Calculus [2nd ed.] 9781319248482, 1319248489

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CALCULUS EARLY TRANSCENDENTALS

SECOND EDITION

Michael Sullivan Chicago State University

Kathleen Miranda State University of New York, Old Westbury

Vice President, STEM: Daryl Fox Program Director: Andrew Dunaway Program Manager: Nikki Miller Dworsky Development Editor: Andrew Sylvester Associate Editor: Andy Newton Editorial Assistant: Justin Jones Senior Marketing Manager: Nancy Bradshaw Marketing Assistant: Savannah DiMarco Executive Media Editor: Catriona Kaplan Assistant Media Editor: Doug Newman Director of Digital Production: Keri deManigold Senior Media Project Manager: Alison Lorber Photo Editor: Christine Buese Photo Research: Donna Ranieri Director of Content Management Enhancement: Tracey Kuehn Managing Editor: Lisa Kinne Director of Design, Content Management: Diana Blume Text Design: Diana Blume Senior Content Project Manager: Edgar Doolan Senior Workflow Manager: Paul Rohloff Illustration Coordinator: Janice Donnola Illustrations: Network Graphics Composition: Lumina Datamatics Ltd. ISBN-13: 978-1-319-24848-2 (epub) © 2019, 2014 by W. H. Freeman and Company All rights reserved. 1 2 3 4 5 6

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What Is Calculus? Calculus is a part of mathematics that evolved much later than other subjects. Algebra, geometry, and trigonometry were developed in ancient times, but calculus as we know it did not appear until the seventeenth century. The first evidence of calculus has its roots in ancient mathematics. For example, In his book A History of π, Petr Beckmann explains that Greek mathematician Archimedes (287– 212 BCE) “took the step from the concept of ‘equal to’ to the concept of ‘arbitrarily close to’ or ‘as closely as desired’…and reached the threshold of the differential calculus, just as his method of squaring the parabola reached the threshold of the integral calculus.”* But it was not until Sir Isaac Newton and Gottfried Wilhelm Leibniz, each working independently, expanded, organized, and applied these early ideas, that the subject we now know as calculus was born. Although we attribute the birth of calculus to Newton and Leibniz, many other mathematicians, particularly those in the eighteenth and nineteenth centuries, contributed greatly to the body and rigor of calculus. You will encounter many of their names and contributions as you pursue your study of calculus. But what is calculus? The simple answer is: calculus models change. Since the world and most things in it are constantly changing, mathematics that explains change becomes immensely useful. Calculus has two major branches, differential calculus and integral calculus. Let’s take a peek at what calculus is by looking at two problems that prompted the development of calculus.

The Tangent Problem—The Basis of Differential Calculus Suppose we want to find the slope of the line tangent to the graph of a function at some point P=(x1, y1). See Figure 1(a). Since the tangent line necessarily contains the point P, it remains only to find the slope to identify the tangent line. Suppose we repeatedly zoom in on the graph of the function at the point P. See Figure 1(b). If we can zoom in close enough, then the graph of the function will look approximately linear, and we can choose a point Q, on the graph of the function different from the point P, and use the formula for slope.

Figure 1 Repeatedly zooming in on the point P is equivalent to choosing a point Q closer and closer to the point P. Notice that as we zoom in on P, the line connecting the points P and Q, called a secant line, begins to look more and more like the tangent line to the graph of the function at the point P. If the point Q can be made as close as we please to the point P, without equaling the point P, then the slope of the tangent line mtan can be found. This formulation leads to differential calculus, the study of the derivative of a function. The derivative gives us information about how a function changes at a given instant and can be used to solve problems involving velocity and acceleration; marginal cost and profit; and the rate of change of a chemical reaction. Derivatives are the subjects of Chapters 2 through 4.

The Area Problem—The Basis of Integral Calculus If we want to find the area of a rectangle or the area of a circle, formulas are available. (See Figure 2.) But what if the figure is curvy, but not circular as in Figure 3? How do we find this area?

Figure 2

Figure 3 Calculus provides a way. Look at Figure 4(a). It shows the graph of y=x2 x=0 to x=1. Suppose we want to find the area of the shaded region.

from

Figure 4 By subdividing the x-axis from 0 to 1 into small segments and drawing a rectangle of height x2 above each segment, as in Figure 4(b), we can find the area of each rectangle and add them together. This sum approximates the shaded area in Figure 4(a). The smaller we make the segments of the x-axis and the more rectangles we draw, the better the approximation becomes. See Figure 4(c). This formulation leads to integral calculus, and the study of the integral of a function.

Two Problems—One Subject? At first differential calculus (the tangent problem) and integral calculus (the area problem)

appear to be different, so why call both of them calculus? The Fundamental Theorem of Calculus establishes that the derivative and the integral are related. In fact, one of Newton’s teachers, Isaac Barrow, recognized that the tangent problem and the area problem are closely related, and that derivatives and integrals are inverses of each other. Both Newton and Leibniz formalized this relationship between derivatives and integrals in the Fundamental Theorem of Calculus. *Beckmann, P. (1976). A History of π (3rd. ed., p. 64). New York: St. Martin’s Press.

Contents | Calculus, Early Transcendentals, Second Edition Preface Applications Index

P Preparing for Calculus P.1 Functions and Their Graphs P.2 Library of Functions; Mathematical Modeling P.3 Operations on Functions; Graphing Techniques P.4 Inverse Functions P.5 Exponential and Logarithmic Functions P.6 Trigonometric Functions P.7 Inverse Trigonometric Functions P.8 Sequences; Summation Notation; the Binomial Theorem

1 Limits and Continuity 1.1 Limits of Functions Using Numerical and Graphical Techniques 1.2 Limits of Functions Using Properties of Limits 1.3 Continuity 1.4 Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions 1.5 Infinite Limits; Limits at Infinity; Asymptotes 1.6 The ε−δ

Definition of a Limit

Chapter Review Chapter Project: Pollution in Clear Lake

2 The Derivative 2.1 Rates of Change and the Derivative 2.2 The Derivative as a Function 2.3 The Derivative of a Polynomial Function; The Derivative of y=ex 2.4 Differentiating the Product and the Quotient of Two Functions; Higher-Order Derivatives 2.5 The Derivative of the Trigonometric Functions Chapter Review Chapter Project: The Lunar Module

3 More About Derivatives 3.1 The Chain Rule 3.2 Implicit Differentiation 3.3 Derivatives of the Inverse Trigonometric Functions 3.4 Derivatives of Logarithmic Functions 3.5 Differentials; Linear Approximations; Newton’s Method 3.6 Hyperbolic Functions Chapter Review Chapter Project: World Population

4 Applications of the Derivative 4.1 Related Rates

4.2 Maximum and Minimum Values; Critical Numbers 4.3 The Mean Value Theorem 4.4 Local Extrema and Concavity 4.5 Indeterminate Forms and L’Hôpital’s Rule 4.6 Using Calculus to Graph Functions 4.7 Optimization 4.8 Antiderivatives; Differential Equations Chapter Review Chapter Project: The U.S. Economy

5 The Integral 5.1 Area 5.2 The Definite Integral 5.3 The Fundamental Theorem of Calculus 5.4 Properties of the Definite Integral 5.5 The Indefinite Integral; Method of Substitution 5.6 Separable First-Order Differential Equations; Uninhibited and Inhibited Growth and Decay Models Chapter Review Chapter Project: Managing the Klamath River

6 Applications of the Integral 6.1 Area Between Graphs 6.2 Volume of a Solid of Revolution: Disks and Washers 6.3 Volume of a Solid of Revolution: Cylindrical Shells 6.4 Volume of a Solid: Slicing 6.5 Arc Length; Surface Area of a Solid of Revolution 6.6 Work 6.7 Hydrostatic Pressure and Force 6.8 Center of Mass; Centroid; The Pappus Theorem Chapter Review Chapter Project: Determining the Amount of Concrete Needed for a Cooling Tower

7 Techniques of Integration 7.1 Integration by Parts 7.2 Integrals Containing Trigonometric Functions 7.3 Integration Using Trigonometric Substitution: Integrands Containing a2−x2,x2+a2, or x2−a2, a>0

7.4 Integrands Containing ax2+bx+c, a≠0 7.5 Integration of Rational Functions Using Partial Fractions; the Logistic Model 7.6 Approximating Integrals: The Trapezoidal Rule, the Midpoint Rule, Simpson’s Rule 7.7 Improper Integrals 7.8 Integration Using Tables and Computer Algebra Systems

7.9 Mixed Practice Chapter Review Chapter Project: The Birds of Rügen Island

8 Infinite Series 8.1 Sequences 8.2 Infinite Series 8.3 Properties of Series; Series with Positive Terms; the Integral Test 8.4 Comparison Tests 8.5 Alternating Series; Absolute Convergence 8.6 Ratio Test; Root Test 8.7 Summary of Tests 8.8 Power Series 8.9 Taylor Series; Maclaurin Series 8.10 Approximations Using Taylor/Maclaurin Expansions Chapter Review Chapter Project: How Calculators Calculate

9 Parametric Equations; Polar Equations 9.1 Parametric Equations 9.2 Tangent Lines 9.3 Arc Length; Surface Area of a Solid of Revolution 9.4 Polar Coordinates 9.5 Polar Equations; Parametric Equations of Polar Equations; Arc Length of Polar Equations 9.6 Area in Polar Coordinates 9.7 The Polar Equation of a Conic Chapter Review Chapter Project: Polar Graphs and Microphones

10 Vectors; Lines, Planes, and Quadric Surfaces in Space 10.1 Rectangular Coordinates in Space 10.2 Introduction to Vectors 10.3 Vectors in the Plane and in Space 10.4 The Dot Product 10.5 The Cross Product 10.6 Equations of Lines and Planes in Space 10.7 Quadric Surfaces Chapter Review Chapter Project: The Hall Effect

11 Vector Functions 11.1 Vector Functions and Their Derivatives 11.2 Unit Tangent and Principal Unit Normal Vectors; Arc Length

11.3 Arc Length as Parameter; Curvature 11.4 Motion Along a Curve 11.5 Integrals of Vector Functions; Projectile Motion 11.6 Application: Kepler’s Laws of Planetary Motion Chapter Review Chapter Project: How to Design a Safe Road

12 Functions of Several Variables 12.1 Functions of Two or More Variables and Their Graphs 12.2 Limits and Continuity 12.3 Partial Derivatives 12.4 Differentiability and the Differential 12.5 Chain Rules Chapter Review Chapter Project: Searching for Exoplanets

13 Directional Derivatives, Gradients, and Extrema 13.1 Directional Derivatives; Gradients 13.2 Tangent Planes 13.3 Extrema of Functions of Two Variables 13.4 Lagrange Multipliers Chapter Review Chapter Project: Measuring Ice Thickness on Crystal Lake

14 Multiple Integrals 14.1 The Double Integral over a Rectangular Region 14.2 The Double Integral over Nonrectangular Regions 14.3 Double Integrals Using Polar Coordinates 14.4 Center of Mass; Moment of Inertia 14.5 Surface Area 14.6 The Triple Integral 14.7 Triple Integrals Using Cylindrical Coordinates 14.8 Triple Integrals Using Spherical Coordinates 14.9 Change of Variables Using Jacobians Chapter Review Chapter Project: The Mass of Stars

15 Vector Calculus 15.1 Vector Fields 15.2 Line Integrals of Scalar Functions 15.3 Line Integrals of Vector Fields; Work 15.4 Fundamental Theorem of Line Integrals 15.5 Green’s Theorem

15.6 Parametric Surfaces 15.7 Surface and Flux Integrals 15.8 The Divergence Theorem 15.9 Stokes’ Theorem Chapter Review Chapter Project: Modeling a Tornado

16 Differential Equations 16.1 Classification of Ordinary Differential Equations 16.2 Separable and Homogeneous First-Order Differential Equations; Slope Fields; Euler’s Method 16.3 Exact Differential Equations 16.4 First-Order Linear Differential Equations; Bernoulli Differential Equations 16.5 Power Series Methods Chapter Review Chapter Project: The Melting Arctic Ice Cap

APPENDIX A Precalculus Used in Calculus A.1 Algebra Used in Calculus A.2 Geometry Used in Calculus A.3 Analytic Geometry Used in Calculus A.4 Trigonometry Used in Calculus

APPENDIX B Theorems and Proofs B.1 Limit Theorems and Proofs B.2 Theorems and Proofs Involving Inverse Functions B.3 Derivative Theorems and Proofs B.4 Integral Theorems and Proofs B.5 A Bounded Monotonic Sequence Converges B.6 Taylor’s Formula with Remainder

APPENDIX C Technology Used in Calculus C.1 Graphing Calculators C.2 Computer Algebra Systems (CAS) Answers Index

Preface To the Student As you begin your calculus course, you may feel anxious about the large number of theorems, definitions, problems, and pages that you see in the book. Don’t worry, your concerns are normal. As educators we understand the challenges you face in this course, and as authors we have crafted a text that gives you the tools to meet them. If you pay attention in class, keep up with your homework, and read and study what is in this book, you will build the knowledge and skills you need to be successful. Here are some ways that you can use the book to your benefit. Read actively. When you are busy, it is tempting to skip reading and go right to the problems. Don’t! This book has examples with clear explanations to help you learn how to logically solve mathematics problems. Reading the book actively, with a pencil in hand, and doing the worked examples along with us will help you develop a clearer understanding than you will get from simple memorization. Read before class, not after. If you do, the classroom lecture will sound familiar, and you can ask questions about material you don’t understand. You’ll be amazed at how much more you get out of class if you read first. Use the examples. Mastering calculus requires a deep understanding of the concepts. Read and work the examples, and do the suggested NOW WORK Problems listed after them. Also do all the problems your instructor assigns. Be sure to seek help as soon as you feel you need it. Use the book’s features. In the classroom, we use many methods to communicate. These methods are reflected in various features that we have created for this book. These features, described on the following pages, are designed to make it easier for you to understand, review, and practice. Taking advantage of the features will help you to master calculus. Engage with the applications. Students come to this course with a wide variety of educational goals. You may be majoring in life sciences or social sciences, business or economics, engineering or mathematics. This book has many applied exercises—some written by students like you—related to courses in a wide variety of majors. Please do not hesitate to contact us with any questions, sug​gestions, or comments that might improve this text. We look forward to hearing from you, and we hope that you enjoy learning calculus and do well—not only in calculus but in all your studies. Best Wishes,

Michael Sullivan and Kathleen Miranda

To the Instructor The challenges facing calculus instructors are daunting. Diversity among students, both in their mathematical preparedness for learning calculus and in their ultimate educational and career goals, is vast and growing. Increased emphasis by universities and administrators on structured assessment and its related data analytics puts additional demands on instructors. There is not enough classroom time to teach every topic in the syllabus, to answer every student’s questions, or to delve into the rich examples and applications that showcase the beauty and utility of calculus. As mathematics instructors, we share these frustrations. As authors, our goal is to craft a calculus textbook that supports your teaching philosophy, promotes student understanding, and identifies measurable learning outcomes. Calculus, Early Transcendentals, Second Edition, is a mathematically precise calculus book that embraces proven pedagogical strategies to increase both student and instructor success. The language of the text is simple, clear, and precise. Definitions and theorems are carefully stated, and titled when appropriate. Numbering of definitions, equations, and theorems is kept to a minimum. The mathematics is rigorous. Proofs of theorems are presented throughout the text in one of three ways: (1) following the statement of a theorem; (2) preceding a theorem followed by “We have proved the following theorem”; or (3) in Appendix B, if the proof is more involved. (When the proof of a theorem is beyond the scope of this book, we note that it may be found in advanced calculus or numerical analysis texts.) The text is adaptable to your class and your teaching style. Proofs can be included or omitted and exercises are paired, grouped, and graduated from easy to difficult. By dividing exercises into Skill Building, Applications and Extensions, and Challenge Problems, we make it easier for you to tailor assignments to the specific needs of your students based on their ability and preparedness. The structural pedagogical features, described on the pages that follow, can be used to design an assessment tool for your students. Learning objectives are listed at the start of each section; the structure of the solved examples and/or the proofs can be used to write rubrics for evaluation; and assessment questions can be drawn from the Now Work Problems, end of section exercises, and chapter review problems. Additional media and supplement resources available to you and your students are described on page xiii. Let us know what you think of the text and do not hesitate to contact us with any suggestions or comments that might improve it. We look forward to hearing from you.

Best Wishes, Michael Sullivan and Kathleen Miranda

Pedagogical Features Promote Student Success Just In Time Review Throughout the text there are margin notes labeled NEED TO REVIEW? followed by a topic and page references. These references point to a previous presentation of a concept related to the discussion at hand. RECALL margin notes provide a quick refresher of key results that are being used in theorems, definitions, and examples.

IN WORDS These notes translate complex formulas, theorems, proofs, rules, and definitions using plain language that provide students with an alternate way to learn the concepts.

CAUTIONS and NOTES provide supporting details or warnings about common pitfalls. ORIGINS give biographical information or historical details about key figures and discoveries in the development of calculus.

Effective Use of Color The text contains an abundance of graphs and illustrations that carefully utilize color to make concepts easier to visualize. Dynamic Figures The text includes many pieces of art that students can interact with through the online e-Book. These dynamic figures, indicated by the icon next to the figure label, illustrate select principles of calculus, including limits, rates of change, solids of revolution, convergence, and divergence.

Learning Objectives Each section begins with a set of Objectives that serve as a broad outline of the section. The objectives help students study effectively by focusing attention on the concepts being covered. Each learning objective is supported by appropriate definitions, theorems, and proofs. One or more carefully chosen examples enhance the learning objective, and where appropriate, an application example is also included. Learning objectives help instructors prepare a syllabus that includes the important topics of calculus, and concentrate instruction on mastery of these topics.

Examples with Detailed and Annotated Solutions Examples are named according to their purpose. The text includes hundreds of examples with detailed step-by-step solutions. Additional annotations provide the formula used, or the reasoning needed, to perform a particular step of the solution. Where procedural steps for solving a type of problem are given in the text, these steps are followed in the solved examples. Often a figure or a graph is included to complement the solution.

Chapter Projects The case studies that open each chapter demonstrate how major concepts apply to recognizable and often contemporary situations in biology, environmental studies, astronomy, engineering, technology, and other fields. At the end of the chapter, there is an extended project with questions that guide students through a solution to the situation.

Immediate Reinforcement of Skills Following most examples there is the statement, NOW WORK. This callout directs students to a related exercise from the section’s problem set. Doing a related problem immediately after working through a solved example encourages

active learning, enhances understanding, and strengthens a student’s ability to use the objective. This practice also serves as a confidence-builder for students.

The Exercises are Divided into Four Categories Concepts and Vocabulary provide a selection of quick fill-in-the-blank, multiple choice, and true/false questions. These problems assess a student’s comprehension of the main points of the section.

Skill Building problems are grouped into subsets, usually corresponding to the objectives of the section. Problem numbers in red are the NOW WORK problems that are correlated to the solved examples. The skill building problems are paired by odd/even and usually progress from easy to difficult.

Applications and Extensions consist of applied problems, questions that require proof, and problems that extend the concepts of the section. (See the Applications Index for a full list of the applied examples and problems in the text.)

Challenge Problems provide more difficult extensions of the section’s material. They are often combined with concepts learned in previous chapters. Challenge problems are intended to be thought-provoking and require some ingenuity to solve—they are an excellent addition

to assignments for exceptionally strong students or for use in group work.

The Chapter Review is in Three Parts Things to Know is a detailed list of important definitions, formulas, and theorems contained in the chapter, and page references to where they can be found.

Objectives Table is a section-by-section list of each learning objective of the chapter (with page references), as well as references to the solved examples and review problems related to that learning objective.

Review Problems provide exercises related to the key concepts of the chapter and are matched to the learning objectives of each section of the chapter.

New to the Second Edition The Second Edition of Calculus, Early Transcendentals, is the culmination of more than 100 reviews of the first edition, more than 20 focus groups, and 4 years of work. Contributions came from a variety of sources: students who studied from the first edition, as well as professors at community colleges, 4-year universities, and research universities, some of whom taught from the first edition and some of whom did not. We also commissioned a team of reputable mathematicians to review every aspect of the revised presentation. Using this invaluable input, as well as insights from our own experiences teaching calculus, we have revised and modified as needed, while preserving the writing style and presentation that students using the first edition praise. We acknowledge the crucial importance of clarity and rigor, particularly considering the investment an instructor makes when adopting a calculus textbook. In preparing for this revision, we analyzed every component of the text: content, organization, examples, exercises, rigor, mathematical accuracy, depth and breadth of coverage, accessibility, and readability. Reviewers suggested how and where to revise, cut, deepen, or expand the

coverage. Many offered suggestions for new exercises or nuances to enhance a student’s learning experience. Wherever new Objectives or Examples were added, corresponding problems in the exercise sets were also added. A brief description of the major changes to the content and organization in the second edition is provided below. It is by no means comprehensive and does not reflect the more subtle, but no less important, changes we made in our quest to increase clarity and deepen conceptual understanding.

Chapter P Preparing for Calculus Sequences; Summation Notation; the Binomial Theorem has been moved from Appendix A to Chapter P, and now includes a full exercise set.

Chapter 2 The Derivative Section 2.2 includes NEW EXAMPLES on obtaining information about a function from the graph of its derivative function and determining whether a (piecewise-defined) function has a derivative at a number.

Chapter 3 More About Derivatives Implicit Differentiation and Derivatives of Inverse Trigo​nometric Functions are now treated in separate sections. The discussion of Taylor Polynomials (previously Section 3.5) is now in Chapter 8 (Infinite Series).

Chapter 4 Applications of the Derivative Section 4.3 includes a NEW EXAMPLE on obtaining information about the graph of a function from the graph of its derivative. Section 4.4 includes NEW EXAMPLES on obtaining information about the graph of a function f from the graphs of f′ and f″.

Chapter 5 The Integral Section 5.2 The Definite Integral has been rewritten and expanded to accommodate the following objectives: Form Riemann sums Define a definite integral as a limit of Riemann sums Approximate a definite integral using Riemann sums Know conditions that guarantee a definite integral exists Find a definite integral using the limit of Riemann sums Form Riemann sums from a table

To improve conceptual understanding, three new interpretations of the definite integral have been included: Section 5.3: Objective 3: Interpret the integral as a rate of change, accompanied by two new examples, and Objective 4: Interpret the integral as an accumulation function, accompanied by a new example. Section 5.4: Objective 4: Interpreting integrals involving rectilinear motion. Section 5.5: The Indefinite Integral and Substitution are now treated in one section. Section 5.6 now covers separable first-order differential equations and growth and decay models (both uninhibited and inhibited).

Chapter 6 Applications of the Integral Section 6.5: NEW Objective 3: Find the surface area of a solid of revolution.

Chapter 7 Techniques of Integration Section 7.1: NEW Objective 2: Find a definite integral using integration by parts. Section 7.2: NEW Summary tables are now included. Section 7.5: NEW Objective 5: Solve the logistic differen​tial equation. Section 7.6: NEW Objective 2: Approximate an integral using the Midpoint Rule. NEW Section 7.9: Mixed Practice provides exercises that “mix up” all the integrals.

Chapter 8 Infinite Series Section 8.3: NEW Objective 5: Approximating the sum of a convergent series. Section 8.4 now includes an expanded form of the Limit Comparison Test. Section 8.10: NEW Objective 1: Approximate a function and its graph using a Taylor Polynomial.

Chapter 9 Parametric Equations; Polar Equations Arc length, formerly in Section 9.2, now appears in Section 9.3.

Chapter 11 Vector Functions Section 11.5 has been expanded to include more examples of both indefinite and definite integrals of vector functions.

Chapter 12 Functions of Several Variables Section 12.3 now includes an example that uses the definition to find the partial

derivatives of a function on two variables, and an example that uses the Chain Rule to find partial derivatives has been added.

Chapter 13 Directional Derivatives, Gradients, and Extrema Section 13.1 has an increased emphasis on the geometric interpretation of the gradient. Section 13.2: NEW Objective 3: Find an equation of a tangent plane to a surface defined explicitly.

Chapter 14 Multiple Integrals The presentation now uses double and triple Riemann sums to define double and triple integrals and to develop applications.

Chapter 15 Vector Calculus The presentation in Sections 15.2, 15.3, and 15.4 now distinguishes between line integrals of scalar functions and line integrals of vector functions.

Chapter 16 Differential Equations Section 16.2 has two new topics Using a Slope Field. Euler’s Method to Approximate Solutions to a First-order Differential Equation.

WebAssign Premium Macmillan’s integrates Sullivan/Miranda’s Calculus into a powerful online system with an interactive e-book, powerful answer evaluator, and easy-to-assign exercise bank that includes algorithmically generated homework and quizzes. Combined with Macmillan’s esteemed content including “CalcTools” (described below), WebAssign Premium offers course and assignment customization to extend and enhance the classroom experience for instructors and students. Student resources available through WebAssign Premium include, Macmillan’s CalcTools Dynamic Figures: Around 100 figures from the text have been recreated in an interactive format for students to manipulate and explore as they master key concepts. Tutorial videos explain how to use the figures. CalcClips: These whiteboard tutorial videos provide a step-by-step walk-through

illustrating key concepts from examples adapted from the book. This powerful, self-paced formative assessment tool provides instant feedback tied to specific sections of the text. Question difficulty level and topic selection adapt based on the individual student’s performance. Tutorial questions that break up questions into segments to help students work through learning a concept. A full, interactive, and easily navigated e-book with highlighting and notetaking features. Additional student resources, including the text’s student solutions manual, MapleTM Manual, and Mathematica® Manual. And for instructors, Over 7,000 exercises culled directly from the end-of-chapter sections of the text, with detailed solutions available to students at your discretion. Ready-to-use Course Pack Assignments created and curated from the full exercise bank greatly decreases your preparation time. A suite of Instructor Resources, including iClicker questions, Instructor’s Manual, PowerPoint lecture slides, a printable test bank, and more.

WeBWorK webwork.maa.org Macmillan Learning offers hundreds of algorithmically generated questions (with full solutions) authored directly from the book through this free, open-source online homework system created at the University of Rochester. Adopters also have access to a shared national library test bank with thousands of additional questions, including questions tagged to the book’s topics.

Additional Resources For Instructors Instructor’s Solutions Manual Contains worked-out solutions to all exercises in the text. ISBN: (complete) 978-1-319-10828-1; (SV) 978-1-319-10836-6; (MV) 978-1-319-10833-5 Test Bank Includes a comprehensive set of multiple-choice test items.

Instructor’s Resource Manual Provides sample course outlines, suggested class time, key points, lecture materials, discussion topics, class activities, worksheets, projects, and questions to accompany the dynamic figures. ISBN: 978-1-319-10828-1 iClicker is a two-way radio-frequency classroom response solution developed by educators for educators. Each step of iClicker’s development has been informed by teaching and learning. To learn more about packaging iClicker with this textbook, please contact your local sales representative or visit www.iclicker.com. Lecture Slides offer a customizable, detailed lecture presentation of concepts covered in each chapter. Image Slides contain all textbook figures and tables. For Students Student Solutions Manual Contains worked-out solutions to all odd-numbered exercises in the text. ISBN: (SV) 978-1-319-06756-4; (MV) 978-1-319-06754-0 Software Manuals MapleTM and Mathematica® software manuals serve as basic introductions to popular mathematical software options.

Acknowledgments Ensuring the mathematical rigor, accuracy, and precision, as well as the complete coverage and clear language that we have strived for with this text, requires not only a great deal of time and effort on our part, but also on the part of an entire team of exceptional reviewers. The following people have also provided immeasurable support reading drafts of chapters and offering insight and advice critical to the success of the second edition: Kent Aeschliman, Oakland Community College Anthony Aidoo, Eastern Connecticut State University Martha Allen, Georgia College & State University Jason Aran, Drexel University Alyssa Armstrong, Wittenberg University Mark Ashbrook, Arizona State University Beyza Aslan, University of North Florida Dr. Mathai Augustine, Cleveland State Community College C. L. Bandy, Texas State University Scott Barnett, Henry Ford College William C. Bauldry, Appalachian State University Nick Belloit, Florida State College at Jacksonville Robert W. Benim, University of Georgia Daniel Birmajer, Nazareth College Dr. Benkam B. Bobga, University of North Georgia, Gainesville Campus Laurie Boudreaux, Nicholls State University Alain Bourget, California State University, Fullerton David W. Boyd, Valdosta State University James R. Bozeman, Ph.D., Lyndon State College Naala Brewer, Arizona State University, Tempe Campus Light Bryant, Arizona Western College Meghan Burke, Kennesaw State University James R. Bush, Waynesburg University Sam Butler, University of Nebraska, Omaha Joseph A. Capalbo, Bryant University Debra Carney, Colorado School of Mines Stephen N. Chai, Miles College David Chan, Virginia Commonwealth University Youn-Sha Chan, University of Houston—Downtown E. William Chapin, Jr., University of Maryland Eastern Shore Kevin Charlwood, Washburn University, Topeka

Jeffrey Cohen, El Camino College Hugh Cornell, University of North Florida Ana-Maria Croicu, Kennesaw State University Charles Curtis, Missouri Southern State University Seth Daugherty, St. Louis Community College Christopher B. Davis, Tennessee Technological University Shirley Davis, South Plains College John C. D. Diamantopoulos, Northeastern State University Roberto Diaz, Fullerton College Tim Doyle, DePaul University Deborah A. Eckhardt, St. Johns River State College Steven Edwards, Kennesaw State University Amy H. Erickson, Georgia Gwinnett College Keith Erickson, Georgia Gwinnett College Karen Ernst, Hawkeye Community College Nancy Eschen, Florida State College at Jacksonville Kevin Farrell, Lyndon State College Ruth Feigenbaum, Ph.D., Bergen Community College Mohammad Ganjizadeh, Tarrant County College Dennis Garity, Oregon State University Tom Geil, Milwaukee Area Technical College Bekki George, University of Houston Dr. Jeff Gervasi, Porterville College Aaron Gibbs, Johnson County Community College William Griffiths IV, Kennesaw State University Mark Grinshpon, Georgia State University Gary Grohs, Elgin Community College Boyko Gyurov, Georgia Gwinnett College Teresa Hales Peacock, Nash Community College Sami M. Hamid, University of North Florida Craig Hardesty, Hillsborough Community College Alan T. Hayashi, Oxnard College Mary Beth Headlee, The State College of Florida—Venice Campus Beata Hebda, University of North Georgia Piotr Hebda, University of North Georgia Shahryar Heydari, Piedmont College Kaat Higham, Bergen Community College Syed Hussain, Orange Coast College Mohamed Jamaloodeen, Georgia Gwinnett College Dr. Christopher C. Jett, University of West Georgia Paul W. Jones II, University of South Alabama

Victor Kaftal, University of Cincinnati Dr. Robin S. Kalder, Central Connecticut State University John Khoury, Eastern Florida State College Minsu Kim, University of North Georgia Wei Kitto, Florida State College at Jacksonville Ashok Kumar, Valdosta State University Hong-Jian Lai, West Virginia University Carmen Latterell, University of Minnesota Duluth Barbara A. Lawrence, Borough of Manhattan Community College Richard C. Le Borne, Tennessee Technological University Glenn Ledder, University of Nebraska, Lincoln Jeffrey Ledford, Virginia Commonwealth University Namyong Lee, Minnesota State University, Mankato Rowan Lindley, Westchester Community College Yung-Way Liu, Tennessee Technological University Doron Lubinsky, Georgia Institute of Technology Frank Lynch, Eastern Washington University Jeffrey W. Lyons, Nova Southeastern University Filix Maisch, Oregon State University Mark Marino, Erie Community College Jeff McGowan, Central Connecticut State University John R. Metcalf, St. Johns River State College Ashod Minasian, El Camino College Matthew Mitchell, Florida State College at Jacksonville Val Mohanakumar, Hillsborough Community College Ronald H. Moore, Florida State College at Jacksonville Tom Morley, Georgia Institute of Technology Catherine Moushon, Elgin Community College Keith Nabb, Moraine Valley Community College Jeffrey Neugebauer, Eastern Kentucky University Shai Neumann, Eastern Florida State College Mike Nicholas, Colorado School of Mines Jon Oaks, Macomb Community College C. Altay Özgener, State College of Florida, Manatee—Sarasota Joshua Palmatier, State University of New York at Oneonta Sam Pearsall, Los Angeles Pierce College Stan Perrine, Georgia Gwinnett College Davidson B. Pierre, State College of Florida, Manatee—Sarasota Cynthia Piez, University of Idaho Daniel Pinzon, Georgia Gwinnett College Katherine Pinzon, Georgia Gwinnett College

Mihaela Poplicher, University of Cincinnati Elise Price, Tarrant County College Brooke P. Quinlan, Hillsborough Community College Mahbubur Rahman, University of North Florida Joel Rappoport, Florida State College at Jacksonville William Rogge, University of Nebraska—Lincoln Richard Rossi, Montana Tech of the University of Montana Bernard Rothman, Ramapo College Daniel Russow, Arizona Western College Kristen R. Schreck, Moraine Valley Community College Randy Scott, Santiago Canyon College Daniel Seabold, Hofstra University Vicki Sealey, West Virginia University Plamen Simeonov, University of Houston–Downtown Derrick Thacker, Northeast State Community College Jen Tyne, The University of Maine Jossy Uvah, University of West Florida William Veczko, St. Johns River State College H. Joseph Vorder Bruegge, Hillsborough Community College M. Vorderbruegge, Hillsborough Community College Kathy Vranicar, University of Nebraska, Omaha Kathryn C. Waddel, Waynesburg University Martha Ellen Waggoner, Simpson College Qing Wang, Shepherd University Yajni Warnapala, Roger Williams University Mike Weimerskirch, University of Minnesota Benjamin Wiles, Purdue University Catalina Yang, Oxnard College Dr. Michael A. Zeitzew, El Camino College Hong Zhang, University of Wisconsin Oshkosh Additional applied exercises, as well as many of the case study and chapter projects that open and close each chapter of this text were contributed by a team of creative individuals. We would like to thank all of our exercise contributors for their work on this vital and exciting component of the text: Wayne Anderson, Sacramento City College (Physics) Allison Arnold, University of Georgia (Mathematics) Kevin Cooper, Washington State University (Mathematics) Adam Graham-Squire, High Point University (Mathematics) Sergiy Klymchuk, Auckland University of Technology (Mathematics)

Eric Mandell, Bowling Green State University (Physics and Astronomy) Eric Martell, Millikin University (Physics and Astronomy) Barry McQuarrie, University of Minnesota, Morris (Mathematics) Rachel Renee Miller, Colorado School of Mines (Physics) Kanwal Singh, Sarah Lawrence College (Physics) John Travis, Mississippi College (Mathematics) Gordon Van Citters, National Science Foundation (Astronomy) We would like to thank the dozens of instructors who provided invaluable input throughout the development of the first edition of this text. Marwan A. Abu-Sawwa, University of North Florida Jeongho Ahn, Arkansas State University Weam M. Al-Tameemi, Texas A&M International University Martha Allen, Georgia College & State University Roger C. Alperin, San Jose State University Robin Anderson, Southwestern Illinois College Allison W. Arnold, University of Georgia Mathai Augustine, Cleveland State Community College Carroll Bandy, Texas State University Scott E. Barnett, Henry Ford Community College Emmanuel N. Barron, Loyola University Chicago Abby Baumgardner, Blinn College, Bryan Thomas Beatty, Florida Gulf State University Robert W. Bell, Michigan State University Nicholas G. Belloit, Florida State College at Jacksonville Mary Beth Headlee, State College of Florida, Manatee-Sarasota Daniel Birmajer, Nazareth College of Rochester Justin Bost, Rowan-Cabarrus Community College-South Campus Laurie Boudreaux, Nicholls State University Alain Bourget, California State University at Fullerton Jennifer Bowen, The College of Wooster David Boyd, Valdosta State University Brian Bradie, Christopher Newport University James Brandt, University of California, Davis Jim Brandt, Southern Utah University Light R. Bryant, Arizona Western College Kirby Bunas, Santa Rosa Junior College Dietrich Burbulla, University of Toronto Joni Burnette Pirnot, State College of Florida, Manatee-Sarasota James Bush, Waynesburg University

Shawna M. Bynum, Napa Valley College Joe Capalbo, Bryant University Mindy Capaldi, Valparaiso University Luca Capogna, University of Arkansas Deborah Carney, Colorado School of Mines Jenna P. Carpenter, Louisana Tech University Nathan Carter, Bentley University Vincent Castellana, Craven Community College Stephen Chai, Miles College Julie Clark, Hollins University Adam Coffman, Indiana University-Purdue University, Fort Wayne William Cook, Appalachian State University Sandy Cooper, Washington State University David A. Cox, Amherst College Mark Crawford, Waubonsee Community College Charles N. Curtis, Missouri Southern State University Larry W. Cusick, California State University, Fresno Alain D’Amour, Southern Connecticut State University Rajendra Dahal, Coastal Carolina University John Davis, Baylor University Ernesto Diaz, Dominican University of California Robert Diaz, Fullerton College Geoffrey D. Dietz, Gannon University Della Duncan-Schnell, California State University, Fresno Deborah A. Eckhardt, St. Johns River State College Karen Ernst, Hawkeye Community College Mark Farag, Fairleigh Dickinson University Kevin Farrell, Lyndon State College Judy Fethe, Pellissippi State Community College Md Firozzaman, Arizona State University Tim Flaherty, Carnegie Mellon University Walden Freedman, Humboldt State University Kseniya Fuhrman, Milwaukee School of Engineering Melanie Fulton Douglas R. Furman, SUNY Ulster Community College Rosa Garcia Seyfried, Harrisburg Area Community College Tom Geil, Milwaukee Area Technical College Jeff Gervasi, Porterville College William T. Girton, Florida Institute of Technology Darren Glass, Gettysburg College Jerome A. Goldstein, University of Memphis

Giséle Goldstein, University of Memphis Lourdes M. Gonzalez, Miami Dade College Pavel Grinfield, Drexel University Mark Grinshpon, Georgia State University Jeffery Groah, Lone Star College Gary Grohs, Elgin Community College Paul Gunnells, University of Massachusetts, Amherst Semion Gutman, University of Oklahoma Teresa Hales Peacock, Nash Community College Christopher Hammond, Connecticut College James Handley, Montana Tech Alexander L. Hanhart, New York University Gregory Hartman, Virginia Military Institute Karl Havlak, Angelo State University LaDawn Haws, California State University, Chico Janice Hector, DeAnza College Anders O.F. Hendrickson, St. Norbert College Shahryar Heydari, Piedmont College Max Hibbs, Blinn College Rita Hibschweiler, University of New Hampshire David Hobby, SUNY at New Paltz Michael Holtfrerich, Glendale Community College Keith E. Howard, Mercer University Tracey Hoy, College of Lake County Syed I. Hussain. Orange Coast College Maiko Ishii, Dawson College Nena Kabranski, Tarrent County College William H. Kazez, The University of Georgia Michael Keller, University of Tulsa Steven L. Kent, Youngstown State University Eric S. Key, University of Wisconsin, Milwaukee Michael Kirby, Colorado State University Stephen Kokoska, Bloomsburg University of Pennsylvania Alex Kolesnik, Ventura College Natalia Kouzniak, Simon Fraser University Sunil Kumar Chebolu, Illinois State University Ashok Kumar, Valdosta State University Geoffrey Laforte, University of West Florida Tamara J. Lakins, Allegheny College Justin Lambright, Anderson University Peter Lampe, University of Wisconsin-Whitewater

Donald Larson, Penn State Altoona Carmen Latterell, University of Minnesota, Duluth Glenn W. Ledder, University of Nebraska, Lincoln Namyong Lee, Minnesota State University, Mankato Denise LeGrand, University of Arkansas, Little Rock Serhiy Levkov and his team Benjamin Levy, Wentworth Institute of Technology James Li-Ming Wang, University of Alabama Aihua Li, Montclair State University Amy H. Lin Erickson, Georgia Gwinnett College Rowan Lindley, Westchester Community College Roger Lipsett, Brandeis University Joanne Lubben, Dakota Weslyan University Matthew Macauley, Clemson University Filix Maisch, Oregon State University Heath M. Martin, University of Central Florida Vania Mascioni, Ball State University Betsy McCall, Columbus State Community College Philip McCartney, Northern Kentucky University Kate McGivney, Shippensburg University Douglas B. Meade, University of South Carolina Jie Miao, Arkansas State University John Mitchell, Clark College Val Mohanakumar, Hillsborough Community College Catherine Moushon, Elgin Community College Suzanne Mozdy, Salt Lake Community College Gerald Mueller, Columbus State Community College Will Murray, California State University, Long Beach Kevin Nabb, Moraine Valley Community College Vivek Narayanan, Rochester Institute of Technology Larry Narici, St. John’s University Raja Nicolas Khoury, Collin College Bogdan G. Nita, Montclair State University Charles Odion, Houston Community College Giray Ökten, Florida State University Nicholas Ormes, University of Denver Chihiro Oshima, Santa Fe College Kurt Overhiser, Valencia College Edward W. Packel, Lake Forest College Joshua Palmatier, SUNY College at Oneonta Chad Pierson, University of Minnesota, Duluth

Cynthia Piez, University of Idaho Jeffrey L. Poet, Missouri Western State University Shirley Pomeranz, The University of Tulsa Vadim Ponomarenko, San Diego State University Elise Price, Tarrant County College, Southeast Campus Harald Proppe, Concordia University Frank Purcell, Twin Prime Editorial Michael Radin, Rochester Institute of Technology Jayakumar Ramanathan, Eastern Michigan University Joel Rappaport, Florida State College at Jacksonville Marc Renault, Shippensburg University Suellen Robinson, North Shore Community College William Rogge, University of Nebraska, Lincoln Yongwu Rong, George Washington University Amber Rosin, California State Polytechnic University Richard J. Rossi, Montana Tech of the University of Montana Bernard Rothman, Ramapo College of New Jersey Dan Russow, Arizona Western College Adnan H. Sabuwala, California State University, Fresno Alan Saleski, Loyola University Chicago John Samons, Florida State College at Jacksonville Brandon Samples, Georgia College & State University Jorge Sarmiento, County College of Morris Ned Schillow, Lehigh Carbon Community College Kristen R. Schreck, Moraine Valley Community College Randy Scott, Santiago Canyon College George F. Seelinger, Illinois State University Andrew Shulman, University of Illinois at Chicago Mark Smith, Miami University John Sumner, University of Tampa Geraldine Taiani, Pace University Barry A. Tesman, Dickinson College Derrick Thacker, Northeast State Community College Millicent P. Thomas, Northwest University Tim Trenary, Regis University Kiryl Tsishchanka and his team Pamela Turner, Hutchinson Community College Jen Tyne, University of Maine David Unger, Southern Illinois University-Edwardsville Marie Vanisko, Carroll College William Veczko, St. Johns River State College

James Vicknair, California State University, San Bernardino Robert Vilardi, Troy University, Montgomery David Vinson, Pellissippi Community College Klaus Volpert, Villanova University Bryan Wai-Kei, University of South Carolina, Salkehatchie David Walnut, George Mason University Lianwen Wang, University of Central Missouri Qing Wang, Shepherd University E. William Chapin, Jr., University of Maryland, Eastern Shore Rebecca Wong, West Valley College Kerry Wyckoff, Brigham Young University Jeffrey Xavier Watt, Indiana University-Purdue University Indianapolis Carolyn Yackel, Mercer University Catalina Yang, Oxnard College Yvonne Yaz, Milwaukee State College of Engineering Hong Zhang, University of Wisconsin, Oshkosh Qiao Zhang, Texas Christian University Qing Zhang, University of Georgia Because the student is the ultimate beneficiary of this material, we would be remiss to neglect their input in its creation. We would like to thank students from the following schools who have provided us with feedback and suggestions for exercises throughout the text, and have written application exercises that appear in many of the chapters: Barry University Boston University Bronx Community College California State University—Bakersfield Catholic University Colorado State University Idaho State University Lamar University Lander University University of Maryland Minnesota State University Millikin University University of Missouri—St. Louis Murray State University University of North Georgia North Park University University of North Texas

University of South Florida Southern Connecticut State University St. Norbert College Texas State University—San Marcos Towson State University Trine University State University of West Georgia University of Wisconsin—River Falls Finally, we would like to thank the editorial, production, and marketing teams at W. H. Freeman, whose support, knowledge, and hard work were instrumental in the publication of this text: Ruth Baruth and Terri Ward, who initially brought our text to Macmillan; Nikki Miller Dworsky, Andrew Sylvester, Katharine Munz, Justin Jones and Andy Newton, who formed the Editorial team; Diana Blume, Edgar Doolan, Paul Rohloff, Christine Buese, and Janice Donnola on the Project Management, Design, and Production teams; and Nancy Bradshaw and Leslie Allen on the Marketing and Market Development teams. Thanks as well to the diligent group at Lumina, lead by Sakthivel Sivalingam and Misbah Ansari, for their expert composition, and to Ron Weickart at Network Graphics for his skill and creativity in executing the art program. Michael Sullivan Kathleen Miranda

Applications Index Note: Italics indicates Example.

