ~ I

27. a. The x-axis

h. The line y

o

X

I

h. The line y = -2/5

x ~

1

:5 'Tr/4

a. Showthatxg(x) = (tan x)', 0:5

~

y

--;:-t---------"'"-~x

= {~~x)'/x, ~: ~

c. The x-axis

In Exercises 27 and 28, use 1he shell method to find 1he volumes of 1he solids generated by revolving 1he shaded regioos about 1he indicated axes.

e. The line y = 8/5

14. Letg(x)

325

h. The line x =-1

c. The x-axis

'Tr.

b. Find the volwne of1he solid generated by revolving the shaded regioo about the y-axis in 1he accompanying figure.

y=

Volumes Using Cylindrical Shells

12(Y' -

y'J

:5 'Tr/4.

h. Find the volwne ofthe solid generated by revolving the

shaded regioo about 1he y-axis in the accompanying figure.

~f'--------L-----c>x

y

'l

tan' x

y

'"

'"

= { --·o<x~x 4 0,

x~

0

28. a. The x-axis

e. The line y

h. The line y = 2

=5

d. The line y = -5/8

y --;:-r--:o--~x

o

'I! 4

(2,2)

Revolution About the x-AxIs Use the shell me1hod to find the volwnes of1he solids generated by revolving the regioos bounded by 1he curves and lines in Exercises 15-22 about the x-axis. 15. x ~

vY,

16. x

y2,

=

x ~ -y,

X = -y, y

y ~ 2 =

17. x = 2y - y', x = 0 19. y =

Ixi.

y = I

18. x = 2y - y', x = Y

20. y = x, y = 2x, y = 2

21. y~~, y~O, y~x-2

22.y~~, 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 e. The line x = -I e. Theliney ~ 7

Choosing the Washer Method or Shell Method For some regioos, both the washer and shell me1hods work well for 1he solid generated by revolving the regioo about 1he coordinate axes, but this is not always the case. When a region is revolved about the y-axis, for example, and washera are used, we must integrate wi1h respect toy. It may not be possible, however, to express the integrand in terms ofy. In sneh 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 each coordinste axis using a. 1he shell me1hod.

h. The line x = 4

d. The x-axis f. The line y ~ -2

h. 1he washer method.

30. Compute the volwne ofthe solid generated by revolving 1he triangu1ar regioo bounded by the lines 2y = x + 4, y = x, and x = 0 about

a. 1he x-axis using 1he washer me1hod.

24. y = x', y = 8, x = 0 a. The y-axis e. The line x = -2 e. Theliney ~ 8

--oot-----:------:;-> x o 2

2, y ~ 0

h. The line x = 3

h. 1he y-axis using 1he shell method.

d. The x-axis f. Theliney~-I

e. 1he line x

~

4 using 1he shell method.

d. 1he line y

~

8 using 1he washer me1hod.

326

Chapter 6: Applications of Definite Integrals

Ie 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 rmd 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 a. the x-axis

e. the line x

b. the y-axis ~

10/3

32. The region bounded by y =

d. the Hoe y

v-;;, y =

1 ~

I

----!'--------cot------o--> x

2, x = 0 about

a. the x-axis

b. the y-axis

e. the line x = 4

d. the Hoe y = 2

33. The region in the rITst quadrant bounded by the curve x = y - y' and the y-axis about

a. the x-axis

b. the Hoe y

~

I

-2

0

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 rmd 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 a. the x-axis

b. the y-axis

e. the line x = I

d. the Hoe y = I

3S. The region bounded by y ~

--;cf"""':....-+.----> x

o

b. the y-axis

x2

and y = x about a. the y-axis b. the Hoe x = I 37. The region in the rITSt quadrant that is bounded above by the curvey = I/x'/"ontheleftbythelinex = 1/16, and below by 36. The region bounded by y = 2x -

the line y

y

v-;; and y ~ x 2/8 about

a. the x-axis

= 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 drill bit ofradius 3.

Find the volume ofthe solid by

a. Find the volume ofthe bead.

a. the washer method. b. the shell method. 38. The region in the rITSt quadrant that is bounded above by the curve y = I/V-;;, on the left by the Hoe x = 1/4, and below by the line y

= 1 is revolved about the y-axis to generate a solid.

Find the volume ofthe solid by

a. the washer 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 ofthe removed portion ofthe sphere.

42. A Bondt cake, well known for having a ringed shape, is formed by revolving aroond the y-axis the region boonded 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 JITCcise 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 many 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 x = a to x = b. Ie order to derive an integral formula for the length ofthe curve, we assume that fhas a continuous derivative at every point of [a, bl. Such a function is called smooth, and its gmph is a smooth eurve because it does not have any breaks, corners, or cusps.

y

327

Arc Length

6.3

lJ.-,

-f---~-,----,----rlf-----'---'------"-----c---"----_X

FIGURE 6.22 The length of the polygonal path PoP,P2 ••• p. approximates the length ofthe curve y = f(x) from point A to pointB. y

AYk

I I

I

L1

r------I AXl

P,

I I I

We partition the interval [a, bj into n subintervals with a = Xo < XI < X2 < ... < x. = b. IfY, = f(x,), then the corresponding point P,(x" y,) 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 ofthe curve (Figure 6.22). If Axk = Xk - Xl-I and ~Yk = Yk - Yk-lo then a representative line segment in the path bas length (see Figure 6.23)

-ccI-cc'--------cL--> X

o

Xl_l

Xk

FIGURE 6.23 The arc P'_IP, ofthe curve y = f(x) is approximated by the straight line segment shown here, which has length L, ~ v'(ax,)2 + (ay,)2.

so the length of the curve is approximated by the sum (1)

We expect the approximation to improve as the partition of [a, bj becomes rmer. Now, by the Mean Value Theorem, there is a point c'" with Xk-I < Ck < X", such that ~Yk =

f'(et) Axk.

With this substitution for ~Y'" the sums in Equation (I) take the form

•

~ Lk

=

•

~ V(Axk)2 + (f'(Ck)~Xk)2

=

•

~ VI + [('(Ck»)' ~Xk'

(2)

Because VI + [('(x»)' is continuous on [a, bj, the limit of the Riemann sum on the righthand side ofEquation (2) exists as the norm of the partition goes to zero, giving lim

±

n_ OO k=1

Lk = lim

±

n_ OO k=l

VI

+

[('(et»)' Axk =

r

Ja

VI

+ [('(x»)' dx.

We derme the value ofthis limiting integral to be the length ofthe curve.

DEFINmON If j' is continuous on [a, bj, then the length (arc length) of the curve Y = f(x) from the point A = (a,j(a» to the pointB = (b,j(b» is the value ofthe integral

L

=

t

VI

+

[{'(x)]'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

4V2 3 x3 / 2 -

I ,O";x";l.

We use Equation (3) with a = 0, b = I, and

4V2 x3/ 2 _

y =

I

~ =

3

l,y

R:

0.89

dy = 4V2 • lx'/2 = 2V2x1/2 dx 3 2

FIGURE 6.24 The length of the curve is slightly larger than the length of the line segment joining points A and B (Example 1).

(:Y

Y

2

(2V2x1/

=

&.

=

The length of the curve over x = 0 to x = I is

L= =

[)1+

(:Ydx=

t· t(I +

&)3/2

[~dx

J: Ii '"

Eq. (3) with a=O,b=1

Let" = 1 + ~. integrate, and replace u by

2.17.

=

1

+ 8.<.

Notice that the length of the curve is slightly larger than the length ofthe straight-line segment joiningthepointsA = (0, -1)andB = (1,4V2/3 - I) on the curve (see Figure 6.24): 2.17

EXAMPLE 2

,!I2 + (1.089)2 '" 2.14

>

Solution

•

Find the length of the gmph of

f(x) y

Decimal approximations

=

x3

I

12 + x'

I ,,; x ,,; 4.

A gmph 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

=

2 4

I + (x _ l)2 2

=

I + (x' _ ! + l )

x'

(x

2

I

x

I

16

=T6+z+ x '= 4+

I X

2

)2

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):

329

6.3 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 =

(4)

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:

y3/2

(~t Raise both sides to the p~ 312.

= ~

2

x = 2y 3/2.

-cc1'------cL------cL--->.

FIGURE 6.26 The graph of y = (./2)2/3 from x = Otox = 2 is also the graph ofx ~ 2y 3/2 from y ~ 0 to Y ~ I (Example 3).

+ [g'(y)]2dy.

Find the length ofthe curve y = (X/2)2/3 from x = 0 to x = 2.

y=

0 1 2

t VI

Sol~

for;t.

From this we see that the curve whose length we want is also the graph of x = 2y 3/2 from = 0 to Y = I (Figure 6.26). The derivative

y

~ = 2(~),'/2 =

3y '/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!+9Ydy

+ 9y)3/2 ]1

0

{7 (IOv'iO - I) ,., 2.27.

Eq. (4) with c=O,d=l. Lctu~ 1 +9y, du/9 ~ dy, integrate, and substitute back..

•

The Differential Formula for Arc Length If y = j(x) and iff' is continuous on [a, b], then by the Fundamental Theorem ofCalcu1us we can defme a new function

1VI x

s(x)

=

+ [f'(t)f dt.

(5)

From Equation (3) and Figure 6.22, we see that this function s(x) is continuous and measures the length along the curve y = j(x) from the initial point Po(a, j(a» to the point Q(x, j(x» for each x E [a, b]. The function s is called the arc length funetlon for y = j(x). From the Fundamental Theorem, the function s is differentiable on (a, b) and :

=

VI + [f'(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

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

o

(aJ y

Ydx' + dy',

=

(7)

which can be integrated between appropriate limits to give the total length ofa curve. From this point ofview, all the arc length fonnulas are simply different expressions for the equation L = tis. Figure 6.27a gives the exact interpretation of tis corresponding to Equation (7). Figure 6.2Th is not strictly accurate, but is to be thought of as a simplified approximation of Figure 6.27a. That is, tis RJ /is.

J

EXAMPLE 4 Find the arc length function for the curve in Example 2 taking A = (1, 13/12) as the starting point (see Figure 6.25). Solution

In the solution to Example 2, we found that

-ccI-------~x

1

o

+

[f'(x»)' = (x; + ~,

r

(b)

Therefore the arc length function is given by FIGURE 6.27 Diagrams for remembering the equation tis = v'fix' + dy'.

s(x)

= [YI+[f'(I»)'dl= 3 1

1

= [ 12 - /

r 1

x

3

[(~+I~)dl

1

To compute the arc length along the curve from A instance, we simply calculate s(4)

11

=12-:X+12'

43

1

=

(1, 13/12) to B

=

(4,67/12), for

11

= 12 - 4 + 12 = 6.

•

TIris is the same result we obtained in Example 2.

Exercises 6.3 findIng Lengths of Curves Find the leogths of the curves in Exercises 1-10. Ifyou have a grapher, you may want to graph these curves to see what they look like. 1. y ~ (l/3)(x' + 2)'/2 from x ~ 0 tox ~ 3 2. y~x'/' from x~Otox~4 3. x = (y'/3) + 1/(4y) from y = I toy = 3 4. x = (y'/2/3) - yl/2 from y = I toy = 9 5. 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»:'/3 - (3/8»:'/3 + 5, I '" x '" 8 8. y = (x'/3) + x' + x + 1/(4x + 4), 0'" x '" 2

9. x = 10. y

D

J.Y v'sec't -Idt,

4 ~ 1\1:3/ -, -=1 dt,

-1f/4 "'y '" 1f/4 -2 '" x '" -I

findIng Integrals for Lengths of Curves In Exercises 11-18, do the following.

a. Set up an integral for the leogth of the curve. b. Graph the curve to see what it looks like. e. Use your grapher's or computer's integral evaluator to find the curve's leogth numerically.

= x', -I '" x '" 2 Y = tanx, -'lr/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 ofradius r centered at the origin. You will learn how to evaluate the integral in Section 8.3.

11. Y

12. 13.

v"J=Y2,

14. x ~ 15. y' + 2y ~ 2x

+

I.'

0

25. If9x' ~ y(y - 3)', show that

-1/2 '" Y '" 1/2 I from (-I, -I) to (7, 3)

16. y = sinx - x cosX',

17. Y

=

18. x

= J.Y V sec' I -

tan I dl,

331

Arc Length

6.3

O:s; x :s;

'1r

, _ (y

ds-

+ I)' 4y

dy. ,

26. If 4x' - y' ~ 64, show that

0 '" x '" 'lr/6

ds' = 4, (5x' - 16) dx'. y

-'lr/3

I dl,

'" Y '"

'lr/4

Theory and Example. 19. a. 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. a. Find a curve through the point (0, I) whose length integral (Equation 4) is L=

27. Is there a smooth (continuously differentiable) curve y

j(x)

28. Using tangent fins to derive the length formula for curves Assume that j is smooth on [a, b] and partition the interval [a, b] in the usual way. In each subinterval [x'-I,x.l, construct the langenlfin at the point (x.-l, j(x._I)) , as shown in the accompanying figure. a. Show that the length ofthe kth tangent rm over the interval

[XH, xii equals '1'(&1:.)'

+ (!'(XH) &1:.)'.

b. Showthat

lim ±(Iengthofkthtangentfiu) =

[~I+ :.dy.

~

whose length over the interval 0 '" x '" a is always v'2a? Give reasons for your answer.

n_ OO k=zl

l'

VI

a

+ (j'(x))'dx,

which is the lengthL of the curve y = j(x) from a to b.

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

from x = 0 to x = 'lr/4.

22. The length of an astroid The graph of the equation x'/3 + y'/3 ~ I is one of a family of curves called astroids (not "asteroids',) because oftheir starlike appearance (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.

(xk_l. !(xk_l» : with slope ----f'!"''-------clix-,--~, f'("_I) I I I

'''''1

"

29. Approximate the arc length of one-quarter of the unit circle (which is~) by computing the length ofthe polygonal approximation with n ~ 4 segments (see accompanying figure).

y y

1 x 2J3

+ y'1J3 = 1

---;.oEO:---c+----:3~,

-1

1

~+-Cc'-c~~'-c-~~,

o

-1

23. Length of a line segment 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 lriangle.

0.25 0.5 0.75 1

30. Distance between two points Assume that the tv.<> points (Xl, Yl) and (x" Y:z) lie on the graph of the straight line y = nIX + b. Use the arc length formula (Equation 3) to rmd the distance between the two points. 31. Find the arc length fimction fur the graph ofj(x) = 2x 3/2 using (0, 0) as the stirling point Wbatis 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)?

COMPUTER EXPLORATIONS In Exercises 33-38, use a CAS to perform the following steps for the given graph of the function over the closed interval. 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.

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.

33. j(x) =~, -1:::; x :::; 1 34. j(x) = x l/3 + x 2/3 , 35. j(x) = sin (1TX 2 ), 36. j(x) = x 2 cosx, 37. j(x) =

x-I 2

4x

38. j(x) = x 3

-

+

0:::;

0:::; x :::; 1

, - -2

1

x 2,

X :::;

0:::; X :::;

2

v2

1T

:::; x :::; 1

-1:::; x :::; 1

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 ofrevolution. 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

~x

y I+-~x-+I

A

B

27Ty X

0

NOT TO SCALE

(b)

(a)

FIGURE 6.28 (a) A cylindrical surface generated by rotating the horizontal line about the x-axis segment AB of length has area 21TyaX. (b) The cut and rolled-out cylindrical surface as a rectangle.

ax

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 dx about the x-axis (Figure 6.28a), we generate a cylinder with surface area 2'7Tydx. This area is the same as that of a rectangle with side lengths dx and 2'7Ty (Figure 6.28b). The length 2'7Ty 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 2'7Ty * 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 2'7TY* (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 == f(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

21TY* Yl

o NOT TO SCALE

(a)

(b)

FIGURE 6.29 (a) The frustum ofa cone generated by rotating the slanted line segment AB of length L about the x-axis has area

Yl + Y2 21TY* L. (b) The area of the rectangle for y* = --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 27ry*L, where y* is the average height of the line segment joining P and Q, and L is its length (just as before). Since ! ~ 0, from Figure 6.32 we see that the average height of the line segment is y*

==

+ !(xk))/2, and the slant length is L == Y(dXk)2 + (dYk)2. Therefore, !(Xk-l) + !(Xk) ~ / 2 · V (dXk)2 + (dYk)2 Frustum surface area == 27r·

(!(Xk-l)

==

7r(!(Xk-l)

+ !(Xk))Y(dxk)2 + (dYk)2.

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

L7r(!(Xk-l)

+ !(Xk))Y(dxk)2 + (dYk)2.

(1)

k=l

FIGURE 6.31 The line segment joining P and Q sweeps out a frustum of a cone.

We expect the approximation to improve as the partition of [a, b] becomes finer. Moreover, if the function! is differentiable, then by the Mean Value Theorem, there is a point (Ck, !(Ck)) on the curve between P and Q where the tangent is parallel to the segment PQ (Figure 6.33). At this point, dYk !'(Ck) == dXk' dYk

p

Segment length: L = V (~xk)2 + (~Yk)2

==

!'(Ck) dXk.

With this substitution for dYk, the sums in Equation (1) take the form n

L7r(!(Xk-l) k=l

+ !(Xk))Y( dxk)2 + (!'(Ck) dXk)2

n

==

L7r(!(Xk-l)

+ !(xk))Y1 + (!'(Ck))2 dXk.

(2)

k=l

These sums are not the Riemann sums of any function because the points Xk-l, Xk, 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

27T!(X)Yl

+ U'(x)f dx.

334

Chapter 6: AppLications of Definite Integrals (ck.f(Ck»

I

Tangent parallel to chord

We therefore define this integral to be the area of the surface swept out by the graph of f from a to b.

DEFINmON 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

y =f(x)

FIGURE 6.33 Iff is smooth, the Mean Value Theorem guarantees the existence of a point Ck where the tangent is parallel to segmentPQ.

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\/.;, 2, about the x-axis (Figure 6.34). We evaluate the formula

l Y)1 (2Y b

S =

+

27T

dx

Eq.(3)

with x

a

= 1,

dy dx

b = 2,

1 \/';'

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 I we calculate the area of this surface.

=

~l + i = ~x ~ 1 = ~.

With these substitutions, we have S =

r 27T· 2\/.; v.;-+l {2 ~ dx = 47TJI v.;-+l dx

JI

]2

2 = 47T· I(x + 1)3/2 I = 8;

(3V3 -

2V2).

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, 0 :::; 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:

x+y= 1

Lateral surface area

1

==

base circumference . 2 X slant heIght

==

'IT

"

V

I::

2.

1

-----

T

To see how Equation (4) gives the same result, we take

I 1

---_01

-I- - _

c

1

==

0,

x

d == 1,

==

dx dy

1 - y,

I 1

x

I

1 FIGURE 6.35 Revolving line segmentAB

and calculate

about the y-axis generates a cone whose lateral surface area we can now calculate in two different ways (Example 2).

s=

i

+

(:Y

d

21TX

-1,

1

=

21TV2 [y -

==

'ITV2.

+ (:

Y

~2I =

1 1

dy

=

21TV2

21T(1 - y)V2 dy

(1 - t) •

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. 0 ~ x ~ 7T'/4;

1. y = tanx,

0 ~

X

~ 2;

x-axis

3. xy = 1,

1

~

y

~

y-axis

0

~

y

4. x = siny,

5. x I/2 6. y

~

7T'; y-axis

+ from (4,I)to(I,4); x-axis + 2 vY = x, 1 ~ y ~ 2; y-axis

7. x =

l

8. y =

IX ~ dt,

tan t dt,

0

~ y ~ n/3; I

y-axis

~ x ~ V5;

x-axis

9. Find the lateral (side) surface area of the cone generated by revolving the line segment y = x/2, 0 ~ x ~ 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) + (1/2), 1 ~ x ~ 3, about the x-axis. Check your result with the geometry formula

X base circumference X slant height.

+ r2)

X slant height.

12. Find the surface area of the cone frustum generated by revolving the line segment y = (x/2) + (1/2), 1 ~ x ~ 3, about the y-axis. Check your result with the geometry formula Frustum surface area = 7T'(ri

+ 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.

13. y = x 3/9, 14. y =

Finding Surface Area

Lateral surface area =

t

Frustum surface area = 7T'(ri

yI /2 = 3 Y

Lateral surface area =

x-axis

2. y = x 2 ,

2;

10. Find the lateral surface area of the cone generated by revolving the line segment y = x/2, 0 ~ x ~ 4, about the y-axis. Check your answer with the geometry formula

v.;,

0 ~ x ~ 2;

3/4 ~ x ~ 15/4;

15. y = v'2x - x 2 , 16. y =~, 17. x = y3/3,

x-axis x-axis

0.5 ~ x ~ 1.5; 1 ~ x ~ 5;

x-axis

x-axis

0 ~ y ~ 1; y-axis

18. x = (1/3)y3 /2 - yI /2,

1 ~ y ~ 3;

y-axis

336

Chapter 6: Applications of Definite Integrals

19. x = 2~,

0 ~ y ~ 15/4;

y (em)

y-axis

~

+ y2

/ __ ~ __ . . . x. 2

y

4'cri /0,~) 2V4"=Y x

I I

I

.....

/

=

16 2 = 256

......

I I

'

=

----i---.-0: . . _-_ I

......

X

I

-16

28. Slicing bread Did you know that if you cut a spherical loaf of

20. x =

\I2.Y--=-1,

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.)

5/8 ~ y ~ 1; y-axis y

y

(1, 1)

x

=

A

V2Y=1"

5 8

o x

_L--

-r

21. y = (x 2/2)

+ (1/2), 0 + 2)3/2, + dy 2 in

22. y = (1/3)(x 2 ds =

S =

Vdx 2

ds =

0 ~

X

~

4

y-axis

\12;

y-axis (Hint: Express

terms of dx, and evaluate the integral

r-

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 27TRh.

Vdx

2

+ 1/(8y 2), 1 ~ y ~ 2; x-axis (Hint: + dy 2 in terms of dy, and evaluate the

Express

~

integral

h

J 27Ty ds with appropriate limits.)

T

24. Write an integral for the area of the surface generated by revolving the curve y = cosx, -7T/2 ~ x ~ 7T/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 47Ta 2 by using Equation (3) to find the area of the surface generated by revolving the y = ~, -a ~ x ~ a, about the x-axis.

curve

R

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

26. Testing the new definition

The lateral (side) surface area of a cone of height h and base radius r should be 7TrW+];i, the semiperimeter 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, 0 ~ x ~ h, about the x-axis.

D 27.

h

J 27TX ds with appropriate limits.)

23. x = (y4/4)

S =

~ x ~ 1;

-+------I...-----L_---L-r_~~x 0 a a + h Q))

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 1 cm3 = 1 mL, so 1 L = 1000 cm3 .)

bottom)?

D

b. Express the answer to the nearest square foot.

~

Ti

45 ft: Center

II/Radius

tl

22.5 ftl

~45ft

1.1-----~ 1

6.5 Work and Fluid Forces 31. An alternative derivation of the surface area formula Assume fis smooth on [a, b] and partition [a, b] in the usual way. In the kth subinterval [x,_" xu, construct the tangent line to the curve at the ntidpointm, = (X'-1 + x,)/2, as in the accompanying figure.

c. Show that the lateral surface area of the frustom of the cone swept out by the tangent line segment as it revolves ahout the x-axis is 2".f(m,)"V' I + (j'(m,»' !1x,. d. Show that the area of the surface generated by revolving y = f(x) about the x-axis over [a, b] is

a. Show that

r, = f(m,) - f'(m,)

t.;'

337

and r, = f(m,) + f'(m,) 11;,.

lim

±

n_OO 1=1

(lateral surface area\ of kth frustum )

~

l'

2".f(x)"V'1

+ (f'(x»'dx.

a

b. Show that the length L, of the tangent line segment in the kth

suhinterval is L, ~ "V'(!1x,)'

+ (f'(m,) !1x,)'.

32. The surface of an astroid Find the area of the surface generated by revolving about the x-axis the portion of the astroid x'/3 + y'/3 = I ahowo in the accompanying figure.

(Hint: Revolve the first-quadrant portion y ~ (I - x'/3)312,

o'" x '"

I, about the x-axis and double your result.) y

_J-...-c_~+-

-1

~I'---l1x,.----->l'I

.........L_x

0

Work and FLuid Forces

6.5

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 (I) we see that the unit OfWOlK in any system is the unit of force multiplied by the unit ofdistance. In SI units (SI 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 (Nom). This combination appears so often it has a special name, the joule. In the British system, the unit of WOIk is the foot-pound, a unit frequently used by engineers.

I

Joules The joule, ahbreviated J and pronounced ''jewel,'' is named after the English physicist James Prescott Joule (181l>-1889). The defming equation is 1joule

~

(I newton)(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 Ib 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 Ib). The total work performed by the jack on the car is 1000 X 1.25 = 1250 ft-Ib. In SI 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 fonnula W = Fd has to be replaced by an integral fonnuia 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 assurne that the magnitode of the force is a continuous 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, b] in the usual way and choose an arbitrary point Cl in each subinterval [Xl- h 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 the same as it wouid be if F were constant and we couid apply Equation (I). The total work done from a to b is therefore approximated by the Riemannsurn

axh

n

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: n

n~ ~ F(cl)

axl =

lb

a F(x) dx.

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) dx.

(2)

The units of the integral arejouies 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=

1'0 I 1

x2

1]'01 =

dx = - X

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 4f--.<'

Work done by F fromx ~ 0 tox ~ 0.25

-o:1"---:c':-:------~x(ft)

0.25

(3)

The constant k, measured in force units per unit length, is a characteristic of the spring, called the fo",. constant (or spring constant) of the spring. Hooke's Law, Equation (3), gives good resuits 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 natorallength of I ft to a length of 0.75 ft if the force constant is k = 16Ib/ft. Solution We picture the uncompressed spring laid out along the x-axis with its movable end at the origin and its f'lXed 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 = l6x. To compress the spring from 0 to 0.25 ft, the force must increase from F(O) = 16·0 = Olb

to

F(0.25) = 16·0.25 = 4 lb.

The work done by F over this interval is r.25

w = Jo

l6x dx = 8x

2 ]0.25 ° = 0.5 ft-lb.

Eq. (2) with

a = 0, b = 0.25, F(x) = 1&

•

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

(c) How far will a 45-N force stretch the spring? SoLution

0.8

1

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

1 2

W =

=

0 m to x

2 m is

=

2 30x dx = l5X ]: = 60 J.

(c) 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-lb.

•

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 57lb/ft3 . How much work does it take to pump the oil to the rim of the tank? 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 + d y has a volume of about

SoLution

10 - y

LlV

= 1T(radius?(thickness) =

Y

1T(~Y

~ y2 Lly fP.

=

Lly

The force F(y) required to lift this slab is equal to its weight, x

FIGURE 6.39 Example 5.

The olive oil and tank in

F(y)

==

57 dV

==

Weight = (weight per unit volume) X volume

4577T' Y 2 dy lb.

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 dW

==

577T' 2 4(10 - y)y dyft-lb.

Assuming there are n slabs associated with the partition of [0, 8], and that y == Yk denotes the plane associated with the kth slab of thickness dYk, we can approximate the work done lifting all of the slabs with the Riemann sum W

~

±5~1T

(10 - Yk)Yk 2 LlYkft-lb.

k=l

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

=

lim

~

n~OO k=l

57 1T (10 - Yk)Yk 2 LlYk 4

=

r

Jo

= 571T 4 ==

57 7T' (10 - y)y2 dy 4

r (10y 2 - y

Jo

577T' [10

4

y3

3

_ y4]8 4 0

3)

~

dy 30 561 ft-1b '

• .

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 Worle and FLuid Forces

I Weight-deD.lity A fluid's weight-density w is i11l weight per unit volume. 1jpical values (lbjft3) are listed below.

-

341

Tbe Pralure-Depth Equation In a fluid that is standing still, the pressure p at depth II is the fluid's weightdensity 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 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

100

Molasses Olive oil

'7

S.......

64

Freshwater

62.4

F - total force - force per unit area X area = pressure X area = pA = whA.

P =

wA: from Eq. (4)

Fluid Force on a Constant-Depth Surface

F=pA=whA FIGURE 15.41

These containers are filled

wi1h water to 1he same dep1h and have 1he same base area. The total force is therefore the same on the bottom ofeach container. The containers' shapes do not matter here.

y •

Surface of fluid

, y

•

....

Submerged vertical.

-t~=L(Y)

-1}.y Strip

(')

For example, the weight-density of freshwater is 62.41b/tP, 80 the fluid force at the bottom ofa 10 ft X 20 ftrectangular swimming pool 3 ft deep is

F- whA. - (62.41b/ft')(3fl)(lO-20ft') = 37,440 lb.

For a flat plate submmged horizontIJUy, 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). Iftbe plate is submetged vertically, however, !ben tbe pressure against it will be different at different depths and Equation (5) no longer is usable in that fonn (because II varies). Suppose we want to know the force exerted by a fluid against one side of a vertical plate submerged in a fluid of weight-density w. To f'md it, we model the plate as a region extending from y - a to y - b in the xy-plane (Figure 6.42). We partition [a, b] in the usual way and imagine the region to be cut into thin horizontal strips by planes perpendicularto the y-axis at the partition points. The typical strip fromy to Y + l!J.y is !JJ.y units wide by L(y) units long. We assume L(y) to be a continuous function ofy. The pressure varies across the strip from top to bottom.. If the strip is narrow enough, however, the pressure will remain close to its bottom-edge value ofw X (strip depth). The force exerted by the fluid against one side ofthe strip will be about

I!J.F = (pressure along bottom edge) X (area)

- w • (,trip depth) - L(y) ay.

Strip leng1h at level J

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 I!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 ODe side ofa thin, f1at horizontal strip is about AF = pressure X area = w X (strip depth) X L(y)Ay.

F ~ '5'(w o (,trip 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 ofthese sums:

,

lim '5'(w-(,tripdepth)"L(y,)ay, ~

n_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 ofweight-deosity w runs from y = a to y = b on the y-axis. Let L(y) be the leogth 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 (tt)

/0 /

l

b

w· (strip depth). L(y) dy.

(7)

EXAMPLE 6 A flat isosceles right-triangular plate with base 6 ft and 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 p1acing 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 leogth 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

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

one here.

F =

t

w·

u:~) 'L(y) dy

Eq. (7)

= 1362.4(5 - y)2y dy 3

= 124.81

5

(5y - y2) dy

= 124.8 [Zy2

y3]30 =

- 3

1684.8 lb.

•

Exercises 6.5 Springs 1. Spring eonstant It took 1800 J of work to s1retch a spring from its natural length of2 m to a length of5 In. Find the spring's force

constant. 2. Stretebing a spring A spring has a natural length of 10 in. An 800-Ib force stretches the spring to 14 in.

4. Stretcbing a spring If a force of 90 N s1retches a spring 1 m beyond its natural length, how much work does it take to stretch the spring 5 m beyond its natural length? 5. Subway ear springs It takes a force of 21,714 Ib to compress a coil spring assembly on a New York City Transit Authority subway car from its free height of8 in. to its fully compressed height of5 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 flISt half inch? the second half inch? Answer to the nearest in.-Ib.

c. How far beyond its natural length will a 1600-Ib force stretch the spring? 3. Stretebing a 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 thc rubber band this far?

6. Bathroom seale A ba1hroom scale is compressed 1/16 in. when a 150-Ib person stands on it. Assuming that the scale behaves like a spring that obeys Hocke'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 lb 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 multistrand cable weighing 4.5 lbI ft. When the car is at the frrst floor, 180 ft ofcable 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 klx 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,

o< a <

b.

11. Leaky bucket Assume the bucket in Example 4 is leaking. It starts with 2 gal of water (16 lb) 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 lb) 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 62.4 Ib/ft3 . 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-Ib/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 10ft (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 62.26 Ib/ft3 ? 62.59 Ib/ft3 ? 10 ft Ground level 0

12 ft

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-Ib/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 62.26 Ib/ft3 ? 62.59 Ib/ft3 ?

o

/

Ground level

10 10 ft 20 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 51.2Ib/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/ft3) 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 ~ x ~ 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/ft3). 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/ft3 . 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/m 3• y

Q

20 y

14. Emptying a cistern The rectangular cistern (storage tank for rainwater) shown has its top 10ft below ground level. The cistern,

343

currently full, is to be emptied for inspection by pumping its contents to ground level.

..t t Ay

y

Work and Fluid Forces

==:-:-r 4m

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 10ft and is full of benzene weighing 56Ib/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

Y x2

+ y2 + z2 =

i

100

Outlet pipe

\10 1 2 f t I f I I

0) __

//

z

--~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 Xl 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

-= --=

vdv dx

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· m2 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 Xl and X2. In physics, the expression (1/2 )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 2

r

6,370,000

. 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 of 280 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 tournament 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?

/ / /

Surface of pool

//

x (ft)

o __-+-_ _~~_1_=_=2 //5

Depth

Iyl

/

~.r--------r---./(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 milksbake 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

y (ft)

Pool surface at Y = 2

-_-:>3,-----::+--"""71:-+. (ft) 3

33. Rectangular plate In a pool filled with water to a depth of 10ft, calculate the fluid force on one side of a 3 ft by 4 ft rectangular plate ifthe plate rests vertically at the bottom of the pool L

on its 4-ft edge

b. on its 3-ft edge.

34. Semldn:ulor plate Calculate the fluid force on one side of a semicin:ular plate of nuIius 5 ft that rests vertically on its diameter at the bottom ofa pool filled with water to a depth of6 it. y Surface of water

6

5

Work and Fluid Forces

345

38. Semicircular plate A semieircular 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.

39. 1iIled plate Calculate the fluid force 00 one side ofa 5 ft by 5 ft square plate if the plate is at the bottom ofa pool ftlled with water to a depth of g ft and L 1yina flat on its 5 ft by 5 ft face. b. restiog vertically on a 5-ft edge. c. restiog on a 5-ft edge and tilted at 45' to the bottom ofthe pool. 40. lilted plate Calculate the fluid force 00 one side of a righttriangular plate with edges 3 ft, 4 ft, and 5 ft if the plate sits at the

bottom of a pool ftlled 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 metal tank shown here has a parabolic gate held in place by bolts and designed to withs1lmd a fluid force of 160 lb without rupturing. The liquid you plan to store has a weightdensity of 50 Ib/tt' . a. What is the fluid force on the gate when the liquid is 2 ft deep? b. What is the maximwn height to which the container can be filled without exceeding the gate's design limitation? y (ft)

_ _L-_ _--"--_ _---l_ _ •

(-1, 1)

(I, I)

35. Triangular plate The isosceles triangular plate shown 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 wculd be the fluid force on one side of the plate if the water were seawater instead of freshwater? Surface level

Parabolic gate

EDlarged view of

plUBbolic gate

42. The end plates of the trough shown here were designed to withs1lmd a fluid force of 6667 lb. How many cubic fi:et of water can the tank hold without exceeding this limitation? Round down to the nearest cubic fool 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 on one face of the plate now?

---cc'of----+. (ft) End view of trough

43. A vertical rectangular plate a uuits long by b uuits wide is sui>metged in a fluid of weight-density w with its long edges parallel to the fluid's surface. Find the average value of the pressure along the vertical dimension ofthe plate. Explain your answer. 37. New EngbmdAquarium The viewing portion of the rectsngular glass 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 force against this portion of the window. The \Wight-density of seawater is 64 Ib/tt' . (In case you were wcndering, the glass is 3/4 in. thick and the tank wal1s extend 4 in. above the water to keep the fISh fromjumpina oul)

44. (Continuation oJfuerciae 43.) Show that the force exerted by the fluid 00 one side of the plate is the average value of the pressure (found in Exercise 43) times the area of the plate.

45. Water pours into the tank shown here at the rate of 4 tt'/min. The tank's cross-sections are 4-ft.
346

Chapter 6: Applications of Definite Integrals 46. Watering trough The vertical eods of a watering trough are sqWl1'OS 3 ft 00 a side.

hole in the hottom at the rate of 5 It'/min. Will the movable eod reach the hole before the tank overllows?

a. Find the fluid force against the eods wheo the trough is full. Movable end

b. How many inches do you have to lower the trough to reduce the fluid force by 25%?

y

~

_or

level in the

hole

Side view

Drain hole

6.6

Moments and Centers of Mass Many structures and mechanical systems hehave as if their masses were concentrated at a single point, called the center ofmass (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 IS.

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

X2

X,

:.&< :2

m,

0

•

>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 exerts a downward force m.g (the weight of m.) 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 m.g by the signed distance x. 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.

m.

System torque = ml8Xt

+

m28X2

+

m38X3

The system will balance if and only ifits torque is zero. Ifwe factor out the g in Equation (I), we see that the system torque is

g

---

. (mixi

+ 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. M o = 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

X2

0

• ml

• /\ m2

L_~~

X3

• m3

~

) 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 ofmass as the center glides in a verticalline.

Suppose that we have a finite collection ofmasses located in the plane, with mass mk at the point (xt, Yk) (see Figure 6.45). The mass ofthe system is

y

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

System mass:

= ~

Moment about x-axis:

Mx

Moment about y-axis:

My = ~ mkXk.

mkYk,

The x-coordinate of the system's center of mass is defined to be FIGURE 6.45 Each mass mk has a moment about each axis.

_

My M

x=-=

(3)

348

Chapter 6: Applications of Definite Integrals

y

With this choice of X, as in the one-dimensional case, the system balances about the line = x (Figure 6.46). The y-coordinate ofthe system's center ofmass is dermed to be

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

ofmasses balances on its center of mass.

y Strip of mass 11m

In many applications, we need to find the center ofmass of a thin, flat plate: a disk of aluminum, say, or a triangular sheet of stee\. 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 .lJ'-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 ofthe 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 1hin strips parallel to the y-axis. The moment exerted by a typical strip about each axis is the moment its mass .11m would exert if concentrated at the strip's center of mass

_

M

x 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, and Center of Mass of a Thin Plate Covering a Region in the xy-Plane

I Density

A material's density is its mass per unit area For wires, rods, and narrow strips, we use mass per unit length.

Moment about the x-axis:

Mx = J'1dm

Moment about the y-axis:

My=Jxdm

Mass:

M=

J

(5)

dm

Center ofmass:

The differential dm is the mass ofthe strip. Assuming the density Il ofthe plate to be a continuous function, the mass differential dm equals the product Il dA (mass per unit area times area). Here dA represents the area ofthe strip. To evaluate the integrals in Equations (5), we picture the plate in the coordinate plane and sketch a strip ofmass parallel to one ofthe coordinate axes. We then express the strip's mass dm and the coordinates (x, '1) ofthe strip's center ofmass in terms ofx or y. Finally, we integrate '1 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/cm2 • 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 ofmass (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 ofmass (c.m.): length: width: area: mass: distance of c.m. from y-axis:

The plate in Example I.

(1,2)

My = Strip c.m. is halfway.

J 1 xdm =

16x2dx

=

2x 3

J:

=

2g·cm.

(b) The plate's mass:

\<, y) ~ (x, x)

J 1 1

M =

dm =

J:

6x dx = 3x 2

=

3 g.

(c) The x-coordinate ofthe plate's center of mass:

--;cf-----,--,---+--'----~x

->Idxr- I

o

x

The moment ofthe strip about the y-axis is xdm = x'6xdx = 6x 2 dx. The moment ofthe plate about the y-axis is therefore

y 2

(x,y) = (x,x) 2x dx dA = 2x dx dm = IldA = 3·2xdx = 6xdx = x

Units in centimeters

FIGURE 6.49 Modeling the plate in Example 1 with vertical strips.

_

My

x

= M =

2g'cm 2 3 g = 3 cm .

By a similar computation, we could find Mx and y = M x/ M.

Method 2: Horizontal Strips

(Figure 6.50)

(8) The moment My: The y-coordinate ofthe center ofmass of a typical horizontal strip is y (see the figore), so

y(em)

y=y. (1,2)

2

Strip c.m. is halfway.

Y

__)= (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):

(y/2) + 1

_

1-2~

x

=

2

y

=

Y+2

1

4+2

=

-4-'

dy (I,y) -oo+-------!---~x(em)

FIGURE 6.50 Modeling the plate in Example 1 with horizootaI strips.

We also have length:

I -

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 strip about the y-axis is

:tdm

=

y+2 2-y -4-" 3· -2-dy

3 8(4 - y2)dy.

=

The moment of the plate about the y-axis is

My = J :tdm =

f~(4

-

y2)dy = H4Y - y;]: =

H~6) = 2g"cm.

(b) The plate's mass:

M= J

dm = (' 1(2 - y) dy = 1 [2Y - y2]2 = 1(4 - 2) = 3 g. 102 22 0 2

(c) The x-coordinate ofthe plate's center ofmass:

_ My 2g"cm 2 x= M = 3g =3 cm. By a similar computation, we could fmd M x 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

EXAMPLE 2 Find the center ofmass 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.

4 y~4-x'

Center of mass (i,y)

=

(x. 4 ~

L

2

2 X )

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 follnwing relevant data.

Solution

~L--~+-,,---"--+,'---~x

-2

0

x

2

---*Idx!

(:t, y)

center ofmass (c.m.):

(x, 4 ~ x 2

=

)

length: 4 - x 2 width: dx area: ciA = (4 - x 2 ) dx mass: dm = 8 ciA = 8(4 -x 2 )dx _ 4 - x2 distance from c.m. to x-axis: y=-2The moment of the strip about the x-axis is

y dm

=

4 - x2 -2-.8(4 - x 2) dx

=

8

2(4 - x 2)2dx.

The moment of the plate about the x-axis is

Mx=JYdm= =

('06x 2

1-2

M=J

dm =

-

f2~(4_x2)'dx=

1-2

8x 4 + x 6 ) dx

1:

8 2c:i5 1

2 8(4 - x ) dx =

= f2 (8x2 _ 2x4) dx

1-2

=

=

2;t

f\2(4-x2)'dx

1-2

1:

2 2 2x (4 - x ) dx

351

Moments and Centers of Mass

6.6

Therefore, _

Y

2048

Mx

M

=

15

8

7'

105 • 256 =

=

The plate's center ofmass is

• Plates Bounded by Two Curves Suppose a plate covers a region that lies between two curves y = g(x) and y = j(x), where j(x) 2: g(x) and a :5 x :5 b. The typical vertical strip (see Figure 6.52) has

y

"Y

=

g(X)

I

I I I

I I I

a

dx

--::-t---'----:::>:I-1o::---:--- x

o

(X, y)

center ofmass (c.m.): length: width: area: mass:

= (x,! [((x)

+ g(x)])

j(x) - g(x)

ax

dA = [((x) - g(x)] ax dm = Il dA = Il[{(x) - g(x)]

ax.

b

The moment ofthe plate about the y-axis is FIGURE 6.52 Modeling the plate bounded by _ curves with vertical strips. The strip

c.m. is halfway, so 'ji

=

t

[f(x)

J 1 b

My =

+ g(x)].

xdm =

xll[{(x) - g(x)]

ax,

+ g(x)]'Il[{(x)

- g(x)]

and the moment about the x-axis is

Jy 1t b

Mx =

dm =

b

=

1 a

2Il [f(x)

[((x)

- g2(x)]

ax

ax.

These moments give the formulas

11

b

X= M

a

Ilx [((x) - g(x)]

ax

(6)

(7)

EXAMPLE 3 Find the center ofmass for the thin plate bounded by the curves g(x) and j(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

(V; _~)2 ax

x2

= x/2

=

t 10

3 (:x:'/2 _ x ) 2

ax

=

[2.7 x 7/2 _ Ix-]! =~. 8 0 56

352

Chapter 6: Applications of Definite Integrals Then from Equations (6) and (7) we get

x= =

y 1

56 9

[x2.{v'~ -~) dx

(x7/2 _~4)

56 [ 9

= 56

[2. x9/2 _

9 9

~ xs]l 10

0

dx 308

405'

and g(x) ~ ~

--o-\------------:--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 I 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. to x-axis:

a c.m.

-a

ds = a dO dm = [; tis = [;a dO y = a sin O.

Mass per unit length times 1ength

Hence,

(0. ~a)

0

x

2 2 (Figure 6.54). The Solution 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 (x, 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 (x, 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 (x, y) subtends an angle of dO radians, so tis = a dO. Thus a typical subarc segment has these relevant data for calculating y:

a

(b)

FIGURE 6.54 The semicircular wire in Example 4. (a) The dimensions and variables used in rmding the center of mass. (b) The center of mass does not lie on the wire.

_ 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 [; = I 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 willleam how to find the center of mass of wires lying along more general smooth curves in the plane (or in space).

6.6

Moments and Centers of Mass

353

Auid Forces and Centroids

Surface level of fluid 11 = centroid depth

Ifwe know the location ofthe 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

The force against one side ofthe plate is w·1i . plate area. FIGURE 6.55

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 ofa swimming pool. (fbis is Example 6, Section 6.5.) Use Equation (8) to find the force exerted by the water against one side ofthe 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 -----~-=---~

In the third century, an Alexandrian Greek named Pappas 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)

c

- - -- -

-~-"-~---~

Centroid

-O:+--------·~x FIGURE 6.56 The region R is to be revolved (once) about the x-axis to generate a solid. A 1700-year-old theorem says that the solid's volume can be calculated by multiplying the region'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,,(sbell

nldius)(shell height) dy

~ 2,,1' y L(y) dy,

(10)

The y-coordinate ofR's centroid is

,

_l'

Distance from Will of revoluUOII to centroid

I

'ji
y =.A.

l'

y L(y) dy

.A.

y-

y, dA. - L(y)dy

so that

, FIGURE &.57 Wi1hPappus'sfrrst theorem, we can fmd the volume of a tonJs wi1hout having to integrate (Example 6).

Substituting Ay for the last integral in Equation (10) gives V = 2'll"j/A. With p equal to y, we have V = 2'1rpA. •

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 ~ 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 = '1ra 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-

, • ~a Ca1troid 3.

-_.'-------.o.J----'a'!f)-u FIGURE &.58 Wi1h Pappus's frrst 1heorem, we can locate 1he centroid of a semicircular region without having to integrate (Example 1).

•

2,,(b)("a') - 2,,'00'.

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 ofthe 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 ofrevolution. 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 ofthe centroid's coordinate y.

EXAM PLE 7

Locate the centroid of a semiciroulsr region of radius a.

Ya

2 Solution We consider the region ~ the semicirele y 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

_ V (4/3)"0' 4 y - 2'1rA - 2'1r(l/2)1Ta2 - 3'1r a.

•

THEOREM 2 'IPPus'S Theorem for Surface Areas If an arc ofa 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 leng1h L ofthe arc time, the ms-e1IlM:led by the ""~ cen1roid during the revolution If p is the distance from the axis ofrevolution to the centroid, then

S = 21TpL.

(13)

6.6 Moments and Centers of Mass

355

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 ofthe surface generated by the arc is S=

o

l

x~b

27T)1

x-a

tis

lx~b

Y

= 2'IT

x-a

tis.

(12)

The y-coordinate ofthe arc's centroid is

_ E:bytis E:bytis btis y L

FIGURE 6.59 Figure for proving Pappus's Theorem for surface area. The arc leogth differential tis is giveo by Equation (6) in Section 6.3.

=

l

x-

J

L = ds is the arc's length and y ~ y.

=

x=a

Hence

l ytis x~b

x~a

=

YL.

Substituting yL for the last integral in Equation (12) gives S we have S = 2'ITpL.

EXAMPLE 8

=

2'ITYL. With p equal to y, •

Use Pappus's area 1heorem tof"md 1he surface area of1he torus in Example 6.

Solution From Figure 6.57, the surface of the torus is generated by revolving a circle of radius a about 1he z-axis, and b ;" a is 1he distance from 1he centroid to 1he axis ofrevolution. The arc leng1h of the smooth curve generating this surface of revolution is the circumference of the circle, so L = 2'ITa. Substituting these values into Equation (II), we f"md 1he surface area of1he torus to be

•

S = 2'IT(b)(2'ITa) = 4".2ba.

Exercises 6.6 10. a. The region cut from the frrst quadrant by the circle x 2 + y2 = 9

Thin Plates with Constant Density In Exercises 1-12, find the ceoter of mass of a 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 eocloscd by the parabolas y ~ x 2

-

3 and y ~ - z,,2

S. The region bounded by the y-axis and the curve x ~ y - y',

O"'y",1 6. The region bounded by the parabola x = y2 - Y and the line y = x 7. The region bounded by the x-axis and the curve y = cos x,

-'IT/2 '" x '" 'IT/2 8. The region between the curve y = sec' x, -'IT/4 '" x '" 'IT/4 and the x-axis 9. The region bounded by the parabolas y ~ z,,2 - 4x and

y=z,,-x 2

b. The region bounded by the x-axis and the semicirele

y=

vI9=7

Compare your answer in part (b) with the answer in part (a). 11. The "ttiangular" region in the frrst quadrant betweeo the cirele x 2 + y2 ~ 9 and the lines x ~ 3 and y ~ 3. (Hint: Use geometry to fmd the area.) 12. The region bounded above by the curve y = I/x', below by the curve y ~ -I/x', and on the left and right by the lines x ~ I and x = a > 1. Also, :fmd lima_ oo j.

Thin Plates with varying Density 13. Find the center of mass of a thin plate covering the region 2

betweeo the x-axis and the curve y = 2/x , I '" x '" 2, if the plate's deosity at the point (x,y) is B(x) = x 2• 14. Find the ceoter of mass of a thin plate covering the region bounded below by the parabola y = x 2 and above by the line y = x if the plate's deosity at the point (x,y) is B(x) = Iz".

356

Chapter 6: Applications of Definite Integrals

15. The region bounded by the curves y = ±4/V; and the lines x ~ 1 and x = 4 is revolved about the y-axis to generate a solid. Find the volume of the solid. b. Find the center of mass of a thin plate covering the region if the plate's density at the point (x,y) is 6(x) = I/x. .. Sketch the plate and show the center ofmass in your skeU:b. L

16. The region between the curve y = 2/x and the x-axis from x to x = 4 is revolved about the x-axis to generate a solid. L

=I

Find the volume of the solid.

15. Vorlahle density Suppose that the density of the wire in Example 4 is 6 = hin8 (kconstant). Find the center of mass. 26. Vuiahle density Suppose that lbe density of the wire in Example4is6 ~ I + klcos81 (kCODStant). Find the center of mass. Plates Bounded by Two Cunes In Exercises 27-30, fmd the centroid of the thin plate bounded by the graphs of the given functions. Use Equations (6) aod (7) with 6 = I and M ~ area of the regiro covered by the plate. 27. g(x) ~ x'

b. Find the center of mass ofa thin plate covering the region if the plate's density at the point (x,y) is 6(x) = V;.

.. Sketch the plate aod show the center of mass in your skeU:b.

Centroids of Triangles 17. The centroid of a triangle lift at the iDtenedion of the triangle's medians Yoo may recall that the point inside a triangle that lies OIIe-third of the way from each side taward the opposite \'CrtCX is the point where the triangle's three medians intersect. Sbow that Ibe centroid lics at the intersection of the medians by showing Ibat it too lies one-third of the way from eacb side toward Ibe opposite \'CrtCX. To do so, take the following steps. I) Stand onc side of the triangle on the x-axis as in part (b) of Ibe accompaoying figure. Express dm in terms of L and dy. il) Use similar triangles to sbow thalL = (b/h)(h - y). Substitute this expression for L in your formula for dm. ill) Showthatji = h/3. Iv) Extend the arguroent to the other sides.

and fIx) ~ x

+ I),

211. g(x) ~ x' (x

29. g(x) ~ x'(x - I) 30. g(x)

= 0,

(Hint:

J

f(;<)

+6

f(x) ~ 2,

and f(x) ~

= 2 + sin x,

xsinxd<

and x = 0

~ sin x -

x

x' = 0,

xcosx

and x

= 211'

+ C.)

Theory and Examples Verify the statemenl8 and formulas in Exercises 31 and 32. 31. The coordinates of the centroid of a differentiable plane curve are

_ J xds

x ~ length'

_

fyds

y = length·

y

y

, \ \

\

dy

\ \

..........

/"

h~

\ ,,"

T

~

,cio~" \

...............

32. Whatever the value of p > 0 in the equation y = x'/(4p), the y-coonlinate of the centroid of the parabolic segment abown here is ji = (3/5}a. y

Use the result in Exercise 17 to fmel the centroids of the triangles wbose vertices appear in Exercises 111-22. Assume a, b > D. 18. (-1,0),(1,0),(0,3) 19. (0,0), (I, 0), (0, I) ZO. (0,0), (a, 0), (0, a) 21. (0,0), (a, 0), (0, b) 11. (0,0), (a, 0), (a/2, b)

Tlrin Wires 23. Constant density Find the moment about Ibe x-axis of a wire of constant density that lies along the curve y = V; from x = 0 tox = 2. 24. Constant density Find the moment about Ibe x-axis ofa wire of constant density that lies along the corve y = x' from x = 0 tox = 1.

Th. Theorems of Pappus 33. The square region with vertices (0, 2), (2, D), (4, 2), and (2, 4) is revolved about the x-axis to generate a solid. Fmel the volume and surface area of the solid. 34. Use a theorem of Pappus to fmd the volume generated by revolving about Ibe line x = 5 Ibe triangular region bounded by the coordinate axes and the line 2x + y = 6 (see Exercise 17). 35. Find lbe volume of lbe torus generated by revolving the circle (x - 2)' + y' = I about lbe y-axis.

Chapter 6 Practice Exercises 36. Use the theorems of Pappus to fmd the lateral surface area and the volume ofa right-circular cone. 37. Use Pappus's Theorem for sorface area and the fact that the surface area ofa sphere ofradius a is 4'1Ta 2 to fmd the centroid ofthe 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 point (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 point (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

ofthe given triangle. Use the fact that the volume of a cone ofradius r and height h is V ~ ~ 'lTr 2h. 43.

y (0, b)

=;;;t----~~x (0,0) (a, 0)

44.

y

(a, c)

41. The region ofExercise 40 is revolved about the line y = x - a to generate a solid. Find the volume ofthe solid.

Chapter

Questions to Guide Your Review

1. How do 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 calculationa by these methods. 3. Describe the method of cylindrical sbells. Give an example. 4. How do you fmd the length of the grapb of a smooth function over a closed interval? 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

~

6. How do you defme and calculate the work done by a variable force directed along a portion of the x-axis? How do you calculate the work it takes to pump a liquid from a tank? Give examples. 7. How do you calculate the force exerted by a liquid against a portion ofa flat vertical wall? Give an example. 8. What is a center ofmass? a centroid? 9. How do you locate the center of mass ofa thin flat plate ofmat<>rial? Give an example. 10. How do 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 ofthe 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-sectiona perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y ~ x 2 to the parabola y ~ ~.

3. The solid lies between planes perpendicular to the.HXis at x = 'IT/4 and x ~ 5'IT/4. The cross-sectiona between these planes are circular disks whose diameters run from the curve y = 2 cos x to the curve y=2sinx.

2. The base of the solid is the region in the fust quadrant between the line y = x and the parabola y = 2 ~. The cross-sections of

4. The solid lies between planes perpendicular to the x-axis at x = 0 and x = 6. The cross-sectiona between these planes

358

Chapter 6: Applications of Definite Integrals y

are squares whose bases run from the x-axis up to the curve x'f2 + y'f2 =

v'6.

y ~ll;t-----:+--+'ll~X

6

-2

2

Lengths of Curves Find the leogths of the curves in Exercises 17-20.

17. y

5. The solid lies betweeo planes perpeodicular to the x-axis at x = D and x ~ 4. The cross-sections of the solid perpeodicular to the x-axis between these planes are circular disks whose diameters run from the curve x' ~ 4y to the curve y' ~ 4x.

= x 'f2

- (1/3)x'f2,

I '" x '" 4

18. x = y'/3, I '" y '" 8 19. y = (5/12)x 6/' - (5/8)x 4/', 20. x ~ (y'/12)

+ (l/y),

I '" y '" 2

Areas of Surfaces of Revolution In Exercises 21-24, fmd the areas ofthe surfaces geoerated hy revolv-

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 equilateral triangle with one edge in the plane. (The triangles all lie on the same side of the plane.)

ing the curves about the given axes.

7. Find the volume of the solid geoerated 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.

24. x =

8. Find the volume of the solid geoerated 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 geoerated by revolving the region bounded on the left by the parabola x ~ y' + I and on the right by the line x = 5 about (a) thex-axis; (b)they-axis; (c) the line x = 5. 10. Find the volume of the solid geoerated by revolving the region bouoded by the parabola y' = 4x and the line 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 geoerated 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 geoerated 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 geoerated by revolving the region be2x about (a) the x-axis; tween the x-axis and the curve y ~ (b) the liney = -I; (c) the line x = 2; (d) the line y = 2.

x' -

14. Find the volume of the solid geoerated 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

ft is 15. Volume of a solid sphere hole A rouod hole of radius bored throngb the cooter of a solid sphere of a radius 2 ft. Find the volume of material removed from the sphere.

_haD

16. Voiome of a The prome of a fuothall resembles the ellipse shown here. Find the football's volume to the nearest cubic inch.

I '" x '" 32

21. y ~~, 0 '" x '" 3; x-axis 22. y = x 3/3, 0:5 x:5 1; x-axis 23. x ~

v'4y -

YY,

y',

I '" y '" 2;

2 '" y '" 6;

y-axis

y-axis

Work 25. Lifting equipmeot A rock climber is about to baul up 100 N (about 22.5 Ib) of equipmeot that bas beeo banging beoeath her on 40 m of rope that weighs 0.8 newtoo per meter. How much work will it take? (Hint: Solve for the rope and equipmeot separately, theo add.) 26. Leaky tank truck You drove an 800-gal tank truck of water from the base of Mt. Washington to the summit and discovered on arrival that the tank was only balffull. You started with a full tank, clin3bed at a steady rate, and accomplisbed the 475D-ft elevation cbange in 50 min. Assuming that the water leaked out at a steady rate, how much work was speat 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 leogth, how much work does it take to stretch the spring this far? An additional lOot? 28. Garage door spring A force of 200 N will stretch a garage door spring 0.8 m beyond its uostressed length. How far will a 300-N force stretch the spring? How much work does it take to stretch the spriog this far from its unstressed length? 29. Pumping a reservoir A reservoir sbaped 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 reservoir (Continuation of Exercise 29.) The reservoir is filled to a depth of5 ft, and the water is to be pumped to the same level as the top. How much work does it take? 31. Pumping a couical tank A right-circular conical tank, point down, with top radius 5 ft and height 10 ft is filled with a liquid whose weight-deosity is 60 Ib/ft' . How much work does it take to pump the liquid to a point 2 ft above the tank? If the pump 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 ofa thin, flat plate covering the region enclosed by the x-axis, the lines x = 2 and x = -2, and the parabola y = x 2•

