ACTIVE CALCULUS 2018 Edition - UPDATED Matthew Boelkins David Austin Steven Schlicker Active Calculus Active Calculus Matthew Boelkins Grand Valley State University Contributing Authors David Austin Grand Valley State University Steven Schlicker Grand Valley State University Production Editor Mitchel T. Keller Morningside College July 2, 2019 Cover Photo: James Haefner Photography Edition: 2018 Updated Website: http://activecalculus.org ©2012–2019 Matthew Boelkins Permission is granted to copy and (re)distribute this material in any format and/or adapt it (even commercially) under the terms of the Creative Commons Attribution-ShareAlike 4.0 International License. The work may be used for free in any way by any party so long as at- tribution is given to the author(s) and if the material is modiﬁed, the resulting contributions are distributed under the same license as this original. All trademarks™ are the registered® marks of their respective owners. The graphic that may appear in other locations in the text shows that the work is licensed with the Cre- ative Commons and that the work may be used for free by any party so long as attribution is given to the author(s) and if the material is modiﬁed, the resulting contributions are dis- tributed under the same license as this original. Full details may be found by visiting https://creativecommons.org/licenses/by-sa/4.0/ or sending a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA. vi Acknowledgements This text began as my sabbatical project in the winter semester of 2012, during which I wrote most of the material for the ﬁrst four chapters. For the sabbatical leave, I am indebted to Grand Valley State University for its support of the project, as well as to my colleagues in the Department of Mathematics and the College of Liberal Arts and Sciences for their en- dorsement of the project. I’m also grateful to the American Institute of Mathematics for their support of free and open textbooks in general, and their support of this one in particular. The beautiful full-color .eps graphics in the text are only possible because of David Austin of GVSU and Bill Casselman of the University of British Columbia. Building on their long- standing eﬀorts to develop tools for high quality mathematical graphics, David wrote a li- brary of Python routines that employ Bill’s PiScript program; David’s routines are so easy to use that even I could generate graphics like the professionals that he and Bill are. I am deeply grateful to them both. The current .html version of the text is possible only because of the amazing work of Rob Beezer and his development of the original Mathbook XML, now known as PreTeXt. My ability to take advantage of Rob’s work is due in large part to the support of the American Institute of Mathematics, which funded me to attend a weeklong workshop in Mathbook XML in San Jose, CA, in April 2016, as well as the support of the ongoing user group. A subsequent workshop in June 2019 has oﬀered further support and more improvements to the text. David Farmer’s conversion script saved me hundreds of hours of work by taking my original LATEX source and converting it to PreTeXt; David remains a major source of ongoing support and advocacy. Alex Jordan of Portland Community College has also been a tremendous help, and it is through Alex’s fantastic work that live WeBWorK exercises are not only possi- ble, but also included from the 2017 version forward. Mitch Keller of Morningside College agreed in early 2018 to serve as the book’s production editor; his technical expertise has contributed to many aspects of the book, including the presence of answers to activities and non- WeBWorK exercises and other supporting material for instructors. For the 2018 edition, Kathy Yoshiwara of the AIM Editorial Board read the entire text and contributed editorial suggestions for every section. In short, she made the prose cleaner, more direct, and simply better. I’m deeply thankful for her time, eﬀort, and insights. Over my more than 20 years at GVSU, many of my colleagues have shared with me ideas and resources for teaching calculus. I am particularly indebted to David Austin, Will Dickin- son, Paul Fishback, Jon Hodge, and Steve Schlicker for their contributions that improved my teaching of and thinking about calculus, including materials that I have modiﬁed and used over many diﬀerent semesters with students. Parts of these ideas can be found throughout this text. In addition, Will Dickinson and Steve Schlicker provided me access to a large num- ber of their electronic notes and activities from teaching of diﬀerential and integral calculus, and those ideas and materials have similarly impacted my work and writing in positive ways, with some of their problems and approaches ﬁnding parallel presentation here. In the summer of 2012, David and Steve each agreed to write a chapter to support the com- pletion of the material on integral calculus. David is the lead author of Chapter 7 and Steve the lead author of Chapter 8. Along with our colleague Ted Sundstrom, Steve has also con- tributed a large number of problem and activity solutions and answers. I’m especially grate- ful for how the work of these friends and colleagues has made the text so much better. Shelly Smith of GVSU and Matt Delong of Marian University both provided extensive com- ments on the ﬁrst few chapters of early drafts, feedback that was immensely helpful in im- proving the text. As more and more people use the text, I am grateful to everyone who reads, edits, and uses this book, and hence contributes to its improvement through ongoing discussion. Finally, I am grateful for all that my students have taught me over the years. Their responses and feedback have helped to shape me as a teacher, and I appreciate their willingness to wholeheartedly engage in the activities and approaches I’ve tried in class, to let me know how those aﬀect their learning, and to help me learn and grow as an instructor. Early on, they also provided useful editorial feedback on this text. Any and all remaining errors or inconsistencies are mine. I will gladly take reader and user feedback to correct them, along with other suggestions to improve the text. viii Contributors A large and growing number of people have generously contributed to the development or improvement of the text. Contributing authors David Austin and Steven Schlicker have each written drafts of at least one full chapter of the text. Production editor Mitchel Keller has been an indispensable source of technological support and editorial counsel. The following contributing editors have oﬀered feedback that includes information about typographical errors or suggestions to improve the exposition. David Austin Patti Hunter GVSU Westmont College Rene Ardila Mitchel Keller GVSU Morningside College Allan Bickle GVSU Sam Kolins Lebanon Valley College David Clark GVSU Dave Kung Will Dickinson St. Mary’s College of Maryland GVSU Paul Latiolais Nate Eldredge Portland State University University of Northern Colorado Charles Fortin Hugh McGuire Champlain Regional College GVSU St-Lambert, Quebec, Canada Martin Mohlenkamp Marcia Frobish Ohio University GVSU Teresa Gonske Ray Rosentrater University of Northwestern - St. Paul Westmont College Luis Sanjuan Amy Stone Conservatorio Profesional de Musica de Avila GVSU Spain Robert Talbert Michael Santana GVSU GVSU Greg Thull Steven Schlicker GVSU GVSU Michael Shulman Sue Van Hattum University of San Diego Contra Costa College Brian Stanley Kathy Yoshiwara Foothill Community College AIM Editorial Board x Active Calculus: Our Goals Several fundamental ideas in calculus are more than 2000 years old. As a formal subdisci- pline of mathematics, calculus was ﬁrst introduced and developed in the late 1600s, with key independent contributions from Sir Isaac Newton and Gottfried Wilhelm Leibniz. The subject has been understood rigorously since the work of Augustin Louis Cauchy and Karl Weierstrass in the mid 1800s when the ﬁeld of modern analysis was developed. As a body of knowledge, calculus has been completely understood for at least 150 years. The disci- pline is one of our great human intellectual achievements: among many spectacular ideas, calculus models how objects fall under the forces of gravity and wind resistance, explains how to compute areas and volumes of interesting shapes, enables us to work rigorously with inﬁnitely small and inﬁnitely large quantities, and connects the varying rates at which quantities change to the total change in the quantities themselves. While each author of a calculus textbook certainly oﬀers their own creative perspective on the subject, it is hardly the case that many of the ideas they present are new. Indeed, the mathematics community broadly agrees on what the main ideas of calculus are, as well as their justiﬁcation and their importance. In the 21st century and the age of the internet, no one should be required to purchase a calculus text to read, to use for a class, or to ﬁnd a coherent collection of problems to solve. Calculus belongs to humankind, not any individual author or publishing company. Thus, a primary purpose of this work is to present a calculus text that is free. See https://activecalculus.org for links to both the .html and .pdf versions of the text. In addition, instructors who are looking for a calculus text should have the opportunity to download the source ﬁles and make modiﬁcations that they see ﬁt; thus this text is open- source. See GitHub for the source. Since August 2013, Active Calculus - Single Variable has been endorsed by the American Institute of Mathematics and its Open Textbook Initiative. In Active Calculus - Single Variable, we actively engage students in learning the subject through an activity-driven approach in which the vast majority of the examples are generated by students. Where many texts present a general theory followed by substantial collections of worked examples, we instead pose problems or situations, consider possibilities, and then ask students to investigate and explore. Following key activities or examples, the presen- tation normally includes some overall perspective and a brief synopsis of general trends or properties, followed by formal statements of rules or theorems. While we often oﬀer plau- sibility arguments for such results, rarely do we include formal proofs. It is not the intent of this text for the instructor or author to demonstrate to students that the ideas of calculus are coherent and true, but rather for students to encounter these ideas in a supportive, leading manner that enables them to begin to understand calculus for themselves. This approach is consistent with the scholarly consensus that calls for students to be interactively engaged in class. Moreover, this approach is consistent with the following goals: • To have students engage in an active, inquiry-driven approach, where learners con- struct solutions and approaches to ideas, with appropriate support through questions posed, hints, and guidance from the instructor and text. • To build in students intuition for why the main ideas in calculus are natural and true. Often we do this through consideration of the instantaneous position and velocity of a moving object. • To challenge students to acquire deep, personal understanding of calculus through reading the text and completing preview activities on their own, working on activities in small groups in class, and doing substantial exercises outside of class time. • To strengthen students’ written and oral communicating skills by having them write about and explain aloud the key ideas of calculus. xii Features of the Text Instructors and students alike will ﬁnd several consistent features in the presentation, in- cluding: Motivating Questions At the start of each section, we list 2–3 motivating questions that pro- vide motivation for why the following material is of interest to us. One overall goal of each section is to answer each of the motivating questions. Preview Activities Each section of the text begins with a short introduction, followed by a preview activity. This brief reading and the preview activity are designed to foreshadow the upcoming ideas in the remainder of the section; both the reading and preview activity are intended to be accessible to students in advance of class, and to be completed by students before the day on which a particular section is to be considered. Activities A typical section in the text has at least three activities. These are designed to engage students in an inquiry-based style that encourages them to construct solutions to key examples on their own, working individually or in small groups. Exercises There are dozens of calculus texts with (collectively) tens of thousands of exer- cises. Rather than repeat standard and routine exercises in this text, we recommend the use of WeBWorK with its access to the Open Problem Library and around 20,000 calculus problems. In this text, each section includes a small number of anonymous WeBWorK exercises, as well as 3–4 challenging problems per section. The WeBWorK exercises are best completed in the .html version of the text, as this provides students with immediate feedback without penalty. Almost every non- WeBWorK exercise has multiple parts, requires the student to connect several key ideas, and expects that the student will do at least a modest amount of writing to answer the questions and explain their ﬁndings. For instructors interested in a more conventional source of exercises, consider the freely available APEX Calculus text by Greg Hartmann et al., available from www.apexcalculus.com. Graphics We strive to demonstrate key fundamental ideas visually, and to encourage stu- dents to do the same. Throughout the text, we use full-color¹ graphics to exemplify and magnify key ideas, and to use this graphical perspective alongside both numerical and algebraic representations of calculus. ¹To keep cost low, the graphics in the print-on-demand version are in black and white. When the text itself refers to color in images, one needs to view the .html or .pdf electronically. Links to interactive graphics Many of the ideas of calculus are best understood dynami- cally; java applets oﬀer an often ideal format for investigations and demonstrations. Relying primarily on the work of David Austin of Grand Valley State University and Marc Renault of Shippensburg University, each of whom has developed a large li- brary of applets for calculus, we frequently point the reader (through active links in the electronic versions of the text) to applets that are relevant for key ideas under con- sideration. Summary of Key Ideas Each section concludes with a summary of the key ideas encoun- tered in the preceding section; this summary normally reﬂects responses to the moti- vating questions that began the section. xiv Students! Read this! This book is diﬀerent. The text is available in three diﬀerent formats: HTML, PDF, and print, each of which is available via links on the landing page at https://activecalculus.org/single/. The ﬁrst two formats are free. If you are going to use the book electronically, the best mode is the HTML. The HTML version looks great in any browser, including on a smartphone, and the links are much easier to navigate in HTML than in PDF. Some particular direct suggestions about using the HTML follow among the next few paragraphs; alternatively, you can watch this short video from the author. It is also wise to download and save the PDF, since you can use the PDF oﬄine, while the HTML version requires an internet connection. A print copy costs about $21 via Amazon. This book is intended to be read sequentially and engaged with, much more than to be used as a lookup reference. For example, each section begins with a short introduction and a Preview Activity; you should read the short introduction and complete the Preview Activity prior to class. Your instructor may require you to do this. Most Preview Activities can be completed in 15-20 minutes and are intended to be accessible based on the understanding you have from preceding sections. There are not answers provided to Preview Activities, as these are designed simply to get you thinking about ideas that will be helpful in work on upcoming new material. As you use the book, think of it as a workbook, not a worked-book. There is a great deal of scholarship that shows people learn better when they interactively engage and struggle with ideas themselves, rather than passively watch others. Thus, instead of reading worked examples or watching an instructor complete examples, you will engage with Activities that prompt you to grapple with concepts and develop deep understanding. You should expect to spend time in class working with peers on Activities and getting feedback from them and from your instructor. You can purchase a separate Activities Workbook from Amazon (Chapters 1-4, Chapters 5-8) in which to record your work on the activities, or you can ask your instructor for a copy of the PDF ﬁle that has only the activities along with room to record your work. Your goal should be to do all of the activities in the relevant sections of the text and keep a careful record of your work. You can ﬁnd answers to the activities in the back matter. Each section concludes with a Summary. Reading the Summary after you have read the section and worked the Activities is a good way to ﬁnd a short list of key ideas that are most essential to take from the section. A good study habit is to write similar summaries in your own words. At the end of each section, you’ll ﬁnd two types of Exercises. First, there are several anony- mous WeBWorK exercises. These are online, interactive exercises that allow you to submit answers for immediate feedback with unlimited attempts without penalty; to submit an- swers, you have to be using the HTML version of the text (see this short video on the HTML version that includes a WeBWorK demonstration). You should use these exercises as a way to test your understanding of basic ideas in the preceding section. If your institution uses WeBWorK, you may also need to log in to a server as directed by your instructor to complete assigned WeBWorK sets as part of your course grade. The WeBWorK exercises included in this text are ungraded and not connected to any individual account. Following the WeB- WorK exercises there are 3-4 additional challenging exercises that are designed to encourage you to connect ideas, investigate new situations, and write about your understanding. There are answers to most of the non-WeBWorK exercises in the back matter. You can ﬁnd additional support for your work in learning calculus from the GVSU Math 201 YouTube Channel and GVSU Math 202 YouTube Channel where there are several short video tutorials for each section of the text, numbered in alignment with the textbook sections. Math 201 corresponds to Chapters 1-4 and Math 202 to Chapters 5-8; there are about 90 videos for each, totally more than 180. The best way to be successful in mathematics generally and calculus speciﬁcally is to strive to make sense of the main ideas. We make sense of ideas by asking questions, interacting with others, attempting to solve problems, making mistakes, revising attempts, and writing and speaking about our understanding. This text has been designed to help you make sense of calculus; we wish you the very best as you undertake the large and challenging task of doing so. xvi Instructors! Read this! This book is diﬀerent. Before you read further, ﬁrst read “Students! Read this!”. Chapters 1-4 are designed to correspond to what is often called diﬀerential calculus. Chap- ters 5-8 correspond roughtly to what is often called integral calculus, including chapters on diﬀerential equations and inﬁnite series. Among the three formats (HTML, PDF, print), the HTML is optimal for display in class if you have a suitable projector. The HTML is also best for navigation, as links to internal and external references are much more obvious. We recommend saving a downloaded version of the PDF format as a backup in the event you don’t have internet access. It’s a good idea for each student to have a printed version of the Activities Workbook, which is available on Amazon (Chapters 1-4, Chapters 5-8) or as a PDF document by direct request to the author (boelkinm at gvsu dot edu); many instructors use the PDF to have coursepacks printed for students to purchase from their local bookstore. The text is written so that, on average, one section corresponds to two hours of class meet- ing time. A typical instructional sequence when starting a new section might look like the following: • Students complete a Preview Activity in advance of class. Class begins with a short debrief among peers followed by all class discussion. (5-10 minutes) • Brief lecture and discussion to build on the preview activity and set the stage for the next activity. (5-10 minutes)� • Students engage with peers to work on and discuss the ﬁrst activity in the section. (15-20 minutes)� • Brief discussion and possibly lecture to reach closure on the preceding activity, fol- lowed by transition to new ideas. (Varies, but 5-15 minutes) • Possibly begin next activity. The next hour of class would be similar, but without the Preview Activity to complete prior to class: the principal focus of class will be completing 2 activities. Then rinse and repeat. We recommend that instructors use appropriate incentives to encourage students to com- plete Preview Activities prior to class. Having these be part of completion-based assign- ments that count 5% of the semester grade usually results in the vast majority of students completing the vast majority of the previews. If you’d like to see a sample syllabus for how to organize a course and weight various assignments, you can request one via email to the author. Note that the WeBWorK exercises in the HTML version are anonymous and there’s not a way to track students’ engagement with them. These are intended to be formative for stu- dents and provide them with immediate feedback without penalty. If your institution is a WeBWorK user, we have existing sets of .def ﬁles that correspond to the sections in the text; these are available upon request to the author. In the back matter of the text, you’ll ﬁnd answers to the Activities and to non-WeBWorK Exercises. Instructors interested in solutions to these should contact the author directly. You and your students can ﬁnd additional resources in the GVSU Math 201 YouTube Chan- nel and GVSU Math 202 YouTube Channel where there are short video tutorials for every section of the text. Math 201 (GVSU’s Calculus I) corresponds to Chapters 1-4 and Math 202 (GVSU’s Calculus II) to Chapters 5-8. The PreTeXt source code for the text can be found on GitHub. If you ﬁnd errors in the text or have other suggestions, you can ﬁle an issue on GitHub, use the Feedback link in the HTML version (found at the bottom left in the main menu), or email the author directly. To engage with instructors who use the text, we maintain both an email list and the Open Calculus blog; you can request that your address be added to the email list by contacting the author. Finally, if you’re interested in a video presentation on using the text, you can see this online video presentation to the MIT Electronic Seminar on Mathematics Education; at about the 17-minute mark, the portion begins where we demonstrate features of and how to use the text. Thank you for considering Active Calculus as a resource to help your students develop deep understanding of the subject. I wish you the very best in your work and hope to hear from you. xviii Contents Acknowledgements vii Contributors ix Active Calculus: Our Goals xi Features of the Text xiii Students! Read this! xv Instructors! Read this! xvii 1 Understanding the Derivative 1 1.1 How do we measure velocity? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 The notion of limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.3 The derivative of a function at a point . . . . . . . . . . . . . . . . . . . . . . . 22 1.4 The derivative function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 1.5 Interpreting, estimating, and using the derivative . . . . . . . . . . . . . . . . 45 1.6 The second derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 1.7 Limits, Continuity, and Diﬀerentiability . . . . . . . . . . . . . . . . . . . . . . 68 1.8 The Tangent Line Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 80 2 Computing Derivatives 89 2.1 Elementary derivative rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 2.2 The sine and cosine functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 2.3 The product and quotient rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 2.4 Derivatives of other trigonometric functions . . . . . . . . . . . . . . . . . . . . 115 Contents 2.5 The chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 2.6 Derivatives of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 2.7 Derivatives of Functions Given Implicitly . . . . . . . . . . . . . . . . . . . . . 139 2.8 Using Derivatives to Evaluate Limits . . . . . . . . . . . . . . . . . . . . . . . . 146 3 Using Derivatives 157 3.1 Using derivatives to identify extreme values . . . . . . . . . . . . . . . . . . . 157 3.2 Using derivatives to describe families of functions . . . . . . . . . . . . . . . . 170 3.3 Global Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 3.4 Applied Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 3.5 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 4 The Deﬁnite Integral 201 4.1 Determining distance traveled from velocity . . . . . . . . . . . . . . . . . . . 201 4.2 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 4.3 The Deﬁnite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 4.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . 244 5 Evaluating Integrals 259 5.1 Constructing Accurate Graphs of Antiderivatives . . . . . . . . . . . . . . . . . 259 5.2 The Second Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . 270 5.3 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 5.4 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 5.5 Other Options for Finding Algebraic Antiderivatives . . . . . . . . . . . . . . . 301 5.6 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 6 Using Deﬁnite Integrals 325 6.1 Using Deﬁnite Integrals to Find Area and Length . . . . . . . . . . . . . . . . . 325 6.2 Using Deﬁnite Integrals to Find Volume . . . . . . . . . . . . . . . . . . . . . . 334 6.3 Density, Mass, and Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . 344 6.4 Physics Applications: Work, Force, and Pressure . . . . . . . . . . . . . . . . . 354 6.5 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 xx 7 Diﬀerential Equations 375 7.1 An Introduction to Diﬀerential Equations . . . . . . . . . . . . . . . . . . . . . 375 7.2 Qualitative behavior of solutions to DEs . . . . . . . . . . . . . . . . . . . . . . 386 7.3 Euler’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 7.4 Separable diﬀerential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 7.5 Modeling with diﬀerential equations . . . . . . . . . . . . . . . . . . . . . . . . 415 7.6 Population Growth and the Logistic Equation . . . . . . . . . . . . . . . . . . . 423 8 Sequences and Series 435 8.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 8.2 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 8.3 Series of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 8.4 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 8.5 Taylor Polynomials and Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . 483 8.6 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 A A Short Table of Integrals 509 B Answers to Activities 511 C Answers to Selected Exercises 565 Index 623 xxi Contents xxii CHAPTER 1 Understanding the Derivative 1.1 How do we measure velocity? Motivating Questions • How is the average velocity of a moving object connected to the values of its position function? • How do we interpret the average velocity of an object geometrically on the graph of its position function? • How is the notion of instantaneous velocity connected to average velocity? Calculus can be viewed broadly as the study of change. A natural and important question to ask about any changing quantity is “how fast is the quantity changing?” We begin with a simple problem: a ball is tossed straight up in the air. How is the ball moving? Questions like this one are central to our study of diﬀerential calculus. Preview Activity 1.1.1. Suppose that the height s of a ball at time t (in seconds) is given in feet by the formula s(t) 64 − 16(t − 1)2 . a. Construct a graph of y s(t) on the time interval 0 ≤ t ≤ 3. Label at least six distinct points on the graph, including the three points showing when the ball was released, when the ball reaches its highest point, and when the ball lands. b. Describe the behavior of the ball on the time interval 0 < t < 1 and on time interval 1 < t < 3. What occurs at the instant t 1? c. Consider the expression s(1) − s(0.5) AV[0.5,1] . 1 − 0.5 Compute the value of AV[0.5,1] . What does this value measure on the graph? What does this value tell us about the motion of the ball? In particular, what are the units on AV[0.5,1] ? Chapter 1 Understanding the Derivative 1.1.1 Position and average velocity Any moving object has a position that can be considered a function of time. When the motion is along a straight line, the position is given by a single variable, which we denote by s(t). For example, s(t) might give the mile marker of a car traveling on a straight highway at time t in hours. Similarly, the function s described in Preview Activity 1.1.1 is a position function, where position is measured vertically relative to the ground. On any time interval, a moving object also has an average velocity. For example, to compute a car’s average velocity we divide the number of miles traveled by the time elapsed, which gives the velocity in miles per hour. Similarly, the value of AV[0.5,1] in Preview Activity 1.1.1 gave the average velocity of the ball on the time interval [0.5, 1], measured in feet per second. In general, we make the following deﬁnition: Average Velocity. For an object moving in a straight line with position function s(t), the average velocity of the object on the interval from t a to t b, denoted AV[a,b] , is given by the formula s(b) − s(a) AV[a,b] . b−a Note well: the units on AV[a,b] are “units of s per unit of t,” such as “miles per hour” or “feet per second.” Activity 1.1.2. The following questions concern the position function given by s(t) 64 − 16(t − 1)2 , considered in Preview Activity 1.1.1. a. Compute the average velocity of the ball on each of the following time inter- vals: [0.4, 0.8], [0.7, 0.8], [0.79, 0.8], [0.799, 0.8], [0.8, 1.2], [0.8, 0.9], [0.8, 0.81], [0.8, 0.801]. Include units for each value. b. On the graph provided in Figure 1.1.1, sketch the line that passes through the points A (0.4, s(0.4)) and B (0.8, s(0.8)). What is the meaning of the slope of this line? In light of this meaning, what is a geometric way to interpret each of the values computed in the preceding question? c. Use a graphing utility to plot the graph of s(t) 64 − 16(t − 1)2 on an interval containing the value t 0.8. Then, zoom in repeatedly on the point (0.8, s(0.8)). What do you observe about how the graph appears as you view it more and more closely? d. What do you conjecture is the velocity of the ball at the instant t 0.8? Why? 2 1.1 How do we measure velocity? feet s 64 B A 56 48 sec 0.4 0.8 1.2 Figure 1.1.1: A partial plot of s(t) 64 − 16(t − 1)2 . 1.1.2 Instantaneous Velocity Whether we are driving a car, riding a bike, or throwing a ball, we have an intuitive sense that a moving object has a velocity at any given moment -- a number that measures how fast the object is moving right now. For instance, a car’s speedometer tells the driver the car’s velocity at any given instant. In fact, the velocity on a speedometer is really an average velocity that is computed over a very small time interval. If we let the time interval over which average velocity is computed become shorter and shorter, we can progress from average velocity to instantaneous velocity. Informally, we deﬁne the instantaneous velocity of a moving object at time t a to be the value that the average velocity approaches as we take smaller and smaller intervals of time containing t a. We will develop a more formal deﬁnition of instantaneous velocity soon, and this deﬁnition will be the foundation of much of our work in calculus. For now, it is ﬁne to think of instantaneous velocity as follows: take average velocities on smaller and smaller time intervals around a speciﬁc point. If those average velocities approach a single number, then that number will be the instantaneous velocity at that point. Activity 1.1.3. Each of the following questions concern s(t) 64 − 16(t − 1)2 , the position function from Preview Activity 1.1.1. a. Compute the average velocity of the ball on the time interval [1.5, 2]. What is diﬀerent between this value and the average velocity on the interval [0, 0.5]? b. Use appropriate computing technology to estimate the instantaneous velocity of the ball at t 1.5. Likewise, estimate the instantaneous velocity of the ball at t 2. Which value is greater? 3 Chapter 1 Understanding the Derivative c. How is the sign of the instantaneous velocity of the ball related to its behavior at a given point in time? That is, what does positive instantaneous velocity tell you the ball is doing? Negative instantaneous velocity? d. Without doing any computations, what do you expect to be the instantaneous velocity of the ball at t 1? Why? At this point we have started to see a close connection between average velocity and instan- taneous velocity. Each is connected not only to the physical behavior of the moving object but also to the geometric behavior of the graph of the position function. We are interested in computing average velocities on the interval [a, b] for smaller and smaller intervals. In order to make the link between average and instantaneous velocity more formal, think of the value b as b a + h, where h is a small (non-zero) number that is allowed to vary. Then the average velocity of the object on the interval [a, a + h] is s(a + h) − s(a) AV[a,a+h] , h with the denominator being simply h because (a + h)− a h. Note that when h < 0, AV[a,a+h] measures the average velocity on the interval [a + h, a]. To ﬁnd the instantaneous velocity at t a, we investigate what happens as the value of h approaches zero. Example 1.1.2 Computing instantaneous velocity for a falling ball. The position function for a falling ball is given by s(t) 16 − 16t 2 (where s is measured in feet and t in seconds). a. Find an expression for the average velocity of the ball on a time interval of the form [0.5, 0.5 + h] where −0.5 < h < 0.5 and h , 0. b. Use this expression to compute the average velocity on [0.5, 0.75] and [0.4, 0.5]. c. Make a conjecture about the instantaneous velocity at t 0.5. Solution. a. We make the assumptions that −0.5 < h < 0.5 and h , 0 because h cannot be zero (otherwise there is no interval on which to compute average velocity) and because the function only makes sense on the time interval 0 ≤ t ≤ 1, as this is the duration of time during which the ball is falling. We want to compute and simplify s(0.5 + h) − s(0.5) AV[0.5,0.5+h] . (0.5 + h) − 0.5 We start by ﬁnding s(0.5 + h). To do so, we follow the rule that deﬁnes the function s. s(0.5 + h) 16 − 16(0.5 + h)2 16 − 16(0.25 + h + h 2 ) 16 − 4 − 16h − 16h 2 12 − 16h − 16h 2 . 4 1.1 How do we measure velocity? Now, returning to our computation of the average velocity, we ﬁnd that s(0.5 + h) − s(0.5) AV[0.5,0.5+h] (0.5 + h) − 0.5 (12 − 16h − 16h 2 ) − (16 − 16(0.5)2 ) 0.5 + h − 0.5 12 − 16h − 16h 2 − 12 h −16h − 16h 2 . h At this point, we note two things: ﬁrst, the expression for average velocity clearly de- pends on h, which it must, since as h changes the average velocity will change. Further, we note that since h can never equal zero, we may remove the common factor of h from the numerator and denominator. It follows that AV[0.5,0.5+h] −16 − 16h. b. From this expression we can compute the average for any small positive or negative value of h. For instance, to obtain the average velocity on [0.5, 0.75], we let h 0.25, and the average velocity is −16 − 16(0.25) −20 ft/sec. To get the average velocity on [0.4, 0.5], we let h −0.1, and compute the average velocity as −16 − 16(−0.1) −14.4 ft/sec. c. We can even explore what happens to AV[0.5,0.5+h] as h gets closer and closer to zero. As h approaches zero, −16h will also approach zero, so it appears that the instantaneous velocity of the ball at t 0.5 should be −16 ft/sec. Activity 1.1.4. For the function given by s(t) 64 − 16(t − 1)2 from Preview Activ- ity 1.1.1, ﬁnd the most simpliﬁed expression you can for the average velocity of the ball on the interval [2, 2 + h]. Use your result to compute the average velocity on [1.5, 2] and to estimate the instantaneous velocity at t 2. Finally, compare your earlier work in Activity 1.1.2. 1.1.3 Summary • For an object moving in a straight line with position function s(t), the average velocity of the object on the interval from t a to t b, denoted AV[a,b] , is given by the formula s(b) − s(a) AV[a,b] . b−a • The average velocity on [a, b] can be viewed geometrically as the slope of the line be- tween the points (a, s(a)) and (b, s(b)) on the graph of y s(t), as shown in Figure 1.1.3. 5 Chapter 1 Understanding the Derivative s s(b)−s(a) m= b−a (b, s(b)) (a, s(a)) t Figure 1.1.3: The graph of position function s together with the line through (a, s(a)) and s(b)−s(a) (b, s(b)) whose slope is m b−a . The line’s slope is the average rate of change of s on the interval [a, b]. • Given a moving object whose position at time t is given by a function s, the average s(b)−s(a) velocity of the object on the time interval [a, b] is given by AV[a,b] b−a . View- ing the interval [a, b] as having the form [a, a + h], we equivalently compute average s(a+h)−s(a) velocity by the formula AV[a,a+h] h . • The instantaneous velocity of a moving object at a ﬁxed time is estimated by consider- ing average velocities on shorter and shorter time intervals that contain the instant of interest. 1.1.4 Exercises 1. Average velocity from position. Consider a car whose position, s, is given by the table t (s) 0 0.2 0.4 0.6 0.8 1 s (ft) 0 0.5 1.4 3.8 6.5 9.6 Find the average velocity over the interval 0 ≤ t ≤ 0.2. Estimate the velocity at t 0.2. 6 1.1 How do we measure velocity? 2. Rate of calorie consumption. The table below shows the number of calories used per minute as a function of an individual’s body weight for three sports: Activity 100 lb 120 lb 150 lb 170 lb 200 lb 220 lb Walking 2.7 3.2 4 4.6 5.4 5.9 Bicycling 5.4 6.5 8.1 9.2 10.8 11.9 Swimming 5.8 6.9 8.7 9.8 11.6 12.7 a) Determine the number of calories that a 200 lb person uses in one half-hour of walk- ing. b) Who uses more calories, a 170 lb person swimming for one hour, or a 220 lb person bicycling for a half-hour? c) Does the number of calories of a person walking increase or decrease as weight in- creases? 3. Average rate of change - quadratic function. Let f (x) 9 − x 2 . a) Compute each of the following expressions and interpret each as an average rate of change: f (1)− f (0) (i) 1−0 f (3)− f (1) (ii) 3−1 f (3)− f (0) (iii) 3−0 b) Based on the graph sketched below, match each of your answers in (i) - (iii) with one of the lines labeled A - F. Type the corresponding letter of the line segment next to the appropriate formula. Clearly not all letters will be used. f (1) − f (0) 1−0 f (3) − f (1) 3−1 f (3) − f (0) 3−0 7 Chapter 1 Understanding the Derivative 4. Comparing average rate of change of two functions. Consider the graphs of f (x) and 1(x) below: For each interval given below, decide whether the average rate of change of f (x) or 1(x) is greater over that particular interval. Interval Which function has GREATER average rate of change? 0≤x≤4 (□ f □ g □ both have an equal rate of change) 0≤x≤8 (□ f □ g □ both have an equal rate of change) 0 ≤ x ≤ 2.2 (□ f □ g □ both have an equal rate of change) 5.2 ≤ x ≤ 6.1 (□ f □ g □ both have an equal rate of change) 5.2 ≤ x ≤ 6.9 (□ f □ g □ both have an equal rate of change) 5. Matching a distance graph to velocity. A car is driven at a constant speed, starting at noon. Which of the following could be a graph of the distance the car has traveled as a function of time past noon? 1. 2. 3. 4. 5. 6. 7. 8. 8 1.1 How do we measure velocity? 6. A bungee jumper dives from a tower at time t 0. Her height h (measured in feet) at time t (in seconds) is given by the graph in Figure 1.1.4. In this problem, you may base your answers on estimates from the graph or use the fact that the jumper’s height function is given by s(t) 100 cos(0.75t) · e −0.2t + 100. 200 s 150 100 50 t 5 10 15 20 Figure 1.1.4: A bungee jumper’s height function. a. What is the change in vertical position of the bungee jumper between t 0 and t 15? b. Estimate the jumper’s average velocity on each of the following time intervals: [0, 15], [0, 2], [1, 6], and [8, 10]. Include units on your answers. c. On what time interval(s) do you think the bungee jumper achieves her greatest average velocity? Why? d. Estimate the jumper’s instantaneous velocity at t 5. Show your work and ex- plain your reasoning, and include units on your answer. e. Among the average and instantaneous velocities you computed in earlier ques- tions, which are positive and which are negative? What does negative velocity indicate? 7. A diver leaps from a 3 meter springboard. His feet leave the board at time t 0, he reaches his maximum height of 4.5 m at t 1.1 seconds, and enters the water at t 2.45. Once in the water, the diver coasts to the bottom of the pool (depth 3.5 m), touches bottom at t 7, rests for one second, and then pushes oﬀ the bottom. From there he coasts to the surface, and takes his ﬁrst breath at t 13. a. Let s(t) denote the function that gives the height of the diver’s feet (in meters) above the water at time t. (Note that the “height” of the bottom of the pool is −3.5 meters.) Sketch a carefully labeled graph of s(t) on the provided axes in Figure 1.1.5. Include scale and units on the vertical axis. Be as detailed as possible. b. Based on your graph in (a), what is the average velocity of the diver between t 2.45 and t 7? Is his average velocity the same on every time interval within [2.45, 7]? 9 Chapter 1 Understanding the Derivative s v t t 2 4 6 8 10 12 2 4 6 8 10 12 Figure 1.1.5: Axes for plotting s(t) Figure 1.1.6: Axes for plotting v(t) in part (a). in part (c). c. Let the function v(t) represent the instantaneous vertical velocity of the diver at time t (i.e. the speed at which the height function s(t) is changing; note that velocity in the upward direction is positive, while the velocity of a falling object is negative). Based on your understanding of the diver’s behavior, as well as your graph of the position function, sketch a carefully labeled graph of v(t) on the axes provided in Figure 1.1.6. Include scale and units on the vertical axis. Write several sentences that explain how you constructed your graph, discussing when you expect v(t) to be zero, positive, negative, relatively large, and relatively small. d. Is there a connection between the two graphs that you can describe? What can you say about the velocity graph when the height function is increasing? decreasing? Make as many observations as you can. 8. According to the U.S. census, the population of the city of Grand Rapids, MI, was 181,843 in 1980; 189,126 in 1990; and 197,800 in 2000. a. Between 1980 and 2000, by how many people did the population of Grand Rapids grow? b. In an average year between 1980 and 2000, by how many people did the popula- tion of Grand Rapids grow? c. Just like we can ﬁnd the average velocity of a moving body by computing change in position over change in time, we can compute the average rate of change of any function f . In particular, the average rate of change of a function f over an interval [a, b] is the quotient f (b) − f (a) . b−a f (b)− f (a) What does the quantity b−a measure on the graph of y f (x) over the inter- val [a, b]? d. Let P(t) represent the population of Grand Rapids at time t, where t is measured in years from January 1, 1980. What is the average rate of change of P on the interval t 0 to t 20? What are the units on this quantity? 10 1.1 How do we measure velocity? e. If we assume the population of Grand Rapids is growing at a rate of approxi- mately 4% per decade, we can model the population function with the formula P(t) 181843(1.04)t/10 . Use this formula to compute the average rate of change of the population on the intervals [5, 10], [5, 9], [5, 8], [5, 7], and [5, 6]. f. How fast do you think the population of Grand Rapids was changing on January 1, 1985? Said diﬀerently, at what rate do you think people were being added to the population of Grand Rapids as of January 1, 1985? How many additional people should the city have expected in the following year? Why? 11 Chapter 1 Understanding the Derivative 1.2 The notion of limit Motivating Questions • What is the mathematical notion of limit and what role do limits play in the study of functions? • What is the meaning of the notation limx→a f (x) L? • How do we go about determining the value of the limit of a function at a point? • How do we manipulate average velocity to compute instantaneous velocity? In Section 1.1 we used a function, s(t), to model the location of a moving object at a given time. Functions can model other interesting phenomena, such as the rate at which an auto- mobile consumes gasoline at a given velocity, or the reaction of a patient to a given dosage of a drug. We can use calculus to study how a function value changes in response to changes in the input variable. Think about the falling ball whose position function is given by s(t) 64 − 16t 2 . Its average velocity on the interval [1, x] is given by s(x) − s(1) (64 − 16x 2 ) − (64 − 16) 16 − 16x 2 AV[1,x] . x−1 x−1 x−1 2 Note that the average velocity is a function of x. That is, the function 1(x) 16−16x x−1 tells us the average velocity of the ball on the interval from t 1 to t x. To ﬁnd the instantaneous velocity of the ball when t 1, we need to know what happens to 1(x) as x gets closer and closer to 1. But also notice that 1(1) is not deﬁned, because it leads to the quotient 0/0. This is where the notion of a limit comes in. By using a limit, we can investigate the behavior of 1(x) as x gets arbitrarily close, but not equal, to 1. We ﬁrst use the graph of a function to explore points where interesting behavior occurs. Preview Activity 1.2.1. Suppose that 1 is the function given by the graph below. Use the graph in Figure 1.2.1 to answer each of the following questions. a. Determine the values 1(−2), 1(−1), 1(0), 1(1), and 1(2), if deﬁned. If the function value is not deﬁned, explain what feature of the graph tells you this. b. For each of the values a −1, a 0, and a 2, complete the following sentence: “As x gets closer and closer (but not equal) to a, 1(x) gets as close as we want to em minus 4.5emem .” c. What happens as x gets closer and closer (but not equal) to a 1? Does the function 1(x) get as close as we would like to a single value? 12 1.2 The notion of limit g 3 2 1 -2 -1 1 2 3 -1 Figure 1.2.1: Graph of y 1(x) for Preview Activity 1.2.1. 1.2.1 The Notion of Limit Limits give us a way to identify a trend in the values of a function as its input variable approaches a particular value of interest. We need a precise understanding of what it means to say “a function f has limit L as x approaches a.” To begin, think about a recent example. In Preview Activity 1.2.1, we saw that as x gets closer and closer (but not equal) to 0, 1(x) gets as close as we want to the value 4. At ﬁrst, this may feel counterintuitive, because the value of 1(0) is 1, not 4. But limits describe the behavior of a function arbitrarily close to a ﬁxed input, and the value of the function at the ﬁxed input does not matter. More formally,¹ we say the following. Deﬁnition 1.2.2 Given a function f , a ﬁxed input x a, and a real number L, we say that f has limit L as x approaches a, and write lim f (x) L x→a provided that we can make f (x) as close to L as we like by taking x suﬃciently close (but not equal) to a. If we cannot make f (x) as close to a single value as we would like as x approaches a, then we say that f does not have a limit as x approaches a. Example 1.2.3 For the function 1 pictured in Figure 1.2.1, we make the following observa- tions: lim 1(x) 3, lim 1(x) 4, and lim 1(x) 1. x→−1 x→0 x→2 When working from a graph, it suﬃces to ask if the function approaches a single value from each side of the ﬁxed input. The function value at the ﬁxed input is irrelevant. This reasoning ¹What follows here is not what mathematicians consider the formal deﬁnition of a limit. To be completely precise, it is necessary to quantify both what it means to say “as close to L as we like” and “suﬃciently close to a.” That can be accomplished through what is traditionally called the epsilon-delta deﬁnition of limits. The deﬁnition presented here is suﬃcient for the purposes of this text. 13 Chapter 1 Understanding the Derivative explains the values of the three limits stated above. However, 1 does not have a limit as x → 1. There is a jump in the graph at x 1. If we approach x 1 from the left, the function values tend to get close to 3, but if we approach x 1 from the right, the function values get close to 2. There is no single number that all of these function values approach. This is why the limit of 1 does not exist at x 1. For any function f , there are typically three ways to answer the question “does f have a limit at x a, and if so, what is the limit?” The ﬁrst is to reason graphically as we have just done with the example from Preview Activity 1.2.1. If we have a formula for f (x), there are two additional possibilities: 1 Evaluate the function at a sequence of inputs that approach a on either side (typically using some sort of computing technology), and ask if the sequence of outputs seems to approach a single value. 2 Use the algebraic form of the function to understand the trend in its output values as the input values approach a. The ﬁrst approach produces only an approximation of the value of the limit, while the latter can often be used to determine the limit exactly. Example 1.2.4 Limits of Two Functions. For each of the following functions, we’d like to know whether or not the function has a limit at the stated a-values. Use both numerical and algebraic approaches to investigate and, if possible, estimate or determine the value of the limit. Compare the results with a careful graph of the function on an interval containing the points of interest. ( ) b. 1(x) sin πx ; a 3, a 0 2 a. f (x) 4−x x+2 ; a −1, a −2 Solution. a. We ﬁrst construct a graph of f along with tables of values near a −1 and a −2. From Table 1.2.5, it appears that we can make f as close as we want to 3 by taking x suf- ﬁciently close to −1, which suggests that limx→−1 f (x) 3. This is also consistent with the graph of f . To see this a bit more rigorously and from an algebraic point of view, 2 consider the formula for f : f (x) 4−xx+2 . As x → −1, (4 − x ) → (4 − (−1) ) 3, and 2 2 (x + 2) → (−1 + 2) 1, so as x → −1, the numerator of f tends to 3 and the denominator tends to 1, hence limx→−1 f (x) 31 3. The situation is more complicated when x → −2, because f (−2) is not deﬁned. If we try to use a similar algebraic argument regarding the numerator and denominator, we observe that as x → −2, (4 − x 2 ) → (4 − (−2)2 ) 0, and (x + 2) → (−2 + 2) 0, so as x → −2, the numerator and denominator of f both tend to 0. We call 0/0 an indeterminate form. This tells us that there is somehow more work to do. From Table 1.2.6 and Figure 1.2.7, it appears that f should have a limit of 4 at x −2. 14 1.2 The notion of limit f x f (x) x f (x) 5 −0.9 2.9 −1.9 3.9 −0.99 2.99 −1.99 3.99 −0.999 2.999 −1.999 3.999 −0.9999 2.9999 −1.9999 3.9999 3 −1.1 3.1 −2.1 4.1 −1.01 3.01 −2.01 4.01 −1.001 3.001 −2.001 4.001 1 −1.0001 3.0001 −2.0001 4.0001 -3 -1 1 Table 1.2.5: Table of Table 1.2.6: Table of f values near x −1. f values near x −2. Figure 1.2.7: Plot of f (x) on [−4, 2]. To see algebraically why this is the case, observe that 4 − x2 lim f (x) lim x→−2 x→−2 x + 2 (2 − x)(2 + x) lim . x→−2 x+2 It is important to observe that, since we are taking the limit as x → −2, we are considering x values that are close, but not equal, to −2. Because we never actually allow x to equal −2, the quotient 2+x x+2 has value 1 for every possible value of x. Thus, we can simplify the most recent expression above, and ﬁnd that lim f (x) lim 2 − x. x→−2 x→−2 This limit is now easy to determine, and its value clearly is 4. Thus, from several points of view we’ve seen that limx→−2 f (x) 4. b. Next we turn to the function 1, and construct two tables and a graph. x 1(x) x 1(x) 2 2.9 0.84864 −0.1 0 g 2.99 0.86428 −0.01 0 2.999 0.86585 −0.001 0 -3 -1 1 3 2.9999 0.86601 −0.0001 0 3.1 0.88351 0.1 0 -2 3.01 0.86777 0.01 0 3.001 0.86620 0.001 0 Figure 1.2.10: Plot of 1(x) on [−4, 4]. 3.0001 0.86604 0.0001 0 Table 1.2.8: Table of Table 1.2.9: Table of 1 values near x 3. 1 values near x 0. First, as x → 3, it appears from the table values that the function is approaching a number between 0.86601 and 0.86604. From the graph it appears that 1(x) → 1(3) as x → 3. The 15 Chapter 1 Understanding the Derivative √ exact value of 1(3) sin( π3 ) is 3 2 , which is approximately 0.8660254038. This is convincing evidence that √ 3 lim 1(x) . x→3 2 As x → 0, we observe that πx does not behave in an elementary way. When x is positive and approaching zero, we are dividing by smaller and smaller positive values, and πx increases without bound. When x is negative and approaching zero, πx decreases without bound. In this sense, as we get close to x 0, the inputs to the sine function are growing rapidly, and this leads to increasingly ( rapid ) oscillations in the graph of 1 betweem 1 and −1. If we plot the function 1(x) sin πx with a graphing utility and then zoom in on x 0, we see that the function never settles down to a single value near the origin, which suggests that 1 does not have a limit at x 0. How do we reconcile the graph with the righthand table above, which seems to suggest that the limit of 1 as x approaches 0 may in fact be 0? The data misleads us because of the special −k nature of the sequence ( ) of input values {0.1, 0.01, 0.001, . . .}. When we evaluate 1(10 ), we π get 1(10−k ) sin 10−k sin(10k π) 0 for each positive integer value of k. But if we take a diﬀerent sequence of values approaching zero, say {0.3, 0.03, 0.003, . . .}, then we ﬁnd that ( ( k ) √ −k π ) 10 π 3 1(3 · 10 ) sin −k sin ≈ 0.866025. 3 · 10 3 2 √ That sequence of function values suggests that the value of the limit is 23 . Clearly the func- tion cannot have two diﬀerent values for the limit, so 1 has no limit as x → 0. An important lesson to take from Example 1.2.4 is that tables can be misleading when de- termining the value of a limit. While a table of values is useful for investigating the possible value of a limit, we should also use other tools to conﬁrm the value. Activity 1.2.2. Estimate the value of each of the following limits by constructing ap- propriate tables of values. Then determine the exact value of the limit by using alge- bra to simplify the function. Finally, plot each function on an appropriate interval to check your result visually. √ 2 −1 (2+x)3 −8 a. limx→1 xx−1 b. limx→0 x c. lim x→0 x+1−1 x Recall that our primary motivation for considering limits of functions comes from our inter- est in studying the rate of change of a function. To that end, we close this section by revisiting our previous work with average and instantaneous velocity and highlighting the role that limits play. 16 1.2 The notion of limit 1.2.2 Instantaneous Velocity Suppose that we have a moving object whose position at time t is given by a function s. We s(b)−s(a) know that the average velocity of the object on the time interval [a, b] is AV[a,b] b−a . We deﬁne the instantaneous velocity at a to be the limit of average velocity as b approaches a. Note particularly that as b → a, the length of the time interval gets shorter and shorter (while always including a). We will write IVta for the instantaneous velocity at t a, and thus s(b) − s(a) IVta lim AV[a,b] lim . b→a b→a b−a Equivalently, if we think of the changing value b as being of the form b a + h, where h is some small number, then we may instead write s(a + h) − s(a) IVta lim AV[a,a+h] lim . h→0 h→0 h Again, the most important idea here is that to compute instantaneous velocity, we take a limit of average velocities as the time interval shrinks. Activity 1.2.3. Consider a moving object whose position function is given by s(t) t 2 , where s is measured in meters and t is measured in minutes. a. Determine the most simpliﬁed expression for the average velocity of the object on the interval [3, 3 + h], where h > 0. b. Determine the average velocity of the object on the interval [3, 3.2]. Include units on your answer. c. Determine the instantaneous velocity of the object when t 3. Include units on your answer. The closing activity of this section asks you to make some connections among average ve- locity, instantaneous velocity, and slopes of certain lines. Activity 1.2.4. For the moving object whose position s at time t is given by the graph in Figure 1.2.11, answer each of the following questions. Assume that s is measured in feet and t is measured in seconds. 17 Chapter 1 Understanding the Derivative 5 s 3 1 t 1 3 5 Figure 1.2.11: Plot of the position function y s(t) in Activity 1.2.4. a. Use the graph to estimate the average velocity of the object on each of the fol- lowing intervals: [0.5, 1], [1.5, 2.5], [0, 5]. Draw each line whose slope represents the average velocity you seek. b. How could you use average velocities or slopes of lines to estimate the instan- taneous velocity of the object at a ﬁxed time? c. Use the graph to estimate the instantaneous velocity of the object when t 2. Should this instantaneous velocity at t 2 be greater or less than the average velocity on [1.5, 2.5] that you computed in (a)? Why? 1.2.3 Summary • Limits enable us to examine trends in function behavior near a speciﬁc point. In partic- ular, taking a limit at a given point asks if the function values nearby tend to approach a particular ﬁxed value. • We read limx→a f (x) L, as “the limit of f as x approaches a is L,” which means that we can make the value of f (x) as close to L as we want by taking x suﬃciently close (but not equal) to a. • To ﬁnd limx→a f (x) for a given value of a and a known function f , we can estimate this value from the graph of f , or we can make a table of function values for x-values that are closer and closer to a. If we want the exact value of the limit, we can work with the function algebraically to understand how diﬀerent parts of the formula for f change as x → a. • We ﬁnd the instantaneous velocity of a moving object at a ﬁxed time by taking the limit of average velocities of the object over shorter and shorter time intervals containing the time of interest. 18 1.2 The notion of limit 1.2.4 Exercises 1. Limits on a piecewise graph. Use the ﬁgure below, which gives a graph of the function f (x), to give values for the indicated limits. (a) lim f (x) x→−1 (b) lim f (x) x→0 (c) lim f (x) x→1 (d) lim f (x) x→4 2. Estimating a limit numerically. Use a graph to estimate the limit sin(6θ) lim . θ→0 θ Note: θ is measured in radians. All angles will be in radians in this class unless other- wise speciﬁed. 3. Limits for a piecewise formula. For the function x 2 − 4, 0≤x<4 f (x) 4, x4 3x + 0, 4 < x use algebra to ﬁnd each of the following limits: lim f (x) x→4+ lim f (x) x→4− lim f (x) x→4 Sketch a graph of f (x) to conﬁrm your answers. 4. Evaluating a limit algebraically. Evaluate the limit x 2 − 49 lim x→−7 x+7 19 Chapter 1 Understanding the Derivative 16−x 4 5. Consider the function whose formula is f (x) x 2 −4 . a. What is the domain of f ? b. Use a sequence of values of x near a 2 to estimate the value of limx→2 f (x), if you think the limit exists. If you think the limit doesn’t exist, explain why. 16−x 4 c. Use algebra to simplify the expression x 2 −4 and hence work to evaluate lim f (x) x→2 exactly, if it exists, or to explain how your work shows the limit fails to exist. Discuss how your ﬁndings compare to your results in (b). d. True or false: f (2) −8. Why? 4 e. True or false: 16−x x 2 −4 −4 − x 2 . Why? How is this equality connected to your work above with the function f ? f. Based on all of your work above, construct an accurate, labeled graph of y f (x) on the interval [1, 3], and write a sentence that explains what you now know about 4 limx→2 16−x x 2 −4 . |x+3| 6. Let 1(x) − x+3 . a. What is the domain of 1? b. Use a sequence of values near a −3 to estimate the value of limx→−3 1(x), if you think the limit exists. If you think the limit doesn’t exist, explain why. |x+3| c. Use algebra to simplify the expression x+3 and hence work to evaluate lim 1(x) x→−3 exactly, if it exists, or to explain how your work shows the limit fails to exist. Discuss how your ﬁndings compare to your results in (b). (Hint: |a| a whenever a ≥ 0, but |a| −a whenever a < 0.) d. True or false: 1(−3) −1. Why? |x+3| e. True or false: − x+3 −1. Why? How is this equality connected to your work above with the function 1? f. Based on all of your work above, construct an accurate, labeled graph of y 1(x) on the interval [−4, −2], and write a sentence that explains what you now know about limx→−3 1(x). 7. For each of the following prompts, sketch a graph on the provided axes of a function that has the stated properties. a. y f (x) such that • f (−2) 2 and limx→−2 f (x) 1 • f (−1) 3 and limx→−1 f (x) 3 • f (1) is not deﬁned and limx→1 f (x) 0 • f (2) 1 and limx→2 f (x) does not exist. 20 1.2 The notion of limit b. y 1(x) such that • 1(−2) 3, 1(−1) −1, 1(1) −2, and 1(2) 3 • At x −2, −1, 1 and 2, 1 has a limit, and its limit equals the value of the function at that point. • 1(0) is not deﬁned and limx→0 1(x) does not exist. 3 3 -3 3 -3 3 -3 -3 Figure 1.2.12: Axes for plotting y f (x) in (a) and y 1(x) in (b). 8. A bungee jumper dives from a tower at time t 0. Her height s in feet at time t in seconds is given by s(t) 100 cos(0.75t) · e −0.2t + 100. a. Write an expression for the average velocity of the bungee jumper on the interval [1, 1 + h]. b. Use computing technology to estimate the value of the limit as h → 0 of the quantity you found in (a). c. What is the meaning of the value of the limit in (b)? What are its units? 21 Chapter 1 Understanding the Derivative 1.3 The derivative of a function at a point Motivating Questions • How is the average rate of change of a function on a given interval deﬁned, and what does this quantity measure? • How is the instantaneous rate of change of a function at a particular point deﬁned? How is the instantaneous rate of change linked to average rate of change? • What is the derivative of a function at a given point? What does this derivative value measure? How do we interpret the derivative value graphically? • How are limits used formally in the computation of derivatives? The instantaneous rate of change of a function is an idea that sits at the foundation of calculus. It is a generalization of the notion of instantaneous velocity and measures how fast a partic- ular function is changing at a given point. If the original function represents the position of a moving object, this instantaneous rate of change is precisely the velocity of the object. In other contexts, instantaneous rate of change could measure the number of cells added to a bacteria culture per day, the number of additional gallons of gasoline consumed by increas- ing a car’s velocity one mile per hour, or the number of dollars added to a mortgage payment for each percentage point increase in interest rate. The instantaneous rate of change can also be interpreted geometrically on the function’s graph, and this connection is fundamental to many of the main ideas in calculus. Recall that for a moving object with position function s, its average velocity on the time interval t a to t a + h is given by the quotient s(a + h) − s(a) AV[a,a+h] . h In a similar way, we make the following deﬁnition for an arbitrary function y f (x). Deﬁnition 1.3.1 For a function f , the average rate of change of f on the interval [a, a + h] is given by the value f (a + h) − f (a) AV[a,a+h] . h Equivalently, if we want to consider the average rate of change of f on [a, b], we compute f (b) − f (a) AV[a,b] . b−a It is essential that you understand how the average rate of change of f on an interval is connected to its graph. 22 1.3 The derivative of a function at a point Preview Activity 1.3.1. Suppose that f is the function given by the graph below and that a and a + h are the input values as labeled on the x-axis. Use the graph in Fig- ure 1.3.2 to answer the following questions. y f x a a+h Figure 1.3.2: Plot of y f (x) for Preview Activity 1.3.1. a. Locate and label the points (a, f (a)) and (a + h, f (a + h)) on the graph. b. Construct a right triangle whose hypotenuse is the line segment from (a, f (a)) to (a + h, f (a + h)). What are the lengths of the respective legs of this triangle? c. What is the slope of the line that connects the points (a, f (a)) and (a+h, f (a+h))? d. Write a meaningful sentence that explains how the average rate of change of the function on a given interval and the slope of a related line are connected. 1.3.1 The Derivative of a Function at a Point Just as we deﬁned instantaneous velocity in terms of average velocity, we now deﬁne the instantaneous rate of change of a function at a point in terms of the average rate of change of the function f over related intervals. This instantaneous rate of change of f at a is called “the derivative of f at a,” and is denoted by f ′(a). Deﬁnition 1.3.3 Let f be a function and x a a value in the function’s domain. We deﬁne the derivative of f with respect to x evaluated at x a, denoted f ′(a), by the formula f (a + h) − f (a) f ′(a) lim , h→0 h provided this limit exists. Aloud, we read the symbol f ′(a) as either “ f -prime at a” or “the derivative of f evaluated 23 Chapter 1 Understanding the Derivative at x a.” Much of the next several chapters will be devoted to understanding, computing, applying, and interpreting derivatives. For now, we observe the following important things. Note 1.3.4 • The derivative of f at the value x a is deﬁned as the limit of the average rate of change of f on the interval [a, a + h] as h → 0. This limit may not exist, so not every function has a derivative at every point. • We say that a function is diﬀerentiable at x a if it has a derivative at x a. • The derivative is a generalization of the instantaneous velocity of a position function: if y s(t) is a position function of a moving body, s ′(a) tells us the instantaneous velocity of the body at time t a. f (a+h)− f (a) • Because the units on h are “units of f (x) per unit of x,” the derivative has these very same units. For instance, if s measures position in feet and t measures time in seconds, the units on s ′(a) are feet per second. f (a+h)− f (a) • Because the quantity h represents the slope of the line through (a, f (a)) and (a + h, f (a + h)), when we compute the derivative we are taking the limit of a collec- tion of slopes of lines. Thus, the derivative itself represents the slope of a particularly important line. We ﬁrst consider the derivative at a given value as the slope of a certain line. When we compute an instantaneous rate of change, we allow the interval [a, a + h] to shrink as h → 0. We can think of one endpoint of the interval as “sliding towards” the other. In particular, provided that f has a derivative at (a, f (a)), the point (a + h, f (a + h)) will approach (a, f (a)) as h → 0. Because the process of taking a limit is a dynamic one, it can be helpful to use computing technology to visualize it. One option is a java applet in which the user is able to control the point that is moving. For a helpful collection of examples, consider the work of David Austin of Grand Valley State University, and this particularly relevant example. For applets that have been built in Geogebra¹, see Marc Renault’s library via Shippensburg University, with this example being especially ﬁtting for our work in this section. Figure 1.3.5 shows a sequence of ﬁgures with several diﬀerent lines through the points (a, f (a)) and (a + h, f (a + h)), generated by diﬀerent values of h. These lines (shown in the ﬁrst three ﬁgures in magenta), are often called secant lines to the curve y f (x). A se- cant line to a curve is simply a line that passes through two points on the curve. For each f (a+h)− f (a) such line, the slope of the secant line is m h , where the value of h depends on the location of the point we choose. We can see in the diagram how, as h → 0, the secant lines start to approach a single line that passes through the point (a, f (a)). If the limit of the slopes of the secant lines exists, we say that the resulting value is the slope of the tangent line to the curve. This tangent line (shown in the right-most ﬁgure in green) to the graph of y f (x) at the point (a, f (a)) has slope m f ′(a). ¹You can even consider building your own examples; the fantastic program Geogebra is available for free down- load and is easy to learn and use. 24 1.3 The derivative of a function at a point y y y y f f f f x x x x a a a a Figure 1.3.5: A sequence of secant lines approaching the tangent line to f at (a, f (a)). If the tangent line at x a exists, the graph of f looks like a straight line when viewed up close at (a, f (a)). In Figure 1.3.6 we combine the four graphs in Figure 1.3.5 into the single one on the left, and zoom in on the box centered at (a, f (a)) on the right. Note how the tangent line sits relative to the curve y f (x) at (a, f (a)) and how closely it resembles the curve near x a. y f x a Figure 1.3.6: A sequence of secant lines approaching the tangent line to f at (a, f (a)). At right, we zoom in on the point (a, f (a)). The slope of the tangent line (in green) to f at (a, f (a)) is given by f ′(a). Note 1.3.7 The instantaneous rate of change of f with respect to x at x a, f ′(a), also mea- sures the slope of the tangent line to the curve y f (x) at (a, f (a)). The following example demonstrates several key ideas involving the derivative of a function. Example 1.3.8 Using the limit deﬁnition of the derivative. For the function f (x) x − x 2 , use the limit deﬁnition of the derivative to compute f ′(2). In addition, discuss the meaning of this value and draw a labeled graph that supports your explanation. Solution. From the limit deﬁnition, we know that f (2 + h) − f (2) f ′(2) lim . h→0 h Now we use the rule for f , and observe that f (2) 2−22 −2 and f (2+ h) (2+ h)−(2+ h)2 . 25 Chapter 1 Understanding the Derivative Substituting these values into the limit deﬁnition, we have that (2 + h) − (2 + h)2 − (−2) f ′(2) lim . h→0 h In order to let h → 0, we must simplify the quotient. Expanding and distributing in the numerator, 2 + h − 4 − 4h − h 2 + 2 f ′(2) lim . h→0 h Combining like terms, we have −3h − h 2 f ′(2) lim . h→0 h Next, we remove a common factor of h in both the numerator and denominator and ﬁnd that f ′(2) lim (−3 − h). h→0 Finally, we are able to take the limit as h → 0, and thus conclude that f ′(2) −3. We note that f ′(2) is the instantaneous rate of change of f at the point (2, −2). It is also the slope of the tangent line to the graph of y x − x 2 at the point (2, −2). Figure 1.3.9 shows both the function and the line through (2, −2) with slope m f ′(2) −3. m = f ′ (2) 1 2 -2 -4 y = x − x2 Figure 1.3.9: The tangent line to y x − x 2 at the point (2, −2). The following activities will help you explore a variety of key ideas related to derivatives. 26 1.3 The derivative of a function at a point Activity 1.3.2. Consider the function f whose formula is f (x) 3 − 2x. a. What familiar type of function is f ? What can you say about the slope of f at every value of x? b. Compute the average rate of change of f on the intervals [1, 4], [3, 7], and [5, 5 + h]; simplify each result as much as possible. What do you notice about these quantities? c. Use the limit deﬁnition of the derivative to compute the exact instantaneous rate of change of f with respect to x at the value a 1. That is, compute f ′(1) using the limit deﬁnition. Show your work. Is your result surprising? d. Without doing any additional computations, what are the values of f ′(2), f ′(π), √ and f ′(− 2)? Why? Activity 1.3.3. A water balloon is tossed vertically in the air from a window. The balloon’s height in feet at time t in seconds after being launched is given by s(t) −16t 2 + 16t + 32. Use this function to respond to each of the following questions. a. Sketch an accurate, labeled graph of s on the axes provided in Figure 1.3.10. You should be able to do this without using computing technology. y 32 16 t 1 2 Figure 1.3.10: Axes for plotting y s(t) in Activity 1.3.3. b. Compute the average rate of change of s on the time interval [1, 2]. Include units on your answer and write one sentence to explain the meaning of the value you found. c. Use the limit deﬁnition to compute the instantaneous rate of change of s with respect to time, t, at the instant a 1. Show your work using proper notation, include units on your answer, and write one sentence to explain the meaning of the value you found. 27 Chapter 1 Understanding the Derivative d. On your graph in (a), sketch two lines: one whose slope represents the average rate of change of s on [1, 2], the other whose slope represents the instantaneous rate of change of s at the instant a 1. Label each line clearly. e. For what values of a do you expect s ′(a) to be positive? Why? Answer the same questions when “positive” is replaced by “negative” and “zero.” Activity 1.3.4. A rapidly growing city in Arizona has its population P at time t, where t is the number of decades after the year 2010, modeled by the formula P(t) 25000e t/5 . Use this function to respond to the following questions. a. Sketch an accurate graph of P for t 0 to t 5 on the axes provided in Fig- ure 1.3.11. Label the scale on the axes carefully. y t Figure 1.3.11: Axes for plotting y P(t) in Activity 1.3.4. b. Compute the average rate of change of P between 2030 and 2050. Include units on your answer and write one sentence to explain the meaning (in everyday language) of the value you found. c. Use the limit deﬁnition to write an expression for the instantaneous rate of change of P with respect to time, t, at the instant a 2. Explain why this limit is diﬃcult to evaluate exactly. d. Estimate the limit in (c) for the instantaneous rate of change of P at the instant a 2 by using several small h values. Once you have determined an accurate estimate of P ′(2), include units on your answer, and write one sentence (using everyday language) to explain the meaning of the value you found. e. On your graph above, sketch two lines: one whose slope represents the average rate of change of P on [2, 4], the other whose slope represents the instantaneous rate of change of P at the instant a 2. 28 1.3 The derivative of a function at a point f. In a carefully-worded sentence, describe the behavior of P ′(a) as a increases in value. What does this reﬂect about the behavior of the given function P? 1.3.2 Summary f (b)− f (a) • The average rate of change of a function f on the interval [a, b] is b−a . The units on the average rate of change are units of f (x) per unit of x, and the numerical value of the average rate of change represents the slope of the secant line between the points (a, f (a)) and (b, f (b)) on the graph of y f (x). If we view the interval as being [a, a+h] instead of [a, b], the meaning is still the same, but the average rate of change is now f (a+h)− f (a) computed by h . • The instantaneous rate of change with respect to x of a function f at a value x a is denoted f ′(a) (read “the derivative of f evaluated at a” or “ f -prime at a”) and is deﬁned by the formula f (a + h) − f (a) f ′(a) lim , h→0 h provided the limit exists. Note particularly that the instantaneous rate of change at x a is the limit of the average rate of change on [a, a + h] as h → 0. • Provided the derivative f ′(a) exists, its value tells us the instantaneous rate of change of f with respect to x at x a, which geometrically is the slope of the tangent line to the curve y f (x) at the point (a, f (a)). We even say that f ′(a) is the “slope of the curve” at the point (a, f (a)). • Limits allow us to move from the rate of change over an interval to the rate of change at a single point. 29 Chapter 1 Understanding the Derivative 1.3.3 Exercises 1. Estimating derivative values graphically. Consider the function y f (x) graphed below. Give the x-coordinate of a point where: A. the derivative of the function is negative B. the value of the function is negative C. the derivative of the function is smallest (most negative) D. the derivative of the function is zero E. the derivative of the function is approxi- mately the same as the derivative at x 2.25 (be sure that you give a point that is distinct from x 2.25!) 2. Tangent line to a curve. The ﬁgure below shows a function 1(x) and its tangent line at the point B (6.8, 2). If the point A on the tangent line is (6.74, 2.05), ﬁll in the blanks below to complete the statements about the function 1 at the point B. 1( ) 1′( ) 30 1.3 The derivative of a function at a point 3. Interpreting values and slopes from a graph. Consider the graph of the function f (x) shown below. Using this graph, for each of the following pairs of numbers decide which is larger. Be sure that you can explain your answer. A. f (6) (□ < □ = □ >) f (8) B. f (6) − f (4) (□ < □ = □ >) f (4) − f (2) f (4)− f (2) f (6)− f (2) C. 4−2 (□ < □ = □ >) 6−2 D. f ′(2) (□ < □ = □ >) f ′(8) 4. Finding an exact derivative value algebraically. Find the derivative of 1(t) 2t 2 + 2t at t 7 algebraically. 5. Estimating a derivative from the limit deﬁnition. Estimate f ′(3) for f (x) 6x . Be sure your answer is accurate to within 0.1 of the actual value. 6. Consider the graph of y f (x) provided in Figure 1.3.12. a. On the graph of y f (x), sketch y and label the following quantities: 4 f • the secant line to y f (x) on the interval [−3, −1] and the secant line to y f (x) on the interval [0, 2]. x • the tangent line to y f (x) at -4 4 x −3 and the tangent line to y f (x) at x 0. b. What is the approximate value of the average rate of change of f on -4 [−3, −1]? On [0, 2]? How are these values related to your work in (a)? c. What is the approximate value of Figure 1.3.12: Plot of y f (x). the instantaneous rate of change of f at x −3? At x 0? How are these values related to your work in (a)? 7. For each of the following prompts, sketch a graph on the provided axes in Figure 1.3.13 of a function that has the stated properties. 31 Chapter 1 Understanding the Derivative 3 3 -3 3 -3 3 -3 -3 Figure 1.3.13: Axes for plotting y f (x) in (a) and y 1(x) in (b). a. y f (x) such that • the average rate of change of f on [−3, 0] is −2 and the average rate of change of f on [1, 3] is 0.5, and • the instantaneous rate of change of f at x −1 is −1 and the instantaneous rate of change of f at x 2 is 1. b. y 1(x) such that 1(3)−1(−2) 1(1)−1(−1) • 5 0 and 2 −1, and • 1 ′(2) 1 and 1 ′(−1) 0 8. Suppose that the population, P, of China (in billions) can be approximated by the func- tion P(t) 1.15(1.014)t where t is the number of years since the start of 1993. a. According to the model, what was the total change in the population of China between January 1, 1993 and January 1, 2000? What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on Janu- ary 1, 2000? Explain and justify, being sure to include proper units on all your answers. b. According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, 2012? c. Write an expression involving limits that, if evaluated, would give the exact in- stantaneous rate of change of the population on today’s date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (in- cluding units) of the value you have found. d. Find an equation for the tangent line to the function y P(t) at the point where the t-value is given by today’s date. 32 1.3 The derivative of a function at a point 9. The goal of this problem is to compute the value of the derivative at a point for several diﬀerent functions, where for each one we do so in three diﬀerent ways, and then to compare the results to see that each produces the same value. For each of the following functions, use the limit deﬁnition of the derivative to compute the value of f ′(a) using three diﬀerent approaches: strive to use the algebraic approach ﬁrst (to compute the limit exactly), then test your result using numerical evidence (with small values of h), and ﬁnally plot the graph of y f (x) near (a, f (a)) along with the appropriate tangent line to estimate the value of f ′(a) visually. Compare your ﬁndings among all three approaches; if you are unable to complete the algebraic approach, still work numerically and graphically. a. f (x) x 2 − 3x, a 2 d. f (x) 2 − |x − 1|, a 1 b. f (x) 1 x, a1 √ π c. f (x) x, a 1 e. f (x) sin(x), a 2 33 Chapter 1 Understanding the Derivative 1.4 The derivative function Motivating Questions • How does the limit deﬁnition of the derivative of a function f lead to an entirely new (but related) function f ′? • What is the diﬀerence between writing f ′(a) and f ′(x)? • How is the graph of the derivative function f ′(x) related to the graph of f (x)? • What are some examples of functions f for which f ′ is not deﬁned at one or more points? We now know that the instantaneous rate of change of a function f (x) at x a, or equiva- lently the slope of the tangent line to the graph of y f (x) at x a, is given by the value f ′(a). In all of our examples so far, we have identiﬁed a particular value of a as our point of interest: a 1, a 3, etc. But it is not hard to imagine that we will often be interested in the derivative value for more than just one a-value, and possibly for many of them. In this section, we explore how we can move from computing the derivative at a single point to computing a formula for f ′(a) at any point a. Indeed, the process of “taking the derivative” generates a new function, denoted by f ′(x), derived from the original function f (x). Preview Activity 1.4.1. Consider the function f (x) 4x − x 2 . a. Use the limit deﬁnition to compute the derivative values: f ′(0), f ′(1), f ′(2), and f ′(3). b. Observe that the work to ﬁnd f ′(a) is the same, regardless of the value of a. Based on your work in (a), what do you conjecture is the value of f ′(4)? How about f ′(5)? (Note: you should not use the limit deﬁnition of the derivative to ﬁnd either value.) c. Conjecture a formula for f ′(a) that depends only on the value a. That is, in the same way that we have a formula for f (x) (recall f (x) 4x − x 2 ), see if you can use your work above to guess a formula for f ′(a) in terms of a. 1.4.1 How the derivative is itself a function In your work in Preview Activity 1.4.1 with f (x) 4x − x 2 , you may have found several patterns. One comes from observing that f ′(0) 4, f ′(1) 2, f ′(2) 0, and f ′(3) −2. That sequence of values leads us naturally to conjecture that f ′(4) −4 and f ′(5) −6. We also observe that the particular value of a has very little eﬀect on the process of computing the value of the derivative through the limit deﬁnition. To see this more clearly, we compute f ′(a), where a represents a number to be named later. Following the now standard process 34 1.4 The derivative function of using the limit deﬁnition of the derivative, f (a + h) − f (a) 4(a + h) − (a + h)2 − (4a − a 2 ) f ′(a) lim lim h→0 h h→0 h 4a + 4h − a − 2ha − h − 4a + a 2 2 2 4h − 2ha − h 2 lim lim h→0 h h→0 h h(4 − 2a − h) lim lim (4 − 2a − h). h→0 h h→0 Here we observe that neither 4 nor 2a depend on the value of h, so as h → 0, (4 − 2a − h) → (4 − 2a). Thus, f ′(a) 4 − 2a. This result is consistent with the speciﬁc values we found above: e.g., f ′(3) 4 − 2(3) −2. And indeed, our work conﬁrms that the value of a has almost no bearing on the process of computing the derivative. We note further that the letter being used is immaterial: whether we call it a, x, or anything else, the derivative at a given value is simply given by “4 minus 2 times the value.” We choose to use x for consistency with the original function given by y f (x), as well as for the purpose of graphing the derivative function. For the function f (x) 4x − x 2 , it follows that f ′(x) 4 − 2x. Because the value of the derivative function is linked to the graph of the original function, it makes sense to look at both of these functions plotted on the same domain. m=0 (0, 4) 4 4 m=2 m = −2 3 3 (1, 2) 2 2 1 1 m=4 m = −4 (2, 0) 1 2 3 4 1 2 3 4 -1 -1 (3, −2) -2 -2 y = f (x) y = f ′ (x) -3 -3 -4 -4 (4, −4) Figure 1.4.1: The graphs of f (x) 4x − x 2 (at left) and f ′(x) 4 − 2x (at right). Slopes on the graph of f correspond to heights on the graph of f ′. In Figure 1.4.1, on the left we show a plot of f (x) 4x−x 2 together with a selection of tangent lines at the points we’ve considered above. On the right, we show a plot of f ′(x) 4 − 2x with emphasis on the heights of the derivative graph at the same selection of points. Notice 35 Chapter 1 Understanding the Derivative the connection between colors in the left and right graphs: the green tangent line on the original graph is tied to the green point on the right graph in the following way: the slope of the tangent line at a point on the lefthand graph is the same as the height at the corresponding point on the righthand graph. That is, at each respective value of x, the slope of the tangent line to the original function is the same as the height of the derivative function. Do note, however, that the units on the vertical axes are diﬀerent: in the left graph, the vertical units are simply the output units of f . On the righthand graph of y f ′(x), the units on the vertical axis are units of f per unit of x. An excellent way to explore how the graph of f (x) generates the graph of f ′(x) is through a java applet. See, for instance, the applets at http://gvsu.edu/s/5C or http://gvsu.edu/s/ 5D, via the sites of Austin and Renault¹. In Section 1.3 when we ﬁrst deﬁned the derivative, we wrote the deﬁnition in terms of a value a to ﬁnd f ′(a). As we have seen above, the letter a is merely a placeholder, and it often makes more sense to use x instead. For the record, here we restate the deﬁnition of the derivative. Deﬁnition 1.4.2 Let f be a function and x a value in the function’s domain. We deﬁne the f (x+h)− f (x) derivative of f , a new function called f ′, by the formula f ′(x) limh→0 h , provided this limit exists. We now have two diﬀerent ways of thinking about the derivative function: 1 given a graph of y f (x), how does this graph lead to the graph of the derivative function y f ′(x)? and 2 given a formula for y f (x), how does the limit deﬁnition of derivative generate a formula for y f ′(x)? Both of these issues are explored in the following activities. Activity 1.4.2. For each given graph of y f (x), sketch an approximate graph of its derivative function, y f ′(x), on the axes immediately below. The scale of the grid for the graph of f is 1 × 1; assume the horizontal scale of the grid for the graph of f ′ is identical to that for f . If necessary, adjust and label the vertical scale on the axes for f ′. When you are ﬁnished with all 8 graphs, write several sentences that describe your overall process for sketching the graph of the derivative function, given the graph the original function. What are the values of the derivative function that you tend to identify ﬁrst? What do you do thereafter? How do key traits of the graph of the derivative function exemplify properties of the graph of the original function? ¹David Austin, http://gvsu.edu/s/5r; Marc Renault, http://gvsu.edu/s/5p. 36 1.4 The derivative function f g x x f′ g′ x x p q x x p′ q′ x x 37 Chapter 1 Understanding the Derivative r s x x r′ s′ x x w z x x w′ z′ x x For a dynamic investigation that allows you to experiment with graphing f ′ when given the graph of f , see http://gvsu.edu/s/8y.