  Acoustics Configuring microphones, 732 Sound level of a leafblower, 240 Vibrating strings, 882 Aerodynamics Air resistance, 1126 Aeronautics Direction of an airplane, 756, 761 Paths of spacecraft, 783 Position of aircraft, 752 Speed and direction of aircraft, 742, 749, 761 Speed and direction of a bird, 761 Agriculture Crop yields and fertilizer, 291 Field enclosure, 945 Irrigation ditches, 937 Rate of a growing fruit on a tree, 163 Archaeology Age of a fossil, 416 Amount of carbon left in ancient man found in ice, 421 Carbon dating of early human fossilized remains, 211 Architecture Window design, 340 Concrete needed to build a cooling tower, 497 Area of a sphere, 724 of an enclosed region, 718–720, 723–724, 731 Art Rate of change in its value, 194 Astronomy Brightness of stars, 887 Density of a star, 1023 Escape speed, 1061

Event Horizon of a Black Hole, 270 Modeling a moon of Saturn, 1013 Motion of planets, 846 Orbit of a comet, 729 Orbit of Mercury, 729 Planetary orbit, 727, 846 Searching for exoplanets, 906 Biology Air flowing through a windpipe, 280 Bacterial growth, 1125 Chemical flow through bodies, 920 Yeast biomass as function of time, 309 Business Cost of a box, 945 of manufacturing, 13, 16, 945 of materials, 936, 942 of electricity, 113 of production, 16 of removing pollutants from a lake, 138 postal rates per ounce, 89, 113 Manufacturing, 270, 933, 945 Maximizing profit, 934, 937, 945, 948 Minimizing cost, 948 Production of ball bearings, 244 Product production, 860, 902 Profit generated by unnamed technology, 309 Revenue of an unnamed commodity, 206 Sales, 162, 1125 Chemistry Chemical reactions, 341 Crystals (structure of), 753, 737 Decomposition, 421 Decomposition reactions, 134, 138–139 Decrease of mass (glucose), 156–157 Energy within a system, 396 Growth of bacteria, 415–416, 420–421 Mixtures, 1141, 1150 Radioactive decay, 420, 425, 1148 Salt solutions, 421 Communications

Rate of texting, 1125 Speed of a telephone call, 341 Computers Graphics, 752 Construction Minimizing cost, 340 of a box, 936 of a dam, 971 of a rain gutter, 279 of building a fence, 339 Supporting beam of a wall, 340 Decorating Christmas lights, 257, 283 Demographics Population growth, 261, 1141–1142 U.S. citizens, no High School diploma, 26 of a rare bird species, 572 Design Maximizing plot of land, 339 of a cylinder, 945 of a play area, 334–335 Ski slope, 471 Distance Between cities, 710 Height of a building, 249 of an unnamed object, 268, 282, 295, 516, 554, 572 Traveled by a bee, 248 Traveled by a dropped ball, 596, 600 Traveled by motorcycle, 356 Economics Cobb-Douglas Model, 882, 902 Demand equations, 163, 194 Margin of productivity, 882 Market penetration of smartphones, 226 Mean earnings for workers 18 years and older, 217 Price of tablets, 220 Production rates of unnamed commodities, 226 Rate of productivity, 877 U.S. Budget, 309 U.S. Economy, 356 Education

Learning curve, 226 Student rating of faculty, 217 Test Scores, 89 Electricity Current in RL circuit, 319 Density of a current in a wire, 195 Drop in voltage, 421 Electrical charge, 771, 860, 1142 Electrical current, 195 Electrical field, 411, 860, 919 Electrical potential, 860 Flow over a cylinder, 1091 Height of a cable, 282 Kirchhoff‘s Law, 1142 Magnetic effect on a TV, 772 Magnetic fields, 860 Magnetism of a cord, 772 of electrical charge, 732 of electrical current, 772 Parallel circuits, 891 Potential, 919, 920 Rates of change for current, 188 Resistance, 217, 891, 905 Electronics Circuitry, 1126 Energy stored in a capacitor, 588 How calculators calculate, 678 Energy Nuclear chain reactions, 588 Solar, 925 Engineering Amount of earth removed for a train to pass, 553 Bridges (St. Louis Arch), 257 Cable length, 471 Distribution of structural forces, 398 Electrical currents, 563 Electrical theory, 563 Length of hanging piece of rope, 471 Measuring ice depths, 949 River management, 426

Suspension, 254, 257 Wearing down of ball bearing, 244 Entertainment Cable subscriptions, 326 Environment Air quality, 1142 Disposal of pollutant, 152 Melting ice caps, 1150 Oil leakage, 269 Pollution in Gulf of Mexico, 1131 Farming Field enclosure, 332 Finance Consumer consumption, 600 Cost of fuel, 282, 937 of health care, 435 of labor, 937 of mailing packages, 936 of mobile phone plans, 859 of removing pollutant from a lake, 194 of water, 101, 113 Diminishing returns, 600 Flow of currency, 1142 Maximizing profits, 356 Present value of an unnamed currency, 563 Price of wine, 340 Projected business revenues, 202 Revenue maximization, 339 Sale of stereos, 356 Sales of an unnamed object, 305 Tax rates, 283 Force (Hydrostatic) Gasoline in a tank, 485, 496 in a swimming pool, 485 in a trapezoid, 496 in a trough, 485 on a dam, 485, 496 on a floodgate, 485 on a submarine, 485 in an oil tank, 485

Resultant, 483–484 Forensic Sciences Change in body temperature following death 421, 426 Geography Longitude and latitude, 1011 Geometry Area of a circle, 165, 174, 397 of a disk, 248 of a rectangle, 268, 341, 356 of a triangle, 202, 496, 771, 891, 892 of an ellipse, 1064, 1110 enclosed by parabola, 1021 of a parallelogram, 768, 770 Centroid, 1083 Cone volume, 341 Cube (volume of), 250, 260, 268 Cylinder (volume of), 248 Density of an unnamed object, 397 Dimensions of a box, 945 Dimensions of a rectangle, 336, 339 Equation of a line, 875, 905 Finding the center of a sphere, 1013 Fractals, 600 Maximizing area of a triangle, 338 Parallelepiped (vectors), 753 Right triangle (area of), 366 of a river channel, 554 Snow ball (volume of), 268 Sphere (radius of), 268 Sphere (volume of), 250, 268 Surface area, 985, 989, 1018, 1019, 1021, 1079–1080 of a balloon, 264, 268 of a bead, 724 of a box, 860 of a circle, 1021 of a cone, 1022 of a cylinder, 1032, 1039 of a dome, 989 of a helicoid, 1080 of a hemisphere, 989

of a paraboloid, 989, 1022, 1110 of a plane, 989 of a plug, 724 of a pond, 554 of a sphere, 268, 724, 988, 990 of a torus, 1078, 1080 of an unnamed solid, 699, 700, 722, 731 Tetrahedron (vectors), 753 Trapezoid (area of), 366 Triangle (area of), 268, 271, 340 Triangle (dimensions of), 268 Volume of area removed from a sphere, 461 of a box, 945 of a circle drilled in sphere, 1006 of a cone, 456–457, 461, 496, 1099 of a cube, 163 of a cylinder, 162, 461, 888, 891, 946, 975, 997, 1022 of a liquid, 970 of a mountain, 1005 of a paraboloid, 994, 1019, 1022 of a parallelepiped, 771, 1099 of a pyramid, 458, 461 of a silo, 891 of a solid, 496, 948, 956–957, 959, 965, 968, 971, 977, 994, 998, 999, 1003, 1005, 1010, 1012, 1019, 1021 of a sphere, 174, 977, 1012, 1022 of a spherical cap, 1013 of a tetrahedron, 971, 1021 of a wedge removed from a cylinder, 461 of an ellipsoid, 999, 1021 of ice, 902 of intersecting pipes, 461, 1006 of water in a bowl, 461 of water in a glass, 461 of water in a tank, 977 Health Body mass index, 249 Human respiration rate, 162 Spread of an epidemic, 1139, 1141 Investment

Price of stocks, 600 Compound interest, 240 in CD’s, 240 Mass center of, 1005, 1012, 1021 intersection of two rods, 1005 of a cone, 1109 of a cube, 999 of a cylinder, 999, 1005 of fluid, 1094–1095 of a plane, 980–981 of a solid, 984, 1005, 1012 of a sphere, 1005, 1011, 1012, 1091 of a square, 983 of a tetrahedron, 993, 999 of a triangle, 979–980, 983 of a washer, 984 of a wire, 1032, 1039 Medicine Drug concentration, 194, 340, 377, 1126 Drug reaction, 563, 936, 937 Radiation from x-rays, 352 Spread of a disease, 308, 601 Meteorology Atmospheric pressure, 173, 217 Heat index, 859 Modeling a tornado, 1111 Temperature zones, 859 Military Path of a falling bomb, 183 Miscellaneous Area of ground illuminated by a light, 342 Baking a potato, 111 Carrying a pipe around a corner, 341 Density of an object, 892 Fashioning a metal box, 333 Human intelligence, 672 Man walking in relation to the height of a tower, 268 Maximizing carry-on luggage, 945 Maximizing the volume of a box, 339, 355 Metal detectors, 937

Minimizing distance between two numbers, 341 Minimizing the surface area of a box, 339 Optimal dimensions for a can, 341 Optimal viewing angle for art on a wall, 341 Optimal wire length, 340 Path between two homes, 339 Weight lifting, 771 Motion of a ball, 26, 190, 194, 352, 835 of a bullet, 216 of a car, 217, 791 of a dropped rock, 352 of a golf ball, 835 of hailstones, 217 of a harmonic spring, 138 of a hydrogen atom, 836 of an elevator, 195 of the Lunar Module, 206 of ocean waves, 202 of a pendulum, 35, 217, 249, 260 of a person walking and jogging, 162 of a piston in an automobile, 218 of a projectile, 283, 471, 693, 842, 843 of a rocket, 411, 1126 of a spring, 8 of a subway train, 162 of a thrown rock, 349 of a train, 689 of an unnamed object, 193, 202, 206, 216, 226, 352, 385 of a wave, 883 of a weight on a spring, 200, 201, 218 of water down a hill, 919 Path of a satellite in orbit, 860 Navigation Steering a boat, 761 Tracking an airplane, 270 Tracking a rocket, 269 Optics Intensity of a light, 283 Light intensity, 340 Light passing through different materials, 338

Mirrors, 588 Photography Camera distance calculation, 139, 250 Physics Air resistance on a falling object, 411 Bodies in free fall, 341 Braking load, 761 Compressibility of gas, 226 Effect of elevation on weight, 249 Flow of liquid, 1088, 1089, 1090, 1091, 1110 Force acting on a car, 850 acting on person in a car, 836 of an object dragged along the ground, 341 on an unnamed object, 553 needed to keep vehicle from skidding, 829, 835 Force fields, 1043, 1046, 1069, 1097 Frictional force, 1126 Gravitation/gravitational fields, 563, 860, 891, 920, 1013, 1028 Gravity (on Europa), 113 Harmonic motion, 283 Inertia, 352, 981, 984, 996, 999, 1004, 1005, 1012, 1013, 1022, 1069 Ideal Gas Law, 860, 881, 892, 901 Kepler’s Laws, 846, 849 Newton’s Law of Cooling, 138 Potential of gravity, 920 Potential of magnets, 564 Proton in a particle collider, 835 Repelling charged particles, 1046 Snell’s Law, 892 Specific gravity of a sphere, 250 Speed of a thrown ball, 352 Speed of light, 672 Thermodynamics, 860, 881 Trajectory of a baseball, 689 of a football, 689 Velocity of a ball, 162 of a box sliding down an inclined plane, 355 of a circular disk, 217

of a falling object, 161 of a falling rock, 157, 158, 163 of a liquid through a tube, 183 of a thrown object, 111 of an automobile, 163, 173 of an unnamed object, 182, 195 Water pressure against the glass of an aquarium, 524 Work done by a chain, 479 done by a pulled rope, 473 done by a pump, 476–477, 479 done by an electrical charge, 480 done by gravity on an unwinding cable, 479 done by horse pulling plow, 762 done by unnamed force, 1110 done to pull ore from a mine, 496 done to raise an anchor, 496 needed to lift a bucket, 479–480 needed to lift a rocket off the Earth, 480 needed to move a piston, 480 of an expanding gas, 480 pulling a cart, 753 pulling a wagon, 753, 762 pulling crates, 762 pushing lawnmower, 760 Ramp, 762 Tension in a cable, 753 to lift an elevator, 479 Tug of war, 753 Using a wrench, 771 Population Growth, 173, 194, 420, 425 of endangered bald eagles, 138 of insects, 138, 420 of the United States, 309 of wolves, 319 Stocking a lake with fish, 588, 600 Probability Coin flips, 601 Production of cylindrical containers, 335, 339

Rate Blood flow through an artery, 183 Change in light intensity on surface, 194 Change in light intensity through sunglasses, 202 Change in luminosity of the Sun, 183, 249 Change in the inclination of a ladder as it slips from a wall, 268 Change of angle of mirror as it slips on a wood floor, 202 Change of density in a diver’s body under water, 194 Change of gas pressure in a can, 194 Falling water level in cone, 268 of a baseball passing third base, 269 of a person rowing a boat, 340 of a person walking, 269 of a pile of sand increasing, 269 of a rumor spreading, 1141, 1150 of a turning searchlight, 271 of an unnamed object, 347 of change in a melting snowball, 354 of change in the angle of a rising balloon, 271 of corporate sales, 377, 385 of flowing water, 1126, 1135 of gas pressure changing, 269 of heated metal plate expansion, 269 of helium leaking from a balloon, 385 of miles per gallon, 339 of string being pulled by a kite, 269 of water being added to a reservoir, 387 of water released from a dam, 425 of weight change of an unnamed object in outer space, 270 Water leaking out of a pool, 183 Speed Kinetic energy of an object approaching the speed of light, 139 of an aircraft, 114, 354, 752, 753 of a boat approaching a pier, 266 of a car 287, 292, 352, 397 of a child’s lengthening shadow, 269 of a decelerating car, 348, 352 of a downhill skier, 352 of a falcon moving away from its trainer, 269 of a person in a parachute, 1134 of a person riding a Ferris wheel, 270

of a rising container pulled by a pulley, 270 of a rotating lighthouse beacon, 354 of a shadow moving along a dome, 271 of a shadow moving along a wall, 271 of a skydiver, 319, 1141 of a train, 271 of a truck, 282 of a turning police searchlight, 271 of an unnamed object, 385, 396, 752, 769, 770, 1141 of elevator and delivery truck moving away from one another, 270 of engine pistons, 270 of falling hailstones, 138 of Mars orbiter, 835 of moving radar beam, 269 of rotation of lighthouse beacon, 267, 269 of satellite in orbit, 828–829 of sound, 882 of two vehicles moving towards one another, 270 of wind, 753 Sports Baseball (bat’s center of mass), 493 Baseball (ERA), 859 Basketball (field goal percentage), 859 Golf (motion of a ball), 282 Golf (radius of water ripple if ball lands in a pond), 263 High jump on the Moon, 352 Mountaineering, 920 Races, 600 Swimming, 739, 761 Temperature Along an unnamed point, 860 Boyle’s Law, 905 Change in, 421, 876, 919, 920 Cooling time of a non–specific object, 138 Cooling time of a pie, 421 Distribution, 881, 882 Kinetic energy of gas, 101 of a metal plate, 915, 918, 948 of a rod, 396 of a sphere, 945 of an unnamed object, 419

Readings on a thermometer, 421 Volume Change in volume of an open box, 271 of a ball, 250 of a balloon, 248 of a paper cup, 248 of a rain gutter, 340 of a tree trunk cut into logs, 547 of a trough, 340 of an unnamed revolving solid object, 507, 516, 523, 540, 554, 558, 562, 572 of expanding gas, 553 of gasoline in a tank, 3 of liquid in a cone, 23 of motor oil in a pan, 249 of water in a reservoir, 554 Rising water level in a cone, 268 Rising water level in a swimming pool, 265, 268 Rising water level in a tank, 268 Weather Rainfall measurement, 396

P Preparing for Calculus

P.1 Functions and Their Graphs P.2 Library of Functions; Mathematical Modeling P.3 Operations on Functions; Graphing Techniques P.4 Inverse Functions P.5 Exponential and Logarithmic Functions P.6 Trigonometric Functions P.7 Inverse Trigonometric Functions P.8 Sequences; Summation Notation; the Binomial Theorem

Until now, the mathematics you have encountered has centered mainly on algebra, geometry, and trigonometry. These subjects have a long history, well over 2000 years. But calculus is relatively new; it was developed less than 400 years ago. In the time since, it has profoundly influenced every facet of human activity from biology to medicine to economics and physics, to name just a few. The year 2019 marks the 50th anniversary of the Apollo 11 mission, which put the first humans on the Moon, an achievement made possible by mathematics, in particular calculus. Calculus deals with change and how the change in one quantity affects other quantities. Fundamental to these ideas are functions and their properties. In this chapter, we discuss many of the functions used in calculus. We also provide a review of techniques from precalculus used to obtain the graphs of functions and to transform

known functions into new functions. Your instructor may choose to cover all or part of this chapter. Regardless, throughout the text, you will see the NEED TO REVIEW? marginal notes. They reference specific topics, often discussed in Chapter P.

P.1 Functions and Their Graphs OBJECTIVES When you finish this section, you should be able to: 1 Evaluate a function 2 Find the difference quotient of a function 3 Find the domain of a function 4 Identify the graph of a function 5 Analyze a piecewise-defined function 6 Obtain information from or about the graph of a function 7 Use properties of functions 8 Find the average rate of change of a function

Often there are situations where one variable is somehow linked to another variable. For example, the price of a gallon of gas is linked to the price of a barrel of oil. A person can be associated to her telephone number(s). The volume V of a sphere depends on its radius R. The force F exerted by an object corresponds to its acceleration a. These are examples of a relation, a correspondence between two sets called the domain and the range. If x is an element of the domain and y is an element of the range, and if a relation exists from x to y, then we say that y corresponds to x or that y depends on x, and we write x→y. It is often helpful to think of x as the input and y as the output of the relation. See Figure 1.

Figure 1 Suppose an astronaut standing on the Moon throws a rock 20 meters up and starts a stopwatch as the rock begins to fall back down. If x represents the number of seconds on the stopwatch and if y represents the altitude of the rock at that time, then there is a relation between time x and altitude y. If the altitude of the rock is measured at x = 1, 2, 2.5, 3, 4, and 5 seconds, then the altitude is approximately y = 19.2, 16.8, 15, 12.8, 7.2, and 0 meters, respectively. This is an example of a relation expressed verbally. The astronaut could also express this relation numerically, graphically, or algebraically. The relation can be expressed by a table of numbers (see Table 1) or by the set of ordered pairs {(0,20),(1,19.2),(2,16.8),(2.5,15),(3,12.8),(4,7.2),(5,0)}, where the first element of each pair denotes the time x and the second element denotes the altitude y. The relation also can be expressed visually, using either a graph, as in Figure 2, or a map, as in Figure 3. Finally, the relation can be expressed algebraically using the formula y=20−0.8x2 TABLE 1 Time, x (in seconds) 0 1 2 2.5 3

Altitude, y (in meters) 20 19.2 16.8 15 12.8

4 5

7.2 0

Figure 2

Figure 3 The NOTE Not every relation is a function. If any element x  in the set X  corresponds to more than one element value, y in the set Y, then the relation is not a function. 0 Time In this example, notice that if X is the set of times from 0 to 5 seconds and Y is (seconds) the set of altitudes from 0 to 20 meters, then each element of X corresponds to one corresponds and only one element of Y. Each given time value yields a unique, that is, exactly one,

altitude value. Any relation with this property is called a function from X into Y. DEFINITION Function

Let X and Y be two nonempty sets.* A function f from X into Y is a relation that associates with each element of X exactly one element of Y. The set X is called the domain of the function. For each element x in X, the corresponding element y in Y is called the value of the function at x, or the image of x. The set of all the images of the elements in the domain is called the range of the function. Since there may be elements in Y that are not images of any x in X, the range of a function is a subset of Y. See Figure 4.

Figure 4 1 Evaluate a Function Functions are often denoted by letters such as f, F, g, and so on. If f is a function, then for each element x in the domain, the corresponding image in the range is denoted by the symbol f(x), read “f of x.” f(x) is called the value of f at x. The variable x is called the independent variable or the argument because it can be assigned any element from the domain, while the variable y is called the dependent variable because its value depends on x. EXAMPLE 1 Evaluating a Function

For the function f defined by f(x)=2x2−3x, a. f(5) b. f(x+h)

find:

c. f(x+h)−f(x) Solution a. f(5)=2⋅52−3⋅5=50−15=35 b. The function f(x)=2x2−3x expand (x+h)2, of 3 and (x+h).

gives us a rule to follow. To find f(x+h), multiply the result by 2, and then subtract the product

c. f(x+h)−f(x)=[2x2+4hx+2h2−3x−3h]−[2x2−3x]=4hx+2h2−3h f open parenthesis xNOW WORK Problem 13. plus h EXAMPLE 2 Finding the Amount of Gasoline in a Tank close parenthesis A Shell station stores its gasoline in an underground tank that is a right circular cylinder lying equals on its side. The volume V of gasoline in the tank (in gallons) is given by the formula 2 open parenthesis V(h)=40h296h−0.608 x where plus h is the height (in inches) of the gasoline as measured on a depth stick. See Figure 5. h close parenthesis squared minus 3 open parenthesis x plus h close parenthesis equals

▪

Figure 5 a. If h=12 b. If h=1

inches, how many gallons of gasoline are in the tank? inch, how many gallons of gasoline are in the tank?

Solution a. We evaluate V when h=12. V(12)=40⋅1229612−0.608=40⋅1448−0.608=57607.392≈15,660

There are about 15,660 in the tank is 12 inches. b. Evaluate V when h=1.

gallons of gasoline in the tank when the height of the gasoline

V(1)=40⋅12961−0.608=4096−0.608=4095.392≈391

There are about 391 tank is 1 inch.

gallons of gasoline in the tank when the height of the gasoline in the

▪

Implicit Form of a Function

In general, a function f defined by an equation in x and y is given implicitly. If it is possible to solve the equation for y in terms of x, then we write y=f(x) and say the function is given explicitly. For example, Implicit FormExplicit Formx2−y=6y=f(x)=x2−6          xy=4y=g(x)=4x

In calculus we sometimes deal with functions that are defined implicitly and that cannot be expressed in explicit form. For example, if the function with independent variable x and dependent variable y is defined by sin(xy)=xy−2y+cos x−sin y, there is no method to solve for y and express the function explicitly. 2 Find the Difference Quotient of a Function An important concept in calculus involves working with a certain quotient. For a given function y=f(x), the inputs x and x+h, h≠0, result in the images f(x) and f(x+h). The quotient of their differences

f(x+h)−f(x)(x+h)−x=f(x+h)−f(x)h with h≠0,

is called the difference quotient of f at x.

DEFINITION Difference Quotient

The difference quotient of a function f at x is given by

f(x+h)−f(x)h   h≠0 The difference quotient is used in calculus to define the derivative, which is used in applications such as the velocity of an object and optimization of resources. When finding a difference quotient, it is necessary to simplify the expression in order to cancel the h in the denominator, as illustrated in the next example. EXAMPLE 3 Finding the Difference Quotient of a Function

Find the difference quotient of each function.

a. f(x)=2x2−3x b. f(x)=x Solution

▪

(a) f open NOW WORK Problem 23. parenthesis x 3 Find the Domain of a Function plus h In applications, the domain of a function is sometimes specified. For example, we might be close interested in the population of a city from 1990 to 2019. The domain of the function is time, parenthesis in years, and is restricted to the interval [1990, 2019]. Other times the domain is restricted by minus the context of the function itself. For example, the volume V of a sphere, given by the f of function V=43πR3, makes sense only if the radius R is greater than 0. But x often the domain of a function f is not specified; only the formula defining the function is all given. In such cases, the domain of f is the largest set of real numbers for which the value over

f(x)

is defined and is a real number.

EXAMPLE 4 Finding the Domain of a Function

Find the domain of each of the following functions: a. f(x)=x2+5x b. g(x)=3xx2−4 c. h(t)=4−3t d. F(u)=5uu2−1 Solution a. Since f(x)=x2+5x domain of f is the set of all real numbers.

is defined for any real number x, the

b. Since division by zero is not defined, x2−4 x≠2. x=−2

cannot be 0, that is, x≠−2

and

The function g(x)=3xx2−4 is defined for any real number except and x=2. So, the domain of g is the set of real numbers {x|x≠−2,x≠2}.

c. Since the square root of a negative number is not a real number, the value of 4−3t must be nonnegative. The solution of the inequality 4−3t≥0 domain of h is the set of real numbers t|t≤43

is t≤43,

so the

or the interval −∞,43.

NEED TO REVIEW? Solving inequalities is discussed in Appendix A.1, pp. A-6 to A-8.

NEED TO REVIEW? Interval notation is discussed in Appendix A.1, p. A-5.

d. Since the square root is in the denominator, the value of u2−1 must be not only nonnegative, it also cannot equal zero. That is, u2−1>0. The solution of the inequality u2−1>0 is the set of real numbers {u|u1} or the set (−∞,−1)∪(1,∞).

▪



If x is in the domain of a function f, we say that f is defined at x, or f(x)

exists. If x is not in the domain of f, we say that f is not defined at x, or f(x) does not exist. The domain of a function is expressed using inequalities, interval notation, set notation, or words, whichever is most convenient. Notice the various ways the domain was expressed in the solution to Example 4. NOW WORK Problem 17.

4 Identify the Graph of a Function In applications, often a graph reveals the relationship between two variables more clearly than an equation. For example, Table 2 shows the average price of gasoline at a particular gas station in the United States (for the years 1985–2016 adjusted for inflation, based on 2014 dollars). If we plot these data using year as the independent variable and price as the dependent variable, and then connect the points (year, price) we obtain Figure 6. TABLE 2 Average Retail Price of Gasoline (2014 dollars) Year 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995

Price 2.55 1.90 1.89 l.81 1.87 2.03 1.91 1.82 1.74 1.71 1.72

Year 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006

Source: U.S. Energy Information Administration

Price 1.80 1.76 1.49 1.61 2.03 1.90 1.76 1.99 2.31 2.74 3.01

Year 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016

Price 3.19 3.56 2.58 3.00 3.69 3.72 3.54 3.43 2.74 2.24

Figure 6 Graph of the average retail price of gasoline (2014 dollars). The graph shows that for each date on the horizontal axis there is only one price on the Graph vertical axis. So, the graph represents a function, although the rule for determining the price of from the year is not given. priceWhen a function is defined by an equation in x and y, the graph of the function is in the set of points (x,y) in the xy -plane that satisfy the equation. dollars versus NEED TO REVIEW? The graph of an equation is discussed in Appendix A.3, pp. A-16 to A-18. years shows But not every collection of points in the xy -plane represents the graph of a function. an Recall that a relation is a function only if each element x in the domain corresponds to up exactly one image y in the range. This means the graph of a function never contains two down points with the same x -coordinate and different y -coordinates. Compare the graphs in moving Figures 7 and 8. In Figure 7, every number x is associated with exactly one number y, curving but in Figure 8 some numbers x are associated with three numbers y. Figure 7 shows the beginning graph of a function; Figure 8 shows a graph that is not the graph of a function. at price 2.50 in 1984 which fluctuates until the year 2002 after

Figure 7 Function: Exactly one y for each x. at most one point.

Every vertical line intersects the graph in

Figure 8 Not a function: x=c has 3 y’s associated with it. The vertical line x=c intersects the graph in three points. For a graph to be a graph of a function, it must satisfy the Vertical-line Test. THEOREM Vertical-line Test

A set of points in the xy -plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. NOTE The phrase “if and only if” means the statements on each side of the phrase are equivalent. That is, they have the same meaning.

EXAMPLE 5 Identifying the Graph of a Function

Which graphs in Figure 9 represent the graph of a function?

Figure 9 First Solution The graphs in Figure 9(a), 9(b), and 9(e) are graphs of functions because every diagram vertical line intersects each graph in at most one point. The graphs in Figure 9(c) and 9(d) are shows not graphs of functions because there is a vertical line that intersects each graph in more than a one point. parabolic curve NOW WORK Problems 31(a) and (b). of function Notice that although the graph in Figure 9(e) represents a function, it looks different from y the graphs in (a) and (b). The graph consists of two pieces plus a point, and they are not equals connected. Also notice that different equations describe different pieces of the graph. x Functions with graphs similar to the one in Figure 9(e) are called piecewise-defined squared, functions. seconds shows 5 Analyze a Piecewise-Defined Function an SSometimes a function is defined differently on different parts of its domain. For example, the shaped absolute value function f(x)=|x | is actually defined by two equations: f(x)=x curve if x≥0 and f(x)=−x if x0 when the y -coordinate of a point (x,y) on the graph of f is positive. This occurs when x is in the set 0,π2∪3π2,5π2∪7π2,4π.

▪

NOW WORK Problems 37, 39, 41, 43, 45, 47, and 49.

EXAMPLE 8 Obtaining Information About the Graph of a Function

Consider the function f(x)=x+1x+2. a. What is the domain of f ? b. Is the point 1,12 c. If x=2, d. If f(x)=2,

on the graph of f ?

what is f(x)

? What is the corresponding point on the graph of f ?

what is x ? What is the corresponding point on the graph of f ?

e. What are the x -intercepts of the graph of f (if any)? What point(s) on the graph of f correspond(s) to the x -intercept(s)? Solution a. The domain of f consists of all real numbers except −2; {x|x≠−2}. The function is not defined at −2.

that is, the set

then

b. When x=1, point 1,12 c. If x=2,

The point 1,23

is on the graph of f ; the

is not on the graph of f. then

The point 2,34 then x+1x+2=2.

d. If f(x)=2,

is on the graph of f.

Solving for x, we find

x+1=2(x+2)x+1=2x+4x=−3 The point (−3,2)

is on the graph of f.

e. The x -intercepts of the graph of f occur when y=0. That is, they are the solutions of the equation f(x)=0. The x -intercepts are also called the real zeros or roots of the function f. The real zeros of the function f(x)=x+1x+2 so x=−1. The only x -intercept is −1, on the graph of f.



▪

Figure 13 shows the graph of f.

satisfy the equation x+1=0, so the point (−1,0) is

Figure 13 f(x)=x+1x+2 NOW WORK Problems 55, 57, and 59.

7 Use Properties of Functions One of the goals of calculus is to develop techniques for graphing functions. Here we review some properties of functions that help obtain the graph of a function. DEFINITION Even and Odd Functions

A function f is even if, for every number x in its domain, the number −x the domain and

is also in

f(−x)=f(x) A function f is odd if, for every number x in its domain, the number −x domain and f(−x)=−f(x) For example, f(x)=x2 f(−x)=(−x)2=x2=f(x)

is an even function since

is also in the

Also, g(x)=x3

is an odd function since

g(−x)=(−x)3=−x3=−g(x) See Figure 14 for the graph of f(x)=x2 and Figure 15 for the graph of g(x)=x3. Notice that the graph of the even function f(x)=x2 is symmetric with respect to the y -axis, and the graph of the odd function g(x)=x3 with respect to the origin.

Figure 14 The function f(x)=x2 respect to the y -axis.

Figure 15 The function g(x)=x3. respect to the origin.

is symmetric

is even. The graph of f is symmetric with

is odd. The graph of g is symmetric with

NEED TO REVIEW? Symmetry of graphs is discussed in Appendix A.3, pp. A-17 to A-18.

THEOREM Graphs of Even and Odd Functions

A function is even if and only if its graph is symmetric with respect to the y -axis. A function is odd if and only if its graph is symmetric with respect to the origin.

EXAMPLE 9 Identifying Even and Odd Functions

Determine whether each of the following functions is even, odd, or neither. Then determine whether its graph is symmetric with respect to the y -axis, the origin, or neither. a. f(x)=x2−5 b. g(x)=4xx2−5 c. h(x)=5x3−13 d. F(x)=|x |

e. H(x)=x2+2x−1(x−5)2 Solution a. The domain of f is (−∞,∞), −x is also in the domain. Replace x by −x

so for every number x in its domain, and simplify.

f(−x)=(−x)2−5=x2−5=f(x) Since f(−x)=f(x), with respect to the y -axis.

the function f is even. So the graph of f is symmetric

b. The domain of g is {x|x≠±5}, is also in the domain. Replace x by −x

so for every number x in its domain, −x and simplify.

g(−x)=4(−x)(−x)2−5=−4xx2−5=−g(x) Since g(−x)=−g(x), with respect to the origin.

the function g is odd. So the graph of g is symmetric

c. The domain of h is (−∞,∞), also in the domain. Replace x by −x

so for every number x in its domain, −x and simplify.

h(−x)=5(−x)3−13=−5x3−13=−(5x3+1)3=−5x3+13

is

Since h(−x)≠h(x) and h(−x)≠−h(x), the function h is neither even nor odd. The graph of h is not symmetric with respect to the y -axis and not symmetric with respect to the origin. d. The domain of F is (−∞,∞), also in the domain. Replace x by −x

so for every number x in its domain, −x and simplify.

is

F(−x)=|−x|=|−1|⋅|x|=|x|=F(x) The function F is even. So the graph of F is symmetric with respect to the y -axis. e. The domain of H is {x|x≠5}. The number x=−5 is in the domain of H, but x=5 is not in the domain. So the function H is neither even nor odd, and

▪

the graph of H is not symmetric with respect to the y -axis or the origin. NOW WORK Problem 61.

Another important property of a function is to know where it is increasing or decreasing. DEFINITION

A function f is increasing on an interval I if, for any choice of x1 and x2 in I, with x10,

then

au⋅av=au+vauav=au−v(au)v=auv(ab)u=au⋅bu(ab)u=aubu

NEED TO REVIEW? The Laws of Exponents are discussed in Appendix A.1, pp. A-8 to A-9.

For example, we can use the Laws of Exponents to show the following property of an exponential function: f(x+1)f(x)=ax+1ax=a(x+1)−x=a1=a EXAMPLE 1 Graphing an Exponential Function

Graph the exponential function g(x)=12x.

Solution We begin by writing 12 as 2−1.

Then

g(x)=12x=(2−1)x=2−x Now use the graph of f(x)=2x shown in Figure 61 and reflect the graph about the y -axis to obtain the graph of g(x)=2−x. See Figure 63.

Figure 63

▪ NOW WORK Problem 19.

The graph of g(x)=12x in Figure 63 is typical of all exponential functions that have a base between 0 and 1. Figure 64 illustrates the graphs of several more exponential functions whose bases are between 0 and 1. Notice that the graphs with smaller bases are steeper when x0, these graphs are closer to the x -axis.

Figure 64 f(x)=ax,00

is x>0.

then y is the exponent in x=ay.

and a≠1,

its inverse

function is f−1(x)=logax. If the base of a logarithmic function is the number e, then it is called the natural logarithmic function, and it is given a special symbol, ln (from the Latin logarithmus naturalis). That is, y=ln xif and only ifx=ey Since the exponential function y=ax, a>0,a≠1 and the logarithmic function y=logax are inverse functions, the following properties hold: loga(ax)=x  alogax=x

for all real numbers x for all x>0

Because exponential functions and logarithmic functions are inverses of each other, the graph of the logarithmic function y=logax, a>0 and a≠1, is the reflection of the graph of the exponential function y=ax, about the line y=x, as shown in Figure 67.

Figure 67 Based on the graphs in Figure 67, we see that When aloga1=0logaa=1a>0anda≠1 is between 0EXAMPLE 3 Graphing a Logarithmic Function and Graph: 1, a. f(x)=log2x both curves b. g(x)=log1⁄3x and the c. F(x)=ln x line y Solution a. To graph f(x)=log2x, graph y=2x and reflect it about the equaly=x. line See Figure 68(a). to x b. To graph g(x)=log1⁄3x, graph y=13x and reflect it about the intersect line See Figure 68(b). at y=x. ac. To graph F(x)=ln x, graph y=ex and reflect it about the line y=x. point See Figure 68(c).

Figure 68

▪

The reflection of NOW WORK Problem 39. logarithm of Since a logarithmic function is the inverse of an exponential function, it follows that x in base Domain of the logarithmic function=Range of the exponential function=(0,∞) 2 about the Range of the logarithmic function=Domain of the exponential function=(−∞,∞) line y equal to x The domain of a logarithmic function is the set of positive real numbers, so the argument of a gives logarithmic function must be greater than zero. the exponential graph EXAMPLE 4 Finding the Domain of a Logarithmic Function 2 Find the domain of each function: raised to a. F(x)=log2(x+3) x. The reflection b. g(x)=ln1+x1−x of c. h(x)=log1⁄2|x | logarithm

Solution a. The argument of a logarithm must be positive. So to find the domain of F(x)=log2(x+3), we solve the inequality x+3>0. The domain of F is {x|x>−3}. NEED TO REVIEW? Solving inequalities is discussed in Appendix A.1, pp. A-5 to A-8.

b. Since ln1+x1−x

requires 1+x1−x>0,

solving the inequality 1+x1−x>0.

we find the domain of g by

Since 1+x1−x

is not defined for x=1,

and the solution to the equation 1+x1−x=0 is x=−1, we use −1 and 1 to separate the real number line into three intervals (−∞,−1), (−1,1), and (1,∞). Then we choose a test number in each interval and evaluate the rational expression 1+x1−x at these numbers to determine if the expression is positive or negative. For example, we chose the numbers −2, 0, and 2 and found that 1+x1−x>0

on the interval (−1,1).

of g(x)=ln1+x1−x Interval

See the table on the left. So the domain

is {x|−10

and a≠1

The domain of f is the set of all positive real numbers; the range is the set of all real numbers.

The x -intercept of the graph of f is 1.

There is no y -intercept.

A logarithmic function is decreasing on the interval (0,∞) increasing on the interval (0,∞) if a>1.

if 00 passes through three points: , and . 10. Multiple Choice The graph of f(x)=log2x (c) neither]. 11. True or False If y=logax,

and a≠1, ,

is [(a) increasing (b) decreasing

then y=ax.

12. True or False The graph of f(x)=logax, a>0 x -intercept equal to 1 and no y -intercept. 13. True or False ln ex=x

.

and a≠1,

has an

for all real numbers.

14. ln e=_______________. 15. Explain what the number e is. 16. What is the x -intercept of the function h(x)=ln(x+1)? Practice Problems

17. Suppose that g(x)=4x+2. a. What is g(−1) b. If g(x)=66, g ?

? What is the corresponding point on the graph of g ? what is x ? What is the corresponding point on the graph of

18. Suppose that g(x)=5x−3. a. What is g(−1) b. If g(x)=122, of g ?

? What is the corresponding point on the graph of g ? what is x ? What is the corresponding point on the graph

In Problems 19–24, the graph of an exponential function is given. Match each graph to one of the following functions: a. y=3−x b. y=−3x c. y=−3−x d. y=3x−1 e. y=3x−1 f. y=1−3x 19.

20.

21.

22.

23.

24.

In Problems 25–30, use transformations to graph each function. Find the domain and range. 25. f(x)=2x+2 26. f(x)=1−2−x⁄3 27. f(x)=413x 28. f(x)=12−x+1 29. f(x)=e−x 30. f(x)=5−ex In Problems 31–34, find the domain of each function. 31. F(x)=log2x2 32. g(x)=8+5ln(2x+3) 33. f(x)=ln(x−1) 34. g(x)=ln x In Problems 35–40, the graph of a logarithmic function is given. Match each graph to one of the following functions: a. y=log3x b. y=log3(−x) c. y=−log3x d. y=log3x−1

e. y=log3(x−1) f. y=1−log3x 35.

36.

37.

38.

39.

40.

In Problems 41–44, a. Find the domain of f. b. Graph f. c. From the graph of f, determine the range of f. d. Find f−1,

the inverse of f.

e. Use f−1

to find the range of f.

f. Graph f−1. 41. f(x)=ln(x+4) 42. f(x)=12log(2x) 43. f(x)=3ex+2 44. f(x)=2x⁄3+4 45. How does the transformation y=ln(x+c), c>0, intercept of the graph of the function f(x)=ln x? 46. How does the transformation y=ecx, c>0, graph of the function f(x)=ex? In Problems 47–62, solve each equation. 47. 3x2=9x 48. 5x2+8=1252x 49. e3x=e2ex 50. e4x⋅ex2=e12 51. e1−2x=4 52. e1−x=5 53. 5(23x)=9 54. 0.3(40.2x)=0.2 55. 31−2x=4x 56. 2x+1=51−2x 57. log2(2x+1)=3 58. log3(3x−2)=2 59. logx18=3

affect the x affect the y -intercept of the

60. logx64=−3 61. ln(2x+3)=2ln3 62. 12log3x=2log32 In Problems 63–66, use technology to solve each equation. Express your answer rounded to three decimal places. 63. log5(x+1)−log4(x−2)=1 64. ln x=x 65. ex+ln x=4 66. ex=x2 67. a. If f(x)=ln(x+4) and g(x)=ln(3x+1), and g on the same set of axes.

graph f

b. Find the point(s) of intersection of the graphs of f and g by solving f(x)=g(x). c. Based on the graph, solve f(x)>g(x). 68. a. If f(x)=3x+1 set of axes.

and g(x)=2x+2,

graph f and g on the same

b. Find the point(s) of intersection of the graphs of f and g by solving f(x)=g(x). Round answers to three decimal places. c. Based on the graph, solve f(x)>g(x). 1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

P.6 Trigonometric Functions OBJECTIVES When you finish this section, you should be able to: 1 Work with properties of trigonometric functions 2 Graph the trigonometric functions

NEED TO REVIEW? Trigonometric functions are discussed in Appendix A.4, pp. A-27 to A-32.

1 Work with Properties of Trigonometric Functions Table 6 lists the six trigonometric functions and the domain and range of each function. TABLE 6 Function

Symbol

Domain

sine

y=sin x

All real numbers

{y|−1≤y≤1}

cosine

y=cos x

All real numbers

{y|−1≤y≤1}

tangent

y=tan x

cosecant

y=csc x

secant

y=sec x

cotangent y=cot x

x|x≠odd integer multiples of π2 {x|x≠integer multiples of π} x|x≠odd integer multiples of π2 {x|x≠integer multiples of π}

All real numbers {y|y≤−1 or  {y|y≤−1 or  All real numbers

NOTE In calculus, radians are generally used to measure angles, unless degrees are specifically mentioned.

An important property common to all trigonometric functions is that they are periodic. DEFINITION Periodic Function

A function f is called periodic if there is a positive number p with the property that whenever x is in the domain of f, so is x+p, and f(x+p)=f(x) If there is a smallest number p with this property, it is called the (fundamental) period of f. The sine, cosine, cosecant, and secant functions are periodic with period 2π

; the

tangent and cotangent functions are periodic with period π. THEOREM Period of Trigonometric Functions

sin(x+2π)=sin xcos(x+2π)=cos xtan(x+π)=tan xcsc(x+2π)=csc xsec(x+2π)=sec xcot(x+π)=cot

Because the trigonometric functions are periodic, once the values of the function over one period are known, the values over the entire domain are known. This property is useful for graphing trigonometric functions. The next result, also useful for graphing the trigonometric functions, is a consequence of the even-odd identities, namely, sin(−x)=−sin x and cos(−x)=cos x. From these, we have tan(−x)=sin(−x)cos(−x)=−sin xcos x= −tan xsec (−x)=1cos(−x)=1cos x=sec xcot(−x)=1tan(−x)=1−tanx= −cot xcsc(−x)=1sin(−x)=1−sin x=−cscx

THEOREM Even-Odd Properties of the Trigonometric Functions

The sine, tangent, cosecant, and cotangent functions are odd, so their graphs are symmetric with respect to the origin. The cosine and secant functions are even, so their graphs are symmetric with respect to the y -axis. NEED TO REVIEW? The values of the trigonometric functions for select numbers are discussed in Appendix A.4, pp. A-29 and A-31.

2 Graph the Trigonometric Functions To graph y=sin x, we use Table 7 on page 54 to obtain points on the graph. Then we plot some of these points and connect them with a smooth curve. Since the sine function has a period of 2π, continue the graph to the left of 0 and to the right of 2π. See Figure 72. TABLE 7

x

y=sin x

(x,y)

0

0

π6

12 1

π2

5π6

12

π

0

7π6

3π2

11π6 2π

−12

(0,0)

π6,12 π2,1

5π6,12 (π,0)

7π6,−12

−1 3π2,−1 −12

11π6,−12 0

(2π,0)

Figure 72 f(x)=sin x Notice the symmetry of the graph with respect to the origin. This is a consequence of f being an odd function. The graph of y=sin x

illustrates some facts about the sine function.

Properties of the Sine Function f(x)=sin x

The domain of f is the set of all real numbers. The range of f consists of all real numbers in the closed interval −1,1. The sine function is an odd function, so its graph is symmetric with respect to the origin. The sine function has a period of 2π. The x -intercepts of f are …, the y -intercept is 0.

