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 August 30, 2007
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Copyright c 2003 David H. Collingwood and K. David Prince. 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 FrontCover, and with no BackCover 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 multistep 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 112; 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 multistep 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 experi
ence and confidence in working with a particular set of mathematical tools.
This text was created to achieve these goals and the 200405 academic year marks the eleventh year in which it has been used. Several thousand students have successfully passed through the course.
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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 several people have led to numerous 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, Don Marshall, Linda Martin, Patrick
vi Perkins, Lisa Peterson, Ken Plochinski, Eric Rimbey, Tim Roberts, Aaron Schlafly, David Schneider, Marilyn Stor, Lukas Svec, Sarah Swearinger, 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 preengineers. 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 is used to maintain and develop this textbook.
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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 AHa! 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 AHa! 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 multistep process such that the “AHa!” 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 welldeveloped 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 multistep 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 stepbystep 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
PSfrag replacements axis
Pictures
Symbols
axis axis
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 . The tube is filled with oil and piston moves to the right 2 in/sec, as indicated. This will cause piston to move to the right in/sec. Assume the oil does not compress; that means the volume of the oil between the two pistons is always the same. PSfrag replacements axis axis
oil
2 in/sec
axis
m in/sec B
A
1. If the diameter of the narrow part of the tube is inch, what is the speed of piston ? 2. If moves the tube?
in , sec
what is the diameter of the narrow part of
Solution. The first thing to do with any story problem is to draw a picture of the problem. In this case, you might resketch the picture so that it looks 3dimensional: 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
PSfrag replacements
Piston B. xA
axis axis
B
Piston A.
axis
Figure 2: A resketch of the original given figure.
1. Let and stand for the change in volumes as piston A moves to the right. 2. Let and represent the diameters of each cylinder. 3. Let and represent the radii of each cylinder. 4. Let and stand for the speeds of each piston. 5. Let and 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: 1. "!
inches . second
2. #$&% inch. After you have studied this problem for a while, you would write down some useful relationships: 1. The volume of any cylinder is
*
')(
*
where is the radius of the cylinder, and is its height or length. From this, you can derive the volume of a cylinder in terms of its diameter, :
'
(+
*
xiii 2. “Distance” write
“Rate” ,
“Time”. In terms of this problem, you would
.0/21 where is the distance your piston moves, and 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:
(4 + (4 7+ 6 8/:9 3 53
and
;<
(4 + =
(> + 6 2/:9 <
2. Since the oil does not compress, at each instant when piston A is moving, you must have 7?) , thus:
(4 + 6 @/:9 (4 + 6 2/29
After canceling ( , / , and 4, you end up with a mathematical model describing this problem that you can use to answer all sorts of interesting questions:
$ A
Using your model, you can compute the following solutions: 1. Given: ' in, B model, you derive:
$ D
&% in, and BC!
in , sec
F D E
from which you can compute
GHIHJ
in sec
2. Likewise, you can use your model to compute
K
D exactly.
! in 1
find . From your
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Contents 1
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Warming Up 1.1 Units and Rates . . . 1.2 Total Change = Rate , 1.3 The Modeling Process 1.4 Exercises . . . . . . .
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Imposing Coordinates 2.1 The Coordinate System . . . . . . . . . . . . . . . 2.1.1 Going from L 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 . . . . . . . . . . . . . . . . . . . . . . . Three Simple Curves 3.1 The Simplest Lines . . 3.2 Circles . . . . . . . . . 3.3 Intersecting Curves I 3.4 Exercises . . . . . . .
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Linear Modeling 4.1 The Earning Power Problem . . . . . . . 4.2 Relating Lines and Equations . . . . . . 4.3 Nonvertical 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 Exercises . . . . . . . . . . . . . . . . . . xv
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CONTENTS
xvi 5
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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 . . . . . . . . . . . . . . .
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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 . . . . . . . . . . . . . . . . . . . . . .
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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 Exercises . . . . . . . . . . . . . . . . . . . . . . . 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 . . . . . . . . . . . . . . . . .
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CONTENTS 9
Three Construction Tools 9.1 A Lowtech Exercise . . . 9.2 Reflection . . . . . . . . . 9.3 Shifting . . . . . . . . . . 9.4 Dilation . . . . . . . . . . 9.5 Vertex Form and Order of 9.6 Summary of Rules . . . . 9.7 Exercises . . . . . . . . .
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10 Arithmetic 10.1 Function Arithmetic . . . . . . . . 10.1.1 What about the domain? 10.2 Graphical Interpretation . . . . . 10.3 Step Functions . . . . . . . . . . . 10.3.1 Building Step Functions . 10.4 Exercises . . . . . . . . . . . . . .
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11 Inverse Functions 11.1 Concept of an Inverse Function . . . . . . . . . 11.1.1 An Example . . . . . . . . . . . . . . . . 11.1.2 A Second Example . . . . . . . . . . . . 11.1.3 A Third Example . . . . . . . . . . . . . 11.2 Graphical Idea of an Inverse . . . . . . . . . . . 11.2.1 Onetoone Functions . . . . . . . . . . 11.3 Inverse Functions . . . . . . . . . . . . . . . . . 11.3.1 Schematic Idea of an Inverse Function 11.3.2 Graphing Inverse Functions . . . . . . 11.4 Trying to Invert a Non onetoone Function . . 11.5 Exercises . . . . . . . . . . . . . . . . . . . . . .
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123 123 124 126 128 131 132 136
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149 149 150 151 151 152 154 155 156 157 157 160
12 Rational Functions 163 12.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13 Measuring an Angle 13.1 Standard and Central Angles . 13.2 An Analogy . . . . . . . . . . . 13.3 Degree Method . . . . . . . . . 13.4 Radian Method . . . . . . . . . 13.5 Areas of Wedges . . . . . . . . 13.5.1 Chord Approximation 13.6 Great Circle Navigation . . . . 13.7 Exercises . . . . . . . . . . . .
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CONTENTS
xviii 14 Measuring Circular Motion 14.1 Different ways to measure Cosmo’s speed . . . . . . . . 14.2 Different Ways to Measure Circular Motion . . . . . . . 14.2.1 Three Key Formulas 14.3 Music Listening Technology 14.4 Belt and Wheel Problems . . 14.5 Exercises . . . . . . . . . . .
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. 187 . 187 . 190 . 193 . 196
15 The Circular Functions 15.1 Sides and Angles of a Right Triangle . . . . . . . . . . 15.2 The Trigonometric Ratios . . . . . . . . . . . . . . . . . 15.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Circular Functions . . . . . . . . . . . . . . . . . . . . . 15.4.1 Are the trigonometric ratios functions? . . . . 15.4.2 Relating circular functions and right triangles 15.5 What About Other Circles? . . . . . . . . . . . . . . . . 15.6 Other Basic Circular Function . . . . . . . . . . . . . . 15.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
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199 199 200 202 204 205 207 208 209 212
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217 217 220 222 223 223 226 227 227 229 229 230
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233 233 236 239 241 244
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247 247 250 253 256
16 Trigonometric Functions 16.1 Easy Properties of Circular Functions . . . 16.2 Identities . . . . . . . . . . . . . . . . . . . . 16.3 Graphs of Circular Functions . . . . . . . . 16.3.1 A matter of scaling . . . . . . . . . . 16.3.2 The sine and cosine graphs . . . . . 16.3.3 The tangent graph . . . . . . . . . . 16.4 Trigonometric Functions . . . . . . . . . . . 16.4.1 A Transition . . . . . . . . . . . . . . 16.4.2 Graphs of trigonometric functions . 16.4.3 Notation for trigonometric functions 16.5 Exercises . . . . . . . . . . . . . . . . . . . .
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17 Sinusoidal Functions 17.1 A special class of functions . . . . . . . . . . . . . . 17.1.1 How to roughly sketch a sinusoidal graph 17.1.2 Functions not in standard sinusoidal form 17.2 Examples of sinusoidal behavior . . . . . . . . . . . 17.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 18 Inverse Circular Functions 18.1 Solving Three Equations . . . . . . . 18.2 Inverse Circular Functions . . . . . . 18.3 Applications . . . . . . . . . . . . . . . 18.4 How to solve trigonometric equations
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185
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CONTENTS
xix
18.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Exponential Functions 19.1 Functions of Exponential Type . . . . . . . 19.1.1 Reviewing the Rules of Exponents 19.2 The Functions MN' PORQTS . . . . . . . . . . . 19.2.1 The case QU$ . . . . . . . . . . . 19.2.2 The case Q VW . . . . . . . . . . . . 19.2.3 The case %YX)QZX[ . . . . . . . . . 19.3 Piano Frequency Range . . . . . . . . . . . 19.4 Exercises . . . . . . . . . . . . . . . . . . .
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20 Exponential Modeling 20.1 The Method of Compound Interest . . 20.1.1 Two Examples . . . . . . . . . . 20.1.2 Discrete Compounding . . . . 20.2 The Number \ and the Exponential Function . . . . . . . . 20.2.1 Calculator drill . . . . . . . . . 20.2.2 Back to the original problem... 20.3 Exercises . . . . . . . . . . . . . . . . .
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260 265 266 267 269 270 271 271 272 274
277 . . . . . . . . . . . 278 . . . . . . . . . . . 279 . . . . . . . . . . . 280 . . . .
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281 283 283 284
21 Logarithmic Functions 21.1 The Inverse Function of MN]\ S . . . 21.2 Alternate form for functions of exponential type . . . . 21.3 The Inverse Function of MN)QAS . . . 21.4 Measuring the Loudness of Sound 21.5 Exercises . . . . . . . . . . . . . . .
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22 Parametric Equations 22.1 Parametric Equations . . . . . . . . 22.2 Motivation: Keeping track of a bug 22.3 Examples of Parametrized Curves . 22.4 Function graphs . . . . . . . . . . . 22.4.1 A useful trick . . . . . . . . 22.5 Circular motion . . . . . . . . . . . 22.5.1 Standard circular motion . 22.5.2 General circular motion . . 22.6 Exercises . . . . . . . . . . . . . . .
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23 Linear Motion 315 23.1 Motion of a Bug . . . . . . . . . . . . . . . . . . . . . . . . 315 23.2 General Setup . . . . . . . . . . . . . . . . . . . . . . . . . 317 23.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
CONTENTS
xx Appendix
325
A
Useful Formulas
327
B
Answers
331
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 . . .
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347 . 348 . 349 . 349 . 350 . 352 . 352 . 352 . 353 . 353 . 353
References
355
Index
355
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 / seconds? PSfrag replacements We can attack these questions from two directions. If axis we knew some basic physics, then we would have equaPath of tossed ball. axis tions for the motion of the ball. Going from these equations axis 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 onemile 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) (ft)
(constant speed) , (elapsed time) (ft/sec) , (sec)
(1.1)
To apply the formula, let / 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 ////E ft 6 !^_!A9 6a` 1 !J%9 mile //// 6 !^c!A9 6d` 1_!J%9 ft
!^_! mile , 6a` 1 !J% ft/mile 9b Formula 1.1 can now be applied:
6 !^c!A9 6a` 1_!J%9 ft g
f/t 6 !^c!A9 d6 ` 1_!J%9 eJ h f/t/sec
i
1&^J ` cjj ef sec
eJ ft f sec ,U/ / i
1&^J ` kjj sec "/ i
So, the runner would complete the race in /Z 1&^J ` kjj seconds. If we wanted this answer in more sensible units, we would go through yet another units conversion:
/
i
g &1 ^i J ` kjj sec , 6 min/^% sec 94, 6 hr/^% min 9 1&^J ` cjj
hr + ^% h !Aaj J 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 1.7) 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 density
def
mass volume
For example, pure water has a density of g f cml . Notice, given any two of the quantities “density, volume or mass,” we can solve for the remaini ing unknown using the formula. For example, if J ` g of an unknown substance has a volume of 2.1 liters, then the density would be
mass volume i J ` g !Ag a Li g
g
g J ` , , L /h + !Aamh % %J gf cml
/ L n1&%%% cml h
Example 1.1.1. A sphere of solid gold has a mass of e%% kg and the density of gold is eokj gf cml . What is the radius of the sphere? Solution. This problem is more involved. To answer this, let be the unknown radius of the sphere in units of cm. The volume of the sphere is Hqp (l . Since the sphere is solid gold, the density of gold is the ratio
l
density of gold
mass of sphere volume of sphere
Plugging in what we know, we get the equation g eokj cml
e%% kg e %% kg r tr p ( l?s p (@ ls l l
Solving for l we find
l $0!nj^&o ``` e^ cml from which we get
i
i
+
i wyxz0!nj cml0{ } l $e% j ` cm
g
e%%% g e %vu g tr kg h p ( lTs l
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: rate
def
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: rate
def
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 time
6 value of quantity at final time 9A
6 value of quantity at initial time 9 6 final time 9T 6 initial time 9
For example, suppose that on June 4 we measure that the temperai ture at 8:00 am is ^ ` F and at 10:00 am it is 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
rate
final value of quantity initial value of quantity final time initial time i >^ ` degrees e%n
c%%Jn
c%% hours j deg/hr
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
rate
final value of quantity initial value of quantity final time initial time i ^ ` degrees e%n
_%%Jn
c%% hours j deg/hr
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 ^% mi/hr. The fact that the speedometer needle indicates a steady speed of ^% 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 nonconstant 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 cm /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 /HJ minutes, Total Surface Area
, Timeg + Rate g + J cm hr , ^ % h hr h
` ^ cm
The formula for the area of a circular region of radius is given at the back of this + text. Using this, the puddle has radius G at time /;.J minutes.
uR kjj ` cm
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 is the number of hours spent on precalculus a given week, it is certainly the case that:
# (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 pages of text to read during the week, then (hours reading text)
` hours ^%
As for homework, if a typical homework problem takes (on average) 25 * minutes and there are homework problems for the week, then (hours doing homework)
! ` * hours ^%
We now have a mathematical model for the weekly time commitment to precalculus:
! * Z ` ` ` hours ^% ^%
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) Which is faster: 100 mph or 150 ft/s? (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) Assume it takes 180 quarter credits to get a baccalaureate degree. If 1 quarter credit counts for one classroom hour of lecture each week of the quarter and you study 2.5 hours for each hour in class, how many hours must you invest to get a degree? (You may assume each quarter has 10 weeks of class and no holidays.) Problem 1.2. A typical scuba diving tank is initially filled with pressurized air at R lb/in . A good diver constantly monitors the pressure in their tank during a dive. (a) Suppose Cherie is using air at a rate of lb/in /min. Assuming she should surface her air pressure reaches R lb/inwhen , how long can she remain underwater?
(b) Dave plans a dive that will require minutes of underwater time. Assume he plans to surface when his air pressure reaches R lb/in . What is the maximum rate at which Dave can use his air, in units of “lb/in /min.” Problem 1.3. A prime number is an integer greater than one such that the only positive integers that divide it evenly (i.e., with remainder zero) are itself and 1. For example, , , , , R , , , are the first few prime numbers. Recently, someone discovered a prime number with zR RR digits. (a) If you were going to write down this number and each digit requires R mm, how long would the number be? (b) If you could speak at the rate of dig hours each day, how its/second for long would it take you to recite the number? (The exact number is ¢¡£¥¤¥¦§R¨A© , which your calculator will not like very much!)
Problem 1.4. Nonresident yearly tuition at the UW in Rª was $ 0 . Suppose nonresident yearly tuition at the UW in R was $ ¥ . (a) What is the rate of change in yearly nonresident tuition? (b) Assume the rate of change in yearly nonresident tuition is constant, then what would be the yearly nonresident tuition in the year R ? Problem 1.5. 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 1.6. The density of lead is 11.34 g/cm « and the density of aluminum is 2.69 g/cm « . Find the radius of lead and aluminum spheres each having a mass of 50 kg. Problem 1.7. 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? (b) Allyson has a pace of 6 min/mile; what is her speed? (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? Problem 1.8. Convert each of the following sentences into “pseudoequations.” For example, suppose you start with the sentence: “The cost of the book was more than $ and the cost of the magazine was $ .” A first step would be these “pseudoequations”: (Book cost) ¬ $10 and (Magazine cost) = $4. (a) John’s salary is $ªe RR pays no taxes.
a year and he
(b) John’s salary is at most $ª R a year and he pays § % of his salary in taxes.
1.4. EXERCISES (c) John’s salary is at least $ª R a year and he pays more than % of his salary in taxes. (d) The number of students taking Math §0 at the UW is somewhere between §0R and each year. (e) The cost of a new red Porsche is more than three times the cost of a new Ford F150 pickup truck.
9 Problem 1.11. During a typical evening in Seattle, Pagliacci receives phone orders for pizza delivery at a constant rate: orders in a typical minute period. How many pies are sold in hours? Assume Pagliacci starts taking orders at ;¸0 pm and the profit is a constant rate of $ on orders. When will phone order profit exceed $ R ?
(f) Each week, students spend at least two but no more than three hours studying for each credit hour.
Problem 1.12. In Jonathan Swift’s book, Gulliver’s Travels, there is the following passage:
(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.
The reader may please to observe, that, in the last article of the recovery of my liberty, the emperor stipulates to allow me a quantity of meat and drink sufficient for the support of 1728 Lilliputians. Some time after, asking a friend at court how they came to fix on that determinate number, he told me that his majesty’s mathematicians, having taken the height of my body by the help of a quadrant, and finding it to exceed theirs in the proportion of twelve to one, they concluded from the similarity of their bodies, that mine must contain at least 1728 of theirs, and consequently would require as much food as was necessary to support that number of Lilliputians. By which the reader may conceive an idea of the ingenuity of that people, as well as the prudent and exact economy of so great a prince.
(h) The difference between Cady’s high and low midterm scores was ¥ %. Her final exam score was %. (i) The vote tally for Gov. Tush was within onehundredth of one percent of onehalf the total number of votes cast. Problem 1.9. Which is a better deal: A 10 inch diameter pizza for $8 or a 15 inch diameter pizza for $16? Problem 1.10. The famous theory of relativity predicts that a lot of weird things will happen when you approach the speed of light ¡U;®I¥¯ m/sec. For example, here is a formula that relates the mass °I± (in kg) of an object at rest and its mass when it is moving at a speed ² :
°
¡
³ ° ± ´©Zµ· ¶ ¶
(a) Suppose the object moving is Dave, who has a mass of ° ± ¡ªRª kg at rest. What is Dave’s mass at z % of the speed of light? At R % of the speed of light? At Re % of the speed of light? (b) How fast should Dave be moving to have a mass of R kg?
Where does this number 1728 come from? Does it seem reasonable?
Problem 1.13. 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 “endtoend”; then the sum total length will be approximately meters.
CHAPTER 1. WARMING UP
10 isolate DNA from nucleus
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 ¿ be the number of trees in the orchard and À the yield per tree.
cell nucleus
lay out end−to−end PSfrag replacements axis axis axis
2m
Assume the human body has ¥¹¨»º cells containing DNA. How many times would the sum total length of DNA in your body wrap around the equator of the earth? Problem 1.14. 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 R cm /hour. (a) Find the area and radius of the puddle after minute, ¼ minutes, hours, day. (b) Is the radius of the puddle increasing at a constant rate? Problem 1.15. During the ¼ s, Seattle was dumping an average of 0 million gallons of sewage into Lake Washington each day. (a) How much sewage went into Lake Washington in a week? In a year? (b) In order to illustrate the amounts involved, imagine a rectangular prism whose base is the size of a football field ( ¥R yards ®½0 yards) with height ¾ yards. What are the dimensions of such a rectangular prism containing the sewage dumped into Lake Washington in a single day? (Note: There are gallons in one cubic foot. Dumping into Lake Washington has stopped; now it goes into the Puget Sound.) Problem 1.16. Dave has inherited an apple orchard on which 60 trees are planted. Under
(a) Find a formula for À in terms of the unknown ¿ . (Hint: Make a table of data with one column representing various values of ¿ and the other column the corresponding values of À . 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 1.17. Congress is debating a proposed law to reduce tax rates. If the current tax rate is Á %, then the proposed rate after ¿ years is given by this formula:
Á ´Â
´Â ¿
Rewrite this formula as a simple fraction. Use your formula to calculate the new tax rate after , , and 0 years. Would tax rates increase or decrease over time? Congress claims that this law would ultimately cut peoples’ tax rates by %. Do you believe this claim? Problem 1.18. (a) The temperature at 7:00 am is 44 Ã F and the temperature at 10:00 am is 50 Ã 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 Ã 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 Ã F and the temperature at 6:15 pm is 26 Ã 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?
1.4. EXERCISES Problem 1.19. (b) Solve for Å :
³
11
(a) Solve for Ä :
´Â Æ¨ ¡Ç
zÄ© ¡ZReÂÄ .
.
È ¿ ÂÅ ¡¹ÅmÂÉ¿ Solve for Ä : ´©ÉÄÊ¬ÉË©DRÄ .
is this:
(c) Solve for ¿ : (d)
Î be a positive integer (i.e. ÎÏ¡ R v ¹ 2 v ªe& ). A useful general formula
(b) Let
.
(e) Write as a single fraction:
© ¿ Ì ¿ ÂÍ Problem 1.20. zÂ ¨ , zÂ ¨
Â ¨ º
(c)
(a) Find the exact value of: , zÂ ¨ Â ¨ Â ¨ , zÂ ¨ Â ¨ Â ¨ Â ¨ .
º
¯
º
¯
¨}¦
´Â Â Â Â ÂÐÐÐÂ 0 Ñ ¡Ò© Ñ « º To five decimal places, what is ËÂ ¨ Â ¨ ¨ ¨ ¨ ¶ Â Ó Â Ô ÂÐÐÐ¼Â Õ ? Solve this inequality (Î is a positive integer):
´Â Â Â Â ÂÐÐÐÂ 0 Ñ ¬UR R « º
12
CHAPTER 1. WARMING UP
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 twodimensional 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 flowPSfrag of ideas underreplacements lying twodimensional coordinate systems in such a way that it easily generalizes to threedimensions. Suppose axis we start with a blank piece of paper and mark two points; axis Ö × let’s label these two points “ Ø ” and “ Ù .” This presents the axis 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 Ø in the plane, the plan is to use these two axes to obtain a pair of real numbers 6 @1 M39 that will give us the exact location of ØT With this in mind, the horizontal axis is often called the axis and the vertical axis is often called the M axis. Remember, a typical real number line (like the axis or the M 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 axis and M axis. Unless we say otherwise, we will always adopt the convention that the positive axis consists of those numbers to the right of the origin on 13
CHAPTER 2. IMPOSING COORDINATES
14
PSfrag replacements PSfrag replacements the axis and the positive M axis consists of those numbers above the axis axis origin on the M axis. We have just described the 2M coordinate system for axis axis the plane: axis
axis
Positive axis Positive axis Negative axis Negative axis
Ú
Positive axis Negative real numbers
Origin
Positive real numbers
Origin
Negative real Negative axis numbers
Positive axis
Zero Positive real numbers
Negative axis
S
Zero
(a) Number line.
S
Ú
(b)
¿Û
coordinate system.
Figure 2.2: Coordinates.
ements
2.1.1 Going from Ü to a Pair of Real Numbers.
axis axis axis
Ú
axis
Ö Ý
Ú
ÝßÞ S Figure
S
axis
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 Ø to a pair of real numbers: 1. First, draw two new lines passing through Ø , one parallel to the axis and the other parallel to the M axis; call these à and àRá , as pictured in Figure 2.3. 2. Notice that à will cross the M axis exactly once; the point on the M axis where these two lines cross will be called “M .” Likewise, the line à á will cross the axis exactly once; the point on the axis where these two lines cross will be called “ .”
3. If you begin with two different points, like Ø and Ù in Figure 2.1, you will see that the two pairs of points you obtain will be different; i.e., if Ù gives you the pair 6 ´á1 Mâá9 , then either H ã á or M]" ã Mâá . 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 Ø . 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 6 81 M39 . Locate the number on the axis and the number M on the M axis. Now draw two lines: a line à parallel to the axis passing through the number M on the M axis and a line à á parallel to the M axis passing
2.2. THREE FEATURES OF A COORDINATE SYSTEM
15
through the number on the axis. The two lines à and àá will intersect in exactly one point in the plane, call it Ø . 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 Ø in the M plane corresponds to a unique pair of real numbers 6 @1§M39 , where is a number on the horizontal axis and M is a number on the vertical M axis; for this reason, we commonly use the notation “ ØF 6 81 M39 ”PSfrag replacements 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.
axis
Ú
axis axis Second Quadrant
First Quadrant
Third Quadrant
Fourth Quadrant
A coordinate system involves scaling, labeling and units on each of the axes.
2.2.1 Scaling Sketch two M coordinate systems. In the first, make the scale on each axis the same. In the second, assume “one unit” on the axis has the same length as “two units” on the M axis. Plot the points 6 n1d9 , 6 Én1d9 , x Ip 1 { , x l 1Ì ä { , x 1Ë p { , x U 1 { , 6 %´1&%9 , x 17 { , x 1Ë p { , x l 1Ì ä { , x p 1 { , 6 n1å9 . u u u u u u u u u u u u u u u u Both pictures illustrate how the points lie on a parabola in the 2M coordinate system, but the aspect ratio has changed. The aspect ratio is defined by this fraction: def
S
axis
Figure 2.4: Quadrants in the ¿0Û plane.
2.2 Three Features of a Coordinate System
aspect ratio
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
. In problem solving, you will often need to make a rough assumption
axis
axis
axis
axis
CHAPTER 2. IMPOSING COORDINATES
16
æ O
O O Ú O ç O O p æ O u O O
axis
O u S
(a) Aspect ratio
¡Z .
axis
O
æ O
OO Oç Ú OO p æ O u O O
axis
O u S
(b) Aspect ratio
¡ ¨
axis
O
.
Figure 2.5: Coordinates.
about the relative axis scaling. This scaling will depend entirely on the 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 represent the week and the variable M represent the gross sales (in thousands of dollars) in week . We can then plot the points 6 @1 M9 in the 2M coordinate system; see Figure 2.6. Notice, the units on the two axes are very different: M axis units are “thousands of dollars” and 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.
17
Ú
axis (Thousands of Dollars)
1,400 1,200 1,000 800 600 400 200
S
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 M 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 18 shows four natural choices of 2M 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 and M coordinates of any point on the path of the ball will be nonnegative. Example 2.3.2. Michael and Aaron are running toward each other, beginning at opposite ends of a e%´1&%%% ft. airport runway, as pictured in Figure 2.8 on page 19. 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 è 6 %´1%9 (the origin) and Aaron is initially located at the point é 6 e%´1&%%%´1%9 . To find the coordinates of Michael after / seconds, we need to think about how distance and time are related. Since Michael is moving at the rate of ` ft/second, then after one second he is located ` feet right of the origin; i.e., at the point 6 ` 1%9 . After ! seconds, Michael has moved an additional ` feet, for a total of j% feet; so he is located at the point 6 j%´1%9 , etc. Conclude Michael has traveled ` / ft. to the right after / seconds; i.e., his location is the point
CHAPTER 2. IMPOSING COORDINATES
18
Ú
Ú
axis
PSfrag replacements
Path of tossed ball.
PSfrag replacements
axis
S
axis axis
axis Path of tossed ball.
axis axis
axis axis
Cliff.
Cliff.
S
(b) Origin at the bottom of the ledge.
(a) Origin at the top of the ledge.
PSfrag replacements
Ú
Path of tossed ball.
Ú
axis
PSfrag replacements
axis
axis
axis
axis
axis
axis Path of tossed ball.
S
axis Cliff.
axis
axis
S
Cliff. axis
(c) Origin at the landing point.
(d) Origin at the launch point.
Figure 2.7: Choices when imposing an
¿Û
coordinate system.
è 6 /:9Ò 6 ` /21%9 . Similarly, Aaron is located 8 ft. left of his starting location after 1 second (at the point 6 o´1&oov!1%9 ), etc. Conclude Aaron has traveled J:/ ft. to the left after / seconds; i.e., his location is the point 6 /:9D 6 e%´1k%%%J:/21%9
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 coordinates of è 6 /:9 and 6 /:9 agree. This is a straight forward algebra problem:
` / !nj/ /
e%´1&%%%êJ:/ + e%´+ 1&%%% i j J
(2.1)
To the nearest tenth + of+ a second, the runners collide after 434.8 i + + + + seconds. Plugging /. j J into either expression for the position: è 6 j &J98 6 `86 j J9z1%9 6 ^´1 ` !!1%9 .
Sfrag replacements axis
axis
axis axis
axis
axis axis
axis
2.4.axis DISTANCE
Michael: Aaron: axis
ft sec ft sec
Michael: Aaron:
Michael: ft
u
ft sec
Aaron:
ç
ft sec
19
ft sec ft sec
ft
Ú
axis axis axis
ORë O¥O¥O
î
Sfrag replacements axis axis
axis
ìPîcò íïî ò
(b) The
NOT TO SCALE!
Michael starts here.
ORë ORð
ð ò u ëóORð ì
after
seconds.
îcò íïî
ð
¿Û
S axis ORë O¥O¥ORëñOð
Oë ORð
ft
(a) The physical picture.
Ú
íïî
ìNíïî
axis
coordinate picture.
ò
Aaron starts here.
ORòë O¥O¥O æ ç ëåORð after
seconds.
S
axis
axis Michael: Aaron:
ft sec ft sec
ft axis
(c) Building a visual model
Michael: 10 mph.
Aaron: 8 mph.
axis
6,522 feet to collision point.
(d) Michael and Aaron’s collision point. Figure 2.8: Michael and Aaron running headon.
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 Ø) 6 1 M 9 and Ùô 6 1 M 9 in the 2M 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 Ø (Denver) and flying to the location Ù (New York) along a straight line
axis
axis
axis
axis
CHAPTER 2. IMPOSING COORDINATES
20
??
segment; see Figure 2.9(a). Now ask yourself this question: To what Begin (start) here. overall extent have the and M coordinates changed?
Ú
End (stop) here.
axis
??
×
??
Ö
End (stop) here.
S
axis
SRö ë Ú ö ð
Ú ö
ø í î P
÷ Ú Ú õ
Begin (start) here.
Beginning point. Ending point.
Ú
íPî
Ö
Seõ ë Ú õ ð
Beginning point.
axis
Seõ ÷
??
(a) Starting and stopping points.
×
S
(b) Coordinates for
Figure 2.9: The meaning of
ûA¿
and
ûÛ
Szö ù
Ending point.
S
and
axis
ú
.
.
To answer this, we introduce visual and notational aides into this figure. We have inserted an “arrow” pointing from the starting position Ø to the ending position Ù ; see Figure 2.9(b). To simplify things, introduce the notation 3 to keep track of the change in the coordinate and 4M to keep track of the change in the M coordinate, as we move from Ø to Ù . Each of these quantities can now be computed:
3
change in coordinate going from Ø to Ù (2.2) ( coord of ending point) ( coord of beginning point)
ü
4M
change in M coordinate going from Ø to Ù (M coord of ending point) (M coord of beginning point)
M NM
We can interpret 3 and 4M using the right triangle in Figure 2.9(b). This means we can use the Pythagorean Theorem to write:
0ý 6 39 6 4M39
that is,
Çÿþ 6 39 6 4M39 1
which tells us the distance from Ø to Ù . In other words, is the distance + + we would fly if we had flown along that line segment connecting the two + As an example, if Ø 6 n19 and Ù 6a` 1 9 , then 7 ` ? , points. 4M 'w"j and Ç ` .
axis axis
2.4. DISTANCE
21
Ú
There is a subtle idea behind the way we defined 3 and 4M : 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 3 and 4M will both be negative and the lengths of the sides of the right triangle are computed by taking the absolute value of 3 and >M . As far as a distance calculation is concerned, the previous formula still works because of this algebra equality:
÷ Ú
axis
Ú õ
Ö ø
S õ ëÚ õ ð
Beginning point.
íPî
Ú ö
SRö ë Ú ö ð Sõ S
Ending point.
× Szö ÷ S
þ 6 3@9 6 4M39 þ 6 7 9 6 4 M 9
íPî
axis
Figure 2.10: A different direction.
We will sometimes refer to 3 and 4M as directed distances in the and M 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 Ø" 6a` 1 9 and Ù 6 n19 , then 3ô; ` , 4M] Hj and  ` Important Fact 2.4.2 (Distance formula). If ØU 6 1§M 9 and Ù$ 6 1§M 9 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
þ 6 3@9 þ 6 ü
6 4M39
9 6 M ½M 9
(2.3)
If your algebra is a little rusty, a very common mistake may crop up when you are using the distance formula. For example,PSfrag replacements
?
?
axis
axis
Notice, you have an impossible situation:
axis
is never equal to .
!!!
þ
CAUTION !!!
CHAPTER 2. IMPOSING COORDINATES
22
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 j% ft/sec. Find the distance (in miles) between the two cars after hour 0! minutes. In addition, determine when the two cars would be exactly mile apart.
ements axis
axis axis
$" %&!
Solution. Begin with a picture of the situation. We have indicated the locations of the two vehicles after / seconds and the distance between them at time / . By the distance formula, the distance between them is
axis (North)
#
þ 6(' %9 6 %wQÌ9
þ ' Q
axis (East)
This formula is a first step; the difficulty is that we have traded the mystery distance for two new unknown numbers ' and Q . To find the coordinate ' for the Eastbound car, we know the car is moving at the rate of j% ft/sec, so it will travel j%:/ feet after / seconds; i.e., ' "j%:/2 Similarly, we find that Q"Bj%:/: Substituting into the formula for we arrive at
"! Figure 2.11: Two departing cars.
þ 6 j%:/:9 6 j%:/29
þ !n/ 6 j%9 j%:*/ ) !A First, we need to convert hour and 0! minutes into seconds so that
our formula can be used: 1 hr 12 min
1 + 12/60 hr 1.2 hr g 60 min 6 _! hr 9 hr h 4,320 sec
g
60 sec min h
+ 1cj2!% sec and recalling that mile = ` 1_!J% feet, we arrive Substituting /< at
0!o´1&^%%+) ! feet
183,282 feet 34.71 miles.
For the second question, we specify the distance being mile and want to find when this occurs. The idea is to set equal to mile and solve for / . However, we need to be careful, since the units for are feet:
j%:/ ) !
` 1 !J%
2.4. DISTANCE
23
Solving for / :
` 1_!J% j% ) !
/
124.45 seconds 2 minutes 4 seconds
The two cars will be mile apart in ! minutes,
+ seconds.
CHAPTER 2. IMPOSING COORDINATES
24
2.5 Exercises Problem 2.1. In the following four cases, let ù be the initial (starting) point and ú the ending point; recall Equation 2.2 and Figure 2.10 on Page 21. Compute , = the distance from ù to ú , ûA¿ and û´Û . Give your answer in exact form; eg. È is an exact answer, whereas 1.41 is an approximation of È . (a) (b) (c) (d)
ù ¡.ñ¹ 0/}úN¡1 R&2/ . ù . ¡ :&" /}úN1 ¡  R©?2/ . ù . ¡ ó©>R & / ¥ú½. ¡ :©?" / . ù3 ¡ »R & /}ú'4 ¡ »@ÂÇzÄ mÂÄ2/ , where Ä
constant.
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 Kingston
is a
Problem 2.2. Start with two points 5B¡6»Å 78/ and Àü¡9:0 Ä2/ in the ¿0Û coordinate system. Let PSfrag , be the distance between these tworeplacements points. Answer these questions and make sure you axis axis can justify your answers:
,;¡<; »Å4©=:2/ Â>?77©YÄ2/ TRUE or FALSE: ,;¡<; »Å4©=:2/ Â>ÄT©@78/ , ¡ ; A:T©Å/ Â>ÄT©@78/ TRUE or FALSE: ; Suppose 5 is the beginning point and À
(a) TRUE or FALSE:
.
(b)
.
(c) (d)
.
is the ending point; recall Equation 2.2 and Figure 2.10 on Page 21. What is ûA¿ ? What is û´Û ?
(e) Suppose À is the beginning point and 5 is the ending point; recall Equation 2.2 and Figure 2.10 on Page 21. What is ûA¿ ? What is û´Û ?
(f) If ûA¿ =0, what can you say about the relationship between the positions of the two points 5 and À ? If û´Û =0, what can you say about the relationship between the positions of the two points 5 and À ? (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?
Edmonds
North Ballard UDub
axis
(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
sec sec Ä
Ferry
Erik
Distance Between
min 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
25 Brooke
(a) Find an expression for the distance between the two cars after Ä seconds.
ocean
(b) Find the distance in miles between the two cars after 3 hours 47 minutes.
5 mi
kayak reaches shore here
(c) When are the two cars 1 mile apart? PSfrag replacements Problem 2.6. Allyson and Adrienne 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 Adrienne moves 8 ft/sec in the directions indicated.
Building
axis axis axis
axis
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 either the point “B ”.
20 ft
rag replacements
axisshore axis
(c) Determine the total time to reach Kono’s if she paddles directly to Kono’s.
Allyson
30 ft Adrian
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”.)
(d) Do you think your answer to (b) or (c) is the minimum time required for Brooke to reach lunch? (e) Determine the total time to reach Kono’s if she paddles directly to a point on the shore half way between point “B ” 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)? Problem 2.8. Nikki and Laura are standing on opposite sides of a hive of killer bees. The bees can fly 7.5 m/sec in still air and the wind is blowing 4.5 m/sec in the indicated direction. Notice, if the bees are flying upwind (against the wind), then the wind speed will decrease their ground speed. Likewise, if the bees are flying downwind (with the wind), then the wind will increase their ground speed. Two bees leave the hive; one is heading for Nikki and the other toward Laura. Assume that both women start running away from the hive at the same instant. wind
nikki laura Problem 2.7. Brooke is locatedPSfrag 5 miles out replacements from the nearest point B along a straight hive axis shoreline in her seakayak. Hunger strikes and axis she wants to make it to Kono’s for lunch; see 20 m 26 m axis picture. Brooke can paddle 2 mph and walk 4 (a) What is the ground speed of each bee? mph.
CHAPTER 2. IMPOSING COORDINATES
26
200 ft
(b) Assume each person runs 3.1 m/sec. Determine who escapes and who doesn’t. If someone can’t escape, determine where and when attack will occur.
hang glider 210 ft
gliderport
(c) Determine the minimum speed each must run in order to avoid being stung. Is it realistic for Nikki to outrun the bees? How about Laura? PSfrag replacements
500 feet 250 ft boat
axis axis axis
Problem 2.9. Using an ¿Û coordinate system, plot the points B"¡4»R"/ , CF¡4 « "/ , DZ¡4 v"/ , E ¡1ó©>R¼©Í"/ , F ¡1å©4R¼© ¨ / , and Gm¡1å©4 0/ .
(a) Use the distance formula to find: (a1) The distance from (a2) The distance from (a3) The distance from
B
to
C
E
to G .
(b) For each number Ä , define a point ùH Ä2/ by the formula ùH Ä2/¡3 8ÂÄ&2/ . Plot the points ùHó0/ , ùHó I/ , ùJ»"/ , ùJó©>"/ , ùHI/ . (c) Likewise, for fine a point
each
number
ú  Ä2/ by the K úK Ä2/¡.ó©>R© ´ÂYÄ2/ . Plot the úKó0/ ¥úLó &/ ¥úL " / , K ú å©4"/}úKI/ .
75 ft Ocean level
300 ft
(a) What are the coordinates of the hang glider? (b) What are the coordinates of the seagull? (c) What are the coordinates of the boat? (d) What are the coordinates of the gliderport?
.
to F .
D
seagull
Ä , deformula points
(d) Use the distance formula to compute the distance between ùJÄ2/ and úKÄ2/ . Your formula will involve Ä . (e) Find a value of Ä so that the distance between ùHÄ2/ and úL Ä2/ is 6. Where are the two points located for this value of Ä ?
Problem 2.10. 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. Impose three different coordinate systems: one with origin at the gliderport, one with the origin at the hang glider and the third with origin at the boat location. Answer these questions for each coordinate system separately.
Problem 2.11. Plot the following four points in the ¿Û plane: BZ¡1 R &/ , CÉ¡1» ©"/ , DÉ¡1å©@: ©"/ , E ¡4å©4R"/ ; each axis will be scaled the same and the units will be feet. Form the region E E bounded by the line segments BMC CND D B . A bug begins at point B and starts moving around the perimeter of the region in a clockwise direction inches/second; i.e., the bug E walks from B to C to D to to B , etc. Also, assume the bug doesn’t slow down at the corners. (a) Sketch the region. (b) Find the perimeter of the region. (c) How long does it take the bug to complete one trip around the region? (d) Where is the bug located after 8 seconds? (e) When will the bug first cross the ¿ axis?
(f) When will the bug first cross the Û axis?
(g) During the bug’s first trip around the reE gion, when is it between D and ? Problem 2.12. A spider is located at the position  R &/ in a coordinate system, where the units on each axis are feet. An ant is located at the position  §v 0/ in the same coordinate system. Assume the location of the spider after Ä minutes is :Ä2/=¡O ´ÂÉRÄ ËÂÉÄ2/ and the location of the ant after Ä minutes is ÅÄ2/¡1 §Ê©ÉRÄ RÄ2/ .
2.5. EXERCISES (a) Sketch a picture of the situation, indicating the locations of the spider and ant at times Ä¡[R : ¹ : minutes. Label the locations of the bugs in your picture, using the notation :ñ*/ , :»"/ ,..., :P &/ , ÅQó0/ , ÅQ»2/ , ..., ÅQ &/ .
(b) When will the ¿ coordinate of the spider equal ? When will the Û coordinate of the ant equal ?
27 (a)
¿¹©Í ¿ÌÂ ¡ ¡ (b)
&¿4ÂïÛH/ ©S&¿Ì©PÛN/
(c) Where is the spider located when its Û coordinate is ? (d) Where is each bug located when the Û coordinate of the spider is twice as large as the Û coordinate of the ant?
(g) Find the speed of each bug along its line of motion; which bug is moving faster? Problem 2.13. 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 ª¼z feet north of the intersection of Broadway and Aloha, at the same time that the Mercedes is R feet east of the intersection. Assume the Mercedes is traveling at the constant speed of z miles/hour. Find the speed of the Ferrari so that a collision occurs in the intersection of Broadway and Aloha. Problem 2.14. 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 2.15. 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).
¡ ¡
&¿ ÂïÛ /¹©¢¿ ©ïÛ
(c)
*&¿4©¢I/d ¿m©Í§
(e) How far apart are the bugs when their ¿ coordinates coincide? Draw a picture, indicating the locations of each bug when their ¿ coordinates coincide. (f) A sugar cube is located at the position óe ªR/ . Explain why each bug will pass through the position of the sugar cube. Which bug reaches the sugar cube first?
¿@Â>å©42/ ¿ÌÂ? ¿ Â ©> ¿ ¿Ì©Í ¡
¡ ¡ ¡
Problem 2.16. Assume stants. Solve for ¿ . (a)
U:¿ÌÂ>Vw¡ X2Y[¨ Z \
(b)
X¨ Â \¨ ¡ Y ¨
(c)
U3Â \ ¨ ¡ Y ¨
Problem 2.17.
&¿m©¢T/å ¿4©?§  ¿4©?§I / z¿Ì©?§ ¿4©?§v U´WV
(a) Find
û
are nonzero con
if
¿¹ ¿ Z ¡ ¿*] (b) Find
û
if
 È I/^H¶_ /¡ ] (c) Factor
Å¹©`7
.
(d) Expand and simplify
»Å4ÂY278/a»Å4©É"78/ .
(e) Simplify as much as possible:
Åm©@7 Æ¨ © b ¨ Problem 2.18. Simplify as far as possible. (a) (b) (c) (d)
»´©DÄ2/ Âc ÊÂYRÄ2/ Ä´ÂÍ2å/ @Âcå©@Ä´©?"/d8©Y d Z ¨ © dfe ¨ (write as a single fraction)
¨ ¨ ;  ÊÂÉÄ2/ ÂÉÄ
28
CHAPTER 2. IMPOSING COORDINATES
Chapter 3 Three Simple Curves Before we discuss graphing, we first want to become ac axis quainted with the sorts of pictures that will arise. This is A typical curve. replacements surprisingly easy to accomplish: ImposePSfrag an M coordinate system on a blank sheet of paper. Take a sharp pencil andaxis axis begin moving it around on the paper. The resulting picaxis ture is what we will call a curve. For example, here is aaxis 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 and M ). We consider the three simplest situations in PSfrag this chapter: horizontal lines, vertical lines andreplacements circles.
3.1 The Simplest Lines
axis
Undoubtedly, the simplest curves in the plane are the horizontal and vertical lines. For example, sketch a line parallel to the axis passing through 2 on the M axis; the result is a horizontal line à , as pictured. This means the line à passes through the point 6 %´1!A9 in our coordinate system. A concise symbolic prescription for ALL of the points on à can be given using “set notation”:
Ú
axis
gihkjmlonqprgtsWlonWp g$nulonWp gtv[lnWp , a typical point.
Ý
Figure 3.2: The points
à>xw 6 81!A9 is any real numbery We read the righthand side of this expression as “the set of all points 6 81!A9 where N any real number.” Notice, the points 6 @1§M39 on the line à are EXACTLY the ones that lead to solutions of the equation M"[! ; i.e., take any point on this line, plug the coordinates into the equation Mê.! 29
S
axis
&¿dÛH/ .
CHAPTER 3. THREE SIMPLE CURVES
30
ments
axis
Ú
axis
z
and you get a true statement. Because the equation does not involve the variable and only constrains M to equal 2, we see that can take on any real value. In short, we see that plotting all of the solutions 6 @1§M39 to the equation MN"! gives the line à . We usually refer to the set of all solutions of the equation MN"! as the graph of the equation M½"! . As a second example, sketch the vertical line passing through j on the axis; this means the line passes through the point 6 jA1%9 in our coordinate system. A con{tu}?~R , a typical point. cise symbolic prescription for ALL of the points on can be given using “set notation”:
{tu}oq
{tu} {tu}A
Gw 6 Aj 1§M39 M is any real numbery¼ Notice, the points 6 @1§M39 on the line
are EXACTLY ö the ones that lead to solutions of the equation .j ; i.e., take any point on this line, plug the coordinates into the equation Ç"j and you get a true statement. Because the Figure 3.3: Stacked points. equation does not involve the variable M and only specifies that 5j , M can take on any real number value. Plotting all of the solutions 6 81§M9 to the equation 'j gives the line . We usually refer to the set of all solutions of the equation )j as the graph of the equation Ç)j . 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 6 @1§M39 of the equation in the 2M coordinate system. These observations work for any horizontal or vertical line.
S
axis
Definition 3.1.1. Horizontal and Vertical Lines: A horizontal line à passing through on the M axis is precisely a plot of all solutions 6 81§M9 of the equation* Mÿ
; i.e., à is the graph of Mÿ
. A vertical line passing through * on the axis is precisely a* plot of all solutions 6 @1§M39 of the equation Ç ; i.e., is the graph of Ç .
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 and M . As a special case of the distance formula (2.3), suppose ØW 6 %´1%9 is the origin and Ù[ 6 @1§M39 is any point in the plane; then
þ 6 Y%9 6 M ê%9
þ M This calculation tells us that a point 6 @1§M39 is of distance from
the origin
. This if and only if þ M or, squaring each side, that M [ distance from Ø to Ù
shows
w 6 @1§M39 distance 6 81§M39 to origin is yÊxw 6 @1§M39 M " y¼
(3.1)
PSfrag replacements
3.2. CIRCLES
31
axis axis
Pencil. What is the lefthand side of Equation 3.1? To picaxis ture all points in the plane of distance from the origin, fasten a pencil to one end of a nonelastic string (a string that will not stretch) of length 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 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 righthand side of Equation
3.1?
A point 6 81§M9 in the righthand set is a solution to the equation NM ) ; i.e., if we plug in the coordinates we get a true statement. For example, in Figure 3.5 we PSfrag replacements 1 æ I , plot eight solutions 6 1%9 , 6 ;1%9 , 6 %´1n9 , 6 %´1z;n9 , $ 1 T , Naxis
axis point is æ 1 æ I , Ï æ 1 I , of the equation. To see that the last
axis æ a solution, here is the sample calculation: I & ö ö
Since the two sides of Equation 3.1 are equal, drawing the circle of radius is the same
as plotting all of the solutions of the equation ?M " . The same reasoning can be used to show that drawing a circle of radius centered * at a point 6 1W89 is the* same as plotting
all of the solutions of the equation: 6 < 9 6 M89 " We usually refer to the set of all solutions of the equation as the graph of the equation.
{tW}o^
{tm}Aq
{i*m}q Figure
{tW}A*^ 3.5: Computing points.
î
*
Definition 3.2.1 (Circles). Let 6 1W89 be a given point in the PSfrag replacements M plane and IV'% a given * positive real number. The circle of radius centered at 6 1W89 is precisely all of the solutions 6 81 M39 of the equation
ý * 6 ¢ 9 6 M 89 "
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.
axis
Ú
axis
î
S ëÚ ð
ëeð
S
axis
Figure 3.6: Defining a circle.
CHAPTER 3. THREE SIMPLE CURVES
32
Be very careful with the minus signs “ ” in the standard form for a circle equation. For example, the equation
ements axis axis
!!!
axis
CAUTION !!!
^
J
is NOT in standard form. We can rewrite it in standard form:
0
^
J
*
so, this equation describes a circle of radius
centered at ¡0J .
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 'M . This circle is called the unit circle and will be used extensively.
*
2. A circle of radius 3 centered at the point 6 1W89Ò 6 n1zÉ9 is the
graph of the
ý equation 6 89 6 M7 6 É99 )j ; or, i equivalently 6 @9 6 M79 j or, equivalently M !¢ !vM 3. The circle of radius ) ` centered at 6 !1zjT9 does not pass through the
origin; this is because 6 %´1%9 is not a solution of the equation 6 <½!A9 6 M üjT9 ` .
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.
placements
axis
PSfrag replacements 3.3. INTERSECTING axis CURVES I
Ú
axis
í
axis
î
axis Vertical Horizontal line: line: . . Horizontal line: axis. Vertical Point: . line: .
Ú
í
ëeð
Point: .
S
33
S
(b) Possible intersections.
(a) Line equations.
Figure 3.7: Circles and lines.
Example 3.3.1. GloTek Industries has designed a new halogen street light fixture for the city of Seattle. According to the product literature, when placed on a ` % foot light pole, the resulting useful illuminated area is a + 0!% feet in diameter. Assume the light pole is located !% feet circular disc east and % 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?
PSfrag replacements 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 coordinateaxis system; we will use units of feet for each axis. The illumi+ region will be a circular disc centered at the point nated 6 !%´1 %9 in the coordinate system; the radius of the disc will be <H^% feet. We need to find the points of intersection Ø , Ù , ¤ , and ¥ of the circle with the axis and the M * axis. The+ equation for the circle with '^% and center 6 1W89Ê 6 !%´1 %9 is
+
6 ¢ !%9 6 M
Ú
axis (Parkside)
Ö
S
×
3.8: Illuminated street.
To find the circular disc intersection with the M axis, we have a system of two equations to work with:
6 ¢ !%9 6 M
+
%9
jA1&^%% %
axis (Wilson)
¢ £
Figure
%9 "j^%%
Illuminated zone.
ý
To find the intersection points we simultaneously solve both equations. To do this, we replace W % in the first equation (i.e., we impose the
CHAPTER 3. THREE SIMPLE CURVES
34
conditions of the second equation on the first equation) and arrive at +
6 % !+ %9 6 M + %9 %%w 6 M + %9 6 M + %9 6 M %9b M
ý jA1&^%% ý jA1&^%% ý jA1_!%% + ¦ ) jA1 !%% % ¦ ) jA1_!%% i i Ée^ ` or o^ `
Notice, we have two solutions. This means that the circle and M axis i i intersect at the points ØU 6 %´1&o^ ` 9 and Ù 6 %´1'e^ ` 9 . Similarly, to find the circular disc intersection with the axis, we have a system of two equations to work with: +
6 Pü!%9 6 MU
%9
M
jA1&^%% %
ý
Replace M q% in the first equation (i.e., we impose the conditions of the second equation on + the first equation) and arrive at
6 Pü!%9 6 % % 9 6 ¢ !%9 9 6 ¢ !%b
ý Aj 1&^%% ý !1&%%% ý ¦@) !1&%%% !L % +¦ ) !1&%%% + i i ;! ! or ^ A!
+ the circle and axis intersect at the points ¥ê Conclude i ¤U 6 ;! !1k%9 .
+
i 6 ^ ! 1k%9 and
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. EXERCISES
35
3.4 Exercises Problem 3.1. This exercise emphasizes the “mechanical aspects” of circles and their equations.
(b) Each of these equations can be put in the ”standard form” above; find the appropriate ” Å ” and ”7 ” to do this:
(a) Find an equation whose graph is a circle of radius centered at ó©Ò¹ I/ . (b) Find an equation whose graph is a circle of diameter ¨ centered at the point  ¹ © ¨¨ / .
«
(c) Find four different equations whose graphs are circles of radius through »R"/ .
(d) Consider the equation &¿T©ï2/dÂ§cÛËÂï"/d>¡ . Which of the following points lie on the graph of this equation: »R"/ , »©Í2/ , »R©É*/ , »´Â È v 0/}Wñ¹¼©?´© È I/ , ñ 0/ .
(i) (ii) (iii) (iv)
¿¹Â¢ÛÌ¡ ¿¹ÂÉÛm¡ ¿¹Â¥Ûm¡ ¿¹ÂD¼Ûm¡Ç
(c) Find where the horizontal line ÛD¡Z and vertical line ¿Ç¡©> intersect the ellipse in (a). 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.
Problem 3.2. Water is flowing from a major broken water main at the intersection of two streets. The resulting puddle of water is cir60 feet cular and the radius Á of the puddle is given by the equation ÁY¡]RÄ feet, where Ä represents time in minutes elapsed since the the operator main broke. PSfrag replacements (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 12 miles per hour. When will the runner’s feet get wet? (b) Suppose, instead, that when the main broke, the runner was 6 miles east, and 500 feet north of the intersection. The runner runs due west at 12 miles per hour. When will the runner’s feet get wet? Problem 3.3. The equation
¨ ¿
Â ¨ Û ¡Z Å © ª 7 ©
defines an ellipse; i.e., the graph of all points ¿n ÛH/ satisfying the equation will be an ellipse in the ¿0Û plane.
(a) If Å¡Ç and 7<¡ , sketch the ellipse with the above equation; do this by plotting all of the points with ¿#¡. , « , «Ò or ÛY¡? , « , «Ò , «Ê .
axis axis axis
24 feet
rider
100 feet ground level 62 ft. tower
(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. 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 nonconformist he cuts off an edge as pictured. John then takes one of the remaining triangular slices. Who has more pizza and by how much?
CHAPTER 3. THREE SIMPLE CURVES
36 John’s part
(f) Find the area of GRASS watered after one hour.
PSfrag replacements axis axis
Lee’s part
Problem 3.7. 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. sailboat
axis
Kingston
Problem 3.6. A crawling tractor sprinkler is located as pictured below, ¥R feet South of a sidewalk. Once the water is turned on, the sprinkler waters a circular disc of radius feet and moves North along the hose at the rate of ¨ inch/second. The hose isPSfrag perpendicular 10 ft. wide sidewalk. Assume replacements to the there is axis grass on both sides of the sidewalk. axis
W
E S sidewalk
PSfrag replacements axis axis
North Ballard UDub
axis
N
hose
Edmonds
tractor sprinkler
axis
(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. (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.
(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) How would you determine where and when the ferry exits the radar zone? (e) How long does the ferry spend inside the radar zone?
3.4. EXERCISES
37
Problem 3.8. Draw the graphs of ¿;¡?©>RdÛÉ¡ and a circle of radius 4 centered at the point (2,3) in the ¿0Û coordinate system. Let ¬ be the point where the horizontal and vertical line intersect.
N W S
(c) At the same instant, a spider starts at the location (1,3) and moves upward along the line ¿¡ ©4 . Assume the position of the spider after Ä seconds is the point úKÄ2/¡Oó©>R©IÒÂYÄ2/ . Plot the spider locations úLó0/ ¥úK»"/}úK&I/}úKóª0/ (d) When will the ant exit the circular zone? When will the spider enter the circular zone? (e) Which bug reaches the point
¬
first?
(f) Use the distance formula to find the distance between the ant and the spider at time Ä seconds. Your answer will involve Ä .
(g) For times Ä>¡F: ¹ , draw a line segment connecting ùJÄ2/ and úKÄ2/ and compute its length; i.e., plot ùJ&/ and úKI/ and connect these by a line segment, etc. Problem 3.9. 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 westernmost edge of the field.
swath cut
combine
(a) Determine where the vertical line and the horizontal line intersect the circle; i.e., find the coordinates of the intersecPSfrag replacements tions. Label all points of intersection in axis your picture. (b) An ant starts at the location (6,2) and moves to the left along the line ÛÉ¡ . Assume the position of the ant after Ä seconds is the point ùHÄ2/Ì¡óªm©?RÄ &/ . Plot the ant locations ùHó0/ }ùH»2/ ùJ&/} ùJ&I/}
E
20 ft
a center wheat field
axis axis
(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? (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 3.10.
(a) Solve for ¿ :
¿¹©Éz¿ÌÂ ¡¿Ì©Y ¿mÂY (b) Solve for ¿ :
¿Ì©É ¡# ¿ÌÂY (c) If ¿¡?©8 , find ALL solutions of the equation
&¿mÂÍ"/ ÂScÛw©?"/ ¡#¥ (d) If Û?¡# , find ALL solutions of the equation
T&¿4Â?"/ ÂYTkÛw©Í"/ ¡Z¥
38
CHAPTER 3. THREE SIMPLE CURVES
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 fulltime 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 YEAR 1970 1987
WOMEN (dollars) $5,616 $18,531
YEAR 1970 1987
MEN (dollars) $9,521 $28,313
(a) Women. (b) Men. Table 4.1: Earning power data.
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 2M coordinate systems, where the axis has units of “year” and the M axis has units of “dollars;” see Figure 4.1. In each coordinate system, the data in our table gives us two points to plot: i In the case of women, the data table gives us the points Ø# 6 eo %´1 ` 1&^e^9 i and Ù 6 eoJ 1eJ´1 ` jT9 . Likewise, for the men, the data table gives us the i i points ¤U 6 eo %´1o´1 ` !A9 and ¥ 6 eoJ 12!J´1cjTjT9 . 39
CHAPTER 4. LINEAR MODELING
40 PSfrag replacements axis 30000 axis 25000 axis 20000
PSfrag replacements
Ú
axis (dollars)
´ ¯ {
axis 30000 axis 25000
} ²a
Ú
axis (dollars)
¸ ¯ {
}  W
õ°^¶m± ö ¶ õ
axis 20000
õµ°^¶m± õ¶ õ
15000
15000
10000
®8¯ {
5000
õ°^±
10000
W} ²(³ ³(
õ S
· ¯ {
axis (year)
5000
õµ°m±
W} ²(³ ³
õ
S
axis (year)
(a) Data points for women.
1995
1990
1985
1980
1975
1970
1995
1990
1985
1980
1975
1970
aca
(b) Data points for men.
Figure 4.1: Visualizing the data.
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 is $M , then the point 6 81§M9 lies on the line connecting Ø and Ù . Likewise, for men, if the earning power in year is $M , then the point 6 81§M9 lies on the line connecting ¤ and ¥ . In the real world, the validity of this kind of assumption would involve a lot of statistical analysis. This kind of assumption leads us to what is called a linear model, since we are demanding that the data points predicted by the model (i.e., the points 6 @1§M39 ) lie on a straight line in a coordinate system. Now that we have made this assumption, our job is to find a way to mathematically describe when a point 6 81§M39 lies on one of the two lines pictured in Figure 4.2. Our goal in the next subsection is to review the mathematics necessary to show that the lines in Figure 4.2 are the socalled graphs of Equations 4.1 and 4.2.
MJ¹»º z½¼f¾
g
M z¿¼À¾
eJ´1 ` jT4 ` 1&^e^ i i ¢Heo eoJ )eo %Bh 6 i 0!1&o ` i 6 Y'eo %9´ ` 1&^e^ g !J´1cjTjwo´1 ` !A i i ¢Heo eoJ )eo % h 6 i i eJ´1 ov! i 6 Y'eo %9´o´1 ` !A i
i
%9´ ` 1&^e^
(4.1)
%9´o´1 ` !A
(4.2)
4.2 Relating Lines and Equations A systematic approach to studying equations and their graphs would begin with the simple cases, gradually working toward the more com
PSfrag replacements
PSfrag replacements
axis
4.2. axis RELATING LINES AND EQUATIONS axis axis
41
axis
}  W
õ°^¶m± ö ¶ õ
~
{$Á[} ~R
15000
on the line: This means men dollars in earn year .
·8¯ { 1970
5000
(a) Linear model for women.
~
W} ²
õµ°m± ° ö õ S axis (year) Á 1995
10000
Á
1990
20000
1995
1970
5000
1990
10000
1985
~
1980
} ^²  õ°a¶m± õ¶ õ {tÁ[} ~R on the line: This ~ means women Á earn dollars in year . ® ¯ { W} ²(³ ³ õ°m± õ axis (year) S Á 1975
15000
axis (dollars)
¸ ¯ {
25000
´ ¯ {
20000
Ú
30000
1985
25000
axis (dollars)
1980
Ú
30000
1975
axis
(b) Linear model for men.
Figure 4.2: Linear models of earning power.
plicated. 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 coordinate; we refer to this line as the graph of the equation Ç . Likewise, a point on the horizontal line in Figure 4.3(b) will always have the same M coordinate; we refer to this line as the graph of the equation MêÂ . Figure 4.3(c) is different, in the sense that neither the nor the M coordinate is constant; i.e., as you move a point along the line, both coordinates of the point are changrag replacements replacements ing. ItPSfrag is reasonable to guess that this line is the graph of some equation PSfrag replacements involving both and M . The question is: What is the equation? axis
Ú
axis graph of
Æ Å0Ç ÈRÉ
ÃÄÅ
axis
Ú
axis
Æ Ã[Ç Ê"É
is a typical point on this line
is a typical point on this line.
location
S Å
axis on
Ã
axis
(a) Vertical line.
graph of location
Ê
on
È
axis
ÈÄÊ
axis
S
axis
(b) Horizontal line. Figure 4.3: Lines in a plane.
Ú
axis
{tÁ[} ~R
is a typical point on this line graph of some equation involving and
Á
~
S
axis
(c) Sloped line.
CHAPTER 4. LINEAR MODELING
42
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 nonvertical lines.
4.3 Nonvertical Lines Assume in this section that à is a nonvertical line in the plane; for example, the line in Figure 4.3(c). If we are given two points ØW 6 1§M 9 and Ùé 6 1§M 9 on a line à , then Equation (2.2) on page 20 defined two quantities we can calculate:
3Ç change in going from Ø to Ù " 4MN change in M going from Ø to Ù[êM ½M We define the slope of the line à to be the ratio of 4M by 3 , which is usually denoted by :
def
slope of àt
>M 3 M M
(4.3)
change in M change in
Notice, we are using the fact that the line is nonvertical to know that this ratio is always defined; i.e., we will never have 7F$% (which would lead to illegal division by zero). There is some additional terminology that goes along with the definition of the slope. The term 4M is sometimes called the rise of à and 3 is called the run of à . For this reason, people often refer to the slope of a line à as “the rise over the run,” meaning slope of à
def
rise of à run of à
4M ' 3
In addition, notice that the calculation of 4M involves taking the difference of two numbers; likewise, the calculation of 7 involves taking the difference of two numbers. For this reason, the slope of a line à is sometimes called a difference quotient.
PSfrag replacements axis
4.3. NONVERTICAL LINES
+
p ëu ð
u
í î Ö ×
÷ Ú
ë ð í ÷
í
S
l
Ì Î
ÎÌ ¡
p
Figure 4.4: Computing the slope of a line.
9 M M w>M 3 w 9
PSfrag replacements We CANNOT reverse the order in just one of the calculations and get the same slope: axis PSfrag replacements ÍÌ ÍÌ ÏÌ ÍÌ
Ë
p
and
Ë
ÎÌ
ÎÌ
axis axis
axis
!!!
>M M M 6 M NM 3 6 ü
í î
axis
For example, suppose Ø 6 n19 + and Ùy 6 1 ` 9 lie on a line à . In this case, the rise '4MN and the run '3Ç)j , so Gqp is the slope of à . l When computing 3 , pay special attention that it is the coordinate of the destination point Ù minus the coordinate of the starting point Ø ; likewise, when computing >M , it is the M coordinate of the destination point Ù minus the M coordinate of the starting point Ø . We can reverse this order in both calculations and get the same slope:
ô
43
axis
CAUTION !!!
axis
axis
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 Ø and Ù . 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 ØâáF 6 á 1§Mâá 9 and right Ù á 6 á 1§M á 9 on à , then we would get two similar triangles: See Figure 4.5. Basic properties of similar triangles tell us ratios of lengths of common sides are equal, so that
Ñ ÁÒ } ~2Ò ¯ ´ Ò ¶ ¶(Ó {tÁ } ~ ¯ ´ ~ H ~ Ô~ Ò J ~2Ò ¶ ¶ ${ Á } ~ ¯Ð® ¶ ¶ _ _ Á Á _ _ ¶ _ ÁÒ ÁÒ ® Ò ¯ Ñ ÁÒ ¶ } ~ÔÒ _ _ _Ó Figure 4.5: Using similar triangles.
M M Mâ á N Mâá ý
á ü á
ô
but this just says the calculation of the slope is the same for any pair of distinct points on à . For example, lets redo the slope calculation when Ø á Ø# 6 1§M 9 and Ù á 6 81§M9 represents an arbitrary point on the line. Then the two ratios of lengths of common sides give us the equation
M ½M
M N M ¢ü 6 ¢
1 9
CHAPTER 4. LINEAR MODELING
44 This can be rewritten as
M
6 Y 9T½M or g M M ¢ 9T½M h 6
M
(4.4) (4.5)
Equation 4.4 is usually called the point slope formula for the line à (since the data required to write the equation amounts to a point 6 1 M 9 on the line and the slope ), whereas Equation 4.5 is called the two point formula for the line à (since the data required amounts to the coordinates of the points Ø and Ù ). In any event, we now see that Important Fact 4.3.1. 6 @1 M39 lies on à if and only if 6 81 M39 is a solution to MNH 6 Y 9´NM .
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 M coordinate system, the line
ements
axis
Important Fact 4.3.2. à>xw 6 81 6 P 9A M 9 is any real numbery¼
axis axis
graph of
~ ¯
Ák
ÔÓ
{i }ÍØ
õ
Ó
Ó_
{Õ³q} {×Ö[} ²W ¶ Ó Ó { } {iq}õ õ Ó_
Figure 4.6: Verifying points on a line.
+ Figure 4.6, through Example 4.3.3. Consider the line à , in + ØB 6 n1d9 and Ù 6 1 ` 9 . Then the slope the two points f:j and à consists of all pairs of points 6 81 M39 of à is such that the coordinates and M satisfy the equation M p 6 9mÿ Letting 5%´1n1^ and É , we conclude that l the following four points lie on the line à : 6 %´1ßp 6 %ïW9
W9 æ l 6 %´1 l 9 , 6 n1ßpl 6 $9Ò 9ï 6 æ n1d9 , 6 ^´1ßpl 6 ^¢$9Ò 9ï 6 ^´1 l l 9 and 6 Én1apl 6 Éw 9Ò[9¢ 6 Én1 l u 9 . By the same reasoning, the point 6 %´1&%9 does not lie on the line à . As a set of points in the plane, we have àÚÙ
g
+
81 6 ¢H9´' is any real number Û h j
Returning to the general situation, we can obtain a third general equation for a nonvertical line. To emphasize what is going on here, plug the specific value ½B% into Equation 4.4 and obtain the point ¤ 6 %´1cQÌ9 on the line, where QUH 6 %3 9FM .<F FM . But, Equation (4.4) can be written
M
6 ¢ T9 M FYF M FYQâ
4.4. GENERAL LINES
45
The point ¤ is important; it is precisely the point where the line à crosses the M axis, usually called the M intercept. The slope intercept equation of the line is the form
M½HFYQm1 where the slope of the line is
and Q is the M intercept of the line.
Summary 4.3.4. Nonvertical Lines: Let à be a nonvertical line in the M plane. There are three ways to obtain an equation whose graph is à , depending on the data provided for à :
6 1 M 9 areg two different points on the line, then M M ¢N 9´#M gives an equation the twopoint formula MN h 6 whose graph is à .
1. If Ø'
6 1 M 9z1Ù
2. If Øq 6 1 M 9 is a point on the line and is the slope of à , then the pointslope formula MNH 6 9 #M gives an equation whose graph is à . 3. If the line à intersects the M axis at the point 6 %´1kQÌ9 and is the slope of the line à , then the slopeintercept formula M FÉêQ gives an equation whose graph is à .
4.4 General Lines Summarizing, any line in the plane is the graph of an equation involving and M and the equation always has the form
ïP AMU
H%´1
for some constants , , . Equations like this are called linear equations. In general, nonvertical 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 nonvertical line in the 2M coordinate system, then its slope will be the rate of change of M with respect to :
4M 3
slope
change in M change in
def
rate of change of M with respect to
CHAPTER 4. LINEAR MODELING
46
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 pointslope 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 M with respect to . An interesting thing to notice is how the units for and M figure into the rate of change calculation. For example, suppose that we have the equation M½$e%´1k%%%Ìï!%%´1k%%% , which relates the value M of a house (in dollars) to the number of years you own it. For example, after 5 years, Ç ` and the value of the house would be M ôe%´1k%%% 6d`
[email protected] !%%´1k%%%# $! ` %´1&%%% In this case, the equation MN$e%´1k%%%Ìw½!%%´1k%%% is linear and already written in slopeintercept form, so the slope can be read as te%´1&%%% . If
4M
we carry along the units in the calculation of , then the numerator 7 involves “dollar” units and the denominator “years” units. That means that carrying along units, the slope is actually ôe%´1k%%% dollars/year. ements In other words, the value of the house is changing at a rate of e%´1k%%% axis dollars/year. axis At the other extreme, if the units for both and M are the same, then axis the units cancel out in the rate of change calculation; in other words, the slope is a unitless 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 Þ ß of a moving object. Suppose an object moves along a speed Ý straight line at a constant speed : See Figure 4.7. reference If we specify a reference point, we can let Q be the startpoint location of the obinitial location ject at time Ü 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 Q , the time / , the conline. stant speed and the location at time / :
ements axis axis axis
(location of object at time / ) (initial location of object) + (distance object travels in time / )
QZ/:
where / is in the same time units used to define the rate . Notice, both Q and would be constants given to us, so this is a linear equation involving the variables and / . We can graph the equation in the / coordinate system: See Figure 4.8. It is important to distinguish between this picture (the à axis graph of Z/ WQ ) and the path of the object in Fig¯ Ý ÜÔá ß graph of Þ ure 4.7. The graph of the equation should be thought of â òÞ as a visual aid attached to the equation Z/7"Q . The â Ü ¯ãâ Þ rate of change general idea is that using this visual aid can help answer ßª¯ Þ intercept â Ü various questions involving the equation, which in turn axis
Figure
4.8: The graph of :Ò¡Ç°ïÄÂ@7 .
4.5. LINES AND RATE OF CHANGE
47
will tell us things about the motion of the object in Figure 4.7. Two other comments related to this discussion are important. First, concerning notation, the speed is often symbolized by ä to denote constant velocity and Q is written as (the subscript “ % ” meaning “time zero”). With these changes, the equation becomes D ä/ , 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 / axis. This was because / represented time, which is always a nonnegative quantity. Example 4.5.1. Linda, Asia and Mookie are all playing frisbee. Mookie is PSfrag replacements 10 meters in front of Linda and always runs 5 m/sec. Asia is 34 meters in front of Linda and axis always runs 4 m/sec. Linda yells “go!” and both axis running directly away from Linda to catch a tossed Mookie and Asia start frisbee. Find linear equations for the distances between Linda, Mookie and axis Asia after / seconds.
ì
Linda
Mookie
Asia
be the distance between Linda and Mookie and the Solution. Let distance ì between Linda and Asia, after / seconds. An application of the above formula tells us
ì
(initial distance between Linda and Mookie) EeEeE (distance Mookie runs in / seconds)
EeEeE
e%w ` /2
Likewise,
ì
(initial distance between Linda and Asia) Ee+EeE + (distance Asia runs in / seconds)
j
EeEeE
/2
If is the distance between Mookie and Asia after / seconds, we comì ì pute
HÍê
6j
+
+
/:9´ 6 e%w ` /29Ò 6 !
+
/:9 meters
In all cases, the distances we computed are given by linear equations of the form w'QZ/ , for appropriate Q and rate .
CHAPTER 4. LINEAR MODELING
48
ements
4.6 Back to the Earning Power Problem
axis axis
Ú
axis
å
axis (dollars)
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:
40000
æ
35000
¢
30000 25000
{tÁ[} ~R
~
S 1980
1975
1970
Á
i 6 eo ´% 1&o´1 ` !A9
¤ªç ¼Wè(éo¾ and
axis (year)
¥ ¼À¾ ø
1995
5000
Á
1990
£
10000
~
on the line: means men earn dollars in year
15000
1985
20000
Figure 4.9: Linear model of Men’s Earning Power.
i 6 eoJ _1 !J´1_jTjT9
We find that
4M
3
!J´1_jTj<o´1 ` !A
eoJ
i eJ´1 vo !
i
i Heo %
i
4M 3
i eJ´1 i ov!
If we apply the “pointslope formula” for the equation of a line, we arrive at the equation:
i i eJ´1 vo ! MN i 6 ¢)eo %9´êo´1 ` !A
(4.6)
The graph of this line will pass through the two points ¤ and ¥ in Figure 4.2. We sketch the graph in Figure 4.9, indicating two new points and ê . 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 qeoo ` up to the graph and label the intersection point ê .
Draw a horizontal line à through ê . The line à crosses the M axis at the point
i i i eJ´1 vo ! i 6 eoo ` 'eo %9´o ` !A)j 1å ` ^
4.7. WHAT’S NEEDED TO BUILD A LINEAR MODEL?
49
The coordinates of the point ê 6 eoo ` 1j i 1å ` ^9 . i Conclude that $j 1d ` ^ is the Men’s Earning Power in eoo ` For another
example, suppose we wanted to know when men’s earning power will equal $jjA1k%%% ? This means we seek a data point on the men’s earning curve whose M coordinate is jjA1&%%% By (4.6), has the form
Z
g
i i eJ´1 vo ! 1 i 6 Y)eo %9´êo´1 ` A! @ h
We want this to be a data point of the form 6 81jjA1k%%%9 . Setting these two points equal and equating the second coordinates leads to an algebra problem:
i i eJ´1 vo ! i 6 ¢Heo %9´o´1 ` !A
jjA1k%%% + eoo_!
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 M"[jjA1k%%% and label the intersection point on the model.
Draw a vertical line à through . The line à crosses the axis at the point 1991.24.
+ The coordinates of the point # 6 eooc! c1 jjA1k%%%9 . In the exercises, you will be asked to show that the women’s earning power model is given by the equation
M½
i 0!1&o ` i 6 ¢Heo %9´ ` 1&^e^
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
50
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 Washingi i ton was $ eJv! in eoJo and $!o% in eoo ` 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 $10,000?
ements
axis 10000
Ú(dollars) axis Solution. If we consider a coordinate system where the axis represents the year and the M axis represents i dollars, we are given two data points: ØF 6 eoJo´1n1&Jv! 9 i and Ù[ 6 eoo ` 1!1&o% 9 . Using the twopoint formula for the equation of line through Ø and Ù , we obtain the equation
8000
Ö
axis S(year) 2020
2010
×
2000
1990
2000
2040
4000
2030
6000
i MN$eJ% 6 P 'eoJo9´'n1&Jv!
Figure 4.10: Linear tuition model.
The graph of this equation gives a line through the given points as pictured in Figure 4.10. i If we let q!%%% , we get M) $jA1&J% , which tells us the tuition in the year !%%% . On the other hand, if we set the equation equal to $ e%´1&%%% , we can solve for :
e%´1&%%% i J´1d + +j !1k%:j
eJ% 6 ï HeoJo9´)n1&Jv! eJ% 6 ï HeoJo9
i
Conclude the tuition is $ e%´1&%%% in the year !%:j `
4.9 Perpendicular and Parallel Lines Here is a useful fact to keep in mind. Important Facts 4.9.1. Two nonvertical lines in the plane are parallel precisely when they both have the same slope. Two nonvertical lines are perpendicular precisely when their slopes are negative reciprocals of one another.
PSfrag replacements
4.10. INTERSECTING CURVES II
51 axis
Example 4.9.2. Let à be a line in the plane passing through the points 6 n19 and 6 ^´1zÉ9 . Find a linear equation whose graph is a line parallel to à passing through 5 on the M axis. + whose graph is perpendicular to à Find a linear equation and passes through 6 1^9 . Solution. Letting Ø point formula”:
M
6 n19 and Ùq 6 ^´1zÉ9 , apply the “two
Ú
axis
given line
ë
The graph of this equation will be à . This equation
is in slope intercept æ form and we can read off that the slope
is u . The desired line ' is parallel to à ; it must have slope and M intercept 5. Plugging into u the “slope intercept form”:
M½
;! Y ` `
The desired line Q is a line perpendicular to à (so its slope is 3í +
æ = ²ö
u ) and passes through the point 6 1^9 , so we can use the “point slope
formula”:
M½ ` 6 ¢ !
+
9T^
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 and M . A useful
tool when working with equations involving squared terms (i.e., or M ), is the quadratic formula . Important Fact 4.10.1. Quadratic Formula: Consider the equation
H Q ð ½ ï b% , where ' 1cQm1iï are constants. The solutions for this equa' î î tion are given by the formula
î
wQñ¦ ) Q ! '
+
' ï
The solutions are real numbers if and only if Q
+
S
axis
ì
Figure 4.11: Graph of perpendicular line.
;! 6 ¢'9A) ` i ! Y ` `
ß
' ï=ò)% .
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.”
axis axis
CHAPTER 4. LINEAR MODELING
52
North flight path
´
East
® South
õaô
²
miles
West
irrigated field
ó
miles
crop duster
Figure 4.12: The flight path of a crop duster.
Example 4.10.2. A crop dusting airplane flying a constant speed of 0!% mph is spotted ! miles South and ` miles East of the center of a circular irrigated field. The irrigated field has a radius of mile. Impose a coordinate system as pictured, with the center of the field the origin 6 %´1k%9 .. The flight path of the duster is a straight line passing over the labeled points Ø and Ù . Assume that the point Ù where the plane exits the airspace above the field is the Westernmost 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 Ø 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 Ù[ 6 Én1%9 and ¥ 6 ` 1z;!A98 duster spotting point. Construct a line through Ù and ¥ . The slope is <%J¢H and the line equation becomes:
MN.<%&Jv¢%J
(4.7)
2. The equation of the boundary of the irrigated region is M Ï . We need to solve this equation AND the line equation MN.<%JvPê%J simultaneously. Plugging the line equation into the unit circle equation gives:
< % v J P % & J 9 + + 6
%^ + ' _!Jv¢%&^ ^ ' _!Jv¢%cj^
%
n1%c!Aeo ` . Conclude Apply the quadratic formula and find É that the coordinate of Ø is %_!Aeo ` . To find the M coordinate, plug i into the line equation and get Mt <%o ` ^ . Conclude that Ø i 6 %_!Aeo ` 1z<%o ` ^9
4.10. INTERSECTING CURVES II 3. Find the distance from Ø to Ù
53 by using the distance formula:
i þ 6 É>%_!Aeo ` 9 6 % 6 <%o ` ^99 ` ^v! miles +
miles i Now, u O mph H%&%j%v! hours 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 slope æ O c! ` . So this perpendicular line has equation ç My c! ` . Intersecting this line with the flight path gives us the point closest to the center of the field. The coordinate of this point is found by setting the two line equations equal and solving:
<%&JvY%&J
_! ` <%cjo%v! +
This means that the closest point on the flight path is 6 <%cjo´1z<% o9 . Apply the distance formula and the shortest distance to the flight path is
+
þ 6 <%kjo9 6 <% o9 % ^v!^:jT
CHAPTER 4. LINEAR MODELING
54
4.11 Exercises Problem 4.1. This exercise emphasizes the “mechanical aspects” of working with linear equations. Find the equation of a line:
»R©#"/
and
ó©>R¼©I&/
with
(b) Find U so that the intersection point in (a) has ¿ coordinate 10.
°y¡
(c) Find U so that the intersection point in (a) lies on the ¿ axis.
(a) Passing through the points å©@: I/ . (b) Passing through the point slope °'¡ . (c) With Û intercept ©8 .
7U¡ê©@
and slope
(d) Passing through the point having slope °)¡Ç .
&:R"/
and
(e) Perpendicular to the line in (a) and passing through »R"/ . (f) Parallel to the line in (b) and having intercept 7 ¡?©4§ .
Û

