TB2011-12 - Precalculus Precalculus David H Collingwood...

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Unformatted text preview: Precalculus Precalculus David H. Collingwood Department of Mathematics University of Washington K. David Prince Minority Science and Engineering Program College of Engineering University of Washington Matthew M. Conroy Department of Mathematics University of Washington September 2, 2011 ii Copyright c 2003 David H. Collingwood and K. David Prince. Copyright c 2011 David H. Collingwood, K. David Prince, and Matthew M. Conroy. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.1 or any later version published by the Free Software Foundation; with no Invariant Sections, with no Front-Cover, and with no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”. Author Note For most of you, this course will be unlike any mathematics course you have previously encountered. Why is this? Learning a new language Colleges and universities have been designed to help us discover, share and apply knowledge. As a student, the preparation required to carry out this three part mission varies widely, depending upon the chosen field of study. One fundamental prerequisite is fluency in a “basic language”; this provides a common framework in which to exchange ideas, carefully formulate problems and actively work toward their solutions. In modern science and engineering, college mathematics has become this “basic language”, beginning with precalculus, moving into calculus and progressing into more advanced courses. The difficulty is that college mathematics will involve genuinely new ideas and the mystery of this unknown can be sort of intimidating. However, everyone in this course has the intelligence to succeed! Is this course the same as high school Precalculus? There are key differences between the way teaching and learning takes place in high schools and universities. Our goal is much more than just getting you to reproduce what was done in the classroom. Here are some key points to keep in mind: • The pace of this course will be faster than a high school class in precalculus. Above that, we aim for greater command of the material, especially the ability to extend what we have learned to new situations. • This course aims to help you build the stamina required to solve challenging and lengthy multi-step problems. • As a rule of thumb, this course should on average take 15 hours of effort per week. That means that in addition to the 5 classroom hours per week, you would spend 10 hours extra on the class. This is only an average and my experience has shown that 12–15 hours iii iv of study per week (outside class) is a more typical estimate. In other words, for many students, this course is the equivalent of a halftime job! • Because the course material is developed in a highly cumulative manner, we recommend that your study time be spread out evenly over the week, rather than in huge isolated blocks. An analogy with athletics is useful: If you are preparing to run a marathon, you must train daily; if you want to improve your time, you must continually push your comfort zone. Prerequisites This course assumes prior exposure to the “mathematics” in Chapters 1-12; these chapters cover functions, their graphs and some basic examples. This material is fully developed, in case you need to brush up on a particular topic. If you have never encountered the concept of a function, graphs of functions, linear functions or quadratic functions, this course will probably seem too advanced. You are not assumed to have taken a course which focuses on mathematical problem solving or multi-step problem solving; that is the purpose of this course. Internet There is a great deal of archived information specific to this course that can be accessed via the World Wide Web at the URL address http://www.math.washington.edu/˜m120 Why are we using this text? Prior to 1990, the performance of a student in precalculus at the University of Washington was not a predictor of success in calculus. For this reason, the mathematics department set out to create a new course with a specific set of goals in mind: • A review of the essential mathematics needed to succeed in calculus. • An emphasis on problem solving, the idea being to gain both experience and confidence in working with a particular set of mathematical tools. This text was created to achieve these goals and the 2004-05 academic year marks the eleventh year in which it has been used. Several thousand students have successfully passed through the course. v Notation, Answers, etc. This book is full of worked out examples. We use the the notation “Solution.” to indicate where the reasoning for a problem begins; the symbol is used to indicate the end of the solution to a problem. There is a Table of Contents that is useful in helping you find a topic treated earlier in the course. It is also a good rough outline when it comes time to study for the final examination. The book also includes an index at the end. Finally, there is an appendix at the end of the text with ”answers” to most of the problems in the text. It should be emphasized these are ”answers” as opposed to ”solutions”. Any homework problems you may be asked to turn in will require you include all your work; in other words, a detailed solution. Simply writing down the answer from the back of the text would never be sufficient; the answers are intended to be a guide to help insure you are on the right track. How to succeed in Math 120. Most people learn mathematics by doing mathematics. That is, you learn it by active participation; it is very unusual for someone to learn the material by simply watching their instructor perform on Monday, Wednesday, and Friday. For this reason, the homework is THE heart of the course and more than anything else, study time is the key to success in Math 120. We advise 15 hours of study per week, outside class. Also, during the first week, the number of study hours will probably be even higher as you adjust to the viewpoint of the course and brush up on algebra skills. Here are some suggestions: Prior to a given class, make sure you have looked over the reading assigned. If you can’t finish it, at least look it over and get some idea of the topic to be discussed. Having looked over the material ahead of time, you will get FAR MORE out of the lecture. Then, after lecture, you will be ready to launch into the homework. If you follow this model, it will minimize the number of times you leave class in a daze. In addition, spread your study time out evenly over the week, rather than waiting until the day before an assignment is due. Acknowledgments The efforts of numerous people have led to many changes, corrections and improvements. We want to specifically thank Laura Acuna, Patrick ˜ Averbeck, Jim Baxter, Sandi Bennett, Daniel Bjorkegren, Cindy Burton, Michael D. Calac, Roll Jean Cheng, Jerry Folland, Dan Fox, Grant Galbraith, Peter Garfield, Richard J. Golob, Joel Grus, Fred Kuczmarski, Julie Harris, Michael Harrison, Teri Hughes, Ron Irving, Ian Jannetty, Mark Johnson, Michael Keynes, Andrew Loveless, Don Marshall, Linda vi Martin, Alexandra Nichifor, Patrick Perkins, Lisa Peterson, Ken Plochinski, Eric Rimbey, Tim Roberts, Aaron Schlafly, David Schneider, Marilyn Stor, Lukas Svec, Sarah Swearinger, Jennifer Taggart, Steve Tanner, Paul Tseng, and Rebecca Tyson. I am grateful to everyone for their hard work and dedication to making this a better product for our students. The Minority Science and Engineering Program (MSEP) of the College of Engineering supports the development of this textbook. It is also authoring additional materials, namely, a student study guide and an instructor guide. MSEP actively uses these all of these materials in its summer mathematics program for freshman pre-engineers. We want to thank MSEP for its contributions to this textbook. We want to thank Intel Corporation for their grant giving us an ”Innovation in Education” server donation. This computer hardware was used to maintain and develop this textbook. Comments Send comments, corrections, and ideas to [email protected] or [email protected] Preface Have you ever noticed this peculiar feature of mathematics: When you don’t know what is going on, it is really hard, difficult, and frustrating. But, when you know what is going on, mathematics seems incredibly easy, and you wonder why you had trouble with it in the first place! Here is another feature of learning mathematics: When you are struggling with a mathematical problem, there are times when the answer seems to pop out at you. At first, nothing is there, then very suddenly, in a flash, the answer is all there, and you sit wondering why you didn’t “see” the solution sooner. We have a special name for this: It’s the “AHa!” experience. Often the difficulty you have in studying mathematics is that the rate at which you are having an A-Ha! experience might be so low that you get discouraged or, even worse, you give up studying mathematics altogether. One purpose of this course is to introduce you to some strategies that can help you increase the rate of your mathematical A-Ha! experiences. What is a story problem? When we ask students if they like story problems, more often than not, we hear statements like: “I hate story problems!” So, what is it about these kinds of problems that causes such a negative reaction? Well, the first thing you can say about story problems is that they are mostly made up of words. This means you have to make a big effort to read and understand the words of the problem. If you don’t like to read, story problems will be troublesome. The second thing that stands out with story problems is that they force you to think about how things work. You have to give deep thought to how things in the problem relate to each other. This in turn means that story problems force you to connect many steps in the solution process. You are no longer given a list of formulas to work using memorized steps. So, in the end, the story problem is a multi-step process such that the “A-Ha!” comes only after lots of intense effort. All of this means you have to spend time working on story problems. It is impossible to sit down and spend only a minute or two working each problem. With story problems, you have to spend much more time working toward a solution, and at the university, it is common to spend vii viii an hour or more working each problem. So another aspect of working these kinds of problems is that they demand a lot of work from you, the problem solver. We can conclude this: What works is work! Unfortunately, there is no easy way to solve all story problems. There are, however, techniques that you can use to help you work efficiently. In this course, you will be presented with a wide range of mathematical tools, techniques, and strategies that will prepare you for university level problem solving. What are the BIG errors? Before we look at how to make your problem solving more efficient, let’s look at some typical situations that make problem solving inefficient. If you want to be ready for university level mathematics, we are sure you have heard somewhere: “You must be prepared!” This means you need to have certain well-developed mathematical skills before you reach the university. We would like to share with you the three major sources of errors students make when working problems, especially when they are working exam problems. Every time we sit down and review solutions with a student who has just taken an exam, and who has lost a lot of points in that exam, we find errors falling pretty much into three categories, and these errors are the major cause of inefficient mathematical problem solving. The first type of error that loses points is algebra. This is an error of not knowing all of the algebraic rules. This type of error also includes mistakes in the selection and use of mathematical symbols. Often, during the problem solving process, you are required to introduce mathematical symbols. But, without these symbols, you cannot make any further progress. Think of it this way: Without symbols, you cannot do any mathematics involving equations! The second error we see in problem solving has to do with visualization. In this case, we’re talking about more than the graphics you can get from a calculator. Graphing and curve sketching are very important skills. But, in doing story problems, you might find it almost impossible to create a solution without first drawing a picture 1 of your problem. Thus, by not drawing a good picture of the problem, students get stuck in their exams, often missing the solution to a problem entirely. Finally, the third big source of error is not knowing mathematical definitions. Actually, this is a huge topic, so we will only touch on some of the main features of this kind of error. The key thing here is that by not knowing mathematical definitions, it becomes very hard to know what to do next in a multi-step solution to a story problem. 1 Whenever we talk about a picture of your problem, we mean not just the drawing itself. In this case, the picture must include the drawing and the labels which clearly signify the quantities related to your problem. ix Here is what it all boils down to: Mathematical definitions, for the most part, provide little cookbook procedures for computing or measuring something. For example, if you did not know the mathematical definition of “speed,” you would not know that to measure speed, you first measure your distance and you simultaneously measure the time it takes to cover that distance. Notice this means you have two measuring instruments working at the same time. The second thing you must do, according to the definition of speed, is divide the distance you measured by the time you measured. The result of your division is a number that you will call speed. The definition is a step-by-step procedure that everyone agrees to when talking about “speed.” So, it’s easy to understand that if you are trying to solve a story problem requiring a speed computation and you did not know the definition or you could not remember the definition of speed, you are going to be “stuck” and no further progress will be possible! What does all of this mean for you? As you study your mathematics, make sure you are the best you can be in these three areas: Algebra, Visualization, and Definitions. Do a little algebra every day. Always draw a picture to go with all your problems. And, know your mathematical definitions without hesitation. Do this and you will see a very large portion of your math errors disappear! Problem Solving Strategies This topic would require another book to fully develop. So, for now, we would like to present some problem solving ideas you can start using right away. Let’s look now at a common scenario: A student reads a story problem then exclaims, maybe with a little frustration: “If I only had the formula, I could solve this problem!” Does this sound familiar? What is going on here, and why is this student frustrated? Suppose you are this student. What are you actually trying to do? Let’s break it down. First, you are reading some descriptive information in words and you need to translate this word information into symbols. If you had the symbolic information, you would be in a position to mathematically solve your problem right away. Unfortunately, you cannot solve anything without first translating your words into symbols. And, going directly from words to symbols is usually very difficult! So, here we are looking for some alternative approach for translating words into symbols. Figure 1 is the answer to this problem solving dilemma. A lot is going on in Figure 1. Let’s consider some of the main features of this diagram. First, it is suggesting that you are dealing with information in three different forms: Words, Pictures, and Symbols. The arrows in this diagram suggest that in any problem solving situation, you are x Words Pictures Symbols Figure 1: Problem solving as a transformation process. actually translating information from one form to another. The arrows also suggest that there are alternative paths you can take to get from one form to another! This is a very, very important point: the idea that there is more than one way to get from words to symbols. Let’s rewind this discussion: You’re reading a story problem. But, now, before giving any thought to what your formula is, that is, before worrying about your symbolic information, you grab a blank sheet of paper and start drawing a picture of your problem. And, to your picture you add symbols denoting the quantities you need in your problem. At this point in your problem solving, you are not trying to write any equations; you are only trying to see what your problem looks like. You are also concentrating on another extremely important step: Deciding what symbols to use in your problem! Now you have a good picture of your problem. It shows not only what the problem looks like, but symbolically shows all the problem’s variables and constants. You can start using this information to mathematically model your problem. The process of creating a mathematical model is actually nothing more than the arrow in the diagram going from pictures to symbols. Mathematical modeling is the jump you make from the visual information you have created to information contained in your formulas. Let’s summarize the problem solving process. You start with a description of a problem that is presented to you mainly in the form of words. Instead of trying to jump directly from words to symbols, you jump from words to pictures. Once you have a good picture, you jump from pictures to symbols. And, all the time, you are relying on mathematical definitions as you interpret the words of your problem; on visualization techniques as you draw pictures related to your problem; and, on your algebra skills as you are formulating the equations you need to xi solve your problem. There is one final thing to notice about the diagram in this section. All of this discussion so far deals with the situation where your direction is from Words =⇒ Pictures =⇒ Symbols. But when you study the diagram you see that the arrows go both ways! So, we will leave you with this to think about: What does it mean, within the context of problem solving, when you have Symbols =⇒ Pictures =⇒ Words ? An Example. Here is a worked example that is taken from a typical homework assignment for Section 1.1 of this book. See if you can recognize the multitude of steps needed to arrive at the equations that allow us to compute a solution. That is, try to identify the specific way in which information is being transformed during the problem solving process. This problem illustrates the principle used to make a good “squirt gun”. A cylindrical tube has diameter 1 inch, then reduces to diameter d. The tube is filled with oil and piston A moves to the right 2 in/sec, as indicated. This will cause piston B to move to the right m in/sec. Assume the oil does not compress; that means the volume of the oil between the two pistons is always the same. oil 2 in/sec A m in/sec B 1. If the diameter of the narrow part of the tube is is the speed of piston B? 1 2 inch, what in 2. If B moves 11 sec , what is the diameter of the narrow part of the tube? Solution. The first thing to do with any story problem is to draw a picture of the problem. In this case, you might re-sketch the picture so that it looks 3-dimensional: See Figure 2. As you draw, add in mathematical symbols signifying quantities in the problem. The next thing is to clearly define the variables in your problem: xii Volume Leaving Cylinder A. Volume Entering Cylinder B. d dB A x B Piston B. x A Piston A. Figure 2: A re-sketch of the original given figure. 1. Let VA and VB stand for the change in volumes as piston A moves to the right. 2. Let dA and dB represent the diameters of each cylinder. 3. Let rA and rB represent the radii of each cylinder. 4. Let sA and sB stand for the speeds of each piston. 5. Let xA and xB stand for the distance traveled by each piston. Now that you have some symbols to work with, you can write the given data down this way: inches 1. sA = 2 second . 2. dA = 1.0 inch. After you have studied this problem for a while, you would write down some useful relationships: 1. The volume of any cylinder is V = πr2 h where r is the radius of the cylinder, and h is its height or length. From this, you can derive the volume of a cylinder in terms of its diameter, d: V= π2 d h. 4 xiii 2. “Distance” = “Rate” × “Time”. In terms of this problem, you would write x = st, where x is the distance your piston moves, and s is the speed of the piston’s motion. Now you are in a position to create a mathematical model that describes what is going on: 1. From the two relationships above, you can derive the volume equations for each cylinder so that the diameters and speed of the pistons are included: VA = and VB = πd2 (sA t) πd2 xA A A =⇒ VA = 4 4 πd2 xB πd2 (sB t) B B =⇒ VB = 4 4 2. Since the oil does not compress, at each instant when piston A is moving, you must have VA = VB , thus: πd2 (sA t) πd2 (sB t) A B = . 4 4 After canceling π, t, and 4, you end up with a mathematical model describing this problem that you can use to answer all sorts of interesting questions: d2 sA = d2 sB . A B Using your model, you can compute the following solutions: in 1. Given: dB = 1 in, dA = 1.0 in, and sA = 2 sec , find sB . From your model, 2 you derive: d2 sA = d2 sB =⇒ sB = A B d2 · sA A d2 B from which you can compute m = sB = 8 in . sec 2. Likewise, you can use your model to compute dB = exactly. 2 in, 11 xiv Contents 1 2 3 4 Warming Up 1.1 Units and Rates . . . . . . . 1.2 Total Change = Rate × Time 1.3 The Modeling Process . . . . 1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 4 6 8 Imposing Coordinates 2.1 The Coordinate System . . . . . . . . . . . . . . . 2.1.1 Going from P to a Pair of Real Numbers. 2.2 Three Features of a Coordinate System . . . . . . 2.2.1 Scaling . . . . . . . . . . . . . . . . . . . . 2.2.2 Axes Units . . . . . . . . . . . . . . . . . . 2.3 A Key Step in all Modeling Problems . . . . . . . 2.4 Distance . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 11 12 13 13 14 15 17 22 . . . . . 25 25 26 28 30 31 . . . . . . . . . . . . 33 33 35 36 39 39 42 43 44 44 45 47 50 Three Simple Curves 3.1 The Simplest Lines . . 3.2 Circles . . . . . . . . . 3.3 Intersecting Curves I 3.4 Summary . . . . . . . 3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear Modeling 4.1 The Earning Power Problem . . . . . . . 4.2 Relating Lines and Equations . . . . . . 4.3 Non-vertical Lines . . . . . . . . . . . . . 4.4 General Lines . . . . . . . . . . . . . . . . 4.5 Lines and Rate of Change . . . . . . . . . 4.6 Back to the Earning Power Problem . . . 4.7 What’s Needed to Build a Linear Model? 4.8 Linear Application Problems . . . . . . . 4.9 Perpendicular and Parallel Lines . . . . . 4.10 Intersecting Curves II . . . . . . . . . . . 4.11 Uniform Linear Motion . . . . . . . . . . 4.12 Summary . . . . . . . . . . . . . . . . . . xv . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS xvi 4.13 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 6 7 8 51 Functions and Graphs 5.1 Relating Data, Plots and Equations 5.2 What is a Function? . . . . . . . . . 5.2.1 The definition of a function (equation viewpoint) . . . . 5.2.2 The definition of a function (conceptual viewpoint) . . . 5.3 The Graph of a Function . . . . . . 5.4 The Vertical Line Test . . . . . . . . 5.4.1 Imposed Constraints . . . . 5.5 Linear Functions . . . . . . . . . . . 5.6 Profit Analysis . . . . . . . . . . . . 5.7 Exercises . . . . . . . . . . . . . . . ............. ............. 55 55 57 ............. 58 . . . . . . . . . . . . . . 61 62 63 64 64 65 68 . . . . . . . . 73 73 73 74 76 76 77 79 82 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graphical Analysis 6.1 Visual Analysis of a Graph . . . . . . . . . . . . 6.1.1 Visualizing the domain and range . . . 6.1.2 Interpreting Points on the Graph . . . . 6.1.3 Interpreting Intercepts of a Graph . . . 6.1.4 Interpreting Increasing and Decreasing 6.2 Circles and Semicircles . . . . . . . . . . . . . . 6.3 Multipart Functions . . . . . . . . . . . . . . . . 6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Quadratic Modeling 7.1 Parabolas and Vertex Form . . . . . . . . . . . . . 7.1.1 First Maneuver: Shifting . . . . . . . . . . 7.1.2 Second Maneuver: Reflection . . . . . . . 7.1.3 Third Maneuver: Vertical Dilation . . . . 7.1.4 Conclusion . . . . . . . . . . . . . . . . . 7.2 Completing the Square . . . . . . . . . . . . . . . 7.3 Interpreting the Vertex . . . . . . . . . . . . . . . 7.4 Quadratic Modeling Problems . . . . . . . . . . . 7.4.1 How many points determine a parabola? 7.5 What’s Needed to Build a Quadratic Model? . . . 7.6 Summary . . . . . . . . . . . . . . . . . . . . . . . 7.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 85 87 89 89 89 90 93 94 98 101 101 103 Composition 8.1 The Formula for a Composition . . . . 8.1.1 Some notational confusion . . 8.2 Domain, Range, etc. for a Composition 8.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 108 111 112 116 . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS 9 xvii Inverse Functions 9.1 Concept of an Inverse Function . . . . . . . . . 9.1.1 An Example . . . . . . . . . . . . . . . . 9.1.2 A Second Example . . . . . . . . . . . . 9.1.3 A Third Example . . . . . . . . . . . . . 9.2 Graphical Idea of an Inverse . . . . . . . . . . . 9.2.1 One-to-one Functions . . . . . . . . . . 9.3 Inverse Functions . . . . . . . . . . . . . . . . . 9.3.1 Schematic Idea of an Inverse Function 9.3.2 Graphing Inverse Functions . . . . . . 9.4 Trying to Invert a Non one-to-one Function . . 9.5 Summary . . . . . . . . . . . . . . . . . . . . . . 9.6 Exercises . . . . . . . . . . . . . . . . . . . . . . 10 Exponential Functions 10.1 Functions of Exponential Type . . . . . . . 10.1.1 Reviewing the Rules of Exponents 10.2 The Functions y = A0 bx . . . . . . . . . . . 10.2.1 The case b = 1 . . . . . . . . . . . 10.2.2 The case b > 1 . . . . . . . . . . . . 10.2.3 The case 0 < b < 1 . . . . . . . . . 10.3 Piano Frequency Range . . . . . . . . . . . 10.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Exponential Modeling 11.1 The Method of Compound Interest . . 11.1.1 Two Examples . . . . . . . . . . 11.1.2 Discrete Compounding . . . . 11.2 The Number e and the Exponential Function . . . . . . . . 11.2.1 Calculator drill . . . . . . . . . 11.2.2 Back to the original problem... 11.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 12 Logarithmic Functions 12.1 The Inverse Function of y = ex . . . 12.2 Alternate form for functions of exponential type . . . . 12.3 The Inverse Function of y = bx . . . 12.4 Measuring the Loudness of Sound 12.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 119 120 121 121 122 124 125 126 127 127 129 130 . . . . . . . . 133 134 135 137 138 138 139 140 142 145 . . . . . . . . . . . 146 . . . . . . . . . . . 147 . . . . . . . . . . . 148 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 151 151 152 153 . . . . . . . . . . . . . 153 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 157 159 163 13 Three Construction Tools 165 13.1 A Low-tech Exercise . . . . . . . . . . . . . . . . . . . . . . 165 13.2 Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 13.3 Shifting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 CONTENTS xviii 13.4 13.5 13.6 13.7 Dilation . . Vertex Form Summary of Exercises . ........ and Order of Rules . . . . ........ ....... Operations ....... ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 173 174 178 14 Rational Functions 181 14.1 Modeling with Linear-to-linear Rational Functions . . . . 185 14.2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 14.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 15 Measuring an Angle 15.1 Standard and Central Angles . 15.2 An Analogy . . . . . . . . . . . 15.3 Degree Method . . . . . . . . . 15.4 Radian Method . . . . . . . . . 15.5 Areas of Wedges . . . . . . . . 15.5.1 Chord Approximation 15.6 Great Circle Navigation . . . . 15.7 Summary . . . . . . . . . . . . 15.8 Exercises . . . . . . . . . . . . 16 Measuring Circular Motion 16.1 Different ways to measure Cosmo’s speed . . . . . . . . 16.2 Different Ways to Measure Circular Motion . . . . . . . 16.2.1 Three Key Formulas 16.3 Music Listening Technology 16.4 Belt and Wheel Problems . . 16.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 192 193 193 196 199 201 202 204 205 207 ................. 207 . . . . . . . . . . 209 209 212 215 218 . . . . . . . . . 221 221 222 224 226 227 229 230 231 234 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 The Circular Functions 17.1 Sides and Angles of a Right Triangle . . . . . . . . . . 17.2 The Trigonometric Ratios . . . . . . . . . . . . . . . . . 17.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Circular Functions . . . . . . . . . . . . . . . . . . . . . 17.4.1 Are the trigonometric ratios functions? . . . . 17.4.2 Relating circular functions and right triangles 17.5 What About Other Circles? . . . . . . . . . . . . . . . . 17.6 Other Basic Circular Function . . . . . . . . . . . . . . 17.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Trigonometric Functions 237 18.1 Easy Properties of Circular Functions . . . . . . . . . . . 237 18.2 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 18.3 Graphs of Circular Functions . . . . . . . . . . . . . . . . 242 CONTENTS xix 18.3.1 A matter of scaling . . . . . . . . . . 18.3.2 The sine and cosine graphs . . . . . 18.3.3 The tangent graph . . . . . . . . . . 18.4 Trigonometric Functions . . . . . . . . . . . 18.4.1 A Transition . . . . . . . . . . . . . . 18.4.2 Graphs of trigonometric functions . 18.4.3 Notation for trigonometric functions 18.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 243 246 247 247 249 249 250 19 Sinusoidal Functions 19.1 A special class of functions . . . . . . . . . . . . . . 19.1.1 How to roughly sketch a sinusoidal graph 19.1.2 Functions not in standard sinusoidal form 19.2 Examples of sinusoidal behavior . . . . . . . . . . . 19.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . 19.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 251 254 257 259 262 263 . . . . . . 267 267 270 273 276 280 281 20 Inverse Circular Functions 20.1 Solving Three Equations . . . . . . . 20.2 Inverse Circular Functions . . . . . . 20.3 Applications . . . . . . . . . . . . . . . 20.4 How to solve trigonometric equations 20.5 Summary . . . . . . . . . . . . . . . . 20.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix 285 A Useful Formulas 287 B Answers 291 C GNU Free Documentation License C.1 Applicability and Definitions . . . . . C.2 Verbatim Copying . . . . . . . . . . . C.3 Copying in Quantity . . . . . . . . . . C.4 Modifications . . . . . . . . . . . . . . C.5 Combining Documents . . . . . . . . C.6 Collections of Documents . . . . . . . C.7 Aggregation With Independent Works C.8 Translation . . . . . . . . . . . . . . . C.9 Termination . . . . . . . . . . . . . . . C.10 Future Revisions of This License . . . 303 304 305 305 306 308 308 308 309 309 309 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References 311 Index 311 xx CONTENTS Chapter 1 Warming Up The basic theme of this book is to study precalculus within the context of problem solving. This presents a challenge, since skill in problem solving is as much an art or craft as it is a science. As a consequence, the process of learning involves an active apprenticeship rather than a passive reading of a text. We are going to start out by assembling a basic toolkit of examples and techniques that are essential in everything that follows. The main ideas discussed in the next couple of chapters will surely be familiar; our perspective on their use and importance may be new. The process of going from equations to pictures involves the key concept of a graph , while the reverse process of going from pictures (or raw data) to equations is called modeling . Fortunately, the study of graphing and modeling need not take place in a theoretical vacuum. For example, imagine you have tossed a ball from the edge of a cliff. A number of natural questions arise: Where and when does the ball reach its maximum height? Where and when does the ball hit the ground? Where is the ball located after t seconds? We can attack these questions from two directions. If we knew some basic physics, then we would have equaPath of tossed ball. tions for the motion of the ball. Going from these equations to the actual curved path of the ball becomes a graphing problem; answering the questions requires that we really Cliff. understand the relationship between the symbolic equaGround level. tions and the curved path. Alternatively, we could approach these questions without knowing any physics. The idea would be to collect some data, keeping track of the Figure 1.1: Ball toss. height and horizontal location of the ball at various times, then find equations whose graphs will “best” reproduce the collected data points; this would be a modeling approach to the problem. Modeling is typically harder than graphing, since it requires good intuition and a lot of experience. 1 CHAPTER 1. WARMING UP 2 1.1 Units and Rates A marathon runner passes the one-mile marker of the race with a clocked speed of 18 feet/second. If a marathon is 26.2 miles in length and this speed is maintained for the entire race, what will be the runner’s total time? This simple problem illustrates a key feature of modeling with mathematics: Numbers don’t occur in isolation; a number typically comes with some type of unit attached. To answer the question, we’ll need to recall a formula which precisely relates “total distance traveled” to “speed” and “elapsed time”. But, we must be VERY CAREFUL to use consistent units. We are given speed units which involve distance in “feet” and the length of the race involves distance units of “miles”. We need to make a judgment call and decide on a single type of distance unit to use throughout the problem; either choice is OK. Let’s use “feet”, then here is the fact we need to recall: (total distance traveled) = (constant speed) × (elapsed time) (ft) = (ft/sec) × (sec) (1.1) To apply the formula, let t represent the elapsed time in seconds and first carry out a “conversion of units” using the conversion factor “5,280 ft/mile”. Recall, we can manipulate the units just like numbers, canceling common units on the top and bottom of a fraction: mile ////· ft mile //// = (26.2)(5,280) ft. 26.2 mile × (5,280 ft/mile) = (26.2)(5,280) Formula 1.1 can now be applied: (26.2)(5,280) ft = 18 ft/sec × t (26.2)(5,280) 18 ft / =t ft/sec / 7,685.33 1 = 7,685.33 sec = t 1/sec So, the runner would complete the race in t = 7,685.33 seconds. If we wanted this answer in more sensible units, we would go through yet another units conversion: t = 7,685.33 sec × (1 min/60 sec) × (1 hr/60 min) 7,685.33 hr = 602 = 2.1348 hr. 1.1. UNITS AND RATES 3 The finish clock will display elapsed time in units of “hours : minutes : seconds”. Two further conversions (see Exercise 1.5) lead to our runner having a time of 2:08:05.33; this is a world class time! Manipulation of units becomes especially important when we are working with the density of a substance, which is defined by def density = mass . volume For example, pure water has a density of 1 g/cm3 . Notice, given any two of the quantities “density, volume or mass,” we can solve for the remaining unknown using the formula. For example, if 857 g of an unknown substance has a volume of 2.1 liters, then the density would be mass volume 857 g = 2.1 L g 857 × = 2.1 L / 3 = 0.408 g/cm . d= × 1/ L 1,000 cm3 Example 1.1.1. A sphere of solid gold has a mass of 100 kg and the density of gold is 19.3 g/cm3 . What is the radius of the sphere? Solution. This problem is more involved. To answer this, let r be the unknown radius of the sphere in units of cm. The volume of the sphere is V = 4 πr3 . Since the sphere is solid gold, the density of gold is the ratio 3 density of gold = mass of sphere volume of sphere Plugging in what we know, we get the equation 19.3 g = cm3 100kg 4 πr3 3 = 100kg 4 πr3 3 Solving for r3 we find r3 = 1236.955516 cm3 from which we get r = 1237 cm3 1/3 = 10.73457 cm 1000 g 1 kg = 105 g 4 πr3 3 CHAPTER 1. WARMING UP 4 1.2 Total Change = Rate × Time We live in a world where things are changing as time goes by: the temperature during the day, the cost of tuition, the distance you will travel after leaving this class, and so on. The ability to precisely describe how a quantity is changing becomes especially important when making any kind of experimental measurements. For this reason, let’s start with a clear and careful definition. If a quantity is changing with respect to time (like temperature, distance or cost), we can keep track of this using what is called a rate (also sometimes called a rate of change ); this is defined as follows: def rate = change in the quantity change in time This sort of thing comes up so frequently, there is special shorthand notation commonly used: We let the Greek letter ∆ (pronounced “delta”) be shorthand for the phrase “change in.” With this agreement, we can rewrite our rate definition in this way: def rate = ∆ quantity ∆ time But, now the question becomes: How do we calculate a rate? If you think about it, to calculate “∆ quantity” in the rate definition requires that we compare two quantities at two different times and see how they differ (i.e., how they have changed). The two times of comparison are usually called the final time and the initial time. We really need to be precise about this, so here is what we mean: ∆ quantity = (value of quantity at final time) − (value of quantity at initial time) ∆ time = (final time) − (initial time). For example, suppose that on June 4 we measure that the temperature at 8:00 am is 65◦ F and at 10:00 am it is 71◦ F. So, the final time is 10:00 am, the initial time is 8:00 am and the temperature is changing according to the ∆ quantity ∆ time final value of quantity − initial value of quantity = final time − initial time 71 − 65 degrees = 10:00 − 8:00 hours = 3 deg/hr. rate = As a second example, suppose on June 5 the temperature at 8:00 am is 71◦ and at 10:00 am it is 65◦ . So, the final time is 10:00 am, the initial 1.2. TOTAL CHANGE = RATE × TIME 5 time is 8:00 am and the temperature is changing according to the ∆ quantity ∆ time final value of quantity − initial value of quantity = final time − initial time 65 − 71 degrees = 10:00 − 8:00 hours = −3 deg/hr. rate = These two examples illustrate that a rate can be either a positive or a negative number. More importantly, it highlights that we really need to be careful when making a rate computation. In both examples, the initial and final times are the same and the two temperatures involved are the same, BUT whether they occur at the initial or final time is interchanged. If we accidently mix this up, we will end up being off by a minus sign. There are many situations where the rate is the same for all time periods. In a case like this, we say we have a constant rate. For example, imagine you are driving down the freeway at a constant speed of 60 mi/hr. The fact that the speedometer needle indicates a steady speed of 60 mi/hr means the rate your distance is changing is constant. In cases when we have a constant rate, we often want to find the total amount of change in the quantity over a specific time period. The key principle in the background is this: Total Change in some Quantity = Rate × Time (1.2) It is important to mention that this formula only works when we have a constant rate, but that will be the only situation we encounter in this course. One of the main goals of calculus is to develop a version of (1.2) that works for non-constant rates. Here is another example; others will occur throughout the text. Example 1.2.1. A water pipe mounted to the ceiling has a leak. It is dripping onto the floor below and creates a circular puddle of water. The surface area of this puddle is increasing at a constant rate of 4 cm2 /hour. Find the surface area and dimensions of the puddle after 84 minutes. Solution. The quantity changing is “surface area” and we are given a “rate” and “time.” Using (1.2) with time t = 84 minutes, Total Surface Area = Rate × Time cm2 84 = 4 × hr hr 60 = 5.6 cm2 . The formula for the area of a circular region of radius r is given at the back of this text. Using this, the puddle has radius r = time t = 84 minutes. 5.6 π = 1.335 cm at CHAPTER 1. WARMING UP 6 1.3 The Modeling Process Modeling is a method used in disciplines ranging from architecture to zoology. This mathematical technique will crop up any time we are problem solving and consciously trying to both “describe” and “predict.” Inevitably, mathematics is introduced to add structure to the model, but the clean equations and formulas only arise after some (or typically a lot) of preliminary work. A model can be thought of as a caricature in that it will pick out certain features (like a nose or a face) and focus on those at the expense of others. It takes a lot of experience to know which models are “good” and “bad,” in the sense of isolating the right features. In the beginning, modeling will lead to frustration and confusion, but by the end of this course our comfort level will dramatically increase. Let’s look at an illustration of the problem solving process. Example 1.3.1. How much time do you anticipate studying precalculus each week? Solution. One possible response is simply to say “a little” or “way too much!” You might not think these answers are the result of modeling, but they are. They are a consequence of modeling the total amount of study time in terms of categories such as “a little,” “some,” “lots,” “way too much,” etc. By drawing on your past experiences with math classes and using this crude model you arrived at a preliminary answer to the question. Let’s put a little more effort into the problem and try to come up with a numerical estimate. If T is the number of hours spent on precalculus a given week, it is certainly the case that: T = (hours in class) + (hours reading text) + (hours doing homework) Our time in class each week is known to be 5 hours. However, the other two terms require a little more thought. For example, if we can comfortably read and digest a page of text in (on average) 15 minutes and there are r pages of text to read during the week, then (hours reading text) = 15 r hours. 60 As for homework, if a typical homework problem takes (on average) 25 minutes and there are h homework problems for the week, then (hours doing homework) = 25 h hours. 60 We now have a mathematical model for the weekly time commitment to precalculus: T =5+ 25 15 r + h hours. 60 60 1.3. THE MODELING PROCESS 7 Is this a good model? Well, it is certainly more informative than our original crude model in terms of categories like “a little” or “lots.” But, the real plus of this model is that it clearly isolates the features being used to make our estimated time commitment and it can be easily modified as the amount of reading or homework changes. So, this is a pretty good model. However, it isn’t perfect; some homework problems will take a lot more than 25 minutes! CHAPTER 1. WARMING UP 8 1.4 Exercises Problem 1.1. (a) Verify that 7685.33 seconds is 2 hours 8 minutes 5.33 seconds. (b) Allyson has a pace of 6 min/mile; what is her speed? (b) Which is faster: 100 mph or 150 ft/s? (c) Adrienne and Dave are both running a race. Adrienne has a pace of 5.7 min/mile and Dave is running 10.3 mph. Who is running faster? (c) Gina’s salary is 1 cent/second for a 40 hour work week. Tiare’s salary is $1400 for a 40 hour work week. Who has a higher salary? (d) Suppose it takes 180 credits to get a baccalaureate degree. You accumulate credit at the rate of one credit per quarter for each hour that the class meets per week. For instance, a class that meets three hours each week of the quarter will count for three credits. In addition, suppose that you spend 2.5 hours of study outside of class for each hour in class. A quarter is 10 weeks long. How many total hours, including time spent in class and time spent studying out of class, must you invest to get a degree? Problem 1.2. Sarah can bicycle a loop around the north part of Lake Washington in 2 hours and 40 minutes. If she could increase her average speed by 1 km/hr, it would reduce her time around the loop by 6 minutes. How many kilometers long is the loop? Problem 1.3. The density of lead is 11.34 g/cm3 and the density of aluminum is 2.69 g/cm3 . Find the radius of lead and aluminum spheres each having a mass of 50 kg. Problem 1.4. The Eiffel Tower has a mass of 7.3 million kilograms and a height of 324 meters. Its base is square with a side length of 125 meters. The steel used to make the Tower occupies a volume of 930 cubic meters. Air has a density of 1.225 kg per cubic meter. Suppose the Tower was contained in a cylinder. Find the mass of the air in the cylinder. Is this more or less than the mass of the Tower? Problem 1.5. Marathon runners keep track of their speed using units of pace = minutes/mile. (a) Lee has a speed of 16 ft/sec; what is his pace? Problem 1.6. Convert each of the following sentences into “pseudo-equations.” For example, suppose you start with the sentence: “The cost of the book was more than $10 and the cost of the magazine was $4.” A first step would be these “pseudo-equations”: (Book cost) > $10 and (Magazine cost) = $4. (a) John’s salary is $56,000 a year and he pays no taxes. (b) John’s salary is at most $56,000 a year and he pays 15% of his salary in taxes. (c) John’s salary is at least $56,000 a year and he pays more than 28% of his salary in taxes. (d) The number of students taking Math 120 at the UW is somewhere between 1500 and 1800 each year. (e) The cost of a new red Porsche is more than three times the cost of a new Ford F-150 pickup truck. (f) Each week, students spend at least two but no more than three hours studying for each credit hour. (g) Twice the number of happy math students exceeds five times the number of happy chemistry students. However, all of the happy math and chemistry students combined is less than half the total number of cheerful biology students. (h) The difference between Cady’s high and low midterm scores was 10%. Her final exam score was 97%. (i) The vote tally for Gov. Tush was within one-hundredth of one percent of onehalf the total number of votes cast. Problem 1.7. Which is a better deal: A 10 inch diameter pizza for $8 or a 15 inch diameter pizza for $16? 1.4. EXERCISES 9 Problem 1.8. The famous theory of relativity predicts that a lot of weird things will happen when you approach the speed of light c = 3 × 108 m/sec. For example, here is a formula that relates the mass mo (in kg) of an object at rest and its mass when it is moving at a speed v: m = mo 1− v2 c2 Problem 1.11. A typical cell in the human body contains molecules of deoxyribonucleic acid, referred to as DNA for short. In the cell, this DNA is all twisted together in a tight little packet. But imagine unwinding (straightening out) all of the DNA from a single typical cell and laying it “end-to-end”; then the sum total length will be approximately 2 meters. . isolate DNA from nucleus (a) Suppose the object moving is Dave, who has a mass of mo = 66 kg at rest. What is Dave’s mass at 90% of the speed of light? At 99% of the speed of light? At 99.9% of the speed of light? cell nucleus (b) How fast should Dave be moving to have a mass of 500 kg? Problem 1.9. During a typical evening in Seattle, Pagliacci receives phone orders for pizza delivery at a constant rate: 18 orders in a typical 4 minute period. How many pies are sold in 4 hours? Assume Pagliacci starts taking orders at 5 : 00 pm and the profit is a constant rate of $11 on 10 orders. When will phone order profit exceed $1,000? Problem 1.10. Aleko’s Pizza has delivered a beautiful 16 inch diameter pie to Lee’s dorm room. The pie is sliced into 8 equal sized pieces, but Lee is such a non-conformist he cuts off an edge as pictured. John then takes one of the remaining triangular slices. Who has more pizza and by how much? John’s part lay out end−to−end 2m Assume the human body has 1014 cells containing DNA. How many times would the sum total length of DNA in your body wrap around the equator of the earth? Problem 1.12. A water pipe mounted to the ceiling has a leak and is dripping onto the floor below, creating a circular puddle of water. The area of the circular puddle is increasing at a constant rate of 11 cm2 /hour. (a) Find the area and radius of the puddle after 1 minute, 92 minutes, 5 hours, 1 day. (b) Is the radius of the puddle increasing at a constant rate? Problem 1.13. During the 1950s, Seattle was dumping an average of 20 million gallons of sewage into Lake Washington each day. (a) How much sewage went into Lake Washington in a week? In a year? Lee’s part (b) In order to illustrate the amounts involved, imagine a rectangular prism whose base is the size of a football field (100 yards × 50 yards) with height h yards. What are the dimensions of such a rectangular prism containing the CHAPTER 1. WARMING UP 10 sewage dumped into Lake Washington in a single day? (Note: There are 7.5 gallons in one cubic foot. Dumping into Lake Washington has stopped; now it goes into the Puget Sound.) Problem 1.14. Dave has inherited an apple orchard on which 60 trees are planted. Under these conditions, each tree yields 12 bushels of apples. According to the local WSU extension agent, each time Dave removes a tree the yield per tree will go up 0.45 bushels. Let x be the number of trees in the orchard and N the yield per tree. (a) Find a formula for N in terms of the unknown x. (Hint: Make a table of data with one column representing various values of x and the other column the corresponding values of N. After you complete the first few rows of the table, you need to discover the pattern.) (b) What possible reason(s) might explain why the yield goes up when you remove trees? Problem 1.15. Congress is debating a proposed law to reduce tax rates. If the current tax rate is r %, then the proposed rate after x years is given by this formula: r 1+ . 1 1+ 1 x Rewrite this formula as a simple fraction. Use your formula to calculate the new tax rate af- ter 1, 2, 5 and 20 years. Would tax rates increase or decrease over time? Congress claims that this law would ultimately cut peoples’ tax rates by 75 %. Do you believe this claim? Problem 1.16. (a) The temperature at 7:00 am is 44◦ F and the temperature at 10:00 am is 50◦ F. What are the initial time, the final time, the initial temperature and the final temperature? What is the rate of change in the temperature between 7:00 am and 10:00 am? (b) Assume it is 50◦ F at 10:00 am and rate of change in the temperature tween 10:00 am and 2:00 pm is same as the rate in part (a). What is temperature at 2:00 pm? the bethe the (c) The temperature at 4:30 pm is 54◦ F and the temperature at 6:15 pm is 26◦ F. What are the initial time, the final time, the initial temperature and the final temperature? What is the rate of change in the temperature between 4:30 pm and 6:15 pm? Problem 1.17. (b) Solve for a: (a) Solve for t: 3t − 7 = 11 + t. 1 a = 3. √ (c) Solve for x: x2 + a2 = 2a + x. 1+ (d) Solve for t: 1 − t > 4 − 2t. (e) Write as a single fraction: 2 1 − x x+1 Chapter 2 Imposing Coordinates You find yourself visiting Spangle, WA and dinner time is approaching. A friend has recommended Tiff’s Diner, an excellent restaurant; how will you find it? Of course, the solution to this simple problem amounts to locating a “point” on a two-dimensional map. This idea will be important in many problem solving situations, so we will quickly review the key ideas. 2.1 The Coordinate System If we are careful, we can develop the flow of ideas underlying two-dimensional coordinate systems in such a way Q that it easily generalizes to three-dimensions. Suppose we start with a blank piece of paper and mark two points; let’s label these two points “P ” and “Q.” This presents the P basic problem of finding a foolproof method to reconstruct the picture. Figure 2.1: Two points in a The basic idea is to introduce a coordinate system for plane. the plane (analogous to the city map grid of streets), allowing us to catalog points in the plane using pairs of real numbers (analogous to the addresses of locations in the city). Here are the details. Start by drawing two perpendicular lines, called the horizontal axis and the vertical axis , each of which looks like a copy of the real number line. We refer to the intersection point of these two lines as the origin . Given P in the plane, the plan is to use these two axes to obtain a pair of real numbers (x,y) that will give us the exact location of P. With this in mind, the horizontal axis is often called the x-axis and the vertical axis is often called the y-axis. Remember, a typical real number line (like the x-axis or the y-axis) is divided into three parts: the positive numbers, the negative numbers, and the number zero (see Figure 2.2(a)). This allows us to specify positive and negative portions of the x-axis and y-axis. Unless we say otherwise, we will always adopt the convention that the positive x-axis consists of those numbers to the right of the origin on 11 CHAPTER 2. IMPOSING COORDINATES 12 the x-axis and the positive y-axis consists of those numbers above the origin on the y-axis. We have just described the xy-coordinate system for the plane: Positive y-axis Negative real numbers Positive real numbers Origin Negative x-axis Zero Positive x-axis Negative y-axis (a) Number line. (b) xy-coordinate system. Figure 2.2: Coordinates. 2.1.1 Going from P to a Pair of Real Numbers. y-axis P ℓ y x-axis x ℓ∗ Figure 2.3: Coordinate pairs. Imagine a coordinate system had been drawn on our piece of paper in Figure 2.1. Let’s review the procedure of going from a point P to a pair of real numbers: 1. First, draw two new lines passing through P, one parallel to the x-axis and the other parallel to the y-axis; call these ℓ and ℓ∗ , as pictured in Figure 2.3. 2. Notice that ℓ will cross the y-axis exactly once; the point on the y-axis where these two lines cross will be called “y.” Likewise, the line ℓ∗ will cross the x-axis exactly once; the point on the x-axis where these two lines cross will be called “x.” 3. If you begin with two different points, like P and Q in Figure 2.1, you will see that the two pairs of points you obtain will be different; i.e., if Q gives you the pair (x∗ ,y∗ ), then either x = x∗ or y = y∗ . This shows that two different points in the plane give two different pairs of real numbers and describes the process of assigning a pair of real numbers to the point P. The great thing about the procedure we just described is that it is reversible! In other words, suppose you start with a pair of real numbers, say (x,y). Locate the number x on the x-axis and the number y on the y-axis. Now draw two lines: a line ℓ parallel to the x-axis passing through the number y on the y-axis and a line ℓ∗ parallel to the y-axis passing through the number x on the x-axis. The two lines ℓ and ℓ∗ will intersect 2.2. THREE FEATURES OF A COORDINATE SYSTEM 13 in exactly one point in the plane, call it P. This procedure describes how to go from a given pair of real numbers to a point in the plane. In addition, if you start with two different pairs of real numbers, then the corresponding two points in the plane are going to be different. In the future, we will constantly be going back and forth between points in the plane and pairs of real numbers using these ideas. Definition 2.1.1. Coordinate System: Every point P in the xy-plane corresponds to a unique pair of real numbers (x, y), where x is a number on the horizontal x-axis and y is a number on the vertical y-axis; for this reason, we commonly use the notation “P = (x,y).” Having specified positive and negative directions on the horizontal and vertical axes, we can now divide our two dimensional plane into four quadrants . The first quadrant corresponds to all the points where both coordinates are positive, the second quadrant consists of points with the first coordinate negative and the second coordinate positive, etc. Every point in the plane will lie in one of these four quadrants or on one of the two axes. This quadrant terminology is useful to give a rough sense of location, just as we use the terminology “Northeast, Northwest, Southwest and Southeast” when discussing locations on a map. y-axis Second Quadrant First Quadrant Third Quadrant Fourth Quadrant Figure 2.4: Quadrants in the xy-plane. 2.2 Three Features of a Coordinate System A coordinate system involves scaling, labeling and units on each of the axes. 2.2.1 Scaling Sketch two xy coordinate systems. In the first, make the scale on each axis the same. In the second, assume “one unit” on the x axis has the same length as “two units” on the y axis. Plot the points (1,1), (−1,1), 39 24 11 11 24 39 4 4 − 5 , 16 , − 5 , 25 , − 5 , 25 , − 5 , 25 , (0,0), 5 , 25 , 5 , 25 , 5 , 25 , 5 , 16 , (1,1). 25 25 Both pictures illustrate how the points lie on a parabola in the xycoordinate system, but the aspect ratio has changed. The aspect ratio is defined by this fraction: def aspect ratio = x-axis length of one unit on the vertical axis . length of one unit on the horizontal axis Figure 2.5(a) has aspect ratio 1, whereas Figure 2.5(b) has aspect ratio 1 . In problem solving, you will often need to make a rough assumption 2 about the relative axis scaling. This scaling will depend entirely on the CHAPTER 2. IMPOSING COORDINATES 14 1.0 y-axis 0.8 0.6 1.0 y-axis 0.8 0.6 0.4 0.2 0.4 0.2 −1.0 −0.5 0.0 0.5 x-axis 1.0 (a) Aspect ratio = 1. −1.0 −0.5 0.0 x-axis 0.5 1.0 1 (b) Aspect ratio = 2 . Figure 2.5: Coordinates. information given in the problem. Most graphing devices will allow you to specify the aspect ratio. 2.2.2 Axes Units Sometimes we are led to coordinate systems where each of the two axes involve different types of units (labels). Here is a sample, that illustrates the power of using pictures. Example 2.2.1. As the marketing director of Turboweb software, you have been asked to deliver a brief message at the annual stockholders meeting on the performance of your product. Your staff has assembled this tabular collection of data; how can you convey the content of this table most clearly? week 1 2 3 4 5 6 7 8 9 10 sales 11.0517 12.214 13.4986 14.9182 16.4872 18.2212 20.1375 22.2554 24.596 27.1828 week 11 12 13 14 15 16 17 18 19 20 TURBOWEB SALES (in $1000’s) sales week sales week 30.0417 21 81.6617 31 33.2012 22 90.2501 32 36.693 23 99.7418 33 40.552 24 110.232 34 44.8169 25 121.825 35 49.5303 26 134.637 36 54.7395 27 148.797 37 60.4965 28 164.446 38 66.8589 29 181.741 39 73.8906 30 200.855 40 sales 221.98 245.325 271.126 299.641 331.155 365.982 404.473 447.012 494.024 545.982 week 41 42 43 44 45 46 47 48 49 50 sales 603.403 666.863 736.998 814.509 900.171 994.843 1099.47 1215.1 1342.9 1,484.131 One idea is to simply flash an overhead slide of this data to the audience; this can be deadly! A better idea is to use a visual aid. Suppose we let the variable x represent the week and the variable y represent the gross sales (in thousands of dollars) in week x. We can then plot the points (x,y) in the xy-coordinate system; see Figure 2.6. Notice, the units on the two axes are very different: y-axis units are “thousands of dollars” and x-axis units are “weeks.” In addition, the aspect ratio of this coordinate system is not 1. The beauty of this picture is the visual impact it gives your audience. From the coordinate plot we can get a sense of how the sales figures are dramatically increasing. In fact, this plot is good evidence you deserve a big raise! 2.3. A KEY STEP IN ALL MODELING PROBLEMS Mathematical modeling is all about relating concrete phenomena and symbolic equations, so we want to embrace the idea of visualization. Most typically, visualization will involve plotting a collection of points in the plane. This can be achieved by providing a “list” or a “prescription” for plotting the points. The material we review in the next couple of sections makes the transition from symbolic mathematics to visual pictures go more smoothly. 15 y-axis (Thousands of Dollars) 1,400 1,200 1,000 800 600 400 200 x-axis (Weeks) 10 20 30 40 50 Figure 2.6: Turboweb sales. 2.3 A Key Step in all Modeling Problems The initial problem solving or modeling step of deciding on a choice of xy-coordinate system is called imposing a coordinate system : There will often be many possible choices; it takes problem solving experience to develop intuition for a “natural” choice. This is a key step in all modeling problems. Example 2.3.1. Return to the tossed ball scenario on page 1. How do we decide where to draw a coordinate system in the picture? Figure 2.7 on page 16 shows four natural choices of xy-coordinate system. To choose a coordinate system we must specify the origin. The four logical choices for the origin are either the top of the cliff, the bottom of the cliff, the launch point of the ball or the landing point of the ball. So, which choice do we make? The answer is that any of these choices will work, but one choice may be more natural than another. For example, Figure 2.7(b) is probably the most natural choice: in this coordinate system, the motion of the ball takes place entirely in the first quadrant, so the x and y coordinates of any point on the path of the ball will be non-negative. Example 2.3.2. Michael and Aaron are running toward each other, beginning at opposite ends of a 10,000 ft. airport runway, as pictured in Figure 2.8 on page 17. Where and when will these guys collide? Solution. This problem requires that we find the “time” and “location” of the collision. Our first step is to impose a coordinate system. We choose the coordinate system so that Michael is initially located at the point M = (0, 0) (the origin) and Aaron is initially located at the point A = (10,000, 0). To find the coordinates of Michael after t seconds, we need to think about how distance and time are related. Since Michael is moving at the rate of 15 ft/second, then after one second he is located 15 feet right of the origin; i.e., at the point (15, 0). After 2 seconds, Michael has moved an additional 15 feet, for a total of 30 feet; so he is located at the point (30, 0), etc. Conclude Michael has traveled 15t ft. to the right after t seconds; i.e., his location is the point CHAPTER 2. IMPOSING COORDINATES 16 y-axis Path of tossed ball. y-axis Path of tossed ball. x-axis Cliff. Cliff. x-axis (a) Origin at the top of the ledge. Path of tossed ball. y-axis (b) Origin at the bottom of the ledge. y-axis Path of tossed ball. x-axis Cliff. Cliff. x-axis (c) Origin at the landing point. (d) Origin at the launch point. Figure 2.7: Choices when imposing an xy-coordinate system. M(t) = (15t, 0). Similarly, Aaron is located 8 ft. left of his starting location after 1 second (at the point (9,992, 0)), etc. Conclude Aaron has traveled 8t ft. to the left after t seconds; i.e., his location is the point A(t) = (10,000 − 8t, 0). The key observation required to solve the problem is that the point of collision occurs when the coordinates of Michael and Aaron are equal. Because we are moving along the horizontal axis, this amounts to finding where and when the x-coordinates of M(t) and A(t) agree. This is a straight forward algebra problem: 15t = 10,000 − 8t 23t = 10,000 t = 434.78 (2.1) To the nearest tenth of a second, the runners collide after 434.8 seconds. Plugging t = 434.78 into either expression for the position: M(434.8) = (15(434.8), 0) = (6,522, 0). 2.4. DISTANCE Michael: 15 17 ft sec Aaron: 8 ft sec y-axis x-axis M = (0,0) 10,000 ft (b) The xy-coordinate picture. (a) The physical picture. y-axis NOT TO SCALE! Aaron starts here. Michael starts here. (0,0) A = (10,000, 0) M(t) = (15t, 0) A(t) = (10,000 − 8t, 0) x-axis M after t seconds. A after t seconds. (c) Building a visual model Michael: 10 mph. Aaron: 8 mph. 6,522 feet to collision point. (d) Michael and Aaron’s collision point. Figure 2.8: Michael and Aaron running head-on. 2.4 Distance We end this Chapter with a discussion of direction and distance in the plane. To set the stage, think about the following analogy: Example 2.4.1. You are in an airplane flying from Denver to New York. How far will you fly? To what extent will you travel north? To what extent will you travel east? Consider two points P = (x1 ,y1 ) and Q = (x2 ,y2 ) in the xy coordinate system, where we assume that the units on each axis are the same ; for example, both in units of “feet.” Imagine starting at the location P (Denver) and flying to the location Q (New York) along a straight line segment; see Figure 2.9(a). Now ask yourself this question: To what overall extent have the x and y coordinates changed? To answer this, we introduce visual and notational aides into this figure. We have inserted an “arrow” pointing from the starting position P to the ending position Q; see Figure 2.9(b). To simplify things, introduce the notation ∆x to keep track of the change in the x-coordinate and ∆y CHAPTER 2. IMPOSING COORDINATES 18 y-axis y-axis Q = (x2 , y2 ) Q y2 End (stop) here. ?? ?? Ending point. d ∆y Begin (start) here. P x-axis ?? (a) Starting and stopping points. y1 P = (x1 , y1 ) Beginning point. x1 x-axis x2 ∆x (b) Coordinates for P and Q. Figure 2.9: The meaning of ∆x and ∆y. to keep track of the change in the y-coordinate, as we move from P to Q. Each of these quantities can now be computed: ∆x = = = ∆y = = = change in x-coordinate going from P to Q (2.2) (x-coord of ending point) − (x-coord of beginning point) x2 − x1 change in y-coordinate going from P to Q (y-coord of ending point) − (y-coord of beginning point) y2 − y1 . We can interpret ∆x and ∆y using the right triangle in Figure 2.9(b). This means we can use the Pythagorean Theorem to write: d2 = (∆x)2 + (∆y)2 ; that is, d= (∆x)2 + (∆y)2 , which tells us the distance d from P to Q. In other words, d is the distance we would fly if we had flown along that line segment connecting the two points. As an example, if P = (1, 1) and Q = (5, 4), then ∆x = 5 − 1 = 4, ∆y = 4 − 1 = 3 and d = 5. There is a subtle idea behind the way we defined ∆x and ∆y: You need to specify the “beginning” and “ending” points used to do the calculation in Equations 2.2. What happens if we had reversed the choices in Figure 2.9? Then the quantities ∆x and ∆y will both be negative and the lengths of the sides of the right triangle are computed by taking the absolute value 2.4. DISTANCE 19 of ∆x and ∆y. As far as a distance calculation is concerned, the previous formula still works because of this algebra equality: d= = (∆x)2 + (∆y)2 (|∆x|)2 + (|∆y|)2 . We will sometimes refer to ∆x and ∆y as directed distances in the x and y directions. The notion of directed distance becomes important in our discussion of lines in Chapter 4 and later when you learn about vectors ; it is also very important in calculus. For example, if P = (5, 4) and Q = (1, 1), then ∆x = y-axis 1 − 5 = −4, ∆y = 1 − 4 = −3 and d = 5. Important Fact 2.4.2 (Distance formula). If P = (x1 , y1) and Q = (x2 , y2) are two points in the plane, then the straight line distance between the points (in the same units as the two axes) is given by the formula d= = (∆x)2 + (x2 − x1 Beginning point. d |∆y| y2 Q = (x2 , y2 ) Ending point. (∆y)2 )2 P = (x1 , y1 ) y1 x2 + (y2 − y1 )2 . (2.3) x-axis x1 |∆x| Figure 2.10: A different direction. If your algebra is a little rusty, a very common mistake may crop up when you are using the distance formula. For example, ?√ = 32 + √ 42 ? = 3+4 = 7. Notice, you have an impossible situation: 5 is never equal to 7. !!! 32 + 42 √ 9 + 16 5 CAUTION !!! CHAPTER 2. IMPOSING COORDINATES 20 Example 2.4.3. Two cars depart from a four way intersection at the same time, one heading East and the other heading North. Both cars are traveling at the constant speed of 30 ft/sec. Find the distance (in miles) between the two cars after 1 hour 12 minutes. In addition, determine when the two cars would be exactly 1 mile apart. Solution. Begin with a picture of the situation. We have indicated the locations of the two vehicles after t seconds and the distance d between them at time t. By the distance formula, the distance between them is y-axis (North) (0,b) d x-axis (East) d= = (a,0) Figure 2.11: Two departing cars. (a − 0)2 + (0 − b )2 a2 + b2 . This formula is a first step; the difficulty is that we have traded the mystery distance d for two new unknown numbers a and b. To find the coordinate a for the Eastbound car, we know the car is moving at the rate of 30 ft/sec, so it will travel 30t feet after t seconds; i.e., a = 30t. Similarly, we find that b = 30t. Substituting into the formula for d we arrive at d= (30t)2 + (30t)2 2t2 (30)2 √ = 30t 2. = First, we need to convert 1 hour and 12 minutes into seconds so that our formula can be used: 1 hr 12 min = 1 + 12/60 hr = 1.2 hr 60 min = (1.2hr) hr = 4,320 sec. 60 sec min Substituting t = 4,320 sec and recalling that 1 mile = 5,280 feet, we arrive at √ d = 129,600 2 feet = 183,282 feet = 34.71 miles. For the second question, we specify the distance being 1 mile and want to find when this occurs. The idea is to set d equal to 1 mile and solve for 2.4. DISTANCE t. However, we need to be careful, since the units for d are feet: √ 30t 2 = d = 5,280 Solving for t: 5,280 √ 30 2 = 124.45 seconds = 2 minutes 4 seconds. t= The two cars will be 1 mile apart in 2 minutes, 4 seconds. 