−2π,

−π,

0, π, 2π,

3π,…;

The maximum value of f is 1 and occurs at x=…,−3π2, π2, 9π2,…

; the minimum value of f is −1 7π2,

5π2,

and occurs at x=…,−π2, 3π2,

11π2,….

The graph of the cosine function is obtained in a similar way. Locate points on the graph of the cosine function f(x)=cos x for 0≤x≤2π. Then connect the points with a smooth curve and continue the graph to the left of 0 and to the right of 2π to obtain the graph of y=cos x. See Figure 73.

Figure 73 f(x)=cos x The graph of y=cos x

illustrates some facts about the cosine function.

Properties of the Cosine Function f(x)=cos x The domain of f is the set of all real numbers.

The range of f consists of all real numbers in the closed interval [−1,1]. The cosine function is an even function, so its graph is symmetric with respect to the y -axis. The cosine function has a period of 2π. The x -intercepts of f are …,−3π2,

−π2,

π2,

3π2,

5π2,…

; the y -intercept is 1. The maximum value of f is 1 and occurs at x=…,−2π, 0, 2π, 6π,… ; the minimum value of f is −1 and occurs at x=…,−π, π, 3π, 5π,….

4π,

Many variations of the sine and cosine functions can be graphed using transformations. EXAMPLE 1 Graphing Variations of f(x)=sin x

Use the graph of f(x)=sin x Solution Notice that g(x)=2 f(x), the graph of f(x)=sin x.

Using Transformations

to graph g(x)=2 sin x. so the graph of g is a vertical stretch of Figure 74 illustrates the transformation.

Figure 74

▪ Notice that the values of g(x)=2 sin x lie between −2 and 2, inclusive. In general, the values of the functions f(x)=A sin x and g(x)=A cos x, where A≠0, will satisfy the inequalities −|A|≤A sin x≤|A|         and         −|A|≤A cos x≤|A|

respectively. The number A g(x)=A cos x.

is called the amplitude of f(x)=A sin x

EXAMPLE 2 Graphing Variations of f(x)=cos x

Use the graph of f(x)=cos x

and of

Using Transformations

to graph g(x)=cos(3x).

Solution The graph of g(x)=cos(3x) is a horizontal compression of the graph of f(x)=cos x. Figure 75 on page 56 shows the transformation.

Figure 75

▪ From the graph, we notice that the period of g(x)=cos(3x)

is 2π3.

NOW WORK Problems 27 and 29.

In general, if ω>0,

the functions f(x)=sin(ωx)

and g(x)=cos(ωx)

have period T=2πω. If ω>1, the graphs of f(x)=sin(ωx) and g(x)=cos(ωx) are horizontally compressed, and the period of the functions is less than 2π. If 00, y≠0. are

The partial derivatives of f

fx(x, y)=1xyfy(x, y)=-ln xy2 Since both partial derivatives exist and are continuous at every point (x0, y0) in the domain of f, the function z=f(x, y) is differentiable at every point (x0, y0) in its domain. The differential dz is dz=1xy dx-ln xy2 dy

▪ NOW WORK Problem 11.

If z=f(x, y) of the differential dz.

is a differentiable function, then Δz

can be expressed in terms

where lim(Δx,Δy)→(0, 0) η1=0    andlim(Δx,Δy)→(0, 0) η2=0.

When the differentials dx=Δx and dy=Δy are close to 0, then η1 and η2 are also close to 0, and the differential dz is approximately equal to Δz . (4) That is, Δz≈dz=fx(x0, y0)dx+fy(x0, y0)dy The differential dz is usually easier to calculate than Δz, making dz a useful approximation to Δz . The error introduced in using dz to approximate Δz equals the expression η1Δx+η2Δy . 3 Use the Differential as an Approximating Tool EXAMPLE 4 Using the Differential in Astronomy

The luminosity L (total power output in watts, W) of a star is given by the formula

L=L(R, T)=4πR2 σT4 where R is the radius of the star (in meters), T is its effective surface temperature (in kelvin, K), and σ is the Stefan-Boltzmann constant. For the sun, Ls=(3.90×1026) W, Rs=(6.94×108) m, and Ts=4800 K. Suppose in a billion years, the changes in the Sun will be ΔRs=(0.08×108) m and ΔTs=100 K. increase in luminosity?

What will be the resulting percent

Solution We begin with L=4πR2σT4.

Then

dL=4πσ(2R)T4 dR+4πσR2(4T3)dT=8πσRT3(TdR+2R dT) The relative error in luminosity is ΔLL≈dLL=8πσRT34πR2 σT4TdR+2RdT=2dRR+2 dTT=2 ΔRR+4  ΔTT=2(0.08×108)6.94×108+4·1004800≈0.106

The percent increase in luminosity will be approximately 10.6%

.

▪

Incidentally, a reasonable, although rough, estimate of how this change would affect Earth’s temperature is ΔTe≈14⋅0.106 Te=(0.0265)(290 K)≈7.69 K That is, the average temperature of Earth’s surface will change from 16.9 ∘C (62.4 ∘F to 24.6 ∘C (76.3 ∘F ). Such a change in temperature would be enough to modify Earth’s climate. EXAMPLE 5 Using the Differential in Error Analysis

A cola company requires a can in the shape of a right circular cylinder of height 10 cm and radius 3 cm . If the manufacturer of the cans claims a percentage error of no more than 0.2% in the height and no more than 0.1% in the radius, what is the approximate maximum variation in the volume of the can?

)

Solution The volume V of a right circular cylinder of height h cm is V=π R2 h cm3 . We find the differential dV.

and radius R cm

dV=∂V∂R dR+∂V∂h dh=2π Rh dR+π R2 dh

The relative error in the radius R is |ΔR|R=|dR|R=0.001,

and the

relative error in the height h is Δhh=|dh|h=0.002 the volume V is

. The relative error in

|ΔV|V≈|dV|V=2π Rh dR+π R2 dhπ R2 h=2 dRR+dhh=2 ΔRR+Δhh≤2  ΔRR+Δhh=2(0.001)+0.002=0.004

The maximum variation in the volume is approximately 0.4%, the container varies as follows:

so the actual volume of

V=π R2 h±0.004(π R2 h)=π R2 h(1±0.004)=90π(1±0.004) cm3 The volume V is between 89.64π≈281.612 cm3

and

90.36π≈283.874 cm3

.

▪

NOW WORK Problem 39.

EXAMPLE 6 Using the Differential dz

to Approximate the Value of a Function z

For the function f(x, y)=x2 y-1, f(1.1, 1.9) .

use the differential dz to approximate

Solution Example 2 shows f is differentiable and fx(x, y)=2xy fy(x, y)=x2

and

.

Let (x0, y0)=(1, 2)

and (x0+Δx, y0+Δy)=(1.1, 1.9) . Then

f(x0, y0)=f(1, 2)=12⋅2-1=1f(x0+Δx, y0+Δy)=f(1.1, 1.9)fx(x0, y0)=2x0y0=2⋅1⋅2=4fy(x0, y0)= (x0)2=1dx=Δx=1.1-1=0.1dy=Δy=1.9-2=-0.1

Now Δz=f(x0+Δx, y0+Δy)-f(x0, y0) statement (4), we have

and from

Δz≈dz=fx(x0, y0)dx+fy(x0, y0)dyf(x0+Δx, y0+Δy)f(x0, y0)≈fx(x0, y0)dx+fy(x0, y0)dyf(x0+Δx, y0+Δy)≈f(x0, y0)+fx(x0, y0)dx+fy(x0, y0)dy

Then f(1.1, 1.9)≈1+4⋅0.1+1⋅(-0.1)=1.3

▪ The actual value of f(1.1, 1.9)=(1.1)2(1.9)-1=1.299, so the error in using the differential dz to approximate z is 0.001.

NOW WORK Problem 33.

Continuity and Differentiability for Functions of Two Variables Differentiable functions of a single variable are necessarily continuous. This result is also true for functions of two variables. THEOREM Differentiability Is Sufficient for Continuity

Let z=f(x, y) be a function of two variables whose domain is D . Let (x0, y0) be an interior point of D . If f is differentiable at (x0, y0), then f is continuous at (x0, y0) . Proof The function z=f(x, y)

is continuous at (x0, y0)

if

lim(x, y)→(x0, y0) f(x, y)=f(x0, y0) This is equivalent to the statement limx→y→(x0, y0)  Δz=0. z=f(x, y) is differentiable at (x0, y0), then Δz

Since can be expressed as

Δz=fx(x0, y0)Δx+fy(x0, y0)Δy+η1Δx+η2Δy where lim(Δx,Δy)→(0, 0) η1=0 and    lim(Δx,Δy)→(0, 0)  η2=0. Then Δz=fx(x0, y0)+η1Δx+fy(x0, y0)+η2Δy Now let Δx=x-x0

and Δy=y-y0 is equivalent to (x, y)→(x0, y0)

. Then (Δx,Δy)→(0, 0) and

lim(x, y)→(x0, y0)Δz= lim(x, y)→(x0, y0){[fx(x0, y0)+η1 ](x−x0)+[fy(x0, y0)+η2 ](y

−y0)}=0 That is, f is continuous at (x0, y0)

.

▪

For functions of two variables, differentiability implies continuity. However, the existence of partial derivatives at a point does not necessarily result in continuity at that point because fx and/or fy might not be continuous at the point. For example, the function

f(x, y)=xyx2+y2if(x, y)≠(0, 0)0if(x, y)=(0, 0) has partial derivatives at (0, 0), continuous at (0, 0) .

but as shown in Example 10 on page 879, f is not

A function that is differentiable at (x0, y0) is continuous at (x0, y0). So, a function that is not continuous at (x0, y0) is not differentiable at (x0, y0) . The following corollary provides a condition for z=f(x, y) to be continuous. COROLLARY Continuity of a Function of Two Variables

Let z=f(x, y) be a function of two variables whose domain is D . Let (x0, y0) be an interior point of D . If the partial derivatives fx and fy exist at each point of some disk centered at (x0, y0), and if fx and fy are each continuous at (x0, y0), then f is continuous at (x0, y0) . Although the precise formulations are given as theorems, the following summary might be helpful: If z=f(x, y), The continuity of the partial derivatives fx and fy implies the differentiability of f . The differentiability of f implies the continuity of f . The continuity of the partial derivatives fx and fy implies the continuity of f . The existence of the partial derivatives fx differentiable.

and fy

does not necessarily mean f is

The existence of the partial derivatives fx continuous.

and fy

does not necessarily mean f is

4 Find the Differential of a Function of Three or More Variables Under suitable conditions, the definitions and theorems involving the differentiability of a function of two variables extend to functions of three or more variables. If w=f(x, y, z) is a function of three variables, the function f is differentiable at a point (x0, y0, z0) if the change Δw in w can be

expressed in the form Δw=fx(x, y, z)Δx+fy(x, y, z)Δy+fz(x, y, z)Δz+η1Δx+η2Δy+η3Δz

where η1, η2,

and η3 are each functions of Δx,Δy,

and Δz

for which

lim(Δx, Δy, Δz)→(0, 0, 0) η1=0   and lim(Δx, Δy, Δz)→(0, 0, 0)  η2=0     and lim(Δx, Δy,  Δz)→(0, 0, 0) η3=0

If w=f(x, y, z) differentials dx, dy,

is differentiable at a point (x0, y0, z0), and dz are defined as

the

dx=Δxdy=Δydz=Δz The differential dw

is defined as

dw=fx(x0, y0, z0) dx+fy(x0, y0, z0) dy+fz(x0, y0, z0) dz

It can be shown that if w=f(x, y, z) is defined within a ball centered at (x0, y0, z0), and if the partial derivatives fx, fy, and fz exist in this ball and are continuous at (x0, y0, z0), then f is differentiable at (x0, y0, z0) . EXAMPLE 7 Finding the Differential of a Function of Three Variables

Find the differential dw

of the function w=f(x, y, z)=3x2 sin2 y cos z .

Solution The function f is defined everywhere in space. We begin by finding the partial derivatives of f. fx(x, y, z)=6x sin2 y cos zfy(x, y, z)=6x2 sin y cos y cos zfz(x, y, z)=-3x2sin2 y sin z

Since the partial derivatives are continuous everywhere, we have dw=fx(x, y, z) dx+fy(x, y, z) dy+fz(x, y, z) dz=6x sin2 y cos z dx+6x2 sin y cos y cos z dy3x2sin2 y sin z dz

▪ NOW WORK Problem 23.

The discussion above extends to functions of more than three variables in an analogous way.

12.4 Assess Your Understanding Concepts and Vocabulary

1. True or False If z=x sin y,

then the change in z from 0.9, 0

to 1,π4

is Δz=sin π4-0.9 sin 0=22. 2. True or False If fx(x, y)=-y21-x

and fy(x, y)=2y ln(1-x),

then f is differentiable at (0, 2) 3. True or False Let z=f(x, y)=ex cos y of z at 0,π2

. . Then the differential dz

is dz=-dy.

4. True or False If the partial derivatives of a function z=f(x, y) fx(x, y)=2x+y

and fy(x, y)=x,

are then f is continuous.

5. True or False If the partial derivatives of a function z=f(x, y) fx(x, y)=2x+y

and fy(x, y)=x,

are then f is differentiable.

6. True or False If w=f(x, y, z)=exyz, dw=yzexdx+xzeydy+xyezdz.

then

Skill Building

In Problems 7–10, for each function z=f(x, y), 7. z=x2+y2 8. z=2x2+xy-y2 9. z=ex ln(xy)

from (1, 3)

find the change in z.

to (1.1, 3.2) from (2,-1)

from (1, 2)

to (2.1,-1.1)

to (0.9, 2.1)

10. z=xyx+y

from (-1, 2)

In Problems 11–22, find the differential dz

to (-0.9, 1.9) of each function.

11. z=x2+y2 12. z=2x2+xy-y2 13. z=x sin y+y sin x 14. z=ex cos y+e-x sin y 15. z=tan-1 yx 16. z=x tan-1 y 17. z=ln yx 18. z=ln(x2+y2) 19. z=exy 20. z=ex2+y 21. z=x2y+ey2 22. z=xy3-ln x2 In Problems 23–28, find the differential dw

of each function.

23. w=xeyz+yexz+zexy 24. w=x2y+y2z+z2x 25. w=ln(x2+y2+z2) 26. w=ln(xy)+ln(xz)+ln(yz) 27. w=xyz 28. w=xyzx+y+z In Problems 29–32, show that the function z=f(x, y) (x0, y0) in its domain by: a. Finding Δz .

is differentiable at any point

b. Finding η1

and η2

so that Δz=fx(x0, y0)Δx+fy(x0, y0)Δy+η1Δx+η2Δy holds.

c. Showing that lim(Δx, Δy)→(0, 0) η1=0 and lim(Δx, Δy)→(0, 0) η2=0 . 29. z=f(x, y)=xy2-2xy 30. z=f(x, y)=3x2+y2 31. z=f(x, y)=yx 32. z=f(x, y)=2xy In Problems 33–36, for each function z=f(x, y) differential dz to approximate the given value of f . 33. Approximate f(2.3,-2.1) .

if z=16x2+y2

34. Approximate f(-0.9, 1.9) .

if z=xyx+y

35. Approximate f(2.03, 0.1) f(x0, y0)=f(2, 0)

if z=x2 y+ey2 .

36. Approximate f(0.9, 2.1) .

if z=exln(xy)

use f(x0, y0)

and the

and f(x0, y0)=f(2,-2)

and f(x0, y0)=f(-1, 2)

and and f(x0, y0)=f(1, 2)

Applications and Extensions

37. Area of a Triangle Approximate the increase in the area of a triangle if its base is increased from 2 to 2.05 cm and its height is increased from 5 to 5.1 cm . 38. Area of a Triangle Approximate the change in the area of a triangle if the base is increased from 5 to 5.1 cm and the height is decreased from 10 to 9.8 cm . 39. Volume and Surface Area of a Cylinder

a. Use a differential to approximate the change in the volume of a right circular cylinder if the height changes from 2 to 2.1 cm

and the radius changes from 0.5

to 0.49 cm

.

b. Approximate the change in the surface area of the cylinder. Assume that the cylinder is closed on the top and on the bottom.

40. Electricity In a parallel circuit, the total resistance R due to two sources of resistance R1 and R2 is given by 1R=1R1+1R2 . If R1=50 Ω with a possible error of 1.2% and R2=75 Ω with a possible error of 1%, what is the approximate maximum variation in the total resistance? 41. Specific Gravity The specific gravity of an object is defined as s=aa-w, where a is the weight of the object in the air and w is its weight in water. If a is found to be 6 lb with a possible error of 1% and w is 5 lb with a possible error of 2%, what is the approximate maximum variation in the specific gravity? 42. Volume A grain silo consists of a hemisphere mounted on a cylinder of the same radius (see the figure). The height and radius of the cylinder are measured as 14 m and 5 m, respectively. However, the device used to make this measurement was found to be in error by 1% . What is the approximate maximum variation in the volume of the silo?

43. Snell’s Law The index of refraction is defined as μ=sin isin r, where i is the angle of incidence and r is the angle of refraction. If i=30∘ and r=60∘, and each measurement is subject to a possible error of 2%, what is the approximate maximum relative error for μ?

44. Ideal Gas Law The equation PV=kT, where k is a constant, relates the pressure P, volume V, and temperature T of one mole of an ideal gas. If P=0.1 g⁄mm2, V=12 mm3, and T=32 ∘C, approximate the change in P if V and T change to 15 mm3 and 29 ∘C, respectively. 45. Determining Density In a laboratory experiment, the average density ρ of a spherical object is computed by measuring the mass m and diameter d of the sphere. In all empirical measurements, there is error caused by the inaccuracy of the tool, the scientist, or both. Suppose the experimental measurements are m=24.0 g±1% and d=1.5 cm±5.0% . a. Find the approximate maximum change in the computed density. b. Find the approximate maximum percentage error in the computed density.

46. Geometry Two sides, b and c, and the included angle α of a triangle are measured by a ruler (for sides) and protractor (for angles), which are subject to errors of 2% and 3%, respectively. The area of the triangle is then computed from the formula A=12 bc sin α a. Show that dAA=dbb+dcc+cot α dα

b. If α=π4,

47.

.

what is the approximate maximum relative variation in the computation of A?

a. Find the partial derivatives fx

and fy

of the function f(x, y)=x2+y2-2x-

6y+10. b. Determine where the partial derivatives are continuous. c. Graph the surface f(x, y)=x2+y2-2x-6y+10. d. Use the graph to give a geometric interpretation to the discontinuity found in (b). 48. If x=r cos θ

and y=r sin θ,

show that x dy-y dx=r2 d θ

. In Problems 49 and 50, show that each function f has partial derivatives at (0, 0). Also show that f is not continuous at (0, 0). Is f differentiable at (0, 0)? Explain. 49. Let f(x, y)=2xyx2+y2if(x, y)≠(0, 0)0if(x, y)=(0, 0)

50. Let f(x, y)=xy(1+y2)x2+y2if(x, y)≠(0, 0)0if(x, y)=(0, 0)

51. Let f(x, y)=xy-1x2+y2-2ifx2+y2≠212ifx2+y2=2

Show that fx(1, 1) (1, 1) .

and fy(1, 1)

each exist, but f is not differentiable at

52. Let f(x, y)=x2y2x4+y4if(x, y)≠(0, 0)0if(x, y)=(0, 0)

Show that fx(0, 0) (0, 0) .

and fy(0, 0)

each exist, but f is not differentiable at

Challenge Problems

53. Electrical Resistance The electrical resistance R of a wire is given by R=ρLA, where L is the length of the wire, A is its cross-sectional area, and ρ is the resistivity of the wire. If the temperature of the wire increases by ΔT, the wire expands so L and A change. The resistivity ρ also changes. The change in the length L is given by ΔL=L0αΔT, the change in the radius r is Δr=r0αΔT, and the change in the resistivity ρ is Δρ=ρ0KΔT, where α and K are constants that depend on the material being heated, L0 is the initial length, r0 is the initial radius, and ρ0 is the initial resistivity. A copper wire with a circular cross-sectional area A has a resistance R of 0.50  Ω at 0∘C . But suppose the temperature of the wire increases to 40 ∘C from the heat generated by the current passing through the wire. For copper, α=5.1×10-

5∘C-1

and K=3.93×10-3∘C-1

.

a. Find the resistance R of the wire at 40 ∘C . b. Find the approximate maximum percentage change in the resistance R of the wire caused by the change in temperature. c. Which effect contributes more to the change in resistance: the expansion of the wire or the increase in the resistivity? 54.

a. Find the differential of f(x, y, z)=xyz b. Find the differential of g(x, y, z)=(xy)z

1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

. .

12.5 Chain Rules OBJECTIVES When you finish this section, you should be able to: 1 Differentiate functions of several variables where each variable is a

function of a single variable 2 Differentiate functions of several variables where each variable is a function of two or more variables 3 Differentiate an implicitly defined function of several variables 4 Use a Chain Rule in a proof 1 Differentiate Functions of Several Variables Where Each Variable Is a Function of a Single Variable Recall that the Chain Rule is used to find the derivative of a composite function. For functions of several variables, there is more than one version of the Chain Rule. The first Chain Rule is used when the independent variables of a composite function are each a function of a single variable t. For example, for the function z=f(x, y) of two variables, if x=x(t) and y=y(t), then the composite function z=f(x(t), y(t)) is a function of a single independent variable t . NEED TO REVIEW? The Chain Rule is discussed in Section 3.1, pp. 208–215.

THEOREM Chain Rule I: One Independent Variable

If x=x(t)

and y=y(t) are differentiable functions of t, and if z=f(x, y) is a differentiable function of x and y, then z=f(x(t), y(t)) is a differentiable function of t . Moreover,

dzdt=∂z∂x dxdt+∂z∂y dydt

Proof Since dzdt=limΔt→0 ΔzΔt, we seek an expression for Δz z=f(x, y) is differentiable, the change Δz is

. Since

Δz=∂z∂xΔx+∂z∂yΔy+η1Δx+η2Δy where η1

and η2

are functions of Δx

and Δy

and lim(Δx, Δy)→(0, 0) η1=0

and lim(Δx, Δy)→(0, 0) η2=0

. Next we divide both

sides by Δt. ΔzΔt=∂z∂x ΔxΔt+∂z∂y ΔyΔt+η1ΔxΔt+η2ΔyΔt Then dzdt=limΔt→0 ΔzΔt=limΔt→0∂z∂x ΔxΔt+∂z∂y ΔyΔt+η1ΔxΔt+η2ΔyΔt

In the right-hand expression, ∂z∂x and ∂z∂y and do not depend on Δt . Also, since x=x(t) differentiable,

are evaluated at (x(t), y(t)) and y=y(t) are

limΔt→0 ΔxΔt=dxdtandlimΔt→0 ΔyΔt=dydt Furthermore, as Δt→0, and η2→0

then (Δx,Δy)→(0, 0), . Putting this all together, we get

so that η1→0

dzdt=∂z∂x limΔt→0 ΔxΔt+∂z∂y limΔt→0 ΔyΔt+limΔt→0 η1⋅limΔt→0 ΔxΔt+limΔt→0  η2⋅limΔt→0 ΔyΔt=∂z∂x dxdt+∂z∂y dydt+0⋅dxdt+0⋅dydt=∂z∂x dxdt+∂z∂y dydt

▪ The tree diagram in Figure 38 may help you to remember the form of Chain Rule I.

Figure 38 dzdt=∂z∂x dxdt+∂z∂y dydt EXAMPLE 1 Differentiating a Function of Two Variables Where Each Variable Is a Function of t

Let z=x2 y-y2 x,

where x=sin t

and y=et

Solution We begin by finding the partial derivatives ∂z∂x derivatives dxdt

. Find dzdt

and ∂z∂y,

.

and the

and dydt.

∂z∂x=2xy-y2∂z∂y=x2-2xydxdt=cos tdydt=et

Then use Chain Rule I to find dzdt.

(1)

Since z is a function of t, we express dzdt

in terms of t.

dzdt=(2et sin t-e2t)(cos t)+(sin2 t-2et sin t)(et)x=sin t,y=et=et[sin(2t)-et cos t+sin2 t-2et sin t]2 sin t cos t=sin(2t)

▪

When dzdt

is expressed in terms of x, y, and t, as in (1), we say it is expressed

in mixed form. When dzdt is expressed in terms of t alone, as in (2), we say it is in final form. When computations become involved, we will leave our answers in mixed form. NOW WORK Problem 3.

EXAMPLE 2 Differentiating a Function of Two Variables Where Each Variable Is a Function of t

Let z=ex sin y, .

where x=et

and y=π3 e-t

Solution Begin by finding the partial derivatives ∂z∂x dxdt

and dydt

and ∂z∂y,

. Find dzdt

at t=0

and the derivatives

.

∂z∂x=ex sin y∂z∂y=ex cos ydxdt=etdydt=-π3 e-t

Then use Chain Rule I to find dzdt.

The above expression for dzdt . When t=0,

is in mixed form. We can use it to evaluate dzdt

then x=e0=1

and y=π3 e0=π3

at t=0

. So, when

t=0, dzdt t=0=e sin π3(1)+e cos π3-π3=e32-πe6=e6(33-π)

▪ NOW WORK Problem 41.

Chain Rule I extends to functions of three or more variables, where each of these variables is a function of a single variable t. If z=f(x1, x2,…,xn) differentiable and each variable xi=xi(t), i=1, 2,…,n, is a differentiable function of t, then

is

dzdt=∂z∂x1 dx1dt+∂z∂x2 dx2dt+⋯+∂z∂xn dxndt

where each of the partial derivatives ∂z∂x1,…,∂z∂xn t.

is expressed in terms of

EXAMPLE 3 Differentiating a Function of Three Variables Where Each Variable Is a Function of t

Find dwdt and z=5t

if w=x2 y+y2 z, .

where x=sin t,

y=cos t,

Solution Use Chain Rule I since each variable is a function of a single variable t . dwdt=∂w∂x dxdt+∂w∂y dydt+∂w∂z dzdt=(2xy)(cos t)+(x2+2yz)(-sin t)+y2(5)

Now since w is a function of t, write the derivative in terms of t alone. dwdt=2 sin t cos2 t-sin3 t-10 t sin t cos t+5 cos2 t

▪ NOTE For functions of a single variable t, it is sometimes easier to express the function in terms of t, and then differentiate it. NOW WORK Example 3 by expressing w as a function of t . NOW WORK Problem 17.

2 Differentiate Functions of Several Variables Where Each Variable Is a

Function of Two or More Variables A second version of the Chain Rule is used for differentiating z=f(x, y), where x and y are each functions of two independent variables u and v . For example, if x=g(u, v) and y=h(u, v), then the composite function z=f(x, y)=f(g(u, v), h(u, v)) is a function of the two variables u and v . We seek the partial derivatives ∂z∂u

and ∂z∂v

.

THEOREM Chain Rule II : Two Independent Variables

Let z=f(g(u, v), h(u, v)) be the composite of z=f(x, y), where x=g(u, v) and y=h(u, v) . If g and h are each continuous and have continuous first-order partial derivatives at a point (u, v) in the interior of the domains of both g and h, and if f is differentiable in some disk centered at the point (x, y)=(g(u, v), h(u, v)), then ∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂uand∂z∂v=∂z∂x ∂x∂v+∂z∂y ∂y∂v

Proof To find ∂z∂u, we hold v fixed. Then x=g(u, v) and y=h(u, v) are functions of u alone, and we can use Chain Rule I. Then ∂z∂u=∂z∂x dxdu+∂z∂y dydu. and replace dydu

with ∂y∂u

Now replace dxdu

with ∂x∂u,

.

A similar argument is used for finding ∂z∂v

.

▪

The tree diagrams in Figures 39 and 40 may help you remember the form of Chain Rule II.

Figure 39 ∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂u

Figure 40 ∂z∂v=∂z∂x ∂x∂v+∂z∂y ∂y∂v EXAMPLE 4 Differentiating a Function of Two Variables Where Each Variable Is a Function of Two Variables

Find ∂z∂u

and ∂z∂v

if z=f(x, y)=x2+xy-y2

and x=e2u + v

and y=ln vu. Solution The function z=f(x, y) is a composite function of two independent variables u and v. So, we use Chain Rule II.

∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂uand∂z∂v=∂z∂x ∂x∂v+∂z∂y ∂y∂v

Since ∂z∂x=2x+y∂z∂y=x-2y∂x∂u=2e2u + v∂y∂u=∂∂u(ln v-ln u)=-1u

we have

Similarly, since ∂x∂v=e2u + v

and ∂y∂v=1v,

we have

▪ Notice that since z is a function of u and v, the final form of ∂z∂u is expressed in terms of only u and v.

and ∂z∂v

NOW WORK Problem 31.

The form of Chain Rule II stays the same if z is a function of m≥3 variables. That is, if z=f(x1, x2,…,xm) is a differentiable function, and if each of the variables x1=g1(u1, u2,…,un), x2=g2(u1, u2,…,un), …,xm=gm(u1, u2,…,un) has continuous first-order partial derivatives, then the composite function z=f(g1(u1, …,un), g2(u1,…,un),…,gm(u1,…,un)) is a function of u1, u2,…,un, and the partial derivatives are found using an extension of Chain Rule II.

∂z∂u1=∂z∂x1 ∂x1∂u1+∂z∂x2 ∂x2∂u1+⋯+∂z∂xm ∂xm∂u1∂z∂u2=∂z∂x1 ∂x1∂u2+∂z∂x2  ∂x2∂u2+⋯+∂z∂xm ∂xm∂u2⋮∂z∂un=∂z∂x1 ∂x1∂un+∂z∂x2 ∂x2∂un+⋯+∂z∂xm ∂xm∂un

These partial derivatives can be written more compactly as

∂z∂ui=∑j = 1m∂z∂xj ∂xj∂uii=1, 2,…,n EXAMPLE 5 Differentiating a Function of Three Variables Where Each Variable Is a Function of Four Variables

Find ∂w∂u, z=u+2v+3s+4t

∂w∂v,

∂w∂s, ∂w∂t where x=uvst, .

for the function w=f(x, y, z)=x2+y2+z2, y=eu + v + s + t, and

Solution We use an extension of Chain Rule II, where each variable x, y, and z is a function of four variables: u, v, s, t . ∂w∂u=∂w∂x ∂x∂u+∂w∂y ∂y∂u+∂w∂z ∂z∂u∂w∂v=∂w∂x ∂x∂v+∂w∂y ∂y∂v+∂w∂z ∂z∂v=(2x) (vst)+(2y)(eu + v + s + t)+(2z)(1)=(2x)(ust)+(2y)(eu + v + s + t)+(2z) (2)=2uv2s2t2+2e2(u + v + s + t)+2(u+2v+3s+4t)=2u2 vs2t2+2e2(u + v + s + t)+4(u+2v+3s+4t)∂ ∂x∂s+∂w∂y ∂y∂s+∂w∂z ∂z∂s∂w∂t=∂w∂x ∂x∂t+∂w∂y ∂y∂t+∂w∂z ∂z∂t=(2x)(uvt)+(2y) (eu + v + s + t)+(2z)(3)=(2x)(uvs)+(2y)(eu + v + s + t)+(2z) (4)=2u2v2 st2+2e2(u + v + s + t)+6(u+2v+3s+4t)=2u2v2s2 t+2e2(u + v + s + t)+8(u+2v+3s+4t)

Again, notice that the partial derivatives of f in Example 5 are expressed in terms of

u, v, s,

and t alone.

▪

NOW WORK Problem 37.

3 Differentiate an Implicitly Defined Function of Several Variables If a differentiable function y=f(x) of one variable is defined implicitly by the equation F(x, y)=0, then F(x, f(x))=0 for all x in the domain of f . That is, F(x, f(x))≡0 . NEED TO REVIEW? Implicit differentiation is discussed in Section 3.2, pp. 219–224.

We can find the derivative dydx

as follows. Let

z=F(u, y)whereu=xandy=f(x) Since u and y are functions of one independent variable x, we use Chain Rule I to find dzdx. dzdx=∂F∂u⋅dudx+∂F∂y⋅dydx Since the composite function z=F(u, y)=F(x, f(x))≡0, derivative dzdx=0.

Also because u=x,

. So, 0=∂F∂x⋅1+∂F∂y⋅dydx

Now if ∂F∂y≠0,

we can solve for dydx.

dydx=-∂F∂x∂F∂y=-FxFy∂F∂y≠0

the ∂F∂u=∂F∂x

and dudx=1

Implicit Differentiation Formula I Suppose F is a differentiable function and y=f(x) implicitly by the equation F(x, y)=0 . Then

is a function defined

(3) dydx=-Fx(x, y)Fy(x, y)

provided ∂F∂y=Fy(x, y)≠0.

EXAMPLE 6 Differentiating an Implicitly Defined Function

Find dydx

if y=f(x)

is defined implicitly by F(x, y)=ey cos x-x-1=0 .

Solution First find the partial derivatives of F . Fx=∂F∂x=-ey sin x-1andFy=∂F∂y=ey cos x

Then use (3). If ey cos x≠0, dydx=-FxFy=--ey sin x-1ey cos x=ey sin x+1ey cos x

▪ NOW WORK Problem 47.

If a differentiable function z=f(x, y) the equation F(x, y, z)=0,

of two variables is defined implicitly by we can find the partial derivatives ∂z∂x

∂z∂y by using Chain Rule II. We begin by letting w=F(u, v, z), where u=x, v=y, z=f(x, y) . Since the composite function w=F(x, y, f(x, y))≡0,

and

and

it follows that ∂w∂x=0 expression for ∂w∂x,

and ∂w∂y=0

. To find an

use Chain Rule II.

∂w∂x=∂F∂u⋅∂u∂x+∂F∂v⋅∂v∂x+∂F∂z⋅∂z∂x=0 Since u=x,

then ∂F∂u=∂F∂x

then ∂F∂v=∂F∂y

and ∂v∂x=0

and ∂u∂x=1,

and since v=y,

. So,

∂F∂x⋅1+∂F∂y⋅0+∂F∂z⋅∂z∂x=0∂F∂x+∂F∂z⋅∂z∂x=0 If ∂F∂z≠0,

it follows that

∂z∂x=-∂F∂x∂F∂z=-Fx(x, y, z)Fz(x, y, z) In a similar way, it can be shown that

∂z∂y=-Fy(x, y, z)Fz(x, y, z) Implicit Differentiation Formulas II If a differentiable function z=f(x, y) F(x, y, z)=0, then

∂z∂x=-Fx(x, y, z)Fz(x, y, z)and∂z∂y=-Fy(x, y, z)Fz(x, y, z)

provided ∂F∂z=Fz(x, y, z)≠0.

is defined implicitly by the equation

EXAMPLE 7 Differentiating an Implicitly Defined Function

Find ∂z∂x

and ∂z∂y

if z=f(x, y)

is defined implicitly by the function

F(x, y, z)=x2z2+y2-z2+6yz-10=0 Solution First find the partial derivatives of F. Fx=∂F∂x=2xz2Fy=∂F∂y=2y+6zFz=∂F∂z=2x2z-2z+6y

Then use (4). If Fz=2x2z-2z+6y≠0,

∂z∂x=-2xz22x2z-2z+6y=-xz2x2z-z+3y and ∂z∂y=-2y+6z2x2z-2z+6y=-y+3zx2z-z+3y

▪ NOW WORK Problem 55.

4 Use a Chain Rule in a Proof Chain Rules are often used in proofs involving functions of two or more variables. EXAMPLE 8 Using a Chain Rule in a Proof

Let p=f(v-w, v-u, u-w)

be a differentiable function. Show that

∂p∂u+∂p∂v+∂p∂w=0 Solution Let x=v-w,

y=v-u,

and z=u-w

. We use an extension of Chain Rule II. Since ∂x∂u=0, and ∂z∂u=1,

we have

. Then p=f(x, y, z) ∂y∂u=-1,

∂p∂u=∂p∂x ∂x∂u+∂p∂y ∂y∂u+∂p∂z ∂z∂u=∂p∂x(0)+∂p∂y(-1)+∂p∂z(1)=-∂p∂y+∂p∂z

Since ∂x∂v=1,

∂y∂v=1,

and ∂z∂v=0,

we have

∂p∂v=∂p∂x ∂x∂v+∂p∂y ∂y∂v+∂p∂z ∂z∂v=∂p∂x(1)+∂p∂y(1)+∂p∂z(0)=∂p∂x+∂p∂y

Since ∂x∂w=-1,

∂y∂w=0,

and ∂z∂w=-1,

we have

∂p∂w=∂p∂x ∂x∂w+∂p∂y ∂y∂w+∂p∂z ∂z∂w=∂p∂x(-1)+∂p∂y(0)+∂p∂z(-1)=-∂p∂x-∂p∂z

Adding these, we get

▪ NOW WORK Problem 71.

12.5 Assess Your Understanding Concepts and Vocabulary

1. True or False If a differentiable function z is defined implicitly by the equation F(x, y, z)=0, provided Fz(x, y, z)≠0. 2. True or False If x=x(t) and if z=f(x, y) dzdt=∂z∂x+∂z∂y.

then ∂z∂x=Fx(x, y, z)Fz(x, y, z),

and y=y(t) are differentiable functions of t is a differentiable function of x and y, then

Skill Building

Answers to Problems 3–39 are given in final form and in mixed form. In Problems 3–14, find dzdt

using Chain Rule I.

3. z=x2+y2,x=sin t,y=cos(2t) 4. z=x2-y2,x=sin(2t),y=cos t 5. z=x2+y2,x=tet,y=te-t 6. z=x2-y2,x=te-t,y=t2e-t 7. z=eu sin v,u=t,v=πt 8. z=eu⁄v,u=t,v=t3+1 9. z=eu⁄v,u=tet,v=et2 10. z=ln(uv),u=t5,v=t+1 11. z=ex2 + y2,x=sin(2t),y=cos t 12. z=ex2 - y2,x=sin(2t),y=cos(2t) 13. z=xyx2+y2,x=sin t,y=cos t 14. z=y ln x+xy+tan y,x=tt+1,y=t3-t

In Problems 15–22, find dpdt

.

15. p=x2+y2-z2,x=tet,y=te-t,z=e2t 16. p=x2-y2-z2,x=te-t,y=t2e-t,z=e-t 17. p=ex sin y cos z,x=t,y=πt,z=t2 18. p=ln(xyz),x=t5,y=t+1,z=t2 19. p=w ln uv,u=tet,v=et2,w=e2t

20. p=weu⁄v,u=t,v=t3+1,w=et 21. p=u2 vw,u=sin t,v=cos t,w=et 22. p=uvw,u=et,v=tet,w=t2e2t

In Problems 23–34, find ∂z∂u

and ∂z∂v

using Chain Rule II.

23. z=x2+y2,x=uev,y=veu 24. z=x2-y2,x=u ln v,y=v ln u 25. z=ex sin y,x=u2 v,y=ln(uv) 26. z=1y ln x,x=uv,y=vu 27. z=ln(x2+y2),x=v2u,y=uv2 28. z=x sin y-y sin x,x=u2 v,y=uv2 29. z=x2+y2,x=sin(u-v),y=cos(u+v) 30. z=ex+y,x=tan -1uv,y=ln(u+v) 31. z=ser,r=u2+v2,s=vu 32. z=s2+r2,s=ln(uv),r=uv 33. z=x y2w3,x=2u+v,y=5u-3v,w=2u+3v

34. z=x2-y2+w,x=eu+v,y=uv,w=vu In Problems 35–40, find each partial derivative. 35. Find ∂f∂u,∂f∂v,∂f∂w if f(x, y, z)=x2+y2+z2,x=uv,y=eu+2v+3w,z=2v+3w .

36. Find ∂f∂u,∂f∂v,∂f∂w

if f(x, y, z)=x-y2+z2,x=u+v,y=(u+w)ln v,z=2-v+3w .

37. Find ∂f∂u,∂f∂v,∂f∂w if f(x, y, z)=x cos yz cos x+x2 yz,x=uvw,y=u2+v2+w2,z=w . 38. Find ∂f∂u,∂f∂v,∂f∂w

if f(x, y, z)=x2+y2,x=sin(u-v),y=cos(u+v),z=uw2 .

39. Find ∂f∂u,∂f∂v,∂f∂w,∂f∂t

if f(x, y, z)=x+2y2-z2, x=ut,y=eu+2v+3w+4t,z=u+12 v+4t .

40. Find ∂f∂u,

∂f∂v,

∂f∂w, ∂f∂t if f(x, y, z)=x2+y2+z, x=sin(u+t), y=cos(v-t), z=uw2 .

In Problems 41–44, for each function z=f(x, y), 41. z=sin(πx + y),

where x=et

42. z=cos(x + πy2),

and y=t2

where x=t3

43. z=1+3x2+y2,

44. z=x2+y2,

find dzdt

In Problems 45 and 46, z=f(x, y)

, x=u(t),

differentiable. Use the given information to find dzdt 45. u(2)=5,u′(2)=12,fx(5, 1)=4

and y=sin t

and y=t

.

; t0=0

and y=et⁄2

where x=cos t

where x=ln t

at t=t0

; t0=0 ; t0=π6

; t0=1 and y=v(t)

at t=t0

.

are

v(2)=1,v′(2)=3,fy(5, 1)=1 t0=2 46. u(1)=-3,u′(1)=2,fx(-3, 0)=4 v(1)=0,v′(1)=-2,fy(-3, 0)=5 t0=1

In Problems 47–52, y is a function of x . Find dydx. 47. F(x, y)=x2 y-y2 x+xy-5=0 48. F(x, y)=x3y2-xy+x2 y-10=0 49. F(x, y)=x sin y+y sin x-2=0 50. F(x, y)=x ey+y ex-xy=0 51. F(x, y)=x1⁄3+y1⁄3-1=0 52. F(x, y)=x2⁄3+y2⁄3-1=0

In Problems 53–58, z is a function of x and y . Find ∂z∂x

and ∂z∂y

53. F(x, y, z)=xz+3yz2+x2y3-5z=0 54. F(x, y, z)=x2 z+y2 z+x3 y-10z=0 55. F(x, y, z)=sin z+y cos z+xyz-10=0 56. F(x, y, z)=x sin y-cos z+x2 z=0 57. F(x, y, z)=x eyz+yexz+xyz=0 58. F(x, y, z)=eyzln x+yexz-yz=0

In Problems 59 and 60, find ∂w∂x, 59. w=(2x+3y)4z 60. w=(2x)3y + 4z

∂w∂y,

and ∂w∂z

.

.

Applications and Extensions

61. Ideal Gas Law One mole of a gas obeys the Ideal Gas Law PV=20T, where P is pressure, V is volume, and T is temperature in kelvin (K). If the temperature T of the gas is increasing at the rate of 5 K⁄s and if, when the temperature is 353 K, the pressure P is 10 N⁄m2 and is decreasing at the rate of 2 Nm2⋅s, find the rate of change of the volume V with respect to time. 62. Melting Ice A block of ice of dimensions l, w, and h is melting. When l=3m, w=2m, h=1m, these variables are changing so that dldt=-1 m/h,

dwdt=-1 m/h,

and dhdt=-0.5 m/h

. a. What is the rate of change in the surface area of the block of ice? b. What is the rate of change in the volume of the block of ice? 63. Wave Equation The one-dimensional wave equation ∂2 f∂x2=1v2 ∂2 f∂t2 describes a wave traveling with speed v along the x -axis. The function f represents the displacement x from the equilibrium of the wave at time t . a. Show that z=f(x, t)=sin(x+vt) b. Show that z=f(x, t)=ex-vt c. Show that z=f(x, t)=sin x+sin(vt) wave equation.

satisfies the wave equation. satisfies the wave equation. does not satisfy the

d. Show that any twice-differentiable function of the form f(x+vt) solution of the wave equation.

is a

64. Economics A toy manufacturer’s production function satisfies a Cobb–Douglas model, Q(L, M)=400L0.3M0.7, where Q is the output in thousands of units, L is the labor in thousands of hours, and M is the machine hours (in thousands). Suppose the labor hours are decreasing at a rate of 4000 h/yr and the machine hours are increasing at a rate of 2000 h/yr. Find the rate of change of production when a. L=19 b. L=21

and M=21 and M=20

65. Related Rates Let y=h(a, b), . Suppose when s=1

where a=h(s, t) and b=k(s, t) and t=3, we know that

∂h∂s=4∂k∂s=-3∂h∂t=1∂k∂t=-5 Also, suppose h(1, 3)=6k(1, 3)=2fa(6, 2)=7fb(6, 2)=2

What are ∂y∂s

and ∂y∂t

at the point (1, 3)?