² Now, recall the opening example of this section: we began with the function y f (x) 4x − ²Marc Renault, Calculus Applets Using Geogebra. 38 1.4 The derivative function x 2 and used the limit deﬁnition of the derivative to show that f ′(a) 4 − 2a, or equivalently that f ′(x) 4−2x. We subsequently graphed the functions f and f ′ as shown in Figure 1.4.1. Following Activity 1.4.2, we now understand that we could have constructed a fairly accurate graph of f ′(x) without knowing a formula for either f or f ′. At the same time, it is useful to know a formula for the derivative function whenever it is possible to ﬁnd one. In the next activity, we further explore the more algebraic approach to ﬁnding f ′(x): given a formula for y f (x), the limit deﬁnition of the derivative will be used to develop a formula for f ′(x). Activity 1.4.3. For each of the listed functions, determine a formula for the derivative function. For the ﬁrst two, determine the formula for the derivative by thinking about the nature of the given function and its slope at various points; do not use the limit deﬁnition. For the latter four, use the limit deﬁnition. Pay careful attention to the function names and independent variables. It is important to be comfortable with us- ing letters other than f and x. For example, given a function p(z), we call its derivative p ′(z). a. f (x) 1 c. p(z) z 2 e. F(t) 1t √ b. 1(t) t d. q(s) s 3 f. G(y) y 1.4.2 Summary f (x+h)− f (x) • The limit deﬁnition of the derivative, f ′(x) limh→0 h , produces a value for each x at which the derivative is deﬁned, and this leads to a new function y f ′(x). It is especially important to note that taking the derivative is a process that starts with a given function ( f ) and produces a new, related function ( f ′). • There is essentially no diﬀerence between writing f ′(a) (as we did regularly in Sec- tion 1.3) and writing f ′(x). In either case, the variable is just a placeholder that is used to deﬁne the rule for the derivative function. • Given the graph of a function y f (x), we can sketch an approximate graph of its derivative y f ′(x) by observing that heights on the derivative’s graph correspond to slopes on the original function’s graph. • In Activity 1.4.2, we encountered some functions that had sharp corners on their graphs, such as the shifted absolute value function. At such points, the derivative fails to exist, and we say that f is not diﬀerentiable there. For now, it suﬃces to understand this as a consequence of the jump that must occur in the derivative function at a sharp corner on the graph of the original function. 39 Chapter 1 Understanding the Derivative 1.4.3 Exercises 1. The derivative function graphically. Consider the function f (x) shown in the graph below. Carefully sketch the derivative function of the given function (you will want to estimate val- ues on the derivative function at diﬀerent x val- ues as you do this). Use your derivative func- tion graph to estimate the following values on the derivative function. at x -3 -1 1 3 the derivative is 2. Applying the limit deﬁnition of the derivative. Find a formula for the derivative of the function 1(x) 4x 2 − 8 using diﬀerence quotients. 3. Sketching the derivative. For the function f (x) shown in the graph below, sketch a graph of the derivative. You will then be picking which of the following is the correct derivative graph, but should be sure to ﬁrst sketch the derivative yourself. Which of the following graphs is the derivative of f (x)? 40 1.4 The derivative function 1. 2. 3. 4. 5. 6. 7. 8. 4. Comparing function and derivative values. The graph of a function f is shown below. At which of the labeled x-values is f (x) least? f (x) greatest? f ′(x) least? f ′(x) greatest? 5. Limit deﬁnition of the derivative for a rational function. Let 1 f (x) x−4 Find (i) f ′(3), (ii) f ′(5), (iii) f ′(6), and (iv) f ′(8). 6. Let f be a function with the following properties: f is diﬀerentiable at every value of x (that is, f has a derivative at every point), f (−2) 1, and f ′(−2) −2, f ′(−1) −1, f ′(0) 0, f ′(1) 1, and f ′(2) 2. a. On the axes provided at left in Figure 1.4.3, sketch a possible graph of y f (x). Explain why your graph meets the stated criteria. b. Conjecture a formula for the function y f (x). Use the limit deﬁnition of the derivative to determine the corresponding formula for y f ′(x). Discuss both graphical and algebraic evidence for whether or not your conjecture is correct. 41 Chapter 1 Understanding the Derivative 3 3 -3 3 -3 3 -3 -3 Figure 1.4.3: Axes for plotting y f (x) in (a) and y f ′(x) in (b). 7. Consider the function 1(x) x 2 − x + 3. a. Use the limit deﬁnition of the derivative to determine a formula for 1 ′(x). b. Use a graphing utility to plot both y 1(x) and your result for y 1 ′(x); does your formula for 1 ′(x) generate the graph you expected? c. Use the limit deﬁnition of the derivative to ﬁnd a formula for p ′(x) where p(x) 5x 2 − 4x + 12. d. Compare and contrast the formulas for 1 ′(x) and p ′(x) you have found. How do the constants 5, 4, 12, and 3 aﬀect the results? 8. Let 1 be a continuous function (that is, one with no jumps or holes in the graph) and suppose that a graph of y 1 ′(x) is given by the graph on the right in Figure 1.4.4. 2 2 -2 2 -2 2 -2 -2 Figure 1.4.4: Axes for plotting y 1(x) and, at right, the graph of y 1 ′(x). 42 1.4 The derivative function a. Observe that for every value of x that satisﬁes 0 < x < 2, the value of 1 ′(x) is constant. What does this tell you about the behavior of the graph of y 1(x) on this interval? b. On what intervals other than 0 < x < 2 do you expect y 1(x) to be a linear function? Why? c. At which values of x is 1 ′(x) not deﬁned? What behavior does this lead you to expect to see in the graph of y 1(x)? d. Suppose that 1(0) 1. On the axes provided at left in Figure 1.4.4, sketch an accurate graph of y 1(x). 9. For each graph that provides an original function y f (x) in Figure 1.4.5, your task is to sketch an approximate graph of its derivative function, y f ′(x), on the axes immediately below. View the scale of the grid for the graph of f as being 1 × 1, and assume the horizontal scale of the grid for the graph of f ′ is identical to that for f . If you need to adjust the vertical scale on the axes for the graph of f ′, you should label that accordingly. 43 Chapter 1 Understanding the Derivative f f x x f′ f′ x x f f x x f′ f′ x x Figure 1.4.5: Graphs of y f (x) and grids for plotting the corresponding graph of y f ′(x). 44 1.5 Interpreting, estimating, and using the derivative 1.5 Interpreting, estimating, and using the derivative Motivating Questions • In contexts other than the position of a moving object, what does the derivative of a function measure? • What are the units on the derivative function f ′, and how are they related to the units of the original function f ? • What is a central diﬀerence, and how can one be used to estimate the value of the derivative at a point from given function data? • Given the value of the derivative of a function at a point, what can we infer about how the value of the function changes nearby? It is a powerful feature of mathematics that it can be studied both as abstract discipline and as an applied one. For instance, calculus can be developed almost entirely as an abstract col- lection of ideas that focus on properties of functions. At the same time, if we consider func- tions that represent meaningful processes, calculus can describe our experience of physical reality. We have already seen that for the position function y s(t) of a ball being tossed straight up in the air, the derivative of the position function, v(t) s ′(t), gives the ball’s velocity at time t. In this section, we investigate several functions with speciﬁc physical meaning, and consider how the units on the independent variable, dependent variable, and the derivative function add to our understanding. To start, we consider the familiar problem of a position function of a moving object. Preview Activity 1.5.1. One of the longest stretches of straight (and ﬂat) road in North America can be found on the Great Plains in the state of North Dakota on state high- way 46, which lies just south of the interstate highway I-94 and runs through the town of Gackle. A car leaves town (at time t 0) and heads east on highway 46; its posi- tion in miles from Gackle at time t in minutes is given by the graph of the function in Figure 1.5.1. Three important points are labeled on the graph; where the curve looks linear, assume that it is indeed a straight line. a. In everyday language, describe the behavior of the car over the provided time interval. In particular, discuss what is happening on the time intervals [57, 68] and [68, 104]. b. Find the slope of the line between the points (57, 63.8) and (104, 106.8). What are the units on this slope? What does the slope represent? c. Find the average rate of change of the car’s position on the interval [68, 104]. Include units on your answer. d. Estimate the instantaneous rate of change of the car’s position at the moment t 80. Write a sentence to explain your reasoning and the meaning of this 45 Chapter 1 Understanding the Derivative value. (104, 106.8) s 100 80 (57, 63.8) 60 (68, 63.8) 40 20 t 20 40 60 80 100 Figure 1.5.1: The graph of y s(t), the position of the car along highway 46, which tells its distance in miles from Gackle, ND, at time t in minutes. 1.5.1 Units of the derivative function As we now know, the derivative of the function f at a ﬁxed value x is given by f (x + h) − f (x) f ′(x) lim , h→0 h and this value has several diﬀerent interpretations. If we set x a, one meaning of f ′(a) is the slope of the tangent line at the point (a, f (a)). df dy We also sometimes write dx or dx instead of f ′(x), and these alternate notations help us see the units (and thus the meaning) of the derivative as the instantaneous rate of change of f with f (x+h)− f (x) respect to x. The units on the slope of the secant line, h , are “units of y per unit of x,” and when we take the limit as h goes to zero, the derivative f ′(x) has the same units: units of y per unit of x. It is helpful to remember that the units on the derivative function are “units of output per unit of input,” for the variables of the original function. For example, suppose that the function y P(t) measures the population of a city (in thousands) at the start of year t (where t 0 corresponds to 2010 AD). We are told that P ′(2) 21.37. What is the meaning of this value? Well, since P is measured in thousands and t is measured in years, we can say that the instantaneous rate of change of the city’s population with respect to time at the start of 2012 is 21.37 thousand people per year. We therefore expect that in the coming year, about 21,370 people will be added to the city’s population. 46 1.5 Interpreting, estimating, and using the derivative 1.5.2 Toward more accurate derivative estimates f (x+h)− f (x) Recall that to estimate the value of f ′(x) at a given x, we calculate a diﬀerence quotient h with a relatively small value of h. We should use both positive and negative values of h in order to account for the behavior of the function on both sides of the point of interest. To that end, we introduce the notion of a central diﬀerence and its role in estimating derivatives. Example 1.5.2 Suppose that y f (x) is a function for which three values are known: f (1) 2.5, f (2) 3.25, and f (3) 3.625. Estimate f ′(2). f (2+h)− f (2) Solution. We know that f ′(2) limh→0 h . But since we don’t have a graph or a formula for the function, we can neither sketch a tangent line nor evaluate the limit alge- braically. We can’t even use smaller and smaller values of h to estimate the limit. Instead, we have just two choices: using h −1 or h 1, depending on which point we pair with (2, 3.25). So, one estimate is f (1) − f (2) 2.5 − 3.25 f ′(2) ≈ 0.75. 1−2 −1 The other is f (3) − f (2) 3.625 − 3.25 f ′(2) ≈ 0.375. 3−2 1 Because the ﬁrst approximation looks backward from the point (2, 3.25) and the second ap- proximation looks forward, it makes sense to average these two estimates in order to account for behavior on both sides of x 2. Doing so, we ﬁnd that 0.75 + 0.375 f ′(2) ≈ 0.5625. 2 The intuitive approach to average the two estimates found in Example 1.5.2 is in fact the best possible way estimate to a derivative when we have just two function values for f on opposite sides of the point of interest. To see why, we think about the diagram in Figure 1.5.3. On the left, we see the two secant f (1)− f (2) lines with slopes that come from computing the backward diﬀerence 1−2 0.75 and from f (3)− f (2) the forward diﬀerence 3−2 0.375. Note how the ﬁrst slope over-estimates the slope of the tangent line at (2, f (2)), while the second slope underestimates f ′(2). On the right, we see the secant line whose slope is given by the central diﬀerence f (3) − f (1) 3.625 − 2.5 1.125 0.5625. 3−1 2 2 Note that this central diﬀerence has the same value as the average of the forward and back- ward diﬀerences (and it is straightforward to explain why this always holds). The central diﬀerence yields a very good approximation to the derivative’s value, because it yields a line closer to being parallel to the tangent line. 47 Chapter 1 Understanding the Derivative 3 3 2 2 1 1 1 2 3 1 2 3 Figure 1.5.3: At left, the graph of y f (x) along with the secant line through (1, 2.5) and (2, 3.25), the secant line through (2, 3.25) and (3, 3.625), as well as the tangent line. At right, the same graph along with the secant line through (1, 2.5) and (3, 3.625), plus the tangent line. The central diﬀerence approximation to the value of the ﬁrst derivative is given by f (a + h) − f (a − h) f ′(a) ≈ . 2h This quantity measures the slope of the secant line to y f (x) through the points (a−h, f (a− h)) and (a + h, f (a + h)). Activity 1.5.2. A potato is placed in an oven, and the potato’s temperature F (in de- grees Fahrenheit) at various points in time is taken and recorded in the following table. Time t is measured in minutes. t 0 15 30 45 60 75 90 F(t) 70 180.5 251 296 324.5 342.8 354.5 Table 1.5.4: Temperature data in degrees Fahrenheit. a. Use a central diﬀerence to estimate the instantaneous rate of change of the tem- perature of the potato at t 30. Include units on your answer. b. Use a central diﬀerence to estimate the instantaneous rate of change of the tem- perature of the potato at t 60. Include units on your answer. c. Without doing any calculation, which do you expect to be greater: F′(75) or F′(90)? Why? d. Suppose it is given that F(64) 330.28 and F′(64) 1.341. What are the units on these two quantities? What do you expect the temperature of the potato to 48 1.5 Interpreting, estimating, and using the derivative be when t 65? when t 66? Why? e. Write a couple of careful sentences that describe the behavior of the temper- ature of the potato on the time interval [0, 90], as well as the behavior of the instantaneous rate of change of the temperature of the potato on the same time interval. Activity 1.5.3. A company manufactures rope, and the total cost of producing r feet of rope is C(r) dollars. a. What does it mean to say that C(2000) 800? b. What are the units of C′(r)? c. Suppose that C(2000) 800 and C′(2000) 0.35. Estimate C(2100), and justify your estimate by writing at least one sentence that explains your thinking. d. Do you think C′(2000) is less than, equal to, or greater than C′(3000)? Why? e. Suppose someone claims that C′(5000) −0.1. What would the practical mean- ing of this derivative value tell you about the approximate cost of the next foot of rope? Is this possible? Why or why not? Activity 1.5.4. Researchers at a major car company have found a function that relates gasoline consumption to speed for a particular model of car. In particular, they have determined that the consumption C, in liters per kilometer, at a given speed s, is given by a function C f (s), where s is the car’s speed in kilometers per hour. a. Data provided by the car company tells us that f (80) 0.015, f (90) 0.02, and f (100) 0.027. Use this information to estimate the instantaneous rate of change of fuel consumption with respect to speed at s 90. Be as accurate as possible, use proper notation, and include units on your answer. b. By writing a complete sentence, interpret the meaning (in the context of fuel consumption) of “ f (80) 0.015.” c. Write at least one complete sentence that interprets the meaning of the value of f ′(90) that you estimated in (a). In Section 1.4, we learned how use to the graph of a given function f to plot the graph of its derivative, f ′. It is important to remember that when we do so, the scale and the units on the vertical axis often have to change to represent f ′. For example, suppose that P(t) 400 − 330e −0.03t tells us the temperature in degrees Fahrenheit of a potato in an oven at time t in minutes. In Figure 1.5.5, we sketch the graph of P on the left and the graph of P ′ on the right. Notice that the vertical scales are diﬀerent in size and diﬀerent in units, as the units of P are ◦ F, while those of P ′ are ◦ F/min. 49 Chapter 1 Understanding the Derivative ◦F ◦ F/min 400 16 300 y = P(t) 12 200 8 y = P′ (t) 100 4 min min 20 40 60 80 20 40 60 80 Figure 1.5.5: Plot of P(t) 400 − 330e −0.03t at left, and its derivative P ′(t) at right. 1.5.3 Summary • The derivative of a given function y f (x) measures the instantaneous rate of change of the output variable with respect to the input variable. • The units on the derivative function y f ′(x) are units of y per unit of x. Again, this measures how fast the output of the function f changes when the input of the function changes. • The central diﬀerence approximation to the value of the ﬁrst derivative is given by f (a + h) − f (a − h) f ′(a) ≈ . 2h This quantity measures the slope of the secant line to y f (x) through the points (a − h, f (a − h)) and (a + h, f (a + h)). The central diﬀerence generates a good approximation of the derivative’s value. 1.5.4 Exercises 1. A cooling cup of coﬀee. The temperature, H, in degrees Celsius, of a cup of coﬀee placed on the kitchen counter is given by H f (t), where t is in minutes since the coﬀee was put on the counter. (a) Is f ′(t) positive or negative? (Be sure that you are able to give a reason for your answer.) (b) What are the units of f ′(35)? 50 1.5 Interpreting, estimating, and using the derivative Suppose that | f ′(35)| 1.5 and f (35) 68. Fill in the blanks (including units where needed) and select the appropriate terms to complete the following statement about the temperature of the coﬀee in this case. At minutes after the coﬀee was put on the counter, its (□ derivative □ temperature □ change in temperature) is and will (□ increase □ decrease) by about in the next 30 seconds. 2. A cost function. The cost, C (in dollars) to produce 1 gallons of ice cream can be ex- pressed as C f (1). (a) In the expression f (100) 250, what are the units of 100? What are the units of 250? (b) In the expression f ′(100) 1.2, what are the units of 100? What are the units of 1.2? (Be sure that you can carefully put into words the meanings of each of these statement in terms of ice cream and money.) 3. Weight as a function of calories. A laboratory study investigating the relationship between diet and weight in adult humans found that the weight of a subject, W, in pounds, was a function, W f (c), of the average number of Calories, c, consumed by the subject in a day. (a) In the statement f (1600) 165 what are the units of 1600? What are the units of 165? (Think about what this statement means in terms of the weight of the subject and the number of calories that the subject consumes.) (b) In the statement f ′(2000) 0, what are the units of 2000? What are the units of 0? (Think about what this statement means in terms of the weight of the subject and the number of calories that the subject consumes.) (c) In the statement f −1 (173) 2400, what are the units of 173? What are the units of 2400? (Think about what this statement means in terms of the weight of the subject and the number of calories that the subject consumes.) (d) What are the units of f ′(c) dW/dc? (e) Suppose that Sam reads about f ′ in this study and draws the following conclusion: If Sam increases her average calorie intake from 2800 to 2840 calories per day, then her weight will increase by approximately 0.8 pounds. Fill in the blanks below so that the equation supports her conclusion. ( ) f′ 4. Displacement and velocity. The displacement (in meters) of a particle moving in a straight line is given by s t 2 − 5t + 16, where t is measured in seconds. (A) (i) Find the average velocity over the time interval [3,4]. 51 Chapter 1 Understanding the Derivative (ii) Find the average velocity over the time interval [3.5,4]. (iii) Find the average velocity over the time interval [4,5]. (iv) Find the average velocity over the time interval [4,4.5]. (B) Find the instantaneous velocity when t 4. 5. A cup of coﬀee has its temperature F (in degrees Fahrenheit) at time t given by the function F(t) 75 + 110e −0.05t , where time is measured in minutes. a. Use a central diﬀerence with h 0.01 to estimate the value of F′(10). b. What are the units on the value of F′(10) that you computed in (a)? What is the practical meaning of the value of F′(10)? c. Which do you expect to be greater: F′(10) or F′(20)? Why? d. Write a sentence that describes the behavior of the function y F′(t) on the time interval 0 ≤ t ≤ 30. How do you think its graph will look? Why? 6. The temperature change T (in Fahrenheit degrees), in a patient, that is generated by a dose q (in milliliters), of a drug, is given by the function T f (q). a. What does it mean to say f (50) 0.75? Write a complete sentence to explain, using correct units. b. A person’s sensitivity, s, to the drug is deﬁned by the function s(q) f ′(q). What are the units of sensitivity? c. Suppose that f ′(50) −0.02. Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a). 7. The velocity of a ball that has been tossed vertically in the air is given by v(t) 16 − 32t, where v is measured in feet per second, and t is measured in seconds. The ball is in the air from t 0 until t 2. a. When is the ball’s velocity greatest? b. Determine the value of v ′(1). Justify your thinking. c. What are the units on the value of v ′(1)? What does this value and the corre- sponding units tell you about the behavior of the ball at time t 1? d. What is the physical meaning of the function v ′(t)? 8. The value, V, of a particular automobile (in dollars) depends on the number of miles, m, the car has been driven, according to the function V h(m). a. Suppose that h(40000) 15500 and h(55000) 13200. What is the average rate of change of h on the interval [40000, 55000], and what are the units on this value? b. In addition to the information given in (a), say that h(70000) 11100. Deter- mine the best possible estimate of h ′(55000) and write one sentence to explain the meaning of your result, including units on your answer. c. Which value do you expect to be greater: h ′(30000) or h ′(80000)? Why? d. Write a sentence to describe the long-term behavior of the function V h(m), 52 1.5 Interpreting, estimating, and using the derivative plus another sentence to describe the long-term behavior of h ′(m). Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing. 53 Chapter 1 Understanding the Derivative 1.6 The second derivative Motivating Questions • How does the derivative of a function tell us whether the function is increasing or decreasing on an interval? • What can we learn by taking the derivative of the derivative (the second derivative) of a function f ? • What does it mean to say that a function is concave up or concave down? How are these characteristics connected to certain properties of the derivative of the function? • What are the units of the second derivative? How do they help us understand the rate of change of the rate of change? Given a diﬀerentiable function y f (x), we know that its derivative, y f ′(x), is a related function whose output at x a tells us the slope of the tangent line to y f (x) at the point (a, f (a)). That is, heights on the derivative graph tell us the values of slopes on the original function’s graph. At a point where f ′(x) is positive, the slope of the tangent line to f is positive. Therefore, on an interval where f ′(x) is positive, the function f is increasing (or rising). Similarly, if f ′(x) is negative on an interval, the graph of f is decreasing (or falling). The derivative of f tells us not only whether the function f is increasing or decreasing on an inter- val, but also how the function f is increasing or decreasing. Look at the two tangent lines shown in Figure 1.6.1. We see that near point A the value of f ′(x) is positive and relatively close to zero, and near that point the graph is rising slowly. By con- trast, near point B, the derivative is negative and relatively large in absolute value, and f is decreas- A ing rapidly near B. B Besides asking whether the value of the deriva- tive function is positive or negative and whether it is large or small, we can also ask “how is the derivative changing?” Because the derivative, y f ′(x), is itself a func- Figure 1.6.1: Two tangent lines on a tion, we can consider taking its derivative — the graph. derivative of the derivative — and ask “what does the derivative of the derivative tell us about how the original function behaves?” We start with an investigation of a moving object. 54 1.6 The second derivative Preview Activity 1.6.1. The position of a car driving along a straight road at time t in minutes is given by the function y s(t) that is pictured in Figure 1.6.2. The car’s position function has units measured in thousands of feet. For instance, the point (2, 4) on the graph indicates that after 2 minutes, the car has traveled 4000 feet. y s 14 10 6 2 t 2 6 10 Figure 1.6.2: The graph of y s(t), the position of the car (measured in thousands of feet from its starting location) at time t in minutes. a. In everyday language, describe the behavior of the car over the provided time interval. In particular, you should carefully discuss what is happening on each of the time intervals [0, 1], [1, 2], [2, 3], [3, 4], and [4, 5], plus provide commen- tary overall on what the car is doing on the interval [0, 12]. b. On the lefthand axes provided in Figure 1.6.3, sketch a careful, accurate graph of y s ′(t). c. What is the meaning of the function y s ′(t) in the context of the given prob- lem? What can we say about the car’s behavior when s ′(t) is positive? when s ′(t) is zero? when s ′(t) is negative? d. Rename the function you graphed in (b) to be called y v(t). Describe the behavior of v in words, using phrases like “v is increasing on the interval . . .” and “v is constant on the interval . . ..” e. Sketch a graph of the function y v ′(t) on the righthand axes provide in Fig- ure 1.6.3. Write at least one sentence to explain how the behavior of v ′(t) is connected to the graph of y v(t). 55 Chapter 1 Understanding the Derivative y y t t 2 6 10 2 6 10 Figure 1.6.3: Axes for plotting y v(t) s ′(t) and y v ′(t). 1.6.1 Increasing or decreasing So far, we have used the words increasing and decreasing intuitively to describe a function’s graph. Here we deﬁne these terms more formally. Deﬁnition 1.6.4 Given a function f (x) deﬁned on the interval (a, b), we say that f is in- creasing on (a, b) provided that for all x, y in the interval (a, b), if x < y, then f (x) < f (y). Similarly, we say that f is decreasing on (a, b) provided that for all x, y in the interval (a, b), if x < y, then f (x) > f (y). Simply put, an increasing function is one that is rising as we move from left to right along the graph, and a decreasing function is one that falls as the value of the input increases. If the function has a derivative, the sign of the derivative tells us whether the function is increasing or decreasing. Let f be a function that is diﬀerentiable on an interval (a, b). It is possible to show that that if f ′(x) > 0 for every x such that a < x < b, then f is increasing on (a, b); similarly, if f ′(x) < 0 on (a, b), then f is decreasing on (a, b). For example, the function pictured in Figure 1.6.5 is increasing on the entire interval −2 < x < 0, and decreasing on the interval 0 < x < 2. Note that the value x 0 is not included in either interval since at this location, the function is changing from increasing to decreasing. 56 1.6 The second derivative A 2 -2 2 y = f (x) -2 B Figure 1.6.5: A function that is decreasing on the intervals −3 < x < −2 and 0 < x < 2 and increasing on −2 < x < 0 and 2 < x < 3. 1.6.2 The Second Derivative We are now accustomed to investigating the behavior of a function by examining its deriv- ative. The derivative of a function f is a new function given by the rule f (x + h) − f (x) f ′(x) lim . h→0 h Because f ′ is itself a function, it is perfectly feasible for us to consider the derivative of the derivative, which is the new function y [ f ′(x)]′. We call this resulting function the second derivative of y f (x), and denote the second derivative by y f ′′(x). Consequently, we will sometimes call f ′ “the ﬁrst derivative” of f , rather than simply “the derivative” of f . Deﬁnition 1.6.6 The second derivative is deﬁned by the limit deﬁnition of the derivative of the ﬁrst derivative. That is, f ′(x + h) − f ′(x) f ′′(x) lim . h→0 h The meaning of the derivative function still holds, so when we compute y f ′′(x), this new function measures slopes of tangent lines to the curve y f ′(x), as well as the instantaneous rate of change of y f ′(x). In other words, just as the ﬁrst derivative measures the rate at which the original function changes, the second derivative measures the rate at which the ﬁrst derivative changes. The second derivative will help us understand how the rate of change of the original function is itself changing. 57 Chapter 1 Understanding the Derivative 1.6.3 Concavity In addition to asking whether a function is increasing or decreasing, it is also natural to in- quire how a function is increasing or decreasing. There are three basic behaviors that an increasing function can demonstrate on an interval, as pictured in Figure 1.6.7: the function can increase more and more rapidly, it can increase at the same rate, or it can increase in a way that is slowing down. Fundamentally, we are beginning to think about how a par- ticular curve bends, with the natural comparison being made to lines, which don’t bend at all. More than this, we want to understand how the bend in a function’s graph is tied to behavior characterized by the ﬁrst derivative of the function. Figure 1.6.7: Three functions that are all increasing, but doing so at an increasing rate, at a constant rate, and at a decreasing rate, respectively. On the leftmost curve in Figure 1.6.7, draw a sequence of tangent lines to the curve. As we move from left to right, the slopes of those tangent lines will increase. Therefore, the rate of change of the pictured function is increasing, and this explains why we say this function is increasing at an increasing rate. For the rightmost graph in Figure 1.6.7, observe that as x increases, the function increases, but the slopes of the tangent lines decrease. This function is increasing at a decreasing rate. Similar options hold for how a function can decrease. Here we must be extra careful with our language, because decreasing functions involve negative slopes. Negative numbers present an interesting tension between common language and mathematical language. For example, it can be tempting to say that “−100 is bigger than −2.” But we must remember that “greater than” describes how numbers lie on a number line: x > y provided that x lies to the right of y. So of course, −100 is less than −2. Informally, it might be helpful to say that “−100 is more negative than −2.” When a function’s values are negative, and those values get more negative as the input increases, the function must be decreasing. Now consider the three graphs shown in Figure 1.6.8. Clearly the middle graph depicts a function decreasing at a constant rate. Now, on the ﬁrst curve, draw a sequence of tangent lines. We see that the slopes of these lines get less and less negative as we move from left to right. That means that the values of the ﬁrst derivative, while all negative, are increasing, and thus we say that the leftmost curve is decreasing at an increasing rate. This leaves only the rightmost curve in Figure 1.6.8 to consider. For that function, the slopes 58 1.6 The second derivative Figure 1.6.8: From left to right, three functions that are all decreasing, but doing so in diﬀerent ways. of the tangent lines are negative throughout the pictured interval, but as we move from left to right, the slopes get more and more negative. Hence the slope of the curve is decreasing, and we say that the function is decreasing at a decreasing rate. We now introduce the notion of concavity which provides simpler language to describe these behaviors. When a curve opens upward on a given interval, like the parabola y x 2 or the exponential growth function y e x , we say that the curve is concave up on that interval. Likewise, when a curve opens down, like the parabola y −x 2 or the opposite of the ex- ponential function y −e x , we say that the function is concave down. Concavity is linked to both the ﬁrst and second derivatives of the function. Figure 1.6.9: At left, a function that is concave up; at right, one that is concave down. In Figure 1.6.9, we see two functions and a sequence of tangent lines to each. On the lefthand plot, where the function is concave up, observe that the tangent lines always lie below the curve itself, and the slopes of the tangent lines are increasing as we move from left to right. In other words, the function f is concave up on the interval shown because its derivative, f ′, is increasing on that interval. Similarly, on the righthand plot in Figure 1.6.9, where the 59 Chapter 1 Understanding the Derivative function shown is concave down, we see that the tangent lines alway lie above the curve, and the slopes of the tangent lines are decreasing as we move from left to right. The fact that its derivative, f ′, is decreasing makes f concave down on the interval. We state these most recent observations formally as the deﬁnitions of the terms concave up and concave down. Deﬁnition 1.6.10 Let f be a diﬀerentiable function on an interval (a, b). Then f is concave up on (a, b) if and only if f ′ is increasing on (a, b); f is concave down on (a, b) if and only if f ′ is decreasing on (a, b). Activity 1.6.2. The position of a car driving along a straight road at time t in minutes is given by the function y s(t) that is pictured in Figure 1.6.11. The car’s position function has units measured in thousands of feet. Remember that you worked with this function and sketched graphs of y v(t) s ′(t) and y v ′(t) in Preview Activ- ity 1.6.1. y s 14 10 6 2 t 2 6 10 Figure 1.6.11: The graph of y s(t), the position of the car (measured in thousands of feet from its starting location) at time t in minutes. a. On what intervals is the position function y s(t) increasing? decreasing? Why? b. On which intervals is the velocity function y v(t) s ′(t) increasing? decreas- ing? neither? Why? c. Acceleration is deﬁned to be the instantaneous rate of change of velocity, as the acceleration of an object measures the rate at which the velocity of the object is changing. Say that the car’s acceleration function is named a(t). How is a(t) computed from v(t)? How is a(t) computed from s(t)? Explain. d. What can you say about s ′′ whenever s ′ is increasing? Why? 60 1.6 The second derivative e. Using only the words increasing, decreasing, constant, concave up, concave down, and linear, complete the following sentences. For the position function s with velocity v and acceleration a, • on an interval where v is positive, s is . • on an interval where v is negative, s is . • on an interval where v is zero, s is . • on an interval where a is positive, v is . • on an interval where a is negative, v is . • on an interval where a is zero, v is . • on an interval where a is positive, s is . • on an interval where a is negative, s is . • on an interval where a is zero, s is . Exploring the context of position, velocity, and acceleration is an excellent way to understand how a function, its ﬁrst derivative, and its second derivative are related to one another. In Activity 1.6.2, we can replace s, v, and a with an arbitrary function f and its derivatives f ′ and f ′′, and essentially all the same observations hold. In particular, note that the following are equivalent: on an interval where the graph of f is concave up, f ′ is increasing and f ′′ is positive. Likewise, on an interval where the graph of f is concave down, f ′ is decreasing and f ′′ is negative. Activity 1.6.3. A potato is placed in an oven, and the potato’s temperature F (in de- grees Fahrenheit) at various points in time is taken and recorded in the following table. Time t is measured in minutes. In Activity 1.5.2, we computed approximations to F′(30) and F′(60) using central diﬀerences. Those values and more are provided in the second table below, along with several others computed in the same way. t F(t) t F′(t) 0 70 0 NA 15 180.5 15 6.03 30 251 30 3.85 45 296 45 2.45 60 324.5 60 1.56 75 342.8 75 1.00 90 354.5 90 NA Table 1.6.12: Select values of F(t). Table 1.6.13: Select values of F′(t). a. What are the units on the values of F′(t)? b. Use a central diﬀerence to estimate the value of F′′(30). c. What is the meaning of the value of F′′(30) that you have computed in (b) in terms of the potato’s temperature? Write several careful sentences that discuss, 61 Chapter 1 Understanding the Derivative with appropriate units, the values of F(30), F′(30), and F′′(30), and explain the overall behavior of the potato’s temperature at this point in time. d. Overall, is the potato’s temperature increasing at an increasing rate, increasing at a constant rate, or increasing at a decreasing rate? Why? Activity 1.6.4. This activity builds on our experience and understanding of how to sketch the graph of f ′ given the graph of f . In Figure 1.6.14, given the respective graphs of two diﬀerent functions f , sketch the corresponding graph of f ′ on the ﬁrst axes below, and then sketch f ′′ on the second set of axes. In addition, for each, write several careful sentences in the spirit of those in Activity 1.6.2 that connect the behaviors of f , f ′, and f ′′. For instance, write some- thing such as f ′ is on the interval , which is connected to the fact that f is on the same interval , and f ′′ is on the interval. but of course with the blanks ﬁlled in. Throughout, view the scale of the grid for the graph of f as being 1 × 1, and assume the horizontal scale of the grid for the graph of f ′ is identical to that for f . If you need to adjust the vertical scale on the axes for the graph of f ′ or f ′′, you should label that accordingly. 1.6.4 Summary • A diﬀerentiable function f is increasing on an interval whenever its ﬁrst derivative is positive, and decreasing whenever its ﬁrst derivative is negative. • By taking the derivative of the derivative of a function f , we arrive at the second deriv- ative, f ′′. The second derivative measures the instantaneous rate of change of the ﬁrst derivative. The sign of the second derivative tells us whether the slope of the tangent line to f is increasing or decreasing. • A diﬀerentiable function is concave up whenever its ﬁrst derivative is increasing (or equivalently whenever its second derivative is positive), and concave down whenever its ﬁrst derivative is decreasing (or equivalently whenever its second derivative is neg- ative). Examples of functions that are everywhere concave up are y x 2 and y e x ; examples of functions that are everywhere concave down are y −x 2 and y −e x . • The units on the second derivative are “units of output per unit of input per unit of input.” They tell us how the value of the derivative function is changing in response to changes in the input. In other words, the second derivative tells us the rate of change of the rate of change of the original function. 62 1.6 The second derivative f f x x f′ f′ x x f ′′ f ′′ x x Figure 1.6.14: Two given functions f , with axes provided for plotting f ′ and f ′′ below. 63 Chapter 1 Understanding the Derivative 1.6.5 Exercises 1. Comparing f , f ′ , f ′′ values. Consider the function f (x) graphed below. For this function, are the following nonzero quantities positive or negative? f (0.5), f ′(0.5), f ′′(0.5) 2. Signs of f , f ′ , f ′′ values. At exactly two of the labeled points in the ﬁgure below, which shows a function f , the derivative f ′ is zero; the second derivative f ′′ is not zero at any of the labeled points. Give the sign for each of f , f ′ and f ′′ at each marked point. 3. Acceleration from velocity. Suppose that an accelerating car goes from 0 mph to 64.1 mph in ﬁve seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: 1 mph is 22/15 ft/sec.) Find the average acceleration of the car over each of the ﬁrst two seconds. t (s) 0 1 2 3 4 5 v(t) (ft/s) 0.00 32.05 55.55 72.64 85.45 94.00 average acceleration over the ﬁrst second = average acceleration over the second second = 64 1.6 The second derivative 4. Rates of change of stock values. Let P(t) represent the price of a share of stock of a corporation at time t. What does each of the following statements tell us about the signs of the ﬁrst and second derivatives of P(t)? (a) The price of the stock is falling slower and slower. The ﬁrst derivative of P(t) is (□ positive □ zero □ negative) The second derivative of P(t) is (□ positive □ zero □ negative) (b) The price of the stock is close to bottoming out. The ﬁrst derivative of P(t) is (□ positive □ zero □ negative) The second derivative of P(t) is (□ positive □ zero □ negative) 5. Interpreting a graph of f ′. The graph of f ′ (not f ) is given below. (Note that this is a graph of f ′, not a graph of f .) At which of the marked values of x is A. f (x) greatest? B. f (x) least? C. f ′(x) greatest? D. f ′(x) least? E. f ′′(x) greatest? F. f ′′(x) least? 6. Suppose that y f (x) is a twice-diﬀerentiable function such that f ′′ is continuous for which the following information is known: f (2) −3, f ′(2) 1.5, f ′′(2) −0.25. a. Is f increasing or decreasing near x 2? Is f concave up or concave down near x 2? b. Do you expect f (2.1) to be greater than −3, equal to −3, or less than −3? Why? c. Do you expect f ′(2.1) to be greater than 1.5, equal to 1.5, or less than 1.5? Why? d. Sketch a graph of y f (x) near (2, f (2)) and include a graph of the tangent line. 65 Chapter 1 Understanding the Derivative 7. For a certain function y 1(x), its derivative is given by the function pictured in Fig- ure 1.6.15. y = g′ (x) 4 2 -3 -1 1 3 Figure 1.6.15: The graph of y 1 ′(x). a. What is the approximate slope of the tangent line to y 1(x) at the point (2, 1(2))? b. How many real number solutions can there be to the equation 1(x) 0? Justify your conclusion fully and carefully by explaining what you know about how the graph of 1 must behave based on the given graph of 1 ′. c. On the interval −3 < x < 3, how many times does the concavity of 1 change? Why? d. Use the provided graph to estimate the value of 1 ′′(2). 8. A bungee jumper’s height h (in feet ) at time t (in seconds) is given in part by the table: t 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 h(t) 200 184.2 159.9 131.9 104.7 81.8 65.5 56.8 55.5 60.4 69.8 t 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 h(t) 81.6 93.7 104.4 112.6 117.7 119.4 118.2 114.8 110.0 104.7 a. Use the given data to estimate h ′(4.5), h ′(5), and h ′(5.5). At which of these times is the bungee jumper rising most rapidly? b. Use the given data and your work in (a) to estimate h ′′(5). c. What physical property of the bungee jumper does the value of h ′′(5) measure? What are its units? d. Based on the data, on what approximate time intervals is the function y h(t) concave down? What is happening to the velocity of the bungee jumper on these time intervals? 66 1.6 The second derivative 9. For each prompt that follows, sketch a possible graph of a function on the interval −3 < x < 3 that satisﬁes the stated properties. a. y f (x) such that f is increasing on −3 < x < 3, concave up on −3 < x < 0, and concave down on 0 < x < 3. b. y 1(x) such that 1 is increasing on −3 < x < 3, concave down on −3 < x < 0, and concave up on 0 < x < 3. c. y h(x) such that h is decreasing on −3 < x < 3, concave up on −3 < x < −1, neither concave up nor concave down on −1 < x < 1, and concave down on 1 < x < 3. d. y p(x) such that p is decreasing and concave down on −3 < x < 0 and is increasing and concave down on 0 < x < 3. 67 Chapter 1 Understanding the Derivative 1.7 Limits, Continuity, and Diﬀerentiability Motivating Questions • What does it mean graphically to say that f has limit L as x → a? How is this con- nected to having a left-hand limit at x a and having a right-hand limit at x a? • What does it mean to say that a function f is continuous at x a? What role do limits play in determining whether or not a function is continuous at a point? • What does it mean graphically to say that a function f is diﬀerentiable at x a? How is this connected to the function being locally linear? • How are the characteristics of a function having a limit, being continuous, and being diﬀerentiable at a given point related to one another? In Section 1.2, we learned how limits can be used to study the trend of a function near a ﬁxed input value. In this section, we aim to quantify how the function acts and how its values change near a particular point. If the function has a limit L at x a, we will consider how the value of the function f (a) is related to limx→a f (x), and whether or not the function has a derivative f ′(a) at x a. Preview Activity 1.7.1. A function f deﬁned on −4 < x < 4 is given by the graph in Figure 1.7.1. Use the graph to answer each of the following questions. Note: to the right of x 2, the graph of f is exhibiting inﬁnite oscillatory behavior similar to the function sin( πx ) that we encountered in the key example early in Section 1.2. f 3 2 1 -3 -2 -1 1 2 3 -1 -2 -3 Figure 1.7.1: The graph of y f (x). a. For each of the values a −3, −2, −1, 0, 1, 2, 3, determine whether or not lim f (x) x→a exists. If the function has a limit L at a given point, state the value of the limit 68 1.7 Limits, Continuity, and Diﬀerentiability using the notation limx→a f (x) L. If the function does not have a limit at a given point, write a sentence to explain why. b. For each of the values of a from part (a) where f has a limit, determine the value of f (a) at each such point. In addition, for each such a value, does f (a) have the same value as limx→a f (x)? c. For each of the values a −3, −2, −1, 0, 1, 2, 3, determine whether or not f ′(a) exists. In particular, based on the given graph, ask yourself if it is reasonable to say that f has a tangent line at (a, f (a)) for each of the given a-values. If so, visually estimate the slope of the tangent line to ﬁnd the value of f ′(a). 1.7.1 Having a limit at a point In Section 1.2, we learned that f has limit L as x approaches a provided that we can make the value of f (x) as close to L as we like by taking x suﬃciently close (but not equal to) a. If so, we write limx→a f (x) L. Essentially there are two behaviors that a function can exhibit near a point where it fails to have a limit. In Figure 1.7.3, at left we see a function f whose graph shows a jump at a 1. If we let x approach 1 from the left side, the value of f approaches 2, but if we let x approach 1 from the right, the value of f tends to 3. Because the value of f does not approach a single number as x gets arbitrarily close to 1 from both sides, we know that f does not have a limit at a 1. For such cases, we introduce the notion of left and right (or one-sided) limits. Deﬁnition 1.7.2 We say that f has limit L1 as x approaches a from the left and write lim f (x) L1 x→a − provided that we can make the value of f (x) as close to L1 as we like by taking x suﬃciently close to a while always having x < a. We call L1 the left-hand limit of f as x approaches a. Similarly, we say L2 is the right-hand limit of f as x approaches a and write lim f (x) L2 x→a + provided that we can make the value of f (x) as close to L2 as we like by taking x suﬃciently close to a while always having x > a. In the graph of the function f in Figure 1.7.3, we see that lim f (x) 2 and lim+ f (x) 3. x→1− x→1 Precisely because the left and right limits are not equal, the overall limit of f as x → 1 fails to exist. 69 Chapter 1 Understanding the Derivative f g 3 3 2 2 1 1 1 Figure 1.7.3: Functions f and 1 that each fail to have a limit at a 1. For the function 1 pictured at right in Figure 1.7.3, the function fails to have a limit at a 1 for a diﬀerent reason. While the function does not have a jump in its graph at a 1, it is still not the case that 1 approaches a single value as x approaches 1. In particular, due to the inﬁnitely oscillating behavior of 1 to the right of a 1, we say that the right-hand limit of 1 as x → 1+ does not exist, and thus limx→1 1(x) does not exist. To summarize, if either a left- or right-hand limit fails to exist or if the left- and right-hand limits are not equal to each other, the overall limit does not exist. A function f has limit L as x → a if and only if lim f (x) L lim+ f (x). x→a − x→a That is, a function has a limit at x a if and only if both the left- and right-hand limits at x a exist and have the same value. In Preview Activity 1.7.1, the function f given in Figure 1.7.1 fails to have a limit at only two values: at a −2 (where the left- and right-hand limits are 2 and −1, respectively) and at x 2, where limx→2+ f (x) does not exist). Note well that even at values such as a −1 and a 0 where there are holes in the graph, the limit still exists. 70 1.7 Limits, Continuity, and Diﬀerentiability Activity 1.7.2. Consider a function that is piecewise-deﬁned according to the formula 3(x + 2) + 2 for −3 < x < −2 2 (x + 2) + 1 for −2 ≤ x < −1 3 f (x) 2 (x + 2) + 1 for −1 < x < 1 3 for x 1 2 4 − x for x > 1 Use the given formula to answer the following questions. 3 2 1 -2 -1 1 2 -1 Figure 1.7.4: Axes for plotting the function y f (x) in Activity 1.7.2. a. For each of the values a −2, −1, 0, 1, 2, compute f (a). b. For each of the values a −2, −1, 0, 1, 2, determine lim− f (x) and lim+ f (x). x→a x→a c. For each of the values a −2, −1, 0, 1, 2, determine limx→a f (x). If the limit fails to exist, explain why by discussing the left- and right-hand limits at the relevant a-value. d. For which values of a is the following statement true? lim f (x) , f (a) x→a e. On the axes provided in Figure 1.7.4, sketch an accurate, labeled graph of y f (x). Be sure to carefully use open circles (◦) and ﬁlled circles (•) to represent key points on the graph, as dictated by the piecewise formula. 71 Chapter 1 Understanding the Derivative 1.7.2 Being continuous at a point Intuitively, a function is continuous if we can draw its graph without ever lifting our pencil from the page. Alternatively, we might say that the graph of a continuous function has no jumps or holes in it. In Figure 1.7.5 we consider three functions that have a limit at a 1, and use them to make the idea of continuity more precise. f g h 3 3 3 2 2 2 1 1 1 Figure 1.7.5: Functions f , 1, and h that demonstrate subtly diﬀerent behaviors at a 1. First consider the function in the left-most graph. Note that f (1) is not deﬁned, which leads to the resulting hole in the graph of f at a 1. We will naturally say that f is not continuous at a 1. For the function 1, we observe that while limx→1 1(x) 3, the value of 1(1) 2, and thus the limit does not equal the function value. Here, too, we will say that 1 is not continuous, even though the function is deﬁned at a 1. Finally, the function h appears to be the most well-behaved of all three, since at a 1 its limit and its function value agree. That is, lim h(x) 3 h(1). x→1 With no hole or jump in the graph of h at a 1, we say that h is continuous there. More formally, we make the following deﬁnition. Deﬁnition 1.7.6 A function f is continuous at x a provided that a. f has a limit as x → a, b. f is deﬁned at x a, and c. limx→a f (x) f (a). Conditions (a) and (b) are technically contained implicitly in (c), but we state them explicitly to emphasize their individual importance. The deﬁnition says that a function is continuous at x a provided that its limit as x → a exists and equals its function value at x a. If a function is continuous at every point in an interval [a, b], we say the function is “continuous on [a, b].” If a function is continuous at every point in its domain, we simply say the function is “continuous.” Thus, continuous functions are particularly nice: to evaluate the limit of a continuous function at a point, all we need to do is evaluate the function. For example, consider p(x) x 2 − 2x + 3. It can be proved that every polynomial is a 72 1.7 Limits, Continuity, and Diﬀerentiability continuous function at every real number, and thus if we would like to know limx→2 p(x), we simply compute lim (x 2 − 2x + 3) 22 − 2 · 2 + 3 3. x→2 This route of substituting an input value to evaluate a limit works whenever we know that the function being considered is continuous. Besides polynomial functions, all exponential functions and the sine and cosine functions are continuous at every point, as are many other familiar functions and combinations thereof. Activity 1.7.3. This activity builds on your work in Preview Activity 1.7.1, using the same function f as given by the graph that is repeated in Figure 1.7.7. f 3 2 1 -3 -2 -1 1 2 3 -1 -2 -3 Figure 1.7.7: The graph of y f (x) for Activity 1.7.3. a. At which values of a does limx→a f (x) not exist? b. At which values of a is f (a) not deﬁned? c. At which values of a does f have a limit, but limx→a f (x) , f (a)? d. State all values of a for which f is not continuous at x a. e. Which condition is stronger, and hence implies the other: f has a limit at x a or f is continuous at x a? Explain, and hence complete the following sentence: “If f at x a, then f at x a,” where you complete the blanks with has a limit and is continuous, using each phrase once. 1.7.3 Being diﬀerentiable at a point We recall that a function f is said to be diﬀerentiable at x a if f ′(a) exists. Moreover, for f ′(a) to exist, we know that the function y f (x) must have a tangent line at the point 73 Chapter 1 Understanding the Derivative (a, f (a)), since f ′(a) is precisely the slope of this line. In order to even ask if f has a tangent line at (a, f (a)), it is necessary that f be continuous at x a: if f fails to have a limit at x a, if f (a) is not deﬁned, or if f (a) does not equal the value of limx→a f (x), then it doesn’t make sense to talk about a tangent line to the curve at this point. Indeed, it can be proved formally that if a function f is diﬀerentiable at x a, then it must be continuous at x a. So, if f is not continuous at x a, then it is automatically the case that f is not diﬀerentiable there. For example, in Figure 1.7.5, both f and 1 fail to be diﬀerentiable at x 1 because neither function is continuous at x 1. But can a function fail to be diﬀerentiable at a point where the function is continuous? In Figure 1.7.8, the function has a sharp corner at a point. For the pictured function f , we observe that f is clearly continuous at a 1, since limx→1 f (x) 1 f (1). f 1 (1, 1) 1 Figure 1.7.8: A function f that is continuous at a 1 but not diﬀerentiable at a 1; at right, we zoom in on the point (1, 1) in a magniﬁed version of the box in the left-hand plot. But the function f in Figure 1.7.8 is not diﬀerentiable at a 1 because f ′(1) fails to exist. One way to see this is to observe that f ′(x) −1 for every value of x that is less than 1, while f ′(x) +1 for every value of x that is greater than 1. That makes it seem that either +1 or −1 would be equally good candidates for the value of the derivative at x 1. Alternately, we could use the limit deﬁnition of the derivative to attempt to compute f ′(1), and discover that the derivative does not exist. Finally, we can see visually that the function f in Figure 1.7.8 does not have a tangent line. When we zoom in on (1, 1) on the graph of f , no matter how closely we examine the function, it will always look like a “V”, and never like a single line, which tells us there is no possibility for a tangent line there. If a function does have a tangent line at a given point, when we zoom in on the point of tangency, the function and the tangent line should appear essentially indistinguishable¹. Conversely, if we zoom in on a point and the function looks like a single straight line, then the function should have a tangent line there, and thus be diﬀerentiable. Hence, a function that is diﬀerentiable at x a will, up close, look more and more like its tangent line at (a, f (a)). Therefore, we say that a function that is diﬀerentiable at x a is locally linear. To summarize the preceding discussion of diﬀerentiability and continuity, we make several ¹See, for instance, http://gvsu.edu/s/6J for an applet (due to David Austin, GVSU) where zooming in shows the increasing similarity between the tangent line and the curve. 74 1.7 Limits, Continuity, and Diﬀerentiability important observations. • If f is diﬀerentiable at x a, then f is continuous at x a. Equivalently, if f fails to be continuous at x a, then f will not be diﬀerentiable at x a. • A function can be continuous at a point, but not be diﬀerentiable there. In particular, a function f is not diﬀerentiable at x a if the graph has a sharp corner (or cusp) at the point (a, f (a)). • If f is diﬀerentiable at x a, then f is locally linear at x a. That is, when a function is diﬀerentiable, it looks linear when viewed up close because it re- sembles its tangent line there. Activity 1.7.4. In this activity, we explore two diﬀerent functions and classify the points at which each is not diﬀerentiable. Let 1 be the function given by the rule 1(x) |x|, and let f be the function that we have previously explored in Preview Activity 1.7.1, whose graph is given again in Figure 1.7.9. f 3 2 1 -3 -2 -1 1 2 3 -1 -2 -3 Figure 1.7.9: The graph of y f (x) for Activity 1.7.4. a. Reasoning visually, explain why 1 is diﬀerentiable at every point x such that x , 0. |h| b. Use the limit deﬁnition of the derivative to show that 1 ′(0) limh→0 h . c. Explain why 1 ′(0) fails to exist by using small positive and negative values of h. d. State all values of a for which f is not diﬀerentiable at x a. For each, provide a reason for your conclusion. e. True or false: if a function p is diﬀerentiable at x b, then limx→b p(x) must exist. Why? 75 Chapter 1 Understanding the Derivative 1.7.4 Summary • A function f has limit L as x → a if and only if f has a left-hand limit at x a, f has a right-hand limit at x a, and the left- and right-hand limits are equal. Visually, this means that there can be a hole in the graph at x a, but the function must approach the same single value from either side of x a. • A function f is continuous at x a whenever f (a) is deﬁned, f has a limit as x → a, and the value of the limit and the value of the function agree. This guarantees that there is not a hole or jump in the graph of f at x a. • A function f is diﬀerentiable at x a whenever f ′(a) exists, which means that f has a tangent line at (a, f (a)) and thus f is locally linear at x a. Informally, this means that the function looks like a line when viewed up close at (a, f (a)) and that there is not a corner point or cusp at (a, f (a)). • Of the three conditions discussed in this section (having a limit at x a, being con- tinuous at x a, and being diﬀerentiable at x a), the strongest condition is being diﬀerentiable, and the next strongest is being continuous. In particular, if f is diﬀer- entiable at x a, then f is also continuous at x a, and if f is continuous at x a, then f has a limit at x a. 1.7.5 Exercises 1. Limit values of a piecewise graph. Use the ﬁgure below, which gives a graph of the function f (x), to give values for the indicated limits. If a limit does not exist, enter none. (a) lim f (x) x→−1 (b) lim f (x) x→0 (c) lim f (x) x→1 (d) lim f (x) x→4 2. Limit values of a piecewise formula. For the function 3x − 2, 0≤x<1 f (x) 5, x1 x 2 − 2x + 2, 1 < x use algebra to ﬁnd each of the following limits: lim f (x) x→1+ lim f (x) x→1− 76 1.7 Limits, Continuity, and Diﬀerentiability lim f (x) x→1 Sketch a graph of f (x) to conﬁrm your answers. 3. Continuity and diﬀerentiability of a graph. Consider the function graphed below. At what x-values does the function appear to not be continuous? At what x-values does the function appear to not be diﬀerentiable? 4. Continuity of a piecewise formula. Find k so that the following function is continuous: { kx if 0 ≤ x < 2 f (x) 5x 2 if 2 ≤ x. 5. Consider the graph of the function y p(x) that is provided in Figure 1.7.10. Assume that each portion of the graph of p is a straight line, as pictured. a. State all values of a for which limx→a p(x) does not exist. b. State all values of a for which p is not continuous at a. c. State all values of a for which p is not diﬀerentiable at x a. d. On the axes provided in Figure 1.7.10, sketch an accurate graph of y p ′(x). 77 Chapter 1 Understanding the Derivative p 3 3 -3 3 -3 3 -3 -3 Figure 1.7.10: At left, the piecewise linear function y p(x). At right, axes for plotting y p ′(x). 6. For each of the following prompts, give an example of a function that satisﬁes the stated criteria. A formula or a graph, with reasoning, is suﬃcient for each. If no such example is possible, explain why. a. A function f that is continuous at a 2 but not diﬀerentiable at a 2. b. A function 1 that is diﬀerentiable at a 3 but does not have a limit at a 3. c. A function h that has a limit at a −2, is deﬁned at a −2, but is not continuous at a −2. d. A function p that satisﬁes all of the following: • p(−1) 3 and limx→−1 p(x) 2 • p(0) 1 and p ′(0) 0 • limx→1 p(x) p(1) and p ′(1) does not exist 7. Let h(x) be a function whose derivative y h ′(x) is given by the graph on the right in Figure 1.7.11. a. Based on the graph of y h ′(x), what can you say about the behavior of the function y h(x)? b. At which values of x is y h ′(x) not deﬁned? What behavior does this lead you to expect to see in the graph of y h(x)? c. Is it possible for y h(x) to have points where h is not continuous? Explain your answer. d. On the axes provided at left, sketch at least two distinct graphs that are possible functions y h(x) that each have a derivative y h ′(x) that matches the provided graph at right. Explain why there are multiple possibilities for y h(x). 78 1.7 Limits, Continuity, and Diﬀerentiability 3 3 2 y = h′ (x) 1 -3 3 -3 -2 -1 1 2 3 -1 -2 -3 -3 Figure 1.7.11: Axes for plotting y h(x) and, at right, the graph of y h ′(x). √ 8. Consider the function 1(x) |x|. a. Use a graph to explain visually why 1 is not diﬀerentiable at x 0. b. Use the limit deﬁnition of the derivative to show that √ ′ |h| 1 (0) lim . h→0 h c. Investigate the value of 1 ′(0) by estimating the limit in (b) using small positive √ |−0.01| and negative values of h. For instance, you might compute 0.01 . Be sure to use several diﬀerent values of h (both positive and negative), including ones closer to 0 than 0.01. What do your results tell you about 1 ′(0)? d. Use your graph in (a) to sketch an approximate graph of y 1 ′(x). 79 Chapter 1 Understanding the Derivative 1.8 The Tangent Line Approximation Motivating Questions • What is the formula for the general tangent line approximation to a diﬀerentiable function y f (x) at the point (a, f (a))? • What is the principle of local linearity and what is the local linearization of a diﬀer- entiable function f at a point (a, f (a))? • How does knowing just the tangent line approximation tell us information about the behavior of the original function itself near the point of approximation? How does knowing the second derivative’s value at this point provide us additional knowledge of the original function’s behavior? Among all functions, linear functions are simplest. One of the powerful consequences of a function y f (x) being diﬀerentiable at a point (a, f (a)) is that, up close, the function y f (x) is locally linear and looks like its tangent line at that point. In certain circum- stances, this allows us to approximate the original function f with a simpler function L that is linear: this can be advantageous when we have limited information about f or when f is computationally or algebraically complicated. We will explore all of these situations in what follows. It is essential to recall that when f is diﬀerentiable at x a, the value of f ′(a) provides the slope of the tangent line to y f (x) at the point (a, f (a)). If we know both a point on the line and the slope of the line we can ﬁnd the equation of the tangent line and write the equation in point-slope form¹. Preview Activity 1.8.1. Consider the function y 1(x) −x 2 + 3x + 2. a. Use the limit deﬁnition of the derivative to compute a formula for y 1 ′(x). b. Determine the slope of the tangent line to y 1(x) at the value x 2. c. Compute 1(2). d. Find an equation for the tangent line to y 1(x) at the point (2, 1(2)). Write your result in point-slope form. e. On the axes provided in Figure 1.8.1, sketch an accurate, labeled graph of y 1(x) along with its tangent line at the point (2, 1(2)). ¹Recall that a line with slope m that passes through (x 0 , y0 ) has equation y − y0 m(x − x0 ), and this is the point-slope form of the equation. 80 1.8 The Tangent Line Approximation Figure 1.8.1: Axes for plotting y 1(x) and its tangent line to the point (2, 1(2)). 1.8.1 The tangent line Given a function f that is diﬀerentiable at x a, we know that we can determine the slope of the tangent line to y f (x) at (a, f (a)) by computing f ′(a). The equation of the resulting tangent line is given in point-slope form by y − f (a) f ′(a)(x − a) or y f ′(a)(x − a) + f (a). Note well: there is a major diﬀerence between f (a) and f (x) in this context. The former is a constant that results from using the given ﬁxed value of a, while the latter is the general expression for the rule that deﬁnes the function. The same is true for f ′(a) and f ′(x): we must carefully distinguish between these expressions. Each time we ﬁnd the tangent line, we need to evaluate the function and its derivative at a ﬁxed a-value. In Figure 1.8.2, we see the graph of a function f and its tangent line at the point (a, f (a)). Notice how when we zoom in we see the local linearity of f more clearly highlighted. The function and its tangent line are nearly indistinguishable up close. Local linearity can also be seen dynamically in the java applet at http://gvsu.edu/s/6J. 1.8.2 The local linearization A slight change in perspective and notation will enable us to be more precise in discussing how the tangent line approximates f near x a. By solving for y, we can write the equation for the tangent line as y f ′(a)(x − a) + f (a) This line is itself a function of x. Replacing the variable y with the expression L(x), we call L(x) f ′(a)(x − a) + f (a) 81 Chapter 1 Understanding the Derivative y y = f (x) y = f (x) (a, f (a)) (a, f (a)) y = f ′ (a)(x − a) + f (a) y = L(x) a x Figure 1.8.2: A function y f (x) and its tangent line at the point (a, f (a)): at left, from a distance, and at right, up close. At right, we label the tangent line function by y L(x) and observe that for x near a, f (x) ≈ L(x). the local linearization of f at the point (a, f (a)). In this notation, L(x) is nothing more than a new name for the tangent line. As we saw above, for x close to a, f (x) ≈ L(x). Example 1.8.3 Suppose that a function y f (x) has its tangent line approximation given by L(x) 3 − 2(x − 1) at the point (1, 3), but we do not know anything else about the function f . To estimate a value of f (x) for x near 1, such as f (1.2), we can use the fact that f (1.2) ≈ L(1.2) and hence f (1.2) ≈ L(1.2) 3 − 2(1.2 − 1) 3 − 2(0.2) 2.6. We emphasize that y L(x) is simply a new name for the tangent line function. Using this new notation and our observation that L(x) ≈ f (x) for x near a, it follows that we can write f (x) ≈ f (a) + f ′(a)(x − a) for x near a. Activity 1.8.2. Suppose it is known that for a given diﬀerentiable function y 1(x), its local linearization at the point where a −1 is given by L(x) −2 + 3(x + 1). a. Compute the values of L(−1) and L′(−1). b. What must be the values of 1(−1) and 1 ′(−1)? Why? c. Do you expect the value of 1(−1.03) to be greater than or less than the value of 1(−1)? Why? d. Use the local linearization to estimate the value of 1(−1.03). e. Suppose that you also know that 1 ′′(−1) 2. What does this tell you about the graph of y 1(x) at a −1? 82 1.8 The Tangent Line Approximation f. For x near −1, sketch the graph of the local linearization y L(x) as well as a possible graph of y 1(x) on the axes provided in Figure 1.8.4. Figure 1.8.4: Axes for plotting y L(x) and y 1(x). From Activity 1.8.2, we see that the local linearization y L(x) is a linear function that shares two important values with the function y f (x) that it is derived from. In particular, • because L(x) f (a) + f ′(a)(x − a), it follows that L(a) f (a); and • because L is a linear function, its derivative is its slope. Hence, L′(x) f ′(a) for every value of x, and speciﬁcally L′(a) f ′(a). Therefore, we see that L is a linear function that has both the same value and the same slope as the function f at the point (a, f (a)). Thus, if we know the linear approximation y L(x) for a function, we know the original function’s value and its slope at the point of tangency. What remains unknown, however, is the shape of the function f at the point of tangency. There are essentially four possibilities, as shown in Figure 1.8.5. Figure 1.8.5: Four possible graphs for a nonlinear diﬀerentiable function and how it can be situated relative to its tangent line at a point. 83 Chapter 1 Understanding the Derivative These possible shapes result from the fact that there are three options for the value of the second derivative: either f ′′(a) < 0, f ′′(a) 0, or f ′′(a) > 0. • If f ′′(a) > 0, then we know the graph of f is concave up, and we see the ﬁrst possibility on the left, where the tangent line lies entirely below the curve. • If f ′′(a) < 0, then f is concave down and the tangent line lies above the curve, as shown in the second ﬁgure. • If f ′′(a) 0 and f ′′ changes sign at x a, the concavity of the graph will change, and we will see either the third or fourth ﬁgure.². • A ﬁfth option (which is not very interesting) can occur if the function f itself is linear, so that f (x) L(x) for all values of x. The plots in Figure 1.8.5 highlight yet another important thing that we can learn from the concavity of the graph near the point of tangency: whether the tangent line lies above or below the curve itself. This is key because it tells us whether or not the tangent line ap- proximation’s values will be too large or too small in comparison to the true value of f . For instance, in the ﬁrst situation in the leftmost plot in Figure 1.8.5 where f ′′(a) > 0, because the tangent line falls below the curve, we know that L(x) ≤ f (x) for all values of x near a. Activity 1.8.3. This activity concerns a function f (x) about which the following infor- mation is known: • f is a diﬀerentiable function deﬁned at every real number x • f (2) −1 • y f ′(x) has its graph given in Figure 1.8.6 y = f ′ (x) 2 2 2 x x x 2 2 2 Figure 1.8.6: At center, a graph of y f ′(x); at left, axes for plotting y f (x); at right, axes for plotting y f ′′(x). Your task is to determine as much information as possible about f (especially near the value a 2) by responding to the questions below. ²It is possible that f ′′ (a) 0 but f ′′ does not change sign at x a, in which case the graph will look like one of the ﬁrst two options. 84 1.8 The Tangent Line Approximation a. Find a formula for the tangent line approximation, L(x), to f at the point (2, −1). b. Use the tangent line approximation to estimate the value of f (2.07). Show your work carefully and clearly. c. Sketch a graph of y f ′′(x) on the righthand grid in Figure 1.8.6; label it ap- propriately. d. Is the slope of the tangent line to y f (x) increasing, decreasing, or neither when x 2? Explain. e. Sketch a possible graph of y f (x) near x 2 on the lefthand grid in Fig- ure 1.8.6. Include a sketch of y L(x) (found in part (a)). Explain how you know the graph of y f (x) looks like you have drawn it. f. Does your estimate in (b) over- or under-estimate the true value of f (2.07)? Why? The idea that a diﬀerentiable function looks linear and can be well-approximated by a lin- ear function is an important one that ﬁnds wide application in calculus. For example, by approximating a function with its local linearization, it is possible to develop an eﬀective algorithm to estimate the zeroes of a function. Local linearity also helps us to make further sense of certain challenging limits. For instance, we have seen that the limit sin(x) lim x→0 x is indeterminate, because both its numerator and denominator tend to 0. While there is no sin(x) algebra that we can do to simplify x , it is straightforward to show that the linearization of f (x) sin(x) at the point (0, 0) is given by L(x) x. Hence, for values of x near 0, sin(x) ≈ x, and therefore sin(x) x ≈ 1, x x which makes plausible the fact that sin(x) lim 1. x→0 x 1.8.3 Summary • The tangent line to a diﬀerentiable function y f (x) at the point (a, f (a)) is given in point-slope form by the equation y − f (a) f ′(a)(x − a). • The principle of local linearity tells us that if we zoom in on a point where a function y f (x) is diﬀerentiable, the function will be indistinguishable from its tangent line. 85 Chapter 1 Understanding the Derivative That is, a diﬀerentiable function looks linear when viewed up close. We rename the tangent line to be the function y L(x), where L(x) f (a) + f ′(a)(x − a). Thus, f (x) ≈ L(x) for all x near x a. • If we know the tangent line approximation L(x) f (a) + f ′(a)(x − a) to a function y f (x), then because L(a) f (a) and L′(a) f ′(a), we also know the values of both the function and its derivative at the point where x a. In other words, the linear approximation tells us the height and slope of the original function. If, in addition, we know the value of f ′′(a), we then know whether the tangent line lies above or below the graph of y f (x), depending on the concavity of f . 1.8.4 Exercises √ √ 1. Approximating x. Use linear approximation to approximate 36.1 as follows. √ Let f (x) x. The equation of the tangent line to f (x) at x 36 can be written in the √ form y mx + b. Compute m and b. Using this ﬁnd the approximation for 36.1. 2. Local linearization of a graph. The ﬁgure below shows f (x) and its local linearization at x a, y 4x − 4. (The local linearization is shown in blue.) What is the value of a? What is the value of f (a)? Use the linearization to approximate the value of f (3.2). Is the approximation an under- or overestimate? 3. Estimating with the local linearization. Suppose that f (x) is a function with f (130) 46 and f ′(130) 1. Estimate f (125.5). 4. Predicting behavior from the local linearization. The temperature, H, in degrees Cel- sius, of a cup of coﬀee placed on the kitchen counter is given by H f (t), where t is in minutes since the coﬀee was put on the counter. (a) Is f ′(t) positive or negative? (b) What are the units of f ′(30)? Suppose that | f ′(30)| 0.9 and f (30) 51. Fill in the blanks (including units where needed) and select the appropriate terms to complete the following statement about the temperature of the coﬀee in this case. At minutes after the coﬀee was put on the counter, its (□ derivative □ temperature □ change in temperature) is and will (□ increase □ decrease) 86 1.8 The Tangent Line Approximation by about in the next 75 seconds. 5. A certain function y p(x) has its local linearization at a 3 given by L(x) −2x + 5. a. What are the values of p(3) and p ′(3)? Why? b. Estimate the value of p(2.79). c. Suppose that p ′′(3) 0 and you know that p ′′(x) < 0 for x < 3. Is your estimate in (b) too large or too small? d. Suppose that p ′′(x) > 0 for x > 3. Use this fact and the additional information above to sketch an accurate graph of y p(x) near x 3. Include a sketch of y L(x) in your work. 6. A potato is placed in an oven, and the potato’s temperature F (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time t is measured in minutes. t F(t) 0 70 15 180.5 30 251 45 296 60 324.5 75 342.8 90 354.5 Table 1.8.7: Temperature data for the potato. a. Use a central diﬀerence to estimate F′(60). Use this estimate as needed in subse- quent questions. b. Find the local linearization y L(t) to the function y F(t) at the point where a 60. c. Determine an estimate for F(63) by employing the local linearization. d. Do you think your estimate in (c) is too large or too small? Why? 7. An object moving along a straight line path has a diﬀerentiable position function y s(t); s(t) measures the object’s position relative to the origin at time t. It is known that at time t 9 seconds, the object’s position is s(9) 4 feet (i.e., 4 feet to the right of the origin). Furthermore, the object’s instantaneous velocity at t 9 is −1.2 feet per second, and its acceleration at the same instant is 0.08 feet per second per second. a. Use local linearity to estimate the position of the object at t 9.34. b. Is your estimate likely too large or too small? Why? c. In everyday language, describe the behavior of the moving object at t 9. Is it moving toward the origin or away from it? Is its velocity increasing or decreasing? 87 Chapter 1 Understanding the Derivative For a certain function f , its derivative is known to be f ′(x) (x − 1)e −x . Note that you 2 8. do not know a formula for y f (x). a. At what x-value(s) is f ′(x) 0? Justify your answer algebraically, but include a graph of f ′ to support your conclusion. b. Reasoning graphically, for what intervals of x-values is f ′′(x) > 0? What does this tell you about the behavior of the original function f ? Explain. c. Assuming that f (2) −3, estimate the value of f (1.88) by ﬁnding and using the tangent line approximation to f at x 2. Is your estimate larger or smaller than the true value of f (1.88)? Justify your answer. 88 CHAPTER 2 Computing Derivatives 2.1 Elementary derivative rules Motivating Questions • What are alternate notations for the derivative? • How can we use the algebraic structure of a function f (x) to compute a formula for f ′(x)? • What is the derivative of a power function of the form f (x) x n ? What is the deriv- ative of an exponential function of form f (x) a x ? • If we know the derivative of y f (x), what is the derivative of y k f (x), where k is a constant? • If we know the derivatives of y f (x) and y 1(x), how do we compute the deriv- ative of y f (x) + 1(x)? In Chapter 1, we developed the concept of the derivative of a function. We now know that the derivative f ′ of a function f measures the instantaneous rate of change of f with respect to x. The derivative also tells us the slope of the tangent line to y f (x) at any given value of x. So far, we have focused on interpreting the derivative graphically or, in the context of a physical setting, as a meaningful rate of change. To calculate the value of the derivative at a speciﬁc point, we have relied on the limit deﬁnition of the derivative, f (x + h) − f (x) f ′(x) lim . h→0 h In this chapter, we investigate how the limit deﬁnition of the derivative leads to interesting patterns and rules that enable us to ﬁnd a formula for f ′(x) quickly, without using the limit deﬁnition directly. For example, we would like to apply shortcuts to diﬀerentiate a function such as 1(x) 4x 7 − sin(x) + 3e x Chapter 2 Computing Derivatives Preview Activity 2.1.1. Functions of the form f (x) x n , where n 1, 2, 3, . . ., are of- ten called power functions. The ﬁrst two questions below revisit work we did earlier in Chapter 1, and the following questions extend those ideas to higher powers of x. a. Use the limit deﬁnition of the derivative to ﬁnd f ′(x) for f (x) x 2 . b. Use the limit deﬁnition of the derivative to ﬁnd f ′(x) for f (x) x 3 . c. Use the limit deﬁnition of the derivative to ﬁnd f ′(x) for f (x) x 4 . (Hint: (a + b)4 a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 . Apply this rule to (x + h)4 within the limit deﬁnition.) d. Based on your work in (a), (b), and (c), what do you conjecture is the derivative of f (x) x 5 ? Of f (x) x 13 ? e. Conjecture a formula for the derivative of f (x) x n that holds for any positive integer n. That is, given f (x) x n where n is a positive integer, what do you think is the formula for f ′(x)? 2.1.1 Some Key Notation In addition to our usual f ′ notation, there are other ways to denote the derivative of a func- tion, as well as the instruction to take the derivative. If we are thinking about the relationship between y and x, we sometimes denote the derivative of y with respect to x by the symbol dy dx which we read “dee-y dee-x.” For example, if y x 2 , we’ll write that the derivative is dy dx 2x. This notation comes from the fact that the derivative is related to the slope of a ∆y ∆y line, and slope is measured by ∆x . Note that while we read ∆x as “change in y over change dy in x,” we view dx as a single symbol, not a quotient of two quantities. We use a variant of this notation as the instruction to take the derivative. In particular, d [□] dx means “take the derivative of the quantity in □ with respect to x.” For example, we may d write dx [x 2 ] 2x. It is important to note that the independent variable can be diﬀerent from x. If we have dy f (z) z 2 , we then write f ′(z) 2z. Similarly, if y t 2 , we say dt 2t. And it is also true that [ ] df d2 f dq [q ] d 2 2q. This notation may also be used for second derivatives: f ′′(z) d dz dz dz 2 . In what follows, we’ll build a repertoire of functions for which we can quickly compute the derivative. 90 2.1 Elementary derivative rules 2.1.2 Constant, Power, and Exponential Functions So far, we know the derivative formula for two important classes of functions: constant functions and power functions. If f (x) c is a constant function, its graph is a horizontal d line with slope zero at every point. Thus, dx [c] 0. We summarize this with the following rule. Constant Functions. For any real number c, if f (x) c, then f ′(x) 0. √ Example 2.1.1 If f (x) 7, then f ′(x) 0. Similarly, dx [ d 3] 0. In your work in Preview Activity 2.1.1, you conjectured that for any positive integer n, if f (x) x n , then f ′(x) nx n−1 . This rule can be formally proved for any positive integer n, and also for any nonzero real number (positive or negative). Power Functions. For any nonzero real number n, if f (x) x n , then f ′(x) nx n−1 . Example 2.1.2 Using the rule for power functions, we can compute the following derivatives. If 1(z) z −3 , then 1 ′(z) −3z −4 . Similarly, if h(t) t 7/5 , then dh dt 5 t 7 2/5 d , and dq [q π ] πq π−1 . It will be instructive to have a derivative formula for one more type of basic function. For now, we simply state this rule without explanation or justiﬁcation; we will explore why this rule is true in one of the exercises. And we will encounter graphical reasoning for why the rule is plausible in Preview Activity 2.2.1. Exponential Functions. For any positive real number a, if f (x) a x , then f ′(x) a x ln(a). Example 2.1.3 If f (x) 2x , then f ′(x) 2x ln(2). Similarly, for p(t) 10t , p ′(t) 10t ln(10). It is especially important to note that when a e, where e is the base of the natural logarithm function, we have that d x [e ] e x ln(e) e x dx since ln(e) 1. This is an extremely important property of the function e x : its derivative function is itself! Note carefully the distinction between power functions and exponential functions: in power functions, the variable is in the base, as in x 2 , while in exponential functions, the variable is in the power, as in 2x . As we can see from the rules, this makes a big diﬀerence in the form of the derivative. 