THOMAS'
CALCULUS Twelfth Edition
Multivariable Based on the original work by
George B. Thomas, Jr. Massachusetts Institute of Technology as revised by
Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis
Addison-Wesley Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
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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 978-0-321-64369-8 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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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
viii
Contents
17
Second-Order Differential Equations 17.1 17.2 17.3 17.4 17.5
online
Second-Order Linear Equations Nonhomogeneous Linear Equations Applications Euler Equations Power Series Solutions
Appendices
AP-1 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.
ix
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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.
Preface
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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
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
12 VECTORS AND THE GEOMETRY OF SPACE OVERVIEW To apply calculus in many real-world situations and in higher mathematics, we need a mathematical description of three-dimensional space. In this chapter we introduce three-dimensional coordinate systems and vectors. Building on what we already know about coordinates in the xy-plane, we establish coordinates in space by adding a third axis that measures distance above and below the xy-plane. Vectors are used to study the analytic geometry of space, where they give simple ways to describe lines, planes, surfaces, and curves in space. We use these geometric ideas later in the book to study motion in space and the calculus of functions of several variables, with their many important applications in science, engineering, economics, and higher mathematics.
Three-Dimensional Coordinate Systems
12.1
z
= constant (0,0, z) (0, y, z) (x, 0, z)
°l / /
---
P(x, y, z)
---_
/ / /
~ y = constant
(x, 0, 0)
x
x = constant
(0, y, 0)
-~y
(x, y, 0)
FIGURE 12.1 The Cartesian coordinate system is right-handed.
660
To locate a point in space, we use three mutually perpendicular coordinate axes, arranged as in Figure 12.1. The axes shown there make a right-handed coordinate frame. When you hold your right hand so that the fingers curl from the positive x-axis toward the positive y-axis, your thumb points along the positive z-axis. So when you look down on the xy-plane from the positive direction of the z-axis, positive angles in the plane are measured counterclockwise from the positive x-axis and around the positive z-axis. (In a left-handed coordinate frame, the z-axis would point downward in Figure 12.1 and angles in the plane would be positive when measured clockwise from the positive x-axis. Right-handed and left-handed coordinate frames are not equivalent.) The Cartesian coordinates (x, y, z) of a point P in space are the values at which the planes through P perpendicular to the axes cut the axes. Cartesian coordinates for space are also called rectangular coordinates because the axes that define them meet at right angles. Points on the x-axis have y- and z-coordinates equal to zero. That is, they have coordinates of the form (x, 0, 0). Similarly, points on the y-axis have coordinates of the form (0, y, 0), and points on the z-axis have coordinates of the form (0, 0, z). The planes determined by the coordinates axes are the xy-plane, whose standard equation is z == 0; the yz-plane, whose standard equation is x == 0; and the xz-plane, whose standard equation is y == o. They meet at the origin (0, 0, 0) (Figure 12.2). The origin is also identified by simply 0 or sometimes the letter o. The three coordinate planes x == 0, y == 0, and z == 0 divide space into eight cells called octants. The octant in which the point coordinates are all positive is called the first octant; there is no convention for numbering the other seven octants. The points in a plane perpendicular to the x-axis all have the same x-coordinate, this being the number at which that plane cuts the x-axis. The y- and z-coordinates can be any numbers. Similarly, the points in a plane perpendicular to the y-axis have a common y-coordinate and the points in a plane perpendicular to the z-axis have a common z-coordinate. To write equations for these planes, we name the common coordinate's value. The plane x == 2 is the plane perpendicular to the x-axis at x == 2. The plane y == 3 is the plane perpendicular to the y-axis
12.1
661
Three-Dimensional Coordinate Systems
z xz-plane: y = 0
\ xy-plane: z = 0 ------
"'"
//
/
/ ~ yz-plane: x =
0
/ /
'" Plane y = 3
/0/ . .
--
: \ ngm : (0, 0, 0)
(0,3,0)
Zy
y
I
x Line x = 2, y = 3
FIGURE 12.2 The planes x = O,y = O,andz = space into eight octants.
°divide
FIGURE 12.3 The planes x = 2, y = 3, and z = 5 determine three lines through the point (2, 3, 5).
at y == 3. The plane z == 5 is the plane perpendicular to the z-axis at z == 5. Figure 12.3 shows the planes x == 2, y == 3, and z == 5, together with their intersection point (2, 3, 5). The planes x == 2 and y == 3 in Figure 12.3 intersect in a line parallel to the z-axis. This line is described by the pair of equations x == 2, y == 3. A point (x, y, z) lies on the line if and only if x == 2 and y == 3. Similarly, the line of intersection of the planes y == 3 and z == 5 is described by the equation pair y == 3, z == 5. This line runs parallel to the x-axis. The line of intersection of the planes x == 2 and z == 5, parallel to the y-axis, is described by the equation pair x == 2, z == 5. In the following examples, we match coordinate equations and inequalities with the sets of points they define in space.
EXAMPLE 1 (a) z
~
We interpret these equations and inequalities geometrically.
0
The half-space consisting of the points on and above the xy-plane.
(b) x == -3
z The circle
x 2 + y2
=
4, z = 3
/
The plane perpendicular to the x-axis at x == - 3. This plane lies parallel to the yz-plane and 3 units behind it.
(c)
z==O,x:::;O,y~O
The second quadrant of the xy-plane.
(d)
x~O,y~O,z~O
The first octant.
(e) -1 :::; y :::; 1
The slab between the planes y included).
(f) y == - 2, z == 2
The line in which the planes y == -2 and z == 2 intersect. Alternatively, the line through the point (0, -2, 2) parallel to the x-axis. _
EXAMPLE 2
FIGURE 12.4 The circle x 2 + the plane z = 3 (Example 2).
y2
= 4 in
1 and y
==
1 (planes
What points P(x, y, z) satisfy the equations and
x
== -
z == 3?
SoLution The points lie in the horizontal plane z == 3 and, in this plane, make up the circle x 2 + y2 == 4.We call this set of points "the circle x 2 + y2 == 4 in the plane z == 3" or, more simply, "the circlex 2 + y2 == 4, z == 3" (Figure 12.4). _
662
Chapter 12: Vectors and the Geometry of Space
Distance and Spheres in Space
Z
The formula for the distance between two points in the xy-plane extends to points in space.
The Distance Between Pt(Xt,Yt, Zt) and P 2(X2,Y2, Z2) is
+
IP I P 2 1 == Y(X2 - XI)2
x
FIGURE 12.5 We find the distance between PI and P 2 by applying the Pythagorean theorem to the right triangles PIAB and P I BP2 •
(Y2 - YI)2
+
(Z2 - ZI)2
Proof We construct a rectangular box with faces parallel to the coordinate planes and the points PI and P 2 at opposite comers of the box (Figure 12.5). If A(X2, YI, ZI) and B(X2, Y2, ZI) are the vertices of the box indicated in the figure, then the three box edges PIA, AB, and BP2 have lengths IABI == IY2-YII,
Because triangles PIBP2 and PIAB are both right-angled, two applications of the Pythagorean theorem give 2 and IPIP212 == IPIBI2 + IBP21 (see Figure 12.5). So IPIP212
==
IPIBI2
+
2 IBP21
IPIAI2
+
IABI
IX2 - xI1 ==
2
(X2 - XI)2
2
+
+ IY2 + (Y2 -
Substitute
2 IBP21
YI1
2
YI)2
IP 1BI 2
+ IZ2 + (Z2 -
=
IP 1AI 2 + IABI 2 •
zI12 ZI)2
Therefore
• EXAMPLE 3
The distance between PI(2, 1,5) and P2( -2,3,0) is IP IP21
==
Y( -2 -
==
Y16
==
V45 ~
2)2
+
+4+
25
(3 - 1)2
+
(0 - 5)2
•
6.708.
We can use the distance formula to write equations for spheres in space (Figure 12.6). A point P(x, Y, z) lies on the sphere of radius a centered at Po(xo, Yo, zo) precisely when IPoPI == aor
Z
The Standard Equation for the Sphere of Radius a and Center (xo,Yo, zo)
o
(x - xO)2
+ (y -
YO)2
+
(z - zO)2
==
a2
y x
EXAMPLE 4
Find the center and radius of the sphere x2
FIGURE 12.6 The sphere of radius a centered at the point (xo, Yo, zo).
+ y 2 + z2 +
3x - 4z
+
1
==
O.
We find the center and radius of a sphere the way we find the center and radius of a circle: Complete the squares on the X-, y-, and z-terms as necessary and write each
SoLution
663
12.1 Three-Dimensional Coordinate Systems
quadratic as a squared linear expression. Then, from the equation in standard form, read off the center and radius. For the sphere here, we have
x2
(X 2 + 3x +
GY)
+ y2 + Z2 + 3x - 4z + I + 3x) + y2 + (Z2 - 4z)
(X 2
+ y2 +
0
= -I
(-;4 y) -I + (~y + (-;4Y
+
(Z2 - 4z
=
=
21 (x+ 23)2 +y2+(z-2)2=-1+ 49 +4=4' From this standard form, we read that Xo = -3/2, Yo = 0, Zo = 2, and a = v21/2. The center is (-3/2,0,2). The radius isv21/2. •
EXAMPLE 5 Here are some geometric interpretations of inequalities and equations involving spheres. (8) x 2
The interior ofthe sphere x 2
+ y2 + z2 < 4 2 x + y2 + z2 :5 4
(h)
+ y2 + z2
= 2
4.
The solid ball bounded by the sphere x + y2 + 4. Alternatively, the sphere x 2 + y2 + z2 = 4 together with its interior.
z2 =
(c) x 2
x2
(d)
+ y2 + z2 > 4 + y2 + z2 = 4, z
The exterior ofthe sphere x 2 :5
+ y2 + z2
= 4.
The lower hemisphere cut from the sphere x 2 + y2 + z2 = 4 by the xy-plane (the plane z = 0).
0
•
Just as polar coordinates give another way to locate points in the xy-plane (Section 11.3), alternative coordinate systems, different from the Cartesian coordinate system developed here, exist for three-dimensional space. We examine two of these coordinate systems in Section 15.7.
Exercises 12.1 Geometric Interpretations of Equations
10 Exercises 1-16, give a geometric description ofthe set of points in space whose coordinates satisfY the given pairs of equations. 1. x
=
2, y
=
3
3. y = 0, z = 0
2. x
=
-1, z
xl
15. Y
=x
2
,
z
=0
16. z = y2, X = 1
0
17.
4. x = 1, Y = 0
+ y2 = 4, z = 0 6. x 2 + y2 7. x 2 + z2 = 4, Y = 0 8. y2 + z2 2 9. x + y2 + z2 = 1, x = 0 10. x 2 + y2 + z2 ~ 25, Y ~ -4 11. x 2 + y2 + (z + 3)' ~ 25, z ~ 0 12. x 2 + (y - 1)2 + z2 ~ 4, Y ~ 0 13. x 2 + y2 = 4, z = Y 14. x 2 + y2 + z2 = 4, y = x S.
=
=
4,
= 1,
Geometric Interpretations of Inequalities and Equations 10 Exercises 17-24, describe the sets of points in space whose coordinates satisJY the given inequalities or combinations of equations and inequalities.
z
=
-2
x= 0
8.
x
~
0, y
~
0, z = 0
18. a. 0 ::s x :s 1 O:s y :s 1,
19. a. x 2 + y2 + z2 :s 1 20.
8.
x
C.
x2
+ y2 :s 1, z = 0
+ y2:5
~
0, y
~
b. 0 :s x :s I,
c. 0 :s x :s 1, 2
b. x
1,
O:s z b.
Xl
b. x
2
0, z = 0 O:s y ::s 1
:s 1
+ y2 + z2 > 1 + y2 :s 1, z =
3
norestrictiononz
21.8.1 :sx 2 +y2+ z 2 S 4 b. x 2
+ y2 + z2 :s
8. x = y, z = 0 23. a. y ~ x2, z ~ 0
22.
24.
8.
z
=
b. z =
I, z ~ 0 b. x
=
y,
b. x
S
y2,
1 - y, no restriction onx y3,
X
=2
no restriction on z 0
S
z
S
2
664
Chapter 12: Vectors and the Geometry of Space
In Exercises 25-34, describe the given set with a single equatioo or
44. P I (3, 4, 5),
P2(2, 3, 4)
with a pair ofequations.
45. PI(O, 0, 0),
P2(2, -2, -2)
46. P I (5, 3, -2),
P2(0, 0, 0)
25. The plaoe perpendicular to the R.
x-axis at (3, 0, 0)
b. y-axis at (0, -I, 0)
c. z-axis at (0, 0, -2) 26. The plaoe tbrougb the point (3, -1,2) perpendicular to the
a. x-axis
b. y-axis
c. z-axis
Sphell!s Find the centers and radii ofthe spheres in Exercises 47-50. 47. (x
+ 2)' + y2 + (z - 2)' = 8
27. The plaoe tbrougb the point (3, -I, I) parallel to the R.
xy-plaoe
b. yz-plaoe
c. xz-plaoe
28. The circle of radius 2 centered at (0, 0, 0) and lying in the R.
xy-plaoe
b. yz-plane
xy-plaoe
b. yz-plane
xy-plaoe
b. yz-plane
&+ t)' +
b. y-axis
c. z-axis
33. The circle in which the plane tbrnugb the point (I, 1,3) perpendicular to 1Ire z-axis meets the sphere ofradius 5 centered at the origin 34. The set of points in space that lie 2 units from 1Ire point (0, 0, I) and, at 1Ire sarae time, 2 units from the point (0, 0, -I) Inequalities to Describe Sets of Points Write inequalities to describe the sets in Exercises 35-40.
x2 + Y+3
( 1)2 + (z-31)2 =9 16
Find equations for the spheres whose centers and radii are given in Exercises 51-54. Center
Radius
51. (1,2,3)
Vi4
52. (0, -1,5)
2
53.
(-I,t,-t)
4 9
54. (0, -7,0)
7
Find the centers and radii ofthe spheres in Exercises 55-58. 55. x 2 + y2 + z2 + 4x - 4z = 0
56. x 2 + y2 57.
+ z2 - 6y + 8z ~ 0 + 2y 2 + 2z2 + X + Y + z ~ 2 3x + 3y 2 + 3z2 + 2y - 2z = 9
2x 2
35. The slab bounded by the planes z = 0 and z = I (planes included)
58.
36. The solid cube in the frrst octant bounded by 1Ire coordioate plaoes and 1Ire planes x = 2, y = 2, and z = 2
Theory and Examples
37. The half-space consisting of the points 00 and below the xy-plane 38. The upper hemisphere of the sphere of radius I centered at the origin 39. The (a) interior and (b) exterior of the sphere of radius I centered at the point (I, I, I) 40. The closed regioo bounded by the spheres ofradius I and radius 2 centered at the origin. (Closed means the spheres are to be included. Had we wanted the spheres left out, we would have asked for the open regioo bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out) Distance In Exercises 41-46, fmd the dislance between points PI and P2.
P 2(3, 3, 0)
42. Pl( -I, 1,5),
P2(2, 5, 0)
43. PI(I, 4, 5),
P2(4, -2,7)
= 25
50.
c. xz-plaoe
32. The set of points in space equidistant from the origin and the point (0, 2, 0)
41. Pl(l, I, I),
+ 3)'
(x - yz)' + (y - yz)2 + (z + yz)' ~ 2
31. The line through the point (I, 3, -I) parallel to the
a. x-axis
(z
c. plane y = 2
30. The circle of radius I centered at (-3,4, I) and lying in a plane parallcl to the R.
+
49.
c. xz-plaoe
29. The circle of radius 2 centered at (0, 2, 0) and lying in the R.
48. (x - I)'
9
59. Find a formula for the distance from the point P(x, y, z) to the
a. x-axis
b. y-axis
c. z-axis
60. Find a formula for the distance from the point P(x, y, z) to the R.
xy-plane
b. yz-plane
c. xz-plane
61. Find the perimeter of the triangle with vertices A( -1,2, I), B(I, -I, 3), and C(3, 4, 5). 62. Show that the point p(3, I, 2) is equidistant from the points A(2, -1,3) andB(4, 3, I). 63. Find an equatioo for the set of all points equidistant from the planesy = 3 andy = -1.
64. Find an equatioo for the set of all points equidistant from the point (0, 0, 2) and the xy-plane. 65. Find the point on the sphere x 2 nearest
a. thexy-plane.
+ (y -
3)'
+ (z + 5)'
~ 4
b. thepoint(O, 7, -5).
66. Find the point equidistant from the points (0, 0, 0), (0,4, 0), (3, 0, 0), and (2, 2, -3).
12.2 Veeto"
12.2
665
I_v_e_ct_or_s- - - - - - - - - - - - - - - - Some of the things we measure are determined simply by their magnitudes. To record mass, length, or time, for example, we need only write down a number and name an appropriate unit ofmeasure. We need more information to describe a force, displacement, or velocity. To describe a force, we need to record the direction in which it acts as well as how large it is. To describe a body's displacement, we have to say in what direction it moved as well as how far. To describe a body's velocity, we have to know where the body is headed as well as how fast it is going. In this section we show how to represent thiogs that have both magnitude and direction in the plane or in space. Terminal point B
FIGURE 12.7 The directed line segment
AB is called a vector.
Component Form A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment (Fignre 12.7). The arrow points in the direction of the action and its length gives the magnitude of the action in terms of a suitably chosen unit. For example, a force vector points in the direction in which the force acts and its length is a measure of the force's strength; a velocity vector points in the direction of motion and its length is the speed of the moving object. Figure 12.8 displays the velocity vector v at a specific location for a particle moving along a path in the plane or in space. (This application of vectors is studied in Chapter 13.) y
y B _______ D
A
C p
F (a) two dimensions
E
FIGURE 12.9 The four arrows io the plane (directed lioe segments) shown here have the same length and direction. They therefore represeot the same vector, and we write
-7f'----~"------y
o
------c:t"~------x
o
o
-;;f--------->x
x
(b) three dimensions
FIGURE 12.8 The velocity vector ofa particle moving along a path (a) io the plane (b) io space. The arrowhead on the path iodicates the direction ofmotion ofthe particle.
AB = cD = OP = EF.
DEFINmONS The vector represented by the directed line segmen.!....AB has initial point A and terminal point B and its length is denoted by lABI. Two vectors are equal if they have the same length and direction.
The arrows we use when we draw vectors are understood to represent the same vector if they have the same length, are parallel, and point in the same direction (Fignre 12.9) regardless ofthe initial point. In textbooks, vectors are usually written in lowercase, boldface letters, for example u, v, and w. Sometimes we use uppercase boldface letters, such as F, to denote a force vector. In handwritten form, it is customary to draw small arrows above the letters, for example ii,
v,w,andF. FIGURE 12.10 A vector PQ io standard position has its initial point ~the origin. The directed lioe segments PQ and v are
parallel and have the same length.
We need a way to represent vectors algebraically so that we can be more precise shout the direction of a vector. Let v = PQ. There is one directed line segment equal to PQ whose initial point is the origin (Fignre 12.10). It is the representative ofv in standard position and is the vector we normally use to represent v. We can specify v by writing the
666
Chapter 12: Vectors and the Geometry of Space
coordinates of its terminal point (v" V2, v,) when v is in standard position. Ifv is a vector in the plane its terminal point (V" V2) has two coordinates.
DEFINmON If v is a two-dimensional vector in the plane equal to the vector with initial point at the origin and terminal point (V" V2), then the component form of v is v = (V" V2).
Ifv is a three-dimensional vector equal to the vector with initial point at the origin and terminal point (V" V2, v,), then the componentform ofv is v = (V" V2, v,).
So a two-dimensional vector is an ordered pair v = (v" V2) of real numbers, and a three-dimensional vector is an ordered triple v = (V" V2, v,) of real numbers. The numV2, and are the components ofv. bers If v = (V" V2, v,) is represented by the directed line segment PQ, where the initial point is P(x"y" Z,) and the terminal point is Q(X2,Y2, Z2), then Xl + VI = X2, Yl + V2 = Y2, and ZI + = Z2 (see Figure 12.10). Thus, VI = X2 - X" V2 = Y2 - YI, and = Z2 - Zl are the components of PQ. In summary, given the points P(x" Y" Zl) and Q(X2,Y2, Z2), the standard position vector v = (v" V2, v,) equal to PQ is
v"
v,
v,
v,
v = (X2 - X"Y2 - y"Z2 - ZI).
If v is two-dimensional with P(X"YI) and Q(X2,Y2) as points in the plane, then v = (X2 - X" Y2 - Yl). There is no third component for planar vectors. With this understanding, we will develop the algebra of three-dimensional vectors and simply drop the third component when the vector is tw<>-dimensional (a planar vector). Two vectors are equal if and only if their standard position vectors are identical. Thus (u" U2, u,) and (V" V2, v,) are equal if and only iful = v" U2 = V2, and u, = v,. The magnitude or length of the vector PQ is the length of any of its equivalent directed line segment representations. In particular, ifv = (X2 - X" Y2 - Y" Z2 - Zl) is the standard position vector for PQ, then the distance formula gives the magnitude or length of v, denoted by the symbollvl orllvll.
The magnitude or length of the vector v =
Ivl
=
Yv? + vo' + vl = Y(X2 -
PQ is the nonnegative number X,)2
+
(Y2 - YIl'
+
(Z2 - zll'
(see Figure 12.10).
The only vector with length 0 is the zero vector 0 = (0, 0) or 0 = (0, 0, 0). This vector is also the only vector with no specific direction.
EXAMPLE 1 Find the
V2 = Y2 - Yl = 2 - 4 = -2,
12.2
Veeto"
667
and V3 =
2 - I = 1.
ZI =
Z2 -
~
The component form of PQ is v = (-2, -2, I). ~
y
(b) The length or magnitude of v = PQ is
t
I
Ivl = Y(-2)2
I
I I I
I
FIGURE 12.11 The force pulling the cart forward is represeoted by the vector F whose horizontal component is the effective force (Example 2).
+
(-2)2
+ (1)2
=
V9 =
•
3.
EXAMPLE 2 A small cart is being pulled along a smooth horizontal floor with a 20-lb force F making a 45° angle to the floor (Figure 12.11). What is the effective force moving the cart forward? Solution
The effective force is the horizontal component ofF = (a, b), given by
a
= IFI cos 45° = (20)('?) "" 14.14 lb.
•
Notice that F is a two-dimensional vector.
Vector Algebra Operations Two principal operations involving vectors are vector addition and scalar multiplication. A scalar is simply a real number, and is called such when we want to draw attention to its differences from vectors. Scalars can be positive, negative, or zero and are used to "scale" a vector by multiplication.
DEFINmONS
Let u = (UI, U2, U3) and v = (VI, V2, V3) be vectors with k a
scalar.
+v
+ VI, U2 + V2, U3 + V3)
Addition:
u
Scalar multiplication:
ku = (kuI, ku2, ku3)
= (Ul
We add vectors by adding the corresponding components of the vectors. We multiply a vector by a scalar by multiplying each component by the scalar. The definitions apply to planar vectors except there are ouly two components, (UI, U2) and (VI, V2). The definition of vector addition is illustrated geometrically for planar vectors in Figure 12.12a, where the initial point of one vector is placed at the terminal point of the other. Another interpretation is shown in Figure 12.12b (called the parallelogram law of y
y
__ ....-.-.-""--I -
I
/
I
I
I I
I
~F---;;----'-------->x
(0)
FIGURE 12.12 vector addition.
o (b)
(0) Geomelric interpretation of the vector .om. (b) The parallelogram law of
668
Chapter 12: Vectors and the Geometry of Space
~
addition), where the swn, called the resultant vector, is the diagonal ofthe parallelogram.
//
/~~
In physics, forces add vectorially as do velocities, accelerations, and so on. So the force
acting on a particle subject to two gravitational forces, for example, is obtained hy adding the two force vectors. Figure 12.13 displays a geometric interpretation of the product ku of the scalar k and vector u. If k > 0, then ku has the same direction as u; if k < 0, then the direction of ku is opposite to that of u. Comparing the lengths of u and ku, we see that
lkul
FIGURE 12.13 Scalar multiples ofu.
=
V(I
= WVU12
+ ui + u,l
Ikllul.
=
The length of ku is the absolute value of the scalar k times the length of u. The vector ( -I)u = -u bas the same length as u but points in the opposite direction. The dilTerenee u - v of two vectors is defmed hy u-v=u+(-v). Ifu = (uI.
U2, U3)
and v =
u - v =
(a)
then
(VI. V2, V3), (U1 -
VI. U2 -
V2, U3 -
V3).
Note that (u - v) + v = u, so adding the vector (u - v) to v gives u (Figure 12.14a). Figure 12.14b shows the difference u - vas the swn u + (-v).
EXAMPLE 3
Y
(a) 2u
+ 3v
Let u = (-1,3, I) and v = (4,7,0). Find the components of (b) u - v
I!ul.
(e)
Solution (a) 2u + 3v = 2(-1,3, I) + 3(4,7,0) = (-2,6,2) + (12,21,0) = (10,27,2)
FIGURE 12.14 (a)Thevector u - Y, when added to v, gives u. (b)u - v
= u + (-v).
(b) u - v= (-1,3,1) - (4,7,0) = (-I - 4,3 -7,1 - 0) = (-5,-4,1) (e)
I!ul I(-!,~,!)I ~(-!y + (~y + (!y !V11. =
=
•
Vector operations have many ofthe properties of ordinary arithmetic.
Properties orVector Operations Let u, v, w be vectors and a, b be scalars. 1.u+v=v+u 3. u + 0 = u 5. Ou = 0 7. a(bu) = (ab)u 9. (a + b)u = au
2. 4. 6. 8.
(u + v) + w = u + (v + w) u + (-u) = 0 lu = u a(u + v) = au + av
+ bu
These properties are readily verified using the definitions ofvector addition and multiplication by a scalar. For instance, to establish Property I, we have u
+v
=
(uI. U2, U3) +
(VI. V2, V3)
= (U1
+ VI. U2 + V2, U3 + V3)
=
+
(VI
UI. V2
= (VI. V2, V3) =
v
+ u.
+
U2, V3
+
U3)
+ (uI. U2, U3)
12.2 Veeto"
669
When three or more space vectors lie io the same plane, we say they are coplanar vectors. For example, the vectors u, v, and u + v are always coplanar.
Unit Vectors A vector v of length I is called a unit vector. The standard unit vectors are i = (1,0,0),
j = (0,1,0),
and
k = (0,0,1).
Aoy vector v = (v!> V2, V3) can be written as a linear combination of the standard unit vectors as follows: v = (v!> V2, V3) = (v!> 0, 0) = vI(I, 0, 0) = VIi
+ v2j
+ (0, V2, 0) + (0,0, V3)
+ V2(0, 1,0) + V3(0,0, 1) + V3k.
We call the scalar (or number) VI the i-component of the vector v, V2 the j-component, and V3 the k- Y!> ZI) to P2(X2,,Y2, Z2) is
x
FIGURE 12.15 The vector from PI to P2 + (Y2 - YI)i +
is P-;P2 ~ (X2 - xl)l (Z2 - z,)k.
(Figure 12.15). Wheneverv oF O,itslength Ivl is not zero and
That is, v/lvl is a unit vector io the direction of v, called the direction of the nonzero vectorv.
EXAMPLE 4 P2(3, 2, 0). Solution
Fiod a unit vector u io the direction of the vector from P,(I, 0,1) to
We divide P-;P2 by its length:
P-;P2 IP-;P21
= (3 -
l)i
= Y(2)2
+
+
(2 - O)j
(2)2
+
+
(0 - l)k = 2i
(-1)2 = Y4
+4+
+ 2j 1=
~
- k
V9 =
3
•
The unit vector u is the direction of P,P2.
EXAMPLE 5 If v = 3i - 4j is a velocity vector, express v as a product of its speed times a unit vector io the direction ofmotion. Solution
Speed is the magnitude (length) of v: Ivl = Y(3j2
HIsTORICAL BIOGRAPHY
Hermann Grassmann (180!l-1877)
+
(-4j2 =
V9+16 =
The unit vector v/ Iv I has the same direction as v: v Ivl
3i-4j 3.4. 5 = 5' - 5 J ·
5.
670
Chapter 12: Vectors and the Geometry of Space
So v = 3i -
4i
5Gi -
=
/ Length
ti).
•
Direction ofmotion
(speed)
In summary, we can express any nonzero vector v in terms of its two importsnt features,
length and direction, by writing v = Iv II; I .
IfvoF 0, then 1. 1;1 is a unit vector in the direction of v; 2. the equation v = Ivll;1 expresses v as its length times its direction.
EXAMPLE 6 A force of 6 newtons is applied in the direction of the vector v = 2i + 2i - k. Express the force F as a product of its magnitude and direction.
Solution
The force vector has magnitude 6 and direction I; I ' so 21_._+-c2c:--i_-_k 2i + 2i - k = 6_ F = 6_v_ = 6 2 2 Ivl Y2 +2 +(-I)' 3 =6(t i +t i
-t
•
k ).
Midpoint of a Line Segment Vectors are often useful in geometry. For example, the coordinates of the midpoint of a line segment are found by averaging. The midpoint M of the line segment joining points PI(x"y"Zt) and P2(X2, Y2, Z2) is the point XI (
+ X2 2 '
YI
+ Y2 Zt + Z2) 2 ' 2 .
To see why, observe (Figure 12.16) that --OM = OPt
+ 21(P-tP2)
I~
=
o
Xt
FIGURE 12.16 The coordinates of the midpoint are the averages ofthe coordinates of PI andP2.
2(OP t
EXAMPLE 7
=
-OPt
1---+ 2(OP2 - OPt)
~
+ OP2)
+ X2 • Yt + Y2 • Zt + Z2 2 1+ 2 I + -2- k .
The midpoint ofthe segment joining P t (3, -2,0) andP2(7, 4, 4) is (
3 + 7 -2 + 2' 2
4' 2 4) 0 +
= (5 1 2) ".
•
12.2 Veeto"
671
Applications An important application of vectors occurs in navigation.
EXAMPLE 8 Ajet airliner, flying due east at 500 mph in still air, encounters a 70-mph tailwind blowing in the direction 60° north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?
N
500
u
NOTTOSCAIE
FIGURE 12.17 Vectors representing the velocities of the airplane u and tailwind v in Example 8.
Solution Ifu = the velocity of the airplane alone and v = the velocity of the tailwind, then lui = 500 and Ivl = 70 (Figure 12.17). The velocity of the airplane with respect to the ground is given by the magnitude and direction of the resultant vector u + v. Ifwe let the positive x-axis represent east and the positive y-axis represent north, then the component fonus of u and v are u = (500,0)
v = (70 cos 60°, 70 sin 60°) = (35, 35v'3).
and
Therefore, u lu
+v
+ vi
= (535, 35v'3) = 535i = \/535
2
+
+
35v'3 j
(35V'3)' '" 538.4
and (J =
tan-I
3~~
'" 6.50
Figure 12.17
The new ground speed of the airplane is about 538.4 mph, and its new direction is about _ 6.5° north of east. Another important application occurs in physics and engineering when several forces are acting on a single object.
40'
EXAMPLE 9 A 75-N weight is suspended by two wires, as shown in Figure 12.18a. Find the forces F 1and F2 acting in both wires. Solution
The force vectors F 1 and F2 have magnitudes IF11 and IF21 and components that are measured in Newtons. The resultant force is the sum F l + F 2 and must be equal in magnitude and acting in the opposite (or upward) direction tu the weight vectur w (see Figure 12.18b). It follows from the figure that
(aJ
Fl = (-IFI Icos 55°, IFdsin55°) Since Fl
/
/
+ F2
and
F2 = (IF2Icos40°, IF2Isin400).
= (0,75), the resultant vectur leads to the system of equations
/ /
+ IF21cos40° IFdsin55° + IF21sin400
-IFl Icos 55°
= 0 = 75.
Solving for IF21 in the 1mt equation and substituting the result into the second equation, we get
w
~
(0, -75)
(bJ
FIGURE 12.18 in Example 9.
IF2 1 =
IFdcos55° cos 40°
. ° IFl lsm55
and
+
IFdcos55°. ° _ cos 40° sm40 - 75.
It follows that
The suspended weight IFll = sin 55°
+
75 cos 55° tan 40° '" 57.67N,
672
Chapter 12: Vectors and the Geometry of Space and
IF 2 I =
75 cos 55° sin 55° cos 4()0 + cos 55° sin 40° 75 cos 55° sin(550 + 400) '" 43.18 N.
•
The force vectors are then F, = (-33.08,47.24) and F2 = (33.08,27.76).
Exercises 12.2 Vectors In the Plane
Geometric Representations
10 Exercises 1--1l, letu = (3, -2) and v = (-2,5). Find the (a) compooeot form and (b) magnitude (Ieogth) of the vector.
10 Exercises 23 and 24, copy vectors U, v, and w head to tail as needed to sketch the indicated vector.
1. 3u
3. u
2. -2v
+v
5.2u-3v
3
23.
4. u - v
6. -2u
4
7. SU + SV
8.
w
+ 5v
5
u
12
-TI u + TIv
v
10 Exercises !l-16, f"md the compooeot form of the vector.
a. u
9. The vector PQ, where P = (1,3) and Q = (2, -I)
oP
10. The vector where 0 is the origin and P is the midpoint of segmeotRS, whereR ~ (2, -I) andS ~ (-4,3)
+v
C. U -
b.u+v+w d.u.-w
v
24.
11. The vector from the point A = (2, 3) to the origin
cD,
where A=(I,-I),B=(2,0), 12. The sam of AB and C = (-1,3),andD = (-2,2) 13. The unit vector that makes an angle 8 ~ 27f/3 with the positive x-axis
u
14. The unit vector that makes an angle 8 = -37f/4 with the positive x-axis 15. The unit vector ohtained by rotating the vector (0, I) 120" counterclockwise about the origin 16. The unit vector obtained by rotating the vector (I, 0) 135" counterclockwise about the origin
b.u-v+w d.u+v+w
a. u - v
Vectors In Space In Exercises 17-22, express each vector in the form v = ",j + v,k.
C. Vii
20. -
V
+ Length and DIrection
17. PJ>2 if P, is the point (5, 7, -I) andPz is the point (2, 9, -2)
10 Exercises 25-30, express each vector as a product of its leogth and
18. PJ>2 if P, is the point (I, 2, 0) and Pz is the point (-3, 0, 5)
direction.
19.
20.
AB ifA is the point (-7, -8, I) andB is the point (-10,8, I) AB ifA is the point (1,0,3) andB is the point( -1,4,5)
25.2i+j-2k
26.9i-2j+6k
27.5k
28.~i+!k
29. _I_ i __I_ __I_ k j
30.
21. 5u - vifu = (I, I, -I) and v = (2,0,3) 22. -2u+ 3vifu = (-1,0,2) and v = (1,1,1)
Vii
Vii
Vii
i
j
k
v'3 + v'3 + v'3
12.2 Veeto" 31. Find the vectors whose lengths snd directions are given. Try to do the calculations without writiog. Length
30'
673
45'
Direction
a. 2 b.
V3
-k
c.
2
1
1j
5
+i
5
k
46. Consider a 50-N weight suspended by two wires as shown in the accompsnying figure. If the magnitude of vector F1 is 35 N, fmd angle a and the magnitude ofvectorF,.
~1-~j+1k
d. 7
777
32. Find the vectors whose lengths snd directions are given. Try to do the calculations without writiog. Length
Direction
a. 7
-j
b.
_1 1 - i
v2 13
k
5
5
3.
4.
12 k
TI'- TIl - TI _1_ + _I_ __I_ 1 j k
c'IT d. a> 0
v2
V3
V6
47. Consider a w-N weight suspended by two wires as shown in the accompsnying figure. Ifthe magnitude of vector F, is 100 N, fmd wand the magnitude of vector Fl.
33. Find a vector of magnitude 7 in the direction ofv = 121 - 5k. 34. Find a vector of magnitude 3 in the direction opposite to the directionofv = (1/2)1 - (1/2)j - (1/2)k.
40'
35'
Di"'ction and Midpoints In Exercises 35-38, find
a. the direction of P-;P, snd b. the midpoint ofline segment P, P,.
35. P,( -I, I, 5)
P,(2, 5, 0)
36. P,(I, 4, 5)
P,(4, -2,7)
37. P,(3, 4, 5)
P,(2, 3, 4)
38. P,(O, 0, 0)
P,(2, -2, -2)
39. If AB 40. IfAB
= I + 4j = -71 +
48. Consider a 25-N weight suspended by two wires as shown in the accompanying figure. If the magnitudes of vectors F1 and F, are both 75 N, then angles a and fJ are equal. Find a. a
fJ
- 2k sndB is the point (5, 1,3), fmdA. 3j + 8ksndAisthepoint(-2,-3,6),fmdB.
TheoIY and Applications 41. Linear combination Let u = 2i + j, v = i + j. and w = i - j. Find scalars a and b such that u = av + bw. 42. Lloear eombloation Let u = I - 2j, v = 2i + 3j, snd w = i + j. Write u = Ul + U2. where UI is parallel to v and U2 is parallel to w. (See Exercise 41.) 43. Velocity An airplsne is flying in the direction 25' west of north at 800 kmfb. Find the component form of the velocity of the air-
plane, assuming that the positive x-axis represents due east and the positive y-axis represents due north. 44. (Continuation of Example 8.) What speed and direction should the jetliner in Example 8 have in order for the resultant vector to be 500 mph due east? 45. Consider a loo-N weight suspended by two wires as shown in the accompsnying figure. Find the magnitudes and components of the force vectors F1 andF,.
49. Location A bird flies from its nest 5 km in the direction 60' north of east, where it stops to rest on a tree. It then flies 10 km in the direction due southeast and lands atop a telephone pole. Place an xy-coordinate system so that the origin is the bird's nest, the
x-axis points east, and the y-axis points north. a. At what point is the tree located? b. At what point is the telephone pole?
SO. Use similar trisngles to fmd the coordinates ofthe point Q that divides the segment from P,(XloY1, Zl) to P,(x"y" z,) into two lengths whose ratio is p/q ~ r. 51. Medions of a triangle Suppose that A, B, and C are the corner points of the thin triangular plate of constant density shown here. a. Find the vector from Cto the midpointM ofsideAB. b. Find the vector from C to the point that lies lwD-thirds of the way from C to M on the median CM.
674
Chapter 12: Vectors and the Geometry of Space
c. Find the coordinates ofthe point in which the medians of !iAJJC intersect Accordiog to Exercise 17, Section 6.6, this
point is the plate~ cooter ofmass. C(I, I, 3)
·c.m.
~y B(l, 3, 0)
M
A(4, 2, 0)
53. Let ABCD be a gooeral, not necessarily planar, quadrilateral in
space. Show that the two segmoots joining the midpoints of opposite sides of ABCD bisect each other. (Hint: Show that the segmoots have the same midpoint.) 54. Vectors are drawn from the cooter of a regular n-sided polygon in the plane to the vertices ofthe polygon. Show that the sum ofthe vectors is zero. (Hint: What happoos to the sum if you rotate the polygon about its cooter?) 55. Suppose that A, B, and C are vertices of a triangle and that a, b, and c are, ~ectively, the midpoints ofthe opposite sides. Show
thatAa + Bb + Co = o. 56. Unit veetors in the plane Show that a unit vector in the plane can be expressed as u ~ (cosO)i + (sinO)j,obtainedbyrotating i through an angle 0 in the counterclockwise direction. Explain why this form gives every unit vector in the plane.
52. Find the vector from the origin to the point of intersection of the medians of the triangle whose vertices are
A(I,-1,2),
12.3
B(2,I,3),
and C(-1,2,-1).
The Dot Product
v
Length ~
IF I cos 9
FIGURE 12.19 The magnitude of the force F in the direction ofvector v is the length IFI cos 0 of the projection ofF ontov.
If a force F is applied to a particle moving along a path, we often need to know the magnitode ofthe force in the direction ofmotion. Ifv is parallel to the tangent line to the path at the point where F is applied, then we want the magnitode ofF in the direction ofv. Figure 12.19 shows that the scalar quantity we seek is the length IFI cosO, where 0 is the angle between the two vectors F and v. In this section we show how to calcolate easily the angle between two vectors directly from their components. A key part of the calcolation is an expression called the dot product. Dot products are also called inner or scalar products because the product results in a scalar, not a vector. After investigating the dot product, we apply it to finding the projection ofone vector onto another (as displayed in Figure 12.19) and to finding the work done by a constant force acting through a displacement.
Angle Between Vectors When two nonzero vectors u and v are placed so their initial points coincide, they form an angle 0 of measure 0 ,,; 0 ,,; 1f (Figure 12.20). If the vectors do not lie along the same line, the angle 0 is measured in the plane containing both of them. If they do lie along the same line, the angle between them is 0 if they point in the same direction and 1f if they point in opposite directions. The angle 0 is the angle between u and v. Theorem I gives a fonnula to detennine this angle.
v
THEOREM 1-Angle Between Two Vedors The angle 0 between two nonzero vectors u = (u," U2, U3) and v = (v," v" V3) is given by _
o-
-I
cos
(UtVt + U2 V2 + U3 V3 ) lullvl
.
FIGURE 12.20 The angle betweeo u and v.
Before proving Theorem 1, we focus attention on the expression UI VI + U2 V2 + U3 V3 in the calcolation for O. This expression is the sum of the products of the corresponding components for the vectors u and v.
12.3
The Dotproduct
675
DEFINmON The dot product u' v ("u dot v') of vectors u = (u!, U2, U3) and v = (v!, V2, V3) is
EXAMPLE 1
+ (-2)(2) + (-1)(-3) -6 - 4 + 3 = -7
(8) (1,-2,-1)'(-6,2,-3) = (1)(-6) =
(b)
(!i +
+
3j
k)
'(4i -
j
+ 2k)
(!)t4) +
=
+
(3)(-1)
(1)(2) =
I
•
The dot product of a pair oftwo-dimensional vectors is dermed in a similar fashion:
(u!, U2)' (v!, V2) = U1 V1
+ U2V2.
We will see throughout the remainder of the book that the dot product is a key tool for many important geometric and physical calculations in space (and the plane), not just for rmding the angle between two vectors. Proof of Theorem 1 Applying the law of cosines (Equation (8), Section 1.3) to the triangle in Figure 12.21, we find that
Iwl 2 = lul 2 21ullvl cosO = lul 2
+ Ivl 2 - 21ullvl cosO +
Law of cosines
Ivl 2 - Iw1 2 .
Becausew = u - v,thecomponentformofwis(U, - V!,U2 - V2,U3 - V3).SO lul 2 = (VU12 + ui + ul)' = U,' + ui + ul Ivl 2 = (Vv,' + vi + vl)' = v,' + vi + vi Iwl 2 = (V(U1 - v1l' + (U2 - v2l' + (U3 - V3)2)'
FIGURE 12.21 The parallelogram law of addition of vectors gives w = u - v.
=
(U1 - v,)2
+ (U2 -
vol' + (U3
- V3)2
and Therefore, 21ullvl cosO = lul 2 + Ivl 2 - Iwl 2 = 2(U1V1 lullvl cos 0 = U1 V1 UtVt
cosO = Since 0
:5
0
<
?T,
+ U2V2 + U3V3)
+ U2V2 + U3V3 +
U2'V2
+
U3V3
lullvl
we have V U2 V2 + U3 V3) 0 -- cos-1 (U1 1 +lullvl .
In the notation of the dot product, the angle between two vectors u and v is
o=
1 cos- C:I'I:I)'
•
676
Chapter 12: Vectors and the Geometry of Space
EXAMPLE 2 Solution
Find the angle between u = i - 2j - 2k and v = 6i
+ 3j + 2k.
We use the fonnula above:
+
u'v = (1)(6)
lui = VO)' + Ivl = V(6)' () =
(-2)(3)
+ (-2)(2) =
(-2)' + (-2)' =
1
v9 = 3
v49 = 7
+ (3)2 + (2)2 =
cos- C:I'I:I) = cos-
6 - 6 - 4 = -4
1
(3)~») '" 1.76radians.
-
The angle fonnula applies to two-dimensional vectors as well. y B(3,5)
EXAMPLE 3 A
Find the angle () in the triangle ABC detennined by the vertices
= (0, O),B = (3,5), and C = (5,2) (Figure 12.22). ~
~
Solution The angle () is the angle between the vectors CA and CB. The component fonns of these two vectors are
eA
=
(-5, -2)
and
cE
=
(-2,3).
First we calculate the dot product and magnitudes ofthese two vectors. FIGURE 12.22 The triangle in Example 3.
eA'cE =
(-5)(-2)
+ (-2)(3)
=
4
+ (-2)' = v29 V(-2)' + (3)' = '\Il3
leAl = V(-5)' IcEl =
Then applying the angle formula, we have
'" 78.1
0
or
1.36 radians.
-
Perpendicular (Orthogonal) Vectors Two nonzero vectors u and v are perpendicular or orthogonal if the angle between them is For such vectors, we have u' v = 0 because cos (1f/2) = O. The converse is also true. If u and v are nonzero vectors with u' v = lullvl cos () = 0, then cos () = 0 and () = cos- 1 0 = 1f/2.
1f/2.
DEFINmON
Vectors u and v are orthogonal (or perpendieular) if and ouly
ifu'v = O.
EXAMPLE 4
To determine if two vectors are orthogonal, caleulate their dot product.
+ (-2)(6) = O. (b) u = 3i - 2j + k and v = 2j + 4k are orthogonal because u' v = (3)(0) + (-2)(2) + (1)(4) = O. (a) u = (3, -2) and v = (4,6) are orthogonal because u' v = (3)(4)
12.3
The Dotproduct
677
(c) 0 is orthogonal to every vector u since
o· u
=
(0,0, O) • (UI. U2, U3)
= (O)(Ul)
+
(0)(U2)
+ (0)(U3)
=0.
•
Dot Product Properties and Vector Projections The dot product obeys many of the laws that hold for ordinary products of real numbers (scalars).
Properties of the Dot Product Ifu, v, and w are any vectors and c is a scalar, then 1. U'v = v'U 2. (cu)'v = u'(cv) = c(u'v) 3. u'(v + w) = U'v + U'w 4. u·u = lul 2 5.0·u=0.
Carl Friedrich Gauss
Proofs of Properties 1 and 3 The properties are easy to prove using the detmition. For instance, here are the proofs of Properties 1 and 3.
(1777-1855)
1.
u·v =
3.
u· (v + w)
HIsTORICAL BIOGRAPHY
UIVI
+ U2V2 + U3V3
=
= (UI. U2, U3)' (VI = Ul(Vl
VIUl
+
+ V2U2 +
WI. V2
V3U3
= v·u
+ W2, V3 + W3)
+ WI) + U2(V2 + W2) + U3(V3 + W3)
Q
/1.,
We now return to the problem of projecting one vector onto another, posed in the opening to this section. The ~ctor projection of u = PQ onto a nonzero vector v = PS (Figure 12.23) is the vector PR determined by dropping a perpendicular from Q to the line PS. The notation for this vector is
• R
FIGURE 12.23 u onto v.
•
=u·v+u·w
s
P
proj. u
The vector projection of
/
("the vector projection ofu onto v').
If u represents a force, then proj. u represents the effective force in the direction of v (Figure 12.24). If the angle Ii between u and v is acute, proj. u has length Iu I cos Ii and direction Iv I (Figure 12.25). If Ii is obtuse, cos Ii < 0 and proj. u has length -I u I cos Ii and diIvl. In both cases, rection
vi
-vi
proj.u = (lui cos Ii) v
proj,.u
FIGURE 12.24 Ifwe pull on the box with force u, the effective force moving the box forward in the direction v is the projection ofo ontov.
1:1
(u.v) v (u.v) Ivl
Ivl 2
Ivl
V.
lui cos 8 ~
1.11.1 cos 8 1'1
.0' 1'1
678
Chapter 12: Vectors and the Geometry of Space
v
v
Length ~ lui cos 9
Leugth ~ -lui cos 9 (b)
(aJ
FIGURE 12.25 The length ofprojyu is (a) lui cos 8 ifcos 8 '" (b)-lui cos 9 if cos 8 < O.
oand
The number Iu I cos II is called the scalar component of u in the direction of v (or of u onto v). To summarize,
The vector projection of u onto v is the vector (u.v) . prOJyU = Ivl 2 v.
(1)
The scalar component of u in the direction of v is the scalar
u·v
lui cos II = lVT =
v
(2)
UOT;!"
Note that both the vector projection of u onto v and the scalar component of u onto v depend only on the direction of the vector v and not its length (because we dot u with v/ Iv I, which is the direction of v).
+ 3j + 2k onto v
EXAMPLE 5 Find the vector projection of u = 61 and the scalar component ofu in the direction ofv. Solution
=
I - 2j - 2k
We find projy u from Equation (I):
.
uov
6-6-4.
proJy u = v ° v v = 1 =
+ 4 + 4 (.
.
- 2) - 2k)
_1(1 - 2j - 2k) = _1 1 + ~j + ~k 9999'
We f"md the scalar component of u in the direction ofv from Equation (2):
lui cos II
= UOI;I =
=
2 - 2 -
(61 + 3j + 2k)·
(t 1 - t j - tk)
4 34 = -3'
•
Equations (I) and (2) also apply to two-dimensional vectors. We demonstrate this in the next example.
EXAMPLE 6 Find the vector projection of a force F = 51 the scalar component ofF in the direction ofv.
+ 2j
onto v = I - 3j and
12.3 The Dotproduct
Solution
679
The vector projection is projv F = =
(~;I~)v 5 - 6 (i - 3j) 1+9
= -
I~ i +
= -
~ (i - 3j) 10
?oj·
The scalar component ofF in the direction ofv is IFI cosO
F'v
5 - 6
=-1-1 =.~ v vi + 9
I
•
---
v'W'
A routine calculation (see Exercise 29) verifies that the vector u - projv u is orthogonal to the projection vector projv u (which has the same direction as v). So the equation . . u ) = (u.v) u = proJvU + (u - proJv Ivl 2 V + ( u - (u.v)) Ivl 2 V Parallel to v
Orthogonal to v
expresses u as a sum of orthogonal vectors.
Work
1----1 I
: I I I
I
p
I
D
In Chapter 6, we calculated the work done by a constant force of magnitude F in moving an object through a distance d as W = Fd. That formula holds only ifthe force is directed along the lioe of motion. If a force F moving an object through a displacement D = has some other direction, the work is performed by the component of F in the direction of D. If 0 is the angle between F and D (Figure 12.26), then
PQ
Q
IFI cos 6
component OfF)(1 gth f ) Work = (scalar in the direction of D en 0 D
FIGURE 12.26 The worl< done by a constant force F during a displacement D is ( IF I cos 8) ID I, which is the dot product
(lFI cosO)IDI F·D.
F·D.
DEFINmON
The work done by a constant force F acting through a displace-
mentD = PQis
W= F·D.
EXAMPLE 7 IflFI = 40N(newtons), IDI F in acting from P to Q is
= 3m,andO = 60°, the work done by
Work = F'D
Ocrmition
= IFIIDI cosO = (40)(3) cos 60° = (120)(1/2) = 60 I (joules).
Given values
•
We encounter more challenging work problems in Chapter 16 when we learn to rmd the work done by a variable force along a path in space.
680
Chapter 12: Vectors and the Geometry of Space
Exercises 12.3 Dot Product and Projections In Exercises 1-8, fmd
a. Showtllat cos a =
a. v' u, lvi, lui
a Fr'
b
cosf:l =
and cos2 a
c. the scalar component ofn in the direction ofv
the direction cosines of v.
2. v = (3/5)1 + (4/5)k,
u = 51 + 12j
3. v = 101 + 11j - 2k,
u = 3j + 4k
4. v
~
21 + 10j - 11k,
5. v = 5j - 3k, 6.v=-I+j, 7. v ~ 51 & v
=
u
~
direction cosines ofv.
16. Water main construction A water main is to be constructed wi1h a 20% grade in 1he nur1h direction and a 10% grade in 1he east direction. Determine 1he angle 8 required in 1he water main for 1he turn from nur1h to east.
21 + 2j + k
u = I+ j + k u=v2I+V3j+2k
+ j, u ~
2i
(:n, VJ)'
+ u
I
Wj
=
c
Fr'
b. Unit vectors are built from direction cosines Show tllat if v = ai + bj + ck is a unit vector, then a, b, and c are the
d. 1he vector proj, u. u = -21 + 4j - VSk
cos 1 =
+ cos2 fj + cos2 'Y = 1. These cosines are called
b. 1he cosine of1he angle between v and u
1. v = 21 - 4j + VSk,
Fr'
I
I I
8
~-------~-----j
(:n,- VJ)
/
East
//
///~,
Angle Between Vecton
D Find 1he angles betweeo 1he vectors in Exercises !l-12 to 1he nearest hundredtll of a radian. 9. u=21+j,
Theory and Examples 17. Sums and differences In 1he accompanying figure, it looks as ifvl + V2 and VI - V2 are orthogonal. Is this mere coincidence, or are 1here circumstances under which we may expect 1he sura of two vectors to be ur1hogona1 to 1heir difference? Give reasons for
v=I+2j-k
10. u = 21 - 2j + k, 11. u = V31 - 7j,
v = 31 + 4k v = V31 + j - 2k
12. u=I+v2j-v2k,
your answer.
v=-I+j+k
13. Triangle Find 1he measures of 1he angles of 1he mangle whose vertices are A = (-1,0), B = (2, I), and C = (I, -2). 14. Rectangle Find 1he measures of 1he angles between 1he diagonals of1he rectangle whose vertices are A ~ (I, O),B ~ (0,3), C = (3,4), andD = (4, I).
15. Direction angles and direction cosines The direction angles a, f3, and 'Y of a vector v = ai + bj + ck are dermed as follows: a is the angle between v and the positive x-axis (0 :s; a :s; '11') f:l is 1he angle betweeo v and 1he positive y-axis (0 '" f:l '" 'If)
1 Is 1he angle between v and 1he positive z-axis (0 '" 1 '" 'If). 18. Orthogonality on a circle Suppose 1ha! AB is 1he diameter of a circle with center 0 and that C is a p-oint on one of the two arcs joiningA andB. Show tllat and CB are ur1hogona1.
cA
A ~-_-:u'---~O'-------;o,---+lB
y
19. Diagonals of a rhombus Show tllat 1he diagonals of a rhombus (parallelogram wi1h sides of equalleng1h) are perpendicular.
12.3 20. Perpendicular diagonals Show that squares arc the only rectangles with perpendicular diagooals. 21. When parallelograms are rectangles Prove that a parallelogram is a rectangle if and only if its diagooals are equal in leogth. (This fact is often exploited by carpeoters.) 22. Diagonal of parallelogram Show that the indicated diagooal of the parallelogram determined by vectors u and v hisects the angle hetweenuandviflnl ~ Ivl.
The Dot Product
681
Equations for Lines in the Plane 31. Line perpendicular to a vector Show that v = ai + bj is perpeodicular to the line ax + by ~ c by establishing that the slope
of the vector v is the negative reciprocal of the slope of the given line. 32. Line parallel to a vector Show that the vector v ~ ai + bj is parallel to the line bx - ay = c by establishing that the slope of the line segmeot represeoting v is the same as the slope of the giveo line. In Exercises 33-36, use the result of Exercise 31 to fmd an equation for the line 1hrough P perpendicular to v. Theo sketch the line. Include v in your sketch as a vector starting at the origin.
23. Projectile motion A guo with muzzle velocity of 1200 ft/sec is fired at an angle of 8' ahove the horizontal. Find the horizontal and vertical componeots of the velocity. 24. Indined plane Suppose that a box is being towed up an inclined plane as shown in the figure. Find the force w needed to make the componeot of the force parallel to the inclined plane equal to 2.5 lb.
~
25. a. Canchy-Schwartz inequality Since u' v ~ Iu II v I cos 9, show that the inequality Iu • v I '" Iu II v I holds for any vectors u and v. h. Under what circumstances, if any, does Iu • v I equal Iu II v I? Give reasons for your answer. 26. Copy the axes and vector shown here. Theo shade in the points (x,y) for which (xi + yj) • v '" O. JustifY your answer.
33.P(2,1),
v=i+2j
35. P( -2, -7),
v
~
-2i + j
34.P(-1,2),
v=-2!-j
36. P(ll, 10),
v
~
2! - 3j
In Exercises 37-40, use the result of Exercise 32 to fmd an equation for the line through P parallel to v. Theo sketch the line. Include v in your sketch as a vector starting at the origin.
37. P(-2, I), 39. P(I,2),
v
~
38. P(O, -2),
i - j
40. P(I, 3),
v = -i - 2j
v
~
2i + 3j
v = 3i - 2j
Work 41. Work along a line Find the work done by a force F = 5i (magnitnde 5 N) in moving an object along the line from the otigin to the point (I, I) (distance in meters).
42. Locomotive The Union Pacific's Big Boy locomotive could pull 6000-ton 1rains with a tractive effort (Pull) of 602,148 N (135,375 Ib). At this level of effort, about how much work did Big Bay do on the (approximately straight) 605-kro journey from
San Francisco to Los Angeles?
43. Indined plane How much work does it take to slide a crate 20 m along a loading dock by pulling on it with a 200 N force at an angle of30' from the horizontal? 44. Sailboat The wind passing over a boat's sail exerted a 1000-lb magnitude force F as shown here. How much work did the wind perform in moving the boat forward I mi? Answer in foot-pounds.
y
60'
27. Orthogonal unit vectors If 0.1 and tors and v = aUt + hU2,f'mdv·ut.
U2
F
are orthogonal unit vec-
28. CanceDation in dot products In real-numher multiplication, if U'VI = Ul/2 and u =F 0, we can cancel the u and conclude that v, = v,. Does the same rule hold for the dot product? That is, if U'VI = U'V2and u =F O,canyouconcludethatvt = v2?Give
Angles Between Lines In the Plane The acute angle between intenecting lines that do not cross at right angles is the same as the angle determined by vectors nonnal to the lines or by the vectors parallel to the lines.
reasons for your answer.
L,
29. Using the definition of the projection ofu onto v, show by direct calculation that (u - proj. u) •proj. u = O. 30. A force F = 2i + j - 3k is applied to a spacecraft with velocity vector v = 3i - j. Express F as a sum of a vectur parallel to v and a vector orthogooal to v.
L, 9
682
Chapter 12: Vectors and the Geometry of Space
Use this fact and the results of Exercise 31 or 32 to f"md the acute angles betweeo the lines in Exercises 45-50. 45. 3x+y~5, 2x-y~4 46. y=
V3x-l,
12.4
y= -V3x+2
47.
V3x - y = -2, x - V3y = I x + V3y = 1, (1 - V3)x + (I
48. 49. 3x-4y=3, x-y=7 SO. 12x
+ 5y
= I,
+ V3)y
= 8
2x - 2y = 3
I_T_h_e_c_ro_ss_p_ro_d_uct
_
In studying lines in the plane, when we needed to describe how a line was tilting, we used the notions of slope and angle of inclination. In space, we want a way to describe how a plane is tilting. We accomplish this by multiplying two vectors in the plane together to get a third vector perpendicular to the plane. The direction of this third vector tells us the ''inclination'' ofthe plane. The product we use to multiply the vectors together is the vector or cross product, the second ofthe two vector multiplication methods. We study the cross product in this section.
The Cross Product of Two Vectors in Space We start with two nonzero vectors u and v in space. Ifu and v are not parallel, they determine a plane. We select a unit vector n perpendicular to the plane by the right-hand rule. TIris means that we choose n to be the unit (normal) vector that points the way your right thumb points when your f'mgers curl through the angle 0 from u to v (Figure 12.27). Then the cross product u X v ("u cross v') is the vector defined as follows.
DEFINmON u X v = (IullvlsinO) n FIGURE 12.27 The construction of
u X v. Unlike the dot product, the cross product is a vector. For this reason it's also called the vector product of u and v, and applies only to vectors in space. The vector u X v is orthogonal to both u and v because it is a scalar multiple of n. There is a straightforward way to calculate the cross product of two vectors from their components. The method does not require that we know the angle between them (as suggested by the definition), but we postpone that calculation momentarily so we can focus first on the properties of the cross product. Since the sines of 0 and 1f are both zero, it makes sense to define the cross product of two parallel nonzero vectors to be O. If one or both of u and v are zero, we also define u X v to be zero. TIris way, the cross product of two vectors u and v is zero ifand only ifu and v are parallel or one or both of them are zero. Parallel Vectors Nonzero vectors u and v are parallel if and only if u X v = O. The cross product obeys the following laws. Properties of the Cross Product If u, v, and w are any vectors and r, s are scalars, then
+
+u Xw +w Xu
1. (ru) X (sv) = (rs)(u X v)
2. u X (v
3. v X u = -(u X v)
4. (v
5.0Xu=O
6. UX(VXw) = (u'w)v-(u'v)w
+ w)
w) = u X v
X u = v X u
12.4
FIGURE 12.28 The construction of v Xu.
The (ross Product
683
To visualize Property 3, for example, notice that when the fingers of your right hand curl through the angle 0 from v to u, your thumb points the opposite way; the unit vector we choose in forming v X u is the negative of the one we choose in forming u X v (Figure 12.28). Property I can be verified by applying the definition of cross product to both sides of the equation and comparing the results. Property 2 is proved in Appendix 8. Property 4 follows by multiplying both sides of the equation in Property 2 by -I and reversing the order of the products using Property 3. Property 5 is a def"mition. As a rule, cross product multiplication is not associative so (u X v) X w does not generally equal u X (v X w). (See Additioual Exercise 17.) When we apply the definition to calculate the pairwise cross products of i, j, and k, we find (Figure 12.29)
i Xj =
-0
X i) = k
j X k = -(k X j) = i
k X i = -(I X k) = j
Diagram for recalliug
these products
and I X I = j X j = k X k = O.
IU X V I Is the Area of a Parallelogram Because n is a unit vector, the magnitode of u X v is
FIGURE 12.29 The pairwise cross products ofi, j, and k
lu X
Area = base' height ~ 01 ~
lui' Ivllsm luxvl
\
u
FIGURE 12.30 The parallelogram detennined by u and v.
vi
lullvl sinO.
lullvl IsinOllnl
This is the area of the parallelogram determined by u and v (Figure 12.30), Iu I being the base of the parallelogram and Iv II sin 0 I the height.
Determinant Formula for u X v Our next objective is to calculate u X v from the components of u and v relative to a Cartesian coordinate system. Suppose that u = Uli
+
u2j
+
and
U3k
v = VIi
+
v2j
+
v3k.
Then the distributive laws and the rules for multiplying I, j, and k tell us that u X v = (Ul1
+
u2j
= Ul VIi X I
+
+
+ Ul V2i
u2v!i X I
+ U3Vlk
U3k) X (VII
X I
+
X j
= (U2V3 - U3V2)1 -
v2j
+ Ul V31
u2v2j X j
+ U3V2k
+
X j
(UIV3 -
+
+
V3k)
X k
U2V3 j X k
+ U3V3k X k U3Vl)j + (UIV2
-
U2 Vl)k.
The component terms in the last line are hard to remember, but they are the same as the terms in the expansion of the symbolic determinant
j
k
684
Chapter 12: Vectors and the Geometry of Space
I Determinant. 2 X 2 and 3 X 3 determinants are evaluated as follows:
I:
~I
= ad - bc
So we restate the calculation in this easy-to-remember form.
Calculating the Cro•• Product as a Determinant Ifu = uji + u2j + U3k and v = vii + v2j + V3k, then
EXAMPLE
I
I_~ ~ ~ (2)(3) -
u X v = (1)( -4)
i
j
k
Ul
U2
"3
VI
V2
V3
=6+4=10
a, b,
a, a, b, b,
C, c, C, - a,
=
EXAMPLE 1
b'c,l
a1 1b> c,
Find u X v and v X u ifu = 2i + j + k and v = -4i + 3j + k.
Solution
IbC, b'l + a,Ib1 b'c,l 1
C3
uXv=
Ct
EXAMPLE -5 3 1
2 -4
1
3
~
-5(1 - 3) - 3(2 + 1(6 + 4)
~I
+ 4)
= 10 - 18 + 10 = 2 (For more information, see the Web site at www.aw.comlthomas.)
3
•
PR
Solution The vector PQ X is perpendicular to the plane because it is perpendicular to both vectors. In terms of components,
PQ =
(2 - I)i + (I + I)j + (-I - O)k = i + 2j - k
PR =
(-I - I)i + (I + I)j + (2 - O)k = -2i + 2j + 2k
R(-I,I,2)
":/:.
11
=
EXAMPLE 2 Find a vector perpendicular to the plane of P(I, -I, 0), Q(2, I, - I), and R( -I, 1,2) (Figure 12.31).
PQXPR=
z
0' ,
1 I
v X u = -(u X v) = 2i + 6j - 10k
1
- (3)1_~ :1 + (I)I-~
k
= -2i - 6j + 10k
1 1 = (-5) 11 11
3
2 -4
.
1J 3
,
:i'-
FIGURE 12.31 The vector PQ X PR is perpendicu1llr to the plane oftriangle PQR (Example 2). The area oftriangle PQR is ba1fof IPQ X PRI (Example 3).
2 =
1
2
-~Ii-I_~ -~Ij+ I_~ ~Ik •
=6i+6k.
EXAMPLE 3 Find the area of the triangle with vertices P(I, -1,0), Q(2, I, -I), and R( -I, 1,2) (Figure 12.31). Solution
"."..,
~ -~2 2
I -2
The area ofthe parallelogram determined by P, Q, andR is
IPQ X PRI
= 16i + 6kl = V(6)' + (6)' =
Values from Example 2
v'2-36 =
6V2.
The triangle's area is half ofthis, or 3V2.
EXAMPLE 4
•
Find a unit vector perpendicular to the plane ofP(I, -1,0), Q(2, I, -I),
andR(-I,I,2).
PR
Solution Since PQ X is perpendicular to the plane, its direction n is a unit vector perpendicular to the plane. Taking values from Examples 2 and 3, we have ~
~
PQ X PR 6i + 6k I. I n= IPQ X PRI = 6V2 = V2' + V2 k .
•
12.4
The Cross Product
685
For ease in calculating the cross product using determinants, we usually write vectors in the form v == VI i + V2j + V3 k rather than as ordered triples v == (VI, V2, V3) .
Torque When we tum a bolt by applying a force F to a wrench (Figure 12.32), we produce a torque that causes the bolt to rotate. The torque vector points in the direction ofthe axis of the bolt according to the right-hand rule (so the rotation is counterclockwise when viewed from the tip of the vector). The magnitude of the torque depends on how far out on the wrench the force is applied and on how much of the force is perpendicular to the wrench at the point of application. The number we use to measure the torque's magnitude is the product of the length of the lever arm r and the scalar component of F perpendicular to r. In the notation of Figure 12.32, Component of F perpendicular to r. Its length is IFI sin ~
Magnitude of torque vector
Ir II F I sin 8,
==
or Ir X F I. If we let D be a unit vector along the axis of the bolt in the direction of the torque, then a complete description of the torque vector is r X F, or Torque vector
(I r II F I sin 8)
==
D.
FIGURE 12.32 The torque vector describes the tendency of the force F to drive the bolt forward.
Recall that we defined u X v to be 0 when u and v are parallel. This is consistent with the torque interpretation as well. If the force F in Figure 12.32 is parallel to the wrench, meaning that we are trying to tum the bolt by pushing or pulling along the line of the wrench's handle, the torque produced is zero.
3 ft bar p L.-------r-~"Q
Figure 12.33 is
EXAMPLE 5
20lb magnitude force
The magnitude of the torque generated by force F at the pivot point P in IPQ X FI
F
FIGURE 12.33 The magnitude of the torque exerted by F at P is about 56.4 ft-Ib (Example 5). The bar rotates counterclockwise around P.
==
IPQIIFI sin 70°
~
(3)(20)(0.94)
~
56.4ft-lb.
In this example the torque vector is pointing out of the page toward you.
•
Triple Scalar or Box Product The product (u X v)· w is called the triple scalar product ofu, v, and w (in that order). As you can see from the formula I(u X v)·wl
==
lu X vllwllcos81,
the absolute value of this product is the volume of the parallelepiped (parallelogram-sided box) determined by u, v, and w (Figure 12.34). The number Iu X v I is the area of the base uXv
v
Area of base ----=Iuxvl
u
Volume = area of base . height = lu X vi IwlIcos 01 = I(u X v)· wi
FIGURE 12.34
The number I (u X v)· wi is the volume ofa parallelepiped.
686
Chapter 12: Vectors and the Geometry of Space parallelogram. The number Iw II cos () I is the parallelepiped's height. Because of this geometry, (u X v)· w is also called the hox product ofu, v, and w. By treating the planes of v and w and of w and u as the base planes of the parallelepiped detennined by u, v, and w, we see that
(11 X v)· w
= (v X w)· u = (w X u)· v.
Since the dot product is commutative, we also have 1
The dot and cross may be interchaoged in a 1riple scalar product without altering its value.
(11 X v)·w
= Il'(v X w).
The triple scalar product can be evaluated as a determinant:
(11 X v)·w
:1
= [I::
1 -I::
Ul
U2
U3
VI
V2
V3
WI
W2
W3
::Ij
+
I::
::Ik]-w
Calculating the Triple Scalar Product as a Dcterminant
(u
EXAMPLE 6 v = -21 Solution
X
v)· w =
Ul
U2
U3
VI
V2
V3
Find the volume ofthe box (parallelepiped) determined by u = 1 + 2j - k, = 7j - 4k.
+ 3k,andw
Using the rule for calculating determinants, we rmd
(u X v), w
=
I -2
2 0
-I 3 = -23.
o
7
-4
The volume is I(u X v)· wi = 23 units cubed.
•
Exercises 12.4 7. u = -81 - 2j - 4k, v = 21 + 2j + k
Cross Product Call:ulations In Exercises 1-8, rmd the length and direction (when dermed) of
uXvandvXu. 1. u=21-2j -k, v=l-k 2. u = 21 3.
+ 3j, v
= -I
u~21-2j+4k,
+j
v~-I+j-2k
4.u=l+j-k, v=O S. u
=
6. u
~
v = -3j 1 X j, v ~ j X k 2i,
8. u = tl -
t j + k,
v = 1 + j + 2k
In Exercises 9-14, sketch the coordinate axes and then include the
vectors u, v, and u X vas vectors starting at the origin.
9.u=i, v=j
10. u = 1 - k, v = j
11.u=I-k, v=j+k
12. u = 2i - j, v = 1 + 2j
13. u=i+j,
14. u
v=i-j
~
j + 2k,
v
~
1
12.4 The (ross Product Triangles in Space In Exercises 15-18, a. Find the area of the triangle detennined by the points P, Q, andR. b. Find a unit vector perpendicular to plane PQR. 15. P(I, -1,2), Q(2,0, -I), R(O, 2, I) 16. P(I, I, I),
Q(2, 1,3), R(3, -I, I)
c. (-u) X v
d. (en)'v
~
~
-(u X v)
u'(ev)
~
c(n'v)
(anynurnbere)
e. c(u X v) = (eu) X v = u X (ev) f.
U'u
= lul
687
(anynurnberc)
2
g. (uXu)'u=O h. (uXv)'u=v'(uXv)
29. Given nonzero vectors n, v, and lV, use dot product and cross product notatioo, as appropriate, to describe the followiog.
17. P(2,-2,1),
Q(3,-1,2), R(3,-I, I)
18. P( -2,2,0),
Q(O, I, -I), R( -1,2, -2)
a. The vector projection of u onto v
b. A vector orthogonal to u and v
Triple Scalar Products In Exercises 1!l-22, verifY that (u X v)· w = (v X w)· u = (w X u)· v and fmd the volume of the parallelepiped (box) determined by n, v, and w. w
2j
2k
20.i-j+k
2i+j-2k
-i
21.2i+j
2i-j+k
i
22.i+j-2k
-i - k
2i+4j-2k
19. 2i
+ 2j
30. Compute (i X j) X j and i X (j X j). What can you conclude about the associativity of the cross product? - k
+ 2k
Theory and Examples 23. Parallel and perpendicular vectoR l.et u = Si - j + k, v = j - 5k,w = -lSi + 3j - 3k.Whichveetors,ifany, are (a) perpendicular'l (b) Parallel? Give reasons for your answers. 24. Parallel and perpendicular vectOR I.ct u = i + 2j - k, v = -i + j + k, w = i + k, r = -("./2)i - ".j + ("./2)k. Which vectors, if any, are
on the bolt atP if pounds.
IPQI
= 8in. and
25.
IFI
= 301b. Answer in foot-
26.
'ii'> ~
p • p
e. A vector orthogonal to u X v and u X w
f. A vector oflength Iu I in the direction of v
v
u
c. A vector orthogonal to u X v and w
d. The volwue of the parallelepiped detennined by u, v, and w
27. Which of the following are always true, and which are not always true? Give reasons for your answers.
31. Let u, v, and w be vectors. Which of the followiog make seose, and which do not? Give reasons for your answers. a. (uXv)'w b. u X (v'w)
c. u X (v X w) d. u'(v'w) 32. Cross products of three vectOR Show that except in degenerate cases, (u X v) X w lies in the plane of u and v, whereas u X (v X w) lies in the plane ofv and w. What are the degenerate cases? 33. Cancellation in cross products If u X v = u X w and u =F 0, then does v = w? Give reasons for your answer. 34. Double cancellation If u ~ 0 and if u X v = u X w and U'v = U· lV, then does v = w? Give reasons for your answer.
Area of a Parallelogram Find the areas of the parallelograms whose vertices are given in Exercises 35-40. 35. A(I,O), B(O, I), C(-1,0), D(O, -I) 36. A(O,O),
B(7,3),
C(9, 8),
37. A( -1,2),
B(2,0),
38. A( -6,0),
B(l, -4),
39. A(O, 0, 0),
B(3, 2, 4),
D(2,5)
C(7, I),
D(4,3)
C(3, I),
D( -4,5)
C(5, 1,4),
D(2, -1,0)
a. lui =~
40. A(I, 0, -I),
b. u'n ~ lui
e.uXv=vXn
Area of a Triangle Find the areas of the triangles whose vertices are given in Exercises 41-47. 41. A(O,O), B( -2,3), C(3, I)
f. u X (v + w) = u X v + u X w
42. A( -I, -I),
c.uXO=Oxu=O d. u X (-n) ~ 0
g.
(uXv),v~O
h. (uXv)'w=u'(vXw) 28. Which of the following are always true, and which are not always true? Give reasons for your answers. a.u·v~v·n
b.uXv~-(vXu)
B(l, 7, 2),
B(3,3),
C(2,4, -I),
C(2, I)
43. A( -5,3),
B(I, -2),
44. A( -6,0),
B(IO, -5),
45. A(I, 0, 0),
B(0,2,0),
46. A(O, 0, 0),
B( -I, I, -I),
47. A(I, -I, I),
B(O, I, I),
C(6, -2) C( -2,4) C(O,O,-I) C(3, 0, 3)
C(I,O, -I)
D(O, 3, 2)
688
Chapter 12: Vectors and the Geometry of Space
48. Find the volume of a parallelepiped if four of its eight vertices are A(O, 0, 0), B(l, 2, 0), C(O, -3,2), and D(3, -4, 5).
50. Triangle area Find a concise formula for the area of a triangle in the xy-plane with vertices (al. az), (bl. 1>,.), and (c" cz).
49. Triangle area Find a formula for the area of the triangle in the xy-plane with vertices at (0,0), (ai, az), and (bl. bz ). Explain
your work.
Lines and Planes in Space
12.5
This section shows how to use scalar and vector products to write equations for lines, line segments, and planes in space. We will use these representations throughout the rest ofthe book.
z
Lines and Line Segments in Space
i Po(xo. Yo. zo)
~ __-:--4~"=~ • I
L
/
..
-
P(x,y,z)
v
o y
In the plane, a line is detennined by a point and a number giving the slope of the line. In space a line is detennined by a point and a veclor giving the direction of the line. Suppose that L is a line in space passing through a point Po(xo, Yo, zo) paralle.!10 a vector v = v,i + v2j + V3k. Then !:...is the set of all points P(x, y, z) for which PoP is parallel to v (Figure 12.35). Thus, PoP = tv for some scalar parameter I. The value of I depends on the location of the point P along the line, and the domain of I is (- 00, 00). The expanded form ofthe equation P--;P = tv is
(x - xo)i + (y - Yo)j + (z - zo)k FIGURE 12.35 A point P lies on L through Po parallel to v if and only if P-;P is a scalar multiple ofv.
= t(v,i
+ V2j + V3k),
which can be rewritten as
xi
+ yj + zk
= xoi
+ yoj + zok + I(vli + v2j + V3k).
(1)
Ifr(l) is the position vector ofa pointP(x,y, z) on the line and ro is the position vector of the point Po(xo, Yo, zo), then Equation (I) gives the following vector form for the equation of a line in space.
Vector Equation for a line A vector equation for the line L through Po(XO, Yo, zo) parallel to v is r(l) = ro
+ tv,
-00 < 1< 00,
(2)
where r is the position vector of a point P(x, y, z) on L and ro is the position vector of Po(xo, Yo, zo).
Equating the corresponding components of the two sides of Equation (I) gives three scalar equations involving the parameter I:
x=xo+tvt,
y=YO+ IV2,
These equations give us the standard parametrization of the line for the parameter interval -00<1<00.
Parametric Equations for a Une The standard parametrization of the line through Po(xo, yo, zo) paraDel to v = v,i + V2j + v3kis
x
=
xo + Iv"
Y
=
Yo + IV2, Z = Zo + IV3,
- 00 < I < 00
(3)
12.5
EXAMPLE 1 v = 2i VO(-2, 0, 4) 4 _
+ 4j
Lines and Planes in Space
Find parametric equations for the line through (-2,0,4) parallel to - 2k (Figure 12.36).
Solution With Po(xo, Yo, zo) equal to (-2,0,4) and v,i 2i + 4j - 2k, Equations (3) become
"\;t=O
x = -2
EXAMPLE 2 Q(l, -1,4).
4
Solution
v=2i+4j-2k
+ 21,
y
+ v2j + V3k
•
z = 4 - 21.
= 41,
equal to
Find parametric equations for the line through P( -3,2, -3) and
The vector
PQ = FIGURE 12.36 Selected points and parameter values on the line in Example I. The arrows show the direction of increasing t.
689
(I - (-3))i
= 4i - 3j
+
+ (-I -
2)j
+
(4 - (-3))k
7k
is parallel to the line, and Equations (3) with (xo, Yo, zo) = (-3,2, -3) give
x = -3
+ 41,
y = 2 - 31,
z = -3
+ 71.
We could have chosen Q( I, -I, 4) as the ''base point" and written
x = I
+ 41,
Y = -I - 31,
Z
= 4 + 71.
These equations serve as well as the first; they simply place you at a different point on the • line for a given value of I. Notice that parametrizations are not unique. Not ouly can the ''base poinf' change, but so can the parameter. The equations x = -3 + 41 3,y = 2 - 31 3,andz = -3 + 71 3 also parametrize the line in Example 2. To parametrize a line segment joining two points, we first parametrize the line through the points. We then find the I-values for the endpoints and restrict I to lie in the closed interval bounded by these values. The line equations together with this added restriction parametrize the segment.
EXAMPLE 3 Parametrize the line segment joining the points P( -3,2, -3) and Q(l, -1,4) (Figure 12.37).
Q(I,-I,4) /
t= 1
Solution We begin with equations for the line through P and Q, taking them, in this case, from Example 2:
x = -3
"'t~O
P(-3, 2, -3)
+ 41,
z
y=2-31,
=
-3
+ 71.
We observe that the point
Y
FIGURE 12.37 Example 3 derives a parametrization ofline segmentPQ. The arrow shows the direction ofincreasing t.
(x,y,z) = (-3
+ 41,2
- 31, -3
+ 71)
on the line passes through P( -3,2, -3) at I = 0 and Q(I, -1,4) at I = 1. We add the restriction 0 :5 I :5 I to parametrize the segment:
x = -3
+ 41,
z
y=2-31,
=
-3 + 71,
0:51:51.
•
The vector fonn (Equation (2)) for a line in space is more revealing if we think of a line as the path of a particle starting at position Po(xo, Yo, zo) and moving in the direction ofvector v. Rewriting Equation (2), we have r(l) = ro = ro
/
Initial
position
+
Iv
+ Ilvl I~I'
! \ "-
Time
Speed
(4)
Direction
690
Chapter 12: Vectors and the Geometry of Space
In other words, the position of the particle at time I is its initial position plus its distance moved (speed X time) in the direction Ivl of its straight-line motion.
vi
EXAMPLE 4 A helicopter is to fly directly from a helipad at the origin in the direction of the point (I, I, I) at a speed of 60 ft/sec. What is the position of the helicopter after 10 sec? Solution We place the origin at the starting position (helipad) ofthe helicopter. Then the unit vector u = _I_ i
v3
+ _1_ j + _I_ k v3
v3
gives the flight direction of the helicopter. From Equation (4), the position of the helicopter at any time I is r(l) = ro + I(speed)u = 0 + 1(60)(_I_ i + _1_ j + _1_ k ) v3 v3 v3 = 20v3I(i + j + k). When I = 10 sec, r(lO) = 200v3 (i + j + k) = (200v3, 200v3, 200v3). After 10 sec of flight from the origin toward (1, I, 1), the helicopter is located at the point (200v3, 200v3, 200v3) in space. It has traveled a distance of (60 ft/secHIO sec) = _ 600 ft, which is the length of the vector r(10).
The Distance from a Point to a Line in Space
s
To f"md the distance from a point S to a line that passes througJI a point P parallel to a vec-
PS
tor v, we find the absolute value of the scalar component of in the direction of a vector normal to the line (Figure 12.38). In the n.Q1ation of the figure, the absolute value of the ~ IPS X
scalar component is
FIGURE 12.38 The distance from S to the line through P parallel to v is IPSI sin 8, where 8 is the angle between
Distance from
IPSI sinO, which is
B
Ivl
vi
Point S to B Line Through P Parallel to v
d=
IPS X vi Ivl
(5)
PSandv. EXAMPLE 5
Find the distance from the point S(I, 1,5) to the line
L:
x = 1
+ I,
y = 3 - I,
z = 21.
Solution We see from the equations for L that L passes through P(I, 3, 0) parallel to v=i-j+2k.With
PS =
(I - I)i + (I - 3)j + (5 - O)k = -2j + 5k
691
12.5 Lines and Planes in Space
omd
J
I PS X v = 0 1
-2 -1
k 5 = i 2
+ 5j + 2)[,
Equation (5) gives
d
~ IPS X vi ~ VI + 25 + 4 ~ v':iO ~ vS Ivl VI + 1 + 4 v'6 .
•
An Equation for I Plane in Space
1FIGURE 12.39 The standard equation for a plane in space is defined in terms ofa vector normal to the plane: A point Plies in the plane through Po nonnal to D. if and only ifn' p-;p - O.
A plane in space is determined by knowing a point on the plane and its "tilf' or orientation. This "tilf' is defmed by specifying a vector that is perpendicular or normal to the plane. Suppose that plane M passes through a point Po(xo,Yo. zo) and is normal to the n~ vector D = AI + Bj + CIL Then M is the set of all P~ts p(x, y, z) for which PoP is orthogonal to D (Figure 12.39). Thus, the dot product D· PoP = O. This equation is equivaleJlt to (AI
+ BJ +
Ck)'[(x - xo)1
+ (y
YoH + (z -
-
zo)k]
~
0
or A(x - ",,)
+ B(y -
Yo)
+ C(Z -
Equation for a Plane The plane tbroughPo(XO,Yo, zo) normal to n = Ai
'0)
~
o.
+ Bj + Ckhas
a·PoP = 0 A(x - ",,) + B(y - Yo) + C(z - zo) Ax + By + Cz = D, where D - Axo + 1Jyo + Czo
Vector equation: Component equation: Component equation aimplified:
~
0
EXAMPLE 6 Find an equation for the plane through poe- 3, 0, 7) perpendicular to D=5i+2j-k,
Solution
The component equation is 5(x - (-3))
+ 2(y -
0)
+ (-I)(z
- 7)
~
O.
Simplifying, we obtain
5x
+
15
+ 2y - z + 7 5x
+ 2y -
=
0
z = -22.
•
Notice in Example 6 how the components of D = 51 + 2J - k became the coefficients of oX, y, and z in the equation 5x + 2y - z = -22. The vector D = AI + BJ + CIt. is normal to the plane Ax + By + Cz = D.
692
Chapter 12: Vectors and the Geometry of Space
EXAMPLE 7
Find an equation for the plane throughA(O, 0,1), B(2, 0, 0), and qo, 3, 0).
SoLution We find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equation for the plane. The cross product
---"
---"
AB X AC = 2
o
j 0 3
k -1 = 3i -1
+ 2j + 6k
is normal to the plane. We substitute the components of this vector and the coordinates of A(O, 0, 1) into the component form of the equation to obtain
3(x - 0)
+ 2(y - 0) + 3x
6(z - 1) = 0
+ 2y +
6z
= 6.
-
Lines of Intersection Just as lines are parallel if and only if they have the same direction, two planes are parallel if and only if their normals are parallel, or 01 = k02 for some scalar k. Two planes that are not parallel intersect in a line.
EXAMPLE 8 Find a vector parallel to the line of intersection of the planes 3x - 6y - 2z = 15 and 2x + y - 2z = 5. SoLution The line of intersection of two planes is perpendicular to both planes' normal vectors 01 and 02 (Figure 12.40) and therefore parallel to 01 X 02. Turning this around, 1 X 02 is a vector parallel to the planes' line of intersection. In our case,
°
01
X
Any nonzero scalar multiple of
FIGURE 12.40 How the line of intersection of two planes is related to the planes' normal vectors (Example 8).
j -6 1
3 2
02 =
°
1
X
02
k -2 = 14i
+ 2j + 15k.
-2
-
will do as well.
EXAMPLE 9 Find parametric equations for the line m which the planes 3x - 6y - 2z = 15 and 2x + Y - 2z = 5 intersect. SoLution We fmd a vector parallel to the line and a point on the line and use Equations (3). Example 8 identifies v = 14i + 2j + 15k as a vector parallel to the line. To find a point on the line, we can take any point common to the two planes. Substituting z = 0 in the plane equations and solving for x andy simultaneously identifies one of these points as (3, -1,0). The line is
x = 3
+
14t,
y = -1
+ 2t,
z = 15t.
The choice z = 0 is arbitrary and we could have chosen z = 1 or z = -1 just as well. Or we could have let x = 0 and solved for y and z. The different choices would simply give _ different parametrizations of the same line. Sometimes we want to know where a line and a plane intersect. For example, if we are looking at a flat plate and a line segment passes through it, we may be interested in knowing what portion of the line segment is hidden from our view by the plate. This application is used in computer graphics (Exercise 74).
12.5
EXAMPLE 10
Lines and Planes in Space
693
Find the point where the line
8 x=3+ 21,
Y = -21,
z = 1
+1
intersects the plane 3x + 2y + 6z = 6.
Solution
The point
(~ +
21, -21, 1 + I)
lies in the plane if its coordinates satisfy the equation ofthe plane, that is, if
3(~ +
21) + 2(-21) + 6(1 + I) = 6 8+61-41+6+61=6 81 = -8 1 = -1.
The point of intersection is
(x,y,z)lt=-l =
(~- 2,2,1
- 1) = (f,2,O).
•
The Distance from a Point to a Plane If P is a point on a plane with normal n, then the distance from any point S to the plane is the length ofthe vector projection of pS onto n. That is, the distance fromS to the plane is
d =
Ips· 1:11
(6)
where n = Ai + Bj + Ck is normal to the plane.
EXAMPLE 11
Find the distance fromS(I, 1,3) to the plane 3x + 2y + 6z = 6.
Solution We find a point P in the plane and calco1ate the length ofthe vector projection of pS onto a vector n normal to the plane (Figure 12.41). The coefficients in the equation 3x+2y+6z=6give n=3i+2j+6k.
FIGURE 12.41 The distance from Sto the plane is the length of the vector projection of liS onto n (Example 11).
694
Chapter 12: Vectors and the Geometry of Space
The points on the plane easiest to find from the plane's equation are the intercepts. If we take P to be the y-intercept (0, 3, 0), then
Ps == (1 ==
101
==
- O)i
+ (1 - 3)j + (3 - O)k
+ 3k, Y(3)2 + (2)2 +
i - 2j
(6)2
==
V49 == 7.
The distance from S to the plane is d =
IPs. 1:11
lengthofprojn PS
I(i - 2j + 3k)·
11 - ± + lil 7
7
(t + t
==
7
i
j
+
*
k)
I
12 7 .
•
Angles Between Planes
The angle between two intersecting planes is defined to be the acute angle between their normal vectors (Figure 12.42). EXAMPLE
12
Find the angle between the planes 3x - 6y - 2z
==
15 and
2x+y-2z==5.
The vectors
SoLution
01 ==
3i - 6j - 2k,
02 ==
2i
+j
- 2k
are normals to the planes. The angle between them is FIGURE 12.42
The angle between two planes is obtained from the angle between their normals.
1
C:l\I::I)
() =
cos-
=
cos-
~
1.38 radians.
1
(241)
•
About 79 deg
Exercises 12.5 9. The line through (0, -7, 0) perpendicular to the plane x + 2y + 2z = 13
Lines and Line Segments
Find parametric equations for the lines in Exercises 1-12.
1. The line through the point P(3, -4, -1) parallel to the vector i +j + k
10. The line through (2, 3, 0) perpendicular to the vectors u = i 2j + 3k and v = 3i + 4j + 5k
2. The line through P(I, 2, -1) and Q( -1,0,1)
11. The x-axis
+
12. The z-axis
3. The line through P( -2,0,3) and Q(3, 5, -2)
4. The line through P(I, 2, 0) and Q(I, 1, -1) 5. The line through the origin parallel to the vector 2j
+k
6. The line through the point (3, - 2, 1) parallel to the line x = 1 + 2t, y = 2 - t, z = 3t
Find parametrizations for the line segments joining the points in Exercises 13-20. Draw coordinate axes and sketch each segment, indicating the direction of increasing t for your parametrization.
13. (0, 0, 0),
(1,1,3/2)
7. The line through (1, 1, 1) parallel to the z-axis
15. (1, 0, 0),
(1,1,0)
8. The
17. (0,1,1), (0, -1, 1) 19. (2, 0, 2),
20. (1,0, -1),
(0, 3, 0)
3x
+
line through 7y - 5z = 21
(2, 4, 5)
perpendicular
to
the
plane
(0,2, 0)
14. (0, 0, 0),
(1,0,0)
16. (1,1,0),
(1,1,1)
18. (0, 2, 0),
(3, 0, 0)
12.5 Planes Find equations for the planes in Exercises 21-26.
Angles Find the aog1es between the planes in Exercises 47 and 48.
21. The plane through Pol 0, 2, -I) normal to n = 3i - 2j - k
47. x + Y = I, 2x + Y - 1. = 2 48. 5x + y-z = 10, x -2y + 3z=-1
22. The plane through (I, -1,3) parallel to the plane
D Use a calculator to fmd the acute angles between the planes in Exer-
3x+y+z=7
cises 49-52 to the nearest hundredth of a radian.
23. The plane through (I, I, -I), (2, 0, 2), and (0, -2, I)
49.2x+2y+1.=3,
24. The plane through (2, 4, 5), (I, 5, 7), and (-1,6,8)
+ t,
+ 31,
Y = 1
2x-2y-z=5
SO. x + Y + z = I, z = 0 (thexy-plane) 51. 2x + 2y - z = 3, x + 2y + z = 2
25. The plane through Po(2, 4, 5) perpendicular to the line x = 5
695
Lines and Planes in Space
z = 4t
52. 4y
26. The plane through A(I, -2, I) perpendicular to the vector from the origin to A 27. Find the point ofintersection ofthe lines x = 21 + I, Y = 31 + 2, z = 4t + 3, and x = • + 2, Y = 2s + 4, z = -48 - I, and then fmd the plane detennined by these lines. 28. Find the point of intersection of the lines x = I, Y = -I + 2, z = t + I, and x = 2s + 2, Y = • + 3, z = 5. + 6, and then fmd the plane determined by these lines.
+
3z
= -12,
3x+ 2y+ 6z
=6
Intersecting Lines and Planes In Exercises 53-56, fmd the point in which the line meets the plane.
53.x=I-I, y=31, z=I+I; 2x-y+3z=6 54. x = 2, Y = 3 + 2t, z = -2 - 2t; fix + 3y - 4z = -12 55.x=I+2t,
56. x
=
y=I+5t,
+ 31,
-1
y = -2,
z=3t;
x+y+z=2
z = 51;
2x - 3z = 7
In Exercises 29 and 30, fmd the plane detennined by the intersecting
Find pararnetrizations for the lines in which the planes in Exercises 57--{j0 interseet.
lines. 29. Ll:x
=
58.3x-6y-1.=3,
L2: x
=
57. x + Y + z
30. LI: x =
+ t,
-1
Y
+ t, 1 + 2s, 2
=
1 - 4s, y = I,
L2: x = 1
Y = 3 - 31,
+ s,
< t < 00 -00 < s < 00
-00
z = 2 - 2s;
< 1 < 00 z = -1 + s; -00 < s < 00
z = -2 -
+ S,
Y= 4
1 - t;
Z =
I;
-00
31. Find a plane tbroughPo(2, I, -I) and perpendicular to the line of inletsection of the planes 2x + Y - z = 3, x + 2y + z = 2. 32. Find a plane through the points PI(I, 2, 3), P2(3, 2, I) and petpendicular to the plane 4x - y + 1. = 7.
59. x - 2y
~
I,
x +Y
+ 4z = 2,
60. 5x-2y= 11,
~
2
2x+y-1.=2
x + Y - 1.
=5
4y-5z= -17
Given two lines in space, either they are parallel, or they inletseet, or they are skew (imagine, for example, the flight paths of two planes in the sky). Exercises 61 and 62 each give three lines. In each exercise, determine whether the lines, tsken two at a time, are parallel, intersect, or are skew. If they intersect, find the point of intersectino.
2 - t;
-00
00
Distances
L2:x= I + 4.,y= I + 2s,z= -3 +48;
-00
<. <
00
In Exercises 33-38, find the distance from the point to the line.
L3: x
-00
00
61. LI: x
x = 4t,
33. (0,0,12); 34. (0,0,0);
x
~
5
35. (2, 1,3);
x
~
2
Y = -2t,
+ 3t, + 2t,
36. (2,1,-1);
x~2t,
37.(3,-1,4);
x~4-t,
38. (-1,4,3);
x = 10
Y ~ 5 Y ~ I
y~
+ 4t,
I +2t, Y = -3,
40. (0,0,0), 41. (0, I, I), 42. (2,2,3), 43. (0, -1,0), 44. (1,0, -I),
~
-3 - 5t
z
~
3
L3: x = 5
~
-I
+ 4t, z
y=3s,
+ 2r, y =
~
-2
+ 2r;
z = 31;
z=l+s;
1 - r,
Z
= 8
-00
< 1<
00
-oo<s
+ 3r;
-00
00
Theory and Examples 63. Use Equatinos (3) to generate a parame1rizatino of the line through P(2, -4,7) parallel to VI = 2i - j + 3k. Then generate another parame1rizatino of the line using the point P2( -2, -2, I) and the vector V2 ~ -i + (I/2)j - (3/2)k.
z~-5+3t
z = 4t
64. Use the component form to generate an equation for the plane through P,(4, 1,5) normal to nl = i - 2j + k. Then generate
+ 2y + 1. = 13 3x + 2y + 6z ~ 6 4y + 3z = -12 2x + Y + 1. = 4 2x + Y + 1. = 4 -4x + Y + z ~ 4 x
45. Find the distance from the plane x x + 2y + 6z = 10.
Y
+ 2r, y = 2 + T, Z = + 21, Y = -I - I,
z
In Exercises 39-44, find the distance from the point to the plane.
39. (2, -3,4),
3
+ 2t,
L2:x=2-s,
z~2t
y~3+2t,
=
3
62. LI: x = I
z = 2t
+ 4t, + 6t,
~
another equatino for the same plane using the point P2(3, -2, 0) and the normal vector n2 ~ - Y2i + 2 V2j - "\12k.
65. Find the points in which the line x ~ I + 2t, Y ~ -I - t, z = 3t meets the coordinste planes. Describe the reasnoing behind your answer.
+ 2y + 6z
46. Find the distance from the line x = 2 z ~ -(1/2) - (I/2)ttotheplanex + 2y + 6z
~
I to the plane
+ t, Y ~
10.
= I
+ t,
66. Find equations for the line in the plane z ~ 3 thst makes an angle of"./6rad with i and an aog1e of"./3 rad withj. Describe the reasnoiog behind your answer. 67. Is the line x 2x + Y - z
I - 2t, Y ~ 2 + 5t, z ~ - 3t parallel to the plane = 8? Give reasons for your answer.
~
696
Chapter 12: Vectors and the Geometry of Space
68. How can you tell when two planes A,x + Bly + Clz = D l and A,x + B,y + C,z ~ D, are parallel? Perpendicular? Give reasons for your answer.
b. Test the formulas obtained for y and z in part (a) by investigating their behavior at Xl ~ and Xl ~ Xo and by seeing what happens as Xo --+ 00. What do you find?
°
69. Find two differeot planes whose intersectioo is the line x = 1 + t,Y = 2 - I, Z = 3 + 2t. Write equations for each plane in the form Ax + By + Cz = D. 70. Find a plane through the origin that is perpendicular to the plane M: 2x + 3y + z ~ 12 in a right angle. How do you know that your plane is perpendicular to M? 71. The graph of (x/a) + (y/b) + (z/c) = I is a plane for any nonzero numbers a, b, and c. Which planes have an equation of this form?
P(O,y, z)
-------"JL---7"''--+----->y
72. Suppose L l and L, are disjoint (noointersectiog) nooparallellines. Is it possible for a nonzero vector to be perpeodicular to both L, and L,.? Give reasons for your answer. 73. Perspective in computer graphics In computer graphics and perspective drawing, we need to represent ol!iecls seeo by the eye in space as images on a two-dimensional plane. Suppose that the eye is at E(xo, 0, 0) as sbown bere and that we want to represent a point Pl(xJ,yJ, Zl) as a point 00 the yz-plane. We do this by projecting PI onto the plane with a ray from E. The point PI will be portrayed as the point p(0, y, z). The problem for us as graphics designers is to rmdy andz givenE andP,.
a. Write a vector equatioo that bolds betweeo BP and BPI' Use the equation to expressy andz in terms OfXO,Xl,Yt. andzi.
12.6
74. Hidden lines in computer graphics Here is another typical problem in computer graphics. Your eye is at (4, 0, 0). You are looking at a triangular plate whose vertices are at (I, 0, I), (I, I, 0), and ( -2, 2, 2). The line segment from (I, 0, 0) to (0, 2, 2) passes through the plate. What portioo of the line segment is hiddeo from your view by the plate? (This is an exercise in rmding intersectioos of lines and planes.)
Cylinders and Quadric Surfaces Up to now, we have studied two special types ofsurfaces: spheres and planes. In this section, we extend our inventory to include a variety of cylinders and quadric surfaces. Quadric surfaces are surfaces defmed by second-degree equations in x, y, and z. Spheres are quadric surfaces, but there are others of equal interest which will be needed in Chapters 14-16.
Cylinders Generating curve (in Ihe yz-plane)
A cylinder is a surface that is generated by moving a straight line along a given planar curve while holding the line parallel to a given fixed line. The curve is called a generating curve for the cylinder (Figure 12.43). In solid geometry, where cylinder means circular cylinder, the generating curves are circles, but now we allow generating curves of any kind. The cylinder in our first example is generated by a parabola
EXAMPLE 1 Find an equation for the cylinder made by the lines parallel to the z-axis that pass through the parabola y = x 2 , z = 0 (Figure 12.44).
----"y
\
lines through
generating curve parallel to x-axis
FIGURE 12.43 A cylinder and generating curve.
The point Po(xo, xo2, 0) lies on the parabola y = x' in the xy-pJane. Then, for Solution any value of z, the point Q(xo, xo2, z) lies on the cylinder because it lies on the line x = xo,y = xo' through Po parallel to the z-axis. Conversely, any point Q(xo,xo2,z) whose y-coordinate is the square of its x-coordinate lies on the cylinder because it lies on the line x = Xo, Y = xo2 through Po parallel to the z-axis (Figure 12.44). Regardless of the value of z, therefore, the points on the surface are the points whose coordinates satisfY the equation y = x 2 • This makes y = x 2 an equation for the cylinder. Because of this, we call the cylinder ''the cylinder y = x 2 ." •
12.6
,
Cylinder.; and Quadric Surfaces
697
As Example 1 suggests, any curve f(x,y) = c in the xy-plane defines a cylinder paralle! to the z-axis whose equation is also f(x,y) = c. For instance, the equation x 2 + y2 = 1 def'mes the circular cylinder made by the lines parallel to the z-axis that pass through the circle x 2 + y2 = 1 in the xy-plane. In a similar way, any curve g(x, z) = c in the xz-plane defines a cylinder parallel to they-axis whose space equation is also g(x,z) = c. Any curve h(y,z) = c def'mes a cylinder parallel to the x-axis whose space equation is also hey, z) = c. The axis of a cylinder need not be parallel to a coordinate axis, however.
Quadric Surfaces A quadric surface is the graph in space of a second-degree equation in x, y, and z. We fucus on the special equation ..(x2
FIGURE 12.44 Everypointofthe cylinder in Example 1 has coordinates of the form (XO' xo2, z). we call it "the cylinder y _ x 2 ."
+ By2 + Cz 2 + Dz -
E,
where A, B, C, D, and E are constants. The basic quadric smfaces are ellipsoids, paraboloids, elliptical cones, and hyperboloid.. Spheres are special cases of ellipsoids. We present a few examples illustrating how to sketch a quadric surface, and then give a S1DDmary table of graphs of the basic types.
EXAMPLE 2
The ellipsoid
(Figure 12.45) cuts the coonlinate axes at (± a, 0, 0), (0, ± b, 0), and (0, 0, ± c). It lies within the rectangular box dermed by the inequalities Ix I :S a, IyI :s h. and Iz I :s c. The surface is symmetric with respect to each of the coordinate planes because each variable in the defining equation is squared.
Elliptical cross-section in the pIane z -lo
,
,
/c
\
"i'
/,
I
l'.:"
~
"
"
, :
I
, ,
,
I
~
---I---~
I....
~ _I'"
... , ,
... -
I I
-
I
'>,t-.:..'--:-:--,;""-1--, , ,,, "
FIGURE 12.45 The ellipsoid :x2
iZ
z2
a
b
c
y
I /
~'f
-+-+--1 2 2 2 in Example 2 has elliptical CJ'08S-sections in each ofthe three coordinate planes.
The curves in which the three coordinate planes cut the surface are ellipses. For example, 2
2
~2 + L2 a
b
=
1
when
z
=
o.
698
Chapter 12: Vectors and the Geometry of Space
= zo, IZo I < c, is the ellipse
The curve cut from the surface by the plane z
x2 ----::-----=----:2,2
a (I - (zoic) )
+
y2 2
2
b (I - (zoic) )
= 1.
If any two of the semiaxes a, b, and c are equal, the surface is an ellipsoid of revolu_ tion. If all three are equal, the surface is a sphere.
EXAMPLE 3
The hyperbolic paraboloid
y2
x2
2
a2
---
b
z c'
c>
0
has symmetry with respect to the planes x = 0 and y = 0 (Figure 12.46). The crosssections in these planes are
x
=
0:
y = 0:
the parabola z = c2 y2 . b
(1)
the parabola z = - c2 x 2 •
(2)
a
In the plane x = 0, the parabola opens upward from the origin. The parabola in the plane y = 0 opens downward. If we cut the surface by a plane z = Zo > 0, the cross-section is a hyperbola,
y2 b2
x2
Zo
a2
C'
---
with its focal axis parallel to the y-axis and its vertices on the parabola in Equation (1). If Zo is negative, the focal axis is parallel to the x-axis and the vertices lie on the parabola in Equation (2).
2
z
2
Part of the hyperbola y 2 - X 2 = 1 b a
~y 122 Part of the hyperbola \
/
The parabola z = - ~ ~
a
/" in the plane z = -c
a
- y2 = 1 b x
in the xz-plane
FIGURE 12.46 The hyperbolic paraboloid (y2/b 2) - (x 2/a 2) = z/c, c > O. The cross-sections in planes perpendicular to the z-axis above and below the xy-plane are hyperbolas. The cross-sections in planes perpendicular to the other axes are parabolas.
Near the origin, the surface is shaped like a saddle or mountain pass. To a person traveling along the surface in the yz-p1ane the origin looks like a minimum. To a person traveling the xz-p1ane the origin looks like a maximum. Such a point is called a saddle point of a surface. We will say more about saddle points in Section 14.7. _ Table 12.1 shows graphs of the six basic types of quadric surfaces. Each surface shown is symmetric with respect to the z-axis, but other coordinate axes can serve as well (with appropriate changes to the equation).
12.6
Cylinder.; and Quadric Surfaces
699
TABLE 12.1 Graphs of Quadric Surfaces
,
,
,
Tho poabolu _ ~J:3
ill 1bI J:3-plmo
" .....
Tho p",,,bolu _ ~:I
___
iJIlbo 1f'1I!IIDO
, ELLIPSOID
,
,
ELLIPTICAL PARABOLOID
,
.,
ELLIPTICAL CONE
HYPERBOLOID OF ONE SHEET
,
. HYPERBOLOID OF 1WO SHEETS
HYPERBOLIC PARABOLOID
% C'
0>0
700
Chapter 12: Vector.; and the Geometry of Space
Exercises 12.6 Mitdrlng Equations with Surfaces
Ie
,
L
In Exercises 1-12, match the equation with the surface it defines. Also, identify each surface by type (paraboloid, ellipsoid, etc.) The
swfaces are labeled (a)--{l). 1. x:1 +y:1 + 4z2 = 10 3. 9y2 +z:1 = 16 5. x_ y 2_ z 2 7.x2 +2z 2 _8 9. X=z2_ y :1 11. x:1
+ 4z:1
.
2.z:1+4y 2_4x:1=4 4.y:1+ z 2=X 2 6.x __ 2_ 2 y z 8.z 2 +x:1_ y 2_1
Drawtng
10. z = _4.1: 2 _ y2
= y2
12. 9x:1
+ 4y:1 + 2z:1
Sketch the surfaces in Exercises 13-44. = 36
,
L
,
CYUNDERS 13. x 2 +y:1=4 15. x2+~_16
14.z=y2_1 16. 4x 2 + y2 _ 36
ELUPSOIDS 17. 9%2 + y2
18. 4x:1
19. 4x:1
20. 9%2
+ z2 = 9 + 9y2 + ~ _ 36
+ 4y:1 + z:1 = 16 + 4y 2 + 36z:1 _ 36
PARABOLOIDS AND CONES
,.
,
d.
~
•
y
•
23. x=4-4y:1-~ 25. x:1+ y :1=z:1
24.y=I-X2 -z2 26. 4x:1 + ~ = 9y2
HYPERBOLOIDS 28.y2+ Z 2_ X2=1
30. (y'/4) - (x'/4) -
ASSORTED
y
34. 4x:1 + 4y:1 = z2 36. 1&2 + 4y 2 _ 1
33. z= 1 +y2_ X:1 35. y _ _ (x:1 + z2) 37. x 2 +y:1_ z 2_4 x2
38.x2+~_y
40. 16y2 + 9z:1 _ 4x 2 42.y:1_ X2_ z 2=1 44.x 2 +y:1=z
+z2
_1 41. z = _(x2 + y2) 43.4y 2+ z 2_4x:1=4
..
~ 1
HYPERBOUC PARABOLOIDS
39.
y
z'
31. y2- x :1_ z
,
f.
22.z=8-x 2 -y:1
27. X:1+ y 2_ z :1= 1 29. z2_ X:1_ y :1= 1
•
,
,.
21. z=x:1+4y:1
h•
Theory and Examples 45. L Express the area A ofthe cross-section cut from. the ellipsoid
, ,
x:1+L+~=l 4 9 by the plane z - c as a function of c. (The area of an ellipse with semiaxesa andbis7rab.)
,
L
J.
b. Use slices perpendicular to the z-axis to fmd the volume of the ellipsoid in part (a).
c. Now find the volume ofthe ellipsoid x2 a:1
y2 b
z:1 c
- + -2+ -2- 1
•
y
.
Does your formula give the volume of a sphere ofradius a if
a - b - c?
Chapter 12
701
b. Express your answer in part (a) in terms of h and the areas A o and A h of the regions cut by the hyperboloid from the planes z = 0 andz = h.
46. The barrel shown here is shaped like an ellipsoid with equal pieces cut from the ends by planes perpendicular to the z-axis. The crosssections perpendicular to the z-axis are circular. The barrel is 2h units high. its midsection radius is R, and its end radii are both r. Find a formula for the barrel's volume. Then check two things. First, suppose the sides of the barrel are straightened to turn the barrel into a cylinder ofradius R and height 2h. Does your formula give the cylinder's volume? Second, suppose r = 0 and h = R so the barrel is a sphere. Does your formula give the sphere's volume?
z
Questions to Guide Your Review
c. Show that the volume in part (a) is also given by the formula
V
=
"6h (A o + 4A m + A h ),
where Am is the area of the region cut by the hyperboloid from the plane z = h/2.
Viewing Surfaces
D Plot the surfaces in Exercises 49-52 over the indicated domains.
If
you can, rotate the surface into different viewing positions.
-2 ~ x ~ 2,
49. z = y2,
-2 ~ x ~ 2,
50. z = 1 - y2, 51. z = x 2 y
52. z
-3
~
x
~
b. -1
~
x
c. -2
~
x
d. -2
~
x
8.
47. Show that the volume ofthe segment cut from the paraboloid x2 y2 z -2+ - = a b2 c
by the plane z = h equals half the segment's base times its altitude. 48.
8.
Find the volume of the solid bounded by the hyperboloid x2 y2 z2 -2+ -2- - = 1 a b c2
and the planes z = 0 and z = h, h
Chapter
>
x
=
-2 ~ y ~ 2
x ~ 3,
-3 ~ y ~ 3
3,
-3
~
y
~
3
~
1,
-2
~
y
~
3
~
2,
-2
~
y
~
2
~
2,
-1
~
y
~
1
COMPUTER EXPLORATIONS Use a CAS to plot the surfaces in Exercises 53-58. Identify the type of quadric surface from your graph. x2
53.
y2
9 + 36
55. 5x 2
o.
+ y2, -3 ~ + 2y 2 over
2
-0.5 ~ y ~ 2
=
9 -
1 - 25
z2 - 3y 2
x2
57.
z2 =
y2
1
=
z2
16 + 2
Questions to Guide Your Review
1. When do directed line segments in the plane represent the same vector? 2. How are vectors added and subtracted geometrically? Algebraically? 3. How do you find a vector's magnitude and direction? 4. If a vector is multiplied by a positive scalar, how is the result related to the original vector? What if the scalar is zero? Negative? 5. Define the dot product (scalar product) of two vectors. Which algebraic laws are satisfied by dot products? Give examples. When is the dot product of two vectors equal to zero? 6. What geometric interpretation does the dot product have? Give examples. 7. What is the vector projection of a vector u onto a vector v? Give an example of a useful application of a vector projection. 8. Define the cross product (vector product) of two vectors. Which algebraic laws are satisfied by cross products, and which are not? Give examples. When is the cross product oftwo vectors equal to zero? 9. What geometric or physical interpretations do cross products have? Give examples.
10. What is the determinant formula for calculating the cross product of two vectors relative to the Cartesian i, j, k-coordinate system? Use it in an example. 11. How do you find equations for lines, line segments, and planes in space? Give examples. Can you express a line in space by a single equation? A plane? 12. How do you find the distance from a point to a line in space? From a point to a plane? Give examples. 13. What are box products? What significance do they have? How are they evaluated? Give an example. 14. How do you find equations for spheres in space? Give examples. 15. How do you find the intersection oftwo lines in space? A line and a plane? Two planes? Give examples. 16. What is a cylinder? Give examples of equations that define cylinders in Cartesian coordinates. 17. What are quadric surfaces? Give examples of different kinds of ellipsoids, paraboloids, cones, and hyperboloids (equations and sketches).
702
Chapter 12: Vectors and the Geometry of Space
Chapter
Practice Exercises
Vedor Calculations In Two Dimensions In Exercises 1-4, let u = (-3,4) and v = (2, -5). Find (a) the componeot form of the vector aod (b) its magnitude.
In Exercises 25 aod 26, fmd (a) the area of the parallelogram determined by vectors u aod v and (b) the volume of the parallelepiped determined by the vectors u, v, and w.
1.3u-4v
2.u+v
25.n~i+j-k,
3. -2u
4. 5v
26.u=i+j,
In Exercises 5-8, fmd the componeot form of the vector.
5. The veetor obtained by rotating (0, I) througb ao angle of 21f/3
radians 6. The unit vector that makes an angle of 1f/6 radiao with the positivex-axis 7. The vector 2 units long in the direction 4i - j
8. The veetor 5 units long in the direction opposite to the direction of(3/5}i + (4/5)j
v~2i+j+k,
v=j,
w=i+j+k
Unes, Planes, and Distances 27. Suppose that n is nonna! to a plane aod that v is parallel to the plane. Describe how you would fmd a veetor n that is both perpeodicu1ar to v aod parallel to the plane. 28. Find a veetor in the plaoe parallel to the line ax + by
9. v2i + v2j
11. Velocityveetorv = (-2sint)i + (2cost)jwheot = 1f/2.
+ (etsint + etcost)j
Vedor Calculations in Three Dimensions Express the veetors in Exercises 13 and 14 in terms of their leogtha and directions. 13. 2i - 3j + 6k
c.
29. (2,2,0); x
=
-t, Y
=
t,
Z =
+t
-1
z = t
31. Parmnetrize the line that passes thruugh the point (I, 2, 3) parallel to the vector v ~ - 3i + 7k. 32. Parmnetrize the line segroeot joining the points p(1, 2, 0) aod
10. -i - j
12. Velocity vector v = (etcost - etsint)i wheot ~ 102.
~
In Exercises 29 aod 30, fmd the distance from the point to the line.
30. (0,4, I); x = 2 + t, Y = 2 + t, Express the vectors in Exercises 9-12 in terms of their leogtha aod directions.
w~-i-2j+3k
14. i + 2j - k
15. Find a vector 2 units long in the direction ofv = 4i - j + 41<. 16. Find a vector 5 units long in the direction opposite to the direction ofv = (3/5) i + (4/5)1<.
Q(I,3, -I). In Exercises 33 aod 34, fmd the distance from the point to the plane.
33. (6,0, -6),
x - Y = 4
34. (3,0, 10),
2x
+ 3y + z
= 2
35. Find an equation for the plaoe that passes through the point (3, -2, I) normal to the veetor n = 2i + j + k. 36. Find an equation for the plane that passes through the point (-1,6,0) perpendicular to the line x ~ -I + t,y ~ 6 - 2t, z = 3t. In Exercises 37 and 38, fmd an equation for the plane thruugh points
P,Q,andR. In Exercises 17 and 18, fmd Ivl,lul,v'u,u'v,vxu,uxv,
37. P(l, -1,2),
Iv
38. P(l, 0, 0),
X u I, the angle between v and u, the scalar component of u in the direction of v, and the vector projection ofo onto v. 17. v = i + j
18. v = i + j + 2k
u~2i+j-2k
u~-i-k
In Exercises 19 aod 20, fmd proj. u.
19. v=2i+j-k u=i+j-5k
Q(2, 1,3),
R( -1,2, -I)
Q(O, 1,0), R(O,O, I)
39. Find the points in which the line x ~ I + 2t, Y z = 31 meets the three coordinate planes.
~
-I - t,
40. Find the point in which the line througb the origin perpeodicu1ar to the plane 2x - Y - z ~ 4 meets the plaoe 3x - 5y + 2z ~ 6. 41. Find the acute angle between the planes x = 7 and x + y = -3.
v2z
20. u = i - 2j
v=i+j+k
42. Find the acute angle betweeo the planes x + y y+z= I.
~
+
I aod
In Exercises 21 and 22, draw coordinate axes and then sketch u, v, and u X v as vectors at the origin.
43. Find parame1ric equations for the line in which the planes x + 2y + z ~ I and x - y + 2z ~ -8 intersect.
21. u=i.,
44. Show that the line in which the plaoes
v=i+j
22. u=i-j,
v=i+j
23. !flvl ~ 2, Iwl ~ 3, aodthe aog1e betweeo v aodw is 1f/3, fmd Iv - 2wl· 24. For what value or values of a will the vectors u = 2i + 4j - 5k aod v ~ -4i - 8j + ak be parallel?
x+2y-2z=5
and
5x-2y-z=0
intersect is parallel to the line
x = -3 + 2t, Y = 3t,
z = I + 4t.
Chapter 12 45. The planes 3x + 6z = I and 2x + 2y - z = 3 intersect in a linc. L
Shaw that the planes are orthogonal.
L (21 - 3J + 3k)' b. x-3-t,
b. Find equations for the line of intersection. 46. Find an equation for the plane that passes through the point (1,2, 3) parallel to u - 2i + 3j + kandv - i - j + 2k. 47. Is v = 2i - 4j + k related in any special way to the plane 2x + Y = 5? Give reasons for your answer. 4&. The equation n' P-;P = 0 represents ~p1ane through Po normal to D. What set does the inequality D' PoP> 0 tepresen.fl
e. (x
«x + 2)1 + (y -
y--Ilt,
+ 2) + l1(y
Practice Exercises
703
I)J + zk) - 0
z-2-3t
- I) - 3z
d. (2i - 3j + 3k) X ((x + 2)i + (y - I)j + zk) = 0 e. (2i - j + 3k) X (-3i + k)·((x + 2)i + (y - I)j + zk) ~O
61. The parallelogram shawn here has vertices at A(2, -1,4), B(I, 0, -I), C(I, 2, 3), andD. Find
,
49. Find the distance from the point P(I, 4, 0) to the pla:ne through A(O, 0, 0), B(2, 0, -I), "'" C(2, -1,0).
vI,
50. Find the distance from the point (2, 2, 3) to the plane 2x + 3y + 5z - o. 51. Find a vector parallel to the plane 2x - y - z - 4 and orthogonaltot+J+k. A(2,-1,4)
51. Find a unit vector orthogonal to A in the plane of B and C if A - 2i - j + k,B - i+ 2j + k,andC - i + j - 2k. 53. Find a vector of magnitude 2 parallel to the line ofintersection of theplane8x + 2y +z - I = oand x -y + 2z+ 7 = O. Sol. Find the point in which the line through the origin perpendicular tothep1ane2x - y - z = 4meetsthep1ane3x - 5y +2z = 6.
55. Find the point in which the line through p(3, 2, I) normal ttl the plane 2x - y + 2z = -2 meets the plane. 56. What angle does the line of intersection of the planes 2x + Y - z = 0 and x + y + 2z = 0 make with the positive
x-axis? 57. The line L:
x=3+2t,
y=2t,
z=t
in1ersects the plane x + 3y - z = -4 in a point P. Find the coordinates of P and find equations for the line in the plane through P perpendicular ttl L. 58. Shaw that :for every real number k the plane
x - 2y+ z + 3 + k(2x - y- z + I) = 0
contains the line of intersection ofthe planes
L
the coordinates of D,
b. the cosine ofthe interior angle atB,
e. the vector projection. of.BA. onto iiC, d. the area of the plml1l.elogram, e. an equatioo for the plane ofthe parallelogram. f. the areas of the orthogonal projections ofthe pamllelogram on the three coordi.na.re planes. 63. Distante betwem JineI Find the distance between the line L, through the points .4(1,0,-1) and B(-I, 1,0) and the line L2 through the points C(3, I, -I) andD(4, 5, -2). The distance is ttl be measured along the line perpendicular tD the two lines. Frrst fmd a vector D perpendicular to both lines. Then project AC onto D. 64. (Colltinuation ojExerci.!e 63.) Find the distance between the line through A(4, 0, 2) and B(2, 4, 1) and the line through C(l, 3, 2)
""'D(2, 2, 4).
x - 2y + z + 3 = 0 and 2x - Y - z + I = O. 59. Find an equation for the plane through A(-2,O, -3) and B(I, -2, I) that lies parallel to the line through
C( -2, -13/', 26/') ""'V(16/', -13/', 0). 60. Is the line x = 1 + 2t, Y = -2 + 31, z = -5t related in any way to the plane -4x - 6y + 10z - 91 Give reasoos for your
""""'.
61. Which of the following are equations for the plane through the points P(I, I, -I), Q(3, 0, 2), andR(-2, I, O)?
Qu_drk Surhces Identi:fy and sketch the smfaces in Exercises 65-76. 65. x 2 + y2 + z2 = 4 66. x 2 + (y - 1)2 + z2 = I 2 67. 4x + 4y 2 + z2 = 4 68. 3fu:2 + 91 2 + 4z 2 = 36 69. z = _(x 2 + y2)
70. Y = _(x 2 + ::2)
71. X2 +y2=z2 73. X2 +y2_ z 2=4 75. y 2_ x 2_ z 2=1
72. x 2 +::2 =y2
74. 4y 2+::2-4¥2=4 76.z 2 _x 2 _ y 2=1
704
Chapter 12: Vectors and the Geometry of Space
Chapter
----
AdditionaL and Advanced Exercises
---
1. Submarine hunting Two surface ships on maneuvers are trying to determine a submarine's course and speed to prepare for an aircraft intercept. As shown here, ship A is located at (4, 0, 0), whereas ship B is located at (0, 5, 0). All coordinates are given in thousands of feet. Ship A locates the submarine in the direction of the vector 2i + 3j - (l/3)k, and ship B locates it in the direction of the vector 18i - 6j - k. Four minutes ago, the submarine was located at (2, -1, -1/3). The aircraft is due in 20 min. Assuming that the submarine moves in a straight line at a constant speed, to what position should the surface ships direct the aircraft?
clockwise when we look toward the origin fromA. Find the velocity v of the point of the body that is at the position B(l, 3, 2).
z
IA~(::l,~l,~l!..)_ - - - - ; , B ( l , 3, 2) I
---
x
(4,0,0)
y
x
:~: : ----.J!___ v: / I
z
Ship A
I
/
I
1/
I~
I /
-----_
I
-------1//
~y
//
5. Consider the weight suspended by two wires in each diagram. Find the magnitudes and components of vectors F I and F 2, and angles a and {3.
a.
-\ Submarine
Nor TO SCALE
2. A helicopter rescue Two helicopters, HI and H 2, are traveling together. At time t = 0, they separate and follow different straight-line paths given by HI:
x = 6
H2:
x = 6
+ 40t, + 11 Ot,
+ lOt, -3 + 4t,
y = -3 y =
b.
+ 2t -3 + t.
z = -3 z =
Time t is measured in hours and all coordinates are measured in miles. Due to system malfunctions, H 2 stops its flight at (446, 13, 1) and, in a negligible amount of time, lands at (446, 13,0). Two hours later, HI is advised of this fact and heads toward H2 at 150 mph. How long will it take HI to reach H2? 3. Torque The operator's manual for the Toro® 21 in. lawnmower says "tighten the spark plug to 15 ft-lb (20.4 N· m)." If you are installing the plug with a 1O.5-in. socket wrench that places the center of your hand 9 in. from the axis of the spark plug, about how hard should you pull? Answer in pounds.
(Hint: This triangle is a right triangle.)
6. Consider a weight ofw N suspended by two wires in the diagram, where TI and T 2 are force vectors directed along the wires.
9in.~
~ '/
4. Rotating body The line through the ongm and the point A( 1, 1, 1) is the axis of rotation of a right body rotating with a constant angular speed of3/2 rad/sec. The rotation appears to be
a. Find the vectors T I and T 2 and show that their magnitudes are w cos (3 ITII = sin (a + (3) and T
-
121-
wcosa sin (a+{3)
Chapter 12 b. For a ftxed {j determine the value of a which minimizes the magnitude IT ,I.
ax+by=cis
Xl -
X
X3 -
x
y, - Y Y2 - Y y, - Y
Zt -
Z
Z2 -
Z
d ~O
Z3 - z
is an equation for the plane through the three noncollinear points Pt(Xt,Yt,Zt), P2(X2,Y2, Z2), and P3(X3,Y3,Z3). b. What set ofpoints in space is described by the equation
x
y
Xl
Yl
X2 X,
Y2 Y,
8. Determinants and line.
d =
I"", + byl
- cl
~'--;=.C===.<~ 2 b2 •
Va
+
12. a. Use vectors to abow that the distance from P,(XhYh Zl) to the plane Ax + By + Cz = Dis
a. Showthat
x, -x
705
11. Use vectors to show that the distance from Pl(XhYl) to the line
e. For a ftxed a determine the value of (j which minimizes the magnitude IT 2 1· 7. Determinants and planes
Additional and Advanced Exercises
z
I I I I
Zl Z2
z,
~ ,-IAx---":l--;+;=B:c:Y~l=+=:c=C=Zl~-"D-,-1
VA 2 +B2 +C2
13. a. Show that the distance between the parallel planes Ax + By + Cz = D, and Ax + By + Cz = D, is
d ~
07
.
b. Find an equation for the spbere that is tangent to the planes x + Y + Z = 3 and X + y + Z = 9 if the planes 2x - y = 0 and 3x - Z = 0 pass througb the center of the spbere.
~
ID I IAi
-
D21 Ckl .
+ Bj +
b. Find the distance between the planes 2x + 3y - z = 6 and 2x + 3y - z = 12. e. Find an equation for the plane parallel to the plane 2x - Y + 2z = -4 ifthe point (3, 2, -I) is equidistant from the two planes.
Show that the lines
d. Write equations for the planes that lie parallel to and 5 units away from the plane x - 2y + z = 3.
and
14. Prove that four points A, B, C, and D are coplanar (lie in a common plane) if and only iLW • (AB X BC) = O. intersect or are parallel if and only if at
Cl
a2
C2
a3
C3
b1-d1 b2 - d2 b3 -d3
15. Tbe projection of a vector on a plane Let P be a plane in space and let v be a vector. The vector projection ofv onto the plane P, projp v, can be defmed informally as follows. Suppose the sun is shining so that its rays are norma1 to the plane P. Then projp v is the "shadow" of v onto P. If P is the plane x + 2y + 6z = 6 and v = i + j + k,fmdprojpv.
O.
=
9. Consider a regular tetrabedron of side length 2.
a. Use vectors to fmd the angle 0 formed by the base of the tetrabedron and any one of its other edges. D
2
-
C
I
B
b. Use vectors to fmd the angle 0 formed by any two adjacent faces of the tetrabedron. This angle is commonly referred to as a dihedral angle. 10. In the ftgure here, D is the midpoint of side AB oftriang1e ABC, and E is one-third of the way between C and B. Use vectors to prove that F is the midpoint of line segment CD.
17. Triple vector products The triple vector products (u X v) X w and n X (v X w) are usua1ly not equal, although the formulas for evaluating them from components are similar: (u X v) X w n X (v X w)
~ D
B
= (n'w)v = (n'w)v -
(v·w)u. (u·v)w.
VerilY each formula for the following vectors by evaluating its two sides and coroparing the results. u
C
A
_~-------"vfJ~Y'-------W
-L
p
2
16. The accompanying figure shows nonzero vectors v, W, and z, with z orthogonal to the line L, and v and w ma1cing equal angles {jwithL.Assuming Ivl = Iwl,fmdwintermsofvandz.
a. 2i
v
w
2j
2k
b. i - j + k
2i+j-2k
-i+2j-k
e. 2i + j
2i-j+k
i + 2k
d. i + j - 2k
-i - k
2i+4j-2k
706
Chapter 12: Vectors and the Geometry of Space
18. Cross and dot products
Show that if u, v, w, and r are any
21. Use vectors to prove that
vectors, then
(a' + b')(c' + d') '" (ac + bd)'
LUX~X~+vX~X~+wX~X~-O
for any four nurohers a, b, c, and d. (Hint: Let u = ai + bj and + dj.)
b. u X v - (u'v X i)i + (u'v X j)j + (u'v X k)k C.(UXv).(wXr)=IU'W
u·r
19. Cross and dot products
V'WI.
v = ci
22. Dot multiplication is positive def"IDite
v"r
20. By forming the cross product of two appropriate vectors, derive the trigonometric identity
23. Show that Iu + v I
'"
Iu I +
Iv I for any vectors u and v.
24. Show that W - Iv Iu + Iu Iv bisects the angle hetweeo n and v. 25. Showthat Ivlu + lulvand Ivlu - lulvare orthogonal.
sin(A - B) = sinAcosB - cosAsinB.
Chapter
!E
~
for every vector u and that u· u = 0 if and only ifu = O.
Prove or disprove the formula
u X (u X (n X v))·w - -Inl'n'v X w.
Show that dot multipli-
cation of vectors is positive definite; that is, show that u . u
Technology Application Projects
Mathematica/Maple Module: Using ~ctors to Represent Lines and Find Distances Parts 1 and 11: Learn the advantages of interprctiog lines as vectors. Part ill: Use vectors to f"md the distance from a point to a line. Putting a Scene in Three Dimensions onto Q 1Wo-Dimensional Caln'llS Use the concept of planes in space to obtain a two-dimensional image.
Getting Started in Plotting in 3D Part I: Use the vector def"mition of lines and planes to generste graphs and equations, and to compare different fonns for the equations of a single line. Part 11: Plot functions that are def"med implicitly.
0
13 VECTOR-VALUED FUNCTIONS AND MOTION IN SPACE OVERVIEW Now that we have learned about vectors and the geometry of space, we can combine these ideas with our earlier study of functions. In this chapter we introduce the calculus of vector-valued functions. The domains of these functions are real numbers, as before, but their ranges are vectors, not scalars. We use this calculus to describe the paths and motions of objects moving in a plane or in space, and we will see that the velocities and accelerations of these objects along their paths are vectors. We will also introduce new quantities that describe how an object's path can turn and twist in space.
13.1
Curves in Space and Their Tangents When a particle moves through space during a time interval I, we think of the particle's coordinates as functions defined on I: x
z
= I(t),
y = get),
z
= h(t),
tEl.
(1)
The points (x,y, z) = (f(t), get), h(t», tEl, make up the curve in space that we call the particle's path. The equations and interval in Equation (1) parametrize the curve. A curve in space can also be represented in vector form. The vector ret) =
x FIGU~ 13.1 The position vector r = OP of a particle moving through space is a function of time.
oF =
I(t)i
+ g(t)j + h(t)k
(2)
from the origin to the particle's position P(f(t), g(t), h(t» at time t is the particle's position vector (Figure 13.1). The functions I, g, and h are the component functions (components) of the position vector. We think of the particle's path as the curve traced by r during the time interval 1. Figure 13.2 displays several space curves generated by a computer graphing program. It would not be easy to plot these curves by hand. z
z
z
x
x
y
x
y r(t) = (sin3t)(cost)i + (sin3t)(sint)j + tk
r(t) = (cost)i
+ (sint)j + (sin2t)k
r(t) = (4
(4
+ sin20t)(cost)i + + sin20t)(sint)j +
(cos20t)k (a)
FIGURE 13.2
(b)
(c)
Space curves are defined by the position vectors ret).
707
708
Chapter 13: Vector-Valued Functions and Motion in Space Equation (2) defines r as a vector function of the real variable t on the interval 1. More generally, a vector-valued function or vector function on a domain set D is a rule that assigns a vector in space to each element in D. For now, the domains will be intervals of real numbers resulting in a space curve. Later, in Chapter 16, the domains will be regions in the plane. Vector functions will then represent surfaces in space. Vector functions on a domain in the plane or space also give rise to "vector fields," which are important to the study of the flow of a fluid, gravitational fields, and electromagnetic phenomena. We investigate vector fields and their applications in Chapter 16. Real-valued functions are called scalar functions to distinguish them from vector functions. The components of r in Equation (2) are scalar functions of t. The domain of a vector-valued function is the common domain of its components.
z
EXAMPLE 1
ret) = (cos t)i + (sin t)j + t k.
)
~ t=
2Tr:
Graph the vector function
......... t
..,
SoLution
The vector function
ret) = (cos t)i + (sin t)j + tk
= 11'
is defined for all real values of t. The curve traced by r winds around the circular cylinder x 2 + y2 = 1 (Figure 13.3). The curve lies on the cylinder because the i- and j-components ofr, being the x- andy-coordinates ofthe tip ofr, satisfy the cylinder's equation: x 2 + y2
= (cos t)2 + (sin t)2 = 1.
The curve rises as the k-component z = t increases. Each time t increases by 21T, the curve completes one turn around the cylinder. The curve is called a helix (from an old Greek word for "spiral"). The equations
y
x = cost,
x
z
z
z
x
------".
y
FIGURE 13.4
z = t
parametrize the helix, the interval -00 < t < 00 being understood. Figure 13.4 shows more helices. Note how constant multiples of the parameter t can change the number of turns per unit of time. _
FIGURE 13.3 The upper half of the helix ret) = (cos t)i + (sin t)j + tk (Example 1).
r(t) = (cos t)i
y = sint,
+ (sin t)j + tk
r(t) = (cos t)i
x
y
Y
+ (sin t)j + O.3tk
r(t) = (cos 5t)i
+ (sin 5t)j + tk
Helices spiral upward around a cylinder, like coiled springs.
Limits and Continuity The way we define limits of vector-valued functions is similar to the way we define limits of real-valued functions.
13.1 Curves in Space and Their Tangents
709
DEFINmON Let r(t) = I(t)i + g(t)j + h(t)k be a vector function with domain D, and L a vector. We say that r has limit Last approaches to and write
lim r(t)
= L
1-/0
if, for every number E for all tED Ir(t) -
IfL = Lji
>
0, there exists a corresponding number lJ
LI <
whenever
E
<
0
It - tol
<
+ L2j + L3k, then it can be shown that lim,...."r(t)
>
0 such that
lJ.
= L precisely when
and We omit the proof. The equation lim r(t) =
1-/0
(lim I(tj\ + (lim g(t)'j + (lim h(t)\ I-I{)
')
I-I{))
(3)
1-10)
provides a practical way to calculate limits ofvector functions.
EXAMPLE 2
rfr(t) = (cost)i
+
(sint)j
lim r(t) = (lim cos t)i
t-w/4
t---1'w/4
+
+
tk, then
(lim sin t)j l-w/4
+
lim t)k ( l-w/4
=V2·+V2·+1T k J 2'
2
•
4'
We derme continuity for vector functions the same functions.
WH:j
we derme continuity for scalar
DEFINmON A vector function r(t) is continuous at a point t = to in its domain if lim,....~ r(t) = r(to). The function is continuous if it is continuous at every point in its domain. From Equation (3), we see that r(t) is continuous at t = to if and only if each component function is continuous there (Exercise 31).