2 --'----~.:::::....------'-~x

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

UNITS IN FEET

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 ofthe trough when the syrup is 10 in. deep? y y=x-2// ..,----------"t--/ ----->-"-2--0+---/-}/2"--- 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 ofmass 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 ofmass if, instead ofbeing constant, the density is 8(x) = x. (Use vertical strips.)

~~~~O

-4

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/nJ) 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).

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.

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 <

l VI

a:::;

+ cos2 0 dO >

b. Generalize the result in part (a).

/

/

I

I

Tr/2,

a

/

/

/

/

Va2 + sin2 a.

360

Chapter 6: Applications of Definite Integrals

Surface Area 7. At points on the curve y = 2 ~, line segments of length 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?

'7

/ /

/

2\13

2~

(3,2\13) ~

"

y=2~

x

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.

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 43.5-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 S 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 a. My - bSA if the plate lies to the right of the line, and b. bSA - My if the plate lies to the left of the line.

14. Find the center ofmass ofa thin plate covering the region bounded by the curve y 2 = 4ax and the line x = a, a = positive constant, if the density at (x,y) is directly proportional to (a) x, (b) IYI. 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?

l a x

Work 9. A particle of mass m starts from rest at time [ = 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([) = [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 C is 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

Mathematica/Maple Modules: Using Riemann Sums to Estimate Areas, 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 EXAMPLE 1 Some functions are one-to-one on their entire natural domain. Other 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) = \IX 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 sine function is one-toone on [0, ,,/2], however, because it is an increasing function on [0, ,,/2] giving distinct outputs for distinct inputs. • The graph of a one-to-one function y = f(x) can intersect a given horizontal line at most once. If the function intersects the line 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 Une Test for One-to-One Fnnctions A function y = fix) is one-to-one if and only if its graph intersects each horizontalline at most once.

y

Same y-value

0.5 --!--'~'---!--

-1

0

I

\

,

5:;

, Inverse Functions

y = sinx (b) Not one-to-one: Graph meets one or more horizonta1lines more than once.

=,'

=

FIGURE 7.1 (e)y andy v'Xare 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 (-00,00).

Since each output of a one-to-one function comes from just one input, the effect of the function can be inverted to send an output back to the input from which it came.

DEFINmON Suppose that f is a one-to-one function on a domain D with rnnge R. The inverse function 1 is defmed by

r

rl(b) = The domain of

r

0

if f(o) = b.

l is R and the rnnge of

r

l

is D.

r

r

l for the inverse of f is read "f inverse." The "-I" in l is not an The symbol exponent; rl(x) does not mean 1/fix). Notice that the domains and ranges of f and l are interchanged.

EXAMPLE 2

r

Suppose a one-to-one function y = fix) 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 interchanging the values in the columns of the table for f:

y

•

rl(y)

r

If we apply f to send an input x to the outputf(x) and follow by applying l to fix) we get right back to x, just where we started Similarly, if we take some number y in the range of f, apply l to it, and then apply f to the resulting value rl(y), we get back the value y with which we began. Composing a function and its inverse has the same effect as

r

doing nothing.

U-l U

0

0

f)(x) = x,

rl)(y) = y,

for all x in the domain of f for all y in the domain of f- l (or range of f)

Only a one-to-one function can have an inverse. The reason is that if f(xI) = y and f(X2) = y for two distinct inputs Xl and X2, then there is no way to assign a value to rl(y) that satisfies both rlU(Xl)) = Xl and r l U(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 i' 0 and flO) = 0, defined on (- 00, 00) and passing the horizontal line test.

Finding Inverses The graphs ofa 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

°

x DOMAINOF

f

(a) To :lind the value off at x, we start at X, go up to the curve, and then over to the y-axis.

°

X RANGE OF

/-1

r'

(b) The graph of ia Ibe graph off, hot 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 ofj-l is the range off, The range ofj-l is the domain off.

y

y ~r'(x)

--L---'O";/01"-_....,I-_ _---->, y /

----onl--""""l~---~x

/ / / / / / /

(c) To draw !he graph ofr' in Ibe

(d) Then we interchange the letters x and y. 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.

FIGURE 7.2 Detennining the graph ofy = r'(x) from the graph ofy = j(x). The graph of is obtained 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

r'

to the y-axis to get the value of (x). Figure 7.2 indicates the relationship between the graphs of 1 and 1 • The graphs are interchanged by reflection through the line y = x. 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 ofpassing from 1 to 1-1 can be summarized as a !w<>-step procedure.

r

1.

2. y

Solve the equation y = I(x) for x. This gives a formula x = r1(y) where x is expressed as a function ofy. Interchange x and y, obtaining a formula y = r1(x) where 1 is expressed in the conventional format with x as the independent variable and y as the dependent variable.

r

y~2x-2

EXAMPLE 3

Find the inverse of y

~ x + I, expressed as a function ofx.

=

Solution 1.

Solve for x in terms ofy:

y =

2y = x

~joo"C~jL-f-----~x

-2

/

/

x

/

// -i /

FIGURE 7.3 Graphingf(x) = (1/2)x + I and r'(x) = 2x - 2 together shows the

2I x +

2.

Interchange x and y:

I

+2

= 2y -

2.

y = 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:

graphs' symmetry with respect to the line

r1(f(x))

=

2(~X + I) - 2 =

I(r (x)) '

=

~(2x -

y = x (Example 3).

2)

+

+2- 2

x

I = x - I

=

x

+

I = x.

;;"

0, expressed as a function

•

y

EXAMPLE 4 ofx. Solution

Find the inverse of the function y

=

x 2, X

We flISt solve for x in terms of y: y = x2

l:tl ~""'-------~x

another (Example 4).

because x

2:

0

We then interchange x andy, obtaining

~

FIGURE 7,4 The functions y ~ Vi and y = xl, X ~ 0, are inverses of one

= x

y= vX. = x 2,x;;"

The inverse ofthe functiony O,isthefunctiony = vX (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 function y = x 2, with no domain restrictions, is not one-to-one (Figure 7.lb) and therefore has _ oo~e.

Derivatives of Inverses of Differentiable Functions If we calculate the derivatives of I(x) = (1/2)x + I and its inverse r1(x) from Example 3, we see that

~/(x)=~(~x+I)=~ ~rl(x)

=

~(2x

- 2)

=

2.

= 2x -

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 # 0, the reflected line has slope I/m.

The slopes are reciprocal: (r')'(b) ~ _I_or (f-l)'(b) ~ I 1'(0) f'(f '(b»

FIGURE 7.5 The graphs of inverse functions hsve reciprocal slopes at corresponding points.

r

The reciprocal relationship between the slopes of I and 1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = I(x) at the point (a, I(a» is j'(a) and j'(a) # 0, then the slope of y = r'(x) at the point (f(a), a) is the reciprocal 1/ j'(a) (Figure 7.5). Ifwe set b = I(a), then

(f-I)'(b)

=

j'~a)

=

j'(f \b»'

r'

If y = I(x) has a horizontal tangent line at (a, I(a» then the inverse function has a is not differentiable vertical tangent line at (f(a), a), and this wmite slope implies that at I(a). Theorem I gives the conditions under which is differentiable in its domain (which is the same as the range of I).

r'

THEOREM I-The Derivative Rule for Inverses and j'(x) exists and is never zero on I, then its domain (the range of f). The value of

r'

If I has an interval I as domain

r is differentiable at every point in (r ')' at a point b in the domain of r ' isthereciprocalofthevalueofj' at the point a = r'(b): 1

I (/ -1)'(b) _ - j'(f '(b»

(1)

or

drl I dx

x~b

-

_1_

d/l dx

x-r'(b)

Theorem I makes two assertions. The tITSt of these has to do with the conditions under which I-I is differentiable; the second assertion is a formula for the derivative of

366

Chapter 7: Transcendental Functions

I -1 when it exists. While we omit the proof of the first assertion, the second one is proved in the following way:

l(r 1(x»

=

x

Inverse function relationship

:x l(r (x»

=

I

Differentiating both sides

=

I

Chain Rule

1

/'(r 1(x»' ~ r dx d

dx

1

(x)

1-1(

I

) _

x-

/'(f

Solving for the derivative

'(x»"

Thefunction/(x) = x 2 ,x;" oand its inverse r'(x) = v'Xhavederivatives/,(x) = 2xand(r 1)'(x) = 1/(2v'X). Let's verify that Theorem I gives the same fannula for the derivative of 1(x):

y

EXAMPLE 5

r

4

(r')'(x)

/'(f \x»

=

3

rex)

I

2(f l(x»

2

~

2x with x replaced

by r'(xl

I

I/J

2(v'X)"

--ccfL----!--~___!---"-----~ x

o

t

2

3

4

FIGURE 7.6 The derivative of r'(x) ~ Vi at the point (4, 2) is the

Theorem I gives a derivative that agrees with the known derivative of the square root function. Let's examine Theorem I at a specific point. We pick x = 2 (the number a) and = 4 (the value b). Theorem I says that the derivative of I at 2, /,(2) = 4, and the derivative ofr l at 1(2), (f-1 )'(4), are reciprocals. It states that

1(2)

reciprocal ofthe derivative of I(x) = x 2 at (2, 4) (Example 5).

(/

- 1)'(4)

_

I

_

I

_ I

- /'(f 1(4» - /,(2) - 2x

I.-2 -_4'I

•

See Figure 7.6.

y

6

y=x 3 -

(2, 6)

2 Slope 3<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 drl/dx without knowing a formula for r 1•

Let I(x) = x 3 finding a fannula for 1(x).

EXAMPLE 6

Reciprocal slope:

lz

1(6,2) I

Solution

-

r

2. Find the value of drl/dx at x = 6 = 1(2) without

We apply Theorem I to obtain the value ofthe derivative of

r

1

at x = 6:

I

ax Ix=2

--!----=1--+------~x

-dl

6

1

dr

dX FIGURE 7.7 The derivative of I(x) ~ x' - 2atx ~ 2 tells us the derivative of at x = 6 (Example 6).

r'

See Figure 7.7.

=

3>:2

= 12

x=2

I x=/(2) =

I

I

dl dx

I

I 12

Eq. (I)

x=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? y y 2. 1. y

W..

= -3x3 X

y 1

o

13. y

y

= intx

14.

y ~f(x)

...

~

y

sin x,

-¥sxs¥

0-0 0-0 0-0

y ~ f(x)

"

0-0 0-0 0-0

6. y=

l x

\

::r..

one-to-one. 7. f(x) =

G,-

x,

x
~

{lxx ++ 6, { 1 - 2' x _x_ x + 2'

xsO

9. f(x) ~

2 2 -x ,

x S 1 x> I

10. f(x)

'2

15.

x

16. y

6

f(x)

f(x) ~ 6 - lx, O=SxS3

~

X+I, -ISxSO 2 { -2 + "3x, 0 < x < 3

-"--c+-----fJ-->x -I 3

--:ct------'::c-->x

o

3

17. a. Graph the functionf(x) ~ ~, 0 S x S I. What symmetry does the graph have?

In Exercises 7-10, determine from its graph if the function is

8. f(x)

"

x

y

y

taox,

2

0

y

~

-¥<x<¥

1

X

5.

__x

~f------l---~x

y

4.

X

y~f(x)~ I-l,x>o

o

I

y

12.

y

~f---j'''-------

y=x 4 _x 2

3.

11.

~

4,

r

x,

v?

b. Show that f is its own inverse. (Rernernher that = x if 0.) 18. a. 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 r l in each case. 19. f(x) = x 2

x>O

+ I, x '" 0

y

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 not necessary to rmd a formula for r l .) Identify the domain and rangeofr'.

r'

20. f(x) = x 2,

x

S

0

y

-----'~---l--~x

--:ct-----:-------> x

o

368

Chapter 7: Transcendental Functions

22. j(x) ~ x 2 - lx

21. j(x) ~ x' - I

+

I, x "" I

40. a. Showthath(x) = x'/4andk(x) = (4x)'/3 are inverses of one

another.

y

y

h. Graph h and k over an x-inrervallarge enough to show the graphs inrersecting at (2, 2) and (-2, -2). Be sure the picture 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). -----jf-c-hl---~x --c-+-----'""-----~x

o

23. j(x) ~ (x

+ 1)2,

X "" -I 24. j(x) ~ x 2/3,

y

x "" 0

y

d. What lines are tangent to the corves at the origin? 41. Let j(x) ~ x' - 3x 2 - I, x "" 2. Find the value of dr'/dx at the point x = -I =j(3). 42. Let j(x) = x 2 - 4x - 5,x > 2. Find the value of dr'/dx at the point x ~ 0 ~ j(5).

43. Suppose that the differentiable function y = j(x) bas an inverse and that the graph of j passes through the point (2, 4) and hss a slope ofl/3 there. Find the value of dr'/dx atx ~ 4. 44. Suppose that the differentiable function y = g(x) bas 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. Inverses of Lines 45. a. Find the inverse of the function j(x) stant different from zero.

~

mx:, where m is a con-

b. What can you conclude about the inverse of a function y ~ j(x) whose graph is a line through the origin with a Derlvattves of Inverse Functtons Each of Exercises 25-34 gives a fannula for a function y = j(x). In each case, fmd r'(x) and identilY the domain and range of r'. Ail a check, showthatJ
25. j(x) = x' I

29. j(x) = l/x 2, X 31. j(x) =

x

x "" 0

26. j(x) = x\

+

27. j(x) = x'

28. j(x) = (1/2)x - 7/2

>0

+3

---=2

x 33. j(x) = x 2 - lx, x '" I (Hint: Complete the square.)

30. j(x) = I/x', 32. j(x) = •

x¢'O

y; r

VX - 3

46. Show that the graph of the inverse of j(x) ~ mx: + h, where m and h are constants and m ¢' 0, is a line with slope I/m and y-intercept -h/m. 47. a. Find the inverse ofj(x) = x + I.Graphjanditsinverse together. Add the line y = x to your sketch, drawing it with dashes or dots for contrast b. Find the inverse of j(x) = x + h(h constant). How is the reJared to the graph of f? graph of

r'

e. What can you conclude about the inverses of functions whose graphs are lines parallel to the line y = x? 48. a. Find the inverse ofj(x) = -x + I. Graph the line y ~ -x + I together with the line y ~ x. At whst angle do

+ 1)'/'

34. j(x) = (lx'

nonzero slope m?

the lines intersect? In Exercises 35-38:

a. Findr'(x). b. Graph j and

b. Find the inverse of j(x) = -x + h (h constant). What angle does the line y ~ -x + h mske with the line y ~ x?

r' together.

e. Evaluaredj/dxatx ~ aanddr'/dxatx ~ j(a) to show that at thesepointsdr'/dx = I/(dj/dx).

+

3,

a = -I

37. j(x) ~ 5 - 4x,

a ~ 1/2

35. j(x) = lx

36. j(x) = (1/5)x + 7, a = -I 38. j(x) ~ lx 2, X "" 0, a ~ 5

39. a. Show that j(x) ~ x' and g(x) ~

'Ii'X

are inverses of one

e. 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 j(X2) ¢' j(x,).

another. b. Graph j and g over an x-inrervallarge enough to show the graphs in_ecting at (I, I) and ( -I, -I). Be sure the pictore shows the required symmetry about the line y = x. e. Find the slopes of the tangents to the graphs of j and g at (I, I) and (-I, -I) (four tangents in all).

d. What lines are tangent to the carves at the origin?

Use the results of Exercise 49 to show that the functions in Exercises 5(}-54 hsve inverses over their domains. Find a formula for dr'/dx using Theorem I.

50. j(x) = (1/3)x

+ (5/6)

52. j(x) = I - 8x' 54. j(x) =

x S/'

51. j(x) = 27x'

53. j(x) = (I - x)'

7.2

Theory and Appl;cations 55. If f(x) is on<>-to-one, can anything be said about g(x) = - f(x)?

Is it also one-to-one? Give reasons for your answer. 56. If f(x) is one-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 dermed. If f and g are one-ro-one, can anything be said about fog? Give reasons for your answer.

Natural Logarithms

369

COMPlITER EXPLORATIONS 10 Exercises 61--{j8, you will explore some functions and their inverses together wilb their derivatives and linear approximating functions at specified points. Perform the followiog steps using your CAS: a. Plot the function y = f(x) together with its derivative over the giveo interval. Explain why you know that f is on<>-to-<me over lbe interval. b. Solve the equationy = f(x) for x as a function ofy, and name lbe resulting inverse function g.

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 f 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 f. 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 tangent line to g at the point (f(Xo), xo) located symmetrically across the 45 0 line y = x (which is the graph ofthe ideotity function). Use Theorem I to find the slope of this tangeot line.

r'

60. Equivalence of the washer and sheIl methods for rmding volume Let f be differeotiable and increasing on the interval a '" x '" b, with a > 0, and suppose that f bas a differeotiable inverse, Revolve about the y-axis the region bounded by the graph of f and the lines x = a and y = f(b) to geoerate a solid. Theo the values of the integrals giveo by the washer and shell methods for the volume have ideotiea1 values:

r '.

i

f('J 1T«r'(y»' -

a') t(y =

f(a)

l'

21TX(f(b) - f(x)) dx.

To prove this equality, derme

i

61. Y

l'

1T«r (y))' - a') t(y '

21TX(f(t) - f(x)) dx.

4x

63. y = - , - - ,

+

I

x'

64. y = - , - - , x + I 65. y

= x3

66. y

=

-

t '" x '" 4,

2,

- 3x+2 62.y-2x-11'

3x 2

Xo = 3

-2< _ x<2 _ , Xo

= 1/2

-I '" x '" I, Xo = 1/2 -I '" x '" I, Xo = 1/2 -

2 - x - x 3,

67. y = eX,

68. y Theo show that the functions W and 8 agree at a point of [a, b] and have ideotical derivatives on [a, b]. As you saw in Section 4.7, Exercise 90, this will guarantee W(t) = 8(t) for all t in [a, b]. 10 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

= V3x -

1, 2 ~ x ~ 5, Xo

f(t)

flu)

8(t) =

e. Plot the functions f and g, the ideotity, the two tangeot lines, and lbe line segroeot joiniog the points (xo, f(xo» and (f(xo), xo). Discuss the symmetries you see across the main diagonal.

x

a

W(t) =

(xo ,f(xo)).

-2:5 x:s; 2, :to

-3 s x s 5,

Xo

= 27 10

="23

= 1

= sinx,

10 Exercises 69 and 70, repeat the steps above to solve for the functions y ~ f(x) and x ~ r'(y) dermed implicitly by lbe giveo equations over the interval.

69. y'/' - I = (x + 2)', 70. cosy = x l / S, 0 S x

S

-5 '" x '" 5, Xo = -3/2 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 natura1logarithm as an integral using the Fundamental Theorem of Calculus. While this indirect approach may at first seem strange, it provides an elegant and rigorous way to obtain the key characteristics of logarithmic and exponential functions.

Definition of the Natural Logarithm Function The natura1logarithm of any positive number x, written as

In x, is defined as an integral.

370

Chapter 7: Transcendental Functions

DEFINmON

The natural logarithm is the function given by Inx =

il("Itdl,

x> O.

(1)

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 < I, 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 = y

il[11t dl =

O.

HO <.< I, then In. ~ jX!dt ~ -j\dt gives the negative of this area.

H.> l,th.DIn. ~ jX!dt gives this area.

1

1

y = Inx

FIGURE 7.8 The graph of y = Inx and its relation to the function y = I/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 Typical2-place values of ln x

x

Inx

o

undefined

0.05 0.5

1 2 3 4 10

-3.00 -0.69

o 0.69 1.10 1.39 2.30

lnx =

("I "dx,

il

with x having two meanings. Thus we change the variable of integration to I. By using rectangles to obtain tmite 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].

DEFINmON

The number e is that number in the domain of the naturalloga-

rithm satisfying In(e) = 1.

7.2 Natural Logarithms

371

So the number e lies within the interval [2, 3] and satisfies ' I ]

1

tdt

1.

=

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 calco1ated 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 Calco1us (Section 5.4),

d

d!: Inx

=

d]'1

d!:

1

tdt

=

x'

=

I x'

For every positive value of x, we have

d d!: Inx

I

and the Chain Rule extenda this formula for positive functions u(x):

d d du -Inu = -Inu'd!: du d!:

u> O.

EXAMPLE 1

(2)

We use Equation (2) to [md derivatives.

dId I I (s) d!: 1n2x = 2x d!: (2x) = 2x (2) = x' (b) Equation (2) with u = x 2

~1n(x2 + 3) d!:

=

x> 0

+ 3 gives _1_.~(x2 + 3) x 2 +3d!:

=

_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 I I -Inbx = -'-(bx) = -(b) = - . d!: bx d!: bx x If x < 0 and b < 0, then bx b = -I we get

(3)

> 0 and Equation (3) still applies. In partico1ar, if x < 0 and d

d!: In (-x)

=

xI

for x < 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.

d

d!:1n Ixl

I x'

x oF 0

(4)

372

Chapter 7: Transcendental Functions

Properties of Logarithms HiSlORICAL BIOGRAPHY

John Napier (155{}-1617)

Logarithms, invented by John Napier, were the single most important improvement in arithmetic calculation before the modern electronic computer. What made them so 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 naturailogarithm satisfies the following rules:

b

1. Product Rule:

In bx = In b + Inx

2. Quotient Rule:

In

3. Reciprocal Rule:

xb = I In x =

4. Power Rule:

Inx T

=

In b - In x -In x

Rule 2 with b ~ 1

rlnx

For r rational

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)

(b) 1n

Product

x + I 2x

_ 3 = In(x + I) - 1n(2x - 3)

(0) Int = -inS

Quotient Reciprocal

= -In23 = -3ln2

•

Power

We now give the proof ofTheorem 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:

d

The argument starts by observing that In bx and In x

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

373

rlnx (assuming r rational) We use the same-derivative argument

again. For all positive values ofx,

l-"-

Eq. (2) with u

=

1 r-l -rx

General Power Rule for derivatives, r rational

=

r·f =

-"- In x' = (x') dx x'dx x'

=

xr

:x (rlnx).

Since In x' and r In x have the same derivative, Inx' = rlnx

+C

for some constant C. Taking x to be 1 identifies C as zero, and we're done. (Exercise 46 in Section 3.7 indicates a proof ofthe 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 ofRule 2, obtained by setting b = 1 and noting that In 1 = O. This covers all cases of Theorem 2. • y

l

y= I

If-----+----=..,.._ 2 O~----+I----~2;--~x


y

We have not yet proved Rule 4 for r irrational; however, the role does hold for all r, rational or irrational. We will show this in the next section after we derme exponential functions and irrational exponents.

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 of x. 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 = l/x and above the interval [I, 2]. In Figure 7.9a, a rectangle ofheigbt 1/2 over the interval [1,2] fits under the graph. Therefore the area under the graph, which is In 2, is greater than the area, 1/2, ofthe rectangle. So 1n2 > 1/2. Knowing this we have

y = lnx

1n2n = nln2 --;;-f-~c_;;;---------+x

>

n(!)

<

-n(!)

=

~

and lnZ-n = -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 interva1l '" 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 = f'(x) dx and

f'(x) f(x) dx

J

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

f'-22x Jo

x

-

5

Ja:.

dx=1--s 1a:=ln l l ]-1-s u~x'-5, =In I-II-In I-51 =Inl -In5 =-In5 u

du, u(2) ~ -1

u(O) ~ -5,

•

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 dx =

+C

=

-In lui

=

I In-- cos I xI

For the cotangent,

J

J

sinx eosx dx =

cot x dx =

J

u ~ cosx > 0 on (-Tr/2, Tr/2), dll = -sinXtU

-du

--..

=

+C

-In leosxl

+C

In Iseexl

+ C.

=

Jeo~x J dx = smx

u

du

coax dx

=

=In lui + C =In Isinxl + C =-In lesexl + C.

J -J Ja:

(seex seex (seex

sec x dx =

sinx,

=

du

U

+ tanx).

To integrate sec x, we multiply and divide by (sec x

=

Reciprocal Rule

In lui

+

+ tanx) + tanx) dx -

C = In Iseex

J

2

see x + seextanx seex + tanx dx

+ tanxl +

u

C

+ tanx (sec x tan x + sec2 x) dx

secx

=

du

=

For esex, we multiply and divide by (esc x + eotx).

J J J-:u esex dx =

=

=

esex

-In lui

(esex (esex

+

+ eotx) dx + eotx)

C = -In lese

=

x+

J

ese2 x + esc x cot x dx esex + eotx

cot xl

+

u

C

=

du

=

cscx + cotx (-cscxcot.:t - csc2 x)dx

Integrals of the tangent, cotangent, secant, and cosecant runctions

J J

tanudu

=

In Isecul

+C

eotudu

=

Inlsinul

+C

J J

+ tanul + C

seeudu

=

In Iseeu

eseudu

=

-Inleseu

+ eotul + C

7.2

Natural Logarithms

375

EXAMPLE 4 r~/6

Jo

r~/3

tan2xdx = =

Jo

d tanu';

I ]~/3 Iln Isecul 0

I r~/3

=

IJo

=

I I(ln2 -Inl)

tanudu =

I Iln2

Substitute u = 2x, dx ~ du/2, u(O) ~ 0, u(7T/6) ~ 7T/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

(x 2 + I)(x + 3)'/2

x> 1.

x-I

y=

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 + I) + In (x + 3)'/2 -In(x - I)

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-I'

x+3

Next we solve for dy/dx:

I

I)

(2x dy dx=y x2 +1 +2x+6-x-1 . Fina11y, we substitute for y from the original equation:

dy = (x 2 + I)(x + 3)'/2 (~+ _1 dx x-I \;2+1 2x+6

1_) 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 logarithms 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 and In 7.

a. In (1/125) b. In 9.8 d. In 1225 e. In 0.056 f. (In 35 + In (1/7))/(1n 25)

c. In 70

376

Chapter 7: Transcendental Functions

Use the properties of logarithms to simplify the expressions in Exercises 3 and 4. 3. a. In sin 8 - In

(S~8)

b. In (3x 2

9x) + In (ix)

-

~1n(4t4) -In2

C.

Finding Derivatives In Exercises 5-36, find. the derivative of y with respect to x, t, or (}, as appropriate.

5. y = In3x

6. y = In kx, k constant

In (t 2 )

= = In x10

In (t 3/2)

8. y

3

9.y= ln :x

49. 51.

4. a. In sec 8 + In cos 8 b. 1n(8x + 4) - 2ln2 c. 31n~ -In(t + I)

7. y =

47.

10. y

Jr/2 6

Jo

1

~

53

.

= Vx(x +

57. y

= ~t ~

61. y

x4 x4 17,y~4lnx-T6

18. y ~ (x 2 lnx)4 • Y

~

~I+lnt

t

22. y ~ I + Inx

23. y = In (lnx)

24. y = In (In (lnx))

25. y = 8(sin (In 8) + cos (In 8)) ~

_ (x + I)')

l + x .y =lln 2 I-x

28

.~

vl-x

lr'/2

= In (VSin8COS8) .y 1+2ln8 (x + I)' (x + 2)20

= In

34. y

36.y= ['lntdt

Jv.

Evaluating Integrals

1

~~

-3

39.

J

41.

Jo

2y
['"

43. 45 •

x

sin t 2 - cost dt

[22lnx~

it

2

40.

10 J

42.

Jo

38.

X(lnX)2

3~

3--=2

-1 x

8r dr

4r 2 -5

{'TT/3

44.

x

/, 4~

6tan3x~

J

secx~

VIn (sec x + tanx)

I

t(t

+ I)(t + 2)

56. y 58.

c-.-~-=

= V(x 2 +

y=~t(t~

62

• Y

~

t(t

I I)(t

+

64.y= 8sin8

65

66. y

=xW+! (x

+

3

(x - 2) x 2 +1

67.y=

1)'/3

I)

60. y ~ (tan8)v'29+1

8 +5 63. y = 8cos8

• y

I)(x - I)'

+ 2)

~ (x

=

+ 1)10

+ I)' + I)(x - 2) 68. y = 2 (x + 1)(2x + 3) (2x

3

x(x

a. In (cos x) on [-1f/4, 1f/3], b. cos (In x) on [1/2, 2].

46 •

4 sin (J

1- 4cos8d8

['~

J2

xlnx

/,16 2

> I.

< xifx > I.

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 ~ -1f/4tox ~ 1f/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 ~ 1f/6 to x ~ 1f/2 is revolved about the x-axis to generate a solid. Find the volume of the solid.

Evaluate the integrals in Exercises 37-54.

37.

~

I)

b. Using part (a), show that Inx

32

\Yr

35. y = {'" In vldt

/12

70. a. Provethatf(x) = x -Inxisincressingforx

2

33. y-In

•

Jor

69. Locate and identifY the absolute extreme values of

= I + Int 29 • y I - Int

31. y = In (sec (In 8))

cottdt

Theory and Applications

In (sec 8 + tan8)

I 27. y=ln . ~ xvx + I

54

v'i+3 sin 8

xlnx

Inx 21. y ~ I + Inx

26. y

3

55. y

16. y

'0

1

~/2

Logarithmic Differentiation

14. y = (lnx)3

-t-

52.

secytany 2 + secy
In Exercises 55-<)8, use logarithmic differeotiation to fmd the derivative ofy with respect to the given independent variable.

15. y = t(lnt)'

~

8 2 cot d8

J

~/4

+ 2x

59. y ~

19. y

50.

J2YX~

12. y = In (28 + 2)

Int

x

48.

tan2~

~/2

11. y = In (8 + I) 13. y = Inx 3

= tv'mt

3seo't + 3tant dt

~

2x~

75. The region between the curve y = l/x 2 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 361f. 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 = 2is L =

D 85.

[~I + :,dx.

82. Use a derivative to show that g(x) = Yx'

86. Use the same-derivative argument, as was done to prove Rules 1 and 4 ofTheorem 2, to prove the Quotient Rule property oflogarithms.

+ lox is one-ro-one.

84.

d'y

dx' ~ sec' x,

y(l) ~ 3

D 87.

and y'(O) ~ I

= 2,

a-<1cpeodeot upper bound for ly'I.) Does the graph of y = Vi - lox, x> 0, bave 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 general logarithmic functions, log.,x.

y y = In-1x

or 7

+ sinx) for a

Exponential Functions

7.3 8

a. Grapb y = sin x and the curves y = In (a 4, 8, 20, and 50 together for 0 ,;; x ,;; 23.

b. Why do the curves flatten as a increases? (Hint: Find an

D 88.

y(O) ~ 0

'" x at x ~ O.

c. Graph 10 (I + x) and x togcther for 0 ,;; x,;; O.5.Use differeot colors, ifavailable. At what points does the approximation ofln (I + x) seem best? !.cast good? By reading coordinstes from the graphs, find as good an upper bound for the error as your grapher will allow.

Solve the initial value problems in Exercises 83 and 84. dy I dx ~ I + x'

+ x)

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 ~ I to x ~ 2. Give the coordinates to two decimal places. b. Sketch the region and show the centroid in your sketch. 80. a. Find the ceoter ofmass ofa thin plate of constant deosity covering the region b _ the curve y = I/Vi and the x-axis from x = ltox = 16. b. Find the ceoter of mass if, instead of being constant, the densityfunctionisS(x) = 4/Vi. 81. Use a derivative to show that f(x) = 10 (x' - I) is one-to-one.

83.

The linearization of In (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. a. Derive the linearization 10 (l

D 79.

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 Figure 7.10, lim In-I x = 00

lim In-I x

and

x_-oo

X--i'OO

= O.

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 deimed tu satislY the equation In (e) = I, so e = exp (I). We can raise the number e to a rational power r in the usual algebraic way:

e 2 =e'e,

e-2

=l e 2'

el/2=~,

and so 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 Thegrapbsofy ~ 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' = exp r

for r rational.

cxp i.In-I .

(1)

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 irrational. So Equation (I) 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.

DEFINmON For every real number x, we define the natural exponential function to be eX = exp x. Typical values of

ex

x

eX (rounded)

-I

0.37

o

I

I

2.72

2

7.39

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 JC

e lnx = x

22026 2.6881 X 1043

In(eX)=x

EXAMPLE 1

(all x

> 0)

(all x)

Solve the equation e2>-6 = 4 for x.

Solution We take the natorallogaritbrn of both sides of the equation and use the second inverse equation:

In (e2>-6)

=

In 4

2x-6=ln4 2x = 6

Invorse n:lationship

+ 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)

x

Inverse n:lationship

!L ax In (eX)

= 1

Differentiate both sides.

:x· :Ix (eX)

= I

Eq. (2), Section 72, with

:x eX y

=

=

iT.

u~ ...

Solve for the derivative.

dyJax

That is, for = eX, we find that = eX so the natural exponential function eX is its own derivative. Moreover, if f(x) = eX, then t' (0) = eO = 1. 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_ .du axe-eax'

(2)

7.3 Exponential Functions

EXAMPLE 2

379

We fmd derivatives of the exponential using Equation (2).

(a) iL(Se x) = SiLex = Sex

ax

ax

(b) iLe-x

ax

= e-xiL(-x) = e-C«-I) = -e-C<

d·

ax

.

d

.

(c) _ e smx = eSlDX-(sinx) = eSlDX·cosx

ax

(d)

~(eV3x+l) = =

ax

e V3x +1 • ~(~)

e V3x+l .1(3x + 2

Eq. (2) with"

~

-x

Eq. (2) with"

~

sin.

Eq.(2)with"~ ~ 3 e V3x+l 2~

1)-1/2. 3 =

•

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

e"du = e"

+

C

EXAMPLE 3 ln2

(a)

1 o

e 3x ax =

11n' 0

1 e"·-du 3

u

1

= 3x, 3du = dx.

14(0)

= O.

"(ln2) ~ 3ln2 ~ 1n2' ~ inS

r' e" du

=

1 3Jo

=

1e"]In' 3 0

=t(S-I)=! (b)

Jo("/2 esinxcosxdx = esinx ],,/2 0

Antiderivative from Example 2c

•

= e ' - eO = 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 = eX I 4. (ex,)'

=

e rx" ifr is mtional

Proof of Law 1 Let Yl = eX, and Y2 = eX,. Then Xl = !nYI Xl

+ X2

X2 = !nY2

and

Inverse equations

+ !nY2

=

!nYI

=

!nYIY2

Product Rule for logarithms

Exponentiate.

=

YlY2

• Proof of Law 4

Let Y = (e"y. Then

!ny = !n (ex,y =

r!n(e")

Power Rule for logarithms, mtional r

=

rxl

lne"

=

"with"

= 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 (elnay = 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.

DEFINmON

For any numbers a

> 0 and X,

the exponential function with

base a is

Theorem 3 is also valid for aX, the exponential function with base a. For example, aXl. aX2

= eXtlna. e X2lna

Def'mition of a~

= eXtlna+x2lna

Law I

= 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 n 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 = 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 n, and x n- 1 all exist.

Proof Differentiating x n with respect to x gives Deflnition of x", x

>0

Chain Rule for eO, Eq. (2)

Dcf'mition and derivative of In x

In short, whenever x

For x < 0, if y

> 0,

=

x n, y' , and x n - 1 all exist, then

Inlyl

=

Inlxl n

=

n InIx!-

Using implicit differentiation (which assumes the existence ofthe derivative y') and Equation (4) in Section 7.2, we have

y'

n

y

x'

Solving for the derivative,

y xn y' = n x = n x = nx n-

1

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

Differentiate f(x) = xx, x

o.

>

Solution We cannot apply the power rule here because the exponent is the variable x rather than being a constant value n (rational or irrational). However, from the def"mition ofthe general exponential function we note that f(x) = XX = exlnx , and differentiation gives /,(x) =

~ (e xInX ) dx

=

e xlnx ~ (x In x) dx

=

e

=

XX (in x + I).

Xlnx ( Inx

Eq. (2) with U

~

.In.

+ x·i)

•

.>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. THEOREM 4-The Number. as a Untit limit

e

Proof Iff(x)

=

=

The number e can be calculated as the

lim (I

x-o

+ x)l/x.

Inx,then/,(x) = I/x,so/,(I) = I. But,bythedef"mitionofderivative,

/,(1) = lim f(1 h-O

=

lim In (I

x-o =

lim In (I

x-o

+ h) - f(l)

=

lim f(1

h

x-o

+ x)

- In I = lim

x-o

X

+ x)!/x

=

In [lim(I

x-o

+ x) - f(l) x

lin (I + x)

lnl~O

X

+ x)!/x]

In is continuous;

use Theorem lOin Chapter 2.

Because /,(1) = I, we have In [lim(1

.-0

+ X)!/x]

=

I

Therefore, exponentiating both sides we get lim (I + x)!/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 = e xlna . 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 ofthe general exponential function.

If a > 0 and u is a differentiable function of x, then a U is a differentiable function ofx and

%x aU

=

aUlna :,.

(3)

The integral equivalent ofthis last result gives the general antiderivative

J

aUdu

y~ F E-----'--

2

a>

1.Asx---+oo,wehaveax---+Oif

O 1.As x-+ -00, we have aX -+ 00 if 0 < a < 1 andax-+Oifa> 1.

aU

Ina

+ C.

(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 d -(aX) dx 2

FIGURE 7.11 Exponential functions decrease if 0 < a < 1 and increase if

= -

=

d -(axIna) dx

=

(Inaj2a x

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

We fmd derivatives and integrals using Equations (3) and (4).

(a)

~3x

(b)

~rx = rX(ln3)~(-x)

dx

=

3x ln3

dx

dx

=

- rx ln3

d· . d . (C) dx 3""X = 3,mx(ln3) dx (sinx) = 3,mx(ln3) cosx

(d) (e)

J J

2x dx =

~x2 + C

2 .... cosxdx =

Eq. (3) with a

~

Eq.(3)witha

~ 3,_ ~-.

... ,_

~

2u du =

2sinx

= In2

+

C

~U2 + C

_

Il

~

•

,in.

Eq.(4)witha

J

3, _

~ 2,_ ~.

~ ,in',d_ ~ cosx
•

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 logax.

384

Chapter 7: Transcendental Functions y

DEFINmON

For any positive number a "!' I, log"x is the inverse function of aX.

The graph of y = log"x can be obtained by reflecting the graph of y = aX across the

~/~//~/-/-//-/"O*(--i'1---;'2~~~-~-X

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 loBax

=

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 natura110garithms ... . .. divided by In. ... . .. gives Rule 1 for base a logarithms.

Quotient Rule: log" ~ = log" x - log" y

Derivatives and Integrals Involving log. x

Reciprocal Rule:

To rmd 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) = Ina'u tU'

d I du -(Iog"u) = -I , tU In a u tU

7.3

Exponential Functions

385

EXAMPLE 6 I

I

(8)

axd loglO(3x +

(b)

r°'6)X ax = ~zJ ~x ax =

axd (3x +

I) = In 10 • 3x + I

log,x

~zJUdU

3

I) = (In 1O)(3x + I)

~ :;

u~lnx, du~itU

I u2 I (lnx)' (In x)' =lnzT+C=lnz-z-+ C = ZlnZ +C

•

Exercises 7.3

1. a.

e--()·3t =

b.

27

2. a. • -
35.

e kl =

b • • tt

1

c.

2

e(lnO.2)t =

= ...L e 10'

.(ln2)t

0.4

= 12

37. 39.

et

41.

Finding Derivatives In Exercises 5-24, rmd the derivative ofy with respect to x, I, or 9, as appropriate. 5. y ~ .-;, 6. Y ~ .2%/3 8. y = e(4v'i+x2) 7. Y = e5 - 7:r:

17.

+ 2).x + cos 9)

= .B(sin 9 Y = cos (.....') Y = 10 (310--<)

19. Y 21. y

23. y

+ 2x).-2% = (9x 2 - fu + 2)." = 10 (39.- B) = 0'.-28 cos 59 = 10 (2.-' sin I)

10. y ~ (I

9.y=xe x -e x 11. y = (x 2 - 2x 15.

12. Y 14. Y 16. Y 18. Y

= 10 (----"'-) 1 + eO = e(cost+1nt) ['"x = Jo sin.'dl

Ioe

= :~) = ."'''(lnI 2 + I)

20. Y 22. y 24. Y

=

In Exercises 25-28, find dyJ dx.

= eY sinx 27•• 2% ~ sin (x + 3y) 25. loy

l.

e4Vi.

Ioldl

26. In xy = e X +Y 28. !any

~

.x + lox

Finding Integrals

31.

f(." 5.~) l .x f +

in,

dx

dx

30.

8."+1) dx

-

-1n2

34.

21 .-t' dl

40.

.'/x 7dx

42.

44.

f f f

J.

-V;

'y;. dr 1'.(") dl

.-l/x' ----;>dx ~j2

+ .""B) esc2 9 d9

(I

~/4

45. fescc'lTtsec7Tttanwtdt

46.

47.

f .~o(~+t) l

ln(~j2)

esc (1r

+ I) cot (1r + I) dl

2evcosevdv

48.

In ('11'/6)

49.

0

f ·' I

j,v'];;;,: 2xeX2cos(eX2)dx

+ .' dr

50.

f

I + dx

.x

Initial Value Problems Solve the initial value problems in Exercises 51-54. 51. 52.

:it = :it =

.'sin(.' - 2), y(102)

.--<see2 (1r.-t ), y(104) = 2/1r

d 2y 53. dx 2 = 2.-x, y(O) = I d 2y

-2

dl

=0

= I -

.2t, y(l)

and y'(O)

= -I

=0

and y'(I) = 0

3.-2%) dx

32..~ dx

1n2

33.

(2.x

38.

r /4 (I + .""'B) sec' 9 d9

54.

f 10 f

y;.dr

4

.4 dx

36. Jo

43. Jo

'b

Evaluate the integrals in Exercises 29-50.

29.

l ·v; f f f

.xj2 tU

1n4

4. e(x~e(2x+1) =

13. y

[In 16

in,

Solving Exponential Equations In Exercises 1-4, solve for I.

2.(2%-1) dx

Differentiation In Exercises 55-82, rmd the derivative of Y with respect to the given independent variable.

55.y=2' 57. Y = 5V;

56.y=3~

58. Y =

2(")

386

Chapter 7: Transcendental Functions

59. y = x 7T

60. y = t 1-

61. y = (cos O)V2

62. y = (In O)7T

63. y = 7sec (J In 7 65. y = 2 sin 3t

64. y = 3tan (J In 3

67. y = log 2 50

68. y = log3(1

66. y =

+ log4x2

69. y = log4x

71. y = x 10glO x

111. y = (x

5-cOS 2t

+

Oln3)

72. y = log3 r · logg r

U

((~ ~

IOg3

In Exercises 111-118, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.

70. y = log2s eX - 10gsvX

3

73. y =

Logarithmic Differentiation

e

( 3x + 2

~)lnS

1n3 )

74. y = logs

76

75. y = 0 sin (lOg7 0)

= 1

• Y

Og7

0 0)

(sin cos e(J2(J

x

84.

{I 2-(J dO

91.

86.

h

Jo

7

1 J

a. Find all absolute extreme values for f.

3 .:.X 3x dx

b. Find all inflection points for f. eX 122. Let f(x) = - - 2. 1+ e X

(O 5-(J dO

L2

~dx

4

2

a. Find all absolute extreme values for f.

dx

88.1

sin t dt

(l)tan 90. Jo "3 sec t dt

)

cost

4 x 2x (1

{7T/4

+ Inx) dx

92.

t

J

2

3x V3 dx

97. 99.

94.

3

(y!2 + l)xv'2dx

o 124. Graph f(x) = (x -

3 )2ex 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.

J 1 1

10glO x

98.

0

x

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 .

e

x(In2)-Idx

1 1 1

126. Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = e X / 2 , below by the curve y = e -x/2 , and on the right by the line x = 2 In 2 .

log2 x

-x-dx

e

dx

100.

21n 10 IOg10 x

x

127. Find a curve through the origin in the xy-plane whose length dx

1

10

210g2(X + 2) x +2 dx

{g 210g lO (x

+

102.

J

dx xloglO x

log 10 (lOx)

x

104. i2 106.

x-I

dx

J

dx x(loggx)2

Evaluate the integrals in Exercises 107-110. {lnx 1 {eX 1 107. il t dt , x > 1 108. il t dt

109.

1x /

1 1

t1 dt, x > 0

110. -1-1

na

1 x

1

1)1 1

{3 210g 2 (x - 1)

dx

from x = 0 to x = 1 is

L =

dx

1/10

1)

x + 1

103. io 105.

125. Find the area of the "triangular" region in the first quadrant that

1

In 210g2x

1

101.

x 0.-1 dx

4

-x-dx

4

J

96.1

b. Find all inflection points for f.

123. Find the absolute maximum value of f(x) = x 2 1n (l/x) and say where it is assumed.

x2

2 _x_dx 2 1 + 2x

Evaluate the integrals in Exercises 93-106.

9s.1

120. Where does the periodic function f(x) = 2e sin (x/2) take on its

121. Let f(x) = xe-x.

J

2

93.

119. Find the absolute maximum and minimum values of f(x) = eX - 2x on [0, 1].

o

87.10. x2(x 89.

Theory and AppLications

--------------If----------~x

5 dx

{7T/2

118. y = (lnx)lnx

82. y = t log3 (e(Sin t)(ln 3))

Evaluate the integrals in Exercises 83-92.

85.

115. y = (sinx)X 117. y = sinx x

y

Integration

J

114. y = tv! 116. y = x sinx

80. y = 3 logg (lOg2 t)

3log2t

81. y = log2 (8t ln2 )

83.

(vty

extreme values and what are these values?

05(J 78. y = 2 - 1ogs 0

79. y =

113. y =

112. y = x 2 + x 2x

+ I)X

t1 dt, x > 0

+ ±eXdx.

128. Find the area of the surface generated by revolving the curve x = (e Y + e-Y )/2, 0 ~ y ~ In2, about they-axis.

t

Y

X=

eY + e-Y --2-

In2[ x

7.4

137. Find the area of the region between the curve y ~ 2x/(I and the interval -2 ~ x ::s; 2 of the x-axis.

In Exercises 129--132, find the length of each curve.

129. Y = k(e' +

e~)fromx

130. y = In(e' - I) -In(e'

= Otox = I

+

o 139. The equation x

131. y = In (cos x) from x = Otox = 'IT/4

o 140. Could x

possibly be the same as 21n ' for x two functions and explain what you see.

b. Find the average value oflnx over [I, e]. ~

~

I/x on [I, 2]. ~

1

+ x at x =

~ 2' at x coefficients to two decinzal places.

a. Find the linearization of f(x)

O.

o b. Estimate to five decimal places the magnitude of the error involved in replacing e' by I + x on the interval [0, 0.2]. o c. Graph e' and I + x together for - 2 x 2. Use different :5

x:5

:5

a. Show that the graph of eX is concave up over every interval ofx-values. b. Show, by reference to the accompanying figure, that if 0< a < hthen

J.

e'dx < e

In.

lna

+e

lnb

'(lnh -Ina).

2

~

O. Then round its

:5 x:5

142. The linearization 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

136. The geometric, logarithmic, and arithmetic mean inequality

lnb

> 01 Graph the

o b. Graph the linearization and function together for -3:5 3and-l 1.

colors, if availahle. On what intervals does the approximation appear to overestimate eX? Underestimate eX?

e(ln.+lnb)/2'(lnh -Ina) <

ln2

141. The linearization oUx

0

a. Derive the linear approximation eX

2

= 2' bas three solutions: x = 2, x = 4, and one other. Estimate the third solution as accurately as you can by graphing.

132. y = In (cscx) from x = 'IT/6 to x = 'IT/4 133. a. ShowthatJlnxdx ~xlnx -x + C.

135. The linearization of eX at x

+ x 2)

138. Find the area of the region between the curve y = 2 1- ' and the interval -1 :5 x :5 1 ofthe x-axis.

I) from x = 1n2tox = 1n3

134. Find the average value of f(x)

387

Exponential Change and Separable Oifferential Equations

o

window

O:5x:5 8and2:5x :54. 143. Which is bigger, 'IT' or eW? Calculators have taken some 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.

a. Find an equation for the line through the origin tangent to the graph of y = In x.

E

B'

[-3,6] by [-3, 31

I

A'

_-=I_ _~_LI~--~:::-__ 'X

Ina+lnb 2

Ina

c. Show that In (x')

Nor TO SCALE

c. Use the inequality in part (b) to conclude that

.rc

h-a a+h vah < Inh - Ina < - 2 - '

This inequality says that the geometric mean of two positive

numbers is less than their logarithmic mean, which in turn is less than their arithmetic mean.

7.4

b. Give an argoment hased on the grapha of y = In x and the tangent line to explain why In x < x/e for all positive x oF e.

< x for all positive x

oF e.

d. Conclude that xt' < eX for all positive x =F e. e. So which is bigger, "r or e""?

o 144. A decimal representation of e

Find e to as many decimal places as your calculator allows by solving 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 witb changes in tbe independent variable. They describe growth or decay in many natural and industrial situations. The variety of models based on tbese functions partly accounts for tbeir importance. We now investigate tbe 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 t. 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 t = 0 is called Yo, then we can find y as a function of t by solving the following initial value problem:

dy dt

Differential equation:

=

y = Yo

Initial condition:

(la)

ky when

t = O.

(lb)

Ify is positive and increasing, then k is positive, and we use Equation (Ia) to say that the rate of growth is proportional to what has already been accumulated. Ify 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'dt=k

J~:dt=Jkdt In

Iyl Iyl Iyl y

= = = =

y =

kt + C e kHC eC'e k' ±eCekt Ae kt •

Integrate with respect to t;

J(I/u)du ~ In lui

+ c.

Exponentiate.

If Iyl ~

r, theny

~

±r.

A is a shorter name for ±e c .

By allowing A to tske 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

t = 0:

Yo

=

Ae k ' O = A.

The solution ofthe initial value problem is therefore y = yoe kt •

(2)

Quantities changing in this way are said to undergo exponential growth if k > 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 ofthe form dy/dx = ky for some nonzero constant k. More generally, suppose we have a differential equation ofthe form

dy dx = f(x,y),

(3)

7.4 Exponential Change and Separable Differential Equations

389

where j 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) = j(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 j can be expressed as a product of a function of x and a function of y. The differential equation then has the fann dy dx = g(x)H(y).

g is a function of:t; His a function of y.

When we rewrite this equation in the form dy dx

g(x) h(y) ,

1 H(y) ~ hey)

its differential fann 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 ofthis 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

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

=

> -1.

+ y)e"

(I + y)e" dx

~=e"dx 1+ Y 1+ Y J ~=Je"dx

In(l+y)=e"+C The last equation gives y as an implicit function of x.

Treat dyldx 88 a quotient of

diffi:rentials and multiply both sides by dx. Divide by (1

+ y).

Integrafe both sides. C represents the combined

constants of integration.

•

390

Chapter 7: Transcendental Functions

EXAMPLE 2 Solution

Solve the equation y(x

+

I):

= X(y2

+

I).

We change to differential form, separate the variables, and integrate:

y(x

+

I)dy = X(y2

ydy y2+1

+

I)dx

xdx x+1

- -

- -

x =F

1 J ~-J (1 __ l+y2-

tln(1

+ y2)

x+1

=x -lnlx

)dx

-1

Dividcxbyx + 1.

+ 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 oftime 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 asswne 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 dy/dt is the birth rate minus the death rate, which is the difference of the two proportioualities under our asswnptions. In other words, dy/ 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 dy/dt = ky models unlimitedpopulation 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/ 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 oft, 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

y = 1O,000e kt •

(5)

7.4

Exponential Change and Separable Differential Equations

391

The value of k. When t = I year, the number of cases will be 80% of its present value, or 8000. Hence, 8000 = 1O,000e.\(1)

e k = 0.8 In (e k ) = In 0.8 k= 1n0.8

Eq. (5) with t ~ I and y ~ 8000 Logs ofboth 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 O.'ll 0.1

e(lno.')1 =

(lnO.8)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 ca,e, 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 ofintermediate 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 = - ky, k > O. It is conventional to use -k, with k > 0, to emphasize thaty is decreasing. If Yo is the number ofradioactive nuclei present at time zero, the number still present at any later time t will be

k> 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 =

e-kl = -kt

!YO

2

!

2

= In! = -In2

2

t = 1n2 k

R
392

Chapter 7: Transcendental Functions

Half-life =

~2

(7)

For example, the half-life for radon-222 is half-life =

~I~

"" 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 caroon-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% ofthe 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-kf = 0.9 e -(!n2/S700)f = 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 ofCooling, although it applies to warming as well. If H is the temperature ofthe object at time t and Hs is the constant surrounding temperatore, then the differential equation is dH (it = -k(H - Hsl·

(8)

7.4 Exponential Change and Separable Differential Equations

393

If we substitute y for (H - Hs ), then dy d dB dt = dt (H - Hs) = at

-

d dt (Hs)

dH _ 0 dt dB dt = -k(H- Hs) = -kyo =

H s is a constant.

Eq. (8) H-Hs~Y

Now we know that the solution of dy/dt = -ky is Y = yoe-k', where y(O) = Yo. Substituting (H - Hs) for y, this says that

H - Hs = (Ho - Hs)e-k',

(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-k ,.

To find k, we use the information that H = 38 when t = 5: 38 = 18 e

-5k

-5k k

=

t

+ 80e-5k

1

="4

ln 1 = -ln4 4

=

ln4 = 0.2ln4

The egg's temperature at time tis H = 18

+

(about 0.28).

80e-(O.2ln4),. Now rmd the time t when

H= 20:

20 = 18

+ 80e-(O.2ln4)'

80e-(O.2ln4)' = 2 e-(O.2ln4), = ~ 40 1 -(0.2ln4)t = In 40 = -ln40

t

=

ln40 13' 0.2ln4 "" mm.

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 fuoction y = j(x) is a solution of the

accompanying differential equation. 1. 2y' + 3y = e~ a. y = e" b. y = e"

+ e-{3/2~

c. y = e ~ + Ce-Of2" 2. y' = y2 I I h. y = - - a. y =-Xx+3 I 3. Y = XlI Tdt, x 2y' + xy = eX

I c. y=-x+C

ret I

4. y =

'\I'l+7

l'

v'1+t'dt, y'

I

+ -2x' -y I + x'

present size?

= I

c. What will be our desceodaots' tooili size 20,000 years from

In Exercises 5-8, show that each function is a solution of the given

initial value problem.

S.y'+y~

I

Initial condition

Solution

+ 4e 2r

y(-In2) ~ 'If.. 2

y ~ e-'tan- 1(2e')

= e-.' -

6. y'

2xy

y(2) = 0

y = (x - 2)e-r'

7. xy'

y(f)

y=-x-

2

+ Y = -sinx,

candidate

cosx

= 0

x"y' ~ xy X> I

y2,

y(e)

~

e

11.

~=

13.

~=

15.

y;, ~

e'-Y 2

vYcos vY

the rate at which the amount of a substance changes with time is

proportional to the amount preseot For the chaage of B-giucono lactoue into gluconic acid, for example,

dy _ 2. r 10. fix - x vy, y > 0

dy

dt

12.

~=

14.

~~=

3x

2

e"'"

dy.,.-------, 17. fix - 2x V I - y-,

I

21. }

3x"y' -

-I < y < I

6>;2

~ = yer' + 2vY er'

dy 20. fix=xy+3x-2y-6

27. Working underwater The intensity L(x) oflightx feet beoeaili the surface of the ocean satisfies the differential equation

tiL

22. :

= eX - Y

+ eX +

-0.6y

called ''inversion'' that changes the sugar's molecular s1ructW'e. Once the process has begun, lhe rate of chaage of lhe amount of raw sugar is proportional to the amount of raw sugar remaining. If 1000 kg ofraw sugar reduces to 800 kg ofraw sugar during lhe IlrSt 10 hours, how much raw sugar will remain after aoother 14 hours?

dy e 2r - y 18. dx = eX+Y 19. y2 ~ =

~

wheo t is measured in hours. If there are 100 grams of B-giucono lactoue present wheo t = 0, how many grams will be left after ilie first hour? 26. The inversion of sngar The processing of raw sugar has a step

x> 0 16. (secx) ~ = e Y+'"

= eY+v',;,

Differential equation: dp/dh = kp (k a constant) Initial condition: p = Po wheo h = 0

c. At what altitode does ilie pressure equal 900 millibars? 25. Fint-order chemical reactions In some chemical reactions,

Separable Differential Equations Solve the differential equation in Exercises g....22. • ,.. dy _ 9. 2 vxy fix - I, x,y > 0

now (as a perceotage of our preseot tooili size)? 24. Atmospheric pressure The earth's atmospheric pressme p is ofteo modeled by assuruing that lhe rate dp/dh at which p chaages wiili the altitode h above sea level is proportional to p. Suppose that ilie pressure at sea level is 1013 millibars (about 14.7 pounds per square inch) aod that the pressure at ao altitode of20 km is 90 millibars. &. Solve lhe initial value problem

to express p in terms of h. Determine ilie values of po aod k from ilie giveo altitude-pressure data. b. What is the atmospheric pressure at h ~ 50 km?

x>O 8.

value of k in the equation y = yoe lt . Then use this value of k

to aoswer the following questions. b. In about how maoy years will human teeth be 90% of their

Initial value Prohlems

Differential equation

23. Human ....Iution continues The aoalysis of tooth shrinkage by C. Loring Brace aod colleagues at the University of Michigao's Museom ofAothropology indicates that human tooth size is continuing to decrease aod that ilie 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. a. If t represents time in years and y represents tooth size, use theconditionthaty ~ 0.99yowheot ~ 1000 to find the

e-Y

+1

Applications and Examples D The answers to most of the following exercises are in terms of logarithms aod exponeotials. A calculator cao be helpful, enabling you to express the aoswers in decimal form.

fix

~

-kL.

As a diver, you know from experieoce that diving to 18 ft in ilie Caribbean Sea cuts ilie intensity in half. You csunot work wiiliout artificial light wheo the intensity falls below on.,..tenth of the surface value. About how deep cao you expect to work wiiliout artificiallight?

7.4 28. Voltage in a discharging capacitor Suppose that electricity is draining 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. Cholera hacteria Suppose that the bactetia in a colony can grow unchecked, by the law of exponential change. The colony starts with I bacterium and doubles every balf-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 an infected perliOll, many bacteris are destroyed, but this example helps explain why a person who 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? 31. The incidence of a disea.e (Continuation ofExample 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 ofcases to less than I? 32. The U.S. population The U.S. Census Bureau keeps a running clock totaling the U.S. population. On March 26, 2008, the tota1 was increasing at the rate of I perlion 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 constant rate, fmd the rate constant for the population's growth (people per 365-day year). b. At this rate, what will the U.S. population be at 2:31 P.M. EST on March 26, 2015? 33. Oil depletion Suppose the amount of oil pumped from one ofthe canyon wells in Whittier, California, decreases at the continuous 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 ordeni, your firm's .ale. department applies a continuous discount that makes the unit price a functionp(x) of the numher 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 unit price P(90) for a 90-unit order. c. The sales department has asked you to find out if it is discounting so much that the fInD's revenue, r(x) ~ x· p(x) , will actually be less for a 100-unit order than, say, for a

Exponential Change and Separable Differential Equations

395

90-unit order. Reassure them by showing that r has its maximum value atx ~ 100. d. Graph the revenue function r(x)

~

xp(x) for 0 :5 x :5 200.

35. Plutooium-239 The half-life of the plutonium 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 half-life of polonium is 139 days, but your sample will not be useful to you after 95% of the radioactive nuclei present 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. The mean life of a radioactive 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 mean life ofa carbon-14 nucleus is more than 8000 years. Show that 95% ofthe radioactive nuclei origiually present in a sample will disintegrate within three mean lifetimes, i.e., by time t = 3/k. Thus, the mean life ofanucleus gives a quick way to estimate how long the radioactivity of a sample will last. 38. Californium-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 westem world since its discovery by Glenn Seaburg in 1950. The balf-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 i. the value of k in the decay equation for this isotope? b. What is the isotope's mean life? (See Exercise 37.) c. How long will it take 95% of a sample's radioactive nuclei to disintegrate? 39. Cooling .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 answer the following questions. a. How much longer would it take the soup to cool to 35'C? b. Instead ofbeing left to stand in the room, the cup of90'C soup is put in a freezer whose temperature is -15°C. How long will it take the .oup 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 temperliturewas 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. Silver cooling in air The temperliture of an ingot of silver is 60'C above room temperliture right now. Twenty minutes ago, it

was 70D e above room temperature. How far above room temperature will the silver be

a. 15 min from now?

b. 2 hours from now?

c. When will the silver be 1DoC 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 contained 44.5% of the carbon-14 found in living matter. About how old is Crater Lake? 44. The sensitivity of earhon-14 dating to measurement To see the effect ofa relatively small error in the estimate ofthe amount of carhon-14 in a sample heing dated, consider this hypothetical situation:

a. A fossilized bone found in centrallllinois in the year A.D. 2000 contains 17% ofits original carhon-14 content. Estimate the year the animal died.

45. Carhon-14 The oldest known frozen humao mummy, discovered in the Schnalstal glacier of the Italian Alps in 1991 and called Olzi, was found wearing straw shoes and a leather coat with goat fur, and holdiog a copper ax and stone dagger. It was estimated that Otzi died 5000 years hefore he was discovered in the melting glacier. How much of the origins! carhon-14 remained in Otzi at the time of his discovery? 46. Art forgery A painting attrihuted to Vermeer (1632-1675), which should contain no more than 96.2% of its originsl carhon14, contains 99.5% instead. Ahouthow old is the forgery?

I>. Repeat part (a) assuming 18% instead of 17%.

e. 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'H6pitai (1661-1704) Johano Bernoulli (1667-1748)

John (Johann) 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'Hopital" Rule, after Guillaume de I'HOpitai. He was a French nobleman who wrote the first introductory differential calculus text, where the rule f"ITSt 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 cannot apply the Quotient Rille 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 undef"med. 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 Rille, 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 cannot evaluate. We use % as a notation for an expression known as an indeterminate form. Other meaningless expressions often occur, such as 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 manipillations. 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

'( ) = li f(x) - f(a) f a m x 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:Hopital's Rule enables us to draw on our success with derivatives to evaluate limits that otherwise lead to indetenninate fonns. THEOREM 5- L'H6pital's Rule Supposetbatf(a) = g(a) = 0, that f and g are differentiable on an open interval I containing a, and that g' (x) oF 0 on fif x oF a. Then lim f(x) g(x)

=

x~.

lim f'(x) g'(x) ,

x~.

assuming that the limit on the right side ofthis equation exists. We give a proof ofTheorem 5 at the end of this section.

I Caution

To apply I'H6pital's Rille to Ilg, divide the derivative of I by the derivative ofg. Do not fall into the trap oftaking the detivative of IIg. The quotient to use is ng', not <JIg)'.

EXAMPLE 1 The following limits involve % indeterminate forms, so we apply I'H5pital's Rule. In some cases, it must be applied repeatedly. (a) lim 3x - sinx = lim 3 - cosx = 3 - cos x l

x-o

x

x-o

1

1

= 2

x=o

1 (b) lim

x-o

(c) lim

yIJh - I

=

x

yIJh -

lim 2y1Jh

x-o

1

I 2

1 - x/2

o

2

x-o

o

X

=

=

xr'/ x-o 2x . -(1/4)(1 + xr lim lim (1/2)(1 +

2

-

32 /

x-a

2

1/2 Still

I 8

=--

Not~; limit is found

o

(d) lim x - sinx

x-o

~; differentiate again.

o

x3

=

lim I - cosx

x-o =

3x 2

lim sinx

x-o 6x

Not~; limit is found. Here is a summary of the procedure we followed in Example I. Using UH6pital's Rule

Tofmd lim f(x) g(x)

x~.

by I'HOpital's Rule, continue to differentiate f 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:H5pital'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+x 2 =

. 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'HOpitai '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 inde• terminate 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 oc and + oc mean the same thing.

Positive for x

o

. sin x (b) Iun-2 .%-0-

o

x

=

> 0

lim cosx x-o- 2x

Indetenninate Fonns

= -00

Negative for x < 0

OI:J / OI:J, OI:J •

0,

OI:J -

•

OI: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 we11 as to 0/0. Iff(x) --+ ±oo andg(x) --+ ±oo asx --+ a, then lim fix) = lim f'(x) g(x) x~. g'(x)

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
lim x-1T/2

Find the limits ofthese 00/00 forms:


secx

1

+ tanx

•

eX


x

Solution


secx

+ tanx

:

lim

from the left

x~("/2r

secxtanx seel x

lim

x~("/2)-

sinx = I

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 1. liz

I

I/~

~

• 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 ofhow we might work with these indeterminate forms.

EXAMPLE 5

Find the limits ofthese

00 • 0

forms:

Solution (8)

lim

X~OO

(x sin i)

=

(-hI sin h) =

lim

h-O+

lim sinh = I

QO'O;Leth

h

11-0+

00 • 0

=

lim x-O+

=

EXAMPLE 6

lim

x-o+

I/x -1/2x 3/ 2

(-2\1'.;)

liz.

converted to

00/00

I'H8pital's Rule

•

= 0

Find the limit ofthis

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

o o

lim x - .sinx

x-a x smx

= lim

I - cosx

Still()

o

x-a sinx + xcosx sin x x_o2cosx-xsinx

= lim

=

Q= 0 2

.

•

400

Chapter 7: Transcendental Functions

Indetenninate Powers Limits that lead to the indeterminate fonns 100, 0°, and 00 0 can sometimes be bandled by first taking the logarithm of the function. We use I'Hllpitai'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

Applyl'Hllpital'sRuletoshowthatlimx~o+(1

+ 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)'/X

+ x)'/X

and

i In (I + x),

=

I'Hllpital's Rule now applies to give

+ x)

lim In j(x) = lim In (I x-+O+

x--+o+

0 0

X

I 1 + x Iim - x--+o+ 1

=+=1. Therefore, lim (I

x-o+

EXAMPLE 8

+ x)l/x =

lim j(x)

x-o+

= lim

x-o+

elnfe-)

= el =

e.

•

Find limx~oo x'/x.

Solution

The limit leads to the indeterminate fonn lim_oo In j(x). Since

00°.

We let j(x) = xl/x and f"md

~x,

Inj(x) = Inxl/x =

I'Hllpital's Rule gives lim In j(x) = lim Inx

x_co

x--+OO

x-+oo

= lim x---+OO

j(x)

= lim

x-co

00

1

¥

o.

elnf(x)

=

=

x---+OO

00

I/x lim-

=

Therefore lim xl/x

X

=

eO

= 1.

•

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, hI and differentiable throughout (a, h) and also suppose g'(x) oF 0 throughout (a, h). Then there exists a number c in (a, h) at which

I'(c) g'(c)

f(h) - f(a) g(h) - 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(h). For ifg(h) did equalg(a), then the Mean Value Theorem would give g'(c)

=

g(h) - g(a) h-a

= 0

for some c between a and h, which cannot happen because g'(x) oF 0 in (a, h). We next apply the Mean Value Theorem to the function

F(x)

=

f(h) - f(a) f(x) - f(a) - g(h) _ g(a) [g(x) - g(a)].

This function is continuous and differentiable where f and g are, and F(h) = F(a) = O. Therefore, there is a number c between a and h for which F' (c) = O. When expressed in terms of f and g, this equation becomes

F'(c)

=

I'(c) -

:~:~ =~~~ [g'(c)1 =

0

so that

I'(c) g'(c)

y f'(c)

slope ~,«c)

p

-""""

A (g(a).j(a)) -;;i-----------~x

o

FIGURE 7.13 There is at least one point P on lhe curve C for which lhe slope oflhe tangent to the curve at P is the same as the slope oflhe secaot lioe joioiog lhe poiots A(g(a), f(a)) aodB(g(b), f(b)).

f(h) - f(a) g(h) - 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),f(a)) and B = (g(h), f(h)). 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 I' /g' evaluated at the number c in the interval (a, h), as asserted in Theorem 6. Because the slope of the secant line joining A andB is

f(h) - f(a) g(h) - 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 I'(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)

f(x)

g'(c)

g(x)"

As x approaches a, c approaches a because it always lies between a and x. Therefore, lim f(x) = lim j'(c) =

.r-a+ g(x)

c-a+ g'(c)

lim j'(x)

.r-a+ g'(x) ,

which establishes I'Hllpital'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, a], x < a. •

Exercises 7.5 FimHng Unrits in Two Ways In Exercises l--{j, use I'Hllpital's Rule to evaluare the limit. Then evaluare the limit using a method studied in Chapter 2,

+2

1. lim x

.x_-2x 2 -

x-o

x

4. lim x' - I x-14x 3 - X - 3

5. lim I - cosx

6. lim

2x

2

+ 3x

In(x 2 + 2x) In %-0+ x Y5y

7. lim x-2 .r_2x 2 -4 9. lim I' - 41 + 15 1--3 t 2 - t - 12 11. lim 5x' - 2x x_OO 7x 3 + 3 ,

2

13. lim sm I t

1-0

8x 2 15. lim=~. x-o cosX' 17 lim 20 - 'If · '~w/2 cos (2'1f 0) 19

lim

• 6-1T/2

I - sinO 1 + cos 26 2

21 lim x • x~o In (sec x) 1(1 - cos I) 23. lim . 1-0

25.

lim

(x - ~2) sec x

' 3"'" - I 27. lun 0

lim x -25 %--5 x+S

10. lim

39

•

41

r

2

12. lim x - 8x %_00 t2x 2 + 5x

+

(lnX)2

~-a

I»

y

lim (_1 x-I

1_) Inx

,

(3X + I__,1_) smx X

42. lim (cscx - cotx %-0+

+ cosx)

43. lim cos 0 - I 8_0e fJ - 8-1

+

t

2

45. lim _e_1_ 14. lim sin 51 1-0 2t

1_00

,

r 30 + 'If . '~'!!;/' sin (0 + ('If/3»

.1

x_1Inx - SIn 7TX

22.

lim

(eX - 1)2 .

x-o x smx

18

x -

t

48. lun

16. lim sinX' - x x-o x 3

20. lim

e' -

49

·

tanO -

0

r

SO. x~

sin3x-3x+x 2 sin x sin 2x

Indeterminate Powers and Products Find lhe limits in Exercise 51--{j6. 51. lim xl/(l-x) 52. lim x!/(>-l)

In (esc x)

x~w/2 (x - ('If/2l)2

lim 0 - sinO cosO ,~o

x-I+

%-1+

53. lim (lnx)l/x

54. lim (In x) 1/(>-.) x-e+

x~oo

24. lim

I sin I 1_01 - cost

56. lim.

X IJlu

X~OO

26.

lim

x-(n/2)-

28. ,~o lim

("-2 -

(1/2)' - I 6

x) tan x

57. lim (1

+ 2x)I/(2lnx)

58. lim (eX x~o

x~oo

59.

li.s x'

x~o

a>O

38. lim (lnx - In sin x) x-o+ 40. lim x-o+

x~+ In (sinx)

• %-1+

t- 3

36. lim

y~O

X~OO

I' - I

1-14t 3 -

(-SInt

x-(7rf2)-

.~o

8.

2

Inx

%-0+

y

37. lim (In 2x - In(x

Applying l'H6pital's Rule Use I'Hllpital's rule to fmd the limits in Exercises 7-50.

34. lim In (eX - I)

+ 25 - 5

y~O

~oox3+x+l

x2

lo~x

33. lim 35. lim

-

32 r log, x • x~log,(x + 3)

31. lim In(x + I) %_00

3. lim 5x - 3x x_OO 7x 2 + 1 ~o

x-o~

2. lim sin 5x

4

2

, 3x - I 30. lun~1

60.

lim

%-0+

+ x)!/x

(1 + l)X X

7.5 Indeterminate Forms and rH6pitars Rule

61. lim %_00

63.

65.

(Xx+- I2)'

lim

62.

2

lim x 1nx

x_oo

X)

lim xtan(" x-o+ 2

X

lim sin x oinx

66.

X~

...

x_ OO

69.

v.;-+I

lim cotx escx

70

lim sec x x-(7ff2t tanx

o 81. oc -

oc Form

a. Estimate the value of X~OO

=x

hy graphing f(x) val ofx-values.

• %-0+

. ZX - 3x 71. x~3x + 4%

82. Find lim

x~oo

(W+i - -...h).

o 83. 0/0 Form

i' 73• x~xex I· e

Estimate the value of lim 2x' - (3x

75. Which one is correct, and which one is wrong? Give reasons for

your answers.

=

b. lim x-3 %-3 x 2 - 3

=Q=o 6

and the limit

. (1+"I)X =e.

your answers.

lim

a. lim x '-2x x_ox2 - sinx

2x ~lim-

x' -

X~OO

2

a. Use I'H6pital's Rule to show that

x-O 2x - cosX'

= lim

lim x_o2

2

+ sin x

2x = lim 2x - 2 sinx x-O 2x cosX'

-2

d.

lim Inx

-00

= x~o+ (I/x) = ()O = -

Db. Graph

f(x)

~

(I +

1

x~"'(1

=

lim x-O+

lim (I +

k--+oo

x)

(I/x)

(-l/x 2)

=

lim (-x)

=0

x-o+

78. Find all values of c thst satisfY the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.

a. f(x) ~ x, g(x) ~ x', (a, b) ~ (-2,0) b. f(x) ~ x, g(x) ~ x', (a, b) arbitrary c. f(x) = x'/3 - 4x, g(x) = x', (a, b) = (0,3) 79. Continuous extension Find a value ofc thst makes the function f(x) =

and g(x)

~

(I

+~)'

85. Show thst

lim x In x ~ lim In/x

X~...

:S

toge1her for x '" O. How does the behavior off compare with thst ofg? Estimate the value oflim~oo f(x). c. Confmn your estimate oflim~oo f(x) by calculatiog it with I'Hllpital's Rule.

b. lim xlnx = 0'(-00) = -00 x-o+ x

X~OO

0 - 1 = 2

%-0+

In

lim

2+0

others wrong? Give reasons for your answers. a. lim xlnx ~ 0·(-00) ~ 0

r

. (1+"I)X =e.

= _2_ = I

77. Only one of these calculations is correct Which one? Why are the

c. x-'?J+ X

I)-...h + 2

X

76. Which one is correct, and which one is wrong? Give reasons for

x_ox2

+

1 hy graphing. Then confmn your estimate with I'Hllpital's Rule. x-I

84. This exercise explores the difference between the limit

lim ~=l x-3 2x 6

a. lim x-3 x-3 x 2 - 3

b.

- Yx' + x over a suitably large inter-

b. Now confirm your estimate hy fmding the limit with I'Hllpital's Rule. Ail the fIrSt step, multiply f(x) hy the fraction (x + ~)/(x + Yx' + x) and simplifY the new numerator.

68. lim -...h x-o+ Ysinx

'\!9x+1

bX) = O?

lim (tan 2x + "- + sin x-a x3 x2 X

lim (x-~)

Theory and Applications I:Hllpital'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

80. For what values of a and b is

x+2

lim (In x)' x-o+

64.

x-o+

(X' + I )'IX

403

9X - 3sin3x, x ~ 0 5x' { c, x= 0

continuous at x = O. Explain why your value of c works.

86. Given that x a. Xl/X b. xiii'

1')' k

=

e'.

> 0, fmd the maximum value, if any, of

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.

(I)

a. y~xtan "

2x

3x + e b. y~ 2x+e'"

e-Vi' 88. Findf'(O) for f(x) ~ { 0, '

x ~ 0 x = O.

404

D 89.

Chapter 7: Transcendental Functions

D 90.

The continuons extension of(sinxf to [0, 'IT]

The function (sin x)...• (Continuation ofExercise 89.)

= (sinx)""'on1heinterva1-7:5X:5 7. How do you account for the gaps in the graph? How wide are the gaps?

a. Graphf(x) ~ (sinxf on the interval 0 :5 x:5 'IT. What value would you assign to f to make it continuous at x = O?

a. Graphf(x)

b. VerilY your conclusion in part (a) by rmding lim,....,. f(x) with I'H6pital~ Rule.

b. Now graph f on the interval 0 :5 x :5 '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)+.)

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 crosse. the x-axis. To simplilY your work, you might want to delete the exponential factor from the expression for f' aod 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:

Y = sin-Ix

'IT'

-1';;x,;;1

Range: -TTI2 :S Y :S 'lT12

_~

1

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.

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 -1 at x = -1t/2 to + 1 at x = 1t/2. By restricting its domain to the interval [-1t/2, 1t/2], we make it one-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

sin x

= sinx Domain: [-'IT/2, 'IT/2]

y

Range: [-1,1]

y = cosx Domain: [0, 'IT] Range: [-1, 1] y

sec x

-o;;t---,,;;---;;,,~x

2

i( y = cotx Domain: (0, 'IT) Range: (-00, 00)

y = tao x Domain: (-'IT/2, 'IT/2) Range: (-00, 00)

y = secx Domain: [0, 'IT/2) U ('IT/2, 'IT] Range: (-00, -1] U [I, 00)

----,;,,;--c;cot---c,,;;-~x

-~

2

y = cscx Domain: [-'IT/2, 0) U (0, 'IT/2] Range: (-00, -1] U [I, 00)

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 and y = cos-1 x for radian angles in the frrst quadrant. For a unit circle, lhe = 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 x, 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

or

y

= cos- 1 x

or

y

= tan-I

x

or

= coC l X y = sec- t x

equation s

whose sine is x."

y = sin-Ix

y y

Y = arcsinx Y = arccosx y = arctanx

or

Y = arccotx y = arcsecx

or

y

or

= csc- l x

= arccscx

These equations are read "y equals the arcsine of x" or "y equals arcsin x' and so OD. Caution The - I in the expressions for the inverse means ''inverse.'' It does not mean reciprocal. For example, the reciprocal of sin x is (sinx)-l = I/sinx = cscx.

y

x2

+ y2 = 1

Arc whose sine is x

~-+--

Angle whose sine is x -+-----::r\---'-~-t:_-~x

o

x

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 tske a closer look at these functions and their derivatives. Domain: -1 =S x =S 1 Range" - J! :s y s '!

Domain: -1:s x:s 1 Range: 0 :S Y :S '1r

y

y

.

1

2

Anglewho!le

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:=$1T,Y=F

I

(c)

Range: -I=SY=SI'Y#:O

Domain: _00 < x < 00 Range: O
y

y

Domain:

x:s -1 or x 2: 1

y

r-__ "

~.. 2 -------

y=sec-1x

-------

-c:-----:-+---.f-----:-~x

-2

-1

~-

(d)

1

2

x -c:-----:-+----;c---:-~x

-2

-1

(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 radians) 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

"

S . y y=smx'-2:=Sx

"2:

Domain: [-,,/2, ,,12] 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- 1 x.

(1)

The graph of y = cos-1 x (Figure 7.17) has no such symmetry. (aJ

EXAMPLE 1

y

Solution

x = siny

Y = sin-Ix Domain: [-1, 11 Range: [-,,/2, ,,/21

"

'2 )

-t;o

1

•


because sin (',,"!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

(bJ

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=

CO!lX,O:SxSTr

Domain: [0, ,,]

"'1/3/2

7f/3

7f/6

[-I,ll

v2/2

7f/4

7f/4

1/2

Range:

1 /

/

°

-1/2

7f/6 -7f/6

7f/3 27f/3

(aJ

-v2/2

-7f/4

37f/4

-"'1/3/2

-7f/3

57f/6

/

/

-1

y x = cosy

y

sin-1V3 2

= '!!'

3

COS(~1T) = -~ (b)

FIGURE 7.17

(b)

The graphs of

(a)y = cosx, 0 ::;; x:5 71',and (h) its 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.

FIGURE 7.18

Values of the arcsine and arccosine functions

(Example 1).

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 e = a + b.

7.6 Inverse Trigonometric Functions

407

Solution From Figure 7.19 and elementary geometry, we see that 180 sin a = 12 and 62 sinb = 12, so

a = sin-I

/io '"

b = sin-I

~~

0.067 radian '" 3.8°

'" 0.195 radian '" 11.2°

•

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- I x + cos-It -x) = 1r,

(2)

or COS-I (-x) =

1r -

COS-I x.

Also, we can see from the triangle in Figure 7.21 that for x sin-I x

+

(3)

>

0,

COS-I x = 1r/2.

(4)

Equation (4) holds for the other values ofx in [-1, 1] as well, but we cannot conclude this from the triangle in Figure 7.21. It is, however, a consequence of Equations (1) 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 ofx is an angle whose cotangent is x. The angles belong to the restricted domains of the tangent and cotangent functions. FIGURE 7.20 COS-I X and cos-It -x) are supplementary angles (so their sum is 'IT).

DEFINmON

y y

= tan-'xisthenumberin(-1r/2,1r/2)forwhichtany = = coC' x is the number in (0, 1r) for which coty = x.

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.l5c). Algebraically this means that tan-I (-x) = -tan-I x; FIGURE 7.21 sin-I x and COS-I x are complementary angles (so their sum is 'IT/2).

the arctangent is an odd function. The graph of y = coCI x has no such symmetry (Figure 7.l5f). Notice from Figure 7.l5c that the graph of the arctangent function has two horizontal asymptotes: one at y = 1r/2 and the other at y = -1r/2.

EXAMPLE 3

Y tan-I_I- = tan-I V3 = 1!

tan-' x

x

The accompanying figures show two values of tan-I x.

---,--~

Y'3 I

Y'3/3 -Y'3/3 -1

-Y'3

'IT/3 'IT/4 'IT/6 -1r/6 -1r/4 -1r/3

v'3

tan 1!

3

6

= _1_

6v'3

The angles come from the first and fourth quadrants because the range of tan-I x is

(-1r/2, 1r/2).

•

408

Chapter 7: Transcendental Functions Domain: Range:

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.

Ixl" 1

OSY:S'1T.y*~ y

3" ---------- ------------2'

~" ---\-----y

------------~

sec-Ix -+--,~_--

=

Caution There is no general agreement 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 (I/x). It also makes sec-I x an increasing function on each interval of its domain. Some 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 fannula for the derivative (our fonnula needs absolute value signs) but fail to satisfY the computational equation sec- I x = cos-I (I/x). From this, we can derive the identity

---------- -------------2 "

~ -" ------------ -------------2' ,/

(5)

by applying Equation (4).

3"

FIGURE 7.22 There are several logical choices for the left-hand hranch of y = sec-1 x. With choice A, sec-I x = COS-I (I/x), a useful identity employed by many calculators. y

, "\

"

2

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 I therefore assures us that the inverse function y = sin-I x is differentiable throughout the interval - I < x < I. We cannot expect it to be differentiable at x = I or x = - I 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 I with j(x) = sin x and rl(x) = sin-I x:

y = sin-Ix Domain: -1:s x:s 1 Range: -'1T12 :s y :S 1Tfl

U-1),(x) =

f'U~I(X))

Theorem 1

I cos (sin 1 x)

/'(Il)

=

cosu

I

VI -

sin2 (sin 1 x)

I The graph of y = sin-I x has vertical tangents at x = -1 and x = I.

sin (sin-1 x)

= X

FIGURE 7.23

If u is a differentiable function of x with Iu I

< I, we apply the Chain Rule to get

d ( . -I ) _ I du dx ' 2 dx sm u - •V ,.------, 1- u

EXAMPLE 4

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)

=

f'(j \x))

Theorem I

I sec2 (tan 1 x)

/,(u) ~ sec' u

I I

+ tan2 (tan 1 x)

sec2 u

=

1 + tan 2 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 < TT/2 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

Diffen:ntiate both sides.

dx

dx

dy

secytany dx = I

dy dx

I secytany

Chain Rule

Since: Ixl > l,ylicsin (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-----c+-~CL--~X

-1

0

!Lsec-l x

1

FIGURE 7.24 TIre slope ofthe curve y = sec-Ix is positive for both x <-1 and x > I.

=

dx

I

x~

1

I

x~

if x > I ifx<-l.

With the absolute value symbol, we can write a single expression that elinJinates the "±" ambiguity:

!L sec-l x

I

=

Ixl~

dx If u is a differentiable function of x with Iu I

.

> I, we have the fonou1a

lui>

EXAMPLE 5

I.

Using the Chain Ru1e and derivative ofthe arcsecant function, we tmd d sec-I ( 4) 5x = 5x 4

I IV(5xI 4 )2 -

dx

d ( 4) I dx 5x 5A"4> 1 >0

4

•

xV25x s - I·

Derivatives of the Other Three We cou1d use the same techniques to find the derivatives ofthe 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- l x

coC I X csc- I x

=

TT/2 -

=

TT/2 - sec-I x

tan-I

x

We saw the tmt of these identities in Equation (4). The others are derived in a similar way. It follows easily that the derivatives ofthe inverse cofunctions are the negatives ofthe

7.6 Inverse Trigonometric Functions

411

derivatives ofthe corresponding inverse functions. For example, the derivative of COS-I x is calculated as follows: d (COS-1) d (1T . -1 x ) dx x =dx 2- sm

Identity

1

Derivative of arcsine

v"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- 1 u)

I

dx

1.

du

d(cos- I u)

2.

1

dx

I

lui <

dx

du

1

du

-----

1

d(sec- 1 u)

+ u 2 dx I

du

dx

lul~dx'

d(csc- 1 u)

1

dx

6.

I

+ u 2 dx

I

d(coC 1 u)

5.

1

- - -

dx

4.

du

~dx'

d(tan- 1 u)

3.

lui <

~dx'

lui>

du

lul~dx'

1

lui>

I

Integration Formulas The derivative formulas in Table 7.3 yield three useful integration formulas in Table 704. 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

1.

2. 3.

Jv'a~u_ J~ J a2

u2 =

du

(*) + ~tan-l (*) +

u 2 = sin-

uv'u2 _ a 2

=

1

a

2

(Valid for u

C

(Valid for all u) (Valid for

2


C

1 sec-II "I + C a

* O.

lui> a > 0)

The derivative formulas in Table 7.3 have a = 1, but in most integrations a the formulas in Table 704 are more useful.

* I, and

412

Chapter 7: Transcendental Functions

EXAMPLE 6
V3 2 /

J

Vo/2

These examples illustrate how we use Table 7.4.

ax = v'1=7

('?) -

1V3 ax- - 11 Va 4x2

-

2

du

2 -

= tsin-I

6 =


1

x in Table 7.4, Formula 1

=

('?)

~-

;

= ;;

~ Y3,u ~ 'lx,anddu/2 ~ fix

•

c

+

=

Table 7.4, Formula I

u

= eX,du = eXdx, dx = duje:I: = du/u,

a2

.~

v6

1

du uVu 2 - a2

= !sec-11*1

EXAMPLE 7

I, u

=

(~) + C


sin-I

u2

= tSin-1 (*)

=

a

Vo/2

= sin-I

(b)

]V3/2

sin- 1 X

+

c

Table 7.4, Formula 3

I sec 1 ( eX ) Vii Vii +c -

•

Evaluate

ax

1

4x 2

(b)

V4x - x 2

~x + 2

+

Solution
-

4x) = -(x 2

-

Then we substitute a = 2, u = x - 2, and du =

1V4:- 1V4 - ~ x2

=

=

1Va 1

(x;

a

c

2)

+ 4x + 2

=

4(x 2

=

4

=

2, u

=

x - 2, anddu

=

Table 704, Fonnula 1

+

C

(b) We complete the square on the binomia14x 2 4x 2

(x - 2)2.

2)'

u2

+

= 4 -

ax to get

du 2 -

= sin- (*)

= sin-1

+ 4) + 4

4x

+ x) + 2

=

(x + tY+ I

+ 4x: 4(X2 =

(2x

+X+

t) + t

+ 1)2 + 1.

2 -

dx

Inverse Trigonometric Functions

7.6

413

Then,

J

~x + 2

4x + 2

= =

J

(2x

+

~)2 + 1

V

"2

=

(II) + C

1. ltan- I 2 a

~a

a

= l,u = a + I, = dx

anddu/2

2

Table 7.4, Fonnula 2

a

a

= I, u = 2x + 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-II

e. tan-I

e. tan-I

3.a.sm. (-I) T 4.Rosm. (I)"2 a. cos- (t) -I

b. sin-I

(~)

-I

b. sin-I

(-=1) v'2

1

5.

6. a. esc-I

v'2

b. cos-I

(-2) V3 sec-I (VJ)

b. esc-I b.

S. a. coC ' (-I)

(-I) v'2

b. coCI (0)

c.

. -I sm

. -I

c. sm C.

cos-I

(JJ) (~)

(-V3) (V3) 2 (V3) 2 -2-

c. csc-1 2

c. cot-I

(-I) V3

Evaluations Find the values in Exercises l}.-12. 9. sin (COS-I

('?))

11. tan (Sin-'(-t))

10. sec (COS-I t)

23. y 25. y 27. y

2S. y

29• y -- sec-11t'

12. cot (Sin-'(-

'?))

14.

lim cos-1 x

x--l+

=

= tan-Iv.;z::-J +

40. y

=

42. y

= In(x2

lim. csc-1 X

Find;ng Derivatives 10 Exercises 21-42, fmd the derivative ofy with respect to the appropriate variable. 21. y = COS-I (x 2 ) 22. y = COS-I (I/x)

sin-Il,

-

vt=l

36. y = COS-I (e-l)

8~ + COS-I 8

tan-I x

+ 4) -

X

3S. Y

=

v.;z::-J - sec-I 8

esc-I x, x > I

tan-I

41. y

= X

sin-Ix + ~

(f)

Evaluating Integrals

43J dx • ~ 45J dx • 17+x 47J dx • xV2Sx' 49.

2

/. 2

51~2 • o8+2t

i-

ely yV4y 2 - I

55J 3dr • VI - 4(r - I)' 57J dx • 2+ I)' 59

·

J

(x -

44J dx • VI-4x 46J 9+3x dx 2 4SJ xVSx'dx -

2

•

' 4 tis /.o ~

-I

X~-OO

=

Evaluate the integrals in Exercises 43--j)6.

53. 20.

30. y

34. y = tan-I (In x)

39. y

Vif2

X~OO

sec-I S8

t

37. y

}

=

32. y = cot-I

= esc-I (el)

15. lim tan-I x

X~OO

O
35. y

coc l

26. y

I), x>O

Vi

graph.)

19. lim csc-1 x

I)

2

Find the limits in Exercises 13-20. (If in doubt, look at the functino" x-l-

24. y = sin-I (I - t)

31. y = coC' 33. y = In (tan-I x)

Llm;ts

13. lim sin- 1 x

= sin-Iv'2 t = sec-I (2s + = esc-I (x 2 + = esc-I f

•

SO.

/.3Vi/4

i2 0

4

tis V9-48 2

52~2

• _24+3t Vi/3 dy 54. -2/3 yV9y2 - I

i-

56J 6dr • V 4 - (r + I)' 5SJ dx • I + (3x + I)'

dx (2x - I)V(2x - I)' - 4

•

414 60 •

Chapter 7: Transcendental Functions

J

Integration Fonnulas

dx (x + 3)V(x + 3)' - 25

61.1~/2

62.

J.~/4

63.

64.

1'<1'

2 cos 0 dO ~/2 I + (sinO)' r In v'3 e dx --

I + e 2r

].

•

1

J · v'l=7

65

esc' x dx ~/6 I + (cotx)'

J .

66

ydy

VerilY 1he integration fonnu1as in Exercises 99-102. tan-I x 1 tan-I x 99. ~dx ~ Inx - IIn(l + x') - - x - +

J

x4 x'cos-1 5xdx = 4COS-15x

4 dl I( I + In'l)

100.

sec'ydy

101. J (sin-I x)' dx

J

~ x(sin-1 x)' -

+ 45J

4 x dx VI - 25x'

2x + 2v'!=7 sin-I X + C

VI - tan'y

102. J In (a' + x')dx = xln(a' + x') - 2x + Evaluate 1he integrals in Exercises 67-1l0.

J 1· 71J , 1 67 ·

V-x'+4x-3

69

6dl

•

dx

-1 V3 - 21 - I'

dy y' - 2y

•

73

•

·

41 - 41'

dy

J

(x

4

+ 2x -

I dx

x' + 9 +

dx l)Vx'

+ 2x

78.

80

·

1-2 dl 1'-61+10

J J

I' - 21' + 31 - 4 dl

+

I'

I

e.m-'r dx

(sin-Ix)'dx

83..~

84.

vI-xl dy

•

(tan 1 y)(1 + y')

87.

sec'(sec-I x) dx • ,,------; 2

V2

89.J

I

x+1

~ = -,- -

dy 105. ~ =

xvx

I, y(O) = I

I

xvx 2

1

-

106 dy = _ I_ _

. dx

I +

2 v'!=7'

x'

-

1

88.

I

dx

107. You are sitting in a classroom next to the wall looking at the blackboard at 1he froot of1he room. TIre blackboard is 12 ft loog aod starts 3 ft from 1he wall you are sitting next to.

a = coC1 -..£ - coC1 .!. 15 3

vI-xl

~dx ,

if you are x ft from 1he froot wall.

b. Find x so 1hat a is as large as possible.

l+x

dy

(sin-1 y)v'!=Y'

l'

cos (sec-I x) dx • ,,------; 2

2;V3 xvx

-

1

T'!i 0

12'

1~

dx

90.J~dx 1- e

3'

-l

1x

I' L'H6pital's Rule Find 1he limits in Exercises 91-98. . -15 91. lim 8m x 92. lim x-a x x-l+ sec

W-=l

93. lim r~OO

xtan-l~

tan-I x 2 x-O x sin- X .

95. l u n - -1-

. -1

2

x x-o+(sin lx)2 lim

sm

You

a

Wall x

'I

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 geocrate a solid. Find 1he volume of1he solid.

1x

94. lim 2 tan-I 3x' x-a 7x 2 96. lim eX tan-I eX x_lXl e 1x + x 98.

y(O) = 2

a. Sbow 1hat your viewing aogle is

J

~(x + 1)((tan-1 ~)' + 9)

'It

Applications and Theory

(x - 2)Vx' - 4x + 3

86.J

x > I; y(2) =

."------;'

82• • ~

vI-xl

J 85J 'J

dy

""

J ero,-'r dx

81..~

y(O) = 0

vI-xl

r.u.

104.

Evaluate 1he integrals in Exercises 81--90.

J

dy I 103. ~ = .~'

r.u.

77. Jx'

C

Solve 1he initial value problems in Exercises 103-106.

6dl

y'+6y+1O

76.J

2atan-l~ +

Initial Value Prnblems

dx

74 2dx " / , x'-6x+1O

8dx

75.JX+4 dx x' + 4

79 •

JV2x - x' l' + J

70 • 1/2 V3

72

+5

x'-2x+2

1

68.

c

y

3"

-co+---~------':-x

o

2

7.6 109. The slant height of the cone shown here is 3 m. How large should the indicated angle be to maximize the cone's volume?

Inverse Trigonometric Functions

415

114. Show that the sum tan-1 x + tan-1 ( 1/x) is constant. 115. Use the identity csc-Iu = ~ - sec-Iu 2

to derive the formula for the derivative of csc-1 u in Table 7.3 from the formula for the derivative of sec-1 u.

116. Derive the formula 1

dy

dx

+ x2

1

for the derivative of y = tan-1 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

1

-1

Ixl > 1.

dx sec x = -lx-I-V-x-2---1' 118. Use the identity 111. Here is an informal proof that tan-II + tan- I 2 + tan- I 3 = Explain what is going on.

cot- I U = ~ - tan- I u 2

'IT.

to derive the formula for the derivative of cot-1 u in Table 7.3 from the formula for the derivative of tan-1 u.

119. What is special about the functions

x ~ 0,

g(x) = 2 tan-IVX?

and

Explain.

120. What is special about the functions

j(x) = sin- I

and g(x) = tan-I}?

1

W+1

Explain.

121. Find the volume of the solid of revolution shown here. 112. Two derivations of the identity sec-1 ( - x) =

1fT -

sec-1 x

a. (Geometric) Here is a pictorial proof that sec-1 ( -x) = 'IT - sec-1 x. See if you can tell what is going on. y

y

=

y

\

1

=

Vi+7

~~\ I:

sec-Ix

---l.

:=.I.7--1j= __ -.J

-\13

-3-

I,'

_

\13

J

17--__

--

Y

_~ x

1 1

-x

-1

o

x

b. (Algebraic) Derive the identity sec- I (-x) = 'IT - sec- I x by combining the following two equations from the text:

cos- I (-x) = 'IT - cos- I X sec- I x = cos- I (l/x)

Eq. (3) Eq. (5)

113. The identity sin-1 x + cos-1 X = 1fT /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).

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.

123. The solid lies between planes perpendicular to the x-axis at x = - 1 and x = 1. The cross-sections perpendicular to the x-axis are a. circles whose diameters stretch from the curve y = -1/v1+7 to the curve y = 1/v1+7. b. vertical squares whose base edges run from the curve y = -1/v1+7 to the curve y = 1/v1+7.

416

Chapter 7: Transcendental Functions

124. The solid lies between planes perpendicular to the x-axis at x ~ - V2/2 and x ~ Y2/2. The cross-sections perpendicular to the x-axis are a. circles whose diameters stretch from the x-axis to the curve y

= 2/~I-x'.

b. squares whose diagonals stretch from the x-axis to the curve y ~ 2/~I-x'.

D 125. Find the values of the following. a. sec-I 1.5

b. csc- I (-1.5) b. csc- I 1.7

b. Y

~

cos (cos-I x)

D Use your graphing utility for Exercises 13(l-I34. 130. Graph y see.

~

sec (sec- 1 x)

~

sec (cos-I(I/x)). Explain what you

131. Newton'. .erpentine Graph y ~ 4x/(x' + 1), Imowo 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?

c. coel (-2)

133. Graph f(x) ~ sin-I x together with its fIrst two derivatives. Comment on the behavior of f and the shape ofits graph in relation to the signs and values of f' and 134. Graph f(x) ~ tan-I x together with its frrst two derivatives. Comment on the behavior of f and the shape ofits graph in relation to the signs and values of f' and

Explain.

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. a. y = tan-I (tanx) b. y = tan (tan- I x) I 128. a. y = sin- (sin x) b. Y = sin (sin- I x)

7.7

~ ooS-1 (cos x)

e. coel 2

D 126. Find the values of the following. a. sec- I(-3)

129. a. y

r. r.

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

and

coshx =

eX

+

2

e-x

.

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 e-X) (eX +2 e-X) e2: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

TABLE 7.5

417

The six basic hyperbolic functions y

Yy

=

y

coshx

eX 3

y

=2 2 h ~ L"

Y

=

sinhx

-2

y~cotbx\

(aJ Hyperbolic sine: .

sinhx

(cJ

(bJ

Hyperbolic tangent:

Hyperbolic cosine:

eX - e-x

= "'---020"'--

coshx

+ e-x

eX

=

tanhx

2

= _sinh __ x cosh x

= .. e~x~---,e ..~~ eX + e-x

Hyperholic cotangent: cosh x

cothx

y

= sinhx =

eX + e-x eX _ e-x

-'""-JcccI-+'~_x

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

+

1 2 y=sechx

(dJ Hyperbolic secant:

sechx = _1_ = 2 cosh x eX + e-x

(eJ Hyperbolic cosecant:

cschx = _._1_ = sinhx

I

sinh2 X = cosh 2x - I 2 2 tanh X = I - sech2 X coth2 X = I + csch2 X

For any real 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 ofthe first identity cosh2 U

Derivatives of hyperboLic functions

TABLE 7.7

:fx (sinh u) =

cosh u::

:fx (cosh u) =

sinh u::

:fx (tanh u) = JL (coth u) dx

sech2

2

eX - e-x

=

1,

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 rational combioations of the differentiable functions

u::

= -csch2 U

sinh2 U

-

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~ du/dx 2

dcsch () dx u = -csch u coth u •dx = coshu

du dx

Delmition of sinh u

Derivative of ell Defmition of cosh u

418

Chapter 7: Transcendental Functions

Integral formulas for hyperbolic functions

TABLE 7.8

J J J J J J

sinhudu

=

coshu

TIris gives the first derivative formula. From the definition, we can calculate the derivative of the hyperbolic cosecant function, as follows:

~(CSChU) = ~ (s~u)

+C

= tanhu + C

csch'udu

= -cothu + C

sechutanhudu = -sechu + C

cschucothudu = -cschu + C

coshu du sinh'u dx

Quotient Rul.

I coshu du sinh u sinh u dx

Rearrange terms.

----

coshudu = sinhu + C sech'udu

Def"tnition of csch u

------

du

= -csch u coth u 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 V1+7)

=

(b)

J

coth5x dx

t Jo

sinh'xdx = =

=

(d)

u

=JCOsh5Xdx=lJdU sinh5x 5 u =

(c)

V1+7. ~ (V1+7) 1 sech' V1+7 V1+7

= sech'

tin

lui

+c

{ ' cosh2x -

Jo

1t

2Jo

su:

{In' 4ex sinh x dx =

Jo

=

tin I

sinh 5xl

=

I dx

Table 7.6

1 [Sinh2x 2

2 -

~

_ x]1

2

0

Evaluate with a calculator.

,., 0.40672

{In' 4ex eX - e-x dx = 2

Jo

[e 2x -

sinh5x, 5 cosh 5x d<

~

+c

2 (cosh2x _ I) dx =

=

du

{In' (2e 2x

Jo

_

2) dx

2x]~' = (e'In' - 2ln2) - (I - 0)

= 4 - 2ln2 -

I,., 1.6137

•

Inverse Hyperbolic Functions The inverses of the six basic hyperbolic functions are very useful in integration (see Chapter 8). Since d(sinhx)/dx = coshx > 0, the hyperbolic sine is an increasing function of x. We denote its inverse by y = sinh-I x.

For every value of x in the interval - 00 < x < 00, the value of y = sinh-I x is the number whose hyperbolic sine is x. The graphs of y = sinh x and y = sinh-I x are shown in Figure 7.25a. The function y = coshx is not one-to-one because its graph in Table 7.5 does not pass the horizontal line test. The restricted function y = cosh x, x ;" 0, however, is oneto-one and therefore has an inverse, denoted by y = COsh-I x.

7.7

419

Hyperbolic Functions

For every value of x ;", I, Y = cosh-1 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-1 x are shown in Figure 7.25b. y

y

y

y=siDhx y=x / /

y

y=x

y = sech-1 x (x = seehy, 3 y '" 0) /

8

/

/

/

= coshx,

X2!=O

// Y = sinh-I X (x = sinhy)

/

7

6 5

2

4

/

/

3 2 I

y=

/

COSh-Ix

123 4 5 6 7 8

x

/ / / / /

/

/

y

/

(x~coshy,y",O)

o

/

/

y=x /

= sechx

X2:0

-:+''----+-----E=~_x

o

2

3

(c)

(b)

(a)

FIGURE 7.25 The graphs of the inverse hyperbolic sine, cosine, and secant ofx. Notice the synunetries about lbe line y = x. Like Y = coshx, the function y = sechx = l/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- 1 x is the nonnegative number whose hyperbolic secant is x. The graphs of y = sech x, x ;", 0, and y = sech-1 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,

= coth- 1 x,

Y

Y

= csch- 1 x.

These functions are graphed in Figure 7.26. y

y ,

,

x=ltanhy

x=cothyi

Y = :tanh-1x

Y = coth-Ix:

, , ,

---;i'---,r.;--+.--> x -I,

, , , ,

y

l ,

x

, ,

, ,, ,

(a)

TABLE 7.9

Identities for inverse

hyperbolic functions sech-1x = cosh-11 x csch-1X = sinh-11 x

(c)

FIGURE 7.26 The graphs oflbe inverse hyperbolic tangent, cotangent, and cosecant of x.

Useful Identities We use the identities in Table 7.9 to calculate the values ofsech- 1 x, csch- I x, and coth- I x on calculators that give only cosh-1 x, sinh-1 x, and tanh-I x. These identities are direct consequences of the defmitions. For example, if 0 < x ,,;; 1, then

420

Chapter 7: Transcendental Functions We also know that sech (sech- I 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- I u)

dx d(cosh- I u)

dx d(tanh- I u)

Derivatives of inverse hyperbolic functions

I du ~dx I du W-=1dx'

u

du 2dx l-u '

lui <

I

I du l-u 2dx '

lui>

I

I

----

dx d(coth- I u)

----

dx d(sech- I u)

dx d(csch- I u)

I

I du u~dx'

0< u < I

du lul~dx'

u",O

I

dx

>

The restrictions Iu I < I and Iu I > I on the derivative formulas for tanh-I u and coth- I 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- I u)/dx. 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 I, then

d ( -I) _ I du ~dx' 2 dx cosh u - .vu - I Solution First we find the derivative of y = COsh-I X for x > I by applying Theorem I with f(x) = coshx andr'(x) = COsh-I x. Theorem I can be applied because the derivative of cosh x is positive for 0 < x.

(r')'(x)

=

The=ml

f'(j \ (x)) I sinh (cosh

1 x)

I Ycosh2 (cosh 1 x) - I I

1'(-)

~

coshl

U -

sinhu

=

sinh_

sinhl

U =

I,

Vcosh2 u - 1

421

7.7 Hyperbolic Functions HIsTORICAL BIOGRAPHY

The Chain Rule gives the final result:

Sonya Kova1evsky (185
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. TABLE 7.11

1.

2.

3.

1 1V2

du = sinh-1 (*) + C, Va' + u'

5.

a>O

du = sh-1 (1l)+c a ' u - a 2 co

1 1V 1 V2 2

2

du

u

a2

u

a

du

EXAMPLE 3

Solution

a

u =

-

-

u>a>O

r-'(')+c ~coth-l

du

a

4.

Integrals leading to inverse hyperbolic functions

a

u2

'

+ u2

a2

u2 > a 2

(*) + C,

u2

<

1 =_1 a sech- (1l)+C a '

O
= _1 CSCh-111l1 + C a a '

u"Oanda>O

Evaluate

The indefinite integral is

1V3

2dx

+ 4x 2

=

1

du Va 2 + u 2

Il =

=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 ofExercises 1--4 gives a value of sinh x or coshx. Use the definitions and the identity cosh' x - sinh' x ~ I to fmd the values ofthe remaining five hyperbolic functions. 1. sinhx =

-i

2. sinhx =

!

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. 2 cosh (lox)

6. sinh (2 In x)

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

a. 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.

sinh2 X = 1.

finding Derivatives

51.

In Exercises 13-24, fmd the derivative of y with respect to the appropriate vatiable.

53.

13. y ~ 6sinhf

14. y

VI

~

k sinh (2x

17. y = In (sinhz)

18. y

= I'tanbt = In (coshz)

19. y ~ sechO(l - InsechO)

20. y

~

21. y = In cosh v - ktanb'v

22. y

= In sinh v

15. y = 2V1tanb

23. y ~ (x'

+

16. y

+ I)

cschO(l - IncschO)

- kcoth'v

55.

J J J J J

sinh2xdt 6 cosh

tanb

(~ -

1n3) dt

~ dt

(x - k) dt

48.

sech

VI::;: VI dl

SO.

csch' (5 -

JorID' tanb 2x dt rID' 54. Jo 4.-8 sinh 0 dO

cosh (tan 0) sec' 0 dO

cosh (In I) I

Jor/2 2 sinh (sin 0) cos 0 dO

56.

58.1 flO 4

dl

8 cosh Vi dt

Vi

I

cosh'

dt

csch (In I) ~oth (In I) dl

~/4

' l 1:,

dO

coth

52.

I

59.

4 cosh (3x - In 2) dt

44.

sech'

~/4

57.

sinhfdt

46.

ID4 r cothx dt JID' ID r' 2.' cosh 0 dO J-1n4

1

J J J ..iJ J x) J

42.

(~) dt

60.

4 sinh'

(~) dt

I) 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 logatithons, as showo here.

In Exercises 25-36, fmd the derivative of y with respect to the appropriate vatiable.

25. y = sinh-I 27. y

Vi

26. y

~

(l - 0) tanb-IO 29. y ~ (l - I) coth- I VI

31. y

=

cos- 1 X

33. y = csch-I

-

X

28. y

= cosh-I 2"\fx+l ~ (0' + 20) tanb- I (0 + I)

30. y ~ (l - I') coth- I I

sech- 1 x

(k)"

32. y ~ Inx 34. y

+ ~sech-Ix

= csch- 1 2'

35. y = sinh-I (tanx)

36. y = cosh-I (sec x),

0

= In (x + ~), cosh-I x = In (x + v'.~2::-1), tanb-I = lin I + x sinh-I X

I-x'

sech-I X

~ In

C+ ~),

-I

_

(Ix + ~)

-I

- llnx

Ixl

+

I

~

Ixl

2

coth x- Z

Integration Formulas

x

x

csch x - In

< x < "If/2

'

00

1

< I

0<x:51 x#O

Ixl

x-I'

<x <

-00

>

I

VetifY the integration formulas in Exercises 37-40.

J J J J J

37. a.

sechx dt = tan-I (sinh x)

+C

b.

sechx dt = sin-I (tanbx)

+C

38. 39. 40.

xsech-Ixdt = xcoth-Ixdt = tanb-Ixdt

Use the formulas in the box here to express the ntanbers in Exercises 61--{j6 in tenns ofnaturallogatithons.

~ sech-Ix - k~ + C

x' ~

I coth-Ix

+~+C

= xtanb- I x + kln(1

- x')

+C

61. sinh-I (-5/12)

62. cosh-I (5/3)

63. tanb- I (-1/2)

64. coth- I (5/4)

65. sech-I (3/5)

66. csch-I (-1/0)

Evaluate the integrals in Exercises 67-74 in terms of

a. inverse hyperbolic functions. b. naturallogarithons.

7.7

67.

69

•

'v3

J., o

v'4+7

1 1 xVI -

68. 70.

5j41-x

fix

1/'

73. J.~

72

J. VI J. l 'xv'4+7 l' xVI

cou fix

fix

81. Arc length

--,

1-

X

•

v'5.

82. Usc thc definitions of the byperholic functions to fmd each of the following lintits.

1

lim tanh

(b)

(c) lim sinhx

(d)

(e) lim sechx

(f) tim cotb x

(a)

+ (In X)'

1

Find thc lengib of the gmph of y = (1/2) cosh 2x

fromx = Otox = In

fix

74

o VI + sin'x

o

The region enclosed by the curve y = sech x, the

V3

+ 9x'

0

x~oo

X~OO

Appllcattons and Examples 75. Show that if a function I is defmed on an interval symmetric about the origin (so that I is defmed at -x whenever it is defined at x), then

_ I(x) + I( -x) ( )Ix 2

+

I(x) - I( -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.

fix

•

16>;'

80. Volume

6 fix

112

----""-2

3/13

71.

113

fix

Hyperbolic Functions

(1)

Then show that (f(x) + I( -x))/2 is even and that (f(x) - I( -x))/2 is odd.

W+i)

76. Derive the fornmla sinh-I 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 hody 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 t sec into the fall satisfies the differential equation

x~oo

lim tanhx

.%_-00

lim sinh x

x_-oo x~oo

(g) lim coth x

x-o+

(1)

lim csch

.%_-00

83. Hanging cables Jn3agine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable's weight per unit lengib is a constant w and thc horizontal tension at its lcmest point is a vector oflength H. Ifwe choose a c0ordinate system for the plane ofthe cable 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 cable lies at the point y = H/w on thc y-axis (see accompanying figure), then it can be shown that the cable lies along the graph ofthe hypetbolic cosine

H

w

y = ",coshnx.

dv , mdt=mg-kv, where k is a constant that depends on the hody's aerodynamic 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 = Owhent = O. b. Find the body's limiting velocity, lim,-.oo v. c. For a160-lb skydiver (mg = 160), with time in seconds and distance in fee!, a typical value for k is 0.005. What is the diver~ limiting velocity?

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning "chain." a. Le!P(x,y) denote an arbitrary point on the cable. The next accompanying figure displays the tension at P as a vector of lengib (magnitude) T, as well as the tension H at the lowest pnintA. Show that the cable's slope atP is

dy

.

78. Accelerations whose magnitudes are proportional 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 coshkt + b sinhkt. Show in hoth cases that the acceleration d's/dt' is proportional to s but that in the flIst case it is directed toward the origin, whereas in the sccond case it is directed away from the origin. 79. Volume A region in the flIS! quadrant is hounded above by the curve y = cosh x, below by the curve y = sinh x , and on the left and right by thc y-axis and thc line x = 2, respectively. Find the volume of the solid generated by revolving the region about the x-axis.

w

tan'" = fix = sinhnx.

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 at P(x,y) is exactly equal to the weight of y units of cable.

84. (Continuation ofExercise 83.) The leogth of arc AP in the Exercise 83 figure is s = (I/o) sinh ax, where 0 = w/H. Show that the coordinates ofP may be expressed in terms of s as

x =

~Sinh-l as,

y =

Another analogy between hyperbolic and circular functions is that the variable u in the coordinates (cosh u, sinh u) for the points of the right-band 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.

a. Sbowthat 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 = (I/o) cosh ax, the coordinate axes, and the line x = b is the sarae as the area of a rectangle of height and leogth s, where s is the length of the curve from x ~ 0 to x = b. Draw a figure illustrating this result.

b. Differentiate both sides of the equation in part (a) with respect to u to show that

1/0

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-band branch of the unit hyperbola, X 2 _ y 2=1.

-1°O""~fix.

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

y x2

+

r =1

P(cos U, sin u) ........,-.~

---t-+--'I---~x

Since cosh2 U - sinh2 U = 1, the point (cosh u, sinh u) lies on the right-band branch of the hyperbola x 2 - y2 = I for every value of u (Exercise 86).

7.8

One of the analogies 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 eventually become and remain positive as x --+ 00 .

y

160 140 120 100

Growth Rates of Functions

80 60 40 20

FIGURE 7.27 2

andx .

The graphs of eX, 2x ,

You may have noticed that exponential functions like'];" and eX seem to grow more rapidly as x gets large than do polynomials and rational functions. These exponentials certainly grow more rapidly than x itself, and you can see'];" 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 em,

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 ceiling if it hasn't done so already). At x = 10 em, the graph is e 'O '" 22,026 em '" 220 m high, higher than mosl buildings. Al x = 24 em, 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 ofe' and 10 x.

x

Rates of Growth as x --+

DEFINmON

0Cl

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

x--+OO

2x =lim2=2 x 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 j.

EXAM PLE 1

Let's compare the growth rates of several common functions.

(a) e' grows faster than x 2 as x --+

e'

Iim -2 x

,.--+00

00/00

lim eX 2x

00

=

x_OO

because

lim eX = 2

00.

Using l'H6pital's Rule twice

x_OO

00/00

(h) 3' grows faster than 2' as x --+

00

because

. 3' . (3)'

Inn 2' = x--i>OO lim -2

x_OO

=

00.

426

Chapter 7: Transcendental Functions (c)

X

2

grows faster than In x as x ---> lim x---+OO

2 X

Inx

00

k

= lim x--+OO

l/x

(d) In x grows slower than x '/n as x --->

because = lim 2>;2 =

00

I'H6pital's Rule

%---+00

00

for any positive integer n because

lim Inx = lim

I/x

,_00 x l / n ,_00 (I/n)x(l/n)-I

!'H6pilal's Rule

= lim --"-= O. x---+oo x 1/ n

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 : log" x

.

.

In x/In a

hm--=hm ,_00 10llbX ,_00 Inx/lnb

Inb =Ina'

•

The limiting ratio is always finite and never zero.

x --->

I

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 ,_00

= L1

and

together imply

If L, andL2 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+ x25 x x_OO

(2VX -

x-oo

1)2 = lim

x

x-oo

(2VX Vi

=1

'

1)2 = lim x-oo

(2 __Vi

1_)2 = 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.

Here we use little-oh notation.

EXAMPLE 3

(a) lnx = o(x) as x ---> (b) x 2 = 0(x 3

+

00

because

lim lnx = 0 x---+ OO X 2

1) as x --->

00

because

•

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 /

Here we use big-oh notation.

EXAMPLE 4 (a) x (b) eX

+

sinx = O(x) as x --->

+ x2

= O(g) ("I is big-oh of g").

=

O( eX) as x --->

00

00

because because

x

+ xsinx ,;;

eX

+x x 2 ~lasx--+oo

.

2 for x sufficiently large.

e

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 rate, thoo I = O(g) and g = OU) (Exercise ll). (c) x = O( eX) as x --->

00

because

Sequential vs. Binary Search Computer sciootists often measure the efficioocy of an algorithm by counting the number of steps a computer must take to execute the algorithm. There can be significant differooces in how efficiently algorithms perform, even if they are designed to accomplish the sarae task. These differooces are often described in big-oh notation. Here is an exarople. Webster. 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 arrangemoot. 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, thoo 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 ) That certainly beats a possible 26,000 steps.

< I.

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 detennine 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(log2 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 Compartsons with the Exponentfal eX

1. Which of the following functions grow faster than eX as x --+ 00 ? Which grow at the same rate as eX? Which grow slower? Lx-3 b.x 3 +sin2x c. v'X c. (3/2Y'

d.

4X

r.

eX/2

g. eX/2

h. log"x

g. In (In x)

2. Which of the following functions grow faster than eX as x --+ Which grow at the same rate as eX? Which grow slower?

a. lOx' + 30x c. vT+7

+

I

00

?

d. (5/2Y' f. xeX"

3. Which ofthe following functions grow faster than x 2 as x --+ 00 ? Which grow at the same rate as x'? Which grow slower? a.x 2 +4x b.x S -x 2 c. v'x' + x 3 d. (x + 3)2 e. xlnx f. 2X" 3 g. x e -x h. &x 2 2

4. Which ofthe following functions grow faster than x as x --+

00

?

Which grow at the same rate as x'? Which grow slower? b. lOx'

r.

(I/IOY' h. x 2 + IOOx

Compartsons with the Logartthm In x 5. Which of the following functions grow faster than In x as x ---+ oo? Which grow at the same rate as In x? Which grow

c. In v'X e. x g.

1/x

Ordering Functions by Growth Rates

a. eX

b.x'

c. (lnxY'

d. e X / 2

a. 2x c. (ln2Y'

d. v'X

r.

5lnx

b. x 2 d. eX

Big-oh and Little-oh; Order

9. True, or false? As x ---+ 00. a. x = o(x) c. x = O(x + 5) e. eX = o(e2%) g. Inx = 0(1n 2x) 10. True, or false? As x

d. log" (x')

b. In 2x

+ 5)

8. Order the following functions from slowest growing to fastest growing as x ---+ 00.

Compartsons with the Power i'

slower? a. 10g)x

h. In (2x

7. Order the following functions from slowest growing to fastest growing as x ---+ 00.

b. x Inx - x

h • ex-I

a. x' + v'X c. x'e ~ e. x 3 - x' g. (l.l Y'

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. log, (x 2 ) b. log" lOx c. 1/v'X d. I/x' f. e-x e. x - 2lnx

a. x

=

d. x = 0(2x)

f. x + Inx = O(x) W+5 = O(x)

h. ---+ 00,

~ 3 = O(})

c. } - :'

b.x=0(x+5)

o(})

b. }

+ :' =

O(})

d. 2

+ cosx

= 0(2)

e. eX + x = O(e X) f. x Inx = 0(x 2 ) g. In (In x) = O(lnx) h. In (x) = 0(1n (x 2 + I)) 11. Show that ifpositive functions f(x) and g(x) grow at the same rate asx--+ 00, theof= O(g) andg = 0(/). 12. Wheo is a polynomial f(x) of smaller order than a polynomial g(x) as x ---+ oo? Give reasons for your answer. 13. Wheo 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 ofrational functions tell us about the relative growth ofpolynomia1sasx~ oo?

D

Other Comparisons DIS. Investigate

lim In (x + I) Inx

.%_00

and

.

11m

];_00

In (x + 999) In x .

X'/I,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. c. Even x '/10 takes a long time to overtake In x. Experiment with a calculator to fmd the value of x at which the graphs of x 1/10 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.

D 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 1han 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

x_OO

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 + I and y';+J 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. a. Suppose you have three different algorithms for solving the same prohlem and each algorithm takes a number of steps that is of the order ofone of the functions listed here: n lo~ n,

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 lJ for any positive integer n, even x 1,000,000 . (Hint: What is the nth derivative of XII?) 20. The function c" outgrows any polynomial Show that e" grows faster as x ---> 00 than any polynomial

21. a. Show that Inx grows slower as x ---> 00 than xii' for any positive integer n, even x 1/1,000,000 •

Db.

n 3j2 , n(lo~ n? .

Which of the algorithms is the most efficient 10 the long roo? Give reasons for your answer.

D 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,

Y;; log, n,

(log, n? .

D 25.

Suppose you are looking for an item in an ordered list one million iterus long. How many steps might it take to fmd that item with a sequential search? A binary search?

D 26.

You are looking for an item in an ordered list 450,000 iterus long (the length of Webster. Third New International Dictionary). How many steps might it take to fmd the item with a sequential search? A binary search?

Although the values ofx'/I,OOll,OOO eventually overtake the values 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

429

Questions to Guide Your Review

1. What functions have inverses? How do you know if two functions f and g are ioverses of one another? Give examples of functions that are (are not) ioverses of one another. 2. How are the domains, raoges, and graphs of functions and their ioverses related? Give an example. 3. How can you sometimes express the inverse ofa function of x as a function ofx? 4. Under what circumstances can you be sure that the ioverse of a function f is differentiable? How are the derivatives of f and related?

r'

5. What is the natoral 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 iotegrals of tan x and cot x? 8. How is the exponential function c' 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 truth is there 10 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 chaoge? How can it he derived from an initial value problem? What are some of the applications of the law? 12. Descrihe 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 - OO? Give examples.

14. How can you sometimes handle limits that lead to indeterminate forms Xl, 00, and 00 001 Give examples.

r

15. How are the inverse trigooometric 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 functioos? How do suhstitutioo and completing the square broaden the applicatioo 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

19. What are the derivatives of the six basic hyperbolic functioos? What are the correspooding integral formulas? What similarities do you see here with the six basic trigonometric functions? 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-' x, and coth-I x using a calculator's keys for cosh- 1 X, sinh-1 x, and tanh-I x? 21. What integrals lead natorally to inverse hyperbolic functioos? 22. How do you compare the growth rates of positive functions as X~OO?

23. What roles do the functioos e% and In x play in growth comparisons? 24. Describe big-oh and little-oh notatioo. Give examples. 25. Which is more efficient---
Practice Exercises

finding Derivatives

In Exercises 1-24, Imd the derivative ofy with respect to the appropti-

28. y

=

29. y

= (sinO)v6

ate vatiable.

Vie y,.

2u2"

W+1 = (lnx)'/(Inx)

30. y

1. y = lOe~/'

2. y =

I"" 3. y -_I"" "4xe - 16 e

4. y = x 22/ e- x

Evaluating Integrals Evaluate the integrals in Exercises 31-78.

5. y ~ In (sin2 0)

6. y ~ In (seC> 0)

31. / e% sin (e") dx

2

7. y = 10!l2 (x /2) 9. y = 8--1 11. y = 5x,·6

12. y = y'2,,-v2

+ 2)"+2

14. y = 2(lnx)"/2

13. y = (x

10. y = 9"

15. y = sin-I~, 16. y

33. / e% sec2 (e% - 7) dx

1 17. y

~

Incos-Ix

18. y~zcos-Iz- ~

39.

>1

41.

sec-lVi

<

0

< 'ff/2

43.

45. Logarithmic DIfferentiation

In Exercises 25-30, use 10garitInnic differentiatioo to Imd the derivative ofy with respect to the approptiate vatiable. 2

_ 2(x + I) 25. y - .,----;:v cos 2x + 1)(1 27. y (I _ 2)(1 + 3)

_((I

I) cot (e"

I»)',

l'

11:dx _ 4

-I

20. y = (I + 12) COe' 21 21. y ~ zsec-' z - '\Iz2=1, z

23. y ~ csc- 1 (sec 0), 0 24. y ~ (I + x 2 )e""-' %

+

seC> (x)e""% dx

35. / 37.

19. y = I tan-I I - tlnl

2v;=!

34. / e"csc(e"

+ I)
0< u< 1

~ sin-I (~), v>

22. y =

32. / e' cos (3e' - 2) dl

8. Y = log, (11: - 7)

26. y

t> 2

=

~3X + 4 2x - 4

47.

J. J.' o

w tan

40.

1~

v

dv

(lnx)-'

/

42.

44.

- x - dx

46.

1 rcsC> (I

48.

1

'3

1'~ 2 cot 'ffX dx

1 W

tan (In v)

/

1'~ - x - dx I

~3 dx

49. / x3%' dx 51.

38.

__ 21_ dl o t 2 - 25

/

csc2 x e OOl %dx

36. /

, xdx

+ Inr) dr

6

cos I dt -wf2 1 - smt /

/

dv vlnv

/

In (x - 5) x 5 dx

/cos(I-lnv)

v

SO. / 2""% seC> x dx 52.

1'2 1

I

5x dx

dv

Chapter7

53.

54.1' (~ -~)

l' (~+ ~)dx

56.1

1

55.1- e --«+1) dx 57.

/.,"5

e'(3e'

59.1' ~ ' l 61.

(I

+ 1)-'/2 dr

+ 7lnxt l/'

58.

dx

60

(In (v + I)? v+1 dv

1

1'/ 1

"

65.

4

'19 -

4x 2

66.

6712~2

68

69 ·

J

dy yY4y 2

-

2/3

dy

70.

I

•

J

72. 74

dx

Y-2x-x 2

1-

•

1

75 •

77 •

-2

J

(t

105. lim

(i _1)t

4 _ 25x2

~

2

+

2dv

76

+ 4v + 5

•

--=-r~== - 16

dt I)Yt 2 + 2t - 8

78

•

:-::,~== Iy IV5y 2 - 3

Y-x 2

(3t

dx +4x-1

3 dv

+ 4v + 4

+

dt I)Y9t 2 + 6t

Solving Equations In Exerci,e, 79-84, solve for y.

79. 3" =

80. 4-> = 3"+2

81. ge'" ~ x 2

82. 3"

= x + Iny

~

84. In (lOlny)

= 1n5x

86.1im x ' - 1 x-I x b - 1

85.limx2+3x-4 x-I x 1

88. lim

x-o

2 8in x

+ smx

lim sec 7x cos 3x

x-n/2-

94. lim(1

93. lim (esc x - eotx)

x_a\;4

x~o

2

95. lim (Yx + X~OO

96

x

tan~

, sinmx 90• Iun . x-o smnx

89. 11m - ( x-o tan x 2) 91.

G'

X

+ I - Yx

lim _x 'x_oo

2-1

,)

x_ x2

+t

108. lim x-o+

2

-

x)

(I

3 )X + _x

= log,x,

g(x) = log,x I g(x) = x + X

e. f(x) ~ x/IOO, d. f(x) ~ x,

g(x)

e. f(x) = ese- l x,

g(x) = I/x

r.

g(x)

f(x)

~

~ xe~

g(x) ~ tan-I x

sinhx,

~

eX

110. Does f grow faster, slower, or at the same rate as g as x -+ oo?

Give reasons for your answers. .. f(x)

~ 3~,

g(x)

b. f(x) ~ 1n2x,

~ 2~

g(x) ~ Inx 2

e. f(x) = lOx' + d. f(x) ~ tan- l(1jx),

2x 2,

g(x) = eX g(x) ~ I/x

e. f(x) = sin- (l/x),

g(x) = l/x 2

r.

g(x) =

f(x) = seehx,

e~

111. True, or false? Give reasons for your answers.

3 Inx

Apply;ng L:H6pUal's Rule Use I'HOpital's Rule to find the limits in Exercises 85-108.

•

+ I )'"x

l

2"+1

83. In (y - I)

y-o+

b. f(x) = x,

dy

-21'15

4v 2

106. lim e-l/Y lny

Comparing Growth Rates of Functions 109. Does f grow faster, slower, or at the same rate as g as x -+ oo? .. f(x)

24dy

yVy 2

-1

+3_x

Give reasons for your answers.

1

V

2

.t_4ex4

eX - 1

x_OO

2

sin ('ll'x)

104. lim

t

107. lim (eX

I

6dx

1Y J I J 1 J

t-O+

102. lim x smx x-o tan3 x

+ 2t)

+ Int)tlntdt

-1/5

•

101. lim 5 - 5 eosx 1 x_oe x x t-O+

z-m" -

x-o eX-l

t - In (I 2 t

-V.M

71. /, :-::-r~= Yz/3ly IV9y2 - I 73

1)1/2 d6

8ln31og, 6 d6 6

{'

100. lim

x_oex-l

· lv'33 + t 2

• _24+3t

98. lim 3' - I 6-0 8

x

103. lim

1/5

6dx

x-o

dx

x~

64'1

-:;:~~

-'/4

l"-I-

97. lim lOX - I 99. lim 2'"" - I

dw

/.,"9 e'(e' _

' l

-6- d6

1

2w

e

62. j,4(1

'IOg., 6

63.

dx

0

-'"2

-2

x2

3x

1

431

Practice Exercises

-~) x2

.. ~2 + ~4 = O(~) 2 x e. x

~

b.

x x o(x + Inx)

~2 + ~4 = O(~) 4

x x d. In (lnx)

e. tan-I x ~ 0(1)

~

x o(lnx)

f. eoshx ~ O(e X )

112. Tme, or false? Give reasons for your answers.

"~~O(~+~) x4 x2 x4

b.~~o(~+~) x4 x2 x4

c. Inx = o(x + I)

d. 1n2x = O(lnx)

e. see-l x = 0(1)

f. sinhx = O(e X )

Theory and Applications 113. Thefimctioof(x) = eX + x,heingdifferen1ishleaodone-to-one, has a differeotiable inverse (x) , Find the value of /dx at the point f(ln 2),

rl

drl

114. Find the inverse of the fimctioof(x) = I + (I/x),x # O.Thea 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

In Exercises 125-128 solve the differential equstioo.

values ofeach functioo 00 the given interval.

dy • , 125. fix = vYcos' vy

126. y' =

127. y.y' = secy' sec' x

128. y cos' x dy

[i., f]

115. y = xlo2x - x, 116. y = Hh:(2 - Inx),

3y(x y

+ I)' 1

+ sin x fix

= 0

(0, .']

117. Area Find the area between the curve y x-axis from x = 1 to x = e.

= 2(lnx)/x and the

118. a. Show that the area hetween 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 aloog the curve y = Inx. Its x-coordinare is increasing at the rate (fix/dt) = VX 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. -xl' . Her y-coordinare is changiog at the rate dy/dt = ( -1/4)~ ft/sec. At approximarely what rate is her x-coordinate changiog when she reaches the bottom ofthe slide at x = 9 ft? (Take.' to be 20 and rouod your answer to the nearest ft/sec.) 121. 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 = e -r. What dimensions give the rectangle its largest area, and what is that area? y

In Exercises 129-132 solve the initial value problem.

.-'-Y-',

129. :

=

dy 130. fix

=-

ylny

I

+x

"

y(O) = -2

=•

y(0)

,

131. xdy - (y

+ vY) fix

132. y-' dyfix =

,f-, y(O) = • +I

= 0, y(l) = I

1

133. What is the age ofa sample ofcharcoal in which 90% of the carbon-14 origioally present has decayed? 134. Cooling a pie A deep-dish apple pie, whose inremal 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 teruperatore was 180°F. How long did it take the pie to cool from there to 70°F? 135. Locating a .olar .tation You are under cootract to build a solar .tation at grouod level on the east-we.t 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

cot

x

C1 50 -

60 - co

x

~.

Then fmd the value of x that maximizes 8.

DO

~+------'---'---=-. 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 dimensions give the rectangle its largest area, and what is that area?

::~

0.2 0.1

o

,_

..

~

~I

II

>x

D 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)/VX b. y ~ .-.' e. y ~ (l + x).~

D 124. Graph f(x)

= x In x. Does the functioo appear to have an absolure minimum value? Cnofmn your answer with calculus.

DD

DO DO

o

x

SOm

136. A rouod underwater transmissioo cable consists ofa core ofcopper wire. surrounded by nonconducting insulation. 1f x denores the ratio of the radius of the core to the thickness of the insulation, it is known that the speed ofthe transmission signal is given by the equation v = x'in (I/x). Ifthe radius ofthe core is I em, what insulation thickness h will allow the greatest transmission speed? Insulation x= !.h

Chapter 7 Additional and Advanced Exercises

'f}

Chapter

Additional and Advanced Exercises

Limns

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.

1. lim

[b

b-l-)O

2. lim} {' tan-I r dt

dx

~

%_00

%-0+

Jo

+ .')'/'

4. lim (x

3. lim (cos VX)I/,

%_00

5. lim (_1_ 11_00 n+ 1

+n+2 +... +~) 2n

• I) g(x

x

6. lim 1 (ellA + n_oon

+ ... + e(n-l)/II + en/II)

volving the region about the x-axis. Find the following limits.

b. lim V(t)/A(t)

1_00

c. lim V(t)/A(t) 1-0+

/-+00

8. Varying a logarithm's base a. Findlimlog.,2aso .... 0+, 1-, 1+, and

0< a

00.

= log., 2 as a function of a over the interval

:5

4.

of the larger area to the smaller?

010. Graph j(x) = tan-I x + tan-l(l/x) for -5

x :5 5. Then use calculus to explain what you see. How would you expect j to behave beyond the interval [-5,5]? Give reasons for your answer. :5

0 does x(r) = (x')'? Give reasons for your

answer.

012.

a. Showthatg(O) = O.

b. Showthatg'(x) = I

+ [g(x)]'.

c. Find g(x) by solving the differential equation in part (b).

17. Center of mass Find the center of mass of a thin plate of constant density covering the region in the firat and fourth quadnlnts enclosed by the curves y ~ 1/(1 + x') and y ~ -1/(1 + x') and by the lines x = 0 and x = I. 18. Solid of revolution The region between the curve y = 1/(2VX) and the x-axis from x ~ 1/4 to x ~ 4 is revolved about the x-axis to generate a solid.

b. Find the centroid ofthe region.

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 = •. What is the ratio

>

=I

h

a. Find the volume of the solid.

Theory and Examples

11. For what x

+ g(y)

il) lim g(h) = 0 iii) lim g(h)

7. LelA(t) be the area ofthe region in the IITst quadrant enclosed by the coordinate axes, the curve y = e -x, and the vertical line x ~ t, t > O. Let V(t) be the volume of the solid generated by rea. lim A(t)

g(x)

= I _ g(x)g(y) forallrealnumbers x,y, and

.~o

_1_

e2/A

+ y)

+ y in the domain ofg

h-O

o b. Graph y

433

Graphj(x) = (sinx)"nxover[O, 31T]. Explain what you see.

13. Findj'(2) if j(x)

~ • .<x) andg(x) ~

],'_t-4 dt. , I +t

14. a. Find dfjdx if j(x) =

19. The best hranching angles for hlood vessels and pipes 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 SOD2C energysaving point of view. We might require, for instance, that energy loss due to friction be minimized aloog the sectionAOB showo in the accompaoylng figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe upstream from B, and 0 is the point where the branchiog occurs. A law due to Poiseui11e states that the loss of energy due to friction in nontorbulent flow is proportional to the length of the path and inversely proportional to the fourth power of the radius. Thus, the loss along AO is (kd l )/R 4 and along OB is (kd,)/r 4 , where k 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:

['2~tdt.

dl

d,

L~k4+k4' R r

b. Find j(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) = ofor allxandh(x) = ofor allx. b. If j(x) = h (x) + j 0 (x) is the sum of an even function h(x) and an odd function jo(x) , then show that

h(x) = j(x)

+ j( -x) 2

and

jo(x) = j(x) - j( -x) 2

c. What is the sigoificance of the result in part (b)?

A l d _

1<---:_ _

a

In our model, we assume that AC Thus we have the relations

~

a and BC

~

b are fixed.

434

Chapter 7: Transcendental Functions

D 20.

so that d,=bcsc8, dl = a - d,cos8 = a - beat8. We can express the total loss L as a function of8:

Urban gardening A vegetable garden 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 tall, where should the garden be placed in order to receive the maximum number ofbours of sunlight exposure? (Hint: Determine the value of x in the accompanying fIgure that maximizes sunlight exposure for the garden.)

L= k(a - bcot8 + bCSC8). R4

7

4

DO DO DO DO DO

a. Show that the critical value of8 for which dL/d8 equals zero is 8e =

-1 COS

r4 R4 " 200 ft tall

b. If the ratio ofthe pipe radii is r/R = 5/6, estimate to the oearest degree the optimal branching angle given in part (a).

The mathematical analysis descrihed here is also used to explain the angles at which arteries branch in an aoimal's body.

West

DO DO DO 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 antiderivatives (or indefinite integrals) for many combinations of functions whose antiderivatives cannot be found using the methods presented before.

Basic integration formulas

TABLE 8.1

1.

3.

4. 5. 6.

7. 8. 9. 10. 11.

I

kdx

= kx + C

I~ = In Ixl

I

eX dx = eX

lax

I I I I I I

dx

=L

+

Ina

+

12.

(n =i' -1)

13. 14.

C

C

+

15.

(a>O,a=i' 1)

C

sinxdx = -cosx cosxdx = sinx 2

(any number k)

sec xdx = tanx

+

+ +

csc2 xdx = -cotx

C

17.

C

18.

C

+

16.

19.

C

20.

I I c I I I I I v'ar:- x (~) I ~ ~tan-l (~) I V ~sec-l I~ I tanxdx = In Isecxl

+

cotxdx = In Isinxl

+

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

+

C

C

sin-

+

+

x2 =

dx x 2 - a2

1

=

C

C

+

C

21.

(a> 0)

22.

(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 j(x)g(x) dx. It is useful when j can be differentiated repeatedly and g can be integrated repeatedly without difficulty. The integrals

and

JxCOSXdx

are such integrals because j(x) = x or j(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 Fonn If j and g are differentiable functions of x, the Product Rule says that

d dx [f(x)g(x)] = j'(x)g(x) + j(x)g'(x). In terms of indermite integrals, this equation becomes

J

:x [f(x)g(x)] dx

=

J [f'(x)g(x)

+ j(x)g'(x)] dx

or

J

:x [f(x)g(x)] dx

=

J j'(x)g(x) dx + J j(x)g'(x) dx.

Rearranging the terms of this last equation, we get

J j(x)g'(x) dx = J

:x [f(x)g(x)] dx -

J j'(x)g(x) dx,

leading to the integration by parts formula

J j(x)g'(x) dx = j(x)g(x) - J j'(x)g(x) dx

(l)

Sometimes it is easier to remember the formula if we write it in differential form. Let u = j(x) and v = g(x). Then du = j'(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 Parts 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 =x, du = dl:,

u dv

xcosxdl:.

=

uv - j v du with

dv = cosxdx, v = sinx.

Simplest atrtiderivativc of cos x

Then

j xcosxdl:

=

x sin x - j sinxdl:

=

x sin x

+ cosx +

C.

•

There are four choices available for u and dv in Example 1:

1. Letu = I anddv = xcosxdl:. Letu = xcosx and dv = dl:.

2. 4.

3.

Letu = xanddv = cosxdl:. Letu = cosxanddv = xdl:.

Choice 2 was used in Example 1. 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)dl:.

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 dl:, 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 arl>itrary 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 Inxdl:. Solution

J udv = u du

Since

uv -

= Inx =

I

xdl:,

J In x dl: can J vduwith

be written as

Simplifies when differentiated

J In x • 1 dl:, dv

= dl:

v =x.

we use the formula

Easy to integrate

Simplest antidcrivative

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 = dx, v = eX, and

1

xeXdx

=

xe X -

1

eXdx

=

xe X - eX +

c.

Using this last evaluation, we then obtain

1

x 2e Xdx

21

=

x 2e x -

=

x 2e x - 2xe X

xeXdx

+

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 eventual1y 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 = cosxdx.Then du = eXdx,v = sinx,and

1

eXcosxdx

=

eX sin x -

1

eXsinxdx.

The second integral is like the frrst 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,

du=exdx.

v = -cosx,

1 1

=

eX sin x - (-ex cosx -

=

eX sin x + eXcosx -

(-cosx)(e Xdx»)

eXcosxdx.

8.1 Integration by Parts

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'cosx +

e'cosxdx = e'sinx

Ct.

,.

Dividing by 2 and renaming the constant of integration give

+' e , cosx dx -- e SfiX 2 e cosx + C .

J EXAMPLE 5

•

Obtain a fannula 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·- l x • cos x. Then we let

Solution

u

=

cosn-

1X

dv

and

cosxdx,

=

so that

du = (n - I) cos·- 2 x(-sinx dx)

v = sinx.

and

Integration by parts then gives

J

cos'xdx = cos·-lxsinx

+

(n -

I)J J I)J

sin2xcos·- 2xdx

= cos·- l x sin x

+ (n - I)

(1 - cos2 x) cos·- 2 X dx

= cos·-lxsinx

+ (n -

cos·- 2xdx - (n -

Ifwe add (n -

I)J

I)J

cos·xdx.

cos'xdx

to both sides of this equation, we obtain

J

n

cos' xdx = cos·- l x sin x

+

J

(n - I)

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 fannula 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 fann having the power reduced. When n is a positive integer, we may apply the fannula 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

2J

+3

= tcos2XSinX

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 ofthe Fundamental Theorem gives

Integration by Parts Formula for Definite Integrals

[j(x)g'(x) dx

=

j(x)g(x)

J: - [f'(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\e-X dx. -1

1

2

3

4

x

Letu = x,dv = e-xdx,v = -e-x,anddu = dx. Then,

1\e-x dx

1

=

-xe -x]~

=

[-4e-4 - (0)] +

=

-4e-4 - e-x]~

=

-4e-4 - e-4 - (-eO)

-

4

( -e-X)

1

dx

4

FIGURE 8.1

The region in Example 6.

e-X dx

= I -

5e-4 '" 0.91.

•

Tabular Integration

J

We have seen that integrals of the fonn j(x)g(x) dx, in which j 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 e"dx.

Solution

With j(x) = x 2 and g(x) = eX, we list:

j(x) and its derivatives

g(x) and its integrals

Integration by Parts

8.1

441

We combine tbe products oftbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain

J

x 2e' dx = x 2e' - 2xe'

+ 2e' + C.

•

Compare tbis witb tbe result in Example 3.

EXAMPLE 8

Evaluate

J

x 3 sinxdx.

Solution

Witb f(x) = x 3 and g(x) = sin x, we list:

f(x) and its derivatives

g(x) and its integrals

x3

(+)

sin x

3x 2

(-)

-cosx

(+) (-)

.... -sinx

6x 6

0

-----------

-------------------

~

cos x

~

sin x

Again we combine tbe products oftbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain

J

x 3 sinxdx = -x 3 cosx

+ 3x 2 sinx + 6xcosx

- 6 sin x

+ C.

•

The Additional Exercises at tbe end of this chapter show how tabular integration can be used when neitber function f nor g can be differentiated repeatedly to become zero.

Exercises 8.1 Integratton by Parts Evaluate the integrals in Exercises 1-24 using integration by parts.

1.

1

xsinIdx

3. Il'eosldl 5.

/,2

xlnxdx

7·lxe'dx 9·lx2 e- X dx 11. 13.

1 1

2. 1

4.

ocos 1rO dO

1

x 2 sinxdx

6. /,e x3lnx dx

8. 10.

Ixe

3x

I(X' - 2x +

l)e 2x dx

1

sin-1 ydy

tan-1ydy

12.

xsec2 xdx

14. 14Xsec'2xdx

p4e -P dp

16.

17. 1 (x' - 5x)e'dx

18.1 (r' + r + l)e'dr

19. 1 x'e'dx

20.

21. 1 e'sinOdO

22.

23.

dx;

1

15. 1 x'e'dx

1

e lx cos 3x dx

1 1

l'e 4t dl

e-Yeosydy

24. 1 e-2x sin2xdx

Using Substitution Evaluate !be integrals in Exercises 25-30 by using a substitution prior to integration by parts.

25.

1

e y,;+o tis

26.

/,lx~dx

442

27.

29.

Chapter 8: Techniques of Integration

1.

~/3

xtan2xdx

0

28.

J

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. JxsecX 2 dx

32.

33. JXQnx)2dx

34.

J~:

dx

36.

37. JX 3 e x'dx

38.

35.

39. 41. 43.

JX3~dx

J Je~

sin 3x cos 2x dx

sin eX fix

45. Jcos 47. 49.

1.

~/2

0

('

40. 42. 44.

~dx

46.

8 2 sin28d8

48.

tsec-1tdt

50.

1_ dx xQnx) 2 (lnx)3

J J J JVX J~eVidx x S e" dx

x 2 sin x 3 dx sin 2x cos 4x dx

e Vi

1. 1/V2x 0

1. 0

dx

3

cos 2x dx

a. about the y-axis. b. about the line x

d. What pattern do you see here? What is the area b-.eo the curve aod the x-axis for mr S x '" (n + I)", n ao arbilnlry

nonnegative integer? Give reasons for your answer.

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. ,,/2 S x S 3,,/2. 5,,/2.

c. 5,,/2 S x S 7,,/2.

1.

55. Finding volume Find the volume of the solid generated by revolving the region in the f11'St quadraot bounded by the coordinate axes and the curve y = cos x, 0 :5 x :s; 71'/2, about b. the line x

~

,,/2.

56. Finding volume Find the volume of the solid generated by revolving the region bounded by the x-axis aod the curve y = x sin x, :5 x:s; '11',about

°

a. the y-axis.

=

7T'.

57. Consider the region bounded by the graphs of y = In x, Y = 0, andx = e.

a. Find the area ofthe region. b. Find the volume ofthe solid formed by revolving this region

about the x-axis.

y 10

~

(See Exercise 51 for a graph.)

1T'

S

54. Finding volume Find the volume of the solid generated by revolving the region in the f11'St quadraot bounded by the coordinate axes, the curve y = e -r, and the line x = 1

b. the line x

~ 'IT.

b. 3,,/2 '" x

53. Finding volume Find the volume of the solid generated by revolving the region in the f11'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 enclosed by the curve y = x sin x aod the x-axis (see the accompaoying figure) for

x

10

J__ J

Theory and Examples

~

y

Jco~dx

~/2

12;v3

a. 0

«~) 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 ~ I.

= tan-I x, y = 0,

a. Find the area ofthe 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.

Find the average value ofy aver the intervaJ. 0

:!;';

,

t

:!;';

2'11".

443

Integration by Parts

8.1

The idea is to take the most complicated part of the integral, in this l(x), and simplify it fIrst For the integral of In x, we get

case

r

J

Inxdx

~

J

ye'dy

+C x + c.

= ye:1 _ e:1

= xlnx -

o

For the integral ofC08-1X we get

,

J

C08- l xdt - x COS-Ix

-I

y

cosydy

= X COS-IX - siny

=

em-I x

+C

Use the formula 60. Avenge value In a mass-spring-dashpot system l:ikc the on.c 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

61.

61.

J Ix.

I

x ncoudt = x"sinx - n sinXIU =

-x·

cosx

+

J

(4)

flY) dy

to evaluate the integrals in Exercises 67-70. Express your lUlSWel'S in terms ob.

:s: t :s: 2'11".

67.

X,,-l sinXIU

nI

r'(x) dx - xr'(x) -

o.

Reduction Formulas In Exercises 61--64, use integration by parts to establish the reduction

-.....

J 69.

J J

68.1

sin-lxdt

tan-lxdt

70·/IO~Xdt

I

sec- xdt

Another way to integrate rl(x) (when course) is to use integration by parts with u rewrite the integral. of 1 as

r

I

x·- l COSXIU

rl(x)dt

=

xrl(x)

r l is integrable, of r 1(x) and dv = dt to

-I x(~rl(x))dt.

(5)

Exm:ises 71 and 72 compare the:results ofusing Equations (4) and (5). 64.

J

(lnx)"dt = x(lnx)· -

nJ

(lnX)"-ldt

65. Show that

71. Equations (4) and (5) give diffen:ul: form.ulas for the integral. of cos-1x: L

J

cos-I xdt - x cos- l X

b.1 66. Use inqration by part9 to obtain the fonnula

2 2 J ~dt=lxVI="?+lJ

1 dx. VI="?

Inbgratfng Invems of functions Integration by parts leads to a rule for integrating inverses that usually gives good results:

J

r'(x)dx

~

J

~ yf(y) -

J J

~ xr'(x) -

flY) dy flY) dy

Tntcgndion by partI, with II

-"dv -I'(,)dy

cos-lxdt = Xcos- l X -

sin(OOS-l x )

+C

VI="? + C

Bq.(4) Bq.(S)

Can both integrations be correct? Explain. 72. Equations (4) and (5) lead to different form.ulas for the integral of tan-I x:

.. I b. J

tan-lxdx = Xtan-I x -Insec(tan-lx) tan-I xdt = Xtan-I x -Inv'1+"7"

+C

+C

Bq.(4) Bq.(5)

Can both integrations be correct? Explain.

, - rl(x), x - fey)
yf'(y)dy

-

Evaluate the integrals in Exercises 73 and 74 with (8) Eq. (4) and (b) Eq. (5). In each case, check your work by differentiating your answer with respect to x.

73.1

sinh-IXIU

74.1

tanh-lxdx

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+'x = (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).

J

If m i. even and n i. odd in sin"' x cos' x dx, we write n as 2k + 1 and use the identity cos2 x = 1 - sin2 X to obtain Case 2

cos'x = cos2k+'x = (COS2X)kcosx = (l- sin2X)kcosx. We then combine the single cos x withdx and set cosx dx equal to d(sinx). Case 3

If both m and n are even in . 2

_ 1 - cos2x

smx-

2

'

J sin"' x cos' x dx, we substitote 2 l+cos2x 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

sin3 xcos2 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

III

(I - COS2 X)cos2 x(-d(cosx))

sin.",,~ -d(co..)

4

(U

-

f

2 U )

du

co. .

Multiplytenns.

Evaluate

f Solution

u~

2 2 U )(U )( -du)

(I -

is odd.

COSS xdx.

This is an example of Case 2, where m = 0 is even and n = 5 is odd.

cossxdx = =

=

f

f f

cos4 xcosxdx =

f

(I - sin2 xj2d(sinx)

cosxdx

(I - u 2 j2du (I - 2u 2

u

=

=

d(sinx)

sinx

+ u 4 ) du

• EXAMPLE 3

Solution

Evaluate

f

sin2 xcos4 xdx.

f e-~OS2x)e if if

This is an example of Case 3.

f

sin2 xcos4 xdx =

=

=

=

For the term involving cos2

f

(I - cos2x)(1

(I

i [x +

+

+

~os2xy dx

m and n both even

+ 2cos2x + cos2 2x)dx

cos2x - cos2 2x - cos3 2x)dx

tsin2x -

f(COS22X + cos3 2x)dx

l

2x, we use

2

cos 2xdx = t f (I = t

+

cos4x)dx

(x + i Sin4X)'

Omitting the constant of

integration until the fInal result

446

Chapter 8: Techniques of Integration

For the cos3 2x term, we have

J

cos3 2xdx = =

J V

u = sin2t,

(I - sin2 2x)cos2xdx 2

(I - u )du

=!

du

t Sin32x ).

(Sin2x -

~

2 cos 2xd<

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"f

+ cos 20)/2 to eliminate a square root.

4

Solution

To eliminate the square root, we use the identity cos20 = 1+ cos20 2

With 0 =

+ cos4xdx.

VI

or

1+ cos20

= 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

r

lo

= Vi

[S~2x

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.

8.2 Trigonometric Integrals

447

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 +

Evaluate

1

x +

•

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 +

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 me dx,

1

sin mx cos me dx,

and

1

cos mx cos me 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 me = ! [cos (m - n)x - cos (m + n)x],

(3)

sin mx cos me = ! [sin (m - n)x + sin (m + n)x] ,

(4)

cos mx cos me =

21 [cos (m

- n)x + cos (m + n)x].

(5)

These identities come from the angie 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 Solution

sin 3x cos 5xdx.

From Equation (4) with m = 3 and n = 5, we get

j sin3xcos5xdx tj [sin (-h) + sinRx]dx tj (sinRx - sinh)dx =

=

_cosRx 16

=

+ cosh + C 4

•

.

Exercises 8.2 Powers of Sines and Co';nes Evaluate the integrals in Exercises 1-22.

13.

J J J J J J J

15.

Jo

1.

3. 5.

7. 9. 11.

cos2xdx

2.

cos3 x sin x dx

4.

3

sin xdx

6.

Jo

Jo

sin3 x cos3 x dx

12.

J

cos2 x dx

14.

10

17.1~ 8sin'xdx

18.

J J

16 8in2 xcos2 xdx

8 cos' 20 sin 20 dO

[~/'

3 cos' 3x dx

cos' 2x sin' 2x dx

[~/2

16.

2

sin xdx

J J

1

8in2 x

VI -

cosx

VI -

- ;osx dx

33. 35.

1.

'~/'

VI -

sin2xdx

~/2

32.

sec' x tan x dx

34.

sec' x tan x dx

36.

sec' x tan2 x dx

38.

0

2

~/'

sec' x dx

40.

41.

J

sec' 0 dO

42.

43.

1.~/2 esc' 0 dO

44.

sin2 20 cos' 20 dO

47.

J J

tanS xdx

48.

49.

1.~/3 cot' x dx

50.

45.

- cos2xdx

cos20dO

J J J

39.1

24.1~ VI

30.

J~ (I

- cos2 t)'/2 dt

Powers of Tangents and Secants Evaluate the integrals in Exercises 33-50.

~/'

23.12~~I

(Hint: Multiply by ~: :: :::.)

cos x dx sinx

Jo OVI -

8 cos' 2'ITx dx

Integrating Square Roots Evaluate the integrals in Exercises 23-32.

28.

[~/2

31.

Jor VI + sinxdx l6

dx

,

~

5,"/6

37.

[~/2

Jo

~/3

7cos7 tdt

20.1~ 8 sin'ycos2yth> 22.

29.

cos' 4x dx

10.

1.

'"/2

3 sffi 3 dx

sin' 2x cos 2x dx

cos3 xdx

7

. x

J J f sins~dx

8.

sin yth>

21.

[~

sinS xdx

[~/2

19.

27.

4tan'xdx

46.

~/'

J J J J J J

sec x tan2 X dx sec' x tan' x dx

sec' x tan2 X dx

eX sec' eX dx

3 sec' 3x dx 6

sec

1

~/'

X

dx

6tan'xdx

~/'

J J

cot' 2x dx

8 cot" tdt

8.3 Products of Sines and Cosines Evaluate the integrals in Exercises 51-56.

J

52.

J

53.

J~ sin3xsin3xdx

54.

J.1T/2 sinxcosxdx

55.

sin 11: cos 2x dx

J

cos 11: cos 4x dx

56.

sin 2x cos 11: dx

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.

63

•

J

sec x dx tan x

8.3

64.

esc x dx

66.

x sin' x dx

68.

J J

449

cotx dx

cos2 x

x cos' x dx

Applications

J~; cos x cos 7x dx

69. Arc length

Find the length of the curve

y

~

In (secx),

0:5 x:5 ",/4.

cos' 28 sin 8 d8

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.

sin' 8 cos 28 d8

71. Volume Find the volume generated by revolving one arch of the curve y = sinx about the x-axis.

J J J

72. Area

VI

sin 8 sin 28 sin 38 d8

J

3

sin

X

cos4 X

+

Find the area between the x-axis and the curve y cos4x,O ~ x ~

~

'IT.

73. Centroid Find the centroid of the region bounded by the graphs ofy = x + cos x andy = o forO ::=:; x::=:; 2'7T.

Assorted Integrations Use any method to evaluate the integrals in Exercises 63--{)8. 3

tan' x

67.

Exercises 57--{)2 require the use of various trigonometric identities before you evaluate the integrals.

57.

J J

65.

51.

Trigonometric Substitutions

74. Volume Find the volume ofthe 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.

dx

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 2 + x 2 , Va' - x 2 , and Vx' - a 2 into integrals we can evaluate directly since they come from the reference right triangles in Figure 8.2. Withx = atanlJ,

a2

+ x2

=

a2

+ a 2 tan2 1J

x'

=

a2

-

= a 2 (1

+ tan2 1J)

= a 2 sec2 1J.

Withx = asinlJ,

a2

-

a 2 sin2 1J

=

a'(l - sin'lJ)

=

a' cos2 1J.

a x=atan8

x=asin8

x=asecO

Va' + x' ~ aisecOI

Va' - x' ~ aleosOI

Vx'- a'~ aitanOI

FIGURE 8.2 Reference triaogles for the three basic substitutions identifying the sides labeled x and a for each substitution.

450

Chapter 8: Techniques of Integration Withx = asecO,

x2 6

"2

-

a2

= a 2 sec'0 -

a2

= a 2 (sec2 0 -

I) = a 2 tan2 0.

We want any substitution we use in an integration to be reversible so that we can change 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 ,;; 2'

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

(~)

"

if ~ ~ 1,

a. 2<0';;" if "'<-I

-2 "

To simplify calculations with the substitution x = a sec 0, we will restrict its use to integralsinwhichx/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 ofabsolute values, provided a > O.

6

.-J "

"

(J =

sec-1 ~

"2 x

-1

0 1

a

FIGURE 8.3 The arctangeot, arcsine, and arcsecant of x/a, graphed as functions of

x/a.

Procedure For a Trigonometric Substitution I. Write down the substitution for x, calculate the differential dx, and specify the selected values of 0 for the substitution. 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, 4

+ x2

= 4

dx = 2 sec2 0 dO,

+ 4tan2 0

= 4(1

_"<0<" 2 2'

+ tan2 0)

= 4 sec2 O.

8.3

21'

Then

JV4+7 dx

2

= =

FIGURE 8.4 Reference triangle for , = 2 tan 8 (Example 1):

=

tan8 = :t: 2

J J

2 sec2 0 dO V4sec2 0 =

J

2

sec 0 dO IsecOI sec9>Ofor-~<8<.!!.-

sec 0 dO

In IsecO

2

From Fig. 8.4

C.

-

sec 8 -

v'4+7 2 .

Notice how we expressed In 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

dx = 3 cosO dO,

x = 3 sinO, x2 =

9 -

9sin2 0 =

9 -

Then

JV9=7 x2dx

FIGURE 8.5 Reference triangle for , ~ 3 sin 8 (Example 2):

sin8

=

.! 3

and 00s8=

2

+ tan0l + C

=ln17 +~I +

and

451

Trigonometric Substitutions

V9=7 3

EXAMPLE 3

=

J J J

9(1 -

_

1r

2

sin2 0) =

<0< 9

2

cos2 O.

2

9sin 0·3cosOdO 13 cosOI 2

= 9

sin 0 dO

= 9

I -

-

cos8>Ofor-~<8<'!!.2 2

~OS20 dO

=

~ (0

=

~(O

=

~ (Sin-It -

=

~Sin-lt - ~V9=7 + c.

Sin 20) 2

+C

- sinO cosO) t·

+

C

sin28

~) + C

We first rewrite the radical as

=

2 sin 8 cos 8

Fig. 8.5

-

Evaluate

x

Solution

1r

2

>5'

452

Chapter 8: Techniques of Integration to put the radicand in the form x 2

x 2 x -

-

a 2 • We then substitute

2

2

dx = Ssec 0 lanO dO,

= Ssec 0

(~)' = =

is

- I) =

25

~lan20 25

~X2 (~)' ~ ItanOI ~lanO. =

-

j

dx

v'25x 2 -

FIGURE 8.6 Ifx = (2/5)sec6, 0< 6 < 'IT/2,then6 ~ sec- 1(5x/2), and we can read the values ofthe other 1rigonometric functions of6 from this right 1riangle (Example 3).

4 = =

=

j

dx

5v'x 2 -

(4/25) =

tj secOdO tin tin I~ + v'25~2

(2/5) sec 0 lan 0 dO 5' (2/5) lanO

IsecO

=

> ofor

tan 6

0< 0 < ",/2

=

With these substitutions, we have

j

<;

2~

sec20 -

~(sec20

0< 0

- 41

+ tanOI +

C

+ C.

Fig. 8.6

•

EXERCISES 8.3 Using Trigonometric Substitutions 19

Evaluate the integrals in Exercises 1-2S.

1

.

1v'9+7 fix

2

.

'/2

7.

1

fix

~

6.

8.

'\!'25=7dt

fix 9. , x 1 V4x 2 - 49 11.

1

>

Vy2 - 49 Y dy, Y

fix 13. • ~, 1 X 2 VX 2 - 1

1VI

3fix

21.

+ 9x 2

2 4.1. fix 2 oS+2x

3.12~2 -24+x 5. 1.0

•

x

-7 2

10.

> 7 12.

>

1

14.

1. ' /2Vi 0

1 1 1 1

2fix

5fix , x V25x 2 - 9

> -3 5

dy, Y

>5 31.

x

>

1

Assorted Integrattons Use any method to evaluate the integrals in Exercises 15-34. Most will require 1rigonome1ric substitutions, but some can be evaluated by other methods. x2 16. --2fix 4+x 9 - x2

15·1~fix . 1W+4

17

x'fix

18.

1 1

29

•

• 2fix ~, 2 X 3 Vx - 1

fix X2~

0

2 fix '/2' (x - 1)

33 ·

1 1 1 1

x

6

Sfix

(4x 2

22.

4x2 fix (1 _ x 2 )'/2

(1 - x 2 )'/2

27.

20·1~dw

w2

100 fix 36 + 25x 2

25.1

v'l=9?dt

y'

Sdw

w2

V3/2

23.

VI _ 4x2

V y 2 - 25

1 V4 1 1. +

x> 1

1

x Vx 2 - 4 fix

1 fix 24. 1.0 (4 _ x2)'/2

26.

1(x,x~~)'/2' 1 1 1 1 (I-x 2 )1/2

fix

1)2

x' fix x2 - 1

28.

30 • 32 •

4

34.

fix

6dt (9t 2 + 1)2

25

xfix 2 + 4x

(1 - r 2 )'/2

2

v dv (I - v 2 )'/2

x

x> 1

r

8

dr

In Exercises 35-4S, use an appropriate substitution and then a 1rigonome1ric substitution to evaluate the integrals.

35 •

1.

ln4

37.

I dt

Ve'"

0

['/

e

4

+9

2 dt

11/12 vi + 4tvl

36

•

lin

(4/')

In ('/4)

t dt

_-"e~~

(I

+ e,")'/2

38 {' dy . yVI + (lny)'

1,

8.4 Integration of RationaL Functions by PartiaL Fractions

.. J J · '\17=1 J · 'V1+? . . J)4 ;x JVx~
40.J~ l+x


xy';f-=l

J JVI

41

x
42 .~
43

x
44.

453

56. Consider 1he region bounded by the graphs of y - V:xtan-1 x and y - 0 for 0 :S X :s L Find the volume of 1he solid formed by revolving this region about the x-axis (see accompanying ftgure). y

(In x)'" xlnx tb

-, _ V:xtall- 1x


~fL-----4x

(Hint: Let x _ u 1.)

o

47.

57. Evaluate

Initial Value Problems Solve the initial value problems in Exercises 49-52 for y as a fimction ofx.

49. x dy_.ry-o dx - vx- - 4, x;;:: 2, y () 2 = 0 SO.

v'x2 -

51. (x 2

9 :

+ 4)

:

= I,

x> 3, y(5) = 1n3

= 3, y(2) = 0

-

W+I,

y(O) - 1

a. integration by pllrts. e. a trigonometric substitution. 58. Path of. water dder Suppose that a boot 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 UIlknown path y = j(x), as shown in the acccwnpanying figure.

Showthatf'(x)

=

AppLiCitions lod Exlmpl.es

V9 -

x'l

b. Solve the equation in pert (a) fur j(;&), using j(30) = O. y Y - f<x) path of mel'

Find the area enclosed by the ellipse

x2

-v'9~

(lIint: Assume that the skier is always pointed diIectIy at the boat and the rope is on a line tangent to the path y = j(x).)

53. Area Find the area of the :region in the first quadrant that is enclosed by the coordinate axes and the curve y = x 213. 54. Area

J x 3 ~_dx using

b. a v-substitution.

L

52. (x 2 + 1)2 :

1

y2

2+2=1.

a b 55. ConsidcIthcregi.on bmmdedbythe graphs ofy = sin-I :e,y = 0, andx = 1/2.

a. Find the area ofthe region. b. Find the centroid ofthe regicm..

8.4

~f-L-_-=';';:;!--=-_x

0];

(30,0)

Integration of Rational Functions by Partial Fractions This section shows how to express a rational function (a quotient ofpolynomials) as a sum

of simpler fractions, called partialfractions, which are easily integrated. For instance, the ,..tiona! function (S.% - 3)/(x' - 2x - 3) can be rewri_ as --,-5~x"c-;-,3"--:o ~ _2_ x'l 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 using certain transfonn methods to solve differential 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

ax =

5x - 3

+ l)(x - 3)

J_2_ ax x + 1

= 2ln

+ J_3_ ax

Ix + 11 +

x - 3

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+l 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 + 1) 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 (1) of fractions and regroup in powers of x, obtaining

5x - 3 = A(x - 3) + B(x + 1) = (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 U(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 2. Let x 2

455

+ px + q be an irreducible quadratic factor of g(x) so that + q bas no real roots. Suppose that (x 2 + px + q)n is the highest

x 2 + px

power of this factor that divides g(x). Then, to this factor, assign the sum of the n partial fractions:

Btx

+

Cl

c--.~-~c-

~+~+~

+

B2X

+

C2

~+~+~

Bnx + Cn + ... + 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

x 2 + 4x

+I

+ I)(x + 3) + B(x - I)(x + 3) + C(x - I)(x 2 = A(x + 4x + 3) + B(x 2 + 2x - 3) + C(x 2 - I) =

A(x

= (A

+B +

C)x 2 + (4A

+ 2B)x +

+ I)

(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 obo:

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 + 2j2

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

6x

= Ax

+ 2) + B + (2A + B)

Multiply both sides by (.

+ 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

6x+7 dx _ j ( _ 6 _ 5 )dx (x + 2)' x+2 (x + 2)'

j c:

=

6

=

6ln

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

(x 2

-2x + 4 + l)(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 + I)(x - 1)2

= Ax

x2

+B +~ + D + I x-I (x - I)"

(2)

8.4 Integration of Rational Functions by Partial Fractions

457

Clearing the equation of fractions gives -2x + 4 = (Ax + B)(x - 1)2 + C(X - 1)(x 2 + I) + D(x 2 + I) = (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 : Coefficients ofx

2

O=A+C

°

= -2A + B - C -2 = A - 2B + C

:

Coefficients of x I: Coefficients ofxo:

+D

4=B-C+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 fIrSt 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 (x 2

-2x + 4 = 2x + I _ _ 2_ + I I)(x - 1j2 x2 + I x - I (x - 1j2 .

+

Finally, using the expansion above we can integrate:

!

(x 2

-2x+4 dx_!(2x+I _ _ 2_+ I )dx I)(x - 1)2 ~2 + I x - I (x - 1j2

+

_!C'~_+_I

-

=

EXAMPLE 5

~2 + I

2_+ I )dx x - I (x - 1)2

10 (x 2 + I) + tan-I x - 210 Ix - II - x

~

I + C. •

Use partial fractions to evaluate

! Solution

x2 + I

x(x 2

a:: 1)2 .

The fonn of the partial fraction decomposition is =4+Bx+C+ Dx+E I x(x 2 + 1)2 x x2 + I (x 2 + 1)2

Multiplying by x(x 2 + I j2, we have I = A(x 2 + 1j2 + (Bx + C)x(x 2 + I) + (Dx + E)x = A(x 4 + 2x2 + 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

= x2 +

du

=

I, 2xdx

I I =Inlxl-zlnlul +2u+ K = In

= In

HIsTORICAL BIOGRAPHY

Oliver Heaviside (185
Ixl Ixl

I 2 -2ln (x

~

+

+

1)

+ (2

I

) 2x + 1

1 2(x2 + 1)

+K

+ K.

•

The Heaviside uCover-upw Method for Linear Factors When the degree ofthe polynomial I(x) is less than the degree of g(x) and

g(x) = (x - r,)(x - r2)··· (x - rn) is a product of n distinct linear factors, each raised to the Irrst 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 2

x + 1

(x - 1)(x - 2)(x - 3) Solution

=_A_+~+~ x - 2

x-I

x - 3·

(3)

If we multiply both sides of Equation (3) by (x - 1) to get x2 + 1 B(x - 1) C(x - 1) -o---"-::-cO---'----;;-c = A + + ----'--~ (x - 2)(x - 3) x - 2 x - 3

and set x = 1, the resulting equation gives the value of A:

(1)' + 1 (I - 2)(1 - 3) = A

+ 0 + 0,

A=1. Thus, the value of A is the number we would have obtained if we had covered the factor (x - 1) in the denominator of the original fraction

x2 + 1 (x - 1)(x - 2)(x - 3)

(4)

and evaluated the rest at x = 1:

(I)' + 1 A = <=1(c=x=-='1)~I"-:(1---2:-:-)(:-:-1---=-:-3)

n

Covor

2

(-1)(-2)

=

1.

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 thefactors (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:

Al

f(r,)

=

(rl - r2)'" (rl - rn) A = f(r2) 2 (r2 - rl)(r2 - 1'3) ... (r2 - rn)

3. Write the partia/fraction 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 dx 3x 2 - lOx .

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 rl = O,r2 = 2,andr3 = -5.Wefind Al =

0 +4 2)(0 + 5)

4 = _~ (-2)(5) 5

=

GJ (0 f

Cover

A2 =

21

2

+4

_6_ (2)(7)

=

I (2 + 5)

(x - 2)

1

=

7

f

Cover

A3 =

-5 + 4 (-5)(-5 - 2) I (x

+

=

5)

1

-I (-5)(-7)

I 35·

=--

f

Cover

Therefore,

x +4 x(x - 2)(x

+

= _~

+ 5)

I

3 7(x - 2)

5x

35(x

+ 5)'

and

J

x+4

x(x _ 2)(x

+

5)dx =

2

3

I 21 - 35 In

-slnlxl + 'lin Ix -

Ix + 51 + C.

•

Other Ways to Determine the Coeffidents 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 + 1)3

_A_ X + I

=

(x

+

B

(x

+ 1)2

+

C

(x

+ 1)3

by clearing fractions, differentiating the result, and substitoting x = -I.

Solution

We fIrst clear fractions: x - I = A(x

+ If + B(x + I) + C.

Substituting x = - I shows C = -2. We then differentiate both sides with respect to x, obtaining

+ I) + B.

I = 2A(x

Substituting x = - I shows B = I. We differentiate again to get 0 = 2A, which shows A = O. Hence,

x - I + 1)3

2

I

(x

(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

+ B(x

3)

= A(x - 2)(x -

- I)(x - 3)

+ C(X

-

I)(x - 2).

Then let x = 1,2,3 successively to f"mdA, B, and C: I = A(-I)(-2) + B(O) + C(O) 2 = 2A A=I (2)2 + I = A(O) + B(I)( -I) + C(O) 5 =-B

x = I:

x

x

(I)'

= 2:

+

B =-5 (3)2 + I = A(O) + B(O) + C(2)(l) 10 = 2C C = 5.

= 3:

Conclusion: 2

x + I =_1 (x - I)(x - 2)(x - 3) x-I

5_+_5_ x - 3.

•

x - 2

Exercises 8.4 Expanding Quotients into Partial Fractions Expand the quotients in Exercises 1-8 by partial fractions.

1 5x-13 • (x - 3)(x - 2)

3.

x+4 , (x + I)

5 z + I 'z'(z-I) 7. , /

/' + 8 - 5t + 6

5x - 7 2. , x -3x+2

4 2x + 2 'x'-2x+1

z

6

11./ 13.

x+4 dx x'+5x-6

/.S ,

15 / •

ydy , y -2y-3 dt t'+t'-2t

8. t'

+9 + 9t'

10./~ x + 2x

12./ l

2x+1 dx x' - 7x + 12

+4 'Y --dy 1/2 y' + Y

14.

x + 3 dx 2x' - 8x

16. /

2

l' _

•

-1

x2

-

2x + 1

~ I)'

2

x2 dx

20 / •

,

+

• , (x

dx 1)(x 2 + I)

y2+ 2Y + I

(y2 + I)' dy

23. /

Repeated Linear Factors 10 Exercises 17-20, express the integrand as a sum of partial fractions and evaluate the integrals. 17 (1-c;-__x,--'__dx _ 18 ~ x,-'__ dx _

· Jo x +Zx+ 1

],

·z3_ z 2_6z

t'

(x2

(x - 1)(x

+ 2x + I)

Irredudble Quadratic Factors 10 Exercises 21-32, express the integrand as a sum ofpartial fractioos and evaluate the integrals. 21

Nonrepeated Linear Factors 10 Exercises 9-16, express the integrand as a sum of partial fractions and evaluate the integrals.

9./~ I - x

19. /

25/ 23+2 tis • (8 2 + 1)(8 - I)' 27. / x

2 -

X

+ 2 dx

x' - I X2

29. /

-,--dx x - I

22.

I

v33t2+t+4 , dt 1 / + t

24./ 26. /

8x2

+ 8x + 2 dx (4x 2 + I)'

8' 2 8(8

+ 81 tis

+ 9)'

28. /_,_I_ dx x +x 30./

2 x +x dx x'-3x 2 -4

2

31. /

26' + 56 + 86 + 4 d6 (8' + 26 + 2)'

32. /

6' - 46' + 28' - 36 + I d6 (8' + I)'

Improper Fractions 10 Exercises 33-38, perform loog c1ivisioo 00 the integrand, write the proper fraction as a sum. of partial fractions, and then evaluate the integral. 2 x' 33./ 2x'-2x +l dx 34. -2--dx x2 - x / x - I

462

Chapter 8: Techniques of Integration

35. /9X'-3X+ I dx 3 x

y' 37.

/

36. /

xl

-

+ y2 - I , dY y +Y

16x'

4x 2 -4x+1

56. The y-axis

dx

y

_

y - (x

2'

II--.

38 y dY • / y' - y2 + Y - I

2

+ 1)(2 x) ......

Evaluating Integrals Evaluate the integrals in Exercises 39-50.

39.

e'dI e + 3e t + 2

40. /

cosydY sin2 y + siny - 6

42. /

U

/ 41. /

-;O+-----~I--+x

2t , e 4t + 2e -edt 2t e + 1 0082 8

o

sinO dO

+ cosB - 2

o

(x - 2)'tan- 1 (2x) - 12x' - 3x 43.

2

(4x + I)(x - 2)

/

(x 44.

dx

2

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.

by

58. Find the x-coordiuate of the centroid ofthis region to two decimal places. y

I)'tan- I (3x)

+

+ 9x' + x 2 dx (9x + I)(x + I) 2

/

45. /

x

'/2 I VXdx x

46. /

(x

(3, 1.83)

1/3 I VXdx - I) x

(5,0.98)

(Hinl: Letx = u'.)

47.

(Hinl: Let x 49./

x(x'

+ I)

x x

9

I x' (x s

+ 4)

o

o

I = u2.)

+

I

~f-h~---lc-----~-_x

48./ ~dx +

/~dx dx

50/

•

dx

by::.)

( Hinl: Multiply

Inltfal Value Problems Solve the initial value problems in Exercises 51-54 for x as a function of I. 51. (1 2 - 31 52. (31' 53. (1 2 54. (I

+ 2):

+ 41 2 +

(I

>

I ) : = 2V3,

+ 21): ~

+ I):

= I

2x

= x2

+2

(I,x

2),

x(3) = 0

0),

x(l)

+ I (I > -I), x(O)

~

5

59. Social diffusion Sociologists s0n3ctimes use the phrase "social diffusion" to describe the way information spreads through a population. The information might be a rumor, a cultural fad, or news about a tecbnica1 innovation. In a sufficiently large population, the nU03ber of people x wbo bave the information is treated as a differentiable function oftime I, and the rate of diffusion, dx/dl, is asSU03ed to be proportional to the number of people who bave the information times the number of people wbo do not. This leads to the equation

dx ~ kx(N - x) dl ' where N is the nmober of people in the population. Suppose I is in days, k = 1/250, and two people start a TUmor at time I = 0 in a population of N = 1000 people.

x(1) = -1fV3/4

>

3

a. Find x as a function of t. I

b. When will balf the population bave heard the rumor? (This is when the rumor will be spreading the fastest.)

o 60. Second-order chemical reactions

= 0

Appllcatfons and Examples In Exercises 55 and 56, fmd the volwne of the solid generated volving the shaded region about the indicated axis.

by re-

Many chemical reactions are the result of the interaction of two molecules that undergo a cbange to produce a new product. The rate of the reaction typically depends on the coocentrations of the two kinds of molecules. If a is the amount of substance A and b is the amount of substance B at time I = 0, and ifx is the amount ofproduct at time I, then the rate of formation of x may be given the differential equation

by

55. The x-axis y~

y

~

k(a - x)(b - x),

or

I dx ~ k (a - x)(b - x) dl '

2

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

o

dx dl

3

Y3x-x 2 (0.5, 2.68) (2.5, 2.68)

0.5

2.5

where k is a constant for the reaction. Integrate both sides of this equation to obtain a relation betweeo x and I
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 BriefTable ofIntegrals is provided at the back ofthe 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 S requires n # - I , for example, and Formula 11 requires n # -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 # 0, and Formulas 13a and 13b cannot be used unless b is positive.

EXAMPLE 1

Find

J

+ Sri dx.

x(2x

We use Formula 23 at the back ofthe book (not 22, which requires n # -I):

Solutton

J

x(ax

+ b)-ldx

~-

=

:2

In

lax + bl + c.

With a = 2andb = S,wehave

J

x(2x

EXAMPLE 2

+

Sri dx =

I - ~ In 12x + SI + c.

•

Find

Jx~' Solutton

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 -

V4

+

+ C.

2

JV

n . -1 x n+ 1 . -1 a x sm axdx--+ Ism ax--+ n n I

J

C = tan-I ~x - 2

Find

We begin by using Formula 106:

Solutton

4

4

x n+ 1 dx 1 _ a 2x 2

'

n # -1.

•

464

Chapter 8: Techniques of Integration Withn = 1 and a = 1, we have

1/

2

. -1 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:

2 -:-;c=x:,;=2=",dx = a sin-I 2 / Va2-x2

(~)

_ lxVa 2 _ x 2 + C. 2

Witha = 1,

x2dx ~

/

1. -1 X - -1'''------'1 =-sm X V I - X -2 2 2

+ C.

The combined result is /

X

sin-1xdx = x; sin-I X

-

2

=

tx~ + C)

t (tsin-1 x -

x 1). -1 ( 2-4 sm x

+ 41 XV1 '''------'1 -x-2 + C' .

•

Reduction Formulas The time required for repeated integrations by parts can sometimes be shortened by applying reduction fonnulas like

tannxdx = _l_ tann -l x - /tan n- 2xdx n- 1

/ /

sinnxcosmxdx=

(lnx)n dx = x(lnx)n - n/ . n-l

m+1

sm mX~O:

1/

X+~~n

(1)

(lnxr l dx

sinn-1xcosmxdx

(2)

(n"" -m).

/

(3)

By applying such a fonnula repeatedly, we can eventually express the original integral in terms ofa power low enough to be evaluated directly. The next example illustrates this procedure.

EXAMPLE 4

Find

5 / tan xdx. We apply Equation (1) with n = 5 to get

Solution

5 4 / tan xdx = ±tan x - /

3 tan xdx.

We then apply Equation (1) again, withn = 3, to evaluate the remaining integral:

/

2 tan3xdx = ttan x - /

2

tanxdx = ttan x

+ In Icosxl + c.

The combined result is

/ tan5xdx = ±tan4 x - ttan2 x -In Icosxl

+ C'.

•

As ilieir 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 algebra 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 integral ofthe function

j(x) = x 2Ya 2 + x 2. Using Maple, you first define or name the function:

> j:=

x A2 * sqrt (a A2

+ x A2);

Then you use the integrate command on j, identifYing the variable of integration:

>

int(j,x);

Maple retorns the answer

t

x(a 2 + x 2)3/2 - ta2xYa2 + x 2 - t a4ln (x + Ya 2 + x 2).

If you want to see ifthe answer can be simplified, enter

>

simplify(%);

Maple retorns

ta2xYa2 + x 2 + tx3Ya2 + x 2 - ta4ln (x + Ya 2 + x 2). If you want the def'mite integralfor 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 4ln(2) 64

32

8

- ta4ln(1T + Y4a 2 +,.2) + l~a4ln(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 + tln(YZ EXAMPLE 6

Solution

•

I).

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

+ I~

cos(xj2sin(x)

+

2 15 sin(x).

•

Compoter algebra systems vary in how they process integrations. 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 [IJ:= 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]'2' 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 ofrandom 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

x

dx ,

f

e (.'j d!:,

f

f~

d!:,

VI - k'sin2 xd!:,

f

In (In x) d!:,

0

f

sinx d!: x '

< k < I,

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 I, because they are continuous. However, none ofthe antiderivatives are elementary. None ofthe 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 8.5 Using Integral Tables

2

Use Ibe table of integrals at Ibe back oflbe book to evaluate Ibe integrals in Exercises 1-26.

1/ ·

2/ · xy';+4

dx x~

38.

Yx 2 -

/

4x

+5

dx

Y5 - 4x - x 2 dx

39. /

x 2 Y2x - x 2 dx

40. /

dx

x

Using Reduction Fonnulas

xdx

4 /

xdx (2x + 3)'/2

Use reduction fonnulas to evaluate Ibe integrals in Exercises 41-50.

5./x~dx

6. /

x(7x + 5)'/2 dx

41. /

7·/7

8 / dx • x 2Y4x - 9

3 /

·

9. /

11 /

·

13.

\Ix=2

•

dx

xY4x -x2 dx

10.

/~dx dx x~

12 /

dx xv7+7-

·

/~ dx

14.

/~dx

15. /

e 2t cos 3t dt

16. /

e-3t sin 4t dt

17. /

X

cos-1 xdx

18. /

xtao-l xdx

19. / 21. /

X

2 tao- l xdx

sin 3x cos 2x dx 8 sin 4t sin

25.

8 8 coS:3 cos 4 d8

/

24•

. t . t d

Sffi

/

3

S1n"6

t

cOS~COS78d8

26. /

31. / 33. /

34./ 36. /

2 sin2 t sec' t dt

45. /4tao'2xdx

46. /8COt'tdt

47. /

2 sec' 7fX dx

48. / 3 sec' 3x dx

49. /

csc' x dx

50. /

e'sec' (e' - I)dt

l' 0

(x 2 + 1)2 dx

28.

l

sin- ...;xdx

...;x

~

dx

cottYI - sin2 tdt,

0

/X2+6x (x2

+ 3)' dx

30. /

cos-:rx...;x dx

32. /

~ dx ...;x


7f/2

dt taotY4-sin2 t

35/

tao-I v'Ydy

37. /

•

16x'(ln x)2 dx

55. /, 1

dy yY3+(lny)'

I dx Yx 2 + 2x + 5 (Hint. Complete Ibe square.)

cs~ d8

52. /

2W+1 dx

54.

2(r2 - 1)'/2 r dr

56.

1Y3/2 dy o (I - y2)'/2

1'

;Y3

0

dt (~

/2

+ If

Applications 57. Surface area Find the area of !be surface geoerated by revolving the curve y = 0 S x S about the x-axis.

W+2,

X'+X+I

29. /

44. /

53.

In Exercises 27-40, use a substitution to change Ibe integral into one you can fmd in Ibe table. Theo evaluate Ibe integral.

/

43. / sin2 28 cos' 28 d8

51. /

Substitution and Integral Tables

27.

8 cos' 27ft dt

sin 2x cos 3x dx

22. /

I dt

23. /

42. /

Evaluate Ibe integrals in Exercises 51-56 by making a substitution (possibly trigonometric) and Iben applying a reduction formula.

m:~' x dx

20. /

sin' 2x dx

58. Arc length

v'2,

Find Ibe lengtb of the curve y = x 2 , 0 S

X

S

v3/2. 59. Centroid Find !be centroid of the region cut from the first quadrant by the curve y = I/~ and Ibe line x = 3. 60. Moment about y-axis A thin plate of constaot deosity 6 = I occupies the region enclosed by the curve y = 36/(2x + 3) and Ibe line x = 3 in the f11"St quadrant. Find Ibe momeot of !be plate about the y-axis.

D 61.

Use the integral table and a calculator to fmd to two decimal places Ibe area of the surface geoerated by revolving Ibe curve y = x2 • -1 :s x :s 1, about the x-axis.

62. Volume The bead of your fmn's accouoting department has asked you to fmd a formula she can use in a computer program to calculate Ibe year-<:nd inveotory ofgasoline in the company's tanks. A typical tank is shsped like a right circular cylinder ofradius r and lengtb L, mounted borizontally, as shown in !be accompanying figure. The data COD3O to the accounting office as deptb measuremeots 1aken wilb 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

l:

V = 2L

r d

+

J

d. What pattern do you see? Predict the formula for x 4 In x dx and then see if you are correct by evaluating it with a CAS. e. What is the formula for x n lnx dx, n ~ I? Check your answer using a CAS.

J

v;:z=yz dy .

66. Evaluate the integrals

b. Evaluate the integral.

a.

y

J~;

dx

J1;3

X

b.

dx

c.

JI;:

dx.

d. What pattern do you see? Predict the formula for - Measuring stick

J

r

lnx dx

x5

and then see if you are correct by evaluating it with a CAS. d = Depth of gasoline

-rFL

e. What is the formula for

J

lnxd n x, x

n

~

2?

Check your answer using a CAS.

63. What is the largest value

67. a. Use a CAS to evaluate [7T/2 sinn x Jo sinn x + cosn x dx

can have for any a and b? Give reasons for your answer.

where n is an arbitrary positive integer. Does your CAS find the result?

64. What is the largest value

l

b

b. In succession, find the integral when n = 1,2, 3, 5, and 7. Comment on the complexity of the results.

xV2x - x 2 dx

can have for any a and b? Give reasons for your answer. COMPUTER EXPLORATIONS In Exercises 65 and 66, use a CAS to perform the integrations.

65. Evaluate the integrals

a.

J

xlnxdx

8.6

b.

J

2

x 1nxdx

c.

J

3

x lnxdx.

c. Now substitute x = ('1T /2) - u and add the new and old integrals. What is the value of 7T/2

1 o

.

sinn x

SInn x

+ cosn X

dx?

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), I/ln x, and Vl+?, have no elementary formulas. When we cannot find a workable antiderivative 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 ofnumerical integration. In this section we study two such methods, the Trapezoidal Rule and Simpsons 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 + YZ)6.x

'1

Yz

YII-l

Yn

FIGURE 8.7 The Tn1pezoidal Rule approximates short s1retches ofthe curve y ~ /(x) with line segments. To approximate the integral of/ from a to b, we add the areas ofthe trapezoids made by joining the ends ofthe segments to the x-axis.

Figure 8.7. It is not necessary for the subdivision points Xo, Xl> 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 ofthe trapezoid that lies above the ith subinterva1 is

a.: (Y'-1 2+ Y')

a.:

=T(Y'-I+Y'),

where Y'_I = /(X'_I) andy, = /(x,). This area is the length a.: ofthe trapezoid's horizontal "altitude" times the average of its two vertical "bases." (See Figure 8.7.) The area below the curve Y = /(x) and above the x-axis is then approximated by adding the areas of all the trapezoids:

T

I

= Z(yo

I

+ yt!a.: + Z(YI + Y2)~x + ...

I I + z(Yn-2 + Yn-I)a.: + z(Yn-l + Yn)a.: =

a.: (!YO + Yl

+ Y2 + ... + Yn-I + !Yn)

a.:

= T(YO + 2Yl + 2Y2 + ... + 2Yn-1 + Yn),

where Yo = /(a),

Yl = /(Xl),

... ,

Yn-I = /(Xn-I),

Yn = /(b).

The Trapezoidal Rule says: Use T to estimate the integral of/ 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 + Yn).

To approximate

=

The y's are the values of j at the partition points XQ

=

where

a,Xl = a + .4x,x2 = a ~ =

+ 2.6.x, ... ,Xn-l

=

a + (n -l)4x,xn = b,

(b - a)/n.

EXAMPLE 1

Use the Trapezoidal Rllie with n = 4 to estimate estimate with the exact value.

o

1

5 6 7 2 "4 "4 "4

FIGURE 8.8

The trapezoidal approximalion ofthe area under the graph of y = x 2 from x = I to x = 2 is a slight overestimate (Example I).

TABLE 8.2

x

y

I

I

5 4 6 4 7 4 2

25 16 36 16 49 16 4

=

J,2 x 2 dx. Compare the

x2

Solution Partition [1, 2] into four subintervals of equallengtb (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 Rllie, 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 ofthe curve with parabolas.

Another rule for approximating the definite integral of a continuous function resllits from using parabolas instead of the straight line segments that produced trapezoids. As before, we partition the interval [a, b] into n subinterva1s of equal length h = ~ = (b - a)/n, but this time we require that n be an even number. On each consecutive pair ofinterva1s 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-lo Yi-,), (Xi, y,), and (Xi+l,Yi+l) on the curve. Let's calcu1ate 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

(0, Yl)

...--r-..... (h,y,)

y = Ax2

y =Ax 2 +Bx+ C

so the area under it from x

-h to x

=

y,

Yo

Ap =

0

h

FIGURE 8.10 By integratiog from -h to h, we fmd the shaded area to be

3h (Yo + 4Yl + Y2).

1:

+ Bx + C,

=

his

2

+ Bx + C)dx

(Ax

2 ]h Ax , =_+Bx +CX 3 2 -h

-_C"-_+---'---~x

-h

471

X2 = h (Figure 8.10), where h = dx = (b - a)ln. The area under the parabola will be the same ifwe shift the y-axis to the left or right. The parabola has an equation of the fann

y (-h,y.,)

Numerical Integration

= 2Ah'

3

+

2Ch =

k (2Ah 2 + 6C). 3

Since the curve passes tbrough the tbree points (-h, Yo), (0, YI), and (h,Y2), we also have

Yo

= Ah

2 - Bh + C,

YI

=

C,

Y2

= Ah

2 + Bh + C,

from which we obtain C 2 Ah - Bh Ah 2 + Bh

=

Y"

=

Yo - Y"

= Y2 - Y" 2 2Ah = Yo + Y2 - 2Yl.

Hence, expressing the area Ap in terms of the ordinates Yo, Yl , and Y2, we have

Now shifting the parabola horizontally to its shaded position in Figure 8.9 does not change the area under it. Thus the area under the parabola tbrough (xo, Yo), (X"YI), and (X2' Y2) in Figure 8.9 is still

Similarly, the area under the parabola tbrough the points (X2, Y2), (x, ,y,), and (X4,Y4) is

Computing the areas under all the parabolas and adding the results gives the approximation

HISTORICAL BIOGRAPHY

Thomas Simpson (172IH761)

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 suhintervals.

472

Chapter 8: Techniques ofIntegration

Simpson's Rule To approximate

1: f(x) dx, use

S

=

dx T(YO

+ 4Yl + 2Y2 + 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 ofthe coefficients in the above rule: 1,4,2,4,2,4,2, ... ,4, 1.

EXAMPLE 2

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

o

0

1 2

5 16 5 405 16 80

1 3

2 2

2

Use Simpson's Rule with n = 4 to approximate f0 5x 4 dx.

S = = =

~ VO + 4Yl + 2Y2 + 4)/3 + Y4)

i (0 + 4(;6) + 2(5) + 4(4I°i) + 80) 32 I

12 .

This estimate differs from the exact value (32) by ouiy 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 ofthe def"mite integral

1:

THEOREM l-Enor 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

'frap<=idal Rille

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 ofthe integral off from a to b for n steps satisfies the inequality

IE I ,;; M(b

- a)5 I80n 4

s

Simpson~ Ru1c

To see why Theorem I 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

f(x) dx

=

b - a T - - - . l"(c)(dx)2 12

8.6

Numerical Integration

473

for some number c between a and b. Thus, as llx 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, bj, 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, bj, so that Ij"(x)l :5 M on [a, bj, then

Ifwe substitute (b - a)Jn for Jlx, we get

a)3

< M(b -

IE T I -

12n

2

To estimate the error in Simpson's rule, we start with a result from advanced calculus that says that ifthe fourth derivative /4) is continuous, then

l

b

j(x) dx

=

a

S _ b - a • f4l(c)(Jlx)4 180

for some point c between a and b. Thus, as llx 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, bj, 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 ofmax I 4l (x) Iand have to replace it with an upper bound. IfM is any upper bound for the values of If4l 1 on [a, bj, then

f

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

To estimate the error, we f'mt 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

Solution

474

Chapter 8: Techniques ofIntegration the constant value f4 l(x) = 120, we take M error estimate for Simpson's Rille gives E

I

I< s -

= 120.

With b - a

M(b - a)5 = 120(2)5 180n4 180'44

= 2

and n

= 4,

the

I 12·

This estimate is consistent with the resillt of Example 2.

•

Theorem I 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 I, if we choose the number of subintervals n to satisfy M(b - a)5 -4 180n

<

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)5 I ---'-'04 <4 180n

10

or, equivalently, n

4>64'10

It follows that

n> 10

4

3.

/ (T64)1 '" 21.5. 4

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 integral

As we saw in Chapter 7, the value of In 2 can be calculated from the 2

1n2 =

1 1

x1 dx .

ft.2

Table 8.4 shows T and S values for approximations of (I/x) dx using various values of n. Notice how Simpson's Rille dramatically improves over the Trapezoidal Rille. TA8LE 8,4 Trapezoidal RuLe approximations (T,) and Simpson's RuLe approximations (5,) of In 2 = (l/x) rJx

J:

II

10 20 30 40 50 100

T.

IError I less than ...

S.

IError I 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

Numerical Integration

475

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 thefour/h. This has a dramatic effect as ~ = (2 - I)/n 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

b-a 2 b-a 2 ET = -----rz f"(c)(~) = -----rz(O)(~x) =

146 ft

Vertical spacing

~

The dimensions ofthe SWllDlp in Example 6.

FIGURE 8.11

20 ft

o.

The Trapezoids! 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 Trapezoids! approximations in theory, it may fail to do so in practice. When ~ is very small, say ~ = 10-5 , 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? Solutfon 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 Estlmattng 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 Trapezoidal Rule a. Estimate the integral with n = 4 steps and f"md an upper bound for IETI. b. Evaluate the integral directly and f"md IETI .

e. Use the fannula ( IETI /( trae value)) X 100 to express IETI as a percentage ofth.e integral's true value.

II. Using Simpson'. Rule

a. Estimate the integral with n bound for IEsl.

= 4 steps and f"md an upper

b. Evaluate the integral directly and f"md IEsl .

e. Use the formula (IEs l/(tme value)) X 100 to express IEsl as a percentage of the integral's trae value.

2.1'

(Z>; - I)
476

Chapter 8: Techniques ofIntegration

3.1:

(x

5./,2 (I'

1

4.1:

(x 2

-

I) dt

+ I)dl

6.1:

(I'

+

I)dl

I

2

7.

+ I) dt

2

2 tis

8.

1 •

J2f'( s -I 1ftis

10./,1

9./,~ sinldl

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 Trapezoidsl Rule and (b) Simpson's Rule. (The integrals in Exercises 11-18 are the integrals from Exereises 1-8.)

12·1'

ll·l'xdt

13.1:

(x

15. Jo['

(I'

1

14.1: 16.1

+ I) dt

2

(I'

-

I) dt

+

I) dl

Yo ~ 1.5 ft, Yl ~ 1.6 ft, Y2 ~ 1.8 ft, Y, ~ 1.9 ft, Y4 = 2.0 ft, Ys = Y6 = 2.1 ft Horizontal spacing

1

I tis 2(.-1)2

18.

'\I';+l dt

' I 20. /, • ~dt o yx + I

+ I) dt

22.1: cos (x

26.

+ 1t) dt

Position (ft)

x

o 5 10 15 20 25

[SO

Jo

Depth (ft) h(x) 6.0 8.2 9.1 9.9 10.5 11.0

30· h(x) dt.

Position (ft)

x 30 35 40

45

50

Depth (ft) h(x)

11.5 11.9 12.3 12.7 13.0

24. Distance traveled The accompanying table shows time-tl>speed dsta 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 corve, but be careful: The time intervals vary in length.)

= 1 ft

on consumption on Patbf"mder Island A diesel generator runs continuously, consuming oil at a gradually increasing rate until it must be temporarily shut down to have the filters replaced. Use the Trapezoidsl Rule to estimate the amount of oil consumed by the generator daring that week.

Estimate. with Numerical Data 23. Volume of water in a swimming pool A rectangular swimming pool is 30 ft wide and 50 ft long. The accOOJpanying 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 Trapezoidsl Rule with n = 10 applied to the integral

V=

2.2 3.2 4.5 5.9 7.8 10.2 12.7 16.0 20.6 26.2 37.1

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/ft' . Estimate the length ofthe tank by SinJpson's Rule.

4

2 tis

21./,2 sin (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

_:::::::::::::=::===~}illffi)

-1

1 •

19. /,'

(x

Time (sec)

25. Wmg de.ign The design of a new airplane requires a gasoline

(2x - I)dt

1

+ I) dl

I

2

17.

2

Speed change

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.Q35

Theory and Example. 27. U.able \'lIIue. of the sine-integral function

The .ine-inlegral

jUnction, Si(x) =

/,% S~I dl,

"Sine intcgml ofx"

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 nornerlcal integration. Although the notation does not show it explicitly, the function being integrated is /(1) =

{

sinl I'

I

~

0

I,

t

=

0,

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 =

x

D 33.

Your metal fabrication company is bidding for a contract to make sheets of corrugated iron roofing like the one shown here. The cross-sections ofthe corrugated sheets are to conform to the curve . 37r y -- sm 20 x,

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

+ cos2 xdx.

27f

7f

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 ( ) =

1'" VI

Estimate L by Simpson's Ru1e with n = 8. t dt

t

__-t--.L

477

AppLications

D 32.

y

Si (x)

NumericaL Integration

Jo sU; 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 function, 2 [X Y; Jo

erf(x) =

e-

I

2

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 antiderivative 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

y = 25 cos (7fx/50)

J:

f(x) dx is a 29. Prove that the sum T in the Trapezoidal Rule for 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 ofthe surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.

b

turns out to be Length = 4a

["'/2

Jo \h -

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 = x 2/4,

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_ dx.

Jo

1

+ x2

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 domain of integration [a, b] be finite. Second, that the range ofthe integrand be f"mite on this

y 0.2 0.1

o

2

3

4

5

6


x

domain. In practice, we may encounter problems that fail to meet one or both ofthese conditions. The integral for the area under the curve y = (lnx)/x 2 from x = 1 to x = 00 is an example for which the domain is inf"mite (Fignre 8.l2a). 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 inf"mite (Fignre 8.l2b). In either case, the integrals are said to be improper and are calcnlated as limits. We will see in Chapter 10 that improper integrals play an important 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 inf"mite 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)

J: =

= J.b e--./2dx = _2e-x / 2

-2e-b/2 + 2

(b)

FIGURE 8.12 Are ilie areas under iliese inf"mite curves finite? We will see that the answer is yes for both curves.

Then f"md the limit of A(b) as b ---> 00 lim A(b)

=

b---+oo

lim (-2e- b/ 2 + 2)

2.

=

b-"J>oo

The value we assign to the area under the curve from 0 to 00 is J.o

oo e--./2dx =

lim J.b e--./2 dx

=

2.

b_oo 0

y

DEnNmON Integrals with inf"mite limits of integration are improper integrals ofType I. 1. If f(x) is continuous on [a, 00), then

---+-----====--=-=.x

1

00

a


f(x) dx = lim

l

b

b-co a

f(x) dx.

2. If f(x) is continuous on (-00, b], then

3. If f(x) is continuous on (- 00, 00), then b (bJ

FIGURE 8.13
J.':,f(X)dx =

J~f(X)dx +

1

00

f(x)dx,

where c is any real number. In each case, ifthe 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 shown 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 bas the tmite value 2. If f ;;,: 0 and the improper integral diverges, we say the area under the curve is infinite.

Is the area under the curve y = (lnx)lx 2 from x = I to x = so, wbat is its value?

EXAMPLE 1

y

00

finite? If

Solutfon We find the area under the curve from x = 1 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 = Inx, dv = th/x 2, du ~ dx/., v ~ -1/.

[~:dx = [(In x) (-i)]: -[ (-i) (i) dx

0.2 0.1

o FIGURE 8.14 The area under this curve is an improper integral (Example I).

The limit ofthe area as b --+

1

00

1

00

is

lnx dx = lim x2 b_oo

Ib

lnx dx x2

1

= lim [_lnb _1+ I] b_oo b b =-[Iimlnb]-O+l

b

b_oo

= - [lim

lib]

b-OO 1

+ I = 0 + 1 = 1.

l'H6pital~ Rule

Thus, the improper integral converges and the area has tmite value I.

EXAMPLE 2

Evaluate dx

1

00

-oo1+x 2 '

Solutfon

According to the detmition (part 3), we can choose c = 0 and write

1

dx = -oo1+x 2

HISTORICAL BIOGRAPHY

Lejeune Dirichlet (1805-1859)

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 =

. hm

1°

0_-00

-dxa 1 + x2

lim tan-I x]O 0 _ - 00

a

•

480

Chapter 8: Techniques of Integration

1"

1

. -dx - - hm o 1 + x 2 - b_oo

0

-dx1 + x2

lim tan-1

=

b-oo y

=

b

x]b 0

lim (tan- t b - tan- 1 0) =

b~""

°

=

7f -

2

7f

2

Thus,

-==--~I---=~x

o

NOT TO SCALE

FIGURE 8.15 The area under this curve is fmite (Example 2).

Since l/(l + 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

If p

I

1

'¢'

bdx -

xP

J,"" dx/x

P

converge? When the in-

1,

=

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' { 00,

P

< 1

because lim _1_ = {O,

b-oo bP - 1

00,

p>1 p<1.

Therefore, the integral converges to the value 1/(p - 1) if p

p<1.

>

1 and it diverges if

8.7 Improper Integrals

481

If p = 1, the integral also diverges:

1 1 ax 1ax 00

00

axP =

x

1

1

X

b

lim

=

b_oo 1

lim lnX]~

=

b_oo

lim (lnb - In 1) =

=

b_oo

•

00.

Integrands with Vertical Asymptotes

y

Another type of improper integral arises when the integrand has a vertical asymplole---an infmite discontinuity----at a limit of integration or at some point between the limits of integration. Ifthe integrand f is positive over the inlerva1 ofintegration, 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 quadrant that lies under the curve y = l/YX from x = 0 10 x = 1 (Figure 8.l2b). First we fmd the area ofthe portion from a 10 1 (Figure 8.16).

y~_l_

Vi

l' ~

~f-L----'---~x

o

X

a

1

FIGURE 8.16 The area under this curve is an example of an improper integral of the second kind.

= 2YX]: = 2 - 2va.

Then we find the limit ofthis area as a ---> 0+:

Therefore the area under the curve from 0 10 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 ofType II. 1. If fix) is continuous on (a, bl and discontinuous at a, then b b fix) ax = lim+i fix) ax.

i i a

C-Q

c

2. If fix) is continuous on [a, b) and discontinuous at b, then b

a

fix)

ax =

c

lim_i fix)

c-b

a

3. If fix) is discontinuous at e, where a [a, c) U (e, bl, then [

fix)

ax =

[f(X)

ax +

<

t

e

ax. < b, fix)

and continuous on

ax.

In each case, ifthe limit is finite we say the improper integral converges and that the limit is the value ofthe 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 ofthe defmition, the integral on the left side ofthe equation converges if both integrals 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

b-l-

o

r_ - x

Jo

I

-I - dx = lim [-In II b-l-

1

b

=

FIGURE 8.17 The area beneath the curve aod above the x-axis for [0, 1) is not a real number (Example 4).

xl]Ob

lim [-In(1 - b) + 0]

b-I-

= DO.

•

The limit is infinite, 80 the integral diverges.

EXAMPLE 5

Evaluate

f'

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

dx Jo (x - 1)2/3

y - (x - 1)213

=

{' dx Jo (x - 1)2/3

1 3

dx

+ 1 (x - 1),/3·

Next, we evaluate each improper integral on the right-band side ofthis equation.

1.' _(x-_-"dx""-I~)'~/3

= = =

o

1

FIGURE 8.18

Example 5 shows that the

area under the curve exists (so it is a real

lim 3(x - 1)1/3]b

b-l-

lim [3(b - 1)'/3 + 3]

b---+l-

1

3c-~'-c-;;", dx . - hm 1 (x - 1)'/3 - c~l+ =

number).

0

1

c

b

b~_lb (x _dxl )2/3

=

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]

~1+

=

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 - I). (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 % instructa 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]=

-".

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

y

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.

EXAMPLE 6 Solutfon

--;cr---:---~~-~x

FIGURE 8.19 The graph of e~' lies below the graph of e ~ for x > I (Example 6).

Does the integral

.hoo 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

b

e-i' 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 ofthe real numbers, discussed in Appendix 6.

484

Chapter 8: Techniques of Integration

The comparison of e --i' 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

diverges if

2. [ " g(x) dx

f

and g be continuous on [a, 00)

converges.

[ " g(x) dx

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

b

> a.

From this it can be argued, as in Example 6, that

to to

f(x) dx

[X'

converges if

Turning this around says that

EXAMPLE 7

g(x)dx

to

diverges if

g(x)dx

converges.

f(x)dx

diverges.

•

These examples illustrate how we use Theorem 2.

2

(a) JOO sin x dx

converges because

x2

1

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 >

0.1 -

1

x

on

OO I

[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

•

8.7 Improper Integrals

485

We omit the more advanced proof ofTheorem 3. Althoogh 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

roo

J, y

converges by comparison with

1,00 (l/x 2 ) ax. Find and compare the two integral values.

The functions fix) ous on [1, 00). Also,

Solutfon

fi x) . lim -

x~oo g(x)

= l/x 2 and g(x) =

=

= --;O;f-----;-------;;2c---;;-+X 3

FIGURE 8.20

ax 1 + x2

. bm

x~oo

lim

x_OO

l/x

2

I/O +

1+ x_2 lim __

x 2)

x2

x~oo

(~+ 1) = Xl

a positive finite limit (Figore 8.20). Therefore,

The functions in

Example 8.

1/(1 + x 2) are positive and continu-

0

+1

=

1,

converges because roo ~ J1roo ~ 1+X J, x

converges. The integrals converge to different values, however: Example 3

and

/. I

00 -ax- - Inn . /.6 -ax1+x 1

- b - OO l

=

EXAMPLE 9 TABLE 8.5 6

b /1 .

2 5

10 100 1000 10000 100000

1 _ e-x

x

ax

0.5226637569 1.3912002736 2.0832053156 4.3857862516 6.6883713446 8.9909564376 11.2935415306

1+x 2

lim [tan-' b - tan-'ll

b~oo

= 11' _ 11' = 11'.

2

roo 1 - x e-

x

Investigate the convergence of J,

4

4

•

ax.

The integrand suggests a comparison of fix) = (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 1<:st we fmd that

Solutfon

which is a positive fmite limit. Therefore,

/.00 1 -xe

-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 !NFINITE LIMITS OF INTEGRATION:

INTEGRAND BECOMES !NFINITE: TYPE IT

TYPE I

4. Upper endpoint

I. Upper limit

fl

10

dx = lim (x - 1)2/3

r (X -dx1)'/3

b-c10

Y Y

\

c+----'-------"c~x

+-'-------+x

o

o

5.

2. Lower limit

1 0

--00

. -dx - - Inn 1

+ x2

-

1 0

6--i> -00

1

3

Lower endpoint

j

-dx-

a

Y~r;: 1

~-I)'"

+ x2

1

3

dx (X - 1)'/3

r

=

e

dx

d~+1d (X - 1)'/3

Y Y~

Y

1

_

1

1 (x -

I)'"

Y- 1 +-<,

c+---;------:--->x

~=-C+----=-"""',,->x

o

6. Interior point

3. Both limits

e

10

dx (x - 1)2/3 =

t

10

dx (x - 1)'/3

+

j3 1

Y Y

1

Y~

1 (x -

I)'"

~O+--~---~x

dx (X - 1)2/3

8.7 Improper Integrals

487

Exercises 8.7 Evaluating Improper Integrals Evaluate lhe integrals in Exercises 1-34 without using tables.

1 •

ax

{I

Jo

5.

1

ax

13

6.

-1 X

ax

8

Jo VI==7

100 ], 100

2ax

-2

-oox-l 2 -2--dv 2 v - v

2xax 2

17

8

Jo

19

ax

roo

· Jo

(I + x)Vi

dv

roo

· Jo (I + v 2)(1 + tan Iv) 21. J~ 8e'd8 23.

J~ e-i>1ax 2

27 •

ax

2

29 •

1, 1 v'fxT 1

tis

~

31 •

•

-1

B'

. 11roo ~ x3 + 1

e

12.],00~ 2

51

roo

ax

52],00

ax

2

100

t

+4

'2~

1

-

+ I tis

Jo v'4=7

18.

roo

Jl

ax

I

+ Xl

22.],00 2e -IJ sin 8 d8 -x'

ax

30 •

34

roo

00

100 1

· Jo

62.

Vex - x

ax

00

~W+1

0

Jl x(inx)"

1>.

2X dx

ax

e'+ e ~

],00 ax

x(inx)"

2

,~oo

roo

J!,/(x)dx

Show that

2xax

Jo x2 + 1 diverges and bence that

00 2xax

1 l' 2xax ~ o. --00

x2

+

1

diverges. Then show lhat lim

b_oo -b x 2

+1

Exercises 67-70 are about lhe infmite region in lhe first quadrant between the curve y = e ---x and the x-axis.

cot8d8

cos 8 d8

Vi

~

66. J~oo/(x)dxmaynotequal lim

{~/2

],1 _e--Vi ax

1

64.

(' ax

~/2 (". - 28) /3

40.

itroo eX _I

I

.~ax

(x + 1)(x + I)

Jo

x

I

],2 inx ax 60. [00 in (in x) ax

59. Jl ~ ax 61.

00

58.

roo ,

a.

ax 2

38.1~/2

],o

2 dt t 3/2-1

'o~

sin8d8

e-1/, ax

+;osx dx

65. Find lhe values ofp for which eacb integral cooverges.

t'\J'12=4 2 ax ],

37. r

-2

54],00 xax • 2 W-=1 00 56.1 I + ~inx ax

Theory and Examples

2

36.

ln2 x

2

/.00

4rdr

tan8d8

39.

57.

63.

Jo v'1-=-7 ],' dt

Jo

~

00

1

J:" 2xe

'2~

xl

,

20. roo 16 tan-I x ax 1

it 55.1

roo ~ 1+ 6

~

x~

10

. Jo W+1

53. roo~ax

xax

16. {2 s

r/2

Jo

· , Vi-I

· Jo

x2

Testing for Convergence In Exercises 35-64, use integration, the Direct Comparison Test, or lhe Limit Coroparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. 35.

/.00 ax

48

50

32

+6

46.1:-xinlxlax

dv

-1

d8 + 58

Jol - x

49],00

--00

00

33

{' ax

44·

o 1- x

47

28. {'

Is~

ax

--2

],

~

26.],1 (-inx) ax

v'4=7

], 0

(Hint:t",=sintfort"'= 0)

10.1 ~

24.

25.],1 xinxax

dt.

t-smt

45. 1: in Ixlax

• ~ (x 2 + 4)3/2

+ I d8 vB' + 28

{I

sint

X

• ]0, rO. 999

14

• ~ (x + I)'

15.

ax

2

-2--

Vi +

2

43.

1;3

-s

1

10

ax

1

dt

Jo

42. {I

~

1

2j3

{I

11.

{'

4· Jo

Vi 1

9.

2-. 1 xl. OOI

+1

x

0

3.

7.

100 ax

ax],00 2

41. {~

1

67. Find the area of the region. 68. Find lhe centroid of the region. 69. Find the volume of the solid generated by revolving the region

about the y-axis.

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 = Tr/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 ofthe 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

• jdt. VTr If it converges, what appears to be its value? You will see how to confirm your estimate in Section 15.4, Exercise 41.

77. Normal probability distribution

1 ~ 2Tr x \j 1 + -;tdx,

f( X )

which measures the surface area of the solid of revolution traced out by revolving the curve y = l/x, 1 ,.;; x, about the x-axis, diverges also. By comparing the two integrals, we see that, for every finite value b > 1,

12Tri~1

b

2Tr

=

The function

- -1- 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

1

b

+ :4dx >

2e-t2

o

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

2

2 -t V;dt,

a. Plot the error function for 0 ,.;; x ,.;; 25.

x2

00

r

Jo

b. Explore the convergence of

2

e- dx numerically.

,

The function erf(x) =

dx without introducing an error ofmagnitude 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.

idx.

1:

y

f(x) dx

=

1.

In what follows, let f.L = 0 and u = 1.

o

D

--...I -_ I

1

---

------------

----

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 ofrevolution 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

> 1, and for b > 1,

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-integraljUnction, 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 COMPUTER EXPLORATIONS In Exercises 79-1l2, use a CAS to explore the integrals for various values ofp (include noointeger values). For what values of p does the integral converge? What is the value of the integral when it does converge? Plot the integraod for various values ofp.

Chapter:

81.

80.

],00 xP 10 x dx

100

x P lox dx

82. /:, xPIo

Ixldx

Questions to Guide Your Review

1. What is the formula for integratioo by parts? Where does it come from? Why might you waut to use it? 2. When applying the formula for integratioo by parts, how do you choose the u aod dv? How cao you apply integratioo hy 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 ofeach case. 4. What substitotioos are made to evaluate integrals of sin mx sin nx, sin no: cos nx, and cos mx cos nx? Give an example ofeach case.

J

5. What suhstitotioos are sOlllelinres 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

79.],'XP InXdx

489

Va

Vx

6. What restrictions cao you place 00 the variables involved in the three basic trigooometric substitotioos to make sure the suhstitotions are reversible (have inverses)?

a. is a product of distinct linear factors? b. consists ofa repeated linear factot?

c. cootains ao irreducible quadratic facto!? What do you do if the degree of f is not less thao the degree ofg? 9. How are integral tables typically used? What do you do if a particular integral you waut to evaluate is not listed in the table? 10. What is a reduction formula? How are reductioo formulas used? Give ao example. 11. You are collaborating to produce a short "how-to" maoual for numerical integration, aod you are writing about the Trapezoidal Rule. (a) What would you say about the rule itself aod how to use it? How to achieve accuracy? (b) What would you say ifyou were writing about Simpsoo ~ Rule instead? 12. How would you compare the relative merits of Simpsoo's Role aod the Trapezoidal Rule?

7. What is the goal ofthe method ofpartial fractioos?

13. What is ao improper integral ofType I? Type II? How are the values ofvarious types ofimproper integrals def"med? Give examples.

8. When the degree of a polyoornia1 f(x) is less thao the degree of a polyoontial g(x), how do you write f(x)/g(x) as a sum ofpartia1 fractions if g(x)

14. What tests are available for determining the convergence aod divergence of improper integrals that caooot be evaluated directly? Give examples oftheir use.

Chapter J

1 Practice Exercises

Integration by Parts Evaluate the integrals in Exercises 1--1l using integratioo by parts. 1. / 10 (x

+

3. /

3x dx

tao-I

2 2. / x 10xdx

I)dx

4. /

(x

7. /

e%cos2xdx

(f)

8. /

sinOdO

cos2 8 + cos 8 - 2

15. /3X

dx

6. / x 2 sin(1 - x)dx

+ I}'e%dx

5. /

cos-1

13 / ·

e-2x sin3xdx

17. / 19 / · 21. /

2

v + 3 dv

2v 3 - 8v

xdx x 2 -3x+2

11 / dx x(x + 1)2 •

2

3

x +x

2

x +X x

3

-

+ 4x

2

dx

2

It may be necessary to use a

23. /

xdx x 2 +4x+3

25 / •

dx x(3v;+l)

12./ x+1 dx x 2(x - I)

27. /

e'

~28.

10 / .

2

dx

4xdx x 3 + 4x (3v - 7)dv 18. / (v - l)(v - 2)(v - 3)

tdl t 4 _t 2 -2

22./ + l X3

x3

24. /

-

x

3

dx

2x + x

2

-

2

21x + 24 dx

x +2x-8

x +4x+3

~I

cosOdO sin2 8 + 8inB - 6

16. /

20. /

dt 1'+412 +3

Partial Fractions Evaluate the integrals in Exercises suhstitotioo 11!St 9 / .

+ 4x + 4 dx +x

x3

14. /

26. /

28. /

dx

x( I + \¥x) 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

•

31.

J

30 •

ydy v'16-y'

J~ 4-

32.

x'

J Jv'4t'=1 xdx

33.

J

34.

35.J~ 9-x

j(x)

tdt

J

36.J~ 9-~

Cv

39. 41.

43.

sin3 x cos4 x dx

38.

tan4 x sec2 x dx

40.

sin 58 cos 68d8

42.

v'1

+ cos (tI2) dt

44.

J J J J

coss x sins x dx 3

tan X sec

3

x dx

= 8.27 +

10-' (26T - 1.87T').

51. Fuel eflkieney An automohile computet gives a digital readout of fuel consumption in gallons per hoor. Doring a trip, a passenger recorded 1he fuel consumption every 5 min for a full hoor of travel

Time

Gai/h

Time

Gai/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

e'v'tan' e'

+ I dt

Numerical Integration 45. According to 1he error-bound funnula for Simpson's Rule, howIIUUJY subintervals should you use to be sure of estimating 1he va1ue of 1n3

~ [~dx

by Simpson's Rule wi1h an error of no more 1han 10-4 in absolute

value? (Remember 1hat for Simpson's Rule, 1he number of subinterva1s has to be even.) 46. A brief calculation shows 1hat if 0 '" x '" I, 1hen 1he second derivative of j(x) ~ v'1+7 lies between 0 and 8. Based on 1his, about how IIUUJY subdivisions would you need to estimate 1he integral of j from 0 to I wi1h an error no greater 1han 10-' in ai>solute value using 1he Trapezoidal Rule?

0 5 10

15 20 25 30

a. Use 1he Trapezoidal Rule to approximate 1he total fuel consumption during 1he hoor. b. If1he automobile covered 60 nti in 1he hour, what was its fuel

efficiency (in ntiles per gallon) for 1hat portion of1he trip? 52. A new parking lot To meet 1he demand for parking, yoor town has allocated the area shown here. As the town engineer, you have been asked by 1he town council to Imd out if1he lot can be built for $11,000. The cost to clear 1he land will be $0.10 a square foot, and 1he lot will cost $2.00 a square foot to pave. Use Simpson's Rule to Imd out if1hejob can be done for $11,000.

47. A direct calculation shows 1hat

Oft

54ft How close do you come to 1his value by using 1he Trapezoidal Rule wi1hn = 6?Simpson'sRulewi1hn = 6?Try1hemandlmdout 48. You are planning to use Simpson's Rule to estimate 1he value of 1he integral

l'

+ 25

Find 1he average value of Cv for 20' '" T '" 675'C and 1he temperature at which it is attained.

Evaluate 1he integrals in Exercises 37-44.

J J J J

(x - 101»)

50. Heat capacity of a gas Heat capacity Cv is 1he amount ofbeat required to raise the temperature of a given mass of gas with constant volume by I 'C, measured in units of calldeg-mol (calories per degree gtan3 molecular weight). The heat capacity of oxygen depends on its temperature T and satisfies 1he formula

( dx ') x9-x

Trigonometric Integrals 37.

U:s

~ 37 sin

for a 365-day year. This is one way to estimate 1he armual mean air tetoperature in Fairbanks, Alaska. The National Wea1her Service's official figure, a numerical average of the daily normal mean air lemperatures for 1he year, is 25.7'F, which is slightly higher 1han 1he average value of j(x).

v'4+x'

Evaluate 1he integrals in Exercises 33-36. -xdx -, 9-x

49. Mean temperature Compute 1he average value of1he temperature function

j(x)dx

wi1h an error magnitude less 1han 10-'. You have detennined 1hat Ij<')(x) I '" 3 throughout 1he interval of integration. How IIUUJY subinletva1s should you use to assure 1he required accarsey? (IWmember 1hat for Simpson's Rule 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 Improper Integrals Evaluate the improper integrals in Exercises 53-j)2.

83

dx 53. /, ' • ~ o v9-x2

85.

•

57.

/, 0

61

56

/,00

00

2du

u2

58.

2u

-

/,00 x2e ~ dx roo

· J-oo

60.

dx 4x 2

10

dO 1)'/5

+

• -2 (0

(y - 1)2/3

3

59.

0

dy

2

55

54. /,' lnxdx

62

+9

/, 1

3v - I '2dv 4v - v

1~ xe" dx

roo

65.

1

00

67.

~

2dx

f

71

•

73.

I

1 1

2 dx

x(x +

75.1

70.

+ Vi

I?

2 - cosx + sin x dx sin2 x

9dv 81 - v'

77.1 ocos (20 79 •

72

+

1

l)dO

dx

sin 20 dO (I + cos20l'

81.1

xdx

~

Chapter

l

lx'

+ 2 dx 4 - x2

· 1V-a 1 dx

~/'

82·1~dv

1

(sin-I x? dx

98.

vr

--2

+x

I

l

2

1

x

vb 1 - x4

113... Show that

+ 4e'dO

dv

4x' - 20x dx IOx 2 + 9 x2 --,dx I +x

102.

I (I

+ x dx + x)' +

104.

Jor

v'f+Xdx

'f2

106.

+ V1=7dx

VI l

108·1x'lnx. ln(lnX)dx

xlDXlnX

x

e'V3

2

+ x dx + x'

110.

--x-dx

111·1

tan'tdt

100.

dx

lnx x+xlnxdx

1

x dx

dx

112.

1

(lnX)'nx

[~ + In ~X)] dx

l~dx

J; f(x) dx = J; f(a -

x) dx.

b. Use part (a) to evaluate ,"/2 /, o

114. 116.

1. 1

sinX' . dx. smx + cosx

sinx

smx

+ cosX'

dx

115.

1

1

2

sin x2 dx

+ 8in x

I - cos x dx 1 + cosX'

Additional and Advanced Exercises

Evaluating Integrals Evaluate the integrals in Exercises I-j). I.

dr

x

x'dx x 2 -a+1

VI + cos4xdx

96

x'

l

1m:~' 1

1 . 1wv=J 1x' 1 1 1VI

sin5tdt

1+

x'e""ldx

94.

+ (cos 5t)2

I

2

•

1

x(x

+

x'

Insmv

-rx. VI + Vi dx 105. 1-rxk dx +

1 J.

1

90.

2 -

491

92.1 cotvdv

Y

103.1

x2

2

~f2

88.

4t 2

Iny - , dy

I

109.

sin 0 dO cos2 (J 00 dx 76. / 2, (x - I )2

80.

tdt e'dt

/,

97.1

107.

74.

78.

100

xdx

86.1

+ 3et + 2

e2t

1V8 - a

+I + 4 )dz

Z

2( 2

z z

101. 1 I

· J~ x 2(1 + eX)

xdx

•

e;,dt

Assorted Integrations Evaluate the integrals in Exercises 69-116. The integrals are listed in random order.

69.1

95

99.

roo

68

e~ + e ~

.

93.1 ebrVxdx

4dx

· 1-00 x 2 + 16

66.

89.

84

dy

y2-2y+2

1

64./,00

/,00 I~Z dz

1 1

87.1 V9 91.

Which of the improper integrals in Exercises 6J--j)8 converge and which diverge? 00 dO e-cosudu 63. / 6.

W+1

•

Additional and Advanced Exercises

I)(x

3·1 xsin-Ixdx

+dx2)"'(x + m) 4·1 sin- I yYdy

492

5.

Chapter 8: Techniques of Integration

Jt-V1=t2

•.

dt

J~ ;t4+4

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-axUt to generate a solid. Find the volume of the solid.

Evaluate the limits in Exercises 7 and 8.

7. lim

1'"

x-->00 -->:

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 lineA" - e.

;tl1c~tdt x-->o+ t

sintdt

8. lim

%

Evaluate the limits in Exercises 9 and 10 by identifying them with definite integrals and evaluating the integrals.

9. lim

±1n41

n-->00 1-1

+~

Applfcattons 11. Finding an: length

10. lim

~ Vn 2I-

.......00 .1;-0

t2

Find the length of the curve

y -1% Vcos2tdt,

O:!;';;t:!;';

'lr/4.

21. Centmid of a region Find the centroid of the region in the plane enclosed by the curves y - ±( I - ;t2t1/2 and the lines ;t - Oand;t - 1. 23. Length of. curve ;t-lto;t-e.

Find the length of the curve y - In;t from

24. Finding mrfaee area Find the area of the smface generated by revolving the curve in Exercise 23 about the y-axis. 25. The.urfaee poenrted by an astroid The gnlph of the equation ;t2/3 + y2/3 _ I is an a.ftroid (see accompanying ftgure). Find the area ofthe surface generatM by revolving the curve about the x-mds.

,

11. Finding an: lengtb Find the length ofthe graph of the fimction y -In(l _;t2), O:!;'; x:!;'; 1/2.

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. 14. FIndIng volume The region in the first quadrant that is enclosed by the".ui>, the """" ~ '/(x~). ud the tim", ~ 1 and ;t = 4 is revolved about the »-axis to gwerate a solid. Find the volume of the solid.

y

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 tile 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 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.

17. Finding volume Let R be the ''t:riangular" 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 volume of the solid generated by revolving R about L the ;t-axUt. 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-lIXis and the curve

is revolved about the »-axis to generate the solid shown here. L

Show that f is continuous at;t = O.

-1

26. Lengthofaclll'ft Find the leug1h of the curve

y= [%v'Vt-ldt.

l:!;';xsI6.

27. For what value or values of a does

[G::,-~)dt converge? Evaluate the corresponding integral(s). 28. For each ;t > O. let G(;t) for each x > O.

10

00

e ~ dt. Prove that ;tG(;t) - 1

29. IDf'Ulite area lIDd (mite wlume What values of p have the following 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 :follow.. iog 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 in:finile but the voluIne of the solid genented by revolving the region about one of the COOIt1ina1J; axes is finite. The S.mm. Function .nd Stlrlfng's Fonnula Euler's gamma ftmction f(;t) ("gamma ofr'; f is a Greek capital g) uses an integral to extend the factorial function ftom. the nonnegative integers to otheI' real. values. The formula is

b. Find the volume of the solid.

,

_1 r 00

o

,

f(;t)

1e-t dt,

;t> O.

For each positive;t. the number f{x) is the integral oft r - 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

"'\Yn! '" ~.

o b. 2

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 refmement of Equation (2) gives

--+-±---+---;;+----:--:-----;;-~x

o

3

-I

f(x) =

-2 -3

(~Y l#e1/(l"')(1

FIGURE 8.21 Euler's gamma function f(x) is a continuous function of> whose value at each positive integer n + 1 is n!' The defIning 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 + I»/x, which is the subject of Exercise 31. 31. H n is a nonnegative integer, f(n

+ 1)

(~Y l#e1/(l"'), (6)

Cmnpare the values given for IO! by your calculator, Stirling's approximation, and Equation (6). Tabular Integratton The technique of tabular integration also applies to integrals of the fonn f(x )g(x) fix when neither function can be differentiated repeatedly to become zero. For example, to evaluate

J

= n!

b. Then apply integration by parts to the integral for f(x show that f(x + I) = This gives

xf(x).

+

J

I) to

eh"cosxdt

f(2)

~

If(l)

~

I

we begin as before with a table listing successive derivatives of e'lx

f(3)

~

2f(2)

~

2

and integrals of cos x:

f(4)

~

3f(3)

~

6

+ I)

= nf(n) = n!

(1)

Use mathematical induction to verify Equation (I) for every nonnegative integer n.

32. Stirling's formula Scottish mathematician James Stirling (1692-1770) showed that

e2" and its

cos x and its

derivatives

integrals

e"'~

coox

2o"'~inX (+) ~

4e2¥

-cosx

<E-

so, for large x, .(x) --+ 0 aox --+

DC.

(2)

f1rst row except for multi-

(~)" VZ;;;

J

~ +(e"'sinx) +

(Stirling's formula).

(3)

a. Stirling's approsimation for n! Use Equation (3) and the fact that n! = nf(n) to show that

n! '"

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 interpret the table as saying

e"'cooxfix

Dropping .(x) leads to the approximation

(21i (ex)X "l/x

Stophere:Rowissameas

plicative constants (4 on the left, -I on the right).

x~('iYfIrf(X)= I,

f(x) '"

.(x»

which tells us that

a. Show that f( I) = I.

c.

+

or f(x) '"

r(n

(5)

(Stirling's approIimation).

(4)

(2o"'(-coox»

J

(4e"')( -cos x) fix.

We take signed products from the diagonal arrows and a signed int.,. gra1 for the last horizontal arrow. Transposing the integral on the righthand side over to the left-hand side now gives

5

J

e 2x cosxdx

=

e2t"sinx

+ 2e2xcosx

494

Chapter 8: Techniques of Integration

or

.

/

. " dx -- e" smX' +5 2" e cosX' e cosx

2z

(9)

smx=--2' I +z

+ C'

Finally, x = 2tan-1 z,so

after dividing by 5 and adding the constant of integration. Use tabular integration to evaluate the integrals in Exercises 33-40. 33. /

e"cos3xdx

34. /

35. / sin 3x sinxca

(to)

e"'sin4xdx

Examples

s/

36. / cos5xsin4xdt

37. /

e~sinbxdx

38. /

e~cosbxdx

39. /

In (ax) dx

40. / x 2 1n (ax) dx

.

I dx~/I+z'~ I + cosx 2 I + z'

=/dz=Z+C = tan

The Substitution z

~

(f)

+ C

tan (x/2)

The substitution

z = tan~ 2

(7)

b. /

I dx = / 2 + sinx

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 accompanying figure

I

+ z2

2dz

2 + 2z + 2z 2 1 + z2 z' +

~+ I = /

(z +

(I/~)2 + 3/4

=/,;~,; =

~tan-l (~)

+ C

= ~tan-12z + I + C y!3 y!3 2 _II + 2 tan (x/2) =y!3tan y!3 +C

Use the substitutions in Equations (7}-{IO) to evaluate the integrals in Exercises 41-48. Integrals like these arise in calculating the average aogu1ar velocity of the output shaft of a uoiversal joint when 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

W/2

43. cosx

~ 2cos2 (f) - I ~ sec2~X/2) 1-

o /.

I

w/2

45. /. 0

2 _1=_2_ _ 1 I + tan2 (x/2) I + z2 cosX'

I

z2

= --2' I +z

(8)

dx.

- smX'

dx . 1 + smx dO 2+cosO

47. / . dt smt - cost

~I~~. __dx"'"-c-_ _ + SInX' + cosX'

42. /

44.

46

l

W/2

,"/3

1

dx 1 - cosX'

2w /3 cos 0 dO • '"/2 sinO cos 0 + sinO

48. /

costdt 1 - cost

Use the substitution z = tan (0/2) to evaluate the integrals in Exercises 49 and 50.

and

. . x x sin (x/2) 2 (x) smx = 2 sm 2cos2 = 2 cos (x/2) • cos 2 = 2tan~. I 2 sec2 (x/2)

2 tan (x/2) I + tan2 (x/2)

49. /

secOdO

50. /

cscOdO

Chapter 8 Technology Application Projects

Chapter

Technology Application Projects

Mathematica/Maple Modules: Riemann, Trapezoidal, and SimpsonApproximations Part I: Visualize the error involved in using Riemann sums to approximate the area under a curve.

Part II: Build a table of values and compute the relative magnitude of the error as a function of!be step size 4>:. Part III: Investigate !be effect of the detivative function on !be error. Part. IV and V: Trapezoidal Rule approximations. PartVI: Simpson'. Rule approximations.

Games o/Chance: Exploring the Monte Carlo Probabilistic Technique/or Numerkal Integration Graphically explore the Monte Carlo method for approximating def"mite integrals. Computing Probabilities with Improper Integral. More explorations of the Monte Carlo method for approximating definite integral•.

495

9 FIRST-ORDER DIFFERENTIAL EQUATIONS OVERVIEW In Section 4.7 we introduced differential equations of the form dyjdx = 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 dyjdx (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 ofx-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 = Cjx :

=

+ 2 gives C

~ (i)

+0

= -

~.

We need to show that the differential equation is satisfied when we substitute into it the expressions (Cjx) + 2 for y, and -Cjx 2 for dy/dx. That is, we need to 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 mther 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 = XC. Thus the gmph of the particular solution passes through the point (xo, Yo) in the ~-plane. 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 to 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 +

I

_lex) 3

=

I _

leX

3'

498

Chapter 9: First-Order Oifferential Equations y

On the right side ofthe equation: y

=

(x

+ 1) -

!e x

3

y - x = (x

+ I) - tex - x

=

I - tex.

-"-----!----"+-----J;\--L-~x

4

The function satisfies the initial condition because

y(O) FIGURE 9.1 Graph ofthe solution to the initial value problem in Example 2.

=

I X] I 2 =1--=-. [(x+I)--e 3 x-o 3 3

The graph ofthe 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' = I(x, y), the solution curve (graph ofthe solution) is required to pass through the point (xo, Yo) and to have slope I(xo, Yo) there. We can picture these slopes graphically by drawing short line segments of slope I(x, y) at selected points (x, y) in the region of the '!y-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 resuiting picture 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 tskes at each point it passes through.

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.

9.1 Solutions, Slope Fields, and Euler's Method

499

(c) y'~ (I-x)y+ ~

y

y

~

L(x)

~

Yo + f(x.. yoJ(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 directions, 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 our 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) = Yo, we can approximate the solution y = y(x) by its linearization

L(x) FIGURE 9.5 The fIrst Euler srep

= L(x,).

approximates y(x,) with y,

Y

Euler approximation (x2.

Yv

=

y(xo)

+ y'(xo)(x - xo)

I ,

I I ,

Troe solution ClU'Ve y ~ y(x) : I I

I I

FIGURE 9.6 Three sreps in the Euler

approximation to the solution ofthe initial value problem y' = f(x, y), y (xo) = Yo. As we take more steps, the errors involved

usually accumulate, but not in the exaggerated way shown here.

L(x)

=

Yo

+ f(xo,yo)(x - xo).

The function L(x) gives a good approximation to the solution y(x) in a short interval about Xo (Figure 9.4). The basis ofEuler'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 (Figure 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 ofthe 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 Oifferential Equations and so 00. We are literally building an approximatioo to ooe of the solutioos by following the directioo ofthe slope field ofthe differential equatioo. The steps in Figure 9.6 are drawn large to illustrate the constroctioo process, so the approximatioo looks crude. In practice, would be small enough to make the red curve 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 = 1. Next we determine the values of x at which the Euler approximations will take place: Xl = Xo + = 0.1, X2 = Xo + 2 = 0.2, and X3 = Xo + 3 = 0.3. Then we find

ax

ax

ax

First:

Second:

Yl

Yo + /(xo, Yo)

=

Yo + (I

=

I

+ (I + 1)(0.1)

1.2

=

+ /(Xl, Yl) ax Yl + (I + Yl) ax 1.2 + (I + 1.2)(0.1)

Y2 = Yl = =

Third:

ax + Yo) ax

=

=

+ /(X2, Y2) ax Yo + (I + Y2) ax 1.42 + (I + 1.42)(0.1)

1.42

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 ofthem, set

=

Xo + Xl +

ax ax

=

Xn-l

+ dx.

Xl X2

=

Xn

Then calculate the approximations to the solutioo,

Yl Yo

= =

Yo + /(xo,yo) ax Yl + /(Xl, Yl) ax

Yn = Yn-l

HISTORICAL BIOGRAPHY

Leonhard Euler (1703-1783)

+

/(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 ofnumerical 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 I, starting at Xo = 0 and taking (a) dx = 0.1 and (b) dx = 0.05. Compare the approximations with the values ofthe exact solution y = 2e x - I. Solution

(a) We used a computer to generate the approximate values in Table 9.1. The "error" column is obtained by subtracting the unrounded Euler values from the unrounded values found using the exact solution. All entries are then rounded to four decimal places.

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 (exact)

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

---oJ--------:--~x

o

FIGURE 9.7 Thegraphofy = 2e' - I superimposed on a scatterplot of the Euler approximations shown in Table 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 approximate 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 Oifferential Equations

Euler solution of y' step size dx = 0.05

TABLE 9.2

x 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

1

=

+ y, y(O)

=

1,

y(Euler)

y (eIllet)

Error

I

I

1.1 1.205 1.3153 1.4310 1.5526 1.6802 1.8142 1.9549 2.1027 2.2578 2.4207 2.5917 2.7713 2.9599 3.1579 3.3657 3.5840 3.8132 4.0539 4.3066

1.1025 1.2103 1.3237 1.4428 1.5681 1.6997 1.8381 1.9836 2.1366 2.2974 2.4665 2.6442 2.8311 3.0275 3.2340 3.4511 3.6793 3.9192 4.1714 4.4366

0 0.0025 0.0053 0.0084 0.0118 0.0155 0.0195 0.0239 0.0287 0.0340 0.0397 0.0458 0.0525 0.0598 0.0676 0.0761 0.0853 0.0953 0.1060 0.1175 0.1300

Exercises 9.1 Slope Fields In Exercises 1-4, match the differential equations with their slope fields, graphed here. y

y

!!WlH I!!!!!!! IIIIIIII IIIIIIII I111I111 III III

........................................................

•

.......