91 Chapter 2 Computing Derivatives Activity 2.1.2. Use the three rules above to determine the derivative of each of the following functions. For each, state your answer using full and proper notation, la- beling the derivative with its name. For example, if you are given a function h(z), you should write “h ′(z) ” or “ dh dz ” as part of your response. a. f (t) π d. p(x) 31/2 f. s(q) q −1 b. 1(z) 7z √ c. h(w) w 3/4 e. r(t) ( 2)t g. m(t) 1 t3 2.1.3 Constant Multiples and Sums of Functions Next we will learn how to compute the derivative of a function constructed as an algebraic combination of basic functions. For instance, we’d like to be able to take the derivative of a polynomial function such as p(t) 3t 5 − 7t 4 + t 2 − 9, which is a sum of constant multiples of powers of t. To that end, we develop two new rules: the Constant Multiple Rule and the Sum Rule. How is the derivative of y k f (x) related to the derivative of y f (x)? Recall that when we multiply a function by a constant k, we vertically stretch the graph by a factor of |k| (and reﬂect the graph across y 0 if k < 0). This vertical stretch aﬀects the slope of the graph, so the slope of the function y k f (x) is k times as steep as the slope of y f (x). Thus, when we multiply a function by a factor of k, we change the value of its derivative by a factor of k as well.¹, The Constant Multiple Rule. For any real number k, if f (x) is a diﬀerentiable function with derivative f ′(x), then ′ dx [k f (x)] k f (x). d In words, this rule says that “the derivative of a constant times a function is the constant times the derivative of the function.” Example 2.1.4 If 1(t) 3 · 5t , we have 1 ′(t) 3 · 5t ln(5). Similarly, −2 dz [5z ] d 5(−2z −3 ). Next we examine a sum of two functions. If we have y f (x) and y 1(x), we can compute a new function y ( f + 1)(x) by adding the outputs of the two functions: ( f + 1)(x) f (x) + 1(x). Not only is the value of the new function the sum of the values of the two known functions, but the slope of the new function is the sum of the slopes of the known functions. Therefore², we arrive at the following Sum Rule for derivatives: ¹The Constant Multiple Rule can be formally proved as a consequence of properties of limits, using the limit deﬁnition of the derivative. ²Like the Constant Multiple Rule, the Sum Rule can be formally proved as a consequence of properties of limits, using the limit deﬁnition of the derivative. 92 2.1 Elementary derivative rules The Sum Rule. If f (x) and 1(x) are diﬀerentiable functions with derivatives f ′(x) and 1 ′(x) respec- d tively, then dx [ f (x) + 1(x)] f ′(x) + 1 ′(x). In words, the Sum Rule tells us that “the derivative of a sum is the sum of the derivatives.” It also tells us that a sum of two diﬀerentiable functions is also diﬀerentiable. Furthermore, because we can view the diﬀerence function y ( f − 1)(x) f (x) − 1(x) as y f (x) + (−1 · 1(x)), the Sum Rule and Constant Multiple Rules together tell us that dx d [ f (x) + (−1 · 1(x))] ′ ′ f (x) − 1 (x), or that “the derivative of a diﬀerence is the diﬀerence of the derivatives.” We can now compute derivatives of sums and diﬀerences of elementary functions. Example 2.1.5 Using the sum rule, dw d (2w + w 2 ) 2w ln(2) + 2w. Using both the sum and constant multiple rules, if h(q) 3q − 4q −3 , then h ′(q) 3(6q 5 ) − 4(−3q −4 ) 18q 5 + 12q −4 . 6 Activity 2.1.3. Use only the rules for constant, power, and exponential functions, to- gether with the Constant Multiple and Sum Rules, to compute the derivative of each function below with respect to the given independent variable. Note well that we do not yet know any rules for how to diﬀerentiate the product or quotient of functions. This means that you may have to do some algebra ﬁrst on the functions below before you can actually use existing rules to compute the desired derivative formula. In each case, label the derivative you calculate with its name using proper notation such as f ′(x), h ′(z), dr/dt, etc. a. f (x) x 5/3 − x 4 + 2x e. s(y) (y 2 + 1)(y 2 − 1) b. 1(x) 14e x + 3x 5 − x √ x 3 −x+2 f. q(x) c. h(z) z + z14 + 5z x √ d. r(t) 53 t 7 − πe t + e 4 g. p(a) 3a 4 − 2a 3 + 7a 2 − a + 12 In the same way that we have shortcut rules to help us ﬁnd derivatives, we introduce some language that is simpler and shorter. Often, rather than say “take the derivative of f ,” we’ll instead say simply “diﬀerentiate f .” Similarly, if the derivative exists at a point, we say “ f is diﬀerentiable at that point,” or that f can be diﬀerentiated. As we work with the algebraic structure of functions, it is important to develop a big picture view of what we are doing. Here, we make several general observations based on the rules we have so far. • The derivative of any polynomial function will be another polynomial function, and that the degree of the derivative is one less than the degree of the original function. For instance, if p(t) 7t 5 − 4t 3 + 8t, p is a degree 5 polynomial, and its derivative, p ′(t) 35t 4 − 12t 2 + 8, is a degree 4 polynomial. • The derivative of any exponential function is another exponential function: for exam- ple, if 1(z) 7 · 2z , then 1 ′(z) 7 · 2z ln(2), which is also exponential. • We should not lose sight of the fact that all of the meaning of the derivative that we developed in Chapter 1 still holds. The derivative measures the instantaneous rate of 93 Chapter 2 Computing Derivatives change of the original function, as well as the slope of the tangent line at any selected point on the curve. Activity 2.1.4. Each of the following questions asks you to use derivatives to answer key questions about functions. Be sure to think carefully about each question and to use proper notation in your responses. √ a. Find the slope of the tangent line to h(z) z + 1z at the point where z 4. b. A population of cells is growing in such a way that its total number in millions is given by the function P(t) 2(1.37)t + 32, where t is measured in days. i. Determine the instantaneous rate at which the population is growing on day 4, and include units on your answer. ii. Is the population growing at an increasing rate or growing at a decreasing rate on day 4? Explain. c. Find an equation for the tangent line to the curve p(a) 3a 4 − 2a 3 + 7a 2 − a + 12 at the point where a −1. d. What is the diﬀerence between being asked to ﬁnd the slope of the tangent line (asked in (a)) and the equation of the tangent line (asked in (c))? 2.1.4 Summary • Given a diﬀerentiable function y f (x), we can express the derivative of f in several d f dy diﬀerent notations: f ′(x), dx , dx , and dx d [ f (x)]. • The limit deﬁnition of the derivative leads to patterns among certain families of func- tions that enable us to compute derivative formulas without resorting directly to the limit deﬁnition. For example, if f is a power function of the form f (x) x n , then f ′(x) nx n−1 for any real number n other than 0. This is called the Rule for Power Functions. • We have stated a rule for derivatives of exponential functions in the same spirit as the rule for power functions: for any positive real number a, if f (x) a x , then f ′(x) a x ln(a). • If we are given a constant multiple of a function whose derivative we know, or a sum of functions whose derivatives we know, the Constant Multiple and Sum Rules make it straightforward to compute the derivative of the overall function. More formally, if f (x) and 1(x) are diﬀerentiable with derivatives f ′(x) and 1 ′(x) and a and b are constants, then d [ ] a f (x) + b1(x) a f ′(x) + b1 ′(x). dx 94 2.1 Elementary derivative rules 2.1.5 Exercises 1. Derivative of a power function. Find the derivative of y x 15/16 . 1 2. Derivative of a rational function. Find the derivative of f (x) . x 19 √ 3. Derivative of a root function. Find the derivative of y x. 4. Derivative of a quadratic. Find the derivative of f (t) 3t 2 − 7t + 2. √ 5. Derivative of a sum of power functions. Find the derivative of y 6t 6 − 9 t + 7t . √ 6. Simplifying a product before diﬀerentiating. Find the derivative of y x(x 3 + 9). x6 + 9 7. Simplifying a quotient before diﬀerentiating. Find the derivative of y . x 8. Finding a tangent line equation. Find an equation for the line tangent to the graph of f at (3, 76), where f is given by f (x) 4x 3 − 4x 2 + 4. 9. Determining where f ′(x) 0. If f (x) x 3 + 6x 2 − 288x + 5, ﬁnd analytically all values of x for which f ′(x) 0. 10. Let f and 1 be diﬀerentiable functions for which the following information is known: f (2) 5, 1(2) −3, f ′(2) −1/2, 1 ′(2) 2. a. Let h be the new function deﬁned by the rule h(x) 3 f (x) − 41(x). Determine h(2) and h ′(2). b. Find an equation for the tangent line to y h(x) at the point (2, h(2)). c. Let p be the function deﬁned by the rule p(x) −2 f (x) + 12 1(x). Is p increasing, decreasing, or neither at a 2? Why? d. Estimate the value of p(2.03) by using the local linearization of p at the point (2, p(2)). 11. Let functions p and q be the piecewise linear functions given by their respective graphs in Figure 2.1.6. Use the graphs to answer the following questions. a. At what values of x is p not diﬀerentiable? At what values of x is q not diﬀeren- tiable? Why? b. Let r(x) p(x) + 2q(x). At what values of x is r not diﬀerentiable? Why? c. Determine r ′(−2) and r ′(0). d. Find an equation for the tangent line to y r(x) at the point (2, r(2)). 95 Chapter 2 Computing Derivatives p 3 2 1 -3 -2 -1 1 2 3 -1 q -2 -3 Figure 2.1.6: The graphs of p (in blue) and q (in green). 12. Consider the functions r(t) t t and s(t) arccos(t), for which you are given the facts that r ′(t) t t (ln(t) + 1) and s ′(t) − √ 1 2 . Do not be concerned with where these 1−t derivative formulas come from. We restrict our interest in both functions to the domain 0 < t < 1. a. Let w(t) 3t t − 2 arccos(t). Determine w ′(t). b. Find an equation for the tangent line to y w(t) at the point ( 12 , w( 21 )). c. Let v(t) t t + arccos(t). Is v increasing or decreasing at the instant t 12 ? Why? 13. Let f (x) a x . The goal of this problem is to explore how the value of a aﬀects the derivative of f (x), without assuming we know the rule for dx d [a x ] that we have stated and used in earlier work in this section. a. Use the limit deﬁnition of the derivative to show that ax · ah − ax f ′(x) lim . h→0 h b. Explain why it is also true that ah − 1 f ′(x) a x · lim . h→0 h c. Use computing technology and small values of h to estimate the value of ah − 1 L lim h→0 h when a 2. Do likewise when a 3. d. Note that it would be ideal if the value of the limit L was 1, for then f would be a particularly special function: its derivative would be simply a x , which would 96 2.1 Elementary derivative rules mean that its derivative is itself. By experimenting with diﬀerent values of a be- tween 2 and 3, try to ﬁnd a value for a for which ah − 1 L lim 1. h→0 h e. Compute ln(2) and ln(3). What does your work in (b) and (c) suggest is true about dx [2 ] and dx [3 ]? d x d x f. How do your investigations in (d) lead to a particularly important fact about the function f (x) e x ? 97 Chapter 2 Computing Derivatives 2.2 The sine and cosine functions Motivating Questions dx [a ] a x ln(a)? d x • What is a graphical justiﬁcation for why • What do the graphs of y sin(x) and y cos(x) suggest as formulas for their re- spective derivatives? • Once we know the derivatives of sin(x) and cos(x), how do previous derivative rules work when these functions are involved? Throughout Chapter 2, we will develop shortcut derivative rules to help us bypass the limit deﬁnition and quickly compute f ′(x) from a formula for f (x). In Section 2.1, we stated the rule for power functions, if f (x) x n , then f ′(x) nx n−1 , and the rule for exponential functions, if a is a positive real number and f (x) a x , then f ′(x) a x ln(a). Later in this section, we will use a graphical argument to conjecture derivative formulas for the sine and cosine functions. Preview Activity 2.2.1. Consider the function 1(x) 2x , which is graphed in Fig- ure 2.2.1. a. At each of x −2, −1, 0, 1, 2, use a straightedge to sketch an accurate tangent line to y 1(x). b. Use the provided grid to estimate the slope of the tangent line you drew at each point in (a). c. Use the limit deﬁnition of the derivative to estimate 1 ′(0) by using small values of h, and compare the result to your visual estimate for the slope of the tangent line to y 1(x) at x 0 in (b). d. Based on your work in (a), (b), and (c), sketch an accurate graph of y 1 ′(x) on the axes adjacent to the graph of y 1(x). e. Write at least one sentence that explains why it is reasonable to think that 1 ′(x) c1(x), where c is a constant. In addition, calculate ln(2), and then discuss how this value, combined with your work above, reasonably suggests that 1 ′(x) 2x ln(2). 98 2.2 The sine and cosine functions 7 7 6 6 5 5 4 4 3 3 2 2 1 1 -2 -1 1 2 -2 -1 1 2 Figure 2.2.1: At left, the graph of y 1(x) 2x . At right, axes for plotting y 1 ′(x). 2.2.1 The sine and cosine functions The sine and cosine functions are among the most important functions in all of mathemat- ics. Sometimes called the circular functions due to their deﬁnition on the unit circle, these periodic functions play a key role in modeling repeating phenomena such as tidal eleva- tions, the behavior of an oscillating mass attached to a spring, or the location of a point on a bicycle tire. Like polynomial and exponential functions, the sine and cosine functions are considered basic functions, ones that are often used in building more complicated func- tions. As such, we would like to know formulas for dx d [sin(x)] and dx d [cos(x)], and the next two activities lead us to that end. Activity 2.2.2. Consider the function f (x) sin(x), which is graphed in Figure 2.2.2 below. Note carefully that the grid in the diagram does not have boxes that are 1 × 1, but rather approximately 1.57 × 1, as the horizontal scale of the grid is π/2 units per box. 1 1 2π −2π −π -1 π −2π −π -1 π 2π Figure 2.2.2: At left, the graph of y f (x) sin(x). π π a. At each of x −2π, − 3π 2 , −π, − 2 , 0, 2 , π, 2 , 2π, 3π use a straightedge to sketch an accurate tangent line to y f (x). b. Use the provided grid to estimate the slope of the tangent line you drew at each point. Pay careful attention to the scale of the grid. 99 Chapter 2 Computing Derivatives c. Use the limit deﬁnition of the derivative to estimate f ′(0) by using small values of h, and compare the result to your visual estimate for the slope of the tangent line to y f (x) at x 0 in (b). Using periodicity, what does this result suggest about f ′(2π)? about f ′(−2π)? d. Based on your work in (a), (b), and (c), sketch an accurate graph of y f ′(x) on the axes adjacent to the graph of y f (x). e. What familiar function do you think is the derivative of f (x) sin(x)? Activity 2.2.3. Consider the function 1(x) cos(x), which is graphed in Figure 2.2.3 below. Note carefully that the grid in the diagram does not have boxes that are 1 × 1, but rather approximately 1.57 × 1, as the horizontal scale of the grid is π/2 units per box. 1 1 −2π −π -1 π 2π −2π −π -1 π 2π Figure 2.2.3: At left, the graph of y 1(x) cos(x). π π a. At each of x −2π, − 3π 2 , −π, − 2 , 0, 2 , π, 2 , 2π, 3π use a straightedge to sketch an accurate tangent line to y 1(x). b. Use the provided grid to estimate the slope of the tangent line you drew at each point. Again, note the scale of the axes and grid. c. Use the limit deﬁnition of the derivative to estimate 1 ′( π2 ) by using small values of h, and compare the result to your visual estimate for the slope of the tangent line to y 1(x) at x π2 in (b). Using periodicity, what does this result sug- gest about 1 ′(− 3π 2 )? can symmetry on the graph help you estimate other slopes easily? d. Based on your work in (a), (b), and (c), sketch an accurate graph of y 1 ′(x) on the axes adjacent to the graph of y 1(x). e. What familiar function do you think is the derivative of 1(x) cos(x)? The results of the two preceding activities suggest that the sine and cosine functions not only have beautiful connections such as the identities sin2 (x)+cos2 (x) 1 and cos(x − π2 ) sin(x), but that they are even further linked through calculus, as the derivative of each involves the other. The following rules summarize the results of the activities¹. ¹These two rules may be formally proved using the limit deﬁnition of the derivative and the expansion identities for sin(x + h) and cos(x + h). 100 2.2 The sine and cosine functions Sine and Cosine Functions. For all real numbers x, d d [sin(x)] cos(x) and [cos(x)] − sin(x). dx dx We have now added the sine and cosine functions to our library of basic functions whose derivatives we know. The constant multiple and sum rules still hold, of course, as well as all of the inherent meaning of the derivative. Activity 2.2.4. Answer each of the following questions. Where a derivative is re- quested, be sure to label the derivative function with its name using proper notation. a. Determine the derivative of h(t) 3 cos(t) − 4 sin(t). sin(x) b. Find the exact slope of the tangent line to y f (x) 2x + 2 at the point where x π6 . c. Find the equation of the tangent line to y 1(x) x 2 + 2 cos(x) at the point where x π2 . d. Determine the derivative of p(z) z 4 + 4z + 4 cos(z) − sin( π2 ). e. The function P(t) 24 + 8 sin(t) represents a population of a particular kind of animal that lives on a small island, where P is measured in hundreds and t is measured in decades since January 1, 2010. What is the instantaneous rate of change of P on January 1, 2030? What are the units of this quantity? Write a sentence in everyday language that explains how the population is behaving at this point in time. 2.2.2 Summary • For an exponential function f (x) a x (a > 1), the graph of f ′(x) appears to be a scaled version of the original function. In particular, careful analysis of the graph of f (x) d 2x , suggests that dx [2x ] 2x ln(2), which is a special case of the rule we stated in Section 2.1. • By carefully analyzing the graphs of y sin(x) and y cos(x), and by using the d limit deﬁnition of the derivative at select points, we found that dx [sin(x)] cos(x) and d dx [cos(x)] − sin(x). • We note that all previously encountered derivative rules still hold, but now may also be applied to functions involving the sine and cosine. All of the established meaning of the derivative applies to these trigonometric functions as well. 101 Chapter 2 Computing Derivatives 2.2.3 Exercises 1. Suppose that V(t) 24 · 1.07t + 6 sin(t) represents the value of a person’s investment portfolio in thousands of dollars in year t, where t 0 corresponds to January 1, 2010. a. At what instantaneous rate is the portfolio’s value changing on January 1, 2012? Include units on your answer. b. Determine the value of V ′′(2). What are the units on this quantity and what does it tell you about how the portfolio’s value is changing? c. On the interval 0 ≤ t ≤ 20, graph the function V(t) 24 · 1.07t + 6 sin(t) and describe its behavior in the context of the problem. Then, compare the graphs of the functions A(t) 24 · 1.07t and V(t) 24 · 1.07t + 6 sin(t), as well as the graphs of their derivatives A′(t) and V ′(t). What is the impact of the term 6 sin(t) on the behavior of the function V(t)? 2. Let f (x) 3 cos(x) − 2 sin(x) + 6. π a. Determine the exact slope of the tangent line to y f (x) at the point where a 4. b. Determine the tangent line approximation to y f (x) at the point where a π. π c. At the point where a 2, is f increasing, decreasing, or neither? d. At the point where a 3π 2 , does the tangent line to y f (x) lie above the curve, below the curve, or neither? How can you answer this question without even graphing the function or the tangent line? 3. In this exercise, we explore how the limit deﬁnition of the derivative more formally shows that dxd [sin(x)] cos(x). Letting f (x) sin(x), note that the limit deﬁnition of the derivative tells us that sin(x + h) − sin(x) f ′(x) lim . h→0 h a. Recall the trigonometric identity for the sine of a sum of angles α and β: sin(α + β) sin(α) cos(β) + cos(α) sin(β). Use this identity and some algebra to show that sin(x)(cos(h) − 1) + cos(x) sin(h) f ′(x) lim . h→0 h b. Next, note that as h changes, x remains constant. Explain why it therefore makes sense to say that cos(h) − 1 sin(h) f ′(x) sin(x) · lim + cos(x) · lim . h→0 h h→0 h c. Finally, use small values of h to estimate the values of the two limits in (c): cos(h) − 1 sin(h) lim and lim . h→0 h h→0 h d. What do your results in (b) and (c) thus tell you about f ′(x)? e. By emulating the steps taken above, use the limit deﬁnition of the derivative to d argue convincingly that dx [cos(x)] − sin(x). 102 2.2 The sine and cosine functions 103 Chapter 2 Computing Derivatives 2.3 The product and quotient rules Motivating Questions • How does the algebraic structure of a function guide us in computing its derivative using shortcut rules? • How do we compute the derivative of a product of two basic functions in terms of the derivatives of the basic functions? • How do we compute the derivative of a quotient of two basic functions in terms of the derivatives of the basic functions? • How do the product and quotient rules combine with the sum and constant multiple rules to expand the library of functions we can diﬀerentiate quickly? So far, we can diﬀerentiate power functions (x n ), exponential functions (a x ), and the two fundamental trigonometric functions (sin(x) and cos(x)). With the sum rule and constant multiple rules, we can also compute the derivative of combined functions. Example 2.3.1 Diﬀerentiate √ f (x) 7x 11 − 4 · 9x + π sin(x) − 3 cos(x) Because f is a sum of basic functions, we can now quickly say that f ′(x) 77x 10 −4·9x ln(9)+ √ π cos(x) + 3 sin(x). What about a product or quotient of two basic functions, such as p(z) z 3 cos(z), or sin(t) q(t) ? 2t While the derivative of a sum is the sum of the derivatives, it turns out that the rules for computing derivatives of products and quotients are more complicated. Preview Activity 2.3.1. Let f and 1 be the functions deﬁned by f (t) 2t 2 and 1(t) t 3 + 4t. a. Determine f ′(t) and 1 ′(t). b. Let p(t) 2t 2 (t 3 + 4t) and observe that p(t) f (t) · 1(t). Rewrite the formula for p by distributing the 2t 2 term. Then, compute p ′(t) using the sum and constant multiple rules. c. True or false: p ′(t) f ′(t) · 1 ′(t). d. Let q(t) t 2t+4t 3 1(t) 2 and observe that q(t) f (t) . Rewrite the formula for q by dividing each term in the numerator by the denominator and simplify to write 104 2.3 The product and quotient rules q as a sum of constant multiples of powers of t. Then, compute q ′(t) using the sum and constant multiple rules. 1 ′ (t) e. True or false: q ′(t) f ′ (t) . 2.3.1 The product rule As part (b) of Preview Activity 2.3.1 shows, it is not true in general that the derivative of a product of two functions is the product of the derivatives of those functions. To see why this is the case, we consider an example involving meaningful functions. Say that an investor is regularly purchasing stock in a particular company. Let N(t) represent the number of shares owned on day t, where t 0 represents the ﬁrst day on which shares were purchased. Let S(t) give the value of one share of the stock on day t; note that the units on S(t) are dollars per share. To compute the total value of the stock on day t, we take the product V(t) N(t) shares · S(t) dollars per share. Observe that over time, both the number of shares and the value of a given share will vary. The derivative N ′(t) measures the rate at which the number of shares is changing, while S′(t) measures the rate at which the value per share is changing. How do these respective rates of change aﬀect the rate of change of the total value function? To help us understand the relationship among changes in N, S, and V, let’s consider some speciﬁc data. • Suppose that on day 100, the investor owns 520 shares of stock and the stock’s current value is $27.50 per share. This tells us that N(100) 520 and S(100) 27.50. • On day 100, the investor purchases an additional 12 shares (so the number of shares held is rising at a rate of 12 shares per day). • On that same day the price of the stock is rising at a rate of 0.75 dollars per share per day. In calculus notation, the latter two facts tell us that N ′(100) 12 (shares per day) and S′(100) 0.75 (dollars per share per day). At what rate is the value of the investor’s to- tal holdings changing on day 100? Observe that the increase in total value comes from two sources: the growing number of shares, and the rising value of each share. If only the number of shares is increasing (and the value of each share is constant), the rate at which which total value would rise is the product of the current value of the shares and the rate at which the number of shares is changing. That is, the rate at which total value would change is given by dollars shares dollars S(100) · N ′(100) 27.50 · 12 330 . share day day 105 Chapter 2 Computing Derivatives Note particularly how the units make sense and show the rate at which the total value V is changing, measured in dollars per day. If instead the number of shares is constant, but the value of each share is rising, the rate at which the total value would rise is the product of the number of shares and the rate of change of share value. The total value is rising at a rate of dollars per share dollars N(100) · S′(100) 520 shares · 0.75 390 . day day Of course, when both the number of shares and the value of each share are changing, we have to include both of these sources. In that case the rate at which the total value is rising is dollars V ′(100) S(100) · N ′(100) + N(100) · S′(100) 330 + 390 720 . day We expect the total value of the investor’s holdings to rise by about $720 on the 100th day.¹ Next, we expand our perspective from the speciﬁc example above to the more general and abstract setting of a product p of two diﬀerentiable functions, f and 1. If P(x) f (x) · 1(x), our work above suggests that P ′(x) f (x)1 ′(x) + 1(x) f ′(x). Indeed, a formal proof using the limit deﬁnition of the derivative can be given to show that the following rule, called the product rule, holds in general. Product Rule. If f and 1 are diﬀerentiable functions, then their product P(x) f (x) · 1(x) is also a diﬀerentiable function, and P ′(x) f (x)1 ′(x) + 1(x) f ′(x). In light of the earlier example involving shares of stock, the product rule also makes sense intuitively: the rate of change of P should take into account both how fast f and 1 are chang- ing, as well as how large f and 1 are at the point of interest. In words the product rule says: if P is the product of two functions f (the ﬁrst function) and 1 (the second), then “the deriv- ative of P is the ﬁrst times the derivative of the second, plus the second times the derivative of the ﬁrst.” It is often a helpful mental exercise to say this phrasing aloud when executing the product rule. Example 2.3.2 If P(z) z 3 · cos(z), we can use the product rule to diﬀerentiate P. The ﬁrst function is z 3 and the second function is cos(z). By the product rule, P ′ will be given by the ﬁrst, z 3 , times the derivative of the second, − sin(z), plus the second, cos(z), times the ¹While this example highlights why the product rule is true, there are some subtle issues to recognize. For one, if the stock’s value really does rise exactly $0.75 on day 100, and the number of shares really rises by 12 on day 100, then we’d expect that V(101) N(101) · S(101) 532 · 28.25 15029. If, as noted above, we expect the total value to rise by $720, then with V(100) N(100) · S(100) 520 · 27.50 14300, then it seems we should ﬁnd that V(101) V(100) + 720 15020. Why do the two results diﬀer by 9? One way to understand why this diﬀerence occurs is to recognize that N ′ (100) 12 represents an instantaneous rate of change, while our (informal) discussion has also thought of this number as the total change in the number of shares over the course of a single day. The formal proof of the product rule reconciles this issue by taking the limit as the change in the input tends to zero. 106 2.3 The product and quotient rules derivative of the ﬁrst, 3z 2 . That is, P ′(z) z 3 (− sin(z)) + cos(z)3z 2 −z 3 sin(z) + 3z 2 cos(z). Activity 2.3.2. Use the product rule to answer each of the questions below. Through- out, be sure to carefully label any derivative you ﬁnd by name. It is not necessary to algebraically simplify any of the derivatives you compute. a. Let m(w) 3w 17 4w . Find m ′(w). b. Let h(t) (sin(t) + cos(t))t 4 . Find h ′(t). c. Determine the slope of the tangent line to the curve y f (x) at the point where a 1 if f is given by the rule f (x) e x sin(x). d. Find the tangent line approximation L(x) to the function y 1(x) at the point where a −1 if 1 is given by the rule 1(x) (x 2 + x)2x . 2.3.2 The quotient rule Because quotients and products are closely linked, we can use the product rule to under- stand how to take the derivative of a quotient. Let Q(x) be deﬁned by Q(x) f (x)/1(x), where f and 1 are both diﬀerentiable functions. It turns out that Q is diﬀerentiable every- where that 1(x) , 0. We would like a formula for Q ′ in terms of f , 1, f ′, and 1 ′. multiplying both sides of the formula Q f /1 by 1, we observe that f (x) Q(x) · 1(x). Now we can use the product rule to diﬀerentiate f . f ′(x) Q(x)1 ′(x) + 1(x)Q ′(x). We want to know a formula for Q ′, so we solve this equation for Q ′(x). Q ′(x)1(x) f ′(x) − Q(x)1 ′(x) and dividing both sides by 1(x), we have f ′(x) − Q(x)1 ′(x) Q ′(x) . 1(x) f (x) Finally, we recall that Q(x) 1(x) . Substituting this expression in the preceding equation, we have f (x) ′ f ′(x) − 1(x) 1 (x) ′ Q (x) 1(x) f (x) ′ f ′(x) − 1(x) 1 (x) 1(x) · 1(x) 1(x) 107 Chapter 2 Computing Derivatives 1(x) f ′(x) − f (x)1 ′(x) . 1(x)2 This calculation gives us the quotient rule. Quotient Rule. f (x) If f and 1 are diﬀerentiable functions, then their quotient Q(x) 1(x) is also a dif- ferentiable function for all x where 1(x) , 0 and 1(x) f ′(x) − f (x)1 ′(x) Q ′(x) . 1(x)2 As with the product rule, it can be helpful to think of the quotient rule verbally. If a function Q is the quotient of a top function f and a bottom function 1, then Q ′ is given by “the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared.” Example 2.3.3 If Q(t) sin(t)/2t , we call sin(t) the top function and 2t the bottom function. By the quotient rule, Q ′ is given by the bottom, 2t , times the derivative of the top, cos(t), minus the top, sin(t), times the derivative of the bottom, 2t ln(2), all over the bottom squared, (2t )2 . That is, 2t cos(t) − sin(t)2t ln(2) Q ′(t) . (2t )2 In this particular example, it is possible to simplify Q ′(t) by removing a factor of 2t from both the numerator and denominator, so that cos(t) − sin(t) ln(2) Q ′(t) . 2t In general, we must be careful in doing any such simpliﬁcation, as we don’t want to execute the quotient rule correctly but then make an algebra error. Activity 2.3.3. Use the quotient rule to answer each of the questions below. Through- out, be sure to carefully label any derivative you ﬁnd by name. That is, if you’re given a formula for f (x), clearly label the formula you ﬁnd for f ′(x). It is not necessary to algebraically simplify any of the derivatives you compute. 3z a. Let r(z) z 4 +1 . Find r ′(z). Find v ′(t). sin(t) b. Let v(t) cos(t)+t 2 . x 2 − 2x − 8 c. Determine the slope of the tangent line to the curve R(x) at the x2 − 9 point where x 0. d. When a camera ﬂashes, the intensity I of light seen by the eye is given by the 108 2.3 The product and quotient rules function 100t I(t) , et where I is measured in candles and t is measured in milliseconds. Compute I ′(0.5), I ′(2), and I ′(5); include appropriate units on each value; and discuss the meaning of each. 2.3.3 Combining rules In order to apply the derivative shortcut rules correctly we must recognize the fundamental structure of a function. Example 2.3.4 Determine the derivative of the function x2 f (x) x sin(x) + . cos(x) + 2 How do we decide which rules to apply? Our ﬁrst task is to recognize the structure of the function. This function f is a sum of two slightly less complicated functions, so we can apply the sum rule² to get [ ] d x2 f ′(x) x sin(x) + dx cos(x) + 2 [ ] d d x2 [x sin(x)] + dx dx cos(x) + 2 Now, the left-hand term above is a product, so the product rule is needed there, while the right-hand term is a quotient, so the quotient rule is required. Applying these rules respec- tively, we ﬁnd that (cos(x) + 2)2x − x 2 (− sin(x)) f ′(x) (x cos(x) + sin(x)) + (cos(x) + 2)2 2x cos(x) + 4x 2 + x 2 sin(x) x cos(x) + sin(x) + . (cos(x) + 2)2 Example 2.3.5 Diﬀerentiate y · 7y .s(y) y2 + 1 The function s is a quotient of two simpler functions, so the quotient rule will be needed. To d begin, we set up the quotient rule and use the notation dy to indicate the derivatives of the numerator and denominator. Thus, [ ] [ ] (y 2 + 1) · d dy y · 7y − y · 7y · d dy y2 + 1 s ′(y) . (y 2 + 1)2 ²When taking a derivative that involves the use of multiple derivative rules, it is often helpful to use the notation d dx [ ] to wait to apply subsequent rules. This is demonstrated in each of the two examples presented here. 109 Chapter 2 Computing Derivatives [ ] Now, there remain two derivatives to calculate. The ﬁrst one, d y · 7 y calls for use of the [ ] dy product rule, while the second, dy d y 2 + 1 needs only the sum rule. Applying these rules, we now have (y 2 + 1)[y · 7 y ln(7) + 7 y · 1] − y · 7 y [2y] s ′(y) . (y 2 + 1)2 While some simpliﬁcation is possible, we are content to leave s ′(y) in its current form. Success in applying derivative rules begins with recognizing the structure of the function, followed by the careful and diligent application of the relevant derivative rules. The best way to become proﬁcient at this process is to do a large number of examples. Activity 2.3.4. Use relevant derivative rules to answer each of the questions below. Throughout, be sure to use proper notation and carefully label any derivative you ﬁnd by name. a. Let f (r) (5r 3 + sin(r))(4r − 2 cos(r)). Find f ′(r). cos(t) b. Let p(t) . Find p ′(t). t 6 · 6t c. Let 1(z) 3z 7 e z − 2z 2 sin(z) + z z 2 +1 . Find 1 ′(z). d. A moving particle has its position in feet at time t in seconds given by the func- 3 cos(t)−sin(t) tion s(t) et . Find the particle’s instantaneous velocity at the moment t 1. e. Suppose that f (x) and 1(x) are diﬀerentiable functions and it is known that f (3) −2, f ′(3) 7, 1(3) 4, and 1 ′(3) −1. If p(x) f (x) · 1(x) and f (x) q(x) , calculate p ′(3) and q ′(3). 1(x) As the algebraic complexity of the functions we are able to diﬀerentiate continues to increase, it is important to remember that all of the derivative’s meaning continues to hold. Regardless of the structure of the function f , the value of f ′(a) tells us the instantaneous rate of change of f with respect to x at the moment x a, as well as the slope of the tangent line to y f (x) at the point (a, f (a)). 2.3.4 Summary • If a function is a sum, product, or quotient of simpler functions, then we can use the sum, product, or quotient rules to diﬀerentiate it in terms of the simpler functions and their derivatives. • The product rule tells us that if P is a product of diﬀerentiable functions f and 1 ac- cording to the rule P(x) f (x)1(x), then P ′(x) f (x)1 ′(x) + 1(x) f ′(x). 110 2.3 The product and quotient rules • The quotient rule tells us that if Q is a quotient of diﬀerentiable functions f and 1 f (x) according to the rule Q(x) 1(x) , then 1(x) f ′(x) − f (x)1 ′(x) Q ′(x) . 1(x)2 • Along with the constant multiple and sum rules, the product and quotient rules enable us to compute the derivative of any function that consists of sums, constant multiples, products, and quotients of basic functions. For instance, if F has the form 2a(x) − 5b(x) F(x) , c(x) · d(x) then F is a quotient, in which the numerator is a sum of constant multiples and the de- nominator is a product. This, the derivative of F can be found by applying the quotient rule and then using the sum and constant multiple rules to diﬀerentiate the numerator and the product rule to diﬀerentiate the denominator. 2.3.5 Exercises 1. Derivative of a basic product. Find the derivative of the function f (x), below. It may be to your advantage to simplify ﬁrst. f (x) x · 13x 2. Derivative of a product. Find the derivative of the function f (x), below. It may be to your advantage to simplify ﬁrst. √ f (x) (x 9 − 3 x)2x 3. Derivative of a quotient of linear functions. Find the derivative of the function z, below. It may be to your advantage to simplify ﬁrst. 2t + 7 z 8t + 7 4. Derivative of a rational function. Find the derivative of the function h(r), below. It may be to your advantage to simplify ﬁrst. r3 h(r) 9r + 13 5. Derivative of a product of trigonometric functions. Find the derivative of s(q) 6 cos q sin q. 6. Derivative of a product of power and trigonmetric functions. Find the derivative of f (x) x 5 cos x 7. Derivative of a sum that involves a product. Find the derivative of h(t) t sin t + tan t 8. Product and quotient rules with graphs. Let h(x) f (x) · 1(x), and k(x) f (x)/1(x). Use the ﬁgures below to ﬁnd the exact values of the indicated derivatives. 111 Chapter 2 Computing Derivatives f (x) 1(x) A. h ′(1) B. k ′(−2) 9. Product and quotient rules with given function values. Let F(4) 4, F′(4) 5, H(4) 4, H ′(4) 5. A. If G(z) F(z) · H(z), then G′(4) B. If G(w) F(w)/H(w), then G′(4) 10. Let f and 1 be diﬀerentiable functions for which the following information is known: f (2) 5, 1(2) −3, f ′(2) −1/2, 1 ′(2) 2. a. Let h be the new function deﬁned by the rule h(x) 1(x) · f (x). Determine h(2) and h ′(2). b. Find an equation for the tangent line to y h(x) at the point (2, h(2)) (where h is the function deﬁned in (a)). 1(x) c. Let r be the function deﬁned by the rule r(x) f (x) . Is r increasing, decreasing, or neither at a 2? Why? d. Estimate the value of r(2.06) (where r is the function deﬁned in (c)) by using the local linearization of r at the point (2, r(2)). 11. Consider the functions r(t) t t and s(t) arccos(t), for which you are given the facts that r ′(t) t t (ln(t) + 1) and s ′(t) − √ 1 2 . Do not be concerned with where these 1−t derivative formulas come from. We restrict our interest in both functions to the domain 0 < t < 1. a. Let w(t) t t arccos(t). Determine w ′(t). b. Find an equation for the tangent line to y w(t) at the point ( 12 , w( 21 )). tt c. Let v(t) arccos(t) . Is v increasing or decreasing at the instant t 12 ? Why? 112 2.3 The product and quotient rules 12. Let functions p and q be the piecewise linear functions given by their respective graphs in Figure 2.3.6. Use the graphs to answer the following questions. a. Let r(x) p(x) · q(x). Determine p r ′(−2) and r ′(0). 3 b. Are there values of x for which 2 r ′(x) does not exist? If so, which values, and why? 1 c. Find an equation for the tangent line to y r(x) at the point (2, r(2)). -3 -2 -1 1 2 3 q(x) -1 d. Let z(x) p(x) . Determine z ′(0) and z ′(2). q -2 e. Are there values of x for which -3 z ′(x) does not exist? If so, which values, and why? Figure 2.3.6: The graphs of p (in blue) and q (in green). 13. A farmer with large land holdings has historically grown a wide variety of crops. With the price of ethanol fuel rising, he decides that it would be prudent to devote more and more of his acreage to producing corn. As he grows more and more corn, he learns eﬃciencies that increase his yield per acre. In the present year, he used 7000 acres of his land to grow corn, and that land had an average yield of 170 bushels per acre. At the current time, he plans to increase his number of acres devoted to growing corn at a rate of 600 acres/year, and he expects that right now his average yield is increasing at a rate of 8 bushels per acre per year. Use this information to answer the following questions. a. Say that the present year is t 0, that A(t) denotes the number of acres the farmer devotes to growing corn in year t, Y(t) represents the average yield in year t (mea- sured in bushels per acre), and C(t) is the total number of bushels of corn the farmer produces. What is the formula for C(t) in terms of A(t) and Y(t)? Why? b. What is the value of C(0)? What does it measure? c. Write an expression for C′(t) in terms of A(t), A′(t), Y(t), and Y ′(t). Explain your thinking. d. What is the value of C′(0)? What does it measure? e. Based on the given information and your work above, estimate the value of C(1). 14. Let f (v) be the gas consumption (in liters/km) of a car going at velocity v (in km/hour). In other words, f (v) tells you how many liters of gas the car uses to go one kilometer if it is traveling at v kilometers per hour. In addition, suppose that f (80) 0.05 and f ′(80) 0.0004. a. Let 1(v) be the distance the same car goes on one liter of gas at velocity v. What is the relationship between f (v) and 1(v)? Hence ﬁnd 1(80) and 1 ′(80). 113 Chapter 2 Computing Derivatives b. Let h(v) be the gas consumption in liters per hour of a car going at velocity v. In other words, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. What is the algebraic relationship between h(v) and f (v)? Hence ﬁnd h(80) and h ′(80). c. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response. 114 2.4 Derivatives of other trigonometric functions 2.4 Derivatives of other trigonometric functions Motivating Questions • What are the derivatives of the tangent, cotangent, secant, and cosecant functions? • How do the derivatives of tan(x), cot(x), sec(x), and csc(x) combine with other de- rivative rules we have developed to expand the library of functions we can quickly diﬀerentiate? One of the powerful themes in trigonometry comes from a very simple idea: locating a point on the unit circle. (x, y) 1 sin(θ ) θ cos(θ ) Figure 2.4.1: The unit circle and the deﬁnition of the sine and cosine functions. Because each angle θ in standard position corresponds to one and only one point (x, y) on the unit circle, the x- and y-coordinates of this point are each functions of θ. In fact, this is the very deﬁnition of cos(θ) and sin(θ): cos(θ) is the x-coordinate of the point on the unit circle corresponding to the angle θ, and sin(θ) is the y-coordinate. From this simple deﬁnition, all of trigonometry is founded. For instance, the Fundamental Trigonometric Identity, sin2 (θ) + cos2 (θ) 1, is a restatement of the Pythagorean Theorem, applied to the right triangle shown in Fig- ure 2.4.1. There are four other trigonometric functions, each deﬁned in terms of the sine and/or cosine functions. sin(θ) • The tangent function is deﬁned by tan(θ) cos(θ) ; cos(θ) • the cotangent function is its reciprocal: cot(θ) sin(θ) . 115 Chapter 2 Computing Derivatives • The secant function is the reciprocal of the cosine function, sec(θ) 1 cos(θ) ; • and the cosecant function is the reciprocal of the sine function, csc(θ) 1 sin(θ) . These six trigonometric functions together oﬀer us a wide range of ﬂexibility in problems involving right triangles. Because we know the derivatives of the sine and cosine function, we can now develop short- cut diﬀerentiation rules for the tangent, cotangent, secant, and cosecant functions. In this section’s preview activity, we work through the steps to ﬁnd the derivative of y tan(x). Preview Activity 2.4.1. Consider the function f (x) tan(x), and remember that sin(x) tan(x) cos(x) . a. What is the domain of f ? b. Use the quotient rule to show that one expression for f ′(x) is cos(x) cos(x) + sin(x) sin(x) f ′(x) . cos2 (x) c. What is the Fundamental Trigonometric Identity? How can this identity be used to ﬁnd a simpler form for f ′(x)? d. Recall that sec(x) 1 cos(x) . How can we express f ′(x) in terms of the secant function? e. For what values of x is f ′(x) deﬁned? How does this set compare to the domain of f ? 2.4.1 Derivatives of the cotangent, secant, and cosecant functions In Preview Activity 2.4.1, we found that the derivative of the tangent function can be ex- pressed in several ways, but most simply in terms of the secant function. Next, we develop the derivative of the cotangent function. Let 1(x) cot(x). To ﬁnd 1 ′(x), we observe that 1(x) cos(x) sin(x) and apply the quotient rule. Hence sin(x)(− sin(x)) − cos(x) cos(x) 1 ′(x) sin2 (x) sin2 (x) + cos2 (x) − sin2 (x) By the Fundamental Trigonometric Identity, we see that 1 ′(x) − sin12 (x) , and recalling that csc(x) 1 sin(x) , it follows that we can express 1 ′ by the rule 1 ′(x) − csc2 (x). 116 2.4 Derivatives of other trigonometric functions Note that neither 1 nor 1 ′ is deﬁned when sin(x) 0, which occurs at every integer multiple of π. Hence we have the following rule. Cotangent Function. For all real numbers x such that x , kπ, where k 0, ±1, ±2, . . ., d [cot(x)] − csc2 (x). dx Notice that the derivative of the cotangent function is very similar to the derivative of the tangent function we discovered in Preview Activity 2.4.1. Tangent Function. (2k+1)π For all real numbers x such that x , 2 , where k ±1, ±2, . . ., d [tan(x)] sec2 (x). dx In the next two activities, we develop the rules for diﬀerentiating the secant and cosecant functions. Activity 2.4.2. Let h(x) sec(x) and recall that sec(x) 1 cos(x) . a. What is the domain of h? b. Use the quotient rule to develop a formula for h ′(x) that is expressed completely in terms of sin(x) and cos(x). c. How can you use other relationships among trigonometric functions to write h ′(x) only in terms of tan(x) and sec(x)? d. What is the domain of h ′? How does this compare to the domain of h? Activity 2.4.3. Let p(x) csc(x) and recall that csc(x) 1 sin(x) . a. What is the domain of p? b. Use the quotient rule to develop a formula for p ′(x) that is expressed completely in terms of sin(x) and cos(x). c. How can you use other relationships among trigonometric functions to write p ′(x) only in terms of cot(x) and csc(x)? d. What is the domain of p ′? How does this compare to the domain of p? Using the quotient rule we have determined the derivatives of the tangent, cotangent, secant, and cosecant functions, expanding our overall library of functions we can diﬀerentiate. Ob- serve that just as the derivative of any polynomial function is a polynomial, and the deriv- 117 Chapter 2 Computing Derivatives ative of any exponential function is another exponential function, so it is that the derivative of any basic trigonometric function is another function that consists of basic trigonometric functions. This makes sense because all trigonometric functions are periodic, and hence their derivatives will be periodic, too. The derivative retains all of its fundamental meaning as an instantaneous rate of change and as the slope of the tangent line to the function under consideration. Activity 2.4.4. Answer each of the following questions. Where a derivative is re- quested, be sure to label the derivative function with its name using proper notation. a. Let f (x) 5 sec(x) − 2 csc(x). Find the slope of the tangent line to f at the point where x π3 . b. Let p(z) z 2 sec(z) − z cot(z). Find the instantaneous rate of change of p at the point where z π4 . tan(t) c. Let h(t) − 2e t cos(t). Find h ′(t). t2 + 1 r sec(r) d. Let 1(r) . Find 1 ′(r). 5r e. When a mass hangs from a spring and is set in motion, the object’s position oscillates in a way that the size of the oscillations decrease. This is usually called a damped oscillation. Suppose that for a particular object, its displacement from equilibrium (where the object sits at rest) is modeled by the function 15 sin(t) s(t) . et Assume that s is measured in inches and t in seconds. Sketch a graph of this function for t ≥ 0 to see how it represents the situation described. Then compute ds/dt, state the units on this function, and explain what it tells you about the object’s motion. Finally, compute and interpret s ′(2). 2.4.2 Summary • The derivatives of the other four trigonometric functions are d d [tan(x)] sec2 (x), [cot(x)] − csc2 (x), dx dx d d [sec(x)] sec(x) tan(x), and [csc(x)] − csc(x) cot(x). dx dx Each derivative exists and is deﬁned on the same domain as the original function. For example, both the tangent function and its derivative are deﬁned for all real numbers x such that x , kπ 2 , where k ±1, ±2, . . .. 118 2.4 Derivatives of other trigonometric functions • The four rules for the derivatives of the tangent, cotangent, secant, and cosecant can be used along with the rules for power functions, exponential functions, and the sine and cosine, as well as the sum, constant multiple, product, and quotient rules, to quickly diﬀerentiate a wide range of diﬀerent functions. 2.4.3 Exercises 1. A sum and product involving tan(x). Find the derivative of h(t) t tan t + cos t 5 tan(x) 2. A quotient involving tan(t). Let f (x) . Find f ′(x) and f ′(4). x tan(x) − 2 3. A quotient of trigonometric functions. Let f (x) . Find f ′(x) and f ′(1). sec(x) 2x 2 tan(x) 4. A quotient that involves a product. Let f (x) . Find f ′(x) and f ′(4). sec(x) 5. Finding a tangent line equation. Find the equation of the tangent line to the curve y 3 tan x at the point (π/4, 3). The equation of this tangent line can be written in the form y mx + b. Find m and b. 6. An object moving vertically has its height at time t (measured in feet, with time in 2 cos(t) seconds) given by the function h(t) 3 + 1.2t . a. What is the object’s instantaneous velocity when t 2? b. What is the object’s acceleration at the instant t 2? c. Describe in everyday language the behavior of the object at the instant t 2. 7. Let f (x) sin(x) cot(x). a. Use the product rule to ﬁnd f ′(x). b. True or false: for all real numbers x, f (x) cos(x). c. Explain why the function that you found in (a) is almost the opposite of the sine function, but not quite. (Hint: convert all of the trigonometric functions in (a) to sines and cosines, and work to simplify. Think carefully about the domain of f and the domain of f ′.) 8. Let p(z) be given by the rule z tan(z) p(z) + 3e z + 1. +1 z 2 sec(z) a. Determine p ′(z). b. Find an equation for the tangent line to p at the point where z 0. c. At z 0, is p increasing, decreasing, or neither? Why? 119 Chapter 2 Computing Derivatives 2.5 The chain rule Motivating Questions • What is a composite function and how do we recognize its structure algebraically? • Given a composite function C(x) f (1(x)) that is built from diﬀerentiable functions f and 1, how do we compute C′(x) in terms of f , 1, f ′, and 1 ′? What is the statement of the Chain Rule? In addition to learning how to diﬀerentiate a variety of basic functions, we have also been developing our ability to use rules to diﬀerentiate certain algebraic combinations of them. Example 2.5.1 State the rule(s) to ﬁnd the derivative of each of the following combinations of f (x) sin(x) and 1(x) x 2 : s(x) 3x 2 − 5 sin(x), p(x) x 2 sin(x), and sin(x) q(x) . x2 Solution. Finding s ′ uses the sum and constant multiple rules, because s(x) 31(x) − 5 f (x). Determining p ′ requires the product rule, because p(x) 1(x) · f (x). To calculate q ′ f (x) we use the quotient rule, because q(x) 1(x) . There is one more natural way to combine basic functions algebraically, and that is by com- posing them. For instance, let’s consider the function C(x) sin(x 2 ), and observe that any input x passes through a chain of functions. In the process that deﬁnes the function C(x), x is ﬁrst squared, and then the sine of the result is taken. Using an arrow diagram, x −→ x 2 −→ sin(x 2 ). In terms of the elementary functions f and 1, we observe that x is the input for the function 1, and the result is used as the input for f . We write C(x) f (1(x)) sin(x 2 ) and say that C is the composition of f and 1. We will refer to 1, the function that is ﬁrst applied to x, as the inner function, while f , the function that is applied to the result, is the outer function. Given a composite function C(x) f (1(x)) that is built from diﬀerentiable functions f and 1, how do we compute C′(x) in terms of f , 1, f ′, and 1 ′? In the same way that the rate of change of a product of two functions, p(x) f (x) · 1(x), depends on the behavior of both f and 1, it makes sense intuitively that the rate of change of a composite function C(x) f (1(x)) will 120 2.5 The chain rule also depend on some combination of f and 1 and their derivatives. The rule that describes how to compute C′ in terms of f and 1 and their derivatives is called the chain rule. But before we can learn what the chain rule says and why it works, we ﬁrst need to be comfortable decomposing composite functions so that we can correctly identify the inner and outer functions, as we did in the example above with C(x) sin(x 2 ). Preview Activity 2.5.1. For each function given below, identify its fundamental alge- braic structure. In particular, is the given function a sum, product, quotient, or com- position of basic functions? If the function is a composition of basic functions, state a formula for the inner function 1 and the outer function f so that the overall compos- ite function can be written in the form f (1(x)). If the function is a sum, product, or quotient of basic functions, use the appropriate rule to determine its derivative. a. h(x) tan(2x ) d. m(x) e tan(x) √ b. p(x) 2x tan(x) e. w(x) x + tan(x) √ c. r(x) (tan(x))2 f. z(x) tan(x) 2.5.1 The chain rule Often a composite function cannot be written in an alternate algebraic form. For instance, the function C(x) sin(x 2 ) cannot be expanded or otherwise rewritten, so it presents no al- ternate approaches to taking the derivative. But some composite functions can be expanded or simpliﬁed, and these provide a way to explore how the chain rule works. Example 2.5.2 Let f (x) −4x + 7 and 1(x) 3x − 5. Determine a formula for C(x) f (1(x)) and compute C′(x). How is C′ related to f and 1 and their derivatives? Solution. By the rules given for f and 1, C(x) f (1(x)) f (3x − 5) − 4(3x − 5) + 7 − 12x + 20 + 7 − 12x + 27. Thus, C′(x) −12. Noting that f ′(x) −4 and 1 ′(x) 3, we observe that C′ appears to be the product of f ′ and 1 ′. It may seem that Example 2.5.2 is too elementary to illustrate how to diﬀerentiate a composite fuction. Linear functions are the simplest of all functions, and composing linear functions yields another linear function. While this example does not illustrate the full complexity of a composition of nonlinear functions, at the same time we remember that any diﬀerentiable function is locally linear, and thus any function with a derivative behaves like a line when viewed up close. The fact that the derivatives of the linear functions f and 1 are multiplied 121 Chapter 2 Computing Derivatives to ﬁnd the derivative of their composition turns out to be a key insight. We now consider a composition involving a nonlinear function. Example 2.5.3 Let C(x) sin(2x). Use the double angle identity to rewrite C as a product of basic functions, and use the product rule to ﬁnd C′. Rewrite C′ in the simplest form possible. Solution. Using the double angle identity for the sine function, we write C(x) sin(2x) 2 sin(x) cos(x). Applying the product rule and simplifying, we ﬁnd C′(x) 2 sin(x)(− sin(x)) + cos(x)(2 cos(x)) 2(cos2 (x) − sin2 (x)). Next, we recall that the double angle identity for the cosine, cos(2x) cos2 (x) − sin2 (x). Substituting this result into our expression for C′(x), we now have that C′(x) 2 cos(2x). In Example 2.5.3, if we let 1(x) 2x and f (x) sin(x), we observe that C(x) f (1(x)). Now, 1 ′(x) 2 and f ′(x) cos(x), so we can view the structure of C′(x) as C′(x) 2 cos(2x) 1 ′(x) f ′(1(x)). In this example, as in the example involving linear functions, we see that the derivative of the composite function C(x) f (1(x)) is found by multiplying the derivatives of f and 1, but with f ′ evaluated at 1(x). It makes sense intuitively that these two quantities are involved in the rate of change of a composite function: if we ask how fast C is changing at a given x value, it clearly matters how fast 1 is changing at x, as well as how fast f is changing at the value of 1(x). It turns out that this structure holds for all diﬀerentiable functions¹ as is stated in the Chain Rule. Chain Rule. If 1 is diﬀerentiable at x and f is diﬀerentiable at 1(x), then the composite function C deﬁned by C(x) f (1(x)) is diﬀerentiable at x and C′(x) f ′(1(x))1 ′(x). As with the product and quotient rules, it is often helpful to think verbally about what the chain rule says: “If C is a composite function deﬁned by an outer function f and an inner function 1, then C′ is given by the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.” ¹Like other diﬀerentiation rules, the Chain Rule can be proved formally using the limit deﬁnition of the deriv- ative. 122 2.5 The chain rule It is helpful to identify clearly the inner function 1 and outer function f , compute their derivatives individually, and then put all of the pieces together by the chain rule. Example 2.5.4 Determine the derivative of the function r(x) (tan(x))2 . Solution. The function r is composite, with inner function 1(x) tan(x) and outer function f (x) x 2 . Organizing the key information involving f , 1, and their derivatives, we have f (x) x 2 1(x) tan(x) f ′(x) 2x 1 ′(x) sec2 (x) f ′(1(x)) 2 tan(x) Applying the chain rule, we ﬁnd that r ′(x) f ′(1(x))1 ′(x) 2 tan(x) sec2 (x). As a side note, we remark that r(x) is usually written as tan2 (x). This is common notation for powers of trigonometric functions: cos4 (x), sin5 (x), and sec2 (x) are all composite functions, with the outer function a power function and the inner function a trigonometric one. Activity 2.5.2. For each function given below, identify an inner function 1 and outer function f to write the function in the form f (1(x)). Determine f ′(x), 1 ′(x), and f ′(1(x)), and then apply the chain rule to determine the derivative of the given func- tion. a. h(x) cos(x 4 ) d. z(x) cot5 (x) √ b. p(x) tan(x) c. s(x) 2sin(x) e. m(x) (sec(x) + e x )9 2.5.2 Using multiple rules simultaneously The chain rule now joins the sum, constant multiple, product, and quotient rules in our collection of techniques for ﬁnding the derivative of a function through understanding its algebraic structure and the basic functions that constitute it. It takes practice to get comfort- able applying multiple rules to diﬀerentiate a single function, but using proper notation and taking a few extra steps will help. 2 +2t Example 2.5.5 Find a formula for the derivative of h(t) 3t sec4 (t). Solution. We ﬁrst observe that h is the product of two functions: h(t) a(t) · b(t), where a(t) 3t +2t and b(t) sec4 (t). We will need to use the product rule to diﬀerentiate h. And 2 because a and b are composite functions, we will need the chain rule. We therefore begin by computing a ′(t) and b ′(t). 2 +2t Writing a(t) f (1(t)) 3t , and ﬁnding the derivatives of f and 1, we have 123 Chapter 2 Computing Derivatives f (t) 3t 1(t) t 2 + 2t f ′(t) 3t ln(3) 1 ′(t) 2t + 2 f ′(1(t)) 3t +2t ln(3) 2 Thus, by the chain rule, it follows that a ′(t) f ′(1(t))1 ′(t) 3t 2 +2t ln(3)(2t + 2). Turning next to b, we write b(t) r(s(t)) sec4 (t) and ﬁnd the derivatives of r and s. r(t) t 4 s(t) sec(t) r ′(t) 4t 3 s ′(t) sec(t) tan(t) r ′(s(t)) 4 sec3 (t) By the chain rule, b ′(t) r ′(s(t))s ′(t) 4 sec3 (t) sec(t) tan(t) 4 sec4 (t) tan(t). Now we are ﬁnally ready to compute the derivative of the function h. Recalling that h(t) 3t +2t sec4 (t), by the product rule we have 2 d d h ′(t) 3t 2 +2t [sec4 (t)] + sec4 (t) [3t +2t ]. 2 dt dt From our work above with a and b, we know the derivatives of 3t +2t and sec4 (t), and there- 2 fore h ′(t) 3t +2t 4 sec4 (t) tan(t) + sec4 (t)3t +2t ln(3)(2t + 2). 2 2 Activity 2.5.3. For each of the following functions, ﬁnd the function’s derivative. State the rule(s) you use, label relevant derivatives appropriately, and be sure to clearly identify your √overall answer. 2 a. p(r) 4 r 6 + 2e r d. s(z) 2z sec(z) b. m(v) sin(v 2 ) cos(v 3 ) cos(10y) 2 c. h(y) e 4y +1 e. c(x) sin(e x ) The chain rule now adds substantially to our ability to compute derivatives. Whether we are ﬁnding the equation of the tangent line to a curve, the instantaneous velocity of a mov- ing particle, or the instantaneous rate of change of a certain quantity, if the function under consideration is a composition, the chain rule is indispensable. Activity 2.5.4. Use known derivative rules, including the chain rule, as needed to answer each of the following questions. √ a. Find an equation for the tangent line to the curve y e x + 3 at the point where x 0. 1 b. If s(t) represents the position function of a particle moving horizon- (t 2 + 1)3 tally along an axis at time t (where s is measured in inches and t in seconds), ﬁnd the particle’s instantaneous velocity at t 1. Is the particle moving to the 124 2.5 The chain rule left or right at that instant? c. At sea level, air pressure is 30 inches of mercury. At an altitude of h feet above sea level, the air pressure, P, in inches of mercury, is given by the function P 30e −0.0000323h . Compute dP/dh and explain what this derivative function tells you about air pressure, including a discussion of the units on dP/dh. In addition, determine how fast the air pressure is changing for a pilot of a small plane passing through an altitude of 1000 feet. d. Suppose that f (x) and 1(x) are diﬀerentiable functions and that the following information about them is known: x f (x) f ′(x) 1(x) 1 ′(x) −1 2 −5 −3 4 2 −3 4 −1 2 Table 2.5.6: Data for functions f and 1. If C(x) is a function given by the formula f (1(x)), determine C′(2). In addition, if D(x) is the function f ( f (x)), ﬁnd D ′(−1). 2.5.3 The composite version of basic function rules As we gain more experience with diﬀerention, we will become more comfortable in simply writing down the derivative without taking multiple steps. This is particularly simple when the inner function is linear, since the derivative of a linear function is a constant. Example 2.5.7 Use the chain rule to diﬀerentiate each of the following composite functions whose inside function is linear: d [ ] (5x + 7)10 10(5x + 7)9 · 5, dx d [tan(17x)] 17 sec2 (17x), and dx d [ −3x ] e −3e −3x . dx More generally, following is an excellent exercise for getting comfortable with the derivative rules. Write down a list of all the basic functions whose derivatives we know, and list the derivatives. Then write a composite function with the inner function being an unknown function u(x) and the outer function being a basic function. Finally, write the chain rule for the composite function. The following example illustrates this for two diﬀerent functions. Example 2.5.8 To determine d [sin(u(x))], dx where u is a diﬀerentiable function of x, we use the chain rule with the sine function as the 125 Chapter 2 Computing Derivatives outer function. Applying the chain rule, we ﬁnd that d [sin(u(x))] cos(u(x)) · u ′(x). dx dx [sin(x)] cos(x). d This rule is analogous to the basic derivative rule that dx [a ] a x ln(a), it follows by the chain rule that d x Similarly, since d u(x) [a ] a u(x) ln(a) · u ′(x). dx dx [a ] a x ln(a). d x This rule is analogous to the basic derivative rule that 2.5.4 Summary • A composite function is one where the input variable x ﬁrst passes through one func- tion, and then the resulting output passes through another. For example, the function h(x) 2sin(x) is composite since x −→ sin(x) −→ 2sin(x) . • Given a composite function C(x) f (1(x)) where f and 1 are diﬀerentiable functions, the chain rule tells us that C′(x) f ′(1(x))1 ′(x). 2.5.5 Exercises 1. Mixing rules: chain, product, sum. Find the derivative of f (x) e 5x (x 2 + 7x ). 2. Mixing rules: chain and product. Find the derivative of v(t) t 6 e −ct . Assume that c is a constant. √ 3. Using the chain rule repeatedly. Find the derivative of y e −5t 2 + 9. 4. Derivative involving arbitrary constants a and b. Find the derivative of the function f (x) axe −bx+12 . Assume that a and b are constants. 5. Chain rule with graphs. Use the graph below to ﬁnd exact values of the indicated derivatives, or state that they do not exist. If a derivative does not exist, enter dne in the answer blank. The graph of f (x) is black and has a sharp corner at x 2. The graph of 1(x) is blue. 126 2.5 The chain rule Let h(x) f (1(x)). Find h ′(1), h ′(2), and h ′(3) or explain why they do not exist. 6. Chain rule with function values. Given F(4) 1, F′(4) 5, F(5) 4, F′(5) 6 and G(1) 3, G′(1) 4, G(4) 5, G′(4) 6, ﬁnd each of the following. A. H(4) if H(x) F(G(x)) B. H ′(4) if H(x) F(G(x)) C. H(4) if H(x) G(F(x)) D. H ′(4) if H(x) G(F(x)) E. H ′(4) if H(x) F(x)/G(x) 7. A product involving a composite function. Find the derivative of f (x) 2x sin(6x). 8. Consider the basic functions f (x) x 3 and 1(x) sin(x). a. Let h(x) f (1(x)). Find the exact instantaneous rate of change of h at the point where x π4 . b. Which function is changing most rapidly at x 0.25: h(x) f (1(x)) or r(x) 1( f (x))? Why? c. Let h(x) f (1(x)) and r(x) 1( f (x)). Which of these functions has a derivative that is periodic? Why? 9. Let u(x) be a diﬀerentiable function. For each of the following functions, determine the derivative. Each response will involve u and/or u ′. a. p(x) e u(x) d. s(x) u(cot(x)) b. q(x) u(e x ) e. a(x) u(x 4 ) c. r(x) cot(u(x)) f. b(x) u 4 (x) 127 Chapter 2 Computing Derivatives 10. Let functions p and q be the piecewise linear functions given by their respective graphs in Figure 2.5.9. Use the graphs to answer the following questions. p 3 2 1 -3 -2 -1 1 2 3 -1 q -2 -3 Figure 2.5.9: The graphs of p (in blue) and q (in green). a. Let C(x) p(q(x)). Determine C′(0) and C′(3). b. Find a value of x for which C′(x) does not exist. Explain your thinking. c. Let Y(x) q(q(x)) and Z(x) q(p(x)). Determine Y ′(−2) and Z′(0). 11. If a spherical tank of radius 4 feet has h feet of water present in the tank, then the volume of water in the tank is given by the formula π 2 V h (12 − h). 3 a. At what instantaneous rate is the volume of water in the tank changing with re- spect to the height of the water at the instant h 1? What are the units on this quantity? b. Now suppose that the height of water in the tank is being regulated by an inﬂow and outﬂow (e.g., a faucet and a drain) so that the height of the water at time t is given by the rule h(t) sin(πt) + 1, where t is measured in hours (and h is still measured in feet). At what rate is the height of the water changing with respect to time at the instant t 2? c. Continuing under the assumptions in (b), at what instantaneous rate is the vol- ume of water in the tank changing with respect to time at the instant t 2? d. What are the main diﬀerences between the rates found in (a) and (c)? Include a discussion of the relevant units. 128 2.6 Derivatives of Inverse Functions 2.6 Derivatives of Inverse Functions Motivating Questions • What is the derivative of the natural logarithm function? • What are the derivatives of the inverse trigonometric functions arcsin(x) and arctan(x)? • If 1 is the inverse of a diﬀerentiable function f , how is 1 ′ computed in terms of f , f ′, and 1? Much of mathematics centers on the notion of function. Indeed, throughout our study of cal- culus, we are investigating the behavior of functions, with particular emphasis on how fast the output of the function changes in response to changes in the input. Because each func- tion represents a process, a natural question to ask is whether or not the particular process can be reversed. That is, if we know the output that results from the function, can we deter- mine the input that led to it? And if we know how fast a particular process is changing, can we determine how fast the inverse process is changing? One of the most important functions in all of mathematics is the natural exponential function f (x) e x . Its inverse, the natural logarithm, 1(x) ln(x), is similarly important. One of our goals in this section is to learn how to diﬀerentiate the logarithm function. First, we review some of the basic concepts surrounding functions and their inverses. Preview Activity 2.6.1. The equation y 95 (x − 32) relates a temperature given in x degrees Fahrenheit to the corresponding temperature y measured in degrees Celcius. a. Solve the equation y 59 (x − 32) for x to write x (Fahrenheit temperature) in terms of y (Celcius temperature). b. Let C(x) 59 (x − 32) be the function that takes a Fahrenheit temperature as input and produces the Celcius temperature as output. In addition, let F(y) be the function that converts a temperature given in y degrees Celcius to the temperature F(y) measured in degrees Fahrenheit. Use your work in (a) to write a formula for F(y). c. Next consider the new function deﬁned by p(x) F(C(x)). Use the formulas for F and C to determine an expression for p(x) and simplify this expression as much as possible. What do you observe? d. Now, let r(y) C(F(y)). Use the formulas for F and C to determine an expres- sion for r(y) and simplify this expression as much as possible. What do you observe? e. What is the value of C′(x)? of F′(y)? How do these values appear to be related? 129 Chapter 2 Computing Derivatives 2.6.1 Basic facts about inverse functions A function f : A → B is a rule that associates each element in the set A to one and only one element in the set B. We call A the domain of f and B the codomain of f . If there exists a function 1 : B → A such that 1( f (a)) a for every possible choice of a in the set A and f (1(b)) b for every b in the set B, then we say that 1 is the inverse of f . We often use the notation f −1 (read “ f -inverse”) to denote the inverse of f . The inverse function undoes the work of f . Indeed, if y f (x), then f −1 (y) f −1 ( f (x)) x. Thus, the equations y f (x) and x f −1 (y) say the same thing. The only diﬀerence be- tween the two equations is one of perspective — one is solved for x, while the other is solved for y. Here we brieﬂy remind ourselves of some key facts about inverse functions. Note 2.6.1 For a function f : A → B, • f has an inverse if and only if f is one-to-one ¹ and onto ²; • provided f −1 exists, the domain of f −1 is the codomain of f , and the codomain of f −1 is the domain of f ; • f −1 ( f (x)) x for every x in the domain of f and f ( f −1 (y)) y for every y in the codomain of f ; • y f (x) if and only if x f −1 (y). The last fact reveals a special relationship between the graphs of f and f −1 . If a point (x, y) that lies on the graph of y f (x), then it is also true that x f −1 (y), which means that the point (y, x) lies on the graph of f −1 . This shows us that the graphs of f and f −1 are the reﬂections of each other across the line y x, because this reﬂection is precisely the geometric action that swaps the coordinates in an ordered pair. In Figure 2.6.2, we see this illustrated by the function y f (x) 2x and its inverse, with the points (−1, 21 ) and ( 12 , −1) highlighting the reﬂection of the curves across y x. To close our review of important facts about inverses, we recall that the natural exponential function y f (x) e x has an inverse function, namely the natural logarithm, x f −1 (y) ln(y). Thus, writing y e x is interchangeable with x ln(y), plus ln(e x ) x for every real number x and e ln(y) y for every positive real number y. 2.6.2 The derivative of the natural logarithm function In what follows, we ﬁnd a formula for the derivative of 1(x) ln(x). To do so, we take advantage of the fact that we know the derivative of the natural exponential function, the inverse of 1. In particular, we know that writing 1(x) ln(x) is equivalent to writing e 1(x) ¹A function f is one-to-one provided that no two distinct inputs lead to the same output. ²A function f is onto provided that every possible element of the codomain can be realized as an output of the function for some choice of input from the domain. 130 2.6 Derivatives of Inverse Functions y = f (x) 2 (−1, 21 ) -2 2 ( 12 , −1) -2 y = f −1 (x) y=x Figure 2.6.2: A graph of a function y f (x) along with its inverse, y f −1 (x). x. Now we diﬀerentiate both sides of this equation and observe that d [ 1(x) ] d e [x]. dx dx The righthand side is simply 1; by applying the chain rule to the left side, we ﬁnd that e 1(x) 1 ′(x) 1. Next we solve for 1 ′(x), to get 1 1 ′(x) . e 1(x) Finally, we recall that 1(x) ln(x), so e 1(x) e ln(x) x, and thus 1 1 ′(x) . x Natural Logarithm. dx [ln(x)] x1 . d For all positive real numbers x, This rule for the natural logarithm function now joins our list of basic derivative rules. Note that this rule applies only to positive values of x, as these are the only values for which ln(x) is deﬁned. Also notice that for the ﬁrst time in our work, diﬀerentiating a basic function of a particular type has led to a function of a very diﬀerent nature: the derivative of the natural logarithm is not another logarithm, nor even an exponential function, but rather a rational one. 131 Chapter 2 Computing Derivatives Derivatives of logarithms may now be computed in concert with all of the rules known to date. For instance, if f (t) ln(t 2 + 1), then by the chain rule, f ′(t) t 21+1 · 2t. There are interesting connections between the graphs of f (x) e x and f −1 (x) ln(x). In Figure 2.6.3, we are reminded that since the natural exponential function y = ex has the property that its derivative is it- 8 self, the slope of the tangent to y e x is equal to the height of the curve at that point. For instance, at the point A B (ln(0.5), 0.5), the slope of the tangent line 4 is m A 0.5, and at B (ln(5), 5), the tan- gent line’s slope is m B 5. B′ At the corresponding points A′ and B′ on A the graph of the natural logarithm func- tion (which come from reﬂecting A and -4 4 8 B across the line y x), we know that A′ the slope of the tangent line is the rec- iprocal of the x-coordinate of the point y = ln(x) (since dxd [ln(x)] x1 ). Thus, at A′ -4 (0.5, ln(0.5)), we have m A′ 0.51 2, and ′ at B (5, ln(5)), m B′ 5 . 1 Figure 2.6.3: A graph of the function y e x along with its inverse, y ln(x), where both functions are viewed using the input variable x. In particular, we observe that m A′ m1A and m B′ m1B . This is not a coincidence, but in fact holds for any curve y f (x) and its inverse, provided the inverse exists. This is due to the reﬂection across y x. It changes the roles of x and y, thus reversing the rise and run, so the slope of the inverse function at the reﬂected point is the reciprocal of the slope of the original function. Activity 2.6.2. For each function given below, ﬁnd its derivative. a. h(x) x 2 ln(x) d. z(x) tan(ln(x)) ln(t) b. p(t) e t +1 c. s(y) ln(cos(y) + 2) e. m(z) ln(ln(z)) 2.6.3 Inverse trigonometric functions and their derivatives Trigonometric functions are periodic, so they fail to be one-to-one, and thus do not have inverse functions. However, we can restrict the domain of each trigonometric function so that it is one-to-one on that domain. For instance, consider the sine function on the domain [− π2 , π2 ]. Because no output of the sine function is repeated on this interval, the function is one-to-one and thus has an inverse. 132 2.6 Derivatives of Inverse Functions Thus, the function f (x) sin(x) with [− π2 , π2 ] and codomain [−1, 1] has an inverse function f −1 such that π π f −1 : [−1, 1] → [− , ]. 2 2 We call f −1 the arcsine (or inverse sine) function and write f −1 (y) arcsin(y). It is especially important to remember that (1, π2 ) π 2 y sin(x) and x arcsin(y) f −1 say the same thing. “The arcsine of y” f ( π2 , 1) means “the angle whose sine is y.” For ex- ample, arcsin( 21 ) π6 means that π6 is the angle whose sine is 12 , which is equivalent − π2 π 2 to writing sin( π6 ) 12 . Next, we determine the derivative of the arcsine function. Letting h(x) arcsin(x), − π2 our goal is to ﬁnd h ′(x). Since h(x) is the an- gle whose sine is x, it is equivalent to write sin(h(x)) x. Figure 2.6.4: A graph of f (x) sin(x) (in blue), restricted to the domain [− π2 , π2 ], along with its inverse, f −1 (x) arcsin(x) (in magenta). Diﬀerentiating both sides of the previous equation, we have d d [sin(h(x))] [x]. dx dx The righthand side is simply 1, and by applying the chain rule applied to the left side, cos(h(x))h ′(x) 1. Solving for h ′(x), it follows that 1 h ′(x) . cos(h(x)) Finally, we recall that h(x) arcsin(x), so the denominator of h ′(x) is the function cos(arcsin(x)), or in other words, “the cosine of the angle whose sine is x.” A bit of right triangle trigonom- etry allows us to simplify this expression considerably. 133 Chapter 2 Computing Derivatives Let’s say that θ arcsin(x), so that θ is the angle whose sine is x. We can picture θ as an angle in a right triangle with hypotenuse 1 and a vertical leg of length x, as shown in Figure 2.6.5. The horizontal leg must be √ 1 − x 2 , by the Pythagorean Theorem. Now, because θ arcsin(x), the expression for 1 x cos(arcsin(x)) is equivalent to cos(θ). From the ﬁgure, √ cos(arcsin(x)) cos(θ) 1 − x 2 . Substituting this expression into our formula, h ′(x) θ 1 √ cos(arcsin(x)) , we have now shown that 1 − x2 1 Figure 2.6.5: The right triangle h ′(x) √ . 1 − x2 that corresponds to the angle θ arcsin(x). Inverse sine. For all real numbers x such that −1 < x < 1, d 1 [arcsin(x)] √ . dx 1 − x2 Activity 2.6.3. The following prompts in this activity will lead you to develop the derivative of the inverse tangent function. a. Let r(x) arctan(x). Use the relationship between the arctangent and tangent functions to rewrite this equation using only the tangent function. b. Diﬀerentiate both sides of the equation you found in (a). Solve the resulting equation for r ′(x), writing r ′(x) as simply as possible in terms of a trigonometric function evaluated at r(x). c. Recall that r(x) arctan(x). Update your expression for r ′(x) so that it only involves trigonometric functions and the independent variable x. d. Introduce a right triangle with angle θ so that θ arctan(x). What are the three sides of the triangle? e. In terms of only x and 1, what is the value of cos(arctan(x))? f. Use the results of your work above to ﬁnd an expression involving only 1 and x for r ′(x). While derivatives for other inverse trigonometric functions can be established similarly, for now we limit ourselves to the arcsine and arctangent functions. 134 2.6 Derivatives of Inverse Functions Activity 2.6.4. Determine the derivative of each of the following functions. a. f (x) x 3 arctan(x) + e x ln(x) b. p(t) 2t arcsin(t) c. h(z) (arcsin(5z) + arctan(4 − z))27 d. s(y) cot(arctan(y)) e. m(v) ln(sin2 (v) + 1) ( ) ln(w) f. 1(w) arctan 1 + w2 2.6.4 The link between the derivative of a function and the derivative of its inverse In Figure 2.6.3, we saw an interesting relationship between the slopes of tangent lines to the natural exponential and natural logarithm functions at points reﬂected across the line y x. In particular, we observed that at the point (ln(2), 2) on the graph of f (x) e x , the slope of the tangent line is f ′(ln(2)) 2, while at the corresponding point (2, ln(2)) on the graph of f −1 (x) ln(x), the slope of the tangent line is ( f −1 )′(2) 12 , which is the reciprocal of f ′(ln(2)). That the two corresponding tangent lines have reciprocal slopes is not a coincidence. If f and 1 are diﬀerentiable inverse functions, then y f (x) if and only if x 1(y), then f (1(x)) x for every x in the domain of f −1 . Diﬀerentiating both sides of this equation, we have d d [ f (1(x))] [x], dx dx and by the chain rule, f ′(1(x))1 ′(x) 1. Solving for 1 ′(x), we have 1 ′(x) f ′ (1(x)) 1 . Here we see that the slope of the tangent line to the inverse function 1 at the point (x, 1(x)) is precisely the reciprocal of the slope of the tangent line to the original function f at the point (1(x), f (1(x))) (1(x), x). To see this more clearly, consider the graph of the function y f (x) shown in Figure 2.6.6, along with its inverse y 1(x). Given a point (a, b) that lies on the graph of f , we know that (b, a) lies on the graph of 1; because f (a) b and 1(b) a. Now, applying the rule that 1 ′(x) 1/ f ′(1(x)) to the value x b, we have 1 1 1 ′(b) ′ ′ , f (1(b)) f (a) which is precisely what we see in the ﬁgure: the slope of the tangent line to 1 at (b, a) is the reciprocal of the slope of the tangent line to f at (a, b), since these two lines are reﬂections of one another across the line y x. 135 Chapter 2 Computing Derivatives m = f ′ (a) y = f (x) (a, b) m = g′ (b) (b, a) y = g(x) Figure 2.6.6: A graph of function y f (x) along with its inverse, y 1(x) f −1 (x). Observe that the slopes of the two tangent lines are reciprocals of one another. Derivative of an inverse function. Suppose that f is a diﬀerentiable function with inverse 1 and that (a, b) is a point that lies on the graph of f at which f ′(a) , 0. Then 1 1 ′(b) . f ′(a) More generally, for any x in the domain of 1 ′, we have 1 ′(x) 1/ f ′(1(x)). The rules we derived for ln(x), arcsin(x), and arctan(x) are all just speciﬁc examples of this general property of the derivative of an inverse function. For example, with 1(x) ln(x) and f (x) e x , it follows that 1 1 1 1 ′(x) . f ′(1(x)) e ln(x) x 2.6.5 Summary dx [ln(x)] x1 . d • For all positive real numbers x, • For all real numbers x such that −1 < x < 1, dx [arcsin(x)] d √ 1 . In addition, for all 1−x 2 dx [arctan(x)] d 1 real numbers x, 1+x 2 . • If 1 is the inverse of a diﬀerentiable function f , then for any point x in the domain of 1 ′, 1 ′(x) f ′ (1(x)) 1 . 136 2.6 Derivatives of Inverse Functions 2.6.6 Exercises 1. Composite function involving logarithms and polynomials. Find the derivative of the function f (t) ln(t 3 + 3). 2. Composite function involving trigonometric functions and logarithms. Find the de- rivative of the function 1(t) cos(ln(t)). 3. Product involving arcsin(w). Find the derivative of the function h(w) 5w arcsin w 4. Derivative involving arctan(x). For x > 0, ﬁnd and simplify the derivative of f (x) arctan x + arctan(1/x). (What does your result tell you about f )? 5. Composite function from a graph. Let (x0 , y0 ) (2, 6) and (x1 , y1 ) (2.1, 6.2). Use the following graph of the function f to ﬁnd the indicated derivatives. If h(x) ( f (x))5 , ﬁnd h ′(2). If 1(x) f −1 (x), ﬁnd 1 ′(6). 6. Composite function involving an inverse trigonometric function. Let ( ) f (x) 7 sin−1 x 3 . Find f ′(x). 7. Mixing rules: product, chain, and inverse trig. If f (x) 8x 4 arctan(3x 3 ), ﬁnd f ′(x). 8. Mixing rules: product and inverse trig. Let f (x) 8 cos(x) sin−1 (x). Find f ′(x). 9. Determine the derivative of each of the following functions. Use proper notation and clearly identify the derivative rules you use. a. f (x) ln(2 arctan(x) + 3 arcsin(x) + 5) b. r(z) arctan(ln(arcsin(z))) c. q(t) arctan2 (3t) arcsin4 (7t) ( ) arctan(v) d. 1(v) ln arcsin(v)+v 2 137 Chapter 2 Computing Derivatives 10. Consider the graph of y f (x) provided in Figure 2.6.7 and use it to answer the fol- lowing questions. a. Use the provided graph to estimate the value of f ′(1). b. Sketch an approximate graph of y f −1 (x). Label at least three distinct points on the graph that correspond to three points on the graph of f . y = f (x) c. Based on your work in (a), what is the value of ( f −1 )′(−1)? Why? Figure 2.6.7: A function y f (x) 11. Let f (x) 14 x 3 + 4. a. Sketch a graph of y f (x) and explain why f is an invertible function. b. Let 1 be the inverse of f and determine a formula for 1. c. Compute f ′(x), 1 ′(x), f ′(2), and 1 ′(6). What is the special relationship between f ′(2) and 1 ′(6)? Why? 12. Let h(x) x + sin(x). a. Sketch a graph of y h(x) and explain why h must be invertible. b. Explain why it does not appear to be algebraically possible to determine a formula for h −1 . c. Observe that the point ( π2 , π 2 + 1) lies on the graph of y h(x). Determine the value of (h −1 )′( π2 + 1). 138 2.7 Derivatives of Functions Given Implicitly 2.7 Derivatives of Functions Given Implicitly Motivating Questions • What does it mean to say that a curve is an implicit function of x, rather than an explicit function of x? dy • How does implicit diﬀerentiation enable us to ﬁnd a formula for dx when y is an implicit function of x? dy • In the context of an implicit curve, how can we use dx to answer important questions about the tangent line to the curve? In all of our studies with derivatives so far, we have worked with functions whose formula is given explicitly in terms of x. But there are many interesting curves whose equations involving x and y are impossible to solve for y in terms of x. x2 + y2 = 16 4 x3 − y3 = 6xy A x -4 4 B -4 Figure 2.7.1: At left, the circle given by x 2 + y 2 16. In the middle, the portion of the circle x 2 + y 2 16 that has been highlighted in the box at left. And at right, the lemniscate given by x 3 − y 3 6x y. Perhaps the simplest and most natural of all such curves are circles. Because of the circle’s symmetry, for each x value strictly between the endpoints of the horizontal diameter, there √ are two corresponding y-values. For instance, in Figure 2.7.1, we have labeled A (−3, 7) √ and B (−3, − 7), and these points demonstrate that the circle fails the vertical line test. Hence, it is impossible to represent the circle through a single function of the form y f (x). But portions of the circle can be represented explicitly as a function of x, such as the highlighted arc that is magniﬁed in the center of Figure 2.7.1. Moreover, it is evident that the circle is locally linear, so we ought to be able to ﬁnd a tangent line to the curve at every dy point. Thus, it makes sense to wonder if we can compute dx at any point on the circle, even though we cannot write y explicitly as a function of x. We say that the equation x 2 + y 2 16 deﬁnes y implicitly as a function of x. The graph of the equation can be broken into pieces where each piece can be deﬁned by an explicit 139 Chapter 2 Computing Derivatives function of x. For the circle, we could choose to take the top half as one function of x, namely √ √ y 16 − x 2 and the bottom half as y − 16 − x 2 . The equation for the circle deﬁnes two implicit functions of x. The righthand curve in Figure 2.7.1 is called a lemniscate and is just one of many fascinating possibilities for implicitly given curves. dy How can we ﬁnd an equation for dx without an explicit formula for y in terms of x? The following preview activity reminds us of some ways we can compute derivatives of functions in settings where the function’s formula is not known. Preview Activity 2.7.1. Let f be a diﬀerentiable function of x (whose formula is not known) and recall that dx d [ f (x)] and f ′(x) are interchangeable notations. Determine each of the following derivatives of combinations of explicit functions of x, the un- known function [ 2 f , ]and an arbitrary constant c. [ ] a. dx x + f (x) d d. dxd f (x 2 ) [ ] b. d dx x 2 f (x) [ ] [ ] c. d dx c + x + f (x)2 e. d dx x f (x) + f (cx) + c f (x) 2.7.1 Implicit Diﬀerentiation We begin our exploration of implicit diﬀerentiation with the example of the circle given by dy x 2 + y 2 16. How can we ﬁnd a formula for dx ? By viewing y as an implicit function of x, we think of y as some function whose formula f (x) is unknown, but which we can diﬀerentiate. Just as y represents an unknown formula, so dy too its derivative with respect to x, dx , will be (at least temporarily) unknown. So we view y as an unknown diﬀerentiable function of x and diﬀerentiate both sides of the equation with respect to x. d [ 2 ] d x + y2 [16] . dx dx On the right, the derivative of the constant 16 is 0, and on the left we can apply the sum rule, so it follows that d [ 2] d [ 2] x + y 0. dx dx Note carefully ] diﬀerent roles being played by x and y. Because x is the independent [ 2the d variable, dx x 2x. But y is the dependent variable and y is an implicit function of x. d Recall Preview Activity 2.7.1, where we computed dx [ f (x)2 ]. Computing dx d [y 2 ] is the same, dy dx [y ] 2y 1 dx . We now have that d 2 and requires the chain rule, by which we ﬁnd that dy 2x + 2y 0. dx 140 2.7 Derivatives of Functions Given Implicitly dy We solve this equation for dx by subtracting 2x from both sides and dividing by 2y. dy 2x x − − . dx 2y y mt = − ba There are several important things to observe (a, b) dy about the result that dx − xy . First, this ex- pression for the derivative involves both x and y. This makes sense because there are two b corresponding points on the circle for each mr = a value of x between −4 and 4, and the slope of the tangent line is diﬀerent at each of these points.Second, this formula is entirely consis- tent with our understanding of circles. The slope of the radius from the origin to the point (a, b) is m r ba . The tangent line to the circle at (a, b) is perpendicular to the radius, and thus has slope m t − ba , as shown in Figure 2.7.2. In particular, the slope of the tangent line is zero at (0, 4) and (0, −4), and is undeﬁned at Figure 2.7.2: The circle given by (−4, 0) and (4, 0). All of these values are con- x 2 + y 2 16 with point (a, b) on the circle dy sistent with the formula dx − xy . and the tangent line at that point, with labeled slopes of the radial line, m r , and tangent line, m t . Example 2.7.3 For the curve given implicitly by x 3 + y 2 − 2x y 2, shown in Figure 2.7.4, ﬁnd the slope of the tangent line at (−1, 1). y 3 x -3 3 -3 Figure 2.7.4: The curve x 3 + y 2 − 2x y 2. 