¿ÌÂÉÛ¢¡ . Crossing the ¿ axis at Å¢¡ and having slope °'¡Z .
(g) Having the equation (h)
(a) Find the point of intersection between the line you have graphed and the line ÛÉ¡FÂ@U:¿ ; your answer will be a point in the ¿Û plane whose coordinates involve the unknown U .
Problem 4.2. A spherical balloon of radius 30 feet is to be anchored to the ground by two lines as pictured below:
Problem 4.4. (a) What is the area of the triangle determined by the lines ÛÉ¡?© ¨ ¿ÌÂ , ÛY¡?ª¿ and the Û axis?
7 ¬' and °öõW , then the line Ûê¡ ° ¿7Â7 cuts off a triangle from the first
(b) If
quadrant. Express the area of that triangle in terms of ° and 7 .
(c) The lines Û ¡°<¿Â , Û ¡ü¿ and the Û axis form a triangle in the first quadrant. Suppose this triangle has an area of ¥ square units. Find ° . Problem 4.5. Complete Table 4.2 on page 55. In many cases there may be several possible correct answers. Problem 4.6. The (average) sale price for single family property in Seattle and Port Townsend is tabulated below:
safety line
100 feet safety line
PSfrag replacements axis axis axis
USBA Safety regulation #TD−83−7856a
The U.S. Balloon Association safety regulations require that the safety lines have a slope of absolute value « and be anchored symmet to the basket location on rically with respect the ground. Find the length of each safety line and where it must be anchored to the ground. Problem 4.3. Sketch an accurate picture of the line having equation ÛU¡½m© ¨ ¿ . Let U be an unknown constant.
YEAR 1970 1990
SEATTLE $38,000 $175,000
PORT TOWNSEND $8400 $168,400
(a) Find a linear model relating the year ¿ and the sales price Û for a single family property in Seattle. (b) Find a linear model relating the year ¿ and the sales price Û 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 ¿÷F . (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 agree and what is this price?
4.11. EXERCISES Equation
55
ø intercept
Slope
Point on the line
Point on the line
ÿ
ÿ
ù>ú1ûRüÍýþ
û
þ
þ
û
ÿ þ
þ
ÿ
ÿ
ÿ
Table 4.2: Linear equations table for Problem 4.5.
(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?
9 cup green
(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?
50 feet
replacements (h) When will the Seattle salesPSfrag price be douaxis ble the Port Townsend sales price?
axis ball
(i) Is the Port Townsend sales price ever double the Seattle sales price?
Problem 4.7. Consider the equation: U:¿v7Â *U ¿Â#w¡ . Find the values of ¿ that make this equation true (your answer will involve U ). Find values of U that make this equation true (your answer will involve ¿ ).
d
Problem 4.8. The cup on the hole of a golf course is located dead center in the middle of a circular green that is feet in diameter. Your ball is located as in the picture below:
ball path
rough
40 feet
axis
The ball follows a straight line path and exits the green at the rightmost edge. Assume the ball travels a constant rate of ¥ ft/sec. (a) Where does the ball enter the green? (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.
CHAPTER 4. LINEAR MODELING
56 Problem 4.9. A car traveling on a freeway consumes gas at a constant rate. When the car has gone 100 miles there are 10 gallons of gas in the tank, and when it has gone 220 miles there are 6 gallons left. (a) Find a formula that expresses the amount G of gas in the tank when the car has gone ¿ miles.
three distinct points need not lie on a line. Never the less, we can try to find a ”best fit line” to three given data points. This problem illustrates the idea. Begin by carefully plotting these three points in an ¿0Û coordinate system: ñ 0/ ,  v */ ,  ¥ / .
(a) Plot the line with equation Û)¡ Â ¨ « « ¿ in the coordinate system containing ºyour three points. Which of the three points lie on this line?
(b) How much gas was in the tank when the car started on the trip?
(b) The line in (a) is called the ”best fit line” for the given three data points. In order to measure the ”goodness” of the fit, we can compute a number called the least squares error. This is the sum:
(c) How far has the car gone when it runs out of gas? Problem 4.10. Allyson and Adrienne 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 Adrienne moves 8 ft/sec in the directions indicated. Adrienne stops moving at time Ä¡ v sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (a) Sketch an accurate picture of the situa seconds. Make sure to tion at time ÄÊ¡ label the locations of Allyson and Adrienne; also, compute the length of the seconds. bungee cord at ÄÒ¡ (b) Where is Allyson when the bungee reaches its maximum length?
ó8©
/ Âc Ê© ¥ / Â> © / ª ª
Compute this number and explain in words and in terms of your picture what the number represents. (c) Plot the line with equation ÛÉ¡ ¿ along with the three points. Using the measurement in (b), is the line a better fit to the three plotted points? Problem 4.12. 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 crosssection, and the top view:
20 ft Building
silo
PSfrag replacements
h
axis axis
r
axis
PSfrag replacements axis axis axis
3D view
Ú
Allyson
30 ft
blind spot
Adrian
axis
line of sight #1 (12,0)
start
dave
PSfrag replacements Problem 4.11. Although we know that two distinct points determine a line in the plane,
crosssection
axis
line of sight #2
TOP VIEW
S
axis
4.11. EXERCISES
57
If Dave is standing next to a silo of cross feet at the indicated posectional radius Ám¡ sition, his vision will be partially obstructed. Find the portion of the Û axis that Dave cannot see. (Hint: Let Å be the ¿ 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 PSfrag replacements through the tangency point. Set these two calaxis culations of the slope equal and solve for Å .)
tiff mary 42 nd St. angela
axis
Problem 4.13. While speaking on the phone to a friend in Oslo, Norway, you learned that the current temperature there was ©@Ã Celsius ( ©8Ã C). After the phone conversation, you wanted to convert this temperature to Fahrenheit degrees Ã F, but you could not find a reference with the correct formulas. You then remembered that the relationship between Ã F and Ã C is linear. (a) Using this and the knowledge that zÃ F ¡"Ã C and v§Ã F ¡WRÃ C, find an equation that computes Celsius temperature in terms of Fahrenheit temperature; i.e., an equation of the form D = “an expression involving only the variable G .” (b) Likewise, find an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form G = “an expression involving only the variable D .” (c) How cold was it in Oslo in Ã F? Problem 4.14. 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 4.15. Angela, Mary and Tiff are all standing near the intersection of University and 42 Ñ streets. Mary and Tiff do not move, but Angela runs toward Tiff at ft/sec along a straight line, as pictured. Assume the roads are feet wide and Tiff is ªz feet north of the nearest corner. Where is Angela located when she is closest to Mary and when does she reach this spot?
University Way
axis
Problem 4.16. (a) Using the first and last data points in Table 4.1, Page 39, show that the women’s earning power linear model has equation
Û
¡
e § &¿m©? 0/:ÂÉªeªe
Sketch the graph modeling the years between and 0R What is women’s earning power in R¼ ? When will women’s earning power be $:R R ? (b) Recall the linear models for men and women’s earning power in Figure 4.2. The “women’s percentage of men’s earnings” is given by the formula
¢¡
women s salary men s salary
®É¥R
What is this in Rª ? When %?percentage ¡ ª is Is the percentage ever ¡ ª %? What is happening to as we look farther and farther into the future? (c) Plot the graphs of the men’s and women’s earning power equations in the same coordinate system. Use this graph to explain whether or not women’s earning power is improving with respect to men’s earning power. What aspect of each equation controls whether women are gaining on men? Problem 4.17. Return to Example 4.10.2 on page 52. When is the duster closest to the center of the field?
CHAPTER 4. LINEAR MODELING
58
Problem 4.18. Draw the graphs of ¿;¡?©4 and Û'¡[ in the ¿0Û coordinate system. An ant starts at the location åªe I/ and moves to the left along the line Û¢¡Ç . Assume the position of the ant after Ä seconds is the point ùH Ä2/2¡ óª©RÄ I/ . At the same instant, a spider starts at the location ó©>R§©ï&/ and moves upward along the line ¿7¡?©> . Assume the position of the spider after Ä seconds is the point úK Ä2/Ò¡4ó©>R©ÇmÂÄ2/ . In Problem 3.8 on page 37, we found the distance 100 ft. PSfrag replacements Ä was given between the ant and spider at time by the formula: ,+ Ä2/¡ È RÄ ©É ÄTÂ 2 When axis is the distance between the two bugs exactly axis units? Where are the bugs located at these axis times?
W 100 ft
circular watered zone
(a) Solve for ¿ :
¿ © m ¿ ÂÍ ¡¹ (b) Solve for Ä : (c)
>¡ ; »´ÂÉÄ2/ Â>»A©YRÄ2/ . Solve for Ä : « ¡<; »TÂÉÄ2/ Â>»´©ÉRÄ2/
(d) Solve for Ä :
(b) Solve for ¿ : (c) (d)
m¡ ; »´ÂÉÄ2/ Â>»A©YRÄ2/ (a) Solve for ¿ :
¾ Å ÂÉ ¾ Åm© ¾ ¡FR Å
. .
¾
:
is constant and solve for :
¾ Å ÂÉ ¾ Åm© ¾ ¡FR
Problem 4.22.
(a) Solve for ¿ :
¿ º ©¢R¿ ÂY¡ (c) Find the equation of the line forming the northern boundary of the sidewalk. (Hint: You can use the properties of right triangles.)
z¿vm¡?ª¿
.
z¿¹Ì¡?ª¿4ÂYv Solve for ¿ : z¿¹Ì¡?ª¿4©Yv ¾ Assume is constant and solve for Å
(e) Assume (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.
S
sidewalk
Problem 4.21. (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.
E
20 ft
Problem 4.20. Problem 4.19. 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 along the hose at the rate of ¨ inch/second.
N
hose
(b) Solve for Û :
Ûw©Y È ÛY¡
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 PSfrag replacements Imagine you are standing high atop an oceanside cliff and spot a seagull hovering in the aircurrent. Assuming the axis gull moves up and down along a vertical line of motion, axis how can we best describe its location at time / seconds? axis There are three different (but closely linked) ways to describe the location of the gull:
gull line of motion
a table of data of the gull’s height above cliff level at various times / ;
a plot of the data in a “time” (seconds) vs. “height”
ocean Figure 5.1: Seagull’s height.
(feet) coordinate system;
an equation relating time / (seconds) and height (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 onesecond time intervals 59
cliff level
PSfrag replacements axis axis
CHAPTER 5. FUNCTIONS AND GRAPHS axis
60
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 6 /21z29 in a time (sec.) vs. height (ft.) coordinate system.
(sec) 0 1 2 3
PSfrag replacements
Gull Height (feet above cliff (sec) (ft) (ft) 20 4 10 6.88 5 8.12 2.5 6 2.5 8.12 7 6.88
level) (sec) 8 9 10
u OO p l OO O æ O
(ft) 20 36.88 57.5
PSfrag replacements
(a) Symbolic data.
axis axis
Feet
p
ç
O
Seconds
(b) Visual data.
axis axis
Figure 5.2: Symbolic versus axisvisual view of data.
axis
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:
u OO p O lO O æ O
Feet
p
(a)
¨
Seconds
ç
O
second intervals.
u OO p l OO O æ O
Feet
p
(b)
º
¨
Seconds
ç
O
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 6 /21z29 in any of these three plots is interpreted to mean that the gull is feet above cliff level at time / 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 / . 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
5.2. WHAT IS A FUNCTION?
61
equation involving the variables / and :
` 64 w / J
+ 9 'e% i
If we plug in /¢ %´1n1!1o´1e% , then we get Y !% , ^JJ , ;!A ` , j^JJ , ` ` , respectively; this was some of our initial tabulated data. This same equation produces+ ALL of the data points for the other two plots, using efv! second and ef 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 / value between 0 and 10 and get out a corresponding height . 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.
axis axis
CHAPTER 5. FUNCTIONS AND GRAPHS
62 yaxis
y x
!
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.
xaxis
Figure 5.4: Graph of a procedure.
A table of data, by its very nature, will relate two
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.
Given a curve in Figure 5.4, consider the “procedure” which associates to each on the horizontal axis the M coordinate of the pictured point Ø on the curve. Given an equation relating two quantities and M , plugging in a particular value and going through the “procedure” of algebra often produces a unique output value M .
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 and M 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
j2Y !vM
+
Ç !
MU
+
M þ
'w$n1
can be rewritten as equivalent equations
MN
+
üj2@9 ! 6
! ü !¢JYM
M½
)
'
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 M “a mathematical expression only involving the variable ,”
5.2. WHAT IS A FUNCTION?
63
which we usually indicate via the shorthand notation M"! 6 @9¼ This equation has the very special property that each time we plug in an value, it produces exactly one (a unique) M value. We call the mathematical expression ! 6 @9 ”the rule”.
A set
of values we are allowed to plug into ”domain” of the function. !
6 9 , called the
The set ¤ of output values ! 6 9 , where varies over the domain, called the ”range” of the function.
Any time we have a function M½#! 6 @9 , we refer to as the independent variable (the “input data”) and M as the dependent variable (the “output PSfrag replacements data”). This terminology emphasizes the fact that we have freedom in the values of we plug in, but once we specify an value, axis the M value is axis uniquely determined by the rule ! 6 @9 . axis
Examples 5.2.3. (i) The equation M b;!Ç)j is in the form M &! 6 9 , where the rule is ! 6 9Ê.;!ïj . Once we specify a domain of values, we have a function. For example, we could let the domain be all real numbers.
yaxis
Graph of ù ú%$
$
xaxis
(ii) Take the same rule ! 6 9q;!Dêj from (i) and let the Figure 5.5: Constant funcdomain be all nonnegative real numbers. This detion. scribes a function. However, the functions ! 6 9Ò.;!A j on the domain of all nonnegative real numbers and ! 6 9>[;!Éj 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 nonnegative real numbers. (iii) The equation M Q , where Q is a constant, defines a function on the domain of all real numbers, where the rule is ! 6 93 Q ; we call these the constant functions. Recall, in Chapter 3, we observed that the solutions of the equation MH Q , plotted in the M coordinate system, will give a horizontal line. For example, if QFH% , you get the horizontal axis. (iv) Consider the equation M b , then the rule ! 6 @9Òb defines a function, S S as long as we do not plug in Z % . For example, take the domain to be the nonzero real numbers.
(v) Consider the equation M ) > . Before we start plugging in values, we want to know the expression under the radical symbol (square root symbol) is nonnegative; this insures the square root is
CHAPTER 5. FUNCTIONS AND GRAPHS
64
a real
number. This amounts to solving an inequality equation: %(' ¢ ; i.e., É)' *' These remarks show that the rule ! 6 9Ò ) defines a function, where the domain of values is É)' *' . Typically, the domain of a function M +! 6 9 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 PSfrag replacements Symbolic axis Notation axis
Description
PSfrag , replacements All numbers ü between ü  $ $ ü and , possibly equal to ,.axis $ either , or axis
Picture
axis
Q '
axis PSfrag replacements
All numbers ü between / $ and $ , ü0ú / , and ü0ú%
,
ü
,21
$ 1
Q
axis
'
axis axis PSfrag , replacements All numbers ü between / $ and $ , ü3ú4 and ü possi ,. ü 1 $ axis bly equal to , axis
All numbers ü between , and $ , ü5ú / , and ü possibly equal to $
,21
ü
Q '

axis $
'
Q
Table 5.1: Interval Notations
We can interpret a function as a “prescription” that takes a given value (in the domain) and produces a single unique M 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
Y M
[
and plug in Ç'% , the equation becomes
M
$
This equation has two solutions, Mê ¦I , so the conclusion is that plugging in F % does NOT produce a single output value. This violates
one of the conditions of our function definition, so the equation É M is NOT a function in the independent variable . Notice, if you were to try
5.2. WHAT IS A FUNCTION?
65
and solve this equation for M in terms of , you’d first write M <) and then take a square root (to isolate M ); but the square root introduces TWO roots, which is just another way of reflecting the fact there can be two M values attached to a single value. Alternatively, you can solve the
equation for in terms of M , getting büM ; this shows the equation does define a function Ç46 6 M39 in the independent variable M .
5.2.2 The definition of a function (conceptual viewpoint)
PSfrag replacements PSfrag replacements Conceptually, you can think of a function as a “process”: An allowable axis axis inputaxis goes into a “black box” and outaxis pops a unique new value denoted by the symbol ! 6 @9 . Compare this with the machine making “hulahoops” axis axis 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. 98798977 98798977 98798977 98798977 98798977 98798977 98798977 98798977 98798977 98798977 98798977 979977 789997788tube 789 789 789 789 789 789 789 789 789 789 79 999877788989877987989877 989877987989877 989877987989877 989877987989877 989877987989877 989877987989877 989877987989877 989877987989877 989877987989877 989877987989877 9977999777 997788987978989877987987 989877987987 989877987987 989877987987 989877987987 989877987987 989877987987 989877987987 989877987987 989877987987 99779797 998778987 989877987 989877987 989877987 989877987 989877987 989877987 989877987 989877987 989877987 in 987 97 987 987 987 987 987 987 987 987 987 987 987989877 979977
tubes
:
î_ò ;
out
hoop
ç ¼ð
(a) A hoop machine as a “process” taking “tubes” to “hoops.”
î
in
S domain
:
out
Ú
Sð
(b) A function as a “process” which takes ¿ to Û .
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 M be the total number of gallons of water used by a household between 12:00am and a particular time / ; we will use time units of “hours.” Given a time / , the household will have used a specific (unique) amount of water, call it ¥ 6 /:9 . Then MH + ¥ 6 /29 defines a function in the independent variable / with dependent + variable M . The domain would be %<'W/='.! and the largest possible value of ¥ 6 /:9 on this domain is ¥ + 6 ! 9 . This tells us that the range would be the set of values %>'M?'ð¥ 6 ! 9 . (ii) The height of the center of a basketball as you dribble, depending on time. Let be the height of the basketball center at time / seconds after you start dribbling. Given a time / , if we freeze the action, the center of the ball has a single unique height above the floor,
CHAPTER 5. FUNCTIONS AND GRAPHS
66
*
call it 6 /2* 9 . So, the height of the basketball center is given by a function Ç 6 /29 . The domain would be a given interval of time you are dribbling the ball; for example, maybe %*'H/='H! (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 be the state tax (in dollars) due on a taxable item that sells for î dollars. Given a taxable item that costs î dollars, the state tax due is a single unique amount, call it @ 6µî 9 . So, ê
[email protected] 6µî 9 is a function, where the independent variable is î . The domain could be taken to be %B' î '$n1&%%%´1k%%% , which would cover all items costing up to onemillion dollars. The range of the function would be the set of all values @ 6 î 9 , as î ranges over the domain. (iv) The speed of a chemical reaction depending on the temperature. Let ä be the speed of a particular chemical reaction and the temperature in Celsius C. Given a particular temperature , one could experimentally measure the speed of the reaction; there will be a unique speed, call it 6 ;9 . So, äÇ) 6 9 is a function, where the independent variable is . The domain could be taken to be %C'[5'Be%% , 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 6 ;9 , as ranges over the domain.
5.3 The Graph of a Function Let’s start with a concrete example; the function ! 6 @9¢ô;!Ç'j on the domain of all real numbers. We discussed this Example 5.2.3. Plug in PSfrag replacements the specific values, where .CÉ , % , , ! and tabulate the resulting î M values of the function: axis
í
axis
S
1 0 1 2 .. .
S
S ëÚ ð
Ú
5 3 1 1 .. .
æ SEDml
î
point (1,5) (0,3) (1,1) (2,1) .. .
S ë æ SEDml ð
(a) Tabulated data.
axis
Ú
axis Graph of
Ú
S
SFD4l . axis
(b) Visual data.
Figure 5.7: Symbolic versus visual view of data.
This tells us that the points 6 %´1jT9 , 6 n19 , 6 !1zÉ9 , 6 Én1 ` 9 are solutions of the equation M Ï;!Ç"j . For example, if M b;!Ç"j , % , M yj ,
5.3. THE GRAPH OF A FUNCTION
67
then jZB;! E %P j (which is true), or if MH ;! j , N ! , MH É , then ÉDq;! E !Yêj (which is true), etc. In general, if we plug in we get out ;!"j , so the point 6 @1 !"jT9 is a solution to the function equation M"A! 6 @9 . We can plot all of these solutions in the 2M coordinate system. The set of points we obtain, as we vary over all in the domain, is called the set of solutions of the equation MN.;!Y j : Solutions xw 6 81z;!¢üjT9
any real numbery¼
Notice that plotting these points produces a line of slope G;! with M intercept 3. In other words, the graph of the function ! 6 @94[;!Éj is the same as the graph of the equation M];! j , as we discussed in
Chapter 4. In general, by definition, we say that a point 6 81 M39 is a solution to the function equation MWG! 6 9 if plugging and M into the equation gives a true statement. How can we find ALL the solutions of the equation M &! 6 @9 ? In general, the definition of a function is “rigged” so it is easy to describe all solutions of the equation M &! 6 9 : Each time we specify an value (in the domain), there is only one M value, namely ! 6 9 . This means the point Ø 6 81H! 6 99 is the ONLY solution to the equation M4! 6 9 with first coordinate . We define the graph of the function MI! 6 @9 to be the plot of all solutions of this equation (in the M coordinate system). It is common to refer to this ” or the “graph as either the “graph of J of J .”replacements PSfrag
Graph xw 6 @1K! 6 @9¼9 in the domainy
axis
(5.1)
axis axis
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 L in the domain of a function M &! 6 9 , then you immediately can plot the point 6 Lw1H! 6 L99 on the graph. On the other hand, if someone gives you a point 6 Lw1aä@9 , it will be on the graph only if ä A! 6 L9 is true. We illustrate this in Example 5.3.2.
+ u Example 5.3.2. The function É 6 /:94t ç 6 / 9 We% defines a function in the independent variable / . If we restrict to the domain %<'H/='ÿe% , 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 M 5.1) all lie on this parabola. On * the other hand, the point 6 %´1k%9 is NOT on the graph, since 6 %9Ò"!% H ã %.
OO
uO p O lO O æ O
à axis
p
ò axis
Figure 5.8:
O ¾ : ¡ Ä2/ . Ò ç
CHAPTER 5. FUNCTIONS AND GRAPHS
68
5.4 The Vertical Line Test There is a pictorial aspect of the graph of a function that is very revealing: Since 6 81H! 6 99 is the only point on the graph with first coordinate equal to , a vertical line passing through on the axis (with in the domain) crosses the graph of M &! 6 @9 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 M plane and specify a set of values. Suppose every vertical line through a value in intersects the curve exactly once. Then the curve is the graph of some function on the domain . If we can find a single vertical line through some value in that intersects the curve more than once, then the curve is not the graph of a function on the domain .
For example, draw any straight line in the plane. By the vertical line test, if the line is not vertical, is the graph of a function. On the other hand, if the line is vertical, then is not the graph of a function. These two situations are illustrated in Figure 5.9. As another example,
n1 whose graph is the unit circle and consider the equation 'M PSfrag replacementsspecify the domain to be É)' *' ; recall Example 3.2.2. The vertical line passing through the point x 1% { will intersect the unit circle twice; axis by the vertical line test, the unit circle is not the graph of a function on axis the domain ÉN' *' . axis
Ú
Ú
axis
z z
S
crosses curve twice
Ú
axis
P
crosses curve twice
P
O
S
axis
axis
axis
S
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 (in feet) of a ball above the ground after / seconds is given by the function
*
6 /29ÒHÉe^:/
+
5.5. LINEAR FUNCTIONS
PSfrag replacements
69
axis
We could look at the graph of the function in the / plane 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 / values to lie between 0 and ; notationally, we would write this constraint as %Q' /R' .
à axis
axis axis
Physically interesting portion of graph.
ò
axis
Figure 5.10: Restricting the domain.
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 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 and M . One of the key conclusions was: Important Fact 5.5.1. A nonvertical line in the plane will be the graph of an equation M ÏFÇ)Q , where is the slope of the line and Q is the M intercept. Notice that any nonvertical 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 and M we already obreplacements tained in Chapter 4: The rule ! 6 9Ê'FFQ PSfrag on some specified domain will have a line of slope and M intercept Q as its graph. We call axis a function axis of this form a linear function. axis
Example 5.5.2. You are driving 65 mph from the Kansas state line (mile marker 0) to Salina (mile marker 130) along I35. Describe a linear function that calculates mile marker after / hours. Describe another linear function that will calculate your distance from Salina after / hours.
Solution. Define a function 6 /:9 to be the mile marker after / hours. Using “distance=rate , time,” we conclude that ^ ` / will be the distance traveled after / hours. Since we started at mile marker 0, 6 /29ÒH^ ` / is the rule for the first function. A reasonable domain would be to take %.' /R' ! , since it takes 2 hours to reach Salina. For the second situation, we need to describe a different function, call it 6 /:9 , that calculates your distance from Salina after / hours. To describe the rule of 6 /:9 we
O Ú O¥O O ç OO
p O
axis
ò O u
u
axis
Figure 5.11: Distance functions.
CHAPTER 5. FUNCTIONS AND GRAPHS
70 can use the previous work:
6 /:9
6 mile marker Salina 9T 6 your mile marker at / hrs. 9 j%;H 6 /:9 j%;ê^ ` /:
For the rule 6 /:9 , the best domain would again be %A'5/&'5! . 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. 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 units is $ e%% 6 4 ) 89 . 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 units. All of the costs involved in developing, supporting, distributing and marketing units are controlled by the expense equation (again in dollars): 6
6 9Ê[ ` \ 6 9Ê[e%% 6 4 ) @9
(gross income function) (expense function)
A profit will be realized on the sale of units whenever the gross income exceeds expenses; i.e., this occurs when 6 6 @9ÌV'\ 6 @9 . A loss occurs on the sale of units when expenses exceed gross income; i.e., when \ 6 94VS6 6 @9 . Whenever the sale of units yields zero profit (and zero loss), we call a breakeven point ; i.e., when \ 6 9Ë46 6 9 . The above approach is “symbolic.” Let’s see how to study profit and loss visually, by studying the graphs of the two functions 6 6 9 and \ 6 9 . To begin with, plot the graphs of the two individual functions in the 2M coordinate system. We will focus on the situation when the sales figures are between 0 to 100 units; so the domain of values is the interval %T'b"' e%% . Given any sales figure , we can graphically relate three things:
on the horizontal axis; a point on the graph of the gross income or expense function; M on the vertical axis.
5.6. PROFIT ANALYSIS
axis (dollars)
u O¥O O¥O¥O u O¥O
î S ë uS ð
Ö
O
pO
Ú axis (dollars) î î ¥ O u O O¥O¥O ¼ PSfrag replacements S ë S ðdð Ou ¥O axis S axis (units sold)
O axis O O ç O O¥O × p expense
gross income
Ú
71
S O axisO (unitsO¥O sold) ç
axis
(a) Gross income graph.
(b) Expenses graph.
Figure 5.12: Visualizing income and expenses.
If q!% units sold, there is a unique point Øê 6 !%´1U6 6 !%993 6 !%´1j+%%9 on i the gross income graph and a unique point Ùy 6 !%´10\ 6 !%994 6 !%´1 ` 9 on the expenses graph. Since the M coordinates of Ø and Ù are the function values at q !% , the height of the point above the horizontal axis is controlled by the function. If we plot both graphs in the same coordinate dollars a`Ô !A! system, we can visually study the distance between V[Y ( V\Z ( points on each graph above on the horizontal axis. V (( In the first part of this plot, the expense graph is ^]u ! ! W ( _]u !A! c X ( above the income graph, showing a loss is realized; Y ( b`P !A! the exact amount of the loss will be \ 6 9:d6 6 9 , which Z ( sold units is the length of the pictured line segment. Further to Z Y W V ( X the right, the two graphs cross at the point labeled “ ”; this is the breakeven point; i.e., expense and Figure 5.13: Modelling profit and income agree, so there is zero profit (and zero loss). loss. Finally, to the right of the income graph is above the expense graph, so there is a profit; the exact amount of the profit will be 6 6 @9¹Z\ 6 @9 , which is the length of the rightmost line segment. Our analysis will be complete once we pin down the breakeven point . This amounts to solving the equation 6 6 @9Ò \ 6 @9 .
` ` ¢)e%%
!! ` üj%%%vPHe%%%% !! ` 'j%%%vPHe%%%%
e%% 6 > ) 9 e%% ) e%%%%v % i
i
Applying the quadratic formula, we get two answers: ÇH% J or ` . 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 i solution ` gives us the break even point. What about the other i i ”solution” at é% J ? Try plugging é% J into the original equation: i i `86 % J9Í ã e%% 6 w ) % J=9 . What has happened? Well, when going from
72
CHAPTER 5. FUNCTIONS AND GRAPHS
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 breakeven point using either function:
i i i i i
6a` U1 6 a6 ` 9
[email protected] 6a` 1 J ` `
[email protected] 6a` 01 \ a6 ` 99
5.7. EXERCISES
73
5.7 Exercises Problem 5.1. In each of (a)(h) decide which equations establish a function relationship between the independent variable ¿ and the dependent variable Û . Write out the rule for any functions.
ÛY¡¿ÌÂ? z¿4©ÉÛÉ¡ ¿m©¢¡ ÛÂÉ¡
a. b. c. d.
e. f. g. h.
¿0Û¢¡ ¿¹¥ÛÉ¡ ¿0Ûm¡ Ûw© È ¿ Â?Ò¡
(a) Solve the equation
Problem 5.4.
¿ Y Â rkÛÂ?2/ ¡F for Û in terms of ¿ . You should get two different functions of ¿ . Describe the
largest possible domain of each function. (b) Let Å and 7 be positive constants. Solve the equation
¿mÂ? Â Åon m
Problem 5.2. For each of the following functions, find the expression for e
&¿4Â
¾ / © ¾
e
&¿/
Simplify each of your expressions far enough ¾ so that plugging in ¡? would be allowed. e
(a) e
(b) e
(c) e
(d) e
(e) e
(f)
&¿/¡I¿¹@©Yz¿ . &¿/ ¡ z¿4ÂÉ &¿/ ¡I¿¹@©É &¿/ ¡IË©É¿ &¿/ ¡©gf2¿¹©>f &¿T / ¡ È ¿Ì©Í . (Hint:
merator)
Ú
Rationalize the nu
axis
(20,60)
g(x)
40
replacements
20
(0,24)
f(x)
(20,20)
(0,4) axis
10
for Û in terms of ¿ . You should get two different functions of ¿ . Describe the largest possible domain of each function. (c) Solve the equation
Problem 5.3. Here are the graphs of two linear functions on the domain ihF¿=hZ0e . Find the formula for each of the rules Û?¡ ¿/ and Ûü¡"jr&¿/ . Find the formula for a NEW function ²+¿/ that calculates the vertical distance between the two lines at ¿ . Explain in terms of the picture what ²+¿/ is calculating. What is ²+ &/ ? What is ²+0*/ ? What are the smallest and largest values of ²Q&¿/ on the domain kh ¿lh0 ? 60
Û©É ¡Z 7 n m
20
S
axis
¿¹ ©YrkÛÂ?2/ ¡F for Û in terms of ¿ . You should get two different functions of ¿ . Describe the
largest possible domain of each function. (d) Let Å and 7 be positive constants. Solve the equation m
¿m©? © Å n m
ÛÂY ¡Z 7 n
for Û in terms of ¿ . You should get two different functions of ¿ . Describe the largest possible domain of each function. Problem 5.5. 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 :F¡ ,+Ä2/ which keeps track of Dave’s distance : from Padelford Hall at time Ä . 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 zR feet long.
CHAPTER 5. FUNCTIONS AND GRAPHS
74 PSfrag replacements axis padelford axis
gould
axis
(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 halfway 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 halfway 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 halfway 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.
have been faster, he dawdled in the beautiful sunshine, and ending up getting to the seaside monastery at exactly 11 AM. (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? (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? (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? Problem 5.7. 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. (e) Your height above the water level in a swimming pool after you dive off the high board. Problem 5.8.e Here is a picture of the graph of the function ¿/¡Ç¿ ©Éz¿m©Y .
(f) Suppose you want to sketch the graph of a new function :Ì¡%jTÄ2/ that keeps track of Dave’s distance : from Gould Hall at time Ä . How would your graphs change in (a)(e)? Problem 5.6. 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 mountaintop monastery at 11 PSfrag replacements AM, spent the rest of the day in meditation, and then slept the night there. In the morning, at 5 AM, he began walking back to the seaside axis monastery. Though walking downhill should
Ú
axis
S
axis
5.7. EXERCISES
75
Recall the procedure 5.3.1 on page 67. (a) Find the ¿ and Û intercepts of the graph. (b) Find the exact coordinates of all points &¿ ÛN/ on the graph which have Û coordinate equal to 5. (c) Find the coordinates of all points &¿ ÛN/ on the graph which have Û coordinate equal to 3. (d) Which of these points is on the graph: »R©Y&/ , ó©>R &/ , v : / ,  È ¹ ©D È &/ . (e) Find the exact coordinates of the point &¿ ÛN/ on the graph with ¿;¡ ; ´Â È . Problem 5.9. A biochemist is investigating the enzyme penicillinase that will “deactivate” penicillin by breaking it into two pieces:
replacements axis axis axis
penicillinase
intact penicillin
deactivatated
penicillinase
these functions for: U?¡ü& v ¹ . Make sure to label each of your graphs with its corresponding function. (c) Suppose the biochemist wants the solution to exhibit a turnover rate of 600 molecules/second. How many grams of inhibitor should be added? (d) Suppose the biochemist wants exactly 6000 molecules of penicillin to be deactivated in a 25 second time period. How many grams of inhibitor should be added?
Problem 5.10. 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. You need to get power out to the island so you can listen to your extensive CD collection. A power substation 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 (a) A vessel contains a solution of water, penicillinase and penicillin. Excable path 1 km perimental calculations PSfrag show that the replacements rate at which penicillinase deactix Power vates penicillin molecules is 2000 axis axis molecules/second; this is called the 4 km turnover rate. Find a linear function axis which computes the number of deac(a) Explain why we can assume the cable tivated penicillin molecules after Ä secfollows the path indicated in the picture; onds. Sketch the graph for the first 100 i.e. explain why the path consists of seconds. How many penicillin molecules two line segments, rather than a weird are deactivated after 10 seconds? How curved path AND why it is OK to assume long does it take to deactivate 50,000 the cable reaches shore to the right of penicillin molecules? the power station and the left of the is(b) Adding an inhibitor to the solution will land. slow down the reaction. Theoretical (b) Let ¿ be the location downshore from the calculations show that if U grams of power station where the e cable reaches inhibitor are added to the solution, the land. Find a function &¿/ in the varithen the new turnover rate is qe p p X p ¨ ¶ able ¿ that computes the cost to lay a molecules/second. Find a linear funccable out to your island. tion which computes the number of deactivated penicillin molecules after Ä sece (c) Make a table of values of &¿/ , where onds. (Your equation will involve the ¿Y¡# , ¨ , , « , , , ¤ ,4. Use these calculavariable “Ä ” AND the constant “ U ”; we tions to estimate the installation of minusually call such an U a parameter.) For imal cost. the first 100 seconds, in a common coordinate system, sketch the graphs of
CHAPTER 5. FUNCTIONS AND GRAPHS
76 Problem 5.11. This problem deals with the “mechanical aspects” of working with the rule of a function. For each of e the functions listed e e e in (a)(c), calculate: , , , ó 0 / ó 8 © & / & ; ¿ ? Â * / sr½/ , e srÂ.t§/ . (a) The function &¿/4¡ ¨ ¿©Í*/ on the do main of all real numbers. e
(b) The function &¿/m¡z¿¹Ì©Uª¿ on the domain of all real numbers. e
(c) The function
e
&¿/¡ Uf
.
Problem 5.13. 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 5.14. Find an EXACT answer for each problem. (a) Solve for
Problem 5.12. An object is described that establishes a relationship between two variables. Identify the variables, decide which relationships are functions and in those cases identify the independent and dependent variables. (a) An itemized receipt from the University Book Store. (b) The index to this text. (c) A bathroom scale. (d) A radio dial. (e) Your math professor’s grade book.
(b)
(c)
¿
¿ ¿mÂÉ Â ¿m© ¡ ¿ ©YR¿m©Év Solve for ¿ È z¿4©¢3¡ ¿ ÂY Solve for ¿ È ¿ÌÂÇÈ ¿Ì©Y04¡#¥
(d) Solve for
Ä
È RÄT©?´Â È zÄTÂÉ>¡Ç
5.7. EXERCISES
77
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(n)
(o)
(p)
PSfrag replacements (m) axis axis axis
Figure 5.14: Curves to consider for Problem 5.13
78
CHAPTER 5. FUNCTIONS AND GRAPHS
Chapter 6 Graphical Analysis We ended the previous section with an indepth 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 67) and the surrounding discussion. The key thing about the graph of a function ! 6 9 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 Øê 6 81 M39 will be on the graph of the function ! 6 9 exactly when M &! 6 9 .
6.1.1 Visualizing the domain and range A function is a package that consists of a rule MÏu! 6 9 , a domain of allowed values and a range of output M values. The domain can be visualized as a subset of the axis and the range as a subset of the M 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 axis, then project these points to the M axis. The collection of all values you obtain on the M axis will be the range of the function. This idea of “projection” is illustrated in the two graphs be79
CHAPTER 6. GRAPHICAL ANALYSIS
80
low. We use arrows “ v ” to indicate going from a domain value, up to the graph, then over to the M axis: PSfrag replacements
PSfrag replacements
axis
axis
axis
axis
axis
Æ ÃuÉ2ÄaÃ
x
graph of
range
Ç
Ä ^Ç z
range w
axis
Ä uÇ y{z
domain w x
Ä
domain
Äw xuÇz
w } ~xz
(a) Domain:
¹»^ .
Æ ÃuÉ2Ä^Ãx
graph of
(b) Domain:
©Ò¹{ .
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 ! 6 9 control the height of the point Ø 6 @9 6 @1K! 6 99 on the graph above the axis; if the function value ! 6 @9 PSfrag replacements is negative, the point Ø 6 9 is below the axis.
î
axis
î íPî Ö
î
î :
Sð
î S
units above axis :
Sð
S ë S ðóð :
Sö S
: ð units below the S S
Ö
î
axis
Ú
íPî SRö ð
î
î  î íïî S ð units above the S axis Ö   :  S ð S ë S ðdð S