~~~~~~~~ ~~~~~~~~ ~~~~~~~~ ~~~~~~~~

\\\\\\H \\\\\\\\ (e)

1.y'=x+y (a)

(b)

3.yl=_~

2.y'=y+1

9.1

In Exercises 5 and 6, copy the slope fields and sketch in some of the

solution curves. 5. y'

Solutions, Slope Fields, and Euler's Method

503

19. Use the Euler method with dx = 0.5 to estimate y(5) if y' = y2/~andy(l) = -I. What is the exact value ofy(5)? 20. Use the Euler method with dx ~ 1/3 to estimate y(2) if y' = xsinyandy(O) = I. What is the exact value ofy(2)?

~(y+2)(y-2)

y

21. Show that the solution of the initial value problem

lllllllt llllllll

y'

~x+y,

y(XO)~Yo

is y = -I - x

x

= y(y +

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)

23. y' = y with

y

a. (0, I) h. (0,2) 24. y' = 2(y - 4) with

11111\1'-.... 11111111-.... -....-...-. ...; .....; ..,..............

-4'"............4 ---.............

a. (0, I)

h. (0,4)

x

a. (0, I) a. (0, I)

27. y' = (y - I)(x

Integral Equations In Exercises 7-10, write an equivalent rust-order differential equation and initial condition for y.

8. y ~ 9. y

l'

c. (0,5)

h. (0, -2)

c. (0, 1/4)

d. (-I, -I)

h. (0,2)

c. (0, -I)

d. (0,0)

c. (0,3)

d. (I, -I)

26. y' ~ y2 with

a. (0, -I)

+

c. (0, -I)

25. y' = y(x + y) with

......-....-....-.,.,.-.................

:./............Z. . . . ---.. . . .4-

IIIIII tt 11I111I1 7. y = -I

(I - y(I»dl

28. y'

+ 2) with h. (0, I)

~~ with 2

x +4 a. (0,2)

h. (0, -6)

c.

29. A logistic equation

~ 2 - / , ' (I + y(l))sinldl

O::s;x::S;4,

y' = y(2 - y), y(O) = 1/2;

O~y~3

30. y' = (sinx)(siny), y(O) = 2;

+ /,' y(l) dl

Using Euler's Method

13. y' 14. y'

DIs. D 16.

y'

y'

= x(1 - y), y(l) = 0, dx = 0.2 = 2xy + 2y, y(O) = 3, dx = 0.2 = y2(l + 2x), y( -I) = I, dx = 0.5 = 2xe", y(O) = 2, dx = 0.1 = ye', y(O) = 2, dx = 0.5

17. Use the Euler method with dx ~ 0.2 to estimate y(l) if y' andy(O) = I. What is the exact value ofy(I)?

-6 '" x '" 6,

-6 '" y '" 6

Exercises 31 and 32 have no explicit solution in tenns of elementary functions. Use a CAS to explore graphically each of the differential equations.

In Exercises 11-16, use Euler's method to calculate the rust three approximations to the given initial value prohlem for the specified increment size. Calculate the exact solution 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

12. y'

(-2V3, -4)

In Exercises 29 and 30, obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to rmd the general solution of the differential equation.

['I Jl t dl

10. y = I

".

22. What integral equation is equivalent to the initial value problem y' = f(x), y(xo) = Yo?

iiifiiu ifiiifif 6. y'

+ (I + Xo + Yo) er

31. y' = cos (2x - y), y(O) = 2;

0 '" x '" 5,

0 '" y '" 5

32. A Gompertz equation y' = y(I/2 - loy), 0:5x:54, Q:5y:53

y(O) = 1/3;

33. Use a CAS to rmd the solutions of y' initial conditiony(O) = 0, if f(x) is

a. 2x

b. sin 2x

c. 3e x/2

+y

~

f(x) subjcctto the

d 2e -x/2 cos 2x.

Graph all four solutions over the interval -2 '" x '" 6 to compare the results. 34. a. Use a CAS to plot the slope field of the differential equation

~

y

18. Use the Euler method with dx = 0.2 to estimate y(2) if y' = y/x and y(l) ~ 2. What is the exact value ofy(2)?

,_3x2 +4x+2 y 2(y - I) over the region -3 :5 x :5 3 and -3 :5 Y :5 3. h. Separate the variables and use a CAS integrator to rmd the general solution in implicit fonn.

504

Chapter 9: First-Order Oifferential Equations

c. Using a CAS implicit function grapher, plot solution CUIveS 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) ~ -I.

e. Find the Euler nurnetical approximation to the solution of the initial value problem with 4 subintervals of the x-interva1 and plot the Euler approximation supetimposed on the graph produced in part (d).

In Exercises 35-38, use Euler~ method with the specified step size to estimate the value ofthe solution at the given point x.... Find the value ofthe exact solution at x" . 35. y' ~ 2xe"', y(O) ~ 2, dx ~ 0.1, x· ~ I 36. y' ~ 2y 2(x - I), y(2) ~ -1/2,

dx ~ 0.1,

g. Find the error (y(exact) - y(Euler)) at the specified point x = b for each ofyour four Euler approximations. Discuss the improvement in the percentage error.

x· ~ 3

37. y' ~ -Vx/y, y> 0, y(O) ~ I, dx ~ 0.1, x· ~ I 38. y' ~ I

+ y2, y(O)

~ 0,

dx ~ 0.1,

f. Repcatpart (c) for 8,16, and 32 subinterva1s. Plot these three Euler approximations supetimposed on the graph from part (e).

x· ~ I

Use a CAS to explore graphically each of the differential equations in Exercises 39-42. Perform the followiog steps to help with your explo-

rations.

39. y' = x b~ I

+ y,

y(O) = -7/10;

40. y' = -x/y, y(O) = 2; 41. y' = y(2 - y),

a. Plot a slope field for the differential equation in the given >;y-wiodow.

-4'" x '" 4,

-3 '" x'" 3,

y(O) = 1/2;

-4'" y '" 4;

-3 '" y '" 3; b = 2

0 '" x '" 4, 0 '" y '" 3;

b~3

b. Find the general solution of the differential equation using your CAS DE solver.

42. y' = (sinx)(siny), y(O) = 2; b ~ 37f/2

-6 '" x '" 6,

-6 '" y '" 6;

c. Graph the solutions for the values of the arbitrary constaot C ~ -2, -I, 0, I, 2 superimposed on your slope field plot.

9.2

First-Order Linear Equations A first-order linear differential equation is one that can be written in the form

dy

dx

+ P(x)y

=

Q(x),

(1)

where P and Q are continuous functions of x. Equation (I) is the linear equation's standard 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 lillear (iny) because y and its derivative dy/ dx occur only to the fJrst power, they are not multiplied together, nor do they appear as the argument ofa function (such as siny,

EXAMPLE 1

e", or V dy/dx).

Put the following equation in standard form:

dy xdx

=

x 2 + 3y

'

x > O.

Solution

Divide by~. Standsrd funn withP(x) and Q(x) ~ x

~

-3/x

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 ofthe 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 ofthe 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)

Multiply by posiliw> v(.<).

v(x)Q(x)

v(x) is chosen to make dy d Vt/x +Pvy~ t/x(v·y).

d

dx (v(x)' y)

=

v(x)' y

=

y

=

Original equation is in standard fonn.

J

Integrate with respect tox.

v(x)Q(x) dx

v(~)

J

v(x)Q(x) dx

(2)

Equation (2) expresses the solution of Equation (1) in terms ofthe 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 ofthe 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 sepamtcd, v > 0

Integrate both sides.

lnv =

J

Pdx

e lnv = e!Pdx v = e!Pdx

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 Oifferential 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) = e! P(x) ax 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 ofthe integrating factor and solution function y because ofthe 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 - -:iY

(1752-1833)

=

x,

so P(x) = -3/x is identified. The integrating factor is

v(x) = e! 1'(><) ax = e!H/x) ax =

Constant of integration is 0, so v is as simple as possible.

e-' ln lxl

• >0

,.

I X

Next we multiply both sides of the standard form by v(x) and integrate:

Y_ly) (d = ~.x x x'dx x'

L

I dy 3 I ----y = -2 4 x'dx x x

~(Iy)=~2 dx \.;,

Left-hand side is

x

~ (v' y).

Integrate both sides.

I I ,y = -x + c. x Solving this last equation for y gives the general solution:

y =x' (-} +

EXAMPLE 3

c)

=

-x 2 +

ex',

x>

Find the particular solution of

3xy' - y = lux 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:

,

lnx

I

Y - 3x Y

=

+

I

3x

Then the integrating factor is given by

v =

eJ -dx/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 = -(lnx

+ 4) + ex!/3.

When x = I and y = -2 this last equation becomes

-2

=

-(0 + 4) + C,

so

C = 2. Substitution into the equation for y gives the particular solution

y = 2x!/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 ofthe 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 ofEquation (I) 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 (I), 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

=

t

0

= - P(x) dx

Q(x) - 0 Sepamting the variables

508

Chapter 9: First-Order Oifferential 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 +,---11---,2> I

6

I

a

Switch

b

di L dt

L

R

+ R'I

= 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)

f:We leave the calculation of the solution for you to do in Exercise 28.) Since R and L are positive, -(R/L) is negative and e-(RILl' --> 0 as t --> 00. Thus,

(1' -

lim i = lim 1_00

o

2~

L R

R

R

1_00

1'e-
- 1"0 = R

1'. R

At any given time, the current is theoretically less than V/ R, but as time passes, the current approaches the steady-state value VIR. According to the equation

3~

R

FIGURE 9.9 The growth ofthe 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 ofthe circuit The current gets to within 5% ofits steady-state value in 3 time constants (Exercise 27).

L di dt

+ R'I

= V

,

1= VIR is the current that will flow in the circuit if either L = 0 (no inductance) or di/dt = 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 -(V/R)e-(RjL)' that tends to zero as

•

t~oo.

Exercises 9.2 First-Qrder Unoar Equations Solve the differeo1ia1 equatioos in Exercises 1-14. l.x:+y=e x , x>O 3. xy'

+ 3y = sm,x, x > x

4. y' + (tanx)y = cos' x,

2. eX :+2eX y= I

5.x

dy I +2y=ldx

x,

6. (I

+ x)y' + y ~ y;

8. e 2x y'

0

dy

-'ff/2

< x < 'ff/2

x>O

+ 2e 2x y cosx

~ 2x

10. x dx = -x- - 2y, x > 0

7. 2y' ~ exf2

+y

9. xy' - Y ~ 2xlnx

9.2 11. (I - I)' ~ 12. (I

~

2s

3(1

13. sin 6 ~;

+ (cos6)r =

14. tan 6 ~;

+r

+ I) + ( I )" 1+ I

tan6,

2

1 > -I

0 < 6 < 1r/2

Ldi+R'=O dt l ,

0 < 6 < 1r/2

= sin 6,

which is Equatioo (5) wilb V = O.

a. Solve the equation to express i as a function of t.

Solving Inmal Value Problems

b. How loog after lbe switch is throwo will it take lbe current to fall to half its original value?

Solve lbe initial value problems in Exercises 15-20.

dy 15. dl

+ 2y

16. 1dy dl

c. Showtbat the value oflbe current wheo 1 = L/R is I/e. (The significance of this time is explained in lbe next exercise.)

y(O) = I

= 3,

, 1 > 0, + 2y - I,

dy . 17. 6 d6 + Y = sm6,

509

26. Current in an open RL circuit If lbe switch is thrown open after lbe current in an RL circuit has built up to its steady-state value I = VIR, the decaying current (see accompanying figure) obeys the equatioo

+ 4(1 - 1)2s ~ 1 + I, 1 > I

+ I) tls + d1

First-Order Linear Equations

Y() 2

=I

6> 0, Y(1r/2) = I

dy _, 18. 6 d6 - 2y - 6 sec 6 tan 6, dy

6> 0,

y(1r/3)

~

2

x'

19. (x + I) tis - 2(x 2 + x)y = x e+ I' x > -I, y(O) = 5 20.

dxdy + xy

~

3!-

x, y(O) ~ -6

R

21. Solve lbe expooeotial growth/decay initial value probleoJ for y as a functioo of 1 by lbinking of lbe differential equatioo as a rrrstorderlineareqoatioowilbP(x) = -k and Q(x) = 0:

dy dl

~

ky

(kconstant),

y(O)

~

Yo

22. Solve lbe following initial value probleoJ for u as a functioo of t.

~~ + ~u

= 0

(k and m positive coostants),

u(O) = Uo

27. TIme constants Engineers call !be number L/R lbe time constant of the RL circuit in Figure 9.9. The sigoificance oflbe time coostant is lbat 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.

a. Find the value of i in Equatioo (7) tbat correspoods to 1 = 3L/R and show tbat it is about 95% of the steady-state value I ~ V/ R. b. Approxiruately what perceotage oflbe 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 tbat the solutioo of lbe equation

Theory and Examples 23. Is eilber oflbe following eqoatioos correct? Give reasoos for your

answers. b.

x/ tdx ~ xlnlxl

+ ex

24. Is eilber oflbe following eqoatioos correct? Give reasoos for your

answers. a.

co~x/ cosX' dx =

1/

b. cosX'

is i ~

tanx

cos x dx = tanx

l' + Ce-(RlL)'

+C

C

+ cosX'

25. Current in a closed BL circuit How many seconds after lbe switch in an RL circuit is closed will it take lbe current i to reach half of its steady-state value? Notice tbat lbe time depeods 00 R and L and not 00 how much voltage is applied.

R

b. Theo use the initial cooditioo i(O) ~ 0 to determine lbe value of C. This will complete lbe detivatioo ofEquatioo (7). c. Showtbati = VIR is a solution ofEqoatioo (6) and tbat i = Ce -{RIL)t satisfies the equation

di dl

R.

+ I'

=

o.

510

Chapter 9: First-Order Oifferential 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

+ P(x)y

~

Q(x)y". Of,

equivalently,

Observe that, if n = 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'

31. xy'

+y

Y -2

32. x"y'

~

_y~xy2

+ 2xy

~ y'

AppLications We now look at four applications of frrst-order differential equations. The frrst application analyzes an object moving along a straight line while subject to a fOrce opposing its motion. The second is a model ofpopulation growth. The third application considers a curve or curves intersecting each curve in a second family ofcurves orthogonally (that is, at right angles). The fmal application analyzes chemical concentrations entering and leaving a container. The various models involve separable or linear frrst-<>rder equations.

Motion with Resistance Proportional to Velocity In sonJe 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 line with position function s and velocity vat time t. From Newton's second law ofmotion, the resisting force opposing the motion is Force = mass 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 mass 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 oftime 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 (I) 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

=

+ C.

0 when I = 0 gives vom 0=--+ C k

vom C=-

and

k

The body's position at time I is therefore () sl

_ m _ vom ( -vom T e -(klm)t +VoT - T I - e -(kim)t) .

(2)

To find how far the body will coast, we fmd the limit ofs(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 k

=

t_OO

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: ifm is large, the body will coast a long way.

lIn 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 ft/sec (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-tllS = e-tllS =

I

EQ.(I)withk~ 1/3.

m

1/11

-1/18

=

In (1/11)

I

=

18ln 11 '" 43 sec.

=

=

6, Va

=

11, v

=

1

-In 11

We answer the second question with Equation (3): . Distance coasted

vom

=

T

=

198 ft.

=

11·6

1/3

•

Inaccuracy of the Exponential Population Growth Model In Section 7.4 we modeled population growth with the Law of Exponential Change: dP di

=

kP,

P(O)

=

Po

whereP is the population at time I, k > 0 is a constant growth mte, and Po is the size ofthe 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 dP;dl = k

(4)

512

Chapter 9: First-Order Oifferential Equations

TABLE

World populatioo (1980-2008)

I

)

30

t

FIGURE 9.10 Notice that the value of the solutionP ~ 4454e°.0 17• is 7169 when t = 28, which is nearly 7% more than the actual population in 2008.

Orthogooal trajectory

9.3 World population (midyear)

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 ofthe 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 '" !1P, 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 orthogonal trajectory intersects the family ofcurves at right angles, or orthogonally.

!1P/P

0.017P,

P(O) = 4454.

The solution to this initial value problem gives the population function P = 4454.°.017/. In 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 affecting 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

-+--+-+~+-+--+~X

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 x 2 + y2 = a 2 , centered at the origin (Figure 9.12). Such mutoa1ly orthogonal systems ofcurves 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.

=

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/ dx values. Differentiating xy = a implicitly gives

FIGURE 9.12 Every straight line through the origin is orthogonal to the family of circles centered at the origin.

dy xdx+y=O

or

dy dx

Y 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 y

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.

Each curve is orthogonal to every curve it meets in the other family (Example 2).

(5)

FIGURE 9.13

where b = 2C is an arbitrary constant. The orthogonal trajectories are the family ofhyper• bolas 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 mixture is kept uniform by stirring and flows out of the container at a known rate. 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 (rate at WhiCh) (rate 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 rate of the chemical at time t is y(t) Departure rate = V{t) -(outflowrate)

=

concentration in ) ( container at time t - (outflow rate) .

(7)

Accordingly, Equation (6) becomes

dy . . y(t) dt = (chenncal'samvalrate) - V(t) . {outflow rate).

(B)

If, say, y is measured in pounds, V in gallons, and t in minutes, the units in Equation (8) are pounds

minutes

pounds pounds gallons minutes - gallons - minutes'

EXAMPLE 3 In an oil refinery, a storage tank contains 2000 gal of gasoline that initially has 100 Ib of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 Ib of additive per gallon is pumped into the tank at a rate of 40 gal/min.

514

Chapter 9: First-Drder Differential Equations

The wt:ll-mixed solution is pumped out at a mte of 45 gal/min. How much of the additive is in the tank 20 min after the pumping process begins (Figure 9.14)? ) 40 ga1Imin containing 2 IbIgal

4~ gaJJmi.n c~taining

2 lblgal V

FIGURE 9.14 The storage tank in Example 3 mixes input liquid with stored liquid to produce an output liquid.

Lety be the 8JD01mt (in pounds) of additive in the tank at time I. We know that y = 100 when I = O. The number of gallons of gasoline and additive in solution in the tank at any time I is

Solution

~ 2000 gal + (40: -

V(I)

~

45 : ) (tmin)

(2000 - 51) gal.

Therefore, Rate out

=

y(1) Y(I)' outflow rate

Eq.(7) OutflOW' rue ia4S pJ/min

- (200: ~

51) 45

muh - 2000 - St.

45y Ib 2000 51 min .

Also, Rate in

~

(2

~)(40 : )

~801b. mID

The differen:ti.al equation modeling the mixture process is

dy 45y dt ~ 80 - 2000 51

Eq. (8)

in pmmds per minute. To solve this differeIltial equation, wt: Ill'St write it in standard linear £ann: dy dt

45

51 Y ~ 80.

+ 2000

Thus, P(I) - 45/(2000 - 51) and Q{I) - 80. The integnUing factnr is v(l)

=

eJPdt

=

eJ~dl

= e-9In(2000-St)

~ (2000 -

5Ir'.

9.3

Applications

515

Multiplying both sides ofthe standard equation by v(l) and integrating both sides gives 9 (2000 - 51r •

(2000 - 51r9

(t + 200~5_ Y) 51

t+

= 80(2000 - 51r

9

45(2000 - 51)-10 Y = 80(2000 - 51r9

: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(2oo0 - 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. Coasting bicycle A 66-kg cyclist on a 7-kg bicycle starts coasting on level ground at 9 m/scc. The k in Equation (I) is about 3.9 kgfsec.

a. About how far will tire cyclist coast before reaching a complete stop?

b. How long will it take tire cyclist~ speed to drop to I m/sec? 2. Coasting battleship Suppose that an Iowa class battleship has mass around 51,000 melric 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 tire ship~ speed to drop to I mfsec? 3. The data io Table 9.4 were collected with a motion detector and a CBLTM by Valerie Sharritts, a mathematics teacher at St. Fraucis DeSaies High School in Columbus, Ohio. The table shows tire dislance s (meters) coasted on io-lioe skates io t sec by her daughter Ashley when she was 10 years old. Fiod a model for Ashley's

516

Chapter 9: First-Order Oifferential Equations

position given by the data in Table 9.4 in the form ofEquation (2). Her initial velocity was Vo ~ 2.75 mlsec, her mass m ~ 39.92 kg

In Exercises 5-10, fmd the orthogonal 1rajectories of the family of

(she weighed 88Ib), and her total coasting distaoce was 4.91 m.

curves. Sketch several members ofeach family.

TABLE 9.4 Ashley Sharritts skating data

t (see)

s(m)

t (see)

s(m)

t (see)

s(m)

0

0

2.24

3.05

4.48

4.77

0.16

0.31

2.40

3.22

4.64

4.82

0.32

0.57

2.56

3.38

4.80

4.84

0.48

0.80

2.72

3.52

4.96

4.86

0.64

1.05

2.88

3.67

5.12

4.88

Orthogonal Trajectories

5.y=mx 7. kx' + y' ~ I 9. y = ce ---x

6.y=cx'

8. 2x' + y' ~ 10. y = e b

11. Show that the curves 2x 2

+ 3y 2

= 5 and y2 =

c'

x' are orthogonal.

12. Find the family of solutions of the given differential equation and the fantily of orthogonal trajectories. Sketch both fantilies.

a. xlix + ydy

=0

b. xdy - 2ylix

=a

MIxture Problems 13. Salt mixture A tank initially contains 100 gal of brine in which 50 Ib of salt are dissolved. A brine containing 2 Ib/gal of salt runs into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and flows out of the tank at the rate of4 gal/min.

0.80

1.28

3.04

3.82

5.28

4.89

0.96

1.50

3.20

3.96

5.44

4.90

1.12

1.72

3.36

4.08

5.60

4.90

1.28

1.93

3.52

4.18

5.76

4.91

a. At what rate (pounda per minute) does salt enter the tank at time I?

1.44

2.09

3.68

4.31

5.92

4.90

b. What is the volume ofbrine in the tank at time t?

1.60

2.30

3.84

4.41

6.08

4.91

1.76

2.53

4.00

4.52

6.24

4.90

c. At what rate (pounda per minute) does salt leave the tank at time t?

1.92

2.73

4.16

4.63

6.40

4.91

d. Write down and solve the initial value problem describing the

2.08

2.89

4.32

4.69

6.56

4.91

4. Coasting to a stop Thhle 9.5 shows the distaoce s (meters) coasU:d on in-line skates in terms of time t (seconds) by Kelly Schmitzer. Find a model for her position in the form ofEquation (2). Her initial velocity was Va ~ 0.80 mfsec, her mass m ~ 49.90 kg (110 Ib), and her total coasting distaoce was 1.32 m.

s(m)

0

0

1.5

0.89

3.1

1.30

0.1

0.07

1.7

0.97

3.3

1.31

t (see)

s(m)

0.3

0.22

1.9

1.05

3.5

1.32

0.5

0.36

2.1

1.11

3.7

1.32

0.7

0.49

2.3

1.17

3.9

1.32

0.9

0.60

2.5

1.22

4.1

1.32

1.1

0.71

2.7

1.25

4.3

1.32

1.3

0.81

2.9

1.28

4.5

1.32

9.4

a. At what time will the tank be full? will it contain?

t (see)

s(m)

14. Mixture problem A 200-gal tank is half full of distilled water. At time t = 0, a solution containing 0.5Ib/gal ofconcen1rate enters the tank at the rate of 5 gal/min, and the well-stirred mixture is withdrawn at the rate 00 gal/min. b. At the time the tank is full, how many pounds ofconcen1rate

TABLE 9.5 Kelly Schmitzer skating data

t (see)

mixing process.

e. Find the concentration ofsalt in the tank 25 min afler the process starts.

15. Fertilizer mixture A tank contains 100 gal offresh water. A solution containing I Ib/gal ofsoluble lawn fertilizer runs into the tank at the rate of I gal/min, and the mixture is pomped out of the tank at the rate 00 gal/min Find the maxinuun amount of fertilizer in the tank and the time required to reach the maxinuun. An executive conference room. of a corporation contains 4500 ft' of air initially free ofcarbon monoxide. Starting at time t = 0, cigarette smoke containing 4% carbon monoxide is blown into the racon at the rate of 0.3 ft'/min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.3 ft'/min. Find the time when the concentration ofcarbon monoxide in the room. reaches 0.01 %.

16. Carbon monOlide pollution

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 graphs is a powerful tool in understanding real-world systems. The starting ideas for a graphical 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 Ifdy/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. ------l@)------------i@~--->,

-1

2

Y

518

Chapter 9: First-Order Oifferential Equations 2.

IdentifY and label the intervals where y' > 0 and y' < O. 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 . .. I~) y

__-_-4@)---_-o-------<@r-__-1 2

Similarly, y' < 0 between y = -I and y = 2, so any solution with a value in this interval will decrease toward y = - 1. 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 = -1. Solution curves between the lines y = -I and y = 2 fall away from y = 2 toward Y = - 1. 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 inJplicit differentiation. y' = (y

+ I)(y - 2)

= y2 - Y -

2

< O. To find y", we differ-

Formula for y' ...

y" = ~(Y') = ~(y2 _ Y _ 2)

dx

y

=

y'>O y">O

dx

differentiated implicitly

2y'y' - y'

= (2y -

I)y'

= (2y -

I)(y

with respect to x

+

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). All predicted in Step 2, solutions in the middle and lower bands approach the equilibrium value y = -1 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 = -1, it tends steadily toward that value; y = - 1 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 slays, but if it is offby any amount, no matter how small, it moves away. (Sometimes an equilibrium value is unSlable 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 ofExample 1). The arrows lead away from y = 2 and, once to the left of y = 2, toward Y = - 1. 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 1.

Newton's Law of Cooling dH>O dt

-

: I

In Section 7.4 we solved analytically the differential equation

dH
~~@l---_-

__~, 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 loog run.

dH>O dt 2

d H <0

dt 2 I

I I I I I I I I

@ 15

dH lit

= -k(H - H s),

k>O

modeling Newton's law of cooling. Here H is the temperature of an object at time t and H s 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 - 15) dt

dH <0 dt

-

(1)

2

d H >0

dt 2

•

'H

FIGURE 9.17 The complete phase line for Newton's law of cooling.

Temperature of SlUTOunding

medium

(minus kto give a negative derivative whenH > 15). Since dH/dt = 0 at H = 15, the temperature 15°C is an equilibrium value. If H > 15, Equation (1) tells us that (H - 15) > 0 and dH/dt < O. If the object is hotter than the room, it will get cooler. Similarly, if H < 15, then (H - 15) < 0 and dH/dt > O. An object cooler than the room will warm up. Thus, the behavior described by Equation (1) 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 = 15 is a Slable equilibrium. We determine the concavity of the solution curves by differentiating both sides of Equation (1) with respect to t:

~ (dJ:) = ~(-k(H-

15»

2 d H = _k dH dt 2 dt'

Initial "'mperalure

FIGURE 9.18 Temperature versus time. Regardless of initial temperature, the ohject's remperature H(t) teods toward 15°C, 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 wben 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 Oifferential 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 object is equal to the net force applied to it. In mathematical terms,

F

=

d

(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 object 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 and Equation (2) takes the simpler form

F= ma,

or

0,

(3)

known as Newton s 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

ypositive

Fp FIGURE 9.19

=

An object falling under the

influence ofgravity with a resistive force assumed to be proportional to the velocity.

=

mg,

the force due to gravity. If, however, we think of a real body falling through the air----1lay, 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 m-=mg-ku dt dv k dt =g- m V '

-

dv>O dt __

I

I I

dv
..----{@~~

(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 zero, is

,v

mg

mg v=T'

k

FIGURE 9.20

Initial phase line for the

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

dr 2

<0

I I I I I I I I

@

mg

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

,

dr 2

>0 'v

T FIGURE 9.21 The completed phase line for the falling body.

v

V=

521

mg k

~:~ = ~ (g - ~ v) = - ~ ~~ . We see that d 2v/dt 2 < 0 when v < mg/k and d 2v/dt 2 > 0 when v > mg/k. Figure 9.21 adds tIris infonnation to the phase 11oe. 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

dP

'JYpical velocity curves for a falling body encountering resistance. The value v = mg/k is the tenninal velocity.

lit

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 ofresources 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 k 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 k 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

@

•M •

dP <0 dt ,P

FIGURE 9.23 The initial phase line for logistic growth (Equation 6).