21 CHAPTER 2. IMPOSING COORDINATES 22 2.5 Exercises Problem 2.1. In the following four cases, let P be the initial (starting) point and Q the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. Compute d = the distance from P to Q, ∆x and ∆y. Give your answer in exact form; √ answer, whereas 1.41 is an eg. 2 is an exact√ approximation of 2. (a) P = (0,0), Q = (1,1). Problem 2.4. Erik’s disabled sailboat is floating at a stationary location 3 miles East and 2 miles North of Kingston. A ferry leaves Kingston heading due East toward Edmonds at 12 mph. At the same time, Erik leaves the sailboat in a dinghy heading due South at 10 ft/sec (hoping to intercept the ferry). Edmonds is 6 miles due East of Kingston. sailboat (b) P = (2,1), Q = (1, − 1). (c) P = (−1,2), Q = (4, − 1). Edmonds Kingston (d) P = (1,2), Q = (1 + 3t,3 + t), where t is a constant. Problem 2.2. Start with two points M = (a,b) and N = (s,t) in the xy-coordinate system. Let d be the distance between these two points. Answer these questions and make sure you can justify your answers: (a) TRUE or FALSE: d = (a − s)2 + (t − b)2 . (c) TRUE or FALSE: d = (s − a)2 + (t − b)2 . Ballard UDub (a − s)2 + (b − t)2 . (b) TRUE or FALSE: d = North (d) Suppose M is the beginning point and N is the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. What is ∆x? What is ∆y? (e) Suppose N is the beginning point and M is the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. What is ∆x? What is ∆y? (f) If ∆x=0, what can you say about the relationship between the positions of the two points M and N ? If ∆y=0, what can you say about the relationship between the positions of the two points M and N? (Hint: Use some specific values for the coordinates and draw some pictures to see what is going on.) Problem 2.3. Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads North, at 6 AM, and walks steadily at 3 miles per hour. Elsie sleeps in, and starts walking West at 3.5 miles per hour starting at 8 AM. When will the distance between them be 25 miles? (a) Compute Erik’s speed in mph and the Ferry speed in ft/sec. (b) Impose a coordinate system and complete this table of data concerning locations (i.e., coordinates) of Erik and the ferry. Insert into the picture the locations of the ferry and Erik after 7 minutes. Time Ferry Erik Distance Between 0 sec 30 sec 7 min t hr (c) Explain why Erik misses the ferry. (d) After 10 minutes, a Coast Guard boat leaves Kingston heading due East at a speed of 25 ft/sec. Will the Coast Guard boat catch the ferry before it reaches Edmonds? Explain. Problem 2.5. Suppose two cars depart from a four way intersection at the same time, one heading north and the other heading west. The car heading north travels at the steady speed of 30 ft/sec and the car heading west travels at the steady speed of 58 ft/sec. 2.5. EXERCISES 23 Brooke (a) Find an expression for the distance between the two cars after t seconds. (b) Find the distance in miles between the two cars after 3 hours 47 minutes. ocean 5 mi kayak reaches shore here (c) When are the two cars 1 mile apart? shore Problem 2.6. Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person’s ankle. The cord is 30 feet long, but can stretch up to 90 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 8 ft/sec in the directions indicated. Kono’s A 6 mi (a) If she paddles along a straight line course to the shore, find an expression that computes the total time to reach lunch in terms of the location where Brooke beaches the boat. (b) Determine the total time to reach Kono’s if she paddles directly to the point “A”. 20 ft (c) Determine the total time to reach Kono’s if she paddles directly to Kono’s. Building (d) Do you think your answer to (b) or (c) is the minimum time required for Brooke to reach lunch? Allyson 30 ft Adrian (e) Determine the total time to reach Kono’s if she paddles directly to a point on the shore half way between point “A” and Kono’s. How does this time compare to the times in parts (b) and (c)? Do you need to modify your answer to part (d)? start (a) Where are the two girls located after 2 seconds? (b) After 2 seconds, will the slack in the bungee cord be used up? (c) Determine when the bungee cord first becomes tight; i.e. there is no slack in the line. Where are the girls located when this occurs? (d) When will the bungee cord first touch the corner of the building? (Hint: Use a fact about “similar triangles”.) Problem 2.7. Brooke is located 5 miles out from the nearest point A along a straight shoreline in her seakayak. Hunger strikes and she wants to make it to Kono’s for lunch; see picture. Brooke can paddle 2 mph and walk 4 mph. Problem 2.8. A spider is located at the position (1,2) in a coordinate system, where the units on each axis are feet. An ant is located at the position (15,0) in the same coordinate system. Assume the location of the spider after t minutes is s(t) = (1 + 2t,2 + t) and the location of the ant after t minutes is a(t) = (15 − 2t,2t). (a) Sketch a picture of the situation, indicating the locations of the spider and ant at times t = 0,1,2,3,4,5 minutes. Label the locations of the bugs in your picture, using the notation s(0), s(1),...,s(5), a(0), a(1), ..., a(5). (b) When will the x-coordinate of the spider equal 5? When will the y-coordinate of the ant equal 5? (c) Where is the spider located when its y-coordinate is 3? (d) Where is each bug located when the y-coordinate of the spider is twice as large as the y-coordinate of the ant? 24 (e) How far apart are the bugs when their x-coordinates coincide? Draw a picture, indicating the locations of each bug when their x-coordinates coincide. (f) A sugar cube is located at the position (9,6). Explain why each bug will pass through the position of the sugar cube. Which bug reaches the sugar cube first? (g) Find the speed of each bug along its line of motion; which bug is moving faster? Problem 2.9. A Ferrari is heading south at a constant speed on Broadway (a north/south street) at the same time a Mercedes is heading west on Aloha Avenue (an east/west street). The Ferrari is 624 feet north of the intersection of Broadway and Aloha, at the same time that the Mercedes is 400 feet east of the intersection. Assume the Mercedes is traveling at the constant speed of 32 miles/hour. Find the speed of the Ferrari so that a collision occurs in the intersection of Broadway and Aloha. Problem 2.10. Two planes flying opposite directions (North and South) pass each other 80 miles apart at the same altitude. The Northbound plane is flying 200 mph (miles per hour) and the Southbound plane is flying 150 mph. How far apart are the planes in 20 minutes? When are the planes 300 miles apart? Problem 2.11. Here is a list of some algebra problems with ”solutions.” Some of the solutions are correct and some are wrong. For each problem, determine: (i) if the answer is correct, (ii) if the steps are correct, (iii) identify any incorrect steps in the solution (noting that the answer may be correct but some steps may not be correct). CHAPTER 2. IMPOSING COORDINATES (a) If x = 1, x2 − 1 x+1 x2 + (−1)1 x+1 x 2 −1 + x 1 x−1 = = = (b) (x + y)2 − (x − y)2 = = (x2 + y2 ) − x2 − y2 0 (c) If x = 4, 9(x − 4)2 3x − 12 = = = 32 (x − 4)2 3x − 12 (3x − 12)2 3x − 12 3x − 12. Problem 2.12. Assume α, β are nonzero constants. Solve for x. 1 αx−β (a) αx + β = + 1 β = 1 x (c) α + 1 β = 1 x (b) 1 α Problem 2.13. Simplify as far as possible. (a) (1 − t)2 + (2 + 2t)2 (b) (t + 1)2 + (−t − 1)2 − 2 (c) (d) 1 t−1 − 1 t+ 1 (write as a single fraction) (2 + t)2 + 4t2 Chapter 3 Three Simple Curves Before we discuss graphing, we first want to become acy-axis quainted with the sorts of pictures that will arise. This is A typical curve. surprisingly easy to accomplish: Impose an xy-coordinate system on a blank sheet of paper. Take a sharp pencil and begin moving it around on the paper. The resulting picx-axis ture is what we will call a curve. For example, here is a sample of the sort of “artwork” we are trying to visualize. A number of examples in the text will involve basic curves in the plane. When confronted with a curve in the plane, the fundamental question we always try to answer is this: Figure 3.1: A typical curve. Can we give a condition (think of it as a “test”) that will tell us precisely when a point in the plane lies on a curve? Typically, the kind of condition we will give involves an equation in two variables (like x and y). We consider the three simplest situations in this chapter: horizontal lines, vertical lines and circles. 3.1 The Simplest Lines Undoubtedly, the simplest curves in the plane are the horizontal and vertical lines. For example, sketch a line parallel to the x-axis passing through 2 on the y-axis; the result is a horizontal line ℓ, as pictured. This means the line ℓ passes through the point (0, 2) in our coordinate system. A concise symbolic prescription for ALL of the points on ℓ can be given using “set notation”: y-axis (−1, 2) (0, 2) (2, 2) ℓ x-axis Figure 3.2: The points (x, y). ℓ = {(x, 2)|x is any real number}. We read the right-hand side of this expression as “the set of all points (x, 2) where x = any real number.” Notice, the points (x, y) on the line ℓ are EXACTLY the ones that lead to solutions of the equation y = 2; i.e., take any point on this line, plug the coordinates into the equation y = 2 25 (x, 2), a typical point. CHAPTER 3. THREE SIMPLE CURVES 26 y-axis m and you get a true statement. Because the equation does not involve the variable x and only constrains y to equal 2, we see that x can take on any real value. In short, we see that plotting all of the solutions (x, y) to the equation y = 2 gives the line ℓ. We usually refer to the set of all solutions of the equation y = 2 as the graph of the equation y = 2. As a second example, sketch the vertical line m passing through 3 on the x-axis; this means the line m passes through the point (3, 0) in our coordinate system. A concise symbolic prescription for ALL of the points on m can (3, y), a typical point. be given using “set notation”: (3, 3) m = {(3, y)|y is any real number}. (3, 0) Notice, the points (x, y) on the line m are EXACTLY the ones that lead to solutions of the equation x = 3; i.e., take any point on this line, plug the coordinates into the equation x = 3 and you get a true statement. Because the Figure 3.3: Stacked points. equation does not involve the variable y and only specifies that x = 3, y can take on any real number value. Plotting all of the solutions (x, y) to the equation x = 3 gives the line m. We usually refer to the set of all solutions of the equation x = 3 as the graph of the equation x = 3. These two simple examples highlight our first clear connection between a geometric figure and an equation; the link is achieved by plotting all of the solutions (x, y) of the equation in the xy-coordinate system. These observations work for any horizontal or vertical line. x-axis (3, −2) Definition 3.1.1. Horizontal and Vertical Lines: A horizontal line ℓ passing through k on the y-axis is precisely a plot of all solutions (x, y) of the equation y = k; i.e., ℓ is the graph of y = k. A vertical line m passing through h on the x-axis is precisely a plot of all solutions (x, y) of the equation x = h; i.e., m is the graph of x = h. 3.2 Circles Another common curve in the plane is a circle. Let’s see how to relate a circle and an equation involving the variables x and y. As a special case of the distance formula (2.3), suppose P = (0, 0) is the origin and Q = (x, y) is any point in the plane; then distance from P to Q = = (x − 0)2 + (y − 0)2 x2 + y2 . This calculation tells us that a point (x, y) is of distance r from the origin if and only if r = x2 + y2 or, squaring each side, that x2 + y2 = r2 . This shows {(x, y)|distance (x, y) to origin is r} = {(x, y)|x2 + y2 = r2 }. (3.1) 3.2. CIRCLES 27 Pencil. What is the left-hand side of Equation 3.1? To picr ture all points in the plane of distance r from the origin, fasten a pencil to one end of a non-elastic string (a string r that will not stretch) of length r and tack the other end to Start. Draw with a the origin. Holding the string tight, the pencil point will tight string. locate a point of distance r from the origin. We could visuFigure 3.4: Drawing a circle. alize all such points by simply moving the pencil around the origin, all the while keeping the string tight. What is the right-hand side of Equation 3.1? A point (x, y) in the right-hand set is a solution to the equation x2 + y2 = r2 ; i.e., if we plug in the coordinates we get a true statement. For example, in Figure 3.5 we r r r − plot eight solutions (r, 0), (−r, 0), (0, r), (0, −r), A = √2 , √2 , B = √2 , √r , 2 C= − − √r , √r 2 2 ,D= − r √r , √ 2 2 , of the equation. To see that the last point is a solution, here is the sample calculation: − √r 2 2 + r √ 2 2 = r2 2 Since the two sides of Equation 3.1 are equal, drawing the circle of radius r is the same as plotting all of the solutions of the equation x2 + y2 = r2 . The same reasoning can be used to show that drawing a circle of radius r centered at a point (h, k) is the same as plotting all of the solutions of the equation: (x − h)2 + (y − k)2 = r2 . We usually refer to the set of all solutions of the equation as the graph of the equation. Definition 3.2.1 (Circles). Let (h, k) be a given point in the xy-plane and r > 0 a given positive real number. The circle of radius r centered at (h, k) is precisely all of the solutions (x,y) of the equation (x − h)2 + (y − k)2 = r2 ; i.e., the circle is the graph of this equation. We refer to the equation in the box as the standard form of the equation of a circle. From this equation you know both the center and radius of the circle described. + r2 2 = r2 . (0, r) D A (−r, 0) (r, 0) C B (0, −r) Figure 3.5: Computing points. y-axis (x, y) r (h, k) x-axis Figure 3.6: Defining a circle. CHAPTER 3. THREE SIMPLE CURVES 28 Be very careful with the minus signs “−” in the standard form for a circle equation. For example, the equation !!! CAUTION !!! ( x + 3) 2 + ( y − 1) 2 = 7 is NOT in standard form. We can rewrite it in standard form: √ (x − (−3))2 + (y − 1)2 = ( 7)2 ; √ so, this equation describes a circle of radius 7 centered at (−3, 1). Examples 3.2.2. Here are some of the ways we can discuss circles: 1. The circle of radius 1 centered at the origin is the graph of the equation x2 + y2 = 1. This circle is called the unit circle and will be used extensively. 2. A circle of radius 3 centered at the point (h, k) = (1, −1) is the graph of the equation (x − 1)2 +(y −(−1))2 = 32 ; or, equivalently (x − 1)2 +(y + 1)2 = 32 ; or, equivalently x2 + y2 − 2x + 2y = 7. √ 3. The circle of radius 5 centered at (2, −3) does not pass through the origin; this is because (0, 0) is not a solution of the equation (x − 2)2 + (y + 3)2 = 5. 3.3 Intersecting Curves I In many problem solving situations, we will have two curves in the plane and need to determine where the curves intersect one another. Before we discuss a general procedure, let’s make sure we really understand the meaning of the word “intersect.” From Latin, the word “inter” means “within or in between” and the word “sectus” means “to cut.” So, the intersection of two curves is the place where the curves “cut into” each other; in other words, where the two curves cross one another. If the pictures of two curves are given to us up front, we can often visually decide whether or not they intersect. This is one good reason for drawing a picture of any physical problem we are trying to solve. We will need a small bag of tricks used for finding intersections of curves. We begin with intersections involving the curves studied in this section. Two different horizontal lines (or two different vertical lines) will never intersect. However, a horizontal line always intersects a vertical line exactly once; Figure 3.7(a). Given a circle and a horizontal or vertical line, we may or may not have an intersection. Looking at Figure 3.7(b), you can convince yourself a given horizontal or vertical line will intersect a circle in either two points, one point or no points. This analysis is all pictorial; how do you find the explicit coordinates of an intersection point? Let’s look at a sample problem to isolate the procedure used. 3.3. INTERSECTING CURVES I y-axis 29 Horizontal line: y = k. Point: (h, k). x-axis Vertical line: x = h. (a) Line equations. (b) Possible intersections. Figure 3.7: Circles and lines. Example 3.3.1. Glo-Tek Industries has designed a new halogen street light fixture for the city of Seattle. According to the product literature, when placed on a 50 foot light pole, the resulting useful illuminated area is a circular disc 120 feet in diameter. Assume the light pole is located 20 feet east and 40 feet north of the intersection of Parkside Ave. (a north/south street) and Wilson St. (an east/west street). What portion of each street is illuminated? y-axis Solution. The illuminated area is a circular disc whose diameter and center are both known. Consequently, we really need to study the intersection of this circle with the two streets. Begin by imposing the pictured coordinate system; we will use units of feet for each axis. The illuminated region will be a circular disc centered at the point (20, 40) in the coordinate system; the radius of the disc will be r = 60 feet. We need to find the points of intersection P, Q, R, and S of the circle with the x-axis and the y-axis. The equation for the circle with r = 60 and center (h, k) = (20, 40) is (Parkside) Illuminated zone. P x-axis (Wilson) R S Q Figure 3.8: Illuminated street. (x − 20)2 + (y − 40)2 = 3600. To find the circular disc intersection with the y-axis, we have a system of two equations to work with: (x − 20)2 + (y − 40)2 = 3,600; x = 0. To find the intersection points we simultaneously solve both equations. To do this, we replace x = 0 in the first equation (i.e., we impose the CHAPTER 3. THREE SIMPLE CURVES 30 conditions of the second equation on the first equation) and arrive at (0 − 20)2 + (y − 40)2 = 3,600; 400 + (y − 40)2 = 3,600; (y − 40)2 = 3,200; √ (y − 40) = ± 3,200 √ y = 40 ± 3,200 = −16.57 or 96.57. Notice, we have two solutions. This means that the circle and y-axis intersect at the points P = (0,96.57) and Q = (0, − 16.57). Similarly, to find the circular disc intersection with the x-axis, we have a system of two equations to work with: (x − 20)2 + (y − 40)2 = 3,600; y = 0. Replace y = 0 in the first equation (i.e., we impose the conditions of the second equation on the first equation) and arrive at (x − 20)2 + (0 − 40)2 = 3,600; (x − 20)2 = 2,000; √ (x − 20) = ± 2,000; √ x = 20 ± 2,000 = −24.72 or 64.72. Conclude the circle and x-axis intersect at the points S = (64.72,0) and R = (−24.72,0). The procedure we used in the solution of Example 3.3.1 gives us a general approach to finding the intersection points of circles with horizontal and vertical lines; this will be important in the exercises. 3.4 Summary • Every horizontal line has equation of the form y = c. • Every vertical line has equation of the form x = c. • Every circle has equation of the form (x − h)2 + (y − k)2 = r2 where (h,k) is the center of the circle, and r is the circle’s radius. 3.5. EXERCISES 31 3.5 Exercises Problem 3.1. This exercise emphasizes the “mechanical aspects” of circles and their equations. (a) Find an equation whose graph is a circle of radius 3 centered at (−3,4). (b) Find an equation whose graph is a circle of diameter 1 centered at the point 2 (3, − 11 ). 3 (c) Find four different equations whose graphs are circles of radius 2 through (1,1). (d) Consider the equation (x − 1)2 + (y + 1)2 = 4. Which of the following points lie on the graph of √ this equation:√1,1), (1, − 1), ( (1, − 3), (1 + 3,0), (0, − 1 − 3), (0,0). Problem 3.2. Find the center and radius of each of the following circles. (a) x2 − 6x + y2 + 2y − 2 = 0 (b) x2 + 4x + y2 + 6y + 9 = 0 (c) x2 + 1 x + y2 − 3 10 3y = (d) x2 + y2 = 3 x − y + 2 127 9 35 16 Problem 3.3. Water is flowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the the main broke. (a) When the main broke, a runner was located 6 miles from the intersection. The runner continues toward the intersection at the constant speed of 17 feet per second. When will the runner’s feet get wet? (b) Suppose, instead, that when the main broke, the runner was 6 miles east, and 5000 feet north of the intersection. The runner runs due west at 17 feet per second. When will the runner’s feet get wet? Problem 3.4. An amusement park Ferris Wheel has a radius of 60 feet. The center of the wheel is mounted on a tower 62 feet above the ground (see picture). For these questions, the wheel is not turning. rider 60 feet 100 feet operator ground level 62 ft. tower 24 feet (a) Impose a coordinate system. (b) Suppose a rider is located at the point in the picture, 100 feet above the ground. If the rider drops an ice cream cone straight down, where will it land on the ground? (c) The ride operator is standing 24 feet to one side of the support tower on the level ground at the location in the picture. Determine the location(s) of a rider on the Ferris Wheel so that a dropped ice cream cone lands on the operator. (Note: There are two answers.) Problem 3.5. A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North along the hose at the rate of 1 inch/second. The hose is perpendicular 2 to the 10 ft. wide sidewalk. Assume there is grass on both sides of the sidewalk. N hose W E S sidewalk tractor sprinkler (a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and find equations of the lines forming the North and South boundaries of the sidewalk. CHAPTER 3. THREE SIMPLE CURVES 32 (b) When will the water first strike the sidewalk? (c) When will the water from the sprinkler fall completely North of the sidewalk? (d) Find the total amount of time water from the sprinkler falls on the sidewalk. (e) Sketch a picture of the situation after 33 minutes. Draw an accurate picture of the watered portion of the sidewalk. (f) Find the area of GRASS watered after one hour. Problem 3.6. Erik’s disabled sailboat is floating stationary 3 miles East and 2 miles North of Kingston. A ferry leaves Kingston heading toward Edmonds at 12 mph. Edmonds is 6 miles due east of Kingston. After 20 minutes the ferry turns heading due South. Ballard is 8 miles South and 1 mile West of Edmonds. Impose coordinates with Ballard as the origin. (e) How long does the ferry spend inside the radar zone? Problem 3.7. Nora spends part of her summer driving a combine during the wheat harvest. Assume she starts at the indicated position heading east at 10 ft/sec toward a circular wheat field of radius 200 ft. The combine cuts a swath 20 feet wide and begins when the corner of the machine labeled “a” is 60 feet north and 60 feet west of the western-most edge of the field. N W E S swath cut combine 20 ft a center wheat field sailboat Kingston Edmonds (a) When does Nora’s rig first start cutting the wheat? (b) When does Nora’s rig first start cutting a swath 20 feet wide? North Ballard UDub (a) Find the equations for the lines along which the ferry is moving and draw in these lines. (b) The sailboat has a radar scope that will detect any object within 3 miles of the sailboat. Looking down from above, as in the picture, the radar region looks like a circular disk. The boundary is the ”edge” or circle around this disc, the interior is the inside of the disk, and the exterior is everything outside of the disk (i.e. outside of the circle). Give a mathematical (equation) description of the boundary, interior and exterior of the radar zone. Sketch an accurate picture of the radar zone by determining where the line connecting Kingston and Edmonds would cross the radar zone. (c) When does the ferry enter the radar zone? (d) Where and when does the ferry exit the radar zone? (c) Find the total amount of time wheat is being cut during this pass across the field. (d) Estimate the area of the swath cut during this pass across the field. Problem 3.8. (a) Solve for x: 2 x − 2x + 1 =x−2 x+5 (b) Solve for x: x−3 =1 x+2 (c) If x = −2, find ALL solutions of the equation (x + 1)2 + (y − 1)2 = 10 (d) If y = 3, find ALL solutions of the equation 5(x + 1)2 + 2(y − 1)2 = 10 Chapter 4 Linear Modeling Sometimes, we will begin a section by looking at a specific problem which highlights the topic to be studied; this section offers the first such vista. View these problems as illustrations of precalculus in action, rather than confusing examples. Don’t panic, the essential algebraic skills will be reviewed once the motivation is in place. 4.1 The Earning Power Problem The government likes to gather all kinds of data. For example, Table 4.1 contains some data on the average annual income for full-time workers; these data were taken from the 1990 Statistical Abstract of the U.S. Given this information, a natural question would be: How can we predict the future earning power of women and men? One way to answer this (a) Women. YEAR 1970 1987 (b) Men. WOMEN (dollars) $5,616 $18,531 YEAR 1970 1987 Table 4.1: Earning power data. MEN (dollars) $9,521 $28,313 question would be to use the data in the table to construct two different mathematical models that predict the future (or past) earning power for women or men. In order to do that, we would need to make some kind of initial assumption about the type of mathematical model expected. Let’s begin by drawing two identical xy-coordinate systems, where the x-axis has units of “year” and the y-axis has units of “dollars;” see Figure 4.1. In each coordinate system, the data in our table gives us two points to plot: In the case of women, the data table gives us the points P = (1970, 5,616) and Q = (1987, 18,531). Likewise, for the men, the data table gives us the points R = (1970, 9,521) and S = (1987, 28,313). To study the future earning power of men and women, we are going to make an assumption: For women, if the earning power in year x is $y, 33 CHAPTER 4. LINEAR MODELING 34 PSfrag y-axis (dollars) 30000 y-axis (dollars) 30000 S = (1987,28313) 25000 25000 20000 20000 Q = (1987,18531) 15000 15000 10000 10000 R = (1970,5616) P = (1970,5616) 5000 5000 x-axis (year) x-axis (year) 1995 1990 1985 1980 1975 1970 1995 1990 1985 1980 1975 1970 aca (b) Data points for men. (a) Data points for women. Figure 4.1: Visualizing the data. then the point (x, y) lies on the line connecting P and Q. Likewise, for men, if the earning power in year x is $y, then the point (x, y) lies on the line connecting R and S. In the real world, the validity of this kind of assumption would involve a lot of statistical analysis. This kind of assumption leads us to what is called a linear model, since we are demanding that the data points predicted by the model (i.e., the points (x, y)) lie on a straight line in a coordinate system. Now that we have made this assumption, our job is to find a way to mathematically describe when a point (x, y) lies on one of the two lines pictured in Figure 4.2. Our goal in the next subsection is to review the mathematics necessary to show that the lines in Figure 4.2 are the so-called graphs of Equations 4.1 and 4.2. y-axis (dollars) 30000 y-axis (dollars) 30000 S = (1987,28313) 25000 25000 20000 Q = (1987,18531) y 15000 (x,y) on the line: This means men earn y dollars in year x. 15000 (x,y) on the line: This means women earn y dollars in year x. P = (1970,5616) 5000 10000 R = (1970,9521) 5000 x-axis (year) x-axis (year) 1975 1970 1995 1990 1985 1980 1975 1970 1980 x x 1995 10000 1990 y 1985 20000 (b) Linear model for men. (a) Linear model for women. Figure 4.2: Linear models of earning power. 4.2. RELATING LINES AND EQUATIONS 35 18,531 − 5,616 (x − 1970) + 5,616 1987 − 1970 12,915 = (x − 1970) + 5,616 17 28,313 − 9,521 (x − 1970) + 9,521 = 1987 − 1970 18,792 = (x − 1970) + 9,521 17 (4.1) ywomen = ymen (4.2) 4.2 Relating Lines and Equations A systematic approach to studying equations and their graphs would begin with the simple cases, gradually working toward the more complicated. Thinking visually, the simplest curves in the plane would be straight lines. As we discussed in Chapter 3, a point on the vertical line in Figure 4.3(a) will always have the same x-coordinate; we refer to this line as the graph of the equation x = h. Likewise, a point on the horizontal line in Figure 4.3(b) will always have the same y-coordinate; we refer to this line as the graph of the equation y = k. Figure 4.3(c) is different, in the sense that neither the x nor the y coordinate is constant; i.e., as you move a point along the line, both coordinates of the point are changing. It is reasonable to guess that this line is the graph of some equation involving both x and y. The question is: What is the equation? y-axis y-axis graph of x = h (h,y) is a typical point on this line. x-axis location h on x-axis (a) Vertical line. y-axis (x,k) is a typical point on this line graph of y = k location k on y-axis x-axis (b) Horizontal line. Figure 4.3: Lines in a plane. (x,y) is a typical point on this line graph of some equation involving x and y x-axis (c) Sloped line. CHAPTER 4. LINEAR MODELING 36 Here is the key geometric fact needed to model lines by mathematical equations: Important Fact 4.2.1. Two different points completely determine a straight line. This fact tells us that if you are given two different points on a line, you can reconstruct the line in a coordinate system by simply lining a ruler up with the two points. In our discussion, we will need to pay special attention to the difference between vertical and non-vertical lines. 4.3 Non-vertical Lines Assume in this section that ℓ is a non-vertical line in the plane; for example, the line in Figure 4.3(c). If we are given two points P = (x1 , y1 ) and Q = (x2 , y2 ) on a line ℓ, then Equation (2.2) on page 18 defined two quantities we can calculate: ∆x = change in x going from P to Q = x2 − x1 . ∆y = change in y going from P to Q = y2 − y1 . We define the slope of the line ℓ to be the ratio of ∆y by ∆x, which is usually denoted by m: ∆y ∆x y2 − y1 = x2 − x1 change in y slope of ℓ = change in x def m= (4.3) Notice, we are using the fact that the line is non-vertical to know that this ratio is always defined; i.e., we will never have ∆x = 0 (which would lead to illegal division by zero). There is some additional terminology that goes along with the definition of the slope. The term ∆y is sometimes called the rise of ℓ and ∆x is called the run of ℓ. For this reason, people often refer to the slope of a line ℓ as “the rise over the run,” meaning def slope of ℓ = ∆y rise of ℓ = = m. run of ℓ ∆x In addition, notice that the calculation of ∆y involves taking the difference of two numbers; likewise, the calculation of ∆x involves taking the difference of two numbers. For this reason, the slope of a line ℓ is sometimes called a difference quotient. 4.3. NON-VERTICAL LINES 37 For example, suppose P = (1, 1) and Q = (4, 5) lie on a line ℓ. In this case, the rise = ∆y = 4 and the run = ∆x = 3, 4 so m = 3 is the slope of ℓ. When computing ∆x, pay special attention that it is the x-coordinate of the destination point Q minus the xcoordinate of the starting point P; likewise, when computing ∆y, it is the y-coordinate of the destination point Q minus the y-coordinate of the starting point P. We can reverse this order in both calculations and get the same slope: m= Q = (4,5) 5 P = (1,1) 1 ∆y = 4 ∆x = 3 1 4 Figure 4.4: Computing the slope of a line. y2 − y1 −(y2 − y1 ) y1 − y2 −∆y ∆y = = = = . ∆x x2 − x1 −(x2 − x1 ) x1 − x2 −∆x We CANNOT reverse the order in just one of the calculations and get the same slope: y2 − y1 x2 − x1 = y2 − y1 x1 − x2 , and m= y2 − y1 x2 − x1 = y1 − y2 x2 − x1 It is very important to notice that the calculation of the slope of a line does not depend on the choice of the two points P and Q. This is a real windfall, since we are then always at liberty to pick our favorite two points on the line to determine the slope. The reason for this freedom of choice is pretty easy to see by looking at a picture. If we were to choose two other points P∗ = (x∗ , y∗ ) and 1 1 Q∗ = (x∗ , y∗ ) on ℓ, then we would get two similar right 2 2 triangles: See Figure 4.5. Basic properties of similar triangles tell us ratios of lengths of common sides are equal, so that m= !!! m= CAUTION !!! . x∗ ,y∗ = Q∗ 22 (x2 ,y2 ) = Q y2 − y1 x2 − x1 x∗ − x∗ 2 1 P ∗ = x∗ ,y∗ 11 Figure 4.5: Using similar triangles. y∗ − y∗ y2 − y1 2 1 =∗ ; x2 − x1 x2 − x∗ 1 but this just says the calculation of the slope is the same for any pair of distinct points on ℓ. For example, lets redo the slope calculation when P∗ = P = (x1 , y1 ) and Q∗ = (x, y) represents an arbitrary point on the line. Then the two ratios of lengths of common sides give us the equation y − y1 , x − x1 = m(x − x1 ). m= y − y1 y∗ − y∗ 2 1 (x1 ,y1 ) = P CHAPTER 4. LINEAR MODELING 38 This can be rewritten as y = m(x − x1 ) + y1 or y2 − y1 (x − x1 ) + y1 . y= x2 − x1 (4.4) (4.5) Equation 4.4 is usually called the point slope formula for the line ℓ (since the data required to write the equation amounts to a point (x1 ,y1 ) on the line and the slope m), whereas Equation 4.5 is called the two point formula for the line ℓ (since the data required amounts to the coordinates of the points P and Q). In any event, we now see that Important Fact 4.3.1. (x,y) lies on ℓ if and only if (x,y) is a solution to y = m(x − x1 ) + y1 . We can plot the collection of ALL solutions to the equation in Fact 4.3.1, which we refer to as the graph of the equation. As a subset of the xycoordinate system, the line Important Fact 4.3.2. ℓ = {(x, m(x − x1 ) + y1 ) |x is any real number}. 1 graph of y = 4 x − 3 3 (6, 23 ) 3 (4,5) (−1, − 5 ) 3 (1,1) 1 (0, − 3 ) Figure 4.6: Verifying points on a line. Example 4.3.3. Consider the line ℓ, in Figure 4.6, through the two points P = (1,1) and Q = (4,5). Then the slope of ℓ is m = 4/3 and ℓ consists of all pairs of points (x,y) such that the coordinates x and y satisfy the equation y = 4 (x − 1) + 1. Letting x = 0, 1, 6 and −1, we conclude that 3 the following four points lie on the line ℓ: (0, 4 (0 − 1) + 1) = 3 4 4 (0, −1 ), (1, 3 (1 − 1) + 1) = (1,1), (6, 3 (6 − 1) + 1) = (6, 23 ) and 3 3 4 (−1, 3 (−1 − 1) + 1) = (−1, −5 ). By the same reasoning, the 3 point (0,0) does not lie on the line ℓ. As a set of points in the plane, we have ℓ= 4 x, (x − 1) + 1 |x is any real number 3 Returning to the general situation, we can obtain a third general equation for a non-vertical line. To emphasize what is going on here, plug the specific value x = 0 into Equation 4.4 and obtain the point R = (0,b) on the line, where b = m(0 − x1 ) + y1 = −mx1 + y1 . But, Equation (4.4) can be written y = m(x − x1 ) + y1 = mx − mx1 + y1 = mx + b. 4.4. GENERAL LINES 39 The point R is important; it is precisely the point where the line ℓ crosses the y-axis, usually called the y-intercept. The slope intercept equation of the line is the form y = mx + b, where the slope of the line is m and b is the y-intercept of the line. Summary 4.3.4. Non-vertical Lines: Let ℓ be a non-vertical line in the xy-plane. There are three ways to obtain an equation whose graph is ℓ, depending on the data provided for ℓ: 1. If P = (x1 ,y1 ), Q = (x2 ,y2 ) are two different points on the line, then y2 − y1 the two-point formula y = (x − x1 ) + y1 gives an equation x2 − x1 whose graph is ℓ. 2. If P = (x1 ,y1 ) is a point on the line and m is the slope of ℓ, then the point-slope formula y = m(x − x1 ) + y1 gives an equation whose graph is ℓ. 3. If the line ℓ intersects the y-axis at the point (0,b) and m is the slope of the line ℓ, then the slope-intercept formula y = mx + b gives an equation whose graph is ℓ. 4.4 General Lines Summarizing, any line in the plane is the graph of an equation involving x and y and the equation always has the form Ax + By + C = 0, for some constants A, B, C. Equations like this are called linear equations. In general, non-vertical lines will be of the most interest to us, since these are the lines that can be viewed as the graphs of functions ; we will discuss this in Chapter 5. 4.5 Lines and Rate of Change If we draw a non-vertical line in the xy coordinate system, then its slope will be the rate of change of y with respect to x: ∆y ∆x change in y = change in x slope = def = rate of change of y with respect to x. 40 CHAPTER 4. LINEAR MODELING We should emphasize that this rate of change is a constant; in other words, this rate is the same no matter where we compute the slope on the line. The point-slope formula for a line can now be interpreted as follows: A line is determined by a point on the line and the rate of change of y with respect to x. An interesting thing to notice is how the units for x and y figure into the rate of change calculation. For example, suppose that we have the equation y = 10,000 x + 200,000, which relates the value y of a house (in dollars) to the number of years x you own it. For example, after 5 years, x = 5 and the value of the house would be y = 10,000 (5) + 200,000 = $250,000. In this case, the equation y = 10,000 x + 200,000 is linear and already written in slope-intercept form, so the slope can be read as m = 10,000. If ∆y we carry along the units in the calculation of , then the numerator ∆x involves “dollar” units and the denominator “years” units. That means that carrying along units, the slope is actually m = 10,000 dollars/year. In other words, the value of the house is changing at a rate of 10,000 dollars/year. At the other extreme, if the units for both x and y are the same, then the units cancel out in the rate of change calculation; in other words, the slope is a unit-less quantity, simply a number. This sort of thing will come up in the mathematics you see in chemistry and physics. One important type of rate encountered is the speed s of a moving object. Suppose an object moves along a b speed m straight line at a constant speed m: See Figure 4.7. reference If we specify a reference point, we can let b be the startpoint location of the obinitial location ject at time t ing location of the moving object, which is usually called of the object the initial location of the object. We can write down an Figure 4.7: Motion along a equation relating the initial location b, the time t, the conline. stant speed m and the location s at time t: s = (location of object at time t) s = (initial location of object) + (distance object travels in time t) s = b + mt. where t is in the same time units used to define the rate m. Notice, both b and m would be constants given to us, so this is a linear equation involving the variables s and t. We can graph the equation in the tscoordinate system: See Figure 4.8. It is important to distinguish between this picture (the graph of s = mt + b) and the path of the object in Figure 4.7. The graph of the equation should be thought of as a visual aid attached to the equation s = mt + b. The general idea is that using this visual aid can help answer various questions involving the equation, which in turn will tell us things about the motion of the object in Figure 4.7. 4.5. LINES AND RATE OF CHANGE 41 Two other comments related to this discussion are important. First, concerning notation, the speed m is often symbolized by v to denote constant velocity and b is written as s◦ (the subscript “0” meaning “time zero”). With these changes, the equation becomes s = s◦ + vt, which is the form in which it would be written in a typical physics text. As a second note, if you return to Figure 4.8, you will notice we only drew in the positive t axis. This was because t represented time, which is always a non-negative quantity. s-axis graph of s = mt + b ∆s ∆t rate of change = b = s-intercept Mookie ∆t t-axis Figure 4.8: The graph of s = mt + b. Example 4.5.1. Linda, Asia and Mookie are all playing frisbee. Mookie is 10 meters in front of Linda and always runs 5 m/sec. Asia is 34 meters in front of Linda and always runs 4 m/sec. Linda yells “go!” and both Mookie and Asia start running directly away from Linda to catch a tossed frisbee. Find linear equations for the distances between Linda, Mookie and Asia after t seconds. Linda ∆s Asia Solution. Let sM be the distance between Linda and Mookie and sA the distance between Linda and Asia, after t seconds. An application of the above formula tells us sM = (initial distance between Linda and Mookie) + · · · · · · + (distance Mookie runs in t seconds) sM = 10 + 5t. Likewise, sA = (initial distance between Linda and Asia) + · · · · · · + (distance Asia runs in t seconds) sA = 34 + 4t. If sMA is the distance between Mookie and Asia after t seconds, we compute sMA = sA − sM = (34 + 4t) − (10 + 5t) = (24 − t) meters. In all cases, the distances we computed are given by linear equations of the form s = b + mt, for appropriate b and rate m. CHAPTER 4. LINEAR MODELING 42 4.6 Back to the Earning Power Problem We now return to the motivating problem at the start of this section. Recall the plot in Figure 4.1(b). We can model the men’s earning power using the first and last data points, using the ideas we have discussed about linear equations. To do this, we should specify a “beginning point” and an “ending point” (recall Figure 2.9) and calculate the slope: y-axis (dollars) 40000 U 35000 T 30000 S 25000 20000 (x,y) on the line: means men earn y dollars in year x y 15000 10000 Rbegin = (1970,9,521) R 5000 and x-axis (year) Send = (1987,28,313). 1995 1990 1985 1980 1975 1970 x Figure 4.9: Linear model of Men’s Earning Power. We find that ∆y ∆x m 28,313 − 9,521 1987 − 1970 ∆y ∆x 18,792 17 18,792 17 If we apply the “point-slope formula” for the equation of a line, we arrive at the equation: y= 18,792 (x − 1970) + 9,521. 17 (4.6) The graph of this line will pass through the two points R and S in Figure 4.2. We sketch the graph in Figure 4.9, indicating two new points T and U. We can use the model in (4.6) to make predictions of two different sorts: (i) predict earnings at some date, or (ii) predict when a desired value for earnings will occur. For example, let’s graphically discuss the earnings in 1995: • Draw a vertical line x = 1995 up to the graph and label the intersection point U. • Draw a horizontal line ℓ through U. The line ℓ crosses the y axis at the point 18,792 (1995 − 1970) + 9521 = 37,156. 17 4.7. WHAT’S NEEDED TO BUILD A LINEAR MODEL? 43 • The coordinates of the point U = (1995, 37,156). Conclude that $37,156 is the Men’s Earning Power in 1995. For another example, suppose we wanted to know when men’s earning power will equal $33,000? This means we seek a data point T on the men’s earning curve whose y-coordinate is 33,000. By (4.6), T has the form T= x, 18,792 (x − 1970) + 9,521 . 17 We want this to be a data point of the form (x, 33,000). Setting these two points equal and equating the second coordinates leads to an algebra problem: 18,792 (x − 1970) + 9,521 = 33,000 17 x = 1991.24. This means men’s earning power will be $33,000 at the end of the first quarter of 1991. Graphically, we interpret this reasoning as follows: • Draw a horizontal line y = 33,000 and label the intersection point T on the model. • Draw a vertical line ℓ through T . The line ℓ crosses the x axis at the point 1991.24. • The coordinates of the point T = (1991.24,33,000). In the exercises, you will be asked to show that the women’s earning power model is given by the equation y= 12,915 (x − 1970) + 5,616. 17 Using the two linear models for the earning power of men and women, are women gaining on men? You will also be asked to think about this question in the exercises. 4.7 What’s Needed to Build a Linear Model? As we progress through this text, a number of different “types” of mathematical models will be discussed. We will want to think about the information needed to construct that particular kind of mathematical model. Why would we care? For example, in a laboratory context, if we knew a situation being studied was given by a linear model, this would effect the amount of data collected. In the case of linear models, we can now make this useful statement: CHAPTER 4. LINEAR MODELING 44 Important Fact 4.7.1. A linear model is completely determined by: 1. One data point and a slope (a rate of change), or 2. Two data points, or 3. An intercept and a slope (a rate of change). 4.8 Linear Application Problems Example 4.8.1. The yearly resident tuition at the University of Washington was $1827 in 1989 and $2907 in 1995. Assume that the tuition growth at the UW follows a linear model. What will be the tuition in the year 2000? When will yearly tuition at the University of Washington be $10,000? y-axis 10000 (dollars) 8000 6000 4000 x-axis 2040 2020 (year) 2010 2000 Q 2030 P 1990 2000 Figure 4.10: Linear tuition model. Solution. If we consider a coordinate system where the x-axis represents the year and the y-axis represents dollars, we are given two data points: P = (1989, 1,827) and Q = (1995, 2,907). Using the two-point formula for the equation of line through P and Q, we obtain the equation y = 180(x − 1989) + 1,827. The graph of this equation gives a line through the given points as pictured in Figure 4.10. If we let x = 2000, we get y = $3,807, which tells us the tuition in the year 2000. On the other hand, if we set the equation equal to $10,000, we can solve for x: 10,000 = 180(x − 1989) + 1,827 8,173 = 180(x − 1989) 2,034.4 = x. Conclude the tuition is $10,000 in the year 2035. 4.9 Perpendicular and Parallel Lines Here is a useful fact to keep in mind. Important Facts 4.9.1. Two non-vertical lines in the plane are parallel precisely when they both have the same slope. Two non-vertical lines are perpendicular precisely when their slopes are negative reciprocals of one another. 4.10. INTERSECTING CURVES II 45 Example 4.9.2. Let ℓ be a line in the plane passing through the points (1, 1) and (6, −1). Find a linear equation whose graph is a line parallel to ℓ passing through 5 on the y-axis. Find a linear equation whose graph is perpendicular to ℓ and passes through (4, 6). Solution. Letting P = (1, 1) and Q = (6, −1), apply the “two point formula”: −2 (x − 1) + 1 5 7 2 = − x+ . 5 5 The graph of this equation will be ℓ. This equation is in slope intercept form and we can read off that the slope is m = −2 . The desired line a is 5 2 parallel to ℓ; it must have slope m = − 5 and y-intercept 5. Plugging into the “slope intercept form”: y= y= −2 x + 5. 5 The desired line b is a line perpendicular to ℓ (so its slope is m ′ = 5 ) 2 −1 −2 5 = and passes through the point (4, 6), so we can use the “point slope formula”: 5 y = (x − 4) + 6. 2 4.10 Intersecting Curves II We have already encountered problems that require us to investigate the intersection of two curves in the plane. Ultimately, this reduces to solving a system of two (or more) equations in the variables x and y. A useful tool when working with equations involving squared terms (i.e., x2 or y2 ), is the quadratic formula . Important Fact 4.10.1. Quadratic Formula: Consider the equation az2 + bz + c = 0, where a,b,c are constants. The solutions for this equation are given by the formula √ −b ± b2 − 4ac z= . 2a The solutions are real numbers if and only if b2 − 4ac ≥ 0. The next example illustrates a typical application of the quadratic formula. In addition, we describe a very useful technique for finding the shortest distance between a “line” and a “point.” CHAPTER 4. LINEAR MODELING 46 North flight path irrigated field West East 2 miles Q P 1.5 miles crop duster South Figure 4.11: The flight path of a crop duster. Example 4.10.2. A crop dusting airplane flying a constant speed of 120 mph is spotted 2 miles South and 1.5 miles East of the center of a circular irrigated field. The irrigated field has a radius of 1 mile. Impose a coordinate system as pictured, with the center of the field the origin (0,0). The flight path of the duster is a straight line passing over the labeled points P and Q. Assume that the point Q where the plane exits the airspace above the field is the Western-most location of the field. Answer these questions: 1. Find a linear equation whose graph is the line along which the crop duster travels. 2. Find the location P where the crop duster enters airspace above the irrigated field. 3. How much time does the duster spend flying over the irrigated field? 4. Find the shortest distance from the flight path to the center of the irrigated field. Solution. 1. Take Q = (−1, 0) and S = (1.5, −2) = duster spotting point. Construct a line through Q and S. The slope is −0.8 = m and the line equation becomes: y = −0.8x − 0.8. (4.7) 2. The equation of the boundary of the irrigated region is x2 + y2 = 1. We need to solve this equation AND the line equation y = −0.8x − 0.8 simultaneously. Plugging the line equation into the unit circle equation gives: x2 + (−0.8x − 0.8)2 = 1 x2 + 0.64x2 + 1.28x + 0.64 = 1 1.64x2 + 1.28x − 0.36 = 0 Apply the quadratic formula and find x = −1, 0.2195. Conclude that the x coordinate of P is 0.2195. To find the y coordinate, plug into the line equation and get y = −0.9756. Conclude that P = (0.2195, −0.9756) 4.11. UNIFORM LINEAR MOTION 47 3. Find the distance from P to Q by using the distance formula: (−1 − 0.2195)2 + (0 − (−0.9756))2 d= = 1.562 miles Now, 1.562 miles 120 mph = 0.01302 hours = 47 seconds. 4. The idea is to construct a line perpendicular to the flight path passing through the origin of the coordinate system. This line will have 1 slope m = − −0.8 = 1.25. So this perpendicular line has equation y = 1.25x. Intersecting this line with the flight path gives us the point closest to the center of the field. The x-coordinate of this point is found by setting the two line equations equal and solving: −0.8x − 0.8 = 1.25x x = −0.3902 This means that the closest point on the flight path is (−0.39, −0.49). Apply the distance formula and the shortest distance to the flight path is (−0.39)2 + (−0.49)2 d= = 0.6263. 4.11 Uniform Linear Motion When an object moves along a line in the xy-plane at a constant speed, we say that the object exhibits uniform linear motion. Its location can be described using a pair of linear equations involving a variable which represents time. That is, we can find constants a, b, c, and d such that, at any time t, the object’s location is given by (x,y), where x = a + bt and y = c + dt. Such equations are called parametric equations of motion. The motion is defined in terms of the parameter t. Since there are four constants to be determined, one needs four pieces of information to determine these equations. Knowing the object’s location at two points in time is sufficient. Example 4.11.1. Bob is running in the xy-plane. He runs in a straight line from the point (2,3) to the point (5, − 4), taking 6 seconds to do so. Find his equations of motion. CHAPTER 4. LINEAR MODELING 48 Solution. We begin by setting a reference for our time parameter. Let’s let t = 0 represent the instant when Bob is at the point (2,3). In this way, t will represent the time since Bob left the point (2,3). When t = 6, we know he will be at the point (5, − 4). This is enough information to determine his equations of motion. We seek constants a, b, c, and d so that at time t, Bob’s location is given by x = a + bt and y = c + dt. When t = 0, we know Bob’s location is (2,3). That is, x = 2 and y = 3. Thus, with t = 0, we have the two equations x = 2 = a + b(0) = a and y = 3 = c + d(0) = c. and so a = 2 and c = 3. We’re half-way done. When t = 6, we know Bob’s location is (5, − 4). Thus, with t = 6, we have the two equations x = 5 = a + b(6) = 2 + 6b and y = −4 = c + d(6) = 3 + 6d which we can solve to find b= 1 7 and d = − . 2 6 So, we arrive at the equations of Bob’s motion 7 1 x = 2 + t and y = 3 − t. 2 6 Notice it is easy to check that these are correct. If we plug in t = 0, we find x = 2, y = 3 as required. If we plug in t = 6, we find x = 5, y = −4, as required. So we know we’ve done it right. Now that we have these equations of motion, it is very easy to calculate Bob’s location at any time. For instance, 30 seconds after leaving the point (2,3), we can find that he is at the point (17, − 32) since 1 7 x = 2 + (30) = 17, y = 3 − (30) = −32. 2 6 Example 4.11.2. Olga is running in the xy-plane, and the coordinate are given in meters (so, for example, the point (1,0) is one meter from the origin (0,0)). She runs in a straight line, starting at the point (3,5) and running along the line y = − 1 x + 6 at a speed of 7 meters per second, heading away 3 from the y-axis. What are her parametric equations of motion? 4.11. UNIFORM LINEAR MOTION 49 Solution. This example differs in some respects from the last example. In particular, instead of knowing where the runner is at two points in time, we only know one point, and have other information given to us about the speed and path of the runner. One approach is to use this new information to find where the runner is at some other point in time: this will then give us exactly the same sort of information as we used in the last example, and so we may solve it in an identical manner. We know that Olga starts at the point (3,5). Letting t = 0 represent the time when she starts, we then know that when t = 0, x = 3 and y = 5. To get another point (and time), we can use the fact that we know what line she travels along, and which direction she runs. We may consider any point on the line in the correct direction: any will do. For instance, the point (6,4) is on the line. We then need to find when Olga reaches this point. To do this, we find the distance from her starting point to the point (6,4), and divide this by her speed. The time she takes to get to (6,4) is thus (6 − 3)2 + (4 − 5)2 = 0.45175395 seconds . 7 At this point, we are now in the same situation as in the last example. We know two facts: when t = 0, x = 3 and y = 5, and when t = 0.45175395, x = 6 and y = 4. As we saw in the last example, this is enough information to find the parametric equations of motion. We seek a, b, c, and d such that Olga’s location t seconds after she starts is (x,y) where x = a + bt and y = c + dt. When t = 0, x = 3, and y = 5, so x = 3 = a + b(0) = a and y = 5 = c + d(0) = c and so a = 3 and c = 5. Also, when t = 0.45175395, x = 6 and y = 4, so x = 6 = a+b(0.45175395) = 3+b(0.45175395) and y = 4 = c+d(0.45175395) = 5+d(0.45175395). Solving these equations for b and d, we find b= 3 −1 = 6.64078311 and d = = −2.21359436. 0.45175395 0.45175395 Thus, Olga’s equations of motion are x = 3 + 6.64078311t, y = 5 − 2.21359436t. CHAPTER 4. LINEAR MODELING 50 4.12 Summary • The equation of every non-vertical line can be expressed in the form y = m(x − h) + k (the point-slope form) and y = mx + b (the slope-intercept form) • A vertical line has an equation of the form x = c. • The shortest distance between a point, P, and a line, l, can be found by determining a line l2 which passes through P and is perpendicular to l. Then the point at which l and l2 intersect is the point on l which is closest to l. The distance from this point to P is the shortest distance between P and l. • The location of an object moving at constant speed along a line can be described using a pair of equations (parametric equations) x = a + bt, y = c + dt. 4.13. EXERCISES 51 4.13 Exercises Problem 4.1. This exercise emphasizes the “mechanical aspects” of working with linear equations. Find the equation of a line: (a) Passing through the points (1, − 1) and (−2,4). (b) Passing through the point (−1, − 2) with slope m = 40. (c) With y-intercept b = −2 and slope m = −2. (d) Passing through the point (4,11) and having slope m = 0. (e) Perpendicular to the line in (a) and passing through (1,1). (f) Parallel to the line in (b) and having yintercept b = −14. (g) Having the equation 3x + 4y = 7. (h) Crossing the x-axis at x = 1 and having slope m = 1. Problem 4.2. Sketch an accurate picture of 1 the line having equation y = 2 − 2 x. Let α be an unknown constant. (a) Find the point of intersection between the line you have graphed and the line y = 1 + αx; your answer will be a point in the xy plane whose coordinates involve the unknown α. (b) Find α so that the intersection point in (a) has x-coordinate 10. (c) Find α so that the intersection point in (a) lies on the x-axis. Problem 4.3. (a) What is the area of the tri1 angle determined by the lines y = − 2 x + 5, y = 6x and the y-axis? (b) If b > 0 and m < 0, then the line y = mx + b cuts off a triangle from the first quadrant. Express the area of that triangle in terms of m and b. (c) The lines y = mx + 5, y = x and the yaxis form a triangle in the first quadrant. Suppose this triangle has an area of 10 square units. Find m. Problem 4.4. Complete Table 4.2 on page 52. In many cases there may be several possible correct answers. Problem 4.5. The (average) sale price for single family property in Seattle and Port Townsend is tabulated below: YEAR 1970 1990 SEATTLE $38,000 $175,000 PORT TOWNSEND $8400 $168,400 (a) Find a linear model relating the year x and the sales price y for a single family property in Seattle. (b) Find a linear model relating the year x and the sales price y for a single family property in Port Townsend. (c) Sketch the graph of both modeling equations in a common coordinate system; restrict your attention to x ≥ 1970. (d) What is the sales price in Seattle and Port Townsend in 1983 and 1998? (e) When will the average sales price in Seattle and Port Townsend be equal and what is this price? (f) When will the average sales price in Port Townsend be $15,000 less than the Seattle sales price? What are the two sales prices at this time? (g) When will the Port Townsend sales price be $15,000 more than the Seattle sales price? What are the two sales prices at this time? (h) When will the Seattle sales price be double the Port Townsend sales price? (i) Is the Port Townsend sales price ever double the Seattle sales price? Problem 4.6. The cup on the 9th hole of a golf course is located dead center in the middle of a circular green that is 70 feet in diameter. Your ball is located as in the picture below: CHAPTER 4. LINEAR MODELING 52 Equation Slope y-intercept Point on the line Point on the line y = 2x + 1 (3, −4) −2 (−1, 7) 1 1 2 (0, 1) 1,000 0 (3, 3) (3, −2) (5, −9) Table 4.2: Linear equations table for Problem 4.4. 9 cup green 50 feet ball ball path rough Problem 4.7. Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person’s ankle. The cord is 30 feet long, but can stretch up to 90 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 8 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. 40 feet The ball follows a straight line path and exits the green at the right-most edge. Assume the ball travels a constant rate of 10 ft/sec. (a) Where does the ball enter the green? (a) Sketch an accurate picture of the situation at time t = 7 seconds. Make sure to label the locations of Allyson and Adrian; also, compute the length of the bungee cord at t = 7 seconds. (b) When does the ball enter the green? (c) How long does the ball spend inside the green? (d) Where is the ball located when it is closest to the cup and when does this occur. (b) Where is Allyson when the bungee reaches its maximum length? 4.13. EXERCISES 53 Find the portion of the y-axis that Dave cannot see. (Hint: Let a be the x-coordinate of the point where line of sight #1 is tangent to the silo; compute the slope of the line using two points (the tangent point and (12,0)). On the other hand, compute the slope of line of sight #1 by noting it is perpendicular to a radial line through the tangency point. Set these two calculations of the slope equal and solve for a.) 20 ft Building Allyson 30 ft start Adrian Problem 4.8. Dave is going to leave academia and go into business building grain silos. A grain silo is a cylinder with a hemispherical top, used to store grain for farm animals. Here is a 3D view, a cross-section, and the top view: silo h 3D-view r Problem 4.9. While speaking on the phone to a friend in Oslo, Norway, you learned that the current temperature there was −23◦ Celsius (−23◦ C). After the phone conversation, you wanted to convert this temperature to Fahrenheit degrees ◦ F, but you could not find a reference with the correct formulas. You then remembered that the relationship between ◦ F and ◦ C is linear. (a) Using this and the knowledge that 32◦ F = 0◦ C and 212◦ F = 100◦ C, find an equation that computes Celsius temperature in terms of Fahrenheit temperature; i.e., an equation of the form C= “an expression involving only the variable F .” (b) Likewise, find an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form F= “an expression involving only the variable C .” (c) How cold was it in Oslo in ◦ F? cross-section y-axis blind spot line of sight #1 (12,0) x-axis Dave line of sight #2 TOP VIEW If Dave is standing next to a silo of crosssectional radius r = 8 feet at the indicated position, his vision will be partially obstructed. Problem 4.10. Pam is taking a train from the town of Rome to the town of Florence. Rome is located 30 miles due West of the town of Paris. Florence is 25 miles East, and 45 miles North of Rome. On her trip, how close does Pam get to Paris? Problem 4.11. Angela, Mary and Tiff are all standing near the intersection of University and 42nd streets. Mary and Tiff do not move, but Angela runs toward Tiff at 12 ft/sec along a straight line, as pictured. Assume the roads are 50 feet wide and Tiff is 60 feet north of the nearest corner. Where is Angela located when she is closest to Mary and when does she reach this spot? CHAPTER 4. LINEAR MODELING 54 Problem 4.13. Margot is walking in a straight line from a point 30 feet due east of a statue in a park toward a point 24 feet due north of the statue. She walks at a constant speed of 4 feet per second. tiff mary 42 nd St. (a) Write parametric equations for Margot’s position t seconds after she starts walking. angela (b) Write an expression for the distance from Margot’s position to the statue at time t. University Way Problem 4.12. The infamous crawling tractor sprinkler is located as pictured below, 100 feet South of a 10 ft. wide sidewalk; notice the hose and sidewalk are not perpendicular. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North 1 along the hose at the rate of 2 inch/second. (a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and find the equation of the line forming the southern boundary of the sidewalk. (b) After 33 minutes, sketch a picture of the wet portion of the sidewalk; find the length of the wet portion of the Southern edge of the sidewalk. (c) Find the equation of the line forming the northern boundary of the sidewalk. (Hint: You can use the properties of right triangles.) W 100 ft Problem 4.14. Juliet and Mercutio are moving at constant speeds in the xy-plane. They start moving at the same time. Juliet starts at the point (0, − 6) and heads in a straight line toward the point (10,5), reaching it in 10 seconds. Mercutio starts at (9, − 14) and moves in a straight line. Mercutio passes through the same point on the x axis as Juliet, but 2 seconds after she does. How long does it take Mercutio to reach the y-axis? Problem 4.15. (a) Solve for x: 1 1 − = 3. x x+1 (1 + t)2 + (1 − 2t)2 . (b) Solve for t: 2 = N hose (c) Find the times when Margot is 28 feet from the statue. E S 20 ft (c) Solve for t: 3 √ 5 (d) Solve for t: 0 = Problem 4.16. (1 + t)2 + (1 − 2t)2 . = (1 + t)2 + (1 − 2t)2 . (a) Solve for x: x4 − 4x2 + 2 = 0 100 ft. sidewalk (b) Solve for y: circular watered zone √ y−2 y=4 Chapter 5 Functions and Graphs Pictures are certainly important in the work of an architect, but it is perhaps less evident that visual aids can be powerful tools for solving mathematical problems. If we start with an equation and attach a picture, then the mathematics can come to life. This adds a new dimension to both interpreting and solving problems. One of the real triumphs of modern mathematics is a theory connecting pictures and equations via the concept of a graph. This transition from “equation” to “picture” (called graphing) and its usefulness (called graphical analysis) are the theme of the next two sections. The importance of these ideas is HUGE and cannot be overstated. Every moment spent studying these ideas will pay back dividends in this course and in any future mathematics, science or engineering courses. 5.1 Relating Data, Plots and Equations Imagine you are standing high atop an oceanside cliff and spot a seagull hovering in the air-current. Assuming the gull moves up and down along a vertical line of motion, how can we best describe its location at time t seconds? There are three different (but closely linked) ways to describe the location of the gull: gull line of motion 1 0 1 0 • a table of data of the gull’s height above cliff level at various times t; • a plot of the data in a “time” (seconds) vs. “height” (feet) coordinate system; ocean Figure 5.1: Seagull’s height. • an equation relating time t (seconds) and height s (feet). To make sure we really understand how to pass back and forth between these three descriptive modes, imagine we have tabulated (Figure 5.2) the height of the gull above cliff level at one-second time intervals 55 cliff level CHAPTER 5. FUNCTIONS AND GRAPHS 56 for a 10 second time period. Here, a “negative height” means the gull is below cliff level. We can try to visualize the meaning of this data by plotting these 11 data points (t, s) in a time (sec.) vs. height (ft.) coordinate system. Feet Gull Height (feet above cliff level) s (ft) t (sec) s (ft) t (sec) 20 4 -10 8 6.88 5 -8.12 9 -2.5 6 -2.5 10 -8.12 7 6.88 t (sec) 0 1 2 3 50 40 30 20 10 s (ft) 20 36.88 57.5 2 4 6 8 10 Seconds −10 (b) Visual data. (a) Symbolic data. Figure 5.2: Symbolic versus visual view of data. We can improve the quality of this description by increasing the number of data points. For example, if we tabulate the height of the gull above cliff level at 1/2 second or 1/4 second time intervals (over the same 10 second time period), we might get these two plots: Feet Feet 50 40 30 20 10 Seconds 4 −10 2 (a) 1 2 6 8 10 second intervals. 50 40 30 20 10 Seconds 4 −10 2 (b) 1 4 6 8 10 second intervals. Figure 5.3: Shorter time intervals mean more data points. We have focused on how to go from data to a plot, but the reverse process is just as easy: A point (t, s) in any of these three plots is interpreted to mean that the gull is s feet above cliff level at time t seconds. Furthermore, increasing the amount of data, we see how the plotted points are “filling in” a portion of a parabola. Of course, it is way too tedious to create longer and longer tables of data. What we really want is a “formula” (think of it as a prescription) that tells us how to produce a data point for the gull’s height at any given time t. If we had such a formula, then we could completely dispense with the tables of data and just use the formula to crank out data points. For example, look at this equation involving the variables t and s: s= 15 (t − 4)2 − 10. 8 5.2. WHAT IS A FUNCTION? 57 If we plug in t = 0, 1, 2, 9, 10, then we get s = 20, 6.88, −2.5, 36.88, 57.5, respectively; this was some of our initial tabulated data. This same equation produces ALL of the data points for the other two plots, using 1/2 second and 1/4 second time intervals. (Granted, we have swept under the rug the issue of “...where the heck the equation comes from...” ; that is a consequence of mathematically modeling the motion of this gull. Right now, we are focusing on how the equation relates to the data and the plot, assuming the equation is in front of us to start with.) In addition, it is very important to notice that having this equation produces an infinite number of data points for our gull’s location, since we can plug in any t value between 0 and 10 and get out a corresponding height s. In other words, the equation is A LOT more powerful than a finite (usually called discrete ) collection of tabulated data. 5.2 What is a Function? Our lives are chock full of examples where two changing quantities are related to one another: • The cost of postage is related to the weight of the item. • The value of an investment will depend upon the time elapsed. • The population of cells in a growth medium will be related to the amount of time elapsed. • The speed of a chemical reaction will be related to the temperature of the reaction vessel. In all such cases, it would be beneficial to have a “procedure” whereby we can assign a unique output value to any acceptable input value. For example, given the time elapsed (an input value), we would like to predict a unique future value of an investment (the output value). Informally, this leads to the broadest (and hence most applicable) definition of what we will call a function : Definition 5.2.1. A function is a procedure for assigning a unique output to any allowable input. The key word here is “procedure.” Our discussion of the hovering seagull in 5.1 highlights three ways to produce such a “procedure” using data, plots of curves and equations. y-axis y P = (x,y) x x-axis Figure 5.4: Graph of a • A table of data, by its very nature, will relate two procedure. columns of data: The output and input values are listed as column entries of the table and reading across each row is the “procedure” which relates an input with a unique output. CHAPTER 5. FUNCTIONS AND GRAPHS 58 • Given a curve in Figure 5.4, consider the “procedure” which associates to each x on the horizontal axis the y coordinate of the pictured point P on the curve. • Given an equation relating two quantities x and y, plugging in a particular x value and going through the “procedure” of algebra often produces a unique output value y. 5.2.1 The definition of a function (equation viewpoint) Now we focus on giving a precise definition of a function, in the situation when the “procedure” relating two quantities is actually given by an equation. Keep in mind, this is only one of three possible ways to describe a function; we could alternatively use tables of data or the plot of a curve. We focus on the equation viewpoint first, since it is no doubt the most familiar. If we think of x and y as related physical quantities (e.g. time and distance ), then it is sometimes possible (and often desirable) to express one of the variables in terms of the other. For example, by simple arithmetic, the equations 3x + 2y = 4 1 x2 − x = y − 4 2 y x2 + 1 = 1, can be rewritten as equivalent equations 1 y = (4 − 3x) 2 2x2 − 2x + 8 = y y= √ 1 . +1 x2 This leads to THE MOST IMPORTANT MATH DEFINITION IN THE WORLD: Definition 5.2.2. A function is a package, consisting of three parts: • An equation of the form y = “a mathematical expression only involving the variable x,” which we usually indicate via the shorthand notation y = f(x). This equation has the very special property that each time we plug in an x value, it produces exactly one (a unique) y value. We call the mathematical expression f(x) ”the rule”. • A set D of x-values we are allowed to plug into f(x), called the ”domain” of the function. 