EXAMPLE 3 (a) All the space curves shown in Figures 13.2 and 13.4 are continuous because their component functions are continuous at every value of t in ( - 00, 00).
(b) The function g(t) = (cost)i
+
(sint)j
+
ltJk
is discontinuous at every integer, where the greatest integer function discontinuous.
l t J is •
Derivatives and Motion Suppose that r(t) = I(t)i + g(t)j + h(t)k is the position vector of a particle moving along a curve in space and that I, g, and h are differentiable functions of t. Then the difference between the particle's positions at time t and time t + !J.t is ~r =
r(t
+ !J.t) - r(t)
710
Chapter 13: Vector-Valued Functions and Motion in Space
(Figure 13.5a). In terms of components. ar = r(1 =
=
>,r---------y (a)
+ M) - r(l)
+ al)i + g(1 + M)j + h(1 + M)k] - [/(I)i + g(1)j + h(l)k] [j(1 + at) - /(I)]i + [g(1 + M) - g(1)]j + [j(1
[h(1
+
M) - h(I)]k.
As al approaches zero. three things seem to happen simultaneously. First. Q approaches P along the curve. Second, the secant line PQ seems to approach a limiting position tangent to the curve atP. Third, the quotient arl M (Figure 13.5b) approaches the limit
lim aral [lim /(1 + al)al -
/(I)li
=
.1.'-0
+
.l.t-O
[lim g(1 + al)al .l.t-O
g(l) l' J
+ [lim h(1 + al) - h(l) lk .1.'-0 =
[d/li dl
M
+ [dgl' + [dh lk dl J dl'
We are therefore led to the following definition. ol"---------~y
(b) x
FIGURE 13.5 As <1/ .... 0. thepointQ approaches the point P along the curve C. In the limit, the vector PQ/ al becomes the tangent vector r'(I) ,
DEFINmON The vector function r(l) = /(I)i + g(1)j + h(l)k has a derivative (is differentiable) at I if/. g. and h have derivatives at I. The derivative is the vector function r
'(I) = dr = I' r(1 dl,J!!!o
+ al) al
- r(l) = d/ . dl I
dl J
dh k dl'
A vector function r is dilTerentiable ifit is differentiable at every point ofits domain. The curve traced by r is smooth if dr/dl is continuous and never O. that is. if /. g. and h have continuous first derivatives that are not simultaneously O. The geometric significance of the definition of derivative is shown in ~ure 13.5. The points P and Q have position vectors r(l) and r(1 + al). and the vector PQ is represented by r(1 + al) - r(I). For al > O. the scalar multiple (1/al)(r(1 + al) - r(I» As al ..... O. this vector approaches a vector points in the same direction as the vector that is tangent to the curve atP (Figure 13.5b). The vector r'(I). when different from O. is defined to be the vector tangent to the curve at P. The tangent line to the curve at a point (f(lo). g(lo). h(lo» is defined to be the line through the point parallel to r'(lo). We require drI dl "" 0 for a smooth curve to make sure the curve has a continuously turning tangent at each point. On a smooth curve. there are no sharp corners or cusps. A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smootb (Figure 13.6). Look once again at Figure 13.5. We drew the figure for al positive. so ar points forward, in the direction of the motion. The vector arlal. having the same direction as ar. points forward too. Had M been negative. ar would have pointed backward. against the direction ofmotion. The quotient arI al. however. being a negative scalar multiple of ar. would once again have pointed forward. No matter how ar points. arlal points forward and we expect the vector dr/dl = lim4t-+O arl M. when different from O. to do the same. This means that the derivative drI dl. which is the rate of change ofposition with respect to time. always points in the direction ofmotion. For a smooth curve. drI dl is never zero; the particle does not stop or reverse direction.
PQ.
FIGURE 13.6 A piecewise smooth curve made up offive smooth curves connected end to end in a continuous fashion. The curve here is not smooth at the points joining the five smooth curves.
+ dg. +
13.1 (urves in Space and Their Tangents
711
DEFINmONS If r is the position vector of a particle moving along a smooth curve in space, then dr V(I) = dl is the particle's velocity vector, tangent to the curve. At any time I, the direction of v is the direction of motion, the magnitude of v is the particle's speed, and the derivative a = dv/dl, when it eJcists, is the particle's acceleration veetor. In
summary, dr v = dl'
1. Velocity is the derivative of position:
2. Speed is the magnitude of velocity:
Speed = Ivl.
3. Acceleration is the derivative of velocity:
a =
dv Iii
d 2r
= d1 2 '
4. The unit vectorv/lvl is the direction of motion at time I.
EXAMPLE 4 Find the velocity, speed, and acceleration of a particle whose motion in space is given by the position vector r(I) = 2 cos I I + 2 sin I j + 5 cos2 I k. Sketch the velocity vector v(7"./4). Solution
The velocity and acceleration vectors at time I are V(I) = r'(I) = -2sinll + 2coslj - 10 cos I sinlk
z
= -2 sin I I + 2 cos I j - 5 sin 21 k, a(l) = r"(I) = -2cosll - 2sinlj - IOcos21k, and the speed is
y t
=
7," 4
IV(I) I = v'(-2sinl)2 + (2 cos 1)2 + (-5 sin 21)2 = v'4 + 25sin2 21. When 1= 7"./4, we have
x
FIGURE 13.7 The curve and the velocity vcctor when I ~ 7"./4 for the motion given in Example 4.
V(7:)=V2I+V2j+5k, A sketch of the curve of motion, and the velocity vector when 1= 7"./4, can be seen in Figure 13.7. • We can express the velocity of a moving particle as the product of its speed and direction: Velocity =
IVIC~I) =
(speed)(direction).
Differentiation Rules Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions.
712
Chapter 13: Vector-Valued Functions and Motion in Space
Differentiation Rules for Vector Functions Let u and v be differentiable vector functions of t, C a constant vector, c any scalar, and / any differentiable scalar function.
d
1. Constant Function Rule:
dtC
2. Scalar Multiple Rules:
:t [cu(t)]
0
=
=
:t [f(t)u(t)]
I When you use the Cross Product Rule,
remember to preserve the order ofthe factors. Ifu comes fIrst on the left side ofthe equation, it must also come first on the right or the signs will be wrong.
cu'(t) I'(t)u(t) + /(t)u'(t)
=
3. Sum Rule:
:t [u(t) + v(t)]
4. Difference Rule:
dt [u(t) - v(t)]
5. Dot Product Rule:
:t [u(t)· v(t)]
6. Cross Product Rule:
dt [u(t) X v(t)]
7. Chain Rule:
:t [u(f(t))]
d
u'(t) + v'(t)
=
u'(t) - v'(t)
u'(t)· v(t) + u(t)· v'(t)
=
d
=
=
=
u'(t) X v(t) + u(t) X v'(t)
I'(t)u'(f(t))
We will prove the product rules and Chain Rille but leave the rules for constants, scalar milltiples, sums, and differences as exercises. Proof of the Dot Product Rule
Suppose that
u = u,(t)1
+ U2(t)j + u3(t)k
and Then d
dt(u'v)
=
d
dt(utVt + U2 V2 + U3 V3)
= UlVI
+ U2V2 +
U3V3
+ UtVl + U2V2 + U3V3.
u' ·v
u·v'
•
Proof of the Cross Product Rule We model the proof after the proof ofthe Product Rille for scalar functions. According to the detmition of derivative, ~(
) _ . u(t dtUXV-~
+ h)
X v(t
+ h) - u(t)
X v(t)
h
To change this fraction into an equivalent one that contains the difference quotients for the derivatives ofu and v, we subtract and add u(t) X v(t + h) in the numerator. Then
:t
(u X v)
",u(,:-t_+--,h.c')_X---,-v(,:-t_+--'h.c')_-----=u("-t)__X---'-v-"(t_+~h )__+----=u",(t,-)_x_v,-,(t__+---,-h,-)_---'u::'(tC'-)_X_v'-'("'-t) ~o h
= lim
=
. [U(t Inn
h~O
= lim h-O
+ h)h - u(t) Xvt+h ( ) ( ) v(t + h) - V(t)] +ut X h
u(t + h) - u(t) X lim v(t + h) + lim u(t) X lim v(t + h) - v(t) h
h-O
h-O
h-O
h
713
13.1 Curves in Space and Their Tangents
The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist (Exercise 32). As h approaches zero, vet + h) approaches v(t) because v, being differentiable at t, is continuous at t (Exercise 33). The two fmct:ions approach the values of du/dt and dv/dt at t. In short,
d ( ) tit uXv
IAs an a1~braic convenience, we
sometimes write the product of a scalar c
and a vector v as vc instead. of cv. Thill permits us, for instance, to write the Chain Rule in a familiar form.:
du
~
dv dt'
•
Proof of the Chain Rule Suppose that u(s) = a(s)i + b(s)j + c(s)k is a differentiable vector function ofs and thats = J(t) is a differentiable scalar function oft. Then a, b, and c are differentiable functions of t, and the Chain Rule for differentiable real-valued functions gives
~[o(,)] ~
dD ds
di-didt' whore,
du
=tit Xv + uX
':,;
+ dj;i +
:k
da da db da d£ da -dadt;+dadli+dadl k
f(t).
~:(~;+:i+~k) ~
~
da du dtda
/'(1)0'(/(1)).
1-
•
f(t)
Vector Functions of Constant Length When we tmck a particle moving on a sphere centered at the origin (Figure 13.8), the position vector has a constant length equal to the radius ofthe sphere. The velocity vector dr/tit, taDgen1 to the path ofmotion, is tangent to the sphere and hence perpendicular to r. This is always the case for a differentiable vector function of constant length: The vector and its Imt derivative are orthogonal. By direct calculation,
r(t)'r(t) =
c
2
~[r(I)'r(I)] ~ 0 r'(t)· ret) + r(t)· r'(t)
= 0
lr(t)l- ciJCODItIIllt. Dtili=IWobolh ..... Rnlc 5 with r(1)
=
D(t)
=
"'(1)
2r'(I)' r(l) ~ O. The vectors r'(t) andr(t) are orthogonal because their dot product is O. In summary, If r is a differentiable vector function of t of constant length, then
FIGURE 13.8 Ifa particle moves on a sphere in such a way that its position r is a differentiable function of time, then r·(drfdt) = O.
dr
(4)
r"dt=O.
We will use this observation repeatedly in Section 13.4. The converse is also true (see
_
.. 21).
Exercises 13.1 Motton in the Plane In Exercises 1--4, r(t) is the position of a particle in the xy-plane at time I. Find an equation inx andy whose graph is the path of the particle. Then fmd the particle's velocity and acceleration vectors at the given value oft.
1. r(l) - (I
+ l)i + (,z - I)j,
t - 1
.. r(t) - _t_ i + II' t - -1/2 t+ 1 t' 3. r(t) - ell
+ je 2tJ, t - ln3
4. r(t) - (oos2t)l
+ (3sin2t)J, t -
0
714
Chapter 13: Vector-Valued Functions and Motion in Space
Exercises 5--8 give the position vectors ofparticles moving along various curves in the xy-plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve.
x' + y' ~ 1
5. Motion on the circle r(t)
~
(sint)i + (cost)j;
~
~
'If/4and'lf/2
i) Does the particle have constant speed? If so, what is its con-
x' + y' ~ 16
6. Motion on the circle r(t)
t
(4 cos!} + (4Sin!)j;
t
~ 'lfand3'1f/2
7. Motion on the cyeloid x = t - sin 1, Y = 1 - cos t r(t)
~
(t - sin t)i + (I - cos t)j;
8. Motion on the parabola y ~
t
~
'If and 3'1f/2
t = -1,0, and I
Motion in Space In Exercises !l-14, r(t) is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Then fmd the particle's speed and direction of motion at the given value oft. Write the particle's velocity at that time as the product ofits speed and direction. 9. r(t) ~ (t + l)i + (t' - I)j + 2tk,
V2 ,
10. r(t) = (I + t)i +
t ~ I
3
j +}-k,
12. r(t)
~
(sect)i + (tant)j + jtk,
13. r(l) = (2 In (I + I))i + I'j +
T,
I = 'If/2
~
t
k,
'If/6
I= I
14. r(t) ~ (e -I)i + (2 cos 3t)j + (2 sin 3t)k,
t ~ 0
In Exercises 15-18, r(t) is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time
t
iiI) Does the particle move clockwise or counterclockwise
around the circle? iv) Does the particle begin at the point (I, OJ?
= O.
b. r(t)
~
cos (2t)i + sin (2t)j,
c. r(t)
~
cos (t - 'If/2)i + sin(t - 'If/2)j,
d. r(t) = (cost)i - (sint)j,
('? I)i+ ('? 1- 161 )j vi'+! k
18. r(t) = ~ (I + t)'f2i + ~ (I - t)3f2j + 1 tk 9 9 3 Tangents to Curves As mentioned in the text, the tangent line to a smooth curve r(t) ~ I(t)i + g(t)j + h(t)k at t ~ to is the line that passes through the point (/(10), g(lo), h(lo)) parallel to V(IO) , the curve's velocity vector at to. In Exercises 19-22, fmd paramettic equations for the line that is tangent to the given curve at the given parameter value t ~ to. cos I)j + e'k, to = 0 2 20. r(t) = t i + (2t - l)j + t 3k, to ~ 2 21. r(l) = Inli +:
t'" 0
2
r(l) = (2i + 2j + k)
k)
describes the motion of a particle moving in the circle ofradius 1 centered at the point (2,2, I) and lying in the plane x + y - 2z ~ 2.
25. Motion along a parabola A particle moves along the top of the parabola y2 ~ 2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point (2, 2). 26. Motion along a .yeloid A particle moves in the xy-plane in such a way that its position at time t is
D a.
2
19. r(l) = (sin I)i + (1 2
t'" 0
24. Motion along a eircle Show that the vector-valued function
r(t)
17. r(l) = (In (t 2 + I))i + (tan-1,)j +
t '" 0
e. r(t) ~ cos (t )i + sin (t )j, t '" 0 2
15. r(t) ~ (3t + I)i + y'3tj + t 2k
16. r(l) =
I'" 0
I. I.) + . (I . + I . + I + cost ( y'z' - y'zJ smt y'3' y'3J y'3
t= I
11. r(l) = (2 cos t)i + (3 sin I)j + 41k,
stant speed? h") Is the particle's acceleration vector always orthogonal to its velocity vector?
a. r(l) = (cosl)i + (sint)j,
x' + 1
r(t) = ti + (t' + I)j;
Theory and Examples 23. Motion along a eircle Each of the following equations in parts (aHe) describes the motion of a particle having the same path, namely the unit circle x 2 + y2 = I. Although the path of each particle in parts (aHe) is the same, the behavior, or "dynamics:' of each particle is different For each particle, answer the following questions.
= (t
- sin t)i + (I - cos t)j.
Graph r(t). The resulting curve is a cycloid.
b. Find the maximum and minimum values ofl v 1and 1a I. (Hint:
Find the extreme values ofl v 12 and I a 12 flISt and take square roots later.) 27. Let r be a differentiable vector function of t. Show that if r' (dr/dt) ~ 0 for all t, thenlrl is constant. 28. Derivative. of triple ••alar product.
a. Show that ifu, v, and w are differentiable vector functions of t, then d du dv dw dt(u'v X w) ~ lit' v X w + n' dt X w + u·v X lit. b. Showthat
-
~ ~j
+ Ilntk,
to = I
22. r(t) = (cost)i + (sint)j + (sin2t)k,
to =
f
~ (r. dr dt
dt
2
X
3
d r) = r' (dr X d r) dt 2 dt dt 3 '
(Hint: Differentiate on the left and look for vectors whose produets are zero.)
13.2 29. Prove the two Scalar Multiple Rules for vector functions. 30. Prove the Sum and Difference Rules for vector functions. 31. Component Test for Continuity at a Point Show that the vector function r defined by r(t) = f(t)i + g(t)j + h(t)k is contiouous att = to if and only if f,g, and h are continuous atto.
32. Limits of cross products of vector functions Suppose that r,(t) = f,(t)i + h(lli + h(t)k, r2(1) = g,(t)i + g2(1)j + g,(t)k, lim,......r'(t) = A,andlim,......r2(t) = B. Use the deterotinant formula for cross products and the Limit Product Rule for
Integrals of Vector Functions; Projectile Motion
35. r(t) ~ (siot - tcost)i + (cost o '" t '" &1r, to = 3orr/2
36. r(t) = V2ti + e'j + e--lk, 37. r(t) ~ (sio2t)i to = orr/4 38. r(t) = (10 (t 2 -3 ::;; t:5 5,
+
(10(1
+ tsiotli + t 2k,
-2 '" t '" 3,
+ t))j + tk,
+ 2))i + (tan-' 3t)j + to
=
715
to = 1
0'" t '" 4orr,
v't'+l k,
3
10 Exercises 39 and 40, you will explore graphically the behavior of the helix
scalar functions to show that
r(t) 33. Differentiable vector functions are continuous Show that if r(t) ~ f(t)i + g(t)j + h(t)k is differentiahle at t ~ to, then it is continuous at to as well.
34. Constant Function Rule Prove that if u is the vector function with the constant value C, then do/dt ~ O.
COMPUTER EXPLORATIONS Use a CAS to perform the followiog steps io Exercises 35-38.
a. Plot the space curve traced out by the position vector r. b. Fiod the components of the velocity vector dr/dt.
c. Evaluate dr/dt at the givenpoiot to and deterotine the equation of the tangent lioe to the curve at r(to).
= (cosat)i +
(sioat)j
+ btk
as you change the values of the constants a and b. Use a CAS to perform the steps in each exercise. 39. Set b ~ I. Plot the helix r(t) together with the tangent lioe to the curve at t = 3orr/2 for a = 1, 2, 4, and 6 over the interval o '" t '" 4orr. Descrihe io your own words what happens to the graph of the helix and the position of the tangent line as a increases through these positive values. 40. Set a ~ I. Plot the helix r(t) together with the tangent lioe to the curve at t = 3orr/2 for b = 1/4, 1/2,2, and 4 over the interval o '" t '" 4orr. Descrihe io your own words what happens to the graph of the helix and the position of the tangent line as b increases through these positive values.
d. Plot the tangent lioe together with the curve over the given interval.
13.2
Integrals of Vector Functions: Projectile Motion In this section we investigate integrals of vector functions and their application to motion along a path in space or in the plane.
Integrals of Vector Functions A differentiable vector function R(I) is an antiderivative of a vector function r(1) on an interval I if dR/ dl = r at each point of 1. If R is an antiderivative of r on I, it can be shown, working one component at a time, that every antiderivative of r on I has the form R + C for some constant vector C (Exercise 41). The set of all antiderivatives of r on I is the iudeUnite integral of r on 1.
DEFINmON The indef"mite integral of r with respect to I is the set of all antiderivatives ofr, denoted by r(l) dl. IfR is any antiderivative ofr, then
J
J
r(l) dl = R(I)
+
C.
The usual arithmetic rules for indefinite integrals apply.
716
Chapter 13: Vector-Valued Functions and Motion in Space
EXAMPLE 1
To integrate a vector function, we integrate each of its components.
J
«cos /)1
+j
(J
- 2lk)dl =
COSld}
(J dl)i - (J
+
- (1 2
+ C,)I + (I + C2)j (sin 1)1 + Ij - 12k + C
= (sin I =
2ldl)k
+ C3)k
(1) (2)
C ~ CII + C,l - C,k
As in the integration of scalar functions, we recommend that you skip the steps in Equations (I) and (2) and go directly to the fmal form. Find an antiderivative for each component and add a constant veclor at the end. _ Defmite integrals of vector functions are best defmed in tenus of components. The definition is consistent with how we compute limits and derivatives of vector functions.
DEFINmON Ifthe components ofr(l) = /(1)1 + g(l)j + h(l)k are integrable over [a, h], then so is r, and the defmlte integral ofr from a to h is
t EXAMPLE 2
r(l)dl =
(t /(I)dl) + (tg(l)d) + (th(l)dl)k.
As in Example I, we integrate each component.
J.w«COSI)1
+j
(J.w
- 2lk)dl = =
COSld}
[sinIJ~ I + [IJ~ j
= [0 - 0]1
+ [1T -
(J.w dl)i -
+ -
(J.w2Idl)k
[12J~ k
0li - [1T 2 - 02]k
1T 2k
= 1Tj -
The Fundamental Theorem of Calculus for continuous vector functions says that
t
r(l)
dl
-
= R(I) J: = R(h) - R(a)
where R is any antiderivative ofr, so that R'(I) = r(l) (Exercise 42).
EXAMPLE 3 Suppose we do not know the path ofa hang glider, but only its acceleration vectora(l) = -(3 cos 1)1 - (3 sin I)j + 2k. We also know that initially (at time I = 0) the glider departed from the point (3, 0, 0) with velocity v(O) = 3j. Find the glider's position as a function ofI. Solution
Our goal is to find r(l) knowing
d2~
dl
= -(3 cos 1)1 - (3 sin l)j
The differential equation:
a =
The initial conditions:
v(O) = 3j
and
r(O) = 31
+
(3 cos I)j
+ 21k +
Ct.
We use v(O) = 3j to find C , : 3j = -(3 sin 0)1 3j = 3j
C, =
o.
+
Ct
+
(3 cos O)j
+
(O)k
+
2k
+ OJ + Ok.
Integrating both sides of the differential equation with respect to I gives V(I) = -(3 sin 1)1
+
C,
13.2 Integrals of Vector Functions; Projectile Motion
717
The glider's velocity as a function of time is
~~
+
= V(I) = -(3 sin 1)1
(3 cos I)j
+ 21k.
Integrating both sides of this last differential equation gives
z
r(l) = (3 cos 1)1
+ (3 sin I)j + 12k + C 2.
We then use the initial condition r( 0) = 31 to f'md C2: 31 = (3 cos 0)1 31 = 31 (3,0,0) x
---"-"y
FIGURE 13.9 The path of the hang glider in Example 3. Although the path spirals
around the z-axis, it is not a helix.
y
+ (O)j
+ (3 sin O)j + + (O)k + C2
(02)k
+ C2
C 2 = O. The glider's position as a function of I is r(l) = (3 cos 1)1
+
(3 sin I)j
+
12k.
This is the path of the glider shown in Figure 13.9. Although the path resembles that of a helix due to its spiraling natore around the z-axis, it is not a helix because of the way it is rising. (We say more about this in Section 13.5.) Nole: It turned out in this example that both of the constant vectors of integration, C, and C 2, are O. Exercises 15 and 16 give examples for which the constant vectors of integration are not O. •
The Vector and Parametric Equations for Ideal Projectile Motion
--~e..,-.J....~--_x
r=Oat/lvolcosai timet = 0 a~
-gJ
(a)
y
A classic example of integrating vector functions is the derivation of the equations for the motion of a projectile. In physics, projectile motion describes how an object f'rred at some angle from an initial position, and acted upon by only the force of gravity, moves in a vertical coordinate plane. In the classic example, we ignore the effects of any frictional drag on the object, which may vary with its speed and altitode, and also the fact that the force of gravity changes slightly with the projectile's changing height. In addition, we ignore the long-distance effects of the Earth torning beneath the projectile, such as in a rocket launch or the f'tring of a projectile from a cannon. Ignoring these effects gives us a reasonable approximation of the motion in most cases. To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projectile during its flight is the constant force of gravity, which always points straight down. We assume that the projectile is launched from the origin at time I = 0 into the f'rrst quadrant with an initial velocity Vo (Figure 13.10). Ifvo makes an angle a with the horizontal, then Vo = (lvolcosa)1
+
(lvolsina)j.
If we use the simpler notation Vo for the initial speed Ivol, then Vo = (vo cos a)1 nI'"-------------~x
+
(vosina)j.
(3)
The projectile's initial position is
I<-----R----->!'I
ro = OJ
+ OJ
=
o.
(4)
Horizontal range (b)
FIGURE 13.10
Newton's second law of motion says that the force acting on the projectile is equal to the projectile's mass m times its acceleration, or m(d2 r/dI 2 ) ifr is the projectile's position vector and I is time. If the force is solely the gravitatiouaI force -mgj, then 2
d r = -mgJ. mdl 2
d d '- r . -gJ dl 2 -
an
718
Chapter 13: Vector-Valued Functions and Motion in Space where g is the acceleration due to gravity. We find r as a function of I by solving the following initial value problem.
d'r
Differential equation:
dl 2 =
r = ro
Initial conditions:
.
-gJ
dr
and
dl = Vo
when I = 0
The first integration gives
~;
-(gt)j
=
+ Vo·
A second integration gives r = _!gt2j
+ vol + roo
Substituting the values ofvo and ro from Equations (3) and (4) gives
+ (vo cos a)li + (vosina)lj + O.
r = -tgl1
vol
Collecting terms, we have Ideal Projectile Motion Eqnation r = (vocosa)li
+
(voSina)1 - !gt2)i.
(5)
Equation (5) is the veclor equalion for ideal projectile motion. The angle a is the projectile's launch angle (firing angle, angle of elevation), and Vo, as we said before, is the projectile's initial speed. The components ofr give the parametric equations
x=(vocosa)1
y=(vosina)I-!gt2,
and
(6)
where x is the distance downnmge and y is the height of the projectile at time I ;;,: O.
EXAMPLE 4 A projectile is fired from the origin over horizontal ground at an initial speed of 500 m/sec and a launch angle of 60°. Where will the projectile be 10 sec later? Solution We use Equation (5) withvo = 500,a = 60°,g = 9.8, and I = 10 to find the projectile's components 10 sec after firing. r = (vocosa)li
+
= (500) (t)(IO)i
'" 25001
+
(voSina)1 - !gt2)i
+
(500)('{3)1O - (!)<9.8)(100))i
3840j
Ten seconds after f"rring, the projectile is about 3840 m above ground and 2500 m downrange from the origin. • Ideal projectiles move along parabolas, as we now deduce from Equations (6). If we substitute I = x/(Vo cos a) from the first equation into the second, we obtain the Cartesiancoordinate equation
y=-(
g 2
2
2vo cos a
TIris equation has the form y = ax 2
2 )x +(tana)x.
+ bx, so its graph is a parabola.
13.2 Integrals of Vector Functions; Projectile Motion
719
A ]ITOjectile reaches its highest point when its vertical velocity component is zero. When fired over horizontal ground, the projectile lands when its vertical component equals zero io Equation (5), and the range R is the distance from the origio to the poiot of impact. We summarize the results here, which you are asked to verify io Exercise 27.
y
Height, Flight Time, and Range for Ideal Projectile Motion For ideal ]ITOjectile motion when an object is launched from the origin over a horizontal surface with initial speed Vo and launch angle a:
Maximum height:
Ymax =
t
Flight time: Range:
R
=
(vosioa)' 2g 2vo sin a
---'~
g
vl .
= gsm2a.
-;o+----------~->·
FIGURE 13.11 The path ofa projectile rITed from ('0, Yo) with ao initial velocity Vo at an angle ofa degrees with the
Ifwe rJre our ideal projectile from the point (xo, Yo) instead of the origin (Figure 13.11), the position vector for the path ofmotion is
horizootal.
r = (xo
+ (vocosa)t)i +
&0 + (vosioa)t -
tgt2)j,
(7)
as you are asked to show io Exercise 29.
Projectile Motion with Wind Gusts The next example shows how to account for another force acting on a ]ITOjectile, due to a gust of wind. We also assume that the path of the baseball io Example 5 lies io a vertical plane.
EXAMPLE 5 A baseball is hit when it is 3 fi above the ground. It leaves the bat with ioitial speed of 152 fi/sec, making an angle of 20° with the horizontal. At the iostant the ball is hit, an instantaneous gust of wind blows io the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of -8.8i (fi/sec) to the ball's ioitial velocity (8.8 fi/sec = 6 mph). (a) Fiod a vector equation (position vector) for the path ofthe baseball.
(h) How high does the baseball go, and when does it reach maximum height? (c) Assuming that the ball is not caught, fiod its range and flight time. Solution
(a) Usiog Equation (3) and accounting for the gust of wind, the ioitial velocity of the baseball is Vo = (vocosa)i
+
= (152cos200)i
(vosioa)j - 8.8i
+
(152sio200)j - (8.8)i
= (152 cos 20° - 8.8)i
The ioitial position is ro = OJ
+
~~
+
(152sio20°)j.
3j. Integration of d'rldt 2 = -gj gives = -(gt)j
+
vo.
720
Chapter 13: Vector-Valued Functions and Motion in Space
A second integration gives
I 2.J r = -2,gt
+ vol + roo
Substituting the values of Vo and ro into the last equation gives the position vector of the baseball. r =
I 2.J + vol + ro -Zgt
= -161 2j
+ (152 cos 20°
= (152 cos 20° - 8.8)11
- 8.8)11
+
(3
+
+ (152 sin 20
0
)lj
+ 3j
(152 sin 20°)1 - 161 2)j.
(b) The baseball reaches its highest point when the vertical component of velocity is zero, or
dy . dl = 152sm20° - 321 = O. Solving for I we find 20° '" I 62 I = 152 sin 32 . sec. Substituting this time into the vertical component for r gives the maximum height
y_ = 3
+ (152 sin 20°)(1.62)
- 16(1.62)2
'" 45.2 ft. That is, the maximum height of the baseball is about 45.2 ft, reached about 1.6 sec after leaving the bat. (0) To find when the baseball lands, we set the vertical component for r equal to 0 and solve for I: 3 + (152 sin 20°)1 - 161 2 = 0
3
+
(51.99)1 - 161 2 = O.
The solution values are about I = 3.3 sec and 1= -0.06 sec. Substituting the positive time into the horizontal component for r, we fmd the range
R = (152 cos 20° - 8.8)(3.3) '" 442 ft. Thus, the horizontal range is about 442 ft, and the flight time is about 3.3 sec.
_
In Exercises 37 and 38, we consider projectile motion when there is air resistance slowing down the flight.
Exercises 13.2 Integrattng Vedor-Valued Funettons Evaluate 1he integrals in Exercises 1-10.
1./,\t'1
+ 7j + (t + I)k] dt
2. [[(6 -
6t)1
+ 3v'tj +
(~)k] dt
~/4
1 t)1 + (1 + t)j + t)k] dt r/ 4. Jo [(secttant)1 + (tant)j + (2sintcost)kjdt
3.
[(sin
cos
~/4
3
5. [[~I + 5~
t j + itk]dt
(sec2
13.2
Integrals of Vector Functions; Projectile Motion
721
6. J . l [ b i +
20. Finding muzzle speed Find the muzzle speed of a gun whose maximum range is 24.5 km.
7. J.'rlet'i + e-t j + k] dl
21. Flight time and height A projectile is fued with an initial speed of 500 mfsec at an angle of elevation of 45'.
V3,k]dl 01-1'1+1
a. When and how far away will the projectile stri1re? b. How high overhead will the projectile he when it is 5 km downrange?
8. j"'\leti+etj+lnlk]dl 9.
J.~/2[COSli
- sin21j + sin'lk]dl
10.
J.~[SeCli +
tan'lj - Isinlk]dl
c. What is the greatest height reached by the projectile? 22. Throwing a hasehall A baseball is thrown from the stands 32 I'l above the fIeld at an angle of 30' up from the horizontal. When and how far away will the ball strike the ground if its initial speed is 32 I'll sec?
initial Value Problems Solve the initial value problems in Exercises 11-16 for r as a vector function of I.
11. Differeotial equation: initial condition: 12. Differeotial equation: initial condition: 13. Differeotial equation: initial condition: 14. Differeotial equation: initial condition: 15. Differeotial equation: initial conditions:
-dr
=
dl
-11. - IJ• - I k
r(O) = i
+ 2j +
~~ = (1801)i + r(O)
~
dr = 1(1 + 1)1/2. + -t· + _I_ k dl2 leJ I+ 1 r(O)
~
k
~~ = (13 r(O)
~
i
+ 41)i + Ij + 21'k
+j
dl' r(O) = lOOk and
dl'
~
-(I
8j
+ j + k)
r(O) = lOi + IOj + 10k and dr dt
b. For the same initial speed, fmd the two fIring angles that make the range 6 m. 24. Beaming eIeetrons An electron in a TV tuhe is beamed horizontally at a speed of 5 X 10' mlsec toward the face of the tube 40 em away. About how far will the electron drop before it hits? 25. Equal-range fIring angles What two angles of elevation will enable a projectile to reach a target 16 km downrange on the same level as the gun ifthe projectile's initial speed is 400 mfsec?
It=O --
a. Show that doubling a projectile's initial speed at a given launch angle multiplies its range by 4. b. By about what percentage should you increase the initial speed to double the height and range?
d'r = -32k
d'r
a. What was the ball's initial speed?
26. Range and height versus speed
I = 8i +
initial conditions:
(1801 - 161')j
looj
dr dt t-O 16. Differeotial equation:
3k
23. Firing golf halls A spring gun at ground level fues a golf ball at an angle of 45'. The balllands 10 m away.
0
27. Verify the results given in the text (following Example 4) for the maximum height, flight time, and range for ideal projectile motion. 28. Colliding marbles The accompanying figure shows an experiment with two marbles. Marble A was launched toward marble B with launch angle a and initial speed vo. At the same instant, marble B was released to fall from rest at R tan a units directly above a spot R units downrange from A. The marbles were found to collide regardless of the value of Vo. Was this mere coinci-
dence, or must this happen? Give reasons for your answer.
Motion Along a Straight Line 17. At time I = 0, a particle is located at the point (I, 2, 3). It travels in a straight line to the point (4, 1,4), bas speed 2 at (I, 2, 3) and constant acceleration 3i - j + k. Find an equation for the position vector r(1) ofthe particle at time I. 18. A particle traveling in a straight line is located at the point (1, -1,2) and has speed 2 at time I = O. The particle moves toward the point (3, 0, 3) with constant acceleration 2i + j + k. Find its position vector r(l) at time I.
Projectile Motion Projectile flights in the following exercises are to he treated as ideal unless stated otherwise. AJllaunch angles are assumed to be measured from the horizontal. All projectiles are assumed to be launched from the origin over a horizontal smface unless stated otherwise. 19. Travel time A projectile is fIred at a speed of 840 mlsec at an angle of60'. How long will it take to get 21 km downrange?
29. Firing from (>:0, Yo)
Derive the equations
x
=
Xo
Y
~ Yo
+ (vo cos a)t, + (vosina)1 -
kg,'
(see Equation (7) in the text) by solving the following initial value problem for a vector r in the plane.
722
Chapter 13: Vector-Valued Functions and Motion in Space d 2r . dt 2 = -gJ
Differential equation:
./
/
r(O) = xa i
Initial conditions:
~~ (0) =
//
+ Yaj
......
......
/
Green:
//
I I
/
(vQ cos a)i + (vQ sin a)j
Pin
/
116 ft/sec
: 45°
I
-~~---------------~
30. Where trajectories crest For a projectile fired from the ground at launch angle a with initial speed va, consider a as a variable and Va as a fixed constant. For each a, 0 < a < 7r/2, we obtain a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse 2
x + 4 (y where x
~
V2)2 4~
33. Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6-ft-high net. It leaves the point of impact with an initial velocity of 35 ft/ sec at an angle of 27° and slips by the opposing team untouched.
d. When is the volleyball 7 ft above the ground? How far (ground distance) is the volleyball from where it will land?
".".-----
/
/
/
I
\
369 ft
NOT TO SCALE
c. Find its range and flight time.
/~,.
,
1
b. How high does the volleyball go, and when does it reach maximum height?
Ellipse
I
Tee I+(------------~)II
a. Find a vector equation for the path of the volleyball.
O. y
..",-
T
45 ft
e. Suppose that the net is raised to 8 ft. Does this change things? Explain.
Parabolic trajectory
34. Shot put In Moscow in 1987, Natalya Lisouskaya set a women's world record by putting an 8 lb 13 oz shot 73 ft lOin. Assuming that she launched the shot at a 40° angle to the horizontal from 6.5 ft above the ground, what was the shot's initial speed?
,
"" ...........
_ _ _ _....._-_---=-~......,.--~=----------~----310.____+
o
X
31. Launching downhill An ideal projectile is launched straight down an inclined plane as shown in the accompanying figure. a. Show that the greatest downhill range is achieved when the initial velocity vector bisects angle AOR. b. If the projectile were fired uphill instead of down, what launch angle would maximize its range? Give reasons for your answer.
R
32. Elevated green A golf ball is hit with an initial speed of 116 ft/ sec at an angle of elevation of 45° from the tee to a green that is elevated 45 ft above the tee as shown in the diagram. Assuming that the pin, 369 ft downrange, does not get in the way, where will the ball land in relation to the pin?
35. Model train The accompanying multiflash photograph shows a model train engine moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fired into the air by a spring in the engine's smokestack. The marble, which continued to move with the same forward speed as the engine, rejoined the engine 1 sec after it was fired. Measure the angle the marble's path made with the horizontal and use the information to find how high the marble went and how fast the engine was moving.
36. Hitting a baseball under a wind gust A baseball is hit when it is 2.5 ft above the ground. It leaves the bat with an initial velocity of 145 ft/ sec at a launch angle of 23 0. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of -14i (ft/sec) to the ball's initial velocity. A 15-fthigh fence lies 300 ft from home plate in the direction ofthe flight. a. Find a vector equation for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height?
13.2 e. Find the range and flight time of the baseball, assuming that the ball is not caught
d. When is tlte baseball 20 It high? How far (ground distance) is the baseball from home plate at that height?
Integrals of Vector Functions; Projectile Motion
Theory and Examples 39. Establish the following properties of integrable vector functions. L
The Co1lJlanJ Scalar Mulilple Rule:
e. Has the batter hit a home run? Explain.
16kr(l) dt
Projectile Motion wlth LInear Drag The main force affecting the motion of a projeetile, other than gravity, is air resistance. This slowing down force is drag force, and it acts in a direction opposlle to !be velocity of the proje
~ k16r(l) dt
1
6 6 (-r(1)) dl = - 1 r(1) dl,
is obtained by taking k = -I. b. The Sum and Dliference RJJ1O$: 1
6 (r,(I) ± r2(1» dt = 16r,(I) dl ± 16r2(1) dt
<. The C01lJlanJ Vector Muiliple
16C'r(I)dt
x
=
Derive the equations
y = va (l - e-kt)(sina) k
+ ~(l 2 k
- kl _ e->'J
Initial conditiOllJ:
d'r. ~
dr
0
1=0
= va = (vo cos a)1
(any constant vector C)
d dr do dl(ur) = u dl + r dl' 41. Antiderivatives ofvectorfunetions
+
(vosina)j
The drag eoefficlent k is a positive oonstanl representing resistance due to air density, va and a are the projeetile's initial speed and launch angle, and g is the acceleration of gravity.
38. Hitting a baseball with linear drag Consider the baseball problem in Example 5 wbeo there is linear drag (see Exercise 37). Assume a drag coefficient k = 0.12, but no gust of wind. L
6 r(l) dl
b. If u and r are both differentiable on [a, b], show that ur is differentiable on [a, b] and that
L
ddtrl
X 1
a. Show that ur is continuous on [a, b] if u and r are continuous on [a, b].
dt' = -gJ - tv = -gl - k dl
r(0)
=C
(any COllJtant vector C)
40. Products of ..alar and vector function. Suppose that the scalar function u(l) and the vector function r(1) are both defmed fora ::s t :s b.
by solving the following initial value problem for a vector r in the plane. Differential equation:
Rules:
~ c·16r(l)dl
and 6 1 C X r(l) dl
~O(l - e-kt)cosa
(any scalar k)
The RJJlejiJr NegativeJ,
J
37. Linear drag
723
From Exercise 37, fmd a vector form for !be path of!be
baseha11. b. How high does the baseha11 go, and when dues it reach maximum height?
Use Corollary 2 of the Mean Value Theorem for scalar functions to show that if two veetor functiOllJ R,(I) and R,(I) have identical derivatives on an interval I, !ben !be functions differ by a consrant veetor value throughout I.
b. Use the resu1t in part (a) to show that ifR(l) is any anJiderivative ofr(l) on I, then any other anJiderivative ofr onI equals R(I) + C for some COllJtant ve
dl'
dl • r(T) dor = r(1)
e. Find the range and flight time of the baseball
d. When is the baseball 30 It high? How far (ground distance) is the baseball from home plate at that height? e. A ID-It-high outfield feoce is 340 It from home plate in the direction of the fligbt of the baseball. The outfielder can jump and cal<:h any ball up to 11 It offthe ground to stop it from going over the feoce. Has the batter hit a home run?
at every point I of (a, b). Then use the eonclusion in part (b) of Exer
1 6
r(l) dl = R(b) - R(a).
724
Chapter 13: Vector-Valued Functions and Motion in Space
43. Bitting a baseball with linear drag under a wind gust Consider again the bsseball problero in Example 5. This time assume a drag coefficient of 0.08 and an instantaneous gust of wind that adds a component of -17.61 (ft/sec) to the initial velocity at the instant the baseball is hit.
e. A 20-ft-high outfield fence is 380 ft from home plate in the direction of the flight of the baseball. Has the batter hit a home run? If"yes." what change in the horizontal component of the bsll's initial velocity would bsve kept the ball in the park? If''no:' what change would bsve allowed it to be a homerun?
a. Find a vector equation for the path of the baseball.
44. Heigbt versus time Show that a projectile attains three-quarters of its maximum height in balf the time it lakes to reach the maximum height.
b. How high does the baseball go, and when does it reach
maximum height? e. Find the range and flight time of the baseball. d. When is the bsseball35 fthigh? How far (ground distance) is the baseball from home plate at that height?
13.3 Basepoint
~
~1
r~
Arc Length in Space In this and the next two sections, we study the mathematical features of a curve's shape that describe the sharpness of its turning and its twisting.
/t '"
"
Arc Length Along a Space Curve
4
s
FIGURE 13.12 Smooth corvos can be sealed like number lines, the coordiaate of each point being its directed distance along the curve from a preselected base point.
One of the features of smooth space and plane curves is that they have a measurable length. This enables us to locate points along these curves by giving their directed distance s along the curve from some base point, the way we locate points on coordinate axes by giving their directed distance from the origin (Figure 13.12). This is what we did for plane curves in Section 11.2. To measure distance along a smooth curve in space, we add a z-term to the formula we use for curves in the plane.
DEFINmON
The length of a smooth curve r(l) = X(I)i
a ,;; I ,;; b, that is traced exactly once as I increases from I
=
+ Y(I)j + Z(I)k,
a to I
=
b, is (1)
Just as for plane curves, we can calculate the length of a curve in space from any convenient parametrization that meets the stated conditions. We omit the proof. The square root in Equation (I) is lvi, the length of a velocity vector dr/dl. This enables us to write the formula for length a shorter way.
Arc Length Formula (2)
EXAMPLE 1 Agliderissoaringupwardalongthehelixr(l) = (cosl)i How long is the glider's path from I = 0 to I = 2fT?
+ (sinl)j + Ik.
13.3 z
t=
I
I
,
I I I
_----t----(/ 0 r "t ;-------''''-P
lblvldt
1 Y2
=
Y(-sint)2
=
This is stands.
!!.. 2
(lfdt
dt = 21TY2 units oflength.
Y2 times the circumference of the circle in the xy-plane over which the helix _
Ifwe choose a base point P{to) on a smooth curve C parametrized by t, each value of t determines a point P{t) == (x{t), y{t), z{t)) on C and a "directed distance"
y x
s(t) =
FIGURE 13.13
+ (costf +
21T
= "
I
=
",.,. 1T
"
:
1
21T
L
_--_ )27T / 21T
725
SoLution The path segment during this time corresponds to one full turn of the helix (Figure 13.13). The length of this portion of the curve is
)
t=
Arc Length in Space
The helix in Example 1,
r(t) = (cos t)i + (sin t)j + tk.
z
t
v( 7) Id7,
ito l
measured along C from the base point (Figure 13.14). This is the arc length function we defined in Section 11.2 for plane curves that have no z-component. If t > to, set) is the distance along the curve from P{to) to pet). If t < to, set) is the negative of the distance. Each value of s determines a point on C and this parametrizes C with respect to s. We call s an arc length parameter for the curve. The parameter's value increases in the direction of increasing t. We will see that the arc length parameter is particularly effective for investigating the turning and twisting nature of a space curve.
Arc Length Parameter with Base Point P{to) s(t) =
P(t)
t Y[x'(7)f + [y'( 7)f + [z'( 7)f d7 i~t lv(7) Id7 =
i~
(3)
Base
7-
POint~
x
P(to)
FIGURE 13.14 The directed distance along the curve from P(to) to any point P(t) is
We use the Greek letter T ("tau") as the variable of integration in Equation (3) because the letter t is already in use as the upper limit. If a curve r( t) is already given in terms of some parameter t and s(t) is the arc length function given by Equation (3), then we may be able to solve for t as a function of s: t == t{s). Then the curve can be reparametrized in terms of s by substituting for t: r == r{t{s)). The new parametrization identifies a point on the curve with its directed distance along the curve from the base point.
EXAMPLE 2 This is an example for which we can actually find the arc length parametrization of a curve. If to == 0, the arc length parameter along the helix r{t)
==
{cos t)i
+ (sin t)j + tk
from to to t is s(t) =
= ==
t
v( 7) Idr
ito l
1
tY2d7
Eq. (3)
Value from Example 1
Y2 t. Y2.
Solving this equation for t gives t == s/ Substituting into the position vector r gives the following arc length parametrization for the helix: r(t(s)) = (cos
~} + (sin ~} + ~k.
•
726
Chapter 13: Vector-Valued Functions and Motion in Space
Unlike Example 2, the arc length parametrization is generally difficult to find ana1ytically for a curve already given in tenns of some other parameter t. Fortunately, however, we rarely need an exact fonnula for s(t) or its inverse I(s). HISTORICAL BIOGRAPHY
Josiah Willard Gibbs (1839--1903)
Speed on a Smooth Curve Since the derivatives beneath the radical in Equation (3) are continuous (the curve is smooth), the Fundamental Theorem of Calculus tells us that s is a differentiable function of t with derivative (4)
Equation (4) says lbat lbe speed with which a particle moves along its path is the magnitude of v, consistent with what we know. Although the base point P(to) plays a role in defining s in Equation (3), it plays no role in Equation (4). The rate at which a moving particle covers distance along its path is independent of how far away it is from the base point. Notice that ds/dt > 0 since, by definition, Ivl is never zero for a smooth curve. We see once again that s is an increasing function of t.
Unit Tangent Vector We already know the velocity vector v = dr/dt is tangent to the curve r(t) and that the
vector
is therefore a unit vector tangent to the (smooth) curve, called the unit tangent vector (Figure 13.15). The unit tangent vector T is a differentiable function of t whenever v is a differentiable function of t. As we will see in Section 13.5, T is one of three unit vectors in a traveling reference frame that is used to describe the motion of objects traveling in three dimensions.
EXAMPLE 3 FIGURE 13.15 We !md lbe unit taogeot vector T by dividing v by Iv I.
Find the unit tangent vector of the curve r(t) = (3 cos t)i + (3 sin t)j + t 2k
representing the path of the glider in Example 3, Section 13.2.
Solution
In that example, we found v =
~~
= -(3 sin t)i + (3 cos t)j + 2tk
and Ivl =
y
Y9 + 4t 2 •
Thus,
T=~=Ivl (1,0) x
3sint i+ 4t 2
Y9 +
3cost j+ 4t 2
Y9 +
For the counterclockwise motion r(t) = (cos t)i + (sin t)j around the unit circle, we see that
FIGURE 13.16 Counterclockwise motion around the unit circle.
v = (-sint)i + (costH is already a unit vector, so T = v (Figure 13.16).
2t
Y9 +
k.
4t 2
•
13.3 Arc Length in Space
727
The velocity vector is the change in the position vector r with respect to time t, but how does the position vector change with respect to arc length? More precisely, what is the derivative dr/ds? Since ds/dt > 0 for the curves we are considering, s is one-to-one and has an inverse that gives t as a differentiable function of s (Section 7.1). The derivative of
the inverse is I ds/dt
dt ds
I
Ivl'
This makes r a differentiable function of s whose derivative can be calculated with the Chain Rule to be
dr
dr dt ds = dt ds =
vTVlI =
v
[Vj
=
(5)
T.
This equation says that dr/ds is the unit tangent vector in the direction ofthe velocity vector v (Figure 13.15).
Exercises 13.3 Finding Tangent Vecton and Lengths In Exercises 1-8, find the curve's unit tangent vector. Also, f'md the length of the indicated portion of the curve.
1. 2. 3. 4.
r(t) r(t) r(t) r(t)
+ (2 sin t)j + V5tk, (6 sin 2t)i + (6 cos 2t)j + 5tk,
= (2 cos t)i =
= ti
+ (2/3)t'/2k,
+ t)i - (I + I)j + (COS'I)j + (sin'I)k,
5. r(l) =
6. r(l) ~ 61'i - 21'j - 31'1<, 7. r(l) = (I cos l)i 8. r(l) ~ (I sin I
+
(I sin I)j
Ik,
0 '" t '" 'IT
0 '" I'" 'IT/2 (2V2/3 )I'/'k,
+ cos I)i + (I cos I
- sin I)j,
0 '" I '" 'IT V2 '" I '" 2
9. Find the point on the curve r(l)
= (Ssinl)i +
(Scosl)j
+
121k
at a distance 267T units along the curve from the point (0, 5, 0) in the direction of increasing arc length. 10. Find the point on the curve r(l) ~ (12 sin I)i - (12 cos I)j
+ 51k
at a distance l3'IT units along the curve from the point (0, -12,0) in the direction opposite to the direction of increasing arc length. Arc Length Parameter In Exercises 11-14, find the arc length parameter along the curve from the point where I ~ 0 by evaluating the integral
s = /.tIV(T)ldT from Equation (3). Then f'md the length of the indicated portion of the curve. 11. r(l) = (4 cos I)i 12. r(l) = (cosl
(4 sin I)j
+
311<,
+ Isinl)i + (sin I -
13. r(l) = (e t cos I)i 14. r(l) = (I
+
+ 21)i
0 '" I '" 'IT/2
lCOSI)j,
+ (e t sin I)j + etk, + (I + 31)j + (6 -
Find the length of the curve
+
r(1) = (V2I)i
(V2I)j
+
(I - I')k
from (0, 0, I) to (V2, V2, 0) . 16. Length of heIiI The length 2'IT V2 of the turn of the helix in Example I is also the length of the diagonal of a square 2'IT units on a side. Show how to ohtain this square by cutting away and flattening a portion of the cylinder around which the helix winds.
0 '" I '" 3
I '" 1"'2
+
15. Arc length
0 '" t '" 'IT
0 '" t '" 8
= (2
Theory and Examples
'IT/2 '" I '" 'IT
17. Ellipse a. Show that the curve r(1) = (cosl)i + (sinl)j + (I - COSI)k, o '" I '" 2'IT, is an ellipse by showing that it is the intersection of a right circu1ar cylinder and a plane. Find equatioos for the cylinder and plane. b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors at I = 0, 'IT/2, 'IT, and 3'IT/2. c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for I ~ 0, 'IT/2, 'IT, and 3'IT/2 to your sketch. d. Write an integral for the length of the ellipse. Do not try to evaluate the integral; it is nonelementary. De. Numerical integrator two decimal places.
Estimate the length of the ellipse to
18. Length is independent of parametrization To illustrate that the length of a smooth space curve does not depend on the parame1rization you use to compute it, calculate the length of one turn of the helix in Example I with the following parame1rizatioos. a. r(l) = (cos41)i
+
(sin41)j
+
+ 41k,
-inA'" I '" 0
b. r(l) = [cos (1/2)]i
61)k,
c. r(l) = (cos I)i - (sin l)j - Ik,
-I'" 1"'0
[sin (1/2)]j
+
0'" I '" 'IT/2 (1/2)k,
0 '" I '" 47T
-2'IT '" I'" 0
728
Chapter 13: Vector-Valued Functions and Motion in Space
19. The involute of a circle If a string wollOd arollOd a f"lxed circle is IlOwouod while held taut in the plaoe of the circle, its cod P traces ao involule of the circle. 10 the accompanying f"lgun:, the circle in question is the circle x 2 + y2 = 1 and the tracing point starts at (I, 0). The IlOWOllOd portion of the string is taogcot to the circle at g, aod I is the radian measure of the aogle from the positive x-axis to segment OQ. Derive the parametric equations x
cost
=
+ t sin t, y
=
t>
sint - teost,
0
20. (Continuation ofExercise 19.) Find the unit taogcot vector to the involute of the circle at the point p(x, y). Show that if u is a unit vector, then the arc leogth parameter along the line r(l) ~ Po + lu from the point Po(Xo, Yo, zo) where I = 0, is I itself.
21. Distance along a line
22. Use Siropson's Rule with n = 10 to approximate 1he leng1h of arc ofr(l) = Ii + 12 j + 13k from the origin to 1he point (2, 4, 8).
of the point P(x, y) for the involute.
P(x,y)
-t----o;+--'--o-*--;;c-x
13.4
Curvature and Normal Vectors of a Curve In this section we study how a curve turns or bends. We look flISt at curves in the coordinate plane, and then at curves in space.
y
Curvature of a Plane Curve As a particle moves along a smooth curve in the plane, T = dr/ ds turns as the curve bends. Since T is a unit vector, its length remains constant and only its direction changes as the particle moves along the curve. The rate at which T turns per unit of length along the curve is called the curvature (Figure 13.17). The traditional symbol for the curvature function is the Greek letter K (''kappa'').
s
"I
Po
T
--ocl-"---------_x
o
FIGURE 13.17 As P moves along the curve in the direction of increasing arc length, the unit tangent vector tums. The value ofldT/dsl atP is called 1he curvature of the curve at P.
DEFINmON
1fT is the unit vector of a smooth curve, the curvature function
of the curve is
Ifl dT/ds Iis large, T turns sharply as the particle passes throughP, and the curvature If IdT/ ds I is close to zero, T turns more slowly and the curvature at P is
at P is large. smaller.
729
13.4 Curvature and Normal Vectors of a Curve
If a smooth curve r(t) is already given in terms of some parameter t other than the arc length parameter s, we can calculate the curvature as K =
=
I~I = I~~ ~I
Chain Rule
Ids~dtll~~1
=I~II~~I.
:~Ivi
Formula for Calculating Curvature If r(t) is a smooth curve, then the curvature is I IdTI K=Ndt'
(1)
where T = v/lvl is the unit tangent vector.
Testing the definition, we see in Examples I and 2 below that the curvature is constant for straight lines and circles.
EXAMPLE 1 A straight line is parametrized by r(t) = C + tv for constant vectors C and v. Thus, r'(t) = v, and the unit tangent vector T = v/lvl is a constant vector that always points in the same direction and has derivative 0 (Figure 13.18). It follows that, for any value of the parameter t, the curvature ofthe straight line is K
EXAMPLE 2
I~II~~I = I~IIOI = o.
=
•
Here we find the curvature ofa circle. We begin with the parametrization
r(t) = (a cos t)i + (a sin t)j Along a straight line, T always points in the same direction. The curvature, IdT/ tis I, is zero (Example I). FIGURE 13.18
of a circle of radius a. Then, v =
~~
= -(a sin t)1
Ivl = Y(-asint)'
+ (a cos t)j
+ (a cos t)2
=
W
= lal = a.
Since a > 0,
lal ~
a.
From this we find T =
~~
I~~I
I~I = -(sint)1 + (cost)j
= -(cost)1 - (sint)j 2
= Ycos t
+
2
sin t = I.
Hence, for any value ofthe parameter t, the curvature ofthe circle is K
=
I~II~~I = ~(l) = ~ = ra~us·
•
Although the formula for calculating K in Equation (I) is also valid for space curves, in the next section we find a computational formula that is usually more convenient to apply.
730
Chapter 13: Vector-Valued Functions and Motion in Space Among the vectors orthogonal to the unit tangent vector T is one of particular significance because it points in the direction in which the curve is turning. Since T has constant length (namely, I), the derivative dT/ds is orthogonal to T (Equation 4, Section 13.1). Therefore, if we divide dT/ tis by its length K, we obtain a unil vector N orthogonal to T (Figure 13.19).
DEFINmON At a point where a smooth curve in the plane is FIGURE 13.19 The vector dT/d>, normal to the curve, always points in the direction in which T is turning. The unit normal vector N is the direction ofdT/d>.
K
oF 0, the principal onit normal vector for
The vector dT/ tis points in the direction in which T turns as the curve bends. Therefore, if we face in the direction ofincreasing arc length, the vector dT/tis points toward the right if T turns clockwise and toward the left if T turns counterclockwise. In other words, the principal normal vector N will point toward the concave side of the curve (Fignre 13.19). If a smooth curve r(l) is already given in terms of some parameter I other than the arc length parameter s, we can use the Chain Rule to calculate N directly:
dT/tis N = IdT/tisl
(dT/dl)(dl/tis) IdT/dll Idl/tis I dT/dl ---
d.
1
IdT/dll'
>
0 cancels.
This formula enables us to find N without having to tmd K and s first
Formula for Calculatiog N If r(l) is a smooth curve, then the principal unit normal is
dT/dl
(2)
N = IdT/dll' where T = v/ Iv Iis the unit tangent vector.
EXAMPLE 3
Find T and N for the circular motion r(l) = (cos 21)i
Solution
+
(sin 21)j.
We tmt find T: v = -(2 sin 21)i Ivl = V4sin2 21
+
(2 cos 21)j
+ 4cos2 21
T = I~I = -(sin21)i
+
= 2
(cos21)j.
From this we find
~~
I~~I
= -(2 cos21)i - (2 sin21)j 2
= V4cos 21
+ 4sin2 21
= 2
13.4 Curvature and Normal Vectors of a Curve
731
and
dT/dt N = IdT/dtl = -(cos Zt)i
- (sin Zt)j.
Eq. (2)
Notice that T' N = 0, verifYing that N is orthogonal to T. Notice too, that for the circular • motion here, N points from r(t) towards the circle's center at the origin.
Circle of Curvature for Plane Curves Circle of curvature
The circle of curvature or osculating circle at a point P on a plane curve where K oF 0 is the circle in the plane ofthe curve that
1.
is tangent to the curve at P (has the same tangent line the curve has)
2. 3.
lies toward the concave or inner side ofthe curve (as in Figure 13.Z0).
has the same curvature the curve has at P
The radius of curvature ofthe curve at P is the radius of the circle of curvature, which, according to Example Z, is Radius of curvature = p = FIGURE 13.20 The osculating circle at P(x, y) lies toward the inner side ofthe curve.
k.
To find p, we find K and take the reciprocal. The center of curvature of the curve at P is the center ofthe circle of curvature.
EXAMPLE 4
Find and graph the osculating circle ofthe parabola y = x 2 at the origin.
Solution We parametrize the parabola using the parameter t = x (Section 11.1, Example 5)
r(t) = ti
+ t 2j.
First we find the curvature of the parabola at the origin, using Equation (I): v = dr = i dt
Ivl =
+ Ztj
VI + 4t 2
so that
From this we find
At the origin, t = 0, so the curvature is
K(O)
=
Ivto),I~~ (0)1 + Zjl
=
_1_ 101
=
(I) YC'02'+:-Z="2
VI
Eq. (I)
=
Z.
732
Chapter 13: Vector-Valued Functions and Motion in Space y
Therefore, the radius of curvature is 1/K == 1/2. At the origin we have t == 0 and T == i, so N == j. Thus the center of the circle is (0, 1/2). The equation of the osculating circle is therefore
&-tY (tY"
O? +
(x -
=
You can see from Figure 13.21 that the osculating circle is a better approximation to the parabola at the origin than is the tangent line approximation y == o. •
o FIGURE 13.21 The osculating circle for the parabola y = x 2 at the origin (Example 4).
Curvature and Normal Vectors for Space Curves If a smooth curve in space is specified by the position vector ret) as a function of some parameter t, and if s is the arc length parameter of the curve, then the unit tangent vector T is dr/ ds == v/ Iv I. The curvature in space is then defined to be (3)
just as for plane curves. The vector dT/ ds is orthogonal to T, and we define the principal unit normal to be 1 dT
EXAMPLE 5 r(t) SoLution
=
(4)
IdT/dtl"
Find the curvature for the helix (Figure 13.22) ==
+
(a cos t)i
+ btk,
(a sin t)j
~
a,b
0,
We calculate T from the velocity vector v:
+ (a cos t)j + bk
v
==
-(a sin t)i
Ivl
==
Va 2 sin2 t + a 2 cos2 t + b 2
T
==
_IvI = yI 21 v
y
dT/dt
KdS
N =
a
+
==
[-(a sin t)i
b2
Va 2 + b 2
+
(a cos t)j
+
bk]"
Then using Equation (3),
x FIGURE 13.22
1 IdTI K==F1dt
The helix
r(t) = (a cos t)i + (a sin t)j + btk, drawn with a and b positive and t (Example 5).
~
0
=
V
a
==
a
-_
a
2
2a
+
1
+
b
b2
IV
a
2
1
+
b2
[-(a cos t)i - (a sin t)j]
I
21-(cost)i-(sint)jl
2 a 2 Y( cos t )2 +b
+ (.SIn t )2 -_ 2 a
a
+b
2.
From this equation, we see that increasing b for a fixed a decreases the curvature. Decreasing a for a fixed b eventually decreases the curvature as well. If b == 0, the helix reduces to a circle of radius a and its curvature reduces to 1/a, as it should. If a == 0, the helix becomes the z-axis, and its curvature reduces to 0, again as it should. •
13.4 Curvature and Normal Vectors of a Curve
EXAMPLE 6 Solution
733
Find N for the helix in Example 5 and describe how the vector is pointing.
We have
dT dt
I
•
va + b 2
dT =
I dt
I
I
Va2
2
I
+ b2
+ (a sin t)j]
[(a cos t)i
Va 2 cos2 t
+ a 2 sin2 t
=
ExampleS
_r=;ja====.i Va2 + b2
dT/dt N = IdT/dt I
Eq. (4)
V rca2---+-b'2 a = -(cost)i -
V
I
a2 + b 2
[(a cos t)i + (a sin t)j]
(sint}j.
•
Thus, N is parallel to the xy-plane and always points toward the z-axis.
Exercises 13.4 7. Normals to plane curves a. Sbowthatn(t) = -g'(t)i + f(t)j and -n(t) = g'(t)if(t)j are both normal to the curve r(t) ~ I(t)i + g(t)j at the
Plane Curves Find T, N, and I( for the plane curves in Exercises 1-4.
+ (lncost)j, -'lf/2 < t < 'lf/2 + tj, -'lf/2 < t < 'lf/2 (2t + 3)i + (S - t 2 )j (cost + tsint)i + (sint - tcost)j, t>
1. r(t) = ti
2. r(t)
~
3. r(t) = 4. r(t) =
point (f(t), g(t)).
(In sec t)i
0
5. A formula for tbe eurvature of Ibe grapb of a function in Ibe xy-plaoe ~ I(x) in the xy-plane automatically bas the parametrization x = x, y = I(x), and the vector formula r(x) = xi + I(x)j. Use this fonnula to show that if I is a twice-
a. The grapb y
I«x)
If"(x) I
~
[I
+ (f'(x))2]'/2
.
b. Use the fonnula for I( in part (a) to fmd the curvature of y = In (cos x), -'If12 < x < 'If12. Compare your answer with the answer in Exercise 1. c. Sbow that the curvature is zero at a point of inflection. 6. A formula for Ibe curvature of a parametrized plane curve
a. Sbow that the curvature of a smooth curve r(t) = 1(I)i + g(t)j defmed by twice-differeutiable functions x = I(t) and y ~ g(t) is giveo by the formula
liy - yxl 1(=
(i:2
+ y2)'/2'
The dots in the formula deoote differeutiation with respect to t, one derivative for eacb dot. Apply the formula to fmd the curvatures of the following curves. b. r(t)
~
ti
+ (In sin t)j,
c. r(l) = [tan-I (sinbl)ji
0
'If
+ (lncosbl)j.
To obtain N for a particular plane curve, we can choose the one of n or -n from part (a) that points toward the concave side of the curve, and make it into a mtit vector. (See Figure 13.19.) Apply this method to fmd N for the following curves. b. r(t) ~ ti
+ eUj
v'4=7
c. r(t) = i + Ij, -2 '" t '" 2 8. (Continuation ofExercise 7.) a. Use the method of Exercise 7 to fmd N for the curve r(t) Ii + (I/3)t' j wheo t < 0; wheo t > O.
=
b. Calculate N for t#'O directly from T using Equation (4) for the curve in part (a). Does N exist at t = O? Graph the curve and explain what is bappeoing to N as t passes from negative to positive values.
Space Curves Find T, N, andK for the space curves in Exercises 9-16. 9. r(t) ~ (3 sin t)i + (3 cos t)j + 4tk 10. r(t) = (cost + t sin t)i + (sint - tcost)j + 3k 11. r(t) ~ (e'cost)i + (e'sint)j + 2k 12. r(t) = (6sin2t)i + (6cos2t)j + Stk 13. r(t) ~ (t'/3)i + (t 2/2)j, t> 0 14. r(t) = (cos't}i + (sin't)j, 0 < t < 'lf/2 15. r(t) ~ ti + (a cosh (tla))j, a > 0 16. r(t) = (cosht)i - (sinbl)j + Ik
More on Curvature 17. Show that the parabola y = ax 2, a #' 0, bas its largest curvature at its vertex and has no minimmn curvature. (Note: Since the curvature ofa curve remains the same if the curve is translated arrotatell, this result is true for any parabola.)
734
Chapter 13: Vector-Valued Functions and Motion in Space
18. Show that the ellipse x = a cos I, y = b sin I, a > b > 0, has its largest cmvature on its major axis and its smallest curvature on its minor axis. (As in Exercise 17, the same is true for any ellipse.) 19. MllIimizing the curvature of a helli In Example 5, we found the curvature of the helix r(l) ~ (a cos I)i + (a sin I)j + blk (a, b '" 0) to be " = a/(a' + b'). What is the largest value" can have for a given value of b? Give reasons for your answer. 20. Total curvature We find the tutal curvature of the portioo of a smooth curve that runs :from s = So to s = 31 > So by integrating K from So to Sl • If the curve has some other parameter, say t, then the total curvature is
K =
1" 1" "tis =
to
110
"dl tis dl =
1"
"Ivldl,
to
where 10 and I, com:spood to So and 51 • Find the tota1 curvatures of
a. The portion of the helix r(l) = (3 cos I)i 0:5 t:5 41T. h. The parabola y = x 2 ,
-00
<X <
+ (3 sin I)j + Ik,
b. Calculate the curvature " of the curve at the given value 10 using the appropriate formula from Exercise 5 or 6. Use the paramelrizatioox ~ I andy ~ f(l) if the curve is given as a function y = f(x).
c. Find the unitnorma1 vectorN at 10. Notice that the sigus ofthe compooents ofN depend 00 whether the unit tsngent vector T is turuing clockwise or counterclockwise at 1= 10. (See Exercise 7.) d. IfC = ai + bj is the vector from the origio to the center (a, b) of the oscu1atiog circle, fmd the center C from the vector equatioo
C = r(lo)
22. Find an equation for the circle of curvature of the curve r(l) = (2lnl)i - [I + (I/I)]j, .-',;; I';; .', at the point (0, -2), where I ~ I.
D The formula
The point P(xo, Yo) r(lo) .
00
+ _(I) N(lo). " 10
the curve is given by the positioo vector
e. Plot implicitly the equation (x - a)' + (y - b)' = I/K'ofthe osculatiug circle. Then plot the curve and osculatiug circle together. You may need to experiment with the size of the viewiog window, but be sure it is square.
+ (5 sin I)j, 0,;; I ,;; 2'ff, 10 = 'ff/4 1)i + (sin3 1)j, 0,;; I ,;; 2'ff, 10 = 'ff/4
27. r(l) = (3 cos I)i ,,(x)
=
28. r(l) =
1f"(x)1
[I + (f'(x))']3/2'
derived in Exercise 5, expresses the curvature K(X) of a twicedifferentiable plane curve y ~ f(x) as a functioo of x. Find the curvature function ofeach ofthe curves in Exercises 23-26. Then graph f(x) together with ,,(x) over the giveo interval. You will fmd some surprises. =
a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like.
00.
21. Find an equation for the circle of curvature of the curve r(l) = Ii + (sinl)j at the point ('ff/2, I). (The curve parametrizes the graph ofy ~ sin x in thexy-plane.)
23. y
COMPUTER EXPLORATIONS In Exercises 27-34 you will use a CAS to explore the osculatiug circle at a pointP 00 a plane curve where" oF O. Use a CAS to perform the following steps:
x 2,
-2:S; x
25. Y = sinx,
13.5
:5
2
O::s x:S 21T
24. y
=
x 4/4,
26. Y = eX,
-2:5 x
:5
2
-1::s x:s 2
(COS
3
29. r(l) ~ 1'1
+ (1 3
30. r(l) = (1 3
-
-
31)j,
21' - I)i
+
-4';; I ,;; 4,
h+
j, -2';; I ,;; 5, I I' (2 - 2 cos I)j, 0,;; I ,;; 3'ff,
31. r(l) ~ (21 - sin I)i + 10 = 3'ff/2 32. r(l) = (e-tcos l)i + (.-tsinl)j,
33. y
=x
2
-
x,
10 ~ 3/5
0,;; I';; 6'ff,
10 = I
10 = 'ff/4
-2 ~ x ~ 5, Xo = 1
34. y ~ x(I - x)'/',
-I';; x ,;; 2,
xo ~ 1/2
Tangential and Normal Components of Acceleration
z
If you are traveling along a space curve, the Cartesian i, j, and k coordinate system for representing the vectors describing your motion is not truly relevant to you. What is meaningful instead are the vectors representative of your forward direction (the unit tangent vector T), the direction in which your path is turning (the unit normal vector N), and the tendency of your motion to "twisf' out of the plaoe created by these vectors in the direction perpendicular to this plane (defined by the unil binormal veclor B = T X N). Expressing the acceleration vector along the curve as a linear combination of this TNB frame of mutual1y orthogonal unit vectors traveling with the motion (Figure 13.23) is particularly revealing of the nature of the path and motion along it.
The TNB Frame FIGURE 13.23 The TNB frame of mutually orthogonal unit vectors traveling along a curve in space.
The binormal vector of a curve in space is B = T X N, a unit vector orthogonal to both T and N (Figure 13.24). Together T, N, and B derme a moving right-handed vector frame that plays a significant role in calculating the paths ofparticles moving through space. It is called the Frenet ("fre-nay") frame (after Jean-Frederic Frenet, 1816-1900), or the TNB frame.
735
13.5 Tangential and Normal Components of Acceleration
Tangential and Nonnal Components of Acceleration B
When an object is accelerated by gravity, brakes, or a combination of rocket motors, we usually want to know how much of the acceleration acts in the direction of motion, in the tangential direction T. We can calculate this using the Chain Rille to rewrite v as
p
_ dr _ dr ds _ T ds dt'
v - dt - ds dt -
FIGURE 13.24 The vectors T, N, and B (in that order) make a right-handed frame ofmutually orIhogooal unit vectors in
Then we differentiate both ends of this string of equalities to get 2 _ dv _ ~ ( ds) _ d s + ds dT a-dt-dt T dt -dt 2T dtdt 2 d s dt 2
=
~:~T + K(:)'N.
space.
DEFINmON
T+ dsdt (dTds ds) dt
=
=
2 ds dt 2
T+ dsdt (KN ds) dt
dT tis
~ KN
Ifthe acceleration vector is written as (1)
then and
(2)
are the tangential and normal scalar components of acceleration.
FIGURE 13.25 The tangential and normal components of acceleration. The
acceleratioo a always lies in the plane ofT and N, orthogonal to B.
FIGURE 13.26 The tangential and normal components ofthe acceleration
ofan ohject that is speeding up as it moves counterclockwise around a circle of radius p.
Notice that the binormal vector B does not appear in Equation (1). No matter how the path of the moving object we are watching may appear to twist and tum in space, the acceleration a always lies in the plane of T and N orthogonal to B. The equation also tells us exactly how much of the acceleration takes place tangent to the motion (d 2s/dt2) and how much takes place normal to the motion [K(ds/dt)'] (Figure 13.25). What information can we discover from Equations (2)? By definition, acceleration a is the rate ofchange ofvelocity v, and in general, both the length and direction ofv change as an o~ect moves along its path. The tangential component of acceleration aT measures the rate of change ofthe length ofv (that is, the change in the speed). The normal component of acceleration aN measures the rate of change of the direction of v. Notice that the normal scalar component ofthe acceleration is the curvatore times the square of the speed. This explains why you have to hold on when your car makes a sharp (large K), high-speed (large Ivl) tum. Ifyou double the speed ofyour car, you will experience four times the normal component of acceleration for the same curvatore. If an object moves in a circle at a constant speed, d 2s/dt 2 is zero and all the acceleration points along N toward the circle's center. If the object is speeding up or slowing down, a has a nonzero tangential component (Figure 13.26). To calculate aN, we usually use the formula aN = Vlal 2 - aT', which comes from solving the equation Ia 12 = a' a = aT' + aN' for aN. With this formilla, we can find UN without having to calcillate K first.
Formula for Calculating the Normal Component ofAcceleration aN =
Vlal 2
-
aT'
(3)
736
Chapter 13: Vector-Valued Functions and Motion in Space
EXAMPLE 1
Without finding T and N, write the acceleration ofthe motion
r(t)
=
(cost
+ tsint)! + (sint - tcost)i,
t >0
in the form a = aTT + lINN. (The path ofthe motion is the involute ofthe circle in Figure 13.27. See also Section 13.3, Exercise 19.) Solution
We use the rmt of Equations (2) to rmd aT: v
-f-----Co,~L-----1~=-~x
o
=
~;
=
(tcost)1 + (tsint)i
(-sint + sint + tcost)! + (cost - cost + tsint)i
=
Ivl =
Vt 2 cos2 t + t 2 sin2 t
aT =
dt
d
d Ivl = dt(t)
=
=
W
It I = t
=
t> 0 Eq. (2)
1.
Knowing aT, we use Equation (3) to find aN: FIGURE 13.27 The tangential and normal components ofthe acceleration of themotionr(l) = (cos I + I sin 1)1 +
a
(sin I - Icos I)j, for I > O. Ifa string
=
while held taut in the plane ofthe circle, its endP traces an involute of the circle
(cost - tsint)!
+ (sint + tcost)i
laI2 =t2 +1 ~2~ ~ aN = Vlal - al
wound around a ftxed circle is unwound
(Example I).
=
V(t2
After some algebra
+ I) - (I)
=
W
t.
=
We then use Equation (I) to rmd a: a
=
aTT + lINN = (I)T + (I)N = T + tN.
•
Torsion How does dB/lis behave in relation to T, N, and B? From the rule for differentiating a cross product, we have dB = d(T X N) = dT X N
lis
lis
Since N is the direction of dT/Iis, (dT/ds) X N dB = 0
lis
+ T X dN
lis
+T
ds' =
0 and
X dN = T X dN.
ds
ds
From this we see that dB/lis is orthogonal to T since a cross product is orthogonal to its factors. Since dB/lis is also orthogonal to B (the latter bas constant length), it follows that dB/lis is orthogonal to the plane of B and T. In other words, dB/ds is parallel to N, so dB/lis is a scalar multiple ofN.1n symbols, dB
ds
=
-'TN.
The negative sign in this equation is traditional. The scalar 'T is called the torsion along the curve. Notice that dB
ds • N
=
-'TN' N
We use this equation for our next definition.
=
-'T(I)
=
-'T.
13.5 Tangential and Normal Components of Acceleration
I
Rectifying plane
Binormal
DEFINmON
Let B
=
737
T X N. The torsion function of a smooth curve is
Normal plane
T=
dB -dS'N.
(4)
Principal normal
< A
Unit tangent
FIGURE 13.28 The names ofthe three planes determined by T, N, and B.
Unlike the curvature K, which is never negative, the torsion T may be positive, negative, or zero. The three planes determined by T, N, and B are named and shown in Figure 13.28. The curvature K = IdT/ ds Ican be thought of as the rete at which the normal plane turns as the point P moves along its path. Similarly, the torsion T = -(dB/ds)' N is the rete at which the osculating plane turns about T as P moves along the curve. Torsion measures how the curve twists. Look at Figure 13.29. If P is a !min climbing up a curved track, the rete at which the headlight turns from side to side per unit distance is the curvature of the tmck. The rete at which the engine tends to twist out of the plane formed by T and N is the torsion. In a more advanced course it can be shown that a space curve is a helix if and only ifit has constant nonzero curvature and constant nonzero torsion.
The tmsion
/ '\ /
The cllIValure at P
~ ~ l(dT/d,)I·
s increases
~-
P
I
8=0
FIGURE 13.29 Every moving body travels with a TNB frame that characterizes the geometry ofits path ofmotion.
Computational Formulas The most widely used formula for torsion, derived in more advanced texts, is
x x T=
y
z
)i
'i
x· y
'z'
Iv X al 2
(ifv X
a.. 0).
(5)
The dots in Equation (5) denote differentiation with respect to I, one derivative for each dot. Thus, x ("x dot") means dx/dl, x ("x double dof') means d 2x/dI 2 , and x ("x triple dot'') means d 3x/dI 3 • Similarly, y = dy/dl, and so on. There is also an easy-to-use formula for curvature, as given in the following summary table (see Exercise 21).
738
Chapter 13: Vector-Valued Functions and Motion in Space
Computation Formulas for Curves In Space v
[Vj
Unit tangeot vector:
T
Principal unit normal vector:
dT/dt N = IdT/dtl
Binormal vector:
B=TXN
=
Curvatore:
x y z x ji "i x "ji ·Z· Iv x al 2
dB --oN =
Torsion:
T =
Tangeotial and normal scalar componeots of acceleration:
a = arT
tis
+ aNN
d ar=dt lvl
aN =
Klvl 2 = Vlal 2 -
al
Exercises 13.5 12. r(t) = (6 sin 2t)1 + (6 cos 2t)j + 5tk
finding Tangential and Normal Components In Exercises 1 and 2, write a in the form a finding T and N.
1. r(t) = (a cos t)1
2. r(t) = (I
= aTT
15. r(t) 16. r(t)
- 2)j - 3tk
In Exercises 3--6, write a in the form a = aTT value oft without f"mding T and N.
3. r(t) ~ (t + 1)1 + 2tj + t 2k,
+ tlNN
t~ 0
6. r(t)
+
+
V2e t k,
t ~ 0
t= 0
finding the TNB Frame In Exercises 7 and 8, f"md r, T, N, and B at the giveo value of t. Theo find equatioos for the osculating, nonnal, and rectifying planes at that value oft. 7. r(t) = (cost)1 + (sint)j - k, 8. r(t) = (cos t)1
+
(sin t)j
t = 'It/4
+ tk, t
= 0
In Exercises !l-16 ofSectioo 13.4, you found T, N, and K. Now, in the following Exercises !l-16, f"md Band T for these space curves.
9. r(t) = (3 sin t)1 + (3 cos t)j + 4tk 10. r(t) = (cost + tsint)1 + (sint - tcost)j + 3k 11. r(t)
~
(e t cos t)1 + (e t sin t)j + 2k
~
0
< t < 'It/2
+ (a cosh (t/a))j,
a> 0 (cosh t)1 - (sinh t)j + tk
tl
PhydcalAppUcaUons 17. The speedometer 00 your car reads a steady 35 mph. Could you be accelerating? Explain.
5. r(t) ~ t'l + (t + (I/3)t')j + (t - (I/3)t')k,
(e t sin t)j
~
t> 0
at the given
t~ I
4. r(t) ~ (t cos t)1 + (t sin t)j + t 2k,
13. r(t) ~ (t'/3)1 + (t 2/2)j, 14. r(t) = (cos't)1 + (sin't)j,
+ (a sin t)j + btk
+ 3t)1 + (t
= (e t cos t)1
+ lINN without
18. Can anything be said shout the acceleration of a particle that is moving at a constant speed? Give reasons for your answer. 19. Can anything be said about the speed of a particle whose acceleration is always orthogonal to its velocity? Give reasons for your
answer. 20. An object of mass m travels aloog the parabola y = x 2 with a coostant speed of 10 mtits/sec. What is the force 00 the object due to its acceleratioo at (0, OJ? at (2 1/2, 2)? Write your answers in terms of! and j. (Remember Newton's law, F = ma.)
Theory and Examples 21. Vector furmula for curvature For a smooth curve, use Equatioo (I) to derive the curvature formula X K =
Iv al Ivl3 .
13.6 22. Show that a moving particle will move in a straight line if the nor-
v'k
mal component of its acceleration is zero. 23. A sometime shortcut to curvature If you already know I""I and lvi, then the formula"" = Klvl' gives a convenient way to fmd the curvature. Use it to fmd the curvature and radius of curvature of the curve
r(t)
~
(cost
+ tsint)i + (sint
0 for all tin [a, b]. Show that h(t) ~ 0 for all tin [a, b]. = d'r/dt' and apply the initial conditions in
28. A formula that calculate. T from B and v If we start with the defmition T = -(dB/tis)· N and apply the Chain Rille to rewrite dB/ds as
dBdt dt ds
dB tis
t> O.
- tcost)j,
dB I dt lvi'
we arrive at the formula
24. Show that Kand T arc both zero for the line
+ At)i + (YO + Htli + (zo + Ct)k
25. What can be said about the torsion of a smooth plane curve r(t) ~ f(t)i + g(t)j? Give reasons for your answer. 26. The torsion of a helix
r(t) = (a cos t)i
The advantage of this formula over Equation (5) is that it is easier to derive and state. The disadvantage is that it can take a lot of work to evaluate without a computer. Use the new fannula to fmd the torsion of the helix in Exercise 26.
Show that the torsion of the helix
+ (a sin t)j + btk, a, b '"
is T ~ b/(a' + b'). What is the largest value given value of a? Give reasons for your answer.
T
0
can have for a
27. Differentiable curves with zero torsion lie in planes That a sufficiently differentiable curve with zero torsion lies in a plane is a special case of the fact that a particle whose velocity remains perpendicular to a fIxed vector C moves in a plane perpendicular to C. This, in turn, can be viewed as the following result. Suppose r(t) = f(t)i + g(t)j + h(t)k is twice differentiable for all t in an interval [a, b], that r ~ 0 when t ~ a, and that
13.6
~
739
(Hint: Start with a reverse order.).
(Take"" andlvl from Example 1.)
r(t) ~ (xo
Velocity and Acceleration in Polar Coordinates
COMPUTER EXPLORATIONS Rounding the answers to four decimal places, use a CAS to find v, a, speed, T, N, B, K, T, and the tangential and normal components ofacceleration for the curves in Exercises 29-32 at the given values of t.
+ (t sin t)j + tk, t = v'3 + (e'sint)j + e'k, t = In2 (t - sint)i + (I - cost)j + v=tk, t = -3". (3t - t')i + (3t')j + (3t + t')k, t = I
29. r(t) = (t cos t)i
30. r(t) = (e'cost)i 31. r(t) = 32. r(t) =
Velodty and Acceleration in Polar Coordinates In this section we derive equations for velocity and acceleration in polar coordinates. These equations are useful for calculating the paths of planets and satellites in space, and we use them to examine Kepler'. three law. of planetary motion.
y
Motion in Polar and Cylindrical Coordinates When a particle at P(r, IJ) moves along a curve in the polar coordinate plane, we express its position, velocity, and acceleration in terms of the moving unit vectors u, = (cos lJ)i
+
(sin lJ)j,
ue = -(sinlJ)i
+
(coslJ)j,
(1)
~
~I-""'-----"'---,f'-------_x
o
FIGURE 13.30 The length ofr is the positive polar coordinate r ofthe point P.
Thus, Un which is rllrl, is also r/r. Equations (I) express u, and "" in terms ofiandj.
shown in Figure 13.30. The vector u, points along the position vector OP, so r = ru,. The vector Ue, orthogonal to u" points in the direction of increa.ing IJ. We fmd from Equations (I) that
~~ =
-(sin lJ)i
~~e =
-(co.lJ)i - (sinlJ)j = -u,.
+
(cos lJ)j = ue
When we differentiate u, and Uo with respect to t to fmd how they change with time, the Chain Rule gives . du,. . u, = di IJ = lJue,
.
Ue =
due·
.
dlJ IJ = -lJu,.
(2)
740
Chapter 13: Vector-Valued Functions and Motion in Space
Y
Hence, we can express the velocity vector in terms of u, and U9 as
v
~t'""''----7'''"'---------X
o
=
=
v
I
v=
+ rU. . +rlJ· Do· r = rUr
(i'u, + ,Ii,) + (riJU9 + rOU9 + riJ~).
When Equations (2) are used to evaluate Ii, and ~ and the components are separated, the equation for acceleration in terms of u, and U9 becomes
'Ur + rlJU9.
s
=
(i' - riJ2)u,
+ (rli + 2riJ)U9.
To extend these equations of motion to space, we add zk to the right-hand side of the equation r = ru,. Then, in these cylindrical coordinates, we have
z
r = TD,
= rUr
rUr
See Figure 13.31. As in the previous section, we use Newton's dot notation for time derivatives to keep the formulas as simple as we can: Ii, means du,/ dt, 6 means dfJ/ dt, and soon. The acceleration is
s = FIGURE 13.31 In polar coordinates, the velocity vector is
. d( ) .
r = dt
+ zk 'u, + r6U9 + zk
r = rUr k
+ zk
u,
"
v=
(3)
s = (i' - rif)lI, + (rO + 2,6)U9 + zk. The vectors 11" U9, and kmake a right-handed frame (Figure 13.32) in which urXuo=k,
FIGURE 13.32 Position vector and basic
Planets Move in Planes
unit vectors in cylindrical coordinates. Noticethatlrl # rifz # o.
Newton's law of gravitation says that if r is the radius vector from the center of a sun of mass M to the center of a planet ofmass m, then the force F of the gravitational attraction between the planet and sun is F = GmM - - - r-
Irl 2 Irl
(Figure 13.33). The mnnber G is the universal gravitational constant. Ifwe measure mass in kilograms, force in newtons, and distance in meters, G is about 6.6726 X 10-11 Nm2 kg-2. Combining the gravitation law with Newlon's second law, F = mi', for the force acting on the planet gives ..
mr
..r
G
F=_GmM~
Irl' Irl
r = - GM ---
Irl
2
1
r l"
The planet is accelerated toward the sun's center ofmass at all times. Since i' is a scalar multiple ofr, we have rXi'=O. From this last equation, t dd~X~=txt+rXi'=rxi'=~ o
FIGURE 13.33 The force of gravity is
~
directed along the line joining the centers ofmass.
GmM r
= -~jrj'
It follows that
rXt=C for some constant vector C.
(4)
13.6 Velocity and Acceleration in Polar Coordinates
741
Equation (4) tells us that r and r always lie in a plane perpendicular to C. Hence, the planet moves in a fixed plane through the center ofits sun (Figure 13.34). We next see how Kepler's laws describe the motion in a precise way.
c= rxr
Keplefs First Law (Ellipse Law) Kepler's first law says that a planet's path is an ellipse with the sun at one focus. The eccentricity of the ellipse is
FIGURE 13.34 A planet that obeys Newton's laws of gravitation and motion travels in the plane through the sun's center ofmassperpendiculartoC = r X
r.
rovo2 e= GM-I
(5)
and the polar equation (see Section 11.7, Equation (5» is
+ e)ro
(I
r
HiSTORICAL BIOGRAPHY
= 1
+
(6)
ecosO·
Here Vo is the speed when the planet is positioned at its minimum distance ro from the sun. We omit the lengthy proof. The sun's mass Mis 1.99 X 10 30 kg.
Johannes Kepler (1571-1630)
Keplefs Second Law (Equal Area Law) Planet
/' Sun
Kepler's second law says that the radius vector from the sun to a planet (the vector r in our model) sweeps out equal areas in equal times (Figure 13.35). To derive the law, we use Equation (3) to evaluate the cross product C = r X C=rXr=rXv
X (hI, + rOD.)
= ru,
FIGURE 13.35 The line joining a planet to its sun sweeps over equal areas in equal times.
r from Equation (4): Eq. (3),
=
ri'(D, X D,) + r(rO)(D, X no)
=
r(rO)k.
o
z~
0
k
(7)
Setting t equal to zero shows that
C = [r(rO)]t~o k = rovok. Substitoting this value for C in Equation (7) gives 2'
rovok = r Ok,
r 2'0
or
=
rovo.
This is where the area comes in. The area differential in polar coordinates is
dA
=
2 dO !r 2
(Section 11.5). Accordingly, dA/dt has the constant value
dA
di
=
I
2'
2r 0
=
I
2 rovo.
(8)
So dA/dt is constant, giving Kepler's second law.
Keplefs Third Law (Time-Distance Law) The time T it takes a planet to go around its sun once is the planet's orbital period. Kepler's third law says that T and the orbit's semimajor axis a are related by the equation
T2 a3
4,.' GM'
Since the right-hand side of this equation is constant within a given solar system, the ratio of T 2 to a 3 is the same for every planet in the system.
742
Chapter 13: Vector-Valued Functions and Motion in Space Here is a partial derivation of Kepler's third law. The area enclosed by the planet's elliptical orbit is calculated as follows:
loT dA
Area =
=
10rl2:rovodt
=
2 Trovo.
Eq. (8)
I
If b is the semiminor axis, the area of the ellipse is "TTab, so
2"TTab T= rovo
Planet
=
2"TTa 2 • ~ rovo vI - e-.
For any ellipse, b~.~
(9)
It remains only to express a and e in terms of ro, vo, G, and M. Equation (5) does this for e. For a, we observe that setting 0 equal to "TT in Equation (6) gives Sun Tmax
FIGURE 13.36 The length of the major axis ofthe ellipse is 2a = TO + Tmax _
= TO
1+ e 1 - e"
Hence, from Figore 13.36,
2a
=
ro + r"""
2ro
= I _
e
=
2roGM 2' 2GM - rovo
(10)
Squaring both sides of Equation (9) and substituting the results of Equations (5) and (10) produces Kepler's third law (Exercise 9).
Exercises 13.6 In Exercises 1-5, f'md the velocity and acceleration vectors in terms of
"'and ....
d8 dl=3
1.r=a(l-cos8) and
2. r = a sin 28 and : 3.
r~e"
4. r = a(1 S. r
~
and
:~2
+ sin I)
2 cos 41
= 21
8. Suppose that r is the position vector of a particle moving along a plane curve and dA/dl is the rate at which the vector sweeps out area. Without introducing coordinates, and assuming the necessary derivatives exist, give a geometric argument based on increments and limits for the validity of the equation
and 8 = I - e-'
and 8
~
21
6. Type of orbit For wbat values of Va in Equation (5) is the orbit in Equation (6) a circle? An ellipse? A parabola? A hyperbola?
9. Kepler's third law Complete the derivation of Kepler's third law (the part following Equation (10)). 10. Find the length of the major axis of Earth's orbit using Kepler's third law and the fact that Earth's orbital period is
365.256 days.
7. Circular orbits Show that a planet in a circular orbit moves with a constant speed (Hint: This is a consequence of one of Kepler~ laws.)
Chapter
Questions to Guide Your Review
1. State the rules for differentiating and integratiog vector functions. Give examples.
3. What is special about the derivatives of vector functions of constsnt length? Give an example.
2. How do you defme and calculate the velocity, speed, direction of motion, and acceleration of a body moving along a sufficiently differentiable space curve? Give an example.
4. What are the vector and parametric equations for ideal projectile motion? How do you fmd a projectile~ maximum height, flight time, and range? Give examples.
Chapter 13 5. How do you derme and calculate the length of a segment of a
smooth space curve? Give an example. What mathematical assumptions are involved in the def'mition? 6. How do you measure distance along a smooth curve in space from a preselected base point? Give an example. 7. What is a differentiable curve's unit tangent vector? Give an example. S. Derme curvature, circle of curvature (osculating circle), center of curvature, and radius of curvature for twice-differentiable curves
in the plane. Give examples. What curves have zero curvature?
Practice Exercises
743
10. How do you define N and K for curves in space? How are these quantities related? Give CXlIDlples.
11. What is a curve's binonnal vector? Give an example. How is this vector related to the curve's torsion? Give an example. 12. What formulas are available for writing a moving body's acceleration as a sum of its tangential and normal components? Give an CXlIDlple. Why ntight one want to write the acceleration this way? What if the body moves at a constant speed? At a constant speed
around a circle? 13. State Kepler's laws.
Constant curvature? 9. What is a plane curve's principal normal vector? When is it deImed? Which way does it point? Give an example.
Chapter
Practice Exercises
Motton In the Plane In Exercises 1 and 2, graph the curves and sketch their velocity and acceleration vectors at the given values of t. Then write a in the form a ~ aTT + ilNN without finding T andN, and rmd the value OfK at the given values of I.
1. r(l) = (4 cos 1)1 + (V2sinl)j, 2. r(l) ~ {V3 sec 1)1
+
b. Find v and a at 1 = 0,1,2, and 3 and add these vectors to your sketch. c. At any given time, what is the forward speed of the t0pn3ost point of the wheel? Of C? y
1= Oand'IT/4
(V3tanl)j,
I~ 0
3. The position of a particle in the plane at time I is r =
I
.
+
1
•
v'J+t2' v'J+t2 J. r
Find the particle's highest speed. 4. Suppose r(l) = (e' cos l)i + (e t sin I)j. Show that the angle between r and a never changes. What is the angle? 5. Finding cnrvatnre At point P, the velocity and acceleration of a particle moving in the plane are v = 31 + 4j and a = 51 + 15j. Find the curvature of the particle's path atP.
6. Find the point on the curve y
=
eX where the curvature is greatest
7. A particle moves around the unit circle in the xy-plane. Its position at time 1 is r = xi + yj, where x and y are differentiable functions of I. Find dy/dl ifv' I ~ y. Is the motion clockwise or
counterclockwise? 8. You send a message through a pneumatic tube that follows the (distance in meters). At the point (3, 3), v' I ~ 4 curve 9y ~ anda'l = -2. Find the values ofv' j anda'j at (3, 3).
x'
9. Characterizing circular motion A particle moves in the plane so that its velocity and position vectors are always orthogonal. Show that the particle moves in a circle centered at the origin. 10. Speed along a cycloid A circular wheel with radius I ft and center C rolls to the right along the x-axis at a half-torn per second. (See the accompanying figure.) At time I seconds, the position vector of the point P on the wheel's circumference is r = ('lT1 - sin'ITI)1
+ (I
c
- cos'ITI)j.
a. Sketch the curve traced by P during the interval 0 '" I '" 3.
\ P
'Tf't
1
o Projectile Motion 11. Shot put A shot leaves the thrower's hand 6.5 ft above the ground at a 45' angle at 44 ft/sec. Where is it 3 sec later? 12. Javelin A javelin leaves the !brower's hand 7 ft above the grouod at a 45' angle at 80 ft/sec. How high does it go? 13. A golf ball is hit with an initial speed Vo at an angle a to the horizontal from a point that lies at the foot of a straight-sided hill that is inclined at an angle to the horizontal, where 0< < a <
'IT
2'
Show that the ball lands at a distance 2vo'cosa . ( ) 2 sma-q" goos measured up the face of the hill. Hence, show that the greatest
range that can be achieved for a given Vo occurs when a = (/2) + ('IT/4), Le., when the initial velocity vector bisects the angle between the vertical and the hill.
744
D 14.
Chapter 13: Vector-Valued Functions and Motion in Space
Javelin In Potsdam in 1988, Petra Felke of(thoo) Baat Gennany set a warnoo's world record by throwing ajavelin 262 ft 5 in.
a. Assuming that Felke launched the javelin at a 4QO angle to the horizonta16.5 ft above the ground, wbat was the javelin's initial speed?
30. Radius of curvature Show that the radius of curvature of a twice-differeotiable plane curve r(l) ~ /(I)i + g(l)j is giveo by the formula p =
x2 + »2 Vx2 + y2 _ .2'
where
• =
d 2 + »2 -Vx dl .
b. How high did the javelin go?
31. An alternative def'mition of curvature in tbe plane An alteruative definition gives the curvature of a sufficiently differeotiable plane curve to be Id/dsl, where is the angle betweeo T and i (Figure 13.37a). Figure 13.37b shows the distance s measured cOl.mterclockwise around the circle x 2 + y2 = a 2 :from the point (a, 0) to a pointP, along with the angle at P. Calculate the circle's curvature using the alteruative definition. (Hint: ~ 8 + '1t/2.)
Motion In Space Find the loogths of the curves in Exercises 15 and 16. 15. r(l) = (2 cos I)i + (2 sin I)j + 12k, 0 '" I '" '1t/4 16. r(l) ~ (3 cos I)i + (3 sin I)j + 21 3/2k, 0 '" I '" 3 In Exercises 17-20, fmd T, N, B, 1(, andT at the givoo value of I.
17. r(l)
~ ~ (I
+ 1)3/2i +
~ (I
- 1)3/2j +
t Ik,
18. r(l) ~ (e'sin21)i + (e'cos21)j + 2e'k, 19. r(l) = Ii + te2'j,
I
~
]
fi
0
I ~ 0
1= 1n2
-c+-----------> X o
20. r(l) = (3 cosh 21)i + (3 sinh 21)j + 61k,
(a)
I = In 2 ]
In Exercises 21 and 22, write a in the form a = aTT + IlNN at I = 0
without fmding T and N. 21. r(l) ~ (2 + 31 + 3( 2)i + (41 + 4( 2)j - (6 cos I)k 22. r(l) ~ (2 + l)i + (I + 2( 2)j + (I + ( 2)k 23. Find T, N, B, 1(, and T as functions of I if
r(l) = (sin I)i + (v'2 cos I)j + (sin I)k. (b)
24. At what times in the interval 0 '" I '" '1t are the velocity and acceleration vectors of the motion r(l) = i + (5 cos I)j +(3 sin I)k orthogonal? 25. The position of a particle moving in space at time t
~
0 is
r(l) = 2i + (4Sint) + (3 - if)k. Find the first time r is orthogonal to the vector i - j. 26. Find equations for the osculating, normal, and rectifYing planes ofthe curve r(l) ~ Ii + 12j + 13k at the point (I, I, I). 27. Find parametric equations for the line that is tangeot to the curve r(l) = e'i + (sinl)j + 1n(1 - I)katl = O. 28. Find parametric equations for the line tangoot to the helix r(l) = (v'2cosI)i + (v'2sinl)j + Ikat the point wbereI = '1t/4.
FIGURE 13.37 Figures for Exercise 31. 32. Tbe view from Skylab 4 What percentage of Earth's surface area could the astronauts see wboo Sky1ab 4 was at its apogee height, 437 Ian above the surface? To find out, model the visible surface as the surface generated by revolving the circular arc GT, shown here, about the y-axis. Then carry out these steps: 1. Use sintilar triangles in the figure to show that Yo/6380 6380/(6380 + 437). Solve for Yo. 2. To four significant digits, calculate the visible area as VA
~ fso2'1tX~1
+
(:Y
dy.
3. Express the result as a percentage of Earth's surface area. ]
Theory and Examples 29. Synchronous curves
x
S (Skylab)
By eliminating a from the ideal projectile 437
equations
~
(vocosa)l, y
~
G
x ~ V(6380)2 _]2
T
(vosina)1 - tgt2,
show that x 2 + (y + gI2/2)' = vo212. This shows that projectiles launched simultaneously from the origin at the same initial speed will, at any giveo instant, all lie on the eirele of radius Vol cootered at (0, -gt2/2), regardless of their launch angle. These circles are the synchronous curves of the launching.
6380
-I---+----+__ x o
=
Chapter 13
Chapter
745
Additional and Advanced Exercises
Appllcattons 1. A frictionless particle P, starting from rest at time t point (a, 0, 0), slides down the helix r(8)
Additional and Advanced Exercises
= (a cos 8)i + (a sin 8)j + b8k
= 0 at the
Motton In Polar and Cylindrical Coordinates 3. Deduce from the orbit equation r=
(a, b > 0)
under the influence ofgravity, as in the aceompanying figure. The 8 in this equation is the cylindrical coordinate 8 and the helix is the curve r = a, Z = b8, (} O?:: 0, in cylindrical coordinates. We assome 8 to be a differentiable function of t for the motion. The law of conservation of energy tells us that the particle's speed after it has fallen straight down a distance z is v'2g., where g is the constant acceleration of gravity.
a. Find the angu\arvelocity d8/dt when 8 = 2".. b. Express the particle's 8- and z-coordinates as functions of t.
c. Express the tangential and normal components of the velocity dr/dt and acceleration d'r/dt' as functions of t. Does the aceeleration have any nonzero component in the direction of the binormal vector B?
that a planet is closest to its sun when 8 at that time.
D 4.
y
r-_"/p
The helix //r=a,z=b8
= 0 and show that r = r.
A Kepler equation The problem oflocating a planet in its orbit at a given time and date eventually leada to solving "Kepler" equations of the form
I(x) = x-I - ksinx = O. a. Show that this particular equation has a solution between x ~ 0 and x ~ 2. b. With your computer or calculator in radian mode, use Newton's method to fmd the solution to as many places as
you can.
5. In Section 13.6, we found the velocity of a particle moving in the plane to be v
x
(1 + .)r. 1 + ecos8
= xi + yj
=
rUT
+ rlJutJ.
a. Express i; and j> in terms of r and riJ by evaluating the dot products v· i and v· j. b. Express r and r iI in terms of i; and j> by evaluating the dot products v • Ur and v· lie.
6. Express the curvatore of a twice-differentiable corve r ~ 1(8) in the polar coordinate plane in terms of 1 and its derivatives. Positive z-axis .-------- points down.
1 z
2. Suppose the curve in Exercise 1 is replaced by the conical helix r ~ a8, z ~ b8 shown in the accompanying figure.
a. Express the angular velocity d8/dt as a function of8. b. Express the distance the particle travels along the helix as a function of8.
7. A slender rod through the origin of the polar coordinate plane r0tates (in the plane) about the origin at the rate of 3 rad/min. A beetle starting from the point (2, 0) crawls along the rod toward the origin at the rate of 1 in/min. a. Find the beetle's acceleration and velocity in polar form when it is halfway to (1 in. from) the origin.
D b.
To the nearest tenth of an inch, what will be the length of the path the beetie has traveled by the time it reaches the origin? 8. Arc length in cylindrical coordinates a. Show that when you express ds' = dr' + ely' + dz'in terms of cylindrical coordinates, you get ds' ~ dr' + r'dB' + dz'. b. Interpret this result geometrically in terms of the edges and a
diagonal of a box. Sketch the box. c. Use the result in part (a) to fmd the length of the corve r = e 6,z = e 6,O::S O::s 8InS. Conical helix = a8,.;:; = be
T
1
Conez= ~r
Positive z-axis points down.
z
9. Unit vecton for position and motion in cylindrical coordinates When the position of a particle moving in space is given in cylindrical coordinates, the unit vectors we use to describe its position and motion are u, = (cos8)i + (sin8)j,
... = -(sin8)! + (cos8)j,
and k (see accompanying figure). The particle's position vector is then r = rUr + zk, where r is the positive polar distance coordinate of the particle's position.
746
Chapter 13: Vector-Valued Functions and Motion in Space b. Sbowtbat dU r
(jif
k
~
110
and
dU(J
(jif
~
-u,.
c. Assuming that the necessary derivatives with respect to t exist, express v = i' and a = r in terms ofuy, 1IB. k, r. and iJ. 10. Conservation of angular mom.ntom Let r(l) denote the position in space of a moving object at time t. Suppose the force acting on the object at time t is
F(I) 9
, (r, 9, 0)
a. Show tbat n,., 110, and k, in this order, fonn a right-handed frame of unit vectors.
Chapter
~ -Ir(~) I' r(l) ,
where c is a constant. In physics the angular momentum. of an object at time 1 is dermed to be L(I) ~ r(l) X mV(I) , where m is the mass of the object and v(l) is the velocity. Prove tbat angular momentum is a conaerved quantity; I.e., prove tbat L(I) is a constant vector, independent of time. Remember Newton's law F = mao (This is a calculus problem, not a physics problem.)
Technology Application Projects
Mathematica/Maple Module: Radar Tracking ofa Moving Object Visualize positi~ velocity, and acceleration vectors to analyze motion. Parametric and Polar Equation. with a Figure Slroter Visualize positi~ velocity, and acceleration vectors to analyze motion. Moving in Three Dimensions Compute distance traveled, speed, curvature, and torsion for motion along a space curve. Visualize and compute the tangential, normal, and
binormal vectors associated with motion along a space curve.
14 PARTIAL DERIVATIVES OVERVIEW Many functions depend on more than one independent variable. For instance, the volume of a right circular cylinder is a function V == 7rr 2h of its radius and its height, so it is a function VCr, h) of two variables rand h. In this chapter we extend the basic ideas of single variable calculus to functions of several variables. Their derivatives are more varied and interesting because of the different ways the variables can interact. The applications of these derivatives are also more varied than for single-variable calculus, and in the next chapter we will see that the same is true for integrals involving several variables.
14.1
Functions of Several Variables In this section we define functions of more than one independent variable and discuss ways to graph them. Real-valued functions of several independent real variables are defined similarly to functions in the single-variable case. Points in the domain are ordered pairs (triples, quadruples, n-tuples) of real numbers, and values in the range are real numbers as we have worked with all along.
DEFINITIONS Suppose D is a set ofn-tuples of real numbers (Xl, X2, ... , x n ). A real-valued function f on D is a rule that assigns a unique (single) real number w
==
f(xl,
X2, ... ,
xn )
to each element in D. The set D is the function's domain. The set of w-values taken on by f is the function's range. The symbol w is the dependent variable of f, and f is said to be a function of the n independent variables Xl to X n . We also call the Xj'S the function's input variables and call w the function's output variable.
If f is a function oftwo independent variables, we usually call the independent variables X and y and the dependent variable z, and we picture the domain of f as a region in the xy-plane (Figure 14.1). If f is a function of three independent variables, we call the independent variables x, y, and z and the dependent variable w, and we picture the domain as a region in space. In applications, we tend to use letters that remind us of what the variables stand for. To say that the volume of a right circular cylinder is a function of its radius and height, we might write V == f(r, h). To be more specific, we might replace the notation fer, h) by the formula that calculates the value of V from the values of rand h, and write V == 7rr 2h. In either case, rand h would be the independent variables and V the dependent variable of the function.
747
748
Chapter 14: Partial Derivatives
f f(a, b) -----jl-~____':-~-c--->z
o
FIGURE 14.1
f(x,y)
An arrow diagram for the functionz = j(x, y).
As usual, we evaluate functions defined by fannulas by substituting the values of the independent variables in the fannula and calculating the corresponding value of the 2 + y2 + z2 at the point dependent variable. For example, the value of j(x,y,z) = (3,0,4) is
Yx
j(3, 0, 4)
=
Y(W + (W + (4j2 = V25 =
5.
Domains and Ranges In defining a function ofmore than one variable, we follow the usual practice of excluding inputs that lead to complex numbers or division by zero. If j(x,y) = x 2,y cannot 2 be less than x . If j(x,y) = I/(xy),xy cannot be zero. The domain of a function is assumed to be the largest set for which the defining rule generates real numbers, unless the domain is otherwise specified explicitly. The range consists of the set of output values for the dependent variable.
Yy -
EXAMPLE 1 (a) These are functions of two variables. Note the restrictions that may apply to their domains in order to obtain a real value for the dependent variable z. Function
z =
Yy -
x2
I z=xy z
=
sinxy
Domain
Range
y ~ x2
[0, 00)
xy '" 0
(-00,0) U (0, 00)
Entire plane
[-1,1]
(b) These are functions of three variables with restrictions on some of their domains. Function
w=
Yx 2 + y2 + z2 I
Domain
Range
Entire space
[0, 00) (0, 00)
w
=~--'o--~
(x,y,z) '" (0,0,0)
W
= xyInz
Half-space z
x2
+ y2 + z2
>
0
(-00,00)
•
Fundions of Two Variables Regions in the plane can have interior points and boundary points just like intervals on the real line. Closed intervals [a, b] include their boundary points, open intervals (a, b) don't include their boundary points, and intervals such as [a, b) are neither open nor closed.
Functions of Several Variables
14.1
749
DEFINmONS A point (xo, YO) in a region (set) R in the .lJ'-plane is an interior point of R if it is the center of a disk of positive radius tbat lies entirely in R (Figure 14.2). A point (xo, YO) is a boundary point of R if every disk centered at (xo, YO) contains points tbat lie outside of R as well as points tbat lie in R. (The boundary point itself need not belong to R.) The interior points of a region, as a set, make up the interior of the region. The region's boundary points make up its boundary. A region is open ifit consists entirely of interior points. A region is closed if it contains all its boundary points (Figure 14.3).
(a) Interior point
y
y
y
, 1
R
---1---0+--+-+ x
o
o
o
(b) Boundary point
FIGURE 14.2 Interior points and boundary points of a plane regioo R. An interior point is necessarily a point of R. A
((x,y)
Ix2 + y2< 1)
Open unit disk.
Every point an interior point
{(x,y)lx2+y2~ 1)
Boundary of unit disk. (The unit circle.)
Ix
((x,y) 2 + y2,,;; 1) Closed unit disk. Contains all boundary points.
boundary point of R need not belong to R. FIGURE 14.3
Interior points and boundary points of the unit disk in the plane.
As with a half-open interval of real numbers [a, b), some regions in the plane are neither open nor closed. If you start with the open disk in Figure 14.3 and add to it some of but not all its boundary points, the resulting set is neither open nor closed. The boundary points tbat are there keep the set from being open. The absence of the remaining boundary points keeps the set from being closed.
DEFINmONS A region in the plane is bounded if it lies inside a disk ofrlXed radius. A region is unbounded if it is not bounded.
y Interior points, wherey - x 2 > 0
/
Examples of bounded sets in the plane include line segments, triangles, interiors of triangles, rectangles, circles, and disks. Examples of unbounded sets in the plane include lines, coordinate axes, the graphs of functions defined on infinite intervals, quadrants, half-planes, and the plane itself.
EXAMPLE 2
Describe the domain of the function f(x,y) =
Yy -
x 2•
Solutton Since f is derIDed only where y - x 2 ;" 0, the domain is the closed, unbounded region shown in Figure 14.4. The parabola y = x 2 is the boundary ofthe domain. The points above the parabola make up the domain's interior. • ---~:-~r-"-:---_x
-1
0
FIGURE 14.4 The domain of fix, y) in Example 2 consists of the shaded regioo and its bounding parabola.
Graphs, Level Curves, and Contours of Functions of Two Variables There are two standard ways to picture the values of a function fix, y). One is to draw and label curves in the domain on which f has a constant value. The other is to sketch the surface z = fix, y) in space.
750
Chapter 14: Partial Derivatives
DEFINITIONS The set of points in the plane where a function f(x, y) has a constant value f(x, y) == c is called a level curve of f. The set of all points (x, y, f(x, y)) in space, for (x, y) in the domain of f, is called the graph of f. The graph of f is also called the surface z = f(x,y). EXAM PLE 3 Graph f(x, y) == 100 - X 2 - Y 2 and plot the level curves f(x, y) f(x,y) == 51,andf(x,y) == 75 in the domain off in the plane.
The surface z =f(x,y)
f(x, y) = 75
= 100 - x 2 _ y2 is the graph off.
X f(x, y) = 51
(a typical ~levelcurvein
/ \ the function's domain) y f(x, y) =
x
a
FIGURE 14.5 The graph and selected level curves of the function j(x, y) in Example 3.
The contour curvef(x, y) = 100 - x 2 - y2 = 75 is the circle x 2 + y2 = 25 in the plane z = 75.
==
0,
SoLution The domain of f is the entire xy-plane, and the range of f is the set of real numbers less than or equal to 100. The graph is the paraboloid z == 100 - x 2 - y2, the positive portion of which is shown in Figure 14.5. The level curve f(x, y) == 0 is the set of points in the xy-plane at which f(x,y) == 100 - x 2 - y2 == 0, or x 2 + y2 == 100,
which is the circle of radius 10 centered at the origin. Similarly, the level curves f(x, y) == 51 and f(x, y) == 75 (Figure 14.5) are the circles f(x,y) == 100 - x 2 - y2 == 51, or x 2 + y2 == 49 f(x,y) == 100 - x 2 - y2 == 75, or x 2 + y2 == 25. The level curve f(x, y) == 100 consists of the origin alone. (It is still a level curve.) If x 2 + y2 > 100, then the values of f(x, y) are negative. For example, the circle 2 x + y2 == 144, which is the circle centered at the origin with radius 12, gives the constant value f(x, y) == -44 and is a level curve of f. • The curve in space in which the plane z == c cuts a surface z == f(x, y) is made up of the points that represent the function value f(x, y) == c. It is called the contour curve f(x, y) == c to distinguish it from the level curve f(x, y) == c in the domain of f. Figure 14.6 shows the contour curve f(x,y) == 75 on the surface z == 100 - x 2 - y2 defined by the function f(x, y) == 100 - x 2 - y2. The contour curve lies directly above the circle x 2 + y2 == 25, which is the level curve f(x, y) == 75 in the function's domain. Not everyone makes this distinction, however, and you may wish to call both kinds of curves by a single name and rely on context to convey which one you have in mind. On most maps, for example, the curves that represent constant elevation (height above sea level) are called contours, not level curves (Figure 14.7).
Functions of Three Variables In the plane, the points where a function of two independent variables has a constant value f(x, y) == c make a curve in the function's domain. In space, the points where a function of three independent variables has a constant value f(x, y, z) == c make a surface in the function's domain.
FIGURE 14.6 A plane z = c parallel to the xy-plane intersecting a surface z = j(x, y) produces a contour curve.
DEFINITION The set of points (x, y, z) in space where a function of three independent variables has a constant value f(x, y, z) == c is called a level surface off· Since the graphs of functions ofthree variables consist of points (x,y, z, f(x,y, z)) lying in a four-dimensional space, we cannot sketch them effectively in our three-dimensional frame of reference. We can see how the function behaves, however, by looking at its threedimensional level surfaces.
EXAMPLE 4
Describe the level surfaces of the function f(x,y,z) == Yx 2 + y2 + z2.
14.1
751
FIGURE 14.7 Contours on Mt. Washington in New Hampshire. (Reproduced by permission from the Appalachian Mountain Club.)
x
FIGURE 14.8 The level surfaces of f(x,y, z) = v'x 2 + y2 + z2 are concentric spheres (Example 4).
z
x
Functions of Several Variables
~Y
The value of f is the distance from the origin to the point (x, y, z). Each level surface Yx 2 + Y 2 + z 2 == c, c > 0, is a sphere ofradius c centered at the origin. Figure 14.8 2 shows a cutaway view of three of these spheres. The level surface + y2 + z2 == 0 consists of the origin alone. We are not graphing the function here; we are looking at level surfaces in the function's domain. The level surfaces show how the function's values change as we move through its domain. If we remain on a sphere of radius c centered at the origin, the function maintains a constant value, namely c. If we move from a point on one sphere to a point on another, the function's value changes. It increases if we move away from the origin and decreases if we move toward the origin. The way the values change depends on the direction we take. The dependence of change on direction is important. We return to it in Section 14.5. • SoLution
Yx
The definitions of interior, boundary, open, closed, bounded, and unbounded for regions in space are similar to those for regions in the plane. To accommodate the extra dimension, we use solid balls of positive radius instead of disks.
(a) Interior point
z
x
~Y
1\
...... ,
\ "' "
............. \
'-_/
(b) Boundary point
FIGURE 14.9 Interior points and boundary points of a region in space. As with regions in the plane, a boundary point need not belong to the space region R.
DEFI NITIO NS
A point (xo, Yo, zo) in a region R in space is an interior point of
R if it is the center of a solid ball that lies entirely in R (Figure 14.9a). A point
(xo, Yo, zo) is a boundary point of R if every solid ball centered at (xo, Yo, zo) contains points that lie outside of R as well as points that lie inside R (Figure 14.9b). The interior of R is the set of interior points of R. The boundary of R is the set of boundary points of R. A region is open if it consists entirely of interior points. A region is closed if it contains its entire boundary.
Examples of open sets in space include the interior of a sphere, the open half-space z > 0, the first octant (where x, y, and z are all positive), and space itself. Examples of closed sets in space include lines, planes, and the closed half-space z ~ O. A solid sphere
752
Chapter 14: PartiaL Derivatives with part of its boundary removed or a solid cube with a missing face, edge, or comer point is neither open nor closed. Functions of more than three independent variables are also important. For example, the temperature on a surface in space may depend not only on the location of the point p(x, y, z) on the swface but also on the time t when it is visited, 80 we would write T = j(x, y, z, t).
computer Graphing Three-dimensional graphing programs for computers and calculators make it possible to graph functions of two variables with only a few keystrokes. We can often get information more quickly from a graph than from a formula. w
EXAMPLE 5 The temperature w beneath the Earth's surface is a function ofthe depth x beneath the swface and the time t of the year. If we measure x in feet and t as the number of days elapsed:from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function w = cos (1.7 X 1O-2t - O.2x)e--o·~.
FIGURE 14.10 This graph shows the seasonal variation of the temperature below ground as a fraction of surface ~ (Exampl, 5).
(The temperature at 0 ft is scaled to vary from +I to -I, so that the variation atx feet can be inteIpreted as a :fraction of the variation at the surface.) Figure 14.10 shows a graph of the function. At a depth of 15 ft, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 25 ft, there is almost no variation during the year. The graph also shows that the temperature 15 ft below the surface is about half a year out of phase with the surface temperature. When the temperature is lOM:st on the surface (late January, say), it is at its highest 15 ft below; Fifteen feet below the ground, the seasons are reversed. _ Figure 14.11 shows computer-generated graphs of a number of functions of two variables together with their level curves.
,
, ,
, ,
(a) z - sin x
FIGURE 14.11
+ 2Ioiny
(b) l
-
(4x2 + y2.)eo-r-r
Computer-generated graphs and leve1 curves of typical functions of two variables.
14.1 Functions of Several Variables
753
Exercises 14.1 Domain, Range, and Level Curves
21. f(x,y)
In Exercises 1-4, find the specific function values.
+ xy'
1. f(x,y) ~ x'
y
b. f(-I, I)
a. f(O,O) e. f(2,3) 2. f(x,y) ~ sin (xy)
= .""",'.y')
27. f(x,y)
= sin-1 (y
28. f(x,y)
= tan-I (~)
b.
29. f(x,y)
= In (x' + y'
b. f(l,t,-i)
Exercises 31-36 show level curves for the functions graphed in (aHf) on the following page. Match each set of curves with the appropriate function.
d.
31.
- x)
- I) 30. f(x,y)
+z
0, -t, 0)
f(2,2, 100)
4. f(x,y,z) ~ V49 - x' - y' - z' a. f(O,O,O) b. f(2,-3,6) e. f(-1,2,3)
d.
y
Ir~ ~v~'
f(~, ~, ~)
In Exercises 5-12, fmd and sketeh the domain for each function.
5. f(x,y) = Vy - x - 2 6. f(x,y) = In (x' + y' - 4) (x - I)(y + 2) 7. f(x,y) ~ (y _ x)(y _ x') ~,
x
sin (xy)
,
+Y -
cos- 1 (y
9. f(x,y) = 10. f(x,y) ~ In(xy
- x' - y')
32. y
8. f(x,y)
= In (9
Matching Surfaees with Level Curves
a. f(3,-1,2) e. f(
x' _ y'
26. f(x,y)
x-y 3. f(x,y,z) = - , - - , y
I 16 -
= In (x' + y')
f( -3, ;;) d. f( -f, -7)
f(,."t)
24. f(x, y)
= y/x' = V~9---x~'---y~'
22. f(x,y)
25. f(x,y)
d. f(-3,-2)
a. f(2,');) e.
23. f(x, y)
= xy = ,/
34.
33.
y
25
- x')
+x -
Y - I)
11. f(x,y) = V(x' - 4)(Y' - 9)
I
12. f(x,y) ~ In (4 _ x' _ y') In Exercises 13-16, fmd and sketch the level curves f(x, y) = c on the same set of coordinate axes for the given values of c. We refer to these level curves as a contour map.
13. 14. 15. 16.
f(x,y) f(x,y) f(x,y) f(x,y)
= x
+Y -
= x' = xy,
I,
+ y', c
c = -3, -2, -I, 0, 1,2,3 = 0, 1,4,9, 16,25
c = -9, -4, -I, 0, 1,4,9
~ V25 - x' - y',
c ~ 0, 1,2,3,4
InExereises 17-30, (8) fmd the function's domain, (b) fmd the function's range, (e) describe the function's level curves, (d) fmd the boundary of the function's don3ain, (e) determine if the don3ain is an open region, a closed region, or neither, and (f) decide if the don3ain is bounded or unbounded
17. f(x,y) = y - x 19. f(x,y) ~ 4x' + 9y'
18. f(x,y) = ~ 20. f(x,y) ~ x' - y'
35.
36. y
y
1111 "lIllf
754
Chapter 14: PartiaL Derivatives
,
L
,
f.
b.
<.
,
Functions of Two variables Display the values of the :functions in Exercises 37--48 in two ways: (a) by sketchingthc surfacez = f(x,y) and (b) by drawing an assortment oflcvel curves in the function's dmnain. Label each level curve with its function value.
Vx y":"cc+c-y,,
37. f(x,y) = y2 39. f(x,y) _ x 2 + y2 41. f(x,y) _ x 2 - y
38. f(x,y) ~ 40. f(x,y) _
43. f(x,y) _ 4x 2 + y2
44. f(x,y) - 6 - 2x - 3y ... f(x,y) - 1 -Ixl -Iyl
42. f(x,y) _ 4 - x 2 _ y2
... f(x,y) - 1 -Iyl 47. f(x,y) = yG x ,i'+'---y'OC+-0 4
48. f(x,y) =
Yx2 + y2
4
FInding Level Curves d.
In Exercises 49-52, fmd an equation for and sketch the graph of the level curve ofthe function f(J:, y) that passes through the given point.
,
... f(x,y) -
16 - x' Vx'=l,
51. f(x,y) =
Yx + y2
4•• f(x,y) ~
2y - x 52. f(x,y) - x + y + I'
y',
(2v2, v2)
(1,0)
3,
(3,-1)
(-1,1)
y
Sketching Level Surflces
••
In Exercises 53-60, sketch a typical level surface for the :functioo.. 53. f(x,y,z) - x 2 + y2 + z2 54. f(x,y,z) _ In(x 2 + y2 + z2)
,
+z 2 57. f(x,y,z) _ x + y2
S5. f(x,y,z) = x
56. f(x,y,z) = z 58. f(x,y,z) _ y2
59. f(x,y,z) - z - x2 _ y2 60. f(x,y,z) = (x 2/25) + (y2/16)
+ z2
+ (z2/9)
y
FInding LAvll Surfaces In Exercises 61-64, fmd an equation for the level surface of the function through the given point.
61. f(x,y,z) = ~ - lnz,
62. f(x,y,z) _In(x 2
(3, -1,1)
+ y+z2), (-1,2,1)
14.2
63. g(x,y,z)
x-y+z 64. g(x,y, z) ~ ,_ + ,(1,0, -2)
y-z
-2". '" Y ",,,.,
In Exercises 65--{j8, f"md and sketch the domain of f. Then find an equation for the level curve or surface of the function passing through the given point.
65. f(x,y)
~
it. ~Y. 00
66.g(x,y,z)~~
(x
n=O y
67. f(x, y) =
(1,2)
1 1
dO
vl-OZ
%
y
68. g(x,y,z) =
dt
--2
%
1
+t
73-76. 73. 4ln (x 2 + y2
+ z2)
+ y2
~ I
76. sin
I " ' (ln4, In9, 2)
n.z
(0, I)
dO (0, I, \13) + /.% .~' 0 v4 - OZ
P("., -".)
Use a CAS to plot the implicitly def"med level surfaces in Exercises
75. x
+ y)"
• r.---:;,'
755
71. f(x,y) ~ sin (x + 2 cosy), -2". '" x'" 2"., -2". '" Y '" 2"., P(".,,,.) 72. f(x,y) ~ e(z'LYlsin (x 2 + y2), 0'" X '" 2".,
= Yx 2 + y2 + Z2, (I, -1, 0) .LA.
Limits and Continuity;n Higher Oimens;ons
-
(~)
3z 2
~
74. x 2 + z2 ~ I
1
- (cosy)Yx 2
+ z2
~2
Parametrized Surfaces Just as you describe corvcs in the plane parametrically with a pair of equations x = f(t),y = g(t) def"med on
some parameter interval I, you can sometimes describe surfaces in
.. Plot the surface over the given rectangle.
space with a1riple of equations x = f(u, v),y = g(u, v),z = h(u, v) defined on some parameter rectangle a ::;; u :s; b, c :s; v :s; d. Many computer algebra systeros permit you to plot such surfaces in parametric mode. (parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises 77-80. Also plot several level corvcs in the xy-plane.
h. Plot several level curves in the rectangle.
77. x
COMPUTER EXPLORATIONS Use a CAS to perfonn the following steps for each of the functions in Exercises 69-72.
c. Plot the level curve of f through the given point.
69. f(x,y) =
xsin~ + ysin2x,
0'" x'" 5".,
0'" Y '" 5".,
0:5 u:S; 2,
78.x=ucosv, y=usinv, z=v, 0:5 V:::S; 2'l'T
0::;; u:s; 2,
79. x
P(3".,3".) 70. f(x,y) ~ (sinx)(cosy)e ,9+1"/', o'" Y '" 5"., P(41T, 4".)
14.2
u,
=
u cos v, y :s 21T
o :5 v =
(2
u sin v, z
+ cosu) cos v, y
o ::s; u :s; 2'17",
0 '" x '" 5".,
=
O:s;
V :s;
=
=
(2
+ cosu) sin v, z
=
sinu,
2'17"
80. x = 2cosucosv, y = 2 cos u sin v, z o :5 u :s; 2'l'T, 0:5 v :s; '11'
=
28inU,
Limits and Continuity in Higher Dimensions This section treats limits and continuity for multivariable functions. These ideas are analogous to limits and continuity for single-variable functions, but including more independent variables leads to additional complexity and important differences requiring some new ideas.
Limits for Functions of Two Variables Iflbe values of f(x, y) lie arbitrnrily close to a f"1xed real number L for all points (x, y) suff"1ciently close to a point (xo, yo), we say that f approaches lbe limit L as (x, y) approaches (xo, Yo). This is similar to lbe informal definition for lbe limit of a function of a single variable. Notice, however, lbat if (xo, YO) lies in lbe interior of !'S domain, (x, y) can approach (xo, yo) from any direction. For lbe limit to exist, lbe same limiting value must be obtained whatever direction of approach is taken. We illustrate Ibis issue in several examples following lbe detmition.
756
Chapter 14: Partial Derivatives
DEFINmON We say that a function fIx, y) approaches the limit L as (x, y) approaches (xo, yo), and write lim
(x, y)-(x. Yo)
f(x,y)
=
L
if, for every number E > 0, there exists a corresponding number 6 for all (x, y) in the domain of f,
If(x,y) - LI <
o<
whenever
E
>
0 such that
V(x - xo)' + (y - Yo)' < 6.
The detmition oflimit says that the distance between fIx, y) and L becomes arbitrarily small whenever the distance from (x, y) to (xo, Yo) is made sufficiently small (but not 0). The definition applies to interior points (xo, Yo) as well as boundary points of the domain of f, although a boundary point need not lie within the domain. The points (x, y) that approach (xo, Yo) are always taken to be in the domain of f. See Figure 14.12.
y
ff d'_-,,<
D
f J(x, y)
(xo,Yo)
--'----I • )------> z o L-eLL+E
---+----'-="'----->x
o
FIGURE 14.12 In the limit defInition, II is the radius ofa disk centered at (xo, Yo). For all points (x, y) within 1his disk, the function values I(x,y) lie inside the corresponding interval (L - E, L + E).
As for functions of a single variable, it can be shown that lim
x=xo
lim
y=Yo
lim
k=k
(x, y)--+(xo.Yo)
(x, y) - (xo, Yo) (x, y) - (xo, Yo)
(any number k).
For example, in the first limit statement above, f(x, y) = x and L = xo. Using the definition of limit, suppose that E > 0 is chosen. Ifwe let 6 equal this E, we see that
o<
V(x -
XO)2
+ (y - YO)2 < 6
= E
implies
V(x - xo)' <
Ix - xol < If(x,y) - xol <
E
(x - xo)' s (x - xo)'
E
W~lal
E
x
~
+ (y - Yo)'
I(x,y)
That is, If(x,y) -
xol <
E
whenever
o<
V(x - xo)' + (y - Yo)' < 6.
14.2
Limits and Continuity in Higher Dimensions
757
So a Il has heen found satisfying the requirement ofthe def"mition, and lim f(x,y) = lim x = XO. (x,y)~(Xo,Yo)
(x,y)~(xo,Yo)
All with single-variable functions, the limit of the sum of two functions is the sum of their limits (when they both exist), with similar results for the limits of the differences, constant multiples, products, quotients, powers, and roots.
THEOREM 1-Properties of Limns of Functions of Two variables
The fol-
lowing rules hold if L, M, and k are real numbers and lim
(x, y)~(x. Yo)
f(x,y) = L
1. Sum Rule:
lim
and
(x,y)~(Xo,Yo)
4. Product Rule:
lim
(f(x,y) - g(x,y)) = L - M
lim
kf(x,y) = kL
lim
(f(x,y)· g(x,y)) = L' M
(x,y)~(xo,Yo) (x,y)~(x.Yo)
(x,y)~(xo,Yo) g(x,y)
6. Power Rule:
M# 0
lim
[f(x,y)]n = L n, n a positive integer
lim
-.;jf(x,y) =
(x,y)~(xo,Yo)
7. Root Rule:
= L
(any number k)
f(x,y) = ~, M
lim
5. Quotient Rule:
+M
(f(x,y)
(x,y)~(xo,Yo)
3. Constant Multiple Rule:
+ g(x,y))
lim
(x,y)~(xo,Yo)
2. Difference Rule:
g(x,y) = M.
(x,y)~(xo,Yo)
-.;ji =
Llfn ,
n a positive integer, and if n is even, we assume that L
> O.
While we won't prove Theorem I here, we give an informal discussion ofwhy it's true. If (x, y) is sufficiently close to (XO,yo), then f(x, y) is close to Land g(x, y) is close to M (from the informal interpretation oflimits). It is then reasonable that f(x, y) + g(x, y) is closetoL + M;f(x,y) - g(x,y) is close to L - M; kf(x,y) is close to kL; f(x,y)g(x,y) is close toLM; and f(x,y)/g(x,y) is close to L/Mif M # O. When we apply Theorem I to polynomials and rational functions, we obtsin the useful result that the limits ofthese functions as (x, y) ---> (xo, Yo) can be calculated hy evaluating the functions at (xo, Yo). The only requirement is that the rational functions be def"med at (xo, Yo).
EXAMPLE 1 In this example, we can combine the three aimple results following the limit def"mition with the results in Theorem I to calculate the limits. We simply substitule the x and y values of the point being approached into the functional expression to f"md the limiting value. (a) (h)
x - xy + 3 2 (x,y)~(O,l) x y + 5xy - y3 . hm
lim
(x,y)~(3,
EXAMPLE 2
-4)
Vx 2 + y2 =
0 - (0)(1) + 3 = -3 (0)2(1) + 5(0)(1) - (1)3
V(3)2
+ (-4)2
=
V25 =
-
xy
Find
x2
. r .r· (x,y)~(O,O) V x VY lim
5
•
758
Chapter 14: Partial Derivatives
VY
Solution Since the denominator v;, approaches 0 as (x,y) --+ (0, 0), we cannot use the Q1!otient Rule from Theorem I. If we multiply numerator and denominator by v;, + V;;;, however, we produce an equivalent fraction whose limit we can find: x2
lim (x, y)~(O,O)
lim
n,
-
-J
v;, - VY
(x,y)~(O,O)
lim
(x,y)~(O,O)
(x 2 - .lJI)( v;, + VY) (v;, - VY)( v;, + VY) x(x - y)(v;, x y
+
VY)
Algebra
Cancel the nonzero fa
=
o( VO + VOl
=
0
We can cancel the factor (x - y) because the path y in the domain ofthe function
Known limit values = x
(along which x - y
=
0) is not
• EXAMPLE 3
· d' Fm lim
x
(x, y)~(O,O)
4.lJ12· ·· 2 I f It exists.
2
+y
Solution We first observe that along the line x = 0, the function always has value 0 when y '" O. Likewise, along the line y = 0, the function has value 0 provided x '" O. So if the limit does exist as (x, y) approaches (0, 0), the value of the limit must be O. To see if this is true, we apply the def'mition of limit. Let e > 0 be given, but arbitrary. We want to f'md a /l > 0 such that 4xy2
1x2 + y2
-
01 <
e
o<
whenever
Yx 2
+ y2 < /l
or
o<
whenever
Yx 2
+ y2 <
/l.
Since y2 ,;; x 2 + y2 we have that 41xly2 • r? ".----. 2 2';; 41xl = 4vx 2 ,;; 4Yx 2 + y2. X +Y
So if we choose /l
=
e/4 and let 0 4.lJ12
Ix2 + y2
-
< Yx 2 + y2 < /l, we get
01 ,;; 4Yx 2 + y2 <
4/l =
4(~) 4
=
e.
It follows from the definition that
lim
4
(x,y)~(O,O) x 2
2
.lJI = + y2
O.
•
14.2 Limits and Continuity in Higher Dimensions
EXAMPLE 4
Jff(x,y)=~,
do..
lim
(X,)')-(O,.O)
759
f(x,y)exist'I
The domain off does not include the y-axis, so we do not consider any points (x,Y) where x - 0 in the approach toward the origin (0, 0). Along the x-axis, the value of the function is I(x, 0) - 0 for all x + O. So if the limit does exist as (x,Y) ~ (0, 0), the value ofthe limit must be L - O. On the other hand, along the line y - x, the value of the function is I(x, x) - x/x - 1 for all x .;. O. That is, the function f approaches the value 1 along the line y - x.This means that for every disk ofradius Bcentered at (0, 0), the disk. will contain points (x, 0) on the x-axis where the value of the function is 0, and also points (x, x) along the line y - x where the value ofthe function is 1. So no matter how small we choose B as the radius of the disk in Figure 14.12, there will be points within the disk. for which the function values differ by 1. Therefore, the limit cannot exist because we can take E to be any nwnber less than 1 in the limit definition and deny that L - 0 or I, or any other real nwnber. The limit does not exist because we have different limiting values along different paIbs approacbing the point (0, 0). •
Solution
Continuity & with functions of a single variable, continuity is dermed in terms of limits.
, DEFlNmDN
1. 2.