=

r(M - P)P = rMP - rP 2 •

(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 Oifferential Equations

dP >0 dt I I

aZp

I

dt 2

@, o

>0 : aZp < 0 I

J'

dt2

I,

I

M

I I I I

ZP) and tiP/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 llrst quadrant of the tP-plane into horizontal bands in which we know the signs of both tiP/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

If P

tit
d P >0 I dt 2 @. '>P I

M

1:

FIGURE 9.24 The completed phase line for logistic growth (Equation 6).

> M, then (M -

P

Limiting

f----====:;:::;~~""'''':::::::'''''....~--population

TIme

FIGURE 9.25

Population corves for logistic growth.

Exercises 9.4 Phase Unes and Solution Curves 10 Exercises 1-8,

a. IdentifY the equilibrium values. Which are stahle and which are unstable? b. Construct a phase line. IdentifY the sigos of y' and y'.

c. Sketch several solution curves. dy dy 2. dx 1. dx = (y + 2)(y - 3)

= y2

- 4

dy 3. dx ~ y' - Y

4. dx ~ y' - 2y

dy

5.y'=vy, y>O 7. y' = (y - I)(y - 2)(y - 3)

6. y' = y 8. y' = y3 _ y2

vy,

y > 0

Models of Population Growth The autonomous diffi:rentia1 equations in Exercises 9--12 represent models for population 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, and which are unstable?

dP 9. dt = I - 2P 11. dP dt ~ 2P(P - 3)

dP 10. dt = P( I - 2P)

(I) 2

12. dP dt ~ 3P{I - P) P -

13. Catastrophic change in logistic growth Suppose that a healthy population of some species is growing in a limited environment

and that the current population Po is fairly close to the carrying capacity Mo. You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddeo1y a catastrophe such as the Mount St Helens volcauic eruption contaminates the lake and destroys a siguificant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity M, considerably less than M o and, in fact, less than the current population Po. Starting at some time before the catastrophe, s1retch a "before-and-after" curve that shows how the fish population responds to the change in environment 14. ControUing a population The fish and game department in a certain state is planning to issue huntiog permits to control the deerpopulation (one deer per permit). It is known that if the deer population falls below a certain level m, the deer will become extioct. It is also known that if the deer population rises above the carrying capacity M, the population will decrease back to M through disease and ma1nutrition.

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),

where P is the population of the deer and r is a positive constant ofproportionality. 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

c. ShowthatifP

> Mforallt,thenlim,_ooP(t) < m for all t?

~

M.

d. What happens if P

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 permits should he issued? Applications and Examples 15. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance 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. enrreot in ao RL-clrcuit The accompanying 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

where k is a constant 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 a160-lb skydiver (mg = 160) and with time in seconds and distance in feet, a typical value of k is 0.005. What is the

Switch

diver's terminal velocity?

y;,

16. Resistance proportional to A body ofmass m is projected vertically downWlltd with initial velocity vo. Assume that the resisting force is proportional to the square root of the velocity and fmd the terminal velocity from a graphical analysis. 17. Sailing A sailboat is runuing along a s1raight course with the wind providing a constant fotWlltd 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 I fr/sec. What is the maximum velocity in feet per second of the boat under this wind? 18. The spread ofinformation Sociologists recognize a phenomenon called social diffitsion, which is the spreading of a piece ofinformation, technological innovation, or cu1toral fad among a population. The members of the population can he divided into _ classes: those who have the information and those who do not In a ftxed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number yet to receive it IfX denotes the number 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 booyant force eqoals 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 descends, complete the following steps. a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Fignrc 9.19. b. Using v(t) for the pearl's velocity as a function 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 UN.

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 hass living in a common pond.

524

Chapter 9: First-Order Oifferential Equations

Phase Planes A general system of two first-order differential 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 y(t) 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), y(t» in the xy-plane starting at some specified point. Therefore the solution functions define a solution curve through the specified point, called a trajectory 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 y(t) 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 differential equations. Several factors affect the mtes 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 bass in the pond:

dx dt = (a - by)x,

(la)

dy dt = (m - nx)y.

(lb)

Here x(t) represents the trout population, y(t) 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 y(t) 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 gmphical 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 population x(t) is constant. Similarly, if x = min, theo Equation (Ib) implies dy/dt = 0, and the bass population y(t) is constant. This information is recorded in Figure 9.26. y Bass

y Bass

y Bass I

.!!

<--~=o, dt

a

"il..

b

b

:(mn'ba)

I -----~-----I

y Bass

I I

~.r--___=-----==--~x

'-------=----->x Trout

!!!

•

(b)

(0)

FIGURE 9.26 "---'----=----->x !!! Trout

•

FIGURE 9.27 To the left ofthe line x = min the trajectories move upward, and to the right they move

dowoward.

y Bass

ff-----.,,::==::=---"------=Trou=--t~X

Trout

(c)

Rest points in the competitive-hunter model giveo by Equations (Ia) and (I b).

In setting up our competitive-hunter model, precise values ofthe constants a, b, m, n will not geoeraJly be known. Nonetheless, we can analyze the system ofEquations (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) repreaeots the number of trout and y(t) the number ofbass at time t, we are thinking ofthe pair ofvalues (x(t), y(t» as a point tracing out a trajectory curve in the phase plane. Wheo dx/dt is positive, x(t) is increasing and the point is moving to the right in the phase plane. If dx/dt is negative, the point is moving to the left. Likewise, the 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 of the 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 the line y = alb the trajectories move

y Bass

to the left, and below it they move to the right. a

b

f--+----oii)--+--

'0 @-

(0,0)

c "!

•

Trout

>X

FIGURE 9.29 Composite grapbical analysis ofthe trajectory directious in the four regious determined by x = min and y = alb.

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. Depeoding on where the trajectory begins, it may move downward into region D, leftward into region A,

526

:

Chapter 9: First-Drder Differential Equations

\~~~

~\. \. _'0, c I D

••

(0, 0)

:---;;;---:>" 'Thuut

m

"

FIGURE 9.30 Motion along the trajectories near the rest points (0, 0) and(m/lI,a/b). Y BRIls

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 9.31. Our graphical analysis leads us to conclude that, under the ll&'WIllp1ions of the campetitive-huntermodel, it is unlikely that both species will reach equilibrium levels. This is because it would be almost impossible for the fish populations to move exactly along one of the two approaching 1mjectories for all time. Fm1:hennore, the initial populations point (XO, YO) detennines which of the two species is likely to swvive over time, and mutual coexistence of the species is highly improbable.

LImitations of the Phase·Plane Analysis Method Unlike the situation for the competitive-hunter model, it is not always possible to determine the behavior oftrajectories near a rest point. For example, suppose we know that the 1mjectories near a rest point, chosen here to be the origin (0, 0), behave as in Figure 9.32. The information provided by Figure 9.32 is not sufficient to distinguish between the three possible 1mjectories shown in Figure 9.33. Even if we could determine that a trajectory near an equilibrium point resembles that of Figure 9.33c, we would still not know how the other 1mjectories behave. It could happen that a 1mjectory 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

---+--«!JJY-- x

y

--+-H++-~·x

FIGURE 9.32 Trajectory direction near the rest point (0, 0). (.j

(b)

(oj

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.

y

Another Type of Behavtor The system

FIGURE 9.34 The solutionx 2 is a limit cycle.

+ y2 - 1

:~Y+X-x(x2+y').

(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 + y 2 = 1, we have dy/dx = -x/y (see Exercise 2). If a trajectory starts inside the unit circle, it spimls outward, asymptotically approaching the circle as t - 00. !fa 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 ofthe 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) sod (2b), show that soy 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 - r') sod -dO/dt ~ -1. 3. Develop a model for the growth of trout sod bass, assuming that in isolation trout demonstrate exponential decay [so that a < 0 in Equations (Ia) sod (Ib)] sod that the bass population grows logistically with a population limit M. Analyze gnJphically the motion in the vicinity ofthe rest points in your model. Is coexistence possible? 4. How might the competitive-hunter model be validated? Include a discussion of how the various constaots a, b, m, sod n might be estimated How could state conservation authorities use the model to ensure the survival of both species? 5. Consider soother coropctitive-hunter model dermed by dx = dt

a(1 - ~) x- bxy k

'

1

t = m(1 - :,)y-

c. Give an economic interpretation of the curves that determine the equilibrium points. 7. Two trajectories approach eqnilibrinm 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) sod (Ib) apply the CbainRuie to derive the following equation: dy (m - nx)y (a - ~)X. dx b. Separate the variables, integrate, sod exponentiate to obtain yfJe-by = Kxme-nJC, where K is a constant ofintegration. e. Letf(y) ~ ya/ebysodg(x) ~ xm/e~. Show that f(y) has a unique maximum of = (a/eb)awheny = a/b as shown in Figure 9.35. Similarly, show that g(x) has a unique maximum M, = (m/en)mwhenx = m/n,alsoshowninFigure9.35.

My

f(y)

nxy,

where x sod y represent trout sod bass populations, respectively. .. What assumptions are implicitly being made about the growth of trout sod hass in the absence of competition?

'----,,!:-----~y

b. Interpret the constants a, b, m, n, kh and k2 in terms oftbe physical problem.

b g(x)

e. Perform a graphical aoalysis: I) Find the possible equilibrium levels. ii) Determine whether coexistence is possible. iii) Pick several typical starting points sod sketch typical tra-

jectories in the phase plsoe. Iv) Interpret the outcomes jnedicted by your graphical soalysis in terms oftbe constants a, b, m, n, kh 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 on the market Let Q be the quantity of the item available on the market Both P sod Q are functions oftime. Ifone considers price sod quantity as two interacting species, the following model might be ]naposed: tiP dt dQ

fit

~ aP(~

Q

_p)'

= cQ(fP - Q),

where a, b, c, and f are positive constaots. Justify sod discuss the

adequacy of the model. a. !fa = I,b = 20,000,c = I,sodf = 30, find the equilibrium points of this system. !fpossible, classify each equilibrium point with respect to its stability. Ifa point cannot be readily classified, give some explaoation. b. Perform a graphical stability aoalysis to determine what will happen to the levels of P sod Q as time increases.

'----cO-------~x

'n"

FIGURE 9.35 Graphs of the functions f(y) = ya/e by sodg(x) = xm/e~.

d. Consider what happens as (x, y) approacbes (m/n, a/b). Take limits in part (b) as x ---+ m/n sod y ---+ a/b to show that either lim

x-min

[(y- a)("-~)] = K x eby

M

y-ajb

or MyjM, ~ K. Thus soy solution trajectory that approaches (m/n, a/b) must satisfy

~~ ehy

(My) (x:). M;; e

e. Show that only one trajectory cso approach (m/n, a/b) from below the line y ~ a/b. Pick Yo < a/b. From Fignre 9.35 you cso see that f(yo) < which implies that

My,

528

Chapter 9: First-Order Oifferential 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 specific values.

This in turn implies that

xm

----;a

e

9. What happens to 1he rabbit population if1here are no foxes present?

< Mxo

10. What happens to the fox population ifthere 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 argument 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 1he meaning ofeach ofthese points. 12. Show, by differentiating, that the function

= a lny(t)

C(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

y:

-~. t

UniquexQ

m n

Trout

unchanged at future times. Each value of the constant C gives a trajectory for 1he autonomous system, and 1hese trajectories close op, rather 1han spiraling inwards or outwards. The rabbit and lOx populations oscillate 1hrough repeated cycles along a fixed trajectory. Figure 9.37 shows several trajectories for the predator-prey system. y

FIGURE 9.36 For any Y < alb only one solution trajectory leads to the rest point (min, alb).

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).

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 ofequations that models the populations of two species, one of which preys on the other. Let x(t) represent 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 encouoters 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 offoxes. The number ofencounters between rabbits and foxes is assumed to be proportional to the product of the two populations. These assumptions lead to the autonomous system dx = (a - bylx dt dy = (-c dt

+ dx))'

o FIGURE 9.37

c d

Rabbit population

Some trajectories along which C is conserved.

13. Using a procedure similar to 1hat in 1he text for the competitiVl>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

values. This lag effect is showe in Figure 9.38, which grapha both x(t) and y(t).

Chapter 9 Practice Exercises

529

14. At some time during a trajectory 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 IIUlXimwn 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.

Questions to Guide Your Review

Chapter •

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?

= f(x, y)?

4. Describe Euler's method for solving the initial value problem y' = f(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 fIrSt-order differential equations? 6. What is an orthogonal trajectory of a family of curves? Deseribe how one is found for a given farnily of curves.

Chapter

1. y' = xe"~

5. y' =

+ xcos

2

ydx ~ 0

~

9. Why is the exponential model unrealistic for predicting long-term population growth? How does the logistic model correct for the deficiency in the exponential model for population growth? What is the logistic differential equation? What is the fonn ofits solution? Desctibe the graph of the logistic solution. 10. What is an autonomous system of differential equations? What is a solution to such a system? What is a trajectory of the system?

xj2 = xe

Initial Value Problems In Exercises 17-22 solve the initial value problem.

2. y' = xye" 4.

a 2 dx -

6. y' =

7. x(x - I) th> - ydx = 0 9. 2y' - y

8. How do you construct the phase line for an autonomous differential equation? How does the 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. secxth>

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?

3vYcscxth> ~ 0

xe~-Ycscy

th>

17.(x+I)dx+2y~x,

dy

+ 2y

18. x dx

_ 2 - x

+ I,

x>-I, x

> 0,

15. 16.

~

y(l)

I

I)x- I

8. y' = (y2 y' 10. 2: + Y = e-xsinx

+ 2y ~ I - x-I 12. xy' - y ~ alnx x (l + eX) th> + (ye + e~) dx ~ 0 e~ th> + (e~y - 4x) dx ~ 0 (x + 3y 2) th> + y dx ~ 0 (Hint: d(xy) ~ y dx + x th» xth> + (3y - x-2 cosx) dx = 0, x> 0

th> 19. dx

+ h). = x 2, y(O)

= -I

11. xy'

13. 14.

y(O)~1

20. xth> 21. xy'

+ (y -

+ (x

22. y dx

cosx)dx = 0,

y(~)

= 0

- 2)y ~ h3e~, y(l) ~ 0

+ (3x

- xy

+ 2) dy

= 0,

y(2) = -I, y < 0

530

Chapter 9: First-Order Oifferential Equations

Euler's Method

023.

+

y' = y

cosx,

0,;; x ,;; 2;

y(O) = 0;

Xo

Massm

mgR 2

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.

F~---

,2

~

I I

= 0

I I I

024. y' ~ (2 - y)(2x + 3), y(-3) ~ I; -3:sx:s -1; Xo =-3

dy

025.c~3; dx ~

dy

026.c=4;

dx

=

x-2y

X+!, x2

x

y(l)

'

center

=I

In Exercises 27 and 28, use Euler's method to solve the initial value problem graphically, starting at Xo ~ 0 with

a. dx

027. dxdy =

~

; +2'

eX Y

x2

dy

b. dx

0.1.

028. dx ~ - eY

~

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

-0.1.

v2

2gR2 =

-s-

+ VOl

2gR.

-

y(O) = -2

Thus, the velocity remains positive as loog as Vo '"

+Y

+ x'

v'2iii

y(O) ~ 0

Slope Fields

29.y'=x

30. y' = I/x

= xy

v'2iii,

The velocity Vo = is the moon~ escape velocity. A body projected upward with this velocity or a greater one will escape from the moon's gravitational pull.

In Exercises 2!1-32, sketcb 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. 31. y'

!

y(O) ~ I

-2y+1

iI

M~'sill

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.

32. y' = I/y

Autonomous Differential Equations and Phase Lines In Exercises 33 and 34:

v'2iii, then

b. Show that if Vo =

s=RI+ (

3vo

2R

)2/3

t.

36. Coasting to a stop Table 9.6 shows the dislaDce s (meters) coasted 00 in-line skates in t sec by Johnathon Krueger. Find a model for his position in the form of Equatioo (2) ofSectioo 9.3. His initial velocity was Vo = 0.86 mlsec, his mass m = 30.84 kg (he weighed 68 Ib), and his total coastiog dislaDce 0.97 m.

a. IdentifY the equilibrium values. Which are stable and which are unstable?

TABLE 9.6 Johnathon Krueger skating data

b. Construct a phase line. IdentifY the signs of y' and y'. c. Sketch a representative selection of solution curves. 33.

dy dx

=y2-1

34.

dy dx

=y_y2

Applications 35. Escape velocity

The gravitationsl attraction F exerted by an

airless moon on a body ofmass 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 figure). The force F is negative because it acts in the direction of decreasing s.

Chapter

I (see)

s(m)

I (see)

s(m)

I (see)

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 ofthe movement ofa dissolved substance across a cell's membrane is described by the equation

dy A = k-(c - y) dt V .

-

In this equation, y is the concentration of the substance inside the cell and dy/dtis the rate at whichy changes over time. The letters

L

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

a. Vetify that y(x) problem y'

substance outside the cell. The equation says that the rate at which the conceotration changes within the cell is proportionsl to the difference between it and the outside concentration.

~

531

Yl (x) - Y2(X) satisfies the initial value

+ P(x)y = 0,

y(xo) = O.

b. For the integrating factor v(x) =

eJP«)"', show that

d tb: (v(x )[Yl (x) - Y2(X)]) = O.

a. Solve the equation for y(t), using Yo to deoote y(O). Conclude that v(x )[Yl (x) - Y2(X)]

b. Find the steady-state conceotration, Iim,-..ooy(t).

2. Height of a rocket If an exteroal force F acts upon a system whose mass varies with time, Newton's law of motion is d(mv) tim ----;j/ = F + (v + u)d/' 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 aystem at the rate dm/dt. Suppose that a rocket of initial mass mo starts from rest, but is driven upward by ruing some of its mass directly backward at the constant rate of dmldt = -b units per second and at constant speed relative to the rocket U = -c. The only external force acting on the rocket is F ~ -mg due to gravity. Under these assumptions, sbow that the height of the rocket above the ground at the end of t seconds (t small compared to molb) is _ [ mo - bt In mo - btl _ y-ct+ b mo

1 2 zgt·

3. a. Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Fundanleotal Theorem of Calculus, Part I to show that soy function y satisfYing the equation v(x)y for v(x) =

=

J

v(x)Q(x) tb:

+C

eJ p(.) '" is a solution to the first-<>rder linear equstion dy tb:

+ P(x)y

= Q(x).

J:'

b. If C ~ Yov(xo) v(t)Q(t) dt, theo sbow that any solution Y in part (a) satisfies the initial condition y(xo) = Yo.

4. (Continuation ofExercise 3.) Assmne the hypotheses ofExercise 3, and assume that Yl (x) and Y2(X) are both solutions to the firstorder linear equation satisfYing the initial condition y(xo) = Yo.

Chapter ID

constant.

a

c. From part (a), we have Yl (xo) - Y2(XO) = O. Since v(x) > 0 fora < x < b, usc part (b) to cs1ablisb that Yl(X) - Y2(x) = 0 on the interval (a, b). ThusYl(X) ~ Y2(X) for all a < x < b.

Homogeneous Equatfons A lITst-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 = ylx. Then, y ~ vxand dy dv tb:=v+xtb:.

Substitution into the original differential equstion and collecting tenus with like vatiables theo gives the separable equstion tb:

dv F(v) = O.

x +v_

Afler solving this separable equstion, the solution ofthe original equation is obtained when we replace v by ylx .

Solve the bomogeoeous equstions in Exercises 5-10. First put the equstion in the funn ofa homogeneous equstion.

5. (x 2 + y 2)tb: + xydy 6. x 2 dy

+ (y2

7. (xeY/·

+ y)tb:

- xy)tb:

=0 =0

- xdy = 0

8. (x +y)dy+ (x -y)tb: y y-x 9. y' = x + cos - x -

10.

~

0

(x sin ~ - ycos~) tb: + xcos ~ dy

Technology Application Projects

Mathematica/Maple Modules: Drug Dosages:Are They Effective?Are They Safe? Formulate and solve an initial value model for the absorption ofa drug in the bloodstream. First-Order Differential Equations and Slope Fields Plot slope fields and solution curves for various initial conditions to selected lltSt-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 X n 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 this 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 ofthe sequence be any integer. In the sequence above, {an} starts with a, while {b n } starts withb 6 • Sequences can be described by writing rules that specify their terms, such as an =

V;,

bn = (_I)n+l{,

en = n ; I,

dn = (_I)n+l,

or by listing !enns: {an} =

{Vi, Vi, V3, ... , V;, ... }

{b n }

{1,-!,t,-i,.. ·,Hr' {, ... }

=

1234

n-I

{en} = { 0'2'3'4'5'''''-n-''''

}

{dn} = {I, -I, I, -I, I, -I, ... , (_I)n+\ ... }.

We also sometimes write

Figure 10.1 shows two ways to represent sequences graphically. The fIrst marks the rlrst few points from a" a2, a3,"" an,'" on the real axis. The second method shows the graph ofthe function defIning the sequence. The function is derIDed only on integer inputs, and the graph consists of some points in the xy-plane located at (I, a,), (2, a2), ... , (n, a,,), .... a" a,

•

"2 03 a4 a ,

1

0

•

• •

2

a" ~

•

••• •

0

a"

a,

'" a2

0

vn

3 2 1

1

1 an =n

~t

_L---L---Ll_ll_~, n

2

3

4

5

a"

a, a.

• •

a,

a, a3 0

•

••

1

aJ! =

FIGURE 10.1

(_1)1I+1~

~f

Sequences can be represented as points on the real line or as

points in the plane where the horizontal axis n is the index number ofthe term and the vertical axis a" 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, V2, V3, ... , V;;, ... } L-E

o

L L+E

•• •••• (•• -I- )

f----------~L

L

have tenDs that get larger than any number as n increases, and sequences like

+E

------------(n,a,,)-~-.--­ f------~~--~L

-

E

{I, -I, I, -I, I, -I, ... , (_I).+I, ... }

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,

c:+---:----:----:-----:':~~---n

o

2 3

N

n

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 {(n, a.)}. In this figure, all the a,/s after aN lie within e of L.

FIGURE 10.2

la. - LI <

=>

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 ofthe sequence (Figure 10.2).

The definition is very similar to the definition ofthe limit ofa function f(x) asx tends to 00 (limx~oof(x) in Section 2.6). We will exploit this connection to calculate limits of HISTORICAL BIOGRAPHY

Nicole Oresme (ca. 132
sequences.

EXAMPLE 1 (aJ

lim

n_OO

Show that

1n =

0

(h) lim k = k n_ OO

(any constant k)

Solution (aJ Let E

>

0 be given. We must show that there exists an integer N such that for all n,

n>N This implication will hold if (I/n) < E or n > I/E. If N is any integer greater than I/E, the implication will hold for all n > N. This proves that lim.~oo(l/n) = O.

(h) Let E

>

0 be given. We must show that there exists an integer N such that for all n,

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

Show that the sequence {I, -I, I, -I, 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 I 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.

Solution

10.1 Sequences

535

It follows that IL - II < 1/2, or equivalently, 1/2 < L < 3/2. Likewise, the number - I appears repeatedly in the sequence with arbitnuily 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 have no overlap. Therefore, no such limit L exists and so the sequence diverges. Note that the same aIgIlIl1ent works for any positive number € smaI1er 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

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 smaI1 as n increases. Nor are we asserting that there is some num-

ber infinity that the sequence approaches. We are merely using a notation that captures the idea that an eventually gets and stays larger than any fIxed number as n gets large (see Figure lO.3a). The terms of a sequence might also decrease to negative infinity, as in Figure 10.3b. (b)

FIGURE 10.3
index N all lie in tlle yellow band above M. (bJ The sequence diverges to - 00 because all terms after some index N lie below any

DEFINmON The sequence {an} diverges to infinity iffor every number M there is an integer N such that for all n larger than N, an > M. If this condition holds we write lim a. = 00 or n~OO

Similarly if for every number m there is an integer N such that for all n > N we have an < m, then we say {a.} diverges to negative infinity and write

lim n~OO

an

= -00

or

an ~ -00.

chosen number m.

A sequence may diverge without diverging to infinity or negative inf"mity, 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 offunctions given in Chapter 2 have 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 a. = A and limn_ oo bn = B. 1. Sum Rule:

lim,,~oo(an

2. Difference Rule:

lim,,~oo(an

+ bn)

= A

+B

3. Constant Multiple Rule:

- bn) = A - B lim,,~oo(k'bn) = k·B (any number k)

4. Product Rule:

lim,,~oo(an'

5. Quotient Rule:

Iim,,~oo b

bn) = A' B an A n

= Jj

The proof is similar to that ofTheorem 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)

= -I- lim

n

11--+ 00

1 n

= -1·0 = 0

ConstanlMultipleRuleandExamplela

I) n - I) = lim ( (b) lim ( -nI = lim I 11---+00 11---+00 11---+00

n

lim

11---+ 00

nI =

I - 0 = I

Difference Rule andExamplela

Product Rule

- 7 6 (4/n 6 ) - 7 0 - 7 (d) lim 4 6 n=lim =--=-7. n~OO n + 3 n~OO I + (3/n 6 ) I +0

Sum and Quotient Rules

•

Be cautious in applying Theorem 1. It does not say, for example, that each of the sequences {an} and ibn} 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 ofTheorem I is that every nonzero multiple of a divergent sequence {an} diverges. For suppose, 10 the contrary, that {can} convetges fur some number e oF o. Then, by1llking k = I/e in the Constant Multiple Rule in Theorem I, we see that the sequence

... .... .. .... a; fn

L

b~.~T~.~~:::f~r/I

•• -

c:f--"----------->.

o

FIGURE 10.4 The terms of sequence {bn } are sandwiched betweeo those of {a,,} and {cn}, forcing them to the same common limit L.

converges. Thus, {ea.} cannot converge unless {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 {en} be sequences ofreal numbers. If a. :5 bn :5 en holds 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 Ibnl :5 en and en --+ 0, then bn ---+ 0 because -en ~ bn ~ Cn " We use this fact in the next example.

EXAMPLE 4

Since I/n --+ 0, we know that because

_l
(b) ~--+O n

because

o <:- ~ <: 1. 2" - n'

(c) (- I)nk--+ 0

because

_1n <: (_l)n1n -

(a)

cosn~O

n

2

<:

-

1n"

•

The application ofTheorems 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.

Solutton We know that (n + I)/n --+ 1. Taking f(x) = gives V(n + I)!n --+ VI = I.

VX and L

= I in Theorem 3

•

EXAMPLE 6 The sequence {I/n} converges to O. By taking an = I/n, f(x) = 2\ and L = 0 in Theorem 3, we see thaI2!/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

•

>x

FIGURE 10.5 As n -+ 00, lin -+ 0 and 2'/' -+ 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 fix) ~ 2".

The next theorem formalizes the connection between Iim"~",, an and limx~"" f(x). It enables us to use I'Hllpital's Ru1e to f"md the limits of some sequences.

THEOREM 4 Suppose thai fix) is a function defined for all x ;;,: no and thai {an} is a sequence ofreal numbers such that an = f(n) for n ;;,: no. Then lim f(x) = L

lim an = L.

x~OO

n~OO

Proof Suppose that lim....."" f(x) = L. Then for each positive number ber M such that for all x,

If(x) - LI <

x>M

E

there is a num-

E.

Lei N be an integer grealer than M and greater than or equal to no. Then

n>N

EXAMPLE 7

=>

a,

=

f(n)

Ian -

and

LI

=

If(n) - LI <

E.

•

Show that

Solutton The function (lnx)/x is defined for all x ;;,: I and agrees with the given sequence at positive integers. Therefore, by Theorem 4, Iim"~,,,,(ln n)/n will equal limx~",,(lnx)/x if the latter exists. A single application ofl'Hllpital's Ru1e shows that lim lnx = lim I/x = Q = x- OO 1 1

o.

x-OO X

Weconcludethatlim,~",,(lnn)/n =

•

O.

When we use I'Hllpital's Ru1e 10 f"md the limil of a sequence, we often !real n as a continuous real variable and differentiate directly with respect to n. This saves us from having to rewrite the formu1a for an as we did in Example 7.

EXAMPLE 8

Does the sequence whose nth lerm is an =

("-±-l)' n1

538

Chapter 10: Infinite Sequences and Series Solution The limit leads to the indeterminate form 100 • We can apply I'HllpitaI's Rule if we first change the form to 00 • 0 by taking the natorallogarithm of an:

lnan

=

In(n + I)n n- I

Then, 00·0 fonn

n~~

In(~)

llform

o

lin

-2/(n 2 - I) n~OO -l/n .

I:Hilpital', Rule: diffi:rentiate

= lim -"--'---c-.---'2 = Since

In an --->

numerator and denominator.

lim~=2. 1

n_ 00 ,,2 -

2 and f(x) = eX is continuous, Theorem 4 tells us that

•

The sequence {an} converges toe 2 •

Commonly Occurring Limits The next theorem gives some limits that arise frequently.

THEOREM 5

The following six sequences converge to the limits listed below:

1. lim lnn = 0 n- OO n 3. lim x ' /n = I

2. lim"%=1 n~OO

(x> 0)

n~OO

5. n_ limOO

In

(1+"-)"=e n

(lxl

n 6. lim "-- = 0

(any x)

n~OO

X

(any x)

< I)

4. limxn=O n_OO

Formulas (3) through (6), x remains rlXed as n --->

n!

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 ~n2)

(b)

W

=

These are examples ofthe limits in Theorem 5.

2 ~n

---> 2. 0 =

0

= n2/n = (n lln )'---> (1)2 = I

(c) '{i3;; = 3 /n(n lln ) ---> 1·1 = I ' (d)

(-t)"---> 0

Formula 1 Formula 2

Formula 3 with x

=

Formula 4 with x

= _!

3 and Formula 2

2

10.1 Sequences

I Factorial Notatioo

(e)

The notation n! (Un factorial") means the product I . 2 . 3 ... n ofthe 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 OJ to be 1. Factorials grow even faster than exponentials, as the table suggests. The values in the table are roooded.

It

I 5 10 20

en 3 148 22,026 4.9 X 108

It!

I 120 3,628,800 2.4 X 10 18

(f)

(n ~ 2)" (I + --rnn~e-2 =

loon

-~O

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 defmed recursively by giving

1. The value(s) of the initial tenn or tenns, and 2. A rule, called a recursion formula, for calculating any later tenn from tenns that precede it.

EXAMPLE 10

a,

(a) The statements = I and a" = an-l + I for n > I define the sequence 1,2,3, ... , n, ... ofpositive integers. With = I, we have a2 = + I = 2, a3 = a2 + I = 3, and so on. (h) The statements = I and an = n'an-l for n> I defme the sequence 1,2,6,24, ... , n!,... of factorials. With a, = I, we have a2 = 2· a, = 2, a3 = 3'a2 = 6,,,. = 4'a3 = 24, and so on.

a,

a,

a,

(c) The statements a, = I, a2 = I, and an+l = a" + an-l for n > 2 define the sequence I, 1,2,3,5, ... of Fibonacci numbers. With = I and a2 = I, we have a3 = I + I = 2, a4 = 2 + I = 3, as = 3 + 2 = 5, and so on.

a,

(d) As we can see by applying Newton's method (see Exercise 133), the statements 2 XQ = landxn+l = Xn -- [(sinxn -- xn )J(cosxn -- 2xn)]forn > o define 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 {an} is bounded from above if there exists a number M such that an ,,; M for all n. The number M is an upper bound for {an}. If M is an upper bound for {an} but no number less than M is an upper bound for {an}, then M is the least upper bound for {a,,}. A sequence {a,,} is bounded from below if there exists a number m such that an ;" 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 {an}, then m is the greatest lower bound for {a,,}. If {an} is bounded from above and below, the {an} is bonnded. If {an} is not bounded, then we say that {an} is an nnbounded sequence. EXAMPLE 11 (a) The sequence 1,2,3, ... , n, ... has no upper bound since it eventnally surpasses every number M. However, it is bounded below by every real number less than or equal to 1. The number m = I is the greatest lower bound ofthe sequence. 123 n . (h) Thesequence Z'3'4"'" n + I"" lSbounded above by every real number greater than or equal to I. The upper bound M = I 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 sequences are bounded

If a sequence {a,,} converges to the number L, then by def"mition there is a number N such that Ia" - L I < 1 if n > N. That is,

L-l
forn>N.

If M is a number larger than L + 1 and all of the f"mitely 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 bounded sequences that fail to converge. One example is the bounded sequence {( _1)n+1) discussed in Example 2. The problem here is that some bounded 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.

DEFINITION A sequence {an} is nondeereasing if an :5 an+1 for all n. That is, at ~ a2 ~ a3 ~ .... The sequence is nonincreasing if an ~ an+ 1 for all n. The sequence {an} is monotonic ifit is either nondecreasing or nonincreasing. FIGURE 10.6 Some bounded sequences bouncc around between thcir bounds and fail to converge to any limiting value.

EXAMPLE 12 (a) The sequence 1,2,3, ... , n, ... is nondecreasing.

123 n . d . (b) The sequence 2' 3' 4' ... , n + 1"" IS non ecreasmg. 111

1

()Th c e sequence 1'2' 4' 8"'" 2n,'"

... 18 nonmcreasmg.

(d) The constant sequence 3, 3, 3, ... ,3, is both nondecreasing and nonincreasing. is not monotonic. _ (e) The sequence 1, -1, 1, -1, 1, -1, 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 bound. 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

MI------'-----y~L

........

L I - - - - - - ' - - .-."' ••".,....-r--

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

o

FIGURE 10.7 If the terms ofa nondecreasing sequence have an upper bound M, they have a limit L :5 M.

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 (1, all, (2, a2), ... , (n, an), ... in the xy-plane. If M is an upper bound of the sequence, all 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 bounded 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).+lJn} 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 FInding Terms of a 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" and /l4. 1. an

I - n

2.

= --,-

n

(-1)'+1 3. a, = 2n I

an

=

+ (-I)'

2' - I

Each of Exercises 7-12 gives the fIrst term or_ ofa sequence along with a recursion formula for the remaining terms. Write out the first ten tenns of the sequence.

= I, a.+1 = a. + (1/2') = I, a.+1 = aJ(n + I) = 2, a.+1 = (- 1).+1a,,/2 ~ -2, a,+l ~ naJ(n + I) = a2 = 1, a n+2 = a n+l + an

11.

at

12.

at =

2, a2

=

-1, all +2

33

.

14. The sequence -1, 1, -1, 1, -1,

I ~ with alternating ,igns

.

15. The sequence I, -4,9, -16,25, ...

16. The sequence I, -

I I

4' 9' -

I - 2n 29·a,,=1+2n

31. a" =

FInding a Sequence's Formula In Exercises 13-26, find a formula for the nth term of the sequence.

I I 16' 25""

.a"

I - 5n 4 4

n + 8n

,

=n'-2n+1

n-I

+ (-I)'

35. a"

~

I

37. a"

~

(n;"

1)(1 -

( _1).+1 39. a" ~ 2n I

+

I ./ 1- 3 vn 2n

32 n = n+3 .a n2 +5n+6

34.an =

n'

I 2 70 - 4n

k) k) 38. a, ~ (2 - i,)(3 + i,) 40. a. ~ (-t)" 36. a,

~ (-I)' (I

~ ~n ~ I

I 42. a, ~ (0.9)'

alternating signs

43. a"

~ Sin(f + k)

44. an

Reciprocals of squares ofthe positive integers, with alternating signs l'owmi of2 divided by

45•

_ sinn n

Squares ofthe positive integers, with

Integers differing by 2 divided by prodocts of consecutive integers

20. The sequence -3, - 2, -I, 0, I, ...

30.a.=

41. a"

1's with alternating signs

multiples of3

19. The sequence 0, 3, 8, 15,24, ...

repeated

Convergence and Divergence Which of the sequences {a.} in Exercises 27--90 converge, and which diverge? Find the limit ofeach convergent sequence. n + (-I)' 27. a" = 2 + (0.1)' 28. a, = n

= an+l/an

13. The sequence 1, -1, 1, -1, 1,

Each positive integer

26. The sequence 0, 1, 1,2,2,3,3,4, ...

6.a,,~~

7. a, 8. a, 9. a, 10. a,

Altemating l's and O's

25. The sequence 1,0, 1,0, 1, ...

I

I n.

4. a" = 2

Cubes ofpositive integers divided by powcn; of 5

I 8 27 64 125 24. 25' 125' 625' 3125' 15,625''''

Squares ofthe positive integCIll diminished by I Inlcgers, beginning with -3

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

Inlcgers differing by 3 divided by factorial,

n'11' cos (n'11')

an -

3'

n 47. a" = 2' 49.

=

-

a.

=

48.an = ) n

In(n+l)

y;,

51. a" ~ 8' /'

53. a"

~

55. a" =

57. a"

~

lnn

50. a. = 1n2n 52. a, ~ (0.03)'/'

(I + i.)"

54.a,~(I-k)"

"(:!iOn

56. a. =

(Ii3)'/'

lnn 59. a" = n /'

'

58. a,

~

%2 (n

+ 4),/(,+4)

60. a, = Inn - In(n + I)

542

Chapter 10: Infinite Sequences and Series

61. a, = ~ 63. a,

(-4)' n.

65. an

=

n! 66. an = 2"'3"

67. an

= ( -nl )1/(ln'J

69. a,

=

68. a, =

1n(1 + k)'

70.a'=(n~S 72. a, =

:s

(I -

(10/ 11 )" 74. a, = (9/10)' + (11112)'

cask)

78. an = n(l -

80. a, = (3' ~

y;;tan

88. a,

~

Write out enough early terms of the sequence to deduce a general formula for X, that holda for n '" 2. 100. A sequence of rational numhers is described as follows:

I)'

(3n + 3n - I

(laX:

75. an

= tanhn

Ir,

79. a,

= y;; sin ~

81. all

= tan-I n

83. a,~

X>O

• 1 _ 1 sm1j

= 2n

n

I 3 7 17 a a + 2b --I' 2' S' 12'···· b' a + b •....

----

n2

77. all

85. all

(In n)'

a. VerilY that X1 2 - 2Y12 ~ -I,xi - 2yi ~ +1 and, more generally, that if a 2 - 2b 2 = -I or + I , then

(a + 2b)2 - 2(a +

I (3I)' + vzn

b)'

X n+l

90. an

1

=

/,' pdx, I

!'(x )· n

Do the sequences converge? If so, to what value? In each case, begin by identifYing the functioo f that generates the sequence.

p>1

1 X

Recursively Deftned Sequences In Exercises 91-98, assume that each sequence converges and fmd its

limit 91. at = 2,

Cln+l = 1

92. at = -1, a ll +1 93. al = -4,

72

+ all

95. al = 5, 96. al ~ 3,

I

an + 6 = all + 2

-

2

X n+l = X n -

~ =

b. Xo = I,

X n +l = X n -

---'0;--

c.

X n+l = XII -

1

Xo =

1,

V8 + Za, V8 + Za,

a,+1 =

I

x n2

a. :to = I,

x"

1

2: + Xn

tanxn - 1

sec2 X n

102. a. Suppose that f(x) is differentisble for all X in [0, I] and that f(O) ~ O. Define sequence {a,} by the rule a, ~ nf(l/n). Show that lim,_ooa, ~ 1'(0). Use the result in part (a) to fmd the limits of the following sequences {a,}.

""+1 = ""+1 = vSa;; ""+1 ~ 12 - v;;.

94. al = 0,

-I,

f(x,)

= Xn -

I

1/,'1xdx

or

101. Newton'. method The following sequences come from the recursioo formula for Newtoo's method,

= --n-

• ,,------; •~ vn 2 - 1 - vn 2 +n

= +1

b. The fractions Tn = xJYn approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that r,2 - 2 ~ ±(lly,)2 and that y, is not less than n.)

87. an=n-~

y;;

89. an = 11

b. an = n tan-I

d. a,

I

97.2,2+ , 2 + - -,2+ I , ... 2 1 2+-2 2+-2+ 1 2 98.

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,.

respectively.

(Inn)"')

84.a,=~ 86. a, =

X n+1 =Xl +X2 +"'+X/I'

=

+ 5,)1/,

1-1

n!

10 611

71. a,

76. a, = sinh (In n)

82• a,

99. The first tenn of a sequence is Xl = 1. Each succeeding term is the sum of all those that come hefore it:

~ nn~ (Hint: Compare with lin.)

all = - - , -

64.

Theory and Examples

62. a, = vt32n + 1

VI, VI + VI, VI + VI + VI, VI + VI + VI + VI, ...

k

c.

an

= n(e 1jn

-

1)

~ nln(1 +~)

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

be, respectively, the integer floor and ceiling for a2/2.

10.1

Sequences

543

2' - I

117. a" ~ ~ 119. a"

l~J

~ «(-I)' + l)(n ~

120. The fl!St tenn of a sequence is Xl = 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.+1 6

121. a" = I

a

a. Show that a 2 + b 2 ~ c 2 • (Hint: Let a ~ 2n band c in tenns ofn.)

+

104. The nth root of n! a. Showthat1im.~oo(2mr)1/(2n)= I and hence, using Stirling's approximation (Chapter 8, Additional Exercise 32a), that

D b.

for large values of n.

Test the approximation in part (a) for n ~ 40,50, 60, ... , as far as your calculator will allow.

b. Prove that lim.~oo(l/n') = 0 if c is any positive constant (Hint: If. = 0.001 andc = 0.04, how large shouldNbeto ensure that II/n' - 01 < • ifn > N?)

106. The zipper theorem Prove the "zipper theorem" for sequences: If {a.} and {b.} both converge to L, then the sequence

126. Uniquene•• of least upper bounds Show that if M 1 and M2 are least upper bounds for the sequence {a.}, then M1 ~ M2.

That is, a sequence cannot have two different least upper bounds. 127. Is it true that a sequence {a.} of positive numbers must converge if it is bounded from above? Give reasons for your answer. 128. Prove that if {a.} is a convergent sequence, then to every positive number E there corresponds an integer N such that for all m

andn,

= I.

108. Prove that limn~ooX1/. = I, (x

> 0).

110. Prove Theorem 3.

In Exercises 111-114, detennine ifthe sequence is monotonic and ifit is bounded.

3n + I n+T

2"311 113. an =-, n.

(2n + 3)! 112. a" = (n + I)!

n>N

~

Ia", -

a" 1<

•.

130. Limits and subsequences If the terms ofone 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 L 1 L 2 ,

*

then {a.} diverges.

132. Prove that a sequence {a.} converges to 0 if and only if the quence of absolure values {I a.l} converges to O.

1I6.a"=n-,,

Sl>-

133. Sequence. generated by Newton'. method Newton's method, applied to a differentiable function f(x), begins with a starting value Xo and constructs from it a sequence of numbers {x.} that under favorable circumstances converges to a zero of /. The

recursion fonnula for the sequence is f(x.) Xn+1

2 I 114. a" = 2 - " - 2'

Which of the sequences in Exercises 115-124 converge, and which diverge? Give reasons for your answers. I I

1I5.a.=I-"

m > N and

131. For a sequence {a,,} the tenns of even index are denored by au and the terms of odd index by au+ 1. Prove that if au ---+ L and a2k+1 ~ L, then an ~ L.

converges to L.

111. a. =

n+ I 122. an = - n -

129. Uniquene•• of limit. Prove that limits of sequences are unique. That is, show that if L 1 and L, are numbers such that a" -+ LI and an -+ L2. then Ll = L,..

if c is any positive constant.

109. Prove Theorem 2.

+..j,2.'

+ I)}.

{x.. cos (n

124. at = 1, an+l = 2a" - 3 125. The sequence {n/(n + I)} bas a least upper houud of 1 Show that if M is a number less than I, then the terms of {n/(n + I)} eventually exceed M. That is, if M < I there is an inreger N such that n/(n + I) > M whenever n > N. Since n/(n + I) < I for every n, this proves that I is a least upper bound for {n/(n + I)}.

105. a. Assuming that limn~oo(l/n') = 0 if c is any positive constant, show that

107. Prove that limn~oo'Cf,;

~ max

I and express

b. By direct calculation, or by appealing to the accompanying figure, fmd

vt,;! ~ ~

I)

D

= Xn -

f'(x )· n

a. Show that the recursion formula for f(x) ~ x 2 - a, a > 0, can be written as X.+1 = (x. + a/x.)/2. b. Starting with Xo = I and a = 3, calculate successive tenns of the sequence until the display begins to repeat. What number is being approximared? Explain.

544

Chapter 10: Infinite Sequences and Series

x,

a. Calculate and theo plot the flISt 25 terms of the sequeoce. Does the sequeoce appear to be bouoded from above or below? Does it appear to converge or diverge? If it does cooverge, what is the limit L? b. If the sequence converges, find an integer N such that Ia" - L I :5 0.01 for n '" N. How far in the sequence do you have to get for the terms to lie within 0.0001 ofL?

0134. Arecursivedefioi1ionof"./2 Ifyou starl with = I anddefme the subsequent terms of {x,} by the rule x, = X,-1 + COSX,-I, you geoerate a sequence that converges rapidly to 'If/2. (a) Try it. (b) Use the accompaoying figure to explain why the convergence is so rapid. y

135. a, =

COSXn_l

%

136. a" = I

= an + 5"

137. at = I,

Qn+1

138.

an+! = an

at =

139. an

=

1,

+ (-2)" 140. Cln

sinn

sinn

COMPUTER EXPLORATIONS

Use a CAS to perform the followiog steps for the sequeoces in Exercises 135-1%.

10.2

(I + 0~5)"

=

n sin

k

Inn

141. a, = -n-

142. a" = 11

143. a, = (0.9999)'

144. a" = (123456)'/'

145. an

=

n 41 1%. a" ~ 19"

8' I

n.

Infinite Series An infinite series is the sum of an infmite sequence of numbers

a, + 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 + ... + an

is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n gets 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

I +

I

I

I

I

2 + 4: + 8 + 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

I

First:

SI =

Second:

s2=1+ 2I

Third:

I I s,=I+-+2 4

nth:

I I I s, =1+-+-+ .. ·+-2 4 2,-1

Suggestive expression for partial sum

2 - I

I 3 2 7 4

2 - -I 2 2 - -I 4

2' - I 2,-1

I22,-1

10.2

Infinite Series

545

Indeed there is a pattern. The partial sums form a sequence whose nth tenn is SrI

=

1 2 - 2n - 1 •

This sequence of partial sums converges to 2 because 1im,,~00(l/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 ofpartial 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.

I

1/4

~

o

III

FIGURE 10.8 As the lengths I,

,/2, ,/4, '/" ...

1/8

2

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 + a2 + ,.. + 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 a, + a2 + ... + 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 shorthand when summ.a.tion 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

ar 2

+ ... + ar n- t + ...

=

L ar n- t 11=1

in which a and r are fixed real numbers and a oF O. The series can also be written as ~:o ar". The ratio r can be positive, as in I I (I)a-l 1+-+-+···+ + ...

2

4

2

r~1/2.a~1

'

or negative, as in

I-t+~- .. ·+

r~-1/3.a~1

( - t f ' +....

If r = I, the nth partial sum of the geometric series is Sa = a + a(l) + a(I)2 + ... + a(I)a-l = na,

and the series diverges because lima~oo Sa = ± 00, depending on the sign of a. If r = -I, the series diverges because the nth partial sums alternate between a and O. If 1r 1 oF I. we can detennine the convergence or divergence ofthe series in the following way:

rSn

=

+ aT + ar 2 + ... + ar",-l aT + ar 2 + ... + arn- 1 + ar

Sn - rSn

=

a - ar"

8n = a

ll

Multiply So by r. Subtract rSn from Sn' Most of the terms on the right cancel.

s.(1 - r) = a(1 - r a) 8n =

Factor.

a(1 - r a ) 1- r '

(r oF I).

We can solve for 811 if r #:- 1.

If 1 rl < I, then r a --> 0 as n --> 00 (as in Section 10.1) and Sa --> a/(I - r). Iflrl then 1ral--> 00 and the series diverges.

> I,

Iflrl < I, the geometric series a + ar + ar2 + ... + ar a- I + ... converges to a/(I - r): 00

~ 11=1

ar n -

t =

_a_, 1- r

Irl <

1.

If I r 1 ;;,: 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 ~:1 ar a- l (or with the index n = 0 ifwe write the series as ~:o ar a).

EXAMPLE 1

The geometric series with a = 1/9 and r = 1/3 is I

I

I

00

I (I )a-l

9 + 27 + Sf + ... = ~ 9 3 EXAMPLE 2

1/9 I - (1/3)

The series 00

(-I)a5 4

5 4

5 16

5 64

~--=5--+---+'" a

a~O

I 6'

•

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+ ...

,-I"

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 ~::'~I a, converges. To see why, let al + a2 + ... + a, the "th partial sum. When " is large, both s, and S,_I are close to S, so their difference, a" is close to zero. More formally,

S represent the series' sum and s, = an = 3 n -

~

3 n -l

S- S =

o.

Diffemtce Rule for sequences

This establishes the following theorem.

I Caution

Theorem 7 does not say that ~;1 an converges if an ---+ o. It is possible for a series to diverge when an ---+ 0.

00

If ~ an converges, then an ----i> O.

THEOREM 7

n=1

Theorem 7 leads to a test for detecting the kind ofdivergence that occurred in Example 6.

The Ilth-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)

L00,,+1. -,,- diverges because -,,+1 , , - --> 1. n=1

00

(c)

L (- 1)'+1 diverges because lim,~oo(- 1)'+1 does not exist.

,-I 00

(d)

-n

.

.

EXAMPLE 8

1

-n

~ 2" + 5 diverges because lim.~oo 2" + 5

= -

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 I 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 1. Sum Rule:

~(an + bn) = ~a. + ~bn = A + B

2. Difference Rule:

~(an

3. Constant Multiple Rule:

~kan

- bn) = ~a. - ~bn = A - B = k~an = kA (anynumberk).

Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section 10.1. To prove the Sum Rule for series, let ~=~+_+ ... +~

~=b,+~+···+~.

Then the partial sums of ~(an + bn) are Sn = (a, + b,) + (a2 + b2) + = (a, + ... + an) + (b, + = An

+ (an + bn) + bn)

+ B n·

Sinee An --+ A and B n --+ B, we have Sn --+ A + B by the Sum Rule for sequences. The proof ofthe 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 ofTheorem 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 diverge.

Caution

~bn

diverges, then ~(an + bn) and ~(a.

Remember that ~(an + bn) can converge when ~a. and ~bn both diverge. I + I + I + "'and~bn = (-I) + (-I) + (-1) +'''diverge, = 0 + 0 + 0 + ... converges to O.

Forexample,~an = whereas ~(an + bn)

EXAMPLE 9 00

(s)

L

n=1

- bn) both

Find the sums of the following series.

3n-' _ 1 6n ,

00

=

L

n=1 00

=

L

11=1

(

1 1 ) zn-' - 6n-'

1 zn-' -

I 1 - (I/Z)

4 = Z -6- = 5

5

00

L

11=1

I 6n-' 1

1 - (1/6)

Difference Rule

Geometric series with a ~ 1 andr ~ 1/2,1/6

550

Chapter 10: Infinite Sequences and Series

(b)

Constant 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

00

Lan

+

= at

a2

+ ... +

ak-l

n=1

+ ~ an" 1I=k

Conversely, if~:;"_ka. converges for any k

>

1, 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 n in the fannula fora, by n - h: 00

00

~ an =

11=1

~ an-h

= at

+

a2

+ Q3 + ....

n=-l+h

To lower the starting value ofthe index h units, replace the n in the fannula for a. by n 00

+ h:

00

:L an = n=l-lt :L an+h = 11=1

at

+

a2

+

Q3

+ ....

We saw this reindexing in starting a geometric series with the index n = 0 instead of the index n = 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

1 1

+ 2 + 4 + ...

as 00

1

~2n,

00

1

~ 2,-5' n=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 Find;ng nth Partial Sums In 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.

Using the nth-Term Test In Exercises 27-34, use the nth-Term Test for divergence to show that

the series is divergent, or state that the test is inconclusive.

1. 2+~+~+l+ ... +~+ ... 3 9 27 3'-1

27.

~n

n + 10

2 ~ + _9_ + _9_ + ... + _9_ + ... • 100 1002 100' 100'

00

I

29.

~~ 11-0 n

3.1_

00

1 + 1 _ 1 +"'+(_1),-1_1_+ ... 2

4

8

2,-1

~

Series with Geometric Terms In Exercises 7-14, write out the fITst few terms ofeach series to show how the series starts. Then fwd the sum of the series. 00 I 00 (-I)'

7. ~-4'

8. ,-2 ~4'

,-0

00 7

9. 11. 13.

00 5 10. ~(-I)' 4'

~4'

,-0

(5 + I) (I + -----sn (-I)') ~ 00

~

2"

00

00

~

14.

2"

m (~Y +

32.

,-I

1&

(tY

e;:I)

+

(~)' + (~)' +

.

(~2 Y+ (~2)' + (~2)' + (~2)' + (~2)' + ...

Repeating Declmals 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 ...

22. O.d ~ O.dddd 23. 0.06

= 0.06666

24. 1.414

.

, where d is a digit .

1.414414414 ...

~

25. 1.24123

~

26. 3.142857

~+-+ e n

11-0

00 34. ~ cosmr

,-0

Telescop;ng Series In 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. 00 35. ~

n-I

(In-~I I) n

00 37. ~ (In v;;-+l

-

36.~(3

,-I \;;2

3) (n + I)'

In y';;)

00 38. ~ (tan (n) - tan (n - I)) 11=1

1.24123 123 123

= 3.142857 142857

. .

00 40. ~

,-I

(v;;-+4 -

~)

Find the sum ofeach series in Exercises 41-48. 41

~. 4 • .;:1 (4n - 3)(4n

43

~ 40n • ,~1 (2n - 1)'(2n + 1)2

.

+ (t)' + (t)4 + (t)' + ...

21. 0.7 = 0.7777

,

(5 I)

16. 1+ (-3) + (-3)2 + (-3)' + (-3)4 + 17. (t) +

~+11-1 n + 3

12.~ 2"-3"

3"

In Exercises 15-18, determine if the geometric series converges or diverges. If a series converges. fmd its sum.

15. 1+

30.

00

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) + ...

,-I (n 00

00 I 31. ~ cos n

4.1- 2 +4 - 8 + ... + (-1)'-12'-1 + ...

n(n + I) + 2)(n + 3)

28.

45.

+ I)

00(1 I) ~ y';; - v;;-+l

42

•

~

~

6 (2n - 1)(2n + I)

44.~2n+1 2 ,~1 n (n

46.

00(1 ~ 21/,

+ 1)2 I) - 21/(,+1)

~. ( I _ I ) '.;:1 In(n+2) In(n+l) 00 48. ~(tan-l (n) - tan-I (n + I)) 47

,-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 00(1)' SO. ~(v'2r 49. ~ • r;:: ,-0 11=0 v2 00 51. ~ (-1)'+1 3, 52. ~ ( -1)'+1 n 11=1 2 11=1 00 00 54. ~ cos:'11' 53. ~ cosmr n=O 5 ,-0

552

Chapter 10: Infinite Sequences and Series

00

00

57. 59.

61.

,-I

2

00

58.

211 _ 1

~3" 00

83. Show by example that ~(aJb,) may diverge even though ~a" and ~b, converge and no b, equals O.

56. ~ In 3'

~ 10' ,-I 00

I

00

e-'" ,-0

55. ~

00 60.~ 00

1

illustrate the fact that toAB.

I

I)'

,

62. ~"IFI n!

~ I~O'

2"+ 3" 63. ~11=1 4"

00

00

64.

21'.

~a"b,

85. Show by example that ~(a,/b,) may cooverge to something other than A/Beven whoo A ~ ~a" B ~ ~b, # 0, and no b, equals O.

I-Ii

(

~a, and B = ~b, that may converge without being equal

84. Find convergoot geometric series A =

I

~"' Ixl> 11=0 X

86. If ~a, converges and a, > 0 for all n, can anything be said about ~(I/a,,)? Give reasons for your answer.

+ 4/1

~311+411

87. What happeos if you add a fmite number of terms to a divergent series or delete a ftnite number of tenns :from a divergent series?

65.~1n( n ~I)

Give reasons for your answer.

11""1

Geometric Sertes with a Vartable 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'"

a. a = 2 b. a = 13/2. 90. Find the value of b for which 1 + e b + e 2JJ + e 3b + ... = 9. 91. For whst values of r does the infinite series

,-0

,-0

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 ~arll-1 that converges to the number 5if

00 (-I)' ( I )' 72. ~-2- 3 + sinx

converge? Find the sum ofthe series when it converges. In Exercises 73-78, find the values of x for which the given geometric senes converges. Also, fmd the sum of the series (as a functioo of x) for those values of x. 00

00

73. ~2'x'

74. ~(-I)Y'"

,-0

,-0

00

75. ~(-I)'(x

,-0

92. Show that the error (L - so) obtained by replacing a convergoot geometric series with one of its partial sums Sn is arn/(l - r). 93. The accompanying figure shows the first five of a sequeuce of squares. The outermost square hss an area of 4 m'. 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.

00 ( - I)"(x - 3)' 76. ~ 11=0 2

+ I)'

00

00

77. ~ sin"x

78. ~(lnx)'

,-

Theory and Examples 79. The series in Exercise 5 can also be written as

~

,-I (n

I

~

d

+ I)(n + 2) an

,--I (n

Write it as a sum beginning with (a) n (c) n = 5.

~

I

+ 3)(n + 4) . -2, (b) n

~

0,

80. The series in Exercise 6 can also be written as 00

5

~ n(n + I)

00

and

5

~ (n + I)(n + 2) .

Write it as a sum beginning with (a) n = -I, (b) n = 3, (c) n ~ 20. 81. Make up an infinite series ofnonzero terms whose sum. is a. 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.

94. Helga von Koch's snowflake enrve Helga voo Koch's snowflake is a curve of infmite leogth that oocloses a regioo of fmite area. To see why this is so, suppose the curve is generated by starting with an equilateral triangle whose sides have leogth I.

a. Find the loogth L, of the nth curve C, and show that

limn.. . . oo Ln =

00.

b. Find the area A, of the region oocloscd by C, and show that lim,~ooA, = (8/5)A 1 •

6000 c,

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 ~~1
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 ofnonnegative terms cooverges if and ouly if its partial sums are bounded from above.

EXAMPLE 1

The series

~1= I +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 I/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 I/Zn+1 is greater than zn/zn+ 1 = I/Z. The sequence of partial sums is not bounded from above: If n = Z1, 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 I/n.

EXAMPLE 2

Does the following series converge?

~~= I +1+1+~+ ... +~+ ... 2 2

n-l

n

4

9

16

n

554

Chapter 10: Infinite Sequences and Series

Solution We determine the convergence of ~:::.t(lln2) by comparing it with 00 (l/x 2 ) dx. To carry out the comparison, we think of the terms of the series as values of the function f(x) = I/x 2 and interpret these values as the areas of rectangles under the curve y = I/x 2 •

y

.!t

(1,f(I))

Graph off(x) ~ ~ I

As Figure 10.10 shows,

(2,f(2)

i'

sn

I I I I =-+-+-+ ... +12 22 32 n2 =

< f(l) +

---J---l=-t-L"t:"........;=~~_x

o

1

+ f(2) + f(3) + ... + f(n)

f(l)

n-ln ...

234

I Caution The series and integral need not have lbe same value in the convergent case. As we noted in Example 2, ~:;;" (I/n') = ".'/6 while 1,00(I/x') We = I.

Rectangle areas sum to less than area under graph.

roo

< I + FIGURE 10.10 The sum oflbe areas oflbe rectangles under !he graph of f(x) ~ I/x' is less than !he area under lbe graph (Example 2).

f" J,dx 11 x

J,dx 11 x < I + I = 2.

As in Section 8.7, Example 3, 1,00(I/x2 ) dx ~ 1.

Thus the partial sums of ~~I (1In 2) are bounded from above (by 2) and the series con_ verges. The sum of the series is known to be ""16 '" 1.64493.

THEOREM 9-Tho Integral Tost Let {an} 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 ~~Nan and the integral f(x) dx both converge or both diverge.

I:

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 rectangles in Figure lO.11a, which have areas al, a2, ... , a", collectively enclose more area than that under the curve y = f(x) from x = I tox = n + I. That is,

y

a, a, 2

0

y

~

1

f(x)

n 1

+ f(x) dx ,,; al

n

3

a" n

+I

+ a2 + ... + an'

x

(a)

In Figure 1O.llb the rectangles have been faced to the left instead ofto the right. Ifwe momentarily disregard the first rectangle of area ai, we see that

y

a2

a,

--::-r--"-----'-_"---_---"-'a".L-_~X

o

2

1 n

f(x)dx.

If we include ai, we have

a,

a,

+ a3 + ... + an";

3

al

n-l n

(h)

Subject to the conditions of the Integral Test, lbe series ~:,a. and 00 the integral (x) We both converge or both diverge.

+ a2 + ... + a,,";

a1

+

1 n

f(x)dx.

Combining these results gives

FIGURE 10.11

f"+1 f(x) dx ,,; al + a2 + ... + a" ,,; a1 + 11f" f(x) dx.

11

11

These inequalities hold for each n, and continue to hold as n --->

00 •

11

If f(x) dx is [mite, the right-hand inequality shows that ~a" is [mite. If .!too f(x) dx is infmite, the left-hand inequality shows that ~an is infmite. Hence the series 00

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, then f(x) = l/xP is a positive decreasing function ofx. Since

1

00

1

-1 dx = xP

1

00

x-P dx = lim

1

converges if p

> I, diverges if p

:5 I.

[ ~x

-p+l ]b ~

-P + 1

b_oo

=

_1_ lim ( _ I1 - I)

=

_1_(0 _ I)

b~oo

1- P

1

bP=

I-p

I ~ 1 I The p-series 1I~1 nP

I.

:5

_1_

p-I'

bP - 1 --+ 00 as b --+ 00 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

foo ~dx =

11

xP

_1_ lim (b1-p - I) 1 - Pb_OO

= 00.

The series diverges by the Integral Test. If p = I, we have the (divergent) hannonic series

1 1 ... +-+ 1 ... 1 +-+-+ 2 3 n We have convergence for p

>

1 but divergence for all other values ofp.

•

The p-series with p = 1 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.) The series ~~~l (1/(n 2 Integral Test. The function f(x) = 1/(x 2 x ~ 1,and

EXAMPLE 4

00

1 1

+ +

I)) is not ap-series, but it converges by the I) is positive, continuous, and decreasing for

b~oo

_1_ dx = lim arctan x b +I [ ],

x2

=

lim [arctan b - arctan I]

b~oo

1T - 1T -

2

4

1T

-

4·

Again we emphasize that 1T/4 is not the sum ofthe 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 -

Sn

= 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

+

cz,,+3

+ ... ;;,:

f(x) fix.

• +1

Similarly, from Figure 10.11b, we fmd an upper bound with

1

00

R.

= cz,,+1

+

a.+2

+

cz,,+3

+ ... ,,;;

f(x) fix.

These comparisons prove the following result giving bounds on the size ofthe 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 ~cz" converges to 8. Then the remainder R. = 8 - s. satisfies the inequalities

1

00

1

00

f(x) fix ,,;; R. ,,;;

n+l

f(x) fix.

(1)

n

Ifwe add the partial sum s. to each side of the inequalities in (I), we get

1

00

s.

+

1

00

f(x) fix ,,;; 8 ,,;; s.

+

n+l

f(x) fix

(2)

n

since s. + R. = 8. 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 ofthe interval containing 8, as given by (2).

EXAMPLE 5 n = 10.

Estimate the sum of the series ~(I/n2) 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

1<8< + IT - -

SIO

+1 10·

+ (1/4) + (1/9) + (1/16) + ... + (1/100)

'" 1.54977, these last in-

1.64068 ,,;; 8,,;; 1.64997. If we approximate the sum 8 by the midpoint of this interval, we find that 00

I

~ 2: '" 1.6453.

n=l n

The error in this approximation is less than half the length of the interval, so the error is less than 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 converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied 00 I

1. ~2 11""1 n

I 3. ~-211""1 n + 4 00

4.

00

In(n+I)=

6. ~--2 n(1n n) 2 00 In (n )

.-2

11=1 00

7.~+IFI n + 4

11=2

2

10.

e"j3

~

2

11-2n

Determining Convergence or Divergence Which of the series in Exercises 11-40 cooverge, and which diverge?

Give reasons for your answers. (When you check an answer, rememher that there may he more than one way to determine the selies' convergence or divergence.) 00 00 00 I 12. ~e~ 11. ~ 10" 00 5 14. ~--I IFI n + 00

17.

~ 00

20.

00

15.

13'~n~1

I - 8'

00

Inn

21.

I

26.

~ y!;,(y!;, + I)

29.

~ (ln2)'

00

27.

~

11=2

~

(I/n)

.-3

(lnn)Yln2 n - I

00 I 33. "5".nsinJi

:.:'I

+

1 1

'1 :xtit = I + Inn.

of the hsnnonic selies 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 oniverse 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? 44. Are there any values of x for which ~:::'I(i/(nx)) converges? Give reasons for your answer. 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 h" < On for every n? Is there a "smallest" divergent series of 46. (Continuation ofExercise 45.) Is there a "largesf' convergent se-

~ ~n

ries of positive numbers? Explain. 47. ~:_,( 1/( y';+l)) diverge.

11=2

a. Use the accompanying graph to show that the partia1 sum

2'

Sso = 00

25.

n y!;,ln

2'

.-1

I)'

00

~

(ln3)'

32.

~

I

._1 n(1 + 1n n)

34.

Y

I

y!;+J

tit

Conclude that 11.5

< Sso < {so. ~ tit. Jo V x + I

< Sso <

12.3.

y

1 2

1

I

00

00

1

I

30.

:.:'I

~;~I (1/( v;;-+l)) satisfies

1"

~~I n

II-I

00 ( 28.~I+n

n

•

~_e_ /I-I 1 e ln

I I I :xtitSI+Z+"'+n

positive numbers? Give reasons for your answers.

ny!;,

~3'

I

00

19.

00

n

00

35.

-8

I 24'~-2 -I /I-I n

-2 23. ~~I

31. £J

• -1

11=1

00

00

16.~~

18. ~n

~y!;, 11-0

3

~ y!;,

1

·+1

1

o b. There is absolutely no empirical evidence for the divergence

n - 4 -2n+l

11=1

2a)

0 0(1 42. ~ -11-3 n - 1 n +1

S I +

8. ~-n-

9. ~-"-11-1

I

00

5. ~ e-2•

00

+4

n

I)

OO ( a- - - 41. ~ /I_I n + 2 n +4

43. a. Draw illustrations lilre those in Figures 10.7 and 10.8 to show that the partial sums of the hsnnonic selies satisfY the inequa1ities

~_I-

11""1

Theory and Examples For what values of a, if any, do the selies in Exercises 41 and 42 converge?

nlan

f(x)~.~

n

vx+ 1

2

I ... I I I --+--'---'---LL-L---'---'---LL-~x

36'~-1+ • II-I e

o

1 2 3 4 5 ···48495051

00

38. 00

39. ~ sechn 11=1

~+II-I n + 1

b. What should n he in order that the partial sum

00

40. ~ sech2 n 11=1

S.

~ ~~_I ( 1/( v'i+l)) satisfY s. > 1000?

558

Chapter 10: Infinite Sequences and Series

48. ~:= 1 (1/,,4) eonverges

a. Use ihe accompanying graph to determine ihe error if S30 ~ ~~~I (1/,,4) is used to estimate ihe value of ~:::'I (1/,,4). y

f(x) ~

00 1 e. ~--,

1 d. ~--, nln(n ) ,-2 n(lnn) 57. Euler's eonstant Graphs like those in Figure 10.11 suggest ihat as n increases there is little change in the difference between the sum (Xl

,-2

I I 1+-+···+-

1, x

Z

C I

29

30

31

n

and the integral 32

... 33

x

Inn

["I dx x .

= J,

To explore this idea, carry out ihe followiog steps. b. Find" so ihat ihe partial sum s, = ~~_I (l/i 4) estimates ihe value of~~, (1/,,4) wiih an error of at most 0.000001.

a. Bytakingj(x)

51. How many terms of the convergent series ~:_I (1/,,1.1) should be used to estimate its value with error at most 0.00001?

54. Use ihe Cauchy condensation test from Exercise 53 to show ihat 00 I a. ~ - - diverges; n=2 nlnn b.

~.~ converges if p

::1.

>

I and diverges if p '" I.

n-

an

converges if and ou1y if p

>

f-I-? ,-2 Give reasons for your answer. 56. (Co"tinuatio" ofExercise 55.) Use ihe 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

10.4

"2I + ... + IiI - Inn

is bounded from below and from above.

b. Showihat I I n +

1'+1 x

<,

1

dx = In (n

b.

+ 1) -

Inn,

and use this result to show ihat the sequence {a,} in part (a)

is decreasing. Since a decreasing sequence that is bounded from below converges, ihe numbers a, def'med in part (a) converge: Inn ---. 'Y.

The number 'Y, whose value is 0.5772 ... ,is calledEuler~ constant. 58. Use the Integral Test to show ihat the series

converges.

59. a. For ihe series ~(I/n'), use the inequalities in Equation (Z) with n = 10 to f'md an interval containing the sum S.

,,(In,,)P .

~n(lnn)

1+

I.

b. What implications does the fact in part (a) have for ihe convergence of the series

a.

=

n-

(p a positive constant)

Inn '" I.

Thus, the sequence

I I I + 2 + ... +

/,00 x(:x)P

I + Inn

I I 0< In (n + I) - Inn '" I + 2 + ... +

55. Logarithmie p-series

a. Show ihat ihe improper integral

I

"'+n'"

or

52. How many tenns of the convergent series ~~4(1/,,(In,,)') should be used to estimate its value with error at most 0.01? 53. The Cauchy eondensation test The Cauchy condensation test says: l.ot {a,} be a nonincreasing sequence (a, '" a,+1 for all II) of positive terms that converges to O. Then ~an converges if and ou1y if ~Z'a2" converges. For example, ~(I/,,) diverges because ~Z" (I/Z') = ~I diverges. Show why ihe test works.

l/xintheproofofTheorem9,showihat

I In(n + I) '" I + 2+

49. Estimatethevalueof~:~l (1/,,') to within 0.01 ofits exact value. 50. Estimate the value of ~~2 (1/(,,2 + 4)) to within 0.1 of its exact value.

~

b. Ail in Example 5, use ihe midpoint of the interval found in part (a) to approximate ihe sum of the series. What is !he

maximum error for your approximation? 60. Repeat Exercise 59 using ihe series ~(I/n4).

~n(lnn)l.01

Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence ofmany more series by comparing their tenns to those of a series whose convergence is known.

10.4 Comparison Tests

Let ~a.. ~c.. and nonnegative terms. Suppose that for some integer N

THEOREM 10-The Comparison Test

~dn

559

be series with

n >N.

for all (a) If ~c" converges, then ~a" also converges.

(b) If

y

~dn

diverges, then

~a"

also diverges.

Proof In Part (a), the partial sums of ~a" are bounded above by 00

M

= at

+

a2

+ ... +

aN

+

:L n=N+l

Cn ·

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 ~an are not bounded from above. Ifthey were, the partial sums for ~dn would be bounded by FIGURE 10.12 If tire total area ~c" ofthe taller en rectangles is f'mite, then so is the total area ~all of the shorter a" rectangles.

00

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.

(a) The series 00

~ 5n

5 - I

diverges because its nth term

5

5n - I

=_1_>1 I n n-

S

HisTORICAL BIOGRAPHY

Albert of Saxony (Co. 1316-1390)

is greater than the nth term of the divergent harmonic series. (b) The series

~l=

I + l + l + l + ... I! 2! 3! converges because its terms are all positive and less than or equal to the corresponding terms of 00 I I I 1+~n=I+I+-+2+"" n~O 2 2 2 The geometric series on the left converges and we have "~O n!

00

I +

I

~ 2n = 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. (C) The series

5+~+1+1+ I + I + I + ... + I + ... 37 2+V1 4+V2 8+V3 2"+Y!;; converges. To see this, we ignore the first three terms and compare the remaining terms with those ofthe convergent geometric series ~:;'~o(l/2n). The term I/{ 2n + V;;) 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 lenD by term we have the comparison

1+

1 + I + I + ... ,,;1+1+1+1+ .... 2+V1 4+0 8+\13 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 tin is a rational function of n.

THEOREM l1-Llmtt Comparison Test all n

2:

Suppose that a,

>

0 and b,

>

0 for

N (N an integer).

1. If lim ban = c n

11_00

>

0, then

2. If lim ban = 0 and 11- 00

11

3. If lim ban =

00

n

11_00

~a, and ~b, hoth converge or hoth diverge.

~b, converges, then ~a, converges.

and ~b, diverges, then

~tIn diverges.

Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 55a and b. Since c/2 > 0, there exists an integer N such that for all n n

Thus, for n

>

>

~ I~ -cl

N

<

Limit defmition with

I.

E = c/2,L = c,and .. replaced by a,fb,

N,

c

an

C

--<--c<2 b, 2' £
If ~b, converges, then ~(3c/2)b, converges and ~tIn converges by the Direct Comparison Test. If ~b, diverges, then ~(c/2)b, diverges and ~tIn diverges by the Direct Comparison Test. _

EXAMPLE 2 (a)

Which of the following series converge, and which diverge?

1+~+l + 4

9

16

(b) 1 + 1 + 1 +

1

3

7

00

00

~ + ... = ~ 2n + 1 = ~ 2n + 1 25 ,~l (n + 1)' ,~l n2 + 2n + 1

~ + ... = 15

5: _12' - 1

,~l

(c) 1 + 2m 2 + 1 + 3m 3 + 1 + 4m 4 + ... =

9

14

21

5: 1n+ +n m5n

,~2

2

10.4

Solution

Comparison Tests

561

We apply the Limit Comparison Test to each series.

(8) Let a" = (Zn + I )/(n 2 + Zn + I). For large n, we expect a" to behave like Znln 2 = Zin since the leading terms dominate for large n, so we let b, = lin. Since 00 00 I ~b, = ~ n diverges

,,-I

n-l

and 2

Zn + n 11-+00 n + 2n + 1

lim a" = lim 11-+00

bn

=

2

Z,

~a, diverges by Part I of the Limit Comparison Test. We could just as well have taken b, = Zln, but lin is simpler.

(b) Let a, = I/(Z' - I). For large n, we expect a, to behave like liZ', so we let b, = liZ'. Since 00 00 I ~bn = ~ 211 converges and lim a, = lim ~ 11 bn 11_00 2 1

11_00

=

lim

,~OO

I I - (liZ')

= 1, ~a,

converges by Part I of the Limit Comparison Test.

(c) Leta" = (I + nlnn)/(n 2 + 5). For large n, we expect a, to behave like (nlnn)ln 2 = (lnn)ln,whichisgreaterthanl/nforn;;" 3,soweletb, = lin. Since 00

00

n=2

11=2

I

~b, = ~

n diverges

and

= 00, ~a, diverges

EXAMPLE 3

by Part 3 of the Limit Comparison Test. 00

Inn

-

Does ~ 3/2 converge? 11=1

n

Solution 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 convergentp-series. To get the p-series, we see that l 4

Inn / I
n

n

/

for n sufficiently large. Then taking a,

/

n'/4

= (In n)ln 3/ 2

and b, = I/n s/., we have

lim a, = lim Inn 11-+00

bn

11-+00

=

=

lim

n1/ 4 lin

,~OO (1/4)n 3/4

lim

;/4

I'HOpiIlll's Rule

= O.

n ~(1/n'/4) is ap-series withp > I, it converges, so ~a, converges by PartZ

Since ~b, = ofthe Limit Comparison Test.

11-+ 00

_

562

Chapter 10: Infinite Sequences and Series

Exercises 10.4 Comparison Test 10 Exercises 1--1l, use the Comparison Test to determine if each series converges or diverges. 00 I ~ n-I 2. £--.-1. ~ ~,.--o--::-::­ II-I n + 30 II-I n + 2

3.}: I ,~, v;. 00

11=2

7.

}:~n/

Umit Comparison Test 10 Exercises ~ 16, use the Limit Comparison Test to determine if each series converges or diverges.

}: 9.

11-1

n3

11=1

n-2 n2 + 3

53

n, ++ I2

.-1

5'

(2n 3)'

00 14 ~. -+ ',;':1 5n + 4

45. ~

11=1 00

48 ")'. sec • ~

n

-I

11=1

n

00

n

I

~ 00

n

"(V,;

52. ~-, 11=1 n

n=lnvn

'~1+2+3+"'+n

~_ co~n

49.

nl.3

1 51. ~ ••,

~n

~

46. ~ tan n

n

SID

54}: 1 + 2

2

• 11=1

1

+ 32 + ... + n 2

integer). If and ~a. converges, can anything be said about ~ b. ? Give reasons for yoor answer.

58. Prove that if ~all is a convergent series of nonnegative tenns, then ~a.'converges. 59. Suppose that an 60. Suppose that a" verges.

Determining Convergence or Divergence Which of the series in Exercises 17-54 converge, and which diverge? Use any method, and give reasons for yoor answers. ~ 1 £- . , ." 11=1 2 V n + Vn

20. }: I + ~osn 11=1 n

~ 18. £-

11=1

3 • ,

n+

V

n

21.}:~ 11=1

23}:

·.-1 n(n +IOnI)(n+ I+ 2)

3n - 1

n ). 26. ~ I 3n + I £ .-1.V ~ n' + 2

00

31~. I ',;':1I+Inn

-f"n + 1

00

29.

,&; 00

32.~

.,

~sin2n 19. £- - 2 ' IFI

22.}: n + I 11=1 nZV";,

24.}: 5n' - 3n ._, n'(n - 2)(n' + 5)

~(

(Inn)' 28. ,&1----;;>

> 0 and b. > 0 for n '" N (N an

lim.~oo(a./b.) ~ 00

(Hint: Limit Comparison with ~::"~I (l/n'»

II-I

42. j l n ( n : 1)

57. Suppose that a"

11=1

~

2"+3"

56. If ~:l an is a convergent series of nonnegative numbers, can anything be said about ~';,(aJn)? Explain.

16. }: In (I + 1,) n

34.

00

40'~3'+4'

Theory and Examples 55. Prove (a) Part 2 and (h) Part 3 of the Limit Comparison Thst.

(Hint: Limit Comparison with ~::"_, (l/n»

00

~ + I .-.L

3"-1 + 1 3'

n

00 13 ~.-• ,;':1 v;. 4' 00 I 15. ~·-In ~ n

• .~, -

~

11=1

n(n

25

38. Y ,;':1

1

44. ~ (2)1 11-1 n + . 00 tan-I n 47. ~. -'-,1-

SO. }:

(Hint: Limit Comparison with ~:::, (1/v;.)) 00 + I) 00 2' 11. ~ , 12. ~-3 .-, (n + I)(n - I) + 4'

17.

(Xl

+

n + 3n 5n oo2"_n

IFI

-

(Hint: Limit Comparison with ~::"~I (I/n'» 10. }: ~

39. }:

II

(Hint: First show that (I/n!) '" (I/n(n - 1» for n '" 2.) 00 (n I)! 00 • 1 00 1

8.}:v;.+1 .~IW+3

4 n +4

11=1

n - n

00 I 6. ~-3' 11=1 n

,

5.~co:n 11=1 n /2

~31l

41. ~-2' 11-1 n 00 I 43. ~., ~ n.

4.}:~+2

I

00

37.~.

I v;. In n In (n + I)

n+I

27.

£- In (In n) ._, 00 (Inn)' 30.,&1 n'/2

~ 1 -.n

,.-1

n2

00

I

00

2'

33'~n~

00

35.

~_I-

36.

~ n ;- •

II_I

n 2

>

0 and lim an

= 00.

.~oo

> 0 and lim n'a" .~oo

Prove that ~an diverges.

= O. Prove that ~a. con-

61. Show that ~::,,~, «In n)q/nP) converges for p>l.

-00


00

and

(Hint: Limit Comparison with ~:;;" I/n' for 1 < r < p.) 62. (Continuation of Exercise 61.) Show that ~';, (lnnY'/n P ) diverges for -00 < q < 00 and 0 < P '" I.

(Hint: Limit Comparison with an appropriate p-series.) 10 Exercises 63-68, use the results of Exercises 61 and 62 to determine if each series converges or diverges. 00

63.

(In n)'

~-.-

/1-2 00

65. ~

11=2

n (In n)'oo, n

1.001

67}: I •• _, nl.l(lnn)'

.-, ~m,.n

64. }:

00 (In n)'/5 66. ~----0:99

11=2 00

68.

n

I

,&;~

10.5

The Ratio and Root Tests

563

70. a. Use Theorem 8 to show that

COMPUTER EXPLORATIONS 69. It is not yet known whether the series

s=f

.-1

converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps.

where S ~ ~:I (l/n

s,~

1

11-1

n3 sin2 n

L

+f(~.-1 n2 n(n 1+ I) )

2),the sum ofa convergentp-series.

b. From Example 5, Section 10.2, show that

a. Derme the sequence of partial sums ,

1 n(n + 1)

1

00

S~I+L

• ~I

.

s,

What happens when you try to fmd the limit of as k --> Does your CAS fmd a closed form answer for this limit?

2

n (n

+ 1)

•

e. Explain why taking the fITst M terms in the series in part (b) gives a better approximation to S than taking the fIrst M terms in the otigina1 series ~::'-1 (I/n 2 ).

00 ?

b. Plot the fITst 100 points (I<; s,) for the sequence of partial soms. Do they appear to converge? What would you estimate the limit to be?

d. The exact value ofSis known to be ".2/6. Which ofthe sums 1000000 1

L --, .~I n

e. Next plot the fITst 200 points (I<; s,). Discuss the behavior in your own words.

1000

or

1

+

1

L --c2~'----c .~I n (n + 1)

gives a better approximation to S?

d. Plot the fITst 400 points (I<; s,). What happens when k = 355? Calculate the nomber 355/113. Explain from your calculation what bappened 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 rate of growth
THEOREM

12-The Ratto Test suppose that

Let ~a. be a series with positive terms and

. a,,+l_ I lTD n. - p.

11_00

-n

Then
>

I or p is in-

Proof
a,,+l/a. must lie within e of p when n is large enough, say for alln ;" N.lo particular, a",+l

a;;- < P + e

=

r,

when n ;" N.

564

Chapter 10: Infinite Sequences and Series

1bat is,

< < <

aN+l aN+2 aN+3

raN, raN+ 1 raN+2

< <

rlaN' 3 r aN'

These inequalities show that the tenns of our series, after the Mh tenn, approach zero more mpidly than the terms in a geometric series with mtio r < I. More precisely, consider the series ~cn, where en = an for n = 1,2, ... , N and CN+l = raN, CN+2 = rlaN' ... ,CN+m = rmaN, ... . Now an :5 en for all n, and 00

~c. = al

n=l

= al

The geometric series

+ a2 + ... + aN-l + aN + raN + r 2aN + ... + a2 + ... + aN-l + aN(I + r + r 2 + ...). I + r + r 2 + ... converges because Ir I < I,

so ~c. con-

verges. Since an :5 en, ~an also converges. (b) 1 < P :S oc. From some index M on, and The terms of the series do not approach zero as n becomes infinite, and the series diverges by the nth-Term Test. (0) p = 1. The two series 00

I

~" .-1

and

show that some other test for convergence must be used when p = I. ~ I For ~ ,,:

a.+l a.

I/{n + I) n I/n = n + I

~ I For ~ n2 :

a.+l a.

I/{n + 1)2 = (_n_)2 --> 2 = l/n 2 n + I I I.

--> I.

In both cases, p = I, yet the first series 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 ofthe following series.

~. Z· 3~ 5

~

Solution

(b)

.~=. {nZ,.nn)'" ~

We apply the Ratio Test to each series.

(a) For the series ~~o {Z·

a.+l a.

(O)~. 4'nln/ ~ (Zn).

+ 5)/3',

{Z·+1 + 5)/3.+ {Z· + 5)/3'

1

=

1. Z·+1 + 5 3 Z· + 5

=

1. (Z + 5'Z--') -->1. l 3 I + 5' Z--. 3 I

=

l 3.

The series converges because p = Z/3 is less than I. This does not mean that Z/3 is the sum of the series. In fact, 00 Z. + 5 ~ -..=0 3

=

(z)' + .=03n5

~ .=0 3 00

~ 00

=

I I - (Z/3)

5

+ ------"'--c--oI - (1/3)

ZI

Z·

10.5 The Ratio and Root Tests

565

(2n)! (2n + 2)! (b) Ifa, = -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) Ifa, = 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 ---> 1.

Because the limit is p = I, we cannot decide from the Ratio Test whether the series converges. When we notice that a.+l/a, = (2n + 2)/(2n + I), we conclude that a.+l is always greater than a. because (2n + 2)/(2n + I) is always greater than 1. Therefore, all tenns are greater than or equal to a, = 2, and the nth tenn 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 tenns

n/2" a, = { 1/2',

nodd

n even.

To investigate convergence we write out several terms ofthe series:

fa=l+l+~+l+~+l+2+ ... 2 3 4 5 6 7

'~1'

2

2 2 ' 113

2

2

15

2

2

17

= 2 + 4 + 8 + 16 + 32 + 64 + 128 + ... Clearly, this is not a geometric series. The nth tenn 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 odd

n+ I 2 '

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

Let

~a,

be a series with

a. ;", 0 for n ;", N,

and suppose that

Then (8) the series converges if p < I, (b) the series diverges if p finite, (0) the test is inconclusive if p = I.

>

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 -xYa: ---> p, the terms -xYa: eventually get closer than e to p. In other words, there exists an index M 2: N such that -xYa:
whenn2:M.

Then it is also true that a"

< (p + e)'

forn 2: 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

,,-1

n-M

converges. (b) 1 < P :S 0Cl. For all indices beyond some integer M, we have -xYa: > 1, so that a. > I for n > M. The terms of the series do not converge to zero. The series diverges by the nth-Term Test. (0) p = 1. The series ~:1 (lin) and ~:;C~1 (1/n 2) show that the test is not conclusive when p = I. The first series diverges and the second converges, but in both cases -xYa: ---> 1. _

/2" {n1I' 2 ,

. the senes · · WI t h tenus a. = Consl'der agam

EXAMPLE 2

nodd 11 even.

Does ~a. converge?

Solution We apply the Root Test, finding that

/2, {'fin1/2,

%: = a.

nodd n even.

Therefore,

t ,;

~.

-xYa: ,;

I

Since 'fin ---> (Section 10.1, Theorem 5), we have Iim"~",, -xYa: = 1/2 by the Sandwich Theorem. The limit is less than 1, so the series converges by the Root Test. _

EXAMPLE 3 ""

(8)

Solution (8) (b)

(0)

2

~ ~.

Which of the following series converge, and which diverge? (b)

~ 2:

n=-1 11

(0)

"" (

~

I ). 1+ n

We apply the Root Ths! to each series.

"" n 2 1f.n2 W L 2' converges because 2' = •• r.:;. n=1 v~

2' . ~2' L"" --, diverges because --, = 11

,,=111

=

('finY

--2- --->

1.

2 2 r), ---> --, > 1. n 1

( •• V

L""(I)' -+1 converges because ~(I)' -+1 = 11 n

,,=1

12

2 <

I

-+1 ---> 0 11

< 1.

-

10.5 The Ratio and Root Tests

567

Exercises 10.5 Using the Ratio Test

00

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. 4

4

~

~n~2

II_I

3

00

21l +1

37. 39.

~­

41

311 +2

6·L11=2 Inn

n2 (n n!

+ 2)!

8 ~ • ,~I

32/1

(20

n5' 3) In (n

+

L

11. ~ (4n + 3)' 3n - 5

13 ~ • ~ (3

15.

L

00

,-I

(

+

8 (l/n))2o

12. 14.

i i

(In(e sin'

+

I)

2

+

k) r'

(~)

I)'"

,~OO

= e")

I

00

~n

1+,

Determining Convergence or Divergence

00

19.

21.

23.

20. ~ InOIl

I.

&1 ~O' 00

~

2

22.

+ (-I)'

24.

1.25'

3)'

,-I

31.

OO(

28.

n

(I I)

~. ~n ~ (n

/I-I

+

/I-I 00

(I I)'

~_ n~n

.-=1

l)!

~n

L

11=1

L n . ,-2 (In n )('/2)

40

(Inn )' n lin n. n n(n + 2)!

3' L--,;; n2 00

42.

11-1

~ (20

(n!? (20)'•

45. al = 2,

44.~

+ 3)(2' + 3) 3'+2

,-I

all +l =

46.

al =

1, °n+l

47.

a, =

3'

48.

al

49.

al =

2,

SO.

a, =

5,

51.

a,

52.

al =

I

= 3,

n

= I,

a"

n

3n - I 2n + 5 ""

n

= n

+ 1 an

I

+ Inn

n

+ Inn

0n+l =

I

2'

an

1 + tan-I n =

""+I = an+!

1 + sinn

Cln+1 =

n

n

+ 10

an

alJ

a, ~ t, ""+1 ~ ~ 54. a, t, ""+I (",,)'+1 =

=

Convergence or Divergence Which of the series io Exercises 55-62 converge, and which diverge? Give reasons for your answers. 00 (3n)! 00 2' I I 55 L~ 56. n!(n + I)!(n + 2)! • ,~I (2o)!

&1

00

57.

(n!)'

00

:.:1 00

,

60.

1.3 .... '(20 - l) 4lJ 2lJn!

62.

~ [2 -4-

1.3

(20-1)

(2n)](3'

+ I)

(n!)'

LlJ-l n(lJ1) 00

~ 00

58.

n( ') 2 lJ

61.

+ 2)

00

L ( ')2 lJ-l n

59. Y

I

n.

,

Recursively Def"med Terms Which of the series ~;l a" defmed by the formulas in Exercises 45-54 converge, and which diverge? Give reasons for your answers.

(Inn)'

30. ~ Ii - n 2

32. I)(n

I)'

3n

L-,n 00

.-=1

33. ~

L-

,-I /1""1

29. ~ li- n 2 00

(-2)' , 3

26·LI--

~ In;

/I-I

(n; 2)' ,-I ~ 00

00 ( 25·LI-"

27.

00

53.

I

00

~n!e-II 00

L L

11=2 00

In Exercises 17-44, use any method tc determine if tire series con-

verges or diverges. Give reasons for your answer. 00 Vi 00 17. ~ nZII 18. ~n2e-

~

+ I)!

3'n.,

38. "5' n;

1-"

(Hint: lim (I + x/n)' 16. "5'.

,

n2'(n

11'-

00

. ,_I

43.

In Exercises 9-16, use the Roo! Ths! tc detennine if each series converges or diverges. 00 7 00 4' 10. (3 )" 9. /I-I (2n + 5 )' /1""1 n

L

oo

36.

3!n!3 11

~ (20 :

00

Using the Root Test

/I-I

+ 3)!

(n

00

. ~n3"-1 00

~:II 00

~ 00

L, 00

35.

,

~ (;,)'

568

Chapter 10: Infinite Sequences and Series

Theory and Examples 63. Neither the Ratio Test nor the Root Thst helps with p-senes. Try

65. Let a, = {

them on

Does I ~n=l nP 00

~an

n/ 2'

1/2':

ifn is a prime numher otherwise.

converge? Give reasons for your answer.

66. Show that ~:_I 2(i'J/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

-2

+

1- 2

1

1

1

2 + 3 - 4 + 5 - ... +

I -

1-

1

1

1

2+4- 8+

+3- 4+5- 6+

(_1)'+1 n

+ ...

(1)

(-1)'4

+ ~ +...

(2)

+ (-1)'+ In + ...

(3)

We see from these examples that the nth term of an alternating series is of the fonn
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 ofthe alternating harmonic 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 Serie. Test (Lotbniz'. Test)

The series

00

~(_1)'+IU, = Ul - U2

+ U3

-

U4

+ ...

n=l

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, some integer N.

;;"

U,+I for all n ;;" N, for

3. u,--+O. Proof Assume N = I. If n is an even integer, say n = 2m, then the sum of the f'mt n terms is S2m = (Ul -

= Ul -

U2) (U2 -

+

(U3 -

U3) -

U4)

+ ... + (U2m-1

(U4 -

us) - ... -

-

U2m)

(U2m-2 -

U2m-l) -

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)n+11 = I n~l n

_1 + 1_1 + ... 2

3

4

clearly satisfies the tbree requirements of Theorem 14 with N = I; it therefore converges. _ Rather tban directly verifying the defInition Un ;;,: Un +1o a second way to show that the sequence {un} is nonincreasing is to define a differentiable function f{x) satisfying f{n) = Un' That is, the values of f 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 f{x) is nonincreasing for x ;;,: N. It follows that f{n) ;;,: f{n + I), or Un ;;,: Un + 10 for n ;;,: N.

EXAMPLE 2 Consider the sequence where Un = IOn/{n 2 + 16). Detme f{x) 2 IOx/{x + 16). Then from the Derivative Quotient Rule,

f '() x

=

2

=

1O{16 - x ) <: 0 (x 2 + 16)' -

whenever x

~

4.

It follows that Un ;;,: Un +1 for n ;;,: 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 ofTheorem 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 tbree conditions of Theorem 14 are satisfied with N = I. Starting from the origin of the x-axis, we lay off the positive distance 51 = U1. To tmd the point corresponding to 52 = 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 Sn and sn+ 1 and hence differs from Sn by an amount less tban Un + 1. Because

IL -

Snl

<

Un +1

forn

~

N,

we can make useful estimates of the sums of convergent alternating series.

570

Chapter 10: Infinite Sequences and Series

THEOREM 15-Tbe Alternating Series Estlmation Tbeonom If the alternating series ~:I (-I)o+l uo satisfies the three conditions of Theorem 14, then for n ~N, So = UI -

U2

+ ... + (- 1)0+ I uo

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 n+ t, and the remainder, L - So, bas the same sign as the first unused term.

We leave the verification of the sign of the remainder for Exercise 61.

EXAMPLE 3

We try Theorem 15 on a series whose sum we know:

~(_I)ol0 = I _ 1 + 1_ 1 + l _ l + l __I_II + _1__ ...

o~O

2

2

4

8

16

32

64

128

256

.

The theorem says that ifwe 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 ofthe first nine terms is S9 = 0.66796875. The sum ofthe 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 ahsolutely (is absolutely convergent) if the corresponding series of absolute values, ~I aol, converges.

The geometric series in Example 3 converges absolutely because the corresponding series of absolute values

~l=I+1+1+1+ ...

o~ ~

2

4

8

converges. The alternating harmonic series does not converge absolutely because the corresponding series of absolute values is the (divergent) harmonic series.

DEFINmON 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, -la.1 ,;; a. ,;; la-I,

so

0,;;

a-

+ la.1 ,;; 2Ia.l·

If ~::'~lla.1 converges, then ~::'~12Ia.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 ~:;;'1 a- as the difference of two convergent series: 00

00

00

La. = L(a. +

n=1

1a-1-la.l) =

n=1

00

L(a.

L la.l· n=1

+ la.l) -

n=1

-

Therefore, ~:1 an converges.

Caution We can rephrase Theorem 16 to say tbat 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 00

(8) For

This example gives two series that converge absolutely.

L (-1)'+ .-1

1~ = 1 - 1 + 1 -

4

n

9

~

+ "', the corresponding series of absolute

16

values is the convergent series

5:l=1+1+1+~+ ... 2

• ~1

n

4

9

16

.

The original series converges because it converges absolutely.

(b) For

~sinn sin1 + sin2 sin3 . b 0 th posItive . . and nega-= -+ - + . . . which contains 2

~

n 1 4 9 ' tive tenDs, the corresponding series of absolute values is 1'1=1

I

5: sin n

.-1

n2

I=

Isin11 + Isin21 + "', 1 4

which converges by comparison with ~::'~1 (1/n 2 ) because Isinnl ,;; 1 for every n. The original 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

nP

1

1

1

=1-2P+Y-4P+"',

converges. If p > 1, the series converges absolutely. If 0 tionally. Conditional convergence: Absolute convergence:

< P ,;;

p>O

1, the series converges condi-

I I 1 -I -+ -- + .. ·

V2

V3

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 ofthe proof).

THEOREM 17-The Reammgement Theorem for Absolutely Convergent Series If ~:::I an converges absolutely, and bj, b2, ... , bn, ... is any arrangement of the sequence {an} , then ~bn converges absolutely and 00

00

Lbn = 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 ofthis 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 (_l)n+1/ 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~ ;?t

(_l)n+1 n

=2(I_l+l_l+l_l+l_l+l_~+~_...) 2

3

4

5

6

7

8

9

10

II

212121212 =2-1+---+---+---+---+--'" 3253749511

.

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 - I) -

! + (t -t) - t + (t -t) - i + (% - t) - i + ... (I -! + t - t + t - i + t - i + ~ - I~ + 1\ _ .. -) 00

L 11=1

(_l)n+1 n

= L.

So by rearranging the terms of the conditionally convergent series ~:::12( _l)n+1/ n, the series becomes ~:;"~1 (_I)n+ I/ 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: ~ l/n" converges if p > 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. ~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 In Exercises 1-14, detennine if the alternating series converges or diverges. Some of the aeries do not satislY the conditions of the Alternating Series Test.

1. 3.

~(_I)'+I_I-

,~1

V;;

~(-I)'+I~ IFI n3

2.

11-1

9.

13.

n,n ++ 45

~(_I)'+I I~n

12. i(-I)'In(1 +

~(_I)'+I

V;; + I n+ 1

14.

Inn

k)

V;;

+

I

Absolute and ConcUtional Convergence Which of the aeries in Exercises 15-48 converge absolutely, which converge, and which diverge? Give reasons for your answers. 00

15. L(_I),+I(O.I)' 17.

18.

V;;

00

n

00

23. L(-I),+l 25.

3 5

+n

+n

~(_I)'+II ~ n

11=1

(-I)' ,~I I + V;; 0 0 ,

19. L(-I)'+I+IFI n +1 00 1 21. ~(-I)' + 3

n-l

~

n

20. L(-I),+I 11=1

n;

00

24. L

s:,n

(-2)'+1 + 5'

11-1 n

26.

~(_I)'+I(\Ylo)

11=1

32. Y(-5t'

:=1

n (-100)'

00

34.

1

n.

L

(_1)'-1

~,c'-='---­ 11-1 n + 2n + 1 00

36. ,~_. CO~n1T ~ 00 (-I)'+I(n!)'

+ I)'

~

(2n)'

00

,(In)!

39.

~(-I) 2"nln

38.

~

40.

~(-I)' (In'+ 1)1

(In)l

00

00

00

11=1

11=1

(nl)'3'

00

+ vn -

43. L(-I)'(Yn /1=1

(-I)'

00

44.

n)

V;;) 00

L.r .~ vn+ vn+ 1

45. L(-I)' sechn

,-I

11=1 00

46. L(-I)'cschn

,-I

47.

I

I

I

4: - 6 + 8 -

48. I

I 10

I

I

+ IT - 14 + ...

1111111

+ 4: - 9 - 16 + 25 + 36 - 49 - 64 + ...

2

00.

22. ~(-I)'

Inn n - Inn

00

37.

n-I

~(-I),_I-

,~1

(0 I)'

00

16. L(-I)'+I---il

n-l

~(_I)'

41. L(-I)'(~ - V;;)42. L(-l)'(~-

~(_I)'+I 3~

:=1

L 11=1

nlnn

11-1

~ I

35. ~_ cos mr ~ nV;; 00 (-I)'(n

~(_I)'+I_I_

11-2

30.

n + 1

00

~(-I)' (n + I)!

10.

~

33.

10"

00

8.

00

,

11=1

10

:=1

:.:1

~(_I)'+I_I-

11=2

00

4. L ( - I ) ' - )(' 11=2 In n 6. L(-I),+I

28.

,-I ~(-I)' ~-1 n

31. Y(-I)'

4

~(_I)'+I (~)'

II-I

11.

~(-I)'+I+ n f2 00

7. ~(_I)'+I n2:

29.

L( -1)'n'(2/3)'

n-I

11=1 00

00

S'L(-I)'--,"IFI n + 1

00

27.

Error Estimation In Exercises 49-52, estimate the magnitude of the error involved in using the sum of the first four tenns to approximate the sum of the en-

tire series.

574

Chapter 10: Infinite Sequences and Series

51. f(-I),+1 (O.~I)"

N. you will see in Section 10.7, the sum is In(I.Ot).

,-I

52. I

~I~

i.(

is the same as the sum of the ftrSt n terms of the series I

I

I

I

I

1·2 + 2·3 + 3·4 + 4'5 + 5·6 + .... -1)'1',

0< I < 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? 63. Show that if~:1 a, diverges, then ~:;"_lla,1 diverges.

In Exercises 53-56, determine how many terms should be used to estimate the sum ofthe entire series with an error ofless than 0.001. 1 53. f(-l)' -2--

55.

54. f(_I),+1

n +3

11=1

IFI

~(_I)'+1 (n + ~y;;y

56.

64. Show that if~:1 a, converges absolutely, then

z"n +1

~(-I)' In(ln(~ + 2))

65. Show that if~:1 a" and ~:1 b, brrth converge shsolutely, then so do the following.

D Approximate the sums in Exercises 57 and 58 with an error of magnitude less than 5 X 10..... n .

11=0

58.

f (-I)' 1,

As you will see in Section 10.9,

the sum. is e-1 .

n.

n-O

D

Theory and Examples 59. a. The series

1_1 +

l _ l + l _ l + ... + l _ l + ... 32942783'2'

does not meet one of the conditions ofTheorem 14. Which

one? b. Use Theorem 17 to find the sum of the series in part (a).

D 60.

c. ~

As you will see in Section 10.9, the sum is cos 1, the cosine oflradian.

1

00

57. ~(-I)'(2 )'

The 1intit L of an alternating series that satisfies the conditions of Theorem 14 lies between the values of any two consecutive par-

tial sums. This suggests using the average Sn

+23,,+1 -_

Sn

+ 12 (-1)'+2an+!

ka" (k any number)

66. Show ~ example that ~~=1 a"bll may diverge even if ~~=1 an and ~,-1 b, brrth converge. 67. In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to -1/2. Start the new arrangement 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 than -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 larget at least three times and fInish at or below it. If 9" is the sum of the fIrst n terms of your new series, plot the points (n, s,) to illustrate how the sums are behaving. 68. OutHne of the proof of the Rearrangement Theorem (Theorem!?)

a. Let e be a positive real number, let L = ~~-1 an, and let St

= ~:=1 an· Show that for some index Nt and for some

index N2

O?::

Nt,

to estimate L. Compute

I

I

s20+2'21 as an approximation to the sum of the alternating harmonic series. The exact sum is ln2 = 0.69314718 ....

61. The sign of the remainder of an alternating series that satisfies the conditions of Theorem 14 Prove the assertion in Theorem IS that whenever an alternating series satisfying the conditions ofTheorem 14 is approximated with one ofits partial sums, then the remainder (sum of the unused terms) has the same sign as the first unused term. (Hinl: Group the remainder's terms in consecutive pairs.)

62. Show that the sum of the flfst 2n terms ofthe series 111111111

1-2+2-J+J-~+~-J+J-~+'"

Since all the terms aI, a2, ... , aN: appear somewhere in the sequence {b,}, there is an index N, '" N2 such that if n ~ N3 , then (:~k-l bk ) - SN: is at most a sum. of terms elm withm O?:: Nt. Therefore,ifn O?:: N3,

:5

~la.1

k-Nt

+ ISN, - LI <

E.

b. The argument in part (a) shows that if~::"_1 a, converges

absolutely then ~: I bn converges and ~: 1 bn = ~: 1 an·

Now show that because ~:lla,1 converges, ~:llb,1 converges to ~::"-ll 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

~cllxn

=

Co

+ CIX + C2X2 + ... + cnx + .. ". ll

(1)

11=0

A power series about x = a is a series ofthe form 00

~c.(x - a)' = Co

+ Cl(X

-

a) +

C2(X -

af + ... + c.(x - a)' + ...

(2)

11=0

in which the center a and the coefficients Co, Ch 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 1 in Equation (1) gives the geometric

power series 00

~x'= I +x+x 2 + .. ·+x·+ .. ·. 11=0

This is the geometric series with [rrst term I and ratio x. It converges to 1/(1 - x) for 1. We express this fact by writing

Ix I <

1 2 + .. ·+x • + ... --=I+x+x I-x '

-1<x
(3)

• I Reciprocal Power Series 1

--~ 1- X

00

LX', 11=0

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 ofthe 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 = 1, or -1, 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 = 1, where it has a vertical asymptote. The approximations do not apply when x 20 1.

576

Chapter 10: Infinite Sequences and Series

1

Y=l-x

5 4

Y2=1+x+~

3

~YI~I+X

2 I -I

Yo

=

1

x

0

FIGURE 10.14 The grnphs of f(x) = 1/(1 - x) in Example I and four of its polynomial approximatioos.

EXAMPLE 2

The power series

I -

(_i)"

1 (x -

2) + 1 (x - 2)' + ... + (x - 2)" 2 4 2

+ ...

(4)

matches Equation (2) with a = 2, Co = I, CI = -1/2, C2 = 1/4, ... , C" = (-1/2)". This is a geometric series with flfst 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 =

(x - 2) 2

I -

+

I)"

(x - 2)"

Series (4) generates useful polynomial approximations of f(x) 2

Po(x)

=

2

12 (x

2/x for values of x near 2:

- 2) = 2 - '"2

P2() x = I - -I( x - 2 ) 2

3

=

0< x < 4.

I

PI (x) = I -

o

+ ...,

+ -I(x 4

2 )2 = 3 - -3x

2

+ -x

2

4'

•

FIGURE 10.15 The graphs of f(x) ~ 2/x and its first three polynomial approxima-

and so on (Figure 10.15).

tioos (Example 2).

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

For what values of x do the following power series converge? n-l

(a) ~(-I) "~I

xn

x2

x3

n=x--+--'" 2 3

10.7

00

(b) L(-I)

.-1

(c)

00

577

x 2n - t X3 X5 --=X--+--'"

11-1

x" L = • ~O n!

Power Series

2n - 1

3

X2

X3

2!

3!

5

+ x + - + - + ...

1

00

(d) Ln!x' = 1

+ x + 2!x 2 + 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 = I, we get the alternating hannonic 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 hannonic series; it diverges. Series (a) converges for - 1 < x ,;; 1 and diverges elsewhere.

o

-1

x2n+l

(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)

>x

o

-1

IU~:'1 I(n + 1)!'x'n! I = X·+I

Ixl

n + 1--+0foreveryx.

.1

(. + I)!

The series converges absolutely for all x.

o

IU~:'1 I

(n

(d)

+

1)!x'+

nix'

l

l=

(n + 1)lxl--+oounlessx

The series diverges for all values of x except x =

•

o

>x

=

o.

o.

•

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 Serfes If the power series co La,x' = ao + a,x + a2x2 + ... converges at x = c oF 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 anc' converges. Then lim,~co anc' = 0 by the nth-Term Test. Hence, there is an integer N such that Ianc'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 ofEquation (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 < Ic I; divergence at x ~ d implies divergence forlxl > Idl. The corollary to Theorem 18 asserts the

existence of a radius ofconvergence R '" O.

Since Ix/c I < 1, it follows that the geometric series ~:o Ix/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 ~an(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 hut 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 coX" 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 cod" diverges. Let S be the set of values of x for which ~:o coX" 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 pcint 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)l" 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 (f"mite)


or

a - R

< x < a + R.

2.

Ifthe interval ofabsolute convergence is finite,

3.

Ifthe interval ofabsolute convergence is a - R < x < a + R, the series diverges for Ix - a I > R (it does not even converge conditionally) because the

interval of convergence.

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.

nth term does not approach zero for those values of x.