141 Chapter 2 Computing Derivatives Solution. We begin by diﬀerentiating the curve’s equation implicitly. Taking the derivative of each side with respect to x, d [ 3 ] d x + y 2 − 2x y [2] , dx dx by the sum rule and the fact that the derivative of a constant is zero, we have d 3 d 2 d [x ] + [y ] − [2x y] 0. dx dx dx For the three derivatives we now must execute, the ﬁrst uses the simple power rule, the second requires the chain rule (since y is an implicit function of x), and the third necessitates the product rule (again since y is a function of x). Applying these rules, we now ﬁnd that dy dy 3x 2 + 2y − [2x + 2y] 0. dx dx dy dy We want to solve this equation for dx . To do so, we ﬁrst collect all of the terms involving dx on one side of the equation. dy dy 2y − 2x 2y − 3x 2 . dx dx dy Then we factor the left side to isolate dx . dy (2y − 2x) 2y − 3x 2 . dx Finally, we divide both sides by (2y − 2x) and conclude that dy 2y − 3x 2 . dx 2y − 2x dy Note that the expression for dx depends on both x and y. To ﬁnd the slope of the tangent dy line at (−1, 1), we substitute the coordinates into the formula for dx , using the notation dy 2(1) − 3(−1)2 1 − . dx (−1,1) 2(1) − 2(−1) 4 This value matches our visual estimate of the slope of the tangent line shown in Figure 2.7.4. Example 2.7.3 shows that it is possible when diﬀerentiating implicitly to have multiple terms dy dy involving dx . We use addition and subtraction to collect all terms involving dx on one side dy dy of the equation, then factor to get a single term of dx . Finally, we divide to solve for dx . We use the notation dy dx (a,b) 142 2.7 Derivatives of Functions Given Implicitly dy to denote the evaluation of dx at the point (a, b). This is analogous to writing f ′(a) when f ′ depends on a single variable. d dy There is a big diﬀerence between writing dx and dx . For example, d 2 [x + y 2 ] dx gives an instruction to take the derivative with respect to x of the quantity x 2 + y 2 , presum- ably where y is a function of x. On the other hand, dy 2 (x + y 2 ) dx means the product of the derivative of y with respect to x with the quantity x 2 + y 2 . Under- standing this notational subtlety is essential. Activity 2.7.2. Consider the curve deﬁned by the equation x y 5 − 5y 3 + 4y, whose graph is pictured in Figure 2.7.5. y a. Explain why it is not possible to 3 express y as an explicit function of x. b. Use implicit diﬀerentiation to ﬁnd a formula for dy/dx. x -3 3 c. Use your result from part (b) to ﬁnd an equation of the line tangent to the graph of x y 5 − 5y 3 + 4y at the point (0, 1). -3 d. Use your result from part (b) to de- termine all of the points at which the graph of x y 5 − 5y 3 + 4y has Figure 2.7.5: The curve a vertical tangent line. x y 5 − 5y 3 + 4y. It is natural to ask where the tangent line to a curve is vertical or horizontal. The slope of a horizontal tangent line must be zero, while the slope of a vertical tangent line is undeﬁned. dy Often the formula for dx is expressed as a quotient of functions of x and y, say dy p(x, y) . dx q(x, y) The tangent line is horizontal precisely when the numerator is zero and the denominator is nonzero, making the slope of the tangent line zero. If we can solve the equation p(x, y) 0 for either x and y in terms of the other, we can substitute that expression into the original equation for the curve. This gives an equation in a single variable, and if we can solve that 143 Chapter 2 Computing Derivatives equation we can ﬁnd the point(s) on the curve where p(x, y) 0. At those points, the tangent line is horizontal. Similarly, the tangent line is vertical whenever q(x, y) 0 and p(x, y) , 0, making the slope undeﬁned. Activity 2.7.3. Consider the curve deﬁned by the equation y(y 2 − 1)(y − 2) x(x − 1)(x − 2), whose graph is pictured in Figure 2.7.6. Through implicit diﬀerentiation, it can be shown that dy (x − 1)(x − 2) + x(x − 2) + x(x − 1) 2 . dx (y − 1)(y − 2) + 2y 2 (y − 2) + y(y 2 − 1) Use this fact to answer each of the following questions. y a. Determine all points (x, y) at which the tangent line to the 2 curve is horizontal. (Use tech- nology appropriately to ﬁnd the needed zeros of the relevant 1 polynomial function.) x b. Determine all points (x, y) at which the tangent line is vertical. 1 2 3 (Use technology appropriately to ﬁnd the needed zeros of the -1 relevant polynomial function.) c. Find the equation of the tangent line to the curve at one of the Figure 2.7.6: points where x 1. y(y 2 − 1)(y − 2) x(x − 1)(x − 2). Activity 2.7.4. For each of the following curves, use implicit diﬀerentiation to ﬁnd dy/dx and determine the equation of the tangent line at the given point. a. x 3 − y 3 6x y, (−3, 3) c. 3xe −x y y 2 , (0.619061, 1) b. sin(y) + y x 3 + x, (0, 0) 2.7.2 Summary • In an equation involving x and y where portions of the graph can be deﬁned by explicit functions of x, we say that y is an implicit function of x. A good example of such a curve is the unit circle. • We use implicit diﬀerentiation to diﬀerentiate an implicitly deﬁned function. We dif- ferentiate both sides of the equation with respect to x, treating y as a function of x by 144 2.7 Derivatives of Functions Given Implicitly dy applying the chain rule. If possible, we subsequently solve for dx using algebra. dy dy • While dx may now involve both the variables x and y, dx still gives the slope of the tangent line to the curve. It may be used to decide where the tangent line is horizontal dy dy ( dx 0) or vertical ( dx is undeﬁned), or to ﬁnd the equation of the tangent line at a particular point on the curve. 2.7.3 Exercises 1. Implicit diﬀerentiaion in a polynomial equation. Find dy/dx in terms of x and y if x 2 y − x − 5y − 11 0. dy 2. Implicit diﬀerentiation in an equation with logarithms. Find in terms of x and y dx if x ln y + y 3 3 ln x. 3. Implicit diﬀerentiation in an equation with inverse trigonometric functions. Find dy/dx in terms of x and y if arctan(x 3 y) x y 3 . 4. Slope of the tangent line to an implicit curve. Find the slope of the tangent to the curve x 3 + x y + y 2 31 at (1, 5). 5. Equation of the tangent line to an implicit curve. Use implicit diﬀerentiation to ﬁnd an equation of the tangent line to the curve 3x y 3 + x y 16 at the point (4, 1). 6. Consider the curve given by the equation 2y 3 + y 2 − y 5 x 4 − 2x 3 + x 2 . Find all points at which the tangent line to the curve is horizontal or vertical. Be sure to use a graphing utility to plot this implicit curve and to visually check the results of algebraic reasoning that you use to determine where the tangent lines are horizontal and vertical. 7. For the curve given by the equation sin(x + y) + cos(x − y) 1, ﬁnd the equation of the tangent line to the curve at the point ( π2 , π2 ). 8. Implicit diﬀerentiation enables us a diﬀerent perspective from which to see why the d rule dx [a x ] a x ln(a) holds, if we assume that dx d [ln(x)] x1 . This exercise leads you through the key steps to do so. a. Let y a x . Rewrite this equation using the natural logarithm function to write x in terms of y (and the constant a). b. Diﬀerentiate both sides of the equation you found in (a) with respect to x, keeping in mind that y is implicitly a function of x. dy c. Solve the equation you found in (b) for dx , and then use the deﬁnition of y to dy write dx solely in terms of x. What have you found? 145 Chapter 2 Computing Derivatives 2.8 Using Derivatives to Evaluate Limits Motivating Questions • How can derivatives be used to help us evaluate indeterminate limits of the form 00 ? • What does it mean to say that limx→∞ f (x) L and limx→a f (x) ∞? ∞ • How can derivatives assist us in evaluating indeterminate limits of the form ∞? Because diﬀerential calculus is based on the deﬁnition of the derivative, and the deﬁnition of the derivative involves a limit, there is a sense in which all of calculus rests on limits. In addition, the limit involved in the deﬁnition of the derivative always generates the indeter- minate form 00 . If f is a diﬀerentiable function, then in the deﬁnition f (x + h) − f (x) f ′(x) lim , h→0 h not only does h → 0 in the denominator, but also ( f (x + h) − f (x)) → 0 in the numerator, since f is continuous. Remember, saying that a limit has an indeterminate form only means that we don’t yet know its value and have more work to do: indeed, limits of the form 00 can take on any value, as is evidenced by evaluating f ′(x) for varying values of x for a function such as f ′(x) x 2 . We have learned many techniques for evaluating the limits that result from the derivative deﬁnition, including a large number of shortcut rules. In this section, we turn the situa- tion upside-down: instead of using limits to evaluate derivatives, we explore how to use derivatives to evaluate certain limits. x 5 +x−2 Preview Activity 2.8.1. Let h be the function given by h(x) x 2 −1 . a. What is the domain of h? x5 + x − 2 b. Explain why lim results in an indeterminate form. x→1 x2 − 1 c. Next we will investigate the behavior of both the numerator and denominator of h near the point where x 1. Let f (x) x 5 + x − 2 and 1(x) x 2 − 1. Find the local linearizations of f and 1 at a 1, and call these functions L f (x) and L 1 (x), respectively. L f (x) d. Explain why h(x) ≈ L 1 (x) for x near a 1. e. Using your work from (c) and (d), evaluate L f (x) lim . x→1 L 1 (x) What do you think your result tells us about limx→1 h(x)? 146 2.8 Using Derivatives to Evaluate Limits f. Investigate the function h(x) graphically and numerically near x 1. What do you think is the value of limx→1 h(x)? 2.8.1 Using derivatives to evaluate indeterminate limits of the form 00 . f Lf Lf ≈ f a a Lg Lg ≈ g g Figure 2.8.1: At left, the graphs of f and 1 near the value a, along with their tangent line approximations L f and L 1 at x a. At right, zooming in on the point a and the four graphs. The idea demonstrated in Preview Activity 2.8.1 — that we can evaluate an indeterminate limit of the form 00 by replacing each of the numerator and denominator with their local lin- earizations at the point of interest — can be generalized in a way that enables us to evaluate f (x) a wide range of limits. We have a function h(x) that can be written as a quotient h(x) 1(x) , where f and 1 are both diﬀerentiable at x a and for which f (a) 1(a) 0. We would like to evaluate the indeterminate limit given by limx→a h(x). Figure 2.8.1 illustrates the sit- uation. We see that both f and 1 have an x-intercept at x a. Their respective tangent line approximations L f and L 1 at x a are also shown in the ﬁgure. We can take advantage of the fact that a function and its tangent line approximation become indistinguishable as x → a. First, let’s recall that L f (x) f ′(a)(x − a) + f (a) and L 1 (x) 1 ′(a)(x − a) + 1(a). Because x is getting arbitrarily close to a when we take the limit, we can replace f with L f and replace 1 with L 1 , and thus we observe that f (x) L f (x) lim lim x→a 1(x) x→a L 1 (x) 147 Chapter 2 Computing Derivatives f ′(a)(x − a) + f (a) lim . x→a 1 ′(a)(x − a) + 1(a) Next, we remember that both f (a) 0 and 1(a) 0, which is precisely what makes the original limit indeterminate. Substituting these values for f (a) and 1(a) in the limit above, we now have f (x) f ′(a)(x − a) lim lim ′ x→a 1(x) x→a 1 (a)(x − a) f ′(a) lim ′ , x→a 1 (a) where the latter equality holds because x−a x−a 1 when x is approaching (but not equal to) a. f ′ (a) Finally, we note that 1 ′ (a) is constant with respect to x, and thus f (x) f ′(a) lim ′ . x→a 1(x) 1 (a) This result holds as long as 1 ′(a) is not equal to zero. The formal name of the result is L’Hôpital’s Rule. L’Hôpital’s Rule. Let f and 1 be diﬀerentiable at x a, and suppose that f (a) 1(a) 0 and that f (x) f ′ (a) 1 ′(a) , 0. Then limx→a 1(x) 1′ (a) . In practice, we typically work with a slightly more general version of L’Hôpital’s Rule, which states that (under the identical assumptions as the boxed rule above and the extra assump- tion that 1 ′ is continuous at x a) f (x) f ′(x) lim lim ′ , x→a 1(x) x→a 1 (x) provided the righthand limit exists. This form reﬂects the basic idea of L’Hôpital’s Rule: if f (x) 1(x) produces an indeterminate limit of form 00 as x → a, that limit is equivalent to the limit f ′ (x) of the quotient of the two functions’ derivatives, 1 ′ (x) . For example, if we consider the limit from Preview Activity 2.8.1, x5 + x − 2 lim , x→1 x2 − 1 by L’Hôpital’s Rule we have that x5 + x − 2 5x 4 + 1 6 lim lim 3. x→1 x2 − 1 x→1 2x 2 By replacing the numerator and denominator with their respective derivatives, we often replace an indeterminate limit with one whose value we can easily determine. 148 2.8 Using Derivatives to Evaluate Limits Activity 2.8.2. Evaluate each of the following limits. If you use L’Hôpital’s Rule, in- dicate where it was used, and be certain its hypotheses are met before you apply it. ln(1+x) 2 ln(x) a. limx→0 x c. limx→1 1−e x−1 cos(x) sin(x)−x b. limx→π x d. limx→0 cos(2x)−1 While L’Hôpital’s Rule can be applied in an entirely algebraic way, it is important to re- member that the justiﬁcation of the rule is graphical: the main idea is that the slopes of the f (x) tangent lines to f and 1 at x a determine the value of the limit of 1(x) as x → a. f m = f ′ (a) m = f ′ (a) a m = g′ (a) a m = g′ (a) g Figure 2.8.2: Two functions f and 1 that satisfy L’Hôpital’s Rule. We see this in Figure 2.8.2, where we can see from the grid that f ′(a) 2 and 1 ′(a) −1, hence by L’Hôpital’s Rule, f (x) f ′(a) 2 lim ′ −2. x→a 1(x) 1 (a) −1 It’s not the fact that f and 1 both approach zero that matters most, but rather the rate at which each approaches zero that determines the value of the limit. This is a good way to f (x) remember what L’Hôpital’s Rule says: if f (a) 1(a) 0, the the limit of 1(x) as x → a is given by the ratio of the slopes of f and 1 at x a. Activity 2.8.3. In this activity, we reason graphically from the following ﬁgure to evaluate limits of ratios of functions about which some information is known. 149 Chapter 2 Computing Derivatives g p 2 2 2 f 1 1 1 q r 1 2 3 4 1 2 3 4 1 2 3 4 -1 -1 -1 s -2 -2 -2 Figure 2.8.3: Three graphs referenced in the questions of Activity 2.8.3. a. Use the left-hand graph to determine the values of f (2), f ′(2), 1(2), and 1 ′(2). f (x) Then, evaluate lim 1(x) . x→2 b. Use the middle graph to ﬁnd p(2), p ′(2), q(2), and q ′(2). Then, determine the p(x) value of lim q(x) . x→2 c. Assume that r and s are functions whose for which r ′′(2) , 0 and s ′′(2) , 0 Use the right-hand graph to compute r(2), r ′(2), s(2), s ′(2). Explain why you r(x) cannot determine the exact value of lim s(x) without further information being x→2 r(x) provided, but that you can determine the sign of lim . In addition, state what x→2 s(x) the sign of the limit will be, with justiﬁcation. 2.8.2 Limits involving ∞ The concept of inﬁnity, denoted ∞, arises naturally in calculus, as it does in much of math- ematics. It is important to note from the outset that ∞ is a concept, but not a number itself. Indeed, the notion of ∞ naturally invokes the idea of limits. Consider, for example, the function f (x) x1 , whose graph is pictured in Figure 2.8.4. We note that x 0 is not in the domain of f , so we may naturally wonder what happens as x → 0. As x → 0+ , we observe that f (x) increases without bound. That is, we can make the value of f (x) as large as we like by taking x closer and closer (but not equal) to 0, while keeping x > 0. This is a good way to think about what inﬁnity represents: a quantity is tending to inﬁnity if there is no single number that the quantity is always less than. Recall that the statement limx→a f (x) L, means that can make f (x) as close to L as we’d like by taking x suﬃciently close (but not equal) to a. We now expand this notation and language to include the possibility that either L or a can be ∞. For instance, for f (x) x1 , we now write 1 lim+ ∞, x→0 x 1 by which we mean that we can make x as large as we like by taking x suﬃciently close (but 150 2.8 Using Derivatives to Evaluate Limits 1 f (x) = x 1 1 Figure 2.8.4: The graph of f (x) x1 . not equal) to 0. In a similar way, we write 1 lim 0, x→∞ x since we can make x1 as close to 0 as we’d like by taking x suﬃciently large (i.e., by letting x increase without bound). In general, the notation limx→a f (x) ∞ means that we can make f (x) as large as we like by taking x suﬃciently close (but not equal) to a, and the notation limx→∞ f (x) L means that we can make f (x) as close to L as we like by taking x suﬃciently large. This notation also applies to left- and right-hand limits, and to limits involving −∞. For example, returning to Figure 2.8.4 and f (x) x1 , we can say that 1 1 lim −∞ and lim 0. x→0− x x→−∞ x Finally, we write lim f (x) ∞ x→∞ if we can make the value of f (x) as large as we’d like by taking x suﬃciently large. For example, lim x 2 ∞. x→∞ Limits involving inﬁnity identify vertical and horizontal asymptotes of a function. If lim f (x) x→a ∞, then x a is a vertical asymptote of f , while if limx→∞ f (x) L, then y L is a horizontal asymptote of f . Similar statements can be made using −∞, and with left- and right-hand limits as x → a − or x → a + . In precalculus classes, it is common to study the end behavior of certain families of functions, by which we mean the behavior of a function as x → ∞ and as x → −∞. Here we brieﬂy examine some familiar functions and note the values of several limits involving ∞. 151 Chapter 2 Computing Derivatives y = ex 8 64 y = sin(x) y = f (x) 1 4 -2 2 10 -4 4 8 -64 y = ln(x) -4 y = g(x) Figure 2.8.5: Graphs of some familiar functions whose end behavior as x → ±∞ is known. In the middle graph, f (x) x 3 − 16x and 1(x) x 4 − 16x 2 − 8. For the natural exponential function e x , we note that limx→∞ e x ∞ and limx→−∞ e x 0. For the exponential decay function e −x , these limits are reversed, with limx→∞ e −x 0 and limx→−∞ e −x ∞. Turning to the natural logarithm function, we have limx→0+ ln(x) −∞ and limx→∞ ln(x) ∞. While both e x and ln(x) grow without bound as x → ∞, the exponential function does so much more quickly than the logarithm function does. We’ll soon use limits to quantify what we mean by “quickly.” For polynomial functions of the form p(x) a n x n + a n−1 x n−1 + · · · a1 x + a0 , the end behavior depends on the sign of a n and whether the highest power n is even or odd. If n is even and a n is positive, then limx→∞ p(x) ∞ and limx→−∞ p(x) ∞, as in the plot of 1 in Figure 2.8.5. If instead a n is negative, then limx→∞ p(x) −∞ and limx→−∞ p(x) −∞. In the situation where n is odd, then either limx→∞ p(x) ∞ and limx→−∞ p(x) −∞ (which occurs when a n is positive, as in the graph of f in Figure 2.8.5), or limx→∞ p(x) −∞ and limx→−∞ p(x) ∞ (when a n is negative). A function can fail to have a limit as x → ∞. For example, consider the plot of the sine function at right in Figure 2.8.5. Because the function continues oscillating between −1 and 1 as x → ∞, we say that limx→∞ sin(x) does not exist. Finally, it is straightforward to analyze the behavior of any rational function as x → ∞. Example 2.8.6 Determine the limit of the function 3x 2 − 4x + 5 q(x) 7x 2 + 9x − 10 as x → ∞. Note that both (3x 2 − 4x + 5) → ∞ as x → ∞ and (7x 2 + 9x − 10) → ∞ as x → ∞. Here ∞ we say that limx→∞ q(x) has indeterminate form ∞ . We can determine the value of this limit through a standard algebraic approach. Multiplying the numerator and denominator each 152 2.8 Using Derivatives to Evaluate Limits 1 by x2 , we ﬁnd that 1 3x 2 − 4x + 5 x2 lim q(x) lim · x→∞ x→∞ 7x 2 + 9x − 10 1 x2 3 − 4 x1 + 5 x12 3 lim x→∞ 7+ 9 x1 − 10 x12 7 since x12 → 0 and x1 → 0 as x → ∞. This shows that the rational function q has a horizontal asymptote at y 73 . A similar approach can be used to determine the limit of any rational function as x → ∞. But how should we handle a limit such as x2 lim ? x→∞ ex Here, both x 2 → ∞ and e x → ∞, but there is not an obvious algebraic approach that enables us to ﬁnd the limit’s value. Fortunately, it turns out that L’Hôpital’s Rule extends to cases involving inﬁnity. L’Hôpital’s Rule (∞). If f and 1 are diﬀerentiable and both approach zero or both approach ±∞ as x → a (where a is allowed to be ∞) , then f (x) f ′(x) lim lim ′ , x→a 1(x) x→a 1 (x) provided the righthand limit exists. (To be technically correct, we need to add the additional hypothesis that 1 ′(x) , 0 on an open interval that contains a or in every neighborhood of inﬁnity if a is ∞; this is almost always met in practice.) 2 To evaluate limx→∞ xe x , we can apply L’Hôpital’s Rule, since both x 2 → ∞ and e x → ∞. Doing so, it follows that x2 2x lim x lim x . x→∞ e x→∞ e ∞ This updated limit is still indeterminate and of the form ∞ , but it is simpler since 2x has replaced x 2 . Hence, we can apply L’Hôpital’s Rule again, and ﬁnd that x2 2x 2 lim lim lim x . x→∞ e x x→∞ e x x→∞ e Now, since 2 is constant and e x → ∞ as x → ∞, it follows that 2 ex → 0 as x → ∞, which shows that x2 lim x 0. x→∞ e 153 Chapter 2 Computing Derivatives Activity 2.8.4. Evaluate each of the following limits. If you use L’Hôpital’s Rule, in- dicate where it was used, and be certain its hypotheses are met before you apply it. x tan(x) a. limx→∞ ln(x) d. limx→ π2 − x− π 2 e x +x b. limx→∞ 2e x +x 2 ln(x) c. limx→0+ 1 x e. limx→∞ xe −x f (x) To evaluate the limit of a quotient of two functions 1(x) that results in an indeterminate form ∞ of ∞ , in essence we are asking which function is growing faster without bound. We say that the function 1 dominates the function f as x → ∞ provided that f (x) lim 0, x→∞ 1(x) f (x) f (x) whereas f dominates 1 provided that limx→∞ 1(x) ∞. Finally, if the value of limx→∞ 1(x) is ﬁnite and nonzero, we say that f and 1 grow at the same rate. For example, we saw that 2 3x 2 −4x+5 limx→∞ xe x 0, so e x dominates x 2 , while limx→∞ 7x 2 +9x−10 7 , so f (x) 3x − 4x + 5 and 3 2 1(x) 7x 2 + 9x − 10 grow at the same rate. 2.8.3 Summary • Derivatives can be used to help us evaluate indeterminate limits of the form 00 through L’Hôpital’s Rule, by replacing the functions in the numerator and denominator with their tangent line approximations. In particular, if f (a) 1(a) 0 and f and 1 are diﬀerentiable at a, L’Hôpital’s Rule tells us that f (x) f ′(x) lim lim ′ . x→a 1(x) x→a 1 (x) • When we write x → ∞, this means that x is increasing without bound. Thus, lim f (x) x→∞ L means that we can make f (x) as close to L as we like by choosing x to be suﬃciently large. Similarly, limx→a f (x) ∞, means that we can make f (x) as large as we like by choosing x suﬃciently close to a. • A version of L’Hôpital’s Rule also helps us evaluate indeterminate limits of the form ∞ ∞ . If f and 1 are diﬀerentiable and both approach zero or both approach ±∞ as x → a (where a is allowed to be ∞), then f (x) f ′(x) lim lim ′ . x→a 1(x) x→a 1 (x) 154 2.8 Using Derivatives to Evaluate Limits 2.8.4 Exercises f (x) 1. L’Hôpital’s Rule with graphs. For the ﬁgures below, determine if lim is positive, x→a 1(x) negative, zero, or undeﬁned when f (x) is shown as the blue curve and 1(x) as the black curve. ln(x/4) 2. L’Hôpital’s Rule to evaluate a limit. Find the limit: lim . x→4 x 2 − 16 3. Determining if L’Hôpital’s Rule applies. Compute the following limits using l’Hôpital’s rule if appropriate. 1 − cos(7x) 4x − 3x − 1 lim lim x→0 1 − cos(3x) x→1 x2 − 1 4. Using L’Hôpital’s Rule multiple times. Evaluate the limit using L’Hopital’s rule. 15x 3 lim x→∞ e 2x 5. Let f and 1 be diﬀerentiable functions about which the following information is known: f (3) 1(3) 0, f ′(3) 1 ′(3) 0, f ′′(3) −2, and 1 ′′(3) 1. Let a new function h be f (x) given by the rule h(x) 1(x) . On the same set of axes, sketch possible graphs of f and 1 near x 3, and use the provided information to determine the value of lim h(x). x→3 Provide explanation to support your conclusion. 6. Find all vertical and horizontal asymptotes of the function 3(x − a)(x − b) R(x) , 5(x − a)(x − c) where a, b, and c are distinct, arbitrary constants. In addition, state all values of x for which R is not continuous. Sketch a possible graph of R, clearly labeling the values of 155 Chapter 2 Computing Derivatives a, b, and c. 7. Consider the function 1(x) x 2x , which is deﬁned for all x > 0. Observe that lim+ 1(x) x→0 is indeterminate due to its form of 00 . (Think about how we know that 0k 0 for all k > 0, while b 0 1 for all b , 0, but that neither rule can apply to 00 .) a. Let h(x) ln(1(x)). Explain why h(x) 2x ln(x). 2 ln(x) b. Next, explain why it is equivalent to write h(x) 1 . x c. Use L’Hôpital’s Rule and your work in (b) to compute limx→0+ h(x). d. Based on the value of limx→0+ h(x), determine limx→0+ 1(x). 8. Recall we say that function 1 dominates function f provided that limx→∞ f (x) ∞, f (x) limx→∞ 1(x) ∞, and limx→∞ 1(x) 0. √ a. Which function dominates the other: ln(x) or x? √ b. Which function dominates the other: ln(x) or n x? (n can be any positive integer) c. Explain why e x will dominate any polynomial function. d. Explain why x n will dominate ln(x) for any positive integer n. e. Give any example of two nonlinear functions such that neither dominates the other. 156 CHAPTER 3 Using Derivatives 3.1 Using derivatives to identify extreme values Motivating Questions • What are the critical numbers of a function f and how are they connected to identi- fying the most extreme values the function achieves? • How does the ﬁrst derivative of a function reveal important information about the behavior of the function, including the function’s extreme values? • How can the second derivative of a function be used to help identify extreme values of the function? In many diﬀerent settings, we are interested in knowing where a function achieves its least and greatest values. These can be important in applications — say to identify a point at which maximum proﬁt or minimum cost occurs — or in theory to characterize the behavior of a function or a family of related functions. Consider the simple and familiar example of a parabolic function such as s(t) −16t 2 +32t + 48 (shown at left in Figure 3.1.2) that represents the height of an object tossed vertically: its maximum value occurs at the vertex of the parabola and represents the greatest height the object reaches. This maximum value is an especially important point on the graph, the point at which the curve changes from increasing to decreasing. Deﬁnition 3.1.1 Given a function f , we say that f (c) is a global or absolute maximum of f provided that f (c) ≥ f (x) for all x in the domain of f , and similarly we call f (c) a global or absolute minimum of f whenever f (c) ≤ f (x) for all x in the domain of f . For instance, in Figure 3.1.2, 1 has a global maximum of 1(c), but 1 does not appear to have a global minimum, as the graph of 1 seems to decrease without bound. Note that the point (c, 1(c)) marks a fundamental change in the behavior of 1, where 1 changes from increasing to decreasing; similar things happen at both (a, 1(a)) and (b, 1(b)), although these points are not global minima or maxima. Deﬁnition 3.1.3 We say that f (c) is a local maximum or relative maximum of f provided that f (c) ≥ f (x) for all x near c, and f (c) is called a local or relative minimum of f whenever Chapter 3 Using Derivatives V y = s(t) (c, g(c)) 40 30 (a, g(a)) 20 (b, g(b)) 10 y = g(x) 1 2 Figure 3.1.2: At left, s(t) −16t 2 + 24t + 32 whose vertex is ( 34 , 41); at right, a function 1 that demonstrates several high and low points. f (c) ≤ f (x) for all x near c. For example, in Figure 3.1.2, 1 has a relative minimum of 1(b) at the point (b, 1(b)) and a relative maximum of 1(a) at (a, 1(a)). We have already identiﬁed the global maximum of 1 as 1(c); it can also be considered a relative maximum. Any maximum or minimum may also be called an extreme value of f . We would like to use calculus ideas to identify and classify key function behavior, including the location of relative extremes. Of course, if we are given a graph of a function, it is often straightforward to locate these important behaviors visually. Preview Activity 3.1.1. Consider the function h given by the graph in Figure 3.1.4. Use the graph to answer each of the following questions. a. Identify all of the values of c such that −3 < c < 3 for which h(c) is a local maximum of h. b. Identify all of the values of c such that −3 < c < 3 for which h(c) is a local minimum of h. c. Does h have a global maximum on the interval [−3, 3]? If so, what is the value of this global maximum? d. Does h have a global minimum on the interval [−3, 3]? If so, what is its value? e. Identify all values of c for which h ′(c) 0. f. Identify all values of c for which h ′(c) does not exist. g. True or false: every relative maximum and minimum of h occurs at a point where h ′(c) is either zero or does not exist. 158 3.1 Using derivatives to identify extreme values h. True or false: at every point where h ′(c) is zero or does not exist, h has a relative maximum or minimum. y = h(x) 2 1 -2 -1 1 2 -1 -2 Figure 3.1.4: The graph of a function h on the interval [−3, 3]. 3.1.1 Critical numbers and the ﬁrst derivative test If a continuous function has a relative maximum at c, then it is both necessary and suﬃcient that the function change from being increasing just before c to decreasing just after c. A continuous function has a relative minimum at c if and only if the function changes from decreasing to increasing at c. (See Figure 3.1.6.) There are only two possible ways for these changes in behavior to occur: either f ′(c) 0 or f ′(c) is undeﬁned. Because these values of c are so important, we call them critical numbers. Deﬁnition 3.1.5 We say that a function f has a critical number at x c provided that c is in the domain of f , and f ′(c) 0 or f ′(c) is undeﬁned. Critical numbers are the only possible locations where the function f may have relative extremes. Note that not every critical number produces a maximum or minimum; in the middle graph of Figure 3.1.6, the function pictured there has a horizontal tangent line at the noted point, but the function is increasing before and increasing after, so the critical number does not yield a maximum or minimum. When c is a critical number, we say that (c, f (c)) is a critical point of the function, or that f (c) is a critical value . The ﬁrst derivative test summarizes how sign changes in the ﬁrst derivative (which can only occur at critical numbers) indicate the presence of a local maximum or minimum for a given function. 159 Chapter 3 Using Derivatives Figure 3.1.6: From left to right, a function with a relative maximum where its derivative is zero; a function with a relative maximum where its derivative is undeﬁned; a function with neither a maximum nor a minimum at a point where its derivative is zero; a function with a relative minimum where its derivative is zero; and a function with a relative minimum where its derivative is undeﬁned. First Derivative Test. If p is a critical number of a continuous function f that is diﬀerentiable near p (except possibly at x p), then f has a relative maximum at p if and only¹ if f ′ changes sign from positive to negative at p, and f has a relative minimum at p if and only if f ′ changes sign from negative to positive at p. Example 3.1.7 Let f be a function whose derivative is given by the formula f ′(x) e −2x (3 − x)(x + 1)2 . Determine all critical numbers of f and decide whether a relative maximum, relative minimum, or neither occurs at each. Solution. Since we already have f ′(x) written in factored form, it is straightforward to ﬁnd the critical numbers of f . Because f ′(x) is deﬁned for all values of x, we need only determine where f ′(x) 0. From the equation e −2x (3 − x)(x + 1)2 0 and the zero product property, it follows that x 3 and x −1 are critical numbers of f . (There is no value of x that makes e −2x 0.) Next, to apply the ﬁrst derivative test, we’d like to know the sign of f ′(x) at inputs near the critical numbers. Because the critical numbers are the only locations at which f ′ can change sign, it follows that the sign of the derivative is the same on each of the intervals created by the critical numbers: for instance, the sign of f ′ must be the same for every x < −1. We create a ﬁrst derivative sign chart to summarize the sign of f ′ on the relevant intervals, along with the corresponding behavior of f . To produce the ﬁrst derivative sign chart in Figure 3.1.8 we identify the sign of each factor of f ′(x) at one selected point in each interval. For instance, for x < −1, we could determine the sign of e −2x , (3 − x), and (x + 1)2 at the value x −2. We note that both e −2x and (x + 1)2 are positive regardless of the value of x, while (3 − x) is also positive at x −2. Hence, each of the three terms in f ′ is positive, which we indicate by writing “+ + +.” Taking the product of three positive terms results in a positive value for f ′, which we denote by the “+” in the ¹Technically, we also have to assume that f is not piecewise constant on any intervals. This is because every point on a horizontal line is a relative maximum (and relative minimum) despite the fact that the derivative doesn’t change sign at any point along the horizontal line. 160 3.1 Using derivatives to identify extreme values interval to the left of x −1. And, since f ′ is positive on that interval, we know that f is increasing, so we write “INC” to represent the behavior of f . In a similar way, we ﬁnd that f ′ is positive and f is increasing on −1 < x < 3, and f ′ is negative and f is decreasing for x > 3. f ′ (x) = e−2x (3 − x)(x + 1)2 +++ +++ +−+ sign( f ′ ) + + − behav( f ) INC −1 INC 3 DEC Figure 3.1.8: The ﬁrst derivative sign chart for a function f whose derivative is given by the formula f ′(x) e −2x (3 − x)(x + 1)2 . Now we look for critical numbers at which f ′ changes sign. In this example, f ′ changes sign only at x 3, from positive to negative, so f has a relative maximum at x 3. Although f has a critical number at x −1, since f is increasing both before and after x −1, f has neither a minimum nor a maximum at x −1. Activity 3.1.2. Suppose that 1(x) is a function continuous for every value of x , 2 (x+4)(x−1)2 whose ﬁrst derivative is 1 ′(x) x−2 . Further, assume that it is known that 1 has a vertical asymptote at x 2. a. Determine all critical numbers of 1. b. By developing a carefully labeled ﬁrst derivative sign chart, decide whether 1 has as a local maximum, local minimum, or neither at each critical number. c. Does 1 have a global maximum? global minimum? Justify your claims. d. What is the value of limx→∞ 1 ′(x)? What does the value of this limit tell you about the long-term behavior of 1? e. Sketch a possible graph of y 1(x). 3.1.2 The second derivative test Recall that the second derivative of a function tells us several important things about the behavior of the function itself. For instance, if f ′′ is positive on an interval, then we know that f ′ is increasing on that interval and, consequently, that f is concave up, so throughout that interval the tangent line to y f (x) lies below the curve at every point. At a point where f ′(p) 0, the sign of the second derivative determines whether f has a local minimum or 161 Chapter 3 Using Derivatives local maximum at the critical number p. In Figure 3.1.9, we see the four possibilities for a function f that has a critical number p at which f ′(p) 0, provided f ′′(p) is not zero on an interval including p (except possibly at p). On either side of the critical number, f ′′ can be either positive or negative, and hence f can be either concave up or concave down. In the ﬁrst two graphs, f does not change concavity at p, and in those situations, f has either a local minimum or local maximum. In particular, if f ′(p) 0 and f ′′(p) < 0, then f is concave down at p with a horizontal tangent line, so f has a local maximum there. This fact, along with the corresponding statement for when f ′′(p) is positive, is the substance of the second derivative test. Figure 3.1.9: Four possible graphs of a function f with a horizontal tangent line at a critical point. Second Derivative Test. If p is a critical number of a continuous function f such that f ′(p) 0 and f ′′(p) , 0, then f has a relative maximum at p if and only if f ′′(p) < 0, and f has a relative minimum at p if and only if f ′′(p) > 0. In the event that f ′′(p) 0, the second derivative test is inconclusive. That is, the test doesn’t provide us any information. This is because if f ′′(p) 0, it is possible that f has a local minimum, local maximum, or neither.² Just as a ﬁrst derivative sign chart reveals all of the increasing and decreasing behavior of a function, we can construct a second derivative sign chart that demonstrates all of the im- portant information involving concavity. Example 3.1.10 Let f (x) be a function whose ﬁrst derivative is f ′(x) 3x 4 − 9x 2 . Construct both ﬁrst and second derivative sign charts for f , fully discuss where f is increasing and de- creasing and concave up and concave down, identify all relative extreme values, and sketch a possible graph of f . Solution. Since we know f ′(x) 3x 4 −9x 2 , we can ﬁnd the critical numbers of f by solving 3x 4 − 9x 2 0. Factoring, we observe that √ √ 0 3x 2 (x 2 − 3) 3x 2 (x + 3)(x − 3), √ so that x 0, ± 3 are the three critical numbers of f . The ﬁrst derivative sign chart for f is ²Consider the functions f (x) x 4 , 1(x) −x 4 , and h(x) x 3 at the critical point p 0. 162 3.1 Using derivatives to identify extreme values given in Figure 3.1.11. √ √ f ′ (x) = 3x2 (x + 3)(x − 3) +−− ++− ++− +++ sign( f ′ ) + − − + behav( f ) INC √ − 3 DEC 0 DEC √ 3 INC Figure 3.1.11: The ﬁrst derivative sign chart for f when f ′(x) 3x 4 − 9x 2 3x 2 (x 2 − 3). √ √ We see that f is increasing on the intervals (−∞, − 3) and ( 3, ∞), and f is decreasing on √ √ (− 3, 0) and (0, 3). By the ﬁrst derivative test, this information tells us that f has a local √ √ maximum at x − 3 and a local minimum at x 3. Although f also has a critical number at x 0, neither a maximum nor minimum occurs there since f ′ does not change sign at x 0. Next, we move on to investigate concavity. Diﬀerentiating f ′(x) 3x 4 − 9x 2 , we see that f ′′(x) 12x 3 − 18x. Since we are interested in knowing the intervals on which f ′′ is positive and negative, we ﬁrst ﬁnd where f ′′(x) 0. Observe that ( ) ( √ )( √ ) 3 3 3 0 12x 3 − 18x 12x x 2 − 12x x + x− . 2 2 2 √ This equation has solutions x 0, ± 32 . Building a sign chart for f ′′ in the exact same way we do for f ′, we see the result shown in Figure 3.1.12. q q f ′′ (x) = 12x x + 32 x − 32 −−− −+− ++− +++ sign( f ′′ ) − + − + behav( f ) CCD q CCU 0 CCD q 3 CCU − 32 2 Figure 3.1.12: The second derivative ( ) sign chart for f when f ′′(x) 12x 3 − 18x 12x 2 x 2 − 32 . 163 Chapter 3 Using Derivatives √ √ Therefore, f is concave down on the intervals (−∞, − 2) 3 and (0, 2 ), 3 and concave up on √ √ (− 2 , 0) 3 and ( 2 , ∞). 3 Putting all of this information together, we now see a complete and accurate possible graph of f in Figure 3.1.13. A B f C D E √ √ √ √ − 3 − 1.5 1.5 3 Figure 3.1.13: A possible graph of the function f in Example 3.1.10. √ √ The point A (− 3, f (− 3)) is a local maximum, because f is increasing prior to A and √ √ decreasing after; similarly, the point E ( 3, f ( 3) is a local minimum. Note, too, that f is concave down at A and concave up at B, which is consistent both with our second derivative sign chart and the second derivative test. At points B and D, concavity changes, as we saw in the results of the second derivative sign chart in Figure 3.1.12. Finally, at point C, f has a critical point with a horizontal tangent line, but neither a maximum nor a minimum occurs there, since f is decreasing both before and after C. It is also the case that concavity changes at C. While we completely understand where f is increasing and decreasing, where f is concave up and concave down, and where f has relative extremes, we do not know any speciﬁc information about the y-coordinates√of points on the curve. For instance, while we know that f has a local maximum √ at x − 3, we don’t know the value of that maximum because we do not know f (− 3). Any vertical translation of our sketch of f in Figure 3.1.13 would satisfy the given criteria for f . Points B, C, and D in Figure 3.1.13 are locations at which the concavity of f changes. We give a special name to any such point. Deﬁnition 3.1.14 If p is a value in the domain of a continuous function f at which f changes concavity, then we say that (p, f (p)) is an inﬂection point (or point of inﬂection) of f . Just as we look for locations where f changes from increasing to decreasing at points where 164 3.1 Using derivatives to identify extreme values f ′(p) 0 or f ′(p) is undeﬁned, so too we ﬁnd where f ′′(p) 0 or f ′′(p) is undeﬁned to see if there are points of inﬂection at these locations. At this point in our study, it is important to remind ourselves of the big picture that de- rivatives help to paint: the sign of the ﬁrst derivative f ′ tells us whether the function f is increasing or decreasing, while the sign of the second derivative f ′′ tells us how the function f is increasing or decreasing. Activity 3.1.3. Suppose that 1 is a function whose second derivative, 1 ′′, is given by the graph in Figure 3.1.15. g′′ 2 1 1 2 Figure 3.1.15: The graph of y 1 ′′(x). a. Find the x-coordinates of all points of inﬂection of 1. b. Fully describe the concavity of 1 by making an appropriate sign chart. c. Suppose you are given that 1 ′(−1.67857351) 0. Is there is a local maximum, local minimum, or neither (for the function 1) at this critical number of 1, or is it impossible to say? Why? d. Assuming that 1 ′′(x) is a polynomial (and that all important behavior of 1 ′′ is seen in the graph above), what degree polynomial do you think 1(x) is? Why? As we will see in more detail in the following section, derivatives also help us to understand families of functions that diﬀer only by changing one or more parameters. For instance, we might be interested in understanding the behavior of all functions of the form f (x) a(x − h)2 + k where a, h, and k are parameters. Each parameter has considerable impact on how the graph appears. 165 Chapter 3 Using Derivatives Activity 3.1.4. Consider the family of functions given by h(x) x 2 + cos(kx), where k is an arbitrary positive real number. a. Use a graphing utility to sketch the graph of h for several diﬀerent k-values, including k 1, 3, 5, 10. Plot h(x) x 2 + cos(3x) on the axes provided. What is the smallest value of k at which you think you can see (just by looking at the graph) at least one inﬂection point on the graph of h? 12 8 4 -2 2 Figure 3.1.16: Axes for plotting y h(x). √ b. Explain why the graph of h has no inﬂection points if k ≤ 2, but inﬁnitely √ many inﬂection points if k > 2. c. Explain why, no matter the value of k, h can only have ﬁnitely many critical numbers. 3.1.3 Summary • The critical numbers of a continuous function f are the values of p for which f ′(p) 0 or f ′(p) does not exist. These values are important because they identify horizontal tangent lines or corner points on the graph, which are the only possible locations at which a local maximum or local minimum can occur. • Given a diﬀerentiable function f , whenever f ′ is positive, f is increasing; whenever f ′ is negative, f is decreasing. The ﬁrst derivative test tells us that at any point where f changes from increasing to decreasing, f has a local maximum, while conversely at any point where f changes from decreasing to increasing f has a local minimum. • Given a twice diﬀerentiable function f , if we have a horizontal tangent line at x p and f ′′(p) is nonzero, the sign of f ′′ tells us the concavity of f and hence whether f has a maximum or minimum at x p. In particular, if f ′(p) 0 and f ′′(p) < 0, then f is concave down at p and f has a local maximum there, while if f ′(p) 0 and f ′′(p) > 0, 166 3.1 Using derivatives to identify extreme values then f has a local minimum at p. If f ′(p) 0 and f ′′(p) 0, then the second derivative does not tell us whether f has a local extreme at p or not. 3.1.4 Exercises 1. Finding critical points and inﬂection points. Use a graph below of f (x) ln(2x 2 + 1) to estimate the x-values of any critical points and inﬂection points of f (x). Next, use derivatives to ﬁnd the x-values of any critical points and inﬂection points exactly. 2. Finding inﬂection points. Find the inﬂection points of f (x) 4x 4 + 55x 3 − 21x 2 + 3. 3. Matching graphs of f , f ′ , f ′′. The following shows graphs of three functions, A (in black), B (in blue), and C (in green). If these are the graphs of three functions f , f ′, and f ′′, identify which is which. 4. This problem concerns a function about which the following information is known: • f is a diﬀerentiable function deﬁned at every real number x • f (0) −1/2 • y f ′(x) has its graph given at center in Figure 3.1.17 167 Chapter 3 Using Derivatives f′ 2 x x x 1 1 1 Figure 3.1.17: At center, a graph of y f ′(x); at left, axes for plotting y f (x); at right, axes for plotting y f ′′(x). a. Construct a ﬁrst derivative sign chart for f . Clearly identify all critical numbers of f , where f is increasing and decreasing, and where f has local extrema. b. On the right-hand axes, sketch an approximate graph of y f ′′(x). c. Construct a second derivative sign chart for f . Clearly identify where f is concave up and concave down, as well as all inﬂection points. d. On the left-hand axes, sketch a possible graph of y f (x). 5. Suppose that 1 is a diﬀerentiable function and 1 ′(2) 0. In addition, suppose that on 1 < x < 2 and 2 < x < 3 it is known that 1 ′(x) is positive. a. Does 1 have a local maximum, local minimum, or neither at x 2? Why? b. Suppose that 1 ′′(x) exists for every x such that 1 < x < 3. Reasoning graphically, describe the behavior of 1 ′′(x) for x-values near 2. c. Besides being a critical number of 1, what is special about the value x 2 in terms of the behavior of the graph of 1? 168 3.1 Using derivatives to identify extreme values 6. Suppose that h is a diﬀerentiable function whose ﬁrst derivative is given by the graph in Figure 3.1.18. a. How many real number solutions can the equation h(x) 0 have? h′ Why? b. If h(x) 0 has two distinct real so- lutions, what can you say about the signs of the two solutions? Why? c. Assume that limx→∞ h ′(x) 3, as appears to be indicated in Fig- ure 3.1.18. How will the graph of y h(x) appear as x → ∞? Why? d. Describe the concavity of y h(x) as fully as you can from the pro- vided information. Figure 3.1.18: The graph of y h ′(x). 7. Let p be a function whose second derivative is p ′′(x) (x + 1)(x − 2)e −x . a. Construct a second derivative sign chart for p and determine all inﬂection points of p. √ 5−1 b. Suppose you also know that x 2 is a critical number of p. Does p have a √ 5−1 local minimum, local maximum, or neither at x 2 ? Why? c. If the point (2, 12 e2 ) lies on the graph of y p(x) and p ′(2) − e52 , ﬁnd the equation of the tangent line to y p(x) at the point where x 2. Does the tangent line lie above the curve, below the curve, or neither at this value? Why? 169 Chapter 3 Using Derivatives 3.2 Using derivatives to describe families of functions Motivating Questions • Given a family of functions that depends on one or more parameters, how does the shape of the graph of a typical function in the family depend on the value of the parameters? • How can we construct ﬁrst and second derivative sign charts of functions that depend on one or more parameters while allowing those parameters to remain arbitrary con- stants? Mathematicians are often interested in making general observations, say by describing pat- terns that hold in a large number of cases. Think about the Pythagorean Theorem: it doesn’t tell us something about a single right triangle, but rather a fact about every right triangle. In the next part of our studies, we use calculus to make general observations about families of functions that depend on one or more parameters. People who use applied mathematics, such as engineers and economists, often encounter the same types of functions where only small changes to certain constants occur. These constants are called parameters. d +a d f (t) = a sin(b(t − c)) + d c c + 2π b Figure 3.2.1: The graph of f (t) a sin(b(t − c)) + d based on parameters a, b, c, and d. You are already familiar with certain families of functions. For example, f (t) a sin(b(t − c)) + d is a stretched and shifted version of the sine function with amplitude a, period 2π b , phase shift c, and vertical shift d. We know that a aﬀects the size of the oscillation, b the rapidity of oscillation, and c where the oscillation starts, as shown in Figure 3.2.1, while d aﬀects the vertical positioning of the graph. As another example, every function of the form y mx + b is a line with slope m and y- intercept (0, b). The value of m aﬀects the line’s steepness, and the value of b situates the line 170 3.2 Using derivatives to describe families of functions vertically on the coordinate axes. These two parameters describe all possible non-vertical lines. For other less familiar families of functions, we can use calculus to discover where key behav- ior occurs: where members of the family are increasing or decreasing, concave up or concave down, where relative extremes occur, and more, all in terms of the parameters involved. To get started, we revisit a common collection of functions to see how calculus conﬁrms things we already know. Preview Activity 3.2.1. Let a, h, and k be arbitrary real numbers with a , 0, and let f be the function given by the rule f (x) a(x − h)2 + k. a. What familiar type of function is f ? What information do you know about f just by looking at its form? (Think about the roles of a, h, and k.) b. Next we use some calculus to develop familiar ideas from a diﬀerent perspec- tive. To start, treat a, h, and k as constants and compute f ′(x). c. Find all critical numbers of f . (These will depend on at least one of a, h, and k.) d. Assume that a < 0. Construct a ﬁrst derivative sign chart for f . e. Based on the information you’ve found above, classify the critical values of f as maxima or minima. 3.2.1 Describing families of functions in terms of parameters Our goal is to describe the key characteristics of the overall behavior of each member of a family of functions in terms of its parameters. By ﬁnding the ﬁrst and second derivatives and constructing sign charts (each of which may depend on one or more of the parameters), we can often make broad conclusions about how each member of the family will appear. Example 3.2.2 Consider the two-parameter family of functions given by 1(x) axe −bx , where a and b are positive real numbers. Fully describe the behavior of a typical mem- ber of the family in terms of a and b, including the location of all critical numbers, where 1 is increasing, decreasing, concave up, and concave down, and the long term behavior of 1. Solution. We begin by computing 1 ′(x). By the product rule, d [ −bx ] d 1 ′(x) ax e + e −bx [ax]. dx dx By applying the chain rule and constant multiple rule, we ﬁnd that 1 ′(x) axe −bx (−b) + e −bx (a). To ﬁnd the critical numbers of 1, we solve the equation 1 ′(x) 0. By factoring 1 ′(x), we ﬁnd 0 ae −bx (−bx + 1). 171 Chapter 3 Using Derivatives Since we are given that a , 0 and we know that e −bx , 0 for all values of x, the only way this equation can hold is when −bx + 1 0. Solving for x, we ﬁnd x 1b , and this is therefore the only critical number of 1. We construct the ﬁrst derivative sign chart for 1 that is shown in Figure 3.2.3. g′ (x) = ae−bx (1 − bx) ++ +− sign(g′ ) + − behav(g) INC DEC 1 b Figure 3.2.3: The ﬁrst derivative sign chart for 1(x) axe −bx . Because the factor ae −bx is always positive, the sign of 1 ′ depends on the linear factor (1−bx), which is positive for x < 1b and negative for x > 1b . Hence we can not only conclude that 1 is always increasing for x < 1b and decreasing for x > 1b , but also that 1 has a global maximum at ( 1b , 1( 1b )) and no local minimum. We turn next to analyzing the concavity of 1. With 1 ′(x) −abxe −bx + ae −bx , we diﬀerentiate to ﬁnd that 1 ′′(x) −abxe −bx (−b) + e −bx (−ab) + ae −bx (−b). Combining like terms and factoring, we now have 1 ′′(x) ab 2 xe −bx − 2abe −bx abe −bx (bx − 2). We observe that abe −bx is always positive, and thus the sign of 1 ′′ depends on the sign of (bx − 2), which is zero when x 2b . Since b is positive, the value of (bx − 2) is negative for x < 2b and positive for x > 2b . The sign chart for 1 ′′ is shown in Figure 3.2.4. Thus, 1 is concave down for all x < 2b and concave up for all x > 2b . Finally, we analyze the long term behavior of 1 by considering two limits. First, we note that ax lim 1(x) lim axe −bx lim . x→∞ x→∞ x→∞ e bx This limit has indeterminate form ∞ ∞ , so we apply L’Hôpital’s Rule and ﬁnd that lim 1(x) 0. x→∞ In the other direction, lim 1(x) lim axe −bx −∞, x→−∞ x→−∞ because ax → −∞ and e −bx → ∞ as x → −∞. Hence, as we move left on its graph, 1 decreases without bound, while as we move to the right, 1(x) → 0. 172 3.2 Using derivatives to describe families of functions g′′ (x) = abe−bx (bx − 2) +− ++ sign(g′′ ) − + behav(g) CCD CCU 2 b Figure 3.2.4: The second derivative sign chart for 1(x) axe −bx . All of this information now helps us produce the graph of a typical member of this family of functions without using a graphing utility (and without choosing particular values for a and b), as shown in Figure 3.2.5. global max a −1 be inflection pt g(x) = axe−bx 1 2 b b Figure 3.2.5: The graph of 1(x) axe −bx . Note that the value of b controls the horizontal location of the global maximum and the inﬂection point, as neither depends on a. The value of a aﬀects the vertical stretch of the graph. For example, the global maximum occurs at the point ( 1b , 1( 1b )) ( 1b , ba e −1 ), so the larger the value of a, the greater the value of the global maximum. The work we’ve completed in Example 3.2.2 can often be replicated for other families of functions that depend on parameters. Normally we are most interested in determining all critical numbers, a ﬁrst derivative sign chart, a second derivative sign chart, and the limit of the function as x → ∞. Throughout, we prefer to work with the parameters as arbitrary constants. In addition, we can experiment with some particular values of the parameters present to reduce the algebraic complexity of our work. The following activities oﬀer several key examples where we see that the values of the parameters substantially aﬀect the behavior of individual functions within a given family. 173 Chapter 3 Using Derivatives Activity 3.2.2. Consider the family of functions deﬁned by p(x) x 3 − ax, where a , 0 is an arbitrary constant. a. Find p ′(x) and determine the critical numbers of p. How many critical numbers does p have? b. Construct a ﬁrst derivative sign chart for p. What can you say about the over- all behavior of p if the constant a is positive? Why? What if the constant a is negative? In each case, describe the relative extremes of p. c. Find p ′′(x) and construct a second derivative sign chart for p. What does this tell you about the concavity of p? What role does a play in determining the concavity of p? d. Without using a graphing utility, sketch and label typical graphs of p(x) for the cases where a > 0 and a < 0. Label all inﬂection points and local extrema. e. Finally, use a graphing utility to test your observations above by entering and plotting the function p(x) x 3 − ax for at least four diﬀerent values of a. Write several sentences to describe your overall conclusions about how the behavior of p depends on a. Activity 3.2.3. Consider the two-parameter family of functions of the form h(x) a(1 − e −bx ), where a and b are positive real numbers. a. Find the ﬁrst derivative and the critical numbers of h. Use these to construct a ﬁrst derivative sign chart and determine for which values of x the function h is increasing and decreasing. b. Find the second derivative and build a second derivative sign chart. For which values of x is a function in this family concave up? concave down? c. What is the value of limx→∞ a(1 − e −bx )? limx→−∞ a(1 − e −bx )? d. How does changing the value of b aﬀect the shape of the curve? e. Without using a graphing utility, sketch the graph of a typical member of this family. Write several sentences to describe the overall behavior of a typical func- tion h and how this behavior depends on a and b. Activity 3.2.4. Let L(t) A 1+ce −kt , where A, c, and k are all positive real numbers. a. Observe that we can equivalently write L(t) A(1 + ce −kt )−1 . Find L′(t) and explain why L has no critical numbers. Is L always increasing or always de- creasing? Why? b. Given the fact that ce −kt − 1 L′′(t) Ack 2 e −kt , (1 + ce −kt )3 174 3.2 Using derivatives to describe families of functions ﬁnd all values of t such that L′′(t) 0 and hence construct a second derivative sign chart. For which values of t is a function in this family concave up? concave down? A A c. What is the value of limt→∞ 1+ce −kt ? limt→−∞ 1+ce −kt ? d. Find the value of L(x) at the inﬂection point found in (b). e. Without using a graphing utility, sketch the graph of a typical member of this family. Write several sentences to describe the overall behavior of a typical func- tion L and how this behavior depends on A, c, and k number. f. Explain why it is reasonable to think that the function L(t) models the growth of a population over time in a setting where the largest possible population the surrounding environment can support is A. 3.2.2 Summary • Given a family of functions that depends on one or more parameters, by investigating how critical numbers and locations where the second derivative is zero depend on the values of these parameters, we can often accurately describe the shape of the function in terms of the parameters. • In particular, just as we can created ﬁrst and second derivative sign charts for a single function, we often can do so for entire families of functions where critical numbers and possible inﬂection points depend on arbitrary constants. These sign charts then reveal where members of the family are increasing or decreasing, concave up or con- cave down, and help us to identify relative extremes and inﬂection points. 3.2.3 Exercises 1. Drug dosage with a parameter. For some positive constant C, a patient’s tempera- ( ) ture change, T, due to a dose, D, of a drug is given by T C2 − D3 D 2 . What dosage maximizes the temperature change? The sensitivity of the body to the drug is deﬁned as dT/dD. What dosage maximizes sensitivity? 2. Using the graph of 1 ′. The ﬁgure below gives the behavior of the derivative of 1(x) on −2 ≤ x ≤ 2. Sketch a graph of 1(x) and use your sketch to answer the following questions. A. Where does the graph of 1(x) have inﬂection points? B. Where are the global maxima and minima of 1 on [−2, 2]? C. If 1(−2) −8, what are possible values for 1(0)? How is the value of 1(2) related to the value of 1(0)? 175 Chapter 3 Using Derivatives Graph of 1 ′(x) (not 1(x)) 3. Consider the one-parameter family of functions given by p(x) x 3 − ax 2 , where a > 0. a. Sketch a plot of a typical member of the family, using the fact that each is a cubic polynomial with a repeated zero at x 0 and another zero at x a. b. Find all critical numbers of p. c. Compute p ′′ and ﬁnd all values for which p ′′(x) 0. Hence construct a second derivative sign chart for p. d. Describe how the location of the critical numbers and the inﬂection point of p change as a changes. That is, if the value of a is increased, what happens to the critical numbers and inﬂection point? e −x 4. Let q(x) x−c be a one-parameter family of functions where c > 0. a. Explain why q has a vertical asymptote at x c. b. Determine limx→∞ q(x) and limx→−∞ q(x). c. Compute q ′(x) and ﬁnd all critical numbers of q. d. Construct a ﬁrst derivative sign chart for q and determine whether each critical number leads to a local minimum, local maximum, or neither for the function q. e. Sketch a typical member of this family of functions with important behaviors clearly labeled. (x−m)2 − 5. Let E(x) e 2s 2 , where m is any real number and s is a positive real number. a. Compute E′(x) and hence ﬁnd all critical numbers of E. b. Construct a ﬁrst derivative sign chart for E and classify each critical number of the function as a local minimum, local maximum, or neither. c. It can be shown that E′′(x) is given by the formula ( ) ′′ − (x−m)2 (x − m)2 − s 2 E (x) e 2s 2 . s4 Find all values of x for which E′′(x) 0. 176 3.2 Using derivatives to describe families of functions d. Determine limx→∞ E(x) and limx→−∞ E(x). e. Construct a labeled graph of a typical function E that clearly shows how impor- tant points on the graph of y E(x) depend on m and s. 177 Chapter 3 Using Derivatives 3.3 Global Optimization Motivating Questions • What are the diﬀerences between ﬁnding relative extreme values and global extreme values of a function? • How is the process of ﬁnding the global maximum or minimum of a function over the function’s entire domain diﬀerent from determining the global maximum or min- imum on a restricted domain? • For a function that is guaranteed to have both a global maximum and global min- imum on a closed, bounded interval, what are the possible points at which these extreme values occur? We have seen that we can use the ﬁrst derivative of a function to determine where the func- tion is increasing or decreasing, and the second derivative to know where the function is concave up or concave down. This information helps us determine the overall shape and behavior of the graph, as well as whether the function has relative extrema. Remember the diﬀerence between a relative maximum and a global maximum: there is a relative maximum of f at x p if f (p) ≥ f (x) for all x near p, while there is a global maximum at p if f (p) ≥ f (x) for all x in the domain of f . For instance, in Figure 3.3.1, we see a func- tion f that has a global maximum at x c and a relative maximum at x a, since f (c) global max is greater than f (x) for every value of x, while f (a) is only greater than the value of f (x) for x near a. Since the function appears to de- relative max crease without bound, f has no global min- imum, though clearly f has a relative mini- relative min mum at x b. f Our emphasis in this section is on ﬁnding the global extreme values of a function (if they ex- a b c ist), either over its entire domain or on some restricted portion. Figure 3.3.1: A function f with a global maximum, but no global minimum. 178 3.3 Global Optimization Preview Activity 3.3.1. Let f (x) 2 + 3 1+(x+1)2 . a. Determine all of the critical numbers of f . b. Construct a ﬁrst derivative sign chart for f and thus determine all intervals on which f is increasing or decreasing. c. Does f have a global maximum? If so, why, and what is its value and where is the maximum attained? If not, explain why. d. Determine limx→∞ f (x) and limx→−∞ f (x). e. Explain why f (x) > 2 for every value of x. f. Does f have a global minimum? If so, why, and what is its value and where is the minimum attained? If not, explain why. 3.3.1 Global Optimization In Figure 3.3.1 and Preview Activity 3.3.1, we were interested in ﬁnding the global minimum and global maximum for f on its entire domain. At other times, we might focus on some restriction of the domain. For example, rather than considering f (x) 2+ 1+(x+1) 3 2 for every value of x, perhaps instead we are only interested in those x for which 0 ≤ x ≤ 4, and we would like to know which values of x in the interval [0, 4] produce the largest possible and smallest possible values of f . We are accustomed to critical numbers playing a key role in determining the location of extreme values of a function; now, by restricting the domain to an interval, it makes sense that the endpoints of the interval will also be important to consider, as we see in the following activity. When limiting ourselves to a particular interval, we will often refer to the absolute maximum or minimum value, rather than the global maximum or minimum. Activity 3.3.2. Let 1(x) 13 x 3 − 2x + 2. a. Find all critical numbers of 1 that lie in the interval −2 ≤ x ≤ 3. b. Use a graphing utility to construct the graph of 1 on the interval −2 ≤ x ≤ 3. c. From the graph, determine the x-values at which the absolute minimum and absolute maximum of 1 occur on the interval [−2, 3]. d. How do your answers change if we instead consider the interval −2 ≤ x ≤ 2? e. What if we instead consider the interval −2 ≤ x ≤ 1? In Activity 3.3.2, we saw how the absolute maximum and absolute minimum of a function on a closed, bounded interval [a, b], depend not only on the critical numbers of the function, but also on the values of a and b. These observations demonstrate several important facts that hold more generally. First, we state an important result called the Extreme Value Theorem. 179 Chapter 3 Using Derivatives The Extreme Value Theorem. If f is a continuous function on a closed interval [a, b], then f attains both an absolute minimum and absolute maximum on [a, b]. That is, for some value x m such that a ≤ x m ≤ b, it follows that f (x m ) ≤ f (x) for all x in [a, b]. Similarly, there is a value x M in [a, b] such that f (x M ) ≥ f (x) for all x in [a, b]. Letting m f (x m ) and M f (x M ), it follows that m ≤ f (x) ≤ M for all x in [a, b]. The Extreme Value Theorem tells us that on any closed interval [a, b], a continuous function has to achieve both an absolute minimum and an absolute maximum. The theorem does not tell us where these extreme values occur, but rather only that they must exist. As we saw in Activity 3.3.2, the only possible locations for relative extremes are at the endpoints of the interval or at a critical number. Note 3.3.2 Thus, we have the following approach to ﬁnding the absolute maximum and minimum of a continuous function f on the interval [a, b]: • ﬁnd all critical numbers of f that lie in the interval; • evaluate the function f at each critical number in the interval and at each endpoint of the interval; • from among those function values, the smallest is the absolute minimum of f on the interval, while the largest is the absolute maximum. Activity 3.3.3. Find the exact absolute maximum and minimum of each function on the stated interval. a. h(x) xe −x , [0, 3] b. p(t) sin(t) + cos(t), [− π2 , π2 ] x2 c. q(x) x−2 , [3, 7] d. f (x) 4 − e −(x−2) , (−∞, ∞) 2 e. h(x) xe −ax , [0, 2a ] (a > 0) f. f (x) b − e −(x−a) , (−∞, ∞), a, b > 0 2 The interval we choose has nearly the same inﬂuence on extreme values as the function un- der consideration. Consider, for instance, the function pictured in Figure 3.3.3. In sequence, from left to right, the interval under consideration is changed from [−2, 3] to [−2, 2] to [−2, 1]. • On the interval [−2, 3], there are two critical numbers, with the absolute minimum at one critical number and the absolute maximum at the right endpoint. • On the interval [−2, 2], both critical numbers are in the interval, with the absolute minimum and maximum at the two critical numbers. • On the interval [−2, 1], just one critical number lies in the interval, with the absolute maximum at one critical number and the absolute minimum at one endpoint. 180 3.3 Global Optimization g g g 2 2 2 -2 3 -2 2 -2 1 Figure 3.3.3: A function 1 considered on three diﬀerent intervals. Remember to consider only the critical numbers that lie within the interval. 3.3.2 Moving toward applications We conclude this section with an example of an applied optimization problem. It highlights the role that a closed, bounded domain can play in ﬁnding absolute extrema. Example 3.3.4 A 20 cm piece of wire is cut into two pieces. One piece is used to form a square and the other to form an equilateral triangle. How should the wire be cut to maximize the total area enclosed by the square and triangle? to minimize the area? Solution. We begin by sketching a picture that illustrates the situation. The variable in the problem is where we decide to cut the wire. We thus label the cut point at a distance x from one end of the wire, and note that the remaining portion of the wire then has length 20 − x As shown in Figure 3.3.5, we see that the x cm of wire that is used to form the equilateral triangle with three sides of length x3 . For the remaining 20 − x cm of wire, the square that results will have each side of length 20−x 4 . x 20 − x x 3 20−x 4 Figure 3.3.5: A 20 cm piece of wire cut into two pieces, one of which forms an equilateral triangle, the other which yields a square. At this point, we note that there are obvious restrictions on x: in particular, 0 ≤ x ≤ 20. In 181 Chapter 3 Using Derivatives the extreme cases, all of the wire is being used to make just one ﬁgure. For instance, if x 0, then all 20 cm of wire are used to make a square that is 5 × 5. Now, our overall goal is to ﬁnd the minimum and maximum areas that can be enclosed. √ Because the height of an equilateral triangle is 3 times half the length of the base, the area of the triangle is √ 1 1 x x 3 A∆ bh · · . 2 2 3 6 ( 20−x ) 2 The area of the square is A□ s 2 4 . Therefore, the total area function is √ ( )2 3x 2 20 − x A(x) + . 36 4 Remember that we are considering this function only on the restricted domain [0, 20]. Diﬀerentiating A(x), we have √ ( )( ) √ ′ 3x 20 − x 1 3 1 5 A (x) +2 − x+ x− . 18 4 4 18 8 2 When we set A′(x) 0, we ﬁnd that x 180 √ ≈ 11.3007 is the only critical number of A in 4 3+9 the interval [0, 20]. Evaluating A at the critical number and endpoints, we see that ( ) √ ( )2 3( 180 2 √ ) 20− 180 √ • A 180 √ 4 3+9 4 + 4 3+9 4 ≈ 10.8741 4 3+9 • A(0) 25 √ 3 √ • A(20) 36 (400) 100 9 3 ≈ 19.2450 Thus, the absolute minimum occurs when x ≈ 11.3007 and results in the minimum area of approximately 10.8741 square centimeters. The absolute maximum occurs when we invest all of the wire in the square (and none in the triangle), resulting in 25 square centimeters of area. These results are conﬁrmed by a plot of y A(x) on the interval [0, 20], as shown in Figure 3.3.6. 182 3.3 Global Optimization 25 20 15 y = A(x) 10 5 5 10 15 20 Figure 3.3.6: A plot of the area function from Example 3.3.4. Activity 3.3.4. A piece of cardboard that is 10 × 15 (each measured in inches) is being made into a box without a top. To do so, squares are cut from each corner of the box and the remaining sides are folded up. If the box needs to be at least 1 inch deep and no more than 3 inches deep, what is the maximum possible volume of the box? what is the minimum volume? Justify your answers using calculus. a. Draw a labeled diagram that shows the given information. What variable should we introduce to represent the choice we make in creating the box? Label the di- agram appropriately with the variable, and write a sentence to state what the variable represents. b. Determine a formula for the function V (that depends on the variable in (a)) that tells us the volume of the box. c. What is the domain of the function V? That is, what values of x make sense for input? Are there additional restrictions provided in the problem? d. Determine all critical numbers of the function V. e. Evaluate V at each of the endpoints of the domain and at any critical numbers that lie in the domain. f. What is the maximum possible volume of the box? the minimum? Example 3.3.4 and Activity 3.3.4 illustrate standard steps that we undertake in almost every applied optimization problem: we draw a picture to demonstrate the situation, introduce one or more variables to represent quantities that are changing, ﬁnd a function that models the quantity to be optimized, and then decide on an appropriate domain for that function. Once that is done, we are in the familiar situation of ﬁnding the absolute minimum and maximum of a function over a particular domain, so we apply the calculus ideas that we have been studying to this point in Chapter 3. 183 Chapter 3 Using Derivatives 3.3.3 Summary • To ﬁnd relative extreme values of a function, we use a ﬁrst derivative sign chart and classify all of the function’s critical numbers. If instead we are interested in absolute extreme values, we ﬁrst decide whether we are considering the entire domain of the function or a particular interval. • In the case of ﬁnding global extremes over the function’s entire domain, we again use a ﬁrst or second derivative sign chart. If we are working to ﬁnd absolute extremes on a restricted interval, then we ﬁrst identify all critical numbers of the function that lie in the interval. • For a continuous function on a closed, bounded interval, the only possible points at which absolute extreme values occur are the critical numbers and the endpoints. Thus, we simply evaluate the function at each endpoint and each critical number in the in- terval, and compare the results to decide which is largest (the absolute maximum) and which is smallest (the absolute minimum). 3.3.4 Exercises 1. Based on the given information about each function, decide whether the function has global maximum, a global minimum, neither, both, or that it is not possible to say with- out more information. Assume that each function is twice diﬀerentiable and deﬁned for all real numbers, unless noted otherwise. In each case, write one sentence to explain your conclusion. a. f is a function such that f ′′(x) < 0 for every x. b. 1 is a function with two critical numbers a and b (where a < b), and 1 ′(x) < 0 for x < a, 1 ′(x) < 0 for a < x < b, and 1 ′(x) > 0 for x > b. c. h is a function with two critical numbers a and b (where a < b), and h ′(x) < 0 for x < a, h ′(x) > 0 for a < x < b, and h ′(x) < 0 for x > b. In addition, limx→∞ h(x) 0 and limx→−∞ h(x) 0. d. p is a function diﬀerentiable everywhere except at x a and p ′′(x) > 0 for x < a and p ′′(x) < 0 for x > a. 2. For each family of functions that depends on one or more parameters, determine the function’s absolute maximum and absolute minimum on the given interval. a. p(x) x 3 − a 2 x, [0, a] (a > 0) b. r(x) axe −bx , [ 2b , b ] (a > 0, b > 1) 1 2 c. w(x) a(1 − e −bx ), [b, 3b] (a, b > 0) [π ] d. s(x) sin(kx), 3k , 6k 5π (k > 0) 184 3.3 Global Optimization 3. For each of the functions described below (each continuous on [a, b]), state the loca- tion of the function’s absolute maximum and absolute minimum on the interval [a, b], or say there is not enough information provided to make a conclusion. Assume that any critical numbers mentioned in the problem statement represent all of the critical numbers the function has in [a, b]. In each case, write one sentence to explain your answer. a. f ′(x) ≤ 0 for all x in [a, b] b. 1 has a critical number at c such that a < c < b and 1 ′(x) > 0 for x < c and 1 ′(x) < 0 for x > c c. h(a) h(b) and h ′′(x) < 0 for all x in [a, b] d. p(a) > 0, p(b) < 0, and for the critical number c such that a < c < b, p ′(x) < 0 for x < c and p ′(x) > 0 for x > c 4. Let s(t) 3 sin(2(t − π6 )) + 5. Find the exact absolute maximum and minimum of s on the provided intervals by testing the endpoints and ﬁnding and evaluating all relevant critical numbers of s. a. [ π6 , 7π 6 ] c. [0, 2π] b. [0, π2 ] d. [ π3 , 6 ] 5π 185 Chapter 3 Using Derivatives 3.4 Applied Optimization Motivating Questions • In a setting where a situation is described for which optimal parameters are sought, how do we develop a function that models the situation and use calculus to ﬁnd the desired maximum or minimum? Near the conclusion of Section 3.3, we considered two optimization problems where deter- mining the function to be optimized was part of the problem. In Example 3.3.4, we sought to use a single piece of wire to build an equilateral triangle and square in order to maximize the total combined area enclosed. In the subsequent Activity 3.3.4, we investigated how the volume of a box constructed from a piece of cardboard by removing squares from each corner and folding up the sides depends on the size of the squares removed. In neither of these problems was a function to optimize explicitly provided. Rather, we ﬁrst tried to understand the problem by drawing a ﬁgure and introducing variables, and then sought to develop a formula for a function that modeled the quantity to be optimized. Once the function was established, we then considered what domain was appropriate. At that point, we were ﬁnally ready to apply the ideas of calculus to determine the absolute minimum or maximum. Throughout what follows in the current section, the primary emphasis is on the reader solv- ing problems. Initially, some substantial guidance is provided, with the problems progress- ing to require greater independence as we move along. Preview Activity 3.4.1. According to U.S. postal regulations, the girth plus the length of a parcel sent by mail may not exceed 108 inches, where by “girth” we mean the perimeter of the smallest end. What is the largest possible volume of a rectangular parcel with a square end that can be sent by mail? What are the dimensions of the package of largest volume? Figure 3.4.1: A rectangular parcel with a square end. a. Let x represent the length of one side of the square end and y the length of the longer side. Label these quantities appropriately on the image shown in 186 3.4 Applied Optimization Figure 3.4.1. b. What is the quantity to be optimized in this problem? Find a formula for this quantity in terms of x and y. c. The problem statement tells us that the parcel’s girth plus length may not ex- ceed 108 inches. In order to maximize volume, we assume that we will actually need the girth plus length to equal 108 inches. What equation does this produce involving x and y? d. Solve the equation you found in (c) for one of x or y (whichever is easier). e. Now use your work in (b) and (d) to determine a formula for the volume of the parcel so that this formula is a function of a single variable. f. Over what domain should we consider this function? Note that both x and y must be positive; how does the constraint that girth plus length is 108 inches produce intervals of possible values for x and y? g. Find the absolute maximum of the volume of the parcel on the domain you established in (f) and hence also determine the dimensions of the box of greatest volume. Justify that you’ve found the maximum using calculus. 3.4.1 More applied optimization problems Many of the steps in Preview Activity 3.4.1 are ones that we will execute in any applied opti- mization problem. We brieﬂy summarize those here to provide an overview of our approach in subsequent questions. Note 3.4.2 • Draw a picture and introduce variables. It is essential to ﬁrst understand what quan- tities are allowed to vary in the problem and then to represent those values with vari- ables. Constructing a ﬁgure with the variables labeled is almost always an essential ﬁrst step. Sometimes drawing several diagrams can be especially helpful to get a sense of the situation. A nice example of this can be seen at http://gvsu.edu/s/99, where the choice of where to bend a piece of wire into the shape of a rectangle determines both the rectangle’s shape and area. • Identify the quantity to be optimized as well as any key relationships among the vari- able quantities. Essentially this step involves writing equations that involve the vari- ables that have been introduced: one to represent the quantity whose minimum or maximum is sought, and possibly others that show how multiple variables in the prob- lem may be interrelated. • Determine a function of a single variable that models the quantity to be optimized; this may involve using other relationships among variables to eliminate one or more variables in the function formula. For example, in Preview Activity 3.4.1, we initially found that V x 2 y, but then the additional relationship that 4x + y 108 (girth plus 187 Chapter 3 Using Derivatives length equals 108 inches) allows us to relate x and y and thus observe equivalently that y 108 − 4x. Substituting for y in the volume equation yields V(x) x 2 (108 − 4x), and thus we have written the volume as a function of the single variable x. • Decide the domain on which to consider the function being optimized. Often the phys- ical constraints of the problem will limit the possible values that the independent vari- able can take on. Thinking back to the diagram describing the overall situation and any relationships among variables in the problem often helps identify the smallest and largest values of the input variable. • Use calculus to identify the absolute maximum and/or minimum of the quantity be- ing optimized. This always involves ﬁnding the critical numbers of the function ﬁrst. Then, depending on the domain, we either construct a ﬁrst derivative sign chart (for an open or unbounded interval) or evaluate the function at the endpoints and critical numbers (for a closed, bounded interval), using ideas we’ve studied so far in Chapter 3. • Finally, we make certain we have answered the question: does the question seek the absolute maximum of a quantity, or the values of the variables that produce the max- imum? That is, ﬁnding the absolute maximum volume of a parcel is diﬀerent from ﬁnding the dimensions of the parcel that produce the maximum. Activity 3.4.2. A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can costs $0.015 per square inch and the material for the lids costs $0.027 per square inch. Suppose that we desire to construct a can that has a volume of 16 cubic inches. What dimensions minimize the cost of the can? a. Draw a picture of the can and label its dimensions with appropriate variables. b. Use your variables to determine expressions for the volume, surface area, and cost of the can. c. Determine the total cost function as a function of a single variable. What is the domain on which you should consider this function? d. Find the absolute minimum cost and the dimensions that produce this value. Familiarity with common geometric formulas is particularly helpful in problems such as the one in Activity 3.4.2. Sometimes those involve perimeter, area, volume, or surface area. At other times, the constraints of a problem introduce right triangles (where the Pythagorean Theorem applies) or other functions whose formulas provide relationships among the vari- ables. Activity 3.4.3. A hiker starting at a point P on a straight road walks east towards point Q, which is on the road and 3 kilometers from point P. Two kilometers due north of point Q is a cabin. The hiker will walk down the road for a while, at a pace of 8 kilometers per hour. At some point Z between P and Q, the hiker leaves the road and makes a straight line towards the cabin through the woods, hiking at a pace of 3 kph, as pictured in Figure 3.4.3. In order to minimize the time to 188 3.4 Applied Optimization go from P to Z to the cabin, where should the hiker turn into the forest? cabin 2 P 3 Z Q Figure 3.4.3: A hiker walks from P to Z to the cabin, as pictured. In more geometric problems, we often use curves or functions to provide natural constraints. For instance, we could investigate which isosceles triangle that circumscribes a unit circle has the smallest area, which you can explore for yourself at http://gvsu.edu/s/9b. Or similarly, for a region bounded by a parabola, we might seek the rectangle of largest area that ﬁts beneath the curve, as shown at http://gvsu.edu/s/9c. The next activity is similar to the latter problem. Activity 3.4.4. Consider the region in the x-y plane that is bounded by the x-axis and the function f (x) 25 − x 2 . Construct a rectangle whose base lies on the x-axis and is centered at the origin, and whose sides extend vertically until they intersect the curve y 25− x 2 . Which such rectangle has the maximum possible area? Which such rectangle has the greatest perimeter? Which has the greatest combined perimeter and area? (Challenge: answer the same questions in terms of positive parameters a and b for the function f (x) b − ax 2 .) Activity 3.4.5. A trough is being constructed by bending a 4 × 24 (measured in feet) rectangular piece of sheet metal. Two symmetric folds 2 feet apart will be made parallel to the longest side of the rec- tangle so that the trough has cross-sections in the shape of a trapezoid, as pictured in Figure 3.4.4. At what angle should the folds be made to produce the trough of maximum volume? 