3.
f
A function f(x, y) is continuous at the point (xo,y,) if
is defined at (xo,Yo), lim f(x, y) exists,
(x,Y)-
lim
(x,Y)-
f(x,y) - f(x,,1o).
A function is continuous if it is continuous at every point of its domain. (.)
& with the definition of limit, the def'mition of continuity applies at boundary points as well as interior points of the domain off. The only requirement is that each point (x, y) near (Xo. Yo) be in the domain off. A consequence ofTheorem. 1 is that algebraic com.binatiD1lS of continuous functions are continuous at every point at which all the functions involved are defmed. This means that sums, differeIlces, constant multiples, products, quotients, and powers of continuous functions are continuous where defined In particular, polynomials and rational functions of two variables are continuous at every point at which they are defmed.
y
o
-0.8 -1
0.8 I
0.8
-0.8
o •
0.8 0
EXAMPLE 5
Show thaI
(b)
FIGURE14.13
f( ;t,y) -
(a) The graph of
,2xy ,. :x + y { 0,
(x,y)
~ (0,0)
(x,y) = (0,0).
The fimction is continuous at every point except the origin. (b) The values off are different constants along each line y - nu, x ':F 0 (Example 5).
f(x,y)
=
{x';Y" 0,
(x,y) " (0,0) (x,y)
=
(0,0)
is continuous at every point except the origin (Figure 14.13). Solution The functionf is continuous at any point (x,y) #- (0,0) because its values lD'e then given by a mtional function ofx and y and the limiting value is obtained by substituting the values ofx andy into the functional expression.
760
Chapter 14: Partial Derivatives
At (0,0), the value of f is defined, butf, we claim, has no limit as (x,y) ~ (0, 0). The reason is that different paths of approach to the origin can lead to different results, as we now see. For every value of m, the function f has a constant value on the "punctured" line y = mx,x # O,because
=
f(x,y) I
y=mx
2 X
2xy
+Y
2
I
y=mx
2mx 2
2x(mx) 2 x + (mx)2
2m
+
m2
•
Therefore, f has this number as its limit as (x, y) approaches (0, 0) along the line: lim
(x, y)~(O,O) alongy=mx
f(x, y) =
lim
(x, y)~(O,O)
[f(X, y) I
] y=mx
=
1
2m
+
m2
.
This limit changes with each value of the slope m. There is therefore no single number we may call the limit of f as (x, y) approaches the origin. The limit fails to exist, _ and the function is not continuous. Examples 4 and 5 illustrate an important point about limits of functions of two or more variables. For a limit to exist at a point, the limit must be the same along every approach path. This result is analogous to the single-variable case where both the left- and right-sided limits had to have the same value. For functions of two or more variables, if we ever find paths with different limits, we know the function has no limit at the point they approach. z
Two-Path Test for Nonexistence of a Limit If a function f(x, y) has different limits along two different paths in the domain of f as (x, y) approaches (xo, Yo), then lim(x,y)~(XO'Yo) f(x, y) does not exist.
EXAMPLE 6
Show that the function
f(x,y) = y
x
2x 2y 4
X
+
Y
2
(Figure 14.14) has no limit as (x, y) approaches (0, 0). The limit cannot be found by direct substitution, which gives the indeterminate form 0/0. We examine the values of f along curves that end at (0, 0). Along the curve y = kx 2 , X # 0, the function has the constant value SoLution
(a)
y
o f(x y) ,
Iy=kx2
=
2 2 2x2y I 2x (kx ) 2kx 4 2k = = 4 = - -2 4 4 2 2 4 2 x + y2 y=kx x + (kx )2 x +k x 1+ k .
Therefore, lim
(x, y)~(O,O) alongy=kx2
o (b)
FIGURE 14.14
f(x,y) =
lim
(x,y)~(O,O)
[f(X,y) I
y=kx 2
]
=
~. 1 + k2
This limit varies with the path of approach. If (x, y) approaches (0, 0) along the parabola = x 2 , for instance, k = 1 and the limit is 1. If (x, y) approaches (0, 0) along the x-axis, = 0 and the limit is O. By the two-path test, f has no limit as (x, y) approaches (0, 0). _
y k
It can be shown that the function in Example 6 has limit 0 along every path y = mx (Exercise 53). We conclude that
(a) The graph of
j(x, y) = 2x 2yj(x 4 + y2). (b) Along each path y = kx 2 the value of j is constant, but varies with k (Example 6).
Having the same limit along all straight lines approaching (xo, yo) does not imply a limit exists at (xo, yo).
14.2
Limits and Continuity in Higher Dimensions
761
Whenever it is correctly detmed, the composite of continuous functions is also continuous. The only requirement is that each function be continuous where it is applied. The proof, omitted here, is similar to that for functions of a single variable (Theorem 9 in Section 2.5).
Continuity of Composites If f is continuous at (xo, YO) and g is a single-variable function continuous at f(xo, yo), then the composite function h = g , f detmed by h(x,y) = g(J(x,y)) is continuous at (xo, Yo).
For example, the composite functions
xy cos--, 2
e x-y ,
+
X
In (I
1
+ X'y2)
are continuous at every point (x, y).
Functions of More Than Two Variables The definitions of limit and continuity for functions of two variables and the conclusions about limits and continuity for sums, products, quotients, powers, and composites all extend to functions of three or more variables. Functions like
In (x
+ Y + z)
ysinz x-I
and
are continuous throughout their domains, and limits like
e x +z
lim P-(I,O,-I) z2
+
cos
v;;,
el- I (_1)2 + cosO
1 2'
where P denotes the point (x, y, z), may be found by direct substitution.
Extreme Values of Continuous Functions on Closed, Bounded Sets The Extreme Value Theorem (Theorem I, Section 4.1) states that a function of a single variable that is continuous throughout a closed, bounded interval [a, b] takes on an absolute maximum value and an absolute minimum value at least once in [a, b]. The same holds true of a function z = f(x, y) that is continuous on a closed, bounded set R in the plane (like a line segment, a disk, or a filled-in triangle). The function takes on an absolute maximum value at some point in R and an absolute minimum value at some pointinR. Similar results hold for functions of three or more variables. A continuous function w = f(x, y, z), for example, must take on absolute maximum and minimum values on any closed, bounded set (solid bailor cube, spherical shell, rectangular solid) on which it is detmed. We will learn how to tmd these extreme values in Section 14.7.
Exercises 14.2 Limm with Two variables Find the limits in Exercises 1-12. 3x 2 - y2 + 5 2. lim ~ lim 1 • (x,y)~(O,O) x 2 + y2 + 2 (x,y)~(O,4)
VY
3.
5.
lim
(x,y)~(3,4)
lim.
v'x 2 + y2 -
(x,y)~(O,,,/4)
sec x tany
I
762 7. 9. 11.
Chapter 14: Partial Derivatives lim
(x,y)-(O,1n2)
e X- Y
8.
lim
(;;,,)_(1,1)
In
11 + xZyZI
Y
e sin x
lim
(r.y)~(O,O)
10.
x xsiny --
lim
+
(x,y)-(l, '1f'/6) xl
lim
(r,y)~(1/Z7."')
cos~
1
(x,y)-(l,I)
(;;,,)_(1,1)
x~
x~
1
x -
x~l
16.
y
lim
(r.y)~(2, -4) x'Y - xy y:;t:-4,xt:-:r?
17.
+ 2~ -
x - y
lim
+4 + 4x 2 -
(r.y)~(O,O)
VY
~ -
x~y
18.
21.
lim
(r.y)~(2,2)
y;:;:y - 2
36. a. f(x,y, z) = Inxyz
b. f(x,y,z) = eX+Ycosz
37. a. h(x,y, z) = xy sin}
b. h(x,y, z) =
38. a. h(x,y,z) = Iyl
2VY
28. 30.
2
+z - 1
b. h(x,y,z) = Ixyl
I
+ Izl
XZ
19.
By considering different patha of approach, show that tlle functions in Exercises 41-48 have no limit as (x,y) ---> (0, 0).
lim
(x,y)-(2,O) 2x-Y'~4
lim
(r.y)~(4,3) x:;t:y+l
x
(x,,)-(O,O)
xl
42. f(x,y)
1
y
sin (x 2 + y2)
lim
4
X
41. f(x,y) =
+ y2
= - 4x- - 2 x +Y
z 22.
lim
(x,y)-(O,O)
z
I - cos (xy) xy
+ y3
--+(r.y)~(l,-l) x Y lim
Limits with Three Variables
27.
I
+ Izi
I
2 X
- yZ - zZ I b. h(x,y,z) = -:---=-r~~C=~==:' 4 - Yx2 + y2 + z2 - 9
x4 _
Find tlle limits in Exercises 25-30.
25.
+ y2 - 1
b. f(x,y, z) = Yx 2
40. a. h(x,y,z) = Y4 -
4x
-"\I'Y+l
~
x3
23.
Z + yZ - 2z z
No Umlt at a PoInt x+y-4
x+y:;t:4
20.
X
39. a. h(x,y, z) ~ In (z - X Z - y2 - I) I b. h(x,y,z) ~ • ~ z - vx 2 + y2
xy-y-2x+2
' lIm
(x,y)-(l,I)
continuous? 35. a. f(x,y, z) =
cosy +, I 12. lim (;;,y)-(,"'/2,O) y - smx
Umlts of Quotients Find tlle limits in Exercises 13-24 by rewriting tlle fractions lIrst. xl _ 2xy + y2 xl _ y2 13. lim x y 14. lim x y 15.
Continuity In Space At what points (x, y, z) in space are tlle functions in Exercises 35-40
lim
P-(1.3.4)
(~+ yl + })
26.
(sin2 x
secZz)
lim
P-(1J",'IT,O)
lim
P-{-lj4,1T/2.).)
lim
P-(2,-3,6)
+ cos2y +
tan-1xyz
InYx 2
29.
lim
P-(I,-l,-l)
a; ++ ~ z
x
43. f(x,y) ~
+Y x-y 45. g(x,y) = x + Y
lim P-('IT,O,3)
In Exercises 49 and 50, show that tlle limits do not exist. xy2_1 xy+1 49. lim I 50. lim -2--2
Y-
51. l.etf(x,y) =
x+y 32. a. f(x,y) ~ x - y
33. a. g(x,y) = sin 34. a. g(x,y)
=
;y
xl
x2
-
b. f(x,y) ~ In(x 2 + y2) b. f(x,y)
3x
~
y -2-x + I x
b. g(x,y) = 2
+ y2 +2
x'Y
= -4--2 x +Y
Theory and Examples
(x,y)-(l,l)
+ y)
48. h(x,y)
ze-zycos2x
+ y2 + z2
sin (x
y
xl -
46. g(x,y) = x - y
xl + y 47. h(x,y) = - y -
At what points (x,y) in tlle plane are tlle functions in Exercises 31-34 ~
xy 44. f(x,y) ~ Ixyl
2
X
Continuity In the Plane continuous? 31. a. f(x,y)
y2
4
+Y
ist a. b.
b. g(x,y) = -2-X - Y
y '" x y'" 0
- Y
otherwise.
Find each oftlle following limits, or explain that tlle limit does not ex-
+ cosx 1
I, I, { 0,
(x,y)-(l,-l)x
4
••
lim
\<.y)~(O.l)
f(x,y)
\<.yl~2,3/(x, y) lim
(x,y)-{O,O)
f(x,y)
14.2
52. Letf(x,y)
2 X ,
x=::O
=,x, {
x
<0
Limits and Continuity in Higher Dimensions
763
59. (Continuation ofExample 5.)
.
a. Reread Example 5. Then substitute m = tan 8 into the formula
Find the following limits. a.
b.
linJ
(x, y)-(3, -2)
f(x,y) and sinJplify the result to show how the value of f varies with the line's angle of inclination.
(X,Yl~2, I/(x,y)
c. (x, Y~o, o/(x, y) 53. Show that the function in Example 6 has limit 0 along every straight line approaching (0, 0). 54. If f(Xo, Yo) = 3, what can you say about linJ
(x,Y)-(xo,Yo)
f(x,y)
if f is continuous at (xo, Yo)? If f is not continuous at (xo, YO)?
Give reasons for your answers. The Sandwich Theorem for functions of two variables states that if g(x, y) '" f(x, y) '" h(x, y) for all (x, y) ¢' (xo, Yo) in a disk centered at (xo, Yo) and if g and h have the san3e fInite limit L as (x,y) --+ (XO,yo), then
b. Use the formula you obtained in part (a) to show that the limit of f as (x, y) --+ (0, 0) along the line y ~ mx varies from -I to I depending on the angle of approach. 60. Continuons enension Derme flO, 0) in a way that extends x2 _ y2 f(x,y) ~ xy '+ ' x y to be continuous at the origin. Changing to Polar Coordinates If you cannot mske any headway with 1in3<x,y)--(O,olf(x,y) in rectangular coordinates, try changing to polar coordinates. Substitute x = r cos (J, y = r sin 8, and investigate the limit of the resulting expression as r --+ O. In other words, try to decide whether there exists a number L satisfying the following crit<>-
non: linJ
(x,yl~(Xo,"')
f(x,y) = L.
Given <
Use this result to support your answers to the questions in Exercises 55-58.
>
0, there exists a 6 Irl < 6
lim
(x,y)-(O,O)
tan-I xy
1--3-<~<1
0 such that for all r and 8,
If(r,8) - LI < <.
(1)
If such an L exists, then
55. Does knowing that
x'y'
~
>
f(x,y)
~
lim f(r cos 8, r sin 8)
r-O
~
L.
For instance,
tell you anything about tan-I
lim
(x,yl~(O,Ol
~?
To verify the last ofthese equalities, we need to show that Equation (I) is satisfled with f(r, 8) = r cos' 8 and L = O. That is, we need to show that given any < > 0, there exists a 6 > 0 such that for all r and 8,
xy
Give reasons for your answer. 56. Does knowing that 21xyl-
Irl<6
X 2y 2
6
< 4 - 4 cos ~ < 21xyl
~
Ircos'8-01«·
Since Ircos'81 = Irllcos'81 '" Irl'l = Irl,
tell you anything about
lim
4 - 4cos~xyl
(x,y)~(O,O)
Ixyl
?
the inJplication holds for all r and 8 ifwe take 6 In contrast,
~
<.
Give reasons for your answer. 57. Does knowing that Isin (I/x) I '" I tell you anything about
. Iun
(x,y)-(O,O)
. I? ysm-. x
Give reasons for your answer. 58. Does knowing that Icos (I /y)l
lim
(x,y)~(O,O)
Give reasons for your answer.
'" I tell you anything about xcos 1 ? y
tskes on all values from 0 to I regardless of how small Ir I is, so that linl(x,y)~(O,O)X'/(x' + y') does not exist. In each of these instances, the existence or nonexistence of the limit as r --+ 0 is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) 8 = constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, f(x,y) ~ (2x'y)/(x' + y') becomes . rcos8 sin 28 f(rcos8,rsm8) ~,. . , rcos8+sm8
764
Chapter 14: Partial Derivatives
for r # O. If we hold 6 constant and let r ---+ 0, the limit is O. On the path y = x 2 , however, we have r sin (} = r 2 cos2 (J and . f(rcos6,rsm6) =
rcos8sin28 2
4
(
v'x 2 + y2 < 8
2)2
r cos (} + Teas (}
27 cos2 (} sin (} 2r 2 cos4 (}
3
63. f(x,y) =
65. f(x,y)
2- xy22 x+y y2
X
2
x+y
2
62. f(x,y)
=
64. f(x,y)
=
cos (x: - y:) x+y 2x
2
x+x+y
2
= tan-I (I~I + I~I) x +y x2 _
66. f(x,y) ~
2 X
= In ( 2 X
3X 2 - x"y2
+Y
xy2 73. f(x,y) =
-2--2
74. f(x,y) =
-2--2
x
+Y
x3
+ y4
x
+Y
andf(O,O)
=
0, •
=
0.04
andf(O,O)
=
0, •
=
0.02
2 y"2--c+-z"2 < y'x ..--c+-
2 X
+Y
+
3y2 )
75. f(x,y, z) = x 2 + y2 76. f(x,y, z) = xyz, 77. f(x,y, z) ~
2
8
~ If(x,y, z)
+ z2, •
•.
0.Ql5
• = 0.008 x+y+z
2
2
2
x+y+z+l
"
~
0.Ql5
+ tan2y + tan2 z, • ~ 0.03 79. Show that f(x,y,z) = x + Y - z is continuous at every point
2
(XO,YO, zo).
80. Showthatf(x,y,z) = x 2 + y2
14.3
=
- f(O, 0, 0) I <
78. f(x,y,z) ~ tao2 x
3x"y
68. f(x,y) ~
• ~ 0.Ql 2 70. f(x,y) ~ yj(x + I), • ~ 0.05 71. f(x,y) = (x + y)j(x 2 + 1), • = 0.Ql 72. f(x,y) = (x + y)j(2 + cos x), • = 0.02
(x,y, z),
2
In Exercises 67 and 68, defme f(O, 0) in a way that extends f to be continuous at the origin. 67. f(x,y)
If(x,y) - f(O, 0)1 < •.
Each ofExercises 75-78 gives a function f(x, y, z) and apositive nwnher •. In each exercise, show that there exists a 8 > 0 such that for all
y2
+Y
~
69. f(x,y) ~ x 2 + y2,
1.
In Exercises 6I--{j6, fmd the limit of f as (x,y) ---+ (0, 0) or show that the limit does not exist.
61. f(x,y) =
Using the Limit Definition
Each of Exercises 69-74 gives a function f(x,y) and apositive number •. In each exercise, show 1hat there exists a 8 > 0 such that for all (x, y),
+ Z2 is continuous at the origio.
Partial Derivatives The calculus of several variables is similar to single-variable calculus applied to several variables one at a time. When we hold all but one ofthe independent variables of a function constsnt and differentiate with respect to that one variable, we get a "partial" derivative. This section shows how partial derivatives are defined and interpreted geometrically, and how to calculate them by applying the rules for differentiating functions ofa single variable. The idea of differentiability for functions of several variables requires more than the existence of the partial derivatives, but we will see that differentiable functions of several variables behave in the same way as differentiable single-variable functions.
Partial Derivatives of a Function of Two Variables If (xo, YO) is a point in the domain of a function f(x, y), the vertical plane y = Yo will cut the surface z = f(x, y) in the curve z = f(x, yo) (Figure 14.15). This curve is the graph of the function z = f(x, YO) in the plane y = Yo. The horizontal coordinate in this plane is x; the vertical coordinate is z. The y-value is held constsnt at Yo, so Y is not a variable. We defme the partial derivative of f with respect to x at the point (xo, YO) as the ordinary derivative of f(x, YO) with respect to x at the point x = XO. To distinguish partial derivatives ftmn ordinary derivatives we use the symbol arather than the d previously used. In the definition, h represents a real number, positive or negative.
14.3
Partial Derivatives
765
Vertieal axis in the plane y
=
Yo
P(xo. yo.!(x.. yOJ) z~!(x.y)
The curve z = f(x, Yo) in the plane y = Yo
o Yo y
(xo. Yo) (Xo + h. Yo)
Horizontal axis in the plane y = Yo
FIGURE 14.15 The intersection ofthe plane y = Yo with the surface z ~ f(x, y), viewed from above the fIrSt quadrant ofthexy-plane.
DEFINmON (xo, yo) is
The partial derivative of f(x, y) with respect to x at the point
of
ox
= lim
1
(xo.Yo)
h-O
f(xo + h,yo) - f(xo,yo) h '
provided the limit exists.
An equivalent expression for the partial derivative is
~ f(x, Yo) x~x; 1
The slope ofthe curve z = f(x, Yo) at the point P(xo, Yo, f(xo, Yo» in the plane y = Yo is the value of the partial derivative of f with respect to x at (xo, Yo). (In Figure 14.15 this slope is negative.) The tangent line to the curve at P is the line in the plane y = Yo that passes through P with this slope. The partial derivative of/ax at (xo, Yo) gives the rate of change of f with respect to x wheny is held fixed at the value Yo. We use several notations for the partial derivative:
az
1
ax <X.y,J'
and
The definition of the partial derivative of f(x, y) with respect to y at a point (xo, Yo) is similar to the definition of the partial derivative of f with respect to x. We hold x fixed at the value Xo and take the ordioary derivative of f(xo, y) with respect to y at Yo .
DEFINmON (xo, Yo) is
of 1 ay <x. y,)
The partial derivative of f(x, y) with respect to y at the point =
~f( dy
provided the limit exists.
xo,y
)1
= y~,.
I'
un
h~O
f(xo,yo
+ h) - f(xo,yo) h
'
766
Chapter 14: Partial Derivatives
The slope of the curve z = /(xo,y) at the point P(xo, Yo, /(xo, Yo)) in the vertical plane x = Xo (Figure 14.16) is the partial derivative of / with respect to y at (xo, Yo)· The tangent line to the curve at P is the line in the plane x = Xo that passes through P with this slope. The partial derivative gives the rate of change of/ with respect to y at (xo, Yo) when x is held fixed at the value xo. The partial derivative with respect to y is denoted the same way as the partial derivative with respect to x:
Vertieal axis in the plane x =xo
z
1
P(xo, Yo,!(xo, Yo»
a/
Yo (Xo,Yo)
(Xo,Yo+
k)
Thecurvez ~!(Xo,Y)
in the plane
x=xo
~/ /
Y
a/
/y(xo, yo),
ay (xo, yo),
z ~!(X,Y)
ay'
Notice that we now have two tangent lines associated with the surface z = /(x, y) at the point P(xo, Yo, /(xo, Yo)) (Figure 14.17). Is the plane they detennine tangent to the surface at P? We will see that it is for the differentiable functions defined at the end of this section, and we wilileam how to fmd the tangent plane in Section 14.6. First we have to learn more about partial derivatives themselves.
~
H'
tal
. onzon
z
. axIS
m the plane x = Xo This tangent line has slope I,(xo, Yo).
The intersection ofthe plane x = Xo with the surface z = /(x, y), viewed from above the frrst quadrant of the xy-plane. FIGURE 14.16
r
P(xo, Yo'/(xo, Yo»
This tangent line has slope fixQ. Yo). The ClU'Ve z = f(x, Yo) in the plane y = Yo
z ~ !(x,Y)
y
=
Yo
x = Xo
(xo,Yo)
FIGURE 14.17 Figores 14.15 and 14.16 combined. The tangeot lines at the point (xo, yo, /(XO,Yo») detennine a plane that, in this picture at least, appears to be tangeot to the surface.
Calculations The definitions of at/ax and allay give us two different ways of differentiating / at a point: with respect to x in the usual way while treating y as a constant and with respect to y in the usual way while treating x as a constant. As the following examples show, the values ofthese partial derivatives are usually different at a given point (xo, yo).
EXAMPLE 1
Find the values ofa//ax anda//ay at the point (4, -5) if
/(x,y) Solution
=
x 2 + 3xy
+Y -
1.
To fmd at/ax, we treaty as a constant and differentiate with respect to x:
a/ ax
=
a
ax (x 2 + 3xy
+Y -
1)
=
2x
The value ofa//axat (4, -5) is 2(4) + 3(-5)
+ 3·1· Y + 0 =
-7.
0
=
2x
+ 3y.
14.3
Partial Derivatives
767
To find 0/1oy, we treat x as a constaot and differentiate with respect to y:
0/
0
ily = oy (x
2
+ 3-':)1 + Y
The value ofo/loy at(4, -5) is 3(4)
EXAMPLE 2 Solution
+ 3 ·x·1 + I - 0
- I) = 0
+
= 3x
+ 1.
I = 13.
•
Fiodo/loyasafunctionif/(x,y) = ysio-':)l.
We treat x as a constaot and / as a product of y and sio -':)I:
0/ 0 ( . ) o. oy = oy y sm-':)l = y ily sm-':)l = (y cos -':)I) ~ (xy)
+ (.sm-':)l ) oy 0 ( ) y
+ sio -':)I
= -':)I cos -':)I
+
•
sio-':)l.
Fiod /x and /y as functions if
EXAMPLE 3
2y
/(x,y) = y
Solution
+ cosx.
We treat / as a quotient. With y held constao!, we get
f
=
~J ox
2y
) = (y
\Y + cosx
x
(y
o
0
+ cos x) ax (2y) - 2y ax (y + (y + cosxj2
+ cosx)(O) - 2y( -siox) (y + cosxj2
cosx)
2ysiox
(y
+
cosxj2'
With x held constao!, we get /
= y
~
(
oy
\Y + cosx
(y
+ cosx)(2) - 2y(l) (y + cosx)2
2y
) = (y
o
0
+ cosx)ay(2y) - 2y ay (y + cosx) (y + cosxj2 2cosx (y + cosx)2'
•
Implicit differentiation works for partial derivatives the way it works for ordinary derivatives, as the next example illustrates.
EXAMPLE 4
Fiod ozlax if the equation
yz - lnz = x
+y
dermes z as a function of the two iodependent variables x and y and the partial derivative
exists. Solution We differentiate both sides ofthe equation with respect to x, holding Y constaot and treating z as a differentiable function of x: o 0 ox ox (yz) - ox lnz = ox
+
oy ox
oz_!oz=I+O Yox Zox
With Y constant,
•
.z
az (yz) ~ y az .
&-})~~=I OZ ox
z yz-I'
•
768
Chapter 14: Partial Derivatives
The plane x == 1 intersects the paraboloid z == x 2 + y2 in a parabola. Find the slope of the tangent to the parabola at (1,2,5) (Figure 14.18).
EXAMPLE 5
z
SoLution
r y
Plane/ x=l
The slope is the value of the partial derivative azjay at (1, 2):
az == ~(x2 + y2)1 == 2yl == 2(2) == 4. ay (1,2) ay (1,2) (1,2)
Surface =x 2 + y2
1
Z
Tangent
~line
As a check, we can treat the parabola as the graph of the single-variable function z == (1)2 + Y 2 == 1 + Y 2 in the plane x == 1 and ask for the slope at y == 2. The slope, calculated now as an ordinary derivative, is
-dz 1 == -d (1 + Y 21 ) == 2y 1 == 4. dy y=2 dy y=2 y=2
y
•
x
FIGURE 14.18 The tangent to the curve of intersection of the plane x = 1 and surface z = x 2 + y2 at the point (1, 2, 5) (Example 5).
Functions of More Than Two Variables The definitions of the partial derivatives of functions of more than two independent variables are like the definitions for functions of two variables. They are ordinary derivatives with respect to one variable, taken while the other independent variables are held constant.
EXAMPLE 6
If x, y, and z are independent variables and
f(x, y, z) == x sin (y
+ 3z),
then
:~
= :z [x
sin (y
+
3z)] = x :z sin (y
+
3z)
a
== x cos (y + 3z) az (y + 3z) == 3x cos (y + 3z).
•
EXAMPLE 7 If resistors of Rl, R 2 , and R 3 ohms are connected in parallel to make an R-ohm resistor, the value of R can be found from the equation
(Figure 14.19). Find the value of aRjaR 2 when R 1 == 30, R 2 == 45, and R 3 == 90 ohms. SoLution To find aRjaR 2, we treat R 1 and R 3 as constants and, using implicit differentiation, differentiate both sides of the equation with respect to R 2 : FIGURE 14.19 Resistors arranged this way are said to be connected in parallel (Example 7). Each resistor lets a portion of the current through. Their equivalent resistance R is calculated with the formula 1
Ii
1
1
1
= R 1 + R 2 + R3·
__ 1 ~==0-_1_+0 R 2 aR2 R 22
aR aR2
R2 R22
When R 1 == 30, R 2 == 45, and R 3 == 90,
6 90
1 15'
14.3 soR =
Partial Derivatives
769
15 and aR = aR2
(15)2 (1)2 19' 45 3 =
=
Thus at the given vaiues, a small change in the resistance R 2 leads to a change in R about _ 1/9th as large.
Partial Derivatives and Continuity A function I(x, y) can have partial derivatives with respect to both x and y at a point without the function being continuous there. This is different from functions of a single variable, where the existence of a derivative implies continuity. If the partial derivatives of I(x, y) exist and are continuous throughout a disk centered at (xo,Yo), however, then I is continuous at (xo,Yo), as we see at the end ofthis section.
z~
EXAMPLE 8
{O,XY#:O I, xy
=
0
Let
z
I(x,y)
{~:
=
xy ,. 0 xy = 0
(Figure 14.20). (8) Find the limit of I as (x, y) approaches (0, 0) along the line y = x. (b) Prove that I is not continuous at the origin.
(c) Show that both partial derivatives allax and allay exist at the origin.
Solution
FIGURE 14.20 The graph of
j( x,y) =
{D,I,
D
xy" xy = D
consists ofthe lines L, and L, and the four open quadrants ofthe xy-plane. The function has partiai derivatives at the origin but is not continuous there
(Example 8).
(8) Since I(x,y) is constantly zero along the line y =
lim
(•• y)~(O,o)
I(x,y)
I
y-.
x (except at the origin), we have
lim
0=
(.,y)~(O,o)
o.
(b) Since 1(0,0) = I, the limit in part (a) proves that I is not continuous at (0, 0). (c) To find allax at (0, 0), we holdy rlXed at y = O. Then I(x,y) = 1 for all x, and the graph of I is the line £1 in Figore 14.20. The slope ofthis line at any x is allax = o. In particular, allax = 0 at (0, 0). Similarly, allay is the slope of line £2 at any y, so allay = 0 at (0, 0). _
Example 8 notwithstanding, it is still true in higher dimensions that differentiability at a point implies continuity. What Example 8 suggests is that we need a stronger requirement for differentiability in higher dimensions than the mere existence ofthe partial derivatives. We derme differentiability for functions of two variables (which is slightly more complicated than for single-variable functions) at the end of this section and then revisit the connection to continuity.
Second-Order Partial Derivatives When we differentiate a function I(x, y) twice, we produce its second-order derivatives. These derivatives are usually denoted by
a21
2
ax
or
a21
lxx,
ay2 or I yy ,
and
a21
a)"lx or Ixy·
770
Chapter 14: Partial Derivatives The defining equations are
:y, :x (:~). =
and so on. Notice the order in which the mixed partial derivatives are taken:
a2f
Differentiate first with respect to y, then with respect to x.
axay
Means the same thing.
HiSTORICAL BIOGRAPHY
EXAMPLE 9
Iff(x,y)
=
x cosy + ye~f"mdthesecond-orderderivatives
a2f
Pierre-Simon Laplace (1749--1827)
ayax' Solution
af ax
a2f
and
axay·
The first step is to calculate both first partial derivatives.
=
a ax (x cosy
=
cosy + ye X
af ay
+ ye X)
=
a ay (x cosy
=
-x siny + eX
+ ye X)
Now we find both partial derivatives of each first partial:
2 af
f ay ax
=
iL (a )
:~ =
:x (:~)
ilyilx
= -siny
=
ye
2 af axay
+ eX
=
f ax iIy
iL (a )
= -siny
:~ = ~ (~) =
x •
-x
+ eX
cosy.
•
The Mixed Derivative Theorem You may have noticed that the ''mixed'' second-order partial derivatives
alf ayax
and
a2f axay
in Example 9 are equal. This is not a coincidence. They must be equal whenever f, fx. fy, fxy, andfyx are continuous, as stated in the following theorem.
THEOREM 2-The Mixed Derivative Theorem If f(x,y) and its partial derivatives f x. fy, f xy, and f yx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then fxy(a, b)
HiSTORICAL BIOGRAPHY
Alexis Clairaut (1713-1765)
=
fyx(a, b).
Theorem 2 is also known as Clairaut's Theorem, named after the French mathematician Alexis Clairaut who discovered it. A proof is given in Appendix 9. Theorem 2 says that to calculate a mixed second-order derivative, we may differentiate in either order, provided the continuity conditions are satisfied. This ability to proceed in different order sometimes simplifies our calculations.
EXAMPLE 10
Find a2wjaxay if
eY w=XY+-2--· y + I
14.3 Partial Derivatives
771
Solutton The symbol a'w/ axay tells us to differentiate first with respect to y and then with respect to x. However, if we interchange the order of differentiation and differentiate flISt with respect to x we get the answer more quickly. In two steps,
aw ax = y
a'w ayax = 1.
and
Ifwe differentiate flISt with respect toy, we obtaina'w/axay = I as well. We can differentiate in either order because the conditions ofTheorem 2 hold for w at all points (xo, Yo).•
Partial Derivatives of Still Higher Order Although we will deal mostly with first- and second-order partial derivatives, because these appear the most frequently in applications, there is no theoretical limit to how many times we can differentiate a function as long as the derivatives involved exist. Thus, we get third- and fourth-order derivatives denoted by symbols like
a31 axay 2 = I yyx,
a'i
ax 2ay 2 = Iyyxx,
and so on. As with second-order derivatives, the order of differentiation is irornaterial as long as all the derivatives through the order in question are continuous. EXAMPLE 11
Find/_ if I(x,y,z)
=
I - 2xy2z + x~.
Solutton We flISt differentiate with respect to the variable y, then x, then y again, and fmaIIy with respect to z: I y = -4xyz + x 2 Iyx = -4yz + 2x Iyry = -4z 1_ = -4
•
Differentiability The starting point for differentiability is not the difference quotient we saw in studying singlevariable functions, but rather the idea of increment. Recall from our work with functions of a single variable in Section 3.9 that ify = I(x) is differentiable atx = xo, then the cbange in the value of I that results from changing x from Xo to Xo + dx is given by an equation of the form ay = !'(xo)ax
+ Edx
in which E ---> 0 as dx ---> O. For functions of two variables, the analogous property becomes the defmition of differentiability. The Increment Theorem (proved in Appendix 9) tells us when to expect the property to hold. THEOREM 3-The Increment Theorem for Functions of Two Variables Suppose that the flfSt partial derivatives of I(x, y) are defined throughout an open region R
containing the point (xo, Yo) and that Ix and I y are continuous at (xo, Yo). Then the change
az =
I(xo
+
dx,yO
+
ay) - I(xo,yo)
in the value of I that results from moving from (xo, Yo) to another point (xo + dx, Yo + ay) in R satisfies an equation of the form az = Ix(xO,yo)dx
in which each of Ej,
E2
+ l,(xo,Yo)ay +
---> 0 as both
dx, ay ---> O.
ElaX
+ E2ay
772
Chapter 14: Partial Derivatives You can see where the epsilons come from in the proof given in Appendix 9. Similar results hold for functions ofmore than two independent variables.
DEFINmON A function z = I(x,y) is differentiable at (xo, Yo) if Ix(xo,yo) and I,(xo, YO) exist and!1z satisfies an equation ofthe form
!1z
=
fx(xo,yo)lix
+ Iy(xo,yo)liy + E1/ix + E2liy
in which each of E" E2 ---> 0 as both /ix, liy ---> O. We call I differentiable if it is differentiable at every point in its domain, and say that its graph is a smooth surface.
Because of this definition, an immediate corollary ofTheorem 3 is that a function is differentiable at (xo, YO) if its first partial derivatives are continuous there.
COROLLARY OF THEOREM 3 If the partial derivatives Ix and/y ofa function I(x, y) are continuous throughout an open region R, then I is differentiable at every point of R.
If z = I(x, y) is differentiable, then the definition of differentiability assures that liz = I(xo + /ix,yO + liy) - I(xo, YO) approaches 0 as /ix and liy approach O. This tells us that a function oftwo variables is continuous at every point where it is differentiable.
THEOREM 4-Dlfferentlablllty ImplIes Continuity If a function I(x, y) is differentiable at (xo, Yo), then I is continuous at (XO,yo).
As we can see from Corollary 3 and Theorem 4, a function/(x,y) must be continuous at a point (xo,Yo) if Ix and I y are continuous throughout an open region containing (xo,yo). Remember, however, that it is still possible for a function oftwo variables to be discontinuous at a point where its first partial derivatives exist, as we saw in Example 8. Existence alone of the partial derivatives at that point is not enough, but continnity of the partial derivatives guarantees differentiability.
Exercises 14.3 Calculating First-Order Partial Derivatives 1. I(x,y)
~
3. I(x,y)
= (x 2 -
2x2
-
3y - 4 I)(y
2. I(x,y)
~ x 2 - x;y
+ y2
+ 2)
4. I(x,y) ~ 5xy - 7x 2 - y2
+ 3x
- 6y
6. I(x,y) = (2x - 3y)'
8. I(x,y) = (x'
11. I(x,y) ~ (x
+ y)/(x;y
10. I(x,y) =
15. I(x,y)
~
in (x
+ (y/2»'/' x/(x 2 + y2)
- I) 12. I(x,y) ~ tan-I (y/x)
13. I(x,y) = e(>+y+1)
+ y)
17. I(x,y) ~ sin2 (x - 3y)
~
1
14. I(x,y) = e~sin(x
+ y)
(g continuous for all t)
g(t) dt
.-0
5. I(x,y) = (x;y - I)'
+ y)
21. I(x,y)
Y
22. I(x,y) = ~(x;y)'
+2
7. I(x,y) = v'x 2 + y2 9. I(x,y) = I/(x
20. I(x,y) = logy x
19. I(x,y) = x Y
In Exercises 1-22, fmd ai/ax andal/ay.
(Ix;yl <
I)
In Exercises 23-34, find Ix. I y , and I,. 23. I(x,y,z) = 1 + x;y2 - 2z2 24. I(x,y,z) = x;y
+ yz + xz
25. I(x,y, z) ~ x - v'y 2 + z2 26. I(x,y, z) = (x 2 + y2 + z2)-I/2
16. I(x,y)
~
eXYiny
27. I(x,y,z) ~ sin- 1 (x;yz)
18. I(x,y)
~
cos2 (3x - y2)
29. I(x,y, z) = in (x
28. I(x,y,z)
+ 2y + 3z)
~
sec-1 (x + yz)
14.3
30. f(x,y,z) = yzln(xy) ~
32. f(x,y,z)
31. f(x,y,z) = e-(>'+y'+rJ
e-"'"
33. f(x,y, z) = tanh (x + 2y + 3z) 34. f(x,y, z) ~ sinh (xy - z2)
37. h(p, t/>, 6) = p sin t/> cos 6 39. Work done by the heart
- x
_
+ 2x -
58. f(x,y) - 4 59. f(x,y)
60. f(x,y)
= =
2
W(P, V, 6, v, g) = PV+ 40. Wilson lot size formula
VIlv zg
(Section 4.5, Exercise 51)
A(c, h, k,m, q)
km
=
+ y + xy
43. g(x,y) ~ x"y
q + em + T
+y + I
44. h(x,y) 46. s(x,y)
= t3J.-1 (y/x)
48. w = ye r -
SO. w
45. r(x,y)
~
In (x
+ y)
47. w=x 2 tan(xy) 49. w = x sin (x"y)
Y
x-y
= -2-X +Y
In Exercises 51-54, vetilY that ~
1n(2x + 3y)
53. w ~ xy2
w"" = w,.. ~
54. w ~
+ xlny + ylnx x siny + y sinx + xy e'
55. Which order of differeotiation will calculate f"" faster: x finlt or y fIrst? Try to answer without writing anything down.
a. f(x,y)
~
h. f(x,y)
= I/x =y +
xsiny
+ eY
(x/y)
= xlnxy
56. The fIfth-order partial derivative a'f/iJx 2ay' is zero for each of the followiog functions. To show this as quickly as possible, which variable would you differeotiate with respect to first: x or y? Try to answer without writing anything down. a. f(x,y) = y'x 4e' + 2 b. f(x,y) ~ y2 + y(sinx - x 4) c. f(x,y) ~ x 2 + 5xy + sinx + 7e'
d. f(x,y)
af ay
sin (x' + y4) x 2 + y2 '
(x,y) # (0,0)
0,
(x,y) = (0,0),
af ay
at
at (-2, I)
at (-2, 3)
(0,0)
surface at the point (-I, I) and lying in the a. plane x = -I b. plane y = 1. Let w ~ f(x, y, z) be a function of three independent variables and write the formal defInition of the partial derivative af/az at (xo,Yo, zo). Use this defirtition to fmdaf/az at (I,2,3)forf(x,y,z) ~x"yz2.
63. Three variables
64. Tbree variables Let w = f(x, y, z) be a function of three independent variables and write the formal defmition of the partial derivative af/ay at (XO,YO, zo). Use this defmition to fmdaf/ay at (-1,0,3) for f(x,y, z) = _2xy2 + yz2
defmes z as a function of the two independent variables x and y and the partial derivative exists. 66. Find the value ofiJx/az at the point (I, -I, -3) if the equation xz + ylnx - x 2 + 4 = 0 defmes x as a function of the two independent variables y and z and the partial derivative exists. Exercises 67 and 68 are about the triangle shown here.
d. f(x,y) ~ y + x"y + 4y' - In (y2 + I) e. f(x,y) = x 2 + 5xy + sinx + 7e' f. f(x,y)
and
af iJy
at(I,2)
xy+z'x-2yz~0
52. w
+ x"y' + x'y4
c. f(x,y)
af ax
and
af iJy
Differentiating Implldtly 65. Find the value ofaz/iJx at the point (I, I, I) if the equation
Mixed Partial Derivatives 51. w
42. f(x,y) = sinxy
+ cosy + ysinx
~ xe Y
and
{
and
af ax
2
3y - xy,
+ 3y - I,
V2x
af ax
- 3x"y,
61. Let f(x, y) ~ 2x + 3y - 4. Find the slope of the line tsngent to this surface at the point (2, -I) and lying in the a. plane x = 2 b. plane y = -1. 62. Let f(x,y) ~ x 2 + y'. Find the slope of the line tsngent to this
hq
Calculating 5econd-order Partial Derivatives Find all the second-order partia1 derivatives of the functions in Exercises 41-50.
41. f(x,y) = x
af ax
+Y
=I
38. g(r, 6, z) = r(l - cos 6) - z (Section 3.9, Exercise 49)
773
Using the Partial Derivative Definition In Exercises 57--{j(), use the limit defmition of partial detivative to compute the partial derivatives ofthe functions at the specifled points.
57. f(x,y)
In Exercises 35-40, fmd the partial derivative of the function with respect to each variable. 36. g(u, v) ~ v 2e(2u/v) 35. f(t, a) ~ cos (2".t - a)
Partial Derivatives
= xe Y'/2
B~ C
b
A
67. Express A iroplicitly as a function of a, b, and c and calculate aA/aa and aA/ab. 68. Express a iroplicitly as a function of A, b, and B and calculate aa/aA and aa/aB. 69. Two dependent variable. Express v, in terms of u and y if the equations x = v In u and y = u In v define u and v as functions of the independent variables x and y, and if v, exists. (Hint: Differeotiate both equations with respect to x and solve for v, by elintinating u,.)
774
Chapter 14: Partial Derivatives
Find ax/au and ay/au if the equations u = x 2 - y2 and v = x 2 - y defme x and y as functions of the independent variables u and v, and the partial derivatives exist (See the bint in Exercise 69.) Then let s = x' + y' and fmd as/au.
70. Two dependent variables
Show that each function in Exercises 73-80 satisfies a Laplace equation. 73. f(x,y, z) = x' + y' - 1.' 74. f(x,y, z) = 2z' - 3(x'
+ y')z
75. f(x, y) = e -" cos 2x y '" 0 0 y< .
y" 71. Let f(x,y) = { , -y,
Find f~ f" f"" and fyx, and state the domain for each partial derivative. 72. Let f(x,y) =
x~0 0 x< . and f yx, and state the domain for each partial
'\h { x."
Find f ~ f" f "Y' derivative.
Theory and Examples
The three-dimensional Laplace equation a'f
a'f
a'f
ax'
ay'
az'
-+-+-=0
76. f(x,y) = InYx' + y' 77. f(x,y) ~ 3x + 2y - 4 78. f(x,y) = tan-I
Y
79. f(x,y, z) = (x' + y' + z')-I/2 80. !(x,y, z)
=
e 3x+4y cos 5z
Tbe Wave Equation If we stand on an ocean shore and take a snapshot ofthe waves, the picture shows a regular pattern ofpeaks and valleys in an instant of time. We see periodic vertical motion in space, with respect to distance. If we stand in the water, we can feel the rise and fall of the water as the waves go by. We see periodic vertical motion in time. In pbysics, this beautiful synunetry is expressed by the
one-dimensional wave equation
a'w
,a'w ax 2 '
-=c-
is satisfied by steady-state temperature distributions T = f(x, y, z) in space, by gravitational potentials, and by electrostatic potentials. The
two-dimensional Laplace equation a'f
a'f
ax
ay
iJI 2
where w is the wave height, x is the distance variable, t is the time variable, and c is the velocity with which the waves are propagated.
2+-2=0,
w
obtained by dropping the a'f/az' term from the previous equation, describes potentials and steady-state temperature distributions in a plane (see the accompanying figure). The plane (a) maybe treated as a thin slice ofthe solid (b) perpendicular to the z-axis.
(a)
In our example, x is the distance across the ocean's surface, but in other applications, x ntight be the distance along a vibrating string, distance through air (sound waves), or distance through space (light waves). The number c varies with the medium and type ofwave. Show that the functions in Exercises 81-87 are all solutions of
the wave equation. 81. w 82. w
= sin (x + el) = cos (2x + 2el)
83. w = sin (x + el) + cos (2x + 2cI) 84. w
~
In (2x
+ 2cl) 85. + 3el) + e x+a
w ~ tan (2x - 2el)
86. w ~ 5 cos (3x (b)
87. w = f(u), where f is a differentiable function of u, and u = a(x + ct), where a is a constant 88. Does a function f(x, y) with continuous ftrst partial derivatives throughout an open region R have to be continuous on R? Give
reasons for your answer.
Boundary temperatures controlled
89. If a function f(x, y) has continuous second partial derivatives throughout an open region R, must the first-order partial derivatives off be continuous on R? Give reasons for your answer.
14.4
775
Show that liO, 0) and 1,(0,0) exist, but I is not differentiable at (0,0). (Hint: Use Theorem 4 and show that I is not cootinuous at
90. The heat equation An important partial differential equatioo that descrihes the distributioo of heat in a regioo at time t can be represented by the one-dimensional heat equation
al at
The Chain Rule
(0,0).)
a'i ax' .
92. Let/(x,y) =
Show that u(x, t) ~ sin (ax)' e-~t satisfies the heat equatioo for coostanls a and (3.
o, x' < y < 2>;' . { l, o t h erwtse.
Show that liO, 0) and ly(O, 0) exist, but I is not differentiable at (0,0).
xy'
91. Let/(x,y) =
14.4
{
x' + y4'
(X,y): (0,0)
0,
(x,y) - (0,0).
I_T_h_e_C_ha_in_R_u_le
_
The Chain Rule for functions of a single variable studied in Section 3.6 says that when w = f(x) is a differentiable function of x and x = g( I) is a differentiable function ofI, w is a differentiable function of I and dw/ dl can be calculated by the formula
dw
dl
dwdx dxdl'
For functions of two or more variables the Chain Rule has several forms. The form depends on how many variables are involved, but once this is taken into account, it works like the Chain Rule in Section 3.6.
Functions of Two Variables The Chain Rule formula for a differentiable function w = f(x,y) when x = X(I) and = y(l) are both differentiable functions of I is given in the following theorem.
y
I
Of:~, Z,f,
Each indicates the partial derivative of I with respect to x.
THEOREM 5-Cha;n Rule for Functions of Two Independent Yariables
If
w = f(x,y) is differentiable and if x = x(I),y = Y(I) are differentiable functions of I, then the composite w = f(x(I), y(I)) is a differentiable function of I and
dw dt
=
fx(X(I),y(I))' X'(I) + fy(x(I),y(I))' y'(I),
or
dw
af dx
af dy
= axdl - - +aydl' -dl
Proof The proof consists of showing that if x and y are differentiable at I = 10, then w is differentiable at 10 and
where Po = (x(lo),y(lo)). The subscripts indicate where each of the derivatives is to be evaluated.
776
Chapter 14: Partial Derivatives Let ~x, ~y, and ~w be the increments that result from changing t from to to to Since f is differentiable (see the definition in Section 14.3),
b
=
+
~t.
to
(~; ~x + (~1, ~y + EI~ + E2~y,
where EJ, E2 --> 0 as ~x, ~y --> O. To find dw/dt, we divide this equation through by M and let ~t approach zero. The division gives
~w
--s;t Letting
~t
=
(aw) ~ ox Po ~t
+
(aw) ~y oy Po ~t
~x ~y + EI ~t + E2 ~t .
approach zero gives
Often we write ow/ox for the partial derivative of/ox, so we can rewrite the Chain Rule in Theorem 5 in the fonn
dw = owdl: dt oxdt
I To remember the Chain Rille picture the diagram below. To fmddw/dt, start atw and read down each route to t, multiplying derivatives along the way. Then add the products. CbalnRuIe Dependent
w ~f(x,y)
variable
aw ax
aw ay y
x dx dt
dw dt
=
Intermediate variables
dy
dt t awdx + awdy ax dt ay dt
Independent
+
owdy oydt'
However, the meaning of the dependent variable w is different on each side of the preceding equation. On the left-hand side, it refers to the composite function w = f(x(t),y(t» as a function of the single variable t. On the right-hand side, it refers to the function w = f(x,y) as a function of the two variables x andy. Moreover, the single derivatives dw/dt, dl:/dt, and dy/dt are being evaluated at a point to, whereas the partial derivatives aw/ox and ow/oy are being evaluated at the point (xo, Yo), with Xo = x(to) and Yo = y(to). With that understanding, we will use both of these forms interchangeably throughout the text whenever no confusion will arise. The branch diagram in the margin provides a convenient way to remember the Chain Rule. The ''true'' independent variable in the composite function is t, whereas x and y are intermediate variables (controlled by t) and w is the dependent variable. A more precise notation for the Chain Rule shows where the various derivatives in Theorem 5 are evaluated:
variable
dw of dl: Ii/(to) = oX (xo, yo) • dt(to)
EXAMPLE 1
+
of dy oy(xo,Yo)' dt(to).
Use the Chain Rule to fmd the derivative of w = xy
with respect to t along the path x = cost,y = sint. What is the derivative's value at
t
=
7t/2?
Solution
We apply the Chain Rule to fmd dw/ dt as follows:
dw = owdl: dt ax dt =
+
awdy iJy dt
o(xy) d o(xy) d . ax . dt (cos t) + ay . dt (sm t)
= (y)(-sint)
+ (x)(cost) + (cost)(cost)
= (sint)(-sint) = -sin2 t = cos2t.
+ cos2 t
14.4 The Chain Rule
777
In this example, we can check the result with a more direct calculation. As a fimction oft,
w = xy
.
I. 2
= costsmt = 2sm t,
so
dw lit
(I. ) 2'I 2
d
sm2t
= dt
=
2 cos 2t = cos2t.
In either case, at the given value of t, = cos (2' TT) = ( dw) dt t~~/2 2
COSTT
•
= -I.
Functions of Three Variables You can probably predict the Chain Rule for functions of three variables, as it only involves adding the expected third term to the two-variable formula.
THEOREM 6-Chain Rule for Functions of Thll!e Independent variables If w = f(x,y, z) is differentiable and x, y, and z are differentiable functions of t, then w is a differentiable function of t and dw = awdx dt ax dt
I
Here we have three routes from w to t instead oftwo, but rmding dw/dt is still the same. Read down eacb route, multiplying derivatives aloog the way; then add.
Dependent
variable
aw ax ow ay x
aw az
EXAMPLE 2
dx
di
dy dt
+ z,
x = cost,
z = t.
Using the Chain Rule for three independent variables, we have
dw = awdx dt ax dt
variables
+
awdy ay dt
= (y)(-sint) =
Independent
=
+ awdz az dt
+ (x)(cost) + (1)(1) (sint)(-sint) + (cost)(cost) + I -sin2 t + cos2 t + I = I + cos2t,
dz
tit
t variable dw=owdx +owdy +owdz dt ax dt oy dt iJz dt
y = sint,
In this example the values ofw(t) are changing along the path ofa helix (Section 13.1) as t changes. What is the derivative's value at t = O? Solutton
Z
az dt'
Find dw/dt if w = xy
Intenuediare
y
ay dt
The proof is identical with the proof ofTheorem 5 except that there are now three intermediate variables instead of two. The branch diagram we use for remembering the new equation is similar as well, with three routes from w to t.
ChainRuI. w ~ f(x. y. z)
+ awdy + awdz
Substitute for the intermediate wriables.
so
(r;;;'),~o =
I
+
cos (0) = 2.
•
For a physical interpretation of change along a curve think of an object whose position is changing with time t. If w = T(x, y, z) is the temperature at each point (x, y, z) along a curve C with parametric equations x = x(t), y = y(t), and z = z(t), then the composite function w = T(x(t),y(t),z(t» represents the temperature relative to t along the curve. The derivative dw/ dt is then the instantaneous rate of change of temperature due to the motion along the curve, as calculated in Theorem 6.
Functions Defined on Surfaces If we are interested in the temperature w = f(x, y, z) at points (x, y, z) on the earth's surface, we might prefer to think of x, y, and z as functions of the variables r and s that give
778
Chapter 14: Partial Derivatives
the points' longitudes and latitudes. If x = g(r, s), y = h(r, s), and z = k(r, s), we could then express the temperature as a function of r and s with the composite function w = f(g(r, s), h(r, s), k(r, s».
Under the conditions stated below, w has partial derivatives with respect to both r and s that can be calculated in the following way.
THEOREM 7-Chain Rule for Two Independent variables and Three Intermed;ate variables Supposetbatw = f(x,y,z),x = g(r,s),y = h(r,s),andz = k(r,s). If all four functions are differentiable, then w has partial derivatives with respect to rands, given by the formulas
+ away + awaz
aw = awax ar ax ar aw = awax as ax as
+
ay ar away ayas
+
az ar awaz azas'
The first ofthese equations can be derived from the Chain Rule in Theorem 6 by holding s fixed and treating r as t. The second can be derived in the same way, holding r fixed and treating s as t. The branch diagrams for both equations are shown in Figure 14.21. w
Dependent
variable
aw ax
-----}----:000: I
Intermediate
I--W-
variables
g
Independent
h
k
x
y
ax ar
y
ay as
ax as
az ar
aw az
aw ay
x
z
ay ar
J
aw ax
aw az
aw ay
I
w ~ fr;;, y, z)
~f(x,y,z)
z
k
as
r, s
variables
s ow=owax +iJwiJy +iJwaz
r
w
~
aw=awax+aw~+aw~
f(g(r, s), h (r, s), k (r, s))
aT
ax iJT
as
ax aT
ax as
(b)
(a)
FIGURE 14.21
iJy ar
iJy as
ax as
(c)
Composite function and branch diagrams for Theorem 7.
EXAMPLE 3
Expressaw/ar andaw/as in tenns ofr ands if w = x
Solution
+ 2y + z2,
y = r2
+ Ins,
z = 2r.
Using the fonnulas in Theorem 7, we fmd aw = awax ar ax ar
+ away + awaz ay ar
=
(I)(i) +
=
1 S
+ 4r + (4r)(2)
aw = awax as ax as =
(2)(2r)
az ar
+ (2z)(2) =
1
s + 12r
Substitute for intermediate Vllriablcz.
+ away + awaz ay as
az as
(1)(-'"-) + (2)(!) + (2z)(O) 82 S
=
~S -
'"82
•
14.4 The Chain Rule
779
If f is a function of two variables instead of three, each equation in Theorem 7 becomes correspondingly one term shorter.
CbainRuie w ~f(x,y)
ow
CJw
ax
ay
Ifw = f(x,y),x = g(r,s), andy = h(r,s), then y
iIw = iIw ax ~
ay
ax ar
+ away
b~
+ away
aw = iIw ax
and
i1y~
~
b~
i1y~'
ar r
aw=awax+oway aT
ax aT
iJy aT
FIGURE 14.22 tire equation
aw ar
~
Branch diagram for
Figure 14.22 shows the branch diagram for the first of these equations. The diagram for the second equation is similar; just replace r with s.
EXAMPLE 4
Express ilwlar andawlas in terms ofr ands if
awax + away ax ar iIy ar .
x = r - s, Solution
y = r
+ s.
The preceding discussion gives the following. iIw = aw ax
ar
+
bar
iIw ay
iIw = iIw ax
ayar
as
=
(2x)(I) + (2y)(I)
=
2(r - s)
=
+ 2(r + s)
ayas
+ (2y)(l) -2(r - s) + 2(r + s)
(2x)(-I)
=
= 4r
+ away
ax as
Substitute for the intcnncdiate variables.
•
= 4s
If f is a function of x alone, our equations are even simpler.
Ifw = f(x) and x = g(r,s),then
aw ar
CbainRuie
and
aw as
dwax dxas'
In this case, we use the ordinary (single-variable) derivative, dwldx. The branch diagram is shown in Figure 14.23.
w ~f(x)
dw
Implicit Differentiation Revisited
dx
The two-variable Chain Rule in Theorem 5 leads to a formula that takes some of the algebra out ofimplicit differentiation. Suppose that
x
,
r
aw ar aw a,
dwax dx ar
dwax
dx aT dwax
1.
The function F(x, y) is differentiable and
2.
The equation F(x, y) = 0 defines y implicitly as a differentiable function of x, say y = h(x).
Since w = F(x, y) = 0, the derivative dwldx must be zero. Computing the derivative ftmn the Chain Rule (branch diagram in Figure 14.24), we find
dxai
FIGURE 14.23 Branch diagram for differentiating f as a composite function of r and s with one intennediate variable.
o=
dw dx dx = Fx dx
= Fx'l
dy
+ Fy dx dy
+ F y ' dx'
Theorem 5 with t arulf ~ F
~x
780
Chapter 14: Partial Derivatives w
~
If Fy = ow/oy .. 0, we can solve this equation for dy/dx to get
F(x,y)
dy dx
iJw = F. x
ax
Fx F' y
We state this result formally. dy
dx
~ h'(x)
THEOREM 8-A Formula for Impllctt Dffferenttatton Suppose that F(x, y) is differentiable and that the equation F(x, y) = 0 defines y as a differentiable function of x. Then at any point where Fy .. 0,
x
dw ~F 'l+F.ilL dx.t Ydx
FIGURE 14.24 Branch diagram for differentiating w ~ F(x, y) with respect tox. Settingdw/dx = Dleads to a simple computational fonnula for implicit differentiation (Theorem 8).
dy dx
EXAMPLE 5 Solutton
Fx F' y
(1)
Use Theorem 8 to find dy/dx if y2 - x 2 - sinxy = O.
Take F(x,y) = y2 - x 2 - sinxy. Then
dy dx
Fx Fy
-2x-ycosxy 2y - x cosxy
2x + ycosxy 2y - xcosxy' This calculation is significantly shorter than a single-variable calculation using implicit differentiation. _ The result in Theorem 8 is easily extended to three variables. Suppose that the equation F(x, y, z) = 0 defines the variable z implicitly as a function z = f(x, y). Then for all (x,y) in the domain of f, we have F(x,y,f(x,y)) = O. Assuming thatF and f are differentiable functions, we can use the Chain Rule to differentiate the equation F(x,y, z) = 0 with respect to the independent variable x:
0= of oX iJx oX = Fx·J
+
oFoy oy oX
+ oFoz oz oX oz
+ Fy'O + Fz ' OX'
y is e<mstant when dilferontiating with respect to x.
so
A similar calculation for differentiating with respect to the independent variable y gives
Fy
OZ
+ Fz oy
=
o.
Whenever Fz .. 0, we can solve these last two equations for the partial derivatives of z = f(x, y) to obtain
oz ox
and
OZ
oy
(2)
14.4
781
The Chain Rule
An important result from advanced calculus, called the Implicit Function Theorem, states the conditions for which our results in Equations (2) are valid If the partial derivatives F., Fy , and Fz are continuous throughout an open region R in space containing the point (xo, Yo, zo), and if for some constant c, F(xo, Yo, zo) = c and F.(xo, Yo, zo) oF 0, then the equation F(x, y, z) = c defines z implicitly as a differentiable function of x and y near (xo, Yo, zo), and the partial derivatives of z are given by Equations (2).
EXAMPLE 6 Solution
Find
~;
~;
and
at (0, 0, 0) if x 3
LetF(x,y,z) = x 3 + z2
Fy
= e
xz
+ z2 + ye"" + zcosy
= 0.
+ ye"" + zcosy. Then -
zsiny,
Fz = 2z
and
+ xye"" + cosy.
Since F(O, 0, 0) = 0, F.(O, 0, 0) = I oF 0, and all rlISt partial derivatives are continuous, the Implicit Function Theorem says that F(x,y, z) = dermes z as a differentiable function ofx andy near the point (0, 0, 0). From Equations (2),
°
e XZ
2z
+ xye"" + cosy
and
2z
-
z
siny
+ xye"" + cosy'
At (0, 0, 0) we rmd and
oz oy
- = - - I= - 1
I
.
•
Functions of Many Variables We have seen several different forms of the Chain Rule in this section, but each one is just a special case of one general formula. When solving particular problems, it may help to draw the appropriate branch diagram by placing the dependent variable on top, the intermediate variables in the middle, and the selected independent variable at the bottom. To rmd the derivative ofthe dependent variable with respect to the selected independent variable, star! at the dependent variable and read down each route ofthe branch diagram to the independent variable, calculating and multiplying the derivatives along each route. Then add the products found for the different routes. In general, suppose that w = f(x, y, ... , v) is a differentiable function ofthe variables x,y, ... , v (a rmite set) and the x,y, ... , v are differentiable functions of p, q, ... , t (another rmite set). Then w is a differentiable function ofthe variables p through t, and the partial derivatives of w with respect to these variables are given by equations of the form
oW
op
=
oW oX + owoy + ... + owov
oxop
oyop
avop'
The other equations are obtained by replacing p by q, ... , t, one at a time. One way to remember this equation is to think of the right-hand side as the dot product oftwo vectors with components and Derivatives ofw with respect to the
intermediate variables
Derivatives of the: intermediate variables with respect to the selected independent variable
782
Chapter 14: Partial Derivatives
Exercises 14.4 Chain Rule: One Independent Variable In Exercises l--{j,
1. w = x 2 + y2, X = cos t, Y = sin t; t = '11' 2. w=x 2 +y2, x=cost+sint, y=cost-sint;
3. w = 4. w
i
+~. x = cos2 t. Y =
In(x 2
=
+ y2 + z2),
X =
8in
2
t,
Z
= 1ft;
t=O
t
6. w
= 2ye' -Inz, ~
=
x x
=
cost, Y
= In(t 2 + t, Y
=
sint, z
=
=
4Vt;
I), y
= tan-It,
z
= e';
Z = e t - 1;
In t,
t =
1
Chain Rule: Two and Three Independent Variables In Exercises 7 and 8,
8. z = tan-I (.:tly), x (u, v) = (1.3, '1f/6)
=
u cos v,
= usinv;
y = u sin v;
In Exercises 9 and 10,
10. w = In(x 2 + y2 z = ue v ; (u, v)
x = u
+ v,
y
=u
- v, z = uv;
+ z2), X = uevsinu, = (-2,0)
y = ue V cos u,
+Y
(x,y,z) ~
- z;
12. u = e qr sin-1p, p = sinx, q = z 2 1ny, (x,y,z) ~ ('1f/4, 1/2, -1/2)
r = lIz;
Using a Branch Diagram In Exercises 13-24, draw a branch diagram and write a Cbain Rule fonuula for each derivative. dz 13. dt forz
= f(x,y),
x
dz 14. dt for z ~ f(u, v, w),
aw
15. au and
aw forw
av z=k(u,v)
~
= g(t),
y
x=h(u,v),
y=k(u,v)
aw aw 18. ax and iJy forw ~ g(u, v),
u ~ h(x,y),
v ~ k(x,y)
u ~ g(t),
h(x,y,z),
az
az
v ~ h(t),
x ~ f(u,v),
= f(x,y),
ay 20. ar for y = f(u),
x
x
= g(p,q),
aw aw 23. ar and iii for w ~ f(x,y),
x ~ g(r),
aw 24. as for w
= h(t, s)
y
= h(p,q),
u ~ h(s, t)
= f(x,y,z,v),
z = j(p, q),
y
u = g(r, s)
aw aw 21. as and at for w ~ g(u), aw 22. ap forw
= g(t, s),
v = k(p, q)
= g(x,y),
x
= h(r, s, t),
y
y ~ h(s)
= k(r, s, t)
Implidt Differentiation Assuming that the equations in Exercises 25-28 defme y as a differentiable function of x, use Theorem 8 to fmd the value of dy/tix at the givoo point.
25.
x' -
2y 2
+ X)'
(I, I) 3x - 3 ~ 0, (-I, I)
+ y2 + X)' + y2 - 7 = 0, xe' + sinX)' + y - 1n2
26. X)' 27.
~ 0,
x2
(1,2) ~
0,
(0, In 2)
Find the values ofazjax and azjay at the points in Exercises 2~32.
30.
+ yz + y' - 2
xI + yI + ZI
- I = 0,
= 0,
(I, I, I)
(2,3,6)
+ y) + sin (y + z) + sin (x + z) = 0, ('1f, '1f, '1f) + ye' + 2lnx - 2 - 3ln2 = 0, (I, In 2, 1n3)
31. sin (x 32. xe'
Finding Partial Derivatives at Specified Points 33. Find aw/ar whoo r ~ I,s ~ -I if w ~ (x x = r - s,y = cos(r + s),z = sin(r + s). 34. Find aw/av x = v 2ju,y =
U
whoo u = -I, v + v,z = cosu.
=2
35. Find aw/av whoo u = 0, v = 0 x = u - 2v + 1, y = 2u + v - 2. 36. Find azjau when u = 0, v = 1 x = u 2 + v 2 ,y = uv.
= h(t)
s ~ h(x,y),
k(x,y)
29. z' - X)'
(v'3,2, I)
r ~ g(x,y),
aw () 17. aw au andavforw =gx,y,
28.
In Exercises II and 12,
r ~x
~
19. at and as for z
I z - sinxy,
t
=3
t
t= 3 5. w
aw aw 16. ax and iJy forw ~ f(r,s,t),
w ~ k(t)
37. Find az/au and az/av when Stan-Ix and x=e"+lnv.
y ~ g(u,v),
38. Find az/au and az/av when u and q = "V'iJTI tan-l u.
u ~
if if
w w
+ y + z'1,
= X)' + Inz,
= x 2 + (y/x),
if z = sinX)'
+ x siny,
= In 2, v = I
if z
I, v
~
-2 if z
~
=
In q
14.4 The Chain Rule Theory and Examples 39. Assume thatw = f("
a
+ I') andj'(x)
=
e'. Find
a
a7 and a;'
_(,.)af af (x,y) -_x'2' ' ax (x,y) -_xy, and ay
47. Extreme values on a heli:J: Suppose that the partial derivatives of a function I(x, y, z) at points on the helix x = cos I, Y = sin t,
z=tare
40. Assume that w - f Is , t
aw
f,
= sint,
fy
Ix=cost,
aw
F'md ai and as'
783
= I'
+ 1 - 2.
At what points on the curve, if any, can f take on extreme values?
41. Changing voltage in a circuit The voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery wellTli out At the same time, the resistance R is increasing as the resistor heats up. Use the equatioo ~
dV dl
aVdI aI dl
+ avdR aR dl
to fmd how the current is changing at the instaot when R =
48. A space corve Let w ~ x'e'Y cos 3z. Find the value of dw/dl at the point (l,ln2,0) 00 the curve x = cost,y = In(t+ 2), z = t. 49. Temperature on a circle Let T = f(x, y) be the temperature at the point (x, y) on the circle x = cos I, Y = sin t, 0 :5 t :5 2'7T and suppose that
600 ohms, I ~ 0.04 amp, dR/dl ~ 0.5 ohm/sec, and dV/dl ~ -om volt/sec.
aT -=8x- 4:y ax '
aT ay=8y-4x.
v
~-
I
a. Find where the maximum and minimum temperatures on the circle occur by examining the derivatives dT/dt and d'T/dt'. b. Suppose that T ~ 4x' - 4xy + 4y'. Find the maximum and minimum values of T on the circle.
Bsttery
I
50. Temperature on an ellipse Let T at the point (x, y) 00 the ellipse
R
42. Changing dimensions in a box The lengths a, b, and c of the edges of a rectaogular box are changing with time. At the instaot in questioo, a = 1 m, b = 2 m, c = 3 m, tla/dl = db/dl = 1 mfsec, and dc/dl ~ -3 mfsec. At what rates are the box's volume V and surface area S changing at that instaot? Are the box's interior diagonals increasing in length or decreasing? 43. If flu, v, w) is differentiable and u w = z - x, show that af af af ax + iIy + az =
~
x - y, v
~
y - z, and
o.
44. Polar coordinates Suppose that we substitute polar coordinates x = r cos 8 and y = r sin 8 in a differentiable function w ~ f(x,y).
x
= 2V2 cos t,
y = V2 sint,
aT ax
=
~
+ fy cos o.
~
x.
Differentiating Integrals Undet mild cootinuity restrictioos, it is true that if
/.b
/.b
g(I,x) dl,
g,(t,x) dt. Using this fact and the Chain Rnle, we
can fmd the derivative of
b. Solve the equatioos in part (a) to express f, and fy in terms of aw/ar and aw/ao.
F(x)
=.
/('<)
g(I, x) dl
/.
c. Showthat
(/,)' + (/y)'
O::=:;t::=:;2w,
a. Locate the maximum and minimum temperatures on the ellipse by examining dT/dl and d'T/dl'. b. Suppose that T = xy - 2. Find the maximum and minimum values of T 00 the ellipse.
then F'(x) =
-f, sin 0
aT ay
y,
F(x) =
}:
be the temperature
and suppose that
a. Showthat
and
~ g (x, y)
~ (~;)' + r\ (~;)'.
45. Laplace equations Show that if w = flu, v) satisfies the Laplace equatioo f .. + fvv = 0 and if u = (x' - y')/2 and v = X)I, then w satisfies the Laplace equation W;g- + wyy = O. 46. Laplace equations Let w ~ flu) + g(v), where u ~ x + iy, v ~ x - iy, and i ~ Y=1. Show that w satisfies the Laplace equation W;a- + w". = 0 if all the necessary functions are differentiable.
by letting
G(u, x)
~ /." g(t, x) dt,
where u = f(x). Find the derivatives of the functioos in Exeteises 51 and 52. 51. F(x)
= 0 " Vt'+7 dl
/.
784
Chapter 14: Partial Derivatives
14.5
Directional Derivatives and Gradient Vectors If you look at the map (Figure 14.25) showing contours along the Hudson River in New York, you will notice that the tributary streams flow perpendicular to the contours. The streams are following paths of steepest descent so the waters reach the Hudson as quickly as possible. Therefore, the fastest instantaneous rate of change in a stream's elevation above sea level has a particular direction. In this section, you will see why this direction, called the "downhill" direction, is perpendicular to the contours.
FIGURE 14.25 Contours along the Hudson River in New York show streams, which follow paths of steepest descent, running perpendicular to the contours.
Directional Derivatives in the Plane We know from Section 14.4 that if l(x,Y) is differentiable, then the rate at which I changes with respect to t along a differentiable curve x = get), Y = h(t) is
Y Line x =xo + sU!,Y = Yo + sU2
~
R
-+-----J'---------~x
o
FIGURE 14.26 The rate of change of fin the direction of u at a point Po is the rate at which f changes along this line at Po .
dl dt
al dx ax dt
al dy ay dt .
-=--+--
At any point Po(xo, Yo) = Po(g(to), h(to», this equation gives the rate of change of I with respect to increasing t and therefore depends, among other things, on the direction of motion along the curve. If the curve is a straight line and t is the arc length parameter along the line measured from Po in the direction of a given unit vector u, then dlIdt is the rate of change of I with respect to distance in its domain in the direction of u. By varying u, we find the rates at which I changes with respect to distance as we move through Po in different directions. We now define this idea more precisely. Suppose that the function I(x, y) is defined throughout a region R in the xy-plane, that Po(xo, Yo) is a point in R, and that u = Ul i + u2i is a unit vector. Then the equations x = Xo
+ SUh
y = Yo
+ SU2
parametrize the line through Po parallel to u. If the parameter S measures arc length from Po in the direction ofu, we find the rate of change of I at Po in the direction ofu by calculating dlIds at Po (Figure 14.26).
14.5
785
Directional Derivatives and Gradient Veeto"
DEFINmON
The derivative of fat Po(Xo,Yo) in the direction of the unit
vector u
+ u2j is the number
= Uli
df ) ( tis uRn
=
lim f(xo
+ su"yo + S"2) - f("o,Yo)
8-0
S
(1) '
provided the limit exists.
The directional derivative defmed by Equation (1) is also denoted by "The derivative of/at Po in the dinlction of a"
The partial derivatives fx(xo, YO) and /y(xo, YO) are the directional derivatives of f at Po in the i andj directions. This observation can be seen by comparing Equation (1) to the definitions of the two partial derivatives given in Section 14.3.
EXAMPLE 1
Using the definition, fmd the derivative of
f(x,y)
x 2 + xy
=
at Po(1, 2) in the direction of the unit vector u = (1/v'2)i Solutton
+
(1/v'2)j.
Applying the definition in Equation (1), we obtain
df ) ( ds n.Po
= lint 8-0
f("o + su"yo + SU2) - f("o,Yo) s
Eq. (1)
f(1 + s._I_, 2 + s._I_) - f(I,2) = lint 8-0
= lint
v'2
v'2 S
S S ( 1+_ )2+(I+_S)(2+_ )-(12+ 1'2) v'2
v'2
8-+0
v'2
s
(1+ ~ + S2) + (2 + ~ + S2) - 3
2 2 . v'2 v'2 = 1m -"------'-~--"----;-----'-~--"--3-+0
S
~+S2 = lint v'2 ,~o S
= lint ,~o
(_5_ + s) _5_. =
v'2
v'2
The rate ofchange of f(x,y) = x 2 + xy atPo(1, 2) in the direction u is 5/V2.
•
Interpretation of the Directional Derivative The equation z = f(x, y) represents a surface S in space. If zo = f(xo, YO), then the point P(xo, Yo, zo) lies on S. The vertical piane that passes through P and Po(xo, YO) parallel to u FIGURE 14.27 The slope ofcurve Cat Po is lim slope (PQ); this is the Q~P
directional dctivative
(~) .,p, = (Duf)p,.
intersects S in a curve C (Figure 14.27). The rate of change of f in the direction ofu is the slope of the tangent to C at P in the right-handed system formed by the vectors u and k. When u = i, the directional derivative at Po is af/ax evaluated at ("0, Yo). When u = j, the directional derivative at Po is af/ay evaluated at (xo, Yo). The directional derivative generalizes the two partial derivatives. We can now ask for the rate of change of f in any direction u, not just the directions i and j .
786
Chapter 14: Partial Derivatives For a physical interpretation of the directional derivative, suppose that T = I(x, y) is the temperature at each point (x, y) over a region in the plane. Then 1("0, YO) is the temperature at the point Po(xo, YO) and (D./)po is the instantaneous rate of change of the temperature at Po stepping offin the direction u.
Calculation and Gradients We now develop an efficient formula to calculate the directional derivative for a differentiable function f. We begin with the line
x =
Xo
+
Y=YO+
SUI,
(2)
SU2,
through Po(xo, yo), parametrized with the arc length parameter s increasing in the direction of the unit vectur u = Uti + u2j.Thenhythe Chain Rule we f"md Chain Rule for differentiable f From Eqs. (2),
dx/d!
= III
and dy/ds
= U2
(3) Gradient of / at Po
Direction a
Equation (3) says that the derivative of a differentiable function I in the direction of u at Po is the dot product ofu with the special vectur called the gradient of I at Po.
DEFINmON
The gradient vector (gradient) of I(x, y) at a point Po(xo, YO)
is the vectur
VI = al i + al. ax
ayl
obtained hy evaluating the partial derivatives of I at Po.
The notation VI is read "grad f' as well as "gradient of f" and "del I:' The symbol V hy itselfis read "del." Another notation for the gradient is grad f.
THEOREM 9-The Directional Derivative Is a Dot Product differentiable in an open region containing Po(xo, yo), then
(:to
the dot product ofthe gradient
=
(Vj)p,u,
(4)
VI at Po and u.
EXAMPLE 2 Find the derivative of I(x,y) = xe Y direction ofv = 3i - 4j. Solution
If I(x,y) is
+ cos (xy)
at the point (2,0) in the
The direction ofv is the unit vector obtained by dividing v by its length: u =
v
v
3.
4.
[Vj = 5 = 5' - 51·
14.5
Directional Derivatives and Gradient Veeto"
787
The partial derivatives of / are everywhere continuous and at (2, 0) are given by
Y
fx(2,0)
= (e Y
- y sin (XY))(2,O) = eO - 0 = I
jj,(2,O) = (xe Y - x sin (XY))(2,O) = 2e o - 2· 0 = 2. The gradient of/at (2, 0) is V/I(2,o) = fx(2,0)1 + /y(2,0)j = 1 + 2j x
(Figure 14.28). The derivative of I at (2, 0) in the direction ofv is therefore
(Duf) 1(2,0)
= V/I(2,o)'U =
FIGURE 14.28 Picture V/ as a vector in the domain of /. The figaro shows a number oflevel corves of/. The rate at which / changes at (2, 0) in the direction Q = (3/5)1 - (4/5)i is V/'Q = -I (Example 2).
Eq.(4)
(i + 21')' (Ii - 15 1,) 5
=
I5 - ~5 =
-I
.
•
Evaluating the dot product in the fannula
Dul = V/'u = IV/llulcosO = IVII cos 0, where 0 is the angle between the vectors u and VI, reveals the following properties.
Properties ofthe Directional Derivative Duf =
VI' u
=
IVI Icos 0
1. The function I increases most rapidly when cos 0 = I or when 0 = 0 and u is the direction of Vf. That is, at each point P in its domain, I increases most rapidly in the direction of the gradient vector VI atP. The derivative in this direction is
Duf = IV/leos(O) = IV/I· 2. Similarly, I decreases most rapidly in the direction of - Vf. The derivative in this direction is Duf = IVII cos ('IT) = -I VII. 3. Any direction u orthogonal to a gradient V/oF 0 is a direction of zero change in I because 0 then equals 'IT/2 and Duf
o.
= IVII cos ('IT/2) = IV/I'O =
As we discuss later, these properties hold in three dimensions as well as two.
EXAMPLE 3
Find the directions in which I(x,y) = (x 2/2) + (y2/2)
(a) increases most rapidly at the point (I, I).
(b) decreases most rapidly at (I, I). (c) What are the directions ofzero change in I at (I, I)? Solution
(a) The function increases most rapidly in the direction of VI at (I, I). The gradient there is (Vf)(l,l) =
(xi
+ yj)(l,l)
=
i
+ j.
Its direction is I+j I+j u = Ii + jl = \/(1)2 + (1)2 =
I.
(b) The function decreases most rapidly in the direction of -u = __I_ i __I_j.
o
0
I. 1.
0' + 0
VI at (I, I), which is
788
Chapter 14: Partial Derivatives (0) The directions ofzero change at (1,1) are the directions orthogonal to Vf: n = -1- ,.
v'2
+ -1J•
1.
1.
-n = - , - - J .
and
v'2
v'2
v'2
•
See Figure 14.29.
-~)~!:i~y / k :
I
/-
/
Most rapid /
x
I
'I
'/
I
decreaseinf
/
Most rapid /
Zero change
inf
...
Gradients and Tangents to Level Curves If a differentiable function f(x, y) has a constant value c along a smooth curve r = g(t)i + h(t)j(making the curve a level curve off), thenf(g(t), h(t» = c. Differentiating both sides of this equation with respect to t leads to the equations
I
d
Vf~ 1 + J
d
d/(g(t), h(t» = dt (c)
increase in!
FIGURE 14.29 The direction in which f(x,y) increases most rapidly at (I, I) is the direction of VII (1,1) = i + j.Itcorresponds to the direction of steepest asceot on the surface at (I, I, I) (Example 3).
f a . ( ax'
af dg af dh 0 oxdt+aydt= g af .) • (d . + dh J.) = 0 • ayJ dt' dt
+
Chain Rule
(5)
dr
VI
dt
Equation (5) says that Vf is normal to the tangent vector dr/dt, so it is normal to the curve. The level curve fIx, y)
~
f(xo, Yo)
At every point (xo, YO) in the domain of a differentiable function f(x, y), the gradient of f is normal to the level curve through {xo, yo) (Figure 14.30).
FIGURE 14.30 The gradieot ofa differeotiable function of two variables at a point is always normal to the function's level curve through that point
Equation (5) validates our observation that streams flow perpendicolar to the contours in topographical maps (see Figure 14.25). Since the downflowing stream will reach its destination in the fastest way, it must flow in the direction of the negative gradient vectors from Property 2 for the directional derivative. Equation (5) tells us these directions are perpendicolar to the level curves. 11Iis observation also enables us to find equations for tangent lines to level curves. They are the lines normal to the gradients. The line through a point Po(xo, YO) normal to a vector N = Ai + Bj has the equation A{x - xo)
+ B{y
- YO) = 0
(Exercise 39). If N is the gradient (Vf)~.Yol = fx{xo,yo)i the tangent line given by
fx{xo,yo)(x - xo)
y
Vf(-2, 1)
~
-I
+ 2J
x-2y=-4
EXAMPLE 4
+ fy{xo,Yo)j,
+ fy{XO,yo)(Y - YO)
= O.
Find an equation for the tangent to the ellipse
4x' + y'
= 2
(Figure 14.31) at the point ( -2, I). --t-;;---:----;ot---;-----;:-+-~x
2V2
Solution
The ellipse is a level curve of the function 2
f(x,y) = FIGURE 14.31 We can fmd the tangeot to the ellipse (x'/4) + y' ~ 2 by treating the ellipse as a level curve of the function f(x,y) ~ (x'/4) + y' (Example 4).
x 4 + y2.
The gradient of f at ( -2, 1) is
Vfl(-2,l)
=
(~i + 2yj ) (-2,1) =
-i + 2j.
the equation is (6)
14.5 Directional Derivatives and Gradient Veeto"
789
The tangent is the line
(-I)(x
+ 2) +
(2)(y - I) = 0
x - 2y
=
Eq. (6)
•
-4.
If we know the gradients of two functions f and g, we automatically know the gradients of their sum, difference, constant multiples, product, and quotient. You are asked to establish the following rules in Exercise 40. Notice that these rules bave the same form as the corresponding rules for derivatives of single-variable functions.
Algebra Rules for Gradients
3. Constant Multiple Rule:
V(j + g) = Vf + Vg V(j - g) = Vf - Vg V(kj) = kVf (any number k)
4. Product Rule:
V(jg) = fVg
1. Sum Rule: 2. Difference Rule:
v(L)
5. Quotient Rule:
EXAMPLE 5
=
g
+ gVf
gVf - fVg g2
We illustrate two of the rules with
f(x,y) = x - y Vf = i - j
g(x,y) = 3y Vg = 3j.
We have
1.
V(j - g) = V(x - 4y) = i - 4j = Vf - Vg
2.
V(jg) = V(3xy - 3y 2) = 3yi + (3x - 6y)j
+ 3yj + (3x - 6y)j j) + (3x - 3y)j j) + (x - y)3j = gVf +
=
3y(i - j)
=
3y(i -
=
3y(i -
Rule 2
fVg
Rulc4 •
Functions of Three Variables For a differentiable function f(x, y, z) and a unit vector u =
U1 i
+ u2j + U3 k in space, we
have
and
Duf = Vf·u =
af ax
-U,
af
af
+ -U2 + -U3. ay az
The directional derivative can once again be written in the form
Daf = Vf'u =
IVfllul cosO
=
IVfl cosO,
so the properties listed earlier for functions of two variables extend to three variables. At any given point, f increases most rapidly in the direction of Vf and decreases most rapidly in the direction of - Vf. In any direction orthogonal to Vf, the derivative is zero.
790
Chapter 14: Partial Derivatives
EXAMPLE 6
= x3 -
xy' - z at Po(l, 1,0) in the direction of
(b) In what directions does I change most mpidly at Po, and what are the mtes of change in these directions? Solution
Ivl = Y(2)' + (-3)' + (6)' = V49 = 7 2.
v
3.+6 k
u=fVl=·p-·p
7'
The partial derivatives of I at Po are
Ix = (3x'
- Y')(l,l,O)
= 2,
Iy = -2xy1 (1,1,0) = -2,
Iz =
-11(1,1,0) =
-1.
The gmdient of I at Po is
VII (1,1,0)
= 21 - 2j - k.
The derivative of I at Po in the direction ofv is therefore (Duf)(l,I,O) = V/b,l,o)"U = (21 - 2j -
k)' (tl - ~j + ~k)
4 6 6 4 =-+---=7 7 7 7' (b) The function increases most mpidly in the direction of VI = 21 - 2j - k and decreases most mpidly in the direction of - Vf. The mtes of change in the directions are, respectively,
IVII
= Y(2)' + (-2)' + (-I)' = V9 = 3
-I VII
and
=
-3.
•
Exercises 14.5 Calculating Gradients In Exercises 1- 6, find the gradient of the function at the given point. Then sketch the gradient together with the level curve that passes through the point. 1. f(x,y) = y - x,
(2, I)
3. g(x,y) = xy', (2, -I) 5. f(x,y) 6. f(x,y)
= V2x + 3y, = tan-I
y;,
2. f(x,y) = In (x' + y'), (1,1) ,
y'
4. g(x,y) = x 2
T' (0, I)
rection of u. 11. f(x,y) = 2xy - 3y', P o(5,5),
12. f(x,y) ~
2x' + y',
x-y 13. g(x,y) = xy + 2'
Po( -I, I), Po(l, -I),
(-1,2) 14. h(x,y) = tan-1 (Ylx) u=31-2j
(4, -2)
VI at the given point. 7. f(x,y,z) ~ x' + y' - Zz' + zlnx, (I, I, I) 8. f(x,y,z) = Zz' - 3(x' + y')z + tan-I xz, (I, I, I)
In Exercises 7-10, fmd
9. f(x,y,z) ~ (x' + y' + z')-l/2 + In (xyz), 10. f(x,y,z) = e'+'cosz
finding Directional Derivatives In Exercises 11-18, fmd the derivative of the function at Po in the di-
+ (y + I) sin-lx,
(-1,2, -2) (0,0,"/6)
u = 41 + 3j U
~ 31 - 4j
u = 121 + 5j
+ v'3 sin-I (xyI2),
15. f(x,y, z) = xy + yz + zx, Po(l, -1,2), 16. f(x,y,z) ~ x' + 2y' - 3z', Po(l, I, I), 17. g(x,y,z) = 3e'cosyz, PolO, 0, 0), 18. h(x,y, z) = cosxy u=I+2j+2k
Po(l, I), u = 31 + 6j - 2k U
u = 2i
~ I + j + k
+j
+ e P + Inzx, Po(l,O, 1/2),
- 2k
14.6 In Exercises 19-24, fmd the directions in which the functions increase and decrease most rapidly at Po. Then rmd the derivatives of the functions in these directions.
20. I(x,y) =
+ >;)' + y', Pot -I, I) x'y + ery siny, Po(1,O)
21. I(x,y, z)
~
19. I(x,y) = x'
(x/y) - yz, Po(4, I, I) 22. g(x,y,z) ~ xe Y + z', P o(1,ln2,1/2) 23. I(x,y, z) ~ In>;)' + Inyz + Inxz, Po(1, I, I) 24. h(x,y, z) ~ In (x' + y' - I) + y + 6z, Po(1, 1,0)
Tangent LInes to Level Curves In Exercises 25-28, sketch the curve I(x, y) = c together with VI and the tangent line at the given point Then write an equation for the tangent line.
x' + y' = 4, (v'2, v'2) 26.x'-y=l, (v'2,I) 25.
27. >;)' = -4,
33. Is there a direction 0 in which the rate of change of I(x, y) ~ 3>;)' + 4y' at p(1, 2) equals 14? Give reasons for your
x' -
answer. 34. Changing temperature along a circle Is there a direction 0 in which the rate of change of the temperature function T(x, y, z) = 2>;y - yz (temperature in degrees Celsius, distance in feet) at P(I, -I, I) is -3°C/It? Give reasons for your answer. 35. The derivative of I(x,y) at Po(I, 2) in the direction ofi + j is 2V2 and in the direction of -2j is -3 . What is the derivative of I in the direction of -i - 2j? Give reasons for yoar answer.
2v'3. a. What is Vf at P? Give reasons for your answer.
(-1,2)
a. Du 1(1, -I) is largest c. Du /(1, -I) ~ 0
e. D u /(1, -I) 30. Let I(x,y) =
~
b. What is the derivative of I at P in the direction ofi
n. 1(1, -I) is smallest d. D./(1, -I) ~ 4
b.
-3
~~ ~ ~~. Find the directions 0
D./(-H) Du/(+~) c. Du/( +~) = 0 a.
is largest
Du/(+~)
b. D.
1(+ ~)
+ j?
37. Directional derivatives and scalar components How is the derivative ofa differentiable function I(x,y, z) at a point Po in the direction of a unit vector u related to the scalar component of (V/)p, in the direction ofo? Give reasons foryoor answer. 38. Directional derivatives and partial derivatives Assuming that the necessary derivatives of I(x, y, z) are dermed, how are Dtl, DJ/, and D.I related to I.. I y , and I.? Give reasons for your
answer.
and the valoes of
39. Lines in the xy-plane Show that A(x - xo) + Bly - Yo) = 0 is an equation for the line in the xy-plane throogh the point (xo, Yo) normal to the vectorN = Ai + Bj.
for which
14.6
32. Zero directiooal derivative In what directions is the derivative of I(x,y) = (x' - y')/(x' + y') atp(l, I) equal to zero?
36. The derivative of I(x, y, z) at a point P is greatest in the direction of v ~ i + j - k. In this direction, the value of the derivative is
Theory and Examples 29. Let I(x,y) = x' - >;)' + y' - y. Find the directions u and the values of D. 1(1, -I) for which
e.
31. Zero directional derivative In what direction is the derivative of I(x, y) ~ >;)' + y' at P(3, 2) equal to zero?
(2, -2)
28.x'->;)'+y'=7,
791
rangent Planes and Differentials
is smallest
d. Du/(-t,~) =-2
40. Tbe algebra rules for gradients gradients
=I
VI = al i
ax
+ a/J' + al k, ay
az
Given a constant k and the
Vg = ag i
ax
+ ag J• + ag k, ay
az
establish the algebra rules for gradients.
Tangent Planes and Differentials In this section we derme the tangent plane at a point on a smooth surface in space. Then we show how to calculate an equation of the tangent plane from the partial derivatives of the function defining the surface. This idea is similar to the dermition ofthe tangent line at a point on a curve in the coordinate plane for single-variable functions (Section 3.1). We then study the total differential and linearization of functions of several variables.
Tangent Planes and Normal Lines
+ h(t)j + k(t)k is a smooth curve on the level surface f(x, y, z) = c of a differentiable function f, then f(g(t), h(t), k(t)) = c. Differentiating both sides of this
If r = g(t)i
792
Chapter 14: Partial Derivatives
equation with respect to t leads to d
-I----
f(x, y, z) = c
FIGURE 14.32 The gradient Vi is orthogonal to the velocity vector of every smooth curve in the surface through Po. The velocity vectors at Po therefore lie in a common plane, which we call the tangent plane at Po.
f a i (~
+ af · + af ~J
~
d
dt f(g(t), h(t), k(t))
==
dt (c)
af dg af dh af dk -+-+-
==
0
==
o.
ax dt g (d i
ay dt
k) . &
az dt
+ dh · + dk ~J &
k)
Chain Rule
(1)
dr/~
Vi
At every point along the curve, Vf is orthogonal to the curve's velocity vector. Now let us restrict our attention to the curves that pass through Po (Figure 14.32). All the velocity vectors at Po are orthogonal to Vf at Po, so the curves' tangent lines all lie in the plane through Po normal to Vf. We now define this plane.
DEFINITIONS face f(x, y, z) to Vflpo.
==
The tangent plane at the point Po(xo, Yo, zo) on the level surc of a differentiable function f is the plane through Po normal
The normal line of the surface at Po is the line through Po parallel to Vf Ipo.
From Section 12.5, the tangent plane and normal line have the following equations:
Tangent Plane to /(x,y, z)
fx(Po)(x - xo) Normal Line to /(x,y, z)
==
cat Po(xo, Yo, zo)
+ fy(Po)(y - Yo) + fz(Po)(z - zo) ==
==
0
(2)
cat Po(xo, Yo, zo) (3)
EXAMPLE 1
Find the tangent plane and normal line of the surface
f(x,y,z)
Po(l, 2, 4)
x 2 + y2
==
+z
- 9 == 0
A circular paraboloid
at the point P o( 1, 2, 4). Nonnalline
I
I
I l~
/l . . .
I I I
')
~
SoLution The surface is shown in Figure 14.33. The tangent plane is the plane through Po perpendicular to the gradient of f at Po . The gradient is
Tangent plane
Vf IPo
== (2xi
+ 2yj +
k)(1,2,4) ==
2i
+ 4j +
k.
-......
/// ---.v? -- ~ /
x
FIGURE 14.33 The tangent plane and normal line to this surface at Po (Example 1).
The tangent plane is therefore the plane y
2(x -
1)
+ 4(y
- 2)
+
(z - 4)
==
0,
2
+ 4t,
or
2x
+ 4y + z
==
14.
The line normal to the surface at Po is
x
==
1
+
2t,
Y
==
z == 4
+ t.
•
To find an equation for the plane tangent to a smooth surface z == f(x, y) at a point == f(xo, Yo), we first observe that the equation z == f(x, y) is
Po(xo, Yo, zo) where Zo
14.6
793
Tangent Planes and Differentials
equivalent to f(x,y) - z = O. The surface z = f(x,y) is therefore the zero level surface ofthe function F(x, y, z) = f(x, y) - z. The partial derivatives of Fare
o
=
Ix -
0 =
Ix
=
h-
0 =
h
=
0 - 1 = -1.
Fx = ox (f(x,y) - z)
o
Fy = oy (f(x,y) - z) Fz
=
o
oz(f(x,y) - z)
The formula
F.(Po)(x - xo) + Fy{Po)(y - Yo) + F.(Po)(z - zo)
=
0
for the plane tangent to the level surface at Po therefore reduces to
fx(xo,yo)(x - xo) + fy(xo,yo)(Y - Yo) - (z - zo)
=
o.
Plane Tangent to a Surface:; = f(x,y) at (xo,Yo, fVco,yo» The plane tangent to the surface z = f(x, y) of a differentiable function f at the point Po(xo, Yo, zo) = (XO,YO, f(xo,yo)) is
fx(xo,yo)(x - xo)
EXAMPLE 2
+ fY{xo,yo)(Y - Yo) - (z - zo)
= O.
(4)
Fiod the plane tangent to the surface z = xcosy - ye X at (0,0,0).
Solutton We calculate the partial derivatives of f(x, y) = x cos y - ye X and use Equation (4):
fx(O,O)
=
(cosy - yeX)(o,o) = 1 - 0·1 = 1
/y(O,O)
=
(-x sioy - eX)(o,o)
0- 1
=
-1.
=
The tangent plane is therefore 1'(x - 0) - 1'(Y - 0) - (z - 0) = 0,
or x - y - z =
EXAMPLE 3
Eq.(4)
o.
•
The surfaces
f(x,y,z)
=
x 2 + y2 - 2
=
0
Acylindcr
and
g(x,y,z)
=
x +z - 4
= 0
A plane
meet in an ellipse E (Figure 14.34). Fiod parametric equations for the lioe tangent to E at the poiot Po(I, 1,3). Solutton The tangent lioe is orthogonal to both Vf and Vg at Po, and therefore parallel tov = Vf X Vg. The components ofv and the coordinates of Po give us equations for the lioe. We have Vfl(l,l~) = (2xi
+ 2yj)(I,l,3)
=
2\
+
k)(I,l,3) = \
+
k
Vgl (1,1~)
FIGURE 14.34 This cylinder and plane intersect in an ellipse E (Example 3).
= (\
v = (2\
+ 2j) X (\ + k)
=
+ 2j
2 1
j 2 0
k 0 = 2\ - 2j - 2k. 1
794
Chapter 14: Partial Derivatives The tangent line is
y=I-2t,
x=I+2t,
•
z=3-2t.
Estimating Change in a Specific Direction The directional derivative plays the role of an ordinary derivative when we want to estimate how much the value of a function / changes if we move a small distance tis from a point Po to another point nearby. If / were a function of a single variable, we would have
d/
=
f'(Po) tis.
Ordinary derivative x increment
For a function oftwo or more variables, we use the formula
d/
=
(V/lp,u) tis,
Directional derivative x increment
where u is the direction ofthe motion away from Po.
Estimating the Change In / In a Direction u To estimate the change in the value of a differentiable function / when we move a small distance tis from a point Po in a particular direction u, use the formula
d/
(V/lp, u)
=
tis
Directional Distao<e
derivative increment
EXAMPLE 4
Estimate how much the value of
/(x,y,z)
=
ysinx + 2yz
will change if the point P(x,y, z) moves 0.1 unit from Po(O, 1,0) straight toward
Pt(2, 2, -2). Solution We first fmd the derivative of / at Po in the direction of the vector P-;;Pt = 2i + i - 2k. The direction of this vector is
The gradient of / at Po is V/I(O,I,O) = «ycosx)i
+
(sinx
+ 2z)j + 2yk)(O,I,O)
= i
+ 2k.
Therefore,
V/lpo·u
= (i
+ 2k)·
(t + ti - t t- t -to i
k) =
=
The change d/ in / that results from moving tis = 0.1 unit away from Po in the direction of u is approximately
d/
= (V/lp,u)(tis) =
(-t)tO.I) '"
-0.067unit.
•
How to Linearize a Function of Two Variables Functions of two variables can be complicated, and we sometimes need to approximate them with simpler ones that give the accuracy required for specific applications without being so difficult to work with. We do this in a way that is similar to the way we fmd linear replacements for functions of a single variable (Section 3.9).
14,6 Tangent Planes and Differentials
Apoint _
(x, y)
near (xQ. Yo)
Suppose the function we wish to approximate is z = I(x, y) near a point (xo, Yo) at which we know the values of I, Ix> and I y and at which I is differentiable. Ifwe move from (xo, Yo) to any nearby point (x, y) by increments Ax = x - xo and ~y = y - Yo (see Figure 14.35), then the defmition of differentiability from Section 14.3 gives the change
I(x,y) - I(xo,yo) = h(xo,yo)Ax A point where f is differentiable
795
+ Iy(xo,yo)~y + EtAx +
E2~y,
where Ej, E2 --> 0 as Ax, ~y --> O. If the increments Ax and ~y are small, the products EjAx and E2~y will eventoaIly be smaller still and we have the approximation
/
I(x,y) "" I(xo,yo)
+ I.(xo,yo)(x
+ /y{xo,yo)(y
- xo)
- Yo).
L(x,y)
FIGURE 14.35 If! is differentiable at (xo, yo), then the value of! at soy point (x, y) nearby is approximately !(xo,Yo) + fx(xo,Yo)4x + !y(xO,yo)4y.
In other words, as long as as the linear function L.
~x and ~y
are small, I will have approximately the same value
DEFINmONS
I
The linearization of a function I(x, y) at a point (xo, Yo) where is differentiable is the function
L(x,y) = l(xo,Yo)
+
+
fx{xO,yo)(x - xo)
/y{XO,yo)(y - Yo).
(5)
The approximation
I(x,y) "" L(x,y) is the standard linear approximation of I at (xo, Yo).
From Equation (4), we fmd that the plane z = L(x,y) is tangent to the surface z = I(x, y) at the point (xo, Yo). Thus, the linearization of a function of two variables is a tangent-plane approximation in the same way that the linearization of a function of a single variable is a tangent-line approximation. (See Exercise 63.)
EXAMPLE 5
Find the linearization of
I(x,y) = x 2 - xy
+
t
y2
+3
at the point (3, 2). Solutton
We first evaluate I, Ix. and I y at the point (xo,Yo) = (3,2):
1(3,2) = (x 2 - xy
fx(3,2)
= :
x
+ 1 y2 + 2
(x 2 - xy
/y{3,2) = aa (x 2 - xy Y
3)
+ -21 y2 +
= (3,2)
3)
+ -21 y2 + 3)
8
(3,2)
(3,2)
= (2x - y)(3,2) =
= (-x
+ Y)(3,2)
4
= -I,
giving
L(x, y) = I(xo, Yo) =
8
+
+ (4)(x
fx{xo, yo)(x - xo) - 3)
+ (-I)(y
+
/y{xo, yo)(y - Yo)
- 2) = 4x - y - 2.
The linearization of I at (3, 2) is L(x,y) = 4x - y - 2.
•
When approximating a differentiable function I(x, y) by its linearization L(x, y) at (xo, Yo), an important question is how accurate the approximation might be.
796
Chapter 14: Partial Derivatives
y
If we can tmd a common upper bound M for If", I, Ifyy I, and If", I on a rectangle R centered at (xo,Yo) (Figure 14.36), then we can bound the error E throughout R by using a simple formula (derived in Section 14.9). The error is detmed by E(x,y) =
f(x,y) - L(x,y).
R --;:i-------~x
o
FIGURE 14.3& The rectangular region R: Ix - >:01 '" h, Iy - yol '" kinthe xy-plane.
The Error in the Standard Linear Approximation If f has continuous first and second partial derivatives throughout an open set containing a rectangle R centered at (xo, YO) and if M is any upper bound for the values of If", I, Ifyyl, and If", Ion R, then the error E(x, y) incurred in replacing f(x, y) on R by its linearization
L(x, y)
=
f(xo, YO) + fx(xo, yo)(x - xo) + fixo, YO)(Y - YO)
satisfies the inequality
IE(x,y) I :s t M(lx - xol + Iy - Yo I)'·
To makeIE(x,y) Ismall for a givenM, we just make Ix - xol andly - yol small.
EXAMPLE 6 Find an upper bound for the error in the approximation f(x, y) '" L(x, y) in Example 5 over the rectangle R:
Iy - 21 :s 0.1.
Ix - 31 '" 0.1,
Express the upper bound as a percentage of f(3, 2), the value of f at the center of the rectangle. Solution
We use the inequality
IE(x,y) I :s t M(lx - xol + Iy - Yo I)' . To tmd a suitable value for M, we calculate f "'.!"', and f yy, tmding, after a routine differentiation, that all three derivatives are constant, with values
If", I = 121
If", I = 1-11
= 2,
Ifyyl
= 1,
= 111 =
1.
The largest of these is 2, so we may safely takeMto be 2. With (xo,Yo) = (3,2), we then know that, throughout R,
IE(x,y) I '" t(2)(lx - 31
+ Iy
- 21)2 = (Ix - 31
+ Iy
- 21)2.
Finally, since Ix - 31 '" 0.1 andly - 21 '" 0.1 onR, we have
IE(x,y) I :s (0.1 + 0.1)2
= 0.04.
As a percentage of f(3, 2) = 8, the error is no greater than
0.~4
X 100 = 0.5%.
•
Differentials Recall from Section 3.9 that for a function of a single variable, y = f(x), we defined the change in f as x changes from a to a + ax by
!:.f
=
f(a
+ ax) - f(a)
and the differential of f as
df = f'(a)ax.
14.6 Tangent Planes and Differentials
797
We now consider the differential of a function oftwo variables. Suppose a differentiable function f(x, y) and its partial derivatives exist at a point (xo, yo). If we move to a nearby point (xo + at,yO + lly), the change in f is
llf
=
f(xo + llx,yo + lly) - f(xo,yo).
A straightforward calculation from the definition of L(x, y), using the notation x - Xo and y - Yo = IIy, shows that the corresponding change in L is llL
=
L(xo + at,yO + lly) - L(xo,yo)
=
fx(xO,yo)at + /y(xo,yo)lly.
=
llx
The differentials dx and dy are independent variables, so they can be assigned any values. = llx = x - Xo, and dy = lly = y - Yo. We then have the following def"mition of the differential or total differential of f.
Often we take dx
DEFINmON
If we move from (xo ,YO) to a point (xo
+ dx,yo + dy)
nearby,
the resulting change
df
=
fx(xo ,YO) dx + /y(xo, YO) dy
in the linearization of f is called the total differential of f.
EXAMPLE 7 Suppose that a cylindrical can is designed to have a radius of I in. and a height of 5 in., but that the radius and height are off by the amounts dr = +0.03 and dh = -0.1. Estimate the resulting absolute change in the volume ofthe can. Solution
To estimate the absolute change in V = '1I'r 2h, we use
With V, =
llV'" dV = V,(ro, ho) dr + Vh(ro, ho) dh. 2'11'rh and Vh = '1I'r 2, we get dV = 2'11'roho dr + '1I'r02 dh = 2'11'(1)(5)(0.03) + '1I'(1)2( -0.1) =
EXAMPLE 8
0.3'11' - 0.1'11'
=
0.2'11' '" 0.63 in'
•
Your company manufactures right circular cylindrical molasses storage
tanks that are 25 ft high with a radius of 5 ft. How sensitive are the tanks' volumes to small variations in height and radius?
Solution With V = '1I'r 2h, the total differential gives the approximation for the change in volume as
dV = V,(5, 25) dr + Vh(5, 25) dh 2 = (2'11'rh)(s,2s) dr + ('1I'r )(s,2s) dh = 250'11' dr + 25'11' dh.
r= 5
--l (oj
r=
25
(bJ
FIGURE 14.37 The volume ofcylinder (0) is more sensitive to 0 small change in r than it is to so equally small change in h. The volume of cylinder (b) is more sensitive to small changes in h than it is to small changes in r (Example 8).
Thus, a I-unit change in r will change V by about 250'11' units. A I-unit change in h will change Vby about 25'11' units. The tank's volume is 10 times more sensitive to a small change in r than it is to a small change of equal size in h. As a quality control engineer concerned with being sure the tanks have the correct volume, you would want to pay special attention to their radii. In contrast, if the values of r and h are reversed to make r = 25 and h = 5, then the total differential in Vbecomes
dV
=
(2'11'rh)(2S,5) dr + ('1I'r 2)(2S,5) dh
=
250'11' dr + 625'11' dh.
Now the volume is more sensitive to changes in h than to changes in r (Figure 14.37). The general rule is that functions are most sensitive to small changes in the variables that generate the largest partial derivatives. •
798
Chapter 14: Partial Derivatives
EXAMPLE 9 The volume V = 7TY'-h of a right circular cylinder is to be calculated from measured values ofr and h. Suppose that r is measured with an error of no more than 2% and h with an error of no more than 0.5%. Estimate the resulting possible percentage error in the calculation of V. Solution
We are told that
I~XIOOI';2
~h
and
1
X 1001 ,; 0.5.
Since
dV V
27Trhdr + 7Tr 2 dh 7Tr 2 h
=
=
2dr r
+ dh h'
we have
Id;1
=
12~ + ~hl
,; 12~1 + I~I ,; 2(0.02)
+
0.005 = 0.045.
•
We estimate the error in the volume calculation to be at most 4.5%.
Functions of More Than Two Variables Analogous results hold for differentiable functions of more than two variables.
1.
The linearization of I(x,y, z) at a point Po(xo, Yo, zo) is
L(x,y,z) = I(Po)
2.
+
+ ly(PoHy
- YO)
+ !z{Po)(z
- zo).
Suppose that R is a closed rectangular solid centered at Po and lying in an open region on which the second partial derivatives of I are continuous. Suppose also that 1/",1, I/)'}'I, Ilrl, 1/",,1, I/ul, and 1/,.1 are all less than or equal to Mthroughout R. Then the error E(x,y, z) = I(x,y,z) - L(x,y, z) in the approximation of Iby L is bounded throughout R by the inequality lEI,;
3.
fx{PoHx - xo)
!M(lx -
xol
+
Iy - yol
+
Iz - zoll2.
If the second partial derivatives of I are continuous and if x, y, and z change from xo,Yo, andzo by small amounts dx, dy, anddz, the total differential dl = fx{Po) dx
+ liPo) dy + IAPo) dz
gives a good approximation of the resulting change in I.
EXAMPLE 10
Find the iinearization L(x,y, z) of
I(x,y,z) = x 2
-
xy
+ 3 sinz
at the point (xo,Yo, zo) = (2, 1,0). Find an upper bound for the error incurred in replacing by L on the rectangle
I
R:
Solution
Ix - 21 ,; 0.01,
Iy - II ,; 0.02,
Izl'; 0.01.
Routine calculations give
1(2, I, 0)
=
2,
1.(2, I, 0)
=
3,
/y(2, 1,0)
=
-2,
IA2, I, 0) = 3.
14.6
Tangent Planes and Differentials
- I)
+ 3(z -
799
Thus, L(x,y, z) = 2
+ 3(x
- 2)
+ (-2)(y
0) = 3x - 2y
+ 3z
- 2.
Since
In =
2,
Iyy =
In
0,
Ixy
= -3 sinz,
Ixz =
= -I,
0,
Iyz =
0,
andl-3 sinzl:5 3 sin 0.01 '" .03, we may take M = 2 as a bound on the second partials. Hence, the error incurred by replacing I by L on R satisfies lEI :5
t
(2)(0.01
+ 0.02 + 0.01)2
•
= 0.0016.
Exercises 14.6 Tangent Planes and Normal Lines to Surfaces 10 Exercises 1--1l, find equations for the
Estimating Change 19. By about how much will j(x,y,z) = IoYx 2
2. x 2 + y2 - z2 = 18, P o(3, 5, -4)
change if the point P(x, y, z) moves from Po(3, 4,12) a distance of ds = 0.1 unit in the direction oni + 6j - 2k? 20. By about how much will
x2
3. 2z = 0, Po(2, 0, 2) 4. x 2 + 2xy - y2 + z2 = 7,
!(x,y,z) Po(l, -I, 3)
5. cos 1TX - x'y + en + yz = 4, Po(O, 1,2) 6. x 2 - xy - y2 - z = 0, Po(l, I, -I)
7. x + y + z = I, PolO, 1,0) 8. x 2 + y2 - 2xy - x + 3y - z
= -4,
P o(2, -3, 18)
given surface at the given point. 9. z = 10 (x 2 + y2), (1,0,0) 10. z = e-("+Y~, (0,0, I) 11. z ~~, (1,2, I) 12. z ~ 4x 2 + y2, (I, I, 5) Tangent Lines to Space Curves 10 Exercises 13-18, fmd parametric equations for the line tangent to the curve of intersection of the surfaces at the given point. Point:
+ y2 + 2z =
4,
x = I
Point: Point:
(I, I, I)
(I, I, 1/2)
Point:
+ 3x'y2 + y' + 4xy + y2 + z2 = 11
- z2 ~ 0,
(I, 1,3)
18. Surfaces: x 2 Point:
y = I
(1/2, I, 1/2)
17. Surfaces: x' x2 Point:
= 2,
+ y2
=
4,
(v'2, v'2, 4 )
xl
+ y2
-
- ysinz
+y
change if the pniot P(x, y, z) moves from P o(2, -1,0) a distance ofds ~ 0.2 unit toward the pointP,(O, 1,2)? 22. By about how much will
h(x,y,z) = cos ('1TXy)
+ xz 2
change if the point P(x, y, z) moves fromPo(-I, -I, -I) a distance of ds = 0.1 unit toward the origin? 23. Temperatore change along a circle Suppose that the Celsius temperature at the point (x,y) in thexy-plane is T(x,y) ~ x sin2y and that distance in the xy-plane is measured in meters. A particle is moving clockwise around the circle of radius 1 m centered at the origin at the constant rate of2 m/sec.
P(I/2, V3/2J!
+ 2y + 2z = 4, y = I
+ y2 + z
+ xcosz
b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?
(I, I, I)
16. Surfaces: x
21. By about how much will
a. How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point
14. Surfaces: xyz = I, x 2 + 2y 2 + 3z 2 = 6 15. Surfaces: x 2
= eXcosyz
change as the point P(x, y, z) moves from the origin a distance of ds ~ 0.1 unit in the direction of 2i + 2j - 2k?
g(x,y,z) = x
10 Exercises l}.-12, fmd an equation for the plane that is tangent to the
13. Surfaces: x
+ y2 + z2
Z =
0
24. Changing temperature along a space curve The Celsius temperature in a region in space is given by T(x,y, z) = 2x 2 - xyz. A particle is moving in this region and its position at time t is given by x = 212, y = 3t, Z = -12, where time is measured in seconds and distance in meters. a. How fast is the temperature experienced by the particle changing in degrees Celsius per meter when the particle is at thepointP(8,6, -4)? b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?
800
Chapter 14: Partial Derivatives
Finding Linearizations
34. f(x,y) ~ (1/2).<2
In Exercises 25-30, f"md the linearization L(x, y) of the funetion at
each point. 25. f(x,y) 26. f(x,y) 27. f(x,y) 28. f(x,y) 29. f(x,y) 30. f(x,y)
= x 2 + y2 + I at ~ (x + y + 2» at ~ 3x - 4y + 5 at = xV at = .'cosyat = .2y-, at
.. (0,0),
b. (I, I)
a. (0,0),
b. (1,2)
a. (0,0), .. (I, I),
b. (0,0)
.. (0,0),
b. (0,,,./2)
.. (0,0),
b. (1,2)
R:
n=
35.74
b. (I, I)
36. f(x,y) = xy2
R:
5 25
10 I
5
0
-5 -11
-5 -10 -16 -22
7
15
9
3
-4 -10 -16 -22 -28
15 19 13 v (mph) 20 17 11
6
0
-7 -13
4
-2
25 16
9
3 -4
30 15
8
I
35 14
7
0 -7
-5
-19
-26 -32
-9 -15 -22
-29 -35
-11
-17 -24 -31
-37
-12 -19 -26 -33
-39
-14 -21
-27
-34 -41
.. Vse the table to f"md W(20, 25), W(30, -10), and W(15, 15). b. Vsethe fannula to f"md W(IO, -40), W(50, -40), and W(60,30).
e. Find the linearizationL(v, point (25, 5). d. Vse L( v, values.
n of1he funetion W(v, n at1he
n in part (c) to estimate the fullowing wind chill
i) W(24, 6)
n
n
32. Find the linearization L( v, of 1he function W( v, in Exercise 31 at 1he point (50, -20). Vse it to estimate the following wind chill values. b. W(53, -19)
c. W(60, -30)
Bounding the Error in Linear Approxlmatlons In Exercises 33-38, find the linearization L(x, y) of 1he function f(x, y) at Po. Then f"md an upper bound for the magnitude lEI of the error in the approximation f(x, y) '" L(x, y) over the rectangle R. 33. f(x,y) ~ x 2 - 3xy + 5 at Po(2, I),
R:
Ix - 21:5 0.1,
Iy - 11:5 0.1
+ Y cos (x
Ix-II:50.I,
R:
- I) at P o(l,2),
ly-21:50.1
Ixl:5 0.1,
lyl:5 0.1
(Vsee':5 1.11 and Icosyl :5 I in estimating E.) 38. f(x,y) = Inx R:
+ Inyat
Ix - II :5 0.2,
Po(l, I),
Iy - II :5 0.2
Linearizations for Thnte Variables
39. f(x,y, z) = xy
+ yz + xz at b. (1,0,0)
.. (I, I, I)
c. (0,0,0)
+ y2 + z2 at b. (0, 1,0)
.. (I, I, I) 41. f(x,y, z) ~ \lx 2
c. (1,0,0)
+ y2 + z2 at b. (I, I, 0)
.. (1,0,0)
e. (1,2,2)
42. f(x,y, z) = (sinxy)/z at
b. (2,0, I)
.. ("./2, I, I)
43. f(x,y, z) = e' + cos (y + z) at b. (0,1',0)
.. (0,0,0)
c.
(O,~,~)
44. f(x,y, z) = tan-1 (xyz) at b. (I, I, 0)
.. (1,0,0)
c. (I, I, I)
In Exercises 45-48, find the linearization L(x, y, z) of 1he function f(x, y, z) at Po. Then f"md an upper bound for the magnitude of the error E in the approximation f(x, y, z) '" L(x,y, z) over the region R.
45. f(x,y, z) = xz - 3yz
R:
Ix - 11:5 0.01,
46. f(x,y, z) = x 2
R: R:
R:
+
at Po(l, 1,2),
Iy - 11:5 0.01, Iy - 11:5 0.01,
2yz - 3xz
Ix - II :5 0.01,
48. f(x,y, z) =
+2
+ xy + yz + (1/4)z2
Ix - 11:5 0.01,
47. f(x,y, z) = xy
il) W(27,2)
iii) W(5, -10) (Explain why this value is much different from 1he value found in 1he table.)
a. W(49, -22)
Po(O,O),
lyl:5 0.2
40. f(x,y, z) ~ x 2
19 13
10 21
Po(2,2),
Find 1he linearizations L(x, y, z) of the funetions in Exercises 39--44 at 1he given points.
T' v°.l6,
T("F)
15
+ 4 at
37. f(x,y) ~ e' cosy at Po(O, 0),
where T is air temperature in OF and v is wind speed in mph. A partial wind chill chart is given.
30 25 20
+ Y + xcosyat
Ixl:5 0.2,
3x - 3y
Iy - 21:5 0.1
(Vselcosyl:5 I andlsinyl:5 I in estimating E.)
+ 0.6215 T - 35.75 v O•16
+ 0.4275
+ xy + (I/4)y2 +
Ix - 21:5 0.1,
35. f(x,y) = I
31. W"md chill factor Wmd chill, a measure of the apparent temperature felt on exposed skin, is a funetion ofair temperature and wind speed. The precise fannula, updated by the National Weather Service in 200 I and based on modern heat transfer theory, a human face model, and skin tissue resistance, is W = W(v,
R:
lyl:5 0.01,
at Po(l, 1,2), Iz - 21 :5 0.08
at Po(l, 1,0),
Iy - II :5 0.01,
v'2 cos x sin (y + z)
Ixl:5 0.01,
Iz - 21 :5 0.02
Izl:5 0.01
at Po(O, 0, "./4),
Iz - "./41 :5 0.01
Estimating Error; Sensitivity to Change 49. Estimatiog muimum error Suppose that T is to be found from 1he fannula T ~ x (e Y + e -Y), where x and y are found to be 2 and In 2 with maximum possible errors of Idx I = 0.1 and It!Y I = 0.02. Estimate the maximum possible error in the computed value of T. 50. Estimatiog volume of a cylinder About how accurately may V = ".,2h be calculated from measurements of, and h that are in error by 1%?
14.6 Tangent Planes and Differentials 51. Consider a closed rectangular box with a square base as shown in the accompanying figure. Ifx is measured with error at most 2% and y is measured with error at most 3%, use a differential to estimate the corresponding percentage error in computing the box's
a. surface area b. volume.
801
55. You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer. 56. a. Around the point (1,0), is f(x,y) = x 2(y + 1) more sensitive to changes in x or to changes in y? Give reasons for your answer. b. What ratio of dx to dy will make df equal zero at (1, O)?
57. Error carryover in coordinate changes
y
y x 4
P(3 ± 0.01,4 ± 0.01)
52. Consider a closed container in the shape of a cylinder of radius 10 cm and height 15 cm with a hemisphere on each end, as shown in the accompanying figure. 15 em
l}lOcm I I I I
I I I
• I I I
The container is coated with a layer of ice 1/2 cm thick. Use a differential to estimate the total volume of ice. (Hint: Assume r is radius with dr = 1/2 and h is height with dh = 0.) 53. Maximum percentage error If r = 5.0 cm and h = 12.0 cm to the nearest millimeter, what should we expect the maximum percentage error in calculating V = trr 2h to be? 54. Variation in electrical resistance The resistance R produced by wiring resistors of R 1 and R 2 ohms in parallel (see accompanying figure) can be calculated from the formula
-+-.-------II...-----....L....------+x
o
3
a. If x = 3 ± 0.01 and y = 4 ± 0.01, as shown here, with approximately what accuracy can you calculate the polar coordinates rand (J of the point P(x, y) from the formulas r 2 = X 2 + Y 2 and (J = tan-1 (y Ix)? Express your estimates as percentage changes of the values that rand (J have at the point (xo, Yo) = (3,4). b. At the point (xo, Yo) = (3, 4), are the values of rand (J more sensitive to changes in x or to changes in y? Give reasons for your answer.
58. Designing a soda can A standard 12-fl-oz can of soda is essentially a cylinder ofradius r = 1 in. and height h = 5 in. a. At these dimensions, how sensitive is the can's volume to a small change in radius versus a small change in height?
a. Show that
b. Could you design a soda can that appears to hold more soda but in fact holds the same 12 fl oz? What might its dimensions be? (There is more than one correct answer.)
59. Value of a 2 X 2 determinant Ifl a Iis much greater than Ib I,Ic I, and Id I, to which of a, b, c, and d is the value ofthe determinant b. You have designed a two-resistor circuit like the one shown to have resistances of R 1 = 100 ohms and R 2 = 400 ohms, but there is always some variation in manufacturing and the resistors received by your firm will probably not have these exact values. Will the value of R be more sensitive to variation in R 1 or to variation in R2? Give reasons for your answer.
c. In another circuit like the one shown you plan to change R 1 from 20 to 20.1 ohms and R 2 from 25 to 24.9 ohms. By about what percentage will this change R?
I(a, b, c, d) =
I:
~I
most sensitive? Give reasons for your answer.
60. Estimating maximum error Suppose that u = xe Y + y sinz and that x, y, and z can be measured with maximum possible errors of ±0.2, ±0.6, and ±7r/180, respectively. Estimate the maximum possible error in calculating u from the measured values x = 2, y = In 3, z = 7r/2. 61. The Wilson lot size formula The Wilson lot size formula in economics says that the most economical quantity Q of goods (radios, shoes, brooms, whatever) for a store to order is given by the formula Q = Y2KMlh, where K is the cost of placing the order, M is the number of items sold per week, and h is the weekly holding cost for each item (cost of space, utilities, security, and so on). To which of the variables K, M, and h is Q most sensitive near the point (Ko, Mo, ho) = (2, 20, 0.05)? Give reasons for your answer.
802
Chapter 14: Partial Derivatives
62. Surveying a triangular field The area of a triangle is (1/2)ab sin C, where a and b are the lengths of two sides of the triangle and C is the measure of the included angle. In surveying a triangular plot, you have measured a, b, and C to be 150 ft, 200 ft, and 60°, respectively. By about how much could your area calculation be in error if your values of a and b are off by half a foot each and your measurement of C is off by 2°? See the accompanying figure. Remember to use radians.
64. Change along the involute of a circle Find the derivative of f(x, y) = X 2 + Y 2 in the direction of the unit tangent vector of the curve
r(t) = (cos t + t sin t)i + (sin t - t cos t)j,
t > O.
65. Change along a helix Find the derivative of f(x, y, z) = 2 X + Y 2 + z 2 in the direction of the unit tangent vector of the helix
r(t) = (cos t)i + (sin t)j + tk a = 150± 1ft 2
at the points where t = -Tr/4, 0, and Tr/4. The function f gives the square of the distance from a point P(x, y, z) on the helix to the origin. The derivatives calculated here give the rates at which the square of the distance is changing with respect to t as P moves through the points where t = -Tr/4, 0, and Tr/4.
Theory and ExampLes 63. The linearization of f(x, y) is a tangent-plane approximation Show that the tangent plane at the point Po(xo, Yo, f(xo, Yo)) on the surface z = f(x, y) defined by a differentiable function f is the plane
66. Normal curves A smooth curve is normal to a surface f(x, y, z) = c at a point of intersection if the curve's velocity vector is a nonzero scalar multiple of Vf at the point. Show that the curve
r(t) =
fx(xo, yo)(x - xo) + fy(xo, yo)(y - Yo) - (z - f(xo, YO)) = 0
Vti + Vtj -
is normal to the surface x 2
or
+ y2
-
±(t + 3)k Z
= 3 when t = 1.
A smooth curve is tangent to the surface at a point of intersection if its velocity vector is orthogonal to Vi there. Show that the curve
67. Tangent curves Thus, the tangent plane at Po is the graph of the linearization of f at Po (see accompanying figure).
r(t) =
Vti + Vtj + (2t -
is tangent to the surface x 2
+ y2
-
Z
l)k
= 1 when t = 1.
z = f(x, y) z
(xo, Yo, f(xo, yo)) I I I I I_I I I I I I I I I I
•I I I I I I I I
..
(xo, Yo)
I
I I I
Z=
L(x, y)
I-I~y
I
I
I I I I I I I
x
14.7
Extreme Values and Saddle Points
HISTORICAL BIOGRAPHY
Simeon-Denis Poisson (1781-1840)
Continuous functions of two variables assume extreme values on closed, bounded domains (see Figures 14.38 and 14.39). We see in this section that we can narrow the search for these extreme values by examining the functions' first partial derivatives. A function of two variables can assume extreme values only at domain boundary points or at interior domain points where both first partial derivatives are zero or where one or both of the first partial derivatives fail to exist. However, the vanishing of derivatives at an interior point (a, b) does not always signal the presence of an extreme value. The surface that is the graph of the function might be shaped like a saddle right above (a, b) and cross its tangent plane there.
14.7
Extreme Values and Saddle Points
803
Derivative Tests for Local Extreme Values To find the local extreme values of a function of a single variable, we look for points where the graph has a horizontal tangent line. At such points, we then look for local maxima, local minima, and points of inflection. For a function f(x, y) oftwo variables, we look for points where the surface z = f(x, y) has a horizontal tangent plane. At such points, we then look for local maxima, local minima, and saddle points. We begin by defining maxima and minima. FIGURE 14.38
z
The function
DEFINmONS
(cosx)(cosy)e-vx2+y2
=
has a maximum value of I and a minimum value of about -0.067 on the square regionlxl ~ 31T/2,IYI ~ 31T/2.
Let f(x, y) be defined on a region R containing the point
(a, b). Then
1. f(a, b) is a local maximum value of f if f(a, b) points (x,y) in an open disk centered at (a, b). 2. f(a, b) is a local minimum value of f if f(a, b) points (x,y) in an open disk centered at (a, b).
~
f(x,y) for all domain
:5
f(x, y) for all domain
z Local maxima correspond to mountain peaks on the surface z = f(x, y) and local minima correspond to valley bottoms (Figure 14.40). At such points the tangent planes, when they exist, are horizontal. Local extrema are also called relative extrema. As with functions of a single variable, the key to identifying the local extrema is a first derivative test.
Local maxima (no greater value offnearby)
!
FIGURE 14.39
z
=
The ''roof surface"
±(llxl-lyll-lxl-lyl)
has a maximum value of 0 and a minimum value of -a on the square region IxI ~ a,
Iyl ~
~
Local minimwn (no smaller value off nearby)
a. FIGURE 14.40 A local maximum occurs at a mountain peak and a local minimum occurs at a valley low point.
z of
I
ox
=
~z =f(x,y)
.---......£
o
(x) =}(x, b) I b
:~
a ..... -- .... _....
-
1//
/ ; ..... X
of oy
THEOREM to-First Derivative Test for Local Extreme Values =
0
If f(x, y) has a local maximum or minimum value at an interior point (a, b) of its domain and if the first partial derivatives exist there, then fxCa, b) = 0 and fy(a, b) = O.
h(y) = f(a, y)
----"y
(a, b, 0)
FIGURE 14.41 Ifa local maximum of f occurs at x = a, y = b, then the fIrst partial derivatives fx(a, b) and fy(a, b) are both zero.
Proof Iff has a local extremum at (a, b), then the function g(x) = f(x, b) has a local extremum at x = a (Figure 14.41). Therefore, g'(a) = 0 (Chapter 4, Theorem 2). Now g'(a) = fx(a, b), so fxCa, b) = O. A similar argument with the function h(y) = f(a,y) shows thatfy(a, b) = O. • Ifwe substitute the values fxCa, b) = 0 and fy{a, b) = 0 into the equation
fxCa, b)(x - a) + fy{a, b)(y - b) - (z - f(a, b» = 0
804
Chapter 14: PartiaL Derivatives
,
fur the tangent plane to the surface z = I(x,y) at (a, b), the equation reduces to
O·(x - a) + O·(y - b) - z + !(a, b)
z
y
~
~
0
!(a, b).
Thus, Theorem 10 says that the surface does indeed have a horizontal tangent plane at a local extremum, provided there is a tangent plane there.
,
DEFlNmON
An interiOl'point ofthe domain ofa function I(x,y) where both
1;< and 1y are zero or where one or both of 1 and Iy do not exist is a critical point J:
off.
Theorem 10 says that the only points where a ftmction I(x,y) can assume extreme values are critical points and boundary points. As with d.ifferentiable ftmctions of a single variable, not every critical point gives rise to a local extremum. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variables might have a saddle point.
FIGURE 14.42 Saddle points at the origin.
,
'-i. x
~--;2-c;,c--~~.Y 2
FIGURE 14.43 The graph ofthe fimction !(x,y) - x 2 + y2 - 4y + 9ma panboloid which has a local minimwn value of 5 at the point (0, 2) (Example 1).
DEFlNmON A differentiable ftmction I(x, y) has a JutDe point at a critical point (a, b) if in every open disk centered at (a, b) there are domain points (x,y) when: !(x,y) > !(a, b) and domain points (x,y) when: !(x,y) < !(a, b). The corresponding point (a, b, I(a, b» on the surface z = I(x,y) is called a saddle point ofthe surface (Figure 14.42).
EXAMPLE 1
Find the local extreme values of I(x,y) = x 2 + y2 - 4y
+ 9.
Solution The domain of 1 is the entire plane (so there are no boundary points) and the partial derivatives 1;< = 2x and Iy = 2y - 4 exist everywhere. Therefore, local extreme values can occur only where
IJ:
=
2x
=
0
and
Iy
=
2y - 4
=
O.
The only poooibility is the point (0, 2), when: the value of! is 5. Smce !(x,y) ~ x 2 + (y - 2)2 + 5 is never less than 5, we see that the critical point (0, 2) gives a local minimum (Figure 14.43). _ EXAMPLE 2
Find the local extreme values (if any) of I(x, y) = y2 - x 2•
Solution The domain of 1 is the entire plane (so there are no boundary points) and the partial derivatives 1;< = -2x and Iy = 2y exist everywhere. Therefore, local extrema can occur only at the origin (0,0) where 1;< = 0 and Iy = O. Along the positive x-axis, however, 1 has the value I(x, 0) = -x2 < 0; along the positive y-axis, 1 has the value I(O,y) = y2 > O. Therefore, every open disk in the xy-plane centered at (0, 0) contains points where the function is positive and points where it is negative. The function has a saddle point at the origin and no local extreme values (Figure 14.44a). FigLD'e 14.44b displays the level curves (they are hyperbolas) of I, and shows the ftmction decreasing and increasing in an alternating fashion among the four groupings ofhyperbolas. _ That 1;< = Iy = 0 at an interior point (a, b) ofR does not guarantee 1 has a local ex1reme value there. If1 and its Imt and second partial derivatives are continuous on R, however, we maybe able to learn more from the following theorem, proved in Section 14.9.
14.7 Extreme Values and Saddle Points
805
z
THEOREM 11-Secnnd Derivative rest for Local Extreme Vallles
Suppose that I(x, y) and its first and second partial derivatives are continuous throughout a disk centered at (a, b) and thatlx(a, b) = /y(a, b) = O. Then
< Oand/",lyy - Ix,? > oat (a, b). > Oand/",lyy - Ix,? > Oat(a,b). has a saddle point at (a, b) if/",lyy - Ix'? < oat (a, b).
i) I has a local maximum at (a, b) if I",
il) I has a local minimum at (a, b) if I", iii) I
iv) the test is inconclusive at (a, b) if 1",lyy - Ix,? = 0 at (a, b). In this case, we must fmd some other way to determine the behavior of I at (a, b).
(a)
The expression 1",lyy - Ix,? is called the discriminant or Hessian of I. It is sometimes easier to remember it in determinant form,
1",lyy -
M = I~: ~:I·
Theorem 11 says that if the discriminant is positive at the point (a, b), then the surface curves the same way in all directions: downward if I", < 0, giving rise to a local maximum, and upward if I", > 0, giving a local minimum. On the other hand, if the discriminant is negative at (a, b), then the surface curves up in some directions and down in others, so we have a saddle point. FIGURE 14.44 (a) The origin is a saddle pointoftllefunctionj(x,y)
= yZ
-
EXAMPLE 3
Find the local extreme values of the function I(x,y) = xy - x 2 - y2 - 2x - 2y + 4.
x'.
There are no local extreme values (Example 2). (b) Level curves for tlle function j 10 Example 2.
Solution The function is defined and differentiable for all x and y and its domain has no boundary points. The function therefore has extreme values only at the points where Ix and I y are simultaneously zero. This leads to
Iy
Ix = y - 2x - 2 = 0,
= x -
2y - 2 = 0,
or
x =y
=
-2.
Therefore, the point ( -2, -2) is the only point where I may take on an extreme value. To see if it does so, we calculate
I",
=
-2,
Iyy = -2,
Ixy = 1.
The discriminant of I at (a, b) = (-2, -2) is
1",lyy - Ixy2 = (-2)(-2) - (1)2 = 4 - I = 3. The combination
I", < tells us that
I
I( -2, -2)
8.
=
EXAMPLE 4 Solution
0
and
has a local maximum at (-2, - 2). The value of
I
at this point is
• Find the local extreme values of I(x,y) = 3y 2 - 2y 3
-
3x 2 + 6xy.
Since I is differentiable everywhere, it can assume extreme values only where
Ix = 6y - 6x = 0
and
I y = 6y - 6y 2
+
6x = O.
806
Chapter 14: Partial Derivatives z
From the first of these equations we find x equation then gives
6x - 6x 2 + 6x
==
==
y, and substitution for y into the second
6x (2 - x)
or
0
==
o.
The two critical points are therefore (0, 0) and (2, 2). To classify the critical points, we calculate the second derivatives: y
Ixx
==
I yy
-6,
==
6 - 12y,
Ixy
==
6.
The discriminant is given by
Ixxlyy - Ixy2
x
FIGURE 14.45 The surface z = 3y 2 - 2y 3 - 3x 2 + 6xy has a saddle point at the origin and a local maximum at the point (2, 2) (Example 4).
==
(-36
+ 72y) -
36
==
72(y - 1).
At the critical point (0, 0) we see that the value of the discriminant is the negative number -72, so the function has a saddle point at the origin. At the critical point (2, 2) we see that the discriminant has the positive value 72. Combining this result with the negative value of the second partial I xx == - 6, Theorem 11 says that the critical point (2, 2) gives a local maximum value of 1(2,2) == 12 - 16 - 12 + 24 == 8. A graph of the surface is shown in Figure 14.45. •
Absolute Maxima and Minima on Closed Bounded Regions We organize the search for the absolute extrema of a continuous function I(x, y) on a closed and bounded region R into three steps.
1.
List the interior points ofR where I may have local maxima and minima and evaluate I at these points. These are the critical points of I.