Operations on Power Series On the intersection of their intervals of convergence, two pcwer series can be added and subtracted term by term just like series of constants (Theorem 8). They can be multiplied just as we multiply pclynomials, 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

~~~oa.x· andB(x) = ~:;;'o b.x· converge absolutely forlxl

=

c. = aob.

If

< R, and

•

= ~akb.-.,

+ a,b.- , + a2b.-2 + ... + an-lb! + 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:

(~x').r~ (-l)'~) \,;~o \,;~o n + I = (I

+ x + x 2 + ...)(x _

~2

+ x; - .. -)

(x _ x; + x; _ .. -) + (x2 _ x; + by!

x

2

5x

~4

Multiplysecondseries ... _ .. -)

+ (x 3

~4 + x;

_

byx 3

x

by

_ .. -) + ...

x'

4

and gather the IllS! four powers.

=x+T+T-6····

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/O - x) = ~:;;'ox' converges absolutely for Ixl < I, it follows from Theorem 20 that 1/(1 - 4x 2 ) = ~~_o(4x2)'convergesabsolutelyforI4x21 < I 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 Theonm

radius of convergence R

>

If ~c.(x - a)' has

0, it defines a function

00

j(x)

= ~c.(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)

= ~nc.(x n=1

- a)'-!,

00

j"(x) = ~n(n - 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 j'(x) and j"(x) if

+ x + x 2 + x 3 + x 4 + ... + x' + ...

f(x) = _1_ = I I - x 00

=

~x',

<x<

-I

I

11=0

Solution

We differentiate the power series on the right term by term:

j'(x) =

I (I -

+ 2x + 3x 2 + 4x 3 + ... + nx·- 1 + ...

= I X)2

00

~nx'-"

j"(x) =

2 (I - x)3

<x<

-I

= n=1

=

2

I;

+ 6x + 12x 2 + ... + n(n - l)x·- 2 + ...

00

= ~n(n

- l)x·- 2 ,

<x<

-I

•

I

n=2

Caution Term-by-term differentiation might not work for other kinds of series. For example, the trigonometric series

~ sin (~!x) n

",-I

converges for allx. But if we differentiate term by term we get the series ~ n!cos (n!x) ~ n=1

n2

'

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) = ~c.(x - a)'

.-0

converges fora - R

< x < a + R (R > 0). Then 00

~Cll converges for a - R

J EXAMPLE 5

n+ 1

< x < a + R and f(x)

fora - R

(x _ a)n+l

ax =

00

~c.

(x - a)·+l n + I

+C

< x < a + R. Identify the function 00

f(x)=~

(-I)' x 2n +1 2n+1

x3 =x-

3

xS +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 + ... , -1<x<1. Thcororn21

This is a geometric series with f"rrst term 1 and ratio -

f'(X)=I_~_X2) We can now integrate f'(x)

!

=

1/(1

f'(X)

1"-4 = tan-II = n_02n+1 ~ (-I)'

f(x)

=

x -

x3

!~ 1

+ x2

so C

= 0,

x5

so

l~x2'

+ x 2) to get

ax =

The series for f(x) is zero when x

2 X ,

= O.

x7

3 + 5 - 7 + ... =

= tan-I

x + C.

Hence _

-1<x<1.

tan 1 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 - -1= 1 - 1 + 1 2 -I 3 + ... 1+1

converges on the open interval -I < 1 < I. Therefore, 1n(I+x)=

1,

2 3 4 1 +1 --+ 1 ' " ]' - -1 d l = I - 01+1 234 0

Thcororn 22

2 4 3 = x -x- +x- - -x+ · · · 234

or

I

00

00 (_1),-1

Iln2 ~ ~

,-I

In (I

n

+ x)

=

(_1),-1

L n=1

x'

n

'

-1<x<1.

It can also be shown that the series converges at x = 1 to the number In 2, but that was not

guaranteed by the theorem.

_

Exercises 10.7 Intervals of Convergence In Exercises 1-36,
00

1. ~x'

2. ~(x ,-0

,-0 00

3. ~(-I)'(4x

,-0 00

5."5'

:=6

(x - 2)'

10'

+ I)'

+ 5)'

00

(3x - 2)'

,-I

n

4. ~ 00

6. ~(2x)'

00

,

7. ~--"'---n-O 00

n+2

,-I

,

9. ~_x_ 11""1 nV";,3 n

00

10. ~ 12.

~ 4n~2n

14. ~

00

n=l

,

15 "5'. ---o~x==

. .t=6W+]

+ 2)'

(x - I)'

y;,

11-0 00

,-I

(-I)'(x n

f 311~n n.

(-I)"x' ,,-0 n., 00

11.~

13.

00

8. ~

(X,l

(x - I)' 3,

n 3 (_l)"x n+l

16. Y~.~,­ :=6 vn+3

10.7

00

17.

~ 00

+ 3)'

n(x

00

18 ~

. :?o

5'

'\,f;,;xn

,

nx

4'(n 2

2 )' 47.~ ~ 00

+ I)

48.

(

11=0

3

Power Series

583

2 )' ~ £-=-! 00

(

11=0

2

00

20. ~ %(2x

19. ~-3'

,-I

,-0

+ 5)'

Theory and Examples 49. For what values ofx does the series

00

21. ~ (2

+ (-I)')-{x + 1),-1

I I I)'(x - 3)" + ... 1- 2(x - 3) + 4(x - 3)' + ... + ( -2

11=1 00

22. "5'.

~

23.

(-I)' 32o(x - 2)' 3 n

~ (I + k)" x'

00

24. ~(lnn)x'

00

00

26. ~n!(x - 4)'

25. Lnllx ll

,-I 00

,-0

(_

1)'+1(x + 2)' n2'

27.

~

29.

~_x_

00

~ I/(n(ln n)') ftum Section 10.3,

,-2 n(lnn)' ,

~_x_

n=2

nlnn

50. 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! -

,-0

Get the information you need about

,

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?

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?

Get the infonnation you need about ~ I/(nln n) from Section 10.3,

b. By replacing x by 2x in the series for sin x, fmd a series that

Exercise 54. 00

32.

~

(J>; + 1),+1 2n+2

converges to sin 2x for all x. c. Using the result in part (a) and series multiplicatioo, calculate the first six tenns ofa series for 2 sin x cos x. Compare your answer with the answer in part (b). 52. The series x 2 x 3 x4 X S e' = I + x + 2! + 3! + 4! + 5! + converges to e~ for all x.

00

36. ~

(v;;-+l -

a. Find a series for (d/tb:)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.

J

y';;)(x - 3)'

11=1

1:.

In Exercises 37-4Q, find the series' radius ofconvergence.

~ n! II 37. ,-I ~ 3'6'9 ... 3nx

2.4'6 ... (2n) 38. L ,-I 2·5·8 .. ·(3n - I) 00 (n!)' 39. &12'(2n)! x' 00

(

53. The series

)2 x"

tanx

ofX'.

00

~

45. }:

42. ~ (e' - 4)'

11"'0

,-0

+ 1)20

(x - 1)20 4'

44. Y

(~ -

46. ~(lnx)'

2

00

:=IJ

(x

9'

00

I)'

+

2x'

15 +

17x' 62x' 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 fll'St 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 ofooovergence and, within this interval, the sum of the series as a function

41. ~ 3'x'

+

a. Find the fll'St five terms of the series for In 1sec x I. For what values ofx should the series converge?

(Hint: Apply the Root Test)

43.

=x

converges to tan x for -'IT/2 < x < 'IT/2.

40'~(n~Jx'

00

Replace x by -x in the series for eXto 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.

_

seox-I+

x2

2

5 4 61. 277 8 + 24 x + 720 x + 8064 x + ...

converges to sec x for -'IT/2 < x < 'IT/2. a. Find lbe 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), thea a" = 0 for every n.

b. Find the ftrst four terms of a series for sec x tan x. For what

values of x should the series converge? e. Check your result in part (b) by multiplying the senes for

56. The sum ofthe series ~::"-o (11'/2") To fmd the sum ofthis series, espress 1/(1 - x) as a georoetric 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 a. Shaw that if two power series ~~=oanxn and ~:o bnx ll are convergent and equal for all values ofx in an open interval (-c, c), thea a, = b, for every n. (Hint: Let f(x) ~ ~::"~oa,x' ~ ~:o b,x'. Differeotiate term by term to show that a, and b, both eqoa1 f(')(O)/(nl).)

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 f(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 f(x) is the sum of a power series 00

f(x)

= ~a,(x =

ao

- a)'

+ al(x - a) + a2(x - af + ... + a,(x - a)n + ...

with a positive radius of convergence. By repeated term-by-term differentiation within the interval of convergence I, we obtain

f'(x)

= al

+ 2a2(X - a) + 3a3(X - a)2 + ... + na,,(x - a)n-l + ..., + 2'3a3(X - a) + 3'4/l4(x - af + "',

!,,(x)

=

1'2a2

r(x)

=

1·2· 3a3

+ 2·3· 4/l4(x - a) + 3·4· 5as(x - a)2 + "',

with the nth derivative, for all n, being

I-n)(x)

=

IIla n + a sum of terms with (x - a) as a factor.

Since these equations all hold at x

!,,(a)

=

a, we have

= I, 2a2,

f"'(a)

=

1'2'3a3,

and, in general,

I-nl(a)

=

nlan·

These formulas reveal a pattern in the coefficients of any power series ~::"-o an(x - a)n that converges to the values of f on I (''represents f on 1"). Ifthere is such a series (still an open question), then there is only one such series, and its nth coefficient is

I-n)(a)

an =

-----,.;r-.

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)2 fOlia)

+ ... + --,-(x - a)" + ....

(l)

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 ofI? 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 (I) is the most important and useful series

we will study in this chapter.

HIsTORICAL BIOGRAPIllES

Brook Taylor (1685-1731) Colin Maclaurin (1698-1746)

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 I(k)(a)

00

al" =

~ ~(x -

/,,(a)

I(a)

+ f'(a)(x - a) + ~(x - a)' I(")(a)

+ ... + - - I-(x - a)" + .... n.

The Maclaurin series generated by I is 00 l(k)(O) ~ ~xk = 1(0)

1"(0)

fOliO)

+ f'(O)x + ~X2 + ... + ----,;r-x" +"',

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) = I/x at a = 2. Where, if anywhere, does the series converge to I/x? We need tu find 1(2), 1'(2), /"(2), .... Taking derivatives we get

Solutton

I(x)

=

x-I,

f'(x)

-x-2, /,,(x)

=

=

2!x-3,

f")(x)

=

(-I)"n!x-("+I),

so that

1( 2)

= Z-I =

-21, 1'(2)

= -

1"(2)

2I2 '

~ =

_ 2 3

=

I 23 '

( -I)" 21'1+1 .

The Taylor series is

1(2)

/,,(2)

/")(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 -

2)/2

2

I (x - 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

P 1(x)

=

I(a) + f'(a)(x - a).

In Section 3.9 we used this linearization to approximate I(x) at values of x 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.

DEFINmON 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) +

/"(a) 2 ! (x

- a)' + ...

fk)(a) .

y

fnl(a) n.

+ -k-'-(x - a)k + ... + - - I-(x - a)n.

3.0

We speak of a Taylor polynomial of order n rather than degree n because In)(a) may be zero. The f'rrst two Taylor polynomials of I(x) = cos x at x = 0, for example, are Po(x) = I and PI (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 polynomiaIs 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) = e'

atx = O. Solution Since fnl(x) = e' and fn)(O) = I for every n = 0, 1,2, ... , the Taylor series generated by I at x = 0 (see Figure 10.17) is ---o'o---!o--~c--~--->x

-0.5

a

0.5

1.0

The graph of f(x) and its Taylor polynomials FIGURE 10.17

PI(x) = I P2(X) = I P,(x) = I

= e'

+x + x + (x 2/2!) + x + (x 2/2!) + (x'/3!).

Notice the very close agreement near the center x = 0 (Example 2).

/"(0)

In)(o)

2!

n!

1(0) + f'(O)x + - - x 2 + ... + - - x n + ... 2

= I

x x" +x+-+ .. ·+-+ .. · 2 n! 00

=

k

L~,· k-O •

This is also the Maclaurin series for e'. In the next section we will see that the series converges to e' at every x.

10.8 Taylor and Maclaurin Series

587

The Taylor polynomial of order n at x = 0 is Xl

Xli

+".+,. n.

P.(x)=I+x+2

EXAMPLE 3

•

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)

/o.)(x) = (-I)" cosx, At x = 0, the cosines are I and the sines are 0, so

/0.+1)(0) =

/0.)(0) = (-I)",

o.

The Taylor series generated by 1 at 0 is

1(0)

+ 1'(0)x + l'i!O) x 2 + f"~~0) x3 + ... + /:~O) x" + ... =

I

+ O'x

00

=

~

x2 - 2!

x4

x 2n

+ 0'x 3 + 4! + ... + (-I)" (2n)! + ...

(-I)kx 2k (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 /0.+1)(0) = 0, the Taylor polynomials oforders 2n and 2n + I are identical:

x2 Po.(x) = Po.+l(X) = I - 2!

x4

x 2n

+ 4! - ... + (-I)" (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. • J

2

C:+---:-~c-:---:'---Y---r=--:---\':--~----:o~x

o

-1

-2

FIGURE 10.18

The polynomials n

Po.(x) = ~

(-I)*,;2>

(2k)!

converge to cos x as n ~ 00. We can deduce the behavior of cos x arbitrarily far away solely from knowing the values ofthe cosine and its derivatives at x = 0 (Example 3).

588

Chapter 10: Infinite Sequences and Series

EXAMPLE 4

It can be shown (though not easily) that

y

f(x) =

o {e~I/x',

x = 0 x i' 0

(Figure 10.19) has derivatives of all orders at x = 0 and that 1')(0) = 0 for all n. This means that the Taylor series generated by f at x = 0 is -----'---~+...L-'---------'----x -2

-I

0

/"(0)

f(O) + /,(O)x + ~ x 2 + ... +

2

/,')(0)

----,;r- x'

+ ...

= 0 + O·x + 0'x 2 + ... + O·x· + ...

The graph ofthe continuous extension of y = e-1/;;- is so flat at the origin that all ofits derivatives there are zero (Example 4). Therefore its Taylor series is not the function itself. FIGURE 10.19

=0+0+···+0+···. The series converges for every x (its sum is 0) but converges to f(x) only at x = O. That is, the Taylor series generated by f(x) in this example is not equal to the function f(x) itselL •

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 ofTaylor in the next section.

Exercises 10.8 + x' + 1,

FInding Taylor Polynomials

25. f(x) = x'

In Exercises 1-10, find the Taylorpolynomia1s of orders 0, 1,2, and 3 generated by f at a.

26. f(x) ~ 3x s - x'

2. f(x) = sinx,

I. f(x) = e"', a = 0 3. f(x) = Inx, a = I

4. f(x) = In (1 + x),

5. f(x) = I/x, a = 2 7. f(x) ~ sinx, a ~ 'If/4 9. f(x) =~,

a= 0

a= 4

6. f(x) = I/(x

8. f(x)

~

tanx,

+

2),

a

~

a= 0 a =0 'If/4

10. f(x) =~, a = 0

FInding Taylnr Series at x = 0 (Maclaurin Series) Find the Maclaurin series for the functions in Exercises 11-22. 11. e-x 12. xe~ 13. 1

I

16. sin ~

17. 7 cos (-x)

18. 5 cos '1fX

19. cosh x

=

eX

+ e--;t"

21. x' - lx' - 5x

+4

22. x

x'+ 1

FInding Taylor and Maclaurin Series In Exercises 23-32, fmd the Taylor senes generated by f at x 23. f(x) = x' - lx + 4, a = 2 24. f(x) = lx'

+ x' +

3x - 8,

a = I

a ~ 1

28. f(x) ~ 1/(1 - x)',

a ~ 0

29. f(x) ~ e', a ~ 2 30. f(x) = T, a = 1

31. f(x) = cos (lx + ('If/2», a = 'If/4 32. f(x) = y';+l, a = 0

Theory and Examples 37. Usc the Thylor senes generated by e' at x = a to show that

. eX-e-;; 20. sinhx = 2

2

a ~ -1

verges absolutely. 33. f(x) = cosx - (2/(1 - x» 34. f(x) = (1 - x + x') e' 35. f(x) = (sin x) In (I + x) 36. f(x) = x sin' x

1- x

15. sin 3x

- 2,

In Exercises 33-36, fmd the first three nonzero terms ofthe Maclaurin series for each function and the values of x for which the series con-

14. 2 + x

+x

27. f(x) ~ I/x',

a = -2

+ lx' + x'

e'

~

a.

= eO

[1 + (x _ a) + (x ;1 a)' + ... ].

38. (Continuation ofExercise 37.) Find the Thylor senes generated by e' at x = I. 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 and its flISt 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. Approximation properties of Taylor polynomials Suppose that f(x) is differentiable on an interval centered at x ~ a and that g(x) = bo + bl(X - a) + ... + bo(x - a)O is a polynomial of degree n with constant coefficients ho, ... , b". Let E(x) = f(x) - g(x). Show that ifwe impose on g the conditions

i) E(a) = 0 ") lim II

The approximation error is zcro at:t

E(x)

x-a (x _ a)"

0,

=

Thus, the Taylor polynomial Po(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)o. Quadratic Approsimations The Taylor polynomial of order 2 generated by a twice-differentiable function f(x) at x = a is called the quadratic approximation of f at x = a. In Exercises 41-46, fmd the (a) linearization (Taylor polynomial of order I) and (b) quadratic approximation off at x ~ O. 42. f(x) = e'"'x 41. f(x) = In (cosx)

a.

The error is neg1igible wheo compared to (x - a)II,

then g(x)

= f(a) + f'(a)(x

43. f(x) = 1/"\11=7 45. f(x) = sinx

j"(a) - a) + 2i(x - a}' +

44. f(x) = coshx 46. f(x) = tanx

f(o)(a) +--,-(x-a)o. n.

10.9

l_co_n_v_erg-",-e_n_C_e_of_l_a~YL_O_rS_e_ri_e_s

_

In the last section we asked when a Taylor series for a function can be expected to converge to that (generating) function. We answer the question in this section with the following theorem.

/0)

If f and its first n derivatives 1', /", ... , 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 Theonom

I(b) = I(a)

+ I'(a)(b

+

/0)

/,,(a)

+ 2 ! (b

- a)

AO)(a)

_l"_ _ (b _

n!

a

)0 +

- a)2

Ao+l)(C)

l"

(n

+ I)!

+ ...

(b _

a

)0+1

.

Taylor's Theorem is a generalization of the Mean Value Theorem (Exercise 45). There is a proof ofTaylor'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 I has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I, I(x) = I(a)

+ I'(a)(x

- a)

/o)(a)

+ --, -(x n.

/,,(a)

+ 2 ! (x 0

a)

- a)2

+ ...

+ Ro(x),

(1)

where

/o+l)(c) + 1)1 (x - a)o+1

R.(x) = (n

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 ) at a point The function R.(x) is detennined by the value of the (n + I)st derivative 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)

=

t'k)(a)

L -k'-(x k-O •

a)k.

Often we can estimate R. without knowing the value ofc, as the following example illustrates.

EXAMPLE 1 Show that the Taylor series generated by f(x) f(x) for every real value of x.

= eX at x =

0 converges to

Solution The function has derivatives of all orders throughout the interval I (-00,00). Equations (I) and (2) withf(x) = eX and a = 0 give

x2 xn e X = I +x+-+···+-+R(x) 2! n! n

=

Polynomial from Section 10.8, Example 2

and

_ eC n+l 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' < 1. 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 + I)!

whenx

~

0,

and

x n+ 1

IR.{x) I < eX (n + I)!

when x

>

O.

Finally, because X ·+

lim

.~OO (n

+

1

I)!

= 0

for every x,

Section 10.1, Theorem 5

(3)

I Tbe Number e as a Series 00

1

/FO

n!

e~ ~-

• We can use the result of Example I with x = I to write

e

=

I

+ I + -2!I + ... + -n!I + Rn (I),

10.9 Convergence of Taylor Series

591

where for some c between 0 and I, C

Rn(l) = e (n

11)1 <

(n ;1)1'

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 l.ln+ 1)(I) I :5 M for all 1 between x and a, inclusive, then the remainder term Rn(x) in Taylor's Theorem satisfies the inequality

IRn(x) I :5 M

Ix - al n+1 (n

+ I)!

If this inequality holds for every n and the other conditions ofTaylor'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)

sin x, - sin x,

=

j"(x) =

cos x,

j'(x) =

j'"(x)

=

- cos x,

so

The series has only odd-powered terms and, for n . x3 smx = x - 31

+

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)1) ---> 0 as k---> 00, whatever the value of x, so R2H 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! + 51 - 7!

t=6

+ ...

(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 x4 x 2k cosx = I - 2! + 4! - ... + (-I)k (2k)! + R2I.(x).

Because the derivatives of the cosine have absolute value less than or equal to I, the Remainder Estimation Theorem with M = 1 gives I R2k(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

I)kx 2k

(2k)!

=

series conveIgeS to cos x for every

x2 x' x. I - 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 oftheir 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. I (8) 3(2x + x cos x)

(b) eX cos x

Solution

2 1 (1X2!2 + X'4! -

1 (8) 3(2x + x cos x) = 3 x + 3 x 2

x3

1

2k x ... + (-I)k (2k)! + ... )

x5

x3

x5

=3 x +3 x - 3! + 3'4! -···=x- 6 + 72 - ... 2

3

•

2

3

4

)

(

2

•

)

(b) eX cos x = ( I + x + "-- + "-- + "-- + . .. • I - "-- + "-- - . . . 2! 3! 4! 2! 4!

x + -x+ ' " x ( l + x +2!- + 3! 4!

= 1

x3 x4 +x----+,.· 3 6

)

Multiply the first series by each term of the second series.

-

•

By Theorem 20, we can use the Taylor series ofthe function j to find the Taylor series of j(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 ofconvergence ofthe 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)2I" (2k)! =

00

cos2x =

~

2 2

=

(2x)'

(2x)4

(2x).

Eq. (5) with

I - 2! + 4! - 6 ! + ...

2xfor x

4 4

2x 2x 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 3

5

sinx = x _ x I + x 3! I 5!

_ ...

after (x 3/3!) is no greater than

I~; I

Ixl

s

120 .

Therefore the error will be less than or equal to 3 X 10-4 if s Ixl

120

<

3 X 10-4

or

Ixl

< ~360

X

10 4 '" 0.514.

Rmmdcd 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 S/120 is positive. Figure 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 ,;; 1. y 2 1 c:-i''-------'----:-+-~c--'------.f::____,'::i.....".,._____:----'~-> x

o

-1

-2

FIGURE 10.20 The polyoomials n

P2n+l(X)

(-1)'x2H1

~ ~ (2k + I)!

converge to sin x as n -+ 00. Notice how closely P3(X) approximates the sine curve for x '" I (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

< b. The proof for a > b is nearly the same.

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 ofthe form K(x - a)n+ I , 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

"'.(x)

=

Pn(x) + K(x - a)n+1

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 = "'.(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+1

(7)

With K defined by Equation (7), the function

F(x)

=

f(x) - "'.(x)

"'n

measures the difference between the original function f and the approximating function 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 CI in (a, b).

Next,becauseF'(a) = F'(cl) = OandbothF'andF" are continuous on [a, ctJ,weknow that

F"(C2)

=

0

for some C2 in (a, CI).

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+1 in (a, cn) such that

F(n)(a)

(8)

Ifwe differentiate F(x) = f(x) - P.(x) - K(x - a)n+1 a total ofn

F(n+I)(X)

=

+I

,n+I)(x) - 0 - (n + l)!K.

times, we get (9)

Equations (8) and (9) together give

K

=

,n+I)(c) (n + I)!

for some numberc = Cn+1 in (a, b).

(10)

10.9

Convergence of Taylor Series

595

Equations (7) and (10) give

f(b)

=

Pn(b)

+

/n+1)(c) + I)! (b - a)n+l.

(n

•

This concludes the proof.

Exercises 10.9 Finding Taylor Series Use substitution (as in Example 4) to find the Taylor series at x the functions in Exercises 1-10. 2. e-x/2

1. e--5x

(~x)

4. sin

7. !n(1

8. tao-I

0 of

3. 5 sin (-x)

5. cos 5x

+ x')

~

2

6. cos

(x'/3/V2)

9 I . 1 + ~x3

(3x 4 )

10. -2-

-x

Use power series operations to fmd the Taylor series at x functions in Exercises 11-28.

.

14. smx - x

+ x' 3!

13.

15. x cos 'll"x

16. x' cos (x')

= (I + cos2x)/2.)

18. sin'x

19. 1- 2x

I (I - x)'

24. sinX" cosx 27.

t!n (I + x')

0 for the

12. x 2 sinx

17. cos'x (Hint: cos'x

21.

x' T

~

x'

I

20. xln(I

+ cosx

+ 2x)

25.ex+I~x + x)

26. cosX' - sinx

e~

sin x

!nO +x) 30.

I - x

2

32. cos x' sinx

40. The estimate ~ = I + (x/2) is used when x is small. Estimate the error wben Ix I < 0.01.

42. (Continuation of Exercise 41.) When x < 0, the series for eX is an alternating series. Use the Alternating Series Estimation TheorOll3 to estimate the error that results from replacing eX by I + x + (x'/2) when -0.1 < x < O. Compare your estimate with the one you obtained in Exercise 41. Theory and Examples 43. Use the identity sin' x ~ (I - cos 2x)/2 to obtain the Maclaurin series for 8in2 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.

45. Taylor's Theorem and the Mean Value Theorem Explain how the Mean Value Theoren3 (Section 4.2, Theorem 4) is a special

31. (tao-1 x)'

46. Linearizations at iDfIection points Show that if the graph of a twicl>-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 taogent lines fit so well at inflection points. 47. The (second) seoond derivative test

34. sin (tan- 1 x)

I(x) ~ I(a) Error Estimates 35. Estimate the error if P,(x) ~ x - (x' /6) is used to estimate the valueofsinxatx = 0.1. 36. Esti:matetheerrorifP4(x) = I + x + (x'/2) + (x'/6) is used to estimate the value of eX at x ~ 1/2.

+ (x 4/24)

37. For approximately what values of x can you replace sin x by x - (x'/6) with an error ofmagnitude no greater tban 5 X 1O-4?

Give reasons for your answer.

< IO-'? For

case ofTaylor's Theorem. -!n(I - x)

Find the fIrst four nonzero terms in the Maclaurin series for the functions in Exercises 29-34. 29.

39. How close is the approximation sinx = x when Ixl which of these values ofx is x < sinx?

44. (Continuation of Exercise 43.) Use the identity cos'x ~ cos 2x + sin2 X to obtain a power series for cos2 x.

2 22. ( I-x )'

28. !n(1

too small? Give reasons for your answer.

41. The approximation eX = I + x + (x'/2) is used when x is small. Use the Remainder Estimation Theorem to estimate the error whenlxl < 0.1.

4

I

38. If cos x is replaced by I - (x' /2) and Ix I < 0.5, what estimate can be made of the error? Does I - (x'/2) tend to be too large, or

Use the equatiou

/"(c,)

+ f'(a)(x - a) + -2-(x - a)'

to establish the followiug test. Let 1 have continuous fust and second derivatives and suppose thatf'(a) = O. Then a.

1 bas a local maximum at a if/"'"

0 throughout an interval

whose interior contains a; b.

1 bas a local minimum at a if f" '" 0 throughout an interval whose interior contains a.

596

Chapter 10: Infinite Sequences and Series

48. A cubic apprmdmation 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. Give an upper bound for the magnitode of the error in the approximation whenlxl :5 0.1. 49. a. 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 ofx in the interval [0, I] will the error in the quadratic approximation be less than 1/100? 50. Improving apprmdmations of 'If that P + sin P gives an approximation correct to 3n decimals. (Hint: LetP = 'If + x.) Try it with a calculator.

51. The Taylor .eries generated by f(x) ~ ~::'~oa.x· is ~:=o GilX' A function defined by a power series ~:o GnX" 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 series generated by f(x) = ~::'-o a.x' is the series ~:o Cl"X n itself. An immediate consequence ofthis is that series like • 2 xsmx=x

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 (a) and (b) for the functions and intervals in Exercises

53-58. Step 1: Plot the function over the specified interval.

a. Let P be an approximation of 'IT accurate to n decimals. Show

D b.

and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of x can the function be replaced by each

4

6

8

3!

5!

7!

x + X- - -X + ... --

and

Step 2: Find the Taylor polynomials PI(x), P2(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 ofc over the specified. interval and estimate its maximum absolute value, M. Step 4: Calculate the remainder R,(x) for each polynomial. Using the estimate M from Step 3 in place of /'+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 series by powers of x, as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent. 52. Taylor serie. for even function. and odd functions (Continuation ofSection 10.7, Exercise 55.) Suppose thatf(x) ~ ~:oa.x' converges for all x in an open interval ( - R, R). Show that

a. Iff is even, then al = a, = a, = ... = 0, Le., the Taylor series for f at x

~

0 contains ouly even powers of x.

b. Iff is odd, then no ~ a2 ~ Cl4 ~ ... ~ 0, Le., 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:5

3

4

-t:5 x :5 2

+-, +

Ixl:5 2 x I (cosx)(sinlx), Ixl:5 2

56. f(x)

~

57. f(x)

~ e~coslx,

Ixl:5 I

58. f(x) ~ e'/3 sin lx,

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

10.10 I--Th_e_B_in_o_rn_i_a_L_Se_n_·e_s_a_n_d_A....:p....:.p_Li_c_ati_·o_n_s_o_f_l_a=-yL_o_r_S_en_·e_s

_

In this section we introduce the binonrial series for estimating powers and roots of binomial expressions (1 + x)m. We also show how series can be used to evaluate nonelementary integrals and limits that lead to indeterminate [onns, 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)m, when m is constant, is

+

m(m - I)(m - 2) 3 3! x

+

m(m - I)(m - 2)'" (m - k k!

+ ... + l)

k

x

This series, called the binomial series, converges absolutely for series, we first list the function and its derivatives:

f(x)

=

+ ...

IxI <

(1)

I. To derive the

(I + x)m

+ x)m-t

j'(x)

= m(1

j"(x)

=

m(m - l)(I

j"'(x)

=

m(m - l)(m - 2)(1

pl(x)

=

m(m - l)(m - 2) .. · (m - k

+ x)m-2

+ x)m-3

+ 1)(1 + x)m-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 equsl to zero, the series stops after (m + I) terms because the coefficients from k = 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 x k • Then apply the Ratio Thst for absolute convergence to see that

UH11Im-kl k + I x --+Ixl IUk

ask--+ 00.

=

Our derivation of the binomial series shows ouly that it is generated by (I + x)m 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 - l) (2 2!'

and

m) = m(m - I)(m - 2) .. ·(m - k (k k!

EXAMPLE 1

+ I)

fork;" 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! - ( I). ( 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 + ... + (-Ilx k + ....

•

k=-!

EXAMPLE 2 We know from Section 3.9, Example I, that ~ 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)(-~)(-%) x

+

4

+ ...

4!

x

X3

xl

5x 4

=1+---+---+··· 2 8 16 128 Substitution for x gives still other approximations. For example, for Ix 2 1 small

~I_IRJI_~ __I_ x

2x

for

8x2

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, 2 x 6 x lO X - 3! + 51

x!4

- 7f + ... =

(x 2)3

2

(x 2)s

(x 2)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 ofthe 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!

- -X

14

7!

+ -x

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

599

1

Estimate Jo sinx 2 dx with an error ofless than 0.001.

EXAMPLE 4

From the indetmite integral in Example 3,

Solution

r.

10

2 I I I I I smx dx=:3-7'3! + 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 ofthe preceding two terms gives

l'

2

sinx dx ""

t - 12 "" 0.310.

With two more terms we could estinlate

l'

sinx 2 dx "" 0.310268

with an error ofless than 10--< . With ouly one term beyond that we have

r.

10

2

smx dx ""

I

I

I

I

I

:3 - 42 + 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-l x

= _1_ = I -

I

dx

+ x2

+ x4

x2

x6

-

+ ...

and then integrating to get

x 3 x 5 x7 tan-1 x=x--+---+'" 3 5 7 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)n+l t 2n+2

+ ... + (-I )nt o. + -

t6

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+1 t o.+2 and ratio r = -t 2 . Integrating both sides of Equation (2) fromt = Otot = x gives -1

tan x = x -

x3

xS

x7

3 + 5 -7 + ... + (-I)

where

-l

X

R.{x) -

D

(_l)n+l t o.+2

I

+t

2

n x 2n +

2n

+

1

I

+ Rn(x),

dt.

The denominator of the integrand is greater than or equal to I; hence

IR.{xJl

:5

fixi o. 2

10

t + dt

Ix1o. +3 =

2n

+

3.

600

Chapter 10: Infinite Sequences and Series If Ix I :5 I, the right side of this inequality approaches zero as n --> lim"~""R"(x) = 0 iflxl :5 I and

(-I)"x 2n + 1 tanx- n=O 2 n+1' -I

_

""

L

3

00.

Therefore

Ixl :5 1.

5

(3)

7

x x x tan-1 x=x--+---+··· 3 5 7 '

Ixl :5 1.

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 = I in Equation (3), we get Leibniz'. formula: ~

I 3

I 5

I 7

I 9

(-If 2n+1

-= 1--+---+--···+--+···

4

Because this series converges very slowly, it is not used in approximating ~ to many decimal places. The series for tan-1 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

f3 = tan-1I 3'

and

then 1

I

tana+tanf3 2+3 ~ tan(a + f3) = 1- tanatanf3 = 1- 1 = I = tan 4 6

and ;

=

a + f3 = tan- 1

t

+ tan- 1

t·

Now Equation (3) may be used with x = 1/2 to evaluate tan- 1 (1/2) and with x = 1/3 to give tan- 1 (1/3). The sum of these resuits, multiplied by 4, gives ~.

Evaluating Indeterminate Forms We can sometimes evaluate indeterminate forms by expressing the functions involved as Taylor series.

EXAMPLE 5

Evaluate

Inx " In n - -

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 = I directly or by replacing x by x - I in the series for In (I + x) in Section 10.7, Example 6. Either way, we obtain Inx

= (x -

I) -

t

(x - 1)2

+ ...,

from which we rmd that

lim

x---+l

EXAMPLE 6

~ = x-I

lim

x-I

(I _I(x - I) + ...) 2

Evaluate lim sinx - tanx

x-o

x3

=

1.

•

10.10 The Binomial Series and Applications of Taylor Series

Solution

x, to terms in xS,are

x

The Taylor series for sin and tan

. smx

=

3 x - -x 3!

5

5!

s

3

+ -x - ...

tanx = x

'

601

+ -x3 + -2x + ... 15

Hence,

sinx-tanx=-x; _x; _... =x3(_!_X; _.. -) and

I 2'

•

If we apply series to calculate limx~o((I/sinx) - (l/x)) , 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

I

sin x

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

-1- - -1 sin x x

~

1 =x x·3!

6

or

1 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

i6 (6 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

iO

Ifl ( 1- -2!

'0

.303

·2.2

.404

.505

.606

= I + '-- + _'rr_ + _'_ + _'_ + _'_ + _'_ + ... I! 2! 3! 4! 5! 6!

6

3

. (0 - -0 + -0-5 .. + -04 - -0 + ...)+, ...) = cosO + ,smO 4! 6! 3! 5! .

This does not prove that e 'O = 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 'o to be consistent with other things we know.

DEFINmON ForanyrealnumberO,e lo = cosO + isinO.

(4)

Equation (4), called Euler's identity, enables us to define ea+hi to be ea. ebl 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 ofthe most important constants in mathematics.

TABLE 10.1 Frequently used TayLor series

I

--= 1

I-x

+x+x 2 +···+x"+···=

00

~xn

~

,

Ixl < 1

I 00 - - = I - x + x 2 - ... + (-x)n + ... = ~(_i)nxn, 1+x n=O Xl

X"

00

X"

e X = I +x+-+ .. ·+-+ .. ·= ""2! n! ~n!' .

x3

Ixl <

x 2n+1

x5

x4

x2n

00

cos x = 1 -2-, +-4,- .. ·+(-I)n(2 )' + ... = ~ .. n. 11=0

x2

x3

00

00

smx = x - 3! + 5! - ... + (-I)n (2n + I)! + ... =

x2

Ixl < I

~

(_I)nx 2n+l (2n + I)! '

I )nx 2n (2 )' '

xn

n.

00

2n+l

5

00

(_

00

Ixl <

(_i)n-lxn

1o(1+x)=x-- +-3 _ ... +(_i)n-ln+"·=~ 2 n=1 3

Ixl <

n

(_l)nX2n+l

'

tan-I x = x - "-- + "-- - ... + (- I)n x + ... = ~ -'--..--'--cc---.3 5 2n+1 n~O 2n+I'

00

-1<x:51

Ixl :5 I

Exercises 10.10 Binomial Series Find the ftrSt four terms of the binomial series for the functions in Ex-

4. (I - z,,)'/2

ercises 1-10. 1. (1 + X)'/2

7. (1

2. (1

+ x)'/3

3. (1 - xt'/2

+ x't'/2

8. (1

+ X2)-I/3

10.10 The Binomial Series and Applications of Taylor Series

9.

(I + X1)'/2

x

x-a 1

vl+x

Find the binomial series for the functions in Exercises 11-14.

+ x)4

11. (I

12. (I

13. (I - 2>;)'

I)'

Approxfmattons and Nonelementary Integrals

0.2 sinx

/.°

0.1 17.

o

/.o

2

42.

16.

v'1+74 dx

18.

/.0.25

1

+x

1

/.0.2 I ° "---i- dx ~

dx

0

21.

/.0.1 s~x dx

20.

0.1 v'1+7 dx /.°

22.

~ dx

4S

X

fo

24. Estimate the error if cos VI is approximated by I -

T+ ~! 4

8

in the

t + ~! 2

,

~!

10 Exercises 25-28, fmd a polynomial that will approximate F(x) throughout the given interval with an error of magnitude less than 10-' .

25. F(x) =

I

28. F(x) ~

xln(1 + I) ° I dl, /.

x-a

31. lim

x

t

1-0

4

'

Y - tan-I Y

y-o

y

33. lun

(b) [0, I] (b) [0, I]

35. lim x 2(e- 1/ x'

y-o

-

x~oo

37. lim

1n(1

+x

2

)

~'------"­

x-a 1

sinB - B + (11'/6) 32. 9-0 lim "' tr

34. lun

cosX'

I)

r

1f's

1T'7

3"5!

3"7!

2

2'

2'

2'

3 - 3"3 + 3"5 - 3"7 x 3 + x 4 + X S + x 6 + ... 34x 4

3'>:2 21 XS

-

+ ...

3'x'

+ 41 - 61 +'" + x 7 - x 9 + XII - •••

2 3 22.x 4 23x S 24x 6 50. x -2>; +21-31+41-'"

51. -I +2>;-3x 2 +4x'-5x 4 + ... 52.1+

x x2 x 3 x 4 + + + +",

2

3

4

5

Theory and Examples 53. Replace x by -x in the Taylor series for In (I + x) to oblain a series for In (I - x). Thea subtract Ibis from the Thylor series for In (I + x) to sbowthatforlxl < I,

+


'

,

1_ + ... 4'24

3"3!

1r

46.

_1 3'2'

I - x

2

I - cos I - (1 2/2)

+ ...

I x ( x+-+-+ x' x' 1n--~2

Indeterminate Forms Use series to evaluate the limits in Exercises 29-40. eX - (I + x) 29. lim

+

- - - - + - - - - - + ...

49. x 3

Jo' cos VI dl.

/.X sin 12 dl, [0, I] 26. F(x) ~ /,x I2e -t'dl, [0, I] 27. F(x) = /.X tan-I dl, (a) [0,0.5]

2'22

. 3

47.

- ~os x dx

+

1 __1_ +

48.1-

23. Estimate ~e error ifcos 12 is approximated by I integral cos 12 dl.

in the integral

+

. 2

/.0.1 e -x' dx /.' I °

(t)' (t)4 (t)' (t)"

2 34- - - -3'- + .. · 4 3 . 123- - - + 4 4 '2! 4 '4! 4"6!

44

Use series to approximate the values of the integrals in Exercises 19-22 with an error ofmagnitude less than IO-s .

19.

_tan_-_'~y_-_sin--"--y , y cosy

36. lim (x + 1) sin _+1I x_OO x x2 -

4 38. x-2 lim In (x- I)

+ x')

x-a x· sinx 2

1 I 1 41. 1 + I + 2! + 3! + 4! + ...

010 Exercises 15-18, use series to estimate the integrals' values with an error of magnitude less than 10-'. (The answer section gives the integrals' values rounded to five decimal places.)

15.

40. lim In (I

cos 2x

Using Table 10.1 10 Exercises 41-52, use Table 10.1 to fmd the sum ofeacb series.

+ x 2)'

14. (I -

. 3 2 sm x

39. lim

10•• ,r.-:-

603

3

5

.. · ) .

54. How many tenns of the Taylor series for In (I + x) sbould you add to be sure of calculatiog In (1.1) with an error of magnitude less than 10-8? Give reasons for your answer. 55. According to the Altematiog Series Estimation Theorem, how many tenns of the Thylor series for tan-I I would 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 f(x)

o 57.

Ixl>

= tan-I x

diverges for

1.

Estimating Pi About how many terms of the Taylor series for tan-I x would you have to use to evaluate each term on the righthaod side of the equation

_ 48

'If -

-1

tao

I + 18

32

-1

tan

I 20 -1 1 57 tan 239

with ao error of magnitude less than IO-
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 rrrst three nonzero terms of the Taylor series for In sec x.

by integrating the series

59. a. Use the binomial series and the fact that

in the first case from x to tox. to generate the fIrst four nonzero terms ofthe Taylor series

for sin-1 x. What is the radius ofconvergence? 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. a. Series for sinh-1 x Taylor series for

Find the rrrst four nonzero terms of the

=

sinh-1 x

r

dt

Jo \!l+7

Euler's Identity 67. Use Equatioo (4) to write the followiog powers of e in the form a + hi.

a.

e-l'll'

68. Use Equation (4) to show that

Taylor series for e ifJ and e- ifJ • 70. Show that

a. coshiO = cosO,

+ x)' from the series for

62. Use the Taylor series for 1/(1 - x') to obtain a series for 22/(1 - x')'Estimating Pi The English mathematician Wallis discovered the fommla ~

2'4'4'6'6'8' 3'3'5'5'7'7'

4

00

69. Estshlish the equations in Exercise 68 by combining the formal

estimation error. 61. Obtain the Taylor series for 1/(1 -1/(1 + x).

and in the second case from -

.

Db. Use the rrrst three terms of the series in part (a) to estimate sinh-1 0.25. Give an upper bound for the magnitude of the

D 63.

00

. .

b. sinhi8

= isin8.

71. By multiplying the Taylor series for eX and sin x, rmd the terms through x S ofthe Taylor series for eX sin x. This series is the imaginary part of the series for eXoeix

=

e(l+i)x.

Use this fact to check your anawer. For what values of x should the series for eX sin x converge? 72. Wheo a and b are real, we derme e(a+ib).< with the equation e(a+ib)x = eaxoeibx =

ea.t"(cosbx + isinbx).

Find ~ to two decin2al places with this formula. Differeotiate 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 '( ) = mf(x)

x

1

b. Defineg(x) = (I

+ x·

-I

<

x

73. Use the dermition of and 62 ,

< I.

a.

+ x)-mf(x) and show that g'(x)

= O.

c. From part (b), show that f(x) = (I

j

001__'~3_'5~'~"_' _.,::(2=n~---c-=I:'.) x 2n + 1 2·4·6· .. · • (2n) 2n + I'

j

tan-Ix = ~ _1 + _1 1_ + ... 2X3x3sxS'

= -~ _1 + _1

where C

Derive the series

1_ + ... 2X3x3sxS'

x>

1

x <-I

,

=

1/e i6 .

ea.t" cos bx dx and

j

e a.t" sin bx

dx

from

.-1

Ixl > 1

b. e-i6

74. 1\vo complex nmnbers a + ib and c + id are equal if and ooly if a = c and b = d. Use thia fact to evaluate

smlx=x+~-

66. Series for tan-I x for

= ke'" holds for k complex as

e" to show that for any real nwnbers 6, 61,

ei61ei~ = ei(61+6i),

+ x)m.

65. Series for sio-1 x Integrate the binomial series for (I - x')-I/2 toshowthatforlxl < I,

tan-Ix

= (a + ib )e(a+ib).<.

Thua the familiar rule (d/dx)e'" well as real.

to show that

. _

e(aHb).<

=

e(U+ib)x dx =

a -

a2

ib e(u+ib)x + C

+ b2

'

Ct + iC2 is a complex constant of integration.

Chapter 10

•

Chapter

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? Give 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 can you estimate the error involved in approximating the

soch a sequence have a limit? Give examples.

sum of an alternating series with ooe of the series' partial sums? What is the reasooing behind the estimate?

3. What theorems are available for calculating limits of sequences?

20. What is absolute coovergeoce? Cooditiooa1 convergeoce? How

Give examples. 4. What theorem sometimes enables us to use I'H6pital's Ru1e to calcu1ate the limit of a sequence? Give an example.

5. 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? Give 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 se-

are the two related?

21. What do you know about rearranging the terms of an absolutely convergeot series? Of a cooditionally convergeot series?

22. What is a power series? How do you test a power series for conyergeoce? 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? c. term-by-term differeotiatioo ofpower series?

ries do you know?

d. term-by-term integratioo ofpower series?

9. What is the nth-Thnn Thst for Divergeoce? What is the idea be-

Give examples.

hind the test?

10. What can be said about term-by-term soms and differeoces of coovergeot series? About coostsot mu1tiples of cooyergeot and di-

24. What is the Taylor series geoerated by a functioo 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 ifyou add a fmite number of terms to a coovergeot 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 functioo? Explain.

12. How do you reindex a series? Why might you WlIDt to do this?

27. What are Thylor polyoomials? Ofwhat use are they?

13. Under what circumstsoces will an infntite series of nonnegative terms converge? Diverge? Why study series ofnonnegative 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 fommla say about the error in a linearizatioo? A quadratic approxlmatioo?

14. What is the Integral Test? What is the reasooing behind it? Give an example of its use. 15. Wheo do p-series cooverge? Diverge? How do you know? Give examples of coovergeot and divergeotp-series.

16. What are the Direct Comparisoo Test and the Limit Comparison Test? What is the reasooing behind these tests? Give examples of their use.

17. What are the Ratio and Root 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 cooverge? How is it used?

30. How can you sometimes use power series to estimate the values ofnooelerneotary definite integrals?

31. What are the Taylor series for 1/(1 - x), 1/(1 + x), eX, sinx, cos x, In (I + x), and tso-I x? How do you estimate the errors involved in replacing these series with their partial soms?

Give examples.

Chapter

TI.

Practice Exerdses

Determining Convergence of Sequences Which of the sequences whose nth terms appear in Exercises 1-18 cooverge, and whicb diverge? Find the limit of each convergeot

sequence.

(-I)1. an = l + -n-

3.

I - 2-

an=~

I - (-I)-

2.

a.

=

4.

a.

= I

y;,

+ (0.9)-

5.

a.

~

. n1r sm

7.

a.

~-n-

6. an

2

In (n 2 )

8.an =

_ n + Inn 9.a,,n 11.

a.

n -

=

5)-

= ( -.-

10. an =

sinn1r 1n(2n+I)

n

In (2n'

n

+

I)

( 1)--

12.a_= 1+"

606

Chapter 10: Infinite Sequences and Series

13. all =

,/¥ 'In

15. a, = n(2'/' - I)

17.

all =

(n + I)!

(3)1/,

14. an

= n

16. a,

= 'ZY2n +

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?

I

(-4)' n.

51 1- 1 + -.L - ... + (-I)'-.L + ... · 4 16 4'

18. all = - , -

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

. ,-I (3n - 1)(3n + 2)

f

-8 • ,_, (4n - 3)(4n + I)

00

,3

00

24. }:(-I) 4'

25.

I

00

}:.t 11=1

26.

vn

1_+_1_ _ ...

v3

Determ;nlng Convergence of Series Which of the series in Exercises 25-40 converge absolutely, whicb converge conditionally, and which diverge? Give reasons for your answers.

-5

00

~n

27.

~

(-I)'

y;,

(-I)' 00 I 29."5'. ( ) 30. "5' - - 2 ::lin n + I ,;':1 n (Inn)

9v3 + (-1)'-'

45v3 I

(2n - 1)( v3)2o-'

Find Taylor series at x

~

+ ...

0 for the functions in Exercises 57-64.

I 57· 1 _ 2x

58 _1_ 'I+x'

59.

60.

sin "X

sin

00

f

In: 11=1 n 00 (-I)' 33. "5' .~ ~nvn2+ 1 31.

35.fn~1

n. (-3)' 37. } : - , 11=1 n. 11=1

~ In (Inn)

00 (-I)'3n2 34. "5' '---c,='--''''-~ n + 1 2 + I) 00 (-I)'(n 36. }: ~2'-=-C'­ 1Ft 2n + n - 1

00

39.

Inn

00

32.

00

211 311

38. } : - , 11-1 n

f ,-I v'n(n + I)(n + 2) I

40.

f I '-2n~

Power Series

.&1

00

.&1 00

(-I)'-'(3x - I)' n2 11

45.}:\ ,,-I n 00

47. }:

,,-0

(n

61. cos (x'/')

62. cos

63. .(=/2)

64. .-x'

~

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) = v'3+7 at x = -I 66. f(x) = 1/(1 - x)

67. f(x)

~

I/(x + I) at x

68. f(x) ~ I/x

at x = 2 at x ~ 3 ~

a

>0

Nonelementary Integrals Use series to approximate the values of the integrals in Exercises 651-72 with an error of magnitude less than 10-' . (The answer section gives the integrals' values rounded to 10 decimal places.)

In Exercises 41-50, (a) fmd the series' radius and interval of convergence. 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)!

~ + ... 54. 1- -' "- + -,,4- - ... + (-1)'--9'2! 81'4! 32o (2n)! (In 2)' (In 2)' 55. I + In2 + ----z! + ... + ----.J + ... 56. _1

,-I

00

~+1

'ITs

w3

5 3 , , - - + - - .. ·+(-1)' · 3! 5! (2n

00

44.

~ 00

+ 1)(2x + I)'

(n

(2n + 1)2' 11

46'}:.Xt

+ l)x 2o -1 3'

00

49. ,&1(cschn)x'

11=1 Vn 00

48. }:

11=0

(-I)'(x - 1)2'+' 2 + I n

1/2.-x' fix 1. lf2 -x-fix tan-I x 71. 69.

J.

o

0

0

VX'

Us;ng Series to F;nd Um;ts In Exercises 73-78: a. Use power series to evaluate the limit D b. Then use a grapher to support your calculation. 7 . , -, 26 73. lim ----""£ 74. lim' - . .x-O e'h -

00

50. ,&1(cnth n)x'

1.1x ,in (x') fix 1. 1/64 tan-I x 72. • t fix 70.

0

75 lim

• 1-0

1

I I) ( 2eost 2 -

- -

(2

9-0

8

8m8

(sinh)/h - cosh 76 lim c-----"----,._ _ • h-O

h2

Chapter 10 Additional and Advanced Exercises

77 lim

.

z~o

,

I - cos, z In (I - z) + sinz

b. Show that the functioo defined by the series satisfies a differeotial equatioo of the form

78. lim ------y'---~­ y-O

cosy - coshy

d'y

tlx' ~

Theory and Examples 79. Use a series representation of sin 3x to find values of rand s for which

x'Y + b

and rmd the values of the coostants a andb.

86. a. Find the Maclaurin series for the functioo x '/(1 b. Does the series converge at x

o 80. Compare

the accuracies of the approximations sinx '" x and sinx '" 6>:1(6 + x') by comparing the graphs off(x) ~ sinx - x andg(x) = sinx - (6):1(6 + Describe what you rmd.

x'».

81. Find the radius of convergeoce of the series

3.S.7

'(2n

+

I? Explain.

87. If ~:1 an and ~~=1 b" are convergent series of nonnegative numbers, can anything be said about ~~I aRb,,? 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

89. Prove that the sequeoce {x,} and the series ~~I (XHI -

Xl)

both converge or both diverge.

90. Prove that ~::;'I (aJ(1 ~~=1 an converges.

82. Find the radius of convergeoce of the series 00

~

+ x).

your answer.

2'S.8 • ... '(3n - I) ~ 2·4·6· .... (2n) x' . ,-I 00

~ 4.9.14

607

1)

+ a,»

cooverges if a"

> 0 for all nand

91. Suppose that at, a2, Q3, • .• , all are positive numbers satisfying the following cooditions:

(Sn _ I) (x - I)'.

83. Find a closed-form formula for the nth partial sum of the series ~::,~,In (I - (lIn'» and use it to determine the convergeoce or

divergence ofthe series.

i)

at ~ a2 ~ Q3 ~ ••• ;

ii) the series a2

+ Cl4 + as + al6 + ... diverges.

Show that the series

84. Evaluate ~:2(I/(k' - I» by rmding the limits as n -->

00

of

the series' nth partial sum. 85. a. Find the interval of convergence ofthe series

y

I I = I + -x' +x ' + ... 6 180

+

I

1·4·7· .. ··(3n - 2) (3n)! x"

•

Chapter

+ ...

Determining Convergence of Serfes

~ (3n

3.

~(-I)'tanhn

_ 2),+(1/2)

2.

~

4.

~ log.~n!)

11-1

11-2

n'

+

a,+1 ~ (n

n(n + I) + 2)(n +

6. al

= a2 = 7,

a,+I

7. al

= a2 = 1,

an+! -

= (n

3) a,

n

_ I)(n

+ I) a"

ifn ~ 2

_ _1_ if >2 1 + all n -

ifn is odd,

Choosing Centers for Taylor Serfes Taylor's formula

f(x) = f(a) + f'(a)(x - a) + /,')(a)

a, = n/3' ifn is eveo

!"(a) 2 ! (x

- a)' +

/,,+I)(c)

+ - -n.I-(x - a)' + (n + I)'. (x - a)'+1

n

(Hint: Write out several terms, see which factors cancel, and then generalize.)

8. a, = 1/3'

n_2 nlnn

diverges.

I

Which of the series ~::;'I a" derIDed by the formulas in Exercises S--1l converge, and which diverge? Give reasons for your answers.

S. al ~ I,

+ ~_I-

Additional and Advanced Exerdses

Which of the series ~: 1 an dermed by the fonnulas in Exercises 1-4 converge, and which diverge? Give reasons for your answers. 00 I 00 (tan-I n)'

1.

diverges. 92. Use the result in Exercise 91 to show that

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 eoough to the values of f we are interested in to make (x - a )'+ 1 so small we can neglect the remainder.

In Exercises 9-14, 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 rlrSt 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 {(a' + b')I/'} 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

-""-- = I + f: +

16. Find the sum of the infmite series I

"n+l

+~+~+ -"-- + ~+~+-"--+ ~ 10

10'

10'

10'

10'

10'

10'

3 7 +-+-+ ... 10'

10'

1'+1

11=011

UI

~

,~I

,

nx

(n + 1)(2x + I)'

+

I) Un'

ofor all n,

28. (Continuation of Exercise 27.) If ~::;'I a" converges, and if I> a" > Ofora!ln,abowthat~:::IIn(1 - a,)eonverges.

lin (I

(Hint: First show that

( 1- eos(a/n»)' n •

a constant,

appear to depend on the value of a? If so, how?

a and b constant, b

- a,) I '" a"j(1 - a,,).) Prove Nicole Oresme's Theorem that

29. Nicole Oresme's Theorem

I

b. Does the value of . ( cos (a/n»)' hm 1bn ' 11_00

~ (2n)(2n

b. Does ~::"~I a,/(I - a,) converge? Explain.

converges absolutely.

,~OO

U,+I

27. If~:lanconverges,andifCln =F 1 andan > a. Show that ~: 1 Q,? converges.

a. Does the value of

.

(2n - I)'

~ I,

Apply Raabe's Test to determine whether the series converges.

00

hm

where If(n)l < K for n '" N, then ~::;'I U, converges ifC > I and diverges if C '" I. Show that the results of Raabe's Test agree with what you know about the series ~::"-l (I/n') and ~::"-l (I/n).

I , dx. -l+x

18. Find all values of x for which

D 19.

n2 •

26. (Continuation ofExercise 25. ) Suppose that the terms of ~::"-l U, are defmed recursively by the formulas

17. Evaluate 00 ~

f(n)

n

¢'

0,

+ 1. 2

2

+ 14 . 3 + ... + --"-+ ... = 4. 2'-1

(Hint: Differentiate both sides of the equation 1/(1 - x) = 1 + ~:_IXn.)

30. a. Show that

appear to depend on the value of b? If so, how? c. Use calculus to coufirm yoor fmdings in parts (a) and (b). 20. Show that if ~::"~l a" converges, then

00

~ n-I

n(n

+ I) X,

2x' =(x - I )'

for Ix I > I by differentiating the identity 00

LXn+

,

1

11=1

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 ofthe power series

b. Use part (a) to fmd the rea! solmion 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 functions sinx, Inx, and e' are not polyoomials? Give reasons for your answer.

23. Find the value of a for which the limit lim sin (ax) - sinx - x

x-o

x3

_1_= 1 I-x

+X+X2+ ... +X"+'"

toobtainaseriesforl/(l-x)'.

11. In one throw of two dice, the probability of getting a roll of? is p = 1/6. Ifyou throw the dice repeatedly, the probability that a 7 will appear for the flt'St time at the nth throw is q,-Ip , where q = I - p = 5/6. The expected number of throws until a 7 ftrSt appears is ~:=l nq"-1p. Find the sum

is fmite and evaluate the limit. 24. Find values of a and b for which . cos(ax)-b hm , x-o 2x

a. Differentiate the series

~-l.

25. Raabe's (or Gauss's) Test The following test, which we state without proof, is an extension of the Ratio Test.

of this series. c. Aa an engineer applying statistical control to an industria! operation, yoo inspect items taken at randem from the assembly line. You classilY each sampled item as either "good" or

Chapter 10 Technology Application Projects ''bad.''lfthe probability of an item's beiog good isp and of an item's beiog bad is q ~ I - p, the probability that the flISt bad item found is the nth one inspected is p"-lq. The average number iospected up to and iocluding the flISt bad itero found is ~:lnpll-lq.Evaluatethis sum, assuming 0 < P < 1. 32. Expected value Suppose that a random variable X may assume the values 1, 2, 3, ... , with probabilities Pt,P2,P3, ... , where Pk is the probability that X equals k (k = I, 2, 3, ... ). Suppose also that Pk '" 0 and that ~: I Pk ~ I. The expected 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.)

R. > (1/2)R. (Source: Prescribing Safe and Efficnve Dosage, B. Horelick and S. Koont, COMAP, Inc., Lexiogton, MA.) 34. Time b_een drug doses (ContinUfJtion ofExercise 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 conccotration that is sale (not above CH ) but effective (not below CL)' See the accompanyiog figure. We therefore want to fmd values for Co and to for which

6

I

P1~

e. Ifk ~ 0.01 h- 1 and to ~ IOh, fmd the smallestn such that

5<--1 b.Pk=1

a. Pk=2.... e.

609

I

C

I

k(k+ 1) ~k- k+ I

Highest safe !evel

o 33. Safe aud effective dosage

t

The concentration io the blood resultiog from a siogle dose of a drug nonnally decreases with time as the drug is eliotioated from the body. Doses may therefore need to be repeated periodically to keep the conccotration from dreppiog below some particular level. One model for the effect of

Co

.j.

I

I

Lowest effective level

~to~

repeated doses gives the residual concentration just before the (n

+

I )st dose as

Rn

Cae-otto

=

+ Coe-2tto + ... + Cae-nkto,

where Co = the change io concentration achievable by a single dose (mgJmL), k ~ the elimination constant (h-I ), and to ~ time between doses (h). See the accompanying figure.

Thus Co = CH - CL . When these values are substituted io the equation for R ohtsioed io part (a) of Exercise 33, the resultiog equation simplifies to

C

C1 = Co + Cae- kto

\

C,

To reach an effective level rapidly, one might administer a "Ioadiog" dose that wuu1d produce a conccotration of CH mg/mL. This could be followed every to hours by a dose that raises the concentration by Co ~ CH - CL mg,ImL. a. VerifY the preceding equation for to.

o

b. If k = 0.05 h- 1 and the highest safe conccotration is e times the lowest effective conceotration, fmd the lcogth of time b _ doses that will assure safe and effective

to Time (h)

a. Write Rn in closed form as a single fraction, and fmd R = litnn-+oo Rn . b. CaicuiateR, and RIO for Co = I mg/mL, k = 0.1 h- I , and to ~ 10 h. How good an estimate of R is RIO?

Chapter

•

concentrations. e. Giveo CH ~ 2mg,1mL, CL ~ 0.5 mgImL, andk ~ 0.02 h- I , determioe a scheroe for administeriog the drug. d. Suppose that k = 0.2 h-I and that the smallest effective concentration is 0.03 mgJmL. A siogle dose that produces a concentration of 0.1 mgfmL 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. ~lor PolynomilllApproxinultions

ofa Function

A graphical animation shows the convergeoce of the Taylor polyoontials to functions 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) = (J(t), g(t» at any time t.

DEFINmON

Ifx andy are given as functions x

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

= f(t),

y

= get)

over an interval loft-values, then the set of points (x, y) = (J(t), get»~ 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

ILl

611

Parametrizations of Plane Curves

interval for a curve, we say that we have parametrized the curve. The equations and interval together constitote 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

+ 1,

-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 11.1. 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 seLected values of t. I

x

y

-3 -2 -1 0 1 2 3

9 4 1 0 1 4 9

-2

t

~

3

(9.4)

t

=0 x

(4, -I)

t

0 I

~-3

FIGURE 11.2 The curve given by the parametric equations x = and y = t

,2

2 3 4

-

_(9,-2)

1=-2

-I

+

1

(Example I).

EXAMPLE 2 Identify geometrically the curve in Example 1 (Figure 11.2) by eliminating the parameter I and obtaining an algebraic equation in x and y. Solution We solve the equation y = I + 1 for the parameter I and substitote the result into the parametric equation for x. This procedure gives I = Y - 1 and

x = I' = (y - I)' = y' - 2y

y

+ 1.

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. _ t

=0

-+~~nf-~-+(I',,;O);-x

t

= 31T 2

FIGURE 11.3 The equations x ~ cos t and y = sin t describe motion on the circle x' + y' = I. The arrow shows the direction of increasing t (Example 3).

EXAMPLE 3

Graph the parametric curves

(a) x = cos I,

y

(b)x=acosl,

Y =

=

sint, asint,

o ,,;; I ,,;; 2".. o ,,;; I ,,;; 2"..

Solution (a) Since

x' + y' = cos'l + sin'l =

I, the parametric curve lies along the unit circle x' + y' = 1. 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) Forx = acost,y = a sinl, 0:5 1:5 2"., we have x' + y' = a'cos'l + a' sin'l = a'. The parametrization describes a motion 1hat begins at 1he point (a, 0) and 1Iaverses 1he circle x' + y' = a' once counterclockwise, returning to (a, 0) at I = 2".. The graph is a circle centered at the origin wi1h 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-plane 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 1he interval t '" 0 describe 1he pa1h of a particle 1hat traces 1he righthand half of1he 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 recognizable 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 enlite 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 flISt quadrant as I increases (Figute 1104). 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

-----'::"!"£-------->x FIGURE 11.5 The pa1h 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 ~-plane as we had in Example 4. However, since the parameter I here can now also be negative, we obtain the lefthand 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 (Figute 11.5). • Notice that a parametrization also specifies when (the value of the parameter) a particle 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 we would obtain by the technique used in Example 5 when I = x . However, both parametrizations give the same line.

•

Parametrizations of Plane Curves

11,1

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 pointP(x,y) if 1 X=I+ t ,

values of t.

t

1ft

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 4.8 9.9

613

1 y=l- t ,

1>0.

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 1. 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

x

+y

=

(I + +) + (I - +)

=

21.

We can then eliminate the parameter I by multiplying these last equations together: y

10

(x - y)(x t ~

+ y)

(t)t21)

=

=

4,

10

~0.1,9.9)

or, multiplying together the terms on the left-hand side, we obtain a standard equation for a hyperbola (reviewed in Section 11.6):

5

x' -

y' = 4.

(1)

Thus the coordinates of all the points P(x, y) described by the parametric equations satisfY Equation (1). However, Equation (1) 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 ofthe hyperbola given by Equation (1), points where the x-coordinate would be negative. For small positive values of I, the path lies in the ~0.1,-9.9) -10 fourth quadrant and rises into the first quadrant as I increases, crossing the x-axis when t ~ 0.1 I = 1 (see Figure 11.6). The parameter domain is (0, 00) and there is no starting point and _ FIGURE 11.6 The curve for x ~ I + (1/1), no terminal point for the path. Y = I - (1/1), I > 0 in Example 7. (The Examples 4,5, and 6 illustrate that a given curve, or portion ofit, can be represented by part shown is for 0.1 '" I '" 10.) different parametrizations. In the case of Example 7, we can also represent the right-hand branch ofthe hyperbola by the parametrization -c+'----+--"c----'"'IO~---x

x

=

v'4+f2,

y

=

t,

-00


00,

which is obtained by solving Equation (1) for x ;" 0 and letting y be the parameter. Still another parametrization for the right-hand branch of the hyperbola given by Equation (1) is x = 2 sec t,

y = 2 tan I,

This parametrization follows from the trigonometric identity sec' I - tan' I

x' HISTORICAL BIOGRAPHY

Christian Huygees (1629-1695)

y'

=

4 sec' I - 4 tan' I

=

4(sec' I - tan' I)

=

=

1, so

4.

As I runs between -71/2 and 71/2, 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 71/2. 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

:;;

?::::" )

... .. Cycloid

The prohlem 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 takes the bob to return to center (its lowest position). This does not happen ifthe 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 gnards that caused it to draw up as it swung away from center (Figure 11.7).

•

FIGURE 11.7 InHuygens'pendulum clock, the bob swings in a cycloid, so the frequency is independent ofthe 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 through which the wheel turns, measured in radians. Figure 11.8 shows the wheel a short while later when its base lies at units from the origin. The wheel's center C lies at (at, 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 6 in terms of t, we observe that t + 6 y

=

+ asin6.

31f/2 in the figure, so that

6 3; - t. =

This makes

o FIGURE 11.9 The cycloid curve x = a(t - sint),y = a(1 - cost), for t ~ O.

cos 6

=

cos

(3; -

t) = -sint,

'6 = sm ,(31f) sm T - t = -cost.

The equations we seek are x = at - asint,

y=a-acost.

These are usually written with the a factored out: x = a(t - sint),

B(a7T,2a)

y

y = a(1 - cost).

Figure 11.9 shows the 1mt arch of the cycloid and part of the next.

To study motion along an upside-down cycloid under the influence of gravity, we tum Figure 11.9 upside down. This points the y-axis in the direction ofthe gravitational force and makes the downward y-coordinates FIGURE 11.10

positive. The equations and parameter

interval for the cycloid are still x y

= a(t - sint), = a(l- cost),

t'" 0,

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 relates to the origin 0 and the point B at the bottom 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 property 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 ("taw-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 only 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,

mgy

tmv' - tm(O)'.

=

Thus, the velocity ofthe bead when it reaches (x, y) has to be v =

yzgy.

That is, da is the arc length differential along the bead.. path.

or ds dt = - - =

yzgy

VI +

(dy/dx)' dx

--~"=-"-----

yzgy

The time Tf it takes the bead to slide along a particular path y = j(x) from 0 to B(a"ll", 2a) is _lx~aw

Tf-

1

x-a

+ (dy/dx)' Z

gy

(3)

dx.

What curves y = j(x) , if any, minimize the value ofthis integral? At {lISt sight, we might guess 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 conld travel a longer path and still reach B {lISt. Indeed, this is the right idea. The solution, from a branch of mathematics known as the calculus ofvariations, is that the original cycloid from 0 to B is the one and only brachistochrone for 0 and B. While the solution ofthe 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/dt divided by the derivative dx/dt. Making the derivative calco1ations and substitoting into Equation (3) (we oruit the details ofthe calco1ations here) gives x~aw

Tcycloid =

=

lx-o

1

+ (dy/dx) Z

gy

dx

,=w a (Z - Zcos t) dt t~O Zga( 1 - cos t)

l

From Equations (2),

tb;/dt = a(1 - cost), dy/dt = a sin I, and y = aO - cost)

Thus, the amount of time it takes the frictionless bead to slide down the cycloid to B after it is released from rest at 0 is "ll" 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 to > O. The bead's velocity at any later point (x, y) on the cycloid is

v

=

VZg(y - YO) = V'Zg-a-(c-c-os-t-o---co-s--c t ).

y = 0(1 - 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

a2(2 - 2 cost) ~lw I - cost dt= dt 2ga(cos to - cost) g ~ cos to - cost

~

-J

~ (W

-

- VgJ,o

2

2 sin (t/2)

fc; (W =

VgJ,o

sin (t/2) dt Vcos2 (to/2)

fc;l'=W

=

Vg '='0

dt

(2cos2 (to/2) - I) - (2cos 2(t/2) - I) - cos2(t/2) u -2 du c

-2du Ve 2 - u 2

= cos ('/2) = sin ('/2) d, = cos ('0/2)

. _IU]' -_~[ g -sm C =W 1=10

~

fc; [ . -I

cos (t/2) ]W cos (to/2) '0

=

'1/ g

=

~ (-sin-1 0 + sin-I I) = '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 oftime 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 ofthe amplitode of the swing.

FIGURE 11.11 Beads released simultaneously 00 the upside-down eycloid at 0, A, and C will reach B at the same time.

Exercises 11.1 Finding Carte';an from Parametric Equations

15. x = sec'l - 1, y = tan I,

Exercises 1-18 give parametric equatioos and parameter intervals for the motioo of a particle in the xy-plane. Identify the particle's path by fmding a Cartesian equatioo for it. Graph the Cartesian equation. (The graphs will vary with the equatioo used.) Indicate the portioo of the graph traced by the particle aod the directioo ofmotioo.

1. x

= 3t,

y = 912,

-00


4.x=3-3t, y=2t,

S. x

= cos 21,

6. x

=

y

= sin 2/,

to.

x


O::s t ::s

sint, Y

=

1 + sint, Y

=

cos2t, =

1-2 12. x=t=1' y= 1+ l' I

13. x

= I,

14. x

="\I't+l,

Y

="\Il=7, y

0:5 1:5".

0:5 1:5 2". 0:5 t:5 2'7T

-f:5 It:5

cost- 2, 0:5

Y = ,6 - 21 4,

11. x = t 2 ,

00

1T'

oos(". - I), Y = sin(". - I),

=

= Vi,

-00


t:5w 00

-I <1< 1 -1:5 1:5 I '"

°

"./2

-"./2 < 1<"./2

17. x = -cosht, Y = sinht, 18. x = 2 sinh I, Y = 2 cosh I,

-00


00

-00 < 1<00

19. Find parametric equations and a parameter interval for the motioo ofaparticle that starts at (a, 0) aod traces the circle x' + y' = a'

0:51:51

7. x = 4 cos I, Y = 2sinl, 8. x = 4sint, y = 5eas!, 9. x

-00

< I<

finding Parametric Equations

00

°

2. x = -Vi, y = I, I '" 3. x = 2t - 5, y = 41 - 7,

16. x = -seCI, Y = tanl,

-"./2

°

a. once clockwise. c. twice clockwise.

b. once counterclockwise.

d. twice counterclockwise.

(There are many ways to do these, so your answers may not be the same as the ooes in the back of the book.) 20. Find parametric equations aod a parameter interval for the motioo of a particle that starts at (a, 0) aod traces the ellipse (x'/a') + (y'/b') = I

a. once clockwise. b. once counterclockwise. c. twice clockwise. d. twice counterclockwise. (As in Exercise 19, there are maoy correct aoswers.) In Exercises 21-26, fmd a parametrizatioo 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 ~ x 2

+ 2x

25. the ray (half line) with initial point (2, 3) that passes through the point (-I, -I)

Parametrizations of Plane Curves

34. Find aparametrizatioo for the circle x 2 + y2 = I starting at (1,0) aod moving counterclockwise to the remtinal point (0, I), using the aogle 8 in the accompaoying 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 starting 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 2 in the following way: beginning at (0, 0) it moves to (3, 9), aod then travels back aod forth from (3, 9) to (-3, 9) infmirelymaoy 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 tangent 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 (0, 0) to the point (x, y). 31. Find a parametrizatioo for the line segment joining points (0, 2) aod (4,0) using the aogle 8 in the accompaoying figure as the parameter. y

617

- rf (1,0) x

-2

35. The witch of Maria Agoesi The bell-shaped witch of Maria Agncsi cao be construcred in the following way. Stsrt wi1h a circle of radius I, cenrered at 1he point (0, I), as shown in the accompaoying figure. Cl100se a paintA 00 the line y ~ 2 and 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 borizootalline through B. The witch is 1he curve traced by P as A moves along 1he line y = 2. Find parametric equations and a panuneter interval for the witch by expressing the coordinares ofP in rerms oft, the radian measure of the angle that segment D A makes with 1he positive x-axis. The following equalities (which you may assome) will help. s. x = AQ 1>. y = 2 - AB sin t c. AR·DA = (AQ)2 y

A

--=-f"---"----------'--+x

o

4

32. Find a parametrizatioo for the curve y = ~ with remtinal point (0, 0) using the aogle 8 in the accompaoying figure as the parameter.

-----~'::t~--L------+x

o

y y~vX

/

/~

--=-fL--'8"----------~x

o

33. Find a parametrizatioo for the circle (x - 2)2 + y2 = I starting at (I, 0) aod moving clockwise 0000 around the circle, using the central aogle 8 in the accompaoying figure as the paramere.. y

o

36. Hypocycloid When a circle rolls 00 1he inside of a fixed circle, any point P 00 the circomferenoo of 1he rolling circle describes a hypocycloid. Let 1he fixed circle be x 2 + y2 = 0 2, let the radius of the rolling circle be b, aod let the initial positioo of the tracing point P be A (0, 0). Find parame1ric equatioos for the hypocycloid, using as the parameter the angle 8 from the positive x-axis to the line joining 1he circles' cenrers. In particular, if b = 0/4, 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 t that the line ON makes with the positive y-axis. y

x

43. Parahola

x

44. Cycloid

y

=

,2 _ 1,

t - sin t, y

=

1 - cost, over

~

~

2t

+ 3,

a. 0 ::s t ::s 21T c. '1T ::s t ::s 31T.

-2:5

t :5

2

h. 0 :5 t :5 4'lf

45. Deltoid

A(O, a)

x

= 200st + cos2t,

y = 28int - 8in2t;

0:5 t::s 21T

Whathappeos if you replace 2 with -2 in the equatioos for x aod

y? Graph the new equatioos and fmd out 46. A nice curve x = 3cost

----~~o:;of'~-----~x

+ cos3t,

y = 3sint - sin3t;

0:5 t:5 2'1f

What happeos if you replace 3 with - 3 in the equatioos for x aod 38. Trochoids 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 ofthe 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 \Wen b ~ a.

y? Graph the new equatioos and fmd out 47. a. Epicycloid

x = 9cost - cos9t, y = 9sint - sin9t; h. Hypocycloid

Distance Us;ng Parametric Equations 39. Find the point 00 the parahola x = t,Y = t 2 , -00 < t < 00, closest to the point (2, 1/2). (Hint: Minimize the square of the distance as a functioo of t.)

x = 8east + 200841, y = 8sint - 2sin4t;

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). (Hint: Minimize the square of the distance as a functioo of t.)

5sin3t;

D GRAPHER EXPLORATIONS If you have a parametric equatioo grspher, grsph the equatioos over the giveu intervals in Exercises 41-48. x = 4 cos t,

41. Ellipse

a. 0 C.

~

t

~

2'17"

Y

= 2 sin t,

over b. 0 :s;; t

~ '11'

0:5

t:5

2w

c. Hypotrochoid

x = cost + 500831, y 48. a. x = 6cost

o :s; t

= 6cost -

+ 5cos3t,

0:5 t:5 2w

Y = 6sint - 5sin3t;

:s; 2w

h. x = 6 cos 2t ::s t ::s 1T

o

+ 5cos6t, Y

= 6sin2t - 5sin6t;

c. x = 6cost + 5cos3t, Y = 6sin2t - 5sin3t; 0:5t:52w

d. x

=

6 cos 2t

+ 5cos6t,

y = 6sin4t - 5sin6t;

O::st:51T

-'1f/2 :5 t :5 '1f/2.

42. Hyperhola hranch x = sec t (euter as I/cos (t)), y ter as sin (t)/cos (t», over a. -1.5 :5 t :5 1.5

0:5 t:5 2'1f

= tan t (eu-

h. -0.5 :5 t :5 0.5

c. -0.1 :5 t :5 0.1.

11.2

Calculus with Parametric Curves In this section we apply calculus to parametric curves. Specifically, we f"md slopes, lengths, and areas associated with parametrized curves.

Tangents and Areas A parametrized curve x = /(t) and y = g(t) is differentiable at t if / and g are differentiable at t. At a point on a differentiable parametrized curve where y is also a differentiable function of x, the derivatives dy/dt, dx/dt, and dy/dx are related by the Chain Rule:

dy dy dx dt=dx°dt'

11.2

619

Calculus with Parametric Curves

If dx/dl oF 0, we may divide both sides of this equation by dx/dl to solve for dy/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. (1) 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

dy'/dl dx/dl'

-

EXAMPLE 1

(2)

- -

Find the tangent to the curve

--ccf---/-+-----'--_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 ofthe hyperbola xl -

y2

=

Solution

The slope ofthe curve at I is

1.

dy dx

dy/dl dx/dl

sec21 sec I tan I

sec t

tan I'

Eq. (1)

Setting I equal to 7T/4 gives dy I sec (7T/4) dx t~"/4 = tan (7T/4) =

'?

=

v2.

The tangent line is y - 1= y = y =

EXAMPLE 2

v2 (x - v2) v2x - 2 + 1 v2x - 1.

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

I Finding tJ'y/dx

2

in Term. of I 1. Expressy' = dy/dx in terms of I.

Differentiate y' with respect to I.

2.

dy' = ~ dl dl

2. Finddy'/dl. 3. Divirkdy'/dlbydx/dl.

3.

(II -

312) = 2 - 61 + 61 2 21 (I - 21)2

Derivative Quotient Rule

Divide dy'/ dl by dx/ dl. (2 - 61 + 61 2)/( I - 21j2 I - 21

EXAMPLE 3

y

2-61+61 2 (I - 21)3

Eq. (2)

•

Find the area enclosed by the astroid (Figure 11.13)

x = cos3 t,

y

=

o ,,;; I

sin3 t,

,,;; 2'IT.

By symmetry, the enclosed area is 4 times the area beneath the curve in the first quadrant wbere 0 ,,;; I ,,;; 'IT/2. We can apply the def"mite integral formula for area studied in Chapter 5, using substitution to express the curve and differential dx in terms of the parameter I. So,

Solution ---J.o~--~+-------'"~~x

-I

A =

-I

Figure 11.13 The astroid in Example 3.

41

1

ydx

("/2

= 4 10

=

121"/2

=

2 10

=

2 10

=

2

3 {"/2 3 {"/2

y

=

c

3 2 sin "3 cos I sin Idl

=

Substitotion fOT y and en

e-~os ye ~os 21

+

21) dl

sin4 t

=

(1 - 2cos 21)'

(I - 2 cos 21 + cos2 21)(1 + cos 21) dl

Expand .quare term.

(I - cos 21 - cos2 21 + cos3 21) dl

Multiply term•.

3 [ {"/2 {"/2 2 {"/2 3 ] 10 (I - cos 21) dl - 10 cos 21 dl + 10 cos 21 dl

~ [(I - t sin 21) -

t (I i +

t

sin 21) +

M(; -0- 0- 0) - t (;

+

(Sin 21 - tSin321)r

0- 0- 0) + t(O - 0- 0+ 0)]

3'IT

Section 8.2, Example 3

Evaluate.

•

8'

Length of a Parametrically Defined Curve P,

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

FIGURE 11.14 The smoo1h curve C defmed parametrica11y by the equations x = f(l) aod y = g(I), a :5 1 :5 b. The length ofthe curve from A to B is approximated by the sum ofthe lengths of the polygonal path (straight line segments) starting atA = Po, then to P, , aod so on, ending at B = p •.

Let C be a curve given parametrically by the equations

x = 1(1)

and

y

=

g(1),

a";;

I";; b.

We assume the functions I 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 corners 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" P2 , ... , P n = B (Figure 11.14). These points correspond to a partition of the interval [a, b] by a = 10 < 11 < 12 < ... < In = b,

11.2

Calculus with Parametric Curves

621

where Pk = (t(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(t,»

Lk = Y(dxk)2

+

(dYk)2

= Y[j(lk) - /(lk_I)]2

--::+----------_x o

+

[g(lk) - g(lk-I)]'

(see Figure 11.15). If M k is small, the length Lk is approxilnately the length of arc Pk-IPk. By the Mean Value Theorem there are numbers I; and I;' in [11c- h Iii such that dxk = /(Ik) - /(Ik-l) = 1'(1;)

FIGURE 11.15 The arc P'-lP, is approximated by the straight line segment shown here, which has length

L, =

y'(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 lenda to zero and the number of segments n ---> 00, is the definite integral

.

lim ~ Y[j' (I;)]'

1I111~O

t;1

+

[g'(I;')]' dlk =

l

b

Y[j'(I)]'

a

+

[g'(I)]' dl.

Therefore, it is reasonable to define the length of the curve from A to B as this integral.

DEFINmON 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 L =

l

b

Y[j'(1)]2 + [g'(1)]2 dl.

A smooth curve C does not double back or reverse the direction of motion over the time interval [a, b] since (t')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 ofthe 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 circumfer-

L=

(dy)2 dt + dt dt. (dx)2

2"

1 o

We find

dxdt

=

dy

.

-rsmt,

dt = rcost

and

So

L =

EXAMPLE 5

f2"

10 W

dt =

r[t ]~" =

•

21Tr.

Find the length of the astroid (Figure 11.13)

x

= cos 3 t,

y =

o ,,;; t

8in3 t,

,,;; 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 = cos 3 t,

(:Y

= [3 cos

(:y

y

3 = 8in

2 t( -sin t)f = 9 cos4 t sin2 t

= [3sin2 t(cost)12 =

=

t

9sin4 tcos2 t

Y9 cos2 t sin2 t

= 31costsintl =

costsint~

3costsint.

ofor

OStS'Tr/2

Therefore, Length of first-quadrant portion =

10f"/2 3 cos t sin t dt 3

f"/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) sin2t

•

11.2

Calculus with Parametric Curves

623

HISTORICAL BIOGRAPHY

Length of a Curve y = f(x)

Gregory St. Vincent (1584-1667)

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 and

x=1

a

y = 1(1),

~

t

~

b,

a special case of what we considered before. Then, ax = I dl

it

and

=

/'(1).

From Equation (I), we have dy dy/dl ax = ax/dl = /'(1),

giving

(:)' +

(it)'

=

1

+

[f'(I)]'

=1+[f'(x)]2.

t~.

Substitution into Equation (3) gives the arc length formula for the graph of y sistent with Equation (3) in Section 6.3.

= I(x),

con-

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(I) =

l'

v'[f'(z)]'

+ [g'(z)]' dz.

Then, by the Fundamental Theorem of Calculus,

tis dz

=

v'[f'(I)]'

+

[g'(1)]2 =

~(ax)2 dz +

(dy)2

dz·

The differential of arc length is

~ ( : ) ' + (it)' dl.

ds =

(4)

Equation (4) is often abbreviated to

tis

=

v'ax 2 + dy 2.

Just as in Section 6.3, we can integrate the differential tis 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 ofthe first-quadrant arc ofthe 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 the line y = x, so X = ment of the curve (Figure 11.16) has mass

dm= l'ds= )(:Y +

y. A typical seg-

From Example 5

(tYdl=3COSISinldl.

The curve's mass is -4--'------=-.....c~x

o

A(I,O)

f"/2

The centroid (c.m.) of the astroid arc in Example 6. FIGURE 11.16

M =

10

dm

f"/2

=

10

3 cos I sin I

dl

3

=

2·

Again from Example 5

The curve's moment about the x-axis is Mx =

Jydm

=

10f"/2 sin3 1 '3coslsinldl

= 3

1

,,/2

o

sin4 I cos I

dl

. 5 ]"/2 3 = 3. sm 1 = -.

5

0

5

It follows that

__ M x _ 3/5 _ I y - M - 3/2 - 5·

•

The centroid is the point (2/5,2/5).

Areas of Surfaces of Revolution In Section 6.4 we found integral formulas for the area of a surface when a curve is revolved about a coordinate axis. Specifically, we found that the surface area is S= tis for revolution about the x-axis, and S = tis for revolution about the y-axis. If the curve is parametrized by the equations x = /(1) and y = g(1), a ,;; I ,;; b, where / and g are continuously differentiable and <1')2 + (g')2 > 0 on [a, b], then the arc length differential tis is given by Equation (4). TIris observation leads to the following formulas for area of surfaces of revolution for smooth parametrized curves.

!21fY

!21fx

Area of Surface of Revolution for Parametrized Curves If a smooth curve x = /(I),y = g(I), a ,;; I';; b, is traversed exactly once as I increases from a to b, then the areas of the surfaces generated by revolving the curve about the coordinate axes are as follows.

1. Revolution about the x-axis (y S=

;a"

[21fY

2. Revolution about the y-axis (x

;a"

0):

""c----o-;"

(:Y + (tYdl

(5)

0): ~~c--~~

S=

[21fX

(:Y + (tYdl

(6)

As with length, we can calculate surface area from any convenient parametrization that meets the stated criteria.

625

11.2 Calculus with Parametric Curves

EXAMPLE 7 The standard parametrization of the circle of radius 1 centered at the point (0, 1) in the xy-plane is

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

S

x

=

l

a

b

y (dx)2 dt + (ddt )2 dt

21TY

Eq. (5) fOT revolution about the x-axis;

y=l+sint;;=:Q

(21r

= Jo 21T( 1 + sin t) V~ -sin tf . + (cos t~2 dt 1

{21r

= 21TJo (1 + sin t) dt FIGURE 11.17 In Example 7 we calculate the area of the surface ofrevolution swept out by this parametrized curve.

•

Exerdses 11.2 + 2x

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 + t, Y + 2t 3 = 2x + t 2 ,

4. x = cos t, Y = 5. x = t,

v3 cos t,

7. x = sect, 8. x =

-v"t+!,

9. x = 2t

2

+ 3, y

= t

4

t = 2Tr/3

,

1

t+1'

14. x = t

+ e t,

Area x = aCt - sin t),

x

t y =~, y = 1 - et,

=

a cos t, y

24. Find the area under y parametrizations.

t= 3 t = -1

+ sint, t

a. x = t 2 , Y = t 6

=

t = 2 t = 0

ImpLidtLy Defined Parametrizations

Vi, yet - 1)

17. x + 2x 3/ 2 = t 2 + t,

=

Vi,

t = 4

yv"t+! + 2tV;; = 4,

=

=

b sin t,

0

~

t

~

2Tr.

x 3 over [0, 1] using the following

b. x = t 3 , Y = t 9

Find the lengths ofthe curves in Exercises 25-30. 25. x = cos t, Y = t + sin t, 0 ~ t ~ Tr 26. x = t3, y = 3t 2/2, 0 ~ t ~ v'3 27. x = t 2/2, Y = (2t + 1)3/2/3, 0 ~ t ~ 4

= Tr/2

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 + 2t 2 = 9, 2y 3 - 3t 2 = 4, t = 2

Y5 -

+ e-t •

Lengths of Curves

+ 3)3/2/3, y = 8 cos t + 8t sin t

28. x = (2t

16. x =

y = a(l - cos t).

23. Find the area enclosed by the ellipse

1 11. x = t - sin t, y = 1 - cos t, t = Tr/3 12. x = cost, Y = 1

Tr

t = 0

x = t - t2, Y = 1

V3t,

+ In t, t

t = t = 1

21. Find the area under one arch of the cycloid

t = Tr/6

y =

10. x = l/t, Y = -2

13. x =

20. t = In (x - t), Y = tet,

Vi,

y = tant,

t sin t - 2t = 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 Y =

= t,

t= 0

29. x y

=

o~

8 sin t - 8t cos t, 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 = 2Vt,

0 '" I '"

\/3;

y

y-axis

2

33. x = I + \/2, Y = (1 /2) + \/21, -\/2 '" I'" \/2; y-axis 34. x = In (sec I + tan I) - sin I, Y = cos I, 0 '" I '" 1f/3; x-axis

3

35. A cone frostom 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 ~ I + 1,0'" I '" I. Checkyourresultwiththegeometry fannula: Area = 1f(rl + ,.,)(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 T. Find the cone's surface area with the parametric equations x = ht, Y = rf, 0 :5 t :s; 1. Check your result with the geometry fannula: Area = 1fr(slant height).

44. The curve with parametric equations x=t, y= I-cost, Ost:52w is called a sinusoid and is shown in the accompanying figure. Find the point (x, y) where the slope of the tangent line is a. largest b. smallest

Centroids 37. Find the coordinates of the centroid of the curve

x = cost + tsint, Y = sint - teost,

0 s t :511'/2.

y

38. Find the coordinates of the centroid of the curve

x = e t cos t, y = e t sin t, 0:5 t s

2

1T'.

39. Find the coordinates of the centroid of the curve x

D 40.

= cos t, Y = t + sin t, 0:5

t :5 'Tl'.

Most centroid calculations for curves are done with a calculator or computer that has an integral evaluation program. As a case in point, fmd, to the nearest hundredth, the coordinates of the centroid ofthe curve

x = I', Y = 312/2,

0 '" I '"

-+-"----------'~---->x

o

D

0.

Theory and Examples 41. Length is independent of parametrization To illustrate the fact that the numbers we get for length do not depend on the way we parametrize our curves (except for the mild resmctions preventing doubling back mentioned earlier), calculate the length of the senticircle y ~ 'V'1=7 with these two different parametrlzations:

a. x = cos 21, y = sin 21, 0 '" I '" 1f/2 b. x = sin 1f1, Y = cos 1fI, -1/2 '" I '" 1/2 42. a. Show that the Cartesian fannula

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 flIst quadrant where the tangent to the curve is horizontal, and find the equations of the two tangents at the origin.

45.

46. y

x =

y

=

y

sint sin2t

x = sin2t

y=sin3t

--+-C~--1'---+---->x

47. Cycloid for the length of the curve x ~ g(y), c '" y '" d (Section 6.3, Equation 4), is a special case of the parametric length fannula

a. 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 ~ I. Use this result to fmd the length of each curve. b. x

c. x

= y'/2, 0 '" y '" 4/3 = y2/3, 0'" y '" I

t

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

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

a. 8 = O.

b. 8 = 1f/2.

c. 8 = 4'1f/3.

COMPUTER EXPLORATIONS In Exercises 49-52, use a CAS to perfann the following steps for the

given curve over the closed interval.

a. Plot the curve together with the polygoual 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 ofthe curve by summing the lengths ofthe line segments.

Polar Coordinates

627

I ,

3

49.x- 3 1, Y=ZI, 0"'1"'1 SO. x ~ 21 3 - 161'

e. Evaluate the length ofthe 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

+ 251 + 5,

Y ~ I'

0"'1"'6 = t - cost, Y = 1 + sint,

51. x 52. x

answer.

= etcost,

y

= etsint,

+I

-1r::S

O::s t::s

- 3,

ts

1T'

'T1'

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 calculating 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 angie from the initial ray tu ray OP.

Polar Coordinates P(r, II) Directed distance from 0 to P

oL----,--'------~x

Initial ray

9=0

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 angie 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 ofpolar 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 = -1l1r/6 (Figure 11.19). In some situations we allow r to be negative. That is why 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 angie of 1r/6 radians with the initial ray, and mark the point (2, 1r/6) (Figure 11.21). We then find the angies 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

6 ±

21r,

7T

6 ±

41r,

7T

6 ±

61r, ....

628

Chapter 11: Parametric Equations and Polar Coordinates

(2. *) ~ (_2._ 5;) ~ (-2. 7;)

For r = -2, the angies are 5'1f 6 '

5". -5'1f - ± 2". _ 5". ± 4'IT -6 ± 6"., .... 6 ' 6 '

etc.

The correapoomng cocmfumtepmm of Pare

(2, ~ + 2n".).

~"---+-rLL~~-;--->X

..,

Initial ray

n

=

0, ±I, ±2, ...

n

=

and

FIGURE 11.21 The point P(2. 'If/6) has inf'mitely many polar coordinate pairs

(Example I).

5". ( - 2'-6

+ 2n".)

'

0, ±I, ±2, ....

When n = 0, the formulas give (2,,,./6) and (-2, -5"./6). When n = I. they give (2,13"./6) and (-2, 7"./6), and so 00. _

Polar Equations and Graphs If we hold r fixed at a constant value r = a oF O. the point P(r, 6) will lie Ia Iunits from the origin O. As 8 varies over any interval ofiength 2"., 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, 6) traces the line through 0 that makes an angle of measure 60 with the initial ray. FIGURE 11.22 The polar equation for a circle is r = a.

Equation

Graph

r=a

Circle ofradius Ia Icentered at 0 Line through 0 making an angle 60 with the initial ray

6 = 60 y

(a)

1=sr:S2.0:S(J:S~

EXAMPLE 2 -----cc+--'---L~X

o

1

2


_

y

(b)

Equations of the form r = a and 6 = 60 can be combined to derme regions, segments, and rays. --~'c--=-~X

EXAMPLE 3

Graph the sets of points whose polar coorilinates satisfy the following

conditions.
0';;6';;;

and

/ /

(e)

2"

y

(h) -3';;r,;;2

3

< 6 < 5". (0) 2". 3 - 6 ----:'\-=---~X

Solution

and

6 = ". 4

(no restriction on r)

The graphs are shown in Figure 11.23.

-

Relating Polar and Cartesian Coordinates FIGURE 11.23 The graphs oftypical inequalities in r and 8 (Example 3).

When we use both polar and Cartesian coormnates in a plane, we place the two origins together and take the initial polar ray as the positive x-axis. The ray 6 = "./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, 6)

Equations Relating Polar and Cartesian Coordinates r

Common origin

y

x = rcosO,

6

FIGURE 11.24 The usua1 way to relate polar and Cartesian coordinates.

tanO =)' x

y = rsinO,

The first two of these equations uniquely determine the Cartesian coordinates x and y given the polar coordinates r and O. 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 0 E [0, 2".) satisfying the first two equations, each then giving a polar coordinate representation ofthe 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 coordinates. Polar equation

Cartesian equivalent

rcosO = 2

x=2

r cos 0 sinO = 4

xy = 4 x 2 _ y2 = I

2

r 2 cos2 0

-

r = I

r 2 sin2 0 =

I

y2-3x 2 -4x-1 =0

+ 2rcosO x 4 + y4

r = I - cosO

+ 2x"y2 + 2x3 + 2xy2

_

y2 = 0

Some curves are more simply expressed with polar coordinates; others are not. y

x'+(y-3)'~9

or

......,....... r

= 6 sin 9

Find a polar equation for the circle x 2

EXAMPLE 5 Solution

= 9 (Figure 11.25).

We apply the equations relating polar and Cartesian coordinates:

x2

+ (y

- 3)2 = 9

x 2 + y2 - 6y + 9 = 9 x 2 + y2 - 6y = 0 r 2 -6rsin6=O

------...."f-"'-------+x o

r = 0

FIGURE 11.25 The circle in Example 5.

+ (y - 3)'

•

or

r - 6 sinO = 0 r = 6 sin 0

Expand(y - 3)'. Cancellation

x2 +y2=r 2 Includes both possibilities

• EXAMPLE 6 Replace the following polar equations by equivalent Cartesian equations and identifY their graphs. (a) rcosO = -4

(b) r 2 = 4rcosO 4 (c) r = 2eDse _ sinO Solution

We use the substitutions r cos 0 = x, r sin 0 = y, r 2 = x 2 + y2.

(a) rcosO = -4 The Cartesian equation: The graph:

rcosO = -4 x =-4

Vertical line through x = -4 on the x-axis

630

Chapter 11: Parametric Equations and Polar Coordinates

(b) r 2 = 4r cos 6 The Cartesian equation:

r 2 = 4r cos 6 x 2 + y2 = 4x x 2 -4x+y2=0

x 2 -4x+4+y2=4 (x - 2)' The graph:

(c) r

=

+ y2

=

Completing the squaro

4

Circle, mdius 2, center (h, k) = (2, 0)

4 2cosO - sinO

The Cartesian equation:

r(2 cos 6 - sin 6) = 4

2rcos6 - rsin6

=

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?

a. (3,0)

b. (-3,0)

c. (2,2'ff/3)

d. (2,7'ff/3)

e. (-3, 'ff)

f. (2, 'ff/3)

g. (-3, 2'ff)

h. (-2, -'ff/3)

2. Which polar coordinate pairs label the same point?

a. (-2, 'ff/3) d. (r,9

+ 'ff)

g. (-r, 9

+ 'ff)

b. (2, -'ff/3)

c. (r,9)

e. (-r, 9)

f. (2, -2'ff/3)

h. (-2, 2'ff/3)

3. Plot the following points (given in polar coordinates). Then rmd all the polar coordinates ofeach point.

a. (2, 'ff/2)

b. (2,0)

c. (-2, 'ff/2)

d. (-2,0)

4. Plot the following points (given in polar coordinates). Then rmd all the polar coordinates ofeach point.

a. (3, 'ff/4)

b. (-3, 'ff/4)

c. (3, -'ff/4)

d. (-3, -'ff/4)

Polar to Cartesian Coordinates 5. Find the Cartesian coordinates of the points in Exercise I. 6. Find the Cartesian coordinates of the following points (given in polar coordinates).

a.

('V2, 'ff/4)

c. (0, 'ff/2)

e. (-3,5'ff/6)

f. (5, tan-I (4/3))

g. (-I, 7'ff)

h. (2V3, 2'ff/3)

Cartesian to Polar Coordinates 7. Find the polar coordinates, 0 :5 9 < 2'ff and r '" 0, of the following points given in Cartesian coordinates.

a. (I, I)

b. (-3,0)

c. (V3, -I)

d. (-3,4)

8. Find the polar coordinates, -'ff :5 9 < 'ff and r '" 0, of the following points given in Cartesian coordinates.

a. (-2, -2)

b. (0, 3)

c. (-V3,I)

d. (5,-12)

9. Find the polar coordinates, 0 :5 9 < 2'ff and r :5 0, of the following points given in Cartesian coordinates.

a. (3, 3)

b. (-I, 0)

c. (-I, V3)

d. (4,-3)

10. Find the polar coordinates, -'ff :5 9 < 2'ff and r :5 0, of the following points given in Cartesian coordinates.

a. (-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.

b. (1,0)

11.r=2

12.0:5r:52

d. (- v'2, 'ff/4)

13.r"'l

14.I:5r:52

631

11.4 Graphing in Polar Coordinates 15. 0 :5 0 :5 '11'/6,

r "" 0

16. 0

~

2'11'/3,

17.0 = '11'/3,

-1:5 r:5 3

18. 0

= 11'11'/4,

19. 0 = '11'/2,

r "" 0

20. 0

= '11'/2,

21. 0 :5 0 :5 '11', r = I 23. '11'/4 :5 0 :5 3'11'/4,

r:5-2

r""-I r:5 0

22. 0 :5 0 :5 '11', r

= -I

43. r 2 45.

7

2

+ 2r 2 cos8sin8 = -4r cos (}

=

1

46. r' = -6rsinO

47. r = 8 sinO 49. r = 2cosO

48. r = 3 cosO

+ 2 sinO

SO. r = 2 cosO - sinO

51.rSin(0+~)~2

0:5 r :5 I

44. cos' 0 ~ sin' 0

52. rsin

(2; -

0)

~

5

24. -'11'/4:5 0 :5 '11'/4,

-1:5 r :5 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. S4.y~1 53. x ~ 7 55. x ~ y

26. 0 :5 0 :5 '11'/2,

1:5

2

Polar to Cartesian Equattons Replace the polar equations in Exercises 27-52 with equivalent Cartesian equations. Then describe or identifY the graph. 27. rcosO ~ 2 28. rsinO ~ -I 29. rsinO

~

30. rcosO

0

31. r=4cscO 33. rcosO

+ rsinO

~

0

32. r = -3 sec 0 ~

I

34. 78inB

=

36. r' ~ 4r sin 0

5 37.r=·0 0 sm - 2 cos

38. r'sin20 = 2

39. r

~

colOcscO

40. r

41. r

=

esc (J eTcos 9

42. rsinO

11.4

~

4tanOsecO ~

Inr

x2

+ In cos 0

57.x'+y'=4

y2

59. 9+4~ I

60.xy~2

61. y' = 4x

62.

63.

rcos8

35.r'~1

S6.x-y=3

x' + (y -

65. (x - 3)'

2)' = 4

+ (y + I)'

58.

x' - y' = I

x' + xy + y' =

I

64. (x-5)'+y'=25 ~ 4

66. (x

+ 2)' + (y

- 5)' ~ 16

67. Find all polar coordinales ofthe origin. 68. Vertical and horizontal line. a. Show that every vertical line in the xy-plane has a polar equationoftheformr = asec8. b. 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 describes techniques for graphing these equations using symmetries and tangents to the graph.

Symmetry Figure 11.26 illustrates the standard polar coordinate tests for symmetry. The following summary says how the symmetric points are related.

Symmetry Tests for Polar Graph.

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 (Figure 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 (Figure 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 (Figure 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) I I I

------;ck--+---~x

o

x = r cos 0 = 1(0) cos 0,

I

I

y = rsinO = 1(0) sinO.

I

(r, -8)

or(-r,'1r- 9)

culate dy/dx from the parametric formula

(a) About the x-axis (r,1r- 8) or (-r, -8)

If I is a differentiable function of 0, then so are x andy and, when dx/dO i' 0, we can cal-

Y (r,8)

dy

dy/dO

Section 11.2, Eq. (1)

dx

dx/dO

witht=(J

10 (j(0)' sin 0)

o

d dO (j(0)' cos 0)

df.

dii sm 0 + 1(0) cos 0

(b) About the y-axis

df

y

.

Product Rule for derivatives

dO cos 0 - 1(0) sm 0 (r,8)

Therefore we see that dy/dx is not the same as dl/dO.

o (-r, 8) or (r, 8

+ 1r)

Slope of the Curve r = 1(0)

(c) About the origin

FIGURE 11.26

dyl !,(O) sinO + 1(0) cosO dx (r,B) = !,(O) cosO - 1(0) sinO ,

Three tests for

symmetry in polar coordinates.

provided dx/dO i' 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

dy dx (0, B,) =

!' (00) sin 00 !' (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 = I - cos O.

The curve is symmetric about the x-axis because

(r, 0) on the graph => r = I - cos 0 => r = I - cos ( -0) => (r, -0) on the graph.

cos 8

~

cos (-8)

11.4 Graphing in Polar Coordinates

(J

As 0 increases from 0 to 1T, cos 0 decreases from I to -I, and r = 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 1 and r decreases from 2 back to O. The curve starts to repeat when 0 = 21T because the cosine bas 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 ofvalues 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-cosO

o

o

'!

1 2

3 17

1

'2

3

217

3

'2

17

2 (a)

Graph the curve r 2 = 4 cos O.

EXAMPLE 2

(17.

2)~---~--,,¥--~

x

Solution The equation r 2 = 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

(r, 0) on the graph

(b)

r = 1-

cos'!.O_~

633

=)

r2

=)

r 2 = 4 cos (-0)

=)

(r, -0) on the graph.

__~

= 4 cos 0

ces 0

~

coo (-0)

The curve is also symmetric about the origin because

(r, 0) on the graph

=)

r 2 = 4 cos 0

=)

(-r)2

=)

(-r, 0) on the graph.

(17. 2)-+---"2--"~-

(c)

FIGURE 11.27 The s1eps in graphing the cardioid r = 1 - cos 6 (Example I). The arrow shows the direction ofincreasing 8.

= 4cosO

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 r 2 = 4 cos 0 gives two values of r: r = ±2Ycos O.

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

0

0 +'!

-6

+'!

cosO r=

Y3 2 1

Y2

+'!

!

-3

+'!

-2

0:2

1

-4

r'~4cesO

±2Vcos61

2 0

=

±1.9

=

±1.7

=

±1.4

0 (a)

(b)

The graph ofr 2 = 4 cos 6. The arrows show the direction of increasing 6. The values of r in the tahle are rounded (Example 2). FIGURE 11.28

_

634 (aJ

Chapter 11: Parametric Equations and Polar Coordinates

A Technique for Graphing

,2

One way to graph a polar equation r = /(6) is to make a tahle of (r, 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

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 first Cartesian graph, even when hastily drawn, shows at a glance where r is positive, negative, and nonexistent, as well as where r is increasing and decreasing. Here's an example.

(b) ,

y~+P_from

~---,-lj------j---'--+'_

o

,~

(cJ

8

squareroots

EXAMPLE 3

Graph the lemniscate curve

r 2 = sin26.

-v'sin28 Solution Here we begin by plotting r 2 (not r) as a function of 6 in the Cartesian r 26-plane. See Figure l1.29a. We pass from there to the graph of r = ± sin 26 in the rIJ-plane (Figure lI.29b), and then draw the polar graph (Figure lI.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. •

V

y

USING TECH NOLOGY Graphing Polar Curves Parametrically FIGURE 11.29 To plot r = /(8) in the Cartesiao r8-plaoe in (b), we IITst plot ,2 = sin 28 in the ,28-plaoe in
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

Graph the lemniscates in Exercises 13-16. What symmetries do these

IdentifY the symmetries of the corves in Exercises 1-12. Then sketch the curves. 1. , = I + cos8 2. ,=2-2c058

curves have?

13. ,2 ~ 4 cos 28 15.

r2 ~

14. r 2 ~ 4 sin 28 16. r 2 ~ -cos 28

-sin 28

3. , = I - sin8

4. , = I + sin8

Slopes of Polar Curves

S. , = 2 + sin8

6.,=1+2sin8

7. , = sin (8/2)

8. , = cos (8/2)

Find the slopes of the corves in Exercises 17-20 at the given points. Sketch the corves along with their taogents at these points. 17. Cardioid ,= -I + cos8; 8 = ±'1T/2

9. r 2 = cosO

11. r 2

= -sin 8

10. ,2

~

12. r 2

= -cos 8

sin8

18. Cardioid r

=

-1

+ sin 8;

8

=

0, '1T'

19. Four-leaved rose r ~ sin28; 8 ~ ±'1T/4, ±3'1T/4 20. FOUl'-leaved rose ,= cos 28; 8 = 0, ±'1T/2, '1T

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 lima_s in Exercise 21. Equatioos for lima~oos have the form r = a ± b cos 8 or r = a ± b sin 8. There are four basic shapes. Lima~on.

21.

with an inner loop

a.r=~+co.6 22. Cardioid.

b. r =

a.7="2+oos 8

26. Sketch the regioo defined by the inequalities 0 and -1T/4 :5 6 :5 1T/4.

D 33. D 34.

:5 r :5

2 sec 6

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=8

b. r = -6 c. A logarithmic spiral: r = eO/lo d. A hyperbolic spiral: r ~ 8/6

e. An equilateral hyperbola: r ~ ± 1O/Y8 (Use different colors for the two branches.)

In Exercise. 27 and 28, sketch the region defmed by the inequality.

11.5

The nephroid of Freeth Graph the nephroid ofFreeth:

r=I+2sin~.

Graphing Polar Regions and Curves 25. Sketch the regioo defmed by the inequalities -I :5 r :5 2 and -1T/2 :5 6 :5 1T/2.

27. O~r~2-2cos8

Which of the following has the same graph as r = cos 26?

2

b. r = -2 + sin6

a. r=2+cos8

D 30.

:3_ .in6

Owllim~on.

24.

Which of the following has the same graph as r ~ I - cos 6?

a. r= -1- cosO b.r=l+cos6 Confmn your answer with algebra.

D D 32.

b.r=-I+sin6

23. Dimpled lim~on. 3

635

.. r = -sin (26 + 1T/2) b. r = -cos (6/2) Confmn your answer with algebra. 31. A rose within a ro.e Graph the equation r ~ I - 2 sin 36.

b.r=~+.in6

a.r=1-cos8

D 29.

Areas and Lengths in Polar Coordinates

28. O~r2~cos8

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

~r~f(e)

The region in Figore 11.30 is bounded by the rays 0 = " and 0 = (3 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 tiroes the area of a circle of

r.

ros.

radius 71, or 9=a

t

o

t

A. = rl fl.(}. = (J(O.))' fl.(}•.

~r"""'---_---L-'--------~.

The area of region

ors is approxiroately

FIGURE 11.30 To derive a formula for the area ofregion ars, we approximate the region with fan-shaped circular sectors.

If f is continuous, we expect the approxiroations 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( 9k») M k

=

1~If--+o IF 1

tt

=

FIGURE 11.31

(1(9»)' d9.

Area of the Fan-Shaped Region Between the Origin and the Curve , = /(0), a ,,; 9 ,,; P P A = t,2d9.

l

The area differential dA

for the curve, = f(9).

This is the integral ofthe area differential (Figure 11.31) y

dA =

-*------'t~x

t,2 t d9

=

(1(9»)' d9.

EXAMPLE 1 , =

Find the area of the region in the plane enclosed by the cardioid 2(1 + cosO).

We graph the cardioid (Figure 11.32) and determine that the radius OP sweeps out the region exactly once as 9 runs from 0 to 2". . The area is therefore

Solution

FIGURE 11.32

The cardioid in Example I.

1

8-2"

1

2,2d9

=

8-0

y

o

y

Upper limit (J

= wj2

10

2(1+2cos9+cos2 0)d9

1

2 "(2 + 4cos9 + 2 1 +

~OS29) d9

=

10f2" (3 + 4 cos 0 + cos 29) d9

=

. 29 [39 + 4sin9 + s~

-of=--------> X

= 1 - cos 8

1 2'4(1 + cosO)2d9

0

=

'1

2 "

f2"

=

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 between '2 and the origin.

1

]2" 0

=

•

6". - 0 = 6"..

To find the area of a region like the one in Figure 11.33, which lies between two polar curves '1 = '1(9) and '2 = '2(9) from 9 = a to 9 = f3, we subtract the integral of (1/2h2 d9 from the integral of (1/2),,' d9. This leads to the following formula.

Area of the Region 0 ,,; '1(0)";,,,; '2(0),a"; 0,,;

A

=

l

P t,,'d9 -

l

P

t,t'd9

=

l

P

p

th

2

-

,t') d9

(1)

-+----+--c;i'o--'--+--> x EXAMPLE 2 Find the area of the region that lies inside the circle, = 1 and outside the cardioid, = I - cos 9. We sketch the region to determine its boundaries and find the limits of integration (Figure 11.34). The outer curve is '2 = 1, the inner curve iS'1 = I - cos 9, and 9 runs from -"./2 to "./2. The area, from Equation (I), is

Solution FIGURE 11.34 The region and limits of integration in Example 2.

11.5 Areas and Lengths in Polar Coordinates

A

=

. 1

/2 1

-TT/2

= 2 0

=

0

/. 0

/.

Z (rl - r12 ) dO

../2 1

/. ../2

=

637

Z (rl - r12 ) dO

Symmetry

(1 - (1 - 2cosO + cos2 0)) dO

../2 (2 cos 0 - cos2 0) dO =

= [2sinO-

/."f2(2 cos 0

SqU8l"e rl.

0 - 1

+ ~os 20) dO

O sm ' 20 ]../2 0 =2-;. 4

•

Z_

The fact that we can represent 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 solution may reveal some intersection points without revealing others, so it is sometimes difficult to find all points ofintersection oftwo 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 fonnula for the length of a curve r = flO), a ,,; 0 ,,; (3, by parametrizing the curve as x = rcosO = flO) cosO,

y = r sin 0 = flO) sin 0,

a ,,; 0 ,,; (3.

(2)

The parametric length fonnu1a, Equation (3) from Section 11.2, then gives the length as L =

{P

J.

y (ax)2 (d )2 dO + dO dO.

This equation becomes

when Equations (2) are substituted for x and Y (Exercise 29).

Length of a Polar Curve

If r = flO) has a continuous first derivative for a ,,; 0 ,,; (3 and if the point P(r,O) traces the curve r = flO) exactly once as 0 runs from a to (3, then the length ofthe curve is

y

(3) 8 --+--~~--'?lol~-->x

EXAMPLE 3

Calculating the lengtll of a cardioid (Example 3). FIGURE 11.35

Find the length ofthe cardioid r = 1 - cos o.

Solution We sketch the cardioid to determine the limits ofintegration (Figure 11.35). The point P(r, 0) traces the curve once, counterclockwise as 0 runs from 0 to 2", so these are the values we take for a and (3 .

638

Chapter 11: Parametric Equations and Polar Coordinates With r = 1 - cosO,

dr . 0 dO=sm,

we have r2

+

(~;

y

=

(I - cos 0)'

=

1 - 2cosO

+ (sin 0)' + cos2 0 + sin2 0

=

2 - 2cosO

and

EXEROSES 11.5 Finding Polar "",as

4. Inside the cardioid r = a( I

Find the areas ofthe regions in Exercises 1-8.

1. Bounded by the spiral r

~

8 for 0

:5

8

:5

cos 8),

S. Inside ooeleafofthe four-Ieavedroser 'If

6. Inside ooe leaf of the three-leaved rose r

Y

r=8

+

a> 0 ~

cos28

= cos 38

y

I

..

( , )

2. Bounded by the circle r = 2 sin 8 for 'If/4

:5

8

:5

'If/2

Y

7. Inside ooe loop of the lemniscate r 2 8. Inside the six-leaved rose

r2

= 4 sin 28

~ 2 sin 38

Find the areas ofthe regions in Exercises 9-16.

___-"".. . __ ~

'!! 4 L~_. (}=

o

= 2 cos 8 and r = 2 sin 8

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

13. Inside the

lemniscate r 2

I and r

~

2 sin 8

2 and the cardioid r ~

2(1

~

+ cos 8) and r

2(1 - cos 8) ~

2(1 - cos 8)

= 6 cos 28 and outside the circle r =

V3

11.6

14. Inside the circle r ~ 3a cos 0 and outside the cardioid r = a(1 + cosO), a > 0 15. Inside the circle r = -2 cos (J and outside the circle r = 1 16. Inside the circle r = 6 above the line r = 3 esc 0 ~

17. Inside the circle r r=sec8

4 cos 0 and to the right of the vertical line

27. Thecurver

= cos3 (0/3), 0'" 0 '" ",/4 = v'1 + sin 20, 0 '" 0 '" ",V2

28. The curve r 29. The length of the curve r = /«(J), a '" (J '" fJ Assuming that the necessary derivatives are continuous, show how the substitutions

(Equations 2 in the text) transform

L=t

19. a. Find the area of the shaded region in the accompanying figure. r=tan9

y

-¥ < 8 < ¥

(O:OY+(:YdO

into L --1~ a r2

---o----"':+""----!-~x

-1

0

1

b. It looks as if the graphofr ~ tan 0, -",/2 < 0 < ",/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

11.

27

[(cosO

+

I? - cos2 0]

0

dO = "'.

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. 21. The spiral r =

0>,

0 '" 0 '"

+

V5

25. Theparabolicsegrnentr ~

(dr)2 dO dO.

30. Circumferences of circles As usual, when faccd with a new formula, it is a good idea to try it on familiar objects to be sure it givcs results consisteot with past experience. Use the length formula in Equation (3) to calculate the circorofereoces of the following circles (a > 0).

a. r=a

b. r=acos8

c. r=asin8

Theory and Examples 31. Average value If/is continuous, the average value of the polar coordinate r over the curvc r = /(8), a '" 8 '" fJ, with respect to 0 is given by the formula roy

= /3 -I

a

1~ a /(8) d8.

a. Thc cardioid r = a( I - cos 8)

0 '" 0 '" '" cos 0

24. The curve r = a sin2 (0/2),

+

Use this formula to find the average value of r with respect to 8 over the following curves (a > 0).

/V2,

22. The spiral r = e' 23. The cardioid r = I

b. The circle r = a

a> 0 6/(1 + cosO), 0'" 0 '" ",/2 0 '" 0 '" "',

26. Theparabolicsegrnentr ~ 2/(1 - cosO),

11.6

639

x = /(0) cosO, y = /(0) sinO

18. Inside the circle r = 4 sin 0 and below the horizontal line r=3csc8

2

Conic Sections

",/2'" 0 '" '"

c. The circle r = a cos 8,

-",/2 '" 8 '" ",/2

32. r = /«(J) .s. r = 2/«(J) Can anything be said about the relative lengths of the curves r ~ /(8), a'" 8 '" a ::=:; (J ::=:; {j? Give reasons for your answer.

/3,

and r ~ 2/(8),

Conic Sections In this seelioo we def'me and review parabolas, ellipses, and hyperbolas geometrically and derive their standard Cartesian equations. These curves are called conic sections or conics because they are formed by cutting a double cooe 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 DEFINmONS A set that coosists of all the points in a plane equidistant from a given f'lXed 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

Parabola: plane parallel

Ellipse: plane oblique to cone axis

Hyperbola: plane cuts both halves of cone

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

=

Vex -

PQ

=

Vex -

of + (y - pf = Vx 2 + (y - p)2 xf + (y - (-p)f = V(y + pf.

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

Conic Sections

641

If the parabola opens downward, with its focus at (0, -p) and its directrix the line y = p, then Equations (I) become

x2

y = -4p

and

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

Direclrix

Direclrix

x=-p

x=p

Vertex

Vertex

----:-+o;-f--x

(b)

(aJ

FIGURE 11.38 (a) The parabola y2 = 4px. (b) The parabolay2 = -4px.

EXAMPLE 1 Solution

Find the focus and directrix of the parabolay 2 = lOx.

We f"md 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.

x =-p

•

or

Ellipses

y b

FIGURE 11.40 The ellipse defined by the equation PF, + 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 axis. 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 FI( -c, 0) andF2(c, 0) (Figure 11.40), and PFt + 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 remaining radical, and square again, 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 ofthis 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

-

(3)

c 2 = b 2 and Equation (2) takes the form x 2 y2 a 2 + b 2 = I.

(4)

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

dy

b 2x

ax

a 2y

Ob1Bin.d from Eq. (4) by implicit differentiation

is zero if x = 0 and inf'mite 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-focus distance of the ellipse. If a = b, the ellipse is a circle. 2 y2 Y x I 16+9~ (0,3)

EXAMPLE 2

The ellipse

Vertex

(5)

(-4,0)

(Figure lIAI) 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 =

Semiminor axis:

b

=

v'9 = 3

'\Il6=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 ofthe 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

> h)

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 plane have a constant difference. The two f"1xed

points are the foci ofthe 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).

Points on the focal axis of

FIGURE 11.42

a hyperbola.

If the foci are F,( -e, 0) and F2(e, 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-hand side, square, isolate the remaining radical, and square again, obtaining (8)

---if-~+--+-
FIGURE 11.43 Hyperbolas bave _ brancbes. For points on the right-band branch ofthe hyperbola shown here, PF, - PF, = 2a. For points on the leftband branch, PF, - PF, = 2a. We then leth ~ \Ie' - 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 oftwo sides of triangle PF,F2 , is less than 2e, the third side. The algebraic stcps 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"1es 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 = 1.

(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 C

2

= a2

+ b2•

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

aOy

is infinite 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 x

x2

2

y2

4-5=

(11)

I

is Equation (10) with a 2 = 4 and b 2 = 5 (Figure 11.44). We have

c = Ya 2 + b 2 = V4+5 = 3

Center-to-focus distance: Foci: Center:

(±c,O) = (±3,0), (0,0) x2

Asymptotes: FIGURE 11.44 The hyperbola and its asymptotes in Example 3.

y2

4 - 5

=

Vertices:

0

or y =

(±a,O)

=

(±2, 0)

Y5

•

±T x .

If we interchange x and y in Equation (II), the foci and vertices of the resuIting 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 =

Ya 2 + b 2

x2

2

b2

a

Center-to-focus distance: Foci:

I

c =

Ya 2 + b 2

(0, ±c)

Vertices:

(0, ±a)

x2 y2 b Asymptotes: - 2 - - 2 = 0 or y = ±-ax a b Notice the difference in the asymptote equations (bla in the imt, alb 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

645 + h and

+ k.

Show that the equation x 2 - 4y 2 perbola. Find its center, asymptotes, and foci.

EXAMPLE 4

Sollltlon follows:

+ 2x + 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

+

+ 2x)

- 4( y 2 - 2y) = 7

1) - 4(y2 - 2y (X

+ 1)2 4

+

1) = 7

+

1- 4

- (y - 1)2 = 1.

This is the standard form Equation (10) of a hyperbola with x replaced by x + I and y replaced by y - 1. The hyperbola is shifted one unit to the left and one unit upward, and it has center x + 1 = oandy - 1 = O,orx = -1 andy = 1. Moreover,

so the asymptotes are the two lines

x2 + -1 - (y - I) = 0 The shifted foci have coordinates ( -I ±

and

x ~ 1 + (y _ I)

=

O.

v's, I).

•

Exercises 11.6 Match each conic section in Exercises 5-8 with one ofthese equations:

Identifying Graphs

Match the parabolas in Exercises 1-4 with the following equations: x 2 = 2y, x 2 = -6y, y2 = 8x, y2 = -4x.

Then f"md each parabola's focus and direc1rix.

1.

y

2.

y

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 You may wish to review Section 1.2 before solving Exercises 3<}-56.

8.

7.

y

y

39. The parabola y2 = 8x is shifted down 2 units and right I unit to generatetheparabola(y + 2? ~ 8(x - 1). a. Find the new parabola's vertex, focus, and directrix. b. Plot the new vertex, focus, and directrix, and sketch in the parabola.

40. The parabola x 2 ~ -4y is shifted left I uoit and up 3 units to generate the parabola (x + I)" = -4(y - 3).

Parabolas Exercises <}-16 give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Ioclude the focus and directrix in your sketch. 9. y2 = 121: 10. x 2 = 6y 11. x 2 = -8y 12. y2 = -21: 15. x

13. Y

= -3y 2

= 4x 2

a. Find the new parabola's vertex, focus, and directrix. b. Plot the new vertex, focus, and directrix, and sketch in the parabola. 41. The ellipse (x 2/16) + (y2/9) = 1 is shifted 4 units to the right and 3 units up to generate the ellipse

14. Y = -8x 2

(x - 4)2

16

16. x = 2y 2

+

(y - 3?

9

= I.

a. Find the foci, vertices, and center of the new ellipse. Ellipses Exercises 17-24 give equations for ellipses. Put each equation in standard form. Then sketch the ellipse. Ioclude the foci in your sketch. 17. 16>;2

+ 25y 2 ~

18. 7x 2 + 16y 2 ~ 112

400

19. 21:2 +y2=2

20. 21: 2 +y2=4

21. 3x2

22. 9x 2 + IOy2 = 90

+ 2y 2 = 6 23. 6>;2 + 9y 2 = 54

24. 169x 2

b. Plot the new foci, vertices, and center, and sketch in the new ellipse. 42. The ellipse (x 2/9) + (y2/25) ~ I is shifted 3 units to 1he left and 2 units down to generate the ellipse

(x

+ 3? 9

+ 25y 2 = 4225

+

(y

+ 2? = I.

25

a. Find the foci, vertices, and center ofthe new ellipse. b. Plot the new foci, vertices, and center, and sketch in the new

Exercises 25 and 26 give information about the foci and vertices of ellipses centered at the origin of the xy-plane. 10 each case, rmd the ellipse's standard-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)

(x - 2)2 y2 16 - 9 ~ I.

26. Foci: (0, ±4) Vertices: (0, ±5)

Hyperbolas Exercises 27-34 give equations for hyperbolas. Put each equation in standard form and rmd the hyperbola's asymptotes. Then sketch the hyperbola. Ioclude the asymptotes and foci in your sketch. 27. x 2 - y2 = I

28. 9x 2 - 16y 2 = 144

29.y2_ X 2=8

30.y2_ X 2=4

31. 8x2 - 2y 2 ~ 16

32. y2 - 3x 2 ~ 3

33. 8y 2 - 21: 2 = 16

34. 64x 2 - 36y 2 = 2304

36. Foci:

(0, ±0)

Asymptotes: 37. Vertices:

y = ±x

(±3, 0)

Asymptotes:

y

~ ±~ x 3

Asymptotes: Y = ±_1_ x

v3

(0, ±2)

Asymptotes: Y

b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola. 44. The hyperbola (y2/4) - (x 2/5) ~ 1 is shifted 2 units down to generate the hyperbola

+ 2? 4

(±2,0)

38. Vertices:

Find the center, foci, vertices, and asymptotes of the new hyperbola.

(y

Exercises 35-38 give information about the fuci, vertices, and asymptotes of hyperbolas centered at 1he origin of 1he xy-plane. 10 each case, rmd the hyperbola~ standard-limn equation from the information given. 35. Foci:

8.

~ ±l2 x

2

5x

~

I.

a. Find the center, foci, vertices, and asymptotes ofthe new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

Exercises 45-48 give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and rmd the new vertex, focus, and directrix. 45. y2

= 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

~

I,

left 2, down I

2

50.

x T + y2

= I,

right 3, up4

51.

3 + T = I,

right2,up3

52.

16 + 25

x2

x2

y2

~

dx =

x2

I, left 4, down 5

y2

16 - 9

55. y2 - x 2 y2

56.

3 -

w

/ix.

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. x2 y2 53. 4 - 5 ~ I, right 2, up2 54.

647

70. Suspension bridge cables hang in parabolas The suspensioo bridge cable shown in the accorupaoying figure supports a uniform load of w pounds per horizootal 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

dy

y2

Conic Sections

= I,

left 2, down I

= I,

left I, down I

x 2 ~ I,

71. The width of a parahola at the fueus 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 asymptotes of (x 2/a') - (y2 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 2Ia 2) - (y 2Ib 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

+ 2y 2

59. x 2 + 2x

-

28x

+ 4y

+

12y

+

- 3 = 0

61. x 2 + 5y 2 + 4x ~ I

Sintilar results hold for the remaining portions of the hyperbola andthelinesy ~ ±(bla)x.

114 = 0 60. y2 - 4y - 8x - 12 = 0 62. 9x 2

+ 6y 2 +

36y ~ 0

63. x 2 + 2y 2 - 2x - 4y = -I 64. 4x 2

+ y2 +

8x - 2y ~-I

65. x 2 _ y 2_2x+4y=4 67. 2x2

-

y2

+ 6y

= 3

66. x 2 _ y 2+4x-6y=6 68. y2 - 4x 2

+

16x = 24

Theory and Examples 69. !flines are drawn parallel to the coordinate axes through a point P on the parabola y2 = kx, k > 0, the parabola partitions the rectsngular region bounded by these lines and the coordinate axes into two amaller regions, A and B. a. If the two amaller regions are revolved about the y-axis, show that they generate solids whose volumes have the ratio 4: I.

b. What is the ratio of the volumes generated by revolving the regions about the x-axis? y p

B ~t-------"--~x

o

73. Area Find the dimensions of the rectaog1e of largest area that can be inscribed in the ellipse x 2 + 4y 2 = 4 with its sides parallel to the coordinate axes. What is the area of the rectangle? 74. Volume Find the volume of the solid generated by revolving the regioo enclosed by the ellipse 9x 2 + 4y 2 = 36 about the
= 4px from

77. Tangents Find equations for the tsngents to the circle (x - 2)2 + (y - 1)2 = 5 at the points where the circle crosses the coordinate axes. 78. Volume The region bounded on the left by the y-axis, on the right by the hyperbola x 2 - y2 = I, and above and below by the lioes y ~ ±3 is revolved about the y-axis to generate a solid. Find the volume of the solid. 79. Centroid Find the centroid of the regioo that is bounded below by the x-axis and above by the ellipse (x 2/9) + (y 2/16) ~ I.

W+1,

V2,

80. Surface area The curve y = 0 '" x '" which is part of the upper branch of the hyperbola y2 - x 2 = I, is revolved about the x-asis 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 the parabolay 2 = 4px. The line L is tangent to the parabola at P. The parabola's focus lies at F(p, 0). The ray L' _ding from P to the right is parallel to the x-axis. We show that light from F to P will be reflected out along L' by showing that f3 equals a. Establish this equality by taking the following steps.

This reflective property of parabolas is used in applications like car headlights, radio telescopes, and satellite TV dishes. y L

L'

a. Show that tan f3 = 2pjyo. b. Show that tan 4> = Yoj(x, -

pl·

c. Use the ideatity tan 4> - tan f3 tana = I + tan 4> tan f3 to show that tan a = 2pjyo.

Since a and f3 are both acute, tan f3 = tan a implies f3 = a.

11.7

Conics in PoLar Coordinates Polar coordinates 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 ecceatricity 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 - b • If we fix for an ellipse, we can still determine c from the equation c = 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, only in this case c equals + b2

cia.

instead of

Va

Va

2

-

2

b , and derme these ratios with the somewhat familiar term eccentricity.

DEFINmON The eccentricity of the ellipse (x 2/a 2)

c e=(i=

+ (y2/b 2) =

Va 2a-

1 (a

> b) is

b2

The eccentricity of the hyperbola (x 2/a 2) - (y2/b 2) = 1 is

c

e=a=

Va 2a+ b 2

The eccentricity of a parabola is e = I.

11.7 Conics in Polar Coordinates

Whereas a parabola has one focus and one directrix, each ellipse has two foci and two directrices. These are the lines perpendicular to the major axis at distances ±a/e from the center. The parabola has the property that

y Directrix 2

Directrix I x

=-~

X=

b

649

~

•

(1)

PF = I'PD --+-++-'--,rt-~:;;-'~I---+_x

D2

-b c=ae a

(2) a

e FIGURE 11.45 The foci and directrices ofthe ellipse (x 2/a 2) + (y2/b 2) ~ I. Directrix I corresponds to focus F, and directrix 2 to focus F2.

Direclrix I x

=-~

Y

DireClrix Z X=

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 shown that the equations that replace Equation (I) are

Here, e is the eccentricity, P is any point on the ellipse, FI and Fz are the foci, and Dl and D2 are the points on the directrices nearestP (Figure 11.45). In both Equations (2) the directrix and focus must correspond; that is, if we use the distance from P to F" we must also use the distance from P to the directrix at the same end of the ellipse. The directrix x = -ale corresponds to Fl( -c, 0), and the directrix x = a/e corresponds to F2(C, 0). Ai; with the ellipse, it can be shown that the lines x = ±a/e act as directrices for the hyperbola and that

~

•

----+r.;;i--++-.---:-x

and

(3)

Here P is any point on the hyperbola, F, and F2 are the foci, and D, and D2 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 distance between the foci to the distance between the vertices (because cia = 2c/2a).

E tri"ty distance between foci ccen c, - distance between vertices FIGURE 11.46 The foci and directrices ofthe hyperbola (X 2/a 2) - (y2/b 2) = I. No matter where Plies 00 the hyperbola,

PF, = e·PD, andPF2 = e·PD2 •

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 I. 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 fJXed point F (the focus) is a constant multiple ofits distance from a fixed line (the directrix). That is, suppose

PF= e'PD,

(4)

where e is the constant of proportiona1ity. Then the path traced by P is (a) aparabola if e = I,

(b) an ellipse of eccentricity e if e < I, and (c) a hJPerbola of eccentricity e if e > 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 shown 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 shown in Figure 11.45.

650

Chapter 11: Parametric Equations and Polar Coordinates y

T 3-6 ~! x2

x=l

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.

D(!,y)

Solution We first use the dimensions shown in Figure 11.46 to find the hyperbola's eccentricity. The focus is

P(x,y)

(c,O) --+-~--O++----~x

(3,0),

=

c = 3.

so

The directrix is the line x

=

ea =

When combined with the equation e FIGURE 11.47 The hyperbola and directrix in Example I.

e

c

a

=

=

3

e'

a

so

1,

=

e.

cia that defines eccentricity, these results give

=

e 2 =3

so

and

e=Vi

Knowing e, we can now derive the equation we want from the equation PF = e' PD. In the notation ofFigure 11.47, we bave PF = e'PD

V(x x2

3)2 + (y - 0)2 -

6x + 9 +

v'3lx - II

=

y2 =

Eq. (4)

3(x 2

-

e~

v'3

2x + I)

2x 2 - y 2=6

x2

y2

•

3-6=\' Polar Equations Direc1rix

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

----
k

and PD=k-FB=k-rcosO.

x=k

The conic's focus-directrix equation PF

=

e' PD then becomes

r = e(k - rcosO),

FIGURE 11.48 Ifa conic section is put in the position with its focus placed at the origin and a direc1rix perpendicular to the initial ray and right ofthe 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

where x

=

r=l+ecostJ' k > 0 is the vertical directrix.

(5)

EXAMPLE 2 Here are polar equations for three conics. The eccentricity values identifying the conic are the same for both polar and Cartesian coordinates. e

I = 2:

ellipse

e

= I:

parabola

e

= 2:

hyperbola

k 2 + cosO k r= I + cosO 2k r= 1+2cosO

r=

•

11.7

Conics in Polar Coordinates

651

You may see variations of Equation (5), depending 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

r=

y Directrixy

7T\

ke l-esin9

\

~k

Focus at origin

117tori~

~Y~-k

(c)

(d)

FIGURE 11.49 Equations for cortic sections with eccentricity e > 0 but different locations of the directrix. The grapbs bere sbow a parabola, so e = I.

EXAMPLE 3 = 2.

Find an equation for the hyperbola with eccentricity 3/2 and directrix

x

Solution

We use Equation (5) withk

=

2ande

2(3/2)

or

r = c1-c+-(O-::3-c/2oc)'--co-s--;CO

EXAMPLE 4

6 r = "2-+~3c-c-o-s~O .

•

Find the directrix of the parabola r = 10

Directtix

x=k

3/2:

=

+

25 10cosO"

Solution We divide the numerator and denominator by 10 to put the equation in standard polar form:

--+---==~==-'e--1f---+--~x

5/2 r = 1

+

cosO'

This is the equation ke

with k FIGURE 11.50 In an ellipse with

semimajor axis a, the focus-directrix distance is k = (a/e) - ea, so Ire = a(1 - e').

=

5/2 and e

=

r=1+ecos6 1. The equation ofthe 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 a k = e - ea.

652

Chapter 11: Parametric Equations and Polar Coordinates

From this, we find that Ice = a(1 - e 2 ). Replacing Ice in Equation (5) by a(1 - e 2 ) gives the standard polar equation for an ellipse. Polar Equation for the Ellipse with Eccentricity e and Semimajor AD. 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 Suppose the perpendicular from the origin to line L meets L at the point Po(ro, 1J0), with ro 2: 0 (Figure 11.51). Then, if P(r, IJ) 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

y

ro = rcos(1J - 1J0).

The Standard Polar Equation for Line. If the point Po(ro, 1J0) 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 (0 - 00 ) from 1he right triaogle

r cos (IJ - 1J0) = ro.

(7)

FIGURE 11.51

For example, iflJo = 1f/3 and ro = Z, we find that

OPoP.

rcos(IJ-;)=Z r(cosIJCOS; + SinIJSin;) = Z !rcoslJ +

~ rsinlJ = Z,

or

x

+

V3y =

4.

Circles To find a polar equation for the circle ofradius a centered at Po(ro, 1J0), we let P(r, IJ) be a point on the circle and apply the Law ofCosines to triangle opoP (Figure I1.5Z). This gives

y

r

a2 = rl + r 2

-

Zrorcos(1J - 1J0)·

Ifthe circle passes through the origin, then ro = a and this equation simplifies to ~=-----'=L-_----~x

o

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 + r 2 - Zarcos (IJ - 1J0) r 2 = Zar cos (IJ - 1J0) r = Za cos (IJ - 1J0).

If the circle's center lies on the positive x-axis, 1J0 = 0 and we get the further simplification r = Za cos IJ.

(8)

11.7

Conics in Polar Coordinates

653

If the center lies on the positive y-axis, 8 = 7T/2,cos(8 - 7T/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 severs! 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

Polar equation

(3,0)

r=6cos8

2

(2,7T/2)

1/2

(-1/2,0)

r = 4sin8 r = -cos8

I

(-1,7T/2)

r=-2sin8

•

Exercises 11.7 =3

Ellipses and Eccentridty

21. 8>;' - 2y' = 16

22. y' - 3x'

Io Exercises 1-8, fmd the eccentricity of the ellipse. Then fmd and graph the ellipse's foci and directrices.

23. 8y' - 2x' ~ 16

24. 64x' - 36y' ~ 2304

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. Io 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

Exercises 13-16 give foci and corresponding directrices of ellipses centered at the origin ofthe 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-fonn equation in Cartesian coordinates. 13. Focus:

( v5, 0)

Directrix:

x =

..:Is

15. Focus: (-4,0) Directrix: x = -16

14. Focus:

(4,0)

Directrix:

x = 16 3

16. Focus: (- V2, 0) Directrix: x = -2 V2

Hyperbolas and Eccentridty Io Exercises 17-24, fmd the eccentricity of the hyperbola. Then fmd

Exercises 25-28 give the eccentricities and the vertices or foci of hyperbolas centered at the origio of the xy-plane. Io each case, fmd the hyperbola's standard-form equation in Cartesian coordinates. 25. Eccentricity: 3 26. Eccentricity: 2 Vertices: (0, ±l) Vertices: (±2,0) 27. Eccentricity: 3 Foci: (±3,0)

28. Eccentricity: 1.25 Foci: (0, ±5)

Eccentridtles and Directrices Exercises 29-36 give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. 29. e = I, x = 2 31. e ~ 5, y ~ -6 33. e

~

1/2, x

~

I

35. e = 1/5, y = -10

30. e = I, y = 2 32. e ~ 2, x ~ 4 34. e

~

1/4, x

~

Parabolas and Ellipses 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. 37. r = I

I

+ cos 8

25 39. r = 10 _ 5cos8 400

6 38. r = 2 + cos 8

4 40. r = 2 - 200sB 12

and grapb the hyperbola's foci and directrices. 18. 9x' - 16y' = 144

41. r ~ 16

+ 8 sin 8

42. r ~ 3

+ 3 sin 8

19. y' - x' ~ 8

43. r = 2

8 2 sin 8

44. r = 2

4 sinB

17. x' - y' = I

20. y' - x' ~ 4

-2

36. e = 1/3, y = 6

654

Chapter 11: Parametric Equations and Polar Coordinates

Lines Sketch the lines in Exercises 45-48 and rmd Cartesian equations for them. 45. rcos (8 -

f)

=

V2

47. rcos (8 - 2:) = 3

46. rces (8

+

3:) = I

48. rces (8

+

f)

73. r

V2x V2y = 6

+ 2 sin 8)

74. r ~ 1/(1

+ 2 ces 8)

a. Show that r = a( I - e) when the planet is closest to the son and that r ~ a(l + e) when the planet is far1hest from the son.

b. Use the data in the tahle 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.

= 2

Find a polar equation in the form r ces (8 - 80) = ro for each of the lines in Exercises 4!1-52.

+ 49. 51. y = -5

~ 1/(1

75. Perihelion and aphelion A planet travels about its son in an ellipse whose semimajor axis has length a. (See accompanyiog figure.)

Aphelion

50. V3x-y= 1 52.x=-4

Perihelion

(farthest

(closest

from sun)

to sun)

Circles Sketch the circles in Exercises 53-56. Give polar coordinates for their centers and identiJY their radii.

53. r = 4cos8 55. r ~ -2cos8

54. r = 6 sin 8 56. r ~ -8 sin 8

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.

=4

57. (x - 6)' + y2 = 36

58. (x + 2)2 + y2

59. x 2 + (y - 5)2 = 25 61. x2+2x+y2~0

60. x 2 + (y + 7)' = 49 62. x 2 - 16x + y2 ~ 0 4 64. x 2 + y2 - :fy = 0

63. x 2 +y2+ y =0

76. Plaoetary orbits Use the data in the table below and Equation (6) to rmd polsr equations for the orbits of the planets.

Examples of Polar Equations

o Graph the lines and conic sections in Exercises 65-74. 3 sec (8 - .../3) 67. r=4sin8 69. r = 8/(4 + cos 8)

66. r ~ 4 sec (8 + .../6) 68. r= -2C058 70. r = 8/(4 + sin8)

71. r = 1/(1 - sin 8)

72. r = 1/(1

65. r

~

Chapter

Planet

Semimajor axis (a.tronomical nnit.)

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 8)

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 = /(x) always have a parametrization? Are parametrizations

7. What i. the arc length function for a smooth parametrized curve? What i. it. arc length differential?

of a curve unique? Give examples. 2. Give some typical parametrizations for lines, circles, parabolas,

8. Under what condition. can you find the area of the surface generated by revolving a curve x ~ /(1), Y ~ g(1), a :5 1 :5 b, about

ellipse., and hyperbola•. How might the parametrized curve differ from the graph ofits Cartesian equation? 3. What i. a cycloid? What are typical parametric equations for cycloids? What physical prupertie. accoont for the importance of cycloids? 4. What i. the formula for the .Iope dy/dx of a parametrized curve x = /(t), y = g(t)? When does the formula apply? When can you expect to be able to rmd d'y/dx 2 as well? Give examples. 5. How can you .ometime. rmd the area bounded by a parametrized

curve and one ofthe coordinate axes? 6. How do you rmd the length of a smooth parametrized curve x ~ /(t), y ~ g(t), a :5 1 :5 b? What doe. smoothne.. have to do with length? What else do you need to koow about the parametrization in order to rmd the curve~ length? Give examples.

the x-axis? the y-axis? Give examples. 9. How do you rmd the centroid of a smooth parametrized curve x ~ /(t), y ~ g(t), a :5 1 :5 b? Give an example. 10. What are polar coordinate.? What equation. relate polar coordinates to Cartesian coordinates? Why might you want to change from one ceordinate system to the other? 11. What consequence does the lack of uniqueness of polsr coordinates have for graphing? Give an example. 12. How do you graph equations in polar coordinates? Include in your discussion syounetry, slope, behavior at the origin, and the use of Cartesian graphs. Give examples. 13. How do you rmd the area of a region 0 :5 r,(8) :5 r :5 r2(8), a :5 8 :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), Of. ' " 6 '" ,B, in the polar coordinate plane? Give an example of a typical calculation. 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

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 15. y = (5/12)x6/s - (5/8)x4/s,

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.

16. x ~ (y'/12)

1. x = 1/2, Y = I

+

I;

-00

2.x=V/, y=I-V/;


4. x = -2 cos t, y = 2 sin t; 6. x =

-cos t, Y 4 cos t, y

= =

00

o::=:;

-'1f/2

+ (I/y),


'1f/2

t ::=:; '1T

I '" x '" 32

I '" y '" 2

17. x = 5cosl - cos 51, Y = 5 sin I - sin51, 18. x = ,3 - 6t 2, y = (3 + 6t 2, 0:5 t :5 1 19. x = 3 cos 6, y = 3 sin 6,

1;;,,0

3. x ~ (1/2) ian I, Y ~ (1/2) sec I; X =

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?

Identifying Parametric Equations in the Plane

S.

Practice Exercises

0'" I '" '1f/2

0'" 6 '" 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.

cos2 t; O:s; t :s; 71' 9sint; 0:5 t:5 2'7T

y

1

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 + 9y 2 = 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 x 2 + y2 = 4 three times clockwise. (There are many ways to do this.)

-~+----l_~----}fC---'------~x

o

In Exercises 9 and 10, fmd an equation for the line in the xy-plane that is iangent to the curve 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;

10. x = I + 1/12, Y = I - 3/1;

2

-1

I = '1f/3

I = 2

11. Eliminate the parameter to express the corve in the form y = f(x). a. x = 41 2, y = ,3 - 1 b. x = cos t, Y = tan t 12. Find parametric equations for the given curve.

a. Line through (I, -2) with slope 3 + (y + 2)2 ~ 9 2 coy=4x -x

Surface Areas Find the areas ofthe surfaces generated by revolving the curves in Exercises 21 and 22 about the indicated axes.

,2/2, Y = 21, 0:5 t :5 VS; x-axis 22. x = 12 + 1/(21), Y = 4Vt, 1/V2 '" I '" I; y-axis

21. x

=

b. (x - 1)2

d.

9x 2

+ 4y 2 =

36

Lengths of Curves Find the lengths of the corves in Exercises 13-19. 13. y = x 1/2 - (1/3)x'/2, I '" x '" 4 14. x = y2/3, 1:5 y :5 8

rolar to Cartestan Equations Sketch the lines in Exercises 23-28. Also, fmd a Cartesian equation for each line. 23. rcos (6

+

f)

= 2V3

25.r=2sec6

27. r

~

-(3/2) esc 6

24.

rcos

26. r =

(6 - 3:) V; =

-\12 sec 6

28. r ~ (3V3) esc 6

656

Chapter 11: Parametric Equations and Polar Coordinates A",a In Polar Coordinates

Find Cartesian equations for the circles in Exercises 29--32. Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations.

Find the areas of the regions in the polar coordinate plane described in Exercises 47-50.

29. r ~ -4sinO

30. r ~ 3Y3 sinO

47. Roclosed by the I~on r = 2 - cos 0

31. r ~ 2V2 cos 0

32. r ~ -6cosO

48. Enclosed by one leaf of the three-Ieaved rose r

Cartestan to Polar Equatfons Find polar equations for the circles in Exercises 33-36. Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations.

33. X2 +y2+5y=0 35. x 2 +y2_3x=0

6 cos 8

Length In Polar Coordinates

38. -4sin0:5 r:5 0

6

b. rcosO = I

d. r = sin20

e. r = 8

c. r= 1-200s0 f. ,2 = cos28

g.r=l+cosO

h. r = I - sinO

L7=

j.

k. r = -sinO

sin28

2

I cosO Lr=2cosO+1

40. Spiral y

y

51. r~ -I + cosO 52. r = 2 sinO + 2 cosO, 53. r ~ 8 sin' (0/3),

a. r = cos28

,2 =

r=2sin8 Find the lengths of the curves given by the polar coordinate equations in Exercises 51-54.

Match each graph in Exercises 39--46 with the appropriate equation (a)-(I). There are more equations than graphs, so some equations will not be matched.

39. Four-leaved rose

49. Inside the "figure eight" r = I + cos 20 and outside the circle r = I 50. Inside the cardioid r ~ 2(1 + sin 0) and outside the circle

36.x 2 +y2+4x=0

Graphs In Polar Coordinates

:5

sin 30

34.X 2 +y2_2y=0

Sketch the regions defmed by the polar coordinate inequalities in Exercises 37 and 38.

37. 0 :s; r

~

0:5 0:5 'If/2

0:5 0 :5 'If/4

54. r = v'1 + cos 20,

-'If/2 :5 0 :5 'If/2

Graphing Conic Sections Sketch the parabolas in Exercises 55-58. Include the focos and directrix in each sketch.

56. x 2 ~ 2y

55. x 2 ~ -4y 57. y2

~ 3x

58. y2 ~ -(8/3}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. lfu 2

+ 7y 2 =

60.x 2 +2y 2=4

112

61.3x 2 - y 2=3

62. 5y 2

-

4x 2 = 20

Exercises 63-68 give equations for conic sections and 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.

LinIa~on

centers, and asymptotes, as appropriate. Ifthe curve is a parabola, fmd the new directrix as well.

42. Lemoiscate

y

63. x 2 ~ -12y, 64. y2 ~ lOx, 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 68. 36 - 64 ~ I,

right 2, up 2V2 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). Ifthe curve is a parabola, fmd its directrix as well.

46. Lemoiscate y

y

69. x 2 -4x-4y 2=0

__ x

-~-i-=-

70.4x 2 - y 2+4y=8 72. x 2 - 2x + 8y = -17

71. y2 - 2y + lfu = -49 73. 9x 2 + 16y 2 + 54x - 64y = -I 74. 25x 2 + 9y2 - IOOx + 54y ~ 44 75. X2 +y2_2x-2y=0

76. X2 +y2+4x+2y= I

Chapter 11

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 ofellipses, for the centers as well.

2

8

77. r = 1 + cos 0 6

78. r = 2 + cos 0 12 80.r=3+sinO

79.r=I-2cosO

86. TIre ''triango!ar'' regioo in the fIrSt quadrant bounded by the x-axis, the line x = 4, and the hyperbola 9x 2 - 4y 2 = 36 is revolved about the x-axis to geoerate a solid. Find the volume ofthe solid. 87. Show that the equatioos x = r cos 0, y = rsin 0 transfonn 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 rsin 0 = 2

82•• = 1,

rsin 0 = -6

Theory and Examples 85. Find the volume of the solid generated by revolving the region enclosed by the ellipse 9x 2 + 4y 2 ~ 36 aboot (a) the x-axis, (b) the y-axis.

Chapter

n

into the Cartesian equation

(1 - .2)x 2 + y2 + 2kex - k 2 = O.

rcosO = -4

84•• = 1/3,

657

Additional and Advanced Exercises

88. Archimedes spirals The grapb 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 widtha between the suc-

cessive turns of such a spiral?

Additional and Advanced Exercises

Finding Conic Sections 1. Find an equatioo for the parabola with focus (4, 0) and directrix x = 3. S1retcb the parabola together with its vertex, focus, and directrix.

10. Show that the line

b 2xxl

-

6x - 12y + 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 00 the segment lies a units from ooe 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 an ellipse of eccentricity 0.5 lie at the points (0, ±2). Where do the foci lie? 6. Find an eqoatioo for the ellipse of eccentricity 2/3 that has the line x = 2 as a directrix and the point (4, 0) as the correspoodiog focus. 7. One focus of a hyperbola lies at the point (0, -7) and the c0rresponding directrix is the line y ~ -1. Find an equation for the hyperbola if its eccentricity is (a) 2, (b) 5.

a7.Y1 - a 2b 2

=0

is tsngent to the hyperbola b"x 2 - a2y2 - a 2b 2 (Xl, YI) 00 the hyperbola.

2. Find the vertex, focus, and directrix ofthe parabola

x2

-

= 0 at the point

Equations and Inequalities What points in the xy-plane satisfY the equations and inequalities in Exercises 11-16? Draw a ftgUl'e for each exercise. 11. (x 2 - y2 - 1)(x 2 + y2 - 25)(x 2 + 4y 2 - 4) ~ 0 12. (x + y)(x 2 + y2 - 1) ~ 0

13. (x 2/9) + (y2/16) 14.

(x 2/9)

15. (9x 2 16. (9x 2

- (y2/16)

+ 4y 2 + 4y 2 -

:5 1 :5 1

36)(4x 2 36)(4x 2

+ +

9y2 - 16) :5 0 9y2 - 16)

>

0

Polar Coordinates 17. a. Find an equatioo in polar coordinates for the curve

x = e 2tcost. y

= e 2J sint;

-00


00.

= 2".. = 2 sin' (0/3), 0 :5 0 :5

b. Find the length of the curve from t = 0 to t

8. Find an equatioo for the hyperbola with foci (0, -2) and (0, 2) that passes through the point (12, 7).

18. Find the length of the curve r the polar coordinate plane.

9. Show that the line

Exercises 19-22 give the eccentricities of conic sections with one focus at the origin of the polar coordinate plane, along with the directrix for that focus. Find a polar equatioo for each conic sectioo.

is tangent to the ellipse b 2x 2 + a2y2 - a 2b 2 = 0 at the point (XI, Yl) 00 the ellipse.

19. e = 2,

21••

~

1/2,

rcos6 = 2 rsinO

~

20. e = 1, 2

22••

~

1/3,

3"., in

rcos8 = -4 rsinO

~

-6

658

Chapter 11: Parametric Equations and Polar Coordinates

Theory and Examples 23. Epieycloid. When a circle rolls externally along the circumference ofa second, ftxed circle, any point P on the circumference of the rolling circle describes an epicycloid, as shown here. Let the

Since l/J = t/> - 0 from (1), tanl/J

= tan(t/>

tant/> - tanO

- 0)

= I + tant/>tanO'

IlXed circle have its center at the origin 0 and have radius a. Furthermore,

y

dy

dy/dO

tant/> = is the slope of the curve at P. Also,

Y tanO=x'

-~'---O_---t''----->x

o

A(a, 0)

Hence

Let the radius of the rolling circle be b and let the initial position of the tracing point P be A(a, 0). Find parametric equations for the epicycloid, using as the parameter the angle 0 from the positive x-axis to the line through the circles' centers. 24. Find the centroid of the region ooclosed by the x-axis and the cy-

cloid arch x

~

a(t - sint), y

~

a(I - cost);

+ l/J,

tanl/J = I

(4)

+ x
The numerator in the last expression in Equation (4) is found from Equations (2) aod (3) to be

0:5 t:5 2"..

The Angle Between the Radiu. Vector 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 measured counterclockwise :from the positive x-axis to the tangent line. In polar coordinates, it is more convenient to calculate the angle l/J from the radius vector to the tangent line (see the accoropanying figore). The angle t/> can thoo be calculated from the relation t/> = 0

dy/dO y
----

(1)

which comes from applying the Exterior Angle Theorem to the triangle in the accompanying figore.

Similarly, the dooontinator is

When we substitute these into Equation (4), we obtain r

tanl/J = dr/dO'

(5)

y

This is the equation we use for fmdiog l/J as a function of O. 25. Show, by reference to a figure, that the aogle f3 between the tangents to two curves at a point of intersection may be found from the formula (6)

o Suppose the equation of the curve is givoo in the form r where /(0) is a differentiable function of O. Thoo x

= rcos8

and y = 7sin8

= /(0), (2)

are differentiable functions of 0 with
+ cos.adr dO'

dy a ' a dr dO ~ rcosu + smu dO'

(3)

When will the two curves intersect at right aogles? 26. Find the value oftanl/J for the curver ~ sin4 (0/4). 27. Find the angle hetweoo the radius vector to the curve r ~ 2a sin 30 and its tangent when 0 = "./6. D 28. a. Graph the hyperbolic spiral rO = I. What appears to happen to l/J as the spira1 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 taogoots are perpendicolar there. 30. Find the aogle at which the cardioid r ~ a(I - cos 0) crosses the ray 0 = "./2.

Chapter 11 Technology Application Projects

Chapter

Technology Application Projects

Mathematica/Maple Module: Radar 1l'acking ofII Moving Object Part I: Convert from polar to Cartesian coordinates. Ptmlmetric tJ1Id Polar EqUtJlio". with II Figure Skater Part I: Visualize positioo, velocity, and acccleratioo to analyze motioo defmed by parametric equatioos. Part II: Find and analyze the equatioos of motioo for a figure skater lIacing a polar plot.

659

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

- 43 == 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 of 9'so 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:

+e< b+e b ~ a - e < b - e band e > 0 ~ ae < be band e < 0 ~ be < ae Special case: a < b ~ -b < -a

< 2. a < 3. a < 4. a <

b

>

0

1. a

5. a

~ a

~

1 7i

>

0

6. If a and b are both positive or both . th en a negatIve,

< b

~

b1 <

1 7i

-2

I

I

o

1

2

I

37T

I

)

4

3

The symbol IR 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 rules 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 define precisely. However, the property is essential to the idea of a 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 calculus 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-1

AP-2

Appendices We distinguish three special subsets ofreal numbers.

1. 2. 3.

The natural numbers, namely 1,2,3,4, ... The integers, namely 0, ± I, ±2, ±3, ... The rational numbers, namely the numbers that can be expressed in the form of a fraction min, where m and n are integers and nolO. Examples are

I 3'

4 9

-4 9

4 -9'

200 13'

57 = 57 I

and

The rational numbers are precisely the real numbers with decimal expansions that are either (a) terminating (ending in an infinite string of zeros), for example,

43

= 0.75000 ... = 0.75

or

(b) eventually repeating (ending with a block of digits that 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 the ending zeros repeat. The set ofrational numbers has all the algebraic and order properties ofthe real numbers but lacks the completeness property. For example, there is no rational number whose square is 2; there is a ''hole'' in the rational line where should be. Real numbers that are not rational are called irrational numbers. They are characterized by having nonterminating and noorepeating decimal expansions. Examples are 1r, and 10glO 3. Since every decimal expansion represents a real number, it should be clear that there are infmitely many irrational numbers. Both rational and irrational numbers are found arbitrarily close 10 any point on the realline. Set notation is very useful for specifying a particular subset ofreal numbers. A set is a collection ofobjects, and these objects are the elements ofthe set. If S is a set, the notation a E S means that a is an element of S, and a I'! S means that a is not an element of S. If S and T are sets, then S U T is their union and consists of all elements belonging either to S or T (or to both S and T). The intersection S n T consists of all elements belonging 10 both S and T. The empty set 0 is the set that contains no elements. For example, the intersection ofthe rational numbers and the irrational numbers is the empty set. Some sets can be described by listing their elements in braces. For instance, the set A consisting ofthe natural numbers (or positive integers) less than 6 can be expressed as

v2

v2, '¥5,

A=

{1,2,3,4,5}.

The entire set of integers is written as {O, ± I, ±2, ±3, ... } . Another way to describe a set is 10 enclose in braces a rule that generates all the elements of the set. For instance, the set

A = {xix is an integer and 0 is the set of positive integers less than 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 -1 and 1 (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 ofits 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 B;l is an infinite interval that is both open and closed Table A.l 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)

B;l (set ofall real numbers)

Both open and closed

:5

:5

b}

b}

Picture )

a

b

a

b

a

b

a

b

)

Half-open

)

)

Half-open

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 !

!

0

1

Solution 4

: 3

-7

!

2x < x + x< 4

]X

0

--;;--~;-

4

Add 1 to both sides.

Subtract x from both sides.

The solution set is the open interval (-00,4) (Figure A.1a).

(b)

0

2x-1<x+3

(a)

(a)

_ _':";--_)X 11

(b)

5

(c)

FIGURE A.1 Solution sets for the inequalities in Example I.

-~ 3

< 2x +

1

-x

<

3

-

6x

+

0< 7x + 3 3 < 7x

-~<x

Multiply both sides by 3.

Add x to both sidcs. Subtract 3 from both sides. Divide by 7.

The solution set is the open interval ( - 3/7, 00) (Figure A.1b). 2: 5 can hold only if x > 1, because otherwise 6/(x - 1) is undermed or negative. Therefore, (x - 1) is positive and the inequality will be preserved if we multiply both sides by (x - 1), and we have

(c) The inequality 6/(x - 1)

_6_>5

x-1-

62:5x-5

Multiply both sides by (x - I).

11 2: 5x

Add 5 to both sides.

11> x T.

Orxs

~1.

The solution set is the half-open interval (1,11/5] (Figure A.1c).

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 < 0.

I-51 = -(-5) = 5,

Hall = lal

°°

-

Geometrically, the absolute value of x is the distance from x to on the real number line. Since distances are always positive or 0, we see that Ixl 2: for every real number x, andlxl = if and ouly if x = O.Also,

°

Ix - yl = the distance between x andy ~1-51~5

I

-5

),

131---->

o

3

I

I

on the real line (Figure A.2). Since the symbol y;. always denotes the nonnegative square root of a, an alternate definition ofIx Iis

+---14 - 11 ~ 11 - 41 ~ 3_

I

I

Ixl=W.

4

FIGURE A.2 Absolute values give distances between points on the number line.

W

W

It is important to remember that = Ia I. Do not write = a unless you already know that a 2: 0. 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.

I<---a----ȴI'--a~

----Q.

-a

I

0 r-Ixl--->l

x

FIGURE A.3 Ixl < between -a and a.

0---------->

a

a means x lies

+ Ibl

Note that I-al '¢' -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 ~

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. Ixl = a

The triangie 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 = 1-31 + 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 dermition 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 "ifand 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 x = 5 The solutions of 12x - 31

EXAMPLE 5 Solution

2x - 3 = -7

Equivalent equations without absolute values

2x =-4

Solve as usual.

x =-2

•

= 7 are x = 5 andx = -2.

Solve the inequality 15 - fl < 1.

We bave 2

<1~-1<5-x<1

~-6

~3 > 1

~3

<

-x2 <-4

f>2 I

<x<2'

Property 6

SubtractS. 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 inequality holds if and ouly if(I/3) < x < (1/2). The solution set is the open interval (1/3, 1/2). •

Exercises A.1 1. Express 1/9 as a repeating decimal, using a bar to indicate lbe repeating digits. What are lbe decimal representations of2/9? 3/9? 8/9? 9/9?

Solve lbe inequalities in Exercises 18-21. Express lbe 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, whichoflbe 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
I I I d. 6<:'<2

e. 1<¥<3

f. Ix - 41 < 2

g. -6 < -x < 2

h. -6 < -x < -2

23. Solve lbe equation Ix - II

5·2x-

la

2 "'7x+

7

4

6. S(x - 2)

6

I

< 3(x

- 6)

9. 18 - 391 =

13.

It-II

s I

16. II - xl> I

A.2

14.1 3 -}I < t 17.lr~

II '"

(2)

~ lal 2 + 21allhl + Ihl 2

(3)

= (Ial + Ihl)'

~

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. 11·11-lls3

(1)

s a 2 + 21allhl + h 2

la + hi sial + Ihl

8·121+51=4

10. Ixl < 2

I-x.

=a 2 +2ab+b 2

Solve the equations in Exercises 7-9. 7. Iyl = 3

~

+ hl 2 = (a + h)'

4.5x-3s7-3x

I

x< 0

24. A proof of the triangle inequality Give lbe reason justifYing eacb of lbe nwnbered steps in lbe following proof of lbe triangle inequality.

In Exercises 3--6, solve the inequalities and show the solution sets on lbe real line.

3. -2x > 4

-

22. Do not fall into lbe 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 Iah I = Ia II h Ifor any nwnbers a and h. 26. If Ixl s 3 and x > -1/2, what can you say about x? 27. Graph lbe inequalitylxl + Iyl s I. 28. For any number a, prove that I-a I ~ Ia I. 29. Let a be any positive nwnber. Prove that Ix I > a if and only if x>aorx<-a. 30.

L

Ifh is any nonzero real nwnher, prove lbatl I/hl = I/Ihl.

~ Prove that

I

bal

l

fbi

= lal

for any numbers a and h '" O.

Mathematical Induction Many formulas, like

I

+ 2 + ... + n

n(n

+ I)

= ='-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 + I.

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 hy Step 2 also for n = 3, and hy Step 2 again for n = 4, and so on. If the first 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 troe implies that the next statement in line is troe. Suppose that we can also show that SI is troe. 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

+

I)

+ 2 + ... + n = -"-~2~-'-

We accomplish the proofhy carrying out the two steps above.

The formula holds for n = I because I = --,I("--I=+_1-,-) 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 + I) + (k + 2

(k

+ I)(k + 2)

I)

= k' + k + 2k + 2

(k

2

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 hy 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 -+-+ ... +-= 1 2

2

Solution 1.

2

2'

I 1--

2'·

We accomplish the proofhy carrying out the two steps ofmathematical induction.

The formula holds for n = I because I

-= 21

I 1-21 .

AP-8

Appendices

2.

If

then

Thus, the original formula holds for n = (k

+

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 =

nl

(the first appropriate integer).

Prove that if the formula holds for any integer n = k = (k + I).

20

n" then it also holds for

n

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!

I I

3"

3

2 2 9

3 6 27

4 24 81

5 120 243

6 720 729

7 5040 2187

> 3" for n 20 7. To be sure, we apply mathematical induction. We take 7 in Step 1 and complete Step 2. Suppose k! > 3k for some k 20 7. Then

It looks as if n! nl =

(k Thus, for k

20

+

I)! = (k

+

I)(k!)

> (k + 1)3k > 7'3 k > 3k+'.

7,

k!

> 3k

implies

(k

+

I)!

>

The mathematical induction principle now guarantees n!

3k+l.

20

3" for all n

Proof of the Derivative Sum Rule for Sums of Finitely Many Functions We prove the statement

20

7.

•

Appendix 2 Mathematical Induction

AP-9

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 + UHI) ~

Call this function v.

Call the fUnction delmed by this sum u.

+ Uk) + dU1

dU2

=dx -++ dx

dUk

dUk+1

dx

d Sum Rule for th (u

dUHI

+dx- + dx

+ v)

Eq. (I)

With these steps verified, the mathematical induction principle now guarantees the Sum Ru1e 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

+ b I :5

Ia I + Ib Iholds

7. Show that 2' > n 2 ifn is large enough.

Ixd + IX21 + ... + Ix,1

for aoy n numbers. 2. Show that if r # I, then

1 - r n+1

1 +r+r2+"'+rn=~--"-­ I r

for every positive integer n.

d

dv

6. Show that n! > n' ifn is large enough.

du

3. Use the Product RnIe, dx (uv) = u dx + v dx' and the fact that ::. (x) = I to show that ::. (x') = nx,-l for every positive inte-

8. Show that 2' '" 1/8 for n '" -3. 9. Sums of squares Show that the sum of the squares ofthe fir>t n positive integers is

+ t}n +

+ I)

3

10. Sums of cubes Show that the sum of the cubes of the 'Irst n positive integers is (n(n + 1)/2}'. 11. Rules for ,mite sums Sbow that the following ,mite sum rules bold for every positive integer n. (See Section 5.2.)

gern.

4. Suppose that a function fIx) has the property that /(X1X2) = /(x,) + /(X2) for aoytwo positive numbers x, andx2' Show that /(X,X2' "X,) = /(X1) + /(X2) + ... + /(x,)

for the product ofany n positive numbers Xl, X2, ••• , Xn_ 5. Showthat 2 2 2 I -+-+"'+-=1-31 32 3/1 3"

for all positive integers n.

(aoy number c)

(if at has the constant value c) 12. Sbow that Ix'i = Ix I' for every positive integer n and every real numberx.

AP-10 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

-------oP(a,b) I

I I

Positive y-axis 3

~

I

I I I

2

1

Negative x-axis

I I I

Ori .

/gJn

I

----:c---o--"--o---cc~----:---,--o-~-o--->x

-3

-2

-1

0

1 \2

-1

a3

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 quadrant (+, +)

2

•

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 Cartesian 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 positive 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 A4). 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 )'-Coordinate (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). Stsrting 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 A5. TIris coordinate system is called the rectangular coordinate system or Cartesian coordinate system (after the sixteenth-century French mathematician Rene Descartes). The coordinate axes oftbis coordinate or Cartesian plane divide the plane into four regions called quadrants, numbered counterclockwise as shown in Figure A5. 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. thrust 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

C(5,6)

5

I

B(2, 5)

4

Ax =

2 I

~X =

~y=S-(-3)=8.

2 - 4 = -2,

From C(S, 6) to D(S, 1) the coordinate increments are

D(5, I)

~O+----"I-+----lC+--'---5'-x

-I

~Y =

Ax=S-S=O,

1 - 6 = -S.

See Figure A6.

-2

-3

X, - Xl.

EXAMPLE 1 In going from the point A( 4, -3) to the point B(2, S) the increments in the x- and y-coordinates are

';';~j

3

~~_A(4,-3)

(2,-3)

AP-11

starting point from the coordinates of the ending point. If x changes from Xl to x" the increment in x is

y

6

Lines, Circles. and Parabolas

/i.x =-2

FIGURE A.6 Coordinate increments may be positive, negative, or zero (Example I).

•

Given two points PI(XIoYI) and P,(x"y,) in the piane, we caIl the increments Ax = x, - Xl and ~Y = Y, - Yl the run and the rise, respectively, between P, and P,. Two such points always determine a unique straight line (usually called simply a line) passing through them both. We call the line PIP,. Any nonvertical line in the plane has the property that the ratio rise ~Y Y, - Yl m=run=Ax=X, Xl

y

has the same value for every choice of the two points P,(XIoYI) and P,(x"y,) on the line (Figure A 7). This is because the ratios of corresponding sides for similar triangles are equal.

liy

(rise)

liy'

DEFINmON

The constant ratio rise

lix

Q(x" y,) I

~Y

m = run = Ax =

Y, - Yl X,

X,

(run)

I I -----------~Q'

is the slope of the nonvertical line PIP,.

lix' --~+_--------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 rapid the rise or faIl. The slope of a vertical line is undejined. Since the run Ax 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 e/> (the Greek letter phi) is the inclination of a line, then o :5 e/> < 180°. The relationship between the slope m of a nonverticaI line and the line's angle ofinclination e/> is shown in Figure A.10:

m

= tane/>.

Straight lines have relatively simple equations. All points on the vertlea/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/a/line 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(XIo Yl) 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 - Yl

=

mix - Xl),

Y =YI + mix -Xl).

or

The equation

Y

= Yl

+ mix -

XI)

is the point-slope equation of the line that passes through the point (XhYl) and has slope m. FIGURE A.8 The slope of L, 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

~"Sx

x increases 4 units. !his

----ct'--'.-~x I

...

l1y

m=.ax=taDt/>

I

-t-------x

Inotthis

//Inotthis

FIGURE A.9 Angles of inclination are measured counterclockwise :from the x-axis.

EXAMPLE 2

FIGURE A.I0 The slope ofanonvertical line is the tangeot ofits angle ofinclination.

Write an equation for the line through the point (2, 3) with slope - 3/2.

Solution We substitute Xl = 2,Yl = 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

3

3 - 2(x - 2),

=

80

the line intersects the y-axis at Y = 6.

or

•

-5

m = -2 _ 3 = -5 = I.

Along !his line,

-3-1--+-'----

=

The line's slope is

Solution

-I - 4

y~3

Y

Write an equation for the line through (-2, -I) and (3,4).

5 4

3

-2x + 6.

Y

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)

-f--'--+---l-L---+ x

o

FIGURE A.II

3

4

The standard equations for the vertical and horizontal lines through (2,3) are X ~ 2 andy ~ 3.

Y = -I

+ lo(x - (-2))

y=-I+x+2

With (X"YU = (3,4)

Y = 4

+ lo(x - 3)

y=4+x-3

y=x+1

/y=x+1 - - - - - - - - - - Same ....u1t

Either way, Y =

X

+

I is an equation for the line (Figure A.12).

•

Appendix 3 Lines, Circles, and Parabolass

AP-13

The y-coordinate ofthe point where a nonverticalline intersects the y-axis is called the y-intercept ofthe line. Similarly, the x-intercept of a nonhorizontailine 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 eqoation

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 general linear 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 b.

Lines that are parallel have equal angles ofinclination, 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 mt m2 = -I, so each slope is the negative reciprocal ofthe other: mt =

Y

I -m2'

To see this, notice by inspecting similar triangles in Figure A.14 that mt = a/h, and m2 = -h/a. Hence, mlm2 = (a/h)( -h/a) = -1.

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 4ADCissimilarto !i.CDB. Hence
Y

This distance is

d

Y2

~ Vrel'-2-_-',~12c-+-cI-Y2-_-Y~112 ~

V ('2-·,)2+ (Y2-y,)2

\

o FIGURE A.15 To calculate the distance between P(x"y,) 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

d

=

Y(llx)2

+ (~y)2

=

Y(X2 - Xl)'

+ (Yo - Yl)'.

EXAMPLE 4

y

(a) The distance between P( -1,2) and Q(3, 4) is

Y(3 - (-I)' + (4 - 2)'

•

=

Y(4)' + (2)'

v20 = v'4=5 =

=

2v'5 .

(b) The distance from the origin to P(x, y) is C(h, k)

• (x - h)'

+ (y -

k)' ~.'

--+_---C_---C---C_---C_ _~ 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.

(I)

Equation (1) is the standard equation ofa circle with center (h, k) and radius a. The circle of radius a = I and centered at the origin is the unit circle with equation

X2 +y2=1.

EXAMPLE 5 (a) The standard equation for the circle of radius 2 centered at (3, 4) is

{X - 3)2 + (y - 4)2

= 22 =

4.

(b) The circle

(X - 1)'

+ (y + 5)'

=

3

hash = l,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 fOTIO. The algebraic technique for doing so is completing the square.

EXAMPLE 6

Find the center and radius of the circle X

2 + 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) (x

2

+ (y2 - 6y)

=

Gather tmms. Move the con_ t to the right-hand side.

3

Add the square ofhalfthe coefficient of x to each side of

+ 4x +

the equation. Do the same for y. The parenthetical expressions on the left-hand side are now perfect squares.

Y Exterior: (x - h)'

+ (y -

k)'

> .' (x 2 + 4x

(x

+ 4) + (y2 - 6y + 9)

+ 2)' +

=

3

+4+9 Write each quadratic as a squared linear expression.

(y - 3)2 = 16

•

The center is (- 2, 3) and the radius is a = 4. The points (x, y) satisfYing the inequality Interior: (x - h)'

+ (y - k)' < .'

H----"---------->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 circle (x - h)' + (y - k)' = .2. y y =x 2 (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 ofequations ofthe 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).

•

----'---'---""'",L---'---"---~ x

-2

-1

FIGURE A.18 (Example 7).

0

2

The gmph of an equation of the form

y

Theparabolay = x 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.

The Graph of y

= ax' + I»c + c,

a*"O

The graph of the equation y = ax 2 + bx + c, a i' 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, then we have y = bx + c, which is an equation for a line. The axis, given by Equation (2), can be found by completing the square.

EXAMPLE 8

Graph the equation y = -

2

-

Comparing the equation with y = ax 2

Solution

I a =--

+ 4.

X

+

bx

+ c we see that c = 4.

b = -I,

2'

Since a line

tx

< 0, the parabola opens downward. From Equation (2) the axis is the vertical b x=--= 2a

(-I) 2(-1/2)

= -I.

When x = -I, we have

FIGURE A.19 Besides determining 1he direction in which 1he parabola y = ax 2

1 (_1)2 - (-I) y = _ -2

opens, 1he number a is a scaling factor. The parabola widens as a approaches zero aod narrows as Ia Ibecomes large.

+4

=

-29.

The vertex is (-1,9/2). The x-intercepts are where y = 0:

_lx 2 Intercept at y

~O,4) y=

~

2

4

-

x+4

=

0

x 2 +2x-8=O (x - 2)(x + 4) = 0 x = 2, x =-4

_~x2 -x+ 4

We plot some points, sketch the axis, and use the direction of opening to complete the •

'\2

-k-;;---;;---;--;;i-~~.---~x 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 I aod 2, a particle moves from A to B in 1he coordinste plaoe. Find 1he increments dx aod 4y in 1he particle's coordinstes. Also f"md 1he distance from A toB.

1. A( -3,2), B( -I, -2)

Plot1he points in Exercises 5 aod 6 aod find 1he slope (if soy) of 1he line 1hey determine. Also f"md 1he corornon slope (if soy) of1he lines perpendicular to line AB. S. A( -1,2),

B( -2, -I)

6. A(2,3),

2. A( -3.2, -2), B( -8.1, -2) In Exercises 7 aod 8, find ao equation for

Describe 1he graphs of1he equations in Exercises 3 aod 4.

3. x 2

+ y2 = I

4. x 2

+ y2

'"

3

B( -1,3)

<0> 1he vertical line aod

(b> 1he horizontal line lhrough 1he given point 7. (-1,4/3) 8. (0, -yz)

Appendix 3 In Exercises 9--15, write an equation for each line described.

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

39. y ~ -x 2, Y ~ 2x2 - I 40. x 2 + y2 ~ I, (x - 1)2

13. Has y-intercept 4 and x-intercept -I 14. Passes through (5, -I) and is parallel to the line 2x

+ 5y =

IS

15. Passes through (4, 10) and is perpendicular to the line 6x - 3y ~ 5.

80'

+ 4y

= 12

17.

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,ay = -6. Find its new position.

20. The coordiuates of a particle change by dx = 5 and ay = 6 as it moves fromA(x,y) to B(3, -3). Find x andy. 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 -o/"plane. Include the circle's center in your skeroh. Also, label the circle's x- and y-intercepts, if any, with their coordinate pairs. 22. C(-1,5), a = v'iO 21. C(0,2), a = 2

23. C( -v3, -2), a

=

+ y2

~

I

41. Insulation By measoring slopes in the figure, estimate the temperature change in degrees per inch for (a) the gypsum wallboard; (b) the fiherglass insulation; (0) the wood sheathing.

In Exercises 16 and 17, fmd the line's x- andy-intercepts and use this information to graph the line. ]x

AP-17

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, I) with slope -I

16.

lines, Circles, and Parabolas

~f

:::H=I 1++1 1++1

70'

/.M..Yi!\ WboYrdU ,

60'

~-WF1Jiism 'bORv"

,

studS

'.~.~ :(. ".: ~

R

~

~

30'

I-+-+-

g

g

ffiu~~lfnan . . +- fiJFEW ~ 40' I:; 1+++ '. +.W-l-I+ .

Air E 50' ~siil

b~

%'

I .~

Airo

:ift~

_~Wt

%

20'

1++=

10'

1++1

j~

:+:W9EFEW I IIII III III III I

0'

o

1

.

,,1.;,

234

5

6

7

Distance tbrough wall (inches)

2

The temperature 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 (ifany) 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 Grapb 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

-

I 30. y ~ 2x2

6x - 5

+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

> I,

x 2 + y2

+ 6y < 0,

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 a1mospheres. Find the pressure at 50 meters. 44. Reflected light A ray oflight 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

/

<4

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 radiua V2, and on or to the right ofthe vertical line through (I, 0).

--:+-----"'---------~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 calculating 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 points 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 different 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 Rallway Civil eogineers calculate the slope of roadbed as the ratio of the distance it rises or falls to the distance it runs horizontally. They call1his ratio the grade of the roadbed, usually wtitteo as a percentage. Along the coast, commercial rallroad grades are usually less than 2%. In the mountains, they may go as high as 4%. Highway grades are usually less than 5%. The steepest part of the Mt. Washington Cog Railway in New Hampshire has an exceptional 37.1 % grade. Along this part of the track, the seats in the front 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 k is the line 2x + ~ = 3 perpendicular to the line 4x + Y ~ I? For what value of k are the lines parallel? 52. Midpoint of a line segment nates Xl (

Show that the point with coordi-

+ X2 2

+

Yl

'

2

Y2)

is the midpoint of the line segment joining P(xj, 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 1-L1mlt Laws

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

lim f(x) = L

and

x--'Joe

lim g(x) = M,

then

x-c

+ g(x))

+M

1. Sum Rule:

lim (j(x) x-c

2. Difference Rule:

lim (j(x) - g(x)) = L - M x-c

3. Constant Multiple Rule:

lim (k' f(x)) = k' L x-c

4. Product Rule:

lim (j(x)' g(x)) = L' M x-c

s.

lim f(x) = £ x-cg(x) M'

Quotient Rule:

= L

M" 0

6. Power Rule:

lim [f(x)]' = L', n a positive integer x-c

7. Root Rule:

lim x-c

i;/.fW = "'?'L = L 11., n a positive integer

(lfn is even, we assume that lim f(x) = L x-c

>

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

AP-19

Suppose then that • is a positive number, and write f(x) and g(x) as

f(x)

L + (j(x) - L),

=

g(x)

=

M + (g(x) - M).

Multiply these expressions together and subtract LM:

f(x)' g(x) - LM

= =

=

(L + (j(x) - L))(M + (g(x) - M)) - LM LM + L(g(x) - M) + M(j(x) - L) + (j(x) - L)(g(x) - M) - LM L(g(x) - M) + M(j(x) - L) + (j(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 114 such that for all x inD

o< o< o<

Ix - cl < III

Ix - cl < 112 Ix - cl < 113 0< Ix - cl < 114

=> => => =>

If(x) - LI < yf;f3 Ig(x) - MI < yf;f3 If(x) - LI < ./(3(1 + IMI)) Ig(x) - MI < ./(3(1 + ILl)).

(2)

lfwe take Il to be the smallest numbers Il, through 1l4 , 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

:;:.~~tquality·PPlied

If(x)' g(x) - LMI

+ IMllf(x) - LI + If(x) - Lllg(x) - MI + ILI)lg(x) - MI + (I + IMI)lf(x) - LI + If(x) - Lllg(x) -

,;; ILIIg(x) - MI ,;; (I

••

rere

< 3 + 3 + V3V3

= •.

MI

Value. from (2)

•

This completes the proof ofthe 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) = lim (f(X)

.~,

g(x)

.~,

._1_) g(x)

= lim

.~,

f(x) • lim

.~,

_1_ g(x)

=

L. l M

=

£

M

by the Limit Product Rule. Let. > 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

< •.

Since IMI > 0, there exists a positive number Il, such that for all x

0< Ix - cl < Il,

=>

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(x)l

Therefore, 0

< Ix - c I <

IMI

(4)

Ig(x)l"

III implies that

I I I Ig(x)-M=

M - g(x) I

1

Mg(x)

I

I

I

"'IMI'lg(x)I'IM-g(x)1

2

< IMI'IMI'IM - g(x) I· Since (1/2) IMI 2E >

(5)

Inequality (4)

0, there exists a number 112 > 0 such that for all x (6)

If we take Il to be the smaller of Il, and 1l2 , the conclusions in (5) and (6) both hold for all x such that 0 < Ix - c I < Il. Combining these conclusions gives

0< Ix - cl < Il

~ Ig(~)

-11

<

E.

•

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 contsining c, except possibly at x = c itself. Suppose also that limx~c g(x) = lim_ c h(x) = L. Then limx~c f(x) = L.

Proof for Right-Hand Limits Suppose limx~c+ g(x) = limx~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

+

E

and

L -

E

< h(x) <

L

+

E.

These inequalities combine with the inequality g(x) ", f(x) ", h(x) to give

L -

E

L -

E

-

E

< g(x) ", f(x) ", h(x) < L + < f(x) < L + E, < f(x) - L < E.

Therefore, for all x, the inequality c

E,

< x < c + 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

+

e

We conclude as before that for all x, c - Il

and

L - e

< h(x) < L +

e.

< x < 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) = L and 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 polyoomial functions andg(c) #' 0, then

1. Suppose that functions h(x), h(x), and h(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 /1(x), h(x), ... , I.(x) have limits LI, L 2, ••• ,Ln as x---+ c, then

.

I(x)

I(c) g(c)'

hm-~­

.~c g(x)

6. Composite. of continuous function. Figure A.21 give. the diagram for a proof that the compo.ite of two cootinuous functions is continuous. Reconstruct the proof from the diagram. The .tat<>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 domain of 1 and that I(c) is an interior point of the dumain of g. This will make the 1iroits 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 n = en for any integer n > 1.

4. Limit. of polynomial. Use the fact that 1iro.~c(k) ~ k for any number k together with the results of Exercises I and 3 to show that 1iro.~cl(x) ~ I(c) for any polynomial function

gof

FIGURE A.21

The diagram for a proof that the composite of two continuous functions

is continuous.

A.5

Commonly Occurring Limits This appendix verifies limits (4)-(6) in Theorem 5 of Section 10.1.

..

< 1, lim x"

We need to show that to each e > 0 there corresponds an integer N so large that Ixnl < e for all n greater than N. Since e lfn --> I, while Ixl < I, there exists an integer N for which e 1/ N > IxI- In other words, Limit 4: If Ixl

"~

=

0

(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 Rille, in which we differentiate with respect to n:

lim n In(1

+ ,x:)

n---+OO

=

n

=

lim In(1 + x/n) n--+ OO lin

. C+\/n)- (- :2) . x

Iw

-lIn 2

n_OO

=

lim

n_OO

1

+ x/n =x .

Apply Theorem 3, Section 9.1, with f(x) = eX to conclude that

• •

Limit 6: For any number x, n_OCl lim iLl = 0 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 fjxl/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 Numbe"

A.6

AP-23

Theory of the Real Numbers A rigorous development ofcalculus is based on properties of the 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.

A4

+ (b + c) = (a + b) + c for all a, b, c. a + 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.

Ml

a(bc) = (ab)cforalla,b,c.

M2

ab = ba for all a, b.

Al

A2 A3

a

M3

There is a number called "I" such that a' I = a for all a.

M4

For each nonzero a, there is a b such that ab = I.

n

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

Ifa:5 bandb:5 athena = b.

03

Ifa:5 bandb:5 c then a :5 c.

04

Ifa:5 bthena

05

Ifa:5 bandO:5 c then ac :5 bc.

+

c:5 b

+

c.

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 = 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 the real numbers correspond to all points on the line, with no ~'holes" or ugaps." 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 ofmissing 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

v'=1

0

AP-24 Appendices negative nwnbers. So is M = 1, showing that 2 is not a least upper bound. The least upper bound for the set ofnegative nwnbers 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 rational nwnbers, the set of nwnbers less than is bounded, but it does not have a rational least upper bound, since any rational upper bound M can be replaced by a slightly smaller rational nwnber that is still larger than V2. So the rationals are not complete. In the real nwnbers, 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 maxin3wn value for a function on a closed interval [a, b], as in Section 4.1. The function y = x - x' has a maximwn value on [0, I] at the point x satisfYing 1 - 3x 2 = 0, or x = Vfl3. If we limited our consideration to functions defined only on rational nwnbers, we would have to conclude that the function has no maximwn, since Vfl3 is irrational (Figure A.22). The Extreme Value Theorem (Section 4.1), which in3plies that continuous functions on closed intervals [a, b] have a maxin3wn value, is not true for functions dermed only on the rationals. The Intermediate Value Theorem in3plies that a continuous function 1 on an interval [a,b]with/(a) < Oand/(b) > omust be zero somewhere in [a, b].The function values canoot jwnp from negative to positive without there being some point x in [a, b] where I(x) = O. The Intermediate Value Theorem also relies on the completeness of the real nwnbers and is false for continuous functions defined only on the rationals. The function I(x) = 3x 2 - 1 has 1(0) = -I and 1(1) = 2, but if we consider 1 only on the rational an nwnbers, it never equals zero. The only value ofx for which I(x) = 0 is x = irrational nwnber. We have captored the desired properties of the reals by saying that the real nwnbers are a complete ordered field. But we're not quite finished. Greek mathematicians in the school of Pythagoras tried to in3pose another property on the nwnbers of the real line, the condition that all nwnbers are ratios of integers. They learned that their effort was doomed when they discovered irrational nwnbers such as \12. How do we know that our efforts to specifY the real nwnbers 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 themselves at the bottom. An engineer trying to build such a staircase would find that no structore realized the plans the architect had drawn. Could it be that our design for the reals contains some subtle contradiction, and that no construction of such a nwnber system can be made? We resolve this issue by giving a specific description of the real nwnbers and verifying 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 inImite decimals,

v'z

y

0.5

---j'--!c-'--o'c-'--o'~--o'o-'--o'~---L-> x

0.1

0.3

0.5 I 0.7 y'jf3

0.9 1

FIGURE A.22 The maximum value of y = , - ,'on [0, I] occurs at the irrational number x = V1j3.

V1l3,

a.dt d2d,d4 • •• In this approach a real nwnber is an integer a followed by a sequence of decimal digits dl, 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 nwnbers. The real meaning of the dots " ... " following these digits requires development of the theory of sequences and series, as in Chapter 10. Each real nwnber is constructed as the limit of a sequence of rational nwnbers given by its rmite decin3al approxin3ations. An inImite decimal is then the same

as a series dl d2 a+-+-+ .. · 10 100 This decimal construction ofthe real nwnbers is not entirely straightforward. It's easy enough to check that it gives nwnbers that satisfy the completeness and order properties,

Appendix 7 Complex Numbe"

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 ax 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) "ax as x varied over a closed interval. While the approximating difference quotients ~y/ ~x were understood much as today, it was the quotient ofinfinitesimal quantities, rather than a limit, that was thought to encapsulate the meaning of the derivative. This way ofthinking 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 ofthe 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 can be performed with these numbers without getting outside the system. That is, the system of positive integers is closed under the operations of addition and multiplication. By this we mean that if m and n are any positive integers, then

m+n

=

P

and

mn

= q

(1)

are also positive integers. Given the two positive integers on the left side of either equation in (I), we can find the corresponding positive integer on the right side. More than this, we can sometimes specify the positive integers m and P and tmd a positive integer n such that m + n = p. For instance, 3 + n = 7 can 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 and 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)

an educated person can always find the missing integer that solves the equation m + n = 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 (I), they are given m and q, they discover that sometimes they can find n and sometimes they cannot. Using their imagination, they may be inspired to invent still more numbers and introduce fractions, which are just ordered pairs min of integers m and n. The number zero has special properties that may bother them for a while, but they ultimately discover that it is handy to have all ratios of integers 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:

1. (a) addition (b) subtrsction

1

FIGURE 0\.23

With a straightedge and

compass, it is possible to construct a

2. (a) multiplication (b) division

on any two numbers in the system, except that they cannot divide by zero because it is meaningless. The geometry of the unit square (Figure A.23) and the Pythagorean theorem showed that they could construct a geometric line segment that, in terms of some basic unit of length, has length equal to Thus they could solve the equation

V2.

x2

segment ofirrationalleogth.

= 2

by a geometric construction. But then they discovered that the line segment representing is an incommensurable quantity. This means that cannot be expressed as the ratio oftwo 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 and q with no common factor other !ban 1, and such that

V2

V2

p2 = 2 q 2.

(3)

Since p and q are integers, p must be even; otherwise its product with itself would be odd. In aymbols, p = 2PI, wbere PI is an integer. This leads to 2pl2 = q2 which says q must be even, say q = 2qj, where ql is an integer. This makes 2 a factor of both p and q, contrary to our choice ofp and q as integers with no common factor other !ban I. Hence there is no rational number whose square is 2.

Appendix 7 Complex Numbe"

AP-27

Although our educated people could not f"md a rational solution of the equation x 2 = 2, they could get a sequence ofrational numbers

I

7

l'

5'

239 169'

41 29'

... ,

(4)

whose squares form a sequence

I I'

49 25'

1681 841'

57,121 28,561'

... ,

(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 ofrational 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 Imagination was called upon at many stages 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 contains the previous system. Each system is also richer than its predecessor in that it permits additional operations 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 ofEquations (6) and (7) and, in addition, 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)

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 + I

=

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, tl) are equal if and only if a = candb = d.

Addition (a + ib) + (c + id) = (a + c) + i(b + tl)

The sum of the two complex numbers (a, b) and (c, tl) is the complex number (a + c, b + d).

Multiplication (a + ib)(c + itl) = (ac - btl) + i(ad c(a

+ ib)

=

ac

The product of two complex numbers (a, b) and (c, tl) 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

+ (c,O)

=

(a

(a, 0)' (c, 0)

=

(ac, 0),

(a,O)

+ c,O),

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, atl) = a(c, d). In particular, the complex number (0, 0) plays the role of zero in the complex number

system, and the complex number (I, 0) plays the role of unity or one. The number pair (0, I), which has real part equal to zero and imaginary part equal to one, has the property that its square,

(0,1)(0, I)

=

(-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, I) wbose square can be added to unity = (1,0) to produce zero = (O,O),thatis,

(0, 1)2 + (1,0)

=

(0,0).

Appendix 7 Complex Numbe"

AP-29

The equation

x2

+

I = 0

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, I) = i than there is about the symbol (I, 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) = 0 + iO. But if a + ib 0, then we may carry out a division as follows:

*

e

+ id

a + ib

(e (a

+ id)(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 O,sincea + ib = (a,b) (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 Z (read "z bar") to denote the complex conjugate ofz; thus

z

=

a + ib,

z=

a - ib.

Multiplying the numerator and denominator ofthe fraction (e + id)/(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. (a) (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 ofthe complex number z = x + iy:

1. 2.

as the point P(x, y) in the xy-plane as the vector

oF from the origin to P.

•

AP-30

Appendices

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

y P(x, y) I I I I I

x

:y

X

X

FIGURE A.24 This Argand diagram represents z = x + iy both as a point

P(x, y) and as a vector liP.

y

rsin8,

=

and

I I I

0

rcosB,

=

+ isinll). (10) We define the absolute value of a complex number x + iy to be the length r of a vector z = x

+ 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

e iO

=

cos 8 + isin8,

called Euler's formula, enables us to rewrite Equation (10) as

z

=

re iO •

This formula, in tum, leads to the following rules for calculating products, quotients, powers, and roots of complex numbers. It also leads to AIgand diagrams for e '•. Since cos II + i sin II is what we get from Equation (10) by taking r = I, we can say that e '• 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

Yei8=cos6+isin8 \

-+----:;;or~----f'-'>x

(a)

(cos 6, sin 6)

-+----:;;or~~-+-->x

(b)

FIGURE A.25 Atgand diagrams for e i8 = cos II vector and (b) as a point.

+ i sin II (a) as a

Products To multiply two complex numbers, we multiply their absolute values and add their angles. Let

(11)

Appendix 7 Complex Numbe"

AP-31

so that

y

Then

and hence IZjZ21 arg (ZjZ2) FIGURE A.26 When z, and Z2 are multiplied,lzjz21 = 7,"72 and arg (ZjZ2) = 8j + 82.

=

=

9j

7t 72 = IZd"I Z21

+ 92

arg Zl

=

(12)

+ arg Z2.

Thus, the product of two complex numbers is represented by a vector whose length is the product ofthe lengths ofthe two factors and whose argument is the sum oftheir 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.

EXAMPLE 2 Let Zl = 1 + i, Z2 = y!3 - 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 -i7f/6 .

I

n11C'-"-r-~=-r---~'---r~x Then

1+v3

Zl Z 2 =

2v'z exp

-I

=

FIGURE A.27

e:-

in

2v'z (cos;;

=

2v'zexpG~)

+ i sin ;;) '" 2.73 + 0.73i.

To multiply two complex

numbers, multiply their absolute values and add their arguments.

•

The notation exp(A) stands fore A •

Quotients Suppose 1'2 oF 0 in Equation (11). Then

Hence

That is, we divide lengths and subtract angles for the quotient of complex numbers.

EXAMPLE 3 1

Letzl

+i _

=

1

+ iandz2 = y!3 - i,asinExample2.Then

v'ze tw/4 _

y!3 _ i - 2e tw/6 '" 0.183

v'z ''''/12 0707( 57T 2 e "'. cos 12

+ 0.683i.

+ I..sm 57T) 12

•

AP-32

Appendices

Powers If n is a positive integer, we may apply the product fonnulas in Equation (12) to [md zll = z·z· ... ·Z.

With z

= re

iO

,

11

factors

we obtain " summands

(13) The length r = Iz Iis raised to the nth power and the angle (J = arg z is multiplied by n. If we take r = I in Equation (13), we obtain De Moivre's Theorem.

De Mcrivre's Theorem (cos (J

+ i sin (J)"

= cos n(J

+ i sin n(J.

(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 n(J and sin n(J as polynomials of degree n in cos (J and sin (J.

EXAMPLE 4

Ifn = 3 in Equation (14), we have (cos(J

+

isin(j)3 = cos3(J

+ i sin 3(J.

The left side ofthis equation expands to cos3 (J

+

3i cos2 (J sin (J - 3 cos (J sin2 (J - i sin3 (J.

The real part ofthis must equal cos 3(J and the imaginary part must equal sin 3(J. Therefore, cos 3(J = cos3 (J - 3 cos (J sin2 (J, sin 3(J = 3 cos2 (J sin (J - sin3 (J.

•

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, ... , W.-I, that are nth roots ofz. To see why, let W = pe la be an nth root ofz = 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 must be equal, we can say that they may differ only by an integer multiple of2'lT. That is,

(J

na =

(J

+ 2k7r,

k

=

0, ±I, ±2, ....

Therefore,

a

(J

2'lT

=,,+ kn.

Appendix 7 Complex Numbe"

AP-33

Hence, all the nth roots of z = rei' are given by

y

k =

----j---~F~-_;;_+_~x

(15)

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

= 0, 1,2, ... ,n -

1.

All the nth roots of rei' 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 = rem.

Wz FIGURE A.28

0, ±l, ±2, ....

The three cube roots of

z=re i8 .

EXAMPLE 5

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 rei'. Here, z = -16, r = + 16, and (I = 7T. One of the fourth roots of l6e i " is 2e i"(4. We obtain others by successive additions of 27T/4 = 7T/2 to the argument ofthis first one. Hence,

y

'¢'16

.

expl1T -

2

.(7T 37T 57T 77T) expl 4'4'4'4 '

and the four roots are x

-----.1____

-16

Wo = 2 [cos

FIGURE A.29 of-16.

The four fourth roots

Wj

= 2 [cos

W2

= 2 [cos

W3

= 2 [cos

~ + isin ~]

=

3; + 3;] 5; + 5;] 7; + 7;]

V2(1

+

i)

+ i)

isin

=

V2(-1

isin

=

V2(-1 - i)

isin

=

V2(l - 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 ofthe 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

anz n +

n

Cln_lZ -

1

+ ... +

alZ

+ ao

=

0,

in which the coefficients ao, aj, ... , an are any complex numbers, whose degree n is greater than or equal to one, and whose leading coefficient an 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 compl"" numbers Find (a, b)· (e, d) = (ae - bd,ad

+ be).

a. (2,3)' (4, -2) c. (-I, -2)'(2, i)

b. (2, -I)· (-2,3)

15. cos 48

(This is bow complex numbers are multiplied by computers.)

+ 4i)'

- 2(x - iy) = x

(I + ~)' + _1_. ~ I+ x+lY

b.

16. sin 48

17. Find lbe tbree cube roots of 1.

2. Solve lbe following equations for lbe real numbers, x andy.

a. (3

Powers and Roots Use De Moivre's Theorem to express the trigonometric functions in Exercises 15 and 16 in tenns of cos 8 and sin 8.

I-I

+ iy

18. Find lbe two square roots of i. 19. Find lbe tbree cube roots of -8i. 20. Find lbe six sixlb roots of 64.

i

23. Find all solntions oflbe equation x' 24. Solve lbe equation

Graphing and Geometry 3. How msy lbe followiog complex numbers be obtained from z = x + iy geomelrically? Sketch.

z

b. (-z)

c. -z

10 Exercises 5-10, graph lbe points z conditions. b. Izl

6. Iz - II = 2

+ II ~ Iz 10. Iz + 11'" Izl 8. Iz

II

<2

~

x

+ iy that satisfy lbe given

+

13. 1 + 1-

'\1'=3)'

iV3 iV3

O.

Theory and Examples

25. Complex numben and vecton in the plane Show with an

9. Iz

+ il

~ Iz -

26. Complex aritbmetic with conjugates Show that lbe conjngate oflbe sum (product, or quotient) of two complex numbers, ZI and z" is lbe same as lbe sum (product, or quotient) of lbeir conjugates. 27. Complex roots of polynomials with real coefficients come in complex-conjugate pairs

>2 7. Iz + 11 = 1

c. Izl

a. Extend lbe results of Exercise 26 to show that j(Z) = j(z) if 11

f(z) = a"zn

+ an_tZ"-1 + "." + alZ + ao

is a polynomial with real coefficients aQ•...• all"

Express the complex numbers in Exercises 11-14 in the form re i6 , wilb r '" 0 and -'IT < 8 '" 'IT. Draw an Argand diagram for each calculation.

11. (1

x' + 1 =

O.

Argand diagram that lbe law for adding complex numbers is lbe same as lbe parallelogram law for adding vectors.

d. liz

4. Sbow that lbe distance b&weeo lbe two points ZI and z, in an Argand diagram is IZI - z,l.

5. a. Izi = 2

+ 2z3 + 4x' + 16 =

22. Find lbe six solutions oflbe equation z·

c. (3 - 2i)(x + iy) = 2(x - 2iy) + 2i - I

a.

+ 4 ~ O. + 2 = O.

21. Find lbe four solutions oftbe equation z· - 2z'

12L±l • 1- i 14. (2

+ 3i)(I

- 2i)

b. Ifzisarootoflbeequationj(z) = O,wherej(z) is a polynomial wilb real coefficients as in part (a), show lbat the conjugate z is also a root of the equation. (Hint: Let j(z) ~ u + iv ~ 0; lben bolb u and v are zero. Use tbe fact lbat j(z) = j(z) = u - iv.)

28. Absolutevalueofaconjugate Showthatlzl = 14 29. When z = i If z and z are equal, what can you say abont the location of the point Z in the complex plane?

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 numbers z, Zl , and Z2. a. z + Z = 2Re(z) b. z -

z=

2iIm(z)

c. IRe(z) I :::; Izl IZI + z21 2 = IZll2 + IZ212 + 2Re(zlz2) e. IZI + z21 :::; IZII + IZ21

d.

AP-35

ANSWERS TO ODD-NuMBERED EXERCISES = {x,

29. (a) f(x)

CHAPTER 1

2 -x + ,

Section 1.1, pp. 11-13

2,

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. (a) Not a function ofx because some values ofx have two values ofy (b) A function ofx because fur every x there is only one possible y

_\/3

9. A -

_

'Ix,2

p - 3x

2

-/Ox +

13. L = v'2Ox

15. (-00,00)

_

d

11. x - \/3'

A ~

2d2,

_d' V-

3\/3

25 17. (-00,00)

,

,

=

(b) f(x)

0:5x<1

0, 1:5 X < 2 2,2::S;x<3

1

0,3::S;x:54

-x

-1:5 x < 0 0 < X :5 I l<x<3 -2:5x:50 (b) f(x) = 0 < X :5 I -I, l<x:53 33. (a) 0 :5 X < I (b) -I < x :5 0 35. Yes 37. Symmetric about the origin 39. Symmetric about the origin

=

31. (a) f(x)

I,' { -~x+~, ~x, -2x + 2,

{

,

g(x) .. -JJXr

-.5-4-3-2-~1

I 2 3 4.5

x

-2

--4~-_-:02-+-~.---c~-X

-2

me. -00

< x < Oaud

O<x
,

19. (-oo,O)U(O,oo) 2

Dec. -00 < x < 00 41. Symmetric about the y-axis

F(t) .. i...

,

1'1 11--'-'--

4321

43. Symmetric about the origin

,

1234

~~

4

21. (-00,-5)U(-5,-3]U[3,5)U(5,00) 23. (a) 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 < x ::s 0; inc.O::S;x
Inc.

-00

<x<

00

45. No symmetry

,

--;--*-~~X

"ok-~~~~x

-1

-2

25.

,

27.

~)=l:x:

OS-tSI

2-x, l<x:S2

--;of----;---'::2-~X

-,

,

-4

->

Dec.O:5x
47. Even 49. Even 51. Odd 53. Even 55. Neither 57. Neither 59. t = 180 61. 8 = 2.4 63. V ~ x(l4 - 2x)(22 - 2x) 65. (a) h (b) f (c) g 67. (a) (-2,0) U (4, 00) 71. C = 5(2 + Y2)h

A-1

A-2

Chapter 1: Answers to Odd-Numbered Exercises

Section 102, ppo 19-22 1. Df : -00 < x < 00, Dg : x

1, Rf : -00 < y < 00, R.: y '" 0, D f +. = D f .• = DIP Rf +.: y '" I, Rf .• : y '" a 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.ff: y '" 1/2 5. (a) 2 (b) 22 (0) x 2 + 2 (d) x 2 + lOx + 22 (e) 5 (f) -2 (g) x + 10 (b) x' - /ix2 + 6

7. 13 - 3x

9.

11. (a) f(g(x))