3.4.2 Summary • While there is no single algorithm that works in every situation where optimization is used, in most of the problems we consider, the following steps are helpful: draw a picture and introduce variables; identify the quantity to be optimized and ﬁnd rela- tionships among the variables; determine a function of a single variable that models the quantity to be optimized; decide the domain on which to consider the function 189 Chapter 3 Using Derivatives 1 1 θ 2 Figure 3.4.4: A cross-section of the trough formed by folding to an angle of θ. being optimized; use calculus to identify the absolute maximum and/or minimum of the quantity being optimized. 3.4.3 Exercises 1. Maximizing the volume of a box. An open box is to be made out of a 10-inch by 18- inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume. 2. Minimizing the cost of a container. A rectangular storage container with an open top is to have a volume of 26 cubic meters. The length of its base is twice the width. Material for the base costs 11 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container. 3. Maximizing area contained by a fence. An ostrich farmer wants to enclose a rectangu- lar area and then divide it into six pens with fencing parallel to one side of the rectangle (see the ﬁgure below). There are 620 feet of fencing available to complete the job. What is the largest possible total area of the six pens? 4. Minimizing the area of a poster. The top and bottom margins of a poster are 8 cm and the side margins are each 6 cm. If the area of printed material on the poster is ﬁxed at 388 square centimeters, ﬁnd the dimensions of the poster with the smallest area. printe d mate rial 5. Maximizing the area of a rectangle. A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y 1 − x 2 . What are the dimensions of such a rectangle with the greatest possible area? 6. A rectangular box with a square bottom and closed top is to be made from two mate- rials. The material for the side costs $1.50 per square foot and the material for the top 190 3.4 Applied Optimization and bottom costs $3.00 per square foot. If you are willing to spend $15 on the box, what is the largest volume it can contain? Justify your answer completely using calculus. 7. A farmer wants to start raising cows, horses, goats, and sheep, and desires to have a rectangular pasture for the animals to graze in. However, no two diﬀerent kinds of animals can graze together. In order to minimize the amount of fencing she will need, she has decided to enclose a large rectangular area and then divide it into four equally sized pens by adding three segments of fence inside the large rectangle that are parallel to two existing sides. She has decided to purchase 7500 ft of fencing. What is the maximum possible area that each of the four pens will enclose? 8. Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched from the top of one pole to some point on the ground between the poles, and then to the top of the other pole. What is the minimum possible length of cable required? Justify your answer completely using calculus. 9. A company is designing propane tanks that are cylindrical with hemispherical ends. Assume that the company wants tanks that will hold 1000 cubic feet of gas, and that the ends are more expensive to make, costing $5 per square foot, while the cylindrical barrel between the ends costs $2 per square foot. Use calculus to determine the minimum cost to construct such a tank. 191 Chapter 3 Using Derivatives 3.5 Related Rates Motivating Questions • If two quantities that are related, such as the radius and volume of a spherical bal- loon, are both changing as implicit functions of time, how are their rates of change related? That is, how does the relationship between the values of the quantities aﬀect the relationship between their respective derivatives with respect to time? In most of our applications of the derivative so far, we have been interested in the instanta- neous rate at which one variable, say y, changes with respect to another, say x, leading us to dy compute and interpret dx . We next consider situations where several variable quantities are related, but where each quantity is implicitly a function of time, which will be represented by the variable t. Through knowing how the quantities are related, we will be interested in determining how their respective rates of change with respect to time are related. For example, suppose that air is being pumped into a spherical balloon so that its volume increases at a constant rate of 20 cubic inches per second. Since the balloon’s volume and ra- dius are related, by knowing how fast the volume is changing, we ought to be able to discover how fast the radius is changing. We are interested in questions such as: can we determine how fast the radius of the balloon is increasing at the moment the balloon’s diameter is 12 inches? Preview Activity 3.5.1. A spherical balloon is being inﬂated at a constant rate of 20 cubic inches per second. How fast is the radius of the balloon changing at the instant the balloon’s diameter is 12 inches? Is the radius changing more rapidly when d 12 or when d 16? Why? a. Draw several spheres with diﬀerent radii, and observe that as volume changes, the radius, diameter, and surface area of the balloon also change. b. Recall that the volume of a sphere of radius r is V 34 πr 3 . Note well that in the setting of this problem, both V and r are changing as time t changes, and thus both V and r may be viewed as implicit functions of t, with respective dt and dt . Diﬀerentiate both sides of the equation V 3 πr with derivatives dV dr 4 3 dV respect to t (using the chain rule on the right) to ﬁnd a formula for dt that depends on both r and dr dt . c. At this point in the problem, by diﬀerentiating we have “related the rates” of change of V and r. Recall that we are given in the problem that the balloon is being inﬂated at a constant rate of 20 cubic inches per second. Is this rate the value of dr dV dt or dt ? Why? d. From part (c), we know the value of dV dt at every value of t. Next, observe that when the diameter of the balloon is 12, we know the value of the radius. In the dt 4πr dt , substitute these values for the relevant quantities and equation dV 2 dr 192 3.5 Related Rates dr solve for the remaining unknown quantity, which is dt . How fast is the radius changing at the instant d 12? e. How is the situation diﬀerent when d 16? When is the radius changing more rapidly, when d 12 or when d 16? 3.5.1 Related Rates Problems In problems where two or more quantities can be related to one another, and all of the vari- ables involved are implicitly functions of time, t, we are often interested in how their rates are related; we call these related rates problems. Once we have an equation establishing the relationship among the variables, we diﬀerentiate implicitly with respect to time to ﬁnd connections among the rates of change. Example 3.5.1 Sand is being dumped by a conveyor belt onto a pile so that the sand forms a right circular cone, as pictured in Figure 3.5.2. h r Figure 3.5.2: A conical pile of sand. Solution. As sand falls from the conveyor belt, several features of the sand pile will change: the volume of the pile will grow, the height will increase, and the radius will get bigger, too. All of these quantities are related to one another, and the rate at which each is changing is related to the rate at which sand falls from the conveyor. We begin by identifying which variables are changing and how they are related. In this problem, we observe that the radius and height of the pile are related to its volume by the standard equation for the volume of a cone, 1 2 V πr h. 3 Viewing each of V, r, and h as functions of t, we diﬀerentiate implicitly to arrive at an equa- tion that relates their respective rates of change. Taking the derivative of each side of the 193 Chapter 3 Using Derivatives equation with respect to t, we ﬁnd [ ] d d 1 2 [V] πr h . dt dt 3 d On the left, dt [V] is simply dV dt . On the right, the situation is more complicated, as both r and h are implicit functions of t. Hence we need the product and chain rules. We ﬁnd that [ ] dV d 1 2 πr h dt dt 3 1 d 1 d πr 2 [h] + πh [r 2 ] 3 dt 3 dt 1 dh 1 dr πr 2 + πh2r 3 dt 3 dt (Note particularly how we are using ideas from Section 2.7 on implicit diﬀerentiation. There dy we found that when y is an implicit function of x, dxd [y 2 ] 2y dx . The same principles are applied here when we compute dt d [r 2 ] 2r dr dt .) The equation dV 1 dh 2 dr πr 2 + πrh , dt 3 dt 3 dt relates the rates of change of V, h, and r. If we are given suﬃcient additional information, we may then ﬁnd the value of one or more of these rates of change at a speciﬁc point in time. Example 3.5.3 In the setting of Example 3.5.1, suppose we also know the following: (a) sand falls from the conveyor in such a way that the height of the pile is always half the radius, and (b) sand falls from the conveyor belt at a constant rate of 10 cubic feet per minute. How fast is the height of the sandpile changing at the moment the radius is 4 feet? Solution. The information that the height is always half the radius tells us that for all values of t, h 12 r. Diﬀerentiating with respect to t, it follows that dh dt 2 dt . These relationships 1 dr enable us to relate dVdt to just one of r or h. Substituting the expressions involving r and dt dr dh for h and dt , we now have that dV 1 1 dr 2 1 dr πr 2 · + πr · r · . (3.5.1) dt 3 2 dt 3 2 dt Since sand falls from the conveyor at the constant rate of 10 cubic feet per minute, the value dt , the rate at which the volume of the sand pile changes, is dt 10 ft /min. We are of dV dV 3 interested in how fast the height of the pile is changing at the instant when r 4, so we substitute r 4 and dVdt 10 into Equation (3.5.1), to ﬁnd 1 2 1 dr 2 1 dr 8 dr 16 dr 10 π4 · + π4 · 4 · π + π . 3 2 dt r4 3 2 dt r4 3 dt r4 3 dt r4 dr Only the value of dt r4 remains unknown. We combine like terms on the right side of the 194 3.5 Related Rates equation above to get 10 8π dr dt r4 , and solve for dr dt r4 to ﬁnd dr 10 ≈ 0.39789 dt r4 8π feet per second. Because we were interested in how fast the height of the pile was changing dt when r 4. Since dt 2 dt for all values of t, it follows at this instant, we want to know dh dh 1 dr dh 5 ≈ 0.19894 ft/min. dt r4 8π Note the diﬀerence between the notations dr dr dt and dt r4 . The former represents the rate of change of r with respect to t at an arbitrary value of t, while the latter is the rate of change of r with respect to t at a particular moment, the moment when r 4. Had we known that h 21 r at the beginning of Example 3.5.1, we could have immediately simpliﬁed our work by writing V solely in terms of r to have ( ) 1 2 1 1 V πr h πr 3 . 3 2 6 From this last equation, diﬀerentiating with respect to t implies dV 1 dr πr 2 , dt 2 dt from which the same conclusions can be made. Our work with the sandpile problem above is similar in many ways to our approach in Pre- view Activity 3.5.1, and these steps are typical of most related rates problems. In certain ways, they also resemble work we do in applied optimization problems, and here we sum- marize the main approach for consideration in subsequent problems. Note 3.5.4 • Identify the quantities in the problem that are changing and choose clearly deﬁned variable names for them. Draw one or more ﬁgures that clearly represent the situation. • Determine all rates of change that are known or given and identify the rate(s) of change to be found. • Find an equation that relates the variables whose rates of change are known to those variables whose rates of change are to be found. • Diﬀerentiate implicitly with respect to t to relate the rates of change of the involved quantities. • Evaluate the derivatives and variables at the information relevant to the instant at which a certain rate of change is sought. Use proper notation to identify when a de- rivative is being evaluated at a particular instant, such as dr dt r4 . When identifying variables and drawing a picture, it is important to think about the dynamic ways in which the quantities change. Sometimes a sequence of pictures can be helpful; for 195 Chapter 3 Using Derivatives some pictures that can be easily modiﬁed as applets built in Geogebra, see the following links,¹ which represent • how a circular oil slick’s area grows as its radius increases http://gvsu.edu/s/9n; • how the location of the base of a ladder and its height along a wall change as the ladder slides http://gvsu.edu/s/9o; • how the water level changes in a conical tank as it ﬁlls with water at a constant rate http://gvsu.edu/s/9p (compare the problem in Activity 3.5.2); • how a skateboarder’s shadow changes as he moves past a lamppost http://gvsu.edu/ s/9q. Drawing well-labeled diagrams and envisioning how diﬀerent parts of the ﬁgure change is a key part of understanding related rates problems and being successful at solving them. Activity 3.5.2. A water tank has the shape of an inverted circular cone (point down) with a base of radius 6 feet and a depth of 8 feet. Suppose that water is being pumped into the tank at a constant instantaneous rate of 4 cubic feet per minute. a. Draw a picture of the conical tank, including a sketch of the water level at a point in time when the tank is not yet full. Introduce variables that measure the radius of the water’s surface and the water’s depth in the tank, and label them on your ﬁgure. b. Say that r is the radius and h the depth of the water at a given time, t. What equation relates the radius and height of the water, and why? c. Determine an equation that relates the volume of water in the tank at time t to the depth h of the water at that time. d. Through diﬀerentiation, ﬁnd an equation that relates the instantaneous rate of change of water volume with respect to time to the instantaneous rate of change of water depth at time t. e. Find the instantaneous rate at which the water level is rising when the water in the tank is 3 feet deep. f. When is the water rising most rapidly: at h 3, h 4, or h 5? Recognizing which geometric relationships are relevant in a given problem is often the key to ﬁnding the function to optimize. For instance, although the problem in Activity 3.5.2 is about a conical tank, the most important fact is that there are two similar right triangles involved. In another setting, we might use the Pythagorean Theorem to relate the legs of the triangle. But in the conical tank, the fact that the water ﬁlls the tank so that that the ratio of radius to depth is constant turns out to be the important relationship. In other situations where a changing angle is involved, trigonometric functions may provide the means to ﬁnd relationships among various parts of the triangle. ¹We again refer to the work of Prof. Marc Renault of Shippensburg University, found at http://gvsu.edu/s/5p. 196 3.5 Related Rates Activity 3.5.3. A television camera is positioned 4000 feet from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. In addition, the auto-focus of the camera has to take into account the increasing distance between the camera and the rocket. We assume that the rocket rises vertically. (A similar problem is discussed and pictured dynamically at http://gvsu.edu/s/9t. Exploring the applet at the link will be helpful to you in answering the questions that follow.) a. Draw a ﬁgure that summarizes the given situation. What parts of the picture are changing? What parts are constant? Introduce appropriate variables to rep- resent the quantities that are changing. b. Find an equation that relates the camera’s angle of elevation to the height of the rocket, and then ﬁnd an equation that relates the instantaneous rate of change of the camera’s elevation angle to the instantaneous rate of change of the rocket’s height (where all rates of change are with respect to time). c. Find an equation that relates the distance from the camera to the rocket to the rocket’s height, as well as an equation that relates the instantaneous rate of change of distance from the camera to the rocket to the instantaneous rate of change of the rocket’s height (where all rates of change are with respect to time). d. Suppose that the rocket’s speed is 600 ft/sec at the instant it has risen 3000 feet. How fast is the distance from the television camera to the rocket changing at that moment? If the camera is following the rocket, how fast is the camera’s angle of elevation changing at that same moment? e. If from an elevation of 3000 feet onward the rocket continues to rise at 600 feet/ sec, will the rate of change of distance with respect to time be greater when the elevation is 4000 feet than it was at 3000 feet, or less? Why? In addition to ﬁnding instantaneous rates of change at particular points in time, we can of- ten make more general observations about how particular rates themselves will change over time. For instance, when a conical tank is ﬁlling with water at a constant rate, it seems obvi- ous that the depth of the water should increase more slowly over time. Note how carefully we must phrase the relationship: we mean to say that while the depth, h, of the water is increasing, its rate of change, dh dt , is decreasing (both as a function of t and as a function of h). We make this observation by solving the equation that relates the various rates for one particular rate, without substituting any particular values for known variables or rates. For instance, in the conical tank problem in Activity 3.5.2, we established that dV 1 dh πh 2 , dt 16 dt and hence dh 16 dV . dt πh 2 dt Provided that dV dh dt is constant, it is immediately apparent that as h gets larger, dt will get smaller but remain positive. Hence, the depth of the water is increasing at a decreasing rate. 197 Chapter 3 Using Derivatives Activity 3.5.4. As pictured in the applet at http://gvsu.edu/s/9q, a skateboarder who is 6 feet tall rides under a 15 foot tall lamppost at a constant rate of 3 feet per sec- ond. We are interested in understanding how fast his shadow is changing at various points in time. a. Draw an appropriate right triangle that represents a snapshot in time of the skateboarder, lamppost, and his shadow. Let x denote the horizontal distance from the base of the lamppost to the skateboarder and s represent the length of his shadow. Label these quantities, as well as the skateboarder’s height and the lamppost’s height on the diagram. b. Observe that the skateboarder and the lamppost represent parallel line seg- ments in the diagram, and thus similar triangles are present. Use similar tri- angles to establish an equation that relates x and s. dx ds c. Use your work in (b) to ﬁnd an equation that relates dt and dt . d. At what rate is the length of the skateboarder’s shadow increasing at the instant the skateboarder is 8 feet from the lamppost? e. As the skateboarder’s distance from the lamppost increases, is his shadow’s length increasing at an increasing rate, increasing at a decreasing rate, or in- creasing at a constant rate? f. Which is moving more rapidly: the skateboarder or the tip of his shadow? Ex- plain, and justify your answer. In the ﬁrst three activities of this section, we provided guided instruction to build a solution in a step by step way. For the closing activity and the following exercises, most of the detailed work is left to the reader. Activity 3.5.5. A baseball diamond is 90′ square. A batter hits a ball along the third base line and runs to ﬁrst base. At what rate is the distance between the ball and ﬁrst base changing when the ball is halfway to third base, if at that instant the ball is traveling 100 feet/sec? At what rate is the distance between the ball and the runner changing at the same instant, if at the same instant the runner is 1/8 of the way to ﬁrst base running at 30 feet/sec? 3.5.2 Summary • When two or more related quantities are changing as implicit functions of time, their rates of change can be related by implicitly diﬀerentiating the equation that relates the quantities themselves. For instance, if the sides of a right triangle are all changing as functions of time, say having lengths x, y, and z, then these quantities are related by the Pythagorean Theorem: x 2 + y 2 z 2 . It follows by implicitly diﬀerentiating with 198 3.5 Related Rates respect to t that their rates are related by the equation dx dy dz 2x + 2y 2z , dt dt dt so that if we know the values of x, y, and z at a particular time, as well as two of the three rates, we can deduce the value of the third. 3.5.3 Exercises 1. Height of a conical pile of gravel. Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 23 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V 13 πr 2 h. 2. Movement of a shadow. A street light is at the top of a 13 foot tall pole. A 6 foot tall woman walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 feet from the base of the pole? 3. A leaking conical tank. Water is leaking out of an inverted conical tank at a rate of 9600.0 cm3 /min at the same time that water is being pumped into the tank at a constant rate. The tank has height 7.0 m and the the diameter at the top is 5.0 m. If the water level is rising at a rate of 22.0 cm/min when the height of the water is 1.5 m, ﬁnd the rate at which water is being pumped into the tank in cubic centimeters per minute. Hint. Let R be the unknown rate at which water is being pumped in. Then you know dt R − 9600.0. Use geometry (similar triangles) to ﬁnd that if V is volume of water, dV the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by 31 πr 2 h 4. A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow. If the rope is being pulled in at a rate of 2 feet per second, how fast is the boat approaching the dock when the length of rope from bow to pulley is 13 feet? 5. A swimming pool is 60 feet long and 25 feet wide. Its depth varies uniformly from 3 feet at the shallow end to 15 feet at the deep end, as shown in the Figure 3.5.5. Suppose the pool has been emptied and is now being ﬁlled with water at a rate of 800 cubic feet per minute. At what rate is the depth of water (measured at the deepest point of the pool) increasing when it is 5 feet deep at that end? Over time, describe how the depth of the water will increase: at an increasing rate, at a decreasing rate, or at a constant rate. Explain. 199 Chapter 3 Using Derivatives 25 60 3 15 Figure 3.5.5: The swimming pool. 6. A baseball diamond is a square with sides 90 feet long. Suppose a baseball player is advancing from second to third base at the rate of 24 feet per second, and an umpire is standing on home plate. Let θ be the angle between the third baseline and the line of sight from the umpire to the runner. How fast is θ changing when the runner is 30 feet from third base? 7. Sand is being dumped oﬀ a conveyor belt onto a pile in such a way that the pile forms in the shape of a cone whose radius is always equal to its height. Assuming that the sand is being dumped at a rate of 10 cubic feet per minute, how fast is the height of the pile changing when there are 1000 cubic feet on the pile? 200 CHAPTER 4 The Deﬁnite Integral 4.1 Determining distance traveled from velocity Motivating Questions • If we know the velocity of a moving body at every point in a given interval, can we determine the distance the object has traveled on the time interval? • How is the problem of ﬁnding distance traveled related to ﬁnding the area under a certain curve? • What does it mean to antidiﬀerentiate a function and why is this process relevant to ﬁnding distance traveled? • If velocity is negative, how does this impact the problem of ﬁnding distance traveled? In the ﬁrst section of the text, we considered a moving object with known position at time t, namely, a tennis ball tossed into the air with height s (in feet) at time t (in seconds) given by s(t) 64 − 16(t − 1)2 . We investigated the average velocity of the ball on an interval s(b)−s(a) [a, b], computed by the diﬀerence quotient b−a . We found that we could determine the instantaneous velocity of the ball at time t by taking the derivative of the position function, s(t + h) − s(t) s ′(t) lim . h→0 h Thus, if its position function is diﬀerentiable, we can ﬁnd the velocity of a moving object at any point in time. From this study of position and velocity we have learned a great deal. We can use the de- rivative to ﬁnd a function’s instantaneous rate of change at any point in the domain, to ﬁnd where the function is increasing or decreasing, where it is concave up or concave down, and to locate relative extremes. The vast majority of the problems and applications we have considered have involved the situation where a particular function is known and we seek information that relies on knowing the function’s instantaneous rate of change. For all these tasks, we proceed from a function f to its derivative, f ′, and use the meaning of the deriv- ative to help us answer important questions. Chapter 4 The Deﬁnite Integral We have also encountered the reverse situation, where we know the derivative of a function, f ′, and try to deduce information about f . We will focus our attention in Chapter 4 on this problem: if we know the instantaneous rate of change of a function, can we ﬁnd the function itself? We start with a more speciﬁc question: if we know the instantaneous velocity of an object moving along a straight line path, can we ﬁnd its corresponding position function? Preview Activity 4.1.1. Suppose that a person is taking a walk along a long straight path and walks at a constant rate of 3 miles per hour. a. On the left-hand axes provided in Figure 4.1.1, sketch a labeled graph of the velocity function v(t) 3. mph miles 8 8 4 4 hrs hrs 1 2 1 2 Figure 4.1.1: At left, axes for plotting y v(t); at right, for plotting y s(t). Note that while the scale on the two sets of axes is the same, the units on the right-hand axes diﬀer from those on the left. The right-hand axes will be used in question (d). b. How far did the person travel during the two hours? How is this distance related to the area of a certain region under the graph of y v(t)? c. Find an algebraic formula, s(t), for the position of the person at time t, assuming that s(0) 0. Explain your thinking. d. On the right-hand axes provided in Figure 4.1.1, sketch a labeled graph of the position function y s(t). e. For what values of t is the position function s increasing? Explain why this is the case using relevant information about the velocity function v. 202 4.1 Determining distance traveled from velocity 4.1.1 Area under the graph of the velocity function In Preview Activity 4.1.1, we learned that when the velocity of a moving object’s velocity is constant (and positive), the area under the velocity curve over an interval of time tells us the distance the object traveled. mph mph 3 3 y = v(t) v(t) = 2 A2 1 A1 1 hrs hrs 1 2 3 1 2 3 Figure 4.1.2: At left, a constant velocity function; at right, a non-constant velocity function. The left-hand graph of Figure 4.1.2 shows the velocity of an object moving at 2 miles per hour over the time interval [1, 1.5]. The area A1 of the shaded region under y v(t) on [1, 1.5] is miles 1 A1 2 · hours 1 mile. hour 2 This result is simply the fact that distance equals rate times time, provided the rate is con- stant. Thus, if v(t) is constant on the interval [a, b], the distance traveled on [a, b] is equal to the area A given by A v(a)(b − a) v(a)∆t, where ∆t is the change in t over the interval. (Since the velocity is constant, we can use any value of v(t) on the interval [a, b], we simply chose v(a), the value at the interval’s left endpoint.) For several examples where the velocity function is piecewise constant, see http:/ /gvsu.edu/s/9T.¹ The situation is more complicated when the velocity function is not constant. But on rela- tively small intervals where v(t) does not vary much, we can use the area principle to esti- mate the distance traveled. The graph at right in Figure 4.1.2 shows a non-constant velocity function. On the interval [1, 1.5], the velocity varies from v(1) 2.5 down to v(1.5) ≈ 2.1. One estimate for the distance traveled is the area of the pictured rectangle, miles 1 A2 v(1)∆t 2.5 · hours 1.25 miles. hour 2 Note that because v is decreasing on [1, 1.5], A2 1.25 is an over-estimate of the actual distance traveled. ¹Marc Renault, calculus applets. 203 Chapter 4 The Deﬁnite Integral To estimate the area under this non-constant velocity function on a wider interval, say [0, 3], one rectangle will not give a good approximation. Instead, we could use the six rectangles pictured in Figure 4.1.3, ﬁnd the area of each rectangle, and add up the total. Obviously there are choices to make and issues to understand: How many rectangles should we use? Where should we evaluate the function to decide the rectangle’s height? What happens if the velocity is sometimes negative? Can we ﬁnd the exact area under any non-constant curve? mph 3 y = v(t) 1 hrs 1 2 3 Figure 4.1.3: Using six rectangles to estimate the area under y v(t) on [0, 3]. We will study these questions and more in what follows; for now it suﬃces to observe that the simple idea of the area of a rectangle gives us a powerful tool for estimating distance traveled from a velocity function, as well as for estimating the area under an arbitrary curve. To explore the use of multiple rectangles to approximate area under a non-constant velocity function, see the applet found at http://gvsu.edu/s/9U.² Activity 4.1.2. Suppose that a person is walking in such a way that her velocity varies slightly according to the information given in Table 4.1.4 and graph given in Fig- ure 4.1.5. a. Using the grid, graph, and given data appropriately, estimate the distance trav- eled by the walker during the two hour interval from t 0 to t 2. You should use time intervals of width ∆t 0.5, choosing a way to use the function consis- tently to determine the height of each rectangle in order to approximate distance traveled. b. How could you get a better approximation of the distance traveled on [0, 2]? Explain, and then ﬁnd this new estimate. ²Marc Renault, calculus applets. 204 4.1 Determining distance traveled from velocity t v(t) mph 0.00 1.500 3 y = v(t) 0.25 1.789 0.50 1.938 2 0.75 1.992 1.00 2.000 1.25 2.008 1 1.50 2.063 1.75 2.211 hrs 2.00 2.500 1 2 Table 4.1.4: Velocity data for the per- son walking. Figure 4.1.5: The graph of y v(t). c. Now suppose that you know that v is given by v(t) 0.5t 3 − 1.5t 2 + 1.5t + 1.5. Remember that v is the derivative of the walker’s position function, s. Find a formula for s so that s ′ v. d. Based on your work in (c), what is the value of s(2) − s(0)? What is the meaning of this quantity? 4.1.2 Two approaches: area and antidiﬀerentiation When the velocity of a moving object is positive, the object’s position is always increasing. (We will soon consider situations where velocity is negative; for now, we focus on the situa- tion where velocity is always positive.) We have established that whenever v is constant on an interval, the exact distance traveled is the area under the velocity curve. When v is not constant, we can estimate the total distance traveled by ﬁnding the areas of rectangles that approximate the area under the velocity curve. Thus, we see that ﬁnding the area between a curve and the horizontal axis is an important exercise: besides being an interesting geometric question, if the curve gives the velocity of a moving object, the area under the curve tells us the exact distance traveled on an interval. We can estimate this area if we have a graph or a table of values for the velocity function. In Activity 4.1.2, we encountered an alternate approach to ﬁnding the distance traveled. If y v(t) is a formula for the instantaneous velocity of a moving object, then v must be the derivative of the object’s position function, s. If we can ﬁnd a formula for s(t) from the formula for v(t), we will know the position of the object at time t, and the change in position over a particular time interval tells us the distance traveled on that interval. For a simple example, consider the situation from Preview Activity 4.1.1, where a person is walking along a straight line with velocity function v(t) 3 mph. On the left-hand graph of the velocity function in Figure 4.1.6, we see the relationship between area and distance 205 Chapter 4 The Deﬁnite Integral mph miles 8 8 s(t) = 3t 4 4 v(t) = 3 s(1.5) = 4.5 A = 3 · 1.25 = 3.75 hrs s(0.25) = 0.75 hrs 1 2 1 2 Figure 4.1.6: The velocity function v(t) 3 and corresponding position function s(t) 3t. traveled, miles A3 · 1.25 hours 3.75 miles. hour In addition, we observe³ that if s(t) 3t, then s ′(t) 3, so s(t) 3t is the position function whose derivative is the given velocity function, v(t) 3. The respective locations of the person at times t 0.25 and t 1.5 are s(1.5) 4.5 and s(0.25) 0.75, and therefore s(1.5) − s(0.25) 4.5 − 0.75 3.75 miles. This is the person’s change in position on [0.25, 1.5], which is precisely the distance traveled. In this example there are profound ideas and connections that we will study throughout Chapter 4. For now, observe that if we know a formula for a velocity function v, it can be very helpful to ﬁnd a function s that satisﬁes s ′ v. We say that s is an antiderivative of v. More generally, we have the following formal deﬁnition. Deﬁnition 4.1.7 If 1 and G are functions such that G′ 1, we say that G is an antiderivative of 1. For example, if 1(x) 3x 2 + 2x, G(x) x 3 + x 2 is an antiderivative of 1, because G′(x) 1(x). Note that we say “an” antiderivative of 1 rather than “the” antiderivative of 1, because H(x) x 3 +x 2 +5 is also a function whose derivative is 1, and thus H is another antiderivative of 1. Activity 4.1.3. A ball is tossed vertically in such a way that its velocity function is given by v(t) 32 − 32t, where t is measured in seconds and v in feet per second. ³Here we are making the implicit assumption that s(0) 0; we will discuss diﬀerent possibilities for values of s(0) in subsequent study. 206 4.1 Determining distance traveled from velocity Assume that this function is valid for 0 ≤ t ≤ 2. a. For what values of t is the velocity of the ball positive? What does this tell you about the motion of the ball on this interval of time values? b. Find an antiderivative, s, of v that satisﬁes s(0) 0. c. Compute the value of s(1) − s( 12 ). What is the meaning of the value you ﬁnd? d. Using the graph of y v(t) provided in Figure 4.1.8, ﬁnd the exact area of the region under the velocity curve between t 21 and t 1. What is the meaning of the value you ﬁnd? ft/sec 24 v(t) = 32 − 32t 12 sec 1 2 -12 -24 Figure 4.1.8: The graph of y v(t). e. Answer the same questions as in (c) and (d) but instead using the interval [0, 1]. f. What is the value of s(2) − s(0)? What does this result tell you about the ﬂight of the ball? How is this value connected to the provided graph of y v(t)? Explain. 4.1.3 When velocity is negative The assumption that its velocity is positive on a given interval guarantees that the movement of an object is always in a single direction, and hence ensures that its change in position is the same as the distance it travels. As we saw in Activity 4.1.3, there are natural settings in which an object’s velocity is negative, and we would like to understand this scenario as well. Consider a simple example where a woman goes for a walk on the beach along a stretch of very straight shoreline that runs east-west. We assume that her initial position is s(0) 0, and that her position function increases as she moves east from her starting location. For instance, s 1 mile represents one mile east of the start location, while s −1 tells us she is 207 Chapter 4 The Deﬁnite Integral one mile west of where she began walking on the beach. Now suppose she walks in the following manner. From the outset at t 0, she walks due east at a constant rate of 3 mph for 1.5 hours. After 1.5 hours, she stops abruptly and begins walking due west at a constant rate of 4 mph and does so for 0.5 hours. Then, after another abrupt stop and start, she resumes walking at a constant rate of 3 mph to the east for one more hour. What is the total distance she traveled on the time interval from t 0 to t 3? What the total change in her position over that time? These questions are possible to answer without calculus because the velocity is constant on each interval. From t 0 to t 1.5, she traveled D[0,1.5] 3 miles per hour · 1.5 hours 4.5 miles. On t 1.5 to t 2, the distance traveled is D[1.5,2] 4 miles per hour · 0.5 hours 2 miles. Finally, in the last hour she walked D[2,3] 3 miles per hour · 1 hours 3 miles, so the total distance she traveled is D D[0,1.5] + D[1.5,2] + D[2,3] 4.5 + 2 + 3 9.5 miles. Since the velocity for 1.5 < t < 2 is v −4, indicating motion in the westward direction, the woman ﬁrst walked 4.5 miles east, then 2 miles west, followed by 3 more miles east. Thus, the total change in her position is change in position 4.5 − 2 + 3 5.5 miles. We have been able to answer these questions fairly easily, and if we think about the problem graphically, we can generalize our solution to the more complicated setting when velocity is not constant, and possibly negative. In Figure 4.1.9, we see how the distances we computed can be viewed as areas: A1 4.5 comes from multiplyimg rate times time (3 · 1.5), as do A2 and A3 . But while A2 is an area (and is therefore positive), because the velocity function is negative for 1.5 < t < 2, this area has a negative sign associated with it. The negative area distinguishes between distance traveled and change in position. The distance traveled is the sum of the areas, D A1 + A2 + A3 4.5 + 2 + 3 9.5 miles. But the change in position has to account for travel in the negative direction. An area above the t-axis is considered positive because it represents distance traveled in the positive direc- tion, while one below the t-axis is viewed as negative because it represents travel in theneg- ative direction. Thus, the change in the woman’s position is s(3) − s(0) (+4.5) + (−2) + (+3) 5.5 miles. 208 4.1 Determining distance traveled from velocity (3, 5.5) mph miles (1.5, 4.5) 4.5 4.5 y = v(t) y = s(t) 3.0 3.0 1.5 1.5 (2, 2.5) A1 = 4.5 A3 = 3 hrs hrs 1 3 1 3 -1.5 -1.5 A2 = 2 -3.0 -3.0 -4.5 -4.5 Figure 4.1.9: At left, the velocity function of the person walking; at right, the corresponding position function. In other words, the woman walks 4.5 miles in the positive direction, followed by two miles in the negative direction, and then 3 more miles in the positive direction. Negative velocity is also seen in the graph of the position function y s(t). Its slope is negative (speciﬁcally, −4) on the interval 1.5 < t < 2 because the velocity is −4 on that interval. The negative slope shows the position function is decreasing because the woman is walking east, rather than west. To summarize, we see that if velocity is sometimes negative, a moving object’s change in po- sition diﬀerent from its distance traveled. If we compute separately the distance traveled on each interval where velocity is positive or negative, we can calculate either the total distance traveled or the total change in position. We assign a negative value to distances traveled in the negative direction when we calculate change in position, but a positive value when we calculate the total distance traveled. Activity 4.1.4. Suppose that an object moving along a straight line path has its velocity v (in meters per second) at time t (in seconds) given by the piecewise linear function whose graph is pictured at left in Figure 4.1.10. We view movement to the right as being in the positive direction (with positive velocity), while movement to the left is in the negative direction. Suppose further that the object’s initial position at time t 0 is s(0) 1. a. Determine the total distance traveled and the total change in position on the time interval 0 ≤ t ≤ 2. What is the object’s position at t 2? b. On what time intervals is the moving object’s position function increasing? Why? On what intervals is the object’s position decreasing? Why? 209 Chapter 4 The Deﬁnite Integral m/sec 4 8 y = v(t) 2 4 sec 2 4 6 8 2 4 6 8 -2 -4 -4 -8 Figure 4.1.10: The velocity function of a moving object. c. What is the object’s position at t 8? How many total meters has it traveled to get to this point (including distance in both directions)? Is this diﬀerent from the object’s total change in position on t 0 to t 8? d. Find the exact position of the object at t 1, 2, 3, . . . , 8 and use this data to sketch an accurate graph of y s(t) on the axes provided at right in Figure 4.1.10. How can you use the provided information about y v(t) to determine the concavity of s on each relevant interval? 4.1.4 Summary • If we know the velocity of a moving body at every point in a given interval and the velocity is positive throughout, we can estimate the object’s distance traveled and in some circumstances determine this value exactly. • In particular, when velocity is positive on an interval, we can ﬁnd the total distance traveled by ﬁnding the area under the velocity curve and above the t-axis on the given time interval. We may only be able to estimate this area, depending on the shape of the velocity curve. • An antiderivative of a function f is a new function F whose derivative is f . That is, F is an antiderivative of f provided that F′ f . In the context of velocity and position, if we know a velocity function v, an antiderivative of v is a position function s that satisﬁes s ′ v. If v is positive on a given interval, say [a, b], then the change in position, s(b) − s(a), measures the distance the moving object traveled on [a, b]. • If its velocity is sometimes negative, a moving object is sometimes traveling in the opposite direction or backtracking. To determine distance traveled, we have to think 210 4.1 Determining distance traveled from velocity compute the distance separately on intervals where velocity is positive or negative, and account for the change in position on each such interval. 4.1.5 Exercises 1. Estimating distance traveled from velocity data. A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded: Time since brakes applied (sec) 0 2 4 6 Velocity (ft/s) 90 46 17 0 Give lower and upper estimates (using all of the available data) for the distance the car traveled after the brakes were applied. On a sketch of velocity against time, show the lower and upper estimates you found above.. 2. Distance from a linear veloity function. The velocity of a car is f (t) 11t meters/ second. Use a graph of f (t) to ﬁnd the exact distance traveled by the car, in meters, from t 0 to t 10 seconds. 3. Change in position from a linear velocity function. The velocity of a particle moving along the x-axis is given by f (t) 12 − 4t cm/sec. Use a graph of f (t) to ﬁnd the exact change in position of the particle from time t 0 to t 4 seconds. 4. Comparing distance traveled from velocity graphs. Two cars start at the same time and travel in the same direction along a straight road. The ﬁgure below gives the ve- locity, v (in km/hr), of each car as a function of time (in hr). The velocity of car A is given by the solid, blue curve, and the velocity of car B by dashed, red curve. (a) Which car attains the larger maximum velocity? (b) Which stops ﬁrst? (c) Which travels farther? 5. Finding average acceleration from velocity data. Suppose that an accelerating car goes from 0 mph to 68.2 mph in ﬁve seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: 1 mph is 22/15 feet per sec = 22/15 ft/s.) Find the average acceleration of the car over each of the ﬁrst two seconds. 211 Chapter 4 The Deﬁnite Integral t 0 1 2 3 4 5 v(t) 0.00 34.09 59.09 77.27 90.91 100.00 6. Change in position from a quadratic velocity function. The velocity function is v(t) t 2 −3t+2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−2, 5]. 7. Along the eastern shore of Lake Michigan from Lake Macatawa (near Holland) to Grand Haven, there is a bike path that runs almost directly north-south. For the purposes of this problem, assume the road is completely straight, and that the function s(t) tracks the position of the biker along this path in miles north of Pigeon Lake, which lies roughly halfway between the ends of the bike path. Suppose that the biker’s velocity function is given by the graph in Figure 4.1.11 on the time interval 0 ≤ t ≤ 4 (where t is measured in hours), and that s(0) 1. mph miles 10 y = v(t) 10 6 6 2 2 hrs hrs -2 1 2 3 4 5 -2 1 2 3 4 5 -6 -6 -10 -10 Figure 4.1.11: The graph of the biker’s velocity, y v(t), at left. At right, axes to plot an approximate sketch of y s(t). a. Approximately how far north of Pigeon Lake was the cyclist when she was the greatest distance away from Pigeon Lake? At what time did this occur? b. What is the cyclist’s total change in position on the time interval 0 ≤ t ≤ 2? At t 2, was she north or south of Pigeon Lake? c. What is the total distance the biker traveled on 0 ≤ t ≤ 4? At the end of the ride, how close was she to the point at which she started? d. Sketch an approximate graph of y s(t), the position function of the cyclist, on the interval 0 ≤ t ≤ 4. Label at least four important points on the graph of s. 8. A toy rocket is launched vertically from the ground on a day with no wind. The rocket’s vertical velocity at time t (in seconds) is given by v(t) 500 − 32t feet/sec. a. At what time after the rocket is launched does the rocket’s velocity equal zero? Call this time value a. What happens to the rocket at t a? b. Find the value of the total area enclosed by y v(t) and the t-axis on the interval 212 4.1 Determining distance traveled from velocity 0 ≤ t ≤ a. What does this area represent in terms of the physical setting of the problem? c. Find an antiderivative s of the function v. That is, ﬁnd a function s such that s ′(t) v(t). d. Compute the value of s(a) − s(0). What does this number represent in terms of the physical setting of the problem? e. Compute s(5) − s(1). What does this number tell you about the rocket’s ﬂight? 9. An object moving along a horizontal axis has its instantaneous velocity at time t in seconds given by the function v pictured in Figure 4.1.12, where v is measured in feet/ sec. Assume that the curves that make up the parts of the graph of y v(t) are either portions of straight lines or portions of circles. y = v(t) 1 1 2 3 4 5 6 7 -1 Figure 4.1.12: The graph of y v(t), the velocity function of a moving object. a. Determine the exact total distance the object traveled on 0 ≤ t ≤ 2. b. What is the value and meaning of s(5) − s(2), where y s(t) is the position func- tion of the moving object? c. On which time interval did the object travel the greatest distance: [0, 2], [2, 4], or [5, 7]? d. On which time interval(s) is the position function s increasing? At which point(s) does s achieve a relative maximum? 10. Filters at a water treatment plant become dirtier over time and thus become less ef- fective; they are replaced every 30 days. During one 30-day period, the rate at which pollution passes through the ﬁlters into a nearby lake (in units of particulate matter per day) is measured every 6 days and is given in the following table. The time t is measured in days since the ﬁlters were replaced. Day, t 0 6 12 18 24 30 Rate of pollution in units per day, p(t) 7 8 10 13 18 35 Table 4.1.13: Pollution data for the water ﬁlters. a. Plot the given data on a set of axes with time on the horizontal axis and the rate 213 Chapter 4 The Deﬁnite Integral of pollution on the vertical axis. b. Explain why the amount of pollution that entered the lake during this 30-day period would be given exactly by the area bounded by y p(t) and the t-axis on the time interval [0, 30]. c. Estimate the total amount of pollution entering the lake during this 30-day period. Carefully explain how you determined your estimate. 214 4.2 Riemann Sums 4.2 Riemann Sums Motivating Questions • How can we use a Riemann sum to estimate the area between a given curve and the horizontal axis over a particular interval? • What are the diﬀerences among left, right, middle, and random Riemann sums? • How can we write Riemann sums in an abbreviated form? In Section 4.1, we learned that if an object moves with positive velocity v, the area between y v(t) and the t-axis over a given time interval tells us the distance traveled by the object over that time period. If v(t) is sometimes negative and we view the area of any region below the t-axis as having an associated negative sign, then the sum of these signed areas tells us the moving object’s change in position over a given time interval. For instance, for the velocity function given in Figure 4.2.1, if the areas of shaded re- gions are A1 , A2 , and A3 as labeled, then y = v(t) the total distance D traveled by the moving object on [a, b] is D A1 + A2 + A3 , while the total change in the object’s posi- A1 A3 tion on [a, b] is s(b) − s(a) A1 − A2 + A3 . a A2 b Because the motion is in the negative di- rection on the interval where v(t) < 0, we subtract A2 to determine the object’s total change in position. Figure 4.2.1: A velocity function that is sometimes negative. Of course, ﬁnding D and s(b) − s(a) for the graph in Figure 4.2.1 presumes that we can actually ﬁnd the areas A1 , A2 , and A3 . So far, we have worked with velocity functions that were either constant or linear, so that the area bounded by the velocity function and the horizontal axis is a combination of rectangles and triangles, and we can ﬁnd the area exactly. But when the curve bounds a region that is not a familiar geometric shape, we cannot ﬁnd its area exactly. Indeed, this is one of our biggest goals in Chapter 4: to learn how to ﬁnd the exact area bounded between a curve and the horizontal axis for as many diﬀerent types of functions as possible. In Activity 4.1.2, we approximated the area under a nonlinear velocity function using rectan- gles. In the following preview activity, we consider three diﬀerent options for the heights of the rectangles we will use. 215 Chapter 4 The Deﬁnite Integral Preview Activity 4.2.1. A person walking along a straight path has her velocity in miles per hour at time t given by the function v(t) 0.25t 3 − 1.5t 2 + 3t + 0.25, for times in the interval 0 ≤ t ≤ 2. The graph of this function is also given in each of the three diagrams in Figure 4.2.2. mph mph mph 3 3 3 y = v(t) y = v(t) y = v(t) 2 2 2 1 A4 1 B4 1 C4 A3 B3 C3 A2 B2 C2 hrs B1 hrs C1 hrs A1 1 2 1 2 1 2 Figure 4.2.2: Three approaches to estimating the area under y v(t) on the interval [0, 2]. Note that in each diagram, we use four rectangles to estimate the area under y v(t) on the interval [0, 2], but the method by which the four rectangles’ respective heights are decided varies among the three individual graphs. a. How are the heights of rectangles in the left-most diagram being chosen? Ex- plain, and hence determine the value of S A1 + A2 + A3 + A4 by evaluating the function y v(t) at appropriately chosen values and observ- ing the width of each rectangle. Note, for example, that 1 1 A3 v(1) · 2 · 1. 2 2 b. Explain how the heights of rectangles are being chosen in the middle diagram and ﬁnd the value of T B1 + B2 + B3 + B4 . c. Likewise, determine the pattern of how heights of rectangles are chosen in the right-most diagram and determine U C1 + C2 + C3 + C4 . d. Of the estimates S, T, and U, which do you think is the best approximation of D, the total distance the person traveled on [0, 2]? Why? 216 4.2 Riemann Sums 4.2.1 Sigma Notation We have used sums of areas of rectangles to approximate the area under a curve. Intuitively, we expect that using a larger number of thinner rectangles will provide a better estimate for the area. Consequently, we anticipate dealing with sums of a large number of terms. To do so, we introduce sigma notation, named for the Greek letter Σ, which is the capital letter S in the Greek alphabet. For example, say we are interested in the sum 1 + 2 + 3 + · · · + 100, the sum of the ﬁrst 100 natural numbers. In sigma notation we write ∑ 100 k 1 + 2 + 3 + · · · + 100. k1 ∑ We read the symbol 100k1 k as “the sum from k equals 1 to 100 of k.” The variable k is called the index of summation, and any letter can be used for this variable. The pattern in the terms of the sum is denoted by a function of the index; for example, ∑ 10 (k 2 + 2k) (12 + 2 · 1) + (22 + 2 · 2) + (32 + 2 · 3) + · · · + (102 + 2 · 10), k1 and more generally, ∑ n f (k) f (1) + f (2) + · · · + f (n). k1 Sigma notation allows us to vary easily the function being used to describe the terms in the sum, and to adjust the number of terms in the sum simply by changing the value of n. We test our understanding of this new notation in the following activity. Activity 4.2.2. For each sum written in sigma notation, write the sum long-hand and evaluate the sum to ﬁnd its value. For each sum written in expanded form, write the sum in ∑sigma notation. a. 5k1 (k 2 + 2) d. 4 + 8 + 16 + 32 + · · · + 256 ∑6 b. i3 (2i − 1) ∑6 c. 3 + 7 + 11 + 15 + · · · + 27 e. 1 i1 2i 4.2.2 Riemann Sums When a moving body has a positive velocity function y v(t) on a given interval [a, b], the area under the curve over the interval gives the total distance the body travels on [a, b]. We are also interested in ﬁnding the exact area bounded by y f (x) on an interval [a, b], 217 Chapter 4 The Deﬁnite Integral a b x0 x1 x2 ··· xi xi+1 ··· xn−1 xn △x Figure 4.2.3: Subdividing the interval [a, b] into n subintervals of equal length ∆x. regardless of the meaning or context of the function f . For now, we continue to focus on ﬁnding an accurate estimate of this area by using a sum of the areas of rectangles. Unless otherwise indicated, we assume that f is continuous and non-negative on [a, b]. The ﬁrst choice we make in such an approximation is the number of rectangles. If we desire n rectangles of equal width to subdivide the interval [a, b], then each rectangle must have width ∆x b−an . We let x 0 a, x n b, and deﬁne x i a + i∆x, so that x 1 x 0 + ∆x, x2 x0 + 2∆x, and so on, as pictured in Figure 4.2.3. We use each subinterval [x i , x i+1 ] as the base of a rectangle, and next choose the height of the rectangle on that subinterval. There are three standard choices: we can use the left endpoint of each subinterval, the right endpoint of each subinterval, or the midpoint of each. These are precisely the options encountered in Preview Activity 4.2.1 and seen in Figure 4.2.2. We next explore how these choices can be described in sigma notation. Consider an arbitrary positive function f on [a, b] with the interval subdivided as shown in Figure 4.2.3, and choose to use left endpoints. Then on each interval [x i , x i+1 ], the area of the rectangle formed is given by A i+1 f (x i ) · ∆x, as seen in Figure 4.2.4. y = f (x) A1 A2 ··· Ai+1 ··· An x0 x1 x2 xi xi+1 xn−1 xn Figure 4.2.4: Subdividing the interval [a, b] into n subintervals of equal length ∆x and approximating the area under y f (x) over [a, b] using left rectangles. 218 4.2 Riemann Sums If we let L n denote the sum of the areas of these rectangles, we see that L n A1 + A2 + · · · + A i+1 + · · · + A n f (x0 ) · ∆x + f (x 1 ) · ∆x + · · · + f (x i ) · ∆x + · · · + f (x n−1 ) · ∆x. In the more compact sigma notation, we have ∑ n−1 Ln f (x i )∆x. i0 Note that since the index of summation begins at 0 and ends at n − 1, there are indeed n terms in this sum. We call L n the left Riemann sum for the function f on the interval [a, b]. To see how the Riemann sums for right endpoints and midpoints are constructed, we con- sider Figure 4.2.5. For the sum with right endpoints, we see that the area of the rectangle on an arbitrary interval [x i , x i+1 ] is given by B i+1 f (x i+1 ) · ∆x, and that the sum of all such areas of rectangles is given by R n B1 + B2 + · · · + B i+1 + · · · + B n f (x1 ) · ∆x + f (x2 ) · ∆x + · · · + f (x i+1 ) · ∆x + · · · + f (x n ) · ∆x ∑ n f (x i )∆x. i1 We call R n the right Riemann sum for the function f on the interval [a, b]. For the sum that uses midpoints, we introduce the notation x i + x i+1 x i+1 2 so that x i+1 is the midpoint of the interval [x i , x i+1 ]. For instance, for the rectangle with area C 1 in Figure 4.2.5, we now have C 1 f (x 1 ) · ∆x. y = f (x) y = f (x) B1 B2 ··· Bi+1 ··· Bn C1 C2 ··· Ci+1 ··· Cn x0 x1 x2 xi xi+1 xn−1 xn x0 x1 x2 xi xi+1 xn−1 xn Figure 4.2.5: Riemann sums using right endpoints and midpoints. 219 Chapter 4 The Deﬁnite Integral Figure 4.2.6: A snapshot of the applet found at http://gvsu.edu/s/a9. Hence, the sum of all the areas of rectangles that use midpoints is M n C1 + C 2 + · · · + C i+1 + · · · + C n f (x1 ) · ∆x + f (x2 ) · ∆x + · · · + f (x i+1 ) · ∆x + · · · + f (x n ) · ∆x ∑ n f (x i )∆x, i1 and we say that M n is the middle Riemann sum for f on [a, b]. Thus, we have two variables to explore: the number of rectangles and the height of each rectangle. We can explore these choices dynamically, and the applet¹ found at http:// gvsu.edu/s/a9 is a particularly useful one. There we see the image shown in Figure 4.2.6, but with the opportunity to adjust the slider bars for the heights and the number of rec- tangles. By moving the sliders, we can see how the heights of the rectangles change as we consider left endpoints, midpoints, and right endpoints, as well as the impact that a larger number of narrower rectangles has on the approximation of the exact area bounded by the function and the horizontal axis. When f (x) ≥ 0 on [a, b], each of the Riemann sums L n , R n , and M n provides an estimate of the area under the curve y f (x) over the interval [a, b]. We also recall that in the context of a nonnegative velocity function y v(t), the corresponding Riemann sums approximate the distance traveled on [a, b] by a moving object with velocity function v. There is a more general way to think of Riemann sums, and that is to allow any choice of where the function is evaluated to determine the rectangle heights. Rather than saying we’ll always choose left endpoints, or always choose midpoints, we simply say that a point x ∗i+1 will be selected at random in the interval [x i , x i+1 ] (so that x i ≤ x ∗i+1 ≤ x i+1 ). The Riemann sum is then given by ∑ n f (x1∗ ) · ∆x + f (x 2∗ ) · ∆x + · · · + f (x ∗i+1 ) · ∆x + · · · + f (x ∗n ) · ∆x f (x ∗i )∆x. i1 ¹Marc Renault, Geogebra Calculus Applets. 220 4.2 Riemann Sums At http://gvsu.edu/s/a9, the applet noted earlier and referenced in Figure 4.2.6, by uncheck- ing the “relative” box at the top left, and instead checking “random,” we can easily explore the eﬀect of using random point locations in subintervals on a Riemann sum. In computa- tional practice, we most often use L n , R n , or M n , while the random Riemann sum is useful in theoretical discussions. In the following activity, we investigate several diﬀerent Riemann sums for a particular velocity function. Activity 4.2.3. Suppose that an object moving along a straight line path has its velocity in feet per second at time t in seconds given by v(t) 29 (t − 3)2 + 2. a. Carefully sketch the region whose exact area will tell you the value of the dis- tance the object traveled on the time interval 2 ≤ t ≤ 5. b. Estimate the distance traveled on [2, 5] by computing L4 , R 4 , and M4 . c. Does averaging L4 and R 4 result in the same value as M4 ? If not, what do you think the average of L4 and R 4 measures? d. For this question, think about an arbitrary function f , rather than the particular function v given above. If f is positive and increasing on [a, b], will L n over- estimate or under-estimate the exact area under f on [a, b]? Will R n over- or under-estimate the exact area under f on [a, b]? Explain. 4.2.3 When the function is sometimes negative For a Riemann sum such as ∑ n−1 Ln f (x i )∆x, i0 we can of course compute the sum even when f takes on negative values. We know that when f is positive on [a, b], a Riemann sum estimates the area bounded between f and the horizontal axis over the interval. y = f (x) y = f (x) y = f (x) A1 A3 A2 a b c d a b c d a b c d Figure 4.2.7: At left and center, two left Riemann sums for a function f that is sometimes negative; at right, the areas bounded by f on the interval [a, d]. 221 Chapter 4 The Deﬁnite Integral For the function pictured in the ﬁrst graph of Figure 4.2.7, a left Riemann sum with 12 subin- tervals over [a, d] is shown. The function is negative on the interval b ≤ x ≤ c, so at the four left endpoints that fall in [b, c], the terms f (x i )∆x are negative. This means that those four terms in the Riemann sum produce an estimate of the opposite of the area bounded by y f (x) and the x-axis on [b, c]. In the middle graph of Figure 4.2.7, we see that by increasing the number of rectangles the approximation of the area (or the opposite of the area) bounded by the curve appears to improve. In general, any Riemann sum of a continuous function f on an interval [a, b] approximates the diﬀerence between the area that lies above the horizontal axis on [a, b] and under f and the area that lies below the horizontal axis on [a, b] and above f . In the notation of Figure 4.2.7, we may say that L24 ≈ A1 − A2 + A3 , where L 24 is the left Riemann sum using 24 subintervals shown in the middle graph. A1 and A3 are the areas of the regions where f is positive, and A2 is the area where f is negative. We will call the quantity A1 − A2 + A3 the net signed area bounded by f over the interval [a, d], where by the phrase “signed area” we indicate that we are attaching a minus sign to the areas of regions that fall below the horizontal axis. Finally, we recall that if the function f represents the velocity of a moving object, the sum of the areas bounded by the curve tells us the total distance traveled over the relevant time interval, while the net signed area bounded by the curve computes the object’s change in position on the interval. Activity 4.2.4. Suppose that an object moving along a straight line path has its velocity v (in feet per second) at time t (in seconds) given by 1 2 7 v(t) t − 3t + . 2 2 a. Compute M5 , the middle Riemann sum, for v on the time interval [1, 5]. Be sure to clearly identify the value of ∆t as well as the locations of t0 , t1 , · · ·, t5 . In addi- tion, provide a careful sketch of the function and the corresponding rectangles that are being used in the sum. b. Building on your work in (a), estimate the total change in position of the object on the interval [1, 5]. c. Building on your work in (a) and (b), estimate the total distance traveled by the object on [1, 5]. d. Use appropriate computing technology² to compute M10 and M20 . What exact value do you think the middle sum eventually approaches as n increases with- out bound? What does that number represent in the physical context of the overall problem? ²For instance, consider the applet at http://gvsu.edu/s/a9 and change the function and adjust the locations of the blue points that represent the interval endpoints a and b. 222 4.2 Riemann Sums 4.2.4 Summary • A Riemann sum is simply a sum of products of the form f (x ∗i )∆x that estimates the area between a positive function and the horizontal axis over a given interval. If the function is sometimes negative on the interval, the Riemann sum estimates the diﬀer- ence between the areas that lie above the horizontal axis and those that lie below the axis. • The three most common types of Riemann sums are left, right, and middle sums, but we can also work with a more general Riemann sum. The only diﬀerence among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. For a left Riemann sum, we evaluate the function at the left endpoint of each subinterval, while for right and mid- dle sums, we use right endpoints and midpoints, respectively. • The left, right, and middle Riemann sums are denoted L n , R n , and M n , with formulas ∑ n−1 L n f (x0 )∆x + f (x1 )∆x + · · · + f (x n−1 )∆x f (x i )∆x, i0 ∑n R n f (x1 )∆x + f (x2 )∆x + · · · + f (x n )∆x f (x i )∆x, i1 ∑n M n f (x 1 )∆x + f (x 2 )∆x + · · · + f (x n )∆x f (x i )∆x, i1 where x 0 a, x i a + i∆x, and x n b, using ∆x b−a n . For the midpoint sum, x i (x i−1 + x i )/2. 223 Chapter 4 The Deﬁnite Integral 4.2.5 Exercises 1. Evaluating Riemann sums for a quadratic function. The rectangles in the graph below −x 2 illustrate a left endpoint Riemann sum for f (x) + 2x on the interval [3, 7]. 4 The value of this left endpoint Riemann sum is , and this Riemann sum is (□ an overestimate of □ equal to □ an underestimate of □ there is ambiguity) the area of the region enclosed by y f (x), the x-axis, and the vertical lines x 3 and x 7. Left endpoint Riemann sum for y −x4 + 2x on [3, 7] 2 The rectangles in the graph below illustrate a right endpoint Riemann sum for f (x) −x 2 + 2x on the interval [3, 7]. 4 The value of this right endpoint Riemann sum is , and this Riemann sum is (□ an overestimate of □ equal to □ an underestimate of □ there is ambiguity) the area of the region enclosed by y f (x), the x-axis, and the vertical lines x 3 and x 7. Right endpoint Riemann sum for y −x4 + 2x on [3, 7] 2 224 4.2 Riemann Sums 2. Estimating distance traveled with a Riemann sum from data. Your task is to estimate how far an object traveled during the time interval 0 ≤ t ≤ 8, but you only have the following data about the velocity of the object. time (sec) 0 1 2 3 4 5 6 7 8 velocity (feet/sec) -4 -2 -3 1 2 3 2 3 4 To get an idea of what the velocity function might look like, you pick up a black pen, plot the data points, and connect them by curves. Your sketch looks something like the black curve in the graph below. Left endpoint approximation You decide to use a left endpoint Riemann sum to estimate the total displacement. So, you pick up a blue pen and draw rectangles whose height is determined by the velocity measurement at the left endpoint of each one-second interval. By using the left end- point Riemann sum as an approximation, you are assuming that the actual velocity is approximately constant on each one-second interval (or, equivalently, that the actual acceleration is approximately zero on each one-second interval), and that the velocity and acceleration have discontinuous jumps every second. This assumption is proba- bly incorrect because it is likely that the velocity and acceleration change continuously over time. However, you decide to use this approximation anyway since it seems like a reasonable approximation to the actual velocity given the limited amount of data. (A) Using the left endpoint Riemann sum, ﬁnd approximately how far the object trav- eled. Using the same data, you also decide to estimate how far the object traveled using a right endpoint Riemann sum. So, you sketch the curve again with a black pen, and draw rectangles whose height is determined by the velocity measurement at the right endpoint of each one-second interval. 225 Chapter 4 The Deﬁnite Integral Right endpoint approximation (B) Using the right endpoint Riemann sum, ﬁnd approximately how far the object trav- eled. 3. Writing basic Riemann sums. On a sketch of y e x , represent the left Riemann sum ∫1 with n 2 approximating 0 e x dx. Write out the terms of the sum, but do not evaluate it. ∫1 On another sketch, represent the right Riemann sum with n 2 approximating 0 e x dx. Write out the terms of the sum, but do not evaluate it. Which sum is an overestimate? Which sum is an underestimate? 4. Consider the function f (x) 3x + 4. a. Compute M4 for y f (x) on the interval [2, 5]. Be sure to clearly identify the value of ∆x, as well as the locations of x 0 , x1 , . . . , x4 . Include a careful sketch of the function and the corresponding rectangles being used in the sum. b. Use a familiar geometric formula to determine the exact value of the area of the region bounded by y f (x) and the x-axis on [2, 5]. c. Explain why the values you computed in (a) and (b) turn out to be the same. Will this be true if we use a number diﬀerent than n 4 and compute M n ? Will L4 or R 4 have the same value as the exact area of the region found in (b)? d. Describe the collection of functions 1 for which it will always be the case that M n , regardless of the value of n, gives the exact net signed area bounded between the function 1 and the x-axis on the interval [a, b]. 5. Let S be the sum given by S ((1.4)2 + 1) · 0.4 + ((1.8)2 + 1) · 0.4 + ((2.2)2 + 1) · 0.4 + ((2.6)2 + 1) · 0.4 + ((3.0)2 + 1) · 0.4. a. Assume that S is a right Riemann sum. For what function f and what interval [a, b] is S this function’s Riemann sum? Why? b. How does your answer to (a) change if S is a left Riemann sum? a middle Riemann sum? 226 4.2 Riemann Sums c. Suppose that S really is a right Riemann sum. What is geometric quantity does S approximate? d. Use sigma notation to write a new sum R that is the right Riemann sum for the same function, but that uses twice as many subintervals as S. 6. A car traveling along a straight road is braking and its velocity is measured at several diﬀerent points in time, as given in the following table. seconds, t 0 0.3 0.6 0.9 1.2 1.5 1.8 Velocity in ft/sec, v(t) 100 88 74 59 40 19 0 Table 4.2.8: Data for the braking car. a. Plot the given data on a set of axes with time on the horizontal axis and the ve- locity on the vertical axis. b. Estimate the total distance traveled during the car the time brakes using a middle Riemann sum with 3 subintervals. c. Estimate the total distance traveled on [0, 1.8] by computing L6 , R 6 , and 12 (L6 +R 6 ). d. Assuming that v(t) is always decreasing on [0, 1.8], what is the maximum possi- ble distance the car traveled before it stopped? Why? 7. The rate at which pollution escapes a scrubbing process at a manufacturing plant in- creases over time as ﬁlters and other technologies become less eﬀective. For this par- ticular example, assume that the rate of pollution (in tons per week) is given by the function r that is pictured in Figure 4.2.9. a. Use the graph to estimate the value of M4 on the interval [0, 4]. tons/week b. What is the meaning of M4 in terms 4 of the pollution discharged by the y = r(t) plant? 3 c. Suppose that r(t) 0.5e 0.5t . Use 2 this formula for r to compute L5 on [0, 4]. 1 d. Determine an upper bound on the weeks total amount of pollution that can escape the plant during the pic- 1 2 3 4 tured four week time period that is accurate within an error of at most one ton of pollution. Figure 4.2.9: The rate, r(t), of pollution in tons per week. 227 Chapter 4 The Deﬁnite Integral 4.3 The Deﬁnite Integral Motivating Questions • How does increasing the number of subintervals aﬀect the accuracy of the approxi- mation generated by a Riemann sum? • What is the deﬁnition of the deﬁnite integral of a function f over the interval [a, b]? • What does the deﬁnite integral measure exactly, and what are some of the key prop- erties of the deﬁnite integral? In Figure 4.3.1, we see evidence that increasing the number of rectangles in a Riemann sum improves the accuracy of the approximation of the net signed area bounded by the given function. y = f (x) y = f (x) y = f (x) A1 A3 A2 a b c d a b c d a b c d Figure 4.3.1: At left and center, two left Riemann sums for a function f that is sometimes negative; at right, the exact areas bounded by f on the interval [a, d]. We therefore explore the natural idea of allowing the number of rectangles to increase with- out bound. In an eﬀort to compute the exact net signed area we also consider the diﬀerences among left, right, and middle Riemann sums and the diﬀerent results they generate as the value of n increases. We begin with functions that are exclusively positive on the interval under consideration. Preview Activity 4.3.1. Consider the applet found at http://gvsu.edu/s/a9¹. There, you will initially see the situation shown in Figure 4.3.2. Note that the value of the chosen Riemann sum is displayed next to the word “rela- tive,” and that you can change the type of Riemann sum being computed by dragging the point on the slider bar below the phrase “sample point placement.” 228 4.3 The Deﬁnite Integral Figure 4.3.2: A right Riemann sum with 10 subintervals for the function x2 f (x) sin(2x) − 10 + 3 on the interval [1, 7]. The value of the sum is R 10 4.90595. Explore to see how you can change the window in which the function is viewed, as well as the function itself. You can set the minimum and maximum values of x by clicking and dragging on the blue points that set the endpoints; you can change the function by typing a new formula in the “f(x)” window at the bottom; and you can adjust the overall window by “panning and zooming” by using the Shift key and the scrolling feature of your mouse. More information on how to pan and zoom can be found at http://gvsu.edu/s/Fl. Work accordingly to adjust the applet so that it uses a left Riemann sum with n 5 subintervals for the function is f (x) 2x + 1. You should see the updated ﬁgure shown in Figure 4.3.3. Then, answer the following questions. a. Update the applet (and view window, as needed) so that the function being considered is f (x) 2x + 1 on [1, 4], as directed above. For this function on this interval, compute L n , M n , R n for n 5, n 25, and n 100. What appears to be the exact area bounded by f (x) 2x + 1 and the x-axis on [1, 4]? b. Use basic geometry to determine the exact area bounded by f (x) 2x + 1 and the x-axis on [1, 4]. c. Based on your work in (a) and (b), what do you observe occurs when we increase the number of subintervals used in the Riemann sum? d. Update the applet to consider the function f (x) x 2 + 1 on the interval [1, 4] (note that you need to enter “x ^ 2 + 1” for the function formula). Use the 229 Chapter 4 The Deﬁnite Integral applet to compute L n , M n , R n for n 5, n 25, and n 100. What do you conjecture is the exact area bounded by f (x) x 2 + 1 and the x-axis on [1, 4]? e. Why can we not compute the exact value of the area bounded by f (x) x 2 + 1 and the x-axis on [1, 4] using a formula like we did in (b)? Figure 4.3.3: A left Riemann sum with 5 subintervals for the function f (x) 2x + 1 on the interval [1, 4]. The value of the sum is L5 16.2. 4.3.1 The deﬁnition of the deﬁnite integral In Preview Activity 4.3.1, we saw that as the number of rectangles got larger and larger, the values of L n , M n , and R n all grew closer and closer to the same value. It turns out that this occurs for any continuous function on an interval [a, b], and also for a Riemann sum using any point x ∗i+1 in the interval [x i , x i+1 ]. Thus, as we let n → ∞, it doesn’t really matter where we choose to evaluate the function within a given subinterval, because ∑ n lim L n lim R n lim M n lim f (x ∗i )∆x. n→∞ n→∞ n→∞ n→∞ i1 That these limits always exist (and share the same value) when f is continuous² allows us to make the following deﬁnition. ¹Marc Renault, Shippensburg University, Geogebra Applets for Calclulus, http://gvsu.edu/s/5p. ²It turns out that a function need not be continuous in order to have a deﬁnite integral. For our purposes, we assume that the functions we consider are continuous on the interval(s) of interest. It is straightforward to see that any function that is piecewise continuous on an interval of interest will also have a well-deﬁned deﬁnite integral. 230 4.3 The Deﬁnite Integral y = f (x) A1 A3 A2 a b c d Figure 4.3.5: A continuous function f on the interval [a, d]. Deﬁnition 4.3.4 The deﬁnite integral of a continuous function f on the interval [a, b], de- ∫b noted a f (x) dx, is the real number given by ∫ b ∑ n f (x) dx lim f (x ∗i )∆x, a n→∞ i1 where ∆x b−a n , x i a + i∆x (for i 0, . . . , n), and x ∗i satisﬁes x i−1 ≤ x ∗i ≤ x i (for i 1, . . . , n). ∫ We call the symbol the integral sign, the values a and b the limits of integration, and the ∫b function f the integrand. The process of determining the real number a f (x) dx is called evaluating the deﬁnite integral. While there are several diﬀerent interpretations of the deﬁnite ∫b integral, for now the most important is that a f (x) dx measures the net signed area bounded by y f (x) and the x-axis on the interval [a, b]. For example, if f is the function pictured in Figure 4.3.5, and A1 , A2 , and A3 are the exact areas bounded by f and the x-axis on the respective intervals [a, b], [b, c], and [c, d], then ∫ b ∫ c f (x) dx A1 , f (x) dx −A2 , a b ∫ d f (x) dx A3 , c ∫ d and f (x) dx A1 − A2 + A3 . a We can also use deﬁnite integrals to express the change in position and the distance traveled by a moving object. If v is a velocity function on an interval [a, b], then the change in position 231 Chapter 4 The Deﬁnite Integral f (x) = 2x + 1 9 3 R4 1 (2x + 1) dx 1 4 Figure 4.3.6: The area bounded by f (x) 2x + 1 and the x-axis on the interval [1, 4]. of the object, s(b) − s(a), is given by ∫ b s(b) − s(a) v(t) dt. a ∫b If the velocity function is nonnegative on [a, b], then a v(t) dt tells us the distance the object traveled. If the velocity is sometimes negative on [a, b], we can use deﬁnite integrals to ﬁnd the areas bounded by the function on each interval where v does not change sign, and the sum of these areas will tell us the distance the object traveled. To compute the value of a deﬁnite integral from the deﬁnition, we have to take the limit of a sum. While this is possible to do in select circumstances, it is also tedious and time- consuming, and does not oﬀer much additional insight into the meaning or interpretation of the deﬁnite integral. Instead, in Section 4.4, we will learn the Fundamental Theorem of Calculus, which provides a shortcut for evaluating a large class of deﬁnite integrals. This will enable us to determine the exact net signed area bounded by a continuous function and the x-axis in many circumstances. For now, our goal is to understand the meaning and properties of the deﬁnite integral, rather than to compute its value. To do this, we will rely on the net signed area interpretation of the deﬁnite integral. So we will use as examples curves that produce regions whose areas we can compute exactly through area formulas. We can thus compute the exact value of the corresponding integral. ∫4 For instance, if we wish to evaluate the deﬁnite integral 1 (2x + 1) dx, we observe that the region bounded by this function and the x-axis is the trapezoid shown in Figure 4.3.6. By the formula for the area of a trapezoid, A 12 (3 + 9) · 3 18, so ∫ 4 (2x + 1) dx 18. 1 232 4.3 The Deﬁnite Integral Activity 4.3.2. Use known geometric formulas and the net signed area interpretation of the deﬁnite integral to evaluate each of the deﬁnite integrals below. ∫1 a. 0 3x dx ∫4 b. −1 (2 − 2x) dx ∫1 √ c. −1 1 − x 2 dx ∫4 d. −3 1(x) dx, where 1 is the function pictured in Figure 4.3.7. Assume that each portion of 1 is either part of a line or part of a circle. y = g(x) 1 -3 -2 -1 1 2 3 4 -1 Figure 4.3.7: A function 1 that is piecewise deﬁned; each piece of the function is part of a circle or part of a line. 4.3.2 Some properties of the deﬁnite integral Regarding the deﬁnite integral of a function f over an interval [a, b] as the net signed area bounded by f and the x-axis, we discover several standard properties of the deﬁnite integral. It is helpful to remember that the deﬁnite integral is deﬁned in terms of Riemann sums, which consist of the areas of rectangles. ∫a For any real number a and the deﬁnite integral a f (x) dx it is evident that no area is en- closed, because the interval begins and ends with the same point. Hence, ∫a If f is a continuous function and a is a real number, then a f (x) dx 0. Next, we consider the result of subdividing the interval of integration. In Figure 4.3.8, we see that ∫ b ∫ c f (x) dx A1 , f (x) dx A2 , a b ∫ c and f (x) dx A1 + A2 , a 233 Chapter 4 The Deﬁnite Integral which illustrates the following general rule. y = f (x) A1 A2 a b c Figure 4.3.8: The area bounded by y f (x) on the interval [a, c]. If f is a continuous function and a, b, and c are real numbers, then ∫ c ∫ b ∫ c f (x) dx f (x) dx + f (x) dx. a a b While this rule is easy to see if a < b < c, it in fact holds in general for any values of a, b, and c. Another property of the deﬁnite integral states that if we reverse the order of the limits of integration, we change the sign of the integral’s value. If f is a continuous function and a and b are real numbers, then ∫ a ∫ b f (x) dx − f (x) dx. b a This result makes sense because if we integrate from a to b, then in the deﬁning Riemann sum we set ∆x b−an , while if we integrate from b to a, we have ∆x n − n , and this a−b b−a is the only change in the sum used to deﬁne the integral. There are two additional useful properties of the deﬁnite integral. When we worked with derivative rules in Chapter 2, we formulated the Constant Multiple Rule and the Sum Rule. Recall that the Constant Multiple Rule says that if f is a diﬀerentiable function and k is a constant, then d [k f (x)] k f ′(x), dx 234 4.3 The Deﬁnite Integral and the Sum Rule says that if f and 1 are diﬀerentiable functions, then d [ f (x) + 1(x)] f ′(x) + 1 ′(x). dx These rules are useful because they allow to deal individually with the simplest parts of certain functions by taking advantage of addition and multiplying by a constant. In other words, the process of taking the derivative respects addition and multiplying by constants in the simplest possible way. It turns out that similar rules hold for the deﬁnite integral. First, let’s consider the functions pictured in Figure 4.3.9. B = 2 f (xi )△x y = 2 f (x) A = f (xi )△x y = f (x) B A a xi xi+1 b a xi xi+1 b Figure 4.3.9: The areas bounded by y f (x) and y 2 f (x) on [a, b]. Because multiplying the function by 2 doubles its height at every x-value, we see that the height of each rectangle in a left Riemann sum is doubled, f (x i ) for the original function, versus 2 f (x i ) in the doubled function. For the areas A and B, it follows B 2A. As this is true regardless of the value of n or the type of sum we use, we see that in the limit, the area of the red region bounded by y 2 f (x) will be twice the area of the blue region bounded by y f (x). As there is nothing special about the value 2 compared to an arbitrary constant k, the following general principle holds. Constant Multiple Rule. If f is a continuous function and k is any real number, then ∫ b ∫ b k · f (x) dx k f (x) dx. a a We see a similar situation with the sum of two functions f and 1. 235 Chapter 4 The Deﬁnite Integral C = ( f (xi ) + g(xi ))△x f +g A = f (xi )△x f B = g(xi )△x C A g B a xi xi+1 b a xi xi+1 b a xi xi+1 b Figure 4.3.10: The areas bounded by y f (x) and y 1(x) on [a, b], as well as the area bounded by y f (x) + 1(x). If we take the sum of two functions f and 1 at every point in the interval, the height of the function f + 1 is given by ( f + 1)(x i ) f (x i ) + 1(x i ). Hence, for the pictured rectangles with areas A, B, and C, it follows that C A + B. Because this will occur for every such rectangle, in the limit the area of the gray region will be the sum of the areas of the blue and red regions. In terms of deﬁnite integrals, we have the following general rule. Sum Rule. If f and 1 are continuous functions, then ∫ b ∫ b ∫ b [ f (x) + 1(x)] dx f (x) dx + 1(x) dx. a a a The Constant Multiple and Sum Rules can be combined to say that for any continuous func- tions f and 1 and any constants c and k, ∫ b ∫ b ∫ b [c f (x) ± k1(x)] dx c f (x) dx ± k 1(x) dx. a a a Activity 4.3.3. Suppose that the following information is known about the functions f , 1, x 2 , and x 3 : ∫2 ∫5 • 0 f (x) dx −3; 2 f (x) dx 2 ∫2 ∫5 • 0 1(x) dx 4; 2 1(x) dx −1 ∫2 ∫5 • 0 x 2 dx 83 ; 2 x 2 dx 117 3 ∫2 ∫5 • 0 x 3 dx 4; 2 x 3 dx 609 4 Use the provided information and the rules discussed in the preceding section to evaluate each of the following deﬁnite integrals. 236 4.3 The Deﬁnite Integral ∫2 ∫5 a. 5 f (x) dx d. 2 (3x 2 − 4x 3 ) dx ∫5 b. 0 1(x) dx ∫5 ∫0 c. 0 ( f (x) + 1(x)) dx e. 5 (2x 3 − 71(x)) dx 4.3.3 How the deﬁnite integral is connected to a function’s average value One of the most valuable applications of the deﬁnite integral is that it provides a way to discuss the average value of a function, even for a function that takes on inﬁnitely many values. Recall that if we wish to take the average of n numbers y1 , y2 , . . ., y n , we compute y1 + y2 + · · · + y n AV G . n Since integrals arise from Riemann sums in which we add n values of a function, it should not be surprising that evaluating an integral is similar to averaging the output values of a function. Consider, for instance, the right Riemann sum R n of a function f , which is given by R n f (x1 )∆x + f (x2 )∆x + · · · + f (x n )∆x ( f (x 1 ) + f (x2 ) + · · · + f (x n ))∆x. Since ∆x b−a n , we can thus write b−a R n ( f (x1 ) + f (x 2 ) + · · · + f (x n )) · n f (x 1 ) + f (x2 ) + · · · + f (x n ) (b − a) . (4.3.1) n We see that the right Riemann sum with n subintervals is just the length of the interval (b −a) times the average of the n function values found at the right endpoints. And just as with our eﬀorts to compute area, the larger the value of n we use, the more accurate our average will be. Indeed, we will deﬁne the average value of f on [a, b] to be f (x1 ) + f (x2 ) + · · · + f (x n ) fAVG[a,b] lim . n→∞ n But we also know that for any continuous function f on [a, b], taking the limit of a Riemann ∫b sum leads precisely to the deﬁnite integral. That is, limn→∞ R n a f (x) dx, and thus taking the limit as n → ∞ in Equation (4.3.1), we have that ∫ b f (x) dx (b − a) · fAVG[a,b] . (4.3.2) a Solving Equation (4.3.2) for fAVG[a,b] , we have the following general principle. 