2.
List the boundary points ofR where I has local maxima and minima and evaluate I at these points. We show how to do this shortly.
3.
Look through the lists for the maximum and minimum values of I. These will be the absolute maximum and minimum values of I on R. Since absolute maxima and minima are also local maxima and minima, the absolute maximum and minimum values of I appear somewhere in the lists made in Steps 1 and 2.
EXAMPLE 5
y
Find the absolute maximum and minimum values of
I(x, y)
B(O, 9)
y=9-x
==
2
+ 2x + 2y - x 2 - y2
on the triangular region in the first quadrant bounded by the lines x
==
0, y
==
0, y
==
9 - x.
SoLution Since I is differentiable, the only places where I can assume these values are points inside the triangle (Figure 14.46) where Ix == I y == 0 and points on the boundary.
x=O (1, 1)
•
-O--.-..........--y-=-O::::-----A...... (9l'---,O-)-~ x
FIGURE 14.46 This triangular region is the domain of the function in Example 5.
(a) Interior points. For these we have
Ix
==
2 - 2x
yielding the single point (x, y)
==
==
0,
h
==
2 - 2y
==
(1, 1). The value of I there is 1(1,1)
==
4.
(b) Boundary points. We take the triangle one side at a time: i) On the segment OA, y
==
O. The function
I(x,y)
==
0,
l(x,O)
==
2
+ 2x - x 2
14.7 Extreme Values and Saddle Points
807
may now be regarded as a function of x dermed on the closed interval 0 :5 x :5 9. Its extreme values (we know from Cbapter 4) may occur at the endpoints
x=O
where
1(0,0) = 2
x=9
where
1(9,0) = 2
+
18 - 81 = -61
and at the interior points where!, (x, 0) = 2 - 2x = O. The only interior point where !,(x,O) = 0 is x = I, where
l(x,O) = 1(1,0) = 3. ti) On the segment OB, x = 0 and
I(x,y) = I(O,y) = 2
+ 2y
- y2.
We know from the symmetry of I in x and y and from the analysis we just carried out that the candidates on this segment are
1(0,0) = 2,
1(0,9) = -61,
1(0,1) = 3.
iii) We have already accounted for the values of I at the endpoints ofAB, so we need only look at the interior points of AB. With Y = 9 - x, we have
I(x,y) = 2
+ 2x + 2(9
- x) - x 2 - (9 -
xf =
-61
+
18x - 2x 2.
Setting !,(x, 9 - x) = 18 - 4x = 0 gives
9
18
x = 4=Z' At this value of x,
9
9
y=9- Z =Z
and
I(x,y)
=
(99) =-2' 41
I Z'Z
Summary We list all the candidates: 4, 2, -61,3, -(41/2). The maxilnum is 4, which I assumes at (I, I). The minimum is-61, which I assumes at (0, 9) and (9, 0). • Solving extreme value problems with algebraic constraints on the variables usually requires the method of Lagrange multipliers introduced in the next section. But sometimes we can solve such problems directly, as in the next example.
EXAMPLE 6 A delivery company accepts only rectangular boxes the sum of whose length and girth (perimeter of a cross-section) does not exceed 108 in. Find the dimensions of an acceptable box of largest volume. Let x, y, and z represent the length, width, and height ofthe rectangular box, respectively. Then the girth is 2y + 2z. We want to maximize the volume V = xyz ofthe box (Figure 14.47) satisfying x + 2y + 2z = 108 (the largest box accepted by the delivery company). Thus, we can write the volume of the box as a function of two variables:
Solution Girth = distance
around here
V(y,z) = (108 - 2y - 2z)yz z
Y=~and
x
~
108 - 2y - 2z
= 108yz - 2y 2z - 2yz 2.
Setting the first partial derivatives equal to zero, x
FIGURE 14.47
y
The box in Example 6.
V,(y,z) = IOSz - 4yz - 2z2 = (108 - 4y - 2z)z = 0 Vz(y,z) = 108y - 2y 2 - 4yz = (108 - 2y - 4z)y = 0,
808
Chapter 14: Partial Derivatives gives the critical points (0, 0), (0, 54), (54, 0), and (18, 18). The volume is zero at (0, 0), (0,54), (54, 0), which are not maximum values. At the point (18, 18), we apply the Second Derivative Test (Theorem II):
Vyy
=
v,. =
-4z,
108 - 4y - 4z.
Then
Vyy V,. - V,.2 = 16yz - 16(27 - Y - z)'. Thus,
Vyy(l8, 18)
= -4(18)
<
0
and
[Vyy V,. -
V,.2](IS,IS) = 16(18)(18) - 16(-9)2
>
0
imply that (18,18) gives a maximum volume. The dimensions of the package are = 108 - 2(18) - 2(18) = 36 in.,y = 18 in., andz = 18 in. The maximum volume is V = (36)(18)(18) = 11,664 in', or 6.75 ft'. •
x
Despite the power ofTheorem II, we urge you to remember its limitations. It does not apply to boundary points ofa function's domain, where it is possible for a function to have extreme values along with nonzero derivatives. Also, it does not apply to points where either I x or I y fails to exist.
Summary of Max-Min Tests The extreme values of I(x, y) can occur only at i)
boundary points ofthe domain of I
il) critical points (interior points where Ix = I y = 0 or points where Ix or I y fails to exist). If the first- and second-order partial derivatives of I are continuous throughout a disk centered at a point (a, b) and fx(a, b) = Iy(a, b) = 0, the natore of I(a, b) can be tested with the Second Derivative Test:
I., < 0 and/.,lyy - Ix,' > 0 at (a, b) => local maximum il) I., > 0 and/.,lyy - Ix,' > 0 at (a, b) => local minimum iii) 1.,lyy - Ix,' < 0 at (a, b) => saddle point iv) 1.,1yy - Ix,' = 0 at (a, b) => test is inconclusive i)
Exercises 14.7 finding Local Extrema
Find all the local maxima, local minima, and saddle points of the functions in Exercises 1-30. 1. I(x,y) ~ x 2 + xy + y2 + 3x - 3y + 4 2. I(x,y) = 2xy - 5x2 - 2y' + 4x + 4y - 4 3. I(x,y) ~ x' + xy + 3x + 2y + 5 4. I(x,y) ~ 5xy - 7x 2 + 3x - 6y + 2
s.
+ 3x + 4 + y' + 6y + 2 ~ 2x' + 3xy + 4y' - 5x + 2y = x' - 2xy + 2y' - 2x + 2y + ~ x' - y' - 2x + 4y + 6 = x' + 2xy
I(x,y) ~ 2xy - x' - 2y'
6. I(x,y) = x' - 4xy 7. I(x,y) 8. I(x,y) 9. I(x,y) 10. I(x,y)
I
14.7
11. I(x,y) = "';56.1: 2
By'-
11. I(x,y) _ 1 - "'¥x 2
+ y'-
13. l(x,y) - ;t3 14. f(x,y) _ x 3 15. j(x,y)
16l:
y3 - 2xy + 3xy + y3 -
31
+1-
[b(6
+6
20. f(x,y) 21. l(x,y) =
has its largest value.
x" - y"
2.
1
+ y 2.
1
22. I(x,y) = :i
- 1
23. l(x,y) = ysinx 25. l(x,y) - e..'+T-.v 27. f(x,y) - e"7(x 2 + y2)
41. Te.mperatorel A flat circular plate has the shape of the region ;t2 + y'l :!"; I. The plate, including the boundaIy where ;t2 + y'l _ l,isheatedsothatthetempemtureatthepoin1:(x,y)is
+ xy + y-1
7l:;t,y) =
Z4. I(x,y) = e2>:cosy 26. f(x,y) - eY - ~ 28. I(x,y) _ r(x2 _ y2)
+ lny - 4x In (x + y) + xl - Y
29. j(x,y) = 2mx 30. J(x,y) =
[b (24 _ 2r _ X2)1/3 dr
x" + y" + 4zy x
f(x,y) ~ '" in the open f1ISt quadnurt (;t on a minimum 1:lu:I'c.
31. I(x,y) = 21: 2 - 4x + y2 - 4y + I on the closed triangular plate bounded by the lines x = 0, y = 2, Y = 2x in the flJ'St quodnmt
31. D(x,y) = x 2 - xy + y2 + Ion the closed triangularplatein the frrst quadnntbounded by the lines x = O,y = 4,1 = x
33. l(x,y) - ;t2 + y2 on the closed triangular plate bounded by the Iin.es x = 0, y = 0, y + 2t = 2 in the first quadIant 34. 7l:;t. y) - x 2 + xy + y2 - 6x on the rectangular plate o :!;'; x :!"; 5, -3 :!;'; Y :!;'; 3
+ .xy + y'l
- 6x"
36. f(;t,y) - 48xy - 32r 3 O:!;';;t:!"; 1,0:!;';y:!;'; I
-
+ 2 on
24y 'l
on
the rectangular plate the
rectangular
x.
-
42. Find the critical point of
y
finding Absolute ExtNma
o :!;'; ;t :!;'; 5, -3 :!;'; Y :!"; 0
x2 + 2 y 2
Find the temperatures at the hottest and coldest points on the plate.
In Exercises 31-38, fmd the absolute maxima and minima of the :functions on the given doma.ins.
35. 7l:;t, y) - ;t2
-;t - ;t'l) dr
has its largest value. 40. Find two munbers a and b with a :!;'; b such that
Lx 3
16. I(x,y) = x 3
17. l(x,y) = 18. l(x,y) 19. f(x,y) _ 4xy -
809
39. Find two munbers a and b with a s b such that
8x
+ 3y 2 + 6xy + y3 + 3.1:2 - 3y 2 - 8 x 3 + 3xy2 - 15x + y3 - 15y 2x 3 + 2y 3 - 9x 2 + 3y 2 - 12y
= 6.:1: 2 -
Exbeme Values and Saddle Points
plate
37. f(;t,y) - (4;t - ;t2) cosy on the rectangular plate I S;t :!;'; 3, -,"/4 Sy:!;'; ,"/4 (see accompanying figure).
In.,,,
+ "" > O,y >
0) and shaw that f takes
Theory and Examples 43. Find the muima, minima, and saddle poin1:s of f(;t, y), if any,
given'" f" = 2r - 4y
L
b.
I" = I" -
2t - 2
and fy = 2y - 4x and
Iy =
2y - 4
!h"2 - 9 and fy - 2y + 4 Describe your reasoning in each case.
c.
44. Thediscriminantlaf», - A/iszero at the origin for each ofthe following functions, so the Second Derivative Test fails there. Determine whether the function has a maximum, a minimlDn, or neither at the origin by imagining what the SUIface z - f(x, y) looks like. Describe your reasoning in each case. L f(;t,y) - ;t1y2 b. f(;t,y) - 1 - x1y'l c. f(;t,y) - xy2 II. f(;t,y) _ ;ty2
e. f(;t,y) - ;ty3 f. f(;t,y) - ;t4y4 45. Shaw that (0, 0) is a critical point of I(x, y) = x 2 + kty + y'l no matter what va1ue the constant k has. (Hint: Consider two cases: k - Oandk.;. 0.) 46. For what values ofthe constant k docs the Second Derivative Test guanurtee that f(x,y) = x 2 + tty + y'l will have a saddle poin1: at (0, O)? A local minjmlDn at (0, O)? For what values of k is the Second Derivative Test inconclusive? Give :reasons for your
""""".
47. Iff,,(a,b) - fy(a,b) - 0. must f have a localmaxinrum or minimmn value at (a, b)? Give reasons for your answer.
,
,
38. I(;t,y) = 4x - 8xy + 2y + lonthetriangularplatebmmdedby the linesx = 0, y = O,;t + Y = 1 in the fllllt quadnurt
4&. Can you. conclude anything about f(a, b) iff and its fmrt and second partial derivatives are continuous throughout a disk centered at the critical point (a, b) and fnl..a, b) andfyy(a, b) differ in sign? Give reasons for your answer.
49. Among all the points on the graph of z - 10 - x 2 - y'l that lie above the plane ;t + 2y + 3z - 0, find the point farthest from the plane.
810
Chapter 14: Partial Derivatives
50. Find the point on the graph of z ~ x 2 + y2 plane x + 2y -z= O. 51. Find the point on the plane 3x origin.
+ 2y + z
~
+
10 nearest the
Curves:
6 that is nearest the
x+y-z~2.
53. Find three numbers whose sum is 9 and whose sum ofsquares is a minimum. 54. Find three positive numbers whose sum is 3 and whose product is a maximum. the
maximum
of s
value
~
xy
+ yz + xz
where
x+y+z=6.
iii) The line segment x = 2t,
57. Find the dimensions of the rectangular box of maximum volume that can be inscribed inside the sphere x 2 + y2 + z2 ~ 4. 58. Among all closed rectangolar boxes of volume 27 em3, what is the smallest surface area? 59. You are to construct an open rectangular box from 12 It' of material. What dimensions will result in a box ofmaximum volume? 60. Consider the function I(x, y) ~ x 2 + y2 over the square 0 s x ::s 1 and 0 s y ::s 1.
I,
+ 2xy
- x - y
+
y = t + I,
0 '" t '" I
a. I(x,y) ~ x 2 + y2
= 1/(x 2 + y2)
b. g(x,y)
Curves: i)Thelinex=t,
y=2-2t 0 '" t '" I
65. Least squares and regression lines When we try to fit a line y = mx + b to a set of numerical data points (XhY'), (X2,Y,), ... , (X.,y.) (Figure 14.48), we usually choose the line
that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means fmding the values of m and b that minimize the value of the function
w = (mx, + b - y,? + ... + (mx. + b - y.)2.
(1)
Show that the values of m and b that do this are
I
(};x.)
m=
I in this square. What is the absolute minimum
(};Y.) -
n};xkY. (2)
(};x.y -n};xl
b. Find the absolute maximum value of I over the square. Ertreme Values on Parametrized Curves To fmd the extreme values of a function I(x, y) on a curve x ~ x(t), y ~ y(t), we treat I as a function of the single variable t and use the Chain Rule to fmd where dl/dt is zero. As in any other single-variable case, the extreme values of I are then found among the values at the
-I '" t '" 0
64. Functions:
a. Show that I has an absolute minimum along the line segment ~
+
il) The line segment x = t, Y = 2 - 2t,
56. Find the minimum distance from the cone z ~ \lx 2 + y2 to the point (-6,4,0).
2x + 2y value?
y=t+1
i)Thelinex=2t,
il) The line segment x = 2t, y = t
52. Find the minimum distance from the point (2, -I, I) to the plane
55. Find
63. Function: I(x, y) ~ xy
b~
*
(};Yk -
m};x.),
(3)
with all sums running from k = I to k = n. Many scientific calculators have these formulas built in, ensbling you to fmd m and b with only a few keystrokes after you have entered the data. The line y ~ mx + b derermined by these values of m and b
a. critical points (points where dI / dt is zero or fails to exist), and
is called the least squares line, regression line, or trend line for
b. endpoints of the parameter domain.
the data under study. Finding a least squares line lets you
1. summarize data with a simple expression, Find the absolute maximum and minimorn values of the following functions on the given curves.
2. predict values ofy for other, experimental1y untried values of x, 3. handle data analytically.
61. Functions:
a. I(x,y) = x + Y b. g(x,y) = xy 2 c. h(x,y) = 2x + y2 Curves: i) The semicircle x 2
+ y2
= 4,
ii) The quarter circle x 2 + y2
y '" 0 4, x ~ 0, y
=
Use the parametric equations x
y
=
2 cos t,Y
=
O?::
0
2 sin t.
62. Functions:
a. I(x,y) = 2x + 3y c. h(x,y) = x 2 + 3y 2
b. g(x,y) = xy
Curves: i) The semiellipse (x 2/9)
+
(y2/4) ~ I, y '" 0
il) The quarter ellipse (x 2/9) + (y2/4) ~ I, x '" 0, y '" 0 Use the parametric equations x = 3 cos t,Y = 2 sin t.
-1-------------> x
o
FIGURE 14.48 To fit a line to noncollinear points, we choose the line that minimizes the sum of the squares of the deviations.
14.8 Lagrange MultiplieT5
In
Exercises 6tH;8, use Equations (2) and (3) 10 rmd the least squares line for each set of data points. Then use the linear equation you obtain to predict the value ofy that would correspond lox = 4. 66. (-2,0), (0,2), (2,3) 67. (-1,2), (0, I), (3, -4)
68. (0, 0), (1,2), (2, 3)
COMPUTER EXPLORATIONS Exercises 6~74, you will explore functions to identifY their local extrema. Use a CAS to perform the followiog steps: a. Plot the function over the given rectangle.
In
b. Plot some level curves in the rectangle.
c. Calculate the function's Hrst partial derivatives and use the CAS equation solver 10 rmd the critical points. How do the critical points relate 10 the level corves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for
d. Calculate the function's second partial derivatives and rmd the discriminant f~f", - f,}. e. Using the max-min tests, classifY the critical points found in part (c). Are your rmdings consistent with your discussion in part (c)? 69. f(x,y) = x 2 + y3 - 3:ry, -5';; x,;; 5, -5';; y';; 5 70. f(x,y) ~ x 3 - 3:ry2 + y2, -2';; x ,;; 2, -2';; Y ,;; 2
71. f(x,y) ~ x 4 + y2 -6 ,;; Y ,;; 6
ax2 - 6y + 16,
-3';; x,;; 3,
72. f(x,y) = 2>;4 + y4 - 2>;2 - 2y 2 + 3, -3/2,;; x,;; 3/2, - 3/2 ,;; Y ,;; 3/2 73. f(x,y) ~ 5x' + lax' - 3Ox 4 + 3Oxy2 - 12Ox 3, -4 ,;; x ,;; 3, -2';; Y ,;; 2 74. f(x,y)
= {~:In(x2 + y2),
-2 ::;; x
your answer.
14.8
:5
2,
-2:::S; y
(x,y) ~ (0,0) (x,y) = (0,0)' 2
:5
I_La-=Q_ra---'nQ=--e_M_U_lti--=."p_li_ers
HIsTORICAL BIOGRAPHY
Joseph Louis Lagrange (1736-1813)
811
_
Sometimes we need to find the extreme values of a function whose domain is constrained to lie within some particular subset of the plane-----
Constrained Maxima and Minima We first consider a problem where a constrained minimum can be found by eliminating a variable.
EXAMPLE 1
FindthepointP(x,y,z)ontheplane2x
+ y -z -
5 =
othat is closest
to the origin. Solution
The problem asks us to rmd the minimum value of the function
IoFl = Y(x - 0)2 + (y = Yx2 + y2 + z2
- 0)2
+
(z - 0)2
subject to the constraint that
2x+y-z-5=0.
oF I has a minimum value wherever the function
Since I
f(x,y, z) = x 2
+ y2 + z2
has a minimum value, we may solve the problem by finding the minimum value of f(x, y, z) subject to the constraint 2x + Y - z - 5 = 0 (thus avoiding square roots). Ifwe regard x and y as the independent variables in this equation and write z as z=2x+y-5, our problem reduces to one of finding the points (x,y) at which the function
h(x, y) = f(x, y, 2x
+Y
- 5) = x 2
+ y2 +
(2x
+Y
- 5)'
812
Chapter 14: Partial Derivatives has its minimum value or values. Since the domain of h is the entire xy-plane, the First Derivative Test of Section 14.7 tens us that any minima that h might have must occur at points where
hx = 2x
+ 2(2x + Y
- 5)(2) = 0,
hy = 2y
+ 2(2x + Y
- 5) = O.
This leads to
lOx + 4y = 20,
4x
+ 4y
=
10,
and the solution
x
5
=
3'
We may apply a geometric argument together with the Second Derivative Thst to show that these values minimize h. The z-coordinate of the corresponding point on the plane z=2x+y-5is
5 6' Therefore, the point we seek is
p(t,i,-i).
Closest point:
•
The distance fromPto the origin is 5jV6 '" 2.04.
Attempts to solve a constrained maximum or minimum problem by substitotion, as we might call the method of Example I, do not always go smoothly. This is one of the reasons for learning the new method of this section.
z
EXAMPLE 2
Find the points on the hyperbolic cylinder x 2 - z2 - I = 0 that are clos-
est to the origin. Solution 1 The cylinder is shown in Figure 14.49. We seek the points on the cylinder closest to the origin. These are the points whose coordinates minimize the value of the function
!(x,y,z) = x 2 + y2
+ z2
Square ofthe distance
subject to the constraint that x 2 - z2 - I = O. Ifwe regard x andy as independent variables in the constraint equation, then
and the values of !(x,y, z) = x 2 + y2 + z2 on the cylinder are given by the function
x
h(x, y) = x 2 + y2 + (x' - I) = 2x 2 + y2 - 1. To f'md the points on the cylinder whose coordinates minimize !, we look for the points in the :ry-plane whose coordinates minimize h. The only extreme value of h occurs where The hyperbolic cylinder I = 0 in Example 2.
FIGURE 14.49
x' - z' -
hx =4x=0
and
hy =2y=0,
that is, at the point (0, 0). But there are no points on the cylinder where both x and y are zero. What went wrong? What happened was that the First Derivative Test found (as it shonld have) the point in the domain ofh where h has a minimum value. We, on the other hand, want the points on the cylinder where h has a minimum value. Although the domain of h is the entire
14.8 The hyperbolic cylinder x 2 On this part,
x=W+1
- Z2 =
1
On this part,
z x=-W+1
813
xy-plane, the domain from which we can select the first two coordinates of the points (x, y, z) on the cylinder is restricted to the "shadow" of the cylinder on the xy-plane; it does not include the band between the lines x == -1 and x == 1 (Figure 14.50). We can avoid this problem if we treat y and z as independent variables (instead of x and y) and express x in terms of y and z as x2
With this substitution, I(x, y, z) k(y, z)
FIGURE 14.50 The region in the xy-plane from which the first two coordinates of the points (x, y, z) on the hyperbolic cylinder 2 x - z2 = 1 are selected excludes the band - 1 < x < 1 in the xy-plane (Example 2).
Lagrange Multipliers
+
(z2
==
x2
==
==
+
z2
1.
+ y2 + z2 becomes 1)
+ y2 + z2
1
==
+ y2 + 2z 2
and we look for the points where k takes on its smallest value. The domain of k in the yzplane now matches the domain from which we select the y- and z-coordinates of the points (x, y, z) on the cylinder. Hence, the points that minimize k in the plane will have corresponding points on the cylinder. The smallest values of k occur where ky == 2y == 0
or where y
==
z
==
and
kz == 4z == 0,
O. This leads to x2
==
z2
+
1
1,
==
x
==
±1.
The corresponding points on the cylinder are (± 1, 0, 0). We can see from the inequality k(y, z)
==
1
+ y2 +
2z 2 ~ 1
that the points (± 1, 0, 0) give a minimum value for k. We can also see that the minimum distance from the origin to a point on the cylinder is 1 unit. SoLution 2 Another way to find the points on the cylinder closest to the origin is to imagine a small sphere centered at the origin expanding like a soap bubble until it just touches the cylinder (Figure 14.51). At each point of contact, the cylinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and cylinder are represented as the level surfaces obtained by setting
/
and
x
/
y
FIGURE 14.51 A sphere expanding like a soap bubble centered at the origin until it just touches the hyperbolic cylinder 2 x - z2 - 1 = 0 (Example 2).
g(x,y,z)
==
x 2 - z2 - 1
equal to 0, then the gradients VI and Vg will be parallel where the surfaces touch. At any point of contact, we should therefore be able to find a scalar A ("lambda") such that
VI
==
AVg,
or 2xi
+
2yj
+
2zk
==
A(2xi - 2zk).
Thus, the coordinates x, y, and z of any point of tangency will have to satisfy the three scalar equations 2x
==
2y
2Ax,
==
0,
2z
==
-2Az.
For what values of A will a point (x, y, z) whose coordinates satisfy these scalar equations also lie on the surface x 2 - z2 - 1 == O? To answer this question, we use our knowledge that no point on the surface has a zero x-coordinate to conclude that x i=- O. Hence, 2x == 2Ax only if 2
==
2A,
or
A==1.
For A == 1, the equation 2z == -2Az becomes 2z == -2z. If this equation is to be satisfied as well, z must be zero. Since y == 0 also (from the equation 2y == 0), we conclude that the points we seek all have coordinates of the form
(x, 0, 0).
814
Chapter 14: Partial Derivatives What points on the surface x 2 points (x, 0, 0) for which
x2
-
(0)'
z2 =
-
=
I have coordinates of this fonn? The answer is the
x2
I,
=
I,
or
x
=
± 1.
The points on the cylinder closest to the origin are the points (±1,0,0).
•
The Method of Lagrange Multipliers In Solution 2 of Example 2, we used the method of Lagrange multipliers. The method says that the extreme values of a function I(x, y, z) whose variables are subject to a constraint g(x,y,z) = 0 are to be found on the surface g = 0 among the points where VI = AVg for some scalar A (called a Lagrange multiplier). To explore the method further and see why it works, we first make the following observation, which we state as a theorem.
THEOREM 12-The Orthogonal Gradient Theorem
Suppose that I(x, y, z) is differentiable in a region whose interior contains a smooth curve C:
r(t)
=
g(t)i
+ h(t)i + k(t)k
If Po is a point on C where I has a local maximum or minimum relative to its values on C, then VI is orthogonal to C at Po.
Proof We show that VI is orthogonal to the curve's velocity vector at Po. The values of I on C are given by the composite I(g(t), h(t), k(t», whose derivative with respect to tis
dl dt At any point Po where curve, dl/dt = 0, so
I
=
al dg ax dt
al dh
al dk
+ ay dt + az dt
= VI' v.
has a local maximum or minimum relative to its values on the
V/'Y
=
o.
•
By dropping the z-terms in Theorem 12, we obtain a similar result for functions of two variables.
COROLLARY OF THEOREM 12
At the points on a smooth curve where a differentiable function I(x, y) takes on its local maxima and minima relative to its values on the curve, VI' Y = 0, where
r(t)
=
g(t)i
Y =
dr/dt.
+ h(t)i
Theorem 12 is the key to the method ofLagrange multipliers. Suppose that I(x, y, z) and = 0 where I has a local maximum or minimum value relative to its other values on the sorCace. We assume also that Vg oF 0 at points on the surface g(x,y,z) = o. Then I takes on a local maximum or minimum at Po relative to its values on every differentiable curve through Po on the surface g(x,y, z) = o. Therefore, Viis orthogonal to the velocity vector of every soch differentiable curve through Po. So is Vg, moreover (because Vg is orthogonal to the level surface g = 0, as we saw in Section 14.5). Therefure, at Po, VI is some scalar multiple A ofVg.
g(x, y, z) are differentiable and that Po is a point on the surface g(x, y, z)
815
14.8 Lagrange MultiplieT5
The Method of Lagrange Multlpllen
Suppose that f(x, y, z) and g(x, y, z) are differentiable and Vg '" 0 when g(x,y,z) = O. To find the local maximum and minimum values of f subject to the constraint g(x, y, z) = 0 (if these exist), f"md the values of x, y, z, and A that simultaneously satisfY the equations
Vf
= AVg
g(x,y, z) =
and
o.
(1)
For functions of two independent variables, the condition is similar, but without the variable z.
Some care must be used in applying this method. An extreme value may not actually exist (Exercise 41). y
EXAMPLE 3 X2
V2
8+
Find the greatest and smallest values that the function
y2
f(x,y) = xy
2~1
takes on the ellipse (Figure 14.52) x
0
2V2 We want to find the extreme values of f(x, y)
Solution
Example 3 shows how to f"md the largest and smallest values ofthe product"" on this ellipse. FIGURE 14.52
g(x,y) =
x2
y2
8 +2 - I
=
=
xy subject to the constraint
O.
To do so, we fiT5t find the values ofx,y, and Afor which
Vf
= AVg
and
g(x,y) = O.
The gradient equation in Equations (I) gives yi
+ xj
=
Ay,
=
~ xi + Ayj,
from which we find y
xy=-2
xy=2
x
so that y
-~~i~;~~~~~~
x
=
0 or A = ±2. We now consider these two cases.
Case 1: If y = 0, then x = y = O. But (0, 0) is not on the ellipse. Hence, y '" O. Casel: If y '" 0, then A = ±2 and x = ±2y. Substituting this in the equation g(x, y) = 0 gives (±2y)' -8-
xy=2/
xy=-2
FIGURE 14.53 When subjected to the constraint g(x, y) = x'/8 + y'/2 - I = 0, the function f(x, y) ~ "" takes on extreme values at the four points (±2, ± I ~ These are the points on the ellipse when Vf (red) is a scalar multiple ofVg (blue) (Example 3).
and
y2
+2
=
1,
4y 2 + 4y 2 = 8
and
y = ±l.
The function f(x,y) = xy therefore takes on its extreme values on the ellipse at the four points (±2, I), (±2, -I). The extreme values are xy = 2 andxy = -2. The Geometry of the Solution The level curves ofthe function f(x, y) = xy are the hyperbolas xy = c (Figore 14.53). The farther the hyperbolas lie from the origin, the larger the absolute value of f. We want to f"md the extreme values of f(x, y), given that the point (x, y) also lies on the ellipse x 2 + 4y 2 = 8. Which hyperbolas intersecting the ellipse lie farthest from the origin? The hyperlJolas that just graze the ellipse, the ones that are tangent to it, are
816
Chapter 14: Partial Derivatives farthest. At these points, any vector nonnai to the hyperbola is nonnai to the ellipse, so VI = yi + xj is a multiple (A = ±2) ofVg = (x/4); + yj. At the point (2, I), for example, VI = ;
+
Vg
2j,
I. + J, . 2'
=
and
VI = 2Vg.
and
VI = -2Vg.
At the point (- 2, I), VI = ; - 2j,
EXAMPLE 4 3x
Vg =
-t; +
j,
•
Find the maximum and minimum values of the function I(x,y) = = I.
+ 4y on the circle x 2 + y2
Solution
We model this as a Lagrange multiplier pmblem with
I(x,y) = 3x
+ 4y,
g(x,y) = x 2 + y2 - I
and look for the values ofx,y, and Athat satisfy the equations VI = AVg:
g(x,y) =
+ 4j = 2xAi + 2yAj 0: x 2 + y2 - 1 = O. 3;
The gradient equation in Equations (I) implies that A
x
3 = 2A'
Y
'¢'
0 and gives
2
=
X'
These equations tell us, among other things, that x and y have the same sign. With these values for x and y, the equation g(x, y) = 0 gives
(iAY + (iY -
I
=
0,
so 9
+
16 = 4A2,
4A2 = 25,
5
and
A = ±2'
Thus,
Y
2
=
4
X = ±5'
and/(x,y) = 3x + 4yhas extreme values at (x,y) = ±(3/5,4/5). By calculating the value of 3x + 4y at the points ±(3/5, 4/5), we see that its maximum and minimum values on the circle x 2 + y2 = 1 are 3x+4y=5
and 3x
+ 4y =-5
FIGURE 14.54 The functionf(x,y) ~ 3x + 4y takes on its largest value on the unitcircleg(x,y) = x 2 + y2 - I = Oat the point (3/5, 4/5) and its smallest value at the point (-3/5, -4/5) (Example 4). At each of these points, Vf is a scalar multiple ofVg. The figure shows the gradients at the Il!St point but not the second.
The Geometry of the Solution The level curves of I(x, y) = 3x + 4y are the lines 3x + 4y = c (Figure 14.54). The farther the lines lie from the origin, the larger the absolute value of f. We want to find the extreme values of I(x,y) given that the point (x,y) also lies on the circle x 2 + y2 = 1. Which lines intersecting the circle lie farthest from the origin? The lines tangent to the circle are farthest. At the points of tangency, any vector nonnai to the line is normal to the circle, so the gradient VI = 3; + 4j is a multiple (A = ±5/2) of the gradient Vg = 2xi + 2yj. At the point (3/5, 4/5), for example, VI = 3;
+
4j,
6. Vg = 5'
+ 8.
5 J,
and
5
VI = 2Vg.
•
14.8
Lagrange Multipliers
817
Lagrange Multipliers with Two Constraints Many problems require us to find the extreme values of a differentiable function I(x, y, z) whose variables are subject to two constraints. If the constraints are and and gl and g2 are differentiable, with Vg 1 not parallel to Vg2, we find the constrained local maxima and minima of I by introducing two Lagrange multipliers A and J.L (mu, pronounced "mew"). That is, we locate the points P(x, y, z) where I takes on its constrained extreme values by finding the values of x, y, z, A, and J.L that simultaneously satisfy the equations
c (2) FIGURE 14.55 The vectors Vg 1 and Vg2 lie in a plane perpendicular to the curve C because Vg 1 is normal to the surface gl = 0 and Vg2 is normal to the surface g2 = o.
z ~_r-_C_ylinder x 2 + y2 = 1
Equations (2) have a nice geometric interpretation. The surfaces gl == 0 and g2 == 0 (usually) intersect in a smooth curve, say C (Figure 14.55). Along this curve we seek the points where I has local maximum and minimum values relative to its other values on the curve. These are the points where VI is normal to C, as we saw in Theorem 12. But Vg 1 and Vg2 are also normal to C at these points because C lies in the surfaces gl == 0 and g2 == o. Therefore, VI lies in the plane determined by Vg 1 and Vg2, which means that VI == AVgl + J.L Vg2 for some A and J.L . Since the points we seek also lie in both surfaces, their coordinates must satisfy the equations gl (x, y, z) == 0 and g2(X, y, z) == 0, which are the remaining requirements in Equations (2). The plane x + y + z == 1 cuts the cylinder x 2 + y2 == 1 in an ellipse (Figure 14.56). Find the points on the ellipse that lie closest to and farthest from the origin.
EXAMPLE 5
SoLution
We find the extreme values of I(x,y,z) == x 2
+ y2 + z2
(the square of the distance from (x, y, z) to the origin) subject to the constraints gl(X,y, z) == x 2 + y2 - 1 == 0 g2(X,y,Z) == x x
+y +z
-
1
==
(3)
O.
(4)
The gradient equation in Equations (2) then gives
2xi 2xi
+ 2yj + + 2yj +
VI
==
2zk
==
2zk
==
+ J.L Vg2 A(2xi + 2yj) + J.L(i + j + k) (2Ax + J.L)i + (2Ay + J.L)j + J.Lk
AVg 1
or 2x == 2Ax FIGURE 14.56 On the ellipse where the plane and cylinder meet, we find the points closest to and farthest from the origin. (Example 5).
+ J.L,
2y == 2Ay
+ J.L,
2z
==
J.L.
(5)
The scalar equations in Equations (5) yield
+ 2z~(1 2Ay + 2z ~ (1
2x == 2Ax
- A)X == z,
2y ==
- A)Y
==
(6)
z.
Equations (6) are satisfied simultaneously if either A == 1 and z == 0 or A i:- 1 and == y == z/(l - A). If z == 0, then solving Equations (3) and (4) simultaneously to find the corresponding points on the ellipse gives the two points (1, 0, 0) and (0, 1, 0). This makes sense when you look at Figure 14.56.
x
818
Chapter 14: Partial Derivatives
If x = y, then Equations (3) and (4) give
+x 2 -I=0
X2
x+x+z- 1 = 0
2x 2 = 1
z=I-2x
x= ± V2
z= 1 'f
2
V2.
The corresponding points on the ellipse are
PI =
('?, '?,
1 -
V2)
and
Here we need to be careful, however. Although PI and P2 both give local maxima of f on the ellipse, P2 is farther from the origin than P,. The points on the ellipse closest to the origin are (1, 0, 0) and (0, 1,0). The point on • the ellipse farthest from the origin is P2 •
Exercises 14.8 Two Independent Varlables with One ConstraInt 1. Extrema on an ellipse Find the points 00 the ellipse x 2 + 2y 2 ~ 1 wbere f(x, y) ~ xy has its extreme values. 2. Extrema on a circle Find the extreme values of f(x, y) subject to the constraint g(x, y) = x 2 + y2 - 10 = 0.
~
xy
3. Maximum on a line Find the maximum value of f(x, y) = 49 - x 2 - y2 00 the line x + 3y ~ 10.
= x7
4. Extrema on a line Find the local extreme values of f(x, y) 00 the
line x
+y
= 3.
S. Constrained minimum nearest the origin.
Find the points on the curve xy2
6. Constrained minimum nearest the origin.
Find the points
00
the curve
=
54
x7 =
2
7. Use the method of Lagrange multipliers to find
a. Minimum on a hyperbola The minimum value of x subject to the coostraints xy = 16, x > 0, y > 0
+ y,
b. Muimum on a line The maximum value ofxy, subject to lbe constraint x + y ~ 16.
Comment 00 lbe geometry of eacb solutioo. 8. EItrema on a curve Find the points on the curve x + xy + y2 = 1 in lbe xy-plane lbat are nearest to ood fartbest from lbe OOgio. 2
9. Minimum surface area with ("lXed volume Find lbe dimensioos of the closed rigbt circular cylindrical cao of smallest surface area whose volume is 16'7T cm.3• 10. Cylinder in a sphere Find lbe radius ood heigbt of the open rigbt circular cylinder oflargest surface area that cao be inscrihed in a sphere ofradius a. What is !be largest surface area? 11. Rectangle of greatest area in an ellipse Use the method of Lagrange multipliers to f"md the dimensions of the rectsngIe of greatest area that coo he inscribed in lbe ellipse x 2/16 + y2/9 = I with sides parallel to the coordioate ases. 12. Rectangle of longest perimeter in an ellipse Find the dimensioos of lbe rectaogle oflargest perimeter that coo be inscrihed in
the ellipse x 2/a 2 + y2/b 2 = I with sides parallel to the coordinate ases. What is lbe largest perimeter? 13. Extrema on a circle Find the maximum ood minimum values of x 2 + y2 subject to the coostraint x 2 - 2x + y2 - 4y = O.
14. Extrema on a circle Find the maximum and minimum values of 3x - y + 6 sul!iect to lbe coostraint x 2 + y2 ~ 4. 15. Ant on a metal plate The temperatore at a point (x, y) on a metal plate is T(x, y) = 4x 2 - 4xy + y2. Ao oot 00 the plate walks around lbe circle of radius 5 centered at lbe OOgio. What are the higbest ood lowest temperatores encountered by the oot? 16. Cheapest storage tank Yoor f"mn has been asked to design a storage tank for liquid petroleum gas. The customer's specificatioos call for a cylindrical tank with hemispherical ends, ood the tank is to hold 8000 m' ofgas. The customer also waats to use the smallest amount of material possible in buildiog the tank. What radius ood heigbt do you recommend for the cylindrical portion of the tank?
Th.... e Independent Variables with One Cons!ra;nt 17. Minimum distance to a point Find !be point x + 2y + 3z = 13 closest to the point (1,1, I).
00
lbe plane
18. Maximum distance to a point Find the point 00 the sphere x 2 + y2 + z2 ~ 4 fartbest from lbe point (I, -1, 1). 19. Minimum distance to the origin Find lbe minimum distaoce from the surface x 2 - y2 - z2 = 1 to the origin. 20. Minimum distance to the origin z = xy + 1 nearest the origin.
Find the point 00 the surface
21. Minimum distance to the origin Find the points 00 the surface z2 = xy + 4 closest to the origin. 22. Minimum distance to the origin face xyz = I closest to the OOgio.
Find the point(s) on the sur-
23. Extrema on a sphere Find the maximum aod minimum values of f(x,y,z)
on the sphere
x2
~
+ y2 + z2
x - 2y
= 30.
+ 5z
14.8 Lagrange Multiplie" 24. Ertrema on a sphere Find the points on the sphere x 2 + y2 + z2 ~ 25 where !(x,y, z) ~ x + 2y + 3z has its
maximum and minimum values. 25. Minjmjzing a sum of squares Find three real numbers whose sum. is 9 and the sum of whose squares is as small as possible. 26. Maximizing a product Find the largest product the positive numbers x, y, andz can have if x + y + z2 = 16. 27. Rectangular h"" of largest volume in a sphere Find the dimensions of the closed rectangular box with maximum volume that can be inseribed in the unit sphere.
28. Bm: with vertex on a plane Find the volume of the largest closed rectangular box in the fll'Sl octant having three faces in the coordinareplanesandavertexontheplanex/a + y/b + z/c = I,where a> O,b > o,andc > O. 29. Hottest point on a space probe A space probe in the shape of the ellipsoid 4x 2 + y2 + 4z 2 ~ 16 enters Earth's atmosphere and its surface begins to heat After 1 hour, the temperature at the point (x,y, z) on the probe's surface is
T(x,y,z) =
ax 2 + 4yz -
16z
+ 600.
Find the hotrest point on the probe's surface.
30. Ertreme temperatores on a sphere Suppose that the Celsius temperature at the point (x, y, z) on the sphere x 2 + y2 + z2 ~ 1 is T = 40Oxyz2 . Locare the highest and lowest temperatures on the sphere. 31. Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts x and y of two capital goods G1 and G2 is sometimes measured by a function U(x, y). For example, Gl and (h might be two chemicals a pharmaceutical coropany needs to have on hand and U(x, y) the gain froro manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If Gl costs a dollars per kilogram, (h costs b dollars per kilogram, and the total amount allocared for the purchase of G1 and G2 together is c dollars, then the company's managers want to maximize U(x, y) given that ax + by = c. Thus, they need to solve a typical Lagrange multiplier problem. Suppose that U(x,y) = xy and that the equation ax
+
by
+ 2x
= c simplifies to
2x + Y = 30. Find the maximum value of U and the corresponding values of x and y subject to this latter constraint 32. Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it where the magnetic field ofthe plauet is weakest. The plauet is spherical, with a radius of 6 units. Based on a coordinare system whose origin is at the center of the plauet, the streugth of the magnetic field is given by M(x, y, z) = 6x - y2 + xz + 60. Where should you Iocare the radio reIescope?
Extnlme val.e. Subject to Two Constraints
33. Maximize the function !(x,y, z) ~ x 2 + 2y - z2 subject to the constraints 2x - y = 0 and y + z = O.
819
34. Minimize the function !(x,y,z) ~ x 2 + y2 + z2 subject to the constraints x + 2y + 3z = 6 and x + 3y + 9z = 9.
35. Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes y + 2z ~ 12 and x + y = 6. 36. Muimum value on line of intersection Find the maximum value that !(x, y, z) ~ x 2 + 2y - z2 can have on the line of intersection of the planes 2x - Y = 0 and y + z = O. 37. Extrema on a curve of intersection Find the extreme values of !(x, y, z) ~ x'yz + 1 on the intersection of the plane z ~ I with the sphere x 2 + y2 + z2 = 10. 38. a. Maximum on line of intersection Find the maximum value of w ~ xyz on the line of intersection of the two planes x + y + z = 40 and x + y - z = O. b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of w.
39. Extrema on a circle ofintenectioD Find the extreme values of the function!(x,y, z) = xy + z2 on the circle in which the plane y - x = 0 intersects the sphere x 2 + y2 + z2 = 4. 40. Minimum distance to the origin Find the point closest to the origio on the curve of intersection of the plane 2y + 4z = 5 and the cone z2 = 4x 2 + 4y 2. Theory and Examples 41. The condition V! = AVgisnotsnllicient Although V! ~ AVg is a necessary condition for the occurrence of an extreme value of !(x, y) subject to the conditions g(x, y) ~ 0 and Vg oF 0, it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to fmd a maximum value of !(x, y) ~ x + y subject to the constraint that xy ~ 16. The method will identifY the two points (4, 4) and (-4, -4) as candidares for the location of extreme values. Yet the sum (x + y) has no maximum value on the hyperbola xy = 16. The farther you go fron2 the origin on this hyperbola in the first quadrant, the larger the sum !(x,y) = x + ybecoroes.
42. A lea.t sqnare. plane The plaue z = Ax "fitted" to the following points (x" Yb z,): (0,0,0),
(0, I, 1),
+ By + C
is to be
(1,0, -1).
(I, I, I),
Find the values of A, B, and C that minimize 4
f-1 (Ax, + By, + C - z,l', the sum of the squares of the deviations.
43. a. Maximum on a sphere Show that the maximum value of a 2b 2c 2 on a sphere of radius r centered at the origin of a Cartesian abc-coordinare systero is (r 2/3)3 b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers a, b, and c,
a (abc ) 1/3 < -
+b +c. 3
'
that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean. 44. Sum of products Let alo a2, ... , an be n positive numbers. Find the maximum of ~?=1 a'Xt subject to the constraint ~?=1 x? = 1.
820
Chapter 14: Partial Derivatives
COMPUTER EXPLORATIONS
46. Minimize x2
In Exercises 45-50, use a CAS to perform the following steps imple-
+ y2
-
j(x, y, z)
~ X}'i"
subject
to
the
constraints
1 = 0 and x - z = O.
47. Maximize j(x, y, z) = x 2 + y2 + z2 sul!iect to the constraints 2y + 4z - 5 ~ 0 and4x 2 + 4y 2 - z2 ~ O. 48. Minimize j(x,y,z) ~ x 2 + y2 + z2 subject to the constraints x 2 - xy + y2 _ z2 - I = 0 andx 2 + y2 - I = O.
menting tire method of Lagrange multipliers for fmding constrained ex1rema:
a. Formthefunctionh = j - AlgI - A2g2,wherejisthefunction to optimize subject to tire constraints gl ~ 0 and g2 ~ O.
49. Minimize j(x,y,z, w) = x 2 + y2 + z2 +w 2 subject to the constraints 2x-y+z-w-I~O and x+y-z+ w - I = O.
h. Determine all the first partial derivatives of h, including the par-
tials with respect to Al andA2, and set them equal to O. c. Solve the system ofequations found in part (b) for all the unknowns, including Al and A2.
SO. Determine the distance from the line y = x + I to the parahola y2 ~ x. (Hint: Let (x, y) be a point on tire line and (w, z) a point on the parabola. You want to minimize (x - w)' + (y - z)'.)
d. Evaluate j at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise.
45. Minimize j(x,y,z) ~ xy + yz sul!iect to the constraints x 2 + y2 - 2 = 0 andx 2 + z2 - 2 = O.
14.9
I_Ta-"-Y_Lo_r's_Fo_rm_uL_a_fo_r_T_w_O_Va_ri_a_b_Le_s
-------
In this section we use Taylor's formula to derive the Second Derivative Test for local extreme values (Section 14.7) and the error formula for linearizations of functions of two independent variables (Section 14.6). The use of Taylor's formula in these derivations leads to an extension of the formula that provides polynomial approximations of all orders for functions of two independent variables.
Derivation of the Second Derivative Test t
~
1 S(a
+ h,b + k)
Let I(x, y) have continuous partial derivatives in an open region R containing a point P( a, b) where Ix = I y = 0 (Figure 14.57). Let h and k be increments small enough to put the point S(a + h, b + k) and the line segment joining it to P inside R. We parametrize the segment PS as
x=a+th, If F(t) = I(a
O:5t:51.
y=b+tk,
+ th, b + tk), the Chain Rule gives , dx F (t) = Ix dt
t= 0 P(a, b)
dy
+ I y dt
=
hlx
+ kly •
Part of open region R
FIGURE 14.57 We begin the derivation of the Second Derivative Test atp(a, b) by parametrizing a typical line segment from P to a point S nearby.
Since Ix and I y are differentiable (they have continuous partial derivatives), F' is a differentiable function of t and
F" = aF'dx axdt = h 21xx
+ aF'dy aydt
=
~(hf + kn'h + ~(hf + kn'k ax x y ay x y t", ~ tyx
+ 2hklxy + elyy .
Since F and F' are continuous on [0, I] and F' is differentiable on (0, I), we can apply Taylor's formula with n = 2 and a = 0 to obtain
F(I) = F(O)
+ F'(O)(I - 0) + F"(c)
F(I) = F(O)
+ F'(O) + F"(c)
t
(l
2
0)' (1)
14.9 Taylor's Formula for Two Variables
821
for some c between 0 and I. Writing Equation (I) in terms of I gives I(a
+ h, b + k)
= I(a, b)
+ hlx(a, b) +
+ ~ (h2lxx +
2hkl"1
kly(a, b)
+ k2lyy) I
.
(2)
(a+ch, b+ck)
Since Ix(a, b) = I,(a, b) = 0, lbis reduces to I(a
I(a
+ h,b + k)
- I(a,b) =
~(h2Ixx + 2hkl"1 + k2lyy) I(a+ch, b+ck).
(3)
The presence of an extremum of I at (a, b) is determined by lbe sigo of k) - I(a, b). By Equation (3), this is lbe same as lbe sigo of
+ h, b +
Q(c) = (h 2lxx
+ 2hkl"1 + k2Iyy)l(a+oh,b+ck)'
Now, if Q(O) oF 0, lbe sigo of Q(c) will be lbe same as lbe sigo of Q(O) for sufficiently small values of h and k. We can predict lbe sigo of
Q(O)
= h 2Ixx(a, b)
+ 2hkl"1(a, b) + k2Iyy(a, b)
(4)
from lbe signs of Ixx and Ixxlyy - 1"12 at (a, b). Multiply both sides of Equation (4) by Ixx and rearrange lbe right-band side to get (5)
From Equation (5) we see that
3.
If Ixx < 0 and Ixxlyy - 1"12 > 0 at (a, b), lben Q(O) < 0 for all sufficiently small nonzero values of h and Ie, and I has a local maximum value at (a, b). If Ixx > 0 and Ixxlyy - 1"12 > 0 at (a, b), lben Q(O) > 0 for all sufficiently small nonzero values of h and Ie, and I has a local minimum value at (a, b). If Ixxlyy - 1"12 < 0 at (a, b), lbere are combinations of arbitrarily small nonzero val-
4.
ues of h and k for which Q( 0) > 0, and other values for which Q( 0) < O. Arbitrarily close to lbe point Po(a, b, I(a, b)) on lbe surface z = I(x,y) lbere are points above Po and points below Po, so I has a saddle point at (a, b). If I xxlyy - 1"12 = 0, anolber test is needed. The possibility that Q(O) equals zero pre-
1. 2.
vents us from drawing conclusions about lbe sigo of Q(c).
The Error Formula for Linear Approximations We want to show lbat lbe difference E(x, y), between lbe values of a function I(x, y), and its linearization L(x, y) at (xo, Yo) satisfies lbe inequality IE(x,y)l:5
~M(lx -
xol
+ Iy
- Yo I)'·
The function I is assumed to have continuous second partial derivatives throughout an open set containing a closed rectangular region R centered at (xo, yo). The number M is an upper bound forl/xxl, I/yyl, and 1/"11 onR. The inequality we want comes from Equation (2). We substitute Xo and Yo for a and b, and x - Xo and y - Yo for h and k, respectively, and rearrange lbe result as I(x,y) = l(xo,Yo)
+ fx{xO,yo)(x
- xo)
+ ly(xo,Yo)(y
- Yo)
linearization L(x, y) 1
+ -2 (x
- xc)'lxx
+ 2(x
- xo)(y - yo)I"1
+ (y
error E(x, y)
- yo)'lyy)l(Xo + c (X-Xo.Yo ) + c (Y-Yo » .
822
Chapter 14: Partial Derivatives TIris equation reveals that
IEI:5 !Ox - xol 2 ll,,,1 + 21x - xolly - Yolll",1 + Iy - YoI 2 1/)'}'!). Hence, if Mis an upper bound for the values ofl I", I, 1/",,1, and II)'}' IonR,
IEI:5 !Ox =
!
xol 2M
M(lx -
+ 21x -
xolly - YolM
+ Iy -
Yo I2M)
xol + Iy - Yo1l2•
Taylor's Formula for Functions of Two Variables The formulas derived earlier for F' and F" can be obtained by applying to I(x, y) the operators
(h~ax + k~) ay
and
(
h~ + k~ ax
2
ay )
2
=
2
h2~2 + 2hk-aax
ax ay
2
+ k2~. ay2
These are the first two instances of a more general formula, F (n)()_~()_(~ t - dtn F1 t - h ax
~)n() + kay I x,y ,
(6)
which says that applying dn/dt n to F(t) gives the same result as applying the operator
+ k~)n (h~ ax ay to I(x, y) after expanding it by the Binomial Theorem.
If partial derivatives of I through order n + I are continuous throughout a rectangular region centered at (a, b), we may extend the Taylor formula for F(t) to
F(t) and take t
=
=
F(O)
F"(O)
+ F'(O)t + ----z! t 2 +
F(n)(o)
+ ----,;y- t(n) +
remainder,
I to obtain F( I)
=
F(0)
F"(O)
F(n)(o)
+ F' (0) + ----z! + ... + ----,;y- +
remainder.
When we replace the first n derivatives on the right of this last series by their equivalent expressions from Equation (6) evaluated at t = 0 and add the appropriate remainder term, we arrive at the following formula.
Taylor'. Formula for I(x,y) at the Point (a, b)
Suppose I(x, y) and its partial derivatives through order n + I are continuous throughout an open rectangular region R centered at a point (a, b). Then, throughout R, I(a
+ h,b + k)
=
I(a, b)
+ (hi" + kly) I(a, b) +
i! (h 2/", + 2hkl"" + k2/)'}')I(a,b)
3)1 I + -3I1. (h 31= + 3h 2klxxy + 3hk2IX)')' + ki m (a,b) + ... +, n.
+
(n
I
+ 1)!
(
h~ + k~ ) n+1I ax
ay
I (a+ch,b+ck)'
(a a)n ha + ka II Y x
(a, b)
(7)
823
14.9 Taylor's Formula for Two Variables
The f"lISt n derivative terms are evaluated at (a, b). The last term is evaluated at some point (a + ch, b + ck) on the line segment joining (a, b) and (a + h, b + k). If (a, b) = (0, 0) and we treat h and k as independent variables (denoting them now by x and y), then Equation (7) assumes the following simpler form.
Taylor's Formula for f(x,y) at the Origin f(x,y) = f(O, 0)
+ +
+ xfx + yf, +
i!
(x 3f=
(n
+
I
1)1
i!
(x 2fxx
+
2xyfxy
+ y2fyy)
+ 3x'yfxxy + 3xy2fxyy + y3fyyy) +
(x~ + y~)n+lfl ax
ay
I +n!
(a x - + ya - )nf ax ay (8)
lox,,,,)
The first n derivative terms are evaluated at (0, 0). The last term is evaluated at a point on the line segment joining the origin aod (x, y). Taylor's formula provides polynomial approximations of two-variable functions. The f"lISt n derivative terms give the polynomial; the last term gives the approximation error. The first three terros of Taylor's formula give the function's linearization. To improve on the linearization, we add higher-power terms.
EXAMPLE 1 Find a quadratic approximation to f(x,y) = sinxsiny near the origin. How accurate is the approximation if Ixl ,,;; 0.1 aod Iyl ,,;; O.I? Solution
We take n = 2 in Equation (8):
Calculating the values ofthe partial derivatives, f(O,O) = sinxsinYI(o,o) = 0,
f",,(O,O) = -sinxsinYI(o,o) = 0,
f,(O,O) = cosxsinYllO,O) = 0,
fxy(O,O) = cosxcosYllo,o) = I,
1,(0,0) = sinxcosYllo,o) = 0,
fyy(O,O) = -sinxsinYllo,o) = 0,
we have the result sinxsiny '" 0
+ 0 + 0 + !(x 2(0) + 2xy(I) + y2(0)),
or
sin x siny '" xy.
The error in the approximation is
The third derivatives never exceed I in absolute value because they are products of sines aod cosines. Also, Ixl ,,;; 0.1 andlyl ,,;; 0.1. Hence IE(x,y)1 ,,;; i«0.I)3
+ 3(0.1)3 + 3(0.1)3 +
(0.1)3) =
~(0.I)3
,,;; 0.00134
(rounded up). The error will not exceed 0.00134 iflxl ,,;; 0.1 aodlyl ,,;; 0.1.
•
824
Chapter 14: Partial Derivatives
Exercises 14.9 Finding Quadratic and Cubic Approximations In Exercises 1-10, use Taylor's formula for f(x,y) althe origin to fmd quadratic and cubic approximations of f near the origin.
1. f(x,y) =
xe'
3. f(x,y) = ysinx 5. f(x,y) = er In (I
+ y) 2 7. f(x,y) = sin (x + y2)
2. f(x,y) = ercosy 4. f(x,y) = sinxcosy
+ y + I) 2 8. f(x,y) = cos (x + y2) 6. f(x,y) = In (2x
9. f(x,y) ~ I
I
I
10. f(x,y) ~ I -x-y
-x-y
+ xy
11. Use Taylor's formula to fmd a quadratic approximation of fix, y) = cos x cos y at the origin. Estimate the error in the approximationiflxl'" 0.1 andlyl '" 0.1. 12. Use Taylor's formula to fmd a quadratic approximation of e r siny at the origin. Estimate the error in the approximation iflxl '" 0.1 andlyl'" 0.1.
14.10 t-p_a_rti_'a_L_D_e_ri_v_ati_·V_e_S_Wl_'_th_C_on_s_t_ra_in_e_d_V_a_"_'a_b_Le_s
_
In finding partial derivatives of functions like w = fix, y), we have assumed x andy to be independent. In many applications, however, this is not the case. For example, the internal energy U ofa gas may be expressed as a function U = f(P, V, 1') ofpressureP, volume V, and temperature T. If the individual molecules of the gas do not interact, however, P, V, and T obey (and are constrained by) the ideal gas law
PV= nRT
(n and R constant),
and fail to be independent. In this section we learn how to find partial derivatives in situations like this, which occur in economics, engineering, and physics.·
Decide Which Variables Are Dependent and Which Are Independent If the variables in a function w = f(x,y, z) are constrained by a relation like the one imposed on x, y, and z by the equation z = x 2 + y2, the geometric meanings and the numerical values of the partial derivatives of f will depend on which variables are chosen to be dependent and which are chosen to be independent. To see how this choice can affect the outcome, we consider the calculation of iJwj ax when w = x 2 + y2 + z2 and z = x 2 + y2.
EXAMPLE 1
Find iJwjax if w = x 2 + y2
+ z2 and z
= x2
+ y~
We are given two equations in the four unknowns x, y, z, and w. Like many such systems, this one can be solved for two of the unknowns (the dependent variables) in terms of the others (the independent variables). In being asked for iJwjax, we are told that w is to be a dependent variable and x an independent variable. The possible choices for the other variables corne down to
Solution
Dependent w,z w,y
Independent x,y
x,z
In either case, we can express w explicitly in terms of the selected independent variables. We do this by using the second equation z = x 2 + Y 2 to eliminate the remaining dependent variable in the flISt equation. *This section is based on notes written for MIT by Arthur P. Mattuck.
14.10 Partial Oerivatives with Constrained Variables
825
In the f'lISt case, the remaining dependent variable is z. We eliminate it from the first equation by replacing it by x' + y' . The resulting expression for w is
+ y' + z' = x' + y' + (x' + y')' x' + y' + x 4 + 2x'y' + y4
w = x' =
and
~;
,
= 2x
+ 4x 3 + 4xy'.
(1)
This is the formula for aw/ ax when x and y are the independent variables. In the second case, where the independent variables are x and z and the remaining dependent variable is y, we eliminate the dependent variable y in the expression for w by replacing y' in the second equation by z This gives
x' .
w = x'
+ y' + z'
=
x'
+
(z - x')
+ z'
= z
+ z'
and
aw = 0 ax .
x
FIGURE 14.58 If P is constrained to lie on the paraboloid z = x 2 + y~ the value ofthe partial derivative of w = x 2 + y2 + z2 with respect to x at P depends on the direction of motion (Example I). (I) Ail x changes, with y = 0, P moves up or down the surface on the parabola z = x 2 in the xz-plane with itw/iix ~ 2x + 4x'. (2)Ailxchanges, with z = 1, P moves on the circle x' + y' = I,z = I,andaw/ax = O.
(2)
This is the formula for aw/ ax when x and z are the independent variables. The formulas for aw/ax in Equations (I) and (2) are genuinely different. We cannot change either formula into the other by using the relation z = x' + y'. There is not just one aw/ax, there are two, and we see that the original instruction to f'md aw/ax was incomplete. Which aw/ ax? we ask. The geometric interpretations of Equations (I) and (2) help to explain why the equations differ. The function w = + y' + z' measures the square of the distance from the point (x, y, z) to the origin. The condition z = x' + y' says that the point (x, y, z) lies on the paraboloid of revolution shown in Figure 14.58. What does it mean to calculate aw/ax at a point P(x, y, z) that can move only on this surface? What is the value of aw/ax when the coordinates ofP are, say, (I, 0, I)? If we take x and y to be independent, then we f'md aw/ ax by holding y fixed (at y = 0 in this case) and letting x vary. Hence, P moves along the parabola z = x' in the xz-piane. As P moves on this parabola, w, which is the square of the distance from P to the origin, changes. We calculate aw/ax in this case (our first solution above) to be
x'
aw ax = 2x
+ 4x 3 + 4xy'.
At the point P(I, 0, I), the value of this derivative is
~;
= 2
+4 +
0 = 6.
If we take x and z to be independent, then we f'md aw/ax by holding z fixed while x varies. Since the z-coordinate of P is I, varying x moves P along a circle in the piane z = 1. As P moves along this circle, its distance from the origin remains constant, and w, being the square of this distance, does not change. That is,
aw = 0 ax ' as we found in our second solution.
•
How to Find iJwjiJx When the Variables in w = I(x, y, z) Are Constrained by Another Equation As we saw in Example I, a typical routine for finding aw/ ax when the variables in the function w = /(x,y, z) are related by another equation has three steps. These steps apply to finding aw/ay and aw/az as well.
826
Chapter 14: Partial Derivatives
1. Decide which variables are to be dependent and which are to be independent. (In practice, the decision is based on the physical or theoretical context of our work. In the exercises at the end of this section, we say which variables are which.)
2. Eliminate the other dependent variable(s) in the expression for w. 3. Differentiate as usual.
Ifwe cannot carry out Step 2 after deciding which variables are dependent, we differentiate the equations as they are and try to solve for awjax afterward. The next example shows how this is done.
EXAMPLE 2
Findawjax at the point (x,y, z) = (2, -I, I) if w = x2
+ y2 + z2,
Z3 - xy
+ yz + y3
= I,
and x and y are the independent variables.
It is not convenient to eliminate z in the expression for w. We therefore differentiate both equations implicitly with respect to x, treating x and y as independent variables and w and z as dependent variables. This gives
Solution
aW=2x+2z az ax ax
(3)
and
2aZ 3z - - y ax
az + yax -+0
=
o.
(4)
These equations may now be combined to express awjax in terms of x, y, and z. We solve Equation (4) for azjax to get
az ax
y y+3z2
and substitute into Equation (3) to get
aw=2x+ ax
2yz y+3z2 '
The value ofthis derivative at (x,y, z) = (2, -I, I) is
aw) = 2(2) + 2(-1)(1) = 4 + -2 = 3. ( ax (2,-1,1) -I + 3(1)2 2
•
HISTORICAL BIOGRAPHY
Notation
Sonya Kovalevsky
To show what variables are assumed to be independent in calculating a derivative, we can use the following notation:
(185{}-1891)
(~;)y
awjax with x andy independent
(:~t
af jay with y, x and t independent
14.10 Partial Derivatives with Constrained Variables
EXAMPLE 3 Solution
Find (iIw/ax)y,zifw = x 2
+ Y - z + sin I and x + y
827
= I.
With x, y, z independent, we have I = x
(~;)y,z =
+ y,
w
= x2
+ Y - z + sin (x + y)
2x + 0 - 0 + cos (x +
y)~(x + y)
•
2x + cos (x + y).
=
Arrow Diagrams In solving problems like the one in Example 3, it often helps to start with an arrow diagram that shows how the variables and functions are related. If w = x2
+Y
- z
+ sinl
x+y=1
and
and we are asked to findilw/ax whenx,y, andz are independent, the appropriate diagram is one like this:
(5)
Independent variables
Intermediate variables
Dependent variable
To avoid confusion between the independent and intermediate variables with the same symbolic names in the diagram, it is helpful to rename the intermediate variables (so they are seen as jUnctions ofthe independent variables). Thus, let u = x, v = y , and s = z denote the renamed intermediate variables. With this notation, the arrow diagram becomes
(;)
m
--->
Independent
variables
w
--->
Intermediate variables and
(6)
Dependent
variable
n:lations u=x v=y ~=z
t
= x +y
The diagram shows the independent variables on the left, the intermediate variables and their relation to the independent variables in the middle, and the dependent variable on the right. The function w now becomes w
=
u2 +
V -
s + sint,
where
u =x,
v =y,
s = z,
and
I =
x
+ y.
828
Chapter 14: Partial Derivatives
To find aw/ax, we apply the four-variable fonn ofthe Chain Rille to w, guided by the arrow diagram in Equation (6):
= (2u)(1)
+
+
(1)(0)
(-1)(0)
+
(cost)(I)
=2u+cost = 2x
+
+ y).
cos (x
Sub,tituting the original independent variables u = ;t and t = x + y
Exercises 14.10 finding Partial Derivatives with Constrained Variables In Exercises 1-3, begin by drawing a diagram that shows the relations among the variables.
1. Ifw
=
2. Ifw
=
+ y2 + z2 andz
x2
+Y
x2
- z
(~;L ~ (~;t
x2
+ y2, fmd
+ sintandx + y t.
L
=
=
Theory and Examples 9. Establish the fact, widely used in bydrodyoamics, that if f(x,y, z) ~ 0, then
t,find
(:t (:t e.
e. (aa7L
f.
(~7t·
= f(P, V, 1') be the internal energy of a gas that obeys the ideal gas law PV ~ nRT (n and R constant). Find
3. Let U
(~~)v
L
each give awjax, depending on which variables are chosen to be dependent and which variables are chosen to be independent IdentilY the independent variables in each case.
t.
(~~)v'
t.
(:)y
(Hint: Express all the derivatives in terms of the formal partial derivatives afjax, afjiJy, and afjaz.) 10. Ifz = x
+ f(u), where u ilz
= xy, show that iJz
x ax - y ay
~
x.
4. Find
(~;)y
L
~ 0 determines z as a differentiable function of the independent variables x and y and that g, O. Show that
11. Suppose that the equation g(x, y, z)
*
at the point (x,y,z) = (0, I,,,) if
ysinz
and
+ zsinx
=
az) agjiJy ( iJy , = - agjaz .
O.
5. Find
12. Suppose thatf(x,y,z, w) = 0 andg(x,y,z, w) = 0 determinez and w as differentiable functions of the independent variables x and y, and suppose that
at the point (w,x,y, z) = (4,2, I, -I) if
w
= x'y' + yz -
z'
and
6. Find (aujay), at the point (u, v) ~ y = uv. 7. Suppose that x 2 nates. Find
+ y2
(v2, I), if x
~ u'
+ v' and
and
•
af ag ax Ow af ag
af ag awax af ag
aw
iJw iJz
-----
-----
iJz
and y 2+4z+t
and
x+2z+t~25.
Ow -=2x-1
afag az ay af ag
afag ayaz af ag . ----iJz ilw ilwaz -----
Show that the equations
ax
aw iJz
Show that
(~:);
8. Suppose that _
iJz iJw
= r 2 and x = r cos 0, as in polar coordi-
(~;), w=x 2
af ag af ag -----*0
x' + y' + z' = 6.
and
aW=2x_2 ax
Chapter 14 Practice Exercises
Chapter
M
Questions to Guide Your Review
1. What is a real-valued function of two independent variables? Three independent variables? Give examples. 2. What does it mean for sets in the plane or in space to be open? Closed? Give examples. Give examples of sets that are neither open nor closed. 3. How can you display the values ofa function f(x,y) of two independent variables graphically? How do you do the same for a function f(x,y, z) of three independent variables? 4. What does it mean for a function f(x, y) to have limit L as (x, y) ---+ (XQ, Yo)? What are the basic properties of limits of functions of two independent variables? 5. When is a function of two (three) independent variables continuous at a point in its domain? Give examples of functions that are
continuous at some points but not others. 6. What can be said about algebraic combinations and composites of
continuous functions? 7. Explain the two-path test for nonexistence of limits.
8. How are the partial derivatives af/ax and af/ay of a function f(x,y) dermed? How are they interpreted and calculated? 9. How does the relation between rITst partial derivatives and continuity of functions of two independent variables differ from the relation between rITst derivatives and continuity for real-valued functions of a single independent variable? Give an example.
10. What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives ofsecond and higher orders? Give examples. 11. What does it mean for a function f(x, y) to be differentiable? What does the Increment Theorem say about differentiability?
12. Haw can you sometimes decide from examjning Ix and /, that a function f(x, y) is differentiable? What is the relation between the differentiability off and the continuity of f at a point? 13. What is the general Chain Rule? What form does it take for functions of two independent variables? Three independent variables? Functions dermed on surfaces? How do you diagram these differ-
14. What is the derivative ofa function f(x,y) at a point Po in the direction of a unit vector n? What rate does it describe? What geometric interpretation does it have? Give examples. 15. What is the gradient vector of a differentiable function f(x, y)? How is it related to the function's directional derivatives? State the analogous results for functions of three independent variables. 16. How do you rmd the tangent line at a point on a level curve of a differentiable function f(x, y)? How do you rmd the tangent plane and normal line at a point on a level surface of a differentiable function f(x, y, z)? Give examples. 17. How can you use directional derivatives to estimate change? 18. How do you linearize a function f(x, y) of two independent variables at a point (xo, Yo)? Why right you want to do this? How do you linearize a function of three independent variables? 19. What can you say about the accuracy of linear approximations of functions of two (three) independent variables? 20. If(x,y) moves from (XQ,Yo) to a point (xo +
row your search for these values? Give examples. 23. How do you rmd the extrema of a continuous function f(x, y) on a closed bounded region of the xy-plane? Give an example. 24. Describe the method ofLagraoge multipliers and give examples. 25. How does Taylor's formula for a function f(x, y) generate polynomial approximations and error estimates? 26. Ifw = f(x,y,z), where the variables x, y, and z are conslrained bY an equation g(x, y, z) ~ 0, what is the meaning of the notation (aw/ax),? How can an arrow diagram help you calculate this par-
tial derivative with constrained variables? Give examples.
ent forms? Give examples. What pattern enables one to remember all the different forms?
Chapter
829
Practice Exercises
DomaIn, Range, and Level Curves In Exercises 1-4, rmd the domain and range of the given function and
7. h(x,y,z) =
identitY its level curves. Sketch a typical level curve.
+'
1. f(x,y) ~ 9x 2 + y2
2. f(x,y) ~ eX
3. g(x,y) ~ I/xy
4. g(x,y) ~ ~
In Exercises 5--1l, rmd the domain and range of the given function and identitY its level surfaces. Sketch a typical level surface.
5. f(x,y, z) = x 2 + y2 - z
6. g(x,y, z) = x 2 + 4y 2 + 9z 2
2 X
8. k(x,y,z) =
I
2
+Y +z
2
I 2 2 2 x+y+z+1
Evaluating Um;ts Find the lirrtits in Exercises 9-14. 9. lim e" cosx 10. (x,y)~(~,ln2)
lim
(X,Y)~(O,O) X
+Y + cosy
2
830
11. 13.
Chapter 14: Partial Derivatives
~ 2
lim
(xJl)-(1,l) X
lim
P-(l,-l,e)
12.
y2
_
14.
Inlx + y + zl
lim
X 3y 3 -
(x~)~(I,I)
lim
P-(I,-l,-I)
xy
1
34. Show that if w = f(') is any differentiahle functioo of. and if .=y+5x,then
1
tan-I (x + Y + z)
By considering different paths of approach, show that the limits in Exercises 15 and 16 do not exist
Implidt Differentiation Assuming that the equatioos in Exercises 35 and 36 defme y as a differentiable functioo of x, fmd the value of dy/dx at point P. 17. Continuous extension Let f(x,y) = (x 2 - y2)/(X 2 + y2) for (x, y) # (0, 0). Is it possihle to defme flO, 0) in a way that makes f cootinuous at the origin? Why? Let
18. Continuous extension
+e
X
+' - 2 =
0,
P(O,I)
P(O, In 2)
In Exercises 37-40, fmd the directions in which f increases and decreases most rapidly at Po and find the derivative of f in each directioo. Also, find the derivative of f at Po in the direction of the
Ixl + Iyl # 0
Ixl + Iyl ' 0,
{
36. 2xy
Directional Derivatives
Sin(X- y ) f(x,y) =
35. I - x - y2 - sinxy = 0,
(x,y) = (0,0),
vector v.
Is f cootinuous at the origin? Why?
37. f(x,y) = cosxcosy,
P o("./4, "./4),
v = 31 + 4j
38. f(x,y) = x 2e-2"
Partial Derivatives In Exercises I<}-24, fmd the partial derivative of the functioo with
respect to each variable. 19. g(r,O) = rcosO
20. f(x,y)
=
tin
+ rsinO
(x 2 + y2) + tan-I
1
21. f(R" R2, R,) = Rl
I R2
+
v=i+j+k
~
41. Derivative in velocity direction Find the derivative of f(x,y, z) = xyz in the directioo of the velocity vector ofthe helix
I
+ R,
r(l) = (cos 31)i + (sin 31)j + 31k
22. h(x,y,z) = sin (2".x + Y - 3z)
= ~T
23. P(n, R, T, V) 24. f(r,/,T,w)
at 1 = "./3. 42. Mnimnm directional derivative What is the largest value that the directional derivative of f(x,y, z) = xyz can have at the point (I, I, I)?
(the ideal gas law)
= 2~/~
Second-Order Partials Find the secood-order partia1 derivatives of the functioos in Exercises 25-28, 25. g(x,y)
=
y
x
+Y
26. g(x,y)
= eX
+ ysinx
= y2
+
- 3xy
cosy
tioo ofu is (f,(xo,Yo)i + f,(xo,yo)j)' u, b. The derivative of f at (xo, Yo) in the directioo of u is a vector.
29. Find dw/dl at 1 = 0 if w In (I + I).
= sin (xy
+ ".),
X
=e, t
and y =
30. Find dw/dl at 1 = I if w = xe' + y sinz - cosz, x = 2vt, y = 1 - I + Inl,andz = ".1. 31. Findaw/arandaw/a.wheor =
+ sins,y
aw/au
=
= ".and. = Oifw = sin(2x
d. At (xo, Yo) , vector Vf is normal to the curve f(x, y) = f(xo, Yo),
Gradients, Tangent Planes, and Normal Lines
and
respecttotonthecurvex
c. The directional derivative off at (xo, yo) has its greatest value in the directioo of Vf '
- y),
rs.
aw/av when u = v = 0 if w = 1nV1+7 - tan-Ix and x = 2e"cosv. Find the value ofthe derivative off(x,y,z) = xy + yz + xzwith
32. Find 33.
r
b. Find the derivative off at (I, 2) in the directioo toward the point (4, 6).
a. If u is a unit vector, the derivative of f at (xo, Yo) in the direc-
+ 7e'
Chain Rule Calculations
x
43. Directional derivatives with given values At the point (I, 2), the functioo f(x, y) has a derivative of 2 in the directioo toward (2, 2) and a derivative of -2 in the directioo toward (I, I), a. Find fx(I, 2) andf,(I, 2).
44. Which of the following Ststeo3eotS are true if f(x, y) is differentiable at (xo, yo) ? Give reasoos for your aoswetli.
27. f(x,y) = x + xy - 5x' + In(x 2 + 1)
28. f(x,y)
Po(I,O), v = 1 + j 39. f(x,y, z) = In (2x + 3y + 6z), P o(-1, -1, 1), v = 21 + 3j + 6k 40. f(x,y, z) = x 2 + 3xy - z2 + 2y + z + 4, PolO, 0, 0),
=
cost,Y
=
sint,z
=
cos2tatt
=
1.
In Exercises 45 and 46, sketch the surface f(x,y, z) = c together with Vf at the given points, 45. x 2 + y + z2 = 0; 46. y2 + z2 = 4;
(0, -I, ±I),
(2, ±2, 0),
(0,0,0)
(2, 0, ±2)
831
Chapter 14 Practice Exercises In Exercises 47 and 48, rmd an equation for the plane tangent to the level surface f(x,y, z) ~ c at the point Po. Also, find parametric equations for the line that is nonnal to the surface at Po. 47. x 2 - y - 5z = 0, Po(2, -I, I) 48. x 2 + y2 + Z = 4, P o(l,1,2) In Exercises 49 and 50, rmd an equation for the plane tangent to the surface z ~ f(x, y) at the given point. 49. z ~ In (x 2 + y2), (0, 1,0) 50. z ~ 1/(x 2
+ y2), (I, I, 1/2)
In Exercises 51 and 52, rmd equations for the lines that are tangent and nonna! to the level curve f(x,y) = c at the point Po. Then sketch the lines and level curve together with Vf at Po. 51. y - sinx ~ I,
Po('1T, I)
52.
y2
x2
2 - 2
~
(I, I, 1/2) Point: 54. Surfaces: x + y2 + Z = 2, y = I (1/2,1,1/2)
Linearizations In Exercises 55 and 56, rmd the linearization L(x, y) of the function f(x, y) at the point Po. Then rmd an upper bound for the magnitude of the error E in the approximation f(x, y) '" L(x, y) over the rectangle R. 55. f(x,y) = sinxcosy, PO('1T/4, '1T/4)
Ix - *1
S
0.1,
Iy - *1
S
0.1
56. f(x,y) ~ :x:y - 3y 2 + 2, Po(l, I) R:
Ix - II
S
0.1, Iy - II
S
0.2
Find the linearizations of the functions in Exercises 57 and 58 at the given points. 57. f(x,y, z) = :x:y + 2yz - 3:x:z at (1,0,0) and (I, 1,0) 58. f(x,y,z) ~ v2cosxsin(y
+ z)at(O, 0, '1T/4) and ('1T/4,'1T/4, 0)
Estimates and Sensitivity to Change 59. Measuring the volume of a pipeline You plan to calculate the volmne inside a stretcb ofpipeline that is about 36 in. in diameter and I mile long. With which measurement should you be more careful, the length or the diameter? Why? 60. Sensitivity to change Is f(x,y) = x 2 - :x:y + y2 - 3 more sensitive to changes in x or to changes in y when it is near the point (I, 2)? How do you know? 61. Change in an electrical circuit Suppose that the current I (amperes) in an electrical circuit is related to the voltage V(volts) and the resistance R (ohms) bY the equation 1= V/R. If the voltage drops from 24 to 23 volts and the resistance drops from 100 to 80 ohms, will I increase or decrease? By about how much? Is the change in I more sensitive to change in the voltage or to change in the resistance? How do you know?
=
u
+ v,
a. Ifu is measured with an error of2% and v with an error of3%, about what is the percentage error in the calculated value ofy? b. Show that the percentage error in the calculated value ofz is less than the percentage error in the value of y.
64. Cardiac index To make different people comparable in studies of cardiac output, researchers divide the measured cardiac oUlput bY the body surface area to rmd the cardiac index C:
3
In Exercises 53 and 54, rmd parametric equations for the line that is tangent to the curve ofintersection of the surfaces at the given point 53. Surfaces: x 2 + 2y + 2z = 4, y = I
R:
63. Error in estimating a product Let y = uv and z where u and v are positive independent variables.
C
2' P o(l,2)
Tangent Unes to Curves
Point:
62. Muimum error in estimating the area of an ellipse If a ~ 10 em and b ~ 16 em to the nearest millimeter, what should you expect the maximmn percentage error to be in the calculated areaA ~ '1Tab of the ellipsex2/a 2 + y2/b 2 ~ I?
cardiac output body surface area .
~ ~-~-'-----
The body surface area B ofa person with weight w and height h is approximated bY the formula B
= 71.84wo.42Sho.72S,
which gives B in square centimeters when w is measured in kilograms and h in centimeters. Vou are about to calculate the cardiac index of a person 180 em tall, weighing 70 kg, with cardiac output of 7 L/min. Which will have a greater effect on the calculation, a I-kg error in measuring the weight or a I-em error in measuring the height?
Local Extrema Test the functions in Exercises 65-70 for local maxima and minima and saddle points. Find each function's value at these points. 65. f(x,y) = x 2 -:x:y + y2 + 2:x: + 2y - 4 66. f(x, y) 67. f(x,y)
= 5x 2 + 4:x:y - 2y 2 + 4x = 2:x:' + 3:x:y + 2y'
68. f(x,y) ~ x' 69. f(x,y) ~ x' 70. f(x, y) ~ x 4
4y
+ y' - 3:x:y + IS + y' + 3x2 - 3y 2 - 8:x;2 + 3y 2 - 6y
Absolute Extrema In Exercises 71-78, rmd the absolute maximmn and minimmn values of f on the region R. 71. f(x,y) = x 2 +:x:y + y2 - 3:x: + 3y R: The triangular region cut from the rmlt quadrant
bY the line
x+y~4
72. f(x,y) ~ x 2 - y2 - 2:x:
+ 4y +
I
R: The rectangular region in the rmlt quadrant bounded coordinate axes and the lines x = 4 and y = 2
73. f(x, y)
= y2 -
bY
the
:x:y - 3y + 2:x:
R: The square region enclosed bY the lines x 74. f(x,y) ~ 2:x: + 2y - x 2 - y2
~
±2 and y
~
±2
R: The square region bounded bY the coordinste axes and the lines x = 2, y = 2 in the rmlt quadrant 75. f(x,y) = x 2 - y2 - 2:x: + 4y R: The triangular region bounded below bY the x-axis, above the line y ~ x + 2, and on the right bY the line x ~ 2
bY
832
Chapter 14: Partial Derivatives
76. j(x,y) ~ 4xy - x' - y' + 16 R: The triangu1llr region bounded below by the line y = -2, above by the line y = x, and on the right by the line x = 2
93. If a and b are constants, ax + by, show that
u3
W =
+ tanh u +
cos U, and u =
77. j(x,y) ~ x 3 + y3 + 3x2 - 3y 2 R: The square region enclosed by the lines x 78. j(x,y) ~ x 3 + 3xy + y3 + I
~
R: The square region enclosed by the lines x
~
± I and y ± I and y
~
~
±I 94. Using the Chain Role lfw = In (x 2 + y2 + 2z), x = r + s, y = r - s, and z = 2r8, fmd Wr and w" by the Chain Rule.
±I
Then check your answer another way.
Lagrange MultipUers 79. Extrema on a circle Find the extreme values of j(x, y) x 3 + y2 on the circle x 2 + y2 = 1. 80. Extrema on a circle Find the extreme values of j(x, y) the circle x 2 + y2 = 1.
=
95. Angle between vectors The equations e" cos v - x = 0 and e" sin v - y ~ 0 def"me u and v as differentiable functions of x and y. Show that the angle between the vectors
= xy on
81. Extrema in a disk Find the extreme values of j(x, y) x 2 + 3y 2 + 2y on the unit disk x 2 + y2 '" I.
=
82. Extrema in a disk Find the extreme values of j(x, y) xl + y2 - 3x - xy on the diskx 2 + y2 ~ 9.
=
83. Extrema on a sphere Find the extreme values of j(x,y, z) x - y + zontheunitspherex 2 + y2 + z2 = 1.
=
OUi+OU j ax ay
is constant. 96. Polar coordinates and second derivatives Introducing polar coordinates x = r cos 8 and y = r sin 8 changes j(x, y) to g{r,8). Find the value of a2g/arf' at the point (r, 8) ~ (2, ",/2), given that
aj aj a2j a2j -=-=-=-=1 2 ax ay ax ay 2
84. Minimum distance to origin Find the points on the surface xl - zy = 4 closest to the origin.
85. Minjmjzjng cost of a bOI A closed rectangular box is to have volume V em3• The cost of the material used in the box is a cents/em2 for top and bottom, b cents/em2 for front and back, and c cents/cm2 for the remaining sides. What dimensions minimize the total cost of materials? 86. Leastvolume Find the plane x/a + y/b + z/e = I that passes through the point (2, I, 2) and cuts off the least volume from the f"lrSt octant.
87. EItrema on curve of intersecting surfaces Find the extreme valuesofj(x,y,z) = x(y + z) on the curve of intersection of the right circular cylinder x 2 + y2 = I and the hyperbolic cylinder xz=l. 88. Minimum. distance to origin on curve of intersecting plane and cone Find the point closest to the origin on the curve of intersection of the plane x + y + z = 1 and the cone z2 = 2x 2 + 2y 2. Partial Derivatives with Constrained Variables In Exercises 89 and 90, begin by drawing a diagram that shows the
relations among the variables. 89. Ifw ~ x 2.". andz ~ x 2 - y2 f"md
and
at that point. 97. Normal line parallel to a plane
(y + z)2 + (z - x?
L
(~~)p
~ (~~)T'
Theory and Examples 91. Let w ~ j(r, 8), r ~ v'x 2 + y2, and 8 ~ tan-I (y/x). Find aw/ax andaw/ay and express your answers in terms ofr and 8. 92. Let z = j(u, v), u = ax + by, and v = ax - by. Express Zx and Zy in terms of fu,lv, and the constants a andh.
= 16
where the normal line is parallel to the yz-plane. 98. Tangent plane parallel to xy-plane face
Find the points on the sur-
xy+yz+zx-x-z 2 =0 where the tangent plane is parallel to the xy-plaoe.
99. When gradient is parallel to position vector Suppose that Vj(x,y, z) is always parallel to the position vector xi + yj + zk. Show that j(O, 0, a) ~ j(O,O, -a) for any a. 100. One-sided directional derivative in all directions, but no gradient The one-sided directional derivative off at P (xo. Yo, zo) in the direction u = Uti + u2j + U3k is the number
lim
lexo + SUllYo + SU2,ZO + SU3)
3-0+
- !(Xo,Yo,zo)
S
•
Show that the one-sided directional derivative of
j(x,y, z) = v'x 2 90. Let U = j(P, V, T) be the internal energy of a gas that obeys the ideal gas law PV = nRT (n and R constant). Find
Find the points on the surface
+ y2 + z2
at the origin equals I in any direction but that j has no gradient vector at the origin. 101. Normal line through origin Show that the line normal to the surfacexy + z = 2 atthe point (I, I, I) passes through the origin. 102. Tangent plane and oormalline a. Sketch the surface x 2 - y2
+ z2
~
4.
~ Find a vector uormal to the surface at
(2, -3,3). Add the
vector to your sketch. c. Find equations for the tangent plane and normal line at
(2, -3,3).
Chapter 14 Additional and Advanced Exercises
Chapter
833
Additional and Advanced Exercises
Partial Derivatives
6. Surface in polar coordinates
1. Function with saddle at the origin If you did Exercise 60 in Section 14.2, you know that the function
sin6r
f(r,
(J) =
{
(x,y) "# (0,0) (x,y) = (0,0)
Let
r"#O
6r' 1,
r
0,
=
where r and (J are polar coordinates. Find
(see the accompanying figure) is continuous at (0, 0). Find fxy(O, 0) andfyx(O, 0).
a. lim f(r, (J) r--->O
b. fr(O,O)
c. fe(r, (J),
r"# O.
z
y
x
Gradients and Tangents
2. Finding a function from second partials Find a function w = f(x,y) whose first partial derivatives are aw/ax = 1 + eX cosy and aw/ay = 2y - eX siny and whose value at the point (In 2, 0) is In 2. 3. A proof of Leibniz's Rule Leibniz's Rule says that if f is continuous on [a, b] and if u(x) and v(x) are differentiable functions ofx whose values lie in [a, b], then V
d l (X) dv du dx f(t) dt = f(v(x» dx - f(u(x» dx . u(x)
7. Properties of position vectors r = Irl.
Let r = xi
+ yj + zk
and let
a. Show that Vr = r/r. b. Show that V(r n) = nr n- 2r. c. Find a function whose gradient equals r. d. Show that r • dr = r dr. e. Show that V(A· r) = A for any constant vector A.
8. Gradient orthogonal to tangent Suppose that a differentiable function f(x, y) has the constant value c along the differentiable curve x = g(t),y = h(t);thatis,
Prove the rule by setting
g(u, v)
=
l
f(g(t), h(t»
v
u
f(t) dt,
=
v
u(x),
=
v(x)
and calculating dg/ dx with the Chain Rule. 4. Finding a function with constrained second partials Suppose that f is a twice-differentiable function of r, that r = x 2 + y2 + z2 , and that
+ fyy + fzz
=
O.
c
for all values of t. Differentiate both sides ofthis equation with respect to t to show that Vf is orthogonal to the curve's tangent vector at every point on the curve.
9. Curve tangent to a surface
V
fxx
=
r(t) = (In t)i
Show that the curve
+ (t In t)j +
tk
is tangent to the surface
Show that for some constants a and b,
f(r)
=
xz 2
ra + b.
-
yz
+ cos xy
= 1
at (0,0, 1). 5. Homogeneous functions A function f(x, y) is homogeneous of degree n (n a nonnegative integer) if f(tx, ty) = tnf(x,y) for all t, x, and y. For such a function (sufficiently differentiable), prove that
af a. x ax
af
+ y ay
2f 2 b. x ( aax 2 )
+
= nf(x,y)
2f 2xy (a axiJy )
10. Curve tangent to a surface
Show that the curve
is tangent to the surface
2f + y2 (aay 2)
x 3 + y3 =
n(n - l)f.
at (0, -1, 1).
+ z3 - xyz
=
0
834
Chapter 14: Partial Derivatives
Extreme Values 11. Exlrema on a .urfa.. Show that the only po••ible maxima and minima of z on the surface z = x 3 + y3 - 9xy + 27 occur at (0, 0) and (3, 3). Show that neither a maximum nor a minimum occurs at (0, 0). Determine whether z has a maximum or a minimum at (3, 3).
12. Maximum in closed f"Irst quadrant Find the maximum value of I(x, y) = fuye -(2%+3y) in the closed first quadrant (includes the nonnegative axes). 13. Minimum volume cut from f"lI'St octant Find the minimum volume for a regioo boonded by the planes x = 0, y = 0, z = 0 2
2
£+L+~=I 2 2 2 a
b
c
a. Among all the possible directions tangentisI to the surface S
at a point in the fIrst octant
14. Minimnm distance from a line In a parabola in "",,"plane By minimizing the functioo I(x,y, u, v) = (x - uj' + (y - vj' subject to the coostraints y = x + I and u = v 2, fmd the minimum distance in the xy-plane from the line y ~ x + I to the parabola y2 = x.
Theory and Examples 15. Boundedne.. of f'rr.t partials implies continuity Prove the following theorero: If I(x,y) is defmed in an opeo regiooR of the xy-plane and if Ix and/y are bounded ooR, theo I(x,y) is cootinuous 00 R. (The assumptioo ofboundedoess is esseotia!.) 16. Suppose that r(l) = g(l)i + h(l)j + h(l)k is a smooth corve in the domain of a differeotiable functioo I(x, y, z). Describe the relatioo betweendl/dl, VI, and v = dr/dl. What can be said aboot VI and v at interior points of the corve where I has extreme values relative to its other values on the curve? Give reasons for your answer.
17. Finding function. from partial derivative. Suppose that and g are functions of x and y such that
al ay
ag ax
and
al
ag
ax
ay'
I
ax
=
0,
l(l, 2)
~
g(I, 2)
~
5
and
1(0, 0)
~
4.
Find I(x,y) andg(x,y). We koow that ifI(x,y) is a function oftwo variables and ifu = ai + hj is a unit vector, then Du/(x,y) = fx(x,y)a + I,(x,y)b is the rate of change of I(x,y) at (x,y) in the directioo ofu. Give a similar formula for the rate of change afthe rale afchange afl(x,y) at (x,y) in the directioo u.
18. Rate of change of the rate of change
at the point (0, 0, 10), which directioo will make the rate of change oftemperatore at (0, 0,10) a maximum? b. Which directioo tangential to S at the point (I, I, 8) will make the rate of change oftemperatore a maximum? 22. DriIIing another borehole On a flat surface ofiand, geologists drilled a borehole straight down and hit a mineral deposit at 1000 ft. They drilled a secood borehole 100 ft to the north of the fIrSt and hit the mineral deposit at 950 ft. A third borehole 100 ft east of the fIrSt borehole struck the mineral deposit at 1025 ft. The geologists have reasoos to believe that the mineral deposit is in the shape of a dome, and for the sake of econoroy, they would like to fmd where the deposit is closest to the surface. Assuming the surface to be the xy-plane, in what directioo from the first borehole would you suggest the geologists dri11 their foorth borehole? The one-
This equation is called the one-dimensional heat equation. The value of the positive coostant c 2 is detemtined by the material from which the wire is made.
and suppose that
al
20. Velocity after a ricochet A particle traveling in a straight line with coostant velocity i + j - 5k passes through the point (0, 0, 30) and hits the surface z = 2x 2 + 3y 2. The particle ricochets off the surface, the aog1e of reflectioo being eqoal to the angle of incidence. Assuming no loss of speed, what is the velocity of the particle after the ricochet? SimplilY your answer. 21. Directional derivative. tangent to a .urface Let S be the surface that is the graph of I(x, y) = 10 - x 2 - y2 . Suppose that the temperatore in space at each point (x, y, z) is 11:x,y,z) ~ x'y + y2z + 4x + 14y + z.
and a plane tangeot to the ellipsoid 2
19. Path of a heat-seeking particle A heat-seeking particle has the property that at any point (x,y) in the plane it moves in the directioo of maximum temperatore increase. !fthe teroperatore at (x,y) is 11:x,y) ~ -e-2y cos x, find an equatiooy ~ I(x) for the path ofa heat-seeking particle at the point (",/4, 0).
23. Find all solutions of the ooe-dimeosionsl heat equatioo of the form w = ert sin '7TX, where r is a constant. 24. Find all solutions of the ooe-dimeosiooa1 beat equatioo that have the form w = e" sin hx and satislY the conditions that w(O, I) = 0 and w(L, I) ~ O. What happeos to these solutions as 1---' oo?
Chapter 14
Chapter
Technology Application Projects
835
Technology Application Projects
Mathematica/Maple Module: Plotting Surfaces
Efficiently generate plots of surfaces, contours, and level curves. Exploring the Mathematics Behind Skateboarding: Analysis ofthe Directional Derivative The path of a skateboarder is introduced, Hrst on a level plaoe, then on a ramp, aod fmally on a paraboloid. Compute, plot, aod aoalyze the directional derivative in terms of the skateboarder. Lookingfor Patterns andApplying the Method ofLeast Squares to Real Data Fit a line to a set of numerical data points by choosing the line that minimizes the sum of the squares of the vertical illstaoces from the points to the line. Logrange Goes Skateboarding: HOHI High Does He Go? Revisit aod aoalyze the skateboarders' adventures for maximum aod minimum heights from both a graphical aod aoa1ytic perspective using Lagraoge multipliers.
15 MULTIPLE INTEGRALS OVERVIEW In this chapter we consider the integral of a function of two variables I(x, y) over a region in the plane and the integral of a function of three variables I(x, y, z) over a region in space. These multiple integrals are defined to be the limit of approximating Riemann sums, much like the single-variable integrals presented in Chapter 5. We illustrate several applications of multiple integrals, including calculations of volumes, areas in the plane, moments, and centers of mass.
Double and Iterated Integrals over Rectangles
15.1
In Chapter 5 we defined the definite integral of a continuous function I(x) over an interval [a, b] as a limit of Riemann sums. In this section we extend this idea to define the double integral of a continuous function of two variables I(x, y) over a bounded rectangle R in the plane. In both cases the integrals are limits of approximating Riemann sums. The Riemann sums for the integral of a single-variable function I(x) are obtained by partitioning a finite interval into thin subintervals, multiplying the width of each subinterval by the value of I at a point Ck inside that subinterval, and then adding together all the products. A similar method of partitioning, multiplying, and summing is used to construct double integrals.
Double Integrals We begin our investigation of double integrals by considering the simplest type of planar region, a rectangle. We consider a function I(x, y) defined on a rectangular region R, R:
d
~Ak /
fiYk
Ie-
C:::; y :::; d.
We subdivide R into small rectangles using a network of lines parallel to the x- and y-axes (Figure 15.1). The lines divide R into n rectangular pieces, where the number of such pieces n gets large as the width and height of each piece gets small. These rectangles form a partition of R. A small rectangular piece of width dx and height dy has area dA == dxdy. If we number the small pieces partitioning R in some order, then their areas are given by numbers dA 1, dA 2 , • •• , dAn, where dA k is the area of the kth small rectangle. To form a Riemann sum over R, we choose a point (Xk, Yk) in the kth small rectangle, multiply the value of I at that point by the area dAk, and add together the products:
Y R
a:::; x :::; b,
-(xk' Yk)
fixk
c --+--....L....--------------L...-+X
o
a
FIGURE 15.1 Rectangular grid partitioning the region R into small rectangles of area aA k = aXk aYk.
836
b
n
Sn
==
L I(Xk, Yk) dA k. k=l
Depending on how we pick (Xk, Yk) in the kth small rectangle, we may get different values for Sn.
15.1
Double and Iterated Integrals over Rectangles
837
We are interested in what happens to these Riemann sums as the widths and heights of all the small rectangles in the partition of R approach zero. The norm of a partition P, written Ilpll, is the largest width or height of any rectangle in the partition. If Ilpll == 0.1 then all the rectangles in the partition of R have width at most 0.1 and height at most 0.1. Sometimes the Riemann sums converge as the norm of P goes to zero, written Ilpll ~ o. The resulting limit is then written as n
L I(Xk, Yk) ilA k.
lim
IIPII~o k=l
As Ilpll ~ 0 and the rectangles get narrow and short, their number n increases, so we can also write this limit as n
lim
n~OO
L I(Xk, Yk) ilAk,
k=l
with the understanding that Ilpll ~ 0, and hence ilAk ~ 0, as n ~ 00. There are many choices involved in a limit of this kind. The collection of small rectangles is determined by the grid of vertical and horizontal lines that determine a rectangular partition of R. In each of the resulting small rectangles there is a choice of an arbitrary point (Xk, Yk) at which I is evaluated. These choices together determine a single Riemann sum. To form a limit, we repeat the whole process again and again, choosing partitions whose rectangle widths and heights both go to zero and whose number goes to infinity. When a limit of the sums Sn exists, giving the same limiting value no matter what choices are made, then the function I is said to be integrable and the limit is called the double integral of lover R, written as
11
f(x,y) dA
or
R
11
f(x, y) dx dy.
R
It can be shown that if I(x, y) is a continuous function throughout R, then I is integrable, as in the single-variable case discussed in Chapter 5. Many discontinuous functions are also integrable, including functions that are discontinuous only on a finite number of points or smooth curves. We leave the proof of these facts to a more advanced text.
Double Integrals as Volumes
z z = fix, y) ",--
• T a
b
x
f(xk' Yk)
i1
Y
'AA k
FIGURE 15.2 Approximating solids with rectangular boxes leads us to define the volumes of more general solids as double integrals. The volume of the solid shown here is the double integral of f(x, y) over the base region R.
When I(x, y) is a positive function over a rectangular region R in the xy-plane, we may interpret the double integral of lover R as the volume of the 3-dimensional solid region over thexy-plane bounded below by R and above by the surface z == I(x,y) (Figure 15.2). Each term I(Xk, Yk)ilA k in the sum Sn == ~ I(Xk, Yk)ilA k is the volume of a vertical rectangular box that approximates the volume of the portion of the solid that stands directly above the base ilAk. The sum Sn thus approximates what we want to call the total volume of the solid. We define this volume to be Volume = lim Sn =
n~OO
JJRff f(x,y) dA,
where ilAk~ 0 as n ~ 00. As you might expect, this more general method of calculating volume agrees with the methods in Chapter 6, but we do not prove this here. Figure 15.3 shows Riemann sum approximations to the volume becoming more accurate as the number n of boxes Increases.
838
Chapter 15: Multiple Integrals
(a)n_16
(e) n. 256
(b)n_64
FIGURE 15.3 As n increases, the Riemann 9UID approximations approach the total volume of the solid shown in Figure 15.2.
Fubini's Theo",m for Cllculating Double Integrals Suppose that we wish to calculate the volume 1Ulder the plane z - 4 - x - y over the rec4
tangular region R: 0 :s; x :s; 2, 0 :s; y :s; 1 in the xy-plane. If we apply the method of slicing from Section 6.1, with slices perpendicular to the x-axis (Figure 15.4), then the volume is
1'-' 1,-0'-' ,=0
A(x) dx,
(1)
where A(x) is the cross-sectional area at x. For each value of x, we may calculate A(x) as the integral
2
,~ A(x).
~
~
A(x)
1
y
J,'-' (4-x-y)dy , -0
FIGURE 15.4 To obtain the crosssectional areaA(x), we holdx f1XCd and integrate with respect toy.
(4 - x - y) dy,
(2)
which is the area under the curve z = 4 - x - Y in the plane of the cross-section at x. In calculatingA(x), x is held fIxed and the integration takes place with respect to y. Combining Equations (1) and (2), we see that the volume ofthe entire solid is Volume
~ 1'-'A(x) dx ~ 1'-'(1'-'(4 - x =
x-o
x-o
x-o
4y-xy-2
,-0
y)dy) dx
y'r,-0- 1'-' (7
1,·,[
1
[~x - 2 J: =
dx=
x-O
)
- - x dx 2
2
=
X
5.
(')
If we just wanted to write a formula for the volume, without carrying out any of the integrations, we could write
The expression on the right, cal1ed an Iterated or repeated Integral, says that the volume is obtained by integrating 4 - x - y with respect to y from y = 0 to Y = 1, holding x f'txed, and then integrating the resulting expression in x with respect to x from x = 0 to x = 2. The limits ofintegration 0 and I are associated withy, so they are placed on the integral closest to dy. The other limits of integration, 0 and 2, are associated with the variable x, so they are placed on the outside integral symbol that is paired with dx.
15.1
Double and Iterated Integrals over Rectangles
839
What would have happened if we had calculated the volume by slicing with planes perpendicular to the y-axis (Figure 15.5)? As a function of y, the typical cross-sectional
'""" is
A(y) ~
1,-,
[' ]'-'
(4 - x - y) d< ~ 4x - "- - xy 2
,,-0
",-0
~ 6 - 2y,
(4)
The volume ofthe entire solid is therefore
Volum, ~
rl
1
y=o
A(y) dy ~
1',-0-1
(6 - 2y) dy ~ [6y - y']: ~ 5,
in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing
FIGURE 15.5 To obtain the cross-sectional. area A(y), we hold y fDWd and integrate with respect to x.
112.(4 1
,
Volume =
x - y) duly.
The expression on the right says we can find the volume by integrating 4 - x - y with respect tax from x - 0 tax - 2 as in Equation (4) and integrating tberesult with respect to y from y - 0 to Y - 1. In this iterated integral, the order of integration is f11'S1: x and then y, the reverse ofthe order in Equation (3). What do these two volume calculations with iterated integrals have to do with the double integral
JJ(4 -
x - y) dA
R H:rsWRICAL BIOGRAPHY
Guido Fubini (1879-1943)
over the rectangle R: 0 ~ x ~ 2, 0 ~ y ~ I? The answer is that both iterated integrals give the value of the double integral. This is what we would reasonably expect, since the double integm1 measures the volume ofthe smne region as the two iterated integrals. A theorem. published in 1907 by Guido Fubini says that the double integral of any continuous function aver a rectangle can be calculated as an itemted integral in either order of integra1ion. (Fubini proved his theorem. in greater generality, but this is what it says in our setting.)
THEOREM l-Fubini's Theorem
JJ
f(x,y) dA
~
I'/.'
f(x,y) d
~
1.'1'
f(x,y) dy d<,
R
Fubini's Theorem. says that double integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time. Fubini's Theorem. also says that we may calculate the double integral by integrating in either order, a genuine convenience. When we calculate a volume by slicing, we may use either planes perpendicular to the x-axis or planes perpendicular to the y-axis.
EXAMPLE 1
Ca1<:ulate IfRf(x,y) dA for
f(x,y) ~ 100 -
6x'y
and
R:
O:Sx:S2,
-l:Sy:Sl.
840
Chapter 15: Multiple Integrals Solution
Figw-e 15.6 displays the volume beneath the surface. By Fubini's Theorem,
~ 1:(200 -
16y) dy
~
[lOlly - 8y'].'.,
~ 400.
Reversing the order of integration gives the same answer: x
1 ,
R
,
{' {\100 - 6<'y)dydY.
Jo J-l
~
3x'y'~::, dY.
('[lOlly -
10
FIGURE 15.6 The double integral f(x, y) dA. gives the volume under this surface aver the rectangular region R
~ /,\(100 -
(Example 1).
- 12200 dx - 400.
DR
3x') - (-100 - 3x')] dY. •
, EXAM PLE 2 z - 10 + x 2
, , x
Find the volume of the region bounded above by the ellipitical paraboloid I,O:S;y:s; 2.
+ 3y 2 andbelowbytherectang1eR:O:S; x:S;
Solution The surface and volume a:re shown in Figure 15.7. The volmne is given by the double integral
1~
if + -l'[
FIGUIE 15.7 The double inWgral DR f(x, y) dA gives the volume under this
V=
(10
11 1
x2
+ 3y 2)dA
=
2
(10
+ x 2 + 3y 2) dydx
R
smfaoe aver the rectangular region R (Example 2).
lOy + x'y + y'~:: dY.
=
Jor (20 + 2x
2
+ 8)dx =
+ ~X3 +
[2Ox
3
ax]l = 86,3 0
•
Exercises 15.1 EYalua~ng IteIiltMl IntegRls In Exercises 1-12, evaluate the iterated integral
1.
3.
J.'110f'
1-tl-t"['
2xydydY.
(x
+ y + l)dxdy
5. ].' [' (4 - y')dydY.
010
y 7. ].0 '].' 0 I +xydxdy
9. ].0 "'11[b' e2>:+y dy rb:
11.
f' (x 01-1
2. ].'
y)dydY.
].'].'(
4. 0 0
1-
f" (x'y 01-2
6. ].'
x' + ') dxdy ~
1-,']"."'
y sin;tduly
12·1'lfr1" (sinx + cosy)dxdy
EVllulling Double Integrals OVIr Rectangles In Exercises 13-20, evaluate the double integral. aver tlu: given regionR.
13. 2xy)dydY.
1/(6y2 -2l;)dA,
R: O:!;';x:!;'; I, 0 :!;';y:!"; 2
R
14.
jj
R
(v:) dA, Y
IS. jjxycosYdA, R
R: O:!"; x :s: 4, 1:S: Y :s: 2
R: -lsxSl, OSyS'1T
15.2 Double Integrals over General Regions
16.
JJ JJ JJ
y sin (x
+ y) dA,
R:
-11':S
x :s 0,
O:s y ::s
1T'
R
17.
e
X -
Y
dA,
D '" x '" In 2,
R:
D '" Y '" In 2
R
18.
xyexY' dA,
0 '" x '" 2, 0 '" Y '" I
R:
20.
JJRff x ,xy3+ I dA,
JJ X'y; +
Volume Beneath a Surface 23. Find the volume of the region bounded above by the paraboloid z = x' + y' and below by the square R: -I '" x '" I, -I "'y"'l. 24. Find the volume of the region bounded above by the ellipitical paraboloid z = 16 - x' - y' and below by the square R:O::S;x::S;2,O::S;y::s; 2. 25. Find the volume of the region bounded above by the plane z = 2 - x - y and below by the square R: 0 '" x '" I,
R
19.
841
0 '" x '" I,
R:
I dA,
R:
0 '" x '" I,
0 '" Y '" 2
0 '" Y '" I
R
In Exercises 21 and 22, integrate f over the given region.
21. Square f(x,y) = 1/(xy) over the square I '" x'" 2, I'" Y '" 2 22. Rectangle f(x,y) ~ y cosxy over the rectangle 0 '" x '" 'IT,
O"'y"'l
o '" Y '" I. 26. Find the volume of the region bounded above by the plane z = y/2 and below by the rectangle R: 0 '" x '" 4,0 '" Y '" 2. 27. Find the volwne of the region bounded above by the surface z = 2 sin x cos y and below by the rectangle R: 0 '" x '" 'IT/2, o'" Y '" 'IT/4. 28. Find the volwne of the region bounded above by the surface z = 4 - y' and below by the rectangle R: 0 '" x '" I, o'" Y '" 2.
z=f(x,y)
Double Integrals over General Regions
15.2
In this section we def'me and evaluate double integrals over bounded regions in the plane which are more general than rectangles. These double integrals are also evaluated as iterated integrals, with the main practical problem being that ofdetermining the limits ofintegration. Since the region of integration may have boundaries other than line segments parallel to the coordinate axes, the limits of integration often involve variables, not just constants.
Double Integrals over Bounded, Nonrectangular Regions
.....
R
~Yk
Ie
.u,
r-zx-t'Yk
/
FIGURE 15.8 A rectangular grid partitioning a bounded nonrectangular region into rectangu1ar cells.
To define the double integral of a function f(x, y) over a bounded, nonrectangular region R, such as the one in Figure 15.8, we again begin by covering R with a grid of small rectangular cells whose union contains all points of R. This time, however, we cannot exactly f'JII R with a f'mite number of rectangles lying inside R, since its boundary is curved, and some of the small rectangles in the grid lie partly outside R. A partition of R is formed by taking the rectangles that lie completely inside it, not using any that are either partly or completely outside. For commonly arising regions, more and more of R is included as the norm of a partition (the largest width or height of any rectangle used) approaches zero. Once we have a partition of R, we number the rectangles in some order from I to n and let aAk be the area ofthe kth rectangle. We then choose a point (Xk, YO) in the kth rectangle and form the Riemann sum
As the norm of the partition forming S. goes to zero, Ilpll --> 0, the width and height of each enclosed rectangle goes to zero and their number goes to inf'mity. If f(x, y) is a continuous function, then these Riemann sums converge to a limiting value, not dependent on any of the choices we made. This limit is called the double integral of f(x,y) over R:
±
lim f(X.,Yk) aA k = IIPII-o Fl
rr f(x,y) dA.
JJR
842
Chapter 15: Multiple Integrals
The nature of the boundary of R introduces issues not found in integrals over an interval. When R has a curved boundary, the n rectangles of a partition lie inside R but do not cover all of R. In order for a partition to approximate R well, the parts of R covered by small rectangles lying partly outside R must become negligible as the norm of the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made from polygons, circles, ellipses, and from continuous graphs over an interval, joined end to end. A curve with a "fractal" type of shape would be problematic, but such curves arise rarely in most applications. A careful discussion of which type of regions R can be used for computing double integrals is left to a more advanced text.
Volumes If f(x, y) is positive and continuous over R, we define the volume of the solid region betweenR and the surface z = f(x,y) to be f(x, y) dA, as before (Figure 15.9). If R is a region like the one shown in the xy-plane in Figure 15.10, bounded "above" and ''below'' by the curves y = g2(X) and y = gl(x) and on the sides by the lines x = a, x = b, we may again calculate the volume by the method of slicing. We first calculate the cross-sectional area
JJR
A(x) =
l
y =g2(X)
f(x,y) dy
y=g,(x)
and then integrate A(x) from x
=
a to x = b to get the volume as an iterated integral:
b
V=
A(x)dx= l a
lblg2(X) a
f(x,y)dydx.
(1)
g,(x)
x
R
Volume
=
lim !fiXk, Yk) aAk =
If f(x, y) dA R
FIGURE 15.9 We define the volumes of solids with curved bases as a limit of approximating rectangular boxes.
FIGURE 15.10 The area of the vertical slice shown here is A(x). To calculate the volume of the solid, we integrate this area from x = a to x = b:
l
a
b
A(x)dx =
lblg2(X) a
g,(x)
f(x,y)dydx.
Similarly, if R is a region like the one shown in Figure 15.11, bounded by the curves h2(y) and x == h1(y) and the lines y == c and y == d, then the volume calculated by slicing is given by the iterated integral
z
x z =f(x,y)
X
A
/
==
Volume
y
==
d:lh
1 C
/ /
2 (y)
f(x,y) dxdy.
(2)
hl(Y)
That the iterated integrals in Equations (1) and (2) both give the volume that we defined to be the double integral of f over R is a consequence of the following stronger form of Fubini's Theorem.
/ /
/'
843
Double Integrals over General Regions
15.2
/
/~
x = h 2(y)
FIGURE 15.11
The volume of the solid
shown here is
d
I
c
A(y)dy =
Id:lh2(Y) j(x,y)dxdy. c
h1(y)
For a given solid, Theorem 2 says we can calculate the volume as in Figure 15.10, or in the way shown here. Both calculations have the same result.
THEOREM 2-Fubini's Theorem (Stronger Form) regionR.
Let f(x, y) be continuous on a
1. If R is defined by a :::; x :::; b, gl (x) :::; y :::; g2(X), with gl and g2 continuous on [a, b], then
rrf(x, y) dA JJ
l Jr b
=
g2
R
f(x, y) dy dx.
(x)
g l(X)
a
2. If R is defined by c :::; y :::; d, h1(y) :::; x :::; h2(y), with hI and h2 continuous on [c, d], then
~~
f(x,y) dA
==
d:lh
1 C
R
2(y)
f(x,y) dxdy.
h 1(y)
EXAMPLE 1 Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y == x and x == 1 and whose top lies in the plane z ==
f(x,y)
==
3 - x - y.
SoLution See Figure 15.12. For any x between 0 and 1,y may vary from y (Figure 15.12b). Hence,
==
11[
l1
X (3-x-y)dydx==
1
V==
o
1 (
o
3
==
2
==
x
y=o
0
3x - ~2) dx
0 to y
y2]Y=X dx 3Y-XY-T
0
1
==
[3~ 2
-
2
£ 3 ]X= 1 2 x=o
==
1.
When the order of integration is reversed (Figure 15.12c), the integral for the volume is
V
=
1 11 1 1 (3 - ~ -
(3 - x - y) dx dy
o
3x - ~2 - xy ]X=l dy
0
x=y
2
1
=
11[
=
y
y - 3y
+ Y2 + y2) dy
= 11(%-4y+ty2)dY= The two integrals are equal, as they should be.
[%y-2y2+y;J::~
1. _
844
Chapter 15: Multiple Integrals z y
x=l
R
-O--tL----y-='----O---------------+ X (b)
y
x=l
x=y ,,~--......
R -O--fL------....I..-.------~x
y=x (c)
(a)
FIGURE 15.12 (a) Prism with a triangular base in the xy-plane. The volume of this prism is defined as a double integral over R. To evaluate it as an iterated integral, we may integrate first with respect to y and then with respect to x, or the other way around (Example 1). (b) Integration limits of
r=l (y=x f(x,y)
dydx.
}x=o }y=o If we integrate first with respect to y, we integrate along a vertical line through R and then integrate from left to right to include all the vertical lines in R. (c) Integration limits of
l
Y =l r=l f(x,y) dx dy.
y=o }x=y
If we integrate first with respect to x, we integrate along a horizontal line through R and then integrate from bottom to top to include all the horizontal lines in R.
Although Fubini's Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen.
EXAMPLE 2
Calculate
11 si~x
dA,
R
where R is the triangle in the xy-plane bounded by the x-axis, the line y x == 1.
==
x, and the line
15.2 Double Integrals over General Regions y x
845
Solution The region ofintegration is shown in Figure 15.13. Ifwe integrate rlISt with respect to y and then with respect to x, we rmd
=1
[
([
s~x dY ) dx =
&s~xIJ
[
= -cos (I)
R ~i"----!--->x
o
+
I
RJ
dx = [SinX dx
0.46.
If we reverse the order of integration and attempt to calculate
1. 11 o s~x dx dy,
1
FIGURE 15.13 The region ofintegration in Example 2.
Y
we run into a problem because !«sinx)/x) dx cannot be expressed in terms of elementary functions (there is no simple antiderivative). There is no general role for predicting which order of integration will be the good one in circumstances like these. If the order you rlISt choose doesn't work, try the other. Some_ times neither order will work, and then we need to use numerical approximations.
y
Finding Limits of Integration ~t--------'!-->x
Using Vertical Cross-sections When faced with evaluating lfR f(x,y)dA, integrating rlISt with respect to y and then with respect to x, do the following three steps:
(a)
y
/
We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works.
Leaves at y~~
1.
Sketch. Sketch the region ofintegration and label the bounding curves (Figure 15.14a).
2.
Find the y-limits ofintegration. Imagine a vertica1line L cutting through R in the direction of increasing y. Mark the y-vaiues where L enters and leaves. These are the y-limits of integration and are usually functions of x (instead of constants) (Figure
3.
Find the x-limits of integration. Choose x-limits that include all the vertical lines
Enters at y=l-x
15.14b).
o
x
throughR. The integral shown here (see Figure 15.14c) is
ff
(b)
f(x,y)dA =
E~IL~~ f(x,y)dydx.
R
y
Leaves at
/y~~ Enters at
y=l-x
Using Horizontal Cross-sections To evaluate the same double integral as an iterated integral with the order ofintegration reversed, use horizontailines instead of vertical lines in Steps 2 and 3 (see Figure 15.15). The integral is
JJrr f(x,y)dA R
1kr~ 1
=
0
f(x,y)dxdy.
y
~t--+-----'!--->x
l
x
/
Smallest x
Largest x
isx=O
isx=l
Largesty Y isy = 1
"1
Enters at
x=l-y
(c) Y-r--~~'K---->
FIGURE 15.14 Finding the limits of integration wben integrating ftrst with respect to y and then with respect to x.
Smallest y isy ~ 0
"
~ Leaves at x
~
v'l=Y'
~t--------'!--~x
o
FIGURE 15.15 Finding the limits of integration when integrating ftrst with respect to x and then with respect to y.
846
Chapter 15: Multiple Integrals
EXAMPLE 3
Sketch the region of integration for the integral
11"'(4x 2
+ 2)dydx
and write an equivalent integral with the order of integration reversed. The region of integration is given by the inequalities x 2 :5 y :5 2x and o :5 x :5 2. It is therefore the region bounded by the curves y = x 2 and y = 2x between x = 0 and x = 2 (Figure 15.16a). To find limits for integrating in the reverse order, we imagine a horizontal line passing from left to right through the region. It enters at x = y/2 and leaves at x = To include all such lines, we lety run from y = 0 to y = 4 (Figure 15.16b). The integral is Solution
...;y.
y
r rv,,(4x +
101Y/2
2) dx dy.
•
The common value ofthese integrals is 8.
Properties of Double Integrals Like single integrals, double integrals of continuous functions bave algebraic properties that are useful in computations and applications.
-;of"~--~----x
If f(x, y) and g(x, y) are continuous on the bounded region R, then the following properties hold
FIGURE 15.16 Region of integration for Example 3.
1. constantMUltiPle:ff cf(x,y)dA = cff f(x,y)dA R
(any number c)
R
2. Sum and Difference: ffU(X,y) ± g(x,y)) dA
ff f(x,y) dA ± ff g(x,y) dA
=
R
R
R
3. Domination: y
(a)
if
ff f(x,y) dA ;" 0
f(x,y) ;" 0 onR
R
(b)
ff f(x,y)dA;" ff g(x,y)dA R
o
II R
f(x, y) dA
~
II
f(x, y) dA
R1
II
+
f(x,y) ;" g(x,y) onR
R
4. Additivity: ff f(x,y) dA --cci-\---,/----------> X
if
R
=
ff f(x,y) dA + ff f(x,y) dA Rl
•
if R is the union of two nonoverlapping regions R, and R2 f(x, y) dA
Rz
FIGURE 15.17 The Additivity Property for rectangular regions holds for regions bounded by continuous curves.
Property 4 assumes that the region of integration R is decomposed into nonoverlapping regions R, and R2 with boundaries consisting of a finite number of line segments or smooth curves. Figure 15.17 illustrates an example of this property.
Double Integrals over General Regions
15.2
847
The idea behind these properties is that integrals behave like sums. If the function f(x, y) is replaced by its constant multiple cf(x, y), then a Riemann sum for f n
Sn
z
==
L f(Xk, Yk) ilAk k=l
is replaced by a Riemann sum for cf n
L k=l
Y
y
x
=
4x - 2
(a)
y
y
=
4x - 2
2
----;~----------'--~x
o
0.5
(b)
FIGURE 15.18 (a) The solid "wedgelike" region whose volume is found in Example 4. (b) The region of integration R showing the order dx dy.
n
C
f(Xk' Yk) ilA k
==
c
L f(Xk' Yk) ilA k == cSn. k=l
Taking limits as n ~ 00 shows that c limn~oo Sn == C ffR f dA and limn~oo cSn == ffR C f dA are equal. It follows that the constant multiple property carries over from sums to double integrals. The other properties are also easy to verify for Riemann sums, and carryover to double integrals for the same reason. While this discussion gives the idea, an actual proof that these properties hold requires a more careful analysis of how Riemann sums converge.
EXAMPLE 4
Find the volume of the wedgelike solid that lies beneath the surface z == 16 - x 2 - y 2 and above the region R bounded by the curve Y == 2\h, the line Y == 4x - 2, and the x-axis. SoLution Figure 15.18a shows the surface and the "wedgelike" solid whose volume we want to calculate. Figure 15.18b shows the region of integration in thexy-plane. Ifwe integrate in the order dy dx (first with respect to y and then with respect to x), two integrations will be required because y varies from y == 0 to y == 2VX for 0 :::; x :::; 0.5, and then varies from y == 4x - 2 to y == 2VX for 0.5 :::; x :::; 1. So we choose to integrate in the order dx dy, which requires only one double integral whose limits of integration are indicated in Figure 15.18b. The volume is then calculated as the iterated integral:
11
(16 - x 2 - y2) dA
R
2 1(Y+2)/4
==
1 o
1
2 [
==
o
1/4
x3 2]x=(Y+2)/4 16x - - - xy dx 3 x=y~4
{2 [
=
Jo
(16 - x 2 - y2)dxdy
4(y
191Y [24
+ 2) -
(y + 2)3 3 • 64
63y2
145y 3
(y
+ 2)y2 4
49y 4
- 4y 2
y5
y6
y7]2
+ ~ - ~ - 768 + 20 + 1344
0
4. 0
~
y
~
1, y
~
x
~
2y
5. 0
~
x
~
1,
eX
~
y
~
dy
20803 = 1680 ~ 12.4.
Exercises 15.2 Sketching Regions of Integration In Exercises 1-8, sketch the described regions of integration. 1. 0 ~ x ~ 3, 0 ~ y ~ 2x 2. -1 ~ x ~ 2, x - I ~ y ~ x 2 3. -2 ~ y ~ 2, y2 ~ X ~ 4
y4]
+ 3. 64 + 4
e
6. 1 ~ x ~ e 2 ,
0 ~ y ~ lnx
7. 0 ~ y ~ 1,
0 ~ x ~ sin- 1 y
•
848
Chapter 15: Multiple Integrals
Finding Limits of Integration In Exercises <}-18, write an iterated integral for lfR dA over the described region R using (a) vertical cross-sections, (b) horizontal cross-
sections.
29.1:1-'2 dp dv
(the pv-plaoe)
['{~ 30. Jo Jo 8t dt tis (the st-plaoe)
10.
9.
y
31.
1
~/'j,E" 3 cos t du dt
-13
32.
(the tu-plaoe)
0
j,o
'/2~4-2' 4
2
~dvdu
(theuv-plane)
v
I
Reversing the Drder of Integration -~.!L---+-_x
In Exercises 33--4Q, sketcb the region of integration and write an equivalent double integral with the order of integration reversed.
[' t-
12.
11.
2x
34. {' {'
dy
33. Jo J2
Jo }Y-2
['iVY
y
36.
35. Jo,
[' {,-
13. Bounded by y
= Vi, y = 0, and x = 9
14. Boundedbyy
~
lanx,x
~
O,andy
~
40. Jo Jo
1v'i=Y' 3y
= e~,y = I, and x = In 3 Bounded by Y = O,X = O,y = I, andy = Inx
44.
xy dy
43. JI Jo
15. Bounded by y 16.
17. Bounded by y
~
3 - 2>:,y
~
x, and x
18. Bounded by y ~ x 2 aody ~ x
~
°
+2
finding Regions of Integration and Double Integrals In Exercises 1<}-24, sketch the region of integration aod evaluate the integral.
21.
-v'k'
~/6L'/2
j,o
6x dy
xy2 dy
sinr
46. Jo
Jo
y';J
In Exercises 47-56, sketch the region of integration, reverse the order of integration, aod evaluate the integral.
47.j,~1~ s~y dy
48.j,2122y2SinXYdy
1 1
51.
itr'r' Jo e +Y dx dy
j,o
{'1/3 {'m,-"
45. j,'l" (x + y)
49.j, '
19.j,~j,x x siny dy
1v'k'
2
42.
{' {IU 1
J"
{' r-I' y
16x dy
1
41.
I-x
38. Jo
{'/2 {'-4x' 39. Jo Jo x=2
lll-X' dy
{1n2 {' dy
37. Jo JI
j,o
{2-.1m3 {-.1m3 ex'
Jo
X
50.j,2j,4-x' 4xe_2yy dy
x 2e""
52. {' {I
}yj2
Jo
lvQ3
.r dy
53. {1/16 {1/2 cos (I61rx')
[' p'
Jo
23. Jo Jo 3y'e"" :, x = 1, and x = 2 26. Triangle I(x, y) = x 2 + y2 over the triaogular region with vertices (0, 0), (1, 0), aod (0,1)
"I!rJ
27. Triangle I(u, v) ~ v over the triangular region cut from the fIrSt quadraot of the uv-plaoe by the line u + v = 1 28. Curved region I(s, t) = e' In t over the region in the fIrSt quadraot of the st-plane that lies above the curve s ~ In t from
t=Itot=2 Each of Exercises 2<}-32 gives an integral over a region in a Cartesiao coordinate plane. Sketch the region and evaluate the integral.
54.
),1/4
'h
j,o
2
~
dy
55. Square region lfR (y - 2>:2) dA where R is the region boundedbythesquarelxl + Iyl = 1 56. Triangular region lfRXY dA where R is the region bounded by thelinesy = x,y = 2x,andx + y = 2
Volume Beneath a Surface z
~
/(x, y)
57. Find the volume of the region bounded above by the paraboloid z ~ x 2 + y2 aod below by the triangle enclosed by the lines y = x,x = 0, and x + y = 2 in thexy-plane. 58. Find the volume of the solid that is bounded above by the cylinder z ~ x 2 aod below by the region enclosed by the parabola y = 2 - x 2 and the line y = x io the xy-plane.
15.2
59. Fmd the volume of the solid whose base is the region in the ;cyplane that is bolDlded hy the parabola Y ~ 4 - x' and the line Y = 3x, while the top of the solid is boIDlded hy the plane • = x + 4. 60. Fmd the volume of the oolid in the first oe1lInt bounded hy the eoonIinate planes, the cylinde< x' + Y' = 4, and the plane
• + Y = 3. 61. Fmd the volume of the oolid in the frn octant bounded hy the coordinate planes, the plane x = 3, IIIld the parabolie cylinder • = 4 _ y'. 62. Fmd the volume of the oolid cut from the fIr.lt octant hy the
surface. = 4 -
x' - y.
63. Fmd the volume of the wedge cut from the f1ISl octant hy the cylinder z = 12 - 3y' and the plane x + y ~ 2. 64. Fmd the volume of the oolid cut from the square eolumn Ixl + Iyl S 1 hytheplaneaz = oand 3x + z = 3. 65. Fmd the volume of the solid that is bounded on the front IIIld baek hy the planes x ~ 2 and x ~ I, on the sides hy the cylinden y = ± I/x, and above IIIld below hy the planes • = x + 1 and • = O.
66. Find the volume of the solid boIDlded on the front and back hy the planes x = ±",/3, on the sides hy the cylinders y = ±seex, above hy the cylinder z ~ I + y', and below hy the xy·plane. In Exercises 67 and 68, sketch the region of integration and the oolid whose volume is given hy the double integral. 67.
68.
t/.2-2'/3 (I - tx -t
y ) elydx
/.o
,!-,ft6-,' V25 -x' -
y 2 dxely
-Vt6-y1
Oouble Integrals over General Regions
849
73. f(x,y) = x + yover the region R bounded above hy the semicir· cle y = -vT=X2 and below hy the x-axis, using the partition x = -I, -1/2,0, 1/4, 1/2, I and y = 0, 1/2, I with (Xl,y,) the Jov",r left corner in the Ath subreetangle (provided the subreetangle lies within R)
74. f(x,y) = x + 2y over the region R inside the circle (x - 2)' + (y - 3)' = I using the partitionx = 1,3/2,2,5/2, 3 andy ~ 2, 5/2, 3, 7/2, 4 with (Xl,Yk) the center (centroid) in the Ath subrectangle (provided the subrectangle lies within R) Theory and Examples 75. Cin:ular sector
V4 - x' over the smaller x' + y2 :s 4 hy the rays 8 = ",/6 and
Integrate f(x, y) =
sector cut from the disk 8 = ",/2.
76. Unbounded regioo Integrate f(x,y) = 1/[(x2 - x)(y - 1f/3] over the infmite rectangle 2 S x < 00,0 :s y :s 2.
77. Nooeircular eyliDder A oolid right (ooocireular) cylioder has its base R in the ;cy-plane and is bounded above hy the paraboloid z = x' + y2. The cyliude<'s volume is 2 2 V = /.'/.'(x' + y2) dx ely + /. -,(x' + y') dx dy.
1
Sketch the base region R and express the cyliuder's volume as a single iterated integral with the order of integration reversed. Then evaluate the integral to fmd the volume. 78. Converting to a double Integral Evaluate the integral /''
tan-I
-
x) dx.
(Hint: Write the integrand as an integral.) 79. M.xjmjzj0g a dou.ble inte&ra1 maximizes the value of
What region R in the xy-plane
Integrals ewer Unbounded Reglo.. Improper double integrals can often be computed similarly to im· proper integrals of one variable. The f1ISl iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable hy taking appropriate limits, as in Section 8.7. Evaluate the improper integrals in Exercises 69-72 as iterated integrals. 69.
1"'/.' 1
70.
-1
-1/~
+
I)elydx
f'" r'" 1 dx ely . J-J- (x' + l)(y2 + I)
/."'1.'"
rninirnius the value of
dA.1
R
(2y
71
72.
R
Give reasons for your answer. 80. Minimizing. double integral What region R in the xy-plane jj(x 2 + y2 - 9)
-,-elydx I
e-' X y
i 'i'~
JJ(4 - x' - 2y') dA.1
xe-l>+") dx ely
Approxlmattng Integrals wltll Finite Sums In Exercises 73 and 74, approximate the double integral of f(x,y) over the region R partitioned hy the given vertiea1lines x = a and borizon· tal lines y = c.1n each subreetangle, use (XbYl) as indicated for your approximation.
Give reasons for your ansvwer. 81. Is it possible to evaluate the integral of a continuous limction fv<, y) over a rectangolar region in the xy·plane and get di1fen:nt answers depending on the order of integration1 Give reasons for your answer. 82 How would you evaluate the double integral of a contiouous fonetionfV<,y) over the regionR in the;cy-planeeoelased hy the trian· gle with vertices (0, 1), (2, 0), and 0, 2)1 Give reasons for your answer. 83. Unbounded region
" r'" .~,-" 1'('-oo.J-oo
Prove that dx ely =
lim
1'1' .~,-"
b_oo -b -b
dx ely
850
Chapter 15: Multiple Integrals
84. Improper double integral Evaluate the improper integral
].1].' x2 o
0
(y - 1)2/'
dydx
Use a CAS double-integral evaluator to fmd the integrals in Exercises 8~94. Thoo reverse the order of integratioo and evaluate, again wilh a CAS.
.
COMPUTER EXPLORATIONS Use a CAS double-integral evaluator to estimate the values of the integrals in Exercises 85-1l8.
[' [X
85. 87.
Jl Jl
].1].1
I xydydx
86.
tan-I xy dy dx
15.3
88.
].1].1
1
].11: eX'
91.
].o2~4v2Y (x"y - xy2) dx dy
dxdy
90. ] . ' [ X cos (y2) dydx
y'
e--(x'+y') dy dx
2
93.
1] . \I'i=:;>
-1
89.
3\h -
x 2 - y2dydx
1:1.
92.
].0
94.
1
X '
_+1 dydx Y
lOX
1
2].4-Y' 0
2~8 y'
eX)!
dx dy
I
Yx2 + y2 dxdy
0
Area by Double Integration In this section we show how to use double integrals to calculate the areas of bounded regions in the plane, and to find the average value of a function oftwo variables.
Areas of Bounded Regions in the Plane If we take !(x, y) = I in the defmition of the double integral over a region R in the preceding section, the Riemarm sums reduce to
S. =
• • L !(XhYk) aAk = L aAk. 1-1 i-I
(1)
TIris is simply the swn of the areas of the small rectangles in the partition of R, and approximates wbat we would like to call the area of R. As the norm of a partition of R approaches zero, the height and width ofall rectangles in the partition approach zero, and the coverage of R becomes increasingly complete (Figure 15.8). We define the area of R to be the limit lim . 1IJ1I~o~
DEFINmON
aAk =
fJif dA.
(2)
R
The area of a closed, bounded plane region R is
A =
11
dA.
R
As with the other defmitions in this cbapter, the definition here applies to a greater variety of regions than does the earlier single-variable definition of area, but it agrees with the earlier defmition on regions to which they both apply. To evaluate the integral in the definition of area, we integrate the constant function !(x, y) = I over R.
EXAMPLE 1 quadrant.
Find the area of the region R bounded by Y = x and Y = x 2 in the frrst
15.3
851
Area by Double Integration
Solution
y
We sketch the region (Figure 15.19), noting where the two curves intersect at the origin and (I, I), and calealate the area as
1
A = = ~'--""'-----+---~l-+X
FIGURE 15.19
The region in Example 1.
10f' (x -
x
2 )
=
[~]:ax
ax = [; 2-
3]'0
~
=
I 6'
ax,
Notice that the single-variable integral Jo'(x - x2 ) obtained from evaluating the inside iterated integral, is the integral for the area between these two curves using the method of Section 5.6. • Find the area ofthe regionR enclosed by the parabolay = x 2 and the line
EXAMPLE 2 y = x + 2. y
!oJ dyax
Solution
Ifwe divide R into the regions R, and R2 shown in Figure 15.200, we may calculate the area as
On the other hand, reversing the order ofintegmtion (Figure 15.20b) gives A =
----=~~~X o
2
{X+2 dy ax.
This second result, which requires only one integral, is simpler and is the only one we would bother to write down in practice. The area is 2
y
1-,J..
A =
1 -,
[Y
]X+2 ax x'
1
2
=
-,
(x + 2 - x 2 )
ax =
•
Average Value
------'~""_1----x
o
(b)
FIGURE 15.20 Calculating this area takes
The avemge value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length ofthe interval. For an integrable function oftwo variables defmed on a bounded region in the plane, the average value is the integral over the region divided by the area of the region. This can be visualized by thinking of the function as giving the height at one instant of some water sloshing around in a tank whose vertical walls lie over the boundary of the region. The average height of the water in the tank can be found by letting the water settle down to a constant height. The height is then equal to the volume of water in the tank divided by the area of R. We are led to define the average value of an integrable function f over a region R as follows:
AveragevaIueof/overR = arealofRjj IdA.