~~~ :

2 y-l=_I• -1

0

3x

•

~o • , .. 1

37.

(d) j(j(x))

39.

y

x

I x

x

a

-2

0

(f)

a

45.

•

),.01-.1"3

47.

y

-I

2

y

2

~ .-2

•

3

-I

0

2

3

-I

•

-2

49.

+ 1)3

y

51.

y

, >,,",]1;3

-2

-2

53.

0

1

,

2

y""

3

1 (.%_1)2

2

•

-I 0

1

2

3

y

J~ t+

2x

y

->

----.hb----->.

•

,"{+2

1 -3 -2 -I

y

"--

3

)'+1"(.1:+1)

-0.81

,

2

y

)' .. ~-l

y

---c"'c-t-~---~.

1

43.

y

+I

~

0

•

21. (a) y = -(x + 7)2 (b) y = -(x - 4)2 23. (a) Position 4 (b) Position I (0) Position 2 (d) Position 3 25. (x + 2)' + (y + 3)2 = 49 27. y + I = (x

31. y

•

•

~~ + I,g(f(x)) = ~

v'x + 0.81

• ,

2

41.

(e) -I

x

~ (1,1)

x

(d)

3

2

+6

1+ 1 x

(0) -2

y

g)(x)

(b) Dfo, = (-00,-1] U (0, oo),D,of = (-1,00) (0) Rfo, = [0, I)U(I, oo),R,of = (0,00) 2x 19. g(x) = x - I

29. y ~

•

2

-I

w-=s

x x - I

(b) 2

y .. ..[i+4

1

~

v;=s

~

17. (a) f(g(x)) =

y

y= Ix-21 (f

~ 3x

(d) _x_ x - I I (e) x - I

15. (a) I

(0) g(g(x))

f(x)

(a) x -7 (b) x + 2 (0) x2

35.

y

(f) h(j(f(x)))

g(x)

(f)

I 33. y - I = - x- I

:

(b) j(g(x))

(e) g(h(f(x)))

13.

~

10

1 2

1

•

55. (a) D: [0,2], R: [2, 3]

4

(b) D: [0,2], R:[-I,O]

y

y

,-f(x)-l

0

0

2

3

•

•

•

•

A-3

Chapter 1: Answers to Odd-Numbered Exercises

,

(c) D: [0,2], R: [0,2]

(d) D: [0,2], R:[-I,O]

,

75.

,

,

77.

-, .. -f(x)

0

,

2

x

2

0

-~_2C-_~''---+--'¥-~2:--->X

x

(e) D: [-2,0], R:[O,I]

(f) D: [1,3], R:[O,I]

,

,

-4

-.

-I

-I

,

79.

,

81.

2 2

--~2-_~,-I-~-c:---:----->x

y-f(x+2)

y=f(x-l)

x

-2

x

0

(g) D: [-2,0], R:[O,I]

-~_2c-_~,-f----:-~2:--->X

(h) D: [-I, I], R:[O,I]

,

,

2

83.

2 ,--f(x+l)+l

y=f(-x)

----*"2

(x

+ 4)'

+

16

63. y =

10

,

---~. 65. y = I - 27x'

,

69.

85. (8) Odd (f) Even

2

(b) Odd (g)Even

(c) Odd (d) Even (h) Even (1) Odd

(e) Even

x

-2 -I

-3 -2 -1

I

2

, , 4

Section 1.3, pp. 28-30 x

1. (8) 8". m

5. 0

-4

, , Y=~-l

71.

,

Center: (-4,3)

(H4i + 0_3)2 _1 8

67.,

4

I

<1+2)2

,

3

~4 - ~

3)'

(x-I?

-2- + -,-_I

The major axis is the line segment between (-8,3) and (0,3).

16

~ 3>;2 -

9

---~'-~O.L--~-->x

~-~-~~x

57. y

(y -

-4

1

-".

sin 8

0

4

,

cosB

-I

2

2

tsnO

0

cot 0

UND

sec 0

-I

esc 8

UND

~

,

73.

4

4

x

x y=-~

3. 8.4 in.

9

,

1 2

(h) 55". m

-2"./3

0

V3

0

2 I 2

V3 I

V3 -2 2

V3

7. cosx

= -4/5, tsnx = -3/4

9. sinx

= - V8 tsnx = -V8 3 '

11. sinx

I 2 = --,COSX =--

V5

V5

"./2

3"./4 I

V2 0

I

V2

0

UND

-I

UND

0

-I

UND

-V2 V2

UND

A-4

Chapter 1: Answers to Odd-Numbered Exercises

15. Period 2

13. Period" y

67. A

y

---f----.lc----.r.--~,

17. Period 6

2 = -",B = 4,C = O,D ="1

--f--.+---\-+---f-+---L~.,

19. Period 2" y

y

Section 1.4, p. 34 1. d 3. d 5. [-3,5]by[-15,40]

7. [-3,6] by [-250, 50]

--;;f---;C--.x---k--~'

'0

21. Period 2"

23. Period"/2, symmetric about the origin

10

y

-_*,-_-;,-t--;--y--{-t;-~,

, , , ,

,

1",~2t i

,

. \'

"'-•...-0l-.~~"'!---~7,...4, -4

"4

4

\

\

4

I

T

, , , , , ,

9. [-3, 3] by [-6, 6]

,,

, , , , , , ,

I

I

11. [-2, 6] by [-5,4]

i

y

\ ~ \

o

, , , ,

I 25. Period 4, symmetric about

\

-10

y

f(Jl)=X~

29. D: (-00, 00), R:y~-I,O,I

they-axis

y , , ,

I='~¥

,_!inx

I,

-211" ......--1(

, , ,

, , , , ,

, , , , ,

, , , , ,

I

I

I

I

39. -cos x

41. -cosx

47.

2+ V2

53.

6'2'6'2

4 '1T '11'

65. A

49.

51T 31T

~

2a"

....."

13. [-2,8] by [-5,10]

15. [-3, 3] by [0, 10]

y

y

-1

i(\2!-1 i(\2! , , , , ,

..

-.''-O..:'~.;-,--.f'-O'''''.;-. ~,I-~,

43.

V6+V2 4

2- v'3 51.

4 ...

45.

V2+V6

" 2" 4?J" 5"

3'3'3'3

~

59. v7 '" 2.65

4

-_.:-_*,-+--!c----!"'c-;;-o\c-~, -,

.,

-.

17. [-10, 10] by [-10, 10] 63. a = 1.464

19. [-4,4] by [0, 3]

y

y

,.,

= 2,B = 2", C = -",D = -I s

y

)1=

'D

x+2

6

y_2sinl;;+1t)_1

10 8 6-4

• , • 6 S

,

"

f(x)=

246810

~+2

"+1

'D

, 1.0 0' -4 -3 -2 -I

1

,,•

,

Chapter 1: Answers to Odd-Numbered Exercises

21. [-10, 10] by [ -6, 6]

23. [-6, 10] by [-6,6]

y

,

, ,• ,

39. y

y

!(x) .. 6.1:

2

-, -,

,

\,

x

10

-, -4

46810:%

,•

2 -1

-, -,

8

3

\

1;5'1;5] by

/

x

• 1 • -,- x-3

/ '4321

1234':%

-,

27. [-IO()".,IOO".]by

[-1.25, 1.25]

y

Practice Exercises, pp. 35-36

y

l.0

y-~(~)

1. A

= ".r', C = 2".r'41T A = C'

3. x

= tanO,Y = tan'O

5. 13. 17. 19. 21. 23. 25. 27. 29.

Origin 7. Neither 9. Even 11. Even Odd 15. Neither (a) Even (b) Odd (0) Odd (d) Even (0) Even (a) Domain: allreals (b) Range: [-2, 00) (a) Domain: [-4,4] (b) Range: [0,4] (a) Domain: all reals (b) Range: (- 3, 00) (a) Domain: allreals (b) Range: [-3, I] (a) Domain: (3, 00) (b) Range: all reals (a) Increasing (b) Neither (0) Decreasing (d) Increasing 31. (a) Domain: [-4,4] (b) Range: [0,2]

-300

-0.' -1.0

~, ~]by

7

-1

[-1.25, 1.25]

29. [-

, S

S 6

4

-4

25. [-

y

. .

•

15% + 6

4x 2 -10%

8

10 8 6

37.

A-5

31.

[-0.25, 0.25]

33. f(x) = {I-X, 2-x,

35. (a) (d)

0 :5x< I 1:5x:52

~ v'2.5 "1/5

(b)_I_=

(o)x,x¢O

I

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 j): [-2,2] (0) Range (j 0 g): (-00,2], range (g 0 f): [0,2] 35.

33.

39.

y

y

., f(;J:) = -tan 2t

,• ,

• ,x -,

-,

-.

-20

41. Replace the portion for x < 0 with mirror image of the portion for x > 0 to make the new grapb synnnetric with respect to the y-axis.

A-6

Chapter 2: Answers to Odd-Numbered Exercises

43. Reflects the portion for y < a across the x-axis 45. Reflects the portion for y < a 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. (a) y = g (x - 3)

(c) y

~

g(-x)

+

t

(b) y = g (x ~

(d) y

-g(x)

+ ~) ~

(0) y

(f) y = g(Sx)

,

51.

, -,

-4 -3 -2-1

-1

1

,

(b) '" SO m/scc or 180 km/h 17. (a)

(b) '" $S6,000Iyear (e) '" $42,000Iyear 19. (a) 0.414213,0.449489,

+h \1'1+1; (\1'1+1; I

-, b = \/3

61. (a) a = I

b

63. (a) a = tanB 65. '" 16.98 m

(b) a = 2\/3/3

67. (b) 4'1f

Additional and Advanced Exerdses, pp. 37-38 1. Yes.Forinstancc:f(x) = I/xandg(x) = I/x,orf(x) = 2x andg(x) ~ x/2, or f(x) ~ eX andg(x) ~ lox. 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, thea g(x) = f(x) - 2 is ,

~f-I------_x

I)/h

I)/h

(b) g(x) =

Vi:

1.01

1.001

1.0001

1.04880

1.004987

1.0004998

1.0000499

0.4880

0.4987

0.4998

0.499

1.000001

1.00000S

1.000000S

0.5

0.5 (c) 0.5

(\1'1+1; -

l.l

1.00001

c = 4\/3/3

a (b) c = sinA

also even.

5.1

,

Y_2cos(x_~)

,

3\/3 -----.r y = 4x -

x

,

5.

(b)

'If

(a) 4 (b) 7 (a) 2 (b) y ~ 2x - 7 (a) 12 (b) y = 12x - 16 13. (a) -9 (b) y = -9x - 2 15. Your estimates may not completely agrec with these. (a) PQ, PQ, PQ, PQ. The appropriate units are m/sec. 43 46 49 SO

,

y_cos2.1:'

(b) I

4

(a) -

57. Period2

,

59.

3. 7. 9. 11.

Sg(x)

Y=~+1

x

55. Period 'If

1. (a) 19

~ ,.

'--Vi +~

Section 2.1, pp. 44-46

2

,

53.

CHAPTER 2

(d) 0.5

21. (a) IS mph, 3.3 mph, 10 mph (e) 20 mph when t = 3.5 br

(b) 10 mph, 0 mph, 4 mph

Section 2.2, pp. 54-57 1. (a) Does not exist As x approacbes 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 --+ I. (b) I (c) a (d) 1/2 3. (a) True (b) True (c) False (d) False (0) False (f) True (g) True 5. Asx approsches a from the left, x/lxl approaches -I. Asx approsches a from the right, x/lxl approaches I. There is no single number L that thc function values all get arbitrarily close to as

_1

x---+

7. 15. 27. 39.

o.

Nothing can be said. 9. No; no; no 11. -9 13.-8 S/8 17. 27 19. 16 21. 3/2 23. 1/10 25.-7 3/2 29. -1/2 31. -I 33. 4/3 35. 1/6 37. 4 1/2 41. 3/2 43. -I 45. I 47. 1/3 49.-y'4-=-;;:

Chapter 2: Answers to Odd-Numbered Exercises

25.

(W, Vn),

~.

'7

(2 _ 0.03 2 m'

29.

(k-~,~+n 8=~

Vi? -

8 =

+

0.03) m'

•

U

A-7

4 ... 0.12

~

0.03 m

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 eodpoint was rounded up and the right eodpoint rounded down.

59. The limit does not exist as x approaelres 3.

section 2.4, pp. 71-73 1. (a) (f) (1<) 3. (a)

True True True

2, I

(c) 3,3 5. (a) No 7. (a)

(b) True (0) False (d) True (g) False (b) False (1) False (I) False (b) No, lim f(x) ¢ lim f(x) x-2+

(d) Yes, 3 (b) Yes, 0 y

(e) True (j) False

.%"-2-

(0) No

(b) I, I

(c) Yes, I

x'"

9. (a) D: 0'" 2,R: 0 < Y '" I andy = 2 (b) (0,1) U (I, 2) (0) x = 2 (d) x = 0 y

•

2

~.

y.

o

j

0<<<1

1,

ISz<2

2,

x=2

2

11. v3 13. I 15. 2/V5 17. (a) I (b)-I 19. (a) I (b) 2/3 21. I 23. 3/4 25. 2 27. 1/2 29. 2 31. 0 33. I 35. 1/2 37. 0 39. 3/8 41. 3 47. 8 = .2, lim v;=s = 0 .%"-5+

51. (a) 400

(b) 399

section 2.5, pp. 82-84 1. No; discontinuous at x Section 2.3, pp. 63-66 1. 8 ~ 2

-'----+)~"

-+( 1

'7

3. 8 = 1/2

)

5. 8 = 1/18

( 419

7. 15. 19. 23.