S
axis
: SRö ë RS ö ðåð
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
6.1. VISUAL ANALYSIS OF A GRAPH
81
plot for the function. We include a “shadow” of the graph in Figure 6.3 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.
eplacements axis positive
axis
positive
negative
negative
S
axis
Figure 6.3: Sign plot.
By moving through a sequence of 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 move from 1 to 5, left to right; we have indicated how the corresponding points on the curve will move eplacementsand how the function values will change.
î
ð :
axis
î
Ú
Ö
axis
î
:
graph of
uð :
í î Ú
lîð :
:
to
×
î î
Ö
ð :
#3: S ð values move like this
#2: points on graph move from
pð
#1:
S
u
×
values move from
to
u
Figure 6.4: Dynamic interpretation of a graph.
S
axis
:
Sð
ements
CHAPTER 6. GRAPHICAL ANALYSIS
82
6.1.3 Interpreting Intercepts of a Graph
axis
The places where a graph crosses the axes are often significant. We isolate each as an important feature to look for when doing graphical analysis. The graph of the func{$Á } {$Á } q axis Ó ¶ {$Á } S tion Mÿ! 6 @9 crosses the M axis at the point 6 %´1! 6 %9¼9 ; so, _ the M intercept of the graph is just ! 6 %9 . The graph of the Á intercepts have the form “ {$Á[} q ” function M"! 6 9 crosses the axis at points of the form Figure 6.5: Intercepts of a 6 Ov1K! 6 O¹99 , where ! 6 O¹9<é% . The values O are called roots PSfrag replacements graph. or zeros of the function ! 6 9 . There can be at most one M intercept, but there can be several intercepts or no intercept: See Figure 6.5 * axis The* graph of a function at the M A! 6 9 crosses the vertical line Z * point 6 1K! 6 99 . To find where the graph of a function MI! 6 @9 crosses the horizontal line M 4 , first solve the equation ?&! 6 @9 for . If the equation FI! 6 9 has solutions , , l , p , then the points of intersection would have the coordinatesPSfrag given.replacements
Ú axis ~ intercept ¯ {iq} f{iqA
Ú
{8R}t{8a{8A {tÁ } " _ Á
axis a{$Áu {$Á } [
graph of
{$Á í } [
¶
_
Á
¶
S
Ó
axis
í Á
{$Á } [
Ó
(a) General curve.
Á
ÚÔ
Ô S
Ú
¯
axis
í
{ }
upper semicircle radius centered at
í
axis
S
ö õ Ú í
ö
öõ Ú
S
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 MN twice if and only if )'3X)j ; the graph will cross the horizontal line M½ once if and only if ?"j . The graph will * not intersect +
M Ï and the graph will cross the vertical line Ç the line if and only if * %Q' ' .
ements
6.1.4 Interpreting Increasing and Decreasing axis
axis
S
up
hi ll
Ú
dow
nh
ill
¯
axis
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 values. Typically, we want to study what is happening to the values ! 6 9 as 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
6.1. VISUAL ANALYSIS OF A GRAPH
83
to the right” along the graph, the function values are increasing if you are walking uphill. Likewise, if you were “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 as a local extrema. PSfragcase replacements People have invested a lot of time (centuries!) and energy (lifetimes!) into axisgraphs. We the study of how to find local extrema for particular function axis will see some basic examples in this course and others will surface in axis future courses once you have the tools of calculus at your disposal. Examples range from business applications that involve optimizing profit to understanding the threedimensional 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 / minutes is given by the graph in Figure 6.8: 1. When is the pilot climbing and descending? 2. When is the pilot at the glider port elevation?
a³ ^ Ö(^ ^
ft above gliderport
ö
^ 0Ö(ö a I³aa
²
õ
minutes
Figure 6.8: Hanglider elevations.
3. How much time does the pilot spend flying level? Solution. 1. Graphically, we need to determine the portions of the graph that are increasing or decreasing. In this example, it is increasing when i %'q/<'q! and 'ÿ/<' o . And, it is decreasing when j('q/<' ` and o.' /'$e% . 2. Graphically, this question amounts to asking when the elevation + horiis 0, which is the same as finding when the graph crosses the zontal axis. We can read off there are four such times: /;'%´1 1J´1e% . 3. Graphically, we need to determine the portions of the graph that are made up of horizontal line segments. This happens when !'W/='Wj i and ` ' /i' . So, our pilot flies level for a total of 3 minutes.
CHAPTER 6. GRAPHICAL ANALYSIS
84
6.2 Circles and Semicircles Back in Chapter 3, we discussed equations whose graphs were circles: We found that the graph of the equation
*
9 6 M 89 )
6 P
(6.1)
*
is a circle of radius centered at the point 6 1W89 . It is possible to manipulate * and become confused. We could rewrite this as
this
equation M6 #89 [ 6 I 9 , then take the square root of each side. However, the resulting equivalent equation would be
*
MN @¦ þ 6 Y 9 and the presence of that ¦
M
M
sign is tricky; it means we have two equations:
* 6 ¢ 9 or *
Y]þ 6 ¢ 9 Y]þ
Each of these two equations defines a function: ! 6
6 9 6 9
* 6 Y 9 or *
D þ 6 Y 9
D.þ
(6.2) (6.3)
So, even though the Equation 6.1 is not a function, we were able to obtain two different functions ! 6 @9 and 6 6 9 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 ! 6 @9 and the lower semicircle is the graph of PSfrag replacements the function 6 6 9 . axis axis axis
ù axis
upper semicircle
ù axis
ÿ
ÿ
ÿ Graph of ù>ú ü
ü axis
lower semicircle
ÿ Graph of ù ú? ü
ü axis
Figure 6.9: Upper and lower semicircles.
6.2. CIRCLES AND SEMICIRCLES
85
Example 6.2.1. A tunnel connecting two portions of a space station has a circular crosssection 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. PSfrag replacements Solution. Impose a coordinate system so that the origin is at the center of the circular cross section of the tunnel; by axis symmetry the walkway is centered about the origin.
With
axis Deck A this coordinate system, the graph of the equation ÉM
í ` $!! ` will be the circular crosssection of the tunnel. S axis In the case of Deck , we basically need to determine Deck B how close to each edge of the tunnel a 6 foot high person ft u can stand without hitting his or her head on the tunnel; PSfrag replacements a similar remark applies to Deck . This means we are really trying to fit two sixfoothigh rectangular safe walk axis (a) Crosssection of axis tunnel. zones into the picture: axis Our job is to find the coordinates of the four points Ø , Ù , ¤ , and ¥ . Let’s denote by , , l ,and p the Deck A coordinates of these four points, then ØÏ 6 1^9 , Ù 6 1^9 , ¤[ 6 l 1z
Intersecting the the upper semicircle with the horizontal line having equation M ^ will determine ! 6 9I and ; the upper semicircle is the graph of
) !! ` ü .
(b) Walk zones. Figure 6.10: Space station tunnels.
Intersecting the lower semicircle with the horizontal line having equation M½.
of 6 6 @9ÒH ) !! ` ü . For Deck , we simultaneously solve the system of equations
M Ù M
) !! ` '^
Û
Plugging in M[b^ into the first equation of the system gives é!! `
i i ^ eJo ;i i.e., ¦ ) eJ o#¦IjT ` This tells us that Ø 6 ÉjT ` 1^9 and Ùy 6 jT ` 1^9 . In a similar way, for Deck , we find ¤ 6 É0!A^o´1z
¥Ç
6 0!A^o´1z
CHAPTER 6. GRAPHICAL ANALYSIS
86
i
In the case of Deck , we would paint a safety stripe jT ` feet to the right and left of the centerline. In the case of Deck , we would paint a safety stripe 0!A^o 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 ! 6 @9 as a nice compact expression in the variable . Sometimes we have to work harder. As an example of what we have in mind, consider the graph in Figure 6.11(a):
Ú
PSfrag replacements
axis
É
É
1
1
axis
2
3
4
S
axis
−1
!
6 9Ò
¡ ¢
É (a) Graphing a multipart function.
if if if if if
%Q' Ç X[ N' Ç X ! !0' ÇX"+j j ' +ÇX £ Ç
(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 %B' *' , this curve will define the graph of + some function ! 6 9 . To see why this is true, imagine a vertical line moving from left to right within the domain %d' ¤' on the 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 ! 6 9 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 X 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 ! 6 jT ` ^9 . First, we would find which of the five cases covers Ç"jT ` ^ , then apply that part of the rule to compute ! 6 jT ` ^98$ .
6.3. MULTIPART FUNCTIONS
87
Our first multipart function example illustrated how to go from a graph in the plane to a rule for ! 6 9 ; we can reverse this process and go from the rule to the graph. Example 6.3.1. Sketch the graph of the multipart function 6
6 @9Ò
¡
> ) >ü ¢
if * ''É N' *' if É if >ò
PSfrag replacements
Solution. The graph of 6 6 9 will consist of three pieces. second part Ú axis The first case consists of the graph of the function M axis of graph 6 6 9>y on the domain '$É , this consists of all points on the horizontal line M [ to the left of and including the S axis point 6 Én1å9 . We have “lassoed” this portion of the graph first part of graph third part of graph in Figure 6.12. Likewise, the third case in the definition ¯ graph of ¥ {$Áu yields the graph of the function MN¦6 6 @9Ò[ on the domain 3òô ; this is just all points on the horizontal line M ô Figure 6.12: Multipart functo the right of and including the point 6 n19 . Finally, we tion jT¿k/ . need to analyze the middle case,
which means we need to look at the graph of ; ) > on the domain É§'q¤' . This is just the upper semicircle of the circle of radius 1 centered at 6 %´1d9 . If we paste these three pieces together, we arrive at the graph of 6 6 @9 . PSfrag replacements axis axis axis
5 4
feet
Example 6.3.2. You are dribbling a basketball and the * function w /6 29 keeps track of the height of the ball’s center above the* floor after / seconds. Sketch a reasonable graph of w 6 /:9 .
3
2 1 1
seconds
2
Solution. If we take the domain to be %£')/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 / axis?
CHAPTER 6. GRAPHICAL ANALYSIS
88
6.4 Exercises Problem 6.1. The absolute value function is defined by the multipart rule:
¿ ¡%© ©¿¿ ¨
if if
¨
h ¿ k ¿Íõ
The graph of the absolute value function is pictured below:
Ú
axis
Problem 6.4. Pizzeria Buonapetito makes a triangularshaped 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 ¿ be the bottom length of Alice’s portion and Û be the length of the cut as shown, answer the following questions:
y= x 
shaded region is the piece Alice takes
PSfrag replacements
S
PSfrag replacements
axis axis
axis
axis axis
(b) (c)
¨
¨
¨
,
¨
¨
,
¨
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.
Problem 6.2.e For each of the following func¨e ¨ tions, graph ¿k/ and jT¿/Ò¡ ¿k/ in the same coordinate system and give the multipart rule for jT¿k/ .
¿/¡?©Ò¹ z¿m©Í ¿/ ¡z¿Ì©É ¿k/ ¡¿ÌÂÉ
e
(a) e
(b) (c)
e
Problem 6.3. Solve each of the following equations for ¿ . (a) (b)
¨
r&¿/¡Z , where Tj ¿/¡ z¿ÌÂY e  ¿/¡ZR where j
e
(c)
&¿/¡ ©
z¿ Ë©¢¿
¾ &¿/n¡?©4 where ¾ &¿/¡ © © ©YR¿ Â « ¨ ¿
if if
¨
¿Ïõ ¿÷
if if
, .
¿lh?©8 ¿;¬©@
, .
y
x
x 10
20
(a) Find a formula for Û as a multipart function of ¿ , for hD¿h . Sketch the graph of this function and calculate the range.
¨
©É . ¨ ¨ ¨ ¨ ¨ ¨ Solve for ¿ : ¿ ¡ ; ¿ ¡ , ¿ ¡?©> . ¨ Sketch the graph of ÛY¡ ¨ ¿TÂ and ÛY¡ ¿
(a) Calculate:
20 y
(b) Find a formula for the area of Alice’s portion as a multipart function of ¿ , for hI¿ªhÇ . (c) If Alice wants her portion to have half the area of the pizza, where should she make the cut?
Problem 6.5. This problem deals with cars traveling between Bellevue and Spokane, which are 280 miles apart. Let Ä be the time in hours, measured from 12:00 noon; for example, ÄÒ¡?©> 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 «( Ä2/ that computes her distance from Bellevue at time Ä . 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 :Ä2/ for his distance from Bellevue at time Ä . Sketch the graph, specify the domain and determine the range.
6.4. EXERCISES
89
(c) Find a function ,+Ä2/ that computes the distance between Joan and Steve at time Ä . Problem 6.6. 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 Ä . (Hint: This will be a multipart function.) Try to sketch a graph of this function.
edgar d(t)
PSfrag replacements axis axis
home plate
axis
Problem 6.7. A course uses four scores to determine your final grade. These scores are numbers between 0 and 100; one each for homework ¬ , quizzes ú , midterm exams 5 and a final exam G . The professor first calculates your course percentage , based on this information:
Homework
Quizzes
Midterms
Final
G
¬
is worth 10%.
5
are worth 20%. are worth 30%.
is worth 40%.
(a) Find an equation involving ¬ , ú , 5 , G for calculating your course percentage . If , yourR scores are ¬ ¡ ª , ú¡ z , 5y¡ G4¡ , what is your course percentage? (b) Assume these facts:
A course grade is either 0 or a decimal number between 0.7 and 4.0.
(c) If your are ¬Z¡?ª , ú ¡?z , R , scores G4¡ what is your final grade?
A course percentage of É¡Íª gives a grade of 4.0 and a course percentage of ½¡0 gives a grade of 0.7. Between these two percentages, the grade is given by a linear function.
5
¡
,
(d) What course percentage is needed to get a grade of 3.7?
Problem 6.8. A polynomial function has the form
90 ft
ú
Find a multipart function that calculates the grade based on the course percentage . Make sure to specify the domain and range of this function.
Û ¡F &¿/
Z ¨ ¿ Ñ ¨ ÂÐÐÐkÂÅ ¿ ÂÅ ¨ ¿2ÂÅ p Ñ Z ¨ Å Ñ are all given con &¿ / is the stants. The degree of a polynomial highest power of ¿ occuring in the expression ÛÍ¡U &¿ / . (Note: Linear functions, which have the standard equation form Û¡H°<¿wc Â 7 , are
Û¢¡D ¿k/¡ZÅ Ñ ¿ Ñ ÂÅ Ñ Z where Å Å ¥Å Å p ¨
degree one polynomial functions. Quadratic functions, which have the standard equation form Û)¡ Åe¿¹ÂS7v¿<Â , are degree two polynomial functions.) The roots of a polynomial function Û½¡ &¿/ are the ¿ intercepts for the graph; these are found by factoring the polynomial &¿/ into degree one terms.
(a) Use factoring to find the ¿ intercepts of the graphs of these quadratic polynomials: ÛÍ¡ ¿¹ËÂ z¿3Â?ª , ÛU¡Z¿¹ËÂz¿3©I and Û"¡qÅ¿¹Â>7v¿<Â , where Å 7 are constants. (b) If ÛY¡
& ¿/ is a polynomial of degree 2 and ÛÉ¡CjT&¿/ is a polynomial of degree 3, what e
are the degrees of the NEW polynomials: e e e &¿/\jr&¿/ , ÛÉ¡ &¿/ Âjr&¿/ , ÛÉ¡1 ¿k/ ÂâT/ .
ÛÉ¡
(c) This question will focus on a degree three polynomial; these are called cubic polynomials. (c1) Sketch an accurate graph of the e polynomial Û ¡ &¿/¡ ¿¹«;©#e ¿ by plotting points of the form ßÅ ßÅ+/À/ , where Å is a multiple of ¨ .
º
ó©@¿/ and (c2) Sketch the graphs of ÛÉ¡ ¨ ¨ Û¢¡ e ¿k/ and write out the function rules explicitly. e
(c3) Find a e constant B so that the roots of ÛY¡ ?BÊ¿/ are ®ß¹W« ¥^¯ .
CHAPTER 6. GRAPHICAL ANALYSIS
90 (c4) Sketch the graph of ÛÉ¡Ç¿vT© and find its roots Á Á . What feature(s) e ¨ of the graph of Û"¡ &¿/ correlate with the roots Á »Á ?
¨
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) When will the ditch be completely full?
Problem 6.9. Pagliacci Pizza has designed a cardboard delivery box from a single piece of cardboard, as pictured.
(b) Find a multipart function that models the vertical crosssection of the ditch. (c) What is the width of the filled portion of the ditch after 1 hour and 18 minutes?
(a) Find a polynomial function ²+&¿/ that computes the volume of the box in terms of ¿ . What is the degree of ² ?
(b) Find a polynomial function Å&¿/ that computes the exposed surface area of the closed box in terms of ¿ . What is the degree of Å ? What are the explicit dimensions if the exposed surface area of the closed box is ªzR sq. inches?
(d) When will the filled portion of the ditch be 42 feet wide? 50 feet wide? 73 feet wide? Problem 6.11. The graph of a function Û"¡ jT&¿/ on the domain ©Ëª±h ¿ªhª consistes of line segments and semicircles of radius 2 connecting the points ó©Ëªe */ oó©@: T/ Wó I/ u&: I/ uóªe 0/ .
remove squares to get
Ú
axis
x x 20 in
PSfrag replacements axis axis
PSfrag replacements
x x 50 in
S
axis
axis
axis
NOT TO SCALE
(a) What is the range of j ? Problem 6.10. The vertical crosssection of a drainage ditch is pictured below:
(b) Where is the function increasing? Where is the function decreasing? (c) Find the multipart formula for ÛY¡§jT¿/ . (d) If we restrict the function to the smaller domain ©8²hI¿ªhÇ , what is the range? (e) If we restrict the function to the smaller domain kh ¿ªhI , what is the range?
PSfrag replacements axis axis
3D−view of ditch
axis
20 ft
PSfrag replacements axis axis axis
Problem 6.12.
20 ft
R
R R
R
vertical cross−section
Here, ° indicates a circle of radius 10 feet and all of the indicated circle centers lie along the
(a) Simply as far as possible
© Å Å ÂÍ ´Â Æ ¨ m Å , 7 , that simultaneously satisfy
(b) Find these three equations:
Å Â@73© m ¹Å4©D"73Â Å[email protected] Â 73Â
¡
¡
¡
©>
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 Mq³! 6 9 whose graph reproduces the path of the ball.
eplacements axis axis axis 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 M 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: 91
CHAPTER 7. QUADRATIC MODELING
92
Ú
PSfrag replacements
axis
II
axis
III
I
IV
S
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 M'! 6 @9 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.
ements
35 30 25 20 15
axis
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 Ú axis the graph of M[y 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 axis are obtained as the graphs of functions having the form S 2
4
Figure 7.3: Graph of ÛÉ¡
6
¿¹
.
MN ' QA¢ï1
7.1. PARABOLAS AND VERTEX FORM
93
for various constants ' , Q , and ï , with ' B ã % . 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
M½ 'â6 ¢
* 9 6Ò1
*
for some constants ' , , and , with ' ã % . 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 MW ' )QA ï form. For
example, suppose we start with the vertex form , so M ! U y 9 ] j 6 * that ' !1 Cn1W j . Then we can rewrite the equation in the form M ' QAP ï as follows:
+
! 6 ¢)9 üjD"! 6 !Y)9A jI ! Y ` 1 + so ' !1¼Q . u1 ïP ` . 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 M Hj2 Y^vÇ , it not at all obvious why the vertex form is obtained by this equality:
j2 ê ^vPH.j 6 ¢ '9 !A
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 axis * with the graph in Figure 7.3 and hori40 + PSfrag replacements zontally shift it units to the right. To be specific, con* * 30 sider the two cases b! and . To visualize this, 20 imagine making a wire model of the graph, set in on top * of the curve, then slide the wire model units to the right. axis 10 axis What you will obtain are the two “dashed curves” in Fig−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
M
! A 9 + + 6 ¢
P 1
CHAPTER 7. QUADRATIC MODELING
94
which is the plot with lowest point 6 !1%9 and +
9 Jv¢He^´1
M
ements
40
30
20
axis
10
−6
−4
−2
6 ¢
+
*
which is the plot with lowest *point 6 1%9 . In general, if is positive, the
graph of the function * M 6 w 9 is the parabola obtained by shifting the
graph of MN" by units * to the right. * Next, if is negative, shifting units to the right is the axis * <^Q+ ' *')^ , same as shifting units left! On the domain * Figure 7.5 indicates this for the cases B;!12 , using “dashed curves” for the
shifted graphs and a solid line for the graph of M)$ . Using a graphing device, we can axis check that the corresponding equations for the dashed graphs would be 2
4
6
Figure 7.5: Shift to the left.
M
6 ¢ 6 ; !A99 6 ¢
!A+ 9 + ¢ 1
which is the plot with lowest point 6 ;!1%9 and +
M
6 ¢ + 6 99 6 ¢
9 Jv¢He^´1
+
*
which is the plot with lowest point * 6 1%9 . In general, if is negative, the M * 6 Y 9 gives the parabola obtained by shifting graph of the function
the graph of MN" by units to the LEFT.
The conclusion thus far is this: Begin with the graph of MC in * Figure 7.3. Horizontally shifting this graph units* to
the right gives a new (standard) parabola whose equation is M 6 ¢ 9 . We can also imagine vertically shifting the graph in Ú axis ements Figure 7.3. This amounts to moving the graph units 40 30 vertically upward. It turns out that this vertically shifted
20 graph corresponds to the graph of the function M½" 1 . 10 We can work out a few special cases and use a graphing axis S axis −4 −6 4 6 −2 2 device to illustrate what all this really means. −10 + + Figure 7.6 illustrates the graphs of M b in the e % . É e % , and , , leading to vertically cases when Figure 7.6: Vertical shifts. shifted graphs. Positive values of lead to the upper two “dashed curves” and negative values of lead to the lower two “dashed
curves”; the plot of M y is again the solid
+ line. The
+ equations giving these graphs would be M W We% , M W , M] and Mê] He% , from bottom to top dashed plot. If we combine horizontal and vertical* shifting, we end up with the
graphs of functions of the form M 6 É 9 . Figure 7.7(a) illustrates
7.1. PARABOLAS AND VERTEX FORM the four cases with corresponding equations MN identify which equation goes with each curve.
6 ª¦ !A9 ¦
+
95 ; as an exercise,
7.1.2 Second Maneuver: Reflection Next, we can reflect any of the curves M G´ 6 @9 obtained by horizontal or vertical shifting across PSfrag the axis. This replacements procedure will produce a new curve which is the graph of the new function M q´ 6 9 . For example, begin with the four dashed curves in the previous figure. Here are the
+reflected parabolas and their equations are MN. 6 ½¦ !A9 ¦ axis axis S −4 : See Figure 7.7(b). −6
7.1.3 Third Maneuver: Vertical Dilation
If ' is a positive number, the graph of M) ' is usually PSfrag called a vertical dilation of the graph of M There are . . replacements
two cases to distinguish here:
If ' VW , we have a vertically expanded graph.
If %YX ' X$ , we have a vertically compressed graph. This is illustrated for ' ! (upper dashed plot) and ' efv! (lower dashed plot): See Figure 7.7(c).
Ú 30 20 10
−2
4
6
30
Ú
axis
20
S
10
axis
−6
−4
−2
axis
gether all three of the operations: shifting, reflection and dilation. This will lead to the graphs of functions that axis have the form:
* 9 6Ò1
2
(a) Combined shifts.
2
−10
4
6
−20 −30
(b) Reflections.
7.1.4 Conclusion PSfrag replacements
Starting with MBG in Figure 7.3, we can combine to
M½ 'â6 ¢ *
axis
40
30
Ú
axis
25 20 15 10
S
−6
5
axis −4
−2
2
4
6
ã % . If you think about it for awhile, for some ' , and , ' ' (c) Vertical dilations. it seems pretty easy to believe that any standard parabola arises from the one in Figure 7.3 using our three geometFigure 7.7: Shifts, reflections, and dilations. ric 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
M +! 6 9ï 'â6 Ç 9 4 , * for some constants ' , , and and ' b ã % . * The vertex of the parabola is 6 1W89 and the axis of symmetry is the line Ç . If ' V.% , then the parabola opens upward; if ' X$% , then the parabola opens downward.
CHAPTER 7. QUADRATIC MODELING
96
ments
Example 7.1.2. Describe a sequence of geometric operations leading from
the graph of MN" to the graph of MN#! 6 9ËHj 6 ¢H9 ü! .
axis axis axis
î Á axis
reflect across
æ l S æ ðö D
¯
horizontal shift by
vertical shift by
 vertical dilate by
¯
õ
ö
(a) What do the symbols of an equation mean?
20
ments
Ú
axis
15 10 5
axis
−6
−4
−2
−5
2
S
4
axis 6
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 MI! 6 9 is 6 n1!A9 , the line Í 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
−10 −15
(b) What does the equation look like?
vertical dilate reflect vertical shift
Figure 7.8(b) illustrates the four curves obtained by applying these successive steps, in this order. As a refer ence, we include the graph of M as a “dashed curve”:
A horizontal shift by * yields the graph of
MN 6 PH9 ; this is the fat parabola opening upward with vertex 6 n1%9 . A dilation by ' "j yields the graph of M "j 6 > 9 ; this is the skinny parabola opening upward with vertex 6 n1%9 . A reflection yields the graph of M j 6 Y.9 ; this is the downward opening parabola with vertex 6 n1%9 . A vertical shift by Í)! yields the graph of M ]j 6 ¢'9 ü! ; this is the downward opening parabola with vertex 6 n1!A9 .
Figure 7.8: Interpreting an equation.
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
97
Example 7.2.1. Find the vertex form of the quadratic function Mü]j2 ^v¢H .
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
* j2 ê ^vPH 'â6 ¢ 9 Ò1 *
for some constants ' , , . Now, expand the
right hand side of this equation and factor out coefficients of and :
j2 ^vY) j2 ^vY)
j2 ^vY) 6 jT9 6 ^9 P 6 É9
'â6 P
' 6m' 9
'â6
* 9 * 6 ü! * ! ' 6 ;! '
*
* 9T * * ' 9 P 6(' 89
If this is an equation, then it must be the case that the coefficients of like powers of match up on the two sides of the equation in* Figure 7.9. Now we have three equations and three unknowns (the ' 1 1W ) and we Equal
eplacements axis axis
axis
¶U· ¸
µ
·¸Uµ¶
»
µ
¶U· J¸
a¹
µ¶U·¸
·
¸Uµ
0º
¹
»
¶
µ
U¹
» ½¼¿¸ ¶U·
Equal Equal Figure 7.9: Balancing the coefficients.
can proceed to solve for these:
j
^ É
'
*
'
;*@! '
The first equation just hands us the value of ' qj . Next, we can plug this value of ' into the second equation, giving us
^
* ;! ' * *! 6 ; jT9 ^ 1
CHAPTER 7. QUADRATIC MODELING
98
*
so . Finally, plug the now known values of ' and equation:
É
*
into the third
*8
j 6 9T j 6Ò1 '
so "! . Our conclusion is then
j2
^vY'w.j 6 P '9 ü !A
Notice, this is the quadratic we studied in Example 7.1.2 on page 96. 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 MqÏ
and M &! 6 @9Ê. ` ¢ü! .
20
ements
Ú
axis
10
−6
axis
−4
−2
2
S
4
axis
Solution. We will go through the algebra to complete the square, then interpret what this all means in terms of graphical + maneuvers. We have
+ ` ¢ ! ! 6 9 ` ¢
6
−10 −20
Figure 7.10: Maneuvering ÛY¡ ¿¹ .
* 9 * * 89¼ ' 6 ;! ' 9 ¢ 6(' 'Ì6 ¢
This gives us three equations: +
`
'
;*8! '
*
! + ' * We conclude that ' 1
8 u ÿ%&^v! ` and N uH¾ jT ` ^v! . So, this tells ç us that we can obtain the graph of M"A! 6 9 from that of M"[ by these steps:
Horizontally shifting by * H%&^v! ` units gives MN 6 ¢%^v! ` + +
Vertically dilate by the factor ' Y%&^v! ` 9 . gives M½ 6 + Reflecting across the axis gives M H 6 Y%^v! ` 9 . + Vertically shifting by ?)jT ` ^v! gives M & ! 6 @9Ò. 6 ¢ê%&^v! `
9 .
9 jT ` ^v! .
Example 7.2.3. A drainage canal has a crosssection 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
PSfrag replacements
99
axis axis axis
ö
feet
centerline Solution. Impose an 2M coordinate system so that the ¿2¿2 ¿ À2 ¿ À2 ¿ À2 ¿ À¿ À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À À2 parabolic crosssection of the canal is symmetric about ¿2¿2 ¿ À2 ¿ À2 ¿ À2 ¿ À¿ À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À À2 feet ¿2¿2 ¿ À2 ¿ õ À2 ¿ À2 ¿ À¿ À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À À2 the M axis and its vertex is the origin. The vertex form of
¿2¿2 ¿ À2 ¿ À2 ¿ À2 ¿ À¿ À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À ¿2 À À2 any such parabola is M"! 6 93 ' , for some ' V % ; this * is because 6 1W89 6 %´1%9 is the vertex and the parabola Figure 7.11: A drainage opens upward! The dimension information given tells us canal. that the points 6 e%´1e%9 and 6 Ée%´1e%9 are on the graph of ! 6 9 . Plugging into the expression for ! , we conclude that e% e%% ' , so
Ï % a ! @ P 9 % a 9 and . Finally, if the water is 5 feet deep, we must ' 6 6
i i solve the equation: ` 6 + %d+ 9 , leading to Ç ¦ ) ` %¢ ¦ % . Conclude the surface of the water is d feet wide when the water is 5 feet deep.
7.3 Interpreting the Vertex
eplacements axis axis axis
æ ç
aÃEÄ
minimum
maximum
value
value
:ÂÁ
: Á
æ ç
bÃÄ
vertex vertex
æ ç
bÃ
æ ç
bÃ
Figure 7.12: The vertex as the extremum of the quadratic function.
If we begin with a quadratic function MüÅ! 6 9Ì ' QAY ï , 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 ' V% , the vertex is the lowest point on the graph; if ' XW% , 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 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 Mÿ ' .QA ï , we can use the technique of completing the square and arrive at a formula for the coordinates of the vertex in terms of ' , Q , and ï . 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
100
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 !
QAP ï1 6 9Ë '
æ ç æ ç the vertex has coordinates Ø x b Ã 1H! x bÃ {e{ . The second coordinate of the vertex will detect the maximum or minimum value of ! 6 9 ; this is often a key step in problem solving. + Example 7.3.2. Discuss the graph of the quadratic function M
;! )0Y .
ements −4
axis axis
2
−2
4
6
8
Solution. We need to place the equation M * I! 6 9 in vertex form. We can simply compute ' ;! , æ bÃ ç ¥ and p ?&! 6 9Ò ç ä , using Fact 7.3.1:
p
−20
−40
!
axis −60
−80
Figure 7.13:e Sketching ¿k/ .
Û.¡
Æ! 6 @9I
ç + 0¢ 6 9ËH;! H
g g + J o H;! ¢ h
This means that the graph of ! 6 @9 is a parabola opening downward with vertex x ¥ 1 ç ä { and axis ¥ ; see Figp ç p 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 / as the important variable and try to find a function M 6 /29 which describes the height of the ball / seconds after the ball is kicked; this would just be the M coordinate of the ball at time / . The function M 6 /29 is a quadratic function. If we had this function in
hand, we could determine when the ball hits the ground by solving the equation %É M 6 /29 , 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 M #! 6 @9 whose graph models the exact path of
the ball. In particular, we could find where the ball hits the ground by solving %¢#! 6 9 , but we would not be able to determine when the ball hits the ground.
7.4. QUADRATIC MODELING PROBLEMS
101
In Chapter 22, we will see these two viewpoints are very closely related. For now, it is most important to obtain some familiarity with the sort of motion problems we are interested in solving using quadratic functions. PSfrag replacements axis axis
Example 7.4.1. Figure 7.14(a) shows a ball is located on axis the edge of a cliff. The ball is kicked and its height (in feet) + level ground is given by the function M 6 /:9; above
the Ée^:/ J:/> ` % , where / 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 M 6 /29 is a quadratic function with a negative leading coefficient, so its graph in the /e PSfrag replacements coordinate system will be a downward opening parabola. We use a graphing device to get the picture in Fig axis axis ure 7.14(b). The vertex is the highest point on the graph, which axis can be found by writing M 6 /:9 in standard form using Fact 7.3.1:
+
M 6 /:9b
Ée^:/ :J /m ` % g j êJ ^ Ée^ /4 ! h
path of kicked ball
cliff ground level
(a) What it looks like physically.
50
−2
−1
1
M 6 /:9b
/
3
4
5
−50
−100
(b) What it looks like graphically. Figure 7.14: Different views of the ball’s trajectory.
The vertex of the graph of M 6 /:9 is x l 1J^ { , so the maximum height of the ball above the level ground is 86 feet, occuring at time /; l . The ball hits the ground when its height above the ground is zero; using the quadratic formula:
2
+
Ée^:/ :J /m ` % + + % + JK¦ þ 6 J 9 6 Ée^9 6d` %9 ! E e^ jTJeJ sec or ê%JeJ 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., M 6 %9Ò ` % feet is the height of the cliff.
CHAPTER 7. QUADRATIC MODELING
102
Here are two items to consider carefully:
ements axis
§ Ç
is NOT the path followed by the ball! Finding the 1. The graph of actual path of the ball is not possible unless additional information is given. Can you see why?
!!!
axis
CAUTION !!!
axis
§ Ç
Ç
is defined for all ; however, in the context of the 2. The function ÇÈÉ problem, there is no physical meaning when . 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
that the path of the balloon follows the graph of M Ê! 6 @9ïÏ O¥O p . u u The 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?
ements axis axis axis
Solution. In the coordinate system indicated, the origin is the takeoff point and the graph of M)Ë! 6 9 is the path of balloon path 200 the balloon. Since ! 6 @9 is a quadratic function with a neg100 ative leading coefficient, its graph will be a parabola which opens downward. The difficulty with this problem is that (ft) lake at any instant during the balloon’s flight, the “height of 500 1000 PSfrag replacements the balloon above the ground” and the “height of the balFigure 7.15: Visualizing. axis loon above the lake level” are different! The picture in axisFigure 7.16 highlights this difference; consequently, two different funcaxistions will be needed to study these two different quantities. height above lake (ft)
balloon
height of balloon above lake level
200 100
lake
Ì Ì
height of balloon above ground level c c
500
height of ground above lake level
1000
Figure 7.16: The height of the balloon
Û
as a function of ¿ .
The function M Í! 6 9 keeps track of the height of the balloon above lake level at a given location on the horizontal axis. The line à with slope O O passing through the origin models the ground level.
7.4. QUADRATIC MODELING PROBLEMS
103
This says that the function
M
e %
keeps track of the height of the ground above lake level at a given location on the horizontal axis. We can determine the maximum height of the balloon above lake level by analyzing the parabolic graph of M¦! 6 9 . Putting ! 6 9 in vertex form, via Fact 7.3.1,
¢ ` %%9 !%% 6 ! ` % % The vertex of the graph of M Å! 6 9 is 6a` %%´1!%%9 . This just tells us that the 6 @9b
!
!
maximum height of the balloon above lake level is 200 feet. To find the landing point, we need to solve the system of equations
Ù
M
M
H ¥O O p u Û O u
As usual, plugging the second equation into the first and solving for , we get +
e %
i J ` i 6 ¢êJ ` 9b
!
i ! ` % % J `
`
% %
eplacements height above lake level (feet)
axis axis axis
vertex (high point above lake)
O¥O O¥O
landing point
takeoff point
u O¥O
O¥O¥O
horizontal distance from launch (feet)
Figure 7.17: Locating the takeoff and landing points.
CHAPTER 7. QUADRATIC MODELING
104
i
From the algebra, we see there are two solutions: % or J ` ; 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 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 i position where F[J ` and to find the M coordinate of this landing point i we plug Í J ` into our function for the balloon height above lake level: i i i i MN#! 6 J ` 98HJ ` feet. So, the landing point has coordinates 6 J ` 1kJ ` 9 . Next, we want to study the height of the balloon above the ground. Let MN¦6 6 @9 be the function which represents the height of the balloon above the ground when the horizontal coordinate is . We find 6
6 9
height of the balloon elevation of ground above lake level with 4Î above lake level with horizontal coordinate ÐÏ horizontal coordinate ÐÏ
Î
6 9T&6 6 9 !
g
!
+
µ
! ` %% ¶U·
` h¸
(balloon above lake level)
!
Y ! ` % % 6
+
g
µ¶U·¸ µ
minus
e ¶U% · h ¸
(ground above lake level)
i j ` 9 ) ` jTa0!1
Notice that 6 6 9 itself is a NEW quadratic function with a negative leading + be a downward opening parabola. coefficient, so the graph of MN¦6 6 9 will i The vertex of this parabola will be 6 j ` 1 ` jTa0!A9 , so the highest elevation of the balloon above the ground is 153.12 feet. We can now sketch the graph of 6 6 9 and the horizontal line determined by Mb ` % 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
H ¥O O ¾ O Ù Û M ` % u + + Plug the second equation i i into the first and apply the quadratic foro^ ` or J ^ This tells us the two possible coormula to get W M
dinates when the balloon is 50 feet above the ground. In terms of the + + where the balloon is original coordinate system imposed, i i the two places i 50 feet above the ground are 6 J ^´1 ` J ` 9 and 6 o^ ` 10!o^9 .
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 equa
7.4. QUADRATIC MODELING PROBLEMS
105
vertex (high point above ground)
eplacements axis axis
graph of height above ground function
O¥O
axis
í line
u O¥O places where
Ú
uO
O¥O¥O uO
feet above ground
Figure 7.18: Finding heights above the ground.
tions for lines in the plane. We could then ask if there is a similar characterization of parabolas. Important Fact 7.4.3. Let Ø 6 1§M 9 , Ù 6 1§M 9 and ¤] 6 1§M 9 be l l three distinct noncollinear points in the plane such that the 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 M &! 6 @9Ò ' QTcï and we can find these coefficients by simultaneously solving the system of three equations and three unknowns obtained by assuming ØT1Ù and ¤ are points on the graph of M½#! 6 9 :
ÑK
' QT ï
êM êM
' l QT l ï
êM l Ó
' QT ï
¡ ¢
Ò
Example 7.4.4. Assume the value of a particular house in Seattle has increased in value according to a quadratic function M ä 6 9 , where the units of M are in dollars and 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 ä 6 9 , 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 M)ä 6 9 .
CHAPTER 7. QUADRATIC MODELING
106
We are going to work in a 2M 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 variable, which represents time, will denote the number of years the house is owned. For the M 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 = $ n1k%%% For example, a house valued at $!nj ` 1k^%% would be worth !nj ` ^ 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: Ø 6 %´1 ` %9 , Ù 6 e%´1J%9 and ¤ 6 !%´1!%%9 . If we plot these points, they do not lie on a common line, so we know there is a unique quadratic function
ä 6 9Ê ' QT¢ ï whose graph (which will be a parabola) passes through these three points. In order to find the coefficients ' , Q , and ï , we need to solve the system of equations:
' %
[email protected]%ï 'â6 e%9 Q 6 e%9 ï 'â6 !%9 Q 6 !%9 ï
¡ ¢
Ò Ñ
` % J% ' !%% Ó
1
which is equivalent to the system ¡
ï
+ e%% ' H e%2Q ï %% ' ü !%2Q ï ¢
Plugging ïP
Ò
Ó
Ñ
` % into the second two equations gives the system
+ e%% ' )e%2Q %% ' !%2Q Ù
` % J% ' !%%
)j% $ ` % Û
Solve the first equation for ' , obtaining ' the second equation to get:
+
g
(7.1)
l O æ ¥O O O(ç , then plug this into
j%He%2Q !%2QU$ ` % ü e%% + h 0!% %2Q ü !%2QU$ ` % j QF. ! Now, plug Q.ô l into the first equation of Equation 7.1 to get e%% ' e% x l { )j% ; i.e., ' ä O We conclude that o j MN ä 6 @9Ê Y ` %´1 !% ! %%
keeping in mind the units here are Ô .
7.5. WHAT’S NEEDED TO BUILD A QUADRATIC MODEL?
107
To find the value of the house on January 1, 1996, we simply note this
ä M ä ! ^ Ò 9 ! ^ 9 of ownership. Plugging in, we get is after Ç"!^ years 6 6 O ý
l !^D ` %ôjT ` _! i.e., the value of the house is $jT ` 1 !%% . To find when the house will be worth $1,000,000, we note that $ n1&%%%´1k%%%P$n1&%%% K and need to solve the equation
e%%%P ä 6 9ô %
o ! % o ! %
!
By the quadratic formula,
+
l ¦ x l { ! x i ` ¦ ) 0!Ac! + + %o ++ i &^ or
`
j
äO {
!
j
Y ` % Yo ` %
x ä O { 6
cjT
+ + Because represents time, we can ignore the negative solution i and so the value of the house will be $ n1&%%%´1k%%% after approximately &^ years of ownership.
7.5 What’s Needed to Build a Quadratic Model? Back in Fact 4.7.1 on page 50, 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 noncollinear 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 ! 6 9 has the form
* 6 @9Ò 'â6 P 9 Ò1 * * where 6 W1 89 is the vertex. If we are given and , together with another point 6 Ov1§MÊO9 on the graph, then plugging in gives this equation: * MÊO3 'â6 OÌ 9 6 8 !
The only unknown in this equation is ' , which we can solve for using algebra. A couple of the exercises will depend upon these observations.
CHAPTER 7. QUADRATIC MODELING
108
7.6 Exercises Problem 7.1. A toy rocket is launched from a position 100 feet from the centerline of a house; see picture below. The highest point on the house roof is 35 feet above the ground and the roof slope is « « . The sides of the house are 10 feet high. Assume coordinates are imposed so that the ¿ axis coincides with ground level and the centerline through the house intersects the ground at the origin. Rocket path is a parabola
where the rocket lands, its maximum height above the ground and whether the rocket strikes the house. If the rocket strikes the house, determine exactly where it hits the house. Problem 7.2. Write the following quadratic functions in vertex form, find the vertex, the axis of symmetry and sketch a rough graph. e
(a) (b)
e
(c)
e
centerline through house
(e) PSfrag replacements axis axis axis
35 feet
10 feet
ground level
100 feet
The path of the rocket is always the graph of a quadratic function, but the actual path will depend on the type and amount of fuel used. Answer these questions: (a) If the path is given by the graph of ÛY¡?© ¨{¨ p ¿ ©?¥¼¿ , determine where the rocket lands, its maximum height above the ground and whether the rocket strikes the house. If the rocket strikes the house, determine exactly where it hits the house. (b) If the path is given by the graph of ÛY¡?© ¨{¨ p ¿¹@Â?¥RR , determine where the rocket lands, its maximum height above the ground and whether the rocket strikes the house. If the rocket strikes the house, determine exactly where it hits the house. (c) If the path is given by the graph of ÛY¡© ¨ &¿¹@ÂIz¼¿4©?¥R0/ , determine where the¨p rocket lands, its maximum height above the ground and whether the rocket strikes the house. If the rocket strikes the house, determine exactly where it hits the house. (d) If the path is given by the graph of ÛY¡?© ¨ ¿¹ÂÍ§0¼¿>ÂY0R0/ , determine p
e
(d) e
¿/¡z¿¹@©Íª¿4Â¢: R ¿/ ¡Ç¿¹@©Í§z¿4© ¿k/ ¡¿¹© « ¿mÂÍ . ¤ ¿/ ¡z¿¹ . ¿k/ ¡ ¨ ¿ .
. .
¨p p
Problem 7.3. 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 the price of the stock is modeled by a quadratic function. (a) Find the quadratic function :Ä2/ giving the stock price after Ä 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)? (c) When will the stock become worthless? graph of of the In the same¨ coordinate¨ system, sketch the graph of Û¡ ¿¹4©Íz¿w©Í , give the multipart rule and label the ¿ and Û intercepts of the graph. Problem
7.4.
Sketch
the
ÛY¡ ¿¹©Yz¿4©É . Label the coordinates ¿ and Û intercepts of the graph.
Problem 7.5. 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 followse is the graph of the quadratic function Û'¡ &¿/ ¡q© º ¿¹Â º ¿ . pÕp from the The land drops at a constant incline plateau at the rate of 1 vertical foot for each 5 horizontal feet. Answer the following questions:
7.6. EXERCISES
109
height above plateau (feet)
Suspended magically in the air are bowls of paint. Each bowl is ft wide at its mouth. If the marble lands in a bowl, it will become the color of the paint in the bowl and magically pass through the bowl, continuing on its way. The location of each bowl and the color of its paint is given in the table. The marble starts out clear. What color is the marble when it hits the ground?
balloon
takeoff
horizontal distance from launch (feet)
replacements
ground incline
axis axis axis
(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 7.6. (a) Suppose ¿/U¡ô¿v<© . Doese the point »R &/ lie on the graph of ÛY¡ &¿/ ? Why or why not? e
(b) If
Paint White Yellow Blue Green
Z
¯ Ä hIÄÖhR ²+ Ä2/¡© ¨ ¨ ÄeÂ §Ä¨{©D p W zhÄ×h0R
is a constant, where does the line Û.¡4ÂF27 intersect the graph of ÛW¡ ¿¹Â`7¹¿mÂ`7 ?
(d)
is a constant, where does the
ÛY¡Í© R ¿ Â¢¿mÂ? white
yellow blue
ag replacements
axis
18 ft
dave’s velocity v(t)
15 10 5
t sec 50
100
150
200
(a) When does Dave have maximum velocity and what is his maximum velocity? What is his “pace” at this time in units of “minutes/mile”? (b) During the first 200 seconds, how much time does Dave spend running at a speed of at least 12 ft/sec?
in the pictured coordinate system.
axis axis
ft/sec 20
Û ¡Z´©Åv intersect the Y graph of PSfrag replacements ÛY¡¿©ÉvÅ¿ÌÂ? ? axis Where does the graph of ÛÉ¡?©8z¿vÂI¿ÌÂ ¥ axis intersect the graph of ÛY¡ ¿vÂY¿Ì©Í ? axis
Problem 7.7. (Magical Marble Toss) A marble is thrown into the air from a ledge so that it starts ft above ground. The path of the marble is given by the graph of
0 ¹&¥ I/  ¹ ª /  v zT/  &"/
Problem 7.8. Dave starts off on a run from the IMA. Assume he runs in a straight line for the first 200 seconds. Here is a picture of the graph of his velocity ²QÄ2/ at time Ä seconds; we will use distance units of FEET and time units of SECONDS. Here is the actual formula for Dave’s velocity:
7
(c) If Å line
Bowl Coordinates, Left Rim
green
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?
CHAPTER 7. QUADRATIC MODELING
110 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. Steve likes to entertain friends at parties with “wire tricks.” Suppose he takes 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 7.14. In each case, find a quadratic function whose graph passes through the given points: (a) (b) (c) (d)
ó 0/ , »&"/ and ñ¹©?"/ . å©4&"/ , »R¼©Y&/ and  v I/ .  v&2/ ,  ¹ &/ and  v"/ . ó"/ , » &"/ and  R */ .
Problem 7.15. The market for summer movies is intense. The movie studios use every form of media to promote their movies. Each time a movie is mentioned either on TV, in the papers, over the web, or in a magazine article is considered an advertisement for the movie. Let Äï¡ represent the first day of the summer movie season. For the movie, Rat Race, the maximum number of advertisements, 0R ,
occurred on the nd day of the summer sea ). On the th day, the number of son (Ä;¡ advertisements was only . (a) Develop a mathematical model, °NÄ2/ , that represents the number of advertisements of Rat Race as a function of the day of the summer season, assuming the function is either zero or a quadratic model. (This will be a multipart function.) Specify the time period when there no advertisements.
(b) Let BLÄ2/¡Í©: ¼:ª¼Â<: ªzÄ=©ï ¼R¹ª¼Ä represents the number of advertisements for the movie, Planet of the Apes. Determine when the number of advertisements for Rat Race was 100 or more than the number of advertisements for Planet of the Apes. Problem e 7.16. If the graph of the quadratic function ¿k/Ì¡ü¿ Â<,¿;Â0, has its vertex on the ¿ axis, what are the possible values of , ? e What if ¿k/¡¿¹ÂD0,¿m© ,ÂÍ ? Problem 7.17. For each of the following equations, find the value(s) of the constant U so that the equation has exactly one solution, and determine the solution for each value. (a) (b) (c) (d)
U:¿ ¢ Â ¿mÂ?Ò¡ ¿ ÂU:¿mÂ?Ò¡ ¿ Â¢¿mÂU ¡? ¿ Â:U ¿mÂ¢& U Â?Ò¡Ç
Problem 7.18.
(a) Solve for
:Ò¡ÇTÄT©?"/ Â?
Ä
¿ ÛÉ¡I¿ ÂÉz¿mÂD
(b) Solve for
Problem 7.19. (a) What ise the maximum value of the function &¿/F¡ô¢© ¿v ? In a grammatically correct sentence, explain why the maximum is that particular value. (b) For what values of Å , 7 and does the equation Å¿¹Â7v¿Â ¡ have two distinct solutions? One solution? No solutions?
¿ º ©YR¿ ÂY¡Ç Find all solutions to ¿º@©YR¿¹@©Y¡Ç
(c) Find all solutions to
.
(d)
.
Chapter 8 PSfrag replacements
Composition
axis axis axis
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. PSfrag replacements To set the stage, let’s look at a simple botany experioxygen rate ment. Imagine a plant growing under a particular steady axis þ ØÙ light source. Plants continually give off oxygen gas to axis ØÚ the environment at some rate; common units would be axis Ø liters/hour. If we leave this plant unbothered, we mea Ø û sure that the plant puts out 1 liter/hour; so, the oxygen hours Ú Ù û þ output is a steady constant rate. However, if we apply a flash of high intensity green light at the time /F and measure the oxygen output of the plant, we obtain the (a) Flash at ÄÒ¡Z . plot in Figure 8.1(a). Using what we know from the previous section on quadratic functions, we can check that a reasonable þ oxygen rate ØÙ model for the graph is this multipart function ! 6 /:9 (on the ØÚ domain %Q' /R' e% ): Ø
Ø
!
6 /298
¡ ¢
/ ç /müj l l
/ ' if i ' R / ' j if N if j£' /
û û
(b) Flash at
hours Ú
þ
Ù
Ä8¡
.
Suppose we want to model the oxygen consumption Figure 8.1: Light flashes. when a green light pulse occurs at time /D ` (instead of time /$ ), 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 ! 6 /29 in hand (the known function) to build the model we want (the new function)? We will return in Exam111
CHAPTER 8. COMPOSITION
112
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 ! and 6 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 ! and 6 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 ! and 6 affect those of the composition; we defer that to the next subsection once we have the mechanics down.
PSfrag replacements axis
;
in
î
è
the “ ” function
S
è ;
ð
:
the “ ” function
in
Ú
í î
S
axis axis
out
in
out
í
Ú
Sð :
out
í î î “composed” function
Ú :
è ;
ðdð
Figure 8.2: Visualizing a composite function ÛY¡
¾ ÜÛ/¡ e
jTÜÛ/f/ .
Here is a very common occurrence: We are handed a function MN#! 6 9 , which means given an value, the rule ! 6 9 produces a new M value. In addition, it may happen that the variable is itself related to a third variable L by some different function equation 'u6 6 L9 . Given L , the rule 6 6 L9 will produce a value of ; from this we can use the rule ! 6 9 to produce a M value. In other words, we can regard M as a function depending on the new independent variable L . 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
113
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? PSfrag replacements How much time must elapse so that the area enclosed by the leading ripple is 300 square feet? axis axis axis
Solution. We know that an object tossed into a pond leading ripple after Ü seconds will generate a series of concentric ripples, which grow leading ripple after seconds ö steadily larger. We are asked questions that relate the area of the circular region bounded by the leading ripple ¯ ({ Ü and time elapsed. leading ripple after õ second Let denote the radius of the leading ripple after / seconds; units of feet. The area of a disc bounded by a Figure 8.3: Concentric ripples. leading ripple will be ÿ[ 6 n9m (@ This exhibits as a function in the variable . However, the radius is changing with respect to time:
w" 6 /29Ò radius after / seconds
g
!Acj
feet / seconds "!Akj/ feet sec h
So, [ 6 /:9 is a function of / . In the expression yÿ 6 9 , replace “ ” by “ 6 /:9z1 ” then
'( 6 !Acj/29 ` c!o2(@/
The new function 'â6 /:98 ` _!o2(8/ gives a precise relationship between area and time.
To answer our first question, 'â6 ^9Ò ` oJkj feet is the area of the region bounded by the leading ripple after 6 seconds. On + the other hand, if
¦ þ j%%+ f 6a` c!o2(49Ì ¦ c! ` . Since / repre'Ì6 /29Ì]j%% ft ;, j%% ` _!o2(8/ , so /P sents time, only the positive solution /; c! ` seconds makes sense.
We can formalize the key idea used in solving this problem, which is familiar from previous courses. Suppose that
M½#! 6 @9 and that additionally the independent variable is itself a function of a different independent variable / ; i.e.,
Ç¦6 6 /:9 Then we can replace every occurrence of “ ” in ! 6 9 by the expression “ 6 6 /:9z1 ” thereby obtaining M as a function in the independent variable / . We usually denote this new function of / :
M½#! 6 6 6 /299¼
CHAPTER 8. COMPOSITION
114
We refer to ! 6 6 6 /:99 as the composition of ! and 6 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 M Ê! 6 9 and Ç¦6 6 /:9 , then Procedure 8.1.2 is what you need. Important Procedure 8.1.2. To obtain the formula for ! 6 6 6 /:99 , replace every occurrence of “ ” in ! 6 9 by the expression “ 6 6 /29 ” Here are some examples of how to do this: Examples 8.1.3. Use the composition procedure in each of these cases. (i) If MN#! 6 @9Ò ! and Ç46 6 /298 !n/ , then ! 6 6 6 /:998#! 6 !n/:98"! . (ii) If MN)j2¢
i
and Ç¦6 6 /29Ò
(iii) If MN#! 6 @9Ò
i
`
6 !n/m' 9 + 6 !n/Ì
H+ 9 ) / /4 A!
þ 6 Y Tj 9 and Ç¦6 6 /:98"!n/ ) , then
! 6 !n/ '9
! ]þ + > 6 !n/ ) üjT9
i ! þ / p 'e^:/ H
(iv) If MN#! 6 @9Ò !ï
6 6 6 /:99Ï
(v) If MN#! 6 @9Ò !
, then ! 6 6 6 /:998&! 6 9Ò"j E
+
!
!
+
' and Ç¦6 6 :/ 98"!n/4) , then
6 6 6 /:99Ï !
+
6 6 6 /:99Ï
and Ç¦6 6 /:98)/4¤Ý , then
6 /4¤Ý 9 6 /m
ÞÝI9
/ !nß/ Ý 5Ý !
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,
MN
*
6 9Ê þ
'
can be written as the composition ! 6 6 6 @99 , where MWG! 6 î 9w ) î and î *
6 6 @9Ò) Ç . Each of the functions * ! and 6 is “simpler” than the original , which can help when studying . Examples 8.1.4. Here we use composite functions to “simplify” a given function.
8.1. THE FORMULA FOR A COMPOSITION î (i) The function M where MN#! 6µî 9Ò
115
can be written as a composition Mê¦! 6 6 6 @99 , S æ l ðö D à and î 46 6 @9Ò"Pj . ö D
+ of radius (ii) The upper semicircle
! centered at (1,2) is the graph of the function M '!P þ 6 Y'9 . This function+ can be written as a composition M &! 6 6 6 99 , where MN#! 6µî 98"! ) î and î ¦6 6 9Ò"Y) .
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 MüÅ! 6 9 and ÍA6 6 /29 , then MÅ! 6 6 6 /299 is the composition. This illustrates that the three variables / , , and M can all be of different types. For example, / might represent time, could be speed and M could be distance. If we are given two functions that involve the same independent variable, like ! 6 9Ë and 6 6 9Ò"!N , then we can still form a new function ! 6 6 6 99 by following the same prescription as in Procedure 8.1.2: Important Procedure 8.1.5. To obtain the formula for every occurrence of “ ” in ! 6 9 by the expression “ 6 6 @9 .” !
For our example, this gives us: !
6 6 6 998&! 6 !P'9Ê 6 !P'9
Here are three other examples:
If ! 6 9Ë ) , 6 6 @9Ò ! H , then ! 6 6 6 99Ò ) ! ' . If ! 6 9Ë , 6 6 9Ò"!Y) , then ! 6 6 6 998 . S S_D
If ! 6 9Ë , 6 6 9ÒI áq , then ! 6 6 6 @9¼9Ê" á ! áq .
6 6 6 99 , replace
axis axis
õ
Ú
116
CHAPTER 8. COMPOSITION
axis
Example 8.1.6. Let ! 6 9Ò" , 6 6 9Ê ïê * and 6 @9Ê" * < . Find the formulas for ! 6 6 6 99 , 6 6 ! 6 @9¼9 , ! 6 6 99 and 6 ! 6 @99 . Discuss the relationship between the graphs of these four functions.
¶³ Ö
ö 0
ö
õ
*
õ ö
S
axis

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 jB' *' j , together with the graph of ! 6 9Ê .
9¼9 66 [email protected] 6 6 ! 6 @ * 9¼9 ! 6 * 6 @9¼9 6 ! 6 @9¼9
!
6 P '9Ò 6 6 9Ê" ! 6 P'9Ò * 6 9Ò"
!
6 Y'9 '
Y ' 9 6
)
We can identify each graph by looking at its vertex:
!
6 9 has vertex (0,0) !
6 6 6 @9¼9 has vertex (1,0)
6 !
6 ! 6 @9¼9 has vertex (0,1) * 6 6 @9¼9 has vertex (1,0)
* 6 ! 6 @9¼9 has vertex (0,1) Horizontal or vertical shifting of the graph of ! 6 @9Ò" graphs: See Figure 8.4.
gives the other four
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 ! 6 6 6 @9¼9 of ! 6 9 and 6 6 9 , the range values for 6 6 @9 must lie within the domain values for ! 6 9 . This may require that you modify the range values of 6 6 9 by changing its domain. The domain values for ! 6 6 6 99 will be the domain values for 6 6 @9 .
eplacements axis8.2.
DOMAIN, RANGE, ETC. FOR A COMPOSITION
axis
117
axis
in
out
in
out
î î domain
è
è
:
the “ ” function
the “ ” function range of domain of
è
range of
:
è ðåð S
:
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 115. Important Procedure 8.2.1. To obtain the formula for ! 6 6 6 99 , replace every occurrence of “ ” in ! 6 9 by the expression “ 6 6 9 .” In addition, if there is a condition on the domain of ! that involves , then replace every occurrence of “ ” in that condition by the expression “ 6 6 9 .” The next example illustrates how to use this principle.
Example 8.2.2. Start with the function M ! 6 9 on the domain É)' *' . Find the rule and domain of MN#! 6 6 6 @99 , where 6 6 @Ò 9 Y' . Solution. We can apply the first statement in Procedure 8.2.1 to find the rule for MN#! 6 6 6 99 :
M
6 6 6 99
! 6 ¢H9Ò ¢ H 9 6
!PH !
To find the domain of M]â! 6 6 6 99 , we apply the second statement in Procedure 8.2.1; this will require that we solve an inequality equation:
É)' É) ' %.'
6 9 ¢'
6
'
'
'
!
The conclusion is that M &! 6 6 6 998"
I!âF on the domain %Q' *' ! .
Example 8.2.3. Let MWâ! 6µî 9 ) î , î ã6 6 9<ÿ[ . What is the largest possible domain so that the composition ! 6 6 6 @9¼9 makes sense?
PSfrag replacements PSfrag replacements 118
axis
axis axis
í î Ú
axis
axis
Ú
range
: à
ä
à
ð
domain
ÛY¡ È
í
î í
axis
à
(a)
CHAPTER 8. COMPOSITION
.
à
è ð S
SED
desired range
æ
S
required domain
(b)
¡ ¿;¡F . ä
Figure 8.6: Finding the largest domain for
e
[jT&¿/f/ .
Solution. The largest possible domain for M 4! 6 î 9 will consist of all nonnegative real numbers; this is also the range of the function ! 6µî 9 : See Figure 8.6(a). To find the largest domain for the composition, we try to find a domain of values so that the range of î ½6 6 9 is the domain of M4! 6µî 9 . So, in this case, we want the range of 6 6 @9 to be all nonnegative real numbers, denoted %Q' î . We graph î å6 6 @9 in the î plane, mark the desired range %B' î on the vertical î axis, then determine which values would lead to points on the graph with second coordinates in this zone. We find that the domain of all values greater or equal to É (denoted É0'[ ) leads to the desired range. In summary, the composition M"! 6 6 6 @9¼97 ) ¢H is defined on the domain of values É)' . 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 /ê and measure the oxygen output of the plant, we obtain the plot below and the mathematical model ! 6 /:9 . !
6 /:9Ê
¡ ¢
/ ç /mj l l
/ ' if R '"/R' j if i if jQ'"/
Now, suppose instead we apply the flash of high intensity green light at + model for this experiment is the time / ` . Verify that the mathematical given by ! 6 6 6 /:99 , where 6 6 /:98)/m .
axis
8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION
replacements Solution. Our expectation is that the plotPSfrag for this new experiment will have the “parabolic dip” shifted over to oc axis cur starting at time / ` instead of at time /;$ . In other axis axis words, we expect the graph in Figure 8.7(b). + Our job is to verify that this graph is obtained from the function ! 6 6 6 /:99 , where 6 6 /29q/ . 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: ! 6 /:9é when /.'y . To get the formula for ! 6 6 6 /:99 , we now appeal to Procedure 8.2.1 and just replace every occurrence of / in ! 6 /:9 by 6 6 /:9 . That gives us this NEW domain condition and function equation:
6 /Ì
+
9Ò$ when /m ' when R / ' `
Second part: ! 6 /:9Ê / ç /âNj when )' /R' j . We now l l appeal to Procedure 8.2.1 and just replace every occurrence of / in this function by 6 6 /:9 . That gives us this NEW !
6 6 6 /:99
+
!
domain condition and function equation:
+
119
O ç O O p O
oxygen rate
hours
p
O ç O O p O
oxygen rate
!
6 /Ì
9Ò$ i
when jB' /m
when
(b) Flash at
+
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. !
6 6 /:9[email protected] 6
¡ ¢
/ J:/m l ¾
l
l
if i / ' ` if ` ' /R' i if ' /
i
O
ç Ä8¡
.
Figure 8.7: Applying light at time Ä .
/:
'
õ
p
this NEW domain condition and function equation:
6 6 6 /:99
ÄÒ¡Z .
hours
+ + + J ! 6 6 6 /:99 ! 6 /Ì 9Ò /4 9 6 /Ì 9Tüj when )' /4 ' j ji 6 j ! j i when ` ' /i' / êJ:/Ì j j j ' / . We now appeal to Procedure 8.2.1 and Third part: ! 6 /:9Ò$ when B just replace every occurrence of / in this function by 6 6 /:9 . That gives us
!
O
ç
(a) Flash at
!
+
õ
CHAPTER 8. COMPOSITION
120
8.3 Exercises Problem 8.1. The volume æ of a sphere of radius Á is given by the formula æ ÁÔ/´¡ º fÁ¼« . A « balloon in the shape of a sphere is being inflated with gas. Assume that the radius of the balloon is increasing at the constant rate of inches per second, and is zero when ÄÒ¡ . (a) Find a formula for the volume balloon as a function of time Ä .
of the æ
(b) Determine the volume of the balloon after seconds. (c) Starting with an empty balloon, suppose that the balloon will burst when its volume is ¥ RR cubic inches. At what time will the balloon burst? (d) Find a formula for the surface area ç of the balloon as a function of time Ä ; recall the surface area formula for a sphere of radius Á is çÁP/¡IafÁ . (e) Determine the surface area of the balloon after ª seconds. (f) What will be the surface area of the balloon when it bursts? Problem 8.2. For this ¨ problem, ¾  Ä2/¡ Ä ¨ . jTÄ2/¡Í©@Ä´©? and
e
 Ä2/¡"ÄÊ©
¾
(a) Compute the multipart rules for ¾ and jrÄ2/f/ and sketch their graphs. e
e
,
 Ä2/À/
¾  Ä2/À/
(b) Compute the multipart rules for ¾ and jT Ä2/f/ and sketch their graphs. (c) Compute the multipart rule for and sketch the graph.
¾  ¾  Ä2/»©3"/
Problem 8.3. Write each of the following functions as a composition of two simpler functions: (There is more than one correct answer.) (a) (b) (c) (d) (e) (f)
ÛY¡1&¿m©?R"/ . ÛY¡ È Ó ´Â¢¿ . ÛY¡T &¿4©É* / ©YT¿m©D*/ Â ¨ &¿4©É*/¹Â? . ÛY¡ Y ¨ e . ¶ « ÛY¡ ; È ¿mÂ? . ÛY¡Ê© ; Ê> ©  z¿m©?2 / .
Problem 8.4. (a) Let &¿/ be a linear funce tion, &¿/¡Ze Åee ¿ÌÂ`7 for constants Å and 7 . Show that  ¿/À/ is a linear function. e
(b) Find a function ª¼¿m© . Problem 8.5. Let
e
r&¿/ j
such that
&¿/¡ ¨ ¿4ÂÉ
r T¿/À/¡ j j
.
(a) Sketch the graphs of ¿/ on the interval ©8²hI¿hU¥ . e
e
e
&¿/f/   ¿/À/À/ e
e
e
(b) Your graphs should all intersect at the point (6,6). The value ¿¡ ª e is called a fixed point of the function ¿k/ since e is, 6 is fixed  it doesn’t åªR/;¡.ª ; that e move when is applied to it. Give an explanation for e why 6e is a fixed point for e e any function   »& &¿/}&i/f/À/ . (c) Linear functions can have at most one fixed point. Quadratic functions can have at most two. Find the fixed points of the function jT¿/¡ ¿ ©Y . (d) Give a quadratic function whose fixed points are ¿¡?©8 and ¿7¡Ç . Problem 8.6. A car leaves Seattle heading east. The speed of the car in mph after ° minutes is given by the function
z° DÎ °ñ/¡ ¥8ÂÉ° e function ° ¡ :Ô/
(a) Find a that converts seconds : into minutes ° . Writee out the formula for the new function DÎ A:2/f/ ; what does this function calculate?
¾
(b) Find a function °é¡#jT / that converts ¾ hours into minutes ° . Write out the ¾ formula for the new function DÎjr /À/ ; what does this function calculate? (c) Find a function ä ¡H²+:Ô/ that converts mph : into ft/sec ä . Write out the formula for the new function ²+ DÎ °ñ/ ; what does this function calculate? Problem 8.7. A contractor has just built a retaining wall to hold back a sloping hillside. To monitor the movement of the slope the contractor places marker posts at the positions indicated in the picture; all dimensions are taken in units of meters.
8.3. EXERCISES
121 hillside profile at time wall is constructed
markers
e
(g) e
(h)
3
5
Problem 8.9. Let Û ¡  ä /N¡ È Ë© ä and . Compute the composition z ¿ Í Â ä ¡"jr&¿/¡ e Û#¡ jT¿/À/ . Find the largest possible domain e of ¿ values so that the composition ÛI¡ [jT¿k/f/ is defined. e
axis1 axis
3
2
axis
3
(a) Find a function Û ¡ ¿/ that models the profile of the hillside. e
Problem 8.10. A plant is growing under a particular steady light source. If we apply a flash of high intensity green light at the time Ä¡) and measure the oxygen output of the plant, e we are led to the mathematical model  Ä2/ .
(b) Assume that the hillside moves as time goes by and the profile is modeled by a function j Ñ &¿/ after Î years. If ÎW¡ , then j p &¿/Í¡ e &¿/ . After one year, the profile is modeled by the function e e j ¨ ¿/A¡  &¿/À/ . After two years, the profile is modeled by the function j &¿/;¡ e e e   &¿/À/f/ . After Î years,e ite is e modeled by the function j Ñ ¿/¡  » &¿//f/ , where we have composed the original function ÎUÂê times. Find a formula for j Ñ &¿/ that does not involve compositions. (Hint: To do this, start by writing out the formulas for Î ¡F v ¹ . You will see a pattern developing. To get the general formula, the following fact will be useful: Given a real number Mõ?Á õ and a positive integer è ,
e
ìí
if if if
Ä2/¡Íë Ä© ¯ Ä´ÂÉ
«
«
(a) Sketch the graph of ÛZ¡ main khÄ×hZ¥ hours.
Ä×hZ hÄ×hÇ × hÄ k e
Ä2/
on the do
(b) Suppose instead we apply the flash of high intensity green light at the time ÄÒ¡ . Verify that the mathematical model for this experiment is given by e jrÄ2/f/ , where jT Ä2/¡Ä©Ç , on the domain hIÄÖhZ¥ hours. Sketch the graph modeling this experiment and write out the multipart rule.
e
(c)
e
(f)
wall
replacements
&¿/¡ R¿¹«@©I¹ jr&¿/¡ È Ó z¿4ÂIª &¿/ ¡R¿mÂ? jT¿/¡ ¿ « . &¿/ ¡Çv jT ¿/ ¡ R¿¹@ÂYz¿4Â? . &¿/ ¡?©@:ÕjT& ¿/ ¡? . e
(e)
Áé ¨ ´ÂYÁTÂYÁ ÂYÁ « ÂDÁ º Â?ÐÐÐÂYÁHé¡ ´ ©Y ´ ©YÁ Your final formula for j Ñ & ¿ / will involve both ¿ and Î .) ¿ / for Sketch the graphs of j Ñ & Î¡R : ¹ : in the same coordinate system.
(c) Suppose you subject the plant to a flash of high intensity green light at the time ÄU¡ and at time ÄU¡ . Sketch the graph modeling this experiment on the domain Ch"Ä.h$¥ hours and find the corresponding multipart function.
(d) What is happening to the marker posts? (e) Estimate when the hillside will start to spill over the retaining wall. Problem e e [jT&¿/f/ , e
(a) (b)
e e
(c) (d)
e
e
8.8. Compute the compositions e ¿k/f/ and jT ¿/À/ in each case:
&¿/¡I¿ ÕjT&¿/¡ ¿mÂÉ . &¿/ ¡F êz¿ jT¿/¡ È ¿ . &¿/ ¡¿4ÂYvÕjT& ¿/¡ ¨ &¿4©Y&/ . £ &¿/ ¡ª¿¹8ÂY: jT ¿/ ¡I¿m©Y .
Û
Problem 8.11. Suppose you have a function ¡ e &¿/ such that the e domain of e &¿/ is ÖhI¿ªh?ª and the range of &¿/ is ©8hDÛÐh .
T¿m©D*/À/ ? What is the range of  T&¿4©D*/f/ ? e / ©É ? What is the domain of & ¿¹ e What is the range of  ¿¹ / ©D ? Can you find constants C and e D so that the domain of C& ¿4©D/À/ is h ¿lh? ?
(a) What is the domain of (b) (c) (d) (e)
e
e
CHAPTER 8. COMPOSITION
122 (f) Can you find constants e E B the range of B &¿/¹Â is
and
E
so that hYÛ.hU ?
e
¿/¡1z¿4Â?"/d
.
e
¿k/¡ È RË©¢¿
.
(b) (c)
Problem 8.12. For each of the given functions e
ÛN¡ ¾ ¿k/ , simplify the following expression so that is no longer a factor in the denomina¾ tor, then calculate the result of setting ¡ in this simplified expression: e
(a)
e
¿mÂ
¾ / © ¾
¿/¡ Y[Z ¨
¨
e
.
¿k/
(a) Write the function ,+Ä2/¡ ; z c Â e ózÒ©Íª¼Ä2/ as a composition ,+ Ä2/¡ [ jT Ä2/À/ , where jT Ä2/¡ZÅÄ@Â@7:ÄÂ is 7 . a quadratic function; find Å =
Problem 8.13.
(b) Compute
rÄ2/¡ÇÄ©? j
when ÜÛ/´¡ î ¨ and . Simplify your answer.
T ¿>ÂÉ¿¹R/f/ j
e
e
Chapter 9 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 manuevers were laid out. In this section, we extend those graphical techniques to general function graphs.
9.1 A Lowtech Exercise
PSfrag replacements axis
í î
This section is all about building new functions from ones axis we already have in hand. This can be approached symbolÚ : Sð axis ically or graphically. Let’s begin with a simple handson exercise involving the curve in Figure 9.1. By the vertical line test, we know this represents the graph of a function MN#! 6 @9 . With this picture and a piece Figure 9.1: 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. PSfrag replacements Another way to build new curves from old ones is to exploit the built in axis reflecting symmetry of the 2M coordinate system. For example, imagine axis the graph of M &! 6 @9 across the axis or the M axis. In all of the above cases, we moved from the original axis 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 9.2: Rotating a curve. to a new curve, but it may not be the graph of a function: See Figure 9.2. The pictures in Figure 9.3 highlight most of what we have to say in 123
CHAPTER 9. THREE CONSTRUCTION TOOLS
124
PSfrag replacements PSfrag replacements this section; the hard work remaining is a symbolic reinterpretation of theseaxis graphical operations. axis axis
axis
axis
axis
left
right
up
PSfrag replacements
down
PSfrag replacements
axis
Horizontal shift.
axis
axis
Vertical shift.
axis
axis
axis pull
PSfrag replacements
pull
push
push
PSfrag replacements expand horizontally
axis axis
axis
Horizontal expansion.
axis
pull
pull
axis
compress horizontally Horizontal compression.
axis push
expand vertically
push
compress vertically
PSfrag replacements axis
PSfrag replacements
Vertical expansion.
axis
Vertical compression.
axis axis
axis reflect across axis
S
Vertical reflection.
Ú
reflect across axis
axis
Horizontal reflection.
Figure 9.3: Shifting, dilating, and reflecting Û¢¡ e
&¿/ .
9.2 Reflection In order to illustrate the technique of reflection, we will use a concrete example: Function: MN3´ 6 @9Ò"!Pü! Domain: !0' C' ! ; ;0 ! 'M?')^ Range: As we know, the graph of M"´ 6 @9 on the domain ;!ï'H%'.! is a line of slope ! with M intercept ! , as pictured in Figure 9.4(a). Now, start with
axis axis axis
9.2. REFLECTION
125
the function equation M'å´ 6 9q!Iê! and replace every occurrence of “M ” by “ 3M .” This produces the new equation 3MN"!3Í! ; or, equivalently,
M½åð 6 9ËH;!Y !A
PSfrag replacements axis
The domain is still ;!Ê' ã' ! , but the range will axis ç Ú axis change; we obtain the new range by replacing “M ” by “ 3M ” axis p in the original range: ;!Þ'Ï3M'é^ ; so ! òqM ò<^ . The S axis graph of this function is a DIFFERENT line; this one has æ æ æ slope ;! and M intercept ;!A We contrast these two curves æ p æ in Figure 9.4(b), where ð 6 9 is graphed as the “dashed æ ç line” in the same picture with the original ´ 6 @9 . Once we do this, it is easy to see how the graph of PSfrag just ð 6 9 is really replacements (a) Graph of &¿/¡z¿mÂY . the original line reflected across the axis. axis Next, take the original function equation MéÆ´ 6 @9Ç axis ç Ú axis !"! and replace every occurrence of “ ” by “ ; .” This axis produces a new equation MN"! 6 ;@9T ! ; or, equivalently, p
æ
M½" 6 @9Ê.;!Y !A The domain must also be checked by replacing “ ” by “ ; ” in the original domain condition: ;!Å'ô;I' ! , so !ò òy;! . It just so happens in this case, the domain is unchanged. This is yet another DIFFERENT line; this one has slope ;! and M intercept 2. We contrast these two curves in Figure 9.4(c), where 6 9 is graphed as the “dashed line” in the same picture with the original ´ 6 @9 . Once we do this, it is easy to see how the graph of 6 @9 is really just the original curve reflected across the M axis. This example illustrates a general principle referred to as the reflection principle. Important Facts 9.2.1 (Reflection). Let M function equation.
!
6 @9 be a
æ æ æ p æ æ ç
æ
p
+¿/¡?©@R¿@©w ñ
Ú ç
S
axis
ÁR&¿/¡?©8z¿mÂÉ
Figure 9.4: Reflecting
(i) We can reflect the graph across the axis and the re ¿k/ . sulting curve is the graph of the new function obtained by replacing “M ” by “ 3M ” in the original equation. The domain is the same as the domain for M &! 6 9 . If the range for M½&! 6 9 is ïÐ'Mò'[ , then the range of 3M &! 6 9 is ï>''3M¤' . In other words, the reflection across the axis is the graph of MN.ª! 6 9 . (ii) We can reflect the graph across the M axis and the resulting curve is the graph of the new function obtained by replacing “ ” by “ ; ” in the original equation. The range is the same as the range for M I! 6 @9 . If the domain for M[Í! 6 @9 is ' ' &'éQ , then the domain of M +! 6 ;9 is ' 'B;Þ' Q . Using composition notation, the reflection across the M axis is the graph of MN#! 6 ;9 .
.
axis
æ æ æ p æ æ ç
(c) Graph of
axis
(b) Graph of
S
Û
.
¡
CHAPTER 9. THREE CONSTRUCTION TOOLS
126
ements axis
Example 9.2.2. Consider the parallelogramshaped region ó with vertices 6 %´1!A9 , 6 %´1z;!A9 , 6 n1%9 , and 6 Én1%9 . Use the reflection principle to find functions whose graphs bound ó .
axis axis
í î Ö ô
ORí ë ð î ô
l
ements ô
axis
× £
p
axis
S
ô
ë ORð axis
Region
axis
Ú
axis
Figure 9.5: The region
í î Ú :
.
MN#! p 6 @9Ê"!P !A
í î Ú
æ Sð
æ ements p
æ
axis
axisFigure
Ú
:
Sð
í î axis p S æ Ú : Sð
axis axis
õ
9.6: Reflecting semicircle.
Solution. Here is a picture of the region ó : First off, using the two point formula for the equation of a line, we find that the line à passing through the points Ø 6 %´1!A9 and Ù 6 n1%9 is the graph of the function MÏö! 6 @9? ;!Z !A By Fact 9.2.1 (i), à is the graph of the equation 3M .;!¢ ! , which we can write as the function MN#! 6 @9Ê"!P !A By Fact 9.2.1 (ii) applied to à , the line à l is the graph of the function M Å! l 6 9â.;!D !A Finally, by Fact 9.2.1 (i) applied to à , the line à is the graph of the l p equation 3M H;!?! , which we can write as the function
the
Figure 9.6 illustrates the fact that we need to be careful about the domain of the original function when using the reflection principle. For example, consider
MN#! 6 @9Ê$4 þ > + 6 ¢üjT9 . The largest possible domain of values is !B'"*' and the graph is an upper semicircle of radius centered at the point 6 jA19 . Reflection across the axis gives the graph of MN.É> þ 4 6 PjT9 on the same domain; reflection + the graph of MHÏ þ > 6 Y jT9 across the M axis gives on the new domain ' *'';! .
9.3 Shifting
+
Let’s start out with the function MÅ! 6 @9â ) 1 which Z has a largest possible domain ;!T'yS'y! . From ChapV÷ ø ter 6, the graph of this equation is an upper semicircle V of radius 2 centered at the origin 6 %´1%9 . Sliding the graph ÷ø back and forth horizontally or vertically (or both), never axis rotating or twisting, we are led to the “dashed curves” beZ V V Z ù ù low (contrasted with the original graph which is plotted ¡ Figure 9.7: Graph of Û with a solid curve). This describes some shifted curves on È Ë©¢¿ . 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. The lower righthand dashed semicircle is of radius 2+ and is centered
at 6 jA1%9 , so the corresponding equation must be M þ 6 Y jT9 . The + 2 and upper lefthand dashed semicircle is of radius centered at 6 %´1jT9 , so
the corresponding equation must be M "j= ) . The upper righthand Z_÷ ø
axis
PSfrag replacements
9.3. SHIFTING
127
axis axis
dashed semicircle is of radius ! + and centered at 6 jA1jT9 , so the correspond axis 6 ¢üjT9 . ing equation must be MN)jw þ Keeping this same example, we can continue this kind u of shifting more generally by thinking about the effect of + the p making
following three replacements in the equation M ) ü : l
ú
M ú 6 and M39ûú
* Y MU * and M 89 6 Y
PSfrag
* ý 6 Y 9 +
ý þ M  and, + * M ÿþ 6 Y 9 M½ÿþ
p
u
axis axis
Upper semicircle with radius Ù 2 and center * 1 %9 . 6 ü
Upper semicircle with radius
ü
ü
Ù 2 and center %´1W89 . 6
Upper semicircle with radius Ù 2 and center * W1 89 . 6
Be careful with the sign (i.e., ¦ ) of * and . In Fig* ure 9.9, if , we horizontally* shift the semicircle unit to the right; whereas, if bÉ , we horizontally shift the semicircle É units to the right. But, shifting É unit to the right is the same as shifting * unit to the left! In other words, if is positive, * * then a horizontal shift units * by will move the graph to the right; if is negative, * * then a horizontal shift will move the graph
l
axis
There are three potentially confusing points with this example:
by
replacementsFigure 9.8: Shifting the upper semicircle.
These substitutions lead to three new equations, each the equation of a semicircle:
+
æ æ
Ú
axis
original curve
negative
curve shifted units left if is negative
S
axis
positive
curve shifted units right if is positive
Figure 9.9: Potentially confusing points.
units to the left.
If is positive, then a vertical shift by will move the graph units up; if is negative, then a vertical shift by will move the graph * units down. These conventions insure that the “positivity” of and match up with “rightward” and “upward” movement of the graph.
When shifting, the domain of allowed values may change. This example illustrates an important general principle referred to as the shifting principle. Important Facts 9.3.1 (Shifting). Let MN#! 6 9 be a function equation.
CHAPTER 9. THREE CONSTRUCTION TOOLS
128
*
(i) If we replace “ ” by “U ” in the original function * equation, then 9 is obtained by the graph of the resulting new function * Mÿ* ! 6 ? ! 6 9 by horizontally shifting the graph of . If is positive, the picture * * shifts to the right units; if is negative, the picture shifts to the left * units. If * the domain of* ! 6 9 is an interval ' ' C'"Q , then the domain 9 is ' 'ÏÍ 'ôQ . The range remains unchanged under of ! 6  horizontal shifting.
ements axis axis axis
(ii) If we replace “M ” by “MÂ ” in the original function equation, then the graph of the resulting new function M û! 6 @9Ëð is obtained by vertically shifting the graph of ! 6 @9 by . If is positive, the picture shifts upward units; if is negative, the picture shifts downward units. If the range of ! 6 @9 is an interval ï'[M¦' , then the range of ! 6 @9Ò4 is ï(' M 4T'G . The domain remains unchanged under vertical shifting.
9.4 Dilation õ ² I³
0Ö
ô±² ô² ôö
ö
~
To introduce the next graphical principle we will look at the function
axis
ö
& ² * ô ö ² & ô ² ô±
ÖÁ ³ axis
MN#! 6 9Ê '
Using a graphing device, we have produced a plot of the graph on the domain <^d' ?'q^ . Figure 9.10 shows the õ curve has a high point ý (like a “mountain peak”) and a Figure 9.10: Graph of B Û ¡ low point þ (like a “valley”). Using a graphing device, we Y e ¿k/¡ Y e . can determine that the high point is ýb 6 n1 9 (it lies on ¶ ¨ the line with equation M ) and the low point is þ x Én1 { (it lies on the line with equation line M ]F ), so the range is F 'MÞ' . Draw two new horizontal lines with equations MüH! E ¦ Â¦I . Grab the high point ý on the curve and uniformly pull straight up, so that the high point now lies on the horizontal line M'Ï at 6 n19 . Repeat this process by pulling þ straight downward, so that the low point is now on the line MBtÉ at 6 Én1zÉ9 . We end up with the ”stretched dashed curve” illustrated in Figure 9.11(a). In terms of the original function equation M½ ö S D , we are S simply describing the graphical effect of multiplying the M coordinate of every point on the curve by the positive number ! . In other words, the
S ! x öS { . dashed curve is the graph of MN SöD S D Next, draw the two horizontal lines with equations M$ E ¦Z ¦Z . p Grab the high point ý on the curve (in Figure 9.10) and uniformly push straight down, so that the high point now lies on the horizontal line M at 6 n1 9 . Repeat this process at the low point þ by pushing the p upward, so that the low point is now on the line M] curvep straight at 6 Én1zU 9 . We end up with the new ”dashed curve” illustrated in Figp p ure 9.11(b). In terms of the original function equation MW ö S , we are
S
D
9.4. DILATION
129
î simply describing the graphical effect of PSfrag multiplying the M coordinate of replacements
every point on the curve by the positive number . In other words, the dashed curve is the graph of M ö S ð . S D We could repeat this process systematically: axis Pick a positive number ï .
Draw the two horizontal lines MNðï E ¦ . If ïDVW , then the graph of Mü is parallel and above the graph of M . On the other hand, if %D X ïYXq , then this new line is parallel and below MNb .
&³
0Ö
õ ² Ú axis ô±² ô² ôö Ö * ²ö ö & ô ö ² S * ô ² ô ±
³ axis
õ
PSfrag replacements Uniformly deform the original graph (in Figure 9.10) (a) Vertical expansion. so that the new curve has it’s high and low points just touching Mt ¦ . This will involve vertically axis stretching or compressing, depending on whether õ ² Ú axis ô±² ZX ï or % XïNXt , respectively. A number of posôÿ ÿ× ÿ× ÿ× ÿ× ÿ×ÿ×² ÿ ô ö sibilities are pictured in Figure 9.11(c). & ³ 0 Ö ö * ²ö Ö ³ & ô ö ² S axis We refer to each new dashed curve as a vertical dilation ô * ² of the original (solid) curve. This example illustrates an ô± õ important principle. Important Facts 9.4.1 (Vertical dilation). Let ïDV % be a positive number and M &! 6 @9 a function equation.
(i) If we replace “ M ” by “ Ú ” in the original equation, then the graph of the resulting new equation is obtained by vertical dilation of the graph of M)! 6 @9 . The domain of values is not affected.
(ii) If ïV[ , then the graph of Ú "! 6 @9 (i.e., M ï! 6 9 ) is a vertically stretched version of the original graph.
(b) Vertical compression.
Ú axis
õ ² ô±² ô²
&
³ 0Ö ô ö
ö * ²ö Ö S & ô ö ² * ô ² ô± õ
(iii) If %ÉX ï¢X[ , then the graph of Ú #! 6 9 (i.e., MNðï! 6 9 ) is a vertically compressed version of the original graph.
(c) Many possibilities. Figure 9.11: Dilating e &¿/ .
If we combine dilation with reflection across the axis, we can determine the graphical relationship between M ! 6 9 and M ï! 6 @9 , for any constant ï . The key observation is that reflection across the axis corresponds to the case ï¢HÉ . Example 9.4.2. Describe the relationship between the graphs of
M½#! 6 @9
ÿþ 4 6 ¢H9 1
M½.ª! 6 @9B. þ > 6 Y'9 1 and + +
M½. ! 6 @9Ê. þ > 6 ¢H9
³
axis
Û
¡
ements axis axis
CHAPTER 9. THREE CONSTRUCTION TOOLS
130
axis
step 1: start with upper semicircle
Á axis step 2: reflect across
PSfrag replacements step 3: stretch curve from
~ ¯ RÖÕa{$Áu
step 2: to get
Solution. The graph of M]Í! 6 @9 is an upper semicircle of + radius 1 centered at the point 6 Én1&%9 . To obtain the picture of the graph of MÿG ! 6 @9 , we first reflect M ! 6 9 across the axis; this gives us the graph of MB ª! 6 9 + . PSfrag replacements Then, we vertically dilate this picture by a factor of ï to get the graph of Ú ª! 6 + 9 , which is the same as the p MN. ! 6 9 . See Figure 9.12. graph of the equation
Figure 9.12: Reflecting axis and dilating a lower semicircle.
axis
Let’s return to the original example M ö S D and inS 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 righthand end of the graph (in Figure 9.10) and pull to the right, while at the same time pulling the lefthand end to the left. We can quantify this by stipulating that the high point ý éxzn1 { of the original curve moves to the new location x}!1 { and the low point þ x Én1z { moves to the new location x ;!1z { .
õ ² ô±² ô² ôö
I³ 0Ö ö
Ú
axis
& ² ö * ô ö ² & ô ² ô±
Ö
S
&³ 0 Ö
³
axis
õ
õ ² Ú axis ô±² ô² ôö Ö ö & ô ö ² ö S & ² & ô ² ô±
³
axis
õ
(a) Stretching.
(b) Compressing.
Figure 9.13: Horizontally dilating ÛY¡
Y Y e
¶ ¨
.
The result will be somewhat analogous to stretching a spring. By the same token, we could push the righthand end to the left and push the lefthand end to the right, like compressing a spring. We can quantify this by stipulating that the high point ý éxzn1 { of the original curve moves to the new location x2 1 { and the low point þ½ x Én1z { moves to the new location x U 1zU { . These two situations are indicated in Figure 9.13. We refer to each of the dashed curves as a horizontal dilation of the original (solid) curve. The tricky point is to understand what happens on the level of the originalî equation. In the case of the stretched graph in Figure 9.13(a), you can use
a graphing device to verify that this looks like the graph S  ; in other words, we replaced “ ” by “fv! ” in the original î of M S  ðö D equation. In the case of the compressed graph in Figure 9.13(b), you can use a graphing device to verify that this looks like the graph of MN ð S ö ; S D in other words, we replaced “ ” by “ S "! ” in the original equation.