237 Chapter 4 The Deﬁnite Integral Average value of a function. If f is a continuous function on [a, b], then its average value on [a, b] is given by the formula ∫ b 1 fAVG[a,b] · f (x) dx. b−a a Equation (4.3.2) tells us another way to interpret the deﬁnite integral: the deﬁnite integral of a function f from a to b is the length of the interval (b − a) times the average value of the function on the interval. In addition, when the function f is nonnegative on [a, b], Equa- tion (4.3.2) has a natural visual interpretation. y = f (x) y = f (x) y = f (x) fAVG[a,b] A2 A1 Rb a f (x) dx (b − a) · fAVG[a,b] a b a b a b Figure 4.3.11: A function y f (x), the area it bounds, and its average value on [a, b]. ∫b Consider Figure 4.3.11, where we see at left the shaded region whose area is a f (x) dx, at center the shaded rectangle whose dimensions are (b − a) by fAVG[a,b] , and at right these two ﬁgures superimposed. Note that in dark green we show the horizontal line y fAVG[a,b] . Thus, the area of the green rectangle is given by (b − a) · fAVG[a,b] , which is precisely the ∫b value of a f (x) dx. The area of the blue region in the left ﬁgure is the same as the area of the green rectangle in the center ﬁgure. We can also observe that the areas A1 and A2 in the rightmost ﬁgure appear to be equal. Thus, knowing the average value of a function enables us to construct a rectangle whose area is the same as the value of the deﬁnite integral of the function on the interval. The java applet³ at http://gvsu.edu/s/az provides an opportunity to explore how the average value of the function changes as the interval changes, through an image similar to that found in Figure 4.3.11. √ Activity 4.3.4. Suppose that v(t) 4 − (t − 2)2 tells us the instantaneous velocity of a moving object on the interval 0 ≤ t ≤ 4, where t is measured in minutes and v is measured in meters per minute. √ a. Sketch an accurate graph of y v(t). What kind of curve is y 4 − (t − 2)2 ? ∫4 b. Evaluate 0 v(t) dt exactly. ³David Austin, http://gvsu.edu/s/5r. 238 4.3 The Deﬁnite Integral c. In terms of the physical problem of the moving object with velocity v(t), what ∫4 is the meaning of 0 v(t) dt? Include units on your answer. d. Determine the exact average value of v(t) on [0, 4]. Include units on your answer. e. Sketch a rectangle whose base is the line segment from t 0 to t 4 on the ∫4 t-axis such that the rectangle’s area is equal to the value of 0 v(t) dt. What is the rectangle’s exact height? f. How can you use the average value you found in (d) to compute the total dis- tance traveled by the moving object over [0, 4]? 4.3.4 Summary • Any Riemann sum of a continuous function f on an interval [a, b] provides an estimate of the net signed area bounded by the function and the horizontal axis on the interval. Increasing the number of subintervals in the Riemann sum improves the accuracy of this estimate, and letting the number of subintervals increase without bound results in the values of the corresponding Riemann sums approaching the exact value of the enclosed net signed area. • When we take the limit of Riemann sums, we arrive at what we call the deﬁnite integral ∫b of f over the interval [a, b]. In particular, the symbol a f (x) dx denotes the deﬁnite integral of f over [a, b], and this quantity is deﬁned by the equation ∫ b ∑ n f (x) dx lim f (x ∗i )∆x, a n→∞ i1 where ∆x b−a n , x i a + i∆x (for i 0, . . . , n), and x ∗i satisﬁes x i−1 ≤ x ∗i ≤ x i (for i 1, . . . , n). ∫b • The deﬁnite integral a f (x) dx measures the exact net signed area bounded by f and the horizontal axis on [a, b]; in addition, the value of the deﬁnite integral is related to ∫b what we call the average value of the function on [a, b]: fAVG[a,b] b−a · a f (x) dx. In 1 ∫b the setting where we consider the integral of a velocity function v, a v(t) dt measures the exact change in position of the moving object on [a, b]; when v is nonnegative, ∫b a v(t) dt is the object’s distance traveled on [a, b]. • The deﬁnite integral is a sophisticated sum, and thus has some of the same natural properties that ﬁnite sums have. Perhaps most important of these is how the deﬁnite integral respects sums and constant multiples of functions, which can be summarized by the rule ∫ b ∫ b ∫ b [c f (x) ± k1(x)] dx c f (x) dx ± k 1(x) dx a a a where f and 1 are continuous functions on [a, b] and c and k are arbitrary constants. 239 Chapter 4 The Deﬁnite Integral 4.3.5 Exercises 1. Evaluating deﬁnite integrals from graphical information. Use the following ﬁgure, which shows a graph of f (x) to ﬁnd each of the indicated integrals. Note that the ﬁrst area (with vertical, red shading) is 18 and the second (with oblique, black shading) is 6. ∫b A. f (x)dx ∫ ac B. b f (x)dx ∫c C. a f (x)dx ∫c D. a | f (x)|dx 2. Estimating deﬁnite integrals from a graph. Use the graph of f (x) shown below to ﬁnd the following integrals. ∫0 A. −4 f (x)dx B. If the vertical red shaded∫ area in the 6 graph has area A, estimate: −4 f (x)dx (Your estimate may be written in terms of A.) 3. Finding the average value of a linear function. Find the average value of f (x) 7x + 1 over [3, 8]. 4. Finding the average value of a function given graphically. The ﬁgure below to the left is a graph of f (x), and below to the right is 1(x). 240 4.3 The Deﬁnite Integral f (x) 1(x) (a) What is the average value of f (x) on 0 ≤ x ≤ 2? (b) What is the average value of 1(x) on 0 ≤ x ≤ 2? (c) What is the average value of f (x) · 1(x) on 0 ≤ x ≤ 2? (d) Is the following statement true? Average( f ) · Average(g) Average( f · 1) 5. Estimating a deﬁnite integral and average value from a graph. Use the ﬁgure below, which shows the graph of y f (x), to answer the following questions. ∫3 A. Estimate the integral: −3 f (x) dx B. Which of the following average values of f is larger? ⊙ Between x −3 and x 3 ⊙ Between x 0 and x 3 6. Using rules to combine known integral values. Suppose ∫ −4.5 ∫ −7.5 ∫ −4.5 f (x) dx 10, f (x) dx 8, and f (x) dx 10. −9 −9 −6 Find ∫ ∫ −6 −7.5 f (x) dx and (10 f (x) − 8) dx. −7.5 −6 241 Chapter 4 The Deﬁnite Integral 7. The velocity of an object moving along an axis is given by the piecewise linear function v that is pictured in Figure 4.3.12. Assume that the object is moving to the right when its velocity is positive, and moving to the left when its velocity is negative. Assume that the given velocity function is valid for t 0 to t 4. a. Write an expression involving deﬁ- ft/sec nite integrals whose value is the to- 2 tal change in position of the object on the interval [0, 4]. 1 b. Use the provided graph of v to sec determine the value of the total change in position on [0, 4]. 1 2 3 4 c. Write an expression involving deﬁ- -1 nite integrals whose value is the to- y = v(t) tal distance traveled by the object on [0, 4]. What is the exact value of -2 the total distance traveled on [0, 4]? d. What is the object’s exact average Figure 4.3.12: The velocity function of a velocity on [0, 4]? moving object. e. Find an algebraic formula for the object’s position function on [0, 1.5] that satisﬁes s(0) 0. 8. Suppose that the velocity of a moving object is given by v(t) t(t − 1)(t − 3), measured in feet per second, and that this function is valid for 0 ≤ t ≤ 4. a. Write an expression involving deﬁnite integrals whose value is the total change in position of the object on the interval [0, 4]. b. Use appropriate technology (such as http://gvsu.edu/s/a9⁴) to compute Rie- mann sums to estimate the object’s total change in position on [0, 4]. Work to ensure that your estimate is accurate to two decimal places, and explain how you know this to be the case. c. Write an expression involving deﬁnite integrals whose value is the total distance traveled by the object on [0, 4]. d. Use appropriate technology to compute Riemann sums to estimate the object’s total distance travelled on [0, 4]. Work to ensure that your estimate is accurate to two decimal places, and explain how you know this to be the case. e. What is the object’s average velocity on [0, 4], accurate to two decimal places? 9. Consider the graphs of two functions f and 1 that are provided in Figure 4.3.13. Each piece of f and 1 is either part of a straight line or part of a circle. ⁴Marc Renault, Shippensburg University. 242 4.3 The Deﬁnite Integral y = g(x) 2 2 y = f (x) 1 1 1 2 3 4 1 2 3 4 -1 -1 -2 -2 Figure 4.3.13: Two functions f and 1. ∫1 a. Determine the exact value of 0 [ f (x) + 1(x)] dx. ∫4 b. Determine the exact value of 1 [2 f (x) − 31(x)] dx. c. Find the exact average value of h(x) 1(x) − f (x) on [0, 4]. d. For what constant c does the following equation hold? ∫ 4 ∫ 4 c dx [ f (x) + 1(x)] dx 0 0 10. Let f (x) 3 − x 2 and 1(x) 2x 2 . a. On the interval [−1, 1], sketch a labeled graph of y f (x) and write a deﬁnite integral whose value is the exact area bounded by y f (x) on [−1, 1]. b. On the interval [−1, 1], sketch a labeled graph of y 1(x) and write a deﬁnite integral whose value is the exact area bounded by y 1(x) on [−1, 1]. c. Write an expression involving a diﬀerence of deﬁnite integrals whose value is the exact area that lies between y f (x) and y 1(x) on [−1, 1]. d. Explain why your expression in (c) has the same value as the single integral ∫1 −1 [ f (x) − 1(x)] dx. e. Explain why, in general, if p(x) ≥ q(x) for all x in [a, b], the exact area between y p(x) and y q(x) is given by ∫ b [p(x) − q(x)] dx. a 243 Chapter 4 The Deﬁnite Integral 4.4 The Fundamental Theorem of Calculus Motivating Questions • How can we ﬁnd the exact value of a deﬁnite integral without taking the limit of a Riemann sum? • What is the statement of the Fundamental Theorem of Calculus, and how do anti- derivatives of functions play a key role in applying the theorem? • What is the meaning of the deﬁnite integral of a rate of change in contexts other than when the rate of change represents velocity? Much of our work in Chapter 4 has been motivated by the velocity-distance problem: if we know the instantaneous velocity function, v(t), for a moving object on a given time interval [a, b], can we determine the distance it traveled on [a, b]? If the velocity function is nonneg- ative on [a, b], the area bounded by y v(t) and the t-axis on [a, b] is equal to the distance ∫b traveled. This area is also the value of the deﬁnite integral a v(t) dt. If the velocity is some- times negative, the total area bounded by the velocity function still tells us distance traveled, while the net signed area tells us the object’s change in position. For instance, for the velocity function in Figure 4.4.1, the total distance D traveled by the moving object on [a, b] is D A1 + A2 + A3 , and the total change in the object’s position is s(b) − s(a) A1 − A2 + A3 . The areas A1 , A2 , and A3 are each given by deﬁnite integrals, which may be computed by limits of Riemann sums (and in special circumstances by geometric formulas). y = v(t) A1 A3 a A2 b Figure 4.4.1: A velocity function that is sometimes negative. 244 4.4 The Fundamental Theorem of Calculus We turn our attention to an alternate approach. Preview Activity 4.4.1. A student with a third ﬂoor dormitory window 32 feet oﬀ the ground tosses a water balloon straight up in the air with an initial velocity of 16 feet per second. It turns out that the instantaneous velocity of the water balloon is given by v(t) −32t + 16, where v is measured in feet per second and t is measured in seconds. a. Let s(t) represent the height of the water balloon above ground at time t, and note that s is an antiderivative of v. That is, v is the derivative of s: s ′(t) v(t). Find a formula for s(t) that satisﬁes the initial condition that the balloon is tossed from 32 feet above ground. In other words, make your formula for s satisfy s(0) 32. b. When does the water balloon reach its maximum height? When does it land? c. Compute s( 12 ) − s(0), s(2) − s( 12 ), and s(2) − s(0). What do these represent? d. What is the total vertical distance traveled by the water balloon from the time it is tossed until the time it lands? e. Sketch a graph of the velocity function y v(t) on the time interval [0, 2]. What is the total net signed area bounded by y v(t) and the t-axis on [0, 2]? Answer this question in two ways: ﬁrst by using your work above, and then by using a familiar geometric formula to compute areas of certain relevant regions. 4.4.1 The Fundamental Theorem of Calculus Suppose we know the position function s(t) and the velocity function v(t) of an object mov- ing in a straight line, and for the moment let us assume that v(t) is positive on [a, b]. Then, as shown in Figure 4.4.2, we know two diﬀerent ways to compute the distance, D, the object travels: one is that D s(b) − s(a), the object’s change in position. The other is the area ∫b under the velocity curve, which is given by the deﬁnite integral, so D a v(t) dt. Since both of these expressions tell us the distance traveled, it follows that they are equal, so ∫ b s(b) − s(a) v(t) dt. (4.4.1) a Equation (4.4.1) holds even when velocity is sometimes negative, because s(b) − s(a),the object’s change in position, is also measured by the net signed area on [a, b] which is given ∫b by a v(t) dt. Perhaps the most powerful fact Equation (4.4.1) reveals is that we can compute the integral’s value if we can ﬁnd a formula for s. Remember, s and v are related by the fact that v is the derivative of s, or equivalently that s is an antiderivative of v. Example 4.4.3 Determine the exact distance traveled on [1, 5] by an object with velocity func- 245 Chapter 4 The Deﬁnite Integral y = v(t) Rb D= a v(t) dt = s(b) − s(a) a b Figure 4.4.2: Finding distance traveled when we know a velocity function v. tion v(t) 3t 2 + 40 feet per second. The distance traveled on the interval [1, 5] is given by ∫ 5 ∫ 5 D v(t) dt (3t 2 + 40) dt s(5) − s(1), 1 1 where s is an antiderivative of v. Now, the derivative of t 3 is 3t 2 and the derivative of 40t is 40, so it follows that s(t) t 3 + 40t is an antiderivative of v. Therefore, ∫ 5 D 3t 2 + 40 dt s(5) − s(1) 1 (5 + 40 · 5) − (13 + 40 · 1) 284 feet. 3 Note the key lesson of Example 4.4.3: to ﬁnd the distance traveled, we need to compute the area under a curve, which is given by the deﬁnite integral. But to evaluate the integral, we can ﬁnd an antiderivative, s, of the velocity function, and then compute the total change in s on the interval. In particular, we can evaluate the integral without computing the limit of a Riemann sum. It will be convenient to have a shorthand symbol for a function’s anti- derivative. For a continuous function f , we will often denote an antiderivative of f by F, so that F′(x) f (x) for all relevant x. Using the notation V in place of s (so that V is an antiderivative of v) in Equation (4.4.1), we can write ∫ b V(b) − V(a) v(t) dt. (4.4.2) a ∫b Now, to evaluate the deﬁnite integral a f (x) dx for an arbitrary continuous function f , we could certainly think of f as representing the velocity of some moving object, and x as the variable that represents time. But Equations (4.4.1) and (4.4.2) hold for any continuous ve- locity function, even when v is sometimes negative. So Equation (4.4.2) oﬀers a shortcut route to evaluating any deﬁnite integral, provided that we can ﬁnd an antiderivative of the 246 4.4 The Fundamental Theorem of Calculus 140 120 y = v(t) 100 80 60 40 D= R5 v(t) dt 1 20 = 284 1 3 5 Figure 4.4.4: The exact area of the region enclosed by v(t) 3t 2 + 40 on [1, 5]. integrand. The Fundamental Theorem of Calculus (FTC) summarizes these observations. Fundamental Theorem of Calculus. ∫b If f is a continuous function on [a, b], and F is any antiderivative of f , then a f (x) dx F(b) − F(a). A common alternate notation for F(b) − F(a) is F(b) − F(a) F(x)| ba , where we read the righthand side as “the function F evaluated from a to b.” In this notation, the FTC says that ∫ b f (x) dx F(x)| ba . a The FTC opens the door to evaluating a wide range of integrals if we can ﬁnd an antideriv- [ 3 x ] x 2 , the FTC tells us that d 1 3 ative F for the integrand f . For instance since dx ∫ 1 1 1 3 x 2 dx x 0 3 0 1 3 1 3 (1) − (0) 3 3 1 . 3 But ﬁnding an antiderivative can be far from simple; it is often diﬃcult or even impossible. While we can diﬀerentiate just about any function, even some relatively simple functions don’t have an elementary antiderivative. A signiﬁcant portion of integral calculus (which 247 Chapter 4 The Deﬁnite Integral is the main focus of second semester college calculus) is devoted to the problem of ﬁnding antiderivatives. Activity 4.4.2. Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly. For each, sketch a graph of the integrand on the relevant interval and write one sentence that explains the meaning of the value of the integral in terms ∫ 4 of the (net signed) area bounded by the curve. ∫1 a. −1 (2 − 2x) dx d. −1 x 5 dx ∫ π 2 b. 0 sin(x) dx ∫1 ∫2 c. 0 e x dx e. 0 (3x 3 − 2x 2 − e x ) dx 4.4.2 Basic antiderivatives The general problem of ﬁnding an antiderivative is diﬃcult. In part, this is due to the fact that we are trying to undo the process of diﬀerentiating, and the undoing is much more diﬃcult than the doing. For example, while it is evident that an antiderivative of f (x) sin(x) is F(x) − cos(x) and that an antiderivative of 1(x) x 2 is G(x) 13 x 3 , combinations of f and 1 can be far more complicated. Consider the functions sin(x) 5 sin(x) − 4x 2 , x 2 sin(x), , and sin(x 2 ). x2 What is involved in trying to ﬁnd an antiderivative for each? From our experience with de- rivative rules, we know that derivatives of sums and constant multiples of basic functions are simple to execute, but derivatives involving products, quotients, and composites of fa- miliar functions are more complicated. Therefore, it stands to reason that antidiﬀerentiating products, quotients, and composites of basic functions may be even more challenging. We defer our study of all but the most elementary antiderivatives to later in the text. We do note that whenever we know the derivative of a function, we have a function-derivative pair, so we also know the antiderivative of a function. For instance, since we know that d [− cos(x)] sin(x), dx we also know that F(x) − cos(x) is an antiderivative of f (x) sin(x). F and f together form a function-derivative pair. Clearly, every basic derivative rule leads us to such a pair, and thus to a known antiderivative. In Activity 4.4.3, we will construct a list of the basic antiderivatives we know at this time. Those rules will help us antidiﬀerentiate sums and constant multiples of basic functions. For example, since − cos(x) is an antiderivative of sin(x) and 13 x 3 is an antiderivative of x 2 , it follows that 4 F(x) −5 cos(x) − x 3 3 is an antiderivative of f (x) 5 sin(x) − 4x 2 , by the sum and constant multiple rules for diﬀerentiation. 248 4.4 The Fundamental Theorem of Calculus Finally, before proceeding to build a list of common functions whose antiderivatives we know, we recall that each function has more than one antiderivative. Because the derivative of any constant is zero, we may add a constant of our choice to any antiderivative. For instance, we know that G(x) 31 x 3 is an antiderivative of 1(x) x 2 . But we could also have chosen G(x) 13 x 3 + 7, since in this case as well, G′(x) x 2 . If 1(x) x 2 , we say that the general antiderivative of 1 is 1 3 G(x) x + C, 3 where C represents an arbitrary real number constant. Regardless of the formula for 1, including +C in the formula for its antiderivative G results in the most general possible antiderivative. Our current interest in antiderivatives is so that we can evaluate deﬁnite integrals by the Fundamental Theorem of Calculus. For that task, the constant C is irrelevant, and we usually omit it. To see why, consider the deﬁnite integral ∫ 1 x 2 dx. 0 For the integrand 1(x) x 2 , suppose we ﬁnd and use the general antiderivative G(x) 3 x + C. Then, by the FTC, 1 3 ∫ 1 1 1 3 x 2 dx x +C 0 3 0 ( ) ( ) 1 3 1 (1) + C − (0)3 + C 3 3 1 +C−0−C 3 1 . 3 Observe that the C-values appear as opposites in the evaluation of the integral and thus do not aﬀect the deﬁnite integral’s value. In the following activity, we work to build a list of basic functions whose antiderivatives we already know. Activity 4.4.3. Use your knowledge of derivatives of basic functions to complete Ta- ble 4.4.5 of antiderivatives. For each entry, your task is to ﬁnd a function F whose derivative is the given function f . When ﬁnished, use the FTC and the results in the table to evaluate the three given deﬁnite integrals. 249 Chapter 4 The Deﬁnite Integral given function, f (x) antiderivative, F(x) k, (k is constant) x n , n , −1 x, x > 0 1 sin(x) cos(x) sec(x) tan(x) csc(x) cot(x) sec2 (x) csc2 (x) ex a x (a > 1) 1 1+x 2 √ 1 1−x 2 Table 4.4.5: Familiar basic functions and their antiderivatives. ∫ 1 ( ) a. x 3 − x − e x + 2 dx 0 ∫ π/3 b. (2 sin(t) − 4 cos(t) + sec2 (t) − π) dt 0 ∫ 1 √ c. ( x − x 2 ) dx 0 4.4.3 The total change theorem Let us review three interpretations of the deﬁnite integral. • For a moving object with instantaneous velocity v(t), the object’s change in position on ∫b ∫b the time interval [a, b] is given by a v(t) dt, and whenever v(t) ≥ 0 on [a, b], a v(t) dt tells us the total distance traveled by the object on [a, b]. ∫b • For any continuous function f , its deﬁnite integral a f (x) dx represents the net signed area bounded by y f (x) and the x-axis on [a, b], where regions that lie below the x-axis have a minus sign associated with their area. • The value of a deﬁnite integral is linked to the average value of a function: for a con- tinuous function f on [a, b], its average value fAVG[a,b] is given by ∫ b 1 fAVG[a,b] f (x) dx. b−a a 250 4.4 The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any deﬁnite integral for which we are able to ﬁnd an antiderivative of the integrand. A slight change in perspective allows us to gain even more insight into the meaning of the deﬁnite integral. Recall Equation (4.4.2), where we wrote the Fundamental Theorem of Cal- culus for a velocity function v with antiderivative V as ∫ b V(b) − V(a) v(t) dt. a If we instead replace V with s (which represents position) and replace v with s ′ (since ve- locity is the derivative of position), Equation (4.4.2) then reads as ∫ b s(b) − s(a) s ′(t) dt. (4.4.3) a In words, this version of the FTC tells us that the total change in an object’s position function on a particular interval is given by the deﬁnite integral of the position function’s derivative over that interval. Of course, this result is not limited to only the setting of position and velocity. Writing the result in terms of a more general function f , we have the Total Change Theorem. Total Change Theorem. If f is a continuously diﬀerentiable function on [a, b] with derivative f ′, then f (b) − ∫b f (a) a f ′(x) dx. That is, the deﬁnite integral of the rate of change of a function on [a, b] is the total change of the function itself on [a, b]. The Total Change Theorem tells us more about the relationship between the graph of a func- tion and that of its derivative. Recall that heights on the graph of the derivative function are equal to slopes on the graph of the function itself. If instead we know f ′ and are seeking information about f , we can say the following: diﬀerences in heights on f correspond to net signed areas bounded by f ′. To see why this is so, consider the diﬀerence f (1) − f (0). This value is 3, because f (1) 3 and f (0) 0, but also because the net signed area bounded by y f ′(x) on [0, 1] is 3. That is, ∫ 1 f (1) − f (0) f ′(x) dx. 0 In addition to this observation about area, the Total Change Theorem enables us to answer questions about a function whose rate of change we know. Example 4.4.7 Suppose that pollutants are leaking out of an underground storage tank at a rate of r(t) gallons/day, where t is measured in days. It is conjectured that r(t) is given by the formula r(t) 0.0069t 3 − 0.125t 2 + 11.079 over a certain 12-day period. The graph of y r(t) 251 Chapter 4 The Deﬁnite Integral (2, 4) 4 4 (3, 3) 3 3 (1, 3) 2 2 3 1 1 1 3 4 (0, 0) (4, 0) 1 2 1 1 2 3 4 -1 -1 3 -2 -2 y = f ′ (x) y = f (x) -3 -3 -4 -4 Figure 4.4.6: The graphs of f ′(x) 4 − 2x (at left) and an antiderivative f (x) 4x − x 2 at right. Diﬀerences in heights on f correspond to net signed areas bounded by f ′. ∫ 10 is given in Figure 4.4.8. What is the meaning of 4 r(t) dt and what is its value? What is the average rate at which pollutants are leaving the tank on the time interval 4 ≤ t ≤ 10? gal/day 12 10 8 y = r(t) 6 4 2 days 2 4 6 8 10 12 Figure 4.4.8: The rate r(t) of pollution leaking from a tank, measured in gallons per day. ∫ 10 Solution. Since r(t) ≥ 0, the value of 4 r(t) dt is the area under the curve on the interval 252 4.4 The Fundamental Theorem of Calculus [4, 10]. A Riemann sum for this area will have rectangles with heights measured in gallons per day and widths measured in days, so the area of each rectangle will have units of gallons · days gallons. day Thus, the deﬁnite integral tells us the total number of gallons of pollutant that leak from the tank from day 4 to day 10. The Total Change Theorem tells us the same thing: if we let R(t) denote the total number of gallons of pollutant that have leaked from the tank up to day t, then R′(t) r(t), and ∫ 10 r(t) dt R(10) − R(4), 4 the number of gallons that have leaked from day 4 to day 10. To compute the exact value of the integral, we use the Fundamental Theorem of Calculus. Antidiﬀerentiating r(t) 0.0069t 3 − 0.125t 2 + 11.079, we ﬁnd that ∫ 10 10 1 4 1 0.0069t 3 − 0.125t 2 + 11.079 dt 0.0069 · t − 0.125 · t 3 + 11.079t 4 4 3 4 ≈ 44.282. Thus, approximately 44.282 gallons of pollutant leaked over the six day time period. To ﬁnd the average rate at which pollutant leaked from the tank over 4 ≤ t ≤ 10, we compute the average value of r on [4, 10]. Thus, ∫ 10 1 44.282 rAVG[4,10] r(t) dt ≈ 7.380 10 − 4 4 6 gallons per day. Activity 4.4.4. During a 40-minute workout, a person riding an exercise machine burns calories at a rate of c calories per minute, where the function y c(t) is given in Figure 4.4.9. On the interval 0 ≤ t ≤ 10, the formula for c is c(t) −0.05t 2 + t + 10, while on 30 ≤ t ≤ 40, its formula is c(t) −0.05t 2 + 3t − 30. a. What is the exact total number of calories the person burns during the ﬁrst 10 minutes of her workout? b. Let C(t) be an antiderivative of c(t). What is the meaning of C(40) − C(0) in the context of the person exercising? Include units on your answer. c. Determine the exact average rate at which the person burned calories during the 40-minute workout. d. At what time(s), if any, is the instantaneous rate at which the person is burning calories equal to the average rate at which she burns calories, on the time interval 0 ≤ t ≤ 40? 253 Chapter 4 The Deﬁnite Integral cal/min 15 y = c(t) 10 5 min 10 20 30 40 Figure 4.4.9: The rate c(t) at which a person exercising burns calories, measured in calories per minute. 4.4.4 Summary • We can ﬁnd the exact value of a deﬁnite integral without taking the limit of a Riemann sum or using a familiar area formula by ﬁnding the antiderivative of the integrand, and hence applying the Fundamental Theorem of Calculus. • The Fundamental Theorem of Calculus says that if f is a continuous function on [a, b] and F is an antiderivative of f , then ∫ b f (x) dx F(b) − F(a). a Hence, if we can ﬁnd an antiderivative for the integrand f , evaluating the deﬁnite integral comes from simply computing the change in F on [a, b]. • A slightly diﬀerent perspective on the FTC allows us to restate it as the Total Change Theorem, which says that ∫ b f ′(x) dx f (b) − f (a), a for any continuously diﬀerentiable function f . This means that the deﬁnite integral of the instantaneous rate of change of a function f on an interval [a, b] is equal to the total change in the function f on [a, b]. 254 4.4 The Fundamental Theorem of Calculus 4.4.5 Exercises 1. Finding exact displacement. The velocity function is v(t) −t 2 + 4t − 3 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−1, 5]. ∫ 4 1 2. Evaluating the deﬁnite integral of a rational function. Find the value of dx. 2 x2 3. Evaluating the deﬁnite integral of a linear function. Evaluate the deﬁnite integral ∫ 9 (4x + 10) dx. 2 4. Evaluating the deﬁnite integral of a quadratic function. Evaluate the deﬁnite integral ∫ 6 (36 − x 2 ) dx. −6 5. Simplifying an integrand before integrating. Evaluate the deﬁnite integral ∫ 8 8x 2 + 3 √ dx. 3 x 6. Evaluating the deﬁnite integral of a trigonometric function. Evaluate the deﬁnite integral ∫ π 8 sin(x) dx. 0 7. The instantaneous velocity (in meters per minute) of a moving object is given by the function v as pictured in Figure 4.4.10. Assume that on the interval 0 ≤ t ≤ 4, v(t) is given by v(t) − 14 t 3 + 32 t 2 + 1, and that on every other interval v is piecewise linear, as shown. a. Determine the exact distance traveled by the object on the time interval 0 ≤ t ≤ 4. b. What is the object’s average velocity on [12, 24]? c. At what time is the object’s acceleration greatest? d. Suppose that the velocity of the object is increased by a constant value c for all val- ues of t. What value of c will make the object’s total distance traveled on [12, 24] be 210 meters? 255 Chapter 4 The Deﬁnite Integral m/min y = v(t) 15 12 9 6 3 min 4 8 12 16 20 24 Figure 4.4.10: The velocity function of a moving body. 8. A function f is given piecewise by the formula −x 2 + 2x + 1, if 0 ≤ x < 2 f (x) −x + 3, if 2 ≤ x < 3 . x 2 − 8x + 15, if 3 ≤ x ≤ 5 a. Determine the exact value of the net signed area enclosed by f and the x-axis on the interval [2, 5]. b. Compute the exact average value of f on [0, 5]. c. Find a formula for a function 1 on 5 ≤ x ≤ 7 so that if we extend the above ∫7 deﬁnition of f so that f (x) 1(x) if 5 ≤ x ≤ 7, it follows that 0 f (x) dx 0. 9. When an aircraft attempts to climb as rapidly as possible, its climb rate (in feet per minute) decreases as altitude increases, because the air is less dense at higher altitudes. Given below is a table showing performance data for a certain single engine aircraft, giving its climb rate at various altitudes, where c(h) denotes the climb rate of the air- plane at an altitude h. h (feet) 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10,000 c (ft/min) 925 875 830 780 730 685 635 585 535 490 440 Let a new function called m(h) measure the number of minutes required for a plane at altitude h to climb the next foot of altitude. a. Determine a similar table of values for m(h) and explain how it is related to the table above. Be sure to explain the units. b. Give a careful interpretation of a function whose derivative is m(h). Describe what the input is and what the output is. Also, explain in plain English what the function tells us. c. Determine a deﬁnite integral whose value tells us exactly the number of minutes required for the airplane to ascend to 10,000 feet of altitude. Clearly explain why 256 4.4 The Fundamental Theorem of Calculus the value of this integral has the required meaning. d. Use the Riemann sum M5 to estimate the value of the integral you found in (c). Include units on your result. 10. In Chapter 1, we showed that for an object moving along a straight line with position function s(t), the object’s “average velocity on the interval [a, b]” is given by s(b) − s(a) AV[a,b] . b−a More recently in Chapter 4, we found that for an object moving along a straight line with velocity function v(t), the object’s “average value of its velocity function on [a, b]” is ∫ b 1 v AVG[a,b] v(t) dt. b−a a Are the “average velocity on the interval [a, b]” and the “average value of the velocity function on [a, b]” the same thing? Why or why not? Explain. 11. In Table 4.4.5 in Activity 4.4.3, we noted that for x > 0, the antiderivative of f (x) x1 is F(x) ln(x). Here we observe that a key diﬀerence between f (x) and F(x) is that f is deﬁned for all x , 0, while F is only deﬁned for x > 0, and see how we can actually deﬁne the antiderivative of f for all values of x. a. Suppose that x < 0, and let G(x) ln(−x). Compute G′(x). b. Explain why G is an antiderivative of f for x < 0. c. Let H(x) ln(|x|), and recall that { −x, if x < 0 |x| . x, if x ≥ 0 Explain why H(x) G(x) for x < 0 and H(x) F(x) for x > 0. d. Now discuss why we say that the antiderivative of 1 x is ln(|x|) for all x , 0. 257 Chapter 4 The Deﬁnite Integral 258 CHAPTER 5 Evaluating Integrals 5.1 Constructing Accurate Graphs of Antiderivatives Motivating Questions • Given the graph of a function’s derivative, how can we construct a completely accu- rate graph of the original function? • How many antiderivatives does a given function have? What do those antiderivatives all have in common? ∫x • Given a function f , how does the rule A(x) 0 f (t) dt deﬁne a new function A? A recurring theme in our discussion of diﬀerential calculus has been the question “Given information about the derivative of an unknown function f , how much information can we obtain about f itself?” In Activity 1.8.3, the graph of y f ′(x) was known (along with the value of f at a single point) and we endeavored to sketch a possible graph of f near the known point. In Example 3.1.7 — we investigated how the ﬁrst derivative test enables us to use information about f ′ to determine where the original function f is increasing and decreasing, as well as where f has relative extreme values. If we know a formula or graph of f ′, by computing f ′′ we can ﬁnd where the original function f is concave up and concave down. Thus, knowing f ′ and f ′′ enables us to understand the shape of the graph of f . We returned to this question in even more detail in Section 4.1. In that setting, we knew the instantaneous velocity of a moving object and worked to determine as much as possible about the object’s position function. We found connections between the net signed area under the velocity function and the corresponding change in position of the function, and the Total Change Theorem further illuminated these connections between f ′ and f , showing that the total change in the value of f over an interval [a, b] is determined by the net signed area bounded by f ′ and the x-axis on the same interval. In what follows, we explore the situation where we possess an accurate graph of the deriv- ative function along with a single value of the function f . From that information, we’d like to determine a graph of f that shows where f is increasing, decreasing, concave up, and concave down, and also provides an accurate function value at any point. Chapter 5 Evaluating Integrals Preview Activity 5.1.1. Suppose that the following information is known about a function f : the graph of its derivative, y f ′(x), is given in Figure 5.1.1. Further, assume that f ′ is piecewise linear (as pictured) and that for x ≤ 0 and x ≥ 6, f ′(x) 0. Finally, it is given that f (0) 1. y = f ′ (x) 3 3 1 1 1 3 5 1 3 5 -1 -1 -3 -3 Figure 5.1.1: At left, the graph of y f ′(x); at right, axes for plotting y f (x). a. On what interval(s) is f an increasing function? On what intervals is f decreas- ing? b. On what interval(s) is f concave up? concave down? c. At what point(s) does f have a relative minimum? a relative maximum? d. Recall that the Total Change Theorem tells us that ∫ 1 f (1) − f (0) f ′(x) dx. 0 What is the exact value of f (1)? e. Use the given information and similar reasoning to that in (d) to determine the exact value of f (2), f (3), f (4), f (5), and f (6). f. Based on your responses to all of the preceding questions, sketch a complete and accurate graph of y f (x) on the axes provided, being sure to indicate the behavior of f for x < 0 and x > 6. 260 5.1 Constructing Accurate Graphs of Antiderivatives 5.1.1 Constructing the graph of an antiderivative Preview Activity 5.1.1 demonstrates that when we can ﬁnd the exact area under the graph of a function on any given interval, it is possible to construct a graph of the function’s anti- derivative. That is, we can ﬁnd a function whose derivative is given. We can now determine not only the overall shape of the antiderivative graph, but also the actual height of the graph at any point of interest. This is a consequence of the Fundamental Theorem of Calculus: if we know a function f and the value of the antiderivative F at some starting point a, we can determine the value of ∫b F(b) via the deﬁnite integral. Since F(b) − F(a) a f (x) dx, it follows that ∫ b F(b) F(a) + f (x) dx. (5.1.1) a ∫b We can also interpret the equation F(b) − F(a) a f (x) dx in terms of the graphs of f and F as follows. On an interval [a, b], diﬀerences in heights on the graph of the antiderivative given by F(b) − F(a) correspond to the net signed area bounded by the original function on the interval [a, b], which is ∫b given by a f (x) dx. Activity 5.1.2. Suppose that the function y f (x) is given by the graph shown in Figure 5.1.2, and that the pieces of f are either portions of lines or portions of circles. In addition, let F be an antiderivative of f and say that F(0) −1. Finally, assume that for x ≤ 0 and x ≥ 7, f (x) 0. y = f (x) 1 1 2 3 4 5 6 7 -1 Figure 5.1.2: At left, the graph of y f (x). a. On what interval(s) is F an increasing function? On what intervals is F decreas- ing? b. On what interval(s) is F concave up? concave down? neither? c. At what point(s) does F have a relative minimum? a relative maximum? d. Use the given information to determine the exact value of F(x) for x 1, 2, . . . , 7. In addition, what are the values of F(−1) and F(8)? 261 Chapter 5 Evaluating Integrals e. Based on your responses to all of the preceding questions, sketch a complete and accurate graph of y F(x) on the axes provided, being sure to indicate the behavior of F for x < 0 and x > 7. Clearly indicate the scale on the vertical and horizontal axes of your graph. f. What happens if we change one key piece of information: in particular, say that G is an antiderivative of f and G(0) 0. How (if at all) would your answers to the preceding questions change? Sketch a graph of G on the same axes as the graph of F you constructed in (e). 5.1.2 Multiple antiderivatives of a single function In the ﬁnal question of Activity 5.1.2, we encountered a very important idea: a function f has more than one antiderivative. Each antiderivative of f is determined uniquely by its value at a single point. For example, suppose that f is the function given at left in Figure 5.1.3, and suppose further that F is an antiderivative of f that satisﬁes F(0) 1. f G 3 2 F 1 1 3 5 2 4 H -1 -2 -3 Figure 5.1.3: At left, the graph of y f (x). At right, three diﬀerent antiderivatives of f . Then, using Equation (5.1.1), we can compute ∫ 1 F(1) F(0) + f (x) dx 0 1 + 0.5 1.5. Similarly, F(2) 1.5, F(3) −0.5, F(4) −2, F(5) −0.5, and F(6) 1. In addition, we can use the fact that F′ f to ascertain where F is increasing and decreasing, concave up and concave down, and has relative extremes and inﬂection points. We ultimately ﬁnd that the graph of F is the one given in blue in Figure 5.1.3. If we want an antiderivative G for which G(0) 3, then G will have the exact same shape as F (since both share the derivative f ), but G will be shifted vertically from the graph of 262 5.1 Constructing Accurate Graphs of Antiderivatives ∫1 F, as pictured in red in Figure 5.1.3. Note that G(1) − G(0) 0 f (x) dx 0.5, just as F(1) − F(0) 0.5, but since G(0) 3, G(1) G(0) + 0.5 3.5, whereas F(1) 1.5. In the same way, if we assigned a diﬀerent initial value to the antiderivative, say H(0) −1, we would get still another antiderivative, as shown in magenta in Figure 5.1.3. This example demonstrates an important fact that holds more generally: If G and H are both antiderivatives of a function f , then the function G − H must be constant. To see why this result holds, observe that if G and H are both antiderivatives of f , then G′ f and H ′ f . Hence, d [G(x) − H(x)] G′(x) − H ′(x) f (x) − f (x) 0. dx Since the only way a function can have derivative zero is by being a constant function, it follows that the function G − H must be constant. We now see that if a function has at least one antiderivative, it must have inﬁnitely many: we can add any constant of our choice to the antiderivative and get another antiderivative. For this reason, we sometimes refer to the general antiderivative of a function f . To identify a particular antiderivative of f , we must know a single value of the antideriv- ative F (this value is often called an initial condition). For example, if f (x) x 2 , its general antiderivative is F(x) 31 x 3 + C, where we include the “+C” to indicate that F includes all of the possible antiderivatives of f . If we know that F(2) 3, we substitute 2 for x in F(x) 31 x 3 + C, and ﬁnd that 1 3 (2)3 + C, 3 or C 3 − 8 3 13 . Therefore, the particular antiderivative in this case is F(x) 13 x 3 + 31 . Activity 5.1.3. For each of the following functions, sketch an accurate graph of the antiderivative that satisﬁes the given initial condition. In addition, sketch the graph of two additional antiderivatives of the given function, and state the corresponding initial conditions that each of them satisfy. If possible, ﬁnd an algebraic formula for the antiderivative that satisﬁes the initial condition. a. original function: 1(x) |x| − 1; initial condition: G(−1) 0; interval for sketch: [−2, 2] b. original function: h(x) sin(x); initial condition: H(0) 1; interval for sketch: [0, 4π] x2 , if 0 < x < 1 c. original function: p(x) −(x − 2)2 , if 1 < x < 2 ; initial condition: P(0) 1; 0 otherwise interval for sketch: [−1, 3] 263 Chapter 5 Evaluating Integrals 5.1.3 Functions deﬁned by integrals Equation (5.1.1) allows us to compute the value of the antiderivative F at a point b, provided that we know F(a) and can evaluate the deﬁnite integral from a to b of f . That is, ∫ b F(b) F(a) + f (x) dx. a In several situations, we have used this formula to compute F(b) for several diﬀerent values of b, and then plotted the points (b, F(b)) to help us draw an accurate graph of F. This suggests that we may want to think of b, the upper limit of integration, as a variable itself. To that end, we introduce the idea of an integral function, a function whose formula involves a deﬁnite integral. Deﬁnition 5.1.4 If f is a continuous function, we deﬁne the corresponding integral function A according to the rule ∫ x A(x) f (t) dt. (5.1.2) a Note that because x is the independent variable in the function A, and determines the end- point of the interval of integration, we need to use a diﬀerent variable as the variable of integration. A standard choice is t, but any variable other than x is acceptable. One way to think of the function A is as the “net signed area from a up to x” function, where we consider the region bounded by y f (t). For example, in Figure 5.1.5, we see a∫ function x f pictured at left, and its corresponding area function (choosing a 0), A(x) 0 f (t) dt shown at right. y = f (t) 3 1 π 2π x 1 A(x) -1 π 2π x Figure 5.1.5: ∫ x At left, the graph of the given function f . At right, the area function A(x) 0 f (t) dt. 264 5.1 Constructing Accurate Graphs of Antiderivatives The function A measures the net signed area from t 0 to t x bounded by the curve y f (t); this value is then reported as the corresponding height on the graph of y A(x). At http://gvsu.edu/s/cz, we ﬁnd a java applet¹ that brings the static picture in Figure 5.1.5 to life. There, the user can move the red point on the function f and see how the corresponding height changes at the light blue point on the graph of A. The choice of a is somewhat arbitrary. In the activity that follows, we explore how the value of a aﬀects the graph of the integral function. Activity 5.1.4. Suppose that 1 is given by the graph at left ∫ xin Figure 5.1.6 and that A is the corresponding integral function deﬁned by A(x) 1 1(t) dt. g 3 3 1 1 1 3 5 1 3 5 -1 -1 -3 -3 Figure 5.1.6: At left, the graph of y ∫ x1(t); at right, axes for plotting y A(x), where A is deﬁned by the formula A(x) 1 1(t) dt. a. On what interval(s) is A an increasing function? On what intervals is A decreas- ing? Why? b. On what interval(s) do you think A is concave up? concave down? Why? c. At what point(s) does A have a relative minimum? a relative maximum? d. Use the given information to determine the exact values of A(0), A(1), A(2), A(3), A(4), A(5), and A(6). e. Based on your responses to all of the preceding questions, sketch a complete and accurate graph of y A(x) on the axes provided, being sure to indicate the behavior of A for x < 0 and x > 6. ∫x f. How does the graph of B compare to A if B is instead deﬁned by B(x) 0 1(t) dt? ¹David Austin, Grand Valley State University 265 Chapter 5 Evaluating Integrals 5.1.4 Summary • Given the graph of a function f , we can construct the graph of its antiderivative F provided that (a) we know a starting value of F, say F(a), and (b) we can evaluate the ∫b integral f (x) dx exactly for relevant choices of a and b. For instance, if we wish to a ∫3 know F(3), we can compute F(3) F(a) + a f (x) dx. When we combine this infor- mation about the function values of F together with our understanding of how the behavior of F′ f aﬀects the overall shape of F, we can develop a completely accurate graph of the antiderivative F. • Because the derivative of a constant is zero, if F is an antiderivative of f , it follows that G(x) F(x) + C will also be an antiderivative of f . Moreover, any two antiderivatives of a function f diﬀer precisely by a constant. Thus, any function with at least one antiderivative in fact has inﬁnitely many, and the graphs of any two antiderivatives will diﬀer only by a vertical translation. ∫x • Given a function f , the rule A(x) a f (t) dt deﬁnes a new function A that measures the net-signed area bounded by f on the interval [a, x]. We call the function A the integral function corresponding to f . 5.1.5 Exercises 1. Deﬁnite integral of a piecewise linear function. Use the graph of f (x) shown below to ﬁnd the following integrals. ∫0 A. −5 f (x)dx B. If the vertical red shaded area in the graph ∫7 has area A, estimate: −5 f (x)dx (Your estimate may be written in terms of A.) 266 5.1 Constructing Accurate Graphs of Antiderivatives 2. A smooth function that starts out at 0. Consider the graph of the function f (x) shown below. A. Estimate the integral B. If F is an antiderivative of the same function f and F(0) 30, estimate F(7). 3. A piecewise constant function. Assume f ′ is given by the graph below. Suppose f is continuous and that f (3) 0. Sketch, on a sheet of work paper, an accurate graph of f , and use it to ﬁnd f (0) and f (7. Then ∫7 ﬁnd the value of the integral: 0 f ′(x) dx. (Note that you can do this in two diﬀerent ways!) 4. Another piecewise linear function. The ﬁgure below shows f . If F′ f and F(0) 0, ﬁnd F(b) for b 1, 2, 3, 4, 5, 6. 267 Chapter 5 Evaluating Integrals 5. A moving particle has its velocity given by the quadratic function v pictured in Fig- ure 5.1.7. In addition, it is given that A1 76 and A2 83 , as well as that for the corre- sponding position function s, s(0) 0.5. a. Use the given information to determine s(1), s(3), s(5), and s(6). b. On what interval(s) is s increasing? On what interval(s) is s decreasing? c. On what interval(s) is s concave up? On what interval(s) is s concave down? d. Sketch an accurate, labeled graph of s on the axes at right in Figure 5.1.7. e. Note that v(t) −2 + 12 (t − 3)2 . Find a formula for s. 3 3 s v 1 A1 t t 1 6 2 4 6 A2 -1 -3 -3 Figure 5.1.7: At left, the given graph of v. At right, axes for plotting s. 6. A person exercising on a treadmill experiences diﬀerent levels of resistance and thus burns calories at diﬀerent rates, depending on the treadmill’s setting. In a particu- lar workout, the rate at which a person is burning calories is given by the piecewise constant function c pictured in Figure 5.1.8. Note that the units on c are “calories per minute.” a. Let C be an antiderivative of c. What does the function C measure? What are its units? b. Assume that C(0) 0. Determine the exact value of C(t) at the values t 5, 10, 15, 20, 25, 30. c. Sketch an accurate graph of C on the axes provided at right in Figure 5.1.8. Be certain to label the scale on the vertical axis. d. Determine a formula for C that does not involve an integral and is valid for 5 ≤ t ≤ 10. 268 5.1 Constructing Accurate Graphs of Antiderivatives cal/min 15 c 10 5 min 10 20 30 10 20 30 Figure 5.1.8: At left, the given graph of c. At right, axes for plotting C. 7. Consider the piecewise linear function f given ∫ x in Figure 5.1.9.∫ xLet the functions A, B, and C be deﬁned by the rules A(x) −1 f (t) dt, B(x) 0 f (t) dt, and C(x) ∫x 1 f (t) dt. a. For the values x −1, 0, 1, . . . , 6, make a table that lists corresponding values of A(x), B(x), and C(x). b. On the axes provided in Figure 5.1.9, sketch the graphs of A, B, and C. c. How are the graphs of A, B, and C related? d. How would you best describe the relationship between the function A and the function f ? 3 3 f 1 1 1 3 5 1 3 5 -1 -1 -3 -3 Figure 5.1.9: At left, the given graph of f . At right, axes for plotting A, B, and C. 269 Chapter 5 Evaluating Integrals 5.2 The Second Fundamental Theorem of Calculus Motivating Questions ∫x • How does the integral function A(x) 1 f (t) dt deﬁne an antiderivative of f ? • What is the statement of the Second Fundamental Theorem of Calculus? • How do the First and Second Fundamental Theorems of Calculus enable us to for- mally see how diﬀerentiation and integration are almost inverse processes? In Section 4.4, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that follows it. Recall that the First FTC tells us that if f is a continuous function on [a, b] and F is any antiderivative of f (that is, F′ f ), then ∫ b f (x) dx F(b) − F(a). a We have used this result in two settings: 1 If we have a graph of f and we can compute the exact area bounded by f on an interval [a, b], we can compute the change in an antiderivative F over the interval. 2 If we can ﬁnd an algebraic formula for an antiderivative of f , we can evaluate the integral to ﬁnd the net signed area bounded by the function on the interval. For the former, see Preview Activity 5.1.1 or Activity 5.1.2. For the latter, we can easily evaluate exactly integrals such as ∫ 4 x 2 dx, 1 since we know that the function F(x) 13 x 3 is an antiderivative of f (x) x 2 . Thus, ∫ 4 4 1 x dx x 3 2 1 3 1 1 1 (4)3 − (1)3 3 3 21. Thus, the First FTC can used in two ways. First, to ﬁnd the diﬀerence F(b) − F(a) for an anti- derivative F of the integrand f , even if we may not have a formula for F itself. To do this, we ∫b must know the value of the integral a f (x) dx exactly, perhaps through known geometric formulas for area. In addition, the First FTC provides a way to ﬁnd the exact value of a deﬁ- nite integral, and hence a certain net signed area exactly, by ﬁnding an antiderivative of the integrand and evaluating its total change over the interval. In this case, we need to know a formula for the antiderivative F. Both of these perspectives are reﬂected in Figure 5.2.1. 270 5.2 The Second Fundamental Theorem of Calculus 20 20 (4, 64 3 ) F(4) − F(1) = 21 10 f (x) = x2 10 F(x) = 31 x3 (1, 13 ) R4 2 1 x dx = 21 1 2 3 4 1 2 3 4 Figure 5.2.1: At left, the graph of f (x) x 2 on the interval [1, 4] and the area it bounds. At right, the antiderivative function F(x) 31 x 3 , whose total change on [1, 4] is the value of the deﬁnite integral at left. The value of a deﬁnite integral may have additional meaning depending on context: as the change in position when the integrand is a velocity function, the total amount of pollutant leaked from a tank when the integrand is the rate at which pollution is leaking, or other total changes if the integrand is a rate function. Also, the value of the deﬁnite integral is connected ∫b to the average value of a continuous function on a given interval: fAVG[a,b] 1 b−a a f (x) dx. ∫x In the last part of Section 5.1, we studied integral functions of the form A(x) c f (t) dt. Figure 5.1.5 is a particularly important image to keep in mind as we work with integral functions, and the corresponding java applet at gvsu.edu/s/cz can help us understand the function A. In what∫ x follows, we use the First FTC to gain additional understanding of the function A(x) c f (t) dt, where the integrand f is given (either through a graph or a formula), and c is a constant. Preview Activity 5.2.1. Consider the function A deﬁned by the rule ∫ x A(x) f (t) dt, 1 where f (t) 4 − 2t. a. Compute A(1) and A(2) exactly. b. Use the First Fundamental Theorem of Calculus to ﬁnd a formula ∫ x for A(x) that does not involve integrals. That is, use the ﬁrst FTC to evaluate 1 (4 − 2t) dt. c. Observe that f is a linear function; what kind of function is A? 271 Chapter 5 Evaluating Integrals d. Using the formula you found in (b) that does not involve integrals, compute A′(x). e. While we have deﬁned f by the rule f (t) 4 − 2t, it is equivalent to say that f is given by the rule f (x) 4 − 2x. What do you observe about the relationship between A and f ? 5.2.1 The Second Fundamental Theorem of Calculus The result of Preview Activity 5.2.1 is not particular to the function f (t) 4 − 2t, nor to the choice of “1” as the lower bound in the∫integral that deﬁnes the function A. For instance, if x we let f (t) cos(t) − t and set A(x) 2 f (t) dt, we can determine a formula for A by the First FTC. Speciﬁcally, ∫ x A(x) (cos(t) − t) dt 2 x 1 sin(t) − t 2 2 2 1 sin(x) − x 2 − (sin(2) − 2) . 2 Diﬀerentiating A(x), since (sin(2) − 2) is constant, it follows that A′(x) cos(x) − x, and thus we see that A′(x) f (x), so A is an antiderivative of f . And since ∫ 2 A(2) f (t) dt 0, 2 A is the only antiderivative of f for which A(2) 0. In general, if f is any continuous function, and we deﬁne the function A by the rule ∫ x A(x) f (t) dt, c where c is an arbitrary constant, then we can show that A is an antiderivative of f . To see why, let’s demonstrate that A′(x) f (x) by using the limit deﬁnition of the derivative. Doing so, we observe that A(x + h) − A(x) A′(x) lim h→0 h ∫ x+h ∫x c f (t) dt − c f (t) dt lim h→0 h ∫ x+h x f (t) dt lim , (5.2.1) h→0 h 272 5.2 The Second Fundamental Theorem of Calculus ∫x ∫ x+h ∫ x+h where Equation (5.2.1) follows from the fact that c f (t) dt + x f (t) dt c f (t) dt. Now, observe that for small values of h, ∫ x+h f (t) dt ≈ f (x) · h, x by a simple left-hand approximation of the integral. Thus, as we take the limit in Equa- tion (5.2.1), it follows that ∫ x+h f (t) dt f (x) · h A′(x) lim x lim f (x). h→0 h h→0 h ∫c Hence, A is indeed an antiderivative of f . In addition, A(c) c f (t) dt 0. The preceding argument demonstrates the truth of the Second Fundamental Theorem of Calculus, which we state as follows. The Second Fundamental Theorem of Calculus. If f is a continuous function and c is any constant, then f has a unique antiderivative ∫x A that satisﬁes A(c) 0, and that antiderivative is given by the rule A(x) c f (t) dt. Activity 5.2.2. Suppose that f is the function given in Figure 5.2.2 and that f is a piecewise function whose parts are either portions of lines or portions of circles, as pictured. y = f (x) 1 1 2 3 4 5 6 7 -1 Figure 5.2.2: At left, the graph of y f (x). At right, axes for sketching y A(x). ∫x In addition, let A be the function deﬁned by the rule A(x) 2 f (t) dt. a. What does the Second FTC tell us about the relationship between A and f ? b. Compute A(1) and A(3) exactly. c. Sketch a precise graph of y A(x) on the axes at right that accurately reﬂects where A is increasing and decreasing, where A is concave up and concave down, and the exact values of A at x 0, 1, . . . , 7. d. How is A similar to, but diﬀerent from, the function F that you found in Activ- ity 5.1.2? 273 Chapter 5 Evaluating Integrals e. With as ∫little additional work as∫ possible, sketch precise graphs of the functions x x B(x) 3 f (t) dt and C(x) 1 f (t) dt. Justify your results with at least one sentence of explanation. 5.2.2 Understanding Integral Functions The Second FTC provides us with a way to construct an antiderivative of any continuous function. In particular, if we are given a continuous function 1 and wish to ﬁnd an anti- derivative of G, we can now say that ∫ x G(x) 1(t) dt c provides the rule for such an antiderivative, and moreover that G(c) 0. Note especially that we know that G′(x) 1(x), or [∫ x ] d 1(t) dt 1(x). (5.2.2) dx c This result is useful for understanding the graph of G. Example 5.2.3 Investigate the behavior of the integral function ∫ x e −t dt. 2 E(x) 0 Solution. E is closely related to the well known error function ¹ in probability and statistics. It turns out that the function e −t does not have an elementary antiderivative. 2 While we cannot evaluate E exactly for any value other than x 0, we still can gain a tremen- dous amount of information about the function E. By applying the rule in Equation (5.2.2) to E, it follows that [∫ x ] ′ d e dt e −x , −t 2 2 E (x) dx 0 so we know a formula for the derivative of E, and we know that E(0) 0. This information is precisely the type we were given in Activity 3.1.2, where we were given information about the derivative of a function, but lacked a formula for the function itself. Using the ﬁrst and second derivatives of E, along with the fact that E(0) 0, we can de- termine more information about the behavior of E. First, we note that for all real numbers x, e −x > 0, and thus E′(x) > 0 for all x. Thus E is an always increasing function. Further, 2 as x → ∞, E′(x) e −x → 0, so the slope of the function E tends to zero as x → ∞ (and 2 similarly as x → −∞). Indeed, it turns out that E has horizontal asymptotes as x increases or decreases without bound. In addition, we can observe that E′′(x) −2xe −x , and that E′′(0) 0, while E′′(x) < 0 for 2 x > 0 and E′′(x) > 0 for x < 0. This information tells us that E is concave up for x < 0 and concave down for x > 0 with a point of inﬂection at x 0. 274 5.2 The Second Fundamental Theorem of Calculus The only thing we lack at this point is a sense of how big E can get as x increases. If we use a midpoint Riemann sum with 10 subintervals to estimate E(2), we see that E(2) ≈ 0.8822; a similar calculation to estimate E(3) shows little change (E(3) ≈ 0.8862), so it appears that as x increases without bound, E approaches a value just larger than 0.886, which aligns with the fact that E has horizontal asymptotes. Putting all of this information together (and using the symmetry of f (t) e −t ), we see the results shown in Figure 5.2.4. 2 R x −t 2 f (t) = e−t 2 E(x) = 0e dt 1 1 -2 2 -2 2 -1 -1 Figure ∫5.2.4: At left, the graph of f (t) e −t . At right, the integral function 2 x E(x) 0 e −t dt, which is the unique antiderivative of f that satisﬁes E(0) 0. 2 Because E is the antiderivative of f (t) e −t that satisﬁes E(0) 0, values on the graph of 2 y E(x) represent the net signed area of the region bounded by f (t) e −t from 0 up to x. 2 We see that the value of E increases rapidly near zero but then levels oﬀ as x increases, since there is less and less additional accumulated area bounded by f (t) e −t as x increases. 2 ∫x Activity 5.2.3. Suppose that f (t) t 1+t 2 and F(x) 0 f (t) dt. a. On the axes at left in Figure 5.2.5, plot a graph of f (t) 1+t t 2 on the interval −10 ≤ t ≤ 10. Clearly label the vertical axes with appropriate scale. b. What is the key relationship between F and f , according to the Second FTC? c. Use the ﬁrst derivative test to determine the intervals on which F is increasing and decreasing. d. Use the second derivative test to determine the intervals on which F is concave up and concave down. Note that f ′(t) can be simpliﬁed to be written in the form ∫x 2 ¹The error function is deﬁned by the rule erf(x) √2 π 0 e −t dt and has the key property that 0 ≤ erf(x) < 1 for all x ≥ 0 and moreover that limx→∞ erf(x) 1. 275 Chapter 5 Evaluating Integrals 1−t 2 f ′(t) (1+t 2 )2 . e. Using technology appropriately, estimate the values of F(5) and F(10) through appropriate Riemann sums. f. Sketch an accurate graph of y F(x) on the righthand axes provided, and clearly label the vertical axes with appropriate scale. Figure 5.2.5: Axes for plotting f and F. 5.2.3 Diﬀerentiating an Integral Function We have seen that the Second FTC enables us to construct ∫x an antiderivative F for any con- tinuous function f as the integral function F(x) c f (t) dt. If we have a function of the ∫x [∫ x ] form F(x) c f (t) dt, then we know that F′(x) d dx c f (t) dt f (x). This shows that integral functions, while perhaps having the most complicated formulas of any functions we have encountered, are nonetheless particularly simple to diﬀerentiate. For instance, if ∫ x F(x) sin(t 2 ) dt, π then by the Second FTC, we know immediately that F′(x) sin(x 2 ). In general, we know by the Second FTC that [∫ x ] d f (t) dt f (x). dx a 276 5.2 The Second Fundamental Theorem of Calculus This equation says that “the derivative of the integral function whose integrand is f , is f .” We see that if we ﬁrst integrate the function f from t a to t x, and then diﬀerentiate with respect to x, these two processes “undo” each other. What happens if we diﬀerentiate a function f (t) and then integrate the result from t a to t x? That is, what can we say about the quantity ∫ x d [ ] f (t) dt? a dt [ ] We note that f (t) is an antiderivative of d dt f (t) and apply the First FTC. We see that ∫ d [ ] x x f (t) dt f (t) a dt a f (x) − f (a). Thus, we see that if we ﬁrst diﬀerentiate f and then integrate the result from a to x, we return to the function f , minus the constant value f (a). So the two processes almost undo each other, up to the constant f (a). The observations made in the preceding two paragraphs demonstrate that diﬀerentiating and integrating (where we integrate from a constant up to a variable) are almost inverse processes. This should not be surprising: integrating involves antidiﬀerentiating, which re- verses the process of diﬀerentiating. On the other hand, we see that there is some subtlety involved, because integrating the derivative of a function does not quite produce the func- tion itself. This is because every function has an entire family of antiderivatives, and any two of those antiderivatives diﬀer only by a constant. Activity 5.2.4. Evaluate each of the following derivatives and deﬁnite integrals. Clearly cite whether [∫ x you use ] the First or Second FTC in so∫ doing. x d [ ] d a. dx 4 e dtt 2 d. 3 dt ln(1 + t 2 ) dt ∫x [ ] [∫ x 3 ] b. d t4 dt e. d dx 4 sin(t 2 ) dt . −2 dt 1+t 4 [∫ 1 ] c. d dx x cos(t 3 ) dt 5.2.4 Summary ∫x • For a continuous function f , the integral function A(x) 1 f (t) dt deﬁnes an anti- derivative of f . • The Second Fundamental Theorem of Calculus is the formal, more general statement ∫ofx the preceding fact: if f is a continuous function and c is any constant, then A(x) c f (t) dt is the unique antiderivative of f that satisﬁes A(c) 0. 277 Chapter 5 Evaluating Integrals • Together, the First and Second FTC enable us to formally see how diﬀerentiation and integration are almost inverse processes through the observations that ∫ x d [ ] f (t) dt f (x) − f (c) c dt and [∫ x ] d f (t) dt f (x). dx c 5.2.5 Exercises ∫x 1. A deﬁnite integral starting at 3. Let 1(x) 0 f (t) dt, where f (t) is given in the ﬁgure below. Find each of the following: A. 1(0) B. 1 ′(1) C. The interval (with endpoints given to the nearest 0.25) where 1 is concave up: D. The value of x where 1 takes its max- imum on the interval 0 ≤ x ≤ 8. ∫ a d 2. Variable in the lower limit. Find the derivative: ln(ln(t)) dt. dx x Approximating a function with derivative e −x /5 . Find a good numerical approxima- 2 3. tion to F(4) for the function with the properties that F′(x) e −x /5 and F(0) 3. 2 4. ∫ x 1 be the function pictured at left in Figure 5.2.6, and let F be deﬁned by F(x) Let 2 1(t) dt. Assume that the shaded areas have values A1 4.29, A2 12.75, A3 0.36, and A4 1.79. Assume further that the portion of A2 that lies between x 0.5 and x 2 is 6.06. Sketch a carefully labeled graph of F on the axes provided, and include a written analy- sis of how you know where F is zero, increasing, decreasing, CCU, and CCD. 278 5.2 The Second Fundamental Theorem of Calculus 6 15 4 10 y = g(t) A2 2 5 A4 A1 1 2 3 4 5 6 -1 1 2 3 4 5 6 -2 -5 A3 -4 -10 Figure 5.2.6: At left, the graph of 1. At right, axes for plotting F. 5. The tide removes sand from the beach at a small ocean park at a rate modeled by the function ( ) 4πt R(t) 2 + 5 sin 25 A pumping station adds sand to the beach at rate modeled by the function 15t S(t) 1 + 3t Both R(t) and S(t) are measured in cubic yards of sand per hour, t is measured in hours, and the valid times are 0 ≤ t ≤ 6. At time t 0, the beach holds 2500 cubic yards of sand. a. What deﬁnite integral measures how much sand the tide will remove during the time period 0 ≤ t ≤ 6? Why? b. Write an expression for Y(x), the total number of cubic yards of sand on the beach at time x. Carefully explain your thinking and reasoning. c. At what instantaneous rate is the total number of cubic yards of sand on the beach at time t 4 changing? d. Over the time interval 0 ≤ t ≤ 6, at what time t is the amount of sand on the beach least? What is this minimum value? Explain and justify your answers fully. 6. When an aircraft attempts to climb as rapidly as possible, its climb rate (in feet per minute) decreases as altitude increases, because the air is less dense at higher altitudes. Given below is a table showing performance data for a certain single engine aircraft, giving its climb rate at various altitudes, where c(h) denotes the climb rate of the air- plane at an altitude h. 279 Chapter 5 Evaluating Integrals h (feet) 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10,000 c (ft/min) 925 875 830 780 730 685 635 585 535 490 440 Table 5.2.7: Data for the climbing aircraft. Let a new function m, that also depends on h, (say y m(h)) measure the number of minutes required for a plane at altitude h to climb the next foot of altitude. a. Determine a similar table of values for m(h) and explain how it is related to the table above. Be sure to discuss the units on m. b. Give a careful interpretation of a function whose derivative is m(h). Describe what the input is and what the output is. Also, explain in plain English what the function tells us. c. Determine a deﬁnite integral whose value tells us exactly the number of minutes required for the airplane to ascend to 10,000 feet of altitude. Clearly explain why the value of this integral has the required meaning. d. Determine a formula for a function M(h) whose value tells us the exact number of minutes required for the airplane to ascend to h feet of altitude. e. Estimate the values of M(6000) and M(10000) as accurately as you can. Include units on your results. 280 5.3 Integration by Substitution 5.3 Integration by Substitution Motivating Questions • How can we begin to ﬁnd algebraic formulas for antiderivatives of more complicated algebraic functions? • What is an indeﬁnite integral and how is its notation used in discussing antideriva- tives? • How does the technique of u-substitution work to help us evaluate certain indeﬁnite integrals, and how does this process rely on identifying function-derivative pairs? In Section 4.4, we learned the key role that antiderivatives play in the process of evaluating deﬁnite integrals exactly. The Fundamental Theorem of Calculus tells us that if F is any antiderivative of f , then ∫ b f (x) dx F(b) − F(a). a Furthermore, we realized that each elementary derivative rule developed in Chapter 2 leads to a corresponding elementary antiderivative, as summarized in Table 4.4.5. Thus, if we wish to evaluate an integral such as ∫ ( ) 1 √ x 3 − x + 5x dx, 0 √ it is straightforward to do so, since we can easily antidiﬀerentiate f (x) x 3 − x+5x . Because one antiderivative of f is F(x) 14 x 4 − 23 x 3/2 + ln(5) 1 5x , the Fundamental Theorem of Calculus tells us that ∫ ( ) 1 1 √ 1 2 1 x x 3 − x + 5x dx x 4 − x 3/2 + 5 0 4 3 ln(5) 0 ( ) ( ) 1 4 2 3/2 1 1 1 0 (1) − (1) + 5 − 0−0+ 5 4 3 ln(5) ln(5) 5 4 − + . 12 ln(5) We see that we have a natural interest in being able to ﬁnd such algebraic antiderivatives. We emphasize algebraic antiderivatives, as opposed to any antiderivative, ∫x since we know by the Second Fundamental Theorem of Calculus that G(x) a f (t) dt is indeed an antiderivative of the given function f , but one that still involves a deﬁnite integral. Our goal in this section is to “undo” the process of diﬀerentiation to ﬁnd an algebraic antiderivative for a given function. 281 Chapter 5 Evaluating Integrals Preview Activity 5.3.1. In Section 2.5, we learned the Chain Rule and how it can be applied to ﬁnd the derivative of a composite function. In particular, if u is a diﬀeren- tiable function of x, and f is a diﬀerentiable function of u(x), then d [ ] f (u(x)) f ′(u(x)) · u ′(x). dx In words, we say that the derivative of a composite function c(x) f (u(x)), where f is considered the “outer” function and u the “inner” function, is “the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function.” a. For each of the following functions, use the Chain Rule to ﬁnd the function’s derivative. Be sure to label each derivative by name (e.g., the derivative of 1(x) should be labeled 1 ′(x)). i. 1(x) e 3x iv. q(x) (2 − 7x)4 ii. h(x) sin(5x + 1) iii. p(x) arctan(2x) v. r(x) 34−11x b. For each of the following functions, use your work in (a) to help you determine the general antiderivative¹ of the function. Label each antiderivative by name (e.g., the antiderivative of m should be called M). In addition, check your work by computing the derivative of each proposed antiderivative. i. m(x) e 3x iv. v(x) (2 − 7x)3 ii. n(x) cos(5x + 1) iii. s(x) 1+4x 1 2 v. w(x) 34−11x c. Based on your experience in parts (a) and (b), conjecture an antiderivative for each of the following functions. Test your conjectures by computing the deriv- ative of each proposed antiderivative. 2 i. a(x) cos(πx) iii. c(x) xe x ii. b(x) (4x + 7)11 5.3.1 Reversing the Chain Rule: First Steps Whenever f is a familiar function whose antiderivative is known and u(x) is a linear func- tion, it is straightforward to antidiﬀerentiate a function of the form h(x) f (u(x)). ¹Recall that the general antiderivative of a function includes “+C” to reﬂect the entire family of functions that share the same derivative. 282 5.3 Integration by Substitution Example 5.3.1 Determine the general antiderivative of h(x) (5x − 3)6 . Check the result by diﬀerentiating. For this composite function, the outer function f is f (u) u 6 , while the inner function is u(x) 5x − 3. Since the antiderivative of f is F(u) 71 u 7 + C, we see that the antiderivative of h is 1 1 1 H(x) (5x − 3)7 · + C (5x − 3)7 + C. 7 5 35 The inclusion of the constant 15 is essential precisely because the derivative of the inner func- tion is u ′(x) 5. Indeed, if we now compute H ′(x), we ﬁnd by the Chain Rule (and Constant Multiple Rule) that 1 H ′(x) · 7(5x − 3)6 · 5 (5x − 3)6 h(x), 35 and thus H is indeed the general antiderivative of h. Hence, in the special case where the outer function is familiar and the inner function is linear, we can antidiﬀerentiate composite functions according to the following rule. If h(x) f (ax + b) and F is a known algebraic antiderivative of f , then the general antiderivative of h is given by 1 H(x) F(ax + b) + C. a It is useful to have shorthand notation that indicates the instruction to ﬁnd an antiderivative. Thus, in a similar way to how the notation d [ ] f (x) dx represents the derivative of f (x) with respect to x, we use the notation of the indeﬁnite inte- gral, ∫ f (x) dx to represent the general antiderivative of f with respect to x. Returning to the earlier exam- ple with h(x) (5x − 3)6 , we can rephrase the relationship between h and its antiderivative H through the notation ∫ 1 (5x − 3)6 dx (5x − 6)7 + C. 35 When we ﬁnd an antiderivative, we will often say that we evaluate an indeﬁnite integral. ∫ Just d as the notation dx [□] means “ﬁnd the derivative with respect to x of □,” the notation □ dx means “ﬁnd a function of x whose derivative is □.” 283 Chapter 5 Evaluating Integrals Activity 5.3.2. Evaluate each of the following indeﬁnite integrals. Check each anti- derivative ∫ that you ﬁnd by diﬀerentiating. ∫ a. sin(8 − 3x) dx d. csc(2x + 1) cot(2x + 1) dx ∫ ∫ b. sec2 (4x) dx e. √ 1 dx 1−16x 2 ∫ ∫ c. 1 11x−9 dx f. 5−x dx 5.3.2 Reversing the Chain Rule: u-substitution A natural question arises from our recent work: what happens when the inner function is not linear? For example, can we ﬁnd antiderivatives of such functions as 2 2 1(x) xe x and h(x) e x ? It is important to remember that diﬀerentiation and antidiﬀerentiation are almost inverse processes (that they are not is due to the +C that arises when antidiﬀerentiating). This almost-inverse relationship enables us to take any known derivative rule and rewrite it as a corresponding rule for an indeﬁnite integral. For example, since d [ 5] x 5x 4 , dx we can equivalently write ∫ 5x 4 dx x 5 + C. Recall that the Chain Rule states that d [ ] f (1(x)) f ′(1(x)) · 1 ′(x). dx Restating this relationship in terms of an indeﬁnite integral, ∫ f ′(1(x))1 ′(x) dx f (1(x)) + C. (5.3.1) Equation (5.3.1) tells us that if we can view a given function as f ′(1(x))1 ′(x) for some appro- priate choices of f and 1, then we can antidiﬀerentiate the function by reversing the Chain Rule. Note that both 1(x) and 1 ′(x) appear in the form of f ′(1(x))1 ′(x); we will sometimes say that we seek to identify a function-derivative pair (1(x) and 1 ′(x)) when trying to apply the rule in Equation (5.3.1). If we can identify a function-derivative pair, we will introduce a new variable u to represent the function 1(x). With u 1(x), it follows in Leibniz notation that du ′ dx 1 (x), so that in ′ terms of diﬀerentials², du 1 (x) dx. Now converting the indeﬁnite integral to a new one in ²If we recall from the deﬁnition of the derivative that du du ′ dx ≈ ∆x and use the fact that dx 1 (x), then we see that ∆u 1 ′ (x) ′ ≈ ∆x . Solving for ∆u, ∆u ≈ 1 (x)∆x. It is this last relationship that, when expressed in “diﬀerential” notation ∆u enables us to write du 1 ′ (x) dx in the change of variable formula. 284 5.3 Integration by Substitution terms of u, we have ∫ ∫ ′ ′ f (1(x))1 (x) dx f ′(u) du. Provided that f ′ is an elementary function whose antiderivative is known, we can easily evaluate the indeﬁnite integral in u, and then go on to determine the desired overall anti- derivative of f ′(1(x))1 ′(x). We call this process u-substitution, and summarize the rule as follows: With the substitution u 1(x), ∫ ∫ f ′(1(x))1 ′(x) dx f ′(u) du f (u) + C f (1(x)) + C. To see u-substitution at work, we consider the following example. Example 5.3.2 Evaluate the indeﬁnite integral ∫ x 3 · sin(7x 4 + 3) dx and check the result by diﬀerentiating. Solution. We can make two algebraic observations regarding the integrand, x 3 ·sin(7x 4 +3). First, sin(7x 4 + 3) is a composite function; as such, we know we’ll need a more sophisticated approach to antidiﬀerentiating. Second, x 3 is almost the derivative of (7x 4 + 3); the only issue is a missing constant. Thus, x 3 and (7x 4 + 3) are nearly a function-derivative pair. Furthermore, we know the antiderivative of f (u) sin(u). The combination of these ob- servations suggests that we can evaluate the given indeﬁnite integral by reversing the chain rule through u-substitution. Letting u represent the inner function of the composite function sin(7x 4 + 3), we have u dx 28x . In diﬀerential notation, it follows that du 28x dx, and thus 7x 4 + 3, and thus du 3 3 x dx 28 du. The original indeﬁnite integral may be slightly rewritten as 3 1 ∫ sin(7x 4 + 3) · x 3 dx, and so by substituting u for 7x 4 + 3 and 1 28 du for x 3 dx, it follows that ∫ ∫ 1 sin(7x + 3) · x dx 4 3 sin(u) · du. 28 Now we may evaluate the easier integral in u, and then replace u by the expression 7x 4 + 3. Doing so, we ﬁnd ∫ ∫ 1 sin(7x 4 + 3) · x 3 dx sin(u) · du 28 ∫ 1 sin(u) du 28 285 Chapter 5 Evaluating Integrals 1 (− cos(u)) + C 28 1 − cos(7x 4 + 3) + C. 28 To check our work, we observe by the Chain Rule that [ ] d 1 1 − cos(7x 4 + 3) − · (−1) sin(7x 4 + 3) · 28x 3 sin(7x 4 + 3) · x 3 , dx 28 28 which is indeed the original integrand. The u-substitution worked because the function multiplying sin(7x 4 + 3) was x 3 . If instead that function was x 2 or x 4 , the substitution process would not have worked. This is one of the primary challenges of antidiﬀerentiation: slight changes in the integrand make tremendous diﬀerences. For instance, we can use u-substitution with u x 2 and du 2xdx to ﬁnd that ∫ ∫ x2 1 xe dx eu · du 2 ∫ 1 e u du 2 1 eu + C 2 1 2 e x + C. 2 However, for the similar indeﬁnite integral ∫ 2 e x dx, the u-substitution u x 2 is no longer possible because the factor of x is missing. Hence, part of the lesson of u-substitution is just how specialized the process is: it only applies to situations where, up to a missing constant, the integrand is the result of applying the Chain Rule to a diﬀerent, related function. Activity 5.3.3. Evaluate each of the following indeﬁnite integrals by using these steps: • Find two functions within the integrand that form (up to a possible missing constant) a function-derivative pair; • Make a substitution and convert the integral to one involving u and du; • Evaluate the new integral in u; • Convert the resulting function of u back to a function of x by using your earlier substitution; • Check your work by diﬀerentiating the function of x. You should come up with the integrand originally given. 286 5.3 Integration by Substitution ∫ ∫ √ x2 cos( x) a. 5x 3 +1 dx c. √ dx x ∫ b. e x sin(e x ) dx 5.3.3 Evaluating Deﬁnite Integrals via u-substitution We have introduced u-substitution as a means to evaluate indeﬁnite integrals of functions that can be written, up to a constant multiple, in the form f (1(x))1 ′(x). This same technique can be used to evaluate deﬁnite integrals involving such functions, though we need to be careful with the corresponding limits of integration. Consider, for instance, the deﬁnite integral ∫ 5 2 xe x dx. 2 Whenever we write a deﬁnite integral, it is implicit that the limits of integration correspond to the variable of integration. To be more explicit, observe that ∫ 5 ∫ x5 x2 2 xe dx xe x dx. 2 x2 When we execute a u-substitution, we change the variable of integration; it is essential to note that this also changes the limits of integration. For instance, with the substitution u x 2 and du 2x dx, it also follows that when x 2, u 22 4, and when x 5, u 52 25. Thus, under the change of variables of u-substitution, we now have ∫ x5 ∫ u25 2 1 xe x dx eu · du x2 u4 2 u25 1 eu 2 u4 1 25 1 4 e − e . 2 2 ∫ 2 Alternatively, we could consider the related indeﬁnite integral xe x dx, ﬁnd the antideriv- 2 ative 21 e x through u-substitution, and then evaluate the original deﬁnite integral. With that method, we’d have ∫ 5 5 2 1 x2 xe x dx e 2 2 2 1 25 1 4 e − e , 2 2 which is, of course, the same result. 287 Chapter 5 Evaluating Integrals Activity 5.3.4. Evaluate each of the following deﬁnite integrals exactly through an appropriate ∫2 u-substitution. x ∫ 4/π cos( 1 ) a. 1 1+4x 2 dx c. 2/π x 2 x dx ∫1 b. 0 e −x (2e −x + 3)9 dx 5.3.4 Summary • To ﬁnd algebraic formulas for antiderivatives of more complicated algebraic functions, we need to think carefully about how we can reverse known diﬀerentiation rules. To that end, it is essential that we understand and recall known derivatives of basic func- tions, as well as the standard derivative rules. ∫ • The indeﬁnite integral provides notation for antiderivatives. When we write “ f (x) dx,” we mean “the general antiderivative of f .” In particular, if we have functions f and F such that F′ f , the following two statements say the exact thing: ∫ d [F(x)] f (x) and f (x) dx F(x) + C. dx That is, f is the derivative of F, and F is an antiderivative of f . ∫ technique of u-substitution helps us to evaluate indeﬁnite integrals of the form • The f (1(x))1 ′(x) dx through the substitutions u 1(x) and du 1 ′(x) dx, so that ∫ ∫ f (1(x))1 ′(x) dx f (u) du. A key part of choosing the expression in x to be represented by u is the identiﬁcation of a function-derivative pair. To do so, we often look for an “inner” function 1(x) that is part of a composite function, while investigating whether 1 ′(x) (or a constant multiple of 1 ′(x)) is present as a multiplying factor of the integrand. 5.3.5 Exercises 1. Product involving 4th power of a polynomial. Find the following integral. ∫ ( )2 t3 t4 − 4 dt. 2. Product involving sin(x 6 ). Find the the general antiderivative F(x) of the function f (x) given below. f (x) 7x 5 sin(x 6 ). 3. Fraction involving ln9 . Find the following integral. ∫ ln8 (z) dz z 288 5.3 Integration by Substitution 4. Fraction involving e 5x . Find the following integral. ∫ e 5x dx 1 + e 5x √ y 5. Fraction involving e 5 . Find the following integral. ∫ √ 2e 4 y √ dy y 6. Deﬁnite integral involving e −cos(q) . Use the Fundamental Theorem of Calculus to ﬁnd ∫ π ( ) e sin(q ) · cos q dq π/2 7. This problem centers on ﬁnding antiderivatives for the basic trigonometric functions other than sin(x) and cos(x). ∫ a. Consider the indeﬁnite integral tan(x) dx. By rewriting the integrand as sin(x) tan(x) cos(x) and identifying an ∫appropriate function-derivative pair, make a u-substitution and hence evaluate tan(x) dx. ∫ b. In a similar way, evaluate cot(x) dx. c. Consider the indeﬁnite integral ∫ sec2 (x) + sec(x) tan(x) dx. sec(x) + tan(x) Evaluate this integral using the substitution u sec(x) + tan(x). d. Simplify the integrand in (c) by factoring the numerator. What is a far simpler way to write the integrand? ∫ e. Combine your work in (c) and (d) to determine sec(x) dx. ∫ f. Using (c)-(e) as a guide, evaluate csc(x) dx. ∫ √ 8. Consider the indeﬁnite integral x x − 1 dx. a. At ﬁrst glance, this integrand may not seem suited to substitution due to the pres- ence of x in separate locations in the integrand. Nonetheless, using the composite √ function x − 1 as a guide, let u x − 1. Determine expressions for both x and dx in terms of u. b. Convert the given integral in x to a new integral in u. √ c. Evaluate the integral in (b) by noting that u u 1/2 and observing that it is now possible to rewrite the integrand in u by expanding through multiplication. 