(3)
R
If I is the temperature of a thin plate covering R, then the double integral of f over R divided by the area of R is the plate's average temperature. If I(x, y) is the distance from the point (x, y) to a flXed point P, then the average value oflover R is the average distance of points inR fromP.
852
Chapter 15: Multiple Integrals
EXAMPLE 3 R: 0 :5 x :5
Find the average value of f(x,y) = xcosxy over the rectsngle 'IT,
0
:5
Y :5 1.
The value of the integral of f over R is
Solution
1"1' o
xcosxy dydx =
0
1"[
sinxy
0
dx ]Y-l y=o
= l"(SinX -
f
xcosxydy
I
0) dx = -cosx
=
= sinxy + C
I
+I
= 2.
The area of R is 7f. The average value of f over R is 2/7f.
•
Exercises 15.3 Area by Double Integrals In Exercises 1-12, sketch the region bounded by the given lines and curves. Then express the region's area as an iterated double integral and evaluate the integral.
1. The coordinate axes and the line x
+y
= 2
2. The lines x = O,y = 2x,andy = 4
3. The parabola x ~ _y2 and the line y ~ x
+2
4. The parabola x = y - y2 and the line y = -x 5. The corve y = eX and the lines y = 0, x = 0, and x = In 2 6. Thecorvesy = loxandy = 210xandthelinex = e,inthefIrSt quadrant
Find;ng Average Values 19. Find the average value of f(x, y) = sin (x
+ y) over a. the rectangle 0 ::s x :s 1r. O:s y :s 1r. b. the rectangle 0 ,;; x ,;; 7f, 0,;; y ,;; 7f/2.
20. Which do you think will be larger, the average value of f(x,y) = xy over the square 0 ,;; x ,;; 1,0 ,;; Y ,;; I, or the average value of f over the quarter circle x 2 + y2 ,;; I in the first quadrant? Calculate them to fmd out 21. Find the average height of the paraboloid z = x 2 + y2 over the square 0 :s x :s 2, 0 :s y :s 2. 22. Find the average value of f(x,y) ~ I/(xy) over the square 102,;; x,;; 2102,102';; y,;; 2102.
7. The parabolas x = y2 and x = 2y _ y2 Theory and Examples
8. The parabolas x ~ y2 - I and x ~ 2y 2 - 2 9. The linesy = x,y =
x/3, andy
= 2
10. The lines y = I - x and y = 2 and the curve y = eX 11. Thelinesy = 2x,y = x/2,andy = 3 - x 12. The lines y ~ x - 2 and y ~ -x and the corve y ~
VX
Identlfytng the Region of Integration The integrals and sums of integrals in Exercises 13-18 give the areas of regions in the xy-plane. Sketch each region, label each bounding curve with its equ.ati~ and give the coordinates of the points where the corves intersect Then find the area ofthe region.
II
15.
17.
/, 61 o
/,o
K
dxdy
y'J./3
,,/41.
00
0%
16.
0
-2;(
/, 2:£0
Ox 2 -4
-x Y+2
dydx
01'-X dydx + /,2:£'-X dydx
1
/,31X(2-X) dydx 0
sinx
-1
18.
2Y
dydx
+
-x/2
/,4:£Vx dydx 00
- y. 1~
dx dy
23. Bacterium population If f(x,y) = (IO,OOOe Y)/(I + IxV2) represents the "population density" of a certain bactetiU03 on the xy-plane, where x and y are measured in centimeters, fmd the total population of bactetia within the rectangle -5 ,;; x ,;; 5 and -2 ,;; Y ,;; O. 24. Regional population If f(x, y) ~ 100 (y + I) represents the population density of a planar region on Earth, where x and y are measmed in miles, find the number of people in the region bounded by the corvesx = y2 and x = 2y _ y2. 25. Average temperature in TeIas According to the Texas Almanac, Texas has 254 counties and a National Weather Service station in each county. Assume that at time to, each of the 254 weather stations recorded the local temperature. Find a formula that would give a reasonable approximation of the average temperature in Texas at time to. Your answer should involve information that you would expect to be readily available in the Texas Almanac. 26. If Y ~ f(x) is a nonnegative continuous function over the closed interval a ,;; x ,;; b, show that the double integral defInition of area for the closed plane region bounded by the graph of f, the vertical lines x = a and x = b, and the x-axis agrees with the defmition for area beneath the curve in Section 5.3.
15.4
15.4
853
Double Integrals in Polar Form
Ir--D_ou_b_Le_I_n_te-=9;....ra_Ls_i_n_p_oL_a_r_Fo_rm
_
Integrals are sometimes easier to evaluate if we change to polar coordinates. This section shows how to accomplish the change and how to evaluate integrals over regions whose boundaries are given by polar equations.
Integrals in Polar Coordinates When we dermed the double integral of a function over a region R in the xy-plane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. These were the natoralshapes to use because their sides have either constant x-values or constant y-values. In polar coordinates, the natural shape is a ''polar rectangle" whose sides have constant r- and 6-values. Suppose that a function fir, 0) is dermed over a region R that is bounded by the rays = a and 0 = {3 and by the continuous curves r = gt
o
r=a
8
~ 'If--"---------------'=----------~8 ~
o
0
FIGURE 15.21 TheregionR:g,(8):5 r s g2(8),a s 8 s ,8,is contained in the fanshaped region Q: 0 :5 r :5 a, a :5 8 :5 ,8. The partition of Q by circular arcs and rays induces a partition of R.
We cover Q by a grid of circular arcs and rays. The arcs are cut from circles centered at the origin, with radii !J.r, 2!J.r, ... , m!J.r, where !J.r = aim. The rays are given by
o=
a,
0
=
a
+ !J.O,
0
=
a
+ 2!J.0,
... ,
0
=
a
+ m'!J.O
= {3,
where !J.O = ({3 - a)lm'. The arcs and rays partition Q into small patches called ''polar rectangles." We number the polar rectangles that lie inside R (the order does not matter), calling their areas !J.A" !J.A 2, ... , !J.A•. We let (rk, Ok) be any point in the polar rectangle whose area is !J.Ak. We then form the sum
•
S. = ~. f( rh Ok) !J.Ak. ~
If f is continuous throughout R, this sum will approach a limit as we refine the grid to make !J.r and !J.O go to zero. The limit is called the double integral of f over R. In symbols, lim S. =
n~OO
JJRrr fir, 0) dA.
854
Chapter 15: Multiple Integrals
To evaluate this limit, we first have to write the sum S. in a way that expresses aA. in terms of!i.r and!i.lI. For convenience we choose r. to be the average of the radii ofthe inner and outer arcs bounding the kth polar rectangle aA•. The radius of the inner arc bounding aA. is then r. - (!i.r/2) (Figure 15.22). The radius of the outer arc is
r.
+ (!i.r/2). The area of a wedge-shaped sector of a circle having radius r and angle II is
A = Large sector
o FIGURE 15.22 aA k =
The observation lbat
1 11 • r 2 2
'
as can be seen by multiplying 7Tr 2, the area of the circle, by 1I/27T, the fraction of the circle's area contained in the wedge. So the areas of the circular sectors subtended by these arcs at the origin are
-
Car::e~) (m:::~r)
leads to lbe formula aA, = r, !i.r !i.O.
I(r. - --f
!i.)2 !i.Ii
Inner mdius:
2
Outer radius:
!i.r)2 !i.1I. 2I ( r. + T
Therefore,
aA.
= area oflarge sector - area of small sector =
y
~II [ (r. + ~r)' -
(r. -
~)'] = ~II (2r. !i.r) =
r. ar!i.li.
Combining this result with the sum defming S. gives
•
S. = ~ i(r., lI.h ar !i.Ii.
~I------~x
As n ---> integral
00
and the values of ar and all approach zero, these sums converge to the double
o
lim S. =
(a)
n_OO
rr
JJ R
y
A version of Fubini's Theorem says that the limit approached by these sums can be evaluated by repeated single integmtions with respect to r and II as
2 "in 8 ~ y ~
i(r,lI) r dr dll.
v2 f-....,.f-~
or Enters at T
r=V2csc9
=
v2 esc 8
8 --:cf-"---------> x
o
Largest 6 is
I.
L
2
/
// Y
=x
V21--+-....:/~/ / / /
/
R
l
'~Pl,~g,J')
8=a
i(r,lI) r dr dll.
r=g,(8)
Finding Limits of Integration The procedure for finding limits of integration in rectangular coordinates also works for polar coordinates. To evaluate lfR i(r, II) dA over a region R in polar coordinates, integmting fITst with respect to r and then with respect to II, take the following steps.
(b) y /
fJif
i(r,lI) dA =
~ Smallest 6 is ~.
1. Sketch. Sketch the region and label the bounding curves (Figure 15.23a). 2. Find the r-limits ofintegration. Imagine a my L from the origin cutting through R in the direction ofincreasing r. Mark the r-values where L enters and leaves R. These are the r-limits of integmtion. They usually depend on the angle II that L makes with the positive x-axis (Figure 15.23b).
/
--:cf---"--------> x
o
(c)
FIGURE 15.23 Finding !be limits of integration in polar coordinates.
3. Find the II-limits of integration. Find the smallest and largest II-values that bound R. These are the II-limits of integmtion (Figure 15.23c). The polar iterated integral is
fJif R
i(r, II) dA =
1
'~"/21'~2
9=7f/4
r=V2csc6
i(r,lI) r dr dll.
15.4
Double Integrals in Polar Form
855
EXAMPLE 1 Find the limits ofintegration for integrating fir, 0) over the region R that lies inside the cardioid r = I + cos 0 and outside the circle r = 1.
y
Solutton
O=-I
L
Enters
at
Leaves at
r=I+0089
1. We flIst sketch the region and label the bounding curves (Figure 15.24). 2. Next we fmd the r-limits ofintegration. A typical ray from the origin enters R where r = I and leaves where r = I + cos 0. 3. Finally we find the O-limits ofintegration. The rays from the origin that intersectR run from 0 = -",/2 to 0 = ",/2. The integral is
1
r=1
-fTj2
FIGURE 15.24 Finding the limits of
integratioo in polar coordinates for the regioo in Example I.
~/211+'os,
•
fir, 0) r dr dO.
1
If fir, 0) is the constant function whose value is I, then the integral of f over R is the areaofR.
Area In Polar Coordinates The area of a closed and bounded region R in the polar coordinate plane is
I
Area Differential in Polar Coordinates dA = rdrd6
A = IJrdrdO. R
This fonnula for area is consistent with all earlier fonnulas, although we do not prove this fact. y
Leaves at r~
!.
v. cas 2e
--l---~~---+--_x
Enters at r= 0
r 2 =400829
FIGURE 15.25 To integrate over the shaded regioo, we run r from 0 to
y'4 cos 26 and 6 from 0 to ",/4 (Example 2).
EXAMPLE 2
Find the area enclosed by the lemniscate r 2
=
4 cos 20.
Solutton We graph the lemniscate to determine the limits ofintegration (Figure 15.25) and see from the symmetry ofthe region that the tota1 area is 4 times the first-quadrant portion.
J.
A=4
~/4foV4,"S26
rdrdO=4
J.~/4 [r2]r~V4'OS2'
00
=
J.
4
0
~/4
2 cos 20 dO
2
0
=
4 sin 20
]~/4 0
dO
r=O
=
4.
•
Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integralffR fix, y) dx dy into a polar integral has twc steps. First substitute x = r cos 0 and y = r sin 0, and replace dx dy by r dr dO in the
Cartesian integral. Then supply polar limits of integration for the boundary of R. The Cartesian integral then becomes I I fix, y) dx dy = I I fir cos 0, r sin 0) r dr dO, R
G
where G denotes the same region ofintegration now described in polar coordinates. This is like the substitution method in Chapter 5 except that there are now twc variables to substitute for instead of one. Notice that the area differential dx dy is not replaced by dr dO but by r dr dO. A more general discussion of changes of variables (substitutions) in multiple integrals is given in Section 15.8.
856
Chapter 15: Multiple Integrals
EXAMPLE 3
Evaluate
11
x2 y2
e + dy dx,
R
where R is the semicircular region bounded by the x-axis and the curve y (Figure 15.26).
==
~
SoLution In Cartesian coordinates, the integral in question is a nonelementary integral x2 y2 and there is no direct way to integrate e + with respect to either x or y. Yet this integral and others like it are important in mathematics-in statistics, for example-and we need to find a way to evaluate it. Polar coordinates save the day. Substituting x == r cos 0, y == r sin 0 and replacing dy dx by r dr dO enables us to evaluate the integral as
y
(}=o
(} = 11'
1
o
-1
x
(7T
FIGURE 15.26 The semicircular region in Example 3 is the region
o~
r
~
1,
0
~ (J ~ 7T.
=
1 Jo 2(e -
l)dO
7T
=
2(e -
I).
r2
The r in the r dr dO was just what we needed to integrate e • Without it, we would have been unable to find an antiderivative for the first (innermost) iterated integral. _
EXAMPLE 4
SoLution
Evaluate the integral
Integration with respect to y gives
an integral difficult to evaluate without tables. Things go better if we change the original integral to polar coordinates. The region of integration in Cartesian coordinates is given by the inequalites 0 :::; y :::; ~ and o :::; x :::; 1, which correspond to the interior of the unit quarter circle x 2 + y2 == 1 in the first quadrant. (See Figure 15.26, first quadrant.) Substituting the polar coordinates x == rcosO,y == rsinO,O:::; 0:::; 7T/2andO:::; r:::; 1, and replacing dx dybyrdrdO in the double integral, we get
z
{I
{~
Jo Jo
(x
2
{7T/2
+ y2) dy dx
=
Jo
{I
Jo
4Jr= 1 _17T/2 1 _ dO 4 dO r=O 0
_17T/2 [r 4 o
2 (r ) r dr dO 7T 8·
Why is the polar coordinate transformation so effective here? One reason is that x 2 + y2 simplifies to r 2 • Another is that the limits of integration become constants. _
EXAMPLE 5 Find the volume of the solid region bounded above by the paraboloid == 9 - x 2 - y2 and below by the unit circle in the xy-plane.
z x
FIGURE 15.27 Example 5.
The region of integration R is the unit circle x 2 + y2 == 1, which is described in polar coordinates by r == 1, 0 :::; 0 :::; 27T. The solid region is shown in Figure 15.27. The volume is given by the double integral SoLution
The solid region in
15.4
Double Integrals in Polar Form
857
• EXAMPLE 6 Using polar integration, find the area of the regionR in the xy-plane enclosed by the circle x 2 + y2 + 4, above the line y = 1, and below the line y = 0x. Solutton A sketch of the region R is showo in Figure 15.28. First we note that the lioe y = 0x has slope 0 = tan 0, so 0 = 7T/3. Next we observe that the line y = 1 intersects the circle x 2 + y2 = 4 when x 2 + 1 = 4, or x = 0. Moreover, the radial line from the origin through the point (0,1) has slope 1/0 = tan 0, giving its angle ofinclination as 0 = 7T/6. This information is showo in Figure 15.28. Now, for the regionR, as 0 varies from 7T/6 to 7T/3, the polar coordinate r varies from the horizontal line y = 1 to the circle x 2 + y2 = 4. Substitutiog r sin 0 for Y in the equation for the horizontallioe, we have r sin 0 = 1, or r = esc 0, which is the polar equation of the line. The polar equation for the circle is r = 2. So in polar coordinates, for 7T/6 :5 0 :5 7T/3, r varies from r = esc 0 to r = 2. It follows that the iterated integral for the area then gives
y
--=-I"'--""---Ll-----!-~x
o
2
ff dA JJR
FIGURE 15.28 The region R in Example 6.
=
f"
Jrr/6
-
1 2
/3
1 [1 ,,/3
'1rj6
rdrdO
esc (J
-r 2 2
]'~2
dO
r-csc9
,,/3 1
=
1
=
Z1 [46 + cot 0],,/ "/6
=
1 (47T +
,,/6
Z[4-csc2 0]dO 3
_1_) _
1(47T + 23026
0)
= 7T -
Exercises 15.4 Regions In Polar Coordinates 10 Exercises 1--1l, describe the given region io polar coordinates.
L
3.
y
1 y
--:------::'1'----:---+ x -1 0 1 -+---+---,-~x
9
----+-'I--+-~x
o
4
4.
0
3'
•
858
Chapter 15: Multiple Integrals Area in Polar Coordinates
6.
5. y
27. Find the area ofthe region cut from the fIrst quadrant by the curve r ~ 2(2 - sin 26)1/2.
y
28. Cardioid overlapping a clrcle Find the area of the region that lies inside the cardioid r = 1 + cos 0 and outside the circle r = 1.
2f------,
29. One leaf of a rose r = 12 cos 36.
30. Snall shen Find the area of the region enclosed by the positive x-axis and spiral r = 46/3,0 '" 6 '" 2". The region looks like a anall shell.
2
7. The region enclosed by the circle x 2 + y2
= lx.
2
8. Theregionenclosedbythesemicirclex + y2 = 2y,y '" O.
In Exercises 9-22, change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.
1 1].~
dy dx
-1
n 13.
~
18.
19.
20.
22.
(x 2 + y2) dx dy
~
dydx
1
1°1° 1 11~
].2].' j2Vi j'v4=Y'
-~ 1+ Yx 2 + y2
-I
2 dydx -~ (I + x 2 + y2f
ln2:£V(1n2)'-"
Y dy dx
34. Average height ofa cone Find the average height of the (single)
dxdy
• r>c>
e vx'+,' dx dy
In(x 2 +y2+I)dxdy
].'ly'i::;>
(x
0,
2].~
°
24.
].° ° r
1
25. ~
35. Average distance from interior of disk to center Find the average distance from a point P(x, y) in the disk x 2 + y2 '" a 2 to
37. Converting to a polar integral Integrate f(x,y) = [In (x 2 + y2)]/yx 2 + y2 over the region I '" x 2 + y2 '" e. 38. Converting to a polar integral Integrate f(x, y) = [In (x 2 + y2)1/(x 2 + y2) over the region I '" x 2 + y2 '" e 2.
+ 2y)dydx
39. Volume of noncircular right cylinder The region that lies inside the cardioid r = 1 + cos 8 and outside the circle r = 1 is the base of a solid right cylinder. The top of the cylioder lies in the
I dydx (x 2 + y2)2
plane z = x. Find the cylinder's volume.
3
whose top is bouoded by the sphere z = ~. Find the 41. Converting to polar integrals
a. The uaual way to evaluate the improper integral 00 I = e --;f1 dx is frrst to calculate its square:
10
I
~/4 ].2"" ].° °
40. Volume of noncircular right cylinder The region enclosed by the lemniscate r 2 ~ 2 cos 26 is the base of a solid right cylinder
cylinder's volume. sin6eos6drd6
~/21=' r 2 cos 6 dr d6
,"/6
eonez ~ Yx 2 + y2 above the diskx 2 + y2 '" a 2 inthexy-plane.
its houndary Find the average value of the square of the distance from the point p(x, y) in the disk x 2 + y2 '" I to the bouodary pointA(I, 0).
In Exercises 23-26, sketch the region of integration and convert each polar integral or sum of integrals to a Cartesian integral or sum of integrals. Do not evaluate the integrals. 23.
Find the average height of x 2 - y2 above the disk
Ya 2 -
Theory and Examples
-v'I=Yl
~/2].1
the hemispherical surface z = x 2 + y2 ~ a 2 in the xy-plane.
the origin.
I 1v'i=Y' I
f(r, 6) r dr d6.
R
36. Average distance squared from a point in a disk to a point in
].
1
JJ
33. Average height of a hemisphere
dydx
-I
°1 °
Are~(R)
0
~ -V",-x'
14.
2
Average values In polar coordinates, the average valu.e of a function over a region R (Section 15.3) is given by
u lal~ dydx
x dx dy
1°1'
-1
21.
].1].v'i=Y' 0
W+y~dxdy ].°2:£v4=Y' °
I
17.
10.
0
].6].'
31. Cardioid in the rrrst qnadrant Find the area of the region cut from the frrst quadrant by the cardioid r = I + sin 6. 32. Overlapping cardioids Find the area of the region connnon to the interiors of the cardioids r ~ I + cos 6 and r ~ I - cos 6.
Evaluating Polar Integrals
9.
Find the area enclosed by one leaf ofthe rose
r' sin2 6 drd6
'"''-'1].3'''''' ~dr~+ 1~/2 ].o 0 tim-It
].4=' 0
~dr~
Evaluate the last integral uaing polar coordinates and solve the resulting equation for 1.
Triple Integrals in Rectangular Coordinates
15.5 b. Evaluate x
lim erf(x) = lim x-HX)
42. Converting to a polar integral
1001 o
1
x-HX)
0
00
0
2 -t 2 ~ e/ dt.
46. Area Suppose that the area of a region in the polar coordinate plane is A
V 7T
=
31T/4:l2 sinO
1 1T/4
Evaluate the integral
1- - d x dYe ---(1 + x 2 + y2)2
43. Existence Integrate the function f(x, y) = 1/( 1 - x 2 - y2) over the disk x 2 + y2 ~ 3/4. Does the integral of f(x, y) over the disk x 2 + y2 ~ 1 exist? Give reasons for your answer.
859
rdrdO. esc 0
Sketch the region and find its area. COMPUTER EXPLORATIONS In Exercises 47-50, use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise.
a. Plot the Cartesian region of integration in the xy-plane. b. Change each boundary curve ofthe Cartesian region in part (a) to its polar representation by solving its Cartesian equation for r and o.
44. Area formula in polar coordinates Use the double integral in polar coordinates to derive the formula
c. Using the results in part (b), plot the polar region of integration in the rO-plane.
d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility.
for the area of the fan-shaped region between the origin and polar curve r = f(O), ex ~ 0 ~ {3. 45. Average distance to a given point inside a disk Let Po be a point inside a circle of radius a and let h denote the distance from Po to the center of the circle. Let d denote the distance from an arbitrary point P to Po. Find the average value of d 2 over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and Po on the x-axis.)
15.5
47.
ee~dydx Jo Jx + x
1l 1
49.
48.
e r/ Jo Jo
50.
Jo
y
Y 3 /
Y
o -y/3 v'x 2 + y2
dxdy
2
+---zdydx x +Y
f112-Y y
~dxdy
Triple Integrals in Rectangular Coordinates Just as double integrals allow us to deal with more general situations than could be handled by single integrals, triple integrals enable us to solve still more general problems. We use triple integrals to calculate the volumes of three-dimensional shapes and the average value of a function over a three-dimensional region. Triple integrals also arise in the study of vector fields and fluid flow in three dimensions, as we will see in Chapter 16.
Triple Integrals
z
j
If F(x, Y, z) is a function defined on a closed, bounded region D in space, such as the region occupied by a solid ball or a lump of clay, then the integral of F over D may be defined in the following way. We partition a rectangular boxlike region containing D into rectangular cells by planes parallel to the coordinate axes (Figure 15.29). We number the cells that lie completely inside D from 1 to n in some order, the kth cell having dimensions dXk by dYk by dZk and volume d Vk == dXkdYkdzk. We choose a point (Xk, Yk, Zk) in each cell and form the sum
D
x
/ ~y
n
Sn
==
L F(Xk, Yk, Zk) d Vk.
(1)
k=l
FIGURE 15.29 Partitioning a solid with rectangular cells of volume a Vk •
We are interested in what happens as D is partitioned by smaller and smaller cells, so that dXk, dYk, dZk and the norm ofthe partition Ilpll, the largest value among dXk, dYk, dZk, all approach zero. When a single limiting value is attained, no matter how the partitions and points (Xk, Yk, Zk) are chosen, we say that F is integrable over D. As before, it can be
860
Chapter 15: Multiple Integrals shown that when F is continuous and the bounding surface of D is fonned from tmitely many smooth surfaces joined together along tmitely many smooth curves, then F is integrable. As Ilpll ---> 0 and the number of cells n goes to 00, the sums S. approach a limit. We call this limit the triple integral of F over D and write
~S. =
•
iiiF(x,y,z)dV
lim S. =
or
11P1I~o
D
rrr F(x,y,z)
JJJ D
dxdy dz.
The regions D over which continuous functions are integmble are those having ''reasonably smooth" boundaries.
Volume of a Region in Space If Fis the constant function whose value is 1, then the sums in Equation (1) reduce to
As dxk' d Yk, and dZk approach zero, the cells d Vk become smaller and more numerous and ml up more and more of D. We therefore define the volume of D to be the triple integml
DEFINmON
The volume of a closed, bounded region D in space is
V= iff dV. D
TIris detmition is in agreement with our previous detmitions ofvolume, although we omit the verification ofthis fact. As we see in a moment, this integral enables us to calculate the volumes of solids enclosed by curved surfaces.
Finding Limits of Integration in the Order dz dy dx We evaluate a triple integml by applying a three-dimensional version of Fubini's Theorem (Section 15.2) to evaluate it by three repeated single integrations. As with double integmls, there is a geometric procedure for finding the limits ofintegmtion for these single integmls. To evaluate
iff F(x,y,z)dV D
over a region D, integrate first with respect to z, then with respect to y, and finally with respect to x. (You might choose a different order ofintegmtion, but the procedure is similar, as we illustmte in Example 2.)
1.
Sketch. Sketch the region D along with its "shadow" R (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces of D and the upper and lower bounding curves of R.
Triple Integrals in Rectangular Coordinates
15.5
861
z
x
2.
Find the z-limits ofintegration. Draw a line M passing through a typical point (x, y) in R parallel to the z-axis. As z increases, M enters D at z == ! 1(x, y) and leaves at z == !2(X, y). These are the z-limits of integration. z
Leaves at
1- . . . . ' 1 " z = f2(x, 1
y:
1 1
D /
//
.".--r--1 1 1
a
b
x
--- --(x, y)
3.
Find the y-limits ofintegration. Draw a line L through (x, y) parallel to the y-axis. As y increases, L enters R at y == gl (x) and leaves at y == g2(X). These are the y-limits of integration. z
D
Enters at y = gl(x) a
y
x b
x
\ L Leaves at y = g2(x)
862
Chapter 15: Multiple Integrals
4.
Find the x-limits ofintegration. Choose x-limits that include all lines through R parallel to the y-axis (x == a and x == b in the preceding figure). These are the x-limits of integration. The integral is
l
x =bl y =g2(X)l z =!2(X,y)
x=a
F(x, y, z) dz dy dx. y=gl(X)
z=!I(x,y)
Follow similar procedures if you change the order of integration. The "shadow" of region D lies in the plane of the last two variables with respect to which the iterated integration takes place. The preceding procedure applies whenever a solid region D is bounded above and below by a surface, and when the "shadow" region R is bounded by a lower and upper curve. It does not apply to regions with complicated holes through them, although sometimes such regions can be subdivided into simpler regions for which the procedure does apply.
EXAM PLE 1 and z
==
SoLution
Find the volume of the region D enclosed by the surfaces z 8 - x 2 - y2. The volume is
v=
iff
==
x2
+
3y 2
dzdydx,
D
the integral of F(x, y, z) == 1 over D. To find the limits of integration for evaluating the integral, we first sketch the region. The surfaces (Figure 15.30) intersect on the elliptical cylinder x 2 + 3y 2 == 8 - x 2 - y2 or x 2 + 2y 2 == 4, z > O. The boundary of the region R, the projection of D onto the xy-plane, is an ellipse with the same equation: x 2 + 2y 2 == 4. The "upper" boundary of R is the curve y == V(4 - x 2)/2. The lower boundary is the curve y == -V(4 - x 2 )/2. Now we find the z-limits of integration. The line M passing through a typical point (x, y) in R parallel to the z-axis enters D at z == x 2 + 3y 2 and leaves at z == 8 - x 2 - y2. M
Leaves at z = 8 - x2
z
- y2
\
The curve of intersection
I I I I I I I I I I
Enters at
z = x 2 + 3y 2
'(-2,0,4)
..J-----
-- I I .... I
Enters at
y = - ~--(4-_-x-2-)/-2
'..J
..
(x,y)
x R
7
Leavesat = Y(4 - x 2 )/2
L
y
y
FIGURE 15.30 The volume of the region enclosed by two paraboloids, calculated in Example 1.
Triple Integrals in Rectangular Coordinates
15.5
863
Next we find the y-limits of integration. The line L through (x, y) parallel to the y-axis enters R at y == - Y(4 - x 2 )/2 and leaves at y == Y(4 - x 2 )/2. Finally we find the x-limits of integration. As L sweeps across R, the value of x varies from x == -2 at (-2, 0, 0) to x == 2 at (2, 0, 0). The volume of Dis
v=
111
dzdydx
D
1 1 1 1 2( 1 2[ ( 1
21Y(4-X2)/218-X2-y2
==
dz dy dx
-2 -Y(4-x 2 )/2 2
==
Y(4-X 2)/2
x 2 +3y 2
(8 - 2x 2
4y 2) dy dx
-
-2 - Y(4-x 2 )/2 2 [
==
4 ]Y=Y(4-X (8 - 2x )y - -3 y3 2
= ==
dx
- 38(4~2)3/2) dx ~ ~r 4~X 2)3/2 _~ (4~X 2)3/2] dx=4~2J_2(4-x2)3/2dx
-2
2(8 -
_2
8
81TV2.
)P
y=-Y(4-x 2
-2
=
2 )/2
~ 2x2\/~
2
After integration with the substitution x = 2 sin u
•
In the next example, we project D onto the xz-plane instead of the xy-plane, to show how to use a different order of integration. z __________ (0, 1,1)
(1,1,0)
x
Finding the limits of integration for evaluating the triple integral of a function defined over the tetrahedron D (Examples 2 and 3). FIGURE 15.31
EXAMPLE 2 Set up the limits of integration for evaluating the triple integral of a function F(x, y, z) over the tetrahedron D with vertices (0, 0, 0), (1, 1,0), (0, 1,0), and (0, 1, 1). Use the order of integration dy dz dx. SoLution We sketch D along with its "shadow" R in the xz-plane (Figure 15.31). The upper (right-hand) bounding surface of D lies in the plane y == 1. The lower (left-hand) bounding surface lies in the plane y == x + z. The upper boundary of R is the line z == 1 - x. The lower boundary is the line z == O. First we find the y-limits of integration. The line through a typical point (x, z) in R parallel to the y-axis enters D at y == x + z and leaves at y == 1. Next we find the z-limits of integration. The line L through (x, z) parallel to the z-axis enters R at z == 0 and leaves at z == 1 - x. Finally we find the x-limits of integration. As L sweeps across R, the value of x varies from x == 0 to x == 1. The integral is
I I1I-Xi l o 0
x+z
F(x, y, z) dy dz dx.
•
EXAMPLE 3 Integrate F(x, y, z) == 1 over the tetrahedron D in Example 2 in the order dz dy dx, and then integrate in the order dy dz dx. SoLution First we find the z-limits of integration. A line M parallel to the z-axis through a typical point (x, y) in the xy-plane "shadow" enters the tetrahedron at z == 0 and exits through the upper plane where z == y - x (Figure 15.32). Next we find the y-limits of integration. On the xy-plane, where z == 0, the sloped side of the tetrahedron crosses the plane along the line y == x. A line L through (x, y) parallel to the y-axis enters the shadow in the xy-plane at y == x and exits at y == 1 (Figure 15.32).
864
Chapter 15: Multiple Integrals
Finally we find the x-limits of integration. As the line L parallel to the y-axis in the previous step sweeps out the shadow, the value of x varies from x == 0 to x == 1 at the point (1, 1, 0) (see Figure 15.32). The integral is
z
{IiIJo
(y-x
Jo For example, if F(x, y, z)
==
x
1, we would find the volume of the tetrahedron to be
{IiIJo li 1t 1 (t - t (y-x
V= Jo
1
(y - x) dy dx
(1,1,0)
FIGURE 15.32 The tetrahedron in Example 3 showing how the limits of integration are found for the order dz dy dx.
dz dy dx
x
1
= x
F(x, y, z) dz dy dx.
=
1 [
y2
xy
-
]Y=l dx
o
y=x
1
=
+ X2 )
x
[Ix - Ix 2 2 2
==
dx
Ix 3]1 6 0
+
1
6· We get the same result by integrating with the order dy dz dx. From Example 2,
ll i l
V==
l
x
l
dydzdx
-
o 0
x+z
{I {l-x
=
Jo Jo
==
1
1 [
o
1
(1 -
1 ]Z=l-X (1 - x)z - - z2 dx 2 z=o
1
=
[(1 - x)2 -
111
== -
2
=
x- z) dz dx
t
(1 - X)2] dx
(1 - x)2 dx
0
-61 (1
- x)3
]1
0
6·1
•
Average Value of a Function in Space The average value of a function F over a region D in space is defined by the formula
Average value of F over D =
vol~e of D
111
F dV.
(2)
D
For example, if F(x,y, z) == Yx 2 + y2 + z2, then the average value of F over D is the average distance of points inD from the origin. If F(x,y, z) is the temperature at (x,y, z) on a solid that occupies a region D in space, then the average value of F over D is the average temperature of the solid.
15.5 Triple Integrals in Rectangular Coordinates
865
EXAMPLE 4 Find the average value of F(x,y, z) = xyz throughout the cubical region D bounded by the coordinate planes and the planes x = 2, Y = 2, and z = 2 in the fIrst octant. Solution We sketch the cube with enough detail to show the limits of integration (Figure 15.33). We then use Equation (2) to calculate the average value of F over the cube. The volume of the region Dis (2)(2)(2) = 8. The value of the integral of F over the cube is
D
2
2 2 222 1 1 xyzdxdydz 1212 [2 ~ yz ]X~2 dydz 112yzdydz 1 2 2[]Y~2 = y2z dz= 1 4zdz= []2 2z2 =8. 1o
/ x
=
000
FIGURE 15.33 in Example 4.
The region ofintegration
=
00
x-O
y=O
00
0
0
With these values, Equation (2) gives Average value of = / .1Yz over the cube vo ume
(-81 )(8)
fff xyz dV = 111 cube
= 1.
In evaluating the integral, we chose the order dx dy dz, but any of the other five possible orders would have done as well. •
Properties of Triple Integrals Triple integrals have the same algebraic properties as double and single integrals. Simply replace the double integrals in the four properties given in Section 15.2, page 846, with triple integrals.
Exercises 15.5 Triple Integrals In Different Iteration Orders 1. Evaluate the integral in Example 2 taking F(x,y, z) = 1 to Imd the volume of the tetrahedron in the order dz dx dy. 2. Volume of rectangular solid Write six different iterated triple integrals for the volume of the rectangular solid in the Itrst octant bounded by the coordiuate planes and the planes x; ~ I, Y ~ 2, and z = 3. Evaluate one of the integrals.
Evaluating Triple Iterated Integrals Evaluate the integrals in Exercises 7-20. 7./,'/,'/,,<x;2
+ y2 + z2) dz ely dx
[12f L:~::-Y'
dz dx ely
9.
3. Volume of tetrabedron Write six different iterated triple int<>grals for the volume of the tetrahedron cut from the IlrSt octant by the plane 6x + 3y + 2z = 6. Evaluate one of the integrals.
10. [' {'-3x {'-3x- dzdydx
11.
4. Volume of solid Write six different iterated triple integrals for the volume of the region in the first octant enclosed by the cylin+ = 4 and the plane y = 3. Evaluate one of the der integrals.
12. {' [' {2(x;
X;2 z2
5. Volume enclosed by paraboloids Let D be the region bounded by the paraboloids = 8 y2 and = + y2. Write six different triple iterated integrals for the volmne of D. Evaluate one of the integrals.
z
X;2 -
z X;2
6. Volume inside paraboloid benealb a plane Let D be the region bounded by the paraboloid z = x;2 + y2 and the plane z = 2y. Write triple iterated integrals in the order dz dx ely and dz ely dx that give the volume ofD. Do not evaluate either integral.
&
y
Jo Jo Jo
i-tio Jo
+ Y + z)elydxdz
13.1'f9~j,vh' dzelydx
ill ~
dx ely dz
{~/' [' {' ysinzdxelydz
Jo Jo J-2
866 19.
Chapter 15: Multiple IntEgrals
fo"/4foltt&_·f~ rdxd'du
fo 1'lov'R
(tw:-~)
r :
,
(pqr--~)
7
2t.
Z4. The resion in the flnt octant bounded by the coordinate p1anel mdtbeplanee:f + Z - 1,1 +:b - 2
I tip dq tir
nncHog~_~
ZL Bc:rc II ~rcaioo ofiotepaticoofthc imqral
1'1, '1.'-' Z'
-1
x
dzdydx.
D
:zs.
The n:eioo in ibe fUst: octant bounded by the ~ plme:s. + Z = 2, and the cylinder.¥ = 04 _ ,2
the p1ue,
,
(-I, 1, 0)
;{
(I, I, 0)
Ilewritc the iotqpl u an equivalent itJ::rmd integral in the mdcI' L
dydzdx
b. dydxdz
Co
dz dy dz
d. b. tfz dy
x
16. The wedge cut &m:1 the cylinder ~
z- -yandz-O
I. ,1-1f'J." D
-
dzdydx.
D
I by the pIaneI
,
e. dzb.dy. 22. Here iI the reglOII of ln1egratiOD. ofthe ~
+ ,2 -
-
,
(0,-1,1)
x 0,-1.1)
27. The Il:tn1hcdroo in the fint oetantboundcdbythc ooordinatc planca "'" the p_ """iDa _ (I, 0, D), (0, 2, 0), "'" (0, 0, 3)
,
o , (1,-1,0)
•
1
Rewrite thl: iDtearal u an equivalent iterated integral in the order
dy tfz dJ:
b. dy tb: tfz
e. dJ: dy dz e. dztb:dy.
d. tb: tfz fly
L
x
.... UIf"il Tltpla _gnU FiDel the volumes ofthc: RlIio:oa in Eu:n:iIIcs 23-36. n The n:poa betwc:c::u tbe cylindc:r z = y2 aDd tbt: .¥)'-pIaDc that is bouIKblbytbopl&Des x = o.x = 1,1 = -l,y = 1 I1ndt"" _
,
28. The rqioo in the flISt octant bounded by the coordinate p1ancI, the pIaDC 1 1 -.x, and the 5DIfiKlc z = COlI (",x/2),
=
Os ... :S1
,
867
15.5 Triple Integrals in Rectangular Coordinates 29. The region common to the interiors of the cylinders x 2 + y2 = 1 and x 2 + z2 = 1, one-eighth of which is shown in the accompanying figure
z
33. The region between the planes x z = 4 in the first octant
+ y + 2z
34. The finite region bounded by the planes z y = 8, andz = 0
=
2 and 2x
= x, x + z = 8, z = y,
35. The region cut from the solid elliptical cylinder x 2 the xy-plane and the plane z = x + 2
xZ+r=l
+ 2y +
+ 4y 2
::::;
4 by
36. The region bounded in back by the plane x = 0, on the front and sides by the parabolic cylinder x = 1 - y2, on the top by the paraboloid z = x 2 + y2, and on the bottom by the xy-plane
Average Values In Exercises 37-40, find the average value of F(x, y, z) over the given region. 37. F(x,y, z) = x 2 + 9 over the cube in the first octant bounded by the coordinate planes and the planes x
x
30. The region in the first octant bounded by the coordinate planes and the surface z = 4 - x 2 - Y
= 2, y = 2, and z = 2
+Y -
38. F(x, y, z) = x
z over the rectangular solid in the first octant bounded by the coordinate planes and the planes x = l,y = 1, andz = 2 39. F(x,y,z) = x 2 + y2 + z2 over the cube in the first octant bounded by the coordinate planes and the planes x = l,y = 1, andz = 1
40. F(x, y, z) = xyz over the cube in the first octant bounded by the coordinate planes and the planes x = 2, y = 2, and z = 2
z
Changing the Order of Integration Evaluate the integrals in Exercises 41-44 by changing the order of integration in an appropriate way. 41. y
42.
x
31. The region in the first octant bounded by the coordinate planes, the plane x + y = 4, and the cylinder y2 + 4z 2 = 16
43.
4~112 4 cos (x • r o 0 2y 2 yz
1
r rl112xzezy2
Jo Jo
1hI x2l
)
dxdydz
dy dx dz
x2
1
o
2
~
ln3
'Tfe
2x
0
r r- r si~2z Jo Jo Jo
sin 7TJ'2
Y
2
dxdydz
2
z
44.
dy dz dx
z
4
Theory and Examples 45. Finding an upper limit of an iterated integral y
114-a-x214-X2-y
1 o
x
32. The region cut from the cylinder x 2 + y2 = 4 by the plane z = 0 and the plane x + z = 3 z
0
Solve for a:
_-.£
dzdydx - 15'
a
46. Ellipsoid For what value of c is the volume of the ellipsoid x 2 + (y/2)2 + (z/c)2 = 1 equal to 8'Tf? 47. Minimizing a triple integral mizes the value of the integral
What domain D in space mini-
JJJ(4x 2 + 4y 2 + z2 - 4) dV? D
Give reasons for your answer. 48. Maximizing a triple integral mizes the value of the integral
What domain D in space maxi-
JJJ(l - x 2 - y2 - z2) dV? y
x
D
Give reasons for your answer.
868
Chapter 15: Multiple Integrals
COMPUTER EXPLOIATIONS In Exercises 49-52, use a CAS integration utility to evaluate the triple integral of the given :function aver the specified solid region.
49. F(x,y,z) = x1y2z over the solid cylinder x 2 + y2 -= 1 and the planesz =- Oandz =- 1
51. F(x,y, z) = the cone z =
52. F(x,y, z) = x" z2 :S
:
2 3/2
over the solid bounded below by
v';(2 + y2 and above by the plane z =
bounded by
SO. F(x, y, z) = IxyzIover the solid bounded below by the paraboloid z = x 2 + y2 and above by the plane z = I
15.6
2
(x +y +z)
+ y2 + i2
1
over the solid sphere x2
+ y2 +
1
I---'-'Mo.:..:m.:..:e.:..:n:.:::ts.:..:a.:..:n:.:::d-=ce.:..:n.:..:te.:..:l'S.:..:.:..:of.:..:M.:..:a.:..:ss=---
_
This section shows how to calculate the masses and moments of two- and threedimensional objects in Cartesian coordinates. Section 15.7 gives the calculations for cylindrical and spherical coordinates. The defmitions and ideas are similar to the single-variable case we studied in Section 6.6, but now we can consider more realistic situations.
Masses and First Moments If B(x, y, z) is the density (mass per unit volwne) ofan object occupying a region D in space, the integral of B over D gives the mall of the object. Th see why, imagine partitioning the object into n mass element!llike the one in Figure 15.34. The object's mass is the limit
,
y
FIGURE 15.34 To def'me an object's we fll8t imagine it to be partiti.oncd into a fmite number of mass elements
tnaSIl,
The first moment of a solid region D about a coordinate plane is defined as the triple integral """ D of the dis1anc. ftum a point (x, y, z) in D '" the plane multiplied by the density
ofthe solid at that point For insta:nce, the f11'Sl: moment about the yz-plane is the integral
<1m.
MY'
~ JJf x8(x,y,z) dV. D
The center ofma33 is found from the f11'Sl: moments. For instance, the x-coordinate of the center ofmass is x - Myz/M. For a two-dimensional object, such as a thin, flat plate, we calculate f11'8t moments about the coordinate axes by simply dropping the z-coordinate. So the first moment about the y-axis is the double integral over the region R fanning the plate of the distance from the axis multiplied by the density, or M,
o
ff
>8(x, y) dA.
R
• y
, FIGURE 15.35 Finding the center of ofa solid (Example I).
tnaSIl
~
Table 15.1 summarizes the formulas. EXAMPLE 1
Find the center ofmass of a solid of constant density B bmmded below
by the disk R: x 2 + y2 :S 4 in the plane z = 0 and above by the paraboloid z = 4 - x 2 - y2 (Figure 15.35).
15.6
TABLE 15.1
Moments and Centers of Mass
869
Mass and first moment formulas
THREE-DIMENSIONAL SOLID
a~a(x,y,z)isthedensityat(x.y.z).
Mass:M= JJJ8dV D
First moments about the coordinate planes:
My< = JJJ x8dV,
Mxy = JJJ z8dV
D
D
Center of mass: _
_ Mxy z=-
M",
Y=M'
M
TWO-DIMENSIONAL PLATE Mass:
M=JJ8dA
a
~
a(x,y) i.the density at (x,y).
R
First moments:
My = JJ x8dA,
Center of mass:
_ My x= M'
R
Solution
By symmetry x = y = 0. To tmd l', we first calculate
Mxy =
~~~
z=o
R
=
z-4-x'-1
z8dzdydx =
~if [z2
2
R
~JJ(4 -
]z-4-x'-18dydx z=o
x 2 - y2)2dydx
R
2810
=-
0
=
"1
2
2
(4-r 2)2 r drdO
r 210
~
Polar coordinate. simplify the integration.
0
2 " [_1(4 _ 6
r2)3]'~2 dO ,~O
= IM!2" dO = 321T8. 3 3
10
A similar calculation gives the mass
M =
rrr-x'-18 dz dy dx
1110
=
81T8.
R
Thereforel' = (Mxy/M) = 4/3 and the cenler ofmass is (x, y, l') = (0,0,4/3).
•
When the density of a solid object or plate is constant (as in Example I), the center of mass is called the centroid of the object. To tmd a centroid, we set 8 equal to I and proceed to find X, y, and l' as before, by dividing first moments by masses. These calculations are also valid for two-dimensional objects.
EXAMPLE 2 Find the centroid ofthe region in the first quadrant that is bounded above by the line Y = x and below by the parabola Y = x 2•
870
Chapter 15: Multiple Integrals
y
SoLution We sketch the region and include enough detail to determine the limits of integration (Figure 15.36). We then set 8 equal to 1 and evaluate the appropriate formulas from Table 15.1:
1 6 -o~--_.....L...-_-+X
FIGURE 15.36 The centroid of this region is found in Example 2.
1
12· From these values of M, Mx , and My, we find _ x
My ==
1/12
1
M == 1/6 ==
and
2
•
The centroid is the point (1/2, 2/5).
Moments of Inertia An object's first moments (Table 15.1) tell us about balance and about the torque the object experiences about different axes in a gravitational field. If the object is a rotating shaft, however, we are more likely to be interested in how much energy is stored in the shaft or about how much energy is generated by a shaft rotating at a particular angular velocity. This is where the second moment or moment of inertia comes in. Think of partitioning the shaft into small blocks of mass dmk and let rk denote the distance from the kth block's center of mass to the axis of rotation (Figure 15.37). If the shaft rotates at a constant angular velocity of w == d8/dt radians per second, the block's center of mass will trace its orbit at a linear speed of
Vk
==
d dt (rk 8 )
==
d8 rk dt
==
rk w .
The block's kinetic energy will be approximately FIGURE 15.37 To find an integral for the amount of energy stored in a rotating shaft, we first imagine the shaft to be partitioned into small blocks. Each block has its own kinetic energy. We add the contributions of the individual blocks to find the kinetic energy of the shaft.
2 1A 2umkVk
==
21 umk rk w )2 A
(
1 2 2 A 2 w rk umk·
==
The kinetic energy of the shaft will be approximately ~ 1 W2 2
£.J
2
A
rk umk·
The integral approached by these sums as the shaft is partitioned into smaller and smaller blocks gives the shaft's kinetic energy:
_J12 w r
KEshaft -
22
_1
2
dm - 2w /
2
r dm.
(1)
The factor 1=
J
2
r dm
is the moment ofinertia of the shaft about its axis of rotation, and we see from Equation (1) that the shaft's kinetic energy is _ 1 KEshaft -
2 fw
2
.
15.6
z
x
FIGURE 15.38 Distances from dVto the coordinate planes and axes.
Moments and Centers of Mass
871
The moment of inertia of a shaft resembles in some ways the inertial mass of a locomotive. To start a locomotive with mass m moving at a linear velocity v, we need to provide a kinetic energy ofKE = (1/2)mv 2 • To stop the locomotive we have to remove this amount of energy. To start a shaft with moment of inertia I rotating at an angular velocity cu, we need to provide a kinetic energy of KE = (l/2)Icu2 • To stop the shaft we have to take this amount of energy back out. The shaft's moment of inertia is analogous to the locomotive's mass. What makes the locomotive hard to start or stop is its mass. What makes the shaft hard to start or stop is its moment of inertia. The moment of inertia depends not only on the mass of the shaft but also on its distribution. Mass that is farther away from the axis of rotation contributes more to the moment of inertia. We now derive a formula for the moment of inertia for a solid in space. If r(x, y, z) is the distance from the point (x,y, z) in D to a line L, then the moment of inertia of the mass tJ.mk = 8(xk,Yk,zk)tJ.Vk about the line L (as in Figure 15.37) is approximately tJ.h = r 2(xk, Yk, Zk) tJ.mk. The moment of inertia about L of the entire object is
If L is the x-axis, then r 2 = y 2
Ix
+ z2 (Figure 15.38) and = JJJ(y
2
+ z2) 8(x, y, z) dV.
D
Similarly, if L is the y-axis or z-axis we have and Table 15.2 summarizes the formulas for these moments of inertia (second moments because they invoke the squares of the distances). It shows the definition of the polar moment about the origin as well.
EXAMPLE 3
z
Find Ix, Iy , Iz for the rectangular solid of constant density 8 shown in
Figure 15.39. SoLution
The formula for Ix gives
c
-~
Ix
= lc/21b/21a/2(y2
+ z2) 8 dx dy dz.
-c/2 -b/2 -a/2 y
Center of block
FIGURE 15.39 Finding Ix, Iy, and Iz for the block shown here. The origin lies at the center of the block (Example 3).
We can avoid some of the work of integration by observing that (y2 + z2)8 is an even function of x, y, and z since 8 is constant. The rectangular solid consists of eight symmetric pieces, one in each octant. We can evaluate the integral on one of these pieces and then multiply by 8 to get the total value.
r/ r/ r/ 3 +
Ix = 8Jo
2 2 r a/2 Jo Jo (y2 2
= 4a8
[y3
Jo
= 4a81c/2
+ z2) 8 dx dy dz =
2 rb/2 Jo (y2
+ z2) dy dz
r=b/2 z2y y=o dz
(~: + z~b )
= 4a8(~3~ +
r/
4a8Jo
~:)
=
dz
ar~8 (b 2 + c 2) = ~ (b 2 + c 2).
M = abc8
872
Chapter 15: Multiple Integrals
TABLE 15.2
Moments of inertia (second moments) formulas
THREE-DIMENSIONAL SOLID
fJJ(y2 + Z2) Il dV
About the x-axis:
Ix
About the y-axis:
I = fJJ(X 2+ z2) Il dV
About the z-axis:
Iz =
fJJ(X 2+ y2) Il dV
About a line L:
h
fJJr 21l dV
=
8
~
8(z,y,.)
y
=
rex, y, z) = distance from the point (:t. y, z) to line L
TWO-DIMENSIONAL PLATE About the x-axis:
Ix
About the y-axis:
Iy =
About a line L: About the origin
=
ff ff
2
8
y 1ldA
~
8(z,y)
dA h = fJr2(X,y) Il dA (;:~j~~L distance from 10 = ff(x 2+ y2)lldA = Ix + I x
2
1l
y
(polar moment):
Similarly,
l = M (a 2 y 12
+
c2)
and
IZ
_ M (2 a + b2) . -T2
•
EXAMPLE 4 A thin plate covers the triangular region bounded by the x-axis and the lines x = I and y = 2x in the first quadrant. The plate's density at the point (x, y) is Il(x, y) = 6x + 6y + 6. Find the plate's moments ofinertia about the coordinate axes and the origin. y 2
(1,2)
Solution
We sketch the plate and put in enough detail to determine the limits ofintegration for the integrals we bave to evaluate (Figure 15.40). The moment of inertia about the
x-axis is
nf----+--~z
FIGURE 15.40 The triangular region covered by the plate in Example 4.
=
[s.x S + 4X4]~
=
12.
15.6
Moments and Centers of Mass
873
Similarly, the moment of inertia about the y-axis is
Beam A
Iy =
Axis
39 10t 10fOx x 2S(x,y) dy dx = S'
Notice that we integrate y2 times density in calculating Ix and x 2 times density to find IySince we know Ix and Iy , we do not need to evaluate an integral to tmd 10; we can use the equation 10 = Ix + Iy from Table 15.2 instead: BeamB
[_G
10 = 12
+
39 = 60
5
+ 5
39
99 5'
•
Axis
The moment of inertia also plays a role in detennining how much a horizontal metal beam will bend under a load. The stiffness ofthe beam is a constant times I, the moment of inertia of a typical cross-section of the beam about the beam's longitudinal axis. The greater the value of I, the stiffer the beam and the less it will bend under a given load. That FIGURE 15.41 The greater the polar is why we use I-beams instead of beams whose cross-sections are square. The flanges at moment ofinertia of the cross-section of a the top and bottom of the beam hold most of the beam's mass away from the longitudinal beam about the beam's longitudinal axis, the axis to increase the value of I (Figure 15.41). stiffer the beam. Beams A and B have the same cross-sectional area, but A is stiffer.
Exercises 15.6 Plates of Constant Density 1. Finding a center of mass Find the center of mass of a thin plate of density B = 3 bounded by the lines x = 0, y = x, and the parabola y = 2 - x 2 in the tITst quadrant 2. Finding moments of inertia Find the moments of inertia about the coordioate axes of a thio rectangular plate of constant density B bounded by the lines x = 3 and y = 3 in the tITst quadrant. 3. Finding a centroid Fiod the centroid of the region in the first quadrant bounded by the x-axis, the parabola y2 = 2x, and the lioex+y~4.
4. Finding a centroid Find the centroid of the triangular region cut from the tITSt quadrant by the line x + y = 3. 5. Finding a centroid Fiod the centroid of the region cut from the tITSt quadrsnt by the circle x 2 + y2 = a 2. 6. Finding a centroid Find the centroid of the region b-'en the x-axis and the archy = sin x, 0:5 x :s; 71'. 7. Finding moments of inertia Find the moment of inertia about the x-axis of a thio plate of deosity B ~ I bounded by the circle x 2 + y2 = 4. Then use your result to tmd 1y and 10 for the plate. 8. Finding a moment of inertia Find the moment of inertia with respect to the y-axis of a thio sheet of constant density B = I bounded by the curve y = (sio2 x)/x 2 and the iotorval 'TT'
:s; X :s; 2w of the x-axis.
9. The centroid of an infinite region Find the centroid of the intmite region io the second quadrant enclosed by the coordioate axes and the curve y = e~. (Use improper integrals in the massmoment formulas.)
10. The fin! moment of an inImite plate Find the tITSt moment about the y-axis of a thio plate of density B(x, y) = I covering the 2 infmite region under the curve y = e -x f2 in the first quadrant.
Plates with Varying Density 11. Finding a moment of inertia Find the moment of inertia about the x-axis of a thio plate bounded by the parabola x ~ y - y2 and the line x + y = if B(x,y) = x + y.
ma..
o
12. Finding Find the mass of a thin plate occupying the smaller region cut from the ellipse x 2 + 4y 2 = 12 by the parabola x ~ 4y 2 if B(x, y) ~ 5x. 13. Finding a center of mass Find the center of mass of a thin triangular plate bounded by the y-axis and the lines y ~ x and y = 2 -xifB(x,y) = 6x+ 3y+ 3. 14. Finding a center of mass and moment of inertia Find the center of mass and moment of inertia about the x-axis of a thin plate bounded by the curves x ~ y2 and x ~ 2y - y2 if the density at the point (x. y) is B(x,y) = y + I. 15. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thin rectangular plate cut from the tITSt quadrsnt by the lioes x ~ 6 and y ~ I if B(x,y) = x + Y + I. 16. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thio plate bounded by the line y = I and the parabola y = x 2 if the density is B(x,y) = y + I. 17. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the y-axis of a thio plate bounded by the x-axis, the lines x = ± 1, and the parabola y = x 2 if B(x,y) ~ 7y + I.
874
Chapter 15: Multiple Integrals
18. Center of mass, moment of inertia Find the center of mass and the moment of inertia about the x-axis of a thin rectangular plate bounded by the linesx = 0, x = 20, Y = -I, and y = 1 if B(x,y) ~ I + (x/20). 19. Center of mass, moments of inertia
Find the center of mass,
the moment of inertia about the coordinate axes, and the polar moment of inertia of a thin triangular plate bounded by the linesy = x,y = -x, andy = 1 ifB(x,y) = Y + 1.
20. Center of mass, moments of inertia Repeat Exercise 19 for B(x,y) ~ 3x 2 + 1.
l-----" y
Solids with Constant Dens;ty 21. Moments of inertia Find the moments of inertia of the rectangular solid shown here with respect to its edges by calculating Ix, [y, and /z.
c I I
/
/ /
!--
-- b
Y
a
22. Moments of inertia The coordinate axes in the figure run through the centroid of a solid wedge parallel to the labeled edges. Find Ix, [y, and I z if a = b = 6 and c = 4.
Centroid
8t(0, 0, 0)
c
~
3
a
23. Center of mass and moments of inertia A solid ''trough'' of constant density is bounded below by the surface z = 4y 2, above by the plane z = 4, and on the ends by the planes x = I and x ~ -1. Find the center of mass and the moments of inertia with respect to the three axes. 24. Center of ma.. A solid of constant density is bounded below by the plane z ~ 0, on the sides by the elliptical cylinder x 2 + 4y 2 = 4, and above by the plane z = 2 - x (see the accompanying figure).
a. Find>' andy. b. Evaluate the integral Mxy =
1'j -
(1j2)~J.2-x
-(1f2)~ 0
zdzdydx
using integral tables to carry out the fmal integration with respect to x. Then divide Mxy by M to verify that z = 5/4.
25. a. Center of mass Find the center of mass of a solid of constant density bounded below by the paraboloid z = x 2 + y2 and above by the plane z ~ 4. b. Find the plane z ~ c that divides the solid into two parts of equal volume. This plane does not pass through the center ofmass. 26. Moments A solid cube, 2 units on a side, is bounded by the planes x = ±I,z = ±I,y = 3, and y = 5. Find the center of
mass and the moments ofinertia about the coordinate axes. 27. Moment of inertia about a line A wedge like the one in Exercise 22 has a = 4, b = 6, and c = 3. Make a quick sketch to check for yourself that the square of the distance from a typical point (x, y, z) of the wedge to the line L: z = O,y = 6 is r 2 = (y - 6)2 + z2. Then calculate the moment of inertia of the wedge about L. 28. Moment of inertia about a line A wedge like the one in Exercise 22 has a = 4, b = 6, and c = 3. Make a quiek sketch to check for yourself that the square of the distance from a typical point (x, y, z) of the wedge to the line L: x = 4, Y = 0 is r 2 ~ (x - 4j2 + y2. Then calculate the moment of inertia of the wedge about L.
Solids with varying Dens;ty In Exercises 29 and 30, fmd
a. the mass of the solid. b. the center of mass. 29. A solid region in the flIst octant is bounded by the coordinate planes and the plane x + y + z = 2. The density of the solid is B(x,y,z) = 2x. 30. A solid in the flIst octant is boWlded by the planes y = 0 and z = 0 andbythesurfacesz = 4 - x 2 andx = y2 (seethe accompanying figure). Its density function is B(x, y, z) = !xy, k a constant
z
4~
15.7 Triple Integrals in Cylindrical and Spherical Coordinates In Exercises 31 and 32, fmd
a. the mass of the solid.
~1
b. the center of mass.
c. the moments ofinertia about the coordinate axes. 31. A solid cube in the fIrSt octant is bounded by the coordinate planes and by the planes x = I,y = I,andz = 1. The density of the cube is 8(x,y,z) ~ x + y + z + 1.
L
34. Ma.. Find the mass of the solid region bounded by the parabolic surfaces z = 16 - 2x 2 - 2y 2 and z = 2x 2 + 2y 2 if the density ofthe solid is 8(x,y,z) = v'x 2 + y2. Theory and Examples The Parallel AD. Theorem Let L,m. be a line through the center of mass of a body ofmass m and let L be a parallel line h units away from Lc.m • The Parallel Axis Theorem says that the moments of inertia Ic:m. and h of the body about L,m. and L satislY the equation
.~xi+yJ
--',-......_.
(x,y, z)
,y
32. A wedge like the one in Exercise 22 bas dimensions a ~ 2, b ~ 6, and c = 3. The density is 8(x,y,z) = x + 1. Notice that if the density is constant, the center ofmass will be (0, 0, 0). 33. Mas. Find the mass of the solid bounded by the planes x + z = I, x - z = -I, y = 0 and the surface y = ;/Z. The density of the solid is 8(x,y,z) ~ 2y + 5.
875
D
(h, 0, 0)
b. To prove the Parallel Axis Theorem, place the body with its center ofmass at the origin, with the line L c.m . along the z-axis and the line L perpendicular to the xy-plane at the point (h, 0, 0). Let D be the region ofspace occupied by the body. Then, in the notation of the figure,
h~ fIJIV - hil
2
dm.
D
Expand the integrand in this integral and complete the proof.
(2)
36. The moment of inertia about a dismeter ofa solid spbere ofconslant density and radius a is (2/5)ma 2, where m is the mass ofthe sphere. Find the moment of inertia about a line tangent to the spbere.
As in the two-dimensional case, the theorem. gives a quick way to
37. The moment of inertia of the solid in Exercise 21 about the z-axis is I, ~ abc(a 2 + b 2 )/3.
calculate one moment when the other moment and the mass are knowo. 35. Proof of the Parallel AD. Theorem a. Sbow that the fIrSt moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the yz-plane. What does the fannulax = M,,/M then tell you?)
a. Use Equation (2) to fmd the moment of inertia ofthe solid about 1he line parallel ro 1he z-axis through the solid's center ofmass. b. Use Equation (2) and 1he result in part (a) to fmd the moment of inertia of1he solid about the line x ~ 0, y ~ 2b.
38. If a ~ b ~ 6 and c ~ 4, 1he moment of inertia of the solid wedge in Exercise 22 about the x-axis is Ix = 208. Find 1he moment of inertia of the wedge about 1he line y ~ 4,z ~ -4/3 (1he edge of the wedge's narrow end).
Triple Integrals in Cylindrical and Spherical Coordinates
15.7
When a calculation in physics, engineering, or geometry involves a cylinder, cone, or sphere, we can often simplify our work by using cylindrical or spherical coordinates, which are introduced in this section. The procedure for transforming to these coordinates and evaluating the resulting triple integrals is similar to the transformation to polar coordinates in the plane studied in Section 15.4. P(r, 8, z)
Integration in Cylindrical Coordinates We obtain cylindrical coordinates for space by combining polar coordinates in the xy-plane with the usual z-axis. This assigns to every point in space one or more coordinate triples of the form (r, fJ, z), as shown in Figure 15.42.
o 8
r
y
-
~y /
FIGURE 15,42 The cylindrical coordinates of a point in space are r, 6, andz.
DEFINmON Cylindrical coordinates represent a point P in space by ordered triples (r, fJ, z) in which 1. r and fJ are polar coordinates for the vertical projection of P on the xy-plane 2. z is the rectangular vertical coordinate.
876
Chapter 15: Multiple Integrals
The values of x, y, r, and () in rectangular and cylindrical coordinates are related by the usual equations. Equations Relating Rectangular (x,y, z) and Cylindrical (r, (J, z) Coordinates x = rcos(),
y = rsin(),
r 2 = x 2 + y2,
z 0=00 ,
randzvary-
z = zo, rand 0 vary
I
tan()
=
y/x
In cylindrical coordinates, the equation r = a describes not just a circle in the xy-plane but an entire cylinder about the z-axis (Figure 15.43). The z-axis is given by r = o. The equation () = ()o describes the plane that contains the z-axis and makes an angle ()o with the positive x-axis. And, just as in rectangular coordinates, the equation z = Zo describes a plane perpendicular to the z-axis. Cylindrical coordinates are good for describing cylinders whose axes run along the z-axis and planes that either contain the z-axis or lie perpendicular to the z-axis. Surfaces like these have equations of constant coordinate value:
r=4 y
() =
Cylinder, radius 4, axis the z-axis
7T
Plane containing the z-axis
3 z = 2.
r = a, oand z vary
FIGURE 15.43 Constant-coordinate equations in cylindrical coordinates yield cylinders and planes.
z = z,
Plane perpendicular to the z-axis
When computing triple integrals over a region D in cylindrical coordinates, we partition the region into n small cylindrical wedges, rather than into rectangular boxes. In the kth cylindrical wedge, r, () and z change by !J.rk, !J.()k, and !J.Zk, and the largest of these numbers among all the cylindrical wedges is called the norm of the partition. We define the triple integral as a limit of Riemann sums using these wedges. The volume of such a cylindrical wedge !J. Vk is obtained by taking the area !J.A k of its base in the rO-plane and multiplying by the height !J.z (Figure 15.44). For a point (rk, ()k, Zk) in the center of the kth wedge, we calculated in polar coordinates that !J.A k = rk !J.rk !J.()k. So !J. Vk = !J.zk rk !J.rk !J.()k and a Riemann sum for f over D has the form n
Sn =
'L f(rk, ()k, Zk) !J.zk rk !J.rk !J.()k. k=!
The triple integral of a function f over D is obtained by taking a limit of such Riemann sums with partitions whose norms approach zero:
I
Volume Differential in Cylindrical Coordinates dV = dzrdrdO
n~~Sn =
Iff D
fdV=
III
fdzrdrd().
D
Triple integrals in cylindrical coordinates are then evaluated as iterated integrals, as in the following example.
z r !:J.r!:J.O
rM
L ~--...,
r!:J.z
~
EXAMPLE 1 Find the limits of integration in cylindrical coordinates for integrating a function fer, (), z) over the region D bounded below by the plane z = 0, laterally by the circular cylinder x 2 + (y - 1)2 = 1, and above by the paraboloid z = x 2 + y2. The base of D is also the region's projection R on the xy-plane. The boundary of R is the circle x 2 + (y - 1f = 1. Its polar coordinate equation is
SoLution
+ (y - 1)2 = 1 2 x + y2 - 2y + 1 = 1 r 2 - 2r sin () = 0 r = 2 sin(). x2
FIGURE 15.44 In cylindrical coordinates the volume of the wedge is approximated by the product a v = az rar ao.
15.7
Top Cartesian: z = x2 Cylindrical: z = ,2
+ y2
D
Triple Integrals in Cylindrical and Spherical Coordinates
877
The region is sketched in Figure 15.45. We find the limits of integration, starting with the z-limits. A line M through a typical point (r, 8) in R parallel to the z-axis enters D at z == 0 and leaves at z == x 2 + y2 == r 2 . Next we find the r-limits of integration. A ray L through (r, 8) from the origin enters R at r == 0 and leaves at r == 2 sin 8. Finally we find the 8-limits of integration. As L sweeps across R, the angle 8 it makes with the positive x-axis runs from 8 == 0 to 8 == 7r. The integral is
111
f(r, 0, z) dV
17r12sinlJlr2 f(r, 0, z) dz r dr dO.
=
•
D i------+y
lJ (r, lJ)
R /
x2
x
i
+ (y - 1 Cartesian: Polar: r = 2 sin lJ
=
L
Example 1 illustrates a good procedure for finding limits of integration in cylindrical coordinates. The procedure is summarized as follows.
1
How to Integrate in Cylindrical Coordinates To evaluate
111
FIGURE 15.45
Finding the limits of integration for evaluating an integral in cylindrical coordinates (Example 1).
f(r, 0, z) dV
D
over a region D in space in cylindrical coordinates, integrating first with respect to z, then with respect to r, and finally with respect to 8, take the following steps. 1.
Sketch. Sketch the region D along with its projection R on the xy-plane. Label the surfaces and curves that bound D and R. z
/-
D ----
y x
r
2.
=
h2(lJ)
Find the z-limits of integration. Draw a line M through a typical point (r, 8) of R parallel to the z-axis. As z increases, M enters D at z == gl (r, 8) and leaves at z == g2(r, 8). These are the z-limits of integration. z M
878
Chapter 15: Multiple Integrals
3.
Find the r-limits of integration. Draw a ray L through (r, 8) from the origin. The ray enters R at r == h 1(8) and leaves at r == h2(8). These are the r-limits of integration. z M
L
4.
Find the 8-limits ofintegration. As L sweeps across R, the angle 8 it makes with the positivex-axis runs from 8 == ex to 8 == f3. These are the 8-limits ofintegration. The integral is
~~if
j(r, 8, z) dV
==
j(r, 8, z) dz r dr d8.
()=a
D
r=h 1(())
z=gl(r, ())
EXAMPLE 2 Find the centroid (8 == 1) of the solid enclosed by the cylinder x 2 + y2 == 4, bounded above by the paraboloid z == x 2 + y2, and bounded below by the xy-plane.
z
4
We sketch the solid, bounded above by the paraboloid z == r 2 and below by the plane z == 0 (Figure 15.46). Its base R is the disk 0 :::; r :::; 2 in the xy-plane. The solid's centroid (x, y, z) lies on its axis of symmetry, here the z-axis. This makes x == y == o. To find z, we divide the first moment M xy by the mass M. To find the limits of integration for the mass and moment integrals, we continue with the four basic steps. We completed our initial sketch. The remaining steps give the limits of integration. The z-limits. A line M through a typical point (r, 8) in the base parallel to the z-axis enters the solid at z == 0 and leaves at z == r 2 • The r-limits. A ray L through (r,8) from the origin enters R at r == 0 and leaves at r == 2. The 8-limits. As L sweeps over the base like a clock hand, the angle 8 it makes with the positive x-axis runs from 8 == 0 to 8 == 27T. The value of M xy is SoLution
Centroid
R
x
l
()=f31r=h2(())lz=g2(r, ())
/\~ y
(r,O)
L
FIGURE 15.46 Example 2 shows how to find the centroid of this solid.
{27T {2 {r
2
Mxy = Jo Jo Jo
-1 1
27T 2
-
o
{27T {2 [ 2]r z dz r dr dO = Jo Jo ~ 0 r dr dO 27T 27T 2
_1
5
r2 dr d8 -
0
[~]2 12 d8
_1
-
000
~ 3 d8 -_
327T 3 .
The value of M is
{27T {2 {r
M= Jo Jo Jo {27T {2
=
Jo Jo
3
{27T {2 [ ]r
2
dz r dr dO =
r dr dO =
Jo Jo
z
2
0
r dr dO
{27T [ 4]2 {27T ~ 0 dO = Jo 4 dO =
Jo
87T.
879
Triple Integrals in Cylindrical and Spherical Coordinates
15.7
Therefore,
z P(p,
_
ep, (J)
M xy
z == M
==
321T 1 4 -3-81T == 3'
•
and the centroid is (0, 0, 4/3). Notice that the centroid lies outside the solid. p
z= x
p cos
ep
--- 7----y (J
r
y
x
--- /
/
FIGURE 15.47 The spherical coordinates 4>, and () and their relation to x,y, z, and r.
Spherical Coordinates and Integration Spherical coordinates locate points in space with two angles and one distance, as shown in Figure 15.47. The first coordinate, p = 1&1, is the point's distance from t~ origin. Unlike r, the variable p is never negative. The second coordinate, ep, is the angle OP makes with the positive z-axis. It is required to lie in the interval [0, 1T]. The third coordinate is the angle () as measured in cylindrical coordinates.
p,
DEFINITION triples (p,
Spherical coordinates represent a point P in space by ordered
ep, ()) in which
1. p is the distance from P to the origin.
2.
ep is the angle & makes with the positive z-axis (0
:::;
ep :::;
1T).
3. () is the angle from cylindrical coordinates (0 :::; () :::; 21T). On maps of the Earth, () is related to the meridian of a point on the Earth and ep to its latitude, while p is related to elevation above the Earth's surface. The equation p == a describes the sphere of radius a centered at the origin (Figure 15.48). The equation ep == epo describes a single cone whose vertex lies at the origin and whose axis lies along the z-axis. (We broaden our interpretation to include the xy-plane as the cone ep == 1T/2.) If epo is greater than 1T/2, the cone ep == epo opens downward. The equation () == (}o describes the half-plane that contains the z-axis and makes an angle (}o with the positive x-axis.
z
ep = epo, p and (J vary
~y
Equations Relating Spherical Coordinates to Cartesian and Cylindrical Coordinates
x
p
=
a,
ep and (J vary
-I
r
==
p sin ep,
x == r cos ()
==
p sin ep cos (),
z
==
p cos ep,
y
r sin ()
==
p sin ep sin (),
p
(J = (Jo,
==
==
VX 2 + Y 2 + z2
==
(1)
V r 2 + z 2.
p and ep vary
FIGURE 15.48 Constant-coordinate equations in spherical coordinates yield spheres, single cones, and half-planes.
EXAMPLE 3 SoLution
Find a spherical coordinate equation for the sphere x 2 + y2
+ (z
- 1)2
We use Equations (1) to substitute for x, y, and z: x2
+ y2 + (z + (p cos ep -
p2 sin2 ep cos2 () + p2 sin2 ep sin2 () p2 sin2 ep( cos2 () + sin2 ()) + p2 cos2 ep
-
1)2
==
1
1)2
==
1
1
==
1
cos2 ep)
==
2p cos ep
p2
==
2p cos ep
p
==
2 cos ep .
2p cos ep
+
Eqs. (1)
1
p2(sin2 ep
+ 1
p
>
0
==
1.
880
Chapter 15: Multiple Integrals
The angle cP varies from 0 at the north pole of the sphere to 1T/2 at the south pole; the angle (J does not appear in the expression for P, reflecting the symmetry about the z-axis (see Figure 15.49). •
z x2
+ y2 + (z
2
- 1)2 = 1 p=2cose/>
EXAMPLE 4
v'x 2 + y2.
SoLution 1 Use geometry. The cone is symmetric with respect to the z-axis and cuts the first quadrant of the yz-plane along the line z = y. The angle between the cone and the positive z-axis is therefore 1T/4 radians. The cone consists of the points whose spherical coordinates have cP equal to 1T/4, so its equation is cP = 1T/4. (See Figure 15.50.)
p
Use algebra. Ifwe use Equations (1) to substitute for x,y, andz we obtain the
SoLution 2
y
FIGURE 15.49
Find a spherical coordinate equation for the cone z =
same result:
The sphere in Example 3.
z
=
v'x 2 + y2 v'p2 sin2 cP
Example 3
=
P sin cP
p
z
=
P cos cP P cos cP
>
0, sine/> ;;, 0
cos cP = sin cP 1T cP = 4·
e/>=J!
4
•
Spherical coordinates are useful for describing spheres centered at the origin, half-planes hinged along the z-axis, and cones whose vertices lie at the origin and whose axes lie along the z-axis. Surfaces like these have equations of constant coordinate value:
y
x
FIGURE 15.50
p=4
Sphere, radius 4, center at origin
cP=1T 3 1T (J = 3.
Cone opening up from the origin, making an angle of 71/3 radians with the positive z-axis
The cone in Example 4.
Half-plane, hinged along the z-axis, making an angle of 71(3 radians with the positive x-axis
When computing triple integrals over a region D in spherical coordinates, we partition the region into n spherical wedges. The size of the kth spherical wedge, which contains a point (Ph cPh (h), is given by the changes aph a(Jh and acPk in P, (J, and cPo Such a spherical wedge has one edge a circular arc of length Pk acPh another edge a circular arc of length Pk sin cPk a(Jh and thickness apk. The spherical wedge closely approximates a cube of these dimensions when aph a(Jh and acPk are all small (Figure 15.51). It can be shown that the volume of this spherical wedge a Vk is a Vk = pl sin cPk apk acPk a(Jk for (Ph cPh (h) a point chosen inside the wedge. The corresponding Riemann sum for a function f(p, cP, (J) is
I Volume Differential in Spherical Coordinates
dV = p2 sin e/> dp de/> dO
z p sin e/> psine/>M
n
Sn
= ~ f(Ph cPh (Jk) Pk2 sin cPk apk acPk a(Jk. k=l
As the norm of a partition approaches zero, and the spherical wedges get smaller, the Riemann sums have a limit when f is continuous:
n~ Sn = y
f(p, cP, (J) dV
D
=
111
2
f(p, cP, (J) p sin cP dp dcP d(J.
D
In spherical coordinates, we have
F
dV = p2 sin cP dp dcP d(J.
x
FIGURE 15.51
dV
111
= =
In spherical coordinates
dp· p de/> • p sin e/> dO p2 sin e/> dp de/> dO.
To evaluate integrals in spherical coordinates, we usually integrate first with respect to p. The procedure for finding the limits of integration is as follows. We restrict our attention to integrating over domains that are solids of revolution about the z-axis (or portions thereof) and for which the limits for (J and cP are constant.
15.7
Triple Integrals in Cylindrical and Spherical Coordinates
881
How to Integrate in Spherical Coordinates To evaluate
iff
f(p,
D
over a region D in space in spherical coordinates, integrating first with respect to p, then with respect to cfJ, and finally with respect to 8, take the following steps. 1.
Sketch. Sketch the region D along with its projection R on the xy-plane. Label the surfaces that bound D.
z
x
2.
Find the p-limits of integration. Draw a ray M from the origin through D making an angle cfJ with the positive z-axis. Also draw the projection of M on the xy-plane (call the projection L). The ray L makes an angle 8 with the positive x-axis. As p increases, M enters D at p == gl (cfJ, 8) and leaves at p == g2( cfJ, 8). These are the p-limits of
integration. z
R
- - - - () =
f3
y L
x
3.
e
cfJ that M makes with the cfJmax. These are the cfJ -limits of integration.
Find the cfJ-limits of integration. For any given 8, the angle
z-axis runs from cfJ
==
cfJmin to cfJ
==
882
Chapter 15: Multiple Integrals
4.
Find the 8-limits ofintegration. The ray L sweeps over R as 8 runs from ex to f3. These are the 8-limits of integration. The integral is
~~if
()= f31¢=¢max
j(p, cP, 8) dV
SoLution over D.
R (J
FIGURE 15.52
¢=¢rnin
P=g2(¢, ())
1
j(p, cP, 8) p2 sin cP dp dcP d8.
p=gl(¢, ())
EXAMPLE 5 Find the volume of the "ice cream cone" D cut from the solid sphere p :::; 1 by the cone cP == 7T/3.
z
Example 5.
1
()=a
D
x
==
L
Y
The ice cream cone in
The volume is V
==
fffn p2 sin cP dp dcP d8,
the integral of j(p, cP, 8)
==
1
To find the limits of integration for evaluating the integral, we begin by sketching D and its projection R on the xy-plane (Figure 15.52). The p-limits ofintegration. We draw a ray M from the origin through D making an angle cP with the positive z-axis. We also draw L, the projection of M on the xy-plane, along with the angle 8 that L makes with the positive x-axis. Ray M enters D at p == 0 and leaves at p == 1. The cP-limits ofintegration. The cone cP == 7T/3 makes an angle of 7T/3 with the positive z-axis. For any given 8, the angle cP can run from cP == 0 to cP == 7T/3. The 8-limits of integration. The ray L sweeps over R as 8 runs from 0 to 27T. The volume is
{27T {7T/3 [p3] 1 {27T {7T/3 1 3 0 sin 4J d4J dO = Jo Jo 3 sin 4J d4J dO
= Jo Jo 27T
==
1 o
[1 - - cos cP ]7T/3 d8 3 0
1
27T
==
0
(-16
+ -1) d8 3
== -1
6
(2 7T)
•
7T. == 3
EXAMPLE 6 A solid of constant density 8 == 1 occupies the region D in Example 5. Find the solid's moment of inertia about the z-axis. SoLution
In rectangular coordinates, the moment is
In spherical coordinates, x 2 + y2 Hence,
==
(p sin cP cos 8)2
+ (p sin cP sin 8)2
==
p2 sin2 cPo
For the region in Example 5, this becomes
1_
{27T {7T/3
= SJo Jo 1_
{27T ( 1
=SJo
2
(1 - cos
4J) sin 4J d4J dO
1 1) -"2+ 1 +24-3
1_
{27T [
=
SJo
1_
{27T 5
dO=SJo
-cos 4J
24 dO
=
cos 3 cP ]7T/3
+ -3-
1 24 (27T)
7T
0
=U·
dO •
15.7 Triple Integrals in Cylindrical and Spherical Coordinates
883
Coordinate Conversion Formulas C'YuNDRICAL TO
SPHERICAL TO
SPHERICAL TO
REcTANGULAR
RECTANGULAR
CYLINDRICAL
rcosO
p sin "'cosO
r=psin'"
y = rsinO
Y = psin"'sinO
z=pcos'"
z=z
z=pcos'"
0=0
X =
X =
Corresponding fonnulas for dV in triple integrals:
dV= dxdydz = dzrdrdO = ;
sin '" dp d'" dO
In the next section we offer a more general procedure for detennining dV in cylindrical and spherical coordinates. The resuits, of course, will he the same.
Exercises 15.7 Evaluating Integrals In CyUndrical Coordinates Evaluate the cylindricsl coordinate integrals in Exercises l--{j.
1.
'WJ.1J~ dzrdrdO
J.o
0
2. J.,wJ.'J,vts=;' dzrdrdO rl~
rOO
3. J.'w f6/'w f3+24,.' dz r dr dO 4. J.w {'Iw
Jo Jo J.'WJ.'Jl~3 o
5.
0
f'~ z dz r dr dO
Jo J-v'h'-
Jo
Changtng the Order of Integration In Cylindrical Coordinates
9. 10.
0
J.
Jo Jo 0 Jo
I
-I 0
(r' cos' 0
J. 'l~J.2w o
r-2
(rsinO
b. dr dz dO
Jo
+ z')r dO dr dz + I)rdOdzdr
Finding Iterated Integrals In CyUndrical Coordinates 13. Give the limits of integration for evaluatiog the integral
ff!
f(r, 0, z) dz r dr dO
14. Convert the integral
[J.VH[(x' +
to an equivalent integral in cylindrical coordinates and evaluate
the result In Exercises 15-20, set up the iterated integral for evaluatiog fffnf(r, 0, z) dz r dr dO over the given regionD.
15. D is the right circular cylinder whose base is the circle r ~ 2 sin 0 in the xy-plane and whose top lies in the plane z = 4 - y.
11. Let D be the region bounded below by the plane z = 0, above by the sphere + y' + = 4, nod on the sides by the cylinder + y' ~ I. Set up the triple integrals in cylindricsl coordinates that give the volume of D using the following orders of integration.
x'
a. dz drdO
y') dz dx dy
z
0
x'
c. dO dzdr
as an iterated integral over the region that is bounded below by the plnoe z = 0, on the side by the cylinder r = cos 0, nod on top by the paraboloid z ~ 3r'.
+ z')dzrdrdO
The integrals we have seen so far suggest that there are preferred orders of integration for cylindricsl coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integrsls in Exercises 7-10. 1 7. J.'w {' f'P r'drdzdO 8. J.'w f1+~"4rdrdOdz
1J.V, f'w
a. dz dr dO
dzrdrdO
6. J.'w f111/2(r'sin'O o -1/2
o
integrals in cylindricsl coordinates that give the volwne of D using the following orders of integration.
z=4-y
z'
b. dr dz dO
c. dO dz dr
12. Let D be the region bounded below by the cone z = y'x' + y' nod above by the paraboloid z = 2 - x' - y'. Set up the triple
/
884
Chapter 15: Multiple Integrals
16. D is the right circular cylinder whose base is the circle r = 3 cos 0 and whose top lies in the plane z = 5 - x.
z
z
2
z=5-x
y
"r =
3 cos ()
x
x
y=x
17. D is the solid right cylinder whose base is the region in the xy-plane that lies inside the cardioid r = 1 + cos 0 and outside the circle r = 1 and whose top lies in the plane z = 4.
z
EvaLuating IntegraLs in SphericaL Coordinates Evaluate the spherical coordinate integrals in Exercises 21-26.
7T {7T {2 sin4J
21.
1io io 1i i 1io io 1 io io 1 1io io
p2 sin c/J dp dc/J dO
0
27T {7T/4 {2
22.
0
0
27T {7T
~
x
r
=
r
=
1
23.
18. D is the solid right cylinder whose base is the region between the circles r = cos 0 and r = 2 cos 0 and whose top lies in the plane z = 3 - y.
{(I-cos 4J)/2
24. 25.
z
\
26.
{I
27T17T/312
o
3
5p3 sin c/J dp dc/J dO
0
3p2 sin 4> dp d4> dO
sec4J
0
27T {7T/4
z=3-y
p2 sin c/J dp dc/J dO
0
37T/2 {7T
1 + cos ()
(p cos c/J) p2 sin c/J dp dc/J dO
0
{sec 4J
(p cos c/J) p2 sin c/J dp dc/J dO
0
Changing the Order of Integration in SphericaL Coordinates The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals in Exercises 27-30.
/\ x
\ r
r
=
=
cos 0
2 cos ()
19. D is the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in theplanez = 2 - y.
z
27.1 ro 2
o J-7T
28. {1T/3
J7T/6
{2 esc q, rz1Tp2 sin c/J dO dp dc/J Jo
Jcsc4J
29. 1111T11T/412P sin3 c/J dc/J dO dp
2 z=2-y
ix I
/
/
30. {7T/2 {
1
7T12
J7T/6 J-7T12
2
y=x
20. D is the prism whose base is the triangle in the xy-plane bounded by the y-axis and the lines y = x and y = 1 and whose top lies in the plane z = 2 - x.
5p4 sin3 c/J dp dO dc/J
esc 4J
31. Let D be the region in Exercise 11. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.
a. dp d4> dO
/ /
x
{1T/2p3 sin 2c/J dc/J dO dp
J7T/4
b. d4> dp dO
32. Let D be the region bounded below by the cone z = v'x 2 + y2 and above by the plane z = 1. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.
a. dp d4> dO
b. d4> dp dO
Finding Iterated IntegraLs in SphericaL Coordinates In Exercises 33-38, (a) find the spherical coordinate limits for the integral that calculates the volume of the given solid and then (b) evaluate the integral. 33. The solid between the sphere p = cos 4> and the hemisphere p = 2,z ~
°
z p = cos
ep
2
885
Triple Integrals in Cylindrical and Spherical Coordinates
15.7
volume of D as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find V. 41. Let D be the smaller cap cut from a solid ball of radius 2 units by a plane 1 unit from the center of the sphere. Express the volume of D as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals. 42. Express the moment of inertia /z of the solid hemisphere x 2 + y2 + z2 ~ 1, z ~ 0, as an iterated integral in (a) cylindrical and (b) spherical coordinates. Then (c) find /z.
P= 2
VoLumes Find the volumes of the solids in Exercises 43-48. 43. x
44.
y
34. The solid bounded below by the hemisphere p = 1, Z above by the cardioid of revolution p = 1 + cos 4>
z
z ~
0, and
z P= 1
y
x
x
y
\
z=-~
z = (x 2 + y2)2 - 1
45. z =-y
f
y
x
46.
z
35. The solid enclosed by the cardioid of revolution p = 1 - cos 4> 36. The upper portion cut from the solid in Exercise 35 by the xy-plane
r
3 cos ()~
=
37. The solid bounded below by the sphere p = 2 cos 4> and above by 2 the cone z = + y2
y
x
"r y
x
v'x
47.
=
-3 cos ()
48.
z
x
y
x
38. The solid bounded below by the xy-plane, on the sides by the sphere p = 2, and above by the cone 4> = '1T/3
r
/ sin () =
y x
y
r
z
=
cos ()
49. Sphere and cones Find the volume of the portion of the solid sphere p ~ a that lies between the cones 4> = '1T /3 and 4> = 2'1T/3. p=2 x
50. Sphere and half-planes Find the volume of the region cut from the solid sphere p ~ a by the half-planes (J = and (J = '1T /6 in the first octant.
°
51. Sphere and plane Find the volume of the smaller region cut from the solid sphere p ~ 2 by the plane z = 1.
Finding TripLe IntegraLs 39. Set up triple integrals for the volume of the sphere p = 2 in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. 40. Let D be the region in the first octant that is bounded below by the cone 4> = '1T /4 and above by the sphere p = 3. Express the
52. Cone and planes Find the volume of the solid enclosed by the cone z = v'x 2 + y2 between the planes z = 1 and z = 2. 53. Cylinder and paraboloid Find the volume of the region bounded below by the plane z = 0, laterally by the cylinder x 2 + y2 = 1, and above by the paraboloid z = x 2 + y2.
886
Chapter 15: Multiple Integrals
54. CyHnder and paraboloids Find 1he volwne ofthe regioo bouoded below by the paraboloid z ~ ",' + y', laterally by the cylinder x' + y' = I,andabovebytheparaboloidz =x' + y' + I.
73. Moment of inertia of solid cone Find the moment of inertia of a right circular cone of base radius I and height I about an axis through the vertex parallel to the base. (Take 8 = I.)
55. Cylinder and cones Find the volume of the solid cut from the thick-walled cylinder 1:5 + y' :5 2 by the cooes z ~
74. Moment of inertia of solid sphere Find the moment of inertia ofa solid sphere of radius a about a diameter. (Take 8 = I.)
x'
±Vx' + y'.
56. Sphere and cylinder
side the sphere
x'
Find 1he volume of the regioo that lies in+ y' + z' ~ 2 and outside the cylinder
x'+y'=1. 57. Cylinder and planes Find the volume ofthe regioo enclosed by the cylinder ",' + y' ~ 4 and the planes z ~ 0 and y + z ~ 4. 58. Cylinder and planes Find the volume of the regioo enclosed by the cylinder + y' = 4 and the planes z = 0 and x + y + z ~ 4.
x'
75. Moment of inertia of solid cone Find the moment of inertia of a right circular cone of base radius a and height h about its axis. (Hint: Place the cone with its vertex at the origin and its axis along the z-axis.) 76. Variable density A solid is bounded on the top by the paraboloid z = r', 00 the bottom by the plane z = 0, and on the sides by the cylinder r = I. Find the center of mass and the moment of inertia about the z-axis if the density is
a. 8(r, 0, z) = z
b. 8(r, 0, z) = r.
59. Region trapped by paraboloids Find the volume of the regioo y' and below by bounded above by 1he paraboloid z = 5 the paraboloid z ~ 4x' + 4y'.
77. Variable density A solid is bounded below by the cone z= x' + y' and above by the plane z = I. Find the center of mass and the moment of inertia about the z-axis if the density is
60. Paraboloid and cylinder Find the volwne of the regioo y', below by the bounded above by the paraboloid z = 9 xy-plane, and lying outside the cylinder ",' + y' ~ I.
78. Variable density A solid ball is bounded by the sphere p Find the moment of inertia about the z-axis if the density is
x' -
x' -
61. Cylinder and sphere Find 1he volume of the regioo cut from the solid cylinder + y' :5 I by the sphere + y' + z' = 4.
x'
x'
62. Sphere and paraboloid Find the volume of the regioo bounded + y' + z' ~ 2 and below by the paraboabove by the sphere loidz = x 2 + y2.
x'
Average Values 63. Find the average value of the functioo f(r, 0, z) = r over the regioo bounded by the cylinder r ~ I between the planes z ~ -I andz = I.
64. Find the average value of the function f(r, 0, z) = r over the solid + z' ~ I. (This is 1he sphere ball bounded by the sphere + y' + = I.)
r'
z'
x'
65. Find the average value of the function f(p, q" 0) = p over the solid ball p :5 I. 66. Find the average value of the functioo f(p, q" 0) the solid upper ball p :5 1,0 :5 q, :5 'If/2.
~
p cos q, over
Masses, Moments, and Centroids 67. Center of mass A solid of cooatant density is bounded below by the plane z = 0, above by the cone z = r, r O?:: 0, and on the sides by the cylinder r = I. Find the center of mass. 68. Centroid Find the centroid of thVCgioo in the fll'St octant that is bounded above by the cone z = + y', below by the plane z = 0, and 00 the sides by the cylinder + y' = 4 and the planes x ~ 0 and y ~ O.
x'
69. Centroid
x'
Find the centroid of the solid in Exercise 38.
70. Centroid Find the centroid of the solid bounded above by the spherep ~ a and below by the cooe q, ~ 'If/4.
V
a. 8(r, 0, z) = z
b. 8(r, 0, z) = z'.
a. 8(p, q" 0) = p'
~
a.
b. 8(p, q" 0) = r = p sin q,.
79. Centroid of solid semiellipsoid Show that the centroid of the solid semiellipsoid ofrevolutioo (r'/a') + (z'/h') :5 I,z '" 0, lies on the z-axis three-eighths of the way from the base to the top. The special case h = a gives a solid hemisphere. Thus, the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base to the top. 80. Centroid of solid cone Show that the centroid of a solid right eircular eone is one-fourth of the way from the base to the vertex. (10 general, the centroid of a solid cone or pyramid is one-fourth of the way from the centroid of the base to the vertex.) 81. Density of eenter of a planet A planet is in the shape of a sphere of radius R and total mass M with spherically symmetric
density distribution that increases linearly as one approaches its center. What is the density at the center of this planet if the density at its edge (surface) is tsken to be zero?
82. Mass of planet's atmosphere A spherical planet of radius R bas an atmosphere whose density is p. = lJ<Je ~', where h is the altitude above the surface of1he planet, IJ<J is the density at sea level, and c is a positive constant Find the mass of1he planet's atmosphere. Theory and Examples 83. Vertical planes in cylindrical coordinates
a. Show that planes perpendicular to the x-axis bave equatiooa of the form r
~
a sec 0 in cylindrical coordinates.
b. Show that planes perpendicular to the y-axis bave equations of the form r = b csc O. 84. (Continuation of Exercise 83.) Find an equation of the form r = f(O) in cylindrical coordinates for 1he plane ax + by = c, c # O.
71. Centroid Find the centroid of 1he regioo that is bounded above by the surface z = 00 the sides by the cylinder r = 4, and below by the xy-plane.
85. Symmetry What symmetry will you fmd in a surface that bas an equatioo of the form r = f(z) in cylindrical coordinates? Give
Find the centroid of the region cut from the solid ball z':5 I by the balf-planes 0 = -'If/3,r '" O,andO = 'If/3,
86. Symmetry What symmetry will you fmd in a surface that bas an equation of the form p = f( q,) in spherical coordinates? Give
v;.,
72. Centroid
r' + r O?:: O.
reasons for your answer.
reasons for your answer.
15.8 Substitutions in Multiple Integrals
15.8
I_SU_b_Sti_·t_u_ti_O_nS_i_n_M_U_Lti~·p_Le_In_te....::g~ra_Ls
o
_(u_ v)
G
887
---------
The goal oftbis section is to introduce you to the ideas involved in coordinate transformations. You will see how to evaluate multiple integrals by substitution in order to replace complicated integrals by ones that are easier to evaluate. Substitutions accomplish this by simplifying the integrand, the limits of integration, or both. A thorough discussion ofmultivariable transformations and substitutions, and the Jacobian, is best left to a more advanced course following a study oflinear algebra.
Substitutions in Double Integrals u ------c+-------o Cartesian uv-plane x
y
~ g(u, 0)
1 y
~
h(u, 0)
The polar coordinate substitution ofSeclion 15.4 is a special case ofa more general substitution method for double integrals, a method that pictures changes in variables as transformations of regions. Suppose that a region G in the uv-piane is transformed one-to-one into the region R in the xy-piane by equations ofthe form
x = g(u, v),
y = h(u, v),
as suggested in Figure 15.53. We call R the image of G under the transformation, and G the preimage of R. Any function j(x, y) defined on R can be thought of as a function j(g(u, v), h(u, v» dermed on G as well. How is the integral of j(x,y) over R related to the integral of j(g(u, v), h(u, v» over G? The answer is: If g, h, and j have continuous partial derivatives and J(u, v) (to be discussed in a moment) is zero only at isolated points, if at all, then Cartesian xy-plane
FIGURE 15.53 The equations x = g(u,v) andy = h(u,v) allow us to change an integral over a region R in the xy-plane into an integral over a region G in the uv-plane by using Equation (I).
ff
j(x,y)dxdy =
ff
(1)
j(g(u,v),h(u,v»IJ(u,v)ldudv.
G
R
The factor J(u, v), whose absolute value appears in Equation (1), is the Jacobian of the coordinate transformation, named after German mathematician Carl Jacobi. It measures how much the transformation is expanding or contracting the area around a point in G as G is transformed into R.
DEFINmON The Jacobian determinant or Jacobian of the coordinate transformationx = g(u, v),y = h(u, v) is
J(u, v) =
ax au
ax av
ay au
ay av
ax ay auav
ay ax au av'
-----
(2)
The Jacobian can also be denoted by
a(x,y) J(u, v) = a(u, v) HIsTORICAL BIOGRAPHY
Carl Gustav Jacob Jacobi (1804-1851)
to help us remember how the determinant in Equation (2) is constrocted from the partial derivatives ofx andy. The derivation of Equation (1) is intricate and properly belongs to a course in advanced calculus. We do not give the derivation here.
EXAMPLE 1 Find the Jacobian for the polar coordinate transformation x = r cos 0, y = r sin 0, and use Equation (1) to write the Cartesian integral fJR j(x, y) dx dy as a polar integral.
888
Chapter 15: Multiple Integrals
o "f--,-,------, 2
G
Solution Figure 15.54 shows how the equations x = r cos 6, y = r sin 6 transfonn the rectangle G: 0 :5 r :5 1,0 :5 0 :5 Tr/2, into the quarter circle R bounded by x 2 + y2 = I in the first quadrant of the ~-plane. For polar coordinates, we have r and 0 in place of u and v. With x = r cos 6 and y = r sin 0, the Jacobian is
J(r,O)
Cartesian rO-plane
=
ax oX or 00 oy
oy
or
00
=
Icoso sinO
11
f(x,y)dxdy
=
R
0=
~
1
r(cos2 0 + sin2 0)
=
Since we assume r ;;,: 0 when integrating in polar coordinates, IJ(r, 0)1 Equation (1) gives
I x=rcos8 ~y=rsin9
y
-rSin°1 rcosO
11
=
=
r.
Irl
=
f(rcos6,rsinO)rdrdO.
r, so that
(3)
G
This is the same fonnula we derived independently using a geometric argument for polar area in Section 15.4. Notice that the integral on the right-hand side of Equation (3) is not the integral of f(r cos 0, r sin 0) over a region in the polar coordinate plane. It is the integral of the product of f(r cos 0, r sin 0) and r over a region G in the Cartesian rO-plane. _ Here is an example of a substitution in which the image of a rectangle under the coordinate transfonnation is a trapezoid. Transformations like this one are called linear transformations.
Cartesian xy-plane
EXAMPLE 2
Evaluate
FIGURE 15.54 The equations x = r cos 8, y = rsin 8 transform G into R.
1o
4:LX~(Y/2)+1
x-y/2
2x - y --dxdy 2
by applying the transfonnation
2x - Y u=--2-'
y v=2
(4)
and integrating over an appropriate region in the uv-plane. Solution We sketch the region R of integration in the xy-plane and identify its boundaries (Figure 15.55).
y
v 2
u=o
v=2
4
u
G
=1
--:ct---;;-~---u
o
v- 0
1
x=u+v y = 2v )
o \
1
y~O
FIGURE 15.55 The equations x = u + v and y = 2v transform G into R. Reversing the transformatioo by the equations u = (2x - y)/2 and v ~ y/2 transforms R into G (Example 2).
15.8 Substitutions in Multiple Integrals
889
To apply Equation (I), we need to fmd the corresponding uv-region G and the Jacobian of the transformation. To find them, we first solve Equations (4) for x andy in terms of u and v. From those equations it is easy to see that
x
u + v,
=
y
=
2v.
(5)
We then find the boundaries of G by substituting these expressions into the equations for the boundaries of R (Figure 15.55). .>y-equations for the boundary of R
x = y/2 x = (y/2) + I y=O y=4
Corresponding uv-equations for the boundary of G
u + v = 2v/2 = v u + v = (2v/2) + I = v 2v = 0 2v = 4
+
Simplified uv-equations
u=o u = I v=O v=2
I
The Jacobian of the transformation (again from Equations (5» is
ax au J(u, v) = ay au
ax av ay av
a au(u + v)
a av(u + v)
a au (2v)
a av (2v)
=
I~ ~I
=
2.
We now have everything we need to apply Equation (I):
:L o 1 4
X
l
-(Y/2 +1
2x - y -2-dxdy
=
1"-21"-'
x=yj2
v
v=O
ulJ(u,v)ldudv
u=O
=f[(U)(2) du dv
v=u
I
=
f
[u
2
]:
dv
=f dv =2.
--c1(--+--~U
• EXAMPLE 3
tl=-2u
Evaluate
r
f'-x
10 '10
-2
~ (y
- 2x)' dy dx.
Solution We sketch the region R ofintegration in the xy-plane and identify its boundaries (Figure 15.56). The integrand suggests the transformation u = x + y and v = y - 2x. Routine algebra produces x and y as functions of u and v:
y
x x+y=l
x=o R
-0+--'\---"-----> X
=
u
v
3 - 3'
y
2u
=
v
3 + 3'
(6)
From Equations (6), we can find the boundaries of the uv-region G (Figure 15.56). .>y-equations for the boundary of R
Corresponding uv-equations for the boundary of G
Simplified uv-equations
y~O
FIGURE 15.56 The equations x ~ (u/3) - (v/3) andy = (2u/3) + (v/3) transform G into R. Reversing the transformation by the equations u ~ x + y and v = y - 2x transforms R into G (Example 3).
x+y=1 x=O y=O
(1-¥)+(2;+¥)=1 '!_"=O 3 3 2u+,,=0 3
3
u
=I
v=u
v
=-2u
890
Chapter 15: Multiple Integrals The Jacobian of the transfonnation in Equations (6) is
J(u, v) =
ax au
ax av
I 3
I 3
ay au
ay av
2 3
I 3
I 3.
Applying Equation (I), we evaluate the integral:
=
1'1" o
11.'
=-
9
u'/2 v 2
-2u
(I)
1.£1 [I
dv du = 3
-
3
ul/2(u 3 +8u 3 )du=
11 0
0
0
u 1/ 2 -v3 3
]V-"
du
v--2u
2]1 29
u7/2du=-u912 9
=-.
0
•
In the next example we illustrate a nonlinear transfonnation ofcoordinates resulting from simplifying the fonn of the integrand Like the polar coordinatea' transfonnation, nonlinear transfonnations can map a straight line boundary of a region into a curved boundary (or vice versa with the inverse transfonnation). In general, nonlinear transfunnations are more complex to analyze than linear ooes, and a complete treatment is left to a more advanced course. y
EXAMPLE 4
Evaluate the integral
2 f--\~-'--~./-
1r ~eVxYdxdY. 2
, }'/Y'/X
1
Solution
The square root terms in the integt'lll!d suggest that we might sinoplify the inteand v = VYf;:. Squaring these equatioos, we readily gratioo by substituting u = have u 2 = xy and v 2 = y/x, which inoply that u 2if = y2 and u 2/if = x 2. So we obtain the transfonnation (in the same ordering ofthe variables as discussed before)
-c-l'-------'-----'-~.x
o
1
2
'\IxY
u x=v
FIGURE 15.57 The region of integration R in Example 4.
y = uv.
and
Let's firat see what happens to the integrand itself under this transfonnatioo. The Jacobian ofthe transfonnation is
u=1
J-----
J(u, v) =
xy=l
/
v=l
1
2
ax av
I v
-u
ay au
ay av
v
u
if
2u v'
If G is the region of integration in the uv-plane, then by Equation (I) the transfonned
I
y=x ,
double integral under the substitution is
I
0
ax au
"
FIGURE 15.58 The boundaries of the region G correspond to those of region R in Figure 15.57. Notice as we move counterclockwise around the region R, we also move counterclockwise around the region G. The inverse transformation equations u ~ v'i,Y. v ~ vY7i prodoce the region G from the region R.
11 ~ R
e VxY dx dy =
11
ve"
~ du dv =
G
11
2ue" du dv.
G
The transfonned integrand function is easier to integrate than the original one, so we proceed to detennine the limits ofintegratioo for the transfonned integral. The regioo of integratioo R of the original integral in the xy-plane is shown in Figore and v = VYf;:, we see that the inoage of 15.57. From the substitution equations u = the left-hand boundary xy = I for R is the vertical line segment u = 1,2 ;;,: v ;;,: I, in G (see Figure 15.58). Likewise, the right-hand boundary y = x of R maps to the horizontal line segment v = I, I :5 U :5 2, in G. Finally, the horizontal top boundary Y = 2 of R
'\IxY
Substitutions in Multiple Integrals
15.8
891
maps to uv == 2, 1 :::; v :::; 2, in G. As we move counterclockwise around the boundary of the region R, we also move counterclockwise around the boundary of G, as shown in Figure 15.58. Knowing the region of integration G in the uv-plane, we can now write equivalent iterated integrals:
11~ 2
Y
1
x e vxy dx dy
11 2
==
l/y
1
2
1
/
U
2ue u dv duo
Note the order of integration.
We now evaluate the transformed integral on the right-hand side,
11 2
2
U
/
u
2ue dv du =
=
21 21
2
vue U
]~:7/u du
2
(2e U
ue U ) du
-
=
21 (2 -
==
2[(2 - u)e U
2
==
2(e
2
-
u)e U du
+
(e
+
e U ]::~
e))
==
Integrate by parts.
2e(e - 2).
•
Substitutions in Triple Integrals The cylindrical and spherical coordinate substitutions in Section 15.7 are special cases of a substitution method that pictures changes of variables in triple integrals as transformations of three-dimensional regions. The method is like the method for double integrals except that now we work in three dimensions instead of two. Suppose that a region G in uvw-space is transformed one-to-one into the region D in xyz-space by differentiable equations of the form
x
==
g(u, v, w),
y
==
h(u, v, w),
z == k(u, v, w),
as suggested in Figure 15.59. Then any function F(x, y, z) defined on D can be thought of as a function F(g(u, v, w), h(u, v, w), k(u, v, w))
H(u, v, w)
==
defined on G. If g, h, and k have continuous first partial derivatives, then the integral of F(x, y, z) over D is related to the integral of H(u, v, w) over G by the equation
iff
F(x,y,z)dxdydz =
D
iii
H(u,v,w)IJ(u,v,w)ldudvdw.
G
z
w x = g(u, v, w) y = h(u, v, w) z = k(u, v, w) )
y
v u
Cartesian uvw-space
x
Cartesian xyz-space
FIGURE 15.59 The equations x = g(u, v, w),y = h(u, v, w), and z = k(u, v, w) allow us to change an integral over a region D in Cartesian xyz-space into an integral over a region G in Cartesian uyw-space using Equation (7).
(7)
892
Chapter 15: Multiple Integrals The factor J(u, v, w), whose absolute value appears in this equation, is the Jacobian determinant
Cube with sides parallel to the coordinate axes
z
J(u,v,w)
==
8
r
ax au ay au
ax av ay av
ax aw ay aw
az au
az av
az aw
a(x, y, z) a(u, v, w)·
Cartesian r8z-space
This determinant measures how much the volume near a point in G is being expanded or contracted by the transformation from (u, v, w) to (x, y, z) coordinates. As in the twodimensional case, the derivation of the change-of-variable formula in Equation (7) is omitted. For cylindrical coordinates, r, 8 , and z take the place of u, v, and w. The transformation from Cartesian r8z-space to Cartesian xyz-space is given by the equations
x = r cos 8 y = r sin 8
1
z=z
z
x
==
r cos 8,
y
==
z==z
r sin 8,
(Figure 15.60). The Jacobian of the transformation is D
8 = constant x
Y
J(r, 8, z)
Cartesian xyz-space
FIGURE 15.60 The equations x = r cos (), y = r sin (), and z = z transform the cube G into a cylindrical wedgeD.
==
ax ar ay ar
ax a8 ay a8
ax az ay az
az ar
az a8
az az
cos 8 sin 8 0
-r sin 8 r cos 8 0
0 0 1
The corresponding version of Equation (7) is
fff
F(x,y,z)dxdydz
fff
=
D
H(r,O,z)lrldrdOdz.
G
We can drop the absolute value signs whenever r ~ o. For spherical coordinates, p, c/J, and 8 take the place of u, v, and w. The transformation from Cartesian pc/J8-space to Cartesian xyz-space is given by
x
==
p sin c/J cos 8,
y
==
p sin c/J sin 8,
z
==
p cos c/J
(Figure 15.61). The Jacobian of the transformation (see Exercise 19) is
J(p, c/J, 8)
ax ap ay ==
ax
ac/J ay
ap
ac/J
az
az
ap
ac/J
ax a8 ay a8
==
p2 sin c/J.
az a8
The corresponding version of Equation (7) is
fff D
F(x, y, z) dx dy dz
=
fff G
H(p,
Substitutions in Multiple Integrals
15.8
(J
893
p = constant
Cube with sides parallel to the coordinate axes
(J =
constant
sin ep cos (J sin ep sin (J z = p cos ep
x y
= p = p
)
G
(J
p
Cartesian pep(J-space
x
Cartesian xyz-space
FIGURE 15.61 The equations x = p sin 4> cos (J, y = p sin 4> sin (J, and z = p cos 4> transform the cube G into the spherical wedge D.
We can drop the absolute value signs because sin cP is never negative for 0 :::; cP :::; 'IT. Note that this is the same result we obtained in Section 15.7. Here is an example of another substitution. Although we could evaluate the integral in this example directly, we have chosen it to illustrate the substitution method in a simple (and fairly intuitive) setting.
w
EXAMPLE 5
Evaluate
1 1 1 4
3
2
)
o
~y
x=u+ y Y = 2y
l
=(Y/2)+1
(2X y - -+ -z) dxdydz 2
x=y/2
u == (2x - y)/2,
z = 3w
z
3
Rear plane:
x D
=
~
2'
==
z/3
(8)
SoLution We sketch the region D of integration in xyz-space and identify its boundaries (Figure 15.62). In this case, the bounding surfaces are planes. To apply Equation (7), we need to find the corresponding uvw-region G and the Jacobian of the transformation. To find them, we first solve Equations (8) for x, y, and z in terms of u, v, and w. Routine algebra gives
ory = 2x
)
x
)
4 /
~ + 1, or y
=
==
u
+ v,
y
==
2v,
z == 3w.
(9)
y
Front plane: =
w
v == y/2,
and integrating over an appropriate region in uvw-space.
3
x
0
by applying the transformation
u
x
X
We then find the boundaries of G by substituting these expressions into the equations for the boundaries of D:
2x - 2
FIGURE 15.62 The equations x = u + v,y = 2v, andz = 3w transform G into D. Reversing the transformation by the equations u = (2x - y)/2, v = y/2, and w = z/3 transforms D into G (Example 5).
xyz-equations for the boundary of D x
==
Corresponding uvw-equations for the boundary of G u
y/2
x == (y/2) y==O
+
1
u
+v
+v ==
==
2v/2 == v
(2v/2)
+
1 == v
+
Simplified uvw-equations
u u
==
0
==
1
2v == 0
v==O
y==4
2v == 4
v==2
z==O
3w == 0
w==O
z == 3
3w == 3
w==1
894
Chapter 15: Multiple Integrals The Jacobian of the transfonnation, again from Equations (9), is
J(u, v, w) =
ax au
ax av
ax aw
ay au
ay av
ay aw
az au
az av
az aw
I 0 0
I 2 0
0 0 = 6. 3
We now bave everything we need to apply Equation (7):
1 1
4 :£X-(Y/2)+ 1
3
(2x---y + -z) dxdydz
23 2 = 1 1 1\u + w)lJ(u,v,w)ldudvdw
00x=y/2 1
=
=
=
1J1\u+W)(6)dudvdw=61J[~2 +uwJ>vdw
61J (! 6[w
+ W2]~
+
w) dvdw
=
61' [¥
+ vw
J: dw
=
61'(1
+ 2w) dw
•
= 6(2) = 12.
Exercises 15.8 Jacoblans and Tnnsfonned Regions In the Plane 1. a. Solve the system
u = x - y,
v=2x+y
for x and y in terms of u and v. Then rmd the value of the Jacobian a(x, y)/a(u, v). b. Find the image under the transformatioo u ~ x - y, v = 2x + Y of the triangularregioo with vertices (0, 0), (I, I), and (I, -2) in the xy-plane. Sketcb the transformed regioo in the uv-plane. 2. a. Solve the system
u = x + 2y,
v=x-y
for x and y in terms of u and v. Then rmd the value of the Jacobian a(x, y)/a(u, v). b. Find the image under the transformatioo u ~ x + 2y, v = x - y of the triangular region in the xy-plane bounded by the lines y = 0, y = x , and x + 2y = 2. Sketch the transformed region in the uv-plane. 3. a. Solve the system
u = 3x
by the x-axis, the y-axis, and the line x transformed region in the uv-p1ane.
+y
= I. Sketch the
4. a. Solve the system
u
= 2x
- 3y,
v = -x + y
for x and y in terms of u and v. Then rmd the value of the Jacobian a(x,y)/a(u, v). b. Find the image uoder the transformation u = 2x - 3y, v ~ -x + y of the parallelogram R in the xy-plane with bouodaties x = -3, x = O,y = x, and y = x + 1. Sketch the transformed regioo in the uv-plane.
Substitutions In Double Integrals 5. Evaluate the integral
4:£X-(Y/2)+1 2x - Y /.o ;;-y/2
--dxdy 2
from Example I directly by integration with respect to x and y to
confmn that its value is 2. 6. Use the transformation in Exercise I to evaluate the integral
+ 2y,
v = x
+ 4y
for x andy in terms ofu and v. Then rmd the value of the Jacobian a(x, y)/a(u, v). b. Find the image under the transformatioo u ~ 3x + 2y, v = x + 4y ofthe triangular region in thexy-plane bounded
JJ(2x 2 - xy - y2) dx dy R
for the regioo R in the rust quadrant bounded by the lines y = -2x + 4,y = -2x + 7,y = x - 2, andy = x + 1.
15.8
7. Vse the transfonnation in Exercise 3 to evaluate the integral
Substitutions in Multiple Integrals
16. Vse the transformation x = u 2
-
895
,}l,y = 2uvto evaluate the in-
tegral
jj(3x 2 + 14xy + 8y 2) dx dy
t
r2~ Jo Jo
R
for the region R in the fIrst quadrant bounded by the lines y = -(3/2)x + I, y = -(3/2)x + 3, y = -(1/4)x, and y = -(1/4)x + 1.
8. Vse the transformation and parailelogramR in Exercise 4 to evaluate the integral
Finding Jacobians
a. x = ucosv, y = usinv
R
9. Let R be the region in the flIst quadrant of the xy-plane bounded by the hyperbolas xy = I, xy = 9 and the lines y = x, y = 4x. Vse the transformation x = u/v,y = uv with u > 0 and v > 0 to rewrite
jj
(Ix +
V;;)dxdy
R
as an integral over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G. 10. a. Find the Jacobian of the transfonnation x = u, y = uv and sketch the region G: 1 ::s; u ::s; 2, 1 ::s; uv ::s; 2, in the uv-plane. b. Then use Equation (I) to transform the integral
2{2y
1J, x 1
12. The area of an ellipse The area '/Tab of the ellipse x2/a 2 + y2/b 2 ~ I can be found by integrating the function j(x, y) = I over the region bounded by the ellipse in the 'o/"plane. Evaluating the integral directly requires a trigonometric substi1ution. An easier way to evaluate the integral is to use the transfor-
au, y
=
+ v2
18. Find the Jacobiana(x,y, z)/a(u, v, w) of the transfonnation
a. x = ucosv, y = usinv, z = w b. x = 2u - I, Y = 3v - 4, z = (1/2)(w - 4). 19. Evaluate the appropriate determinant to sbow that the Jacobian of the transformation from Cartesian p. 20. Substitutions in single integrals How can substi1utions in single defInite integrals be viewed as transfonnations of regions? What is the Jacobian in such a case? lllus1rate with an example.
Substitutions In Triple Integrals x,y,andz.
11. Polar moment of inertia of an elliptical plate A thin plate of constant density covers the region bounded by the ellipse x 2/a 2 + y2/b 2 = I, a > 0, b > 0, in the xy-plane. Find the frrst moment of the plate about the origin. (Hint: V se the transformationx = arcos8,y = brsin8.)
=
b. x = usinv, Y = ucosv.
21. Evaluate the integral in Example 5 by integrating with respect to
dydx
into an integral over G, and evaluate bo1h integrals.
mation x
(Hint Show that the image of the triangular region G with vertices (0, 0), (I, 0), (I, I) in the uv-plane is the region of integration R in the xy-plane defmed by the limits of integration.)
17. Find the Jacobiana(x,y)/a(u, v) of the transformation
jj2(X - y)dxdy.
the disk G: u 2
Yx 2 + y 2 dydx.
bv and evaluate the transformed integral over I in the uv-plane. Find the area this way.
22. Volume of an ellipsoid
Find the volume of the ellipsoid
= au, y = 00, and z = cwo Then find the volume of an appropriate region in uvw-space.)
(Hint: Let x 23. Evaluate
over the solid ellipsoid
'"
x2 y2 z2 2+2+2~1.
13. Vse the transformation in Exercise 2 to evaluate the integral
abc
2/3 [2-2y
o Jy ].
(x + 2y)e(y-r) dx dy
(Hint: Let x = au, y = bv, and z appropriate region in uvw-space.)
by flIst writing it as an integral over a region G in the uv-plane.
14. Vse the transformation x
~
u
+ (l/2)v,y
~
24. Let D be the region in xyz-space defmed by the inequalities
v to evaluate the 1
integral 2 {(y+4)/2 y'(2x
_ y)e(2r-y )' dx dy
].
o }yf2
by lust writing it as an integral over a region G in the uv-plane. 15. Vse the transformation x ~ u/v,y ~ uv to evaluate the integral
sam
11 2
1
Y
l(y
= cw. Then integrate over an
::s;
x ::s; 2,
O::s; xy
::s;
2,
O::s; z
~
1.
Evaluate
ffj(x2y
+ 3xyz) dx dy dz
D
by applying the transfonnation
(x 2 + y2) dx dy
+ 1414(Y (x 2 + y2) dx dy. 2
y(4
u
= x,
v = xy,
w = 3z
and integrating over an appropriate region G in uvw-space.
896
Chapter 15: Multiple Integrals
25. Centroid of a solid semieIIipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base toward the top, show, by traosforming the appropriate integrals, that the center of mass of a solid semiellipsoid (x 2Ia 2) + (y2Ib 2) + (z 2/e 2) '" I, z '" 0, lies on the z-axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)
Chapter
26. Cylindrieal sheIls In Section 6.2, we learned how to rmd the volume of a solid ofrevolution using the shell method; namely, if the region between the curve y = f(x) and the x-axis from a to b (0 < a < b) is revolved about the y-axis, the volume of the 2'lfxf(x) fix. Prove that rmding volumes by resulting solid is using triple integrals gives the same result. (Hint: Use cylindrical coordinates with the roles ofy andz cbanged.)
1:
Questions to Guide Your Review
1. Derme the double integral of a function of two variables over a bounded region in the coordinate plane. 2. How are double integrals evaluated as iterated integrals? Does the order of integration matter? How are the limits of integration determined? Give examples.
3. How are double integrals used to calculate areas and average values. Give examples. 4. How can you change a double integral in rectangular coordinates into a double integral in polar coordinates? Why might it be worthwhile to do so? Give an example. 5. Derme the triple integral of a function f(x, y, z) over a bounded region in space.
7. How are double and triple integrals in rectangular coordinates used to calculate volumes, average values, masses, moments, and
centers of mass? Give examples. 8. How are triple integrals dermed in cylindrical and sphetical coordinates? Why might one prefer working in one of these coordinate systems to worldng in rectangular coordinates? 9. How are triple integrals in cylindrical and sphetical coordinates evaluated? How are the limits ofintegration found? Give examples.
10. How are substitutions in double integrals pictured as transformations of two-dimensional regions? Give a sample calculation.
11. How are substitutions in triple integrals pictured as transformations of three-dimensional regions? Give a sample calculation.
6. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example.
Chapter
Practice Exercises
Evaluating Double Iterated Integrals
Areas and Volumes Using Double Integrals
In Exercises 1-4, sketch the region of integration and evaluate the double integral.
13. Area between line and parabola Find the area of the region enclosed by the line y ~ 2x + 4 and the parabola y ~ 4 - x 2 in the xy-plane.
['0 {IIY
1.
3.
JI Jo
/,o
ye" fix dy
,/2:£v9=4t' tdsdt -V9-4t 2
2./,1/," 4.
/,o
eY/r dyfix
IJ2-v,; xyfixdy v,;
In Exercises 5-8, sketch the region of integration and write an equivalent integral with the order of integration reversed. Then evaluate both integrals. 5.
7.
4:£(;;-4)/2 /,o -V4=Y fixdy /,o
,/2:£V9=4Y' yfixdy -V9-4,,'
6.
{".Ix dy fix Jo lx ['
2
{2
8.
r-r'
Jo Jo
2xdyfix
9. [' {'4
h~
11.
/, sh o
2
dyfix ~ y4 + 1
fix dy
10.
12.
{2 (I er' fix dy h~
/,'h'
o"o/Y
angular" region in the xy-plane that is bounded on the right by the parabola y = x 2 , on the left by the line x + y = 2, and above by the line y ~ 4. 15. Volume of the region under a paraboloid Find the volume under the paraboloid z = x 2 + y2 above the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xy-plane.
16. Volume of the region under parabolic cylinder Find the volume under the parabolic cylinder z = x 2 above the region enclosed by the parabola y ~ 6 - x 2 and the line y ~ x in the xy-plane.
Average Values
Evaluate the integrals in Exercises 9-12. cos (x 2 )
14. Area bonnded by lines and parabola Find the area of the ''tri-
2'If sm'lfX . 2 fix dy x2
Find the average value of f(x, y) and 18.
= xy over the regions in Exercises 17
17. The square bounded by the lines x quadrant
~
1, Y
~
1 in the first
18. The quarter circle x 2 + y2 '" 1 in the Itrst quadnlnt
Chapter 15 PoLar Coordinates Evaluate the integrals in Exercises 19 and 20 by changing to polar coordinates.
19. 20.
1
11~
-1 1
b. First quadrant
The triangle with vertices (0, 0), (1, 0),
The first quadrant of the xy-plane.
24.
0
2
l
1
0
32. Rectangular to cylindrical coordinates (a) Convert to cylindrical coordinates. Then (b) evaluate the new integral.
11~ -~
1
VoLumes and Average VaLues Using TripLe IntegraLs 27. Volume Find the volume of the wedge-shaped region enclosed on the side by the cylinder x = -cosy, -7r/2 ~ Y ~ 7r/2, on the top by the plane z = - 2x, and below by the xy-plane.
2
(x
J-v'x 2+y2
dzdydx
34. Rectangular, cylindrical, and spherical coordinates Write an iterated triple integral for the integral of j(x, y, z) = 6 + 4yover the region in the first octant bounded by the cone z = v'x 2 + y2 , the cylinder x 2 + y2 = 1, and the coordinate planes in (a) rectangular coordinates, (b) cylindrical coordinates, and (c) spherical coordinates. Then (d) find the integral of j by evaluating one of the triple integrals. 35. Cylindrical to rectangular coordinates Set up an integral in rectangular coordinates equivalent to the integral
{7T/2 {v'3 (vi4=?
io il il =
2
+y ) 21xy2 dz dy dx _(x 2+y2)
-1
Z
x
r ~ 0
to (a) rectangular coordinates with the order of integration dz dx dy and (b) spherical coordinates. Then (c) evaluate one of the integrals.
e -~ 1 11~
e(x+y+z) dz dy dx
In4
eix1z2Y 3dydzdx 1
3 dz r dr dO,
r
33. Rectangular to spherical coordinates (a) Convert to spherical coordinates. Then (b) evaluate the new integral.
cos (x + y + z) dx dy dz
{I {X {x+Y 25. Jo Jo Jo (2x - y - z) dz dy dx 26.
Convert
{27T (Vijvl4=? Jo Jo
o
7T
ln71ln2llns
In6
of j(x, y, z) =
by the coordinate planes and the planes x = 1, y = 3, z = 1.
1
EvaLuating TripLe Iterated Integrals Evaluate the integrals in Exercises 23-26.
111 1
value
CyLindricaL and SphericaL Coordinates 31. Cylindrical to rectangular coordinates
+ y2 + 1) dx dy
a. Triangular region and (1, \13).
23.
average
2
21. Integrating over lemniscate Integrate the function j(x, y) = 1/(1 + x 2 + y2)2 over the region enclosed by one loop of the lemniscate (x 2 + y2)2 - (x 2 - y2) = O. 22. Integrate j(x,y) = 1/(1 + x 2 + y2)2 over
7T
value Find the
w+y over the rectangular solid in the first octant bounded
30. Average value Find the average value of p over the solid sphere p ~ a (spherical coordinates).
-1
7T
30xz
897
2dydx
_~ (1 + x 2 + y2)2
l 1V!=.? In (x -V!=.?
29. Average
Practice Exercises
3
r (sin 0 cos O)z2 dz dr dO.
Arrange the order of integration to be z first, then y, then x.
-cosy
36. Rectangular to cylindrical coordinates The volume of a solid is
1
o
1T
x
2
y
28. Volume Find the volume of the solid that is bounded above by the cylinder z = 4 - x 2, on the sides by the cylinder x 2 + y2 = 4, and below by the xy-plane. z
21~1-v'4-~-Y2 0
--v'4-x 2_y2
dz dy dx.
a. Describe the solid by giving equations for the surfaces that form its boundary. b. Convert the integral to cylindrical coordinates but do not evaluate the integral.
37. Spherical versus cylindrical coordinates Triple integrals involving spherical shapes do not always require spherical coordinates for convenient evaluation. Some calculations may be accomplished more easily with cylindrical coordinates. As a case in point, find the volume of the region bounded above by the sphere x 2 + y2 + z2 = 8 and below by the plane z = 2 by using (a) cylindrical coordinates and (b) spherical coordinates. Masses and Moments 38. Finding I z in spherical coordinates Find the moment of inertia about the z-axis of a solid of constant density S = 1 that is bounded above by the sphere p = 2 and below by the cone cf> = 7r/3 (spherical coordinates).
898
Chapter 15: Multiple Integrals
39. Moment orinertia of a "thick" sphere Find the moment of inertia of a solid of constant density 8 bounded by two concentric spheres ofradii a and b (a < b) about a diameter.
40. Moment of inertia of an apple Find the moment of inertia about the z-axis of a solid of density 8 ~ I enclosed by the spberical coordinate surface p = I - cos t/> • The solid is the red curve rotated about the z-axis in the accompanying figure.
z
bounded by the line y = x and the parabola y = x 2 in the xy-plane ifthe density is 8(x, y) = x + I . 47. Plate with variable density Find the mass and first moments about the coordinate axes of a thin square plate bounded by the linesx ~ ± I, Y ~ ± I in the xy-plane if the density is 8(x, y) ~ x 2 + y2 + 1/3. 48. Triangles with same inertial moment Find the moment of inertia about the x-axis of a thin triangular plate of constant density 8 whose base lies along the interval [0, b] on the x-axis and whose vertex lies on the line y = h above the x-axis. As you will see, it does not matter where on the line this vertex lies. All such triangles have the same moment of inertia about the x-axis. 49. Centroid Find the centroid ofthe region in the polar coordinate plane defined by the inequalities 0 :5 r :5 3, -7f/3 :5 9 :5 7f/3. 50. Centroid Find the centroid of the region in the firat quadrant bounded by the rays 9 = 0 and 9 = 7f/2 and the circles r = I andr ~ 3.
51. a. Centroid Find the centroid of the region in the polar coordinate plane that lies inside the cardioid r = 1 + cos (} and outside the circle r ~ 1. 41. Centroid Find the centroid of the "triangular" region bounded by the lines x = 2, Y = 2 and the hyperbola xy = 2 in the xy-plane. 42. Centroid Find the centroid of the region between the parabola x + y2 - 2y = 0 and the line x + 2y = 0 in the xy-plane. 43. Polar moment Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 = 3 bounded by the y-axis and the lines y = 2x and y = 4 in the xy-plane. 44. Polar moment Find the polar moment of inertia about the center of a thin rectangular sheet of constant density 8 = I bounded by the lines
a. x = ±2, y = ± I in the xy-plane ~ ±a, y ~ ±b in the xy-plane.
b. x
(Hint: Find Ix. Then use the formula for Ix to fmd Iy and add the two to find 10). 45. Inertial moment Find the moment of inertia about the x-axis of a thin plate of constant density 8 covering the triangle with vertices (0, 0), (3, 0), and (3, 2) in thexy-plane. 46. Plate with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin plate
Chapter
b. Sketch the region and show the centroid in your sketch.
52. a. Centroid Find the centroid of the plane region defined by the polar coordinate inequalities 0 ~ r ~ a, -a ::=:; (J ~ a (0 < a::;;
'7T).Howdoesthecentroidmoveasa~'7T-?
b. Sketch the region fora ~ 57f/6 and show the centroid in your sketch.
Substitutions 53. Show that if u = x - y and v = y, then
54. What relationship must hold between the constants a, b, and c to mske
(Hint: l.et s = ax + ,By and t = yx + 8y, where (a8 - ,By)' ac - b 2. Thenax2 + 2bxy + cy2 = 8 2 + t 2.)
=
Additional and Advanced Exerdses
Volumes 1. Sand pile: double and triple integrals The base of a sand pile covers the region in the xy-plane that is bounded by the parabola x 2 + y = 6 and the line y = x . The height of the sand above the point (x, y) is x 2 • Express the volume of sand as (a) a double integral, (b) a triple integral. Then (c) fmd the volume.
3. Solid cylindrical region between two plane. Find the volume ofthe portion of the solidcylinderx 2 + y2:5 I that lies between the planes z ~ o and x + y + z ~ 2.
2. Water in a hemispherical bowl A hemispherical bowl of radius 5 em is filled with water to within 3 em of the top. Find the volume of water in the bowl.
5. Two paraboloids Find the volume of the region bounded above by the paraboloid z = 3 - x 2 - y2 and below by the paraboloid z ~ 2x 2 + 2y 2.
4. Spbere and paraboloid Find the volume of the region bounded above by the sphere x 2
loidz
= xl
+ y2 + z2
= 2 and below by the parabo-
+ y2,
Chapter 15 Additional and Advanced Exercises 6. Spherical coordinates Find the volume of the region enclosed by the spherical coordinate surface p = 2 sin c/J (see accompanying figure).
p=2sinq,
Similarly, it can be shown that
ll
x
V
u
em(x-t) f(t) dt du dv = lx (x - t)2 em(x-t) f(t) dt. 2 l0 0 0 0
14. Transforming a double integral to obtain constant limits Sometimes a multiple integral with variable limits can be changed into one with constant limits. By changing the order of integration, show that y
1 1
X
f(x) ( l g(x - y)f(y) dY ) dx
1 (1 1
=
7. Hole in sphere A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is
=
1
f(y)
g(x - y)f(x) dx) dy
Ire Jo
g( Ix - yl )f(x)f(y) dx dy.
"2Jo
Masses and Moments
[2" [V3 [~
Jo Jo Jl
V= 2
899
r dr dz dO.
a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral.
8. Sphere and cylinder Find the volume of material cut from the solid sphere r 2 + z2 ::::; 9 by the cylinder r = 3 sin o. 9. Two paraboloids Find the volume ofthe region enclosed by the surfaces z = x 2 + y2 and z = (x 2 + y2 + 1)/2.
15. Minimizing polar inertia A thin plate of constant density is to occupy the triangular region in the first quadrant of the xy-plane having vertices (0, 0), (a, 0), and (a, l/a). What value of a will minimize the plate's polar moment of inertia about the origin?
16. Polar inertia of triangular plate Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 = 3 bounded by the y-axis and the lines y = 2x and y = 4 in the xy-plane.
10. Cylinder and surface z = xy Find the volume of the region in the first octant that lies between the cylinders r = I and r = 2 and that is bounded below by the xy-plane and above by the surfacez = xy.
17. Mass and polar inertia of a counterweight The counterweight of a flywheel of constant density I has the form of the smaller segment cut from a circle of radius a by a chord at a distance b from the center (b < a). Find the mass of the counterweight and its polar moment of inertia about the center of the wheel.
Changing the Order of Integration
18. Centroid of boomerang Find the centroid of the boomerangshaped region between the parabolas y2 = -4(x - I) and y2 = -2(x - 2) in thexy-plane.
11. Evaluate the integral
1
00
e-ax -
o
dx. Theory and Examples
(Hint: Use the relation
e -ax - e -bx x
e-bx
x
=
l
19. Evaluate b
e-XYdy
a
to form a double integral and evaluate the integral by changing the order of integration.)
12. a. Polar coordinates that
l
o
Show, by changing to polar coordinates,
aSinf3j~ In(x 2 + y2)dxdy
(I) -"2 '
= a 2{3 Ina
ycotf3
where a
where a and b are positive numbers and
20. Show that
> 0 and 0 < (3 < 'TT' /2.
b. Rewrite the Cartesian integral with the order of integration reversed.
over the rectangle Xo ::::; x ::::; xl, Yo ::::; Y ::::; y[, is
13. Reducing a double to a single integral By changing the order of integration, show that the following double integral can be reduced to a single integral: l
x l
u
em(x-t) f(t) dt du = l
x (x - t)em(x-t) f(t) dt.
21. Suppose that f(x,y) can be written as a product f(x, y) = F(x)G(y) of a function of x and a function of y. Then
900
Chapter 15: Multiple Integrals
the integral off over the rectangle R: a
~ x ~ b, c be evaluated as a product as well, by the formula
ff
dx)
f(x,y) dA = ( [ F(x)
(t
~
y
~
d can
a. If you have not yet done Exercise 41 in Section 15.4, do it now to show that
(1)
G(y) dy).
R
The argument is that
ii
f(x,y) dA
~
R
= =
t ([ dx) t (G(Y{ dx) t ([ dx )G(Y) F(x)G(y)
F(x)
b. Substitute y = Vi in Equation (2) to show that f( 1/2) = 2I = Y-ii.
dy
(i)
dy
F(x)
dy
(ii) (iii)
= ([F(X) dx)tG(y) dy.
(iv)
a. Give reasons for steps (i) through (iv).
24. Total electrical charge over circular plate The electrical charge distribution on a circular plate of radius R meters is O'(r, 0) = kr(1 - sinO) coulomb/m2 (k a constant). Integrate 0' over the plate to fmd the total charge Q. 25. A parabolic rain gauge A bowl is in the shape of the graph of z = x 2 + y2 from z = 0 to z = 10 in. You plan to calibrate the bowl to make it into a rain gauge. What height in the bowl would correspond to 1 in. ofrain? 3 in. ofrain? 26. Water in a satemte disb A parabolic satellite dish is 2 m wide and 1/2 m deep. Its axis of symmetry is tilted 30 degrees from the vertical.