66. Related Rates Let z=f(x, y), . Suppose when s=2

where x=g(s, t) and y=h(s, t) and t=-1, we know that

∂x∂s=5∂y∂s=2∂x∂t=-3∂y∂t=-2 Also, suppose g(2,-1)=3h(2,-1)=4fx(3, 4)=12fy(3, 4)=7

What are ∂z∂s 67. If z=f(x, y),

and ∂z∂t

at the point (2,-1)?

where x=g(u, v)

expressions for ∂2 z∂u2, ∂2 z∂u∂v, 68. If z=f(x, y)

and y=h(u, v),

find

and ∂2 z∂v2.

and x=r cos θ, y=r sin θ,

show that

∂z∂r2+1r2 ∂z∂ θ2=∂z∂x2+∂z∂y2 69. If z=f(x, y), +v cos θ,

where x=u cos θ-v sin θ with θ a constant, show that

and y=u sin θ

∂f∂u2+∂f∂v2=∂f∂x2+∂f∂y2 70. If z=f(u-v, v-u), x=u-v and y=v-u

show that ∂z∂u+∂z∂v=0 .

. Hint: Let

71. If z=vf(u2-v2), 72. If w=f(u)

show that v ∂z∂u+u ∂z∂v=uzv. and u=x2+y2+z2,

show that

∂w∂x2+∂w∂y2+∂w∂z2=dwdu2 73. If z=fxy,

show that x ∂z∂x+y ∂z∂y=0

74. Show that if z=fuv,vw,

.

then u ∂z∂u+v ∂z∂v+w ∂z∂w=0 .

75. Suppose we denote the expression ∂2∂x2+∂2∂y2 where x=r cos θ and y=r sin θ,

by Δ . If z=f(x, y), show that

Δf=∂2 f∂r2+1r⋅∂f∂r⋅+1r2⋅∂2 f∂θ2 76. Prove that if F(x, y, z)=0

is differentiable, then

∂z∂x⋅∂x∂y⋅∂y∂z=-1 Challenge Problems

77. a. Suppose that F(x, y) has continuous second-order partial derivatives and F(x, y)=0 defines y as a function of x implicitly. Show that d2 ydx2=-Fy2Fxx-2FxFyFxy+Fx2FyyFy3Fy≠0

b. Use the result from (a) to find d2 ydx2 78. f(t, x)=∫0x⁄(2λt)e-u2 du 1. = NOW WORK problem = Graphing technology recommended

for x3+3xy-y3=6 . Show that ft=λfxx

. .

= Computer Algebra System recommended

Chapter Review THINGS TO KNOW 12.1 Functions of Two or More Variables and Their Graphs

Definitions: Function z=f(x, y) Function w=f(x, y, z)

of two variables (p. 852) of three variables (p. 852)

Contour line: The curve formed by the intersection of a surface z=f(x, y) and the plane z=c, where c is a constant (p. 854) Level curve: the projection of a contour line onto the xy -plane (p. 855) 12.2 Limits and Continuity

Definitions: A δ -neighborhood of P0 P0 (p. 861)

: the set of all points P that lie within a distance δ of

In the plane, a δ -neighborhood of P0 P0. In space, a δ -neighborhood of P0

is called a disk of radius δ centered at

is called a ball of radius δ centered at P0.

Limit of a function f of two variables (p. 861) Limit of a function f of three variables (p. 862) Interior point, boundary point, open set, closed set (p. 867) Continuity at a point (p. 868) Theorems: Properties of limits (p. 863) Properties of continuous functions (p. 869) 12.3 Partial Derivatives

First-order partial derivatives of a function z=f(x, y) fx(x, y)=∂f∂x=∂z∂x=limΔx→0 f(x+Δx, y)

:

−f(x, y)Δxfy(x, y)=∂f∂y=∂z∂y=limΔy→0 f(x, y+Δy)−f(x, y)Δy

provided the limits exist. (p. 872) Mixed partials: ∂2 z∂x∂y=fyx(x, y)

and ∂2 z∂y∂x=fxy(x, y)

(p. 878) Equality of mixed partials (Clairaut’s Theorem) (p. 878) 12.4 Differentiability and the Differential

Change in z : Δz=f(x0+Δx, y0+Δy)-f(x0, y0) (p. 884) Differentiability of z=f(x, y) Differentials: dx=Δx,

at a point (x0, y0) dy=Δy,

(p. 885)

dz=fx(x0, y0) dx+fy(x0, y0) dy

(p. 886) Summary:

(p. 890) If z=f(x, y),

The continuity of the partial derivatives fx of f .

and fy

implies the differentiability

The differentiability of f implies the continuity of f . The continuity of the partial derivatives fx

and fy

implies the continuity of f .

The existence of the partial derivatives fx is differentiable.

and fy

does not necessarily mean f

The existence of the partial derivatives fx is continuous.

and fy

does not necessarily mean f

12.5 Chain Rules

Chain Rule I: dzdt=∂z∂x dxdt+∂z∂y dydt Chain Rule II: ∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂u

(p. 893)

∂z∂v=∂z∂x ∂x∂v+∂z∂y ∂y∂v Implicit differentiation formula I: If F(x, y)=0, provided Fy≠0.

(p. 896)

dydx=-Fx(x, y)Fy(x, y) (p. 898)

Implicit differentiation formula II: If F(x, y, z)=0, ∂z∂x=-Fx(x, y, z)Fz(x, y, z)

provided Fz≠0.

and ∂z∂y=-Fy(x, y, z)Fz(x, y, z)

(p. 899)

OBJECTIVES Section

You should be able to …

Example

12.1

1 Work with functions of two or three variables (p. 852) 2 Graph functions of two variables (p. 854) 3 Graph level curves (p. 854) 4 Describe level surfaces (p. 857) 1 Define the limit of a function of several variables (p. 861) 2 Find a limit using properties of limits (p. 863) 3 Examine when limits exist (p. 865) 4 Determine where a function is continuous (p. 867) 1 Find the partial derivatives of a function of two variables (p. 871) 2 Interpret partial derivatives as the slope of a tangent line (p. 874) 3 Interpret partial derivatives as a rate of change (p. 875) 4 Find second-order partial derivatives (p. 878) 5 Find the partial derivatives of a function of n variables (p. 879)

1–3 4 5, 6 7, 8

Review Exercises 1–9 10–13 14–16 17, 18

1 2–4 5–8 1–4 5 6–8 9, 10 11, 12

19, 20 21, 22 23–25 26–31, 41 32, 33 34(a), (b) 35, 36 37–40

1 Find the change in z=f(x, y) (p. 884) 2 Show that a function of two variables is differentiable (p. 885) 3 Use the differential as an approximating tool (p. 887) 4 Find the differential of a function of three or more variables (p. 890) 1 Differentiate functions of several variables where each variable is a function of a single variable (p. 893) 2 Differentiate functions of several variables where each variable is a function of two or more variables (p. 895)

1

42

2, 3 4–6 7

43 44, 45, 48–51 46, 47

1–3

34(c), 52, 53

4, 5

54, 55

3 Differentiate an implicitly defined function of several variables (p. 898) 6, 7

56, 57

12.2

12.3

12.4

12.5

4 Use a Chain Rule in a proof (p. 900)

REVIEW EXERCISES

8

58

In Problems 1–3, evaluate each function. 1. f(x, y)=ex ln y a. f(1, 1) b. f(x+Δx, y) c. f(x, y+Δy) 2. f(x, y)=2x2+6xy-y3 a. f(1, 1) b. f(x+Δx, y) c. f(x, y+Δy) 3. f(x, y, z)=exsin -1(y+2z) a. fln 3,12,14 b. f1, 0,14 c. f(0, 0, 0) In Problems 4–7, find the domain of each function and graph the domain. Use a solid curve to indicate that the domain includes the curve, and a dashed curve to indicate that the domain excludes the curve. 4. z=f(x, y)=ln(x2-3y) 5. z=f(x, y)=9-x2-4y2

6. z=f(x, y)=25xy5-y2

7. z=f(x, y)=(y-3x)2x+2y 8. Find the domain of w=f(x, y, z)=y2+z2x2. 9. Find the domain of w=f(x, y, z)=ex+y ln z

.

In Problems 10–13, graph each surface. 10. z=f(x, y)=x-y+5 11. z=f(x, y)=sin x 12. z=f(x, y)=ln x 13. z=f(x, y)=ey 14. For z=f(x, y)=x2-2y, c=-4, -1, 0, 1, 4.

graph the level curves corresponding to

15. For z=f(x, y)=x2+y2, c=0, 1, 4, 9.

graph the level curves corresponding to

16. For z=f(x, y)=e4x2+y2, c=1, e, e4, e16

graph the level curves corresponding to .

17. Describe the level surfaces associated with the function w=f(x, y, z)=4x2+y2+z2 . 18. Describe the level surfaces associated with the function w=f(x, y, z)=x+y+2z . 19. Find lim(x, y)→(π⁄2, 0) sin x cos yx

.

20. Find lim(x, y)→(1, 2) 4x2+y22x+y 21. Let f(x, y)=3x y2x2+y4

. .

a. Show that lim(x, y)→(0, 0) f(x, y)=0 . b. Find lim(x, y)→(0, 0) f(x, y)

along the lines y=mx

along the parabola x=y2

c. What can you conclude? 22. Show that lim(x, y)→(0, 0) 2y2-x2x2+y2 23. Determine where the function f(x, y)=2x2 y-y2+3

does not exist. is

.

continuous. 24. Determine where the function R(x, y)=xyx2-y2 25.

is continuous.

a. Determine where the function f(x, y)=tan -1 1x2+y2 continuous.

is

b. Find lim(x, y)→(0, 1) tan-1 1x2+y2. 26. Find fx

and fy

for z=f(x, y)=ex2+y2 sin(xy)

27. Find fx

and fy

for z=f(x, y)=x+yy

28. Find ∂z∂x

and ∂z∂y

for z=f(x, y)=x-2y2

29. Find ∂z∂x

and ∂z∂y

for z=f(x, y)=ex ln(5x+2y)

30. For f(x, y)=x2-y2,

find fx(2, 1)

31. For F(x, y)=ex sin y,

find Fx0,π6

. . . . and fy(2,-1) and Fy0,π6

. .

32. Find an equation for the tangent line to the curve of intersection of the ellipsoid x224+y212+z26=1, and z is positive.

and the plane y=1,

where x=4

33. Find an equation for the tangent line to the curve of intersection of the surface z=4x2y2+7 : a. With the plane y=-2 b. With the plane x=1

at the point (1,-2, 7)

.

at the point (1,-2, 7)

.

34. Boyle’s Law The volume V of a gas varies directly with the temperature T and inversely with the pressure P . a. Find the rate of change of the volume V with respect to the temperature T. b. Find the rate of change of the volume V with respect to the pressure P. c. If T and P are functions of t, what is dVdt

?

In Problems 35 and 36, find the second-order partial derivatives fxx, fxy, fyx,  and fyy . 35. z=f(x, y)=(x+y2)e3x 36. z=f(x, y)=sec(xy) In Problems 37–40, find fx, fy,

and fz .

37. w=f(x, y, z)=exyz 38. w=f(x, y, z)=zexy 39. f(x, y, z)=ex sin y+ey sin z 40. w=f(x, y, z)=z tan-1yx 41. For the function z=x3y2-2x y4+3x2y3, ∂z∂y=5z

show that x ∂z∂x+y 

.

42. Find the change Δz in z=f(x, y)=x y2+2 from (x0, y0) to (x0+Δx, y0+Δy) . Use the answer to calculate the change in z from (1, 0) to (0.9, 0.2) . 43. Show that the function z=f(x, y)=xy-5y2 point (x, y) in its domain by: a. Finding Δz

.

b. Finding η1

and η2

is differentiable at any

so that Δz=fx(x0, y0)Δx+fy(x0, y0)Δy+η1Δx+η2Δy .

c. Show that lim(Δx,Δy)→(0, 0) η1=0 η2=0

and lim(Δx,Δy)→(0, 0) 

.

In Problems 44 and 45, find the differential dz of each function. 44. z=x1+y2 45. z=sin-1xy, y>0

In Problems 46 and 47, find the differential dw

of each function.

46. w=zexy 47. w=ln(xyz) 48. Use differentials to estimate the change in z=x1+y2 (4.1, 0.1).

from (4, 0)

49. Use differentials to estimate the change in z=sin-1 xy (0.1, 1.1).

from (0, 1)

to

to

50. Use the differential of f(x, y)=y2 cos x to approximate the value of f(0.05, 1.98) . (Compare your answer with a calculator result.) 51. Electricity The electrical resistance R of a wire is R=k LD2 where L is the length of the wire, D is the diameter of the wire, and k is a constant. If L has a 1% error and D has a 2% error, what is the approximate maximum percentage error in the computation of R ? 52. Find dzdt .

if z=sin(xy)-x sin y,

53. Find dwdt

if w=xy+yz+zx,

and z=1t3 54. Find ∂w∂u

55. Find ∂u∂r

where x=et

and y=tet

where x=1t, y=1t2,

. and ∂w∂v and z=v and ∂u∂s

if w=xy+yz-xz, where x=u+v, y=uv, . Express each answer in terms of u and v.

if u=x2+y2+z2,

x=r cos s, y=r sin s, answer in terms of r and s . 56. Find ∂z∂x and ∂z∂y if z=f(x, y) F(x, y, z)=x2+y2-2xyz=0

where

and z=r2+s2

. Express each

is defined implicitly by .

57. Find ∂z∂x and ∂z∂y if z=f(x, y) F(x, y, z)=2x sin y+2y sin x+2xyz=0 58. If z=uf(u2+v2),

is defined implicitly by

show that 2v ∂z∂u-2u ∂z∂v=2vzu.

1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

.

CHAPTER 12 PROJECT Searching for Exoplanets

To find an exoplanet generally requires powerful telescopes, but some exoplanets can be observed by amateur astronomers with smaller telescopes and careful observation using the transit method. When using less powerful telescopes, very small errors can greatly affect the results of the observation since the change in apparent brightness of the star because of the transit is at most a few percent. Atmospheric scintillation, the main source of error, refers to variations, due to differences in the index of refraction of Earth’s atmosphere, in the apparent brightness (magnitude) of an extrasolar object. When a star is viewed through a lens, the error E in magnitude due to atmospheric scintillation is estimated by (1) E(D, Z, h, T)=0.09D-2⁄3(2T)-1⁄21cos Z1.75e(-h⁄8000)

where D Z h T

is the diameter, in centimeters, of the lens of the telescope is the zenithal angle, in degrees, of the planet is the altitude, in meters above sea level, of the telescope is the exposure time in seconds.

The zenithal angle is defined to be the angle from the highest point in the sky (Z=0∘ ) to an object in the heavens, and it is important because there is less atmosphere to get in the way of viewing a star that is higher in the sky. To obtain observations accurate enough to detect the small fluctuation that would indicate the transit of an exoplanet, the fluctuation in magnitude due to scintillation needs to be smaller than 0.001 . 1. Suppose the diameter of the telescope lens is 7.1 cm,

the zenithal angle is 20∘,

and the altitude of the telescope is 3000 m . What is the minimum exposure time needed to keep the error E in magnitude below 0.001 ? Round the answer to the nearest second. 2. Repeat Problem 1 for h=4000 m and h=5000 m . Comment on the effect of altitude on the error. Explain why your result makes sense in terms of the effects of atmospheric scintillation. 3. Find limZ→0 E(7.1, Z,3000, 180)

and

limZ→90 E(7.1, Z,3000, 180). zenithal angle on the error. 4. Find ∂E∂D

. Interpret ∂E∂D

5. Find ∂E∂Z

. Interpret ∂E∂Z

6. Find ∂E∂h

. Interpret ∂E∂h

7. Find ∂E∂T

. Interpret ∂E∂T

8. Find the differential dE

Comment on the effect of the

. . . .

. Interpret dE

.

9. Suppose D=7.1, Z=20, h=3000, and T=180 . Use the differential to determine which of the following reduces the error by a greater amount: a. using a telescope with a lens of diameter of 7.3 cm,

or

b. using a platform to increase the altitude of the telescope to 3200 m

.

10. Suppose you are interested in astronomy and searching for exoplanets. You research the cost of telescopes with different-size lenses and the resulting effects that the different lenses will have on your observations. Do a cost benefit analysis, and then write a position paper supporting your decision to purchase the telescope you chose.

13 Directional Derivatives, Gradients, and Extrema

13.1 Directional Derivatives; Gradients 13.2 Tangent Planes 13.3 Extrema of Functions of Two Variables 13.4 Lagrange Multipliers Chapter Review Chapter Project

Measuring Ice Thickness on Crystal Lake Minnesota’s nickname is the Land of 10,000 Lakes, and in the winter many of those lakes freeze over. While this might put an end to some water sports like swimming, it opens up a host of other activities for Minnesotans to enjoy—ice skating, ice fishing, ice hockey, ice golf, ice bowling, and ice driving, to name just a few. The annual Art Shanty Projects festival on Medicine Lake even features shacks (complete with radio and heaters), sculptures, printing presses, and concerts, all comfortably resting atop the lake’s frozen surface. Of course, a frozen lake can only host those people and events that its ice will support. Specific estimates vary, but the recommended ice depth that will safely support one ice skater is about 4 inches. A dozen or more ice skaters might

require closer to 7 inches of ice. A small truck? That would need about 10.5 inches. But how do you determine the depth of the ice on a given lake? Trial and error is prohibitively risky. The Minnesota Department of Natural Resources suggests that you ask at local bait shops or resorts. In the chapter project, we examine another method for measuring ice depth. CHAPTER 13 PROJECT In the Chapter Project on page 949, we use two hypothetical models to measure the ice depth in a frozen lake.

In Chapter 12 we defined the partial derivatives ∂ z∂ x=fx(x, y) ∂z∂y=fy(x, y)

of a function z=f(x, y)

and of two variables. One

interpretation of fx(x, y) and fy(x, y) is the rate of change of z in the direction of the positive x -axis and in the direction of the positive y -axis, respectively. Here we generalize these ideas to obtain the rate of change of z in any direction, a directional derivative. Directional derivatives lead us to the gradient, a vector that is used in finding tangent planes to a surface. The gradient also provides information about paths of quickest ascent (or descent) on a surface. Finally, we revisit optimization. Just as derivatives are used to find extreme values of functions of a single variable, partial derivatives are used to find extreme values of functions of several variables. We identify techniques for locating extreme values on a surface and investigate applications involving optimization of functions of two variables.

13.1 Directional Derivatives; Gradients OBJECTIVES When you finish this section, you should be able to: 1 Find the directional derivative of a function of two variables 2 Find the gradient of a function of two variables 3 Work with properties of gradient 4 Find the directional derivative and gradient of a function of three

variables 1 Find the Directional Derivative of a Function of Two Variables In Chapter 2, we saw that an important interpretation of the derivative of a function y=f(x) is the rate of change of f. In Chapter 12, the partial derivatives of a function z=f(x, y) at the point (x0,y0) are defined as fx(x0,y0)=limΔx→0 f(x0+Δx, y0)-f(x0,y0)Δxfy(x0,y0)=limΔy→0 f(x0,y0+Δy)-f(x0,y0)Δy

where (x0,y0) fx(x0,y0)

is an interior point of the domain of f. The partial derivative equals the rate of change of f at (x0,y0) in the direction of the

positive x -axis, and the partial derivative fy(x0,y0) equals the rate of change of f at (x0,y0) in the direction of the positive y -axis. But what if we want the rate of change of f in a direction other than the positive x -axis or the positive y -axis? To lay the groundwork consider a function z=f(x, y) whose graph is a surface in space. In Figure 1, P0=(x0,y0) is an interior point in the domain of f, and u=cos θi+sin θj is a unit vector in the xy -plane with its initial point at P0 making an angle θ, 0≤θ0

and A>0,

then f has a local minimum at (x0,y0).

If AC-B2>0

and A0 and A=fxx(4,-2)=2>0, Partial Derivative Test states that f has a local minimum at (4,-2),

▪

is the local minimum value. See Figure 17.

the Second and f(4,-2)=-6

Figure 17 z=f(x, y) =x2+xy+y2-6x+6 NOW WORK Problems 13 and 17. NOTE To use the Second Partial Derivative Test for Local Extrema we must solve the system of equations fx=0 and fy=0. If the system is nonlinear, no general method of solution is available and a CAS is generally used. If solving by hand, care must be taken because extraneous roots are sometimes introduced.

EXAMPLE 4 Using the Second Partial Derivative Test

Find any local maxima, local minima, and saddle points for z=f(x, y)=x3+y2+2xy-4x-3y+5 Solution We begin with the system of equations

{ fx (x, y)=3x2 +2y −4=0fy(x, y)=2y +2x −3=0 We solve the system using substitution. Solving for y in fy(x, y)=0,

we obtain

y=12(3-2x). result is

Then if we substitute for y in fx(x, y)=0,

the

3x2+2[12(3−2x)]−4=03x2−2x−1=0(3x+1)(x−1)=0x=−13  or  x=1

When x=-13,

When x=1,

then y=12(3-2x)=12 3+23=116.

then y=12(3-2x)=12.

The critical points are -13,116 and 1,12. The second-order partial derivatives of f are fxx(x, y)=6xfxy(x, y)=2fyy(x, y)=2 The requirements of the Second Partial Derivative Test are met at each critical point. At -13,116, A=fxx -13,116=-2B=fxy -13,116=2C=fyy -13,116=2AC-B2=-4-4=-80,

f has a local minimum at 1,12,

and the local minimum value is f1,12=74

▪



A graph of z=f(x, y)=x3+y2+2xy-4x-3y+5 shown in Figure 18.

is

Figure 18 z=f(x, y) =x3+y2+2xy-4x-3 y+5 NOW WORK Problems 19 and 23.

3 Find the Absolute Extrema of a Function of Two Variables For a function f of one variable, the Extreme Value Theorem states that if f is continuous on a closed interval [a,b], then f has an absolute maximum value and an absolute minimum value on that interval. For functions of two variables, there is a similar theorem. But first we need to define the set in the plane that is analogous to a closed interval on the real line. Recall that a set D is closed if it contains all its boundary points. We say that D is bounded if D can be enclosed by some disk (if D is in the plane) or by some ball (if D is in space). With this definition, we can state the following two-variable version of the Extreme Value Theorem:

THEOREM Extreme Value Theorem for Functions of Two Variables

Let z=f(x, y) bounded set D,

be a function of two variables. If f is continuous on a closed, then f has an absolute maximum and an absolute minimum on D.

NEED TO REVIEW? Open and closed sets are discussed in Section 12.2, p. 867.

To find the absolute extrema of a function satisfying the criteria of the Extreme Value Theorem for Functions of Two Variables, we note that if f has an extreme value at (x0,y0), then (x0,y0) either is a critical point of f or is a boundary point of D. As a result, we have the following test: THEOREM Test for Absolute Maximum and Absolute Minimum

Let z=f(x, y) be a function of two variables defined on a closed, bounded set D. If f is continuous on D, then the absolute maximum value and the absolute minimum value of f are, respectively, the largest and smallest values found among the following: The values of f at the critical points of f in D The values of f on the boundary of D

EXAMPLE 5 Finding Absolute Extrema

Find the absolute maximum and the absolute minimum of z=f(x, y)=2x-2xy+y2 whose domain is the region defined by 0≤x≤4

and 0≤y≤3.

Solution The function f is continuous on its domain, which is a closed, bounded set. Then the Extreme Value Theorem for Functions of Two Variables guarantees that f has an absolute maximum and an absolute minimum on its domain. To find them, we need the critical points of f. Since fx and fy exist at every point in the domain of f, the critical points of f are the solutions of the system of equations

fx(x, y)=2-2y=0fy(x, y)=-2x+2y=0 Solving, we find that the only critical point is (1, 1).

The value of f at (1, 1)

is

f(1, 1)=2⋅1-2⋅1⋅1+12=1 The domain of f is the set 0≤x≤4, 0≤y≤3. domain of f consists of the four line segments L1, L2, Figure 19. We evaluate f on each line segment.

The boundary of the L3, and L4 shown in

Figure 19 z=f(x, y)=2x-2xy+y2,

0≤x≤4,

On L1

: x is in the interval 0,4

0≤y≤3

and y=0.

The function f(x, y)=f(x,0)=2x is increasing on 0≤x≤4, so its extreme values occur at the endpoints 0 and 4. At x=0, f(0, 0)=0andatx=4,f(4, 0)=8 On L2

: x=4

and y is in the interval [0,3].

The function f(x, y)=f(4, y)=8-8y+y2. To find the extreme values of f on [0,3], begin by finding the critical number(s) of the function g(y)=8-8y+y2. That is, we find where g′(y)=0.

g′(y)=-8+2y=0y=4 Since 4 is not in the interval [0,3], and 3.

the extreme values occur at the endpoints 0

At y=0,f(4, 0)=8andat y=3, f(4, 3)=-7 On L3

: x is in the interval [0,4]

and y=3.

The function f(x, y)=f(x,3)=2x-6x+9=-4x+9 is decreasing on 0≤x≤4, so its extreme values occur at the endpoints 0 and 4. At x=0, f(0, 3)=9andat x=4, f(4, 3)=-7 On L4

: x=0

and y is in the interval [0,3].

The function f(x, y)=f(0, y)=y2 is increasing on 0≤y≤3, so its extreme values occur at the endpoints 0 and 3. At y=0,f(0, 0)=0andat y=3,f(0, 3)=9 The values of f at the critical point (1, 1) boundary are Point

(1, 1)

Value

(0, 0)

1

0

and at the extreme values on the

(4, 0) 8

(4, 3)

(0, 3)

-7

9

The absolute maximum value of f is f(0, 3)=9 f(4, 3)=-7.

▪



NOW WORK Problem 37.

EXAMPLE 6 Finding Absolute Extrema

Find the absolute maximum and the absolute minimum of z=f(x, y)=x2+y2-2x+2y-5 whose domain is the disk x2+y2≤9.

; the absolute minimum value is

Solution The function f is continuous on its domain, a closed, bounded set. The Extreme Value Theorem for Functions of Two Variables guarantees that f has an absolute maximum and an absolute minimum on its domain. To find them, find the critical points of f. Since fx and fy exist at every point in the domain of f, the critical points of f are the solutions of the system of equations

{ fx (x, y)=2x−2=0fy(x, y)=2y+2=0 The only solution is x=1, y=-1, which is an interior point of the domain of f. The value of f at the critical point is f(1,-1)=-7. The boundary of the domain of f is the circle x2+y2=9. Express the boundary using the parametric equations x=x(t)=3 cos t, y=y(t)=3 sin t, 0≤t≤2π, and evaluate f on its boundary. z=f(x,y)=f(3 cos t, 3 sin t)=9 cos2 t+9 sin2 t−6 cos t+6 sin t−5=4−6 cos t +6 sin t

Now determine where dzdt=0.

dzdt=6 sin t+6 cos t=0tan t=-1t=3π4ort=7π4 At t=3π4,

the value of f is

f3 cos 3π4,3 sin 3π4=4-6 cos 3π4+6 sin 3π4=4-6-22+6⋅22=4+62

At t=7π4,

the value of f is

f3 cos 7π4,3 sin 7π4=4-6 cos 7π4+6 sin 7π4=4-6⋅22+6-22=4-62

The absolute maximum value of f, which occurs on the boundary of f, is 4+62≈12.485. The absolute minimum value of f, which occurs at an interior point, is -7.



▪

Figure 20 shows the graph of z=f(x, y)=x2+y2-2x+2y-5 with the absolute extrema marked.

Figure 20 NOW WORK Problem 39.

4 Solve Optimization Problems Now that we have the tools for finding the extrema of functions of two variables, we can investigate applied problems that involve optimization. EXAMPLE 7 Solving an Optimization Problem: Minimizing Resources Used

A manufacturer wants to make an open rectangular box of volume V=500 cm3 using the least possible amount of material. Find the dimensions of the box. Solution Let x and y be the dimensions of the base of the box, and let z be the height of the box. Then V=500=xyz cm3x>0y>0z>0 The manufacturer wants to minimize the amount of material used, which equals the surface area S of the box. Since the box is open, S=xy+2xz+2yz If we solve the equation 500=xyz for z and substitute z=500xy into the formula for the surface area S of the box, we can express S as a function of two variables. S=S(x, y)=xy+2x⋅500xy+2y⋅500xy=xy+1000y+1000x

This is the function to be minimized. The partial derivatives of S are Sx=y-1000x2andSy=x-1000y2 Since x>0

and y>0,

the critical points satisfy the system of equations

y-1000x2=0x-1000y2=0 Using substitution, we find x=y=10001/3=10. the only critical point of S. The second-order partial derivatives of S are

So the point (10, 10) is

Sxx=∂∂x y-1000x2=2000x3Sxy=∂∂y y-1000x2=1Syy=∂∂y x-1000y2=2000y3

At the critical point (10, 10),

A=Sxx(10, 10)=2000103=2B=Sxy(10, 10)=1C=Syy(10, 10)=2000103=2AC-B2=3>0

From the Second Partial Derivative Test, S has a local minimum at (10, 10). But, we want to know where S attains its absolute minimum. Since the physical properties of the problem require that the absolute minimum exists, the local minimum is also the absolute minimum. Consequently, the dimensions (in centimeters) of the open box of volume V=500 cm3 that uses the least amount of material are x=10 cmy=10 cmz=500100=5 cm

▪ In Example 7, the conditions of the Extreme Value Theorem for Functions of Two Variables were not met, since the domain of S is not bounded. Fortunately, in applied problems (such as Example 7), it is often possible to argue from physical or geometric properties that an absolute extremum must exist and that it must occur at one of the local extrema. NOW WORK Problem 49.

EXAMPLE 8 Solving an Optimization Problem: Maximizing Profit

The demand functions for two products are p=12-2xandq=20-y where p and q are the respective prices (in thousands of dollars) of each product, and x and y are the respective amounts (in thousands of units) of each sold. Suppose the joint cost function is C(x, y)=x2+2xy+2y2 a. Find the revenue function R=R(x, y)

and the profit function P=P(x, y).

b. Determine the prices and amounts that will maximize profit. c. What is the maximum profit? Solution a. Revenue is the amount of money brought in. That is, revenue is the product of price and quantity sold. The revenue function R=R(x, y) is the sum of the

revenues from each product. R(x, y)=xp+yq=x(12-2x)+y(20-y) Profit is the difference between revenue and cost. The profit function P=P(x, y) is P(x, y)=R(x, y)-C(x, y)=[x(12-2x)+y(20-y)]-[x2+2xy+2y2]=12x-2x2+20y-y2-x2-2xy2y2=-3x2-3y2-2xy+12x+20y

b. The partial derivatives of P are Px(x, y)=∂∂x (-3x2-3y2-2xy+12x+20y)=-6x-2y+12Py(x, y)=∂∂y (-3x2-3y22xy+12x+20y)=-6y-2x+20

Find the critical points by solving the system of equations: -6x-2y+12=0-2x-6y+20=0 We solve the first equation for y and substitute the result into the second equation. Since y=6-3x, then -2x-6(6-3x)+20=0x=1 and y=6-3⋅1=3, so (1, 3) The second-order partial derivatives are

is the only critical point.

Pxx(x, y)=-6Pxy(x, y)=-2Pyy(x, y)=-6 At the critical point (1, 3), Pxx(1, 3)=-60 From the Second Partial Derivative Test, P has a local maximum at (1, 3). The domains of the demand functions p=12-2x and q=20-y

are

0≤x≤6 and 0≤y≤20, respectively. Since the domain of the profit function P is 0≤x≤6, 0≤y≤20, the domain is closed and bounded, so a maximum profit exists. From the test for absolute maximum and absolute minimum, the maximum profit is found at a critical point or on the boundary of the domain. Figure 21 shows the boundary of the profit function P, and Table 1 gives the maximum value of P on its boundary and at the critical point ( 1, 3 ).

Figure 21 The boundary of P TABLE 1 Maximum L1:0≤x≤6,y=0

P(x,0)=-3x2+12x

P(2, 0)=12

L2:x=6,x≤y≤20

P(6,y)=-3y2+8y-36

P(6,y)0, (0, 0) if A0, then z=f(x, y) (x0,y0) ; if A0, has a local maximum at (x0,y0) then z=f(x, y)

* If AC-B2=0,

Show that if A>0 has a local minimum at then z=f(x, y) ; and if AC-B20 and y>0. We express this as a problem in one variable by solving xy=4 z=12x+3y.

for y and substituting y=4x

in the expression

Then dzdx=12-12x2=12x2-12x2x>0 The critical numbers of z are -1 examine x=1, for x=1,

and 1. We exclude x=-1,

since x>0.

To

use the Second Derivative Test. Since d2 zd x2=24x3>0 z has a minimum when x=1

and y=4x=4.

The minimum

value of z is 24

at the point (1, 4)

▪

on the hyperbola.

NEED TO REVIEW? The Second Derivative Test is discussed in Section 4.4, pp. 302–304.

In Example 1, we solved the problem by eliminating the variable y and treating the problem as a minimum problem in one variable. Sometimes, however, it is not easy, or perhaps it is impossible, to eliminate a variable. Then the Method of Lagrange Multipliers is useful. Before we use the method, let’s look at a geometric rationale. A Geometric Rationale for the Method of Lagrange Multipliers Consider Example 1 again. We want to minimize the function z=f(x, y)=12x+3y, subject to the constraint g(x, y)=xy-4=0. The graph of g(x, y)=xy-4=0 is a curve (a hyperbola) in the xy -plane. We are interested in finding the smallest value of z=f(x, y), where (x, y) is a point in the first quadrant on the graph of g(x, y)=0. Figure 22 shows several level curves of the plane z=12x+3y. On each level curve, the value of z is fixed. We want the smallest z that coincides with the hyperbola xy=4. Figure 22 shows that the minimum value of z occurs precisely at the point where the line 12x+3y=24 is tangent to the graph of the hyperbola xy=4. At this point, the two curves have a common tangent line, so they must have the same normal line. Since the gradient gives the direction of the normal line to a curve, it follows that ∇f is parallel to ∇g. That is, there is a scalar λ≠0, so that ∇f=λ∇gandg(x, y)=0

Figure 22 Minimum value of z=12x+3y

subject to xy=4

is z=24.

This discussion leads to the following result. THEOREM Lagrange’s Theorem

Let f and g be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve g(x, y)=0. Suppose that f, when restricted to points on the curve g(x, y)=0, has a local extremum at the point (x0,y0) and that ∇g(x0,y0)≠0. Then there is a number λ for which ∇f(x0,y0)=λ∇g(x0,y0) Proof Express the smooth curve C defined by g(x, y)=0 as a parametric equation in vector form r(t)=x(t)i+y(t)j. Define the function s(t)=f(x(t),y(t)) and choose t0 so that x0=x(t0), y0=y(t0). Now since f has a local extremum at the point (x0,y0), the extreme value is given by s(t0)=f(x(t0),y(t0))=f(x0,y0)

Then s′(t0)=0.

By Chain Rule I,

s′(t0)=fx(x0,y0) x′(t0)+fy(x0,y0) y′(t0)dsdt=∂s∂x dxdt+∂s∂y dydt=∇f(x0,y0)⋅r′(t0)Definition of the gradient=0

If ∇f(x0,y0)≠0, it is orthogonal to r′(t0), which is a tangent vector to C at (x0,y0). Since the gradient is normal to the level curve at (x0,y0), the vector ∇g(x0,y0) is also orthogonal to r′(t0). If ∇f(x0,y0)=0, then let λ=0. Then, in every case, there is a number λ for which ∇f(x0,y0)=λ∇g(x0,y0).

▪



ORIGINS Joseph-Louis Lagrange (1736–1813), a.k.a. Giuseppe Lodovico Lagrangia, was a self-taught mathematician and astronomer. He was born and began his career teaching in Turin, Italy, but later moved first to Berlin and then to Paris, where he held prestigious positions at the Berlin Academy and Académie des Sciences. After the French Revolution (1789), Lagrange was named the first professor of analysis at the École Polytechnique, in Paris. Lagrange is one of 72 names of mathematicians, engineers, and scientists inscribed on the Eiffel Tower.

The number λ in the equations ∇f(x, y)=λ∇g(x, y) is called a Lagrange multiplier. Since ∇f=λ∇g, we have fxi+fyj=λ(gxi+gyj), so that fx(x, y)=λ gx(x, y)

and

fy(x, y)=λ gy(x, y). These equations and the equation g(x, y)=0 form a system of three equations in the three unknowns x,y, and λ.

fx(x, y)=λ gx(x, y)fy(x, y)=λ gy(x, y)g(x, y)=0

A solution (x0,y0) of the system of equations is called a test point, since it is a candidate for the desired extrema. The maximum and minimum values, if they exist, are found by choosing the largest and smallest values of z=f(x, y) at the test points. 1 Use Lagrange Multipliers for an Optimization Problem with One Constraint EXAMPLE 2 Solving Example 1 Using Lagrange Multipliers

Find the point in the first quadrant on the hyperbola xy=4, where the value of z=12x+3y is a minimum. What is the minimum value? Solution We can restate the problem to read as follows: Find the minimum value of z=f(x, y)=12x+3y subject to the condition xy=4, where x>0 and y>0. The test points satisfy the equations ∇f(x, y)=λ∇g(x, y)g(x, y)=xy-4=0x>0y>0

(1)

where λ is a number. Since ∇f(x, y)=12i+3j and ∇g(x, y)=yi+xj, the equations in (1) lead to the following system of three equations containing three variables: 12=λyfx(x, y)=λ gx(x, y)3=λxfy(x, y)=λ gy(x, y)xy-4=0g(x, y)=0

We solve this system by eliminating λ from the first two equations. This results in y=4x, so the third equation becomes 4x2-4=0x=1orx=-1 Since x>0, point is (1, 4).

▪



we ignore x=-1. When x=1, then y=4, The corresponding minimum value of z=12x+3y

so the only test is z=24.

NOTE In eliminating λ,

we use the fact that λ≠0

since ∇f≠0.

NOW WORK Problem 5.

Steps for Using Lagrange Multipliers

Step 1 Express the problem in the following form: Find the maximum (or minimum) value of z=f(x, y) subject to the constraint g(x, y)=0. Step 2 Check that the functions f and g satisfy Lagrange’s Theorem. Step 3 Solve the equations ∇f(x, y)=λ∇g(x, y) and g(x, y)=0 for the test points by solving the system of equations

fx(x, y)=λ gx(x, y)fy(x, y)=λ gy(x, y)g(x, y)=0 Step 4 Evaluate z=f(x, y) at each test point found in Step 3. Choose the maximum (or minimum) value of z.

EXAMPLE 3 Using Lagrange Multipliers

Find the maximum and minimum values of z=f(x, y)=3x-y+1 subject to the constraint 3x2+y2-9=0. Solution Follow the steps for using Lagrange multipliers: Step 1 The problem is expressed in the required form. Step 2 The functions f(x, y)=3x-y+1 and g(x, y)=3x2+y2-9=0 each have continuous partial derivatives. Step 3 ∇f(x, y)=3i-j

and ∇g(x, y)=6xi+2yj

The equations ∇f(x, y)=λ∇g(x, y) lead to the system of equations

and g(x, y)=0

3=6λxfx(x, y)=λ gx(x, y)-1=2λyfy(x, y)=λ gy(x, y)3x2+y2-9=0g(x, y)=0

From the first two equations, we find that x=12λ

and y=-12λ

from which x=-y.

Substituting into the third equation, we have

4x2=9x=±32

Then x=±32,

y=∓32,

and the test points are 32,-32

and

-32,32. Step 4 The values of z corresponding to the test points are z=92+32+1=7 and z=-92-32+1=-5. of z is 7 and the minimum value is -5.

The maximum value

▪



NOW WORK Problem 9.

EXAMPLE 4 Using Lagrange Multipliers to Find Absolute Extrema

Find the absolute maximum and absolute minimum of f(x, y)=x2+y2+4x-4y+3 subject to x2+y2≤2. Solution From the Extreme Value Theorem for Functions of Two Variables, f has both an absolute maximum and an absolute minimum on the closed, bounded set x2+y2≤2. Begin by finding any critical points of f.

fx=2x+4=0fy=2y-4=0x=-2y=2 The point (-2, 2) lies outside the disk x2+y2≤2, so f has no relevant critical points. The extrema must occur on the boundary of the domain, that is, on the curve x2+y2=2. To find the extrema, we use Lagrange multipliers. Step 1 Using g(x, y)=x2+y2-2=0 expressed in the required form. Step 2 The functions f(x, y)=x2+y2+4x-4y+3 g(x, y)=x2+y2-2=0 derivatives. Step 3 The equations ∇f(x, y)=λ∇g(x, y) lead to the system of equations

as the constraint, the problem is and each have continuous partial and g(x, y)=0

2x+4=2λxfx(x, y)=λ gx(x, y)2y-4=2λyfy(x, y)=λ gy(x, y)x2+y2=2g(x, y)=0

Eliminate λ from the first two equations. 2x+4=2xλ2y-4=2yλλ=2x+42xλ=2y-42yλ=1+2xλ=1-2y

Then 1+2x=1-2y,

so y=-x.

If we substitute y=-x 2x2=2. Then x=±1 (1,-1) and (-1, 1). Step 4 Evaluate z=f(x, y)

into the third equation x2+y2=2, we find and y=∓1. There are two test points: at each test point.

f(1,-1)=13Maximum valuef(-1,1)=-3Minimum value

▪ The cylinder x2+y2=2

and the elliptic paraboloid z=(x+2)2+(y-2)2-5 are shown in Figure 23.

Figure 23 NOW WORK Problem 17.

Lagrange multipliers can be used for functions of three variables. The extreme values of w=f(x, y,z) subject to the constraint g(x, y,z)=0, if they exist, occur at the solutions (x, y,z) of the system of equations. ∇f(x, y,z)=λ∇g(x, y,z)andg(x, y,z)=0 These equations form a system of four equations in the four unknowns x, y, z, and λ.

EXAMPLE 5 Using Lagrange Multipliers with a Function of Three Variables

A container in the shape of a rectangular box is open on top and has a fixed volume of 12 m3 . The material used to make the bottom of the container costs $3 per square meter, while the material used for the sides costs $1 per square meter. What dimensions will minimize the cost of material?

Solution Let x be the width of the container, y be its depth, and z be its height. Then the volume of the box is xyz=12. The cost function C=C(x, y,z), which is to be minimized, is C(x, y,z)=3xy+2xz+2yz We use Lagrange multipliers. Step 1 We want to minimize C subject to the constraint g(x, y,z)=xyz-12=0. Step 2 The functions C and g each have continuous partial derivatives. and g(x, y, z)=0

Step 3 The equations ∇C(x, y,z)=λ∇g(x, y,z) lead to the system of equations 3y+2z=λyzCx(x, y,z)=λ gx(x, y,z)(1)3x+2z=λxzCy(x, y,z)=λ gy(x, y,z) (2)2x+2y=λxyCz(x, y,z)=λ gz(x, y,z)(3)xyz-12=0g(x, y,z)=0(4)

Now use the facts that x>0, equations for λ, obtaining

y>0,

and z>0

λ=3z+2y(5)λ=3z+2x(6)λ=2y+2x(7)

From these, we find that y=xFrom equations (5) and (6)z=32 xFrom equations (5) and (7)

to solve the first three

Substituting into xyz-12=0,

we find

xyz-12=0x⋅x⋅32 x-12=032 x3=12x=2 Then y=2, z=3, and the test point is (2,2,3). Step 4 The dimensions of the container that minimize the cost are 2 m by 2 m by 3 m. The minimum cost is C(2,2,3)=3⋅2⋅2+2⋅2⋅3+2⋅2⋅3=$36.

▪

NOW WORK Problem 33.

2 Use Lagrange Multipliers for an Optimization Problem with Two Constraints Lagrange multipliers can be used to solve problems involving more than one constraint. In the two-constraint situation, we seek the extreme values of a function w=f(x, y,z) subject to two constraints g(x, y,z)=0 and h(x, y,z)=0. The extreme values of f, if they exist, occur at solutions (x, y,z) of the system of five equations containing the five variables x,y,z,λ1, and λ2. ∇f(x, y,z)=λ1∇g(x, y,z)+λ2∇h(x, y,z)g(x, y,z)=0h(x, y,z)=0

where λ1

and λ2

are Lagrange multipliers.

EXAMPLE 6 Using Lagrange Multipliers with Two Constraints

Find the points of intersection of the ellipsoid x2+y2+9z2=25 and the plane x+3y-2z=0 that are farthest from the origin. Also find the points that are closest to the origin. Solution See Figure 24.

Figure 24 Let C be the curve formed by intersecting the ellipsoid and the plane, and let P=(x, y,z) be a point on C. We use the steps for Lagrange multipliers. Step 1 The function to be optimized is the distance d from P to the origin: d=f(x, y,z)=x2+y2+z2 subject to the constraints g(x, y,z)=x2+y2+9z2-25=0The point must be on the ellipsoid.h(x, y,z)=x+3y-2z=0The point must be on the plane.