289 Chapter 5 Evaluating Integrals ∫ √ ∫ √ d. Evaluate each of the integrals x 2 x − 1 dx and x x 2 − 1 dx. Write a para- graph to discuss the similarities among the three indeﬁnite integrals in this prob- lem and the role of substitution and algebraic rearrangement in each. ∫ 9. Consider the indeﬁnite integral sin3 (x) dx. a. Explain why the substitution u sin(x) will not work to help evaluate the given integral. b. Recall the Fundamental Trigonometric Identity, which states that sin2 (x) + cos2 (x) 1. By observing that sin3 (x) sin(x) · sin2 (x), use the Fundamental Trigonometric Identity to rewrite the integrand as the product of sin(x) with another function. c. Explain why the substitution u cos(x) now provides a possible way to evaluate the integral in (b). ∫ d. Use your work in (a)-(c) to evaluate the indeﬁnite integral sin3 (x) dx. ∫ e. Use a similar approach to evaluate cos3 (x) dx. 10. For the town of Mathland, MI, residential power consumption has shown certain trends over recent years. Based on data reﬂecting average usage, engineers at the power com- pany have modeled the town’s rate of energy consumption by the function r(t) 4 + sin(0.263t + 4.7) + cos(0.526t + 9.4). Here, t measures time in hours after midnight on a typical weekday, and r is the rate of consumption in megawatts³ at time t. Units are critical throughout this problem. a. Sketch a carefully labeled graph of r(t) on the interval [0,24] and explain its mean- ing. Why is this a reasonable model of power consumption? ∫ 24 b. Without calculating its value, explain the meaning of 0 r(t) dt. Include appro- priate units on your answer. c. Determine the exact amount of energy Mathland consumes in a typical day. d. What is Mathland’s average rate of power consumption in a given 24-hour period? What are the units on this quantity? ³The unit megawatt is itself a rate, which measures energy consumption per unit time. A megawatt-hour is the total amount of energy that is equivalent to a constant stream of 1 megawatt of power being sustained for 1 hour. 290 5.4 Integration by Parts 5.4 Integration by Parts Motivating Questions • How ∫ do we evaluate∫indeﬁnite integrals that involve products of basic functions such as x sin(x) dx and xe x dx? • What is the method of integration by parts and how can we consistently apply it to integrate products of basic functions? • How does the algebraic structure of functions guide us in identifying u and dv in using integration by parts? ∫ of u-substitution for evaluating indeﬁnite integrals. In Section 5.3, we learned the technique For example, the indeﬁnite integral x 3 sin(x 4 ) dx is perfectly suited to u-substitution, be- cause one factor is a composite function and the other factor is the derivative (up to a con- stant) of the inner function. Recognizing the algebraic structure of a function can help us to ﬁnd its antiderivative. Next we consider integrands with a diﬀerent elementary algebraic structure: a product of basic functions. For instance, suppose we are interested in evaluating the indeﬁnite integral ∫ x sin(x) dx. The integrand is the product of the basic functions f (x) x and 1(x) sin(x). We know that it is relatively complicated to compute the derivative of the product of two functions, so we should expect that antidiﬀerentiating ∫ a product should be similarly involved. Intuitively, we expect that evaluating x sin(x) dx will involve somehow reversing the Product Rule. To that end, in Preview Activity 5.4.1 we refresh our understanding of the Product Rule and then investigate some indeﬁnite integrals that involve products of basic functions. Preview Activity 5.4.1. In Section 2.3, we developed the Product Rule and studied how it is employed to diﬀerentiate a product of two functions. In particular, recall that if f and 1 are diﬀerentiable functions of x, then d [ ] f (x) · 1(x) f (x) · 1 ′(x) + 1(x) · f ′(x). dx a. For each of the following functions, use the Product Rule to ﬁnd the function’s derivative. Be sure to label each derivative by name (e.g., the derivative of 1(x) should be labeled 1 ′(x)). i. 1(x) x sin(x) iv. q(x) x 2 cos(x) ii. h(x) xe x iii. p(x) x ln(x) v. r(x) e x sin(x) 291 Chapter 5 Evaluating Integrals b. Use your work in (a) to help you evaluate the following indeﬁnite integrals. Use diﬀerentiation to check your work. ∫ ∫ i. xe x + e x dx iv. x cos(x) + sin(x) dx ∫ ii. e x (sin(x) + cos(x)) dx ∫ ∫ iii. 2x cos(x) − x 2 sin(x) dx v. 1 + ln(x) dx c. Observe that the examples in (b) work nicely because of the derivatives you were asked to calculate in (a). Each integrand in (b) is precisely the result of diﬀerentiating one of the products of basic functions found in (a). To see what happens when an integrand is still a product but not necessarily the result of diﬀerentiating an elementary product, we consider how to evaluate ∫ x cos(x) dx. i. First, observe that d [x sin(x)] x cos(x) + sin(x). dx Integrating both sides indeﬁnitely and using the fact that the integral of a sum is the sum of the integrals, we ﬁnd that ∫ ( ) ∫ ∫ d [x sin(x)] dx x cos(x) dx + sin(x) dx. dx In this last equation, evaluate the indeﬁnite integral on the left side as well as the rightmost indeﬁnite integral on the right. ii. ∫In the most recent equation from (i.), solve the equation for the expression x cos(x) dx. iii. For which product of basic functions have you now found the antideriva- tive? 5.4.1 Reversing the Product Rule: Integration by Parts Problem (c) in Preview Activity 5.4.1 provides a clue to the general technique known as Integration by Parts, which comes from reversing the Product Rule. Recall that the Product Rule states that d [ ] f (x)1(x) f (x)1 ′(x) + 1(x) f ′(x). dx Integrating both sides of this equation indeﬁnitely with respect to x, we ﬁnd ∫ ∫ ∫ d [ ] f (x)1(x) dx f (x)1 ′(x) dx + 1(x) f ′(x) dx. (5.4.1) dx 292 5.4 Integration by Parts On the left side of Equation (5.4.1), we have the indeﬁnite integral of the derivative of a function. Temporarily omitting the constant that may arise, we have ∫ ∫ f (x)1(x) f (x)1 ′(x) dx + 1(x) f ′(x) dx. (5.4.2) We solve for the ﬁrst indeﬁnite integral on the left to generate the rule ∫ ∫ f (x)1 ′(x) dx f (x)1(x) − 1(x) f ′(x) dx. (5.4.3) Often we express Equation (5.4.3) in terms of the variables u and v, where u f (x) and v 1(x). In diﬀerential notation, du f ′(x) dx and dv 1 ′(x) dx, so we can state the rule for Integration by Parts in its most common form as follows: ∫ ∫ u dv uv − v du. To apply integration by parts, we look for a product of basic functions ∫ that we can identify as u and dv. If we can ∫ antidiﬀerentiate dv to ﬁnd v, and evaluating v du is not more diﬃcult than evaluating u dv, then this substitution usually proves to be fruitful. To demonstrate, we consider the following example. Example 5.4.1 Evaluate the indeﬁnite integral ∫ x cos(x) dx using integration by parts. Solution. When we use integration by parts, we have a choice for u and dv. In this problem, we can either let u x and dv cos(x) dx, or let u cos(x) and dv x dx. While there is ∫ a universal rule for how to choose u and dv, a good not ∫ guideline is this: do so in a way that v du is at least as simple as the original problem u dv. This leads us to choose¹ u x and dv cos(x) dx, from which it follows that du 1 dx and v sin(x). With this substitution, the rule for integration by parts tells us that ∫ ∫ x cos(x) dx x sin(x) − sin(x) · 1 dx. ∫ All that remains to do is evaluate the (simpler) integral sin(x) · 1 dx. Doing so, we ﬁnd ∫ x cos(x) dx x sin(x) − (− cos(x)) + C x sin(x) + cos(x) + C. Observe that when we get to the ﬁnal stage of evaluating the last remaining antiderivative, it is at this step that we include the integration constant, +C. ¹Observe that if we considered the alternate choice, and let u cos(x) and dv x dx, then du − sin(x) dx 293 Chapter 5 Evaluating Integrals The general technique of integration by parts involves trading the problem of integrating the product of two functions for the problem of integrating ∫ the product of two related ∫ func- tions. That is, we convert the problem of evaluating u dv to that of evaluating v du. This clearly ∫ shapes our choice of u and v. In Example 5.4.1, the original integral to evaluate was x cos(x) dx, and through ∫ the substitution provided by integration by parts, we were in- stead able to evaluate sin(x) · 1 dx. Note that the original function x was replaced by its derivative, while cos(x) was replaced by its antiderivative. Activity 5.4.2. Evaluate each of the following indeﬁnite integrals. Check each anti- derivative ∫ that you ﬁnd by diﬀerentiating. ∫ a. te −t dt c. z sec2 (z) dz ∫ ∫ b. 4x sin(3x) dx d. x ln(x) dx 5.4.2 Some Subtleties with Integration by Parts Sometimes integration by parts is not an obvious choice, but the technique is appropriate nonetheless. Integration by parts allows us to replace one function in a product with its derivative while replacing the other with its antiderivative. For instance, consider evaluating ∫ arctan(x) dx. Initially, this problem seems ill-suited to integration by parts, since there does not appear to be a product of functions present. But if we note that arctan(x) arctan(x)·1, and realize that we know the derivative of arctan(x) as well as the antiderivative of 1, we see the possibility for the substitution u arctan(x) and dv 1 dx. We explore this substitution further in Activity 5.4.3. ∫ In a related problem, consider t 3 sin(t 2 ) dt. Observe that there is a composite function present in sin(t 2 ), but there is not an obvious function-derivative pair, as we have t 3 (rather than simply t) multiplying sin(t 2 ). In this problem we use both u-substitution and integra- tion by parts. First we write t 3 t · t 2 and consider the indeﬁnite integral ∫ t · t 2 · sin(t 2 ) dt. We let z t 2 so that dz 2t dt, and thus t dt 12 dz. (We are using the variable z to perform a “z-substitution” ﬁrst so that we may then apply integration by parts.) Under this z-substitution, we now have ∫ ∫ 1 t · t · sin(t ) dt 2 2 z · sin(z) · dz. 2 ∫ ∫ and v 12 x 2 , from which we would write x cos(x) dx 12 x 2 cos(x) − 12 x 2 (− sin(x)) dx. Thus we have replaced the problem of integrating x cos(x) with that of integrating 12 x 2 sin(x); the latter is clearly more complicated, which shows that this alternate choice is not as helpful as the ﬁrst choice. 294 5.4 Integration by Parts The resulting integral can be evaluated by parts. This, too, is explored further in Activ- ity 5.4.3. These problems show that we sometimes must think creatively in choosing the variables for substitution in integration by parts, and that we may need to use substitution for an additional change of variables. Activity 5.4.3. Evaluate each of the following indeﬁnite integrals, using the provided hints. ∫ a. Evaluate arctan(x) dx by using Integration by Parts with the substitution u arctan(x) and dv 1 dx. ∫ b. Evaluate ln(z) dz. Consider a similar substitution to the one in (a). ∫ c. Use the substitution z t 2 to transform the integral t 3 sin(t 2 ) dt to a new integral in the variable z, and evaluate that new integral by parts. ∫ 3 d. Evaluate s 5 e s ds using an approach similar to that described in (c). ∫ e. Evaluate e 2t cos(e t ) dt. You will ﬁnd it helpful to note that e 2t e t · e t . 5.4.3 Using Integration by Parts Multiple Times Integration by parts is well suited to integrating the product of basic functions, allowing us to trade a given integrand for a new one where one function in the product is replaced ∫ by its derivative, ∫ and the other is replaced by its antiderivative. The goal in this trade of u dv for v du is that the new integral be simpler to evaluate than the original one. Sometimes it is necessary to apply integration by parts more than once in order to evaluate a given integral. ∫ Example 5.4.2 Evaluate t 2 e t dt. Solution. Let u t 2 and dv e t dt. Then du 2t dt and v e t , and thus ∫ ∫ t e dt t e − 2 t 2 t 2te t dt. The integral on the right side is simpler to evaluate than the one on the left, but it still requires integration by parts. Now letting u 2t and dv e t dt, we have du 2 dt and v e t , so that ∫ ( ∫ ) t 2 e t dt t 2 e t − 2te t − 2e t dt . (Note the parentheses, ∫ which remind us to distribute the minus sign to the entire value of the integral 2te t dt.) The ﬁnal integral on the right is a basic one; evaluating that integral and distributing the minus sign, we ﬁnd ∫ t 2 e t dt t 2 e t − 2te t + 2e t + C. 295 Chapter 5 Evaluating Integrals Of course, even more than two applications of integration by parts may be necessary. In the preceding example, if the integrand had been t 3 e t , we would have had to use integration by parts three times. Next, we consider the slightly diﬀerent scenario. ∫ Example 5.4.3 Evaluate e t cos(t) dt. Solution. We can choose to let u be either e t or cos(t); we pick u cos(t), and thus dv e t dt. With du − sin(t) dt and v e t , integration by parts tells us that ∫ ∫ e cos(t) dt e cos(t) − t t e t (− sin(t)) dt, or equivalently that ∫ ∫ e t cos(t) dt e t cos(t) + e t sin(t) dt. (5.4.4) The new integral has the same algebraic structure as the original one. While the overall situation isn’t necessarily better than what we started with, it hasn’t gotten worse. Thus, we proceed to integrate by parts again. This time we let u sin(t) and dv e t dt, so that du cos(t) dt and v e t , which implies ∫ ( ∫ ) e cos(t) dt e cos(t) + e sin(t) − t t t t e cos(t) dt . (5.4.5) We seem to be back where we∫started, as two applications of integration by parts has led us back to the original problem, e t cos(t) dt. But if we look closely at Equation (5.4.5), ∫ we see that we can use algebra to solve for the value of the desired integral. Adding e t cos(t) dt to both sides of the equation, we have ∫ 2 e t cos(t) dt e t cos(t) + e t sin(t), and therefore ∫ 1( t ) e t cos(t) dt e cos(t) + e t sin(t) + C. 2 Note that since we never actually encountered an integral we could evaluate directly, we didn’t have the opportunity to add the integration constant C until the ﬁnal step. Activity 5.4.4. Evaluate each of the following indeﬁnite integrals. ∫ a. x 2 sin(x) dx ∫ b. t 3 ln(t) dt ∫ c. e z sin(z) dz ∫ d. s 2 e 3s ds 296 5.4 Integration by Parts ∫ e. t arctan(t) dt (Hint: At a certain point in this problem, it is very helpful to note t2 that 1+t 2 1− 1 1+t 2 .) 5.4.4 Evaluating Deﬁnite Integrals Using Integration by Parts We can use the technique of integration by parts to evaluate a deﬁnite integral. Example 5.4.4 Evaluate ∫ π/2 t sin(t) dt. 0 Solution. One option is to ﬁnd an antiderivative (using indeﬁnite integral notation) and then apply the Fundamental Theorem of Calculus to ﬁnd that ∫ π/2 π/2 t sin(t) dt (−t cos(t) + sin(t)) 0 0 ( π π π ) − cos( ) + sin( ) − (−0 cos(0) + sin(0)) 2 2 2 1. Alternatively, we can apply integration by parts and work with deﬁnite integrals throughout. With this method, we must remember to evaluate the product uv over the given limits of integration. Using the substitution u t and dv sin(t) dt, so that du dt and v − cos(t), we write ∫ π/2 π/2 ∫ π/2 t sin(t) dt − t cos(t) − (− cos(t)) dt 0 0 0 π/2 π/2 − t cos(t) + sin(t) 0 0 ( π π π ) − cos( ) + sin( ) − (−0 cos(0) + sin(0)) 2 2 2 1. As with any substitution technique, it is important to use notation carefully and completely, and to ensure that the end result makes sense. 5.4.5 When u-substitution and Integration by Parts Fail to Help Both integration techniques we have discussed apply in relatively limited circumstances. It is not hard to ﬁnd examples of functions for which neither technique produces an antideriv- ative; indeed, there are many, many functions that appear elementary but that do not have an elementary algebraic antiderivative. For instance, neither u-substitution nor integration 297 Chapter 5 Evaluating Integrals by parts proves fruitful for the indeﬁnite integrals ∫ ∫ 2 e x dx and x tan(x) dx. While there are other integration techniques, some of which we will consider brieﬂy, none of 2 them enables us to ﬁnd an algebraic antiderivative for e x or x tan(x). We do know from the Second Fundamental Theorem∫ x 2 of Calculus that we can construct2 an integral antiderivative ∫x for each function; F(x) 0 e t dt is an antiderivative of f (x) e x , and G(x) 0 t tan(t) dt is an antiderivative of 1(x) x tan(x). But ﬁnding an elementary algebraic formula that doesn’t involve integrals for either F or G turns out not only to be impossible through u- substitution or integration by parts, but indeed impossible altogether. Antidiﬀerentiation is much harder in general than diﬀerentiation. 5.4.6 Summary • Through the method of integration by parts, ∫ we can evaluate∫ indeﬁnite integrals that involve products of basic functions such as x sin(x) dx and x ln(x) dx. Using a sub- stitution enables us to trade one of the functions in the product for its derivative, and the other for its antiderivative, in an eﬀort to ﬁnd a diﬀerent product of functions that is easier to integrate. • If ∫ the algebraic structure of an integrand is a product of basic functions in the form f (x)1 ′(x) dx, we can use the substitution u f (x) and dv 1 ′(x) dx and apply the rule ∫ ∫ u dv uv − v du ∫ to evaluate the original integral f (x)1 ′(x) dx by instead evaluating ∫ ∫ v du f ′(x)1(x) dx. • When deciding to integrate by parts, we have to select∫ both u and dv. That selection is guided by the overall∫principle that the new integral v du not be more diﬃcult than the original integral u dv. In addition, it is often helpful to recognize if one of the functions present is much easier to diﬀerentiate than antidiﬀerentiate (such as ln(x)), in which case that function often is best assigned the variable u. In addition, dv must be a function that we can antidiﬀerentiate. 5.4.7 Exercises 1. Choose which method to use. For each of the following integrals, indicate whether integration by substitution or integration by parts is more appropriate, or if neither method is appropriate. Do not evaluate the integrals. ∫ 1. x sin x dx 298 5.4 Integration by Parts ∫ x4 2. 1+x 5 dx ∫ 5 3. x 4 e x dx ∫ 4. x 4 cos(x 5 ) dx ∫ 5. √ 1 dx 5x+1 2. Product involving cos(5x). Use integration by parts to evaluate the integral. ∫ 3x cos(4x) dx. 3. Product involving e 8z . Find the integral ∫ (z + 1) e 5z dz. 4. Deﬁnite integral of te −t . Evaluate the deﬁnite integral. ∫ 5 te −t dt. 0 ∫x 5. Let f (t) te −2t and F(x) 0 f (t) dt. a. Determine F′(x). b. Use the First FTC to ﬁnd a formula for F that does not involve an integral. c. Is F an increasing or decreasing function for x > 0? Why? ∫ 6. Consider the indeﬁnite integral given by e 2x cos(e x ) dx. a. Noting that e 2x e x ·e x , use the substitution z e x to determine a new, equivalent integral in the variable z. b. Evaluate the integral you found in (a) using an appropriate technique. ∫ c. How is the problem of evaluating e 2x cos(e 2x ) dx diﬀerent from evaluating the integral in (a)? Do so. d. Evaluate each of the following integrals as well, keeping in mind the approach(es) used earlier in this problem: ∫ • e 2x sin(e x ) dx ∫ • e 3x sin(e 3x ) dx ∫ 2 2 2 • xe x cos(e x ) sin(e x ) dx 7. For each of the following indeﬁnite integrals, determine whether you would use u- substitution, integration by parts, neither∗, or both to evaluate the integral. In each case, write one sentence to explain your reasoning, and include a statement of any sub- stitutions used. (That is, if you decide in a problem to let u e 3x , you should state that, as well as that du 3e 3x dx.) Finally, use your chosen approach to evaluate each 299 Chapter 5 Evaluating Integrals integral. (∗ one of the following problems does not have an elementary antiderivative and you are not expected to actually evaluate this integral; this will correspond with a choice of “neither” among those given.) ∫ a. x 2 cos(x 3 ) dx ∫ b. x 5 cos(x 3 ) dx (Hint: x 5 x 2 · x 3 ) ∫ c. x ln(x 2 ) dx ∫ d. sin(x 4 ) dx ∫ e. x 3 sin(x 4 ) dx ∫ f. x 7 sin(x 4 ) dx 300 5.5 Other Options for Finding Algebraic Antiderivatives 5.5 Other Options for Finding Algebraic Antiderivatives Motivating Questions • How does the method of partial fractions enable any rational function to be antidif- ferentiated? • What role have integral tables historically played in the study of calculus and how ∫ √ can a table be used to evaluate integrals such as a 2 + u 2 du? • What role can a computer algebra system play in the process of ﬁnding antideriva- tives? We have learned two antidiﬀerentiation techniques: u-substitution and integration by parts. The former is used to reverse the chain rule, while the latter to reverse the product rule. But ∫ we have seen that each works only in specialized ∫ circumstances. For example, while x 2 x xe dx may be evaluated by u-substitution and xe dx by integration by parts, neither ∫ 2 method provides a route to evaluate e x dx, and in fact an elementary algebraic antideriv- 2 ative for e x does not exist. No antidiﬀerentiation method will provide us with a simple algebraic formula for a function F(x) that satisﬁes F′(x) e x . 2 In this section of the text, our main goals are to identify some classes of functions that can be antidiﬀerentiated, and to learn some methods to do so. We should also recognize that there are many functions for which an algebraic formula for an antiderivative does not exist, and appreciate the role that computing technology can play in ﬁnding antiderivatives of other complicated functions. Preview Activity 5.5.1. For each of the indeﬁnite integrals below, the main question is to decide whether the integral can be evaluated using u-substitution, integration by parts, a combination of the two, or neither. For integrals for which your answer is aﬃrmative, state the substitution(s) you would use. It is not necessary to actually evaluate any of the integrals completely, unless the integral can be evaluated imme- diately using a familiar basic antiderivative. ∫ ∫ ∫ ∫ a. x 2 sin(x 3 ) dx, x 2 sin(x) dx, sin(x 3 ) dx, x 5 sin(x 3 ) dx ∫ ∫ ∫ ∫ 1 x 2x+3 ex b. 1+x 2 dx, 1+x 2 dx, 1+x 2 dx, 1+(e x )2 dx, ∫ ∫ ln(x) ∫ ∫ c. x ln(x) dx, x dx, ln(1 + x 2 ) dx, x ln(1 + x 2 ) dx, ∫ √ ∫ ∫ ∫ d. x 1 − x 2 dx, √ 1 2 dx, √ x 2 dx, √1 2 dx, 1−x 1−x x 1−x 301 Chapter 5 Evaluating Integrals 5.5.1 The Method of Partial Fractions The method of partial fractions is used to integrate rational functions. It involves reversing the process of ﬁnding a common denominator. Example 5.5.1 Evaluate ∫ 5x dx. x2 −x−2 Solution. If we factor the denominator, we can see how R might be the sum of two fractions A of the form x−2 + x+1 B , so we suppose that 5x A B + (x − 2)(x + 1) x − 2 x + 1 and look for the constants A and B. Multiplying both sides of this equation by (x − 2)(x + 1), we ﬁnd that 5x A(x + 1) + B(x − 2). Since we want this equation to hold for every value of x, we can use insightful choices of speciﬁc x-values to help us ﬁnd A and B. Taking x −1, we have 5(−1) A(0) + B(−3), so B 53 . Choosing x 2, it follows 5(2) A(3) + B(0), so A 10 3 . Thus, ∫ ∫ 5x 10/3 5/3 dx + dx. x −x−2 2 x−2 x+1 This integral is straightforward to evaluate, and hence we ﬁnd that ∫ 5x 10 5 dx ln |x − 2| + ln |x + 1| + C. x2 −x−2 3 3 It turns out that we can use the method of partial fractions to rewrite any ratinal function P(x) R(x) Q(x) where the degree of the polynomial P is less than¹ the degree of Q as a sum of simpler rational functions of one of the following forms: A A Ax + B Ax + B , , , or x − c (x − c)n x 2 + k (x 2 + k) n where A, B, and c are real numbers, and k is a positive real number. Because we can an- tidiﬀerentiate each of these basic forms, partial fractions enables us to antidiﬀerentiate any rational function. ¹If the degree of P is greater than or equal to the degree of Q, long division may be used to write R as the sum of a polynomial plus a rational function where the numerator’s degree is less than the denominator’s. 302 5.5 Other Options for Finding Algebraic Antiderivatives A computer algebra system such as Maple, Mathematica, or WolframAlpha can be used to ﬁnd the partial fraction decomposition of any rational function. In WolframAlpha, entering partial fraction 5x/(x^2-x-2) results in the output 5x 10 5 + . x 2 − x − 2 3(x − 2) 3(x + 1) We will use technology to generate partial fraction decompositions of rational functions, and then evaluate the integrals using established methods. Activity 5.5.2. For each of the following problems, evaluate the integral by using the partial fraction decomposition provided. ∫ 1/4 1/4 a. 1 x 2 −2x−3 dx, given that 1 x 2 −2x−3 x−3 − x+1 ∫ x 2 +1 x 2 +1 b. x 3 −x 2 dx, given that x 3 −x 2 − x1 − 1 x2 + 2 x−1 ∫ −x+2 c. x−2 x 4 +x 2 dx, given that x−2 x 4 +x 2 1 x − 2 x2 + 1+x 2 5.5.2 Using an Integral Table Calculus has a long history, going back to Greek mathematicians in 400-300 BC. Its main foundations were ﬁrst investigated and understood independently by Isaac Newton and Gottfried Wilhelm Leibniz in the late 1600s, making the modern ideas of calculus well over 300 years old. It is instructive to realize that until the late 1980s, the personal computer did not exist, so calculus (and other mathematics) had to be done by hand for roughly 300 years. In the 21st century, however, computers have revolutionized many aspects of the world we live in, including mathematics. In this section we take a short historical tour to precede discussing the role computer algebra systems can play in evaluating indeﬁnite integrals. In particular, we consider a class of integrals involving certain radical expressions. As seen in the short table of integrals found in Appendix A, there are many forms of integrals √ √ that involve a 2 ± w 2 and w 2 − a 2 . These integral rules can be developed using a technique known as trigonometric substitution that we choose to omit; instead, we will simply accept the results presented in the table. To see how these rules are used, consider the diﬀerences among ∫ ∫ ∫ √ 1 x √ dx, √ dx, and 1 − x 2 dx. 1−x 2 1−x 2 The ﬁrst integral is a familiar basic one, and results in arcsin(x) + C. The second integral can be evaluated using a standard u-substitution with u 1 − x 2 . The third, however, is not familiar and does not lend itself to u-substitution. In Appendix A, we ﬁnd the rule ∫ √ (u) u√ 2 a2 (h) a 2 − u 2 du a − u2 + arcsin + C. 2 2 a 303 Chapter 5 Evaluating Integrals Using the substitutions a 1 and u x (so that du dx), it follows that ∫ √ x√ 1 1 − x 2 dx 1 − x 2 − arcsin(x) + C. 2 2 Whenever we are applying a rule in the table, we are doing a u-substitution, especially when the substitution is more complicated than setting u x as in the last example. Example 5.5.2 Evaluate the integral ∫ √ 9 + 64x 2 dx. Solution. Here, we want to use Rule (c) from the table, but we now set a 3 and u 8x. We also choose the “+” option in the rule. With this substitution, it follows that du 8dx, so dx 18 du. Applying the substitution, ∫ √ ∫ √ ∫ √ 1 1 9 + 64x 2 dx 9 + u2 · du 9 + u 2 du. 8 8 By Rule (c), we now ﬁnd that ∫ √ ( ) 1 u√ 2 9 √ 9+ 64x 2 dx u + 9 + ln |u + u 2 + 9| + C 8 2 2 ( √ √ ) 1 8x 9 64x 2 + 9 + ln |8x + 64x 2 + 9| + C . 8 2 2 Whenever we use a u-subsitution in conjunction with Appendix A, it’s important that we not forget to address any constants that arise and include them in our computations, such as the 18 that appeared in Example 5.5.2. Activity 5.5.3. For each of the following integrals, evaluate the integral using u- ∫ √ and/or an entry from the table found∫ in Appendix substitution A. a. x 2 + 4 dx c. √ 2 dx 2 16+25x ∫ ∫ b. √ x dx d. √ 1 dx x 2 +4 x 2 49−36x 2 5.5.3 Using Computer Algebra Systems A computer algebra system (CAS) is a computer program that is capable of executing sym- bolic mathematics. For example, if we ask a CAS to solve the equation ax 2 + bx + c 0√for the b −4ac variable x, where a, b, and c are arbitrary constants, the program will return x −b± 2a 2 . Research to develop the ﬁrst CAS dates to the 1960s, and these programs became publicly available in the early 1990s. Two prominent examples are the programs Maple and Mathemat- ica, which were among the ﬁrst computer algebra systems to oﬀer a graphical user interface. Today, Maple and Mathematica are exceptionally powerful professional software packages 304 5.5 Other Options for Finding Algebraic Antiderivatives that can execute an amazing array of sophisticated mathematical computations. They are also very expensive, as each is a proprietary program. The CAS SAGE is an open-source, free alternative to Maple and Mathematica. For the purposes of this text, when we need to use a CAS, we are going to turn instead to a similar, but somewhat diﬀerent computational tool, the web-based “computational knowl- edge engine” called WolframAlpha. There are two features of WolframAlpha that make it stand out from the CAS options mentioned above: (1) unlike Maple and Mathematica, WolframAlpha is free (provided we are willing to navigate some pop-up advertising); and (2) unlike any of the three, the syntax in WolframAlpha is ﬂexible. Think of WolframAlpha as being a little bit like doing a Google search: the program will interpret what is input, and then provide a summary of options. If we want to have WolframAlpha evaluate an integral for us, we can provide it syntax such as integrate x^2 dx to which the program responds with ∫ x3 x 2 dx + constant. 3 While there is much to be enthusiastic about regarding CAS programs such as WolframAlpha, there are several things we should be cautious about: (1) a CAS only responds to exactly what is input; (2) a CAS can answer using powerful functions from very advanced mathematics; and (3) there are problems that even a CAS cannot do without additional human insight. Although (1) likely goes without saying, we have to be careful with our input: if we enter syntax that deﬁnes the wrong function, the CAS will work with precisely the function we deﬁne. For example, if we are interested in evaluating the integral ∫ 1 dx, 16 − 5x 2 and we mistakenly enter integrate 1/16 - 5x^2 dx a CAS will (correctly) reply with 1 5 x − x3. 16 3 But if we are suﬃciently well-versed in antidiﬀerentiation, we will recognize that this func- 1 tion cannot be the one that we seek: integrating a rational function such as 16−5x 2 , we expect the logarithm function to be present in the result. ∫ 1 Regarding (2), even for a relatively simple integral such as 16−5x 2 dx, some CASs will in- voke advanced functions rather than simple ones. For instance, if we use Maple to execute the command 305 Chapter 5 Evaluating Integrals int(1/(16-5*x^2), x); the program responds with ∫ √ √ 1 5 5 dx arctanh( x). 16 − 5x 2 20 4 While this is correct (save for the missing arbitrary constant, which Maple never reports), the inverse hyperbolic tangent function is not a common nor familiar one; a simpler way to express this function can be found by using the partial fractions method, and happens to be the result reported by WolframAlpha: ∫ 1 1 ( √ √ ) dx √ log(4 5 + 5x) − log(4 5 − 5x) + constant. 16 − 5x 2 8 5 Using sophisticated functions from more advanced mathematics is sometimes the way a CAS says to the user “I don’t know how to do this problem.” For example, if we want to evaluate ∫ e −x dx, 2 and we ask WolframAlpha to do so, the input integrate exp(-x^2) dx results in the output ∫ √ −x 2 π e dx erf(x) + constant. 2 The function “erf(x)” is the error function, which is actually deﬁned by an integral: ∫ x 2 e −t dt. 2 erf(x) √ π 0 So, in producing output involving an integral, the CAS has basically reported back to us the very question we asked. Finally, as remarked at (3) above, there are times that a CAS will actually fail without some additional human insight. If we consider the integral ∫ √ (1 + x)e x 1 + x 2 e 2x dx and ask WolframAlpha to evaluate int (1+x) * exp(x) * sqrt(1+x^2 * exp(2x)) dx, the program thinks for a moment and then reports 306 5.5 Other Options for Finding Algebraic Antiderivatives (no result found in terms of standard mathematical functions) But in fact this integral is not that diﬃcult to evaluate. If we let u xe x , then du (1 + x)e x dx, which means that the preceding integral has form ∫ √ ∫ √ (1 + x)e x 1 + x 2 e 2x dx 1 + u 2 du, which is a straightforward one for any CAS to evaluate. So, we should proceed with some caution: while any CAS is capable of evaluating a wide range of integrals (both deﬁnite and indeﬁnite), there are times when the result can mislead us. We must think carefully about the meaning of the output, whether it is consistent with what we expect, and whether or not it makes sense to proceed. 5.5.4 Summary • We can antidiﬀerentiate any rational function with the method of partial fractions. Any polynomial function can be factored into a product of linear and irreducible quadratic terms, so any rational function may be written as the sum of a polynomial plus rational A Bx+C terms of the form (x−c) n (where n is a natural number) and x 2 +k (where k is a positive real number). • Until the development of compute algebra ∫ √ systems, integral tables enabled students of calculus to evaluate integrals such as a 2 + u 2 du, where a is a positive real number. A short table of integrals may be found in Appendix A. • Computer algebra systems can play an important role in ﬁnding antiderivatives, though we must be cautious to use correct input, to watch for unusual or unfamiliar advanced functions that the CAS may cite in its result, and to consider the possibility that a CAS may need further assistance or insight from us in order to answer a particular question. 5.5.5 Exercises 1. Partial fractions: linear over diﬀerence of squares. Calculate the integral below by partial fractions and by using the indicated substitution. Be sure that you can show how the results you obtain are the same. ∫ 2x dx x 2 − 36 First, rewrite this with partial fractions: ∫ ∫ ∫ 2x x 2 −36 dx dx + dx + +C. Next, use the substitution w x 2 − 36 to ﬁnd the integral. 307 Chapter 5 Evaluating Integrals 2. Partial fractions: constant over product. Calculate the integral: ∫ 1 dx. (x + 6)(x + 8) 3. Partial fractions: linear over quadratic. Calculate the integral ∫ 8x + 6 dx. x 2 − 3x + 2 4. Partial fractions: cubic over 4th degree. Consider the following indeﬁnite integral. ∫ 9x 3 + 6x 2 + 100x + 75 dx x 4 + 25x 2 The integrand has partial fractions decomposition: a b cx + d + + x 2 x x 2 + 25 where a b c d Now integrate term by term to evaluate the integral. 5. Partial fractions: quadratic over factored cubic. The form of the partial fraction de- composition of a rational function is given below. ( ) − 4x 2 + x + 32 A Bx + C + 2 (x − 4)(x 2 + 9) x−4 x +9 A B C Now evaluate the indeﬁnite integral. ∫ ( ) − 4x 2 + x + 32 dx. (x − 4)(x 2 + 9) 6. For each of the following integrals involving rational functions, (1) use a CAS to ﬁnd the partial fraction decomposition of the integrand; (2) evaluate the integral of the resulting function without the assistance of technology; (3) use a CAS to evaluate the original integral to test and compare your result in (2). ∫ x 3 +x+1 a. x 4 −1 dx ∫ x 5 +x 2 +3 b. x 3 −6x 2 +11x−6 dx 308 5.5 Other Options for Finding Algebraic Antiderivatives ∫ x 2 −x−1 c. (x−3)3 dx 7. For each of the following integrals involving radical functions, (1) use an appropriate u-substitution along with Appendix A to evaluate the integral without the assistance of technology, and (2) use a CAS to evaluate the original integral to test and compare your result in (1). ∫ a. √ 1 dx x 9x 2 +25 ∫ √ b. x 1 + x 4 dx ∫ √ c. e x 4 + e 2x dx ∫ d. √ tan(x) dx 9−cos2 (x) 8. Consider the indeﬁnite integral given by ∫ √ √ x + 1 + x2 dx. x a. Explain why u-substitution does not oﬀer a way to simplify this integral by dis- cussing at least two diﬀerent options you might try for u. b. Explain why integration by parts does not seem to be a reasonable way to proceed, either, by considering one option for u and dv. c. Is there any line in the integral table in Appendix A that is helpful for this integral? d. Evaluate the given integral using WolframAlpha. What do you observe? 309 Chapter 5 Evaluating Integrals 5.6 Numerical Integration Motivating Questions ∫1 • How do we accurately evaluate a deﬁnite integral such as 0 e −x dx when we can- 2 not use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative? Are there ways to generate accurate estimates without using extremely large values of n in Riemann sums? • What is the Trapezoid Rule, and how is it related to left, right, and middle Riemann sums? • How are the errors in the Trapezoid Rule and Midpoint Rule related, and how can they be used to develop an even more accurate rule? When we ﬁrst explored ﬁnding the net signed area bounded by a curve, we developed the concept of a Riemann sum as a helpful estimation tool and a key step in the deﬁnition of the deﬁnite integral. Recall that the left, right, and middle Riemann sums of a function f on an interval [a, b] are given by ∑ n−1 L n f (x 0 )∆x + f (x 1 )∆x + · · · + f (x n−1 )∆x f (x i )∆x, (5.6.1) i0 ∑n R n f (x1 )∆x + f (x2 )∆x + · · · + f (x n )∆x f (x i )∆x, (5.6.2) i1 ∑n M n f (x 1 )∆x + f (x 2 )∆x + · · · + f (x n )∆x f (x i )∆x, (5.6.3) i1 where x0 a, x i a + i∆x, x n b, and ∆x b−a n . For the middle sum, we deﬁned x i (x i−1 + x i )/2. A Riemann sum is a sum of (possibly signed) areas of rectangles. The value of n determines the number of rectangles, and our choice of left endpoints, right endpoints, or midpoints determines the heights of the rectangles. We can see the similarities and diﬀerences among these three options in Figure 5.6.1, where we consider the function f (x) 20 1 (x−4)3 +7 on the interval [1, 8], and use 5 rectangles for each of the Riemann sums. While it is a good exercise to compute a few Riemann sums by hand, just to ensure that we understand how they work and how varying the function, the number of subintervals, and the choice of endpoints or midpoints aﬀects the result, using computing technology is the best way to determine L n , R n , and M n . Any computer algebra system will oﬀer this capability; as we saw in Preview Activity 4.3.1, a straightforward option that is freely available online is the applet¹ at http:// gvsu.edu/s/a9. Note that we can adjust the formula for f (x), the window of x- and y-values of interest, the number of subintervals, and the method. (See Preview Activity 4.3.1 for any needed reminders on how the applet works.) In this section we explore several diﬀerent alternatives for estimating deﬁnite integrals. Our ¹Marc Renault, Shippensburg University 310 5.6 Numerical Integration y = f (x) y = f (x) y = f (x) 1 LEFT 8 1 RIGHT 8 1 MID 8 Figure 5.6.1: Left, right, and middle Riemann sums for y f (x) on [1, 8] with 5 subintervals. main goal is to develop formulas to estimate deﬁnite integrals accurately without using a large numbers of rectangles. Preview Activity 5.6.1. As we begin to investigate ways to approximate deﬁnite inte- grals, it will be insightful to compare results to integrals whose exact values we know. ∫3 To that end, the following sequence of questions centers on 0 x 2 dx. a. Use the applet at http://gvsu.edu/s/a9 with the function f (x) x 2 on the window of x values from 0 to 3 to compute L3 , the left Riemann sum with three subintervals. b. Likewise, use the applet to compute R 3 and M3 , the right and middle Riemann sums with three subintervals, respectively. c. Use the Fundamental Theorem of Calculus to compute the exact value of I ∫3 0 x 2 dx. d. We deﬁne the error that results from an approximation of a deﬁnite integral to be the approximation’s value minus the integral’s exact value. What is the error that results from using L 3 ? From R 3 ? From M3 ? e. In what follows in this section, we will learn a new approach to estimating the value of a deﬁnite integral known as the Trapezoid Rule. The basic idea is to use trapezoids, rather than rectangles, to estimate the area under a curve. What is the formula for the area of a trapezoid with bases of length b 1 and b 2 and height h? f. Working by hand, estimate the area under f (x) x 2 on [0, 3] using three subin- tervals and three corresponding trapezoids. What is the error in this approxi- mation? How does it compare to the errors you calculated in (d)? 311 Chapter 5 Evaluating Integrals 5.6.1 The Trapezoid Rule So far, we have used the simplest possible quadrilaterals (that is, rectangles) to estimate areas. It is natural, however, to wonder if other familiar shapes might serve us even better. An alternative to L n , R n , and M n is called the Trapezoid Rule. Rather than using a rectangle to estimate the (signed) area bounded by y f (x) on a small interval, we use a trapezoid. For example, in Figure 5.6.2, we estimate the area under the curve using three subintervals and the trapezoids that result from connecting the corresponding points on the curve with straight lines. y = f (x) D1 D2 D3 x0 x1 x2 x3 ∫b Figure 5.6.2: Estimating a f (x) dx using three subintervals and trapezoids, rather than rectangles, where a x 0 and b x 3 . The biggest diﬀerence between the Trapezoid Rule and a Riemann sum is that on each subin- terval, the Trapezoid Rule uses two function values, rather than one, to estimate the (signed) area bounded by the curve. For instance, to compute D1 , the area of the trapezoid on [x0 , x 1 ], we observe that the left base has length f (x0 ), while the right base has length f (x 1 ). The height of the trapezoid is x1 − x0 ∆x b−a 3 . The area of a trapezoid is the average of the bases times the height, so we have 1 D1 ( f (x0 ) + f (x1 )) · ∆x. 2 Using similar computations for D2 and D3 , we ﬁnd that T3 , the trapezoidal approximation ∫b to a f (x) dx is given by T3 D1 + D2 + D3 1 1 1 ( f (x 0 ) + f (x1 )) · ∆x + ( f (x1 ) + f (x 2 )) · ∆x + ( f (x 2 ) + f (x3 )) · ∆x. 2 2 2 312 5.6 Numerical Integration Because both left and right endpoints are being used, we recognize within the trapezoidal approximation the use of both left and right Riemann sums. Rearranging the expression for T3 by removing factors of 12 and ∆x, grouping the left endpoint and right endpoint evalua- tions of f , we see that 1[ ] 1[ ] T3 f (x0 ) + f (x 1 ) + f (x2 ) ∆x + f (x1 ) + f (x 2 ) + f (x3 ) ∆x. (5.6.4) 2 2 We now observe that two familiar sums have arisen. The left Riemann sum L3 is L3 f (x 0 )∆x + f (x1 )∆x + f (x2 )∆x, and the right Riemann sum is R 3 f (x1 )∆x + f (x2 )∆x + f (x 3 )∆x. Substituting L3 and R 3 for the corresponding expressions in Equation (5.6.4), it follows that T3 12 [L3 + R 3 ]. We have thus seen a very important result: using trapezoids to estimate the (signed) area bounded by a curve is the same as averaging the estimates generated by using left and right endpoints. The Trapezoid Rule. ∫b The trapezoidal approximation, Tn , of the deﬁnite integral a f (x) dx using n subin- tervals is given by the rule [ ] 1 1 1 Tn ( f (x0 ) + f (x 1 )) + ( f (x 1 ) + f (x2 )) + · · · + ( f (x n−1 ) + f (x n )) ∆x. 2 2 2 ∑ n−1 1 ( f (x i ) + f (x i+1 ))∆x. 2 i0 Moreover, Tn 1 2 [L n + R n ]. Activity 5.6.2. In this activity, we explore the relationships among the errors gener- ated by left, right, midpoint, and trapezoid approximations to the deﬁnite integral ∫2 1 1 x2 dx. ∫2 1 a. Use the First FTC to evaluate 1 x2 dx exactly. b. Use appropriate computing technology to compute the following approxima- ∫2 1 tions for 1 x2 dx: T4 , M4 , T8 , and M8 . c. Let the error that results from an approximation be the approximation’s value minus the exact value of the deﬁnite integral. For instance, if we let ET,4 repre- sent the error that results from using the trapezoid rule with 4 subintervals to estimate the integral, we have ∫ 2 1 ET,4 T4 − dx. 1 x2 Similarly, we compute the error of the midpoint rule approximation with 8 subin- tervals by the formula ∫ 2 1 E M,8 M8 − 2 dx. 1 x 313 Chapter 5 Evaluating Integrals Based on your work in (a) and (b) above, compute ET,4 , ET,8 , E M,4 , E M,8 . d. Which rule consistently over-estimates the exact value of the deﬁnite integral? Which rule consistently under-estimates the deﬁnite integral? e. What behavior(s) of the function f (x) 1 x2 lead to your observations in (d)? 5.6.2 Comparing the Midpoint and Trapezoid Rules We know from the deﬁnition of the deﬁnite integral that if we let n be large enough, we can make any of the approximations L n , R n , and M n as close as we’d like (in theory) to the ∫b exact value of a f (x) dx. Thus, it may be natural to wonder why we ever use any rule other than L n or R n (with a suﬃciently large n value) to estimate a deﬁnite integral. One of the primary reasons is that as n → ∞, ∆x b−a n → 0, and thus in a Riemann sum calculation with a large n value, we end up multiplying by a number that is very close to zero. Doing so often generates roundoﬀ error, because representing numbers close to zero accurately is a persistent challenge for computers. Hence, we explore ways to estimate deﬁnite integrals to high levels of precision, but without using extremely large values of n. Paying close attention to patterns in errors, such as those observed in Activity 5.6.2, is one way to begin to see some alternate approaches. To begin, we compare the errors in the Midpoint and Trapezoid rules. First, consider a function that is concave up on a given interval, and picture approximating the area bounded on that interval by both the Midpoint and Trapezoid rules using a single subinterval. T1 M1 M1 ∫b Figure 5.6.3: Estimating a f (x) dx using a single subinterval: at left, the trapezoid rule; in the middle, the midpoint rule; at right, a modiﬁed way to think about the midpoint rule. As seen in Figure 5.6.3, it is evident that whenever the function is concave up on an inter- val, the Trapezoid Rule with one subinterval, T1 , will overestimate the exact value of the deﬁnite integral on that interval. From a careful analysis of the line that bounds the top of the rectangle for the Midpoint Rule (shown in magenta), we see that if we rotate this line segment until it is tangent to the curve at the midpoint of the interval (as shown at right in 314 5.6 Numerical Integration Figure 5.6.3), the resulting trapezoid has the same area as M1 , and this value is less than the exact value of the deﬁnite integral. Thus, when the function is concave up on the interval, M1 underestimates the integral’s true value. M1 ∫b Figure 5.6.4: Comparing the error in estimating a f (x) dx using a single subinterval: in red, the error from the Trapezoid rule; in light red, the error from the Midpoint rule. These observations extend easily to the situation where the function’s concavity remains consistent but we use larger values of n in the Midpoint and Trapezoid Rules. Hence, when- ∫b ever f is concave up on [a, b], Tn will overestimate the value of f (x) dx, while M n will ∫b a underestimate a f (x) dx. The reverse observations are true in the situation where f is con- cave down. Next, we compare the size of the errors between M n and Tn . Again, we focus on M1 and T1 on an interval where the concavity of f is consistent. In Figure 5.6.4, where the error of the Trapezoid Rule is shaded in red, while the error of the Midpoint Rule is shaded lighter red, it is visually apparent that the error in the Trapezoid Rule is more signiﬁcant. To see how much more signiﬁcant, let’s consider two examples and some particular computations. ∫1 If we let f (x) 1 − x 2 and consider 0 f (x) dx, we know by the First FTC that the exact value of the integral is ∫ 1 1 x3 2 (1 − x 2 ) dx x − . 0 3 0 3 Using appropriate technology to compute M4 , M8 , T4 , and T8 , as well as the corresponding errors E M,4 , E M,8 , ET,4 , and ET,8 , as we did in Activity 5.6.2, we ﬁnd the results summarized ∫2 1 in Table 5.6.5. We also include the approximations and their errors for the example 1 x2 dx from Activity 5.6.2. 315 Chapter 5 Evaluating Integrals ∫1 ∫2 Rule 0 (1 − x 2 ) dx 0.6 error dx 0.5 1 1 x2 error T4 0.65625 −0.0104166667 0.5089937642 0.0089937642 M4 0.671875 0.0052083333 0.4955479365 −0.0044520635 T8 0.6640625 −0.0026041667 0.5022708502 0.0022708502 M8 0.66796875 0.0013020833 0.4988674899 −0.0011325101 Table 5.6.5: Calculations of T4 , M4 , T8 , and M8 , along with corresponding errors, for the ∫1 ∫2 deﬁnite integrals 0 (1 − x 2 ) dx and 1 1 x2 dx. ∫b For a given function f and interval [a, b], ET,4 T4 − a f (x) dx calculates the diﬀerence between the approximation generated by the Trapezoid Rule with n 4 and the exact value of the deﬁnite integral. If we look at not only ET,4 , but also the other errors generated by using Tn and M n with n 4 and n 8 in the two examples noted in Table 5.6.5, we see an evident pattern. Not only is the sign of the error (which measures whether the rule generates an over- or under-estimate) tied to the rule used and the function’s concavity, but the magnitude of the errors generated by Tn and M n seems closely connected. In particular, the errors generated by the Midpoint Rule seem to be about half the size (in absolute value) of those generated by the Trapezoid Rule. That is, we can observe in both examples that E M,4 ≈ − 12 ET,4 and E M,8 ≈ − 21 ET,8 . This property of the Midpoint and Trapezoid Rules turns out to hold in general: for a function of consistent concavity, the error in the Midpoint Rule has the opposite sign and approximately half the magnitude of the error of the Trapezoid Rule. Written symbolically, 1 E M,n ≈ − ET,n . 2 This important relationship suggests a way to combine the Midpoint and Trapezoid Rules to create an even more accurate approximation to a deﬁnite integral. 5.6.3 Simpson’s Rule When we ﬁrst developed the Trapezoid Rule, we observed that it is an average of the Left and Right Riemann sums: 1 Tn (L n + R n ). 2 If a function is always increasing or always decreasing on the interval [a, b], one of L n and ∫b R n will over-estimate the true value of a f (x) dx, while the other will under-estimate the integral. Thus, the errors that result from L n and R n will have opposite signs; so averaging L n and R n eliminates a considerable amount of the error present in the respective approx- imations. In a similar way, it makes sense to think about averaging M n and Tn in order to generate a still more accurate approximation. We’ve just observed that M n is typically about twice as accurate as Tn . So we use the weighted average 2M n + Tn S2n . (5.6.5) 3 316 5.6 Numerical Integration The rule for S2n giving by Equation (5.6.5) is usually known as Simpson’s Rule.² Note that we use “S2n ” rather that “S n ” since the n points the Midpoint Rule uses are diﬀerent from the n points the Trapezoid Rule uses, and thus Simpson’s Rule is using 2n points at which to evaluate the function. We build upon the results in Table 5.6.5 to see the approximations generated by Simpson’s Rule. In particular, in Table 5.6.6, we include all of the results in Table 5.6.5, but include additional results for S8 2M43+T4 and S16 2M83+T8 . ∫1 ∫2 Rule 0 (1 − x 2 ) dx 0.6 error 1 1 x2 dx 0.5 error T4 0.65625 −0.0104166667 0.5089937642 0.0089937642 M4 0.671875 0.0052083333 0.4955479365 −0.0044520635 S8 0.6666666667 0 0.5000298792 0.0000298792 T8 0.6640625 −0.0026041667 0.5022708502 0.0022708502 M8 0.66796875 0.0013020833 0.4988674899 −0.0011325101 S16 0.6666666667 0 0.5000019434 0.0000019434 Table 5.6.6: Table 5.6.5 updated to include S8 , S16 , and the corresponding errors. ∫2 The results seen in Table 5.6.6 are striking. If we consider the S16 approximation of 1 x12 dx, the error is only ES,16 0.0000019434. By contrast, L8 0.5491458502, so the error of that estimate is EL,8 0.0491458502. Moreover, we observe that generating the approximations for Simpson’s Rule is almost no additional work: once we have L n , R n , and M n for a given value of n, it is a simple exercise to generate Tn , and from there to calculate S2n . Finally, note ∫1 that the error in the Simpson’s Rule approximations of 0 (1 − x 2 ) dx is zero!³ ∫1 e −x dx, for 2 These rules are not only useful for approximating deﬁnite integrals such as 0 which we cannot ﬁnd an elementary antiderivative of e −x , but also for approximating def- 2 inite integrals when we are given a function through a table of data. Activity 5.6.3. A car traveling along a straight road is braking and its velocity is mea- sured at several diﬀerent points in time, as given in the following table. Assume that v is continuous, always decreasing, and always decreasing at a decreasing rate, as is suggested by the data. ²Thomas Simpson was an 18th century mathematician; his idea was to extend the Trapezoid rule, but rather than using straight lines to build trapezoids, to use quadratic functions to build regions whose area was bounded by parabolas (whose areas he could ﬁnd exactly). Simpson’s Rule is often developed from the more sophisticated perspective of using interpolation by quadratic functions. ³Similar to how the Midpoint and Trapezoid approximations are exact for linear functions, Simpson’s Rule approximations are exact for quadratic and cubic functions. See additional discussion on this issue later in the section and in the exercises. 317 Chapter 5 Evaluating Integrals v seconds, t Velocity in ft/sec, v(t) 0 100 0.3 99 0.6 96 0.9 90 1.2 80 1.5 50 t 1.8 0 0.3 0.6 0.9 1.2 1.5 1.8 Table 5.6.7: Data for the braking car. Figure 5.6.8: Axes for plotting the data in Activity 5.6.3. a. Plot the given data on the set of axes provided in Figure 5.6.8 with time on the horizontal axis and the velocity on the vertical axis. b. What deﬁnite integral will give you the exact distance the car traveled on [0, 1.8]? c. Estimate the total distance traveled on [0, 1.8] by computing L3 , R 3 , and T3 . Which of these under-estimates the true distance traveled? d. Estimate the total distance traveled on [0, 1.8] by computing M3 . Is this an over- or under-estimate? Why? e. Using your results from (c) and (d), improve your estimate further by using Simpson’s Rule. f. What is your best estimate of the average velocity of the car on [0, 1.8]? Why? What are the units on this quantity? 5.6.4 Overall observations regarding L n , R n , Tn , M n , and S2n . As we conclude our discussion of numerical approximation of deﬁnite integrals, it is impor- tant to summarize general trends in how the various rules over- or under-estimate the true value of a deﬁnite integral, and by how much. To revisit some past observations and see some new ones, we consider the following activity. Activity 5.6.4. Consider the functions f (x) 2 − x 2 , 1(x) 2 − x 3 , and h(x) 2 − x 4 , all on the interval [0, 1]. For each of the questions that require a numerical answer in 318 5.6 Numerical Integration what follows, write your answer exactly in fraction form. a. On the three sets of axes provided in Figure 5.6.9, sketch a graph of each function on the interval [0, 1], and compute L1 and R 1 for each. What do you observe? ∫ 1 ∫ 1 b. Compute M1 for each function to approximate f (x) dx, 1(x) dx, and ∫ 1 0 0 h(x) dx, respectively. 0 c. Compute T1 for each of the three functions, and hence compute S2 for each of the three functions. ∫1 ∫1 ∫1 d. Evaluate each of the integrals 0 f (x) dx, 0 1(x) dx, and 0 h(x) dx exactly us- ing the First FTC. e. For each of the three functions f , 1, and h, compare the results of L1 , R 1 , M1 , T1 , and S2 to the true value of the corresponding deﬁnite integral. What patterns do you observe? 2 2 2 1 1 1 Figure 5.6.9: Axes for plotting the functions in Activity 5.6.4. The results seen in Activity 5.6.4 generalize nicely. For instance, if f is decreasing on [a, b], ∫b L n will over-estimate the exact value of a f (x) dx, and if f is concave down on [a, b], M n will over-estimate the exact value of the integral. An excellent exercise is to write a collection of scenarios of possible function behavior, and then categorize whether each of L n , R n , Tn , and M n is an over- or under-estimate. Finally, we make two important notes about Simpson’s Rule. When T. Simpson ﬁrst devel- oped this rule, his idea was to replace the function f on a given interval with a quadratic function that shared three values with the function f . In so doing, he guaranteed that this new approximation rule would be exact for the deﬁnite integral of any quadratic polyno- mial. In one of the pleasant surprises of numerical analysis, it turns out that even though it was designed to be exact for quadratic polynomials, Simpson’s Rule is exact for any cubic ∫5 polynomial: that is, if we are interested in an integral such as 2 (5x 3 − 2x 2 + 7x − 4) dx, S2n will always be exact, regardless of the value of n. This is just one more piece of evidence that shows how eﬀective Simpson’s Rule is as an approximation tool for estimating deﬁnite 319 Chapter 5 Evaluating Integrals integrals.⁴ 5.6.5 Summary ∫1 • For a deﬁnite integral such as 0 e −x dx when we cannot use the First Fundamental 2 Theorem of Calculus because the integrand lacks an elementary algebraic antideriva- tive, we can estimate the integral’s value by using a sequence of Riemann sum approx- imations. Typically, we start by computing L n , R n , and M n for one or more chosen values of n. ∫b • The Trapezoid Rule, which estimates a f (x) dx by using trapezoids, rather than rec- tangles, can also be viewed as the average of Left and Right Riemann sums. That is, Tn 21 (L n + R n ). • The Midpoint Rule is typically twice as accurate as the Trapezoid Rule, and the signs of the respective errors of these rules are opposites. Hence, by taking the weighted ∫b average S2n 2Mn3+Tn , we can build a much more accurate approximation to a f (x) dx by using approximations we have already computed. The rule for S2n is known as Simpson’s Rule, which can also be developed by approximating a given continuous function with pieces of quadratic polynomials. 5.6.6 Exercises Two notes about how Exercise 5.6.6.1 is coded: (i) as explained in the problem header, you need responses to every single entry before you can get individual parts marked as correct; if you enter only an answer for (a) and submit, (a) will be marked wrong regardless. And (ii), in this problem, the notation ”SIMP(2)” is actually what we have called ”SIMP(4)” in our previous work. Diﬀerent authors use diﬀerent notation, and the author of this WeBWorK exercise chooses to write ”SIMP(n)” where we have written ”SIMP(2n)” in Section 5.6. ∫4 1. Various methods for e x numerically. (a) What is the exact value of 0 e x dx? (b) Find LEFT(2), RIGHT(2), TRAP(2), MID(2), and SIMP(2); compute the error for each. (c) Repeat part (b) with n 4 (instead of n 2). (d) For each rule in part (b), as n goes from n 2 to n 4, does the error go down approximately as you would expect? Explain by calculating the ratios of the errors: Error LEFT(2)/Error LEFT(4) = Error RIGHT(2)/Error RIGHT(4) = Error TRAP(2)/Error TRAP(4) = Error MID(2)/Error MID(4) = ⁴One reason that Simpson’s Rule is so eﬀective is that S2n beneﬁts from using 2n + 1 points of data. Because it combines M n , which uses n midpoints, and Tn , which uses the n + 1 endpoints of the chosen subintervals, S2n takes advantage of the maximum amount of information we have when we know function values at the endpoints and midpoints of n subintervals. 320 5.6 Numerical Integration Error SIMP(2)/Error SIMP(4) = (Be sure that you can explain in words why these do (or don’t) make sense.) 2. Comparison of methods for increasing concave down function. Using the ﬁgure ∫3 showing f (x) below, order the following approximations to the integral 0 f (x) dx and its exact value from smallest to largest. Enter each of ”LEFT(n)”, ”RIGHT(n)”, ”TRAP(n)”, ”MID(n)” and ”Exact” in one of the following answer blanks to indicate the correct ordering: < < < < 3. Comparing accuracy for two similar functions. Using a ﬁxed number of subdivisions, we approximate the integrals of f and 1 on the interval shown in the ﬁgure below. (The function f (x) is shown in blue, and 1(x) in black). For which function, f or 1 is LEFT more accurate? For which function, f or 1 is RIGHT more accurate? For which function, f or 1 is MID more accurate? For which function, f or 1 is TRAP more accurate? 4. Identifying and comparing methods. Consider the four functions shown below. On ∫b the ﬁrst two, an approximation for a f (x) dx is shown. 321 Chapter 5 Evaluating Integrals 1. 2. 3. 4. 1. For graph number 1, Which integration method is shown? ⊙ midpoint rule ⊙ left rule ⊙ right rule ⊙ trapezoid rule Is this method an over- or underestimate? 2. For graph number 2, Which integration method is shown? ⊙ right rule ⊙ midpoint rule ⊙ left rule ⊙ trapezoid rule Is this method an over- or underestimate? 3. On a copy of graph number 3, sketch an estimate with n 2 subdivisions using the midpoint rule. Is this method an over- or underestimate? 322 5.6 Numerical Integration 4. On a copy of graph number 4, sketch an estimate with n 2 subdivisions using the trapezoid rule. Is this method an over- or underestimate? ∫1 5. Consider the deﬁnite integral 0 x tan(x) dx. a. Explain why this integral cannot be evaluated exactly by using either u-substitution or by integrating by parts. b. Using appropriate subintervals, compute L4 , R 4 , M4 , T4 , and S8 . c. Which of the approximations in (b) is an over-estimate to the true value of ∫1 0 x tan(x) dx? Which is an under-estimate? How do you know? 6. For an unknown function f (x), the following information is known. • f is continuous on [3, 6]; • f is either always increasing or always decreasing on [3, 6]; • f has the same concavity throughout the interval [3, 6]; ∫6 • As approximations to 3 f (x) dx, L4 7.23, R 4 6.75, and M4 7.05. a. Is f increasing or decreasing on [3, 6]? What data tells you? b. Is f concave up or concave down on [3, 6]? Why? ∫6 c. Determine the best possible estimate you can for 3 f (x) dx, based on the given information. 7. The rate at which water ﬂows through Table Rock Dam on the White River in Branson, MO, is measured in cubic feet per second (CFS). As engineers open the ﬂoodgates, ﬂow rates are recorded according to the following chart. seconds, t 0 10 20 30 40 50 60 ﬂow in CFS, r(t) 2000 2100 2400 3000 3900 5100 6500 Table 5.6.10: Water ﬂow data. a. What deﬁnite integral measures the total volume of water to ﬂow through the dam in the 60 second time period provided by the table above? b. Use the given data to calculate M n for the largest possible value of n to approxi- mate the integral you stated in (a). Do you think M n over- or under-estimates the exact value of the integral? Why? c. Approximate the integral stated in (a) by calculating S n for the largest possible value of n, based on the given data. 1 d. Compute 60 S n and 2000+2100+2400+3000+3900+5100+6500 7 . What quantity do both of these values estimate? Which is a more accurate approximation? 323 Chapter 5 Evaluating Integrals 324 CHAPTER 6 Using Deﬁnite Integrals 6.1 Using Deﬁnite Integrals to Find Area and Length Motivating Questions • How can we use deﬁnite integrals to measure the area between two curves? • How do we decide whether to integrate with respect to x or with respect to y when we try to ﬁnd the area of a region? • How can a deﬁnite integral be used to measure the length of a curve? Early on in our work with the deﬁnite integral, we learned that for an object moving along an axis, the area under a non-negative velocity function v between a and b tells us the distance the object traveled on that time interval, and that area is given precisely by the deﬁnite in- ∫b ∫b tegral a v(t) dt. In general, for any nonnegative function f on an interval [a, b], a f (x) dx measures the area bounded by the curve and the x-axis between x a and x b. Next, we will explore how deﬁnite integrals can be used to represent other physically im- portant properties. In Preview Activity 6.1.1, we investigate how a single deﬁnite integral may be used to represent the area between two curves. Preview Activity 6.1.1. Consider the functions given by f (x) 5 − (x − 1)2 and 1(x) 4 − x. a. Use algebra to ﬁnd the points where the graphs of f and 1 intersect. b. Sketch an accurate graph of f and 1 on the axes provided, labeling the curves by name and the intersection points with ordered pairs. c. Find and evaluate exactly an integral expression that represents the area be- tween y f (x) and the x-axis on the interval between the intersection points of f and 1. d. Find and evaluate exactly an integral expression that represents the area be- tween y 1(x) and the x-axis on the interval between the intersection points of Chapter 6 Using Deﬁnite Integrals f and 1. e. What is the exact area between f and 1 between their intersection points? Why? 6 4 2 1 2 3 Figure 6.1.1: Axes for plotting f and 1 in Preview Activity 6.1.1 6.1.1 The Area Between Two Curves In Preview Activity 6.1.1, we saw a natural way to think about the area between two curves: it is the area beneath the upper curve minus the area below the lower curve. Example 6.1.2 Find the area bounded between the graphs of f (x) (x − 1)2 + 1 and 1(x) x + 2. 6 6 6 4 4 4 g g g 2 2 2 f f 1 2 3 1 2 3 1 2 3 Figure 6.1.3: The areas bounded by the functions f (x) (x − 1)2 + 1 and 1(x) x + 2 on the interval [0, 3]. Solution. In Figure 6.1.3, we see that the graphs intersect at (0, 2) and (3, 5). We can ﬁnd 326 6.1 Using Deﬁnite Integrals to Find Area and Length these intersection points algebraically by solving the system of equations given by y x + 2 and y (x − 1)2 + 1: substituting x + 2 for y in the second equation yields x + 2 (x − 1)2 + 1, so x + 2 x 2 − 2x + 1 + 1, and thus x 2 − 3x x(x − 3) 0, from which it follows that x 0 or x 3. Using y x + 2, we ﬁnd the corresponding y-values of the intersection points. On the interval [0, 3], the area beneath 1 is ∫ 3 21 (x + 2) dx , 0 2 while the area under f on the same interval is ∫ 3 [(x − 1)2 + 1] dx 6. 0 Thus, the area between the curves is ∫ 3 ∫ 3 21 9 A (x + 2) dx − [(x − 1)2 + 1] dx −6 . (6.1.1) 0 0 2 2 We can also think of the area this way: if we slice up the region between two curves into thin vertical rectangles (in the same spirit as we originally sliced the region between a single curve and the x-axis in Section 4.2), we see (as shown in Figure 6.1.4) that the height of a typical rectangle is given by the diﬀerence between the two functions, 1(x) − f (x), and its width is ∆x. Thus the area of the rectangle is Arect (1(x) − f (x))∆x. 6 4 g g g(x) − f (x) 2 f f △x x 1 2 3 Figure 6.1.4: The area bounded by the functions f (x) (x − 1)2 + 1 and 1(x) x + 2 on the interval [0, 3]. 327 Chapter 6 Using Deﬁnite Integrals The area between the two curves on [0, 3] is thus approximated by the Riemann sum ∑ n A≈ (1(x i ) − f (x i ))∆x, i1 and as we let n → ∞, it follows that the area is given by the single deﬁnite integral ∫ 3 A (1(x) − f (x)) dx. (6.1.2) 0 In many applications of the deﬁnite integral, we will ﬁnd it helpful to think of a “representa- tive slice” and use the deﬁnite integral to add these slices. Here, the integral sums the areas of thin rectangles. Finally, it doesn’t matter whether we think of the area between two curves as the diﬀerence between the area bounded by the individual curves (as in (6.1.1)) or as the limit of a Riemann sum of the areas of thin rectangles between the curves (as in (6.1.2)). These two results are the same, since the diﬀerence of two integrals is the integral of the diﬀerence: ∫ 3 ∫ 3 ∫ 3 1(x) dx − f (x) dx (1(x) − f (x)) dx. 0 0 0 Our work so far in this section illustrates the following general principle. If two curves y 1(x) and y f (x) intersect at (a, 1(a)) and (b, 1(b)), and for all x ∫b such that a ≤ x ≤ b, 1(x) ≥ f (x), then the area between the curves is A a (1(x) − f (x)) dx. Activity 6.1.2. In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a repre- sentative slice, and (iv) state the area of the representative slice. Then, state a deﬁnite integral whose value is the exact area of the region, and evaluate the integral to ﬁnd the numeric value of the region’s area. √ a. The ﬁnite region bounded by y x and y 14 x. b. The ﬁnite region bounded by y 12 − 2x 2 and y x 2 − 8. c. The area bounded by the y-axis, f (x) cos(x), and 1(x) sin(x), where we consider the region formed by the ﬁrst positive value of x for which f and 1 intersect. d. The ﬁnite regions between the curves y x 3 − x and y x 2 . 328 6.1 Using Deﬁnite Integrals to Find Area and Length 6.1.2 Finding Area with Horizontal Slices At times, the shape of a region may dictate that we use horizontal rectangular slices, instead of vertical ones. Example 6.1.5 Find the area of the region bounded by the parabola x y 2 − 1 and the line y x − 1, shown at left in Figure 6.1.6. x = y2 − 1 x = y2 − 1 x = y2 − 1 2 2 2 1 1 1 △y 1 2 3 1 2 3 1 2 3 -1 y = x−1 -1 y = x−1 -1 x = y+1 Figure 6.1.6: The area bounded by the functions x y 2 − 1 and y x − 1 (at left), with the region sliced vertically (center) and horizontally (at right). Solution. By solving the second equation for x and writing x y + 1, we ﬁnd that y + 1 y 2 − 1. Hence the curves intersect where y 2 − y − 2 0. Thus, we ﬁnd y −1 or y 2, so the intersection points of the two curves are (0, −1) and (3, 2). If we attempt to use vertical rectangles to slice up the area (as in the center graph of Fig- ure 6.1.6), we see that from x −1 to x 0 the curves that bound the top and bottom of the rectangle are one and the same. This suggests, as shown in the rightmost graph in the ﬁgure, that we try using horizontal rectangles. Note that the width of a horizontal rectangle depends on y. Between y −1 and y 2, the right end of a representative rectangle is determined by the line x y + 1, and the left end is determined by the parabola, x y 2 − 1. The thickness of the rectangle is ∆y. Therefore, the area of the rectangle is Arect [(y + 1) − (y 2 − 1)]∆y, and the area between the two curves on the y-interval [−1, 2] is approximated by the Rie- mann sum ∑ n A≈ [(y i + 1) − (y i2 − 1)]∆y. i1 Taking the limit of the Riemann sum, it follows that the area of the region is ∫ y2 A [(y + 1) − (y 2 − 1)] dy. (6.1.3) y−1 329 Chapter 6 Using Deﬁnite Integrals We emphasize that we are integrating with respect to y; this is because we chose to use horizontal rectangles whose widths depend on y and whose thickness is denoted ∆y. It is a straightforward exercise to evaluate the integral in Equation (6.1.3) and ﬁnd that A 92 . Just as with the use of vertical rectangles of thickness ∆x, we have a general principle for ﬁnding the area between two curves, which we state as follows. If two curves x 1(y) and x f (y) intersect at (1(c), c) and (1(d), d), and for all y such that c ≤ y ≤ d, 1(y) ≥ f (y), then the area between the curves is ∫ yd A (1(y) − f (y)) dy. yc Activity 6.1.3. In each of the following problems, our goal is to determine the area of the region described. For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a repre- sentative slice, and (iv) state the area of the representative slice. Then, state a deﬁ- nite integral whose value is the exact area of the region, and evaluate the integral to ﬁnd the numeric value of the region’s area. Note well: At the step where you draw a representative slice, you need to make a choice about whether to slice vertically or horizontally. a. The ﬁnite region bounded by x y 2 and x 6 − 2y 2 . b. The ﬁnite region bounded by x 1 − y 2 and x 2 − 2y 2 . c. The area bounded by the x-axis, y x 2 , and y 2 − x. d. The ﬁnite regions between the curves x y 2 − 2y and y x. 6.1.3 Finding the length of a curve We can also use the deﬁnite integral to ﬁnd the length of a portion of a curve. We use the same fundamental principle: we slice the curve up into small pieces whose lengths we can easily approximate. Speciﬁcally, we subdivide the curve into small approximating line segments, as shown at left in Figure 6.1.7. We estimate the length L slice of each portion of the curve on a small interval of length ∆x. We use the right triangle with legs parallel to the coordinate axes and hypotenuse connecting the endpoints of the slice, as seen at right in Figure 6.1.7. The length, h, of the hypotenuse approximates the length, Lslice , of the curve between the two selected points. Thus, √ Lslice ≈ h (∆x)2 + (∆y)2 . Next we use algebra to rearrange the expression for the length of the hypotenuse into a form that we can integrate. By removing a factor of (∆x)2 , we ﬁnd √ Lslice ≈ (∆x)2 + (∆y)2 330 6.1 Using Deﬁnite Integrals to Find Area and Length y f h △y Lslice x x0 x1 x2 x3 △x Figure 6.1.7: At left, a continuous function y f (x) whose length we seek on the interval a x0 to b x3 . At right, a close up view of a portion of the curve. √ ( ) (∆y)2 (∆x)2 1 + (∆x)2 √ (∆y)2 1+ · ∆x. (∆x)2 ∆y dy Then, as n → ∞ and ∆x → 0, we have that ∆x → dx f ′(x). Thus, we can say that √ Lslice ≈ 1 + f ′(x)2 ∆x. Taking a Riemann sum of all of these slices and letting n → ∞, we arrive at the following fact. Given a diﬀerentiable function f on an interval [a, b], the total arc length, L, along the curve y f (x) from x a to x b is given by ∫ b √ L 1 + f ′(x)2 dx. a Activity 6.1.4. Each of the following questions somehow involves the arc length along a curve. a. Use the deﬁnition and appropriate computational technology to determine the arc length along y x 2 from x −1 to x 1. 331 Chapter 6 Using Deﬁnite Integrals √ b. Find the arc length of y 4 − x 2 on the interval −2 ≤ x ≤ 2. Find this value in two diﬀerent ways: (a) by using a deﬁnite integral, and (b) by using a familiar property of the curve. c. Determine the arc length of y xe 3x on the interval [0, 1]. d. Will the integrals that arise calculating arc length typically be ones that we can evaluate exactly using the First FTC, or ones that we need to approximate? Why? e. A moving particle is traveling along the curve given by y f (x) 0.1x 2 +1, and does so at a constant rate of 7 cm/sec, where both x and y are measured in cm (that is, the curve y f (x) is the path along which the object actually travels; the curve is not a “position function”). Find the position of the particle when t 4 sec, assuming that when t 0, the particle’s location is (0, f (0)). 6.1.4 Summary • To ﬁnd the area between two curves, we think about slicing the region into thin rec- tangles. If, for instance, the area of a typical rectangle on the interval x a to x b is given by Arect (1(x) − f (x))∆x, then the exact area of the region is given by the deﬁnite integral ∫ b A (1(x) − f (x)) dx. a • The shape of the region usually dictates whether we should use vertical rectangles of thickness ∆x or horizontal rectangles of thickness ∆y. We want the height of the rectangle given by the diﬀerence between two curves: if those curves are best thought of as functions of y, we use horizontal rectangles, whereas if those curves are best viewed as functions of x, we use vertical rectangles. • The arc length, L, along the curve y f (x) from x a to x b is given by ∫ b √ L 1 + f ′(x)2 dx. a 6.1.5 Exercises 1. Area between two power functions. Find the area of the region between y x 1/2 and y x 1/5 for 0 ≤ x ≤ 1. 2. Area between two trigonometric functions. Find the area between y 8 sin x and y 9 cos x over the interval [0, π]. Sketch the curves if necessary. ¹This integral is actually ”improper” because the integrand is undeﬁned at the endpoints, x ±2. We will learn how to evaluate such integrals in Section 6.5. 332 6.1 Using Deﬁnite Integrals to Find Area and Length 3. Area between two curves. Sketch the region enclosed by x + y 2 56 and x + y 0. Decide whether to integrate with respect to x or y, and then ﬁnd the area of the region. √ 4. Arc length of a curve. Find the arc length of the graph of the function f (x) 2 x 3 from x 2 to x 5. 5. Find the exact area of each described region. a. The ﬁnite region between the curves x y(y − 2) and x −(y − 1)(y − 3). b. The region between the sine and cosine functions on the interval [ π4 , 4 ]. 3π c. The ﬁnite region between x y 2 − y − 2 and y 2x − 1. d. The ﬁnite region between y mx and y x 2 − 1, where m is a positive constant. 6. Let f (x) 1 − x 2 and 1(x) ax 2 − a, where a is an unknown positive real number. For what value(s) of a is the area between the curves f and 1 equal to 2? 7. Let f (x) 2 − x 2 . Recall that the average value of any continuous function f on an ∫b interval [a, b] is given by 1 b−a a f (x) dx. √ a. Find the average value of f (x) 2 − x 2 on the interval [0, 2]. Call this value r. b. Sketch a graph of y f (x) and y r. Find their intersection point(s). √ c. Show that on the interval [0, 2], the amount of area that lies below y f (x) and above y r is equal to the amount of area that lies below y r and above y f (x). d. Will the result of (c) be true for any continuous function and its average value on any interval? Why? 333 Chapter 6 Using Deﬁnite Integrals 6.2 Using Deﬁnite Integrals to Find Volume Motivating Questions • How can we use a deﬁnite integral to ﬁnd the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular axis? • In what circumstances do we integrate with respect to y instead of integrating with respect to x? • What adjustments do we need to make if we revolve about a line other than the x- or y-axis? Just as we can use deﬁnite integrals to add the areas of rectangular slices to ﬁnd the exact area that lies between two curves, we can also use integrals to ﬁnd the volume of regions whose cross-sections have a particular shape. In particular, we can determine the volume of solids whose cross-sections are all thin cylin- ders (or washers) by adding up the volumes of these individual slices. We ﬁrst consider a familiar shape in Preview Activity 6.2.1: a circular cone. Preview Activity 6.2.1. Consider a circular cone of radius 3 and height 5, which we view horizontally as pictured in Figure 6.2.1. Our goal in this activity is to use a deﬁnite integral to determine the volume of the cone. y 3 x x 5 ∆x Figure 6.2.1: The circular cone described in Preview Activity 6.2.1 a. Find a formula for the linear function y f (x) that is pictured in Figure 6.2.1. b. For the representative slice of thickness ∆x that is located horizontally at a lo- 334 6.2 Using Deﬁnite Integrals to Find Volume cation x (somewhere between x 0 and x 5), what is the radius of the repre- sentative slice? Note that the radius depends on the value of x. c. What is the volume of the representative slice you found in (b)? d. What deﬁnite integral will sum the volumes of the thin slices across the full horizontal span of the cone? What is the exact value of this deﬁnite integral? e. Compare the result of your work in (d) to the volume of the cone that comes from using the formula Vcone 31 πr 2 h. 6.2.1 The Volume of a Solid of Revolution A solid of revolution is a three dimensional solid that can be generated by revolving one or more curves around a ﬁxed axis. For example, the circular cone in Figure 6.2.1 is the solid of revolution generated by revolving the portion of the line y 3 − 35 x from x 0 to x 5 about the x-axis. Notice that if we slice a solid of revolution perpendicular to the axis of revolution, the resulting cross-section is a circle. We ﬁrst consider solids whose slices are thin cylinders. Recall that the volume of a cylinder is given by V πr 2 h. Example 6.2.2 Find the volume of the solid of revolution generated when the region R bounded by y 4 − x 2 and the x-axis is revolved about the x-axis. Solution. First, we observe that y 4 − x 2 intersects the x-axis at the points (−2, 0) and (2, 0). When we revolve the region R about the x-axis, we get the three-dimensional solid pictured in Figure 6.2.3. y y = 4 − x2 x ∆x Figure 6.2.3: The solid of revolution in Example 6.2.2. 335 Chapter 6 Using Deﬁnite Integrals We slice the solid into vertical slices of thickness ∆x between x −2 and x 2. A repre- sentative slice is a cylinder of height ∆x and radius 4 − x 2 . Hence, the volume of the slice is Vslice π(4 − x 2 )2 ∆x. Using a deﬁnite integral to sum the volumes of the representative slices, it follows that ∫ 2 V π(4 − x 2 )2 dx. −2 It is straightforward to evaluate the integral and ﬁnd that the volume is V 15 π. 512 For a solid such as the one in Example 6.2.2, where each slice is a cylindrical disk, we ﬁrst ﬁnd the volume of a typical slice (noting particularly how this volume depends on x), and then integrate over the range of x-values that bound the solid. Often, we will be content with simply ﬁnding the integral that represents the volume; if we desire a numeric value for the integral, we typically use a calculator or computer algebra system to ﬁnd that value. This method for ﬁnding the volume of a solid of revolution is often called the disk method. The Disk Method. If y r(x) is a nonnegative continuous function on [a, b], then the volume of the solid of revolution generated by revolving the curve about the x-axis over this interval is given by ∫ b V πr(x)2 dx. a A diﬀerent type of solid can emerge when two curves are involved, as we see in the following example. Example 6.2.4 Find the volume of the solid of revolution generated when the ﬁnite region R that lies between y 4 − x 2 and y x + 2 is revolved about the x-axis. Solution. First, we must determine where the curves y 4 − x 2 and y x + 2 intersect. Substituting the expression for y from the second equation into the ﬁrst equation, we ﬁnd that x + 2 4 − x 2 . Rearranging, it follows that x 2 + x − 2 0, and the solutions to this equation are x −2 and x 1. The curves therefore cross at (−2, 0) and (1, 1). When we revolve the region R about the x-axis, we get the three-dimensional solid pictured at left in Figure 6.2.5. Immediately we see a major diﬀerence between the solid in this example and the one in Example 6.2.2: here, the three-dimensional solid of revolution isn’t “solid” because it has open space in its center along the axis of revolution. If we slice the solid perpendicular to the axis of revolution, we observe that the resulting slice is not a solid disk, but rather a washer, as pictured at right in Figure 6.2.5. At a given location x between x −2 and x 1, 336 6.2 Using Deﬁnite Integrals to Find Volume the small radius r(x) of the inner circle is determined by the curve y x + 2, so r(x) x + 2. Similarly, the big radius R(x) comes from the function y 4 − x 2 , and thus R(x) 4 − x 2 . y R(x) x r(x) Figure 6.2.5: At left, the solid of revolution in Example 6.2.4. At right, a typical slice with inner radius r(x) and outer radius R(x). To ﬁnd the volume of a representative slice, we compute the volume of the outer disk and subtract the volume of the inner disk. Since πR(x)2 ∆x − πr(x)2 ∆x π[R(x)2 − r(x)2 ]∆x, it follows that the volume of a typical slice is Vslice π[(4 − x 2 )2 − (x + 2)2 ]∆x. Using a deﬁnite integral to sum the volumes of the respective slices across the integral, we ﬁnd that ∫ 1 V π[(4 − x 2 )2 − (x + 2)2 ] dx. −2 Evaluating the integral, we ﬁnd that the volume of the solid of revolution is V 5 π. 108 This method for ﬁnding the volume of a solid of revolution generated by two curves is often called the washer method. 337 Chapter 6 Using Deﬁnite Integrals The Washer Method. If y R(x) and y r(x) are nonnegative continuous functions on [a, b] that satisfy R(x) ≥ r(x) for all x in [a, b], then the volume of the solid of revolution generated by revolving the region between them about the x-axis over this interval is given by ∫ b V π[R(x)2 − r(x)2 ] dx. a Activity 6.2.2. In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a deﬁnite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you ﬁnd. √ a. The region S bounded by the x-axis, the curve y x, and the line x 4; revolve S about the x-axis. √ b. The region S bounded by the y-axis, the curve y x, and the line y 2; revolve S about the x-axis. √ c. The ﬁnite region S bounded by the curves y x and y x 3 ; revolve S about the x-axis. d. The ﬁnite region S bounded by the curves y 2x 2 + 1 and y x 2 + 4; revolve S about the x-axis. √ e. The region S bounded by the y-axis, the curve y x, and the line y 2; revolve S about the y-axis. How is this problem diﬀerent from the one posed in part (b)? 6.2.2 Revolving about the y-axis When we revolve a given region about the y-axis, the representative slices now have thick- ness ∆y, which means that we must integrate with respect to y. Example 6.2.6 Find √ the volume of the solid of revolution generated when the region R that lies between y x and y x 4 is revolved about the y-axis. Solution. These two curves intersect when x 1, hence at the point (1, 1). When we re- volve the region R about the y-axis, we get the three-dimensional solid pictured at left in Figure 6.2.7. Note that the slices are cylindrical washers only if taken perpendicular to the y-axis. We slice the solid horizontally, starting at y 0 and proceeding up to y 1. The thickness of a representative slice is ∆y, so we√must express the integrand in terms of y. The inner radius is determined by the curve y x, so we solve for x and get x y 2 r(y). In the same way, we solve the curve y x 4 (which governs the outer radius) for x in terms of y, and hence 338 6.2 Using Deﬁnite Integrals to Find Volume √4 x y. Therefore, the volume of a typical slice is √ Vslice π[R(y)2 − r(y)2 ] π[( 4 y)2 − (y 2 )2 ]∆y. y R(y) r(y) x Figure 6.2.7: At left, the solid of revolution in Example 6.2.6. At right, a typical slice with inner radius r(y) and outer radius R(y). We use a deﬁnite integral to sum the volumes of all the slices from y 0 to y 1. The total volume is ∫ y1 [√ ] V π ( 4 y)2 − (y 2 )2 dy. y0 It is straightforward to evaluate the integral and ﬁnd that V 15 π. 7 Activity 6.2.3. In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a deﬁnite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you ﬁnd. √ a. The region S bounded by the y-axis, the curve y x, and the line y 2; revolve S about the y-axis. √ b. The region S bounded by the x-axis, the curve y x, and the line x 4; revolve S about the y-axis. c. The ﬁnite region S in the ﬁrst quadrant bounded by the curves y 2x and y x 3 ; revolve S about the x-axis. 339 Chapter 6 Using Deﬁnite Integrals d. The ﬁnite region S in the ﬁrst quadrant bounded by the curves y 2x and y x 3 ; revolve S about the y-axis. e. The ﬁnite region S bounded by the curves x (y − 1)2 and y x − 1; revolve S about the y-axis 6.2.3 Revolving about horizontal and vertical lines other than the coordi- nate axes It is possible to revolve a region around any horizontal or vertical line. Doing so adjusts the radii of the cylinders or washers involved by a constant value. A careful, well-labeled plot of the solid of revolution will usually reveal how the diﬀerent axis of revolution aﬀects the deﬁnite integral. Example 6.2.8 Find the volume of the solid of revolution generated when the ﬁnite region S that lies between y x 2 and y x is revolved about the line y −1. Solution. Graphing the region between the two curves in the ﬁrst quadrant between their points of intersection ((0, 0) and (1, 1)) and then revolving the region about the line y −1, we see the solid shown in Figure 6.2.9. Each slice of the solid perpendicular to the axis of revolution is a washer, and the radii of each washer are governed by the curves y x 2 and y x. But we also see that there is one added change: the axis of revolution adds a ﬁxed length to each radius. The inner radius of a typical slice, r(x), is given by r(x) x 2 + 1, while the outer radius is R(x) x + 1. y x Figure 6.2.9: The solid of revolution described in Example 6.2.8. Therefore, the volume of a typical slice is [ ] Vslice π[R(x)2 − r(x)2 ]∆x π (x + 1)2 − (x 2 + 1)2 ∆x. 340 6.2 Using Deﬁnite Integrals to Find Volume Finally, we integrate to ﬁnd the total volume, and ∫ 1 [ ] 7 V π (x + 1)2 − (x 2 + 1)2 dx π. 0 15 Activity 6.2.4. In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a deﬁnite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you ﬁnd. For each prompt, use the ﬁnite region S in the ﬁrst quadrant bounded by the curves y 2x and y x 3 . a. Revolve S about the line y −2. c. Revolve S about the line x −1. b. Revolve S about the line y 4. d. Revolve S about the line x 5. 6.2.4 Summary • We can use a deﬁnite integral to ﬁnd the volume of a three-dimensional solid of revo- lution that results from revolving a two-dimensional region about a particular axis by taking slices perpendicular to the axis of revolution which will then be circular disks or washers. • If we revolve about a vertical line and slice perpendicular to that line, then our slices are horizontal and of thickness ∆y. This leads us to integrate with respect to y, as opposed to with respect to x when we slice a solid vertically. • If we revolve about a line other than the x- or y-axis, we need to carefully account for the shift that occurs in the radius of a typical slice. Normally, this shift involves taking a sum or diﬀerence of the function along with the constant connected to the equation for the horizontal or vertical line; a well-labeled diagram is usually the best way to decide the new expression for the radius. 6.2.5 Exercises 1. Solid of revolution from one function about the x-axis. The region bounded by y e x , y 0, x −2, x −1 is rotated around the x-axis. Find the volume. 2. Solid of revolution from one function about the y-axis. Find the volume of the solid obtained by rotating the region in the ﬁrst quadrant bounded by y x 6 , y 1, and the y-axis around the y-axis. 3. Solid of revolution from two functions about the x-axis. Find the volume of the solid obtained by rotating the region in the ﬁrst quadrant bounded by y x 6 , y 1, and the y-axis around the x-axis. 4. Solid of revolution from two functions about a horizontal line. Find the volume of the solid obtained by rotating the region in the ﬁrst quadrant bounded by y x 6 , y 1, 341 Chapter 6 Using Deﬁnite Integrals and the y-axis about the line y −5. 5. Solid of revolution from two functions about a diﬀerent horizontal line. Find the volume of the solid obtained by rotating the region bounded by the curves y x6 , y1 about the line y 5 . 6. Solid of revolution from two functions about a vertical line. Find the volume of the solid obtained by rotating the region bounded by the given curves about the line x −6 y x2 , x y2 3 7. Consider the curve f (x) 3 cos( x4 ) and the portion of its graph that lies in the ﬁrst quadrant between the y-axis and the ﬁrst positive value of x for which f (x) 0. Let R denote the region bounded by this portion of f , the x-axis, and the y-axis. a. Set up a deﬁnite integral whose value is the exact arc length of f that lies along the upper boundary of R. Use technology appropriately to evaluate the integral you ﬁnd. b. Set up a deﬁnite integral whose value is the exact area of R. Use technology appropriately to evaluate the integral you ﬁnd. c. Suppose that the region R is revolved around the x-axis. Set up a deﬁnite integral whose value is the exact volume of the solid of revolution that is generated. Use technology appropriately to evaluate the integral you ﬁnd. d. Suppose instead that R is revolved around the y-axis. If possible, set up an in- tegral expression whose value is the exact volume of the solid of revolution and evaluate the integral using appropriate technology. If not possible, explain why. 8. Consider the curves given by y sin(x) and y cos(x). For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice. a. Sketch the region R bounded by the y-axis and the curves y sin(x) and y cos(x) up to the ﬁrst positive value of x at which they intersect. What is the exact intersection point of the curves? b. Set up a deﬁnite integral whose value is the exact area of R. c. Set up a deﬁnite integral whose value is the exact volume of the solid of revolution generated by revolving R about the x-axis. d. Set up a deﬁnite integral whose value is the exact volume of the solid of revolution generated by revolving R about the y-axis. e. Set up a deﬁnite integral whose value is the exact volume of the solid of revolution generated by revolving R about the line y 2. f. Set up a deﬁnite integral whose value is the exact volume of the solid of revolution generated by revolving R about the line x −1. 