When it applies, Equation (I) can be a time-saver. Use it to evaluate the following integrals. b.
{1n2 {~/2 eX cosy dy dx
c.
Jo Jo
hold. (Hint: Put your coordinate system so that the satellite dish is in "standsrd position" and the plane of the water level is slanted) (Caution: The limits of integration are not "nice.")
{2 (I x dx dy
11
i-I Y2
22. Let Duf denote the derivative of f(x,y) = (x 2 direction ofthe unit vector u = "1 i + U2j.
+ y2)/2
in the
a. Finding average value Find the average value of Duf over the triangular region cut from the fIrSt quadrant by the line x+y=l. b. Average value and centroid Show in general that the average value of Duf over a region in the >;y-plane is the value of D.f at the cenlroid of the region. 23. The value ofr(1/2)
a. Set up, but do not evaluate, a triple integral in rectangular coordinates that gives the amount ofwaler the satellite dish will
b. What would be the antallest tilt ofthe satellite dish so that it holds no water? 27. An infinite balf-eylinder Let D be the interior of the infmite right circular half-cylinder ofradius I with its single-end face suspended 1 unit above the origin and its axis the ray from (0, 0, I) to 00. Use cylindrical coordinates to evaluate
iff
The gamma function,
1 ,..-1
z(r 2
+ z 2 t'/2 dV.
D
00
f(x) =
e-t dt,
extends the factorial function from the noonegative integers to other real values. Of particular interest in the theory of differential equations is the number
Chapter
J: dx
28. Hypenolnme We have learned that 1 is the length of the interval [a, b] on the number line (one-dimensional space), !fR 1 dA is the area ofregionR in thexy-plane (two-;yz-space). We could contioue: If Q is a region in 4-space (xyzw-space), then g I dV is the ''bypervolume'' of Q. Use your generalizing abilities and a Cartesian coordinate system of 4-space to find the bypervolume inside the unit 3-dimensional spherex2 + y2 + z2 + w2 = 1.
Technology Application Projects
Mathematica/Maple Module: 1lIke Your Chances: 1l'y the MOllte Carlo Techllique for Numerical llItegratioll ill Three Dimemioll' Use the Monte Carlo technique to integrate numerically in three dimensions. Means alld MomeIrts and Exploring New Plotting Techlliques, Part II Use the method ofmoments in a form that makes use of geometric symmetry as well as multiple integration.
I1ff
16 INTEGRATION IN VECTOR FIELDS OVERVIEW In this chapter we extend the theory of integration to curves and surfaces in space. The resulting theory of line and surface integrals gives powerful mathematical tools for science and engineering. Line integrals are used to find the work done by a force in moving an object along a path, and to find the mass of a curved wire with variable density. Surface integrals are used to find the rate of flow of a fluid across a surface. We present the fundamental theorems of vector integral calculus, and discuss their mathematical consequences and physical applications. In the final analysis, the key theorems are shown as generalized interpretations of the Fundamental Theorem of Calculus.
16.1
Line Integrals
z
To calculate the total mass of a wire lying along a curve in space, or to find the work done by a variable force acting along such a curve, we need a more general notion of integral than was defined in Chapter 5. We need to integrate over a curve C rather than over an interval [a, b]. These more general integrals are called line integrals (although path integrals might be more descriptive). We make our definitions for space curves, with curves in the xy-plane being the special case with z-coordinate identically zero. Suppose that I(x, y, z) is a real-valued function we wish to integrate over the curve C lying within the domain of I and parametrized by r(t) == g(t)i + h(t)j + k(t)k, a :::; t :::; b. The values of I along the curve are given by the composite function I(g(t), h(t), k(t)). We are going to integrate this composite with respect to arc length from t == a to t == b. To begin, we first partition the curve C into a finite number n of subarcs (Figure 16.1). The typical subarc has length J).Sk. In each subarc we choose a point (Xh Yk, Zk) and form the sum n
Sn
== 'LI(Xk, Yk, Zk) J).Sk, k=l
x
FIGURE 16.1 The curve r(t) partitioned into small arcs from t = a to t = b. The length of a typical subarc is aSk.
which is similar to a Riemann sum. Depending on how we partition the curve C and pick (Xh Yh Zk) in the kth subarc, we may get different values for Sn. If I is continuous and the functions g, h, and k have continuous first derivatives, then these sums approach a limit as n increases and the lengths J).Sk approach zero. This limit gives the following definition, similar to that for a single integral. In the definition, we assume that the partition satisfies J).Sk~ 0 as n ~ 00.
If I is defined on a curve C given parametrically by ret)
DEFINITION g(t)i
+
h(t)j
+
==
k(t)k, a :::; t :::; b, then the line integral of f over Cis
Jcr !(x, y, z) ds =
lim ±!(Xk'Yk, Zk) Llsk,
n~OO
(1)
k=l
provided this limit exists.
901
902
Chapter 16: Integration in Vector Fields If the curve C is smooth for a :5 t:5 b (so v = dr/dt is continuous and never 0) and the function f is continuous on C, then the limit in Equation (1) can be shown to exist. We can then apply the Fundamental Theorem of Calculus to differentiate the arc length equation,
l'
s(t) =
Eq. (3) of Section 13.3
IV(T) I dT,
withto=a
to express tis in Equation (1) as tis = Iv(t) I dt and evaluate the integral oflover C as
l
!c1(x,y,z)dS =
b
I(g(t), h(t), k(t» Iv(t) I dt.
(2)
Notice that the integral on the right side of Equation (2) is just an ordinary (single) definite integral, as defined in Chapter 5, where we are integrating with respect to the parameter t. The formula evaluates the line integral on the left side correctly no matter what parametrization is used, as long as the parametrization is smooth. Note that the parameter t defines a direction along the path. The starting point on C is the position r(a) and movement along the path is in the direction of increasing t (see Figure 16.1).
How to Evaluate a Line Integral To integrate a continuous function I(x, y, z) over a curve C: 1. Find a smooth parametrization of C,
r(t) = g(t)i
+ h(t)j + k(t)k,
a
:5
t
:5
b.
2. Evaluate the integral as
!c I(x,y, z) tis =
t
=
l
b
I(g(t), h(t), k(t))lv(t) I dt.
If 1 has the constant value 1, then the integral of 1 over C gives the length of C from = b in Figure 16.1.
a to t
EXAMPLE 1 Integrate I(x,y,z) = x - 3y 2 origin to the point (1, 1, 1) (Figure 16.2).
+ z over the line segment C joining the
(I, I, 1) I I I I
: --cA~-
\ \ \ \ \
X
FIGURE 16.2
r(t) = ti +
---)oy
I I I I I I
~
(1, 1,0)
Example I.
We choose the simplest parametrization we can think of:
Solution
ti +
tk,
The components have continuous first derivatives and Iv(t)1 = Ii 2
2
V1 + 1 over Cis
2
+1
=
+ i + kl
=
V3 is never 0, so the parametrization is smooth. The integral of 1
The integration path in
!c I(x,y, z) ds =
=
=
l' l'
I(t, t, t)( V3) dt
(t - 3t 2
V3.Jo{' (2t -
Eq. (2)
+ t)V3 dt 3t
2
)
dt =
V3 [t 2 -
1
t 3 ]0 =
o.
•
16.1
Line Integrals
903
Additivity Line integrals have the useful property that if a piecewise smooth curve C is made by joining a finite number of smooth curves C" C2 , ••• , Cn end to end (Section 13.1), then the integral of a function over C is the sum ofthe integrals over the curves that make it up:
{Ids= {Ids+ {Ids+ ... + (Ids.
J~
Jc (1, 1, 1)
J~
(3)
J~
EXAMPLE 2 Figure 16.3 shows another path from the origin to (I, I, I), the union of line segments CI and C2. Integrate I{x,y, z) = x - 3y 2 + Z over C, U C2. Solution We choose the simplest parametrizations for C, and C2 we can find, calculating the lengths ofthe velocity vectors as we go along:
+ Ij, 0:5 I :5 I; Ivl = VI 2 + 12 = Vi i + j + Ik, 0:5 1:5 I; Ivl = V02 + 02 + 12 = 1.
C,:
r{l) = Ii
C2:
r{l) =
With these parametrizations we fmd that FIGURE 16.3
Example 2.
The path afinlegration in
{
I{x,y,z)ds
=
}C,UC2
=
=
=
{
lei
l' l'
I{x,y,z)ds + ( I(x,y,z)ds
Eq. (3)
JC2
l'
I{I, I, O)Vi dl +
I{I, 1,1)(1) dl
{I - 31 2 + O)Vidl +
Vi [1
2
2
l'
2
_ / 3 ]1
+ [1 - 21]1
0
0
2
(I - 3
= _
Eq. (2)
+ 1)(I)dl
Vi _ 1. 2
2
•
Notice three things about the integrations in Examples I and 2. First, as soon as the components ofthe appropriate curve were substituted into the fannula for I, the integration became a standard integration with respect to I. Second, the integral oflover c, U C2 was obtained by integrating I over each section ofthe path and adding the results. Third, the integrals oflover C and C, U C2 had different values.
The value of the line integral along a path joining two points can change if you change the path between them.
We investigate this third observation in Section 16.3.
Mass and Moment Calculations We treat coil springs and wires as masses distributed along smooth curves in space. The distribution is described by a continuous density function Il(x, Y, z) representing mass per unit length. When a curve C is parametrized by r{l) = X{I)i + Y{I}j + Z{I)k, a :5 I :5 b, thenx,y, andz are functions ofthe parameter I, the density is the function Il{X{I), y(1), z{I», and the arc length differential is given by
ds=~(:Y + (itY + (~Ydl.
904
Chapter 16: Integration in Vector Fields (See Section 13.3.) The spring's or wire's mass, center of mass, and moments are then calculated with the formulas in Table 16.1, with the integrations in terms of the parameter I over the interval [a, b]. For example, the formula for mass becomes
M =
t
8(x(I),y(I), z(I))
~ ( : ) ' + (~)' + (:)' dl.
These formulas also apply to thin rods, and their derivations are similar to those in Section 6.6. Notice how alike the formulas are to those in Tables 15.1 and 15.2 for double and triple integrals. The double integrals for planar regions, and the triple integrals for solids, become line integrals for coil springs, wires, and thin rods.
Mass and moment formuLas for coil springs, wires, and thin rods lying aLong a smooth curve Cin space
TABLE 16.1
Mass:
M =
fc 8 tis
B
~ B(x, y, z) is the density ,,(x, y, z)
First moments about the coordinate pIanes:
Myz = fc x 8 tis,
Mxy = fc z 8 tis
Coordinates of the center of mass:
y
= Mxz/M,
Moments of inertia about axes and other lines:
Ix
=
fc (y2
+ z2)8 tis,
h= fcr 28dS
r(x,y, z) = distance from the point (x,y, z) to line L
Notice that the element of mass dm is equal to 8 tis in the table rather than 8 dV as in Table 15.1, and that the integrals are taken over the curve C.
EXAMPLE 3 A slender metal arch, denser at the bottom than top, lies along the semicircle y2 + z2 = I, z ;" 0, in theyz-plane (Figure 16.4). Find the center ofthe arch's mass if the density at the point (x, y, z) on the arch is 8(x, y, z) = 2 - z.
z
c.m.
-1
Solution We know that x = 0 and y = 0 because the arch lies in the yz-plane with its mass distributed symmetrically about the z-axis. To find z, we parametrize the circle as r(l) = (cos l)j
FIGURE 16.4 Example 3 shows how to find the center ofmass of a circular arch of
variabledeosity.
+
(sin I)k,
o :5 I
:5 1f.
For this parametrization,
IV(I)I = so tis =
Ivldl =
(:)'
dl.
+
(~)' +
(:)'
=
Y(Of
+
(-sinl)2
+
(cos 1)2 = I,
16.1
905
Line Integrals
The formulas in Table 16.1 then give M= fc8ds= fc(Z-Z)ds= !."(Z-Sint)dt=Z1T-Z Mxy = fc z8 ds = fc z(Z - z) ds = !."(Sint)(Z - sint) dt =
!."(ZSint - sin2 t) dt
_ Mxy Z= M
=
=
8 - 1T I -Z-'Z1T Z
8;
1T
8 41T
1T
4'" 0.57.
With z to the nearest hundredth, the center ofmass is (0, 0, 0.57).
•
Line Integrals in the Plane There is an interesting geometric interpretation for line integrals in the plane. If C is a smooth curve in the :>ry-plane parametrized by r(t) = x(t)i + y(t)j, a ,;; t ,;; b, we generate a cylindrical surface by moving a straight line along C orthogonal to the plane, holding the line parallel to the z-axis, as in Section IZ.6. Ifz = f(x,y) is a nonnegative continuous function over a region in the plane containing the curve C, then the graph of f is a surface that lies above the plane. The cylinder cuts through this surface, forming a curve on it that lies above the curve C and follows its winding nature. The part of the cylindrical surface that lies beneath the surface curve and above the :>ry-plane is like a "winding wall" or "fence" standing on the curve C and orthogonal to the plane. At any point (x, y) along the curve, the height ofthe wallis f(x, y). We show the wall in Figure 16.5, where the "top" of the wall is the curve lying on the surface z = f(x, y). (We do not display the surface formed by the graph of f in the figure, ouiy the curve on it that is cut out by the cylinder.) From the definition
heightj(x, y)
y
t
=b
FIGURE 16.5 The line integral
Ie f
tis
gives the area ofthe portion ofthe cylindrical surface or "wall" beneath z = f(x,y) '" O. where !isk~ 0 as n ~ shown in the figure.
00,
we see that the line integral Ief ds is the area of the wall
Exercises 16.1 Graphs of Vector Equations
Match the vector equations in Exercises 1-8 with the graphs (a)-(h) given here. a.
c.
b.
z
z
d.
( 2,2,2)
2
-1
' r2M z
z
,
1
y
y
2
x
y
2
:~y
~~-
I
I
I---J,/
/
/
906
Chapter 16: Integration in Vector Fields
e.
z
f.
z 2 (0,0, 1)
r""
1~ __
I--
x
(0,0,0) ~
-1
(1, 1, 1)
(0,0,0)
~y
~
y x
y
-2
(1,1,-1)
(b)
(a)
h.
g.
The paths of integration for Exercises 15 and 16.
z
z
16. Integrate f(x, y, z) = x + vY - z' over the path from (0, 0, 0) to (I, I, I) (see accompanying figure) given by
y
~
+ (I
Ii
- l)j,
+ j + Ik,
2. r(l) = I
3. r(l) = (2 cos 1)1 4. r(l)
~
Ii,
5. r(l)
~
Ii
~
0:5 I :5 I
20. Evaluate
Ie f
11. Evaluate e (xy + Y + z) tis along the curve r(l) ~ 21i Ij + (2 - 21)k, 0 :5 I :5 1.
+
Ie \Ix' + y' tis along the curve r(l) ~ (4 cos 1)1 + + 31k, -2". :5 I :5 2"..
13. Find the line integral of f(x,y, z) = x + Y line segment from (I, 2, 3) to (0, -I, I).
2'17".
21. Find the line integral of f(x,y) ~ ye'" along the curve r(l) = 4ti - 3tj, -I :5 I :5 2. 22. Find the line integral of f(x, y) = x - y r(l) ~ (cos 1)1 + (sin I)j, 0 :5 I :5 2".. 23.
Evaluatel:~3 tis, eY
+ I'j,
I + j + Ik,
2
0:5 I :5 I
3
,
for
1 :s; t :s; 2.
24. Find the line integral of f(x,y) ~ 0/x along the curve r(l) = 1'1 + I'j, 1/2 :5 I :5 1. 25. Evaluate figure.
f e(x + 0) tis where C is given in the accompanying y
+ y' + z') over
__ x
---=--='~=-----~-
0:5 I :5 I
+ 3 along the curve
where C is the curve x = t , Y = t
15. Integrate f(x, y, z) ~ x + vY - z' over the path from (0, 0, 0) to (I, I, I) (see accompanying figure) given by
~
~
Ie \Ix + 2y tis, where Cis
+ z over the straight-
14. Find the line integral of f(x,y, z) = 'II3/(x' the curve r(l) ~ II + Ij + Ik, I :5 I :5 00.
r(l)
o~ t
h. C1 U C,; C1 is the line segment from (0, 0) to (I, 0) and C, is the line segment from (I, 0) to (I, 2).
".
(x - y + z - 2) tis where C is the straight-line seg10. Evaluate mentx ~ I,y ~ (I - I),z ~ I, from (0, I, l)to(I,O, I).
C,:
+ (a sin I)k,
.. the straight-line segment x = I, Y = 41, from (0, 0) to (I, 4).
x ~ I,y ~ (I - I),z ~ 0, from (0, I, 0) to (I, 0, 0).
r(l) ~ II
+ z'overthecircle
h. the parabolic curve x ~ I,y ~ I', from (0,0) to (2, 4).
Evaluating Line Integrals over Space Curves 9. Evaluate Ie (x + y) tis where C is the straight-line segment
Cl:
0:5 I :5 I
Ii
.. the straight-line segment x = I, Y = 1/2, from (0, 0) to (4, 2).
0:5 I :5 2
+ 21k, -1:5 1:5 I (2 cos 1)1 + (2 sin I)k, 0:5 I :5
12. Evaluate (4 sin I)j
r(l)
Line Integrals over Plane Curves 19. Evaluate fex tis, where Cis
0:5 I :5 2".
7. r(l) = (I' - I)j 8. r(l) =
r(l) = Ij + k,
r(l) = (a cos I)j
-1:5 I :5 I
6. r(l) = Ij + (2 - 21)k,
C,:
18. Integratef(x,y,z) ~ -\Ix'
-1:5 I :5 I
+ Ij + Ik,
0:5 I :5 I
+ j + k, 0:5 I :5 I 17. Integratef(x,y,z) = (x + Y + z)/(x' + y' + z') over the path r(l) ~ II + Ij + Ik, 0 < a :5 I :5 b.
0:5 I :5 1
+ (2 sin I)j,
r(l) = Ik,
C,:
y
1. r(l)
C1:
(0,0)
16.2
26. Evaluate
1
2
eX
1
+ y2 + I
tis where C is given in the accompanying
figure.
yz-plane. Find the moments of inertia of the rod about the three coordinate axes. 39. Two springs of constant density A spring of constant density 8 lies along the helix
y
o :s t :s 2w.
r(t) = (cos t)1 + (sin t)j + tk, (0, I)'+--f-~(I, I)
a. Find /z'
---:c:-:;+---+--:04:::-~x
(0,0)
(1,0)
In Exercises 27-30, integrate! over the given curve.
27. !(x,y) = x 3/y, C: y = x 2/2, 0 '" x'" 2 28. !(x,y) = (x + y2)/vT+7, C: y = x 2/2 from (I, 1/2) to (0,0)
29. !(x,y) = x + y,
C:
907
Vector Fields and Line Integrals: Work, Circulation, and Flux
x 2 + y2 = 4 in the first quadrant from
(2, 0) to (0, 2)
30. !(x,y) = x 2 - y, C: x 2 + y2 = 4 in the first quadrant from (0, 2) to (v'2, v'2) 31. Find the area of one side of the "winding wall" standing orthogonally on the curve y = x 2• 0 :s x :s 2. and beneath the curve on the surface !(x, y) = x +
b. Suppose that you have another spring of constant density 8 that is twice as long as the spring in part (a) and lies along the helix for 0 '" t '" 4". Do you expect I. for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calcularing I. for the longer spring.
40. Wire of constant density A wire of constant density 8 along the curve
r(t) ~ (tcost)1 + (tsint)j + (2V2/3)t 312k,
Masses and Moments 33. Mass of a wire Find the mass of a wire that lies along the curve r(t) ~ (12 - I)j + 2tk, 0 '" t '" I, if the density is 8 ~ (3/2)t. 34. Center of mass of a euned wire A wire of density 8(x,y, z) = IS'\1'.Y+2 lies along the curve r(t) = (t 2 - I)j + 2tk, -I '" t '" I. Find its cenlet of mass. Then sketch the curve and cenlet ofmass together.
0'" t '" I.
Find. and I•. 41. The arch in Eilimple 3
Find I. for the arch in Example 3.
42. Center of mass and moments of inertia for wire with variable density Find the cenlet of mass and the moments of inertia about the coordinate axes of a thin wire lying along the curve
r(t) = tl + 2 r t 312j +
v'Y.
32. Find the area of one side of the "wall" standing orthogona1ly on the curve 2x + 3y = 6, 0 '" x '" 6, and beneath the curve on the surface !(x, y) = 4 + 3x + 2y.
= I lies
if the density is 8
= I/(t +
~ k,
0 '" t '" 2,
I).
COMPUTER EXPLORATIONS In Exercises 43-46, use a CAS to perfOt03 the following steps to evaluate the line integrals. a. Find tis = Iv(t) I dtforthepathr(t) = g(t)i
+ h(t)j +
k(t)k. b. Express the integrand !(g(t), h(t), k(t)) Iv(t) I as a function of the parameter t.
e. Evaluate
Ie ! tis using Equation (2) in the text
35. Mass of wire with variable density Find the mass of a thin wire lying along the curve r(t) ~ V2ti + V2tj + (4 - t 2)k, o'" t '" 1, if the density is (a) 8 = 3t and (b) 8 = I.
43. !(x,y,z) =
36. Center of mass of wire with variable density Find the cenlet of mass of a thin wire 1ying along the curve r(t) ~ ti + 2tj + (2/3)t 312k, 0 '" t '" 2, ifthe density is 8 = 3'V"5+t.
44. !(x,y,z) =
37. Moment of Inertia of wire hoop A circulat wire hoop of constant density 8 lies along the circlex 2 + y2 ~ a 2 in thexy-plane. Find the hoop's moment of inertia about the z-axis.
45. !(x,y,z) ~ xyY - 3z 2 ; r(t) ~ (cos2t)1 + (sin2tJi + Stk, o :5 t :5 2w
38. Inertia of a slender rod A slender rod of constant density lies along the line segment r(t) ~ tj + (2 - 2t)k, 0 '" t '" I, in the
16.2
r(t) = ti + t 2j + 3t 2k,
VI
+ 3Ox 2 +
VI
+ x 3 + Sy3; r(t)
lOy;
0:51:52
= ti
+
tt
2
j
+ Vlk,
0:51:52
46. !(x,y,z) t'12k,
~
(I
+ ~zl/3
)'/4;
r(t)
~ (cos2t)1 + (sin2t)j +
0 '" t '" 2"
Vector Fields and Line Integrals: Work, Circulation, and Flux Gravitational and electric forces have both a direction and a magnitude. They are represented by a vector at each point in their domain, producing a vector field. In this section we show how to compute the work done in moving an object through such a field by using a line integral involving the vector field. We also discuss velocity fields, such as the vector
908
Chapter 16: Integration in Vector Fields
field representing the velocity of a flowing fluid in its domain. A line integral can be used to find the rate at which the fluid flows along or across a curve within the domain.
Vector Fields
FIGURE 16.6 Velocity vectors of a flow around an airfoil in a wind tunnel.
Suppose a region in the plane or in space is occupied by a moving fluid, such as air or water. The fluid is made up of a large number of particles, and at any instant of time, a particle has a velocity v. At different points of the region at a given (same) time, these velocities can vary. We can think of a velocity vector being attached to each point of the fluid representing the velocity of a particle at that point. Such a fluid flow is an example of a vector field. Figure 16.6 shows a velocity vector field obtained from air flowing around an airfoil in a wind tunnel. Figure 16.7 shows a vector field of velocity vectors along the streamlines of water moving through a contracting channel. Vector fields are also associated with forces such as gravitational attraction (Figure 16.8), and to magnetic fields, electric fields, and also purely mathematical fields. Generally, a vector field is a function that assigns a vector to each point in its domain. A vector field on a three-dimensional domain in space might have a formula like
F(x,y,z)
FIGURE 16.7 Streamlines in a contracting channel. The water speeds up as the channel narrows and the velocity vectors increase in length.
==
M(x,y,z)i
+ N(x,y,z)j + P(x,y,z)k.
The field is continuous if the component functions M, N, and P are continuous; it is differentiable if each of the component functions is differentiable. The formula for a field of two-dimensional vectors could look like
F(x,y)
==
M(x,y)i
+ N(x,y)j.
We encountered another type of vector field in Chapter 13. The tangent vectors T and normal vectors N for a curve in space both form vector fields along the curve. Along a curve ret) they might have a component formula similar to the velocity field expression
vet)
==
I(t)i
+ g(t)j + h(t)k.
If we attach the gradient vector VI of a scalar function I(x, y, z) to each point of a level surface of the function, we obtain a three-dimensional field on the surface. If we attach the velocity vector to each point of a flowing fluid, we have a three-dimensional field defined on a region in space. These and other fields are illustrated in Figures 16.9-16.15. To sketch the fields, we picked a representative selection of domain points and drew the
FIGURE 16.8 Vectors in a gravitational field point toward the center of mass that gives the source of the field.
/
........
//1 ~
x/ : /
//
""'"
~
/
x
~
I
I I
I
I~ ~
FIG URE 16.9 A surface, like a mesh net or parachute, in a vector field representing water or wind flow velocity vectors. The arrows show the direction and their lengths indicate speed.
16.2
Vector Fields and Line Integrals: Work, Circulation, and Flux
y
\ i i
\
1t !
'\
909
y
,/ ~
!
f(x,y,z) = c \
FIGURE 16.10 The field of gradient vectors V! on a surface!(x,y,z) = c.
z
1
FIGURE 16.11 The radial field F = xi + yj of position vectors of points in the plane. Notice the convention that an arrow is drawn with its tail, not its head, at the point where F is evaluated.
FIGU RE 16.12 unit vectors
A "spin" field of rotating
F = (-yi
+ xj)/(x 2 + y2)1/2
in the plane. The field is not defined at the origin.
vectors attached to them. The arrows are drawn with their tails, not their heads, attached to the points where the vector functions are evaluated.
Gradient Fields The gradient vector of a differentiable scalar-valued function at a point gives the direction of greatest increase of the function. An important type of vector field is formed by all the ~,-.
FIGURE 16.13 The flow of fluid in a long cylindrical pipe. The vectors v = (a 2 - r 2 )k inside the cylinder that have their bases in the xy-plane have their tips on the paraboloid z = a 2 - r 2• y
(
I
--+-----.-+ X
o
WIND SPEED, FIGURE 16.14 The velocity vectors v(t) of a projectile's motion make a vector field along the trajectory.
o
2
4
6
8
Mis
10 12 14 16+
FIGURE 16.15 NASA's Seasat used radar to take 350,000 wind measurements over the world's oceans. The arrows show wind direction; their length and the color contouring indicate speed. Notice the heavy storm south of Greenland.
910
Chapter 16: Integration in Vector Fields gradient vectors of the function (see Section 14.5). We define the gradient field of a differentiable function f(x, y, z) to be the field of gradient vectora
Vf =
of.
of.
of
ax' + ayJ + az k .
At each point (x, y, z), the gradient field gives a vector pointing in the direction ofgreatest increase of f, with magnitude being the value ofthe directional derivative in that direction. The gradient field is not always a force field or a velocity field.
EXAMPLE 1
Suppose that the temperature T at each point (x, y, z) in a region of space
is given by
T = 100 - x 2 - y2 - z2, and that F(x, y, z) is defined to be the gradient of T. Find the vector field F.
Solution The gradient field F is the field F = VT = -2xi - 2yj - Zzk. At each point in space, the vector field F gives the direction for which the increase in temperature is greatest. _
Line Integrals of Vector Fields In Section 16.1 we dermed the line integral of a scalar function f(x,y, z) over a path e. We turn our attention now to the idea of a line integral of a vectur field F along the curve
e.
Aasurne that the vectur field F = M(x,y, z)i + N(x,y, z)j + P(x,y, z)k has continuous components, and that the curve C has a smooth parametrization r(l) = g(l)i + h(l)j + k(1)k, a ,,; I ,,; b. Aa discussed in Section 16.1, the parametrization r(l) defines a direction (or orientation) along C which we call the forward direction. At each point along the path C, the tangent vector T = dr/lis = v/ Iv Iis a unit vector tangent to the path and pointing in this forward direction. (The vector v = dr/ dl is the velocity vectur tangent to C at the point, as discussed in Sections 13.1 and 13.3.) Intuitively, the line integral ofthe vector field is the line integral of the scalar tangential component of F along e. This tangential component is given by the dot product
dr lis'
F'T = F ' -
so we have the following formal definition, where 16.1.
f
= F' T in Equation (I) of Section
DEFINmON Let F be a vectur field with continuous components defined along a smooth curve C parametrized by r(I), a ,,; I ,,; b. Then the line integral of F along Cis
We evaluate line integrals of vectur fields in a way similar to how we evaluate line integrals of scalar functions (Section 16.1).
16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux
Evaluating the Line Integral of F = Mi C: r(t) = g(t)i + h(t)j + k(t)k
911
+ Hj + Pk along
1. Express the vector field F in terms of the parametrized curve Cas F(r(l)) by substituting the components x = g(1),y = h(I), z = k(1) ofr into the scalar components M(x,y,z), N(x,y,z), P(x,y, z) ofF. 2. Find the derivative (velocity) vector dr/dl. 3. Evaluate the line integral with respect to the parameter I, a :5 I :5 b, to obtain
EXAMPLE 2
Evaluate feF • dr, where F(x,y, z) = zi + ~ j - y 2k along the curve C
given by r(l) = l'i + tj + Solution
VI k, 0 :5 I :5 1.
We have F(r(I)) =
VIi + 13j
- 12k
and dr=2ti+j+_l_k.
dl
2V1
Thus,
• Line Integrals With Respect to the xyz Coordinates It is sometimes useful to write a line integral ofa scalar function with respect to one of the coordinates, such as fe M dx. This integral is not the same as the arc length line integral feM tis we defined in Section 16.1. To define the new integral for the scalar function M(x,y,z), we specify a vector field F = M(x,y,z)i over the curve C parametrized by r(l) = g(l)i + h(l)j + k(1)k, a :5 I :5 b. With this notation we have x = g(l) and dx = g'(I) dl. Then, dr F'dr = F' dldl = M(x,y,z)g'(I)dl = M(x,y,z)dx. So we define the line integral of M over C with respect to the coordinate x as
fc M(x,y,z) dx = fc F'dr,
where
F = M(x,y,z)i.
In the same way, by defming F = N(x,y,z)j, or F = P(x,y, z)k, we obtain the integrals feN dy and feP dz. Expressing everything in terms of the parameter I, we have the following formulas for these integrals:
912
Chapter 16: Integration in Vector Fields
fc M(x,y, z) dx
=
fc N(x,y,z) dy
=
fc P(x,y, z) dz
=
l l l
b M(g(/), h(/), k(1)) g'(/) dl
(1)
N(g(/), h(/), k(1» h'(/) dl
(2)
P(g(/), h(/), k(1)) k'(/) dl
(3)
b
b
It often happens that these line integrals occur in combination, and we abbreviate the nota-
tion by writing
fc
M(x,y,z) dx
EXAMPLE 3 r(/) = (cos 1)1
+
fc
N(x,y,z) dy
+
Evaluate the line integral
+ (sin I)j + Ik, 0
fc
P(x,y,z) dz =
fc
M dx
+ N dy + Pdz.
Je -y dx + z dy + 2x dz, where C is the helix
:5 I :5 21T.
Solution We express everything in terms of the parameter I, so x = cos I, Y = sin I, z = I, and dx = - sin I dl, dy = cos I dl, dz = dl. Then,
fc -y dx + zdy + 2x dz
1 1 2
=
=
"
2 "
[(-sin/l( -sin I)
[2 cos I
= [18inl =
[0
=
1T.
+ I cos 1+
2 cos I] dl
+ I cos I + sin2 I] dl
+ (I sin I + cos I) +
(t - S~2/)
+ (0 + I) + (1T - 0)] - [0 + (0 +
r
I)
+ (0 - 0)]
•
Work Done by a Force over a Curve in Space Suppose that the vector field F = M(x,y,z)1 + N(x,y,z)j + P(x,y,z)k represents a force throughout a region in space (it might be the force of gravity or an electromagnetic force of some kind) and that r(/) = g(/)1
FIGURE 16.16 The work done along the subarc shown here is approximately F." T• .1sh where F. = F(XhY., z.) and T. = T(Xh Yh z.).
+ h(/)j + k(1)k,
a
:5
t
:5
b,
is a smooth curve in the region. The formula for the work done by the force in moving an object along the curve is motivated by the same kind ofreasoning we used in Chapter 6 to derive the formula W = Jab F(x) dx for the work done by a continuous force ofmagnitude F(x) directed along an interval of the x-axis. For a curve C in space, we define the work done by a continuous force field F to move an object along C from a point A to another point B as follows. We divide C into n subarcs P.-IP. with lengths fl.sh starting at A and ending at B. We choose any point (Xk, Yh Zk) in the subarc Pk--lPk and let T(Xh Yk, Zk) be the unit tangent vector at the chosen point. The work Wk done to move the object along the subarc Pk-jPk is approximated by the tangential component of the force F(Xk, Yk, Zk) times the arclength fl.sk approximating the distance the object moves along the subarc (see Figure 16.16).
913
16,2 Vector Fields and Line Integrals: Work. Circulation. and Flux
The tota1 work done in moving the object from point A to point B is then approximated by summing the work done along each of the subarcs, so n
W RJ
n
L Wk k-l L F(Xk, Yh Zk) • T(Xh Yh Zk) /isk. k-l RJ
For any subdivision of C into n subarcs, and for any choice of the points (Xk, Yk, Zk) within each subarc, as n --> 00 and /isk --> 0, these sums approach the line integral
fc F·TdI-. This is just the line integral of F along C, which is defined to be the total work done.
DEFINmON Let Cbe a smooth curve parametrized by r(t), a :5 t:5 b, and F be a continuous force field over a region containing C. Then the work done in moving an object from the point A = r(a) to the pointB = r(b) along Cis
T F
W
A t= a
The worl< done by a force is the line integral ofthe scalar
FIGURE 16.17 F
component F . T over the smooth curve fromA toB.
=
lcr F' T ds =
l
a
b
F(r(t))· dr dt dt.
The sign of the number we calculate with this integral depends on the direction in which the curve is traversed. If we reverse the direction of motion, then we reverse the direction ofT in Fignre 16.17 and change the sign ofF' T and its integral. Using the notations we have presented, we can express the work integral in a variety of ways, depending upon what seems most suitable or convenient for a particular discussion. Table 16.2 shows five ways we can write the work integral in Equation (4). TABLE 16.2 Different ways to write the work integral for F = Mi the curve C: r(t) = g(t)i + h(t)j + k(t)k, a :5 t:5 b
=
=
=
z =
(1, 1, 1)
~
I
I
~i~y
;/
I I
__
I
-_
---
X
r(t) = ti
I I
+ P:i + t 3k--- 1/
FIGURE 16.18
//
/
/
/ /
(1, 1,0)
The curve in Example 4.
fc F ' dr
l l
b
F·
a b
a
Vector differential form
dr dt dt
dx (M dt
+ Nj + Pk over
The definition
W= fcF'TdI-
(0,0,0)
(4)
Parametric vector evaluation
dy
+ N dt + pdz)dt dt
Parametric scalar evaluation Scalar differential form
fcMdx+NdY+Pdz
Find the work done by the force field F = (y - x 2 )i + (z _ y 2)j + (x - z2)k along the curve r(t) = Ii + t 2j + t'k, :5 t :5 I, from (0, 0, 0) to (I, I, I) (Fignre 16.18).
EXAMPLE 4
Solution
°
First we evaluate F on the curve r(t): F = (y - x 2 )i + (z - y 2)j + (x - z2)k =
(t 2
-
o
t 2 )i
+ (t' - t 4 )j + (t -
t 6 )k.
Substitute x Y = ,2, Z =
=
rJ.
t,
914
Chapter 16: Integration in Vector Fields Then we find dr/ dt,
~~
~ (ti +
=
t 2j + t 3k)
= 1+
2tj + 3t 2k.
Finally, we find F . dr/ dt and integrate from t = 0 to t = I: F' dr = [(t 3 - t 4 )j + (t - t 6 )k]' (I + 2tj + 3t 2k)
dt
=
(t 3 - t 4 )(2t)
+ (t - t 6 )(3t 2 )
=
2t 4
-
2t S
+ 3t 3 - 3t 8
so, Work =
l'
(2t 4
-
2t S + 3t 3 - 3t 8) dt
• EXAMPL! 5 Find the work done by the force field F = xi + yj + zk in moving an object along the curve C parametrized by r(t) = C08 ('ITt) 1 + t 2j + sin ('ITt) k, 0 :5 t :5 1. Solution
We begin by writing F along C as a function of t, F(r(t)) = cos ('ITt) 1 + t 2j
+ sin ('ITt) k.
Next we compute dr/ dt,
~~
= -'IT sin ('ITt) 1 + 2tj
+ 'IT cos ('ITt) k.
We then calculate the dot product, F(r(t)).: =
-'IT
sin ('ITt) cos ('ITt)
+ 2t 3 + 'IT sin ('ITt) cos('lTt)
=
2t 3.
The work done is the line integral
l
a
b
F(r(t)). dr dt =
dt
f' 2t3 dt =
10
4
t ]' = 1. 2 0 2
•
Row Integrals and Circulation for Velocity Fields Suppose that F represents the velocity field of a fluid flowing through a region in space (a tidal basin or the turbine cbamber of a hydroelectric generator, for example). Under these circumstances, the integral ofF' T along a curve in the region gives the fluid's flow along the curve.
DEFINmONS Ifr(t) parametrizes a smooth curve C in the domain of a continuous velocity field F, the flow along the curve from A = rea) to B = reb) is Bow =
fc
F·Tds.
(5)
The integral in this case is called a flow integral. If the curve starts and ends at the same point, so that A = B, the flow is called the circulation around the curve.
915
16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux
The direction we travel along C matters. If we reverse the direction, then T is replaced by -T and the sign of the integral changes. We evaluate flow integrals the same way we evaluate work integrals.
EXAMPLE 6 A fluid's velocity field is F = xi + zj + yk. Find the flow along the helix r(t) = (cost)i + (sint)j + tk,O:5 t:5 7T/2. Solution
We evaluate F on the curve,
F = xi
+ zj + yk
= (cos t)i
+ tj +
(sin t)k
Substitute x
=
cos t, Z
=
t, Y
=
sin t.
and then rmd dr/ dt: :
=
(-sint)i + (cost)j + k.
Then we integrate F' (dr/dt) from t = 0 to t = ; :
F' :
= (cost)(-sint) =
+ (t)(cost) + (sint)(I)
-sintcost + tcost + sint
so, Flow =
y
- "/ / - - ~. //. / ~/-y-t--+--+:-~~x /~..---c
I I
I
=
l~~b F' :dt =
1"/2
(-sintcost
[CO~2t + tSintr =(0 + ;) -
+ tcost + sint)dt
(t + 0) =; - t.
-
EXAMPLE 7 Find the circulation of the field F = (x - y)i + xj around the circle r(t) = (cos t)i + (sin tH, 0 :5 t :5 27T (Figure 16.19).
,~
Solution
Onthecircle,F = (x - y)i
FIGURE 16.19 The vector field F and curve r(t) in Example 7.
:
=
+ xj
=
(cost - sint)i + (cost)j,and
(-sint)i + (cost)j.
Then F'
~~
-sintcost + sin2 t + cos2 t
=
1
gives
Jor " F' drdt dt = Jor " (I 2
Circulation = Simple, not closed
=
[t -
2
s~2tr =
sin t cos t) dt
27T.
As Figure 16.19 suggests, a fluid with this velocity field is circulating counterclockwise around the circle. _ Not simple, not closed.
Not simple,
closed
FIGURE 16.20 Distinguishing curves that are simple or closed. Closed curves are also called loops.
Aux Across a Simple Plane Curve A curve in the xy-plane is simple if it does not cross itself (Figure 16.20). When a curve starts and ends at the same point, it is a closed curve or loop. To find the rete at which a fluid is entering or leaving a region enclosed by a smooth simple closed curve C in the xy-plane,
916
Chapter 16: Integration in Vector Fields
we calculate the line integral over C of F' n, the scalar component of the fluid's velocity field in the direction of the curve's outward-pointing nonnal vector. The value of this integral is thejlux ofF across C. Flux is Latin forjlow, but many flux calculations involve no motion at all. If F were an electric field or a magnetic field, for instance, the integral of F n would still be called the flux of the field across C. 0
DEFINITION If C is a smooth simple closed curve in the domain of a continuous vector field F = M(x, y)i + N(x, y)j in the plane, and if n is the outwardpointing uuit normal vector on C, the flux of F across Cis Flux ofF across C =
z
1
F' n
tis.
(6)
Notice the difference between flux and circulation. The flux of F across C is the line integral with respect to arc length ofF n, the scalar component ofF in the direction of the outward nonnal. The circulation of F around C is the line integral with respect to arc length of F • T, the scalar component of F in the direction of the unit tangent vector. Flux is the integral of the normal component of F; circulation is the integral of the tangential component ofF. To evaluate the integral for flux in Equation (6), we begin with a smooth parametrization 0
For clockwise motion, k X T points outward. y
c x = g(t),
z For counterclockwise
motion, T X k points outward.
y x
c
y = h(t),
a :5 t :5 b,
that traces the curve C exactly once as t increases from a to b. We can f'md the outward uuit normal vector n by crossing the curve's uuit tangent vector T with the vector k. But which order do we choose, T X k or k X T? Which one points outward? It depends on which way C is traversed as t increases. If the motion is clockwise, k X T points outward; if the motion is counterclockwise, T X k points outward (Figure 16.21). The usual choice is n = T X k, the choice that assumes counterclockwise motion. Thus, although the value of the integral in Equation (6) does not depend on which way C is traversed, the fonnulas we are about to derive for computing n and evaluating the integral assume counterclock-
wise motion. In terms of components,
FIGURE 16.21 To [md an outward unit normal vector for a smooth simple curve C in the xy-plane that is traversed counterclockwise as t increases, 'We take n = T X k. For clockwise motion, we taken ~ k X T.
n = T X k = (tb; i
ds
IfF = M(x,y)i
+ dy j) ds
X k = dy i _ tb; j.
tis
tis
+ N(x,y)j, then Fon = M(x,y)
dy
tb;
tis - N(x,y) ds'
Hence, lFonds=
l(M~-N:)tIs= fMdY-Ntb;. c
We put a directed circle 0 on the last integral as a reminder that the integration around the closed curve C is to be in the counterclockwise direction. To evaluate this integral, we express M, dy, N, and tb; in terms of the parameter t and integrate from t = a to t = b. We do not need to know n or tis explicitly to find the flux.
16.2
917
Vector Fields and Line Integrals: Work, Circulation, and Flux
Calculating FIlII Across a Smooth Closed Plane Curve (Flux ofF = Mi
+ Nj across C)
=
f
ax
M dy - N
(7)
C
The integral can be evaluated from any smooth parametrizatioo x = g(t),y = h(t), a :5 t :5 b, that traces C counterclockwise exactly ooce.
EXAMPLE 8 Find the flux ofF = (x - y)i + xi across the circle x' + y' = I in the xy-plane. (The vector field and curve were shown previously in Figure 16.19.) Solution The parametrizatioo r(t) = (cost)i + (sint)j,O:5 t:5 2'IT, traces the circle counterclockwise exactly ooce. We can therefore use this parametrizatioo in Equatioo (7). With
M = x - y = cost - sint,
dy = d(sin t) = cos t dt
N=x=cost,
ax =
d(cost) = -sintdt,
we find Flux = =
1M
dy - N
{'W cos' t dt
Jo
l'w
ax = =
(cos' t - sintcost
{'W I +
Jo
cos 2t dt =
2
[1 + 2
+
cos t sin t) dt
sin 2t]'W =
'IT.
a
4
Eq.(7)
The flux of F across the circle is 'IT. Since the answer is positive, the net flow across the curve is outward A net inward flow would have given a negative flux. •
Exercises 16.2 Vector Fields Find ilie gradient fields ofilie functions in Exercises 1-4. 1. !(x,y,z) ~ (x'
+ y' + z')-l/2
2. !(x,y,z) ~ InVx' + y' + z' 3. g(x,y, z) = e' - In (x' + y')
xy + yz + xz S. Give a fannula F = M(x, y)i
4. g(x,y, z)
~
+ N(x, y)j for ilie vector field in the plane iliat has ilie property iliat F points toward the origin wiili magnitude inversely proportional to ilie square of ilie distance from (x, y) to ilie origin. (The field is not defined at (0, 0).)
.. The slraight-line paili Cl: ret)
= Ii + tj + tk,
7. F = 3yi + 2xj + 4zk 9. F = Vii - 2xj + yYk
8. F = [I/(x' + l)]j 10. F = xyi + yzj + xzk
11. F = (3x' - 3x)i + 3zj + k 12. F
= (y + z)i +
(z
+ x)j + (x + y)k z
6. Give a fannula F = M(x, y)i + N(x, y)j for ilie vector field in ilie plane iliat has ilie properties iliat F ~ 0 at (0, 0) aod iliat at any oilier point (a, b), F is tangent to ilie circle x' + y' ~ + b' aod points in the clockwise direction wiili magnitode IFI = Va' + b'.
a'
~y /
Line Integnlls of Vector Fields In Exercises 7-12, fmd ilie line integrals of F from (0, 0, 0) to (I, I, I) over each ofilie following pailis in ilie accompanying figure.
0:5 t :5 I
1>. The curved paili C,: ret) = ti + t'j + t 4k, 0 '" t '" I e. The paili C, U C4 coosisling ofilie line segment from (0, 0, 0) to (1, 1,0) followed by ilie segment from (1, 1,0) to (1, 1, 1)
/ / /
-
/ /
/
(1. 1.0)
918
Chapter 16: Integration in Vector Fields
Une Integrals with Respect to x, y, and z
Une Integrals in the Plane
In Exercises 13-16, fmd the line integrals along the given path C.
23. Evaluate Ie xy d< + (x + y) dy along the curve y ~ x' from (-I,l)to(2,4).
13.1
14. 15.
(x - y)d<,wbereC:x
1'J; 1
~ t,y ~ 2t +
l,forO:5 t:5 3
dy, where C: x = t, Y = I', for I :5 t :5 2
(x' + y') dy, where C is given in the accompanying figure.
24. Evaluate Ie (x - y) d< + (x + y) dy counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, I). 25. Evaluate Ie F • T ds for the vector field F ~ x'i - yj along the curve x = y' from (4, 2) to (I, -I). 26. Evaluate Ie F • dr for the vector field F = yi - xj counterclockwise alnog the unit circle + y' ~ I from(I,O)to(O, I).
x'
y
Work, Cin:ulatlon, and Flux in the Plane (3,3)
c (0,0)
1\fx+Y
28. Work Find the work done by the gradient of f(x, y) = (x + y)' counterclockwise around the circle + y' = 4 from (2, 0) to itself.
x'
---;;=t--~-;:;~-~x
16.
27. Work Find the work done by the force F = xyi + (y - x)j over the straight line from (I, I) to (2, 3).
(3,0)
29. Circulation and filII
F1 d<, where C is given in the accompanying figure.
= xi + yj
and
F2
= -yi + xj
around and across each of the following curves.
a. The circle r(t) y
Find the circulation and flux of the fields
~
(cos t)i + (sin t)j,
0:5 t :5 2".
h. The ellipse r(t) = (cos t)i + (4 sin I)j,
c
30. FIlII across a circle
0:5 t :5 21f
Find the flux of the fields
F 1 = 2xi - 3yj
F, = 2xi + (x - y)j
and
across the circle r(l)
o ::s t ::s 21T.
= (a co. t)i + (a.in I)j,
~~---~x
(0,0)
17. Along the curve r(t) = Ii - j + t'k, 0 :5 t :5 I, evaluate each of the following integrals.
1(X + h. 1(X + a.
c.
1
(x
y -z)d<
- z) dz
1
xz d<
h.
1
xz dy
c.
1
xyz dz
Work In Exercises 1~22, fmd the work done by F over the curve in the
direction ofincreasing t. 19. F ~ xyi + Y.i - yzk r(l) = Ii + I'j + Ik,
32. F = x'i + y'j
33. F ~ -yi + xj
34. F ~ -y'i + x'j
a. The upper half of the circle x' + y' ~ I
18. Alnog the curve r(l) = (cos t)i + (sin I)j - (cos t)k, 0 :5 I :5 1f, evaluate each of the following integrals.
a.
line segment r,(t) = Ii, -a :5 t :5 a. 31. F = xi + yj
35. Flow integral. Find the flow of the velocity field F ~ (x + y)i - (x' + y')jalnogeachofthefollowingpathsfrom(I,O) to (-1,0) in thexy-plane.
y -z)dy
+Y
In Exercise. 31-34, fmd the circulation and flux of the fieldF around and across the closed semicircular path that consists of the semicircular arch rl(t) = (a cos I)i + (a sin I)j, 0 :5 t :5 1f, followed by the
h. The line segment from (I, 0) to (-1,0) c. The line segment from (I, 0) to (0, -I) followed by the line segment from (0, -I) to (-1,0) 36. FIlII aero•• a triangle Find the flux of the field F in Exercise 35 outward across the triangle with vertices (I, 0), (0, I), (-1,0). 37. Find the flow of the velocity field F = y'i + 2xyj along each of the following paths from (0, 0) to (2, 4).
a.
21. F=zi+xj+yk r(t) ~ (sin t)i + (cos t)j + tk,
(2, 4)
}=x2
0:5 t:5 2".
0:5 t :5 2".
22. F ~ 6zi + y'j + 12xk r(l) = (sin I)i + (cos t)j + (1/6)k,
y
(2,4)
0:5 I :5 I
20. F ~ 2yi + 3xj + (x + y)k r(t) ~ (cost)i + (sint)j + (t/6)k,
h.
y
0:5 I :5 21f
I I I I X
(0, 0)
2
X
(0, 0)
2
c. Use any path from (0, 0) to (2, 4) different from parts (a) and (b).
16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux 38. Find the circulation of the field F = yi + (x + 2y)j around each of the following closed paths.
a.
y
(-1,1)
(1, 1)
(-I, -1)
(1, -1)
919
43. Unit vectors pointing toward lbe origin Find a field F = M(x, y)1 + N(x, y)j in the xy-plane with the property that at each point (x,y) # (0,0), F is a unit vector pointing toward the origin. (The field is undefmed at (0, 0).)
44. Two "central" fields Find a field F ~ M(x, y)1 + N(x, y)j in the >;y-plane with the property that at each point (x,y) # (0,0), F points toward the origin and IF I is (a) the distance from (x, y) to the origin, (h) inversely proportional to the distance from (x, y) to the origin. (The field is undefined at (0, 0).)
45. Work and area Suppose that I(t) is dilfereotiable and positive for a :5 t :5 b. Let C be the path r(t) ~ ti + I(t)j, a :5 t :5 b, and F = yl. Is there any relation between the value of the work integral
b.
y
and the area of the region bounded by the t-axis, the graph of I, and the lines t ~ a and t ~ b? Give reasons for your answer. -+---+--+~X
46. Work done by a radial force with constant magnitnde A particle moves along the smooth curve y = I(x) from (a, I(a)) to (b, I(b)). The force moving the particle has constant magoitude k and always points away from the origin. Show that the work done by the force is
fc F'Tda
c. Use any closed path dilfereot from parts (a) aod (h).
Vector fields In the Plane Draw the spin field
39. Spin field
= k[(b 2 + (f(b)2)1j2 - (a 2 + (f(a)j2)'/2].
Flow Integrals In Space
In Exercises 47-50, F is the velocity field ofa fluid flowing through a F=-
Y 1+ x j v'x2 + y2 v'x2 + y2
region in space. Find the flow along the given curve in the direction of increasing t.
-4xyi + 8yj + 2k 0:5 t :5 2 48. F = x2j + yzj + y 2k r(t) ~ 3tj + 4tk, 0:5 t :5 I 49. F ~ (x - z)i + xk r(t) = (cos t)1 + (sin t)k, 0:5 t :5 'iT
47. F (see Figure 16.12) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y2 = 4.
40. Radial field Draw the radial field F=xl+yj (see Figure 16.11) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y2 = 1. 41. A field oftangent vectors
a. Find a field G = P(x, y)1 + Q(x, y)j in the >;y-plane with the property that at any point (a, b) # (0, 0), G is a vector of
v'
magoitude a 2 + b 2 taogent to the circle x 2 + y2 = a 2 + b 2 and pointing in the counterclockwise direction. (The field is undefmed at (0, 0).)
~
r(t)
= ti + t 2j + k,
SO. F=-yi+xj+2k r(t) ~ (-2 cos t)1 + (2 sin t)j + 2tk,
0 :5 t :5 2'iT
51. Circulation Find the circulation of F = 2xi + 2zj + 2yk around the closed path consisting of the following three curves traversed in the direction ofincreasing t. C,:
r(t) = (cost)1 + (sint)j + tk,
C2:
r(t) = j + ('iT/2)(1 - t)k,
C,:
r(t)
~
ti
+ (I
- t)j,
b. How is Grelated to the spin fieldF in Figure 16.12?
0:5 t:5 'iT/2
0:5 t:5 I
0:5 t :5 I
(0,1,
¥)
42. A field of tangent vectors
a. Find a field G = P(x, y)1 + Q(x, y)j in the >;y-plane with the property that at any point (a, b) # (0,0), G is a unit vector tangent to the circle x 2 + y2 = a 2 + b 2 and pointing in the clockwise direction.
b. How is Grelated to the spin fieldF in Figure 16.12?
C2
(1, o'7o)_:::::.._---"'~(0, 1,0) C,
y
920
Chapter 16: Integration in Vector Fields
52. Zero circulation Let C be the ellipse in which the plane 2x + 3y - z = 0 meets the cylinder x 2 + y2 = 12. Show, without evaluating either line integral directly, that the circulation of the field F = xi + yj + zk around C in either direction is zero. 53. Flow along a curve The field F = xyi + yj - yzk is the velocity field of a flow in space. Find the flow from (0, 0, 0) to (1, 1, 1) along the curve of intersection of the cylinder y = x 2 and the plane z = x . (Hint: Use t = x as the parameter.)
z=x
+ h(t)j + k(t)k.
1
F · dr.
+ 2)j; r(t) = (2 cos t)i + (sin t)j, r(t) = (cost)i + (sint)j,
O~t~7T
57. F = (y + yz cosxyz)i + (x 2 + xz cosxyz)j + (z + xy cos xyz )k; r(t) = (2 cos t)i + (3 sin t)j o ~ t ~ 27T
L_ y
=
58. F = 2xyi - y2j + zexk;
x2
x
r(t) = -ti +
+ k,
Ytj + 3tk,
1~t~4
54. Flow of a gradient field
Find the flow ofthe field F = V(xy 2z 3):
a. Once around the curve C in Exercise 52, clockwise as viewed from above b. Along the line segment from (1, 1, 1) to (2, 1, -1).
16.3
c. Evaluate
56. F = - 3 -2i + ~j; l+x l+y
(1, 1, 1)
I I
\
a. Find dr for the path r(t) = g(t)i
b. Evaluate the force F along the path.
55. F = xy 6 i + 3x(xy 5 o ~ t ~ 27T
z
_1
COMPUTER EXPLORATIONS In Exercises 55-60, use a CAS to perform the following steps for finding the work done by force F over the given path:
59. F = (2y + sinx)i + (z2 r(t) = (sin t)i + (cos t)j
+ (1/3)cosy)j + x 4 k; + (sin 2t)k, -7T/2 ~ t
~
7T/2
60. F = (x2y)i + tx3j + xyk; r(t) = (cos t)i + (sin t)j + (2 sin2 t - 1)k, 0 ~ t ~ 27T
Path Independence, Conservative Fields, and Potential Functions A gravitational field G is a vector field that represents the effect of gravity at a point in space due to the presence of a massive object. The gravitational force on a body of mass m placed in the field is given by F == mG. Similarly, an electric field E is a vector field in space that represents the effect of electric forces on a charged particle placed within it. The force on a body of charge q placed in the field is given by F == qE. In gravitational and electric fields, the amount of work it takes to move a mass or charge from one point to another depends on the initial and final positions of the object-not on which path is taken between these positions. In this section we study vector fields with this property and the calculation of work integrals associated with them.
Path Independence If A and B are two points in an open region D in space, the line integral of F along C from A to B for a field F defined on D usually depends on the path C taken, as we saw in Section 16.1. For some special fields, however, the integral's value is the same for all paths from A toB.
DEFINITIONS Let F be a vector field defined on an open region D in space, and suppose that for any two points A and B in D the line integral F • dr along a path C from A to B in D is the same over all paths from A to B. Then the integral F · dr is path independent in D and the field F is conservative on D.
Ie
Ie
The word conservative comes from physics, where it refers to fields in which the principle of conservation of energy holds. When a line integral is independent of the path C from
16.3 Path Independence, Conservative Fields, and Potential Functions
point A to point B, we sometimes represent the integral by the symbol usual line integral symbol
J:
921
rather than the
Je' This substitution helps us remember the path-independence
property. Under differentiability conditions nonnally met in practice, we will show that a field F is conservative if and only if it is the gradient field ofa scalar function I-that is, if and only if F = VI for some j. The function I then has a special name.
DEFINmON If F is a vector field defined on D and F = VI for some scalar function I on D, then I is called a potential function for F.
A gravitatioual potential is a scalar function whose gradient field is a gravitatioual field, an electric potential is a scalar function whose gradient field is an electric field, and so on. As we will see, once we have found a potential function I for a field F, we can evaluate all the line integrals in the domain of F over any path between A and B by
1 B
1 B
F'dr =
V/'dr = I(B) - I(A).
(1)
If you think of VI for functions of several variables as being something like the derivative j' for functions of a single variable, then you see that Equation (1) is the vector calculus analogue ofthe Fundamental Theorem of Calculus fannula
l
b
j'(x) dx = I(b) - I(a).
Conservative fields have other remarkable properties. For example, saying that F is conservative on D is equivalent to saying that the integral ofF around every closed path in D is zero. Certain conditions on the curves, fields, and domains must be satisfied for Equation (I) to be valid. We discuss these conditions next.
Assumptions on Curves, Vector Fields, and Domains In order for the computations and results we derive below to be valid, we must assume certain properties for the curves, surfaces, domains, and vector fields we consider. We give these assumptions in the statements of theorems, and they also apply to the examples and exercises unless otherwise stated. The curves we consider are piecewise smooth. Such curves are made up of fmitely many smooth pieces connected end to end, as discussed in Section 13.1. We will treat vector fields F whose components have continuous first partial derivatives. The domains D we consider are open regions in space, so every point in D is the center of an open ball that lies entirely in D (see Section 13.1). We also assume D to be connected. For an open region, this means that any two points in D can be joined by a smooth curve that lies in the region. Finally, we assume D is simply connected, which means that every loop in D can be contracted to a point in D without ever leaving D. The plane with a disk removed is a two-dimensioual region that is 1101 simply connected; a loop in the plane that goes around the disk cannot be contracted to a point without going into the ''hole'' left by the removed disk (see Figure 16.22c). Similarly, if we remove a line from space, the remaining region D is 1101 simply connected. A curve encircling the line cannot be shrunk to a point while remaining inside D.
922
Chapter 16: Integration in Vector Fields Connectivity and simple connectivity are not the same, and neither property implies the other. Think of connected regions as being in "one piece" and simply connected regions as not having any "Ioop-catching holes." All of space itself is both connected and simply connected. Figure 16.22 illustrates some of these properties.
y
-+---=====-, Simply connecred
Caution Some of the results in this chapter can fail to hold if applied to situations where the conditions wtl've imposed do not hold In particular, the component test for conservative fields, given later in this section, is not valid on domains that are not simply connected (see ExampleS).
(.)
Line Integrals in Conse",.tive Fields Gradient fJ.elds F are obtained by different:iatiDg a scalar function I. A theorem analogous to the Fun.danJental. Theorem of Calculus gives a way to evaluate the line integmls of gradient fJ.elds.
THEOREM 1-Fund.mentaL ThUMm of Line Integr1lls Let C be a smooth curve joining the point A to the point B in the plane or in space and parametrized by r(t). Let I be a differentiable function with a continuous gradient vector F = VI on a domain D containing C. Then (h)
fc F' dr -
y
o.
/(B) - /(4).
Like the Fundamental Theorem, Theorem. I gives a way to evaluate line integmls without having to take limits of Riemann sums or Imding the line integral by the procedure used in Section 16.2. Before proving Theorem I, we give an example.
Not simply connected
-+---------,
EXAMPLE 1
Suppose the force field F = VI is the gradient ofthe function
(.)
I(x,y,z) = -
2 X
~------
,,,"" ,, / ,
/ -" '
....
',-----_ "
1
2
+Y +z
2'
Find the work done by F in moving an object along a smooth curve C joining (I, 0, 0) to (0, 0, 2) that does not pass through the origffi.
,
,,
~i"_====,______. y Not simply connected
,
(d)
FIGURE Ui.22 Four connected regions. In (a) and (b), the regions are simply cOIllltlCted. In (c) and (d), the regions are not simply connected becllllSe tlu: curves Ct and C2 cannot be contracted to a point inside the regions containing them.
Solution An application ofTheore:m 1 shows that the work done by F along any smooth curve C joining the two points and not passing through the origin is
fc
F ' dr - /(0,0,2) - /(1,0,0) -
-i -
(-1) -
t·
•
The gravitational force due to a planet, and the electric force associated with a chaIged particle, can both be modeled by the fJ.eld F given in Example I up to a constant
that depends on the units of measurement. Proof of Theorem 1 Suppose that A and B are two points in region D and that C: r(t) = g{t)i + h(t)j + k{t)k, a ~ t ~ b, is a smooth curve in D joining A to B.
16.3
923
Path Independence, Conservative Fields, and Potential Functions
We use the abbreviated form r(t) = xi + yj + zk for the parametrization of the curve. Along the curve, I is a differentiable function of t and Chain Rule in Section 14.4 with x ~ g(,), y ~ h('), z ~ k(,)
dl al dx al dy al dz -=--+--+-dt ax dt ay dt az dt
=V/·(dxi+dYj+dzk)=V/·dr=F.dr ~
~
~
~~'
Therefore,
1 e
F·dr = =
l
j
bdl '=b F·--'"dt d = -dt '=a dt a dt
I(g(t), h(t), k(t))
J:
=
Because F
r(o)
~
~
Vf
A, r(b)
~
B
I(B) - I(A).
•
So we see from Theorem I that the line integral of a gradient field F = VI is straightforward to compute once we know the function f. Many important vector fields arising in applications are indeed gradient fields. The next result, which follows from Theorem I, shows that any conservative field is of this type. THEOREM 2-Conservati"" Fields are Gradient Fields Let F = Mi + Nj + Pk be a vector field whose components are continuous throughout an open connected region D in space. Then F is conservative if and only if F is a gradient field VI for a differentiable function f.
Theorem 2 says that F = VI if and only iffor any two points A andB in the regionD, the value ofline integral IcF' dr is independent of the path CjoiningA toB inD.
Proof of Theorem 2 If F is a gradient field, then F = VI for a differentiable function I, and Theorem I shows that F' dr = I(B) - I(A). The value of the line integral does not depend on C, but only on its endpoints A and B. So the line integral is path independent and F satisfies the definition of a conservative field. On the other hand, suppose that F is a conservative vector field. We want to f'md a function I on D satisfYing VI = F. First, pick a point A in D and set I(A) = O. For any F • dr, where C is any smooth path in D from A other point BinD define I(B) to equal to B. The value of I(B) does not depend on the choice of C, since F is conservative. To show that VI = Fweneedto demonstrate that ai/ax = M,al/ay = N,andaflaz = P. Suppose that B has coordinates (x, y, z). By def'mition, the value of the function I at F • dr, where Co is any path from A to Bo. We a nearby point Bo located at (xo,y, z) is take a path C = Co U L from A to B formed by first traveling along Co to arrive at Bo and then traveling along the line segment L from Bo to B (Figure 16.23). When Bo is close to B, the segment L lies in D and, since the value I(B) is independent of the path from A toB,
Ie
Ie
Xo
-~~~~---1~~) I
X _~~~~~~~~~/
Ie,
/
x
FIGURE 16.23 The function I(x, y, z) in the proof ofTheorem 2 is computed by a line integral f e• F . dr ~ I(Bo) fromA to BOo plus a line integral F' dr along a line segment L parallel to tire x-axis and joining Bo to B located at (x,y, z). The value of 1 atA is I(A) = O.
.h
I(x,y,z) = Differentiating, we have
1. 1 (1. 1
;x/(X,y,z) = ;x
F'dr +
F'dr +
F·dr.
F'dr).
924
Chapter 16: Integration in Vector Fields Only the last tenn on the right depends on x, so
a
a1
ax f(x,y, z) = ax
L
F' dr.
Now parametrize L as r(t) = Ii + yj + zk, Xo :5 t :5 x. Then dr/dt = i, F' dr/dt = M, and fL F' dr = f:' M(t, y, z) dt. Substitution gives
a a axf(x,y,z) = ax
x
Ix,
M(t,y,z)dt = M(x,y,z)
by the Fundamental Theorem of Calculus. The partial derivatives af/ay = N and af/az = P follow similarly, showingthatF = Vf. •
EXAMPLE 2
Find the work done by the conservative field F = yzi
+ xzj + xyk
=
Vf,
where
f(x, y, z) = xyz,
along any smooth curve Cjoining the pointA( -1,3,9) toB(I, 6, -4). With f(x,y, z) = xyz, we have
Solution
1 1 B
F'dr
=
=
F ~ VI and padJ independeoce
Vf'dr
f(B) - f(A)
Theorem 1
= XYZ I('.6.-4j - XYZI(-1~.9) =
(1)(6)(-4) - (-1)(3)(9)
=
-24 + 27
=
•
3.
A very useful property of line integrals in conservative fields comes into play when the for integration path of integration is a closed curve, or loop. We often use the notation C around a closed path (discussed with more detail in the next section).
f
THEOREM 3-Loop Property of Conservative Fields are equivalent. 1.
f
The following statements
F' dr = 0 around every loop (that is, closed curve
q
in D.
C
2. The field F is conservative on D.
Proof that Part 1 => Part 2 We want to show that for any two points A and Bin D, the integral of F • dr has the same value over any two paths C, and C2 from A to B. We reverse the direction on C2 to make a path -C2 fromB toA (Figure 16.24). Together, C, and -C2 make a closed loop C, and by assumption,
r F' dr - Jczr F' dr leIr F' dr + J-cz r F' dr Jcr F' dr
lei FIGURE 16.24 Ifwe have two paths from A to B, one of them cao he reversed to make a loop.
=
=
=
o.
Thus, the integrals over C, and C2 give the same value. Note that the definition of F' dr shows that changing the direction along a curve reverses the sign ofthe line integral.
16.3
925
Path Independence, Conservative Fields, and Potential Functions
Proof that Part 2 => Part 1 We want to show that the integral of F • dr is zero over any closed loop C. We pick two points A and B on C and use them to break C into two pieces: Cl from A to B followed by C2 from B back to A (Figure 16.25). Then
f
F ' dr =
c
r
lei
F'dr
+
r
lcz
F.dr =
r
r
F'dr -
JA.
F'dr = O.
JA-
•
The following diagram summarizes the results ofTheorems 2 and 3. A
A
Theorem 2
If A andB lie on a loop, we can reverse part of the loop to make two patha from A to B. FIGURE16.25
F = VIon D
=
Theorem 3
= fF'dr
F conservative onD
=
0
c
over any loop in D
Two questions arise:
1.
How do we know whether a given vector field F is conservative?
2.
If F is in fact conservative, how do we f"md a potential function I (so that F = V/)?
Finding Potentials for Conservative Fields The test for a vector field being conservative involves the equivalence of certain partial derivatives of the field components.
Component Test for Conservative Fields Let F = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k be a field on a connected and simply connected domain whose component functions have continuous first partial derivatives. Then, F is conservative ifand only if
oP oy
Proof that Equations that
ON
oM
OP
oz'
iJz
ax'
oN oX
and
oM oy'
(2)
(2) hold if F b conservative There is a potential function I such
F = Mi
+ Nj + Pk
01
= oX
i
+
01
oyj
+
01
oz k
Hence,
(0
0
oP _ ~ 1) _ 1 oy - oy oz - oyoz 2
all
Mixed Derivative Th~. Section 14.3
ozoy
=:z (:;)
=
~~.
The others in Equations (2) are proved similarly.
•
The second half of the proof, that Equations (2) imply that F is conservative, is a consequence of Stokes' Theorem, taken up in Section 16.7, and requires our assumption that the domain of F be simply connected.
926
Chapter 16: Integration in Vector Fields
Once we know that F is conservative, we usually want to find a potential function for F. This requires solving the equation Vj = F or
oj i + oj" + oj k ox oyJ OZ
=
Mi
+ N" + Pk J
for j. We accomplish this by integrating the three equations
oj oX =M,
oj oy
oj oz =P,
= N,
as illustrated in the next example.
EXAMPLE 3 Show that F = (eX cosy + yz)i + (xz - eX siny)j + (xy + z)k is conservative over its natural domain and find a potential function for it. Solution The natoral domain of F is all of space, which is connected and simply connected. We apply the test in Equations (2) to M = eXcosy
+ yz,
N = xz - eXsiny,
P=xy+z
and calculate OP oN oy =x= oz'
oM
iiZ
= y =
OP ox'
oN oX
-exsiny + z
=
=
oM. oy
The partial derivatives are continuous, so these equalities tell us that F is conservative, so there is a function j with Vj = F (Theorem 2). We fmd j by integrating the equations
oj oX
-
=
oj
eXcosy + yz '
oj oz
oy = xz - eX siny,
=
xy + z.
(3)
We integrate the first equation with respect to x, holding y and z fIXed, to get
j(x,y, z)
eX cosy + xyz + g(y, z).
=
We write the constant of integration as a function of y and z because its value may depend ony andz, though not onx. We then calculate ojjoy from this equation and match it with the expression for ojjoy in Equations (3). This gives
-eX siny + xz + so ogjoy
=
:~ =
xz - eX siny,
O. Therefore, g is a function of z alone, and
j(x,y, z)
=
eX cosy + xyz + h(z).
We now calculate ojjoz from this equation and match it to the formula forojjoz in Equations (3). This gives dh
xy+ dz =xy+z,
or
so
h(z)
Z2
=
2 + c.
dh dz
=
z,
16.3
Path Independence, Conservative Fields, and Potential Functions
927
Hence,
!(X,y, Z) =
eX cosy
+
lryZ
+
Z2
2
+
c.
-
We have infinitely many potential functions ofF, one for each value of C.
EXAMPLE 4 Solution
Show that F = (h - 3)1 - zj + (cosz)k is not conservative.
We apply the component test in Equations (2) and find immediately that
oP 0 oy = oy (cosz) = 0,
oN = -(-z) 0 = -1.
-
OZ
OZ
-
The two are unequal, so F is not conservative. No further testing is required.
EXAMPLE 5
Show that the vector field F =
2 -y 2 1+ 2 X 2j + Ok x +y x +y
satisfies the equations in the Component Test, but is not conservative over its natural domain. Explain why this is possible. Solution We have M = -y/(x 2 + y2), N = x/(x 2 + y2), and P = O. If we apply the Component Test, we find
oP=O=oN oy oz'
oP=O=oM ax
oM oy
and
iJz'
y2 - x 2 (x 2 + y2)2
oN oX .
So it may appear that the field F passes the Component Test. However, the test assumes that the domain of F is simply connected, which is not the case. Since x 2 + y2 cannot equal zero, the natural domain is the complement of the z-axis and contains loops that cannot be contracted to a point. One such loop is the unit circle C in the lry-plane. The circle is parametrized by r(l) = (COSI)I + (sinl)j,O :5 I :5 21T. This loop wraps around the z-axis and cannot be contracted to a point while staying within the complement of the z-axis. To show that F is not conservative, we compute the line integral F • dr around the loop C. First we write the field in terms of the parameter I: C
f
F =
- y2.1 + 2 x 2.J = . 2-sin I__2 1. + . 2 cosl _2 J. = ( -sml . ). + ( ). 2 1 cosl l· x + Y x + y sm I + cos-I sm I + co.-I
Next we f'md dr/dl = (-sinl)1 + (COSI)j, and then calculate the line integral as
f C
F·dr =
f F'~~ 1 dl
=
2 2 2 " (sin + COS
1
1) dl =
21T.
C
Since the line integral ofF around the loop C is not zero, the field F is not conservative, bY Theorem 3. _ Example 5 shows that the Component Test does not apply when the domain of the field is not simply connected. However, if we change the domain in the example so that it is restricted to the ball of radius I centered at the point (2, 2, 2), or to any sirnilar ball-shaped region which does not contain a piece ofthe z-axis, then this new domain D is simply connected Now the partial derivative Equations (2), as well as all the assumptions of the Component Test, are satisfied In this new situation, the field F in Example 5 is conservative on D.
928
Chapter 16: Integration in Vector Fields Just as we must be careful with a function when determining if it satisfies a property throughout its domain (like continuity or the intermediate value property), so must we also be careful with a vector field in determining the properties it mayor may not have over its assigned domain.
Exact Differential Forms It is often conveuient to express work and circulation integrals in the differential form
!cMax+Ndy+Pdz discussed in Section 16.2. Such line integrals are relatively easy to evaluate if
Max + N dy + P dz is the total differential of a function t and C is any path joining the two points from A to B. For then {
( at at at ax ax + ay dy + az dz
Jc M ax + N dy + P dz = Jc
1 8
=
Vtodr
= t(B) - t(A).
Vfiscooservative.
The= I
Thus,
1 8
df= t(B) - t(A),
just as with differentiable functions of a single variable.
DEFINmONS Any expression M(x,y,z)ax + N(x,y,z)dy + P(x,y,z)dz is a dilTerential form. A differential form is exact on a domain D in space if af
at
at
Max+Ndy+Pdz=~ax+~dy+~dz=~
for some scalar function
t throughout D.
Notice that if M ax + N dy + P dz = dt on D, then F = Mi + Nj + Pk is the gradient field of t on D. Conversely, if F = Vf, then the form Max + N dy + P dz is exact. The test for the form's being exact is therefore the same as the test for F being conservative.
Component Test for Exactness of M dx + N dy + P dz The differential form Max + N dy + P dz is exact on a connected and simply connected domain if and only if
ap ay
aN az'
aM
ap
iJz
ax'
and
This is equivalent to saying that the field F = Mi
aN ax
aM ay·
+ Nj + Pk is conservative.
16.3
EXAMPLE 6
929
Path Independence, Conservative Fields, and Potential Functions
+ x dy + 4 dz is exact and evaluate the integral
Show that y dx
<2,3,-')
1
ydx + xdy + 4dz
(1,1,1)
over any path from (I, I, I) to (2, 3, - I). Solution
We let M = y, N = x, P = 4 and apply the Test for Exactness:
aN=I=aM ax ay'
+ x dy + 4 dz is exact, so
These equalities tell us that y dx
y dx + x dy + 4 dz
dl
=
for some function I, and the integral's value is 1(2, 3, -I) - 1(1, I, I). We f'md 1 up to a constant by integmting the equations
al ax = y,
al ay = x,
al az = 4.
(4)
From the first equation we get
I(x,y,z)
xy + g(y,z).
=
The second equation tells us that
al ag ay=x+ay=x,
ag
or
iJy =
o.
Hence, g is a function of z alone, and
I(x,y, z)
=
xy
+ h(z).
or
h(z)
The third of Equations (4) tells us that
al az
=
0
+ dh
= 4 dz'
=
4z + C.
Therefore,
I(x,y,z)
=
xy
+ 4z +
C.
The value of the line integral is independent of the path taken from (I, I, I) to (2, 3, -I), and equals
1(2,3, - I) - I(I, I, I)
=
2 + C - (5 + C)
=
•
-3.
Exercises 16.3 resting for Conservative fields
8. F
Which fields in Exercises 1--6 are conservative, and which are not? 1. F =yzi +xzj +xyk 2. F = (y sinz)i + (x sinz)j + (xy cosz)k 3. F = yi + (x + z)j - yk 4.F=-yi+xj S. F ~ (z + y)i + zj + (y + x)k 6. F ~ (eX cosy)i - (eX siny)j + zk
9. F ~ eY+"(i
finding Potential Functions 10 Exercises 7-12, rmd a poteotial function f for the fieldF. 7. F=2xi+3yj+4zk
~
(y + z)i + (x + z)j + (x + y)k
+ xj + 2xk)
10. F = (y sinz)i + (x sinz)j + (xy cosz)k 11. F ~ (lux + sec2 (x + y))i +
~)j + ~k ( sec'(x + y) + Y +z y +z 12.F=
I + YX 2 y 2 i+( 1+ xX 2 y 2 + VI _z y2 z 2 )j+ (
Y
VI - y2 z 2
+
l)k z
930
Chapter 16: Integration in Vector Fields
Exact Diffell!ntial Fonns In Exercises 13-17, show that the differential fanns in the integrals are exact Then evaluate the integrals. (2,3,-<)
13.
1
2xtb:
L
+ 2ydy + 2zdz
1,
yztb:+xzdy+:rydz (1~'3)
1
2xy tb:
+ (x'
16.
1
(0,0,0) (0,1,1)
17.
1,
+
~
(sin I)j
z
+
~
1
(1/2'IT)k, 0 '" I '" 2'IT
z
- z') dy - 2yz dz
(0.0.0) (3.3.1)
= 0, 0
c. The x-axis from (I, 0, 0) to (0, 0, 0) followed by the parabola z = x 2 ,y = 0 from (0, 0, 0) to (I, 0, I)
(1,1,')
15.
The line segment x = l,y
b. The helix r(l) = (cos 1)1
(0,0,0) (3,5,0)
14.
29. Work along different paths Find the work done by F = (x' + y)1 + (y2 + x)j + ze'k over the following paths from (1,0,0) to (I, 0, I).
4
2xdx - y'dy - - - d z I + z' sinycosxdx + cosy sin x dy + dz
",--,,,(0:,.,"O'C::0,-)_ - . " y
(1,0,0)
finding Potential Functions to Evaluate LIne Integrals Although they are not defmed on all of space R 3, the fields associated with Exercises 18-22 are simply connected and the Component Test can be used to show they are conservative. Find a potential function for each field and evaluate the integrals as in Example 6.
18.
1
(1,7/2,')
2 cosy dx
(I
)
I
+ y - 2xsiny dy + zdz
(0,',1)
19.
1,
(1~'3)
1,
(2,1,1)
,
3x'tb:
(1,1,1)
20.
+~
dy
a. The line segroentx = I,y = 'lT1/2,z = I - 1,0 '" I '" I
z
+ 2z loy dz
(2xloy - yz)dx +
(
(1,2,1)
21.
30. Work along different path. Find the work done by F = e"'l + (xze'" + zcosy)j + (:rye'" + siny)k over the following paths from (1,0, I) to (I, 'IT/2, 0).
2
)
~ - xz dy -:rydz
I (I x ) 1, Y-dx + z - - dy - -dz 2xdx + 2ydy + 2zdz x + + 1 (2~~)
'If
2
Y
(1,1,1)
..P.....-----/+~y
z2
y2
r;~o)
(2~~)
22.
2
(-I, -1, -1)
y2
z2
Applications and Examples 23. Revisiting Example 6
Evaluate the integral
(2,3.-1)
1,
ydx
+ xdy + 4dz
x
1
b. The line segroent from (I, 0, I) to the origin followed by the line segroent from the origin to (I, 'IT/2, 0)
z
0,1,1)
from Example 6 by fmding parame1ric equations for the line segment from (I, I, I) to (2,3, -I) and evaluating the line integral ofF ~ yi + xi + 4k along the segroent Since F is conservative, the integral is independent of the path.
'If
2 )i>i,"-"=----/+~y
r;~o)
24. Evaluate
fc x 2 dx + yz dy + (y2/2) dz along the line segroent C joining (0, 0, 0) to (0, 3, 4). Independence of path Show that the values of the integrals in Exercises 25 and 26 do not depend on the path taken from A toB.
25.1" z 2 dx 26.
l
A
z
+ 2ydy + 2xzdz
1
+ ydy + zdz --:-r~==:'=~ \/x' + y2 + z2
"xdx
e;/)i,
In Exercises 27 and 28, fmd a potential function for F.
27. F = :;' I 28. F
+
c. The line segroent from (I, 0, I) to (I, 0, 0), followed by the x-axis from (I, 0, 0) to the origin, followed by the parabola y = 'lTx 2/2,z = o from there to (I, 'IT/2, 0)
((x,y): y > 0)
~ (e'loy)1 + (~ + Sinz)j + (ycosz)k
16.4 31. Evaluating a work integral two ways Let F ~ V(x'y2) and let C be the path in the xy-plane from (-I, I) to (I, I) that consists of the line segment from (-I, I) to (0, 0) followed by the line segment from (0, 0) to (I, I). Evaluate leF'dr in two ways.
a. Find parametrizations for the segments that make up C and evaluate the integral. b. Use!(x,y) ~ x'y2 as a potential function forF. 32. Integral along different path. Evaluate the line integral Ie 2x cos y dx - x 2 sin y dy along the following paths C in the xy-plane. a. The parabola y = (x - I)' from (1,0) to (0, I) b. The line segment from (-I,,,.) to (I, 0) c. Thex-axi. from (-I, 0) to (I, 0) d. The astroid r(l) = (cos'l)i + (sin' I)j, 0 ,;; I';; 2"., counterclockwise from (I, 0) back to (I, 0) y
Green's Theorem in the Plane
34. Gradient of a line integral vative vector field and g(x,y,z) =
Suppose that F
ic
~
931
V! is a conser-
«3X)
F·dr.
(0,0,0)
ShowthatVg = F. 35. Path of lea.t work You have been asked to find 1he pa1h along which a force field F will perform 1he least work in moving a particle between two locations. A quick calculation on your part shows F to be conservative. How should you respond? Give reasons for your answer. 36. A revealing esperiment By experiment, you fmd that a force field F performs only half as much work in moving an object along pa1h C, from A to B as it does in moving 1he object along path C2 from A to B. What can you conclude about F? Give reasons for your answer. 37. Work by a coostant foree Show that 1he work done by a constant force field F ~ oi + bj + ck in moving a particle along aoypa1hfromA toB i. W = F·AB.
(0, I)
38. Gravitational field (-1,0)
(1,0)
a. Find a potential function for the gravitational field
'---ooEc--+----c:lM~x
F = -GmM
_~n,,-'_+-,y~j_+~zk~ (x2 + y2 + z2)'/2
(0, -I)
(G, m, andM are constants).
33. a. Exact differential form How are 1he constants related if the following differential form is exact?
0,
b, and c
(ay2 + 2czx) dx + y(bx + cz) dy + (ay2 + ex 2) dz
For what values of b and c will F = (y2 + 2=)i + y(bx + cz)j + (y2 + ex 2 )k be a gradient field?
b. Gradient field
16.4
s,
b. Let P, and P2 be points at distance and S2 from the origin. Show that 1he work done by 1he gravitational field in part (a) in moving a particle from PI to Pz is
GmM(~~). 82 Sl
Green's Theorem in the Plane If F is a conservative field, then we know F = Vt for a differentiable function t, and we can calculate the line integral of F over any path C joining point A to B as fcF'dr = t(B) - t(A). In this section we derive a method for computing a work or flux integral over a closed curve C in the plane when the field F is not conservative. This method, known as Green's Theorem, allows us to convert the line integral into a double integral over the region enclosed by C. The discussion is given in terms of velocity fields of fluid flows (a fluid is a liquid or a gas) because they are easy to visualize. However, Green's Theorem applies to any vector field, independent of any particular interpretation of the field, provided the assumptions of the theorem are satisfied. We introduce two new ideas for Green's Theorem: divergence and circulation density arowd an axis perpendicular to the plane.
Divergence Suppose that F(x, y) = M(x, y)1 + N(x, y)i is the velocity field of a fluid flowing in the plane and that the first partial derivatives of M and N are continuous at each point of a region R. Let (x, y) be a point in R and let A be a small rectangle with one corner at (x, y) that, along with its interior, lies entirely in R. The sides of the rectangle, parallel to the coordinate axes, have lengths of ax and Ily. Assume that the components M and N do not
932
Chapter 16: Integration in Vector Fields
(x,y
+ Oy)
(x
+ Dx,Y + Oy)
F'j>O
,I' F ·(-1) < 0
'"
A
!t Oy
F·I>O F·(-j)
(x
+ Ox, y)
k-I,- ----ox------->l,1 FIGURE 16.26 The rate at which the fluid leaves the rectangular region A across the bottom edge in the direction of the outwardnormal-j is approximately F(x, y). (-j) At, which is negative for the vector field F shown here. To approximate the flow rate at the point (x, y), we calculate the (approximate) flow rates across eacb edge in the directions of the red arrows, sum these rates, and then divide the sum by the area ofA. Thking the limit .. At --+ 0 and ay--+ 0 gives the flow rate per unit area.
change sign tbroughout a small region containing the rectangle A. The rate at which fluid leaves the rectangle across the bottom edge is approximately (Figure 16.26)
F(x,y)'(-j) ax
=
-N(x,y)ax.
TIris is the scalar component ofthe velocity at (x,y) in the direction ofthe outward normal times the length of the segment. If the velocity is in meters per second, for example, the flow rate will be in meters per second times meters or square meters per second. The rates at which the fluid crosses the other three sides in the directions of their outward nonnals can be estimated in a similar way. The flow rates may be positive or negative depending on the signs of the components of F. We approximate the net flow rate across the rectangular boundary of A by summing the flow rates across the four edges as defined by the following dot products. Fluid Flow Rates:
F(x,y + ay)' j ax = N(x,y + ay)ax F(x,y)'(-j) ax = -N(x,y)ax
Top: Bottom:
F(x + ax,y)'i ay = M(x + ax,y)ay F(x,y)' (-i) ay = -M(x,y)ay.
Right: Left: Summing opposite pairs gives Top and bottom:
(N(x,y
Right and left:
(M(x
+
(~~ ay)ax
+ ay) - N(x,y»ax '" ax,y) - M(x,y))ay '"
(~~ ax)ay .
Adding these last two equations gives the net effect of the flow rates, or the
ax + ON) oy axay.
Flux across rectangle boundary '" ( oM
We now divide by axay to estimate the total flux per unit area or flux density for the rectangle: Flux across rectangle boundary '" rectangle area
(OM + ON). oX
oy
16.4
Source: div F (xo. Yo) > 0 A gas expanding at the point (xo. Yo).
~r/ /1""-
<--
-----
SInk: div F (xo, Yo)
Green's Theorem in the Plane
933
ax
Finally, we let and lly approach zero to define the flux density ofF at the point (x,y). In mathematics, we call the flux density the divergence of F. The symbol for it is div F, pronounced "divergence ofF' or "div F."
DEFINmON The divergence (flux density) of a vector field F = MI the point (x, y) is
div F
<0
=aM+aN ax ily'
+ Nj at
(1)
A gas compressing at the point (xo, Yo).
".1/
7r~ FIGURE 16.27 Ifa gas is expanding at a point (Xo, Yo), the lines of flow have positive divergence; if the gas is compressing, the divergence is negative.
A gas is compressible, unlike a liquid, and the divergence of its velocity field measures to what extent it is expanding or compressing at each point. Intuitively, if a gas is expanding at the point (xo,Yo), the lines of flow would diverge there (hence the name) and, since the gas would be flowing out of a small rectangle about (xo, Yo), the divergence of F at (xo, Yo) would be positive. If the gas were compressing instead of expanding, the divergence would be negative (Figure 16.27).
EXAMPLE 1 The following vector fields represent the velocity of a gas flowing in the xy-plane. Find the divergence of each vector field and interpret its physical meaning. Figure 16.28 displays the vector fields.
(8) Uniform expansion or compression: F(x, y) = cxi
(b) Uniform rotation: F(x,y) = -cyi
+ cxj
(c) Shearingflow: F(x,y) = yi
(d) Whirlpool effect: F(x, y) =
2 -y 2 I + 2 X 2j x +y x +y
+ cyj
934
Chapter 16: Integration in Vector Fields Solution
<
+ ~ (cy)
divF =
>
2c: If c
0, the gas is undergoing unifonn expan-
0, it is undergoing unifonn compression.
(b) div F = :x (-cy)
=
:x (y)
+ ~ (ex)
= 0: The
gas is neither expanding nor compressing.
= 0: The gas is neither expanding nor compressing.
~ ( -y ) + ~( x ) oX x2 + y2 oy x2 + y2
2xy 2xy = O. (x2 + y2)2 (x2 + y2)2 . Again, the divergence is zero at all points in the domain ofthe velocity field. =
• Cases (b), (c), and (d) of Figure 16.28 are plausible models for the two-dimensional flow of a liquid. In fluid dynamics, when the velocity field of a flowing liquid always bas divergence equal to zero, as in those cases, the liquid is said to be incompressihle.
Spin Around an Axis: The k-Component of Curl The second idea we need for Green's Theorem bas to do with measuring how a floating paddle wheel, with axis perpendicular to the plane, spins at a point in a fluid flowing in a plane region. This idea gives some sense of how the fluid is circulating around axes located at different points and perpendicular to the region. Physicists sometimes refer to this as the circulation density of a vector field F at a point. To obtain it, we retorn to the velocity field F(x,y) = M(x,y)i
+
N(x,y)j
and consider the rectangle A in Figure 16.29 (where we assume both components ofF are positive). (x,y
+ Oy)
(x
+ OX,y + Oy)
l
I
F·(-I)
1 F'(-j)
f
A
Oy
F(x,y)
] J +
OP
,
F·I>O (x,y)
I,
(x OX
Ox,y)
,I
FIGURE 16.29 The rate at which a fluid flows along the bottom edge ofa rectangular region A in the direction i is approximately F(x, y) • i ax, which is positive for the vector field F shown here. To approximate the rate ofcirculation at the point (x, y), we calcn1ate the (approximate) flow rates along each edge in the directions of the red arrows, sum. these rates, and. then divide the sum by the area of A. Taking the limit as ax ---+ 0 and ay ---+ 0 gives the rate of the circulation per unit area.
The circulation rate of F around the boundary of A is the sum of flow rates along the sides in the tangential direction. For the bottom edge, the flow rate is approximately F(x,y)'iAx = M(x,y)Ax.
16.4
This is the scalar component of the velocity F(x, y) in the tangent direction i times the length ofthe segment. The flow rates may be positive or negative depending on the components ofF. We approximate the net circulation rate around the rectangular boundary of A by summing the flow rates along the four edges as dermed by the following dot products.
Vertical axis
kt
eb
Top: Bottom: Right: Left:
----
CurlF (xo,Yo)· k > 0 Counterclockwise circulation
Top and bottom:
c6
Right and left:
-
F(x,y + lly) (-i) llx = -M(x,y + lly)llx F(x,y) °i llx = M(x,y)llx F(x + llx,y) j lly = N(x + llx,y)lly F(x,y)' (-j) lly = -N(x,y)lly. 0
0
We sum opposite pairs to get
Vertical axis
kt
935
Green's Theorem in the Plane
-(M(x,y (N(x
+ lly) - M(x,y»llx '" -
+ llx,y) - N(x,y))lly '"
(~~ llY)llx
(~~ llx )lly.
Adding these last two equations gives the net circulation relative to the counterclockwise orientation, and dividing by llxlly gives an estimate of the circulation density for the rectangle:
CurlF (xo, Yo) • k < 0 Clockwise circulation
aN aM Circulation around rectangle '" ax -a,u. rectangle area J
FIGURE 16.30 In the flow ofan incompressible fluid over a plane region, the k-component of the curl measures the rate of the fluid's rotation at a point. The k-component ofthe curl is positive at points where the rotation is counterclockwise and negative where the rotation is clockwise.
We let llx and lly approach zero to define the circulation density of F at the point
(x,y). If we see a counterclockwise rotation looking downward onto the xy-plane from the tip ofthe unit k vector, then the circulation density is positive (Figure 16.30). The value of the circulation density is the k-component of a more general circulation vector field we derme in Section 16.7, called the curl of the vector field F. For Green's Theorem, we need only this k-component.
DEFINmON The circulation density of a vector field F = Mi point (x, y) is the scalar expression aN ax
+ Nj at the
aM ay'
---
(2)
This expression is also called the k-component of the curl, denoted by (curl F) k. 0
If water is moving about a region in the :ry-plane in a thin layer, then the k-component ofthe curl at a point (xo, YO) gives a way to measure how fast and in what direction a small paddle wheel spins if it is put into the water at (xo, YO) with its axis perpendicular to the plane, parallel to k (Figure 16.30).
EXAMPLE 2 Find the circulation density, and interpret what it means, for each vector field in Example I. Solution (a) Uniform expansion: (curl F) k = 0
at very small scales.
:x (0') -
~ (ex)
= O. The gas is not circulating
936
Chapter 16: Integration in Vector Fields y
--- --- --- ---
(b) Rotation: (curl F)' k = :x (ex) -
indicates rotation at every point. If c the rotation is clockwise.
(0) Shear: (curl F)' k = -
:0---
----
~ (-cy)
~ (y)
>
= 2c. The constant circulation denaity
0, the rotation is counterclockwise; if c
<
0,
= -1. The circulation density is constant and negative,
so a paddle wheel floating in water undergoing such a shearing flow spins clockwise. The rate ofrotation is the same at each point. The average effect of the fluid flow is to push fluid clockwise around each of the small circles shown in Figure 16.31. (d) Whirlpool:
FIGURE 16.31 A shearing flow pushes the fluid clockwise around each point (Example 2c).
(curlF)'k=~( ax
x )_~( -y )= y2_ x 2 _ y2_ x 2 =0 x 2 + y2 ay x2 + y2 (x2 + y2)2 (x2 + y2)2 .
The circulation density is 0 at every point away from the origin (where the vector field is undefined and the whirlpool effect is taking place), and the gas is not circulating at any point for which the vector field is defined _
Two Forms for Green's Theorem In one form, Green's Theorem says that under suitable conditions the outward flux of a vector field across a simple closed curve in the plane equals the double integral of the divergence of the field over the region enclosed by the curve. Recall the formulas for flux in Equations (3) and (4) in Section 16.2 and that a curve is simple if it does not cross itseIf.
THEOREM 4-Green's Theorem (flUX-DIvergence or Normal Form) Let C be a piecewise smooth, simple closed curve enclosing a region R in the plane. Let F = Mi + Nj be a vector field with M and N having continuous first partial derivatives in an open region containing R. Then the outward flux of F across C equals the double integral of div F over the region R enclosed by C.
f
F . nds
C Outward flux
=
fMdY-Ndx= JJ(aa~+~~)dxdy C
We introduced the notation
(3)
R
Divergence integral
fc in Section 16.3 for integration around a closed curve.
We elaborate further on the notation here. A sirople closed curve C can be traversed in two possible directions. The curve is traversed counterclockwise, and said to be positively oriented, if the region it encloses is always to the left of an object as it moves along the path. Otherwise it is traversed clockwise and negatively oriented. The line integral of a vector field F along C reverses sign if we change the orientation. We use the notation
f
F(x,y)'dr
c
for the line integral when the sirople closed curve C is traversed counterclockwise, with its positive orientation. A second form of Green's Theorem says that the counterclockwise circulation of a vector field around a sirople closed curve is the double integral of the k-<:omponent of the curl of the field over the region enclosed by the curve. Recall the defming Equation (2) for circulation in Section 16.2.
937
16.4 Green's Theorem in the Plane
THEOREM 5-Green's Theorem (Circulation-Curl or Tangential Form) Let C be a piecewise smooth, simple closed curve enclosing a region R in the plane. Let F = Mi + Nj be a vector field with M and N having continuous first partial derivatives in an open region containing R. Then the counterclockwise circulation of F around C equals the double integral of (curl F) • k over R.
f
f Mdx
F ' Tds =
C
+ Ndy
=
ff(~~-~~)dxdY
(4)
R
C
Counterclockwise circulation
Curl integral
The two forms of Green's Theorem are equivalent. Applying Equation (3) to the field + Mj gives
G, = Ni - Mj gives Equation (4), and applying Equation (4) to G, = -Ni Equation (3).
Both forms of Green's Theorem can be viewed as two-dimensional generalizations of the Net Change Theorem in Section 5.4. The outward flux of F across C, dermed by the line integral on the left-hand side of Equation (3), is the integral of its rate of change (flux density) over the region R enclosed by C, which is the double integral on the right-hand side ofEquation (3). Likewise, the counterclockwise circulation ofF around C, dermed by the line integral on the left-hand side of Equation (4), is the integral of its rate of change (ciIculation density) over the region R enclosed by C, which is the double integral on the right-hand side of Equation (4).
EXAMPLE 3
Verify both forms of Green's Theorem for the vector field
F(x,y) = (x - y)i
+ xj
and the region R bounded by the unit circle C:
Solution
ret)
= (cost)i
+
(sint)j,
o ,,; t ,,; 21T.
Evaluating F(r(t» and differentiating components, we have M= cost - sint,
dx
N = cost,
dy = d(sin t) = cos t dt,
aM = I ax '
aM =-1 ay ,
=
d( cos t)
aN = I
ax
= -sin t dt,
'
aN = 0
ay
.
The two sides of Equation (3) are
f
C
Mdy -
Ndx 1 =
'~2" 1=0
2"
=
JJ (~~ + ~~) dxdy
=
R
0 /.
(cost - sin t)(costdt) - (cost)(-sintdt)
cos2 tdt
= 1T
ff (I + O)dxdy R
=
ff dx dy R
= area inside the unit circle = 1T.
938
Chapter 16: Integration in Vector Fields y
The two sides of Equation (4) are
f Mdx + Ndy
=
c =
JJ (~~ - ~~)
dxdy
=
R
FIGURE 16.32 The vector field in Example 3 has a counterclockwise circulatioo of2". around the unit circle.
1~~2w (cost -
12w JJ
sint)(-sintdt) + (cos t)(cos tdt)
(-sintcost + I) dt
(I - (-I» dxdy
=
R
2".
=
2JJ dxdy
2"..
=
R
•
Figure 16.32 displays the vector field and circulation around C.
Using Green's Theorem to Evaluate Line Integrals If we construct a closed curve C by piecing together a number of different curves end to end, the process of evaluating a line integral over C can be lengthy because there are so many different integrals to evaluate. If C bounds a region R to which Green's Theorem applies, however, we can use Green's Theorem to change the line integral around C into one double integral over R.
EXAMPLE 4
Evaluate the line integral
f xydy - y 2 dx,
c where C is the square cut from the first quadrant by the lines x = I and y = I.
Solution
We can use either form of Green's Theorem to change the line integral into a double integral over the square. 1.
With the Normal Form Equation (3): Taking M = xy, N = y2, and C and R as the square's boundary and interior gives fXYdy - y2 dx= JJ(y + 2y)dxdy= l'l'3Y dxdy C
R
=
2.
1 1
o
[]X~1 3xy dy x-o
=
l'
3ydy
=
With the Tangential Form Equation (4): Taking M
=
3]' 3
_y2 2
0
_y2 and N
=-.
0
2
=
xy gives the same
=~.
•
result:
f_ y 2 dx + xydy C
=
JJ(y - (-2y))dxdy R
EXAMPLE 5 Calculate the outward flux of the vector field F(x,y) = xi the square bounded by the lines x = ± I and y = ± 1.
+ y2j
across
16.4 Green's Theorem in the Plane
939
Solution
Calculating the flux with a line integral would take four integrations, one for each side of the square. With Green's Theorem, we can change the line integral to one double integral. WithM = x, N = y2, Cthe square, andR the square's interior, we have
Flux = f
Fond< = f Mdy - Ndx
C
C
= ff (aM + aN) dxd JJ ax ay Y
1 11 1-1 -1 +
Omen's Theorem
R
=
(I
1
2y) dxdy
1
=
(2 + 4y)dy
=
11-1 [ + X
[2Y + 2y2 ]1
=
-1
2xy
=
]X=1
x=-I
dy
•
4.
-1
Y
Proof of Green's Theorem for Special Regions Let C be a smooth simple closed curve in the ;ry-plane with the property that lines parallel ,
R
,, , ,
,
:C.
that M, N, and their fIrst partial derivatives are continuous at every point of some open region containing C and R. We want to prove the ciroulation-curl fonn of Green's Theorem,
,
,I
f Mdx + Ndy
--;ot-~---~--;--~x
o
a
11 (~~ -~~)dxdY.
_,(),P,(x.f,Vc))
11'Y-JIXI ,
to the axes cut it at no more than two points. Let R be the region enclosed by C and suppose
x
b
=
C
(5)
R
Figure 16.33 shows Cmade up of two directed parts:
FIGURE 16.33 The boundary curve Cis made up ofC" the graph ofy ~ j,(x), and C2, the graph of y = f,(x).
C1: y = h(x),
a:5
x
C2 : y = h(x),
b,
:5
b
~
x
~
Q.
For any x between a and b, we can integrate aM/ay with respect to y from Y = h(x) to = j,(x) and obtain
Y
h (X) aM
1
ady
f,«)
y
=
M(x,y)
r=
!z(x)
=
M(x,j,(x» - M(x,h(x)).
y= f,(x)
We can then integrate this with respect to x from a to b:
ll b
a
h
«) adydx aM = 1I(x) ~
= Y d
Ci: x ~ 8'(Y) ----\--~--_.........
l -l b
[M(x,j,(x)) - M(x,h(x))]dx
a
a
M(x, j,(x» dx -
l
b
M(x, h(x)) dx
=-fMdx-f Mdx
lei
JC'l
Y
=-fMdx. C
c
Therefore
--;cr------------> x
o
FIGURE 16.34 The boundary curve Cis made up ofCl. the graphofx = g,(y), and C;, the graph of x ~ g2(Y).
f Mdx C
=
11 (-~~)dxdy.
(6)
R
Equation (6) is half the result we need for Equation (5). We derive the other halfby integrating
aN/ax f'rrst with respect to x and then with respect to y, as suggested by Figure 16.34.
940
Chapter 16: Integration in Vector Fields
y
TIris shows the curve C of Figure 16.33 decomposed into the two directed parts C\: x = gt(Y), d ;;,: Y ;;,: c and Co: x = g2(Y), c ,,; Y ,,; d. The result of this double integration is
e
f
~
N dy =
e
II ~~
(7)
dx dy.
R
Summing Equations (6) and (7) gives Equation (5). TIris concludes the proof.
o
•
Green's Theorem also holds for more general regions, such as those shown in Figures 16.35 and 16.36, but we will not prove this result here. Notice that the region in Figure 16.36 is not simply connected. The curves C , and Ch on its boundary are oriented so that the region R is always on the left-hand side as the curves are traversed in the directions shown. With this convention, Green's Theorem is valid for regions that are not simply connected. While we stated the theorem in the xy-p1ane, Green's Theorem applies to any region R contained in a pIane bounded by a curve C in space. We will see how to express the double integral over R for this more general form of Green's Theorem in Section 16.7.
(a)
y b
a
y
--..Of---L..... ...J,b-~ x a (b)
FIGURE 16.35 Other regions to which Green's Theorem applies.
-+-+---..t--.:i-+'-I --+ x
FIGURE 16.36 Green's Theorem may be applied to the aonular region R by sununiog the line integrals along the houodaries C, and Ch in the directions shown.
Exercises 16.4 Verifying Green's Theorem
10 Exercises 1-4, verifY the conclosion ofGreen's Theorem by evaluating both sides ofEquations (3) and (4) for the field F = Mi + Nj. Take the domains of iotegration io each case to be the disk R: x 2 + y2 ,,; a 2 and its bouodiog circle C: r = (a cos I)i + (a sio I)j, I ,,; 2".
9. F ~ (xy
+ y2)i + (x
- y)j
10. F
~
+ 3y)i + (2x -
(x
y)j
y
y
°,,;
l.F~-yi+xj
2.F~yi
3. F = 2rl - 3yj
4. F = -x'yi + xy2j
---cr---+--+-~x
-2
2
---;:;;-;;;+-~------+x
(0,0)
Clrculatton and Flux 10 Exercises 5-14, use Green's Theorem to fmd the couoterclockwise circolation and outward flux for the field F and corve C. S. F
c:
~
(x - y)i
+ (y -
x)j
11. F = x 3y 2 i
The square bouoded by x = o,X = l,y = o,y = I
-I
+ kX4yj
12. F = ~i+ (tan-1y)j l+y
y
y
(x 2
6. F = + 4y)i + (x + y2)j C: The square bouoded by x = O,X = l,y = O,y = I 7. F ~ (y2 - x 2)i + (x2 + y2)j C: The triangle houoded by y = 0, x
= 3 , and y = x
8. F ~ (x + y)i - (x 2 + y2)j C: The triangle houoded by y = 0, x
= 1 , and y = x
1
x 2 +y2=1
(2,2)
1
-I
---;:o-:;;~-7'-------+x
(0,0)
-I
x
Green's Theorem in the Plane
16.4
13. F
~
(x
+ eX siny)i + (x + eX cosy)j
C: The right-hand loop of the lemniscste r 2 ~ cos 28 14. F
Green'. Theorem Area Formula
~ (tan-I~} + In(x 2 + y2)j
AreaofR = k f xdy - ydx
c
C: The boundary of the region defmed by the polar coordinate inequalities 1 :s r :s 2, 0 ::s 8 ::s 11'
15. Find the counterclockwise circulation and outward flux of the field F ~ xyi + y2j around and over the boundary of the region enclosed by the curves y = x 2 and y = x in the frrst quadrant
The reason is that by Equation (3), run backward, AreaofR =
16. Find the counterclockwise circulation and the outward flux of the field F ~ (-siny)i + (xcosy)j around and over the square cut from the frrst quadrant by the lines x = ".12 and y = "./2.
F =
+ cos 8), a > O.
18. Find the counterclockwise circulation of F = (y + eX In y)i + (eXIy)j around the boundary of the region that is bounded above by the curve y = 3 - x 2 and below by the curve y = x' + I.
Work In Exercises 19 and 20, fmd the work done by F in moving a particle
~
+ (2x -
C: The circle (x - 2»
+
f kxdy - kydx.
25. The circle r(t) ~ (a cos t)i 26. The ellipse r(t) = (a cos t)i 27. The astroid r(t)
+ (a sin t)j,
0 '" t '" 2".
+ (b sin t)j,
= (cos3 t)i +
(sin3 t)j,
0 '" t '" 2". 0 '" t '" 2".
28. One arch of the cycloid x = t - sint, y = 1 - cost 29. Let C be the boundary of a region on which Green's Theorem holds. Use Green~ Theorem to calculate
.. f f
f(x) dx + g(y) dy
ky dx
+ hx dy (k and h constants).
c 30. Integral depeodent only on area
Show that the value of
4y)j
f xy2 dx + (x2y + 2x) dy
(y - 2)2 ~ 4
c
Using Green's Theorem Apply Green~ Theorem to evaluate the integrals in Exercises 21-24.
21. f (y2 dx
+ k)dydx
R
Use the Green's Theorem area formula given above to fmd the areas of the regions enclosed by the curves in Exercises 25-28.
b.
C: The boundary of the ''triangular'' region in the frrst quadrant enclosed by the x-axis, the line x = 1, and the curve y = x 3
(4x - 2y)i
II (k
c
once counterclockwise around the given curve. 19. F = lxy':i + 4x2y2j
~
dydx =
c
(3xy - ~)i + (eX + tan-I y)j 1 +Y
aerossthe cardioidr = a(1
JJ R
17. Find the outward flux ofthe field
20. F
941
+ x 2 dy)
around any square depends only on the area of the square and not on its location in the plane. 31. What is special about the integral
f
4x 3y dx
+ x' dy?
c
C
C: The mangle bounded by x = 0, x
+y
= I,y = 0
Give reasons for your answer. 32. What is special about the integral
22. f (3ydx
+ 2xdy)
f
c C: The boundary of 0 '" x",,,., 0 '" Y '" sinx
23. f
(6y
C: The circle (x - 2» 24. f
(2x
+ y2) dx +
+ (y -
(lxy
c Give reasons for your answer. 33. Area as a nne integral Show that if R is a region in the plane bounded by a piecewise smooth, simple closed curve C, then
+ x)dx + (y + 2x)dy
c
_y3dy+x 3 dx?
3»
=4
+ 3y) dy
c C: Any simple closed curve in the plane for which Green's Theorem holds Calculating Area with Green" Tbeorem If a simple closed curve C in the plane and the region R it encloses satisfy the hypotheses of Green's Theorem, the area of R is given by
AreaofR = f xdy = - f ydx.
c
c
34. Definite integral as a nne integral Suppose that a nonnegative functiony = f(x) bas a continuous frrstderivative on [a, b]. Let C be the boundary of the region in the xy-plane that is bounded below by the x-axis, above by the grapb of f, and on the sides by the lines x = a and x = b. Show that
/.b
f(x)dx
=-
f ydx.
c
942
Chapter 16: Integration in Vector Fields
35. Area and the centroid Let A be the area and xthe x-coordinate of the centroid of a region R that is bounded by a piecewise
a. Let f(x,y) ~ in (x 2 + y2) and let Coo lbe circle x 2 + y2 = a 2. Evaluate the flux integral
smoolb, simple closed curve C in lbe ~-p1llne. Show that
f
Vf·ntis.
c 36. Moment of inertia Let I y be the moment of loertia about lbe y-axis oflbe region 10 Exercise 35. Show that
37. Green's Theorem and Laplace's equation Assuming that all the necessary derivatives exist and are continuous, show that if f(x,y) satisfies lbe Laplace equation
a2f a2f '+-2=0. ax ay lben
b. Let Kbe an arbitrary smoolb, simple closed curve in lbe plane that does not pass through (0, 0). Use Green's Theorem to show that
has two possible values, depending on whelber (0, 0) lies inside K or outside K.
40. Bendi::l:son's criterion The streamlines of a planar fluid flow are lbe smoolb curves traced by lbe fluid's individual particles. The vectors F ~ M(x, y)i + N(x, y)j of lbe flow's velocity field are the tsngent vectors of lbe streamlines. Show that if the flow takes place over a simply connected region R (no holes or misslog points) and lbat if M, + Ny ¢ 0 throughout R, lben none of the streamlines in R is closed In other words, no particle of fluid ever has a closed trajectory in R. The criterion M, + Ny ¢ 0 is called Bendb:son's criterion for the nonexistence of closed
trajectories. 41. Estshlish Equation (1) to f"mish lbe proof of lbe special case of for all closed curves C to which Green's Theorem applies. (The converse is also true: Ifthe line integral is always zero, !ben f satisfies lbe Laplace equation.)
38. Mnjrnjzing work Among all smooth, simple closed curves in the plane, oriented counterclockwise, find the one along which lbe work done by
Green's Theorem. 42. Curl component of conservative fields
Can anytblog be said
about the curl component of a conservative two-dimensional vector field? Give reasons for your answer.
COMPUTER EXPLORATIONS In Exercises 43-46, use a CAS and Green's Theorem to f"md lbe counterclockwise circulation of the field F around the simple closed curve C. Perform lbe following CAS steps.
a. Plot C in lbe xy-p1llne. is greatest. (Hint: Where is (curl F) • k positive?) 39. Regions with many holes Green's Theorem holds for a region R wilb any f"mite number of holes as long as lbe bounding curves
are smooth, simple, and closed and we integrate over each component oflbe boundary 10 lbe direction that keeps R on our immediate left as we go along (see accompanyiog figure).
b. Determine lbe integrand (aN/ax) - (aM/ay) forlbe curl form of Green's Theorem.
c. Deterntine lbe (double integral) limits of integration from your plot in part (a) and evaluate lbe curl integral for lbe circulation. 43. F = (2x - y)i
+ (x + 3y)j,
44. F = (2x' - y')i
45. F = x-1e"i
C: The ellipse x 2
+ (x' + y')j,
+ 4y 2
2 C: The ellipse ~
= 4
y2
+9
= 1
+ (e"lnx + 2x)j,
C: The boundary of lbe region def"med by y = 1 andy ~ 2 (ahove) 46. F ~ xeYi + (4x 2 iny)j,
+ x' (below)
C: The triangle wilb vertices (0, 0), (2, 0), and (0, 4)
16.5
I
16.5
Surfaces and Area
943
Surfaces and Area We have dermed curves in the plane in three different ways:
II _
•
constant
_ _+,::-::;_--"11 _ constant
(II, v)
~
fuplicitform:
y
j(x)
Implicitform:
F(x,y)
Parame1ric vector form:
r(t) ~ j(t~ + g(t)J,
~
0
a" t "
b.
We have analogous def'mitions of surfaces in space:
R
;;1---+----. o
~
Explicit form:
z
Implicit form:
F(x,y,z) -
j(x,y)
o.
There is also a parametric form for surfaces that gives the position of a point on the surface as a vector function of two variables. We discuss this new form in this section and apply the form to obtain the area of a surface as a double integral. Double integral formulas for areas of surfaces given in implicit and explicit forms are then obtained as special cases of the more general parametric formula. "-
Curve II %
constaDt
_
r(lI,lI) .. 1(11, 11)1 + g(lI, v)J + 11(11,11)11:, Position vector to surface point
~y FIGURE 16.37
A parametrized swfacc S
~SllCd as a vector function of two variables defmed on a region R.
Parametrizatlons of Surfaces Suppose
r(a,v) - j(a,v)1
EXAMPLE 1
y = g(u, v),
z = h(u, v).
Find a parametrization ofthe cone
z = Yx 2
I
I-W+]2
-,
(1)
is a continuous vector function that is dermed on a region R in the uv-plane and one-toone on the interior of R (Figure 16.37). We call the range of r the surface S defined or traced by r. Equation (1) together with the domain R constitute a parametrization of the surface. The variables u and v are the parameters, and R is the parameter domain. To simplify our discussion, we ta1re R to be a rectangle def'med by inequalities of the form a SUS b, C S v S d. The requirement that r be one-to-one on the interior of R ensures that S does not cross itself. Notice that Equation (1) is the vector equivalent of three parametric equations: x = /(u, v),
Cone:
+ g(a,v)J + h(a,v)k
+ y2,
0
S Z S
1.
Solution Here, cylindrical coordinates provide a parametrization. A ~t (x,y, z) on the cone (Figure 16.38) has x = rcos6,y = rsin6, and z = Yx 2 + y2 = r, with S r S I and 0 S 6 S 2v. Thking u = r and v = 6 in Equation (1) gives the parametrization
o
fJi'x:::1
r(r,6) - (r 00!l + (r sin fJJ + rll:
.
,
r(r, 6)
I
,,
"
~
(, ces 6)1
+ (r sin 6)J + rk,
o :S r :S
1,
O:S 8
:S
21T.
The parametrization is one-to-one on the interior of the domain R, though not on the boundary tip ofits cone where r = O. •
I
y
FIGURE lfi.38 The cone in Example I can be parametrized using cylindrical coordinates.
EXAMPLE 2
Find a parametrization ofthe sphere x 2
+ y2 + z2
= a 2.
Solution Spherical coordinates provide what we need. A typical point (x, y, z) on the sphere (Figure 16.39) has x=asin~cos8, y=asin~sin8, and z=acos~,
944
Chapter 16: Integration in Vector Fields
o ~ t/J ~ 7T, 0 ~ () ~ 27T. Taking U = t/J 8IId v
= () in Equation (1) gives the parame-
1rization r(,9) - (a sin eo, 9)i
+ (asin,in9)J + (ao..
Again, the parametrization is one-to-one on the interior of the domain R, though not on its
boundary "poles" where t/J EXAMPLE 3
- 0 or t/J -
+ (y -
Cylindcr: x Z + (y - 3f = 9
x 2 + (y 2
r
6y
-
r=6sinB,
+ 9) - 9
}f;2+ y l=r1, y=rllinO
6rsinO = 0
o :s B :s 7T.
A typical point on the cylinder therefore has
x - rcosB - 6sinBcosB - 3sin2B Y = rsinB = 6sin2 B
r(6,z)
*= ""---
r~(r·,·~ z " , ,: ........ (311i:l1.26,6sin O,zJ
,
FIGURE 18.40 The cylinder in Example 3 can be parametrized using cylindrical coordinates.
5.
or
r=6ain6
\
Z :S
Solution In cylindrical coordinates, a point (x, Y, z) has x - rcosB,Y - rsinB, 8IId z - z. For points on the cylinder x 2 + (y - 3)2 - 9 (Figure 16.40), the equation is the same as the polar equation for the cylinder's base in the xy-plane:
oc
,, ,
0~
3)2 = 9,
r2 z
•
Find a parametrization ofthe cylinder x2
FIGURE 18.39 The sphere in Example 2 can be parametrized using gpherica1 coordinates.
7T.
z = z.
TakiDg u - B and v - z in Equation (I) gives the one-to-one parametrization
r(B,z)-(3sin28)i+(6sinZB)J+zk,
O~B:S7T,
O:Sz:S5.
•
Surface Area Our goal is to fwd a double integral for calculating the area of a curved surface S based on the parametrization
r(u, v) = f(u, v)i
+ g(u, v)j + h(u, v)k,
a
~
u
~
b,
c
~
v
~
d.
We need S to be smooth for the construction we are about to carry out The def'tnition of smoothness involves the partial derivatives of r with respect to II and v:
ar
af
ag
ah
af
ag
ah
r ll
-
au - au i + au J + auk
r"
=
au
iIr =
au i + avJ + auk.
DEFlNmDN AJlll"IDetrized_r(u,v) ~ f(u,v)i + g(u,v)j H(u,v)k is smooth if r II and r" are continuous and r II X r" is never zero on the interior of the parameter domain. The condition that r" X r" is never the zero vector in the defmition of smoothness means that the two vectors r v and r v are nonzero and never lie along the same line, so they always determine a plane tangent to the surface. We relax this condition on the boundary of the domain, but this does not affect the area computations.
16.5
945
Surfaces and Area
Now consider a small rectangle J1A. lIU inR with sides on the lines u - "0," - Uo + .6.u, v - Vo, and v - 110 + &v (Figure 16.41). Each side of dA rw maps to a curve on the surface S, and together these four curves bound a "curved patch element" dUIIV " In the notation of the figure, the side v - Vo maps to curve Ct. the side u - Uo maps to C2, and their common vertex (uo. vol maps to Po.
Parametrization
v
~
d
va + Av
"", 0
~~~~~
----R
•
"
f
Uo+AIl'
b
,
x
y
'.
FIGURE 16.42 A magnified view ofa surface patch element 40'wu.
FIGURE 11.41 A rectangular area element l1A..., in the uv-plane maps onto a curved
patch element .&0'In' on S.
Figure 16.42 shows an enlmged view of .6.0'111/' The partial derivative vector rll(uo, vol is tangent to Cl at Po. Likewise, ru(UO, vol is tangent to C2 atPo. The cross product r u X r" is normal to the surface at Po- (Here is where we begin to use the assmnption that S is smooth. We want to be sure that r 1/ X r 1/ '# 0.) We next approximate the surface patch element b.uvv by the parallelogram on the tangent plane whose sides are determined by the vectors .6.urll and bovr" (Figure 16.43). The area oftbis parallelogram is
l.6.urv X .6.vrvl - Irll
x
I"", A'C' y
FIGURE Ui.43 The area of the parallelogram determined by the vector!
X
r"l.6.u avo
(2)
A partition of the region R in the llV-p1mJ.e by rectangular regions ':u1II/ induces a partition of the surface S into surface patch elements 40'111/" We defme the area of each s1D'face patch element 40'1IV to be the pamllelogram area in Equation (2) and sum these areas together to obtain an approximation of the surface area of S: (')
As .4" and .4v approach zero independently, the number of area elements PI tends to 00 and the continuity of r ll and rIO guarantees that the sum in Equation (3) approaches the double integral !cd );,b Irll X rIO I du dv. This double integral over the regionR def'mes the area of the surface S.
Aur. and .&vr"is def"med to be the area of the surface patch element Au.....
DEFlNmON «u, vJ
~
The area of the smooth surface f(u, v~
+ g(u, v)i + h(u, v)"
a~u~b.
c~v~d
is A
~ JJlr. X r.1 dA ~ R
1'1' Ir,
X
r.1
dudv.
(4)
946
Chapter 16: Integration in Vector Fields
We can abbreviate the integral in Equation (4) by writing du for Ir. X rvl du dv. The surface area differential du is analogous to the arc length differential tis in Section 13.3.
Surface Area Differential for a Parametrized Surface
11
du = Ir. X rvl dudv
(5)
du
s
Surl'ace area differential
EXAMPLE 4
Differential fonnula for surface area
Find the surface area of the cone in Example I (Figure 16.38).
In Example I, we found the parametrization
Solution
r(r, 0)
=
(r cos O)i
+
(r sin O)j
+ rk,
0 ,,;; r ,,;; I,
0,,;; 0 ,,;; 2"..
To apply Equation (4), we first fmd r, X r.:
r, X r9
=
i cos 8
-rsinO =
Thus, Ir,
X
r.1
1"1' 1"1'
o
-(rcosO)i - (rsinO)j
Yr 2 cos2 0
=
k I
j sinO rcosO
+ r 2 sin2 0 + r 2
+ (rcos2 0 + rsin2 0)k. ,
Ir, X r.1 drdO
2
=
EXAMPLE 5 Solution
Eq.(4)withu
1 '(j 2
v'2r drdO
=
"
W
=
2
A =
dO =
,
,
v'2r.Theareaoftheconeis
=
~ r,v ~.
'(j (2".)
=
".v'2 units squared.
Find the surface area ofa sphere ofradius a.
We use the parametrization from Example 2: r(, 0)
=
(asincosO)i + (asinsinO)j o ,,;; ";;,,., 0,,;; 0 ,,;; 2"..
+ (a cos
For r", X r., we get
r", X r.
i
j
k
=
a cos cos 0 -a sin sin 0
a cos sin 0 a sin cos 0
-a sin 0
=
(a 2 sin2 cosO)i
+
(a 2 sin2 sinO)j
+ (a 2 sin cos
Thus, Ir", X r.1
=
Ya 4 sin4 cos2 0
+ a 4 sin4 sin2 0 + a4 sin2 cos2
=
Ya 4 sin4
+ a4 sin2 cos2
=
a 2 y sin2
=
a 2 sin ,
=
Ya 4 sin2 (sin2
+ cos2
•
16.5
since sin cP
~
11 1 I 1 27r
7r
2
a sin 4J d4J dO
27r
27r
2
2a 2 dO = 41Ta 2
dO =
cos 4J
units squared.
•
EXAMPLE 6 Let S be the "football" surface formed by rotating the curve x == cos z, y == 0, -7T/2 :::; z :::; 7T/2 around the z-axis (see Figure 16.44). Find a parametrization for S and compute its surface area.
I
/
r = cos z is the radius of a circle
~ ~, at height z ~ -T~ (x, y, z)
..... (J
--.. .
.... .... x
[ -a
This agrees with the well-known formula for the surface area of a sphere.
z
: :
947
0 for 0 :::; cP :::; 7T. Therefore, the area of the sphere is
A =
=
Surfaces and Area
y
SoLution Example 2 suggests finding a parametrization of S based on its rotation around the z-axis. If we rotate a point (x, 0, z) on the curve x == cos z, y == 0 about the z-axis, we obtain a circle at height z above the xy-plane that is centered on the z-axis and has radius r == cosz (see Figure 16.44). The point sweeps out the circle through an angle of rotation 8, 0 :::; 8 :::; 27T. We let (x, y, z) be an arbitrary point on this circle, and define the parameters u == z and v == 8. Then we have x == r cos 8 == cos u cos v, y == r sin 8 == cos u sin v, and z == u giving a parametrization for S as
r (u, v )
==
cos u cos V
"+ cos u. SIn v J"+
uk,
I
_7T
0 :::; v:::; 27T.
Next we use Equation (5) to find the surface area of S. Differentiation of the parametrization gives FIGURE 16.44 The "football" surface in Example 6 obtained by rotating the curve x = cos z about the z-axis.
ru
sin u cos v i-sin u sin v j
== -
+
k
and rv
==
-cosusinvi
+
cosucosvj
Computing the cross product we have
ru X rv
==
- SIn u cos v - cos u SIn v
j
k
- sIn u sIn v
1 0
cos u cos v
-cos u cos v i-cos u sin v j - (sin u cos u cos2 V
+
cos u sin u sin2 v)k.
Taking the magnitude of the cross product gives
Ir u
X
r v I ==
==
Y cos2 U (cos2 V + sin2 v) + sin2 u cos2 u y cos2 u (1 + sin2 u) cos u
Yl
+
sin2 u.
cos u ~ 0 for _J! 2 -< u -< J! 2
From Equation (4) the surface area is given by the integral 27T
A
==
1 o
17T/2 -7T/2
cos u
Y1 + sin2 u du dv.
948
Chapter 16: Integration in Vector Fields
To evaluate the integml, we substitute w = sin u and dw = cos u du, - 1 ~ w ~ 1. Since the smface S is symmetric across the xy-plane, we need only integrate with respect to w from 0 to 1, and multiply the result by 2. In summary, we have A- 2
Jof" Jot VI+w2 dwdv
21 [;VI +w2+t1n(w+ VI ~ f 2[t Vi +tln(l +Vi)]dV. 2w
=
~
+w2
)]:av •
2"'[Vi + In(l + Vi)].
Implidt Surfaces Surfaces are often presented as level sets of a function, described by an equation such as F(x, y, z)
Surface F~, y, .t) '"' c
p
l The vatical projection or "Ilhadow" of S on a coordinate plane
FIGURE 18.4!1 At. we soon see, the area of a surface S in space can be calculated by evaluating a relatM double integral over the vertical projection or "shadow" of S on a coordinate plane. The unit vector p is nonna1 to the plane.
=
c,
for some constant c. Such a level surface does not come with an explicit parametrization, and is called an implicitly defined surface. Implicit surfaces arise, for example, as equipotential surfaces in electric or gravitational fields. Figure 16.45 shows a piece of such a surface. It may be difficult to fmd Cllplicit formulas for the functions j, g, and h that describe the surface in the form r{u, v) = j(u, v)1 + g(u, v)J + h(u, v)k. We now show how to compute the surface area differential du for implicit surfaces. Figure 16.45 shows a piece ofan implicit surface S that lies above its "shadow" region R in the plane beneath it. The surface is defined by the equation F(x,y, z) = c and p is a unit vector normal to the plane region R. We assume that the surface is smooth (F is differentiable and VF is nonzero and continuous on S) and that VF· P =F 0, so the surface never folds back over itself Assume that the normal. vector p is the unit vector k, 80 the region R in Figure 16.45 lies in the xy-plane. By assumption, we then have VF· P = VF· k = Fz =I- 0 on S. An advanced calculus theorem. called the Implicit Function Theorem implies that S is then the graph of a differentiable function z = h(x, y), although the function h(;K, y) is not Cllplicitly known. Define the parameters u and v by u = x and v = y. Then z = h(u, v) and r(u, v) = ui
+ vj + h(u, v)k
(6)
gives a parametrization ofthe surface S. We use Equation (4) to f"md the area of S. Calculating the partial derivatives of r, we f"md ru - I
ah + auk
ru - j
ah + avk.
Applying the Chain Rule for implicit differentiation (see Equation (2) in Section 14.4) to F(x,y, z) = c, where x = u,y = v, andz = h(u, v), we obtain the partial derivatives
ah
au
F, ~
F,
and
ah
au
F,
F;
~
Substitution of these derivatives into the derivatives of r gives and
.
F,
r u = J - F k. z
16.5
Surfaces and Area
949
From a routine calculation of the cross product we find
=
1 F,. (F;rl
=
VF =
+ Fyj +
F.k)
VF
F.
VF-k
VF - VF·p·
p->
Therefore, the surface area differential is given by IVFI
du ~ Ir. X r.1 dudv ~ IVF'pldxdy.
w-"andv·,.
We obtain similar calculations if instead the vector p = j is normal to the xz-plane when F y =# 0 on S, or if P = i is normal to the yz-plane when F x =I- 0 on S. Combining these results with Equation (4) then gives the following general formula.
Formala for the SlIl1'aeeArea of an ImpUclt Surface = c over a closed and bounded plane region R is
The area ofthe surface.F(x,y,z)
Surface area ~
rr IVF'pl IVFI dA,
ff
(7)
R
wherep - i,j,orkisnormaltoRand VF'p =I- O.
Thus, the area is the double integral over R of the magnitude of VF divided by the magnitude of the scalar component of VF normal to R. We reached Equation (7) under the assumption that VF- P =j:. 0 throughoutR and that VF is continuous. Whenever the integral exists, however, we derme its value to be the area of the portion of the surface F(x,y,z) = c that lies over R. (Recall that the projection is asswned to be one-to-one.)
, EXAMPLE 7 x2
+ y2
-
Z
Find the area of the surface cut from the bottom. of the pamboloid = Obytheplanez = 4.
4
Solution We sketch the surface S and the region R below it in the xy-p1ane (Figure 16.46). The surface S is part of the level surface F(x,Y, z) - x 2 + y 2 - Z - 0, and R is the disk. x 2 + y2 :s; 4 in the xy-plane. To get a unit vector normal to the plane of R, we can take p - k. At any point (x,y, z) on the surface, we have F(x,y,z) = x 2 + y2 - z
VF-2x1+2yJ-k IVFI ~ "11(2<)' + (2y)' + ( 1)' FIGURE 1lUlI The area ofthis parabolic surface is calculated in Example 7.
=
V4x 2
+ 4y 2 +
1
IVF'pl ~ IVF·kl ~ 1-11 ~ 1.
950
Chapter 16: Integration in Vector Fields In the region R, dA = dx dy. Therefore,
ff
Surface area =
IVFI IVF' P I dA
11 R
11
2 V4x + 4y 2 + I dxdy
r+;"4 f2" f'
=
10 10
=
f2"
Eq. (7)
2 V4r + IrdrdlJ
[l
Polar coordinates
(4r2 + 1)3/2]2 dlJ
10 12 0 f''' = 10 1~(173/2_I)dlJ=
~(I7V17-I).
•
Example 7 illustrates how to find the surface area for a function z = I(x, y) over a region R in the xy-plane. Actually, the surface area differential can be obtained in two ways, and we show this in the next example.
EXAMPLE 8 Derive the surface area differential du of the surface z = I(x, y) over a region R in the ~-plane (a) parametrically using Equation (5), and (b) implicitly, as in Equation (7). Solution (a) We parametrize the surface by tsking x = u, y = v, and z = I(x, y) over R. This gives the parametrization r(u, v) = ui + vj + I(u, v)k Computing the partial derivatives gives r.
r.
X
rv
=
+ I + I. k, r v
-1.1 - lui + k.
= j
I I
J
k
0
o
I
f. f.
+ Iv k and
Thenlr. X rvldudv = V/. 2 + Iv 2 + I dudv. Substitutingforuandvthengives the surface area differential
du
=
VI.' + 1/ + I dxdy.
(b) We define the implicit function F(x, y, z) = I(x, y) - z. Since (x, y) belongs to the regionR, theunitnonnal to the plane ofR is p = k. Then VF = Ix I + Iyj - k so that
IVF'pl = I-II = I,IVFI = V/x 2 + face area differential is again given by
du
=
1/ + l,andIVFI/IVF'pl = IVF~ Thesur-
VI.' + 1/ + I dxdy.
•
The surface area differential derived in Example 8 gives the following fonnula for calculating the surface area ofthe graph of a function deIfied explicitly as z = I(x, y). Formula for the SurfaeeArea oca Graph z = I(x, y) For a graph z = I(x, y) over a region R in the ~-plane, the surface area fonnula is
A
=
11 R
VI,' +
f/ + Idxdy.
(8)
16.5
951
Surfaces and Area
Exercises 16.5 Finding Parametrizations In Exercises 1-16, f"md a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.) 1. The paraboloid z = x 2
+ y2, Z
2. Theparaboloidz = 9 - x 2
-
~ 4
y2,z ~ 0
3. Cone frustum The first-octant portion of the cone z Y x' + y'/2 between the planes z = 0 and z = 3
=
4. Cone frnstnm The portion of the cone z = 2Yx' + y' between the planes z ~ 2 and z ~ 4 5. Spherical cap Ve cap cut from the sphere bytheconez= x 2 +y2
x' + y' + z' =
9
x'
6. Spherical cap The portion of the sphere + y' + z' = 4 in the fIrst octant between the xy-plane and the conez = Yx' + y'
x'
7. Spherical band The portion of the sphere + y' + z' ~ 3 between the planes z = v'3/2 and z = - v'3/2 8. Spherical cap The upper portion cut from the sphere + y' + z' = 8 by the plane z = -2
x'
9. Parabolic cylinder between planes The surface cut from the parabolic cylinder z ~ 4 - y' by the planes x ~ 0, x ~ 2, and z=O 10. Parabolic cylinder between planes The surface cut from the parabolic cylinder y ~ by the planes z ~ 0, z ~ 3, and y ~ 2
x'
11. CIrcular cylinder baud The portion ofthe cylinder y' + z' ~ 9 between the planes x = 0 and x = 3
x' + z' =
12. CIrcular cylinder baud The portion ofthe cylinder above the xy-plane between the planes y ~ - 2 and y
4
x'
x'
x'
x' -
by the cone z = Yx' + y' 24. Parabolie baod The portion of the paraboloid z = between the planes z ~ I and z ~ 4
x' + y'
25. Sawed-off sphere The lower portion cut from the sphere + y' + z' = 2 by the cone z = Yx' + y'
x'
26. Spherical baud The portion of the sphere between the planes z ~ -I and z ~ v'3
x' + y' +
z' = 4
Planes Tangent to Parametrized Surfaces The tangent plane at a point Po(f("", vol, g(uo, vol, h(uo, vo)) on a parametrized surface r(u, v) ~ f(u, v)i + g(u, v)j + h(u, v)k is the plane througbPo normal to the vectorr.(uo, vol X r.(uo, vol, the cross product of the tangent vectors r.(uo, vol and r.(uo, vol at Po. In Exercises 27-30, f"md an equation fur !he plane tangent to !he surface at Po. Then f"md a Cartesian equation fur the surface and sketch the surface and tangent plane together.