Step 2 The functions f,g, and h are each functions of three variables, and each has continuous partial derivatives at every point of some open set containing the smooth curve defined by g(x, y,z)=0 and h(x, y,z)=0. Step 3 We introduce two Lagrange multipliers and solve the system of equations ∇f(x, y,z)=λ1∇g(x, y,z)+λ2∇h(x, y,z), g(x, y,z)=0, and h(x, y,z)=0. xx2+y2+z2=2 λ1x+λ2fx(x, y,z)=λ1 gx(x, y,z)+λ2 hx(x, y,z)yx2+y2+z2=2 λ1y+3  λ2fy(x, y,z)=λ1 gy(x, y,z)+λ2 hy(x, y,z)zx2+y2+z2=18 λ1z-2  λ2fz(x, y,z)=λ1 gz(x, y,z)+λ2 hz(x, y,z)x2+y2+9z2-25=0g(x, y,z)=0x+3y-2z=0h(x, y,z)=0

Eliminating λ1 and λ2 from the first three equations, we have two possibilities: z=0 or y=3x. Using each of these equations with the constraints x2+y2+9z2-25=0 and x+3y-2z=0 results in the following: z=0:-1510,510,05 units from the origin1510,-510,05 units from the originy=3x:5235,15235,25235875235≈1.930 units from the origin5235,-15235,-25235875235≈1.930 units from the origin

The points -1510,510,0

and 1510,-510,0

farthest from the origin, and the points 5235,15235,25235

are and

-5235,-15235,-25235 are closest to the origin. Since the intersection of the ellipsoid and the plane is an ellipse, the points we have found are the

▪

endpoints of the major and minor axes of this ellipse. NOW WORK Problem 19.

13.4 Assess Your Understanding

Concepts and Vocabulary

1. Multiple Choice The number λ in the equations ∇f(x, y)=λ∇g(x, y) is called a [(a) factor (b) Lagrangian (c) Lagrange multiplier]. 2. True or False Extreme values of the function z=f(x, y) subject to the constraint g(x, y)=0 are found as solutions of the system of equations ∇f(x, y)=λ∇g(x, y) and g(x, y)=0. 3. True or False Lagrange multipliers can be used only for functions of two variables. 4. True or False When using Lagrange multipliers to find the extreme values of a function, the number of Lagrange multipliers introduced depends on the number of variables in the function. Skill Building

In Problems 5–16, use Lagrange multipliers to find the maximum and minimum values of f subject to the constraint g(x, y)=0. 5. f(x, y)=3x+y,

g(x, y)=xy-8=0

6. f(x, y)=3x+y+4,

g(x, y)=xy-1=0

7. f(x, y)=3x+y+4,

g(x, y)=x2+4y2-1=0

8. f(x, y)=3x+y, 9. f(x, y)=x-2y2, 10. f(x, y)=x2+4y3,

g(x, y)=x2+y2-4=0 g(x, y)=x2+y2-1=0 g(x, y)=x2+2y2-2=0

11. f(x, y)=2xy,

g(x, y)=x2+y2-2=0

12. f(x, y)=xy,

g(x, y)=9x2+4y2-36=0

13. f(x, y)=x2-4xy+4y2,

g(x, y)=x2+y2-4=0

14. f(x, y)=9x2-6xy+y2,

g(x, y)=x2+y2-25=0

15. f(x, y,z)=4x-3y+2z,

g(x, y,z)=x2+y2-6z=0

16. f(x, y,z)=x2+2y2+z2,

g(x, y,z)=2x-3y+z-6=0

17. Find the absolute maximum and the absolute minimum of the function f(x, y)=x2+y2+4xy subject to the constraint x2+y2≤2. 18. Find the absolute maximum and the absolute minimum of the function f(x, y)=2x2+y2 subject to the constraint x2+y2≤4. 19. Find the absolute maximum and the absolute minimum values of the function f(x, y,z)=x2+y2+z2 subject to the constraints z2=x2+y2 and x+y-z+1=0. 20. Find the absolute minimum value of w=x2+y2+z2 constraints 2x+y+2z=9 and 5x+5y+7z=29.

subject to the

Applications and Extensions

21. Minimizing Distance Find the point on the line x-3y=6 origin.

that is closest to the

22. Minimizing Distance Find the point on the plane 2x+y-3z=6 closest to the origin.

that is

23. Find the maximum product of two numbers x and y subject to x+2y=21. 24. Find the minimum quotient of two positive numbers x and y subject to 4x+2y=100. 25. a. Find the points on the intersection of 4x2+y2+z2=16 plane 3-y=0 that are farthest from the origin.

and the

b. Find the points that are closest to the origin. 26. Find the point on the intersection of the sphere x2+2x+y2+z2=16 and the plane 3x+y-z=0 that is farthest from the origin. Also find the point that is closest to the origin. 27. At which points on the ellipse x2+2y2=2

is the product xy a maximum?

28. Maximizing Volume Find the dimensions of an open-topped box that maximizes volume when the surface area is fixed at 48 square centimeters. 29. Manufacturing and Design Find the optimal dimensions for a can in the shape of a

right circular cylinder of fixed volume V. That is, find the height h and the radius r of the can in terms of V so that the surface area is minimized. Assume the can is closed at the top and at the bottom. Source: Problem submitted by the students at Minnesota State University 30. Minimizing Materials A manufacturer receives an order to build a closed rectangular container with a volume of 216 m3. What dimensions will minimize the amount of material needed to produce the container? 31. Cost of a Box An open-topped box has a volume of 12 m3 and is to be made from material costing $1 per square meter. What dimensions minimize the cost? 32. Cost of a Box A rectangular box is to have a bottom made from material costing $2 per square meter, while the top and sides are made from material costing $1 per square meter. If the volume of the box is to be 18 m3, what dimensions will minimize the cost of production? 33. Carry-On Luggage Requirements The linear measurements (length+width+height ) for luggage carried onto a Delta Airlines plane must not exceed 45 inches. Find the dimensions of the rectangular suitcase of greatest volume that meets this requirement.

Source: Luggage measurements from Delta Airlines, 2017. 34. Extreme Temperature Suppose that T=T(x, y,z)=100x2yz is the temperature (in degrees Celsius) at any point (x, y,z) on the sphere given by x2+y2+z2=1. Find the points on the sphere where the temperature is greatest and least. What is the temperature at these points? 35. Fencing A farmer has 340 m of fencing for enclosing two separate fields, one of which

is to be a rectangle twice as long as it is wide and the other is a square. The square field must enclose at least 100 m2, and the rectangular one must enclose at least 800 m2. a. If x is the width of the rectangular field, what are the maximum and minimum values of x ? b. What is the greatest number of square meters that can be enclosed in the two fields? 36. Fencing in an Area A Vinyl Fence Co. prices its Cape Cod Concave fence, which is 3 ft tall, at $26.62 per linear foot. A home builder has $5000 available to spend on enclosing a rectangular garden. What is the largest area that can be enclosed? Source: A Vinyl Fence Co. San Jose, California, 2017. 37. Joint Cost Function Let x and y be the number of units (in thousands) of two products manufactured at a factory, and let C=18x2+9y2 in thousands of dollars be the joint cost of production of the products. If x+y=5400, find x and y that minimize production cost. 38. Production Function The production function of a company is P(x, y)=x2+3xy-6x, where x and y represent two different types of input. Find the amounts of x and y that maximize production if x+y=40. 39. Economics: The Cobb-Douglas Model Use the Cobb-Douglas production model P=1.01K0.25 L0.75 as follows: Suppose that each unit of capital K has a value of $175 and each unit of labor L has a value of $125. a. If there is a total of $175,000 to invest in the economy, use Lagrange multipliers to find the units of capital and the units of labor that maximize the total production in the manufacturing sector of the economy. b. What are the maximum units of production that the manufacturing sector of the economy could generate under these conditions? 40. The surface xyz=-1 is cut by the plane x+y+z=1, resulting in a curve C. Find the points on C that are nearest to the origin and farthest from the origin. 41. Find the points on the curve of intersection of the cylinder x2+y2=4 and the plane 2x+y+z=2 that are closest to and furthest from the origin. 42. Maximizing Volume A closed rectangular box of fixed surface area and maximum volume is a cube. Use Lagrange multipliers to confirm this fact. 43. Maximizing Volume A closed cylindrical can of fixed surface area and maximum volume has a height equal to the diameter of its base. Use Lagrange multipliers to

confirm this fact. 44. Find the points of intersection on the plane x+y+z=1 x2+y2-z2=1 nearest the origin.

and the hyperboloid

45. At what points on the union of the two curves x2+y2=1 is the function f(x, y)=x4+y4+4 points is it a minimum?

and x3+y3=1 a maximum? At what

46. Find the extreme values of f(x, y,z)=xyz

on the sphere x2+y2+z2=1.

Challenge Problems

47. Minimize x4+y4+z4

subject to the constraint Ax+By+Cz=D.

48. Use Lagrange multipliers to show that the triangle of largest perimeter that can be inscribed in a circle of radius R is an equilateral triangle. 49. Find the point on the paraboloid z=2-x2-y2 (1,1,2). 50. What points on the surface xy-z2-6y+36=0 origin? 1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

that is closest to the point are closest to the

Chapter Review THINGS TO KNOW 13.1 Directional Derivative; Gradient

The directional derivative of a differentiable function z=f(x, y) point (x0,y0) in the direction of the unit vector u=cos θ i+sin θ j is Du f(x0,y0)=fx(x0,y0) cos θ+fy(x0,y0) sin θ.

at the

(p. 909) The directional derivative Du f(x0,y0) of a differentiable function f equals the slope of the tangent line to the curve C at the point (x0,y0,f(x0,y0)) on the surface z=f(x, y), where C is the intersection of the surface with the plane perpendicular to the xy -plane and containing the line through (x0,y0) in the direction u. (p. 910) The gradient of a differentiable function f of two variables: ∇f(x, y)=fx(x, y)i+fy(x, y)j 911) of three variables: ∇f(x, y,z)=fx(x, y,z)i+fy(x, y,z)j+fz(x, y,z)k

(p.

(p. 917) The directional derivative Du f(x, y)=∇f(x, y)⋅u, u=cos θ i+sin θ j. (p. 911) Du f(x, y,z)=∇f(x, y,z)⋅u, β j+cos γ k.

where where u=cos αi+cos 

(p. 917)

Properties of the Gradient: The gradient ∇f(x0,y0)≠0 (x0,y0). (p. 912) If ∇f(x0,y0)=0 direction u. (p. 913)

is normal to the level curve of f at P0=

,then Du f(x0,y0)=0

for any

If ∇f(x0,y0)≠0, then the directional derivative of f at (x0,y0) is a maximum when u is in the direction of ∇f(x0,y0). The maximum value of Du f(x0,y0) is ∇f(x0,y0). (p. 913)

If ∇f(x0,y0)≠0, then the directional derivative of f at (x0,y0) is a minimum when u is in the direction of -∇f(x0,y0). The minimum value of Du f(x0,y0) is -∇f(x0,y0). (p. 913) z=f(x, y) z=f(x, y)

increases most rapidly in the direction of ∇f(x0,y0)≠0. (p. 914) decreases most rapidly in the direction of -∇f(x0,y0)≠0. (p. 914)

The value of z=f(x, y) ∇f(x0,y0)≠0.

remains the same for directions orthogonal to (p. 914)

13.2 Tangent Planes

Definition of a tangent plane to a surface (p. 921) Equation of a tangent plane to a surface (p. 922) Equations of the normal line to the surface F(x, y,z)=0 (x0,y0,z0) :

at the point P0=

vector equation: r(t)=r0+t∇F(x0,y0,z0), r0=x0i+y0j+z0k is the position vector of P0 position vector of any point P on the normal line. (p. 922) parametric equations: x=x0+at, where a=Fx(x0,y0,z0), and c=Fz(x0,y0,z0).

y=y0+bt,

symmetric equations: x-x0a=y-y0b=z-z0c (p. 923) If ∇f(x0,y0)=0, at (x0,y0) the xy -plane. (p. 923)

where and r is the

and z=z0+ct, b=Fy(x0,y0,z0), (p. 923) if abc≠0.

then the tangent plane to the surface z=f(x, y) is z=z0. That is, the tangent plane is parallel to

13.3 Extrema of Functions of Two Variables

Definitions: Local maximum; local minimum (p. 926) Absolute maximum; absolute minimum (p. 926) Critical point (p. 927)

Saddle point (p. 928) A Necessary Condition for Local Extrema

Let z=f(x, y) be a function of two variables whose domain is an open set D. If f has a local extremum at (x0,y0), then (x0,y0) is a critical point of f. (p. 927) Second Partial Derivative Test: (p. 929)

Let z=f(x, y) be a functionof two variables for which the first- and second-order partial derivatives are continuous in some disk containing the point (x0,y0). Suppose that fx(x0,y0)=0 and fy(x0,y0)=0. Let A=fxx(x0,y0)B=fxy(x0,y0)C=fyy(x0,y0) If AC-B2>0 minimum at (x0,y0).

and fxx(x0,y0)>0,

then f has a local

If AC-B2>0 maximum at (x0,y0).

and fxx(x0,y0)0,

c>0.

31. Minimizing Cost The base of a rectangular box costs five times as much as the other five sides. Use Lagrange multipliers to find the proportions of the dimensions for the cheapest possible box of volume V. 32. Find the extreme values of f(x, y,z)=xyz x2+y2=1 and y=3z.

subject to the constraints

CHAPTER 13 PROJECT Measuring Ice Thickness on Crystal Lake

In this project we examine safety issues involving frozen fresh-water lakes. When a lake freezes over, the depth of the ice varies from place to place on the lake. If we can construct a model that gives the depth of ice at any position on a lake, then we can determine whether a given position is safe or not for a particular load/activity. Based on studies of ice depth, Table 1 lists minimum safe ice depths for various loads and activities on clear lake ice. TABLE 1 Load/Activity Cross-country skiing (one person) Ice fishing or skating (one person) One snowmobile Ice boating Group activities One car Several snowmobiles Light truck (2.5 tons) Medium truck (3.5 tons) 10-ton load 25-ton load

Minimum Safe Ice Depth (in inches) 3 4 5 6 7 8 9 10.5 12 15 20

Data from: U.S. Army Corps of Engineers, Minnesota Department of Natural Resources, Pennsylvania Fish and Boat Commission, and The Old

Farmer’s Almanac Our study focuses on Crystal Lake in Otter Tail, Minnesota. Crystal Lake is roughly circular in shape, with a radius of 0.838 mi. 1. Place the center of the lake at the origin (0, 0) of a rectangular coordinate system. Using rectangular coordinates (x, y) measured in miles, write an equation for the shoreline of Crystal Lake. What are the minimum and maximum values of x and y? Draw a graph of the lake with the direction north pointing up. A simple model for determining the depth d of ice is d=d(x, y)=α+α sin(3πxy) where d is measured in inches, (x, y) are the coordinates of a point on the lake, as given in Problem 1, and α, measured in inches, is a nonnegative constant related to various conditions such as temperature, lake depth, currents, and so on. Suppose on a certain day in December, the value of α for Crystal Lake is α=9 in. 2. What is the domain of d=d(x, y)? minimum and maximum values of d?

What is the range? What are the

3. What is the depth of the ice at the center of the lake? 4. Graph the level curves of d(x, y)=c, and 18.

for c=0, 3, 6, 12, 15,

5. a. Using the model, determine whether a snowmobile can safely cross the lake along the path y=x. b. Using the model, determine whether a car can safely cross the lake along the path y=-x. 6. a. Find the rate of change of d at the point (0.4,-0.1) east (the direction of i).

in the direction

b. Find the rate of change of d at the point (0.4,-0.1) north (the direction of j). c. Interpret the results obtained in (a) and (b).

in the direction

7. a. Find the direction from the point 0.4, -0.1 increases most rapidly.

for which the ice depth

b. Travel a short distance, say 1 ft, in the direction found in (a). At the new point, recalculate the direction for which the ice depth increases most rapidly.

c. Explain the results obtained in (a) and (b). 8. a. Find the direction from the point (0.4,-0.1) in ice depth d.

that results in no change

b. Travel a short distance, say 1 ft, in the direction found in (a), and at the new point, recalculate the direction that again results in no change in ice depth. c. Explain the results obtained in (a) and (b). Another model for the depth of ice on Crystal Lake is given by h=h(x, y)=α|x+y|+β|x-y| where h is measured in inches and α, β, measured in inches per mile, are nonnegative constants determined by local conditions. Suppose that on a certain day in January, the value of α=β=9 in/mi for Crystal Lake. 9. Sketch the level curves of h(x, y)=c, 15.

for c=0, 3, 6, 12,

10. a. Using this model, find the direction from the point (0.4,-0.1) the ice depth increases most rapidly.

and for which

b. Travel a short distance, say 1 ft, in the direction found in (a). At the new point, recalculate the direction for which the ice depth increases most rapidly. c. Explain the results obtained in (a) and (b). 11. a. Using this model, find the direction from the point 0.4, -0.1 results in no change in ice depth h.

that

b. Travel a short distance, say 1 ft, in the direction found in (a), and at the new point, recalculate the direction for no increase in ice depth. A year earlier, the values of α and β were different. For that day in January, α=5 and β=10. 12. Sketch the level curves of h(x, y)=5|x+y|+10|x-y|, c=0,3,6,12, and 15. 13. a. Find the direction from the point 0.4,-0.1 increases most rapidly.

for for which the ice depth

b. Travel a short distance, say 1 ft, in the direction found in (a). At the new point, recalculate the direction for which the ice depth increases most rapidly. c. Explain the results obtained in (a) and (b). 14. a. Find the direction from the point (0.4,-0.1) ice depth.

that results in no change in

b. Travel a short distance, say 1 ft, in the direction found in (a), and at the new point, recalculate the direction for no increase in ice depth. b. Explain the results obtained in (a) and (b). 15. What conditions might have influenced the difference in α and β from one year to the next? 16. Compare the results obtained from the two models. Which would you choose to model the ice depth of Crystal Lake? Write a brief report that supports your decision.

14 Multiple Integrals

14.1 The Double Integral over a Rectangular Region 14.2 The Double Integral over Nonrectangular Regions 14.3 Double Integrals Using Polar Coordinates 14.4 Center of Mass; Moment of Inertia 14.5 Surface Area 14.6 The Triple Integral 14.7 Triple Integrals Using Cylindrical Coordinates 14.8 Triple Integrals Using Spherical Coordinates 14.9 Change of Variables Using Jacobians Chapter Review Chapter Project

The Mass of Stars Some of the most massive objects in the universe are not solid, but are large balls of gas—stars. The nearest star, the Sun, has a mass of 1.99×1030 kg and is about 333,000 times more massive than Earth. Stars are often depicted on a Hertzsprung–Russell diagram, or HR diagram, which compares the luminosity (brightness) and color (temperature) of stars. As

centuries pass, the composition of a star changes because of the nuclear reactions at the center that generate the energy we see as light. Over billions of years changes in composition alter the star’s structure and the star evolves, tracing a unique path in the HR diagram determined primarily by its mass. CHAPTER 14 PROJECT The Chapter Project on page 1023 investigates models for determining the mass of a star. In Chapter 5, we defined the definite integral of a function (of a single variable). When extended to functions of two or more variables, definite integrals are called multiple integrals. The integral of a function of two variables is called a double integral, and the integral of a function of three variables is a triple integral. For consistency, we refer to the integral of a function of one variable as a single integral. In Chapter 6, we used a single integral to find the area of a region, the volume and surface area of a solid, arc length, and the work done by a variable force. Two of the many uses of multiple integrals are to find areas and volumes of more general regions than those considered in Chapter 6.

14.1 The Double Integral over a Rectangular Region OBJECTIVES When you finish this section, you should be able to: 1 Find double Riemann sums of z=f(x,y)

over a closed rectangular

region 2 Find the value of a double integral defined on a closed rectangular region 3 Find the volume under a surface and over a rectangular region Historically, the investigation of the area problem led to the definite integral. Similarly, the search for methods to find the volume of an irregularly shaped solid led to the development of multiple integrals. 1 Find Double Riemann Sums of z=f(x,y) Region

over a Closed Rectangular

Since the development of multiple integrals parallels that of a single integral, we begin by reviewing the process that led to the definition of a single integral. To define a single integral, we begin with a function y=f(x) interval [a,b]. Then we:

defined on a closed

Partition [a,b]

into n subintervals of lengths Δx1, Δx2,…,Δxi,…,Δxn. (Recall that the lengths of the subintervals need not be equal and the length of the largest subinterval is the norm, max Δxi, of the partition.) Choose a number ui in each subinterval. Evaluate f at ui,

multiply f(ui)

by Δxi,

and form the Riemann sums

∑i=1n f(ui)Δxi.

If limmax Δxi→0 ∑i=1n f(ui)Δxi integral of f from a to b and is denoted by

exists, it is called the definite

∫ab f(x) dx=limmax Δxi→0 ∑i=1n f(ui)Δxi For a function f of two variables, the double integral of f is defined similarly. We

begin with a function z=f(x,y) of two variables defined on a closed rectangular region R defined by a≤x≤b and c≤y≤d. Partition the interval [a,b] into n subintervals of length Δxi, i=1,2,…,n, and the interval [c,d] into m subintervals of length Δyj, j=1,2, …,m. By drawing lines parallel to the coordinate axes through the endpoints of the subintervals, we form a rectangular partition P of the region R into nm subrectangles Rij. Now we define the norm, ∥Δ∥, of the rectangular partition as the largest of the lengths of the subintervals Δxi or Δyj. That is, ∥Δ∥=max(Δxi,Δyj). See Figure 1.

Figure 1 A rectangular partition of the region R.

The norm of the partition, ∥Δ∥,

the largest of the lengths of the subintervals Δxi

or Δyj.

In each subrectangle Rij, choose a point (uij,vij), z=f(x,y) at (uij,vij), multiply f(uij,vij) Δyj, and form the sums ∑j=1m∑i=1n f(uij,vij)ΔAij=∑j=1m∑i=1n f(uij,vij)Δxi Δyj

is

evaluate the function by the area ΔAij=Δxi 

These sums are called the double Riemann sums of f for the partition P. Although the subintervals Δxi, i=1, 2,…,n need not be equal, we usually partition [a,b] into n subintervals each of the same length Δx=b-an. Similarly, we usually partition [c,d]

into m subintervals each of the

same length Δy=d-cm. EXAMPLE 1 Finding Double Riemann Sums

Let f(x,y)=x2y be a function defined on the square having its lower left corner at (1, 0) and its upper right corner at (5, 4), as shown in Figure 2. a. Find a double Riemann sum of f over this region by partitioning the square into four congruent subsquares. Choose the lower left corner of each subsquare as (uij,vij). b. Find a double Riemann sum of f over this region by partitioning the square into eight congruent rectangles with sides of length Δxi=2, i=1, 2, and Δyj=1, j=1, 2, 3, 4. Choose the lower right corner of each rectangle as (uij,vij).

Figure 2 f(x,y)=x2y

is defined on the square 1≤x≤5, 0≤y≤4.

Solution a. Begin by partitioning the square in Figure 2 into four congruent subsquares. We form four subsquares by partitioning the interval [1, 5] into two subintervals each of

length Δxi=5-12=2, i=1, 2,

and by partitioning the interval [0, 4]

into two subintervals each of length Δyj=4-02=2,j=1, 2. So, Δxi Δyj=2⋅2=4 for i=1, 2 and j=1, 2. Now draw lines parallel to the coordinate axes through the endpoints of each subinterval. Figure 3 shows the lower left corner of each subsquare.

Figure 3 The square is partitioned into four congruent subsquares. The double Riemann sum for which (uij,vij) subsquare is

is the lower left corner of each

∑j=12∑i=12 f(uij,vij)Δxi Δyj=f(1, 0)⋅4+f(3, 0)⋅4+f(1, 2)⋅4+f(3, 2)⋅4= [f(1, 0)+f(3, 0)+f(1, 2)+f(3, 2)]⋅4=[0+0+2+18]⋅4=80f(x,y)=x2 y

b. Now partition the square in Figure 2 into eight congruent rectangles with sides Δxi=2, i=1, 2, and Δyj=1, j=1, 2, 3, 4, as shown in Figure 4. The double Riemann sum, for which (uij,vij) is the lower right corner of each subrectangle, is ∑j=14∑i=12 f(uij,vij)Δxi Δyj= [f(3, 0)+f(5, 0)+f(3, 1)+f(5, 1)+f(3, 2)+f(5, 2)+f(3, 3)+f(5, 3)]⋅2⋅1= [0+0+9+25+18+50+27+75]⋅2=408

Figure 4 The square is partitioned into eight congruent subrectangles.

▪ NOW WORK Problem 5.

As Example 1 demonstrates, the double Riemann sums of a function z=f(x,y) over a region R depend both on the choice of the partition, which depends on Δxi Δyj, and on the choice of the points (uij,vij), i=1,2,…,n, j=1,2,…,m, in R. If

and

lim∥Δ∥→0 ∑j=1m∑i=1n f(uij,vij)Δxi Δyj=lim∥Δ∥→0 ∑j=1m∑i=1n f(uij,vij)ΔAij=I

exists and does not depend on the choice of the partition or on the choice of uij or vij, then the number I is called the double integral of z=f(x,y) over the region R and is denoted by ∬Rf(x,y) dA. DEFINITION Double Integral

Let f be a function of two variables defined on a closed rectangular region R.

Then the

denoted by ∬Rf(x,y)dA,

double integral of f over R,

is defined by

∬Rf(x,y) dA=lim∥Δ∥→0 ∑j=1m∑i=1n f(uij,vij)ΔAij

provided this limit exists. In this case, f is said to be integrable on R. Since Δxi Δyj=Δyj Δxi=ΔAij, integral of f over R are

other symbols for the double

∬Rf(x,y) dx dyand∬Rf(x,y) dy dx In Chapter 5, an important theorem states that if a function f is continuous on a closed interval [a,b], two variables.

then ∫ab f(x) dx

exists. A similar result holds for functions of

THEOREM Existence of the Double Integral

If a function z=f(x,y) the double integral ∬Rf(x,y) dA

is continuous on a closed, rectangular region R,

then

exists.

A proof can be found in most advanced calculus books. 2 Find the Value of a Double Integral Defined on a Closed Rectangular Region To find the value of a double integral of a function f of two variables whose domain is a closed rectangle without using Riemann sums, we need the concept of partial integration. Partial integration is the process of finding the single integral of a function f of two variables with respect to one of the variables, while treating the other variable as a constant. The symbol ∫ab f(x,y) dx means treat y as a constant and integrate f with respect to the variable x. The result is a function of y alone. Similarly, ∫cd f(x,y) dy means treat x as a constant and integrate f with respect to the variable y. This result is a function of x alone. IN WORDS Partial Integration is the reverse of partial differentiation.

EXAMPLE 2 Using Partial Integration

Use partial integration to find: a. ∫x=1x=2x3y2 dx b. ∫y=0y=4x3y2 dy Solution a. For ∫x=1x=2x3y2 dx, respect to x :

treat y as a constant and integrate with

∫x=1x=2x3y2 dx=y2 ∫x=1x=2x3 dx=y2 x44x=1x=2=y24-14=154 y2

b. For ∫y=0y=4x3y2 dy,

treat x as a constant and integrate with respect to y :

∫y=0y=4x3y2 dy=x3 ∫y=0y=4y2 dy=x3 y33y=0y=4=64x33

▪ NOW WORK Problem 11.

EXAMPLE 3 Using Partial Integration

Use partial integration to find:

a. ∫y=0y=4 ∫x=1x=2x3y2 dx dy b. ∫x=1x=2 ∫y=0y=4x3y2 dy dx Solution a. Begin by using partial integration to find the integral within the bracket. Then integrate the result with respect to y.

b.

▪ NOW WORK Problem 29.

Integrals of the form ∫x=ax=b ∫y=cy=d f(x,y) dy dxand∫y=cy=d ∫x=ax=b f(x,y) dx dy

are called iterated integrals. In the iterated integral on the left, we first integrate the function f partially with respect to y from c to d. The result is a function of x that we then integrate with respect to x from a to b. In the iterated integral on the right, we first integrate f partially with respect to x from a to b. The result is a function of y that we then integrate with respect to y from c to d. In writing iterated integrals, we will omit the variables x and y from the upper and lower limits of integration. Then the iterated integrals in Example 3 are written as ∫04  ∫12 x3y2 dx dy and ∫12 ∫04 x3y2 dy dx. Notice that the two iterated integrals in Example 3 are equal. This is a consequence of Fubini’s Theorem, which provides a practical way to find certain double integrals. THEOREM Fubini’s Theorem

If a function z=f(x,y) is continuous on a closed rectangular region R defined by a≤x≤b and c≤y≤d, then ∬Rf(x,y) dA=∫ab ∫cd f(x,y) dy dx=∫cd ∫ab f(x,y) dx dy

Fubini’s Theorem states that an iterated integral can be integrated in either order provided the function z=f(x,y) is continuous on the closed rectangular region R. ORIGINS Guido Fubini (1879–1943) was an Italian mathematician born in Venice. His father was a mathematics teacher who encouraged Guido to study mathematics. A short man, Fubini made many contributions in diverse areas of math, earning him the nickname “the little giant.” In 1939 the anti-Semitic

policies of Mussolini caused Fubini and his family to immigrate to the United States. He accepted a position at the Institute for Advanced Study in Princeton, NJ, and also taught at Princeton University for several years. He died in New York in 1943.

EXAMPLE 4 Finding the Value of a Double Integral Using Fubini’s Theorem

Find the value of ∬R(2x+y) dA by 1≤x≤5 and 0≤y≤4. Solution The function f(x,y)=2x+y

if R is the closed rectangular region defined is continuous at every point (x,y

the xy -plane. Then by Fubini’s Theorem, the double integral ∬R(2x+y) dA equals either of the iterated integrals

) in

(1)

∫15 ∫04(2x+y) dy dxor∫04 ∫15(2x+y) dx dy

We use the integral on the left in (1) to find the double integral.

▪ NOW WORK Example 4 using the integral on the right in (1). NOW WORK Problem 41.

3 Find the Volume Under a Surface and over a Rectangular Region The double integral ∬Rf(x,y)dA over a rectangular region R has a geometric interpretation. If the function z=f(x,y) is nonnegative on R, then f(uij,vij) represents the height of z at the point (uij,vij), and the product f(uij,vij)Δxi Δyj equals the volume of a small, rectangular cylinder of height f(uij,vij) and a base of area Δxi Δyj as shown in Figure 5. The sum of the volumes of these small cylinders, ∑j=1m∑i=1n f(uij,vij)Δxi Δyj,

is an approximation for the volume of the solid under the graph of z=f(x,y) and over R, as illustrated in Figure 6. If f is continuous on R, then lim∥Δ∥→0 ∑j=1m∑i=1n f(uij,vij)Δxi Δyj=lim∥Δ∥→0 ∑j=1m∑i=1n f(uij,vij)ΔAij exists, and ∬Rf(x,y)dA can be interpreted as the volume of the solid under the graph of z=f(x,y) over the region R.

Figure 5

Figure 6 Volume Under a Surface

Let z=f(x,y) be a function of two variables that is continuous and nonnegative on a closed, rectangular region R. The volume V under the surface z=f(x,y) and over the region R is given by

V=∬Rf(x,y) dA

NOTE It can be proved that this formula for volume is consistent with the formulas for volume given in Chapter 6.

EXAMPLE 5 Finding the Volume of a Solid

Find the volume V under the paraboloid z=f(x,y)=4-x2-y2 over the rectangular region R defined by -1≤x≤1 and 0≤y≤1. Solution Begin by graphing z=f(x,y) in Figure 7.

and the rectangular region R,

and as shown

Figure 7 z=f(x,y)=4-x2-y2 -1≤x≤1, 0≤y≤1 Since z=f(x,y) is continuous and z≥0 on the rectangular region -1≤x≤1, 0≤y≤1, the volume V under the surface z=f(x,y) and over R is given by

V=∬Rf(x,y) dA=∬R(4-x2-y2) dA Using Fubini’s Theorem, we have V=∫−11∫01(4-x2-y2) dydx=∫−114y-x2y-y3301dx=∫−114-x2-13  dx=∫−11113-x2 dx=113 xx33-11=113-13--113+13=203 cubic units

▪

NOW WORK Example 5 by finding ∫01 ∫-11(4-x2-y2) dx dy. NOW WORK Problem 61.

14.1 Assess Your Understanding Concepts and Vocabulary

1. True or False ∫02 ∫01 xy2 dy dx=∫02 x dx⋅∫01y2 dy.

2. True or False Fubini’s Theorem states that if a function z=f(x,y) continuous on a closed rectangular region R defined by a≤x≤b then ∫cd ∫ab f(x,y) dx dy=∫ab ∫cd f(x,y) dx dy.

is and c≤y≤d,

3. Multiple Choice The result of integrating ∫14 x2y dy is [(a) a number (b) a function of y (c) a function of x and y (d) a function of x ]. 4. True or False If a function z=f(x,y) region R,

is continuous on a closed, rectangular

then the double integral ∬Rf(x,y) dA

exists.

Skill Building

5. Let f(x,y)=x(3-y)

be defined on the region R shown in the figure.

a. Find the double Riemann sum of f over R by partitioning the region into nine congruent subsquares with sides Δxi=1, i=1,2,3, and Δyj=1,j=1,2,3. Choose the lower right corner of each subsquare as (uij,vij), i=1,2,3 and j=1,2,3. b. Find the double Riemann sum of f over the partition used in (a) but choose the upper left corner of each subsquare as (uij,vij), i=1,2,3 and j=1,2,3.

6. Let f(x,y)=3xy2 0≤x≤6, 0≤y≤4.

be defined on the rectangular region R defined by

a. Find the double Riemann sum of f over R by partitioning the region into six congruent subsquares, with sides Δxi=2, i=1,2,3, and Δyj=2, j=1,2. Choose the lower left corner of each subsquare as (uij,vij). b. Find the double Riemann sum of f over the partition used in (a) but choose the lower right corner of each subsquare as (uij,vij). 7. Let f(x,y)=x2+y 1≤x≤5, 2≤y≤4.

be defined on the rectangular region R defined by

a. Find the double Riemann sum of f over R by partitioning the region into four congruent subrectangles with sides Δxi=2, i=1,2, and Δyj=1, j=1,2. Choose the lower left corner of each subrectangle as (uij,vij). b. Find the double Riemann sum of f over the partition used in (a) but choose the upper right corner of each subrectangle as (uij,vij). 8. Let f(x,y)=x 1-y 0≤x≤3, 0≤y≤4.

be defined on the rectangular region R defined by

a. Find the double Riemann sum of f over R by partitioning the region into six congruent subrectangles with sides Δxi=1, i=1,2,3, and Δyj=2, j=1,2. Choose the lower left corner of each subrectangle as (uij,vij).

b. Find the double Riemann sum of f over the partition used in (a) but choose the upper left corner of each subrectangle as (uij,vij). 9. Find the double Riemann sum for Example 1 if R is divided into four congruent subsquares. Choose (uij,vij) as the center of the ijth subsquare. 10. Find the double Riemann sum for Example 1 if R is divided into eight congruent subrectangles with Δxi=1, i=1,2,3,4, and Δyj=2, j=1,2. Choose (uij,vij) as the center of the ijth subrectangle. In Problems 11–16, find the indicated partial integral. 11. ∫1e xy dy 12. ∫02 xy dx 13. ∫0π/2 x sin y dx 14. ∫0π/2  x sin y dy 15 ∫01ex dx 16. ∫01ex dy In Problems 17–36, find each iterated integral.

17. ∫01 ∫02x2y dy dx 18. ∫02 ∫04x2y dx dy 19. ∫03 ∫023xy dy dx 20. ∫02 ∫013xy dx dy

21. ∫-11 ∫013x2y2 dx dy 22. ∫01 ∫023x2y2 dy dx 23. ∫02 ∫-112x y2 dx dy 24. ∫-11 ∫122x y2 dy dx 25. ∫0π/4 ∫02 x cos y dx dy

26. ∫02 ∫0π/3 x cos y dy dx

27. ∫03 ∫0π/3(4x-3)2 sin y dy dx 28. ∫0π/3 ∫03(4x-3)2 sin y dx dy

29. ∫01 ∫0π/2ex cos y dy dx 30. ∫0π/4 ∫01ex cos y dx dy

31. ∫02 ∫-π/2π/2  sin y2x+1 dy dx 32. ∫0π/2  ∫02 sin y2x+1 dx dy 33. ∫0π/2 ∫02 xex sin y dx dy

34. ∫01 ∫0π/2 xex sin y dy dx

35. ∫12 ∫0π/2 x cos xy dx dy 36. ∫0π/3 ∫12 x cos xy dy dx In Problems 37–46, use Fubini’s Theorem to find each double integral over the rectangular region R.

37. ∬R(2x+y2) dA,0≤x≤2, 0≤y≤3

38. ∬R(x2-3y) dA,-1≤x≤2, 0≤y≤1

39. ∬Rx(x2+5) dA,0≤x≤2, -1≤y≤1

40. ∬Ry(3x2+x) dA,0≤x≤2, 0≤y≤4

41. ∬R2xey dA,-3≤x≤2, 0≤y≤1

42. ∬R e2x+y dA,0≤x≤1, -1≤y≤1

43. ∬Rx sec2y dA,0≤x≤3, 0≤y≤π4

44. ∬R y2 sec x tan x dA,0≤x≤π3, -1≤y≤3

45. ∬R x2y+3 dA,0≤x≤2, 0≤y≤1

46. ∬R y2x-3 dA,4≤x≤5, 0≤y≤3 Applications and Extensions

In Problems 47–52, use Fubini’s Theorem to find each double integral over the given rectangular region.

47. ∬Rx sin(xy) dA,0≤x≤π2, 0≤y≤1

48. ∬Ry cos(xy) dA,0≤x≤1, π2≤y≤2π

49. ∬R x3 cos(x2y) dA,0≤x≤π2, 0≤y≤1

50. ∬R x3 sin(x2y) dA,0≤x≤π2, 0≤y≤1

51. ∬R y2(1+x y2)3 dA,0≤x≤2, 0≤y≤3

52. ∬R(x3+x)(x2y+y)dA,0≤x≤1, 0≤y≤1 In Problems 53 and 54: a. Use Fubini’s Theorem to find each double integral over the given rectangular region. b. Change the order of integration and use a CAS to find the double integral, showing that both orders of integration yield the same results.

53. ∬R y5 ex y3 dA,0≤x≤1, 0≤y≤2

54. ∬R x5y ex3y2 dA,0≤x≤2, 0≤y≤1 In Problems 55–64, find the volume under the surface z=f(x,y) rectangular region. 55. f(x,y)=x+2y,0≤x≤1, 0≤y≤2 56. f(x,y)=2x+3y,0≤x≤2, 0≤y≤3

and over the given

57. f(x,y)=x2+y2,0≤x≤2, 0≤y≤1 58. f(x,y)=4x2+3y2,0≤x≤3, -2≤y≤1 59. f(x,y)=sin x,0≤x≤π2, 0≤y≤1 60. f(x,y)=cos y,0≤x≤1, 0≤y≤π2 61. f(x,y)=x2+y2xy1≤x≤2, 1≤y≤2 62. f(x,y)=y2x2,1≤x≤2, 1≤y≤2 63. f(x,y)=ex+y,0≤x≤2, 0≤y≤1 64. f(x,y)=ex-y,0≤x≤2, 1≤y≤2 65. Volume Find the volume of the solid below the paraboloid z=x2+y2 and over the square in the xy -plane enclosed by the lines x=±1 and y=±1. 66. Volume Find the volume of the solid under the elliptic paraboloid z=9-x2-3y2 and over the rectangle in the xy -plane enclosed by the coordinate axes and the lines x=2 and y=1. In Problems 67–70, use the following discussion: Suppose a function P is defined as P(x,y)=f(x,y)for x and y in a closed rectangular region R0elsewhere

Then P is a joint probability density function of the continuous random variables X and Y if f(x,y)≥0,

for all x, y

in R and ∬Rf(x,y) dA=1.

The probability Pr((X,Y)∈D)=∬Df(x,y) dA, a subset of R.

where D is

67. Probability The joint probability density function P for the random variables X and Y is given by P(x,y)=23(2x+y)on R={(x,y)|0≤x≤1,0≤y≤1}0elsewhere

a. Graph the region R. b. Verify that P is a joint probability density function by showing that P(x,y)≥0 on R,

and ∬RP(x,y) dA=1.

c. Find the probability Pr((X,Y)∈D)

where D is the region

14≤X≤1, 0≤Y≤12. 68. Probability The joint probability density function P for the random variables X and Y is given by P(x,y)=xyonR={(x,y)|0≤x≤2,0≤y≤1}0elsewhere

a. Graph the region R. b. Verify that P is a joint probability density function by showing that P(x,y)≥0 on R and ∬RP(x,y) dA=1. c. Find the probability that Pr((X,Y)∈D)

where D is the region

0≤X≤1, 12≤Y≤1. 69. Probability a. Find the number c that makes the function P a joint probability density function for the random variables X and Y. P(x,y)=c x2y0≤x≤1,0≤y≤10elsewhere b. Use the probability density function from (a) to find Pr 0≤X≤12,14≤Y≤1 70. Probability a. Find the number c that makes the function P a joint probability density function for the random variables X and Y. P(x,y)=cx(1+y2)0≤x≤1,0≤y≤30elsewhere

b. Use the probability density function from (a) to find Pr(0≤X≤1,1≤Y≤2).

71. Show that ∬RdA=∫cd ∫ab dx dy=(b-a)(d-c). That is, show that the volume of a solid with height 1, ∬RdA, defined over a rectangular region a≤x≤b, c≤y≤d is numerically equal to the area of the rectangle. Challenge Problems

72. Average Value of a Function Suppose that z=f(x,y) is integrable over a closed, rectangular region R in the xy -plane. Let P be a partition of R into nm subrectangles of equal area ΔA. Evaluate f at the center (uij,vij) of the ijth subrectangle (i=1,2,…,n and j=1,2,…,m), and let AVG be the average of these nm values. a. Show that AVG=1A∑j=1m∑i=1n f(uij,vij)Δxi Δyj, where A is the area of R. b. Explain why

lim∥Δ∥→0AVG=1A ∬Rf(x,y) dA (This is called the average value of f over R.

)

In Problems 73 and 74, find the average value of each function over the given rectangular region. 73. f(x,y)=xyx2+1,0≤x≤1, 0≤y≤2 74. f(x,y)=y cos x,0≤x≤π, 1≤y≤5 1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

14.2 The Double Integral over Nonrectangular Regions OBJECTIVES When you finish this section, you should be able to: 1 Use Fubini’s Theorem for an x

-simple region 2 Use Fubini’s Theorem for a y -simple region 3 Use properties of double integrals 4 Use double integrals to find area In Section 14.1, we defined the double integral of a function f over a closed rectangular region. Although single integrals are always integrated over an interval, for double integrals the region of integration is not always rectangular. In this section, we extend the definition of a double integral to functions defined over closed, bounded regions. RECALL A region R in the plane is closed if it contains all its boundary points. A region R can be enclosed by some circle, or equivalently, some rectangle in the plane.

is bounded if it

We begin with a function z=f(x,y) defined on a closed, bounded region R. See Figure 8(a). Then R can be enclosed by a rectangle D defined by a≤x≤b and c≤y≤d, as shown in Figure 8(b). Now partition the interval [a,b] into n subintervals of length Δxi, i=1,2,…,n, and the interval [c,d] into m subintervals of length Δyj, j=1,2,…,m, as shown in Figure 8(c). This partitions the rectangle D into mn subrectangles Dij. The norm ∥Δ∥ of this partition is defined as the largest of the lengths of the subintervals Δxi or Δyj.

Figure 8 Now we introduce a function P=P(x,y)

whose domain is the rectangular

region D.

We define P as follows:

P(x,y)=f(x,y)if(x,y) is in R0if(x,y) is in D , but not in R

In other words, P(x,y)=f(x,y) for points in D, but not in R.

for points in R,

and P(x,y)=0

Choose a point (uij,vij) in each subrectangle Dij and evaluate P(uij,vij). If Δxi Δyj is the area Aij of the subrectangle Dij, form the double Riemann sums ∑j=1m∑i=1n P(uij,vij)Δxi Δyj=∑j=1m∑i=1n P(uij,vij)ΔAij.

function z=P(x,y)

Then the double integral of the over the rectangle D is

(1)

∬DP(x,y)dA=lim∥Δ∥→0 ∑j=1m∑i=1n P(uij,vij)Δ Aij

provided the limit exists and does not depend on the partition or on the choice of (uij,vij).

If ∬DP(x,y)dA

exists, we define the double integral of f over R as

∬Rf(x,y)dA=∬DP(x,y)dA Notice that this definition makes sense because for points (uij,vij) R, P(uij,vij)=0. So such points do not contribute to ∑j=1m∑i=1n P(uij,vij)ΔAij. f(x,y)≥0 on R.

in D,

but not in

Figure 9(a) and (b) illustrate this for

Figure 9 The limit of the double sums in (1) exists under certain conditions on the function z=f(x,y) and the boundary of the region R. Before giving a theorem which guarantees the existence of a double integral, we need the following definition. Recall that a curve C is smooth if the parametric equations x=x(t), y=y(t) defining C have derivatives dxdt and dydt that are continuous and are never simultaneously zero. If a curve C is not smooth, but consists of a finite number of smooth curves that are joined end to end, then C is a piecewise-smooth curve. NEED TO REVIEW? Smooth curves are discussed in Section 11.2, p. 809.