342 6.2 Using Deﬁnite Integrals to Find Volume 9. Consider the ﬁnite region R that is bounded by the curves y 1 + 12 (x − 2)2 , y 21 x 2 , and x 0. a. Determine a deﬁnite integral whose value is the area of the region enclosed by the two curves. b. Find an expression involving one or more deﬁnite integrals whose value is the volume of the solid of revolution generated by revolving the region R about the line y −1. c. Determine an expression involving one or more deﬁnite integrals whose value is the volume of the solid of revolution generated by revolving the region R about the y-axis. d. Find an expression involving one or more deﬁnite integrals whose value is the perimeter of the region R. 343 Chapter 6 Using Deﬁnite Integrals 6.3 Density, Mass, and Center of Mass Motivating Questions • How are mass, density, and volume related? • How is the mass of an object with varying density computed? • What is is the center of mass of an object, and how are deﬁnite integrals used to compute it? Studying the units on the integrand and variable of integration helps us understand the meaning of a deﬁnite integral. For instance, if v(t) is the velocity of an object moving along an axis, measured in feet per second, and t measures time in seconds, then both the deﬁnite integral and its Riemann sum approximation, ∫ b ∑ n v(t) dt ≈ v(t i )∆t, a i1 have units given by the product of the units of v(t) and t: (feet/sec) · (sec) feet. ∫b Thus, a v(t) dt measures the total change in position of the moving object in feet. Unit analysis will be particularly helpful to us in what follows. Preview Activity 6.3.1. In each of the following scenarios, we consider the distribu- tion of a quantity along an axis. a. Suppose that the function c(x) 200 + 100e −0.1x models the density of traﬃc on a straight road, measured in cars per mile, where x is number of miles east of a ∫2 major interchange, and consider the deﬁnite integral 0 (200 + 100e −0.1x ) dx. i. What are the units on the product c(x) · ∆x? ii. What are the units on the deﬁnite integral and its Riemann sum approxi- mation given by ∫ 2 ∑ n c(x) dx ≈ c(x i )∆x? 0 i1 ∫2 ∫2( ) iii. Evaluate the deﬁnite integral 0 c(x) dx 0 200 + 100e −0.1x dx and write one sentence to explain the meaning of the value you ﬁnd. b. On a 6 foot long shelf ﬁlled with books, the function B models the distribution of the weight of the books, in pounds per inch, where x is the number of inches from the left end of the bookshelf. Let B(x) be given by the rule B(x) 0.5 + 1 (x+1)2 . 344 6.3 Density, Mass, and Center of Mass i. What are the units on the product B(x) · ∆x? ii. What are the units on the deﬁnite integral and its Riemann sum approxi- mation given by ∫ 36 ∑ n B(x) dx ≈ B(x i )∆x? 12 i1 ∫ 72 ∫ 72 ( ) iii. Evaluate the deﬁnite integral 0 B(x) dx 0 0.5 + 1 (x+1)2 dx and write one sentence to explain the meaning of the value you ﬁnd. 6.3.1 Density The mass of a quantity, typically measured in metric units such as grams or kilograms, is a measure of the amount of the quantity. In a corresponding way, the density of an object measures the distribution of mass per unit volume. For instance, if a brick has mass 3 kg and volume 0.002 m3 , then the density of the brick is 3kg kg 3 1500 3 . 0.002m m As another example, the mass density of water is 1000 kg/m3 . Each of these relationships demonstrate the following general principle. For an object of constant density d, with mass m and volume V, m d , or m d · V. V But what happens when the density is not constant? The formula m d · V is reminiscent of two other equations that we have used in our work: for a body moving in a ﬁxed direction, distance = rate · time, and, for a rectangle, its area is given by A l · w. These formulas hold when the principal quantities involved, such as the rate the body moves and the height of the rectangle, are constant. When these quantities are not constant, we have turned to the deﬁnite integral for assistance. By working with small slices on which the quantity of interest (such as velocity) is approximately constant, we can use a deﬁnite integral to add up the values on the pieces. For example, if we have a nonnegative velocity function that is not constant, over a short time interval ∆t we know that the distance traveled is approximately v(t)∆t, since v(t) is almost constant on a small interval. Similarly, if we are thinking about the area under a nonnegative function f whose value is changing, on a short interval ∆x the area under the curve is approximately the area of the rectangle whose height is f (x) and whose width is ∆x: f (x)∆x. Both of these principles are represented visually in Figure 6.3.1. In a similar way, if the density of some object is not constant, we can use a deﬁnite integral to compute 345 Chapter 6 Using Deﬁnite Integrals ft/sec y y = v(t) y = f (x) v(t) f (x) sec x △t △x Figure 6.3.1: At left, estimating a small amount of distance traveled, v(t)∆t, and at right, a small amount of area under the curve, f (x)∆x. the overall mass of the object. We will focus on problems where the density varies in only one dimension, say along a single axis. Let’s consider a thin bar of length b whose left end is at the origin, where x 0, and assume that the bar has constant cross-sectional area of 1 cm2 . We let ρ(x) represent the mass density function of the bar, measured in grams per cubic centimeter. That is, given a location x, ρ(x) tells us approximately how much mass will be found in a one-centimeter wide slice of the bar at x. x ∆x Figure 6.3.2: A thin bar of constant cross-sectional area 1 cm2 with density function ρ(x) g/cm3 . The volume of a thin slice of the bar of width ∆x, as pictured in Figure 6.3.2, is the cross- sectional area times ∆x. Since the cross-sections each have constant area 1 cm2 , it follows that the volume of the slice is 1∆x cm3 . And because mass is the product of density and volume, we see that the mass of this slice is approximately g massslice ≈ ρ(x) · 1∆x cm3 ρ(x) · ∆x g. cm3 346 6.3 Density, Mass, and Center of Mass The corresponding Riemann sum (and the integral that it approximates), ∑ n ∫ b ρ(x i )∆x ≈ ρ(x) dx, i1 0 therefore measure the mass of the bar between 0 and b. (The Riemann sum is an approxi- mation, while the integral will be the exact mass.) For objects whose cross-sectional area is constant and whose mass is distributed relative to horizontal location, x, it makes sense to think of the density function ρ(x) with units “mass per unit length,” such as g/cm. Thus, when we compute ρ(x) · ∆x on a small slice ∆x, the resulting units are g/cm · cm = g, which thus measures the mass of the slice. The general principle follows. For an object of constant cross-sectional area whose mass is distributed along a single axis according to the function ρ(x) (whose units are units of mass per unit of length), the total mass, M, of the object between x a and x b is given by ∫ b M ρ(x) dx. a Activity 6.3.2. Consider the following situations in which mass is distributed in a non-constant manner. a. Suppose that a thin rod with constant cross-sectional area of 1 cm2 has its mass distributed according to the density function ρ(x) 2e −0.2x , where x is the dis- tance in cm from the left end of the rod, and the units on ρ(x) are g/cm. If the rod is 10 cm long, determine the exact mass of the rod. b. Consider the cone that has a base of radius 4 m and a height of 5 m. Picture the cone lying horizontally with the center of its base at the origin and think of the cone as a solid of revolution. i. Write and evaluate a deﬁnite integral whose value is the volume of the cone. ii. Next, suppose that the cone has uniform density of 800 kg/m3 . What is the mass of the solid cone? iii. Now suppose that the cone’s density is not uniform, but rather that the cone is most dense at its base. In particular, assume that the density of the cone is uniform across cross sections parallel to its base, but that in each such cross section that is a distance x units from the origin, the density of the cross section is given by the function ρ(x) 400 + 1+x200 2 , measured in 3 kg/m . Determine and evaluate a deﬁnite integral whose value is the mass of this cone of non-uniform density. Do so by ﬁrst thinking about the mass of a given slice of the cone x units away from the base; remember that in such a slice, the density will be essentially constant. 347 Chapter 6 Using Deﬁnite Integrals c. Let a thin rod of constant cross-sectional area 1 cm2 and length 12 cm have its mass be distributed according to the density function ρ(x) 251 (x − 15)2 , mea- sured in g/cm. Find the exact location z at which to cut the bar so that the two pieces will each have identical mass. 6.3.2 Weighted Averages The concept of an average is a natural one, and one that we have used repeatedly as part of our understanding of the meaning of the deﬁnite integral. If we have n values a1 , a2 , . . ., a n , we know that their average is given by a1 + a2 + · · · + a n , n and for a quantity being measured by a function f on an interval [a, b], the average value of the quantity on [a, b] is ∫ b 1 f (x) dx. b−a a As we continue to think about problems involving the distribution of mass, it is natural to consider the idea of a weighted average, where certain quantities involved are counted more in the average. A common use of weighted averages class grade grade points credits is in the computation of a student’s chemistry B+ 3.3 5 GPA, where grades are weighted ac- calculus A- 3.7 4 cording to credit hours. Let’s consider history B- 2.7 3 the scenario in Table 6.3.3. psychology B- 2.7 3 Table 6.3.3: A college student’s semester grades. If all of the classes were of the same weight (i.e., the same number of credits), the student’s GPA would simply be calculated by taking the average 3.3 + 3.7 + 2.7 + 2.7 3.1. 4 But since the chemistry and calculus courses have higher weights (of 5 and 4 credits respec- tively), we actually compute the GPA according to the weighted average 3.3 · 5 + 3.7 · 4 + 2.7 · 3 + 2.7 · 3 3.16. 5+4+3+3 The weighted average reﬂects the fact that chemistry and calculus, as courses with higher credits, have a greater impact on the students’ grade point average. Note particularly that in the weighted average, each grade gets multiplied by its weight, and we divide by the sum of the weights. In the following activity, we explore further how weighted averages can be used to ﬁnd the balancing point of a physical system. 348 6.3 Density, Mass, and Center of Mass Activity 6.3.3. For quantities of equal weight, such as two children on a teeter-totter, the balancing point is found by taking the average of their locations. When the weights of the quantities diﬀer, we use a weighted average of their respective locations to ﬁnd the balancing point. a. Suppose that a shelf is 6 feet long, with its left end situated at x 0. If one book of weight 1 lb is placed at x1 0, and another book of weight 1 lb is placed at x2 6, what is the location of x, the point at which the shelf would (theoretically) balance on a fulcrum? b. Now, say that we place four books on the shelf, each weighing 1 lb: at x 1 0, at x2 2, at x3 4, and at x4 6. Find x, the balancing point of the shelf. c. How does x change if we change the location of the third book? Say the locations of the 1-lb books are x1 0, x2 2, x 3 3, and x4 6. d. Next, suppose that we place four books on the shelf, but of varying weights: at x1 0 a 2-lb book, at x2 2 a 3-lb book, at x3 4 a 1-lb book, and at x4 6 a 1-lb book. Use a weighted average of the locations to ﬁnd x, the balancing point of the shelf. How does the balancing point in this scenario compare to that found in (b)? e. What happens if we change the location of one of the books? Say that we keep everything the same in (d), except that x3 5. How does x change? f. What happens if we change the weight of one of the books? Say that we keep everything the same in (d), except that the book at x3 4 now weighs 2 lbs. How does x change? g. Experiment with a couple of diﬀerent scenarios of your choosing where you move one of the books to the left, or you decrease the weight of one of the books. h. Write a couple of sentences to explain how adjusting the location of one of the books or the weight of one of the books aﬀects the location of the balancing point of the shelf. Think carefully here about how your changes should be considered relative to the location of the balancing point x of the current scenario. 6.3.3 Center of Mass In Activity 6.3.3, we saw that the balancing point of a system of point-masses¹ (such as books on a shelf) is found by taking a weighted average of their respective locations. In the activity, we were computing the center of mass of a system of masses distributed along an axis, which is the balancing point of the axis on which the masses rest. ¹In the activity, we actually used weight rather than mass. Since weight is proportional to mass, the computations for the balancing point result in the same location regardless of whether we use weight or mass. The gravitational constant is present in both the numerator and denominator of the weighted average. 349 Chapter 6 Using Deﬁnite Integrals Center of Mass (point-masses). For a collection of n masses m 1 , . . ., m n that are distributed along a single axis at the locations x1 , . . ., x n , the center of mass is given by x1 m1 + x2 m2 + · · · + x n m n x . m1 + m2 + · · · + m n Now consider a thin bar over which density is distributed continuously. If the density is constant, it is obvious that the balancing point of the bar is its midpoint. But if density is not constant, we must compute a weighted average. Let’s say that the function ρ(x) tells us the density distribution along the bar, measured in g/cm. If we slice the bar into small sections, we can think of the bar as holding a collection of adjacent point-masses. The mass m i of a slice of thickness ∆x at location x i , is m i ≈ ρ(x i )∆x. If we slice the bar into n pieces, we can approximate its center of mass by x 1 · ρ(x 1 )∆x + x2 · ρ(x2 )∆x + · · · + x n · ρ(x n )∆x x≈ . ρ(x 1 )∆x + ρ(x2 )∆x + · · · + ρ(x n )∆x Rewriting the sums in sigma notation, we have ∑n i1 x i · ρ(x i )∆x x≈ ∑ n . (6.3.1) i1 ρ(x i )∆x The greater the number of slices, the more accurate our estimate of the balancing point will be. The sums in Equation (6.3.1) can be viewed as Riemann sums, so in the limit as n → ∞, we ﬁnd that the center of mass is given by the quotient of two integrals. Center of Mass (continuous mass distribution). For a thin rod of density ρ(x) distributed along an axis from x a to x b, the center of mass of the rod is given by ∫b xρ(x) dx x ∫a b . a ρ(x) dx Note that the denominator of x is the mass of the bar, and that this quotient of integrals is simply the continuous version of the weighted average of locations, x, along the bar. Activity 6.3.4. Consider a thin bar of length 20 cm whose density is distributed ac- cording to the function ρ(x) 4 + 0.1x, where x 0 represents the left end of the bar. Assume that ρ is measured in g/cm and x is measured in cm. a. Find the total mass, M, of the bar. b. Without doing any calculations, do you expect the center of mass of the bar to be equal to 10, less than 10, or greater than 10? Why? 350 6.3 Density, Mass, and Center of Mass c. Compute x, the exact center of mass of the bar. d. What is the average density of the bar? e. Now consider a diﬀerent density function, given by p(x) 4e 0.020732x , also for a bar of length 20 cm whose left end is at x 0. Plot both ρ(x) and p(x) on the same axes. Without doing any calculations, which bar do you expect to have the greater center of mass? Why? f. Compute the exact center of mass of the bar described in (e) whose density func- tion is p(x) 4e 0.020732x . Check the result against the prediction you made in (e). 6.3.4 Summary • For an object of constant density D, with volume V and mass m, we know that m D · V. • If an object with constant cross-sectional area (such as a thin bar) has its density dis- tributed along an axis according to the function ρ(x), then we can ﬁnd the mass of the object between x a and x b by ∫ b m ρ(x) dx. a • For a system of point-masses distributed along an axis, say m 1 , . . . , m n at locations x1 , . . . , x n , the center of mass, x, is given by the weighted average ∑n xi mi x ∑i1 n . i1 mi If instead we have mass continuously distributed along an axis, such as by a density function ρ(x) for a thin bar of constant cross-sectional area, the center of mass of the portion of the bar between x a and x b is given by ∫b xρ(x) dx x ∫a b . a ρ(x) dx In each situation, x represents the balancing point of the system of masses or of the portion of the bar. 6.3.5 Exercises 1. Center of mass for a linear density function. A rod has length 4 meters. At a distance x meters from its left end, the density of the rod is given by δ(x) 5 + 2x g/m. 351 Chapter 6 Using Deﬁnite Integrals (a) Complete the Riemann sum for the total mass of the rod. (b) Convert the Riemann sum to an integral and ﬁnd the exact mass. 2. Center of mass for a nonlinear density function. A rod with uniform density (mass/ unit length) δ(x) 8 + sin(x) lies on the x-axis between x 0 and x π. Find the mass and center of mass of the rod. 3. Interpreting the density of cars on a road. Suppose that the density of cars (in cars per mile) down ( a 20-mile ( √ stretch))of the Pennsylvania Turnpike is approximated by δ(x) 250 2 + sin 4 x + 0.125 , at a distance x miles from the Breezewood toll plaza. Sketch a graph of this function for 0 ≤ x ≤ 20. (a) Complete the Riemann sum that approximates the total number of cars on this 20- mile stretch. (b) Find the total number of cars on the 20-mile stretch. Number = 4. Center of mass in a point-mass system. A point mass of 1 grams located 7 centimeters to the left of the origin and a point mass of 4 grams located 8 centimeters to the right of the origin are connected by a thin, light rod. Find the center of mass of the system. 5. Let a thin rod of length a have density distribution function ρ(x) 10e −0.1x , where x is measured in cm and ρ in grams per centimeter. a. If the mass of the rod is 30 g, what is the value of a? b. For the 30g rod, will the center of mass lie at its midpoint, to the left of the mid- point, or to the right of the midpoint? Why? c. For the 30g rod, ﬁnd the center of mass, and compare your prediction in (b). d. At what value of x should the 30g rod be cut in order to form two pieces of equal mass? 6. Consider two thin bars of constant cross-sectional area, each of length 10 cm, with re- spective mass density functions ρ(x) 1+x −0.1x . 2 and p(x) e 1 a. Find the mass of each bar. b. Find the center of mass of each bar. c. Now consider a new 10 cm bar whose mass density function is f (x) ρ(x) + p(x). (i) Explain how you can easily ﬁnd the mass of this new bar with little to no additional work. ∫ 10 (ii) Similarly, compute 0 x f (x) dx as simply as possible, in light of earlier com- putations. (iii) True or false: the center of mass of this new bar is the average of the centers of mass of the two earlier bars. Write at least one sentence to say why your conclusion makes sense. 7. Consider the curve given by y f (x) 2xe −1.25x + (30 − x)e −0.25(30−x) . a. Plot this curve in the window x 0 . . . 30, y 0 . . . 3 (with constrained scaling so 352 6.3 Density, Mass, and Center of Mass the units on the x and y axis are equal), and use it to generate a solid of revolution about the x-axis. Explain why this curve could generate a reasonable model of a baseball bat. b. Let x and y be measured in inches. Find the total volume of the baseball bat generated by revolving the given curve about the x-axis. Include units on your answer. c. Suppose that the baseball bat has constant weight density, and that the weight density is 0.6 ounces per cubic inch. Find the total weight of the bat whose volume you found in (b). d. Because the baseball bat does not have constant cross-sectional area, we see that the amount of weight concentrated at a location x along the bat is determined by the volume of a slice at location x. Explain why we can think about the function ρ(x) 0.6π f (x)2 (where f is the function given at the start of the problem) as being the weight density function for how the weight of the baseball bat is dis- tributed from x 0 to x 30. e. Compute the center of mass of the baseball bat. 353 Chapter 6 Using Deﬁnite Integrals 6.4 Physics Applications: Work, Force, and Pressure Motivating Questions • How do we measure the work accomplished by a varying force that moves an object a certain distance? • What is the total force exerted by water against a dam? • How are both of the above concepts and their corresponding use of deﬁnite integrals similar to problems we have encountered in the past involving formulas such as “dis- tance equals rate times time” and “mass equals density times volume”? y y = f (x) y = v(t) ρ(x) △x f (x) v(t) t a △x b a △t b Figure 6.4.1: Three settings where we compute the accumulation of a varying quantity: the area under y f (x), the distance traveled by an object with velocity y v(t), and the mass of a bar with density function y ρ(x). We have seen several diﬀerent circumstances where the deﬁnite integral enables us to mea- sure the accumulation of a quantity that varies, provided the quantity is approximately con- stant over small intervals. For instance, to ﬁnd the area bounded by a nonnegative curve y f (x) and the x-axis on an interval [a, b], we take a representative slice of width ∆x that has area Aslice f (x)∆x. As we let the width of the representative slice tend to zero, we ﬁnd that the exact area of the region is ∫ b A f (x) dx. a In a similar way, if we know the velocity v(t) of a moving object and we wish to know the distance the object travels on an interval [a, b] where v(t) is nonnegative, we can use a def- inite integral to generalize the fact that d r · t when the rate, r, is constant. On a short time interval ∆t, v(t) is roughly constant, so for a small slice of time, dslice v(t)∆t. As the width of the time interval ∆t tends to zero, the exact distance traveled is given by the 354 6.4 Physics Applications: Work, Force, and Pressure deﬁnite integral ∫ b d v(t) dt. a Finally, if we want to determine the mass of an object of non-constant density, because M D ·V (mass equals density times volume, provided that density is constant), we can consider a small slice of an object on which the density is approximately constant, and a deﬁnite integral may be used to determine the exact mass of the object. For instance, if we have a thin rod whose cross sections have constant density, but whose density is distributed along the x axis according to the function y ρ(x), it follows that for a small slice of the rod that is ∆x thick, Mslice ρ(x)∆x. In the limit as ∆x → 0, we then ﬁnd that the total mass is given by ∫ b M ρ(x) dx. a All three of these situations are similar in that we have a basic rule (A l · w, d r · t, M D · V) where one of the two quantities being multiplied is no longer constant; in each, we consider a small interval for the other variable in the formula, calculate the approximate value of the desired quantity (area, distance, or mass) over the small interval, and then use a deﬁnite integral to sum the results as the length of the small intervals is allowed to approach zero. It should be apparent that this approach will work eﬀectively for other situations where we have a quantity that varies. We next turn to the notion of work: from physics, a basic principle is that work is the product of force and distance. For example, if a person exerts a force of 20 pounds to lift a 20-pound weight 4 feet oﬀ the ground, the total work accomplished is W F · d 20 · 4 80 foot-pounds. If force and distance are measured in English units (pounds and feet), then the units of work are foot-pounds. If we work in metric units, where forces are measured in Newtons and distances in meters, the units of work are Newton-meters. Of course, the formula W F · d only applies when the force is constant over the distance d. In Preview Activity 6.4.1, we explore one way that we can use a deﬁnite integral to compute the total work accomplished when the force exerted varies. Preview Activity 6.4.1. A bucket is being lifted from the bottom of a 50-foot deep well; its weight (including the water), B, in pounds at a height h feet above the water is given by the function B(h). When the bucket leaves the water, the bucket and water together weigh B(0) 20 pounds, and when the bucket reaches the top of the well, B(50) 12 pounds. Assume that the bucket loses water at a constant rate (as a function of height, h) throughout its journey from the bottom to the top of the well. a. Find a formula for B(h). b. Compute the value of the product B(5)∆h, where ∆h 2 feet. Include units on your answer. Explain why this product represents the approximate work it took to move the bucket of water from h 5 to h 7. 355 Chapter 6 Using Deﬁnite Integrals c. Is the value in (b) an over- or under-estimate of the actual amount of work it took to move the bucket from h 5 to h 7? Why? d. Compute the value of the product B(22)∆h, where ∆h 0.25 feet. Include units on your answer. What is the meaning of the value you found? e. More generally, what does the quantity Wslice B(h)∆h measure for a given value of h and a small positive value of ∆h? ∫ 50 f. Evaluate the deﬁnite integral 0 B(h) dh. What is the meaning of the value you ﬁnd? Why? 6.4.1 Work Because work is calculated by the rule W F · d whenever the force F is constant, it follows that we can use a deﬁnite integral to compute the work accomplished by a varying force. For example, suppose that a bucket whose weight at height h is given by B(h) 12 + 8e −0.1h is being lifted in a 50-foot well. In contrast to the problem in the preview activity, this bucket is not leaking at a constant rate; but because the weight of the bucket and water is not constant, we have to use a deﬁnite integral to determine the total work done in lifting the bucket. At a height h above the water, the approximate work to move the bucket a small distance ∆h is Wslice B(h)∆h (12 + 8e −0.1h )∆h. Hence, if we let ∆h tend to 0 and take the sum of all of the slices of work accomplished on these small intervals, it follows that the total work is given by ∫ 50 ∫ 50 W B(h) dh (12 + 8e −0.1h ) dh. 0 0 While it is a straightforward exercise to evaluate this integral exactly using the First Fun- damental Theorem of Calculus, in applied settings such as this one we will typically use ∫ 50 computing technology. Here, it turns out that W 0 (12 + 8e −0.1h ) dh ≈ 679.461 foot- pounds. Our work in Preview Activity 6.4.1 and in the most recent discussion above employs the following important general principle. For an object being moved in the positive direction along an axis with location x by a force F(x), the total work to move the object from a to b is given by ∫ b W F(x) dx. a 356 6.4 Physics Applications: Work, Force, and Pressure Activity 6.4.2. Consider the following situations in which a varying force accom- plishes work. a. Suppose that a heavy rope hangs over the side of a cliﬀ. The rope is 200 feet long and weighs 0.3 pounds per foot; initially the rope is fully extended. How much work is required to haul in the entire length of the rope? (Hint: set up a function F(h) whose value is the weight of the rope remaining over the cliﬀ after h feet have been hauled in.) b. A leaky bucket is being hauled up from a 100 foot deep well. When lifted from the water, the bucket and water together weigh 40 pounds. As the bucket is being hauled upward at a constant rate, the bucket leaks water at a constant rate so that it is losing weight at a rate of 0.1 pounds per foot. What function B(h) tells the weight of the bucket after the bucket has been lifted h feet? What is the total amount of work accomplished in lifting the bucket to the top of the well? c. Now suppose that the bucket in (b) does not leak at a constant rate, but rather that its weight at a height h feet above the water is given by B(h) 25+15e −0.05h . What is the total work required to lift the bucket 100 feet? What is the average force exerted on the bucket on the interval h 0 to h 100? d. From physics, Hooke’s Law for springs states that the amount of force required to hold a spring that is compressed (or extended) to a particular length is propor- tionate to the distance the spring is compressed (or extended) from its natural length. That is, the force to compress (or extend) a spring x units from its natural length is F(x) kx for some constant k (which is called the spring constant.) For springs, we choose to measure the force in pounds and the distance the spring is compressed in feet. Suppose that a force of 5 pounds extends a particular spring 4 inches (1/3 foot) beyond its natural length. i. Use the given fact that F(1/3) 5 to ﬁnd the spring constant k. ii. Find the work done to extend the spring from its natural length to 1 foot beyond its natural length. iii. Find the work required to extend the spring from 1 foot beyond its natural length to 1.5 feet beyond its natural length. 6.4.2 Work: Pumping Liquid from a Tank In certain geographic locations where the water table is high, residential homes with base- ments have a peculiar feature: in the basement, one ﬁnds a large hole in the ﬂoor, and in the hole, there is water. For example, in Figure 6.4.2 we see a sump crock¹. A sump crock pro- vides an outlet for water that may build up beneath the basement ﬂoor to prevent ﬂooding the basement. ¹Image credit to www.warreninspect.com/basement-moisture. 357 Chapter 6 Using Deﬁnite Integrals In the crock we see a ﬂoating pump. This pump is activated by elevation, so when the water level reaches a particular height, the pump turns on and pumps water out of the crock, hence relieving the water buildup beneath the foundation. One of the ques- tions we’d like to answer is: how much work does a sump pump accomplish? Figure 6.4.2: A sump crock. Example 6.4.3 Suppose that a sump crock has the shape of a frustum of a cone, as pictured in Figure 6.4.4. The crock has a diameter of 3 feet at its surface, a diameter of 1.5 feet at its base, and a depth of 4 feet. In addition, suppose that the sump pump is set up so that it pumps the water vertically up a pipe to a drain that is located at ground level just outside a basement window. To accomplish this, the pump must send the water to a location 9 feet above the surface of the sump crock. How much work is required to empty the sump crock if it is initially completely full? y+ (0, 1.5) ∆x (4, 0.75) x+ Figure 6.4.4: A sump crock with approximately cylindrical cross-sections. Solution. It is helpful to think of the depth below the surface of the crock as being the independent variable, so we let the positive x-axis point down, and the positive y-axis to the right, as pictured in the ﬁgure. Because the pump sits on the surface of the water, it makes sense to think about the pump moving the water one “slice” at a time, where it takes a thin slice from the surface, pumps it out of the tank, and then proceeds to pump the next slice below. Each slice of water is cylindrical in shape. We see that the radius of each slice varies according to the linear function y f (x) that passes through the points (0, 1.5) and (4, 0.75), where x is the depth of the particular slice in the tank; it is a straightforward exercise to ﬁnd that 358 6.4 Physics Applications: Work, Force, and Pressure f (x) 1.5 − 0.1875x. Now we think about the problem in several steps: a. determining the volume of a typical slice; b. ﬁnding the weight² of a typical slice (and thus the force that must be exerted on it); c. deciding the distance that a typical slice moves; d. and computing the work to move a representative slice. Once we know the work it takes to move one slice, we use a deﬁnite integral over an appro- priate interval to ﬁnd the total work. Consider a representative cylindrical slice at a depth of x feet below the top of the crock. The approximate volume of that slice is given by Vslice π f (x)2 ∆x π(1.5 − 0.1875x)2 ∆x. Since water weighs 62.4 lb/ft3 , the approximate weight of a representative slice is Fslice 62.4 · Vslice 62.4π(1.5 − 0.1875x)2 ∆x. This is also the approximate force the pump must exert to move the slice. Because the slice is located at a depth of x feet below the top of the crock, the slice being moved by the pump must move x feet to get to the level of the basement ﬂoor, and then, as stated in the problem description, another 9 feet to reach the drain at ground level. Hence, the total distance a representative slice travels is dslice x + 9. Finally, the work to move a representative slice is given by Wslice Fslice · dslice 62.4π(1.5 − 0.1875x)2 ∆x · (x + 9). We sum the work required to move slices throughout the tank (from x 0 to x 4), let ∆x → 0, and hence ∫ 4 W 62.4π(1.5 − 0.1875x)2 (x + 9) dx. 0 When evaluated using appropriate technology, the integral shows that the total work is W 3463.2π foot-pounds. The preceding example demonstrates the standard approach to ﬁnding the work required to empty a tank ﬁlled with liquid. The main task in each such problem is to determine the volume of a representative slice, followed by the force exerted on the slice, as well as the distance such a slice moves. In the case where the units are metric, there is one key diﬀerence: in the metric setting, rather than weight, we normally ﬁrst ﬁnd the mass of a slice. For instance, if distance is measured in meters, the mass density of water is 1000 kg/ ²We assume that the weight density of water is 62.4 pounds per cubic foot. 359 Chapter 6 Using Deﬁnite Integrals m3 . In that setting, we can ﬁnd the mass of a typical slice (in kg). To determine the force required to move it, we use F ma, where m is the object’s mass and a is the gravitational constant a 9.81 N/kg. That is, in metric units, the weight density of water is 9810 N/m3 . Activity 6.4.3. In each of the following problems, determine the total work required to accomplish the described task. In parts (b) and (c), a key step is to ﬁnd a formula for a function that describes the curve that forms the side boundary of the tank. y+ x+ Figure 6.4.5: A trough with triangular ends, as described in Activity 6.4.3, part (c). a. Consider a vertical cylindrical tank of radius 2 meters and depth 6 meters. Sup- pose the tank is ﬁlled with 4 meters of water of mass density 1000 kg/m3 , and the top 1 meter of water is pumped over the top of the tank. b. Consider a hemispherical tank with a radius of 10 feet. Suppose that the tank is full to a depth of 7 feet with water of weight density 62.4 pounds/ft3 , and the top 5 feet of water are pumped out of the tank to a tanker truck whose height is 5 feet above the top of the tank. c. Consider a trough with triangular ends, as pictured in Figure 6.4.5, where the tank is 10 feet long, the top is 5 feet wide, and the tank is 4 feet deep. Say that the trough is full to within 1 foot of the top with water of weight density 62.4 pounds/ft3 , and a pump is used to empty the tank until the water remaining in the tank is 1 foot deep. 6.4.3 Force due to Hydrostatic Pressure When building a dam, engineers need to know how much force water will exert against the face of the dam. This force comes from water pressure. The pressure a force exerts on a region is measured in units of force per unit of area: for example, the air pressure in a tire is often measured in pounds per square inch (PSI). Hence, we see that the general relationship is given by F P , or F P · A, A 360 6.4 Physics Applications: Work, Force, and Pressure where P represents pressure, F represents force, and A the area of the region being consid- ered. Of course, in the equation F PA, we assume that the pressure is constant over the entire region A. We know from experience that the deeper one dives underwater while swimming, the greater the pressure exerted by the water. This is because at a greater depth, there is more water right on top of the swimmer: it is the force that “column” of water exerts that determines the pressure the swimmer experiences. The total water pressure is found by computing the total weight of the column of water that lies above a region of area 1 square foot at a ﬁxed depth. At a depth of d feet, a rectangular column has volume V 1 · 1 · d ft3 , so the corre- sponding weight of the water overhead is 62.4d. This is the amount of force being exerted on a 1 square foot region at a depth d feet underwater, so the pressure exerted by water at depth d is P 62.4d (lbs/ft2 ). Because pressure is force per unit area, or P F A, we can compute the total force from a variable pressure by integrating F PA. Example 6.4.6 Consider a trapezoid-shaped dam that is 60 feet wide at its base and 90 feet wide at its top, and assume the dam is 25 feet tall with water that rises to within 5 feet of the top of its face. Water weighs 62.4 pounds per cubic foot. How much force does the water exert against the dam? Solution. First, we sketch a picture of the dam, as shown in Figure 6.4.7. Note that, as in problems involving the work to pump out a tank, we let the positive x-axis point down. 45 y+ x−5 y = f (x) x △x x+ (25, 30) Figure 6.4.7: A trapezoidal dam that is 25 feet tall, 60 feet wide at its base, 90 feet wide at its top, with the water line 5 feet down from the top of its face. Pressure is constant at a ﬁxed depth, so we consider a slice of water at constant depth on the face, as shown in the ﬁgure. The area of this slice is approximately the area of the rectangle pictured. Since the width of that rectangle depends on the variable x, we ﬁnd a formula for the line that represents one side of the dam. It is straightforward to ﬁnd that y 45 − 35 x. 361 Chapter 6 Using Deﬁnite Integrals Hence, the approximate area of a representative slice is 3 Aslice 2 f (x)∆x 2(45 − x)∆x. 5 At any point on this slice, the depth is approximately constant, so the pressure can be con- sidered constant. Because the water rises to within 5 feet of the top of the dam, the depth of any point on the representative slice is approximately (x − 5). Now, since pressure is given by P 62.4d, we have that at any point on the slice Pslice 62.4(x − 5). Knowing both the pressure and area, we can ﬁnd the force the water exerts on the slice. Using F PA, it follows that 3 Fslice Pslice · Aslice 62.4(x − 5) · 2(45 − x)∆x. 5 Finally, we use a deﬁnite integral to sum the forces over the appropriate range of x-values. Since the water rises to within 5 feet of the top of the dam, we start at x 5 and take slices all the way to the bottom of the dam, where x 25. Hence, ∫ x25 3 F 62.4(x − 5) · 2(45 − x) dx. x5 5 Using technology to evaluate the integral, we ﬁnd F 848640 pounds. Activity 6.4.4. In each of the following problems, determine the total force exerted by water against the surface that is described. y+ x+ Figure 6.4.8: A trough with triangular ends, as described in Activity 6.4.4, part (c). a. Consider a rectangular dam that is 100 feet wide and 50 feet tall, and suppose that water presses against the dam all the way to the top. b. Consider a semicircular dam with a radius of 30 feet. Suppose that the water rises to within 10 feet of the top of the dam. 362 6.4 Physics Applications: Work, Force, and Pressure c. Consider a trough with triangular ends, as pictured in Figure 6.4.8, where the tank is 10 feet long, the top is 5 feet wide, and the tank is 4 feet deep. Say that the trough is full to within 1 foot of the top with water of weight density 62.4 pounds/ft3 . How much force does the water exert against one of the triangular ends? Although there are many diﬀerent formulas involving work, force, and pressure, the funda- mental ideas behind these problems are similar to others we’ve encountered in applications of the deﬁnite integral. We slice the quantity of interest into more manageable pieces and then use a deﬁnite integral to add them up. 6.4.4 Summary • To measure the work done by a varying force in moving an object, we divide the prob- lem into pieces on which we can use the formula W F · d, and then use a deﬁnite integral to sum the work done on each piece. • To ﬁnd the total force exerted by water against a dam, we use the formula F P · A to measure the force exerted on a slice that lies at a ﬁxed depth, and then use a deﬁnite integral to sum the forces across the appropriate range of depths. • Because work is computed as the product of force and distance (provided force is con- stant), and the force water exerts on a dam can be computed as the product of pressure and area (provided pressure is constant), problems involving these concepts are sim- ilar to earlier problems we did using deﬁnite integrals to ﬁnd distance (via “distance equals rate times time”) and mass (“mass equals density times volume”). 6.4.5 Exercises 1. Work to empty a conical tank. A tank in the shape of an inverted right circular cone has height 5 meters and radius 4 meters. It is ﬁlled with 3 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ 1070 kg/m3 . Your answer must include the correct units. 2. Work to empty a cylindrical tank. A fuel oil tank is an upright cylinder, buried so that its circular top is 10 feet beneath ground level. The tank has a radius of 4 feet and is 12 feet high, although the current oil level is only 10 feet deep. Calculate the work required to pump all of the oil to the surface. Oil weighs 50 lb/ft3 . 3. Work to empty a rectangular pool. A rectangular swimming pool 50 ft long, 30 ft wide, and 8 ft deep is ﬁlled with water to a depth of 6 ft. Use an integral to ﬁnd the work required to pump all the water out over the top. (Take as the density of water δ 62.4lb/ft3 .) 4. Work to empty a cylindrical tank to diﬀering heights. Water in a cylinder of height 11 ft and radius 4 ft is to be pumped out. The density of water is 62.4 lb/ft3 . Find the 363 Chapter 6 Using Deﬁnite Integrals work required if (a) The tank is full of water and the water is to be pumped over the top of the tank. (b) The tank is full of water and the water must be pumped to a height 6 ft above the top of the tank. (c) The depth of water in the tank is 6 ft and the water must be pumped over the top of the tank. 5. Force due to hydrostatic pressure. A lobster tank in a restaurant is 1.25 m long by 0.5 m wide by 90 cm deep. Taking the density of water to be 1000kg/m3 , ﬁnd the water forces on the bottom of the tank. on each of the larger sides of the tank. on each of the smaller sides of the tank. 3 6. Consider the curve f (x) 3 cos( x4 ) and the portion of its graph that lies in the ﬁrst quadrant between the y-axis and the ﬁrst positive value of x for which f (x) 0. Let R denote the region bounded by this portion of f , the x-axis, and the y-axis. Assume that x and y are each measured in feet. a. Picture the coordinate axes rotated 90 degrees clockwise so that the positive x- axis points straight down, and the positive y-axis points to the right. Suppose that R is rotated about the x axis to form a solid of revolution, and we consider this solid as a storage tank. Suppose that the resulting tank is ﬁlled to a depth of 1.5 feet with water weighing 62.4 pounds per cubic foot. Find the amount of work required to lower the water in the tank until it is 0.5 feet deep, by pumping the water to the top of the tank. b. Again picture the coordinate axes rotated 90 degrees clockwise so that the positive x-axis points straight down, and the positive y-axis points to the right. Suppose that R, together with its reﬂection across the x-axis, forms one end of a storage tank that is 10 feet long. Suppose that the resulting tank is ﬁlled completely with water weighing 62.4 pounds per cubic foot. Find a formula for a function that tells the amount of work required to lower the water by h feet. c. Suppose that the tank described in (b) is completely ﬁlled with water. Find the total force due to hydrostatic pressure exerted by the water on one end of the tank. 7. A cylindrical tank, buried on its side, has radius 3 feet and length 10 feet. It is ﬁlled completely with water whose weight density is 62.4 lbs/ft3 , and the top of the tank is two feet underground. a. Set up, but do not evaluate, an integral expression that represents the amount of work required to empty the top half of the water in the tank to a truck whose tank lies 4.5 feet above ground. b. With the tank now only half-full, set up, but do not evaluate an integral expression that represents the total force due to hydrostatic pressure against one end of the tank. 364 6.5 Improper Integrals 6.5 Improper Integrals Motivating Questions • What are improper integrals and why are they important? • What does it mean to say that an improper integral converges or diverges? • What are some typical improper integrals that we can classify as convergent or di- vergent? Another important application of the deﬁnite integral measures the likelihood of certain events. For instance, consider a company that manufactures incandescent light bulbs. Based on a large volume of test results, they have determined that the fraction of light bulbs that fail between times t a and t b of use (where t is measured in months) is given by ∫ b 0.3e −0.3t dt. a For example, the fraction of light bulbs that fail during their third month of use is given by ∫ 3 3 0.3e −0.3t dt −e −0.3t 2 2 −e −0.9 + e −0.6 ≈ 0.1422. Thus about 14.22% of all lightbulbs fail between t 2 and t 3. Clearly we could adjust the limits of integration to measure the fraction of light bulbs that fail during any time period of interest. Preview Activity 6.5.1. A company with a large customer base has a call center that receives thousands of calls a day. After studying the data that represents how long callers wait for assistance, they ﬁnd that the function p(t) 0.25e −0.25t models the time customers wait in the following way: the fraction of customers who wait between t a and t b minutes is given by ∫ b p(t) dt. a Use this information to answer the following questions. a. Determine the fraction of callers who wait between 5 and 10 minutes. b. Determine the fraction of callers who wait between 10 and 20 minutes. c. Next, let’s study the fraction who wait up to a certain number of minutes: 365 Chapter 6 Using Deﬁnite Integrals i. What is the fraction of callers who wait between 0 and 5 minutes? ii. What is the fraction of callers who wait between 0 and 10 minutes? iii. Between 0 and 15 minutes? Between 0 and 20? d. Let F(b) represent the fraction of callers who wait between 0 and b minutes. Find a formula for F(b) that involves a deﬁnite integral, and then use the First FTC to ﬁnd a formula for F(b) that does not involve a deﬁnite integral. e. What is the value of the limit limb→∞ F(b)? What is its meaning in the context of the problem? 6.5.1 Improper Integrals Involving Unbounded Intervals In view of the above examples, we see that we may want to integrate over an interval whose upper limit grows without bound. For example, to ﬁnd the fraction of light bulbs that fail eventually, we wish to ﬁnd ∫ b lim 0.3e −0.3t dt, b→∞ 0 for which we will also use the notation ∫ ∞ 0.3e −0.3t dt. (6.5.1) 0 Such an integral can be interpreted as the area of an unbounded region, as pictured at right in Figure 6.5.1. y y ··· t b t Figure 6.5.1: At left, the area bounded by p(t) 0.3e −0.3t on the ﬁnite interval [0, b]; at right, the result of letting b → ∞. By “· · ·” in the righthand ﬁgure, we mean that the region extends to the right without bound. 366 6.5 Improper Integrals We call an integral for which the interval of integration is unbounded improper. For instance, the integrals ∫ ∞ ∫ 0 ∫ ∞ 1 1 e −x dx 2 dx, dx, and 1 x 2 −∞ 1 + x 2 −∞ are all improper because they have limits of integration that involve ∞. To evaluate an im- proper integral we replace it with a limit of proper integrals. That is, ∫ ∞ ∫ b f (x) dx lim f (x) dx. 0 b→∞ 0 ∫b We ﬁrst attempt to evaluate 0 f (x) dx using the First FTC, and then evaluate the limit. Is it even possible for the area of an unbounded region to be ﬁnite? The following activity explores this issue and others in more detail. ∫∞ ∫∞ 1 1 Activity 6.5.2. In this activity we explore the improper integrals 1 x dx and 1 x 3/2 dx. ∫∞ 1 a. First we investigate 1 x dx. ∫ 10 ∫ 1000 1 1 i. Use the First FTC to determine the exact values of dx, dx, and ∫ 100000 1 x 1 x 1 1 dx. Then, use your computational device to compute a decimal x approximation of each result. ∫b 1 ii. Use the First FTC to evaluate the deﬁnite integral 1 x dx (which results in an expression that depends on b). iii. Now, use your work from (ii.) to evaluate the limit given by ∫ b 1 lim dx. b→∞ 1 x ∫∞ 1 b. Next, we investigate 1 x 3/2 dx. ∫ 10 ∫ 1000 1 1 i. Use the First FTC to determine the exact values of dx, dx, ∫ 100000 1 x 3/2 1 x 3/2 1 and 1 x 3/2 dx. Then, use your calculator to compute a decimal approx- imation of each result. ∫b 1 ii. Use the First FTC to evaluate the deﬁnite integral 1 x 3/2 dx (which results in an expression that depends on b). iii. Now, use your work from (ii.) to evaluate the limit given by ∫ b 1 lim dx. b→∞ 1 x 3/2 c. Plot the functions y and y x 3/2 on the same coordinate axes for the values 1 x 1 x 0 . . . 10. How would you compare their behavior as x increases without bound? What is similar? What is diﬀerent? ∫∞ ∫∞ d. How would you characterize the value of 1 x1 dx? of 1 x 3/2 1 dx? What does this tell us about the respective areas bounded by these two curves for x ≥ 1? 367 Chapter 6 Using Deﬁnite Integrals 6.5.2 Convergence and Divergence ∫b Activity 6.5.2 suggests that limb→∞ 1 f (x) dx is either ﬁnite or inﬁnite (or it doesn’t exist). With these possibilities in mind, we introduce the following terminology. ∫∞ If f (x) is nonnegative for x ≥ a, then we say that the improper integral a f (x) dx converges provided that ∫ b lim f (x) dx b→∞ a ∫∞ exists and is ﬁnite. Otherwise, we say that a f (x) dx diverges. We will restrict our interest to improper integrals for which the integrand is nonnegative. ∫ ∞that limx→∞ f (x) 0, for if f does not approach 0 as x → ∞, then it is Also, we require impossible for a f (x) dx to converge. Activity 6.5.3. Determine whether each of the following improper integrals converges or diverges. ∫∞ For each integral that converges, ﬁnd∫its exact value. ∞ a. 1 x12 dx d. 4 (x+2)3 5/4 dx ∫∞ ∫∞ b. e −x/4 dx e. 0 xe −x/4 dx 0 ∫∞ 1 ∫∞ f. dx, where p is a positive real 1 xp 9 c. 2 (x+5)2/3 dx number 6.5.3 Improper Integrals Involving Unbounded Integrands An integral is also called improper if the integrand is unbounded on the interval of integra- tion. For example, consider ∫ 1 1 √ dx. 0 x Because f (x) √1x has a vertical asymptote at x 0, f is not continuous on [0, 1], and the integral represents the area of the unbounded region shown at right in Figure 6.5.2. We address the problem of the integrand being unbounded by replacing the improper integral ∫1 with a limit of proper integrals. For example, to evaluate √1 dx, we replace 0 with a and 0 x let a approach 0 from the right. Thus, ∫ 1 ∫ 1 1 1 √ dx lim+ √ dx. 0 x a→0 a x ∫1 We evaluate the proper integral a √1x dx, and then take the limit. We will say that the im- proper integral converges if this limit exists, and diverges otherwise. In this example, we 368 6.5 Improper Integrals y y f (x) = √1 f (x) = √1 x x x x a 1 1 Figure 6.5.2: At left, the area bounded by f (x) √1x on the ﬁnite interval [a, 1]; at right, the result of letting a → 0+ , where we see that the shaded region will extend vertically without bound. observe that ∫ 1 ∫ 1 1 1 √ dx lim+ √ dx 0 x a→0 a x √ 1 lim+ 2 x a a→0 √ √ lim+ 2 1 − 2 a a→0 2, ∫1 so the improper integral √1 dx converges (to the value 2). 0 x We have to be particularly careful with unbounded integrands, for they may arise in ways that may not initially be obvious. Consider, for instance, the integral ∫ 3 1 dx. 1 (x − 2)2 At ﬁrst glance we might think that we can simply apply the Fundamental Theorem of Cal- 2 to get − x−2 and then evaluating from 1 to 3. Were we to do 1 1 culus by antidiﬀerentiating (x−2) so, we would be erroneously applying the FTC because f (x) (x−2)1 2 fails to be continuous throughout the interval, as seen in Figure 6.5.3. Such an incorrect application of the FTC leads to an impossible result (−2), which would itself suggest that something we did must be wrong. Instead, we must address the vertical asymptote at x 2 by writing ∫ 3 ∫ a ∫ 3 1 1 1 dx lim− dx + lim dx. 1 (x − 2)2 a→2 1 (x − 2)2 b→2+ b (x − 2)2 We then evaluate two separate limits of proper integrals. For instance, doing so for the 369 Chapter 6 Using Deﬁnite Integrals y 1 y= (x−2)2 1 2 3 x Figure 6.5.3: The function f (x) 1 (x−2)2 on an interval including x 2. integral with a approaching 2 from the left, we ﬁnd ∫ 2 ∫ a 1 1 dx lim− dx 1 (x − 2)2 a→2 1 (x − 2)2 a 1 lim− − a→2 (x − 2) 1 1 1 lim− − + a→2 (a − 2) 1 − 2 ∞, ∫2 since 1 a−2 → −∞ as a approaches 2 from the left. Thus, the improper integral 1 1 (x−2)2 dx ∫3 1 diverges; similar work shows that 2 (x−2)2 dx also diverges. From either of these two results, ∫3 1 we can conclude that that the original integral, 1 (x−2)2 dx diverges, too. Activity 6.5.4. For each of the following deﬁnite integrals, decide whether the integral is improper or not. If the integral is proper, evaluate it using the First FTC. If the integral is improper, determine whether or not the integral converges or diverges; if the integral ∫1 converges, ﬁnd its exact value. ∫2 1 a. 0 x 1/3 dx d. −2 x12 dx ∫2 ∫ π/2 b. 0 e −x dx e. 0 tan(x) dx ∫4 ∫1 c. √1 dx f. √ 1 dx 1 4−x 0 1−x 2 370 6.5 Improper Integrals 6.5.4 Summary ∫b • An integral a f (x) dx can be improper if at least one of a or b is ±∞, making the interval unbounded, or if f has a vertical asymptote at x c for some value of c that satisﬁes a ≤ c ≤ b. One reason that improper integrals are important is that certain probabilities can be represented by integrals that involve inﬁnite limits. • When we encounter an improper integral, we work to understand it by replacing the improper integral with a limit of proper integrals. For instance, we write ∫ ∞ ∫ b f (x) dx lim f (x) dx, a b→∞ a and then work to determine whether the limit exists and is ﬁnite. For any improper in- tegral, if the resulting limit of proper integrals exists and is ﬁnite, we say the improper integral converges. Otherwise, the improper integral diverges. • An important class of improper integrals is given by ∫ ∞ 1 dx 1 xp where p is a positive real number. We can show that this improper integral converges whenever p > 1, and diverges whenever 0 < p ≤ 1. A related class of improper ∫1 integrals is 1 0 xp dx, which converges for 0 < p < 1, and diverges for p ≥ 1. 6.5.5 Exercises 1. An improper integral on a ﬁnite interval. Calculate the integral below or explain why it diverges. ∫ 3 9 √ dx 0 x x 2. An improper integral on an inﬁnite interval. Calculate the integral below, if it con- verges. ∫ ∞ 3x 2 e −x dx 3 2 3. An improper integral involving a ratio of exponential functions. Calculate the inte- gral, if it converges. ∫ 3 e 2x dx −∞ 1 + e 2x 4. A subtle improper integral. Calculate the integral, if it converges. ∫ 3 1 dv −3 v 371 Chapter 6 Using Deﬁnite Integrals 5. An improper integral involving a ratio of trigonometric functions. Find the area un- der the curve y tan(t) between t 0 and t π/2. 6. Determine, with justiﬁcation, whether each of the following improper integrals con- verges or diverges. ∫∞ ln(x) a. e x dx ∫∞ 1 b. e x ln(x) dx ∫∞ 1 c. e x(ln(x))2 dx ∫∞ 1 d. e x(ln(x))p dx, where p is a positive real number ∫1 ln(x) e. 0 x dx ∫1 f. 0 ln(x) dx 7. Sometimes we may encounter an improper integral for which we cannot easily ∫ ∞ evalu- 1 ate the limit of the corresponding proper integrals. For instance, consider 1 1+x 3 dx. 1 While it is hard (or perhaps impossible) to ﬁnd an antiderivative for 1+x 3 , we can still determine whether or not the improper integral converges or diverges by comparison to a simpler one. Observe that for all x > 0, 1 + x 3 > x 3 , and therefore 1 1 < 3. 1 + x3 x It therefore follows that ∫ ∫ b b 1 1 dx < dx 1 1 + x3 1 x3 ∫∞ for every b > 1. If we let b → ∞ so as to consider the two improper integrals 1 dx ∫∞ 1 1+x 3 1 and 1 dx, we know that the larger of the two improper integrals converges. And x3 thus, since the smaller one lies below∫a convergent integral, it follows that the smaller ∞ 1 one must converge, too. In particular, 1 1+x 3 dx must converge, even though we never explicitly evaluated the corresponding limit of proper integrals. We use this idea and similar ones in the exercises that follow. ∫∞ a. Explain why x 2 + x + 1 > x 2 for all x ≥ 1, and hence show that 1 dx ∫∞ 1 x 2 +x+1 1 converges by comparison to 1 x2 dx. b. Observe that for each x > 1, ln(x) < x. Explain why ∫ b ∫ b 1 1 dx < dx 2 x 2 ln(x) ∫b for each b > 2. Why must it be true that 1 2 ln(x) dx diverges? 372 6.5 Improper Integrals √ c. Explain why x x+1 4 4 > 1 for all x > 1. Then, determine whether or not the im- proper integral ∫ √ ∞ 1 x4 + 1 · dx 1 x x4 converges or diverges. 373 Chapter 6 Using Deﬁnite Integrals 374 CHAPTER 7 Diﬀerential Equations 7.1 An Introduction to Diﬀerential Equations Motivating Questions • What is a diﬀerential equation and what kinds of information can it tell us? • How do diﬀerential equations arise in the world around us? • What do we mean by a solution to a diﬀerential equation? In previous chapters, we have seen that a function’s derivative tells us the rate at which the function is changing. The Fundamental Theorem of Calculus helped us determine the total change of a function over an interval from the function’s rate of change. For instance, an object’s velocity tells us the rate of change of that object’s position. By integrating the velocity over a time interval, we can determine how much the position changes over that time interval. If we know where the object is at the beginning of that interval, we have enough information to predict where it will be at the end of the interval. In this chapter, we introduce the concept of diﬀerential equations. A diﬀerential equation is an equation that provides a description of a function’s derivative, which means that it tells us the function’s rate of change. Using this information, we would like to learn as much as possible about the function itself. Ideally we would like to have an algebraic description of the function. As we’ll see, this may be too much to ask in some situations, but we will still be able to make accurate approximations. Preview Activity 7.1.1. The position of a moving object is given by the function s(t), where s is measured in feet and t in seconds. We determine that the velocity is v(t) 4t + 1 feet per second. a. How much does the position change over the time interval [0, 4]? b. Does this give you enough information to determine s(4), the position at time t 4? If so, what is s(4)? If not, what additional information would you need to know to determine s(4)? Chapter 7 Diﬀerential Equations c. Suppose you are told that the object’s initial position s(0) 7. Determine s(2), the object’s position 2 seconds later. d. If you are told instead that the object’s initial position is s(0) 3, what is s(2)? e. If we only know the velocity v(t) 4t + 1, is it possible that the object’s position at all times is s(t) 2t 2 + t − 4? Explain how you know. f. Are there other possibilities for s(t)? If so, what are they? g. If, in addition to knowing the velocity function is v(t) 4t + 1, we know the initial position s(0), how many possibilities are there for s(t)? 7.1.1 What is a diﬀerential equation? A diﬀerential equation is an equation that describes the derivative, or derivatives, of a func- tion that is unknown to us. For instance, the equation dy x sin x dx describes the derivative of a function y(x) that is unknown to us. As many important examples of diﬀerential equations involve quantities that change in time, the independent variable in our discussion will frequently be time t. In the preview activity, we considered the diﬀerential equation ds 4t + 1. dt Knowing the velocity and the starting position of a moving object, we were able to ﬁnd its position at any later time. Because diﬀerential equations describe the derivative of a function, they give us information about how that function changes. Our goal will be to use this information to predict the value of the function in the future; in this way, diﬀerential equations provide us with something like a crystal ball. Diﬀerential equations arise frequently in our every day world. For instance, you may hear a bank advertising: Your money will grow at a 3% annual interest rate with us. This innocuous statement is really a diﬀerential equation. Let’s translate: A(t) will be amount of money you have in your account at time t. The rate at which your money grows is the de- rivative dA/dt, and we are told that this rate is 0.03A. This leads to the diﬀerential equation dA 0.03A. dt 376 7.1 An Introduction to Diﬀerential Equations dt 4t + 1. This diﬀerential equation has a slightly diﬀerent feel than the previous equation ds In the earlier example, the rate of change depends only on the independent variable t, and we may ﬁnd s(t) by integrating the velocity 4t + 1. In the banking example, however, the rate of change depends on the dependent variable A, so we’ll need some new techniques in order to ﬁnd A(t). Activity 7.1.2. Express the following statements as diﬀerential equations. In each case, you will need to introduce notation to describe the important quantities in the statement so be sure to clearly state what your notation means. a. The population of a town grows continuously at an annual rate of 1.25%. b. A radioactive sample loses mass at a rate of 5.6% of its mass every day. c. You have a bank account that continuously earns 4% interest every year. At the same time, you withdraw money continually from the account at the rate of $1000 per year. d. A cup of hot chocolate is sitting in a 70◦ room. The temperature of the hot choco- late cools continuously by 10% of the diﬀerence between the hot chocolate’s temperature and the room temperature every minute. e. A can of cold soda is sitting in a 70◦ room. The temperature of the soda warms continuously at the rate of 10% of the diﬀerence between the soda’s temperature and the room’s temperature every minute. 7.1.2 Diﬀerential equations in the world around us Diﬀerential equations give a natural way to describe phenomena we see in the real world. For instance, physical principles are frequently expressed as a description of how a quantity changes. A good example is Newton’s Second Law, which says: The product of an object’s mass and acceleration equals the force applied to it. For instance, when gravity acts on an object near the earth’s surface, it exerts a force equal to m1, the mass of the object times the gravitational constant 1. We therefore have ma m1, or dv 1, dt where v is the velocity of the object, and 1 9.8 meters per second squared. Notice that this physical principle does not tell us what the object’s velocity is, but rather how the object’s velocity changes. Activity 7.1.3. Shown below are two graphs depicting the velocity of falling objects. On the left is the velocity of a skydiver, while on the right is the velocity of a meteorite entering the Earth’s atmosphere. 377 Chapter 7 Diﬀerential Equations 6 6 v v 5 5 4 4 3 3 2 2 1 1 t t 1 2 3 1 2 3 Figure 7.1.1: A skydiver’s velocity. Figure 7.1.2: A meteorite’s velocity. a. Begin with the skydiver’s velocity and use the given graph to measure the rate of change dv/dt when the velocity is v 0.5, 1.0, 1.5, 2.0, and 2.5. Plot your values on the graph below. You will want to think carefully about this: you are plotting the derivative dv/dt as a function of velocity. 378 7.1 An Introduction to Diﬀerential Equations 1.5 dv dt 1.0 0.5 v 1 2 3 4 5 -0.5 -1.0 -1.5 b. Now do the same thing with the meteorite’s velocity: use the given graph to measure the rate of change dv/dt when the velocity is v 3.5, 4.0, 4.5, and 5.0. Plot your values on the graph above. c. You should ﬁnd that all your points lie on a line. Write the equation of this line being careful to use proper notation for the quantities on the horizontal and vertical axes. d. The relationship you just found is a diﬀerential equation. Write a complete sen- tence that explains its meaning. e. By looking at the diﬀerential equation, determine the values of the velocity for which the velocity increases. f. By looking at the diﬀerential equation, determine the values of the velocity for which the velocity decreases. g. By looking at the diﬀerential equation, determine the values of the velocity for which the velocity remains constant. The point of this activity is to demonstrate how diﬀerential equations model processes in the real world. In this example, two factors inﬂuence the velocities: gravity and wind resis- tance. The diﬀerential equation describes how these factors inﬂuence the rate of change of the velocities. 7.1.3 Solving a diﬀerential equation A diﬀerential equation describes the derivative, or derivatives, of a function that is unknown to us. By a solution to a diﬀerential equation, we mean simply a function that satisies this 379 Chapter 7 Diﬀerential Equations description. For instance, the ﬁrst diﬀerential equation we looked at is ds 4t + 1, dt which describes an unknown function s(t). We may check that s(t) 2t 2 + t is a solution because it satisﬁes this description. Notice that s(t) 2t 2 + t + 4 is also a solution. If we have a candidate for a solution, it is straightforward to check whether it is a solution or not. Before we demonstrate, however, let’s consider the same issue in a simpler context. Suppose we are given the equation 2x 2 − 2x 2x + 6 and asked whether x 3 is a solution. To answer this question, we could rewrite the variable x in the equation with the symbol □: 2□2 − 2□ 2□ + 6. To determine whether x 3 is a solution, we can investigate the value of each side of the equation separately when the value 3 is placed in □ and see if indeed the two resulting values are equal. Doing so, we observe that 2□2 − 2□ 2 · 32 − 2 · 3 12, and 2□ + 6 2 · 3 + 6 12. Therefore, x 3 is indeed a solution. We will do the same thing with diﬀerential equations. Consider the diﬀerential equation dv 1.5 − 0.5v, or dt d□ 1.5 − 0.5□. dt Let’s ask whether v(t) 3 − 2e −0.5t is a solution¹. Using this formula for v, observe ﬁrst that dv d□ d [3 − 2e −0.5t ] −2e −0.5t · (−0.5) e −0.5t dt dt dt and 1.5 − 0.5v 1.5 − 0.5□ 1.5 − 0.5(3 − 2e −0.5t ) 1.5 − 1.5 + e −0.5t e −0.5t . −0.5t , we have indeed found dt and 1.5 − 0.5v agree for all values of t when v 3 − 2e Since dv a solution to the diﬀerential equation. ¹At this time, don’t worry about why we chose this function; we will learn techniques for ﬁnding solutions to diﬀerential equations soon enough. 380 7.1 An Introduction to Diﬀerential Equations Activity 7.1.4. Consider the diﬀerential equation dv 1.5 − 0.5v. dt Which of the following functions are solutions of this diﬀerential equation? a. v(t) 1.5t − 0.25t 2 . c. v(t) 3. d. v(t) 3 + Ce −0.5t where C is any b. v(t) 3 + 2e −0.5t . constant. This activity shows us something interesting. Notice that the diﬀerential equation has in- ﬁnitely many solutions, which are parametrized by the constant C in v(t) 3 + Ce −0.5t . In Figure 7.1.3, we see the graphs of these solutions for a few values of C, as labeled. 6 v 5 3 4 2 1 3 0 −1 2 −2 1 −3 t 1 2 3 Figure 7.1.3: The family of solutions to the diﬀerential equation dv dt 1.5 − 0.5v. Notice that the value of C is connected to the initial value of the velocity v(0), since v(0) 3 + C. In other words, while the diﬀerential equation describes how the velocity changes as a function of the velocity itself, this is not enough information to determine the velocity uniquely: we also need to know the initial velocity. For this reason, diﬀerential equations will typically have inﬁnitely many solutions, one corresponding to each initial value. We have seen this phenomenon before: given the velocity of a moving object v(t), we cannot uniquely determine the object’s position function unless we also know its initial position. If we are given a diﬀerential equation and an initial value for the unknown function, we say that we have an initial value problem. For instance, dv 1.5 − 0.5v, v(0) 0.5 dt is an initial value problem. In this problem, we know the value of v at one time and we know 381 Chapter 7 Diﬀerential Equations how v is changing. Consequently, there should be exactly one function v that satisﬁes the initial value problem. This demonstrates the following important general property of initial value problems. Initial value problems that are “well behaved” have exactly one solution, which exists in some interval around the initial point. We won’t worry about what “well behaved” means—it is a technical condition that will be satisﬁed by all the diﬀerential equations we consider. To close this section, we note that diﬀerential equations may be classiﬁed based on certain characteristics they may possess. You may see many diﬀerent types of diﬀerential equations in a later course in diﬀerential equations. For now, we would like to introduce a few terms that are used to describe diﬀerential equations. A ﬁrst-order diﬀerential equation is one in which only the ﬁrst derivative of the function occurs. For this reason, dv 1.5 − 0.5v dt is a ﬁrst-order equation while d2 y −10y dt 2 is a second-order equation. A diﬀerential equation is autonomous if the independent variable does not appear in the description of the derivative. For instance, dv 1.5 − 0.5v dt is autonomous because the description of the derivative dv/dt does not depend on time. The equation dy 1.5t − 0.5y, dt however, is not autonomous. 7.1.4 Summary • A diﬀerential equation is simply an equation that describes the derivative(s) of an un- known function. • Physical principles, as well as some everyday situations, often describe how a quantity changes, which lead to diﬀerential equations. • A solution to a diﬀerential equation is a function whose derivatives satisfy the equa- tion’s description. Diﬀerential equations typically have inﬁnitely many solutions, pa- rametrized by the initial values. 382 7.1 An Introduction to Diﬀerential Equations 7.1.5 Exercises 1. Matching solutions with equations. Match the solutions to the diﬀerential equations. If there is more than one solution to an equation, select the answer that includes all solutions. (a) dy 4y A. y sin(4x) or y 4 sin(x) dx d2 y B. y sin(4x) (b) dx 2 16y C. y e −4x or y e 4x dy (c) dx −4y D. y e 4x d2 y (d) dx 2 −16y E. y e −4x F. y 4 sin(x) 2. Finding constant to complete solution. Find a positive value of k for which y sin(kt) satisﬁes d2 y + 9y 0. dt 2 3. Choosing solution of dy/dt k(1 − Ay). Let A and k be positive constants. dy Which of the given functions is a solution to dt −k(y + A)? ⊙ y −A + Ce −kt ⊙ y A + Ce −kt ⊙ y −A + Ce kt ⊙ y A−1 + Ce Akt ⊙ y A + Ce kt ⊙ y A−1 + Ce −Akt 4. Suppose that T(t) represents the temperature of a cup of coﬀee set out in a room, where T is expressed in degrees Fahrenheit and t in minutes. A physical principle known as Newton’s Law of Cooling tells us that dT 1 − T + 5. dt 15 a. Supposes that T(0) 105. What does the diﬀerential equation give us for the dt |T105 ? Explain in a complete sentence the meaning of these two facts. value of dT b. Is T increasing or decreasing at t 0? c. What is the approximate temperature at t 1? d. On the graph below, make a plot of dT/dt as a function of T. 383 Chapter 7 Diﬀerential Equations 5 dT 4 dt 3 2 1 T 30 60 90 120 -1 -2 -3 e. For which values of T does T increase? For which values of T does T decrease? f. What do you think is the temperature of the room? Explain your thinking. g. Verify that T(t) 75 + 30e −t/15 is the solution to the diﬀerential equation with initial value T(0) 105. What happens to this solution after a long time? 5. Suppose that the population of a particular species is described by the function P(t), where P is expressed in millions. Suppose further that the population’s rate of change is governed by the diﬀerential equation dP f (P) dt where f (P) is the function graphed below. dP dt P 1 2 3 4 a. For which values of the population P does the population increase? b. For which values of the population P does the population decrease? 384 7.1 An Introduction to Diﬀerential Equations c. If P(0) 3, how will the population change in time? d. If the initial population satisﬁes 0 < P(0) < 1, what will happen to the population after a very long time? e. If the initial population satisﬁes 1 < P(0) < 3, what will happen to the population after a very long time? f. If the initial population satisﬁes 3 < P(0), what will happen to the population after a very long time? g. This model for a population’s growth is sometimes called “growth with a thresh- old.” Explain why this is an appropriate name. 6. In this problem, we test further what it means for a function to be a solution to a given diﬀerential equation. a. Consider the diﬀerential equation dy y − t. dt Determine whether the following functions are solutions to the given diﬀerential equation. i. y(t) t + 1 + 2e t ii. y(t) t + 1 iii. y(t) t + 2 b. When you weigh bananas in a scale at the grocery store, the height h of the ba- nanas is described by the diﬀerential equation d2 h −kh dt 2 where k is the spring constant, a constant that depends on the properties of the spring in the scale. After you put the bananas in the scale, you (cleverly) observe that the height of the bananas is given by h(t) 4 sin(3t). What is the value of the spring constant? 385 Chapter 7 Diﬀerential Equations 7.2 Qualitative behavior of solutions to DEs Motivating Questions • What is a slope ﬁeld? • How can we use a slope ﬁeld to obtain qualitative information about the solutions of a diﬀerential equation? • What are stable and unstable equilibrium solutions of an autonomous diﬀerential equation? In earlier work, we have used the tangent line to the graph of a function f at a point a to approximate the values of f near a. The usefulness of this approximation is that we need to know very little about the function; armed with only the value f (a) and the derivative f ′(a), we may ﬁnd the equation of the tangent line and the approximation f (x) ≈ f (a) + f ′(a)(x − a). Remember that a ﬁrst-order diﬀerential equation gives us information about the derivative of an unknown function. Since the derivative at a point tells us the slope of the tangent line at this point, a diﬀerential equation gives us crucial information about the tangent lines to the graph of a solution. We will use this information about the tangent lines to create a slope ﬁeld for the diﬀerential equation, which enables us to sketch solutions to initial value problems. Our aim will be to understand the solutions qualitatively. That is, we would like to understand the basic nature of solutions, such as their long-range behavior, without precisely determining the value of a solution at a particular point. Preview Activity 7.2.1. Let’s consider the initial value problem dy t − 2, y(0) 1. dt a. Use the diﬀerential equation to ﬁnd the slope of the tangent line to the solution y(t) at t 0. Then use the initial value to ﬁnd the equation of the tangent line at t 0. Sketch this tangent line over the interval −0.25 ≤ t ≤ 0.25 on the axes provided in Figure 7.2.1. b. Also shown in Figure 7.2.1 are the tangent lines to the solution y(t) at the points t 1, 2, and 3 (we will see how to ﬁnd these later). Use the graph to measure the slope of each tangent line and verify that each agrees with the value speciﬁed by the diﬀerential equation. c. Using these tangent lines as a guide, sketch a graph of the solution y(t) over the interval 0 ≤ t ≤ 3 so that the lines are tangent to the graph of y(t). 386 7.2 Qualitative behavior of solutions to DEs d. Graph the solution you found in (d) 3 on the axes provided, and compare it y to the sketch you made using the tan- gent lines. 2 1 t 1 2 3 -1 -2 Figure 7.2.1: Grid for plotting partial tangent lines. 7.2.1 Slope ﬁelds Preview Activity 7.2.1 shows that we can sketch the solution to an initial value problem if we know an appropriate collection of tangent lines. We can use the diﬀerential equation to ﬁnd the slope of the tangent line at any point of interest, and hence plot such a collection. dy Let’s continue looking at the diﬀerential equation dt t − 2. If t 0, this equation says that dy/dt 0 − 2 −2. Note that this value holds regardless of the value of y. We will therefore sketch tangent lines for several values of y and t 0 with a slope of −2, as shown in Figure 7.2.2. 3 3 y y 2 2 1 1 t t 1 2 3 1 2 3 -1 -1 -2 -2 Figure 7.2.2: Tangent lines Figure 7.2.3: Adding at points with t 0. tangent lines at points with t 1. Let’s continue in the same way: if t 1, the diﬀerential equation tells us that dy/dt 1 − 2 −1, and this holds regardless of the value of y. We now sketch tangent lines for several values of y and t 1 with a slope of −1 in Figure 7.2.3. 387 Chapter 7 Diﬀerential Equations Similarly, we see that when t 2, dy/dt 0 and when t 3, dy/dt 1. We may therefore add to our growing collection of tangent line plots to achieve Figure 7.2.4. 3 3 y y 2 2 1 1 t t 1 2 3 1 2 3 -1 -1 -2 -2 Figure 7.2.4: Adding Figure 7.2.5: A completed tangent lines at points slope ﬁeld. with t 2 and t 3. In Figure 7.2.4, we begin to see the solutions to the diﬀerential equation emerge. For the sake of even greater clarity, we add more tangent lines to provide the more complete picture shown at right in Figure 7.2.5. Figure 7.2.5 is called a slope ﬁeld for the diﬀerential equation. It allows us to sketch solutions just as we did in the preview activity. We can begin with the initial value y(0) 1 and start sketching the solution by following the tangent line. Whenever the solution passes through a point at which a tangent line is drawn, that line is tangent to the solution. This principle leads us to the sequence of images in Figure 7.2.6. 3 3 3 y y y 2 2 2 1 1 1 t t t 1 2 3 1 2 3 1 2 3 -1 -1 -1 -2 -2 -2 Figure 7.2.6: A sequence of images that show how to sketch the IVP solution that satisﬁes y(0) 1. In fact, we can draw solutions for any initial value. Figure 7.2.7 shows solutions for several dy diﬀerent initial values for y(0). Just as we did for the equation dt t − 2, we can construct a slope ﬁeld for any diﬀerential equation of interest. The slope ﬁeld provides us with visual information about how we expect solutions to the diﬀerential equation to behave. 388 7.2 Qualitative behavior of solutions to DEs 3 y 2 1 t 1 2 3 -1 -2 dy Figure 7.2.7: Diﬀerent solutions to dt t − 2 that correspond to diﬀerent initial values. Activity 7.2.2. Consider the autonomous diﬀerential equation dy 1 − (y − 4). dt 2 dy a. Make a plot of dt versus y on the axes provided in Figure 7.2.8. Looking at the graph, for what values of y does y increase and for what values of y does y decrease? b. Next, sketch the slope ﬁeld for this diﬀerential equation on the axes provided in Figure 7.2.9. c. Use your work in (b) to sketch (on the same axes in Figure 7.2.9.) solutions that satisfy y(0) 0, y(0) 2, y(0) 4 and y(0) 6. d. Verify that y(t) 4 + 2e −t/2 is a solution to the given diﬀerential equation with the initial value y(0) 6. Compare its graph to the one you sketched in (c). e. What is special about the solution where y(0) 4? 7 y 6 3 5 dy dt 4 2 3 1 2 y 1 -1 1 2 3 4 5 6 7 t -1 -1 1 2 3 4 5 6 7 -1 -2 Figure 7.2.9: Axes for plotting the dy dy Figure 7.2.8: Axes for plotting dt versus y. slope ﬁeld for dt − 12 (y − 4). 389 Chapter 7 Diﬀerential Equations 7.2.2 Equilibrium solutions and stability As our work in Activity 7.2.2 demonstrates, ﬁrst-order autonomous equations may have solutions that are constant. These are simple to detect by inspecting the diﬀerential equation dy/dt f (y): constant solutions necessarily have a zero derivative, so dy/dt 0 f (y). dy For example, in Activity 7.2.2, we considered the equation dt f (y) − 12 (y − 4). Constant solutions are found by setting f (y) − 21 (y − 4) 0, which we immediately see implies that y 4. dy Values of y for which f (y) 0 in an autonomous diﬀerential equation dt f (y) are called equilibrium solutions of the diﬀerential equation. Activity 7.2.3. Consider the autonomous diﬀerential equation dy 1 − y(y − 4). dt 2 dy a. Make a plot of dt versus y on the axes provided in Figure 7.2.10. Looking at the graph, for what values of y does y increase and for what values of y does y decrease? 6 y 5 3 4 dy dt 3 2 2 1 y 1 t -1 1 2 3 4 5 6 7 -1 1 2 3 4 5 6 7 -1 -1 -2 -2 Figure 7.2.10: Axes for plotting dy/dt Figure 7.2.11: Axes for plotting the dy dy vs y for dt − 12 y(y − 4). slope ﬁeld for dt − 12 y(y − 4). b. Identify any equilibrium solutions of the given diﬀerential equation. c. Now sketch the slope ﬁeld for the given diﬀerential equation on the axes pro- vided in Figure 7.2.11. d. Sketch the solutions to the given diﬀerential equation that correspond to initial values y(0) −1, 0, 1, . . . , 5. e. An equilibrium solution y is called stable if nearby solutions converge to y. This means that if the initial condition varies slightly from y, then limt→∞ y(t) y. Conversely, an equilibrium solution y is called unstable if nearby solutions are 390 7.2 Qualitative behavior of solutions to DEs pushed away from y. Using your work above, classify the equilibrium solutions you found in (b) as either stable or unstable. f. Suppose that y(t) describes the population of a species of living organisms and that the initial value y(0) is positive. What can you say about the eventual fate of this population? g. Now consider a general autonomous diﬀerential equation of the form dy/dt f (y). Remember that an equilibrium solution y satisﬁes f (y) 0. If we graph dy/dt f (y) as a function of y, for which of the diﬀerential equations repre- sented in Figure 7.2.12 and Figure 7.2.13 is y a stable equilibrium and for which is y unstable? Why? dy dy dt = f (y) dt = f (y) y y y y dy dy Figure 7.2.12: Plot of dt as a function of Figure 7.2.13: Plot of dt as a diﬀerent y. function of y. 7.2.3 Summary • A slope ﬁeld is a plot created by graphing the tangent lines of many diﬀerent solutions to a diﬀerential equation. • Once we have a slope ﬁeld, we may sketch the graph of solutions by drawing a curve that is always tangent to the lines in the slope ﬁeld. • Autonomous diﬀerential equations sometimes have constant solutions that we call equilibrium solutions. These may be classiﬁed as stable or unstable, depending on the behavior of nearby solutions. 391 Chapter 7 Diﬀerential Equations 7.2.4 Exercises 1. Graphing equilibrium solutions. Consider the direction ﬁeld below for a diﬀerential equation. Use the graph to ﬁnd the equilibrium solutions. 2. Sketching solution curves. Consider the two slope ﬁelds shown, in ﬁgures 1 and 2 below. ﬁgure 1 ﬁgure 2 On a print-out of these slope ﬁelds, sketch for each three solution curves to the diﬀer- ential equations that generated them. Then complete the following statements: For the slope ﬁeld in ﬁgure 1, a solution passing through the point (4,-3) has a (□ pos- itive □ negative □ zero □ undeﬁned) slope. For the slope ﬁeld in ﬁgure 1, a solution passing through the point (-2,-3) has a (□ pos- itive □ negative □ zero □ undeﬁned) slope. For the slope ﬁeld in ﬁgure 2, a solution passing through the point (2,-1) has a (□ pos- itive □ negative □ zero □ undeﬁned) slope. For the slope ﬁeld in ﬁgure 2, a solution passing through the point (0,3) has a (□ pos- itive □ negative □ zero □ undeﬁned) slope. 392 7.2 Qualitative behavior of solutions to DEs 3. Matching equations with direction ﬁelds. Match the following equations with their direction ﬁeld. While you can probably solve this problem by guessing, it is useful to try to predict characteristics of the direction ﬁeld and then match them to the picture. Here are some handy characteristics to start with -- you will develop more as you prac- tice. A. Set y equal to zero and look at how the derivative behaves along the x-axis. B. Do the same for the y-axis by setting x equal to 0 C. Consider the curve in the plane deﬁned by setting y ′ 0 -- this should correspond to the points in the picture where the slope is zero. D. Setting y ′ equal to a constant other than zero gives the curve of points where the slope is that constant. These are called isoclines, and can be used to construct the direction ﬁeld picture by hand. (a) y ′ 2x y + 2xe −x 2 y (b) y ′ + 3 cos(2x) x (c) y ′ y + xe −x + 1 (d) y ′ 2 sin(3x) + 1 + y A B 393 Chapter 7 Diﬀerential Equations C D 4. Describing equilibrium solutions. Given the diﬀerential equation x ′(t) −x 4 − 9x 3 − 19x 2 + 9x + 20. List the constant (or equilibrium) solutions to this diﬀerential equation in increasing order and indicate whether or not these equations are stable, semi-stable, or unstable. (It helps to sketch the graph. (□ □ stable □ unstable □ semi-stable) (□ □ stable □ unstable □ semi-stable) (□ □ stable □ unstable □ semi-stable) (□ □ stable □ unstable □ semi-stable) 5. Consider the diﬀerential equation dy t − y. dt 4 a. Sketch a slope ﬁeld on the axes at y right. 3 b. Sketch the solutions whose initial 2 values are y(0) −4, −3, . . . , 4. 1 c. What do your sketches suggest is t the solution whose initial value is y(0) −1? Verify that this is -4 -3 -2 -1 1 2 3 4 -1 indeed the solution to this initial value problem. -2 d. By considering the diﬀerential -3 equation and the graphs you have sketched, what is the relationship -4 between t and y at a point where a solution has a local minimum? 394 7.2 Qualitative behavior of solutions to DEs 6. Consider the situation from problem 2 of Section 7.1: Suppose that the population of a particular species is described by the function P(t), where P is expressed in millions. Suppose further that the population’s rate of change is governed by the diﬀerential equation dP f (P) dt where f (P) is the function graphed below. dP dt P 1 2 3 4 a. Sketch a slope ﬁeld for this diﬀerential equation. You do not have enough infor- mation to determine the actual slopes, but you should have enough information to determine where slopes are positive, negative, zero, large, or small, and hence determine the qualitative behavior of solutions. b. Sketch some solutions to this diﬀerential equation when the initial population P(0) > 0. c. Identify any equilibrium solutions to the diﬀerential equation and classify them as stable or unstable. d. If P(0) > 1, what is the eventual fate of the species? if P(0) < 1? e. Remember that we referred to