9.5. VERTEX FORM AND ORDER OF OPERATIONS
131
The process just described leads to a general principle. Important Facts 9.4.3 (Horizontal dilation). Let ï]V number and M &! 6 @9 a function equation.
% be a positive
(i) If we replace “ ” by “ S ” in the original function equation, then the graph of the resulting new function MN&! x S { is obtained by a horizontal dilation of the graph of M Å! 6 9 . If the domain of ! 6 9 is ' '"'HQ , then the domain of MN#! x S { is ' ' S '"Q . (ii) If ïDVW , then the graph of MN#! x S { is a horizontal stretch. (iii) If %ÉX
ï¢X[ , then the graph of MN#!x S { is a horizontal compression.
9.5 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 9.5.1. The problem is to describe a sequence of geometric ma
neuvers that transform the graph of MN" into the graph of M Hj 6 ¢'9 ! .
Solution. The idea is to rewrite MN.j 6 ÌU9 I! as a composition of MN" 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: Reflect ·¸Uµ¶
µ¶U·¸
Horizontal shift by ·¸U µ¶
Ì
Dilate by 3
µ¶Uº ·¸
Vertical shift by
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 M a specific name to make things definite:
6 9 ä 6 9ô 6 9 !
Y ! j2@
*
6 9Ë ¢H 6 @9ÒH;
CHAPTER 9. THREE CONSTRUCTION TOOLS
132 Now, verify that
*
*,+
g
ä !"4 Ù $ ! # 6 @9&% Û(')  ä
* /.!10 ¢'3254 h

ä76ß w 6 ¢H9 y 98
ä:#ó 6 j 6 ¢'9 ;9 %
ä0cj 6 ¢'9 2
.j 6 ¢H9 !
.j2 ^vY)
9.6 Summary of Rules
ements axis axis
axis
æ Figure
9.14: A multipart function.
For quick reference, we summarize the consequence of shifting and expanding symbolically and pictorially. The running example for Tables 9.1, 9.2, and 9.3 will be a multipart function M ! 6 @9 whose graph, seen in Figure 9.14, consists of a line segment and a quarter circle on the domain ;!£' C' ! :
!
+ ! Pü 6 9Ë Ù )
if ;!£' *')% if %.' *' !
9.6. SUMMARY OF RULES
133
Reflection Symbolic Change
New Equa Graphical PSfrag replacements tion Consequence
Picture
axis
Replace ; .
with
MN#! 6 ; 9
axis
Replace with ª! 6 9 . !
6 9
MN.ª! 6 9
axis A reflection axis across the
M axis. PSfrag replacements æ
A reflection across the axis.
Table 9.1: Reflecting ÛY¡ e
¿/ .
axis
æ
axis
æ
CHAPTER 9. THREE CONSTRUCTION TOOLS
134
Shifting (Assume <>[email protected]? ) replacements New EquaPSfrag Graphical tion Consequence axis
Symbolic Change
axis
Replace 6 ¢ ï=9 .
with
axis
M½&! 6 Y ï=9
Picture
A shift to the right ï units. PSfrag replacements
p
axis axis
Replace 6 ¢ ï=9 .
with
M½&! 6 Yï=9
A shift to the left ï units.
axis
PSfrag replacements
æ p
æ p
axis ! 6 @9 Replace with 6 ! 6 9Aï=9 .
axis
M½&! 6 9T ï
A shift up ï units.
axis
PSfrag replacements
æ
axis
Replace ! 6 @9 with 6 ! 6 9Aï=9 .
axis
M½&! 6 9T ï
A shift down ï units.
Table 9.2: Shifting Û¢¡ e
¿k/ .
Z axis ù Z
Z
PSfrag replacements
9.6. SUMMARY OF RULES
135
axis axis axis
PSfrag replacements
Dilation Symbolic Change
New Equation
axis axis
Graphical axis Picture í Consequence
PSfrag replacements If Z ï VB , replace with x S { .
MN#!ã ï
A horizontal expansion.
axis axis axis
If ïZVB , replace ! 6 @9 with 6 ï! 6 99 .
If % X F ï Xô , replace ! 6 9 with 6 ï! 6 99 .
MN#! ï
A horizontal compression.
öõ
axis axis axis
A vertical expansion.
A vertical compression.
Table 9.3: Dilating ÛY¡ e
&¿/ .
BBABA l
GDFGDFGDF GDFGDFGDF GFGFGF GFDGFDGDFGDF p GFGF
l
æ pæ læ æ í
MNðï! 6 9
p
A
æ p æ líæ æ O l p
MNðï! 6 9
l
æ pæ læ æ O l p í
PSfrag replacements
ï Xô , reIf % X F place with x S { .
p
öõ
CDCDCDCCC EDEDEDO EEE p
l
l p
HDHDJJJDHH KJKKJJ æ p æ l æ æ IDIDO II l p
CHAPTER 9. THREE CONSTRUCTION TOOLS
136
9.7 Exercises E
for some constants B7 C[email protected] . Find these constants, describe the precise order of graphical operations involved in going ¨ ¨ frome the graph of ÛÇ¡ ¿ to the graph of ÛÉ¡ &¿/ (paying close attention to the order), write out the multipart rule, sketch the graph and calculate the coordinates of the “vertex” of the graph.
Problem 9.1. On a single set of axes, sketch a picture of the graphs of the following four equations: Û¢¡Í©¿=Â È , Û¢¡Í©¿=© È , ÛÉ¡ ¿=Â È , and Û¡)¿© È . 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.
&¿/ Problem 9.2. The graph of a function ÛD¡ is pictured with domain ©42h¿§hW . Sketch the graph of the new function e
y π f(x)
axis axis
x
1
(a) Each of the six functions below can be written in the “standard form”
Problem 9.3. e
¿k/
¨
e
(a5) (a6)
e
(b) Solve the following inequalities using your work in the previous part of this problem:
(c)
¨
¨
¿Ì©É hÇ ¨ ¨ (b2) h ¿ÌÂD hI ¨ ¨ (b3) ÛY¡Ç z¿4©? ÂÉÍ÷U¥ ¨ ¨ The graphs of Û ¡ z¿ê© Â ¨ ¨ and Û¢¡?© ¿Ì©É Â?¥ intersect to form a (b1)
bounded region of the plane. Find the vertices of this region and sketch a picture.
e
axis
Û ¡
e
(a4)
¨
Problem 9.4. Consider the function with multipart definition
π/2
−1
PSfrag replacements
e
(a3)
ÛY¡§jT¿/¡ f e  z¿k/v©D v Find the largest possible domain of the function ÛY¡ ; jT&¿/ .
e
(a2)
(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 centered at å©@:©U"/ sits tangentially. Use the techniques of this section to help.
¨
¿/¡ ¿m©Y ¨ ¨ ¿/ ¡ ¿ÌÂÉ ¨ ¨ ¿/ ¡ R¿Ì©? ¨ ¨ ¿/ ¡ r& ¿Ì© "/ ¨ ¨ ¿/ ¡Ç z¿m©Í ÂY ¨ ¨ ¿/ ¡?©@ ¿mÂÉ ©Í e
(a1)
¨
ÛÉ¡ B CJ¿Ì©D/ Â
E
¿mÂÉ ¿/¡ L z@ © ¿ÌÂY LL ì
íë
if if if if
(a) Sketch the graph of
Û ¡
&¿/ e
¿ªh?©4 ©4Ö hIª ¿ h hIl ¿ h ¿÷
ÛY¡
e
&¿/ .
(b) Is ÛÇ¡ ¿/ an even function? Is Ûe ¡ ¿/ an odd function? (A e function Û¡ &¿/ is e å©¿e / for all ¿ in the called even if &¿/¡ Û¢¡ &¿/ is called odd domain. A function e e if ó©@¿/¡?© &¿/ for all ¿ in the domain.) e
e
(c) Sketch the reflection of the graph across the ¿ axis and Û axis. Obtain the resulting multipart equations for these reflected curves. (d) Sketch the vertical dilations e and ÛY¡ ¨ &¿/ .
Û]¡ÿ e & ¿/
9.7. EXERCISES
137
(e) Sketch the horizontal dilations e and ÛÉ¡  ¨ ¿/ . (f)
Û ¡ e
 z¿/ 4
Find a number ? ¬ so that the highest pointe on the graph of the vertical dilation ÛY¡ &¿/ has Û coordinate 11.
2
(g) Using horizontal dilation, find a number ¬ so that the function values e  Y· / are nonzero for all © õ¿¿õ .
4
2
4
axis axis
cool air in heat exchanger tubing
replacements axis axiswarm air out axis
4
2
(h) Using horizontal dilation, find positive i,"¬ÿ so that the function numbers e values  · ¨ ¿4©>,k/À/ are nonzero precisely PSfrag replacements when õÇ¿ÏõZ . Problem 9.5. A typical home gas furnace contains a heat exchanger which consists of a bunch of cylindrical metal tubes through which air passes as the flames heat the tube. As a furnace ages, these tubes can deform and crack, allowing deadly carbon monoxide gas to leak into the house. Suppose the crosssection of a heat exchanger tube is a circle of radius cm when
2
6
axis
Problem 9.6. An isosceles triangle has sides of length ¿ , ¿ and Û . In addition, assume the triangle has perimeter 12. (a) Find the rule for a function that computes the area of the triangle as a function of ¿ . Describe the largest possible domain of this function. (b) The graph of ä ¡ ÅQ&¿/ from part (a) is given below. Sketch the graph and find the rule for the function ä ¡¹ÅQñ¿=Â*/Â< ; make sure to specify the domain and range of this new function.
flames heating tube...
10
Ú
axis
8 it is brand new. As the furnace ages, assume the bottom semicircular crosssection 6 deforms according to the vertical dilation principle 9.4.1 with a deformation constant of 4 PSfrag replacements
¡ Ä2/¡#´Â Ä ÂYÄ2/ 0R
after Ä years. The top semicircular crosssection does not deform. (a) Sketch an accurate picture of the crosssection of the heat exchanger tubing after 10 and 15 years. (b) Suppose the metal in the tubing will crack if it deforms more than 3 cm out of the original shape. When will the exchanger crack? (c) When will the crosssection of the heat exchanger tubing look like the picture?
2 axis
2
4
6
8
S
axis
10
(c) Assume that the maximum value of the function ÅQ&¿/ in (a) occurs when ¿ ¡ . Find the maximum value of ä ¡FÅQ&¿/ and ä ¡Ç¹ÅQ z¿mÂÉ&/vÂ? . Problem 9.7. Describe how each graph differs from that of ÛÉ¡ ¿¹ . (a) (b)
ÛÉ¡ z¿¹ Û ¡I¿@©Y É
CHAPTER 9. THREE CONSTRUCTION TOOLS
138 (c) (d) (e)
ÛY¡1&¿m©¢I/d ÛY1 ¡  z¿m©?&/å ÛY¡T  z¿m©?& /åÊ©Y
Problem 9.8. Let Å 7 be positive constants. We will work with the ellipse equation
¨ ¿ ¨ Û Åª© Â 7© ¡ZR
(b) By plotting points, sketch the graphs of the functions in (a) when Å;¡Z and 7<¡ ; or Å;¡ and 7<¡F . (c) Describe how to use horizontal and vertical dilation to obtain the pictures in the previous part, starting from the unit circle (the circle of radius 1 centered at the origin).
Å 7 i,
¨
¨
Û ¡ &¿/Ë¡ ¿ and Û#¡&jT¿k/â¡  I¿Â # 2/À/0Â , sketch the graphs of Û¢¡§jr&¿/ and ÛDd ¡ jT[jT&¿ /f/ and write out their multipart
(b) If
e
e
rules.
(a) Solve the equation for Û in terms of ¿ ; you should get two different functions.
Problem 9.9. Let Å 7 both positive.
graph of Û]¡ÍjT[jT¿k/f/ using the graphical operations of shifting and dilation in the horizontal and vertical directions. Make sure to carefully specify the order in which you should do the graphical operations.
be constants, with
ÛH¡ e &¿/ and let ÛÉ¡jr&¿/¡#Å ?7ª&¿4© f/ /vÂ , . Describe how e to go from the graph of Û ¡ ¿k/ to the
(a) Start with ae function
Problem 9.10.¨ ¨ In each case, start with the function ÛH¡ ¿ 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.
Chapter 10 Arithmetic There is yet another way to build new functions from known functions, this time relying on what we know about the arithmetic operations for PSfrag replacements numbers. We begin with a simple example from cellular biology. axis
PSfrag replacements
axis
axis
axis
mV
axis
probe A sensor
msec
axis 1
2
3
4
5
main cell body probe B
−70
resting potential voltmeter
(a) Resting potential.
(b) Probing the cell.
Figure 10.1: A cellular biology experiment.
A biologist has isolated a single nerve cell (a neuron ) and plans to make some experimental measurements. It turns out that there is always a voltage difference between the inside and outside of a living cell. We will focus on an experiment involving the measurement of this voltage, as a function of time. To begin with, the measurement of voltage in this experiment is in units of volts ; the number tells us the tendency of a single unit of positive charge to move from the inside to the outside of the cell. In any event, if we do nothing to this nerve cell, the volti age measured turns out to be % mV; “mV” stands for “millivolts,” or i %F,)e% æ l volts. The fact that we have a minus sign means a positive charge will move inward rather than outward. The plot below indicates a steady constant voltage (called the resting potential ), where we will use msec=milliseconds for time units. Our nerve cell is attached to a touch sensor near the surface of an animal and the biologist has set up an 139
CHAPTER 10. ARITHMETIC
140
apparatus that will activate two different probes that will push on the sensor with equal force upon command. PSfrag replacements
PSfrag replacements
axis
axis mV
axis axis
1
2
3
4
5
mV
axis
msec
msec
axis
6
−50 −70
1
2
3
4
5
6
−50 −70
(a) Experiment #1.
(b) Experiment #2.
Figure 10.2: Raw experimental data.
In experiment #1, the biologist has probe A touch the sensor at time /I and the voltage recordings for the cell are given in the plot shown in Figure 10.2(a). In experiment #2, the biologist has probe B touch the sensor at time / ! and the voltage recordings for the cell are given in the plot shown in Figure 10.2(b). What we want to indicate is how each of the plots in experiments #1 and #2 could have been obtained by an “addition process.” For example, the plot in experiment #1 can be obtained by adding together the two plots, as pictured in Figure 10.3.
PSfrag replacements axis axis
mV
mV
mV
msec
axis 1
2
3
4
5
msec
6
1
2
3
4
5
msec
6
1
−50
−50
−50
−70
−70
−70
2
3
4
5
6
Figure 10.3: Adding experimental voltages.
10.1 Function Arithmetic We want to use the arithmetic operations for numbers: , M 1 to combine two functions. On a symbolic equations level, this is easy to do. If we begin with two functions M #! 6 9 and M å6 6 9 , we can form the
10.1. FUNCTION ARITHMETIC new functions:
6 !Ì&69 69 6 !Ì& 6 g ! E 69
ý 6 9T#6 6 9 ý ! 6 9T#6 6 9 ý ! 6 H 9 6 6 9 ! 6 @9 as long as the denominator 6 6 9;' ã %. 6 6 9
6 @9b 6 @9b 6 @9b
!
9 Òh 6 6
141
!
Examples 10.1.1.
+
i
(10.1)
+
(i) Let M"! 6 9m l> ^v üD and MË6 6 9Ì. with the domain +all real numbers. Then a direct + + å6A9 6 9É calculation shows that 6 ! i
l ` êI + Aj 1 6 !>Å6A9 6 97 l3 ]n1 6 ! E 6A9 6 97 u3 ^v p :
i l Tj 0 ¢ ! J´1 è 6 9Ê p S æ ö S ö DS æ ¾ S D p : Notice that è is defined for all , since the denominator is never going
to be zero.
common (ii) Let MüÅ! 6 @9ËHj and MüA6 6 9â ) >ü . The largest possible
domain is É)' *' ; why? We get > ü 1 ! ( 6A9 6 98 ! ( A 6 9 Ê 9 " > j ) 6 6
: 6 j ) 1 6 ! E 6A9 6 9Ò)j ) 1 6 è 9 6 9Ê læ ö . Notice that the domain S of allowed values is É'BT' for each of the first three functions and É
(ii) Suppose M.qj2 ) H . Find simpler functions ! 6 9 and M½#î ! 6 @9q6 6 9 . To do this, let ! 6 9Ë"j2 and 6 6 @9Ò ) ' .
î
î
6
6 9 so that
æ
(iii) Suppose M[ S Ö . Find simpler functions ! 6 9 and 6 6 @9 so that M$ : * S D ð è ð S ð . To do this, let ! 6 9Ò"Y) , 6 6 @9Ò" p and 6 9Ê$ .
S
D
S
+
+ (iv) Suppose MN 6 9 ! 6 9 . Find simpler functions
! 6 9 and 6 6 @9 + so that MN#! 6 @9q6 6 9 . To do this, first factor MN 6 F!@9 6 F@9 , then let ! 6 @9Ò ! and 6 6 @9Ò" .
10.1.1 What about the domain? When you start combining functions as in Equation (10.1), you need to assume that you are using the same domain for each function. For
example, suppose ! 6 9mH and 6 6 @9Ìô . We can take the domain of ! 6 9
S
ements
CHAPTER 10. ARITHMETIC
142
axis axis
to be all real numbers, but the largest possible domain of 6 6 9 would be the nonzero numbers. When we form the new functions ! 6 9>6 6 98"
S and ! 6 9e§6 6 9Ë , the largest possible domain would be the nonzero S numbers.
axis
PSfrag replacements 10.2
Á á units ö ~ ¯ Á á ö  units õ unit ~ ¯
axis
PSfrag replacements
Graphical Interpretation axis
{$Á[} Á á ö
axis
What do these function arithmetic operations mean on axis the level of their graphs? In the case of the addition and õ axis subtraction operations, this is easy to illustrate by way of { } {tÁ[} õ õ õ õ example. If ! 6 9Ò"@P! and 6 6 9Ê[ , Figure 10.4 illustrates e what is happening graphically at Ç$ and general . Figure 10.4: ¿k/ and jr&¿/ . In Figure 10.5(a), we sketch ! 6 9n 6 6 @98 ;Fj , together with graphical interpretation at Íq and general ; Figure 10.5(b) deals with ! 6 9T#6 6 9Ê ¢H .
{ } W
axis
~
add coordinates of these to get coordinate of is
´
~
®
1
Á
Á á  {$Áu ¯ ¥
a{tÁu áÖ¥ t{ Áu ¯ Á á ~õ subtract of ~ ® coordinates these points ~ to get the ´ ´ coordinate ® is Á á of ö {tÁu ¯ õ ¥ õ a{$Áu ¯ Á á ö 1 Á ~
coordinates of these points to get the coordinate of is
Ö
f{tÁu ¯ Á á ö
a{$Áu á ¥ ${ Áu ¯ Á á  ® ~ ´ add
~
õ
(a) Adding two graphs.
subtract coordinates of these to get coordinate is of
(b) Subtracting two graphs.
Figure 10.5: Using function arithmetic.
This simple example illustrates how you can “add” or “subtract” two graphs: Note that 6 @1H! 6 @99 is the point on the graph of ! 6 9 above (on the axis) and 6 81U6 6 99 is on the point on the graph of 6 6 @9 above (on the axis). The point on the graph of ! 6 9H¦Å6 6 9 above on the axis is obtained by combining the M coordinates:
6 @1H! 6 @9 ¦Å6 6 99
10.3 Step Functions Now we are in a position to describe somewhat complicated functions in terms of simpler ones, as suggested in Figure 10.3 on page 140. For example, the measurement of voltage as a function of time often leads to plots which involve jumps in voltage over a short time period. This was illustrated in the experimental graphs in Figure 10.2. In practice, it is
10.3. STEP FUNCTIONS
143
PSfrag replacements PSfrag replacements common to model these sorts of plots with what are called step functions as sketched in Figure 10.6(b). axis axis axis
axis
axis 30 20 10
axis
volts
30 20 10 1
2
3
4
time
volts
1
(a) The actual voltage plot.
2
3
4
time
(b) A step function model.
Figure 10.6: A mathematical description of an experiment.
Roughly speaking, if we were to interpret the plot in Figure 10.6(a), in the time interval between /;$ and /"j , the voltage is 20 volts; otherwise it is 10 volts. Of course, this is not quite accurate really close to /Çô or /?yj , but those portions of the graph are nearly vertical and rather insignificant when you look at the scale of the entire graph. We could, to a first approximation, replace the plot (Figure 10.6(a)) with the graph in Figure 10.6(b). Notice, this is the graph of the multipart function:
M 6 /:98
¡ ¢
e % !% e%
if . % ' /X$ if ) ' /')j if jD" X /
This is an example of what is usually called a step function. PSfrag replacements
10.3.1 Building Step Functions
We can begin with a basic step function L 6 /:9 and show how to build all others using our function building tools:
M½3L 6 /:9Ò
¡ ¢
% %
if ; / XH% if . % ' /'$ if w X"/
axis
Ú
ò
axis
axis −2
(10.2)
−1
1
2
Figure 10.7: A basic step function Û»Ä2/ .
Example 10.3.1. Let L 6 /:9 be the basic step function in Equation (10.2). Compute the multipart formula for L 6 /4 !A9 PSfrag and sketch the graph. replacements axis
Solution. To start, notice this is really the calculation of a composition of two functions. Moreover, by Fact 9.3.1, this is a horizontal shift of the basic step function in Equation (10.2). The complication is that one of these functions is a multipart function. The key is to study how
Ú
ò
axis
axis 1
2
Figure 10.8:
3
»ÄT©Y&/ . Û
4
CHAPTER 10. ARITHMETIC
144
the multipart rule is effected for each of the multipart cases of L 6 /29 seperately. Here is the procedure and the resulting graph: See Figure 10.8. ¡
MN3L 6 m / ü!A9Ò
% ¢
/ ü!DXH% if Ì % ' /4 !0' if > if X)/4 !
/ X ! if ; if !B' /R'"j j X)/ if É
%
¡
% ¢
%
Example 10.3.2. Let L 6 /:9 be the basic step function in Equation (10.2). Compute the multipart rule for
g
MN
!
+
/4 ^ 6
L
9
j h
and sketch the graph. If this was modeling the voltage in an experiment, interpret what the graph is telling you.
ements axis
Solution. As in the preceding example, we study how the multipart rule is effected for each of the multipart cases of L 6 /29 separately. We will start by finding the effect of the horizontal shift and dilation:
axis
4 3 2
N axis
1 4
+
10
Figure 10.9: Interpreting experimental voltage.
MN3L
g
/m ^ 6
+
9
h
% ¢
MN !
L
g
/4 ^ 6
+
h
¡
jÉ
¢
¡
if %.'é
% % ¢
%
9 X'+ % Ê 6 /Ì + 9' 6 /m 9
if wX + if +/;X if ' /R'[e% if e%YX"/
+
9
¡
if 6 /4
¡
¢
6 e fv!A9Ì,#% if +/X efv! if ' /R' e% 6 efv!A9Ì,#% + if e%YX)/ j if +/;X
¾ if ' /R'[e% j if e%YX"/
Ò
Ó
Ñ
üj
+ follows: If this graph represents a voltage plot, the interpretation is as The voltage is a constant 3 volts except during the time interval ' /R' e% , during which time it jumps up to a constant 3.5 volts.
10.3. STEP FUNCTIONS
Ú
eplacements
145
Ú
axis
ò
axis
1
Ú
axis
ò
axis
ò
1
1
axis
axis
axis
2
1
4
Figure 10.10: The graphical components of
ÛY¡.Û»Ä2/¹ÂNÛ»ÄT©YI/¹©2Û»Ä´©YI/ .
+ Example 10.3.3. Let L 6 /:9 be the basic step function in Figure 10.7 on page 143. Sketch the graph of M !L 6 /:9´ü!L 6 /Ì !A9T !L 6 /4 9. +
Solution. This is the same as M ! 6 L 6 /:9R)L 6 /Y!A9R)L 6 /n 9¼9 . In Figure 10.10 + + M "L 6 /:9?L 6 /3 we graphically examine the components of the equation !A9A(L 6 /Ì 9 . Then, we graph M TL 6 /29´(L 6 /4 !A9T L 6 /m 9 which gives the + by a factor of 2 lefthand plot in Figure 10.11. Finally, we vertically dilate to get the righthand plot of M !L 6 /:9ü!L 6 /m !A9´ü!L 6 /Ì 9.
eplacements axis
2
Ú
2
1 −1
ò
axis
4
axis
vertically dilate by a factor of
1 2 3
axis
1 −1
−2
Ú
ò axis
1 2 3 4
−2 akg
Figure 10.11: Bringing it all together.
CHAPTER 10. ARITHMETIC
146
10.4 Exercises Problem 10.1. Use function arithmetic to solve these problems.
º . Find simpler funcÛ¡ ¿ ¿©D e tions &¿/ , jTk ¿ / , and ¾ & ¿/ so that e ÛÉ¡ &¿/Â ¾ jT&k¿¿ //
Å ¬½ , a function of the form ¿k/Ò¡ ¨ has a graph that looks like a line ¿mÂ Å¿ with a bend in it. For what value(s) of Å is the bend a right angle?
(a) Suppose
(b)
¿º . Find simpler funcSuppose ÛD¡ ¿ ©D ¾ e tions &¿/ ÕjT&¿/ and & ¿/ so that e ÛÉ¡ jTk¿ &/ ¿ ¾ / k¿ /
(c) Suppose ÛY¡Ç¿Q&¿ ©D &/v©Iª¿ « e simpler functions &¿/ and
ÛÉ¡
&¿/\jr&¿/ e
"/Â T¿ (d) Suppose Û¡ZR¿Q&¿ÂF e simpler functions &¿/ and ÛÉ¡
 ¿ ©I*/ . Find jT¿/ so that
e
« ÂI¿ / .
T¿/ j
Find so that
&¿/\jr&¿/
Problem 10.2. In each of thee followinge cases, e find a formula for  Â£jI/a&¿/ ,  ©£j&/^&¿/ ,  Ðaj&/^&¿/ and 5OPr/a&¿/ . In the division case, comment on any conditions required for the domain values.
e
(a) e
(b) e
(c) (d)
e
¿/¡z¿Ì©Y and jr&¿/¡Z ¿¹ÂI¿ . ¿/ ¡Ò© ; Ê©>&¿m©É*/ and jr&¿/¡ ¿mÂ? . ¿k/ ¡Çv ¿¹Â;z¿©Y and jT¿/¡ ¿¹©¿©; . ¿/ ¡ and jr& ¿/ ¡ ¿¹«@Â¢¿ .
Problem¨ ¨ 10.3. Let &¿/Í¡t , jr&¿/Í¡¿ and ¾ &¿/¡ ¿ . Write out the multipart rule and graph each of these functions: e
(a) (b) (c)
ÛY¡ e &¿/Â£jr&¿/¹Â ¾ &¿/ . Û ¡§jT¿/ ¾ ¿/ . Y RQ YTS ÛY¡ P Q YTS , when ¿V¡U .
Problem 10.4. (a) Determine the multipart rule and sketch the graph of the func¨ ¨ tion jT¿/¡ ¿ÌÂ ¿ . (b) Determine the multipart rule and sketch ¨ ¨ ¾ the graph of the function &¿/¡ ¿mÂ R¿ .
e
(c) For ¨
Problem 10.5. For each of the following functions, (1) find the multipart rule, and (2) sketch the graph. (a) (b) (c) (d)
¨
¨
¨
¨
 ¿/¡ ¿m©É Â ¿mÂ¢ . ¨ ¨ jr&¿/¡ ¿¹© ¾ & ¿/ ¡ ¨ ¿¹© ¨ Â¢¿ «& ¿ / ¡Ç°&¿/¹ÂYÎ ¿k/ where e
° ¿/¡© ¿z¿m©Í
if if
¿ÏõÍ ¿÷
and
z ¿ Î ¿k/¡%© ¿m Â Í
if if
¿Ïõ# ¿÷U
Problem 10.6. Let ÛÎ Ä2/ be as in 10.3.1 on Page 143. Compute the multipart rules and graphs d for ÛÎ ¨ Äâ©½"/f/ and Û» © "/ . Explain how to Û»Ä2/ by applying the obtain each function from operations of Chapter 9; be very specific about the order in which you apply the operations. Problem 10.7. Let Û»Ä2/ be the basic step function in 10.3.1 on Page 143. Form the NEW function
ÛÉ¡Y²+ Ä2/¡ B×Û E
m
 Ä´© DJ/ Â E C n
where B3 C[email protected] are all nonnegative E constants. ¡' is just (For example, B[¡ C¡[ , D¡ ÛÎ Ä2/ ; if B ¡ÿ êRvfC)¡[ªe(DH¡.: E ¡W , we get the function in Example 10.3.2 on Page 144.) Consider the rectangle formed by the graph of ÛY¡É²QÄ2/ , the vertical line Ä8¡cD , the vertical line Äw¡3DÉÂ C , and the Ä axis. Sketch this E region and compute its area in terms of B7 C[email protected] . Note: This problem is related to a topic in cellular biology called neuronal arithmetic. The combination of electrical signals arriving at a nerve cell can be modeled by a sum of functions of this type. Adding up the corresponding areas is of physical significance since it is a measurement of “work” being performed.
10.4. EXERCISES
147 decibels
Problem 10.8. Let ÛÎ Ä2/ be the basic step function:
Î Ä2/¡ Û
ìí ë
if if if
60
ÄJõ? , kh ÄÖhU , ÄÒ¬U .
30 Find the multipart rule for each of the following functions. Sketch the graph of each function. PSfrag replacements (a) (b) (c) (d) (e) (f) (g)
ÎñzÄAÂY&/ Û
¨ Û»RÄT©¢T/ ÛÎ 2 Ä /¹Â2Û»Ä´©Y&/ d ÛÎ RÄA©?" /vN © ÛÎ / Ä ÛÎ 2Ä / ÄTÂY& / ÛÎ ÄT©É& / ÛÎ R2 Ä /vÂÍ ÛÎ ÄT©É* /¹Â
Problem 10.9. Mike is a Pearl Jam fanatic and goes into a daze every time he hears the first track of their latest CD. The loudness (in decibels) of this 4 minute tracke is modeled by the graph of the function ÛF¡ ¨  Ä2/ at time Ä minutes; the graph is given below:
axis axis
minutes 4
axis
Suppose Mike makes three successive 4 minute recordings of this e track,e after the orige inal 4 minute track. Let Ä2/ , Ä2/ , and Ä2/ « º represent the loudness (in decibels) of each of e the 4 e minute recordings, with Ä2e / 15% louder e e than Ä2/ , Ä2/ 15% louder than Ä2/ and Ä2/ e ¨ « º 15% louder than  Ä2/ .
« Xe W
YW
Ä2/ is of the formW Ä2/T¾ ¡ (a) Each We function ¾ W / ; find the B ¨  Ä© constants B and explicitly. e
(b) Sketch the graph of
° Ä2/¡ e
e e e ¨ Ä2/Â Ä2/¹Â «  Ä2/¹Â º  Ä2/
What does this graph represent?
W
148
CHAPTER 10. ARITHMETIC
Chapter 11 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, PSfrag replacements maybe you have just mixed together some chemical reactants in a vesaxis remaining, sel. As time goes by, you measure the fraction of reactants tabulate your results, then sketch a graph as indicated in axis Figure 11.1. Viewing the input value as “time” and the output value axis fraction as “fraction of product,” we could find a function M4! 6 /:9 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 O time of reactants exists. In other words, we would like a new Figure 11.1: Fraction of refunction that allows us to input a “fraction of reactants” actants 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 M S! 6 /:9 ? The answer is yes and depends upon the general theory of inverse functions.
11.1 Concept of an Inverse Function Suppose you are asked to solve the following three equations for . How do you proceed?
6 ¢ A! 9 6 ¢ !A9 6 ¢ !A9 l
+
^+
^
^+
In the first equation, you add “ ;! ” to each side, then obtain ÇH^v! . In the third equation,+ you take the cube root of both sides of the equation, giving you ê!Z , then subtract 2 getting ! . In second equation, you take a square root of both sides, BUT you need to remember both the 149
CHAPTER 11. INVERSE FUNCTIONS
150
ements
positive and negative results when doing this. So, you are reduced down to ¢N! ð¦PJ or that ÇHÉe% or ^ Why is it that in two of these cases you axis obtain a single solution, while in the remaining case there are two difaxis ferent answers? We need to sort this out, since the underlying ideas will surface when we address the inverse circular functions in Chapter 18. î Let’s recall the conceptual idea of a function: A function is a process which takes a number and outputs a In Out : ð S S new number ! 6 9 . So far, we’ve only worked with this process from “left to right;” i.e., given , we simply put it into The function a symbolic rule and out pops a new number ! 6 @9 . This is all pretty mechanical and straightforward. Figure 11.2: A function as a axis
process.
11.1.1 An Example
Let’s schematically interpret what happens for the specific concrete exPSfrag replacementsample M ³! 6 9YÏj2Ç , when É , , % , , , ! : See Figure 11.3. axis axis axis
æ æ õ ö O
in
in
in
¥l S æ ¥l S æ ¥l S æ
out
out
out
in
æ p
öõ
æ ² ö
1
æ
in
in
Figure 11.3: Function process
l S æ l S æ l S æ
out
öõ
out
out
u
ÛY¡z¿m©? .
We could try to understand the function process in this example in “reverse order,” going “right to left;” namely, you might ask what 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 some given M values, then we can define a “reverse process” ÍG 6 MF.91 l which returns the value required so that ! 6 @9ÊêM : See Figure 11.4.
11.1. CONCEPT OF AN INVERSE FUNCTION eplacements î
151
î
axis axis axis
Ú D ð l
æ
æ p
Ú D ð l
öõ
î æ õ ö
î
î
O
öõ
Ú D ð l
æ ² ö
1
Ú D ð l
æ
î
Figure 11.4: The reverse process
Ú D ð l Ú D ð l
u
¿¡ « ¨ kÛÂ?2/ .
11.1.2 A Second Example ã % , then we If we begin with a linear function MN#! 6 9Ê'Fw½Q , where ' can always find a “reverse process” for the function. To find it, you must solve the equation MN#! 6 9 for in terms of M :
M M Q
FYüQ F
M QÌ9 6 So, if G"j and QF.É , we just have the first example above. For another example, suppose M H<%Jv¢ ! ; then ô.<%&J and QU"! . In this case, the reverse process is Éc! `86 MZ !A9;q . If we are given the value M½$ , we simply compute that Ç.É_! ` ; i.e., ! 6 É_! ` 9Ê[ .
11.1.3 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 MN#! 6 9Ê 6 7 9 . Figure 11.5 is a schematic of how the function works when we plug in ZqÉ , , % , , , ! ; 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 M values, then we can try to define a “reverse process” Ç[Z\Z\Z which returns an value required so that ! 6 9ÒêM . Unfortunately, there is no way to obtain a single formula for this reverse process; Figure 11.6(a) shows what happens if you are given M½)j and you try to solve for .
axis axis axis
CHAPTER 11. INVERSE FUNCTIONS
152
in V ù
in ù
in
^`_badcDe Z f c
out
^`_badcDe Z f c
out
^`_badcDe Zgf c
out
öõ ø
in
]
[x
1
Z
ments Figure 11.5: Function process
axis
in
in
ÛY¡ e
6 ¢)9 '
6 ¢)9 '
6 ¢)9 '
&¿/¡1¿m©?2/d@ÂÍ
out
Ö²
out
1
out
.
axis axis
á h ö õi
ments õ
hö
axis axis axis
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; in other words, each of these “reverse processes” defines reverse process jk a function. Á ¯ h ~  õ So, given M j , there are TWO possible values, namely Ç[ ¦ ) ! , so that ! 6 4 ) !A9Ò"j and ! 6 4 ) !A9Ò"j . In other words, the reverse process is not given by a single equation; there are TWO POSSIBLE reverse processes.
(a) Reverse process but not a function.
õá h ö õ h ö
11.2 Graphical Idea of an Inverse
reverse process

Á ¯ á h ~ª õ reverse process
Á ¯ h ~ª õ 
(b) Two new reverse processes that are functions. Figure 11.6: What to do if a reverse process is not a function.
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 ! 6 @9 as in Figure 11.7(a). What input values result in an output value of 3? This involves finding all such that ! 6 @9m.j . Graphically, this means we are trying to find points on the graph of ! 6 9 so that their M coordinates are jT The easiest way to to do this is to draw the line M ]j and find where it intersects the graph. In Figure 11.7(b) we can see the points of intersection are 6 ` 1jT9 , 6 Én1jT9 , and 6 o´1jT9 . That means that Ç. ` , É , o produce the output value j ; i.e., ! 6 ` 9Ò#! 6 É9Ò#! 6 o9Ò)j .
axis axis
PSfrag replacements axis 11.2. GRAPHICAL IDEA OF AN INVERSE
í î
axis
153
í î
axis
Ú
axis
:
Sð
Ú ø
u æ u
æ u
:
í ð S
Ú O u
ù
ø ù
ø
(a) Given the function e ÛY¡ ¿k/ .
l
V
ø
(b) Whate values of ¿k/¡ ?
¿
give
Figure 11.7: Using the horizontal line Û¢¡Ç to find values on the ¿ axis.
This leads to our first important fact about the “reverse process” for a function: Important Fact 11.2.1. Given a number ï , the values such that ! 6 9Ë ï can be found by finding the coordinates of the intersection points of the graphs of M &! 6 @9 and M½ðï . Example 11.2.2. Graph M &! 6 @9Ò" when ïP)jA1n1^ .
and discuss the meaning of Fact 11.2.1 PSfrag replacements
Solution. We graph M)$ and the lines M' , M) j and M) ^ . Let’s use ïÍ ^ as an example.
We need to simultaneously solve the equations MW and M]y^ . ++Putting
these++ together, we get ô^ or 'ö¦ ) ^ml ¦
í
axis axis axis
Ú
Sö
í í
Ú í Ú l Ú
Figure 11.8: Graph of Ûy¡ The pictures so far indicate another very important e &¿/¡ ¿¹ . piece of information. For any number ï , we can tell exactly “how many” input values lead to the same output value ï , just by counting the number of times the graphs of M½#! 6 @9 and M ï intersect.
Important Fact 11.2.3. For any function ! 6 9 and any number ï , the number of values so that ! 6 @94 ï is the number of times the graphs of M ï and MN#! 6 9 intersect.
CHAPTER 11. INVERSE FUNCTIONS
154 Examples 11.2.4.
(i) If ! 6 9 is a linear function ! 6 9 FUWQm1 ã % , then the graph of ! 6 9 intersects a given horizontal line M ï EXACTLY once; i.e., the equation ïP#! 6 9 always has a unique solution.
(ii) If ! 6 93y is a constant function and ïy , then every input value in the domain leads to the output value ï . On the other hand, if ïU[ ã , then no input value will lead to the output value ï . For example, if ! 6 9 G and ïF , then every real number can be input to produce PSfrag replacements an output of 1; if ï¢)! , then no input value of will lead to an output of !A axis
î
axis
í
axis
Ú
zí
Sð :
axis
Ú
SED ç Ú
axis
î
:
S
S
axis
í Sð
axis
any of these inputs the only input which leads to an output of
linear functions
Figure 11.9: Does a horizontal line ÛÉ¡
leads to an output of
constant functions
intersect a curve once or more than once?
11.2.1 Onetoone Functions For a specified domain, onetoone functions are functions with the property: Given any number ï , there is at most one input value in the domain so that ! 6 @9É ï . Among our examples thus far, linear functions
(degree 1 polynomials) are always onetoone. However, ! 6 @9ÿ is not onetoone; we’ve already seen that it can have two values for some of its inverses. By Fact 11.2.3, we can quickly come up with what’s called the horizontal line test. Important Fact 11.2.5 (Horizontal Line Test). On a given domain of values, if the graph of some function ! 6 9 has the property that every horizontal line crosses the graph at most only once, then the function is onetoone on this domain.
11.3. INVERSE FUNCTIONS
155
PSfrag replacements Example 11.2.6. By the horizontal line test, it is easy to axis see that ! 6 9Iôl is onetoone on the domain of all real axis numbers. axis Although it isn’t common, it’s quite nice when a function is onetoone because we don’t need to worry as much about the number of input values producing the same output M value. In effect, this is saying that we can define a “reverse process” for the function M ¦! 6 9 which will also be a function; this is the key theme of the next section.
yaxis
xaxis
lines
Figure 11.10:e A onetoone function &¿/¡I¿¹« .
11.3 Inverse Functions Let’s now come face to face with the problem of finding the “reverse process” for a given function M' ! 6 @9 . 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 11.11 shows the function ! 6 9Db with three different domains specified and the corresponding range values.
eplacements axis
range
range
axis
range
axis domain
domain
horizontal
domain
Figure 11.11: 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 and M for the domain (input) and range (output) variables of ! 6 9 and the “reverse process” is going to reverse these roles. It then seems natural to simply write M (instead of ï ) for the input values of the “reverse process” and for its output values.
CHAPTER 11. INVERSE FUNCTIONS
156
Important Fact 11.3.1. Suppose a function ! 6 9 is onetoone on a domain of values. Then define a NEW FUNCTION by the rule
æ 6 M39Ò the value so that ! 6 9ÒêM3 æ The domain of M values for the function ! 6 M39 is equal to the range of the function ! 6 9 . !
The rule defined here is the “reverse process” for the given function. It æ is referred to as the inverse function and we read ! 6 M39 as “...eff inverse of M ...”.
ements axis
À
!!!
axis
and the “rule” J have equal influence on Both the “domain” of J whether the inverse rule is a function. Keep in mind, you do NOT get an inverse function automatically from functions that are not onetoone!
CAUTION !!!
axis
ements axis
11.3.1 Schematic Idea of an Inverse Function æ Suppose that ! 6 9 is onetoone, so that ! 6 M39 is a function. As a result, æ we can model ! 6 M39 as a black box. What does it do? If we put in M in the input side, we should get out the such that ! 6 @9ÒêM .
axis axis
in
out
n  !
n such ! that
Figure 11.12: e A new funcZ tion ¿7¡ ¨PcÛH/ .
Now, let’s try to unravel something very special that is happening on a symbolic level. What would happen if we plugged ! 6(' 9 into the inverse function for some number ' ? æ Then the inverse rule ! 6 ! 6(' 99 tells us that we want to find some so that ! 6 @9Ç ! 6(' 9 . But, we already know  ' works and since ! æ 6 M39 is a function (hence gives us æ unique answers), the output of ! 6 ! 6(' 99 is just ' . Symbolically, this means we have Fact 11.3.2.
PSfrag replacements Important Fact 11.3.2. For every ' value in the domain of ! 6 @9 , we have axis
!
axis axis
æ 6 ! 6(' 998 '
(11.1)
This is better shown in the black box picture of Figure 11.13.
î Ã
in :
î Sð
out :
Ã
î ð
in
Figure 11.13: Visualizing
e
:
Z
out
õ Sð
¨2 e ßÅ+/À/¡FÅ
Ã
.
A good way to get an idea of what an inverse function is doing is to æ æ remember that ! 6 M39 reverses the process of ! 6 @9 . We can think of ! 6 M39 as a “black box” running ! 6 9 backwards.
11.4. TRYING TO INVERT A NON ONETOONE FUNCTION
157
11.3.2 Graphing Inverse Functions
How can we get the graph of an inverse function? The idea is to manipulate the graph of our original onetoone function in some prescribed æ way, ending up with the graph of ! 6 M39 . 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 M &! 6 @9 looks like 6 81H! 6 99 . Now æ let’s take a look at the inverse function Zå! 6 M39 . Given a number M in æ the domain of ! 6 M39 , M &! 6 9 for some in the domain of ! 6 @9 ; i.e., we are æ using the fact that the domain of ! equals the range of ! . The function æ M39 takes the number ! 9 and sends it to , by Fact 11.3.2. So when ! 6 6 ! 6 9 is the input value, becomes the output value. Conclude a point on æ the graph of ! 6 M9 looks like 6 ! 6 9z19 . It’s similar to the graph of M S! 6 9 , only the and M coordinates have reversed! What does that do to the graph? Essentially, you reorient the picture so that the positive axis eplacementsand positive M axis are interchanged. Figure 11.14 shows the process for æ M39  M . We ! axisthe function MHG! 6 @9ÿ l and its inverse function 6 ) axisplace some o symbols on the graph to help keep track of what is happenaxising.
Úæ D
axis
D S æ
*
axis rotate O ä clockwise
*
* *
2S æ D
D
axis
* *
Ú æ
2S æ
axis
D
axis
* *
¿¡ e
Z
Ú æ
axis
*
flip across horiz axis
Figure 11.14: Graphically finding
D
¨2cÛH/¡ È Ó Û
.
11.4 Trying to Invert a Non onetoone Function
Suppose we blindly try to show that ) M is the inverse function for MN" , without worrying about all of this onetoone stuff. We’ll start out with i æ the number . If ! 6 M39Ë ) M , then we know that !
+ i i æ æ 6 ! 6 998&! 6 o 9Ò
On the other hand, the formula in Fact 11.3.2 tells us that we must have !
i i æ 6 ! 6 998H 1
i
i
æ
so we have just shown ! So clearly ! 6 M39U ã ) M . Even if we try æ M39Ò. M , we produce a contradiction. It seems that if you didn’t have ! 6 )
CHAPTER 11. INVERSE FUNCTIONS
158
to worry about negative numbers, things would be all right. Then you æ could say that ! 6 M39Ò ) M . Let’s try to see what this means graphically. Let’s set ! 6 9I , but only for nonnegative values. That means that we want to erase the graph to the left of the M axis (so remember  no negative values allowed). The graph would then look like Figure 11.15.
í
PSfrag replacements axis D
axis
Ú Ú
axis
Sö S D
inverse function
2S
Ú D
D
domain nonnegative
S
domain nonnegative
Ú
Ú
Figure 11.15: Restricting the domain: No negative ¿ values.
This is a now a onetoone function! And now, one can see
that its inverse function is ) M . Similarly, we could have taken ! 6 9Ì' but only æ for the nonpositive values. In that case, ! 6 M39ïÏ ) M . In effect, we have split the graph of Mê] into two parts, each of which is the graph of a onetoone function; Figure 11.16. PSfrag replacements í axis axis axis
Ú
Sö
D
Ú
S D
S D
domain nonpositive
domain nonnegative
Ú
S
D
Ú
inverse function
æ Ú
Figure 11.16: Restricting the domain: No positive ¿ values.
It is precisely this splitting into two cases that leads us to multiple solutions of an equation like ` . We obtain U ) ` and U$ ) ` ; one
11.4. TRYING TO INVERT A NON ONETOONE FUNCTION
159
solution comes from the side of the graph to the left of the M axis, and the other from the right of the M axis. This is because we have separate inverse functions for the left and right side of the graph of MN" .
CHAPTER 11. INVERSE FUNCTIONS
160
11.5 Exercises on the largest Problem 11.1. Let &¿/Ë¡ « Y[Z º makes domain for which the formula sense. e
(a) Find the domain and range of sketch the graph.
e
¿k/ , then
(b) Find the domain, range and rule for e Z the inverse function ¨ , then sketch its graph. Problem 11.2. Find the inverse function of the each of the following functions. Specify the domains of the inverse functions. (a) (b) (c) (d) (e)
¿/¡ e
¾ &¿/n¡ r&¿/¡ «& ¿/n¡ èQ ¿ / ¡ j
¿ÌÂ ¿ÌÂI eÈ 8©É¿Ì© È ¿4Â È ¿Ì©? ; ª@©¢¿ k h ¿lh
Problem 11.5. Show that the functions Ûê¡ ¿« , ÛU¡ z¿7ÂBf and ÛU¡©8z¿ are onetoone by checking their graphs. Problem 11.6. Which of the following graphs are onetoone? If they are not onetoone, section the graph up into parts that are onetoone. A
B
C
D
PSfrag replacements
Problem 11.3. For this problem, Û¡ ¿k/7¡ z¿¹A©Y¿â©Ç on the domain of all real numbers. e
(a) Sketch the function graph and find the coordinates of the vertex ù<¡ ßÅ 78/ .
(b) Explain why Û ¡ &¿/ does not have an inverse function on the domain of all real numbers. e
&¿/ to the domain ®§ÅIh (c) Restrict Û ¡ ¿¯ and find the formula for the inverse e Z ¨2cÛH/ . What are the domain function and range of the inverse function? e
(d) Restrict ÛB¡ ¿/ to the domain ®å¿4h the formula for the inverse Å¯ and find e Z function ¨2cÛH/ . What are the domain and range of the inverse function?
axis axis axis
Problem 11.7. Show that, for every value of Å , the function e
¿/¡FÅmÂ
¿m©?Å
is its own inverse.
e
Problem 11.4. A function is called onto if every horizontal line will intersect the graph at least once. (a) Show that onto.
¿¹«
is onto and that
¿¹
is not
(b) Draw pictures of a graph (you do not need an algebraic formula  just the graph) of a function with domain all real numbers which is (i) onetoone and onto (don’t use ¿¹« ) (ii) onetoone but not onto (iii) not onetoone but onto (iv) neither onetoone nor onto (don’t use ¿v )
Problem 11.8. 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 ¿ axis extending horizontally, the path of the rocket is described by the formula ÛY¡?©8z¿vÂÍ§0¿ .
¾
(a) Give a function ¡ &¿/ relating the ¾ height of the rocket above the sloping ground to its ¿ coordinate. e
(b) Find the maximum height of the rocket above the sloping ground. What is its ¿ coordinate when it is at its maximum height?
11.5. EXERCISES
161
¾
(c) Clovis measures its height of the rocket above the sloping ground while it ¾ is going up. Give a function ¿<¡%jr / relating the ¿ coordinate of the rocket to ¾ .
(a) Give a function r ¡  Ä2/ relating the width r of the surface of the water to the time Ä , in hours. Make sure to specify the domain and compute the range too.
(d) Does this function still work when the rocket is going down? Explain.
(b) After how many hours will the surface of the water have width of 6 feet?
Problem 11.9. For each of the following functions: (1) sketch the function, (2) find the inverse function(s), and (3) sketch the inverse function(s). In each case, indicate the correct domains and ranges. (4) Finally, make sure you test each of the functions you propose as an inverse with the following compositions:
(c) Give a function ÄY¡ ¨Ôsr / relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too.
e
and (a)
e
e
(b)
e e
(c) (d) (e)
e
e
Z