The next theorem, proved in most advanced calculus texts, is very similar to the theorem stated in Section 14.1 guaranteeing the existence of a double integral over a closed, rectangular region. THEOREM Existence of a Double Integral

If a function z=f(x,y)

is continuous on a closed, bounded region R whose

boundary is a piecewise-smooth curve, then ∬Rf(x,y)dA

When ∬Rf(x,y)dA

exists.

exists, we say f is integrable on R.

As before, other representations for ∬Rf(x,y)dA

are

∬Rf(x,y)dxdy  and  ∬Rf(x,y)dydx If z=f(x,y)≥0

on a closed, bounded region R whose boundary is a

piecewise-smooth curve, then ∬Rf(x,y)dA equals the volume of the solid under the surface z=f(x,y) and over the region R. Refer back to Figure 9. Volume Under a Surface and over a Region R

Let z=f(x,y) be a function of two variables that is continuous and nonnegative on a closed, bounded region R whose boundary is a piecewise-smooth curve. The volume V under the surface z=f(x,y) and over the region R is given by the double integral

V=∬Rf(x,y) dA Certain types of regions have boundaries that make it possible to express a double integral as an iterated integral. One such type of region is called x -simple. DEFINITION x -Simple Region

Let R be a closed, bounded region. R is an x-simple region if the boundary of R consists of two smooth curves, y=g1(x) and y=g2(x), each defined on a closed interval [a,b], where g1(x)≤g2(x) on [a,b]. The boundary may also consist of a portion of the vertical lines x=a and x=b. Figure 10 illustrates some typical x -simple regions. Notice in each case the top and bottom boundaries of R are smooth curves that can be expressed as functions of x, as the name x -simple implies. Also notice that the boundary of an x -simple region is a piecewise-smooth curve.

Figure 10 Examples of x -simple regions Fubini’s Theorem can be extended to regions that are x -simple. 1 Use Fubini’s Theorem for an x -Simple Region THEOREM Fubini’s Theorem for an x -Simple Region

If a function z=f(x,y) is continuous on a closed, bounded region R that is x simple, then the double integral of f over R is given by

∬Rf(x,y) dA=∫ab [∫g1(x)g2(x) f(x,y) dy] dx where y=g1(x)

and y=g2(x), g1(x)≤g2(x), a≤x≤b, are two smooth curves that form the boundary of R.

The proof of Fubini’s Theorem can be found in advanced calculus books. Here we give a geometric argument for Fubini’s Theorem for x -simple regions when f(x,y)≥0 is continuous on R. Suppose we choose a number xi in the closed interval [a,b] and let A(xi) equal the area of the intersection of the plane x=xi and the solid under the surface z=f(x,y) and over R. See Figure 11. Using slicing, the volume V of the solid under z=f(x,y) and over R from x=a to x=b is V=∫ab A(x) dx Since the volume V is also given by V=∬Rf(x,y) dA,

we have (2)

V= ∬Rf(x,y) dA=∫ab A(x) dx But A(xi) y=g1(xi)

is the area of the plane region under the surface z=f(xi,y) to y=g2(xi). That is,

from

A(x)=∫g1(x)g2(x) f(x,y) dy where x is held constant. By substituting ∫g1(x)g2(x) f(x,y) dy in the volume formula (2), we find

for A(x)

Figure 11 V=∫ab A(x) dx

A(x)=∫g1(x)g2(x) f(x,y) dy NEED TO REVIEW? Finding the volume of a solid using slicing is discussed in Section 6.4, pp. 456–460.

EXAMPLE 1 Using Fubini’s Theorem for an x -Simple Region

a. Find ∬Rxy dA

if R is the region enclosed by y=x2

b. Verify that ∬Rxy dA

and y=x.

represents the volume of a solid. Describe the solid.

Solution a. We begin by graphing the region R. See Figure 12. Observe that R is a closed, bounded region. Also notice that the bottom and top boundaries of R are smooth curves expressed as functions of x : g1(x)=x2, g2(x)=x, 0≤x≤1. That is, R is x -simple. Since f(x,y)=xy is continuous on R, we can use Fubini’s Theorem. ∬Rxy dA=∫01 ∫x2x xy dy dx=∫01 x∫x2x y dy dx=∫01 x y22x2x dxIntegrate partially with respect toy.=∫01 xx-x42 dx=12 ∫01(x2-x5) dx=12 x33-x6601=112

b. Since f(x,y)=xy integral ∬Rxy dA f(x,y)=xy

is continuous and nonnegative on the region R,

the double

can be interpreted as the volume of the solid under the surface and over the region R.

The volume equals 112

cubic unit.

Figure 12 The region R is x -simple.

▪ NOW WORK Problems 19 and 25.

2 Use Fubini’s Theorem for a y -Simple Region Regions whose left and right boundaries consist of smooth curves expressed as functions of y are referred to as y -simple regions. DEFINITION y -Simple Region

Let R be a closed, bounded region. Then R is a y -simple region if the boundary of R consists of two smooth curves, x=h1(y) and x=h2(y), each defined on the closed interval c≤y≤d, where h1(y)≤h2(y) on c, d. The boundary may also consist of a portion of the horizontal lines y=c and y=d.

Figure 13 illustrates some typical y -simple regions.

Figure 13 Examples of y -simple regions Fubini’s Theorem also applies to closed, bounded regions that are y -simple. THEOREM Fubini’s Theorem for a y -Simple Region

If a function z=f(x,y) is continuous on a closed, bounded region R that is y simple, the double integral of f over R is given by

∬Rf(x,y) dA=∫cd ∫h1(y)h2(y) f(x,y) dx dy where x=h1(y)

and x=h2(y), h1(y)≤h2(y), c≤y≤d, are two smooth curves that form the boundary of R.

Figure 14 provides justification for Fubini’s Theorem for a y -simple region R when z=f(x,y)≥0 is continuous on R. The volume V of the solid under z=f(x,y) from y=c to y=d is given by V=∫cd A(y) dy

Figure 14 V=∫cd A(y) dy, A(y)=∫h1(y)h2(y) f(x,y) dx

Since this volume V is also given by V=∬Rf(x,y) dA,

we have

(3)

V=∬Rf(x,y) dA=∫cd A(y) dy But A(yi) x=h1(yi)

is the area of the plane region under the surface z=f(x,yi) to x=h2(yi). That is,

from

A(y)=∫h1(y)h2(y) f(x,y) dx where y is held constant. By substituting ∫h1(y)h2(y) f(x,y) dx in the volume formula (3), we find

EXAMPLE 2 Using Fubini’s Theorem for a y -Simple Region

for A(y)

Find ∬R3x2y dA if R is the region bounded by the smooth curves x=y and x=-y, and the line y=1. Solution Always begin by graphing the region R. See Figure 15. Observe that R is a closed, bounded region that is y -simple, where h1(y)=-y and h2(y)=y, 0≤y≤1. Using Fubini’s Theorem for a y -simple region, we have ∬R3x2y dA=∫01 ∫-yy3x2y dx dy=∫01 y ∫-yy3x2 dx dy=∫01 yx3 yy dy=∫01 y(y3/2+y3) dy=∫01(y5/2+y4) dy=y7/27/2+y5501=27+15=1735

Figure 15 The region R is y -simple.

▪ NOW WORK Problem 15.

When the region R is both x -simple and y -simple, we can choose the order of integration. In some cases, integration in one order uses simpler techniques than the techniques needed to integrate in the opposite order. EXAMPLE 3 Finding a Double Integral over a Region That Is Both x -Simple and y -Simple

Find ∬R ey2 dA,

where R is the region shown in Figure 16.

Figure 16 The region R is both x -simple and y -simple. Solution The region R is both x -simple and y -simple. If we treat R as x -simple, then the bottom boundary of R is y=2x, boundary is y=2, and x varies from 0 to 1. Then by Fubini’s Theorem,

the top

∬R ey2 dA=∫01 ∫2x2 ey2 dy dx Since the integral ∫ey2 dx

cannot be expressed in terms of elementary functions, we

cannot find ∫2x2 ey2 dy using the Fundamental Theorem of Calculus. So, we treat the region R as y -simple. Then the left boundary of R is x=0, the right boundary is x=y2,

and y varies from 0 to 2. Then by Fubini’s Theorem,

∬R ey2 dA=∫02 ∫0y/2 ey2 dx dy=∫02 ey2∫0y/2 dx dy=∫02 ey2 x 0y/2 dy=∫02 y2 ey2 dy=12  ∫02 y ey2 dy=12 ey2202=14(e4-1)

▪ NEED TO REVIEW? Elementary functions are discussed in Section 7.6, pp. 541–542. NOW WORK Problem 39.

EXAMPLE 4 Using a Double Integral to Find the Volume of a Solid

Find the volume V of the solid in the first octant enclosed by the plane x+y+z=1. Solution Figure 17(a) shows the solid in the first octant enclosed by the plane x+y+z=1. The volume V of the solid lies under the plane z=f(x,y)=1-x-y and over the region R in the xy -plane bounded by the lines x=0, y=0, and x+y=1. See Figure 17(b). The region R is x -simple and y -simple. If we treat R as x -simple, then y varies from y=0 to y=1x, 0≤x≤1. The volume V of the solid is V=∬Rf(x,y) dA=∫01 ∫01-x(1-x-y) dy dx=∫01 (1-x)y-y2201-x dx=∫01 (1-x)2-(1-x)22 dx=12  ∫01 (1-x)2 dx=12 x-x2+x3301=16cubic unit

Figure 17

▪ NOW WORK Problem 55.

EXAMPLE 5 Using a Double Integral to Find the Volume of a Solid

Find the volume V common to the two cylinders x2+y2=1

and x2+z2=1.

Solution Each cylinder has radius 1 ; their axes are perpendicular to each other and lie on the z -axis and y -axis, respectively. Figure 18(a) shows the portion of the solid lying in the first octant. If we find the volume of this portion of the solid, then since the volume in each octant is the same, the volume V is eight times the volume in the first octant. That is, if

z=f(x,y)=1-x2,

then

V=8∬Rf(x,y) dA=8∬R (1-x2)1/2 dA where R is the region in the first quadrant inside the circle x2+y2=1. Figure 18(b).

See

Figure 18 Consider the integrand. It is simpler to integrate with respect to y first. (Do you see why?) So, we treat R as an x -simple region. (If this approach fails, we would try treating R as a y -simple region.) If R is x -simple, then y varies from 0 to (1x2)1/2, 0≤x≤1.

The volume V is

V=8∬R (1-x2)1/2 dy dx=8 ∫01 ∫01-x2 (1-x2)1/2 dy dx=8 ∫01 (1-x2)1/2y01-x2 dx=8 ∫01(1-

x2) dx=8 x-x3301=163 cubic units

▪ NOW WORK Problems 53 and 59.

3 Use Properties of Double Integrals Double integrals have properties very similar to properties of single integrals. NEED TO REVIEW? Properties of single integrals are discussed in Section 5.4, pp. 387–390.

THEOREM Properties of Double Integrals

Suppose f and g are functions of two variables that are continuous on a closed, bounded region R, whose boundary is a piecewise-smooth curve, so that both ∬Rf(x,y) dA and ∬Rg(x,y) dA

exist. Then

∬R[f(x,y)±g(x,y)] dA=∬Rf(x,y) dA±∬Rg(x,y) dA.

∬Rcf(x,y) dA=c∬Rf(x,y) dA, constant.

where c is a

If R consists of two subregions, R1 and R2, that have no points in common except for points lying on portions of their common boundary, then ∬Rf(x,y) dA=∬R1 f(x,y) dA+∬R2 f(x,y) dA

EXAMPLE 6 Using Properties of Double Integrals

If R is a closed, bounded region, then a. ∬R(x2y+sin x cos y) dA=∬R x2y dA+∬Rsin x cos y dA

b. ∬R8(x2+y2) dA=8∬R(x2+y2) dA

▪ NOW WORK Problem 47.

The third property of double integrals is useful when the closed, bounded region R is neither x -simple nor y -simple. In such cases, it may be possible to partition R into subregions, each of which is x -simple or y -simple. Figure 19 illustrates how this can be done.

Figure 19 In Figure 19(a), R is neither x -simple nor y -simple. It is not x -simple because of the corner at the point P. It is not y -simple because of the corners at Q1 and Q2. However, by drawing a vertical line through P, we partition R into two nonoverlapping subregions R1 and R2, each of which is x -simple. In Figure 19(b), the region R is neither x -simple nor y -simple. However, by drawing a vertical line that passes through the corner P of the left boundary, we partition R into three nonoverlapping subregions R1, R2, and R3, each of which is x simple. EXAMPLE 7 Using Properties of Double Integrals

Find ∬R x2y dA, x+2,

where R is the region enclosed by the lines y=x, y=x+2, y= and y=-x+4.

Solution Begin by graphing the region R as shown in Figure 20(a). Observe that R is a closed, bounded region, but it is neither x -simple nor y -simple. It is not x -simple because of the corners at (1, 3) and (1, 1). It is not y -simple because of the corners at (0, 2) and (2, 2). But we can partition R into two subregions R1 and R2, both of which are x -simple, by drawing a vertical line from (1, 3) to (1, 1). See Figure 20(b). The subregion R1 is x -simple with g1(x)=-x+2 ; g2(x)=x+2, 0≤x≤1. The subregion R2 is x simple with h1(x)=x ; h2(x)=-x+4, 1≤x≤2. Then ∬R x2y dA=∬R1 x2y dA+∬R2 x2y dA=∫01 ∫-x+2x+2 x2y dy dx+∫12 ∫xx+4 x2y dy dx=∫01 x2 y22-x+2x+2 dx+∫12 x2 y22x-x+4 dx=12 ∫01 x2(x+2)2--x+22 dx+12  ∫12 x2-x+42-x2 dx=12 ∫018x3 dx+12 ∫12(-8x3+16x2) dx=4 x4401-4 x4412+8 x3312=115+563=143

Figure 20

▪ In Example 7, we could have also partitioned R into two subregions, both of which are y -simple.

NOW WORK Example 7 using y -simple subregions. NOW WORK Problem 29.

4 Use Double Integrals to Find Area Double integrals can be used to find the area of a closed, bounded region R. To find area using a double integral, let f(x,y)=1 and use the theorem for finding the volume under a surface. The volume under the surface f(x,y)=1 and over the region R is numerically equal to the area of the region R. That is, Area of the region R=∬RdA EXAMPLE 8 Using a Double Integral to Find the Area of a Region

Use a double integral to find the area A of the region R in the first quadrant enclosed by the parabola y=6x-x2 and the line y=4x-8. Solution The graph of the region R is shaded tan in Figure 21. Notice that R is neither x -simple nor y -simple. It is not x -simple because of the corner at (2, 0) ; it is not y -simple because of the corner at (4, 8). If we partition R by drawing a vertical line through the point (2, 0), we obtain two x -simple subregions R1 and R2. Then the area A of the region R is A=∬RdA=∬R1dA+∫∬R2dA=∫02 ∫06x-x2 dy dx+∫24 ∫4x-86x-x2 dy dx=∫02(6x-x2) dx+∫24(x2+2x+8) dx=3x2-x3302+-x33+x2+8x24=563 square units

Figure 21 Regions R1

and R2

are each x -simple.

▪ NOW WORK Problems 31 and 51.

14.2 Assess Your Understanding Concepts and Vocabulary

1. A closed, bounded region R that has a boundary consisting of two smooth curves, y=g1(x) and y=g2(x), defined on a≤x≤b, where g1(x)≤g2(x) on [a,b] and possibly a portion of the vertical lines x=a and x=b, is called a(n) ________ region. 2. True or False The double integral of a function that is continuous on a closed, bounded, y -simple region R can be expressed as an iterated integral. 3. True or False The region R enclosed by the parabola y=x2+1 y=3x+1 is both x -simple and y -simple. 4. Explain why ∫01 ∫0x f(x,y) dy dx≠∫01 ∫0y f(x,y) dx dy.

and the line

Skill Building

In Problems 5–8, express ∬Rx y2 dA

as an iterated integral in two ways:

a. one with the integration with respect to x first; b. the other with the integration with respect to y first; c. find each double integral over the region R. 5.

6.

7.

8.

In Problems 9–24, find each iterated integral. Identify and graph the region R associated with each integral.

9. ∫01 ∫x2x dy dx 10. ∫01 ∫y2y x dx dy 11. ∫-12 ∫y2y+2 dx dy 12. ∫01 ∫x2x y dy dx 13. ∫01 x∫x3x y dy dx

14. ∫01 y∫yy x2 dx dy

15. ∫01 ∫1ey yx dx dy

16. ∫01 ∫0x2 x ey dy dx 17. ∫02 ∫y2y xy dx dy

18. ∫24 ∫1y2 yx2 dx dy 19. ∫01 ∫x2xx y dy dx 20. ∫01 ∫x2xx dy dx 21. ∫01 ∫x1 1x2+1 dy dx 22. ∫01 ∫yy(x2+y2) dx dy 23. ∫12 ∫0ln x x ey dy dx

24. ∫23 ∫01/y ln y dx dy In Problems 25–30, find each double integral. Start by graphing the region R. Pay particular attention to the boundary, since it determines the correct limits of integration. 25. ∬R(x+y) dA, y2=8x 26. ∬R(x2-y2) dA,

where R is the region enclosed by y=x2

where R is the region enclosed by y=x

and

and

y=x2 27. ∬R y2 dA,

where R is the region enclosed by y=2-x

28. ∬Rxy dA,

where R is the region enclosed by y2=x+1

and y=x2

and y=1-x

29. ∬R x2y dA

where R is the triangular region enclosed by y=2x, y=-x, and y=2

30. ∬R x2y dA

where R is the triangular region enclosed by y=2x, y=-x, and x=1

In Problems 31–38, use double integration to find the area of each region. 31. Enclosed by the graphs of y=x3 32. Enclosed by the graphs of y=2x 33. Enclosed by the graphs of y=x2-9 34. Enclosed by the graphs of x2+y2=16 35. Enclosed by the graph y=1x-1, x=5 36. Enclosed by the graphs of y=x3/2 37. Enclosed by the line x+y=3

and y=x2 and y=x24 and y=9-x2 and y2=6x the x -axis, and the lines x=2

and

and y=x and the hyperbola xy=2

38. Enclosed by the hyperbola xy=3 and the circle x2+y2=4, in the first quadrant only In Problems 39–46, change the order of integration of each iterated integral to obtain an equivalent expression.

39. ∫01 ∫0x f(x,y) dy dx

40. ∫02 ∫y22y f(x,y) dx dy

41. ∫0a ∫0a2-y2 f(x,y) dx dy

42. ∫0223 [∫x2/4x f(x,y) dy] dx

43. ∫016 ∫y/8y1/4 f(x,y) dx dy 44. ∫25 ∫x2-6x+9x-1 f(x,y) dy dx 45. ∫12 ∫0ln y f(x,y) dx dy 46. ∫1e ∫ln x1 f(x,y) dy dx In Problems 47–50, the functions f and g are continuous on a closed, bounded region R,  ∬Rf(x,y) dA=8 and ∬Rg(x,y) dA=-6. properties of double integrals to find each double integral.

Use

47. ∬R[f(x,y)-g(x,y)]dA

48. ∬R4f(x,y) dA

49. ∬R[3f(x,y)+4 g(x,y)] dA

50. ∬R[2f(x,y)+5 g(x,y)] dA Applications and Extensions

51. Area Find the area of the region R in the first quadrant enclosed by the graphs of

y=6x-x2 and y=4x-8 by partitioning R with a horizontal line through the point (4, 8). Refer to Figure 21. 52. Area Use double integration to find the area in the first quadrant enclosed by the parabola x2=9y, the y -axis, and the circle x2+y2=10. In Problems 53 and 54, set up, but do not evaluate, an iterated double integral to find the volume under each surface over the given region. 53. Volume Find the volume of the solid under the surface z=ex2+y2 the region enclosed by the graphs of y=x+4 and y=x2-2x.

and over

54. Volume Find the volume of the solid under the surface z=ex2+ey2 enclosed by the graphs of x=y2-6y+5 and y=x+1. Volume In Problems 55–64, find the volume of each solid. 55. The tetrahedron enclosed by the plane x+2y+z=2 planes

and the coordinate

56. Enclosed by the elliptic paraboloid 4x2+9y2=36 and z=1

and the planes z=0

57. Below the paraboloid z=x2+y2 and above the triangular region R formed by the x - and y -axes and the line x+y=1, as shown in the figure below.

58. Enclosed by the cylinder z=9-y2 z=0 59. Enclosed by the surface z=4-x2-y2

and the planes x=0, x=4, and z=0

and

60. Enclosed by the surfaces x2+y2=1, z=|x|,

and z=1

61. The solid bounded from above by the graph of z=x2+y2, xy -plane, and on the sides by the cylinder x2+y2=1, figure.

62. Enclosed by a portion of the parabolic cylinder z=x22 y=0, y=x, x=2, and z=0 63. Enclosed by the coordinate planes, the plane y=3, 64. Enclosed by the surfaces z=x ey,

from below by the as shown in the

and the four planes and the surface z=y+1-x2

z=0, y=0, x=1,

and y=x2

In Problems 65–72, (a) graph the region associated with each iterated integral, (b) reverse the order of integration, and (c) find the new iterated integral.

65. ∫0π/2 ∫yπ/2 sin x2 dx dy 66. ∫01/2 ∫2x1 ey2 dy dx

67. ∫01 ∫01-x2 11-y2 dy dx

68. ∫01 ∫y1 sin xx dx dy 69. ∫01 ∫y12+x2 dx dy

70. ∫01 [∫x311+y4 dy] dx

71. ∫01 ∫y1 ey/x dx dy

72. ∫01 ∫sin-1yπ/2 ecos x dx dy 73. Properties of Double Integrals Let R be the region enclosed by y=1, y=-1, x=0, and x=1 ; let R1 and R2 be the subregions of R in the first and fourth quadrants, respectively. Suppose f is continuous on R and ∬R3f(x,y) dA-2∬R1f(x,y) dA=∬R2f(x,y) dA,∬R25f(x,y) dA-2∬R1 f(x,y) dA=18

Find ∬Rf(x,y) dA. 74. Properties of Double Integrals Let R be the region enclosed by y=0, y=2, x=-2, and x=1 ; let R1 and R2 be the subregions of R in the first and second quadrants, respectively. Suppose f is continuous on R and ∬R23 f(x,y) dA−∬R1f(x,y) dA=2∬Rf(x,y) dA,∬R14 f(x,y) dA+∬R22 f(x,y) dA=7

Find ∬Rf(x,y) dA.

75. Properties of Double Integrals Let R be the region enclosed by y=x, y=-1, and x=2 ; let R1 and R2 be the subregions of R above the x -axis and below the x -axis, respectively. Suppose f is continuous on R and 2∬R2f(x,y) dA−7∬R1f(x,y) dA=17∬R2f(x,y) dA−2∬R1f(x,y) dA=7

Find ∬Rf(x,y) dA. 76. Properties of Double Integrals Let R be the region enclosed by y=x2+1, y=0, x=-1, and x=2 ; let R1 and R2 be the subregions of R in the first and second quadrants, respectively. Suppose f is continuous on R and ∬R26 f(x,y) dA−3∬R1f(x,y) dA=−12∬R24 f(x,y) dA+2=∬R1f(x,y) dA

Find ∬Rf(x,y) dA. In Problems 77–80, each of the iterated integrals represents the volume of a solid. Describe the solid.

77. ∫01 ∫01-x21-x2-y2 dy dx

78. ∫02 ∫04-y2(4-x2-y2) dx dy 79. ∫04 ∫013 dy dx 80. ∫01 ∫022 dy dx

81. Find: a. ∬Rx dA, where R is the region enclosed by the circle of radius 1 centered at the origin. b. 4∬R1 x dA, where R1 is the region in the first quadrant enclosed by the circle of radius 1 centered at the origin. (This shows that although the region over which we are integrating is symmetric about the x -axis and the y -axis, we must also consider properties of the integrand on this region before we use symmetry.) 82.

a. Find

∫1e 1x ∫0ln x dy dx b. Reverse the order of integration and find the resulting iterated integral.

83.

a. Find

∫0π/2 sin x∫0cos x dy dx b. Reverse the order of integration and find the resulting iterated integral.

84. Volume of a Liquid A tank in the shape of a half-cylinder is lying on its flat side. The radius of the tank is 1 m and its length is 4 m. If the tank is filled with liquid to a depth of a meters, what is the volume of the liquid? Hint: Position the half cylinder as shown in the figure below.

85. Volume of a Liquid Repeat Problem 84 if the tank is buried so its rounded base is 1 m under the ground and its flat side is at ground level. See the figure below.

86. Damming a Valley A dam is built in a V-shaped valley. The top of the dam is 200 m wide and the lowest point in the V is 150 m below the top, as shown in the figure. The pressure P due to the water at a depth d meters below the surface is P=ρgd, where ρ=1000 kg⁄m3 and g=9.8 m/s2. What is the total outward force F from the water that this dam must be able to withstand? Express your answer in both newtons and pounds using the fact that 1 N≈0.2248 lb. F=PA.)

(Recall that pressure is P=FA,

so the force

In Problems 87–90, refer to the discussion of joint probability density functions on page 959. 87. Probability The joint probability density function P for the random variables X and Y is given by P(x,y)=23(2x+y)0≤x≤1,0≤y≤10elsewhere

a. Find Pr(X+Y≤1). b. Find Pr(Y≤X). 88. Probability The joint probability density function P for the random variables X and Y is given by P(x,y)=xy0≤x≤2,0≤y≤10elsewhere a. Find Pr(X≤Y). b. Find Pr12≤X≤Y. 89. Probability a. Find the number c that makes the function P a joint probability density function for the random variables X and Y. P(x,y)=cxy0≤x≤1,0≤y≤x0elsewhere

b. Use the probability density function from (a) to find Pr12≤Y≤X. 90. Probability a. Find the number c that makes the function P a joint probability density function for the random variables X and Y. P(x,y)=cx(1+y)0≤x≤y,0≤y≤10elsewhere

b. Use the probability density function from (a) to find Pr19≤X≤Y. 91. Explain if the following equality is true or false without finding the exact value of the two integrals: ∫-33 ∫-9-y29-y2(7-x) dx dy=4 ∫03 ∫09-y2(7-x) dx dy

92. Volume Find the volume of the tetrahedron enclosed by the coordinate planes and the plane xa+yb+zc=1. 93. Find ∬Rxy dA,

(Assume that a,b,

and c are positive.)

where R is the region enclosed by y=4-x2

and

y=x2

to the right of the y -axis.

94. Find ∬Rxy dA, where R is the region enclosed by x=4-y2 x=y2 above the x -axis. 95. Find ∬R ex dA,

and

where R is the region enclosed by y=x, x=1, y=0.

96. a. Graph the region R defined by 0≤a≤y≤b, 0≤a≤x≤y. b. If z=f(x)

is continuous on R,

show that ∬Rf(x) dA=∫ab(b-x)f(x) dx

97. Properties of Double Integrals Using properties of double integrals, show that if the functions f and g are integrable on R and f(x,y)≤g(x,y) for all (x,y)

in R,

then ∬Rf(x,y) dA≤∬Rg(x,y) dA

Challenge Problem

98. Volume Find the volume of the solid under the graph of the elliptic paraboloid z=8-2x2y2 and above the first quadrant of the xy -plane. Hint: Use Wallis’s formula from Section 7.1, Problem 85. 1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

14.3 Double Integrals Using Polar Coordinates OBJECTIVES When you finish this section, you should be able to: 1 Find a double integral using polar coordinates 2 Find area and volume using polar coordinates

Suppose the function z=f(x,y) is continuous on a closed, bounded region R whose boundary is a piecewise-smooth curve. Then the double integral ∬Rf(x,y) dA

A of R.

exists, and dA=dx dy=dy dx is the differential of the area If the boundary of R consists of rays (lines through the origin) and parts of

circles, it is often easier to find ∬Rf(x,y) dA by converting to polar coordinates. We convert a function z=f(x,y) to polar coordinates (r,θ) using the equations x=r cos θ, y=r sin θ. It remains to find the differential dA in polar coordinates. NEED TO REVIEW? Polar coordinates are discussed in Section 9.4, pp. 701–708.

Suppose the closed, bounded region R is enclosed by the rays θ=α and θ=β, where 0≤α0, and on the right by the circle r=2a cos θ if its mass density ρ is inversely proportional to the distance from the y -axis. 30. Mass and Center of Mass Find the mass and center of mass of a lamina in the shape of the smaller region cut from the circle r=6 by the line r cos θ=3 if its mass density is ρ=ρ(r,θ)=cos2 θ. 31. Mass Find the mass of the lamina enclosed by y=x2

and y=x3;

the mass

density is ρ=ρ(x,y)=xy. 32. Center of Mass Find the center of mass of the lamina inside r=4 cos θ and outside r=23 if the mass density ρ=ρ(r,θ) is inversely proportional to the distance from the origin. 33. Center of Mass Find the mass and the center of mass of the lamina enclosed by x=y-2 and x=-y2, if the mass density is ρ=ρ(x,y)=x2. 34. Center of Mass Find the center of mass of the lamina enclosed by y=x2 2y+1=0, if the mass density is ρ=ρ(x,y)=2x+8y+2.

and x-

35. Center of Mass Find the center of mass of the homogeneous lamina enclosed by bx2=a2y and ay=bx, a>0,b>0. 36. Center of Mass Find the center of mass of the homogeneous lamina enclosed by y=ln x, the x -axis, and the line x=e2. 37. Moment of Inertia A homogeneous lamina is in the shape of a square with sides s. Find the moment of inertia about a. a side b. a diagonal c. the centroid. 38. Moment of Inertia A homogeneous lamina is in the shape of an equilateral triangle with each side measuring s. Find the moment of inertia about a. a side b. an altitude.

39. Moment of Inertia The mass density at each point of a circular washer of inner radius a and outer radius b is inversely proportional to the square of the distance from the center. a. Find the mass of the washer. Then discuss its behavior as a→0+. (Note that the mass density becomes unbounded as we approach the center of the washer.) b. Find the moment of inertia of the washer about its center, and show that (unlike the mass) it remains finite as a→0+. 40. Moment of Inertia a. Graph the limaçon r=3+2 cos θ

and the circle r=2.

b. Set up the integral for the moment of inertia about the x -axis of the lamina inside the limaçon and outside the circle from (a) if the mass density ρ of the lamina is inversely proportional to the square of the distance from the origin. c. Find the integral in (b). 41. Show that the center of mass of a rectangular homogeneous lamina lies at the intersection of its diagonals. 42. A homogeneous lamina is in the shape of the region enclosed by a right triangle of base b and height h. Show that its moment of inertia about the base is 16 m h2, where m is the mass of the lamina. Hint: Position the triangle so that the base lies on the positive x -axis from (0, 0) to (b, 0). Challenge Problems

43. Show that the center of mass of the region enclosed by a triangular homogeneous lamina lies at the point of intersection of its medians. Hint: Position the triangle so that its vertices are at (a, 0), (b, 0), and (0, c), where a0, and c>0 44. Suppose that a homogeneous lamina occupies a region R in the xy -plane. Show that the average value of f(x,y)=x over R is x, the first coordinate of the center of mass of the lamina. What is the average value of g(x,y)=y over R ? Hint: The average value of f over a region R is defined to be the number 1A  ∬Rf(x,y) dA. 45. a. Show that the moment of inertia about the z -axis of the thin flat plate in the figure

equals the sum of its moments of inertia about the x - and y -axes.

b. Given that the moment of inertia of a homogeneous disk about an axis through its center and perpendicular to its plane is m R22 (where m is mass and R is the radius), use (a) to find its moment of inertia about a diameter. c. What is the moment of inertia of a disk about an axis tangent to its edge? 1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

14.5 Surface Area OBJECTIVE When you finish this section, you should be able to: 1 Find the area of a surface that lies above a region R

So far we have used a double integral to find the volume under a surface z=f(x,y)≥0 and above a closed, bounded region R of the xy -plane. In this section, we use a double integral to find the surface area of z=f(x,y)≥0 above a closed, bounded region R whose boundary is a piecewise-smooth curve. In developing the formula for surface area, we use tangent planes to approximate surface area. 1 Find the Area of a Surface That Lies Above a Region R We begin with the simple case where the surface is a plane z=ax+by+c and the region R is a rectangle with sides of lengths Δx and Δy. The surface area Sp is the area of the part of the plane that lies above the rectangle, in this case, a parallelogram, as shown in Figure 35. Notice that this parallelogram is the intersection of the plane and the cylinder formed by projecting the perimeter of the rectangle with sides Δx and Δy vertically upward.

Figure 35 The adjacent sides of the parallelogram are given by the vectors u and v. Both u

and v have the same initial point (x0, y0, ax0+by0+c) on the parallelogram. The terminal point of u is (x0+Δx, y0, a(x0+Δx)+by0+c) and the terminal point of v is (x0, y0+Δy, ax0+b(y0+Δy)+c). So, u and v are given by u=Δxi+aΔxk,v=Δyj+bΔyk The area of the parallelogram is ∥u×v∥=Δx(i+ak)×Δy(j+bk)=(i+ak)×(j+bk) Δx Δy=-ai-bj+k Δx Δy=a2+b2+1Δx Δy

NEED TO REVIEW? Vectors are discussed in Section 10.2, pp. 738–741, and properties of the cross product are discussed in Section 10.5, pp. 764–769.

Let R be a closed rectangular region in the xy -plane with sides of length Δx and Δy. If the perimeter of R is projected upward parallel to the z -axis, the result is a cylinder. Let Sp denote the area cut from the plane z=ax+by+c by this (1) cylinder. Then the area Sp is given by Sp=a2+b2+1Δx Δy We now develop a formula for the surface area of the graph of a nonnegative function z=f(x,y) that is defined over a closed, rectangular region R. We assume that f has first-order partial derivatives that are continuous at every point of the region R. See Figure 36(a). Using lines parallel to the x - and y -axes, we partition R into subrectangles and consider the subrectangle Rij,i=1,2,…,n, j=1,2,…,m having area Aij=Δxi Δyj. Now select a point (uij,vij) in Rij and construct the tangent plane to the surface at the corresponding point Pij=(uij,vij,f(uij,vij)) on the surface. See Figure 36(b). The equation of this tangent plane can be written as (2) z=fx(uij,vij)(x-uij)+fy(uij,vij)(y-vij)+f(uij,vij)

Figure 36 The sides of the rectangle Rij, when they are projected upward parallel to the z axis, result in a cylinder. Let ΔSij denote the area cut from the tangent plane at the point Pij by the cylinder. Then ΔSij is an approximation to the surface area of the part of the surface that lies above the rectangle Rij. By adding all the areas ΔSij and taking the limit of the sums as the norm ∥Δ∥ of the partition approaches 0, we obtain the (3) surface area S of the part of the surface z=f(x,y) that lies above the region R.

S=lim∥Δ∥→0 ∑j=1m∑i=1n ΔSij To obtain a formula for ΔSij, the surface at the point Pij,

we combine (2), the equation of the tangent plane to

z=fx(uij,vij)(x-uij)+fy(uij,vij)(y-vij)+f(uij,vij)Equation (2)

and formula (1) for the surface area Sp Sp=a2+b2+1 Δx ΔyFormula (1) The result is ΔSij=[fx(uij,vij)]2+[fy(uij,vij)]2+1 ΔAij

of a plane z=ax+by+c

Now substitute this expression for ΔSij into (3) to obtain the surface area S of the part of the surface that lies above the region R. Then S=lim∥Δ∥→0 ∑j=1m∑i=1n[fx(uij,vij)]2+[fy(uij,vij)]2+1 Δxi Δyj

Since the first-order partial derivatives of f are continuous on R, is a double integral.

the limit exists and S

S=∬Rfx(x,y)2+fy(x,y)2+1 dA It can be shown that the surface area S of the graph of a nonnegative function z=f(x,y) over a closed, bounded region R whose boundary is a piecewise-smooth curve can be found using a double integral provided the partial derivatives fx and fy are continuous on R. Surface Area Above a Region R Let z=f(x,y) be a function of two variables defined on a closed, bounded region R whose boundary is a piecewise-smooth curve. If fx and fy are continuous on R, then the surface area S of the part of the surface that lies above R is given by

S=∬R[fx(x,y)]2+[fy(x,y)]2+1 dA

EXAMPLE 1 Finding Surface Area

Find the surface area of the part of the surface z=f(x,y)=23(x3/2+y3/2)

x=0, x=1, y=0,

that lies above the region R, and y=2.

a rectangle enclosed by the lines

Solution Figure 37 shows the surface z=f(x,y)=23(x3/2+y3/2) and the surface area S above the rectangle R. We begin by finding the partial derivatives of z=f(x,y) : fx(x,y)=x1/2fy(x,y)=y1/2

Since fx and fy are continuous on R,

the surface area S above R is

S=∬R[fx(x,y)]2+[fy(x,y)]2+1 dA=∬R(x1/2)2+(y1/2)2+1 dA=∫01  ∫02x+y+1 dy dx=∫0123 (x+y+1)3/202 dx=23 ∫01(x+3)3/2-(x+1)3/2 dx=23⋅25⋅(x+3)5/2(x+1)5/2 01=415(32-93-42+1)=415(33-93-42)

Figure 37 z=23(x3/2+y3/2)

▪ NOW WORK Problem 5.

EXAMPLE 2 Finding Surface Area

Find the surface area of the part of the paraboloid z=f(x,y)=1-x2-y2 that lies above the xy -plane. Solution Figure 38 shows the part of the surface z=1-x2-y2 that lies above the xy -plane and its projection onto the xy -plane, the region R enclosed by the circle x2+y2=1. We begin by finding the partial derivatives of z=f(x,y). fx(x,y)=-2xfy(x,y)=-2y Since fx and fy are continuous on R, paraboloid z=f(x,y)=1-x2-y2

the surface area S of the part of the that lies above R is

S=∬R[fx(x,y)]2+[fy(x,y)]2+1 dA=∬R(-2x)2+(-2y)2+1 dA=∬R4(x2+y2)+1 dA

Since both the region R (a circle) and the integrand involve x2+y2, we use polar coordinates (r,θ). For the region R, we have 0≤r≤1 and 0≤θ≤2π. Then S=∬R4(x2+y2)+1 dA=∫02π  ∫014r2+1r dr dθx2+y2=r2d A=r dr dθ=∫02π 112(4r2+1)3/2 01 dθLet u=4r2+1, then du=8r dr.=1 1)∫02π dθ=112(55-1)2π=π6(55-1)

Figure 38 z=1-x2-y2,z≥0

▪ NOW WORK Problem 9.

EXAMPLE 3 Finding Surface Area

Find the surface area of the part of the sphere x2+y2+z2=a2 the xy -plane and is contained within the cylinder x2+y2=ax, a>0.

that lies above

Solution Figure 39 shows the sphere and the cylinder. The equation of the surface in explicit form is z=f(x,y)=a2-x2-y2 The partial derivatives of z =f(x,y)

are

fx(x,y)=-xa2-x2-y2fy(x,y)=-ya2-x2-y2

The projection of the cylinder onto the xy -plane is the region R enclosed by the circle x2+y2=ax, a>0. Then the surface area S is S=∬Rx2a2-x2-y2+y2a2-x2-y2+1 dA=∬Ra2a2-x2-y2 dA

Since the integrand involves the expression x2+y2 and the boundary of R is the circle x2+y2=ax, we use polar coordinates (r,θ). We begin by finding the limits of integration. Since

x2+y2=axr2=ar cos θr=a cos θ we find that r varies from 0 to a cos θ.

Figure 39 x2+y2+z2=a2, z≥0 Since half the area is to the right of the xz

-plane and half is to the left of the plane, we

can let θ vary from 0 to π2 and double the area. Then the surface area S is

▪ NOW WORK Problem 13.

14.5 Assess Your Understanding Concepts and Vocabulary

1. True or False Suppose R is a closed, rectangular region in the xy -plane with sides of length Δx and Δy. If the perimeter of R is projected upward parallel to the z -axis, the result is a cylinder. If Sp is the area cut from the plane z=ax+by+c by this cylinder, then Sp=a2+b2+1 Δx Δy. 2. True or False Suppose z=f(x,y) is a function of two variables defined on a closed, bounded region R. If fx and fy are continuous on R, then the area S of the part of the surface that lies over R is given by S=∬R fx(x,y)+fy(x,y)+1dA.

Skill Building

3. Find the surface area of the part of the plane 2x+2y+z=6 the region in the xy -plane bounded by x2+y2=4, -axis, as shown in the figure.

4. Find the surface area of the part of the plane 4z-2x-y=8 the region enclosed by the triangle with vertices (0,0,0), (1,0,0), (1,1,0), as shown in the figure.

that lies above the x -axis, and the y

that lies over and

In Problems 5–16, find the surface area described. 5. The part of the surface z=f(x,y)=9-x2 rectangle enclosed by the lines x=0, x=2, y=-1, y=4

that lies above the

6. The part of the surface z=f(x,y)=8-y2 rectangle enclosed by the lines x=0, x=5, y=0, y=2

that lies above the

7. The part of the surface z=23(x3/2+y3/2) enclosed by the lines x=0, y=0,

that lies above the triangle and 2x+3y=6

8. The part of the surface z=23(x3/2+y3/2) enclosed by x=0, y=0, and 3x+y=3 9. The part of the paraboloid z=4-x2-y2 10. The part of the cylinder z=a2-x2 -12 a≤x≤12 a

that lies above the xy

-plane

that lies above the square defined by

and -12 a≤y≤12 a, a>0

11. The part of the cone z=x2+y2

that lies inside the cylinder x2+y2=2x

12. The part of the sphere x2+y2+z2=4z x2+y2=2z 13. The part of the surface z=xy x2+y2=a2, a>0

that lies above the triangle

that lies within the paraboloid

in the first octant that lies within the cylinder

14. The part of the surface z=x2-y2 cylinder x2+y2=4

in the first octant that lies within the

15. The part of the sphere x2+y2+z2=4z and z=3

that lies between the planes z=1

16. The part of the sphere x2+y2+z2=4a2 x2+y2=2ax, a>0

that lies inside the cylinder

Applications and Extensions

17. Surface Area Find the surface area cut from the hyperbolic paraboloid y2-x2=6z by the cylinder x2+y2=36. 18. Surface Area Find the surface area of the part of z=4-y2 octant and is enclosed by z=0, x=0, x=y, 19. Surface Area Find the surface area of the part of x2=y under the plane x+z=3. Hint: Project the surface onto the xz

that lies in the first and y=2. that lies in the first octant

-plane.

20. Surface Area Find the surface area of the part of the paraboloid z=9-x2-y2 that lies between the planes z=0 and z=8. 21. Surface Area Find the surface area of the part of the hemisphere x2+y2+z2=16, z≥0, that lies inside the cylinder bounded by x2+y2=4. See the figure.

22. Surface Area of the Pantheon The Pantheon in Rome is the largest unreinforced concrete dome in the world. Its inner surface is a hemisphere of diameter 43.3 m

with an open circular ocular 9.1 m surface area of the dome.

in diameter cut from its apex. Find the inner

23. Surface Area a. Graph the surface z=10e-(x2+y2)

that lies inside the cylinder x2+y2=4.

b. Find the surface area of the surface graphed in (a).

24. Surface Area a. Set up, but do not evaluate, the integral to find the surface area of the part of the surface z=1-x4-y2 that lies above the xy

-plane.

b. Graph the surface. 25. Derive the formula S=πa a2+h2 circular cone of base radius a and altitude h.

for the lateral surface area of a right

26. Show that the area of the first-octant portion of the plane xa+yb+zc=1 (where a,b, and c are positive) is S=12a2 b2+b2 c2+c2 a2.

27. Use a double integral to derive the formula for the surface area of a sphere of radius R. Challenge Problems

28. Let F denote a function of three variables that possesses continuous first-order partial derivatives at each point of its domain. Suppose also that Fz(x,y,z) is never 0. If S is the area of the part of the surface F(x,y,z)=0 that lies over the closed, bounded region R, show that S=∬R [Fx(x,y,z)]2+[Fy(x,y,z)]2+[Fz(x,y,z)]2|Fz(x,y,z)| dA

29. Surface Area The center of a sphere of radius R is on the surface of a right circular cylinder of base radius R2. See the figure.

Find the surface area of the sphere inside the cylinder.

30. For the plane surface F(x,y,z)=Ax+By+Cz-D=0, C≠0, if z=f(x,y),

show that the

surface area S=∬R[fx(x,y)]2+[fy(x,y)]2+1 dA can be written as

S=∬Rsec γ dA where γ is the positive acute angle between the normal n=Ai+Bj+Ck to the plane and k.