+
The portion of the plane
28. Hemisphere The hemisphere surface r(, 8) ~ (4 sin cos 8)i + (4sinsin8)j + (4 cos '" 1T/2,0 '" 8 '" 21T, at the point Pot 2v'3) corresponding to (,8) = (1T/6,1T/4)
2
a. Inside the cylinder x' + y' ~ 9 b. Inside the cylinder y' + z' = 9
v'2, v'2,
v'2, v'2,
29. CIrcular cylinder The circular cylinder r(8, z) = (J sin 28)i + (6sin'8)j +zk,O '" 8 '" 1T, at the point Po(3Y3/2,9/2,0) corresponding to (8, z) = (1T/3, 0) (See Example 3.)
x-y+2z=2 a. Inside the cylinder x' + z' = 3 b. Inside the cylinder y' + z' = 2
15. Circular cylinder baud The portion of the cylinder (x - 2)' + z' = 4 between the planes y = 0 and y = 3 16. Circular cylinder baud The portion of the cylinder y' (z - 5)' ~ 25 between the planes x ~ 0 and x ~ 10
+
Surface Area of Parametrized Surfaces In Exercises 17-26, use a parametrization to espress the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the sarne values, however.) 17. TIlted plane inside eylinder The portion of the plane y + 2z = 2 inside the cylinder + y' = I
x'
x'
27. Cone The cone r(r, 8) = (r cos 8)i + (rsin 8)j + rk, r '" 0, o'" 8 '" 21T at !he point Pot 2) corresponding to (r,8) = (2,1T/4)
~
13. Tilted plane inside cylinder The portion of the plane x + y z= I
14. TIlted plane inside cylinder
18. Plane inside cylinder The portion of the plane z ~ -x inside the cylinder + y' = 4 + y' 19. Cone frustum The portion of the cone z = 2 Y between the planes z = 2 and z = 6 20. Cone frnstum The portion of the cone z = Yx' + y'/3 between the planes z = I and z = 4/3 21. CIrcular cylinder baud The portion of the cylinder + y' ~ I between the planes z = I and z = 4 + z' = 22. CIrcular cylinder baud The portion of the cylinder 10 between the planes y = -I and y = I y' 23. Parabolic cap The csp cut from the parsholoid z = 2 -
30. Parabolic cylinder The parsholic cylinder surface r(x, y) = xi + yj - x 2k, -00 < X < 00, -00 < y < 00, at the point P o(i,2,-I)correspondingto(x,y) = (1,2)
More Parametrlzatlons of Surfaces 31. a. A torus ofrevolution (doughnut) is obtained by rotating a circle e in the xz-plane about the z-axis in space. (See the accompanying fIgure.) If e has radius r > 0 and center (R, 0, 0), show that a parametrization of the torus is r(u,v) = «R + rcosu)cosv)i
+ «R where 0 fIgure.
S
+ rcosu)sinv)j + (rsinu)k,
u :s 2'1T and 0 :s v :s 21T are the angles in the
952
Chapter 16: Integration in Vector Fields
b. Show that the surface area ofthe torusis.A = ~Rr.
34. Hyperboloid of one meet L
c
c.;;\----+--f-J.l---
Find a parametrization for the hyperboloid ofone sheet ;t'l + y2 - z'l _ I in terms of the angle II associated with the circle;t'l + y2 _ ,2 and the hyperbolic parameter u associated with the hypeIbolic function,2 - z2 - 1. (Hint. cosh 2 u - sinh 2 u -= I.)
b. Generalize the result in part (a) to the hyperboloid (xl/a'l) + (y'l/b'l) - (z2/ c'l) _ I.
35. (Contbwation ofExercise 34.) Find a Cartesian equation for the plane tangent to the hyperboloid;t'l + y2 - z'2 - 2S at the point (::to,Yo, 0), where;to2 + yo'l - 25. 36. Hyperboloid of two IheN Find a parametrization of the hyperboloid of two sheets (z2/c'l) - (;t'l/a 2) - (y2/b 2) - I.
• •
.j
•
, 32. Parametrization of a mrfaee of revolutioD Suppose that the parametrized curve C: (f(u), g(u» is revolved about the ;t-axis, whereg(u) > ofora :!;'; u ~ b. .. Showthat r(u, v) - f(u)i
+ (g(U)C08V)j + (g(u)sinv)k
is a paranu:trizatio of tlu: :resulting surface of revolution, where 0 :!;'; v :!;'; 2'1T is 1he angle from thexy-plane to the point r(1I, v) on the sur6we. (See the accompanying figure.) Notice that f(u):measures distance aJongthe axis ofrevolution and g(u) measures distancefrom the axis ofrevolution.
,
(/(11), gM, 0)
r(II. v)
/ c
\
"-feu)
g(u)
.......-:::
b. Find a parametrization for the surface obtained by revolving the curve:x - y2,y ~ 0, about the;t-axis.
33. .. Parametriation of an el6p1Oid Recall the parametrization ;t - a C06 lI,y - bsinll,O:!;'; lI:!;'; 2'11'frrtheellipse(;t2/a 'l) + (y2/b 2) -= I (Section 3.9, Example S). Using the angles II and 4J in spherical coordinates, shaw that r(lI,4J) - (acos8cos4J)i + (bsinllc064J)j + (csin4J)k isaparametrization oftheeUipsoid (xl/a'l) (z'l/c 2) = 1.
+ (y2/b 2) +
b. Write an integral for the surface area ofthe ellipsoid, but do not evaluate the integral.
Surface Ani for Implktt and ExpUdt Fonns 37. Find the area of the surl'ace cut from. the paraboloidx 2 + y'l - Z = Obytlu: planez = 2. 38. Find the area of the band cut from. the paraboloid x'l + y2 - Z = Obytlu: planes z = 2andz = 6. 39. Find the area of tlu: region cut from the plane x + 2y + 2z - 5 by the cylinder whose wallure x = y'landx = 2 - y~ 40. Find the area of the portion of the surface ;t'l - 2z - 0 that lies abave the 1riangle bounded by the lines ;t -= V3,y - 0, and y = x in the xy-plane. 41. Find the area of the surface x'l - 2y - 2z - 0 that lies above the tria:ngl.e bounded by the lines x -= 2, y - 0, and y - 3:x" in the xyplane. 41. Find the areaoftlu: cap cut from the spbere;t'l + y'l + z'l _ 2by the canez -= y:x'l + y2. 43. Find the area of1he ellipse cut from the plane z - cr (c a constant) by the cylinder x 2 + y'l - 1. 44. Find the area of the upper portion of the cylinder x 2 + z'l = I that lies between the planes x -= ±1/2 andy - ±1/2. 45. Find the area of the portion of the paraboloid :x _ 4 - y2 - z2 that lies above the ring I :!;'; y2 + z2 :!;'; 4 in 1he yz-plane. 46. Findtheareaofthesurfacecutfumtheparaboloidx2 + y + z2_2 by the plane y - O. 47. Find the area of the surface x 2 - 2Inx + z - 0 abave tlu: square R: I :!;'; x :!;'; 2, 0 :!"; Y :!;'; I, in the xy-plane. 48. Find the area of the surface 2:x 3/2 + 2y3/2 - 3z - 0 above the square R: 0 :!"; :x :!"; I, 0 :!;'; Y :!"; I, in the xy-plane.
Wy -
Find the area ofthe SUIfaces in Exercises 49--54. 49. The surface cut from. the bottom. of the paraboloid z _ ;t2 + y2 by tlu: plane z - 3 SO. The surface cut from the ''nose'' of 1he paraboloid ;t - I y2 _ z'2 by the yz-plane
51. The portion of the cane z - V;t2 + y'l that lies over the region between 1he circle x'l + y'l _ I and 1he ellipse 9:%"2 + 4y 'l - 36 in the xy-plane. (Hint: Use formulas from geometry to fmd the area ofthe region.) 51. The tria:ngl.e cut from tlu: plane 2:x + 6y + 3z - 6 by the bounding planes of the fmt octant. Calculate the area three ways, using di:ffurent explicit forms. 53. The surface in the fust octant cut from the cylinder y - (2/3)z3/2 by tlu: planes x - I and y -= 16/3
16.6 Surface Integrals
953
,
54. The portion ofthe plane Y + z = 4 thatli.es above the region aut from the fll8t quadrant of the xz-plane by the parabola x=4-z 2 55. Use the parametrization
r(x, z) = xi
+
f(x, z)j
+ zk
and Equation (5) to derive a furmula fur du associated with the explicit form y = f(x, z). 56. Let S be the surface obtained by rotating the smooth cmrve y = fC!'), a :s x :s b, about the x-axis, when: f(x) 2: O. L
Shaw that the vector function
r(x,9) =xi
+ f(x)cos8j + f(x)sin8k
b. Use Equation (4) to shaw that the surface area oftbis surface ofrevolution is given by
is a parametrization of S, when: 8 is the angle of rotation around the x-axis (see the accompanying figure).
16.6
A
~
I.'
2,,!(x)v'1
+
[f'(x)fd<.
I_S_u_rfa_C_e_ln_te--'9'--ra_ls
_
To compute quantities such as the flow ofliquid across a curved membrane or the upward furce on a falling parachute, we need to integrate a function over a curved surface in spa.ce. This concept of a surface integral is an extension of the idea of aline integml for integrating over a curve.
Surfa.. Integrals
r(u,v) = f(u,v)i
+ g(u,v)j + h(u,v)k,
(u,v)eR.
In Figure 16.47, we see how a subdivision of R (considered as a rectangle for simplicity) divides the surface S into corresponding curved surfa.ce elements, or patches, of area FIGURE 16.47 The area ofthe patch 4Uk is the area ofthe tangent paral.lelogram determined by the vectors 4u r"and 4v r". The point (XDYb Z}) lies on the surface patch, beneath the paral.lelogram shawn hm:.
tJ.ullU
Rl
Ir" X rvl
dudv.
As we did for the subdivisions when defming double integrals in Section 15.2, we number the surface element patches in some order with their areas given by tJ.UI, tJ.uz, . . " tJ.ull . To form a Riemann sum over S, we choose a point (X},yl., z.t) in the kth patch, multiply the value of the function G at that point by the area tJ.Uh and add together the products:
•
~
G(X,b Yh
z.t} tJ.u,t.
Depending on how we pick (X,b Y,b z,V in the kth patch, we may get different values for this Riemann sum. Then we take the limit as the number of surface patches increases, their areas shrink to zero, and both 4u --+ 0 and tJ.v --+ O. This limit, whenever it exists independent of all choices made, delmes the .urface integral of G over the lurface
80s (1)
954
Chapter 16: Integration in Vector Fields Notice the analogy with the definition of the double integral (Section 15.2) and with the line integral (Section 16.1). IfS is a piecewise smooth surface, and G is continuous over S, then the surface integral defined by Equation (I) can be shown to exist. The formula for evaluating the surface integral depends on the manner in which S is described, parametrically, implicitly or explicitly, as discussed in Section 16.5.
Formulas for a Surface Integral 1. For a smooth surface S defined parametrically as r(u, v) = I(u, v)i + g(u, v)j + h(u, v)k, (u, v) ER, and a continuous function G(x, y, z) defined on S, the surface integral of Gover S is given by the double integral over R,
11 G(x,y,z)drr
=
11
S
G(/(u, v),g(u, v),h(u, v))
Irv X rvl
dudv.
(2)
R
2. For a surface S given implicitly by F(x, y, z) = c, where F is a continuously differentiable function, with S lying above its closed and hounded shadow region R in the coordinate plane beneath it, the surface integral of the continuous function G over S is given by the double integral over R,
JJ
G(x,y,z)drr
S
=
JJ G(x,y,Z)I~;~~1
(3)
dA,
R
where p is a unit vector normal to R and VF' p "!'
o.
3. For a surface S given explicitly as the graph of z = I(x, y), where I is a continuously differentiable function over a region R in the xy-plane, the surface integral of the continuous function G over S is given by the double integral over R,
11
G(x, y,z) drr
S
=
11
G(x,y,f(x,y)) V/.
2
+ I; + I dxdy.
(4)
R
The surface integral in Equation (I) takes on different meanings in different applications. If G has the constant value I, the integral gives the area of S. If G gives the mass density of a thin shell of material modeled by S, the integral gives the mass of the shell. If G gives the charge density of a thin shell, then the integral gives the total charge.
EXAMPLE 1
Integrate G(x, y, z) = x 2 over the cone z =
Vx 2 + y2, 0 :5 z
:5
1.
Solutlon Using Equation (2) and the calculations from Example 4 in Section 16.5, we have Ir, X rei = Vir and
J. WJ. ' (r 2cos2(J)(Vir)drd(J 2
11x2drr= S
.~,cos9
16.6 Surface Integrals
955
Surface integrals behave like other double integrals, the integral of the sum of two functions being the sum oftheir integrals and so on. The domain Additivity Property takes the form
11
Gdu
11
=
s
Gdu +
11
~
11
Gdu + ... +
~
Gdu.
~
When S is partitioned by smooth curves into a finite number of smooth patches with nonoverlapping interiors (i.e., if S is piecewise smooth), then the integral over S is the sum of the integrals over the patches. Thus, the integral of a function over the surface of a cube is the sum of the integrals over the faces of the cube. We integrate over a turtle shell of welded plates by integrating over one plate at a time and adding the results.
EXAMPLE 2 Integrate G(x,y, z) == xyz over the surface of the cube cut from the first octant by the planes x == 1, y == 1, and z == 1 (Figure 16.48).
z
1
Side A
SoLution We integrate xyz over each of the six sides and add the results. Since xyz == 0 on the sides that lie in the coordinate planes, the integral over the surface of the cube reduces to
:/ I I I I I
11
1-_ /0 1 // / x
/
---
1
:::~y SideC
SideB
FIGURE 16.48
11
xyz du =
xyz du
11
+
Side A
Cube surface
xyz du
11
+
SideB
xyz du.
SideC
Side A is the surface f(x, y, z) == z == 1 over the square region Rxy : 0 :::; x :::; 1, y :::; 1, in the xy-plane. For this surface and region,
o :::;
p
==
k,
Vf
==
k,
IVfl
==
IVfl
The cube in Example 2.
IVf·pl
1,
==
Ik·kl
==
1
1
du = IVf.pl dA = Tdxdy = dxdy xyz == xy( 1) == xy
and
Symmetry tells us that the integrals of xyz over sides Band C are also 1/4. Hence,
JJff
1
xyz du
=
1
1
3
•
4 + 4 + 4 = 4·
Cube surface
VI -
EXAMPLE 3
Integrate G (x, y, z) == formed by rotating the curve x == cos z, y
==
x 2 - y2 over the "football" surface S 0, -7r/2 :::; z :::; 7r/2, around the z-axis.
The surface is displayed in Figure 16.44, and in Example 6 of Section 16.5 we found the parametrization
SoLution
X ==
cos u cos v,
y == cos u sin v,
z
==
u,
7r -2:::;
u :::;
27r
and
0:::; v :::;
27r,
where v represents the angle of rotation from the xz-plane about the z-axis. Substituting this parametrization into the expression for G gives
The surface area differential for the parametrization was found to be (Example 6, Section 16.5) dO" == cos u
VI
+ sin2 u du dv.
956
Chapter 16: Integration in Vector FieLds
These calculations give the surface integral
{{ \II -
JJS
x2
-
y 2 du
=
{27T17T/2Isinul cos u
Jo
/ {27T {7T 2
=2
_1Jo 1Jo 2
27T
-
\II + sin2 ududv
-7T/2
o
sin u cos u
I d d VW W v
2 3/2 = 21T· 3 w
+
sin2 u du dv
w = 1 + sin2 u
~
1
VI
dw = 2 sin u c;s u du Whenu=Ow=l. ' When u = Tr/2, w = 2.
]2 = 4; (2Y2 - 1).
•
1
Orientation
FIGURE 16.49 Smooth closed surfaces in space are orientable. The outward unit normal vector defines the positive direction at each point. c
d
a
b
Start
We call a smooth surface S orientable or two-sided if it is possible to define a field n of unit normal vectors on S that varies continuously with position. Any patch or subportion of an orientable surface is orientable. Spheres and other smooth closed surfaces in space (smooth surfaces that enclose solids) are orientable. By convention, we choose n on a closed surface to point outward. Once n has been chosen, we say that we have oriented the surface, and we call the surface together with its normal field an oriented surface. The vector n at any point is called the positive direction at that point (Figure 16.49). The Mobius band in Figure 16.50 is not orientable. No matter where you start to construct a continuous unit normal field (shown as the shaft of a thumbtack in the figure), moving the vector continuously around the surface in the manner shown will return it to the starting point with a direction opposite to the one it had when it started out. The vector at that point cannot point both ways and yet it must if the field is to be continuous. We conclude that no such field exists.
Surface Integral for Flux
FIGURE 16.50 To make a Mobius band, take a rectangular strip of paper abed, give the end be a single twist, and paste the ends of the strip together to match a with e and b with d. The Mobius band is a nonorientable or one-sided surface.
Suppose that F is a continuous vector field defined over an oriented surface S and that n is the chosen unit normal field on the surface. We call the integral of F • n over S the flux of F across S in the positive direction. Thus, the flux is the integral over S of the scalar component of F in the direction of n.
DEFINmON The flux of a three-dimensional vector field F across an oriented surface S in the direction of n is
Flux =
11
F·ndu.
(5)
s
The definition is analogous to the flux of a two-dimensional field F across a plane curve C. In the plane (Section 16.2), the flux is
1
F·nds,
the integral of the scalar component ofF normal to the curve. If F is the velocity field of a three-dimensional fluid flow, the flux of F across S is the net rate at which fluid is crossing S in the chosen positive direction. We discuss such flows in more detail in Section 16.7.
16.6
EXAMPLE 4
z
y
=
x
2
,
0
:5 X
Surface Integrals
957
Find the flux of F = yzi + xj - z 2 k through the parabolic cylinder :5 1,0 :5 z :5 4, in the direction n indicated in Figure 16.51.
On the surface we have x = x, y = x 2 , and z = z, so we automatically have the parametrization rex, z) = xi + x 2j + zk,O :5 x :5 1,0 :5 z :5 4. The cross product of tangent vectors is SoLution
k
1
j 2x
o
0
1
I
n""-
rx X rz , y
Finding the flux through the surface of a parabolic cylinder (Example 4).
o = 2xi -
j.
The unit normal vectors pointing outward from the surface as indicated in Figure 16.51 are rx X rz
n =
FIGURE 16051
=
On the surface, y = x
2
,
IrxXrzl
2xi - j
= -----;====;::=====
Y4x 2
+
l'
so the vector field there is
Thus, F n 0
=
V
((x 2z)(2x)
1 2
4x + 1 2x 3z - X Y4x 2 + 1 .
+
+
(x)( -1)
(-z2)(O))
The flux of F outward through the surface is
r e 'V4x +x1 Irx X rzl dxdz = r e 2x z - x Y4x + 1 dx dz Jo Jo Y4x + 1 3z
{{Fonda- = JJ Jo Jo
s
2
-
3
41 141 1
2
2
1
= =
(2x 3z - x) dxdz =
o
0
14 [1 0
1]4
(z - 1) dz = - (z - 1)2 o 2 4 -
1
1
-x 4z - -x 2 ]X=1 dz 2 2 x=o
0
1
= 4(9) - 4(1) = 2. If S is part of a level surface g(x, y, z) = fields
• C,
then n may be taken to be one of the two
Vg n = ±IVgl'
(6)
depending on which one gives the preferred direction. The corresponding flux is Flux =
11
Fonda-
s
=
{{ ( ±Vg) IVgl JJ F0 1Vg1 IVgopl dA
Eqs. (6) and (3)
R
=
{{ ±Vg JJ FOIVgopl dA. R
(7)
958
Chapter 16: Integration in Vector FieLds
,'+,'~1
I;
Rxjt
EXAMPLE 5 Find the flux of F = yzj + z 2k outward through the surface S cut from the cylinder y2 + z2 = 1, z 2:: 0, by the planes x = and x = 1.
°
SoLution The outward normal field on S (Figure 16.52) may be calculated from the gradient of g(x,y, z) = y2 + z2 to be S
(1, -1, 0)
(1, 1,0)
Vg +IVgl
n =
y
With P
=
=
2yj + 2zk • ~ r: = Yl + zk. 2vl
k, we also have
x
FIGURE 16.52 Calculating the flux ofa vector field outward through the surface S. The area of the shadow region Rry is 2 (Example 5).
2yj + 2zk Y4y 2 + 4z 2
du =
IVgl IVg.kl
2 1 dA = 12z1 dA = zdA.
We can drop the absolute value bars because z The value of F • n on the surface is
2::
°on S.
=
+ z 2k) • (yj + zk) y2z + z3 = z(y2 + z2)
=
z.
F • n = (yzj
The surface projects onto the shadow region Rxy, which is the rectangle in the xy-plane shown in Figure 16.52. Therefore, the flux ofF outward through S is
• Moments and Masses of Thin Shells Thin shells of material like bowls, metal drums, and domes are modeled with surfaces. Their moments and masses are calculated with the formulas in Table 16.3. The derivations are similar to those in Section 6.6. The formulas are like those for line integrals in Table 16.1, Section 16.1. TABLE 16.3 Mass and moment formulas for very thin shells
Mass:
M =
11
8 du
8 = 8(x,y,z) = density at (x,y,z) as mass per unit area
s
First moments about the coordinate planes: Myz =
11
x8du,
M xz =
s
11
y 8 du,
s
Coordinates of center of mass:
Moments of inertia about coordinate axes:
rex, y, z)
=
distance from point (x,y, z) to line L
16.6 Surface Integrals
,
x1+r+:?-=a2
II
(0.0. ~)
"
EXAMPLE 6
Find the center ofmass of a thin hemispherical shell ofradius a and COIl-
""'" density 6.
We model the shell with the hemisphere
Solution
s
R/
~y
a
x?-+r_a2
x
FIGURE 11S.53 Thecentl:rofmassofa thin hemispherical shell ofconstant density lies on the axis of symmetry haI:fway from the base to the ~ (Example 6).
959
f(x,y,z)
=
x2 + y2 +
a 2,
z2 =
z;=:
0
(Figure 16.53). The symmetry ofthe surface about the z-axis tells us thatj mains only to findi from the fonnulai = Mxy/M. The mass ofthe shell is M -
JJs
JJs
6 du - 6
Y=
O. Itre-
du - (6)(are. of S) - 2.-0'6.
To evaluate the integral for MJiY' we take p
=
k and calculate
IV/I ~ 12xi + 2yj + 2m1 ~ IV/'pl
=
2V,'
+ y' + z' ~ 20
IV/'kl ~ IZzI ~ Zz IV/I 0 IV/'pl dA ~ zdA.
~
da ~ Then
JJ z6 du - if z~dA - 8aJJ dA - &(7Ta2) - wa 3 z------M 21Ta 8
M;r;y -
6
S
_
Mxy
R
'1ra 3l) 2
R
a 2'
•
The shell's C"",", afDUlllll is the porn! (0, 0, 0/2),
,
EXAMPLE 7
------,!-. r
cone z
:t-Vr+r
=
Find the center of mass of a thin shell of density 8 Vx 2 + y2 by the planes z = 1 andz = 2 (Figure 16.54).
=
Solution The symmetry of the surface about the z-axis tells us that x = z = Mxy/M. Working as in Example 4 of Section 16.5, we have
1
r(r,6) = (rc088)i
+ (rsine)j + rk.
I
~
r
:S
2,
O:s
and
Ir, X rol - Vir. FIliURE 16.54 The cone frustmn formed 2 + y 2 iscutby whentheconez the planes z - 1 lIIld z - 2 (Example 7).
I/z2 cut from the
Therefon:,
yx:
{"
{"
- Y2Jo [In r];d6 - Y2Jo ~ 2.-Y21n2,
1n2d6
y = O. We :fmd (J
:s 217',
960
Chapter 16: Integration in Vector Fields
Mxy =
ff
Bzdu =
1
2 '1r12 :2rV2rdrd6
S
~ Vi},f'Oj' 1 drdIJ ~ _
Vi},f'O dO ~ 2..Vi, M,.
z~-=
M
2..Vi
1
Vi In 2
=-
1n2'
2..
•
The shell's center ofmass is the point (0, 0, ljln2).
Exercises 16.6 Surface Integrals In Exercises 1--8, integrare the given function over the given surface.
1. Parabolic cy6nder G(x,y,z) = x, over the parabolic cylinder y=x 2,O SX s2,O Szs 3
(0, I, 2)
1
G(x,y,z) = z, aver the cylindrical surface y2 +z2 = 4,z:2!: 0,1 Sx s4 3. Sphere G(x,y,z) = x 2, over the unit sphen:: x 2 + y2 +z2 = 1 4. Hemisphere G(x,y,z) = z2, aver the hemispheIe x 2 + y2 + z2=a 2,z2:0 Z. Cireular cylinder
(0, 1, 1)
,,, , ,,, ,,
5. Portion of plane F(x,y, z) = z, aver the portion of the plane x + y + z = 4 that lies abaw: the square 0 S X S I, o s y S I, in the xy-plane 6. CODe F(x,y, z) = z - x, aver the cone z = Yx 2 + y2,
1(0,1,0) 1
Oszsl
1
yS - 4z, over the parabolic
H(x,y,z) - x 2 domez - 1 - x 2 - y2,z:2!: 0
7. Parabolic dome
y
x
8. SpherieaJ. eap H(x,y,z) - yz, over the part of the sphere x2 + y2 + z2 _ 4 that lies above the cone z _ Yx2 + y2 9. Integrate G(x,y,z) - x + Y + z over the surface ofthe cube cut from the fust octant by the planes x - a, y =- a, z - a.
+ z aver the surface of the wedge in the frrst octant bolDlded by the coordinate planes and the planes
10. lIltqp'ate G(x,y,z) = y
x-2andy+z-1.
(I, 1, 0)
16. IntegrateG(x,y,z) = xavertlu: surfacegivcnby z=x 2 +y for OsxSl, -ISysl. 17. IntegrateG(x,y,z) - xyz awr the triangular surface with vertices (I, 0, 0), (0, 2, 0), om! (0, 1, 1).
,
11. lIltqp'ate G(x,y,z) = X)'% over the surface of the rectangular solid cut from the frrst octant by the planes x - a, y - b, and
(0, 1, 1)
1
z - c. 12. lIltqp'ate G(x,y,z) = X)'% over the surface of the rectangular solid bounded by the planes x - ±a, y - ±b, and z - ±c. 13. lIltqp'ate G(x,y,z) = x + Y + z aver the portion of the plane b + 2y + z = 2 that lies in the fust octant
14. ~ G(x,y, z) = XVy 2 + 4 aver the surface cut from the parabolic cylinder y2 + 4z = 16 by the planes x = 0, x = 1,
).. -
--
""-~.Y
(0, 2, 0)
x
(1,0,0)
andz = O. 15.
~ G(x,y,z) = z - x over the portion of the graph of z = x + y2 above the triangle in the xy-plane having vertices (0. 0, 0), (I, 1,0), om! (0, I, 0). (Soo "",ompanying figure.)
18. Integrate G(x,y,z) =- x - y - z aver tlu: portion of the pllllle x + Y = 1 in the fust octant between z = 0 and z = 1 (sec the accompanying figure).
16.6
Surface Integrals
961
+k + zyj + z 2k xi + yj + zk xi + yj + zk V2 2 x + y2 + z
33. F(x,y,z) = yi - xj
z
34. F(x,y,z) = zxi 35. F(x,y,z)
=
(1,0, 1)
36. F(x,y,z) = -~""'----+y
x
Finding FLux Across a Surface In Exercises 19-28, use a parametrization to find the flux ffsF· n du across the surface in the given direction. 19. Parabolic cylinder F = z 2i + xj - 3zk outward (normal away from the x-axis) through the surface cut from the parabolic cylinder z = 4 - y2 by the planes x = 0, x = 1 , and z = 0 20. Parabolic cylinder F = x 2j - xzk outward (normal away from the yz-p1ane) through the surface cut from the parabolic cylinder y = x 2 , -1 :::; x :::; 1 , by the planes z = 0 and z = 2 21. Sphere F = zk across the portion of the sphere x 2 + y2 + z2 = a 2 in the first octant in the direction away from the origin zk across the sphere x 2
22. Sphere F = xi + yj + in the direction away from the origin
25. Cone F = xyi - zk outward (normal away from the z-axis) through the cone z = Yx 2 + y2, 0 :::; Z :::; 1 26. Cone F = y 2i + xzj - k outward (normal away from the z-axis) through the cone z = 2Yx 2 + y2, 0 :::; Z :::; 2 27. Cone frustum F = -xi - yj + z 2k outward (normal away from the z-axis) through the portion of the cone z = Yx 2 + y2 between the planes z = 1 and z = 2 28. Paraboloid F = 4xi + 4yj + 2k outward (normal away from the z-axis) through the surface cut from the bottom of the paraboloid z = x 2 + y2 by the plane z = 1 In Exercises 29 and 30, find the flux of the field F across the portion of the given surface in the specified direction. -i
+ 2j + 3k
S: rectangular surface z = 0, direction k 30. F(x, y, z) = yx 2i - 2j + xzk S: rectangular surface y = 0, direction - j
0:::; x :::; 2,
0:::; y :::; 3,
=
zk
32. F(x, y, z)
=
-yi
+ xj
z 1--_
R yZ
/
I'
/
"
/ 1~
x
S
~y 1'1'
" 40. Let S be the portion of the cylinder y = In x in the first octant whose projection parallel to the y-axis onto the xz-plane is the rectangle Rxz: 1 :::; x :::; e, 0 :::; z :::; 1. Let n be the unit vector normal to S that points away from the xz-plane. Find the flux of F = 2yj + zk through S in the direction ofn. 41. Find the outward flux of the field F = 2xyi + 2yzj + 2xzk across the surface of the cube cut from the first octant by the planes x = a,y = a,z = a. 42. Find the outward flux of the field F = xzi + yzj + k across the surface of the upper cap cut from the solid sphere x 2 + y2 + z2 :::; 25 by the plane z = 3. Moments and Masses 43. Centroid Find the centroid of the portion of the sphere x 2 + y2 + z2 = a 2 that lies in the first octant. 44. Centroid Find the centroid of the surface cut from the cylinder y2 + z2 = 9, z ;;:0: 0, by the planes x = 0 and x = 3 (resembles the surface in Example 5).
-1:::; x :::; 2,
2:::; z :::; 7,
In Exercises 31-36, find the flux of the field F across the portion of the sphere x 2 + y2 + z2 = a 2 in the first octant in the direction away from the origin.
31. F(x,y,z)
39. Let Sbe the portion of the cylinder y = eX in the first octant that projects parallel to the x-axis onto the rectangle Ryz: 1 :::; y :::; 2, o :::; z :::; 1 in the yz-plane (see the accompanying figure). Let n be the unit vector normal to S that points away from the yz-plane. Find the flux of the field F(x, y, z) = -2i + 2yj + zk across S in the direction of n.
I'
24. Cylinder F = xi + yj + zk outward through the portion ofthe cylinder x 2 + y2 = 1 cut by the planes z = 0 and z = a
=
38. Find the flux of the field F(x,y,z) = 4xi + 4yj + 2k outward (away from the z-axis) through the surface cut from the bottom of the paraboloid z = x 2 + y2 by the plane z = 1.
+ y2 + z2 = a 2
23. Plane F = 2xyi + 2yzj + 2xzk upward across the portion of the plane x + y + z = 2a that lies above the square o :::; x :::; a, 0 :::; y :::; a , in the xy-p1ane
29. F(x,y, z)
37. Find the flux of the field F(x,y, z) = z2j + xj - 3zk outward through the surface cut from the parabolic cylinder z = 4 - y2 by the planes x = O,X = 1, andz = O.
45. Thin shell of constant density Find the center of mass and the moment of inertia about the z-axis of a thin shell of constant density {) cut from the cone x 2 + y2 - z2 = 0 by the planes z = 1 andz = 2. 46. Conical surface of constant density Find the moment of inertia about the z-axis of a thin shell of constant density {) cut from the cone 4x 2 + 4y 2 - z2 = 0, Z ;;:0: 0, by the circular cylinder x 2 + y2 = 2x (see the accompanying figure).
962
Chapter 16: Integration in Vector FieLds 47. Spherical shells
z
a. Find the moment of inertia about a diameter of a thin spherical shell of radius a and constant density 8. (Work with a hemispherical shell and double the result.) b. Use the Parallel Axis Theorem (Exercises 15.6) and the result in part (a) to find the moment of inertia about a line tangent to the shell. 48. Conical Surface Find the centroid of the lateral surface of a solid cone of base radius a and height h (cone surface minus the base). y X2
x
16.7
+ y2 = 2x
or r=2cosO
Stokes' Theorem
-lCurlF
«"(/) FIGURE 16.55 The circulation vector at a point (x, y, z) in a plane in a threedimensional fluid flow. Notice its right-hand relation to the rotating particles in the fluid.
As we saw in Section 16.4, the circulation density or curl component of a two-dimensional field F = Mi + Nj at a point (x, y) is described by the scalar quantity (aN/ax - aM/ay). In three dimensions, circulation is described with a vector. Suppose that F is the velocity field of a fluid flowing in space. Particles near the point (x,y, z) in the fluid tend to rotate around an axis through (x,y, z) that is parallel to a certain vector we are about to define. This vector points in the direction for which the rotation is counterclockwise when viewed looking down onto the plane ofthe circulation from the tip of the arrow representing the vector. This is the direction your right-hand thumb points when your fingers curl around the axis of rotation in the way consistent with the rotating motion of the particles in the fluid (see Figure 16.55). The length ofthe vector measures the rate of rotation. The vector is called the curl vector and for the vector field F = Mi + Nj + Pk it is defined to be curl F =
(ap + (aM _ ap)j + (aN _ aM)k. ay _ aN)i az az ax ax ay
(1)
This information is a consequence of Stokes' Theorem, the generalization to space ofthe circulation-curl form of Green's Theorem and the subject of this section. Notice that (curl F)· k = (aN/ax - aM/ay) is consistent with our definition in Section 16.4 when F = M(x,y)i + N(x,y)j. The formula for curl F in Equation (1) is often written using the symbolic operator
V
.a +.a =lax Jay
+k
a az'
(The symbol V is pronounced "del.") The curl of F is V X F : j
k
a VXF= ax
a a ay az M N P
_ aN)i + (aM _ ap). + (aN _ aM)k ( ap ay az az ax J ax ay =
curl F.
(2)
16.7
Stokes'Theorem
curl F = V X F
EXAMPLE 1 SoLution
Find the curl ofF = (x 2 - z)i
963
(3)
+ xeZj + xyk.
We use Equation (3) and the determinant form, so curlF = V X F j
k
a
a
a
ax 2 x - z
ay
az
xe z
xy
- ~(xeZ))i - (~(xy) - ~(x2 (~(xy) ay az ax az + (~(xeZ) - ~(x2 - Z))k ax ay
z))j
= (x - xeZ)i - (y + l)j + (e Z - O)k =
x(l - eji - (y
+
l)j
+
•
e~
As we will see, the operator V has a number of other applications. For instance, when applied to a scalar function f(x, y, z), it gives the gradient of f:
Vf =
af. ax.
af.
af
+ ayJ + azk.
It is sometimes read as "del f' as well as "grad j."
Stokes' Theorem
s
~-
....------ - n'/
ci
J
FIGURE 16.56 The orientation ofthe bounding curve C gives it a right-handed relation to the normal field n. If the thumb of a right hand points along n, the fingers curl in the direction of C.
Stokes' Theorem generalizes Green's Theorem to three dimensions. The circulation-curl form of Green's Theorem relates the counterclockwise circulation of a vector field around a simple closed curve C in the xy-plane to a double integral over the plane region R enclosed by C. Stokes' Theorem relates the circulation of a vector field around the boundary C of an oriented surface S in space (Figure 16.56) to a surface integral over the surface S. We require that the surface be piecewise smooth, which means that it is a finite union of smooth surfaces joining along smooth curves.
THEOREM 6-Stokes'Theorem
Let S be a piecewise smooth oriented surface having a piecewise smooth boundary curve C. Let F = Mi + Nj + Pk be a vector field whose components have continuous first partial derivatives on an open region containing S. Then the circulation of F around C in the direction counterclockwise with respect to the surface's unit normal vector n equals the integral of V X F· n over S.
f
F·dr =
c
Counterclockwise circulation
11
V X F·nda-
s
Curl integral
(4)
964
Chapter 16: Integration in Vector Fields Notice from Equation (4) that if two different oriented surfaces Sl and S2 have the same boundary C, their curl integrals are equal:
Both curl integrals equal the counterclockwise circulation integral on the left side of Equation (4) as long as the unit normal vectors oland 02 correctly orient the surfaces. If C is a curve in the xy-plane, oriented counterclockwise, and R is the region in the xy-plane bounded by C, then du == dx dy and
(V
x F)· 0
==
(V X F)· k
==
aN ( -ax
-
-aM) •
ay
Under these conditions, Stokes' equation becomes
f
11 (~~ -~~)
Fodr =
C
dxdy,
R
which is the circulation-curl form of the equation in Green's Theorem. Conversely, by reversing these steps we can rewrite the circulation-curl form of Green's Theorem for two-dimensional fields in del notation as
cD
f
Curl
Circulation
Fodr =
C
11
V X FokdA.
(5)
R
See Figure 16.57.
Stokes:
Evaluate Equation (4) for the hemisphere S: x 2 + y2 its bounding circle C: x 2 + y2 == 9, z == 0, and the field F == yi - xj.
EXAMPLE 2
FIGURE 16.57 Comparison of Green's Theorem and Stokes' Theorem.
+ z2
==
9, z ~ 0,
SoLution The hemisphere looks much like the surface in Figure 16.56 with the bounding circle C in the xy-plane (see Figure 16.58). We calculate the counterclockwise circulation around C (as viewed from above) using the parametrization r(O) == (3 cos O)i + (3 sin O)j, 0 :::; 0 :::; 27T:
f
+ (3 cos 0 dO)j
dr
==
(-3 sin 0 dO)i
F
==
F· dr
==
yi - xj == (3 sin O)i - (3 cos O)j -9 sin2 0 dO - 9 cos2 0 dO == -9 dO
F ° dr
=
1
27T
-9 dO
-187T.
=
C
For the curl integral ofF, we have n
VX F
/
==
(ap _ aN)i ay az
==
(0 - O)i
az
-
+ (0 xi + yj + zk Vx 2 + y2 + z2
du
==
3 zdA
V X Fonda
=
-~tdA
y 0-
x
FIGURE 16.58 A hemisphere and a disk, each with boundary C (Examples 2 and 3).
+ (aM _ ap)_ + (aN _ aM)k ax J
+ (-1 - l)k xi + yj + zk -----O)j
-
ax ==
ay
-2k Outer unit normal
3 Section 16.6, Example 6, with a = 3
=
-2dA
16.7 Stokes'Theorem and
II
II
V X F'nder =
s
965
-2dA = -18,..
.r+y2.s9
The circulation around the circle equais the integral of the curl over the hemisphere, as it should. _ The surface integral in Stokes' Theorem can be computed using any surface having boundary curve C, provided the surface is properly oriented and lies within the domain of the field F. The next example illustrates this fact for the circulation around the curve C in Example 2.
EXAMPLE 3 Calculate the circulation around the bounding circle C in Example 2 using the disk ofradius 3 centered at the origin in the xy-plane as the surface S (instead of the hemisphere). See Figure 16.58. Solution Ai; in Example 2, V X F = - 2k. For the surface being the described disk in the xy-plane, we have the normal vector n = k so that V X F'nder = -2k'kdA = -2dA and
II
II
V X F'nder =
s
-2dA = -18,.,
.r+y2.s9
a simpler calculation than before. z
_
Find the circulation of the field F = (x 2 - y)i + 4zj + x 2k around the curve C in which the plane z = 2 meets the cone z = Vx 2 + y2, counterclockwise as viewed from above (Figure 16.59).
EXAMPLE 4
C:X 2 +y2=4, z=2
/
Solution Stokes' Theorem enables us to fmd the circulation by integrating over the surface of the cone. Traversing C in the counterclockwise direction viewed from above corresponds to tskiog the inner normal n to the cone, the normal with a positive k-component. We parametrize the cone as r(r, 0) = (r cos O)i + (r sin O)j + rk,
s: r(t) ~ (r cos 6)1 + (r sin 6)J + rk x
FIGURE 16.59
o :5 r :5 2,
0:5 0 :5 2,..
We then have
"'-. y
n=
The curve C and cone S
in Example 4. =
Vz der
=
rr X r. Irr X
-(rcosO)i - (rsinO)j + rk
r.1
rV2
Section 16.5, Example 4
(-(COSO)i - (sinO)j + k)
rV2 drdO
Section 16.5, Example 4
VXF=-4i-2xj+k
Example 1
= -4i - 2rcosOj + k.
x =
Accordingly, V X F'n =
=
Vz Vz
(4 cosO + 2r cos 0 sinO + 1)
(4 cosO + rsin20 + 1)
roosO
966
Chapter 16: Integration in Vector Fields and the circulation is
f
F' dr =
e
II
V X F' n
du
Stokes' Theorem, Eq. (4)
S =
/.2"/.2 Vz
(4 cos Ii
+ rsin21i + 1)(rV2drdli)
= 41T.
•
EXAMPLE 5 The cone used in Example 4 is not the easiest surface to use for calculating the circulation around the bounding circle Clying in the plane z = 3. If instead we use the flat disk of radius 3 centered on the z-axis and lying in the plane z = 3, then the normal vector to the surface S is n = k. Just as in the computation for Example 4, we still bave V X F = -4i - 2xj + k. However, now we get V X F' n = I, so that
II S
V X F' n du =
II
The shadow is the disk ofradius 2 in the .JY""plane.
I dA = 41T.
x2+y2S4
TIris result agrees with the circulation value found in Example 4.
Paddle Wheel Interpretation of V
•
X F
Suppose that F is the velocity field of a fluid moving in a region R in space containing the closed curve C. Then
f
F·dr
e
is the circulation of the fluid around C. By Stokes' Theorem, the circulation is equal to the flux of V X F through any suitably oriented surface S with boundary C:
f
F·dr
=
II
V X F·ndu.
s
e
Suppose we fIX a point Q in the region R and a direction u at Q. Take C to be a circle ofradius p, with center at Q, whose plane is normal to u. If V X F is continuous at Q, the average value of the u-component of V X F over the circular disk S bounded by C approaches the u-component of V X F at Q as the radius p ---> 0:
(V X F'u)g = lim p-O
-l, rr V X F·udu. trp JJ s
If we apply Stokes' Theorem and replace the surface integral by a line integral over C, we get CurlF _ _
(V X F'u)g = lim
l..2f F·dr.
p-o 7rf'
(6)
e
The left-hand side of Equation (6) has its maximum value when u is the direction of V X F. When p is small, the liruit on the right-hand side of Equation (6) is approximately
-1-fF'dr 7f' 2 ' p
FIGURE 16.60 The paddle wheel interpretation of curl F.
e
which is the circulation around C divided by the area of the disk (circulation density). Suppose that a small paddle wheel of radius p is introduced into the fluid at Q, with its axle directed along u (Figure 16.60). The circulation of the fluid around C affects the rate
16.7
Stokes'Theorem
967
of spin of the paddle wheel. The wheel spins fastest when the circulation integral is maximized; therefore it spins fastest when the axle of the paddle wheel points in the direction of V X F.
EXAMPLE 6 A fluid of constant density rotates around the z-axis with velocity F = w( -yi + xj), where w is a positive constant called the angular velocity ofthe rotation (Figure 16.61). Find V X F and relate it to the circulation density. SoLution
+
With F = -wyi
wxj, we find the curl
P(x,y, z)
V X F
F=w(-yi+xj)
P(x,y, O)
(aM _ ap). az ax J
+
(ap _ aN)i ay az
(aN _ aM)k ax ay
=
(0 - O)i + (0 - O)j + (w - (-w»k
=
2wk.
By Stokes' Theorem, the circulation of F around a circle C ofradius p bounding a disk S in a plane normal to V X F , say the xy-p1ane, is
/
o
+
=
f
e
y
x
F·dr =
11
V X F·odu =
s
11 s
Thus solving this last equation for 2w, we have
FIGURE 16.61 A steady rotational flow parallel to the xy-plane, with constant angular velocity w in the positive (counterclockwise) direction (Example 6).
2wk·kdxdy = (2w)( 1Tp2).
(V X F)· k = 2w =
-l-f 1Tp2
F· dr,
e
•
consistent with Equation (6) whenu = k.
Ie
EXAMPLE 7 Use Stokes' Theorem to evaluate F· dr, ifF = xzi + xyj + 3xzk and C is the boundary of the portion of the plane 2x + y + z = 2 in the first octant, traversed counterclockwise as viewed from above (Figure 16.62). z
SoLution
y
The plane is the level surface I(x, y, z) = 2 of the function I(x, y, z) = 2x
+
+ z. The unit normal vector o =
(0,0,2)
VI IVI I =
(2i 2i
1
+ j + k) 1 (. . ) + j + kI = V6 21 + J + k
I
is consistent with the counterclockwise motion around C. To apply Stokes' Theorem, we find j
k
a ax
a ay
a az
xz
xy
3xz
V X F = (x - 3(2 - 2x - y»j
+ yk
=
(7x
6 + y)
=
~ (7X + 4y -
cur1F = V X F =
x y
FIGURE 16.62 Example 7.
=
(x - 3z)j
+ yk.
On the plane, z equals 2 - 2x - y, so
The planar surface in
+
3y - 6)j
+ yk
and
V X F· 0
=
~ (7X + 3y -
The surface area element is du =
IV/I IV/.kl dA
V6
=
-l- dxdy .
6)'
968
Chapter 16: Integration in Vector Fields The circulation is
f
ff
F'dr =
C
V X Fonda
StW:I''T11ccRm,Eq.(4)
S
11 11
1 2 2x -
=
~ (7X + 4y -
1 2 -'b:(7X
=
+ 4y -
6)V6 dydx
•
6)dydx =-1.
Let the surface S be the ellipitica1 paraboloid z = x 2 + 4y 2lying beneath the plane z = 1 (Figure 16.63). We define the orientation of Sby taking the innernonnal vector n to the surface, which is the normal having a positive k-component. Find the flux of the curl V X F across S in the direction D for the vector field F = yI - xzj + xz2JL
EXAM PLE 8
Solution We use Stokes'Theorem to calculate the curl integral by tindiDg the equivalent counterclockwise circulation of F around the cmve of intersection C of the paraboloid z - x 2 + 4y 2 and the plane z - I, as shown in Figure 16.63. Note that the orientation of S is consistent with tntversiDg C in a counterclockwise direction around the z-axis. The cmve Cis the ellipse x 2 + 4y 2 - 1 in the plane z ... 1. We can parametrize the ellipse by x - cos t,y sin t, z - 1 for 0 :s; t :s; 2v, so C is given by
-1
y
FIGURE 16.63 The portion of the ellipitical paraboloid in Example 8, showing itll curve ofintersection C with the pl.anc z = I and itll inner normal orientation by n.
r(t)
~
(cost)1
t
To compute the circulation integral iF dr, we evaluate F along C and fmd the velocity vectDr dr/ dt: 0
P(r(t» -
t
(sin t)1 - (cost)J
-:;;
~
-(sin ')1
and
•
o :S t :S 21T.
+ (sin tll + k,
Thco,
f
P'dr -
c D
=
+
,
t
+ (cost)k
(cos Ill·
dr P(r('»' dt dt
1" 1) 1,"(1-"lsin t - "lco;t 0
2
0
-t1
dt
271
=
c
B
(.)
dJ = -1T.
Therefore the flux of the curl across S in the direction n for the field F is
ff s
V X P'od"
~ -v.
•
Proof of Stokes' lbeorem for Polyhedral Surfaces Let S be • polybednU surlilce <:<JDmsting of • 'milo moobee of plane regions '" filceo. (See Figure 16.64 for examples.) We apply Green's Theorem to each separate face of S. There are two types of faces: (b)
FIGURE 16.64 (a) Part ofa polyhedral surface. (b) Other polyhedral. surfaces.
1. 2.
Those that are surrounded on all sides by other faces. Those that have one or more edges that are not adjacent to other faces.
969
16.7 Stokes'Theorem
The boundary II of S consists of those edges of the type 2 faces that are not adjacent to other faces. In Figure 16.64a, the triangles EAB, BCE, and CDE represent a part of S, with ABCD part of the boundary ll. We apply a generalized tangential form of Green's Theorem to the three triangles of Figure 16.64a in lttm and add the results to get
In the generalized form, the line integral of F around the curve enclosing the plane region 0 equals the double integral of (curl F) • n over R. The three line integrals on the left-hand side of Equation (7) combine into a single line integral taken around the periphery ABCDE because the integrals along interior segments cancel in pairs. For example, the integral along segment BE in triangle ABE is opposite in sign to the integral along the sarne segment in triangle EBC. The sarne holds for segment CEo Hence, Equation (7) reduces to
R normal to
f
F·dr
=
ABCDE
II
VX
F·ndO".
ABCDE
When we apply the generalized form of Green's Theorem to all the faces and add the results, we get
fF'dr= IlvxF.ndO". s
~
This is Stokes' Theorem for the polyhedral surface S in Figure 16.64a. More general polyhedral surfaces are shown in Figure 16.64b and the proof can be extended to them. General smooth surfaces can be obtained as limits of polyhedral surfaces.
Stokes' Theorem for Surfaces with Holes Stokes' Theorem holds for an oriented surface S that has one or more holes (Figure 16.65). The surface integral over S of the normal component of V X F equals the sum of the line integrals around all the boundary curves of the tangential component of F, where the curves are to be traced in the direction induced by the orientation of S. For such surfaces the theorem is unchanged, but C is considered as a union of simple closed curves.
An Important Identity The following identity arises frequently in mathematics and the physical sciences.
curl grad f = 0
FIGURE 16.65 Stokes'Theorem also holds for oriented surfaces with holes.
or
V X Vf= 0
(8)
This identity holds for any function f(x, y, z) whose second partial derivatives are continuous. The proof goes like this:
V X Vf
=
i
j
k
a ax af ax
a ay af ay
a az af az
=
{fzy - fyz)i - {fzx - fxz)j + {fyz - fzy)k.
If the second partial derivatives are continuous, the mixed second derivatives in parentheses are equal (Theorem 2, Section 14.3) and the vector is zero.
970
Chapter 16: Integration in Vector Fields
Conservative Fields and Stokes' Theorem In Section 16.3, we found that a field F being conaervative in an open region D in space is equivalent to the integral of F around every closed loop in D being zero. This, in turn, is equivalent in simply connected open regions to saying that V X F = 0 (which gives a test for determining if F is conservative for such regions). THEOREM 7-Curl F ~ 0 Related to the Closed-Loop Property If V X F = 0 at every point of a simply connected open region D in space, then on any piecewisesmooth closed path C in D,
(sJ
f
F'dr = O.
c
•
Sketch of a Proof Theorem 7 can be proved in two steps. The first step is for simple closed curves (loops that do not cross themselves), like the one in Figure l6.66a. A theorem from topology, a branch of advanced mathematics, states that every smooth simple closed curve C in a simply connected open region D is the boundary of a smooth two-sided surface S that also lies in D. Hence, by Stokes' Theorem,
f
(b)
FIGURE 16.66
F'dr =
c
11
V X F'ndCT
=
O.
s
The second step is for curves that cross themselves, like the one in Figure l6.66b. The idea is to break these into simple loops spanned by orientable surfaces, apply Stokes' Theorem one loop at a time, and add the results. _ The following diagram summarizes the results for conservative fields defined on connected, simply connected open regions. Theorem 2, Section 16.3
F conservative on D Theorem 3,
Section 16.3
fonD
~
Vector identity (Eq. S)
'1/
\I'
over any closed
F
II
""
(continuous second partial derivatives)
fcF.dr~o pathinD
<=:::>
<==
xF~Othro~oomD
Theorem 7
Domain's simple connectivity and
Stokes' Theorem
Exercises 16.7 Using Stokes' Theorem to Find Line Integrals In Exercises l--{j, use the surface integral in Stokes' Theorem to calculate the circulation of the field F around the curve C in the indicated
direction. 1. F=x'i+2xj +z'k C: The ellipse 4x'
+ y'
when viewed from above
=
4 in the xy-plane, counterclockwise
2. F ~ 2yi + 3xj -z'k C: The circle x 2 + y2
=
9 in the xy-plane, counterclockwise
wben viewed from above
3. F=yi+xzj +x'k C: The boundary ofthe triangle em from the plane x + y + z by the first octant, counterclockwise wben viewed from above
=I
971
16.7 Stokes'Theorem 4. F ~ (y' + z')1 + (x' + z')j + (x' + y')k C: The boundary of the triangle cut from the plane x + y + z ~ 1 by the IITSt octant, counterclockwise when viewed from above 5. F = (y' + z')1 + (x' + y')j + (x' + y')k C: The sqUllIe bounded by the lines x = ± 1 and y = ± 1 in the
xy-plane, counterclockwise when viewed from above 6. F~x'y31+j+zk C: The intersectioo of the cylinder + y' = 4 and the hemisphere xl + y2 + z2 = 16, z ~ 0, cmmterclockwise when viewed
x'
from above 7. Let n be the ooter unit normal of the elliptical shell
+ 9y' + 36z'
~ 36,
z '" 0,
and let F = yi + x'j + (x' + y4)3/2 sin e v.;,;; k. Find the value of
II
°
8. Let n be the ooter unit normal (nonnal away from the origin) of the parabolic shell
4x'+y+z'=4,
y"'O,
and let
~
(-z
+ 2
Find the value of
~ x} +
II
1
(tan- y)j +
(x
+ 4
~ z)k.
s
°
V X (yi)'ndu,
where S is the hemisphere xl 11. FlU][ of curl F Show that
JJs
o
r(, 8)
= (\13 sin cos 8)1 + (\13 sinsin8)j Os s "./2,
+ y2 + z2
=
1, z
O?::
Os 8
S
+
2".
+ z'j + xk S: r(, 8) = (2 sin cos 8)1 + (2 sin sin8)j + (2cos s "./2, s 8 S 2".
18. F ~ y'l
°
Theory and Examples 19. Zero circulation Use the identity V X VI ~ 0 (Equatioo (8) in the rext) and Stokes' Theorem to show that the eirculatioos of the followiog fields around the boundary of any smooth orientable surface in space are zero. a. F = 2xi + 2yj + 2zk h. F = V(xy'z3) ~F~VX~+~+~
VX F·ndu.
9. Let S be the cylinder x' + y' = a', s z S h, together with its top, xl + y2 ::;; a 2, z = h. Let F = -yi + xj + x 2k. Use Stokes' Theorem to Imd the flux of V X F outward 1hrougb S. 10. Evaluate
JJs
o
(\I3cos
(Hint: One paramettizatioo of the ellipse at the hase ofthe shell Is x = 3cost,y = 2sint, s t s 2"..)
F
S:
S:
VX F·ndu.
S
S:
+ 3xj + 5yk r(r,8) = (r cos 8)1 + (rsin8)j + (4 - r')k, S r :s 2, O:s 8 s 21T 14. F = (y - z)i + (z - x)j + (x + z)k S: r(r,8) = (r cos 8)1 + (rsin 8)j + (9 - r')k, S r :s 3, O:s 8 :s 21T
13. F = 2zi
15. F = x'yi + 2y 3zj + 3zk S: r(r,8) = (r cos 8)1 + (rsin 8)j + rk, o :s r :s I, O:s 8 :s 21T 16. F = (x - y)1 + (y - z)j + (z - x)k S: r(r,8) = (r cos 8)1 + (rsin8)j + (5 - r)k, o :s r :s 5, O:s 8 :s 21T 17. F = 3yi + (5 - 2x)j + (z' - 2)k
Flux of the Curl
s: 4x'
Stokes' Tbeoll!m for Parametrized Surfaces In Exercises 13-18, use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field F across the surface S in the direction of the outward unit normal D.
~F~~
20. Zero circulation Let I(x, y, z) = (x' + y' + z')-I/2. Show that the clockwise circulatioo of the field F ~ VI around the circle x 2 + y2 = a2 in the xy-plane is zero a. by taking r = (a cos t)1 + (a sin t)j, s t S 2"., and inregratiog F • dr over the circle. h. by applying Stokes' Theorem. 21. Let C be a simple closed smooth curve in the plane 2x + 2y + z ~ 2, orienred as shown here. Show that
°
f
O.
2ydr
+ 3zdy -
xdz
c z
V X F'ndu
2x+2y+z=2
has the same value for all orienred surfaces S that span C and that
induce the same positive direction on C. 12. Let F be a differentiable vector field deImed 00 a regioo cootaining a smooth closed orienred surface S and its inrerior. Let n be the unit normal vector field 00 S. Suppose that S is the unioo of two surfaces SI and S, joined aloog a smooth simple closed curve e. Can anything be said about
JJs
V X F'ndu?
Give reasons for your answer.
~1 x
depends ooly 00 the area of the regioo enclosed by C and not 00 the positioo or shape of C. 22. Show that ifF = xi + yj + zk, then V X F = 0.
972
Chapter 16: Integration in Vector Fields
23. Find a vector field with twice-differentiable compooents whose curl is xi + yj + zk or prove that no such field exists. 24. Does Stokes'Theorem say anything special about circulatioo in a field whose curl is zero? Give reasons for your answer. 25. Let R be a regloo in the xy-plane that is bouoded by a piecewise smooth simple closed curve C and suppose that the moments of inertia of R about the x- and y-axes are koowo to be 1, and 1yo
26. Zero cur~ yet field not conservative Show that the curl of F =
Y - 2-2 I. + - 2x - -· 2+ J k z x+y
x+y
is zero but that
Evaluate the integral
f C
16.8
V(r 4)'nd>,
is not zero if C is the circle x 2 + y 2 ~ I in the xy-plane. (Theorem 7 does not apply here because the domain of F is not simply connected. The field F is not defmed along the z-axis so there is no way to contract eto a point without leaving the domain ofF.)
The Divergence Theorem and a Unified Theory The divergence form of Green's Theorem in the plane states that the net outward flux of a vector field across a simple closed curve can be calculated by integrating the divergence of the field over the region enclosed by the curve. The corresponding theorem in three dimensions, called the Divergence Theorem, states that the net outward flux of a vector field across a closed surface in space can be calculated by integrating the divergence ofthe field over the region enclosed by the surface. In this section we prove the Divergence Theorem and show how it simplifies the calculation of flux. We also derive Gauss's law for flux in an electric field and the continuity equation ofhydrodynamics. Finally, we unify the chapter's vector integral theorems into a single fundamental theorem.
Divergence in Three Dimensions The divergence ofa vector fieldF = M(x,y,z)i function divF = V'F = aM ax
+ N(x,y,z)j + P(x,y,z)k is the scalar + aN + ap. ay
az
(1)
The symbol ~'div F" is read as "divergence ofF" or "div F." The notation V· F is read udel dot F." Div F bas the same physical interpretation in three dimensions that it does in two. If F is the velocity field of a flowing gas, the value of div F at a point (x, y, z) is the rate at which the gas is compressing or expanding at (x, y, z). The divergence is the flux per unit volume or flux density at the point.
EXAMPLE 1 The following vector fields represent the velocity ofa gas flowing in space. Find the divergence of each vector field and interpret its physical meaning. Figure 16.67 displays the vector fields. (a) Expansion: F(x, y, z) = xi + yj + zk (b) Compression: F(x,y,z) = -xi - yj - zk
16.8
The Divergence Theorem and a Unified Theory
(c) Rotation about z-axis: F(x,y,z) = -yi
973
+ xj
(d) Shearing along horizontal planes: F(x, y, z) = zj
z
z
.'
...-...-...-+ ...-...-...-...-Y
-- ..:=---"--
(d)
--
FIGURE 16.67 Velocity fields ofa gas flowing in space (Example 1).
Solutton (a) divF = ;x(x)
+ ~(y) + ;z(z)
= 3: The gas is undergoingunifonn expansion at all
points. (b) divF = :X(-x)
+ ~(-y) + ;z(-z)
=
-3: The gas is undergoing unifonn com-
pression at all points. (c) div F = ;x (-y)
+ ~ (x)
= 0: The gas is neither expanding nor compressing at any
point. (d) div F =
~ (z)
= 0: Again, the divergence is zero at all points in the domain ofthe ve-
locity field, so the gas is neither expanding nor compressing at any point.
•
Divergence Theorem The Divergence Theorem says that under suitable conditions, the outward flux of a vector field across a closed surface equals the triple integral of the divergence of the field over the region enclosed by the surface.
974
Chapter 16: Integration in Vector Fields
THEOREM 8-Divergence Theorem
Let F be a vector field whose components have continuous first partial derivatives, and let S be a piecewise smooth oriented closed surface. The flux of F across S in the direction of the surface's outward unit normal field n equals the integral of V• F over the region D enclosed by the surface:
ff Foodo- fJJvoFdv.
(2)
=
D
S
Outward flux
EXAMPLE 2 F
==
xi
Divergence integral
Evaluate both sides of Equation (2) for the expanding vector field 2 == a (Figure 16.68).
+ yj + zk over the sphere x 2 + y2 + z2
SoLution The outer unit normal to S, calculated from the gradient of f(x, y, z) y2 + z2 - a 2, is
==
x2
+
Hence,
Therefore, FIGURE 16.68 A uniformly expanding vector field and a sphere (Example 2).
The divergence of F is
V·F
==
a ax
a ay
a az
-(x) + -(y) + -(z)
==
3
'
so
• EXAMPLE 3 Find the flux ofF == xyi + yzj + xzk outward through the surface of the cube cut from the first octant by the planes x == 1, y == 1, and z == 1. SoLution Instead of calculating the flux as a sum of six separate integrals, one for each face of the cube, we can calculate the flux by integrating the divergence
v· F
==
a ax (xy)
+
a ay (yz)
+
a az (xz) == y
+z +x
over the cube's interior: Flux =
ff Foodo- fff VoFdV =
Cube surface
Cube interior
{I {I {I
=
Jo Jo Jo
The Divergence Theorem
(x
+
3
Y+ z)dxdydz ="2"
Routine integration
•
16.8 The Divergence Theorem and a Unified Theory
975
Proof of the Divergence Theorem for Special Regions To prove the Divergence Theorem, we take the components of F to have continuous flI'St partial derivatives. We flI'St assume that D is a convex region with no holes or bubbles, such as a solid ball, cube, or ellipsoid, and that 8 is a piecewise smooth surface. In addition, we assume that any line perpendicular to the xy-plane at an interior point of the region Rzy that is the projection ofD on the xy-plane intersects the surface 8 in cuctly two points, producing surfaces
FIGURE 16.69 We prove the Divergence Theorem for the kind of three-dimensional region shown here.
81:
z = ft(x,y),
(x,y) inRzy
8'l:
z
(x,y) inRzy,
=
h(x,y),
with ft :S h. We make similar assumptions about the projection of D onto the other coordinate planes. See Figure 16.69. The components of the unit normal vector a = nIl + n2J + n3k are the cosines of the angles a,/3, and 'Y that D makes with i, i, 8IId k (Figure 16.70). This is true because all the vectors involved are unit vectors. We have nl
,
a'l
=
lalill cosu
=
cosu
., ~ -'i
~
1-llilc08/l
~
co./l
o'k
=
lallklcos'Y
=
cos'Y.
=
n3 =
8, Thus,
D
= (cos ali
+ (oos13)i + (cos 'Y)k
and
+ Ncos/3 + Pcos'Y.
F'. = Mcosa
In component form, the Divergence Theorem states that
x
FIGURE 16.70 The components orn arc P, and 'Y that it
the cosines of the angles a, makes with i, J, and k.
ff s
(Mcosa
+ Ncos13 + P cos 'Y) dO' .
=
.
iff (a:: + a:: + ~) ~dydz. D'
F·o
•
divF
We prove the theorem by proving the three following equalities:
, t:=h~J)
D
s,
JJ
M co. a do-
~ JJp:: dx dy dz
S
JJ Nco./l do- ~ JJJa::
dxdydz
(4)
~ JJJa;: dx dy dz
(5)
S
du
-
x
dA. _
d~dJ
FIGURE 16.71 TheregionDenclosedby the surfaces 81 and /h projects vertically onto Rry in thexy-plane.
JJ
(')
D
D
P C081 do-
S
D
Proof 01 Equation (5) We prove Eqcatioo (5) by coovertiog 1he surface integml 00 lhe left to a double integral over the projection Rzy of D on the xy-plane (Figure 16.71). The surface 8 consists of an upper part s,. whose equation is z = h(x,y) and a lower part 81 whose equation is z = fl(X,y). ODs,., the outer normal 0 has a positive k-component and cos 'Y do' = dx dy
because
dxdy ~
cos 'Y'
976
Chapter 16: Integration in Vector Fields
See Figure 16.72. On 810 the outer normal Dhas a negative k-eomponent and
cos'Ydu - -dxdy. Therefore,
ff PC081du ~ ff PcC" dU + ff PCC'1 du ~ ff P(x,y,j,(x,y»dxdy - ff p(x,y,j,(x,y»dxdy ~ ff [p(x,y,j,(x,y» - p(x,y,!I(x,y))] dxdy
Here 'Y is acute, so du _ dx tty/cos 'Y.
S
8:1
St
R.,
<:]01'-- Here 'Y is obtuse, so du ,., -dx dykOlJ 1.
R.,
R.,
~ ff R.
FIGURE 16.72 An enlarged view of the area patches in Figure 16.71. The relations du = ±dx dyjcos "I come from Eq. (7) in Section 16.S.
u.:
l
a:: dz] dx dy ~ JJJ a:: dz dx dy. D
This proves Equation (5). The proofs for Equations (3) and (4) follow the same pattern; or just permute x,y, z; M, N, P; a, P, 'Y, in order, and get those results :from Equation (5). This proves the Divergence Theorem for these special regions. •
Divergence Theorem for Other Regions
,
y
The Divergence Theorem can be extended to regions that can be partitioned into a f'mite number of simple regions of the type just discussed and to regions that can be defined as limits of simpler regions in certain ways. For an example of one step in such a splitting process, suppose that D is the region between two concentric spheres and that F has continuously d:ifferentiable components throughout D and on the bounding surfaces. Split D by an equatorial plane and apply the Divergence Theorem to each half separately. The bottom half, D}, is shown in Figure 16.73. The surface 8\ that bounds DI consists ofan outer hemisphere, a plane washer-shaped base, and an inner hemisphere. The Divergence Theorem says that (6)
FIGURE 18.73 Thelowerhalfofthe solid region between two concentric
The unit normal 0\ that points outward :from Dl points away from the origin along the outer surface, equals k along the flat base, and points toward the origin along the inner surface. Next apply the Divergence Theorem to D7., and its surface 82 (Figure 16.74):
spheres.
,
ff F'n,du, ~ fff V·FdY,.
1
s,
& we follow 02 over s,., pointiDg outward from. Dz , we see that 02 equals - k along the washer-shaped base in the xy-p1ane, poiDts away :from the origin on the outer sphere, and points toward the origin on the inner sphere. When we add Equations (6) and (7), the integrals over the flat base cancel because of the opposite signs ofo l and 02. We thus arrive at the result
., y
ff F'nda ~ fff V'FdY, S
FIGURE 115.74 Theupperhalfofthe solid region between two concentric ",here.
(7)
D,
D
with D the region between the spheres, 8 the boundary of D consisting of two spheres, and o the unit normal to 8 directed outward from. D.
16.8
EXAMPLE 4
The Divergence Theorem and a Unified Theory
Find the net outward flux of the field xi
+ yj + zk
F ==
3
'
P
across the boundary of the region D: 0 SoLution
FIGURE 16.75 Two concentric spheres in an expanding vector field.
977
< a2
:::;
x2
+ y2 + z2
:::;
b 2 (Figure 16.75).
The flux can be calculated by integrating V· F over D. We have
and 2 -aM -_ -a (xp-3) -_ p-3 - 3xp-4 -ap -_ -1 - -3x ax ax p3 p5 . ax
Similarly,
aN
and
ay
ap
az
Hence,
and
fff
V·FdV
=
o.
V·F
=
divF
D
So the integral of V • F over D is zero and the net outward flux across the boundary of D is zero. There is more to learn from this example, though. The flux leaving D across the inner sphere Sa is the negative of the flux leaving D across the outer sphere Sb (because the sum of these fluxes is zero). Hence, the flux of F across Sa in the direction away from the origin equals the flux of F across Sb in the direction away from the origin. Thus, the flux of F across a sphere centered at the origin is independent of the radius of the sphere. What is this flux? To find it, we evaluate the flux integral directly. The outward unit normal on the sphere of radius a is n ==
+ yj + zk Yx + y2 + z2 xi
-;:::::::=======~
2
xi
+ yj + zk a
Hence, on the sphere,
F·n
xi ==
+ yj + zk a
xi
3
+ yj + zk a
and
ff Sa
F· n da =
a\Jf
da =
:2 (47m 2) = 47T.
Sa
The outward flux of F across any sphere centered at the origin is 4'7T.
•
978
Chapter 16: Integration in Vector FieLds
Gauss's Law: One of the Four Great Laws of Electromagnetic Theory There is still more to be learned from Example 4. In electromagnetic theory, the electric field created by a point charge q located at the origin is
q
1
E(x,y, z)
= 47n:0
~
(r) TrT
q = 47n:o
r
~
q = 47n:o
xi+yj+zk p3 '
where EO is a physical constant, r is the position vector of the point (x, y, z), and p = Ir I = x 2 + y2 + z2. In the notation of Example 4,
V
q E =-4-F. 7TEO
The calculations in Example 4 show that the outward flux ofE across any sphere centered at the origin is q/ EO, but this result is not confined to spheres. The outward flux of E across any closed surface S that encloses the origin (and to which the Divergence Theorem applies) is also q/Eo. To see why, we have only to imagine a large sphere Sa centered at the origin and enclosing the surface S (see Figure 16.76). Since
V·E
=
V·-q-F
=
47TEo
-q-V·F 47TEo
=
°
when p > 0, the integral of V· E over the region D between S and Sa is zero. Hence, by the Divergence Theorem,
11
E·ndu = 0,
Boundary ofD
FIGURE 16.76 A sphere Sa surrounding another surface S. The tops of the surfaces are removed for visualization.
and the flux of E across S in the direction away from the origin must be the same as the flux of E across Sa in the direction away from the origin, which is q/ EO. This statement, called Gauss s Law, also applies to charge distributions that are more general than the one assumed here, as you will see in nearly any physics text.
11
Gauss's Law:
E· n du
=
:0
s
Continuity Equation of Hydrodynamics Let D be a region in space bounded by a closed oriented surface S. Ifv(x, y, z) is the velocity field of a fluid flowing smoothly through D, 8 = 8(t, x, y, z) is the fluid's density at (x, y, z) at time t, and F = 8v, then the continuity equation ofhydrodynamics states that
V·F
T
h = (v M).n
___L
a8
+ at
=
0.
If the functions involved have continuous first partial derivatives, the equation evolves naturally from the Divergence Theorem, as we now see. First, the integral
11
F·ndu
s
s FIGURE 16.77 The fluid that flows upward through the patch au in a short time !1t fills a "cylinder" whose volume is approximately base X height =
v· n au
at.
is the rate at which mass leaves D across S (leaves because n is the outer normal). To see why, consider a patch of area au on the surface (Figure 16.77). In a short time interval at, the volume a v of fluid that flows across the patch is approximately equal to the volume of a cylinder with base area au and height (vat) . n, where v is a velocity vector rooted at a point of the patch:
a v ~ v· n au at.
16.8 The Divergence Theorem and a Unified Theory
979
The mass of this volume of fluid is about
Ilm "" Bv' n Ilu Ilt, so the rate at which mass is flowing out of D across the patch is about
This leads to the approximation
as an estimate of the average rate at which mass flows across S. Finally, letting Ilu --+ 0 and Ilt --+ 0 gives the instantaneous rate at which mass leaves D across S as
~7 = which for oUI particular flow is
~7 =
11 11
Bv' n du,
s
F·ndu.
s
Now let B be a solid sphere centered at a point Q in the flow. The average value of V'FoverBis
vOI~e of B
lff
V' F dV.
B
It is a consequence of the continuity of the divergence that V' F actually takes on this value at some point Pin B. Thus,
11
I of B ( V'F )p = volume
F'ndu
lJifJ V'FdV= volume s of B B
rate at which mass leaves B across its surface S volume ofB
(8)
The last term ofthe equation describes decrease in mass per unit volume. Now let the radius of B approach zero while the center Q stays tIXed. The left side of Equation (8) converges to (V, F)g, the right side to (-aBjat)g. The equality of these two limits is the continuity equation
The continuity equation "explains" V' F: The divergence of F at a point is the rate at which the density of the fluid is decreasing there. The Divergence Theorem
11 S
F'ndu =
111
V'FdV
D
now says that the net decrease in density of the fluid in region D is accounted for by the mass tnmsported across the surface S. So, the theorem is a statement about conservation of mass (Exercise 31).
980
Chapter 16: Integration in Vector Fields
Unifying the Integral Theorems If we think of a two-dimensional field F = M(x, y)i + N(x, y)i as a three-dimensional field whose k-component is zero, then V' F = (aM/ax) + (aN/ay) and the normal form of Green's Theorem can be written as
f
11(aa~+~~)dxdy=
11
F ' nds = V ' FdA . e R R Similarly, V X F' k = (aN/ax) - (aM/ay), can he written as
80
the tangential form of Green's Theorem
11 (~~ -~~)
f
11
F·dr = dxdy = V X F·kdA. e R R With the equations of Green's Theorem now in del notation, we can see their relationships to the equations in Stokes' Theorem and the Divergence Theorem.
Green'. Theorem and Its GeneraIization to Three Dimensions
f
Normal form of Green's Theorem:
11
Divergence Theorem:
f f
F' n drT =
F' dr =
n =-1
n !
a
=I •
F' dr =
dimensional space.
V X F' n drT
S
Notice how Stokes' Theorem generalizes the tangential (curl) form of Green's Theorem from a flat surface in the plane to a surface in three-dimensional space. In each case, the integral of the normal component of curl F over the interior of the surface equals the circulation of F around the boundary. Likewise, the Divergence Theorem generalizes the normal (flux) form of Green's Theorem from a two-dimensional region in the plane to a three-dimensional region in space. In each case, the integral of V• F over the interior ofthe region eqnals the total flux of the field across the boundary. There is still more to be learned here. All these results can be thought of as forms of a single fUndamental theorem. Think back to the Fundamental Theorem of Calculus in Section 5.4. It says that if f(x) is differentiable on (a, b) and continuous on [a, bl, then
b
FIGURE 16.78 The outward unit normals atthe boundary of [a, b] in onl>-
V X F' k dA
R
C
I
V' F dV
D
C
Stokes' Theorem:
V' F dA
R
s
Tangential form of Green's Theorem:
11 111 11 11
F' n ds =
C
l
a
bdf dx dx
field n normal to the boundary of [a,
= =
f(b) - f(a).
bl, then (df/dx) = V' F. Ifwe define the unit vector bl to be i at band -i at a (Figure 16.78), then f(b)i' (i) + f(a)i' (-i) F(b)'n + F(a)'n total outward flux of F across the boundary of [a, bI.
If we let F = f(x)i throughout [a, f(b) - f(a) =
=
981
16.8 The Divergence Theorem and a Unified Theory
The Fundamental Theorem now says that
F(b)'n + F(a)'n
=
J
V·Fdx.
[a,b]
The Fundamental Theorem of Calculus, the normal form of Green's Theorem, and the Divergence Theorem all say that the integral of the differential operator V' operating on a field F over a region equals the sum of the normal field components over the boundary of the region. (Here we are interpreting the line integral in Green's Theorem and the surface integral in the Divergence Theorem as "sums" over the boundary.) Stokes' Theorem and the tangential form of Green's Theorem say that, when things are properly oriented, the integral of the normal component of the curl operating on a field equals the sum of the tangential field components on the boundary of the surface. The beauty of these interpretations is the observance of a single unifYing principle, which we might state as follows.
A Unifying Fundamental Theorem The integral of a differential operator acting on a field over a region equals the sum of the field components appropriate to the operator over the boundary of the region.
Exercises 16.8 9. Portion ofsphere F ~ x'i - 2xyj
Calculating Divergence In Exercises 1-4, find the divergence of the field.
I. The spin field in Figure 16.12 2. The radial field in Figure 16.11 3. The gravitational field in Figure 16.8 and Exercise 38a in Section 16.3 4. The velocity field in Figure 16.13
10. Cylindrical can F ~ (6);2 + 2xy)i + (2y + x 2z)j + 4x:Y'k D: The region cut from the fITSt octant by the cylioder x 2 + y2 = 4 and the plane z ~ 3 F = 2xzi - xyj - z 2k D: The wedge cut from the fITst octant by the plane y andtheellipticalcylinder4x 2 + y2 = 16
11. Wedge
Calculating Flux Us;ng the Divergence Theorem In Exercises 5-16, use the Divergence Theorem to ftnd the outward flux ofF across the boundary of the regionD.
5. Cube F = (y - xli + (z - y)j + (y - x)k D: The cube bounded by the planes x = ± I, y = ± I, and z ~ ±I 6. F=x'i+y2j+z 2k a. Cube D: The cube cut from the first octant by the planes x = I,y = I, andz = I b. Cube D: The cube bounded by the planes x y ~ ±I, andz ~ ±I
+ 3xzk
D: The region cut from the fITst octant by the sphere x 2 + y2 + z2 = 4
= ± I,
c. Cylindrical can D: The region cut from the solid cylinder x 2 + y2 '" 4 by the planes z = 0 and z~ I 7. Cylinder and paraboloid F = yi + xyj - zk D: The region inside the solid cylinder x 2 + y2 '" 4 between the plane z = 0 and the paraboloid z = x 2 + y2 8. Sphere F = x'i + xzj + 3zk D: The solid sphere Xl + y2 + z2 :s 4
+z
~
4
+ y'j + z'k The solid sphere Xl + y2 + z2 ::s a 2
12. Sphere F ~ x':i D:
13. Tbick sphere F ~ Vx 2 + y2 + z2 (xi D: The region 1 :s x 2 + y2 + z2 :s 2 14. Tbick sphere F
= (xi + yj + zk)/Vx 2 + y2 + z2
D: The region 1 :s x 2
IS. Tbick sphere F (5z' + e' cosz)k
+ yj + zk)
+ y2 + z2 :s
= (5x' +
12xy2)i
4
+ (y' + e'sinz)j +
D: The solid region between the spheres x 2 + y2 x 2 + y2 +z2 = 2
16. Tbickcylinder F = In(x 2 + y2)i zVx 2 + y 2 k
+ z2 = I and
(~tan-I~)j +
D: The thick-walled cylioder I '" x 2 + y2 '" 2,
-I '" z '" 2
982
Chapter 16: Integration in Vector Fields
Properties of CurL and Divergence 17. div (curl G) is zero a. Show that if the necessary partial derivatives of the components of the field G = Mi + Nj + Pk are continuous, then V·V X G = O. b. What, if anything, can you conclude about the flux of the field V X G across a closed surface? Give reasons for your answer. 18. Let F land F2 be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities. a. V·(aF l
+
bF2) = aV·Fl
+
bV·F2
b. V X (aFl + bF2) = aV X Fl + bV X F2 c. V· (F l X F 2) = F 2 · V X F l - F l · V X F 2 19. Let F be a differentiable vector field and let g(x, y, z) be a differentiable scalar function. Verify the following identities. a. V·(gF) = gV·F
+
Vg·F
b. V X (gF) = gV X F
+
Vg X F
+ Nj + Pk is a differentiable vector field, we define the notation F · V to mean a a a M ax + N ay + P az .
20. If F = Mi
23. a. Show that the outward flux of the position vector field F = xi + yj + zk through a smooth closed surface S is three times the volume of the region enclosed by the surface.
b. Let n be the outward unit normal vector field on S. Show that it is not possible for F to be orthogonal to n at every point ofS. 24. Maximum flux Among all rectangular solids defined by the inequalities 0 ~ x ~ a, 0 ~ y ~ b, 0 ~ z ~ 1, find the one for which the total flux of F = (-x 2 - 4xy)i - 6yzj + 12zk outward through the six sides is greatest. What is the greatest flux? 25. Volume of a solid region Let F = xi + yj + zk and suppose that the surface S and region D satisfy the hypotheses of the Divergence Theorem. Show that the volume of D is given by the formula
+
(V· F 2)F l -
(V· F l )F 2 b. V(F l · F 2) = (F l · V)F2 F 2 X (V X F l )
+
(F2 · V)F l
+
F l X (V X F 2)
+
Theory and ExampLes 21. Let F be a field whose components have continuous first partial derivatives throughout a portion of space containing a region D bounded by a smooth closed surface S. If IF I ~ 1, can any bound be placed on the size of
fff
26. Outward flux of a constant field Show that the outward flux of a constant vector field F = C across any closed surface to which the Divergence Theorem applies is zero. 27. Harmonic functions A function I(x, y, z) is said to be harmonic in a region D in space if it satisfies the Laplace equation 2 a1 V21 = V· VI = - 2 ax
2
a
+-
1 + -a21 =
ay 2
az 2
0
throughout D. a. Suppose that I is harmonic throughout a bounded region D enclosed by a smooth surface S and that n is the chosen unit normal vector on S. Show that the integral over S of VI · n, the derivative of I in the direction ofn, is zero. b. Show that if I is harmonic on D, then
ff
fVfondO" =
fff
IVfl 2 dV,
D
S
28. Outward flux of a gradient field Let S be the surface of the portion of the solid sphere x 2 + y2 + z2 ~ a 2 that lies in the first octant and let I(x, y, z) = In x 2 + Y 2 + z2. Calculate
VoFdV?
V
D
ff
Give reasons for your answer. 22. The base of the closed cubelike surface shown here is the unit square in the xy-plane. The four sides lie in the planes x = 0, x = 1, y = 0, and y = 1. The top is an arbitrary smooth surface whose identity is unknown. Let F = xi - 2yj + (z + 3)k and suppose the outward flux of F through Side A is 1 and through Side B is - 3. Can you conclude anything about the outward flux through the top? Give reasons for your answer.
(VI · n is the derivative of I in the direction of outward normal n.) 29. Green's first formula Suppose that I and g are scalar functions with continuous first- and second-order partial derivatives throughout a region D that is bounded by a closed piecewise smooth surface S. Show that
ff
f Vgon dO" =
Top
fff
(j V 2g + Vfo Vg) dV.
(9)
D
S
!
VfondO".
S
z
1
FondO",
s
For differentiable vector fields F land F 2, verify the following identities. a. V X (F l X F 2) = (F2 · V)F l - (F l · V)F 2
t ff
Volume ofD =
Equation (9) is Green's first formula. (Hint: Apply the Divergence Theorem to the field F = I Vg.)
30. Green's second formula
(Continuation ofExercise 29.) Interchange I and g in Equation (9) to obtain a similar formula. Then subtract this formula from Equation (9) to show that
/
x
Side A
~y
•
'" SideB
(1,1,0)
ff S
(j Vg - gVf) n dO" = 0
fff
(j V 2g - gV 2f) dV.
D
This equation is Green's second formula.
(10)
Chapter 16 31. Conservation of mass Let v(l, x, y, z) be a continuously differentiable vector field over the region D in space and let p(l, x,y, z) be a continuously differentiable scalar function. The variable 1represents the time domain. The Law of Conservation of Mass asserts that
~JJJp(l,x,y,Z)dV=
-JJpv'ndCT,
D
S
where S is the surface enclosing D.
a. Give a physical interpretation of the conservation ofmass law ifv is a velocity flow field and p represents the density of the fluid at point (x,y, z) at time I.
D
32. Tbe beat _ o n equation Let 1'(1, x, y, z) be a function with continuous second derivatives giving the temperature at time t at the point (x, y, z) of a solid occupying a region D in space. If the solid's heat capacity and mass density are denoted by the constants c and p, respectively, the quantity cpT is called the solid's heat energy per unit volume. a. Explain why - VT points in the direction ofbeat flow. b. Let -kVT denote the energy flux veetor. (Here the constant
k is called the conductivity.) Assuming the Law of Conservation of Mass with -kVT = v and cpT = p in Exercise 31, derive the diffusion (heat) equation
JJJ?t dV,
aT = Kv-T, ..." at
D
where K = k/(cp) > 0 is the difjUsivity constant. (Notice that if 1'(1, x) represents the temperature at time 1at position x in a unifunn conduc1ing rod wi1b perfectly insulated sides, then V'T = a2T/ax 2 and the diffusion equation reduces to the onedimensional beat equation in Chspter l4's Additional Exercises.)
to show that the Law of Conservation of Mass is equivalent to the continuity equation,
V·pv +
Chapter
2!rn'
983
(In the flIst term V' pv, the variable 1is held fixed, and in the second term ap/al , it is asswned that the point (x, y, z) in Dis held fixed.)
b. Use the Divergence Theorem and Leibniz's Rule,
~JJJp(l,x,y,Z)dV=
Practice Exercises
ap
at = o.
Questions to Guide Your Review
1. What are line integrals? How are they evaluated? Give examples. 2. How can you use line integrals to f'md the centers of mass of springs? Explain.
3. What is a vector field? A gradient field? Give examples. 4. How do you calculate the work done by a force in moving a particle along a curve? Give an example. 5. What are flow, circulation, and flux? 6. What is special about pa1b independent fields? 7. How can you tell when a field is conservative? 8. What is a potential function? Sbow by example how to fmd a poteotial function for a conservative field. 9. What is a diffurential form? What does it mean for such a form to be exact? How do you test for exactness? Give examples. 10. What is the divergence ofa vector field? How can you interpret it? 11. What is the corl of a vector field? How can you interpret it?
13. How do you calculate the area ofa parame1rized surface in space? Of an implicitly defmed surface F(x, y, z) ~ O? Of the surface which is the graph of z = f(x, y)? Give examples. 14. How do you integrate a function over a parametrized surface in space? Of surfaces that are defmed implicitly or in explicit form? What can you calculate with surface integrals? Give examples. 15. What is an otiented surface? How do you calculate the flux of a three-dimensional vector field across an otiented surface? Give an example. 16. What is Stokes' Theorem? How can you interpret it? 17. Summarize the chapter's results on conservative fields. 18. What is the Divergence Theorem? How can you interpret it? 19. How does the Divergence Theorem generalize Green's Theorem? 20. How does Stokes' Theorem generalize Green's Theorem? 21. How can Green's Theorem, Stokes' Theorem, and the Divergence Theorem be thought of as forms ofa single fundamental theorem?
12. What is Green's Theorem? How can you interpret it?
Chapter
•
Practice Exerdses
Evaluating LIne Integrals 1. The accompanying figure shows two polygonal paths in space joining the otigin to 1be point (I, I, I). Integrate f(x,y, z) ~ 2x 3y 2 - 2z + 3 over each path.
(0,0,0)
,
O'O'O)
(I, I, 1)
,\,,------, , ,
~
\1
(I, 1,0)
Path 1
y x
z
~ /
(1,1,1)
~y (I, 1,0)
Path 2
984
Chapter 16: Integration in Vector Fields
2. The accompanying figure shows three polygonal paths joining the origin to the point (1, 1, 1). Integrate /(x, y, z) = x 2 + Y - z over each path.
z
b. Find the area of the portion of the cylinder that lies inside the hemisphere. (Hint: Project onto the xz-plane. Or evaluate the integral Jh ds, where h is the altitude of the cylinder and ds is the element of arc length on the circle x 2 + y2 = 2x in the xy-plane.)
z
(0,0,0)
(1, 1, 1)
(1, 1, 1)
~y (1,1,0)
x
x
14. a. Hemisphere cut by cylinder Find the area of the surface cut from the hemisphere x 2 + y2 + z2 = 4, z ~ 0, by the cylinder x 2 + y2 = 2x.
~y
z
Hemisphere
(1,1,0)
j
z=~
z
/
(0,0, 1) (0,0,0)
0'
y
7
x
3. Integrate /(x, y, z) =
v'x 2 + z2 over the circle
r(t) = (a cos t)j + (a sin t)k,
o~
t
~
+ t sin t)i + (sin t - t cos t)j,
0 ~ t ~
\13.
Evaluate the integrals in Exercises 5 and 6. (4,-3,0) dx + dy + dz 5. -;:::::====~ (-1,1,1) Yx + y + z
1
6.
«IO,3,3)
dx -
~dY
-
y
27T.
4. Integrate /(x,y, z) = v'x 2 + y2 over the involute curve r( t) = (cos t
~
Cylinder r = 2 cos ()
~dZ
J(1,I,1)
7. Integrate F = -(y sinz)i + (x sinz)j + (xy cosz)k around the circle cut from the sphere x 2 + y2 + z2 = 5 by the plane z = - 1, clockwise as viewed from above. 8. Integrate F = 3x 2yi + (x 3 + l)j + 9z 2k around the circle cut from the sphere x 2 + y2 + z2 = 9 by the plane x = 2. Evaluate the integrals in Exercises 9 and 10. 9. [8X siny dx - 8y cosx dy C is the square cut from the first quadrant by the lines x = 7T/2 andy = 7T/2. 10. [y2 dx + x 2 dy C is the circle x 2 + y2 = 4.
Finding and EvaLuating Surface Integrals 11. Area of an elliptical region Find the area of the elliptical region cut from the plane x + y + z = 1 by the cylinder x 2 + y2 = 1. 12. Area of a parabolic cap Find the area of the cap cut from the paraboloid y2 + z2 = 3x by the plane x = 1.
13. Area of a spherical cap Find the area of the cap cut from the top of the sphere x 2 + y2 + z2 = 1 by the plane z = V2/2.
15. Area of a triangle Find the area of the triangle in which the plane (x/a) + (y/b) + (z/c) = 1 (a, b, c > 0) intersects the first octant. Check your answer with an appropriate vector calculation. 16. Parabolic cylinder cut by planes
Integrate
yz
a. g(x,y,z) =
Y
+
1
b. g(x,y,z) =
v'4Y~ +
1
4y 2
over the surface cut from the parabolic cylinder y2 - z = 1 by the planes x = 0, x = 3 , and z = O. 17. Circular cylinder cut by planes Integrate g(x, y, z) = x 4y(y2 + z2) over the portion of the cylinder y2 + z2 = 25 that lies in the first octant between the planes x = 0 and x = 1 and above the plane z = 3. 18. Area of Wyoming The state of Wyoming is bounded by the meridians 111 °3' and 104°3' west longitude and by the circles 41 ° and 45° north latitude. Assuming that Earth is a sphere ofradius R = 3959 mi, find the area ofWyoming.
Parametrized Surfaces Find parametrizations for the surfaces in Exercises 19-24. (There are many ways to do these, so your answers may not be the same as those in the back of the book.) 19. Spherical band The portion of the sphere x 2 + y2 + z2 = 36 between the planes z = - 3 and z = 3
V3
20. Parabolic cap The portion of the paraboloid z = -(x 2 + y2)/2 above the plane z = - 2 21. Cone The cone z = 1 + v'x 2 + y2, Z ~ 3 22. Plane above square The portion of the plane 4x + 2y + 4z = 12 that lies above the square 0 ~ x ~ 2, 0 ~ y ~ 2 in the first quadrant
Chapter 16 23. Portion of paraboloid The portion of the paraboloid y = 2(x 2 + z2), Y :5 2, that lies above the xy-plane
24. Portionofbemispbere Theportionof1bebemispherex 2 + y2 z2 = 10, y '" 0, in the Ilrst octant
+
25. Surface area Find the area of the surface r(u, v) = (u + v)i + (u - v)j + vk,
o :s; u :5
1, 0
:s;
38. Flow along different paths V(x 2zeY )
Practice Exercises
985
Find the flow of the field F =
a. once around the ellipse C in which the plane x + y + z ~ I intersects the cylinder x 2 + z2 = 25, clockwise as viewed from 1be positive y-axis. b. along the curved boundsry of the helicoid in Exercise 27 from (1,0,0) to (I, 0, 21T).
v :s; 1.
26. Surface integral Integrate /(x,y,z) = xy - z2 over the surface in Exercise 25.
In Exercises 39 and 40, use the surface integral in Stokes' Theorem to Imd the circulation of the field F around the curve C in the indicated direction.
27. Area of a helicoid
39. Circulation around an ellipse
Find the surface area of the helicoid
r(r, 6) = (r cos 6)i + (rsin 6)j + 6k, in thc accompanying figure.
0:5 6 :5 21T, 0:5 r :5 1,
F = y2;, - yj + 3z2k
C: The ellipse in which the plane 2x + 6y - 3z ~ 6 meets 1be cylinder x 2 + y2 = I, counterclockwise as viewed from above 40. Circulation around a circle (4y 2 - z)k
F = (x 2 + y)i + (x + y)j +
C: The circle in which the plane z ~ -y meets the sphere + y2 + z2 = 4, counterclockwise as viewed from above
x2
Masses and Moments 41. Wirc with different densities Find the mass of a thin wire lying along the curve r(l) = v2ti + v2tj + (4 - 12 )k, o :5 I :5 I, if the density at I is
y
x
28. Surface integral Evaluate the integral ffs v'x 2 + y2 + 1 du, where S is the helicoid in Exercise 27.
Conservative Fields Which of the fields in Exercises 29-32 are conservative, and which arena!? 30. F = (xi + yj + zk)/(x 2 + y2 + z2)'/2
31. F = xeYi + ye%j + ~
Find the center of mass and 1be moments of inertia about the coordioate axes of a thin wire lying along the curve r(l) = Ii + 2V2 1'/2j + f k, 3 2 if the density altis B = 1/(1+ 1).
29. F=xi+yj+zk
32. F
43. Wirc with variable density
44. Center of mass of an arch A slender metal arch lies along 1be semicircle y ~ v'a 2 - x 2 in 1be xy-plane. The density at 1be point (x, y) on 1be arch is B(x,y) = 2a - y. Find the center of mass.
ze~
(i + zj + yk)/(x + yz)
Find potential functions for thc ficlds in Exerciscs 33 and 34. 33. F = 2i + (2y + z)j + (y + l)k 34. F = (zcosxz)i + eYj + (xcosxz)k
Work and Cbeulation In Exercises 35 and 36, Imd the work done by each field along the paths from (0, 0, 0) to (1,1,1) in Exercise I. 2
2
35.F=2xyi+j+x k
36.F=2xyi+x j+k
37. Finding work in two ways
Find the work done by
xi + yj F = C-;;-----':,,= (x2 + y2)'/2 over the plane curve r(l) ~ (e' cos I)i + (e' sin I)j from the point (1,0) to the point (e 2w , 0) in two ways:
a. By using the parame1rization of the curve to evaluate the work integral. b. By evaluating a potential function for F.
45. Wirc with constant density A wire of constant density B ~ 1 lies along the curve r(l) = (e' cos I)i + (e' sin I)j + e'k, 0:5 I :5 In 2. Find ii and I•. 46. Helical wire with constant density Find the mass and center of mass of a wire of constant density B that lies along 1be helix r(l) ~ (2 sin I)i + (2 cos I)j + 31k, 0 :5 I :5 21T.
47. Inertia and center of mass of a shen Find I. and 1be center of mass of a thin shell of density B(x, y, z) = z cut from the upper portion of the spbere x 2 + y2 + z2 ~ 25 by the plane z ~ 3. Find the moment of inertia about the z-axis of the surface of 1be cube cut from 1be Ilrst octant by theplanesx~ I,y~ I,andz~ lifthedensityisB~ I.
48. Moment of inertia of a cnbe
Flux Across a Plane Curve or Surface Use Green's Theorem to Imd the counterclockwise circu1ation and outward flux for the fields and curves in Exercises 49 and 50. 49. Square F = (2xy + xli + (xy - y)j C: The squsre bounded by x = O,x = I,y = O,y = 1
986
Chapter 16: Integration in Vector Fields
50. Triangle
F ~ (y - fu:')i
+ (x + y')j
55. Spherical cap
f
Show that
cosy lnxsinyt(y - -x-dx = 0
c
for any closed curve C to which Green~ Theorem applies.
pianes
52. a. Outward nm: and area Show that tire outward nux of the position vector field F = xi + yj across any closed curve to which Greco ~ Theorem applies is twice the area of the region enclosed by the curve. b. Let n he the outward unit nonnal vector to a closed curve to which Greco's Theorem applies. Show that it is not possihle for F = xi + yj to he orthogonal to n at every point of C.
In Exercises 53-56, fmd tire outward flux of F across tire houodary ofD. 53. Cuhe
F = 2xyi
+
2yzj
+ 2x2k
D: The cuhe cut from the first octant by tire pianes x z =1
54. Spherical cap
F
~
I,y
~
I,
= X2i + yzj + k
D: The entire surface of the upper cap cut from the solid sphere + y' + z':5 25 by the piane z ~ 3
x'
Chapter
•
+ zk
D: The upper region cut from the solid sphere x' + y' + z' :5 2 by the paraholoid z ~ x' + y' 56. Cone and oylinder F = (fu: + y)i - (x + z)j + 4yzk D: The region in the ftrst octant houoded by the cone z = v'x' + y', tire oylinder x' + y' = I, and the coordinate
C: The triangle made by the lines y = 0, y = x , and x = I 51. Zero line integral
F = -2xi - 3yj
57. Hemisphere, oylinder, and plane Let S he tire surface that is bounded on tire leftby the hemisphere x' + y' + z' = a',y:5 0, in lbe middle by the oylinder x' + z' = a', 0 :5 Y :5 a, and on the right by the plane y = a. Find the flux of F = yi + zj + xk outward across S. 58. Cylinder and planes Find the outward flux of the fteld F = 3X2'i + yj - z'k across the surface of the solid in the first octant that is bounded by the oylinder + 4y' ~ 16 and the pianesy = 2z,x = 0, andz = O.
x'
59. Cylindrical can Use the Divergence Theorem to fmd the flux ofF ~ xy'i + x'yj + yk outward through the surface of tire region enclosed by the cylinder + y' = I and the pianes z = I andz ~ -I.
x'
60. Hemisphere Find the flux ofF ~ (3z + I)kupwardacrossthe hemisphere x 2 + y2 + z2 = a 2• z ~ 0 (a) with the Divergence Theorem and (b) by evaluating the flux integral directly.
Additional and Advanced Exercises
FInding Areas with Green's Theorem Use the Greeo~ Theorem area formula in Exercises 16.4 to fmd the areas of the regions enclosed by the curves in Exercises 1-4.
3. The eight curve x = (1/2)sin21,y = sin I, 0 :5 1:5 ".(oneloop) y
I. The lima""" x = 2 cos 1 - cos 21, Y = 2 sin 1 - sin 21, OSI:521T y
-t-----=-+-~-~x
o
4. The teardrop x 2. The deltoid x = 2 cos 1 + cos 21, Y = 2 sin 1 - sin 21, OSt:521T y
-+--ccf----::::~~-.
3
= 2a cos t
-
a sin 2t, y = b sin t, 0 :s t :s 2'17" y
b
Chapter 16 Additional and Advanced Exercises Theory and Applications 5. a. Give an example of a vector field F (x, y, z) that has value 0 at only one point and such that curl F is nonzero everywhere. Be sure to identify the point and compute the curl. h. Give an example ofa vector fieldF (x,y, z) that has value 0
on precisely one line and such that curl F is nonzero everywhere. Be sure to identify the line and compute the curl. c. Give an example ofa vector fieldF (x,y, z) that has value 0 on a surface and such that curl F is nonzero everywhere. Be sure to identify the surface and compute the curl. 6. Find all points (a, b, c) on the sphere x 2 + y2 + z2 ~ R2 where the vector field F = yz'i + xz2j + 2xyzk is nonnal to the surface and F(a, b, c) '¢' O. 7. Find the mass of a spherical shell of radius R such that at each point (x, y, z) on the surface the mass density ~(x,y, z) is its distance to some fixed point (a, b, c) of the surface.
8. Find the mass of a helicoid
r(r, 8) 0:5
r:5
1,0:5
~
(r cos 8)1
(J:5
+ (r sin 8)j + Ok,
2w, if the density function is 8(x,y,z)
=
2Vx 2 + y2. See Practice Exercise 27 for a figure. 9. Among all rectango\ar regions 0 :5 x :5 a, 0 :5 Y :5 b, fmd the one for which the total outward flux of F = (x 2 + 4xy)1 - 6yj across the four sides is least. What is the least flux? 10. Find an equation for the plane through the origin such that the circulation of the flow field F = zi + xj + yk around the circle of intersection of the plane with the sphere x 2 + y2 + z2 = 4 is a maxinnun. 11. A string lies along the circle x 2 + y2 = 4 from (2, 0) to (0, 2) in
the fl!St quadrant. The density of the string is p (x,y)
= xy.
a. Partition the string into a fmite number of subarcs to show that the work done by gravity to move the string straight dowo to the x-axis is given by Work
~
lim
±
n_ OO i-I
g x.t)'.2.1s.
~
lc{ g xy2 ds,
where g is the gravitational constant.
13. Archimedes' principle If an object such as a hall is placed in a liquid, it will either sink to the hottom, float, or sink a certain distance and remain suspended in the liquid. Suppose a fluid has constant weight density w and that the fluid's surface coincides with the plane z = 4. A spherical hall remains suspended in the fluid and occopies the region x 2 + y2 + (z - 2)' :5 I.
a. Show that the surface integral giving the magnitude of the tota1 force on the hall due to the fluid's pressure is Force = lim ±w(4 - z.) .1er. = ({ w(4 - z) der.
12. A thin sheet lies along the portion of the plane x + y + z = 1 in the fl!St octant. The density of the sheet is ~ (x,y, z) = xy. a. Partition the sheet into a fmite numher of suhpieces to show that the work done by gravity to move the sheet straight dowo to the xy-plane is given by Work
~
lim
±
n_OO k=l
g x.t)'.z• .1er.
~
{{ g xyz der,
JJS
where g is the gravitational constant. h. Find the tota1 work done hy evaluating the surface integral in
part (a). c. Show that the total work done equals the work required to move the sheet's center of mass (x,y, z) straight dowo to the xy-plane.
JJs
n-OO.t_l
h. Since the ball is not moving, it is being held up by the buoyant force ofthe liquid. Show that the magnitude of the buoy-
ant force on the sphere is Buoyantforce
JJ
=
w(z - 4)k' n der,
s where n is the outer unit normal at (x, y, z). This illustrates Archimedes' principle that the magnitude of the buoyant force on a submerged solid equals the weight of the displaced fluid.
c. Use the Divergence Theorem to fmd the magnitude ofthe buoyant force in part (b). 14. Fluid force on a curved surface A cone in the shape ofthe stn'face z = Vx 2 + y2, 0 :5 z :5 2 is filled with a liquid of constant weight density w. Assuming the xy-plane is "ground level," show that the tota1 force on the portion of the cone from z = 1 to z ~ 2 due to liquid pressure is the surface integral F
~
JJ
w(2 - z)der.
s Evaluate the integral. 15. Faraday's Law If E(t,x,y,z) and B(t,x, y, z) represent the electric and magnetic fields at point (x,y, z) at time t, a hasic principle of electromagoetic theory says that V X E = -aB/at. In this expression V X E is computed with t held fixed and aB/at is calculated with (x, y, z) fIXed. Use Stokes' Theorem to derive Faraday~ Law,
f
h. Find the tota1 work done hy evaluating the line integral in
part (a). c. Show that the total work done equals the work required to move the string's center of mass (x, y) straight dowo to the x-axis.
987
E ' dr =
-f, JJ B'nder, s
c
where C represents a wire loup thrnugh which current flows counterclockwise with respect to the surface's unit nonna! n, giving rise to the voltage
f
E ' dr
c
around C. The surface integral on the right side of the equation is called the magnetic flux, and S is any oriented surface with boundaryC. 16. Let F=_GmM r 3 Irl be the gravitatioual force field defined for r '¢' O. Use Gauss's
Law in Section 16.8 to show that there is no continuously differentiable vector field H satisfying F = V X H.
988
Chapter 16: Integration in Vector Fields surface. Also, the magnitude du = Idul is the element of surface area (by Equation 5 in Section 16.5). Derive the identity
17. If j(x, y, z) and g(x, y, z) are continuously differentiable scalar functions defmed over the oriented surface S with boundary curve
e, prove that jJ(Vj
X Vg)'ndu
s
=
f
du = (EG - F 2)1/2 du dv where
jVg·dr.
c
E ~ Ir.1 2 ,
18. Suppose that V· F 1 = V· F 2 and V X F 1 = V X F 2 over a re-
gion D enclosed by the oriented surface S with outward unit normal n and that F 1 • n = F 2 • nonS. Prove that F 1 = F 2 through-
F ~ r.·r.,
and
G ~ Ir.1 2 •
21. Sbow that the volume V of a region D in space enclosed by the oriented surface S with outward normal n satisfies the identity
outDo 19. Prove or disprove that if V • F
~
0 and V X F
~
0, then F
~
O.
20. Let S be an oriented surface parame1rized by r(u, v). Defme the notation du = ru du X rv dv so thatdu is a vectornonnal to the
Chapter
•
v=tjJr.ndU,
s where r is the position vector of the point (x,y, z) inDo
Technology Application Projects
Mathematica/Maple Module: Work in ConservatiPe and NOlU:onse1'Vlltive Force Fields Explore integration over vector fields and experiment with conservative and nonconservative force functions along different paths in the field.
How CO" You VISualize Gree" 's Theo,.",? Explore integration over vector fields and use parame1rizations to compute line integrals. Both fonns of Green's Theorem are explored. VISualizing and I"terpreting the Diverge"ce Theorem VerilY the Divergence Theorem by fonnulating and evaluating certain divergence and surface integrals.
SECOND-ORDER DIFFERENTIAL EQUATIONS OVERVIEW In this chapter we extend oUI study of differential equations to those of second order. Second-order differential equations arise in many applications in the sciences and engineering. For instance, they can be applied to the study of vibrating springs and electric circuits. You willleam how to solve such differential equations by several methods in this chapter.
11III
Second-Order Linear Equations An equation of the form
P(x)y"(x) + Q(x)y'(x) + R(x)y(x)
=
G(x),
(1)
which is linear in y and its derivatives, is called a second-order linear differential equation. We assume that the fimctions P, Q, R, and G are continuous throughout some open interval 1. If G(x) is identically zero on I, the equation is said to be homogeneous; otherwise it is called nonhomogeneous. Therefore, the form of a second-order linear homogeneous differential equation is
P(x)y" + Q(x)y' + R(x)y
=
O.
(2)
We also assume thatP(x) is never zero for any x EI. Two fimdamental results are important to solving Equation (2). The first of these says that if we know two solutions Yl and yz of the linear homogeneous equation, then any linear combination y = cm + czyz is also a solution for any constants Ct and Cz.
THEOREM 1-The Superposition Prindple If Yt(x) and y,(x) are two solutions to the linear homogeneous equation (2), then for any constants Ct and cz, the fimction
y(x)
=
CtY1(x) + C2Yz(x)
is also a solution to Equation (2).
17-1
17-2
Chapter 17: Second-Order Oifferential Equations Proof Substituting Y into Equation (2), we have
P(x)y"
+ Q(x)y' + R(x)y = P(X)(CIYl + C2Y2)" + Q(X)(CIYl + C2Y2)' + R(X)(CIYI + C2Y2) = P(X)(CIY!" + C2Y2") + Q(X)(CIY1' + C2Y2') + R(X)(CIYl + C2YV = CI(P(X)YI" + Q(X)YI' + R(X)Yl) + C2(P(X)y2" + Q(X)Y2' + R(X)Y2) = 0, =
Cl(O)
Therefore, Y = CIYI
YI is a solution
=
0, Y2 is a solution
+ C2(0) = O. + C2Y2 is a solution ofEquation (2).
•
Theorem I immediately establishes the following facts concerning solutions tu the linear homogeneous equation.
+ Y2
1.
A sum of two solutions Yl C2 = I.)
2.
A constant multiple kYl of any solution Yl to Equation (2) is also a solution. (Choose Cl = k and C2 = 0.) The trivial solution y(x) = 0 is always a solution tu the linear homogeneous equation. (Choose CI = C2 = 0.)
3.
tu Equation (2) is also a solution. (Choose CI =
The second fundamental result about solutions to the linear homogeneous equation concerns its general solution or solution containing all solutions. This result says that there are two solutions Yl and Y2 such that any solution is some linear combination ofthem for suitable values of the constants Cl and C2. However, not just any pair of solutions will do. The solutions must be linearly independent, which means that neither YI nor Y2 is a constant multiple of the other. For example, the functions f(x) = eX and g(x) = xe x are linearly independent, whereas f(x) = x 2 and g(x) = 7x 2 are not (so they are linearly dependent). These results on linear independence and the following theorem are proved in more advanced courses.
THEOREM 2 If P, Q, and R are continuous over the open interval I and P(x) is never zero on I, then the linear homogeneous equation (2) has two linearly independent solutions YI and Y2 on T. Moreover, if YI and Y2 are any two linearly independent solutions of Equation (2), then the general solution is given by y(x) = CIY1(X)
+ C2Y2(X),
where Cl and C2 are arbitrary constants.
We now turn our attention tu rmding two linearly independent solutions tu the special case of Equation (2), where P, Q, andR are constant functions.
Constant-Coefficient Homogeneous Equations Suppose we wish to solve the second-order homogeneous differential equation
ay"+by'+cy=O,
(3)
17-3
17.1 Second-Order Linear Equations
where a, b, and c are constants. To solve Equation (3), we seek a function which when multiplied by a constant and added to a constant times its first derivative plus a constant times its second derivative sums identically to zero. One function that behaves this W
+ bre + ce rx TX
=
o.
Since the exponential function is never zero, we can divide this last equation through by en<. Thus, y = en< is a solution to Equation (3) if and ouly if r is a solution to the algebraic equation
ar 2 + br + c
=
O.
(4)
Equation (4) is called the auxiliary equation (or characteristic equation) ofthe differential equation ay" + by' + cy = O. The auxiliary equation is a quadratic equation with roots
-b
+ Vb 2 - 4ac 2a
and
'2 =
There are three cases to consider which depend on the value ofthe discriminant b 2 Case 1: b2
-
4ac.
4QC > O. In this case the auxiliary equation has two real and unequal roots e'" and Y2 = e'" are two linearly independent solutions to Equation (3) because e'" is not a constant multiple of e'" (see Exercise 61). From Theorem 2 we conclude the following result. -
rl and r2. Then YI =
THEOREM 3 If rl and r2 are two real and unequal roots to the auxiliary equation ar 2 + br + c = 0, then
is the general solution to ay"
EXAMPLE 1
+ by' + cy
= O.
Find the general solution of the differential equation
y" - y' - 6y
=
O.
Solution Substitution of y = en< into the differential equation yields the auxiliary equation which factors as (r - 3)(r
+ 2)
=
O.
The roots are r1 = 3 and r2 = - 2. Thus, the general solution is y =
clel<
+ C2e-2x.
•
17-4
Chapter 17: Second-Order Oifferential Equations
Case 2: b2 - 4ac = O. In this case rl = r2 = -bI2a. To simplify the notation, let r = -bI2a. Then we have one solution YI = e'" with 2ar + b = O. Since multiplication of e'" by a constant fails to produce a second linearly independent solution, suppose we try multiplying by ajUnction instead. The simplest such function would be u(x) = x, so let's see if Y2 = xe'" is also a solution. Substituting Y2 into the differential equation gives
a)l2" + by! +
a(2re'" + r 2xe"') + b(e'" + rxe"') + exe'" = (2ar + b)e'" + (ar 2 + br + c)xe'"
C)'2 =
= O(e"') + (O)xe'"
=
O.
The first term is zero because r = -bI2a; the second term is zero because r solves the auxiliary equation. The functions Y' = e'" and Y2 = xe'" are linearly independent (see Exercise 62). From Theorem 2 we conclude the following result.
THEOREM 4 ar2 + br + c
=
If r is the only (repeated) real root to the auxiliary equation 0, then
is the general solution to ay"
EXAMPLE 2
+ by' +
C)' =
o.
Find the general solution to
y" + 4y' + 4y
=
O.
Solution The auxiliary equation is
r 2 +4r+4=0, which factors into
(r
+ 2)2
=
O.
Thus, r = -2 is a double root. Therefore, the general solution is
y
=
c,e-:Ix +
C2 xe -:lx.
•
2
Case 3: b - 4ac < O. In this case the auxiliary equation has two complex roots rl = a + if3 and r2 = a - if3, where a and f3 are real numbers and i 2 = -I. (These real numbers are a = -bl2a and f3 = Y4ac - b 2/2a.) These two complex roots then give rise to two linearly independent solutions
y, = e(·+ip)< = e""(cos f3x + i sin f3x)
and Y2 = e(·-ip)< = e""(cos f3x - i sin f3x).
(The expressions involving the sine and cosine terms follow from Euler's identity in Section 9.9.) However, the solutions y, and Y2 are complex valued rather than real valued Nevertheless, because of the superposition principle (Theorem I), we can obtain from them the two real-valued solutions
0 Y3 -_21 Y, +21 Y2_ - " e " cos ",x
an
d
1_""'0 Y4 -_I2i Y, - 2i Y2 - e sm ",x.
The functions Y3 and Y4 are linearly independent (see Exercise 63). From Theorem 2 we conclude the following result.
17.1 Second-Order Linear Equations
17-5
THEOREM 5 If rl = a + if3 and r2 = a - if3 are two complex roots to the auxiliary equation ar 2 + br + c = 0, then
Y = is the general solution to ay"
EXAMPLE 3
.""(Cl
cos f3x
+ by' + cy
+ =
C2Sin
f3x)
o.
Find the general solution to the differential equation
y" - 4y'
+
5y = O.
Solution The auxiliary equation is
r 2 -4r+5=0. The roots are the complex pair r = (4 ± VI6 - 20)/2 or Thus, a = 2 and f3 = I give the general solution
rl =
2
+i
and r2 = 2 - i.
• Initial Value and Boundary Value Problems To determine a unique solution to a {lIllt-order linear differential equation, it was sufficient to specify the value ofthe solution at a single point. Since the general solution to a secondorder equation contains two arbitrary constants, it is necessary to specify two conditions. One way of doing this is to specify the value of the solution function and the value of its derivative at a single point: y(xo) = Yo and y'(xo) = YI. These conditions are called initial conditions. The following result is proved in more advanced texts and guarantees the existence of a unique solution for both homogeneous and nonhomogeneous second-order linear initial value problems.
THEOREM 6 If P, Q, R, and G are continuous throughout an open interval I, then there exists one and only one function y(x) satisfying both the differential equation P(x)y"(x)
+
Q(x)y'(x)
+ R(x)y(x)
= G(x)
on the interval I, and the initial conditions
y(xo) = Yo
and
y'(xo) = Yl
at the specified point Xo E I.
It is important to realize that any real values can be assigned to Yo and Yl and Theorem 6 applies. Here is an example of an initial value problem for a bomogeneous equation.
17-6
Chapter 17: Second-Order Oifferential Equations
EXAMPLE 4
Find the particular solution to the initial value problem
y(O) = I, y'(0) = -1.
y"-2y'+y=0, Solution The auxiliary equatioo is
r2
y
2r + I = (r - 1)2 = O.
-
The repeated real root is r = I, giving the general solutioo
Y =
CteX
+ C2 xex•
Then,
y' = cle x
-4
+ C2(X +
I)e x •
From the initial cooditioos we have -6 -8
FIGURE 17.1 Particularsolutioncurve for Example 4.
and Thus, C1 = I and C2 = -2. The unique solutioo satisfying the initial cooditioos is
y = eX - 2xe x • The solutioo curve is shown in Figure 17.1.
•
Another approach to determine the values of the two arhitrary constants in the general solutioo to a secood-order differential equatioo is to specify the values of the solutioo functioo at two different points in the interval 1. That is, we solve the differential equatioo subject to the boundary values and where Xl and X2 both beloog to 1. Here again the values for Yl and Y2 can be any real numbers. The differential equatioo together with specified boundary values is called a boundary value problem. Unlike the result stated in Theorem 6, boundary value problems do not always possess a solutioo or more than ooe solution may exist (see Exercise 65). These problems are studied in more advanced texts, but here is an example for which there is a unique solution.
EXAMPLE 5
Solve the boundary value problem
y"
+ 4y
=
0,
y(O) = 0, y(;;) = 1.
Solution The auxiliary equatioo is r 2 + 4 = 0, which has the complex roots r = ±2i. The general solution to the differential equation is
y = Cl cos 2x + C2 sin 2x. The boundary cooditioos are satisfied if
y(0)=c l ol+C20 0 =0
It follows that C1 = 0 and C2 = 2. The solution to the boundary value problem is
y = 2 sin2x.
•
17.1
17-7
Second-Order Linear Equations
EXERCISES 17.1 In Exercises 1-30, fmd the general solution of the given equation.
1. y' - y' - 12y
~
0
+ 3y'
~
0
3. y'
- 4y
2. 3y' - y'
4. y" - 9y
5.y·-4y=0 7. 2y" - y' - 3y = 0
19. y'
+ 9y = 0 + 25y = 0 - 2y' + 5y + 2y' + 4y + 4y' + 9y
21. y"
~
13. y' 15. y' 17. y'
43. 4y"
0
18.
~
0
20. 22.
23. dx2+4dx+4y=0 0
47.
+ 12y = 0 + 4y' + 5y = 0 y" + Y = 0 y" + 16y ~ 0 y" - 2y' + 3y ~ 0 4y" - 4y' + By ~ 0 y" + 8y' + 16y ~ 0 d~
d~ ~ - 12 dx2 dx
~
27.4dx2+4dx+y~0
28.4dx2-4dx+y~0
d~ 29. 9 dx 2
~ 6 dx
d~ ~ 30. 9 dx 2 - 12 dx
+Y
= 0
d~
~
58. 0
~
+ 4y
= 0
In Exercises 31-40, fmd the uoique solution of the second-order ioitial value problero.
+ 6y' + 5y ~ 0, y(O) ~ 0, y'(O) ~ 3 + 16y ~ 0, y(0) ~ 2, y'(O) ~ -2 y" + 12y ~ 0, y(0) ~ 0, y'(O) ~ I 12y' + 5y' - 2y = 0, y(O) = I, y'(O) = y" + 8y = 0, y(0) = -I, y'(O) = 2 y" + 4y' + 4y = 0, y(O) = 0, y'(O) = I y" - 4y' + 4y = 0, y(O) = I, y'(O) = 0 4y" - 4y' + Y = 0, y(O) = 4, y'(O) = 4
31. y" 32. y" 33. 34. 35. 36. 37. 38.
54. 4y'
+
16y'
+ 52y =
0
0
+ 2y' + y ~ 0, y(0) ~ I, y'(O) ~ I 4y" - 4y' + y ~ 0, y(O) ~ -I, y'(O) ~ 2 3y" + y' - 14y ~ 0, y(O) ~ 2, y'(O) ~ -I 4y" + 4y' + 5y ~ 0, y(,,') ~ I, Y'(1T) ~ 0
60.
61. Prove that the two solution fimctions io Theorem 3 are lioearly iodependent. 62. Prove that the two solution fimctions io Theorem 4 are lioearly iodependent. 63. Prove that the two solution fimctions io Theorem 5 are lioearly iodependent. 64. Prove that if Yl and Y2 are lioearly iodependent solutions to the homogeneous equation (2), then the fimctions Y3 = Yl + Y2 and y, ~ Yl - Y2 are also lioearly iodependent solutions. 65. a. Show that there is no solution to the bouodary value problero
-I
d~ ~ ~ 39.4 dx 2 + 12 + 9y = 0, y(0) = 2, dx(O) = I dx d~ ~ 40.9 dx 2 - 12 dx
44. 9y' + 12y' + 4y ~ 0 46. y' + 2y' + 2y = 0 48. 6y' + By' - 5y = 0 50. y" + 4y' + 6y = 0 52. 6y' - 5y' - 6y = 0
57. y" 59.
+ 9y
~
0
56. y" - 2y' + 2y = 0, y(0) = 0, y'(O) = 2
~
d~
+
+ 4y' + y ~ 0 + 20y = 0 25y" + lOy' + Y = 0 4y" + 4y' + 5y = 0 16y" - 24y' + 9y = 0 9y" + 24y' + 16y = 0
~
In Exercises 56-60, solve the ioitial value problero.
24. dx2-6dx+9y=0 26. 4
42. 6y' - y' - y
0
45. 4y"
53. 55. 6y" - 5y' - 4y
~
~
~
49. 51.
14. 16.
~
0
10. 3y" - 2Oy'
0
d~~ 25. dx 2 + 6 + 9y dx
41. y" - 2y' - 3y
12. y" ~
0
d~
~
In Exercises 41-55, fmd the general solution.
0
6.y"-64y=0 8. 9y" - y = 0
9. 8y" - lOy' - 3y = 0 11. y'
~
~ + 4y = 0, y(0) = -I, dx(O) = I
y" + 4y = 0, y(O) = 0, Y(1T) = 1. b. Show that there are iofmitely many solutions to the bouodary value problero
y" + 4y = 0, y(O) = 0, Y(1T) = O. 66. Show that if a, b, and c are positive constants, then all solutions of the homogeneous differential equation
ay"+by'+cy=O
approach zero as x ~
00.
17-8
Chapter 17: Second-Order Oifferential Equations
~
Nonhomogeneous Linear Equations
~I------=-------'-------In this section we study two methods for solving second-order linear nonhomogeneous differential equations with constant coefficients. These are the methods of undetermined coefficients and variation ofparameters. We begin by considering the form of the general
solution.
Form of the General Solution Suppose we wish to solve the nonhomogeneous equation ay"
+ by' + ey
(1)
= G(x),
where a, b, and c are constants and G is continuous over some open interval 1. Let Ye = CtYt + C2Y2 be the general solution to the associated complementary equation ay"
+ by' + cy
=
o.
(2)
(We learned how to find Ye in Section 17.1.) Now suppose we could somehow come up with a particular function YP that solves the nonhomogeneous equation (I). Then the sum y = Ye
+ YP
(3)
also solves the nonhomogeneous equation (I) because a(ye
+ Yp)" + b(ye + Yp)' + c(ye + Yp) = (aye" + bye' + eye) + (ayp" + byp' + eyp) y, solves Eq. (2) and YP sol"'s Eq. (t) = 0 + G(x) = G(x).
Moreover, if Y = y(x) is the general solution to the nonhomogeneous equation (I), it must have the form ofEquation (3). The reason for this last statement follows from the observation that for any function YP satisfying Equation (I), we have a(y - Yp)"
+ b(y = (ay"
+ c(y + by' + ey)
- Yp)'
= G(x) - G(x) =
- Yp) - (ayp"
+ byp' + eyp)
o.
Thus, Ye = Y - YP is the general solution to the homogeneous equation (2). We have established the following result.
THEOREM 7 The general solution y = y(x) to the nonhomogeneous differential equation (I) has the form y = Ye
+ YP'
where the complementary solution Ye is the general solution to the associated homogeneous equation (2) and Yp is any particular solution to the nonhomogeneous equation (I).
17.2
Nonhomogeneous Linear Equations
17-9
The Method of Undetennined Coeffidents This method for finding a particular solution YP to the nonhomogeneous equation (I) applies to special cases for which G(x) is a sum of terms of various polynomials p(x) multiplying an exponential with possibly sine or cosine factors. That is, G(x) is a sum of terms of the following forms:
P2(x)e UX cos f3x,
P3(x)e UX sin f3x.
For instance, 1 - x, e2x , xe x, cos x, and Sex - sin 2x represent functions in this category. (Essentially these are functions solving homogeneous linear differential equations with constant coefficients, but the equations may be of order higher than two.) We now present several examples illustrating the method.
EXAMPLE 1
Solve the nonhomogeneous equation y" - 2y' - 3y = I - x 2.
Solution The auxiliary equation for the complementary equation y" - 2y' - 3y = 0 is r2
-
2r - 3 = (r
+ I)(r - 3)
=
O.
It has the roots r = - I and r = 3 giving the complementary solution
Now G(x) = I - x 2 is a polynomial of degree 2. It would be reasonable to assume that a particular solution to the given nonhomogeneous equation is also a polynomial ofdegree 2 because if y is a polynomial of degree 2, then y" - 2y' - 3y is also a polynomial of degree 2. So we seek a particular solution of the form
yp=Ax 2 +Bx+C. We need to determine the unknown coefficients A, B, and C. When we substitote the polynomial yp and its derivatives into the given nonhomogeneous equation, we obtain
2A - 2(2Ax + B) - 3(Ax 2 + Bx + C) = I - x 2 or, collecting terms with like powers of x,
-3Ax 2
+ (-4A
- 3B)x
+ (2A -
2B - 3C) = I - x 2.
This last equation holds for all values of x if its two sides are identical polynomials of degree 2. Thus, we equate corresponding powers of x to get
-3A = -I,
-4A - 3B = 0,
and
2A - 2B - 3C = I.
These equations imply in turn that A = 1/3, B = -4/9, and C = 5/27. Substituting these values into the quadratic expression for our particular solution gives
By Theorem 7, the general solution to the nonhomogeneous equation is
_
y-Yc
+ yp-cle _ -. +
C2e
3.
+ :3I x 2 -94 x + 27' 5
•
17-10
Chapter 17: Second-Order Oifferential Equations
EXAMPLE 2
Find a particular solution of y" - y' = 2 sin x.
Solution If we try to find a particular solution of the fonn Yp=Asinx and substitute the derivatives of Yp in the given equation, we f"md that A must satisfy the equation
-A sin x
+ A cos x
= 2 sin x
for all values of x. Since this requires A to equal both -2 and 0 at the saroe time, we conclude that the nonhomogeneous differential equation has no solution of the form A sin x. It turns out that the required fonn is the sum
+ B cosx.
Yp = A sin x
The result of substitoting the derivatives of this new trial solution into the differential equation is
-A sin x - B cos x - (A cos x - B sin x) = 2 sin x or (B - A) sin x - (A
+ B) cos x
= 2 sinx.
This last equation must be an identity. Equating the coefficients for like terms on each side then gives
B-A=2
A
and
+B
=
O.
Simultaneous solution of these two equations gives A = - I and B = I. Our particu1ar solution is
YP
EXAMPLE 3
=
•
cosx - sinx.
Find a particular solution of y" - 3y'
+
2y = 5e".
Solution If we substitute
YP
=
Ae"
and its derivatives in the differential equation, we f"md that
Ae" - 3Ae" + 2Ae"
=
5e"
or
0= 5e". However, the exponential function is never zero. The trouble can be traced to the fact that
Y = e" is already a solution of the related homogeneous equation y" - 3y'
+
2y = O.
The auxiliary equation is
r2
-
3r
+2
=
(r - 1)(r - 2) = 0,
which has r = I as a root. So we would expect Ae" to become zero when substitoted into the left-hand side ofthe differential equation. The appropriate way to modify the trial solution in this case is to multiply Ae" by x. Thus, our new trial solution is
YP
=
Axe".
17.2
Nonhomogeneous Linear Equations
17-11
The result of substituting the derivatives ofthis new candidate into the differential equation is (Axe X + ZAe") - 3(Axe X + Ae") + ZAxe x = 5e x or Thus, A
=
-5 gives our sought-after particular solution YP
EXAMPLE 4
=
•
-5xe x. = e 3x •
Find a particular solution of y" - 6y' + 9y
Solutton The auxiliary equation for the complementary equation r2
-
6r
+9
=
(r - 3)2 = 0
has r = 3 as a repeated root. The appropriate cboice for Yp in this case is neither Ae 3x nor Axe 3x because the complementary solution contains both of those terms already. Thus, we choose a term containing the next higher power of x as a factor. When we substitute Ax2e 3x and its derivatives in the given differential equation, we get (9Ax 2e 3X + lZAxe 3x + ZAe~ - 6(3Ax 2e 3x + ZAxe 3") YP =
+ 9Ax 2e 3x
=
e 3x
or Thus, A
=
1/2, and the particular solution is YP =
tx2e~
•
When we wish to fmd a particular solution of Equation (1) and the function G(x) is the sum of two or more terms, we choose a trial function for each term in G(x) and add them.
EXAMPLE 5
Find the general solution to y" - y'
=
5e x
-
sin 2x.
Solutton We fITSt check the auxiliary equation r2
Its roots are r = 1 and r mogeneous equation is
=
-
r = O.
O. Therefore, the complementary solution to the associated hoYc
= cle
x
+
C2·
We now seek a particular solution Yp. That is, we seek a function that will produce 5e x - sin 2x when substituted into the left-hand side of the given differential equation. One part of YP is to produce 5e x , the other -sin 2x. Since any function ofthe form CteX is a solution ofthe associated homogeneous equation, we choose our trial solution YP to be the sum Yp = Axex
+ B cos 2x + C sin 2x,
x
including xe where we might otherwise have included only eX. When the derivatives of YP are substituted into the differential equation, the resulting equation is (Axe X + ZAe x - 4B cos 2x - 4C sin 2x) - (Axe X + Aex
-
2B sin 2x
+ 2C cos 2x)
=
5e x
-
sin 2x
17-12
Chapter 17: Second-Order Oifferential Equations or
Ae' - (4B + 2C) cos 2x + (2B - 4C) sin 2x
5e' - sin 2x.
=
TIris equation will hold if
A=5,
2B-4C=-I,
4B+2C=0,
or A = 5, B = -1/10, and C = 1/5. Our particular solution is I .2x I 2x+ Ssm. Yp - 5 xe' - TOcOS The general solution to the differential equation is
Y =Yc + Yp =
Cle'
+
C2
+ 5xe' -
l~cos2x
+ tsin2x.
•
You may find the following table helpful in solving the problems at the end of this section.
TABLE 17.1
The method of undetermined coefficients for selected equations of the form
oy" If G(x) has a term that is a constant multiple of ...
e'"
+ by' + cy
=
And if
G(x). Then include this expression in the trial function for Ypo
r is not a root of the auxiliary equation
Ae'"
r is a single root of the auxiliary equation
Axe'"
r is a double root of the auxiliary equation sin kx, cos kx
ki is not a root of the auxiliary equation
Bcoskx + Csinkx
px 2 +qx+m
ois not a root of the
Dx 2 +Ex+F
auxiliary equation
ois a single root of the
Dx'
+ Ex 2 + Fx
Dx'
+ Ex' + Fx 2
auxiliary equation
ois a double root of the auxiliary equation
The Method of Variation of Parameters TIris is a general method for finding a particular solution of the nonhomogeneous equation (I) once the general solution of the associated homogeneous equation is known. The method consists of replacing the constants c, and C2 in the complementary solution by functions VI = VI(X) and V2 = V2(X) and requiring (in a way to be explained) that the
17.2
Nonhomogeneous Linear Equations
17-13
resulting expression satisfy the nonhomogeneous equation (I). There are two functions to be detennined, and requiring that Equation (I) be satisfied is only one condition. As a second condition, we also require that
00
~~+~~=Q
Then we have
+
Y = VIYI
y' = Y" =
V2Y2,
+ V2Y2', VIYl" + V2Y2" + VIYl'
Vt'Yl'
+ V2'Y2'.
lfwe substitote these expressions into the left-hand side of Equation (I), we obtain vI(aYI"
+ bYI' +
CYI)
+
+
v2(aY2"
+ CY2) + a(vI'YI' + ViY2')
by2'
= G(x).
The first two parenthetical terms are zero since YI and Y2 are solutions of the associated homogeneous equation (2). So the nonhomogeneous equation (I) is satisfied if, in addition to Equation (4), we require that
+ v2'yi)
a(vI'YI'
= G(x).
(5)
Equations (4) and (5) can be solved together as a pair Vt'Yt
+
V2'Y2
= 0,
for the unknown functions VI' and~'. The usual procedure for solving this simple system is to use the method ofdeterminants (also known as Cramer ~ Rule), which will be demonstrated in the examples to follow. Once the derivative functions v,' and V2' are known, the two functions V, = v,(x) and ~ = V2(X) can be found by integration. Here is a summary ofthe method.
Variation of Parameters Procedure To use the method of variation of parameters to find a particular solution to the nonhomogeneous equation ay"
+ by' + cy
= G(x),
we can work directly with Equations (4) and (5). It is not necessary to rederive them. The steps are as follows.
1.
Solve the associated homogeneous equation ay"+by'+cy=O
2.
to tmd the functions YI and Y2. Solve the equations Vt'Yt
Vt'Yt'
3. 4.
+
Vl'Y2 =
+ 1l2'y2'
0, G(x)
=
-a-
simultaneously for the derivative functions VI' and V2'. Integrate VI' and V2' to find the functions VI = VI(X) and V2 = V2(X). Write down the particular solution to nonhomogeneous equation (I) as YP = Vl)'l
+
~Y2·
17-14 Chapter 17: Second-Order Oifferential Equations EXAMPLE 6
Find the general solution to the equation
y" + Y
= tanx.
Solution The solution of the homogeneous equation
y" + Y
=
0
= Cl COS X
+
Cz
is given by
Yc
sinx.
Since y,(x) = cos x and Y2(X) = sin x, the conditions to be satisfied in Equations (4) and (5) are Vt' -Vt'
+
Vl'
sinx = 0,
sinx +
V2'
cosx = tanx.
cos x
a
= 1
Solution of this system gives
Vt' =
0 sin x I tan x cosx I -c'--co-,,-x--S-in--'XC1
-smx
-sin2 x
-tan x sin x 2
cos
X
+
---
cosx·
2 8in X
cos X 1
Likewise,
I c~sx Vl' =
-smx
I c~sx -smx
~xl sinx cos x
I
=
sinx.
After integmting VI' and V2', we have VI(X) =
j
= =
and
V2(X) =
-Sin2 x
cos x dx
j
(sec x - cos x) dx
-In Isecx
j
+ tanxl + sinx,
sin x dx = -cos x.
Note that we have omitted the constants of integration in detennining VI and V2. They would merely be absorbed into the arbitrary constants in the complementary solution. Substitutiog V, and V2 into the expression for YP in Step 4 gives
+ tanxl + sin x] cos x + (-cos x) sin x (-cos x) In lsecx + tanxl.
YP = [-In Isecx =
The general solution is
Y = c, cos x
+ C2 sin x -
(cos x) In Isecx
+ tanxl.
•
17.2
EXAMPLE 7
Nonhomogeneous Linear Equations
17-15
Solve the nonhomogeneous equation
y" + y' - 2y
xe x.
=
Solution The auxiliary equation is r2
+r
+ 2)(r
- 2 = (r
- I) = 0
giving the complementary solution
Yc
Cte-2x
=
+
cze x .
The conditions to be satisfied in Equations (4) and (5) are Vt'e- 2x -2vt'e-lx
+
v2'e x
+
v2'e x = xe x.