).t

-1/2

-712 -3

) ). 112

4n

8 = 0.1 9. 8 = 7/16 11. 8 = V5 - 2 17. (-0.19,0.21), (3.99,4.01), 8 = 0.01 (3,15), 8 ~ 5 21. (10/3,5), 8 ~ 2/3 (- v'4.5, - viS), 8 = v'4.5 - 2 ... 0.12

13. 8 = 0.36 8 = 0.19

(c) The limit does not exist

= 2; not defined at x = 2 (b) Yes (0) Yes (d) Yes 3. Continuous 5. (a) Yes 7. (a) No (b) No 9. 0 11. I, nonremovable; 0, removable 13. All x except x = 2 15. All x except x = 3,x = I 17. Allx 19. All x except x ~ 0 21. All x except x = norr/2, n any integer 23. All x except norr/2, n an odd integer 25. All x '" -3/2 27. Allx 29. Alb 31. 0; continuousatx = 71' 33. 1; continuous aty = 1 35. V2/2; continuous at t = 0 37. g(3) = 6 39. f(l) = 3/2 41. a = 4/3 43. a = -2,3 45. a ~ 5/2, b ~ -1/2 69. x ... 1.8794, -1.5321, -0.3473 71. 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. (a) (f) 3. (a) (b) 15. (a) 21. (a)

31. 43. 53. 55. 57. 59. 61. 63.

00 00 (a) (a) (a) (a) (a)

a

(b) -2 (c) 2 (d) Does not exist (e)-I 00 (g) Does not exist (b) I (i) a -3 (b) -3 5. (a) 1/2 (b) 1/2 7. (a) -5/3 -5/3 9. a 11. -I 13. (a) 2/5 (b) 2/5 a (b) a 17. (a) 7 (b) 7 19. (a) a (b) a -2/3 (b) -2/3 23. 2 25. 00 27. a 29. I 33. I 35. 1/2 37. 00 39. -00 41.-00 45. (a) 00 (b) -00 47. 00 49. 00 51.-00 00 (b) -00 (c) -00 (d) 00 - 00 (b) 00 (c) a (d) 3/2 -00 (b) 1/4 (c) 1/4 (d) 1/4 (e) It will be -00. - 00 (b) 00 00 (b) 00 (c) 00 (d) 00 65.

81. 0 83. -3/4 85. 5/2 93. (a) For every positive real number B tbere exists a corresponding number S > 0 sucb !bat for all x Xo - S

Xo

, __1_

,

,-I

=> f(x) > B.

Xo

<x<

Xo

+ S => f(x) < -B.

(c) For every negative real number -B!bere exists a corresponding number S > 0 sucb !bat for all x Xo - S < x < Xo => f(x) < -B. 99.

101.

, ,

,

10

<x<

(b) For every negative real number -B!bere exists a corresponding number S > a sucb !bat for all x

6

,

:

4

I

, ,

,

2

I I

~

~

I( x2 I / I Y= .%_I=x+l+ x-I

,_,1;+ 1//

I

-~~-;-~;-/~~-;/":;_"':-k-;'i1-;2c-*,--74-,;--.,

103.

lOS.

,

-10

, ,

:x--2 67.

", \

~'..

.5

,

---c-+----I''---+----,

-,.x+3

_

,=-';~

/////// y=:r?-;l

,

_~""L

2

-,.X,,/

,

,

x--2:" ~

,

69. Here is one possibility.

--. ... _--_.

x+2

c---":\.---c:--"or--~,

-2

-----i"':.

y= -,+2 \

x=2

,,

,

107.

71. Here is one possibility.

,

!

73. Here is one possibility.

,

y=f~) -2

-,

109. At 00: 00, at -00: 0

Practice Exerdses, pp. 97-98 1. Atx=-I: 75. Here is one possibility.

,

h(x).-':!"', x¢O

1'1

79. At most one

lim f(x)=I,so

%--}+

lim f(x) = I = f(-I);contiuuousatx = -I.

x-+-l

Atx = 0:

11----o

lim f(x) =

%--}-

Atx

~

I:

lim f(x) = lim f(x) = 0, so lim f(x) = O.

x-lr

*

x-O+

x-O

However, f(O) 0, so f is discontinuous at x = O. The discontinuity cau be removed by redefming flO) to be O. lim f(x) ~ -laud lim f(x) ~ I,so lim f(x) x-lx-l+ x-I does not exist. The function is discontinuous at x = I, aud !be discontinuity is not removable.

Chapter 3: Answers to Odd-Numbered Exercises

, _ _ _L ,

-

(b) 49 I (h) -7

(g) -7

7. (a) (-00,+00) (d) (0,00)

11.

2

13. 2x

-------!,ch¥---~,

(0) 0

(d) I

(e) I

(I) 7

5. 4

(-2,-8)

(h) [0,00)

9. (a) Does not exist

(b) 0 I

15. - 4

17. 2/3

21. I

19. 2/'IT

23. 4

x--l

31. Yes, f does have a continuous extension, to a = 1 with 1(1) ~ 4/3.

37. 2/5

39. 0

47. (a) x = 3 (b) x = I

-,

(0) (-00, 0) and (0, 00)

25. 2 27. 0 29. No in both cases, because lim I(x) does not exist, and lim I(x) does not exist. ,~I

33. No

,

y.I2¥+16

--*-~-~,

3. (a) -21

I

9. Y = 12x + 16

y=f(x)

41. -00

43. 0 45. 1

11. m = 4,y - 5 = 4(x - 2) 13. m ~ -2,y - 3 ~ -2(x - 3) 15. m = 12, Y - 8 = 12(1 - 2) I 1 17. m=4,y-2=4(x-4) 19. m = -10 21. m = -1/4 23. (-2,-5) 25. Y = -(x + I),y = -(x - 3) 27. 19.6 m/sec 29. 6'1T 33. Yes 35. Yes 37. (a) Nowhere 39. (a) At x = 0 41. (a) Nowhere 43. (a) At x = 1 45. (a) Atx ~ 0

section 3.2, pp. 112-115

(0) x = -4

1. -2x,6,0,-2

AddUional and Advanced Exercises, pp. 98-100 3. 0; the Ielt-band limit was needed because the function is undefmed for v > c. 5. 65 < 1 < 75; within 5°F 13. (a) B (h) A (0) A (d) A 21. (a) lim r+(a) ~ 0.5, lim r+(a) ~ I a-O

a--l+

(h) lim r_(a) does not exist, lim r_(a) = 1 a-O

25. 0

Q_-l+

27. 1

31. y = 2x

29. 4

33. y = x, y = -x

5 _3_ _ 3_ 1 _3_ 7 6x' • 2y'36' 20' 2' 2 v 2 '

11.

-1 2(q

• r---:-c

+ 1) v q + 1

= 0,1,4

,

CHAPTER 3 I,P,:

y~2x+5

,

m,

= 5

'2 ~

86-420~:X: 3. P,:

m,

7. Y

~

= 5/2, P,: x + 1

m,

= -1/2

, , .. ,;+1

35. (a) i) 1.5 of/hr ii) 2.9 of/hr iii) O°F/hr iv) -3.7°F/hr (h) 7.3 DF/hrat 12 P.M., -11 °F!hrat6p.M. Slopo

, ,

('F1hr) 6

---.f o lLLLLU\C-Uc4' (!mo)

-, -.-,

-12

"'""'8-

I

~o~ .• r.cc-'-~*c~'

-0.7

-3.3

(0)

~of----+-----;,~__;,c___;.-~,

"

f' on (-4, 6)

I

Section 3.1, pp. 105-106

15. 31' - 21, 5

33.

" ~

m, =

x

-4 y-4=-1(x-6) 19.6 (x - 2)v;=2' 2 -1 -1 21. 1/8 23. (x + 2)' 25. (x _ I)' 27. b 29. d

(h)

1. P,:

9

13. 1 - 2' 0

9 1 • (21 + I)'

17.

31. (a) x

5.

2 1 2 3. -1',2'-4'- 3 0

84

e-:J

0--0

87 88

A-9

A-tO

Chapter 3: Answers to Odd-Numbered Exercises

37 S'

l'

mceh~

•

flO + h) - flO) ~ I h

while lim flO

+~ -

41. (a) 13

flO) = 0,

~

lim

flO

+ h) - flO) h

.~o

'~o'

I, . f(l -2' j (I) ~ lim

h-O

43. 45. 47.

+ h) - f(l)

h-O

h

does not exist and j(x) is not differentiable at x = I. Since j(x) is not continuous at x = 0, j(x) is not differentiable atx = O. (c) None (a) -3 '" x '" 2 (b) None (b) None (c) x = 0 (a) -3 "'x < 0,0 <x '" 3 (b) x = 0 (0) None (a) -I '" x < 0, 0 < x '" 2

Section 3.3, pp. 122-124

dy_, d'y_ 5. ds - 4x - I, ds' - 8x 7 dw=_~+~d'w=~_~ • dz z3 z2' dz2 z4 z3 = 12x - 10 + IOx-',

dr = -2 • tis 38 3

Z

= 12 - 3Ox....

+~

d'r = ~ _ ~ 2s 2 ' ds 2 84 83

13. y' = -5x' + 12x' - 2x - 3 , I -19 15. y' = 3x + lOx + 2 - 17. y' = ---'"''---0 x' (Jx - 2)' 19.

~ x' + x + 4 (x + 0.5)'

'(x) g

~

23. I'(s)

dv 21 • dt

I

25. v'

\h(\h + I)'

t'-2t-1 (I + t')'

~ '--:c~~,-cc

~

63. (a) (b)

:x (uvw)

+ uv'w + u'vw

= uvw'

= UlU2U3U4'

=

=

(c)

~ (Ul ••• Un) = "1"2'" ... + UI'U2"'U" _ _ nRT +

+ 4x -

yen) = Oforn

33. y'

=

O?::

S,y"

35 dr = 38.... d'r = 'd8 'dB' 37. :

~

=

dpi 39. - ~ -q dq 6

2

+ 4,ym

=

6,

+

14x-3

~ 1.-'

I + -q-' + q-'

6

6x

=

-12~'

-z-' - I, ~'v;

+ "1"2'"

U"-2U,,-I'U,,

+

2an' V'

(V-nb)'

1. (a) -2 m, -I m/sec (b) 3 m/sec, 1m/sec; 2 m/sec', 2 m/sec' (0) Changes direction at t = 3/2 sec 3. (a) -9m, -3 m/sec (b) 3 m/sec, 12 m/sec; 6 m/sec', -12 rn/sec' (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' (c) No change in direction 7. (a) a(l) = -6 m/sec', a(3) = 6 rn/sec' (b) v(2) = 3 m/sec (c) 6 m 9. Mars: '" 7.5 sec, Jupiter: '" 1.2 sec 11. g, = 0.75 m/sec' 13. (a) v ~ -32t,lvl ~ 32tft/sec,a ~ -32ft/sec' (b) t '" 3.3 sec (c) v '" -107.0 ft/sec (b) 3 '" t '" 6 15. (a) t = 2, t = 7 (c) (d) Ivl (m/!CC)

,,"

dv a=-

d,

,

S,"",

0

4

2x - 7x-2 ,y"

Un-IU,,'

~

1

a for n '" 5

3x 2

+ UtU2U3'U4 + UI U2'U3"4 +

Ul'U2 U3 U4

, ,

6

,

10

29. y' = 2x' - 3x - I,y' = 6x' - 3,ym = 12x,y(') = 12, y(,)

(0) (2, 6)

+ (n - I)a.-lx·-' + ... + 2a,x + al

~ (UIU2U3U4)

-1, + 2x-'/2 x

27 '= -4x'-Jx'+1 • Y (x' - I)'(x' + x + I)'

31. y'

+2

Section 3.4, pp. 132-135

3. dsd = 1St' - 1St" d'S, = 30t - 60t' t dt

11

53. (a) y = 2x

51. (0, 0), (4, 2)

61. The Product Ru1e is then the Constant Mu1tiple Ru1e, so the latter is a special case of the Product Ru1e.

. dV-

9. :

+ 17

57. a = -3

65 dP

1 ely = -2x d'y = -2 'ds 'ds'

(d) 20

(b) m = -4 at (0, I)

47.a=l,b=l,c=0

59. P'(x) = na"x·- 1

f(l) = 2 while

. f(1 + h) - f(1) lim_ h

41.

49. (2,4)

55. 50

+~ -

(0) 7/25

i

45.y=4x,y=2 does not exist andj(x) is not dif-

ferentiable at x = O. 39. Since lim f(1

-~ +

(0) Y = 8x - 15,y = 8x

.~o

1'(0)

(b) -7

43. (a) y =

d'PII ~ - - -q.... - 5q-< ' dq' 6 2

t(sec)

0 -1

-,-,

1 2 3 4 S

7 8 9 10 ~

~

-4

17. (a) (d) (f) (g)

190 ft/sec (b) 2 sec (0) 8 sec, 0 ft/sec (e) 2.8 sec 10.8 sec, 90 ft/sec Greatest acceleration happens 2 sec after launch Constant acceleration between 2 and 10.8 sec, -32 ft/sec'

19. (a)

~ sec, 280 em/sec

(b) 560 em/sec, 980 em/sec'

(0) 29.75 flashes/sec

21. C = position, A = velocity, B = acceleration (b) $80 (0) $79.90 23. (a) SUO/machine 25. (a) b'(O) ~ 10' hacteria/b (b) b'(5) ~ 0 bacteriajb (0) b'(IO) ~ -10' bacteria/b

A-11

Chapter 3: Answers to Odd-Numbered Exercises

dy I 27. (8) dl ~ IT

-

I

(0) The

dt is 0 mfh wheo = 12 aod the smallest value of dt is -I mfh wheo = O.

(b) The largest value of

I

V15 6 ±3 sec.

. speeds up on (6-V15 (d) The o~ect 3 , 2 ) U (6+V15 3 ,4 ]

t

(0)

o~ect cbaoges direction at I ~

aod slows down on [0, 6 - 3V15) U ( 2, 6 + 3V15) .

y

~ a aod I ~ 4 wheo it is . 7 ' / dl movmg umts sec an s owest at t = 6±V15 3 sec.

(e) The o~ect is moving fastest at I

V15 (f) Wheo t = 6 + 3 the object is at position

8 '"

-6.303

units aod farthest from the origin.

section 3.5, pp. 139-142 29. 1= 25 sec

1. -10 - 38inx

D = 6250 m 9

5. -cscxcotx -

31.

11.

-esc' x 2 (I + ootx)

3. 2xcosx -x 2 sinx

~

7. sinxsec2 x + sinx

13. 4 tan x sec x - csc2x

17. 3x 2 sinxcosx

+X

3

2 COS X

9.0

15. x 2 cosx

-x 3 sin2 x

21. -2csclcott 23. -0 (OoosO + 2 sinO) (I - esc I)' 25. sec 0 esc 0 (tao 0 - cot 0) = sec' 0 - esc' 0 19. sec'1 - I

(8) V = 0 wheo 1= 6.25 sec. (b) v > 0 wheo 0 :5 t < 6.25 => the object moves up; v < 0 wheo 6.25 < I :5 12.5 => the object moves down. (0) The object cbaoges 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 eodpoints I = 0 aod I = 12.5 wheo it is traveling 200 It/sec. It's moving slowest at t ~ 6.25 wheo the speed is O. (f) Wheo I = 6.25 the object is 8 = 625 m from the origin aod farthest away.

33.

(8) v = 0 wheo I =

(b)v
6+

~<

t :5

0 when 0 :::s; t

<

6 - V15 3 or

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. " (q - I) 33. (8) 2 esc' x - cscx (b) 2 sec' x - secx 35. y

'If

41. No

A-12

Chapter 3: Answers to Odd-Numbered Exercises

41. 2'ffsin ('ffl - 2)cos('ff1 - 2)

43.

45. 101 10 tan' I sec' 1+ 101' tan lO I

47.

8 sin (21) , (I + cos 21) -31' (1 2 + 4) (I' _ 41)'

49. -2 cos (cos (21 - 5» (sin (21 - 5» _'c-~~--./<-~~-",'c2-· , ,

2 ) VI + cos (12)

53.

45. 47. 55. 57. 61.

57. 3(21 2 - 5)'(181 2 - 5)


5. -sin (sinx) cosx ,dy

dyd.

•

= (z" + I),y =. : dx = d. dx = 5•• 2 =

9. Witb •

Section 3.7, pp. 153-155

10(z" + I)' -7 dy = ( I - (x / 7» ,y =. : dx =

11. Witb •

dy d. d. dx

1.

=

4.'·

dy

dydu

= «(x 2/8) + x - (I/x»,y = .': dx = d. dx =

(~ +

15. Witb •

~

I + :2) = tanx,y

~

4(~2 +

sec.:

X -

})'(~ +

11.

19.

J--=t

21.

15. -

sin 51)

23.

23. y' =

t8C~ 8

00

~c

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 ). x3 4 - z,,2

)2

33. xsecxtanx + secx 2'\17 + x sec x

35.

2sin8 2 (I + cos 8)

39. (2(::

37. -2 sin (8') sin 28 + 28 cos (28) cos (8')

1~'j2) cos (vf+-J)

~

17.

25. -2 29.
YSin(~) _ cos~) + xy

-r

y2 - (x + 1)2 y'

vY , y' = -~"----I vY + I 2(vY + I)' 27. (-2, I): m ~ -I, (-2, -I):m 7 I 4 29 ~4x ~

3x

- 2

+6

6

(b) y ~

6

2'ff x

~

-7 x +7 I

8

(b) y ~

-67 x -67 2

2

+ 'ff

(b) Y = 'if x - 'if +

37.
x I (b) Y = - 2'ff + 2'ff

39. Points: (- V?, 0) and

(V?, 0), Slope:

41.

I

(b)Y~-3x+3

y~7x+7

35.
9. cos2y

7 I 'y(x+1)2 _y2

8

x +I " 21.y'--y-, y

=.':: = : : : = 3. 2 cosx =

~ (cos 31 -

-ooS2(xy) - y x

I + :2)

dy dy d. dx ~ du dx ~

2 3-1"

I - 2y 3'z,,+2y-1

13.

(sec.tan.)(sec'x) = sec(tanx)tan(tanx)sec2x 17. Witb • = sinx,y 3 sin2X (cosx)

-2xy - y2 x 2 +2xy

-z,,' + J>;2y - xy2 + x 5. x'y-x'+y

-7·-'-(-t) = (l-tY" 13. Witb.

+~)

59. ;, (I + })(I

61. 2csc'(J>; - l)cot(J>; - I) 63. 16(z" + 1)2(5x + I) 65. 5/2 67. -'ff/4 69. a 71.-5 73.
Section 3.6, pp. 147-149 1. Iz,,' 3. 3 cos (3x + I) 7. 10sec'(Hb; - 5)

55. 6 tan (sin' I) sec2 (sin' I) sin2 I cos I

I sin (1

2'ff

-2

m~-Iat (~,~). m~V3at (~,t)

Chapter 3: Answers to Odd-Numbered Exercises

43. (-3,2):m = _2;; (-3, -2): m = 2;; (3,2): m = 2;; (3, -2): m

51. dx

17.

27

= - If

45. (3, -I) dy y'

+ 2xy = x2 + 3xy2'

dx dy

dx dy

I dy/dx

Section 3.8, pp. 160-163 dA

1. dt

dr 21Tr dt

=

3. 10 5. -6 11.
9. 31/13 13.
dJ,

= 1Tr

dV

Iii =

2 ~;

15.
dR

<e) dt =

(b)

2dh 1Tr dt

+

dJ,

7. -3/2 (b) -135m'/min

= 21Thr:

dr 21Thr dt

(b) -

(

3x 2

-

(b) 1.003

-

3)

2y',;

2

dx

I-y 21• • r dx 3vy +x

19 2-2x2 dx • (I + x 'f 5.r

23• • rcos(5Vx)dx

2vx

25. (4X 2 )Sec'(X;)dx

27.

~(cSC(l

29. 31. 33. 37. 41.


- 2y',;)cot(l - 2y',;))dx

45.
3I amp/sec

dI)

I (dV V dt - I dt

I

15.
A-13


x

17.
x

Yx2 +

dt

dx

Practice Exercises, pp. 176-180

+ y2 dt

dx y2 dt

+

Y dy Yx2 + y2 dt

dx ydy
dA

I d8 Z ab cos 8 dt dAI d81.da (b) dt = zab cos 8 dt + Zbsm8 dt dA I ~ I . • I . • <e) dt = Zabcos8 dt + Zbsm8 dt + Zasm8 dt 21.
19.
Iii =

<e) -14/13 em/sec, deereasiog (b) -59.5 ft'/sec 23.
~; ~

29.
2~~ m/min

<e) :

11.19 em/min

(b)


~~

14.92 cm/min

(b) r = Y26y - y2 m

~ - 28~1T m/min

31. I fi/min, 401T ft'/min 33. 11 fi/sec 35. loereasiog at 466/1681 L/min2 37. -5 m/sec 39. -1500 ft/sec 41

Section 3.9, pp. 173-175 9. -5

3. L(x) = 2 I 4 11. 12 x + 3

-2 sio8 2 (cos8 - I) -3x 2 - 4y

y

43.
1. L(x) = lOx - 13

35.

+2 41. - - + 3 x

5. / . 10. 2/ . • 721Tm. mm,Tm mm

7. 2x

1. 5x 4 - 0.25x + 0.25 3. 3x(x - 2) 5. 2(x + 1)(2x 2 + 4x + I) 7. 3(8' + sec 8 + 1)2(28 + sec 8 tao 8) I

5. L(x) = x - 1T

13. /(0) ~ I. Also, I'(x) ~ k (I + x)H, so 1'(0) ~ k. This meaos the linearization at x = 0 is L(x) = I + kx.

47.

43.

I 2y(x+lj'

4x - 4y

49. dp

+2

45.

1/'

-xY

6q - 4p 3p 2+4q

dq

dr 51. ds ~ (2r - 1)(tao2s)

d'y -2xy' - 2x 4 53.
59.

v'3


-k

d'y _2xy2 - I (b) dx2 = 4 , xy (e) 12

(d) 1/4

63.

-2 (2t

+ I)

2

(f) 9/2

A-14

Chapter 3: Answers to Odd-Numbered Exercises

65.
67.
,

117. (a) L(x) = 2x

1f - 2 +2-

-Oao¥----+-'!2-~ ,

(b) Yes

jl;l:)...I

~,-IS.x
l-~.


(H)

(0) No

OSl:<1

(0) Yes

G,-±)

and

71. (-1,27)and(2,0)

73.
(b) (0,20),(1,7)

,

-1

y=-!x-j-l

119. L(x) 77.

I

4

123. (a) 4%

79. 4

81. Tangeot: y ~

I 9 -4'" + 4' normal:

y ~ 4", - 2

83. Tangeot: y = 2x - 4,normai: y = 85. Tangeot: y 87. (I, I): m

~

~ - ~x + 6, normal: I -2:; (I, -I):

y

-t + i x

~ ~x

- 11 5

mnotdefined

89. B = graph of t, A = graph off'

91.

,

121. tIS =

= 1.5x + 0.5

1frho

\lr 2 + ha2

dh

(0) 12%

(b) 8%

Additional and Advanced Exerdses, pp. 180-182 1.
9

2:' a

5v'5

= -2-

7.
93. (a) 0, 0 (b) 1700 rabbits, '" 1400 rabbits 95. -I 97. 1/2 99. 4 101. I 103. To ma1re g continuous at the origin, derme g(0) = I. dS dr dS dh 105. (a) dt = (41fr + 21fh) dt
tIS

(c) dt = (41fr

dr

+ 21fh) dt +

dh 21fr dt

11.
dr r dh (d) dt = - 2r + h dt

107. -40 m 2/aec 113. (a) r = 115. (a)

2. h 5

109. 0.02 ohm/sec (b) -

111. 22 mfaec

125 ft/min 144'lr

~ krn/aec or 600 m/aec

a

(b)~ rpm

~

b

-'if

(b) 12.31 sec, 393.85 ft

(b) m

3 9 17.
~

-I,b

~ 1f

19. todd=> f' is eveo

23. h' is defmed but not continuous at x = 0; k' is deIrned and continuous at x = O. 27.
A-1S

Chapter 4: Answers to Odd-Numbered Exercises

31. Absolute maximum: I; absolute minimum: -1

29. Absolute maximum: 2;

CHAPTER 4

absolute minimum: 0 y

Section 4.1, pp. 189-191 y .. .J4-x'-

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

--+--~-.t--7---~%

11. (c) 13. (d) 17. Absolute maximum at x = 2; no absolute minimum

y

(-2, 0) -1 Ab. min

0

y .. g(.x)

y

12

Absmin

-1

1.0 0.8

0.6

%

AblI max

AblI max

(013, 21"')

(2<13, 21"')

y=escx

('JT/2, 1) 1/"/3 :S x:!li: 211"/3 Ab.

37. locreasing on (0,8), decreasing on (-1,0); absolute maximum: 16 at x = 8; absolute minimum: 0 at x = 0

y

21. Absolute maximum: - 3 ; absolute minimum: -19/3

-I 0 -I

39. locreasing on ( - 32, 1); absolute maximum: I at 8 absolute minimum: -8 at 8 = -32 41. x=3 43. x=l,x=4 45. x = I 47. x~Oandx~4 49. Minimum value is I at x = 2.

23. Absolute maximum: 3; absolute minimum: -I

y

51. Local maximum at (-2, 17); local minimum at

1

2

,

(2, 3) AblI

%

59.

-2:Sx:S3

x=O

-7

61. 27. Absolute maximum: 2; absolute minimum: -}

y

-I

-, -4

y

E-

Abs max

y--~,O.S:sx:S2

(0.5, --4) Absmin

(~, - i~ )

Critical point Derivative Extremum or endpoint 4 x =-5

,=jx-s

25. Absolute maximum: -0.25; absolute minimum: -4

-2

1;

' "2I at x = I ' , val' 1 at x 57. Maxim:om va1ue IS ; tl111111D.UJn De IS -"2

(3,-3)

-,

~

53. Minimum value is 0 at x = -I and x = 1. 55. There is a local minimum at (0, I).

y

-2

0

min

~0+----"'-"c'~-0'=/2~2c'ol3~%

19. Absolute maximum at x ~ '/T/2; absolute minimum at x = 3'/T/2

(-2,-19/3) Abo min

35. Absolute maximum: 2; absolute minimum: -I

~

004 02

-2

-I (-11'/2,-1)

33. Absolute maximum: 2/V3; absolute minimum: 1

y

-_~I--'¥'-----:---;2c-~

(11'/2, 1) Abs IDlIX

-2:SxS 1 ....--t......_

7. No absolute minimum; no absolute maximum 9. Absolute maximum at (0,5) 15. Absolute minimum at x = 0; no absolute maximum

y (0., 2)AbI max

%

2

1

y=~

-

-1. _<:.::%:;.<::..'_ - - - - (8,2) ~ Abs

-1:--1r-:,---:2---:-,---;-,-':-C6;--:7---:'~ -I

(-1,-1) Absmin

%

63.

0

Local max

= - I.

Value 12 10'/3 '" I 034 25 .

Undef'med Local min

0

Critical point or endpoint

Derivative

EItremum

Value

x ~ -2 x =-0 x~ 0 x=2

Undef'med 0 0 Undef'med

Local max Minimum Local min

0 -2 2 0

Critical point or endpoint

Derivative

Extremum

Value

x=I

Undef'med

Minimum

2

Maximmn

A-16 65.

Chapter 4: Answers to Odd-Numbered Exercises

Critical point

or endpoint x ~ -I x = I x = 3

Derivative

EItremum

Value

a

Maximum LocaImin Maximum

5 I 5

UndeImed

a

67. (8) No a for all x '" 2. (0) No, because (-00,00) is not a closed interval.
2

b- a b - a

=

_~~c2(a c2

-b) a_b~c

ab

=

=

Viib.

59. f(x) must be zero at least once b-.eo a and b by the Iotermediate Value Theorem. Now suppose that f(x) is zero twice between a and b. Theo, by the Mean Value Theorem, f' (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 giveo that f'(x) '" 0 on this interval. Therefore, f(x) is zero once and only once between a and b.

69. 1.09999 '" f(O.l) '" l.l

+ 80

75. 2g

77. Maximum value is 11 at x

5; minimum value is 5 on the in-

=

terval [-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

+

0) '" 1.22,t(1

0) '" -0.549

-

7. Does not; f is not differentiable at the interior domsin point x ~ O. 9. Does 11. Does not; f is not differentiable at x = -I. 15. (8) i) • • •

o

-2

ii) - -

-,•

2

---"

• -3•

-4

•1 •0

iii)

•

iv)

•4

0

27. Yes 31. (8)

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. 51. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have beeo going that speed at least once during the trip. 55. The conclusion of the Mean Value Theorem yields I I

29. (8) 4

x2

T +C

I 33. (8) X

+

35. (8) -tcoS21 (0) -tcoS21

37. f(x) = x 2

41. 8 = 4.91 2

T +C

-

x

I

+X+

+C

•

•

24

18

(0) 3

x3

(b) x

C

•

9

(b) 3

(b)

•2


(b) 2 sin

x4

4 +C I (0) 5x - X

I+C

I

+ 51 + 10 2 45. 8 = 161 + 201 + 5

43. 8 =

1.
7. (8) (b) (0)

0, I Iocreasing on (-00,0) and (1,00); decreasing on (0, I) Local maximum at x ~ 0; local minimum at x ~ I

-2, I Iocreasing on (-2, I) and (1,00); decreasing on (-00, -2) No local maximum; local minimum atx =-2

-2, I, 3 Iocreasing on (-2, I) and (3,00); decreasing on (-00, -2) and (1,3) Local maximum atx = I; local minima atx = -2,3 0,-1 Iocreasing on (-00, -2) and (1,00); decreasing on (-2,0) and (0, I) Local minimum atx = I

9. (8) -2,2 (b) Iocreasing on (-00, -2) and (2,00); decreasing on (-2, 0) and (0, 2) (0) Localmaximumatx ~ -2; local minimum atx ~ 2 11.
+

C

and

e:,

4:}decreaSing on

+ cot6

- 21t - I

I - cos (1ft) 1t

47.8= sin(21) - 3

49. If T(I) is the temperature of the thermometer at time I, theo T(O) = -19'CandT(14) = 100 'C. From the MeanValue T(l4) - T(O) Theorem, there existsa 0 < 10 < 14 such that 14 _ 0

x =

(O,T). (T'

2:).

(~,21t)

(c) Local maximum. at x

+ 2Sin + C 39. r(6) = 86

Section 4.3, pp. 201-203

21t

3

=

0 and x

=

41t lal ·· T; ac m1D.1DlUIn at

and x = 21t

15.
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. <s) Increasing on (-00, -1.5); decreasing on (-1.5, 00) (b) Local maximum: 5.25 all = -1.5; absolute maximum: 5.25 atl ~ -1.5 21. <s) Decreasing on (-00,0); increasing on (0, 4/3); decreasing on (4/3,00) (b) Local minimum atx ~ o(0, 0); local maximum at

x ~ 4/3 (4/3, 32/27); no absolute extrema 23. <s) Decreasing on (-00,0); increasing on <0, 1/2); decreasing on (1/2,00) (b) Local minimum at 6 = 0 (0, 0); local maximum at 6 = 1/2 (1/2, 1/4); no absolute extrema 25. <s) Increasing on (-00,00); never decreasing (b) No local extrema; no absolute extrema 27. <s) Increasing on (-2,0) and (2,00); decreasing on (-00, -2) and (0, 2) (b) Local maximum: 16 at x ~ 0; local minimum: 0 at x ~ ±2; no absolute maximum; absolute minimum: 0 at x = ±2 29. <s) Increasing on (-00, -I); decreasing on (-1,0); increasing on (0, I); decreasing on (I, 00) (b) Local maximum: 0.5 atx ~ ±I; local minimum: 0 atx ~ 0; absolute maximum: 1/2 at x = ± I; no absolute minimum 31. <s) Increasing on (10, 00); decreasing on (I, 10) (b) Local maximum: I atx = I; local minimum: -8 atx = 10; absolute minimum: -8 at x = 10 33. <s) Decreasing on (-2V2, -2); increasing on (-2,2); (b)

35. <s)

(b)

37. <s)

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. <s) 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;locai minimum: -24V'2/77/6 '" -3.12 atx ~ 2/v7; no

absolute extrema 41. <s) (b) 43. <s) (b) 45. <s) (b) 47. <s) (b) 49. <s)

Local maximum: I at x ~ I; local minimum: 0 at x ~ 2 Absolute maximum: I at x ~ I; no absolute minimum Local maximum: I at x = I; local minimum: 0 at x = 2 No absolute maximum; absolute minimum: 0 atx = 2 Local maxima: -9 at I ~ -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. <s) Local maximum: 2 at x = 0;

5

=

local minimum:.j3 at x = 2 - V3 4 3 - 6 (b) No absolute maximum; an absolute minimum at x~2-V3 53. <s) 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: V3 at x = 2",; local minimum: -2 atx ~ 7",/6; local minimum: V3 at x = 0 57. <s) Local minimum: (",/3) - V3 atx = 2",/3; local maximum: 0 at x = 0; local maximum: '7T at x = 2'17" 59. <s) Local minimum: 0 at x ~ ",/4 61. Local maximum: 3 at 6 = 0; local minimum: - 3 at 6 ~ 2", 63.

,16

V~' -~ b '

l¥'

,=1(;<)

I

/ \

,=f")

"

,

\!:.I(;<)

0

I

1 (;<)

"

\,

01010101

(s)

65. <s)

(b)

,

, 2

0

(c)

(b)

(d)

,

~ ~

"

2

2

0

2

69. a = -2,b = 4

section 4.4, pp. 211-213 1. Local maximum: 3/2atx ~ -I; local minimum: -3 atx ~ 2; point ofinflection 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: 0 at x = ± I; points ofinflection at

(-V3, 3f) and (V3, 3f}

rising on (-1,0) and (1,00); falling on (-00, -I) and <0, I); concave up on (-00, -V3) and (V3, 00); concave down on

(-V3, V3)

5. Localmaxima:

-i'" + ~

~ -2"'/3,f + ~ at ~ atx = -",/3, 2; - ~

atx

x = ",/3; local minims: - f -

at x = 2",/3; points ofinflection 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 ('fr/3, 2'fr/3), concave up 00 (-'fr/ 2, 0) and ('fr/2, 2'fr/3); coocave down 00 (-2'fr/3, -'fr/2)and (0, 'fr/2) 7. Local maxima: 1 at x ~ -'fr/2 and x ~ 'fr/2, a at x ~ - 2'fr and x = 2'lr, local minima: -I atx = -3'fr/2andx = 3'fr/2, at x = 0; points ofinflectioo at (-'fr, 0) and ('fr, 0); rising 00 (- 3'fr/2, -'fr/2), (0, 'fr/2), and (3'fr/2, 2'fr); falling 00 ( -2'fr, - 3'fr/2), (-'fr/2, 0), and ('fr/2, 3'fr/2); coocave up 00 ( -2'fr, -'fr) and ('fr, 2'fr); concave down 00 (-'fr, 0) and (0, 'fr) 11. , 9.

Absmax

(m'4,112)

o

,

,

27.

'(0",MO)'f-----=",;;---::1k--c;:7.-~.~~'

""' ....

(31rl4,-112) Ab!min

-I

,

29.

,

31.