5.2. WHAT IS A FUNCTION? 59 • The set R of output values f(x), where x varies over the domain, called the ”range” of the function. Any time we have a function y = f(x), we refer to x as the independent variable (the “input data”) and y as the dependent variable (the “output data”). This terminology emphasizes the fact that we have freedom in the values of x we plug in, but once we specify an x value, the y value is uniquely determined by the rule f(x). Examples 5.2.3. (i) The equation y = −2x + 3 is in the form y = f(x), where the rule is f(x) = −2x+3. Once we specify a domain of x values, we have a function. For example, we could let the domain be all real numbers. y-axis Graph of y = b b x-axis Figure 5.5: Constant func(ii) Take the same rule f(x) = −2x + 3 from (i) and let the tion. domain be all non-negative real numbers. This describes a function. However, the functions f(x) = −2x+ 3 on the domain of all non-negative real numbers and f(x) = −2x + 3 on the domain of all real numbers (from (i)) are different, even though they share the same rule; this is because their domains differ! This example illustrates the idea of what is called a restricted domain. In other words, we started with the function in (i) on the domain of all real numbers, then we “restricted” to the subset of non-negative real numbers. (iii) The equation y = b, where b is a constant, defines a function on the domain of all real numbers, where the rule is f(x) = b; we call these the constant functions. Recall, in Chapter 3, we observed that the solutions of the equation y = b, plotted in the xy coordinate system, will give a horizontal line. For example, if b = 0, you get the horizontal axis. 1 1 (iv) Consider the equation y = x , then the rule f(x) = x defines a function, as long as we do not plug in x = 0. For example, take the domain to be the non-zero real numbers. √ (v) Consider the equation y = 1 − x2 . Before we start plugging in x values, we want to know the expression under the radical symbol (square root symbol) is nonnegative; this insures the square root is a real number. This amounts to solving an inequality equation: 0 ≤ 1 − x2 ; i.e., −1 ≤ x ≤ 1. These remarks show that √ the rule f(x) = 1 − x2 defines a function, where the domain of x values is −1 ≤ x ≤ 1. CHAPTER 5. FUNCTIONS AND GRAPHS 60 Typically, the domain of a function y = f(x) will either be the entire number line, an interval on the number line, or a finite union of such intervals. We summarize the notation used to represent intervals in Table 5.1. Common Intervals on the Number Line Description Symbolic Notation All numbers x between a and b, x possibly equal to either a or b a≤x≤b All numbers x between a and b, x = a and x = b a<x<b All numbers x between a and b, x = b and x possibly equal to a All numbers x between a and b, x = a and x possibly equal to b Picture a b a b a≤x<b a b a<x≤b a b Table 5.1: Interval Notations We can interpret a function as a “prescription” that takes a given x value (in the domain) and produces a single unique y value (in the range). We need to be really careful and not fall into the trap of thinking that every equation in the world is a function. For example, if we look at this equation x + y2 = 1 and plug in x = 0, the equation becomes y2 = 1. This equation has two solutions, y = ±1, so the conclusion is that plugging in x = 0 does NOT produce a single output value. This violates one of the conditions of our function definition, so the equation x + y2 = 1 is NOT a function in the independent variable x. Notice, if you were to try and solve this equation for y in terms of x, you’d first write y2 = 1 − x and then take a square root (to isolate y); but the square root introduces TWO roots, which is just another way of reflecting the fact there can be two y values attached to a single x value. Alternatively, you can solve the equation for x in terms of y, getting x = 1 − y2 ; this shows the equation does define a function x = g(y) in the independent variable y. 5.2. WHAT IS A FUNCTION? 61 5.2.2 The definition of a function (conceptual viewpoint) Conceptually, you can think of a function as a “process”: An allowable input goes into a “black box” and out pops a unique new value denoted by the symbol f(x). Compare this with the machine making “hula-hoops” in Figure 5.6. While you are problem solving, you will find this to be a useful viewpoint when a function is described in words. 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 tube in tubes out hoop f(tube) (a) A hoop machine as a “process” taking “tubes” to “hoops.” x domain in out y f ( x) (b) A function as a “process” which takes x to y. Figure 5.6: Viewing a function as a “process.” Examples 5.2.4. Here are four examples of relationships that are functions: (i) The total amount of water used by a household since midnight on a particular day. Let y be the total number of gallons of water used by a household between 12:00am and a particular time t; we will use time units of “hours.” Given a time t, the household will have used a specific (unique) amount of water, call it S(t). Then y = S(t) defines a function in the independent variable t with dependent variable y. The domain would be 0 ≤ t ≤ 24 and the largest possible value of S(t) on this domain is S(24). This tells us that the range would be the set of values 0 ≤ y ≤ S(24). (ii) The height of the center of a basketball as you dribble, depending on time. Let s be the height of the basketball center at time t seconds after you start dribbling. Given a time t, if we freeze the action, the center of the ball has a single unique height above the floor, call it h(t). So, the height of the basketball center is given by a function s = h(t). The domain would be a given interval of time you are dribbling the ball; for example, maybe 0 ≤ t ≤ 2 (the first 2 seconds). In this case, the range would be all of the possible heights attained by the center of the basketball during this 2 seconds. (iii) The state sales tax due on a taxable item. Let T be the state tax (in dollars) due on a taxable item that sells for z dollars. Given a taxable item that costs z dollars, the state tax due is a single unique CHAPTER 5. FUNCTIONS AND GRAPHS 62 amount, call it W (z). So, T = W (z) is a function, where the independent variable is z. The domain could be taken to be 0 ≤ z ≤ 1,000,000, which would cover all items costing up to one-million dollars. The range of the function would be the set of all values W (z), as z ranges over the domain. (iv) The speed of a chemical reaction depending on the temperature. Let v be the speed of a particular chemical reaction and T the temperature in Celsius ◦ C. Given a particular temperature T , one could experimentally measure the speed of the reaction; there will be a unique speed, call it r(T ). So, v = r(T ) is a function, where the independent variable is T . The domain could be taken to be 0 ≤ T ≤ 100, which would cover the range of temperatures between the freezing and boiling points of water. The range of the function would be the set of all speeds r(T ), as T ranges over the domain. 5.3 The Graph of a Function Let’s start with a concrete example; the function f(x) = −2x + 3 on the domain of all real numbers. We discussed this Example 5.2.3. Plug in the specific x values, where x = −1, 0, 1, 2 and tabulate the resulting y values of the function: x -1 0 1 2 . . . x y 5 3 1 -1 . . . −2x + 3 point (x,y) (-1,5) (0,3) (1,1) (2,-1) . . . (x, − 2x + 3) (a) Tabulated data. y-axis Graph of y = −2x + 3. x-axis (b) Visual data. Figure 5.7: Symbolic versus visual view of data. This tells us that the points (0, 3), (1, 1), (2, −1), (−1, 5) are solutions of the equation y = −2x + 3. For example, if y = −2x + 3, x = 0, y = 3, then 3 = −2 · 0 + 3 (which is true), or if y = −2x + 3, x = 2, y = −1, then −1 = −2 · 2 + 3 (which is true), etc. In general, if we plug in x we get out −2x + 3, so the point (x, − 2x + 3) is a solution to the function equation y = f(x). We can plot all of these solutions in the xy-coordinate system. The set of points we obtain, as we vary over all x in the domain, is called the set of solutions of the equation y = −2x + 3: Solutions = { (x, −2x + 3) | x any real number}. Notice that plotting these points produces a line of slope m = −2 with y-intercept 3. In other words, the graph of the function f(x) = −2x + 3 is 5.4. THE VERTICAL LINE TEST 63 the same as the graph of the equation y = −2x + 3, as we discussed in Chapter 4. In general, by definition, we say that a point (x,y) is a solution to the function equation y = f(x) if plugging x and y into the equation gives a true statement. How can we find ALL the solutions of the equation y = f(x)? In general, the definition of a function is “rigged” so it is easy to describe all solutions of the equation y = f(x): Each time we specify an x value (in the domain), there is only one y value, namely f(x). This means the point P = (x, f(x)) is the ONLY solution to the equation y = f(x) with first coordinate x. We define the graph of the function y = f(x) to be the plot of all solutions of this equation (in the xy coordinate system). It is common to refer to this as either the “graph of f(x)” or the “graph of f.” Graph = {(x,f(x)) | x in the domain} (5.1) Important Procedure 5.3.1. Points on a graph. The description of the graph of a function gives us a procedure to produce points on the graph AND to test whether a given point is on the graph. On the one hand, if you are given u in the domain of a function y = f(x), then you immediately can plot the point (u, f(u)) on the graph. On the other hand, if someone gives you a point (u, v), it will be on the graph only if v = f(u) is true. We illustrate this in Example 5.3.2. Example 5.3.2. The function s = h(t) = 15 (t − 4)2 − 10 de8 fines a function in the independent variable t. If we restrict to the domain 0 ≤ t ≤ 10, then the discussion in Chapter 7 tells us that the graph is a portion of a parabola: See Figure 5.8. Using the above procedure, you can verify that the data points discussed in the seagull example (in §5.1) all lie on this parabola. On the other hand, the point (0,0) is NOT on the graph, since h(0) = 20 = 0. 60 50 40 30 20 10 −10 s-axis t-axis 4 2 6 8 10 Figure 5.8: s = h(t). 5.4 The Vertical Line Test There is a pictorial aspect of the graph of a function that is very revealing: Since (x, f(x)) is the only point on the graph with first coordinate equal to x, a vertical line passing through x on the x-axis (with x in the domain) crosses the graph of y = f(x) once and only once. This gives us a decisive way to test if a curve is the graph of a function. Important Procedure 5.4.1. The vertical line test. Draw a curve in the xy-plane and specify a set D of x-values. Suppose every vertical line through a value in D intersects the curve exactly once. Then the curve is CHAPTER 5. FUNCTIONS AND GRAPHS 64 the graph of some function on the domain D. If we can find a single vertical line through some value in D that intersects the curve more than once, then the curve is not the graph of a function on the domain D. For example, draw any straight line m in the plane. By the vertical line test, if the line m is not vertical, m is the graph of a function. On the other hand, if the line m is vertical, then m is not the graph of a function. These two situations are illustrated in Figure 5.9. As another example, consider the equation x2 + y2 = 1, whose graph is the unit circle and specify the domain D to be −1 ≤ x ≤ 1; recall Example 3.2.2. The vertical line passing through the point 1 , 0 will intersect the unit circle twice; by 2 the vertical line test, the unit circle is not the graph of a function on the domain −1 ≤ x ≤ 1. y-axis y-axis y-axis l m m x-axis l crosses curve twice C x-axis x-axis Figure 5.9: Applying the vertical line test. 5.4.1 Imposed Constraints In physical problems, it might be natural to constrain (meaning to “limit” or “restrict”) the domain. As an example, suppose the height s (in feet) of a ball above the ground after t seconds is given by the function s = h(t) = −16t2 + 4. s-axis Physically interesting portion of graph. t-axis Figure 5.10: Restricting the domain. We could look at the graph of the function in the tsplane and we will review in Chapter 7 that the graph looks like a parabola. The physical context of this problem makes it natural to only consider the portion of the graph in the first quadrant; why? One way of specifying this quadrant would be to restrict the domain of possible 1 t values to lie between 0 and 2 ; notationally, we would write this constraint as 0 ≤ t ≤ 1 . 2 5.5 Linear Functions A major goal of this course is to discuss several different kinds of functions. The work we did in Chapter 4 actually sets us up to describe one 5.6. PROFIT ANALYSIS 65 very useful type of function called a linear function. Back in Chapter 4, we discussed how lines in the plane can be described using equations in the variables x and y. One of the key conclusions was: Important Fact 5.5.1. A non-vertical line in the plane will be the graph of an equation y = mx + b, where m is the slope of the line and b is the y-intercept. Notice that any non-vertical line will satisfy the conditions of the vertical line test, which means it must be the graph of a function. What is the function? The answer is to use the equation in x and y we already obtained in Chapter 4: The rule f(x) = mx + b on some specified domain will have a line of slope m and y-intercept b as its graph. We call a function of this form a linear function. Example 5.5.2. You are driving 65 mph from the Kansas state line (mile marker 0) to Salina (mile marker 130) along I-35. Describe a linear function that calculates mile marker after t hours. Describe another linear function that will calculate your distance from Salina after t hours. Solution. Define a function d(t) to be the mile marker after t hours. Using “distance=rate×time,” we conclude that 65t will be the distance traveled after t hours. Since we started at mile marker 0, d(t) = 65t is the rule for the first function. A reasonable domain would be to take 0 ≤ t ≤ 2, since it takes 2 hours to reach Salina. For the second situation, we need to describe a different function, call it s(t), that calculates your distance from Salina after t hours. To describe the rule of s(t) we can use the previous work: y-axis 120 100 80 60 40 20 0.5 1 1.5 2 t-axis Figure 5.11: Distance functions. s(t) = (mile marker Salina) − (your mile marker at t hrs.) = 130 − d(t) = 130 − 65t. For the rule s(t), the best domain would again be 0 ≤ t ≤ 2. We have graphed these two functions in the same coordinate system: See Figure 5.11 (Which function goes with which graph?). 5.6 Profit Analysis Let’s give a first example of how to interpret the graph of a function in the context of an application. CHAPTER 5. FUNCTIONS AND GRAPHS 66 Example 5.6.1. A software company plans to bring a new product to market. The sales price per unit is $15 and the expense to produce and market √ x units is $100(1 + x). What is the profit potential? Two functions control the profit potential of the new software. The first tells us the gross income, in dollars, on the sale of x units. All of the costs involved in developing, supporting, distributing and marketing x units are controlled by the expense equation (again in dollars): g(x) = 15x e(x) = 100(1 + √ (gross income function) (expense function) x) A profit will be realized on the sale of x units whenever the gross income exceeds expenses; i.e., this occurs when g(x) > e(x). A loss occurs on the sale of x units when expenses exceed gross income; i.e., when e(x) > g(x). Whenever the sale of x units yields zero profit (and zero loss), we call x a break-even point ; i.e., when e(x) = g(x). The above approach is “symbolic.” Let’s see how to study profit and loss visually, by studying the graphs of the two functions g(x) and e(x). To begin with, plot the graphs of the two individual functions in the xycoordinate system. We will focus on the situation when the sales figures are between 0 to 100 units; so the domain of x values is the interval 0 ≤ x ≤ 100. Given any sales figure x, we can graphically relate three y-axis (dollars) 1500 1500 expense gross income y-axis (dollars) 1000 (x, 15x) 500 P 20 1000 500 (x, e(x)) Q x-axis (units sold) x-axis (units sold) 40 60 80 100 (a) Gross income graph. 20 40 60 80 100 (b) Expenses graph. Figure 5.12: Visualizing income and expenses. things: • x on the horizontal axis; • a point on the graph of the gross income or expense function; • y on the vertical axis. If x = 20 units sold, there is a unique point P = (20, g(20)) = (20, 300) on the gross income graph and a unique point Q = (20, e(20)) = (20, 547) on the expenses graph. Since the y-coordinates of P and Q are the function values at x = 20, the height of the point above the horizontal axis is controlled by the function. 5.6. PROFIT ANALYSIS 67 dollars y If we plot both graphs in the same coordinate (x, g(x)) 1400 system, we can visually study the distance between 1200 points on each graph above x on the horizontal axis. 1000 (x, e(x)) 800 (x, e(x)) In the first part of this plot, the expense graph is B 600 above the income graph, showing a loss is realized; 400 (x, g(x)) 200 sold units x the exact amount of the loss will be e(x) − g(x), which is the length of the pictured line segment. Further to 20 40 60 80 100 the right, the two graphs cross at the point labeled Figure 5.13: Modelling profit and loss. “B”; this is the break-even point; i.e., expense and income agree, so there is zero profit (and zero loss). Finally, to the right of B the income graph is above the expense graph, so there is a profit; the exact amount of the profit will be g(x) − e(x), which is the length of the right-most line segment. Our analysis will be complete once we pin down the break-even point B. This amounts to solving the equation g(x) = e(x). √ 15x = 100(1 + x) √ 15x − 100 = 100 x 225x2 − 3000x + 10000 = 10000x 225x2 − 13000x + 10000 = 0. Applying the quadratic formula, we get two answers: x = 0.78 or 57. Now, we face a problem: Which of these two solutions is the answer to the original problem? We are going to argue that only the second solution x = 57 gives us the break even point. What about the other ”solution” at x = 0.78? Try plugging x = 0.78 into the original equation: √ 15(0.78) = 100(1 + 0.78). What has happened? Well, when going from the second to the third line, both sides of the equation were squared. Whenever we do this, we run the risk of adding extraneous solutions. What should you do? After solving any equation, look back at your steps and ask yourself whether or not you may have added (or lost) solutions. In particular, be wary when squaring or taking the square root of both sides of an equation. Always check your final answer in the original equation. We can now compute the coordinates of the break-even point using either function: B = (57, g(57)) = (57, 855) = (57, e(57)). CHAPTER 5. FUNCTIONS AND GRAPHS 68 5.7 Exercises Problem 5.1. For each of the following functions, find the expression for f(x + h) − f(x) . h Simplify each of your expressions far enough so that plugging in h = 0 would be allowed. (a) f(x) = x2 − 2x. (b) f(x) = 2x + 3 (c) f(x) = x2 − 3 (d) f(x) = 4 − x2 (e) f(x) = −πx2 − π2 √ (f) f(x) = x − 1. (Hint: Rationalize the numerator) Problem 5.2. Here are the graphs of two linear functions on the domain 0 ≤ x ≤ 20. Find the formula for each of the rules y = f(x) and y = g(x). Find the formula for a NEW function v(x) that calculates the vertical distance between the two lines at x. Explain in terms of the picture what v(x) is calculating. What is v(5)? What is v(20)? What are the smallest and largest values of v(x) on the domain 0 ≤ x ≤ 20? y-axis 60 (20,60) g(x) 40 20 (0,24) f(x) (20,20) (0,4) x-axis 10 20 Problem 5.3. Dave leaves his office in Padelford Hall on his way to teach in Gould Hall. Below are several different scenarios. In each case, sketch a plausible (reasonable) graph of the function s = d(t) which keeps track of Dave’s distance s from Padelford Hall at time t. Take distance units to be “feet” and time units to be “minutes.” Assume Dave’s path to Gould Hall is along a straight line which is 2400 feet long. gould padelford (a) Dave leaves Padelford Hall and walks at a constant speed until he reaches Gould Hall 10 minutes later. (b) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute. He then continues on to Gould Hall at the same constant speed he had when he originally left Padelford Hall. (c) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to figure out where he is. Dave then continues on to Gould Hall at twice the constant speed he had when he originally left Padelford Hall. (d) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Dave gets confused and stops for 1 minute to figure out where he is. Dave is totally lost, so he simply heads back to his office, walking the same constant speed he had when he originally left Padelford Hall. (e) Dave leaves Padelford heading for Gould Hall at the same instant Angela leaves Gould Hall heading for Padelford Hall. Both walk at a constant speed, but Angela walks twice as fast as Dave. Indicate a plot of “distance from Padelford” vs. “time” for both Angela and Dave. (f) Suppose you want to sketch the graph of a new function s = g(t) that keeps track of Dave’s distance s from Gould Hall at time t. How would your graphs change in (a)-(e)? Problem 5.4. At 5 AM one day, a monk began a trek from his monastery by the sea to the monastery at the top of a mountain. He reached the mountain-top monastery at 11 AM, spent the rest of the day in meditation, 5.7. EXERCISES 69 and then slept the night there. In the morning, at 5 AM, he began walking back to the seaside monastery. Though walking downhill should have been faster, he dawdled in the beautiful sunshine, and ending up getting to the seaside monastery at exactly 11 AM. y-axis (a) Was there necessarily a time during each trip when the monk was in exactly the same place on both days? Why or why not? x-axis (b) Suppose the monk walked faster on the second day, and got back at 9 AM. What is your answer to part (a) in this case? Recall the procedure 5.3.1 on page 63. (a) Find the x and y intercepts of the graph. (c) Suppose the monk started later, at 10 AM, and reached the seaside monastery at 3 PM. What is your answer to part (a) in this case? (b) Find the exact coordinates of all points (x,y) on the graph which have ycoordinate equal to 5. (c) Find the coordinates of all points (x,y) on the graph which have y-coordinate equal to -3. Problem 5.5. Sketch a reasonable graph for each of the following functions. Specify a reasonable domain and range and state any assumptions you are making. Finally, describe the largest and smallest values of your function. (a) Height of a person depending on age. (b) Height of the top of your head as you jump on a pogo stick for 5 seconds. (c) The amount of postage you must put on a first class letter, depending on the weight of the letter. (d) Distance of your big toe from the ground as you ride your bike for 10 seconds. (d) Which of these points √ on the graph: is √ (1, − 2), (−1,3), (2.4,8), ( 3,7 − 3 3). (e) Find the exact coordinates of the point √ (x,y) on the graph with x = 1 + 2. Problem 5.7. After winning the lottery, you decide to buy your own island. The island is located 1 km offshore from a straight portion of the mainland. There is currently no source of electricity on the island, so you want to run a cable from the mainland to the island. An electrical power sub-station is located 4 km from your island’s nearest location to the shore. It costs $50,000 per km to lay a cable in the water and $30,000 per km to lay a cable over the land. ocean your island cable path (e) Your height above the water level in a swimming pool after you dive off the high board. power 1 km x 4 km Problem 5.6. Here is a picture of the graph of the function f(x) = 3x2 − 3x − 2. (a) Explain why we can assume the cable follows the path indicated in the picture; CHAPTER 5. FUNCTIONS AND GRAPHS 70 i.e. explain why the path consists of two line segments, rather than a weird curved path AND why it is OK to assume the cable reaches shore to the right of the power station and the left of the island. (b) Let x be the distance downshore from the power sub-station to where the cable reaches the land. Find a function f(x) in the variable x that computes the cost to lay a cable out to your island. (c) Make a table of values of f(x), where 7 1 x = 0, 2 ,1, 3 ,2,. . . , 2 ,4. Use these calcula2 tions to estimate the installation of minimal cost. Problem 5.8. This problem deals with the “mechanical aspects” of working with the rule of a function. For each of the functions listed in (a)-(c), calculate: f(0), f(−2), f(x + 3), f(♥), f(♥ + △). 1 (a) The function f(x) = 2 (x − 3) on the domain of all real numbers. (b) The function f(x) = 2x2 − 6x on the domain of all real numbers. (c) The function f(x) = 4π2 . Problem 5.9. Which of the curves in Figure 5.14 represent the graph of a function? If the curve is not the graph of a function, describe what goes wrong and how you might “fix it.” When you describe how to “fix” the graph, you are allowed to cut the curve into pieces and such that each piece is the graph of a function. Many of these problems have more than one correct answer. Problem 5.10. Find an EXACT answer for each problem. (a) Solve for x 5 30 x + =2 x+3 x−7 x − 4x − 21 (b) Solve for x √ x 5x − 4 = + 2 2 (c) Solve for x √ √ x + x − 20 = 10 (d) Solve for t √ √ 2t − 1 + 3t + 3 = 5 5.7. EXERCISES 71 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) Figure 5.14: Curves to consider for Problem 5.9 72 CHAPTER 5. FUNCTIONS AND GRAPHS Chapter 6 Graphical Analysis We ended the previous section with an in-depth look at a “Profit Analysis Problem.” In that discussion, we looked at the graphs of the relevant functions and used these as visual aids to help us answer the questions posed. This was a concrete illustration of what is typically called “graphical analysis of a function.” This is a fundamental technique we want to carry forward throughout the course. Let’s highlight the key ideas for future reference. 6.1 Visual Analysis of a Graph A variety of information can be visually read off of a function graph. To see this, we ask ourselves the following question: What is the most basic qualitative feature of a graph? To answer this, we need to return to the definition of the graph (see Equation (5.1) on page 63) and the surrounding discussion. The key thing about the graph of a function f(x) is that it keeps track of a particular set of points in the plane whose coordinates are related by the function rule. To be precise, a point P = (x,y) will be on the graph of the function f(x) exactly when y = f(x). 6.1.1 Visualizing the domain and range A function is a package that consists of a rule y = f(x), a domain of allowed x-values and a range of output y-values. The domain can be visualized as a subset of the x-axis and the range as a subset of the y-axis. If you are handed the domain, it is graphically easy to describe the range values obtained; here is the procedure: Important Procedure 6.1.1. Look at all points on the graph corresponding to domain values on the x-axis, then project these points to the y-axis. The collection of all values you obtain on the y-axis will be the range of the function. This idea of “projection” is illustrated in the two graphs be73 CHAPTER 6. GRAPHICAL ANALYSIS 74 low. We use arrows “→” to indicate going from a domain x-value, up to the graph, then over to the y-axis: range = [1,9] graph of f (x) = −2x + 3 domain = [3,5] domain = [−3,1] range = [−7, − 3] graph of f (x) = −2x + 3 (b) Domain: [−3, 1]. (a) Domain: [3, 5]. Figure 6.1: Example projections. 6.1.2 Interpreting Points on the Graph We can visually detect where a function has positive or negative values: Important Fact 6.1.2. The function values f(x) control the height of the point P(x) = (x,f(x)) on the graph above the x-axis; if the function value f(x) is negative, the point P(x) is below the x-axis. f (x) units above x-axis f(x3 ) units above the x-axis P (x) = (x,f (x)) P (x3 ) = (x3 ,f (x3 )) x2 x x3 |f(x)| units below the x-axis x-axis P(x2 ) = (x2 ,f(x2 )) y-axis Figure 6.2: Interpreting points on a graph. In Figure 6.2 we can now divide the domain (in this case the whole number line) into segments where the function is above, below or crossing the axis. Keeping track of this information on a number line is called a sign plot for the function. We include a “shadow” of the graph in Figure 6.3 6.1. VISUAL ANALYSIS OF A GRAPH 75 positive positive negative negative x-axis Figure 6.3: Sign plot. to emphasize how we arrived at our “positive” and “negative” labeling of the sign plot; in practice we would only provide a labeled number line. By moving through a sequence of x values we can investigate how the corresponding points on the graph move “up and down”; this then gives us a dynamic visual sense of how the function values are changing. For example, in Figure 6.4, suppose we let x move from 1 to 5, left to right; we have indicated how the corresponding points on the curve will move and how the function values will change. y-axis f(1) P #2: points on graph move from P to Q f(2) #3: f(x) values move like this graph of y = f(x) f(3) f(5) Q f(4) 1 5 x-axis #1: x values move from 1 to 5 Figure 6.4: Dynamic interpretation of a graph. CHAPTER 6. GRAPHICAL ANALYSIS 76 6.1.3 Interpreting Intercepts of a Graph The places where a graph crosses the axes are often significant. We isolate each as an important feature to look y-intercept = (0,f(0)) for when doing graphical analysis. The graph of the func(x3 ,0) x-axis (x2 ,0) tion y = f(x) crosses the y-axis at the point (0,f(0)); so, (x1 ,0) the y-intercept of the graph is just f(0). The graph of the x-intercepts have the form “(x,0)” function y = f(x) crosses the x-axis at points of the form Figure 6.5: Intercepts of a (x0 ,f(x0 )), where f(x0 ) = 0. The values x0 are called roots graph. or zeros of the function f(x). There can be at most one y-intercept, but there can be several x-intercepts or no x-intercept: See Figure 6.5 The graph of a function y = f(x) crosses the vertical line x = h at the point (h,f(h)). To find where the graph of a function y = f(x) crosses the horizontal line y = k, first solve the equation k = f(x) for x. If the equation k = f(x) has solutions x1 , x2 , x3 , x4 , then the points of intersection would have the coordinates given. y-axis y-axis upper semicircle radius r = 2 centered at (2,1) y-axis (h,(f(h)) y=k graph of f(x) (x1 ,k) (x3 ,k) (x2 ,k) (x4 ,k) y=k x1 x2 x3 x=h x4 y= x-axis x=h (a) General curve. 1 2 x-axis (b) Semicircle. Figure 6.6: Crossing horizontal and vertical lines. As another example, the graph in Figure 6.6(b) above will cross the horizontal line y = k twice if and only if 1 ≤ k < 3; the graph will cross the horizontal line y = k once if and only if k = 3. The graph will not intersect 1 the line y = 2 and the graph will cross the vertical line x = h if and only if 0 ≤ h ≤ 4. 6.1.4 Interpreting Increasing and Decreasing y-axis up hi ll x-axis dow nh ill = local extrema Figure 6.7: Graphically interpreting increasing and decreasing. We use certain terms to describe how the function values are changing over some domain of x values. Typically, we want to study what is happening to the values f(x) as x moves from “left to right” in some interval. This can be linked graphically with the study of “uphill” and “downhill” portions of the function graph: If you were “walking to the right” along the graph, the function values are increasing if you are walking uphill. Likewise, if you were 6.2. CIRCLES AND SEMICIRCLES 77 “walking to the right” along the graph, the function values are decreasing if you are walking downhill. Once we understand where the graph is moving uphill and downhill, we can isolate the places where we change from moving uphill to downhill, or vice versa; these “peaks” and “valleys” are called local maxima and local minima. Some folks refer to either case as a local extrema. People have invested a lot of time (centuries!) and energy (lifetimes!) into the study of how to find local extrema for particular function graphs. We will see some basic examples in this course and others will surface in future courses once you have the tools of calculus at your disposal. Examples range from business applications that involve optimizing profit to understanding the three-dimensional shape a of biological molecule. Example 6.1.3. A hang glider launches from a gliderport in La Jolla. The launch point is located at the edge of a 500 ft. high cliff over the Pacific Ocean. The elevation of the pilot above the gliderport after t minutes is given by the graph in Figure 6.8: ft above gliderport 600 400 200 minutes 5 10 −200 1. When is the pilot climbing and descending? −400 −600 2. When is the pilot at the glider port elevation? 3. How much time does the pilot spend flying level? Figure 6.8: Hanglider elevations. Solution. 1. Graphically, we need to determine the portions of the graph that are increasing or decreasing. In this example, it is increasing when 0 ≤ t ≤ 2 and 7 ≤ t ≤ 9. And, it is decreasing when 3 ≤ t ≤ 5 and 9 ≤ t ≤ 10. 2. Graphically, this question amounts to asking when the elevation is 0, which is the same as finding when the graph crosses the horizontal axis. We can read off there are four such times: t = 0, 4, 8, 10. 3. Graphically, we need to determine the portions of the graph that are made up of horizontal line segments. This happens when 2 ≤ t ≤ 3 and 5 ≤ t ≤ 7. So, our pilot flies level for a total of 3 minutes. 6.2 Circles and Semicircles Back in Chapter 3, we discussed equations whose graphs were circles: We found that the graph of the equation (x − h)2 + (y − k)2 = r2 (6.1) CHAPTER 6. GRAPHICAL ANALYSIS 78 is a circle of radius r centered at the point (h, k). It is possible to manipulate this equation and become confused. We could rewrite this as (y − k)2 = r2 − (x − h)2 , then take the square root of each side. However, the resulting equivalent equation would be r2 − (x − h)2 y=k± and the presence of that ± sign is tricky; it means we have two equations: y = k+ r2 − (x − h)2 y = k− or r2 − (x − h)2 . Each of these two equations defines a function: f(x) = k + r2 − (x − h)2 g(x) = k − or (6.2) r2 − (x − h)2 . (6.3) So, even though the Equation 6.1 is not a function, we were able to obtain two different functions f(x) and g(x) from the original equation. The relationship between the graph of the original equation and the graphs of the two functions in (6.2) and (6.3) is as follows: The upper semicircle is the graph of the function f(x) and the lower semicircle is the graph of the function g(x). y-axis upper semicircle y-axis (h,k) (h,k) lower semicircle x-axis Graph of y = f(x) x-axis Graph of y = g(x) Figure 6.9: Upper and lower semicircles. Example 6.2.1. A tunnel connecting two portions of a space station has a circular cross-section of radius 15 feet. Two walkway decks are constructed in the tunnel. Deck A is along a horizontal diameter and another parallel Deck B is 2 feet below Deck A. Because the space station is in a weightless environment, you can walk vertically upright along Deck A, or vertically upside down along Deck B. You have been assigned to paint “safety stripes” on each deck level, so that a 6 foot person can safely walk upright along either deck. Determine the width of the “safe walk zone” on each deck. 6.3. MULTIPART FUNCTIONS Solution. Impose a coordinate system so that the origin is at the center of the circular cross section of the tunnel; by symmetry the walkway is centered about the origin. With this coordinate system, the graph of the equation x2 + y2 = 152 = 225 will be the circular cross-section of the tunnel. In the case of Deck A, we basically need to determine how close to each edge of the tunnel a 6 foot high person can stand without hitting his or her head on the tunnel; a similar remark applies to Deck B. This means we are really trying to fit two six-foot-high rectangular safe walk zones into the picture: Our job is to find the coordinates of the four points P, Q, R, and S. Let’s denote by x1 , x2 , x3 ,and x4 the xcoordinates of these four points, then P = (x1 , 6), Q = (x2 , 6), R = (x3 , −8), and S = (x4 , −8). To find x1 , x2 , x3 , and x4 , we need to find the intersection of the circle in Figure 6.10(b) with two horizontal lines: 79 Deck A x-axis Deck B r = 15ft (a) Cross-section of tunnel. Deck A Deck B = Safe walk zone (b) Walk zones. Figure 6.10: Space station tunnels. • Intersecting the the upper semicircle with the horizontal line having equation y = 6 will determine x1 and x2 ; the upper semicircle is the graph of f(x) = √ 225 − x2 . • Intersecting the lower semicircle with the horizontal line having equation y = −8 will determine x3 and x4 ; the lower semicircle is the graph of g(x) = √ − 225 − x2 . For Deck A, we simultaneously solve the system of equations √ y = 225 − x2 . y =6 Plugging in y = 6 into the first equation of the system gives x2 = 225 − √ 62 = 189; i.e., x = ± 189 = ±13.75. This tells us that P = (−13.75, 6) and Q = (13.75, 6). In a similar way, for Deck B, we find R = (−12.69, −8) and S = (12.69, −8). In the case of Deck A, we would paint a safety stripe 13.75 feet to the right and left of the centerline. In the case of Deck B, we would paint a safety stripe 12.69 feet to the right and left of the centerline. 6.3 Multipart Functions So far, in all of our examples we have been able to write f(x) as a nice compact expression in the variable x. Sometimes we have to work harder. CHAPTER 6. GRAPHICAL ANALYSIS 80 As an example of what we have in mind, consider the graph in Figure 6.11(a): y-axis 1 1 2 3 4 x-axis −1 (a) Graphing a multipart function. −1 1 −1 f(x) = 1 −1 if if if if if 0≤x<1 1≤x<2 2≤x<3 3≤x<4 x=4 (b) Writing a multipart function. Figure 6.11: A multipart function. The curve we are trying to describe in this picture is made up of five pieces; four little line segments and a single point. The first thing to notice is that on the domain 0 ≤ x ≤ 4, this curve will define the graph of some function f(x). To see why this is true, imagine a vertical line moving from left to right within the domain 0 ≤ x ≤ 4 on the x-axis; any one of these vertical lines will intersect the curve exactly once, so by the vertical line test, the curve must be the graph of a function. Mathematicians use the shorthand notation above to describe this function. Notice how the rule for f(x) involves five cases; each of these cases corresponds to one of the five pieces that make up the curve. Finally, notice the care with the “open” and “closed” circles is really needed if we want to make sure the curve defines a function; in terms of the rule, these open and closed circles translate into strict inequalities like < or weak inequalities like ≤. This is an example of what we call a multipart function. The symbolic appearance of multipart functions can be somewhat frightening. The key point is that the graph (and rule) of the function will be broken up into a number of separate cases. To study the graph or rule, we simply “home in” on the appropriate case. For example, in the above illustration, suppose we wanted to compute f(3.56). First, we would find which of the five cases covers x = 3.56, then apply that part of the rule to compute f(3.56) = 1. Our first multipart function example illustrated how to go from a graph in the plane to a rule for f(x); we can reverse this process and go from the rule to the graph. Example 6.3.1. Sketch the graph of the multipart function if x ≤ −1 1 √ 2 if −1 ≤ x ≤ 1 g(x) = 1+ 1−x 1 if x ≥ 1 6.3. MULTIPART FUNCTIONS 81 second part y-axis Solution. The graph of g(x) will consist of three pieces. of graph The first case consists of the graph of the function y = g(x) = 1 on the domain x ≤ −1, this consists of all points x-axis on the horizontal line y = 1 to the left of and including the third part of graph first part of graph point (−1,1). We have “lassoed” this portion of the graph = graph of g(x) in Figure 6.12. Likewise, the third case in the definition yields the graph of the function y = g(x) = 1 on the domain Figure 6.12: Multipart funcx ≥ 1; this is just all points on the horizontal line y = 1 tion g(x). to the right of and including the point (1, 1). Finally, we need to analyze the middle case, which means we need to √ look at the graph of 1 + 1 − x2 on the domain −1 ≤ x ≤ 1. This is just the upper semicircle of the circle of radius 1 centered at (0,1). If we paste these three pieces together, we arrive at the graph of g(x). 5 4 feet Example 6.3.2. You are dribbling a basketball and the function s = h(t) keeps track of the height of the ball’s center above the floor after t seconds. Sketch a reasonable graph of s = h(t). 3 2 1 1 seconds 2 Solution. If we take the domain to be 0 ≤ t ≤ 2 (the first 2 Figure 6.13: Dribbling. seconds), a reasonable graph might look like Figure 6.13. This is a multipart function. Three portions of the graph are decreasing and two portions are increasing. Why doesn’t the graph touch the t axis? CHAPTER 6. GRAPHICAL ANALYSIS 82 6.4 Exercises Problem 6.1. The absolute value function is defined by the multipart rule: |x| = x −x Problem 6.4. (a) Let f(x) = x + |2x − 1|. Find all solutions to the equation if 0 ≤ x if x < 0 f(x) = 8. The graph of the absolute value function is pictured below: (b) Let g(x) = 3x − 3 + |x + 5|. Find all values of a which satisfy the equation y-axis g(a) = 2a + 8. y = |x| (c) Let h(x) = |x| − 3x + 4. Find all solutions to the equation h(x − 1) = x − 2. x-axis Problem 6.5. Express the area of the shaded region below as a function of x. The dimensions in the figure are centimeters. (a) Calculate: |0|, |2|, | − 3|. (b) Solve for x: |x| = 4; |x| = 0, |x| = −1. (c) Sketch the graph of y = 1 x + 2 and y = |x| 2 in the same coordinate system. Find where the two graphs intersect, label the coordinates of these point(s), then find the area of the region bounded by the two graphs. 5 Problem 6.2. For each of the following functions, graph f(x) and g(x) = |f(x)|, and give the multipart rule for g(x). 3 (a) f(x) = −0.5x − 1 (b) f(x) = 2x − 5 x (c) f(x) = x + 3 6 Problem 6.3. Solve each of the following equations for x. (a) g(x) = 17, where g(x) = |3x + 5| (b) f(x) = 1.5 where 2x 4−x f(x) = if x < 3, if x ≥ 3. (c) h(x) = −1 where h(x) = −8 − 4x 1 + 1x 3 if x ≤ −2, if x > −2. Problem 6.6. Pizzeria Buonapetito makes a triangular-shaped pizza with base width of 30 inches and height 20 inches as shown. Alice wants only a portion of the pizza and does so by making a vertical cut through the pizza and taking the shaded portion. Letting x be the bottom length of Alice’s portion and y be the length of the cut as shown, answer the following questions: 6.4. EXERCISES 83 20 y x 10 20 20 y x Problem 6.8. Arthur is going for a run. From his starting point, he runs due east at 10 feet per second for 250 feet. He then turns, and runs north at 12 feet per second for 400 feet. He then turns, and runs west at 9 feet per second for 90 feet. Express the (straight-line) distance from Arthur to his starting point as a function of t, the number of seconds since he started. Problem 6.9. A baseball diamond is a square with sides of length 90 ft. Assume Edgar hits a home run and races around the bases (counterclockwise) at a speed of 18 ft/sec. Express the distance between Edgar and home plate as a function of time t. (Hint: This will be a multipart function.) Try to sketch a graph of this function. (a) Find a formula for y as a multipart function of x, for 0 ≤ x ≤ 30. Sketch the graph of this function and calculate the range. (b) Find a formula for the area of Alice’s portion as a multipart function of x, for 0 ≤ x ≤ 30. (c) If Alice wants her portion to have half the area of the pizza, where should she make the cut? Problem 6.7. This problem deals with cars traveling between Bellevue and Spokane, which are 280 miles apart. Let t be the time in hours, measured from 12:00 noon; for example, t = −1 is 11:00 am. (a) Joan drives from Bellevue to Spokane at a constant speed, departing from Bellevue at 11:00 am and arriving in Spokane at 3:30 pm. Find a function j(t) that computes her distance from Bellevue at time t. Sketch the graph, specify the domain and determine the range. (b) Steve drives from Spokane to Bellevue at 70 mph, departing from Spokane at 12:00 noon. Find a function s(t) for his distance from Bellevue at time t. Sketch the graph, specify the domain and determine the range. (c) Find a function d(t) that computes the distance between Joan and Steve at time t. 90 ft Edgar d(t) home plate Problem 6.10. Pagliacci Pizza has designed a cardboard delivery box from a single piece of cardboard, as pictured. (a) Find a polynomial function v(x) that computes the volume of the box in terms of x. What is the degree of v? (b) Find a polynomial function a(x) that computes the exposed surface area of the closed box in terms of x. What is the degree of a? What are the explicit dimensions if the exposed surface area of the closed box is 600 sq. inches? CHAPTER 6. GRAPHICAL ANALYSIS 84 (a) When will the ditch be completely full? x (b) Find a multipart function that models the vertical cross-section of the ditch. x 20 in (c) What is the width of the filled portion of the ditch after 1 hour and 18 minutes? (d) When will the filled portion of the ditch be 42 feet wide? 50 feet wide? 73 feet wide? x x 50 in remove shaded squares and fold to get: Problem 6.12. The graph of a function y = g(x) on the domain −6 ≤ x ≤ 6 consistes of line segments and semicircles of radius 2 connecting the points (−6,0),(−4,4), (0,4), (4,4), (6,0). y-axis x-axis Problem 6.11. The vertical cross-section of a drainage ditch is pictured below: (a) What is the range of g? (b) Where is the function increasing? Where is the function decreasing? (c) Find the multipart formula for y = g(x). (d) If we restrict the function to the smaller domain −5 ≤ x ≤ 0, what is the range? (e) If we restrict the function to the smaller domain 0 ≤ x ≤ 4, what is the range? 3D−view of ditch 20 ft 20 ft R R R R vertical cross-section Here, R indicates a circle of radius 10 feet and all of the indicated circle centers lie along the common horizontal line 10 feet above and parallel to the ditch bottom. Assume that water is flowing into the ditch so that the level above the bottom is rising 2 inches per minute. (a) Simply as far as possible Problem 6.13. 1 1+ 1 a − a . a+1 (b) Find a, b, c that simultaneously satisfy these three equations: a+b−c = 2a − 3b + c = a+b+c = 5 4 −1 Chapter 7 Quadratic Modeling If you kick a ball through the air enough times, you will find its path tends to be parabolic. Before we can answer any detailed questions about this situation, we need to get our hands on a precise mathematical model for a parabolic shaped curve. This means we seek a function y = f(x) whose graph reproduces the path of the ball. ground level Figure 7.1: Possible paths for a kicked ball are parabolic. 7.1 Parabolas and Vertex Form OK, suppose we sit down with an xy-coordinate system and draw four random parabolas; let’s label them I, II, III, and IV: See Figure 7.2. The relationship between these parabolas and the fixed coordinate system can vary quite a bit: The key distinction between these four curves is that only I and IV are the graphs of functions; this follows from the vertical line test. A parabola that is the graph of a function is called a standard parabola. We can see that any standard parabola has three basic features: 85 CHAPTER 7. QUADRATIC MODELING 86 y-axis II III I IV x-axis not graphs of functions Figure 7.2: Relationship between a fixed coordinate system and various parabolas. • the parabola will either open “upward” or “downward”; • the graph will have either a “highest point” or “lowest point,” called the vertex ; • the parabola will be symmetric about some vertical line called the axis of symmetry. Our first task is to describe the mathematical model for any standard parabola. In other words, what kind of function equations y = f(x) give us standard parabolas as their graphs? Our approach is geometric and visual: • Begin with one specific example, then show every other standard parabola can be obtained from it via some specific geometric maneuvers. • As we perform these geometric maneuvers, we keep track of how the function equation for the curve is changing. 35 30 25 20 15 10 5 −6 −4 −2 This discussion will amount to a concrete application of a more general set of tools developed in the following section of this chapter. Using a graphing device, it is an easy matter to plot y-axis the graph of y = x2 and see we are getting the parabola pictured in Figure 7.3. The basic idea is to describe how we can manipulate this graph and obtain any standard parabola. In the end, we will see that standard parabolas x-axis are obtained as the graphs of functions having the form 2 4 6 Figure 7.3: Graph of y = x2 . y = ax2 + bx + c, 7.1. PARABOLAS AND VERTEX FORM 87 for various constants a, b, and c, with a = 0. A function of this type is called a quadratic function and these play a central role throughout the course. We will divide our task into two steps. First we show every standard parabola arises as the graph of a function having the form y = a(x − h)2 + k, for some constants a, h, and k, with a = 0. This is called the vertex form of a quadratic function. Notice, if we were to algebraically expand out this equation, we could rewrite it in the y = ax2 + bx + c form. For example, suppose we start with the vertex form y = 2(x − 1)2 + 3, so that a = 2, h = 1, k = 3. Then we can rewrite the equation in the form y = ax2 + bx + c as follows: 2(x − 1)2 + 3 = 2(x2 − 2x + 1) + 3 = 2x2 − 4x + 5, so a = 2, b = −4, c = 5. The second step is to show any quadratic function can be written in vertex form; the underlying algebraic technique used here is called completing the square. This is a bit more involved. For example, if you are simply handed the quadratic function y = −3x2 + 6x − 1, it not at all obvious why the vertex form is obtained by this equality: −3x2 + 6x − 1 = −3(x − 1)2 + 2. The reason behind this equality is the technique of completing the square. In the end, we will almost always be interested in the vertex form of a quadratic. This is because a great deal of qualitative information about the parabolic graph can simply be “read off” from this form. 7.1.1 First Maneuver: Shifting Suppose we start with the graph in Figure 7.3 and horiy-axis 40 zontally shift it h units to the right. To be specific, con30 sider the two cases h = 2 and h = 4. To visualize this, 20 imagine making a wire model of the graph, set in on top of the curve, then slide the wire model h units to the right. 10 What you will obtain are the two “dashed curves” in Figx-axis −6 −4 −2 2 4 6 ure 7.4. We will call the process just described a horizontal shift. Since the “dashed curves” are no longer the Figure 7.4: Shift to the right. original parabola in Figure 7.3, the corresponding function equations must have changed. Using a graphing device, you can check that the corresponding equations for the dashed graphs would be y = (x − 2)2 = x2 − 4x + 4, CHAPTER 7. QUADRATIC MODELING 88 which is the plot with lowest point (2, 0) and y = (x − 4)2 = x2 − 8x + 16, 40 30 20 10 −6 −4 −2 which is the plot with lowest point (4, 0). In general, if h is positive, the graph of the function y = (x − h)2 is the parabola obtained by shifting the graph of y = x2 by h units to the right. Next, if h is negative, shifting h units to the right is the y-axis same as shifting |h| units left! On the domain −6 ≤ x ≤ 6, Figure 7.5 indicates this for the cases h = −2, − 4, using “dashed curves” for the shifted graphs and a solid line for the graph of y = x2 . Using a graphing device, we can check that the corresponding equations for the dashed x-axis graphs would be 2 4 6 Figure 7.5: Shift to the left. y = (x − (−2))2 = (x + 2)2 = x2 + 4x + 4, which is the plot with lowest point (−2, 0) and y = (x − (−4))2 = (x + 4)2 = x2 + 8x + 16, which is the plot with lowest point (−4, 0). In general, if h is negative, the graph of the function y = (x − h)2 gives the parabola obtained by shifting the graph of y = x2 by |h| units to the LEFT. The conclusion thus far is this: Begin with the graph of y = x2 in Figure 7.3. Horizontally shifting this graph h units to the right gives a new (standard) parabola whose equation is y = (x − h)2 . We can also imagine vertically shifting the graph in y-axis Figure 7.3. This amounts to moving the graph k units 40 30 vertically upward. It turns out that this vertically shifted 20 graph corresponds to the graph of the function y = x2 + k. 10 We can work out a few special cases and use a graphing x-axis −4 −6 4 6 −2 2 device to illustrate what all this really means. −10 Figure 7.6 illustrates the graphs of y = x2 + k in the cases when k = 4, 10 and k = −4, −10, leading to vertically Figure 7.6: Vertical shifts. shifted graphs. Positive values of k lead to the upper two “dashed curves” and negative values of k lead to the lower two “dashed curves”; the plot of y = x2 is again the solid line. The equations giving these graphs would be y = x2 − 10, y = x2 − 4, y = x2 + 4 and y = x2 + 10, from bottom to top dashed plot. If we combine horizontal and vertical shifting, we end up with the graphs of functions of the form y = (x − h)2 + k. Figure 7.7(a) illustrates 7.1. PARABOLAS AND VERTEX FORM 89 the four cases with corresponding equations y = (x±2)2 ±4; as an exercise, identify which equation goes with each curve. 7.1.2 Second Maneuver: Reflection Next, we can reflect any of the curves y = p(x) obtained by horizontal or vertical shifting across the x-axis. This procedure will produce a new curve which is the graph of the new function y = −p(x). For example, begin with the four dashed curves in the previous figure. Here are the reflected parabolas and their equations are y = −(x ± 2)2 ± 4: See Figure 7.7(b). y-axis 40 30 20 10 x-axis −4 −6 −2 2 4 6 (a) Combined shifts. 7.1.3 Third Maneuver: Vertical Dilation y-axis 30 2 If a is a positive number, the graph of y = ax is usually called a vertical dilation of the graph of y = x2 . There are two cases to distinguish here: 20 10 x-axis −6 −4 −2 2 4 6 −10 −20 • If a > 1, we have a vertically expanded graph. −30 • If 0 < a < 1, we have a vertically compressed graph. (b) Reflections. This is illustrated for a = 2 (upper dashed plot) and a = 1/2 (lower dashed plot): See Figure 7.7(c). y-axis 30 25 20 7.1.4 Conclusion 15 10 Starting with y = x2 in Figure 7.3, we can combine together all three of the operations: shifting, reflection and dilation. This will lead to the graphs of functions that have the form: y = a(x − h)2 + k, 5 x-axis −6 −4 −2 2 4 6 (c) Vertical dilations. Figure 7.7: Shifts, reflections, and dilations. for some a, h and k, a = 0. If you think about it for awhile, it seems pretty easy to believe that any standard parabola arises from the one in Figure 7.3 using our three geometric maneuvers. In other words, what we have shown is that any standard parabola is the graph of a quadratic equation in vertex form. Let’s summarize. Important Fact 7.1.1. A standard parabola is the graph of a function y = f(x) = a(x − h)2 + k, for some constants a, h, and k and a = 0. The vertex of the parabola is (h, k) and the axis of symmetry is the line x = h. If a > 0, then the parabola opens upward; if a < 0, then the parabola opens downward. CHAPTER 7. QUADRATIC MODELING 90 Example 7.1.2. Describe a sequence of geometric operations leading from the graph of y = x2 to the graph of y = f(x) = −3(x − 1)2 + 2. reflect across x-axis horizontal shift by h = 1 −3(x − 1)2 + 2 vertical shift by k = 2 vertical dilate by 3 (a) What do the symbols of an equation mean? 20 y-axis 15 10 5 −6 −4 −2 x-axis 2 4 6 −5 −10 −15 (b) What does the equation look like? Figure 7.8: Interpreting an equation. Solution. To begin with, we can make some initial conclusions about the specific shifts, reflections and dilations involved, based on looking at the vertex form of the equation. In addition, by Fact 7.1.1, we know that the vertex of the graph of y = f(x) is (1, 2), the line x = 1 is a vertical axis of symmetry and the parabola opens downward. We need to be a little careful about the order in which we apply the four operations highlighted. We will illustrate a procedure that works. The full explanation for the success of our procedure involves function compositions and we will return to that at the end of Chapter 8. The order in which we will apply our geometric maneuvers is as follows: horizontal shift ⇒ vertical dilate ⇒ reflect ⇒ vertical shift Figure 7.8(b) illustrates the four curves obtained by applying these successive steps, in this order. As a reference, we include the graph of y = x2 as a “dashed curve”: • A horizontal shift by h = 1 yields the graph of y = (x − 1)2 ; this is the fat parabola opening upward with vertex (1, 0). • A dilation by a = 3 yields the graph of y = 3(x − 1)2 ; this is the skinny parabola opening upward with vertex (1, 0). • A reflection yields the graph of y = −3(x − 1)2 ; this is the downward opening parabola with vertex (1, 0). • A vertical shift by k = 2 yields the graph of y = −3(x − 1)2 + 2; this is the downward opening parabola with vertex (1, 2). 7.2 Completing the Square By now it is pretty clear we can say a lot about the graph of a quadratic function which is in vertex form. We need a procedure for rewriting a given quadratic function in vertex form. Let’s first look at an example. 7.2. COMPLETING THE SQUARE 91 Example 7.2.1. Find the vertex form of the quadratic function y = −3x2 + 6x − 1. Solution. Since our goal is to put the function in vertex form, we can write down what this means, then try to solve for the unknown constants. Our first step would be to write −3x2 + 6x − 1 = a(x − h)2 + k, for some constants a, h, k. Now, expand the right hand side of this equation and factor out coefficients of x and x2 : −3x2 + 6x − 1 −3x2 + 6x 1 −3x2 + 6x − 1 (−3)x2 + (6)x + (−1) = = = = a ( x − h) 2 + k a(x2 − 2xh + h2 ) + k ax2 − 2xah + ah2 + k (a)x2 + (−2ah)x + (ah2 + k). If this is an equation, then it must be the case that the coefficients of like powers of x match up on the two sides of the equation in Figure 7.9. Now we have three equations and three unknowns (the a, h, k) and we Equal (−3) x2 + (6) x + (−1) = (a) x2 + (−2ah) x + (ah2 + k) Equal Equal Figure 7.9: Balancing the coefficients. can proceed to solve for these: −3 = a 6 = −2ah −1 = ah2 + k The first equation just hands us the value of a = −3. Next, we can plug this value of a into the second equation, giving us 6 = −2ah = −2(−3)h = 6h, CHAPTER 7. QUADRATIC MODELING 92 so h = 1. Finally, plug the now known values of a and h into the third equation: −1 = ah2 + k = −3(12 ) + k = −3 + k, so k = 2. Our conclusion is then −3x2 + 6x − 1 = −3(x − 1)2 + 2. Notice, this is the quadratic we studied in Example 7.1.2 on page 90. The procedure used in the preceding example will always work to rewrite a quadratic function in vertex form. We refer to this as completing the square. Example 7.2.2. Describe the relationship between the graphs of y = x2 and y = f(x) = −4x2 + 5x + 2. 20 y-axis 10 x-axis −6 −4 −2 2 4 6 −10 −20 Solution. We will go through the algebra to complete the square, then interpret what this all means in terms of graphical maneuvers. We have −4x2 + 5x + 2 = a(x − h)2 + k (−4)x2 + 5x + 2 = ax2 + (−2ah)x + (ah2 + k). This gives us three equations: Figure 7.10: Maneuvering y = x2 . −4 = a 5 = −2ah 2 = ah2 + k. 57 We conclude that a = −4, h = 5 = 0.625 and k = 16 = 3.562. So, this tells 8 us that we can obtain the graph of y = f(x) from that of y = x2 by these steps: • Horizontally shifting by h = 0.625 units gives y = (x − 0.625)2. • Vertically dilate by the factor a = 4 gives y = 4(x − 0.625)2. • Reflecting across the x-axis gives y = −4(x − 0.625)2. • Vertically shifting by k = 3.562 gives y = f(x) = −4(x − 0.625)2 + 3.562. Example 7.2.3. A drainage canal has a cross-section in the shape of a parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the top. If the water depth in the ditch is 5 feet, how wide is the surface of the water in the ditch? 7.3. INTERPRETING THE VERTEX 93 20 feet centerline Solution. Impose an xy-coordinate system so that the parabolic cross-section of the canal is symmetric about 10 feet the y-axis and its vertex is the origin. The vertex form of any such parabola is y = f(x) = ax2 , for some a > 0; this is because (h, k) = (0, 0) is the vertex and the parabola Figure 7.11: A drainage opens upward! The dimension information given tells us canal. that the points (10, 10) and (−10, 10) are on the graph of f(x). Plugging into the expression for f, we conclude that 10 = 100a, so a = 0.1 and f(x) = (0.1)x2. Finally, if the water is 5 feet deep, we must √ solve the equation: 5 = (0.1)x2, leading to x = ± 50 = ±7.07. Conclude the surface of the water is 14.14 feet wide when the water is 5 feet deep. 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 7.3 Interpreting the Vertex minimum −b value f 2a maximum −b value f 2a vertex vertex −b 2a −b 2a Figure 7.12: The vertex as the extremum of the quadratic function. If we begin with a quadratic function y = f(x) = ax2 +bx+c, we know the graph will be a parabola. Graphically, the vertex will correspond to either the “highest point” or “lowest point” on the graph. If a > 0, the vertex is the lowest point on the graph; if a < 0, the vertex is the highest point on the graph. The maximum or minimum value of the function is the second coordinate of the vertex and the value of the variable x for which this extreme value is achieved is the first coordinate of the vertex. As we know, it is easy to read off the vertex coordinates when a quadratic function is written in vertex form. If instead we are given a quadratic function y = ax2 + bx + c, we can use the technique of completing the square and arrive at a formula for the coordinates of the vertex in terms of a, b, and c. We summarize this below and label the two situations (upward or downward opening parabola) in the Figure 7.12. Keep in CHAPTER 7. QUADRATIC MODELING 94 mind, it is always possible to obtain this formula by simply completing the square. Important Fact 7.3.1. In applications involving a quadratic function f(x) = ax2 + bx + c, the vertex has coordinates P = −b , f −b . The second coordinate of the 2a 2a vertex will detect the maximum or minimum value of f(x); this is often a key step in problem solving. Example 7.3.2. Discuss the graph of the quadratic function y = f(x) = −2x2 + 11x − 4. −4 2 −2 4 6 8 −20 Solution. We need to place the equation y = f(x) in vertex form. We can simply compute a = −2, h = −b = 11 and 2a 4 k = f( 11 ) = 89 , using Fact 7.3.1: 4 8 f(x) = −2x2 + 11x − 4 −40 −60 = −2 x − −80 Figure 7.13: Sketching y = f(x). 11 4 2 + 89 . 8 This means that the graph of f(x) is a parabola opening downward with vertex 141 , 89 and axis x = 11 ; see Fig8 4 ure 7.13. 7.4 Quadratic Modeling Problems The real importance of quadratic functions stems from the connection with motion problems. Imagine one of the three kicked ball scenarios in Figure 7.1 and impose a coordinate system with the kicker located at the origin. We can study the motion of the ball in two ways: • Regard time t as the important variable and try to find a function y(t) which describes the height of the ball t seconds after the ball is kicked; this would just be the y-coordinate of the ball at time t. The function y(t) is a quadratic function. If we had this function in hand, we could determine when the ball hits the ground by solving the equation 0 = y(t), but we would not be able to determine where the ball hits the ground. • A second approach is to forget about the time variable and simply try to find a function y = f(x) whose graph models the exact path of the ball. In particular, we could find where the ball hits the ground by solving 0 = f(x), but we would not be able to determine when the ball hits the ground. 7.4. QUADRATIC MODELING PROBLEMS 95 Example 7.4.1. Figure 7.14(a) shows a ball is located on the edge of a cliff. The ball is kicked and its height (in feet) above the level ground is given by the function s = y(t) = −16t2 + 48t + 50, where t represents seconds elapsed after kicking the ball. What is the maximum height of the ball and when is this height achieved? When does the ball hit the ground? How high is the cliff? Solution. The function y(t) is a quadratic function with a negative leading coefficient, so its graph in the tscoordinate system will be a downward opening parabola. We use a graphing device to get the picture in Figure 7.14(b). The vertex is the highest point on the graph, which can be found by writing y(t) in standard form using Fact 7.3.1: path of kicked ball cliff ground level (a) What it looks like physically. 50 −2 −1 1 2 3 4 5 −50 −100 (b) What it looks like graphically. y(t) = −16t2 + 48t + 50 3 = −16 t − 2 2 + 86. Figure 7.14: Different views of the ball’s trajectory. 3 The vertex of the graph of y(t) is 2 , 86 , so the maximum height of the ball above the level ground is 86 feet, occur3 ing at time t = 2 . The ball hits the ground when its height above the ground is zero; using the quadratic formula: y(t) = −16t2 + 48t + 50 =0 −48 ± (48)2 − 4 (−16) (50) t= 2 · 16 = 3.818 sec or − 0.818 sec Conclude the ball hits the ground after 3.818 seconds. Finally, the height of the cliff is the height of the ball zero seconds after release; i.e., y(0) = 50 feet is the height of the cliff. Here are two items to consider carefully: 2. The function y(t) is defined for all t; however, in the context of the problem, there is no physical meaning when t < 0. !!! 1. The graph of y(t) is NOT the path followed by the ball! Finding the actual path of the ball is not possible unless additional information is given. Can you see why? CAUTION !!! CHAPTER 7. QUADRATIC MODELING 96 The next example illustrates how we must be very careful to link the question being asked with an appropriate function. Example 7.4.2. A hot air balloon takes off from the edge of a mountain lake. Impose a coordinate system as pictured in Figure 7.15 and assume 2 that the path of the balloon follows the graph of y = f(x) = − 2500 x2 + 4 x. The 5 land rises at a constant incline from the lake at the rate of 2 vertical feet for each 20 horizontal feet. What is the maximum height of the balloon above lake level? What is the maximum height of the balloon above ground level? Where does the balloon land on the ground? Where is the balloon 50 feet above the ground? Solution. In the coordinate system indicated, the origin is the takeoff point and the graph of y = f(x) is the path of balloon path 200 the balloon. Since f(x) is a quadratic function with a negative leading coefficient, its graph will be a parabola which 100 opens downward. The difficulty with this problem is that at any instant during the balloon’s flight, the “height of (ft) lake 500 1000 the balloon above the ground” and the “height of the balFigure 7.15: Visualizing. loon above the lake level” are different! The picture in Figure 7.16 highlights this difference; consequently, two different functions will be needed to study these two different quantities. height above lake (ft) balloon height of balloon above lake level 200 A A 100 B lake B 500 height of balloon above ground level height of ground above lake level 1000 Figure 7.16: The height of the balloon y as a function of x. The function y = f(x) keeps track of the height of the balloon above lake level at a given x location on the horizontal axis. The line ℓ with slope 1 2 m = 20 = 10 passing through the origin models the ground level. This says that the function 1 y= x 10 keeps track of the height of the ground above lake level at a given x location on the horizontal axis. We can determine the maximum height of the balloon above lake level by analyzing the parabolic graph of y = f(x). Putting f(x) in vertex form, via Fact 7.3.1, 2 (x − 500)2 + 200. f(x) = − 2500 7.4. QUADRATIC MODELING PROBLEMS 97 The vertex of the graph of y = f(x) is (500, 200). This just tells us that the maximum height of the balloon above lake level is 200 feet. To find the landing point, we need to solve the system of equations 2 4 y = − 2500 x2 + 5 x 1 y = 10 x . As usual, plugging the second equation into the first and solving for x, we get 1 x 10 x2 x2 − 875x x(x − 875) 224 x+ x 2500 5 = 875x =0 =0 =− height above lake level (feet) vertex (high point above lake) 200 100 landing point horizontal distance from launch (feet) takeoff point 500 1000 Figure 7.17: Locating the takeoff and landing points. From the algebra, we see there are two solutions: x = 0 or x = 875; these correspond to the takeoff and landing points of the balloon, which are the two places the flight path and ground coincide. (Notice, if we had divided out x from the last equation, we would only get one solution; the tricky point is that we can’t divide by zero!) The balloon lands at the position where x = 875 and to find the y coordinate of this landing point we plug x = 875 into our function for the balloon height above lake level: y = f(875) = 87.5 feet. So, the landing point has coordinates (875,87.5). Next, we want to study the height of the balloon above the ground. Let y = g(x) be the function which represents the height of the balloon above CHAPTER 7. QUADRATIC MODELING 98 the ground when the horizontal coordinate is x. We find height of the balloon above lake level with horizontal coordinate x g(x) = − elevation of ground above lake level with horizontal coordinate x = f(x) − g(x) = − 224 x+ x 2500 5 (balloon above lake level) =− 1 x 10 − minus (ground above lake level) 2 (x − 437.5)2 + 153.12, 2500 Notice that g(x) itself is a NEW quadratic function with a negative leading coefficient, so the graph of y = g(x) will be a downward opening parabola. The vertex of this parabola will be (437.5, 153.12), so the highest elevation of the balloon above the ground is 153.12 feet. We can now sketch the graph of g(x) and the horizontal line determined by y = 50 in a common coordinate system, as below. Finding where the balloon is 50 feet above the ground amounts to finding where these two graphs intersect. We need to now solve the system of equations 2 y = − 2500 x2 + y = 50 7 x 10 . Plug the second equation into the first and apply the quadratic formula to get x = 796.54 or 78.46. This tells us the two possible x coordinates when the balloon is 50 feet above the ground. In terms of the original coordinate system imposed, the two places where the balloon is 50 feet above the ground are (78.46, 57.85) and (796.54, 129.6). 