e
Z
¨ &¿/À/q¡ p ¿
¨  e ¿/À/ ¡ p ¿
&¿/¡¿4©Y &¿/ ¡ ¨ ¿ÌÂY &¿/ ¡©@¿¹ÂD¹ß¿ã÷Ç &¿/ ¡I¿¹@ÂYz¿4ÂYv»¿ªh?©4 &¿/ ¡ È Ë©¢¿ , hD¿h
e
e
Z
Problem 11.11. A biochemical experiment involves combining together two protein extracts. Suppose a function t  Ä2/ monitors the amount (nanograms) of extract A remaining at time Ä (nanoseconds). Assume you know these facts: 1. The function t is invertible; i.e., it has an inverse function. 2.
Problem 11.10. 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.
t ñ*/7¡'ª , t » "/7¡" , t  &/7¡" , t  */¡[ , t &T/¡Ç , t »0/¡Ç .
(a) At what time do you know there will be 3 nanograms of extract A remaining? (b) What is you?
t Z ¨Pñ &/
and what does it tell
(c) (True or False) There is exactly one time when the amount of extract A remaining is 4 nanograms.
water
t ut Z P¨ »"/f/¡ Z Calculate t ¨2`t åªR/À/=¡
(d) Calculate
replacements axis axis axis
(e)
5 ft crosssection of trough
(f) What is the domain and range of
t
?
162
CHAPTER 11. INVERSE FUNCTIONS
Chapter 12 Rational Functions î î ð
A rational function is a function of the form ! 6 9 v w S ð Ú axis 6 S where ´ 6 9 and ð 6 @9 are polynomials. The PSfrag easiestreplacements example 4 is ! 6 9Ë which we graph in Figure 12.1. 2 S S axis10 An asymptote is a line that the graph of a function −5 5 −10 −2 gets very close to, without ever quite reaching it. Note axis −4 that as gets very large, the graph of MG becomes −6 S very close to the axis. The line Mé % is a horizontal Figure 12.1: The graph of asymptote for the function. When is close to 0, the e k ¿ / ¡ Y¨ . graph is very close to the M axis. The line ÇHî % is a vertical î asymptote for this function. Note that the graph goes up the vertical asymptote on the right, and down it on the left. Now let’s ð consider a general rational function ! 6 @94 v w S ð . Our first question about S any function is “What are its zeros?”
*
öæ æ S ö æ S . Describe the zeros for this function. S * Solution. We wish to solve %Y 6 9 for . * % 6 9 ! ü¢'
' % E 6 '9 ! ü¢' % 6 ¢H9 6 !PH9¼ Example 12.0.1. Let
6 9Ê
It appears that * there are two answers, #B and Z . But looking closer we see that 6 9 is undefined, because the bottom is zero. So N # is really the only answer. î
î
The problem in the last example can be avoided if we reduce our rað tional function to lowest terms. In general, let ! 6 9P v w S ð be a rational S function which is reduced to lowest terms. Then the zeros of ! 6 9 are the values where ´ 6 @9Ê.% . Furthermore, the vertical asymptotes of M #! 6 9 occur at the values where ð 6 @9ÒH% . 163
CHAPTER 12. RATIONAL FUNCTIONS
164
Example 12.0.2. Put the rational function Describe the zeros and asymptotes.
*
6 9â
öæ æ S ö æ S into lowest form. S
Solution.
*
6 Y'9 6 !¢)9 6 ¢H9 6 ¢)9 !Y' Y' Thus we have a zero at ÇHZ and the line ÇHÉ is a vertical asymptote. 6 9
To find the horizontal asymptote for this function, note that the value of the fraction does not change if we multiply top and bottom by .
*
S
!Y' S Y' E S !ï S S * When is very large, of M is very small and the value 6 9 is very close *
S to . Thus M ! is the horizontal asymptote for 6 @9 . Example 12.0.3. Let 6 6 @9Ò _S ö D æ . Describe the zeros and asymptotes. S ä Solution. As before, the value of 6 6 9 for ã % does not change if we multiply both the top and the bottom by ö . So, we have, for H ã %, S Y ' Sö 6 6 9 E o ö S ö S S 5ä ö S When is very large, the numerator is close to 0 but the denominator is close to 1. Thus the ratio is close to 0. We conclude that M é% is a horizontal asymptote. Note that 6 6 9 _S ö D æ is in lowest terms. The top S vanishes at . É . This is the only zeroä of 6 6 @9 . The bottom is zero ¦
Now we are almost ready to graph MN¦6 6 @9Ò S^ö D æ . But first it is useful S ä to figure out where the function is positive and where it is negative. Of course, the graph can cross the axis at the zero of the function. But it can also jump across at the vertical asymptotes. The function 6 6 9 has one zero, yÉ , and two vertical asymptotes, j and ÿj . Mark these three values on a number line; see Figure 12.2. These points divide + the number line into four intervals. Check the function at a point in each interval. Here I use ÿ , ;! , 0 and 4. Other values would be just as good. I conclude that 6 6 9 is negative when X j or ÉZX Xj and positive when jDX ?XHÉ or Vj .
axis axis
î
î
è æ ðyxO p −4
î
è æ ðTz;O −3
negative
−2 positive
î
è OðyxO −1
0
2
1
3
negative
Figure 12.2: Sign of
165
è ðTz;O p 4
S
axis
positive
T&¿/ . j
PSfrag replacements
Ú
axis
Now let’s graph M$ 6 6 @9¢ S_ö D æ . First draw a pair of S ä Wtj and axes and sketch in the vertical asymptotes ÍHj . We know that the axis is a horizontal asymptote. axis S axis Note that 6 Én1k%9 is the only place where the graph crosses the axis. Á ¯ * Á ¯  Coming in from the right we have VWj and so 6 6 @9V % . Thus the graph must come in above the horizontal Figure 12.3: Starting the asymptote and go up Ç)j on the right. Ye graph of jT&¿/¡ Y Z ¨ . Next, we are in the interval É X ½Xqj so 6 6 9 is neg¶ £ ative. Thus the graph must go down HGj on the left. PSfrag replacements Continue this reasoning to get the rest of the graph. A rational function is called lineartolinear if both the Ú axis top and the bottom have degree one. This is a very important class of functions, with many applications. In axis general, these look like alo
!
' ¢Q 6 @9Ò ï¢H
where ' , Q , ï , and are real numbers. In fact, we can always take ï¢$ . Here’s an example that shows how.
{Õ } q
S
õ
Á ¯ *
axis
Á ¯  alo
Figure 12.4: The graph of jr&¿/ on the right of ¿¡Ç .
CHAPTER 12. RATIONAL FUNCTIONS
166
ements
Ú
Example 12.0.4. Let ! 6 @9Ò
axis
axis !
{i } q
S
õ
Á ¯ 0
axis
Á ¯ 
Figure 12.5: The final graph of jT¿/ .
6 9
!Yüj i E u ` Y
u Y l u u Y ¾ u
S^Dnl . Then u¥S æ ¾
This can be useful when we are trying to model something using lineartolinear rational functions.
Example 12.0.5. Clyde makes extra money scalping 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 $50 each. How much will he charge for a ticket if he holds 20 tickets?
Solution. We want to give a lineartolinear rational function relating the price of a ticket M to the number of tickets that Clyde is holding. As we saw above, we can assume the function is of the form
Q ¢ MN ' ¢' where ' , Q and are numbers. Note that M ' is the horizontal asymptote. When is very large, M is close to 50. This means the line Mü ` % is a horizontal asymptote. Thus ' ` % and %v¢Q MN ` Y) Next we plug in the point 6 n1de%%9 to get a linear equation in Q and . % >Q e%% ` E >' e%% E 6 >H@9b ` %Q ` % Q)e%% Similarly, plugging in 6 e%´1&J%9 and doing a little algebra (do it now!) gives another linear equation j%%'5QN[J% . Solving these two linear equations simultaneously gives  0!A ` and QZ$j%% . Thus our function is
%v¢'j%% MN ` ¢'0!A `
and, if Clyde holds 20 tickets, he will charge
% !%Hj%% i ii MN ` E $ % !%;H0!A `
per ticket.
12.1. EXERCISES
167
12.1 Exercises Problem 12.1. Give the domain of each of the following functions. Find the zeros. Sketch a graph and indicate any vertical or horizontal asymptotes. Give equations for the asymptotes. e Y Ye (a) ¿k/¡ Y[ Z (b) jr&¿/¡ « Y"Z Y ¶ ZI¨ Y"Z Y ¶ e Y e ¨ ¾ eY"Z ¦ (c) ¿/¡ (d) &¿/¡ Y YZ e ¾ ¶ (e) ¿/¡ Y Z Y"« Z (f) ¿k/¡ Y e Y e Q Y ¶ e « S Q Y eº S ¶ Y «e (g) èQ&¿/¡ Q Y e S Q Y e ¨ S (h) ° &¿/¡ Q Y"Z S Q ºY"Z S
«
¦
Problem 12.2. The multipart function Û¢¡ is defined below: e
ìí
&¿/¡Gë
Y¶Z Y"Z ¨ Â Y ¨
¨
if if if
(a) Sketch the graph of
¿V¡Ç U ¿¡Ç ¿¡F
Û¢¡
e
and
e
º
¿/
¿V¡Z U
(a) Give a linear model relating the price of the Martin uke to the year Ä . Take ÄÌ¡F in 1975. (b) Give a linear model relating the price ñ of the Kamaka uke to the year Ä . Again take ÄÒ¡ in 1975. (c) When is the value of the Martin twice the value of the Kamaka? (d) Give a function  Ä2/ which gives the ratio of the price of the Martin to the price of the Kamaka. e
¿/ .
&¿/ to the domain õ ¿Bh (b) Restrict ÛF¡ and find the formula for the inverse e Z function ¨2kÛN/ . e
Problem 12.4. 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.
(c) Restrict ÛÉ¡ &¿/ to the domain h ¿ and find the formula for the inverse function e Z ¨ÔkÛN/ .
(e) In the long run, what will be the ratio of the prices of the ukuleles?
e
ó©@¿k/ and (d) Sketche the graphs of Û ¡ Û ¡)© å©¿/ and describe their relationship with the original function graph in (a). e
Problem 12.3. 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 lineartolinear rational model relating the meter reading Û to how many feet ¿ 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 12.5. 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 Û to the number of tapes sold ¿. (b) Divide the profit function in part (a) by the number of tapes sold ¿ to get a model relating average profit r per tape to the number of tapes sold. (c) How many tapes must she sell in order to make $1.20 per tape in net profit? Problem 12.6. Find the lineartolinear function whose graph passes through the points  R&2/ , v &/ and 0¹ */ . What is its horizontal asymptote? Problem 12.7. Find the lineartolinear function whose graph has Û"¡$ª as a horizontal asymptote and passes through ó&0/ and  ¹ / .
168
CHAPTER 12. RATIONAL FUNCTIONS
Problem 12.8. 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 lineartolinear function of the number of hours that you study. If you want a score of 80%, how long do you need to study?
Problem 12.10. For each of e the following find the linear to linear function ¿k/ satisfying the given requirements:
Problem 12.9. A street light is 10 feet North of a straight bike path that runs EastWest. Olav is bicycling down the path at a rate of 15 MPH. At noon, Olav is 33 feet West of the point on the bike path closest to the street light. (See the picture). The relationship between the intensity D of light (in candlepower) and the distance , (in feet) from the light source is given by DÉ¡ é , where è is a constant depending on the light ¶ source. (a) From 20 feet away, the street light has an intensity of 1 candle. What is è ?
e
ó0/¡¹ »0/¡Z  00/¡#§ e
ó0/¡Z  &/¡ 2 0/¡Ç e
 ¥0/;¡)0  0/w¡'Rv and the graph of ¿/ has Û?¡# as its horizontal asymp
e
(a)
e
(b) (c)
e
e
e
e
tote
Problem 12.11. 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.
(b) Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds.
(a) Finde a linear to linear rational function ÛÇ¡ &¿/ that calculates the number Û of customers/day if $¿ is spent on advertising.
(c) When will the light on Olav have maximum intensity?
(b) How much must the shop spend on advertising to have 300 customers/day.
(d) When will the intensity of the light be 2 candles?
(c) Sketch the graph of the function ÛÉ¡ on the domain khI¿hI0R .
(Top View−looking down)
PSfrag replacements axis axispath axis
10ft
olav 33ft
e
&¿/
(d) Find the rule, domain and range for the inverse function from part (c). Explain in words what the inverse function calculates.
PSfrag replacements
Chapter 13
axis axis axis
Measuring an Angle So far, the equations we have studied have an algebraic character, involving the variables and M , 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:
ÿ
ÿ
Cosmo motors
{
 X~
feet
around
the circle
Figure 13.1: Cosmo the dog walking a circular path.
ÿ
Natural Questions 13.0.1. How can we measure the angles g
,
, and
? How can we measure the arc lengths , , arc and arc ? 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 , , and ? Turning this around, if we know how to compute the coordinates of , , and , can we then measure the angles g
,
, and ? 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 problemsolving power of these functions will become apparent in our discussion of sinusoidal functions in Chapter 17 and parameterized motion in Chapter 22. The 2M 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 twodimensional 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. 169
}
CHAPTER 13. MEASURING AN ANGLE
170
13.1 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 PSfrag replacementsby rotating a single ray from one position to another, sweeping counteraxisclockwise or clockwise : See Figure 13.2. We often insert a curved arrow axisto indicate the direction in which we are sweeping out the angle. The ray is called the initial side and à the terminal side of the angle (É . axisà
(terminal side) ô
ö
START
ô
õ
vertex
SWEEP CLOCKWISE
vertex
ô
õ (initial side)
SWEEP COUNTERCLOCKWISE
Figure 13.2: Angle
NBC
(initial side) ô
õ
(terminal side) ô
ö
.
Working with angles, we need to agree on a standard frame of reference for viewing them. Within the usual 2M coordinate system, imagine a model of 7 @É in Figure 13.2 constructed from two pieces of rigid wire, welded at the vertex. Sliding this model around inside the 2M 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 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 13.3 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 ' 0ï 6 I >9 . 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 axis. For example, ïÙÉØ¥ in the Figure beginning this Chapter is a central angle which is not in standard position.
eplacements 13.2. AN ANALOGY axis
axis
171
c
arc subtended arc(
c Ì
arc(
Ì
c
axis c
)
)
vertex Ì
axis
COUNTERCLOCKWISE
vertex Ì
axis
CLOCKWISE
Figure 13.3: Standard angles and arcs.
13.2 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.
13.3 Degree Method
Begin by drawing a circle of radius , call it , centered at the origin. Divide this circle into 360 equal sized pie shaped wedges, beginning with the the point 6 n1&%9 on the circle; i.e. the place where the circle crosses the axis. We will refer to the arcs along the outside edges of these wedges as onedegree 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
PSfrag replacements axis axis
CHAPTER 13. MEASURING AN ANGLE
172
axis
circle etc.
O
î
a total of 360 equal sized pie shaped wedges inside this disk
ë Oð
typical wedge
this angle is DEFINED to have measure 1 degree
etc.
***NOT TO SCALE***
Figure 13.4: Wedges as
Ã
arcs.
defined to have a measure of one degree, denoted . We can now use onedegree arcs to measure any angle: Begin by sliding the angle (É into standard central position, as in Figure 13.3. Piece together consecutive onedegree arcs in a counterclockwise or clockwise direction, beginning from the initial side and working toward the terminal side, approximating the angle (É to the nearest degree. If we are allowed to divide a onedegree arc into a fractional portion, then we could precisely determine the number of onedegree arcs which consecutively fit together into the given arc. If we are sweeping counterclockwise from the initial side of the angle, is defined to be the degree measure of the angle. If we sweep in a clockwise direction, then < is defined to be the degree measure of the angle. So, in Figure 13.3, the lefthand angle has positive degree measure while the righthand angle has negative degree measure. Simple examples would be angles like the ones in Figure 13.5. Notice, with our conventions, the rays determining an angle with measure Éj ` sit inside the circle in the same position as those for an angle of measure !! ` ; the minus sign keeps track of sweeping the positive axis clockwise (rather than counterclockwise). We can further divide a onedegree arc into 60 equal arcs, each called a one minute arc. Each oneminute 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 +í and one second, denoted +í í :
^ % minutes j^%% seconds.
13.3. DEGREE METHOD
173
çO
ä O
O ¾
eplacements axis axis axis
pu
æ l¥u
l u
Figure 13.5: Examples of common angles.
For example, an angle of measure # 5 degrees 23 minutes 18 seconds is usually denoted ` !nj í eJ í í . We could express this as a decimal of degrees:
` !nj í eJ í í
g
!nj eJ ` ^% j^%% h ` kjJJ:j
In degrees!
i
+
As another example, suppose we have an angle with measure ` ` ^ and + + we+ wish to convert this into degree/minute/second units. First, since i i ` ` ^ ` D% ` ^ , we need to write % ` ^ in minutes by the calculation:
+
i minutes % ` ^ degree ,Z^% "! kj^ í degree + i i i i i This tells us that ` ` ^ ` ! cj^ªí ` ! ví %kj^ í . Now we need to write %kj^ªí in seconds via the calculation: %kj^ minutes ,F^% seconds/minute !A^ í í + i i i In other words, ` ` ^ ` ! í¼!A^ í í .
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 13.3 and recall the situation
CHAPTER 13. MEASURING AN ANGLE
174
where an arc ' Rï 6 D >9 is subtended by the central angle 7 @É . In this situation, the arc length of ' 0ï 6 D >9 , commonly denoted by the letter , 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 o% , i eJ% , ! % , ` , Éj ` , and jT ` . If the circle is of radius 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 13.3.1. (length of a part) (fraction of the part) , (length of the whole) For example, the circumference of the entire circle of radius is !( ; this is the “length of the whole” in the general principle. The arc subO tended by a 90 angle is ä O of the entire circumference; this is the l p principle. The boxed formula implies: “fraction of the part” in the general
ements
axis
Þ ¯
distance along the arc
axis axis
arc length of the 90 arc
degrees
g
+ h
!(<
( !
Similarly, a eJ% angle subtends an arc of length D a jT ` angle subtends an arc of length ¢ôx l uO { !(@Y l¼ etc. In general, we arrive at this formula: ¾
(P , p
,
Important Fact 13.3.2 (Arc length in degrees). Start with a central angle of measure degrees inside a circle of radius . Then this angle will subtend an arc of length
Figure 13.6: The definition of arc length.
g
! ( j^% h
13.4 Radian Method The key to understanding degree measurement was the description of a “basic wedge” which contained an interior angle of measure ; 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 13.3.2. However, the formula for the length of an arc subtended by an angle measured in degrees is sort of cumbersome, involving the curi ous factor O . Our next goal is to introduce an alternate angle measurel ment 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.
13.4. RADIAN METHOD
175
eplacements axis
axis axis
î
equilateral wedge
ë ORð
circle
O
this angle is DEFINED to have measure 1 radian
Figure 13.7: Constructing an equilateral wedge.
As before, begin with a circle of radius . Construct an equilateral wedge with all three sides of equal length ; see Figure 13.7. 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 radian is obtained by symmetrically dividing 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 radian, 2 radian, 3 radian, radian and radian, the length of the arcs subtended by these angles are , ! , j2 , p , and . In other words, a p between the length pattern emerges that gives a very simple relationship of an arc and the radian measure of the subtended angle: PSfrag replacements Important Fact 13.4.1 (Arc length in radians). Start with a central angle of measure radians inside a circle of radius . Then this angle will subtend an arc of length < . These remarks allow us to summarize the definition of the radian measure of (É inside a circle of radius by the formula:
à
I Ù à
if angle is swept counterclockwise if angle is swept clockwise
Þ ¯
axis axis axis
distance along the arc
radians
Figure 13.9: Defining arc length when angles are measured in radians.
CHAPTER 13. MEASURING AN ANGLE
176
2r
PSfrag replacements
r
axis axis
í
í¡
3r
3 radians
2 radians
r
r
r
í
axis
r
r 1 radian r
öõ r
radian
Öõ
radian
r
ements
í¡
í¡
r r
p
Figure 13.8: Measuring angles in radians.
The units of are sometimes abbreviated as rad. It is important to appreciate the role of the radius of the circle when using radian measure of an angle: An angle of radian measure will subtend an arc of length on the radian unit circle. In other words, radian measure of angles is ¢ S axis exactly the same as arc length on the unit circle; we couldn’t hope for a better connection! circle radius The difficulty with radian measure versus degree measure is really one of familiarity. Let’s view a few common Figure 13.10: Arc length afangles in radian measure. It is easiest to start with the ter imposing a coordinate case of angles in central standard position within the unit system. circle. Examples of basic angles would be fractional parts of one complete revolution around the unit circle; for example, revolution, revolution, revolution, revolution, revolution and l revoç lution. One revolution around the p unit circle describes an arc of p length !( and so the subtended angle (1 revolution) is !( radians. We can now easily find the radian measure of these six angles. For example, rev rad. Similarly, olution would describe an angle of measure 6 9 !( rad= the other five angles pictured below have measures rad, rad, rad, ( l p rad and l rad. All of these examples have positive radian measure. For an angle with negative radian measure, such as ' radians, we would locate
by rotating revolution clockwise, etc. From these calculations and our
axis
Ú
axis
{ ¯ Þ
length of arc
p
13.5. AREAS OF WEDGES
177
eplacements axis axis
axis
p
õõö
¶õ
revolution
Öõ
revolution
l ³õ
revolution
l öõ
revolution
revolution
Ö
revolution
Figure 13.11: Common angles measured in radians.
previous examples of degree measure we find that
eJ% degrees )( radians
(13.1)
Solving this equation for degrees or radians will provide conversion formulas relating the two types of angle measurement. The formula
also helps explain the origin of the curious conversion factor O l O ç in Fact 13.3.2.
13.5 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. “pieshaped regions”). This is a key reason why we will almost always prefer to work with radian measure. Example 13.5.1. 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 13. MEASURING AN ANGLE
178 So, for our pizza: (area one slice)
(area whole pie) , (fraction of pie)
ements
axis
£
O
6 J (49 e^2( j
g
0 ! h
area of a Let’s apply the same reasoning to find the circular sector. We know the area of the circular disc bounded by a circle of radius is ( . Let ¤ be the “pie shaped wedge” cut out by an angle 7 @É with positive measure radians. Using the above principle
axis axis
Figure 13.12: Finding the area of a “pie shaped wedge”.
area 6 ¤
9Ò
6 area of disc bounded by 9
, 6 portion of disc accounted for by ¤ 9
g 6 ( 9 A !( h ! For example, if j in. and # rad, then the area of the pie shaped wedge is ä ( sq. in. p ç
Important Fact 13.5.2 (Wedge area). Start with a central angle with positive measure radians inside a circle of
radius . The area of the “pie shaped region” bounded by the angle is .
Example 13.5.3. A water drip irrigation arm 1200 feet long rotates around a pivot Ø 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 !( radians, so after / hours the arm sweeps out an angle 6 /:9 given by
6 /:9Ò
!( radians ( ,U/ hours / radians 0! hours ^
Consequently, by Important Fact 13.5.2 the area 6 /:9 of the irrigated region after / hours is
( 6 /:9Ê 0!%%9 6 /29Ò 0!%%9 ; / [0!%´1&%%%2(@/ square feet. 6 6 ! ! ^ i After 1 hour, the irrigated area is 6 970!%´1&%%%2(ê j ^´1&oo sq. ft. Likewise, after 37 minutes, which is Hl O¾ hours, the area of the irrigated region is 6 Hl O¾ 9ïC0!%´1k%%%2( 6 Hl O¾ 9ïy!nj2!1 ` %% square feet. To answer the final ques the equation /29>ée%%% ; i.e., 0!%´1k%%%2(8/ïée%%% , so tion,g we need to solve 6 g j^%% seconds hours , /; 'o `` seconds. 0!%2( h h hour
13.5. AREAS OF WEDGES
179
13.5.1 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 ¤ and walks counterclockwise to location ¥ . Furthermore, let’s suppose you are standing at the center of the circle determined by the tether and you measure the angle from ¤ to ¥ to be ` ; see the lefthand figure. Because the angle is small, notice that the straight line distance from ¤ to ¥ is approximately the same as the arc length subtended by the angle ¤@Ø¥ ; the righthand picture in Figure 13.13 is a blowup:
eplacements axis axis
¢
axis
¢
Ö 20 feet
u
Ö
u £
20 feet
Figure 13.13: Using the arc length
à
ø £
:
to approximate the chord , .
Example 13.5.4. Estimate the distance from ¤ to ¥ .
+ + ` Solution. We first convert the angle into radian measure via (13.1): i i %%J j radians. By Fact 13.3.2, the arc has length ` feet !%o inches. ¤ to ¥ , since the angle This is approximately equal to the distance fromreplacements PSfrag is small. axis
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. As the size of the angle decreases, the accuracy of the arc length approximation will improve.
¢
axis axis
à chord
£
Important Fact 13.5.5 (Chord Approximation). In Figure 13.14, if the central angle is small, then 9l ¤¥ . Figure
13.14: Chord proximation.
ap
CHAPTER 13. MEASURING AN ANGLE
180
13.6 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 yjA1&o^% miles. We could slice the earth with a twodimensional plane £O which is both perpendicular PSfrag replacementsto a line connecting the North and South poles and passes through the axiscenter of the earth. Of course, the resulting intersection will trace out a axiscircle of radius ¢WjA1&o^% miles on the surface of the earth, which we call axisthe equator. We call the plane £O the equatorial plane. Slicing the earth with any other plane £ parallel to £7O , we can consider the right triangle pictured below and the angle :
ç
line of latitude North Pole
North Pole
equatorial plane
Ö
ç
equator equator center of earth South Pole
center of earth
äO æ
South Pole
Figure 13.15: Measuring latitude.
Essentially two cases arise, depending on whether or not the plane £ is above or below the equatorial plane. The plane £ 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 North line of latitude (denoted N) or the South line of latitude (denoted S). Notice, the radius Q of a line of latitude can vary from a maximum of jA1&o^% miles (in the case of Z % ), to a minimum of 0 miles, (when Ho% ). When QFH% , we are at the North or South poles on the earth. In a similar spirit, we could imagine slicing the earth with a plane ¤ 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 jA1ko^% 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.
PSfrag replacements axis axis
13.6. GREAT CIRCLE NAVIGATION
axis
181
Longitudes are measured using angles East or West of Greenwich. Pictured below, the longitude of is . Because is east of Greenwich, measures longitude East, typically written E; west longitudes would be denoted as W. All longitudes are between % and eJ% . The meridian which is eJ% West (and eJ% 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 Greenwich International meridian Date Line by providing the longitude and latitude of the point. The + usual convention is to list longitude first. For example, i Equator Seattle has coordinates 0!!A%:jjj W, &^ 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 13.16: The Internahere. This means that Seattle is on the line of longitude tional Date Line. + 0!!A%:jjj West i of+ the Greenwich meridian and on the line of latitude ^ North of the equator. In the figure below, we indicate the i &^ and C0!!A%:jjj by inserting the three indicated key angles ¥t radial line segments. eplacements axis axis
not great circle
§
axis
great circles
Seattle, WA
¦
¢
Greenwich Meridian not great circle
Figure 13.17: Distances along great circles.
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.
CHAPTER 13. MEASURING AN ANGLE
182
Important Fact 13.6.1 (Great Circles). The shortest distance between two points on the earth is measured along a great circle connecting them. Example 13.6.2. What is the shortest distance from the North Pole to Seattle, WA ?
ements axis axis axis
¨ © ¢ ª ¸
equator
Greenwich Meridian
Figure 13.18: Distance between the North Pole and Seattle, Washington.
Solution. The line of longitude 0!!A%:jjj W is a great circle connecting the North Pole and Seattle. So, the shortest distance will be the arc length subtended by the angle + in Figure 13.18. Since the + latitude of «¬[email protected] pictured i i + + Seattle is &^ , the angle ¡[email protected] has measure ^ . Since V« is a right angle (i.e., o% ), ¡«®[email protected] has measure !A . By Fact 13.4.1 and Equation 13.1,+ + +
6 jo+ ^% miles 9 6 !A 9 6 %&% i !o jT miles 1
i
` radians/degree9
which is the shortest distance from the pole to Seattle.
13.7. EXERCISES
183
13.7 Exercises Problem 13.1. Let sure ¯ .
NBMC
be an angle of mea
(a) Convert ¯'¡t¹ ± into degrees/ minutes/ seconds and into radians.
(b) Convert dians.
¯3¡Z ± c
into degrees and ra
(c) Convert ¯ï¡½ radian into degrees and degrees/ minutes/ seconds. Problem 13.2. A nautical mile is a unit of distance frequently used in ocean navigation. It is defined as the length of an arc : along a great circle on the earth when the subtending angle has measure ¡ “one minute”. 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 13.3. The rear window wiper blade on a station wagon has a length of 16 inches. The wiper blade is mounted on a inch arm, ª inches from the pivot point.
6" 16"
(c) Suppose bug B lands on the end of the blade farthest from the pivot. Assume the wiper turns through an angle of R Ã . In one cycle (back and forth) of the wiper blade, how far has the bug traveled? (d) Suppose bug C lands on the end of the wiper blade closest to the pivot. Assume the wiper turns through an angle of R0Ã . In one cycle of the wiper blade, how far has the bug traveled? (e) Suppose bug D lands on an intermediate location of the wiper blade. Assume the wiper turns through an angle of R¥Ã . If bug D travels 28 inches after one cycle of the wiper blade, determine the location of bug D on the wiper blade. Problem 13.4. 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 13.5. Astronomical measurements are often made by computing the small angle formed by the extremities of a distant object and using the estimating technique in 13.5.1. In the picture below,± the full moon is shown to form an angle of ¨ when the distance indi Estimate the diameter cated is 248,000 miles. of the moon. moon
replacements axis axis
248,000 miles
axis
(a) If the wiper turns through an angle of R¥Ã , how much area is swept clean? PSfrag replacements (b) Through how much of an angle would the wiper sweep if the area cleaned was 10 square inches?
o 1/2
axis axis axis
earth
CHAPTER 13. MEASURING AN ANGLE
184 Problem 13.6. 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 13.8. 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 ¯ is less than 90 degrees, what is the value of the radius, Á ?
r
satellite plane
PSfrag replacements
θ
axis axis
earth
r
axis
PSfrag replacements axis axis axis
(a) When is the plane directly over a loca tion with latitude eÃß N for the first time?
Problem 13.9. Let D ¦ be the circle of radius 6 inches centered at the origin in the ¿0Û coordinate 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
(b) How fast is the satellite moving?
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? C
C
6
Problem 13.7. Find the area of the sector of a circle of radius 11 inches if the measure ¯ of a central angle of this sector is: (a) (b) (c) (d) (e) (f)
6 y=x
¯3¡IÃ
¯3¡ ± ¯3¡
radians
¯3¡ v ª
radians
¯3¡ ± ¯3¡z ± z &
PSfrag replacements axis axis axis
y=−(1/4)x + 2 C
6
Chapter 14
PSfrag replacements axis axis
Measuring Circular Motion
axis
If Cosmo begins at location ¤ 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 ã¤Ø¥ , this is a “static question”, which we answered in the previous section.
Cosmo moves counterclockwise, maintaining a tight tether.
¸
®
Figure 14.1: How fast Cosmo moving?
14.1 Different ways to measure Cosmo’s speed If Cosmo starts at location ¤ and arrives at location ¥ after some amount of time, we could study
°
measure ¤Ø¥ time required to go from ¤ to ¥
The funny Greek letter “ ° ” on the left of side of the equation is pronounced “ohmega”. We will refer to this as an angular speed. Typical { ”, etc. For example, if the angle ”, “ degrees ”, “ x radians units are “ degrees minute second minute + swept out by Cosmo after 8 seconds is % , then Cosmo’s angular speed u { . Using (13.1), we can convert to radian units and get is x 8 seconds p O { bx sec ° rad rad . This is a new example of a rate and we can ask to u ( min sec l l find the total change, in the spirit of (1.2). If we are given ° in units of deg rad { ”, we have “ x time ” or “ time
I ° /1
which computes the measure of the angle swept out after time / (i.e. the total change in the angle). Angular speed places emphasis upon the 185
· 20 feet
is
CHAPTER 14. MEASURING CIRCULAR MOTION
186
“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 ° H rad means he sec rad is moving clockwise 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 ý 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 ¤ØËè or rev/min. For example, if Cosmo completes one trip around the circle every 2 minutes, then Cosmo is moving at a rate of ¤ØËè . If instead, Cosmo completes one trip around the circle every 12 seconds, then we could first express Cosmo’s speed in units of revolutions/second as rev/second, then convert to ¤ØËè units:
g
rev sec ` RPM e^% 0 ! sec h min
As a variation, if we measure that Cosmo completed l of a revolution in ¾ 2 minutes, then Cosmo’s angular speed is computed by
l rev ¾
! min
j+
RPM
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 14.1, the circumference of the circle of motion is !( 6 !%9> %2( 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 RPM, then he has a linear speed of
g
äÇ
1 rev 2 min h
g +
%2( ft
rev h
"!%2(
Likewise, if Cosmo is moving
äÇ
g
( rad i
sec h
g
rev ! ( rad h
g +
¾
ft min
rad , sec
%2( ft
rev h
then
!%2( i
ft sec
14.2. DIFFERENT WAYS TO MEASURE
CIRCULAR MOTION
187
Important Fact 14.1.1. 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
v
g
ft min h
14.2 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 ° to denote angular speed and ä for linear speed. If an object is moving around a circle of radius at a constant rate, then we can measure it’s speed in two ways:
The angular speed
°
“revolutions” “degrees swept” “radians swept” or or “unit time” “unit time” “unit time”
The linear speed äÇ
“distance traveled” “per unit time”
Important Facts 14.2.1 (Measuring and converting). We can convert between angular and linear speeds using these facts:
1 revolution = 360 = !( radians;
The circumference of a circle of radius units is !(@ units.
14.2.1 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.
axis axis
CHAPTER 14. MEASURING CIRCULAR MOTION
188
STATIC QUANTITIES
Ö
DYNAMIC QUANTITIES
¢
1. arc length
¢
Þ
5. angular speed
£
2. angle swept in time
ò
3. radius
Ö
4. elapsed time
...take a “snapshot” after time ...
£
6. linear speed
²
³
Ü ...see what happens per unit time...
Figure 14.2: Measuring linear and angular speed.
We now know two general relationships for circular motion:
" , where =arclength (a linear distance), =radius of the circular (i) path and =angle swept in RADIAN measure; this was the content of Fact 13.4.1 on page 175.
(ii) I ° / , where is the measure of an angle swept, ° = angular speed and / represents time elapsed. This is really just a consequence of units manipulation. Notice how the units work in these formulas. If =20 feet and cj radians, then the arc length Íy!% 6 kjT9 feet= 26 feet; this is the length of the arc of radius 20 feet that is subtending the angle . If ° j rad rad/second and /; ` seconds, then I)j seconds , ` seconds $ ` radians. ° If we replace “ ” in w of (i) with / in (ii), then we get
" ° /:
ò
This gives us a relationship between arclength (a distance) and time / . Plug in the fact that the linear speed is defined to be äÍ “distance” and we get
° / äÇ " ° / /
All of these observations are summarized below. Important Facts 14.2.2 (Three really useful formulas). If we measure angles in RADIANS and ° in units of radians per unit time, we have these three formulas:
ä
° / °
(14.1) (14.2) (14.3)
14.2. DIFFERENT WAYS TO MEASURE
CIRCULAR MOTION
189
PSfrag replacements + bike Example 14.2.3. You are riding a stationary exercise axis and the speedometer reads a steady speed of % MPH axis (miles per hour). If the rear wheel is 28 inches in diameaxis ter, determine the angular speed of a location on the rear * tire. A pebble becomes stuck to the tread of the rear tire. pebble sticks to tread here Describe the location of the pebble after 1 second and 0.1 second. Figure 14.3: Where is the pebble after
Solution. The tires will be rotating in+ a counterclockwise direction and the radius  !JU inches. The other given quantity, “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
g +
ä
g
miles % + hr h i in % sec
Now, the angular speed
°
i
+%
+
! 6 9(
°
in 0 ! ft h
g
1 hr 60 min h
g
min 60 sec h
of the wheel will be
inches second inches revolution
revolution second J% RPM
+J
g
ft ` !J% mile h
It is then an easy matter to convert this to
g
°
revolution J second h degrees !1&JJ% second
g
j^%
degrees revolution h
If the pebble begins at the “6 o’clock” position (the place the tire touches the ground on the wheel), then after second the pebble will go through J revolutions, so will be in the “ ^ o’clock” position again. After %a seconds, the pebble will go through
J
rev 6 %a sec 9 sec
%&J rev
6 %J rev 9 !JJ
g
j^%
deg rev h
Ä
seconds?
CHAPTER 14. MEASURING CIRCULAR MOTION
190
Keeping in mind that the rotation is counterclockwise, we can view the location of the pebble after 0.1 seconds as pictured below:
PSfrag replacements
counterclockç¥ç wise rotation
axis axis axis after 0.1 second
í
*
located here at time sec
O
*
starts here
pebble sticks to tread in 6 o’clock position
Figure 14.4: Computing the pebble’s position after
Ä@¡¹&
sec.
We solved the previous problem using the “unit conversion method”. There is an alternate approach available, which uses one of the formulas + you could proceed: First, as above, we know in Fact 14.2.2. Here is how i the linear speed is äÇ % in/sec. Using the “äÇ ° ” formula, we have
i %
+
° 6
in sec
°
+
in 9 rad
` %c!J sec
Notice how the units worked out in the calculation: the “time” unit comes from ä and the “angular” unit will always be radians. As a comparison with the solution above, we can convert ° into RPM units:
°
g
rad ` %_!J sec h rev J sec
g
rev ! ( rad h
All of the problems in this section can be worked using either the “unit conversion method” or the “ä ° method”.
14.3 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
14.3. MUSIC LISTENING TECHNOLOGY
191
“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 jjÌ RPM. An arm on a pivot (called the tone arm) has a needle mounted l 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 sidetoside, 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!
LP turning at
l¥l  õ
amp
RPM
eplacements
speakers
axis axis axis
tonearm
needle
Figure 14.5: Reproducing music using analogue technology.
This is known as analogue technology and is based upon the idea of maintaining a constant angular speed of jj + RPM for the storage medium i l (our þ´Ø ). (Older analogue technologies used ` RPM and J RPM records. However, jjm RPM became the consumer standard for stereo music.) l With an LP, the beginning of the record (the leadin 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 leadin groove, the needle gradually works its way to the exit groove. However, whereas the angular speed of the LP is a constant jj> RPM, the linear l speed at the needle can vary quite a bit, depending on the needle location.
192
CHAPTER 14. MEASURING CIRCULAR MOTION
1
Example 14.3.1 (Analogue LP’s). The “leadin 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 “leadin 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.
ements axis axis axis 6
Figure
14.6: Leadin exit grooves.
and
Solution. This is a straightforward application of Fact 14.2.1. Let ä (resp. ä ) 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,
ä
g
jj
rev
j min h
g
! 6 ^9 (
inches rev h
i in 0! ` i min in { min { x 0! ` min x ^% hour ft { x ` !J% mile x 0! inft { aeo MPH Similarly, ä y%_! MPH. To answer the remaining question, let be the radial distance from the center of the þ´Ø to the needle location on the record. If ä $ MPH:
mile hour
ä
g
g
g
g
g
in min ft mile jj !e( ^% j min h + 0! in h ` !J% ft h rev h hr h So, when the needle is < ` &% inches from the center, the linear speed is 1 MPH.
ements laser
axis
laser support arm moves back and forth
axis axis
spinning CD
Figure 14.7: Reproducing music using digital technology.
rev
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.
14.4. BELT AND WHEEL PROBLEMS
193
The technology of <íñ differs from that of þ´Ø¿í& 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 y!1&J:j ` minute speed of c! meters . Secondly, the beginning location of the sec PSfrag replacements 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 axis of the CD is changing, as the laser position changes. axis axis
Example 14.3.2 (Digital CD’s). What is the angular speed (in RPM) of a CD if the laser is at the beginning, located l inches from the center of the disc ? What is the p speed (in RPM) of a CD if the laser is at the end, loangular cated ! 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 14.2.1. Let ° l §p be the angular speed at the start and ° the angular speed at the end of the CD ; the subscript is keeping track of the laser distance from the CD center. ° 6 !J:j ` inches/min 9 !! ` ^ RPM 6 ! 6 !A9 ( inches/rev 9
ö ´´
RÔÓ ´ ´
Start of CD
Figure 14.8: Computing the angular speed of a CD.
!J:j ` inches/min 9 i '^%^ RPM 6 ! 6 % ` 9 ( inches/rev9 To answer the remaining question, let be the radial distance from
°
l §p
the center of the CD to the laser location on the CD. If the angular speed ° at this location is j ` % RPM, we have the equation
°
j ` % RPM c!Jo inches
End of CD
inches { x !1kJ:j ` minute inches { x !( revolution n
So, when the laser is 1.289 inches from the center, the CD is moving j ` %Ï¤ØËè .
14.4 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.
axis
CHAPTER 14. MEASURING CIRCULAR MOTION
194
¼ µ·¶ O
¸
¹
º
»
Figure 14.9: Two typical connected wheel scenarios.
í
ments axis axis
front sprocket radius
axis
ments
5 inches
í rear sprocket radius
2 inches
axis axis axis
(a) A stationary exercise bike. radius
¯
14 inches
¯ radius
5 inches
radius 2 inches
¯
(b) A model of the bike’s connected wheels. Figure 14.10: Visualizing the connected wheels of an exercise bike.
Example 14.4.1. 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 (which is 40 MPH) and we need to find the angular speed of wheel =front sprocket. Denote by ä , ä , and ä O the linear speeds of each of the wheels , , and , respectively. Likewise, let ° , ° , and ° O denote the angular speeds of each of the wheels ,
, and , respectively. In addition, the chain connecting the wheels and will have a linear speed, which we will denote by ä Ã é¾ . The strategy is broken into a sequence of steps which leads us from the known linear speed ä= to the angular speed ° O of wheel :
Step 1: Given ä , find ° . Use the fact ° #¾½ . Step 2: Observe ° ü ° ; this is because the wheel and rear sprocket are both rigidly mounted on a common axis of rotation.
Step 3: Given ° , find ä . Use the fact äD"e ° D"e ° # ä . Step 4: Observe ä´É ä Ã oé ¾ ä O ; this is because the chain is directly connecting the two sprockets and assumed not to slip.
Step 5: Given ä O , find ° O
. Use the fact
° O
½ ½ ä>
14.4. BELT AND WHEEL PROBLEMS
195
Saying that the speedometer reads 40 MPH is the same as+ saying % MPH. that the linear speed of a location on the rear wheel is ä= + out in our solution to Converting this into angular speed was carried J% RPM. This completes Example 14.2.3 above; we found that+ ° y + and so by Step 2, ° ] ° J% RPM. For Step 3, we convert Step 1 ° É J% RPM into linear speed following Fact 14.2.1:
g +
g
revolution inches ä J% ! 6 !A9 (49 6 minute h revolution h inches ^´1k%:j2! minute By Step 4, conclude that the linear speed of wheel is ä O H^´1&%:j2! inches/min. Finally, to carry out Step 5, we convert the linear speed into angular speed:
°
O
{ x§^´1k%:j2! inches min inches { x ! 6d` 9 ( rev eov! RPM rev jTc!
sec In conclusion, the bike rider must pedal the front sprocket at the rate of rev jTc! sec . This example indicates the basic strategy used in all belt/wheel problems. Important Facts 14.4.2 (Belt and Wheel Strategy). Three basic facts are used in all such problems:
Using “unit conversion” or Fact 14.2.2 allows us to go from linear speed ä to angular speed
°
, 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.
196
14.5 Exercises
CHAPTER 14. MEASURING CIRCULAR MOTION
Problem 14.1. An irrigation drip arm is feet long and pivots about a center one revolution each hour. A “drip hole” ¬ is located feet from the pivot point. (a) What is the linear speed of ¬ in units of feet feet inches feet hour , minute , second , and second . drips (b) What is the required drip rate (in second ) if we want at least 4 drips for every 1 inch ¬ travels. (The drip rate must be a counting number; i.e. R v v : )
(c) Suppose drips ¡. cc (cubic centimeter). How many liters flow from ¬ after minutes, assuming the drip rate computed in b.? ( liter ¡FR RR cc)
(d) How much water flows from ¬ after the arm has made complete revolutions (assuming the drip rate computed in b.)? Problem 14.2. 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) You are standing on the equator of the earth (radius 3960 miles). What is your linear and angular speed? (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?
Problem 14.5. 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?
(c) How far does a person sitting by the window move in 100 minutes if the radius of the restaurant is 21 meters?
(d) Find the area of the pie shaped sector enclosed by Lee’s path.
Problem 14.3. You are riding a bicycle along a level road. Assume each wheel is 26 inches in diameter, the rear sprocket has radius 3 inches and the front sprocket has radius 7 inches. How many revolutions per minute of the front sprocket produces a speed of 35 mph?
Problem 14.6. A laser video disc (LVD), was a predecessor to the DVD. Similar to a DVD, an LVD is a thin plastic disc with a silver coating on one side, but they have a diameter of 12 inches. Information is read from the disc using E is designed to spin at a cona laser. The Àßæ stant linear speed below the laser; this speed is R meters/sec ¡ 2835 inches/minute. The laser begins « inch from the center of the spinning disc andº works it way out to the end (5.5 inches from the center).
Problem 14.4. 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 ¿ in rad/sec and the linear speed ² in ft/sec? (b) A wheel of radius 8 in. is rotating ± /sec. What is the linear speed ² , the angular speed in RPM and the angular speed in rad/sec?
E
(a) Find the angular speed of the Àßæ at the E beginning and end of the À æ ; i.e. when the laser is « inch and : inches from the center. º (b) Describe the location of the laser if the angular speed is 100 RPM.
14.5. EXERCISES
197
Problem 14.7. 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 § 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.
(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 ç as pictured. What is the angle ùÁÖç in degrees? What is the angle ùÁÖç in radians? How many feet has Michael traveled?
12 RPM θ P
replacements axis axis
S
θ
4 feet
axis
P
O
(a) Find the radius of the ferris wheel.
(b) Once the wheel is built, John suggests PSfrag replacements that Tiff should take the first ride. The axis wheel starts turning when Tiff is at the axis location ù , which makes an angle ¯ with axis the horizontal, as pictured. It takes her 1.3 seconds to reach the top of the ride. (e) Assume Michael has traveled 88 feet Find the angle ¯ . from the position P to a new position Q. (c) Poor engineering causes Tiff’s seat to fly How many seconds will this take? What off in 6 seconds. Describe where Tiff is will be the angle swept out by Michael? located (an angle description) the instant she becomes a human missile. length of arc (PQ) is 88 ft
Problem 14.8. Michael and Aaron are on the “ULTossum” ride at Funworld. This is a merrygoround of radius 20 feet which spins counterclockwise 60 RPM. The ride is driven by a belt connecting the outer edge PSfrag of the ride replacements to a drive wheel of radius 3 feet:
axis axis
Aaron
axis
Drive wheel radius 3 ft
O
P
Q
Michael O
replacements axis axis
drive belt
P main ride radius 20 ft
axis
(a) Assume Michael is seated on the edge of the ride, as pictured. What is Michael’s linear speed in mph and ft/sec?
Problem 14.9. You are riding a bicycle along a level road. Assume each wheel is inches in diameter, the rear sprocket has radius inches and the front sprocket has radius Á inches. Suppose you are pedaling the front sprocket rev and your forward speed is at the rate of R sec R mph on the bike. What is the radius of the front sprocket?
CHAPTER 14. MEASURING CIRCULAR MOTION
198
Problem 14.10. Your car’s speedometer is geared to accurately give your speed using a certain size tire. Suppose your car has ,P¡ 14 inch diameter wheels and the height of the tire ¾ is ¡ 4.5 inches.
d PSfrag replacements axis axis axis
h tire wheel
¾
(b) You are furious over the speeding ticket and return to the tire dealer, demanding new tires which are the correct size. The dealer only has “low profile tires” in ¾ ¡.¹ inches high. stock, which are If you accept these and drive away from the dealer with your speedometer reading 35 mph, how fast are you really going? Problem 14.11. 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 E D and are rigidly fastened to E the same axle. What is the radius Á of wheel ?
(a) You buy a new set of tires with ¡ : inches and ,¡' inches. On a trip to Spokane, you maintain a constant speed PSfrag replacements of 65 mph, according to your speedometer. However, as luck would have it, you axis are stopped for speeding. Explain how axis this could happen. What did the radar axis gun display as your true speed?
A
D C
B
Chapter 15 The Circular Functions PSfrag replacements axis axis
Â Suppose Cosmo begins at location ¤ and walks in a coun axis Â terclockwise 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: Â Z Ã feet The measure of ã¤Ø¥ exactly pins down a location on x the circle of radius 20 feet. But, we really might prefer Â PSfrag replacements a description of the horizontal and vertical coordinates of ] Cosmo; this would tie in better with the coordinate system axis we typically use. Solving this problem will require NEW axis Figure 15.1: Cosmo moves functions, called the circular functions. axiscounterclockwise maintaining a tight tether. Cosmo?
Where’s
15.1 Sides and Angles of a Right Triangle Example 15.1.1. You are preparing to make your final shot at the British Pocket Billiard World Championships. The position of your ball is as in Figure 15.2, and you must play the ball off the left cushion into the lowerright corner pocket, as indicated by the dotted path. For the big money, where should you aim to hit the cushion?
The billiard table layout.
Ö
find this location
The billiard table layout
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 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 common sides are equal: æ p If we solve this equation ui S S for , we obtain ÇÏ u )jT ` feet.
p
199
ft
²T Á
Ö
Á
²
this pocket for the big money
ft
õö
³
ft
ft
Mathmatically modeling the bank shot.
Figure 15.2: A pocket billiard banking problem.
CHAPTER 15. THE CIRCULAR FUNCTIONS
200
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 ? 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.
ements axis axis axis
15.2 The Trigonometric Ratios
hypotenuse
From elementary geometry, the sum of the angles of any triangle will equal eJ% . Given a right triangle áÍ I , since one of the angles is o% , the remaining two angles must be acute angles ; i.e., angles of measure between % and o% . If we specify one of the acute angles in a right triangle áÍ D
, say angle , we can label the three sides using this terminology. We then consider the following three ratios of side lengths, referred to as trigonometric ratios :
side opposite O
side adjacent
Figure 15.3: Labeling the sides of a right triangle.
length of side opposite length of hypotenuse
length of side adjacent length of hypotenuse
length of side opposite length of side adjacent to
9 sin 6 A
def
9 cos 6 A
def
9 tan 6 A
def
(15.1)
(15.2)
(15.3)
For example, we have three right triangles in Figure 15.4; you can verify that the Pythagorean Theorem holds in each of the cases. In the lefthand triangle, sin 6 A9I u , cos 6 A9I , tan 6 A9I u . In the middle l l . In the righthand triangle, sin 6 A9 , cos 6 A93 , tan 6 A93y triangle,
sin 6 A93 , cos 6 A9 l , tan 6 A93 . The symbols “sin”, “cos”, and “tan” l are abbreviations for the words sine, cosine and / 'ÅÄ 6´\ Ä / , 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 . This follows from the properties of similar triangles. Consider Figure 15.5. Notice áÍ D
and áÍ @Æ O are similar. If we use áÍ D
to compute cos 6 A9 , then we find cos 6 A9Ë A . On the other
µ ¶ . Since the ratios of O mon sides of similar triangles must agree, we have cos 6 A9< A
hand, if we use Í á §Ç , we obtain cos 6 A 9>
com
µ¶ ,
axis axis axis
15.2. THE TRIGONOMETRIC RATIOS
201
l
u
l
Figure 15.4: Computing trigonometric ratios for selected right triangles.
PSfrag replacements axis
which is what we wanted to be true. The same argumentaxis can be used to show that sin 6 A9 and tan 6 A9 can be computed using anyaxis right triangle with acute angle . µ Except for some “rigged” right triangles, it is not easy to calculate the trigonometric ratios. Before the 1970’s, approximate values of sin 6 A9 , cos 6 A9 , tan 6 A9 were listed in long tables or calculated using a slide rule. Today, a scientific calculator saves the day on these computations. Most scientific calculators will give an approximation for O the values of the trigonometric ratios. However, it is good ¶ to keep in mind we can compute the EXACT values of the Figure 15.5: Applying trigonometric ratios to any trigonometric ratios when ] + % , , , , radians or, l p right triangle. equivalently, when IH% , j% , ` , ^% , o% . Angle
È
Deg
Rad
%
%
+
`
o%
È
cos
%
j%
^%
sin
Trigonometric Ratio
p
È
%
l
%
tan
l
l
È
l
) j Undefined
Table 15.1: Exact Trigonometric Ratios
CHAPTER 15. THE CIRCULAR FUNCTIONS
202
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:
ements axis
!!!
axis
When computing values of cos È , sin È , and tan È on your calculator, make sure you are using the correct “angle mode” when entering È ; i.e. “degrees” or “radians”.
CAUTION !!!
axis
i
+ % &% ` , and For example, if W , then cos 6 9? %oooJ , sin 6 9Í i tan 6 9# %+ % ` . In contrast, if+ ÿ radians, then cos 6 # 9 % ` :% j , i sin 6 9ÒH%J ` , and tan 6 98$ `` .
ements axis axis axis
15.3 Applications
cos
sin
{
{
ì
tan
{
ì
Figure 15.6: What do these ratios mean?
When confronted with a situation involving a right triangle where the measure of one acute angle 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: Important Facts 15.3.1 (Trigonometric ratios). Given a right triangle, the trigonometric ratios relate the lengths of the sides as shown in Figure 15.6.
Example 15.3.2. To measure the distance across a river for a new bridge, surveyors placed poles at locations , and . The length D e%% feet and the measure of the angle 7 I
is jT eJ í . Find the distance to span the river. If the measurement of the angle D
is only accurate within ¦w!í , find the possible error in .
ements axis axis axis
Solution. The trigonometric ratio relating these two sides would be the tangent and we can convert into decimal form, arriving at:
O
O¥O
ø A
l çÉ Figure 15.7: The distance spanning a river.
tan 6 jT eJ í 9Ò tan 6 jTkj 9Ê therefore
Ò
e%%
Ç'^%J 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 ;! í , then tan 6 jT !% í»9U tan 6 jTcjjjj 9 %^%JJ and the span is 60.88 ft. On the other hand, if the measurement of the angle was in error by ;!Îí , then i i tan 6 jT e^ í»9Ò tan 6 jT_!^^ 9ÊH%^% ! and the span is 60.72 ft.
PSfrag replacements axis axis
15.3. APPLICATIONS
203
axis
Example 15.3.3. A plane is flying 2000 feet above sea level toward a mountain. The pilot observes the top of the mountain to be eJ above the horizontal, then immediately flies the plane at an angle of !% 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?
æ
Ö
O¥O¥O
Í¶
ft
ÊËÊËÌËÊÌËÊ ÌËÊÌËÊ ÌÊÌÊ
¢
sealevel
Figure 15.8: Flying toward a mountain.
Solution. We can compute the hypotenuse of áCþ´Ø by using the speed and time information about the plane:
! ` miles j
Ø 6 e%% mph 9 6a` minutes 9 6 hourf^% minutes 9Ò
The definitions of the trigonometric ratios show:
! ` sin 6 !% 9Ê"!A&J ` % miles, and j ! ` i Ø þ &J:jT miles cos 6 !% 9Ò j
þ
With this data, we can now find
þ
Ø þ
tan 6 eJ 9Ò"!A `
++
þ
: PSfrag replacements axis
miles
axis
++
axis
The height of the plane above the peak is Î þ þ "!AJ ` %½!A ` %kj%^ miles n1k^e^ feet. The elevation of the peak level is above sea + + + given by: Peak elevation plane altitude þ ¥´Ø þ 5!1&%%% 6 !A ` 9 6d` 1_!J%9[ ` 1 j2! feet.
Example 15.3.4. A Forest Service helicopter needs to de+ termine the width of a deep canyon. While hovering, they J at position (see picture), then measure the angle Ô) descend 400 feet to position and make two measurements of Õ ÿj (the measure of § ), ÖZ ` j (the mea sure of § ). Determine the width of the canyon to the nearest foot.
p O¥O O
ft
Ñ
Ò Ó ¶
ÏËÐËÏ ÐÏ
canyon
Figure 15.9: Finding width of a canyon.
Solution. We will need to exploit three right triangles in the picture: , , and . Our goal is to compute áZ
áÍ áÍ , which suggests more than one right triangle will come into play.
µ the
CHAPTER 15. THE CIRCULAR FUNCTIONS
204
The first step is to use áZ and áÍ 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,
6 % %; 9 tan 6 J 9 tan 6a` j 9¼
Plugging the second equation into the first and rearranging we get
+
+
%% tan 6 J 9 + !1&% ` j feet tan 6a` j 9T tan 6 J 9
Plugging this back into the second equation of the system gives
i
+
6 !% ` jT9 tan 6a` j 9Ê"! ! feet
The next step is to relate áÍ and áÍ , which can now be done + in an effective way using the calculations above. Notice that the measure % . We have of $ is Ö ×Õ#
+
+
i 9 [n1 !nj feet 6 !% ` jT9 tan 6 % Ê + i i W!1 ! W n1 !njyn1&%% feet is the width As noted above,
tan 6 % 9Ò
of the canyon.
ements
15.4 Circular Functions íFî
axis axis axis
¢
O
S ëÚ ð Ú
Ö S
£
If Cosmo is located somewhere in the first quadrant of Figure 15.1, represented by the location ¥ , we can use the trigonometric ratios to describe his coordinates. Impose the indicated 2M coordinate system with origin at Ø and extract the pictured right triangle with vertices at Ø and ¥ . The radius is 20 ft. and applying Fact 15.3.1 gives
¥ Figure 15.10: Cosmo on a circular path.
6 @1§M39Ò 6 !% cos 6 A9z1!% sin 6 A99
Unfortunately, we run into a snag if we allow Cosmo to wander into the second, third or fourth quadrant, since then the angle is no longer acute.
15.4. CIRCULAR FUNCTIONS
PSfrag replacements
205
axis axis 15.4.1 Are the trigonometric ratios functions?
í
axis
Recall that sin 6 A9 , cos 6 A9 , and tan 6 A9 are defined for acute Ö A unit circle with angles inside a right triangle. We would like to say that radius these three equations actually define functions where the . variable is an angle . Having said this, it is natural to ask if these three equations can be extended to be defined for £ ANY angle . For example, we need to explain how sin x { l is defined. To start, we begin with the unit circle pictured in the M coordinate system. Let ¤Ø¢Ø be the angle in standard central position shown in Figure 15.11. If is posFigure 15.11: Coordinates of points on the unit circle. itive (resp. negative), we adopt the convention that is swept out by counterclockwise (resp. clockwise) rotation of the initial side @¤ . The objective is to find the coordinates of the point Ø in this figure. Notice that each coordinate of Ø (the coordinate and the M coordinate) will depend on the given anglereplacements . For this reason, we PSfrag need to introduce two new functions involving the variable . axis
Definition 15.4.1. Let be an angle in standard central position inside the unit circle, as in Figure 15.11. This angle determines a point Ø on the unit circle. Define two new functions, cos 6 A9 and sin 6 A9 , on the domain of all values as follows: cos 6 A 9
9 sin 6 A
def
horizontal coordinate of Ø on unit circle
vertical M coordinate of Ø on unit circle
def
axis axis
í
ÛËÛ
ÛËÜËÜÛ ÛÜÜÛ ËÚÙËÚÙ ÚËÙÚËÙ ÚËÙÚËÙ ÚÙÚÙ kilometer Ë ÜÜ
O O u ÞßËì ;Þ àÝ Michael starts here
Figure 15.12: A circular We refer to sin 6 A9 and cos 6 A9 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 PSfrag replacements will be on your calculator. Again, BE CAREFUL to check the angle mode axis setting on your calculator (“degrees” or “radians”) before doing a calculaaxis tion. axis
Example 15.4.2. Michael is test driving a vehicle counterclockwise around a desert test track which is circular of radius kilometer. He starts at the location pictured, traveling %%v! ` rad . Impose coordinates as pictured. Where is sec Michael located (in M coordinates) after eJ seconds? Solution. Let è 6 /:9 be the point on the circle of motion representing Michael’s location after / seconds and 6 /29 the angle swept out the by Michael after / seconds. Since we are given the angular speed, we get
6 /:98H%%v! ` /
radians
Ú
ìPî_ò íPî î_ò axis
î_ò
î_ò ð
S ðåë Ú ðdð
O O u ÞßËì ;Þ àÝ ð S axis Michael starts here
Figure 15.13: Modeling Michael’s location.
CHAPTER 15. THE CIRCULAR FUNCTIONS
206
Since the angle 6 /:9 is in central standard position, we get
è 6 /29Ò 6 cos 6 6 /:99z1 sin 6 6 /:9998 6 cos 6 %&%v! ` /29z1 sin 6 %%v! ` /:99 So, after 18 seconds Michael’s location will be è
ments ements axis axis axis
CAUTION !!!
starts here ments
~
axis axis axis
Angela starts here
¯
axis
õ
² Y&ä áãå;âæ ô ö Á axis
kilometer
a  Y&ä áãå;âæ ô
Michael starts here
(a) Angela and Michael on the same test track.
~ Angela starts here
è a  Yä&áãå;âæ ô
+
6 eJ9Ò 6 %&o%% 1% j ` %9 .
Interpreting the coordinates of the point L cos È 0¡ sin È * in Figure 15.11 only works if the angle È 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.
!!!
axis axis axis
+
ç
axis
(t)
(t)
M(t) (t)
² Y&ä áå&âæ ô ö Á axis
Michael starts here
Example 15.4.3. 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 15.14(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 2M coordinates) after 18 seconds? Solution. Let è 6 /29 be the point on the circle of motion representing Michael’s location after / seconds. Likewise, let 6 /29 be the point on the circle of motion representing Angela’s location after / seconds. Let 6 /29 be the angle swept out the by Michael and Õ 6 /:9 the angle swept out by Angela after / seconds. Since we are given the angular speeds, we get
6 /:9b
%%v! ` / radians, and %%:j/ radians
Õ 6 /:9b
A(t)
From the previous Example 15.4.2,
(b) Modeling the motion of Angela and Michael. Figure 15.14: Visualizing motion on a circular track.
è 6 /:9 è 6 eJ9
6 cos 6 %+ &%v! ` + /291 sin 6 %&%v! ` /2991 and 6 %o%% 1% j ` %9 Angela’s angle Õ 6 /29 is NOT in central standard position, so we must observe that Õ 6 /:9n½( é Ö 6 /29 , where Ö 6 /29 is in central standard position: See Figure 15.14(b). We conclude that
6 /:9
6 <%&J `
ii
6 cos 6 Ö 6 /2991 sin 6 Ö 6 /:999 6 cos 6 (?%%:j/:9z1 sin 6 (%&%:j/299¼
+ So, after 18 seconds Angela’s location will be 6 eJ9 1z<% ` 9 .
PSfrag replacements
15.4. CIRCULAR FUNCTIONS
207
axis axis
axis 15.4.2 Relating circular functions and right triangles
If the point Ø on the unit circle is located in the first quadrant, then we can compute cos 6 A9 and sin 6 A9 using trigonometric ratios. In general, it’s useful to relate right triangles, the unit circle and the circular functions. To deeplacementsscribe this connection, given we place it in central standard position in the unit circle, where ¤Ø¢ØHê . Draw axis a line through Ø perpendicular to the axis, obtaining axis an inscribed right triangle. Such a right triangle has hyaxis potenuse of length 1, vertical side of length labeled Q and horizontal side of length labeled ' . There are four cases: See Figure 15.16.
O
a
P
P
b
b
ìì ëìë
ëa
R
O
R
a b
í! í í! í íí ! î! î î! î!Oîî
unit circle (radius P
O
CASE II
O
R
ðïð òñò ! ï! ï ñ a ï
bR P
CASE III
Figure 15.16: Possible positions of
¯
CASE IV
on the unit circle.
Case I has already been discussed, arriving at cos 6 A9Ò ' and sin 6 A9Ê Q . In Case II , we can interpret cos 6 A9wy ' 1 sin 6 A9yQ . We can reason
similarly in the other Cases III and IV, using Figure 15.16, and we arrive at this conclusion:
Important Facts 15.4.4 (Circular functions and triangles). View as in Figure 15.16 and form the pictured inscribed right triangles. Then we can interpret cos 6 A9 and sin 6 A9 in terms of these right triangles as follows: Case I Case II Case III Case IV
cos 6 A9Ë '
cos 6 A9ËH
cos 6 A9ËH
cos 6 A9Ë ' 1 1
' 1 ' 1
sin 6 sin 6 sin 6 sin 6
A9Ê'Q A9Ê'Q
A9Ê.wQ A9Ê.wQ
{
{
Figure 15.15: The point the first quadrant.
õ)
R
cos
P CASE I
sin
¯
ù
in
ements axis axis
CHAPTER 15. THE CIRCULAR FUNCTIONS
208
axis
15.5 What About Other Circles?
T
ó
P O O
Figure
R
S
unit circle
15.17: Points other circles.
ements
on
general circles:
axis
What happens if we begin with a circle with radius (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 centered at the origin in the same picture with unit circle and the angle in standard central position for each circle. As pictured, we can view Nô¤¢Ø õK¥öï . If Ø 6 @1 M9 is our point on the unit circle corresponding to the angle , then the calculation below shows how to compute coordinates on
ØF 6 81 M39 6 cos 6 A91 sin 6 A99g÷
axis axis
ø
ø ø ø
M
6 0@9
[
M
6 RM39 6 @1RM39 6 cos 6 A91 sin 6 A99g÷
Important Fact 15.5.1. Let be a circle of radius centered at the origin and ùK¥öï an angle in standard central position for this circle, as in Figure 15.17. Then the coordinates of Z 6 cos 6 A9z1 sin 6 A9¼9 .
~
è
U
B
ç ¯ûú è ¯ ö ô ° Ö õ ³
axis
R T
Q S
P O
A
Á
axis
¯ circle radius circle radius circle radius
¯ õ ¯ ö
Figure 15.18: Coordinates of points on circles.
Examples 15.5.2. Consider the picture below, with IH%&J radians and Õê %_! radians. What are the coordinates of the labeled points? Solution. The angle is in standard central position; Õ is a central angle, + but it is not in standard position. Notice, Ö)Ï(#¾Õ.!A&o e^ is an angle in standard central position which locates the same points êÇ1¼1u¥ as the angle Õ . Applying Definition 15.4.1 on page 205:
ØU Ù ¤ ¥Ç Z ê]
6 cos 6 %&J9z1 sin 6 %&J99 6 ! cos 6 %J91! sin 6 %J99 6 j cos 6 %+ J91j sin 6 %J9+ 9 6 cos 6 !A&o e+ ^9z1 sin 6 !A&o e^+ 99 6 ! cos 6 !Ao + e^91! sin 6 !Ao + e^99 6 j cos 6 !Ao e^91j sin 6 !Ao e^99
+ i i 6 %&^o^ + 1% + + 9 i 6 cjo:j 1 j 9 6 !A&%o%n1!Aa ` !A9 i 6 <%&oJ%n1%aeoJ 9 i 6 É&o+^%v!1%kjo jT9 6 ;!A&o %:jA1% ` o^9 i
15.6. OTHER BASIC CIRCULAR FUNCTION
209
PSfrag replacements Example 15.5.3. Suppose Cosmo begins at the position ¤ axis in the figure, walking around the circle of radius 20 feet axis with an angular speed of p RPM counterclockwise. After 3 axis u minutes have elapsed, describe Cosmo’s precise location.
S
Solution. Cosmo has traveled j p revolutions. If is
u u radians the
angle traveled after 3 minutes, x u rev{ x !( rev { p radians ô
` %J radians. By (15.5.1), we have x !u% cos p rad{ Ée^aeJ feet and M !% sin x p rad { i u Conclude that Cosmo is located u + ++ ^ feet. at the point i ¥Z 6 Ée^aeJ´1 ^9 . Using (13.1), U J^ [! 6 j^% 9W ; ++ this means that Cosmo walks counterclockwise around the circle two complete revolutions, plus .
P
15.6 Other Basic Circular Function
any angle , our constructions offer a concrete link between the and sine functions and right triangles inscribed inside the unit axiscircle: See Figure 15.20. axisGiven
axiscosine
P
R
O
R
O
R
O
R
O P
P CASE I
CASE II
CASE III
CASE IV
Figure 15.20: Computing the slope of a line using the function tan ü¯T/ .
The slope of the hypotenuse of these inscribed triangles is just the slope of the line through ¢Ø . Since ØF 6 cos 6 A9z1 sin 6 A99 and 6 ´ % 1%9 : Slope
sin 6 A9 ý 4M 3 cos 6 A9
ã % . This calculation motivates a this would be valid as long as cos 6 A9?Ï new circular function called the tangent of by the rule tan 6 A 9Ò
sin 6 A9 1 cos 6 A9
provided cos 6 A 9;H ã %
feet
R
Figure 15.19: Where is Cosmo after 3 minutes?
eplacements
P
O
CHAPTER 15. THE CIRCULAR FUNCTIONS
210
The only time cos 6 A9 % is when the corresponding point Ø on the unit circle has coordinate 0. But, this only happens at the positions 6 %´19 and 6 %´1zÉ 9 on the unit circle, corresponding to angles of the form
ê¦ 1[¦ l 1[¦ u 1 EeEeE . These are the cases when the inscribed right triangle would “degenerate” to having zero width and the line segment ¢Ø becomes vertical. In summary, we then have this general idea to keep in mind: Important Fact 15.6.1. The slope of a line tan 6 A9 , where is the angle the line makes with the xaxis (or any other horizontal line) Three other commonly used circular functions come up from time to time. The cotangent function M cot 6 A9 , the secant function M" sec 6 A9 and the cosecant function MN csc 6 A9 are defined by the formulas:
ments axis
9 sec 6 A
def
axis axis
9 cos 6 A
1
9 csc 6 A
def
9 sin 6 A
1
9 cot 6 A
def
9 tan 6 A
Just as with the tangent function, one needs to worry about the values of for which these functions are undefined (due to division by zero). We will not need these functions in this text.
ments axis
North Alaska
Northwest
axis axis
õ§õ
West
²ý
²mþý East
ö
SeaTac South
þý
Delta
(a) The flight paths of three airplanes. Q
Á ¯ m
Alaska Q
Northwest
N
~ ¯
P W
Á ¯ *²m
ö
R
E
Delta
S
(b) Modeling the paths of each flight. Figure 15.21: Visualizing and modeling departing airplanes.
Example 15.6.2. Three airplanes depart SeaTac Airport. A NorthWest flight is heading in a direction ` % counterclockwise from East, an Alaska flight is heading ` counterclockwise from East and a Delta flight is heading !% 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 15.21(a), where “East” (resp. “North”) points along the positive axis (resp. positive M 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 ;!% . 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 6d` % 9 aeo´1 + ` 9 W;!Aa + 1 and slope Alaska line tan 6 slope Delta line tan 6 ;!% 9ÒW<%kj^
15.6. OTHER BASIC CIRCULAR FUNCTION
211
All three flight paths pass through the origin 6 %´1&%9 of our coordinate system, so the equations of the lines through the flight paths will be: NW flight
MN$aeov@+ 1
MN.;!Aa +@1 Alaska flight
MN.<%kj^ Delta flight The Northwest flight is 20 miles North of SeaTac when M)$!% ; plugging into the equation of the line of flight gives !%Ntdeov , so Ce^J and the plane location will be Ø. 6 e^&Jn1!%9 . Similarly, the Alaska flight is + Í] ` % ; plugging into the equation of the 50 miles West of SeaTac when i line of flight gives M] 6 ;!Aa 9 6 ` %9;ôe% and the plane location will be i Ù 6 ` %´1e% 9 . Finally, check that the Delta flight is at ¤ 6 j%´1zÉe%&ov!A9 when it is 30 miles East of SeaTac.
CHAPTER 15. THE CIRCULAR FUNCTIONS
212
15.7 Exercises Problem 15.1. Answer the following questions. (a) Using the circular functions, compute the distance , (to four decimal places) in 13.5.4 on Page 179; compare this to the arc length : . (b) Go back to Chord Approximation in 13.5.1 and give a formula for the EXACT length of the chord in terms of the arc length : . (c) Return to Exercise 13.5 on Page 183 and compute the EXACT diameter of the moon, to the nearest mile.
Problem 15.3. (a) Find the equation of a line passing through the point (1,2) and ± making an angle of with the ¿ axis. (Note: There are two answers; find them both.) (b) Find the equation of a line making an angle of ± with the Û axis and passing through the point  R&2/ . (Note: There are two answers; find them both.) Problem 15.4. The crew of a helicopter needs to land temporarily in a forest and spot a flat horizontal piece of ground (a clearing in the forest) as a potential landing site, but are uncertain whether it is wide enough. They make two measurements from B (see picture) finding U¡½R ± and VÇ¡½z ± . They rise vertically 100 ± . Determine the feet to C and measure ¡Í width of the clearing to the nearest foot.
Problem 15.2. 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 B simply has a vision and has told John these constraints on her dream: (i) the wheel should 100 feet γ rotate counterclockwise with an angular speed A of § RPM; (ii) the linear speed of a rider should be 200 mph; (iii) the lowest point on the ride PSfrag replacements α β should be 4 feet above the level ground. Recall, axis we worked on this in Exercise 14.7. axis C axis
12 RPM θ
PSfrag replacements axis axis axis
P
E
clearing
D
Problem 15.5. A merrygoround 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.
4 feet
(a) Impose a coordinate system and find the coordinates ÿ Ä2/<¡
¿QÄ2/ ÛÄ2/f/ of Tiff at time Ä seconds after she starts the ride. (b) Find Tiff’s coordinates the instant she becomes a human missile.
rotating 3 RPM
θ
jump on here
(c) Find the equation of the tangential line PSfrag replacements along which Tiff travels the instant she axis becomes a human missile. Sketch a pic axis ture indicating this line and her initial axis direction of motion along it when the seat detaches. (a) If ¯é¡ ± , then what are your coordinates after 4 minutes?
¿0Û