,--'{T.i

2

2

~~~---c'-,If-~~~~~'

-4-3-2-

\1234 I -2

-2

(2,-1)

-2

Absmin

-+---C_I:-+---C==~-~'

,

13.

,

33.

,

15.

,

(2, S) Lac max

4

,=,'''G-')

(0,0)

-_-:,~t-'-~,'--'-Y":--:--~·,

-I' ....._~

""' .... (1,-1)

2

(1,1)

I

-_2:--_1~0+--jr--:-~-~4-->'

---C-'l~-_~'-+(O",~O)~'-~-\--:-->'

c...

-I

""' .... (0,-3)

,

17.

19.

,

,

35. y=2x_3x 213 Cusp. Lac IDlIX

2

InfI

(0,0) InfI

,

37.

39.

""' .... , (0. 4) Abs max

4

""'(-,,12,0)

----<>---+---~-~, (-4. 0) (4,0)

---,---;;---;--f----:-----;2'(2~"_=-, o'~->'

AlII min

Ab! min

""' ....

23. ,

, ,

41.

y ...

~-3 ,-2

4

,

43.

V

Ab._ (2"', "') r -.... (2,2)

: (3, 6) Loc min

-'1nfI

:(1. 2) Lac max

I-ro, -"') ---;0;f-'==~.------;2~.~'

25.

,

-, -8

Infl /

5m3,S-fJ"'lT13-1) Lac min

(3'11/2. 3..J3'IT12)

1'/2,~"/2)

-'oil-;r:",;;--.~-C;:';;---;;2.~~' (0,-2) Ab!min

-4

InfL

(-2, _2;-2

Ab' min

Chapter 4: Answers tc Odd-Numbered Exercises

,

45.

,

47.

y=~

65. y" =

A-19

-~ (x + 1)-'/3

-_~4-_~3~_2!-_~,--1f'-!-C:--~3C---4~X

-2

2

(-1,0)

(1,0)

Abo_

Abo_

x

49. y" = 1 -2>;

51. y"

~

3(x - 3)(x - 1)

xO

-2 69. y" = { 2,'

Loo_

,

71.

Loo_

x_o

---Y---.f---1--_ x

x=-1

53. y" = 3(x - 2)(x

+ 2)

Loo_

\

~ InflX~2/

~~'-2

~

73.

x=2..J3

;p,,-2..J3

55. y" = 4(4 - x)(5x 2

-

,

16x + 8)

Loo_ .%= 8/S

75.

57. y" = 2 sec2 x tan x

59. y"

77.

,

= -tcSC2~,

, , ,

,

0<6<2".

":',

4

8='It __........~Ab5mu:

""" ,,

3 )'=2

y=~+x-l

IL

~-I

, ,

- -2

--'-\~=X~l

~ __~_'"'_cc_c" __-_-_\-,'oit-\++~~x

':r \

,

-2

'

" :.;=1

,'

, '

2

61. y" = 2tan6sec 6.

-f < 6 < f

':

79.

,

,

81.

__

, ,_ ..'-1

="'-----;--t----;----'=-.x

, __,_ ..'-1 ='"-'-~f--i-+-~,e-\~-'-'''_'>X

'... :, ,,

---;---

63. y" = -sin t. 0 ::s t ::s 21T x= -1

~ : x-I

" "

A-20

Chapter 4: Answers to Odd-Numbered Exercises

,

83.

85.

,

x--l

··v/ "

I L

y. ;'-x+ 1 -~~~+'¥'"-~~--.,

-4 -3 - 2 :

"1

2

3

,

87.

: \,,.'

,-1

.//",;,,.X-l

,/

''.

,

,--;"

"

,

.

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.

V'2;

,"1=_'_ x-I

,

===c-c-+--;,---c2;-----+'

,

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(l - x) (e)

,, , :.1:=1

11.9XI8in.

Point p

Q R

y' -

y"

+ +

a

a

s

-

T

,

95.

+

7

-

4

105. h ~ -3 109. -1,2 111. a = I, h = 3, C = 9

(e) Maximum volume'" 1309.95 in' when x '" 3.39 in. (d) V'(x) ~ 24x2 + 864, so the critical point is at x ~ 7 which confIrms the result in part (c). (e) x = 2 in. or x = 5 in. 19. '" 2418.40 21. (a) h ~ 24,w ~ 18 (b) v (24. 10368)

'\Il3,

336x

em'

Abs IIUIX 8000 6000

a are extrema and points of

inflection, respectively. Inflection at x

"""«-')

X = 3.3944487 Y .. I309.9547

10000

113. The zeros of y' = 0 and y' =

"=

(b) Domain: (0, 9)

----001--+2-4;-~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

,

15.h:r=8:1t

-

1600

(6.7)

97. (a) Towards origin: a '" t < 2 and 6 '" t '" 10; away from origin: 2 '" t '" 6 and 10 '" t '" 15 (b) t ~ 2, t ~ 6, t ~ 10 (e) t ~ 5, t ~ 7, t ~ 13

x = 0, local minimum at x

13.f

17. (a) V(x) ~ 2x(24 - 2x)(l8 - 2x) v

-------"ot--+----., 93.

t

1·m 2450., 7. 80,000 m2 ,' 400 m by 200 m 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 consideratious such as structural requirements.

y=81~+4)

2

square units, I by

5 14 X 35 X

,

91.

t

= 4.

=

3, local maximum. at

4000 2000

oJ,L~,--+.1Oc-7"~"'"o-2:5fc-~30~,,4. 23. Ifr is the radius of the hemisphere, h the height of the cylinder, f3 and V the volume, then r = 81t and h = 'iT .

( 3V)'

25. (b) x = 51

8

27. Radius =

(3 V)'1'

(e) L '" 11 in.

v2 m, height =

I m, volume = 2:; m'

A-21

Chapter 4: Answers to Odd-Numbered Exercises

29. I

31. 9+

3 33. 2 X 2

9~

m, triangle,

3w

b~

9+

35.
m, circle

3w

(b)-I

37.
53.
l":

2

(b)

l":

59. M=

57.4 X 4 X 3 ft, $288

~

(b) The minimum distance is from the point (3/2,0) to the point (I, I) on the graph of y = Vi, and this occurs at the value x ~ I, where D<x), the distance squared, has its minimum value.

x,

x,

section 4.7, pp. 236-239 3

1.
(b)

3

x T

<e)

3.
(b) -tx-'

I 5.
5 (b) -:x

x T -

x2 +

X

<e) -tx-' + x 2 + 3x

5
7.
(b) Vi

<e) 2W + 2Vi 3 9.
13.
(b) 2 tan

65.
" di .v5 67<)Th • a e nnmmwn 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.
15.
(b)

x T

+

X

3

1 x 23. -:x - T

D(x)=x2-2t+~

27.

~x'/2

31. x 2 +

3

+ C

~x4/'

+

~+

C

C

+ C

o

f + s~ 41 + C

<e) 2 esc ( ,;) + C

21.

2 x4 T - 2s.x

+ 7x + C

3 25. 2x2/' + C

~y'/4

29. 4y 2 -

~

+ C

39. 3eotx + C

43. 4 sec x - 2tanx + C 47.

t 4

33. 2Vt -

37. -21 cos 3 + C

-~tan (~)

2

19. I +

- 3X +


t esc (5x)

2

17.

(t)

+ C 35. -2 sin I + C

I 41. -2eseO + C

I 45. -2cos2x + eotx + C

49. tan 0 + C

51. -cotx -

X

+ C

53. -cos 0 + 0 + C 61.
Section 4.6, pp. 228-230 1.

7. 9.

X2

5 13 = -3' 21

3.

X2

51 5763 = -31' 4945

5.

X2

2387 = 2000

x" and all later approximations will equalxo.

. xsmx

:ix (x; sinx + C) = ~ sinx + ~2 eosx = x2

+ 2cOSX

(b) Wrong:

:ix (-x eosx +

C) = -eosx + x sinx

y


x·

cosx = x sin x . ~ (2x + I)' ) _ 3(2x + 1)'(2) _ 63.
y_J ..rz ,x
2(2x

-l-r-;,.t
x'

11. The points of in1:erBection of y = 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.
+ I)'

(b) Wrong: 6(2x


:ix «2x +

I)' + C) = 3(2x + 1)'(2) =

+ I)'

:ix «2x +

I)' + C)

~

6(2x + 1)2

A-22

Chapter 4: Answers to Odd-Numbered Exercises 33. (a) Local maximum at x = 4; local minimum at x = -4; inflection point at x ~ a (b) x='

65. Right 67. (b)

69. y = x' - 7x

+ 10

I x' I 71. Y=-X+T-2

73. y ~ 9x 1/'

+4

75. s ~ t

77. r = cos ('lT6) - I

+ sin t + 4

79. v =

81. y=-x'+x'+4x+1 85. y = x' - 4x'

2I sect + 2I x",-4

83.r=1+2t-2

35. (a) Local maximum at x = 0; local minima at x = -I sod x = 2; inflection points at x = (I ± v7)/3 (b) Loo_

+5

87. y ~ -siot+ cost

+ t'-I

89. y=2x'/'-50 91.y=x-x'/'+t 93. y = -sinx - cosx - 2

95. (a) (i) 33.2 units, ("Ii) 33.2 units, (ill) 33.2 units 97. t~ 88/k,k~ 16 99. (a) v = IOt'/' - 6t 1/2 (b) S = 4t'/' - 4t'/2

(b) True

37. (a) Localmaximumatx = -V2;localminimumatx = V2; inflection points at x = ± I sod a (b)

Practice Exercises, pp. 240-242 1. No 3. No minimum; absolute maximum: f(l) points: x = I sod 11/3 5. Yes,exceptatx = 7. No

x='

x=-1

~

Loo_

16; critical

a

11. (b) One

13. (b) 0.8555996772

19. GlObaiminimumvalueortatx = 2

(b) t = 3,9 21. (a) t = 0,6,12 (d) a < t < 6, 12 < t < 14

43. 25.

23.

< t < 12

(c) 6

45.

'_X+l_J~·L ,

y

x-3

,

, ,

y""x2;l 5

x-3:

, -----i---

3

.. .;+ {

:

1

:

------ I

-~~~~+-~~~~_x

--=~_"'I.Jc-~-+-O-----!,--_x

y

47.

-v

49.

,

,

,,

..

,'~ ~ :' ,~/

"

y

y

:

-, 51. (a) 0,36

)'=1

-_7,-_cC,-\;--+-;;f---!--cI:;;-,c:,----;,_x

~~!:_~

"

~-4

- - - - T -,,--. .-_-_ _

1

3 y_ x +2 ..

:

i'= x2-3 :x=~

x=-V3

-,

-1

31.

, ,

2

~-yp;;t'~1~,---c,~x

29.

1 2 3 4

,'-I

/.·A~~

-, 27.

-4 -3 -2

-,

.t. +! ,

x

(b) 18,18

53. 54 sqUll1"e units

55. Height = 2, radius = V2

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

~

2 in.;

Chapter 5: Answers to Odd-Numbered Exercises 4

63.

x 4

67. - r

~

61. x, = 2.195823345 65. 21'/2 71.

t

t+ C

5 + 2x2 - 7x + C 5 + C

5•

69. (82 + 1)'/2 + C

· SID'T1' -

. 8m

." + sm .."

2

11. ~ k 17. 19. 21. 29. 31. 33.

73. 10 tan :0 + C

75. - JzCSC V28 + C

77.

kX - Sin I + C

81. r = 41 s/2 + 41'/2 - 81

79. y = x - } - I

=

V32-

2

7• All 0 fthem

9. b

6 4 5

~~ i'

13.

15.

~ (_I)k+Ll

k (a) -15 (b) I (e) I (d) -11 (e) 16 (a) 55 (b) 385 (0) 3025 -56 23. -73 25. 240 27. 3376 (a) 21 (b) 3500 (0) 2620 (a) 4n (b) en (0) (n 2 - n)/2 (a) (b) kal

(I + x 4)'/4 + C

3

A-23

i-I

k-l

y

AddUional and Advanced Exercises, pp. 243-245

y

,

(2,3)

1. The function is constant on the interval. 3. The extreme points will not be at the end of an open inrerval. 5. (a) A local minimwn atx = -1;pointsofinf1ectionatx = 0 and x = 2 (b) A local maximum at x = 0 and local minima

~)"';J;2_1, O:Sx:S2

(~')

...

J~)=x'l-I, : OSl::S2

Right-hand

Left·.... 2

2

,, ,,

. finf1' atx=- I and x= 2 ;pomtso ectionatx= I ± 3v'7

9. No 11. a = I, b = 0, e = I 15. Dri11 the hole at y = h/2. 17. r =

2(H~ R) for H >

19. (a) e ;. b (d) e

13. Yes

2R, r = R ifH '" 2R

(b) e ; b

-I

2

(0) b - 2be:;, e

2

+ 4ae

(c)

y

,

+b +1 2

21. ma = 1 -

I

7i' ml

23. (a) k = -38.72 25• Yces,y = x

=

I 7j

2

(b) 25 ft

--

(2.3)

f~)_x2_1, OS,;:S2

27• Vo = 2V2 3 b'/4

o

c,

c2

I

J c3

c4

",

CHAPTER 5

-1

Section 5.1, pp. 253-255 (a) 0.125 (b) 0.21875 (0) 0.625 (d) 0.46875 (a) 1.066667 (b) 1.283333 (0) 2.666667 (d) 2.083333 0.3125,0.328125 7. 1.5,1.574603 (a) 87 in. (b) 87 in. 11. (a) 3490 ft (b) 3840 ft (a) 74.65 ft/sec (b) 45.28 ft/sec (0) 146.59 ft 31 15. 16 17. I

1. 3. 5. 9. 13.

y f~)

(0) 8

.. IIin::t,

-'lTSxSw

Left·....

•

19. (a) Upper ~ 758 gal, lower ~ 543 gal (b) Upper ~ 2363 gal,lower ~ 1693 gal (0) '" 31.4h, '" 32.4 h 21. (a) 2

(b)

35. (a)

(0)

(b) 2V2 '" 2.828

Sin(~) '" 3.061

(d) Each area is less than the arca of the circle, .". As n increases, the polygon area approaches 'T1' •

Section 5.2, pp. 261-262

37. 1.2

6(1) 6(2) 1'T+1+2+1~7 3. cos(l)." + cos(2)." + cos(3)." + cos(4)."

~

0

43 .

39.

t -in - 6~2'

~+6n+1

6

6n 2

5 '

6

45.

1

~

41. 12 + 27;": 9,

I

I 2n2 '

2 + Ii +

I

2

12

A-24

Chapter 5: Answers to Odd-Numbered Exercises

Section 5.3, pp. 270-274 1. ]"X'dx

3.1'(x' - 3x) dx

°

7.1

('

5.

-7

I

J,

I-xdx

0

secxdx

--rr/4

9. 11. 13. 17. 21. 27. 35. 45. 51.

(a) a (b) -8 (c) -12 (d) 10 (e) -2 (I) 16 (a) 5 (b) 5\13 (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' (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. a Using n subintervals oflcogth lu = bin aod right-<mdpoint

values: Area

65. (a) True. Since j is cootinuous, g is dilfercotiable by Part I of the Fundamcotal Theorem of Calculus. (b) True: g is cootinuous because it is dilfercotiable. (c) True, sinceg'(I) ~ j(l) ~ O. (d) False, sinceg°(l) = 1'(1) > O. (e) True, since g'(I) ~ a aodg°(l) ~ 1'(1) > o. (I) False: gO (x) = I' (x) > 0, so g" never cbaages sign. (g) True, since g' (I) = j(l) = 0 aod g' (x) = j(x) is ao increasing func1ioo ofx (because I'(x) > 0).

Section 5.5, pp. 290-291

5. 11 (3x 2 + 4x)' + C 0

13. t (x'/2 - I) 53. Using n subintervals oflcogth lu values:

= bin aod right-<mdpoint 15. (a) -

7. -tcoS3x + C

11. -6(1 - r')'/2 + C

9. t sec 21 + C

~ ].']X' dx ~ b'

3. -t(x2 + 5)-' + C

1. i(2x + 4)' + C

i

4I (cot' 28)

sin (2x'/2 - 2) + C + C

(b) -

17. -t(3 - 2x)'/2 + C 55. 61. 63. 71. 73.

19.

av(f) ~ a 57. av(f) ~ -2 59. av(f) ~ I (a) av(g) = -1/2 (b) av(g) = I (c) av(g) = 1/4

21. (-2/(1 + Vi» + C

e(b - a) 65. b'/3 - a'/3 67. 9 a ~ a aod b ~ I maximize the integral. Upperbound = I, lower bound = 1/2

25. tSin'

69. b'/4 - a'/4

~a

79. Upper bound

~ 1/2

Section 5.4, pp. 282-284 3. -10/3

5. 8 ".

4

13. -"./4

15. I -

21. -3/4

23.

29. (cosVi)

C~)

v'2 -

35. -tx-l/2SinX 43. 1/2

45.".

49. d,sincey'

~

17.

7. I 2-

~+ I 31. 41'

v'2

11.

33.

a

19. -8/3

25. -I

39. I

37. 0 47.

9. 2\13 4

. (I) + C

~

I (I 16

I (x - 1)12 43 . IT

(~: - I)' + C I 31. 2 cos (21 + I) + C 35. -

sin' (1/8) 2 + C

1)'/2 + C 39. 32(2 - :x

+C

47. 1(x2 + 1)'/2 _1(x 2 + 1)'/2 + C 5 3

49.

~'x +

~'x +

51. (a) - 2 +

41.28/3

C

(b) - 2 +

-3

57. s = 41 - 20in (21

+ ~) + 9

f)

+ 100/+ I

S

-I + C 4(x2 -4)2 C

6 , + C 2+taox

53. isin Y3(2r - 1)2 + 6 + C ~

59.

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

+ C

I 4 2 45. -g(l - x)s + 7(1 - x)7 - 3(1 - x)' + C

(c) -

2 55. 3 bh

8')'/'

I + rr(x - I) 11 + C

51. b,sincey' = secxaody(O) = ]'°seCldl+ 4 = 4

r 53. y = J2 sec 1dl + 3

-

27. 16

V1+7

J.~ I ~ tdl - 3

44 + C + I)

2( 3)'/2 41. 27 I - x'

YJ:'"

I :xaody(".)

27.

33. -sm t - I 37.

1. 6

C

+

-~(I

+ C

I 23. 3tao(3x + 2) + C

2 ('/2 ) 29. -3cOS x + I + C

75. For example, ].' sin (x') dx '" ].' dx = I 77. J.'j(X)dx '" J.'Odx

(t)

4I (csc2 28)

= sin (21-

55. s

~

t(31' - I)' - 5

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

Chaptet 6: Answers to Odd-Numbered Exetcises

17. 27. 35. 45. 51.

25. 16/3 19. 3'/2 - I 21. 3 23. 'IT/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

3/4 2'/2 5/6 8/3 18

61. 56/15

4

63. 4

4

65':3 - 1f

y

,

67. 'IT/2 _~'-~_II---+-4---!,~X

69. 2 71. 1/2 73. I 75. (a) (±VC, c) (b) c ~ 4'/3 77. 11/3 79. 3/4

(0) c ~ 4'/3

17. 1/2

Practice Exercises, pp. 301-303 1. (a) About 680 ft

15. 13/3

25. (a) 0 (e) y

(b) h (feet) 700

23. (b) 'lTr'

(d) x = I -I,x ~ 2

(b) -I (c) -'IT 2x + 2 - 'IT (I) sin4y siny 29. V;I - 2V;1

~

27. 2/x

600

21. /" f (x) dx

19. 1/6

X ~

(g) [-2'IT,Oj

500 400 300 200 100

----"0t-""-.;----7-;---;-~ t (sec)

3. (a) -1/2 5.

l'(2x -

9. (a) 4

11. 8/3

23.

(b) 31

1)-1/2 dx = 2

(b) 2

'? -

15. I I

25. 4

/,X(S~I)dt - 3

53. 27V3/160

61. -I 63.2 65. -2 71. (a) b (b) b 75. 25'F

83.

77.

V~2-+-c-o~s"x

3I'

41.

4 + t + C

49. I

79.

~ 3 +x

x

~

5. (a) 1/4

(b)

13· y

19. 3;'IT

43. 8'IT

-.

0

-.

, y=-4

x

(b) s'h

21. 36'IT

51.8

49. (a)

2'IT 35. T

45.

7'IT

6

I6'IT

15

53. (a) V =

15. 2:;

(b)

29. 2'IT

117'IT 37. -5-

56'IT

15

(0)

32'IT

'lTh'(3a - h)

3

64'IT

15

(b)

31. 3'IT

39. 'IT ('IT - 2)

(b) -5-

(0)

4'IT 41. T

8'IT

T

,

51. V = 2a b'IT

(d)

,

I

TW1T m/sec

59. 4 - ~ + a

\o/l2

.=

.

1

£-

3 2

11.

i~

13. (b) 4'IT

15.

19. 4:;

I~

17. 8:;

'IT

5 I4'IT

7 8

. 'IT

'3

21.

23. (a) I6'IT (d) 24'IT

(b) 32'IT (e) 60'1T

(0) 28'IT (I) 48'IT

y

25. (a) 27'IT

(b) 27'IT

(0) 72'IT

2

27. (a) 6'IT 5

2

(b) 4'IT 5

9 5'IT

'6

I~; (3\12

Z -1f

I

(b) 36

23. 'IT

27. 2'IT

47. (a) 8'IT

~ 3308 em'

2

-s

13. (a) s'h

7. (a) 60

section 6.2, pp. 324-326

9. y ~ x' + 2x - 4

11. 36/5

(b) 8

81. Yes

Additional and Advanced Exerdses, pp. 304-307 7. f(x) ~

17. 4 - 'IT

69. \12 - I

85. Cost '" $10,899 using a lower sum estimate

(b) No

11. 10

57. V

-vT+?

1. (a) Yes

9. 8'IT

--' - 2'IT 33. .,,-

59. 6V3 - 2'IT

67. I

3.

25. 'IT(f + 2\12 - 11) 3

47.2

57. V3

• r-: 5. (a) 2 V 3

316

1. 16

7

37. -4(oosx)'/2 + C

45.16

55. 'IT/2

21. 9/8

31. 6/5

39. IJ' + 6 + sin (26 + 1) + C 43. -tcos(2I'/2) + C

19. 18

8~ -

27.

= 2

(e) 8/5

(d) -2'IT

17. 1/6

29. Min: -4, max: 0, area: 27/4 35. y =

section 6.1, pp. 316-319

(d) 0

J~COSIdx

7.

(c) -2

13. 62

~+

(0) 13

CHAPTER 6

5

(0) 2'IT

+ 5)

(d) 108'IT

5

(d) 2'IT

29. (a) About lbe x-axis: V =

~~; about lbe y-axis: V = ~

(b) About lbe x-axis: V =

~~; about lbe y-axis: V = ~

224'IT

15

A-26

Chapter 6: Answers to Odd-Numbered Exercises

31.
(b) 4;

33.
i~

35.
~".

37.
16

5.

(b) ,

,

4~'"

(b) (b)


~~

(b)

9".


9".

16

39. Disk: 2 integrals; washer: 2 intcgrals; shcll: I integral 41.
(b)

u:".

.~--;----;;---"',~.'

/,7/3 U:

Section 6.3, pp. 330-332 1. 12

53

6

3.

11. ';1 13.
/,7

VI

15.
[7/'

17.
5.

123

32

+ 4x2 dx

7.

(y

sccxdx

7.
9. 2

(b)

+ 1)2 t!Y


0.'

9. 4".VS


1. 5. 7. 15.

25. 31. 35. 39.

400 N/m 3. 4 em, 0.08 J
85.1 ft-Ib 27. 1684.81b 33. 1164.8 Ib 37.
41.

15. 2".

21. (2"'/3)(2V2 - I)

Section 6.5, pp. 342-346


~ VI + y-4 t!Y

13. 98"./81

19. 35".VS/3

27. Order 226.2litcrs of each color.

23. 253"./20

17. 37,306 ft-Ib

3.
tJmtdJ

11. 3".VS

17. ".(v'8 - 1)/9

,

2

L'

o '--coO'o-.l-COO~2-COO~3-con'o-,-COn~, -COo.'o-.-COO.~7 -->,

Section 6.4, pp. 335-337


,1;"


31. {7 <103/2 - I)

[7/'

,


= ±X + C wbere C is any real number

1.

o.s

19.
tan t dt) sec Y dy


+ cos2 Y t!Y +

99

8

21. 7,238,229.47 ft-Ib

98.35 ft-Ib 29. 91.32 in.-oz
(b) 3 ft

43.

~b

45. No. The tank will overflow because the movable end will bave movcd only ft by thc timc thc tank is full.

3!


, 2

Section 6.6, pp. 355-357 1. x = O,y = 12/5 3. x = I,y 5. 9.

13. ';-
15.

= -3/5

x ~ 16/105,y ~ 8/15 7. x ~ O,y ~ "./8 x = l,ji = -2/5 11. x = Y = --24 ". x ~ 3/2,y ~ 1/2
Chapter 7: Answers to Odd-Numbered Exercises

,


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"'---±-

."

(2,0)

I

0

r

4

y.--±."

-4

19.

j

= Y = 1/3

21.

= a/3, y = b/3

j

27.

j

~

,

,

23. 136/6

025 . x= ,y

mr =4 1/2,y ~ 4

17.
6

29.

j

~

35.

= 32"., S = 32v'z". 4".' 37. j = O,y =~

41.

v'z".a 3(4 + 3".)/6

6/5,y

~

8/7

33. V

43.

39.

j

=

O,y

=

:~

~ ~,y ~ ~

j

'"-----~--c-...6~r

Practice Exerdses, pp. 357-359 9". 72". 1. 280

3.".'

7.
5.

19. r'(x)

23. r'(x)

35

25. r'(x)

(b)".
27. r'(x) range:

11. "'(3V3 - ".)/3 13.
(b) 8"./5

17. 10 3


29. r'(x)


19. 2:5

= v;=J 21. r'(x) = ~ =~- I = ~; domain: -00 < x < 00; range: -00 < y < = 5v;=J; domain: -00 < x < 00; <

-00

range:

<

5 -".'300

7. 28/3

3. I(x) =

~ x + a, where C '"

OO,y

range:

-00

35.
y';+1; domain: -I < y :5 I

~

I -~

>

0

oo,x

~ I;

=F 2

33. r'(x) ~ I -

:5

x<

00;

37. (a) r'(x)

~ -~ + ~

(b)

,

Additional and Advanced Exerdses, pp. 359-360

~2x ;; a

0; range: y

+ 13 ; domain: -00 < x <

x-00 < Y <

(b)

1. I(x) =

00

~ ~; domain: x>

31. r'(x) = 2x

28".v'z/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

Y

I

9 4h\l3mh • 3

~ O,y ~ 2n':- I' (0, 1/2) 15.
j

(b) (2a/".,2a/".) 17. "'2329.6Ib


CHAPTER 7 Section 7.1, pp. 367-369 1. One-to-one 3. Not one-ro-one 7. Not one-to-one 9. One-ro-one

s.

One-to-one

(c) -4, -1/4

,

00

A-28

Chapter 7: Answers to Odd-Numbered Exercises

(0) Slope off at (I, I) : 3; slope ofgat (1, 1): 1/3; slope off at

(-1, -1): 3; slope ofg at (-1, -I): 1/3 (d) Y = 0 is tangent to Y = x' at x = 0; x y~~atx~O. 41. 1/9 43. 3 45. (a) rl(x) =

=0

8+5[1 I ] 63'8cos88+5-i1+tan8

is tangent to

65 xW+! • (x + 1)2/3

,k x

(b) The graphofr l is the line through the origin with slope l/m. 47. (a) rl(x) ~ x-I

(b) rl(x) = x-b. The graph of r l is a line parallel to the grapb of f. The graphs of f and rllie 00 opposite sides of the line y = x and are equidistant from that line. (0) Their graphs will be parallel to ooe another and lie 00 opposite sides of the line y = x equidistant from that line.

53. Decreasing, therefore one-to-one; dr1/dx = -tx-2/'

2x) 67 -I ..jx(x2 - 2) (1 - + 1- - -2"3 x +1 x x-2 x +1 69. (a)Max = Oatx = 0, min = -In2atx = 1t/3 (b)Max = I atx = 1, min = cos (In 2) atx = 1/2 and x 71. In 16 73. 41t In 4 75. 1t In 16 77. (a) 6 + 1n2 (b) 8 + 1n9 79. (a)x '" 1.44,y '" 0.36 (b) ,

Section 7.2, pp. 375-377

83. Y ~ x + Inlxl + 2

3. (a) In 5 5. l/x

(b) In (x - 3)

7. 2/t

9. -I/x

15. 2(lnt) + (lntl'

21.

27.

I x(l + Inx?

3x+2 2x(x + I)

l 23. x In x

43. (ln2? 49. In 2

45.

+1

13. 3/x

I

~ -lOln3

t

11.

x 2e'

~

-2

35. 2xlnlxl- xln

!xr.c VL

I

29. te" -

5e~ + C

~e~~'

+ C

57.

G~J5Y;

73. (x + I)(x _ I)

47. 1n16 + 3tantl + C

55.

(-2 )v'X(X + I)(~ + -I-I) = x +

57.

(t))t ~ I (+ -

59.

'\1'8+3 (sin 8) (2(8 ~ 3)

1

(11)

61. 1(1 + 1)(1 + 2)[+ +

t

+ C

77. 1n\0

43. e

53. y = 2(e-' + x) - I

1tx(~-l)

55.2' 1n2

61. -V2cos8(v2-I)sin8

1)'/2

= 312 +

.

I

75. sm (log7 8) + In 7 cos (log7 8)

Iog 79. +(Iog,3)3 "

81. +

I I 6 as' 2ln2 87' ln2 89' ln7 3x(V3+1) . r. 91. 32760 93•• r.c + C 95. 3 v2+1 v3 + 1 ln x 97. 1n\0 2 ?) + C 99. 2(ln2)" 101. 3

e

+ cot 8)

11 + 112]

41. -ell' + C

5' 83· ln5 +C

2x + I 2v'x(x + I)

= 2Vt(/+

59.

35. 2

49. In (I + e') + C

47. 1

51. Y = I - cos(e' - 2)

33. 8e(>+I) + C

63. 7'" '(In 7?(seo 8 tan 8) 65. (3cos31)(2""')ln2 I 3 x2 2 67. 8ln2 69. xln4 71. In 10 + 3x 10glO x

41. 1n3

'\."x)

31. I

39. -e-l' + C

-2

53. In (I +

15. 28e -8' sin (e-8')

1

2

~4

9. xe'

19. 1/(1 + e') 21. e""'(1 - Isint) yeY cos x 20" - cos (x + 3y) 23. (sinx)/x 25 27 _::-----;---c-~-'-''l-yeYsinx • 3 cos (x + 3y) 17. 1

45.

tan (In 8) 31. 8

() inA 0 1~1n.2

7. -7e(5-7,)

13. 2e' cos 8

37. 2e y; + C

t

(b) I ~ _In2 k

5. -5e-'"

lnt

25. 2 cos (In 8)

39. Inly2 - 251 + C

51. In 27

(t 2 )

19. I

2 29. 1(1 _ In t?

10x 33. 2 + (I ) x +1 21-x 37. In(t)

11. 8

17. x'lnx

(0) -In2

(f) t(3ln3 -In2)

(0) In

85. (b) 0.00469

Section 7.3, pp. 385-387

3. 4(lnx)2

(b) 2(ln2 -In3)

(e) 1n3 + tln2

=2

ccol-----!'=='-!--,

1. (a)

(d) tln3

_x_ _ 2 ] x 2+ I 3(x + I)

= ~x-2/3

51. Increasing, therefore ooe-tl>-ooe; dr1/dx

1. (a) In3 - 2ln2

[1X+

103. In 10 61 + 2

105. (In 10) Inllnxl + C

107. In (In x), x > I

109. -Inx

~2

Chapter 7: Answers to Odd-Numbered Exercises

111. (x + I)'

(X ~

115. (sinx)'(lnsinx + xcatx) ~

119. Maximum: I at x

(~I + k)

113. ("vt)'

I + In (x + I))

~

87. (a) y = I

~

(b) y = 0, y =

89. (a) We shauld ..siga the value I to f(x) = (sin x)' ta make it continuous at x = o.

117. casxx.x-'(l + Inx)

0, minimum: 2 - 2 10 2 at x

83. -I

A-29

10 2

y

121. (a) Absmax: }atx = I

(b) (2, :')

123. Abs max afl/(2e) assumed at x = 1/"';; 125. 2 127. y = e x/' - I

e' 2~

I 131. In (\/2 + I) d I 133. (a) dx(xlnx -x + C) =x'x + Inx - I + 0 = Inx

129.

I (b) e - I

(0) The maximum value af f(x) is clase to I near the point x '" 1.55 (see the graph in part (a».

135. (b) Ierror I '" 0.02140 (0) L(x) ~ x + I never overestimates eX.

section 7.6, pp. 413-416 1. 3. 5. 7.

(a) (a) (a) (a)

'tt/4 (b) -'tt/3 -'tt/6 (b) 'tt/4 'tt/3 (b) 3'tt/4 3'tt/4 (b) 'tt/6

9. 1/\/2 19.0 137. 21n 5

139. x '" -0.76666

141. (a) L(x)

~

25.

I + (102).< '" 0.69x + I

29.

9.

1

=C

y 'I' - x'!,

11. e" - eX

13. -x + 2taovY = C

17. y = sin (x' + C)

21. 41o(yY 23. 25. 31. 33. 39. 41.

+ 2) ~

=C

15. e-Y + 2ey'; = C 47.

(a) -0.00001 (b) 10,536 years (0) 82% 54.88 g 27. 59.8 ft 29. 2.8147498 X 10 14 (a) 8 years (b) 32.02 years

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

29. In 2

31. In 2

41. -1/2 53. I 65. +1

75. (b) is correct

43. -I

55. I/e

31.

-I 2Vt (I + I)

33. I 45. I

35. 1/2 47. 0

11. 5/7

13. 0

25. -I 37. In 2

-1

43. sin-I

55.

I (tao Ix)(1 + x') -2s 1J 37. ~ 39. 0

v?'=l

~sec-II~I

~+ C

+ C

45.• V

~ tao-I • ~ + C 17

V

17

51. 'tt/16

49.2'tt/3

~sin-12(r - I) + C 59.tsec-112X;II+c

63. 'tt/12

65. k sin-I y' + C

69. 'tt

71. ktao-I

f;

I) + C

67. sin-I (x - 2) + C

73. 2'tt

1 1n(x'+4)+2tao-I ":+C 2

2

77. x + 10 (x' + 9) -

79. sec-I Ix +

11 +

83. t(sin- I x)' + C

49.2

57. e !' 59. I 61. e' ' 67. 3 69. I 71. 0 73. 00 27 -I 77. (d) is correct. 79. c ~ TO 81. (b) T

17. 'tt/2

33.

61. 'tt

75.

I

-I

vJ=t2

57. ,?tao-,(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%

15. 'tt/2

-2x (x' + I)Yx' + 2x'

27.

I

12s + II~

53. -'tt/12

+C

13. 'tt/2

~ 23.~ I-x' 1-21'

21.

41. sin-I x

19. tlnly' - 21 = x' + C ex'

11. -1/Y3

-e' 35. le'IY(e')' _ I

Section 7.4, pp. 394-396

(0) 'tt/6 (0) -'tt/3 (0) 'tt/6 (0) 2'tt/3

89. 1tao-1 97. I

(tao-~

C

13°tao-l~ + C 81. e""-'x + C 85. Inltao-I yl + C

Vi) + C

91. 5

87. Y3 - I

93. 2

95. I

103. y ~ sin-I x 105. y ~ sec-I x + 2;,x

>I

A-30

Chapter 7: Answers to Odd-Numbered Exercises

107. (b) y = 3YS

109. 8 = COS-I

(~)

3. sinhx = 8/15, tanh x = 8/17, cothx = 17/8, sechx = 15/17,

"" 54.7'

cschx = 15/8

121. w2/2 123. (a) w2/2 (b) 2'IT 125. (a) 0.84107 (b) -0.72973 (0) 0.46365 127. (a) Domain: all real numbers except those having the fonn

f + k'IT where !is an integer Range:

'IT

I 5. x + x

" 7. 0 " 9. 0

r. tanhVt 15. sech v t + Vt 2.

-2
25.

35. -00

<x <

00;

21. tanh'v

RJmge:

-00


00

lsecxl

23. 2

27. I +1 8 - tanh-I 8

I 2Vx(1 + x)

1n2

31. -sech-I x

29 _1_ - coth-IVt • 2Vt

(b) Domain:

3

17. cothz

19. (In sech 8)(sech 8 tanh 8)

'IT

13. 2cosh x

33.

CO~2x + C

41.

(f -ln3) + 45. 7 In I + 0~/71 + 47. tanh (x - ~) + 49. -2 sechVt + 51. In % 43. 12 sinh

C

0'/7

C

53. {2 + 1n2

61. 1n(2/3)

55. 63.

00;

(b) Domain: - I '" x '" I; RJmge: -I '" Y '" I

y

y

2

0-

-~3

1

)'=005(008-1 x)

(b)

0;)

57. 3/4

59.

~

+ 1nv2

65. 1n3

(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

(h)

(~) In(t)

(~)

- (12/13)2) (I + VI(4/5)- (4/5)2) -In (I + VI(12/13) + In -InG) + In In (4/3) (2) =

=

-2

73. (a) 131. The grnphs are identical.

133.

y

77. (b)

y

a

n

(b)

a (c) 80YS "" 178.89 I'l/sec

79. 2'IT

81.

%

Section 7.8, pp. 428-429

y.-'-

(~)'

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 (c) Slower (d) Faster (0) Slower (f) Slower (g) Same (b) Slower 3. (a) Same (b) Faster (c) Same (d) Same (0) Slower (f) Faster (g) Slower (b) Same 5. (a) Same (b) Same (c) Same (d) Faster (0) Faster (f) Same (g) Slower (b) Faster 7. d, a, c, b 9. (a) False (b) False (c) True (d) True (0) True (f) True (g) False (b) True 13. When the degree of f is less than or equal to the degree of g. 15. I, I 21. (b) (ol"lOOtlOO) ~ 17,000,000 < (0 11'10')1/10' ~ 0 17 "" 24,154,952.75 (0) x "" 3.4306311 X 10" (d) Theycrossatx"" 3.4306311 X 10"-

~

-5/3,

A-31

Chapter 8: Answers to Odd-Numbered Exercises

23. (8) TIre algorithm that takes 0 (n log, n) StepS (b)

117. I 119. I/e m/sec 121. 1/v2 units loog by I/y';; Wlits high, A = 1/y'2;, '" 0.43 units2 123. (8) Absolute maximum of 2/e at x = e 2; inflectioo point (e'/3, (S/3)e-4/3); coocave up 00 (e'I', 00); coocave down 00 (0, e'/3)

, 2500 2000

ISOO

,

1000

'OO~~~~~;';:'~",",,~nn 20

40

60

80

0.'

100

-0.5

-I

25. It could take ooe millioo for a sequential search; at most 20 steps for a binary search.

Practice Exerdses, pp. 430-432 1.

-2e~/'

3. xo'"

13. (x

+ 2)"+2(1n (x + 2) +

-I 17. .,.-------,2 vl-x coS-1X 21.

sin

11. 18x 2.•

9. -S-t(ln S)

~+ z2 -

1

I)

+ tan 2x ]

(t + I)(t 27. 5 [(t _ 2)(t + 3)

[I

1)]' t+T +

VB(SinO)V.(In~ +C

33. tan (eX - 7) 39. inS

49.

2~3

37.

-~

53.15/16

57. 1/6

59. 9/14

+

I)

1n2 79. Y ~ 1n(3/2)

+

C

75. ",,/2

-2

-4

43. -[lnlcos(lnv)ll

69. sec-1 12yl + C

----c:--;;----I::p-~ ......, 3 2 1 2 3

31. -cos eX + C

51.3ln7

",,/"1/3

73. sin-I (x

0)

+C

61. t[(ln4)' - (In 2)'] ort(ln2)' 67.

y=(l+J:)l!"""

I I I] t- I - t - 2 - t+3

+C

C

--"',--..,,2"'--!,---'----!,~2:--,'!---+'

,

+ Inr) + C

(3"')

55. e - I

~

47. -cot(l

45. -k(lnx)-2

,

(c) Absolute maximum of I at x = 0; inflectioo point (I, 2/e); coocave up 00 (I, 00); concave down 00 ( - 00, I)

+ ocot

35. e""x

41. 1n(9/25)

+

15.

23.-1

2(x + I) ~ 2x 25. • ~ -2-vcos2x +I

29.

(J

(b) AbsoluteIIUJXimumofi atx = 0; inflectioopoints (±I/v2, I/y';;); coocaveup 00 (-00, -1/v2) U (l/v2, 00); coocave down 00 (-I/v2,I/v2)

19. tan-I(t) + _ t _ - ~ l+t 22t

sec-Iz

2

2 7. (ln2)x

5. 2 sin 02COS 0 = 2cotO

-1.5

-.

7

+C 125. Y

+ 1n2

~

(tan-I

129. y = -2

(x ~ C)Y

+ In (2

133. IS,935 years 63.

9~2

65.""

71. ",,/12

77.

81. Y ~ Inx - 1n3

t sec-I (t ~ I) + C I 83. Y ~ I _ eX

85. 5 87. 0 89. I 91. 3/7 93. 0 95. I 105. I 107. I 97. In 10 99. In 2 101. 5 103. -00 109. (8) Same rate (b) Sarae rate (c) Faster (d) Faster (e) Saraerate (f) Samerate 111. (8) True (b) False (c) False (d) True (e) True (f) True 113. 1/3 115. Absoluternaximum = 0 at x = e/2, ahsoluteminimum = -0.5 atx = 0.5

- e~)

127. y

2

x

= sin-I (2 tan + C)

131. y = 4x - 4~

135. 20(5 -

+

I

vl7) m

Additional and Advanced Exercises, pp. 433-434 1. ",,/2 3. I/y';; I I 9. 1n2' 2ln2' 2: I 4 17. x = ll.:r , ji = 0

5. In 2

7. (8) I

11. x = 2

(b) ",,/2

13.2/17

19. (b) 61'

CHAPTER 8 Section 8.1, pp. 441-443 1. -2x cos (x/2) + 4 sin (x/2) + C 3 3. t 2 sint + 2tcost - 2sint + C 5. 1n4 - 4 7. xo X - eX + C 9. -(x 2 + 2x - 2) e~ + C

(c)""

A-32 11. 13. 15. 17. 19.

Chapter 8: Answers to Odd-Numbered Exercises

ytan- I (y) - 1n\!'l+7 + C xtanx + In leosxl + C (x' - 3x 2 + 6x - 6)e' + C (x 2 - 7x + 7)e' + C (x' - 5x 4 + 20.' - 60.2 + 120. - 120)e' + C

33. ktan2x + C

43.4/3 45. 2tan2 x - 21n(l + tan2x) + C 1 1 47. 4tan4x - :2tan2x + In lsecxl + C

e'"

49.

23. T3(3sin3x + 2eos3x) + C

t (~e

2~

".\13 _ In (2) _ ".2 3 18

y'j;+o -

ey'j;+o)

37. ttan'x + C

41. ttan8 + t,ee2 8tan8 + C

39. 2\13 + In (2 + \13)

21. k(-e'eos8 + e'sin8) + C

25.

35. t,ee'x + C

.r. 34-Inv3

1. 55• "2 smx

+C

1 1 51. -TOeos5x - :2eo,x + C

53.".

1·7 C + 14 sm x +

- ~sin58 + C . 16 ,in 38 _lsin8 4 20

57

59.

29. k[-xeos(lnx) + x sin (lnx)] + C

-~

eo,'8 + C

61. ± cos 8 - ioeos58 + C

63. ,eex - In lesex + eOlxl + C 65. eosx + ,ecx + C 1 2 1. 1 69. In (l + V2) 67. 4x -4xsm2x-geo,2x+C

1

31. :21n Iseex2 + tanx 2 1 + C 33. kx2(lnx)2 - kx2lnx + ±x2 + C

71. ".2/2

37. 1 e'" +C 4 2 2 2 2 39. t x (x + 1)'/2 - 15 (x + 1)'/2 + C

73 - _ 4". - _ 8".2 + 3 • x - 3'y 12".

1 1 35. -xlnx-x+C

41.

-~sin3xsin2x

".2;; 4

C

7. 2i

45. 2~sin~ + 2 cos ~ + C

43. -eose' + C 47.

~eos3xeos2x +

-

Section 8.3, pp. 452-453 1. 1n1v'9+7 + xl + C 3. "./4

49. 5". -9 3\13

9.

51. (a)". (b) 3". (c) 5". (d) (2n + 1)". 53. 2".(l - 1n2) 55. (a) ".(". - 2) (b) 2". 57. (a) 1 (d)

59.

(b) (e - 2""

j

=

l' (e

+ 9)

x", dv

19. 61. u = x', dv = eosx fix 67. x sin-1 x

e flX dx

=

69. x sec-Ix -Inlx + ~I + C

71. Yes

3. -±cos4 x + C -

tcos5 X

11. ±Sin4 x - t,sin6 x + C

15. 16/35

+C

1

5. 3COS3 X -

cosx + C

9. sinx - tsin3x

Il_ ~3

17. 3".

29

±

/ (1)'/2 _1Il35 _ ~7 (1)7 2

. 5 2

2

31.

~+

,ee-I(~) ]

V2

17.

t(x

+ C 2

13.~ + C

+ 4)'/2 - 4W+4 + C

21. 10 tan-I ~ + C 3 x +C

~ + C

11

+"\Ilo) C

+C

29. 2 tan-I 2x +

+ C

4x + C (4x 2 + I)

1( v )' + C 33. 3"V'J-=-;;z

37. "./6

39. sec-1lxl + C

43. klnlvT+7 + x 2 1 + C

V; + ~~ + C 47. ± ,in- ~ ±~ ~ 45. 4 sin-I

1

13. kx + ±Sin2x + C

19. -4 sin x cos3 x + 2 cosX' sin x + 2x + C 21. -eos4 28 + C 23. 4 25. 2 27 • \j:2

+C

41.

1

25.

(~)'

35. 1n9 - In(l

49

+C

1 2 +:21n 1 Ix 2 31. :2x

Section 8.2, pp. 448-449

+ tcos3 X

-2~

27. --51

(b) X sinh-Ix - (1 + X 2 )1/2 + C

7. -CDSX'

- 49 7

2

23. 4\13 - 4:

+ cos (sin-1 x) + C

73. (a) x sinh-I x - cosh (Sinh-I x) + C

1. ksin2x + C

Vy

15. -~ + C

= ±(e2 + I),)' = k(e - 2)

2~ (1 - e-2~)

63. u

(c)

2

t~ + C

'in-'(t) +

klnl~ + V4X~ -

11. {

5. "./6

-

(1 - 2x)

+C

Chapter 8: Answers to Odd-Numbered Exercises

I~ ('IT

55.
x=

57.
-t

3\13 - 'IT , Y = .,:> + 12\13'IT - 72 4('IT + 6\13 - 12) 12('IT + 6\13 - 12) 2 x 2 (I - x 2)3/2 - 15 (I - x 2)'/2 + C

+

3

15 (I

(0)

section 8.5, pp. 467-468

+ 6\13 - 12)

(b) _1 (I - x 2)3/2

1 (I - x 2)'/2 5

13 (I

- x 2)'/2 -

- x 2)3/2

+C

+C

1.

-h (tan-I~X

; 3) + C

• ,------;:: (2(X - 2)

+ I)

5.

(2x - 3)3/2(X 5

7.

-~ x

9.

(x + 2)(2x - 66)~ +4sm. (x-2- 2) +C

7.1+~+-12 t-3 1-2

5.-2+2+_2_ Z z2 z-l

e

-

2lnl2 +

-~lnlll

+ ilnll + 21 +

l-

x l 19. 1ln l + 4 x - I

x

2(x 2 - I)

17.

13. (lnI5)/2

x;

+

y

I

tin II +C

II + C

17. 3ln2 - 2

29.

31.

2

3 1n

21. ('IT + 2 In 2)/8

+ C 27.

i

In

I

+ + 6 1n Ix 2 + x + II - •v r.:3 tan-I (2x\13

11

Ix -

I) + C

+ 28 + 2

2

33. x + Inlx

~

41.

y2

2

+ In (0'- + 28 + 2) - tan-I (8 + I) + C II + C

21

39. In(e't e

I) C

+ + +2

+C

45. In IYX - II + C

YX+I

In

47. 2\1'lh + In Iv'XTI - II + C v'XTI+I

Ix/~ II + C

[Sin (71/2) sin (91/2)] 7 9 + C

~sin(78/12) + C

) ZI In (x 2 + I +

29. (x -

23. 8

x I-I 2(1 + x 2) + ztan x + C

k) sin-I~ + ~v'x

- x2 + C

ye!------=sin~2'1 I +

C

35. In Ilny + v'3 + (lny)' I + C

57. 1.10

41 •

sin4 2x cos 2x 10

~

sin-I (x ; 2) + C

28m2 2x cos 2x _ 4 cos 2x 15 15

+C

sin3 28 cos228 + sin3 28 + C 10 15 2 45. tan 2x - 2 Inlsee 2x1 + C 43.

47.

I (sec 'lTX) (tan 'lTX) 'IT + 1T Inl sec 'lTX + tan 'lTX I + C

49.

-esc' x cot x 4

51.

ZI [sec (e'

3esexcotx 8

3

I

I

Sin esex + eotx + C

- I) tan (e' - I) +

Inlsec (e ' - I) + tan(e ' - 1)1l + C

51. x ~ Inll - 21 - Inll - II + 1n2

55. 3'IT In 25

cos 5x eosx -----ro - -2- + C

39. x ; 2 v'5 _ 4x _ x2 +

I

I ISin sin siny + 3

i

I

37. In Ix + I + v'x 2 + 2x + 51 + C

(tan- I 2x)2 6 43. 4 -3Inlx-21+x_2+C

49.

C

sml

-Inlyl + ZIn(1 + y2) + C Y

2

33. v'1 - sin2I _ In II +

35. 9x + 2lnlxl +} + 7lnlx - II + C 37.

ix~ +

31. sin-I~ - v'x - x 2 + C

I~ ~ : I + ~ tan-I x + C

0'- -I

+ C

19. ~ tan-Ix - ~ + 61n(1 + x 2 ) + C 21.

25. -(8 - If2 + (8 - 1)-1 + tan-I 8 + C 27.

~I

COS-Ix + isin-Ix -

25. 6 sin (8/12) +

23. tan-I Y - -2-1-

+C

2t

3

15.

f7+71

15. 13(2 cos 31 + 3 sin 31) + C

9. Z[lnll + xl-Inll - xl] + C 11. tlnl(x + 6)2(X - 1)'1 + C

+C

v9-4x+3

-I

13. v'4 - x 2

I

+C

21~-31 - -3 In • ~

11. - 0lnlV7 +

3 _1_+ 3 •x + I (x + I)'

+4 ) +C

3

3. vx-2

Section 8.4, pp. 461-462 1 _2_+_3_ • x - 3 x - 2

A-33

59.
~

53. x

~

1000e". 499 + e I

61 1+2 - I (b) 1.55 days

53. V2 + 1n(V2 + I)

55. 'IT/3

57. 2'IT\13 + 'lTV21n ("V2 + \13) 59.

x ~ 4/3, Y ~ 1nV2

61. 7.62

63. 'IT/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. 1: (8) 1.5, a (b) 1.5, a (0) 0% 11: (8) 1.5, a (b) 1.5, a (0) 0% 3. 1: (8) 2.75, 0.08 (b) 2.67, 0.08 (0) 0.0312 '" 3% 11: (8) 2.67, a (b) 2.67,0 (0) 0% 5. 1: (8) 6.25,0.5 (b) 6,0.25 (0) 0.0417 '" 4% 11: (8) 6, a (b) 6, a (0) 0% 7. 1: (8) 0.509, 0.03125 (b) 0.5,0.009 (0) 0.018 '" 2% 11: (8) 0.5, 0.002604 (b) 0.5, 0.0004 (0) 0% 9. 1: (8) 1.8961,0.161 (b) 2,0.1039 (0) 0.052 '" 5% 11: (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. (8) 76 (b) 12 21. (8) 82 (b) 8 23. 15,990 ft' 25. '" 10.63 ft 27. (8) "'0.00021 (b) "'1.37079 (0) "'0.015% 31. (8) "'5.870 (b) IETI '" 0.0032 33. 21.07 in. 35. 14.4

15.

V3

3. 2

5. 6

17. 'fT

7. 'fT/2

19. 1n(1 +

1')

9. In 3

11. In 4

21. -I

13.

13. _lin looso - II + C 300s0+2 15. 4lnlxl- kin (x' + I) + 4 tan-I X + C 17.

a

23. I

y

I.S I.,

0'

I.' I.'

Si{t)=

I

/,,'

o. o.

.

smIdt

'

o.s 0.' 0.' 0.'

0' 0

0'101'

(b) 'fT/2 77. (8)

20

2S

,

-0.'

"

25.

llnl~

4 2 + 3 1n Ix + 21 + 31n Ix - II + C

- II + C ~+I

3

X

tan-I (3x) - iln(I + 9x2 ) + C

5. (x + I)'e' - 2(x + l)e X + 2e x + C

27. Inll - e-'I + C

29. -V16 - y' + C

31. -kin 14 - x'i + C

33. In

35. lin Ix + 31 + C 6 x-3

I

~

+ C

'7

37. - oo~ x + co; x + C

39.

41. co;o - co~~IO + C 45. 51. 53. 63.

Allessl 16

¥

,

+ C

43. 4VI - cos (t/2) + C

47. T = 'fT, S = 'fT 49. 25'F (b) "'24.83 mi/gal 57. In 3 59. 2 61. 'fT/6 65. Diverges 67. Cooverges

(8) "'2.42 gal 'fT/2 55. 6

Diverges

.

2~ -

2 In ( ~ + I) + C

~ 1_ 12 ( W+1 X )2 + W+1

C

73. -2 colx - Inlcsex + colxl + csex + C

~lnI3+vl 12 3 - v

osin (20 2

81. 2(

3.

+ C

x' - zln 9 3 T Ix + 31 + zln Ix + II + C

77.

1. (x + 1)(1n (x + I» - (x + I) + C

VJ

~ tan-I

23.

Y

Practice Exerdses, pp. 489-491

+C

x' T

75.

(b) "'0.683, "'0.954, "'0.997

v'

21.

71 Inl 20

+ 2)1

I(v - 2)'(v

69. 2x;f2 - x +

,

,_ !lint

,

I

16 1n

19. ktan-l t -

25. -1/4 27. 'fT/2 29. 'fT/3 31. 6 33. In 2 35. Diverges 37. Cooverges 39. Cooverges 41. Cooverges 43. Diverges 45. Cooverges 47. Cooverges 49. Diverges 51. Cooverges 53. Cooverges 55. Diverges 57. Cooverges 59. Diverges 61. Cooverges 63. Cooverges 65. (8) Cooverges wben p < I (b) Coovcrges wben p > I 67. I 69. 2'fT 71. In 2 73. (b) "'0.88621 75. (8) y

+ C

I 11. In Ixl - In Ix + II + x + I + C

Section 8.7, pp. 487-489 1. 'fT/2

11

9. 2lnlx - 21 - Inlx -

+ I)

+ltan-I1J +C 6 3 +

cos (20 + I) 4 + C

(~)3 _ 2~) + C

85.

i ln Izi -1z - i [k ln

87.

-iv'9=4t'

93. tx3f2 + C

+ C 95.

-t

ZI x 2 -

I

2

I

4 sec' 0

89. In(:: :

D

+ C

tan-I cos (5t) + C

Zln (x + I) + C

+ C

83. tan-l(y - I) + C

l (Z2 + 4) + k tan-

97. 2y';. - 21n (I + y';.) + C

99.

79.

m] + C 91. 1/4

A-35

Chapter 9: Answers to Odd-Numbered Exercises 2 101. 31n Ix

I 2 + II + 61n Ix

7. y'

+ II +

- x

_I_ tan- (2x - I) + C 103.

v'3

*

(I

+ -rx)7/2

l-rx +

105. 21n

107. Inx -In II 109.

kx'u +

113. (b)

~

115. x -

9. y'

~

-(I + y)sinx;y(O)

~

2

I-

15. y(exact)

19. y '" -0.2272, exact value is 1/( I - 2VS) '" -0.2880.

kin II -~I Jz

-I

17. y '" 2.48832, exact value is e.

+ Inxl + C 111.

~

= ~, Yl = -0.25, Y2 = 0.3, Y3 = 0.75 = 3e*+2>, YI = 4.2, Y2 = 6.216, y, = 9.697 ~ or' + I, Yl ~ 2.0, Y2 ~ 2.0202, y, ~ 2.0618

13. y(exact)

- ~ (I + -rx)'/2 + ~ (I + -rx)'/2 + C

~I + C

C

x - y;y(I)

11. y(exact)

l

v'3

~

23.

+C

(V2 tan x ) + C

tan-I

Additional and Advanced Exerdses, pp. 492-494 1. x(sin-l x)'

+ 2(sin- 1 x)~ -

2x + C

25.

x 2 sin-1 x x~ - sin-Ix 3. 2 + 4 +C 5.

H (t - V1=t') In

sin-l

7. 0 9. In (4) - I 11. I 17. (a) 1r (b) 1r(2e - 5) 8(ln2)2 16(ln2) 19. (b) 1r ( - 3 - - - 9 -

t)

13. 321r/35 16) + 27

15. 21r

(e- 4+-I' -e-2 -2) 2

21.

vT+7 -In(~ +~) - V2 + 1n(1 + V2)

25.

121r 5

_1_In2 4

e'" sin3x 33. 13(3

37. 41.

e~

a

2

+b

1

<

29. 2 < P - I

+ 2 cos 3x) + C

35. cosxsin3x -8 3 sin x cos 3x

+C

2(asinbx-bcosbx)+C

2 ( / ) l-tanx2

+C

43. I

45.

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

v3931r

eX + C 1. y=-x-' x>O

5·

I I y =I-x+

9. y ~ x(lnx)2

V2

+ +

I

+ tan (8/2) I

47. -II n Itan(t/2) tan (t/2) 49.

In II

_ tan (8/2)

V21 + V2

I I

+C

+C

11.9=

C X

2'

3.y=

x> 0

t'

(I - 1)4

+

r'

XI -

x e+ 1

21. y

23. (b) is correct, but (a) is not.

Section 9.1, pp. 502-504

27. (a) i =

, ill}lIH IHHIH

7. y

= ~xeX/2 + Ce x/2

C

(I - 1)4

~ - ~e-'

I 29. y ~ I + Ce~

=

= yoe kt 25. t =

~ In 2 sec

~(I - e-') '" 0.95~amp

31. y' ~ I + Cx-'

section 9.3, pp. 515-516

HHHH

x>O

13. r = (csc 8)(lnlsec 81 + C), 0 < 8 < 1r/2 _31_2'1 _I 1r 15. y -IIe 7. y - -Ocos8 + 28

CHAPTER 9 3. (a)

C - cosX' " x

+ ex

3(t - 1)4

19. y = 6e

1. (d) 5.

,

+C

23.

27. a - 2'

27.

1. (a) 168.5 m (b) 41.13 sec 3. 9(t) ~ 4.91(1 - ,-(22.36/39.92)')

(b) 86%

A-36

Chapter 9: Answers to Odd-Numbered Exercises 3. y' ~ y' - y ~ (y + I)y(y - I) (8) Y = -I and y = I are unstable equilibria and y = 0 is a stable equilibrium. (b) y" = (3 y 2 - l)y'

5.X 2 +y2=C y

y

~ 3( y +

1)( Y + I/V3}( y

,'<0

-I.'

I/YJ)(y -

-

,'<0

,>0

,>0

1.,

-1

:,'>0

,'<0

I)

Y

y'>O

,

1

1

-;r, 9. y = ±\!2x

+C

;r,

y

(e) 1.,

1'<0.,'<0

1'>0.,,'>0

1'<0.,'<0

-1.5

13. (8) IOlb/min

(b) (100 + t)gal 4y

dy

(e) 4(10:+ t)lb/min

(b) y" =

(d) dt = 10 - 100 + t' y(O) = 50, y = 2(100 + t) - (

ISO

t 1+ 100

5. y' = v'Y,y > 0 (8) There are no equilibrium values.

12 ,'>0

)4

o

2

,">0

. y (25) 188.6 (e) Cooeen1ralloo ~ am!. brine in tank ~ 125

'"

(e)

y

1.5lb/gal

15. y(27.8) '" 14.8Ib, t '" 27.8 min

Section 9.4, pp. 522-523 1. y' ~ (y + 2)(y - 3) (8) Y = -2 is a stable equilibrium value and y = 3 is an unstable equilibrium.

(b) y"

~ 2(y + 2)& -

t)
,>0 ,

-4

~, , ,

(

,

,

0

2

,'>0

2

4

6

7. y' ~ (y - I)(y - 2)(y - 3) (8) Y = I and y = 3 are unstable equilibria and y = 2 is a stable equilibrium. (b) y" = (3 y 2 - 12y + ll)(y - I)(y - 2)(y - 3) =

,>0

)"<0

l ,'<0

-2

,"<0

4)

!, , , ,

'Y

,'>0

3(y - I) (y -

6 - 3YJ) (y

- 2) (y

0.'

(e)

,'<0

y

,

0

4

,>0,,">0

(

,"<0

,'>0

1~

, ,

:,'>0:

,

, ,

l ,"<0

~"1.42

--"-----'''1.''''-''---"-----'--------'--> , 1.,

___J-.....,;:::::::li;;;;;;;:....

!:~~(-~~

2~

,'<0 (

y'>O

,

, ,

, ,

,>0

3~

l ,'>0

:,'<0:

~"'2.S8

,

4

y

A-37

Chapter 9: Answers to Odd-Numbered Exercises

. d2

,

(c)

ConCavIty: dt v2 = -2 (8) ~>o

b

ft

= -2

(k) k 2) m (g - m V

V

(b)

~
I

v

(.2" tJt2>o

~ r--~""--~-

v

villi"!!

= I - 2P has a stable equilibrium at P =

d 2p

V

~
.2,,)

_~,-_L-_~-~2----'314'

9.

(k) m
t;

dP

(c)

dt 2 ~ -2 dt ~ -2(1 - 2P). 17. F

p

Vt=mmol

= Fp

-

(l6O = ~ o:oos = 178.9 ft/sec = 122 mph F,;ma

= 50

dv - 51vl; dt

maximum velocity occurs when ~~

1.,

I = m(50

~

0 or v

- 51vl). The

~

10 ft/sec.

19. Phase line: 0 dt J

p'>o,p' <0 0.5

0.75

1.25

1.5

1.75

d;' dt

P'>O

O~5)

1"<0

"

1">0

:

l!S 2

.~

J

f

lfthe switch is closed at t = 0, then i(O) = 0, and the graph of the solution looks like this:

dt

1"<0

:

I dt

dP >0 ieq"

11. dP = 2P(P - 3) has a stable equilibrium at P = 0 and an 00dt 2 stableequi1ibriumatP ~ 3; ~ 2(2P - 3) dP ~ 4P(2P - 3)(P - 3).

dii dP <0

0


+

2~5

1">0

~3~5)4P

,"<0

P'>O

As t~

00,

i(t) ---+ ilteadylta.te

=

~.

P 4

F:::::---===:::='~P'>O'p· >0 2

-,

0.4

0.5

0.6

0.7 P'>O,P' <0

Section 9.5, pp. 527-529 1. Seasonal variations, nonconfonnity ofthe environments, effects of other interactions, unexpected disasters, etc. 3. This model assumes that the number ofinteractions is proportional to the product ofx and y: fix

dt = (a - by)X, a < 0,

-2

13. Before !be catastrophe, the population exhibits logistic growth and P(t) increases toward M o, the stable equilibrium. After !be catastrophe, the population declines logistically and p(t) decreases toward M" the new stable equilibrium. p

p

M. P, , , ,

--~

----t----

,

dv _ k 2 15. dt - g - mV ' Equilibrium' dv .~

----

~. ,

'--

g,k,m > Oandv(t) '" 0

=g

_

~ v 2 = 0 ~ v = fmi m ~~

dy =

m(1 _l')y - nxy y(m - my - we) =

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 dominates the competition when either population is small, 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 ofcompetition to the trout n: intensity ofcompetition to the bass k,: environmental carrying capacity for the trout k2: environmental carrying capacity for !be hass

~: growth versus competition or net growth oftrout

W: relative survival of bass

A-38

Chapter 10: Answers to Odd-Numbered Exercises

dx (e) dt ~ 0

when x ~ 0

or y ~

a

b-

a bk, x,

33. (a) y = -I is stable and y = I is unstable.

d"y

)".,-1

Bypieking alb > k2 andm/n > k" we insure that an equilibrium point exists inside the flIst quadrant

~>o

dx

~ -In(C - ~(x -

: ~>o : d'-,
(c)

2)s/2 -1(x - 2)'/2)

x2

-

2x + C 2x 2

13. Y =

2

17 y = •

o~

+C

I +

0'

15. '9' + y' = C

I ' 19. y = -3 (I - 4e~)

2x'+3x +6 6(x + I)'

, 2

3. !any = -x sin x - cosx + C 5. (y + l)o" = -Inlxl + C 2 x-I 7. y = C9. y = x40'/2 + Co'/2 x11. y =

I

•

Practice Exerdses, pp. 529-530 1. y

Y

-2

21. y ~ 0~(3x' - 3x 2 )

23.

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

l.l 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

1.6241 1.8319 2.0513 2.2832 2.5285 2.7884 3.0643 3.3579 3.6709 4.0057

1. (a) y = c + (YO - c)o-k(AjY)1 (b) Steady-state solution: yoo ~ c 5. x 2(x 2 + 2y 2) = C 7. In Ixl + 0"/' = C 9. In Ixl -In lsec(y/x - I) + tan(y/x - 1)1 = C

CHAPTER 10 Section 10.1, pp. 541-544 1. al = 0, a2 = -1/4, a, = -2/9, a., = -3/16 3. al = l,a2 = -1/3,a, = 1/5, a., = -1/7 5. al = 1/2, a2 = 1/2, a, = 1/2, a., = 1/2

25. y(3) '" 0.8981 27.

3 7 15 31 63 127 255 511 1023 7. 1'1' 4' 8' 16' 32' 64' 128' 256' 512 IIIIIIII 9. 2, I, -1'-4' 8'16'- 32' - 64' 128' 256

('l~

11. 1,1,2,3,5,8,13,21,34,55 (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 solutionisy = -2 + 1n(2 - o~),whichhas an asymptote atx = -In 2 '" -0.69. Obviously, the Euler approximations are misleading for x :5 -0.7.)

~ -

~

..

31. y(exact)

123 .1 IX -I;y(2) '" 0.4; exact value IS I'

= = -0(>'-1)/2;y(2)

-0'/2 '" -4.4817.

15. a. = (-l)·+1(n)2,n '" I

19. an = n 2

'" -3.4192; exact value is

-

1, n ~ 1

_ 3n + 2 > 23.an n! ,n_l

27. 33. 39. 45. 51. 57.

63.

H,O.2] by [-la, 2]

( ) 29. yexact

13. a.

= (-l)'+',n

'" I

2n - 1

[ 0.2, 4.5] by [ 2.5,O.S]

69. 73. 79. 85. 91.

17. a. = 3(n + 2)'n '" I

21. an = 4n - 3, n ~ 1 I + (_1)·+1 25.an = 2 ,n~l

29. Converges, -I 31. Converges, -5 Diverges 37. Converges, 1/2 41. Converges, v'2 43. Converges, I 47. Converges, 0 49. Converges, 0 53. Converges, 07 55. Converges, I 59. Diverges 61. Converges, 4 Converges, 0 65. Diverges 67. Converges, e-1 Converges, 02/' 71. Converges, x (x > 0) Converges, 0 75. Converges, I 77. Converges, 1/2 Converges, I 81. Converges, 'It/2 83. Converges, 0 Converges, 0 87. Converges, 1/2 89. Converges, 0 8 93. 4 95. 5 97. I + v'2 99. x. = 2.- 2 Converges, 2 Diverges 35. Converges, 0 Converges, 0 Converges, I Converges, I

Chapter 10: Answers to Odd-Numbered Exercises

101.
17. Converges; geometric series, r 19. Diverges; Integral Test 21. Converges; geometric series, r

2(1 - (1/3)') 1-(1/3),3

7

~

2/3 < I

25. Diverges; lim 2+' I "!' 0 n_oon 27. Diverges; lim,_oo (y';;/In n) "!' 0

29. Diverges; geometric series, r

31. 35. 39. 43.

Converges; Integral Thst Converges; Integral Test Converges; Integral Thst
=

~2 >

1

33. Diverges; nth-Thrm Thst 37. Converges; Integral Test 41. a = I

I - (-1/2)' 3.s,= 1_(_1/2),2/3

I II 5. s, = 2 - n + 2' 2 9.

i
23. Diverges; Integral Test

Section 10.2, pp. 551-552 1. S,=

=

A-39

7. I -

1114 64 + "', 5

4: + 16 -

7 7 7 16 + 64 + ..., 3

4: +

G t)

(t + ~) + (i + i7) + "', 2i 13. (I + I) + (1- 1) + (1 + ~) + (1- _1_) + ... 17 2 5 4 25 8 125 ' 6 11. (5+ I) +

+

+

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

35. s, = I - n

39.

s, f =

41. I

~

oos-1

43. 5

37. s, = In

I; converges, I

(n ~

45. I

2}

21. 7/9

v;;-+l; diverges

converges,-~ ~2

47. -

49. Converges, 2 +

v'2

55. Converges, - ,e'- e - I 57. Converges, 2/9 59. Converges, 3/2 61. Diverges 63. Converges, 4 65. Diverges 67. Converges, '11' ~ e 51. Converges, 1

53. Diverges

69. a ~ I, r ~ -x; converges to 1/(1 + x) forlxl < I 71. a ~ 3, r ~ (x - 1)/2; converges to 6/(3 - x) for x in (-1,3) I I I 73.lxl<2'1_2x 75. -2<x<0'2+x 77. x "!' (2k + I) 00

79.
.~, 00

<e)

~

f, k an integer; I

1 (n + 4)(n + 5)

I sinx 00


~

(b) "" 41.55 45. True 47. (b) n '" 251,415 s I 49. Ss ~ ~ --, "" 1.195 51. 10 60 n-l n 59.
section 10.4, pp. 562-563 1. Converges; compare with ~ (I/n') 3. Diverges; compare with ~ (I/y';;) 5. Converges; compare with ~ (l/n 3/l) . 7. Converges; compare WIth

1 (n + 2)(n + 3)

9. Converges

~

~n+4n -4-- ~ n +0

11. Diverges; limit comparison with ~
13. Diverges; limit comparison with ~(I/y';;)

I (n - 3)(n - 2)

17. Diverges; limit comparison with ~ ( I/y';;)

89.
19. Converges; compare with ~(1/2') 21. Diverges; nth-Term Test 23. Converges; compare with ~ (I/n')

Section 10.3, pp. 557-558

25. Converges; (3n :

1. Converges 9. Converges

3. Converges

.rI V 5 ~ --,-., n f~

5. Converges

7. Diverges

11. Converges; geometric series, r = 11 < I 0 13. Diverges; lim +n I = I "!' 0 15. Diverges; p-series, p < I n_oon

27. 29. 31. 33.

I)" < (3:)"

~

(t)"

Diverges; direct comparison with ~(I/n) Diverges; limit comparison with ~(I/n) Diverges; limit comparison with ~(I/n) Converges; compare with ~(I/n3/l)

A-40

Chapter 10: Answers to Odd-Numbered Exercises

35. Converges; n~' ,;;

i,

37. Converges; 3'

I i

+

< _I_ 1

3,,-1

39. Converges; comparison with ~ (l/5n') 41. Diverges; comparison with

~ n(n ~

I

I)

n'

~(I/n)

n'

2 Zen n + ~ = -2n-e e e +1 20' 2 from . . -,- = n' a term a convergent geometric senes.

45. Converges absolutely; sech n =

1. 9. 17. 21.

Converges 3. Diverges 5. Converges 7. Converges Converges 11. Diverges 13. Converges 15. Converges Converges; Ratio Test 19. Diverges; Ratio Test Converges; Ratio Thst 23. Converges; compare with ~(3/(1.25)')

~)' = 0-3 '¢' 0

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/n2) 43. Converges; Ratio Test 45. Converges; Ratio Test 47. Diverges; Ratio Test 49. Converges; Ratio Test I )(1/,0 51. Converges; Ratio Test 53. Diverges; a, =:3 ---> I (

57. Diverges; Root Test 61. Converges; Ratio Test

1. Converges by Theorem 16 3. Converges; Alternating Series Thst 5. Converges; Alternating Series Test 7. Diverges; an ...,.,. 0 9. Diverges; all ...,.,. 0 11. Converges; Alternating Series Test

13. Converges by Theorem 16 15. Converges absolutely. Series of absolute values is a convergent geometric series.

~ diverges.

19. Converges absolutely; compare with ~:,(l/n2). 21. Converges conditionally; I/(n + 3) ---> 0 but ~:, n

+ + nn ~ I

0

~ (-I) 2(n ~ I) converges by Alternating 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 (b) -1/2 59. (a) a, '"

Section 10.7, pp. 582-584 1. (a) I, -I < x < I (b) -I < x < I (c) nonc 3. (a) 1/4, -1/2 < x < 0 (b) -1/2 < x < 0 (c) none 5. (a) 10, -8 < x < 12 (b) -8 < x < 12 (c) none (b) -I < x < I (c) none 7. (a) I, -I < x < I 9. (a) 3, - 3 ,;; x ,;; 3

(b) -3 ,;; x ,;; 3 (e) none (b) forallx (e) none 13. (a) 1/2, -1/2 ,;; x < 1/2 (b) -1/2 < x < 1/2 11. (a) oc,forallx

(e) -1/2 65. Yes

Section 10.6, pp. 573-574

· 3 23• DIverges; 5

<

a,.,

27. 29. 31. 35. 39.

diverges (compare with ~:~l(l/n».

I = I(-nl!~+11 = n~/2

diverges (compare with ~(l/y';,)). 43. Diverges, a, ---> 1/2 '¢' 0

0'

17. Converges conditionally; I/y';, ---> 0 but ~:,

31. Diverges; a" -1+ 0

+ ~) ---> 0, but series of absolute values

I/(y';,

Section 10.5, pp. 567-568

55. Converges; Ratio Thst 59. Converges; Root Test

+ n)/n 2 > I/n

37. Converges absolutely by Root Thst 39. Diverges; a, ---> oc 41. Converges conditionally; ~ - y';, =

53. Converges; limit comparison with ~(I/n') 63. Converges 65. Converges 67. Converges

,~OO

---> 0 but (I

(convergentp-series)

49. Converges; compare with ~(l/n') 51. Diverges; limit comparison with ~(I/n)

25. Diverges; lim (I -

k)

35. Converges absolutely since c:V;;'

or limit comparison with ~ (I/n')

45. Diverges; limit comparison with tan-I n 1f/2 47. Converges; -,-,- < --u

+

27. Converges absolutely; Ratio Test 29. Converges absolutely by Integral Test 33. Converges absolutely by Ratio Test

~ (I/n)

43. Converges; comparison with

25. Converges conditionally; C2

~

15. 17. 19. 21. 23. 25. 27. 29. 31.

(a) I, -I ,;; x < (a) 5, -8 < x < (a) 3, - 3 < x < (a) I, -2 < x < (a) I, -I < x < (a)

(a)

(a) (a) 33. (a) 35. (a)

37. 3 43. -I

(e) x = -I (b) -I < x < I (b) -8 < x < 2 (e) none (b) -3 < x < 3 (e) none (b) -2 < x < 0 (e) none I (b) -I < x < I (e) none O,X = 0 (b) x = 0 (e) none 2, -4 < x ,;; 0 (b) -4 < x < 0 (e) x = 0 I, -I ,;; x ,;; I (b) -I ,;; x ,;; I (e) none 1/4, I ,;; x ,;; 3/2 (b) I ,;; x ,;; 3/2 (e) none oc,forallx (b) forallx (e) none I, -I ,;; x < I (b) -I < x < I (e)-I 39. 8 41. -1/3 < x < 1/3, 1/(1 - 3x) < x < 3, 4/(3 + 2x - x') I

2 3 0

45. 0 < x < 16, 2/{4 -

47. - v'2 < x < v'2, 3/(2 - x 2 ) 49. I < x < 5, 2/(x - I), I < x < 5, -2/(x - I)' x2

3

-rx) x4

x6

51. (a) cosx = I - 2! + 4! - 6! converges for all x (b) Same answer as part (c) 23x 3 ZSx s 2'x' (e) 2x - 3! + 51 - 7!

x8

x lO

+ 8! - IO! + ... ; 29x 9

Zllx ll

+ 9! - IT! + ...

A-41

Chapter 10: Answers to Odd-Numbered Exercises 6 X 17x 8 3lx l0 '11' '11' 45 + 2520 + 14175' -2 < x < 2 2x 4 17x' 62x' 'IT 'IT (b) I +x 2 + - + - + - + ... --<x<3 45 315 '2 2

x2 53.
X

4

IT +

5.

Section 10.8, pp. 588-589 1. Po(x) ~ I, P,(x) ~ I + 2x, P2(x) ~ I + 2x + 2x 2, 4 P,(x) ~ I + 2x + 2x 2 + 3x'

7. 9.

3. Po(x) = O,P,(x) = x - I,P2(X) = (x - I) - t(x - If, P,(x) = (x - I) _l(x - I)' + l(x - I)' 2 3 I I I 5. Po(x) ~ I,PI(X) ~ I - 4(x - 2), P2(X) =

t - ±(X -

15.

P2(X)

P,(x) =

17. I + Y.

V2 + V2 (x _ "'-) 2 2 4

_

P2(X) = 2 + P,(x)

~

2 +

4I (x

00

19. x2~(2x)' ~ x 2 + 2x' + 4x 4 + ...

V2 (x _ ",-)2 4 4

21. ~nx'-I = I + 2x + 3x 2 + 4x' + ...

11=0 00

,-I

I + 4(x -

25.

4),

~ (x

- 4)2 + 5:2 (x - 4)'

(-x)"

13. ~(-I)"x' ~ I - x + x 2 - x'

+ ...

2 1)'3'"+ 'x'"+ 1 00 (-I )"x ' 00 '" ~ (2n +I)'. 17. 7~ )' 19. ~ (;,,)! 11=0 11-0 (2n. x 4 - 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 00

15. 21. 23. 25.

00

11=0

~(- 1),+1
(x - f

00

r

2

",(x - 2)' 11=0 n.

29. ~

33. -1-2x- 5 x 2 - "',-1 <x< I I 2 I, I 4 35. x -IX +6x + "',-1 <x< I 41. L(x) ~ 0, Q(x) ~ -x 2/2 43. L(x) ~ I, Q(x) ~ I 45. L(x) = x, Q(x) = x

Section 10.9, pp. 595-596 00

xS

-

~x4 + 23 x 6 3

45

_

I

4

I 104 '41

<

2

33.I+x+Ix -gX

6

9

-

...

...

+ ... ~x8 + ... 105

+ ...

(_

27. ~(-I)'(n + I)(x - I)' 31.

x2

I

11=0

3

x3

·

xU xIS + - - - + ... 5 7

~---+--

3n

29. x+x2 +T - 30 31

00

~ • ,~I

- -x'3

+ (-I)')X' = 2 + :lx2 _1 x ' + 25 x 4 2 6 24 x3 xS x'

n. (_I)II-lx 2n+1

00

27

x2 x3 x4 11 ~--=I-x+---+--··· • 11=0 n! 2! 3! 4! 00

(1,

~

11=0

I

4) -

x4ll-1

00

23. ~ (-1)'+1_- ~x' ,~I 2n - I

- 4) - 64 (x - 4)2,

±(x -

=

V2 (x _ ",-)2 4 4'

- ~ (x -f)' 9. Po(x) = 2,P,(x) = 2

()

2' 2n 1

(2x)' (2x)4 (2x)' (2x)' 1- 2'2! + 2.41 - 2'6! + 2.81 - ...

(x -

_ "'-) _ 4

(-I)'(2x)'"

::1

'? + '? f),

~ V2 + V2 (x 2 2

~

00

P,(x) ~l_l(X - 2) +l(x- 2)2 -~(X - 2)' 2 4 8 16 7. Po(x) = '?,Pd x ) =

11.

13.

2) + t(X - 2)2,

5(-I)'(-x)'"+1 00 5( -1)'+1x ,"+1 (2n + I)! = ~ (2n + I)! s 7 5x' 5x 5x = - 5 x + - - - + - + ... 3! 5! 7! 2 00 ( - 1)'(5x )," 25x4 625x' ~ (2n)' = I - 2! + 41 - ... 11=0' • . 00 x2n x4 x6 x8 ~ (-I)'+'n~x2- - + - - - + ... ,~I 2 3 4 2 00 3 x6 - 3'x 9 + ... ~ (-1)" (3)' - x 311 = 1 - -3x 3 + ,-0 4 4 42 4' 00 X"+l x3 x4 xS ~ - , = x + x 2 + -2' + -3' + -4' + ... 11=0 n. ... 00 (-I)"x 211 x4 x6 x8 x lO ~ (2n)1 = 4! - 61 + 8! - IO! + ... 7T'x' 'lT4X' 'IT'X 7 00 (-I)'A,"+1 x - -2' + -4' - -6' + ... = 11=0 ~ (2n)'. • • . 00

3.

(-5x)'

1. "Y--,- = I - 5x :=6 n.

52,;2 2.

5'x'

+ - , - - , + ... 3.

35. IError I

:5

4.2 X 10-6

37. Ixl < (0.06)'/' < 0.56968 39. IError I < (10-')'/6 < 1.67 X 10-10, -10-' < x < 0 41. IError I < (3°. 1)(0.1)'/6 < 1.87 X 10-4 k(k - I) 2 49.
section 10.10, pp. 602-604

+ x 2/2

2 1. I + -x - -x + -x' 2 8 16 3x 2 x 3 5.I-x+T-2 9 · 11. 13. 15. 23.

I + -x 3 2 + -x' 5 + ... 3 I + -x . 2 8 16 x 3 3x 6 5x 9 + 7.12 8 - 16 I+~ __I_+_I2x 8x 2 16>;' (I + X)4 = I + 4x + 6>;2 + 4x' + x 4 (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!

A-42

Chapter 11: Answers to Odd-Numbered Exercises

27.
x2

x4 12

x2

x4

T T -

13. With a = 22'lf, oosx = I -

3.4

+

29. 1/2

31. -1/24

39. 3/2

41. e

49.

I

x'

59.
x8

-I

x3

+ ... + (-I)" 31.32 35. -I

37. 2

45. V3/2

47. I

33. 1/3

(I + X)2 3x s 5x 7

•

55. 500 terms

x3

~x

23. a

.

,-I

71 x + xl + lx 3 . 3

(1/v2)(l + z)

19. 27. 31. 35. 39. 41. 43. 45. 47.
(b) -V3

(0) None 49.
<x <e

<x<e

~

(x

(b) -e

< x < V3

+

61.

3(x

+,

55. e ,ln2,2

00

+ I)!

I) 65. 2 - - 2 II . .

+ 1)2

2 '2!

~ +

(-I)'x'Onj3 (In)!

9(x

s

+ I)'

2 '3!

,-0

Section 11.1, pp. 616-618 1.

3.

,

63.

~

83. In (n ;;, I); the seties converges to In

(b) a = I, b = 0

5.

,

7.

,

«'lfx)/2)'

2

n!

,

~2+-i_l

2

+ ...

l_l(x 2

85.
(0) I/q

(b) 6

---~oofL-~---->,

57. ~2'x' 00

3) +l(x - 3)' _l(x - 3)' 4 4 4' 44 71. 0.4872223583 73. 7/2 69. 0.4849171431 77. -2 79. r = -3, S = 9/2 81. 2/3

67.

27. (b) Yes

(0) Empty set

X

53. sinx,'11',O

i),,.z.+IX2n+l

(2n

I

±S

21. b =

00

I 4

51. 1 +x'4'5 59.

17. 'If/2

CHAPTER 11

3. Converges to -I 5. Diverges 9. Converges to I 11. Converges to e-s 15. Converges to In 2 17. Diverges 1/6 21. 3/2 23. e/(e - I) 25. Diverges Converges conditionally 29. Converges conditionally Converges absolutely 33. Converges absolutely Converges absolutely 37. Converges absolutely Converges absolutely (0) x = -7
(_

22'lf)4

~x5 + ...' forallx 30

-

1. Converges to I 7. Converges to 0 13. Converges to 3

00

1! (x -

(0) 7

(o)-j

Practice Exercises, pp. 605-607

I

+

33. (a) R, = Coe-"'(I - e...... )/(I - e-Ito ), R ~ Co(e -"')/(1 - e -Ito) ~ Co/(e lto - I) (b) R , = I/e '" 0.368, RIO = R(I - e- lO ) '" R(0.9999546) '" 0.58195; R '" 0.58198; 0 < (R - RIO)/R < 0.0001

= I

61.1-2x+3x 2 -4x'+"· (b)

-7/6

31. (a) ~nx'-1

2 - x - 6 - 40 - ill

67.
~

2,£

(x - 22'lf?

00

5x 7

3x 5

~

t

+ ,..

15. Converges, limit = b

57. 4 terms

+ 6 + 40 + ill' radius of convergence

'IT

I - 6! (x - 22'lf)'

x 32

5.6 - 7.8

43. cos ~

51

+ x2

x6

-2

75. 1/12

t=O

-1

,_w

2

,

0

-1

-2

(t).

87. It converges.

t=¥

, L.L. 16 4

9.

,

11.

,

Additional and Advanced Exerdses, pp. 607-609 1. Converges; Comparison Test

3. Diverges; nth-Term Test

5. Converges; Comparison Test 9. With a = 'If/3, cosx =

+

Y.f

(x - 'If/3)'

I

2-

7. Diverges; nth-Term Test V3 I -2- (x - 'If/3) - 4: (x - 'If/3?

+ ... 2

11. With a = 0, e' = I

3

+ x + x2! + x3! + ,..

--:-f---+---\-~~,

----7<~'------_+_~,

,

Chapter 11: Answers to Odd-Numbered Exercises

13.

15. y

3

t=O

-, y=·b -? " , \

\

19. I 29.

•

21. 3a'1T

23. ab1T

31. 8~

521T 33. -3-

~ (~ -

:,,:, -

,

'1r

,

-,

,

\

d"y

fix' = 108

13. Y = 9x - I,

y

37. (j,y) -f"t
39. (j,y) = (t,1T

-1)

-3

17.

,

y

.. •

,

,

I

-I

I

\ \

(0) x =

,/ ,,

\

45.

\

,, ,,

(~,I).

33. Possible answer: x = 2 - cos t, Y = sin t,

5. Y = x

5. O:s; t :s; 2'7T

0 S I

< 1T/2

39. (1,1)

9. y=x-4,

d"y tb;2

= -2

(b)~

V2

I

2:

11.y~0x-1Tr+2, Z~-4

3;

-,

(~O)

+ 2n1T) and (-2,

(a) (3,0)

(b) (-3,0)

(e) (3,0)

(C)

7. (a)

(V2, f) 1~1T)

(0)

(I, v'3)

3;

(-I, v'3)

(g) (-3,0)

d"y

fix' =

11.

(I, v'3) (-I, v'3)

(d)

(b)

(d)

(5, 1T -

tan-I

1)

(b)

(0)

0,

+ (2n + 1)1T).

(b) (3, 1T)

(a)

-4

7. Y = 2x -

3.

9. (-3V2,5:) (-1,0) (-2, 5;) (-5, 1T -

+ 2V2, - , ~ -V2

I d"y + 'I' fix'

y = -2xall = 1T

(b) (2,2n1T)and(-2,(2n + 1)1T),naninteger

(0) (2,

d"y

dy 20 - I fix = 0 - 2

n an integer (d) (2,(2n + 1)1T) and (-2, 2n1T),n an integer

Section 11.2, pp. 625-627 d'y

fix' =

0,

Y = 2xall =

dy = 0 fix

(a) (2'f + 2n1T) and (-2,f + (2n + 1)1T),naninleger

O
37. x ~ a sin' I tan I, y ~ a sin'l,

-2 x + 2V2,

(b) x = 0, y = 3,

(-~~)

(0) (2,

Os 8 < 1T/2 aniix = O,y = 2 if8 = 1T/2

3. y =

(b) 1T

(-~O)

4 tan 8 31. Possible answer: x = I + 2 tan 8' Y = I + 2 tan 8'

fix

2)

~~-+-~--,

4

I

35. 31T ..v"5

3- 0 2 '

,y =

1. a,e;b,g;c,h;d,f

-00<1<00

1. Y ~ -x

27. 12

41. (a) 1T

I

17.-6

section 11.3, pp. 630-631

19. (a) x = a cos I, y = -a sin I, 0 SIS 21T (b) X = a cost, y = asint, O:s; t:S; 2'l'T (c) X = a cost, y = -asint, O:s; t:S; 4'1T (d) x = a cos I, y = a sin I, 0 SIS 41T 21. Possible answer: x ~ -I + 51, Y ~ -3 + 41, 0 SIS I 23. Possible answer: x = I' + I, Y = I, ISO 25. Possible answer: x = 2 - 31, Y = 3 - 41, I '" 0 27. Possible answer: x ~ 2 cos I, Y ~ 2 Isin II, 0 SIS 41T . -at a 29. PossIble answer: x = .. ~' y = .. ~' vl+I' vl+I'

35. x=2cott, y=2sin2 t,

2

47. (a) 8a

, \

o -

25. 4

dy I fix = 2

43. (a) x = I, Y = 0,

3

-16

IS.

A-43

(d)

Y

13.

tan-I

~) Y

-30

.~

+-c+--+--~,

I

--+-+---+~~.,

•

A-44

Chapter 11: Answers to Odd-Numbered Exercises

15.

,

,

17.

7. x-axis,y-aJds,origin

5. y-axis

,

6=¥

-ISTS3

,_!lin (8/2)

-~o1"""'----~' -I

,

19.

2

,

21. 8-

i

,.1

,

0

,

9. x-axis, y-axis, origin

~S6S~

,

,

0

,

25.

11. y-axis, x-axis, origin

,

Os 0""11'

r~O

23.

_~It-----3*E----,lf-+'

--~--I--:E---'

_'It-----+-----,-~,

OSrSI

,

-I

13. x-axis, y-axis, ongm 15. Origin 17. The slope at(-I,'ff/2) is -I, at(-I, -'ff/2)is I.

0

,

27. x ~ 2, vertical line through (2, 0) 29. y ~ 0, the x-axis 31. y = 4, horizootalline through (0, 4) 33. x + y = 1,line,m = -l,b = 1 35. x 2 + y2 ~ I, circle, qo, 0), radius 1 37. y - 2x = 5,line,m = 2,b = 5 39. y2 ~ x, parabola, vertex (0, 0), opens right 41. y = eX, graph ofnatura1 expooential functioo 43. x + y = ±I,twostraightlinesofslope-I,y-inter<:eptsb = ±I 45. (x + 2)2 + y2 ~ 4, circle, q-2, 0), radius 2 47. x 2 + (y - 4)2 ~ 16, circle, qo, 4), radius 4 49. (x - 1)2 + (y - 1)2 ~ 2, circle, C(I, I), radius v'2

51. 57. 61. 65.

- - * - - - - - t2 - ,

19. The slope at (I, 'ff/4) is -I, at (-I, -'ff/4) is I, at (-I, 3'ff/4) is l,at(l, -3'ff/4)is -I.

,

v3y +x ~ 4 53. rcosO ~ 7 55. 0 ~ 'ff/4 r ~ 20rr ~ -2 59. 4r 2 cos2 0 + 9r 2 sin2 0 ~ 36 rsi02 0 ~ 4cosO 63. r ~ 4 sinO r 2 = 6rcosO - 2> sinO - 6 67. (O,O),where Oisany angle

Section 11.4, pp. 634-635 1. x-axis

,

3. y-axis

, 21.
r=l+cosO

,

(b)

, ... I-sinS

*----*~,

1E--);-----+~~' 2

-I

-2

,

A-45

Chapter 11: Answers to Odd-Numbered Exercises 23. (8)

(b)

y

-,, ~ 2

13.

y

15.

y

y

1 2

1

,

x=~

,

2

12

1'''~-sin8

0

-,

1/4

0

5

dirmrix)'

25.

27.

12

,

"'-h

,

17.

y

,

I

19.

,

OSrs2-2cosO

'i2

4

~-----~o~~, -5 -2 -4

29. (8)

,

21.

23.

,

Section 11.5, pp. 638-639 I•

I,

(; 'IT

3. 18'IT

13. 3v3 19. (8) 25.

15.

'IT

2:3 - 4'IT

5.

'IT 8

f + '?

21. 19/3

3(YZ + In(1

7. 2

+ yz))

9.

'IT 2 -

17. 8:

I

11.

5'IT -

8

+ v3

23. 8 27.

'IT

3

8 +8

x2

25.

y2

4+T=

I

27. Asymptotes: y

,

Section 11.6, pp. 645-648 1. y2 ~ 8>;, F(2, 0), directrix: x ~ -2

~

±X

29. Asymptotes: y

,

~

±X

3. x 2 ~ -6y, F(O, -3/2), directrix: y ~ 3/2 x 2 y2 5. 4 - 9 = I, F(±v13,o), V(±2,0),

asymptotes: y = x2 7. T

9.

+ y2 ~

±~ x

I, F(±I,O),

,

---='---;+;-,-------> , ,, ,,

,,

V(±YZ,O)

,

11.

.% __ 3

)'=2

2

--_,O+----;;+--;;j>;;o--,

2

-~..-.+-_2_--> ,

31. Asymptotes: y

,

~

±2x

33. Asymptotes: y

,

,,

,,

,

~ ±X/2

A-46

Chapter 11: Answers to Odd-Numbered Exercises

35.

x2 = I

y2 -

37.

x2

57. C(-2,O), 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

y2

9 - 16 =

I

39. (a) Vertex: (I, -2); focus: (3, -2); directrix: x = -I (b) ,

V(v'5 - 2,O)andV(-v'5 - 2,0)

F(-4,O),

. (x - 1)2 63. Ellipse: 2 2

F(O,I),

-O;o+-+',f!.2-c,;--------+' -2

65. Hyperbola: (x - 1)2 - (y - 2)2 = I,

"'- eF(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)

2

+ (y - I) = I, C(I, I), F(2, V(v'2 + 1,1) and V( -v'2 + 1,1)

Vi, 3); vertices: (8, 3) and (0, 3); center: (4,3)

,

+ 3)andV(O,-V6 + 3); v'2x + 3 or y = - v'2x + 3

V(O, V6

F(O,O),

asymptotes: y =

69. (b) I: I 73. Length ~ 2v'2, width ~ v'2, area ~ 4 75. 24". 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

±~ (x

(-2,0); asymptotes: y = (b)

1. e ~

3 S'

. directrices are x =

- 2)

3. e

F(±3,O);

25

±T'

~

I

y'z; F(O, ± I);

directrices arey

~

±2.

,

,

-5

---'-~-_,;\--+---t,-L--"->,

______-_-12

5. e 45. (y + 3)2 ~ 4(x + 2), V(-2, -3), F(-I, -3), directrix: x = -3 47. (x - 1)2 ~ 8(y + 1), 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

(x - 2)'

4

-

I y'3; F(O, ±I);

directrices arey

,

= ±3.

7. e

~ ~;

F(±v3,O);

directrices are x = ±3v3.

,

C(-2, -I)

(y - 3)2

+ 2 V(±v3 + 2,3),

~

= I,

F(3, 3) andF(I, 3), --13

C(2,3)

(y - 2)2 5

= I,

--16

C(2,2),

F(5, 2) and F( -1,2),

V(4, 2) and V(O, 2); asymptotes: (y - 2)

~

v'5 ± 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 =

V2;

F(±V2,O);

. directrices are x

~

19. e =

1

± v'2 .

V2;

F(O, ±4);

directrices arey

~

47. y

±2.

~ ~ x + 2V3

y

y

y

2 --------~

-- 2

,

x+,.",2

-./8-

-~-c:__---'*-__!_-__o_-x __~ ~,~/,../ ,,_ 2 4 ---2 --./8-

-4 F,

49. rcos(o -

= V5;

F( n/iii, 0); . 2 directrices are x = ± 'VTIl"

21. e

23. e =

V5;

F(O,

±vio);

53.

*) ~

3

51. rcos(o

+

f) ~

55.

y

5

y

directrices . are y = ± y'j(j' 2

r=-20058

---I----'+'--4-_x 2

c_2ol---->--+----~x Radiu!

=1


---~----

-~-~___c'1'""-__!_---;--x -4 __ ::~~i~~: ~~~,_~___ 4

RadiIllI = 2

57. r

,

,

-2

~

59. r

12 cosO

~

IOsinO

y

•

y

r=

to lin 9

J-+(y-sl-2S (;<-6(+1-" r=12COl18

2 25.y2_ x =1

27.x2

8

y2 -

8

=1

2 29. r= 1 + cosO

1 33. r ~ 2 + cosO

30 31. r ~ 1 - 5 sinO

------"'-f-"----_x

10

35. r ~ 5 - sinO

39.

37. y

y

x='

,=-'1+0058 ---;---c----;;PHc---->x

61. r = -2 cos 0

63. r

= -sin (J y

y

x=-s

;+~+t)2 ..

l;.;+li+l=l

-+-"",___c___c+-~+-~,,--,,----+(,:", ...o~) x

, __200s6

(t· o)

t

r_ -sinO

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

----~h~---~.x

65.

67.

43.

41. y

y

y

y 4

(s,". f),

400

\ /[\."+'....

r",4sinO

------1(-;~__l;----x

---___c(-;,o~T----->x

------"'-f-"-----x

A-48

Chapter 11: Answers to Odd-Numbered Exercises

69.

71.

,

±lxl'/2

,

13. 10 3 +~--_-!,--f--;-~.

23.

15. 2:5

17. 10

±~

19. 9;

21.

7~

v'3

Y~Tx-4

,

,

-I

---~_I!.±cE----->·

,=-'4+Wl!I9

(b) y=

11. (a) y = - 8 - - 1

x=2

73.

---f---

27. y =

,

~-~.

-~

29.

x' + (y + 2)' =

4

,

,

r- _ 1 1 2sin8

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

•

,

:l-+ (, + 2r--= 4

r_ -4!in8

•

2

1=-1 31.

(x - v'2)' + y' = 2

33. r = -58in8

,

,

r __SsinS r=

Practice Exercises, pp.6SS-6S7 1. 3.

,

Moo! 8

;.+~+tY=-¥

,

• 41-4.'=1

1 1=0

2

-----;;o+----~·

35.r=3c088

37.

,

, O",,,6COl18

,

5. \

\ \

\

----+----,c;,>-c,-+->.

", ,,

,_.1: I

-~_IC---"::t""'~--;---·

7. x

= 3 cos t, v'3

9. y ~2x

Y = 4 sin t, 1 1

+4'4

o::=:;

t ::=:;

2'1T

39. d

41. 1

51. 8

53.

'IT -

43. k

3

45. i

----;;if----+,-.

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

I.

(~, 0).

directrix is x =

4 81.r=I+2cos6

-l

85. (a)

2 83.r=2+sin6

(b) 16,,-

24,,-

y

y

Additional and Advanced Exercises, pp. 657-658

y.1

7

1. x-

61.

0 ~

y2

y

2 =T

2; the asymptotes

arey ~

y

±V3x. 3.

3x2+3y2_8y+4~O

7. (a)

(y - 1)2 x2 16 - 48 = I

5. F(O, ±I)

&+~y - -x-2 = I

(b)

(i~)

,

-t----.t---+-/7c--'

11. y

Y

,'-,,'-1 ••

\

-4

63. (x - 2)2 = -12(y - 3), V(2, 3), F(2, 0), directrix is y = 6. (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 -

2Y2)' 8

(x - 2)2 2 ~ I, C( 2, 2Y2),

F( 2, 2Y2 ± ViO), v( 2, 4Y2) and V(2, 0), the asymptotes arey = 2x - 4 + 2Y2andy = -2x + 4 + 2Y2.

15.

y

9X2+4y2-36=O

69. Hyperbola: C(2, 0), V(O, 0) and V(4, 0), the foci are

F( 2 ± v5, 0) and the asymptotes are y

""

= ± x ; 2.

71. Parabola: V(-3, 1),F(-7,I)andthedirectrixisx = I. 73. Ellipse: C(-3,2),F(-3 ± 0,2), V(I,2)andV(-7,2) 75. Circle: C(I, I) and radius = v'2 77. V(I,O) 79. V(2,,,.) and V(6,,,.) y

y

17. (a) r ~

,=_2_ l+cosO

----.t-+.--;;o---~,

(1,0)

,.-'1-20019

0 28

(b)

4 19.r=I+2cos6

23. x

~ (a +

v5 (0"" 2

I)

2 21.r=2+sin6

b)cos6 - bCOS(a; b 6),

y = (a + b)sin6 - bSin(a; b 6)

27.

(~)

13.

x2+4,r-4-0

65.

A-49

".

2

INDEX

a,logarithms with base, 38J-384 Abscissa, AP-l 0 Aboolu'" change, 170 Absolute cODvetgence, 570 Absolute ConveIgence Test, 571 Aboolu'" extrema, fmding, 188-189 grapbo of, 183 Aboolu'" (global) maximum, 184, 187, 189, 806-808 Aboolu'" (global) minimum, 184, 187,189, 806-808 Aboolu'" value, AP-4-AP-6, AP-30 properties of, AP-5 Absolute wIue function, as piecewise-defined function, 5 Acceleration, 127, 129 derivative of Oork), 127 as derivative ofvelocity, 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 of vectors, 667-668 Addition fannulas, trigonometric, 26 Additivity, 903 fur defini'" in"'grals, 266 double in"'grals and, 846 Albert of Saxony, 559 Algebra, Fundamental Theorem of, AP-3J-AP-34 Algebra operations, vector, 667-669 Algebra roles for f'mit.e smns, 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 Angle convention, 23 Angle of elevation, 718

Angle of inclination, AP-II Angles, 22-23 between planes, 694 direction, 680 in standard position, 23 between vectors, 674-676 Angular velocity ofrotation, 967 Antiderivative linearity roles, 232 Antiderivatives, 230-235 defutition of, 230 and indefuti'" in"'grals, 235-236 motion and, 233-234 ofvector function, 715 Antidifferentistion, 230 Applied optimization, 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 ofbox, 214 Approximations, differential, error in, 169-170 by differential~ 168 error analysis of, 472--475 lioear, error formula for, 796, 821--1l22 Newton's Method for roots, 226 for roots and powers, 166-167 by Simpson~ Rule, 472-475 stuuhudlmear, 165,795 tangent line, 795 by Taylor polynomials, 586 trapezoidal, 469-470 by Trapezoldal Rule, 474, 475 using parabolas, 470-472 AJbitraIy constan~ 231 AJclength,326-330 along curve in space, 724-726 differential fomutia for, 329-330 ofa function, 327, 726 and line mtegrals, 901-902 AJc length differential, 623-624 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 Arctangen~ 599-600 Area, 246-248 ofbounded regions mplane, 850-851 cross-sectional, 308, 309 uoder curve or graph, 247 between curves, 294-296 substitution and, 291-297 as defuti'" in"'graI, 246 defutition of, 268 by double in"'gration, 85Q-j!52 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~ Theorem, 940 infuti"', 479 optimizing, ofrectang1es, 216-217 of parallelogram, 945 vector cross product as, 683 in polar coordinates, 855 of smooth surface, 945-946 surfaces and, 334, 624-625, 943-953 of surfaces of revolution, 332-335, 624-625 toUU,247,28Q-282 Area differential, 636 Argand diagrams, AP-29-AP-30 Argumen~ AP-30 Arrow diagram fur a function, 2, 827-828 Associative laws, AP-23 Astroid, 620 length of, 622 Asymptotes, 9 fmding,209 horizontal,84,86-88,92 ofbyperbolas,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 iT, 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

ofrnultivariablefunctions,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, 3011-315 spin around, 934-936

Base, a, logarithms with, 383-384 of cylinder, 308 of exponential functioo, 380 Bemoolli, Danie~ 185 Bemoolli, Iohann, 145 Binary search, 427-428 Binomial selies, 596-598 Binomial vector, 738 Bolzano, BemanI, 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 noorecumgular, 841-1l42

Bounded sequences, 539-540 Box prodnc~ 684-686

Brachistochrones, 614-616 Branch diagram(s), for multivariahle Chain

niles, 776,777,778, 779,780 Cable, hanging, II Calculators, to estimate !intits, 51-53 graphlng mth, 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 coordinates,855-857 in plane, AP-IO related to cylindrical and spherical coordi-

nates, 879 related to cylindrical coordinates, 876 related to polar cnordinate~ 623--030 three-dimensional. See lbree-dimensional coordinate systems

triple integraia in, 859-865

Cartesian integrals, changing into polar integnds,855-857

CAS. See Computer algebra systems CAST rule, 24 Catenary, 11,423 Cancby, Augnstio-Loui~ 401

Cauchy's Mean Value Theorem, 400-401 Cavalieri, Bonaventura, 310 Cavalieri's principle, 310

Center ofcurvature, for plane curves, 731 Center ofmass, 347 centroid, 352 coordinstes of, 3411-349, 904, 958 moments and, 346-355, 8611-873

of solid, 869 of thin flat plate, 3411-351 of thin shell, 959-960

of wire or spring, 904 Centtoids,352,869-870 fluid forces and, 353 Pappua" 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 "outside-inaide" rule and, 144 mth powers offunctiou, 145-146 proofof,I70

repeated use of, 145 Substitotion Rule and, 285-286 for two independent variablea and three

intermediate variables, 77S-779 for vector functions, 712, 713 Change, ofbase in a logarithm, 384 estimating, in special direction, 794 expunential,387-388

rares of, 39-44,103-104,124-125 sensitivity to, 131, 170-172 Charge, electrical, 978 Circle of curvature, for plane curves, 731-732

Circle ofmdius,AP-14 Circles, length of, 622 oaculating, 731 in plane, AP-B-AP-15 polar equation for, 652--053

standard equation for, AP-14, 18 Circulation, flux versus, 916 Cireulation density, 934-936

Complex conjugate, AP-29 Complex numbers, AP-25-AP-34 imaginary part of, AP-28 roal 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 Compoaite functions, 15-16

continuity of, 77, 76] def'mition of, ]5 derivative of, 142-143 !intit of, 78 Compreasing a graph, 16 Compreasiun ofa gas, uniform, 934 Computational formulaa, for torsioo, 737 Computer algebra systems (CAS), in evalnatiun of improper integraia, 482-483 integnd tablea and, 463-464 integrate command, 465 integration m!h, 465-466 Computer graphing, 30-34 offunctions of two variables, 752 Computers, to estimate limits, 51-53 Concave down graph, 203 Concave np 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, eccentricity of, 648-650 in polar coordinates, 639, 648-653 polar equations of, 650-652 Connected regloo, 921

Connectedness, 80 Conservative fields, component test for, 925,927 fmding potentials for, 925-928 aa gradient fielda, 923

Circulation fOT velocity fields, 914-915

line integrals in, 927

Clairau~ Alexis,

loop property of, 924

770

Claimut's Theorem, 770 Closed curve, 915 Closed regloo, 749, 751

Coefl1cienm,bdnoncrUU,597 determination for partial fractions, 460-461 ofpolynomial, 11-9

ofpower series, 575 undetemtioed, 454

Combining functions, 14-22 Combining series, 549-550 Common functions, 7-11 Common logarithm functioo, 384 Commutativity laws, AP-23

Comparison tests, for convergence of improper integraia, 484-485

for convergence of series, 55S-561 Competitive-hunter model, 524-526 Complete ordered field, AP-23-AP-24 Completeness property ofreal numbers, AP-23 Completing the sqnare,AP-I4-AP-15

potentials for, 925-928 and Stokes' theorem, 970 231

Consum~ arhllnlry,

nonzero,232 rate, 388

spring,338 Conaumt-
Constant Function Rule, 7] 2 Constant functions, def'mition of, 7 derivative of, 116 limi' of, 48

Constant Multiple Rules, for antiderivatives, 232,235 furdctivativea,117-118

and divergent series, 549 for f'mite sums, 257 for gradients, 789 fur integnds, 266, 846

Index fur limits, 49 for limits of functions of two variables, 757 for limits of sequences, 536 for series, 549 Constrained maximum, 811--1l14

Constrained minimum, 811-814 Constrained variables, 824-828 Construction ofreals, AP-24-AP-25

Continuity, 39-100 See also Discontinuity ofconnpomtes, 761 differentiability and, 111, 772

ofa function at a point, 74-76 at interior point, 74 on an interval, 76 of inverse functions, 364 alleft endpoin~ 74 furmuitivariable functions, 75~761 partial derivatives and, 769 of vector functions, 708-709 Continuity equation ofhydrodynamics, 978-979

Continuity Test, 75 Continuous extension, 79-80 Continuous function theorem for sequences, 536

Continuous functions, 76-77 absolute extrema of, 188-189,761 average wIue of, 269-270, 851, 864 nonnegative, 252-253 definition of, 76, 709 differoutiability 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 mdius of, 578-579

ofsequence, 533-535 of series, 554 geometric, 546 power, 575-579 ofTaylorSeties,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 Coordiuste conversion formulas, 883 Coordiuste frame, left-banded, 660 right-banded, 660 Coordiuste pair, AP-IO

Coordinate planes, 660 f'rrst moments about, 904, 958 Coordinate systems, three-dimensional. See Three-dimensional coordinate systems

Coordinates, of center ofmass, 348-349, 904, 948 polar, integrals in, 853--1l54 xyz, line integrals and, 911-912 Coplausr vectors, 669 Corner, 111

Cosecant, 23 Cosecant function, integral of, 374 inverse of, 407 Cosine(s), 23 integrals ofproducts 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, fIXed, 130

variable, 130 Cotangent function, 23 integral of, 374 Courant, 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 role for, 682 of two vectors in space, 682--683 Cross Product Rule for derivatives of vector functions, 712-713

Cross-sections, 308-315 horizontal, limits of integration and, 845--1l46 vertical, lintits of integration and, 845 Cube, integral over surface of, 955 Cube root function, 8

Cubic functions, 9 Curl, k-component of, 934-936 Corl 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,788-789

negatively oriented, 936 parametric, 610-611

polar, graphing of, 624 leogth of, 637--u38

positively oriented, 936 secant to, 41 sigmoid shape, 522 ....hiog, 203-210

slope of, def'Inition of, 41-43, 102 fUldiog,42,103,632 mnooili,3-4,326-327,620-621

curvature of, 728-729 leogth of, 724 speed on, 726

torsion of, 737 in space, 707-713 an: leogth along, 724-726

binonnals to, 734 fonnulas fur, 738

normals to, 730 parnmetric equations for, 7fJ7 vector equations for. See Vector functions tangent line to, 102 tangents to, 3~, 726, 788-789 terminal point of, 610 work done by force over, 912-914 y ~ fIx), leogth of, 327 Cusp, 111 Cycloids,614 Cylinder(s),69_7

base of, 308 pamholic, flux through, 957 parametrization of, 944 slicing with,

31~321

volume of, 308 Cylindrical coordiustes, definition of, 875 integration of, 877--1l79

motion in, 739-740 parametrization by, 944 to rectangular coonlinates, 876, 883 frmn sphetical coordiustes, 883 lriple 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 Decreasing function, 6, 19~200 Dedekiud, Richard, 295, 550, AP-25 Definite integrals, 262-270

applications of, 308-360 definition of, 246, 262-263, 285 evaluation of, by parts, 440

existence of, 262-264 Mean Value Throrem fur, 274-275

pammelrically deliued, leogth of, 620-622 pammelrized, 611, 624-625

notation for, 263

piecewise smooth, 710

substitution in, 291-293

plane, curvature of, 728-733

flux across, 915-917 lengths of, 327, 329, 621--024

parametrizations of, 610-616, 707 plates bouuded by two, 351-352 points of inflection of, 204-206

1-3

properties of, 265-267 of symme1ric functions, 293-294 of vector function, 716 Definite integration by parts, 440

by substitution, 292 Degree,ofpolynonrial,9 DeI(V),786,966-968

1-4

Index

Density, 348 cir<:ulation, 934-936

flux,933 Dependent variable of function, I, 747, 824-1l25 Dctivativo product role, 119-120, 520

Derivative quotientmle, 120-121, 145,569

Derivative rule for inverses, 365 Derivative sumru1e,1l8-119,AP-8

Derivative tests, for local extreme values, 187-188,635-{;38

Derivatives, of absolute value function, 146 alternate formula for, 107 applications of, 184-245 calculation from def'mition, 107 of composite function, 142-143,712 of constant function, 116 CODStantmultipleroie for, 117-118 of cosine function, 136-137 Cross Product Rnle, 712 def'mition of, 106 directional. See Directional derivatives Dot Product Rule, 712 in economics, 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, 108-109 higher-order, 121-122, 152 of hyperbolic functions, 417-418 ofintegnd,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 one-sided, 109-110

partial. See Partial derivatives alpoin~

102-104, 110-111 of power series, 580 as rate of clnmge, 124-131 of reciprocal function, 107 righ~handed, 109-110

second-order, 121-122 of sine function, 135-136 of square root function, 108 symbols for, 122

of tangent vector, 730 third, 122

of trigonometric functions, 135-139 of vector function, 709-711

Difference Rules, 115-122, 235 for combining series, 549

jump, 75 oscillating, 75

for derivativea, 115-116 for gradieo~ 789 for integnds, 266

point of, 74 removable, 75 D~criminant (Hessian) of function, 805

for limits, 49 for limits of functions with two variables, 757 for limits of sequences, 536, 548 for vector functions, 712

Diskmetbod, 311-312 Diaplacemeo~ 125 def"mition of, 251-252, 279 versus distance traveled, 254 Diaplay window, 30-33

Differentiability, 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 grapb of, 203 on interval, 109-110 roles for, 115-121, 143, 712

Taylor's fonnula for, 586 Differential approximation, error in, 169-170 Differential equations, autonomous, 517

D~tance, 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 Diatance formula, 662, AP-14 Diatance traveled, 248-250 versus displacement, 251-252, 279

estimating with f'Inite sums, 249-250 total,251-252

f"rrst-order,496-531 applications of, 510-515

D~tributive Law, AP-23

solutions, 496-498 initial value problema and, 233-234

for vector cross products, 682 Divergence, of improper integrals, 478,

separable, 388-390 system(s) of, 523-526 trajectory of, 524 Differential forms, 928-929 Differential formula, sbort form of arc lengtb, 329-330 Differentials, 164,167-168,796-798

definition of, 167 estimating with, 168-169

surface area, for parametrized surface, 946 total, 797, 798 Differentiation, 102-183 of function, 107 implicit, 149-153, 779-781

and integration, as inverse processes, 280 logaritbmic, 375

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 cbange in, 794 along a path, 61Q-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

as velocity, 125,711 ofy = lnx, 371 ofy = sec- 1 u, 409-410 ofy = sin-I u, 408 ofy = tan-I u, 409 Descartes, Rene, AP-1O

properties of, 787 Directrix (directrices), ofellipse, 649

Delemtinant(s), calculating the crosa produc~ 686 Iacobian, 887,889,890,892

ofbyperbola, 649 ofparabola, 649, 651 Diricbl~ Lejeune, 479

Difference quotient, 103, 107 forms for, 107 limit of, 104

D~amtinw~lll SuahoContinw~

interpretstion of, 785-786 in plane, 784-785

in dy/dx, 328-329 infInite, 75

proof of, AP-35-AP-36

480-481

nth-term test for, 548 of sequence, 533-535 to infmity, 535

to negative inf'mity, 535 of series, 545, 548 tests for, 483-485, 572-573

of vector field, 931-934, 972 Divergence Theorem, 974 fur 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"Inite integmls, 266 Dot prodoc~ 674-<>79

def'mition of, 675 directional derivative as, 786 of ortbogonal vectors, 517 properties of, 677--u79

Dot Product Rule for vector functions, 712 Double-angle fonnulas, trigonometric, 26 Double integrals, Fubini's theorem. for calculating,838-840

over bounded nonrectangular regions, 841-842

over rectangles, 836-840 in polar form, 853-857 properties of, 846-847

substitutions in, 887--891 as volumes, 837, 838 Double integration, area by, 850-852 Dummy variable in integmls, 264

Index e, deImition ofnumber, 370 .. limi~ 382 natura1 exponential and, 377-378

as series, 590-591 Eccen1ricity,

64~53

of ellipse, 648-649 ofhyperbola, 648 ofparabola, 648

polar equation for conic with, 650 ~onrics,derivativesin, 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 ffilipae Law (Kepler's Firat 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 ffilipaoids,697--{j98,699

ofrevo1ution, 698 ffiliptical cooes, 697, 699 ffiliptical paraboloids, 699 Empty~AP-2

Endpoint extreme values, 188 Endpoint values of function, 74, 188, 189 Energy, conversion ofmass to, 172 Equal An:a Law (Kepler's Secood Law), 741 Equations, autonomous, graphical solutions of,

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 identity, 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, 499-502

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, unifonn, of a gas, 934 Exponeotisl cbange (growth or deray), 387-388 Exponeotisl fonctiooa, 377-385

deImition of, 377 derivatives of, 378-379 description of, 10 geoeral antiderivative of, 379 general a:f, 380 natura1, 377 Exponeotisl populatioo growth model, 511-512 Exponeots, Lawa of, 379-380 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 eodpoints, 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 Factorial notation, 539

differential. See Differential equations

Falling body, eocountering reaiatalwe, 520-521 Fan-shaped region in polar coordinates, area of,

for ellipaes, 643, 652 Eoler's identity, 601-602 fucua-directrix, 649 for hyperbolas, 644

Fermat, Pierre de, 42 Fermat's principle in optics, 217 Fibonacci numbers, 539

516-522

ideal projectile motion and, 718 inverse, 384 linear,AP-13 parametric. See Parametric equations forplaoe in space, 691--{j92 point-slope, AP-12 polar for circles, 652 polar for lioea, 652 pressure-deptb, 341

relating polar and Cartesian coordinates, 629

relating rectangular and cylindrical coordi-

nates, 876 relating spherical coordinates to Cartesian and cyliodricsl coordinates, 879 Equilibria, stable and unstable, 519 EquilibriUll1 values, 517, 524-525 E1Tor analysis for numerical integration, 472-475 for linear approximation, 798 in standard linear approximation, 169,796

1-5

636

Ficlds,~tive,920,922-928,970

electric, 920 gradieo~ 923 gravitatiooal, 920

number, AP-23 ordered, AP-23

vector, 910 Finite (bounded) interva1a, 6, AP-3 Finite soms, algebra rulea for, 257-260 estimating with, 246-253 limits of, 258-259

and sigma notation, 256-258 Firing angle, 718 Firat Derivative Tes~ 198-201,207,803, 812-813

First derivative theorem for local extreme values,187-188

First moments, 346-348, 868-869 aboot coordinate axes, 348, 869

about coordinate planes, 869, 958 Flat plate, center ofmass of, 348-351, 869 vertical, fluid force and, 342

calculatioo of, 936, 957--958

versus circulation, 916 dermitioo of, 916, 956 surface integral for, 956-958 Flux density (divergence), of vector field, 933,972

Foci. 639-641 Forces, addition of, 667--668 constan~

337, 338, 339

field of, 912 fluid presaurea and, 340-342

variable along line, 338 work done by, over curve in space, 912-914 through displacemeo~ 679 Free fall, Gallilco's law fur, 39, 127-128 Freo~ Jean-Frederic, 734 Freoet frame, 734

computational formulas, 737 torsion in, 736-737 Fubini, Guido, 839 Fubini's theorem for dooble integrals, 838-840, 843-845,854,860

Functions, 1-38 absolute value, 5

addition of, 14-15 algebraic, 10, 361

arcsine and arccosine, 405-407 arrow diagram of, 2, 748 combiniog of, 14-22 common, 7-11 component, 707 composite. See Composite functions cooatao~ 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 cuberoo~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 diacoutinuity of, 74-75, 759

domain of, 1-3, 14,747,748 even. 6-7

1-6

Index

Functions (continued) exponential. See Exponential functions extreme values of, 184-189,803-808, 814,817 general sine, AP-7 gradient of, 786 graphs of, 3-4, 750 scaling of, 16-18 greatest integer,S growth rates of, 424-426 Hessian offunction of two variables, 805 hyperbolic. See Hyperbolic functions identity, 7, 48, 969 implieitlydef"med, 150-151,779 increasing, 6 independent variable 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 DlllJl!inal eos~ 129-130

square root, 8

of functions, 3-4, 14-22

synnnebic,6-7,293-294,631

offunctions with several variables,

of three variables, 750-752, 777, 789-790 total area under graph of, 281

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

synunetry rests for, 631 teclntique for, 634 ofpolar curves, 634 of sequence, 533 surface area of, 950 symmetric about origin, 6, 631 symmetric about x-axis, 6, 631 symmetric about y-axis, 6, 631

value of, 2 vector. See Vector functions

trigonometric, transformations of, 27 oftrigonometric functions, 25, 32-34 ofy ~ f(x), s1nltegyfor, 198-200

total cost, 130 tr.mscendental, II, 361-428

velocity, 196,251-252,711 vertical line test for, 4-5 Fundamental Theorem ofAlgebra, AP-3J-AP-34 Funtlsmental Theororn of Calculus, 274-282, 329,435,623,921,937,981 for line integrals, 902 Part I (detivative ofintegra1), 275-276 proof of, 276-277 Part 2 (Evaluation Theorem), 277-278 Net Change Theorem, 279 proof of, 277-278

Fundamental Theorem afLine Integrals, 922

maximum and minimum values of, 184-186, 188,200,206,806-807 198-201 of more than twowriables, 761, 768-769, 798-799 multiplication of, 14 natural exponential, def'mition of, 378 naluIallogarithm,369-371 nondUf~tiab1e. 110

Galileo Galilei, 39

IDODOkHriC,

free-fall fonnula, 39, 127-128 law of, 39 Gsuss, Carl Ftiedrieb, 258, 677 Gsuss's Law, 978 General linear equation, AP-13 General Power Rule for derivatives, 381 General sine funetion, 27

nonintegrable, 264 nonnegative, area under graph of, 268-269

General solution of differential equation,

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 funetions 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 roal-valued, 2, 708, 747 :representation as power series, 584 right-continuous, 74 sealar,708 seatterplot of, 4 of several variables, 747-752 sbift fonnulas for, 16 sine, 23, 135-136 in space, average value of, 864-865

747-752

offunctions with three variables, 750-751 offunctions with two variables, 749-750 ofpammetric equations, 634 in polar coordinates, 628, 631--634

233,389 Geometric series, 546-548 convergence of, 546 Gibbs, Josiah Williud, 726 Global (ahsolure) maxUmun, 184, 806-808 Global (absolure) minimum, 184,806-808 Gradient Theorem., Orthogonal, for constrained extrerna,814 Gradient vector fields, 909-910

conservative fields as, 923 Gradient vectors, 784-787 algebra rules for, 789 ourlof,969 definition of, 786 to level curves, 788-789 Graphical solutions, of autonomous equations, 516-522 Graphing, willi calculators and eomputers, 30-34 consequence for, 80 Graphing windows, 30-33 Graphs, asymptotes of, 84-93

Grnssmano, Hennann, 669

Gravitation, Newton's Law of, 740 Gravitstional eonatan~ 740 Gravitstional field, 920 veetors in, 908

Greatest integer function,S as piecewise-defmed function. 5 Green's Theorem, area by, 940 cireulation eurl ortangential fonn, 937, 939, 964,980 comparison with Divergence Theorem, 972, 980 comparison willi Stokes' Theorem, 963, 980 divergenee or normal fonn of, 936, 972, 980 to evaluate line inregra!s, 938-939 funns for, 936-938 generalization in three dimensions, 980 and llie Net Change Theororn, 937 in plane, 931-940 proof of, for special regions, 939-940 Growth, exponential, 388 logistic, 521, 522 Growth ..res, 388, 424-428.

relative, 511-512 Half-angle formulas, trigonometric, 26 Half-life, 391-392 Halley, Edmund, 200 Harmonic series, 533, 553 alternating,568-569 Hest equation, 775 Hest 1ransfer, 392-393 Heaviside, Oliver, 458

Heaviside "cover-up" method, for linear factors, 458-460

Height, maximum in projectile motion, 719 Helix,708

Hessian of function. 805 Higher-orderdetivatives, 121-122,152,771

Hooke's law of springs, 338-339 Horizontalasymptores, 84, 86-88, 92, 209

Horizontal scaling and reflecting formulas, 17

Horizontal shift of function. 16

of common functions, 7-11

Horizontal strips, 349-350

connectedness and, 80 of derivatives, 108-109 of equation, AP-IO

Hydrodynamics, continuity equation of,

Huygens,~tian,613,614

978-979

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 vertices 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-com.ponent ofvector, 669 Ideal gas law, 824 Identity function, 7, 48, 969 Image, 887 Implicit differentiation, 149-153,779-781 furmuia for, 780 Implicit Function Theorem, 781, 948 Implicit surfaces, 948-950 Improper integrals, 478-486 approximations to, 485 ofType I, 478, 486 of1YPe II, 481, 486 Increasing function, 6, 199-200 Increment Theorem for Functions of1\vo Variables, 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, 60lJ-{)()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) interw1a, 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 Iofinitesimala, 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, 610 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, power rule for, 116-117 starting, AP-8 Integrable functions, 264-265, 837, 859 Integral furm, product role 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 Integral1heorems for vector fields, 980-981 Integrals, approximation of, by Riemann sum, 261 by lower swns, 248 by midpoint role, 248, 469 by Simpson's Rule, 470-472 by Trapezoidal Rule, 469-470 by upper sums, 248 BriefTable of, 435 def'mite. See Defmite integrals double. See Double integrals evaluated with inverse trigonometric functions, 411-412 l(lfu) du, 373-374 for fluid force against vertical flat plate, 342 ofhyperbolic functions, 417-418 improper, 478-486 approximations to, 485 ofType I, 478, 486 ofType II, 481, 486 indeImite, 235-236,284-285 involving IOilax, 384-385 iterated, 838 line. See Line integrals logaritlun dermed as, 370 multiple, 836-1l37 substitution:in. 887-894 nonelemeotary, 466, 598-599 polar, changing Cartesian integrals into, 855--1l57 in polar coordinates, 853-854 of powers of tan x and sec x, 446-447 ofrate, 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 worl<, 337-338, 727-728, 912-914 Integraods, 235 mtb vertical asymptote~ 481-482 Integrate connnand (CAS), 465 Integrating factor, 505-507 Integration, 246-307 basic fannulas, 435 mtb CAS, 465-466 in cylindrical coordinates, 875-879 and diffi:rentiation, relationabip between, 280 formulas,411-413 limits of. See Limits, of integration numerical,468-475 by porta, 436-441 by porta 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--1l83

tabular, 440-441 techniques of, 435-495 term-by-term forpower selies, 581 by trigonometric snbatitution, 449-452 variable of, 235, 264 in vector fields, 901-981 ofvector function, 716-717 Interior 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, 8Q-8 I, 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, 61 (J-{j 11 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 Imding, 36~364 byperl>olic,418-419,420-421 one-to-one, 361 trigonometric. See Inverse trigonometric functions Inverse trigonometric functions, 404-413 cofunction identities, 41 Q-411 def'mition of, 404-405 derivatives of, 411 integrals evaluated mtb, 411-412 Inverses, def'ming of, 404-405 integration and differentiation operations, 280 oflnx and number e, 377-378 of tan x, cotx, sec x, and csc x, 407-408

1-8

Index

Inational numbers, AP-2 as exponents, 377-378 Irreducible quadratic polynomial. 454-455 lterat
j-<:omponent of vector, 669 Jacobi, Carl Gustav Jacob, 887 Jacobian determinan~ 887. 889. 890. 892 Jerk, 96. 97 Joule, James Prescott, 337 Joules. 337 Jmnp discontinuity, 75 k-eompooeot of curl. 934-936 k-<:omponent of vector, 669 Kepler. Jobannes. 741 Kepler's First Law (Ellipse Lsw). 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. Josepb-Louis. 193. 811 Lagrange multipliers. 807. 81 1--1l 18 method of. 814-816 with two constraints, 817-818

Laplace, Pierre-Simon, 770 Laplace equation, 774 Launch angle. 718 Law of cooling. Newton's. 392-393 Law of cosines, 26-27, 675 Law of Exponential Change. 511 Law of refraction. 218 Laws of exponents. 37l}-380 Laws oflogarithms, 384 Least integer function, 5 Lesst upper bound, 539. AP-2:l-AP-24 Left-continuous functions, 74 Left-band derivatives. 10l}-110 Left-band limits. 67 defInition of, infanna!, 66-67

precise, 57--63, 68 proofof.AP-21 Left-banded coordinate ftmne. 660 Legendre, Adrien Marie, 506 Leibniz. Gottfried, AP-25 Leibniz's forrnula, 600 Leibniz's no1ation, 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 ofcurves, 326-328, 621-624

of pammetrically def'med curve, 620-622 of polar coordinate curve, 637--638 of vector (magnitude). 665. 667-668 Lenses. light entering. 152 Level curves, of functions of two variables, 750

Level surface, of functions oftbree variables, 750 I'H6pitaI. Guillaume de. 396 I'H6pitaI's Rule. 396. 397-399. 400-402 Imding limits of sequences by. 537-538 proof of. 40Q-402

Limit circle, 526 Limit Comparison 1l:s~ 483. 484. 560-561 Limit Laws. 46-54. 757 lbeorern, 49 Limit Power Rule, 49 Limit Product Rule, proof of, AP-I8-AP-19 Limit Quotient Rule. proof of. AP-Il}-AP-20 Limit Root Rule, 49 Limit theorems, proofs of, AP-I8-AP-21 Limiting population. 521 Limits.3l}-IOO commouly occurring. 538-539. AP-21-AP-22 of continuous functions, 78

definition of, informal, 46-47 precise, 57--63 proving lbeorems wilb. 62--03 testing of. 5lHiO del..... Imding algebraically. 60-62 of difference quotient, 104 estimation of, calculators and computers for, 51-53 e (lbe number) ss. 382 off"mite sums, 258-259 offunction values, 46-54 for functions of two variables, 755-759 indetenninate forms of. 396-399 inImite. 8l}-90 precise def"mitions of, 90-91 at inf"mity, 84-93

of integration, for cylindrical coordinates, 877-879 for definite integrals. 292-293

f"mding of for multiple integrals, 845-846, 854-855.860-862.877-878.881-882 inImite. 478-480. 486 for polar coordinates. 854-855 for rectangular coordinates, 859, 860-864 for spherical coordinates. 881--1l82 left-band See Left-band limits

nonexistence of, two-path test for functions of two variables, 760 one-sided. See One-sided limits ofpolynomials. 50

power rule for. 49 ofmtional functions, 50 of Riemann sums, 262-264 right-band. See Right-band limits

root rule for, 49 Sandwich Theorem. 53-54 of sequences, 534, 536 of(ain 0)/9. 6l}-71 two-sided, 66 of vector-valued functions. 708-709 Lin. integrals. 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 standaJd, 79>'-796 Linear equations, AP-13 Ill"St-<>rder differential. 504-508 staodaId 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, of intersection. for planes, 692--693 masses along, 346-347 motion along, 125-129 nonnal, 152 tangent planes and, 791-793 parallel. AP-13 parame1ric equations for. 68lHi89 perpendicular, AP-13 and planes, in space, 688--694 polar equation for, 652 secant, 41 straight line equations, 688, AP-IO-AP-13 taogen~ 43-44. 102 vector equations for. 688-{)90 vertical, shell formula for revolction ~ 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 mug. of. 373 integral of. 437

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 f"1I'St derivative theorem. for, 187 second derivative test for, 206 Local extreme values, def"med, 186, 803 derivative tests for, 187, 803-804, 805 Local (relative) lIlllJ
Machine diagnun of function. 2 Maclaurin, Colin, 585

Index Maclaurin series, 585-586, 587 Magnitude (length) ofvector, 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 energy, 172 distributed over plane region, 347-348 fonnulas for, 348, 351, 869, 904 by line integral, 904 rneasurementof,511 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 Inequality Rule for MIDi'" in"'gra!s, 266,267,274 ~-~1Csm,200,206,803,805,808

Maximum, sbsolute (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 Mende~ Gregor Johann, 131 Mesh size, 469 Midpoint of line segment in space, f'mding with vectors, 670 Midpoint role, 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 of inertia, 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 Moootooic 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

Newton"ecood law of, 171-172, 510, 520 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 scalar, of vectors, 667--668 Multipller (Lagrange), 807, 811-818

Napier, John, 372 Natora1 domain offunctioo, 2 Natora1 expooeotial functioo, 378 derivative of, 378 grapb of, 377 power series for, 586, 590 Natural logarithm function, algebraic properties of, 372 deflDitioo of, 369-371 power series for, 582, 600 Natora1logatitbms,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-Rapbsoo me1hod, 225-228 Newton~ law ofcooling, 392-393, 519-520 Newton~ lawofgravitatioo, 740 Newton~ me1hod, 225-228 applying, 226-227 convergence of approximations, 228 procedure for, 225-226 Newton's second law of motion, 171-172,510, 520 Noodecressing partial sums, 545 Nondecreasing sequences, 540 Nondifferentiable function, 110 Nooelemeo1ary 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 nib psrtial sum, 544-545 nth-tenn test for divergence, 548 Numerical integrstioo, 468-475 Numerical melbod, 499 Numerical representation of functions, 4 Numerical solution, 499 Oblique (alaot) asymptote, 88-89

Octaots, 660 Odd functions, 6-7

Oh-notatioo, 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

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-23-AP-24 Oresme, Nicole, 534 Otieotable sorface, 956 Otigio, of coordina'" sys_, AP-IO moment of system about, 347 in polar coordinates, 627 Otihogonsl gradieot Ibeorem, 814 Otihogonsl trajectories, 512-513 Ortbogonal vectors, 676-677 Oscillating discontinuities, 75 Osculating circle, 731 Osculating plane, 737 Output variable of function, 747 Outside-Inside interpretation ofchain rule, 144-145

4>-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 spproximatioos by, 470-472 axis of, 640, AP-15 deflDitioo of, 639 directrix of, 639, 641, 651 eccentricity of, 648, 649 focalleogth of, 640 focus of, 639, 641 parametrization of, 611--612 semicubicsl, 155 slope of, 42 vertex of, 640, AP-15 Paraboloids, 697 elliptical, 699 hyperbollc, 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, ares of, 683 hnoofoWfition, 667-668, 675 Paonn_ domain, 610, 943 Parameter interval, 610-611 Param~,610,943

Parametric curve, 611, 612 arc leogth of, 620-622, 724-725 calculus with, 618-625 differentiable, 618 graphing of, 634 Parametric equations, 610-613 ofcircle, 611, 622 for curves in space, 700 ofcycloid, 614 ofbyperhola, 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 ofcurves, 610-616, 707 of cylinder, 944 ofline, 681Hi89 of sphere, 943-944 and surface area, 944-948 of~aces.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 higher-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 ofseries, 5#-545 sequence of, 545 Particular solution, of differential equation, 233 Partitions, 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 ofparticle, 707 Peak voltage, 289 Pendulum clock, 614 Percentage change, 170 Periodicity, of trigonometric functions, 25 Perpendicular lines, AP-13 Perpendicular (orthogonal) vectora, 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, 5 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 :masses distributed over, 347-348 Plane tangent to surface, 792, 793 Planes, angles betwoen, 694 Cartesian coordinates in, AP-10 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 ofintersection for, 692-693 motion of planets in, 740-741 nonnal,737 osculating, 737 parallel, 692 rectifying, 737 in space, 688-694 Planetary motion, Kepler's Firat Law (Ellipse Law) of, 741 Kepler's Second Law (Equal Area Law) of, 741 Kepler's Third Law (Time-Distance Law) of, 741-742 as planar, 740-741 Plate(s), thin fla~ center ofmaas of, 34S-35 I two-dimensional, 869, 872 Point-slope equation, AP-12 Pointa, boundary, 751 of discontinuity, def'mition of, 74 ofirrflection, 188,204-206 interior, 751 in three-dimensional Cartesian coordinate system., distance to plane, 693--694 ~seuiUe,Iean, 171 Poisson, Simeon-Denis, 802 Polar coordinate pair, 627 Polar coordinates, 627-630 ares in, 855 area ofpoIarregion, 636 conics in, 639, 648-653 definition of, 627 graphing in, 628, 631-634 symmetry testa for, 631 initial ray of, 627 integrals in, 853-1154 length ofpolar 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, ofcircles, 652--653 of conic sections, 650-652 graphing of, 628 of lines, 652 Polybedml surfaces, 968-969 Polynomial functions, def'mition of, 8 Polynomials, coefficients of, 8-9 degree of, 9 derivative of, 11 &-1 19 limita 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

Power Cbain Rule, 145-146 Power functions, 7-8 Power Rule, for derivatives, 115-116 for limita, 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 teating 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 indetenninate, 400 of sines and cosines, products of, 444-446 Preimage, 887 Pressure-dep!h equation, 341 Pressures, fluid, and forces, 340-342 Principal unit normal vector, 732, 738 Product Rule, for derivatives, 119-120 fur gradien~ 789 in integral form, integration 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 Producta, of complex numbers, AP-30-AP-31 ofpowers 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 gusta, 719-720 Projection, of vectors, 677--679 Proportionality relationship, 7 Pumping liquids from containers, 340 Pyramid, volume of, 309 Pythagorean theorem., 25, 27, 28, AP-13, AP-26 Quadnmta, of coordinate sys_, AP-I 0 Quadratic approximations, 586 Quadratic polynomial, irreducible, 454-455 Quadric surfaces, 697--699 Quotient Rule, 51 for dctivatives, 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 Quotienta, AP-31 producta and, dctivatives of, 119-120

p-limits of integration, :ftnding 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 ofco~ence,579

ofpower series, 578-579 of curvature, for plane curves, 731-732 Radius units, 22 ~,of~ction,1-3,747, 748 in projectile motion, 719, 720 Rate constant, exponential change, 388 ~s,ofchange,39-44

average, 41 instantaneous, derivative as, 43-44 Ratio, in geometric series, 546 Ratio Tes~ 563-565, 576-577, 579, 597

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 ofreals and, AP-24-AP-25 development of, AP-26-AP-27

properties of, algebraic,AP-1, AP-23 completeness, AP-1, AP-23 order,AP-1,AP-23 andrealline,AP-1 theory of, AP-23-AP-25 Real-valued func1ioos, 2, 747, 546

Remainder, estimating of, in Taylor's Theorem, 590,591-592 in integral teat, 555-556

of order n, def'mition for Taylor's fonnu1a, 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

Rest mass, 172 Reatpoints, 517, 524-525

unstable, 526 Reaultant veetor, 667--068

Revolution, about y-axis, 334-335 areas of surfaces of, 624-625 ellipsoid of, 698 solids of, disk method, 311-313 wasbermethod, 314-315 Riemano, Georg Friedrieb BemIuud, 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-band limits, 67

defInition of, 68 proof of, AP-20 Right-banded coordinate frame, 660 Right-banded derivatives, 1Ol}-110

Reals, construction of, AP-24-AP-25 Rearrangement theorem. for absolutely convergent series, 572

RL circuits, 508

Reeiproeal Rule for natorallogaritlans, 372

Rolle, Michel, 192

Rectangles, approximating area of, 247-248 def'ming Riemann sums, 260-261

Rolle's Theorem, 192-193 Root Imding, 80 Root rule, for limits, 49

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 Regioos, bounded, 749 elosed, 749, 751 eoooeeted, 921

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 Related rates, 155-160

Relative change, 170 Relative growth rate, 511-512 Relative (lueal) extrema, 186, 187,803

Rise,AP-ll

for limits of functions of two variables, 757 RootTea~ 565-566, 579 Roots, AP-32-AP-33

binomial series for, 597-598 Imding by Newton's Method, 225-228 and Intermediate Value Theorem, 80-8 I Rotatioo, 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 ofvectors, 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 ofsecond-order solutions form ofsecond-order solutions j;rced vibrations general solution for linear equations homogeneous equations linear combination linearly independent solutions linearity method ofundetermined coefficients nonhomogeneous equations power series solutions uniqueness 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-coejJicient second-order linear equations superposition principle theorem on general solution j;rm variation ofparameters Second-order partial derivatives, 769-770 Separable differential eqnatioos, 3811-389

Sequences, 532-541 bounded, 53l}-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 smus, 545 nondecreasing, 540 recursively def'med, 539 Sequential search, 427-428

Series, absolutely convergent, 570 adding or delc1iog terms, 550 alternating,5611-570 harmonic, 5611-569 binomial,596-598 combining,54l}-550 cooditiooally coovergen~ 570

convergence of, comparison tests for, 5511-560

1-12

Index

Series (continued) convetgen~ 545, 570 divergen~ 545, 548 geometric, 546-548 harrnanic, 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, 572 ",indexing, 550 representations, of functions of power, 584-585 sum. of, 544-545 Thylor, 585-586, 589, 592-593 tests, for absolute convergence, 571 slternating, 568 comparison, 559 convergence, 558-560 integral, 553-555 limit comparison, 560-561 ratio, 563-565 roo~ 565-566 sommary of, 572-573 Set,AP-2 Shearing flow, 934 Shell fonnola for revolution, 322 Shell method, 321-324 Shells, thin, masses and moments of, 958-960 Shift formulas for functions, 16 Shifting, of fun<:tion graph, 16 Short dilfcrcntisl formula, arc lcngjh, 329-330 SI anits, 337 Sigma notation, 256-258 Simple harmonic motion, 137-138 Simply coonccted region, 921 ShnpsOD,11lomas,471 Simpson's Rule, approximations by, 47ll-472 error analysis and, 472-475

Sinc(s), 23, 24 integrals of prod_ of, 447-448 integrals of prod_ ofpowers 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 cylindcra, 319-321 by parallel planes, 309-310 volume by, 30&-314 Slope,ofcurve,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 Suell van Royen, Willebrord, 217 Snell's Law, 218 Solids, Cavalieri's principle of, 310 cross-section of, 308

three-dimensional, masses and moments, 869,872

volume of, calculation of, 309 by disk method, 311-313 by double integrala, 837, 838, 842-845 by method of slicing, 30&-315 by triple integrala, 859-860 by washer method, 314-315 Solids ofrevolution, by disk method, 311-313 bywasbermethod, 314-315 Solution, of differential equation, 233, 389 general, 496-498 particular, 497-498 Solution curves, 498 Space, curve in, wmk done by force over, 912-914 Speed, 126-127, 158 along smooth curve, 726 average, 39-41 over short time intervals, 40 instantaneous, 39-41 ofparticle 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 Spring coostan~ 338 Springs, Hooke's law for, 33&-339 mass of, 903-904 work to stretch, 339 Square root function, 8 derivative of, 108 Square roots, elimination of, in integrals, 446 Squeeze Theorem. See Sandwich Theorem Sl Vincen~ Gregory, 623 Standard linear approximation, 165,795 Standard anit voctors, 669 Steady-sta.. value, 508 Step size, 469 Stokes'11leorcm, 925, 962-970, 980,981 comparison with Green's Theorem, 962, 963, 964

conservative fields and, 970 for polyhednsl surfaces, 96&-969 surface integral in, 965 for surfaces with holes, 969 Stretching a grapb, 17 Substitution, 285-289 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

roctangn1ar to cylindrics! coordinates, 876 rectangn1ar to spherics! coordinates, 880 Substitution fonnula for definite integrals, 291-293 Substitution Rule, 286-287, 292 evaluation of indefinite integrals with, 284-290

Subtraction, of vectors, 668 Sum Rule, for antiderivatives, 235 for combining series, 549 for Mlnite integrals, 266 derivative, 11 &-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

:f"tnite,572 estimation with, 246-253 lintits of, 25&-259 lower, 247-248 partial, sequence of, 545 Riemann, 259-261 upper, 247, 248 Surfilce area, defming of, 332-334 dilfcrcntisl for pannnetrized surface, 946 of explicit surface, 954 of graph, 950 of implicit surface, 949-950, 954 parametrization of, 944-948 for spbere, 947 Surfilce integrals, 953-960, 965 computation of, 955-956 fur tlus, 956-958 formulas for, 954 Surfilces, and area, 334, 624-625, 943-953 constant depth, fluid force on planar, 341 fun<:tions defmed on, 777-779 with holes, 969 implici~ 94&-950 implicitly deImed, 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 tw<>-sided, 956 of two-variable functions, 750 Symmetric functions, 6 defntite integrals of, 293-294 graphs of, 6-7 Symmetry tests, for graphs in polar coordinates, 631 System. torque, systems ofmasses, 346 8-lintits of integration, fmdiog of, 854-855, 878 Thble ofintegrals, 463-464, T-I-T-6 Tabular integration, 440-441 Tan x and cot x, derivatives of, 138 integrala 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 tographoffunetion, 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 Tangent planes, horizontal, 803-1104 and normal linea, 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, 59lH;02

convergence of, 589-595 freqoenOy 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 Algebnric Properties ofNatural Logarilhrn, 372

Alternating Series Estimation, 570, 593 angle between two vectora, 674 Chain Rule, 142-146 for functions of three variables, 777 for functions of two variables, 775 for two independent variahles 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 related to loop property, 970 De Moivre's, AP-32

Derivative Rule for Inverses, 365 Differentiability implies cootinuity, III, 772 Direct Comparison Tes~ 560

Green's, 937 Implicit FuncOon, 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'H6pital'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 corollllIY I, 195 corollllIY 2, 195

for def'mite integrals, 274-275 Mixed Derivative, 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 ofPappus, 353-354

Properties of continuous functions, 76 Properties of limits of functions of two variables, 757 Ratio Test. 563 Rearrangement, for Ahsolutely Coovergent Series, 572 Remainder Eatimatioo, 591, 593 Rolle's, 192-193 RootTes~ 565-566

Sandwich, 53, 69, 70, 87, 536, AP-20 Second derivative test for local extreme values, 206, 805

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 baaic, 23-24 'nigonometric identities, 25 'nigonometric substitutions, 449-452 Triple integrals, in cylindrical coordinates, 875-879 properties of, 859-1160, 865 in rectangular coordinates, 859-865

in spherical coordinates, 879-883 substitutions in, 891-894 Triple scalar product (box product), 685--086 liming fork dsta, 4 lWo-
Unbounded region, 749 Unbounded sequence, 539

Undetermined coefficients, 454 Unified tIreory, 980 Union of set. AP-2

Stokes',963

Unit hinormal vector, 734

Substitution in def'mite integrals, 292 Substitution Rule, 286

Unit circle, AP-14 Unit nonnal vector, 730 Unit step functions,limits and, 49

Taylor's,589,594-595

Term-by-Term Differentiation, 580

Unit tangent vector, 726-727, 738

Term-hy-Term Integratioo, 581

Unit vectors, defmition of, 669 writing vectors in terms of, 669-670 Universal gravitational constant, 740

Thickness variable, 322-323 Thin shells, moments and masses of, 958-960 Three-dimensional coordinate systems, 66Q-{j63 Cartesian, 660

coordinate planes, 660 cylindrical, 877-1179 right- and left-handed, 660 spherical, 881-1182

Threo-dimensional solid, 869, 872 Three-dimensional vectors, component funn of, 666

1-13

Upper hoood, 267, AP-23 Upper sums, 247, 248

Value(s), absolute, AP-4-AP-6, AP-30 average,252 CKUen1e, 184-188,802-808 of function, 2-3, 864-865 ofimproperintegntl, 478, 481

local maximmn, 186, 803 local minimum, 186,803

Time-Distance Law (Kepler's Third Law),

Variable force, along curve, 912-913

Evaluation, 277-278 Exactness of differential forms, 928 Extreme Value, 185-186, AP-24

741-742 TNB frame, 734 Torque, 346-347,685 Torsion, 736-737, 738

Variable of integration, 235, 264 Variables, constrained, partial derivatives with,

First derivative test for local extreme values,

Toros,951

187,803 Formula for Implicit Differentiatioo, 780

Total diffi:rential, 797, 798 Transcendental functiooa, 11,361-428

FubDri~,839,843-845

Transformations, Jacobian of, 889, 890

Divergence, 974

Fundamental, of Calcolus, part I, 275-277 part 2, 277-278 Fundamental, ofAlgehra, AP-33-AP-34 of Line Integntia, 922

linear, 888-889 oftrigonometric graphs, 27

Transitivity law for real numbers, AP-23 Trapezoid, area of, 269

along line, 338

824-828 dependen~ I,

824-825 funcOooa ofseveral, 747-752, 761, 768-769, 781 independent, 1,775,777,824-825 inpu~ 747

intennediate, 776-777 outpu~ 747 proportiooal, 7

1-14

Index

Variables (continued) tlrickness,322-323 three, functions of, 75(}-752, 78~790 Chain Rule for, 777 two, functions of, 748-749, 752, 771 Chain Rule for, 775-777 independen~ and three intennediate, 778 limits for, 755-759 linearization of, 794-796, 798 partial derivatives of, 764-766 Thylor's formola for, 82(}-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

conservative, 920,924-925 continuous, 908 curl of, 962

direction of, 669 dot product, def'mition of, 674 eqoality of, 665 and geome1ry in space, 66(}-699 gmdien~ 786 in gravitational field, 908 i-com.ponent of, 669 initial point of, 665 j-compooent of, 669 k-component of, 669 length (magnilode) of, 665, 667--{j68 in navigation, 671

Pappus's Theorom for, 353-354

parallel, 682 perpendicolar (orthogonal), 676-677 in physics and engineering, 671--{j72 position, standard, 665--{j66 priocipal onit oormal, 73(}-73I, 738

by slicing, 308-315

scalar multiplication of, 667-fJ68 standard position, for a point, 665-666 standard unit, 669

line integrals of, def'mition, 910

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 detennineo~ 683--{j85 right-band rule for, 682 of two vectors in space, 682--683 curl, 962--963 defmition of, 665

313

double integrals as, 837, 838 by iterated integrals, 842--1l43

ofregion in space, 860

eleelric, 920 flux density of, 933

and line integrals, 907--917 potential function for, 921 Vector functions, 707-713 antiderivatives of, 715 of constsotlength, 713 continuity of, 708-709 defmite integral of, 716-717 derivatives of, deImition of, 710

876

in spherical coordinates, 880 by disks for rotation about axis, 311,

normal,ofcurve, 732-733

projection of, 677-fJ79

gravitstiOllll1, 920 integration in, 901--981

differential, in cylindrical coordinates,

notation for, 665

differentiable, 908 divergence of, 933-934

gradien~ 9~lO, 922--923

Vertical slrip, 319, 349 Vertical tangents, 111 Viewing windows, 3(}-33 Voltsge, peak, 289 Volwne, of cylinder, 308

resoltso~

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

667--{j68

subtraction (differew:e) of, 668 tsogen~ ofcorve, 710 tenninal point of, 665 three-dimensional, 666 torque, 685 triple scalar product of, 685-686 rwo-dllnerunooal,666,676 oni~ definition of, 66~70 derivative in direction of, 785 writing vectors in tenns of, 669-670 onit binormal, 734 unit normal, 732 onit tsogen~ 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, 51(}-511 tenninal,521

Velocity fields, circulation for, 914-915 flow integral, 914-915 Velocity function, 251-252, 711 acceleration and, 196, 711 Vertical asyrnptote~ 84, 91-92. See also Asymptotes definition of, 91

Wasberme1hod, 314-315, 324 Wave equation, 774 Weierstrass, Karl, 484 Weight-density, of fluid, 34(}-341 Whirlpool effe~ 934 Wmdows, graphing, 3(}-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 corve, 913 by variable force along line, 338

x-coonlinate, AP-10 x-intercept, AP-13 x-limits ofinregration, 862, 864 xy-plane, 524, 660 solution curves in, 518 xz-plane, 660

y, integration with respect to, 296-297 Y ~ f(x), graphing of, 206-208

y y y

length of, 326-328, 623 inX', derivative of, 371 = sec-I u, derivative of, 409-410 =

1 = sin- u, derivative of, 408 y = tan-I u, derivative of, 409 y-axis, revolution about, 334-335 y-coordinate, AP-10

y-intercept, AP-13 y-limits of integration, 861, 863 )"-plane, 660

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

z-limits of integration, 861, 862, 863, 877 Zero denomina_, algebraical elimination of, 51 Zero vector, 666 Zero Width Interval Rule, 266

A BRIEF TABLE OF INTEGRALS Basic Fonns 1.

3. 5.

7. 9. 11.

J JrJ; J J J J

kdx

=

In Ixl

+C

~xa

+C

aX dx =

17.

19.

22.

J J JVa

In Isinxl

eosbxdx = sinbx

dx a2 + x2

J J

(ax

dx 2

=

> 0, a

+x

2

6. J sinxdx

"" I)

+

+C

+C

16. C

18.

(a> 0)

20.

+ W 2dx

=

(Vax

+ b) dx

+ C, n ""

J

seextanxdx = seex

J JV JVx

• / 2dx va

x

x2

-

dx

x2

dx

2

-

-

a2

a

2

+C +C

+C

=. sm-1 ~ a + = =

C

~ see-1 I~ I + C eosh- 1 ~

+C

(x

> a > 0)

-1

+ C, n "" 24.

:2 [ In Iax + b I + ax : b] + C 2 (Vax

=

tanx + C

b

~In lax + bl + C

=

=

14. J sinbxdx = eosbx

b] - (ax + b)·+1 [ax + b a2 n +2 - n+ 1

•

-eosx + C

=

12. J tanxdx = In Iseexl

+ b) dx -

•

J

+C

(ax + b)·+1 a(n + 1)

=

+ bt 1dx

25. J x(ax

10.

sinb-l~ + C

=

n + 1 + C (n "" -1)

=

8. J see2 xdx

ltan-l~ + a a

+ b)' dx

x(ax

23. J(ax

27.

(a

esexeolxdx = -esex =

J

xn+l

x· dx

4. JeXdx=eX+C

ese2 xdx = -eolx + C

Forms Involving ax 21.

2.

+C

eosxdx = sinx

13. J eotxdx 15.

+ C (anynumberk)

kx

=

a

n

+ b(2

+2

+ C, n ""

-2

-1,-2

Jx(ax+b)-ldx=~- :21nIax+bl +C

26. J 28.

x(ax~ b) = tin lax: bl + C

J~ dx=2~+bJ xVaxdx + b T-1

T-2 29.

30.

A BriefTable of Integrals

J

(a)

dx

J~ x2

34.

J J

a

2

dx

+x

Vb

- Vb I + c

(b)

Vax+b+Vb

~ +'!J dx +C X 2 x~

dx=-

- I tan-1 x a a+ C

33

dx

Va + x

2

36. JX 2Va 2 + x 2 dx

dx (a 2 + x 2)2

~(a2 + 2x 2)Va 2 + x 2 -

=

f In

2 38. J Va: 2+ x dx

=

In (x + Va 2 + x 2) _ Va / x + C

2

J

dx

xVa 2 + x 2

2a 2(a 2 + x 2)

-"-J dx +C 2b xVax + b

+ _I_tan-l~ + C 2a3

a

2

2 -!In la + V: + x21 + C

=

2

42J . a2 dx _ x2

Ilnlx+al+c 2a x- a

=

44. J Va':- x 2 =

41.

J

2

dx

x 2Va 2 + x 2

-

43J . (a2 dx _ x2)2

sin-l~ + C

46. JX 2Va 2 - x 2 dx =

fsin-l~ -

kXVa2 - x 2 (a 2 - 2x2)

x 2Va 2 - x 2

=

Forms Involving Xl -

JV.

dx

Xl -

a2

=

Vx 2 - a2 dx

Va 2 - x 2 2 +C ax 0

2

In Ix + Vx 2 - a 2 + C 1

=

2

IVx2 - a2 -

=

2a2(a2x _ x2)

+_I_Inlx+al+c 4a3 x- a

~21n Ix + Vx 2 -

a2

1

~2 sin-l~ + C

+C

50.

dx

,+

Va x +C ax

45. J Va 2 - x 2 dx = IVa2 - x 2 +

2 2 2 47. JVa x -x dx=Va2_x2_alnla+V:2-x21+c 48. JVa: -x 2

J

= _

Xl

0

53. J

C

2 x dx=_a21n(x+Va2+x2)+xVa2+x2+c Va2 + x2 2 2

Forms Involving

52.

~ bx x

=

+

2 + V: + x21 + c

Va 2 + x 2 _ a In Ia

.

= -

b

b

+ Va 2 + x 2) + C

(x

=

51

Vb

~1n(x+Va2+x2)+C

2 2 37. J Va x + x dx

40.

dx x 2Vax + b

J

~tan-l~ax -

=

sinh-l~ + C = In (x + Va 2 + x2 ) + C

=

35. JVa2+x2dx=IVa2+x2+

39.J

.

dx

x~

.

2 -

2

J

31 J

+ Xl

Forms Involving Ql 32.

_I_In I~

=

x~

+C

J

dx=-sin-l~-

2 2 Va x -x +C

2 -!In I a + V: - x21 + C xVa 2 - x 2 dx

=

A BriefTable ofIntegrals

54.

(Vx 2 - a 2)" dx

J

55J dx • (Vx 2 -a 2)"

X(VX2 - a 2)n n+ I -

=

(v' 2 =xx-a

(2-n)a 2

2

=

57. J x 2Vx 2 - a 2 dx

i(2x 2

=

J

2

(Vx 2 - a2)"-2 dx,

I

_n-3J dx (n-2)a 2 (Vx 2 -a 2)"

+ C, n"!'-2

-

a 2)Vx 2 - a 2 - as'1n Ix + Vx 2 - a 2 1 + C

a2

-

IXI a

=

Yx 2

2 59. J VX: 2- a dx

=

In Ix

60. J

2 x dx Vx2 - a2

=

a 1n Ix + Vx 2 - a 2 1 + "-Vx 2 - a2 + C 2 2

61·

VXldxX

58.

J

J

a2

=

-

a sec-I

+ Vx 2 _

a2

+ C

1- VX 2x-

a

2

+C

2

1a sec-11"-1 a

+ C = 1 a cos

-11~1 x

68•

J J

sin" ax

~ - s~~ax + C

ax =

smn-I : cos ax

J

(b)

J

(c)

72. 74.

75.

76.

.

' ax + n dx -- cosn axIsm cosax na

69. (a)

70.

64.

C

=

65. J sin2axdx 67.

-~cosax +

=

J

. J

J J

=

+ C

a2

-

x

C

sin2ax

4

b2

r

71.

I -a In Isin axl + C

sinax I cosaxdx = -aln Icosaxl + C

sinn ax cosmax dx =

~sinax +

cosn-2 ax dx

2 2 a "!' b

4a

2 2 Vx 2- a + C a :x

sin,,-2 ax dx

sin(a - b)x sin(a + b)x 2(a _ b) + 2(a + b) + C,

cos2ax

2

=

cosaxdx-2+~+C

2 2 a "!'b

cosaxcosbxdx =

a2

cos ax dx =

.. sin(a - b)x sin(a + b)x smaxsmbxdx= 2(a-b) - 2(a+b) +C,

cos ax -.--dx smax

J J J

n

66.

cos(a + b)x cos(a - b)x . smaxcosbxdx = - 2(a + b) - 2(a _ b) + C,

sm ax cos ax dx = -

J

IJ "--=---IJ

+n;

VXldx Xl

62. J

+ C

Trigonometric Forms 63. JSinaxdx

n"!' -I

n

(VX2 - a 2)"+2 n+2

2 56. JX(VX - a )" dx

2 Vx2x - a dx

2t

nn~

.

73 .

J J

n-I m+l I sm ax cos ax + n aim + n) m+ n

sinn+l ax cosm- 1ax m- 1 sinn ax cosmax dx = aim + n) + m+ n

. n-2

SID

J.

1 n - sin"+ ax sm axcosaxdx - (n + I)a + C,

J

n"!'-I

_ cosn+l ax n· cosaxsmax dx --(n+l)a + C, n "!' - I

m

axcos ax

dx

sinn ax cosm-2 ax dx,

,

n "!' -m

m "!' -n

(reduces sinn ax) (reduces cosmax)

T-3

T-4

A BriefTable of Integrals

77. J b + :inax

=

78. J

=

dx. b + csmax

79. J 1 + 7inax 81. J 82.J

aVb~2_ c 2 tan-I [~~ ~ ~ tan(: -1

aVc2 _ b2

dx

=

dx b + ceosax

=

b+ceosax

85. J x sin ax dx

80. J 1

[~bb+c - ctanax] 2

lnlc+beosax+vc2-b2sinaxl+c b + c cos ax ' 84. J 1

C

~eosax + ~J

b2
%sax

-~eoto; + C

=

C

86. J xeosaxdx = :2 cos ax +

~sinax +

cos ax dx

88. J x' cos ax dx

=

~J

~ln Isinaxl

87. J x' sinaxdx

=

89. J tanaxdx

~In Isecaxl

+ C

90. J cotaxdx

x + C

92. J eot'axdx

=

94. J cot" ax dx

= -

=

_x: cos ax +

ltan(1T + ax) + C a 4 2

=

b2 > c 2

1

~tan 0; +

+ C '

b2 < c2

ttx.smax

tan-I

a\ sin ax -

=

V.

2

aVb2-c2 aVc2 _ b2

=

2 2 b > c

+ C,

c2 In Ic + bsinax + - b2eosaxl + C, b + csmax

-~tan(: - 0;) + C

=

83. J 1+ %sax

T)]

91. J tan2 ax dx

=

~tanax -

93. J tan' ax dx

=

:'~ ':) -

95. Jseeaxdx

99. 100.

101.

=

" J sec ax dx -

,

J

-

esc axdx -

J,

J tan,-2 ax dx,

~ln Isec ax + tanaxl

=

97. J see2 axdx

X,-l

~tanax +

n" 1

96. Jeseaxdx

+ C

2J

sec,,-2 ax tan ax n a(n _ 1) + n _ 1

,-2

sec

n" 0

sin ax dx

-~cotax -

x + C

:(:~ ':) -

J eot"-2 ax dx,

-~eolax +

eotaxl + C

C

ax dx, n" 1

esc"-2 ax col ax n - 2J -2 a(n _ 1) + n _ 1 esc" axdx,

-

=

X,-I

+ C

-~ln leseax +

=

98. J ese2 axdx

C

sec"ax + C, sec ax tan ax dx - ----..,,-

=

x: sin ax -

C

n" 1

102.

J'

-

esc"ax + C, esc ax cot ax dx - - ----..,,-

n"O

n" 1

A BriefTable ofIntegrals

Exponential and Logarithmic Forms 109.

j

e

QX

~eQX + C

dx =

110.

111. jxeQXdx = : : (ax - I)

114. je

QX

b

IbQX

a In b + C,

b

> 0, b oF 1

~xneQX - ~jxn-leQXdx

> O,b oF 1

2e

a +b

2

(a sinbx - b cos bx)

+C

/QX 2(acosbx+bsinbx)+C

116.jlnaxdx=Xlnax-x+c

a +b

117. jxn(lnax)mdx = xn+1(lnaxr - ----"'----jxn(lnax)m-'dx

n+1

j

dx =

QX sinbx dx =

115.jeQXcosbxdx=

118.

j

QX

112. jxneQXdx =

-a~bjxn-lbQXdx,

~i:,:

113. jxnbQXdx =

+C

b

n+1

(lnax)m+l x l(lnax)mdx = m + 1 + C,

n oF-I

'

m oF -I

119.

j

dx xlnax =

lnllnaxl

+ C

Forms Involving Y2ax - xl, a> 0 120. 121.

j Y2ax - x j Y2ax dx

sin-1 (x

=

-

. j (Y2ax -

125. 126. 127.

+

C

(x - a)( Y2ax 2n x ) dx = n+ 1

j j Y2~ j Y2a;2j Y2ax -

x 2)"

=

-2 ~2a ;

2

x dx = =

(x_~_a) + C

na 2 j ( + n + 1 Y2ax n - 3

j

x

2)n-2

dx

dx

+ ----'"-~2

(Y2ax _ x2)"

(n - 2)a

2 x +

-

~ sin-1 (x ~ a)

2

+ C

Y2ax _ x 2 + a sin- 1 (x ~ a) + C

2 x dx =

x2

2 x )"

~a)Y2ax -

xY2ax _ x 2 dx = (x + a)(2x -

xdx

-

(x - a)(Y2ax - x 2)2-n (n - 2)a 2

dx

124.

~ a)

x 2 dx = x_;_a y2ax - x 2 + 'fsin-1

122. j ( Y2ax 123

2

a sin-1 (x

~

x - sin- 1 a) -

(9) +

Y2ax -

C

x2 + C

128.

j x Y2ax -x dx

= 2

_1~2a a

- x +C x

Hyperbolic Forms 129. 131. 133.

j sinh j j sinh

ax dx =

sinh2 axdx = n

~ cosh ax + C S~2ax 1

n

ax dx = sinh

-

I+C

~ cosh ax

- n -;; 1

j sinh

n - 2 ax

j cosh j cosh axdx n °

130.

ax dx =

132.

2

dx,

oF

~ sinh ax + C =

s~;ax + I

+ C

T-5

T-6

A BriefTable of Integrals

134. J

cosh" ax ax = COsh"-l:, sinh ax

135.Jx sinh ax ax = ~ cosh ax 137.Jx" sinh ax ax = 139.]tanh ax ax

n-;; 1J

:2 sinh ax

141.J tanh2 ax ax =

+C

- ~J x"-l cosh ax ax

x; cosh ax

~ tanh ax + C

x -

143.] tanh" ax ax

=

~~-;;;:

+

144.] coth" ax ax

=

(~th~-;;;:

+

=

J J

tanh"-2 ax ax,

noll

coth"-2 ax ax,

noll

~sin-l(tanhax) + C

147.J sech2 ax ax = ~ tanh ax + C

J

" ax sech ax

149.

cosh"-2 ax ax,

~ln(coshax) + C

=

145.] sechaxax

+

--

sech"-2 ax tanh ax (n _ l)a

+

" ax = - csch"-2 ax coth ax 150J . esch ax (n1 )a

n n -_ 21

e 154.Je ax

sinh bx ax =

ax

cosh bx ax =

e; e: e; e: [a

b -

[a

b

f(n) =

(n -

J x cosh ax ax = ~ sinh ax -

:2 cosh ax

138.

J x" cosh ax ax =

140.

Jcothaxax

- ~J x"-l sinh ax ax

142.

J

146.

J

148.

J

h"-2 ax sec ax,

n-

ae~bxb]

152. a2

*b

+ C, a 2

*b

=

x; sinh ax

~ln Isinh axl

coth2 ax ax = x -

cschaxax =

+C

~ coth ax + C

~ln Itanh~1

csch2 ax ax = -

X

w/2

1

157. o

sin" xax =

1W

/2

0

1)1,

cos" xax =

n> 0

4

41 n...-

J

csch" ax coth ax ax = -

csc~= ax + C, nolO

2

2

156.1""e-ax '

1.3.5 ..... (n - 1) 7r •2·4· 6' .... n 2' 2 • 4 • 6 ••••• (n - 1) 3. 5' 7' ...• n '

1

+C

~ coth ax + C

Some Definite Integrals

155.1""x"-le- ax =

+C

n...- 1

- n- 2J ax 1 esch"-2 ax,

ae~bxb] + C,

+

136.

J

151.J sech" ax tanh ax ax = sec~= ax + C, nolO 153.J

nolO

ax =

t.J%,

if n is an even integer ;,,2 ifn is an odd integer;"3

a > 0

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, AP/Wide 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 Cenler, Inc.; Page 361, photo, Getty Images; Page 435, photo, Getty Images; Page 496, photo, Getty Images; Page 532, photo, Getty Images; Page 547, Figure 10.9h, 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 o/Film 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

Trigonometry Formulas

lanA - lanE Ian (A - E) ~ 1 + lanA lanE

y

1. Definitions and Fundamental Identities sinO ~ ~ ~ _I_ T esc f)

Sine:

x r

1 sec f)

y

1 eolO

cos ()

Cosine:

x

Ian 0

Tangent:

P(x, y)

Sin(A - ; ) 0

~

~

~

Sin(A + ; ) sin A sinE

~

eosA,

COS(A - ; ) eos(A + ; )

~

~

sinA

-sinA

±eos (A - E) - ±eos (A + E)

1 1 eosA easE ~ 2eos (A - E) + 2eos (A + E) -sin 0,

sin2 0 + eos2 0 ~ 1, sin20

-eosA,

x

x

2. Identities sin (-0)

lanA + lanE 1 - lanA lanE

Ian (A + E)

~

eos (-0)

eos 0

see2 0 ~ 1 + lan2 0,

2sinOeosO,

2n_1+eos20 COSu2 '

cos 2fJ

=

cos2 f)

-

~A=E~±~~-m+±~~+m

ese2 0 ~ 1 + eofO sin2 f)

sin2 0 ~ 1 - eos 20 2

sin (A + E)

smA cos B + cosA sinE

sin (A - E)

smA cos B - cosA sinE

eos (A + E)

=

cosA cos B

eos (A - E)

=

cosA cos B + sin A sinB

sin A sinB

sin A + sin E

~

2 sin ± (A + E) eos ± (A - E)

sinA - sinE

~

2 eos±(A + E) sin ± (A - E)

eos A + eos E

~

1 1 2 eos 2 (A + E) eos 2 (A - E)

eosA - easE

~

-2sin±(A + E)sin±(A - E)

Trigonometric Functions Radian Measure

Degrees

Radians

L n 5

V2

45

,

,

2

~

90

4

,

'f,

Domain: (_00, 00) Range: [-1,1]

~

y

2

Domain: (_00, 00) Range: [-1,1]

y y=tanx

y=secx

Domain: All real numbers except odd integer multiples of 7T12 Range: (_00,00)

or 180 0 =

1T

radians.

() =

s

y'

y

y y=cscx

The angles of two common triangles, in degrees and radians.

,U -

~O

(\

Domain: All real numbers except odd integer multiples of 7T12 Range: (_00, -1] U [1, 00)

"2

h

II

Domain: x *- 0, ±7T, ±27T,. RaIge: (_00, -1] U [1, co)

y=cotx

x

Domain: x *- 0, ±7T, ±27T, . Range: (_00,00)

SERIES

Tests for Convergence of Infinite Series 5. Series with some negative terms: Does LI an Iconverge? If yes, so does Lan since absolute convergence implies convergence.

1. The nth-Term Test: Unless an ~ 0, the series diverges. 2. Geometric series: Lar n converges if Irl < 1; otherwise it diverges.

2: IjnI' converges if p >

3. p-series:

6. Alternating series: Lan converges ifthe series satisfies the conditions of the Alternating Series Test.

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.

Taylor Series 00

1+

-x

x2

+ ... +

Ixl <

I

=

~ ~(-I)"xn,

+ (-x)" + ...

-

Ixl <

n=O

xn +-+ n!

x2 =I+x+2T+

. smx

~xn,

+ ...

00

- x + x2

I +x X

xn

n=O

I

e

+

x

n

00

~;,

n=O

Ixl <

n.

x2n+ 1

x3 x5 x - - + -3! 5!

I

00

00

(_

1 )nx 2n+ 1

+ (- I)" -c(2"'n-+----I),--! + ... ~ ~ "'(2,--n"'+----I)-! x 2n

00

(-I )"x 2n

~-'--;-:-'...,-;­ + (-I)" (2n)! + ... ~ n~O (2n)! 2

In(1 + x)

~

x -

3

x x 2 + :3 - ... +

1+ I - x

X -1 x~2 ( In--~2tanh

tan

-1

x ~ x -

x3

X

(-I)

3

3

5

5

:3 + 5 - ... +

+

X

2n + 1

00

-1 <x:S;

)

5

x x x+-+-+'"

Ixl <

00

n

n

n-1 x

Ixl <

+

(-I)" 2n + I +

. 00

=

00

2n+1

2~2:

(-I)"x2n+!

~ ~ -'--,2'--n'--+"---;IC-

Ixl <

+ l'

Ixl s

I

I

Binomial Series

m(m - l)x 2

+ mx + ---'-2"'!C-'--- +

Ixl <

m(m - I)(m - 2)x 3

-'----~3"!----'---

+ ... +

m(m - I)(m - 2) .. · (m - k + I)x k k!

I,

where

m(m - I) 2!

m(m - I) .. · (m - k + I) k!

for k '" 3.

+

BASIC ALGEBRA FORMULAS Arithmetic Operations a(b + c)

~

a c

ab + ac,

b'd

ac

~ bd

alb a d cld ~ b' c

!!:..+!:..=ad+bc b d bd

Laws of Signs ~

-(-a) Zero

-a b

a,

a a b-b

Division by zero is not defined.

* 0: 7ia ~ 0,

If a

aO ~ I,

For any nwnber a: a· 0

O· a

=

a

on ~ =

0

Laws of Exponents

If a

* 0, am _ m-n -;;n-a ,

+ b)n

=

-m _

1

- am'

For any positive integer n,

The Binomial Theorem (a

a

+ nan-1b +

an

+

n(n - I)

1.2

n(n - I)(n - 2)

1.2.3

a

a n- 2 b 2

n-3 3

b

+ ... +nab n -

I

+b n .

For instance, (a +

bf

~ a2

(a - b f ~ a 2

+ 2ab + b 2 ,

-

2ab + b 2

(a - b)3 ~ a 3 - 3a 2 b + 3ab 2

(a + b)' ~ a 3 + 3a 2b + 3ab 2 + b 3,

-

Factoring the Difference of Like Integer Powers, n > 1 an - b n

=

(a - b)(a n-

l

+ a n- 2 b + a n- 3 b 2 + ... + ab n- 2 + b n- l )

For instance, a2

-

b 2 ~ (a- b)(a + b),

a 3 - b 3 ~ (a - b )(a 2 + ab + b 2 ), a4

-

Completing the Square ax 2 + bx + c

The Quadratic Formula

b 4 ~ (a - b)(a 3 + a 2 b + ab 2 + b 3).

If a

~ If a

* 0,

au 2 + C

* a and ax x~

(u 2

~x

+ (bI2a), C

+ bx + c ~ 0, then

-b ± Vb 2 2a

-

4ac

~

c - ::)

b3

GEOMETRY FORMULAS A = area, B = area of base, C = circumference, S = lateral area or surface area, V ~ volume

Triangle

Similar Triangles

Pythagorean Theorem

a b'

a'

c'

Z;=c

a

Trapezoid

Circle

b A

~ ~(a + b)h

Any Cylinder or Prism with Parallel Bases

Right Circular Cylinder

v= S

Any Cone or Pyramid

Right Circular Cone

v= S

= 1Trs =

!1TTlh 3

Area of side

=

21Trh

Sphere

=

1Tr 2h

Area of side

LIMITS General Laws

Specific Formulas If P(x) = anx n + an_IX n- 1 + ... + ao, then

If L, M, c, and k are real numbers and lim f(x)

~

L

lim g(x)

and

x~c

~

M,

then

lim P(x) ~ P(c) ~ ancn + an_1Cn-1 + ... + aO. x~c

x~c

lim(j(x) + g(x))

Sum Rule:

~

L +M

~

L - M

x~c

lim(j(x) - g(x))

Difference Rule:

If P(x) and Q(x) are polynomials and Q(c) i' 0, then

x~c

~

lim (j(x) , g(x))

Product Rule:

L, M

.

lim(k' f(x))

Constant Multiple Rule:

~

P(x) Q(x)

P(c) Q(c)'

hm--~-­

x~c

x~c

k, L

x~c

M i' 0

Quotient Rule:

If f(x) is continuous at x

The Sandwich Theorem

lim g(x) then limx~c f(x)

~

c, then

lim f(x)

~

lim hex)

~

~

f(c).

x~c

If g(x) s; f(x) s; h(x) in an open interval containing c, except possibly at x ~ c, and if

x----+c

~

L,

x----+c

lim smx x---+O

L.

x

=

and

lim 1 - cosx x---+O

=

0

X

Inequalities

L'Hopital's Rule

If f(x) s; g(x) in an open interval containing c, except possibly

If f(a) ~ g(a) ~ 0, both l' and g' exist in an open interval I containing a, and g' (x) i' 0 on I if x i' a, then

at x

=

c, and both limits exist, then

lim f(x) s; lim g(x) . x----+c

x----+c

Continuity

asswning the limit on the right side exists.

If g is continuous at L and limx~c f(x)

limg(f(x)) x~c

~

~

L, then

gel).

DIFFERENTIATION RULES General Formulas

Inverse Trigonometric Functions

Asswne u and v are differentiable flUlctions ofx.

A.. (sin-I x) ~

Constant:

d dx (e) ~ 0

Sum:

d du dv dx(u+v)~dx+dx

Difference:

d du dv dx(u-v)~dx-dx

dx

Constant Multiple:

A..(cu) ~ e du

Product:

A.. (uv)

dx

dx

A.. (tan-I x) dx

A.. (eoCI x) dx

A.. (COS-I x)

1

~

dx

~ _

~ _1_ 1 + x2

A.. (sec-I x) ~

~ _ _1_ 1 + x2

A.. (esc-I x)

dx

1

~

1

Ixl~

~ _

dx

1

Ixl~

dx

~ u dv + v du

d ()

dx

dx

du

dv

vd; - ud;

Quotient:

dx¥i~

Power:

.!:!....x n = nx n-

Chain Rule:

:x (j(g(x)) ~ f'(g(x))' g'(x)

v2 1

dx

Hyperbolic Functions

:x (sinh x) ~ eoshx

:x (cosh x) ~ sinhx

dx (tanh x) ~ seeh2 x

d

dx (seehx) ~ -seehxtanhx

d

d dx (esehx) ~ -esehxeothx

dx (eothx) ~ -eseh2 x

d

Inverse Hyperbolic Functions Trigonometric Functions

:x (sin x) ~ eosx

:x (cos x) ~ -sinx

d dx(tarlX) ~ see 2 x

d dx(seex) ~ seextanx

d dx(eotx) ~ -ese2 x

d dx(esex) ~ -esexeotx

Exponential and Logarithmic Functions .!:!....e X = eX

d 1 -lnx = -

.!!....ax=axlna dx

d 1 -d (loga x ) ~ -In x x a

dx

dx

x

A.. (seeh-I x) ~

_

A.. (eseh-I x) ~

_

dx

A.. (eoth-I x) dx

~ _1_ 1 - x2

dx

1

x~

Parametric Equations If x

~

jet) and y

y'

dy dx

~

get) are differentiable, then dy/dt dxjdt

and

d 2y dx 2

1

Ixl~

dy'/dt dxjdt'

INTEGRATION RULES General Formulas Zero:

Order a/Integration:

[f(X) dx

~

a

l

~

- [f(x) dx

af

(x) dx

~

l\f(X) dx

Constant Multiples:

[-f(X)dX~

k1

bf

(x) dx

(Any number

(k~-I)

-[f(X)dX

Sums and Differences:

1

Additivity:

[f(x) dx + [f(X) dx

b (j(X) ± g(x)) dx

~

1

bf

k)

(x) dx ±

1

b g(x) dx

~ [f(X) dx

Max-Min Inequality: If max f and min f are the maximum and minimum values of f on [a, b], then

1 1 1 1

minf'(b - a)

Domination:

f(x) '" g(x) f(x) '"

a

[a, b]

on

[a, b]

on

bf

implies

bf

implies

bf

S;

(x)dx

(x) dx '"

(x) dx '"

S;

maxf'(b - a).

bg

(x) dx

a

The Fundamental Theorem of Calculus Part 1 If f is continuous on [a, b], then F(x) ~ f:f(t) dt is continuous on [a, b] and differentiable on (a, b) and its derivative is f(x);

~

F'(x)

fx[f(t) dt

~

f(x).

Part 2 Iff is continuous at every point of [a, b] and F is any antiderivative of f on [a, b], then

1

bf

(x) dx

~ F(b)

- F(a).

Substitution in Oefinite Integrals b

1

f(g(x))' g'(x) dx

a

~

Integration by Parts 19(b)

g{a)

feu) du

[f(x)g'(x) dx

~ f(x)g(x) J: - [j'(x)g(x) dx