7.4.1 How many points determine a parabola? We all recall from elementary geometry that two distinct points in the plane will uniquely determine a line; in fact, we used this to derive equations for lines in the plane. We could then ask if there is a similar characterization of parabolas. Important Fact 7.4.3. Let P = (x1 , y1), Q = (x2 , y2 ) and R = (x3 , y3 ) be three distinct non-collinear points in the plane such that the x-coordinates are all different. Then there exists a unique standard parabola passing through these three points. This parabola is the graph of a quadratic function y = f(x) = ax2 + bx + c and we can find these coefficients by simultaneously 7.4. QUADRATIC MODELING PROBLEMS 99 vertex (high point above ground) graph of height above ground function 100 line y = 50 500 1000 places where 50 feet above ground Figure 7.18: Finding heights above the ground. solving the system of three equations and three unknowns obtained by assuming P, Q and R are points on the graph of y = f(x): ax2 + bx1 + c = y1 1 2 ax2 + bx2 + c = y2 . ax2 + bx3 + c = y3 . 3 Example 7.4.4. Assume the value of a particular house in Seattle has increased in value according to a quadratic function y = v(x), where the units of y are in dollars and x represents the number of years the property has been owned. Suppose the house was purchased on January 1, 1970 and valued at $50,000. In 1980, the value of the house on January 1 was $80,000. Finally, on January 1, 1990 the value was $200,000. Find the value function v(x), determine the value on January 1, 1996 and find when the house will be valued at $1,000,000. Solution. The goal is to explicitly find the value of the function y = v(x). We are going to work in a xy-coordinate system in which the first coordinate of any point represents time and the second coordinate represents value. We need to decide what kind of units will be used. The x-variable, which represents time, will denote the number of years the house is owned. For the y-variable, which represents value, we could use dollars. But, instead, we will follow a typical practice in real estate and use the units of K, where K = $1,000. For example, a house valued CHAPTER 7. QUADRATIC MODELING 100 at $235,600 would be worth 235.6 K. These will be the units we use, which essentially saves us from drowning in a sea of zeros! We are given three pieces of information about the value of a particular house. This leads to three points in our coordinate system: P = (0, 50), Q = (10, 80) and R = (20, 200). If we plot these points, they do not lie on a common line, so we know there is a unique quadratic function v(x) = ax2 + bx + c whose graph (which will be a parabola) passes through these three points. In order to find the coefficients a, b, and c, we need to solve the system of equations: = 50 a02 + b0 + c a(10)2 + b(10) + c = 80 , a(20)2 + b(20) + c = 200 which is equivalent to c 100a + 10b + c 400a + 20b + c the system = 50 = 80 . = 200 Plugging c = 50 into the second two equations gives the system 100a + 10b = 30 400a + 20b = 150 (7.1) Solve the first equation for a, obtaining a = the second equation to get: 400 30−10b , 100 then plug this into 30 − 10b + 20b = 150 100 120 − 40b + 20b = 150 3 b=− . 2 3 Now, plug b = − 2 into the first equation of Equation 7.1 to get 100a + 9 10 − 3 = 30; i.e., a = 20 . We conclude that 2 92 3 x − x + 50, 20 2 keeping in mind the units here are K. To find the value of the house on January 1, 1996, we simply note this 9 is after x = 26 years of ownership. Plugging in, we get y = v(26) = 20 (26)2 − 3 26 + 50 = 315.2; i.e., the value of the house is $315,200. To find when 2 the house will be worth $1,000,000, we note that $1,000,000 = 1,000 K and need to solve the equation y = v(x) = 92 x− 20 92 x− 0= 20 1000 = v(x) = 3 x + 50 2 3 x − 950. 2 7.5. WHAT’S NEEDED TO BUILD A QUADRATIC MODEL? 101 By the quadratic formula, x= 3 2 −3 2 ± √ 2 −4 2 9 20 9 20 (−950) 1712.25 0.9 = 47.64 or − 44.31. = 1.5 ± Because x represents time, we can ignore the negative solution and so the value of the house will be $1,000,000 after approximately 47.64 years of ownership. 7.5 What’s Needed to Build a Quadratic Model? Back in Fact 4.7.1 on page 44, we highlighted the information required to determine a linear model. We now describe the quadratic model analog. Important Facts 7.5.1. A quadratic model is completely determined by: 1. Three distinct non-collinear points, or 2. The vertex and one other point on the graph. The first approach is just Fact 7.4.3. The second approach is based on the vertex form of a quadratic function. The idea is that we know any quadratic function f(x) has the form f(x) = a(x − h)2 + k, where (h, k) is the vertex. If we are given h and k, together with another point (x0 , y0 ) on the graph, then plugging in gives this equation: y0 = a(x0 − h)2 + k. The only unknown in this equation is a, which we can solve for using algebra. A couple of the exercises will depend upon these observations. 7.6 Summary • A quadratic function is one of the form f(x) = ax2 + bx + c. where a = 0. CHAPTER 7. QUADRATIC MODELING 102 • The graph of a quadratic function is a parabola which is symmetric about the vertical line through the highest (or lowest) point on the graph. This highest (or lowest) point is known as the vertex of the graph; its location is given by (h,k) where h=− b and k = f(h). 2a • If a > 0, then the vertex is the lowest (or minimum) point on the graph, and the parabola ”opens upward”. If a < 0, then the vertex is the highest (or maximum) point on the graph, and the parabola ”opens downward”. • Every quadratic function can be expressed in the form f(x) = a(x − h)2 + k where (h,k) is the vertex of the function’s graph. 7.7. EXERCISES 103 7.7 Exercises Problem 7.1. Write the following quadratic functions in vertex form, find the vertex, the axis of symmetry and sketch a rough graph. (a) f(x) = 2x2 − 16x + 41. (b) f(x) = 3x2 − 15x − 77. (c) f(x) = x2 − 3 x + 13. 7 (d) f(x) = 2x2 . (e) f(x) = 1 2 100 x . Problem 7.2. In each case, find a quadratic function whose graph passes through the given points: (a) (0,0), (1,1) and (3, − 1). (b) (−1,1), (1, − 2) and (3,4). (c) (2,1), (3,2) and (5,1). (d) (0,1), (1,1) and (1,3). Problem 7.3. (a) Sketch the graph of the function f(x) = x2 − 3x + 4 on the interval −3 ≤ x ≤ 5. What is the maximum value of f(x) on that interval? What is the minimum value of f(x) on that interval? (b) Sketch the graph of the function f(x) = x2 − 3x + 4 on the interval 2 ≤ x ≤ 7. What is the maximum value of f(x) on that interval? What is the minimum value of f(x) on that interval? (c) Sketch the graph of the function g(x) = −(x + 3)2 + 3 on the interval 0 ≤ x ≤ 4. What is the maximum value of g(x) on that interval? What is the minimum value of g(x) on that interval? Problem 7.4. If the graph of the quadratic function f(x) = x2 + dx + 3d has its vertex on the x-axis, what are the possible values of d? What if f(x) = x2 + 3dx − d2 + 1 ? Problem 7.5. The initial price of buzz.com stock is $10 per share. After 20 days the stock price is $20 per share and after 40 days the price is $25 per share. Assume that while the price of the stock is not zero it can be modeled by a quadratic function. (a) Find the multipart function s(t) giving the stock price after t days. If you buy 1000 shares after 30 days, what is the cost? (b) To maximize profit, when should you sell shares? How much will the profit be on your 1000 shares purchased in (a)? Problem 7.6. Sketch the graph of y = x2 − 2x − 3. Label the coordinates of the x and y intercepts of the graph. In the same coordinate system, sketch the graph of y = |x2 − 2x − 3|, give the multipart rule and label the x and y intercepts of the graph. Problem 7.7. A hot air balloon takes off from the edge of a plateau. Impose a coordinate system as pictured below and assume that the path the balloon follows is the graph of the 4 quadratic function y = f(x) = − 2500 x2 + 4 x. 5 The land drops at a constant incline from the plateau at the rate of 1 vertical foot for each 5 horizontal feet. Answer the following questions: height above plateau (feet) balloon takeoff horizontal distance from launch (feet) ground incline (a) What is the maximum height of the balloon above plateau level? (b) What is the maximum height of the balloon above ground level? (c) Where does the balloon land on the ground? (d) Where is the balloon 50 feet above the ground? Problem 7.8. (a) Suppose f(x) = 3x2 − 2. Does the point (1,2) lie on the graph of y = f(x)? Why or why not? (b) If b is a constant, where does the line y = 1 + 2b intersect the graph of y = x2 + bx + b? 104 (c) If a is a constant, where does the line y = 1 − a2 intersect the graph of y = x2 − 2ax + 1? (d) Where does the graph of y = −2x2 + 3x + 10 intersect the graph of y = x2 + x − 10? Problem 7.9. Sylvia has an apple orchard. One season, her 100 trees yielded 140 apples per tree. She wants to increase her production by adding more trees to the orchard. However, she knows that for every 10 additional trees she plants, she will lose 4 apples per tree (i.e., the yield per tree will decrease by 4 apples). How many trees should she have in the orchard to maximize her production of apples? Problem 7.10. Rosalie is organizing a circus performance to raise money for a charity. She is trying to decide how much to charge for tickets. From past experience, she knows that the number of people who will attend is a linear function of the price per ticket. If she charges 5 dollars, 1200 people will attend. If she charges 7 dollars, 970 people will attend. How much should she charge per ticket to make the most money? Problem 7.11. A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 24 feet. What should its dimensions be in order to maximize the area of the window and, therefore, allow in as much light as possible? Problem 7.12. Jun has 300 meters of fencing to make a rectangular enclosure. She also wants to use some fencing to split the enclosure into two parts with a fence parallel to two of the sides. What dimensions should the enclosure have to have the maximum possible area? Problem 7.13. You have $6000 with which to build a rectangular enclosure with fencing. The fencing material costs $20 per meter. You also want to have two parititions across the width of the enclosure, so that there will be three separated spaces in the enclosure. The material for the partitions costs $15 per meter. What is the maximum area you can achieve for the enclosure? Problem 7.14. Steve likes to entertain friends at parties with “wire tricks.” Suppose he takes CHAPTER 7. QUADRATIC MODELING a piece of wire 60 inches long and cuts it into two pieces. Steve takes the first piece of wire and bends it into the shape of a perfect circle. He then proceeds to bend the second piece of wire into the shape of a perfect square. Where should Steve cut the wire so that the total area of the circle and square combined is as small as possible? What is this minimal area? What should Steve do if he wants the combined area to be as large as possible? Problem 7.15. Two particles are moving in the xy-plane. The move along straight lines at constant speed. At time t, particle A’s position is given by x = t + 2, y = 1 t−3 2 and particle B’s position is given by 1 x = 12 − 2t, y = 6 − t. 3 (a) Find the equation of the line along which particle A moves. Sketch this line, and label A’s starting point and direction of motion. (b) Find the equation of the line along which particle B moves. Sketch this line on the same axes, and label B’s starting point and direction of motion. (c) Find the time (i.e., the value of t) at which the distance between A and B is minimal. Find the locations of particles A and B at this time, and label them on your graph. Problem 7.16. Sven starts walking due south at 5 feet per second from a point 120 feet north of an intersection. At the same time Rudyard starts walking due east at 4 feet per second from a point 150 feet west of the intersection. (a) Write an expression for the distance between Sven and Rudyard t seconds after they start walking. (b) When are Sven and Rudyard closest? What is the minimum distance between them? Problem 7.17. After a vigorous soccer match, Tina and Michael decide to have a glass of 7.7. EXERCISES 105 their favorite refreshment. They each run in a straight line along the indicated paths at a speed of 10 ft/sec. Parametrize the motion of Tina and Michael individually. Find when and where Tina and Michael are closest to one another; also compute this minimum distance. (200,300) soy milk (−50,275) Find values of α that make this equation true (your answer will involve x). Problem 7.19. For each of the following equations, find the value(s) of the constant α so that the equation has exactly one solution, and determine the solution for each value. (a) αx2 + x + 1 = 0 (b) x2 + αx + 1 = 0 beet juice (c) x2 + x + α = 0 (400,50) Tina (d) x2 + αx + 4α + 1 = 0 Problem 7.20. (a) Solve for t Michael s = 2(t − 1)2 + 1 Problem 7.18. Consider the equation: αx2 + 2α2 x + 1 = 0. Find the values of x that make this equation true (your answer will involve α). (b) Solve for x y = x2 + 2x + 3 106 CHAPTER 7. QUADRATIC MODELING Chapter 8 Composition A new home takes its shape from basic building materials and the skillful use of construction tools. Likewise, we can build new functions from known functions through the application of analogous mathematical tools. There are five tools we want to develop: composition, reflection, shifting, dilation, arithmetic. We will handle composition in this section, then discuss the others in the following two sections. 1 To set the stage, let’s look at a simple botany experioxygen rate hr 1 ment. Imagine a plant growing under a particular steady 0.8 light source. Plants continually give off oxygen gas to 0.6 the environment at some rate; common units would be 0.4 liters/hour. If we leave this plant unbothered, we mea0.2 hours sure that the plant puts out 1 liter/hour; so, the oxygen 2 4 6 8 10 output is a steady constant rate. However, if we apply a flash of high intensity green light at the time t = 1 and (a) Flash at t = 1. measure the oxygen output of the plant, we obtain the 1 plot in Figure 8.1(a). oxygen rate hr 1 Using what we know from the previous section on 0.8 quadratic functions, we can check that a reasonable 0.6 model for the graph is this multipart function f(t) (on the 0.4 domain 0 ≤ t ≤ 10): 0.2 2 1 if t ≤ 1 22 8 t − 3 t + 3 if 1 ≤ t ≤ 3 f( t ) = 3 1 if 3 ≤ t 4 (b) Flash at t = 5. Figure 8.1: Light flashes. Suppose we want to model the oxygen consumption when a green light pulse occurs at time t = 5 (instead of time t = 1), what is the mathematical model? For starters, it is pretty easy to believe that the graph for this new situation will look like the new graph in Figure 8.1(b). But, can we somehow use the model f(t) in hand (the known function) to build the model we want (the new function)? We will return in Exam107 6 hours 8 10 CHAPTER 8. COMPOSITION 108 ple 8.2.4 to see the answer is yes; first, we need to develop the tool of function composition. 8.1 The Formula for a Composition The basic idea is to start with two functions f and g and produce a new function called their composition. There are two basic steps in this process and we are going to focus on each separately. The first step is fairly mechanical, though perhaps somewhat unnatural. It involves combining the formulas for the functions f and g together to get a new formula; we will focus on that step in this subsection. The next step is of varying complexity and involves analyzing how the domains and ranges of f and g affect those of the composition; we defer that to the next subsection once we have the mechanics down. in in out out y x u the “f” function y = f(x) the “g” function x = g(u) in out “composed” function y = f(g(u)) Figure 8.2: Visualizing a composite function y = h(u) = f(g(u)). Here is a very common occurrence: We are handed a function y = f(x), which means given an x value, the rule f(x) produces a new y value. In addition, it may happen that the variable x is itself related to a third variable u by some different function equation x = g(u). Given u, the rule g(u) will produce a value of x; from this x we can use the rule f(x) to produce a y value. In other words, we can regard y as a function depending on the new independent variable u. It is important to know the mechanics of working with this kind of setup. Abstractly, we have just described a situation where we take two functions and build a new 8.1. THE FORMULA FOR A COMPOSITION 109 function which “composes” the original ones together; schematically the situation looks like this: Example 8.1.1. A pebble is tossed into a pond. The radius of the first circular ripple is measured to increase at the constant rate of 2.3 ft/sec. What is the area enclosed by the leading ripple after 6 seconds have elapsed? How much time must elapse so that the area enclosed by the leading ripple is 300 square feet? Solution. We know that an object tossed into a pond leading ripple after t seconds will generate a series of concentric ripples, which grow leading ripple after 2 seconds steadily larger. We are asked questions that relate the area of the circular region bounded by the leading ripple and time elapsed. r = r(t) leading ripple after 1 second Let r denote the radius of the leading ripple after t secFigure 8.3: Concentric riponds; units of feet. The area A of a disc bounded by a ples. 2 leading ripple will be A = A(r) = πr . This exhibits A as a function in the variable r. However, the radius is changing with respect to time: r = r(t) = radius after t seconds = 2.3 feet t seconds = 2.3t feet. sec So, r = r(t) is a function of t. In the expression A = A(r), replace “r” by “r(t),” then A = π(2.3t)2 = 5.29πt2. The new function a(t) = 5.29πt2 gives a precise relationship between area and time. To answer our first question, a(6) = 598.3 feet2 is the area of the region bounded by the leading ripple after 6 seconds. On the other hand, if a(t) = 300 ft2 ;, 300 = 5.29πt2, so t = ± 300/(5.29π) = ±4.25. Since t represents time, only the positive solution t = 4.25 seconds makes sense. We can formalize the key idea used in solving this problem, which is familiar from previous courses. Suppose that y = f(x) and that additionally the independent variable x is itself a function of a different independent variable t; i.e., x = g(t ). Then we can replace every occurrence of “x” in f(x) by the expression “g(t),” thereby obtaining y as a function in the independent variable t. We usually denote this new function of t: y = f(g(t)). CHAPTER 8. COMPOSITION 110 We refer to f(g(t)) as the composition of f and g or the composite function. The process of forming the composition of two functions is a mechanical procedure. If you are handed the actual formulas for y = f(x) and x = g(t), then Procedure 8.1.2 is what you need. Important Procedure 8.1.2. To obtain the formula for f(g(t)), replace every occurrence of “x” in f(x) by the expression “g(t).” Here are some examples of how to do this: Examples 8.1.3. Use the composition procedure in each of these cases. (i) If y = f(x) = 2 and x = g(t) = 2t, then f(g(t)) = f(2t) = 2. (ii) If y = 3x − 7 and x = g(t) = 4, then f(g(t)) = f(4) = 3 · 4 − 7 = 5. (iii) If y = f(x) = x2 + 1 and x = g(t) = 2t − 1, then f(g(t)) = f(2t − 1) = (2t − 1)2 + 1 = 4t2 − 4t + 2. (iv) If y = f(x) = 2 + 1 + (x − 3)2 and x = g(t) = 2t2 − 1, then f(g(t)) = f(2t2 − 1) = 2+ 1 + (2t2 − 1 − 3)2 = 2+ 4t4 − 16t2 + 17. (v) If y = f(x) = x2 and x = g(t) = t + ♥, then f(g(t)) = f(t + ♥) = (t + ♥)2 = t2 + 2t♥ + ♥2 . It is natural to ask: What good is this whole business about compositions? One way to think of it is that we can use composite functions to break complicated functions into simpler parts. For example, y = h(x) = x2 + 1 √ can be written as the composition f(g(x)), where y = f(z) = z and z = g(x) = x2 + 1. Each of the functions f and g is “simpler” than the original h, which can help when studying h. Examples 8.1.4. Here we use composite functions to “simplify” a given function. 8.1. THE FORMULA FOR A COMPOSITION 111 (i) The function y = (x−31)2 +2 can be written as a composition y = f(g(x)), where y = f(z) = z21 2 and z = g(x) = x − 3. + (ii) The upper semicircle of radius 2 centered at (1,2) is the graph of the function y = 2 + 4 − (x − 1)2 . This function can be written as a com√ position y = f(g(x)), where y = f(z) = 2 + 4 − z2 and z = g(x) = x − 1. 8.1.1 Some notational confusion In our discussion above, we have used different letters to represent the domain variables of two functions we are composing. Typically, we have been writing: If y = f(x) and x = g(t), then y = f(g(t)) is the composition. This illustrates that the three variables t, x, and y can all be of different types. For example, t might represent time, x could be speed and y could be distance. If we are given two functions that involve the same independent variable, like f(x) = x2 and g(x) = 2x + 1, then we can still form a new function f(g(x)) by following the same prescription as in Procedure 8.1.2: Important Procedure 8.1.5. To obtain the formula for f(g(x)), replace every occurrence of “x” in f(x) by the expression “g(x).” For our example, this gives us: f(g(x)) = f(2x + 1) = (2x + 1)2 . Here are three other examples: √ √ • If f(x) = x, g(x) = 2x2 + 1, then f(g(x)) = 2x2 + 1. • If f(x) = 1 , g(x) = 2x + 1, then f(g(x)) = x 1 . 2x+1 • If f(x) = x2 , g(x) = △ − x, then f(g(x)) = △2 − 2x△ + x2 . 112 CHAPTER 8. COMPOSITION y-axis Example 8.1.6. Let f(x) = x2 , g(x) = x + 1 and h(x) = x − 1. Find the formulas for f(g(x)), g(f(x)), f(h(x)) and h(f(x)). Discuss the relationship between the graphs of these four functions. 10 8 6 4 2 x-axis −3 −2 −1 1 2 3 Figure 8.4: Sketching composite functions. Solution. If we apply Procedure 8.1.5, we obtain the composition formulas. The four graphs are given on the domain −3 ≤ x ≤ 3, together with the graph of f(x) = x2 . f(g(x)) g(f(x)) f(h(x)) h(f(x)) = = = = f( x + 1) = ( x + 1) 2 g(x2 ) = x2 + 1 f( x − 1) = ( x − 1) 2 h(x2 ) = x2 − 1. We can identify each graph by looking at its vertex: • f(x) has vertex (0,0) • f(g(x)) has vertex (-1,0) • g(f(x)) has vertex (0,1) • f(h(x)) has vertex (1,0) • h(f(x)) has vertex (0,-1) Horizontal or vertical shifting of the graph of f(x) = x2 gives the other four graphs: See Figure 8.4. 8.2 Domain, Range, etc. for a Composition A function is a “package” consisting of a rule, a domain of allowed input values, and a range of output values. When we start to compose functions, we sometimes need to worry about how the domains and ranges of the composing functions affect the composed function. First off, when you form the composition f(g(x)) of f(x) and g(x), the range values for g(x) must lie within the domain values for f(x). This may require that you modify the range values of g(x) by changing its domain. The domain values for f(g(x)) will be the domain values for g(x). 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION in domain g out in the “g” function 113 out the “f” function range of f(g(x)) range of g domain of f Figure 8.5: What is the domain and range of a composite function? In practical terms, here is how one deals with the domain issues for a composition. This is a refinement of Procedure 8.1.5 on page 111. Important Procedure 8.2.1. To obtain the formula for f(g(x)), replace every occurrence of “x” in f(x) by the expression “g(x).” In addition, if there is a condition on the domain of f that involves x, then replace every occurrence of “x” in that condition by the expression “g(x).” The next example illustrates how to use this principle. Example 8.2.2. Start with the function y = f(x) = x2 on the domain −1 ≤ x ≤ 1. Find the rule and domain of y = f(g(x)), where g(x) = x − 1. Solution. We can apply the first statement in Procedure 8.2.1 to find the rule for y = f(g(x)): y = f(g(x)) = f( x − 1) = ( x − 1) 2 = x2 − 2x + 1. To find the domain of y = f(g(x)), we apply the second statement in Procedure 8.2.1; this will require that we solve an inequality equation: −1 ≤ g(x) ≤ 1 −1 ≤ x − 1 ≤ 1 0≤ x ≤2 The conclusion is that y = f(g(x)) = x2 − 2x + 1 on the domain 0 ≤ x ≤ 2. √ Example 8.2.3. Let y = f(z) = z, z = g(x) = x + 1. What is the largest possible domain so that the composition f(g(x)) makes sense? CHAPTER 8. COMPOSITION 114 y z range z domain (a) y = z = g ( x) = x + 1 y = f(z) √ z. desired range −1 x required domain (b) z = x = 1. Figure 8.6: Finding the largest domain for f(g(x)). Solution. The largest possible domain for y = f(z) will consist of all nonnegative real numbers; this is also the range of the function f(z): See Figure 8.6(a). To find the largest domain for the composition, we try to find a domain of x-values so that the range of z = g(x) is the domain of y = f(z). So, in this case, we want the range of g(x) to be all non-negative real numbers, denoted 0 ≤ z. We graph z = g(x) in the xz-plane, mark the desired range 0 ≤ z on the vertical z-axis, then determine which x-values would lead to points on the graph with second coordinates in this zone. We find that the domain of all x-values greater or equal to −1 (denoted −1 ≤ x)√ leads to the desired range. In summary, the composition y = f(g(x)) = x + 1 is defined on the domain of x-values −1 ≤ x. Let’s return to the botany experiment that opened this section and see how composition of functions can be applied to the situation. Recall that plants continually give off oxygen gas to the environment at some rate; common units would be liters/hour. Example 8.2.4. A plant is growing under a particular steady light source. If we apply a flash of high intensity green light at the time t = 1 and measure the oxygen output of the plant, we obtain the plot below and the mathematical model f(t). if t ≤ 1 1 22 8 f( t ) = t − 3 t + 3 if 1 ≤ t ≤ 3 3 1 if 3 ≤ t Now, suppose instead we apply the flash of high intensity green light at the time t = 5. Verify that the mathematical model for this experiment is given by f(g(t)), where g(t) = t − 4. 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION Solution. Our expectation is that the plot for this new experiment will have the “parabolic dip” shifted over to occur starting at time t = 5 instead of at time t = 1. In other words, we expect the graph in Figure 8.7(b). Our job is to verify that this graph is obtained from the function f(g(t)), where g(t) = t − 4. This is a new terrain for us, since we need to look at a composition involving a multipart function. Here is how to proceed: When we are calculating a composition involving a multipart function, we need to look at each of the parts separately, so there will be three cases to consider: First part: f(t) = 1 when t ≤ 1. To get the formula for f(g(t)), we now appeal to Procedure 8.2.1 and just replace every occurrence of t in f(t) by g(t). That gives us this NEW domain condition and function equation: 115 oxygen rate 1 0.8 0.6 0.4 0.2 hours 2 22 t 3 8 t 3 4 6 8 10 (a) Flash at t = 1. oxygen rate 1 1 hr 0.8 0.6 0.4 0.2 hours 2 f(g(t)) = f(t − 4) = 1 when t − 4 ≤ 1 = 1 when t ≤ 5. 1 hr 4 6 8 10 (b) Flash at t = 5. Figure 8.7: Applying light at time t. Second part: f(t) = − + 3 when 1 ≤ t ≤ 3. We now appeal to Procedure 8.2.1 and just replace every occurrence of t in this function by g(t). That gives us this NEW domain condition and function equation: 8 2 f(g(t)) = f(t − 4) = (t − 4)2 − (t − 4) + 3 when 1 ≤ t − 4 ≤ 3 3 3 22 73 = t − 8t + when 5 ≤ t ≤ 7. 3 3 Third part: f(t) = 1 when 3 ≤ t. We now appeal to Procedure 8.2.1 and just replace every occurrence of t in this function by g(t). That gives us this NEW domain condition and function equation: f(g(t)) = f(t − 4) = 1 when 3 ≤ t − 4 = 1 when 7 ≤ t. The multipart rule for this composition can now be written down and using a graphing device you can verify its graph is the model for our experiment. if t ≤ 5 1 22 t − 8t + 73 if 5 ≤ t ≤ 7 f(g(t)) = 3 3 1 if 7 ≤ t CHAPTER 8. COMPOSITION 116 8.3 Exercises Problem 8.1. For this problem, f(t) = t − 1, g(t) = −t − 1 and h(t) = |t|. (a) Compute the multipart rules for h(f(t)) and h(g(t)) and sketch their graphs. (b) Compute the multipart rules for f(h(t)) and g(h(t)) and sketch their graphs. Problem 8.5. A car leaves Seattle heading east. The speed of the car in mph after m minutes is given by the function C(m) = 70m2 . 10 + m2 (c) Compute the multipart rule for h(h(t)−1) and sketch the graph. (a) Find a function m = f(s) that converts seconds s into minutes m. Write out the formula for the new function C(f(s)); what does this function calculate? Problem 8.2. Write each of the following functions as a composition of two simpler functions: (There is more than one correct answer.) (b) Find a function m = g(h) that converts hours h into minutes m. Write out the formula for the new function C(g(h)); what does this function calculate? (a) y = (x − 11)5 . √ (b) y = 3 1 + x2 . (c) y = 2(x − 3)5 − 5(x − 3)2 + 1 (x − 3) + 11. 2 (d) y = (e) y = (c) Find a function z = v(s) that converts mph s into ft/sec z. Write out the formula for the new function v(C(m); what does this function calculate? 1 . x2 +3 √ x + 1. (f) y = 2 − 5 − (3x − 1)2 . Problem 8.3. (a) Let f(x) be a linear function, f(x) = ax + b for constants a and b. Show that f(f(x)) is a linear function. (b) Find a function g(x) such that g(g(x)) = 6x − 8. Problem 8.4. Let f(x) = 1 x + 3. 2 (a) Sketch the graphs of f(x),f(f(x)),f(f(f(x))) on the interval −2 ≤ x ≤ 10. (b) Your graphs should all intersect at the point (6,6). The value x = 6 is called a fixed point of the function f(x) since f(6) = 6; that is, 6 is fixed - it doesn’t move when f is applied to it. Give an explanation for why 6 is a fixed point for any function f(f(f(...f(x)...))). (c) Linear functions (with the exception of f(x) = x) can have at most one fixed point. Quadratic functions can have at most two. Find the fixed points of the function g(x) = x2 − 2. (d) Give a quadratic function whose fixed points are x = −2 and x = 3. Problem 8.6. Compute the compositions f(g(x)), f(f(x)) and g(f(x)) in each case: (a) f(x) = x2 , g(x) = x + 3. √ (b) f(x) = 1/x, g(x) = x. (c) f(x) = 9x + 2, g(x) = 1 (x − 2). 9 (d) f(x) = 6x2 + 5, g(x) = x − 4. √ (e) f(x) = 4x3 − 3, g(x) = 3 2x + 6 (f) f(x) = 2x + 1, g(x) = x3 . (g) f(x) = 3, g(x) = 4x2 + 2x + 1. (h) f(x) = −4, g(x) = 0. √ 4 − z2 and Problem 8.7. Let y = f(z) = z = g(x) = 2x + 3. Compute the composition y = f(g(x)). Find the largest possible domain of x-values so that the composition y = f(g(x)) is defined. Problem 8.8. Suppose you have a function y = f(x) such that the domain of f(x) is 1 ≤ x ≤ 6 and the range of f(x) is −3 ≤ y ≤ 5. (a) What is the domain of f(2(x − 3)) ? (b) What is the range of f(2(x − 3)) ? (c) What is the domain of 2f(x) − 3 ? (d) What is the range of 2f(x) − 3 ? (e) Can you find constants B and C so that the domain of f(B(x − C)) is 8 ≤ x ≤ 9? 8.3. EXERCISES (f) Can you find constants A and D so that the range of Af(x) + D is 0 ≤ y ≤ 1? 117 this simplified expression: f(x + h) − f(x) . h (a) f(x) = Problem 8.9. For each of the given functions y = f(x), simplify the following expression so that h is no longer a factor in the denominator, then calculate the result of setting h = 0 in 1 x−1 . (b) f(x) = (2x + 1)2 . √ (c) f(x) = 25 − x2 . 118 CHAPTER 8. COMPOSITION Chapter 9 Inverse Functions The experimental sciences are loaded with examples of functions relating time and some measured quantity. In this case, time represents our “input” and the quantity we are measuring is the “output.” For example, maybe you have just mixed together some chemical reactants in a vessel. As time goes by, you measure the fraction of reactants remaining, tabulate your results, then sketch a graph as indicated in Figure 9.1. Viewing the input value as “time” and the output value fraction 1 as “fraction of product,” we could find a function y = f(t) modeling this data. Using this function, you can easily compute the fraction of reactants remaining at any time in the future. However, it is probably just as interesting to know how to predict the time when a given fraction 0 time of reactants exists. In other words, we would like a new Figure 9.1: Fraction of reacfunction that allows us to input a “fraction of reactants” tants as a function of time. and get out the “time” when this occurs. This “reverses” the input/output roles in the original function. Is there a systematic way to find the new function if we know y = f(t)? The answer is yes and depends upon the general theory of inverse functions. 9.1 Concept of an Inverse Function Suppose you are asked to solve the following three equations for x. How do you proceed? (x + 2) = 64 (x + 2)2 = 64 (x + 2)3 = 64. In the first equation, you add “−2” to each side, then obtain x = 62. In the third equation, you take the cube root of both sides of the equation, giving you x + 2 = 4, then subtract 2 getting x = 2. In second equation, you take a square root of both sides, BUT you need to remember both the 119 CHAPTER 9. INVERSE FUNCTIONS 120 positive and negative results when doing this. So, you are reduced down to x + 2 = ±8 or that x = −10 or 6. Why is it that in two of these cases you obtain a single solution, while in the remaining case there are two different answers? We need to sort this out, since the underlying ideas will surface when we address the inverse circular functions in Chapter 20. Let’s recall the conceptual idea of a function: A function is a process which takes a number x and outputs a In Out x f ( x) new number f(x). So far, we’ve only worked with this process from “left to right;” i.e., given x, we simply put it into The function a symbolic rule and out pops a new number f(x). This is Figure 9.2: A function as a all pretty mechanical and straightforward. process. 9.1.1 An Example Let’s schematically interpret what happens for the specific concrete example y = f(x) = 3x − 1, when x = −1, − 1 , 0, 1 , 1, 2: See Figure 9.3. 2 2 out in −1 in −5 2 1 −1 2 out 3x − 1 in 0 −4 3x − 1 1 −2 in 1 2 1 2 3x − 1 in out 3x − 1 out out 3x − 1 in 2 out 3x − 1 5 Figure 9.3: Function process y = 3x − 1. We could try to understand the function process in this example in “reverse order,” going “right to left;” namely, you might ask what x value can be run through the process so you end up with the number 11? This is somewhat like the “Jeopardy” game show: You know what the answer is, you want to find the question. For our example, if we start out with 1 some given y values, then we can define a “reverse process” x = 3 (y + 1), which returns the x value required so that f(x) = y: See Figure 9.4. 9.1. CONCEPT OF AN INVERSE FUNCTION placements (y+1) 3 −4 −1 2 (y+1) 3 0 (y+1) 3 −1 121 1 2 (y+1) 3 1 2 −5 2 1 (y+1) 3 2 −1 2 (y+1) 3 5 Figure 9.4: The reverse process x = 1 (y + 1). 3 9.1.2 A Second Example If we begin with a linear function y = f(x) = mx + b, where m = 0, then we can always find a “reverse process” for the function. To find it, you must solve the equation y = f(x) for x in terms of y: y = mx + b y − b = mx 1 (y − b) = x m So, if m = 3 and b = −1, we just have the first example above. For another example, suppose y = −0.8x + 2; then m = −0.8 and b = 2. In this case, the reverse process is −1.25(y − 2) = x. If we are given the value y = 11, we simply compute that x = −11.25; i.e., f(−11.25) = 11. 9.1.3 A Third Example The previous examples hide a subtle point that can arise when we try to understand the “reverse process” for a given function. Suppose we begin with the function y = f(x) = (x − 1)2 + 1. Figure 9.5 is a schematic of how 1 1 the function works when we plug in x = −1, − 2 , 0, 2 , 1, 2; what is being illustrated is a “forward process”, in that each input generates a unique output. For this example, if we start out with some given y values, then we can try to define a “reverse process” x = ??? which returns an x value required so that f(x) = y. Unfortunately, there is no way to obtain a single formula for this reverse process; Figure 9.6(a) shows what happens if you are given y = 3 and you try to solve for x. CHAPTER 9. INVERSE FUNCTIONS 122 in (x − 1) + 1 −1 in 2 5 in (x − 1)2 + 1 in 13 4 2 2 (x − 1)2 + 1 1 out (x − 1)2 + 1 0 1 2 out (x − 1) + 1 −1 2 in out 2 in (x − 1)2 + 1 out 5 4 out 1 out 2 Figure 9.5: Function process y = f(x) = (x − 1)2 + 1. The conclusion is that the “reverse process” has two outputs. This violates the rules required for a function, so this is NOT a function. The solution is to create two new “reverse processes.” Each of these “reverse processes” has a unique output; √ 1+ 2 reverse in other words, each of these “reverse processes” defines process a function. √ x= − y−1 3 + So, given y √ 3, there are √ = TWO possible x √ values, √ 1− 2 namely x = 1 ± 2, so that f(1 + 2) = 3 and f(1 − 2) = 3. In other words, the reverse process is not given by a single (a) Reverse process but not a equation; there are TWO POSSIBLE reverse processes. function. reverse process 1+ √ 2 x =+ √ y−1 3 reverse process 1− √ 2 x =− √ y−1 3 (b) Two new reverse processes that are functions. Figure 9.6: What to do if a reverse process is not a function. 9.2 Graphical Idea of an Inverse We have seen that finding inverses is related to solving equations. However so far, the discussion has been symbolic; we have pushed around a few equations and in the end generated some confusion. Let’s use the tools of Chapter 6 to visualize what is going on here. Suppose we are given the graph of a function f(x) as in Figure 9.7(a). What input x values result in an output value of 3? This involves finding all x such that f(x) = 3. Graphically, this means we are trying to find points on the graph of f(x) so that their y-coordinates are 3. The easiest way to to do this is to draw the line y = 3 and find where it intersects the graph. In Figure 9.7(b) we can see the points of intersection are (−5, 3), (−1, 3), and (9, 3). That means that x = −5, −1, 9 produce the output value 3; i.e., f(−5) = f(−1) = f(9) = 3. 9.2. GRAPHICAL IDEA OF AN INVERSE 123 y = f ( x) y = f ( x) 5 5 y=3 −5 5 10 −5 (a) Given the function y = f(x). −5 5 10 −5 (b) What values of x give f(x) = 3? Figure 9.7: Using the horizontal line y = 3 to find values on the x-axis. This leads to our first important fact about the “reverse process” for a function: Important Fact 9.2.1. Given a number c, the x values such that f(x) = c can be found by finding the x-coordinates of the intersection points of the graphs of y = f(x) and y = c. Example 9.2.2. Graph y = f(x) = x2 and discuss the meaning of Fact 9.2.1 when c = 3, 1, 6. Solution. We graph y = x2 and the lines y = 1, y = 3 and y = 6. Let’s use c = 6 as an example. We need to simultaneously solve the equations y = x2 and y = 6. Putting √ these together, we get x2 = 6 or x = ± 6 ≈ ±2.449; i.e., f(±2.449) = 6. If c = 3, we get x = ±1.732; i.e., f(±1.732) = 3. Finally, if c = 1, we get x = ±1; i.e., f(±1) = 1. y = x2 y=6 y=3 y=1 Figure 9.8: Graph of y = The pictures so far indicate another very important f(x) = x2 . piece of information. For any number c, we can tell exactly “how many” input x values lead to the same output value c, just by counting the number of times the graphs of y = f(x) and y = c intersect. Important Fact 9.2.3. For any function f(x) and any number c, the number of x values so that f(x) = c is the number of times the graphs of y = c and y = f(x) intersect. CHAPTER 9. INVERSE FUNCTIONS 124 Examples 9.2.4. (i) If f(x) is a linear function f(x) = mx + b, m = 0, then the graph of f(x) intersects a given horizontal line y = c EXACTLY once; i.e., the equation c = f(x) always has a unique solution. (ii) If f(x) = d is a constant function and c = d, then every input x value in the domain leads to the output value c. On the other hand, if c = d, then no input x value will lead to the output value c. For example, if f(x) = 1 and c = 1, then every real number can be input to produce an output of 1; if c = 2, then no input value of x will lead to an output of 2. y-axis f(x) = mx + b y-axis y=c f ( x) = 1 x-axis x-axis any of these inputs the only input which leads to an output of 1 leads to an output of c linear functions constant functions Figure 9.9: Does a horizontal line y = c intersect a curve once or more than once? 9.2.1 One-to-one Functions For a specified domain, one-to-one functions are functions with the property: Given any number c, there is at most one input x value in the domain so that f(x) = c. Among our examples thus far, linear functions (degree 1 polynomials) are always one-to-one. However, f(x) = x2 is not one-to-one; we’ve already seen that it can have two values for some of its inverses. By Fact 9.2.3, we can quickly come up with what’s called the horizontal line test. Important Fact 9.2.5 (Horizontal Line Test). On a given domain of x-values, if the graph of some function f(x) has the property that every horizontal line crosses the graph at most only once, then the function is one-to-one on this domain. 9.3. INVERSE FUNCTIONS 125 Example 9.2.6. By the horizontal line test, it is easy to see that f(x) = x3 is one-to-one on the domain of all real numbers. Although it isn’t common, it’s quite nice when a function is one-to-one because we don’t need to worry as much about the number of input x values producing the same output y value. In effect, this is saying that we can define a “reverse process” for the function y = f(x) which will also be a function; this is the key theme of the next section. y-axis x-axis lines Figure 9.10: A one-to-one function f(x) = x3 . 9.3 Inverse Functions Let’s now come face to face with the problem of finding the “reverse process” for a given function y = f(x). It is important to keep in mind that the domain and range of the function will both play an important role in this whole development. For example, Figure 9.11 shows the function f(x) = x2 with three different domains specified and the corresponding range values. range range range domain domain horizontal domain Figure 9.11: Possible domains for a given range. These comments set the stage for a third important fact. Since the domain and range of the function and its inverse rule are going to be intimately related, we want to use notation that will highlight this fact. We have been using the letters x and y for the domain (input) and range (output) variables of f(x) and the “reverse process” is going to reverse these roles. It then seems natural to simply write y (instead of c) for the input values of the “reverse process” and x for its output values. CHAPTER 9. INVERSE FUNCTIONS 126 Important Fact 9.3.1. Suppose a function f(x) is one-to-one on a domain of x values. Then define a NEW FUNCTION by the rule f−1 (y) = the x value so that f(x) = y. The domain of y values for the function f−1 (y) is equal to the range of the function f(x). The rule defined here is the “reverse process” for the given function. It is referred to as the inverse function and we read f−1 (y) as “...eff inverse of y...”. !!! CAUTION !!! Both the “domain” of f(x) and the “rule” f(x) have equal influence on whether the inverse rule is a function. Keep in mind, you do NOT get an inverse function automatically from functions that are not one-to-one! 9.3.1 Schematic Idea of an Inverse Function Suppose that f(x) is one-to-one, so that f−1 (y) is a function. As a result, we can model f−1 (y) as a black box. What does it do? If we put in y in the input side, we should get out the x such that f(x) = y. Now, let’s try to unravel something very special that is in out happening on a symbolic level. What would happen if we f−1 (y) plugged f(a) into the inverse function for some number a? x such that y Then the inverse rule f−1 (f(a)) tells us that we want to f ( x) = y find some x so that f(x) = f(a). But, we already know x = a works and since f−1 (y) is a function (hence gives us Figure 9.12: A new function unique answers), the output of f−1 (f(a)) is just a. Symbolx = f−1 (y). ically, this means we have Fact 9.3.2. Important Fact 9.3.2. For every a value in the domain of f(x), we have f−1 (f(a)) = a. (9.1) This is better shown in the black box picture of Figure 9.13. out in a f ( x) in f(a) out f − 1 ( x) a Figure 9.13: Visualizing f−1 (f(a)) = a. A good way to get an idea of what an inverse function is doing is to remember that f−1 (y) reverses the process of f(x). We can think of f−1 (y) as a “black box” running f(x) backwards. 9.4. TRYING TO INVERT A NON ONE-TO-ONE FUNCTION 127 9.3.2 Graphing Inverse Functions How can we get the graph of an inverse function? The idea is to manipulate the graph of our original one-to-one function in some prescribed way, ending up with the graph of f−1 (y). This isn’t as hard as it sounds, but some confusion with the variables enters into play. Remember that a typical point on the graph of a function y = f(x) looks like (x, f(x)). Now let’s take a look at the inverse function x = f−1 (y). Given a number y in the domain of f−1 (y), y = f(x) for some x in the domain of f(x); i.e., we are using the fact that the domain of f−1 equals the range of f. The function f−1 (y) takes the number f(x) and sends it to x, by Fact 9.3.2. So when f(x) is the input value, x becomes the output value. Conclude a point on the graph of f−1 (y) looks like (f(x), x). It’s similar to the graph of y = f(x), only the x and y coordinates have reversed! What does that do to the graph? Essentially, you reorient the picture so that the positive x-axis and positive y-axis are interchanged. Figure 9.14 shows the process for √ the function y = f(x) = x3 and its inverse function x = f−1 (y) = 3 y. We place some ∗ symbols on the graph to help keep track of what is happening. +y−axis +x−axis * * +x−axis * rotate 900 * clockwise +y−axis +x−axis ** +y−axis * * flip across horiz axis Figure 9.14: Graphically finding x = f−1 (y) = * √ 3 y. 9.4 Trying to Invert a Non one-to-one Function √ Suppose we blindly try to show that y is the inverse function for y = x2 , without worrying about all of this one-to-one stuff. We’ll start out with √ the number −7. If f−1 (y) = y, then we know that f−1 (f(−7)) = f−1 (49) = 7. On the other hand, the formula in Fact 9.3.2 tells us that we must have f−1 (f(−7)) = −7, √ so we have just shown 7 = −7! So clearly f−1 (y) = y. Even if we try √ f−1 (y) = − y, we produce a contradiction. It seems that if you didn’t have CHAPTER 9. INVERSE FUNCTIONS 128 to worry about negative numbers, things would be all right. Then you √ could say that f−1 (y) = y. Let’s try to see what this means graphically. Let’s set f(x) = x2 , but only for non-negative x-values. That means that we want to erase the graph to the left of the y-axis (so remember - no negative x-values allowed). The graph would then look like Figure 9.15. +y y = x2 +x inverse function +x √ y +y domain non-negative y domain non-negative x Figure 9.15: Restricting the domain: No negative x-values. This is a now a one-to-one function! And now, one can see that its √ inverse function is y. Similarly, we could have taken f(x) = x2 but only √ for the non-positive x-values. In that case, f−1 (y) = − y. In effect, we have split the graph of y = x2 into two parts, each of which is the graph of a one-to-one function; Figure 9.16. +y y = x2 +x domain non-negative y +x domain non-positive x +y inverse function √ −y Figure 9.16: Restricting the domain: No positive x-values. It is precisely this splitting into two cases that √ leads us to multiple √ 2 solutions of an equation like x = 5. We obtain x = 5 and x = − 5; one 9.5. SUMMARY 129 solution comes from the side of the graph to the left of the y-axis, and the other from the right of the y-axis. This is because we have separate inverse functions for the left and right side of the graph of y = x2 . 9.5 Summary • Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in the domain of f and the domain of g. • A function f is one-to-one if every equation f(x) = k has at most one solution. If there is a value of k such that the equation f(x) = k has more than one solution, then f is not one-toone. • A function is one-to-one if every horizontal line intersects the function’s graph at most once. • A function has an inverse if the function is one-to-one, and every one-to-one function has an inverse. • The domain of a function is the range of its inverse, and the range of a function is the domain of its inverse. • The graph of a function and its inverse are mirror images of each other across the line y = x. CHAPTER 9. INVERSE FUNCTIONS 130 9.6 Exercises Problem 9.1. Let f(x) = 3x2 4 on the largest − domain for which the formula makes sense. A B (a) Find the domain and range of f(x), then sketch the graph. (b) Find the domain, range and rule for the inverse function f−1 , then sketch its graph. C D Problem 9.2. Find the inverse function of each of the following functions. Specify the domains of the inverse functions. (a) f(x) = 1 x+8 5 5 x+3 √ (c) g(x) = 4 3 − x − 7 √ √ (d) j(x) = x + x − 1 (b) h(x) = (e) k(x) = 16 − x2 , 0 ≤ x ≤ 4 Problem 9.3. For this problem, y = f(x) = 2x2 − 3x − 1 on the domain of all real numbers. (a) Sketch the function graph and find the coordinates of the vertex P = (a,b). (b) Explain why y = f(x) does not have an inverse function on the domain of all real numbers. (c) Restrict y = f(x) to the domain {a ≤ x} and find the formula for the inverse function f−1 (y). What are the domain and range of the inverse function? (d) Restrict y = f(x) to the domain {x ≤ a} and find the formula for the inverse function f−1 (y). What are the domain and range of the inverse function? Problem 9.4. Which of the following graphs are one-to-one? If they are not one-to-one, section the graph up into parts that are one-toone. Problem 9.5. Show that, for every value of a, the function f(x) = a + 1 x−a is its own inverse. Problem 9.6. Clovis is standing at the edge of a cliff, which slopes 4 feet downward from him for every 1 horizontal foot. He launches a small model rocket from where he is standing. With the origin of the coordinate system located where he is standing, and the x-axis extending horizontally, the path of the rocket is described by the formula y = −2x2 + 120x. (a) Give a function h = f(x) relating the height h of the rocket above the sloping ground to its x-coordinate. (b) Find the maximum height of the rocket above the sloping ground. What is its x-coordinate when it is at its maximum height? (c) Clovis measures its height h of the rocket above the sloping ground while it is going up. Give a function x = g(h) relating the x-coordinate of the rocket to h. (d) Does this function still work when the rocket is going down? Explain. Problem 9.7. For each of the following functions: (1) sketch the function, (2) find the inverse function, and (3) sketch the inverse function. In each case, indicate the correct domains and ranges. (4) Finally, make sure you 9.6. EXERCISES 131 test each of the functions you propose as an inverse with the following compositions: ? (c) Give a function t = f−1 (w) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too. f(f−1 (x)) = x and (b) After how many hours will the surface of the water have width of 6 feet? ? f−1 (f(x)) = x. (a) f(x) = 3x − 2 (b) f(x) = 1 x + 5 2 Problem 9.9. A biochemical experiment involves combining together two protein extracts. Suppose a function φ(t) monitors the amount (nanograms) of extract A remaining at time t (nanoseconds). Assume you know these facts: (c) f(x) = −x2 + 3, x ≥ 0 (d) f(x) = x2 + 2x + 5, x ≤ −1 √ (e) f(x) = 4 − x2 , 0 ≤ x ≤ 2 Problem 9.8. A trough has a semicircular cross section with a radius of 5 feet. Water starts flowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. 1. The function φ is invertible; i.e., it has an inverse function. 2. φ(0) = 6, φ(1) = 5, φ(2) = 3, φ(3) = 1, φ(4) = 0.5, φ(10) = 0. (a) At what time do you know there will be 3 nanograms of extract A remaining? water (b) What is φ−1 (0.5) and what does it tell you? 5 ft cross-section of trough (a) Give a function w = f(t) relating the width w of the surface of the water to the time t, in hours. Make sure to specify the domain and compute the range too. (c) (True or False) There is exactly one time when the amount of extract A remaining is 4 nanograms. (d) Calculate φ(φ−1 (1)) = (e) Calculate φ−1 (φ(6)) = (f) What is the domain and range of φ? 132 CHAPTER 9. INVERSE FUNCTIONS Chapter 10 Exponential Functions If we start with a single yeast cell under favorable growth conditions, then it will divide in one hour to form two identical “daughter cells”. In turn, after another hour, each of these daughter cells will divide to produce two identical cells; we now have four identical “granddaughter cells” of the original parent cell. Under ideal conditions, we can imagine how this “doubling effect” will continue: TIME cells t=0 hours t=1 hours t=2 hours t=3 hours Figure 10.1: Observing cell growth. The question is this: Can we find a function of t that will predict (i.e. model) the number of yeast cells after t hours? If we tabulate some data (as at right), the conclusion is that the formula N (t ) = 2t predicts the number of yeast cells after t hours. Now, let’s make a very slight change. Suppose that instead of starting with a single cell, we begin with a population of 3 × 106 cells; a more realistic situation. If we assume 133 Total hours Number of yeast cells 0 1 2 3 4 5 6 1=20 2=21 4=22 8=23 16=24 32=25 64=26 Table 10.1: Cell growth data. CHAPTER 10. EXPONENTIAL FUNCTIONS 134 that the population of cells will double every hour, then reasoning as above will lead us to conclude that the formula N(t) = (3 × 106 )2t gives the population of cells after t hours. Now, as long as t represents a non-negative integer, we know how to calculate N(t). For example, if t = 6, then N(t) = (3 × 106 )26 = (3 × 106 )(2 · 2 · 2 · 2 · 2 · 2) = (3 × 106 )64 = 192 × 106. The key point is that computing N(t) only involves simple arithmetic. But what happens if we want to know the population of cells after 6.37 hours? That would require that we work with the formula N(t) = (3 × 106 )26.37 and the rules of arithmetic do not suffice to calculate N(t). We are stuck, since we must understand the meaning of an expression like 26.37 . In order to proceed, we will need to review the algebra required to make sense of raising a number (such as 2) to a non-integer power. We need to √ understand the precise meaning of expressions like: 26.37 , 2 5 , 2−π , etc. 10.1 Functions of Exponential Type x this is a fixed positive integer this is a variable y=b b y=x this is a fixed number Exponential Picture this is a variable Monomial Picture Figure 10.2: Viewing the difference between exponential and monomial functions. On a symbolic level, the class of functions we are trying to motivate is easily introduced. We have already studied the monomials y = xb , where x was our input variable and b was a fixed positive integer exponent. What happens if we turn this around, interchanging x and b, defining a new rule: y = f(x) = bx . (10.1) 10.1. FUNCTIONS OF EXPONENTIAL TYPE 135 We refer to x as the power and b the base. An expression of this sort is called a function of exponential type. Actually, if your algebra is a bit rusty, it is easy to initially confuse functions of exponential type and monomials (see Figure 10.2). 10.1.1 Reviewing the Rules of Exponents To be completely honest, making sense of the expression y = bx for all numbers x requires the tools of Calculus, but it is possible to establish a reasonable comfort level by handling the case when x is a rational number. If b ≥ 0 and n is a positive integer (i.e. n = 1, 2, 3, 4, . . . ), then we can try to solve the equation tn = b. (10.2) 3 2 n=2 n=4 n=6 1 0 -1.5 -1 -0.5 0 0.5 1 1.5 -1 A solution t to this equation is called an nth root of b. This leads to complications, depending on whether n is even or odd. In the odd case, for any real number b, notice that the graph of y = b will always cross the graph of y = tn exactly once, leading to one solution of (10.2). On the other hand, if n is even and b < 0, then the graph of y = tn will miss the graph of y = b, implying there are no solutions to the equation in (10.2). (There will be complex solutions to equations such as t2 = −1, involv√ ing the imaginary complex numbers ±i = ± −1, but we are only working with real numbers in this course.) Also, again in the case when n is even, it can happen that there are two solutions to (10.2). We do not want to constantly worry about this even/odd distinction, so we will henceforth assume b > 0. To eliminate possible ambiguity, we will single out a particular nth -root; we define the symbols: √ 1 n b = b n = the largest real nth root of b. -2 -3 3 2 1 0 -1.5 -1 -0.5 0 √ q p b = √ q 1 1.5 -2 -3 Figure 10.3: Even and odd monomials. (10.3) Important Facts 10.1.1 (Working with rational exponents). For all positive integers p and q, and any real number base b > 0, we have p 0.5 -1 √ Thus, whereas ±1 are both 4th -roots of 1, we have defined 4 1 = 1. In order to manipulate y = bx for rational x, we need to recall some basic facts from algebra. bq = n=3 n=5 n=7 bp . For any rational numbers r and s, and for all positive bases a and b: CHAPTER 10. EXPONENTIAL FUNCTIONS 136 n y=t y-axis n even = solution y-axis n y=t n odd y=b y=b t-axis y=b* t-axis no solution or two solutions exactly one solution Figure 10.4: How many solutions to tn = b?. 1. Product of power rule: br bs = br+s 2. Power of power rule: (br )s = brs 3. Power of product rule: (ab)r = ar br 4. Zero exponent rule: b0 = 1 5. Negative power rule: b−r = 1 br These rules have two important consequences, one theoretical and the other more practical. On the first count, recall that any rational number p r can be written in the form r = q , where p and q are integers. Consequently, using these rules, we see that the expression y = bx defines a function of x, whenever x is a rational number. On the more practical side of things, using the rules we can calculate and manipulate certain expressions. For example, 2 √ 3 27 5 √ 3 8 27 3 = 8− 3 = 2 −5 = 32 = 9; = 2−5 = 1 1 =. 5 2 32 The sticky point which remains is knowing that f(x) = bx actually defines a function for all real values of x. This is not easy to verify and we are simply going to accept it as a fact. The difficulty is that we need the fundamentally new concept of a limit, which is the starting point of a Calculus course. Once we know the expression does define a function, we can also verify that the rules of Fact 10.1.1 carry through for all real 10.2. THE FUNCTIONS Y = A0 BX 137 exponent powers. Your calculator should have a “y to the x key”, allowing √ you to calculate expressions such as π 2 involving non-rational powers. Here are the key modeling functions we will work with in this Chapter. Definition 10.1.2. A function of exponential type has the form A(x) = A0 bx , for some b > 0, b = 1, and A0 = 0. We will refer to the formula in Definition 10.1.2 as the standard exponential form. Just as with standard forms for quadratic functions, we sometimes need to do a little calculation to put an equation in standard form. The constant A0 is called the initial value of the exponential function; this is because if x represents time, then A(0) = A0 b0 = A0 is the value of the function at time x = 0; i.e. the initial value of the function. Example 10.1.3. Write the equations y = 83x and y = 7 exponential form. 1 2x−1 2 in standard Solution. In both cases, we just use the rules of exponents to maneuver the given equation into standard form: y = 83x = (83 )x = 512x and 1 2 2x−1 y=7 2x =7 1 2 =7 1 2 = 14 1 4 −1 1 2 2 x 2 x 10.2 The Functions y = A0bx We know f(x) = 2x defines a function of x, so we can study basic qualitative features of its graph. The data assembled in the solution of the “Doubling Effect” beginning this Chapter, plus the rules of exponents, produce a number of points on the graph. This graph exhibits four key qualitative features that deserve mention: CHAPTER 10. EXPONENTIAL FUNCTIONS 138 Point on the graph of y = 2x . . . . . . . . . -2 1/4 (-2, 1/4) -1 1/2 (-1, 1/2) 0 1 (0, 1) 1 2 (1, 2) 2 4 (2, 4) 3 8 (3, 8) . . . . . . . . . x 2x (a) Data points from y = 2x . y=2 x (3,8) (2,4) (−1,1/2) (1,2) (0,1) (−2,1/4) −1 1 (b) Graph of y = 2x . Figure 10.5: Visualizing y = 2x . • The graph is always above the horizontal axis; i.e. the function values are always positive. • The graph has y-intercept 1 and is increasing. • The graph becomes closer and closer to the horizontal axis as we move left; i.e. the x-axis is a horizontal asymptote for the lefthand portion of the graph. • The graph becomes higher and higher above the horizontal axis as we move to the right; i.e., the graph is unbounded as we move to the right. The special case of y = 2x is representative of the function y = bx , but there are a few subtle points that need to be addressed. First, recall we are always assuming that our base b > 0. We will consider three separate cases: b = 1, b > 1, and 0 < b < 1. 10.2.1 The case b = 1 In the case b = 1, we are working with the function y = 1x = 1; this is not too exciting, since the graph is just a horizontal line. We will ignore this case. 10.2.2 The case b > 1 If b > 1, the graph of the function y = bx is qualitatively similar to the situation for b = 2, which we just considered. The only difference is the exact amount of “concavity” in the graph, but the four features highlighted above are still valid. Figure 10.6(a) indicates how these graphs 10.2. THE FUNCTIONS Y = A0 BX 139 compare for three different values of b. Functions of this type exhibit what is typically referred to as exponential growth ; this codifies the fact that the function values grow rapidly as we move to the right along the x-axis. y-axis y-axis all graphs pass through (0,1) all graphs pass through (0,1) x-axis x-axis (a) Graph of y = bx , b > 1. (b) Graph of y = bx , 0 < b < 1. Figure 10.6: Visualizing cases for b. 10.2.3 The case 0 < b < 1 We can understand the remaining case 0 < b < 1, by using the remarks above and our work in Chapter 13. First, with this condition on b, notice 1 1x that b > 1, so the graph of y = b is of the type in Figure 10.6(a). Now, using the rules of exponents: y= 1 b −x = 1 b −1 x = bx . −x 1 By the reflection principle, the graph of y = b is obtained by reflect1x ing the graph of y = ( b ) about the y-axis. Putting these remarks together, if 0 < b < 1, we conclude that the graph of y = bx will look like Figure 10.6(b). Notice, the graphs in Figure 10.6(b) share qualitative features, mirroring the features outlined previously, with the “asymptote” and “unbounded” portions of the graph interchanged. Graphs of this sort are often said to exhibit exponential decay, in the sense that the function values rapidly approach zero as we move to the right along the x-axis. Important Facts 10.2.1 (Features of Exponential Type Functions). Let b be a positive real number, not equal to 1. The graph of y = bx has these four properties: 1. The graph is always above the horizontal axis. 2. The graph has y-intercept 1. 3. If b > 1 (resp. 0 < b < 1), the graph becomes closer and closer to the horizontal axis as we move to the left (resp. move to the right); this says the x-axis is a horizontal asymptote for the left-hand portion of the graph (resp. right-hand portion of the graph). CHAPTER 10. EXPONENTIAL FUNCTIONS 140 4. If b > 1 (resp. 0 < b < 1), the graph becomes higher and higher above the horizontal axis as we move to the right (resp. move to the left); this says that the graph is unbounded as we move to the right (resp. move to the left). If A0 > 0, the graph of the function y = A0 bx is a vertically expanded or compressed version of the graph of y = bx . If A0 < 0, we additionally reflect about the x-axis. 10.3 Piano Frequency Range A sound wave will cause your eardrum to move back and forth. In the case of a so-called pure tone, this motion is modeled by a function of the form d(t) = A sin(2πft), where f is called the frequency, in units of “periods/unit time”, called “Hertz” and abbreviated “Hz”. The coefficient A is related to the actual displacement of the eardrum, which is, in turn, related to the loudness of the sound. A person can typically perceive sounds ranging from 20 Hz to 20,000 Hz. A# C# D# F# G# A# C# D# F# G#A# C#D# F# G# A# C# D# F# G#A# C# D# F# G# A# C# D# F# G# A# C# D# F# G#A# ABCDE FGABCDEF GABCDEF GABCDEF GABCDEF GABCDEF GABCDEF GABC 220 Hz middle C Figure 10.7: A piano keyboard. A piano keyboard layout is shown in Figure 10.7. The white keys are labelled A, B, C, D, E, F, and G, with the sequence running from left to right and repeating for the length of the keyboard. The black keys fit into this sequence as “sharps”, so that the black key between A and B is “A sharp”, denoted A# . Thus, starting at any A key, the 12 keys to the right are A, A# , B, C, C# , D, D# , E, F, F# , G, and G# . The sequence then repeats. Notice that between some adjacent pairs of white keys there is no black key. A piano keyboard is commonly tuned according to a rule requiring that each key (white and black) has a frequency 21/12 times the frequency of the key to its immediate left. This makes the ratio of adjacent keys always the same (21/12 ), and it means that keys 12 keys apart have a ratio of frequencies exactly equal to 2 (since (21/12 )12 = 2). Two such keys are 10.3. PIANO FREQUENCY RANGE 141 said to be an o ctave apart. Assuming that the key A below middle C has a frequency of 220 Hz, we can determine the frequency of every key on the keyboard. For instance, the A# to the right of this key has frequency 220 × 21/12 = 220 × 1.059463094... ≈ 233.08188Hz. The B to the right of this key has frequency 233.08188 × 21/12 ≈ 246.94165Hz. CHAPTER 10. EXPONENTIAL FUNCTIONS 142 10.4 Exercises Problem 10.1. Let’s brush up on the required calculator skills. Use a calculator to approximate: (a) 3π (b) 42+ √ 5 (c) ππ (d) 5− √ 3 (d) Anja, a third member of your lab working with the same yeast cells, took these two measurements: 7.246 × 106 cells after 4 hours; 16.504 × 106 cells after 6 hours. Should you be worried by Anja’s results? If Anja’s measurements are correct, does your model over estimate or under estimate the number of yeast cells at time t? 2 (e) 3π √ (f) 11π−7 Problem 10.4. middle C. (a) Find the frequency of (b) Find the frequency of A above middle C. Problem 10.2. Put each equation in standard exponential form: (a) y = 3(2−x ) (b) y = 4−x/2 (c) y = ππx (d) y = 1 (e) y = x 1 3+ 2 3 5 0.3452x−7 (f) y = 4(0.0003467)−0.4x+2 Problem 10.3. A colony of yeast cells is estimated to contain 106 cells at time t = 0. After collecting experimental data in the lab, you decide that the total population of cells at time t hours is given by the function y = 106 e0.495105t . (c) What is the frequency of the lowest note on the keyboard? Is there a way to solve this without simply computing the frequency of every key below A220? (d) The Bosendorfer piano is famous, due in part, to the fact it includes additional keys at the left hand end of the keyboard, extending to the C below the bottom A on a standard keyboard. What is the lowest frequency produced by a Bosendorfer? Problem 10.5. You have a chess board as pictured, with squares numbered 1 through 64. You also have a huge change jar with an unlimited number of dimes. On the first square you place one dime. On the second square you stack 2 dimes. Then you continue, always doubling the number from the previous square. (a) How many cells are present after one hour? (a) How many dimes will you have stacked on the 10th square? (b) (True or False) The population of yeast cells will double every 1.4 hours. (b) How many dimes will you have stacked on the nth square? (c) Cherie, another member of your lab, looks at your notebook and says : ...that formula is wrong, my calculations predict the formula for the number of yeast cells is given by the function (c) How many dimes will you have stacked on the 64th square? y = 106 (2.042727)0.693147t . Should you be worried by Cherie’s remark? (d) Assuming a dime is 1 mm thick, how high will this last pile be? (e) The distance from the earth to the sun is approximately 150 million km. Relate the height of the last pile of dimes to this distance. 10.4. EXERCISES 143 63 64 calculates the fraction of hemoglobin saturated with oxygen at a given pressure p. (a) The graphs of M(p) and H(p) are given below on the domain 0 ≤ p ≤ 100; which is which? fraction 1 0.8 0.6 10 9 0.4 1 2 3 8 0.2 20 Problem 10.6. Myoglobin and hemoglobin are oxygen carrying molecules in the human body. Hemoglobin is found inside red blood cells, which flow from the lungs to the muscles through the bloodstream. Myoglobin is found in muscle cells. The function p Y = M(p) = 1+p calculates the fraction of myoglobin saturated with oxygen at a given pressure p torrs. For example, at a pressure of 1 torr, M(1) = 0.5, which means half of the myoglobin (i.e. 50%) is oxygen saturated. (Note: More precisely, you need to use something called the “partial pressure”, but the distinction is not important for this problem.) Likewise, the function Y = H(p) = p2.8 262.8 + p2.8 40 60 80 100 p (b) If the pressure in the lungs is 100 torrs, what is the level of oxygen saturation of the hemoglobin in the lungs? (c) The pressure in an active muscle is 20 torrs. What is the level of oxygen saturation of myoglobin in an active muscle? What is the level of hemoglobin in an active muscle? (d) Define the efficiency of oxygen transport at a given pressure p to be M(p) − H(p). What is the oxygen transport efficiency at 20 torrs? At 40 torrs? At 60 torrs? Sketch the graph of M(p) − H(p); are there conditions under which transport efficiency is maximized (explain)? 