15.7. EXERCISES
213
±
(b) If ¯¡ , then what are your coordinates after 45 minutes?
¿0Û
downriver

(c) If ¯]¡G©> ± , then what are your ¿0Û coordinates after 6 seconds? Draw an accurate picture of the situation.
dam 0.9
¯ ¡t©8v&
PSfrag replacements rad, then what are your ¿0Û coordinates after 2 hours and 7 sec axis onds? Draw an accurate picture of the axis situation. axis
(d) If
(e) If ¯#¡.v rad, then what are your ¿0Û coordinates after 5 seconds? Draw an accurate picture of the situation. Problem 15.6. In the picture, find a circle of radius Á so that the vertical coordinate of the is 13. point
axis W θ=0.8 P
13
O
axis
replacements circle radius=1 circle radius=r
axis
In the picture, find a circle of radius the horizontal coordinate of is 7.
Ú
7
α=0.2
Z
axis
S
so that
S
O
replacements circle radius=1
axis
:
axis
1.2 a
155 ft
Problem 15.8. A radio station obtains a permit to increase the height of their radio tower on Queen Anne Hill by no more than ¥ 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 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 RÃ above level. You move R feet away from the original measurement and observe that the old top of the tower now makes an angle of eÃ 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 0Ã 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 15.9. Michael is running 8 mph clockwise around a circular track of radius 200 feet. Michael begins at the Northernmost point on the track. Aaron is located 200 feet West and 200 feet South of the center of the circular track.
circle radius=s P
Problem 15.7. 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 PSfrag replacements a point 155 feet farther down river is 0.9 radians; assume the two angle measurements are axis taken at the same elevation above sea level. axis axis How high is the dam?
Michael starts here (8 mph)
N
200 ft.
E
W
O
S Q Aaron
CHAPTER 15. THE CIRCULAR FUNCTIONS
214
(a) If ú is Michael’s location after 45 seconds, what are the coordinates of ú ? (b) What is the distance from Aaron’s location to ú ?
(c) Aaron starts running toward ú (constant speed in a straight line) at the instant Michael starts running toward ú . Aaron plans to tackle Michael the instant he arrives at ú . How fast should Aaron run?
(d) As in c., where is Michael located when Aaron first crosses the circular track?
(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?
Problem 15.12. You are defending your title as the US Billiard Champion. Your final shot requires playing off the left and bottom cushions into the top right corner pocket, as indicated by the dotted path. For the big money, where should you aim to hit the left cushion and where will the ball strike the bottom cushion?
Problem 15.10. 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 PSfrag replacements each other after 11 seconds.
N
this pocket for the big money
4 ft 6 ft
5 ft
axis axis
12 ft
axisfind these locations
PSfrag replacements
George
Paula
axis axis axis
After running for 3 minutes, how far east of his starting point is George?
Problem 15.13. A United flight departs ± clockwise from SeaTac airport heading ªz South. A crop duster leaves a rural airstrip located 100 miles due North of SeaTac and is ± heading R clockwise from South. Where will the two lines of flight cross? rural airstrip
Problem 15.11. 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.
55 o 100 miles
duster
kite
SeaTac
PSfrag replacements PSfrag replacements axis axis axis
o 42 4 feet
axis axis
ground level
axisUnited
60 o
South
15.7. EXERCISES
k¿ / ¡ ¿ sin  z¿Y©ê" / , Â What is the maximum value of the function T© ¨ sin k ¿ / ? Do not use a calculator your answer. and explain
Problem 15.14. e Æ what is (b)
215
(a) If ¨ /?
e
(c) What is the minimum value of the function sin  ¿/ ? Do not use a calculator and explain your answer. cos  Ä2/ , &¿/4¡¿¹ and jT Ä2/4¡ e jrÄ2/À/ and jT ¿/À/ evaluated at
(d) If e
e
(c) Find the coordinates of the bug at time Ä. (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
º
what are ?
ω=4π/9 rad/sec 1.2 rad bug lands here
Problem 15.15. 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:
bug lands here
ω= − 4π/9 rad/sec 1.2 rad
¯ p Â ¿>Ä . Plug
to explicitly
bug lands here
2 in
2 in
(a) Find an angle ¯ in standard central pop sition 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.) PSfrag replacements (b) The location angle of the bug at time Ä is given by the formula ¯kÄ2/¡ in the values for ¯ and ¿ p obtain a formula for ¯kÄ2/ .
2 in
2 in
axis axis axis
ω=4π/9 rad/sec
ω= − 4π/9 rad/sec
ω=4π/9 rad/sec o 40 2 in
bug lands here 0.5 rad
2 in bug lands here
216
CHAPTER 15. THE CIRCULAR FUNCTIONS
Chapter 16 PSfrag replacements Trigonometric Functions axis
Our definitions of the circular functions are based upon the unit circle. This makes it easy to visualize many of their properties.
16.1 Easy Properties of Circular Functions {tWÚ } axis { { õ ® ball moves sin { õ } q counterclockwise {Õ } q õ { S axis cos
How can we determine the range of function values for cos 6 A9 and sin 6 A9 ? To begin with, recall the abstract definition for the range of a function ! 6 A9 :
1
! xw ! 6 A 9Ò
1 is in the domainy¼ Range of
{iq} Using the unit circle constructions of the basic circular õ functions, it is easy to visualize the range of cos 6 A9 and Figure 16.1: Visualizing the sin 6 A9 . Beginning at the position 6 n1%9 , imagine a ball range of sin  ¯T/ and cos  ¯T/ . moving counterclockwise around the unit circle. If we PSfrag replacements “freeze” the motion at any point in time, we will have swept out an angle PSfrag replacements and the corresponding position Ø 6 A9 on the circle will have coordinates Ø 6 A9 6 cos 6 A9z1 sin 6 A99 . UNIT CIRCLE
axis
¶
axis
Ú
ball moves from 0 to
S
axis
Ú
axis
light source
¶
ball moves from 0
radians around unit
to
circle
around unit circle
S
light source
axis
radians
axis
(b) What do you see on the ¿ axis?
(a) What do you see on the Û axis?
Figure 16.2: Projecting the coordinates of points onto the Û axis and the ¿ axis.
By studying the coordinates of the ball as it moves in the first quadrant, we will be studying cos 6 A9 and sin 6 A9 , for %¦' Ë' (Òfv! radians. 217
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
218
We can visualize this very concretely. Imagine a light source as in Figure 16.2(a); then a shadow projects onto the vertical M axis. The shadow locations you would see on the M axis are precisely the values sin 6 A9 , for %Q'[£'"(Òfv! radians. Similarly, imagine a light source as in Figure 16.2(b); then a shadow projects onto the horizontal axis. The shadow locations you would see on the axis are precisely the values cos 6 A9 , for %>'[0'"(Òfv! PSfrag replacements radians. PSfrag replacements There are two visual conclusions: First, the function values of sin 6 A9 vary from 0 to 1 as varies from 0 to (Òfv! . Secondly, the function values of cos axis 0 to (Êfv! . Of course, we can go 6 A9 vary from 1 to 0 as varies from axis ahead and continue analyzing the motion axis as the ball moves into the second, third and fourth quadrant, ending up back at the starting position 6 n1%9 . See Figure 16.3.
Ú #2
#3
light source
axis #1
#4
öú
ball moves from 0 to radians around unit circle
S
axis
ú radians öaround nit circle
ball moves from 0 #2 #3
to
#1 #4
S
axis
light source
(a) What do you see on the Û axis?
(b) What do you see on the ¿ axis?
Figure 16.3: Analyzing the values of the sine and cosine functions.
The conclusion is that after one complete counterclockwise rotation, the values of sin 6 A9 and cos 6 A9 range over the interval 0_Én132 . 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 16.3(a). The values of the sine function vary from % up to while the ball moves through the first quadrant (arrow labeled #1), then from down to % (arrow labeled #2), then from % down to É (arrow labeled #3), then from É up to % (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 6 A9 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 6 n1%9 . 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 nonnegative, beginning with slope % (the degenerate right triangle
16.1. EASY PROPERTIES OF CIRCULAR FUNCTIONS
Ú
Ú
axis 8
7
219
axis
6 5
PSfrag replacements
PSfrag replacements 4
3
2 1
S
axis
axis
1 2
S
axis
3
axis
4 5 7
(a) What happens to the slopes of these triangles?
6
(b) What happens to the slopes of these triangles?
Figure 16.4: Analyzing the values of the tangent function.
when Z % ) then increasing. In fact, as the angle approaches (Òfv! radians, the ball is getting closer to the position 6 %´19 and the hypotenuse is approaching a vertical line. This tells us that as varies from 0 to (Êfv! (but not equal to (Òfv! ), these slopes attain all possible nonnegative values. In other words, the range of values for tan 6 A9 on the domain %' FXq(Òfv! will be %d' î X . Similar reasoning shows that as the ball moves in the fourth quadrant, the slopes of the hypotenuses of the triangles are always nonpositive, varying from 0 to ANY negative value. In other words, the range of values for tan 6 A9 on the domain w(Òfv!ZX ('q% X î ')% . will be On your calculator, you can verify the visual conclusions we just es tablished by studying the values of tan 6 A9 for close (but not equal) to radians = o% :
i
9 ` c!o tan 6 Jo Ê i ! o^ tan 6 Joo 9Ê ` A i ! o ` J tan 6 Jooo 9Ê ` .. .
i
9 . ` _!o tan 6
The fact that the values of the tangent function become arbitrarily large as we get close to ¦ (Òfv! radians means the function output values are unbounded. Important Fact 16.1.1 (Circular function values). For any angle , we always have ÉÐ' ï ´ 6 A9' and É>' sin 6 A9'q . On domain %0'¾C'H!( , the range of both cos 6 A9 and sin 6 A9 is ÉT' î ' . In contrast, on the domain of all values for which tangent is defined, the range of tan 6 A9 is all real numbers.
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
220
For the sine and cosine functions, if the domain is not %5' #'é!( , then we need to consider the “periodic qualities” of the circular functions to determine the range. This is discussed below.
16.2 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 ØZ 6 cos 6 A9z1 sin 6 A99 on the unit circle, corresponding to the standard central
position angle , then recall the equation of the unit circle tells us M [ . But, since the coordinate is cos 6 A9 and
the M coordinate is sin 6 A9 , we have 6 cos
6 A9¼9 6 sin 6 A99 ÿ It is common notational practice to write 6 cos 6 A99 cos 6 A9 and 6 sin 6 A99 sin 6 A9 . This leads to the most important of all trigonometric identities: Important Fact 16.2.1 (Trigonometric identity). For any angle , we
have the identity cos 6 A9´ sin 6 A9Ë[ . Adding any multiple of !( radians (or j^% ) 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 6 A9 and sin 6 A9 on the domain %Q'[£' !( 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.
9 is called Definition 16.2.2 (Periodic function). For ïIV % , a function ! 6 A ï periodic if two things are true: (i) !
6 ï=98&! 6 A9 holds for all ;
(ii) There is no smaller , %ÉX .
?X ï , such that ! 6 H@9â#! 6 A9 holds for all
We usually call ï the period of the function. Using this new terminology, we conclude that the sine and cosine circular functions are !( periodic. In the case of the tangent circular function, it is also true that tan 6 A9¢ tan 6 I)!( Ä 9 , for every integer Ä . However, referring back to the unit circle definitions of the circular functions, we have tan 6 A94 tan 6 Y Ä (49 , for all integers Ä . If you take Ä y , then this tells us that the tangent circular function is ( periodic. We summarize this information below. Important Fact 16.2.3 (Periodicity identity). For any angle and any integer Ä '%´1[¦ n1[¦w!1[¦ jA1ee , we have cos 6 A9Ò cos 6 P!( Ä 9 , sin 6 A9Ò sin 6 !( Ä 9 , and tan 6 A9Ê tan 6 ï Ä (49 .
axis axis
16.2. IDENTITIES
221
its negative in the same Next, we draw an angle and unit circle picture in standard central position. We have the points Ø and Ø æ used to define the circuindicated lar functions. It is clear from the picture in Figure 16.5 that Ø and Ø æ have the same coordinate, but the M coordinates are negatives of one another. This gives the next identity:
{ cos { } sin { A ¯Ð®
{ cos {Õ } sin {i A ¯Ð®
j
¢
unit circle
·
Figure 16.5: Visualizing a trigonometric identity.
Important Fact 16.2.4 (Even/Odd identity). For any angle , sin 6 bA9Ë sin 6 A9 , and cos 6 bA9Ë cos 6 A9 . PSfrag replacements axis axis
We can use the terminology of even and odd functionsaxis 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 and ? ]( in the same unit circle picture in standard central position. We have in on the unit dicated the corresponding points Ø and Ø D funccircle and their coordinates in terms of the circular tions: From the picture in Figure 16.6, the coordinate of and Ø must be the “negative” of the coordinate of Ø D similarly, the M coordinate of Ø must be the “negative” of . This gives us the next identity: the M coordinate of Ø
® Q¯ { cos { } sin { A unit circle
á ú
Figure
k · ® ¯ { { á ú }{ { á ú A
cos
16.6: Visualizing Fact 16.2.5.
D
Important Fact 16.2.5 (Plus identity). For any angle sin 6 ï(>9ÒH sin 6 A9 , and cos 6 ï(49ÊH cos 6 A9 .
, we have
Important Fact 16.2.6. For any angle , we have sin 6 (D A 9Ë sin 6 A 9 and cos 6 ( A9Ê. cos 6 A9 .
sin
For example, we have sin x u { sin x { . This calculation leads to a computational observation: Combining Table 15.1 with the previous two + 6 A9 , and tan 6 A9 identities we can compute the EXACT value of cos 6 A9 , sin at an angle which is a multiple of j% radians or ` radians. p Here are some sample calculations together with a reference as to “why” each equality is valid:
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
222 Example 16.2.7.
+
` 9Ò cos 6 ` 9 ) ! ! + ! ! ` 9Ò sin 6 ` + 'eJ% 9 (ii) sin 6 H sin 6 ` 9 ) ! H ! g !( ( (iii) cos cos ¹ ( j½h j ( H cos j ( H cos j H !
(i)
cos 6
+
+
Fact e^_!A on page !!A Table ` a on page !%
! !% Fact e ^_!Akj on page ! !A Fact e ^_!A ` on page Table ` a on page !% Fact e^_!Akj on page !!% Fact e^_!A ` on page !!A
+
Fact e^_!A on page !!A Table ` a on page !%
16.3 Graphs of Circular Functions We have introduced three new functions of the variable 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 axis. We will label the vertical axis, which corresponds to the dependent variable, the î axis. With these conventions, beginning with any of the circular functions î sin 6 A9 , î cos 6 A9 , or î tan 6 A9 , the graph will be a subset of the î coordinate system. Precisely, given a circular function î S! 6 A9 , the graph consists of all pairs 6 1H! 6 A99 , where 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 “M ” 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 M coordinate system, the coordinates (resp. M coordinates) of points on the unit circle are computed by cos 6 A9 (resp. sin 6 A9 ).
16.3. GRAPHS OF CIRCULAR FUNCTIONS
eplacements
axis
î
íPî Ú
à
axis
223
Ö
cos
î ðåë sin ðdð
axis
S
Coordinate system used to GRAPH the circular functions.
Figure 16.7: The ä
Coordinate system used to DEFINE the circular functions.
¯
versus
¿0Û
coordinate systems.
16.3.1 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 16.3.1. 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 axis so that the length of 1 radian on the 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: j^% = !( radians.
16.3.2 The sine and cosine graphs Using Fact 16.1.1, we know that Éï' sin 6 A9='ô and Éï' cos 6 A92'ô . Pictorially, this tells us that the graphs of î sin 6 A9 and î cos 6 A9 lie between the horizontal lines î and î É ; i.e. the graphs lie inside the darkened band pictured in Figure 16.8. By Fact 16.2.3, we know that the values of the sine and cosine repeat themselves every !( radians. Consequently, if we know the graphs of the + domain %Q'[£' !( , then the picture will repeat for sine and cosine on the the interval !( '[0' ( , ;!( '[0')% , etc.
PSfrag replacements 224
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
í
axis axis
PSfrag replacements à
axis axis
axis
à
axis
í
axis
æ à
Figure 16.8: Visualizing the range of sin ü¯T/ and cos 
í
Ú æ p
Repeat
æ
à
í
O
æ
¯T/ .
axis
Repeat
Ú
axis
Repeat
p
Repeat
picture in here repeats each
axis
units
Figure 16.9: On what intervals will the graph repeat?
! ( can be roughly Sketching the graph of î sin 6 A9 for %' '
, 1 l , and achieved by plotting points. For example, x 1 { , 1
l p { x 1 lie on the graph, as do x l 1zÉ { , u 1z l , ¾ 1z , x ¥ 1zU { , and PSfrag replacements l p !(>1%9 , etc. If we return to our analysis of the range of values for the 6 axis sine function in Figure 16.2, it is easy to see where sin 6 A9 is positive or axis negative; combined with Chapter 4, this tells us where the graph is above and below the horizontal axis (see Figure 16.10).
axis
à
æ
æ
axis
l
axis
Figure 16.10: Where 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 16.11).
axis axis axis
16.3. GRAPHS OF CIRCULAR FUNCTIONS à
1
æ
225
axis
æ
−1
l
axis
one period
Figure 16.11: The graph of
¡
sin 
¯T/ .
ä PSfrag replacements
axis
We could repeat this analysis to arrive at the graph of the cosine. Instead, we will utilize an identity. Given an angle , place it in central standard position in the unit circle, as one of the four cases of Figure 15.16. For example, we have pictured Case I in this figure. Since the sum of the angles in a triangle is eJ% ô( radians, we know that , (Òfv! , and are the three angles of the inscribed right triangle. From the picture in Figure 16.12, it then follows that side adjacent to hypotenuse side opposite to x hypotenuse
9 cos 6 A
sin
(
!
axis
unit circle
{
Using the same reasoning this identity is valid for all . This gives us another useful identity:
9Ê Important Fact 16.3.2 (Conversion identity). For any angle , cos 6 A sin x × { , and sin 6 A9Ê cos x { . This identity can be used to sketch the graph of the cosine function. First, we do a calculation using our new identity:
(
6 bA9 ! ( sin ï !
sin
+
Fact e^c!A on page !!A Fact e^cjTc! (above) Since: 6(' QÌ9Ò
ú
ö
ö
Figure 16.12: Visualizing the conversion identity.
×
cos 6 A 9Ò cos 6 bA9
ú æ
axis
6(' 6 wQÌ99Ò 6 Q 6 ' 99
By the horizontal shifting principle in Fact 9.3.1 on page 127, the graph of î cos 6 A9 is obtained by horizontally shifting the graph of î sin 6 A9 by units to the left. Here is a plot of the graph of the cosine function: See Figure 16.13.
axis axis axis
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
226
à
1
æ
ements
axis
æ
axis
−1
axis
Figure 16.13: The graph of
¶
16.3.3 The tangent graph
¶
¶ axis
+¯
¶
axis graph heads this way, getting close to vertical line as gets close to
¶
¡ ä
cos 
¯T/ .
î
î the sine and cosine circuAs we have already seen, unlike lar functions, the tangent function is NOT defined for all sin ð values of . Since tan 6 A9Ê cos ð , here are some properties we can immediately deduce:
graph heads this way, getting close to vertical line as gets close to
axis
one period
axis
Q¯
l
¶
The function î tan 6 A9 is undefined if and only if I ( , where ?'%´1[¦ n1[w ¦ !1[¦
Figure 16.14: The behavior of tan  ¯T/ as ¯ approaches asymptotes.
The function î tan 6 A9 % if and only if . where ?'%´[1 ¦ n1[¦w!1[¦ jA1 EeEeE
( ,
By Fact 16.2.3, the tangent function is ( periodic, so the picture of the graph will repeat itself every ( units and it is enough to understand the graph when X X . On the domain % ' X , tan 6 A9§òé% ; on the domain XêÞ'y% , 9 ')% . tan 6 A ( , where In the î coordinate system, the vertical lines Ï N %´[1 ¦In[1 ¦w!1 EeEeE will be vertical asymptotes for the graph of the tangent function. Using our slope interpretation in Figure 16.4, what becomes
clear is this: As the values of get close to , the graph is getting close to the vertical line ü AND becoming farther and farther away from the horizontal axis: To this numerically, first understand suppose is { { slightly smaller than , say x %d , x %% , and x %%% { . Then the calculation of tan 6 A9 involves dividing a number very close to 1 by a
very small positive number: tan tan tan
( !
(
( !
%a
%%
oo^^^´1 i oooo^ 1 and
%%%
e%%%
!
16.4. TRIGONOMETRIC FUNCTIONS
tan tan tan
( !
(
( !
%d
%&%
ê%&%%
Ée%%%
!
¶
Q¯
Conclude that as “approaches from below”, the values of tan 6 A9 are becoming larger and larger. This says that the function values become “unbounded”. Likewise, is slightly bigger than , say imagine the case when x %a { , x %% { , and x %%% { . Then the calculation of tan 6 A9 involves dividing a number very close to É by a very small positive number:
227 axis
one period
ú
ö
¶
etc.
etc.
ú
+¯ Ø
ú
Q¯ Ó ¶
ú +¯
¶
öú
+¯ Ó ¶
axis
Figure 16.15: The graph of ä ¡ ÄRÅeÎ ü¯r/ .
Conclude that as “approaches from above”, the values of tan 6 A9 are becoming negative numbers of increasingly larger magnitude: Again, this tells us the function values are becoming “unbounded”. The graph of î tan 6 A9 for can be roughly achieved by comXé X bining the calculations as in Example 16.2.7 and the qualitative features highlighted. Figure 16.15 shows a software plot.
16.4 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 / ; 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 / ; 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.
16.4.1 A Transition Given a real number / , is there a sensible way to define cos 6 /29 and sin 6 /:9 ? The answer is yes and depends on the ideas surrounding radian measure of angles. Given the positive real number / , we can certainly imagine an angle of measure / radians inside the unit circle (in standard position) and we know the arc subtended by this angle has length / (this is why we use the unit circle).
PSfrag replacements axis
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
axis228 axis feet
O D u O D Du O æ u æ O æ u æ O D
ò (time)
ements axis axis

axis
Figure 16.16: A periodic function with input variable Ä .
Æ É"Ä Æ Ã[Ç ÈRÉ arclength
rads
Ä
We already know that cos 6 / radians 9 and sin 6 / radians 9 compute the and M coordinates of the point Ø 6 /29 . In effect, we are just using the measure of the angle / to help us locate the point Ø 6 /:9 . An alternate way to locate Ø 6 /29 is to move along the circumference counterclockwise, beginning at 6 n1%9 , until we have an arc of length / ; that again puts us at the point Ø 6 /29 . In the case of an angle of measure / radians, the point Ø 6 /:9 can be located by moving along the circumference clockwise, beginning at 6 n1%9 , until we have an arc of length / .
Æ mÇ ~qÉ
unit circle
Figure 16.17: The circular functions with input variable Ä.
Definition 16.4.1 (Trigonometric functions). Let / be a real number. We DEFINE the sine function M sin 6 /:9 , the cosine function M cos 6 /29 and the tangent function MN tan 6 /29 by the rules sin 6 /:9 M coordinate of Ø 6 /29 sin 6 / radians 9 def
cos 6 /:9 ) coordinate of Ø 6 /:9Ì cos 6 / radians 9 def sin 6 /29 tan 6 /:9 tan 6 / radians 9 cos 6 /:9 def
We refer to these as the basic trigonometric functions. If we are working with radian measure and / is a real number, then there is no difference between evaluating a trigonometric function at the real number / and evaluating the corresponding circular function at the angle of measure / radians. Example 16.4.2. Assume that the number of hours
of daylight in Seattle i during 1994 is given by the function 6 /:98)jT Ä x 6 /mJ% ` 9 { 0!1 where
l¥
16.4. TRIGONOMETRIC FUNCTIONS
229
/ represents the day of the year and /H% 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 jT¹!JwjTvwj%D; i +jT st day of the year. So, there will be 6 jT9Ê'jT sin 6 ! 6a` % ` 9 (Òf:j^^9'0! Jv! hours of daylight on May 11.
16.4.2 Graphs of trigonometric functions The graphs of the trigonometric functions MÏ sin 6 /:9 , Mb cos 6 /:9 , and M tan 6 /29 will look just like Figures 16.11, 16.13, and 16.15, except that the horizontal axis becomes the / axis and the vertical axis becomes the M axis.
16.4.3 Notation for trigonometric functions
!!!
In many texts, you will find the sine function written as MN sin / ; i.e. the parenthesis around the “/ ” 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 M sin 6 /:9 , 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
M sin 6 / É!n/n 9 , it is crystal clear that the sine function is applied to the expression
“/ !n/ ”; using the alternate notation yields the expression M sin / !n/mH , which is interpreted to mean MN 6 sinPSfrag / 9T 6 !nreplacements /m'9 . , cos , axis As a rule, whenever you see an expression involving sin , we assume “ or tan ” is in units of RADIANS, unless otherwise axis noted. When computing values on your calculator, MAKE SURE YOU axis ARE USING RADIAN MODE!
CAUTION !!!
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
230
16.5 Exercises Problem 16.1. On your calculator you can check that cos åª ± /¢¡ 0 and sinreplacements óª ± /¢¡ PSfrag R .
K/¡ R 0 ± ]¥R ± R K/¡ 0 L/¡¹ ªz ± L/¡¹ 0 ¥R ± K/¡ 0 so that cos .
(a) Find four other angles ¹ 0 and sin  L/¡Ç¹ (b) Find an angle that cos
,
h
, so .
Ëh
and sin
(c) How many angles of measure between will have cos ? ± and (d) How many angles of measure between ¥ ± and will have cos ? Problem 16.2. Work the following problems without using ANY calculators. (a) Sketch Û¢¡ sin &¿/ .
(b) Sketch Û¢¡ sin (c) Sketch Û¢¡
&¿/ .
e e
(b) (c)
e
 Ä2/¡  Ä2/ ¡  Ä2/ ¡
¨
sin Ä2/ . ¨
cos Ä2/ .
Problem 16.6. Start with the equation sin ü¯r/>¡ cos  ¯T/ . Use the unit circle interpretation of the circular functions to find the solutions of this equation; make sure to describe your reasoning. Problem 16.7. Answer the following questions:
& ¿w / ¡ [jT¿k/f/ ?
sin &¿/ and jT¿/w¡Wz¿ , what is e e What is jr&/f/ ? What is jr ¿k/f/ ? e What is jr  &/À/ ? e
(a) If e
(b) If ¿/¡ e sin &¿/ and jr&¿/ ¡.¿¹>ÂU¿<©U , e [jT¿k/f/ ? What is [jT»2/À/ ? What is what is e e jr ¿k/f/ ? What is jT »"/f/ ?
&¿/
.
Problem 16.3. Sketch the graphs of these functions: (a)
axis
e
sin
´Â
axis axis
jT¿/7¡ &¿/ , what is [jT¿k/f/ ? (c) If ¿k/¡¦ sin e e What is [jT "/À/ ? Is jr&¿/À/¡% sin &¿/~ó ? e
e
¨
Problem 16.8. Dave is replumbing his house and needs to carry a copper pipe around the corner of a hallway. As he cheerfully walks down the hall and rounds the corner, the pipe becomes stuck, as pictured. Assume Dave must always hold the pipe level; i.e. he can’t tilt it up or down.
¨
¨
¨
tan Ä2/ .
(d)
T Ä2/¡© j
¨
¨
© sin¨ Ä2/ sin RÄTÂ?2/ ÂÉ
(e) Solve the equation as in (d).
rÄ2/¡I j
¨
if if
Ä õ Ä÷
, where
TÄ2/ j
is
pipe
PSfrag replacements Problem 16.4. Solve the following: (a) If cos  ¯T/¡ º , what are the two possible  ¯T/ ? values of sin
sin ü¯r/¡é©Ò¹
(b) If and ¯ is in the third quadrant of the ¿0Û plane, what is cos ü¯T/ ? (c) If sin ü¯T/¡
«
¤
, what is sin 
©Æ¯T/ ?
Problem 16.5. These graphs represent periodic functions. Describe the period in each case.
θ
4 ft
θ
axis axis axis
3 ft
−TOP VIEW−
(a) Find a formula for the function [ü¯T/ which computes the length of the longest pipe that will fit with the pictured angle ¯ . (b) Describe how you could use the function in a. to find the LONGEST pipe Dave can carry around the corner.
16.5. EXERCISES (c) Assume the length of the pipe is mini mized when ¯<¡ 0v Ã ; how long is the pipe?
Problem 16.9. For each situation described, which could possibly be described using a periodic function? (a) The population of rabbits in a large wooded area as a function of time.
231 (b) As you pedal a bike, the height of your big toe above the ground as a function of time. (c) As you pedal a bike, the distance you have traveled as a function of time. (d) The score for the UW womens Bball team as a function of time elapsed during a game. (e) Tuition at the UW as a function of time.
232
CHAPTER 16. TRIGONOMETRIC FUNCTIONS
PSfrag replacements axis axis axis
Chapter 17 Sinusoidal Functions
depth
² A migrating salmon is heading up a portion of the Ö Columbia River. It’s depth 6 /29 (in feet) below the water  surface is measured and plotted over a 30 minute period, ö as a function of time / (minutes). What is the formula for õ 6 /29 ? ² ² ² m õ ö ö õ time In order to answer the question, we need to introduce eplacements an important new family of functions called the sinusoidal Figure 17.1: The depth of a axis salmon as a function of time. functions. These functions will play a central role in modaxis eling any kind of periodic phenomena. The amazing fact axis is that almost any function you will encounter can be approximated by a sum of sinusoidal functions; a result that has farreaching implications in all of our lives.
17.1 A special class of functions Beginning with the trigonometric function M sin 6 9 , what is the most general function we can build using the graphical techniques of shifting and stretching?
horizontally shift
horizontally dilate
ú
¶
õ
~
õ
axis
¶
~ ¯
ú
sin
{$Áu Á axis
vertically
vertically
dilate
shift
Figure 17.2: Visualizing the geometric operations available for curve sketching.
The graph of M sin 6 9 can be manipulated in four basic ways: horizontally shift, vertically shift, horizontally dilate or vertically dilate. Each of these “geometric operations” corresponds to a simple change in the “symbolic formula” for the function, as discussed in Chapter 9. 233
CHAPTER 17. SINUSOIDAL FUNCTIONS
234
If we vertically shift the graph by units upward, the resulting curve would be the graph of the function MN sin 6 9:c ; see Facts 9.3.1. Recall, the effect of the sign of : If is negative, the effect of shifting units upward is the same as shifting units downward. Notice, the function H M PSfrag replacements sin 6 98ð is still a periodic function, having the same period !( as MN sin 6 @9 . Notice, whereas the graph of the function MN sin 6 @9 oscillates axis between the horizontal lines M ð¦ , the graph of M½ sin 6 @9O oscillates axis between M" ¦ . For this reason, we sometimes refer to the constant axis as the mean of the function M sin 6 9T . In Figure 17.3, notice that the graph of M sin 6 @9e is symmetrically split by the horizontal “mean” line M½ .
È
axis
~ ¯
sin
È
{$Áu
~ ¯
axis
Ã
~ ¯ axis shift
sin
{$Áu á
Ã
axis
units
Figure 17.3: Interpreting the mean.
Next, consider the effect of horizontally shifting the graph of M½ sin 6 9 by units to the right. By Facts 9.3.1, the new curve is the graph of the function M sin 6 ? 9 . Also, recall the effect of the sign of : If is negative, the effect of shifting units right is the same as shifting units left. If the domain of sin 6 9 is % ' T' !( , then the domain of '!( , again by Facts 9.3.1. Rewriting this, the 9 is %Þ'Ï sin 6  9 is 'ÏI'Ï!(½ and the graph will go through PSfrag replacementsdomain of sin 6  precisely one period on this domain. In other words, the new function axis 9 is still !( periodic. The constant is usually called the phase sin 6 Y axis shift of MN sin 6 7 9 . Looking at Figure 17.4, it is possible to interpret axis graphically: will be a point where the graph crosses the horizontal axis on its way up from a minimum to a maximum.
È
È
~ ¯ Þ {$Áu
axis
Ã
axis
shift
axis
~ ¯ Ã
sin
{$Ák
axis
units
Figure 17.4: Interpreting the phase shift.
Vertically dilating the graph, either by vertical expansion or compression, leads to a new curve. The graph of this vertically dilated curve is Mé5 sin 6 @9 , for some positive constant . Furthermore, if V ,
17.1. A SPECIAL CLASS OF FUNCTIONS
235
the graph of M y sin 6 @9 is a vertically expanded version of M] sin 6 9 , whereas, if %?X$ BXy , then the graph of M $ sin 6 9 is a vertically compressed version of M sin 6 9 . Notice, the function M . sin 6 @9 is still !( eplacementsperiodic. What has changed is the band of oscillation: whereas the graph of the function M sin 6 @9 stays between the horizontal lines Mê ¦ , the axis graph of Mê$ sin 6 9 oscillates between the horizontal lines M ¦ . We axis usually refer to as the amplitude of the function M ) sin 6 @9 . axis
È
axis
~ ¯
È
{$Áu sin Ã
~ ¯
axis
sin
{tÁu
Ã
axis
axis stretch
units
Figure 17.5: Interpreting the amplitude.
Finally, horizontally dilating the graph, either by horizontal expansion or compression, leads to a new curve. The equation of this horizontally dilated curve is M) sin 6 ï¹9 , for some constant ïFV[% . We know that M) sin 6 @9 is a !( periodic function and observe that horizontally dilation still results in a periodic function, but the period will typically NOT be !( . For future purposes, it is useful to rewrite the equation for the horizontally stretched curve in a way more directly highlighting the period. To begin eplacements with, once the horizontal stretching factor ï is known, we could rewrite axis axis
!( 1
ïP
axis
~
axis
for some )' ã %
~ ¯
sin
Á stretch
~
{tÁu
axis
axis
~ ¯
sin
{A{
¶
iÁu Á
axis
stretch
Figure 17.6: Interpreting the period.
Here is the point of this yoga with the horizontal dilating constant: If
we let the values of range over the interval 0 %´1 2 , then will range over the interval 0_%´1 !(Î2 . In other words, the function
MN sin x { is periodic and we can read off the period of MH sin x { by viewing the constant in this mysterious way. The four constructions outlined lead to a new family of functions.
CHAPTER 17. SINUSOIDAL FUNCTIONS
236
Definition 17.1.1 (The Sinusoidal Function). Let , , and be fixed constants, where and are both positive. Then we can form the new function
PSfrag replacements axis
g
!( ¢ 9 U1 6
h
MN' sin
axis axiswhich
is called a sinusoidal function. The four constants can be interpreted graphically as indicated:
È
È
axis
axis
Ã axis ~ ¯
sin
Ã
all four
{$Áu
operations
axis
~ ¯
sin
{{
¶
{$Ár à A á
Figure 17.7: Putting it all together for the sinusoidal function.
17.1.1 How to roughly sketch a sinusoidal graph Important Procedure 17.1.2. Given a sinusoidal function in the standard form
MN' sin
g
!( ¢ 9 U1 6
h
once the constants , , , and 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 , , , and . Here is a “five step procedure” one can follow, assuming we are given , , , and . It is a good idea to follow Example 17.1.3 as you read this procedure; that way it will seem a lot less abstract. 1. Draw the horizontal line given by the equation M
; this line will “split” the graph of M' sin x 6 P 9 { into symmetrical upper and lower halves. 2. Draw the two horizontal lines given by the equations MN < ¦I . These two lines determine a horizontal strip inside which the graph of the
17.1. A SPECIAL CLASS OF FUNCTIONS
237
sinusoidal function will oscillate. Notice, the points where the sinusoidal function has a maximum value lie on the line M ]ê . Likewise, the points where the sinusoidal function has a minimum value lie on the line M ðH . 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 , we know these important facts: (1) The period is the horizontal distance between two successive maxima (peaks) in the graph. Likewise, the period 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 .
4. Plot the point 6 1[U9 . This will be a place where the graph of the sinusoidal function will cross the mean line M 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 from 6 1[U9 by horizontally shifting by any integer multiple of . For example, here are three places the graph crosses the mean line:
6 1[U9z1 6
1[U9z1 6 1[U9
5. Finally, midway between 6 1[ 9 and 6 1[U9 there will be a maxima (peak); i.e. at the point 6 1[$" 9 . Likewise, midway between p will be a minima (valley); i.e. at the 6 1[U9 and 6 H 1[U9 there point 6 l 1ÕU 9 . It is now possible to roughly sketch the graph on p '[ò' . by connecting the points described. Once the domain this portion of the graph is known, the fact that the function is periodic tells us to simply repeat the picture in the intervals # ò' *' ?! , ò' *' , etc. To make sense of this procedure, let’s do an explicit example to see how these five steps produce a rough sketch.
CHAPTER 17. SINUSOIDAL FUNCTIONS
238
PSfrag replacements Example 17.1.3. The temperature (in ) of AdriN’s dorm room varies axis
during the day according to the sinusoidal function 6 /:9ÒH^ sin x 6 /4'9 { axis graph of eo´1 where / represents hours after midnight. Roughly sketch the axis 6 /:9 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:
g ( !( 6 /:9ÒH^ sin 6 /4'9 ' eoY) sin /m 9 U 6 0 !
h + Conclude that ô^ , $b! , , e o . Following the first four steps of the procedure outlined, we can sketch the lines Mé eo , MH ¦' Ï eo ¦.^ and three points where the graph crosses the mean line (see Figure17.8).
m ² ö
Ý {Ü
graph will oscillate inside this strip
ö
~ ¯ { } õ§õ õ° ²
õ
{ u} ö õ°
~ ¯
{t^²W} õ°
Ö
³
ö
Ö
¶ õ
³
Ö
³
õ ö õ õ õ¶ ö ö§ö ö ö ö ¶
Figure 17.8: Sketching the mean
E
m 
and amplitude
aÖ m³§
ö B
õ°
~ ¯
õ
²
ö 
õ
¶
.
According to the fifth step in the sketching procedure, we can plot the i maxima 6 1[ 98 6 1! ` 9 and the minima 6 l 1[Í Ç98 6 !o´1jT9 . p p domain )'"/R' We then “connect the dots” to get a rough sketch on the j ` .
m
Ý {Ü
ö
graph
will oscillate inside this
{ maxima } ²W õÀ± ö
²
ö õ õ
{ W} ö õ°
{ } õ§õ õ° ²
strip
Ö
³
¶ õ
{$^²u} õµ°
{ } W ö° õ minima
ö
~ ¯
Ö
Ö
³
õ ö õ õ ³ õ¶ ö ö öPö ö ö ¶
^ 
ö
aÖ¿(³= ¶
Figure 17.9: Visualizing the maximum and minimum over one period.
~ ¯ ~ ¯
²
ö õ
õ° 
17.1. A SPECIAL CLASS OF FUNCTIONS
239
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.
~ axis m maxima { } ²W {imaxima } ²W ² Àõ ± ö ± ö ö { } õ§õ õ° Õ{ u} {Õ ö } õ õ° õõ ² õ° {$²u} W õ õ minima Ö ³ õ õ Ö ³ õ ¶ô ö§ö Ö õ ö ¶ &³ RÖ ¶ õö õ ö õ õ ö ö
{×Ö } ²W
maxima
{ W} {ta²u} õ° ö õµ°
~ ¯
õ ö
{$Ö } ± õµ°
{ } W ö° õ minima Ö ³ m  a Ö ( ³  ¶ ö ö ö¶ ö
{t²^W} W
õ
²
ö
~ ¯ õ ° {$² } ° õ° ~ ¯ 
minima
õ
Ü axis
Figure 17.10: Repeat sketch for every full period.
+
We restrict the picture to the domain %§' /.' ! and obtain the computer generated graph pictured in Figure 17.11; as you can see, our eplacementsrough + + is very accurate. The temperature at 2:00 p.m. is just graph 9Ò"!njTc! C. From the graph, the maximum value of the function will axis 6 axisbe ] [ ! ` C and the minimum value will be ] [[j C. axis
temp
!
u
O u
O u
O
ò (hours)
u
O
Figure 17.11: The computer generated solution.
17.1.2 Functions not in standard sinusoidal form Any time we are given a trigonometric function written in the standard form
M½' sin
g
!( ¢ 9 F1 6 h
for constants , , , and (with and positive), the summary in Definition 17.1.1 tells us everything we could possibly want to know
CHAPTER 17. SINUSOIDAL FUNCTIONS
240
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 or might be negative. + For example, MN.;! sin 6 !A i are examples that fail to be in 94[j and M j sin xU Y) { standard form.
We might use the cosine function in place of the sine function. For example, something like M.! cos 6 j2É.9A! 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 17.1.1? 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 17.1.4.
i
(i) Start with MN.;! sin 6 !m 9j , then here are the steps with reference to the required identities to put the equation in standard form:
i MN.;! sin 6 !¢ 9´j i "! 6 sin 6 !¢ 99´j i "! sin 6 !¢ (49´ 6 jT9 Fact 16.2.5 on page 221 g g i !( ( 8 "! sin Y 6 6 jT9 h
This function is now in the standard form of Definition 17.1.1, so it is æ a sinusoidal function with phase shift ¾ y&o:j , mean qj , amplitude ["! and period ½'( .
+
(ii) Start with MN)j sin 6 F ÌZ9e , then here are the steps with reference to the required identities to put the equation in standard form:
g
+
MN)j sin ¢H h g !g + )j sin ¢H hh ! g g + Fact 16.2.4 on page 221 )j sin ¢H ! h h g + )j sin Y'>( Fact 16.2.5 on page 221 h g ! + +!( )j sin ¢ 0 !<ü!(Î2¼9 ( 6 h
17.2. EXAMPLES OF SINUSOIDAL BEHAVIOR
241
This function is now in the standard form of Definition 17.1.1, so + it + G!Ç'!( , mean , is a sinusoidal function with phase shift ( . amplitude ])j and period (iii) Start with MN"! cos 6 j2ÌZ9e ! , then here are the steps to put the equation in standard form. A key simplifying step is to use the identity: cos 6 /:9Ò sin 6 ü/:9 .
MN"! cos 6 j2PH9´ü! ( "! sin üj2Y) ! "! sin j2Y # É> g ! ( "! sin r Y x { l
!
j
(
!
% 6 ;!A9 (
# É>
%
! h s
6 ;!A9
This function is now in the standard form of Definition 17.1.1, so it is 0_É; 2 , mean é;! , a sinusoidal function with phase shift l amplitude ]"! and period .
l
17.2 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
M½' sin
g
!( ¢ 9 6
h
and asked various questions. The answers typically require either direct calculation or interpretation of the constants. Example 17.1.3 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 , , , and . This is a typical scenario in a “mathematical modeling problem”: the process of observing data, THEN obtaining a mathematical formula. To find , take half the difference between the largest and smallest values of ! 6 @9 . The period
is most easily found by measuring the distance between two successive maxima (peaks) or minima (valleys) in the graph. The mean is the average of the largest and smallest values of ! 6 @9 . The phase shift (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 MW on its way up from a minimum to a maximum. The funny thing is that the phase shift is NOT unique; there are an infinite number of correct choices. One choice that will work
CHAPTER 17. SINUSOIDAL FUNCTIONS
242
is 6 coordinate of a maximum 9A . Any other choice of p from this one by a multiple of the period .
axis
axis axis
!
distance between two successive peaks (or valleys)
coordinate of a maximum max value min value !
will differ
max value min value
ements
+
axis (hours of daylight)
Example 17.2.1. Assume that the number of hours of daylight in Seattle is given by a sinusoidal function 6 /29 of time. ò 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 6 /:9 for the number of hours of daylight on the / day of the year.
"
Solution. Because the function 6 /:9 is assumed to be si nusoidal, it has the form MN' sin x 6 /m 9 { c , for con stants , , , and . 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 :
²
õ ² õ ö ô õ² ±² ô
²^ ^ ²m ^ ²m m^ ^²m õ õ Ü axis ö (days) ö
i ` ê Jcj ÿ [0! !
Figure 17.12: Hours of daylight in Seattle in 1994.
i i ` Jkj )jT !
To find the period, we need to compute the time between two successive maximum values of 6 /29 . To find this, we can simply double the time length of onehalf period, which would be the length of time between successive maximum and minimum values of 6 /:9 . This gives us the equation
"! 6 days between June 21 and December 21 9Ë ! 6 eJ:jT98)j^^
Locating the final constant requires the most thought. Recall, the longest day of the year is June 21, which is day 172 of the year, so
6 day with max daylight 9T
+
i j+^^ [ ! 'J% `
In summary, this shows that
i
g
!( 6 /:9Ò)jT sin /mJ% ` 9 H0!A h j^^ 6 A rough sketch, following the procedure outlined above, gives this graph on the domain %?'y/'éj^^ ; we have included the mean line M$ 0 ! for reference.
17.2. EXAMPLES OF SINUSOIDAL BEHAVIOR
243
We close with the example that started this section. Example 17.2.2. The depth of a migrating salmon below the water surface changes according to a sinusoidal function of time. The fish varies PSfrag replacements between 1 and 5 feet below the surface of the water. It takes the fish axis maximum + 1.571 minutes to move from its minimum depth to its successive axis depth. It is located at a maximum depth when /Í What c!J ` minutes. axis is the formula for the function 6 /:9 that predicts the depth of the fish after / 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 ' ÿ ` jT ) ! ! max depth min depth
u depth
9¼9 , for apSolution. We know that 6 /:9Ê) sin 6 6 ï propriate constants , , , and . We need to use the given information to extract these four constants. The amplitude and mean are easily found using the above formulas:
l
p
p
+
½ ! 6 ` [email protected]"jTa ! Finally, to find
we
(time of maximum depth)
+
+
+
c!J `
jTa+ !
)jT ` %
The formula is now
g
i !( + 6 /:98"! sin /müjT ` 9 jÉ"! sin 6 !n/m 9´üj 6 h jTd ! The depth of the salmon when it was first spotted is just
6 % 9Ò"! sin 6
i
ç
O
Figure 17.13: 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 17.13):
i
time
9TjÉ$&^J^ feet.
+ determining how many Finally, graphically, the last question amounts to times the graph of 6 /:9 crosses the line M on the domain [0,10]. This can be done using Figure 17.13. 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 17. SINUSOIDAL FUNCTIONS
244
17.3 Exercises Problem 17.1. Find the amplitude, period, a phase shift and the mean of the following sinusoidal functions. (a) (b) (c) (d) (e) (f) (g)
ÛY¡ sin  z¿Ì©>fª/Â? Û ¡?ª sin  f2¿/©? Y ÛY¡Ç sin ¿ÌÂDv /¹ÂÉv ÛY¡v ª sin « ¿Ì© ©?§:& Y ÛY¡v sin Â¢: ©e ÛY¡Ç¹  sin R: k &¿>ÂÍ /À/v©ÍR"/ ÛY¡ZRR sin &¿4©Ie & / Â
$# # #
%
#
%
%
%
Problem 17.2. A weight is attached to a spring suspended from a beam. At time ÄU¡ , 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 ¾ height Ä2/ is a sinusoidal function of time Ä . It first reaches a maximum height 0.6 seconds after starting. (a) Follow the procedure outlined in this ¾ section to sketch a rough graph of Ä2/ . 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. (d) Write down a formula for the function ¾ Ä2/ in standard sinusoidal form; i.e. as in 17.1.1 on Page 236. (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.)
machine begins recording a plot of volume per breath versus time (in seconds). Let 7ª Ä2/ be a function of time Ä that tells us the volume (in liters) of a breath that starts at time Ä . 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 7ª Ä2/ 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 17.4. Suppose the high tide in Seattle occurs at ¸ a.m. and ¸ R p.m. at which time the water is ¥ feet above the height of low tide. Low tides occur ª hours after high tides. Suppose there are two high tides and two low tides every day and the height of the tide varies sinusoidally.
¾
(a) Find a formula for the function Û¡  Ä2/ that computes the height of the tide above low tide at time Ä . (In other words, ÛÉ¡Ç corresponds to low tide.) (b) What is the tide height at
R¸ R
a.m.?
Problem 17.5. Use the method of Section 17.1.1 to roughly sketch the graph of the given sinusoidal function: Û[¡ R sin  R _fÇÂ UfÄ2/¹ÂÉ . Problem 17.6. Your seat on a Ferris Wheel is at the indicated position at time Ä8¡ .
Start
Problem 17.3. A respiratory ailment called PSfrag “CheyneStokes Respiration” causes thereplacements volaxis ume per breath to increase and decrease in a axis sinusoidal manner, as a function of time. For axis one particular patient with this condition, a
53 feet
17.3. EXERCISES
245
Let Ä 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 ,+Ä2/ be your height above the ground at time Ä . (a) Argue that ,+ Ä2/ is a sinusoidal function, describing the amplitude, phase shift, period and mean. (b) When are the first and second times you PSfrag replacements are exactly 28 feet above the ground? (c) After 29 seconds, how many times will you have been exactly 28 feet above the ground?
Problem 17.7. The angle of elevation of the ± on sun above the horizon at noon is December 21, the 355th day of the year. The el ± evation is at noon on June 21, the 172nd day of the year. These elevations are the minimum and maximum elevations at noon for the year. Assume this particular year has 366 days and that the elevation on day Ä is given by a sinusoidal function F Ä2/ . (a) Follow the procedure outlined in this section to sketch a rough graph of F Ä2/ . Draw at least two complete cycles of the oscillation, indicating where the maxima and minima occur. What are the mean, amplitude, phase shift and period for this function? (b) Give four different possible values for the phase shift. (c) Write down a formula for the function F Ä2/ . Confirm it is consistent with the graph in (a). (d) What is the elevation at noon February 7? (January has 31 days and we’re counting January 1 as the first day of the year.)
Problem 17.8. Here is a graph of the function ÛY¡ sin ¿k/Â sin R¿k/}
axis axis axis
Is this function sinusoidal? Give a reason.
Problem 17.9. In Exercise 15.15, 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 ùH Ä2/ of the bug at time Ä . Pick three of the scenarios below and answer these two questions:
(a) Both coordinates of ùHÄ2/ are sinusoidal functions in the variable Ä ; i.e. ùJÄ2/7¡ &¿Q Ä2/ ÛJ Ä2/À/ . Put ¿QÄ2/ and Û Ä2/ in standard sinusoidal form. Find the amplitude, mean, period and phase shift for each function.
(b) Sketch a rough graph of the functions ¿QÄ2/ and Û Ä2/ in (a) on the domain hIÄÖh .
CHAPTER 17. SINUSOIDAL FUNCTIONS
246 ω=4π/9 rad/sec bug lands here
ω=4π/9 rad/sec 1.2 rad bug lands here 2 in
2 in
bug lands here
ω= − 4π/9 rad/sec 1.2 rad
bug lands here
2 in
2 in
ω=4π/9 rad/sec
ω= − 4π/9 rad/sec
ω=4π/9 rad/sec
PSfrag replacements axis axis axis
o 40 2 in
bug lands here 0.5 rad
2 in bug lands here
Chapter 18 PSfrag replacements
Inverse Circular Functions
axis axis
An aircraft is flying at an altitude e% miles above the elevation of an airport. If the airplane begins a steady descent 100 miles from the airport, what is the angle of descent? The only natural circular function we can use is î tan 6 A9 , leading to the equation: tan 6 A 9Ò
10 miles
axis
100 miles
Figure 18.1: An aircraft decending toward an airport.
e % e%% e %
The problem is that this equation does not tell us the value of . Moreover, none of the equation solving techniques at our disposal (which all amount to algebraic manipulations) will help us solve the equation for . æ What we need is an inverse function Ê! 6µî 9 ; then we could use the æ fact that tan 6 tan 6 A99Ò and obtain:
I tanæ 6 tan 6 A9¼9Ê tanæ
g
e % h
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 i æ find tan Ìx O { ` . Punch this into your calculator and verify it!
18.1 Solving Three Equations Example 18.1.1. Find all values of true: sin 6 A9ËÏ .
(an angle) that make this equation
Solution. We begin with a graphical reinterpretation: the solutions correspond to the places where the graphs of î sin 6 A9 and î intersect in the î coordinate system. Recalling Figure 16.11, we can picture these two graphs simultaneously as below: 247
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
248
í cross
cross à
æ
æ î í
Graph of à
sin
cross
axis
1
Graph of
ð
−1
cross
l
à
öõ axis
one period
Figure 18.2: Where does sin 
¯T/
cross ä
¡ ¨
?
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 18.1.1! 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 “ ” and “ ”, then all other crossing points are obtained by adding multiples of !( to either “ ” or “ ”. By Table 15.1, radians is a special angle where we computed sin x { , which tells us that the crossing point labeled “A” is the point 6 1 9 . Using the identities in Facts 16.2.4 and 16.2.5, notice that
g
sin
` ( sin ¹ ( ü ( ^#h ^ ( H sin ^ ( H sin ^ ( sin ^ !
So, W u is the only other angle between 0 and !( such that Ex ample 18.1.1 holds. This corresponds to the crossing point labeled “B”, which has coordinates x u 1 { . In view of the remarks above, the crossing points come in two flavors:
g
g
( ^
ü!(>1
1
! h
` ( ü !(>1 1 ^ !8h
H%´1[¦In1[¦
Taking this example as a model, we can tackle the more general problem: For a fixed real number ï , describe the solution(s) of the equation
18.1. SOLVING THREE EQUATIONS
249
ïN+! 6 A9 for each of the circular functions î ã! 6 A9 . Studying solutions
of these equations will force us to come to grips with three important issues:
For what values of ï does ! 6 A9Ò ï have a solution? For a given value of ï , how many solutions does ! 6 A 9Êðï have? Can we restrict the domain so that the resulting function is onetoone?
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 18.1.2. None of the circular functions is onetoone on the domain of all values. The equations ï sin 6 A9 and ïÇ cos 6 A9 have a solution if and only if É=' ïB' ; if ï is in this range, there are infinitely many solutions. The equation ï tan 6 A9 has a solution for any value of ï and there are infinitely many solutions.
PSfrag replacements Example 18.1.3. If two sides of a right triangle have lengths 1 and ) j as pictured below, what are the acute axis angles Õ and Ö ? axis Solution. By the Pythagorean Theorem the remaining side has length
K > &) j
Ó
axis
Ò
l
!A
Since tan 6 Õ49< ) j , we need to solve this equation for Õ . Figure 18.3: What are the Graphically, we need to determine where î ) j crosses values for U and V ? the graph of the tangent function: From Fact 18.1.2, there will be infinitely many solutions to our equa tion, but notice that there is exactly one solution in the interval å 1 and we can find it using Table 15.1:
& '
tan
Õ So, )
( j
tan 6 ^% 9Ò
sin 6 ^% 9 cos 6 ^% 9
l
x2 {
) jT
Ö radians y^% is the only acute angle solution and
^% o% l)j% .
eJ%
PSfrag replacements axis axis
axis250
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
@í æ ú
à
ö
axis
í
one period
²ú
î
ö
only place graphs cross in this period is
ú ë ð l
etc.
etc.
æ
ó æ
í
axis
l
í
@í æ ú ö
ú
ú
ö
ö
Figure 18.4: Where does the line ä
¡UÈ
cross ä
¡
tan ¯ ?
18.2 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 15.1.) 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
æ
✘ sin µ6 î ï 9 solutions æ Ë ✘ cos 6µî 9 solutions
of the equation î sin 6 A9 of the equation î cos 6 A9
æ ✘ tan 6µî 9Ò solutions of the equation î tan 6 A9
(18.1)
There are two main problems with these rules as they stand. First, to æ æ have a solution in the case of sin 6µî 9 and cos 6µî 9 , we need to restrict î so that É£' î 'é . Secondly, having made this restriction on î in the first two cases, there is no unique solution; rather, there are an infinite æ æ æ number of solutions. This means that the rules sin , cos , and tan 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 values on which it becomes onetoone, then we can appeal to Fact 11.3.1 and conclude the inverse function makes sense.
18.2. INVERSE CIRCULAR FUNCTIONS
251
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 î A! 6 A9 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 ! 6 A9 should include the angles between 0 and , since these are the possible acute angles in a right triangle.
On the restricted domain, the function ! 6 A9 should take on all possible values in the range of ! 6 A9 . In addition, the function should be onetoone on this restricted domain.
The function ! 6 A 9 should be “continuous” on this restricted domain;
eplacements
i.e. the graph on this domain could be traced with a pencil, without lifting it off the paper.
In the case of î sin 6 A9 , the principal domain '[0' satisfies our criteria and the picture is given below. Notice, we would not want to take axis the interval %Q'0'"( , since î sin 6 A9 doesn’t achieve negative values on this domain; in addition, it’s not onetoone there. à
æ
æ ú
Ú
axis 1
ú
ö −1
axis
ö
axis
principal domain on the unit circle
S
axis
sine restricted to principal domain
Figure 18.5: Principal domain for sin ü¯T/ .
In the case of î cos 6 A9 , the principal domain %<' ï'W( satisfies our criteria and the picture is given below. Notice, we would not want to take ' ï' , since î cos 6 A9 doesn’t achieve negative values the interval on this domain; in addition, it’s not onetoone there. In the case of î tan 6 A9 , the principal domain X X satisfies our criteria and the picture is given below. Notice, we would not want to take the interval %§' ï' ( , since î tan 6 A9 does not have a continuous graph on this interval; in other words, we do not include the endpoints since tan 6 A9 is undefined there. Important Facts 18.2.1 (Inverse circular functions). Restricting each æ circular function to its principal domain, its inverse rule ! 6 î 9F will define a function.
axis
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
252 à
æ ú
æ
Ú
axis 1
axis
ú
ö
ö
ú
ö
principal domain on the unit circle
axis
S
axis
cosine restricted to principal domain
−1
¯T/ .
Figure 18.6: Principal domain for cos 
æ , then sin 6µî 9 is the unique angle in the principal with the property that sin 6 A9Ò î . ' B' æ (ii) If ÉT' î ' , then cos µ6 î 9 is the unique angle in the principal domain %>'[0'"( with the property that cos 6 A9Ë î . æ (iii) For any real number î , tan µ6 î 9 is the unique angle in the principal domain X X with the property that tan 6 A9Ê î . (i) If É#' î domain
'
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:
ï
ï ï
cos 6 A9 sin 6 A9 ý tan 6 A9
these are called the principal solutions. We also can refer to these as the æ principal values of the inverse circular function rules &! 6µî 9 .
ements axis axis
!!!
axis
CAUTION !!!
As usual, be careful with “radian mode” and “degree mode” when Ì making calculations. For example, if your calculator is in “degree” » ”, the answer given is “86.82”. This mode and you type in “tan 8 º has tan Ì 8 º If your means that an angle of measure È ÕÉ calculator is in “radian” mode and you type in “sin ”, the answer º radians given is “1.12”. É This means that an angle of measure È has sin º
( ) ) ) +*
(
) )
)
There is a key property of the inverse circular functions which is useful in equation solving; it is just a direct translation of Fact 11.3.2 into our current context: Important Facts 18.2.2 (Composition identities). We have the following equalities involving compositions of circular functions and their inverses:
eplacements 18.3. APPLICATIONS
253
axis à
axis
Ú
axis
principal domain on the unit circle
1
æ ú
æ
ö
ú −1
ö
S
axis
axis
tangent function restricted to principal domain
Figure 18.7: Principal domain for tan ü¯T/ .
[£'
(a) If '
æ
, then sin 6 sin 6 A 99Ò .
æ
(b) If %Q'[£'"( , then cos 6 cos 6 A 99Ò
(c) If X X
æ
.
9¼9ÊA , then tan 6 tan 6 A
We have been very explicit about the allowed values for the equations in Fact 18.2.2. This is important and an Exercise will touch on this issue.
18.3 Applications As a simple application of Fact 18.2.2, we can return to the beginning of this section and justify the reasoning used to find the angle of descent of the aircraft:
I tanæ 6 6 tan 6 A99Ò tanæ
g
i i H%&%oo^ rad ` e % h
Let’s look at some other applications. Example 18.3.1. Find two acute angles is satisfied:
! ` x cos 6 A9 { cos 6 A9 e^
so that the following equation
o
+
9 and Solution. Begin by multiplying each side of the equation by cos 6 A rearranging terms: +o %
! ` x > cos 6 A9 { cos 6 A9 e^
! ` ! ` +o cosp 6 A9T cos 6 A9T e^ e ^
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
254
To solve this equation for , 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 î cos 6 A9 :
! ` ! ` +o %Y î î e^ e ^
Applying the quadratic formula, we obtain
K
ö î + ö u u xä { p H%&o:jJ^ or %&%^j^
öð î u
We now use the fact that î cos 6 A9 and note the cosine of an acute angle
ö u ¦
is nonnegative to conclude that
cos 6 A9Ò'%o:jJ^ %o^JJ cos 6 A9Ê'
or cos 6 A9ÒH%%+ ^j^ ii 9 '%c! or cos 6 AÒ
Finally, we use the inverse cosine function to arrive at our two acute angle solutions: + æ I cos 6 %o+^JJ98[ kj ` cos 6 A9Ê'%o+ ^JJ cos 6 A 9Ê'%c!
ii
I
æ
cos 6 %c!
ii
98
i
` ^^
ements axis axis axis