14.6 The Triple Integral OBJECTIVES When you finish this section, you should be able to: 1 Find a triple integral of a function defined in a closed box 2 Find a triple integral of a function defined in a more general solid 3 Find the volume of a solid 4 Find the mass, center of mass, and moments of inertia of a solid 5 Find a triple integral of a function defined in an xz

-simple or a yz -

simple solid Just as single integrals are used to integrate over closed intervals [a,b], and double integrals are used to integrate over two-dimensional, closed, bounded regions R, triple integrals are used to integrate over three-dimensional, closed, bounded solids E. For example, we have seen that if ρ=ρ(x) is the mass density of a long, thin wire, then the single integral ∫ab ρ(x) dx models the mass of the wire from a to b. Also if ρ=ρ(x,y) is the mass density of a lamina, then the double integral ∬Rρ(x,y) dA models the mass M of the lamina over R. Similarly, a triple integral can be used to find the mass of a solid with mass density ρ=ρ(x,y,z). We begin with a function f of three variables defined in a box-shaped region E of three-dimensional space. Partition E into smaller rectangular boxes Eijk, i=1,2,…,n, j=1,2, …,m, and k=1,2,…,l, by drawing planes parallel to the three coordinate planes. See Figure 40.

Figure 40 The norm of the partition, denoted by ∥Δ∥, is defined as the largest of the lengths of the sides of the boxes Eijk. If Δxi, Δyj, and Δzk denote the length, width, and height, respectively, of the ijk th box, then its volume is ΔVijk=Δxi Δyj Δzk. In each box Eijk we arbitrarily select a point (uijk,vijk,wijk), evaluate f(uijk,vijk,wijk), multiply f(uijk,vijk,wijk) by Δxi Δyj Δzk, and form the sums

∑k=1l∑j=1m∑i=1n f(uijk,vijk,wijk)Δxi Δyj Δzk Sums of this form are referred to as triple Riemann sums of f for the partition. If lim∥Δ∥→0 ∑k=1l∑j=1m∑i=1n f(uijk,vijk,wijk)Δxi Δyj Δzk=lim∥Δ∥→0  ∑k=1l∑j=1m∑i=1n f(uijk,vijk,wijk)ΔVijk

exists and does not depend on the choice of the partition or on the choice of (uijk,vijk,wijk), then this limit is called the triple integral of f over E. DEFINITION Triple Integral

Let f be a function of three variables defined in a closed, bounded, box-shaped solid E. The triple integral of f over E is defined as ∭Ef(x,y,z) dV=lim∥Δ∥→0 ∑k=1l∑j=1m∑i=1n f(uijk,vijk,wijk)ΔVijk

provided the limit exists. As with single integrals and double integrals, it can be shown that if f is continuous in E,

then ∭Ef(x,y,z) dV exists. There are six different orders of integration for the triple integral of f in E.

They are

∭Ef(x,y,z) dx dy dz∭Ef(x,y,z) dx dz dy∭Ef(x,y,z) dy dx dz∭Ef(x,y,z) dy dz dx∭ Ef(x,y,z) dz dx dy∭Ef(x,y,z) dz dy dx

Fubini’s Theorem for triple integrals defined in a closed box-shaped solid E,

uses iterated

integrals to find ∭Ef(x,y,z)dV. THEOREM Fubini’s Theorem for Triple Integrals

Let f be a function of three variables defined in the closed box E, x1≤x≤x2, y1≤y≤y2, and z1≤z≤z2. continuous in E, then

where If f is

∭Ef(x,y,z) dV=∫x1x2 ∫y1y2 ∫z1z2 f(x,y,z) dz dy dx

The iterated integral on the right is shown first with respect to z, then y, and finally x. To find ∫z1z2 f(x,y,z) dz, we treat x and y as constants and integrate partially with respect to z. As is the case with iterated double integrals, the braces and brackets are usually omitted.

There are actually six different forms of Fubini’s Theorem for Triple Integrals. Here we state the theorem in the order dz dy dx. But the same result is obtained no matter what order is used. 1 Find a Triple Integral of a Function Defined in a Closed Box EXAMPLE 1 Finding a Triple Integral of a Function Defined in a Closed Box

a. If E is the closed box 0≤x≤1, 0≤y≤2, shown in Figure 41, express ∭E4x yz dV integration.

and 0≤z≤3 using six different orders of

b. Find ∭E4x yz dV.

Figure 41 The closed box E Solution a. Since f(x,y,z)=4xyz is continuous in E, we can find the triple integral using Fubini’s Theorem for Triple Integrals. If we use Fubini’s Theorem in the order dx dy dz, then the limits of integration from left to right are 0≤z≤3, 0≤y≤2, 0≤x≤1, and the iterated integral is ∭E4xyz dV=∫03 ∫02 ∫014xyz dx dy dz=∫03 ∫02 ∫014x dx y dy z dz

Notice how the integrals are nested: ∫014x dx

is on the inside, ∫02…y dy

is in the middle, and ∫03…z dz is on the outside. Using Fubini’s Theorem in the order dy dx dz, the limits of integration from left to right are 0≤z≤3, 0≤x≤1, 0≤y≤2 and the iterated integral is ∭E4x yz dV=∫03 ∫01∫024x yz dy dx dz=∫03 ∫01 ∫024y dy x dx z dz

Following the pattern, in the order dz dy dx,

we obtain

∭E4x yz dV=∫01∫02 ∫034x yz dz dy dx=∫01 ∫02 ∫034z dz y dy x dx

In the order dz dx dy ∭E4x yz dV=∫02 ∫01∫034x yz dz dx dy=∫02 ∫01 ∫034z dz x dx y dy

In the order dx dz dy ∭E4x yz dV=∫02 ∫03 ∫014x yz dx dz dy=∫02 ∫03 ∫014x dx z dz y dy

In the order dy dz dx ∭E4x yz dV=∫01∫03 ∫024x yz dy dz dx=∫01 ∫03 ∫024y dy z dz x dx

b. Using Fubini’s Theorem in the order dx dy dz, ∭E4x yz dV=∫03 ∫02  ∫014x yz dx dy dz=∫03 ∫02 ∫014x dx y dy z dz=∫03  ∫02 2x201 yz dy dz=∫03 ∫022yz dy dz=∫03 y202 z dz=∫034z dz=2z203=18

▪ NOW WORK Problem 9.

2 Find a Triple Integral of a Function Defined in a More General Solid Consider a solid E in space enclosed on the top by a surface z=z2(x,y) and on the bottom by a surface z=z1(x,y), where z1≤z2. The sides of E are a cylinder whose intersection with the xy -plane forms a closed, bounded region R whose boundary is a piecewise-smooth curve. Also the functions z1 and z2 are continuous on R. See Figure 42. Since the projection of E onto the xy -plane results in a closed, bounded region R, and E lies between the two surfaces z1 and z2, the solid E is called xy -simple. The next theorem, which we do not prove, provides a way to find triple integrals defined over solids that are xy -simple.

Figure 42 E is an xy

-simple solid.

THEOREM Triple Integral over an xy

-Simple Solid

Let E be an xy -simple solid that lies between the two surfaces z=z1(x,y) and z=z2(x,y), where z1≤z2, and whose projection onto the xy plane is a closed, bounded region R whose boundary is a piecewise-smooth curve. If f is a function of three variables that is continuous in E, then ∭Ef(x,y,z) dV=∬R∫z1(x,y)z2(x,y) f(x,y,z) dz dA

If the region R is x -simple, and R is enclosed by y=y1(x) where y1≤y2 and a≤x≤b, then ∭Ef(x,y,z) dV= ∬R ∫z1(x,y)z2(x,y) f(x,y,z) dz dA=∫ab ∫y1(x)y2(x)  ∫z1(x,y)z2(x,y) f(x,y,z) dz dy dx

and y=y2(x),

If the region R is y -simple, and R is enclosed by x=x1(y) where x1≤x2 and c≤y≤d, then

and x=x2(y),

∭Ef(x,y,z) dV=∬R∫z1(x,y)z2(x,y) f(x,y,z) dz dA=∫cd ∫x1(y)x2(y)  ∫z1(x,y)z2(x,y) f(x,y,z) dz dx dy

EXAMPLE 2 Finding a Triple Integral of a Function Defined in an xy

Find the triple integral ∭E4x dV planes and the plane 2x+3y+4z=12.

-Simple Solid

over the tetrahedron formed by the coordinate

Solution The solid E is shown in Figure 43(a). A typical plane section of E using a plane perpendicular to the x -axis reveals that the upper surface is z=z2(x,y)=14(12-2x-3y) and the lower surface is z=z1(x,y)=0. The solid E lies between the two surfaces z1 and z2 and the projection of E onto the xy -plane is a closed, bounded region R on which z1 and z2 are continuous. So E is xy -simple. The region R in the xy -plane is the triangle enclosed by the x -axis, the y -axis, and the line y=13(12-2x)=4-2x3, as shown in Figure 43(b). The triangle R is both x -simple and y -simple. If we treat R as an x -simple region, so that R is bounded by y1(x)=0, y2(x)=4-2x3, we have

and 0≤x≤6,

∭E4x dV=∫ab ∫y1(x)y2(x) ∫z1(x,y)z2(x,y)4x dz dy dx=∫06 ∫04-2x/3 ∫0(1/4)(12-2x3y)4x dz dy dx=∫06 ∫04-2x/34x z 0(1/4)(12-2x-3y) dy dx=∫06 ∫04-2x/34x 12-2x3y4 dy dx=∫06 ∫04-2x/3(12x-2x2-3xy) dy dx=∫06 (12x-2x2)y-3x y2204-2x/3 dx=∫06 24x8x2+23 x3 dx=12x2-83 x3+16 x406=72

Figure 43

▪ In Example 2 we used Fubini’s Theorem in the order dz dy dx.

If we treat R as a y

-simple region, then R is bounded by x1(y)=0, x2(y)=6-32 y, and 0≤y≤4. Then Fubini’s Theorem in the order dz dx dy

results in

∭E4x dV=∫cd ∫x1(y)x2(y) ∫z1(x,y)z2(x,y)4x dz dx dy=∫04 ∫06-3y/2 ∫0(1/4)(12-2x3y)4x dz dx dy

The result is the same, as you should verify. NOW WORK Problems 11 and 21.

3 Find the Volume of a Solid If z=f(x,y)=1

on R,

in E,

the double integral ∬Rf(x,y) dA=∬RdA

is numerically equal to the area of the region R. then the triple integral ∭Ef(x,y,z) dV=∭EdV 

If f(x,y,z)=1

is numerically equal to the volume of the solid E.

That is,

Volume V of a solid E:V = ∭EdV EXAMPLE 3 Finding the Volume of a Solid

Find the volume V of the solid E in the first octant that is enclosed by the paraboloid z=16-4x2-y2 and the xy -plane. Solution Figure 44(a) shows the solid E. The upper surface is the paraboloid z=z2(x,y)=16-4x2-y2 and the lower surface is the plane z=z1(x,y)=0. The region R in the xy -plane is enclosed by the x -axis, the y -axis, and part of the ellipse 4x2+y2=16, x≥0, y≥0, and z1 and z2 are continuous on R. So E is xy -simple. The region R is both x -simple and y -simple. We choose to use the iterated integral for an x -simple region, so 0≤y≤16-4x2 Figure 44(b). The volume V of E is given by

and 0≤x≤2.

See

V=∭E dV=∫ab ∫y1(x)y2(x) ∫z1(x, y)z2(x, y) dz dy dx=∫02 ∫016-4x2 ∫016-4x2-y2 dz dy dx=∫02  ∫016-4x2(16-4x2-y2) dy dx=∫02 (16-4x2)y-y33016-4x2 dx=∫02 (16-4x2)3/2-(16-

4x2)3/23 dx=163 ∫02 (4-x2)3/2 dx

We use the Table of Integrals, Integral 67 with a=2.

Then

V=163 ∫02 (4-x2)3/2 dx=163 x4 (4-x2)3/2+3x24-x2+6 sin-1 x202=163⋅6⋅π2=16π

Figure 44

▪ NOTE Alternatively, we could use a CAS to find the integral or we could use the trigonometric substitution x=2 sin θ, -π2≤θ≤π2.

NOW WORK Problem 37.

4 Find the Mass, Center of Mass, and Moments of Inertia of a Solid

Triple integrals can be used to find the mass, center of mass, and moments of inertia of solids. The formulas are obtained in a manner similar to those using double integrals, so we only state them here. The mass M of a solid E of volume V with mass density ρ=ρ(x,y,z) that is continuous in E is given by

M=∭Eρ(x,y,z) dV The moments of the solid about the coordinate planes are Mxy=∭Ezρ(x,y,z) dVMxz=∭Eyρ(x,y,z) dVMyz=∭Exρ(x,y,z) dV

The center of mass of the solid is the point (x,ȳ,z)

whose coordinates are given

by x̄=MyzM=∭Exρ(x,y,z) dV∭Eρ(x,y,z) dVȳ=MxzM=∭Eyρ(x,y,z) dV∭ Eρ(x,y,z) dVz̄=MxyM=∭Ezρ(x,y,z) dV∭Eρ(x,y,z) dV

The moment of inertia about an axis a is

Ia=∭E r2ρ(x,y,z)dV where r is the distance from the point (x,y,z) the moment is to be found.

of the solid to the axis a about which

EXAMPLE 4 Finding the Mass and Center of Mass of a Solid

a. Find the mass M of a solid in the shape of a tetrahedron cut from the first octant by the plane x+y+z=1 if the mass density ρ=ρ(x,y,z) is proportional to the distance from the yz -plane. b. Find the center of mass (x,ȳ,z)

of the tetrahedron.

Solution Figure 45(a) shows the solid E. It is enclosed by the surfaces z=z1(x,y)=0 and z=z2(x,y)=1-x-y, and its projection onto the xy -plane is the closed, bounded x -simple region R defined by the lines y=0 and y=1-x, where 0≤x≤1. See Figure 45(b). Since z1 and z2 are continuous on R, the solid E is xy -simple. a. Since the mass density ρ=ρ(x,y,z) yz -plane, we have ρ(x,y,z)=kx, proportionality. Then the mass M is

is proportional to the distance from the where k is the constant of

M=∭Eρ(x,y,z) dV=∭Ekx dV=k∫01∫01-x ∫01-x-y x dz dy dx

Using a CAS, M=k24. b. The center of mass (x,ȳ,z)

is

x̄=∭Exρ(x,y,z) dVM=∭Exkx dVk24=24∭E x2 dV =24∫01∫01-x ∫01-x-

y x2 dz dy dx=25Use a CAS. Similarly, ȳ=∭Eyρ(x,y,z) dVM=∭Eykx dVk24=24∫01∫01-x ∫01-x-y xy dz dy dx=15

z̄=∭Ezρ(x,y,z) dVM=∭Ezkx dVk24=24∫01∫01-x ∫01-x-y xz dz dy dx=15

Figure 45

The center of mass of the tetrahedron is located at the point 25,15,15.



▪

NOTE The integrals in (a) and (b) can be found by hand, but the integration is somewhat tedious. NOW WORK Problem 49.

EXAMPLE 5 Finding the Moment of Inertia About the z -axis of a Solid

Find the moment of inertia about the z -axis of the homogeneous solid of mass density ρ in the first octant enclosed by the surface z=4xy and the planes z=0, x=3,

and y=2. Solution The moment of inertia about the z -axis is

Iz=∭E r2ρ dV where r=x2+y2

is the distance of the point (x,y,z)

from the z -axis.

Figure 46(a) shows the solid E. The upper surface is z=z2(x,y)=4xy and the lower surface is the plane z=z1(x,y)=0. The region R in the xy -plane is enclosed by the x -axis, the y -axis, the line x=3, and the line y=2, as shown in Figure 46(b). Since z1 and z2 are continuous on R, the solid E is xy simple.

Figure 46 The region R is both x -simple and y -simple. Iz=∭E(x2+y2)ρ dV=ρ ∫03 ∫02 ∫04xy(x2+y2) dz dy dx=ρ ∫03 ∫02(x2+y2) z 04xy dy dx=ρ ∫03  ∫02(4x3y+4x y3) dy dx=ρ ∫03 2x3y2+x y4 02 dx=ρ  ∫03(8x3+16x) dx=ρ2x4+8x2 03=ρ(162+72)=234ρ

▪ NOW WORK Problem 51.

5 Find a Triple Integral of a Function Defined in an xz Simple Solid

-Simple or a yz

-

Triple integrals on certain types of regions in space are best found by integrating first with respect to x or y rather than z. Suppose a solid E in space is enclosed by the two surfaces y=y1(x,z) and y=y2(x,z), where y1≤y2. If the projection of E onto the xz -plane is a closed, bounded region R, and y1 and y2 are continuous on R, then E is called an xz -simple solid. See Figure 47.

Figure 47 An xz

-simple solid; R lies in the xz-plane.

THEOREM Triple Integral over an xz

-Simple Solid

If f is a function of three variables that is continuous in an xz

-simple solid E,

then

∭Ef(x,y,z) dV=∬R∫y1(x, z)y2(x, z) f(x,y,z) dy dA

Similarly, suppose a solid E in space is enclosed by the two surfaces x=x1(y,z) and x=x2(y,z), where x1≤x2. If the projection of E onto the yz -plane is a closed, bounded region R, and x1 and x2 are continuous on R, then E is called a yz -simple solid. See Figure 48.

Figure 48 A yz-simple solid; R lies in the yz -plane. THEOREM Triple Integral over an yz

-Simple Solid

If f is a function of three variables that is continuous in a yz -simple solid E, ∭Ef(x,y,z) dV=∬R∫x1(y, z)x2(y, z) f(x,y,z) dx dA

then

EXAMPLE 6 Finding the Volume of a Solid That Is yz

-simple

Find the volume V of the solid E that is enclosed by the cylinder y2+z2=4, and the planes x=0, z=0, and x+z=5. Solution Figure 49(a) shows that the solid E is yz -simple. The front surface is the plane x2(y,z)=5-z and the back surface is the plane x1(y,z)=0. The region R in the yz -plane is enclosed by the y -axis and the semi-circle z=4-y2; it is y -simple. So, 0≤z≤4-y2 Figure 49(b). The volume V of E is given by

and -2≤y≤2.

See

V=∭EdV=∬R∫x1(y, z)x2(y, z) dx dz dy=∫-22 ∫04-y2 ∫05-z dx dz dy=∫-22 ∫04-y2(5z) dz dy=∫-22 5z-z2204-y2 dy=∫-22 54-y2-4-y22 dy=5 ∫-224-y2 dy-∫-22 2-y22 dy=5⋅2π-2y-

y36-22=10π-163

Figure 49

▪ NOTE ∫-224-y2 dy

equals the area of a semi-circle of radius 2,

NOW WORK Example 2 (p. 993) treating E

as an xz

namely 2π.

-simple region.

14.6 Assess Your Understanding Concepts and Vocabulary

1. True or False A solid E in space is called xy -simple if it is enclosed on the top by a surface z=z2(x,y) and on the bottom by a surface z=z1(x,y),

where z1≤z2 and the sides of E are a cylinder whose intersection with the xy -plane forms a closed, bounded region R, whose boundary is a piecewise-smooth curve. 2. True or False The region R resulting from the projection onto the xy xy -simple solid E is both x -simple and y -simple.

-plane of an

3. True or False The mass M of a solid E of volume V with mass density ρ=ρ(x,y,z) that is continuous in E is given by M=∭Eρ(x,y,z) dV.

4. True or False The triple integral ∭EdV can be interpreted geometrically as the area of the region R that is the projection of E onto the xy -plane. Skill Building

In Problems 5–10, (a) express each triple integral in six different ways. (b) Find the integral. 5. ∭Ex y2 dV,

E is the closed box 0≤x≤1, 0≤y≤2, 0≤z≤4.

6. ∭E x2yz dV, E

is the closed box 0≤x≤2, 0≤y≤1, 0≤z≤3.

7. ∭E(x2+y2+z2) dV, E

is the closed box 0≤x≤1, 0≤y≤2, 0≤z≤3.

8. ∭E(x2-y2+z2) dV, E

is the closed box 0≤x≤3, 0≤y≤2, 0≤z≤1.

9. ∭E ez sin x cos y dV, E 0≤y≤π2,

is the closed box 0≤x≤π4,

0≤z≤1.

10. ∭E e-z cos x cos y dV, E 0≤y≤π3,

is the closed box 0≤x≤π2, 0≤z≤1.

In Problems 11–18, find each iterated triple integral. 11. ∫02 ∫02-3x ∫0x+y x dz dy dx 12. ∫01∫0-x ∫02x+y z dz dy dx 13. ∫03 ∫zz+2 ∫yy+z(2x+1) dx dy dz 14. ∫02 ∫yy+2 ∫yx+y(4z-1) dz dx dy 15. ∫02 ∫01∫0y ex dx dy dz 16. ∫12 ∫0z ∫0y x ex2 dx dy dz

17. ∫0π/2 ∫yπ/2 ∫0xy sin zy dz dx dy 18. ∫0π/2 ∫xπ/2 ∫0xy cos zx dz dy dx

In Problems 19–24, find ∭Exy dV. 19. E is the solid enclosed by the surfaces z=0 and z=5-x-y, whose projection onto the xy -plane is the region enclosed by the rectangle 0≤x≤1 and 0≤y≤3. 20. E is the solid enclosed by the surfaces z=0 and z=16-x2-y2, whose projection onto the xy -plane is the region enclosed by the rectangle 0≤x≤2 and 0≤y≤1. 21. E is the solid enclosed by the surfaces z=0 and z=xy, whose projection onto the xy -plane is the region enclosed by the triangle with vertices (0,0,0), (0,1,0), and (1,0,0). 22. E is the solid enclosed by the surfaces z=0 and z=x2+y2, whose projection onto the xy -plane is the region enclosed by the triangle with vertices (0,0,0), (1,0,0), and (0,2,0).

23. E is the solid enclosed by the surfaces z=0 and z=3-x-y, whose projection onto the xy -plane is the region enclosed by the triangle with vertices (0,0,0), (1,1,0), and (1,-1,0). 24. E is the solid enclosed by the surfaces z=0 and z=3+2y, whose projection onto the xy -plane is the region enclosed by the triangle with vertices (-1,0,0),(0,1,0), and (2,0,0). In Problems 25 and 26, use a CAS

to approximate each integral.

25. ∫01∫1x2 ∫0x yex dz dy dx 26. ∫34 ∫23 ∫01 ln(x2+y2+z2) dx dy dz Applications and Extensions

In Problems 27–30, find each triple integral. 27. ∭Ex dV, if E is the solid enclosed by the tetrahedron having vertices at (0,0,0), (1,1,0), (1,0,0), (1,0,1). 28. ∭E(x2+z2) dV, if E is the solid enclosed by the tetrahedron having vertices at (0,0,0), (1,1,0), (1,0,0), (1,0,1). 29. ∭E(xy+3y) dV, if E is the solid enclosed by the cylinder x2+y2=9 and the planes x+z=3, y=0, and z=0. 30. ∭Ex yz dV, if E is the solid enclosed by the cylinders x2+y2=1 and x2+z2=1. See the figure.

In Problems 31–34, (a) describe the solid whose volume is given by each integral. b. Graph the solid.

31. ∫-22 ∫04-y2∫01 dz dx dy

32. ∫02 ∫04-y2 ∫04-x2-y2 dz dx dy

33. ∫01∫0x2 ∫0y dz dy dx 34. ∫01∫y21 ∫01-x dz dx dy 35. a. Set up the triple integral ∭EdV over the tetrahedron formed by the coordinate planes and the plane x+2y+3z=6 using six different orders of integration. b. Find the volume of the tetrahedron. 36. a. Set up the triple integral ∭EdV over the tetrahedron formed by the coordinate planes and the plane x+y+z=3 using six different orders of integration. b. Find the volume of the tetrahedron. 37. Volume of a Solid Find the volume V of the solid enclosed by y2=z, x=0, and x=y-z.

38. Volume of a Solid Find the volume V of the solid enclosed by z=4-y2, z=9-x, x=0, and z=0. 39. Set up, but do not find, ∭Exy dV, where E is the solid enclosed by the surfaces z=1-x-y and z=3-x-y, whose projection onto the xy -plane is the region enclosed by the circle x2+y2=1. 40. Set up, but do not find, ∭ExydV, surfaces z=0 and z=x2+y, region enclosed by the circle x2+y2=4.

where E is the solid enclosed by the whose projection onto the xy -plane is the

In Problems 41–46, set up, but do not find, an iterated triple integral that equals the volume of the solid, 41. Enclosed by z=x2+y2

and z=16-x2-y2

42. Enclosed by z=x2+y2

and z=4-x2-y2

43. Enclosed by z=x2+y2

and z=2-x

44. Enclosed by z=x2+y2

and z=3-y

45. Enclosed by z2=4x

and x2+y2=2x

46. Enclosed by z2=4y

and x2+y2=2y

47. Mass Set up, but do not find, the integral that equals the mass M of an object in the shape of a tetrahedron cut from the first octant by the plane x+y+z=1, as shown in the figure, if its mass density is proportional to the product of the distances from the three coordinate planes.

48. Mass Set up, but do not find, the integral that equals the mass M of an object in the shape of a right circular cylinder of height h and radius a, if its mass density is proportional to the square of the distance from the axis of the cylinder. 49. Mass Find the mass M of an object in the shape of a cube of edge a if its mass density is proportional to the square of the distance from one corner. 50. Mass A cylindrical bar of radius R and length 2L is positioned with its axis along the x -axis and its center of mass at the origin. The mass density of the bar is given by ρ(x,y,z)=k z2. Show that the mass M of the bar is M=12 πk R4 L.

51. Moments of Inertia Set up, but do not find, the integrals that equal the moments of inertia Ix and Iy for the solid region enclosed by the hemisphere z=9-x2-y2 and the xy -plane, if the mass density is proportional to the distance from the xy -plane. 52. Set up, but do not find, the integral of the function f(x,y,z)=x2yz the solid enclosed by the cone 3x2+3y2=z2, z≥0, z=3.

over and the plane

53. Volume of an Ellipsoid Show that the volume of the ellipsoid x2a2+y2b2+z2c2=1 is 43 πabc. (Assume that a, b, What does this formula reduce to if a=b=c ?

and c are positive.)

Challenge Problems

54. Volume of a Solid Find the volume of the solid in the first octant enclosed by the coordinate planes, a2y=b(a2-x2), and a2 z=c(a2-x2), a>0, b>0, c>0. 55. Volume of a Solid Find the volume V of the region enclosed by z=0, z=1-x2, and z=1-y2. 56. Average Value of a Function The average value of z=f(x,y) over a region R that is not necessarily rectangular is defined to be the number 1A ∬Rf(x,y) dA, where A is the area of R. a. In single variable calculus the average value of a function y=f(x)

over the

closed interval [a,b]

is defined to be the number 1b-a ∫ab f(x) dx.

In what sense is this a special case of the above definition of the average value of z=f(x,y) over R ? b. Let w= f(x,y,z) be integrable over a solid E in space. What definition would you give for the average value of f in E ? 57. Show that the following integrals represent the same volume. Do not find the integrals.

a. 4 ∫04 ∫016-x2 ∫(x2+y2)/44 dz dy dx

b. 4 ∫04 ∫02z ∫04z-x2 dy dx dz

c. 4 ∫04 ∫y2/44 ∫04z-y2 dx dz dy 58. Show that: ∫ab ∫az ∫ay f(x) dx dy dz=∫ab (b-x)22 f(x) dx

1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

14.7 Triple Integrals Using Cylindrical Coordinates OBJECTIVES When you finish this section, you should be able to: 1 Convert rectangular coordinates to cylindrical coordinates 2 Find a triple integral using cylindrical coordinates

We have seen that there are instances where it is easier to find a double integral using polar coordinates than to find it using rectangular coordinates. For triple integrals, we give two alternatives to integration in rectangular coordinates: One uses cylindrical coordinates and the other uses spherical coordinates (Section 14.8). 1 Convert Rectangular Coordinates to Cylindrical Coordinates If the rectangular coordinates of a point P in three-dimensional space are (x,y,z) and if (r,θ,0) are the polar coordinates for the projection of P onto the xy -plane, then the point P can be located by the ordered triple (r,θ,z), called the cylindrical coordinates of the point P. Figure 50 shows how to graph a point P=(r,θ,z). The algebraic relationship between the cylindrical coordinates (r,θ,z) and the rectangular coordinates (x,y,z) of a point P is given by the formulas x=r cos θy=r sin θz=z

Figure 50 EXAMPLE 1 Converting Between Cylindrical Coordinates and Rectangular Coordinates

a. Find the rectangular coordinates of a point P whose cylindrical coordinates are 6,  π3, -2. b. Find the cylindrical coordinates of a point P whose rectangular coordinates are (3, 1, 5 ). Solution a. We use the equations x=r cos θ and θ=π3.

and y=r sin θ

with r=6

Then

x=6 cos π3=3y=6 sin π3=33z=-2 In rectangular coordinates, P=(3, 33,-2). Then cos 

b. In cylindrical coordinates r=x2+y2=3+1=2. θ=xr=32

and sin θ=yr=12

so θ=π6.

the same, the point P in cylindrical coordinates is (r,θ,z)=2, π6, 5.

Since z remains

▪



NOTE The numbers r,θ,

and z are called cylindrical coordinates because if r is a constant, then the

surface traced out by the set of points (r,θ,z)

is a cylinder.

NOW WORK Problems 5 and 13.

Table 1 lists the equations of several surfaces expressed in rectangular coordinates and in cylindrical coordinates. Figure 51 illustrates the surfaces. TABLE 1 Surface

Rectangular

Cylindrical

(a) Half-Plane

y=x tan k

(b) Plane

z=k

z=k

(c) Cylinder

x2+y2=a2

r=a,a>0

(d) Sphere

x2+y2+z2=R2

r2+z2=R2

(e) Circular cone

x2+y2=a2 z2

r=az,a>0

(f) Circular paraboloid

x2+y2=az

r2=az,a>0

θ=k,-π20

is a constant

38. y=ln(C-e-x),dydx=e-(x+y),C 39. Find the values of n so that y=enx

is a constant is a solution of y″+y′-6y=0

40. Find a first-order differential equation that has y=ex+e-x

as a solution.

41. Schrödinger Equation In quantum mechanics, the time-independent Schrödinger equation in one dimension can be written as -h22m d2Y(x)dx2+U(x) Y(x)=E Y(x). What are the degree and order of this differential equation? * It can be shown that this is the general solution of the differential equation.

1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

16.2 Separable and Homogeneous First-Order Differential Equations; Slope Fields; Euler’s Method OBJECTIVES When you finish this section, you should be able to: 1 Solve a separable first-order differential equation 2 Identify a homogeneous function of degree k 3 Use a change of variables to solve a homogeneous first-order differential

equation 4 Find orthogonal trajectories 5 Use a slope field to represent the solution of a first-order differential equation 6 Use Euler’s method to approximate a particular solution of a first-order differential equation In this section, we investigate first-order differential equations. A first-order differential equation is usually written in the form

dydx=f(x,y) where f is continuous on its domain. If we treat dy write dydx=f(x,y)

and dx

as differentials, we can

in differential form as

M(x,y) dx+N(x,y) dy=0 We begin our study of first-order differential equations by reviewing separable differential equations, which were discussed in Section 5.6, pp. 413–414. 1 Solve a Separable First-Order Differential Equation DEFINITION Separable First-Order Differential Equation

A first-order differential equation is said to be separable if it can be written in the form M(x) dx+N(y) dy=0

where M is a function of x alone, N is a function of y alone, and both M are continuous on their domains.

and N

The following steps are used to solve separable first-order differential equations: Steps for Solving a Separable First-Order Differential Equation

Step 1 Express the given equation in the differential form M(x) dx+N(y) dy=0 Step 2 Integrate to obtain the general solution ∫ M(x) dx+∫ N(y) dy=C where C is the constant of integration.

EXAMPLE 1 Solving a Separable First-Order Differential Equation

Solve dydx=2exy2 Solution Use the steps for solving a separable differential equation. Step 1 Express dydx=2exy2

in the differential form:

2exdx-y2dy=0 Step 2 Integrate to obtain the general solution. ∫2exdx-∫y2dy=Cso that2ex-y33=C where C is a constant. This solution is expressed implicitly. To obtain the explicit form, solve for y. y3=6ex+3Cso thaty=6ex+3C3

▪

where C is a constant. NOW WORK Problems 7 and 17.

Some first-order differential equations are not separable, but can be made separable by a change of variable. This is true for differential equations of the form dydx=f(x,y),

where f is a homogeneous function. 2 Identify a Homogeneous Function of Degree k DEFINITION Homogeneous Function of Degree k

A function f(x,y) is said to be a homogeneous function of degree k in x and y if, and only if, for t>0, f(tx,ty)=tk f(x,y) for some real number k.

EXAMPLE 2 Identifying a Homogeneous Function of Degree k

a. f(x,y)=3x2-xy+y2

is a homogeneous function of degree 2, since

f(tx,ty)=3(tx)2-(tx)(ty)+(ty)2=t23x2-t2xy+t2y2=t2(3x2-xy+y2)=t2 f(x,y)for all t>0

b. f(x,y)=x+4y

is a homogeneous function of degree 12,

since

f(tx,ty)=tx+4(ty)=t(x+4y)=t x+4y=t1⁄2 f(x,y)for all t>0

c. f(x,y)=xx2-y2

is a homogeneous function of degree 0, since

f(tx,ty)=tx(tx)2-(ty)2=txt2(x2-y2)=txtx2-y2=t0 f(x,y)for all t>0

d. f(x,y)=x-y2

is not a homogeneous function, since

f(tx,ty)=tx-(ty)2=t(x-t y2)≠tk(x-y2) for all t>0

and some k.

▪

NOW WORK Problem 23.

3 Use a Change of Variables to Solve a Homogeneous First-Order Differential Equation DEFINITION Homogeneous First-Order Differential Equation

A first-order differential equation of the form M(x,y) dx+N(x,y) dy=0 is said to be homogeneous if M degree.

and N are homogeneous functions of the same

NOTE The word “homogeneous” is used in two different ways: to describe a property of a function and to identify a certain type of first-order differential equation. Context will make it clear which meaning applies.

To solve first-order homogeneous differential equations, we use the following change of variables theorem. THEOREM Change of Variables for Homogeneous First-Order Differential Equations

If M(x,y) dx+N(x,y) dy=0 is a homogeneous first-order differential equation, then it can be transformed into a first-order differential equation whose variables are separable by using the substitution y=x v(x) where v=v(x)

is a differentiable function of x.

NOTE For simplicity of notation, we usually write v instead of v(x).

Proof If M(x,y) dx+N(x,y) dy=0 differential equation, use the substitution y=xv, differentiable function of x. Then dy=x dv+v dx

is a homogeneous first-order where v=v(x) is a and

M(x,y) dx+N(x,y) dy=M(x,xv) dx+N(x,xv)(x dv+v dx)=0 Since M and N are each homogeneous functions of degree k, it follows that for some number k

xk M(1,v) dx+xk N(1,v)(x dv+v dx)=0M(x,xv)=xk M(1,v), N(x,xv)=xk N(1,v)xkM(1,v) dx+N(1,v) (x dv+v dx)=0M(1,v) dx+N(1,v) x dv+N(1,v) v dx=0Divide out xk. [M(1,v)+vN(1,v)]dx+N(1,v) x dv=0dxx+N(1,v)M(1,v)+vN(1,v) dv=0

provided neither denominator is 0. This first-order differential equation is separable.

▪

Steps for Solving a Homogeneous First-Order Differential Equation M(x,y) dx+N(x,y) dy=0

Step 1 Confirm that the functions M degree.

and N are homogeneous functions of the same

Step 2 Let y=xv. Substitute for y and dy=x dv+v dx. Step 3 Express the new equation in differential form. The variables will be separable. Step 4 Integrate to obtain the general solution and replace v by yx.

EXAMPLE 3 Solving a Homogeneous First-Order Differential Equation

Solve the differential equation (x2-3y2) dx+2xy dy=0. Solution Follow the steps for solving a homogeneous first-order differential equation. Step 1 Both x2-3y2 see why?

and 2xy

are homogeneous functions of degree 2. Do you

Step 2 Let y=xv. Then dy=x dv+v dx. differential equation:

Substitute these into the

(x2-3y2) dx+2xy dy=0[x2-3(xv)2] dx+2x(xv)(x dv+v dx)=0dy=x dv+vdx(x2x2 v2) dx+2x3 v dv=0x2(1-v2) dx+2x3 v dv=0

Step 3 Then for x≠0 x3(1-v2).

and v≠±1,

we can separate the variables by dividing by

x2(1-v2) dxx3(1-v2)+2x3 v dvx3(1-v2)=0dxx+2v dv1-v2=0

Step 4 Integrate.

∫dxx-∫2v dvv2-1=C1ln x-ln v2-1=C1ln xv2-1=C1 Since C1 Then

is a constant, we can write C1=ln C2,

where C2

is a constant.

ln |xv2−1|=ln C2|xv2−1|=C2|xy2x2−1|=C2v=yx|x3y2−x2|=C2x3=±C2(y2−x2)

Since ±C2 x3=C(y2-x2),

may be either positive or negative, the general solution can be written as where C≠0

▪

is a constant.

NOW WORK Problem 39.

EXAMPLE 4 Solving a Homogeneous First-Order Differential Equation

Solve the differential equation x dy+(2xey⁄x-y) dx=0 when x=1.

if y=0

Solution Follow the steps for solving a homogeneous first-order differential equation: Step 1 Both x and 2xey⁄x-y

are homogeneous of degree 1.

Step 2 Let y=xv. Then dy=x dv+v dx. differential equation.

Substitute these into the

x dy+(2xey⁄x-y) dx=0x(x dv+v dx)+(2xev-xv) dx=0x2dv+xv dx+2xevdx-

xv dx=0x2dv+2xevdx=0 Step 3 To separate the variables, divide by x2 ev.

Then for x≠0,

2dxx+dvev=0 Step 4 Integrate. ∫2dxx+∫dvev=C2∫dxx+∫e-vdv=C2 ln x-e-v=C2 ln x-e-y⁄x=Cv=yx

This is the general solution to the differential equation. To find the particular solution, substitute x=1 and y=0 to find C. 2 ln 1-e0=CC=-1 The particular solution is 2 ln x-e-y⁄x=-1

▪ NOW WORK Problem 43.

4 Find Orthogonal Trajectories Consider the one-parameter family of circles (x-1)2+(y-2)2=CC>0 with center at the point (1,2) and radius C. Figure 1 shows some members of this family of circles. If the equation describing the circles is differentiated with respect to x, we obtain

(1) 2(x-1)+2(y-2)y′=0y′=-x-1y-2

which represents the differential equation of the family of circles.

Figure 1 (x-1)2+(y-2)2=C, For a specific point (x,y), y≠2,

C>0 on any one of the circles, the differential

equation y′=-x-1y-2 gives the slope of the tangent line to the circle at the point (x,y). Now consider another example. The family of nonvertical lines passing through the point (1,2) satisfies the equation y-2=m(x-1) where m is the slope of a particular member of the family of lines. The differential equation for this family of lines is

(2) y′=m=y-2x-1

Figure 2 shows some members of this family of lines.

Figure 2 y-2=m(x-1) If we compare the differential equations (1) and (2), we see that the right side of (1) is the negative reciprocal of the right side of (2). From this, we conclude that if (x,y) is a point of intersection of one of the circles and one of the lines from each family, then the line and the circle are perpendicular (orthogonal) to each other at the point of intersection. Each line in the family y-2=m(x-1) is an orthogonal trajectory of the family of circles, and conversely, each circle in the family (x-1)2+(y-2)2=C is an orthogonal trajectory of the family of lines. In general, if F(x,y,C)=0 and G(x,y,K)=0 are oneparameter families of curves, in which each member of one family intersects the members of

the other family at a right angle, then the two families are said to be orthogonal trajectories of each other. Orthogonal trajectories occur naturally. For example, iron filings sprinkled on a pane of glass over a bar magnet arrange themselves along a family of curved lines, indicating the direction of magnetic force. A family of orthogonal trajectories can be determined by locating all the points in the plane of the glass with the same potential energy.

To find a family G(x,y,K)=0 of orthogonal trajectories for a given family F(x,y,C)=0, we use the steps below: Steps for Finding a Family of Orthogonal Trajectories

Step 1 Find the differential equation for the given family. Step 2 Replace y′ in this differential equation by -1y′; the resulting equation is the differential equation for the family of orthogonal trajectories. Step 3 Find the solution of the differential equation for the family of orthogonal trajectories obtained in Step 2.

EXAMPLE 5 Finding a Family of Orthogonal Trajectories

Find the family of orthogonal trajectories for the one-parameter family y2=Cx3. See Figure 3 for several graphs in the family.

Figure 3 y2=Cx3 Solution Step 1 Differentiate y2=Cx3

with respect to x.

2y y′=3Cx2 Then eliminate C using this equation and the equation y2=Cx3 family. Since C=y2x3,

of the

we find

2y y′=3 y2x3 x2y′=3y2x Step 2 Now replace y′ with -1y′, orthogonal trajectories.

to obtain the differential equation for the family of

-1y′=3y2xy′=-2x3y Step 3 The differential equation y′=dydx=-2x3y written as

is separable and can be

2x dx+3y dy=0 The general solution is x2+32 y2=K The orthogonal trajectories for the family y2=Cx3 x2+32 y2=K,

Figure 4 y2=Cx3

is the family of ellipses

as shown in Figure 4.

; x2+32 y2=K

▪ NOW WORK Problems 57 and 63.

5 Use a Slope Field to Represent the Solution of a First-Order Differential Equation In a first-order differential equation dydx=f(x,y), the function f is continuous on its domain, a set of points in the xy -plane. At each point (x0,y0) in the domain of f, we can evaluate f(x0,y0) and determine dydx=f(x0,y0), the slope of the tangent line to the graph of the solution to the differential equation at (x0,y0). If we draw small line segments representing these tangent lines at every point in the

domain of f, the resulting display, called a slope field, gives a picture of the family of curves that make up the solution of the differential equation.

EXAMPLE 6 Using a Slope Field to Represent the Solution of dydx=y

a. Use a slope field to represent the solution of the differential equation dydx=y. b. Use the slope field representation to draw some graphs of the family of solutions of the differential equation. Solution a. Since dydx=y does not depend on x, the derivatives dydx at all points of the form (x,y0) will be equal. For example, at all points (x,2) , the slopes of the tangent lines to the graphs of the solution to the differential equation will equal 2. Since the points (x,2) all lie on the horizontal line y=2, the slope field along y=2 will consist of parallel tangent lines with slope 2. See Figure 5(a). Similarly, along the horizontal line y=1, the slope field will consist of parallel tangent lines with slope 1; along y=-1, the slope field consists of parallel tangent lines with slope -1. Figure 5(b) illustrates a slope field of the differential equation dydx=y. b. Pick a point in the xy -plane, say (0,2) . Going left and then right from this point, follow the tangent lines and insert a smooth curve. Repeat this process using other points. See Figure 5(c), which shows four of the graphs of the family of solutions.

Figure 5

▪

NOW WORK Problem 45.

EXAMPLE 7 Analyzing a Slope Field

Which of the following differential equations could have the slope field shown in Figure 6? a. dydx=-x b. dydx=x+y c. dydx=x d. dydx=x-y

Figure 6 Solution The tangent lines that lie on a vertical line all have the same slope. That is, for a fixed number x, the slopes dydx

of the tangent lines to the graphs of the solutions of

the differential equation are the same. This means dydx does not depend on y, so the differential equation does not contain y and choices (b) and (d) can be eliminated. Points in quadrants II and III give rise to negative slopes and points in quadrants I and IV give rise to positive slopes. The differential equation in (c) is the only choice that satisfies these conditions.

▪

NOW WORK Problem 51.

6 Use Euler’s Method to Approximate a Particular Solution of a First-Order Differential Equation Euler’s method is a numerical algorithm for approximating a particular solution y=y(x) of a first-order differential equation dydx=f(x,y) with the boundary condition y=y0 when x=x0. It begins with the boundary condition (x0,y0) and a small arbitrary increment h in x. The first approximation to y=y(x) is given by y1=y0+hf(x0,y0) The second approximation uses y1 from the first approximation and x1=x0+h. Then y2=y1+hf(x1,y1)x1=x0+h Each successive approximation follows the same pattern. The third approximation uses y2 and x2=x0+2h, and so on. y3=y2+hf(x2,y2)x2=x1+h=x0+2hy4=y3+hf(x3,y3)x3=x2+h=x0+3h⋮yn=yn-1+hf(xn-1,yn-

1)xn-1=x0+(n-1) h Each successive approximation follows the tangent line to the graph of the solution to the differential equation. That is, it follows the slope field. See Figure 7.

Figure 7 y1=y0+hf(x0,y0) EXAMPLE 8 Using Euler’s Method

Suppose y=y(x)

is the solution to the differential equation

dydx=xy+x2 with the boundary condition y=2 y(1.75) using h=0.25

when x=1. Use Euler’s method to approximate as the increment.

Solution Begin the first approximation with the boundary condition x0=1, and the increment h=0.25. Then x1=x0+h=1.25 y(x1)=y(1.25)≈y1=y0+hf(x0,y0) =2+0.25(1⋅2+12)f(x,y)=xy+x2=2.75

For the second approximation, we use x1=1.25 Then x2=x1+0.25=1.5, and

and y1=2.75.

y(x2)=y(1.5)≈y2=y1+hf(x1,y1)=2.75+0.25[(1.25)(2.75)+1.252]=4

and y0=2 and

For the third approximation y3, x3=x2+0.25=1.75,

we use x2=1.5 and

and y2=4.