144 CHAPTER 10. EXPONENTIAL FUNCTIONS Chapter 11 Exponential Modeling Example 11.0.1. A computer industry spokesperson has predicted that the number of subscribers to geton.com, an internet provider, will grow exponentially for the first 5 years. Assume this person is correct. If geton.com has 100,000 subscribers after 6 months and 750,000 subscribers after 12 months, how many subscribers will there be after 5 years? Solution. The solution to this problem offers a template for many exponential modeling applications. Since, we are assuming that the number of subscribers N(x), where x represents years, is a function of exponential type, N(x) = N◦ bx , for some N◦ and b > 1. We are given two pieces of information about the values of N(x): N(0.5) = 100,000; i.e., N0 b0.5 = 100,000, and N(1) = 750,000; i.e., N0 b = 750,000. We can use these two equations to solve for the two unknowns N◦ and b as follows: If we divide the second equation by the first, we get b1 = 7.5 b0.5 √ b0.5 = b1/2 = b = 7.5 ∴ b = 56.25. Plugging this value of b into either equation (say the first one), we can 100,000 solve for N◦ : N◦ = (56.25)0.5 = 13,333. We conclude that the number of geton.com subscribers will be predicted by N(x) = 13,333(56.25)x. In five years, we obtain N(5) = 7,508,300,000,000 subscribers, which exceeds the population of the Earth (which is between 5 and 6 billion)! 145 CHAPTER 11. EXPONENTIAL MODELING 146 Q 800000 600000 There are two important conclusions we can draw from this problem. First, the given information provides us with two points on the graph of the function N(x): 400000 200000 P 0.2 0.4 0.6 0.8 1 Figure 11.1: Finding the equation for N(x) = N0 bx . P = (0.5, 100,000) Q = (1, 750,000). More importantly, this example illustrates a very important principal we can use when modeling with functions of exponential type. Important Fact 11.0.2. A function of exponential type can be determined if we are given two data points on its graph. !!! CAUTION !!! When you use the above strategy to find the base b of the exponential model, make sure to write down a lengthy decimal approximation. As a rule of thumb, go for twice as many significant digits as you are otherwise using in the problem. 11.1 The Method of Compound Interest You walk into a Bank with P0 dollars (usually called principal), wishing to invest the money in a savings account. You expect to be rewarded by the Bank and paid interest, so how do you compute the total value of the account after t years? The future value of the account is really a function of the number of years t elapsed, so we can write this as a function P(t). Our goal is to see that P(t) is a function of exponential type. In order to compute the future value of the account, the Bank provides any savings account investor with two important pieces of information: r = annual (decimal) interest rate n = the number of compounding periods per year The number n tells us how many times each year the Bank will compute the total value P(t) of the account. For example, if n = 1, the calculation is done at one-year intervals; if n = 12, the calculation is done each month, etc. The bank will compute the value of your account after a r typical compounding period by using the periodic rate of return n . For example, if the interest rate percentage is 12% and the compounding period is monthly (i.e., n = 12), then the annual (decimal) interest rate is 0.12 and the periodic rate is 0.12 = 0.01. 12 !!! CAUTION !!! The number r always represents the decimal interest rate, which is a decimal between 0 and 1. If you are given the interest rate percentage (which is a positive number between 0 and 100), you need to convert to a decimal by dividing by 100. 11.1. THE METHOD OF COMPOUND INTEREST 147 11.1.1 Two Examples Let’s consider an example: P0 = $1,000 invested at the annual interest percentage of 8% compounded yearly, so n = 1 and r = 0.08. To compute the value P(1) after one year, we will have P(1) = P0 + (periodic rate)P0 = P0 + rP0 = P0 (1 + r) = $1,000(1 + 0.08) = $1,080. To compute the value after two years, we need to apply the periodic rate to the value of the account after one year: P(2) = P(1) + (periodic rate)P(1) = P0 (1 + r) + rP0 (1 + r) = P0 (1 + r)2 = $1,000(1 + 0.08)2 = $1,166.40. Notice, the amount the Bank has paid after two years is $166.40, which is slightly bigger than twice the $80 paid after one year. To compute the value after three years, we need to apply the periodic rate to the value of the account after two years: P(3) = P(2) + (periodic rate)P(2) = P0 (1 + r)2 + rP0 (1 + r)2 = P0 (1 + r)3 = $1,000(1 + 0.08)3 = $1,259.71. Again, notice the amount the Bank is paying after three years is $259.71, which is slightly larger than three times the $80 paid after one year. Continuing on in this way, to find the value after t years, we arrive at the formula P(t) = P0 (1 + r)t = $1,000(1.08)t. In particular, after 5 and 10 years, the value of the account (to the nearest dollar) will be $1,469 and $2,159, respectively. As a second example, suppose we begin with the same $1,000 and the same annual interest percentage 8%, but now compound monthly, so n = 12 and r = 0.08. The value of the account after one compounding period is P(1/12), since a month is one-twelfth of a year. Arguing as r before, paying special attention that the periodic rate is now n = 0.08 , we 12 have P(1/12) = P0 + (periodic rate)P0 .08 = P0 1 + 12 = $1,000(1 + 0.006667) = $1,006.67. CHAPTER 11. EXPONENTIAL MODELING 148 After two compounding periods, the value is P(2/12), P(2/12) = P(1/12) + (periodic rate)P(1/12) .08 0.08 .08 = P0 1 + + P0 1 + 12 12 12 2 .08 = P0 1 + 12 = $1,000(1 + 0.006667)2 = $1,013.38. Continuing on in this way, after k compounding periods have elapsed, k the value will be P 12 , which is computed as P(k/12) = P0 1 + .08 12 k . It is possible to rewrite this formula to give us the value after t years, noting that t years will lead to 12t compounding periods; i.e., set k = 12t in the previous formula: P(t) = P0 .08 1+ 12 12t For example, after 1, 5 and 10 years, the value of the account, to the nearest dollar, would be $1,083, $1,490, and $2,220. 11.1.2 Discrete Compounding The two examples above highlight a general formula for computing the future value of an account. Important Fact 11.1.1 (Discrete compounding). Suppose an account is opened with P0 principal. If the decimal interest rate is r and the number of compounding periods per year is n, then the value P(t) of the account after t years will be P(t) = P0 1 + r n nt . Notice, the future value P(t) is a function of exponential type; the base r is the number 1 + n , which will be greater than one. Since P0 > 0, the graph will be qualitatively similar to the ones pictured in Figure 10.6(a). Example 11.1.2. At birth, your Uncle Hans secretly purchased a $5,000 U.S. Savings Bond for $2,500. The conditions of the bond state that the U.S. Government will pay a minimum annual interest rate of r = 8.75%, compounded quarterly. Your Uncle has given you the bond as a gift, subject to the condition that you cash the bond at age 35 and buy a red Porsche. 11.2. THE NUMBER E AND THE EXPONENTIAL FUNCTION 149 On your way to the Dealer, you receive a call from your tax accountant informing you of a 28% tax on the capital gain you realize through cashing in the bond; the capital gain is the selling price of the bond minus the purchase price. Before stepping onto the showroom floor, compute how much cash will you have on hand, after the U.S. Government shares in your profits. Solution. The value of your bond after 35 years is computed by the formula in Fact 11.1.1, using P0 = $2,500, r = 0.0875, n = 4, and t = 35. Plugging this all in, we find that the selling price of the bond is P(35) = $2,500 1 + 0.0875 4 4(35) = $51,716.42. The capital gain will be $51,716.42 - $2,500 = $49,216.42 and the tax due is $(49,216.42)(0.28) = $13,780.60. You are left with $51,716.42 $13,780.60 = $37,935.82. Better make that a used Porsche! 11.2 The Number e and the Exponential Function What happens to the future value of an investment of P0 dollars as the number of compounding periods is increased? For example, return to our earlier example: P0 = $1000 and an annual interest percentage of 8%. After 1 year, the table below indicates the value of the investment for various compounding periods: yearly, quarterly, monthly, weekly, daily, and hourly. Value after 1 year (to nearest dollar) n Compounding Period 1 yearly 4 quarterly $1,000(1 + 0.08)1 = $1,080.00 $1,000 1 + 0.08 4 = 4 $1,082.43 12 monthly $1,000 1 + 0.08 12 12 = $1,083.00 52 weekly $1,000 1 + 0.08 52 52 = $1,083.22 0.08 365 365 = $1,083.28 0.08 8,760 8,760 =$1,083.29 365 daily 8,760 hourly $1,000 1 + $1,000 1 + CHAPTER 11. EXPONENTIAL MODELING 150 We could continue on, considering “minute” and “second” compounding and what we will find is that the value will be at most $1,083.29. This illustrates a general principal: Important Fact 11.2.1. Initially increasing the number of compounding periods makes a significant difference in the future value; however, eventually there appears to be a limiting value. Let’s see if we can understand mathematically why this is happening. The first step is to recall the discrete compounding formula: r nt . n If our desire is to study the effect of increasing the number of compounding periods, this means we want to see what happens to this formula as n gets BIG. To analyze this, it is best to rewrite the expression using a r substitution trick: Set z = n , so that n = rz and n = 1 . Plugging in, we r z have r nt P(t) = P0 1 + n rzt 1 = P0 1 + (11.1) z P(t) = P0 1 + = P0 1+ 1 z z rt . So, since r is a fixed number and z = n , letting n get BIG is the same r as letting z become BIG in (11.1). This all means we need to answer z this new question: What happens to the expression 1 + 1 as z becomes z large? On the one hand, the power in the expression is getting large; at the same time, the base is getting close to 1. This makes it very tricky to make quick predictions about the outcome. It is best to first tabulate z some numerical data for the values of y = g(z) = 1 + 1 and look at a z plot of this function graph on the domain 0.01 ≤ z ≤ 100: See Figure 11.2. z You can see from this plot, the graph of y = 1 + 1 approaches the z “dashed” horizontal asymptote, as z becomes BIG. We will let the letter “e” represent the spot where this horizontal line crosses the vertical axis and e ≈ 2.7182818. This number is only an approximation, since e is known to be an irrational number. What sets this irrational number apart from the √ ones you are familiar with (e.g. 2, π, etc.) is that defining the number e requires a “limiting” process. This will be studied a lot more in your Calculus course. The new number e is a positive number greater than 1, so we can study the function: y = ex . (11.2) Since e > 1, the graph will share the properties in Figure 10.6(a). This function is usually referred to as THE exponential function. Scientific calculators will have a key of the form “ exp(x) ” or “ ex ”. 11.2. THE NUMBER E AND 1+ z 1 2 3 4 20 100 1000 109 1 z THE EXPONENTIAL FUNCTION 151 z 2 2.25 2.37037 2.4414 2.65329 2.70481 2.71692 2.71828 2.5 2 1.5 1 0.5 20 (a) Data points for z 1+ 1 . z 40 60 80 (b) The graph of 1 + 100 1z . z Figure 11.2: What happens when z get very large? 11.2.1 Calculator drill Plugging in x = 1, you can compute an approximation to e on your calculator; you should get e = 2.7183 to four decimal places. Make sure you √ can compute expressions like e3 , eπ , and e = e1/2 ; to four decimal places, you should get 20.0855, 23.1407, and 1.6487. 11.2.2 Back to the original problem... We can now return to our future value formula (11.1) and conclude that as the number of compounding periods increases, the future value is approaching a limiting value: P(t) = P0 1 1+ z z rt =⇒ P0 ert . The right hand limiting formula Q(t) = P0 ert computes the future value using what is usually referred to as continuous compounding. From the investors viewpoint, this is the best possible scheme for computing future value. Important Fact 11.2.2 (Continuous compounding). The future value of P0 dollars principal invested at an annual decimal interest rate of r under continuous compounding after t years is Q(t) = P0 ert ; this value is alr nt ways greater than the value of P0 1 + n , for any discrete compounding scheme. In fact, P0 ert is the limiting value. CHAPTER 11. EXPONENTIAL MODELING 152 11.3 Exercises Problem 11.1. In 1968, the U.S. minimum wage was $1.60 per hour. In 1976, the minimum wage was $2.30 per hour. Assume the minimum wage grows according to an exponential model w(t), where t represents the time in years after 1960. (a) Find a formula for w(t). (b) What does the model predict for the minimum wage in 1960? (a) Sketch rough graphs of these two functions. (b) The graph of the equation x2 − y2 = 1 is shown below; this is called the unit hyperbola. For any value a, show that the point (x,y) = (cosh(a), sinh(a)) is on the unit hyperbola. (Hint: Verify that [cosh(x)]2 − [sinh(x)]2 = 1, for all x.) y (c) If the minimum wage was $5.15 in 1996, is this above, below or equal to what the model predicts. Problem 11.2. The town of Pinedale, Wyoming, is experiencing a population boom. In 1990, the population was 860 and five years later it was 1210. (−1,0) (1,0) x (a) Find a linear model l(x) and an exponential model p(x) for the population of Pinedale in the year 1990+x. (b) What do these models estimate the population of Pinedale to be in the year 2000? Problem 11.3. In 1989, research scientists published a model for predicting the cumulative number of AIDS cases reported in the United States: a(t) = 155 t − 1980 10 3 (thousands) , where t is the year. This paper was considered a “relief”, since there was a fear the correct model would be of exponential type. Pick two data points predicted by the research model a(t) to construct a new exponential model b(t) for the number of cumulative AIDS cases. Discuss how the two models differ and explain the use of the word “relief”. Problem 11.4. Define two new functions: x y = cosh(x) = and y = a cosh( x−h ) + C, a for appropriate constants a, h and C. The constant h depends on how the coordinate system is imposed. A cable for a suspension bridge hangs from two 100 ft. high towers located 400 ft. apart. Impose a coordinate system so that the picture is symmetric about the y-axis and the roadway coincides with the x-axis. The hanging cable constant is a = 500 and h = 0. Find the minimum distance from the cable to the road. towers −x e +e 2 x (c) A hanging cable is modeled by a portion of the graph of the function −x e −e . 2 These are called the basic hyperbolic trigonometric functions. cable d y = sinh(x) = 100 ft roadway 400 ft Chapter 12 Logarithmic Functions If we invest P0 = $1,000 at an annual rate of r = 8% compounded continuously, how long will it take for the account to have a value of $5000? The formula P(t) = 1,000e0.08t gives the value after t years, so we need to solve the equation: 5,000 = 1,000e0.08t 5 = e0.08t . Unfortunately, algebraic manipulation will not lead to a further simplification of this equation; we are stuck! The required technique involves the theory of inverse functions. Assuming we can find the inverse function of f(t) = et , we can apply f−1 (t) to each side of the equation and solve for t: f−1 (5) = f−1 (e0.08t ) = 0.08t (12.5)f−1(5) = t The goal in this section is to describe the function f−1 , which is usually denoted by the symbol f−1 (t) = ln(t) and called the natural logarithm function. On your calculator, you will find a button dedicated to this function and we can now compute ln(5) = 1.60944. Conclude that the solution is t = 20.12 years. 12.1 The Inverse Function of y = ex If we sketch a picture of the exponential function on the domain of all real numbers and keep in mind the properties in Fact 10.2.1, then every horizontal line above the x-axis intersects the graph of y = ex exactly once: See Figure 12.1(a). The range of the exponential function will consist of all possible y-coordinates of points on the graph. Using the graphical techniques of Chapter 6, we can see that the range of will be all POSITIVE real numbers: See Figure 12.1(b). 153 CHAPTER 12. LOGARITHMIC FUNCTIONS 154 PSfrag y=e x y = ex range = all positive numbers graphs of y=c, c>1 cross exponential graph exactly once. −1 1 these horizontal lines miss graph of exponential function. −1 1 domain = all real numbers (a) Horizontal line test for (b) The domain and range for y = ex . y = ex . Figure 12.1: Properties needed to find the inverse of f(x) = ex . By the horizontal line test, this means the exponential function is oneto-one and the inverse rule f−1 (c) will define a function the unique solu tion of the equation , if c > 0 −1 c = ex (12.1) f (c ) = (undefined), if c ≤ 0. y=e x reflecting line y = x y=ln (x) −1 1 Figure 12.2: Visualizing the y = ln(x). This inverse function is called the natural logarithm function, denoted ln(c). We can sketch the graph of the the natural logarithm as follows: First, by Fact 9.2.1, the domain of the function ln(y) = x is just the range of the exponential function, which we noted is all positive numbers. Likewise, the range of the function ln(y) = x is the domain of the exponential function, which we noted is all real numbers. Interchanging x and y, the graph of the natural logarithm function y = ln(x) can be obtained by flipping the graph of y = ex across the line y = x: Important Facts 12.1.1 (Graphical features of natural log). The function y = ln(x) has these features: • The largest domain is the set of positive numbers; e.g. ln(−1) makes no sense. • The graph has x-intercept 1 and is increasing. • The graph becomes closer and closer to the vertical axis as we approach x = 0; i.e. the y-axis is a vertical asymptote for the graph. • The graph is unbounded as we move to the right. Any time we are working with an inverse function, symbolic properties are useful. Here are the important ones related to the natural logarithm. Important Facts 12.1.2 (Natural log properties). We have the following properties: 12.1. THE INVERSE FUNCTION OF Y = EX 155 (a) For any real number x, ln(ex ) = x. (b) For any positive number x, eln(x) = x. (c) ln(bt ) = t ln(b), for b > 0 and t any real number; (d) ln(ba) = ln(a) + ln(b), for all a, b > 0; (e) ln b a = ln(b) − ln(a), for all a, b > 0. The properties (c)-(e) are related to three of the rules of exponents in Facts 10.1.1. Here are the kinds of basic symbolic maneuvers you can pull off using these properties: Examples 12.1.3. (i) ln(83 ) = 3 ln(8) = 6.2383; ln(6π) = ln(6) + ln(π) = 2.9365; ln ln(3) − ln(5) = −0.5108. (ii) ln √ 3 5 = 1 x = ln x1/2 = 2 ln(x); ln x2 − 1 = ln((x − 1)(x + 1)) = ln(x − 1) + ln(x + 1); ln x5 x2 +1 = ln x5 − ln x2 + 1 = 5 ln(x) − ln x2 + 1 . Examples 12.1.4. Given the equation 3x+1 = 12, we can solve for x: 3x+1 = 12 ln 3x+1 = ln(12) (x + 1) ln(3) = ln(12) ln(12) x= − 1 = 1.2619. ln(3) Example 12.1.5. If $2,000 is invested in a continuously compounding savings account and we want the value after 12 years to be $130,000, what is the required annual interest rate? If, instead, the same $2,000 is invested in a continuously compounding savings account with r = 6.4% annual interest, when will the exact account value be be $130,000? Solution. In the first scenario, 130,000 = 2,000e12r 65 = e12r ln(65) = ln e12r ln(65) = 12r ln(65) = 0.3479. r= 12 CHAPTER 12. LOGARITHMIC FUNCTIONS 156 This gives an annual interest rate of 34.79%. In the second scenario, we study the equation 130,000 = 2,000e(0.064)t 65 = e(0.064)t ln(65) = 65.22. t= 0.064 So, it takes over 65 years to accumulate $130,000 under the second scheme. 12.2 Alternate form for functions of exponential type The standard model for an exponential function is A(t) = A0 bt , for some b > 0, b = 1, and A0 = 0. Using the properties of the natural logarithm function, bt = eln(b) t = et ln(b) . This means that every function as in Definition 10.1.2 can be re-written using the exponential function et . Another way of saying this is that you really only need the function keys “ et ” and “ ln(t) ” on your calculator. Important Fact 12.2.1 (Observation). A function of exponential type can be written in the form A(t) = A0 eat , for some constants A0 = 0 and a = 0. By studying the sign of the constant a, we can determine whether the function exhibits exponential growth or decay. For example, given the function A(t) = eat , if a > 0 (resp. a < 0), then the function exhibits exponential growth (resp. decay). Examples 12.2.2. (a) The function A(t) = 200 (2t ) exhibits exponential growth and can be re-written as: A(t) = 200 et ln(2) = 200e0.69315t (b) The function A(t) = 4e−0.2t exhibits exponential decay and can be rewritten as: A(t) = 4e−0.2t = 4 e−0.2 t = 4 0.81873t . 12.3. THE INVERSE FUNCTION OF Y = BX 157 12.3 The Inverse Function of y = bx For some topics in Chemistry and Physics (e.g. acid base equilibria and acoustics) it is useful to have on hand an inverse function for y = bx , where b > 0 and b = 1. Just as above, we would show that f(x) = bx is one-to-one, the range is all positive numbers and obtain the graph using ideas in Figure 12.2. We will refer to the inverse rule as the logarithm function base b, denoted logb (x), defined by the rule: logb (c) = the unique solution of the equation , if c > 0 c = bx (undefined), if c ≤ 0. We will need to consider two cases, depending on the magnitude of b: The important qualitative features of the logarithm function y = logb (x) mirror Fact 12.1.1: x y=b reflecting line y = x x y=b reflecting line y = x y=logb(x) −1 1 −1 1 y=log (x) b The case b > 1 The case 0 < b < 1 Figure 12.3: Cases to consider for b. Important Facts 12.3.1 (Graphical features of general logs). The function y = logb (x) has these features: • The largest domain is the set of positive numbers; e.g. logb (−1) is not defined. • The graph has x-intercept 1 and is increasing if b > 1 (resp. decreasing if 0 < b < 1). • The graph becomes closer and closer to the vertical axis as we approach x = 0; this says the y-axis is a vertical asymptote for the graph. CHAPTER 12. LOGARITHMIC FUNCTIONS 158 • The graph is unbounded as we move to the right. Important Facts 12.3.2 (Log properties). Fix a positive base b, b = 1. (a) For any real number x, logb (bx ) = x. (b) For any positive number x, blogb (x) = x. (c) logb (rt ) = t logb (r), for r > 0 and t any real number; (d) logb (rs) = logb (r) + logb (s), for all r, s > 0; r (e) logb ( s ) = logb (r) − logb (s), for all r, s > 0. It is common to simplify terminology and refer to the function logb (x) as the log base b function, dropping the longer phrase “logarithm”. Some scientific calculators will have a key devoted to this function. Other calculators may have a key labeled “log(x)”, which is usually understood to mean the log base 10. However, many calculators only have the key “ln(x)”. This is not cause for alarm, since it is always possible to express logb (x) in terms of the natural log function. Let’s see how to do this, since it is a great application of the Log Properties listed in Fact 12.3.2. Suppose we start with y = logb (x). We will rewrite this in terms of the natural log by carrying out a sequence of algebraic steps below; make sure you see why each step is justified. y = logb (x) by = x ln(by ) = ln(x) y ln(b) = ln(x) ln(x) y= . ln(b) We have just verified a useful conversion formula: Important Fact 12.3.3 (Log conversion formula). For x a positive number and b > 0, b = 1 a base, logb (x) = ln(x) . ln(b) For example, ln(5) = 0.699 ln(10) ln(11) log0.02 (11) = = −0.613 ln(0.02) ln( 1 ) 1 2 = log20 = −0.2314 2 ln(20) log10 (5) = 12.4. MEASURING THE LOUDNESS OF SOUND 159 The conversion formula allows one to proceed slightly differently when solving equations involving functions of exponential type. This is illustrated in the next example. Example 12.3.4. Ten years ago, you purchased a house valued at $80,000. Your plan is to sell the house at some point in the future, when the value is at least $1,000,000. Assume that the future value of the house can be computed using quarterly compounding and an annual interest rate of 4.8%. How soon can you sell the house? Solution. We can use the future value formula to obtain the equation 1,000,000 = 80,000 1 + 0.048 4 4t 12.5 = (1.012)4t Using the log base b = 1.012, log1.012 (12.5) = log1.012 (1.012)4t log1.012 (12.5) = 4t ln(12.5) t= = 52.934. 4 ln(1.012) Since you have already owned the house for 10 years, you would need to wait nearly 43 years to sell at the desired price. 12.4 Measuring the Loudness of Sound As we noted earlier, the reception of a sound wave by the ear gives rise to a vibration of the eardrum with a definite frequency and a definite amplitude. This vibration may also be described in terms of the variation of air pressure at the same point, which causes the eardrum to move. The perception that rustling leaves and a jet aircraft sound different involves two concepts: (1) the fact that the frequencies involved may differ; (2) the intuitive notion of “loudness”. This loudness is directly related to the force being exerted on the eardrum, which we refer to as the intensity of the sound. We can try to measure the intensity using some sort of scale. This becomes challenging, since the human ear is an amazing instrument, capable of hearing a large range of sound intensities. For that reason, a logarithmic scale becomes most useful. The sound pressure level β of a sound is defined by the equation β = 10 log10 I I0 , (12.2) where I0 is an arbitrary reference intensity which is taken to correspond with the average faintest sound which can be heard and I is the intensity CHAPTER 12. LOGARITHMIC FUNCTIONS 160 of the sound being measured. The units used for β are called decibels, abbreviated “db”. (Historically, the units of loudness were called bels, in honor of Alexander Graham Bell, referring to the quantity log10 II0 .) Notice, in the case of sound of intensity I = I0 , we have a sound pressure level of β = 10 log10 I0 I0 = 10 log10 (1) = 10(0) = 0. We refer to any sound of intensity I0 as having a sound pressure level at the threshold of hearing. At the other end of the scale, a sound of intensity the maximum the eardrum can tolerate has an average sound pressure level of about 120 db. The Table 12.4(a) gives a hint of the sound pressure levels associated to some common sounds. Source of Noise Threshold of pain Riveter Busy Street Traffic Ordinary Conversation Quiet Auto Background Radio Whisper Rustle of Leaves Threshold of Hearing Sound Pressure Level in db 120 95 70 65 50 40 20 10 0 (a) Sources of noise levels. pain threshold 120 100 Zone of Hearing 80 db 60 40 20 0 20 100 hearing threshold 1000 10,000 20,000 Hz (b) Graphing noise levels. Figure 12.4: Considering noise levels. It turns out that the above comments on the threshold of hearing and pain are really only averages and depend upon the frequency of the given sound. In fact, while the threshold of pain is on average close to 120 db across all frequencies between 20 Hz and 20,000 Hz, the threshold of hearing is much more sensitive to frequency. For example, for a tone of 20 Hz (something like the ground-shaking rumble of a passing freight train), the sound pressure level needs to be relatively high to be heard; 100 db on average. As the frequency increases, the required sound pressure level for hearing tends to drop down to 0 db around 2000 Hz. An examination by a hearing specialist can determine the precise sensitivities of your ear across the frequency range, leading to a plot of your “envelope of hearing”; a sample plot is given in Figure 12.4(b). Such a plot would differ from person to person and is helpful in isolating hearing problems. Example 12.4.1. A loudspeaker manufacturer advertises that their model no. 801 speaker produces a sound pressure level of 87 db when a reference test tone is applied. A competing speaker company advertises that 12.4. MEASURING THE LOUDNESS OF SOUND 161 their model X-1 speaker produces a sound pressure level of 93 db when fed the same test signal. What is the ratio of the two sound intensities produced by these speakers? If you wanted to find a speaker which produces a sound of intensity twice that of the no. 801 when fed the test signal, what is its sound pressure level? Solution. If we let I1 and I2 refer to the sound intensities of the two speakers reproducing the test signal, then we have two equations: 87 = 10 log10 93 = 10 log10 I1 I0 I2 I0 Using log properties, we can solve the first equation for I1 : I1 = 10 log10 (I1 ) − 10 log10 (I0 ) I0 log10 (I1 ) = 8.7 + log10 (I0 ) 87 = 10 log10 10log10 (I1 ) = 108.7+log10 (I0 ) I1 = 108.7 10log10 (I0 ) = 108.7 I0 . Similarly, we find that I2 = 109.3 I0 . This means that the ratio of the intensities will be I2 109.3 I0 = 8.7 = 100.6 = 3.98. I1 10 I0 This means that the test signal on the X − 1 speaker produces a sound pressure level nearly 4 times that of the same test signal on the no. 801 speaker. To finish the problem, imagine a third speaker which produces a sound pressure level β, which is twice that of the first speaker. If I3 is the corresponding intensity of the sound, then as above, I3 = 10(β/10) I0 . We are assuming that I3 = 2I1 , so this gives us the equation log10 1 I1 = I3 2 1 108.7 I0 = 10(β/10) I0 2 1 (β/10) 10 108.7 = log10 2 1 8.7 = log10 + log10 10(β/10) 2 β 8.7 = −0.30103 + 10 90 = β 162 CHAPTER 12. LOGARITHMIC FUNCTIONS So, the test signal on the third speaker must produce a sound pressure level of 90 db. 12.5. EXERCISES 163 12.5 Exercises Problem 12.1. These problems will help you develop your skills with logarithms. (a) Compute: log5 3, loge 11, log√2 π, log2 10, log10 2. (b) Solve for x: 35 = ex , log3 x = e, log3 5 = xe3 . (c) Solve each of these equations for x in terms of y: y = 10x , 3y = 10x , y = 103x . Problem 12.2. As light from the surface penetrates water, its intensity is diminished. In the clear waters of the Caribbean, the intensity is decreased by 15 percent for every 3 meters of depth. Thus, the intensity will have the form of a general exponential function. (a) If the intensity of light at the water’s surface is I◦ , find a formula for I(d), the intensity of light at a depth of d meters. Your formula should depend on I◦ and d. (b) At what depth will the light intensity be decreased to 1% of its surface intensity? Problem 12.3. Rewrite each function in the form y = A◦ eat , for appropriate constants A◦ and a. (a) y = 13(3t ) (b) y = 2( 1 )t 8 (c) y = −7(1.567)t−3 (d) y = −17(2.005)−t (e) y = 3(14.24)4t Problem 12.4. (a) If you invest Po dollars at 7% annual interest and the future value is computed by continuous compounding, how long will it take for your money to double? (b) Suppose you invest Po dollars at r% annual interest and the future value is computed by continuous compounding. If you want the value of the account to double in 2 years, what is the required interest rate? (c) A rule of thumb used by many people to determine the length of time to double an investment is the rule of 70. The rule says it takes about t = 70 years to dour ble the investment. Graphically compare this rule to the one isolated in part b. of this problem. Problem 12.5. The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) = 200 (1 − 0.956 e−0.18t ) where L(t) is the length (in centimeters) of a fish t years old. (a) What is the length of a new-born halibut at birth? (b) Use the formula to estimate the length of a 6–year–old halibut. (c) At what age would you expect the halibut to be 120 cm long? (d) What is the practical (physical) significance of the number 200 in the formula for L(t)? Problem 12.6. A cancerous cell lacks normal biological growth regulation and can divide continuously. Suppose a single mouse skin cell is cancerous and its mitotic cell cycle (the time for the cell to divide once) is 20 hours. The number of cells at time t grows according to an exponential model. (a) Find a formula C(t) for the number of cancerous skin cells after t hours. (b) Assume a typical mouse skin cell is spherical of radius 50 × 10−4 cm. Find the combined volume of all cancerous skin cells after t hours. When will the volume of cancerous cells be 1 cm3 ? Problem 12.7. Your Grandfather purchased a house for $55,000 in 1952 and it has increased in value according to a function y = v(x), where x is the number of years owned. These questions probe the future value of the house under various mathematical models. CHAPTER 12. LOGARITHMIC FUNCTIONS 164 (a) Suppose the value of the house is $75,000 in 1962. Assume v(x) is a linear function. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? (b) Suppose the value of the house is $75,000 in 1962 and $120,000 in 1967. Assume v(x) is a quadratic function. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? (c) Suppose the value of the house is $75,000 in 1962. Assume v(x) is a function of exponential type. Find a formula for v(x). What is the value of the house in 1995? When will the house be valued at $200,000? Problem 12.8. Solve the following equations for x: (a) log3 (5) = log2 (x) (b) 10log2 (x) = 3 x (c) 35 = 7 (d) log2 (ln(x)) = 3 (e) ex = 105 (f) 23x+5 = 32 Problem 12.9. A ship embarked on a long voyage. At the start of the voyage, there were 500 ants in the cargo hold of the ship. One week into the voyage, there were 800 ants. Suppose the population of ants is an exponential function of time. (a) How long did it take the population to double? (b) How long did it take the population to triple? (c) When were there be 10,000 ants on board? (d) There also was an exponentially-growing population of anteaters on board. At the start of the voyage there were 17 anteaters, and the population of anteaters doubled every 2.8 weeks. How long into the voyage were there 200 ants per anteater? Problem 12.10. The populations of termites and spiders in a certain house are growing exponentially. The house contains 100 termites the day you move in. After 4 days, the house contains 200 termites. Three days after moving in, there are two times as many termites as spiders. Eight days after moving in, there were four times as many termites as spiders. How long (in days) does it take the population of spiders to triple? Problem 12.11. In 1987, the population of Mexico was estimated at 82 million people, with an annual growth rate of 2.5%. The 1987 population of the United States was estimated at 244 million with an annual growth rate of 0.7 %. Assume that both populations are growing exponentially. (a) When will Mexico double its 1987 population? (b) When will the United States and Mexico have the same population? Problem 12.12. The cities of Abnarca and Bonipto have populations that are growing exponentially. In 1980, Abnarca had a population of 25,000 people. In 1990, its population was 29,000. Bonipto had a population of 34,000 in 1980. The population of Bonipto doubles every 55 years. (a) How long does it take the population of Abnarca to double? (b) When will Abnarca’s population equal that of Bonipto? Chapter 13 Three Construction Tools Sometimes the composition of two functions can be understood by graphical manipulation. When we discussed quadratic functions and parabolas in the previous section, certain key graphical maneuvers were laid out. In this section, we extend those graphical techniques to general function graphs. 13.1 A Low-tech Exercise This section is all about building new functions from ones we already have in hand. This can be approached symboly = f ( x) ically or graphically. Let’s begin with a simple hands-on exercise involving the curve in Figure 13.1. By the vertical line test, we know this represents the graph of a function y = f(x). With this picture and a piece Figure 13.1: Start with some of bendable wire we can build an INFINITE number of new curve. functions from the original function. Begin by making a “model” of this graph by bending a piece of wire to the exact shape of the graph and place it right on top of the curve. The wire model can be manipulated in a variety of ways: slide the model back and forth horizontally, up and down vertically, expand or compress the model horizontally or vertically. Another way to build new curves from old ones is to exploit the built in symmetry of the xy-coordinate system. For example, imagine reflecting the graph of y = f(x) across the x-axis or the y-axis. In all of the above cases, we moved from the original rotate wire model of our function graph to a new curve that (by the vertical line test) is the graph of a new function. The big caution in all this is that we are NOT ALLOWED to NOT a function graph rotate or twist the curve; this kind of maneuver does lead Figure 13.2: Rotating a to a new curve, but it may not be the graph of a function: curve. See Figure 13.2. The pictures in Figure 13.3 highlight most of what we have to say in 165 CHAPTER 13. THREE CONSTRUCTION TOOLS 166 this section; the hard work remaining is a symbolic reinterpretation of these graphical operations. left right up down Horizontal shift. pull pull expand horizontally push push compress horizontally Horizontal expansion. pull Vertical shift. Horizontal compression. pull push push compress vertically expand vertically Vertical expansion. Vertical compression. reflect across y-axis reflect across x-axis Vertical reflection. Horizontal reflection. Figure 13.3: Shifting, dilating, and reflecting y = f(x). 13.2 Reflection In order to illustrate the technique of reflection, we will use a concrete example: Function: Domain: Range: y = p(x) = 2x + 2 −2 ≤ x ≤ 2 −2 ≤ y ≤ 6 As we know, the graph of y = p(x) on the domain −2 ≤ x ≤ 2 is a line of slope 2 with y-intercept 2, as pictured in Figure 13.4(a). Now, start with 13.2. REFLECTION 167 the function equation y = p(x) = 2x + 2 and replace every occurrence of “y” by “−y.” This produces the new equation −y = 2x + 2; or, equivalently, y = q(x) = −2x − 2. The domain is still −2 ≤ x ≤ 2, but the range will change; we obtain the new range by replacing “y” by “−y” in the original range: −2 ≤ −y ≤ 6; so 2 ≥ y ≥ −6. The graph of this function is a DIFFERENT line; this one has slope −2 and y-intercept −2. We contrast these two curves in Figure 13.4(b), where q(x) is graphed as the “dashed line” in the same picture with the original p(x). Once we do this, it is easy to see how the graph of q(x) is really just the original line reflected across the x-axis. Next, take the original function equation y = p(x) = 2x + 2 and replace every occurrence of “x” by “−x.” This produces a new equation y = 2(−x) + 2; or, equivalently, 8 6 4 2 −2 −2 −4 −6 −8 1 8 6 4 2 2 −2 y-axis −1 −2 −4 −6 −8 1 x-axis 2 (b) Graph of q(x) = −2x − 2. 8 6 4 2 −2 −1 −2 −4 −6 −8 y-axis x-axis 1 2 (c) Graph of r(x) = −2x + 2. Figure 13.4: Reflecting y = p(x). Important Facts 13.2.1 (Reflection). Let y = f(x) be a function equation. (i) We can reflect the graph across the x-axis and the resulting curve is the graph of the new function obtained by replacing “y” by “−y” in the original equation. The domain is the same as the domain for y = f(x). If the range for y = f(x) is c ≤ y ≤ d, then the range of −y = f(x) is c ≤ −y ≤ d. In other words, the reflection across the x-axis is the graph of y = −f(x). (ii) We can reflect the graph across the y-axis and the resulting curve is the graph of the new function obtained by replacing “x” by “−x” in the original equation. The range is the same as the range for y = f(x). If the domain for y = f(x) is a ≤ x ≤ b, then the domain x-axis (a) Graph of p(x) = 2x + 2. y = r(x) = −2x + 2. The domain must also be checked by replacing “x” by “−x” in the original domain condition: −2 ≤ −x ≤ 2, so 2 ≥ x ≥ −2. It just so happens in this case, the domain is unchanged. This is yet another DIFFERENT line; this one has slope −2 and y-intercept 2. We contrast these two curves in Figure 13.4(c), where r(x) is graphed as the “dashed line” in the same picture with the original p(x). Once we do this, it is easy to see how the graph of r(x) is really just the original curve reflected across the y-axis. This example illustrates a general principle referred to as the reflection principle. −1 y-axis CHAPTER 13. THREE CONSTRUCTION TOOLS 168 of y = f(−x) is a ≤ −x ≤ b. Using composition notation, the reflection across the y-axis is the graph of y = f(−x). Example 13.2.2. Consider the parallelogram-shaped region R with vertices (0, 2), (0, −2), (1, 0), and (−1, 0). Use the reflection principle to find functions whose graphs bound R. P = (0,2) l3 l 2 Q = (1,0) x-axis l l 4 1 Region R y-axis Figure 13.5: The region R. y-axis y = f(−x) −4 −2 y = f ( x) 2 4 x-axis −y = f(x) Figure 13.6: Reflecting the semicircle. Solution. Here is a picture of the region R: First off, using the two point formula for the equation of a line, we find that the line ℓ1 passing through the points P = (0, 2) and Q = (1, 0) is the graph of the function y = f1 (x) = −2x + 2. By Fact 13.2.1 (i), ℓ2 is the graph of the equation −y = −2x + 2, which we can write as the function y = f2 (x) = 2x − 2. By Fact 13.2.1 (ii) applied to ℓ2 , the line ℓ3 is the graph of the function y = f3 (x) = −2x − 2. Finally, by Fact 13.2.1 (i) applied to ℓ3 , the line ℓ4 is the graph of the equation −y = −2x − 2, which we can write as the function y = f4 (x) = 2x + 2. Figure 13.6 illustrates the fact that we need to be careful about the domain of the original function when using the reflection principle. For example, consider y = f(x) = 1 + 1 − (x − 3)2 . The largest possible domain of x-values is 2 ≤ x ≤ 4 and the graph is an upper semicircle of radius 1 centered at the point (3, 1). Reflection across the x-axis gives the graph of y = −1 − 1 − (x − 3)2 on the same domain; reflection across the y-axis gives the graph of y = 1 + 1 − (x + 3)2 on the new domain −4 ≤ x ≤ −2. 13.3 Shifting √ Let’s start out with the function y = f(x) = 4 − x2 , which has a largest possible domain −2 ≤ x ≤ 2. From Chap2 ter 6, the graph of this equation is an upper semicircle 1.5 1 of radius 2 centered at the origin (0, 0). Sliding the graph back and forth horizontally or vertically (or both), never 0.5 x-axis rotating or twisting, we are led to the “dashed curves” be−2 −1 0 1 2 low (contrasted with the original graph which is plotted Figure 13.7: Graph of y = with a solid curve). This describes some shifted curves on √ 4 − x2 . a pictorial level, but what are the underlying equations? For this example, we can use the fact that all of the shifted curves are still semicircles and Chapter 6 tells us how to find their equations. 2.5 y-axis 13.3. SHIFTING 169 The lower right-hand dashed semicircle is of radius 2 and is centered at (3, 0), so the corresponding equation must be y = 4 − (x − 3)2. The upper left-hand dashed semicircle is of radius 2 and centered at (0, 3), so √ the corresponding equation must be y = 3+ 4 − x2 . The upper right-hand dashed semicircle is of radius 2 and centered at (3, 3), so the corresponding equation must be y = 3 + 4 − (x − 3)2 . Keeping this same example, we can continue this kind 5 of shifting more generally by thinking about the effect of 4 making the following three replacements in the equation √ 3 y = 4 − x2 : 2 1 x y (x and y) x−h y−k (x − h and y − k). −2 −1 1 2 3 4 5 Figure 13.8: Shifting the upper semicircle. These substitutions lead to three new equations, each the equation of a semicircle: y= 4 − ( x − h) 2 ; y−k= 4 − x2 ; and, y−k= 4 − ( x − h) 2 . ⇐ ⇐ ⇐ Upper semicircle with radius 2 and center (h, 0). Upper semicircle with radius 2 and center (0, k). Upper semicircle with radius 2 and center (h, k). There are three potentially confusing points with this example: • Be careful with the sign (i.e., ±) of h and k. In Figure 13.9, if h = 1, we horizontally shift the semicircle 1 unit to the right; whereas, if h = −1, we horizontally shift the semicircle 1 units to the right. But, shifting −1 unit to the right is the same as shifting 1 unit to the left! In other words, if h is positive, then a horizontal shift by h will move the graph |h| units to the right; if h is negative, then a horizontal shift by h will move the graph |h| units to the left. • If k is positive, then a vertical shift by k will move the graph |k| units up; if k is negative, then a vertical shift by k will move the graph |k| units down. These conventions insure that the “positivity” of h and k match up with “rightward” and “upward” movement of the graph. y-axis original curve x-axis h negative h positive curve shifted |h| units left if h is negative curve shifted h units right if h is positive Figure 13.9: Potentially confusing points. CHAPTER 13. THREE CONSTRUCTION TOOLS 170 • When shifting, the domain of allowed x-values may change. This example illustrates an important general principle referred to as the shifting principle. Important Facts 13.3.1 (Shifting). Let y = f(x) be a function equation. (i) If we replace “x” by “x − h” in the original function equation, then the graph of the resulting new function y = f(x − h) is obtained by horizontally shifting the graph of f(x) by h. If h is positive, the picture shifts to the right h units; if h is negative, the picture shifts to the left h units. If the domain of f(x) is an interval a ≤ x ≤ b, then the domain of f(x − h) is a ≤ x − h ≤ b. The range remains unchanged under horizontal shifting. (ii) If we replace “y” by “y − k” in the original function equation, then the graph of the resulting new function y = f(x) + k is obtained by vertically shifting the graph of f(x) by k. If k is positive, the picture shifts upward k units; if k is negative, the picture shifts downward k units. If the range of f(x) is an interval c ≤ y ≤ d, then the range of f(x) + k is c ≤ y − k ≤ d. The domain remains unchanged under vertical shifting. 13.4 Dilation 1 To introduce the next graphical principle we will look at the function x y = f(x) = 2 . x +1 y-axis 0.75 0.5 0.25 −6 −4 −2 2 −0.25 4 6 x-axis Using a graphing device, we have produced a plot of the graph on the domain −6 ≤ x ≤ 6. Figure 13.10 shows the −1 curve has a high point H (like a “mountain peak”) and a Figure 13.10: Graph of y = x low point L (like a “valley”). Using a graphing device, we f(x) = x2 +1 . 1 can determine that the high point is H = (1, 2 ) (it lies on 1 the line with equation y = 2 ) and the low point is L = −1, − 1 (it lies on 2 1 the line with equation line y = − 1 ), so the range is − 2 ≤ y ≤ 1 . Draw two 2 2 new horizontal lines with equations y = 2 · ± 1 = ±1. Grab the high point 2 H on the curve and uniformly pull straight up, so that the high point now lies on the horizontal line y = 1 at (1, 1). Repeat this process by pulling L straight downward, so that the low point is now on the line y = −1 at (−1, −1). We end up with the ”stretched dashed curve” illustrated in Figure 13.11(a). In terms of the original function equation y = x2x 1 , we + are simply describing the graphical effect of multiplying the y-coordinate of every point on the curve by the positive number 2. In other words, the 2x dashed curve is the graph of y = x2 +1 = 2 x2x 1 . + −0.5 −0.75 13.4. DILATION 171 1 Next, draw the two horizontal lines with equations y = 1 · ± 1 = ± 4 . 2 2 Grab the high point H on the curve (in Figure 13.10) and uniformly push straight down, so that the high point now lies on the horizontal 1 1 line y = 4 at (1, 4 ). Repeat this process at the low point L by pushing the 1 curve straight upward, so that the low point is now on the line y = − 4 1 at (−1, − 4 ). We end up with the new ”dashed curve” illustrated in Figure 13.11(b). In terms of the original function equation y = x2x 1 , we are + simply describing the graphical effect of multiplying the y-coordinate of 1 every point on the curve by the positive number 2 . In other words, the x dashed curve is the graph of y = 2(x2 +1) . We could repeat this process systematically: 1 y-axis 0.75 0.5 • Pick a positive number c. • Draw the two horizontal lines y = c · ± 1 . If c > 1, then 2 c the graph of y = 2 is parallel and above the graph of 1 y = 2 . On the other hand, if 0 < c < 1, then this new 1 line is parallel and below y = 2 . • Uniformly deform the original graph (in Figure 13.10) so that the new curve has it’s high and low points just touching y = ± c . This will involve 2 vertically stretching or compressing, depending on whether 1 < c or 0 < c < 1, respectively. A number of possibilities are pictured in Figure 13.11(c). We refer to each new dashed curve as a vertical dilation of the original (solid) curve. This example illustrates an important principle. 0.25 −6 −4 −2 2 −0.25 y c (i) If we replace “ y ” by “ ” in the original equation, then the graph of the resulting new equation is obtained by vertical dilation of the graph of y = f(x). The domain of x-values is not affected. (ii) If c > 1, then the graph of y = f(x) (i.e., y = cf(x)) is a c vertically stretched version of the original graph. 6 x-axis −0.75 −1 (a) Vertical expansion. y-axis 1 0.75 0.5 11 11 00 00 11 11 00 00 11 11 00 00 −6 −4 1111 0000 1111 0000 0.25 1111 0000 2 111 000 111 000 −0.25 111 000 −2 4 6 x-axis −0.5 111 000 111 000 111 000 −0.75 −1 (b) Vertical compression. 1 0.75 Important Facts 13.4.1 (Vertical dilation). Let c > 0 be a positive number and y = f(x) a function equation. 4 −0.5 1111 0000 1111 0000 y-axis 1111 0000 0.5 0.25 11 00 111 11 000 00 11 00 −4 00 111 −2 −6 000 11 11 00 111 11 000 00 2 −0.25 4 6 x-axis −0.5 −0.75 −1 (c) Many possibilities. Figure 13.11: Dilating y = f(x). (iii) If 0 < c < 1, then the graph of y = f(x) (i.e., y = cf(x)) is c a vertically compressed version of the original graph. If we combine dilation with reflection across the x-axis, we can determine the graphical relationship between y = f(x) and y = cf(x), for any constant c. The key observation is that reflection across the x-axis corresponds to the case c = −1. CHAPTER 13. THREE CONSTRUCTION TOOLS 172 Example 13.4.2. Describe the relationship between the graphs of y = f(x) = 1 − (x + 1)2 , y = −f(x) = − 1 − (x + 1)2, and y = −4f(x) = −4 1 − (x + 1)2 . Solution. The graph of y = f(x) is an upper semicircle of radius 1 centered at the point (−1,0). To obtain the picture of the graph of y = −4f(x), we first reflect y = f(x) across the x-axis; this gives us the graph of y = −f(x). Then, we vertically dilate this picture by a factor of c = 4 to get the graph of y = −f(x), which is the same as the 4 graph of the equation y = −4f(x). See Figure 13.12. step 1: start with upper semicircle x-axis step 2: reflect across step 3: stretch curve from step 2: to get y = −4f(x) Figure 13.12: Reflecting and dilating a lower semicircle. Let’s return to the original example y = x2x 1 and in+ vestigate a different type of dilation where the action is taking place in the horizontal direction (whereas it was in the vertical direction before). Grab the right-hand end of the graph (in Figure 13.10) and pull to the right, while at the same time pulling the left-hand end to the left. We can quantify this by stipulating that the 1 1 high point H = 1, 2 of the original curve moves to the new location 2, 2 1 1 and the low point L = −1, − 2 moves to the new location −2, − 2 . 1 y-axis 1 0.75 0.5 −4 0.5 0.25 111 000 −6 111 000 y-axis 0.75 0.25 −2 2 −0.25 4 6 x-axis 11 00 −4 111 −6 000 11 00 111 000 −2 2 −0.25 −0.5 −0.75 −1 6 x-axis −0.5 −0.75 4 −1 (a) Stretching. (b) Compressing. Figure 13.13: Horizontally dilating y = x2x 1 . + The result will be somewhat analogous to stretching a spring. By the same token, we could push the right-hand end to the left and push the left-hand end to the right, like compressing a spring. We can quantify this by stipulating that the high point H = 1, 1 of the original curve moves 2 11 to the new location 2 , 2 and the low point L = −1, − 1 moves to the new 2 1 location − 1 , − 2 . These two situations are indicated in Figure 13.13. We 2 refer to each of the dashed curves as a horizontal dilation of the original (solid) curve. 13.5. VERTEX FORM AND ORDER OF OPERATIONS 173 The tricky point is to understand what happens on the level of the original equation. In the case of the stretched graph in Figure 13.13(a), you can use a graphing device to verify that this looks like the graph x/2 of y = (x/2)2 +1 ; in other words, we replaced “x” by “x/2” in the original equation. In the case of the compressed graph in Figure 13.13(b), you can use a graphing device to verify that this looks like the graph of y = (2x2x+1 ; )2 x in other words, we replaced “x” by “ 1/2 = 2x” in the original equation. The process just described leads to a general principle. Important Facts 13.4.3 (Horizontal dilation). Let c > 0 be a positive number and y = f(x) a function equation. (i) If we replace “ x” by “ x ” in the original function equation, then the c graph of the resulting new function y = f x is obtained by a horizonc tal dilation of the graph of y = f(x). If the domain of f(x) is a ≤ x ≤ b, then the domain of y = f x is a ≤ x ≤ b. c c (ii) If c > 1, then the graph of y = f x c (iii) If 0 < c < 1, then the graph of y = f is a horizontal stretch. x c is a horizontal compression. 13.5 Vertex Form and Order of Operations Using the language of function compositions we can clarify our discussion in Example 7.1.2. Let’s revisit that example: Example 13.5.1. The problem is to describe a sequence of geometric maneuvers that transform the graph of y = x2 into the graph of y = −3(x − 1)2 + 2. Solution. The idea is to rewrite y = −3(x − 1)2 + 2 as a composition of y = x2 with four other functions, each of which corresponds to a horizontal shift, vertical shift, reflection or dilation. Once we have done this, we can read off the order of geometric operations using the order of composition. Along the way, pay special attention to the exact order in which we will be composing our functions; this will make a big difference. To begin with, we can isolate four key numbers in the equation: Horizontal shift by h = 1 Reflect − 3 Dilate by 3 (x − 1 )2 + 2 Vertical shift by k = 2 We want to use each number to define a new function, then compose these in the correct order. We will also give our starting function y = x2 a specific name to make things definite: f(x) = x2 v(x) = x + 2 d(x) = 3x. h(x) = x − 1 r(x) = −x CHAPTER 13. THREE CONSTRUCTION TOOLS 174 Now, verify that v r d f h(x) = v r d f[x − 1] = v r d{(x − 1)2} = v r(3(x − 1)2 ) = v[−3(x − 1)2 ] = −3(x − 1)2 + 2 = −3x2 + 6x − 1. 13.6 Summary of Rules 2 −2 2 Figure 13.14: A multipart function. For quick reference, we summarize the consequence of shifting and expanding symbolically and pictorially. The running example for Tables 13.1, 13.2, and 13.3 will be a multipart function y = f(x) whose graph, seen in Figure 13.14, consists of a line segment and a quarter circle on the domain −2 ≤ x ≤ 2: f(x) = x+2 if −2 ≤ x ≤ 0 √ 2 if 0 ≤ x ≤ 2 4−x 13.6. SUMMARY OF RULES 175 Reflection Symbolic Change Replace x with −x. Replace f(x) with −f(x). New Equation Graphical Consequence y = f(−x) A reflection across the y-axis. y = −f(x) A reflection across the x-axis. Table 13.1: Reflecting y = f(x). Picture 2 −2 2 −2 2 −2 CHAPTER 13. THREE CONSTRUCTION TOOLS 176 Shifting (Assume c > 0 ) Symbolic Change New Equation Graphical Consequence y = f( x − c ) A shift to the right c units. Picture 2 Replace x with (x − c ). 2 Replace x with (x + c ). y = f( x + c ) 4 2 A shift to the left c units. −4 −2 4 Replace f(x) with (f(x) + c). y = f(x) + c A shift up c units. 2 −2 Replace f(x) with (f(x) − c). y = f(x) − c A shift down c units. 2 −2 2 2 Table 13.2: Shifting y = f(x). 13.6. SUMMARY OF RULES 177 Dilation Symbolic Change New Equation Graphical Consequence Picture c=2 If c > 1, replace x with x . c y=f x c A horizontal expansion. −4 −3 −2 −1 c= If 0 < c < 1, replace x with x . c y=f x c A horizontal compression. 1 2 4 3 2 1 01234 4 3 2 1 11 00 −4 −3 −2 −1 00 1 0 11 c=2 If c > 1, replace f(x) with (cf(x)). y = cf(x) A vertical expansion. −4 −3 −2 −1 If 0 < c < 1, replace f(x) with (cf(x)). c= y = cf(x) A vertical compression. 1 2 23 4 111 000 111 000 4 111 000 111 000 111 000 3 2 1 11 00 0 11 00 1234 4 3 2 1 0 1 1 0 1 0 111 −4 −3 −2 −1 000 1 2 3 4 0 111 000 Table 13.3: Dilating y = f(x). CHAPTER 13. THREE CONSTRUCTION TOOLS 178 13.7 Exercises Problem 13.1. On a single set of axes, sketch a picture of the graphs of the √ following four √ √ equations: y = −x + 2, y = −x − 2, y = x + 2, √ and y = x − 2. These equations determine lines, which in turn bound a diamond shaped region in the plane. (a) Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once. (b) Find the intersection points between the unit circle and each of the four lines. (c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help. Problem 13.2. The graph of a function y = f(x) is pictured with domain −2.5 ≤ x ≤ 3.5. y π f(x) π/2 −1 1 x Problem 13.4. (a) Each of the six functions y = f(x) below can be written in the “standard form” y = A|B(x − C)| + D, for some constants A,B,C,D. Find these constants, describe the precise order of graphical operations involved in going from the graph of y = |x| to the graph of y = f(x) (paying close attention to the order), write out the multipart rule, sketch the graph and calculate the coordinates of the “vertex” of the graph. (a1) f(x) = |x − 2| (a2) f(x) = 2|x + 3| Sketch the graph of each of the new functions listed below. (a3) f(x) = |2x − 1| (a) g(x) = 2f(x + 1) (a4) f(x) = |2(x − 1)| (b) h(x) = 1 f(2x − 1) 2 (a5) f(x) = 3|2x − 1| + 5 (c) j(x) = 5f( 1 x + 2) − 2 3 (a6) f(x) = −2|x + 3| − 1 Problem 13.3. The graph of a function y = f(x) is pictured with domain −1 ≤ x ≤ 1. Sketch the graph of the new function y = g(x) = 1 f(3x) − 0.5. π Find the largest possible domain of the function y = g(x). (b) Solve the following inequalities using your work in the previous part of this problem: (b1) |x − 2| ≤ 3 (b2) 1 ≤ 2|x + 3| ≤ 5 (b3) y = 3|2x − 1| + 5 ≥ 10 13.7. EXERCISES (c) The graphs of y = 3|2x − 1| + 5 and y = −|x − 3| + 10 intersect to form a bounded region of the plane. Find the vertices of this region and sketch a picture. Problem 13.5. Consider the function y = f(x) with multipart definition if x ≤ −1 0 2x + 2 if −1 ≤ x ≤ 0 f(x) = −x + 2 if 0 ≤ x ≤ 2 0 if x ≥ 2 179 (c) The graph of z = a(x) from part (a) is given below. Sketch the graph and find the rule for the function z = 2a(3x + 3)+ 1; make sure to specify the domain and range of this new function. 10 y-axis 8 6 4 2 x-axis (a) Sketch the graph of y = f(x). (b) Is y = f(x) an even function? Is y = f(x) an odd function? (A function y = f(x) is called even if f(x) = f(−x) for all x in the domain. A function y = f(x) is called odd if f(−x) = −f(x) for all x in the domain.) (c) Sketch the reflection of the graph across the x-axis and y-axis. Obtain the resulting multipart equations for these reflected curves. (d) Sketch the vertical dilations y = 2f(x) 1 and y = 2 f(x). (e) Sketch the horizontal dilations y = f(2x) and y = f( 1 x). 2 (f) Find a number c > 0 so that the highest point on the graph of the vertical dilation y = cf(x) has y-coordinate 11. (g) Using horizontal dilation, find a number c > 0 so that the function values f( x ) are c 5 non-zero for all − 2 < x < 5. (h) Using horizontal dilation, find positive numbers c,d > 0 so that the function values f( 1 (x − d)) are non-zero precisely c when 0 < x < 1. Problem 13.6. An isosceles triangle has sides of length x, x and y. In addition, assume the triangle has perimeter 12. 2 4 6 8 10 Problem 13.7. Describe how each graph differs from that of y = x2 . (a) y = 2x2 (b) y = x2 − 5 (c) y = (x − 4)2 (d) y = (3x − 12)2 (e) y = 2(3x − 12)2 − 5 Problem 13.8. In each case, start with the function y = |x| and perform the operations described to the graph, in the order specified. Write out the resulting rule for the function and sketch the final graph you obtain. (a) (1) horizontally compress by a factor of 2; (2) horizontally shift to the left by 2; (3) vertically stretch by a factor of 7; (4) vertically shift up 2 units. (b) (1) horizontally stretch by a factor of 2; (2) horizontally shift to the right by 2; (3) vertically compress by a factor of 7; (4) vertically shift down 2 units. (c) (1) horizontally shift to the right 2 units; (2) horizontally compress by a factor of 3. (a) Find the rule for a function that computes the area of the triangle as a function of x. Describe the largest possible domain of this function. Problem 13.9. (a) Begin with the function y = f(x) = 2x . (b) Assume that the maximum value of the function a(x) in (a) occurs when x = 4. Find the maximum value of z = a(x) and z = 2a(3x + 3) + 1. (a1) Rewrite each of the following functions in standard exponential form: f(2x), f(x − 1), f(2x − 1), f(2(x − 1)), 3f(x), 3f(2(x − 1)). 180 CHAPTER 13. THREE CONSTRUCTION TOOLS (a2) Is the function 3f(2(x − 1))+ 1 a function of exponential type? cal dilation, horizontal shifting, horizontal dilation) to the standard exponential model y = Ao bx . For which of the four operations is the resulting function still a standard exponential model? (a3) Sketch the graphs of f(x), f(2x), f(2(x− 1)), 3f(2(x − 1)) and 3f(2(x − 1)) + 1 in the same coordinate system and explain which graphical operation(s) (vertical shifting, vertical Problem 13.10. Begin with a sketch of the dilation, horizontal shifting, horigraph of the function y = 2x on the domain zontal dilation) have been carried of all real numbers. Describe how to use the out. “four tools” of Chapter 13 to obtain the graphs of these functions: y = −2x , y = 2−x , y = 3(2x ), (b) In general, explain what happens when y = 1 (2x ), y = 3 + 2x , y = 2x − 2, y = 2x−2 , you apply the four construction tools 3 y = 2x+2 , y = 23x , y = 2x/3 . of Chapter 13 (vertical shifting, verti- Chapter 14 Rational Functions p A rational function is a function of the form f(x) = q(x) where p(x) and q(x) ( x) are polynomials. For example, the following are all rational functions. f(x) = x x2 + 1 4x5 − 4x2 − 8 x6 g(x) = h(x) = 3 j(x) = 8 3x + 4 3x − 5 x + x2 − x + 1 x + 5x − 1 2 There is a huge variety of rational functions. In this course, we will concentrate our efforts exclusively toward understanding the simplest type of rational functions: linear-to-linear rational functions. Linearto-linear rational functions are rational functions in which the numerator and the denominator are both linear polynomials. The following are linear-to-linear-rational functions. k(x) = 4x + 3 5x − 6 0.34x − 0.113 x 4 m(x) = n(x) = p(x) = 5 3x + 4 2x + 1 x−1 x − 1.117 8 The simplest example of a linear-to-linear rational 1 function is f(x) = x whose graph is shown in Figure 14.1. This is an important example for the study of this class of functions, as we shall see. 1 Let’s consider the graph of this function, f(x) = x . We first begin by considering the domain of f. Since the nu1 merator of x is a constant, and the denominator is just x, the only way we can run into difficulty when evaluating this function is if we try to divide by zero; that is, the only value of x not in the domain of this function is zero. The function is defined for all x except x = 0. At x = 0, there must be a gap, or hole, in the graph. Figure 14.1: The graph of 1 To get the graph started we might simply give ourselves f(x) = x . a point on the graph. For instance, we see that f(1) = 1/1 = 1, so the point (1,1) is on the graph. Then, if we use larger values of x, we see that 1/x becomes smaller as x grows. For instance, f(2) = 1/2, f(10) = 1/10, and f(1000) = 1/1000. In addition, we see that, other than the fact that 1/x > 0 for positive x, there is nothing preventing us from 10 5 0 -10 -5 0 -5 -10 181 5 10 CHAPTER 14. RATIONAL FUNCTIONS 182 making 1/x as small as we want simply by taking x big enough. Want 1/x to be less than 0.001? Just pick x bigger than 1000. Want 1/x to be less than 0.000001? Just use x bigger than 1,000,000. What this means graphically is that as x gets bigger (starting from x = 1), the curve y = 1/x gets closer and closer to the x-axis. As a result, we say that the x-axis is a horizontal asymptote for this function. We see the same behavior for negative values of x. If x is large, and negative (think -1000, or -1000000), then 1/x is very small (i.e., close to zero), and it gets smaller the larger x becomes. Graphically, this means that as x gets large in the negative direction, the curve y = 1/x gets closer and closer to the x-axis. We say that, in both the positive and negative directions, y = 1/x is asymptotic to the x-axis. A similar thing happens when we consider x near zero. If x is a small positive number (think 1/2, or 1/10, or 1/10000), then 1/x is a large positive number. What’s more, if we think of x as getting closer to zero, 1/x gets bigger and bigger. Plus, there is no bound on how big we can make 1/x simply by taking x as close to zero as we need to. For instance, can 1/x be as big as 10000? All you need to do is pick a positive x smaller than 1/10000. Graphically, what this means is that as x approaches zero from the positive side, the y value gets larger and larger. As a result, the curve approaches the y-axis as y gets larger. We say that the y-axis is a vertical asymptote for the curve y = 1/x. We see the same phenomenon as x approaches zero from the negative side: y = 1/x gets larger in the negative direction (i.e., it gets more and more negative). The curve gets closer and closer to the negative y-axis as y becomes more and more negative. Again, we say that the y − axis is a vertical asymptote for the curve y = 1/x. It turns out that every linear-to-linear rational function has a graph that looks essentially the same as the graph of y = 1/x. Let’s see why. +b Consider the linear-to-linear rational function f(x) = ax+d . If we divide cx cx + d into ax + b, the result is f(x) = ax + b a b − ad c =+ cx + d c cx + d which we can rewrite as f(x) = a 1 1 1 ad a ad a bc − ad · · . · + b− = + b− =+ d 2 c c cx + d c c c c c (x + c ) x+ d c If we now let A= bc − ad c2 a ,B= c then we can write f(x) = A + B 1 . x+C , and C = d , c 183 If we let g(x) = 1/x then we have shown that f(x) = A + Bg(x + C), and so the graph of the function f is just a horizontally shifted, vertically shifted and vertically dilated version of the graph of g. Also, if B turned out to be negative the graph would be vertically flipped, too. Why is that useful? It means that the graph of a linear-to-linear rational function can only take one of two forms. Either it looks like this: Or like this: We can sketch an accurate graph of a linear-to-linear rational function by sketching the asymptotes and then sketching in just one point on the graph. That will be enough information to nail down a decent sketch. We can always plot more points to give us more precision, but one point is enough to capture the essence of the graph. Given a linear-to-linear function that we wish to graph, we must first determine the asymptotes. There will be one horizontal asymptote and one vertical asymptote. The vertical asymptote will be a vertical line with equation x = k where k is the one x value which is not in the domain of the function. That is, find the value of x which makes the denominator zero and that will tell you the vertical asymptote. The horizontal asymptote is a little more involved. However, we can quickly get to a shortcut. The essence of a horizontal asymptote is that it describes what value the function is approximately equal to for very large values of x. To study what a linear-to-linear function is like when x is very large, we can perform the following algebraic manipulation: f(x) = a+ b ax + b ax + b 1/x x = · = cx + d cx + d 1/x c+ d x While dividing by x is troublesome if x equals zero, here we are assuming x is very large, so it is certainly not zero. Now, consider this last expression in the above equation. If x is very large, then b ≈0 x CHAPTER 14. RATIONAL FUNCTIONS 184 (where ≈ means ”is approximately”) and likewise d ≈ 0. x Hence, when x is very large, f(x) ≈ a a+0 =. c+0 c We can interpret this by saying that when x is very large, the function f(x) is is close to a constant, and that constant is a . Thus, the horizontal c ax + b is the horizontal line y = a . asymptote of f(x) = c cx + d Example 14.0.1. Sketch the graph of the function f(x) = 3x−1 . 2x+7 Solution. We begin by finding the asymptotes of f. The denominator is equal to zero when 2x + 7 = 0, i.e., when x = −7/2. As a result, the vertical asymptote for this function is the vertical line x = −7/2. By taking the ratio of the coefficients of x in the numerator and denominator, we can find that the horizontal asymptote is the horizontal line 3 y= . 2 10 5 0 -10 -5 0 5 10 -5 -10 Figure 14.2: The graph of f(x) = 3x−1 2x+7 . We then sketch these two asymptotes. The last thing we need is a single point. For instance, we may evaluate f(0): f( 0) = −1 7 14.1. MODELING WITH LINEAR-TO-LINEAR RATIONAL FUNCTIONS185 and so the point (0, −1/7) is on the graph. With this information, we know that the curve lies below the horizontal asymptote to the right of the vertical asymptote, and consequently the curve lies above the horizontal asymptote to the left of the vertical asymptote. We graph the result in Figure 14.2. 