è ö
Ý
// 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 .,., // 00 .,., // 00 // 00 // // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 // 00 //
Example 18.3.2. A 32 ft ladder leans against a building (as shown below) making an angle Õ with the wall. OSHA (Occupational Safety and Health Administration)++specifies a “safety range” for the angle Õ to be ` ' Õ3' . If the base of the ladder is ü e% feet from the house, is this a safe placement? Find the highest and lowest points safely accessible.
O
Solution. If ôe% , then sin 6 Õ49w , so the principal soæ O l lution is Õ sin x { eJc!A ; this lies within the safety l zone. From the picture, it is clear that the highest point safely reached ++ the height will occur precisely when ÕNÿ ` and as this angle increases, * decreases until we reach the lowest safe height when Õ# . We need * + + + + to solve two right triangles. If Õ.G `= , then j2! cos 6 ` 9ïyj%o ft. If * , then cos 6 9Ê"!njT&%v! ft. Õ ) 2 j !
Figure 18.8: A ladder problem.
Example 18.3.3. 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?
18.3. APPLICATIONS
255 PSfrag replacements
Solution. A picture of the situation is shown in Fig axis ure 18.9(a). After a look at the picture, three right triangles pop out and beg to be exploited. We highlight these in the Figure 18.9(b), by imposing a coordinate system and labeling the various sides of our triangles. We will work in radian units and label to be the required intercept heading. We will first determine the sides Y½M and L# of the large right triangle. To do this, we have
43
Ç[ ` M [e% é[ ` L[e%
3
¹
distance traveled time elapsed
+
! _ ! ` %&% `
+
+
² radQ¯ ô South
West
²
õ
õ
miles
rad
miles
East prop plane spotted here
East intercept heading
(a) The physical layout.
axis
Ý ² ô
+
;
õ
Jet
Ú
Jet
+ principal soluD We now have tan 6 A9â Ú W%&J % ` , so the ^D S tion is '%^oo radians, which is about %% ` . This is the 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 + + big right triangle: of the hypotenuse
of the i þ 6 eJ ` ^ 9 6 ` &^%^9 G! c! ` miles. You need to travel this distance in the same amount of time it takes the O prop plane to travel 10 miles at 200 mph; i.e. O¥O %% ` hours. Thus, the intercept speed is w
find speed to intercept
+
cos 6 % ` 98[jT+ ae^ miles 1 cos 6 %98 ` %:j miles 1 i sin 6 % ` 9w a+ eo miles 1 and sin 6 &%9w'J ` miles
Intercept point
North
rad
õ
10
15
Á
1
Intercept point rad
Á á ~
~
S
2 1á2
prop plane spotted here
axis
(b) Modeling the problem. Figure 18.9: Visualizing the Coast Guard problem.
J ` mph
Later you will use an alternative approach to this problem using velocity vectors. In certain applications, knowledge of the principal solutions for the equations ï cos 6 A9 , ï sin 6 A9 , and ï tan 6 A9 is not sufficient. Here is a typical example of this, illustrating the reasoning required. Example 18.3.4. 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?
axis axis axis
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
256
Ò ßp
ft
leaving ground
ements
ft
Ó
ft
airborne
returning to ground
Figure 18.10: Various angles of a vaulter’s pole.
í Ú axis Ò Ú Ó ±
axis
×
Solution. From the obvious right triangles in the picture, we are interested in finding angles where sin 6 A9Ë l . ¾ ßp The idea is to proceed in three steps:
Ö S
Find the principal solution of the equation sin 6 A9Ê l ;
axis
Find all solutions of the equation sin 6 A9Ê l ;
¾
¾
Use the constraints of the problem to find Õ and Ö
Figure 18.11: Modeling the problem with a unit circle.
among the set of all solutions.
Solving the equation sin 6 A9Ê l involves finding the points ¾ on the unit circle with M coordinate equal to l . From the ¾ picture, we see there are two such points, labeled Ø and Ù . The coordinates of these points will be ØF x cos 6 Õ49z1 l { and Ù x cos 6 ÖÊ9z1 l { . ¾ ¾ Notice, Õ is the principal solution of our equation sin 6 A9 l , since æ ¾ %T' Õ 'ôo% ; so Õq sin x l { ! ` kjJ . In general, the solutions come ¾ in two basic flavors:
or
where q
Õ? ! 6 eJ% 9
! ` kjJ ! 6 eJ% 9z1
ÖI ! 6 eJ% 91
%´18¦yn18¦]!18¦[jA1ee . To find the angle properties of the circular functions:
Ö
, we can use basic
Ö 9Ò sin 6 4 Õ 9Ò. sin 6 $Õ>9Ò sin 6 eJ% dÕ49Ò sin 6 ` sin 6 Ê This tells us
+
+
^v! 9
Ö $ ` &^v! .
18.4 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
18.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS
257
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 ïP sin 6 A9z1uïP cos 6 A9 , and ïP tan 6 A9 :
The first strategy is summarized in Procedure 18.4.1. 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 Ex
ample 18.4.2 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 17 or using a graphing device).
Each approach has its merits as you will see in the exercises. Important Procedure 18.4.1. To find ALL solutions to the equations ïI sin 6 A9z1uïï cos 6 A9 , and ïP tan 6 A9 , we can lay out a foolproof strategy.
í
Step 1. Find principal solution
í
í
65 65 8797:7
ORë ë ë
4. Obtain general principal solutions 5. Obtain general symmetry solutions
õ ðî
sin
æ
í
î
Cosine case
2. Find symmetry solution 3. Write out multiples of period
í
î
Sine case
í
í
æ
æ
í
õ ðî D
sin
õ ð
cos
í î
sin
õ ðî
cos
õ ðD
sin
õ ðD D
æ
õ ðî D
cos
î
õ ð
tan
not applicable
í
î cos
í
Tangent case
õ ðD
î tan
õ ðD
not applicable
Example 18.4.2. Assume that the number of hours of daylight in your
i hometown during 1994 is given by the function 6 /29Ò"jT sin x 6 /mJ% ` 9 { l ¥ 0!1 where / 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 M G 6 /29 on the domain %%'q/B'qj^^ . If you apply the graphing procedure discussed in Chapter 17, you obtain the sinusoidal graph below on the larger doi j2!A (The reason we use a larger domain is so that main j^^å' /T' the “pattern” that will arise in the strategy described below is more evident. Ultimately, we will restrict our attention to the smaller domain %('ÿ/C'ÿj^^ .) To determine + when there will be 14 hours of daylight, we need to solve the equation $ 6 /:9 . Graphically, this amounts to finding
PSfrag replacements axis258
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
Ú
axis
u
u O ò
¼ u ¾
æ O¥O
O¥O
axis
p O¥O
Figure 18.12: Where does the sinusoidal function
O¥O ,+Ä2/
cross the line ÛY¡Z .
+
the places where the line M'Ï intersects the graph of 6 /:9 . As can be seen in Figure 18.12, 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 $ 6 /:9 ): 1. Principal Solution. We will find one solution by using the inverse sine function. If we start with the function M½ sin 6 /:9 on its principal æ domain ' /T' , then we can compute the domain of 6 /:9Í i jT sin x 6 /4J% ` 9 { )0! :
l¥
! ( ( /mJ% ` 9Ö' 6 ! j^^ i ! É)'./R' !A
(
'
+ sine function we can find the principal soluNow, using the inverse tion to the equation [ 6 /:9 :
+
+
i
+
% ` % ` i % ` e%J
/
g
jT sin g !( sin j^^ æ sin 6 % ` jTJ
! ( /4J% ` 9 H0! h j^^ 6 +
(18.2)
6 /4J% ` 9 h + ( % ` 98 /ÌêJ% ` 9 eJ:j 6 i
Notice, this answer is in the domain É('Ï/d' ! . In effect, we + have found THE ONLY SOLUTION on this domain. Conclude that there will be about 14 hours of daylight on the th day of the year. 1
You can get the actual data from the naval observatory at this world wide web address: õ http://tycho.usno.navy.mil/time.html ¬
18.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS
eplacements axis
259
+
2. Symmetry Solution. To find another solution to the equation 6 /:9 , we will use symmetry properties of the graph of MB 6 /:9 . This is where having the graph of M) 6 /29 is most useful. We know the i i a maxima on the graph occurs at the point è 6 !1 ` 9 ; review Example 17.2.1 for a discussion of why this is the case. From the graph, we can see there are two symmetrically located intersection + of è . The principal solution gives the intersecpoints on either side tion point 6 jT&J´1 9 . This point is 58.2 horizontal units to the left of + + è ; see the picture i below. So a symmetrically positioned intersection point will be 6 !< ` J_!1 [email protected] 6 !nj%_!1 9 .
ì
Ú
axis
u
u O ò
¼ u ¾
æ O¥O
O¥O
p O¥O
O¥O
axis
Figure 18.13: Finding the symetry solution.
In other words, /Iÿ!nj%c! is a second solution to the equation 6 /:9 . We call this the symmetry solution.
+
+
3. Other Solutions. To find all other solutions of 6 /:9 , we add integer multiples of the period bj^^ to the / coordinates of the principal and symmetric intersection points. On the domain j^^Q' i /*' j2! we get these six intersection points; refer to the picture of the graph:
+
+
+
i
+
6 ;! ` A! _!1 + 9¼1 6 jTJ´1 + 91 6 o J´1 + 91 6 Éj ` &J´1 ¼9 1 6 !nj%c!1 9 1 6a` o^ c!1 9¼ i So, on the + domain j^B ^ ' i / ' j2! we have these six solutions to the equation $ 6 /:9 : + i /H;! ` !Ac!1zÉj ` J´1jT&J´1!nj%_!1 o&J´1 ` o^_!T To conclude the problem, we only are interested in solutions in the domain %§'[/.'[j^^ , so the answers are /YéjTJ´1!nj%_! ; i.e. on days 114 and 230 there will be about 14 hours of daylight.
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
260
18.5 Exercises Problem 18.1. 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:
Z
(a1) sin ¨ Z « ,v
&¿/ ,
for
¨¥¨ Z
(a2) cos ¨2&¿/ , for Z « ,v
¨¥¨ Z
(a3) tan ¨ Z « ,v
¿/ ,
for
¿q¡C , ,©4 , « , ª¼
,
¿]¡G , ,©4 , « , ª¼
,
, ,©4 , « , ª¼
,
¿ ¡
¨¥¨
(b) Redo part (a) with your calculator set in “degree mode”. (c) Find four solutions of sin z¿vÂP¿â©Ç"/¡ .
(d) Find four solutions of ¿ :
2/¡
tan R¿¹ÌÂ
¿3©
Problem 18.4. Suppose
ÿMÄ2/¹¡
axis axis
(b) On what days of the year will there be approximately 10 hours of daylight?
axis
e
¿/
r&¿/
Sketch the graphs of same set of axes.
Set &¿/¡§jT¿/ and find the principal and symmetry solutions.
Indicate the other solutions of the equae tion ¿/¡"jr&¿/ on your graph and describe their relationship to the principal and symmetry solutions.
and
on the
j
e
# ¿Ì© %% #ó¿ÌÂ4¦
(a)
e
(b)
e e
(c) (d)
e
, jr&¿/¡ ¨ . ¿/¡ sin « , jr&¿/¡?©> . ¿/ ¡ sin ¿k/ ¡ sin z¿4©?"/ , jT&¿/¡ ¨ . º ¿/ ¡Z¥ cos  z¿mÂÍ"/¹©É , jr&¿/¡©>
.
bf  Ä© / Â ªRª n z
Problem 18.5. Suppose that you want to run a cable from a power source ù out to an offshore buoy C . The power source is located on land, 200 meters from a straight section of the shoreline, and the buoy is located in the water, 100 meters out from a point 600 meters downshore from ù ; see picture. The cable will run in a straight segment from ù to the point ç on the shoreline, and then from ç in another straight segment out to C .
(a) Find the number of hours of daylight on January 1, May 18 and October 5.
Problem 18.3. For each part of the problem below:
sin
is the temperature (in degrees Fahrenheit) at time Ä , where Ä is measured in hours after midnight on Sunday. You paint the exterior door to your house at p.m. on Monday. The paint information states that 48 hours of Ã F drying time is required; i.e., you can only count time periods when the temperature is at least Ã F. When will the door be dry?
Problem 18.2. Assume that the number of hours of daylight in New Orleans in 1994 is E given by the function &¿/¡ ¤ sin ¿ Â « « of «¼days ¦ « P where x represents the number after March 21. PSfrag replacements200m
# %
m
θ
S
shoreline 100m B
(a) Find a formula for the total length of the cable as a function of the pictured angle ¯ . (b) What value of ¯ will minimize the total length? Note: You do not need to use a graphing calculator (or calculus) to minimize the length. Use common sense instead. (Hint: What is the shortest distance between two points?)
Problem 18.6. 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.
18.5. EXERCISES
261
r= 100 yards
satellite
Earth
0.025 rad/sec
Tiff starts here
horizon circle
Michael starts here
eplacements
center of Earth
axis 0.03 rad/sec axis axis
(a) Where is each runner located (in PSfrag replacements xycoordinates) after 8 seconds?
Earth
axis axis
(b) How far has each runner traveled after 8 seconds?
axis
(c) Find the angle swept out by Michael after Ä seconds.
satellite
α α t CROSS−SECTION
U
in terms of Ä .
(d) Find the angle swept out by Tiffany after Ä seconds.
(a) Find a formula for
(e) Find the xycoordinates of Michael and Tiffany after Ä seconds.
(b) If Ä>¡F RR 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?
(f) Find the first time when Michael’s xcoordinate is 50. (g) Find the first time when Tiffany’s xcoordinate is 50. (h) Find when Tiffany passes Michael the first time.
(c) If Ä;¡)R RR 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?
(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 18.7. Let ÛÉ¡
e
&¿/¡F¥
;º &¿v©Ò*/À/aÂÊª
sin 
.
&¿/ on the do(a) Sketch the graph of ÛF¡ main ªhÍ¿=h . Specify the amplitude, a phase shift, period and mean for this function.
(b) Sketch the graph of ÛÇ¡ main hD¿h . Is ÛY¡ function?
e
¨e
¨e
(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 Ä ?
¨
 ¿k¨ / on the do¿/ a sinusoidal
Problem 18.8. A communications satellite orbits the earth Ä 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 twodimensional crosssectional picture we can use to study the size of the horizon slice:
Problem 18.9. Flowing water causes a water wheel to turn in a CLOCKWISE direction. The wheel has radius 10 feet and has an angular speed of ¿ ¡?©80 rad/sec. Impose a coordinate system with the center of the wheel as the origin. The profile of the flowing water is pictured below. At time Ä¡# , an unlucky salmon gets caught in the wheel at the pictured location. Distance units will be “feet” and time units will be “seconds”.
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
262
Ú
axis
(d) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is at least 410 ± G .
water wheel (radius 10 feet)
20 rad/sec
θ
PSfrag replacements
9 degrees
(25,0)
S
axis
(f) During the first 20 minutes of baking, calculate the total amount of time the ± oven temperature is between 410 G and ± 425 G .
water profile unlucky salmon swimming upstream gets stuck
axisright here
(a) How long does it take the salmon to complete one revolution? (b) What is the salmon’s linear speed? (c) What is the equation of the line modeling the profile of the flowing water? (d) Where (¿0Û coordinates) does the salmon get stuck on the wheel? (e) What is the initial angle
¯
pictured?
(f) Let ùJÄ2/;¡ &¿QÄ2/} Û Ä2/À/ be the location of the salmon at time Ä seconds; find the formulas for ¿QÄ2/ and ÛÄ2/ .
(g) Where (¿0Û coordinates) is the salmon located after 0.12 seconds?
(h) When is the salmon first 4 feet to the right of the Û axis? Problem 18.10. 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 Ä minutes after inserting the scones is given by
ÛÉ¡ :P Ä2/¡Z§
m
sin
f
(e) During the first 20 minutes of baking, calculate the total amount of time the ± oven temperature is at most 425 G .
Ä ´© _ f n
ÂY2§
(a) Find the amplitude, phase shift, period and mean for :P Ä2/ , then sketch the graph on the domain kh ÄÖhI0 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.
Problem 18.11. 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 Ã F and a minimum temperature of 240 Ã . Once the temperature hits 300 Ã , it takes 20 minutes before it is 300 Ã again. Gavin’s cake needs to be in the oven for 30 minutes at temperatures at or above 280 Ã . He puts the cake into the oven when it is at 270 Ã and rising. How long will Gavin need to leave the cake in the oven? Problem 18.12. 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. In the 24 hours after Maria began observing Elasticman, how much of the time will Elasticman be less than 6 feet tall? Problem 18.13. The angle of elevation of the sun above the horizon at noon is approximately a sinusoidal function of the day of the year. The amplitude of this function is 23.452 Ã ; this is the angle at which the earth’s axis is tipped. The period is 365.25 days. The maximum elevation occurs on June 21, the 172nd day of the year. If the latitude where the observation is made is ¯ degrees, then the mean elevation of the sun is zÊ©@¯ degrees. (a) Find a sinusoidal function F Ä2/ which approximates the elevation of the sun on day Ä of the year. (Your answer will involve the unknown latitude ¯ .)
(b) At noon on the 74th day of the year, you measure the elevation of the sun to be 39 Ã 39’ on the UW campus. Use your answer to part (a) to determine the latitude ¯ of the UW campus. Give your answer in degrees and minutes.
18.5. EXERCISES
263
(c) If the latitude ¯ ¡ 40 Ã , find the time(s) when the angle of elevation of the sun at noon is RÃ . (d) Imagine you are way up north in Alaska with a latitude ¯3¡ 70 Ã . Sketch the graph of the function F  Ä2/ . Find the day(s) when the sun never rises. Problem 18.14. Answer the following questions:
hÍ¿Nh (a) If Û ¡ sin ¿/ on the domain © E , ° what is the domain and range of ÛY¡Ç sin  z¿T©¢"/¼Â ? How many solutions does the equation 3¡ sin  ¿:©"/ßÂÊ have E on the domain and what are they?
© hUÄ²h , (b) If ÛÇ¡ sin  Ä2/ on the domain E ° what is the domain and range of ÛY¡ sin  ¨ Ä©â¹ */f/»Â3 How many solutions does the equation R¡ sin   Ä© */À/§Âw have on the domain E and¨ what are they?
=<
;<
Ûê¡ sin Ä2/ on the domain © h'ÄBh E , what is the domain and range ° I/À/8Â# . How of Ûq¡y sin   Ä>© ¼ « ¥ ¦ ¦ many solutions does the equation ¡ sin  E « ¦¦ Ä@© &/À/nÂ have on the domain and what are they?
(c) If
(d) If Û½¡ cos &¿/ on the domain Qh ¿*h?f , E what is the domain and range ° of ÛÉ¡I cos  z¿Â 2/¥©3 ? How many solutions does the equation ©4¡ cos  z¿3ÂZ"/n©Ç E have on the domain and what are they?
(e) If Û ¡ tan ¿/ on the domain © E õÍ° ¿ofõ ÛÉ ¡ , what is the domain and range tan ó©@¿Â&/ÂZ ? How many solutions does the equation ¥R4¡ tan ó©@¿Â&/zÂÉ E have on the domain and what are they? Problem 18.15. Let’s make sure we can handle the symbolic and mechanical aspects of working with the inverse trigonometric functions:
> Â ¿© 2/¡ ¨ z¿ Â¢¿m©?"/¡?
(a) Find four solutions of tan z¿ (b) Solve for ¿ : tan
Z
Problem 18.16. Find the angle the pictures in Figure 18.16.
¯
.
in each of
Problem 18.17. Find the unknown sides and angles in each of the triangles in Figure 18.17.
CHAPTER 18. INVERSE CIRCULAR FUNCTIONS
264
r=3
r=1
r=22
θ
θ
θ y=  0.2
x=10
slope= 5 perpendicular to radial line
r=2
r=5
r=20
θ
θ θ
ag replacements axis axis axis
y=3x
12
x=4 Figure 18.14: Circles for Exercise 18.16.
10
x
β
α
4
α
9
α
axis
y
x
4 β y
x
16
β
α
x 20
ag replacements
1 rad
12
6
α
x
γ
axis axis
Figure 18.15: Triangles for Exercise 18.17.
20 o 6
5
Chapter 19 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
eplacements axis
t=2 hours
axis axis
t=3 hours
Figure 19.1: Observing cell growth.
The question is this: Can we find a function of / that will predict (i.e. model) the number of yeast cells after / hours? If we ò tabulate some data (as at right), the conclusion is that the formula
« 6 /:98"! predicts the number of yeast cells after / hours. Now, let’s make a very slight change. Suppose that instead of starting with a single cell, we begin with a population of j#,"e% cells; a more realistic situation. If we assume 265
Total hours 0 1 2 3 4 5 6 Table
Number of yeast cells
O
1=û 2=û 4=û 8=û l 16=û p 32=û u 64=û
19.1: Cell data.
growth
CHAPTER 19. EXPONENTIAL FUNCTIONS
266
ò cells will double every hour, then that the population of reasoning as above will lead us to conclude that the formula
« 6 /298 6 jÇ,üe% 9 ! gives the population of cells after / hours. Now, as long as / represents a nonnegative integer, we know how to calculate « 6 /:9 . For example, if /;.^ , then
« 6 /298 6 jÇ,üe% 9 !
6 jÇ,üe% 9 6 ! + E ! E ! E ! E ! E !A9 6 jÇ,üe% 9§^ [eov!?,üe%
The key point is that computing « 6 /:9 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
« 6 /298 6 jÇ,üe% 9 ! lH¾ and the rules of arithmetic do not suffice to calculate « 6 /:9 . We are stuck, since we must understand the meaning of an expression like ! l ¾ In order to proceed, we will need to review the algebra required to make sense of raising a number (such as 2) to a noninteger power. Weæ need to understand the precise meaning of expressions like: != lH¾ , ! u , ! , etc.
19.1 Functions of Exponential Type
PSfrag replacements axis
y=b
x
this is a variable this is a fixed number
y=x
b
this is a fixed positive integer this is a variable
axis axis
Exponential Picture
Monomial Picture
Figure 19.2: Viewing the difference between exponential and monomial functions.
On a symbolic level, the class of functions we are trying to motivate is ç easily introduced. We have already studied the monomials M ) , where was our input variable and Q was a fixed positive integer exponent. What happens if we turn this around, interchanging and Q , defining a new rule:
MN#! 6 9Ê)Q S
(19.1)
19.1. FUNCTIONS OF EXPONENTIAL TYPE
267
19.1.1 Reviewing the Rules of Exponents
'QÌ ò
3
PSfrag replacements
¾
/
axis
but it is possible to establish a reasonable comfort level + by handling the case when is a rational number. If Q1ò)% and Ä is a positive integer (i.e. Ä [ , ! , j , , ee ), then we can try to solve the equation
axis
axis
To be completely honest, making sense of the expression
M ÏQTS for all numbers requires the tools of Calculus,
PSfrag replacements
We refer to as the power and Q 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 19.2).
2
n=2 n=4 n=6
1 0
(19.2)
1.5
1
0.5
0
0.5
1
1.5
1
2
ò
the largest real
Thus, whereas ¦ are both In order to manipulate M basic facts from algebra.
+
Ä
root of QÌ
axis
3
3 2
0 1.5
1
0.5
0
A
0.5
1
1.5
1 2 3
Ö
Figure 19.3: Even and odd monomials.
(19.3)
roots of 1, we have defined ) w[ . QTS for rational , we need to recall some
Important Facts 19.1.1 (Working with rational exponents). For all positive integers ´ and ð , and any real number base Q½V % , we have
@?A
n=3 n=5 n=7
1
>
) QU'Q õ
axis
axis
A solution / to this equation is called an Ä root of Q . This leads to complications, depending on whether Ä is even or odd. In the odd case, for any real number Q , notice that the graph of M .Q will always cross the graph ¾ of M "/ exactly once, leading to one solution of (19.2). On the other hand, if Ä is even and Q$X % , then the ¾ graph of MN)/ will miss the graph of M )Q , implying there are no solutions to the equation in (19.2). (There will be complex solutions to equations such as / É , involving the imaginary complex numbers ¦ Y ¦ ) É , but we are only working with real numbers in this course.) Also, again in the case when Ä is even, it can happen that there are two solutions to (19.2). We ò do not want to constantly worry about this even/odd distinction, so we will henceforth assume QV % . To eliminate possible ambiguity, we ò will single out a particular Ä root; we define the symbols:
A
Q Y ) Q v ) Q v For any rational numbers and , and for all positive bases ' and Q :
CHAPTER 19. EXPONENTIAL FUNCTIONS
268
Ú
n y=t n even
axis
Ú
= solution
axis
y=t
n n odd
y=b
ò
axis
y=b*
PSfrag replacements axis
y=b
ò axis
no solution or two solutions exactly one solution
Figure 19.4: How many solutions to
à
1. Product of power rule: Q Q
à
)Q
2. Power of power rule: 6 Q
9 'Q
3. Power of product rule: 6m' Qâ9
O $ æ 5. Negative power rule: Q ç
Ä Ñ ¡Î
?.
à D
à ' Q
4. Zero exponent rule: Q
These rules have two important consequences, one theoretical and the other more practical. On the first count, recall that any rational number can be written in the form ? v w , where ´ and ð are integers. Consequently, using these rules, we see that the expression M]éQ S defines a function of , whenever is a rational number. On the more practical side of things, using the rules we can calculate and manipulate certain expressions. For example,
i ! ö ² J æ 

i
ý
) ! )j o ' æ  ) J u " ! æ u
! u
2j !
The sticky point which remains is knowing that ! 6 9U QS actually defines a function for all real values of . 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 19.1.1 carry through for all real
B
DC
19.2. THE FUNCTIONS Z' PO
269
exponent powers. Your calculator should have a “M to the key”, allowing you to calculate expressions such as ( involving nonrational powers. Here are the key modeling functions we will work with in this Chapter. Definition 19.1.2. A function of exponential type has the form
[email protected] 9Ò' ïORQ S 1 ã , and POYH ã %. for some Q½V % , Q$ We will refer to the formula in Definition 19.1.2 as the standard exponential form. Just as with standard forms for quadratic and sinusoidal functions, we sometimes need to do a little calculation to put an equation in standard form. The constant ¢O is called the initial value of the exponential function; this is because if represents time, then 6 %9Ë. PORQ O . PO is the value of the function at time ÍW% ; i.e. the initial value of the function. Example 19.1.3. Write the equations M½HJ:lS and M½ exponential form.
i
æ x { S in standard
Solution. In both cases, we just use the rules of exponents to manuever the given equation into standard form:
M½HJ lS 6J l 9S ` 0! S and
M½
g i
æ S
! h g æ i S ! h ! h
S g i ! r !Êh s + g + S $ h g
19.2 The Functions
E F G I% H
We know ! 6 9ïB!S defines a function of , 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 19. EXPONENTIAL FUNCTIONS
270
J
Point on the º graph of .. .. .. . . . 2 1/4 (2, 1/4) 1 1/2 (1, 1/2) replacements PSfrag 0 1 (0, 1) axis 1 2 (1, 2) axis 2 4 (2, 4) axis 3 8 (3, 8) .. .. .. . . .
J
º
(a) Datapoints from Û¢¡Ç
Y
.
Figure 19.5: Visualizing
y=2
x
(3,8)
(2,4) (−1,1/2) (1,2) (0,1)
(−2,1/4)
−1
(b) Graph of
ÛY¡Ç Y
1
Û¢¡ Y
.
.
The graph is always above the horizontal axis; i.e. the function values are always positive.
The graph has M intercept 1 and is increasing. The graph becomes closer and closer to the horizontal axis as we move left; i.e. the 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.
We can use these features to argue that MWy!nS is not the graph of any function we have studied thus far. For example, the graph in Figure 19.5 is not the graph of any polynomial, since a polynomial graph never becomes asymptotic to a horizontal axis. It cannot be a sinusoidal function, since the values are not bounded, etc. The special case of M '!S is representative of the function M .QS , but there are a few subtle points that need to be addressed. First, recall we are always assuming that our base Q V % . We will consider three separate cases: QU$ , QNV. , and %YX)QFX[ .
19.2.1 The case
KML N
In the case QZ , we are working with the function MN S ; this is not too exciting, since the graph is just a horizontal line. We will ignore this case.
B DC The case KOPN
19.2. THE FUNCTIONS Z' PO
19.2.2
271
If Q]Vb , the graph of the function M Q S is qualitatively similar to the situation for Q [! , 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 19.6(a) indicates how these graphs compare for three different values of Q . 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 axis.
PSfrag replacements
Ú
PSfrag replacements axis
axis
all graphs pass through (0,1)
S (a) Graph of ÛY¡ñ7
Y
,
Ú
axis
all graphs pass through (0,1)
axis
S
axis
7ï¬
.
(b) Graph of ÛY¡7 Mõ7ãõ# .
Y
axis
,
Figure 19.6: Visualizing cases for 7 .
19.2.3 The case
QRSKMR N
We can understand the remaining case %?X$QüXB , by using the remarks above and our work in Chapter 9. First, with this condition on Q , notice that ç V$ , so the graph of MüÏxnç { S is of the type in Figure 19.6(a). Now, using the rules of exponents:
M½
g
Q h
æ S
r
g
æ S 'Q S Q h s
æ
By the reflection principle, the graph of MG xç { S is obtained by reflecting the graph of Mq 6 ç 9»S about the M axis. Putting these remarks together, if % XtQ X , we conclude that the graph of MyCQS will look like Figure 19.6(b). Notice, the graphs in Figure 19.6(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 axis. Important Facts 19.2.1 (Features of Exponential Type Functions). Let Q be a positive real number, not equal to 1. The graph of M]éQ S has these four properties:
CHAPTER 19. EXPONENTIAL FUNCTIONS
272
1. The graph is always above the horizontal axis. 2. The graph has M intercept 1. 3. If QVÿ (resp. %?X QüX ), the graph becomes closer and closer to the horizontal axis as we move to the left (resp. move to the right); this says the axis is a horizontal asymptote for the lefthand portion of the graph (resp. righthand portion of the graph). 4. If Q VW (resp. %DXHQ#Xq ), 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 POUVé% , the graph of the function M Ï PORQTS is a vertically expanded or compressed version of the graph of Ï M Q S . If P O X % , we additionally reflect about the axis.
19.3 Piano Frequency Range A sound wave will cause your eardrum to move back and forth and this can be modeled using sinusoidal functions. In the case of a socalled pure tone, this motion is modeled by a single sinusoidal function of the form
6 /:9Ò' sin 6 !(!R/291 where ! is called the frequency, in units of “periods/unit time”, called “Hertz” and abbreviated “Hz”. The coefficient 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.
PSfrag replacements
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#
axis axis axis
A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C D E F G A B C 220 Hz
middle C
Figure 19.7: A piano keyboard.
A piano keyboard layout is shown in Figure 19.7. 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
19.3. PIANO FREQUENCY RANGE
273
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 ! R times the frequency of the key to its immediate left. This makes the ratio of adjacent keys
always the same (! R ), and it means that keys
12 keys apart have a ratio of frequencies exactly equal to 2 (since 6 ! }z 9 $! ). Two such keys are said to be an octave 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 A# to the right of this key has frequency the keyboard. For instance,
!!%, ! z $!!%?,ê% ` o ^:j%o ad
lÿ !njjT&%+ JeJ+ J ý î . The B to the right of this key has frequency !njjT%JeJJD,U! z l'! ^o e^ ` ý î .
CHAPTER 19. EXPONENTIAL FUNCTIONS
274
19.4 Exercises Problem 19.1. Let’s brush up on the required calculator skills. Use a calculator to approximate: (a)
(b)
(c)
f
(d) (e) (f) (g) (h)
e
Z « ¶ È R Z ¤  Ó ¥«0/ sin  ¨¨ ª « f / Z Z sin ¨  /
Problem 19.4. According to the Merck Man¾ ual, the relationship between the height (in inches), weight r (in pounds) and surface area ç (in square meters) for a human is approximated by the equation:
¾ Fix
Problem 19.2. Put each equation in standard exponential form: (a) (b) (c) (d) (e) (f)
ÛY¡ÇI ZIY / ÛY¡ ZIY ÛYB ¡ f Y e ÛY¡Z « ¨ « ¶ j ÛY¡ p «º ¶ e ÛY¡ r ó Rzª / Z p º Y
T
DU # % WV < U
Problem 19.3. e
ÛY¡
&¿/¡
(a) Begin with the function .
 z¿/ ,  ¿3©Z"/ , R¿7©Z"/ , r&¿3©Z"/f/ , e  ¿/ , e  r&¿4©Í"/f/ e Is the function  T& ¿n©7"/À/ Â a func
(a2)
e
meter
p
to be your height, in inches. Let ç¡£ç r / be the resulting function computing your surface area as a function of your weight r . What is your surface area? How much weight must you gain for your surface area to increase by 5%? Decrease by 2%?
T
(a1) Rewrite each of the following functions in standard exponential form: e
< < ¤§
¡ ¥¼ ¾ p º r ç
Problem 19.5. Begin with Y a sketch of the graph of the function Ûê¡H on the domain of all real numbers. Describe how to use the “four tools” of Chapter 9 to obtain the graphs Y ZIY Y of these functions: ÛÉ¡U©@ , ÛD¡ , ÛD¡?I / , Y Y Y Y[Z Ûê¡ Y « ¨ e  / , Û¡HY 7Â? ,Y Û¡' © , Û¡' , ÛY¡ , ÛY¡0« , ÛÉ¡ « .
<
Y
(b) In general, explain what happens when you apply the four construction tools of Chapter 9 (vertical shifting, vertical dilation, horizontal shifting, horizontal dilation) to the standard exponential model Û ¡ B ± 7 Y . For which of the four operations is the resulting function still a standard exponential model?
e
e
tion of exponential type?
(a3) Sketch the graphs of &¿/ z¿/} r&¿v© "/f/ e T¿<©N"/À/ and e  T&¿w©N"/À/=Â in the same coordinate system and explain which graphical operation(s) (vertical shifting, vertical dilation, horizontal shifting, horizontal dilation) have been carried out. e
e
e
Problem 19.6. A colony of yeast cells is estimated to contain ¥¦ cells at time Ä>¡ . After collecting experimental data in the lab, you decide that the total population of cells at time Ä hours is given by the function
YX < º¼£
Û¢¡Z¥ ¦
p
d
¨p
(a) How many cells are present after one hour? (b) (True or False) The population of yeast cells will double every 1.4 hours. (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
<
d ÛÉ¡Z¥ ¦ v ¼ / p ¦¥£§«¨»º¼¤
Should you be worried by Cherie’s remark?
19.4. EXERCISES
275
(d) Anja, a third member of your lab working with the same yeast cells, took these two measurements: zª®7¥¦ cells after 4 hours; ªe 0¼ï®D¥0¦ 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 Ä?
Problem 19.7. middle C.
(a) Find the frequency of PSfrag replacements
(b) Find the frequency of A above middle C.
axis axis
1
2
3
63
64
10
9 8
axis
(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 BËR ? (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 19.8. 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 dimes will you have stacked on the 10th square? (b) How many dimes will you have stacked on the Î th square? (c) How many dimes will you have stacked on the 64th square? (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.
Problem 19.9. Myoglobin and hemoglobin are oxygen carrying molecules in the human body. The function
Z
¡c5.k H/¡ ´Âï
calculates the fraction of myoglobin saturated with oxygen at a given pressure torrs. For example, at a pressure of 1 torr, 51»"/>¡) , 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
Z
< < <
¡0¬Ík N/¡ ª Âï¯ ¯ ¯
calculates the fraction of hemoglobin saturated with oxygen at a given pressure . Hemoglobin is found inside red blood cells, which flow from the lungs to the muscles through the bloodstream. Myoglobin is found in muscle cells. (a) The graphs of 5.k H/ and ¬Íc H/ are given below on the domain h BhN¥ ; which is which?
CHAPTER 19. EXPONENTIAL FUNCTIONS
276 fraction 1
(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?
0.8 0.6
PSfrag replacements
0.4
axis 0.2 axis axis
20
40
60
80
100
p
(b) If the pressure in the lungs is 100 torrs, what is level of oxygen saturation of the hemoglobin in the lungs?
(d) Define the efficiency of oxygen transport at a given pressure to be 51c H/:©B¬Íc H/ . What is the oxygen transport efficiency at 20 torrs? At 40 torrs? At 60 torrs? Sketch the graph of 51c H/@©(¬Ík H/ ; are there conditions under which transport efficiency is maximized (explain)?
Chapter 20 Exponential Modeling Example 20.0.1. 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 « 6 9 , where represents years, is a function of exponential type,
« 6 9Ò[« Q S 1 for some « and Q V values of « 6 9 : ý
« 6 % ` 98$e%%´1k%%% ý i « 6 9Ò ` %´1k%%%
. We are given two pieces of information about the
O
i.e., «DORQ u [e%%´1&%%%´1 and i i.e., «IORQU ` %´1&%%%
We can use these two equations to solve for the two unknowns as follows: If we divide the second equation by the first, we get
«
and Q
i Q OQ u `
i Q O u )Q  ) U Q ` QZ ` ^c! ` î Plugging this value of Q into either equation (say the first one), we can solve for « : « O¥OR ë O¥ð O O ô ² jA1cjjj . We conclude that the number of u u
[
geton.com subscribers will be predicted by
« 6 9Ò$jA1_jjj 6a` ^c! ` 9 S i
1 ` %J´1cj%%´1k%%%´1&%%% subscribers, which exIn five years, we obtain « 6d` 9D ceeds the population of the Earth (which is between 5 and 6 billion)! 277
CHAPTER 20. EXPONENTIAL MODELING
278
ements axis
600000
axis
400000
axis
200000
0.4
0.6
0.8
1
Figure 20.1: Finding equation for À &¿/¡?À p
7
More importantly, this example illustrates a very important principal we can use when modeling with functions of exponential type.
Y the .
Important Fact 20.0.2. A function of exponential type can be determined if we are given two data points on its graph.
ements !!!
axis
ØU 6 % ` 1 e%%´1k%%%9 i Ù 6 n1 ` %´1&%%%9
P
0.2
axis
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 « 6 9 :
Q
800000
CAUTION !!!
axis
\
When you use the above strategy to find the base 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.
20.1 The Method of Compound Interest You walk into a Bank with ØO 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 / years? The future value of the account is really a function of the number of years / elapsed, so we can write this as a function Ø 6 /:9 . Our goal is to see that Ø 6 /:9 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:
annual (decimal) interest rate Ä the number of compounding periods per year The number Ä tells us how many times each year the Bank will compute the total value Ø 6 /:9 of the account. For example, if Ä , the calculation is done at oneyear intervals; if Ä 0! , the calculation is done each month, etc. The bank will compute the value of your account after a typical compounding period by using the periodic rate of return ¾ . For example, if the interest rate percentage is 12% and the compounding period is monthly (i.e., Ä 0! ), then the annual (decimal) interest rate is O 0.12 and the periodic rate is '%% .
ements axis axis
!!!
axis
CAUTION !!!
]
The number 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.
20.1. THE METHOD OF COMPOUND INTEREST
279
20.1.1 Two Examples Let’s consider an example: ØOÍ $ n1&%%% invested at the annual interest percentage of J % compounded yearly, so Ä ÿ and P]%&%J . To compute the value Ø 6 9 after one year, we will have
Ø 6 9.ØAOÒ (periodic rate)ØO .ØAOÒ ¹ØAO>HØAO 6 > n9 $ n1&%%% 6 m%%J9Ò $ n1k%J% To compute the value after two years, we need to apply the periodic rate to the value of the account after one year:
Ø 6 !A9.Ø 6 9 (periodic rate)Ø 6 9
.ØAO 6 > 9A ¹ØAO 6 > 9ÊHØAO 6 + > n9
$ n1&%%% 6 m%%J9 $ n1åe^^ % 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:
Ø 6 jT9.Ø 6 !A9 (periodic rate)Ø 6 !A9
.ØAO 6 > 9 ¹ØAO 6 > 9 .ØAO 6 > n9 l i $ n1&%%% 6 m%%J9 l $ n1 ! ` o 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. ò way, to find the value after / years, we arrive at the Continuing on in this formula ò
Ø 6 :/ 9A.ØAO 6 > n 9 $ n1&%%% 6 &%J9 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 J %, but now compound monthly, so Ä 0! and D[%%J . The value of the account after one compounding period is Ø 6 ef0!A9 , since a month is onetwelfth of a year. Arguing as O O ç , we before, paying special attention that the periodic rate is now ¾ have
Ø 6 ef0!A9A.ØAOÊ (periodic rate) ØO g %J .ØAO > 0! h i i $ n1&%%% 6 mê%%%^^^ 9 $ n1&%%^^
CHAPTER 20. EXPONENTIAL MODELING
280
After two compounding periods, the value is Ø 6 !f0!A9 ,
Ø 6 ! f0!A9HØ 6 e f0!A9 (periodic rate) Ø 6 ef0!A9 g g g &%J %%J %J HØAO > ØAO 0!Ph 0!)h 0!Ph g &%J HØAO > 0!Ph i $ n1k%%% 6 m%&%%^^^ 9 $ n1&%jTcjJ Continuing on in this way, after compounding periods have elapsed, the value will be Ø x { , which is computed as
Ø 6 ´ f0!A9HØAO
g
&%J > 0! h
It is possible to rewrite this formula to give us the value after / years, noting that / years will lead to 0!n/ compounding periods; i.e., set U 0!n/ ò in the previous formula:
Ø 6 /29.ØAO
g
%J 0!ïh
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.
20.1.2 Discrete Compounding The two examples above highlight a general formula for computing the future value of an account. Important Fact 20.1.1 (Discrete compounding). Suppose an account is opened with ØAO principal. If the decimal interest rate is and the number of ò year is Ä , then the value Ø 6 /:9 of the account after compounding periods per / years will be
Ø 6 2/ 9.ØAO
Ä
¾
Notice, the future value Ø 6 /:9 is a function of exponential type; the base is the number x ¾ { , which will be greater than one. Since ØOYV % , the graph will be qualitatively similar to the ones pictured in Figure 19.6(a). Example 20.1.2. 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 i U.S. Government will pay a minimum annual interest rate of GJ ` %, 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.
20.2. THE NUMBER
AND
THE EXPONENTIAL FUNCTION
281
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.
+ by the forSolution. The value of your bond after 35 years is computed i , and /ZGj ` . mula in Fact 20.1.1, using ØOÍ î $!1 ` %% , N %%J ` , Ä Plugging this all in, we find that the selling price of the bond is Ø 6 j ` 9A $!1 ` %%
g
i %%+ J ` h
p l¥u ð
i
+
$` n1 e^ !A
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!
20.2 The Number ^ and the Exponential Function What happens to the future value of an investment of Ø@O dollars as the number of compounding periods is increased? For example, return to our earlier example: ØO> $ e%%% and an annual interest percentage of J %. 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)
Compounding Period 1 yearly 4 quarterly
12 monthly 52 weekly 365 daily 8,760 hourly
$ n1k%%% 6 mê%&%J9 = $1,080.00
O O ç { p = $1,082.43 p
O O
{ $ n1&%%% x ç = $1,083.00 O O $ n1&%%%Yx ç { u = $1,083.22 u O O $ n1&%%% x ç { l§u = $1,083.28 lu ë O O O $ n1&%%% x ë ç O { ç ¾ =$1,083.29 ç ¾¥ $ n1k%%% x >
CHAPTER 20. EXPONENTIAL MODELING
282
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 20.2.1. Initially increasing the number of compounding periods makes a significant difference in the future value; however, eventually there appears to be a limiting value.
ò Let’s see if we can understand mathematically why this is happening. The first step is to recall the discrete compounding formula: Ø 6 /29.ØAO
Ä
¾
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 Ä gets BIG. To analyzeò this,¾ it is best to rewrite the expression using a substitution trick: Set î , so that Ä $ î and ¾ à . Plugging in, we have
Ø 6 /:9AWØAO > g WØAO > g]g WØAO
Ä
à
î h
>
¾ ò
î h
ò (20.1)
à
h
¾
So, since is a fixed number and î , letting Ä get BIG is the same as letting î become BIG in (20.1). This all means we need to answer à this new question: What happens to the expression xz> à { as î becomes 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 à some numerical data for the values of M]³6 6µî 9w x 4 à { and look at a plot of this function graph on the domain %&%R' î ' e%% :à See Figure 20.2. You can see from this plot, the graph of M. x > à { approaches the “dashed” horizontal asymptote, as î becomes BIG. We will let the letter “ \ ” represent the spot where this horizontal line crosses the vertical axis and i \ l'!A eJv!JeJ . This number is only an approximation, since \ is known to be an irrational number. What sets this irrational number apart from the ones you are familiar with (e.g. ) ! , ( , etc.) is that defining the number \ requires a “limiting” process. This will be studied a lot more in your Calculus course. The new number \ is a positive number greater than 1, so we can study the function:
MN \ S
(20.2)
Since \#Vb , the graph will share the properties in Figure 19.6(a). This function is usually referred to as THE exponential function. Scientific calculators will have a key of the form “ exp 6 @9 ” or “ \ S ”.
AND
20.2. THE NUMBER
THE EXPONENTIAL FUNCTION
283
aWbdc e Ì f e `
1 2 2 2.25 3 2.37037 PSfrag replacements 4 2.4414 axis 20 2.65329 axis 100 2.70481 axis 1000 2.71692 þ ä 2.71828
# g%g
(a) Datapoints for ´Â ¨ .
2.5
2
1.5
1
0.5
20
40
60
(b) The graph of
Figure 20.2: What happens when
80
100
# ´Â g¨ % g
.
get very large? ä
20.2.1 Calculator drill Plugging in Nb , you can compute an approximation to \ on your cali culator; you should get \ !A eJ:j to four decimal places. Make sure
+ + you can compute expressions like \ l , \ , and ) \\  ; to four decimal i i places, you should get !%&%J `` , !njTa % , and ^ J .
20.2.2 Back to the original problem... We can now return to our future value formula (20.1) and conclude that ò as the number of compounding periods increases, the future value is ò approaching a limiting value:
Ø 6 :/ 9A.ØAO
g g
>
à
î h h
ò
ØAO\
The right hand limiting formula Ù 6 /:9éØO0\ 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.
ò
Important Fact 20.2.2 (Continuous compounding). The future value ò of ØAO dollars principal invested at an annual decimal interest rate of unò der continuous compounding after / years is Ù 6 /:94ÿØO0\ ; this value is al ¾ ways greater than the value of ØO x > ¾ { , for any discrete compounding scheme. In fact, ØO\ is the limiting value.
CHAPTER 20. EXPONENTIAL MODELING
284
20.3 Exercises Problem 20.1. 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 r Ä2/ , where Ä represents the time in years after 1960. (a) Find a formula for
r§Ä2/ .
(b) What does the model predict for the minimum wage in 1960? (c) If the minimum wage was $5.15 in 1996, is this above, below or equal to what the model predicts. Problem 20.2. The town of Pinedale, Wyoming, is experiencing a population boom. In 1990, the population was 860 and five years later it was 1210.
h
(a) Find a linear model &¿/ and an exponential model J¿/ 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 20.3. Return to the Earning Power Problem in 4.1. Use the data in Table 4.1 to obtain exponential models 51&¿/ and 3¿/ for Men’s and Women’s Earning power in the year 4ÂÇ¿ , respectively. What will be the earnings in 1997? In 2010? In 2100? From these calculations, what can you say about whether women are gaining on men?
Problem 20.4. Tiffany is a model rocket enthusiast. She has been working on a pressurized rocket filled with laughing gas. According 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 Z ¨{p pounds/sq.in. a formula ½¡ » 2 / for ¾ miles above sea the atmospheric pressure level. Assume that the rocket is launched at an angle of U 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 Ä seconds is given by the equation Û Ä2/T¡ ©4ªÄÂÍ R sin  U /óÄ .
X T
(a) If the angle of launch is U)¡é§ ± , determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair?
±
(b) If the angle of launch is U)¡ , determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? (c) Assume that the maximum elevation to avoid premature explosion is 3.8526 miles. Find the largest launch angle U so that the rocket will not prematurely explode.
Problem 20.5. Given the initial information ù Ã Á and Î , banks will sometimes provide an additional percentage rate called the effective yield. This is understood to be an annual interest rate : % which would yield the same amount at the end of one year under one compounding as the given data under Î compoundings. (a) If ù dollars is invested at an annual rate Ã of % compounded quarterly, what is the effective yield? (b) If ù Ã dollars is invested at an annual rate of 6.75% continuously compounded, what is the effective yield?
Problem 20.6. In 1989, research scientists published a model for predicting the cumulative number of AIDS cases reported in the United States:
ÅQ Ä2/¡Z§R
m
Ä©Í « ¥ n
 thousands /
where Ä 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 ÅQÄ2/ to construct a new exponential model 7ªÄ2/ for the number of cumulative AIDS cases. Discuss how the two models differ and explain the use of the word “relief”.
20.3. EXERCISES
285
X Y Â X ZIY X Y © X ZIY
Problem 20.7. Define two new functions:
and
ÛY¡
cosh &¿/¡
ÛY¡
sinh ¿/¡
(c) A hanging cable is modeled by a portion of the graph of the function
¾ Ì ¿ © ÛY¡FÅ cosh  Å ¹/ Â[email protected] ¾ appropriate constants Å ¾
for and D . The constant 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 Û axis and the roadway coincides with the ¿ axis. The hanging cable constant is Å ¡'0R ¾ and ¡ . Find the minimum distance from the cable to the road.
These are called the basic hyperbolic trigonometric functions. (a) Sketch rough graphs of these two functions.
(b) The graph of the equation ¿¹©?Û ¡ÿ is shown below; this is called the unit hyperbola. For any value Å , show that the point &¿ ÛH/@¡  cosh ßÅ+/ sinh ßÅ/f/ is on the unit hyperbola. (Hint: Verify that cosh &¿/~ ©C sinh ¿k/ ¡# , for all ¿ .)
y
towers
PSfrag replacements axis axis axis
(−1,0)
Sfrag replacements axis axis axis
(1,0)
x
cable d 400 ft
100 ft roadway
286
CHAPTER 20. EXPONENTIAL MODELING
Chapter 21 Logarithmic Functions ò
If we invest ØO; $ n1k%%% at an annual rate of Y[J % compounded continuously, how long will it take for the account to have a value of $5000? O O The formula Ø 6 /29Ê ò n1&%%%\ ç gives the value after / years, so we need to solve the equation:
ò
O O ` 1&%%%P$n1k%%%\ ç O O ` \ ç Unfortunately, algebraic manipulation will not lead to a further simplifiò cation of this equation; we are stuck! The required technique involves the theory of inverse functions. ò Assuming we can find the inverse function of æ ! 6 /:98 \ , we can apply ! 6 /:9 to each side of the equation and solve for / :
æ 9Ò&! æ \ O O ç 9ÒH%%J:/ a6 ` 6 æ 9 "/ 6 0!A ` 9 ! 6a` Ò !
æ
The goal in this section is to describe the function ! , which is usually æ denoted by the symbol ! 6 /:9 ln 6 /:9 and called the natural logarithm ++ function. On your calculator, you will find a button dedicated to this function and we can now compute ln 6a` 9 &^%o . Conclude that the solution is /; !%a0! years.
21.1 The Inverse Function of
E F ^
If we sketch a picture of the exponential function on the domain of all real numbers and keep in mind the properties in Fact 19.2.1, then every horizontal line above the axis intersects the graph of M)y\S exactly once: See Figure 21.1(a). The range of the exponential function will consist of all possible M 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 21.1(b). 287
CHAPTER 21. LOGARITHMIC FUNCTIONS
288 PSfrag replacements
PSfrag replacements
y=e
axis
graphs of y=c, c>1 cross exponential graph exactly once.
axis axis −1
these horizontal lines miss graph of exponential function.
1
y = ex
x
axis
range = all positive numbers
axis axis −1
1 domain = all real numbers
X
X
(a) Horizontal line test for ÛY¡ Y .
(b) The domain and Y range for ÛÉ¡ .
Figure 21.1: Properties needed to find the inverse of
e
&¿/¡
XY
.
By the horizontal line test, this means the exponential function is oneæ toone and the inverse rule ! 6 ï=9 will define a function
the
!
y=e
x
æ ï=9Ò 6 ¡
¢
reflecting line y = x
ements y=ln (x)
axis axis
−1
1
axis
Figure 21.2: Visualizing the Û¢¡ ln &¿/ .
unique
solu
r tion of the equation 1 if ïDV % s ïP \ S 6 undefined 9z1
(21.1)
if ï>')%
This inverse function is called the natural logarithm function, denoted ln 6 ï9 . We can sketch the graph of the the natural logarithm as follows: First, by Fact 11.2.1, the domain of the function ln 6 M39wÿ is just the range of the exponential function, which we noted is all positive numbers. Likewise, the range of the function ln 6 M39mW is the domain of the exponential function, which we noted is all real numbers. Interchanging and M , the graph of the natural logarithm function M" ln 6 9 can be obtained by flipping the graph of M ]\ S across the line M :
Important Facts 21.1.1 (Graphical features of natural log). The function M½ ln 6 @9 has these features:
The largest domain is the set of positive numbers; e.g. ln 6 É9 makes no sense.
The graph has intercept 1 and is increasing. The graph becomes closer and closer to the vertical axis as we approach ÇH% ; i.e. the M 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.
B i C
21.1. THE INVERSE FUNCTION OF #
289
Important Facts 21.1.2 (Natural log properties). We have the following properties:
î
(a) For any real number , ln 6 \Sn9Ò" .
ò
(b) For any positive number , \
ln
S ð " .
(c) ln 6 Q 9Ò"/ ln 6 QÌ9 , for Q V % and / any real number; (d) ln 6 Q ' 9Ò ln 6m' 9T ln 6 QÌ9 , for all ' 1¼Q½V % ;
ç (e) ln x Ã { ln 6 Qâ9´ ln 6(' 9 , for all ' 1¼Q#V % .
The properties (c)(e) are related to three of the rules of exponents in Facts 19.1.1. Here are the kinds of basic symbolic maneuvers you can pull off using these properties: Examples 21.1.3. (i) ln 6 Jvlv9Z j ln 6 J9U ^ c!njJ:j ; ln 6 ^2(>9F ln 6 jT9T ln 6d` 9ÊH< % ` e% J .
ln 6 ^94
ln 6 (49F
!Ao:j^ ` ; ln x l { u
(ii) ln x ) { ln x  { Ï ln 6 @9 ; ln x ) { ln 6 6 wê9 6 ê9¼9Ê ln 6 wê9 ²
ln 6 ¢'9 ; ln ö S ln x u { ln x ) { ` ln 6 9T ln x H { .
S
D
Examples 21.1.4. (i) Given the equation j2S^D $0! , we can solve for :
j ^S D $0! ln x j ^S D { ln 6 0!A9 6 Y)9 ln 6 jT9Ò ln 6 0!A9 ln 6 0!A9 î )w[c!^eo Ç ln 6 jT9 î (ii) Given the equation \ tan S ð ! , we can solve for : î \ tan S ð ! ð ln x \ tan S { ln 6 !A9 tan 6 9Ë ln 6 !A9ÒH%&^o:jT '%^%^m (>1 where is any integer %´1¦Wn1[¦w!1ee . Example 21.1.5. 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 I[^ % annual interest, when will the exact account value be be $130,000?
CHAPTER 21. LOGARITHMIC FUNCTIONS
290
Solution. In the first scenario,
j%´1k%%%¢ !1&%%%\
^ ` ]\
ln 6 ^ ` 98 ln x¼\ { ln 6 ^ ` 98[0! + i ln 6 ^ ` 9 H%cj o 0!
This gives an annualî interest rate of 34.79%. In the second scenario, we ò study the equation î ò
j%´1k%%%¢ !1&%%%\ O O p ð ^ ` ]\ O O p ð ln 6 ^ ` + 9 /; H^ ` _!!A %&%^
So, it takes over 65 years to accumulate $130,000 under the second scheme.
21.2 Alternate form for functions of exponential type ò
The standard model for an exponential function is 6 /:9m] YORQ , for some ò PO# Q'Vq% , ò Q b ã , î and ò ã î % . Using the properties of the natural logarithm function,
x\ Q
ln
ç¹ð {
]\
ln
ç¹ð
ò
ò
This means that every function as in Definition 19.1.2 can be rewritten using the exponential function \ . Another way of saying this is that you really only need the function keys “ \ ” and “ ln 6 /29 ” on your calculator. Important Factò 21.2.1 (Observation). A function of exponential type can be written in the form
6 /:9Ê) PO\
Ã
1
ã % and ' H ã %. for some constants POÉ' ò
By studying the sign of the constant ' , we can determine whether the function exhibits exponential growth or decay. For example, given the Ã function 6 /29 \ , if ' VG% (resp. ' X % ), then the function exhibits exponential growth (resp. decay). Examples 21.2.2.
B jC
ò
21.3. THE INVERSE FUNCTION OF #W
291
(a) The function 6 /29ïò éî !%% 6 ! 9 exhibits ò exponential growth and can be rewritten as:
ð ò { "!%%\ O ä l u +
ò \ æ O exhibits ò (b) The function 6 /:9ò exponential decay and can be re 6 /:9Ê !%% x \
written as:
6 /:9Ê
+
ln
+ +
i æ O æ O x\ { x %JeJ j { \
21.3 The Inverse Function of
E F H
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 M GQS , where Q)Vq% and Q[ ã . Just as above, we would show that ! 6 @9; QS is onetoone, the range is all positive numbers and obtain the graph using ideas in Figure 21.2. We will refer to the inverse rule as the logarithm function base b, denoted logç 6 9 , defined by the rule:
the
logç 6 ï=9Ò
unique
solu
r tion of the equation 1 if ïDV % s ï¢)Q S
¡ ¢
6 undefined 9z1
if ïÐ')%
We will need to consider two cases, depending on the magnitude of Q : P The important qualitative features of the logarithm function M'
6Qç 6 9 mirror Fact 21.1.1:
x y=b
reflecting line y = x
x y=b
reflecting line y = x
y=logb(x)
eplacements
−1
1
−1
1
axis
y=log (x) b
axis axis The case b > 1
The case 0 < b < 1
Figure 21.3: Cases to consider for 7 .
CHAPTER 21. LOGARITHMIC FUNCTIONS
292
Important Facts 21.3.1 (Graphical features of general logs). The function M½ logç 6 @9 has these features:
The largest domain is the set of positive numbers; e.g. logç 6 É9 is not defined.
The graph has intercept 1 and is increasing if Q V[ (resp. decreasing if %ÉX)QFX[ ). The graph becomes closer and closer to the vertical axis as we approach ÿ % ; this says the M axis is a vertical asymptote for the graph.
The graph is unbounded as we move to the right. ã . Important Facts 21.3.2 (Log properties). Fix a positive base Q , Q[ î (a) For any real number , logç 6 QTS29Ë . ò ß ð (b) For any positive number , Q log S " .
(c) logç 6 9Ò)/ logç 6 9 , for ¢V % and / any real number; (d) logç 6 29Ë logç 6 n9T logç 6 29 , for all 1z V % ;
(e) logç 6 à 9Ë logç 6 n9´ logç 6 29 , for all n1zPV % . It is common to simplify terminology and refer to the function logç 6 9 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 6 @9 ”, which is usually understood to mean the log base 10. However, many calculators only have the key “ln 6 9 ”. This is not cause for alarm, since it is always possible to express logç 6 9 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 21.3.2. Suppose we start with M logç 6 @9 . 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.
M logç 6 @9 Q Ú ln 6 Q Ú 9Ò ln 6 @9 M ln 6 QÌ98 ln 6 @9 ln 6 9 M ln 6 QÌ9 We have just verified a useful conversion formula:
B jC
21.3. THE INVERSE FUNCTION OF #W
293
Important Fact 21.3.3 (Log conversion formula). For a positive numã a base, ber and Q V"%´1¼Q$ logç 6 9Ë
ln 6 9 ln 6 QÌ9
For example, log O 6a` 9Ò
logO O 6 9Ò
log O
g
! h
ln 6a` 9 %^oo H ln 6 e%9 ln 6 9 <%^j . ln 6 %&%v!A9 + ln 6 9 <%c!njT . ln 6 !%9
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 21.3.4. 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 J %. How soon can you sell the house?
ò
Solution. We can use the future value formula to obtain the equation
n1&%%%´1k%%%ïHJ%´1&%%%
gò
+
>
%% + J
0!A ` 6 &%0!A9 p Using the log base QZ[%0! ,
h
p ò
9 log log O 6 0 !A ` 8 + O x 6 %0!A9 p { 9 / log O 6 0 !A ` 8 + + ln 6 0!A ` 9 / ` !Ao:j ln 6 &%0!A9 Since you have already owned the house for 10 years, you would need to wait nearly 43 years to sell at the desired price. Let’s try our hand at a problem that uses both the inverse trigonometric functions and the inverse of the exponential function. We will use î_ò preliminary graphical reasoning as an aide. Example 21.3.5. The voltage output of a circuit at time / is given by the ð function MNS! 6 /:98 ` \ sin . During the first 10 seconds, when is the voltage equal to 8 volts?
CHAPTER 21. LOGARITHMIC FUNCTIONS
294
Ú
ements
axis
Solution. If we look at a software plot of MN#! 6 /:9 and M 'J in the same coordinate system, we see the two graphs cross four times on the domain % 'y/' e% . This means there will be four solutions to the equation J ! 6 /:9 , corresponding to the / coordinates of the four intersection îcò points: See Figure 21.4. The computation îcò is easy to begin:
12
axis
ò
10 8 6 4
axis
2 2
4
6
8
10
X
Figure 21.4: TheQid graph of Û¢¡ sin S .
ð ` \ sinîcò 'J ð ln x ` \ sin { ln 6 J9 9 ln x \ sin ð { ln+ 6 J9 ln a6 ` ´ i sin 6 /:9Ê'%
We now use the technique explained in Chapter 18 to solve this equation. This requires we find the principal solution using the inverse sine function:
+ + i æ /; sin 6 % 8 9 '% Jo:jT
Next, we find the symmetry solution:
+ i æ /;H sin 6 % 9 (#"!A&^ ` !njT
Finally, we conclude EVERY solution has the form:
+
/; 6 % Jo:jw !(49 or 6 !A&^ ` !njw !(49 where ?'%´1[¦ n1[¦w!1[¦ jA1 EeEeE We need to find which of these solutions are between 0 and 10. This is just a calculation:
+
/;'% Jo:jA1!A&^ ` !njA1^
ii
! ` 1Jko:j ``
21.4 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
21.4. MEASURING THE LOUDNESS OF SOUND
295
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 Ö of a sound is defined by the equation
ÖÍ[e%
k
lk kO h g
log O
1
(21.2)
k
where O is an arbitrary reference intensity which is taken to correspond with the average faintest sound which can be heard and is the intensity of the sound being measured. The units used for Ö are called decibels, » abbreviated “db”. (Historically, the units of loudness were called bels » , in honor of Alexander Graham Bell, referring to the quantity log O .) Notice, in the case of sound of intensity O , we have a sound pressure level of
k k
ÖÍ[e%
dk O k Oh g
log O
[e% log O 6 9Ê[e% 6 %9ÒH%
k
We refer to any sound of intensity O 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 21.5(a) gives a hint of the sound pressure levels associated to some common sounds. Sound Pressure Level in db Threshold of pain 120 Riveter 95 PSfrag 70 replacements Busy Street Traffic Ordinary Conversation 65 axis Quiet Auto 50 axis Background Radio 40 axis Whisper 20 Rustle of Leaves 10 Threshold of Hearing 0 Source of Noise
(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 21.5: 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
CHAPTER 21. LOGARITHMIC FUNCTIONS
296
of 20 Hz (something like the groundshaking 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 21.5(b). Such a plot would differ from person to person and is helpful in isolating hearing problems. Example 21.4.1. 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 their model X1 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?
k
k
Solution. If we let and refer to the sound intensities of the two speak ers reproducing the test signal, then we have two equations:
i J
$e% log O
o:jÉ$e% log O
k k O h gmk kO h g
Using log properties, we can solve the first equation for
dk g
i
k
k
k
k
:
[e% log O î [ e% log O 6 9THe% log O 6 O¹9 O h i » » î log O 6 9Ò'J log O 6 O¹9 ð log e% õ õ [e% çz ¾D log õ » ð çz ¾ log ð [ e% çz ¾ O¹ [e% e% õ îJ
k
k
k
k
k
k
Similarly, we find that $e% ä l O . This means that the ratio of the intensities will be
k k
k
k
e% ä l O $e% O )jToJ e% ç ¾ O
n
This means that the test signal on the ü] 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 Ö , which is twice that of the first speaker. If
k
l
21.4. MEASURING THE LOUDNESS OF SOUND
k
k
k
î
is the corresponding intensity of the sound, then as above, [e% l We are assuming that "! , so this gives us the equation
k
k
l
297
k
Ó RO ð z O
.
î ! l î Ó e% ç ¾ O7 e% z O ð O ! g Ó î log O xe% ç ¾z{ log O e% z Oð ! h g Ó i J log O log O x e% z O ð { ! h g i Ö J .<%kj%e%:jw e% h o%¢ Ö
k
k
So, the test signal on the third speaker must produce a sound pressure level of 90 db.
CHAPTER 21. LOGARITHMIC FUNCTIONS
298
21.5 Exercises Problem 21.1. These problems will help you develop your skills with logarithms. (a) Compute: log log .
X
¨p
(b) Solve for ¿ : ¿ «.
zÉ¡
o
, log ´R , log
XY
, log
« ¿Ç¡
X
f
, log , log
¥
,
« É¡
(c) Solve each of these for Y¿ in Y equations Y terms of Û : ÛY¡#¥ , ÛY¡#¥ , ÛY¡Z¥R« .
Problem 21.2. 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.
p
p
(a) If the intensity of light at the water’s surface is , find a formula for (µ,k/ , the inÃ tensity of light at a depth of , meters. Your formula should depend on Ã and , .
p
(b) At what depth will the light intensity be decreased to 1% of its surface intensity?
X
Problem 21.3.d Rewrite each function in the Æ , for appropriate constants B form Û#¡B Ã Ã and Å . (a) (b) (c) (d) (e)
d
ÛY¡ZIñ / d ÛY¡T  ¯¨ / d ÛY¡?© »R ª / Z « d ÛY¡?©> : Rz&/ Z d ÛY¡ÇI »§: RI /óº
(c) Immediately after the February 28, 2001 Seattle earthquake, there were various reports concerning the actual Richter scale reading. These ranged from 6.5 to 7.2 on the Richter scale. How much more intense is a 7.2 earthquake compared to a 6.5 earthquake. Problem 21.5. (a) If you invest ù=± dollars at % 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 ù ± dollars at Á % 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 Ä4¡ ¤ p years to double the investment. Graphically compare this rule to the one isolated in part b. of this problem.
t
Problem 21.6. Recall Exercise 20.7 on Page 285. Potomac Power Co. needs to install a new powerline from the top of a cliff to the level ground below, as pictured. If we impose a coordinate system so that the origin is at the base of pole “A”, the hanging cable is modeled by a portion of the graph of e
Problem 21.4. The Richter Scale is used to measure the strength of an earthquake. The typical unit is °¡ log  / , where Ã is the ¨p minimal intensity one can detect and is the quake intensity.
rq6q s
pp
¿/¡Z¥R
cosh 
¿4©Y0 /© ¹ ¥R
At the position labeled “P”, the powerline is exactly 20 feet above the ground, which is the minimum distance from the ground to the cable.
(a) Discuss the ratio of the intensity between earthquakes measuring 6 and 7 on the Richter scale. (b) Suppose an earthquake is measured 8.2 replacements on the Richter scale and PSfrag an aftershock is onethird the intensity. What is the axis measurement of the aftershock on the axis axis Richter scale?
B
A
50 ft P
50 ft
100 ft
21.5. EXERCISES
299
(a) Find the height of poles “A” and “B”. (b) What portion of the powerline is at least 25 feet above the level ground? (c) What portion of the cable is above the level of the cliff top? (d) What is the horizontal distance from the base of pole “B” to the cable?
Problem 21.7. The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form
X <
d ÀÄ2/¡0R  ´©D ¼ª Z p &¨}¯ /
where À Ä2/ is the length (in centimeters) of a fish Ä years old. (a) What is the length of a newborn 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 À Ä2/ ?
(a) Suppose the value of the house is $75,000 in 1962. Assume ²+&¿/ is a linear function. Find a formula for ²+&¿/ . 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 ²Q&¿/ is a quadratic function. Find a formula for ²+¿/ . 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 ²+&¿/ is a function of exponential type. Find a formula for ²Q&¿/ . What is the value of the house in 1995? When will the house be valued at $200,000? Problem 21.10. Solve the following equations for ¿ :
«  &Q YT/S ¡ log &¿/ ¥ log ¶ ¡ ¡ log  ln &¿/À/¡ Y ¡#¥ Ye 0« ¡ Q T Y S sin ¡ ¨
(a) log (b) (c) (d) (e) (f) (g)
U
X X
Problem 21.8. 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 Ä grows according to an exponential model.
Problem 21.11. 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) Find a formula DÎÄ2/ for the number of cancerous skin cells after Ä hours.
(a) How long did it take the population to double?
(b) Assume a typical mouse skin cell is Z spherical of radius 0®Ë º cm. Find the combined volume of all cancerous skin cells after Ä hours. When will the volume of cancerous cells be 1 cm « ? Problem 21.9. Your Grandfather purchased a house for $55,000 in 1952 and it has increased in value according to a function Û ¡ ²+¿/ , where ¿ is the number of years owned. These questions probe the future value of the house under various mathematical models.
(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 exponentiallygrowing 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?
CHAPTER 21. LOGARITHMIC FUNCTIONS
300
Problem 21.12. 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 21.13. The voltage output(in volts) of an electrical circuit at time Ä seconds is given by the function
ãÄ"/¹¡ « æ
Q d Z sin
e « S ¨
Problem 21.14. A contractor has just built a retaining wall to hold back a sloping hillside. To monitor the movement of the slope the contractor places marker posts at the positions indicated in the picture; all dimensions are taken in units of meters. Assume that the hillside moves as time goes by and the hillside profile is modeled by a function j Ñ &¿/ after Î years. In Exercise 8.7 on Page 120, we showed
Û¢¡§j
Ñ e ¨ e ´ D © Ñ Ñ ¿k/¡ ñ / ¨ ¿mÂ ´ ©D
Find when the hillside first starts to spill over the retaining wall.
(a) What is the initial voltage output of the circuit?
hillside profile at time wall is constructed
markers
(b) Is the voltage output of the circuit ever equal to zero? Explain.
Yu Qid S
(c) The function æ Ä2/8¡ , where J Ä2/4¡ sin bfnÄÌ© _fª/TÂü . Put the sinusoidal 3 function Ä2/ in standard form and PSfrag replacements sketch the graph for ²hÇÄhZ . Label the coordinates of the extrema on the graph. axis1
wall
(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. (f) A picture of the graph of ÛI¡0æãÄ2/ on the domain h Ä hF is given; label the coordinates of the extrema on the graph. volts 15
5
axis axis
2
3
3
Problem 21.15. 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?
12.5 10
(b) When will the United States and Mexico have the same population?
7.5
PSfrag replacements
5
axis 2.5 axis axis
0.2
0.4
0.6
0.8
1
t axis
(g) Restrict the function æãÄ2/ to the domain ¹&Rh ÄRhü ; explain why this function has an inverse and find the formula for the inverse rule. Restrict the function æãÄ2/ to the domain ¹ Þh$Ädh ¹ ; explain why this function has an inverse and find the formula for the inverse rule.
Problem 21.16. 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.
21.5. EXERCISES
301
(a) How long does it take the population of Abnarca to double? (b) In what year will Abnarca’s population equal that of Bonipto? Problem 21.17. The average tenure of a Professor at the University of Washington is 31.6 years. The administration believes that a Professor’s salary after 31.6 years of service should be 2.5 times his/her “hiring in” salary. Assume the Professor’s salary grows with continuous compounding according to this constraint. (a) What is the annual rate Á of salary growth (to the nearest 0.1%)? (b) Assume inflation grows at an annual rate of ÁP¡ ¹ %, compounded continuously. If a Professor is hired at $30,000, what is the inflation adjusted buying power of his/her salary at retirement after 40 years of service? Problem 21.18. Return to the Earning Power Problem in 20.3. Using the data in Table 4.1, you obtained exponential models 3¿/$¡ ªª*» &/ Y and 51&¿/ï¡q¼RvÔ R ªRª¼0 / Y for
Women’s and Men’s Earning power in the year ;Â ¿ , respectively. Is there a time when women and men have equal earning power? If so, when? Problem 21.19. Due to the tremendous start of the 0R season for the Mariners, people are jumping on the bandwagon by joining the teams fan club. After the third game, there were 0 RR members in the Mariners fan club. After games, there were members of the fan club. (a) Develop an exponential model describing the number of members of the fan club as function of the number of games played in the season.
(b) At the allstar break, after games, how many members are in the fan club? (c) If there are ¹§v ª members of the Yankees fan club, during which game will the Mariners have as many members in the fan club as the Yankees?
Problem 21.20. Complete Table 21.1 on page 302.
CHAPTER 21. LOGARITHMIC FUNCTIONS
302
ÿ
vxwzy{~}v+ ò vxwy{~}
v+ vxwz{ ò ú O O l ÿ ú p
Point on ø
Ù
a
aþ
ÿ
}
vxwzy{ Point on ø
} vxwzy{ a f f a û f f a ÿ þf
ÿ þWû
û
Table 21.1: Complete this table for Exercise 21.20.
þ
Chapter 22 Parametric Equations
Imagine a car is traveling along the highway and you look down at the eplacementssituation from high above: axis axis
highway
axis
car
curve (static)
moving point (dynamic)
Figure 22.1: The dynamic motion of a car on a static highway.
We can adopt at least two different viewpoints: We can focus on the entire highway all at once, which is modeled by a curve in the plane; this is a “static viewpoint”. We could study the movement of the car along the highway, which is modeled by a point moving along the curve; this is a “dynamic viewpoint”. The ideas in this chapter are “dynamic”, involving motion along a curve in the plane; in contrast, our previous work has tended to involve the “static” study of a curve in the plane. We will combine our understanding of linear functions, quadratic functions and circular functions to explore a variety of dynamic problems.
303
ments
CHAPTER 22. PARAMETRIC EQUATIONS
304
axis axis
22.1 Parametric Equations
axis
{ ^W} maq
Gatorade
ö
PowerPunch
{i W} ³fq
ö ö
{×Ö(^W} ²mq
ments
tim
axis
michael
axis
(a) Tim and Michael running toward refreshments.
axis
{ ^W} maq
{Õ W} ³aq
ö
ö ö
® ¯ { ²u}
õ ö õ¶^±ô
²W {$Ö(aW} ²mq
Tim’s line of travel
{tW} q
Michael’s line of travel
(b) Modeling Tim and Michael as points moving on a path. Figure
22.2: Visualizing moving points.
Example 22.1.1. After a vigorous soccer match, Tim and Michael decide to have a glass of their favorite refreshment. They each run in a straight line along the indicated paths at a speed of 10 ft/sec. Will Tim and Michael collide? Solution. As a first step, we can model the lines along which both Tim and Michael will travel:
j 1 and 6 9Ò ! 8 + Tim’s line of travel: 6 6 9ÊH ¢ %%9´ ` % ! 6
Michael’s line of travel:
!
It is an easy matter to determine where these two lines cross: Set ! 6 9w+6 6 @9 and solve for , getting üb0! ` , so i the lines intersect at ØU 6 0! ` 1eJ ` 9 . Unfortunately, we have NOT yet determined if the runners collide. The difficulty is that we have found where the two lines of travel cross, but we have not worried about the individual locations of Michael and Tim along the lines of travel. In fact, if we compute the distance from the starting point of each person to Ø , we find:
i 6 %;H0! ` 9 6 %'eJ ` 9 "!! ` + kj ` feet
i dist(Tim, Ø ) ÿþ 6 + %%;)0! ` 9 a6 ` %;HeJ ` 9 i )j% ^ feet
dist(Mike, Ø ) ÿþ
Since these distances are different and both runners have the same speed, Tim and Michael do not collide!
ements
22.2 Motivation: Keeping track of a bug Imagine a bug is located on your desktop. How can you best study its motion as time passes? Let’s denote the location of the bug when you first obS axis served it by Ø . If we let / represent time elapsed since first Figure 22.3: A bug on your spotting the bug (say in units of seconds), then we can let desktop. Ø 6 /29 be the new location of the bug at time / . When /wW% , which is the instant you first spot the bug, the location Ø 6 %9.Ø is the initial location. For example, the path followed by the bug might look something like the dashed path in the next Figure; we have indicated the bug’s explicit position at four future times: / X"/ X)/ X)/ .
axis
Ú
axis
Ö
l
p
PSfrag replacements
22.3. EXAMPLES OF PARAMETRIZED CURVES
axis
How can we describe the curve in Figure 22.4? To start, lets define a couple of new functions. Given a time / , we have the point Ø 6 /:9 in the plane, so we can define:
 axis
8
305
Æ É 
 