=
0, a
= 1
Solving the above system for VI' and vi gives
Ix~x ::1 -_-xe 3e
2x _ -x - -
VI' = C-~e'--c2x;::--e~x:C-1 1 -2e-
2x
I
3 xe
3x
.
eX
Likewise,
Integrating to obtain the parameter functions, we have
Vj(x)
=
J-t
=
xe 3x dx
_t(X~3X
- Je;x dx )
_ I( )3x -27 1 - 3xe , and
Therefore,
The general solution to the differential equation is y =
Cte-lx
+
C2ex -
~xex +
k
x2eX ,
where the term (1/27)e X in YP has been absorbed into the term C2ex in the complementary solution. •
17-16 Chapter 17: Second-Order Oifferential Equations
EXERCISES 17.2 Solve the equations in Exercises 1-16 by the method of undetermined coefficients. 1. y" - 3y' - lOy = -3 2. y" - 3y' - lOy = 2>: - 3 4. y" + 2y' + Y
3. y" - y' = sinx 5. y"
+Y
= cos3x
7. y" - y' - 2y = 20 cos x
9. y" - y = eX
+x
2
+ 3y' + 2y
38. y"
8. y"
39. y" - 8y' =
= e'lx + Y = a+ 3ex + 2y' + Y = 6 sin 2x
+
e-2x -
dy,
d'y
+ 5 dx
13. dx'
dy
= 15x
14. dx' - dx = -8x
d'y dy 15. dx' - 3dx ~ e" - 12>:
d'y
16. dx'
dy
+ 7 dx ~
+3
42>:'
+ 5x +
+ y'
18. y"
+y
I
19. y"
+ Y = sinx + 2y' + Y =
21. y"
= x =
_,!!
tanx,
2
< x < '!!
25. y"
+Y
d'y 28. dx 2
-
=
= secx,
dy dx
=
= eX
24. y" - y = sinx
+ 4y' + Sy
d'y 27. dx'
+ 2y' + Y
22. y" - y = x
23. y" - y = eX
10
26. y" - y' 1f
=
2x
eXcosx, x> 0
+ Bxe~
+ y'
+ sinx, Yp = A cos x + B sinx 2 cosx + sinx, YP = Ax cosx + Bx sinx
- 2y = xe x , YP = Ax 2e x
33 - - = eX • dx' dx
+ e-r
d'y dy 35. dx,-4 -5y =ex +4
dx
_,!!
secxtanx,
=
+
2
< x < '!! 2
2y = eX - e'h
48. y'
+ 4y = x
49. y' - 3y = 5e"
SO. y'
+y
d'y 51. dx'
+Y
= sec
d'y 52. dx'
+y
= e"';
,1f
d'y
53. y"
+ y'
54. y"
+y
dy
34. - 4dx' dx
+ 4)'
= sinx
x,
1f
-2 < x < 2;
y(O) = 0, y'(O) =
y(O) = y'(O) = I 2
5
=
x' T
=
+Y +Y
a
4el{cosx - sinx),
2e x cos x, y(0)
57. y" - 2y' 58. y" - 2y'
x, y(0) = 0, y'(O) = 0
x, YP = 2 sin x + x, y(O) = 0, y'(O) =
ty" + y' + Y ~
YP =
~
~
0, y' (0)
~
I - 2>:, YP
=
2e x, YP = x 2e x, y(O) = 1, y'(O) = 0
=
x -leX, x
>
~
I
x - I, y(O)
~
0, y'(O)
~
1
0,
YP = xexlnx, y(1) = e, y'(I) =
a
In Exercises 59 and 60, two linearly independent solutions YI and Y2 are given to the associated homogeneous equation of the variable= 2e'"
d'y dy 9x 36. dx' - 9 = ge
dx
= x,
56. y" - y' - 2y
+ Bxe x
coefficients.
dy
44. y"
47. y' - 3y = e lO
YP
In Exercises 33-36, solve the given differential equations (8) by variation of parameters and (h) by the method of undetermined d'y
e~
-
The method of undetermined coefficients can sometimes be used to solve flISt-order ordinary differential equations. Use the method to solve the equations in Exercises 47-50.
55.
30. y" - y' = cosx
32. y"
+Y
2y' = x 2
isfYing the equation and the given initial conditions.
In each of Exercises 29-32, the given differential equation has a particular solution YP of the form given. Determine the coefficients in Ypo
31. y" + Y =
45. y"
42. y"
In Exercises 53-58, verify that the given function is a particular solution to the specified nonhomogeneous equation. Find the general solution and evaluate its arbitrary constants to f'md the unique solution sat-
1f
-2 < x < 2
Then solve the differential equation. 29. y" - 5y' = xe~. YP = Ax2e~
+
+ 4y = sinx + 4y' + 5y = x + 2 + 9y = 9x - cosX'
40. y"
e!b:"
Solve the differential equations in Exercises 51 and 52 su!!iect to the given initial conditions.
2
20. y"
e-x
43. y"
46. y" - 3y'
Solve the equations in Exercises 17-28 by variation of parameters. 17. y"
= escX',
41. y" - y' = x 3
x d'y
< x < '11' o<x<w
cotx, 0
=
6. y" + Y 10. y" e-x
=
+Y +Y
37. y"
= x2
11. y" - y' - 6y = e-x - 7 cosx 12. y"
Solve the differential equations in Exercises 37-46. Some of the equations can he solved by the method of undetermined coefficients, hut others cannot.
coefficient nonhomogeneous equation. Use the method of variation of parameters to fmd a particular solution to the nonhomogeneous equation. Assmne x > 0 in each exercise. 59. x7" + 2xy' - 2y = x 2, Yl = x- 2 , Y2 = x 60. x7" +xy'-y=x, Yt =x-1,n=x
17.3 Applications
II1II
17-17
AppLications In this section we apply second-order differential equations to the study of vibrating springs and electric circuits.
Vibrations A spring has its upper end fastened to a rigid support, as shown in Figure 17.2. An object of mass m is suspended from the spring and stretches it a length s when the spring comes to rest in an equilibrium position. According to Hooke's Law (Section 6.5), the tension force in the spring is ks, where k is the spring constant. The force due to gravity pulling down on the spring is mg, and equilibrium requires that massm
ks = mg.
at equilibrium
y
FIGURE 17.2
Massm stretches a spring by length s to the equilibrium position at y = O.
(1)
Suppose that the object is pulled down an additionsl amount Yo beyond the equilibrium position and then released We want to study the object's motion, that is, the vertical position of its center ofmass at any future time. Let y, with positive direction downward, denote the displacement position of the object away from the equilibrium position y = 0 at any time t after the motion has started. Then the forces acting on the object are (see Figure 17.3) the propulsion force due to gravity,
F p = mg,
Fs
= k(s
F,=ll
+ y),
the restoring force of the spring's tension,
dy , dt
a frictional force assumed proportional to velocity.
The frictional force tends to retard the motion ofthe object. The resultant of these forces is F = F p - F, - F" and by Newton's second law F = ma, we must then have
a'ly dy m dt 2 = mg - ks - ky - Il dt' By Equation (I), mg - ks = 0, so this last equation becomes
dly m dt 2
+
dy Il dt
+ ky
=
(2)
0,
subject to the initial conditions y(O) = Yo and y' (0) = O. (Here we use the prime notation
y
to denote differentiation with respect to time t.)
Yo
start
position y
FIGURE 17.3 Thepropulsion force (\Wight) F p pulls Ihe mass downward, but !he spring restoring force F, aod frictional force F, pullihe mass upward. The motion starts at Y = Yo with Ihe mass vihratiog up aod down.
You might expect that the motion predicted by Equation (2) will be oscillatory about the equilibrium position y = 0 and eventually dsmp to zero because of the retarding frictional force. This is indeed the case, and we will show how the constants m, Il, and k determine the nature of the dsmping. You will also see that if there is no friction (so Il = 0), then the object will simply oscillate indefinitely.
Simple Harmonic Motion Suppose first that there is no retarding frictional force. Then Il = 0 and there is no dsmping. Ifwe substitute w = Vkj;. to simplify our calculations, then the second-order equation (2) becomes
y"
+ l1ly
=
0,
with
y(O) = Yo
and
y'(O) = O.
17-18
Chapter 17: Second-Order Oifferential Equations
The auxiliary equation is ,2
+ if
=
0,
having the imaginary roots, = ±wi. The general solution to the differential equation in (2) is
Y =
Cl
cos lJJt
+ C2 sin wt.
(3)
To fit the initial conditions, we compute
y' =
-ClCd sin lJJt
+ C2(JJ cos wt
and then substitute the conditions. This yields c, = Yo and C2 =
o. The particular solution
y=yocoswt
(4)
describes the motion of the object. Equation (4) represents simple harmonic motion of amplitude Yo and period T = 27T/W. The general solution given by Equation (3) can be combined into a single term by using the trigonometric identity
sin(wt
+ cf»
+ sin wt cos cf>.
= cos wt sin cf>
To apply the identity, we take (see Figure 17.4)
c,
c,
=
C sin cf>
and
C2 =
C cos cf>,
where C2
FIGURE 17.4 c, = C sin 4> aod = C cos 4>.
C2
_
and
-1 Cl
cf> - tan
C2·
Then the general solution in Equation (3) can be written in the altenoative form
y = C sin (wt
+ cf».
(5)
Here C and cf> may he taken as two new arbitrary constants, replacing the two constants c, and C2. Equation (5) represents simple harmonic motion of amplitude C and period T = 27T/W. The angle wt + cf> is called the phase angle, and cf> may be interpreted as its initial value. A graph of the simple harmonic motion represented by Equation (5) is given in Figure 17.5.
y
Period
~T=7;:~
C Csin."
I
I
o -C y = C sin(." +.,,)
FIGURE 17.5 Simple hannooic motion of amplitude C aod period T with initial phase angle 4> (Eqoation 5).
17-19
17.3 Applications
Damped Motion Assume now that there is friction in the spring system, so [; oF O. If we substitute W = ~ and 2b = [;/m, then the differential equation (2) is y"
+
2by'
+ .ly
= O.
(6)
The auxiliary equation is r 2 +2br+w2 =O,
with roots r = -b ± yCb"2-_-w'''. Three cases now present themselves, depending upon the relative sizes of b and w. Case 1: b = Cd. The double root of the auxiliary equation is real and equals r general solution to Equation (6) is y = (c,
= w. The
+ c2t)e-".
This situation of motion is called critical damping and is not oscillatory. Figure 17.6a shows an example ofthis kind of damped motion. Case 2: b > Cd. The roots of the auxiliary equation are real and unequal, given by r, = -b + Vb 2 - w' and r2 = -b - Vb 2 - w'. The general solution to Equation (6) is given by y = c,e(-b+v'b'-w'Jt
+
C2 e (-b-v'b'-w'Jt.
Here again the motion is not oscillatory and both r, and r2 are negative. Thus y approaches zero as time goes on. This motion is referred to as overdamping (see Figure 17.6b). Case 3: b < Cd. The roots to the auxiliary equation are complex and given by r = -b ± tVw2 - b 2• The general solution to Equation (6) is given by y = e-bl(c, cosVw' - b 2 t
+ C2SinVw'
- b 2 t).
This situation, called underdamping, represents dam\,ed oscillatory motion. It is analogous to simple harmonic motion of period T = 21T/V w' - b 2except that the amplitude is not constant but damped by the factor e -bl. Therefore, the motion tends to zero as t increases, so the vibrations tend to die out as time goes on. Notice that the period T = 21T/Vw2 - b 2 is larger than the period To = 21T/W in the friction-free system. Moreover, the larger the value of b = [;/2m in the exponential damping factor, the more quickly the vibrations tend to become unnoticeable. A curve illustrating underdamped motion is shown in Figure 17.6c. y
y
y
t
o
y = e-t sin(St + 11/4)
y=(l +t)e-t (a) Critical damping
(b) Overdamping
(c) Underdamping
FIGURE 17.6 Three examples ofdamped vibra1xny motion for a spring system with
friction, so B # O.
17-20
Chapter 17: Second-Order Oifferential Equations An external force F(t) can also be added to the spring system modeled by Equation (2). The forcing function may represent an external disturbance on the system. For instance, if the equation models an automobile suspension system, the forcing function might represent periodic bumps or potholes in the road affecting the performance of the suspension system; or it might represent the effects of winds when modeling the vertical motion of a suspension bridge. Inclusion of a forcing function results in the second-order nonhomogeneous equation
d 2y
dy
m dt 2 + lJ dt + ky
= F(t).
(7)
We leave the study of such spring systems tu a more advanced course.
Elecbic Circuits The basic quantity in electricity is the charge q (analogous to the idea ofmass). In an electric field we use the flow of charge, or current I = dq/dt, as we might use velocity in a gravitational field. There are many similarities between motion in a gravitational field and the flow of electrons (the carriers of charge) in an electric field. Consider the electric circuit shown in Figure 17.7. It consists of four components: voltage source, resistor, inductor, and capacitor. Think of electrical flow as being like a fluid flow, where the voltage source is the pump and the resistor, inductor, and capacitor tend to block the flow. A battery or generator is an example of a source, producing a voltage that causes the current to flow through the circuit when the switch is closed. An electric light bulb or appliance would provide resistance. The inductance is due to a magnetic field that opposes any change in the current as it flows through a coil. The capacitance is normally created by two metal plates that alternate charges and thus reverse the current flow. The following symbols specify the quantities relevant to the circuit: q: charge at a cross section ofa conductor measured in coulombs (abbreviated c);
I: current or rate of change of charge dq/dt (flow of electrons) at a cross section of a conductor measured in amperes (abbreviated A);
E: electric (potential) source measured in volts (abbreviated V); V: difference in potential between two points along the conductor measured in volts (V). R, Resistor
Voltage E soun:e
L,lnd..-
C, Capacitor
FIGURE 17.7 An eleelric circuit. Ohm observed that the current I flowing through a resistor, caused by a potential difference across it, is (approximately) proportional to the potential difference (voltage drop). He named his constant of proportionality 1/R and called R the resistance. So Ohm ~ law is
1 I=li V,
17.3
Applications
17-21
Similarly, it is known from physics that the voltage drops across an inductor and a capacitor are
q C'
and
where L is the inductance and C is the capacitance (with q the charge on the capacitor). The German physicist Gustav R. Kirchhoff (1824-1887) formulated the law that the sum of the voltage drops in a closed circuit is equal to the supplied voltage E(t). Symbolically, this says that
dI RI + L dt
q
+C
=
E(t).
Since I = dq/dt, Kirchhoff's law becomes
d 2q dq I Ldt2+Rdt+Cq=E(t).
(8)
The second-order differential equation (8), which models an electric circuit, has exactly the same form as Equation (7) modeling vibratory motion. Both models can be solved using the methods developed in Section 17.2.
Summary The following chart summarizes our aualogies for the physics of motion of an object in a spring system versus the flow of charged particles in an electrical circuit.
linear Second-Order Constant-Coefficient Models Electrical System
Mechanical System
my" + 8y' + ley
=
F(t)
y: y':
displacement velocity
y": m:
acceleration mass
8: k: F(t):
damping constant spring constant forcing function
Lq"
+ Rq' + lq C
=
E(t)
q:
charge
q': q": L:
current change in current inductance resistance where C is the capacitance voltage source
R:
1/C:
E(t):
EXERCISES 17.3 1. A 16-lb weight is attached to the lower cod of a coil spring suspcoded from the ceiling and having a spring constaot of I lb/ft. The resistance in the spring-mass system is numerically equal to the instaotaoeous velocity. At t ~ 0 the weight is set in motion from a position 2 ft below its eqnilibrium position by giving it a downward velocity of2 ft/sec. Wrilc an initial value problem that models the givco situation.
2. An 8-lb weight stretches a spring 4 ft. The spring-mass system resides in a medium offering a resistance to the motion that is numerically equal to 1.5 times the instantaoeous velocity. If the weight is released at a position 2 ft above its equilibrium position with a downward velocity on ft/sec, write an initial value problem modeling the givco situation.
17-22
Chapter 17: Second-Order Oifferential Equations
3. A 2
circuit. Mechanical units in the British and metric systems may be helpful in doing the following problema. Unit
British System
MKSSystem
Distance
Feet (fi) Slugs Seconds (see) Pouods (lb) 32 ft/seC'
Meters (m) Kilograms (kg) Seconds (sec) Newtoos(N) 9.81 m/scC'
Mass
Time Force g(earth)
7. A 16-Ib weight is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of I Ib/ft. The resistance in the spring-mass system is numerica1ly equal to the instantaneous velocity. At t = a the weight is set in motion from a position 2 fi below its equi1ibrium position by giving it a downward velocity of 2 fi/see. At the cod of 'II' sec, determine whether the mass is above or below the equilibrium position and by what distance. 8. An 8-Ib weight stretches a spring 4 ft. The spring-mass system resides in a medium offering a n:sislmce to the motion equal to 1.5 times the instantaneous velocity. If the weight is released at a position 2 ft shove its equi1ibrium positioo with a downward velocity of 3 ft/see. find its position relative to the equilibrium positioo 2 sec later. 9. A 2
10. A mass of I slug is attached to a spring wbose eoostaot is 25/4 lb/ll. Initially the mass is released I fi above the equilibrium position with a d~ velocity of 3 fi/see. and the subsequent motion takes place in a medium that offers a dsmping force DUmerica1ly equal to 3 times the instantaneous velocity. An external force /(1) is driving the system, but assume that initially /(1) ~ O. Formulate and solve an initial value problem that models the given system. Interpret your results.
11. A I
12. A weight stretches a spring 6 in. It is set in motioo at a point 2 in. below its eqUIlibrium positioo with a ~ velocity of2 in./sec.
a. When does the weight return to its stsrting position? b. When does it reach its highest point'l
e. Show that the maximum velocity is 2v'2g+1 in./sec. 13. A weight of 10 Ib stretches a spring lOin. The weight is drawn down 2 in. below its equilibrium position and given an initial velocity of 4 in./sec. An identical spring has a different weight attached to it. This second weight is drawn down from its equilibrium position a distance equal to the amplitode ofthe fU"llt motion and then given an initial velocity of 2 fi/sec. If the amplitude of the second motion is twice that of the frrs!, what weight is attached to the second spring?
14. A weight stretches one spring 3 in. and a secood weight stretches another spring 9 in. If both weights are simultaneously pulled
down 1 in. below their respective equilibrium positions and then released, fmd the frrst tim. aller t = 0 when their velocities are equal.
15. A weight of 16 lb stretches a spring 4 ft. The weight is pulled down S fi below the equilibrium position and then released. What initial velocity vo given to the weight would have the effect of doubling the amplitude of the vibration?
16. A mass weighing 8 Jb stretches a spring 3 in. The spring-mass system resides in a medium with a damping constant of 2 llHec/ft. If the mass is released from its equib'brium position with a velocity of 4 in./see in the downward direction, fmd the time required for the mass to retum to its equilibrium position for the fU"llt time. 17. A weight suspended from a spring executes dsmped vibratioos with a period of2 sec. If the dsmping factor decreases by 90".4 in 10 sec, find the acceleration of the weight when it is 3 in. below its equilibrium positioo and is moving upward with a speed of2 ft/sec. 18. A I
17.4 Euler Equations 19. An LRC circuit is set up wi1h an inductance of 1/5 henry, a resistance of! ohm, and a capacitance of5/6 farad. Assuming 1he initial charge is 2 coolombs and 1he initial current is 4 amperes, fmd 1he solution fimction descnbing 1he charge on 1he capacitor at any time. What is 1he charge on 1he capacitor after a loog period oftime? 20. An (opoo) electrical circuit coosists of an inductor, a resistor, and a capacitor. There is an initial charge of2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is preseot but no external voltage is being applied. In 1his circuit 1he voltage drops at 1hree points are numerically related as follows: across the capacitor. 10 times the charge; across the resistor, 4 times the instantaneous change in the charge; and across the inductor, 2 times 1he instantaneous change in 1he current. Find 1he charge on the capacitor as a function of time. 21. A 16-lb weight stretches a spring 4 It This spring-mass system is in a medium wi1h a dsmping constant of 4.5Ib-sec/ft, and an external force giveo by /(t) = 4 + e- 2t (in pounds) is being applied. What is 1he solution fimction describing 1he position of 1he mass at any time if1he mass is released from 2 ft below 1he equilibrium position wi1h an initial velocity of 4 ft/ sec downward? 22. A IG-kg mass is a_hed to a spring hsving a spring coostant of 140 N/m. The mass is started in motion from 1he equilibrium position wi1h an initial velocity of I mf sec in 1he upward direction and wi1h an applied external force giveo by /(t) = 5 sin t (in newtons). The mass is in a viscous medium. with a coefficient of resistance eqoal to 90 N-sec/m. Formulate an initial value problem iliat models 1he givoo system; solve 1he model and intetpret 1he results. 23. A 2-kg mass is a_hed to 1he lower ood of a coil spring suspooded ftom 1he ceiling. The mass comes to rest in its equilibrium
~
17-23
position 1hereby stretcbing 1he spring 1.96 m. The mass is in a viscous medium. that offers a resistance in newtons numerically equal to 4 times the instantaneous velocity measmed in meters per second. The mass is 1hoo pulled down 2 m below its equilibrium position and released wi1h a downward velocity of3 mfsec. At this same instant an external force givoo by /(t) ~ 20 cos t (in newtons) is applied to 1he system. At 1he ood of 1T sec detennine if1he mass is above or below its equilibrium position and by how much. 24. An 8-lb weight stretches a spring 4 ft. The spring-mass system rt>sides in a medium offering a resistance to the motion equal to 1.5 times the instantaneous velocity, and an external force given by /(t) = 6 + e -, (in pounds) is being applied. If 1he weight is rt>leased at a position 2 ft above its equilibrium position wi1h downward velocity of 3 ft/sec, fmd its position relative to 1he equilibrium after 2 sec hsve elapsed. 25. Suppose L ~ 10 heorys, R ~ 10 ohms, C ~ 1/500 farads, E = 100 volts, q(O) = 10 coolombs, and q'(O) = i(O) = O. Formulate and solve an initial value problem iliat models 1he givoo LRC circuit. Intetpret your results. 26. A series circuit consisting of an inductor, a resistor, and a capacitor is opeo. There is an initial charge of2 coulombs on 1he capacitor, and 3 amperes ofcurrent is present in the circuit at the instant 1he circuit is closed. A voltage giveo by E(t) ~ 20 cos t is applied. In this circuit 1he voltage drops are numerically eqoal to 1he following: across the resistor to 4 times the instantaneous change in the charge, across the capacitor to 10 times the charge, and across the inductor to 2 times the instantaneous change in the current. Find 1he chsrge on 1he capacitor as a fimction of time. Determine 1he chsrge on 1he capacitor and 1he current at time t = 10.
Euler Equations
. . . . . .1 - - - - . . . . : . . . . - - - - - - - - In Section 17.1 we introduced the second-order linear homogeneous differential equation
P(x)y"(x)
+ Q(x)y'(x) + R(x)y(x)
= 0
and showed how to solve this equation when the coefficients P, Q, and R are constants. If the coefficients are not constant, we cannot generally solve this differential equation in tenns of elementary functions we have studied in calculus. In this section you will learn how to solve the equation when the coefficients have the special forms
P(x) = ax 2,
Q(x) = bx,
R(x) = c,
and
where a, b, and c are constants. These special types of equations are called Euler equations, in honor of Leonhard Euler who studied them and showed how to solve them. Such equations arise in the study of mechanical vibrations.
The General Solution of Euler Equations Consider the Euler equation
ax"y"
+ bxy' + cy
= 0,
x
>
o.
(I)
17-24
Chapter 17: Second-Order Oifferential Equations To solve Equation (I), we rlISt make the change of variables
z = Inx
y(x) = Y(z).
and
We next use the chain rule to find the derivatives y'(x) and y"(x):
y'(x) =
:x
Y(z) =
~Y(z):'
= Y' (z){
and
-~2 Y'(z) + !
y"(x) = !£y'(x) = !£Y'(z)! x = dx dx
x
x
Y"(z)dz = dx
-~2 Y'(z) + ~2 Y"(z). x
x
Substituting these two derivatives into the left-hand side of Equation (I), we rmd
ax"y"
+ bxy' +
:2
cy = ax 2 ( = aY"(z)
Y'(z)
+ (b
:2
+
Y"(Z))
- a)Y'(z)
+ bX({ Y'(Z)) +
cY(z)
+ cY(z).
Therefore, the substitutions give us the second-order linear differential equation with constant coefficients
aY"(z)
+ (b
- a)Y'(z)
+ cY(z)
=
o.
(2)
We can solve Equation (2) using the method of Section 17.1. That is, we find the roots to the associated auxiliary equation
ar 2
+ (b
- a)r
+C
= 0
(3)
to find the general solution for Y(z). After rmding Y(z), we can detennine y(x) from the substitutionz = Inx.
EXAMPLE 1
Find the general solution ofthe equation x 2y"
+
2xy' - 2y =
o.
Solution This is an Euler equation with a = I, b = 2, and C = -2. The auxiliary equation (3) for Y(z) is r2
+ (2
+ 2)
- I)r - 2 = (r - I)(r
=
0,
with roots r = -2 and r = I. The solution for Y(z) is given by
Y(z) = Ct e -2z
+ C2ez.
Substituting z = In x gives the general solution for y(x): y(x) = c,e-2lnx + c2e lnx = c,x-2 + C2X
EXAMPLE 2
Solve the Euler equation x"y" - 5xy'
+
•
9y = O.
Solution Since a = I, b = -5, and C = 9, the auxiliary equation (3) for Y(z) is r2
+ (-5
- I)r
+9
= (r -
3)2 = O.
The auxiliary equation has the double root r = 3 giving Y(z) = c,e 3z + c2ze3~ Substituting z = In x into this expression gives the general solution y(x) = c,e 3lnx + c2lnxe3lnx = CtX3 + C2x3lnx
•
17.4
EXAMPLE 3 Find the particular solution to x"y" - 3xy' initial conditions y(1) = 0 andy'(I) = 1. Solution
Euler Equations
+ 68y
17-25
= 0 that satisfies the
Here a = I, b = -3, and c = 68 substitoted into the auxiliary equation (3)
gives
r 2 -4r+68=0. The roots are r = 2
+
8i and r = 2 - 8i giving the solution
Y(z) = e 2z(c, cos Sz
+ C2 sin Sz).
Substitoting z = In x into this expression gives y(x) = e 21nx (c, cos (8 In x) From the initial condition y(l) = 0, we see that
c,
=
0 and
y(x) = C2x2 sin (8 In x).
y
To fit the second initial condition, we need the derivative
10
y'(x) = c2(8x cos (8 In x) 5
+ 2xsin(8lnx»).
Since y'(I) = I, we immediately obtain C2 = 1/8. Therefore, the particular solution satisfying both initial conditions is
o
2
10
y(x) =
2
-5
+ C2 sin (8 In x»).
t
x 2 sin (8 In x).
y= "ssin(Slnx)
Since -1 :5 sin (8 In x) :5 1, the solution satisfies -10
x2
-8 :5 y(x) FIGURE 17.8 Graph of the solution to Example 3.
:5
x2
8'
Agrapb of the solution is shown in Figure 17.8.
EXERCISES 17.4 In Exercises 1-24, rmd the general solution to the given Euler
equation. Assume x
>
0 throughout
1. x"y" + 2xy' - 2y ~ 0
2. x"y" + xy' - 4y ~ 0
3.x"y"-6y=0
4.x"y"+xy'-y=0
5. xV - 5xy'
+ 8y
~ 0
7. 3xV + 4xy' = 0 9. x"y" - xy' + y = 0 11. x"y" - xy' + 5y = 0
+ 3xy' + lOy ~ 0 15. 4xV + 8xy' + 5y = 0 17. x"y" + 3xy' + y ~ 0 19. xV + xy' ~ 0 13. x"y"
+ 7xy' + 2y ~ 0 8. xV + 6xy' + 4y = 0 10. x"y" - xy' + 2y = 0 12. x"y" + 7xy' + 13y = 0 14. x"y" - 5xy' + lOy ~ 0 16. 4x"y' - 4xy' + 5y = 0 18. x"y" - 3xy' + 9y ~ 0 20. 4x"y" + y ~ 0 6. 2xV
+ 15xy' + y ~ 0 22. 16xV - 8xy' + 9y = 0 23. 16xV + 56xy' + 25y ~ 0 24. 4x"y' - 16xy' + 25y = 0 21. 9x"y'
In Exercises 25-30, solve the given initial value problem.
25. xV 26. 6x"y' 27. xV
+ 3xy' - 3y ~ 0, y(l) ~ 1, y'(I) ~ -I + 7xy' - 2y = 0, y(1) = 0, y'(I) = I - xy' + y = 0, y(l) = 1, y'(I) = I
28. xV + 7xy' + 9y = 0, y(l) = 1, y'(I) = 0 29. xV - xy' + 2y = 0, y(1) = -1, y'(I) = 1 30. xV + 3xy' + 5y = 0, y(l) = 1, y'(I) = 0
•
17-26
Chapter 17: Second-Order Oifferential Equations
~
Power-Series SoLutions
. . . .1 - - - - - - - - - In this section we exteod our stody of second-order linear homogeoeous equations with variable coefficieots. With the Euler equations in Section 17.4, the power of the variable x in the nonconstant coefficieot had to match the order of the derivative with which it was paired: x 2 with yO, Xl with y', and xO(=I) with y. Here we drop that requiremeot so we ean solve more geoeral equations.
Method of Solution The powe.....erie. method for solving a second-order homogeneous differential equation consists of finding the coefficieots of a power series 00
y(x) = ~ c.x· = Co
+ c'x + C2x2 + ...
(1)
n=O
which solves the equation. To apply the method we substitote the series and its derivatives into the differential equation to determine the coefficieots Co, C" C2, .... The technique for finding the coefficieots is similar to that used in the method of uodetermined coefficieots preseoted in Section 17.2. In our first example we demonstrate the method in the setting of a simple equation whose general solution we already know. This is to help you become more comfortable with solutions expressed in series form.
EXAMPLE 1
Solve the equation y"
+Y
=
0 by the power-series method.
Solution We assume the series solution takes the form of
and calculate the derivatives 00
y'
=
~
00
ncll x n -
1
y"
and
= ~
11=1
n(n - 1)c.x·-2 •
11=2
Substitotion of these forms into the second-order equation gives us 00
00
~ n(n - l)c.x·- 2
11-2
+
~ c.x· =
,,-0
o.
Next, we equate the coefficients of each power of x to zero as summarized in the followiog table.
Powerofx
Coefficient Equation
xO
2(I)C2
+ Co
=
0
or
I C2 = -Ico
x'
3(2)C3
+ CI
=
0
or
I C3 = -3-2 c,
x2
4(3)C4
+ C2
=
0
or
x3
5(4)C5
+ C3
=
0
or
x4
6(5)C6
+ C4
=
0
or
I C4 = -4'3 C2 I C5 = -5'4 C3 I C6 = -6'5 C4
+ C.-2
=
0
or
en =
X,,-2
n(n - I)c.
I n(n - I) C.-2
17.5 Power-Series Solutions
17-27
From the table we notice that the coefficients with even indices (n = Zk, k = I, Z, 3, ... ) are related to each other and the coefficients with odd indices (n = Zk + I) are also interrelated. We treat each group in turn. Even indices: Here n = Zk, so the power is x 2k--2. From the last line ofthe table, we have
Zk(Zk - I)cu
+ CU-2
=
0
or
Cu
=
I Zk(Zk _ I) CU-2·
From this recursive relation we find
Cu
= [
Zk(Z1-
(-1)1 (Zk)!
=
1)][
(Zk - ZJ(Zk -
3)]-" [- 4(~)][-!]co
co·
Odd indices: Here n = Zk line of the table yields (Zk
+
+
I, so the power is XU-I. Substituting this into the last
1)(Zk)cu+ I
+ CU-I
=
0
or
CU+ I =
(Zk
I
+ I )(Zk) CU-I·
Thus,
CU+I =
=
[
(Zk+ \)(Zk)] [ (Zk -
(_1)1 (Zk + I)!
1)~Zk -
zJ.. ht4)] htz) Je,
CI·
Writing the power series by grouping its even and odd powers together and substituting for the coefficients yields
00
=
~
00
C2kx2k
k-O 00
t=t,
_ '" - Co
+ ~ C2k+l X2H1 1-0
(-1)1 (Zk)! X U
00
(_1)1
+ CI t=t, '" (Zk + I)! x 2.<+1.
From Table 9.1 in Section 9.10, we see that the 1mt series on the right-hand side ofthe last equation represents the cosine function and the second series represents the sine. Thus, the general solution to y" + Y = 0 is
y = Co cos x
+ Ct sinx.
•
17-28
Chapter 17: Second-Order Oifferential Equations
EXAMPLE 2
+ ~' + Y
Find the general solution to y"
o.
=
Solution We assume the series solution form 00
y = ~CnXll n=O
and calculate the derivatives 00
y'
L ncn x
=
.-1
ll
00
t
-
y"
and
=
Ln(n - l)c.x·- 2•
.-2
Substitution of these forms into the second-order equation yields 00
00
~ n(n - 1)cnx n- 2
+
n=2
00
L ncnxn + L cnx
11=1
= o.
ll
11=0
We equate the coefficients of each power of x to zero as summarized in the following table.
Powerofx
CoeDicleotEquation
+ Co 3(2)C3 + Cl + CI 4(3)C4 + 2C2 + C2 5(4)cs + 3C3 + C3 6(5)C6 + 4C4 + C4 2(I)c2
x'
(n
+ 2)(n + l)c.+2 + (n + l)c.
= 0
0 = 0 = 0 = 0 =
=
0
= = = = =
or or or
C3
or
Cs
or
C6
or
Cn+2 =
C2
C4
-!co -let -lc2
-lC3 -~C4 -
I
n + 2 en
From the table notice that the coefficients with even indices are interrelated and the coefficients with odd indices are also interrelated. Even indices: Here n = 2k - 2, so the power is x 2k- 2• From the last line in the table,
we have I C2k = - 2k C2k-2· From this recurrence relation we obtain
= (2)(4)(6) ... (2k)
co·
Odd indices: Here n = 2k - 1, so the poweris x 2k- I • From the last line in the table,
we have C2k+l = -
2k
1
+
1 C2k-I·
From this recurrence relation we obtain C2k+l =
=
(-2k
~ 1)(-2k ~
(-I)k (3)(5)··· (2k
+ I) Cl.
J.. (-t)(
-t)Cl
17.5 Power-Series Solutions
17-29
Writing the power series by grouping its even and odd powers and substituting for the coefficients yields
Y
=
00
00
1=0
1=0
L C2kx2k + L C2k+ 1X 2k+ 1 (-I)k
00
- 0&'~0 (2)(4)"'(2k) X -
EXAMPLE 3
C
2k
+C
(-I)k
00
1&'~0 (3)(5)"'(2k + 1) X
2k+ 1
•
.
Find the general solution to (1 -
X
2
Ixl <
)y" - 6xy' - 4y = 0,
1.
Solution Notice that the leading coefficient is zero when x = ± 1. Thus, we assume the solution interval I: -1 < x < 1. Substitution ofthe series form
and its derivatives gives us
(1 -
X
2)
00
L
n=2 00
L
n=2
n(n - 1)cnx n- 2 - 6
00
n(n - 1)cnx n- 2 -
L
00
L
00
ncnx n - 4
n=l
n=2
L
cnx n = 0,
n=O
00
n(n - 1)cnx n - 6
L 00
ncnx n - 4
n=l
L
cnx n = O.
n=O
Next, we equate the coefficients of each power of x to zero as summarized in the following table.
Coefficient Equation
Powerofx
- 4co = 0 - 6(1 )Cl - 4Cl = 0
or
4(3)C4 - 2(I)c2 - 6(2)C2 - 4C2 = 0 5(4)cs - 3(2)C3 - 6(3)C3 - 4C3 = 0
or
2(I)c2 3(2)C3
xn
(n
+ 2)(n + I)Cn+2 - [n(n - 1) + 6n + 4]cn = (n + 2)(n + l)cn+2 - (n + 4)(n + l)cn =
or
or
0 0
or
Cn+2
n+4 = n + 2 en
Again we notice that the coefficients with even indices are interrelated and those with odd indices are interrelated Even indices: Here n = 2k - 2, so the power is x 2k• From the right-hand column and last line ofthe table, we get
=
(2k
= (k
+
2k
2) (~)
+ l)co.
2k-2
2) ... &(1)
(2k 2k-4
4 2 Co
17-30
Chapter 17: Second-Order Oifferential Equations Odd indices: Here n = 2k - I, so the power is x 2k+1. The right-hand column and last line of the table gives us
=
=
(2k + 3)(~)(~) 2k+1 2k-1 2k-3
... l(i) 53 Cl
2k + 3 --3- C1 •
The general solution is 00
Y
=
L cnx" .-0
=
L C2kx2k + ~ C2k+ 1x 2k+ 1
00
00
k~O
k~O
00
= CO~
(k
00
~O
EXAMPLE 4
2k+3
•
+ l)x 2k + CI~ -3-X2k+1. ~O
Find the general solution to y" - 2>y'
+Y
=
o.
Solution Assuming that 00
y =
:L CnX", 11=0
substitution into the differential equation gives us 00
00
~ n(n - l)c.x·- 2
-
n=2
00
2 ~ nc.x· + ~ c.x· n=1
=
o.
11=0
We next determine the coefficients, listiog them in the following table.
Powerofx
Coefficient Equation
Xl
2(I)c2
+
3(2)C3 - 2cI 4(3)C4 - 402 5(4)C5 - 6C3
x·
(n
+ 2)(n + l)c.+2 -
I
0
or
C2 = -2co
+ Cl
= 0
or
+ C2
=
0
or
=3 2 c1 3 c4=4 3 c2
Co =
+ C3
=
0
(2n - I)c. = 0
c3
I 0
0
5
or
cS=5 4 c3
or
7 c6=6 5 c4
0
0
2n - I
or
C.+2 =
(n
+ 2)(n + I) c.
17.5 Power-Series Solutions
17-31
From the recursive relation
2n - I
+ 2)(n +
en +2 = (n
I) en>
we write out the first few terms of each series for the general solution: Y =
eO(1 - lx 2
2
lx' -
-
4'
~! x
+ e, (x +
3
n6! x• -
+ ;, x5 +
i~ x7 +
EXERCISES 17.5 In Exercises 1-18, use power series to f"md!he general solution of!he
differential equation.
1. y"
2. y"
+ 2y' ~ 0 + 2y' + y ~
0
11. (x 2
-
2
-
14. y" -
-
+ 2xy' x"y
- 2y ~ 0
~ 0
l)y" - 6y ~ 0
+ 2)y' + 2y = 0 l)y" + 4'>;y' + 2y = 0 2xy' + 4y = 0
15. y" -2xy' +3y=O
6.y"-'>;y'+y=O
7. (I + x)y" - y = 0 8. (I - x 2)y" - 4'>;y' + 6y
l)y"
+ y'
13. (x
+ 2y = 0 2xy' + 2y = 0
4. y" - 3y'
xV -
-
10. y"
12. .>;y" - (x
3.y"+4y~O
5.
9. (x 2
16. (I - x 2)y" - .>;y'
=0
17. y" - xy' + 3y 18.
xV -
4'>;y'
=
+ 6y
+ 4y 0 ~ 0
... )
= 0
... ).
•
ApPENDICES Real Numbers and the Real Line
A.1
This section reviews real numbers, inequalities, intervals, and absolute values.
Real Numbers Much of calculus is based on properties of the real number system. Real numbers are numbers that can be expressed as decimals, such as
- 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
c. IRe(z) I :::; Izl IZI + z21 2 = IZll2 + IZ212 + 2Re(zlz2) e. IZI + z21 :::; IZII + IZ21
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.
d.
a. z + Z = 2Re(z) b. z -
z=
A.8
AP-35
2iIm(z)
The Distributive Law for Vector Cross Products In this appendix we prove the Distributive Law u X (v
+ w)
==
+u
u X v
X w,
which is Property 2 in Section 12.4.
Proof To derive the Distributive Law, we construct u X v a new way. We draw u and v from the common point 0 and construct a plane M perpendicular to u at 0 (Figure A.30). We then project v orthogonally onto M, yielding a vector v' with length IvIsin (). We rotate v' 90° about u in the positive sense to produce a vector v" . Finally, we multiply v" by the length of u. The resulting vector Iu Iv" is equal to u X v since v" has the same direction as u X v by its construction (Figure A.30) and
Iu II v" I == Iu II v' I == Iu II vIsin ()
==
Iu
X
vI·
M'
M
FIGURE A.30
As explained in the text, U X
v = IU Iv" .
Now each of these three operations, namely,
1.
projection onto M
2.
rotation about u through 90°
3.
multiplication by the scalar Iu I
when applied to a triangle whose plane is not parallel to u, will produce another triangle. If we start with the triangle whose sides are v, w, and v + w (Figure A.31) and apply these three steps, we successively obtain the following:
+ w) satisfying the vector equation v' + w' == (v + W)' 2. A triangle whose sides are v", w" , and (v + w)" satisfying the vector equation v" + w" == (v + w)" 1.
A triangle whose sides are v', w' , and (v
I
AP-36
Appendices
(the double prime on each vector has the same meaning as in Figure A.30)
M
FIGURE A.31 The vectors, v, w, v + w, and their projections onto a plane perpendicular to u.
3. A triangle whose sides are lulv",lulw", andlul(v + w)" satisfying the vectorequation
+
lulv"
lulw" = lul(v
+
w)".
Substituting lulv" = u X v, lulw" = u X w, and lul(v from our discussion above into this last equation gives
+u
u X v
X w = u X (v
+ w)"
= u X (v
+ w)
+ w),
•
which is the law we wanted to establish.
A.9
The Mixed Derivative Theorem and the Increment Theorem This appendix derives the Mixed Derivative Theorem (Theorem 2, Section 14.3) and the Increment Theorem for Functions ofTwo Variables (Theorem 3, Section 14.3). Euler rlISt published the Mixed Derivative Theorem in 1734, in a series of papers he wrote on hydrodynamics.
THEOREM 2-Tbe Mixed Derivative Theorem If I(x, y) and its partial derivatives Ix. Iy,I"", and Iyx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then
I",,(a, b)
=
Iyx(a, b).
Proof The equality of 1",,( a, b) and Iyx( a, b) can be established hy four applications of the Mean Value Theorem (Theorem 4, Section 4.2). By hypothesis, the point (a, b) lies in the interior of a rectangle R in the xy-plane on which I, Ix. Iy,I"", and Iyx are all defined. We let h and k be the numbers such that the point (a + h, b + k) also lies in R, and we consider the difference
a
=
F(a + h) - F(a),
(1)
I(x, b + k) - I(x, b).
(2)
where
F(x)
=
Appendix 9 The Mixed Derivative Theorem and the Increment Theorem
AP-37
We apply the Mean Value Theorem to F, which is continuous because it is differentiable. Then Equation (1) becomes
a
=
hF'(e,),
(3)
+ h. From Equation (2), F'(x) = Ix(x,b + k) - Ix(x,b),
where e, lies between a and a
so Equation (3) becomes
a
=
h[f.(e1, b + k) - Ix(e1, b)].
(4)
Now we apply the Mean Value Theorem to the functiong(y) = f.(e"y) and have
g(b
+ k) - g(b)
kg'(d,),
=
or
Ix(e1, b + k) - Ix(e1, b) for some d, between b and b
=
kl"1(e1, d,)
+ k. By substituting this into Equation (4), we get
a
=
hkl"1(e1, d,)
(5)
for some point (e1, d,) in the rectangle R' whose vertices are the four points (a, b), (a + h, b), (a + h,b + k), and (a, b + k). (See FigureA.32.) By substituting from Equation (2) into Equation (1), we may also write
y
R
•
(a, b)
a=
R'
k
= =
h
I(a + h,b + k) - I(a + h,b) - I(a,b + k) + I(a,b) [f(a + h,b + k) - I(a,b + k)] - [/(a + h,b) - I(a,b)] cf>(b + k) - cf>(b),
(6)
where -0;;+------------> X
FIGURE A.32 The key to proving fry(a, b) = f",(a, b) is that no matter how small R' is, fry and f", tske on equal values somewhere ioside R' (although not necessarily at the same point).
cf>(y)
=
I(a + h,y) - I(a,y).
(7)
The Mean Value Theorem applied to Equation (6) now gives
a for some d2 between b and b
=
k¢'(d2 )
(8)
+ k. By Equation (7),
cf>'(y)
=
/y{a + h,y) - Iy(a,y).
(9)
Substitoting from Equation (9) into Equation (8) gives
a=
k[fy(a + h, d2) - Iy(a, d2)].
Finally, we apply the Mean Value Theorem to the expression in brackets and get (10)
for some e2 between a and a + h. Together, Equations (5) and (10) show that
1"1(e1, d,)
=
1",(e2, d2),
(11)
where (e1, d,) and (e2, d2) both lie in the rectangle R' (Figure A.32). Equation (11) is not quite the result we want, since it says only that 1"1 has the same value at (e1, d,) that Iyx has at (e2, d2). The numbers h and k in our discussion, however, may be made as small as we wish. The hypothesis that 1"1 and Iyx are both continuous at (a, b) means that 1"1(e1,d,) = 1"1(a,b) + €,and/yx(e2,d2) = Iyx(a,b) + €2,whereeachof€1,€2--->Oas both h, k---> O. Hence, if we let hand k---> 0, we have 1"1(a, b) = Iyx(a, b). • The eqnality of I "1( a, b) and I yx( a, b) can be proved with hypotheses weaker than the ones we assumed. For example, it is enough for I, Ix> and I y to exist in R and for 1"1 to be continuous at (a, b). Then Iyx will exist at (a, b) and eqnal 1"1 at that point.
AP-38
Appendices
THEOREM 3-The Increment Theorem for Functions of Two variables Suppose that the first partial derivatives of f(x, y) are defined throughout an open region R containing the point (xo, YO) and that fx and fy are continuous at (xo, yo). Then the change
=
f(xo + At,yO + ay) - f(xo,Yo)
in the value of f that results from moving from (xo, YO) to another point (xo + At,yO + ay) inR satisfies an equation of the fonn
=
in which each of Elo
fx(xO,yo)ax + fy(xo,yo)ay + ElAt + E2ay E2 --->
0 as both At, o.
qxo + ax, Yo + ay)
,~~. B(xo
/
1
+ ax, Yo)
Proof We work within a rectangle T centered at A(xo, YO) and lying within R, and we assume that At and
+
At, Yo) - f(xo, Yo)
is the change in the value of f from A to B and T
=
f(xo + At,yO +
is the change in the value of f from B to C (Figure A.34). On the closed interval ofx-values joining xo to xo + At, the function F(x) = f(x,yo) is a differentiable (and hence continuous) function of x, with derivative
F' (x)
=
fx(x, Yo) .
By the Mean Value Theorem (Theorem 4, Section 4.2), there is an x-value c between xo and xo + At at which
F(xo +
=
F'(c) At
or
f(xo +
fx(c,Yo)
=
or
(12)
Similarly, G(y) = f(xo + At,y) is a differentiable (and hence continuous) function ofy on the closed y-interval joining Yo and Yo +
G'(y)
=
fixo + At,y).
Hence, there is ay-value dbetween Yo and Yo
G(yO
+
+
=
G'(d)
or
f(xo +
=
fixo +
or
(13)
AP-39
Appendix 9 The Mixed Derivative Theorem and the Increment Theorem
,
1
,,)--,
,' Q. :=f(~Y)~PO
,'
-----____
pi1
1
,
,
0
1
,
1
,
1
•
1
1
1
1 1
4.1:1
.4.1:
4: 1
'~., 1
::
1__ 1
,, "~
,, Yo
Q
l i p '
1A(.%j).yo)
1
I
::
: :
1
'lli -m s
, ,
Bi
1
~
,
Y
1
: , ,
yo+.4y
I
~,C(",+"-,,y,+.y)
FIGURE "-34 Part ofthc surface z = f(x,y) near Po(Xo, Yo, f(xo,)'o)). The points Po, P', and p w have the same height Zo - f(xo,)'o) abave the .xy-plane. The change in z is 4z - P'S. The change
4z1 = f(xo
+ Ax, Yo) - f(xo, Yo),
shawnasPWQ - P'Q', is ClIllSed by cbangingx from.xotoxo holdingyequal to)'o. Then, with x held equal toX(l + .:ix,
+ Axwbile
... - f(", + "-',Yo + ,,"y) - f(", + "-',Yo) is the change in z caused by changing Yo from Yo + liy, wbicb. is represented by Q'S. The total changeinzis the sum. of 4z1 and 4zz.
Now, 118 both dx and !i.y- 0, we know that c -+ XO and d -+ Yo. Therefore, since h and h are continuous at (xo,Yo) , the quantities E1 ~
f,(e,Yo) - f'("",Yo),
E, ~ j,(XO + """, d) -
(14)
f,(x"Yo)
both approach zero as both dx and !i.y-+ O.
Finally, !i.z
=
!i.z1
+ !i.zz
+ f,(x, + """, d),,"y [f,("",Yo) + Ell""" + [f,(x"Yo) + E,I""Y fJ...xo,yo)dx + fy(xo,yo)!i.y + E 1!i.x + Ez!i.y,
~ j,(e,Yo)"""
=
FromEqa. (12) IIId (13) From Eq. (14)
where both El and EZ - 0 as both!i.x and!i.y -+ 0, which is what we set out to prove.
•
Analogous results hold for functions of any finite number of independent variables. Suppose that the first partial derivatives ofw - f(x,Y, z) are def'med throughout an open region containing the point (Xo. Yo. zo) and that f,;, fy, and fz are continuous at (Xo. Yo. zo). Then !i.w
f(xo + !i.x,Yo + !i.y,zO + !i.z) - f(xo,yo,zo) = f,;!i.x + fy!i.y + fzIM. + l!"l!i.x + EZ!i.y + E3IM., =
(15)
AP-40
Appendices
where £1, E2, £3 ~ 0 as .6.x, .6.y, and t:.z ~ O. The partial derivatives fx> fy, fz in Equation (15) are to be evaluated at the point (xo, Yo, zo) . Equation (15) can be proved by treating aw as the sum oftbree increments,
aWl
=
aW2 aW3
= =
f(xo + ax,yo,zo) - f(xo,yo,zo) f(xo + ax,yO + ay,zo) - f(xo + ax,yO,zo) f(xo + ax,yO + ay,zO + az) - f(xo + ax,yO
(16) (17)
+ ay,zo),
(18)
and applying the Mean Value Theorem to each ofthese separately. Two coordinates remain constant and only one varies in each of these partial increments aWl, aW2, aW3. In Equation (17), for example, only y varies, since x is held equal to xo + ax and z is held equal to zoo Since f(xo + ax,y, zo) is a continuous function ofy with a derivative fy, it is subject to the Mean Value Theorem, and we have
aW2 for some Yl between Yo and Yo
=
+ ay.
fy(xo + ax,yl,Zo) ay
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
A-50
Chapter 12: Answers to Odd-Numbered Exercises 23. The vector v is horizontal and 1 in. long. The vectors u and w
:~ in. loog.
CHAPTER 12
are
Section 12.1, pp. 663-664
horizootal. All vectors must be drawn to scale.
1. The line through the point (2, 3, 0) parallel to the z-axis 3. The x-axis 5. The circle x 2 + y2 = 4 in the xy-plane 2 7. The circle x + z2 = 4 in the xz-plane 9. The circle y2 + z2 = I in the yz-plane 11. The circle x 2 + y2 = 16 in the xy-plane 13. The ellipse formed by the intersection of the cylinder x 2 + y2 ~ 4 and the plane z ~ y 15. The parabola y = x 2 in the xy-plane 17.
39.
< I
51.
(x - 1)2 + (y - 2)2 + (z - 3)' = 14
&-t)'
53. (x + I)' +
55. C(-2,0,2),a = VB 59.
+ (z + t)' = 57.
~~
c(-i,-i,-i).a = 5~
(b) v'x 2 + z2
(0)
63. Y ~ I (b) (0, 5, -5)
25. 3(ti+ tj - t k ) 29.
9. (I, -4) 15. (21. 31
(b)
11. (-2, -3)
1,-t)
+ 5j - 8k
3.
(b)
(t, (-t, 1)
7.
~
••__
27.5(k)
~ (VJI - VJj -VJk)
31.
~k
3
(0) 10 j +
(b) -V3k
33. 17 (121 - 5k) 3 35.
(b) <1/2,3,5/2)
VJI - VJj - VJk
39. A(4, -3, 5)
3 2' b
41. a =
=
G,H)
(b)
2I
43. "'(-338.095,725.046) 1()() cos 45° 45. IFd = sin 75' '" 73.205 N,
IF21
~ I~;~~~O'
'" 89.658 N,
FI ~ (-IFd cos 30', IFd sin 30') '" (-63.397,36.603), F 2 = (IF2 Icos45',IF2 Isin45') '" (63.397,63.397) 100 sin 75' 47. w ~ cos 40' '" 126.093 N,
ws:;;~'
IFd =
51.
v505
-w
'" 106.933 N
(t,
5r)
(b) (5 cos 60' + 10 cos 315', 5 sin 60' + IOsin315')
Section 12.2, pp. 672-674
5.
w
~~
49.
(0) v'x 2 + y2
+6
(b) 3v'i3
u+v+w
(
1.
(b)
•
I 41. 3 43. 7 45. 2V3 47. C( -2, 0, 2), a = 2V2 49. C( V2, V2, - V2), a ~ V2
w is vertical and u makes a 45' angle with the
17. -31 + 2j - k
14) 5
=
5 + 1OV2 5V3 - 1OV2)
2
tl t +
'
2
j - 3k
(b) 1+ j - 2k
(0) (2,2, I)
v'iO
(b)~
19. -31 + 16j
Section 12.3, pp. 680-682 1.
(b) -I
3.
(b) t
5.
\/34, V3
1\
(5j - 3k)
(b)
(0) -5 (0)
t
V3~ 3
34
~ (101 + (0). ~ v34
llj - 2k)
Chapter 12: Answers to Odd-Numbered Exercises
+ vTI, V26, V21
7. (a) 10 (0) 10
(b) 10
9.
+ vTI
~
11.
,
V546 vTI 10 +26 (5i + j)
A-51
(d) 26 9. 0.75 rad 11. 1.77 rad 13. AngleatA = cos-1
G) '"
B
~ cos-1
C
~ cos-1 (~)
(~)
63.435degrees,aogleat
'"
53.130 degrees, aogle at '"
13.
63.435 degrees.
23. Horizootal compooeot '" 1188 fi/sec, vertical componeot: '" 167 fi/sec 25. (a) Since Icos 81 '" I,wehavelu'vl = lullvllcos81 '" lullvl(1) = lullvl· (b) We have equality precisely wheo Icos 81 ~ I or wheo ooe or lxrth ofu and v are O. In the case of nonzero vectors, we have equality wheo 8 ~ 0 or '1T , that is, wheo the vectors are parallel. 27. a 35. -2x + y = -3 33.x+2y=4
,
,
17. (a)
-21+ J i+2j
-~-z;--~of----*~-~,
z .x+2y=4
37. x+y=-I
,
P(l,2)
1'(-2, 1)
,
, -I -
x+)'=-1 -1
41. 5J
43. 3464 J
~
'1T
4
3,directiooisti +
45.
t
49. IfA
2J
47.
tj
'1T
6
+ tk;lv x ul
47. ~
~
~
ali + azj aod B
hli + hzj, theo
i
j
al
a2
hi
b.,
k 0 0
A
X
BI ~ ±tI::
: 1·
The applicable sign is (+ ) ifthe acute angle from A to B runs conoterclockwise in the xy-plane, and (- ) if it runs clockwise.
3,
7.IUXVI~6V5,directinois~i- Vsk;lvxUI~6V5, V5
1
vp
tl
- ~k
+2- k
(I) lu l :
aod the triangle's area is
49. 0.14
3 3 3 3. lu x vi = 0, no directino; Iv X ul = 0, no directino 5. lu X vi ~ 6, directioo is -k; Iv X ul ~ 6, directioo is k
·· I. directLonlS--1 V5
(i - j)
(e) (u X v) X (u X w)
AXB=
45.
directioois-~i _Ij
Vz
19. 8 21. 7 23. (a) None (b) u aod w 25. 1OV3 fi-Ib 27. (a) True (b) Not always true (0) True (d) True (e) Not always true (f) True (g) True (b) True u·v 29. (a) projyu= v'vv (b) ±uXv (0) ±(uXv)Xw
I-J
Section 12.4, pp. 686-688 1. lu x vi
(b) ±
l 31. (a) Yes (b) No (0) Yes (d) No 33. No, v need not equal w. For example, i + j '" -i + j, but i X (i + j) = i X i + i X j = 0 + k = k and i X (-i + j) = -i X i + i X j = 0 + k = k. • ,.;;:;;, 11 25 43. T 35. 2 37. 13 39. v 129 41. T
39. 2x - y = 0
-z
+ j + k)
v6
(d) I(u X v)'wl
-3
,
'?
(b) ± _1_ (21
-2;+y .. -3
,
•
0
15. (a) 2v6
Section 12.5, pp. 694-696 1. 3. 5. 7.
x=3+1, y=-4+1, z=-I+I x = -2 + 51, y = 51, z = 3 - 51 x=O, y=21, z=1 x = I, y = I, z = I + I 9. x = I, Y = -7 + 21, z = 21 11. x = t, Y = 0, z = 0
A-52
Chapter 12: Answers to Odd-Numbered Exercises
13. X=4
Z=~I,
y=I,
z = 0,
OSlsl
,
,,,(ull ,,, _ _->" :, // '_~
(0,0.0)
+ I,
15. x = I, Y = I -}
S t
17.
4i'."'.4i'.,. ,
19.
,
SO
-7t'-----r---+1 /
1
/ /
/
,
J"-:c-=-'~--
,
/
_____ J/
21.
23.
25.
27.
-,
, 17. x
= 0,
Z
= I,
=I a SIS
y
= Z - 24 y = ZI, z=Z-ZI, OSISI
21,
19. x
I
, <0.-1,1)
M2),//1
«(1,1,1)
:
,,
----L----;<''--------'---+1
,
--;<''----='=(0,"!~'":0),-->1
,
21. 3x - Zy - z = -3 23. 7x - 5y - 4z = 6 25. x + 3y + 4z = 34 27. (I, Z, 3), -ZOX + IZy + z = 7 29. Y + z = 3 31. x - Y + z = a 33. zv'3ii 35. a 1
37. 9'(j2 47. -rr/4
39. 3
41. 19/5
49. 1.38 nul
43. 5/3
51. 0.8Z rad
53.
45. 9/V41
G, -~,
t)
55. (I, 1, 0) 57. x = 1 - I, Y ~ 1 + I, z ~ -I 59. x = 4, Y = 3 + 64 z = I + 31 61. LI interseetsLZ; LZ is parallel toL3; LI and L3 are skew. 63. x ~ Z + 21, y = -4 - t, z ~ 7 + 3t; x ~ -Z - 4 y = -Z + (I/Z)I, z = 1 - (3/Z)1
65.
,
29.
31.
i'-"- y'= I
(0,-t,-~).(-I,O,-3),(I,-I,O)
69. Many possible answ=. One possibility: x + y = 3 and Zy + z = 7. 71. (x/a) + (Y/b) + (z/e) ~ I descn1les all planes excepl those
,
through the origin or paral1ello a coordinate axis.
section 12.6, pp. 70G-701 1. (d), ellipsoid 7. (b), cylinder
r
13.' by'.<
33.
35.
3. (a), cylinder 5. (I), hyperbolic paraboloid 9. (k), byperbolieparaboloid 11. (b), cone
IS.
, 1
,
Chapter 12: Answers to Odd-Numbered Exercises
37.
23. 2v7 25. (a) vT4 (b) I 29. v7s/3 31. x = I - 3t, y = 2, z = 3 + 7t 33. \/2 35.2x+y+z=5 37. -9x + y + 7z = 4
39.
39. y
41.
43. 45. 47. 49.
43.
2'IT(9 - e 2 ) 9
(b) S'IT
(e) 4'ITabc
(o,-t,-~),(-I,O,-3),(I,-I,O)
41. 'IT/3
x = -5 + 5t, Y = 3 - t, z = -3t (b) x = -12t, Y = 19/12 + 1St, z = 1/6 + 6t Yes; v is parallel to the plaoe. 3 51. -3j + 3k
53.
JJs
55.
ei,296,_~)
y
45. (a)
(5i - j - 3k)
+ 3t,
57. (I, -2, -I); x ~ I - 5t, Y 59.2x+7y+2z+IO=O 61. (a) No (b) No (e) No 63. ll/Y107
~
65.
4~+41+r: .. 4
x2+r+?=4
,
67.
-2
(d) No
3
z
~
-I
(e) Yes
,
2
Practice Exerdses, pp. 702-703
Vi3i3
1. (a) (-17,32)
(b) (b) 10
3. (a) (6, -S) 5. (7.
'?'-
t) [assuming counterclockwise1
(~,- In)
9• Length =
~'--' . I . I. 2 , uu
11. v ('IT/2) = 2(-i) 13 • Length
=
~'--' . 2.1 -'p 3. 7,UUl:lctionlS"7
+ '1' 6k
15. _S_i - _2_ j + _S_k
v'33
17. Ivl = uX
v'33 \/2, lui =
v~ 2i -
lulcos8 19.
~(2i
v'33
3,v'u = u·v = 3,v X u = -2i + 2j - k,
2j + k,lvx
ul~ 3,8 ~ cos-I (Jz) ~*,
~ ~,projvU ~ ~(i +
j)
+j - k)
21.uxv=k
I
x (i+J)=k
---'f---+---y
Additional and Advanced Exercises, pp. 704-706 1. (26,23,-1/3)
3.IFI=20Ib
5. (a) IFd = SO Ib, IF21 = 60 Ib, F, = (-4S,64), I.J
A-53
F2 = (4S, 36), a =
tan-I~, f3
= tan-I
t
+ 4t
A-54
Chapter 13: Answers to Odd-Numbered Exercises
(b)
2400 IFd = 13 ""
1000 IF21 = 13 ""
184.615Ib,
/ -12,000 28,800) F, = \ 169 '169
76.9231b,
9. (a) 0 = tan-I \/2 "" 54.74'
"" (-71.006,170.414),
v'i4
v(l) =
(b) 0 = tan-I 2\/2 "" 70.53'
+ 2x =
(e) 2x - y
13. (b) _6_
8
" 32. 23 • 13 k I ,. 41' + 41 1 - 41
17. (a) 0,0 (b) -IOi - 2j + 6k, -9i - 2j + 7k (e) -4i - 6j + 2k, i - 2j - 4k (d) -IOi - 10k, -12i - 4j - 8k 19. The fonnula is always true.
CHAPTER 13 Section 13.1, pp. 713-715
~X2,
V
17. 1r/2
= -1,
z
=1+
21. x = t, Y
t
= "3I t,
z = t
23. (a) (i): It bas constant speed I (ii): Yes (iii): Counterclockwise (iv): Yes (b) (i): It has constant speed 2 (ii): Yes (iii): Counterclockwise (iv): Yes (e) (i): It bas constant speed I (ii): Yes (iii): Counterclockwise (iv): It starts at (0, -I) instead of (I, 0) (d) (i): It bas constant speed I (ii): Yes (iii): Clockwise (iv): Yes (e) (i): It bas variable speed (ii): No (iii): Counterclockwise (iv): Yes 25. v ~ 2VSi + VSj
a=2j 1. (1/4)i + 7j + (3/2)k
a = 3i + 8j
" 1r v =\/2. \/2. ,. t =4: T'-T I , a = -\/222 " - \/222 1",
t = 1r/2: v = -j,
15. 1r/2
Section 13.2, pp. 720-724
v=i+2j, = 3i + 4j,
2tj + tk; a = (
V6(~i + ~j + ~k)
19. x = t, Y
(d) x - 2y + z = 3 + 5V6 and x - 2y + z = 3 - 5V6
1. y=x 2 -2x,
(~I)i + t +
-2 2)i + 2j + k; (t + I) .f06dire· I, 2. I ction: V6' + V6 1 + V6 k ; speed: v 0;
/ 12,000 5000) \169' 169 "" (71.006, 29.586)
F2 =
3. y =
13. v =
a = -i
3. (1r + ;\/2} + 2k
5. (ln4)i + (ln4)j + (ln2)k 7 e-I.+e-l.+ k 2' e 1 . 9• ,• - 1·+1r 4k
y
11. r(t) =
(-;>
+ I); +
(-;>
+ 2)j +
(-;>
+ 3)k
13. r(t) = «t + 1)3/2 - I)i + (-e- t + I)j + (In(t + I) + I)k
15. r(t) = 8ti + 8tj + (-16t 2 + loo)k 2
17. r(t) = (tt + .Jut +
= 'IT: v
7. t
= 2i,
a = -j;
t =
3:: v = i - i, a
+(l
t 2 + _2_ t + 3)k = 2 v'il +(i + 2j + 3k)
=-i
y
r,. (I-!in 1)1 + (1- 008 t)J
o 9. v
~i
• + 2tj + 2k; a
~
2j; speed: 3; direction: t i + t j +
tk;v(l) = 3(t i + tj + t k ) 11. v ~ (-2 sin t)i + (3 cos t)j + 4k; a = (-2 cos t)i - (3 sint)j; speed: 2VS; direction: ( -1/VS)i + (2/VS)k; v(1r/2) = 2VS[( -1/VS)i + (2/VS)k]
(I
2
2
(tt + Jut-2}
t2 +
~)(3i v'il
j + k)
50 sec (a) 72.2 sec; 25,510 m (b) 4020 m (e) 6378 m (a) Vo "" 9.9 m/sec (b) a "" 18.4' or 71.6' 39.3' or 50.7' (b) Vo would bisect LAOR. (a) (Assuming that "x" is zero at the point of impact) r(t) = (x(t))i + (y(t))j, where x(t) = (35 cos 27')t and y(t) = 4 + (35 sin 27')t - 16t 2 • (b) At t "" 0.497 sec, it reacbes its maximum beigbt of about 7.945 ft. (e) Range"" 37.45 ft; fligbt time "" 1.201 sec (d) At t "" 0.254 and t "" 0.740 sec, when it is"" 29.554 and "" 14.396 ft from wbere it will land (e) Yes. It changes things because the ball won't clear the net. 35. 4.00 ft, 7.80 ft/ sec 19. 21. 23. 25. 31. 33.
2
I}-
Chapter 13: Answers to Odd-Numbered Exercises ~
43. (a) r(l) x(l)
(x(l))i
+ (y(l))j; where
9. T ~ 3c~sli - 3S~lj + ~k, N ~ (-sinl)i - (COSI)j,
~ (0.~8)(I- e-
17.6) and
K ~
y(l) = 3 + (152)(1 - e-
+(~)(I 2
(b)
(c) (d) (e)
- 0.081 - e-0.08l) 0.08 At I '" 1.527 sec it reaches a maximum height ofabout 41.893 feet. Range'" 351.734 ft; flight time '" 3.181 sec At I '" 0.877 and 2.190 sec, when it is about 106.028 and 251.530 ft from home plate No
Section 13.3, pp. 727-728 1. T
I
V1+t
i+
•
VI k,52 V1+t 3
(Sin!1+1 + teast). I
• I.
N
~
(-costv2- Sint). + (-Sintv2+ cost).I,
~
I
K=--
e'v2 13. T=
.~i+ .~j, 2 2 V
K
1 + I
V
1 + I
Ij
.~ 2
N=
V
1 + I
I
= ---cc-'---c= 2 1(1 + 1)'/2
(tanh ~)j, ~ (-tanh~} + (sech~)j, ~)1+
2i = lsech a a
21.
(x - 'I)' +
y2 = I
25. K(X) = Isinxl/(I + cos2 x)'/2
section 13.5, pp. 738-739 ~
51, L
1. a
5'IT T
3V1
• I.
(COSI~Sinl),
1
23. K(X) = 2/(1 + 4x2)'/2
I
11. s(l)
(cos I.vzSin I} +
K
+ (v2I /2)k,"'> + 1+1 2 'IT 9. (0, 5, 24'IT)
~
19. 1/(2b)
(cost1+1 - Ismt).1 +
=
11. T
N
5. T = -cos Ij + sin Ik, ~ 7 T
3
25
15. T = (seCh
~ (-~Sinl} + (~COSI)i + ~k,3'IT
3. T=
A-55
13. s(l) = v3e' - v3, L = -4-
15. v2 + In(1 + v2)
17. (a) Cylinderisx 2 + y2 ~ I ,piane isx + z ~ I. (b) and (c)
~
laiN
3. a(l)
~
4 2v'S IT + -3-N
7 r(1!) ~ v2 i + v2 1· _ k,T(1!) 42 2 4 •
N(~") = -
5. a(O)
~_
~ 2N
v2 i + v2 1· 2 2'
,?i - '?j,B(f) = k;oscu1atingplane:
z = -I; nonna! plane: -x + y = 0; rectifYing plane:
x+y~ v2 9• B = (icos')i - (iSin')j - l k , T =-~ 5 5 5 25 11. B = k, T = 0 13. B = -k, T = 0 15. B = k, T = 0 17. Yes. If the car is moving on a curved path (K # 0), then tIN ~ Klvl 2 # o and a # O.
\0.,1,1)
,
(d)L~
J.
2~
0
2
VI+sin ldl
t,p
(e) L '" 7.64
Section 13.4, pp. 733-734 1. T
~
(cosl)i - (sinl)j,
3 T ~
I
.
v'J+t2 I
j
v'J+t2 '
K
i -
~
N
~
(-sinl)i - (COSI)j, N ~
j
I
v'J+t2 '
-I
v'J+t2
I
2( v'J+t2)'
5. (b) cosx 7. (b) N
~
-2e 21 VI + 4e
(0) N
4t
i
+
I
VI + 4e 4t
~ -t(v'4=7i + Ij)
j
i-
K ~
cos I
23. K = =I 29. Components of v: -1.8701,0.7089,1.0000 Components of a: -1.6960, - 2.0307 , 0 Speed: 2.2361; Components ofT: -0.8364,0.3170,0.4472 Components ofN: -0.4143, -0.8998, -0.1369 Components ofB: 0.3590, -0.2998, 0.8839; Curvature: 0.5060 Torsion: 0.2813; Tangential component ofacceleration: 0.7746 Nonna! component ofacceleration: 2.5298 31. Components ofv: 2.0000, 0, -0.1629 Components of a: 0, -1.0000, -0.0086; Speed: 2.0066 Components ofT: 0.9967, 0, -0.0812 Components ofN: -0.0007, -1.0000, -0.0086 Components ofB: -0.0812, 0.0086, 0.9967; Curvature: 0.2484 Torsion: 0.0411; Tangential component of acceleration: 0.0007 NomJal component ofacceleration: 1.0000
A-56
Chapter 14: Answers to Odd-Numbered Exercises
Section 13.6, p. 742
(0) V(I)
1. v ~ (3a sin Oju, + 3a(1 - cos O)ne a = 9a(2 cos 0 - I)u, + (18a sin O)ne 3. v = 2ae a6u r + 2ea9u6 a = 4e a8(a 2 - I)u, + 8ae a8u, 5. v ~ (-8sin41)U, + (4cos41)ne a = (-40 cos 41)u, - (32 sin 41)0,
d2r
-~
dl 2
x2
y2
2
gbl Va 2 + b 2
T;
bg T+a ( 2bgt 2 )2N Va2 + b2 a +b
There is no component in the direction ofB. 5. (a) :
= ;'cosO - r8sinO,
." (b) d dt =rXCOSu
Practice Exerdses, pp. 743-744 1. 16 +
~
d:
= ;'sinO
,,"dO + YSInu,r dt
+ r8cosO
"" " = -xStnu +. ycosu
7. (a) aO) ~ -9u, - 6u" vO) ~ -0,. + 30, (b) 6.5 in. 9. (0) v = ;'0,. + r8u, + zk, a = (r - riP)u, + (rii + 2;'6)0, + ilk
~ I y
2
.Jjt-=~ 1
o
-4
«0) •
.(0)
x
CHAPTER 14
-1
Section 14.1, pp. 753-755 Atl
= O:aT = a,,,," = 4,K = 2;
At t
=
1t
4: aT
=
7
3' tlN
=
4\/2 -3-, K
=
4\/2 ------rr-
3. Ivl~ ~ I 5. K ~ 1/5 7. dy/dl ~ -x; clockwise 11. Shot put is 00 the grouml, about 66 ft 3 in. from the stopboard. 15. Leogth
~ f~1+ ~ + In(f + ~I + ~)
17. T(O) =
~I - ~i + tk;
B(O)
3\/2
19. T(102) = • b l + • ~i; vl7 vl7 B(102) 21. a(O) 23. T
y
1/1
y_~ /.f(I,d
///,'
,,/ ,.',-r3
3\/2
K ~ \/2. 7 3'
~-~",,""+--'"'''------->!x
~1 6
N(102) = - • ~I + • bi; vl7 vl7
I
~7"'---+----~x
9. Domain: all points (x,y)
satisfyingx 2
17W
~ ( .vz cos r) - (sin I)i + ( .vz cost)k;
_1-k;K = _1_;7 = 0 \/2 \/2 \/2 25. 1t/3 27. x ~ I + I, Y ~ I, z ~ -I
31. K ~ I/a
11. Dnmain: all points (x, y) for whicb (x - 2)(x + 2)(y - 3)(y + 3) '" 0 3 y=3
Additional and Advanced Exerdses, pp. 745-746 1. (a) dJdO
I
'0=211'
1tgb ~ 2~ a+b 2 2
gbl2 (b) 0 = 2(a 2 + b 2)'
Z
gb 212 = 2(a2 + b 2)
--_2,j---+---f,;-2-~x J=-3
-3
/
/
/1<~,-I)
= lOT + 6N
= _1_ 1 -
" ",/'/
l"o"'I_ _
= k;K = _8_;7 = 0
N = (- .vzSinl} - (cosl)i - (.vzSinl)k; B
y
N(O) = .vzl + .vzi;
~ _ _1_ 1 + _I- i + _4-k; 3\/2
1. (a) 0 (b) 0 (0) 58 (d) 33 3. (a) 4/5 (b) 8/5 (0) 3 (d) 0 5. Domain: all points (x,Y) on 7. Domain: all points (x,Y) or above line y = x + 2 not lying 00 the graph of y=xory=x 3
""
~.I
-I";;y
,,;;x 2
+I
Chapter 14: Answers to Odd-Numbered Exercises
13.
15.
39. (a)
A-57
(b)
y
x+y-l=c ~_,*,~~~~K:'~k~·%
,
-,
03
1
o
,
-1
-, -3
17. (a) (0) (e) 19. (a) (0)
(d) (f) 21. (a) (0)
23.
25.
27.
29.
31. 37.
(d) (f) (a) (0) (d) (e) (a) (0) (d) (e) (a) (b) (0) (d) (e) (a) (b) (0) (d) (e) (f) (a)
All points in the xy-plane (b) Allreals The lines y - x ~ c (d) No boundary points Both opeo and closed (f) Unbounded (b) z '" 0 All points in the xy-plane Farj(x,y) = 0, the origin; for j(x, y) 0, ellipses with the ccoter (0,0), andmajar andminar axes along the x- and y-axes respectively No boundary points (e) Both opeo and closed Unbounded All points in the xy-plane (b) All reals Far j(x,y) = 0, the x- andy-axes; far j(x,y) 0, hyperbolas with the x- and y-axes as asymptotes No boundary points (e) Both opeo and closed Unbounded All (x,y) satisfying x' + y' < 16 (b) z '" 1/4 Circles ccotered at the origin with radii r < 4 + y' = 16 Boundary is the circle Open (f) Bounded (x,y) (0,0) (b) All reals The circles with ccoter (0, 0) and radii r > 0 Boundary is the single point (0, 0) Open (f) Unbounded All (x, y) satisfying -I :5 Y - x :5 I -'IT/2 :5 z :5 'IT/2 Straight lines of the farm y - x ~ c where -I :5 c :5 I Boundary is two straight lines y ~ I + x and y ~ -1 + x Closed (f) Unbounded Domain: all points (x, y) outside the circle x' + y' ~ 1 Range: all reals Circles ccotered at the origin with radii r > 1 + y' ~ I Boundary: Open (f) Unbounded 33. (a) 35. (d) (b) y
(b)
41. (a)
z=-3
,_-'1 %",-1
*
z=o
,_1
,=, ,_3
*
43. (a)
(b)
, 4
z .. 16
x'
*
45. (a)
(b)
,
(0.,0,1)
---
,
~
,-0
, 47. (a)
(b)
0 ... ~,:1 t=
,
,
%",-1
,-0
----,0.t-~,~-~1-~%
x'
-----j---,-,
,
z--1
,
, 1
, ----71'------->,
-4
A-58 49.
Chapter 14: Answers to Odd-Numbered Exercises
x2+y2~ 10
51.x+y2~4
,
,
>ro
->ro
2
>ro x
x
4
-2
->ro 53.
,
55.
37. 39. 41. 43. 45. 47. 49. 51. 55. 59. 63. 69. 75.
(a) All (x,y,z)withz oF 0 (b) All (x,y,z) withx 2 + z2 oF I (a) All points (x, y) satisfYing z > x 2 + y2 + I Coosider paths aloog y = x, x > 0, and aloog y = x, x < O. Coosider the paths y ~ kx 2 , k a coostaot. Coosider the paths y ~ »IX, m a coostaot, m oF -I. Coosiderthepathsy = kx 2 ,kacoostaot,k oF 0, Coosider the pathsx ~ I and y ~ x, (a) I (b) 0 (0) Docs not exist 57. The lintit is O. The limit is I. (a) f(x,y)I,~~ = sin 20 where tao 0 = m 61. 0 Docs not exist 65. 7f/2 67. flO, 0) ~ In 3 8 ~ 0.1 71. 8 ~ 0.005 73. 8 ~ 0,04 8 = Y 0.Q15 77. 8 = 0.005
Section 14.3, pp. 772-775
,
af _ af _ 1. ax -4x,ay - -3
/
~,y,iJ
af _ 3. ax - z,,(y
af af 5. ax = 2y(xy - I), ay = z,,(xy - I)
-,;+%-1
7 af = x af • ax Yx2 + y2'ay
9 af = -I af • ax (x + y)" ay af _y2 - I af 11 - ~ • ax (xy - i)2'ay
,
61. ~ - Inz ~ 2 63. x 2 + y2 + z2 ~ 4 65. Domain: all points (x,y) 67. Domain: all points (x,y) satisfYing Ixl < Iyl satisfYing -I :5 x :5 I and -1:5y:51
,
y-x
y Yx2 + y2 -I
(x + y)' -x 2 - I ~ ----''-----'0 (xy - I)'
13 af = .x+,+1 af = .x+,+1 15 af 'ax 'ay 'ax af 17. ax = 2 sin (x - 3y) cos (x - 3y),
,
y
af _ 2 -x - I
+ 2)'ay
1
af ay af
~
= _1_
af x+y'ay
= _1_ x+y
-6 sin (x - 3y) cos (x - 3y)
ax = yxY
-1
af , ay
af
af
23. fx 25. fx
= x'lnx 21. ax = -g(x), ay = g(y) = y2, f, = 2xy, f, = -4z = I, f, = _y(y2 + z 2t '/2, f, = -z(y2 + z2)-I/2
27. fx
=
19.
yz
,f, V I - x'y2z 2
=
xz
,f, V I - x'y2z 2
xy V I - x'y2z 2
= -:r~~~
29f=
1
1
1
level curve: y = z"
level curve: sin-1 y - sin-1 X = '!! 2
Section 14.2, pp. 761-764 1. 15. 27. 33. 35.
5/2 3. 2\16 5. I 7. 1/2 9. I 11. 1/4 13. 0 -I 17. 2 19. 1/4 21. I 23. 3 25. 19/12 (b) All (x, y) except (0, 0) 2 29. 3 31. (a) All (x, y) (a) All (x,y) except where x ~ 0 or y ~ 0 (b) All (x,y) (a) All (x,y,z) (b) All (x,y,z) except the interior of the cylinder x 2 + y2 = I
I f= 2 f= 3 x+2y+3z' y x+2y+3z' % x+2y+3z 2 2 z2 31 • f x = _2xe-W+ + ) , f y = _Zye-{x + +z ) , f z = _2ze-{x 2+y2+ z2) •
x
r
r
33. fx = sech2(x + 2y + 3z),f, = 2 sech2(x + 2y + 3z), f, = 3 sech2 (x + 2y + 3z) af af 35. iii = -27f sin (27ft - a), aa = sin (27ft - a) ah 37• iJp
=
'.I. oah Sffi'l'cos 'iJq,
39. Wp(P, V, 8, v, g)
~
=
.I. P cos." cos oah 'iJO
=
'.1.'0 -psm."sIn
V, Wy(P, V, 8, v, g) ~ P
2 + 8v 2g'
Vv2( )V8v Wa(P,V,6,v,g ) =z.g,Wv P,V,6,v,g =g. V8v 2 Wg(P, V, 8, v, g) = - 2g2
Chapter 14: Answers to Odd-Numbered Exercises
af 41. ax = I
af
+ y, iJy
a'f a2f = 0, -2 = 0, aX ay
a'f a2f ayax = axay = I
av
•.
2
43. ax - 2xy + Y cosx, ay - x - smy + smx,
a2g -
.
211 - ysmx
iJx2:r
ay2
,
a2g
a'g
ayax ~ axay ~ 2x 45. -ar
ar -l-
ax
47.
--
--
-1
ayax
axiJy
(x + y)'
a'w axay
~
9. (8) au
(x
-I
a'r
+ y)"
iJy2
+ 2x tan (xy), ~;
-I
(x
+ y)"
= x' sec2 (xy),
2x'y sec2 (xy) tan (xy)
a'w ay 2 49. :
~
= sin (x'y)
+ 2x'y cos (x'y), ~;
~ 6xy cos (x'y)
a'w
= -xs sin (x'y)
51 aw = 2 aw • ax 2x + 3y' ay 53.
'av
au
au
13 dz = azdx • dt axdt
"
"ay
= x' cos (x'y),
"
."
a'w axiJy
•
" 17. aw = awax + awiJy aw = awax + away au ax au iJy au' av ax av ay avo
••
•
" •
"•• a.
dw lii
~
0,
(b)
dw lii('1r)
~
3. (8)
dw lii =
I,
(b)
dw lii (3)
= I
0
,
•
•
"
.,
di
"
ay
a•
,
•
1. (8)
'i
."
f",,(x,y) = f",(x,y) = 0 forallpoints(x,y) 89. Yes
"
di
.." . , .•• ." •
••ay
.•• .
71. f.(x,y) = 0 fora1lpoints(x,y), 3y2, y '" 0 f, (x,y) = { -2y, y < 0'
Section 14.4, pp. 782-783
,
,
(f) Y flISt
57. J.(1,2) = -13, f,(I, 2) = -2 59. f.(-2, 3) ~ 1/2,f,(-2,3) ~ 3/4 61. (8) 3 (b) 2 aA a aA ccosA - b 63. 12 65. -2 67. -a b . A' ab ~ bcsm . A a ~CSIn lnv 69. v. = (lnu)(ln v) - I
-y (z - y)'
aydt
-6 (2x + 3y)'
a'w = a'w = 2 + 6xy2 + 12x2..' ay ax ax iJy iY y 55. (8) x flISt (b) Y flISt (0) x flISt (e) y flISt
2
•
~; = y2 + 2xy' + 3x'y', ~; = 2xy + 3x'y2 + 4x'y',
(d) x flISt
_1
15. aw = awax + awiJy + awaz au ax au iJy au az au' aw = awax + away + awaz av ax av ay av az av
- 4x'y2 sin (x'y)
3 a'w 2x + 3y' ayax
= -2V2 (ln2 - 2)
+ azdy
•
a'wa'w 22.. ,. 2.. ayax = axay = 3x cos(Xy) - 2xysm(xy)
ay 2
au
a"; ax
+ 2), :~
(b) au ~ 0 au ~ I au ~ -2 ax 'iJy 'iJz
2x' sec2 (xy) tan (xy)
I
+ 4 cos v,
ziJu 11. (8) ax = 0, iJy = (z _ y)" az
+ ]x2 sec' (xy)
a"; = 4xy sec' (xy) + 2x'y' sec' (xy) tan (xy) + 2 tan (xy) ax
~ '1r +
2 = 2u + 4uv, iJw av = -2v + 2u
(b) aw = 3 aw =
x+y'ax 2
= x'y sec2 (xy)
iJw
'
+ cosx
a2r
~
-cosy
=
Ia'r ---
x+y'ay
a'w ayax
(b): (I)
I,
smv
(b) :: = V2 (ln2
cPr
~;
+
= -4u sin vln (u sin v) + 4u~OS2v
iJZ
ag_
=
4ttan-1 t
7. (8) :: = 4 cos vln (u sin v)
ag_
a2g -
~
5. (8) :
+ x, 2
= I
•
.. di
.,
ay
,
""
A-59
A-60
21.
Chapter 14: Answers to Odd-Numbered Exercises
aw as •
dwau aw
dwau
---
7. Vf = 31 + 2i - 4k
au at
du os' at
• dw
dii
19•
dii
0
0
II
13. 21/13 15. 3 17. 2 1.+I·(Di) ·r.:2 = - y'2' y'21' DJ P, = v,;
21.11=
23.
•
,.T<
a-
ay
,
,
"
-II
I. I. = y'2' - y'21,
~~ ~
37.
::~2,:~~1
39.
aa~
-I,
= 2t
~; ~
29.
-I
.".P, ~;
ay
,
33. 12
.".P
Yj, Yj) and (Yj,-Yj} minimum at ( Yj, Yj) and (- Yj, -Yj)
51. 2xv'x s +
6,min
x' +
~
2
(" Jo 2
kdt t4
,
Vi--I+J
,
d
29.
II
=
~I
(b)
II
=
-~i+ ~i,Duf(l, -I)
(0) II
=
15 1 + i5'i u
~
<)e
-
• -I, u
~i,Duf(l, -I)
~
=
= 5
=-5
_15 1 _ i5 j
24.7. 25 ' - 25 I
. 7.+24. u=-l,u=25' 251
31. u ~ _7_ 1 - _2_ , -u ~ _ _ 7_ 1 + _2_ j i
v'53
v'53
v'53
33. No,themaximumrateofchangeisv'i8s 35. -7/YS
,
v'53
<
14.
1.
17. x
~
I
+ 90t,y
~
I - 90t, z
9 19. df = 11 ,830 '" 0.0008
23.
-----+--""-~r 2
,
27. VI - 2...J2i + 2'/2j
Section 14.6, pp. 799-802 3.
5.
(I + j + k), (D-uf)p, = -2Y3
+x 3
Section 14.5, pp. 790-791 ,
VJ
25.
35.-7
49.
~
= -
(I + i + k), (D,J)p, = 2Y3;
t, ~; ~ -~
41. -0.00005 amps/sec 47. (cos I, sin I, I) and (cos( -2), sin( -2), -2)
VJ
""
~~ ~
= 3s 2
=
-II
"
27. -4/5
II
.1r.:1- .5r.: i - .1r.:k,(D,J)p, =3Y3; 3v3 3v3 3v3 = __ 1_ 1 + _5_ i + _I_k, (D-uf)p = -3Y3 3Y3 3Y3 3Y3 '
a-
~ .. o
"
31.
•
r
~ .. o
'r
1.
;~ k
(D_,J)p, = -y'2
-II
25. 4/3
-
'0
'0
,.T<
;~ i+ ;~i
11. -4
I I ,I " ,I " dw
9. Vf = -
~ sin Y3 -
~
3
21. dg =
a
kcos v'3 '" 0.935'C/ft
(b) Y3 sin v'3 - cos Y3 '" 1.87'C/sec (b) L(x,y) = 2x + 2y - I 25.
+5
Chapter 14: Answers to Odd-Numbered Exercises
29. (a) L(x,y) = I
+x
(b) L(x,y) = -y
+
T
31. (a) W(20,25) = 11°F, W(30, -10) = -39°F, W(15, 15) = OaF (b) W(IO, -40) "" -65.5°F, W(50, -40) "" -88°F, W(60,30) "" 10.2°F (c) L(v, 1) "" -0.36 (v - 25) + 1.337(T - 5) - 17.4088 (d) i) L(24,6) "" -15.7°F il) L(27,2) "" -22.l°F iii) L(5, -10) "" -30.2°F 33. L(x,y) = 7 + x - 6y; 0.06 35. L(x,y) = x + Y + I; 0.08 37. L(x,y) = I + x; 0.0222 39. (a) L(x,y,z) = 2x + 2y + 2z - 3 (b) L(x,y,z) = y + z (c) L(x,y,z) = 0 I I 41. (a) L(x,y,z) ~ x (b) L(x,y,z) ~ v2 x + v2 y
43. (a) L(x,y,z) (b) L(x,y,z)
=2 +X =x - y
2." + ." z+2 +
- z+
(c) L(x,y,z) ~ x - y -
I I
L(x, y, z) ~ 2x - 6y - 2z + 6, 0.0024 L(x,y,z) = x + Y - z - 1,0.00135 Maximum crror (estimate) "'0.31 inmagnilude (a) ±5% (b) ±7% "" ±4.83% Pay more attention to the smaller of the two dimensions. It will generate the larger partia1 derivative. 57. (a) 0.3% 59. f is most sensitive to a change in d. 61. Q is most sensitive to a change in h. 45. 47. 49. 51. 53. 55.
1T'
11'
11'
1T'
65. At- 4 ,- 2v2;atO,0;at 4 '2v2
Section 14.7, pp. 808-811 1. f(-3,3) = -5, local minimum
5.
f(3,~)
3. f(-2,1),saddlepoint
= 1;,localmaximum
7. f(2,-I) = -6, local minimum 11.
f (\6, 0)
~
9. f(I,2),saddiepoint
_1 6, local maximum 7
13. f(O, 0), saddle point; f (
-~, ~)
=
(4, -2) 37. Absolute maximum: 4 at (2, 0); absolute minimum:
3~ at
(3,-f). (3,f). (I,-f).and (I'f)
39. a
~
-3, b
41. Hottest is 2
2
~
r
-t, '?)
at (
and (
-t, -'?}
coldest is
-f (t, 0). at
43. (a) f(O, 0), saddle point (b) f(l, 2), local minimum (c) f(l, -2), local minimum; f( -I, -2), saddle point 49.
(E, 3;:)
51.
(t,H)
53.3,3,3
55. 12
59. 2 ft X 2 ft X I ft
61. (a) On the semicircle, max f = 2v2atl = .,,/4,minf =-2 atl = .".00 the quarter circle, maxf = 2v2 att = .,,/4, minf ~ 2atl ~ 0,.,,/2. (b) On the semicircle, maxg = 2 att = .,,/4, ming = -2 at 1= 3.,,/4. 00 the quarter circle, maxg = 2atl = .,,/4, ming ~ a at I ~ 0, .,,/2. (c) On the semicircle, max h = 8 at I = 0,.,,; min h = 4 at I = .,,/2. 00 the quarter circle, max h = 8 at I = 0, minh ~ 4atl ~ .,,/2. 63. i) minf = -1/2atl = -1/2; no max il) maxf = Oat I ~ -1,O;minf ~ -1/2att ~ -1/2 iiI") maxf ~ 4at 1= I;minf = Oatt = 0 9 71 20 67. y ~ -13 x + 13' yl.~4 ~-13
section 14.8, pp. 818-820 1. (±
Jz,t).( Jz,-t) ±
3.39
5. (3, ±3v2)
7. (a) 8 (b) 64 9. r ~ 2 em, h ~ 4 em 11. Length ~ 4v2, width ~ 3v2 13. f(O,O) = 0 is minimum,f(2, 4) = 20 is maximum. 15. Lowest = 0°, highest = 125°
17.
G, V 2,
19. I
21. (0, 0, 2), (0, 0, -2)
23. f(l, -2, 5) = 30 is maximum, f( -1,2, -5) = -30 is
I;~, local maximum
15. f(O, 0) = 0, local minimum; f(l, -I), saddle point 17. f(O, ±V5), saddle points; f(-2, -I) = 30, local maximum; f(2, I) ~ - 30, local minimum 19. f(O, 0), saddle point; fO, I) = 2, f(-I, -I) = 2, local maxima 21. f(O, 0) = -I, local maximum 23. f(n.", 0), saddle points, for every integer n 25. f(2, 0) = e -4, local minimum 27. f(O, 0) = 0, local minimum; f(O, 2), saddle point 29.
35. Absolute maximum: II at (0, -3); absolute minimum: -10 at
444 57. \/3 X \/3 X \/3
122 (c) L(x,y,z) ~ :3 x + JY + :3 z
A-61
f(t, I) = men - 3, local maximum
31. Absolute maximum: I at (0,0); absolute minimum: -5 at (1,2) 33. Absolute maximum: 4 at (0, 2); absolute minimum: 0 at (0, 0)
minimwn.
25. 3, 3, 3
VJ VJ VJ
27.
by
by
29. (H/3, -4/3, -4/3)
33. f(2/3, 4/3, -4/3) 37. Maximum is I
~
noits
31. U(8, 14) 4
:3
~
$128
35. (2,4,4)
+ 6\/3 at (± v6, \/3, I), minimum is
1- 6\/3 at (±v6, -\/3, I). 39. Maximum is 4 at (0, 0, ±2), minimumis2 at (±v2, ±v2, 0).
section 14.9, p. 824 " x lQuadr • atic.
I, + xy,' b cu ic.' x + xy + Z-xy
3. Quadratic: xy; cubic: xy
A-62
Chapter 14: Answers to Odd-Numbered Exercises
5. Quadratic: y cubic: y
t
+
(2xy - y2);
t i t (2x 2 + 2y 2) = x 2 + y2;
+ (2xy - y2) + (3x'y -
7. Quadratic:
9. Quadratic: I + (x + y) + (x + y)2; cubic: I + (x + y) + (x + y)' + (x
3xy 2 + 2y 3) cubic: x 2 + y2
,
+ y)3
. 11 1 2 ;E( ) 11. Quadratic: ZX 2 - zy x,y '" 0.00134 9. -2 11. 1/2 13. I 15. Let y = kx 2 , k # I 17. No; 1im(x3)_(o,O) f(x,y) does not exist. ag ag 19. or = cosB + sin 8, a8 = -r8in8 + reas8
Section 14.10, p. 828 1. (8) 0
(b) I
+ 2z
(c) I
V) 3. (8) au ap + au( aT nR 5. (8) 5
7.
+ 2z
(b) au (nR) ap V
+ au aT
(b) 5
21.
I af --R1 2 ' aR2
af aRl
I af --R," aR3
I R,'
23 ap = RT ap = nT ap = nR ap = _nRT • an v'aR v'aT v'av V2 alg 2x a2g a2g I a2g 25. -~O-~--~-~--
(:).=ooS6
ax2
'ay 2
y3'oyax
axay
y2
a2f 2 - 2x2 a2f alf a2f 27 - 2 = -3Ox + 2 - = 0 -- = -- = I • ax 'ay ax ax ay (x + I)" ay2
Practice Exerdses, pp. 829-832 1. Domain: all points in the xy-plane; range: z '" O. Level curves
are ellipses with major axis along the y-axis and minor axis along the x-axis.
,
31.
~to =-1 ~wl =2,~wl
33.
ddfl
29.
r
t
(r,,,)-('II',O)
S (7,,)-('IT,0)
= -(sin I
+ cos2)(sinl) +
- 2(sin I
3. Domain: all (x,y) such that x # o andy # O;range:z # O. Level curves are hyperbolas with the x- and y-axes as asymptotes.
,
(cos I
+ cos2)(cosl)
t-I
z=9
35.
=2-".
dyl
fix (x.,)~(O,l)
~
+
cos 1)(sin2)
-I
37. Increases most rspidly in the direction U
decreasesmostrapidlyinthedirection-u
yz
yz
"? "? ~ ,,?i + "?j;
~-
i -
j;
7
Duf = Z;D-uf = --2-;D.,/ = -TOwhereul = z=1
39. Increases most rspidly in the direction U
5. Domain: all points in xyz-space; range: all real numbers. Level
surfaces are paraboloids ofrevolution with the z-axis as axis. 1--+--1f(Z-y,t:)=l:]+I-z .. -l m
7. Domain: all (x,y,z) such that (x,y, z) # (0,0,0); nmge: positive real numbers. Level surfaces are spheres with center (0, 0, 0) and radius r > O.
t
decreases most rapidly in the direction -u = -
~i
D.f
v TVi
~
7; D-..f
~
-7; D.,!
41. "./YZ 43. (8) fil,2) = f,(I, 2) = 2 , 45.
.t=?+I+ 1
,
~ ~i +
,
~
7 where Ul (b) 14/5
~
j
-
v
TVi
+ ~ k;
~j
- §: k;
777
Chapter 15: Answers to Odd-Numbered Exercises 47. Tangent: 4x - y - 5z = 4; normal line: x=2+4t,y= -l-t,z= 1-51 49. 2y - z - 2 = 0 51. Tangent: x + y ~ 'IT + I; normal line: y Y
17. f(x,y) =
~ + 4,g(x,y)
=
A-63
I +~
19. y = 21nlsinxl + 1n2 ~
x -
'IT
b
+ I
(21 + 7j) 21. (8) • v53 23. w = e -C1'lJ"lt sin 1TX
Y"-'; + '11"+ I
(b)
~ (981 29,097
127j + 58k)
CHAPTER 15 53. x~ 1-21,y~ I,z~ 1/2+21 55. Answers will depend 00 the upper bound used for
57. 59. 61.
63. 67. 69.
71. 73. 75. 77.
79. 81.
Section 15.1, pp. 840-841
Ifni, Ifryl, If"l· WithM = v2/2, lEI'" 0.0142. With M = I, lEI'" 0.Q2. L(x,y,z) ~ y - 3z, L(x,y, z) ~ x + Y - z - I Be more careful with the diameter. dJ = 0.038, % change in 1= 15.83%, more sensitive to voltage change (8) 5% 65. Local minimum of -8 at( -2, -2) Saddle point at (0, O),f(O, 0) ~ 0; local maximum of 1/4 at (-1/2, -1/2) Saddle point at (0, 0), f(O, 0) ~ 0; local minimum of -4 at <0,2); local maximum of4 at (-2,0); aaddle point at (-2, 2), f(-2,2) = 0 Absolute maximum: 28 at (0, 4); absolute minimum: -9/4 at <3/2,0) Absolute maximum: 18 at (2, -2); absolute minimum: -17/4 at (-2,1/2) Absolute maximum: 8 at (-2, 0); absolute minimum: -I at (I, 0) Absolute maximum: 4 at (I, 0); absolute minimum: -4 at (0, -I) Absolute maximum: I at (0, ±I) and <1,0); absolute minimum: -I at(-I,O) Maximum: 5 at (0, I); minimum: -1/3 at (0, -1/3)
VJ'-VJ' VJ): (- VJ' VJ'- VJ)
83. Maximum:
85. Width
v'3 at (
minimum: -
t 0' 0' at (
minimum+t (89. (8) (2y + x'z)e'"
v2) and (-
0' 0'-
(a:t
0'-0'0'-0'
v2):
v2) and (
(b) x'e"'& -
~
Section 15.2, pp. 847-850 1.
~) . oaw
sm aT
v2)
3.
,
6
2
-2
5.
7.
,
, 1
------7"~~
Iy"'li' I I I
--+-------!-~x
<8>
--+'----------'--~x
2,x 3 :5y:5 8 (b) 0 '" y '" 8,0 '" x '" ylj3 11. (8) 0 '" X '" 3, x' '" y '" 3x O:5x:5
(b) 0 '" y '" 9, ~ '" x '"
VY
13. (8) 0 '" X '" 9,0 '" y '" ~ (b) 0 '" y '" 3,y' '" x '" 9 15. (8) 0 '" X '" In 3, e~ '" y '" I I (b) 3 '" y '" I, -Iny '" x '" In 3 17. (8) 0 "'x '" I,x "'y '" 3 - 2x 3-y (b) 0 '" y '" 1,0 '" x '" y U I '" y '" 3,0 '" x '" - 2 -
+
cosoaw r
iJ6
19.
~ T
,
+ 2
21. 8ln8 - 16 + e
,
ln'
('11",'11")
AddUional and Advanced Exercises, pp. 833-834 1. fry(O,O) = -I, f,.(O, 0) = I 13. V =
-----------
I I I I
•
+ y' + z')
-----------
-+--------+.-4~X
97. (I, -I ± 4,1),larealnumber
7.
,
-----
9.
~ (c::t,depth ~ (b::t,beigbt ~
87. Maximum:
v'3 at
1. 24 3. I S. 16 7. 21n2 - I 9. (3/2) (5 - e) 11. 3/2 13. 14 15. 0 17. 1/2 19. 2ln2 21. (ln2)' 23. 8/3 25. I 27. v2
~abc
(In In 8, In 8)
zmlny
0
•
x 0
lnln'
A-64
Chapter 15: Answers to Odd-Numbered Exercises
23. e - 2
,
e;
49.
,
r----7(1.1)
.1: ..
, (1,1)
•
25. ~ln2
---coif'----+--·
o
27. -1/10
2
53. 1/(80".)
•
v
-2
v __p
v_p
P
2
55. -2/3
,
29. 8 (-11'13,2)
,
(O.S,o.~
x+,-1
-x+y-I
-2
(-2,-2)
/,, ,
51.2
,2
0
33.
2
(~-'l)
4/.(4-Y)/2 dxdy
,•
-,•
o
•
0.'
;1;-,.=1
-x-y= 1
35. lL'dydx
0
•
-1
,
-1
(1,1)
4
y=4-1t 2
57. 4/3 67.
y=x
--.o+-f-----~·
39. Jo Jo
11 1my
,+ ,-e"
65. 2(1
+ In2)
, II I I I
z=l-
tx-t>'
,L-----
16>; dx dy
,
(l,t)
63. 20
2
['[(~
37. [' [' dx dy
61. 16
,
,-'"
(1,2)
59. 625/12
/
, 2
/ /
---1----------'""c-->.
/
/ /
9
/
3
(1,1)
-01--+-->. o ~
---::0+---;--->'
41.1' [~3Ydydx -1)0
,
69. 1
2
43. /.'
,
L'
77. xy dx dy
71.".2
73. -
/.',1'-' o
2
;2
75. 20r
(x 2 + y2)dydx = 4 -
x
3
~.
~"
,.+
---''--------,+-----'--~.
-1
1
0
---1----f.----...L_. 45.
,
[" [' (x+y)dydx
it Jinx
47.2
, (w,w)
'IT
o
---1+.--------0--.
"
•
•
79. Risthe set of points (x,y) suchthatx 2 + 2y 2 < 4. 81. No, by Fubini's Theorem, the two orders of integration must give the same result. 85. 0.603 87. 0.233
Chapter 15: Answers to Odd-Numbered Exercises
Section 15.3, p. 852 1.
/,,/,,-r
13. 12
dydx = 2 or
3
1'1-Y'
'-2y-2
/"/"-Y dx dy
dxdy=Jt 2
-.
,
Y=COBX
-12
(12,6)
,2=3%
(11'/4, -/212)
2
y=~
,
0
y=l!inx
12
0
r
r
x=-y'J
{'"'r 5. Jo Jo dy dx ~ 1
7.
o
r
•
1 dxdy = 4
or
/, 11'Y o 1/2
,
19. (8) a (b) 4/~ 21. 8/3 23. 40,000(1 - .-')In (7/2) '" 43,329
I
,_2
I I
,
1 dydx + Idxdy+
f::=:;
(J ::=:;
21T,O ::=:; r
~
9
3.
*" ~ (} ~ 3:. 0 ::=:; r ::=:; esc 8
~, 1 '" r '" 2 v'3 sec 0; 671'
r
/,'l,-r
3 or 1 dydx = 2
/,'l'-Y
3 Idxdy~2
1
;;/2
1
1/2
7.
~
-2 '" 0
11. 2~
y=ll:
y_r
,
o x/2
' \ (2,-1)
Y"-~
5. 0 '" 0 '"
2
/, 'i"
---*,~"'cO,-_r
1.
{'Jx/3r 1 dydx + j,')x/3(' 1 dydx = 4 ,
,
Section 15.4, pp. 857-859
--;cor--:"ln",-----+r
Jo
3 2
dxdy-3
(1, I)
Y
17.
/, '1,y-r _1 ,
/,'1'Y
-,
r
~
0
NOf TO SCALE
,-x+2 (-4,-2)
11.
,
(-1,1)
,
9.
15.V2-1
,
,
,
= 2
A-65
'"
~
2' a '" r '" 2 cos 0
13. 36
15. 2 -
19. (2 In 2 - I) (~/2) 23.
v'3
/,'/,vR
or
0
25.
,
2
,=3-x or x=3-y
17. (I - In 2) ~
2(1 + V2) 21. 3
,
/,o'/,~ xydydx
-=''''---"----~,o--''l,-r
~
2
-+-------'---+r
, ,
9.
xy dx dy
;;=2
------------
y=x
-71L-------~,-r
/,,/,r
y'(x' + y')dydx or
27. 2(~ - I)
29. 12~
/,'1\'
31. (3~/8)
+
1
(x' + y')dxdy
33.
32a
35.
32a
A-66
Chapter 15: Answers to Odd-Numbered Exercises
vr.)
37. 2".(2 -
43. ".In 4, no
45.
39.!
+ 5:
t
+ 2h 2 )
(a 2
y;
41. (a) -2-
(b) I
13. x = 15. x = 17. x = 19. x =
23.
].'].1-'13].2-4-413
dy dz dz,
00
/2].1-112-1'/3 dzdzdy,
dy dz dz,
0
2].3-31 ]. o
0
0
'].2-2z/3].1- -1'/3 dzdydz. 112
].o
0
25. 27. 29. 31. 33.
21S-Y'1~ -\f'8=;
1 -vs::; -vs-,-,' ls1VS::;1~ 4
Y'
-2
_~ I dz dz dy,
15. 7/6 21. (a)
17. 0
4
19.
23. 2/3
13. 18
(b)
Xl
{'-'1 Jo Jo {I
VY
-VY
(e)
15.
].11vT=; {1-' dydzdz o -~Jr
l11 1
dzdydz
{11 VY {1- dz dz dy
(d)
o
1 Y -
0
VY
-VY
5.
12
2
29. 16/3
31. 8". -
x = 5/14,y = 38/35 3. x = 64/35,y = 5/7 x = Y = 4a/(3".) 7. I, = Iy = 4,,-,10 = 8". x = -I,y = 1/4 11. I, = 64/105
3.
I~'"
5.
".(6V2 -
32 T
8)
7.
~~
].2~].1].~ r dz dr d8
[~t'J rdrdzd8 + [~[,].Vh' rdrdzd8
(0)
1.']'~].2~ r d8 dz dr
1~/2 -w/2
].""'/"" 0
].~].2
f(r, 8, z) dz r dr d8
0
....].4-,....
19.
f(r, 8, z) dz r dr d8
f(r, 8, z) dz r dr d8
Jo
-,r/2JI
].~/4].~'].2-'''''
25. 5".
Jo -v'YJo
27. I
~a2 + b 2
f(r,8,z)dzrdrd8
dzdzdy
y
25. 20/3
=
(b)
31. (a)
27. 2".
29. (
8
-
i V2)".
].2~].~/6].2'?-sinq,dPdd8 +
].o2~J.~/2].=~p2 sin q, dp dq, d8 1r/6
(b)
Section 15.6, pp. 873-875
9.
2 2 abe(a + b ) 12 ' R,.m
17.1~/2 (1+""•• r
33. 2 35. 4,,- 37. 31/3 39. I 41. 2 sin 4 43.4 45. a = 3 or a = 13/3 47. The domain is the set of all points (x,y, z) such that 4x2 + 4y 2 + z2 ~ 4.
1.
=
3
11. (a)
13.
t -f
11].1-"11-. dydzdz
= I y = I. = 11/6
I,
9. "./3
r1Vil~ I dydzdz + Jo -Vi _~I dydzdz.
5 (2 - yIJ)
11.
-I 0
(0)
x = 4/5,y = Z = 2/5 x = y = z = 8/15 (0)
(b) (b)
4,,-(V2 - I)
12 r1~ -2iz 2 _~1 dy dz dx,
+
The value of all six integrals is 16,,-.
9. 6
2V2
Section 15.7, pp. 883-886 1.
-,-y\ dz dy dz + Jor1Vi1~ I dz dy dz, -yz-~
_\I'8_1'_x11 dy dz dx
=
3
-VS-z-y l
21s-"1~
-2 4
I dz dz dy,
X 1 +y2
12141~
-vs-.-y' I dz dz dy +
s vs 1VS::;1
7. I
(a) 4/3 (a) 5/2
2
-2 _ ~
-2 4
(b) c
_ abc(a + 7b ) _ ~a2 + 7b 2 (b) h 3 ,RL 3
s
2
4
0, z = 8/3
= 1386
0
5. 1 1v'h'l -"-Y'ldzdzdy, -2 -~ ;(2+y 1 121v'4=r18-x:z-y2
l
x= y =
(a)
h
37. (a) 1,=
The value of all six integrals is I.
1
x = Y = 0, z = 12/5, I, = 7904/105 '" 75.28, I y = 4832/63 '" 76.70,1, = 256/45 '" 5.69
3. ].1].2-4].,-",-,y/2 dzdydz, ].2].I-Y/2].'-"'-'Y/2 dzdzdy,
].0 0 0
= 6/5
- M _ M (2 _ M (2 21 • Ir 3 (b2 +c 2),1'-3 a +c 2) ,1%-3 a +b 2)
Section 15.5, pp. 865-868 1. 1/6
1].'-"'].2-4-413
3/8,y = 17/16 11/3,y = 14/27,1y = 432 O,y = 13/31,1y = 7/5 O,y = 7/10; I, = 9/IO,Iy = 3/10,10
].
0
f'].... Jl
2~
0
- '(l/P)
0
p2 sinq,dq,dpd8
+
].2~1.'].~/2P' sin q, dq, dp d8 33.
].2~].~/2 o
0
{2 p2 sinq,dpdq,d8
JcostP
= 3~'"
21.".2
23. "./3
A-67
Chapter 15: Answers to Odd-Numbered Exercises
35.
37.
['7 [7 [l-~~';sinq,dpdq,d8 = 8'IT
10 10 Jo
['7J.7/2 ['~'~ p' sin q, dp dq, d8 =
Jo
Jo
,"/4
1 1 1 3.
3 "'-
3
11.
'lTab(a' + b') 4
8j,7/2j,'j,Vh'rdzdrd8
(b)
41.
9.j,'j,'(u + v)2:: dudv ~ 8 +
7.64/5
8j,7/2j,7/'j,'';Sinq,dPdq,d8
39.
'j,y'4:7j,"'4-%'-Y' j,
17.
13.
-USinvl ucasv
1(
3
=
1
3)
+ .' '"
ucos2 v
2u, and
~
5;In2
0.4687
+ usin2 v
15. =
225
16
u
dzdydx
°° °
(b)
Practice Exercises, pp. S96-S9S
1v'-v'33=? ["'4-%'-Y' =? Jl dz dy dx 1 v3 -v3
43. 8'IT/3 53. 'IT/2 61.
71.
47. 3'IT ; 4 49. 1 4(2\12 - 1)'IT 55. 3 57. 16'IT 45. 9/4
4'IT( 8- 3v'3) 3
63. 2/3
0, z ~ 3/8
69. (x,y, z)
x= y =
0, Z = 5/6
73. 1% = 'IT/4
~
y
77.
2~a'
51. 5'IT/3
~
(0,0,3/8) 75.
a:~'IT
-,
1°1,
4
5.
-<' dydx~-4 3
7.
1-, °
'j,(1/2)~
ydydx
9
~-2
y
~
axis. (f(~.'.~
(M.O+1r,z)
N.l
'+7
2
-2 2;;+4
'lTR' 85. The surface's equationr ~ /(z) tells us that the point (r, 8, z) = (/(z), 8, z) will lie on the surface for 0118. In particular, (/(z), 8 + 'IT, z) lies on the surface wheoever (/(z), 8, z) lies on the surface, so the surface is symmetric with respect to the z-
%
,
10 (lila, 10)
(1.1)
81. 3M
j(t)
3. 9/2
--:Cof-~lN-OT-ro-SCALE--%
),1% = ~
(0, 0, ~),1% =
,
59. 5'IT/2
65. 3/4
~
67. x
1. 9. - 9
,
Section 15.S, pp. S94-S96 u+v v-2uI 1.
~
0, v
~
0, and
-,
°
,
%
17. 1/4 19. 'IT 2(31 - 3'/2) 29. 3
A-68
Chapter 16: Answers to Odd-Numbered Exercises 81f~(bS
41. x
Section 16.2, pp. 917-920
- as)
15
39. I, =
~ ji ~
1. Vf ~ - (xi
2 -\n4
43. 10
~
104
~ 2~
45. 1%
49.x=3~,ji=0
47. M=4,M%=O,My =0
- - 151f + 32 - - 0 51 . () a x - 61f + 48 ,y (b)
3. Vg
5. F 7. 9. 11. 13. 19. 29.
y
+ yi + zk)(x 2 + y2 + z2)-3/2
~ - G2 ~ y2 }
=
(x2
la
G2 ~
-
+ e'k
y2)i
ky /2i,anyk> 0 (x2 + y2)3 (0) 9/2 (0) 0
/21 -
+ y2)3
(a) 9/2 (b) 13/3 (a) 1/3 (b) -1/5 (a) 2 (b) 3/2 (0) 1/2 -15/2 15. 36 17. (a) -5/6
(b) 0 (0) -7/12 1/2 21. -1f 23. 69/4 25. -39/2 27. 25/6 (a) Cire, = 0, eire, = 21f, flux, = 21f, flux2 = 0 (b) Cirel = 0, cirC2 = 8'l'T, flUXI = 8w, flUX2 = 0 31. Cire = 0, flux = a 21f 33. Cire = a 21f, flux = 0
35. (a) -}'
-,
(b) 0
39.
(0) I
37. (a) 32
(b) 32
(0) 32
y
Additional and Advanced Exerdses, pp. 898-900
11-%' 2
1. (a)
11-%'].%'
6
2
x 2 dy dx
(b)
-3x
6
-3x
2
%
dz dy dx
0
(0) 125/4
3. 21f
(b) 4V31f
17.
9. 1f/4
In(£)
11.
Mass = a oos-'(£) - bVa2 2
10 19.
7. (a) Hole radius = I, sphere radius = 2
5. 31f/2
= ~4 oos-'(£)
';b (eo'!>' -
25. h =
I)
~ Va2
_
15.
I/~
43. F b 2,
_ b2 _ b: (a2 _ b2)3/2
21. (b) I
(0) 0
V20 in., h = V60 in.
41. (a) G ~ -yi
xi+n
= •~ vx2
(t) '-?]
1. Graph (c) 9. Vi 17. V3ln 21.
~~ (e 16
3. Graph (g)
11.~
(£) -
13. 3\1114
t
15.
19. (a) 4Y5
e 64 )
23.
25. t(5 3/2 - 7Vi - I) 31.
5. Graph (d)
(173/2 - I)
35. (a) 4Vi - 2 39. (a) 1% ~ 21fVi~
(b)
17 (40 /2 3
27.
t
7. Graph (f)
(5Y5
+ 9)
I~ (173/2 -
+
I 53'-2
~
5. Not conservative
+C
+ Ian (x + y) + tln(y2 + z2) + C
15. -16 17. I 19.9ln2 21.0 23.-3 2 ; I) 29.
v(x
31.
+ b(yb
JF'
- ya)
+ c(zb
- za) ~
F·AB. Section 16.4, pp. 940-942
2
29. 8
In (I + Vi)
(b) 1% ~ 41fVi~
51. 0
+2+2z +C
11. f(x,y,z) = xlnx - x
I)
33. 2Vi - I (b) Vi
49. 1f
3. Not conservative 3y 2 2
feB) - f(A) ~ a(xb - xa)
13 3/2)
1O~ -
2
9. f(x, y, z) = xeY+:z.
27. F
Section 16.1, pp. 90S-907
+ y2
1. Conservative
13. 49
CHAPTER 16
47. 48
+ y 2F
Section 16.3, pp. 929-931 7. f(x,y,z)=x
27. 21f[t -
(b) G ~ Vx 2
+ xi
37. 1% ~ 21f~a3
41. 1% ~ 21f - 2
1. 5. 9. 13. 17. 29.
Flux = 0, eire = 2'7Ta 2 3. Flux = -'l'Ta 2 , eire = 0 Flux = 2, eire = 0 7. Flux = -9, eire = 9 Flux = -11/60, eire = -7/60 11. Flux = 64/9, eire = 0 Flux = 1/2, eire = 1/2 15. Flux = 1/5, eire = -1/12 0 19. 2/33 21. 0 23. -161f 25. 1fa 2 27. 31f/8
Chapter 16: Answers to Odd-Numbered Exercises
Section 16.5, pp. 951-953 1. r(r, 0) ~ (r cas O)i
+ (r sin O)j + r'k, 0
s
0::=:;6::=:;217
3. r(r, 0) ~ (r cas O)i o :5 0 :5 'Tr/2
+ (r sin O)j + (r/2)k, 0 :5
' ' 5. IlzdU = 1 1 (4 - u - v)V3dvdu = 3V3
r :5 6,
5. r(r, 0) = (r cas O)i + (r sin O)j + v'9-=-? k, 0:5 r:5 3Y2/2,0:5 0:5 2'Tr;Alsa:r(4),0) = (3 sin 4> casO)i + (3sin4>sinO)j + (3 cas4»k, 0 :5 4>:5 'Tr/4, o :s (J :s 217 7. r(4), 0) = (V3 sin 4> casO)i + (V3sin4>sinO)j +
s (forx = u,y = v)
7. II x'Y5 - 4zdu =
9. r(x,y) = xi + yj + (4 - y')k, 0 :5 X :5 2, -2 :5 Y :5 2 11. r(u, v) = ui + (3 cas v)j + (3 sin v)k, 0 :5 u :5 3, o s v ::s 21T 13. (8) r(r, 0) = (r cas O)i + (r sin O)j + (I - rcosO - r sin O)k, 0:5 r :5 3,0:5 0 :5 2'Tr (b) r(u, v) = (I - ucasv - usinv)i + (ucasv)j + (usinv)k, 0:5 u:5 3,0:5 v:5 2w 15. r(u,v) = (4cas'v)i + uj + (4 cas v sinv)k,0 :5 u:5 3, -('Tr/2):5 v:5 ('Tr/2); anatherway: r(u, v) ~ (2 + 2cosv)i + uj + (2 sin v)k, 0 :5 U :5 3,0 :5 V :5 2'Tr
[IV;
~ 'Trr
rdrdO
+ Idvdu= [' ['~ u'(4u' +
uY4u'
~~
23. 25.
['~ ['
Jo Jo
['~J.~ 2 sin 4> d4> dO ~
Jo
(4
'Tr
19. -32
+ 6V6)
,
21. 'Tr;
,
35. 'Tr; 43.
37. -32
(%'%'%)
+ be)
23. 13a 4/6
25. 2'Tr/3
,
,
31. 'Tr;
~
33. 'Tr: 41. 3a 4
39. -4
45. (x,y,z)
8'Tr 4. Ta u
13. 2
17. V6/30
29. 18
27. -73'Tr/6
ae
(0,0, 1:),1,
~ 15";Y2 8
201r 4 (b) -3-a 8
section 16.7, pp. 970-972 13. 12'Tr
section 16.8, pp. 981-983
+ 2Y2)'Tr
1. 0 3. 0 5. -16 7. -8'Tr 9. 3'Tr 11. -40/3 13. 45'Tr 15. 12'Tr(4Y2 - 1) 21. The integral ~ value never exceeds the sorface area af S.
,"/4
27.
1 15. 30 (Y2
abe 4 (ab +
I) cos' v dvdu = 11'Tr 12
1. 4'lr 3. -5/6 5. 0 7. -6'Tr 9. 2'Tra' 15. -'Tr/4 17. -15'Tr 25. 161, + 16I,
I)
(5VS uY4u' + I du dv = 6
11.
9. 9a'
47. (8)
19.1'~1' rVS drdO ~ 8'TrVS 21.1'~14 I dudv ~ 6'Tr
1'1'~ u'cas'v'Y4u' + I·
s
(V3 cos 4»k, 'Tr/3 :5 4> :5 2'Tr/3, 0 :5 0 :5 2'Tr
17.
1'~1~ sin' 4> cas' 0 d4> dO = ~
3. II x' du =
:5 r :5 2,
A-69
29.
Practice Exercises, pp. 983-986 I I~
,
33. (b) A
'TrW+!
49.
~(13v'i3 - I)
+
11. 'TrV3
+ ~2 + ~2
17. 50 b c ~ (6sin4>casO)i + (6sin4>sinO)j 2
~)
+ (6cas4»k,
4> :5 2;,0 :5 0 :5 2'Tr
21. r(r, 0) 23. r(u, v)
39. 4
-
51.5'TrY2
41. 6V6 - 2Y2
5VS)
53.
47. 3
t(5VS -
+ 210 2
I)
- I JJ[[ xdu ~ Jo['Jo[' uY4u' + I dudv ~ 17v'i7 4 s
"aI~
~ :5 b'c'cas4 4> cos' 0
Section 16.6, pp. 960-962 1.
13. 2'Tr(1 -
9. 0
5. 0
= (r cas O)i + (r sin O)j + (I + r)k, 0 :5
r :5 2,
0:58:52'l'T
37. 13'Tr/3
45. ~ (17v'i7
43.
7. 8'Tr sin (I)
2 19. r(4),O)
= 1'~1~[a'b'Sin'4>cas'4> +
+ YoY = 25
3. 4a'
15. abe
a'e' cos4 4> sin' 0]'/2 d4> dO 35. xox
1. Path I: 2V3; path 2: I + 3Y2
= (u cas v)i + 2u'j +
(u sin v)k, 0 :5 u :5 I,
O:svsw
25. V6 27. 'Tr[ Y2 + 10 (I + Y2)] 29. Conservative 31. Natconservative 33. /(x,y,z) = y' + yz + 2x + z 35. Path I: 2; path 2: 8/3 37. (8) I - e-'~
41. (8) 4Y2 - 2
43 (- - -) • x,y,z
=
(I '15'3;;; 16 2) I
_ 3 7V3 45. z = 2,1, = -3-
(b) Y2 =
232 I 45' Y
+ 10(1 + Y2) =
64
15' I z
=
956
47. (x,y, z) = (0,0,49/12), I, = 64O'Tr
Appendices: Answers to Odd-Numbered Exercises
A-70
49. Flux: 3/2; eire: -1/2 57. 0
53. 3
7. (a) x = -I (b) y = 4/3 9. y = -x 5 11. y ~ -'Ix + 6 13. y ~ 4x + 4 15. y
sv'2)
55. 2: (7 -
59.".
17. x-intercept = v'3,y-intereept = -
Additional and Advanced Exerdses, pp. 986-988 1. 6". 3. 2/3 5. (a) F(x,y,z) ~ zi (c) F(x,y, z) = zi 7.
16'lTR' 3
9. a
11. (b) 16 g 3 (c) Work =
4 13. (c) } 'lTW
~
(1
+ xj + yk 2, b
~
gxy tis
(b) F(x,y,z)
~
zi
1
-2x + 12
v'2
y
+ yk
I. The mioimum flux is -4.
)or = g xy2 tis =
~
-, 19. (3, -3) x' + (y - 2)' = 4
23.
21.
16 g 3
(x + v'3)' + (y + 2)' =
4
y
(-"'.0)
19. False ifF = yi + xj
---CC-7-~+~" -4
APPENDICES Appendix I, p. AP-6 1. 0.1,0.2, 0.3, 0.8, 0.9 or I 3. x < -2
25.
x' + (y -
, 7. 3, -3
,
9. 7/6,25/6
3/2l' ~ 25/4
27.
(3,0) C(O.3I2)
(2.0)
-,
"
4
15. (-00,-2]U[2,00)
,
-,
--'=''-'''l!.-"--+---,--+="-->,
13. 0 :5 z :5 10
11. -2:5 t :5 4
)
o
10
17. (-00,-3]U[I,00)
"
-,
1
,
1
3
4
-I (0.-1)
-2
f + (y- 3/Zl = 25/4
,, 29.
19. (-3,-2)U(2,3) 21. (0,1) 23. (-00 ,I] 27. The graph of Ixl + Iyl :5 I is the interior and boundary of the
y y __ f_fi.%_5
V(-3,4)
4
"diamond-shaped" region. y
(0.-5)
-1
0, centered at the origin + y' = I andx' + y' = 4 33. The washer between the circles (points with distaoce from the origin between I and 2) 35. (x + 2l' + (y - 1)2 < 6
31. Exterior points ofa circle of radius
x'
Appendix 3, pp. AP-16-AP-18 1. 2, -4; 2\IS I 5. m~ ~-}
3. Unit cirele
37. y
(Vs' Vs). (- Vs,- Vs)
39. (-
VJ,-t). (VJ,-t)
41. (a) '" -2.5 degrees/inch (c) '" -S.3 degrees/inch -----Y'--'---+~.,
(b) '" -16.1 degrees/inch 43. 5.97 atln
;l;
A-71
Appendices: Answers to Odd-Numbered Exercises
45. Ves: C = F = -40' c
3. (a) By reflecting z across the real asis (b) By reflecting z across the imaginary asis (c) By reflecting z in the real asis and then multiplying the length of the vector by 1/lzl' 5. (a) Points 00 the circle + y' = 4 (b) Points inside the circle + y' ~ 4 (c) Points ootside the circle + y' = 4 7. Points 00 a circle ofradius I, center (-I, 0) 11. 4e',,;/3 13. le'w'j3 9. Points 00 the line y = -x
x'
51. k
~
-8, k
~
1/2
Appendix 7, pp. AP-34-AP-35 1. (a) (14,8)
(b) (-1,8)
(c) (0, -5)
x' x'
I,-k ± ~ ';6± '?i,- ';6± ,?i
15. cos'O - 6 cos' 0 sin'O
+ sin'O
19.2i,-v3-i,v3-i 23. I ± v3i, -I ± v3i
21.
17.
i
ANSWERS TO ODD-NuMBERED EXERCISES
Section 17.1, p. 17-7
+ c2e 4x 3. y = cte-4x + c2e x cte-2x + c2e'b: 7. Y = Cte---.t" + C2e 3x/2 3x cte---x/4 + c2e /2 11. Y = Ct cos 3x + C2 sin 3x c, cos 5x + c, sin 5x 15. Y = "(Cl cos 2x + c, sin 2x)
1. y = cte-3x
5. y
=
9. y = 13. y =
I + c2e---.t" + rex
23. y = cte X
CHAPTER 17
17. Y = '-'(Cl cos v'3x
+ c,sin v'3x)
25. y = e-2t(ct cosx + c2sinx) + 2 27. y = Acosx + Bsinx + xsinx + cos x In (cos x) _
29. y- Cl+ C,.
.
33. Y = Cl
33. y
~
.1,- sin2V3x 2v3
35. Y = -cos 2
v'2x +
37. Y = (I - 2x)e2x
Vz
sin 2
v'2x
39. Y = 2(1
+ 2x).-"/2
+ c2e 3x 43. y = cte-x/2 + c2xe-x/2 = Ct cos V5x + C2 sin V5x 47. Y = Cte ---x15 + C2xe ---x/S = .-'/2(Cl cosx + c, sinx) 51. y = C,."/4 + c,xe"/4 = cte-4xj3 + c2xe-4xj3 55. y = cte---x/2 + c2e 4xj3 ~ (1 + 2x)e-' 59. Y = +
41. y = cte---.t"
45. Y 49. y
53. y 57. y
:;.-7,/' g."
1. Y = c,.'"
+ c,.-2x + 130
3• Y =
Ct
+
C2e
5• Y =
Ct
cosx
+I
I
-
smx
I 3x gcas
7. y = Cte2% + C2e-x - 6 cosx - 2 sin x
+ c2e---.t"
" 11• Y -_ Cte
+ C2e
- x2 -
2 + txex
-2x -"4e 1_,49 7. + SOcosx + 50smx
~
C1
+ c,.-'" + x' + ~' -
15. Y =
Ct
+ C2e3x +
13. y
17. Y = Ct
+
c2e-x
2x2
tx2 -
+ C2 sinx
Ex
+ 1-x + txe3X x
!x + C2X).~ + tx"-'
19. y = ct cosx
21. Y = (Cl
+
41. Y = Cl
+ c2e x - x 4/4 - x 3
C1
cot x)]
8
+ c,.-2x -
-
3x 2 - 6x
t.' + x'/6 - x'/4 + x/4
45. Y = Cl COS X + C2 sin x + (x - tan x) cosx - sin x In (COS x) = Cl cosx + C2' sinx + x cosx - (sinx) In (cos x) 47. Y = celt" _lex
2
49. Y = ce lx + 5xe lx 51. Y = 2cosx + sinx - 1
+ sinxIn(secx + tanx) 2 53. Y = -e-x + 1 + lx - x 2 55. Y = 2(e X - e--x:l cosx - 3e-x sinx 57. y = (I-x +x')e'
Section 17.3, pp. 17-21 to 17-23
I .
COSX -
. + C2SmX
9. y = Clex
+ C2eRx + lxe Rx
~
X
59. Yp ~~, 4
Section 17.2, p. 17-16
,I
39. Y = Cl
43. Y
.
+ x smx
Ix + C2e -x - ge -"54 C, cos x + c, sin x - (sinx)[ln (cscx +
37. y =
-te-.n te-J:
I
X
21. Y = 29. y =
I
+ C2e + Ie-x + xe
35. Y = cle S.l'
25. y =
1 2.5.t" 1 S.l' +lOx • -25 xe
31. y = Cl COS X + C2 smx - IX cosx
19. Y = • -2x( C, cos V5x
+ c, sin V5x) Ct + C2X 23. Y = cte-2f + c2xe-2x cte-3x + c2xe-3x 27. y = cte-x/2 + C2xe-x/2 cte---xj3 + c2xe---xj3 31. y = +
.5.t"
-
cosx
1. my"
+ y' + Y
~
0,
y(O)
~
25" 3. 32 Y
2, y'(O)
~
2
5,
Vo
+ 40y ~ 0, y(0) ~ 12' y (0) ~ 12 5. 2q" + 4q' + 10q = 20 cos I, q(O) = 2, q'(O)
= 3
7. 0.0864 ft (above equilibrium) Vo
.
= 0.2917 cos (7.15521) + 85.8623 sm (7.15521) (in feet), or y = 3.5 cos (7.15521) +
9. y(l)
0.~;98 sin (7.15521) (in inches). 11. 0.308 sec
13. 8.3341b 15. 24.4949 ft/sec 17. -1.56 ft/sec' (acceleratioo upward)
+ 10.-2t,
19. q(l)
~
21. y(l)
= I + 2.-t -
-8. -3t
te-
2t _
lim q(l) ~ t___
0
~-St
23. y(".) = -2 m (above equilibrium) 25. q() l -- 5I
+
(49"VT99 . "VT99 995 sm 2 I + 49 5cOS "VT99) 2 I .-t/2 A-1
A-2
Chapter 17: Answers to Odd-Numbered Exercises
Section 17.4, p. 17-25
c,
1. y = 2 +
3. y=co(l-2x2 _
Ct
3. y - 2 +
C2 X
X
C2 X
3
5. Y = CtX 2 + C2X 4 7. Y = CtX- 1j3 + 9. Y ~ x(c, + C2lnX) 11. Y ~ x[c, cos (2 In x) + C2 sin (2 In x)] 13. Y
C2
~ ~[c, cos (3 In x) + C2 sin (3 In x)]
15. Y =
~C, cos (In x) + C2 sin (In x)]
1 17. Y = i(c,
21. Y
=
X
+ C2lnX)
~ ,k(c, + C2lnX)
1 x 25· y =2x'+I
19. Y
= c, + C2lnX
23. Y
~ x- S/ 4 (c, + C2 lnX)
27.y=x
29. Y = x[ -cos (In x)
+ 2 sin (In x)]
Section 17.5, p. 17-31 1. Y = Co
+ Cl(X
=co-
- x2 + ~3 -
Ct -2%
2e
... )
5. Y =
2x , + ... + ···)+c, (x3
)
cocos2x + clsin2x CtX
+ C2X 2
7. Y = CO(I
+ t x2 - ix' + ... ) + c, (x +
9. Y = Co ( 1 - x 2
ix' + ... )
5 4 - ... ) + c'x + 12x
11. Y = co(l - 3x2 + ... ) + c,(x - x') 13. y = CO(I
t
+ x 2 + x4 + ... )
+ Ct (x + x3 + ~xs + ... ) 15. Y = CO(I -
~X2 +
17. y = co( 1 -
~x2 + kx4 +
... ) + c, (x - tx' + ... ... ) +
Ct
)
(x - kX3)
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.
+
VECTOR OPERATOR FORMULAS (CARTESIAN FORM) Formulas for Grad, Diy, Curl, and the Laplacian Cartesian (x, y, z) i, j, and k are llllit vectors in the directions of increasing x, y, and z. M, N, and P are the scalar components of F(x, y, z) in these directions. Gradient
Vf ~ af i + af, + af k ax ayl az
Divergence
V'F~-+-+-
aM ax
aN ay
The Fundamental Theorem of Line Integrals
1.
Let F ~ Mi + Nj + Pk be a vector field whose components are continuous throughout an open connected region D in space. Then there exists a differentiable function f such that F
~ Vf ~ af i + af. + af k ax
ayl
az
if and only if for all points A and E in D the value of pendent of the path j oining A to E in D. 2.
f::
F' dr is inde-
If the integral is independent of the path from A to E, its value is
ap az
Green's Theorem and Its Generalization to Three Dimensions i VXF~
Cnrl
j
k
a
a
a
ax
iIy
az
M
N
P
Normal form of Green's Theorem:
iF.ndS
Divergence Theorem: a2 f a2 f a2 f VZf~ + + ax 2 ay 2 az 2
Laplacian
R
/ /F' n d
Tangential form of Green's Theorem:
iF' dr
u X (v X w)
(v X w)'u ~
~
Stokes' Theorem:
iF • dr
(w X u)'v
V' F dV
D
~
/ / V X F' k dA R
C
~
~ ///
S
Vector Triple Products (u X v)'w
~ //V'FdA
C
~
C
//
V X F' n d
S
(u'w)v - (u'v)w
Vector Identities Iu the identities here, constants.
f
and g are differentiable scalar functions, F, F j , and F 2 are differentiable vector fields, and a and b are real
VX(Vf)~O
V(Jg)
fVg + gVf
~
V'(gF)
~
V X (gF)
gV'F + Vg'F ~
gV X F + Vg X F
V'(aF j + bF2 )
~
V X (aF j + bF2 ) V(F j • F 2 )
~
aV'F j + bV'F2 ~
aV X F j + bV X F 2
(F j • V)F2 + (F2 ' V)F j +
F j X (V X F 2 ) + F 2 X (V X F j )
V X (F j X F 2 ) ~ (F2 ' V)F j (V, F 2 )F j - (V, F j )F2
-
(F j • V)F2 +
V X (V X F) ~ V(V' F) - (V, V)F ~ V(V' F) - VZF (V X F) X F
~
(F' V)F - tV(F'F)
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
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