14.1 Modeling with Linear-to-linear Rational Functions As we have done with other sorts of functions, such as linear and quadratic, we can also model using linear-to-linear rational functions. One reason for using this type of function is their asymptotic nature. Many changing quantities in the world continually increase or decrease, but with bounds on how large or small they can get. For instance, a population may steadily decrease, but a population can never be negative. Conversely, a population may steadily increase, but due to environmental and other factors we may hypothesize that the population will always stay below some upper bound. As a result, the population may ”level off”. This leveling off behavior is exemplified by the asymptotic nature of the linear-to-linear rational functions, and so this type of function provides a way to model such behavior. Given any linear-to-linear rational function, we can always divide the numerator and the denominator by the coefficient of x in the denominator. In this way, we can always assume that the coefficient of x in the denominator of a function we seek is 1. This is illustrated in the next example. Example 14.1.1. Let f(x) = f(x) = = 2x + 3 · 5x − 7 2x+3 . 5x−7 Then 1 5 1 5 3 +5 . x− 7 5 2 x 5 Thus, in general, when we seek a linear-to-linear rational function, we will be looking for a function of the form f(x) = ax + b x+c and thus there are three parameters we need to determine. Note that for a function of this form, the horizontal asymptote is y = a and the vertical asymptote is x = −c. Since these functions have three parameters (i.e., a, b and c), we will need three pieces of information to nail down the function. CHAPTER 14. RATIONAL FUNCTIONS 186 There are essentially three types of modeling problems that require the determination of a linear-to-linear function. The three types are based on the kind of information given about the function. The three types are: 1. You know three points the the graph of the function passes through; 2. You know one of the function’s asymptotes and two points the graph passes through; 3. You know both asymptotes and one point the graph passes through. Notice that in all cases you know three pieces of information. Since a linear-to-linear function is determined by three parameters, this is exactly the amount of information needed to determine the function. The worst case, in terms of the amount of algebra you need to do, is the first case. Let’s look at an example of the algebra involved with this sort. Example 14.1.2. Find the linear-to-linear rational function f(x) such that f(10) = 20, f(20) = 32 and f(25) = 36. Solution. Since f(x) is a linear-to-linear rational function, we know f(x) = ax + b x+c for constants a, b, and c. We need to find a, b and c. We know three things. First, f(10) = 20. So f(10) = 10a + b = 20, 10 + c which we can rewrite as 10a + b = 200 + 20c. (14.1) Second, f(20) = 32. So f(20) = 20a + b = 32, 20 + c which we can rewrite as 20a + b = 640 + 32c. Third, f(25) = 36. So f(25) = 25a + b = 36, 25 + c (14.2) 14.1. MODELING WITH LINEAR-TO-LINEAR RATIONAL FUNCTIONS187 which we can rewrite as 25a + b = 900 + 36c. (14.3) These three numbered equations are enough algebraic material to solve for a, b, and c. Here is one way to do that. Subtract equation 14.1 from equation 14.2 to get 10a = 440 + 12c (14.4) and subtract equation 14.2 from equation 14.3 to get 5a = 260 + 4c (14.5) Note that we’ve eliminated b. Now multiply this last equation by 2 to get 10a = 520 + 8c Subtract equation 14.4 from this to get 0 = 80 − 4c which easily give us c = 20. Plugging this value into equation 14.4, we can find a = 68, and then we can find b = −80. Thus, f(x) = 68x − 80 . x + 20 We can check that we have done the algebra correctly by evaluating f(x) at x = 10, x = 20 and x = 25. If we get f(10) = 20, f(20) = 32 and f(25) = 36, then we’ll know our work is correct. Algebraically, this was the worst situation of the three, since it required the most algebra. If, instead of knowing three points, we know one or both of the asymptotes, then we can easily find a and/or c, and so cut down on the amount of algebra needed. However, the method is essentially identical. Let’s now apply these ideas to a real world problem. Example 14.1.3. Clyde makes extra money selling tickets in front of the Safeco Field. The amount he charges for a ticket depends on how many he has. If he only has one ticket, he charges $100 for it. If he has 10 tickets, he charges $80 a piece. But if he has a large number of tickets, he will sell them for $50 each. How much will he charge for a ticket if he holds 20 tickets? CHAPTER 14. RATIONAL FUNCTIONS 188 Solution. We want to give a linear-to-linear rational function relating the price of a ticket y to the number of tickets x that Clyde is holding. As we saw above, we can assume the function is of the form y= ax + b x+c where a, b and c are numbers. Note that y = a is the horizontal asymptote. When x is very large, y is close to 50. This means the line y = 50 is a horizontal asymptote. Thus a = 50 and y= 50x + b . x+c Next we plug in the point (1,100) to get a linear equation in b and c. 50 · 1 + b 1+c 100 · (1 + c) = 50 + b 50 = b − 100c 100 = Similarly, plugging in (10,80) and doing a little algebra (do it now!) gives another linear equation 300 = b − 80c. Solving these two linear equations simultaneously gives c = 12.5 and b = 1300. Thus our function is y= 50x + 1300 x + 12.5 and, if Clyde holds 20 tickets, he will charge y= 50 · 20 + 1300 = $70.77 20 + 12.5 per ticket. 14.2 Summary • Every linear-to-linear rational function has a graph which is a shifted, scaled version of the curve y = 1/x. As a result, they have one vertical asymptote, and one horizontal asymptote. • Every linear-to-linear rational function f can be expressed in the form f(x) = ax + b . x+c This function has horizontal asymptote y = a and vertical asymptote x = −c. 14.3. EXERCISES 189 14.3 Exercises Problem 14.1. Give the domain of each of the following functions. Find the x- and yintercepts of each function. Sketch a graph and indicate any vertical or horizontal asymptotes. Give equations for the asymptotes. (a) f(x) = (c) h(x) = (e) k(x) = 2x x−1 x+1 x−2 8x+16 5x− 1 2 3x+2 2x−5 (d) j(x) = 4x−12 x+8 9x+24 (f) m(x) = 35x−100 (b) g(x) = Problem 14.2. Oscar is hunting magnetic fields with his gauss meter, a device for measuring the strength and polarity of magnetic fields. The reading on the meter will increase as Oscar gets closer to a magnet. Oscar is in a long hallway at the end of which is a room containing an extremely strong magnet. When he is far down the hallway from the room, the meter reads a level of 0.2. He then walks down the hallway and enters the room. When he has gone 6 feet into the room, the meter reads 2.3. Eight feet into the room, the meter reads 4.4. (a) Give a linear-to-linear rational model relating the meter reading y to how many feet x Oscar has gone into the room. (b) How far must he go for the meter to reach 10? 100? (c) Considering your function from part (a) and the results of part (b), how far into the room do you think the magnet is? Problem 14.3. In 1975 I bought an old Martin ukulele for $200. In 1995 a similar uke was selling for $900. In 1980 I bought a new Kamaka uke for $100. In 1990 I sold it for $400. (a) Give a linear model relating the price p of the Martin uke to the year t. Take t = 0 in 1975. (b) Give a linear model relating the price q of the Kamaka uke to the year t. Again take t = 0 in 1975. (c) When is the value of the Martin twice the value of the Kamaka? (d) Give a function f(t) which gives the ratio of the price of the Martin to the price of the Kamaka. (e) In the long run, what will be the ratio of the prices of the ukuleles? Problem 14.4. Isobel is producing and selling casette tapes of her rock band. When she had sold 10 tapes, her net profit was $6. When she had sold 20 tapes, however, her net profit had shrunk to $4 due to increased production expenses. But when she had sold 30 tapes, her net profit had rebounded to $8. (a) Give a quadratic model relating Isobel’s net profit y to the number of tapes sold x. (b) Divide the profit function in part (a) by the number of tapes sold x to get a model relating average profit w per tape to the number of tapes sold. (c) How many tapes must she sell in order to make $1.20 per tape in net profit? Problem 14.5. Find the linear-to-linear function whose graph passes through the points (1,1), (5,2) and (20,3). What is its horizontal asymptote? Problem 14.6. Find the linear-to-linear function whose graph has y = 6 as a horizontal asymptote and passes through (0,10) and (3,7). Problem 14.7. The more you study for a certain exam, the better your performance on it. If you study for 10 hours, your score will be 65%. If you study for 20 hours, your score will be 95%. You can get as close as you want to a perfect score just by studying long enough. Assume your percentage score is a linear-tolinear function of the number of hours that you study. If you want a score of 80%, how long do you need to study? Problem 14.8. A street light is 10 feet above a straight bike path. Olav is bicycling down the path at a rate of 15 MPH. At midnight, Olav is 33 feet from the point on the bike path directly below the street light. (See the picture). The relationship between the intensity C of light (in candlepower) and the distance d (in feet) from k the light source is given by C = d2 , where k is a constant depending on the light source. CHAPTER 14. RATIONAL FUNCTIONS 190 (a) From 20 feet away, the street light has an intensity of 1 candle. What is k? (b) Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds. (c) When will the light on Olav have maximum intensity? (d) When will the intensity of the light be 2 candles? 10ft olav path Problem 14.10. The number of customers in a local dive shop depends on the amount of money spent on advertising. If the shop spends nothing on advertising, there will be 100 customers/day. If the shop spends $100, there will be 200 customers/day. As the amount spent on advertising increases, the number of customers/day increases and approaches (but never exceeds) 400 customers/day. (a) Find a linear to linear rational function y = f(x) that calculates the number y of customers/day if $x is spent on advertising. 33ft (b) How much must the shop spend on advertising to have 300 customers/day. Problem 14.9. For each of the following find the linear to linear function f(x) satisfying the given requirements: (a) f(0) = 0, f(10) = 10, f(20) = 15 (b) f(0) = 10, f(5) = 4, f(20) = 3 (c) f(10) = 20, f(30) = 25, and the graph of f(x) has y = 30 as its horizontal asymptote (c) Sketch the graph of the function y = f(x) on the domain 0 ≤ x ≤ 5000. (d) Find the rule, domain and range for the inverse function from part (c). Explain in words what the inverse function calculates. Chapter 15 Measuring an Angle So far, the equations we have studied have an algebraic character, involving the variables x and y, arithmetic operations and maybe extraction of roots. Restricting our attention to such equations would limit our ability to describe certain natural phenomena. An important example involves understanding motion around a circle, and it can be motivated by analyzing a very simple scenario: Cosmo the dog, tied by a 20 foot long tether to a post, begins walking around a circle. A number of very natural questions arise: S motors around the Q circle P 20 feet R Figure 15.1: Cosmo the dog walking a circular path. Natural Questions 15.0.1. How can we measure the angles ∠SPR, ∠QPR, and ∠QPS? How can we measure the arc lengths arc(RS), arc(SQ) and arc(RQ)? How can we measure the rate Cosmo is moving around the circle? If we know how to measure angles, can we compute the coordinates of R, S, and Q? Turning this around, if we know how to compute the coordinates of R, S, and Q, can we then measure the angles ∠SPR, ∠QPR, and ∠QPS ? Finally, how can we specify the direction Cosmo is traveling? We will answer all of these questions and see how the theory which evolves can be applied to a variety of problems. The definition and basic properties of the circular functions will emerge as a central theme in this Chapter. The full problem-solving power of these functions will become apparent in our discussion of sinusoidal functions in Chapter 19. The xy-coordinate system is well equipped to study straight line motion between two locations. For problems of this sort, the important tool is the distance formula. However, as Cosmo has illustrated, not all two-dimensional motion is along a straight line. In this section, we will describe how to calculate length along a circular arc, which requires a quick review of angle measurement. 191 Cosmo CHAPTER 15. MEASURING AN ANGLE 192 15.1 Standard and Central Angles An angle is the union of two rays emanating from a common point called the vertex of the angle. A typical angle can be dynamically generated by rotating a single ray from one position to another, sweeping counterclockwise or clockwise : See Figure 15.2. We often insert a curved arrow to indicate the direction in which we are sweeping out the angle. The ray ℓ1 is called the initial side and ℓ2 the terminal side of the angle ∠AOB. (terminal side) l2 B SWEEP CLOCKWISE O A l1 vertex O vertex l1 A (initial side) O A (initial side) l1 (terminal side) START SWEEP COUNTERCLOCKWISE B l2 Figure 15.2: Angle ∠AOB. Working with angles, we need to agree on a standard frame of reference for viewing them. Within the usual xy-coordinate system, imagine a model of ∠AOB in Figure 15.2 constructed from two pieces of rigid wire, welded at the vertex. Sliding this model around inside the xy-plane will not distort its shape, only its position relative to the coordinate axis. So, we can slide the angle into position so that the initial side is coincident with the positive x-axis and the vertex is the origin. Whenever we do this, we say the angle is in standard position. Once an angle is in standard position, we can construct a circle centered at the origin and view our standard angle as cutting out a particular “pie shaped wedge” of the corresponding disc. Notice, the shaded regions in Figure 15.3 depend on whether we sweep the angle counterclockwise or clockwise from the initial side. The portion of the “pie wedge” along the circle edge, which is an arc, is called the arc subtended by the angle. We can keep track of this arc using the notation arc(AB). A central angle is any angle with vertex at the center of a circle, but its initial side may or may not be the positive x-axis. For example, ∠QPS in the Figure beginning this Chapter is a central angle which is not in standard position. 15.2. AN ANALOGY 193 l2 y-axis y-axis B arc subtended l2 B arc(AB) arc(AB) vertex O A l1 x-axis COUNTERCLOCKWISE vertex O x-axis A l1 CLOCKWISE Figure 15.3: Standard angles and arcs. 15.2 An Analogy To measure the dimensions of a box you would use a ruler. In other words, you use an instrument (the ruler) as a standard against which you measure the box. The ruler would most likely be divided up into either English units (inches) or metric units (centimeters), so we could express the dimensions in a couple of different ways, depending on the units desired. By analogy, to measure the size of an angle, we need a standard against which any angle can be compared. In this section, we will describe two standards commonly used: the degree method and the radian method of angle measurement. The key idea is this: Beginning with a circular region, describe how to construct a “basic” pie shaped wedge whose interior angle becomes the standard unit of angle measurement. 15.3 Degree Method Begin by drawing a circle of radius r, call it Cr , centered at the origin. Divide this circle into 360 equal sized pie shaped wedges, beginning with the the point (r,0) on the circle; i.e. the place where the circle crosses the x-axis. We will refer to the arcs along the outside edges of these wedges as one-degree arcs. Why 360 equal sized arcs? The reason for doing so is historically tied to the fact that the ancient Babylonians did so as they developed their study of astronomy. (There is actually an alternate system based on dividing the circular region into 400 equal sized wedges.) Any central angle which subtends one of these 360 equal sized arcs is CHAPTER 15. MEASURING AN ANGLE 194 circle Cr etc. a total of 360 equal sized pie shaped wedges inside this disk r (r,0) typical wedge r this angle is DEFINED to have measure 1 degree etc. ***NOT TO SCALE*** Figure 15.4: Wedges as 1◦ arcs. defined to have a measure of one degree, denoted 1◦ . We can now use one-degree arcs to measure any angle: Begin by sliding the angle ∠AOB into standard central position, as in Figure 15.3. Piece together consecutive one-degree arcs in a counterclockwise or clockwise direction, beginning from the initial side and working toward the terminal side, approximating the angle ∠AOB to the nearest degree. If we are allowed to divide a one-degree arc into a fractional portion, then we could precisely determine the number m of one-degree arcs which consecutively fit together into the given arc. If we are sweeping counterclockwise from the initial side of the angle, m is defined to be the degree measure of the angle. If we sweep in a clockwise direction, then −m is defined to be the degree measure of the angle. So, in Figure 15.3, the left-hand angle has positive degree measure while the right-hand angle has negative degree measure. Simple examples would be angles like the ones in Figure 15.5. Notice, with our conventions, the rays determining an angle with measure −135◦ sit inside the circle in the same position as those for an angle of measure 225◦; the minus sign keeps track of sweeping the positive x-axis clockwise (rather than counterclockwise). We can further divide a one-degree arc into 60 equal arcs, each called a one minute arc. Each one-minute arc can be further divided into 60 equal arcs, each called a one second arc. This then leads to angle measures of one minute, denoted 1 ′ and one second, denoted 1 ′′ : 1◦ = 60 minutes = 3600 seconds. 15.3. DEGREE METHOD 195 180◦ 90◦ 270◦ 45◦ 315◦ −135◦ Figure 15.5: Examples of common angles. For example, an angle of measure θ = 5 degrees 23 minutes 18 seconds is usually denoted 5◦ 23 ′ 18 ′′ . We could express this as a decimal of degrees: ◦ ′ 5 23 18 ′′ 23 18 = 5+ + 60 3600 = 5.3883◦. ◦ ←In degrees! As another example, suppose we have an angle with measure 75.456◦ and we wish to convert this into degree/minute/second units. First, since 75.456◦ = 75◦ + 0.456◦, we need to write 0.456◦ in minutes by the calculation: 0.456 degree × 60 minutes = 27.36 ′. degree This tells us that 75.456◦ = 75◦ 27.36 ′ = 75◦ 27 ′ + 0.36 ′. Now we need to write 0.36 ′ in seconds via the calculation: 0.36 minutes × 60 seconds/minute = 21.6 ′′ . In other words, 75.456◦ = 75◦ 27 ′ 21.6 ′′. Degree measurement of an angle is very closely tied to direction in the plane, explaining its use in map navigation. With some additional work, it is also possible to relate degree measure and lengths of circular arcs. To do this carefully, first go back to Figure 15.3 and recall the situation CHAPTER 15. MEASURING AN ANGLE 196 where an arc arc(AB) is subtended by the central angle ∠AOB. In this situation, the arc length of arc(AB), commonly denoted by the letter s, is the distance from A to B computed along the circular arc; keep in mind, this is NOT the same as the straight line distance between the points A and B. For example, consider the six angles pictured above, of measures 90◦ , 180◦ , 270◦, 45◦ , −135◦ , and 315◦ . If the circle is of radius r and we want to compute the lengths of the arcs subtended by these six angles, then this can be done using the formula for the circumference of a circle (on the back of this text) and the following general principle: Important Fact 15.3.1. (length of a part) = (fraction of the part) × (length of the whole) For example, the circumference of the entire circle of radius r is 2πr; this is the “length of the whole” in the general principle. The arc sub90 1 tended by a 90◦ angle is 360 = 4 of the entire circumference; this is the “fraction of the part” in the general principle. The boxed formula implies: s = arc length of the 90◦ arc = s = distance along the arc r θ degrees πr 1 2πr = . 4 2 Similarly, a 180◦ angle subtends an arc of length s = πr, 315 a 315◦ angle subtends an arc of length s = 360 2πr = 7πr , 4 etc. In general, we arrive at this formula: Important Fact 15.3.2 (Arc length in degrees). Start with a central angle of measure θ degrees inside a circle of radius r. Then this angle will subtend an arc of length r Cr Figure 15.6: The definition of arc length. s= 2π rθ 360 15.4 Radian Method The key to understanding degree measurement was the description of a “basic wedge” which contained an interior angle of measure 1◦ ; this was straightforward and familiar to all of us. Once this was done, we could proceed to measure any angle in degrees and compute arc lengths as in Fact 15.3.2. However, the formula for the length of an arc subtended by an angle measured in degrees is sort of cumbersome, involving the curi2π ous factor 360 . Our next goal is to introduce an alternate angle measurement scheme called radian measure that begins with a different “basic wedge”. As will become apparent, a big selling point of radian measure is that arc length calculations become easy. 15.4. RADIAN METHOD 197 r r r r equilateral wedge r (r,0) circle Cr r this angle is DEFINED to have measure 1 radian Figure 15.7: Constructing an equilateral wedge. As before, begin with a circle Cr of radius r. Construct an equilateral wedge with all three sides of equal length r; see Figure 15.7. We define the measure of the interior angle of this wedge to be 1 radian. Once we have defined an angle of measure 1 radian, we can define an angle of measure 2 radians by putting together two equilateral wedges. Likewise, an angle of measure 1 radian is obtained by symmetrically di2 viding an equilateral wedge in half, etc. Reasoning in this way, we can piece together equilateral wedges or fractions of such to compute the radian measure of any angle. It is important to notice an important relationship between the radian measure of an angle and arc length calculations. In the five angles pictured above, 1 1 radian, 2 radian, 3 radian, 2 radian and 1 radian, the length of the arcs 4 1 1 subtended by these angles θ are r, 2r, 3r, 2 r, and 4 r. In other words, a pattern emerges that gives a very simple relationship between the length s of an arc and the radian measure of the subtended angle: Important Fact 15.4.1 (Arc length in radians). Start with a central angle of measure θ radians inside a circle of radius r. Then this angle will subtend an arc of length s = θr. These remarks allow us to summarize the definition of the radian measure θ of ∠AOB inside a circle of radius r by the formula: θ= s r if angle is swept counterclockwise − s if angle is swept clockwise r s = distance along the arc r θ radians r Cr Figure 15.9: Defining arc length when angles are measured in radians. CHAPTER 15. MEASURING AN ANGLE 198 3r 2r r 3 radians = θ 2 radians = θ r r r r r 1 radian = θ r 1 2 r 1 4 radian = θ radian = θ r 2 r r r 4 r Figure 15.8: Measuring angles in radians. The units of θ are sometimes abbreviated as rad. It is important to appreciate the role of the radius of the circle B Cr when using radian measure of an angle: An angle of radian measure θ will subtend an arc of length |θ| on the θ radian unit circle. In other words, radian measure of angles is O A x-axis exactly the same as arc length on the unit circle; we couldn’t hope for a better connection! circle radius r The difficulty with radian measure versus degree measure is really one of familiarity. Let’s view a few common Figure 15.10: Arc length angles in radian measure. It is easiest to start with the after imposing a coordinate system. case of angles in central standard position within the unit circle. Examples of basic angles would be fractional parts 1 of one complete revolution around the unit circle; for example, 12 revolu1 3 tion, 1 revolution, 6 revolution, 1 revolution, 1 revolution and 4 revolution. 8 4 2 One revolution around the unit circle describes an arc of length 2π and so the subtended angle (1 revolution) is 2π radians. We can now easily 1 find the radian measure of these six angles. For example, 12 revolution 1 would describe an angle of measure ( 12 )2π rad= π rad. Similarly, the other 6 five angles pictured below have measures π rad, π rad, π rad, π rad and 4 3 2 3π rad. 2 All of these examples have positive radian measure. For an angle with negative radian measure, such as θ = − π radians, we would locate B 2 1 by rotating 4 revolution clockwise, etc. From these calculations and our y-axis length of arc(AB) = s 15.5. AREAS OF WEDGES 199 placements π 4 π 6 1 12 π 2 1 4 1 8 revolution 1 6 revolution revolution 3 4 revolution 3π 2 π revolution π 3 1 2 revolution Figure 15.11: Common angles measured in radians. previous examples of degree measure we find that 180 degrees = π radians. (15.1) Solving this equation for degrees or radians will provide conversion formulas relating the two types of angle measurement. The formula π 2π also helps explain the origin of the curious conversion factor 180 = 360 in Fact 15.3.2. 15.5 Areas of Wedges The beauty of radian measure is that it is rigged so that we can easily compute lengths of arcs and areas of circular sectors (i.e. “pie-shaped regions”). This is a key reason why we will almost always prefer to work with radian measure. Example 15.5.1. If a 16 inch pizza is cut into 12 equal slices, what is the area of a single slice? This can be solved using a general principle: (Area of a part) = (area of the whole) × (fraction of the part) CHAPTER 15. MEASURING AN ANGLE 200 So, for our pizza: O θ (area one slice) = (area whole pie) × (fraction of pie) 1 = (82 π) 12 16π . = 3 Let’s apply the same reasoning to find the area of a circular sector. We know the area of the circular disc B bounded by a circle of radius r is πr2 . Let Rθ be the “pie shaped wedge” cut out by an angle ∠AOB with positive Rθ measure θ radians. Using the above principle A Cr Figure 15.12: Finding the area of a “pie shaped wedge”. area(Rθ ) = (area of disc bounded by Cr ) × (portion of disc accounted for by Rθ ) 1 θ = r2 θ. = (πr2 ) 2π 2 For example, if r = 3 in. and θ = 9 the pie shaped wedge is 8 π sq. in. π 4 rad, then the area of Important Fact 15.5.2 (Wedge area). Start with a central angle with positive measure θ radians inside a circle of radius r. The area of the “pie shaped region” bounded by the angle is 1 r2 θ. 2 Example 15.5.3. A water drip irrigation arm 1200 feet long rotates around a pivot P once every 12 hours. How much area is covered by the arm in one hour? in 37 minutes? How much time is required to drip irrigate 1000 square feet? Solution. The irrigation arm will complete one revolution in 12 hours. The angle swept out by one complete revolution is 2π radians, so after t hours the arm sweeps out an angle θ(t) given by θ(t) = π 2π radians × t hours = t radians. 12 hours 6 Consequently, by Important Fact 15.5.2 the area A(t) of the irrigated region after t hours is 1 1 π A(t) = (1200)2θ(t) = (1200)2 t = 120,000πt square feet. 2 2 6 After 1 hour, the irrigated area is A(1) = 120,000π = 376,991 sq. ft. Like37 wise, after 37 minutes, which is 60 hours, the area of the irrigated region 37 37 is A( 60 ) = 120,000π( 60 ) = 232,500 square feet. To answer the final question, we need to solve the equation A(t) = 1000; i.e., 120,000πt = 1000, so 3600 seconds 1 = 9.55 seconds. hours × t= 120π hour 15.5. AREAS OF WEDGES 201 15.5.1 Chord Approximation Our ability to compute arc lengths can be used as an estimating tool for distances between two points. Let’s return to the situation posed at the beginning of this section: Cosmo the dog, tied by a 20 foot long tether to a post in the ground, begins at location R and walks counterclockwise to location S. Furthermore, let’s suppose you are standing at the center of the circle determined by the tether and you measure the angle from R to S to be 5◦ ; see the left-hand figure. Because the angle is small, notice that the straight line distance d from R to S is approximately the same as the arc length s subtended by the angle ∠RPS; the right-hand picture in Figure 15.13 is a blow-up: S S 5◦ P 20 feet d R s 5◦ P 20 feet R Figure 15.13: Using the arc length s to approximate the chord d. Example 15.5.4. Estimate the distance from R to S. Solution. We first convert the angle into radian measure via (15.1): 5◦ = 0.0873 radians. By Fact 15.3.2, the arc s has length 1.745 feet = 20.94 inches. This is approximately equal to the distance from R to S, since the angle is small. CHAPTER 15. MEASURING AN ANGLE 202 S s chord R O We call a line segment connecting two points on a circle a chord of the circle. The above example illustrates a general principal for approximating the length of any chord. A smaller angle will improve the accuracy of the arc length approximation. Important Fact 15.5.5 (Chord Approximation). In Figure 15.14, if the central angle is small, then s ≈ |RS|. Figure 15.14: Chord approximation. 15.6 Great Circle Navigation A basic problem is to find the shortest route between any two locations on the earth. We will review how to coordinatize the surface of the earth and recall the fact that the shortest path between two points is measured along a great circle. View the earth as a sphere of radius r = 3,960 miles. We could slice the earth with a two-dimensional plane P0 which is both perpendicular to a line connecting the North and South poles and passes through the center of the earth. Of course, the resulting intersection will trace out a circle of radius r = 3,960 miles on the surface of the earth, which we call the equator. We call the plane P0 the equatorial plane. Slicing the earth with any other plane P parallel to P0 , we can consider the right triangle pictured below and the angle θ: line of latitude North Pole North Pole b b θ r r r θ equatorial plane P0 r 90◦ − θ◦ equator equator center of earth South Pole center of earth South Pole Figure 15.15: Measuring latitude. Essentially two cases arise, depending on whether or not the plane P is above or below the equatorial plane. The plane P slices the surface of the earth in a circle, which we call a line of latitude. This terminology is somewhat incorrect, since these lines of latitude are actually circles on the surface of the earth, but the terminology is by now standard. Depending on whether this line of latitude lies above or below the equatorial plane, we refer to it as the θ◦ North line of latitude (denoted θ◦ N) or the θ◦ South line of latitude (denoted θ◦ S). Notice, the radius b of a line of latitude can vary from a maximum of 3,960 miles (in the case of θ = 0◦ ), 15.6. GREAT CIRCLE NAVIGATION 203 to a minimum of 0 miles, (when θ = 90◦ ). When b = 0, we are at the North or South poles on the earth. In a similar spirit, we could imagine slicing the earth with a plane Q which is perpendicular to the equatorial plane and passes through the center of the earth. The resulting intersection will trace out a circle of radius 3,960 miles on the surface of the earth, which is called a line of longitude. Half of a line of longitude from the North Pole to the South Pole is called a meridian. We distinguish one such meridian; the one which passes through Greenwich, England as the Greenwich meridian. Longitudes are measured using angles East or West of Greenwich. Pictured below, the longitude of A is θ. Because θ is east of Greenwich, θ measures longitude East, typically written θ◦ E; west longitudes would be denoted as θ◦ W. All longitudes are between 0◦ and 180◦ . The meridian which is 180◦ West (and 180◦ E) is called the International Date Line. Introducing the grid of latitude and longitude lines on Greenwich, England the earth amounts to imposing a coordinate system. In North Pole line of longitude other words, any position on the earth can be determined A Greenwich International meridian Date Line by providing the longitude and latitude of the point. The r θ r usual convention is to list longitude first. For example, Equator Seattle has coordinates 122.0333◦ W, 47.6◦ N. Since the lacenter of earth South Pole bels “N and S” are attached to latitudes and the labels “E and W” are attached to longitudes, there is no ambiguity Figure 15.16: The International Date Line. here. This means that Seattle is on the line of longitude 122.0333◦ West of the Greenwich meridian and on the line of latitude 47.6◦ North of the equator. In the figure below, we indicate the key angles ψ = 47.6◦ and θ = 122.0333◦ by inserting the three indicated radial line segments. not great circle N great circles Seattle, WA Ψ r θ Greenwich Meridian S not great circle Figure 15.17: Distances along great circles. CHAPTER 15. MEASURING AN ANGLE 204 Now that we have imposed a coordinate system on the earth, it is natural to study the distance between two locations. A great circle of a sphere is defined to be a circle lying on the sphere with the same center as the sphere. For example, the equator and any line of longitude are great circles. However, lines of latitude are not great circles (except the special case of the equator). Great circles are very important because they are used to find the shortest distance between two points on the earth. The important fact from geometry is summarized below. Important Fact 15.6.1 (Great Circles). The shortest distance between two points on the earth is measured along a great circle connecting them. Example 15.6.2. What is the shortest distance from the North Pole to Seattle, WA ? Solution. The line of longitude 122.0333◦ W is a great circle connecting the North Pole and Seattle. So, the shortest distance will be the arc length s subtended by the angle ∠NOW pictured in Figure 15.18. Since the latitude of Seattle is 47.6◦, the angle ∠EOW has measure 47.6◦ . Since ∠EON is a right angle (i.e., 90◦ ), ∠NOW has measure 42.4◦ . By Fact 15.4.1 and Equation 15.1, N W O E equator S s = (3960 miles)(42.4◦)(0.01745 radians/degree) = 2943.7 miles, Greenwich Meridian Figure 15.18: Distance between the North Pole and Seattle, Washington. which is the shortest distance from the pole to Seattle. 15.7 Summary • 360◦ = 2π radians • A circular arc with radius r and angle θ has length s, with s = rθ when θ is measured in radians. • A circular wedge with radius r and angle θ has area A, with 1 A = r2 θ 2 when θ is measured in radians. 15.8. EXERCISES 205 15.8 Exercises Problem 15.1. Let ∠AOB be an angle of measure θ. (a) Convert θ = 13.4o into degrees/ minutes/ seconds and into radians. (b) Convert θ = 1o 4 ′ 44 ′′ into degrees and radians. (c) Convert θ = 0.1 radian into degrees and degrees/ minutes/ seconds. Problem 15.2. A nautical mile is a unit of distance frequently used in ocean navigation. It is defined as the length of an arc s along a great circle on the earth when the subtending angle has measure 1 ′ = “one minute” = 1/60 of one degree. Assume the radius of the earth is 3,960 miles. (a) Find the length of one nautical mile to the nearest 10 feet. (b) A vessel which travels one nautical mile in one hours time is said to have the speed of one knot ; this is the usual navigational measure of speed. If a vessel is traveling 26 knots, what is the speed in mph (miles per hour)? (c) If a vessel is traveling 18 mph, what is the speed in knots? Problem 15.3. The rear window wiper blade on a station wagon has a length of 16 inches. The wiper blade is mounted on a 22 inch arm, 6 inches from the pivot point. 6" 16" (a) If the wiper turns through an angle of 110◦ , how much area is swept clean? (b) Through how much of an angle would the wiper sweep if the area cleaned was 10 square inches? (c) Suppose bug A lands on the end of the blade farthest from the pivot. Assume the wiper turns through an angle of 110◦ . In one cycle (back and forth) of the wiper blade, how far has the bug traveled? (d) Suppose bug B lands on the end of the wiper blade closest to the pivot. Assume the wiper turns through an angle of 110◦ . In one cycle of the wiper blade, how far has the bug traveled? (e) Suppose bug C lands on an intermediate location of the wiper blade. Assume the wiper turns through an angle of 110◦ . If bug C travels 28 inches after one cycle of the wiper blade, determine the location of bug C on the wiper blade. Problem 15.4. A water treatment facility operates by dripping water from a 60 foot long arm whose end is mounted to a central pivot. The water then filters through a layer of charcoal. The arm rotates once every 8 minutes. (a) Find the area of charcoal covered with water after 1 minute. (b) Find the area of charcoal covered with water after 1 second. (c) How long would it take to cover 100 square feet of charcoal with water? (d) How long would it take to cover 3245 square feet of charcoal with water? Problem 15.5. Astronomical measurements are often made by computing the small angle formed by the extremities of a distant object and using the estimating technique in 15.5.1. In the picture below, the full moon is shown 1o to form an angle of 2 when the distance indicated is 248,000 miles. Estimate the diameter of the moon. CHAPTER 15. MEASURING AN ANGLE 206 moon (a) θ = 45◦ (b) θ = 80o 248,000 miles (c) θ = 3 radians (d) θ = 2.46 radians (e) θ = 97o 23 ′ 3 ′′ o 1/2 (f) θ = 35o 24 ′ 2 ′′ earth Problem 15.6. An aircraft is flying at the speed of 500 mph at an elevation of 10 miles above the earth, beginning at the North pole and heading South along the Greenwich meridian. A spy satellite is orbiting the earth at an elevation of 4800 miles above the earth in a circular orbit in the same plane as the Greenwich meridian. Miraculously, the plane and satellite always lie on the same radial line from the center of the earth. Assume the radius of the earth is 3960 miles. Problem 15.8. Matilda is planning a walk around the perimeter of Wedge Park, which is shaped like a circular wedge, as shown below. The walk around the park is 2.1 miles, and the park has an area of 0.25 square miles. If θ is less than 90 degrees, what is the value of the radius, r? r θ r satellite plane earth Problem 15.9. Let C6 be the circle of radius 6 inches centered at the origin in the xycoordinate system. Compute the areas of the shaded regions in the picture below; the inner circle in the rightmost picture is the unit circle: y=x y=−x (a) When is the plane directly over a location with latitude 74◦ 30 ′ 18 ′′ N for the first time? C C 6 (b) How fast is the satellite moving? 6 y=x (c) When is the plane directly over the equator and how far has it traveled? (d) How far has the satellite traveled when the plane is directly over the equator? y=−(1/4)x + 2 Problem 15.7. Find the area of the sector of a circle of radius 11 inches if the measure θ of a central angle of this sector is: C 6 Chapter 16 Measuring Circular Motion If Cosmo begins at location R and walks counterclockwise, always maintaining a tight tether, how can we measure Cosmo’s speed? This is a “dynamic question” and requires that we discuss ways of measuring circular motion. In contrast, if we take a snapshot and ask to measure the specific angle ∠RPS, this is a “static question”, which we answered in the previous section. Cosmo moves counterclockwise, maintaining S a tight tether. R P Figure 16.1: How fast is Cosmo moving? 16.1 Different ways to measure Cosmo’s speed If Cosmo starts at location R and arrives at location S after some amount of time, we could study ω= measure ∠RPS . time required to go from R to S The funny Greek letter “ω” on the left of side of the equation is pronounced “oh-meg-a”. We will refer to this as an angular speed. Typical units are “ degrees ”, “ degrees ”, “ radians ”, etc. For example, if the angle minute second minute swept out by Cosmo after 8 seconds is 40◦ , then Cosmo’s angular speed 40◦ 5◦ is 8 seconds = sec . Using (15.1), we can convert to radian units and get π 5 rad ω = 36 rad = 3 π min . This is a new example of a rate and we can ask to sec find the total change, in the spirit of (1.2). If we are given ω in units of deg rad “ time ” or “ time ”, we have θ = ωt, which computes the measure of the angle θ swept out after time t (i.e. the total change in the angle). Angular speed places emphasis upon the 207 20 feet CHAPTER 16. MEASURING CIRCULAR MOTION 208 “size of the angle being swept out per unit time” by the moving object, starting from some initial position. We need to somehow indicate the direction in which the angle is being swept out. This can be done by indicating “clockwise” our “counterclockwise”. Alternatively, we can adopt the convention that the positive rotational direction is counterclockwise, then insert a minus sign to indicate rotation clockwise. For example, saying that Cosmo is moving at an angular speed of ω = − π rad means he 2 sec is moving clockwise π rad . 2 sec Another way to study the rate of a circular motion is to count the number of complete circuits of the circle per unit time. This sort of rate has the form Number of Revolutions ; Unit of Time we will also view this as an angular speed. If we take “minutes” to be the preferred unit of time, we arrive at the common measurement called revolutions per minute, usually denoted RPM or rev/min. For example, if Cosmo completes one trip around the circle every 2 minutes, then Cosmo is moving at a rate of 1 RPM. If instead, Cosmo completes one trip around 2 the circle every 12 seconds, then we could first express Cosmo’s speed in 1 units of revolutions/second as 12 rev/second, then convert to RPM units: 1 rev 12 sec 60 sec = 5 RPM. min As a variation, if we measure that Cosmo completed 3 of a revolution in 7 2 minutes, then Cosmo’s angular speed is computed by 3 rev 7 2 min = 3 RPM. 14 The only possible ambiguity involves the direction of revolution: the object can move clockwise or counterclockwise. The one shortcoming of using angular speed is that we are not directly keeping track of the distance the object is traveling. This is fairly easy to remedy. Returning to Figure 16.1, the circumference of the circle of motion is 2π(20) = 40π feet. This is the distance traveled per revolution, so we can now make conversions of angular speed into “distance traveled per unit time”; this is called the linear speed. If Cosmo is moving 1 RPM, 2 then he has a linear speed of v= 1 rev 2 min 40π ft rev = 20π Likewise, if Cosmo is moving v= π rad 7 sec 1 rev 2π rad ft . min π rad , 7 sec 40π ft rev then = 20π ft . 7 sec 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION 209 Important Fact 16.1.1. This discussion is an example of what is usually called “units analysis”. The key idea we have illustrated is how to convert between two different types of units: rev min converts to −→ ft min 16.2 Different Ways to Measure Circular Motion The discussion of Cosmo applies to circular motion of any object. As a matter of convention, we usually use the Greek letter ω to denote angular speed and v for linear speed. If an object is moving around a circle of radius r at a constant rate, then we can measure it’s speed in two ways: • The angular speed ω= “revolutions” “degrees swept” “radians swept” or or . “unit time” “unit time” “unit time” • The linear speed v= “distance traveled” . “per unit time” Important Facts 16.2.1 (Measuring and converting). We can convert between angular and linear speeds using these facts: • 1 revolution = 360◦ = 2π radians; • The circumference of a circle of radius r units is 2πr units. 16.2.1 Three Key Formulas If an object begins moving around a circle, there are a number of quantities we can try to relate. Some of these are “static quantities”: Take a visual “snapshot” of the situation after a certain amount of time has elapsed, then we can measure the radius, angle swept, arc length and time elapsed. Other quantities of interest are “dynamic quantities”: This means something is CHANGING with respect to time; in our case, the linear speed (which measures distance traveled per unit time) and angular speed (which measures angle swept per unit time) fall into this category. CHAPTER 16. MEASURING CIRCULAR MOTION 210 STATIC QUANTITIES DYNAMIC QUANTITIES S S 1. arc length s 5. angular speed ω 2. angle swept in time r 6. linear speed v θ P P R R 3. radius r 4. elapsed time t ...take a “snapshot” after time t... ...see what happens per unit time... Figure 16.2: Measuring linear and angular speed. We now know two general relationships for circular motion: (i) s = rθ, where s=arclength (a linear distance), r=radius of the circular path and θ=angle swept in RADIAN measure; this was the content of Fact 15.4.1 on page 197. (ii) θ = ωt, where θ is the measure of an angle swept, ω= angular speed and t represents time elapsed. This is really just a consequence of units manipulation. Notice how the units work in these formulas. If r=20 feet and θ = 1.3 radians, then the arc length s = 20(1.3) feet= 26 feet; this is the length of the arc of radius 20 feet that is subtending the angle θ. If ω = 3 rad rad/second and t = 5 seconds, then θ = 3 seconds × 5 seconds = 15 radians. If we replace “θ” in s = rθ of (i) with θ = ωt in (ii), then we get s = rωt. This gives us a relationship between arclength s (a distance) and time t. Plug in the fact that the linear speed is defined to be v = “distance” and we t get s rωt = = rω. t t All of these observations are summarized below. v= Important Facts 16.2.2 (Three really useful formulas). If we measure angles θ in RADIANS and ω in units of radians per unit time, we have these three formulas: s = rθ θ = ωt v = rω (16.1) (16.2) (16.3) 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION Example 16.2.3. You are riding a stationary exercise bike and the speedometer reads a steady speed of 40 MPH (miles per hour). If the rear wheel is 28 inches in diameter, determine the angular speed of a location on the rear tire. A pebble becomes stuck to the tread of the rear tire. Describe the location of the pebble after 1 second and 0.1 second. 211 * pebble sticks to tread here Figure 16.3: Where is the pebble after t seconds? Solution. The tires will be rotating in a counterclockwise direction and the radius r = 1 28 = 14 inches. The other given quantity, 2 “40 MPH”, involves miles, so we need to decide which common units to work with. Either will work, but since the problem is focused on the wheel, we will utilize inches. If the speedometer reads 40 MPH, this is the linear speed of a specified location on the rear tire. We need to convert this into an angular speed, using unit conversion formulas. First, the linear speed of the wheel is miles hr in = 704 . sec v= 5280 40 ft mile 12 in ft 1 hr 60 min 1 min 60 sec Now, the angular speed ω of the wheel will be inches 704 second ω= inches 2(14)π revolution revolution =8 second = 480 RPM It is then an easy matter to convert this to revolution second degrees = 2,880 . second ω= 8 360 degrees revolution If the pebble begins at the “6 o’clock” position (the place the tire touches the ground on the wheel), then after 1 second the pebble will go through 8 revolutions, so will be in the “6 o’clock” position again. After 0.1 seconds, the pebble will go through 8 rev (0.1 sec) = 0.8 rev sec = (0.8 rev) 360 = 288◦ . deg rev CHAPTER 16. MEASURING CIRCULAR MOTION 212 Keeping in mind that the rotation is counterclockwise, we can view the location of the pebble after 0.1 seconds as pictured below: 288◦ counterclockwise rotation after 0.1 second * located here at time = 0.1 sec starts here * pebble sticks to tread in 6 o’clock position Figure 16.4: Computing the pebble’s position after t = 0.1 sec. We solved the previous problem using the “unit conversion method”. There is an alternate approach available, which uses one of the formulas in Fact 16.2.2. Here is how you could proceed: First, as above, we know the linear speed is v = 704 in/sec. Using the “v = ωr” formula, we have 704 in = ω(14 in) sec rad ω = 50.28 . sec Notice how the units worked out in the calculation: the “time” unit comes from v and the “angular” unit will always be radians. As a comparison with the solution above, we can convert ω into RPM units: ω= =8 50.28 rad sec 1 rev 2π rad rev . sec All of the problems in this section can be worked using either the “unit conversion method” or the “v = ωr method”. 16.3 Music Listening Technology The technology of reproducing music has gone through a revolution since the early 1980’s. The “old” stereo long playing record (the LP ) and the 16.3. MUSIC LISTENING TECHNOLOGY 213 “new” digital compact disc (the CD ) are two methods of storing musical data for later reproduction in a home stereo system. These two technologies adopt different perspectives as to which notion of circular speed is best to work with. Long playing stereo records are thin vinyl plastic discs of radius 6 inches onto which small spiral grooves are etched into the surface; we can approximately view this groove as a circle. The LP is placed on a flat 12 inch diameter platter which turns at a constant angular speed of 1 33 3 RPM. An arm on a pivot (called the tone arm) has a needle mounted on the end (called the cartridge), which is placed in the groove on the outside edge of the record. Because the grooves wobble microscopically from side-to-side, the needle will mimic this motion. In turn, this sets a magnet (mounted on the opposite end of the needle) into motion. This moving magnet sits inside a coil of wire, causing a small varying voltage; the electric signal is then fed to your stereo, amplified and passed onto your speakers, reproducing music! amp LP turning at 33 1 RPM 3 speakers tonearm needle Figure 16.5: Reproducing music using analogue technology. This is known as analogue technology and is based upon the idea of 1 maintaining a constant angular speed of 33 3 RPM for the storage medium (our LP). (Older analogue technologies used 45 RPM and 78 RPM records. However, 33 1 RPM became the consumer standard for stereo music.) With 3 an LP, the beginning of the record (the lead-in groove ) would be on the outermost edge of the record and the end of the record (the exit groove ) would be close to the center. Placing the needle in the lead-in groove, the needle gradually works its way to the exit groove. However, whereas the angular speed of the LP is a constant 33 1 RPM, the linear speed at the 3 needle can vary quite a bit, depending on the needle location. 214 CHAPTER 16. MEASURING CIRCULAR MOTION 1 Example 16.3.1 (Analogue LP’s). The “lead-in groove” is 6 inches from the center of an LP, while the “exit groove” is 1 inch from the center. What is the linear speed (MPH) of the needle in the “lead-in groove”? What is the linear speed (MPH) of the needle in the “exit groove”? Find the location of the needle if the linear speed is 1 MPH. 6 Figure 16.6: Lead-in and exit grooves. Solution. This is a straightforward application of Fact 16.2.1. Let v6 (resp. v1 ) be the linear speed at the lead in groove (resp. exit groove); the subscript keeps track of the needle radial location. Since the groove is approximately a circle, 1 rev inches 2(6)π 3 min rev in = 1257 min min in 1257 min 60 hour = ft 5280 mile 12 in ft = 1.19 MPH v6 = 33 Similarly, v1 = 0.2 MPH. To answer the remaining question, let r be the radial distance from the center of the LP to the needle location on the record. If vr = 1 MPH: mile 1 = vr hour 1 rev in min 1 ft 1 mile = 33 2rπ 60 3 min rev hr 12 in 5280 ft So, when the needle is r = 5.04 inches from the center, the linear speed is 1 MPH. laser laser support arm moves back and forth spinning CD Figure 16.7: Reproducing music using digital technology. In the early 1980’s, a new method of storing and reproducing music was introduced; this medium is called the digital compact disc, referred to as a CD for short. This is a thin plastic disc of diameter 4.5 inches, which appears to the naked eye to have a shiny silver coating on one side. Upon microscopic examination one would find concentric circles of pits in the silver coating. This disc is placed in a CD player, which spins the disc. A laser located above the spinning disc will project onto the spinning disc. The pits in the silver coating will cause the reflected laser light to vary in intensity. A sensor detects this variation, converting it to a digital signal (the analogue to digital or AD conversion ). This is fed into a digital to analogue or DA conversion device, which sends a signal to your stereo, again producing music. 16.4. BELT AND WHEEL PROBLEMS 215 The technology of CD ′ s differs from that of LP ′ s in two crucial ways. First, the circular motion of the spinning CD is controlled so that the target on the disc below the laser is always moving at a constant linear inches speed of 1.2 meters = 2,835 minute . Secondly, the beginning location of the sec laser will be on the inside portion of the disc, working its way outward to the end. In this context, it makes sense to study how the angular speed of the CD is changing, as the laser position changes. Example 16.3.2 (Digital CD’s). What is the angular speed (in RPM) of a CD if the laser is at the beginning, located 3 4 inches from the center of the disc ? What is the angular speed (in RPM) of a CD if the laser is at the end, located 2 inches from the center of the disc ? Find the location of the laser if the angular speed is 350 RPM. Solution. This is an application of Fact 16.2.1. Let ω3/4 be the angular speed at the start and ω2 the angular speed at the end of the CD ; the subscript is keeping track of the laser distance from the CD center. (2835 inches/min) ω2 = = 225.6 RPM (2(2)π inches/rev) 2 ′′ 3 ′′ 4 Start of CD End of CD Figure 16.8: Computing the angular speed of a CD. 2835 inches/min) = 601.6 RPM (2(0.75)π inches/rev) To answer the remaining question, let r be the radial distance from the center of the CD to the laser location on the CD. If the angular speed ωr at this location is 350 RPM, we have the equation ω3/4 = 350 RPM = ωr inches 2,835 minute = inches 2rπ revolution 1.289 inches = r. So, when the laser is 1.289 inches from the center, the CD is moving 350 RPM. 16.4 Belt and Wheel Problems The industrial revolution spawned a number of elaborate machines involving systems of belts and wheels. Computing the speed of various belts and wheels in such a system may seem complicated at first glance. The situation can range from a simple system of two wheels with a belt connecting them, to more elaborate designs. We call problems of this sort belt and wheel problems, or more generally, connected wheel problems. Solving problems of this type always uses the same strategy, which we will first highlight by way of an example. CHAPTER 16. MEASURING CIRCULAR MOTION 216 J E C D G F A H B I Figure 16.9: Two typical connected wheel scenarios. front sprocket radius = 5 inches rear sprocket radius = 2 inches (a) A stationary exercise bike. radius A = 14 inches A C radius C = 5 inches B radius B = 2 inches (b) A model of the bike’s connected wheels. Figure 16.10: Visualizing the connected wheels of an exercise bike. Example 16.4.1. You are riding a stationary exercise bike. Assume the rear wheel is 28 inches in diameter, the rear sprocket has radius 2 inches and the front sprocket has radius 5 inches. How many revolutions per minute of the front sprocket produces a forward speed of 40 MPH on the bike (miles per hour)? Solution. There are 3 wheels involved with a belt (the bicycle chain) connecting two of the wheels. In this problem, we are provided with the linear speed of wheel A (which is 40 MPH) and we need to find the angular speed of wheel C=front sprocket. Denote by vA , vB , and vC the linear speeds of each of the wheels A, B, and C, respectively. Likewise, let ωA , ωB , and ωC denote the angular speeds of each of the wheels A, B, and C, respectively. In addition, the chain connecting the wheels B and C will have a linear speed, which we will denote by vchain . The strategy is broken into a sequence of steps which leads us from the known linear speed vA to the angular speed ωC of wheel C: • Step 1: Given vA , find ωA . Use the fact ωA = vA . rA • Step 2: Observe ωA = ωB ; this is because the wheel and rear sprocket are both rigidly mounted on a common axis of rotation. • Step 3: Given ωB , find vB . Use the fact vB = rB ωB = rB rB ωA = rA vA . • Step 4: Observe vB = vchain = vC ; this is because the chain is directly connecting the two sprockets and assumed not to slip. 16.4. BELT AND WHEEL PROBLEMS • Step 5: Given vC , find ωC . Use the fact ωC = vB rC = rB rA rC 217 vC rC = vA . Saying that the speedometer reads 40 MPH is the same as saying that the linear speed of a location on the rear wheel is vA = 40 MPH. Converting this into angular speed was carried out in our solution to Example 16.2.3 above; we found that ωA = 480 RPM. This completes Step 1 and so by Step 2, ωA = ωB = 480 RPM. For Step 3, we convert ωB = 480 RPM into linear speed following Fact 16.2.1: revolution inches (2(2)π) minute revolution inches = 6,032 . minute By Step 4, conclude that the linear speed of wheel C is vC = 6,032 inches/min. Finally, to carry out Step 5, we convert the linear speed into angular speed: vB = ωC = 480 6,032 inches min 2(5)π inches rev = 192 RPM rev . = 3.2 sec In conclusion, the bike rider must pedal the front sprocket at the rate of rev 3.2 sec . This example indicates the basic strategy used in all belt/wheel problems. Important Facts 16.4.2 (Belt and Wheel Strategy). Three basic facts are used in all such problems: • Using “unit conversion” or Fact 16.2.2 allows us to go from linear speed v to angular speed ω, and vice versa. • If two wheels are fastened rigidly to a common axle, then they have the same angular speed. (Caution: two wheels fastened to a common axle typically do not have the same linear speed!) • If two wheels are connected by a belt (or chain), the linear speed of the belt coincides with the linear speed of each wheel. CHAPTER 16. MEASURING CIRCULAR MOTION 218 16.5 Exercises Problem 16.1. The restaurant in the Space Needle in Seattle rotates at the rate of one revolution per hour. (a) Through how many radians does it turn in 100 minutes? (b) How long does it take the restaurant to rotate through 4 radians? (c) How far does a person sitting by the window move in 100 minutes if the radius of the restaurant is 21 meters? Problem 16.2. You are riding a bicycle along a level road. Assume each wheel is 26 inches in diameter, the rear sprocket has a radius of 3 inches and the front sprocket has a radius of 7 inches. How fast do you need to pedal (in revolutions per minute) to achieve a speed of 35 mph? front wheel rear wheel rear sprocket Problem 16.4. Lee is running around the perimeter of a circular track at a rate of 10 ft/sec. The track has a radius of 100 yards. After 10 seconds, Lee turns and runs along a radial line to the center of the circle. Once he reaches the center, he turns and runs along a radial line to his starting point on the perimeter. Assume Lee does not slow down when he makes these two turns. (a) Sketch a picture of the situation. (b) How far has Lee traveled once he returns to his starting position? (c) How much time will elapse during Lee’s circuit? (d) Find the area of the pie shaped sector enclosed by Lee’s path. Problem 16.5. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the worlds greatest ferris wheel. Tiff isn’t into math; she simply has a vision and has told John these constraints on her dream: (i) the wheel should rotate counterclockwise with an angular speed of 12 RPM; (ii) the linear speed of a rider should be 200 mph; (iii) the lowest point on the ride should be 4 feet above the level ground. front sprocket 12 RPM Problem 16.3. Answer the following angular speed questions. (a) A wheel of radius 22 ft. is rotating 11 RPM counterclockwise. Considering a point on the rim of the rotating wheel, what is the angular speed ω in rad/sec and the linear speed v in ft/sec? (b) A wheel of radius 8 in. is rotating 15o /sec. What is the linear speed v, the angular speed in RPM and the angular speed in rad/sec? θ P 4 feet (a) Find the radius of the ferris wheel. (c) You are standing on the equator of the earth (radius 3960 miles). What is your linear and angular speed? (b) Once the wheel is built, John suggests that Tiff should take the first ride. The wheel starts turning when Tiff is at the location P, which makes an angle θ with the horizontal, as pictured. It takes her 1.3 seconds to reach the top of the ride. Find the angle θ. (d) An auto tire has radius 12 inches. If you are driving 65 mph, what is the angular speed in rad/sec and the angular speed in RPM? (c) Poor engineering causes Tiff’s seat to fly off in 6 seconds. Describe where Tiff is located (an angle description) the instant she becomes a human missile. 16.5. EXERCISES 219 Problem 16.6. Michael and Aaron are on the “UL-Tossum” ride at Funworld. This is a merry-go-round of radius 20 feet which spins counterclockwise 60 RPM. The ride is driven by a belt connecting the outer edge of the ride to a drive wheel of radius 3 feet: (e) Assume Michael has traveled 88 feet from the position P to a new position Q. How many seconds will this take? What will be the angle swept out by Michael? length of arc (PQ) is 88 ft Aaron Drive wheel radius 3 ft Michael O P O main ride radius 20 ft drive belt (a) Assume Michael is seated on the edge of the ride, as pictured. What is Michael’s linear speed in mph and ft/sec? (b) What is the angular speed of the drive wheel in RPM? (c) Suppose Aaron is seated 16 feet from the center of the ride. What is the angular speed of Aaron in RPM? What is the linear speed of Aaron in ft/sec? (d) After 0.23 seconds Michael will be located at S as pictured. What is the angle ∠POS in degrees? What is the angle ∠POS in radians? How many feet has Michael traveled? S P Q Problem 16.7. You are riding a bicycle along a level road. Assume each wheel is 28 inches in diameter, the rear sprocket has radius 3 inches and the front sprocket has radius r inches. Suppose you are pedaling the front sprocket rev at the rate of 1.5 sec and your forward speed is 11 mph on the bike. What is the radius of the front sprocket? Problem 16.8. You are designing a system of wheels and belts as pictured below. You want wheel A to rotate 20 RPM while wheel B rotates 42 RPM. Wheel A has a radius of 6 inches, wheel B has a radius of 7 inches and wheel C has a radius of 1 inch. Assume wheels C and D are rigidly fastened to the same axle. What is the radius r of wheel D? θ O P D A C B 220 CHAPTER 16. MEASURING CIRCULAR MOTION Chapter 17 The Circular Functions Suppose Cosmo begins at location R and walks in a counterclockwise direction, always maintaining a tight 20 ft long tether. As Cosmo moves around the circle, how can we describe his location at any given instant? In one sense, we have already answered this question: The measure of ∠RPS1 exactly pins down a location on the circle of radius 20 feet. But, we really might prefer a description of the horizontal and vertical coordinates of Cosmo; this would tie in better with the coordinate system we typically use. Solving this problem will require NEW functions, called the circular functions. S1 S2 P 20 feet S3 R S4 Figure 17.1: Cosmo moves counterclockwise maintaining a tight tether. Where’s Cosmo? 17.1 Sides and Angles of a Right Triangle Example 17.1.1. You are preparing to make your final shot at the British Pocket Billiard World Championships. The position of your ball is as in Figure 17.2, and you must play the ball off the left cushion into the lower-right corner pocket, as indicated by the dotted path. For the big money, where should you aim to hit the cushion? The billiard table layout. 4 ft find this location 5 ft this pocket for the big money 6 ft 12 ft Solution. This problem depends on two basic facts. First, the angles of entry and exit between the path the cushion will be equal. Secondly, the two obvious right triangles in this picture are similar triangles. Let x represent the distance from the bottom left corner to the impact point of the ball’s path: Properties of similar triangles tell us that the ratios of 4 common sides are equal: 5−x = 12 . If we solve this equation x for x, we obtain x = 15 = 3.75 feet. 4 221 4 5−x θ θ x Mathmatically modeling the bank shot. Figure 17.2: A pocket billiard banking problem. CHAPTER 17. THE CIRCULAR FUNCTIONS 222 This discussion is enough to win the tourney. But, of course, there are still other questions we can ask about this simple example: What is the angle θ? That is going to require substantially more work; indeed the bulk of this Chapter! It turns out, there is a lot of mathematical mileage in the idea of studying ratios of sides of right triangles. The first step, which will get the ball rolling, is to introduce new functions whose very definition involves relating sides and angles of right triangles. 17.2 The Trigonometric Ratios From elementary geometry, the sum of the angles of any triangle will equal 180◦ . Given a right triangle △ABC, since one of the angles is 90◦ , the remaining two angles must be acute angles ; i.e., angles of measure between 0◦ and 90◦ . If we specify one of the acute angles in a right triangle △ABC, say angle θ, we can label the three sides using this terminology. We then consider the following three ratios of side lengths, referred to as trigonometric ratios : B hypotenuse side opposite θ A θ C side adjacent θ Figure 17.3: Labeling the sides of a right triangle. def sin(θ) = def cos(θ) = def tan(θ) = length of side opposite θ length of hypotenuse (17.1) length of side adjacent θ length of hypotenuse (17.2) length of side opposite θ . length of side adjacent to θ (17.3) For example, we have three right triangles in Figure 17.4; you can verify that the Pythagorean Theorem holds in each of the cases. In the 5 5 left-hand triangle, sin(θ) = 13 , cos(θ) = 12 , tan(θ) = 12 . In the middle 13 1 1 triangle, sin(θ) = √2 , cos(θ) = √2 , tan(θ) = 1. In the right-hand triangle, √ 1 sin(θ) = 1 , cos(θ) = 23 , tan(θ) = √3 . The symbols “sin”, “cos”, and “tan” 2 are abbreviations for the words sine, cosine and tangent, respectively. As we have defined them, the trigonometric ratios depend on the dimensions of the triangle. However, the same ratios are obtained for any right triangle with acute angle θ. This follows from the properties of similar triangles. Consider Figure 17.5. Notice △ABC and △ADE are similar. If we use △ABC to compute cos(θ), then we find cos(θ) = ||AC|| . On the other AB hand, if we use △ADE, we obtain cos(θ) = |AE| . |AD| Since the ratios of com- mon sides of similar triangles must agree, we have cos(θ) = |AC| |AB| = |AE| , |AD| 17.2. THE TRIGONOMETRIC RATIOS √ 13 223 2 2 5 1 θ 1 θ 12 θ 1 √ 3 Figure 17.4: Computing trigonometric ratios for selected right triangles. which is what we wanted to be true. The same argument can be used to show that sin(θ) and tan(θ) can be computed using any right triangle with acute angle θ. Except for some “rigged” right triangles, it is not easy D to calculate the trigonometric ratios. Before the 1970’s, B approximate values of sin(θ), cos(θ), tan(θ) were listed in long tables or calculated using a slide rule. Today, a scientific calculator saves the day on these computations. θ Most scientific calculators will give an approximation for A C E the values of the trigonometric ratios. However, it is good Figure 17.5: Applying to keep in mind we can compute the EXACT values of the π π π π trigonometric ratios to any trigonometric ratios when θ = 0, 6 , 4 , 3 , 2 radians or, right triangle. equivalently, when θ = 0◦ , 30◦ , 45◦ , 60◦ , 90◦ . Angle θ Trigonometric Ratio Deg Rad sin(θ) cos(θ) tan(θ) 0◦ 0 0 1 0 30◦ π 6 1 2 √ 3 2 1 √ 3 45◦ π 4 √ 2 2 1 60◦ π 3 90◦ π 2 √ 2 2 √ 3 2 1 0 √ 1 2 3 Undefined Table 17.1: Exact Trigonometric Ratios CHAPTER 17. THE CIRCULAR FUNCTIONS 224 Some people make a big deal of “approximate” vs. “exact” answers; we won’t worry about it here, unless we are specifically asked for an exact answer. However, here is something we will make a big deal about: !!! When computing values of cos(θ), sin(θ), and tan(θ) on your calculator, make sure you are using the correct “angle mode” when entering θ; i.e. “degrees” or “radians”. CAUTION !!! For example, if θ = 1◦ , then cos(1◦ ) = 0.9998, sin(1◦ ) = 0.0175, and tan(1◦ ) = 0.0175. In contrast, if θ = 1 radians, then cos(1) = 0.5403, sin(1) = 0.8415, and tan(1) = 1.5574. 17.3 Applications h h sin(θ) θ a tan(θ) θ h cos(θ) When confronted with a situation involving a right triangle where the measure of one acute angle θ and one side are known, we can solve for the remaining sides using the appropriate trigonometric ratios. Here is the key picture to keep in mind: a Figure 17.6: What do these ratios mean? Important Facts 17.3.1 (Trigonometric ratios). Given a right triangle, the trigonometric ratios relate the lengths of the sides as shown in Figure 17.6. Example 17.3.2. To measure the distance across a river for a new bridge, surveyors placed poles at locations A, B and C. The length |AB| = 100 feet and the measure of the angle ∠ABC is 31◦ 18 ′ . Find the distance to span the river. If the measurement of the angle ∠ABC is only accurate within ±2 ′ , find the possible error in |AC|. C Solution. The trigonometric ratio relating these two sides would be the tangent and we can convert θ into decimal form, arriving at: d B 100 A tan(31◦18 ′ ) = tan(31.3◦) = 310 18 ′ Figure 17.7: The distance spanning a river. |AC| d = 100 |BA| therefore d = 60.8 feet. This tells us that the bridge needs to span a gap of 60.8 feet. If the measurement of the angle was in error by +2 ′ , then tan(31◦20 ′ ) = tan(31.3333◦) = 0.6088 and the span is 60.88 ft. On the other hand, if the measurement of the angle was in error by −2 ′ , then tan(31◦16 ′ ) = tan(31.2667◦) = 0.6072 and the span is 60.72 ft. 17.3. APPLICATIONS 225 Example 17.3.3. A plane is flying 2000 feet above sea level toward a mountain. The pilot observes the top of the mountain to be 18◦ above the horizontal, then immediately flies the plane at an angle of 20◦ above horizontal. The airspeed of the plane is 100 mph. After 5 minutes, the plane is directly above the top of the mountain. How high is the plane above the top of the mountain (when it passes over)? What is the height of the mountain? T E P 2000 ft L 111 000 S sealevel 111 000 111 000 Figure 17.8: Flying toward a mountain. Solution. We can compute the hypotenuse of △LPT by using the speed and time information about the plane: |PT | = (100 mph)(5 minutes)(1 hour/60 minutes) = 25 miles. 