Then

y(x3)=y(1.75)≈y3=y2+hf(x2,y2)=4+0.25[(1.5)(4)+1.52]=6.0625

▪ NOW WORK Problem 53.

16.2 Assess Your Understanding Concepts and Vocabulary

1. The general solution of the differential equation dydx=xy

is __________.

2. True or False If f(tx, ty)=tk f(x,y), for all t>0, where k is a real number, then the function f is said to be homogeneous of degree k in x and y. 3. If F(x,y,C)=0 and G(x,y,K)=0 are one-parameter families of curves in which each member of one family intersects the members of the other family orthogonally, then the two families are said to be orthogonal __________ of each other. 4. If M(x,y) dx+N(x,y) dy=0 is a homogeneous differential equation, then it can be transformed into an equation whose variables are separable by using the substitution __________, where __________ is a differentiable function of x. 5. A(n) __________ gives a picture of the family of curves that make up the solution to a first-order differential equation. 6. True or False Euler’s method is used to approximate a particular solution of a firstorder differential equation. Skill Building

In Problems 7–16, solve each differential equation.

7. dydx=x sec y 8. cos y dydx=2x 9. dydx=ey-x 10. x dydx+2y=5 11. dydx=cos x dydx-y sin x 12. dydx=x+y dydx 13. dydx+xy=x 14. (3x+1) dx+ex+ydy=0 15. ln x dxdy=xy 16. dydx=x+22-y In Problems 17–22, obtain a particular solution of each differential equation. (Use the general solutions obtained for Problems 7–12.) 17. dydx=x sec y,y=π4 18. cos y dydx=2x,y=π3 19. dydx=ey-x;y=0 20. x dydx+2y=5;y=1 21. dydx=cos x dydx-y sin x;y=12

when x=1 when x=-1 when x=0 when x=-1 when x=π

22. dydx=x+y dydx,y=1

when x=2

In Problems 23–32, determine if each function is homogeneous. If it is, find its degree. 23. f(x,y)=2x2-3xy-y2 24. f(x,y)=x3-xy2+y3 25. f(x,y)=x3-xy+y3 26. f(x,y)=x2-xy2+y2 27. f(x,y)=2x+x2+y2 28. f(x,y)=x+y 29. f(x,y)=tan 3xy 30. f(x,y)=ex⁄y 31. f(x,y)=ln xy 32. f(x,y)=x ln x-x ln y In Problems 33–44, solve each homogeneous differential equation. Follow the steps on p. 1118. 33. (x-y) dx+x dy=0 34. (x+y) dx+x dy=0 35. (x2+y2) dx+(x2-xy) dy=0 36. xy dx+(x2+y2) dy=0 37. dydx=y-xy+x 38. dydx=x+2y2x+y 39. x dydx=x+y

40. x(x2-y2) dydx-y(x2+y2)=0 41. dydx=2xyx2+y2 42. dydx=x+2y2x-y 43. x2 dydx=x2+xy-4y2;y=1

when x=-1

44. x(x2-y2) dydx-y(x2+y2)=0;y=-2

when x=-1

In Problems 45–50, a. Use a slope field to represent the solution of each differential equation. b. Use the slope field representation to draw some graphs of the family of solutions to the differential equation. 45. dydx=-2y 46. dydx=2x 47. dydx=x2 48. dydx=y2 49. dydx=x+y 50. dydx=x-y 51. Which of the following differential equations could have the slope field shown below? a. dydx=-x b. dydx=x+y

c. dydx=x d. dydx=x-y

52. Which of the following differential equations could have the slope field shown below? a. dydx=-x b. dydx=x2 c. dydx=2x+1 d. dydx=-x2

In Problems 53–56, use Euler’s method to approximate the particular solution to the differential equation with the given boundary condition. Assume y=y(x) is a solution of each differential equation. 53. Approximate y(0.4) if dydx=x2-y Use h=0.2 as the increment.

and y=1

when x=0.

54. Approximate y(0.2) if dydx=y2-x Use h=0.1 as the increment.

and y=1

when x=0.

55. Approximate y(1.4) Use h=0.2

if dydx=xy2-x as the increment.

and y=2

when x=1.

56. Approximate y(1.2) Use h=0.1

if dydx=xy2-x as the increment.

and y=2

when x=1.

In Problems 57–60, find the orthogonal trajectories of each family of curves. Graph the members of each family that contain the point (2,1). 57. xy=C 58. y2=2(C-x) 59. y=Cx2 60. x2=Cy3

Applications and Extensions

In Problems 61 and 62, solve each differential equation by letting x=u+h and y=v+k, where h and k are constants chosen to eliminate the constant terms on the right side. 61. dydx=y-x-3y+x+4 62. dydx=x+2y-32x-y+1

; y=4

when x=3

63. Orthogonal Families Find the orthogonal trajectories of the family of parabolas y2=4Cx. 64. Orthogonal Trajectories Find the orthogonal trajectories of all circles tangent to the x -axis at (3,0). 65. Sales Growth The annual sales of a new company are expected to grow at a rate that is proportional to the difference between the sales and an upper limit of $20 million. Sales are $0 initially and are $3 million for the second year. a. Determine the annual sales at any time t. b. According to the model, what will sales be during the eighth year of operations? c. How long will it take for sales to reach $12

million?

d. Graph the annual sales of the company for its first 20

years of operation.

66. Bacteria Growth A culture of bacteria is growing in a medium that can support a maximum of 1,000,000 bacteria. The rate of growth of the population at time t is proportional to the difference between 1,000,000 and the number present at time t. The culture contains 1000 bacteria initially, and after 1 h there are 1500 bacteria. a. How many bacteria will there be after 5 h b. When will the culture contain 9950

?

bacteria?

67. Learning Curve The number of words per minute W that a person can text increases with practice. Assume that the rate of change of W is proportional to the difference between an upper limit of 150 words per minute and W. That is, the rate of change is proportional to 150-W. Suppose a beginner cannot text at all, that is, W=0 when t=0, and can text 30 words per minute after 10 h of practice. a. How many words per minute can he text after 25 h

of practice?

b. How many hours of practice will be required in order for him to text 100 per minute? c. Graph the number of words per minute W hours of practice.

words

a beginner can text as a function of

68. Drug Concentration A drug is injected into a patient’s bloodstream at a constant rate of r mg/s. Simultaneously, the drug is removed from the bloodstream at a rate proportional to the amount y(t) present at time t. a. Find a differential equation that models this problem. b. Find a solution of the differential equation in (a). c. Find the particular solution that satisfies the initial condition y(0)=0. 69. Friction The frictional force f on an object sliding over a surface depends on the speed v of the object and can be modeled by the function f=-Av, where A is a positive constant that depends on the two surfaces. Note: The force is negative because the direction of the friction is opposite to the direction of the velocity. a. Express the frictional force f as a differential equation by using Newton’s Second Law, F=ma. b. Solve the differential equation found in (a) for v(t). over the surface with an initial speed v0.

Assume the object slides

c. Suppose that the frictional force on a 25-kg sled is 150 N when the speed of the sled is 10 m/s. Determine the value of the constant A. d. How long will it take the sled to slow down to 10%

of its initial speed?

e. How far will the sled travel while slowing down to 1.0 m/s? initial speed of 10 m/s.

Assume an

70. Friction The frictional force f on an object sliding over a surface depends on the speed v of the object and can be modeled by the function f=-Bv, where B is a positive constant that depends on the two surfaces. Note: The force is negative because the direction of the friction is opposite to the direction of the velocity. a. Express the frictional force f as a differential equation by using Newton’s Second Law, F=ma. b. Solve the differential equation found in (a) for v(t). over the surface with an initial speed v0. c. Suppose that the frictional force on a 25-kg

Assume the object slides

sled is 150 N

when the

speed of the sled is 10 m/s.

Determine the value of the constant B.

d. How long will it take the sled to slow down to 10%

of its initial speed?

e. How far will the sled travel while slowing down to 1.0 m/s? initial speed of 10 m/s.

Assume an

71. Electronics: Discharging a Capacitor In a circuit, a capacitor C carries an initial charge q0 and discharges through a resistor R. Using Kirchhoff’s rules with this circuit, we obtain the differential equation dqdt+qRC=0 where q is the charge, in Coulombs, on the capacitor at any time t, and R and C are constants. a. Solve the differential equation for the charge q as a function of time t. b. In a typical laboratory circuit, R=150 Ω (ohms) and C=5.0 μF (microfarads). How long will it take the discharging capacitor to lose 99% its initial charge? Hint: The product (1Ω)(1F)=1 second.

of

72. Air Resistance At high speeds, air resistance (drag) can be modeled by Fdrag=-Av2, where A is a positive constant that depends on the shape of the object, its surface texture, and the conditions of the air. Note: The force is negative because the direction of the drag is opposite to the direction of the velocity. Suppose, in a test facility, a projectile of mass m is launched with initial horizontal speed v0 on a frictionless horizontal track. The only force opposing the motion of the projectile is the drag. a. Express the drag Fdrag Law, F=ma.

as a differential equation by using Newton’s Second

b. Solve the differential equation found in (a) for v(t). c. What happens to the speed of the projectile as t becomes arbitrarily large? Challenge Problems

73. Orthogonal Families Find the orthogonal trajectories of the family of hyperbolas x2C2-y24-C2=1,

where 00v=Ce2x-ex

Step 5 Using v=y1/2,

write the solution in terms of x and y.

y1/2=Ce2x-exy=(Ce2x-ex)2 NOTE Since e-2x v+e-x>0,

then C>0.

▪ NOW WORK Problem 25.

Application: Logistic Functions In the mid-nineteenth century, the Belgian mathematical biologist Pierre F. Verhulst (18041849) used the differential equation

dydt=ky(M-y) where k and M are positive constants, to predict the human population of various countries. Verhulst referred to the model as logistic growth. Because of this, the equation is known as the logistic differential equation and its solutions are called logistic functions. Rewrite the logistic differential equation as

dydt=kMy-ky2 where the first term on the right side, kMy, is a growth term and the second term, ky2, is an “inhibition” or “competition” term that impedes growth. The logistic differential equation dydt=kMy-ky2 differential equation:

is a Bernoulli

dydt-kMy=-ky2 where P(t)=-kM, Q(t)=-k, steps for solving a Bernoulli equation. Step 1 Multiply by y-n=y-2. y-2 dydt-kM y-1=-k

and n=2.

To solve it, follow the

Step 2 Let v=y-1.

Then dvdt=-y-2 dydt.

Step 3 Substitute v and dvdt

into y-2 dydt-kM y-1=-k.

-dvdt-kMv=-kdvdt+kMv=k Step 4 Multiply by the integrating factor e∫ P(t) dt=e∫ kM dt=ekMt. ekMt dvdt+ekMt kMv=k ekMtddt[ekMt v]=k ekMtekMt v=k∫ekMt dt=1M ekMt+C

Step 5 Using v=y-1,

we rewrite the solution in terms of y.

ekMt 1y=1M ekMt+Cv=y-1=1yy(t)=ekMt1M ekMt+C=M ekMtekMt+MC=M1+MCe-kMt

If we assume that the initial condition is y(0)=R, 0≤R0 Find I as a function of t if I=I0 are constants. 1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

when t=0.

Here R,I,E0,

and ω

16.5 Power Series Methods OBJECTIVES When you finish this section, you should be able to: 1 Use power series to solve a linear differential equation

Up to now, we have concentrated on first-order differential equations for which exact solutions could be found. But many first-order differential equations lead to integrals that cannot be expressed in terms of elementary functions. Moreover, most higher-order differential equations cannot be solved exactly. For these reasons, considerable emphasis is placed on methods for approximating solutions of differential equations. One such method, power series, is introduced here. 1 Use Power Series to Solve a Linear Differential Equation Power series methods assume that a solution y=y(x) equation has a power series expansion of the form

of a given differential

y=y(x)=∑k=0∞ak xk A power series method consists of finding power series for the terms in the differential equation, as well as the power series for y, y′, y″, and so on. These power series expansions are substituted into the differential equation to obtain relationships among the coefficients. NEED TO REVIEW? Power series are discussed in Section 8.8, pp. 640–650.

We begin with a simple example to outline the basic idea. EXAMPLE 1 Using Power Series to Solve a Linear Differential Equation

Use power series to solve the differential equation y′=y. Solution Assume that the solution of the differential equation can be expressed as the power series

(1) y(x)=∑k=0∞ak xk

Then

y′(x)=∑k=0∞k ak xk-1 Since y′=y,

this leads to

∑k=1∞k ak xk-1=∑k=0∞ak xk To obtain relationships among the coefficients, we write out the terms. a1 x0+2a2x+3a3x2+4a4x3+⋯=a0x0+a1x+a2x2+a3x3+⋯ Because the coefficients of corresponding powers of x are equal, we have a1=a02a2=a13a3=a24a4=a3…nan=an-1… Now express these relationships recursively. a1=a0a2=12 a1=12! a0a3=13 a2=13⋅2 a0=13! a0a4=14 a3=14! a0…an=1n an-1=1n! a0⋯

The power series (1) takes the form y(x)=∑k=0∞akxk=∑k=0∞1k! a0xk=a0 ∑k=0∞xkk!

which is y(x)=a0ex.



▪

NOW WORK Problem 1.

RECALL As we learned in Section 8.9, ex=∑k=0∞xkk!.

Although this example led to a power series solution we recognized, most solutions will not. The next examples illustrate typical power series solutions.

EXAMPLE 2 Using Power Series to Solve a Linear Differential Equation

Use power series to solve the differential equation y′-2y=e-x. Solution Assume that a solution y=y(x) expressed as the power series y(x)=∑k=0∞ak xk.

of the differential equation can be Then y′(x)=

∑k=0∞k ak xk-1. Since e-x=∑k=0∞(-1)kk! xk, differential equation can be written as

the

(2)

To obtain relationships among the coefficients, write out the terms of the power series. a1+2a2x+3a3x2+4a4x3+5a5x4+⋯-2a0+a1x+a2x2+a3x3+a4x4+⋯=1-x+12!x213!x3+14!x4+⋯

Now combine the coefficients of corresponding powers of x to get (a1-2a0)+(2a2-2a1) x+(3a3-2a2)x2+(4a4-2a3)x3+⋯=1-x+12! x2-13! x3+14! x4-⋯

Since the coefficients of corresponding powers of x are equal, we have a1-2a0=1or equivalently,a1=2a0+12a2-2a1=-1or equivalently,a2=2a1-12=22a0+112=2a0+123a3-2a2=12!or equivalently,a3=2a2+12!3=232a0+12+13⋅2!=43 a0+33!4a42a3=-13!or equivalently,a4=2a3-13!4=12 a3-14!=1243 a0+33!-14!=23 a0+54!5a52a4=14!or equivalently,a5=2a4+14!5=25 a4+15!=2523 a0+54!+15!=415 a0+115!

This recursive formula (n+1)an+1=2an+(-1)n 1(n)! used to find an in terms of a0, as we did for a1, a2, a3, a4, The power series representation of the general solution of the equation is

can be and a5.

y(x)= ∑k=0∞ ak xk=a0+ (2a0+1) x+2a0+12!x2+43 a0+33!x3+23 a0+54!x4+415 a0+115!x5+⋯=a01+2x+2x2+43 x3+2 +x+12! x2+33! x3+54! x4+115! x5+⋯

▪

where a0 is arbitrary.

There is a more direct method for obtaining the recursion formula in Example 2. If we change the index of summation from k to k+1 in the first power series on the left side of Equation (2) and distribute the -2 inside the second power series, we have

∑k=0∞(k+1)ak+1 xk+∑k=0∞(-2)ak xk=∑k=0∞(-1)kk! xk

Now add the two series on the left.

∑k=0∞[(k+1)ak+1-2ak]xk=∑k=0n(-1)kk! xk

Then (n+1)an+1-2an=(-1)nn!,

as before.

EXAMPLE 3 Using Power Series to Solve a Linear Differential Equation

Use power series to find the general solution of the linear differential equation y″+xy′+2y=0

Solution If y(x)=∑k=0∞ak xk then

is a solution to the differential equation,

y′(x)=∑k=1∞k ak xk-1andy″(x)=∑k=2∞k(k-1)ak xk-2

Now substitute these power series into the differential equation. ∑k=2∞k(k-1)akxk-2+x∑k=1∞kakxk-1+2∑k=0∞akxk=0y″+x y′+2y=0∑k=2∞k(k-1)akxk2+∑k=1∞kakxk+∑k=0∞2akxk=0Move x and 2 into the summation.

Next adjust the indexes of summation so that xk appears in each series. Here, only the first series needs modification. If we replace k with k+2 in the first series, we obtain

The index in the first and third series begins at k=0, and the index in the second series begins at k=1. By writing the k=0 term of the first and third series separately, each summation starts at k=1.

Equating the coefficients of corresponding powers of x (the coefficients on the right side are all 0), we have 2a2+2a0=0(n+2)(n+1)an+2+(n+2)an=0a2=-a0an+2=-an(n+1)n=1,2,3,…

If n=1, we find a3=-a12. Then we use the recursion formula on the left to obtain all of the coefficients in terms of a0 and a1. That is, a2=-a0a4=-a23=a03a6=-a45=-a03⋅5a8=-a67=a03⋅5⋅7a3=-a12a5=-a34=a12⋅4a7=-a56=a12⋅4⋅6

and so on. Since a0 and a1 can be chosen arbitrarily, the power series representation for the general solution is y(x)=a01-x2+x43-x63⋅5+x83⋅5⋅7-⋯+a1x-x32+x52⋅4-x72⋅4⋅6+⋯

▪ NOW WORK Problem 7.

A second type of series solution method involves a differential equation with initial conditions and makes use of a Maclaurin series.

NEED TO REVIEW? Maclaurin series are discussed in Section 8.9, pp. 654–660.

EXAMPLE 4 Using a Maclaurin Series to Solve a Linear Differential Equation

(3)

a. Use a Maclaurin series to find the solution of y″=x2 y+ex y′ given the initial conditions y(0)=1

and y′(0)=1.

b. Use the first five terms of the series to approximate values of y=y(x) 0≤x≤1. c. Use a numeric differential equation solver with 10,000

for

equally spaced numbers in

the interval [0,1] to solve the differential equation in (a). Then construct a table showing the values of y for x=0,0.1,0.2,…,0.9,1. Solution a. We assume that the solution of the differential equation is given by the Maclaurin series y(x)=∑k=0∞y(k)(0)k! xn=y(0)+y′(0) x+y″(0)2! x2+y‴(0)3! x3+⋯

Substitute the initial conditions, y(0)=1

and y′(0)=1,

into (3). Then

y″(0)=02⋅1+e0⋅1=1x=0;y(0)=1;y′(0)=1;y″(x)=x2 y+ex y′

Now differentiate y″=x2 y+ex y′

with respect to x to find y‴(0).

y‴(x)=(x2 y′+2xy)+(ex y″+ex y′)y‴(0)=(02⋅1+2⋅0⋅1)+(e0⋅1+e0⋅1)=2

Continue differentiating and evaluating the derivative at x=0. y(4)(x)=x2 y″+4xy′+2y+ex y‴+2ex y″+ex y′y(4)(0)=02⋅1+4⋅0⋅1+2⋅1+e0⋅2+2⋅e0⋅1+e0⋅1=7

and so on. The Maclaurin series then becomes y(x)=1+x+12! x2+23! x3+74! x4+⋯

b. We construct Table 1 that uses the first five terms of the series to approximate select values of y in the interval 0≤x≤1. c. Using every thousandth term from the numeric solution, we construct Table 2. From the results of Tables 1 and 2, we can see how the five-term approximation to the series solution of the differential equation deteriorates as we move away from the center of

▪

convergence 0.

TABLE 1 Maclaurin Series Approximation of y Using Five Terms of the Series x

  0.0

  0.1

  0.2

  0.3

  0.4

  0.5

y 1.0

1.1054

1.2231

1.3564

1.5088

1.6849

  0.6

1.8898

TABLE 2 Select Values of y Using a Numeric Equation Solver x

0.0

0.1

0.2

0.3

0.4

0.5

y 1.0

1.1054

1.2232

1.3569

1.5114

1.6933

NOW WORK Problem 15.

16.5 Assess Your Understanding Skill Building

In Problems 1–12, use power series to solve each differential equation. 1. y′+3xy=0 2. y′-x+3xy=0 3. y″+y=0 4. y″+xy=0 5. y″+x2 y=0 6. y″-2xy=0 7. y″+x2 y′+xy=0 8. y″+3xy′+3y=0 9. y‴+y=0 10. y‴-xy=0

0.6

1.9127

11. (1+x2)y″-4xy′+6y=0 12. (x2+2)y″-3xy′+4y=0 In Problems 13–22: a. Use a Maclaurin series to find the solution of each differential equation using the given initial conditions. b. Use the first five nonzero terms of the series to approximate values of y for 0≤x≤1. Use Table 1 as a guide. 13. y″+xy′+y=0

; y(0)=1,

14. y″-2xy′+y=0

y′(0)=0

; y(0)=2,

15. y″-(sin x) y=0

y′(0)=1

; y(0)=0,

16. y″+(cos x) y=0

y′(0)=1

; y(0)=0,

y′(0)=1

17. y″+y′+ex y=0;y(0)=2,y′(0)=1 18. y″+(3+x) y=0

; y(0)=1,

19. y″+x2 y=0

; y(0)=0,

y′(0)=0 y′(0)=2

20. y″-3x2 y′+2xy=0

; y(0)=1,

y′(0)=1,

21. y(4)-ln(1+x) y=0 ″(0)=0, y‴(0)=0

; y(0)=1,

y′(0)=1,

22. y‴+4y″+2y′-x3 y=0 y″(0)=0

; y(0)=1,

y

y′(0)=1,

Applications and Extensions

23. Exact and Series Solutions a. Use the first six terms of a Maclaurin series to approximate the solution of the differential equation y″+4y′+4y=0 with the initial conditions y(0)=1

and y′(0)=2.

b. Solve the differential equation from (a) using a CAS. c. Graph both the series solution from (a) and the exact solution from (b) on the interval [-1,2]. d. Comment on the graphs. e. Repeat (a) using the first eight terms of a Maclaurin series. f. Add the solution from (e) to the graph in (c). What is happening?

24. Exact and Series Solutions a. Use the first six nonzero terms of a Maclaurin series to approximate the solution of the differential equation y ′=xy(1-y2)

with the initial condition y(0)=2.

b. Solve the differential equation from (a) using a CAS. c. Graph the exact solution from (b) and the series solutions using two, four, and six nonzero terms on the interval [-2,2]. Challenge Problem

25. Age of the Earth’s Crust. Uranium has a half-life of 4.5×109 years. The decomposition sequence is very complicated, producing a very large number of intermediate radioactive products, but the final product is an isotope of lead with an atomic weight of 206, called uranium lead. a. Assuming that the change from uranium to lead is direct, show that u=u0 e-kt, l=u0(1-e-kt), where u and l denote the number of uranium and uranium lead atoms, respectively, present at time t. That is, assume dudt=-ku,

where k>0

is a constant, and l=u0-u.

We can measure the ratio r=lu in a rock, and if it is assumed that all the uranium lead came from decomposition of the uranium originally present in the rock, we can obtain a lower bound for the age of Earth’s crust. Currently, r≈0.054. b. Show that this lower bound is given by t=1k ln(1+r)=1kr-r22+r33-⋯>rk c. What is this lower bound? 1. = NOW WORK problem = Graphing technology recommended = Computer Algebra System recommended

Chapter Review THINGS TO KNOW 16.1 Classification of Ordinary Differential Equations

Ordinary differential equation (p. 1113) Order of a differential equation (p. 1113) Degree of a differential equation (p. 1113) Linear differential equation of order n (p. 1113) General solution; particular solution (p. 1114) 16.2 Separable and Homogeneous First-Order Differential Equations; Slope Fields; Euler’s Method

Separable first-order differential equations (p. 1116) Steps for Solving a Separable Differential Equation (p. 1116) Homogeneous function of degree k in x and y, where k is some real number. (p. 1117) Homogeneous first-order differential equations (p. 1118) Steps for Solving a Homogeneous First-Order Differential Equation (p. 1118) Orthogonal trajectories (p. 1121) Steps for finding orthogonal trajectories (p. 1121) Slope field (p. 1122) Euler’s method (p. 1123) 16.3 Exact Differential Equations

A differential equation of the form M(x,y) dx+N(x,y) dy=0 is an exact differential equation at every point (x,y) in a simply connected open set R if and only if ∂M∂y=∂N∂x in R. (p. 1127) Integrating factor (p. 1130)

for all (x,y)

16.4 First-Order Linear Differential Equations; Bernoulli Differential Equations

First-order linear differential equation: dydx+P(x) y=Q(x), where P and Q are continuous on their domains (p. 1132) If Q(x)=0,

the first-order linear differential equation is separable. (p. 1132)

If Q(x)≠0, multiply by the integrating factor e∫ P(x) dx. Bernoulli equation: dydx+P(x) y=Q(x)yn,n≠0,n≠1,

(p. 1136)

and P and Q are functions of x. (p. 1136) Steps for Solving a Bernoulli Equation (p. 1136) 16.5 Power Series Methods

Assumes a solution y=y(x)

of a differential equation has a power series

expansion of the form y=y(x)=∑k=0∞akxk. (p. 1143) A Maclaurin series can be used to find a particular solution to a differential equation. (p. 1146) OBJECTIVES Section

You should be able to …

Examples

16.1

1 Classify ordinary differential equations (p. 1113) 2 Verify the solution of an ordinary differential equation (p. 1113)

1

Review Exercises 1–4

2, 3

5–8

16.2

1 Solve a separable first-order differential equation (p. 1116) 1 2 3, 4

21, 26, 28, 33

5

41

6, 7

39

8

43

16.3

2 Identify a homogeneous function of degree k (p. 1117) 3 Use a change of variables to solve a homogeneous firstorder differential equation (p. 1118) 4 Find orthogonal trajectories (p. 1120) 5 Use a slope field to represent the solution of a first-order differential equation (p. 1122) 6 Use Euler’s method to approximate a particular solution of a first-order differential equation (p. 1123) 1 Identify and solve an exact differential equation (p. 1128)

13, 19, 29, 30, 32, 34 9–12

1–3

16.4

1 Solve a first-order linear differential equation (p. 1132)

1–3

14, 17, 20 15, 16, 18, 22, 25, 27, 31, 40 23, 24, 42

16.5

2 Find the general solution of a Bernoulli equation (p. 1136) 4, 5 1 Use power series to solve a linear differential equation (p. 1–4 1143)

35–38

REVIEW EXERCISES In Problems 1–4, state the order and degree of each differential equation. Determine whether the equation is linear or nonlinear. 1. dydx+x2 y=sin x

2. d2 ydx2-5 dydx+3y=xex 3. d3 ydx3+4x d2 ydx2+sin2 x dydx+3y=sin x 4. drds3-r=1 In Problems 5–8, verify that the given function is a solution of the differential equation. 5. y=ex,d2 ydx2-y=0 6. y=ln xx2,d2 ydx2+5xdydx+4x2 y=0 7. y=e-3x,d2 ydx2+2 dydx-3y=0 8. y=x2+3x,x dydx-y=x2 In Problems 9–12, determine if each function is homogeneous. If it is, state the degree. 9. f(x,y)=x3+2x2 y+2xy2+y3 10. f(x,y)=x2-x2y2+y2 11. f(x,y)=xey⁄x 12. f(x,y)=ln yx In Problems 13–24: a. Identify each differential equation as separable, homogeneous, exact, linear, or Bernoulli. b. Solve each differential equation. 13. x(y2+1) dx+y(x2-x) dy=0 14. (2y+e2x) dx+(2x+e2y) dy=0

15. x dydx+y=x6 16. dydx-2y=e3x 17. (2xy3+y2 cos x-2x) dx+(3x2y2+2y sin x) dy=0

18. dydx+x2 y-y=x2-1 19. y sin x dx-cos x dy=0 20. (2x sin y-ln y) dx+x2 cos y-xy+3y2dy=0

21. x-y tan yxdx+x tan yx dy=0 22. x dydx-2y=x+1 23. dydx+2x y=y3x2 24. dydx-5y+2y2=0 In Problems 25–34, find the particular solution of each differential equation. 25. x dydx+2y=6,y(2)=8 26. xy dy=y2+xx2+y2 dx,y(1)=1 27. dydx+yx=2,y=0

when x=1

28. (x2+xy+y2) dx-x2dy=0,y=1

when x=1

29. dydx=y-1x+3,y(-1)=0 30. (1-cos x) dydx+y sin x=0,

yπ4=1

31. dydx-2yx=x2 cos x,yπ2=3 32. 2x-1y dx+x⋅1-xy2 dy=0,y(-1)=3 33. (x2+y2) dx-xy dy=0,y=2

when x=1

34. y2 sin x dx+1x-yxdy=0,y=1

when x=π

In Problems 35 and 36, find a power series solution for each differential equation. 35. (x-2)y′+y=0 36. y″+y′-2y=0 In Problems 37 and 38, a. Use a Maclaurin series to find a solution for each differential equation. b. Use the first five terms of the series to approximate values of y=y(x) 0≤x≤1.

for

37. y″+(sin x) y=0, y(0)=1; y′(0)=1 38. y″+exy′+y=0, y(0)=1; y′(0)=1 39. a. Draw a slope field to represent the solution of the differential equation dydx=x2+y2. b. Use the slope field representation to graph the solution that satisfies the initial condition that y=1

when

x=0.

40. Flow in Mixtures A large tank contains 1000 L of water in which 500 mg of chlorine are dissolved. Water containing 2 mg of dissolved chlorine per liter flows into the tank at the rate of 100 L⁄min. The mixture, kept uniform by stirring, runs out of the tank at the rate of 80 L⁄min. How much chlorine is in the tank after an hour? 41. Orthogonal Trajectories Find the orthogonal trajectories of the family of hyperbolas xy=c. Graph a member of each family that contains the point (2,3). 42. Spread of Rumors After a large chip manufacturer posts poor quarterly earnings, the rumor of widespread layoffs spreads among the employees. The rumor spreads throughout the 100,000 employees at a rate that is proportional to the product of

the number of employees who have heard the rumor and those who have not. If 10 employees started the rumor, and it has spread to 100 employees after 1 h, how long will it take for 25% of the employees to have heard the rumor? 43. Use Euler’s method with h=0.1

to approximate the solution y=y(1.3)

to the differential equation dydx=y+xy2 y=2 when x=1.

with the boundary condition that

CHAPTER 16 PROJECT The Melting Arctic Ice Cap

In this project, we study the changes in the thickness of sea ice that forms in the Arctic Ocean. There are many factors that affect the behavior of sea ice: abiotic factors such as temperature, both that of the underlying water and air; wind, sea currents, and the motion of the Earth, and biotic factors such as plant and animal life. We examine only one factor: temperature. We begin with a principle of thermodynamics: The direction of heat transfer is from a warmer to a cooler medium. Fourier’s Law of Conduction states that the rate of heat flow dQdt through a homogeneous solid is directly proportional to the area A of the solid orthogonal to the direction of the heat flow, and to the change ΔT in the temperature T, and is inversely proportional to the thickness x of the solid. See Figure 10. The constant of proportionality k depends on the homogeneous solid and is called the thermal conductivity of the solid.

Figure 10 Fourier’s Law of Conduction

1. Write Fourier’s Law of Conduction as a differential equation. A simple model depicting sea ice consists of a layer of air, a layer of sea ice, and a layer of sea water. A closer look at this model actually shows another layer directly below the ice. See Figure 11. It is this thin layer of super-cold, almost freezing water that is of interest in this project.

Figure 11 Model of sea ice

In this model heat flows from the warmer water beneath the ice, through the ice, to the colder air above the ice. 2. Using k=2.1 W⁄mK (the thermal conductivity for sea ice), A=1m2 as the surface area of the ice, x m as the thickness of the ice, Tw=-1.8∘C as the near freezing temperature of the layer of sea water directly beneath the ice, and Ta=-20∘C as the (average winter) temperature of Arctic air, write Fourier’s Law of Conduction. Then as the thin layer of sea water directly below the ice loses heat Q, it begins to freeze. The amount of water that freezes satisfies the equation: Q=334m where 334 J⁄g is the heat needed to freeze 1 g of the thin layer of water that freezes.

of water, and m is the mass

3. Find an expression for the mass m of the layer of water directly below the ice that

freezes. (The mass density ρ of freezing sea water is 1.025 g⁄ml.) that the surface area of the ice is A=1m2.

Recall

4. Find the rate of change in heat Q with respect to the thickness x of the ice. 5. Combine the result from Problem 4 with Fourier’s Law of Conductivity (Problem 2) to find the rate of change in thickness of the ice with respect to time. 6. Solve the differential equation found in Problem 5. Then use the initial condition that at time t=0, the thickness of the ice is x=x0. 7. Interpret the solution of the differential equation in Problem 6. Is the rate of change of the thickness of the ice increasing or decreasing over time? 8. Discuss whether this simple model offers a possible explanation for NASA’s finding that the multi-year ice is melting faster than seasonal and other perennial ice. 9. Investigate some of the other factors that affect the behavior of sea ice and write a position paper on the impact of at least one these other factors on the depth of ice in the Arctic. To read more and to see how the Arctic Ice cap has changed since 1980, visit http://www.nasa.gov/topics/earth/features/thick-melt.html Sources: http://www.nasa.gov/topics/earth/features/thick-melt.html http://www.nasa.gov/topics/earth/features/arctic-antarctic-ice.html http://www.windows2universe.org/earth/Water/density.html http://plus.maths.org/content/maths-and-climate-change-melting-arctic

Appendix A Precalculus Used in Calculus A.1 Algebra Used in Calculus A.2 Geometry Used in Calculus A.3 Analytic Geometry Used in Calculus A.4 Trigonometry Used in Calculus

The topics reviewed here are not exhaustive of the precalculus used in calculus. However, they do represent a large body of the material you will see in calculus. If you encounter difficulty with any of this material, consult a textbook in precalculus for more detail and explanation.

A.1 Algebra Used in Calculus OBJECTIVES When you finish this section, you should be able to: 1 Factor and simplify algebraic expressions 2 Complete the square 3 Solve equations 4 Solve inequalities 5 Work with exponents 6 Work with logarithms

1 Factor and Simplify Algebraic Expressions EXAMPLE 1 Factoring Algebraic Expressions

Factor each expression completely: a. 2(x+3)(x-2)3+(x+3)2(3)(x-2)2 b. 43x1⁄3(2x+1)+2x4⁄3 Solution a. In expression (a), (x+3) found in each term. Factor them out.

and (x-2)2

are common factors, factors

2(x+3)(x-2)3+(x+3)2(3)(x-2)2=(x+3)(x-2)2[2(x-2)+3(x+3)]Factor out (x+3)(x-2)2.=(x+3)(x2)2(5x+5)Simplify.=5(x+3)(x-2)2(x+1)Factor out 5.

b. Begin by writing the term 2x4⁄3

as a fraction with a denominator of 3.

43x1⁄3(2x+1)+2x4⁄3=4x1⁄3(2x+1)3+6x4⁄33=4x1⁄3(2x+1)+6x4⁄33Add the two fractions.=2x1⁄3[2(2x+1)+3x]3Factor out the common factors 2 and x1⁄3.=2x1⁄3(7x+2)3Simplify.

▪ EXAMPLE 2 Simplifying Algebraic Expressions

a. Simplify (x2+1)(3)-(3x+4)(2x)(x2+1)2. b. Write the expression (x2+1)1⁄2+x⋅12(x2+1)-1⁄2⋅2x as a single quotient in which only positive exponents appear. Solution a.

b. (x2+1)1⁄2+x⋅12(x2+1)-1⁄2⋅2x=(x2+1)1⁄2+x2(x2+1)1⁄2= (x2+1)1⁄2(x2+1)1⁄2(x2+1)1⁄2+x2(x2+1)1⁄2=(x2+1)+x2(x2+1)1⁄2=2x2+1(x2+1)1⁄2

▪ 2 Complete the Square We complete the square in one variable by modifying an expression of the form x2+bx to make it a perfect square. Perfect squares are trinomials of the form x2+2ax+a2=(x+a)2orx2-2ax+a2=(x-a)2

For example, x2+6x+9

is a perfect square because x2+6x+9=(x+3)2. And p2-12p+36

is a perfect square because p2-

12p+36=(p-6)2. To make x2+6x a perfect square, we must add 9. The number to be added is chosen by dividing the coefficient of the first-degree term, which is 6, by 2 and squaring the result 622=9. EXAMPLE 3 Completing the Square

Determine the number that must be added to each expression to complete the square. Then factor. Start

Add

y2+8y

Result y2+8y+16

12⋅82 =16 12⋅(-20)2 a2-20a

a2-20a+100 =100 12⋅(-5)2 =254

p2-5p

2x2+6x=2(x2+3x)

p2-5p+254

12⋅32 =94

2x2+3x+94

▪ CAUTION The original expression x2+bx

and the perfect square x2+bx+b22

are not equal. So when completing the square within an equation or an inequality, we must not only add b22,

we must also subtract it. That is,

x2+bx=x2+bx+(b2)2−(b2)2=0=(x+b2)2−(b2)2

3 Solve Equations To solve a quadratic equation ax2+bx+c=0, a≠0, formula can be used.

the quadratic

THEOREM Quadratic Formula

Consider the quadratic equation ax2+bx+c=0a≠0 If b2-4ac0. An ellipse is oval-shaped. The line segment dividing the ellipse in half the long way is the major axis; its length is 2a. The two points of intersection of the ellipse and the major axis are the vertices of the ellipse. The midpoint of the vertices is the center of the ellipse. The line segment through the center of the ellipse and perpendicular to the major axis is the minor axis. Along the major axis c units from the center of the ellipse, where c2=a2b2, c>0, are two points, called foci, that determine the shape of the ellipse. See Figure 39.

Figure 39

NOTE if a=b, the origin.

then x2a2+y2a2=1

the equation of a circle with radius a and center at

EXAMPLE 12 Graphing an Ellipse with Center at the Origin

Graph the equation 9x2+4y2=36. Solution To put the equation in standard form, we divide each side by 36. x24+y29=1 The graph of this equation is an ellipse. Since the larger number is under y2, the major axis is along the y -axis. The center is at the origin (0, 0). It is easiest to graph an ellipse by finding its intercepts: x-intercepts: Let y=0y-intercepts: Let x=0x24+029=102+y29=1x2=4y2=9x=-2 and x=2y=-3 and y=3

The points (-2, 0), (2, 0), (0,-3), (0, 3) ellipse. See Figure 40.

are the intercepts of the

▪

Figure 40 9x2+4y2=36 A hyperbola is the graph of an equation in one of the two forms:

x2a2-y2b2=1y2a2-x2b2=1 where a>0 and b>0. Two points, called foci, determine the shape of the hyperbola. The midpoint of the line segment containing the foci is called the center of the hyperbola. The foci are located c units from the center, where c2=a2+b2, c>0. The line containing the foci is called the transverse axis. The vertices of a hyperbola are its intercepts. See Figure 41. The graph of a hyperbola consists of two branches.

For the hyperbola x2a2-y2b2=1, right.

the branches of the graph open left and

For the hyperbola y2a2-x2b2=1, down.

the branches of the graph open up and

Figure 41 EXAMPLE 13 Graphing a Hyperbola with Center at the Origin

Graph the equation y24-x25=1. Solution The graph of y24-x25=1 is a hyperbola. The hyperbola consists of two branches, one opening up, the other opening down, like the graph in Figure 41(b). The hyperbola has no x -intercepts. To find the y -intercepts, we let x=0 and solve for y.

y24=1y2=4y=-2ory=2

The y -intercepts are -2 and 2, so the vertices are (0,-2) and (0, 2). transverse axis is the vertical line x=0. To graph the hyperbola, let y=±3 any numbers ≥2 or ≤-2 ). Then

The (or

y24-x25=194-x25=1y=±3x25=54x2=254x=-52orx=52

The points -52,3, -52,-3, 52,3, hyperbola. See Figure 42 for the graph.

Figure 42 y24-x25=1

and 52,-3

▪

are on the

A.4 Trigonometry Used in Calculus OBJECTIVES When you finish this section, you should be able to: 1 Work with angles, arc length of a circle, and circular motion 2 Define and evaluate trigonometric functions 3 Determine the domain and the range of the trigonometric functions 4 Use basic trigonometry identities 5 Use sum and difference, double-angle and half-angle, and sum-to-product

and product-to-sum formulas 6 Solve triangles using the Law of Sines and the Law of Cosines 1 Work with Angles, Arc Length of a Circle, and Circular Motion A ray, or half-line, is the portion of a line that starts at a point V on the line and extends indefinitely in one direction. The point V of a ray is called its vertex. See Figure 43.

Figure 43 If two rays are drawn with a common vertex, they form an angle. We call one ray of an angle the initial side and the other ray the terminal side. The angle formed is identified by showing the direction and amount of rotation from the initial side to the terminal side. If the rotation is in the counterclockwise direction, the angle is positive; if the rotation is clockwise, the angle is negative. See Figure 44.

Figure 44 An angle θ is in standard position if its vertex is at the origin of a rectangular coordinate system and its initial side is on the positive x -axis. See Figure 45 on page A-26.

Figure 45 Suppose an angle θ is in standard position. If its terminal side coincides with a coordinate axis, we say that θ is a quadrantal angle. If its terminal side does not coincide with a coordinate axis, we say that θ lies in a quadrant. See Figure 46.

Figure 46 Angles are measured by determining the amount of rotation needed for the initial side to coincide with the terminal side. The two commonly used measures for angles are degrees and radians. The angle formed by rotating the initial side exactly once in the counterclockwise direction until it coincides with itself (one revolution) measures 360 degrees, abbreviated 360∘. See Figure 47.

Figure 47 1 revolution counterclockwise is 360∘. A central angle is a positive angle whose vertex is at the center of a circle. The rays of a central angle subtend (intersect) an arc on the circle. If the radius of the circle is r and the arc subtended by the central angle is also of length r, then the measure of the angle is 1 radian. See Figure 48.

Figure 48 1 Radian THEOREM Arc Length

For a circle of radius r, a central angle of θ radians subtends an arc whose length s is s=rθ See Figure 49.

Figure 49 s=rθ Consider a circle of radius r. A central angle of one revolution will subtend an arc equal to the circumference of the circle, as shown in Figure 50. Because the circumference of a circle equals 2πr, we use s=2πr in the formula for arc length to find the radian measure of an angle of one revolution. s=rθ2πr=rθFor one revolution: s=2πrθ=2π radiansSolve for θ.

Figure 50 1 revolution =2π

radians

Since one revolution is equivalent to 360∘,

we have 360∘=2π

radians so

that 180∘=π radians In calculus, radians are generally used to measure angles, unless degrees are specifically mentioned. Table 5 lists the radian and degree measures for some common angles. TABLE 5 Radians Degrees 0∘

0

π6 30∘

π4 45∘

π3 60∘

π2 90∘

π 2π3 120∘

3π4 135∘

5π6 150∘

3π2 180∘

270∘

2π 360∘

NOTE If the measure of an angle is given as a number, it is understood to mean radians. If the measure of an angle is given in degrees, then it is marked either with the symbol ∘ or the word “degrees.”

THEOREM Area of a Sector

The area A of the sector of a circle of radius r formed by a central angle of θ radians is

A=12r2θ See Figure 51.

Figure 51 A=12r2θ DEFINITION Linear Speed, Angular Speed

Suppose an object moves around a circle of radius r at a constant speed. If s is the distance traveled along the circle in time t, then the linear speed v of the object is v=st The angular speed ω (the Greek letter omega) of an object moving at a constant speed around a circle of radius r is

ω=θt where θ is the angle (measured in radians) swept out in time t. See Figure 52.

Figure 52 v=st; ω=θt Angular speed is used to describe the turning rate of an engine. For example, an engine idling at 900 rpm (revolutions per minute) rotates at an angular speed of 900revolutionsminute=900revolutionsminute⋅2πradiansrevolution=1800πradiansminute

There is an important relationship between linear speed and angular speed:

So, v=rω where ω is measured in radians per unit time. NOTE In the formula s=rθ

for the arc length of a circle, the angle θ must be measured in radians.

When using the equation v=rω, remember that v=st has the dimensions of length per unit of time (such as meters per second or miles per hour), r has the same length dimension as s, and ω has the dimensions of radians per unit of time. If the angular speed is given in revolutions per unit of time (as is often the case), be sure to convert it to radians per unit of time, using the fact that one revolution =2π radians, before using the formula v=rω.

2 Define and Evaluate Trigonometric Functions There are two common approaches to trigonometry: One uses right triangles, and the other uses the unit circle. We suggest you first review the approach you are most familiar with and then read the other approach. The two approaches are given side-by-side on pages A-28 and A-29. Right Triangle Approach Suppose θ is an acute angle; that is, 0