3 The definitions of the trigonometric ratios show: 25 sin(20◦ ) = 2.850 miles, and 3 25 cos(20◦ ) = 7.831 miles. |PL| = 3 |TL| = With this data, we can now find |EL|: |EL| = |PL| tan(18◦ ) = 2.544 miles. The height of the plane above the peak is |TE| = |TL| − |EL| = 2.850 − 2.544 = 0.306 miles = 1,616 feet. The elevation of the peak above sea level is given by: Peak elevation = plane altitude + |EL| = |SP| + |EL| = 2,000 + (2.544)(5,280) = 15,432 feet. Example 17.3.4. A Forest Service helicopter needs to determine the width of a deep canyon. While hovering, they measure the angle γ = 48◦ at position B (see picture), then descend 400 feet to position A and make two measurements of α = 13◦ (the measure of ∠EAD), β = 53◦ (the measure of ∠CAD). Determine the width of the canyon to the nearest foot. B 400 ft C γ A β α E 111 000 111 000 canyon 111 000 D Figure 17.9: Finding the width of a canyon. Solution. We will need to exploit three right triangles in the picture: △BCD, △ACD, and △ACE. Our goal is to compute |ED| = |CD| − |CE|, which suggests more than one right triangle will come into play. CHAPTER 17. THE CIRCULAR FUNCTIONS 226 The first step is to use △BCD and △ACD to obtain a system of two equations and two unknowns involving some of the side lengths; we will then solve the system. From the definitions of the trigonometric ratios, |CD| = (400 + |AC|) tan(48◦) |CD| = |AC| tan(53◦ ). Plugging the second equation into the first and rearranging we get |AC| = 400 tan(48◦ ) = 2,053 feet. tan(53◦) − tan(48◦ ) Plugging this back into the second equation of the system gives |CD| = (2053) tan(53◦) = 2724 feet. The next step is to relate △ACD and △ACE, which can now be done in an effective way using the calculations above. Notice that the measure of ∠CAE is β − α = 40◦ . We have |CE| = |AC| tan(40◦ ) = (2053) tan(40◦ ) = 1,723 feet. As noted above, |ED| = |CD| − |CE| = 2,724 − 1,723 = 1,001 feet is the width of the canyon. 17.4 Circular Functions S = (x,y) 20 y θ P x R If Cosmo is located somewhere in the first quadrant of Figure 17.1, represented by the location S, we can use the trigonometric ratios to describe his coordinates. Impose the indicated xy-coordinate system with origin at P and extract the pictured right triangle with vertices at P and S. The radius is 20 ft. and applying Fact 17.3.1 gives S = (x, y) = (20 cos(θ), 20 sin(θ)). Figure 17.10: Cosmo on a circular path. Unfortunately, we run into a snag if we allow Cosmo to wander into the second, third or fourth quadrant, since then the angle θ is no longer acute. 17.4. CIRCULAR FUNCTIONS 227 17.4.1 Are the trigonometric ratios functions? Recall that sin(θ), cos(θ), and tan(θ) are defined for acute A unit circle with angles θ inside a right triangle. We would like to say that P radius = 1. these three equations actually define functions where the 1 variable is an angle θ. Having said this, it is natural to ask θ if these three equations can be extended to be defined for π O R ANY angle θ. For example, we need to explain how sin 23 1 is defined. To start, we begin with the unit circle pictured in the xy-coordinate system. Let θ = ∠ROP be the angle in stanFigure 17.11: Coordinates dard central position shown in Figure 17.11. If θ is posof points on the unit circle. itive (resp. negative), we adopt the convention that θ is swept out by counterclockwise (resp. clockwise) rotation of the initial side OR. The objective is to find the coordinates of the point P in this figure. Notice that each coordinate of P (the x-coordinate and the y-coordinate) will depend on the given angle θ. For this reason, we need to introduce two new functions involving the variable θ. Definition 17.4.1. Let θ be an angle in standard central position inside the unit circle, as in Figure 17.11. This angle determines a point P on the unit circle. Define two new functions, cos(θ) and sin(θ), on the domain of all θ values as follows: 11 00 11 00 1111 11 1111 0000 11 r 0000 00 = 1 kilometer 1111 00 0000 00 1111 0000 11 0.025 rad sec Michael starts here def cos(θ) = horizontal x-coordinate of P on unit circle def sin(θ) = vertical y-coordinate of P on unit circle. Figure 17.12: A circular We refer to sin(θ) and cos(θ) as the basic circular funcdriving track. tions. Keep in mind that these functions have variables which are angles (either in degree or radian measure). These functions will be on your calculator. Again, BE CAREFUL to check the angle mode setting on your calculator (“degrees” or “radians”) before doing a calculation. Example 17.4.2. Michael is test driving a vehicle counterclockwise around a desert test track which is circular of radius 1 kilometer. He starts at the location pictured, traveling 0.025 rad . Impose coordinates as pictured. Where is sec Michael located (in xy-coordinates) after 18 seconds? Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds and θ(t) the angle swept out the by Michael after t seconds. Since we are given the angular speed, we get θ(t) = 0.025t radians. y-axis M(t) = (x(t),y(t)) 0.025 θ(t ) rad sec x-axis Michael starts here Figure 17.13: Modeling Michael’s location. CHAPTER 17. THE CIRCULAR FUNCTIONS 228 Since the angle θ(t) is in central standard position, we get M(t) = (cos(θ(t)), sin(θ(t))) = (cos(0.025t), sin(0.025t)). So, after 18 seconds Michael’s location will be M(18) = (0.9004, 0.4350). !!! Interpreting the coordinates of the point P = (cos(θ), sin(θ)) in Figure 17.11 only works if the angle θ is viewed in central standard position. You must do some additional work if the angle is placed in a different position; see the next Example. CAUTION !!! y-axis Angela starts here 0.025 rad sec r = 1 kilometer x-axis Michael starts here 0.03 rad sec (a) Angela and Michael on the same test track. y-axis M(t) Angela starts here β(t) α(t) 0.03 rad sec 0.025 rad sec θ(t) x-axis Michael starts here A(t) (b) Modeling the motion of Angela and Michael. Figure 17.14: Visualizing motion on a circular track. Example 17.4.3. Both Angela and Michael are test driving vehicles counterclockwise around a desert test track which is circular of radius 1 kilometer. They start at the locations shown in Figure 17.14(a). Michael is traveling 0.025 rad/sec and Angela is traveling 0.03 rad/sec. Impose coordinates as pictured. Where are the drivers located (in xy-coordinates) after 18 seconds? Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds. Likewise, let A(t) be the point on the circle of motion representing Angela’s location after t seconds. Let θ(t) be the angle swept out the by Michael and α(t) the angle swept out by Angela after t seconds. Since we are given the angular speeds, we get θ(t) = 0.025t radians, and α(t) = 0.03t radians. From the previous Example 17.4.2, M(t) = (cos(0.025t), sin(0.025t)), and M(18) = (0.9004, 0.4350). Angela’s angle α(t) is NOT in central standard position, so we must observe that α(t) + π = β(t), where β(t) is in central standard position: See Figure 17.14(b). We conclude that A(t) = (cos(β(t)), sin(β(t))) = (cos(π + 0.03t), sin(π + 0.03t)). So, after 18 seconds Angela’s location will be A(18) = (−0.8577, −0.5141). 17.4. CIRCULAR FUNCTIONS 229 17.4.2 Relating circular functions and right triangles If the point P on the unit circle is located in the first quadrant, then we can compute cos(θ) and sin(θ) using trigonometric ratios. In general, it’s useful to relate right triangles, the unit circle and the circular functions. To describe this connection, given θ we place it in central standard position in the unit circle, where ∠ROP = θ. Draw a line through P perpendicular to the x-axis, obtaining an inscribed right triangle. Such a right triangle has hypotenuse of length 1, vertical side of length labeled b and horizontal side of length labeled a. There are four cases: See Figure 17.16. P O θ a b sin(θ) θ O R 1 0 1 0 a 1 0 O Figure 17.15: The point P in the first quadrant. θ θ a R b 11 00 O 11 00 11 00 R θ 11 00 11 11 00 00 11 11 00 00 a O bR P P CASE I CASE II CASE III CASE IV Figure 17.16: Possible positions of θ on the unit circle. Case I has already been discussed, arriving at cos(θ) = a and sin(θ) = b. In Case II , we can interpret cos(θ) = −a, sin(θ) = b. We can reason similarly in the other Cases III and IV, using Figure 17.16, and we arrive at this conclusion: Important Facts 17.4.4 (Circular functions and triangles). View θ as in Figure 17.16 and form the pictured inscribed right triangles. Then we can interpret cos(θ) and sin(θ) in terms of these right triangles as follows: Case I : Case II : Case III : Case IV : cos(θ) = a, cos(θ) = −a, cos(θ) = −a, cos(θ) = a, sin(θ) = b sin(θ) = b sin(θ) = −b sin(θ) = −b R cos(θ) P b unit circle (radius = 1) P CHAPTER 17. THE CIRCULAR FUNCTIONS 230 17.5 What About Other Circles? T P θ O R Cr S unit circle Figure 17.17: Points on other circles. What happens if we begin with a circle Cr with radius r (possibly different than 1) and want to compute the coordinates of points on this circle? The circular functions can be used to answer this more general question. Picture our circle Cr centered at the origin in the same picture with unit circle C1 and the angle θ in standard central position for each circle. As pictured, we can view θ = ∠ROP = ∠SOT . If P = (x,y) is our point on the unit circle corresponding to the angle θ, then the calculation below shows how to compute coordinates on general circles: P = (x,y) = (cos(θ), sin(θ)) ∈ C1 ⇔ ⇔ ⇔ ⇔ x2 + y2 = 1 r2 x2 + r2 y2 = r2 (rx)2 + (ry)2 = r2 T = (rx, ry) = (r cos(θ), r sin(θ)) ∈ Cr . Important Fact 17.5.1. Let Cr be a circle of radius r centered at the origin and θ = ∠SOT an angle in standard central position for this circle, as in Figure 17.17. Then the coordinates of T = (r cos(θ), r sin(θ)). y-axis β = π − α = 2.9416 R U α B T Q S P O Examples 17.5.2. Consider the picture below, with θ = 0.8 radians and α = 0.2 radians. What are the coordinates of the labeled points? θ A x-axis circle radius = 1 circle radius = 2 circle radius = 3 Figure 17.18: Coordinates of points on circles. Solution. The angle θ is in standard central position; α is a central angle, but it is not in standard position. Notice, β = π − α = 2.9416 is an angle in standard central position which locates the same points U, T, S as the angle α. Applying Definition 17.4.1 on page 227: P = (cos(0.8), sin(0.8)) = (0.6967, 0.7174) Q = (2 cos(0.8), 2 sin(0.8)) = (1.3934, 1.4347) R = (3 cos(0.8), 3 sin(0.8)) = (2.0901, 2.1521) S = (cos(2.9416), sin(2.9416)) = (−0.9801, 0.1987) T = (2 cos(2.9416), 2 sin(2.9416)) = (−1.9602, 0.3973) U = (3 cos(2.9416), 3 sin(2.9416)) = (−2.9403, 0.5961). 17.6. OTHER BASIC CIRCULAR FUNCTION 231 Example 17.5.3. Suppose Cosmo begins at the position R in the figure, walking around the circle of radius 20 feet 4 with an angular speed of 5 RPM counterclockwise. After 3 minutes have elapsed, describe Cosmo’s precise location. 4 Solution. Cosmo has traveled 3 5 = 12 revolutions. If θ is 5 the angle traveled after 3 minutes, θ = 152 rev 2π radians = rev 24π radians = 15.08 radians. By (15.5.1), we have x = 5 20 cos 24π rad = −16.18 feet and y = 20 sin 24π rad = 5 5 11.76 feet. Conclude that Cosmo is located at the point S = (−16.18, 11.76). Using (15.1), θ = 864◦ = 2(360◦) + 144◦ ; this means that Cosmo walks counterclockwise around the circle two complete revolutions, plus 144◦. S P Given any angle θ, our constructions offer a concrete link between the cosine and sine functions and right triangles inscribed inside the unit circle: See Figure 17.20. P θ θ O R O θ R O R R θO P P CASE I CASE II CASE III CASE IV Figure 17.20: Computing the slope of a line using the function tan(θ). The slope of the hypotenuse of these inscribed triangles is just the slope of the line through OP. Since P = (cos(θ), sin(θ)) and O = (0, 0): Slope = sin(θ) ∆y = ; ∆x cos(θ) this would be valid as long as cos(θ) = 0. This calculation motivates a new circular function called the tangent of θ by the rule tan(θ) = sin(θ) , cos(θ) provided cos(θ) = 0. R Figure 17.19: Where is Cosmo after 3 minutes? 17.6 Other Basic Circular Function P 20 feet CHAPTER 17. THE CIRCULAR FUNCTIONS 232 The only time cos(θ) = 0 is when the corresponding point P on the unit circle has x-coordinate 0. But, this only happens at the positions (0, 1) and (0, −1) on the unit circle, corresponding to angles of the form θ = ± π , ± 3π , ± 5π , · · · . These are the cases when the inscribed right tri2 2 2 angle would “degenerate” to having zero width and the line segment OP becomes vertical. In summary, we then have this general idea to keep in mind: Important Fact 17.6.1. The slope of a line = tan(θ), where θ is the angle the line makes with the x-axis (or any other horizontal line) Three other commonly used circular functions come up from time to time. The cotangent function y = cot(θ), the secant function y = sec(θ) and the cosecant function y = csc(θ) are defined by the formulas: def sec(θ) = 1 , cos(θ) def csc(θ) = 1 , sin(θ) def cot(θ) = 1 . tan(θ) Just as with the tangent function, one needs to worry about the values of θ for which these functions are undefined (due to division by zero). We will not need these functions in this text. North Alaska Northwest 500 1150 West East 200 SeaTac South Delta (a) The flight paths of three airplanes. x = 30 Alaska Q Northwest N P y = 20 W E R Delta x = −50 S (b) Modeling the paths of each flight. Figure 17.21: Visualizing and modeling departing airplanes. Example 17.6.2. Three airplanes depart SeaTac Airport. A NorthWest flight is heading in a direction 50◦ counterclockwise from East, an Alaska flight is heading 115◦ counterclockwise from East and a Delta flight is heading 20◦ clockwise from East. Find the location of the Northwest flight when it is 20 miles North of SeaTac. Find the location of the Alaska flight when it is 50 miles West of SeaTac. Find the location of the Delta flight when it is 30 miles East of SeaTac. Solution. We impose a coordinate system in Figure 17.21(a), where “East” (resp. “North”) points along the positive x-axis (resp. positive y-axis). To solve the problem, we will find the equation of the three lines representing the flight paths, then determine where they intersect the appropriate horizontal or vertical line. The Northwest and Alaska directions of flight are angles in standard central position; the Delta flight direction will be −20◦ . We can imagine right triangles with their hypotenuses along the directions of flight, then using the tangent function, we have these three immediate conclusions: slope NW line = tan(50◦ ) = 1.19, slope Alaska line = tan(115◦ ) = −2.14, and slope Delta line = tan(−20◦ ) = −0.364. 17.6. OTHER BASIC CIRCULAR FUNCTION 233 All three flight paths pass through the origin (0,0) of our coordinate system, so the equations of the lines through the flight paths will be: NW flight : y = 1.19x, Alaska flight : y = −2.14x, Delta flight : y = −0.364x. The Northwest flight is 20 miles North of SeaTac when y = 20; plugging into the equation of the line of flight gives 20 = 1.19x, so x = 16.81 and the plane location will be P = (16.81, 20). Similarly, the Alaska flight is 50 miles West of SeaTac when x = −50; plugging into the equation of the line of flight gives y = (−2.14)(−50) = 107 and the plane location will be Q = (−50, 107). Finally, check that the Delta flight is at R = (30, −10.92) when it is 30 miles East of SeaTac. CHAPTER 17. THE CIRCULAR FUNCTIONS 234 17.7 Exercises Problem 17.1. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the world’s greatest ferris wheel. Tiff isn’t into math; she simply has a vision and has told John these constraints on her dream: (i) the wheel should rotate counterclockwise with an angular speed of 12 RPM; (ii) the linear speed of a rider should be 200 mph; (iii) the lowest point on the ride should be 4 feet above the level ground. Recall, we worked on this in Exercise 16.5. forest) as a potential landing site, but are uncertain whether it is wide enough. They make two measurements from A (see picture) finding α = 25o and β = 54o . They rise vertically 100 feet to B and measure γ = 47o . Determine the width of the clearing to the nearest foot. B γ 100 feet A β C α E clearing D 12 RPM θ P 4 feet (a) Impose a coordinate system and find the coordinates T (t) = (x(t),y(t)) of Tiff at time t seconds after she starts the ride. (b) Tiff becomes a human missile after 6 seconds on the ride. Find Tiff’s coordinates the instant she becomes a human missile. (c) Find the equation of the tangential line along which Tiff travels the instant she becomes a human missile. Sketch a picture indicating this line and her initial direction of motion along it when the seat detaches. Problem 17.2. (a) Find the equation of a line passing through the point (-1,2) and making an angle of 13o with the x-axis. (Note: There are two answers; find them both.) (b) Find the equation of a line making an angle of 8o with the y-axis and passing through the point (1,1). (Note: There are two answers; find them both.) Problem 17.3. The crew of a helicopter needs to land temporarily in a forest and spot a flat horizontal piece of ground (a clearing in the Problem 17.4. Marla is running clockwise around a circular track. She runs at a constant speed of 3 meters per second. She takes 46 seconds to complete one lap of the track. From her starting point, it takes her 12 seconds to reach the northermost point of the track. Impose a coordinate system with the center of the track at the origin, and the northernmost point on the positive y-axis. (a) Give Marla’s coordinates at her starting point. (b) Give Marla’s coordinates when she has been running for 10 seconds. (c) Give Marla’s coordinates when she has been running for 901.3 seconds. Problem 17.5. A merry-go-round is rotating at the constant angular speed of 3 RPM counterclockwise. The platform of this ride is a circular disc of radius 24 feet. You jump onto the ride at the location pictured below. rotating 3 RPM θ jump on here 17.7. EXERCISES 235 (a) If θ = 34o , then what are your xycoordinates after 4 minutes? (b) If θ = 20o , then what are your xycoordinates after 45 minutes? (c) If θ = −14o , then what are your xycoordinates after 6 seconds? Draw an accurate picture of the situation. (d) If θ = −2.1 rad, then what are your xy-coordinates after 2 hours and 7 seconds? Draw an accurate picture of the situation. (e) If θ = 2.1 rad, then what are your xycoordinates after 5 seconds? Draw an accurate picture of the situation. Problem 17.6. Shirley is on a ferris wheel which spins at the rate of 3.2 revolutions per minute. The wheel has a radius of 45 feet, and the center of the wheel is 59 feet above the ground. After the wheel starts moving, Shirley takes 16 seconds to reach the top of the wheel. How high above the ground is she when the wheel has been moving for 9 minutes? Problem 17.7. The top of the Boulder Dam has an angle of elevation of 1.2 radians from a point on the Colorado River. Measuring the angle of elevation to the top of the dam from a point 155 feet farther down river is 0.9 radians; assume the two angle measurements are taken at the same elevation above sea level. How high is the dam? downriver dam 0.9 1.2 relatively flat area nearby the tower (not necessarily the same altitude as the bottom of the tower), and standing some unknown distance away from the tower, you make three measurements all at the same height above sea level. You observe that the top of the old tower makes an angle of 39◦ above level. You move 110 feet away from the original measurement and observe that the old top of the tower now makes an angle of 34◦ above level. Finally, after the new construction is complete, you observe that the new top of the tower, from the same point as the second measurement was made, makes an angle of 40◦ above the horizontal. All three measurements are made at the same height above sea level and are in line with the tower. Find the height of the addition to the tower, to the nearest foot. Problem 17.9. Charlie and Alexandra are running around a circular track with radius 60 meters. Charlie started at the westernmost point of the track, and, at the same time, Alexandra started at the northernmost point. They both run counterclockwise. Alexandra runs at 4 meters per second, and will take exactly 2 minutes to catch up to Charlie. Impose a coordinate system, and give the x- and y-coordinates of Charlie after one minute of running. Problem 17.10. George and Paula are running around a circular track. George starts at the westernmost point of the track, and Paula starts at the easternmost point. The illustration below shows their starting positions and running directions. They start running toward each other at constant speeds. George runs at 9 feet per second. Paula takes 50 seconds to run a lap of the track. George and Paula pass each other after 11 seconds. N 155 ft a Problem 17.8. A radio station obtains a permit to increase the height of their radio tower on Queen Anne Hill by no more than 100 feet. You are the head of the Queen Anne Community Group and one of your members asks you to make sure that the radio station does not exceed the limits of the permit. After finding a George Paula After running for 3 minutes, how far east of his starting point is George? CHAPTER 17. THE CIRCULAR FUNCTIONS 236 Problem 17.11. A kite is attached to 300 feet of string, which makes a 42 degree angle with the level ground. The kite pilot is holding the string 4 feet above the ground. (d) What are the coordinates of the bug after 1 second? After 0 seconds? After 3 seconds? After 22 seconds? ω=4π/9 rad/sec bug lands here kite 1.2 rad o 42 4 feet 2 in ground level (a) How high above the ground is the kite? (b) Suppose that power lines are located 250 feet in front of the kite flyer. Is any portion of the kite or string over the power lines? bug lands here Problem 17.12. In the pictures below, a bug has landed on the rim of a jelly jar and is moving around the rim. The location where the bug initially lands is described and its angular speed is given. Impose a coordinate system with the origin at the center of the circle of motion. In each of the cases, answer these questions: (a) Find an angle θ0 in standard central position that gives the bugs initial location. (In some cases, this is the angle given in the picture; in other cases, you will need to do something.) (b) The location angle of the bug at time t is given by the formula θ(t) = θ0 + ωt. Plug in the values for θ0 and ω to explicitly obtain a formula for θ(t). (c) Find the coordinates of the bug at time t. 2 in ω=4π/9rad/sec ω= 4π/9 rad/sec bug lands here 0.5 rad 2 in Chapter 18 Trigonometric Functions Our definitions of the circular functions are based upon the unit circle. This makes it easy to visualize many of their properties. 18.1 Easy Properties of Circular Functions How can we determine the range of function values for cos(θ) and sin(θ)? To begin with, recall the abstract definition for the range of a function f(θ): y-axis (0,1) P (θ) sin(θ) 1 θ (−1,0) (1,0) cos(θ) Range of f = {f(θ) : θ is in the domain}. ball moves counterclockwise x-axis UNIT CIRCLE (0, − 1) Using the unit circle constructions of the basic circular Figure 18.1: Visualizing the functions, it is easy to visualize the range of cos(θ) and range of sin(θ) and cos(θ). sin(θ). Beginning at the position (1, 0), imagine a ball moving counterclockwise around the unit circle. If we “freeze” the motion at any point in time, we will have swept out an angle θ and the corresponding position P(θ) on the circle will have coordinates P(θ) = (cos(θ), sin(θ)). y-axis light source ball moves from 0 to π radians around 2 ball moves from 0 to π radians 2 unit circle x-axis around unit circle light source x-axis y-axis (b) What do you see on the (a) What do you see on the x-axis? y-axis? Figure 18.2: Projecting the coordinates of points onto the y-axis and the x-axis. By studying the coordinates of the ball as it moves in the first quadrant, we will be studying cos(θ) and sin(θ), for 0 ≤ θ ≤ π/2 radians. 237 CHAPTER 18. TRIGONOMETRIC FUNCTIONS 238 We can visualize this very concretely. Imagine a light source as in Figure 18.2(a); then a shadow projects onto the vertical y-axis. The shadow locations you would see on the y-axis are precisely the values sin(θ), for 0 ≤ θ ≤ π/2 radians. Similarly, imagine a light source as in Figure 18.2(b); then a shadow projects onto the horizontal x-axis. The shadow locations you would see on the x-axis are precisely the values cos(θ), for 0 ≤ θ ≤ π/2 radians. There are two visual conclusions: First, the function values of sin(θ) vary from 0 to 1 as θ varies from 0 to π/2. Secondly, the function values of cos(θ) vary from 1 to 0 as θ varies from 0 to π/2. Of course, we can go ahead and continue analyzing the motion as the ball moves into the second, third and fourth quadrant, ending up back at the starting position (1, 0). See Figure 18.3. light source y-axis #2 #1 ball moves from 0 to 2π radians around unit circle x-axis #3 #4 ball moves from 0 #1 #2 #3 #4 to 2π radians around nit circle x-axis light source (a) What do you see on the y-axis? (b) What do you see on the x-axis? Figure 18.3: Analyzing the values of the sine and cosine functions. The conclusion is that after one complete counterclockwise rotation, the values of sin(θ) and cos(θ) range over the interval [−1, 1]. As the ball moves through the four quadrants, we have indicated the “order” in which these function values are assumed by labeling arrows #1 — #4: For example, for the sine function, look at Figure 18.3(a). The values of the sine function vary from 0 up to 1 while the ball moves through the first quadrant (arrow labeled #1), then from 1 down to 0 (arrow labeled #2), then from 0 down to −1 (arrow labeled #3), then from −1 up to 0 (arrow labeled #4). What about the tangent function? We have seen that the tangent function computes the slope of the hypotenuse of an inscribed triangle. This means we can determine the range of values of tan(θ) by investigating the possible slopes for these inscribed triangles. We will maintain the above model of a ball moving around the unit circle. We look at two cases, each starting at (1, 0). In the first quadrant, the ball moves counterclockwise and in the fourth quadrant it moves clockwise: In the first quadrant, we notice that these hypotenuse slopes are always non-negative, beginning with slope 0 (the degenerate right triangle when θ = 0) then increasing. In fact, as the angle θ approaches π/2 radians, the ball is getting closer to the position (0, 1) and the hypotenuse 18.1. EASY PROPERTIES OF CIRCULAR FUNCTIONS y-axis y-axis 8 7 239 6 5 4 3 2 1 x-axis 1 x-axis 2 3 4 5 6 7 (a) What happens to the slopes of these triangles? (b) What happens to the slopes of these triangles? Figure 18.4: Analyzing the values of the tangent function. is approaching a vertical line. This tells us that as θ varies from 0 to π/2 (but not equal to π/2), these slopes attain all possible non-negative values. In other words, the range of values for tan(θ) on the domain 0 ≤ θ < π/2 will be 0 ≤ z < ∞. Similar reasoning shows that as the ball moves in the fourth quadrant, the slopes of the hypotenuses of the triangles are always non-positive, varying from 0 to ANY negative value. In other words, the range of values for tan(θ) on the domain −π/2 < θ ≤ 0 will be −∞ < z ≤ 0. On your calculator, you can verify the visual conclusions we just established by studying the values of tan(θ) for θ close (but not equal) to π 2 radians = 90◦ : tan(89◦ ) = 57.29 tan(89.9◦) = 572.96 tan(89.99◦) = 5729.58 . . . tan(−89◦ ) = −57.29 tan(−89.9◦) = −572.96 tan(−89.99◦) = −5729.58 . . . The fact that the values of the tangent function become arbitrarily large as we get close to ±π/2 radians means the function output values are unbounded. Important Fact 18.1.1 (Circular function values). For any angle θ, we always have −1 ≤ cos(θ) ≤ 1 and −1 ≤ sin(θ) ≤ 1. On domain 0 ≤ θ ≤ 2π, the range of both cos(θ) and sin(θ) is −1 ≤ z ≤ 1. In contrast, on the domain of all θ values for which tangent is defined, the range of tan(θ) is all real numbers. For the sine and cosine functions, if the domain is not 0 ≤ θ ≤ 2π, then we need to consider the “periodic qualities” of the circular functions to determine the range. This is discussed below. 240 CHAPTER 18. TRIGONOMETRIC FUNCTIONS 18.2 Identities There are dozens of formulas that relate the values of two or more circular functions; these are usually lumped under the heading of Trigonometric Identities. In this course, we only need a couple frequently used identities. If we take the point P = (cos(θ), sin(θ)) on the unit circle, corresponding to the standard central position angle θ, then recall the equation of the unit circle tells us x2 + y2 = 1. But, since the x coordinate is cos(θ) and the y coordinate is sin(θ), we have (cos(θ))2 + (sin(θ))2 = 1. It is common notational practice to write (cos(θ))2 = cos2 (θ) and (sin(θ))2 = sin2 (θ). This leads to the most important of all trigonometric identities: Important Fact 18.2.1 (Trigonometric identity). For any angle θ, we have the identity cos2 (θ) + sin2 (θ) = 1. Adding any multiple of 2π radians (or 360◦ ) to an angle will not change the values of the circular functions. If we focus on radians for a moment, this says that knowing the values of cos(θ) and sin(θ) on the domain 0 ≤ θ ≤ 2π determines the values for any other possible angle. There is something very general going on here, so let’s pause a moment to make a definition and then an observation. Definition 18.2.2 (Periodic function). For c > 0, a function f(θ) is called c-periodic if two things are true: (i) f(θ + c) = f(θ) holds for all θ; (ii) There is no smaller d, 0 < d < c, such that f(θ + d) = f(θ) holds for all θ. We usually call c the period of the function. Using this new terminology, we conclude that the sine and cosine circular functions are 2π-periodic. In the case of the tangent circular function, it is also true that tan(θ) = tan(θ + 2πn), for every integer n. However, referring back to the unit circle definitions of the circular functions, we have tan(θ) = tan(θ + nπ), for all integers n. If you take n = 1, then this tells us that the tangent circular function is π-periodic. We summarize this information below. Important Fact 18.2.3 (Periodicity identity). For any angle θ and any integer n = 0, ±1, ±2, ±3, . . . , we have cos(θ) = cos(θ + 2πn), sin(θ) = sin(θ + 2πn), and tan(θ) = tan(θ + nπ). 18.2. IDENTITIES Next, we draw an angle θ and its negative in the same unit circle picture in standard central position. We have indicated the points Pθ and P−θ used to define the circular functions. It is clear from the picture in Figure 18.5 that Pθ and P−θ have the same x-coordinate, but the ycoordinates are negatives of one another. This gives the next identity: 241 unit circle (cos(θ), sin(θ)) = Pθ θ O R −θ (cos(−θ), sin(−θ)) = P−θ Figure 18.5: Visualizing a trigonometric identity. Important Fact 18.2.4 (Even/Odd identity). For any angle θ, sin(−θ) = − sin(θ), and cos(−θ) = cos(θ). We can use the terminology of even and odd functions here. In this language, this result says that the cosine function is an even function and the sine function is an odd function. Next, draw the angles θ and θ + π in the same unit circle picture in standard central position. We have indicated the corresponding points Pθ and Pθ+π on the unit circle and their coordinates in terms of the circular functions: From the picture in Figure 18.6, the x-coordinate of Pθ must be the “negative” of the x-coordinate of Pθ+π and similarly, the y-coordinate of Pθ must be the “negative” of the y-coordinate of Pθ+π . This gives us the next identity: Pθ = (cos(θ), sin(θ)) θ R θ+π unit circle Figure Pθ+π = (cos(θ + π), (sin(θ + π)) 18.6: Visualizing Fact 18.2.5. Important Fact 18.2.5 (Plus π identity). For any angle θ, we have sin(θ + π) = − sin(θ), and cos(θ + π) = − cos(θ). Important Fact 18.2.6. For any angle θ, we have sin(π − θ) = sin(θ) and cos(π − θ) = − cos(θ). π 1 For example, we have sin 56 = sin π = 2 . This calculation leads to a 6 computational observation: Combining Table 17.1 with the previous two identities we can compute the EXACT value of cos(θ), sin(θ), and tan(θ) at an angle θ which is a multiple of 30◦ = π radians or 45◦ = π radians. 6 4 Here are some sample calculations together with a reference as to “why” each equality is valid: CHAPTER 18. TRIGONOMETRIC FUNCTIONS 242 Example 18.2.7. (i) (ii) cos(−45◦ ) = cos(45◦ ) √ 2 = 2 Fact 18.2.4 on page 241 Table 17.1 on page 223 sin(225◦ ) = sin(45◦ + 180◦ ) Fact 18.2.3 on page 240 = − sin(45◦ ) Fact 18.2.5 on page 241 √ 2 Table 17.1 on page 223 =− 2 (iii) cos 2π 3 π +π 3 π = − cos − 3 π = − cos 3 1 =− 2 = cos − Fact 18.2.3 on page 240 Fact 18.2.5 on page 241 Fact 18.2.4 on page 241 Table 17.1 on page 223 18.3 Graphs of Circular Functions We have introduced three new functions of the variable θ and it is important to understand and interpret the pictures of their graphs. To do this, we need to settle on a coordinate system in which to work. The horizontal axis will correspond to the independent variable, so this should be the θaxis. We will label the vertical axis, which corresponds to the dependent variable, the z-axis. With these conventions, beginning with any of the circular functions z = sin(θ), z = cos(θ), or z = tan(θ), the graph will be a subset of the θz-coordinate system. Precisely, given a circular function z = f(θ), the graph consists of all pairs (θ, f(θ)), where θ varies over a domain of allowed values. We will record and discuss these graphs below; a graphing device will painlessly produce these for us! There is a point of possible confusion that needs attention. We purposely did not use the letter “y” for the dependent variable of the circular functions. This is to avoid possible confusion with our construction of the sine and cosine functions using the unit circle. Since we viewed the unit circle inside the xy-coordinate system, the x-coordinates (resp. ycoordinates) of points on the unit circle are computed by cos(θ) (resp. sin(θ)). 18.3. GRAPHS OF CIRCULAR FUNCTIONS z 243 y P = (cos(θ), sin(θ)) θ x θ Coordinate system used to GRAPH the circular functions. Coordinate system used to DEFINE the circular functions. Figure 18.7: The zθ versus xy coordinate systems. 18.3.1 A matter of scaling The first issue concerns scaling of the axes used in graphing the circular functions. As we know, the definition of radian measure is directly tied to the lengths of arcs subtended by angles in the unit circle: Important Fact 18.3.1. An angle of measure 1 radian inside the unit circle will subtend an arc of length 1. Since length is a good intuitive scaling quantity, it is natural to scale the θ-axis so that the length of 1 radian on the θ-axis (horizontal axis) is the same length as 1 unit on the vertical axis. For this reason, we will work primarily with radian measure when sketching the graphs of circular functions. If we need to work explicitly with degree measure for angles, then we can always convert radians to degrees using the fact: 360◦ = 2π radians. 18.3.2 The sine and cosine graphs Using Fact 18.1.1, we know that −1 ≤ sin(θ) ≤ 1 and −1 ≤ cos(θ) ≤ 1. Pictorially, this tells us that the graphs of z = sin(θ) and z = cos(θ) lie between the horizontal lines z = 1 and z = −1; i.e. the graphs lie inside the darkened band pictured in Figure 18.8. By Fact 18.2.3, we know that the values of the sine and cosine repeat themselves every 2π radians. Consequently, if we know the graphs of the sine and cosine on the domain 0 ≤ θ ≤ 2π, then the picture will repeat for the interval 2π ≤ θ ≤ 4π, −2π ≤ θ ≤ 0, etc. CHAPTER 18. TRIGONOMETRIC FUNCTIONS 244 z-axis z=1 θ-axis z = −1 Figure 18.8: Visualizing the range of sin(θ) and cos(θ). y=1 Repeat Repeat Repeat 0 −2π −4π z-axis Repeat 2π θ-axis 4π 6π y = −1 picture in here repeats each 2π units Figure 18.9: On what intervals will the graph repeat? Sketching the graph of z = sin (θ) for 0 ≤ θ ≤ 2π can be roughly √ √ 1 achieved by plotting points. For example, π , 2 , π , 22 , π , 23 , and 6 4 3 √ π ,1 2 √ π π π lie on the graph, as do 32 , −1 , 53 , − 23 , 74 , − 22 , 11π , − 1 , and 6 2 (2π, 0), etc. If we return to our analysis of the range of values for the sine function in Figure 18.2, it is easy to see where sin(θ) is positive or negative; combined with Chapter 4, this tells us where the graph is above and below the horizontal axis (see Figure 18.10). z-axis θ-axis −2π −π π 2π 3π Figure 18.10: Where is the graph positive or negative? We now include a software plot of the graph of sine function, observing the three qualitative features just isolated: bounding, periodicity and sign properties (see Figure 18.11). 18.3. GRAPHS OF CIRCULAR FUNCTIONS 245 z-axis 1 θ-axis −2π −π π 2π 3π −1 one period Figure 18.11: The graph of z = sin(θ). We could repeat this analysis to arrive at the graph of the cosine. Instead, we will utilize an identity. Given an angle θ, place it in central standard position in the unit circle, as one of the four cases of Figure 17.16. For example, we have pictured Case I in this figure. Since the sum of the angles in a triangle is 180◦ = π radians, we know that θ, π/2, and π − θ are the three angles of the 2 inscribed right triangle. From the picture in Figure 18.12, it then follows that π 2 θ unit circle π 2 Figure 18.12: Visualizing the conversion identity. side adjacent to θ hypotenuse side opposite to π − θ 2 = hypotenuse π −θ . = sin 2 cos(θ) = Using the same reasoning this identity is valid for all θ. This gives us another useful identity: Important Fact 18.3.2 (Conversion identity). For any angle θ, cos(θ) = sin π − θ , and sin(θ) = cos π − θ . 2 2 This identity can be used to sketch the graph of the cosine function. First, we do a calculation using our new identity: cos(θ) = cos(−θ) π − (−θ) = sin 2 π = sin θ − − 2 −θ Fact 18.2.4 on page 241 Fact 18.3.2 (above) Since: (a + b) = (a − (−b)) = (b − (−a)) By the horizontal shifting principle in Fact 13.3.1 on page 170, the graph of z = cos(θ) is obtained by horizontally shifting the graph of z = sin(θ) by π units to the left. Here is a plot of the graph of the cosine function: 2 See Figure 18.13. CHAPTER 18. TRIGONOMETRIC FUNCTIONS 246 z-axis 1 θ-axis −2π −π π 2π 3π −1 one period Figure 18.13: The graph of z = cos(θ). 18.3.3 The tangent graph θ = −π 2 graph heads this way, getting close to vertical line as θ gets close to π 2 −π 2 π 2 z-axis θ-axis graph heads this way, getting close to vertical line as θ gets close to − π 2 θ= π 2 Figure 18.14: The behavior of tan(θ) as θ approaches asymptotes. As we have already seen, unlike the sine and cosine circular functions, the tangent function is NOT defined for all sin values of θ. Since tan(θ) = cos(θ) , here are some properties (θ) we can immediately deduce: • The function z = tan(θ) is undefined if and only if θ = π + kπ, where k = 0, ±1, ±2, ±3, · · · . 2 • The function z = tan(θ) = 0 if and only if θ = kπ, where k = 0, ±1, ±2, ±3, · · · . • By Fact 18.2.3, the tangent function is π-periodic, so the picture of the graph will repeat itself every πunits and it is enough to understand the graph when −π < θ < π. 2 2 • On the domain 0 ≤ θ < π , tan(θ) ≥ 0; on the domain 2 − π < θ ≤ 0, tan(θ) ≤ 0. 2 In the θz-coordinate system, the vertical lines θ = π + kπ, where 2 k = 0, ±1, ±2, · · · will be vertical asymptotes for the graph of the tangent function. Using our slope interpretation in Figure 18.4, what becomes clear is this: As the values of θ get close to π , the graph is getting close 2 to the vertical line θ = π AND becoming farther and farther away from 2 the horizontal axis: To understand this numerically, first suppose θ is slightly smaller than π , say θ = π − 0.1 , π − 0.01 , and π − 0.001 . Then 2 2 2 2 the calculation of tan(θ) involves dividing a number very close to 1 by a very small positive number: π − 0.1 2 π − 0.01 tan 2 π tan − 0.001 2 tan = 9.9666, = 99.9967, and = 1000. 18.4. TRIGONOMETRIC FUNCTIONS Conclude that as θ “approaches π from below”, the val2 ues of tan(θ) are becoming larger and larger. This says that the function values become “unbounded”. Likewise, imagine the case when θ is slightly bigger than − π , say 2 θ = − π + 0.1 , − π + 0.01 , and − π + 0.001 . Then the cal2 2 2 culation of tan(θ) involves dividing a number very close to −1 by a very small positive number: π tan − + 0.1 2 π tan − + 0.01 2 π tan − + 0.001 2 247 θ = − π z-axis 2 one period θ = 5π 2 etc. etc. θ-axis −2π −π π θ = − 3π 2 θ= π 2 2π 3π θ = 3π 2 Figure 18.15: The graph of z = tan(θ). = −9.9666, = −99.9967, and = −1000. Conclude that as θ “approaches − π from above”, the values of tan(θ) are 2 becoming negative numbers of increasingly larger magnitude: Again, this tells us the function values are becoming “unbounded”. The graph of z = tan(θ) for − π < θ < π can be roughly achieved by com2 2 bining the calculations as in Example 18.2.7 and the qualitative features highlighted. Figure 18.15 shows a software plot. 18.4 Trigonometric Functions To become successful mathematical modelers, we must have wide variety of functions in our toolkit. As an illustration, the graph below might represent the height of the tide above some reference level over the course of several days. The curve drawn is clearly illustrating that the height of the tide is “periodic” as a function of time t; in other words, the behavior of the tide repeats itself as time goes by. However, if we try to model this periodic behavior, the only weapon at our disposal would be the circular functions and these require an angle variable, not a time variable such as t; we are stuck! Modeling the tide graph requires the trigonometric functions, which lie at the heart of studying all kinds of periodic behavior. We have no desire to “reinvent the wheel”, so let’s use our previous work on the circular functions to define the trigonometric functions. 18.4.1 A Transition Given a real number t, is there a sensible way to define cos(t) and sin(t)? The answer is yes and depends on the ideas surrounding radian measure of angles. Given the positive real number t, we can certainly imagine an angle of measure t radians inside the unit circle (in standard position) and we know the arc subtended by this angle has length t (this is why we use the unit circle). CHAPTER 18. TRIGONOMETRIC FUNCTIONS 248 feet +20 +15 +10 +5 0 t (time) −5 −10 −15 −20 Figure 18.16: A periodic function with input variable t. P (t) = (x,y) We already know that cos(t radians) and sin(t radians) compute the x and y coordinates of the point P(t). In effect, we are just using the measure of the angle t to help us locate the point P(t). An alternate way to locate P(t) is to move along the circumference counterclockwise, beginning at (1, 0), until we have an arc of length t; that again puts us at the point P(t). In the case of an angle of measure −t radians, the point P(−t) can be located by moving along the circumference clockwise, beginning at (1, 0), until we have an arc of length t. arc length = t t rads (1,0) unit circle Figure 18.17: The circular functions with input variable t. Definition 18.4.1 (Trigonometric functions). Let t be a real number. We DEFINE the sine function y = sin(t), the cosine function y = cos(t) and the tangent function y = tan(t) by the rules def sin(t) = y-coordinate of P(t) = sin(t radians) def cos(t) = x-coordinate of P(t) = cos(t radians) def sin(t) tan(t) = = tan(t radians) cos(t) We refer to these as the basic trigonometric functions. If we are working with radian measure and t is a real number, then there is no difference between evaluating a trigonometric function at the real number t and evaluating the corresponding circular function at the angle of measure t radians. Example 18.4.2. Assume that the number of hours of daylight in Seattle 2π during 1994 is given by the function d(t) = 3.7sin 366 (t − 80.5) + 12, where 18.4. TRIGONOMETRIC FUNCTIONS 249 t represents the day of the year and t = 0 corresponds to January 1. How many hours of daylight will there be on May 11? Solution. To solve the problem, you need to consult a calendar, finding every month has 31 days, except: February has 28 days and April, June, September and November have 30 days. May 11 is the 31+28+31+30+11 = 131st day of the year. So, there will be d(131) = 3.7 sin(2(50.5)π/366) + 12 = 14.82 hours of daylight on May 11. 18.4.2 Graphs of trigonometric functions The graphs of the trigonometric functions y = sin(t), y = cos(t), and y = tan(t) will look just like Figures 18.11, 18.13, and 18.15, except that the horizontal axis becomes the t-axis and the vertical axis becomes the y-axis. 18.4.3 Notation for trigonometric functions In many texts, you will find the sine function written as y = sin t; i.e. the parenthesis around the “t” are omitted. A similar comment applies to all of the trigonometric functions. We will never do this and the reasoning is simply this: Maintaining the parenthesis, as in y = sin(t), emphasizes the fact that we are dealing with a function and the “input values” are located between the parenthesis. For example, if we write the function y = sin(t2 + 2t + 1), it is crystal clear that the sine function is applied to the expression “t2 + 2t + 1”; using the alternate notation yields the expression y = sin t2 + 2t + 1, which is interpreted to mean y = (sin t2 ) + (2t + 1). !!! As a rule, whenever you see an expression involving sin(· · · ), cos(· · · ), or tan(· · · ), we assume “ · · · ” is in units of RADIANS, unless otherwise noted. When computing values on your calculator, MAKE SURE YOU ARE USING RADIAN MODE! CAUTION !!! CHAPTER 18. TRIGONOMETRIC FUNCTIONS 250 18.5 Exercises Problem 18.1. Work the following problems without using ANY calculators. (a) Sketch y = sin(x). (b) If sin(θ) = −0.8 and θ is in the third quadrant of the xy plane, what is cos(θ)? (c) If sin(θ) = 3 , what is sin( π − θ)? 7 2 (b) Sketch y = sin2 (x). (c) Sketch y = 1 . 1 + sin2 (x) Problem 18.4. These graphs represent periodic functions. Describe the period in each case. Problem 18.2. Sketch the graphs of these functions: (a) f(t) = | sin(t)|. (b) f(t) = | cos(t)|. (c) f(t) = | tan(t)|. Problem 18.3. Solve the following: 24 (a) If cos(θ) = 25 , what are the two possible values of sin(θ)? Problem 18.5. Start with the equation sin(θ) = cos(θ). Use the unit circle interpretation of the circular functions to find the solutions of this equation; make sure to describe your reasoning. Chapter 19 Sinusoidal Functions depth 5 A migrating salmon is heading up a portion of the 4 Columbia River. It’s depth d(t) (in feet) below the water 3 surface is measured and plotted over a 30 minute period, 2 as a function of time t (minutes). What is the formula for 1 d(t)? 5 10 15 20 25 30 time In order to answer the question, we need to introduce Figure 19.1: The depth of a an important new family of functions called the sinusoidal salmon as a function of time. functions. These functions will play a central role in modeling any kind of periodic phenomena. The amazing fact is that almost any function you will encounter can be approximated by a sum of sinusoidal functions; a result that has far-reaching implications in all of our lives. 19.1 A special class of functions Beginning with the trigonometric function y = sin(x), what is the most general function we can build using the graphical techniques of shifting and stretching? y-axis 1 y = sin(x) horizontally shift vertically x-axis horizontally dilate −π π 2 −π 2 π vertically dilate shift −1 Figure 19.2: Visualizing the geometric operations available for curve sketching. The graph of y = sin(x) can be manipulated in four basic ways: horizontally shift, vertically shift, horizontally dilate or vertically dilate. Each of these “geometric operations” corresponds to a simple change in the “symbolic formula” for the function, as discussed in Chapter 13. 251 CHAPTER 19. SINUSOIDAL FUNCTIONS 252 If we vertically shift the graph by D units upward, the resulting curve would be the graph of the function y = sin(x)+D; see Facts 13.3.1. Recall, the effect of the sign of D: If D is negative, the effect of shifting D units upward is the same as shifting |D| units downward. Notice, the function y = sin(x) + D is still a periodic function, having the same period 2π as y = sin(x). Notice, whereas the graph of the function y = sin(x) oscillates between the horizontal lines y = ±1, the graph of y = sin(x) + D oscillates between y = D ± 1. For this reason, we sometimes refer to the constant D as the mean of the function y = sin(x) + D. In Figure 19.3, notice that the graph of y = sin(x) + D is symmetrically split by the horizontal “mean” line y = D. y-axis y = sin(x) + D y-axis y = sin(x) y=D D x-axis x-axis shift D units Figure 19.3: Interpreting the mean. Next, consider the effect of horizontally shifting the graph of y = sin(x) by C units to the right. By Facts 13.3.1, the new curve is the graph of the function y = sin(x − C). Also, recall the effect of the sign of C: If C is negative, the effect of shifting C units right is the same as shifting |C| units left. If the domain of sin(x) is 0 ≤ x ≤ 2π, then the domain of sin(x − C) is 0 ≤ x − C ≤ 2π, again by Facts 13.3.1. Rewriting this, the domain of sin(x − C) is C ≤ x ≤ 2π + C and the graph will go through precisely one period on this domain. In other words, the new function sin(x − C) is still 2π-periodic. The constant C is usually called the phase shift of y = sin(x − C). Looking at Figure 19.4, it is possible to interpret C graphically: C will be a point where the graph crosses the horizontal axis on its way up from a minimum to a maximum. y-axis y-axis C y = sin(x − C) y = sin(x) x-axis x-axis shift C units Figure 19.4: Interpreting the phase shift. Vertically dilating the graph, either by vertical expansion or compression, leads to a new curve. The graph of this vertically dilated curve is y = A sin(x), for some positive constant A. Furthermore, if A > 1, 19.1. A SPECIAL CLASS OF FUNCTIONS 253 the graph of y = A sin(x) is a vertically expanded version of y = sin(x), whereas, if 0 < A < 1, then the graph of y = A sin(x) is a vertically compressed version of y = sin(x). Notice, the function y = A sin(x) is still 2πperiodic. What has changed is the band of oscillation: whereas the graph of the function y = sin(x) stays between the horizontal lines y = ±1, the graph of y = A sin(x) oscillates between the horizontal lines y = ±A. We usually refer to A as the amplitude of the function y = A sin(x). y = A sin(x) y-axis y-axis y = sin(x) A x-axis x-axis stretch A units Figure 19.5: Interpreting the amplitude. Finally, horizontally dilating the graph, either by horizontal expansion or compression, leads to a new curve. The equation of this horizontally dilated curve is y = sin(cx), for some constant c > 0. We know that y = sin(x) is a 2π-periodic function and observe that horizontally dilation still results in a periodic function, but the period will typically NOT be 2π. For future purposes, it is useful to rewrite the equation for the horizontally stretched curve in a way more directly highlighting the period. To begin with, once the horizontal stretching factor c is known, we could rewrite c= 2π , B for some B = 0. y-axis y-axis y = sin(x) y = sin(( 2π )x) B x-axis x-axis stretch Figure 19.6: Interpreting the period. Here is the point of this yoga with the horizontal dilating constant: If we let the values of x range over the interval [0, B], then 2π x will range over B π the interval [0,2π]. In other words, the function y = sin 2B x is B-periodic 2π and we can read off the period of y = sin B x by viewing the constant in this mysterious way. The four constructions outlined lead to a new family of functions. CHAPTER 19. SINUSOIDAL FUNCTIONS 254 Definition 19.1.1 (The Sinusoidal Function). Let A, B, C and D be fixed constants, where A and B are both positive. Then we can form the new function y = A sin 2π (x − C) + D, B which is called a sinusoidal function. The four constants can be interpreted graphically as indicated: B y-axis y-axis A D x-axis x-axis all four y = sin(x) C operations y = A sin(( 2π )(x − c)) + D B Figure 19.7: Putting it all together for the sinusoidal function. 19.1.1 How to roughly sketch a sinusoidal graph Important Procedure 19.1.2. Given a sinusoidal function in the standard form y = A sin 2π (x − C) + D, B once the constants A, B, C, and D are specified, any graphing device can produce an accurate graph. However, it is pretty straightforward to sketch a rough graph by hand and the process will help reinforce the graphical meaning of the constants A, B, C, and D. Here is a “five step procedure” one can follow, assuming we are given A, B, C, and D. It is a good idea to follow Example 19.1.3 as you read this procedure; that way it will seem a lot less abstract. 1. Draw the horizontal line given by the equation y = D; this line will π “split” the graph of y = A sin 2B (x − C) + D into symmetrical upper and lower halves. 2. Draw the two horizontal lines given by the equations y = D ± A. These two lines determine a horizontal strip inside which the graph of the 19.1. A SPECIAL CLASS OF FUNCTIONS 255 sinusoidal function will oscillate. Notice, the points where the sinusoidal function has a maximum value lie on the line y = D + A. Likewise, the points where the sinusoidal function has a minimum value lie on the line y = D − A. Of course, we do not yet have a prescription that tells us where these maxima (peaks) and minima (valleys) are located; that will come out of the next steps. 3. Since we are given the period B, we know these important facts: (1) The period B is the horizontal distance between two successive maxima (peaks) in the graph. Likewise, the period B is the horizontal distance between two successive minima (valleys) in the graph. (2) The horizontal distance between a maxima (peak) and the successive minima (valley) is 1 B. 2 4. Plot the point (C, D). This will be a place where the graph of the sinusoidal function will cross the mean line y = D on its way up from a minima to a maxima. This is not the only place where the graph crosses the mean line; it will also cross at the points obtained 1 from (C, D) by horizontally shifting by any integer multiple of 2 B. For example, here are three places the graph crosses the mean line: 1 (C, D), (C + 2 B, D), (C + B, D) 1 5. Finally, midway between (C, D) and (C+ 2 B, D) there will be a maxima 1 (peak); i.e. at the point (C + 4 B, D + A). Likewise, midway between (C + 1 B, D) and (C + B, D) there will be a minima (valley); i.e. at the 2 3 point (C + 4 B,D − A). It is now possible to roughly sketch the graph on the domain C ≤ x ≤ C + B by connecting the points described. Once this portion of the graph is known, the fact that the function is periodic tells us to simply repeat the picture in the intervals C + B ≤ x ≤ C + 2B, C − B ≤ x ≤ C, etc. To make sense of this procedure, let’s do an explicit example to see how these five steps produce a rough sketch. CHAPTER 19. SINUSOIDAL FUNCTIONS 256 eplacements Example 19.1.3. The temperature (in ◦ C) of Adri-N’s dorm room varies π during the day according to the sinusoidal function d(t) = 6 sin 12 (t − 11) + 19, where t represents hours after midnight. Roughly sketch the graph of d(t) over a 24 hour period.. What is the temperature of the room at 2:00 pm? What is the maximum and minimum temperature of the room? Solution. We begin with the rough sketch. Start by taking an inventory of the constants in this sinusoidal function: d(t) = 6 sin π 2π (t − 11) + 19 = A sin (t − C) + D. 12 B Conclude that A = 6, B = 24, C = 11, D = 19. Following the first four steps of the procedure outlined, we can sketch the lines y = D = 19, y = D ± A = 19 ± 6 and three points where the graph crosses the mean line (see Figure19.8). 30 d(t) graph will oscillate inside this strip 25 y = 25 20 y = 19 (11, 19) (23, 19) (35, 19) 15 y = 13 10 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.8: Sketching the mean D and amplitude A. According to the fifth step in the sketching procedure, we can plot the 3 maxima (C + 1 B, D + A) = (17, 25) and the minima (C + 4 B, D − A) = (29, 13). 4 We then “connect the dots” to get a rough sketch on the domain 11 ≤ t ≤ 35. 30 d(t) graph will oscillate inside this maxima (17, 25) 25 y = 25 (23, 19) 20 y = 19 (11, 19) strip (35, 19) 15 y = 13 (29, 13) minima 10 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.9: Visualizing the maximum and minimum over one period. 19.1. A SPECIAL CLASS OF FUNCTIONS 257 Finally, we can use the fact the function has period 24 to sketch the graph to the right and left by simply repeating the picture every 24 horizontal units. y-axis 30 maxima (−7,25) maxima (17,25) maxima (41,25) y = 25 25 (11,19) 20 (−13,19) (23,19) (35,19) y = 19 (−1,19) 15 (47,19) (59,19) y = 13 (5,13) minima 10 (29,13) minima (53,13) minima t-axis −12 −8 −4 −14 −10 −6 −2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.10: Repeat sketch for every full period. We restrict the picture to the domain 0 ≤ t ≤ 24 and obtain the computer generated graph pictured in Figure 19.11; as you can see, our rough graph is very accurate. The temperature at 2:00 p.m. is just d(14) = 23.24◦ C. From the graph, the maximum value of the function will be D + A = 25◦ C and the minimum value will be D − A = 13◦ C. temp (C) 25 20 15 10 5 10 15 20 t (hours) Figure 19.11: The computer generated solution. 19.1.2 Functions not in standard sinusoidal form Any time we are given a trigonometric function written in the standard form y = A sin 2π (x − C) + D, B for constants A, B, C, and D (with A and B positive), the summary in Definition 19.1.1 tells us everything we could possibly want to know CHAPTER 19. SINUSOIDAL FUNCTIONS 258 about the graph. But, there are two ways in which we might encounter a trigonometric type function that is not in this standard form: • The constants A or B might be negative. For example, y = −2 sin(2x− 7) − 3 and y = 3 sin − 1 x + 1 + 4 are examples that fail to be in 2 standard form. • We might use the cosine function in place of the sine function. For example, something like y = 2 cos(3x + 1) − 2 fails to be in standard sinusoidal form. Now what do we do? Does this mean we need to repeat the analysis that led to Definition 19.1.1? It turns out that if we use our trig identities just right, then we can move any such equation into standard form and read off the amplitude, period, phase shift and mean. In other words, equations that fail to be in standard sinusoidal form for either of these two reasons will still define sinusoidal functions. We illustrate how this is done by way of some examples: Examples 19.1.4. (i) Start with y = −2 sin(2x − 7) − 3, then here are the steps with reference to the required identities to put the equation in standard form: y = −2 sin(2x − 7) − 3 = 2 (− sin(2x − 7)) − 3 = 2 sin(2x − 7 + π) + (−3) 2π 7−π = 2 sin x− π 2 Fact 18.2.5 on page 241 + (−3). This function is now in the standard form of Definition 19.1.1, so it is a sinusoidal function with phase shift C = 7−π = 1.93, mean D = −3, 2 amplitude A = 2 and period B = π. 1 (ii) Start with y = 3 sin(− 2 x + 1) + 4, then here are the steps with reference to the required identities to put the equation in standard form: 1 y = 3 sin − x + 1 + 4 2 1 x−1 +4 = 3 sin − 2 1 = 3 − sin x−1 +4 Fact 18.2.4 on page 241 2 1 x−1+π +4 Fact 18.2.5 on page 241 = 3 sin 2 2π (x − [2 − 2π]) + 4 = 3 sin 4π 19.2. EXAMPLES OF SINUSOIDAL BEHAVIOR 259 This function is now in the standard form of Definition 19.1.1, so it is a sinusoidal function with phase shift C = 2 − 2π, mean D = 4, amplitude A = 3 and period B = 4π. (iii) Start with y = 2 cos(3x + 1) − 2, then here are the steps to put the equation in standard form. A key simplifying step is to use the identity: cos(t) = sin( π + t). 2 y = 2 cos(3x + 1) − 2 π + 3x + 1 − 2 = 2 sin 2 π = 2 sin 3x − −1 − + (−2) 2 2π 1 π = 2 sin x− −1 − 2π 3 2 3 + (−2) This function is now in the standard form of Definition 19.1.1, so it is a sinusoidal function with phase shift C = 1 [−1 − π ], mean D = −2, 3 2 amplitude A = 2 and period B = 2π . 3 19.2 Examples of sinusoidal behavior Problems involving sinusoidal behavior come in two basic flavors. On the one hand, we could be handed an explicit sinusoidal function y = A sin 2π (x − C) + D B and asked various questions. The answers typically require either direct calculation or interpretation of the constants. Example 19.1.3 is typical of this kind of problem. On the other hand, we might be told a particular situation is described by a sinusoidal function and provided some data or a graph. In order to further analyze the problem, we need a “formula”, which means finding the constants A, B, C, and D. This is a typical scenario in a “mathematical modeling problem”: the process of observing data, THEN obtaining a mathematical formula. To find A, take half the difference between the largest and smallest values of f(x). The period B is most easily found by measuring the distance between two successive maxima (peaks) or minima (valleys) in the graph. The mean D is the average of the largest and smallest values of f(x). The phase shift C (which is usually the most tricky quantity to get your hands on) is found by locating a “reference point”. This “reference point” is a location where the graph crosses the mean line y = D on its way up from a minimum to a maximum. The funny thing is that the phase shift C is NOT unique; there are an infinite number of correct choices. One choice that will work CHAPTER 19. SINUSOIDAL FUNCTIONS 260 is C = (x-coordinate of a maximum) − B . Any other choice of C will differ 4 from this one by a multiple of the period B. max value − min value 2 B = distance between two successive peaks (or valleys) B C = x-coordinate of a maximum − 4 max value + min value D= . 2 A= d-axis (hours of daylight) Example 19.2.1. Assume that the number of hours of daylight in Seattle is given by a sinusoidal function d(t) of time. During 1994, assume the longest day of the year is June 21 with 15.7 hours of daylight and the shortest day is December 21 with 8.3 hours of daylight. Find a formula d(t) for the number of hours of daylight on the tth day of the year. 15 12.5 10 7.5 5 50 100 150 200 250 300 350 t-axis (days) Solution. Because the function d(t) is assumed to be siπ nusoidal, it has the form y = A sin 2B (t − C) + D, for constants A, B, C, and D. We simply need to use the given information to find these constants. The largest value of the function is 15.7 and the smallest value is 8.3. Knowing this, from the above discussion we can read off : Figure 19.12: Hours of daylight in Seattle in 1994. D= 15.7 + 8.3 = 12 2 A= 15.7 − 8.3 = 3.7. 2 To find the period, we need to compute the time between two successive maximum values of d(t). To find this, we can simply double the time length of one-half period, which would be the length of time between successive maximum and minimum values of d(t). This gives us the equation B = 2(days between June 21 and December 21) = 2(183) = 366. Locating the final constant C requires the most thought. Recall, the longest day of the year is June 21, which is day 172 of the year, so C = (day with max daylight) − B 366 = 172 − = 80.5. 4 4 In summary, this shows that d(t) = 3.7 sin 2π (t − 80.5) + 12. 366 A rough sketch, following the procedure outlined above, gives this graph on the domain 0 ≤ t ≤ 366; we have included the mean line y = 12 for reference. 19.2. EXAMPLES OF SINUSOIDAL BEHAVIOR 261 We close with the example that started this section. Example 19.2.2. The depth of a migrating salmon below the water surface changes according to a sinusoidal function of time. The fish varies between 1 and 5 feet below the surface of the water. It takes the fish 1.571 minutes to move from its minimum depth to its successive maximum depth. It is located at a maximum depth when t = 4.285 minutes. What is the formula for the function d(t) that predicts the depth of the fish after t minutes? What was the depth of the salmon when it was first spotted? During the first 10 minutes, how many times will the salmon be exactly 4 feet below the surface of the water? max depth − min depth 5−1 A= = =2 2 2 5+1 max depth + min depth = = 3. D= 2 2 5 depth Solution. We know that d(t) = A sin( 2π (x − C)) + D, for apB propriate constants A, B, C, and D. We need to use the given information to extract these four constants. The amplitude and mean are easily found using the above formulas: 4 3 2 1 2 4 B = 2(1.571) = 3.142 Finally, to find C we B 3.142 = 4.285 − = 3.50. 4 4 The formula is now d(t) = 2 sin 8 10 Figure 19.13: Depth of a migrating salmon. The period can be found by noting that the information about the time between a successive minimum and maximum depth will be half of a period (look at the picture in Figure 19.13): C = (time of maximum depth) − 6 time 2π (t − 3.5) + 3 = 2 sin(2t − 7) + 3 3.142 The depth of the salmon when it was first spotted is just d(0) = 2 sin(−7) + 3 = 1.686 feet. Finally, graphically, the last question amounts to determining how many times the graph of d(t) crosses the line y = 4 on the domain [0,10]. This can be done using Figure 19.13. A simultaneous picture of the two graphs is given, from which we can see the salmon is exactly 4 feet below the surface of the water six times during the first 10 minutes. CHAPTER 19. SINUSOIDAL FUNCTIONS 262 19.3 Summary • A sinusoidal function is one of the form f(t) = A sin 2π (t − C) + D B where A, B, C, and D are constants. – A is the amplitude of the function; this is half the vertical distance between a high point and a low point on its graph. – B is the period of the function; this is the horizontal distance between two consecutive high points (or low points) on its graph. – C is the phase shift of the function; it is multi-valued, but one choice for C is a value of t at which the function is increasing and equal to D. – D is the mean value of the function; it is the y-value of the horizontal line about which the graph of the function is balanced. • The graph of a sinusoidal function is a shifted, scaled version of the graph of y = sin t. 19.4. EXERCISES 263 19.4 Exercises Problem 19.1. Find the amplitude, period, a phase shift and the mean of the following sinusoidal functions. (a) y = sin(2x − π) + 1 (b) y = 6 sin(πx) − 1 (c) y = 3 sin(x + 2.7) + 5.2 (d) y = 5.6 sin (e) y = 2.1 sin 2 3x x π − 7 − 12.1 + 44.3 − 9.8 (f) y = 3.9 (sin(22.34(x + 18)) − 11) (g) y = 11.2 sin 5 π (x − 9.2) + 8.3 Problem 19.2. A weight is attached to a spring suspended from a beam. At time t = 0, it is pulled down to a point 10 cm above the ground and released. After that, it bounces up and down between its minimum height of 10 cm and a maximum height of 26 cm, and its height h(t) is a sinusoidal function of time t. It first reaches a maximum height 0.6 seconds after starting. (a) Follow the procedure outlined in this section to sketch a rough graph of h(t). Draw at least two complete cycles of the oscillation, indicating where the maxima and minima occur. (b) What are the mean, amplitude, phase shift and period for this function? (c) Give four different possible values for the phase shift. machine begins recording a plot of volume per breath versus time (in seconds). Let b(t) be a function of time t that tells us the volume (in liters) of a breath that starts at time t. During the test, the smallest volume per breath is 0.6 liters and this first occurs for a breath that starts 5 seconds into the test. The largest volume per breath is 1.8 liters and this first occurs for a breath beginning 55 seconds into the test. (a) Find a formula for the function b(t) whose graph will model the test data for this patient. (b) If the patient begins a breath every 5 seconds, what are the breath volumes during the first minute of the test? Problem 19.4. Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. at which time the water is 10 feet above the height of low tide. Low tides occur 6 hours after high tides. Suppose there are two high tides and two low tides every day and the height of the tide varies sinusoidally. (a) Find a formula for the function y = h(t) that computes the height of the tide above low tide at time t. (In other words, y = 0 corresponds to low tide.) (b) What is the tide height at 11:00 a.m.? Problem 19.5. Your seat on a Ferris Wheel is at the indicated position at time t = 0. (d) Write down a formula for the function h(t) in standard sinusoidal form; i.e. as in 19.1.1 on Page 254. (e) What is the height of the weight after 0.18 seconds? (f) During the first 10 seconds, how many times will the weight be exactly 22 cm above the floor? (Note: This problem does not require inverse trigonometry.) Problem 19.3. A respiratory ailment called “Cheyne-Stokes Respiration” causes the volume per breath to increase and decrease in a sinusoidal manner, as a function of time. For one particular patient with this condition, a Start 53 feet Let t be the number of seconds elapsed after the wheel begins rotating counterclockwise. You find it takes 3 seconds to reach the top, which is 53 feet above the ground. The wheel is rotating 12 RPM and the diameter of the wheel is 50 feet. Let d(t) be your height above the ground at time t. 264 CHAPTER 19. SINUSOIDAL FUNCTIONS ω=4π/9 rad/sec (a) Argue that d(t) is a sinusoidal function, describing the amplitude, phase shift, period and mean. bug lands here 1.2 rad 2 in (b) When are the first and second times you are exactly 28 feet above the ground? bug lands here (c) After 29 seconds, how many times will you have been exactly 28 feet above the ground? 2 in ω=4π/9rad/sec ω= 4π/9 rad/sec bug lands here 0.5 rad 2 in Problem 19.6. In Exercise 17.12, we studied the situation below: A bug has landed on the rim of a jelly jar and is moving around the rim. The location where the bug initially lands is described and its angular speed is given. Impose a coordinate system with the origin at the center of the circle of motion. In each of the cases, the earlier exercise found the coordinates P(t) of the bug at time t. For each of the scenarios below, answer these two questions: Problem 19.7. The voltage output(in volts) of an electrical circuit at time t seconds is given by the function V (t) = 23 sin(5πt−3π)+1 . (a) What is the initial voltage output of the circuit? (a) Both coordinates of P(t) = (x(t),y(t)) are sinusoidal functions in the variable t. Sketch a rough graph of the functions x(t) and y(t) on the domain 0 ≤ t ≤ 9. (b) Use the graph sketches to help you find the the amplitude, mean, period and phase shift for each function. Write x(t) and y(t) in standard sinusoidal form. (b) Is the voltage output of the circuit ever equal to zero? Explain. (c) The function V (t) = 2p(t) , where p(t) = 3 sin(5πt − 3π) + 1. Put the sinusoidal function p(t) in standard form and sketch the graph for 0 ≤ t ≤ 1. Label the coordinates of the extrema on the graph. (d) Calculate the maximum and minimum voltage output of the circuit. (e) During the first second, determine when the voltage output of the circuit is 10 volts. 19.4. EXERCISES 265 (f) A picture of the graph of y = V (t) on the domain 0 ≤ t ≤ 1 is given; label the coordinates of the extrema on the graph. free to move back and forth along the x-axis. The point A is at (2,0) at time t = 0, and the wheel rotates counterclockwise at 3 rev/sec. 2 A volts A -2 15 2 B B 4 6 8 2 4 6 8 -2 12.5 time t > 0 time t=0 10 (a) As the point A makes one complete revolution, indicate in the picture the direction and range of motion of the point B. 7.5 5 2.5 0.2 0.4 0.6 0.8 1 t axis (g) Restrict the function V (t) to the domain 0.1 ≤ t ≤ 0.3; explain why this function has an inverse and find the formula for the inverse rule. Restrict the function V (t) to the domain 0.3 ≤ t ≤ 0.5; explain why this function has an inverse and find the formula for the inverse rule. Problem 19.8. A six foot long rod is attached at one end A to a point on a wheel of radius 2 feet, centered at the origin. The other end B is (b) Find the coordinates of the point A as a function of time t. (c) Find the coordinates of the point B as a function of time t. (d) What is the x-coordinate of the point B when t = 1? You should be able to find this two ways: with your function from part (c), and using some common sense (where is point A after one second?). (e) Is the function you found in (c) a sinusoidal function? Explain. 266 CHAPTER 19. SINUSOIDAL FUNCTIONS Chapter 20 An aircraft is flying at an altitude 10 miles above the elevation of an airport. If the airplane begins a steady descent 100 miles from the airport, what is the angle θ of descent? The only natural circular function we can use is z = tan(θ), leading to the equation: tan(θ) = 10 miles Inverse Circular Functions θ 100 miles Figure 20.1: An aircraft decending toward an airport. 10 1 =. 100 10 The problem is that this equation does not tell us the value of θ. Moreover, none of the equation solving techniques at our disposal (which all amount to algebraic manipulations) will help us solve the equation for θ. What we need is an inverse function θ = f−1 (z); then we could use the fact that tan−1 (tan(θ)) = θ and obtain: θ = tan−1 (tan(θ)) = tan−1 1 10 . Computationally, without even thinking about what is going on, any scientific calculator will allow us to compute values of an inverse circular function and leads to a solution of our problem. In this example, you will 1 find θ = tan−1 10 = 5.71◦. Punch this into your calculator and verify it! 20.1 Solving Three Equations Example 20.1.1. Find all values of θ (an angle) that make this equation 1 true: sin(θ) = 2 . Solution. We begin with a graphical reinterpretation: the solutions correspond to the places where the graphs of z = sin(θ) and z = 1 intersect in 2 the θz-coordinate system. Recalling Figure 18.11, we can picture these two graphs simultaneously as below: 267 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 268 cross cross cross z-axis cross 1 Graph of z = A B 1 2 θ-axis −2π −π π 2π 3π −1 Graph of z = sin(θ) one period Figure 20.2: Where does sin(θ) cross z = 1 2 ? The first thing to notice is that these two graphs will cross an infinite number of times, so there are infinitely many solutions to Example 20.1.1! However, notice there is a predictable spacing of the crossing points, which is just a manifestation of the periodicity of the sine function. In fact, if we can find the two crossing points labeled “A” and “B”, then all other crossing points are obtained by adding multiples of 2π to either “A” or “B”. By Table 17.1, θ = π radians is a special angle where we 6 computed sin π = 1 , which tells us that the crossing point labeled “A” 6 2 is the point ( π , 1 ). Using the identities in Facts 18.2.4 and 18.2.5, notice 62 that sin 5π 6 π +π 6 π = − sin − 6 π = − − sin 6 π = sin 6 1 =. 2 = sin − So, θ = 5π is the only other angle θ between 0 and 2π such that Ex6 ample 20.1.1 holds. This corresponds to the crossing point labeled “B”, π which has coordinates 56 , 1 . In view of the remarks above, the crossing 2 points come in two flavors: π 1 , k = 0, ±1, ±2, ±3, · · · , and + 2kπ, 6 2 5π 1 , k = 0, ±1, ±2, ±3, · · · . + 2kπ, 6 2 Taking this example as a model, we can tackle the more general problem: For a fixed real number c, describe the solution(s) of the equation 20.1. SOLVING THREE EQUATIONS 269 c = f(θ) for each of the circular functions z = f(θ). Studying solutions of these equations will force us to come to grips with three important issues: • For what values of c does f(θ) = c have a solution? • For a given value of c, how many solutions does f(θ) = c have? • Can we restrict the domain so that the resulting function is one-toone? All of these questions must be answered before we can come to grips with any understanding of the inverse functions. Using the graphs of the circular functions, it is an easy matter to arrive at the following qualitative conclusions. Important Fact 20.1.2. None of the circular functions is one-to-one on the domain of all θ values. The equations c = sin(θ) and c = cos(θ) have a solution if and only if −1 ≤ c ≤ 1; if c is in this range, there are infinitely many solutions. The equation c = tan(θ) has a solution for any value of c and there are infinitely many solutions. Example 20.1.3. If two sides of a right triangle have √ lengths 1 and 3 as pictured below, what are the acute angles α and β? β Solution. By the Pythagorean Theorem the remaining side has length 1+ √ 2 3 √ 3 α = 2. 1 √ Since tan(α) = 3, we need to solve this equation for α. Figure 20.3: What are the √ Graphically, we need to determine where z = 3 crosses values for α and β? the graph of the tangent function: From Fact 20.1.2, there will be infinitely many solutions to our equation, but notice that there is exactly one solution in the interval − π , π 22 and we can find it using Table 17.1: ◦ π sin(60 ) tan = tan(60◦ ) = = 3 cos(60◦) √ 3 2 1 2 = √ 3. So, α = π radians = 60◦ is the only acute angle solution and β = 180◦ − 3 60◦ − 90◦ = 30◦ . CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 270 z-axis θ = −π 2 one period θ= 5π 2 only place graphs cross √ this period in is ( π , 3) 3 etc. etc. θ-axis −2π θ = − 3π 2 −π π θ= π 2 2π θ= Figure 20.4: Where does the line z = 3π 3π 2 √ 3 cross z = tan θ? 20.2 Inverse Circular Functions Except for specially chosen angles, we have not addressed the serious problem of FINDING values of the inverse rules attached to the circular function equations. (Our previous examples were “rigged”, so that we could use Table 17.1.) To proceed computationally, we need to obtain the inverse circular functions. If we were to proceed in a sloppy manner, then a first attempt at defining the inverse circular functions would be to write $ sin−1 (z) = solutions θ of the equation z = sin(θ). $ cos−1 (z) = solutions θ of the equation z = cos(θ). (20.1) $ tan−1 (z) = solutions θ of the equation z = tan(θ). There are two main problems with these rules as they stand. First, to have a solution θ in the case of sin−1 (z) and cos−1 (z), we need to restrict z so that −1 ≤ z ≤ 1. Secondly, having made this restriction on z in the first two cases, there is no unique solution; rather, there are an infinite number of solutions. This means that the rules sin−1 , cos−1 , and tan−1 as they now stand do not define functions. Given what we have reviewed about inverse functions, the only way to proceed is to restrict each circular function to a domain of θ values on which it becomes one-to-one, then we can appeal to Fact 9.3.1 and conclude the inverse function makes sense. 20.2. INVERSE CIRCULAR FUNCTIONS 271 At this stage a lot of choice (flexibility) enters into determining the domain on which we should try to invert each circular function. In effect, there are an infinite number of possible choices. If z = f(θ) denotes one of the three circular functions, there are three natural criteria we use to guide the choice of a restricted domain, which we will call a principal domain : • The domain of f(θ) should include the angles between 0 and π , since 2 these are the possible acute angles in a right triangle. • On the restricted domain, the function f(θ) should take on all possible values in the range of f(θ). In addition, the function should be one-to-one on this restricted domain. • The function f(θ) should be “continuous” on this restricted domain; i.e. the graph on this domain could be traced with a pencil, without lifting it off the paper. In the case of z = sin(θ), the principal domain − π ≤ θ ≤ π satisfies our 2 2 criteria and the picture is given below. Notice, we would not want to take the interval 0 ≤ θ ≤ π, since z = sin(θ) doesn’t achieve negative values on this domain; in addition, it’s not one-to-one there. z-axis 1 y-axis principal domain on the unit circle θ −π π 2 −π 2 −1 θ-axis π x-axis sine restricted to principal domain Figure 20.5: Principal domain for sin(θ). In the case of z = cos(θ), the principal domain 0 ≤ θ ≤ π satisfies our criteria and the picture is given below. Notice, we would not want to take the interval − π ≤ θ ≤ π , since z = cos(θ) doesn’t achieve negative values 2 2 on this domain; in addition, it’s not one-to-one there. In the case of z = tan(θ), the principal domain − π < θ < π satisfies 2 2 our criteria and the picture is given below. Notice, we would not want to take the interval 0 ≤ θ ≤ π, since z = tan(θ) does not have a continuous graph on this interval; in other words, we do not include the endpoints since tan(θ) is undefined there. Important Facts 20.2.1 (Inverse circular functions). Restricting each circular function to its principal domain, its inverse rule f−1 (z) = θ will define a function. CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 272 z-axis 1 y-axis principal domain on the unit circle θ −π π 2 −π 2 −1 π 3π 2 θ-axis x-axis cosine restricted to principal domain Figure 20.6: Principal domain for cos(θ). (i) If −1 ≤ z ≤ 1, then sin−1 (z) is the unique angle θ in the principal domain − π ≤ θ ≤ π with the property that sin(θ) = z. 2 2 (ii) If −1 ≤ z ≤ 1, then cos−1 (z) is the unique angle θ in the principal domain 0 ≤ θ ≤ π with the property that cos(θ) = z. (iii) For any real number z, tan−1 (z) is the unique angle θ in the principal domain − π < θ < π with the property that tan(θ) = z. 2 2 We refer to the functions defined above as the inverse circular functions. These are sometimes referred to as the “arcsine”, “arccosine” and “arctangent” functions, though we will not use that terminology. The inverse circular functions give us one solution for each of these equations: c = cos(θ) c = sin(θ) c = tan(θ); these are called the principal solutions. We also can refer to these as the principal values of the inverse circular function rules θ = f−1 (z). !!! CAUTION !!! As usual, be careful with “radian mode” and “degree mode” when making calculations. For example, if your calculator is in “degree” mode and you type in “tan−1 (18)”, the answer given is “86.82”. This means that an angle of measure θ = 86.82◦ has tan(86.82) = 18. If your calculator is in “radian” mode and you type in “sin−1 (0.9)”, the answer given is “1.12”. This means that an angle of measure θ = 1.12 radians has sin(1.12) = 0.9. There is a key property of the inverse circular functions which is useful in equation solving; it is just a direct translation of Fact 9.3.2 into our current context: Important Facts 20.2.2 (Composition identities). We have the following equalities involving compositions of circular functions and their inverses: 20.3. APPLICATIONS 273 z-axis y-axis 1 −π −π 2 −1 principal domain on the unit circle θ π 2 θ-axis x-axis π tangent function restricted to principal domain Figure 20.7: Principal domain for tan(θ). (a) If − π ≤ θ ≤ π , then sin−1 (sin(θ)) = θ. 2 2 (b) If 0 ≤ θ ≤ π, then cos−1 (cos(θ)) = θ. (c) If − π < θ < π , then tan−1 (tan(θ)) = θ. 2 2 We have been very explicit about the allowed θ values for the equations in Fact 20.2.2. This is important and an Exercise will touch on this issue. 20.3 Applications As a simple application of Fact 20.2.2, we can return to the beginning of this section and justify the reasoning used to find the angle of descent of the aircraft: θ = tan−1 ((tan(θ)) = tan−1 1 10 = 0.09967 rad = 5.71◦. Let’s look at some other applications. Example 20.3.1. Find two acute angles θ so that the following equation is satisfied: 252 9 1 − cos2 (θ) . = 4 cos2 (θ) 16 Solution. Begin by multiplying each side of the equation by cos2 (θ) and rearranging terms: 9 252 1 − cos2 (θ) cos2 (θ) = 4 16 252 252 9 0= cos4 (θ) − cos2 (θ) + . 16 16 4 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 274 To solve this equation for θ, we use what is called the technique of substitution. The central idea is to bring the quadratic formula into the picture by making the substitution z = cos2 (θ): 9 252 2 252 z− z+ . 16 16 4 Applying the quadratic formula, we obtain 0= z= 252 16 ± 2 252 16 −4 2(252 ) 16 252 16 9 4 = 0.9386 or 0.06136. We now use the fact that z = cos2 (θ) and note the cosine of an acute angle is non-negative to conclude that cos2 (θ) = 0.9386 or cos(θ) = 0.9688 or cos2 (θ) = 0.06136 cos(θ) = 0.2477. Finally, we use the inverse cosine function to arrive at our two acute angle solutions: cos(θ) = 0.9688 cos(θ) = 0.2477 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 111 000 111 000 111 000 111 000 h α 32 d ⇒ θ = cos−1 (0.9688) = 14.35◦ ⇒ θ = cos−1 (0.2477) = 75.66◦ Example 20.3.2. A 32 ft ladder leans against a building (as shown below) making an angle α with the wall. OSHA (Occupational Safety and Health Administration) specifies a “safety range” for the angle α to be 15◦ ≤ α ≤ 44◦ . If the base of the ladder is d = 10 feet from the house, is this a safe placement? Find the highest and lowest points safely accessible. Solution. If d = 10, then sin(α) = 10 , so the principal so32 10 lution is α = sin−1 32 = 18.21◦; this lies within the safety zone. From the picture, it is clear that the highest point safely reached will occur precisely when α = 15◦ and as this angle increases, the height h decreases until we reach the lowest safe height when α = 44◦ . We need to solve two right triangles. If α = 15◦ , then h = 32 cos(15◦ ) = 30.9 ft. If α = 44◦ , then h = 32 cos(44◦) = 23.02 ft. Figure 20.8: A ladder problem. Example 20.3.3. A Coast Guard jet pilot makes contact with a small unidentified propeller plane 15 miles away at the same altitude in a direction 0.5 radians counterclockwise from East. The prop plane flies in the direction 1.0 radians counterclockwise from East. The jet has been instructed to allow the prop plane to fly 10 miles before intercepting. In what direction should the jet fly to intercept the prop plane? If the prop plane is flying 200 mph, how fast should the jet be flying to intercept? 20.3. APPLICATIONS 275 Intercept point North Solution. A picture of the situation is shown in Figure 20.9(a). After a look at the picture, three right triangles pop out and beg to be exploited. We highlight these in the Figure 20.9(b), by imposing a coordinate system and labeling the various sides of our triangles. We will work in radian units and label θ to be the required intercept heading. We will first determine the sides x + y and u + w of the large right triangle. To do this, we have find speed to intercept Jet West 10 miles 1 rad East prop plane spotted here 15 miles East 0.5 rad θ = intercept heading South (a) The physical layout. Intercept point y-axis 1 rad u x = 15 cos(0.5) = 13.164 miles, y = 10 cos(1.0) = 5.403 miles, w = 15 sin(0.5) = 7.191 miles, and u = 10 sin(1.0) = 8.415 miles. d u+w prop plane w spotted here 15 0.5 rad Jet 10 θ x y x-axis x+y (b) Modeling the problem. w+u x+y We now have tan(θ) = = 0.8405, so the principal soluFigure 20.9: Visualizing the tion is θ = 0.699 radians, which is about 40.05◦. This is the Coast Guard problem. only acute angle solution, so we have found the required intercept heading. To find the intercept speed, first compute your distance to the intercept point, which is the length of the hypotenuse of the big right triangle: d = (18.567)2 + (15.606)2 = 24.254 miles. You need to travel this distance in the same amount of time T it takes the 10 prop plane to travel 10 miles at 200 mph; i.e. T = 200 = 0.05 hours. Thus, the intercept speed s is s= distance traveled 24.254 = = 485 mph. time T elapsed 0.05 Later you will use an alternative approach to this problem using velocity vectors. In certain applications, knowledge of the principal solutions for the equations c = cos(θ), c = sin(θ), and c = tan(θ) is not sufficient. Here is a typical example of this, illustrating the reasoning required. Example 20.3.4. A rigid 14 ft pole is used to vault. The vaulter leaves and returns to the ground when the tip is 6 feet high, as indicated. What are the angles of the pole with the ground on takeoff and landing? CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 276 14 ft α 6 ft leaving ground β 6 ft airborne returning to ground Figure 20.10: Various angles of a vaulter’s pole. y-axis 1 Q y= 3 7 P β α 1 x-axis Solution. From the obvious right triangles in the picture, 6 3 we are interested in finding angles θ where sin(θ) = 14 = 7 . The idea is to proceed in three steps: 3 • Find the principal solution of the equation sin(θ) = 7 ; • Find all solutions of the equation sin(θ) = 3 ; 7 Figure 20.11: Modeling the problem with a unit circle. • Use the constraints of the problem to find α and β among the set of all solutions. 3 Solving the equation sin(θ) = 7 involves finding the points 3 on the unit circle with y-coordinate equal to 7 . From the picture, we see there are two such points, labeled P and Q. 3 The coordinates of these points will be P = cos(α), 3 and Q = cos(β), 7 . 7 Notice, α is the principal solution of our equation sin(θ) = 3 , since 7 0 ≤ α ≤ 90◦ ; so α = sin−1 3 = 25.38◦. In general, the solutions come 7 in two basic flavors: θ = α + 2k(180◦ ) = 25.38◦ + 2k(180◦ ), or θ = β + 2k(180◦), where k = 0, ± 1, ± 2, ± 3, . . . . To find the angle β, we can use basic properties of the circular functions: sin(β) = sin(α) = − sin(−α) = sin(180◦ − α) = sin(154.62◦). This tells us β = 154.62◦. 20.4 How to solve trigonometric equations So far, our serious use of the inverse trigonometric functions has focused on situations that ultimately involve triangles. However, many trigonometric modeling problems have nothing to do with triangles and so we 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 277 need to free ourselves from the necessity of relying on such a geometric picture. There are two general strategies for finding solutions to the equations c = sin(θ), c = cos(θ), and c = tan(θ): • The first strategy is summarized in Procedure 20.4.1. This method has the advantage of offering a “prescription” for solving the equations; the disadvantage is you can lose intuition toward interpreting your answers. • The second strategy is graphical in nature and is illustrated in Example 20.4.2 below. This method usually clarifies interpretation of the answers, but it does require more work since an essential step is to roughly sketch the graph of the trigonometric function (following the procedure of Chapter 19 or using a graphing device). Each approach has its merits as you will see in the exercises. Important Procedure 20.4.1. To find ALL solutions to the equations c = sin(θ), c = cos(θ), and c = tan(θ), we can lay out a foolproof strategy. Step Sine case Cosine case Tangent case 1. Find principal solution θ = sin−1 (c) θ = cos−1 (c) θ = tan−1 (c) 2. Find symmetry solution θ = − sin−1 (c) + π θ = − cos−1 (c) not applicable 3. Write out multiples of period k = 0, ± 1, ± 2, · · · 2kπ 2kπ kπ 4. Obtain general principal solutions θ = sin−1 (c) + 2kπ θ = cos−1 (c) + 2kπ θ = tan−1 (c) + kπ 5. Obtain general symmetry solutions θ = − sin−1 (c) + π + 2kπ θ = − cos−1 (c) + 2kπ not applicable Example 20.4.2. Assume that the number of hours of daylight in your 2π hometown during 1994 is given by the function d(t) = 3.7 sin 366 (t − 80.5) + 12, where t represents the day of the year. Find the days of the year during which there will be approximately 14 hours of daylight?1 Solution. To begin, we want to roughly sketch the graph of y = d(t) on the domain 0 ≤ t ≤ 366. If you apply the graphing procedure discussed in Chapter 19, you obtain the sinusoidal graph below on the larger domain −366 ≤ t ≤ 732. (The reason we use a larger domain is so that the “pattern” that will arise in the strategy described below is more evident. Ultimately, we will restrict our attention to the smaller domain 0 ≤ t ≤ 366.) To determine when there will be 14 hours of daylight, we need to solve the equation 14 = d(t). Graphically, this amounts to finding CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 278 y-axis 15 12.5 10 7.5 t-axis −200 200 400 600 Figure 20.12: Where does the sinusoidal function d(t) cross the line y = 14. the places where the line y = 14 intersects the graph of d(t). As can be seen in Figure 20.12, there are several such intersection points. We now outline a “three step strategy” to find all of these intersection points (which amounts to solving the equation 14 = d(t)): 1. Principal Solution. We will find one solution by using the inverse sine function. If we start with the function y = sin(t) on its principal domain −π ≤ t ≤ π , then we can compute the domain of d(t) = 2 2 2π 3.7 sin 366 (t − 80.5) + 12: 2π π π ≤ (t − 80.5) ≤ 2 366 2 −11 ≤ t ≤ 172. − Now, using the inverse sine function we can find the principal solution to the equation 14 = d(t): 14 = 3.7 sin 0.54054 = sin 2π (t − 80.5) + 12 366 (20.2) 2π (t − 80.5) 366 0.57108 = sin−1 (0.54054) = π (t − 80.5) 183 t = 113.8 Notice, this answer is in the domain −11 ≤ t ≤ 172. In effect, we have found THE ONLY SOLUTION on this domain. Conclude that there will be about 14 hours of daylight on the 114th day of the year. 1 You can get the actual data from the naval observatory at this world wide web address: <http://tycho.usno.navy.mil/time.html> 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 279 2. Symmetry Solution. To find another solution to the equation 14 = d(t), we will use symmetry properties of the graph of y = d(t). This is where having the graph of y = d(t) is most useful. We know the a maxima on the graph occurs at the point M = (172, 15.7); review Example 19.2.1 for a discussion of why this is the case. From the graph, we can see there are two symmetrically located intersection points on either side of M. The principal solution gives the intersection point (113.8, 14). This point is 58.2 horizontal units to the left of M; see the picture below. So a symmetrically positioned intersection point will be (172 + 58.2, 14) = (230.2, 14). y-axis M 15 12.5 10 7.5 t-axis −200 200 400 600 Figure 20.13: Finding the symetry solution. In other words, t = 230.2 is a second solution to the equation 14 = d(t). We call this the symmetry solution. 3. Other Solutions. To find all other solutions of 14 = d(t), we add integer multiples of the period B = 366 to the t-coordinates of the principal and symmetric intersection points. On the domain −366 ≤ t ≤ 732 we get these six intersection points; refer to the picture of the graph: (−252.2, 14), (113.8, 14), (479.8, 14), (−135.8, 14), (230.2, 14), (596.2, 14). So, on the domain −366 ≤ t ≤ 732 we have these six solutions to the equation 14 = d(t): t = −252.2, −135.8, 113.8, 230.2, 479.8, 596.2. To conclude the problem, we only are interested in solutions in the domain 0 ≤ t ≤ 366, so the answers are t = 113.8, 230.2; i.e. on days 114 and 230 there will be about 14 hours of daylight. CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 280 20.5 Summary • The inverse sine function, sin−1 x, is defined as the inverse of the sine function, sin x, restricted to the domain −π/2 ≤ x ≤ π/2. • Inverse cosine and inverse tangent are defined similarly. • To find the solutions to an equation of the form f(t) = A sin 2π (t − C) + D = k, B proceed as follows: 1. Sketch the graph of the function f(t), including a few periods and t = C. 2. Use algebra and the inverse sine function (sin−1 ) to find one solution. This is the principal solution; it is the solution nearest to t = C. Call this solution P. 3. Use the principal solution and your knowledge of the graph of the function to find the symmetry solution. Call this solution S. 4. All solutions are then of the form P, P ± B, P ± 2B, P ± 3B, . . . or S, S ± B, S ± 2B, S ± 3B, . . . . 20.6. EXERCISES 281 20.6 Exercises Problem 20.1. Let’s make sure we can handle the symbolic and mechanical aspects of working with the inverse trigonometric functions: (a) Set your calculator to “radian mode” and compute to four decimal places: (a1) sin−1 (x), for x = 0,1,−1, −3 11 ,2. (a2) cos−1 (x), for x = −3 11 ,2. √ 3 2 ,0.657, √ 0,1,−1, 23 ,0.657, (a3) tan−1 (x), for x = 0,1,−1, −3 11 ,2. √ 3 2 ,0.657, (b) Redo part (a) with your calculator set in “degree mode”. (c) Find four values of x that satisfy the equation 5 sin(2x2 + x − 1) = 2. (d) Find four solutions to the equation 5 tan(2x2 + x − 1) = 2 Problem 20.2. For each part of the problem below: • Sketch the graphs of f(x) and g(x) on the same set of axes. • Set f(x) = g(x) and find the principal and symmetry solutions. • Calculate at least two other solutions to f(x) = g(x). Indicate them on your graph. (a) f(x) = sin x − π 2 1 , g(x) = 3 . (b) f(x) = sin x + π 6 , g(x) = −1. (c) f(x) = sin (2x − 1), g(x) = 1 . 4 (d) f(x) = 10 cos (2x + 1) − 5, g(x) = −1. Problem 20.3. Assume that the number of hours of daylight in New Orleans in 1994 is 7 2π given by the function D(x) = 3 sin 365 x + 35 , 3 where x represents the number of days after March 21. (a) Find the number of hours of daylight on January 1, May 18 and October 5. (b) On what days of the year will there be approximately 10 hours of daylight? Problem 20.4. Hugo bakes world famous scones. The key to his success is a special oven whose temperature varies according to a sinusoidal function; assume the temperature (in degrees Fahrenheit) of the oven t minutes after inserting the scones is given by y = s(t) = 15 sin 3π π t− 5 2 + 415 (a) Find the amplitude, phase shift, period and mean for s(t), then sketch the graph on the domain 0 ≤ t ≤ 20 minutes. (b) What is the maximum temperature of the oven? Give all times when the oven achieves this maximum temperature during the first 20 minutes. (c) What is the minimum temperature of the oven? Give all times when the oven achieves this minimum temperature during the first 20 minutes. (d) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is at least 410o F. (e) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is at most 425o F. (f) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is between 410o F and 425o F. Problem 20.5. The temperature in Gavin’s oven is a sinusoidal function of time. Gavin sets his oven so that it has a maximum temperature of 300◦ F and a minimum temperature of 240◦ . Once the temperature hits 300◦ , it takes 20 minutes before it is 300◦ again. Gavin’s cake needs to be in the oven for 30 minutes at temperatures at or above 280 ◦ . He puts the cake into the oven when it is at 270◦ and rising. How long will Gavin need to leave the cake in the oven? Problem 20.6. Maria started observing Elasticman’s height at midnight. At 3 AM, he was at his shortest: only 5 feet tall. At 9 AM, he was at his tallest: 11 feet tall. Elasticman’s height is a sinusoidal function of time. CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 282 In the 24 hours after Maria began observing Elasticman, how much of the time will Elasticman be less than 6 feet tall? Problem 20.7. Suppose T (t) = 23 sin 2π (t − 7) + 66 24 is the temperature (in degrees Fahrenheit) at time t, where t is measured in hours after midnight on Sunday. You paint the exterior door to your house at 5 p.m. on Monday. The paint information states that 48 hours of 75◦ F drying time is required; i.e., you can only count time periods when the temperature is at least 75◦ F. When will the door be dry? Problem 20.8. Tiffany and Michael begin running around a circular track of radius 100 yards. They start at the locations pictured. Michael is running 0.025 rad/sec counterclockwise and Tiffany is running 0.03 rad/sec counterclockwise. Impose coordinates as pictured. r= 100 yards 0.025 rad/sec (i) Find where Tiffany passes Michael the first time. (j) Find when Tiffany passes Michael the second time. (k) Find where Tiffany passes Michael the second time. Problem 20.9. A communications satellite orbits the earth t miles above the surface. Assume the radius of the earth is 3,960 miles. The satellite can only “see” a portion of the earth’s surface, bounded by what is called a horizon circle. This leads to a two-dimensional cross-sectional picture we can use to study the size of the horizon slice: satellite Earth horizon circle center of Earth Earth satellite α α Tiff starts here t Michael starts here CROSS−SECTION 0.03 rad/sec (a) Find a formula for α in terms of t. (a) Where is each runner located (in xy-coordinates) after 8 seconds? (b) How far has each runner traveled after 8 seconds? (c) Find the angle swept out by Michael after t seconds. (d) Find the angle swept out by Tiffany after t seconds. (e) Find the xy-coordinates of Michael and Tiffany after t seconds. (f) Find the first time when Michael’s x-coordinate is -50. (g) Find the first time when Tiffany’s x-coordinate is -50. (h) Find when Tiffany passes Michael the first time. (b) If t = 30,000 miles, what is alpha? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? (c) If t = 1,000 miles, what is alpha? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? (d) Suppose you wish to place a satellite into orbit so that 20% of the circumference is covered by the satellite. What is the required distance t? Problem 20.10. Answer the following questions: 20.6. EXERCISES (a) If y = sin(x) on the domain − π ≤ x ≤ π , 2 2 what is the domain D and range R of y = 2 sin(3x − 1) + 3? How many solutions does the equation 4 = 2 sin(3x−1)+3 have on the domain D and what are they? (b) If y = sin(t) on the domain − π ≤ t ≤ π , 2 2 what is the domain D and range R of 2π y = 8 sin( 1.2 (t−0.3))+18. How many solu2π tions does the equation 22 = 8 sin( 1.2 (t − 0.3))+ 18 have on the domain D and what are they? (c) If y = sin(t) on the domain − π ≤ t ≤ 2 π 2 , what is the domain D and range R 2π of y = 27 sin( 366 (t − 80.5)) + 45. How many solutions does the equation 40 = 2π 27 sin( 366 (t − 80.5)) + 45 have on the domain D and what are they? (d) If y = cos(x) on the domain 0 ≤ x ≤ π, what is the domain D and range R of y = 4 cos(2x + 1)− 3? How many solutions does the equation −1 = 4 cos(2x + 1) − 3 have on the domain D and what are they? (e) If y = tan(x) on the domain − π < x < π , 2 2 what is the domain D and range R of y = 2 tan(−x + 5) + 13? How many solutions does the equation 100 = 2 tan(−x + 5) + 13 have on the domain D and what are they? Problem 20.11. Tiffany is a model rocket enthusiast. She has been working on a pressurized rocket filled with laughing gas. Ac- 283 cording to her design, if the atmospheric pressure exerted on the rocket is less than 10 pounds/sq.in., the laughing gas chamber inside the rocket will explode. Tiff worked from a formula p = (14.7)e−h/10 pounds/sq.in. for the atmospheric pressure h miles above sea level. Assume that the rocket is launched at an angle of α above level ground at sea level with an initial speed of 1400 feet/sec. Also, assume the height (in feet) of the rocket at time t seconds is given by the equation y(t) = −16t2 + 1400 sin(α)t. (a) At what altitude will the rocket explode? (b) If the angle of launch is α = 12o , determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? (c) If the angle of launch is α = 82o , determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? (d) Find the largest launch angle α so that the rocket will not explode. Problem 20.12. Let’s make sure we can handle the symbolic and mechanical aspects of working with the inverse trigonometric functions: (a) Find four solutions of tan(2x2 + x − 1) = 5. (b) Solve for x: tan−1 (2x2 + x − 1) = 0.5 284 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix 285 286 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix A Useful Formulas Abbreviations inch = in liter = L feet = ft Joule = J yard = yd calorie = cal mile = mi atmosphere = atm millimeter = mm Coulomb = C centimeter = cm radian = rad meter = m degree = deg =◦ kilometer = km second = sec = s minute = min miles per hour = mi/hr = mph hour = hr year = yr feet per second = ft/sec = ft/s ounce = oz pound = lb gram = g meters per second = m/sec = m/s kilogram = kg quart = qt revolutions = rev/min per minute = RPM gallon = gal milliliter = ml 287 APPENDIX A. USEFUL FORMULAS 288 Conversion Factors Length Energy 1 in = 2.54 cm 1 J = 1 kg m2 /s2 1 ft = 0.3048 m 1 cal = 4.184 J 1 mi = 1.609344 km Mass Volume 1 gal = 3.7854 L 1 oz = 28.3495 g 1 qt = 0.946353 L 1 lb = 0.453592 kg Formulas from Plane and Solid Geometry Rectangle r • Perimeter = 2ℓ + 2w • Area = ℓw w Rectanglular prism • Surface Area = 2(ℓw + ℓh + wh) ℓ • Volume = ℓhw Triangle h • Perimeter = a + b + c • Area = w 1 bh 2 ℓ a c h b Right circular cylinder • Surface Area = 2πr2 + 2πrh • Volume = πr2 h Circle • Perimeter = 2πr 2 • Area = πr r h 289 Sphere Right circular cone • Surface Area = 4πr2 • Surface Area = πr2 + πrs 4 • Volume = 3 πr3 • Volume = 1 πr2 h 3 r r Constants Avogadro’s number = N = 6.022142 × 1023 speed of light = c = 2.99792 × 108 m/s2 density of water = 1 g/cm3 mass of earth = 5.9736 × 1024 kg earth’s equatorial radius = 3,960 mi = 6.38 × 106 m acceleration of gravity at earth’s surface = 32 ft/sec2 = 9.8 m/s2 Algebra • ax ay = ax+y • (ax )y = axy • a0 = 1 • a b x = ax bx ax = ax−y ay √ • n a = a1/n • • (a + b)(c + d) = ac + ad + bc + bd • (a + b)2 = a2 + 2ab + b2 • (a + b)3 = a3 + 3a2 b + 3ab2 + b3 2 • Quadratic Formula: If ax + bx + c = 0, then x = −b ± √ b2 − 4ac 2a APPENDIX A. USEFUL FORMULAS 290 b • Completing the Square: ax + bx + c = a x + 2a 2 2 − b2 +c 4a Trigonometry • sin(−x) = − sin x • cos(π − x) = − cos x • cos(−x) = cos x • sin(π + x) = − sin x • sin = cos x • cos(π + x) = − cos x = sin x • sin(x + y) = sin x cos y + cos x sin y π −x 2 π • cos −x 2 π +x • sin 2 π • cos +x 2 = cos x = − sin x • sin(π − x) = sin x • cos(x + y) = cos x cos y − sin x sin y • sin2 x + cos2 x = 1 • sin 2x = 2 sin x cos x • cos 2x = cos2 x − sin2 x Appendix B Answers Answer 1.1 (b) 150 ft/sec. (c) Gina. (d) 6300 hours. Answer 1.14 (a) N = 12 + 0.45(60 − x). (b) Reduced competition for resources (eg. water, nutrients, etc.). Answer 1.2 68.4444 km. Answer 1.3 r = 10.172 cm for lead; r = 16.433 cm for aluminum. Answer 1.4 The mass of the air is 9.740 million kg, more than the mass of the Tower. Answer 1.5 (a) 5.5 min/mi = 5:30 pace. (b) 14 2 ft/sec. (c) 3 Adrienne. Answer 1.6 (a) (John’s Salary) = $56000 and (taxes) = $0. (b) (John’s Salary) ≤$56000 and (taxes) = 0.15× (John’s Salary). (c) (John’s Salary) ≥ $56000 and (John’s taxes) > 0.28× (John’s Salary). (d) 1500 ≤(number of 120 students each year) ≤ 1800. (e) (cost red Porsche) > 3× (cost F-150 pickup). (f) 2 hours ≤ (weekly study time per credit hour) ≤ 3 hours (g) 2× (number of happy math students) > 5× (number of happy chemistry students). (happy math students) + (happy chemistry students) < 1 × (number of 2 cheerful biology students). (h) (Cady’s high score) - (Cady’s low score) = 10%. (Cady’s final exam score) = 97%. (i) 1 (0.9999) × 2 (total votes cast)≤ (Tush votes) ≤ (1.0001) × 1 2 (total votes cast). ( Answer 1.15 Formula simplifies to r11+x) . Rates are +2x about 0.67r, 0.6r, 0.54r and 0.51r at the indicated times. Rates decrease over time, but will never be less than 0.5r. Answer 1.16 (a) Initial time = 7am, Initial temperature = 44◦ F, Final time = 10am, Final temperature = 50◦ F, rate of change = 2◦ F/hr. (b) 58◦ F. (c) Initial time = 4.5pm, Initial temperature = 54◦ F, Final time = 6.25pm, Final temperature = 26◦ F, rate of change = -16◦ F/hr. Answer 1.17 (a) t = 9. (b) a = 1/8. (c) x = −3a/4. (d) t > 3. x+ 2 (e) x(x+1) . √ √ Answer 2.1 (a) d = 2, ∆x = 1 = ∆y = 1. (b) d = 5, √ ∆x = −1, ∆y = −2. (c) d = 34, ∆x = 5, ∆y = −3. (d) √ d = 10t2 + 2t + 1, ∆x = 3t, ∆y = 1 + t. Answer 2.2 (a-c) True. (d) ∆x = s − a, ∆y = t − b. (e) ∆x = a − s, ∆y = b − t. (f) ∆x = 0 means the points line on the same vertical line; ∆y = 0 means the points line on the same horizontal line. Answer 2.3 Just after 12:29 PM that afternoon. Answer 1.10 Lee has 2.265 in2 more pie. Answer 2.4 (a) Erik= 6.818 mph, Ferry = 17.6 ft/sec. (b) Impose coordinates with Kingston the origin and units of miles on each axis; then Edmonds is located at (6,0) and Erik’s sailboat is at (3,2). The table rows have these entries: (0,0), (0.1,0), (1.4,0), (12t,0). (3,2), (3,1.9432), (3,1.2045), (3,2 − 6.818t) 3.606, 3.491, 2.003, (12t − 3)2 + (2 − 6.818t)2 (c) Use coordinates as in (b), then when the ferry reaches (3,0), Erik is at (3,0.296). (d) CG vessel does not catch the ferry before Edmonds. Answer 1.11 About 5 × 106 times around the equator. Answer 2.5 (a) d(t) = (65.3)t (b) 227 minutes, 168.4 miles (c) t = 80.86 seconds. Answer 1.7 Go for the 15 inch pie. Answer 1.8 (a) m90 = 151kg,m99 = 468kg,m99.9 = 1476kg. m (b) v = 2.974 × 108 sec . Answer 1.9 1080 pizzas sold in 4 hours. Profit reaches $1000 at about 8 : 22 pm. Answer 1.12 (a) Radii are about r = 0.2416, 2.317, 4.184 and 9.167 cm at the indicated times. (b) No. Answer 1.13 (a) 140 million gallons per week; about 7300 million gallons per year. (b) 100 yds × 50 yds ×20 yds. Answer 2.6 (a) Allyson’s coordinate position: (0 ft,20 ft). Adrienne’s coordinate position: (−16 ft,0 ft). (b) After 2 seconds there will still be slack in the bungee cord. (c) t ≈ 2.34. Use this time to find where Allyson and Adrienne are located. (d) t = 5.5 seconds. 291 APPENDIX B. ANSWERS 292 Answer 2.7 Impose a coordinate system with origin at A and the shore the horizontal x axis; let x be the location on the shore where Brooke beaches. Equation √ 1 25 + x2 + 1 (6 − x) gives time to reach Kono’s. T (0) = 4 2 4 hr, T (6) = 3.9 hr. Neither time will be the minimum time. 7 Answer 2.8 (b) t = 2; t = 5 . (c) (3,3) (d) spider=( 3 , 8 ), 2 3 ant=( 41 , 4 ). (e) 1.5 feet. (f) Spider reaches (9,6) when t = 4; 33 √ ant reaches (9,6) when t = 3. (g) spider speed is 5 ft/sec; √ ant speed is 8 ft/sec. Answer 3.7 (a) 6.92 seconds. (b) 7.67 seconds. (c) 38.16 seconds. (d) 7332 ft2 < area < 7632 ft2 . Answer 3.8 (a) x = 11/5. (b) no solutions. (c) (−2, − 2) and (−2,4). (d) (−1 ± 2/5,3). 5 Answer 4.1 (a) y = − 3 (x − 1) − 1 (b) y = 40(x + 1) − 2 (c) y = −2x − 2 (d) y = 11 (e) m = 3 , y = 3 (x − 1) + 1 (f) 5 5 y = 40x − 14 (g) y = − 3 x + 7 (h) y = x − 1 4 4 Answer 2.9 49.92 mph (exactly). Answer 2.10 141.46 miles. They are 300 miles apart at time 0.826 hr = 49.6 minutes. Answer 2.11 (a) Final answer is correct, but second equality is wrong. (b) Final answer should be 4xy; key fact is that (x + y)2 = x2 + 2xy + y2 , etc. (c) Answer and steps correct. Answer 2.12 (a) x = ± 1+β2 α2 (b) x = αβ α+ β 1+4α 2 Answer 4.2 (a)( 1+2α , 1+2α ). (b) α = −2/5. (c) α = −1/4. Answer 4.3 (a) area = 25 13 2 b (b) area = − 2m (c) m = − 1 4 Answer 4.4 In some cases, answers in this problem are unique. In other cases, the answers are not unique. Here is a possible solution set: β αβ+1 (c) x = Eqn Answer 2.13 (a) 5t2 + 6t + 5 (b) 2t2 + 4t (c) √ 5t2 + 4t + 4 2 t2 −1 Answer 3.1 (a) (x+3)2 +(y−4)2 = 9 (b) (x−3)2 + y + 11 3 2 = (c) Draw a vertical and horizontal line through (1, 1). On the vertical line, a circle of radius 2 will have a center at either (1, 3) or (1, −1). Likewise, the horizontal line will have circles at either (−1, 1) or (3, 1). C(h,k) 1. 2. 3. 4. (3, 1) (1, 3) (−1, 1) (1, −1) (d) (1,1), (1, − 3), (1 + √ (x − h)2 + (y − k)2 = r2 (x − 3)2 (x − 1)2 (x + 1)2 (x − 1)2 + (y − 1)2 + (y − 3)2 + (y − 1)2 + (y + 1)2 3,0), (0, − 1 − √ y-int Pt on line Pt on line y = 2x + 1 1 16 Eqn Slope 2 1 (0, 1) (− 1 , 0) 2 y = − 11 x + 17 4 4 − 11 4 17 4 (3, − 4) (−1,7) y = −2x + 1 −2 1 (0, 1) ( 1 , 0) 2 y = 1x + 1 2 1 2 1 (0, 1) (−2, 0) y = 1,000 0 1,000 (0, 1,000) (0, 1,000) y=0 0 0 (0, 0) (1, 0) x=3 Undef None (3, 3) (3, −2) y = x − 14 1 −14 (5, − 9) (0, −14) (d) =4 =4 =4 =4 3) √ Answer 3.2 (a) center (3,-1), radius 2 3 (b) center (-2,-3), radius 2 (c) center (-1/6,5/3), radius 203/12 (d) center √ (3/4, -1/2), radius 3 Answer 3.3 (a) Wet in 24.0 minutes. 25.4154041 minutes. Answer 4.5 (a) y = 6,850(x − 1970) + 38,000 (b) y = 8,000(x − 1970) + 8,400 (d) Tabularize your answer: (b) Wet in Answer 3.4 (a) Imposing with x-axis along ground and yaxis along tower, wheel modeled by x2 + (y − 62)2 = 602 . (b) 46.43 ft. to right of tower. (c) (−24,7) and (−24,117). Answer 3.5 (a) Impose a coordinate system so that the tractor is at the origin at t = 0 seconds. With this coordinate system, the south edge of the sidewalk is modeled by ys = 100; the north edge is modeled by, yn = 110. (b) t = 32 minutes. (c) t = 52 minutes. (d) 20 minutes. Answer 3.6 (a)] The equation for eastward travel from Kingston is y = 8. Southward travel along x = −1. (b) Boundary: (x +2)2 +(y−10)2 = 9; Interior:(x +2)2 +(y−10)2 < 9; Exterior:(x + 2)2 + (y − 10)2 > 9. (c) 3.82 minutes. (d) Plug x = −1 into circle equation to find exit point. Use this to find when the ferry exits the radar zone. Be careful to reference all times relative to when the ferry departed Kingston. (e) 20.32 minutes. Year Seattle Port Townsend 1983 1998 $127,050 $229,800 $112,400 $232,400 (e) 1995.74, $214,313 (f) 1982.70, Sea = $124,965 (g) 2008.78, Sea = $303,661 (h) 1972.32, Sea = $53,892 (i) No. Answer 4.6 For these answers, we take the hole to be the origin of our coordinate system. Then, the ball’s line of travel is y = (0.667)(x − 35). (a) (−13.46, −32.31) (b) 3.19 seconds (c) 5.82 seconds (d) (10.78, −16.17) at about t = 6.1 sec. Answer 4.7 (a) Allyson at (0,70); Adrian at (−44,0). Bungee is 83.1401 ft. long. (b) Occurs at time t = 7.7546 seconds and Allyson is at (0,77.546). Allyson’s final location is 77.546 ft. from her starting point. 293 Answer 4.8 The lines of tangency have equations y = 2 ± √ (x − 12). The non-visible portion of the y-axis is 5 √ − 245 5 ≤y≤ √ 24 5 . 5 Answer 4.9 C temperature is = 5 (F − 9 −9.4◦ F. 32) and F = Answer 4.10 At the closest 26.22471828 miles from Paris. 9 C 5 + 32. In Oslo, the point, she will 5 1 Answer 5.8 (a) − 3 ,− 2 , x , 2 (♥ − 3). (b) 0, 20, 6x + 2x2 , 2 2 2 − 6♥, 2(♥ + △ − 3)(♥ + △). (c) Always 4π2 . 2♥ Answer 5.9 Use the vertical line test. For example, (a) is not a function, by the vertical line test; you can split it into two function graphs by slicing the ellipse symmetrically into upper and lower halves. On the other hand, (o) is a function, by the vertical line test; etc. be Answer 5.10 (a) x = 5. √ 242− Answer 4.11 Impose coordinates with Angela’s initial location as the origin. Angela is closest to Mary at (18.8356,41.4383); this takes approximately 3.8 seconds. Answer 4.12 (a) Impose coordinates with sprinkler initial location as the origin (0,0). Line equation becomes 1 y = − 5 x + 100. (b) Sprinkler is located at (0,82.5) at time t = 33 minutes. Circular boundary of watered zone hits southern edge of sidewalk at the points (−6.708,101.3416) 1 and (13.4386,97.3123). (c) y = − 5 x + 110.198. | − 0.5x − 1| = if x ≤ −2 if x > −2 −0.5x − 1 0.5x + 1 y 3 2.5 2 1.5 1 0.5 √ Answer 4.15 (a) (−1 ± 7/3)/2 = (−3 ± 21)/6. (b) t = − √0.463325, 0.863325; or the exact answer would be (1 ± 11)/5. (c) t = 0.2. (d) No real number solutions. 2+ √ 2 Answer 5.1 (a) −2 + h + 2x. (b) 2. (c) h + 2x. (d) −h − 2x. (e) −π(h + 2x) (f) √h+x−1 +√x−1 . 1 -4 Answer 5.3 For example, in (a), suppose Dave has constant speed v ft/min. Then the function s = d(t) = vt will compute the distance Dave travels in t minutes. The graph would be a line with s-intercept 0 and slope v; the domain would be 0 ≤ t ≤ 2400 ; etc. v Answer 5.4 -2 4 2 x Answer 6.3 (a) x = 4 and x = − 22 (b) x = 0.75 (c) the 3 equation has no solutions. Answer 6.4 (a) x = −7 and x = 3 (b) a = 3 (c) x = 8/3 Answer 6.5 For 0 ≤ x ≤ 6, the area is 1 x( x 6 + 18). Answer 6.6 (a) The rule is 2x −x + 30 y= Answer 5.2 g(x) = 9 x + 24, f(x) = 4 x + 4, v(x) = x + 20, 5 5 v(5) = 25, minimum v(x) = 20, maximum v(x) = 40. if 0 ≤ x ≤ 10 if 10 ≤ x ≤ 30 and the range is 0 ≤ y ≤ 20. (b) The rule for the area function a(x) is a(x) = 1 − 2 x2 x2 + 30x − 150 if 0 ≤ x ≤ 10 if 10 ≤ x ≤ 30 (c) x = 12.6795 inches. Answer 6.7 (a) Answer 5.5 Several possible answers for each one. j( t ) = √ Answer 5.6 (a) x-intercepts= 1 (3 ± √ √6 1 (b) ( 6 (3 − 93),5) and ( 1 (3 + 93),5). 6 1+ √ 2, − 2 − 3 (d) Answer 6.2 (a) Answer 4.14 18.923076923 seconds No, Yes. (e) ( (c) x = 36. Answer 6.1 (a) 0, 2, 3. (b) x = ±4, x = 0, no solution. (c) Intersect at (4,4) and (− 4 , 4 ). Area is 16 . 33 3 Answer 4.13 (a) With the origin set at the statue, x = 30 − 3.123475t, y = 2.49878t. (b) The distance to the statue is √ 16t2 − 187.4085t + 900 feet. (c) Margot will be 28 feet from the statue after 0.655672 seconds, and after 11.057360 seconds (assuming she continues past the point due north of the statue). Answer 4.16 (a) There are four answers: x = ± √ √ and x = ± 2 − 2 (b) y = 6 + 2 5. (b) x = 4, 8. 56400 2 33). y-intercept = −2. (c) None. (d) Yes, No, √ √ 1 + 2 + 3(1 + 2)). Answer 5.7 (b) f(x) = 30000x + 50000 (4 − x)2 + 1. (c) The table of values (x,f(x)) will be: (0., 206155), (0.5, 197003), (1, 188114), (1.5, 179629), (2,171803), (2.5, 165139), (3., 160711), (3.5, 160902), (4,170000). Minimal cost occurs for some 2.5 < x < 3.5; the exact answer is x = 13 , but we can4 not solve this in our class since it requires the tools of Calculus. 0 62.22(t + 1) if t < −1 if −1 ≤ t ≤ 3.5 (b) s(t) = 280 280 − 70t if t < 0 if 0 ≤ t ≤ 4 (c) 280 280 − 62.22(t + 1) 217.78 − 132.22t d(t) = 132.22t − 217.78 70t if if if if if t < −1 −1 ≤ t < 0 0 ≤ t < 1.6471 1.6471 ≤ t < 3.5 3.5 ≤ t < 4 APPENDIX B. ANSWERS 294 Answer 6.8 The distance to his starting point t seconds after he starts is 10t 2502 + (12(t − 25))2 d(t) = 4002 + (250 − 9(t − 175 ))2 3 if 0 ≤ t ≤ 25, if 25 ≤ t ≤ 175 , 3 if 175 3 ≤t≤ 205 . 3 Answer 6.9 y = d(t) = if if if if 18t − 5)2 902 + (90 − 18(t − 10))2 90 − 18(t − 15) 902 + 182 (t 0≤x≤5 5 ≤ x ≤ 10 10 ≤ x ≤ 15 15 ≤ x ≤ 20 Here is the graph of d(t) 2 Answer 7.2 (a) y = − 3 x2 + 5 x. (b) y = 1.125x2 − 1.5x − 1.625. 3 2 + 3.5x − 4. (d) No solution. (c) y = −0.5x Answer 7.3 (a) maximum value is 22; minimum value is 1.75 (b) maximum value is 32; minimum value is 2 (c) maximum value is -6; minimum value is -46 Answer 7.4 In the first case, d = 0 or d = 12. In the second case, d = ± √2 . 13 5 1 Answer 7.5 (a) Multipart function: s(t) = − 160 t2 + 8 t + 10 for 0 ≤ t ≤ 114.031, s(t) = 0 for t ≥ 114.031.; $23,125. (b) Sell at time t = 50 days; $2500. Answer 7.6 The parabola has x intercepts at −1, 3 and y intercept at -3. The vertex of the parabola is (1, − 4). ft 140 120 100 80 60 40 20 2 |x − 2x − 3| = x2 − 2x − 3 −x2 + 2x + 3 x2 − 2x − 3 if x ≤ −1 if −1 < x < 3 if x ≥ 3 Answer 7.7 (a) 100 ft. (b) 156.25 ft. (c) (625, − 125). (d) When x = 54.81 ft. or x = 570.19 ft.; i.e. at (54.81,39.04) and (570.19, − 64.04). 5 15 10 20 t sec Answer 6.10 (a) v(x) = 2x3 − 70x2 + 500x; degree 3. (b) a(x) = 1000 − 50x − 2x2 , degree 2. To get 600 sq. in. dimensions are: 6.3746 × 7.2508× 18.6254. Answer 6.11 (a) 2 hours. (b) Impose coordinates with x axis the bottom of the ditch and the y axis the pictured centerline. y= 10 + 10 − 10 − 10 + if if if if if if if x ≤ −40 −40 ≤ x ≤ −30 −30 ≤ x ≤ −20 −20 ≤ x ≤ 20 20 ≤ x ≤ 30 30 ≤ x ≤ 40 40 ≤ x Answer 6.12 (a) 0 ≤ y ≤ 6. (b) Increasing: −6 ≤ x ≤ −2 and 2 ≤ x ≤ 4. Decreasing: −2 ≤ x ≤ 2 and 4 ≤ x ≤ 6. (c) 4+ y= 4− Answer 7.9 She should have 225 trees in the orchard. Answer 7.10 She should charge $7.72 to make the most money. Answer 7.11 The radius of the circular part should be 20 100 − (x + 40)2 100 − (x + 20)2 0 100 − (x − 20)2 100 − (x − 40)2 20 √ (c) 2(40 − 91) feet. (d) 42 ft. wide: 0.3008 minutes. 50 ft. wide: 8.038 minutes. 73 ft. wide: 116.205 minutes. Answer 7.8 (a) No, since f(1) = 1 = 2. (b) The points (1,1 + 2b) and (−(1 + b),1 + 2b). (c) Only the point (a,1 − a2 ). (d) The points (−2.2701, − 7.1168) and (2.9368,1.5612). 2x + 12 4 − (x + 2)2 4 − (x − 2)2 −2x + 12 if if if if −6 ≤ x ≤ −4 −4 ≤ x ≤ 0 0≤x≤4 4≤x≤6 (d) 2 ≤ y ≤ 6. (e) 2 ≤ y ≤ 4. Answer 6.13 (a) 0. (b) a = 13 , 5 b = − 3 , c = −3. 5 Answer 7.1 (a) 2(x − 4)2 + 9, Vertex: (4, 9), Axis: x = 4. (b) 3(x − 5/2)2 − 383/4, Vertex: (5/2, -383/4), Axis: x = 5/2. (c) (x − 3/14)2 + 2539/196, Vertex: (3/14, 2539/196), Axis: x = 3/14. (d)2(x − 0)2 − 0, Vertex: (0, 0), Axis: x = 0. (e) (1/100)(x − 0)2 − 0, Vertex: (0, 0), Axis: x = 0. 24 ≈ 3.36059492 feet. 4+π The short side of the rectangular part should also be equal to this. Answer 7.12 The enclosure should be 50 meters by 75 meters. Answer 7.13 The maximum possible area is 4285.71 m2 . Answer 7.14 To achieve the minimal combined area, cut the wire so that the pieces have lengths 26.394 in. and 33.606 in.; bend the 26.394 inch piece into a circle. Answer 7.15 (a) y = 3.8681948. 1 x−4 2 1 (b) y = 4 + 6 x (c) t = 1350/349 ≈ Answer 7.16 (a) The distance between Sven and Rudyard is 41t2 − 2400t + 36900 feet, where t is the number of seconds after they start moving. (b) They will be closest together after they have been moving for 29.268292 seconds, and will be 42.166916 feet apart at that time. Answer 7.17 Michael M(t) = (5.547t,8.32t), Tina T (t) = (400 − 8.944t,50 + 4.472t). Study the distance SQUARED from M(t) to T (t). Michael and Tina will be closest when t = 26.641 sec; they are 54.3 ft at that instant. 295 Answer 8.5 (a) f(s) = Answer 7.18 x= −2α2 ± α= −x2 ± √ 4α4 − 4α 2α C(f(s)) computes mph when you input seconds s. g(h) = 60h; x4 − 8x 4x Answer 7.19 (d) There are two possible values for α: If √ , α = 8 + 2 17√then the unique solution of the equation will √ , then the unique solution be x = −4 − 17. If α = 8 − 2 17√ of the equation will be x = −4 + 17. 1 (2 ± 2 70s2 10(602 ) + s2 C(f(s)) = √ Answer 7.20 (a) t = s ; 60 (b) 252000h2 10 + 3600h2 C(g(h)) = C(g(h)) computes mph when you input hours h. (c) v(s) = 22 s; 15 √√ √ 2 s − 1). (b) x = −1 ± y − 2. 308m2 30 + 3m2 v(C(m)) = v(C(m)) computes ft/sec when you input minutes m. Answer 8.1 (a) Answer 8.6 (a) f(g(x)) = (x + 3)2 , f(f(x)) = x4 , g(f(x)) = x2 + 1 1 3. (b) f(g(x)) = √x , f(f(x)) = x, g(f(x)) = √x . (c) f(g(x)) = x, h(f(t)) = −t + 1 t−1 if t ≤ 1 if 1 ≤ t h(g(t)) = −t − 1 t+1 if t ≤ −1 if −1 ≤ t f(f(x)) = 81x + 20, g(f(x)) = x. (d) f(g(x)) = 6(x − 4)2 + 5, f(f(x)) = 6(6x2 + 5)2 + 5, g(f(x)) = 6x2 + 1. (e) f(g(x)) = 8x + 21, f(f(x)) = 4(4x3 − 3)3 − 3, g(f(x)) = 2x. (f) f(g(x)) = 2x3 + 1, f(f(x)) = 4x + 3, g(f(x)) = (2x + 1)3 . (g) f(g(x)) = 3, f(f(x)) = 3, g(f(x)) = 43. (h) f(g(x)) = −4, f(f(x)) = −4, g(f(x)) = 0. (b) Answer 8.7 −5/2 ≤ x ≤ −1/2 f(h(t)) = −t − 1 t−1 if t ≤ 0 if 0 ≤ t g(h(t)) = t−1 −t − 1 if t ≤ 0 if 0 ≤ t 7 Answer 8.8 (a) 2 ≤ x ≤ 6. (b) −3 ≤ y ≤ 5. (c) 1 ≤ x ≤ 6. 3 (d) −9 ≤ y ≤ 7. (e) B = 5, C = 39 . (f) A = 1 , D = 8 . 5 8 Answer 8.9 (a) −1 , (x−1)(x+h−1) 4 + 4h + 8x, set h = 0 to get 4 + 8x. (c) (c) −1 (b) (x−1)2 −h−2x √ √ , 25−x2 + 25−(x+h)2 set h = 0 to get −x . 25−x2 set h = 0 to get √ −t − 1 t+1 h(h(t) − 1) = −t + 1 t−1 if if if if t ≤ −1 −1 ≤ t ≤ 0 0≤t≤1 1≤t Answer 9.1 (a) domain of f ={x|x = (b) Answer 9.3 (c) f−1 (y) = − 17 }. 8 2.5 range=f= {y|y = 0}. 3+ √ 17+8y 4 on the domain {y|y ≥ Answer 9.5 1.5 1 Answer 9.6 (a) h = f(x) = −2x2 + 124x. (c) x = g(h) = √ 1 31 − 2 3844 − 2h. 0.5 -1 4 }; 3 Answer 9.4 Only (B) is one-to-one on the entire domain. 2 -2 = 2+4y . 3y Answer 9.2 y 3 -3 f−1 (y) 1 2 3 x Answer 9.7 (a) (1) -0.5 y -1 Answer 8.2 (a) y = f(g(x)), if f(x) = x5 , g(x) = x − 11. (b) √ y = f(g(x)), if f(x) = 3 x, g(x) = 1 + x2 . (c) y = f(g(x)), if 5 − 5x2 + (1/2)x + 11, g(x) = x − 3. (d) y = f(g(x)), if f(x) = 2x √ f(x) = (1/x), g(x) = x2 + 3. (e) y = f(g(x)), if f(x√ = x, ) √ g ( x) = x + 1. (f) y = f(g(x)), if f(x) = 2 − 5 − x2 , g(x) = 3x − 1. Answer 8.4 (c) x = −1 and x = 2 x (2) x = f−1 (y) = bers. (3) y+2 3 has domain and range all real num- APPENDIX B. ANSWERS 296 x Answer 11.4 (c) The curve is modeled by y = x 500cosh( 500 ) − 440.536 and the minimum height is 59.46 feet. Answer 12.1 (a) 0.6826; 2.3979; 3.3030; 3.3219;0.3010. (b) 3.555; 19.8; 0.0729. (c) x = log10 y; x = log10 (3y); x = (1/3) log10 (y). y Answer 12.2 (a) I(d) = I◦ (0.94727)d . (b) 85 meters. (4) f(f−1 (y)) = f( y+2 ) = 3( y+2 ) − 2 = y + 2 − 2 = y; 3 3 f−1 (f(x)) = f−1 (3x − 2) = 3x−2+2 = x. 3 t Answer 9.8 (a) w = f(t) = 2 25 − ( 6 − 5)2 . Domain: 0 ≤ t ≤ 30 hours; Range: 0 ≤ w ≤ 10 ft. (b) t = 6 hours. (c) t = f−1 (w) = 30 − 6 25 − ( w )2 . Domain: 0 ≤ w ≤ 10 ft.; 2 Range: 0 ≤ t ≤ 30 hours. Answer 9.9 (a) 2 nanoseconds. (f) domain φ={0 ≤ t ≤ 10}; range φ={0 ≤ y ≤ 6}. Answer 10.1 (a) 31.5443; (b) 355.1134; (c) 36.4622; (d) 0.0616; (e) 51,168; (f) 0.009794; Answer 10.2 (a) y = 3( 1 )x (b) y = ( 1 )x . (d) y = 2 2 1√x (1). 27 3 Answer 10.3 (a) 1.64 × 106 cells. (b) True. (c) The two formulas are identical. Answer 10.4 (a) 261.31 Hz. (b) 440 Hz. (c) 27.5 Hz. (d) 16.35 Hz. Answer 10.5 (a) 29 = 512. (b) 2n−1 . (c) 263 = 9.2 × 1018 . (d) 9.2 × 1015 meters. Answer 10.6 (a) M(p) is the higher curve. (d) Answer 12.3 (a) y = 13e1.09861t . (b) y = 2e−2.0794t . Answer 12.4 (a) 9.9 years. (b) 34.66%. Answer 12.5 (a) 8.8 cm. (b) 135 cm. (c) 4.84 yrs. (d) Maximum possible length of the halibut. Answer 12.6 (a) C(t) = 1.03526t . Answer 12.7 (c) v(x) $200,000 during 1993. 55000(1.0315)x . = Valued at Answer 12.8 (a) x = 2.7606. (b) 1.392. (c) 0.3552. (d) x = e8 . (e) x = 11.513. (f)-0.61. (g) sin(x) = −0.6931; x = −0.7658 + 2kπ or x = 3.9074 + 2kπ. Answer 12.9 (d) 8.617 weeks. Answer 12.10 31.699250014 days Answer 12.11 (a) 2015 (b) 2048. Answer 12.12 (a) 46.701735 years (b) 137.3113631016 years after 1980 Answer 13.1 The graph is given below: fraction 0.8 y 2 0.6 1.5 0.4 1 0.2 20 40 60 80 $1.113(1.046408)t . Answer 11.1 (a) w(t) = is below; should be $5.70 by the model. 100 0.5 p -2 -1.5 -1 -0.5 0.5 (b) $1.11. (c) It -0.5 Answer 11.2 (a) p(x) = 860(1.070674)x , l(x) = 70x + 860. (b) p(10) = 1702, l(10) = 1560. -1 -1.5 Answer 11.3 If we use 1989 and 2000 in a(t), we get two data points and a corresponding exponential model: E(t) = 15.918(1.243301)( t − 1980). The exponential model grows faster than the cubic model and eventually exceeds a(t). -2 Answer 13.2 (a) 1 1.5 2 x 297 y 30 6 25 4 20 2 15 10 0 -3 -2 -1 0 1 2 5 -2 -1 -2 1 2 4 3 x -4 Answer 13.5 (a) -6 y 4 (b) 3 2 1.5 1 1 -4 -3 -2 -1 1 2 4 3 x 0.5 -1 0 -0.5 0 0.5 1 1.5 2 -0.5 (b) No. (c) 0 x+2 y = f(−x) = −2x + 2 0 -1 -1.5 if if if if x ≤ −2 −2 ≤ x ≤ 0 0≤x≤1 x≥1 y 4 (c) 3 2 1 10 -4 -3 -2 -1 5 -10 -8 -6 -4 -2 0 2 4 -5 0 −2x − 2 y = −f(x) x−2 0 -4 Answer 13.3 3 4 x if if if if x ≤ −1 −1 ≤ x ≤ 0 0≤x≤2 x≥2 y 1 -10 −1 3 2 -1 0 -12 1 ≤ x ≤ 0. -3 -2 -1 1 2 3 4 x -1 -2 Answer 13.4 (a5) (1) horizontally dilate (compress) by a 1 factor of 2; (2) horizontal shift right by 2 ; (3) vertical dilate (expand) by a factor of 3; vertical shift up by 5. f ( x) = −6x + 8 6x + 2 if x < if x ≥ 1 2 1 2 -3 -4 (d) Here are the graphs of y = 2f(x) and y = tively: 1 f ( x) , 2 respec- APPENDIX B. ANSWERS 298 y 4 Answer 13.10 y = 1 (2x ) is obtained by vertically dilating 3 y = 2x ; it is vertically compressed. y = 2x/3 is a horizontal x ; it is horizontally stretched. dilation of y = 2 3 2 Answer 14.1 (a) domain={x|x = 1}; range={y|y = 2}; zero at x = 0; horizontal asymptote y = 2; vertical asymptote x = 1; graph below: 1 -4 -3 -2 -1 1 2 3 4 x y 15 -1 10 y 4 5 -4 -2 3 -5 2 -3 -2 -10 1 -4 -15 -1 1 2 3 4 x -1 1 (e) Here are the graphs of y = f(2x) and y = f( 2 x), respectively: Answer 14.2 (a) y = ft. (c) 10 ft. Answer 14.5 f(x) = 2 asymptote is y = 1 -2 -1 (b) x = 9.143 ft.; x = 9.916 Answer 14.3 (a) m(t) = 35t + 200. (b) k(t) = 30t − 50. (c) 1987. (d) r(t) = 35t+200 . (e) 7 . 30t−50 6 3 -3 0.2x−10.4 . x−10 Answer 14.4 (a) y = 0.03x2 − 1.1x + 14 (b) w = 0.03x − 1.1 + 14 x (c) 70 or 6 2 3 y 4 -4 x 4 2 1 2 3 4 x 41 . 11 Answer 14.6 f(x) = 41 x 11 35 11 65 11 + x+ = 41x + 35 . The horizontal 11x + 65 6x + 10 x+1 -1 Answer 14.7 You should study for 11.25 hours. y 4 Answer 14.8 (a) k = 400 (b) I(t) = t = 1.5 (d) t = 1.05 and 1.95. 400 484t2 −1452t+1189 (c) 3 Answer 14.9 2 Answer 14.10 (a)y = f(x) = 1 (d)x = -4 -2 2 4 x -1 (f) c = 11 . 2 (g) c = 5 . 2 (h) c = d = 1 . 3 √ Answer 13.6 (a) a(x) = 2(6 − x) 3x − 9 with domain 3 ≤ √ x√ 6. (b) Max of a(x) is 4 3 and max of 2a(3x + 3) + 1 is ≤ √ 1 8 3 + 1 (when x = 3 ). (c) 1 + 36(1 − x) x on the domain √ 0 ≤ x ≤ 1. The range is all y values between 1 and 8 3 + 1. Answer 13.7 (c) Horizontally shift the graph of y = x2 to the right 4 units. 20000 = 200y−−y . 400 −1 = {0 ≤ x ≤ f 400x+20000 . x+200 Domain f−1 (b) x = $400. = {100 ≤ y ≤ 388.46}, Range 5000}. The inverse function takes the number of customers per day as an input value and gives the amount the shop spent on advertising as an output value. Answer 15.1 (a) 13o 24 ′ or 0.233874 rads. (b) 1.0788 degs or .01882 rads. (c) 5.7296 degs or 5o 43 ′ 46.5". Answer 15.2 (a) 6080 ft. (b) 29.95 mph. (c) 15.63 knots. Answer 15.3 (a) 430 sq. in. (b) 2.56◦ . (c) 84.47 in. (d) 23.04 in. (e) 7.29 in. Answer 15.4 (a) 1413.7 sq. ft. (c) 4.244 sec. Answer 15.5 2164.208272472 miles. Answer 13.8 (a) y = 7|2x + 4| + 2. Answer 13.9 (a1) f(2x) = 4x . f(2x − 1) = f−1 (y) 1x 4. 2 Answer 15.6 (a) 2.147 hrs. (b) 1103 mph. (c) 12.47 hrs., 6236 miles. (d) 13760 miles. 299 Answer 17.12 Top right scenario: (a) θ◦ = 1.2 rad. (b)θ(t) = 1.2 + 4πt (c) b(t) = (2 cos(1.2 + 4πt ),2 sin(1.2 + 4πt )). 9 9 9 (d) b(1) = (−1.710,1.037). b(0) = (0.725,1.864). b(3) = (1.252, − 1.560). b(22) = (1.753,0.962). Answer 15.7 (b) 84.47 sq. in. (c) 181.5 sq. in. Answer 15.8 0.685078 miles. Answer 15.9 Middle picture: shaded area= 12.537 sq. in. Answer 18.1 (b) sin2 (x) = Answer 16.1 (a) 10π/3 rad = 10.47 radians. (b) 4/2π rev * 1 hour/rev = .64 hours = 38.2 minutes. (c) Using (2.2.2), (21)(10π/3) = 219.91 meters. 1 (1 2 − cos(2x)). 2 1.5 1 Answer 16.2 194 RPM. 0.5 11π 30 Answer 16.3 (a) ω = rad/sec, v = v = 2.094 in/sec, ω = 2.5 RPM. 121π 15 ft/sec. (b) -6 -4 2 4 6 2 4 6 2 4 6 2 -2 4 6 -0.5 -1 Answer 16.4 (b) 700 ft. (c) 70 sec. (d) 15000 sq. ft. -1.5 -2 Answer 16.5 (a) r = 233.427 ft. (b) θ = 0.06283 rad. (c) 2π/5 rad counterclockwise from P. (c) Answer 16.6 (a) 40π ft/sec; 85.68 mph. (b) 400 RPM. (c) 60 RPM; 32π ft/sec. (d) 1.445 rad = 82.8◦ ; 28.9 ft. (e) 0.7 sec; 1.4π rad. 1.5 Answer 16.7 4.4 inches. 0.5 2 1 -6 Answer 16.8 r = 2.45 inches. -4 -2 -0.5 -1 Answer 17.1 (a) If you impose coordinates with the center of the wheel at (0,237.427), then ground level coincides with the x-axis. (a) T (t) = (x(t),y(t)), where x(t) = 233.427 cos( 2π t − 0.06283) and y(t) = 233.427 sin( 2π t − 5 5 0.06283) + 237.427. (b) T (6) = (85.93,454.45). (c) First nd the slope of a radial line from the wheel center out to Tiff’s launch point. -1.5 -2 Answer 18.2 (a) 2 Answer 17.2 (a) y = ±0.2309(x + 1) + 2 1.5 Answer 17.3 290 ft. 0.5 1 -6 Answer 17.4 (a) (-21.91218, -1.498834) (b) (-5.92564, 21.14892) (c) (19.07064, -10.89497) Answer 17.5 (a) (19.9,13.42). (−1.674,23.942). (d) (23.882,2.375). (b) (22.55,8.21). -4 -2 -0.5 -1 -1.5 (c) Answer 17.6 101.496936 feet above the ground. -2 (b) Answer 17.7 The dam is 383 feet high. 2 1.5 Answer 17.8 108 ft. 1 0.5 Answer 17.9 With the center of the track at the origin, and the northernmost point on the positive yaxis, Charlie’s location after one minue of running is (59.84016,4.37666). -6 -4 -2 -0.5 -1 -1.5 Answer 17.10 105.2718216 feet Answer 17.11 (a) 204.74 ft. (b) no. -2 (c) APPENDIX B. ANSWERS 300 x y 2 15 1 4 2 6 t 8 10 -1 -2 5 0 -6 -4 -2 0 2 4 Answer 18.3 (a) 7/25 or -7/25. (b) -0.6. (c) ± √ 40 . 7 Answer 18.4 In the first case, 6 Answer 19.7 (a)2 volts. (b) Never zero since 2x is positive 3 2π for all x. (c)p(t) = 3 sin( 2/5 (t − 5 ) + 1, so A = 3, D = 1, B = 2/5, C = 3/5. (d) 0.25 ≤ V (t) ≤ 16. (e)t = 0.65635 + k(0.4) and t = 0.74365 + k(0.4), k = 0, ± 1, ±2, . . . are ALL solutions. Four of these lie in the domain 0 ≤ t ≤ 1. (f) Maxima have coordinates (0.3 + k(0.4),16) and minima have coordinates (0.1 + k(0.4),0.25), where k = 0, ± 1, ±2, . . . . (g) If we restrict V (t) to 0.5 ≤ t ≤ 0.7, the inverse function has rule: t= y arcsin( ln(3 )−ln)(2) ) + 3π ln(2 5π . If we restrict V (t) to 0.5 + k(0.4) ≤ t ≤ 0.7 + k(0.4), the inverse function has rule: one period t = k(0.4) + arcsin( ln(y)−ln(2) ) 3 ln(2) + 3π . 5π In particular, if we restrict V (t) to 0.1 ≤ t ≤ 0.3, the inverse function has rule: Answer 18.5 π 4 + 2kπ and Answer 19.1 (a) 1, π, π , 2 5π 4 + 2kπ, k = 0, ±1, ±2, ±3, . . . . t = −0.4 + y arcsin( ln(3 )−ln)(2) ) + 3π ln(2 5π 1. (b) 6, 2, 0, -1. . If we restrict V (t) to 0.7 ≤ t ≤ 0.9, the inverse function has rule: ln(y)−ln(2) − arcsin( 3 ln(2) ) + 4π t= . 5π 2π Answer 19.2 (d) h(t) = 8 sin( 1.2 (t − 0.3)) + 18. 2π Answer 19.3 (a) b(t) = 0.6 sin( 100 (t − 30)) + 1.2. Answer 19.4 (a) h(t) = 5 sin( π (t −10)+5, where t indicates 6 hours after midnight. (b) 7.5 ft. above low tide. Answer 19.5 (a) A = 25, B = 5 seconds, C = 1.75, D = 28. (b) t=1.75 and 4.25 seconds 2π Answer 19.6 First scenario: x(t) = 2 sin( 4.5 (t − Cx )), π 9 2π where Cx = −(1.2 + 2 )( 4π ); y(t) = 2 sin( 4.5 (t − Cy )), where 9 Cy = −(1.2)( 4π ). Plots are below: If we restrict V (t) to 0.7 + k(0.4) ≤ t ≤ 0.9 + k(0.4), the inverse function has rule: t = k(0.4) + ln(y)−ln(2) ) 3 ln(2) + 4π . 5π In particular, if we restrict V (t) to 0.3 ≤ t ≤ 0.5, the inverse function has rule: t = −0.4 + x 2 − arcsin( y − arcsin( ln(3 )−ln)(2) ) + 4π ln(2 5π . Answer 19.8 (c) On domain t ≥ 0, 1 B = B(t) = 2 cos(6πt) + 2 4 6 8 36 − 4(sin(6πt))2 . t (e) 0.1023, 0.2310. -1 -2 Answer 20.1 (a1) 0, 1.5708, −1.5708, 1.0472, 0.7168, −0.2762, not defined. 301 Answer 20.2 (a) Principal solution: x = 1.9106, symmetry solution: x = 4.3726; graph below with these two solutions graphically indicated: y 2 1.5 1 0.5 -10 -5 5 10 Answer 20.9 (a)α = arcsin[3960/(3960 + t)]. (b) α = 6.696degs. The interior angle is 166.608 degs and so one satellite covers 46% of the circumference. Thus you need 3 (not two-point-something) satellites to cover the earth’s circumference. (c)α = 52.976deg. The interior angle is 74.047 degs and so one satellite covers 20.57% of the circumference. Thus you need 5 satellites to cover the earth’s circumference. (d) Get an equation for the interior angle in terms of t. Solve 2(90 − arcsin[3960/(3960 + t)]) = 20% of 360 degs = 72 degs. You’ll get t = 934.83 miles. x -0.5 -1 -1.5 Answer 20.10 (a) Domain: 1 ≤ y ≤ 5. One solution: x = 1 3 1 3 1 − π ≤ x ≤ 3 + π ; Range: 6 6 π + 18 ; graph is below: -2 8 Answer 20.3 (a) 9.39, 13.63, 11.09. (b) Feb. 3, Nov. 4. 6 Answer 20.4 (a) A = 15, D = 415, B = 10, C = 15/2. Note that C = 10k + 15/2, k = 0, ±1, ±2, . . . are also all valid choices for the phase shift. (b) maximum temperature= 430o F. (c) minimum temperature= 400o F. (d) 12.1635 minutes. (e) 14.6456 minutes. (f) 6.80907 minutes. Answer 20.5 The cake should be in the oven for 67.572141357 minutes. 4 2 -2 -1.5 -1 -0.5 0.5 1 1.5 2 x -2 Answer 20.6 6.42529 hours Answer 20.7 8.9286 hours of dry time each day. Answer 20.8 The key fact to use over and over is this: M(t)=M’s location after t seconds = (100 cos(0.025t),100 sin(0.025t)); T (t)= T’s location after t seconds = (100 cos(0.03t + π),100 sin(0.03t + π)). Answer 20.11 (a) 3.852624 miles (a) 14.34lb/in2 ; no explosion (b) 8.32lb/in2 ; explosion (c) 54.6 degrees Answer 20.12 (a) Many possible answers; for example: −1.9293, −1.3677, 0.8677, 1.4293. 302 APPENDIX B. ANSWERS Appendix C GNU Free Documentation License Version 1.1, March 2000 Copyright c 2000 Free Software Foundation, Inc. 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. Preamble The purpose of this License is to make a manual, textbook, or other written document “free” in the sense of freedom: to assure everyone the effective freedom to copy and redistribute it, with or without modifying it, either commercially or noncommercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modifications made by others. 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Index x-axis, 11 x-axis,positive, 11 xy-coordinate system, 11, 12 y-axis, 11 y-axis,positive, 12 y-intercept, 39 nth root, 135 adjacent, 222 amplitude, 253 analogue LP’s, 213 angle, 192 angle,central, 192 angle,initial side, 192 angle,standard position, 192 angle,terminal side, 192 angle,vertex, 192 angular speed, 207 arc,length, 199 arc,subtended, 192 arccosine function, 272 arcsine function, 272 arctangent function, 272 area,sector, 199 aspect ratio, 13 axis scaling, 13 axis units, 14 axis,horizontal, 11 axis,vertical, 11 belt/wheel problems, 215 CD’s, 215 central angle, 192 chord, 202 circle, 25 circles, 26, 77 circles,circular function, 230 circles,great, 204 circles,point coordinates on, 230 circles,unit, 28 circular function, 229 circular function,triangles, 229 circular function, 191, 221, 226, 227 circular function,circles, 230 circular function,inverse, 267 circular function,special values, 223 circular motion, 209 compound interest, 146, 148 compounding periods, 146 continuous compounding, 151 converting units, 1 coordinates,imposing, 11 cosecant function, 232 cosine function, 222, 248 cotangent function, 232 curves,intersecting, 28 db, 160 decibel, 160 decreasing function, 76 degree, 194 degree method, 193 degree,minute, 194 degree,second, 194 density, 3 dependent variable, 59 difference quotient, 36 digital compact disc, 215 dilation, 170 dilation,horizontal, 173 dilation,vertical, 171 directed distance, 19 distance, directed, 19 distance,between two points, 17, 19 311 INDEX 312 domain, 58 e, 149, 150 envelope of hearing, 160 equation,quadratic, 45 equatorial plane, 202 even function, 241 exponential decay, 139 exponential function, 149 exponential growth, 139 exponential modeling, 145 exponential type, 139 graphing, 1 great circle, 202, 204 horizon circle, 282 horizontal line, 26 horizontal axis, 11 identity,composition, 272 identity,even/odd, 241 identity,key, 240 identity,periodicity, 240 imposing coordinates, 11, 15 independent variable, 59 interest, 146 function, 58 function,circular, 191, 221, 226, 227, intersecting curves, 28 intervals, 60 229 inverse circular function, 267, 270, function,cosine, 222, 248 271 function,decreasing, 76 inverse function, 267 function,even, 241 function,exponential, 149 function,exponential type, 139 function,logarithm base b, 157 function,logarithmic, 153 function,multipart, 79 function,natural logarithm, 154 function,odd, 241 function,periodic, 240 function,picturing, 55, 57 function,rational, 181 function,sine, 222, 248 function,sinusoidal, 191, 247, 251 function,tangent, 222, 248 function,trigonometric, 247 function,cos(θ), 222 function,cos(x), 248 function,cos−1(z), 271 function,sin(θ), 222 function,sin(x), 248 function,sin−1 (z), 271 function,tan(theta), 222 function,tan(x), 248 function,tan−1(z), 271 graph, 1, 26, 27, 38 graph,circular function, 242 graph,sin(θ), 244 knot, 205 latitude, 202 line,horizontal, 26 line,vertical, 26 linear speed, 208 linear functions, 64 linear modeling, 33 lines, 33, 39 lines,horizontal, 25 lines,parallel, 44 lines,perpendicular, 44 lines,point slope formula, 38 lines,slope intercept formula, 39 lines,two point formula, 38 lines,vertical, 25 logarithm conversion formula, 158 logarithm function base b, 157 logarithmic function, 153 longitude, 203 loudness of sound, 159 LP’s, 213 mean, 252 meridian, 203 meridian,Greenwich, 203 modeling, 1 INDEX modeling,exponential, 145 modeling,linear, 33 modeling,sinusoidal, 251 motion,circular, 209 mulitpart function, 80 multipart functions, 79 natural logarithm, 153 natural logarithm function, 154 natural logarithm function, properties, 154 nautical mile, 205 navigation, 202 odd function, 241 origin, 11 parabola,three points determine, 98 parametric equations, 47 period, 253 periodic, 240, 247 periodic rate, 146 phase shift, 252 piano frequency range, 140 picturing a function, 55, 57 positive x-axis, 11 positive y-axis, 12 principal, 146 principal domain, 271 principal domain, cosine, 271 principal domain, sine, 271 principal domain, tangent, 271 principal solution, 270 Pythagorean Theorem, 18 quadrants, 13 quadratic formula, 45 radian, 198 radian method, 196 range, 59 rate, 4, 39 rate of change, 4 rational function, 181 reflection, 166 restricted domain, 59 right triangles, 229 313 RPM, 208 rules of exponents, 135 scaling, 13 secant function, 232 sector,area, 199 semicircles, 77 shifting, 168 shifting,principle, 170 sign plot, 75 sine function, 222, 248 sinusoidal function, 247 sinusoidal function, 191, 251 sinusoidal modeling, 251 slope, 36 solve the triangle, 267 sound pressure level, 159 speed,angular, 207 speed,circular, 209 speed,linear, 208 standard position, 192 standard angle, 192 standard form, 27 tangent function, 222, 248 triangle,sides, 221 trigonometric function, 247 trigonometric ratios, 223 uniform linear motion, 47 unit circle, 28 units, 1 vertical axis, 11 vertical line test, 63 vertical lines, 26 ...
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TB2011-12 - Precalculus Precalculus David H Collingwood...

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