Æ ~qÉ
6 :/ 9Ò" coordinate of Ø 6 :/ 9 at time /21 M 6 /:98êM coordinate of Ø 6 2/ 9 at time /2
Æ É

Æ É
Æ É
axis
Figure 22.4: A bug’s path.
In other words, the point Ø 6 /29 is described as
Ø 6 :/ 9A 6 6 /:9z1§M 6 /:99 We usually call ' 6 /29 and M M 6 /:9 the coordinate functions of Ø 6 /29 . Also, it is common to call the pair of functions ] 6 /:9 and M"WM 6 /:9 the
parametric equations for the curve. Any time we describe a curve using parametric equations, we usually call it a parametrized curve. Given parametric equations Z 6 /:9 and Mê.M 6 /29 , the domain will be the set of / values we are allowed to plug in. Notice, we are using the same set of / values to plug into both of the equations. Describing the curve in Figure 22.4 amounts to finding the parametric equations 6 /:9 and M 6 /29 . In other words, we typically want to come up with “formulas” for the functions 6 /:9 and M 6 /29 . Depending on the situation, this can be easy or very hard.
22.3 Examples of Parametrized Curves We have already worked with some interesting examples of parametric PSfrag replacements equations. axis axis
Example 22.3.1. A bug begins at the location (1,0) on the axis PSfrag unit circle and moves counterclockwise with anreplacements angular ° bug starts movspeed of ! rad/sec. What are the parametric equa axis ing at ö rad sec tions for the motion of the bug during the first 5 seconds? axis Indicate, via “snapshots”, the location of the bug at 1 sec axisFigure 22.5: A circular path. ond time intervals.
Solution. We can use Fact 14.2.2 to find the angle swept out after / seconds: Í ° /P !n/ radians. The parametric equations are now easy to describe:
" 6 /29 cos 6 !n/29 Ç M½êM 6 :/ 9Ò sin 6 n! /:9
®{ õ ® {i ® {$ÖW ® {tq
® {
ö
® {$²W
Figure 22.6: Six snapshots.
If we restrict / to the domain 0 %´1 ` 2 , then the location of the bug at time / is given by Ø 6 /:9â 6 cos 6 !n/291 sin 6 !n/:99 . We locate the bug via six onesecond snapshots:
CHAPTER 22. PARAMETRIC EQUATIONS
306
When modeling motion along a curve in the plane, we would typically be given the curve and try to find the parametric equations. We can turn this around: Given a pair of functions Ç" 6 /:9 and MNêM 6 /29 , let
Ø 6 2/ 9 6 6 /:9z1 M 6 /29¼9z1
(22.1)
which assigns to each input / a point in the 2M plane. As / ranges over a given domain of allowed / values, we will obtain a collection of points in the plane. We refer to this as the graph of the parametric equations w} 6 /:9z1§M 6 /29fy . Thus, we have now described a process which allows us to obtain a picture in the plane given a pair of equations in a common single variable / . Again, we call curves that arise in this way parametrized curves. The terminology comes from the fact we are describing the curve using an auxiliary variable / , which is called the describing “parameter”. In applications, / often represents time.
ements
axis
Ú
® { ¯ i{ ³q} W ö ® { ¯ {tW} ö ® {Õ ¯ {i*W} õ ® {i ¯ {iI³qõ } S axis
ö
axis
õ
Figure 22.7: Observing the motion of ù .
Example 22.3.2. The graph of the parametric equations 6 /29Çtj/ and M 6 /:9t/ïÿ on the domain ;!¦'C/¤'G! is pictured; it is a line segment. As we let / increase from ;! to ! , we can observe the motion of the corresponding points on the curve.
22.4 Function graphs It is important to realize that the graph of every function can be thought of as a parametrized curve. Here is the reason why: Given a function MN#! 6 @9 , recall the graph consists of points 6 @1K! 6 99 , where runs over the allowed domain values. If we define
" 6 /:93)/ Ç MNêM 6 /29Ò#! 6 /:9z1 then plotting the points Ø 6 /29¢ 6 6 /:9z1§M 6 /:99D 6 /21H! 6 /299 gives us the graph of ! . We gain one important thing with this new viewpoint: Letting / increase in the domain, we now have the ability to dynamically view a point Ø 6 /:9 moving along the function graph. See how this works in Example 22.4.1.
Example 22.4.1. Consider the function MêW on the domain
;!§'H%'H! . As a parametrized curve, we would view the graph of M . as all points
of the form Ø 6 /:9Ë 6 /1/ 9 , where ;!§'./Ð''! . If / increases from ;! to ! , the corresponding points Ø 6 /29 move along the curve as pictured: Solution. For example, Ø 6 ;!A9m etc.
+
9 6 n19 , 6 ;!1 6 ;!A9 93 6 ;!1 9 , Ø 6 9Ì 6 n1
axis
22.4. FUNCTION GRAPHS
Ú
Ö
axis
Ö
í
æ static graph
Ú
S
S
axis
307
î î Ú
î æ ð æ ¾ u ð î Ö æ
æ ð
Ö î
î Ö
Ö î ð
axis
Ö ORð
ð
¾ u ð S
axis
motion along curve
PSfrag replacements Figure 22.8: Visualizing dynamic motion along a static curve.
Ú Not every parametrized curve is the graph of a function. For example, consider these possible curves in the plane: The second curve from the left is the graph of a function; the other curves violate the vertical line test.
axis
axis
S
axis
Figure 22.9: Some curves that are not functions.
22.4.1 A useful trick There is an approach to understanding a parametrized curve which is sometimes useful: Begin with the equation ½ 6 /:9 . Solve the equation F 6 /29 for / in terms of the single variable ; i.e., obtain /¢ 6 6 @9 . Then substitute /¦6 6 9 into the other equation MNêM 6 /:9 , leading to an equation involving only the variables and M . If we were given the allowed / values, we can use the equation Ç 6 /29 to determine the allowed values, which will be the domain of values for the function MNêM 6 6 6 99 . This may be a function with which we are familiar or can plot using available software. Example 22.4.2. Start with the parametrized curve
given by the equations FW 6 /29>]!n/ ` and MêH MPSfrag /6 294[replacements / , when %B''/)'$e% . Find a function M Å! 6 @9 whose graph gives this parametrized curve.
100
Ú
axis
80 60 40
Solution. Following the suggestion, we begin by solving axis [C!n/ï ` for / , giving /ü 6 ` 9 . Plugging this into S axis
the second equation gives My x 6 ¢ ` 9 { x { 6 ` 9 . Figure 22.10: Finding the p curve if and Conclude that 6 81§M9 is on the parametrized
path equation.
only if the equation M)½! 6 9 x 6 ¢ ` 9 { x { 6 I ` 9 is p graph will be an upward satisfied. This is a quadratic function, so the opening parabola with vertex (5,0). Since the / domain is %3'Ï/%' e% , we get a new inequality for the domain: %¤' x: { 6 Ç ` 9='Ge% . Solving this, we get %?'BÇ ` 'B! 6 e%9 , so 20
−10
10
20
CHAPTER 22. PARAMETRIC EQUATIONS
308
of the parametrized curve is the graph ' ')! ` . This means the graph ` '
{ of the function M xv 6 ï ` 9 , with the domain of values 0 ` 1! ` 2 . Here is ! 6 @9 ; the thick portion is the parametrized curve a plot of the graph ofp MN# we are studying.
ements
axis
22.5 Circular motion Ú
axis
®O¯ starting location
r {$Á æ } ~ æ ® { Ü ¯ {tÁ"{ Ü } ~0{ Ü A ¯ location at time Ü
S
axis
Figure 22.11: Circular motion. ements
We can describe the motion of an object around a circle using parametric equations. This will involve the trigonometric functions. The general setup to imagine is pictured: An object moving around a circle of radius centered at a point 6 1§M 9 in the 2M plane. The path traced out is the circle. However, the location of the object at time / will depend on a number of things:
The starting location Ø of the object; The angular speed ° of the object;
The radius and the center 6 1§M 9 .
We will build up to the general solution by considering two cases, the first being a special case of the second.
axis
22.5.1 Standard circular motion Ú
® {
ö
axis
ö²
²Ü ® { Ü ¯ ${ Á"{ Ü } ~*{ Ü A Figure
² S
® {
õ
axis
®8¯ t{ m} q ¯Ð® {tq ¯ starting
As a first case to consider, assume that the center of the circle is 6 %´1%9 and the starting location Ø' 6 1%9 , as pictured below. If the angular speed is ° , then the angle swept out in time / will be Z ° / ; this requires that the time units in ° agree with the time units of / ! We denote by Ø 6 /:9m 6 6 /291§M 6 /:99 the M coordinates of the object at time / . At time / , we can compute the coordinates of Ø 6 /29 6 6 /:9z1§M 6 /:99 using the circular functions:
" 6 /:93" cos 6 ° /:9 Ç 22.12: Standard cirMNêM 6 2/ 9Ò" sin 6 ° :/ 9 cular motion. This parametrizes motion starting at Øê location
6 n1%9 . Using the shifting tech
nology of Chapter 9, we are led to a general description of this type of circular motion, which involves a circle of radius centered at a point 6 1§M 9 ; we refer to this situation as standard circular motion.
Important Fact 22.5.1 (Standard circular motion). Assume an object is moving around a circle of radius centered at 6 1§M 9 with a constant angular speed of ° . Assume the object begins at Ø] 6 '1§M 9 . Then the location of the object at time / is given by the parametric equations: Ç" 6 /:9Ò" cos 6 ° /:9 and MNêM 6 /29ÒêM sin 6 ° /:9 .
axis axis
22.5. CIRCULAR MOTION
Example 22.5.2. Cosmo the dog is tied to a 20 foot long tether, as in Figure 14.1. Assume Cosmo starts at the location “ ¤ ” in the Figure and maintains a tight tether, moving around the circle at a constant angular speed ° rau dians/second. Parametrize Cosmos motion and determine where the dog is located after 3 seconds and after 3 minutes.
309
Cosmo motors
{

} X~
feet
around
the circle
Figure 22.13: Cosmo on a
Solution. Impose a coordinate system so that the pivot running on a circular path. ° V %, point of the tether is 6 =O:1§MÊOe9] 6 %´1%9 . Since Cosmo is walking counterclockwise around the circle. By (4.1.6), the location of Cosmo after / seconds is Ø 6 /:9 6 6 /29z1§M 6 /29¼9 x !% cos x / { 1!% sin x / {{ . After 3 seconds, Cosmo is located at Ø 6 jT9 x !% cos x l u { 1!% sin x l u {e{ 6 <^deJ´1eo%v!A9 . After 3 minutes = 180 seconds, O O u u the location of the dog will be Ø 6 eJ%9 x !% cos x ç { 1!% sin x ç {e{ 6 !%´1k%9 , u PSfragu replacements which is the original starting point.
22.5.2 General circular motion
axis
The circular motion of an object can begin at any location ~ rel axis Ú axis Ø on the circle. To handle the general case, we follow an ®8¯ starting position earlier idea and introduce an auxiliaryrelative coordinate ý system : The rel M rel coordinates are obtained by drawing Á rel axis lines parallel to the M axis and passing through 6 1§M 9 . We are using the subscript “rel” to stand for “relative”. initial angle {$Á æ } ~ æ This new relative coordinate system has origin 6 1§M 9 and S axis allows us to define the initial angle O , which indexes the Figure 22.14: Initial angle starting location Ø , as pictured below: and auxiliary axis. Assume the object starts at Ø and is moving at a constant angular speed ° around the pictured circle of radius . Then after time / has elapsed, the location of the object is indexed by sweeping out an angle ° / , starting from O . In other words, the location after / units of time is going to be determined by the central standard angle nO; ° / with initial side the positive rel axis. This means that if Ø 6 /:9 6 6 /291§M 6 /:99 is the location of the object at time / ,
" 6 2/ 9" cos 6 OÌ ° :/ 9 Ç M½êM 6 :/ 9ÒêM sin 6 Oâ ° /:9
Notice, the case of standard circular motion is just the scenario when nOH% and these parametric equations collapse to those of Fact 22.5.1. Important Fact 22.5.3 (General circular motion). Assume an object is moving around a circle of radius centered at 6 1§M 9 with a constant angular speed of ° . Assume the object begins at the location Ø with initial
CHAPTER 22. PARAMETRIC EQUATIONS
310
ements
angle n O , as in Figure 22.14. The object location at time / is given by: Ç" 6 /:9Ò" cos 6 nOâ ° /:9 and MNêM 6 /:9ÒêM ü sin 6 nOâ ° /:9 .
axis axis axis
r
rad sec


Ä
X~ y
ft
Ticket Booth
Example 22.5.4. A rider jumps on a merrygoround of radius 20 feet at the pictured location. The ride rotates at the ° constant angular speed of G radians/second. The ¾ center of the platform is located 50 feet East and 50 feet North of the ticket booth for the ride. What are the parametric equations describing the location of the rider? Where is the rider after 18 seconds have elapsed? How far from the ticket booth is the rider after 18 seconds have elapsed?
:
axis (north) q
Æ ~qÉ
X~ y
ft
axis (east)
Figure 22.15: A rider jumps on a merrygoround.
Solution. In this example, since ° X % , the rotation is clockwise. Since the angular speed is given in radians, we need to convert the initial angle to radians as well: !AeJ jTJ% ` radians. Impose a coordinate system so that the center of the ride is 6a` %´1 ` %9 and its radius is 20 feet. By Fact 22.5.3, the parametric equations for the rider are given by q 6 /:9F ` %ÇW!% cos x}jTJ% ` / { and ¾ MNêM 6 /:9Ò ` % !% sin x jT&J% ` / { . The location after 18 seconds will be ¾
Ø 6 e J9 6 6a+` 6
+ 6 eJ9z1§M 6 eJ99 + i i %;ü + !% cos 6 _! jT9z1 ` % !% sin 6 _! jT99 o´1^Jde%9
The distance from Ø 6 eJ9 to the origin is
+
+
Çÿþ 6 + o9 6 ^Jae%9 i i o feet
22.6. EXERCISES
311
22.6 Exercises
(a) (b) (c) (d)
¿QÄ2/¡ZRaÛÄ2/¡ Ë©ÉRÄ h Ä h . ¿Q2Ä /¡Ä´©?RaJÛ  Ä2/¡ :}h Ä h . ¿+ 2Ä /¡Ä´©?Rd Û  2Ä /¡ Ë©YRÄ}hIÄÖh ¿Q2Ä /¡ Ë©YÄa Û 2Ä /¡RÄT©Iª hIÖ Ä h
Ú
axis
250
.
200
.
150
100 Problem 22.2. The population of caribou and wolves in a remote Alaskan valley is modeled 50 by PSfrag replacements
¿+ Ä2/ ¡ ÛJ Ä2/ ¡
¥R
where
Ä
¥R
m
sin m
sin
b f Ä´© ªR/ Y Â 0R n b f Ä´© / Y n Â 0R
100
150
200
250
S
axis
axis
 wolves /}
(a) Sketch the graphs of both population models on the same axes, for times between 1970 and 2000. Clearly label the axes on your graphs. (b) The wolves can be regarded as predators and the caribou as prey. In a., the predator graph “lags” the prey graph. Describe what this means in words. (c) Using sketches you made above, how often will the predator population exceed the prey population? (d) Sketch the parametric equations BÍ Ä2/¡ &¿QÄ2/}dÛÄ2/f/ in the ¿0Û coordinate system for the time domain h Ä h .
Problem 22.3. Return to Example 22.3.1 on Page 305. What are the coordinates of the bug in the six “snapshots” in the solution? When will the bug first cross from the first to second quadrant? When will the bug first have ¿ coordinate ¡©Ê ? Problem 22.4. Lee has been boasting all year about his basketball skill and Allyson can’t stand it anymore. So, she has challenged Lee to a game of oneonone. With 2 seconds to go, Allyson is leading 1412 and Lee fires a desperation three. Impose coordinates with Lee’s feet at the origin and use units of feet on each axis. The path of the ball is described by the parametric equations: ¿QÄ2/ ¡[ ¼RÄ , ÛÄ2/¡?©4ª¼Ä@ÂÉ: RÄÂIª .
(e) Using your parametric sketch from the previous question, plot the point BLñ0/¡ &¿Qñ*/ ÛJó0/À/ . Describe the motion ofreplacements BLÄ2/ PSfrag and around the plot for %hHÄ
50
 caribou /
represents years since 1970.
graph of B PÄ2/¡ &¿ Ô Ä2/ Û Ô Ä2/À/ is as pictured below. Explain what is happening. (Note: BLñ0/ is the rightmost endpoint of the curve.)
Problem 22.1. Sketch the curve represented by the parametric equations. Choose four specific Ä values and indicate the corresponding points on the curve. As Ä moves from left to right in the domain, indicate how the corresponding points on the curve are moving.
axis
43 feet
10 feet lee
allyson 16 feet
(a) How high is the ball when Lee fires his shot? (b) When is the ball directly over Allyson?
CHAPTER 22. PARAMETRIC EQUATIONS
312
(c) Allyson can jump and block a ball 9.5 feet above the floor. Can Allyson block Lee’s shot?
100 ft
(d) Where and when is the ball 22 feet above the floor? PSfrag replacements (e) Where and when does the ball reach it’s highest point above the floor?
axis axis
(f) Who wins the game?
axis
(g) Solve the equation ¿;¡ ¿QÄ2/ for Ä in terms of ¿ , then plug this into the equation for ÛD¡ÛÄ2/ . Sketch the graph of the resulte &¿/ ; this is the path of ing function Û?¡ Lee’s shot. Problem 22.5. A six foot long rod is attached at one end B to a point on a wheel of radius 2 feet, centered at the origin. The other end C is free to move back and forth along the ¿ axis. The point B is at v 0/ at time Äw¡ê , and the wheel rotates counterclockwise at 3 rev/sec.
Q
44 ft high pole ground level
(a) What is the angular speed ¿ ? How fast is the rider moving in mph? (b) Find parametric equations for the motion of a rider on the wheel, assuming the rider begins at the lowest point on the wheel. (c) Due to a malfunction, the ride abruptly stops 35 seconds after it began. To the nearest foot, how high above the ground is the rider?
2
PSfrag replacements
A A 2
axis axis axis
2
B
B 4
6
2
8
2
4
6
8
time t > 0
time t=0
(a) As the point B makes one complete revolution, indicate in the picture the direction and range of motion of the point C . (b) Find the coordinates of the point function of time Ä . (c) Find the coordinates of the point function of time Ä .
B
Problem 22.7. Cherie is running clockwise around a circular track of radius 300 feet. She starts at the location pictured running with an angular speed of RPM clockwise. Julie stands at the pictured£ location, 100 feet from the center of the circle.
C
as a 300 ft.
as a
Q=julie
(d) What is the ¿ coordinate of the point C when Ä3¡ê ? 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?). PSfrag (e) Find the first two times when the ¿ replacements coordinate of the point
C
is 5.
Problem 22.6. A ferris wheel of radius 100 feet is rotating at a constant angular speed ¿ counterclockwise. Using a stopwatch, the rider finds it takes 3.4 seconds to go from the lowest point on the ride to a point ú , which is level with the top of a 44 ft pole. Assume the lowest point of the ride is 3 feet above ground level.
axis axis axis
P=cherie 5/9 RPM
(a) When will Cherie first reach the location on the track closest to Julie? (b) Where is Cherie located in 40 seconds? (i.e. find her coordinates). Indicate this point in your picture and label it as ùJ&*/ .
22.6. EXERCISES (c) How far has Cherie traveled (distance she has run) in 40 seconds? (d) Let ,+Ä2/ be the function that calculates the distance between Cherie and Julie at time Ä seconds. (d1) If the domain is taken to be the time required for Cherie to complete one revolution, what is the domain and range of ,+ Ä2/ ?
313 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, find parametric equations describing the location of the bug at time Ä seconds. ω=4π/9 rad/sec bug lands here
ω=4π/9 rad/sec 1.2 rad bug lands here
(d2) Write down a formula to calculate ,+Ä2/ .
2 in
(e) Assume that Julie is instead located at the position ókÅ+/ , where òhÅThqR is a given constant. (So part (d) was the case when Å?¡H¥R . ) Find a function , Æ Ä2/ that computes the distance ω=4π/9 rad/sec between Julie and Cherie at time Ä . Your formula will involve Ä and Å and should collapse to the formula for ,+ Ä2/ in (d) if you set Å;¡ZR . PSfrag replacements
2 in
ω=4π/9 rad/sec
bug lands here
Problem 22.8. 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
axis axis axis
o −40 2 in
2 in bug lands here
ω= − 4π/9 rad/sec
ω= − 4π/9 rad/sec
bug lands here
bug lands here 1.2 rad 0.5 rad
2 in
2 in
314
CHAPTER 22. PARAMETRIC EQUATIONS
Chapter 23 Linear Motion The simplest example of a parametrized curve arises when studying the motion of an object along a straight linePSfrag in thereplacements plane. We will start by studying this kind of motion when the starting and ending locations are axis known. axis PSfrag replacements axis axis
23.1 Motion of a Bug
íPî
axis
Example 23.1.1. A bug is spotted at Ø 6 !1 ` 9 in the M  axis plane. The bug walks in a straight line from Ø to Ù 6 ^´1jT9 at a constant speed . It takes the bug 5 seconds to reach Ù . Assume the units of our coordinate system are feet. What is the speed of the bug along the line connecting Ø and Ù ? Compute the horizontal and vertical speeds of the bug and show they are both constant. Solution. A standard technique in motion problems is to analyze the and M motion separately. This means we look at the projection of the bug location onto the and M axis separately, studying how each projection moves. We can think of these projections as “shadows” cast by a flashlight onto the two axes: For the motion, we study the “shadow” on the axis which starts at “2” and moves toward “6” on the axis. For the M motion, we study the “shadow” on the M axis which starts at “5” and moves toward “3” along the M axis. In general, speed is computed by dividing distance by time elapsed, so
w
dist 6 ØT1ÙÇ9 ft sec ` ! ) ` feet ` sec
þ 6a` üjT9 6 !<ê^9 ft sec `
(23.1)
P axis
¢
S
íPî Q
ël ð
axis
(a) A bug walking from to ú .
~ ~
axis
shadow on axis

²
ù
Á ¯ ${ Á ?} ~ ¯ { } ²W _ _ ~ P ö ¯ {$Á }?~ ¯ {i³qS} W ¶ Á ¶ Q ö shadow³ on Á axis axis
light on bug casts shadow for motion R light on bug casts shadow for motion
(b) How to view motion in the coordinates. Figure 23.1: Visualizing the model for the bug problem.
This is the speed of the bug along the line connecting Ø and Ù . 315
¡¡ ¡ ¡ ë u ð¡¡ ¡ ¡ ¡
CHAPTER 23. LINEAR MOTION
316
The hard part of this problem is to show that the speed of the horizontal and vertical shadows are also constant. This might seem obvious when you first think about it. In order to actually show it, let’s take two intermediary positions ¤ 6 1§M 9 and ¥ü 6 1§M 9 along the bugs path. We are going to relate the horizontal speed ä of the bug between and S , the vertical speed ä Ú of the bug between M and M and the speed of the bug from ¤ to ¥ . Actually, because there are positive or negative directions for the and M axes, we will allow horizontal and vertical “speed” to be a ¦ quantity, with the obvious meaning. If it takes seconds for the bug to travel from ¤ to ¥ , then is the elapsed time for the horizontal motion from to and also the elapsed time for the vertical motion from M to M . The horizontal speed ä S is the directed distance 7 6 U 9 divided by the time elapsed , whereas the vertical speed ä Ú is the directed distance 4M 6 M M 9 divided by the time elapsed . We want to show that ä and ä Ú are both constants! To do this, we S have these three equations:
3 4M M ½ M ä Ú distance 6 ¤â1Õ¥@9 w ä S
þ 3
4M
þ 6 9 6 M NM 9
Now, square
each
side of
the three equations and combine
them to conclude: ä 9ä Ú . We can multiply through by and that gives S us the key equation:
ä . ä Ú S
(23.2)
On the other hand, the ratio of the vertical and horizontal speed gives
÷ x æÚ { >M ÷ 3 x æS {
M M “slope of line connecting Ø and Ù ” H ! ä Ú H ä S ! Solving for ä Ú in terms of ä we can write S ä Ú H ä S ! ä Ú ä S
(23.3)
[
Since ä
S is positive,
g
) ` ) ` ý ä S 1 ä Ú ä ä S ä ä S ! h ! S ! S
(23.4)
23.2. GENERAL SETUP that is,
!
+
! ) ` feet '%J sec ` s ` ) ` ) ` + ft
. This shows the speed in both the horiBy (23.4), ä Ú txU { p <% sec u zontal and vertical directions is constant as the bug moves from ¤ to ¥ . Since ¤ and ¥ were any two intermediary points between Ø and Ù , the ä S
w
!
317
r
horizontal and vertical bug speeds are constant.
Example 23.1.2. Parametrize the motion in the previous problem.
+
S '%J feet/sec and ä Ú H<% feet/sec, so 6 coordinate of the bug at time /:9
Solution. We have shown ä
6 :/ 9
6 beginning coordinate 9T
! %&J:/2
M 6 :/ 9b
distance traveled in r the direction in / s seconds
6 M coordinate of the bug at time /29
PSfrag replacements distance traveled in r the M direction axis in / s seconds axis
6 beginning M coordinate 9T + ` % /2 + axis + As a check, notice that Ø 6 %9Ê 6 !ï 6 %&J9§%´1 ` 6 % 9§%9â 6 !1 ` 9 is the starting ^ 1jT9 is the ending location. location and Ø 6d` 9 6 !7 `86 %&J9z1 ` `86 % 998 6 ´
23.2 General Setup
íïî
Given two points ØZ 6 1§M 9 and Ù 6 1§M 9 in the plane, Ú axis location ending we can study motion of an object along the line connecting starting í location î Q Ø and Ù . In so doing, you need to first specify the starting SRö ë Ú ö ð axisÖ S location and the ending location of the object; lets say we moving from to at a constant speed ë ð P start at Ø and proceed to Ù . Fix the distance units used Sõ Ú õ Figure 23.2: Typical scein the coordinate system (feet, inches, miles, meters, etc.) × nario. and the time units used (seconds, hours, years, etc.). As highlighted in the solution of Example 23.1.2, the key is to analyze the motion and M motion separately. We will be imposing an assumption that the speed along the line connecting Ø and Ù in Figure 23.1(b) is a constant. There is a crucial observation we need to make, a special case of which was the content of Example 23.1.1. Important Facts 23.2.1 (Linear motion). Assume that an object moves from Ø to Ù along a straight line at a constant speed , as in Figure 23.2.
CHAPTER 23. LINEAR MOTION
318
Then the speed ä in the direction and the speed ä Ú in the M direction are S both constant. We also have two useful formulas:
ä 1 ä Ú 1 S slope of line of travel, ä Ú Î when the line is nonä S vertical.
ements axis
!!!
axis
CAUTION !!!
This fact is established using the same reasoning as in Example 23.1.1. Let’s make a few comments. To begin with, if the line of travel is either vertical or horizontal, then either ä y% or ä Ú é% and Fact 23.2.1 isn’t S really saying anything of interest. The formulas in Fact 23.2.1 only work for linear motion.
axis
PSfrag replacements
Ú
axis axis axis
Ú
axis
axis
Ö × ×
Ú
no motion
S
S axis
Ö
axis
S
no motion
Figure 23.3: Horizontal or vertical motion.
For any other line of travel, we can use the reasoning used in Example 23.1.1. Pay attention that the horizontal speed ä and the vertical S speed ä Ú are both directed quantities ; i.e. these can be positive or negative. The sign of ä will indicate the direction of motion: If ä is positive, S S then the horizontal motion is to the right and if ä is negative, then the S horizontal motion is to the left. Similarly, the sign of ä Ú tells us if the vertical motion is upward or downward. Returning to Figure 23.2, to describe the motion, two pieces of information are needed: the starting location (in the direction) and the constant speed ä in the direction. So,
S
6 :/ 9Ê 6 coordinate of the object at time /:9
distance
traveled
in
6 beginning coordinate 9T r the direction in / 1ä S E /2
time units
s
23.2. GENERAL SETUP
319
If we are not given the horizontal velocity directly, rather the time required to travel from Ø to Ù , then we could compute ä using the fact S that the object starts at Ç" and travels to Ç" :
6 directed horizontal distance traveled 9 6 time required to travel this distance 9 ending coordinate 9T 6 starting coordinate 9 6 6 time required to travel this distance 9 7 ä S To describe the M motion in Figure 23.2, we proceed similarly. We will denote by ä Ú the constant vertical speed of the object, then after / time units the object has traveled ä Ú E / units. So,
ä S
[
M 6 /:9b
6 M coordinate of object at time /:9
distance
traveled
in
6 beginning M coordinate 9T r the M direction in /
M ä Ú E /2
s
time units
In summary, Important Fact 23.2.2 (Linear motion). Suppose an object begins at a point Ø 6 1§M 9 and moves at a constantPSfrag speed replacements along a line connecting Ø to another point Ù$ 6 1§M 9 . Then the motion of this objectaxis will trace out axis a line segment which is parametrized by the equations:
M
6 /29Ò" . ä S E /21 M 6 /:9ÒêM ä Ú E /2
axis
~
Example 23.2.3. Return to the linear motion problem studied in Example 23.1.1 and 23.1.2. However, now assume that the point Ù 6 ^´1jT9 is located at the center of a circular region of radius 1 ft. When and where does the bug enter this circular region? Solution. The parametric equations for the linear motion of the bug are given by:
axis
£¤¤£ ¤£¤£ ö ¤£¤£ ¤£¤£ ¤£¤£ ¤£ ¤£
®8¯ { } ²W
location E where bug enters region
´ ¯ {i³q} W
circular region radius ft.
Figure 23.4: A bug crosses a circular boundary.
" 6 /29"! ê Ç %&+J:/1 M½êM 6 /:9Ò ` ê % /2 The equation of the boundary of the circular region centered at Ù by
6 ¢^9 6 M üjT9 [
õ Á axis
is given
CHAPTER 23. LINEAR MOTION
320
To find where and when the bug crosses into the circular region, we determine where and when the linear path and the circle equation have a simultaneous solution. To find such a location, we simply plug Ç" 6 /:9 and M M 6 /:9 into the circle equation:
^ 9 M : / ´ 9 T j 9 + 6 6 /:9´ 6 6
% /mjT9 + %J:/Ì+ ^ 9 + 6a` 6 !<ê e^^ /m%^ / ' &^:/mHde^:/ %J:/ J:/m'eo
%
Notice, this is an equation in the single variable / . Finding the solution of this equation will tell us when the bug crosses into the region. Once we know when the bug crosses into the region, we can determine the location by plugging this time value into our parametric equations. By the quadratic formula, we find the solutions are
/;
J¦ þ ^
+
+
&^
6 eo9 6 %&J9 H^d0! or jT&JJ
We know that the bug reaches the point Ù in 5 seconds, so the second solution / jT&JJ seconds is the time when the bug crosses into the circular region. (If the bug had continued walking in a straight line directly through Ù , then the time when the bug leaves the circular region would correspond to the other solution /] ^d0! seconds.) + bug when it crosses into the region is Finally, the location of the F 6 6 jTJJ9z1§M 6 jTJJ99 6a` de%´1jT ` 9 .
23.3. EXERCISES
321
23.3 Exercises Problem 23.1. 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. (a) Write parametric equations for Margot’s position Ä seconds after she starts walking. (b) Write an expression for the distance from Margot’s position to the statue at time Ä . (c) Find the times when Margot is 28 feet from the statue.
(b) Find the speed and direction of the bug along the line connecting ú to ù ? How about the horizontal and vertical speeds? (c) How far is the bug from the origin after 2 seconds? Problem 23.5. The cup on the th hole of a golf course is located dead center in the middle of a circular green which is 70 feet in diameter. Your ball is located as in the picture below. Assume that you putt the ball at a constant speed in a straight line into the cup. Assume it takes 6 seconds to land in the cup. 9
PSfrag replacements Problem 23.2. 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 Ä seconds after they start walking. (b) When are Sven and Rudyard closest? What is the minimum distance between them?
Û
Problem 23.3. Juliet and Mercutio are moving at constant speeds in the ¿0Û plane. They start moving at the same time. Juliet starts at the point ñ¹0©ÍªR/ and heads in a straight line toward the point » I/ , reaching it in 10 seconds. Mercutio starts at åe©§I/ and moves in a straight line. Mercutio passes through the same point on the ¿ axis as Juliet, but 2 seconds after she does. How long does it take Mercutio to reach the axis? Problem 23.4. Return to Example 23.1.1 on Page 315, but assume the bug begins at ú and walks toward ù . Again assume the bug walks at a constant speed, requiring 5 seconds to traverse the distance. (a) Describe the motion of the bug pictorially and via parametric equations.
Cup
axis axis
Green
axis
72 Feet
Ball
Rough
32 Feet
(a) How far does the ball travel? (b) Find parametric equations for the motion of the ball. (c) How far is the ball from the cup after 3.5 seconds? (d) Where and when does the ball cross from the rough onto the green? Problem 23.6. Return to Example 23.1.2 on Page 317. How far is the bug from the origin at time ÄÒ¡Z ? Where and when will the bug be closest to the point  ¹ I/ ? When is the bug 6 feet from the origin?
d
Problem 23.7. The cup on the hole of a golf course is located 4 feet south and 4 feet east of the center of a circular green which is 54 feet in DIAMETER. Your ball is located as in the picture below. Assume that you mistakenly putt the ball which travels at a constant speed in a straight line toward the center of the green. It takes 7 seconds to reach the center of the green. Use distance units of FEET and time units of SECONDS in this problem.
CHAPTER 23. LINEAR MOTION
322 30 ft
4 ft
ball green
enters green
18
24 ft
cup
PSfrag replacements axis axis
6 ft
5 ft
PSfrag replacements
leaves green rough
axis axis
12 ft
this pocket for the big money
axis
axis
(a) Find parametric equations for the motion of the ball. (b) How far is the ball from the cup after 5 seconds? (c) Where and when does the ball cross into the green and out of the green? (d) Find when and where the ball is closest to the cup. (Hint: Construct a line thru the cup perpendicular to the path of the ball, then find where these two lines intersect.)
Problem 23.9. Two practicing hockey players located at positions B and C in the picture. The player at position B strikes a puck, which we will assume moves at a constant speed along the line connecting B and D ; assume it takes 2 seconds for the puck to go from B to D . At the same instant, the player at position C strikes a puck, which we will assume moves at a conE stant speed along the line connecting C and ; also assume it takes 2 seconds for the puck to E go from C to .
(e) Find a function of Ä that computes the distance from the ball to the cup at any time Ä . (f) When and where is the ball 8 feet from the cup?
30 ft D
30 ft C
50 ft
50 ft centerline
Problem 23.8. Recall the Billard problem in 15.1.1 on Page 199. Assume you strike the ball and it travels a constant speed of 20 in/sec along the dotted path. PSfrag replacements (a) Find parametric equations for the motion of the ball. (Hint: Both the coor axis dinate functions will be multipart func axis tions.) axis (b) Where is the ball located after 2 seconds? (c) Where is the ball located after 5 seconds? (d) When does the ball strike the cushion? (e) When does the ball land in the pocket? (f) When is the ball 2 feet from the pocket?
30 ft
40 ft B
A
rink 20 ft
30 ft
centerline (a) Describe parametric equations for the motion of each puck. How fast is each puck moving? (b) Describe the location of each puck after 1.4 seconds. (c) Compute the distance between the two pucks after 1.4 seconds.
23.3. EXERCISES
323
(d) Write down a function ,¡ ,+Ä2/ which computes the distance between the two pucks at time Ä
speed of 10 ft/sec. Parametrize the motion of Tim and Michael individually. Find when and where Tim and Michaelî are closest to one anî other; also compute this minimum distance. (e) When is the distance between the two PSfrag replacements pucks 40 feet?
(f) When is the distance between the two pucks 5 feet?
axis axis
O¥ORë l O¥ORð
æ u RO ë ¾ u ð
soy milk
î
axis beet juice
(g) Where do the paths of the two pucks cross? Problem 23.10. After a vigorous soccer match, Tina and Michael decide to have a glass of their favorite refreshment. They each run in a straight line along the indicated paths at a
p ¥O ORë u ORð Tina Michael
324
CHAPTER 23. LINEAR MOTION
Appendix
325
326
CHAPTER 23. LINEAR MOTION
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
kilometer km second sec s minute min
deg
miles per hour
hour hr year yr
feet per second
ounce oz pound lb gram g
mi/hr mph
ft/sec ft/s
meters per second
kilogram kg quart qt
revolutions per minute
gallon gal milliliter ml
rev/min
327
m/sec m/s
RPM
APPENDIX A. USEFUL FORMULAS
328
Conversion Factors Length
Energy
+
J $ kg m /s + + cal deJ J
in "!A ` + cm ft H%cj% J m++ mi [^%o:j km Volume
Mass
+
i gal )jT + J ` L
+
oz !Jkj o ` g + lb '% ` j ` vo ! kg
qt H%&o ^:j ` j L
Formulas from Plane and Solid Geometry PSfrag replacements
Rectangle
axis
¥3
3
Area 'à PSfrag replacements
axis
Perimeter "!àÊü!
axis
axis axis
¹
axis
Ý
Rectanglular prism
3
axis
Triangle
3
¦3
Surface Area ! 6 à $à * PSfrag replacements * Volume axis )à
Perimeter ' Q ï PSfrag replacements Area Ï Q * axis
axis
*
¹
Ý
axis axis Ã
ç
Circle Perimeter "!( Area '(
Right circular cylinder Surface Area !(@ ü!( * Volume )(@ * PSfrag replacements axis axis axis
9
329
Sphere Surface Area
+
Right circular cone
(
Surface Area '( (@ Volume Ï ( * l
Volume p @ ( 0l l PSfrag replacements
PSfrag replacements
§
axis axis
axis axis
axis
§
axis
Constants
Avogadro’s number N H^&%v!?, e% l speed of light c "jT% , e% ç m f s
density of water $ gf cml
mass of earth ` oJI, e% p kg
earth’s equatorial radius "jA1&o^% mi H^cjJI,üe% m
acceleration of gravity at earth’s surface "j2! ft/sec HoJ m/s
Algebra ' S ' Ú ' S_D Ú 6(' S9 Ú ' S Ú O [ '
' S ' S Q Q S ' S ' Sæ Ú ' Ú ) ' ' ¾ 6(' Qâ9 6 ïH@9Ò (6 ' Qâ9 ' ! 6(' Qâ9 l ' lmüj
' ï; ' YüQï;Qâ ' Q üQ
' QÇüj ' Q
QAl
wQñ¦ ) Q Quadratic Formula: If ' QAP ï¢H% , then Ç ! '
+
' ï
APPENDIX A. USEFUL FORMULAS
330
Completing the Square: '
QAP ïP '
g
¢
Q
! ' h
+Q '
ï
Trigonometry sin 6 ;@9Ê sin
cos 6 ;9Ë cos sin ( cos ! cos ( sin ! sin ( cos ! cos ( . sin ! sin 6 (? @9Ê sin
cos 6 (Çü9ÊH cos sin 6 (Çü9Ê. sin
cos 6 (Çü9ÊH cos
sin 6 <ZM39Ê sin cos M? cos sin M
cos 6 ÍM39Ò cos cos M sin sin M sin Y cos Ç[ sin !Ç"! sin cos cos !Ç cos Y sin
Appendix B Answers í ßp O
(b) 150 ft/sec. (c) Gina. (d) 6300 hours.
Answer 1.1
Answer 1.2 (a) About
l¥u O
u
u p
minutes. (b)
l¥u
ö
§
lb/in /min
Answer 1.3 (a) meters long; about 0.84 miles. (b) hours to recite the number.
¥O O p çÕ¾ dollars/year. (b) Tuition would be uí í Answer 1.5 km. ¥ç p¥p¥pp O ¾ cm for lead; p ll cm for aluAnswer 1.6 minum. Answer 1.4 (a) $ in .
»p ë l
Answer 1.7 (a) 5.5 min/mi = 5:30 pace. (b) (c) Adrienne.
¨
»p ö
©
ft/sec.
Answer 1.8 (a) (John’s Salary) = $56000 and (taxes) = $0. (b) (John’s Salary) $56000 and (taxes) = 0.15 (John’s Salary). (c) (John’s Salary) $ and (John’s taxes) (John’s Salary). (d) 1500 (number of 120 students each year) 1800. (e) (cost red Porsche) (cost F150 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) (number of cheerful biology students). (h) (Cady’s high score)  (Cady’s low score) = 10%. (Cady’s final exam score) = 97%. (i) (total votes cast) (Tush votes) (total votes cast).
zIO ç zî
©¨
©
¨
ª ¨
© ¨
u OO¥O
©
O ä¥ä¥ä¥ä ð õ ö
©
©
x õ î ö
¨
í
í
z l
í
¨
©
O¥O¥O ð ö õ í
©p O ¶
ßp ¾
½
¾
u kg ë m °^°
m sec
sold in O ç O ¥ pizzas p pm. ç
Answer 1.11 $ at about
O¥O¥O
ä
°
«
Answer 1.12 Answer 1.13 About
u
©
O
³
p ¥ç kg ë m°^°qô °
hours. Profit reaches
í times around the equator.
O 0p l ¾ , p ç p
Answer 1.14 (a) Radii are about , and cm at the indicated times. (b) No.
ä q¾
O¥O
©
© O
O p u O æ S ð
uO
§l O¥O
. (b) Reduced comAnswer 1.16 (a) D petition for resources (eg. water, nutrients, etc.).
({ á uÁ Á õ áõ ö
Answer 1.17 Formula simplifies to . Rates are and at the indicated times. about 0.67 , 0.6 , . Rates decrease over time, but will never be less than
O u p
O u
O u
Answer 1.18 (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.
ò
òËí
Á"{$Á Á á á ö
Answer 1.19 (a) . (e) .
z l
õ
ä.
í
(b)
Ã
í
ç.
(c)
S
æ lÃp
. (d)
ª
í í í í í ¾ í l  ë ¾  p ë uí  ç ë l z (b)í 1.99609. í (c) ¾ í ò ò í ò í ò ÷ ÷ ø ø , Answer 2.1 (a) ,ø .÷ (b) Ú ÷ ÷ ÷ S u í Ç í ò ø S í æ O ,ö Ú í æ , .÷ (c)ò , ÷ Ú lí p , S . u , Ú æ l . (d) D D S l í D ÷ à æ Ã ë÷ æ ç (e) Answer 2.2 ÷ (ac) True. (d) ÷ Ã ÷ à ë Ú ç æ . (f) S O means Ú the points. line æ ÷ S O means the points line on S on the same vertical line; Ú í
Answer 1.20 (a) .
the same horizontal line.
î î Answer 2.4 (a) Erik= 6.818 mph, Ferry = 17.6 ft/sec. (b) Impose with Kingston the origin and units of î coordinates î axis;î then î ò milesî on each Edmonds is located î î î ò at these
ë ORð and Erik’s sailboat is at ë ð . î The table rows have enò î ò l tries: î
ORë ORð , î O ë ORð , p ë ORð , ë ORð . l ë O ð ,, l ë ä p , l Oð ,O l , ë O p u ð æ , l ð ë æ ç æ ç ð ð (c) Use lR lR p¼äz l l ö D ç ç ö coordinates as in (b), then when the ferry reaches ë ORð , l _ î ò í î ò ð . (d) CG vessel does not catch the ferry Erik is at ë O l ä n ò í before Edmonds. ø ð ð (b) 227 minutes, 168.4 Answer 2.5 (a) Oç ç¥ seconds.uR l miles (c) Answer 2.3 Just after 12:29 PM that afternoon.
Answer 1.9 Go for the 15 inch pie. Answer 1.10 (a) m kg (b)
î
Answer 1.15 (a) million gallons per week; about ¾ million gallons per year. (b) yds yds yds.
331
¬
332
î î
ò
òí
O ë O ð
Answer 2.6 (a) Allyson’s coordinate position: ft ft . Adrienne’s coordinate position: ft ft . (b) After 2 seconds there will still be slack in the bungee cord. (c) . Use this time to find where Allyson and Adrienne seconds. are located. (d)
l p
æ ëO ð
u u î Impose a
î í
î í
Answer 2.7 coordinate system with origin at and the shore the horizontal axis; let be the location on the shore where Brooke beaches. Equation gives time to reach Kono’s. D D hr, hr. Neither time will be the minimum time.
S
Ö æ ð ö õ æ u mð S ö õ S l ä
S
æ Oð
p
Answer 2.8 (a) 12 m/s and 3 m/s. (b) Nikki escapes. Laura is eventually stung after 2.921 seconds when she is 35.055 meters to the right of the hive. (c) Nikki must exceed 3 m/sec, which is about 6.7 mph; feasible. However, Laura must exceed 12 m/sec, which is about 27 mph, which would be world record pace of about 8.3 seconds for a 100 meter dash!
î
í î
î
¬
ò
íPî
ò=í
, O . (d) ö . (e)î , Ö , ç ¾  ßp î » p ð D ßp ë î ð , ßp ð æ ëî æ D D> ç ßp ð . î î î î î
Answer 2.10 î If the îorigin is the gliderport: (a) O¥ORë ORð , î ORë æ p u ð , (c) l O× ORë æ u O¥ORð , (d) ORë Oð . If the origin is (b) u u ORë ORð , (b) l¥u ORë æ §lu ð , (c) O¥Oë æ ¾ Oð , the hang
O¥ORë glider:
Oð . (a) æ æ (d) If the is the boat: (a) î æ OORë ORð ,(b)
Oë ð , (c) ORë ORð , (d) æ origin ¾ ¥ O O ë ¥ O R O u ¾u l u ð.
ò@. í (c) sec. î (d) ë O Õ¾ ð î (e) Answer 2.11 (b) ò8í FD u § ç l seconds. î (f) p seconds. (g) l¼ seconds to p¼ä p îseconds. òÊí î òËí ²
ë ð ± ë ¶ ð , Answer (b) ; Ö 2.12 Ö ö . (c) l l (d) spider= ëä ð when ant=  õ ë  ð . (e) 1.5 feet. (f) Spider reaches p ; ant reaches ä ë ð when l . (g) spider speed is ft/sec; ant speed is ft/sec. u ç
Answer 2.13 p¼ä ä mph (exactly). Answer 2.9 (a)
Answer 2.14 141.46 miles. They are 300 miles apart at time 0.826 hr = 49.6 minutes.
î
í
Answer 2.15 (a) Final answer is correct, but second equality is wrong. (b) Final answer should be ; key fact is that , etc. (c) Answer and D D D steps correct.
S
Answer 2.16 (a)
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Answer 2.17 (a) Ã Ã . (e)
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Answer 2.18 (a)
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Answer 4.2 If the basket is at the origin of our coordinate system and we use feet units on each axes, then the lines anchor at the points . Each line is 120.185 ft. long.
î
ä (b) S æ î l ð ö D î ³ D õ õ (c) Draw a vertical and horizontal line ð . On the vertical line, a circle of radius 2 through will have a center at either ë ð or ë æ ð . Likewise, the l horizontal line will have circles at either æ ë ð or ë ð . l
Answer 3.1 (a)
îÚ æ p ðö î
D
ç èçá õ
î î î B. ANSWERS í î APPENDIX î î í îO î î í î ë eð S î æ ð ö D î Ú æ eð ö í ö Eqn î ë ð î æ ð ö îÚ æ ðö í 1. l S æ l ðö D Ú æ ðö p ël ð S î ðö D Ú æ l ðö p î î 2. î æ ë ð 3. S D D E p 4. ëæ ð S æ ðö D Ú D ðö p (d) ë ð , ë æ l ð , D l ë ORð , ORë æ æ l ð í í Answer 3.2 (a) Wet in 27.4048 minutes. (b) Wet in í í í í 27.4446 minutes. í í î î Ã ç l gives a circle Answer 3.3 (b) (i) 3 cen of radius Ã
,ç Ã A ë ; (iii) tered at the origin; (ii) Ã
(c) ë ð  and î æ ë ä í ç ð . l ; ël ç (iv) l p  lî î l l Answer 3.4 (a) Imposing with axis along ground and Ú S Ú æ
ð ö O ö . axis along tower, wheel modeled by ö S D (b) 46.43 ft. to right of tower. (c) æ p ë ¾ ð and æ p ë ¥ ¾ ð . òÒí Answer 3.5 Lee has 2.265 in ö more pie. í í Answer 3.6 (a) Impose system so that the òí òí a coordinate O seconds. With this coortractor is at the origin at dinate system, the south edge of the sidewalk is modeled by Ú Þ is modeled
minutes.
by, Ú í ¥ O . (b) O¥O ; theí (c)north edge minutes. (d) O minutes. î î í î î l u î î Answer 3.7 equation for eastward travel from í (a)] The . Southward travel along Kingston is Ú
ç S ð ö æ . Ú (b)æ æ ð R O ð Boundary: ; Interior: Ú ö ö
ð ö Ú æ ORä ð ö z . (c) 3.82 S D F D S D minutes. F D x ä ; Exterior: S D D (d)ORð öPlug ä æ to find exit point. Use S into circle equation this to find when the ferry exits the radar zone. Be careful í to î reference all times í relative to when the ferry departed Kingston. (e) 20.32 minutes. î ò>í Ò ò í Answer 3.8 then intersection points are òæ í ë ð .(a) IfIf Ú S æ , ò, í then î_ò í ò are ò î
l ë ¾ ð . (d)
î O points î Antî exits at timeintersection Spi D u u at. time u î at time î æ . (e) Ant reaches der enters
and spider ø ¾ Ö at time u . (f) ð Ö ¾ ðdp ë æ lç ð D4u ö . ¾  Ö d ð ë ð (g) distance( ) = , distance( l l )= u, ðp åë p ð ) = . ç distance( í î Answer 3.9 (a) 6.92 (b) 7.67 seconds. × seconds. × (c) 38.16
î î x x seconds. (d) 7332 × ftö area ¾§l ftö . î . (b) no solutions. í (c)î æ ë æ ð Answer 3.10 (a) í S u ¥ 
í u ël ð . í í î and í æ ë ð . (d) æ p í  í í æ ² S æ ð æ  (b) Ú  p O
ðæ Answer 4.1 (a) Ú S D
z æ S æ (d) Ú  (e) ² ,Ú ² S æ ð D (f) (c) Ú Ú p O S æ ßp (g) Ú æ Ö î SE ¥D Ö± (h) Ú S æ
Ò í
í Ö í í öõá ö è ë õõ áá ö èè ð (b) æ  u . (c) æ  p ² ö õ  (b) area æ ö ß Ý ¶ (c) z æ Ö õ (a) area
Answer 4.3 (a) Answer 4.4
Ò í
.
Answer 4.5 In some cases, answers in this problem are unique. In other cases, the answers are not unique. Here is a possible solution set:
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Answer 4.14 At the closest 26.22471828 miles from Paris.
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ç ç§l¥u¼ ë p0 p l¼ç§l ð
Answer 4.7
õ
ö
{tW} Öu í
õ
Port Townsend
Ò
5 h ö ² ÖS æ ² ð Ú²¨ ö ² h
Ú æ ö Ö ²h ²
Answer 4.12 The lines of tangency have equations . The visible portion of the axis is
í
.
O
Answer 4.13 temperature is
æ ä p
î
² ¸ æ ð ° ¸ . l
í
¸ and
Ú
²° O Dâl
î í î P
í î
òí
òí î
Answer 4.17 72.62 seconds after initial spotting.
æ ö p Ö æ p í
SÒ í]î î æ Sö SÖ æ çS î î p S òí O ð æ l¥u ð . (a) Answer 4.8 Line of travel is Ú í î seconds (c) Õ¾ S seconds æ lR p ë æ l l ð (b) (d) l ä uR ç O ¾ç ë æ ¾ ð at about sec. ¸ æ mõ æ îO¥ORð O . (b)  õ gallons. î (c) Answer 4.9 (a) S D l òTí 400 miles. î Answer 4.10 (a) Allyson at ORë ORð ; Adrienne at æ ¾ p¥p ë Oð . Bungee is 83.1401 ft. long. (b) Occurs at time ¾§u p ð . Allyson’s final¾¼locaseconds and Allyson is at Oë ¾ ¾¼ u p í tion is 77.546 ft. from her starting point. î î î í Ý Answer 4.11 (a) None of the points are on . (b) Sum= ±³ . õthe line Ú (c) The analogous sum of squares based on
æ æ æ O R O ð ð R O ð is ö D l ö D ç ö u ; a worse fit to the threeí S î which is visually apparent as well. points,
5
î í í#î î Ö
Ö Answer
ð , 4.18 At ð timesæ ë æ O ¥ u and Ru p ç§ð u . æ ¥ u ë ð ð , ë ð . Ru p ç§u ¾ ¾ ð æ ¥ ë u ð . î ç¥çu p ä ¾ p ç§u î í u p ç§u òÊí Answer 4.19 with ini× (a) Impose coordinates î sprinkler tial as the origin í ORë ORð . Line equation becomes × location î
Ú æ ² õ S D O¥O . (b) Sprinkler is located at ORë ç u ð at time boundary of watered zone hits l¥l minutes. Circular î î æ O ë O ð southern edge of sidewalk at the points
ò í lR p lç ë ä ¾¼ l l ð . (c) Ú æ ² õ S D ¥ O }ä ç. ¾ ç l p0 and î òí ð
= æ Answer 4.20 (a) æ l ð  exact l would  . (b) í answer æ O p §ll u ëñO ç§l¥ í l u ; or¾  the be î ¥ ð  u . (c) O . (d) No î real number solutions. î
æ O ë O ë Answer 4.21 (a) S (b) S (d) ¾ ¾¥ç æ p ¾ ¾ç solutions. (c) No real number í Ã p Ã
 í õÖ Ã Ã ö æ ç D p D ö ðö ð. ö í D ç D p ð  (e) õ p D í D Answer 4.22 (a) Thereî areí four answers: î S
æ and S î í (b) Ú D u . î í : í ð S D . (b) Yes, : S ð ö S æ ² . Answer 5.1 (a) Yes , S î í í (c) No, fails vertical line test. : í (d) Yes,: í ð Ö horizontal line. S u (e) Yes, î ð í Á , S: ð Á Ö S O .O (f) Yes, , S S . öh Á ¶ , S O , fails vertical line test. (g) No, Ú : î (h) Yes, S ð S ö D .
. (b)î .í (c) . î (d) í æ æ . Answer 5.2 (a)î æ í D D S D S ð îS (e)î æ í (f) á Áî õ á Á . h D S õ h õ ð í ²° SD p , : Sî ð Ö² SD p , ½ S ð SD O , Answer 5.3 è S ð ð ½ u í ð u , minimum î ½ S = 20, maximum ½ S = 40.í î æ Ö õ æ í Á ¶ ð , æ l î S l . (b) Answer 5.4 (a) Ú ° Á Ú Lç æ á ì õ ð ö , æ Ã æ S Ã æ . (c) Ú æ Ö õ Á ¶ æ ð , S l or S æ l . (d) Ú æ p ¿ç Á ì õ ð ö æ , Ã ° Ã í î_ò í ò S D or S æ . ò Answer 5.5 For example, ò in (a), suppose Dave ø hasð constant speed ft/min. Then the function à î
õ
$ $
5
be
lR ç
v
ö
í
¥ ¾ ëë O uO OO í $$ ¥ ëë p O¥O¥OO í l p ä ç
ë l l (f) ä ç ¾ O , Sea $ p ë ä §u (g) ¥O O zç ¾ç , (e) ä¥ä u ¾ p , $ ßp (h) Sea $ O ë l l ä ¾ l , Sea $u¥l ë ç ä (i) No. í Ò Ò Ò 1983 1998
will
Answer 4.16 (a) $24,609; 1990.25. (b) 66.3%; 2077. No, never exceeds 68.73%, but gets closer and closer as time goes by. (c) The slopes of the lines involved.
ë O S æ }ä ¾ ORð DÉl¼ç ë O¥O¥O (b) Ú Ú ëç OO¥O S æ } ä ¾ RO ð D>ç ë p O¥O (d)§ç u Tabularize your answer: Seattle
she
Answer 4.15 Impose coordinates with Angela’s initial location as the origin. Angela is closest to Mary at ; this takes approximately seconds.
Answer 4.6 (a)
Year
point,
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. In Oslo, the Answer 5.6
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í
î î
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Answer 5.7 Several possible answers for each one.
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î î ð ³ õ . Ú intercept = æ ³õ l æ S ä í l dð ë u ð ò ³õ lDl ä l l¥ðdl ë u ð . (c) None. (d) í No, Yes. (e) Yes, No, Ò í D ë æ Ò æ í l D Dl D ðåð .
î seconds. Answer 5.9 a^ (a) Ú î O¥O¥O í . 20,000 molecules.
O g. 25 ö õ á è Ü . (c) u ç î g. (d)î (b) Ú ¾ ç î î ¶ î î î î î : ð î OO¥O¥O O¥O¥î O¥O p æ S ð ö D . Answer 5.10 (b) : S l S l Du (c) The í table of values S ë S ðåð will be: O ë O u¥O u ðð ,,
O u ëë ä ¾ O¥O l ðð ,, ë ë O çç ¥ ßð p , ð , ë u O ë O ¾ ä ð , ä ð ë , O¥O¥O¥ORë ð .¾ çMinl u §u l ä lR ¾ ¥ lR u x x ä ; the p ¾ exact answer imal cost occurs for some u î S thislRin u our class since it õ Ö , but we cannot solve is S î î requires the tools of Calculus.  æ ² Á æ ð . (b) O , 20, S D S ö , Answer 5.11 (a) æ
ö æ , öæ , ð ö , ö , ö õ ð . (c) l Always D l D p ö. Answer 5.8 (a) . (b)
intercepts and
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µ
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Answer 5.12 (a) Relates items purchased and their cost; the input (independent variable) would be the item and the output (dependent variable) is the cost. This sets up a function relationship. (b) Let be a word in the index and represents a page number. The index does NOT set up a function relationship between and , since a single item may be referenced to several different pages and a single page may contain several different items being referenced. (d) Independent variable is the dial setting and the dependent variable is the radio station you hear. This sets up a function relationship.
S
Ú
Ú
S
Answer 5.13 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.
í
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Answer 6.4 (a) The rule is
æ SDml SO í is O >Ú O . and the î range Ú
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p ëp ð
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ø
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S
2
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(b)
S
x æ æò ò
if if
¨ ¨ lR u xO Oò¨ ¨ p ò ò
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Á
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Answer 6.7 (a) (b)
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æ S ð æ S  DâS is obtained by reflecting Ú í S ð across the Ú axis. æ S  DmS if S æ æ O  æ if æ æ SS  DmSS if O S S S S S æ S if S  æ and the graph of S S is given below:
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¼ l u ë æl O ë ð O ð æ Oë ORð î p u l u æî ¾ u ë u ð u î î æ ð , Vertex: (4, 9), Axis: = 4. (b) Answer (a) î æ î 7.2 S p ö D ä (5/2, 383/4), Axis: S = 5/2. l(c)S æ u  ð ö æ ð ö lç§l §p , Vertex: S (3/14, 2539/196), Axis: S= 3/14. l z ßp (d) D æ ul ORä¥ð zö äæ ,O ,Vertex: î_ò 0),í Axis: ò S = 0.ò (e) Vertex: (0, S O¥Oð ò æ Oð ö Sæ O , Vertex: c î ò í ò (0, 0), Axis: = 0. S S z ò@í ò8í à ð æ ³aõ ö D ² D O Answer 7.3 (a) Multipart function: à ¶ (b) O for ¥ ßp O l ;õ $23,125. for O ¥ ßp O l O , days;ð $2500. Sell at time (c) On day it is u î ¥ u worthless.
ª
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í
õ . (c4) Roots of lS ö æ areí S These h õ  . plot õ coordinates of ”hilltop” and ”valley” in the of î i.e. í roots î of the î auxillary quadratic Ú î S S í  æ S in (c1);  æ detected the extrema of the graph of Ú S S. ð õ O æ S ð O æ S ð , degree 3. (b) Answer 6.9 (a) Ã ðS O O¥O æ u ½ O S S æ S ö ö S , udegree 2. To
get 600 sq. in. dimensions are: l ¾ p ¾¼ u O ç ç u p . Answer 6.10 (a) 2 hours. (b) Impose coordinates with axis the bottom of the ditch î and the Ú axis the picturedS centerline. î
O if æ O í S p î OO æD O¥O¥OO ææ î ES D p ORORðð öö ifif ææ p OO S ææ l OO S D O if æ l O S O E Ú S
OO æ O¥O¥OO ææ S ææ ORORðð öö ifif OO S l OO î S p O if l O S p D p S
(c) p O æ äz ð feet. (d) 42 ft. wide: 0.3008 minutes. 50 ft. wide: 8.038 minutes. 73 ft. wide: 116.205 minutes. æ S æ Answer 6.11 (a) O Ú . (b) Increasing:
æ and and î S p . Decreasing: S p S . (c) í î
if æ æ p D S S æ æ ð O if p Dæ p æ ES Dæ ð öö if O p S Ú
S p p p æ S S D if p S
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Answer 7.4 The parabola has intercepts at intercept at 3. The vertex of the parabola is
2
axis axis axis
æ ² ,
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Answer 7.1 (a) Rocket hits left portion of roof at position . Maximum height of the rocket oc¾ curs at position . (b) Rocket misses house, and has maximum height at the posilands at . (c) Rocket hits right portion of the roof at tion the position . Maximum height of the rocket ¾ ¾ occurs at position . (d) Rocket misses house, and has maximum height at the position lands at .
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ë æ ð . (d) Answer 7.5 (a) 100 ft. (b) 156.25 ft. (c) u uë O ð î í í When = 54.81 ft. or = 570.19 ft.; i.e. at u p îç l ä p î uHS¾ O ä ë æ î p î O p ð . S and î î : ð . (b) TheÃ points Answer 7.6 (a) No, since Ã ë D ç¹ð and æ D çvðåë D ç¹ð . (c) Only the point ë æ ö ð.
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Answer 7.8 (a) Dave’s maximum velocity is O ft/sec at time u O sec; pace is 4.4 min/mile. (b) 82.4813 seconds. ö æ æ S S l
if if if
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Answer 7.9 She should have 225 trees in the orchard.
¼
Answer 7.10 She should charge $ ¾ ¾ money.
to make the most
Answer 7.11 The radius of the circular part should be 1.787739 feet and the long side of the rectangular part should be 7.404087 feet. Answer 7.12 The enclosure should be 50 meters by 75 meters.
í
í
Answer 7.13 Cut so the pieces have lengths 26.394 in. and 33.606 in.; bend the 26.394 inch piece into a circle.
í
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Answer 7.14 (a) . (c)
u
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Answer 7.15 (a) The quadratic portion is .
æ íl ö
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. In the ø O or ø Answer 7.16 In the first case, Ò ø í ö second case, Ò h õ . Ò í í í Answer 7.17 There are two possible values for : If
,(d)then ofí the equation î unique æ solution
, then ç D æ ¾ æ ò í the . If the unique will be S ç æ ¾ p ¾ solution of the equation will be S p D î ¾ .
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in ¥ö ç . u ò ö inò ö . (e) Hu ¾ inö . (f) p¥p §u î î_ò í ò ò Answer 8.2 (a) î : î_ò ðdð í æ ò ò D if ò ò æ if æ æ if æ è ðdð î î_ò í ò ò D if ò æ ò (b) ò ò : î î_ò ðdð í æ æ ò æ ifif O O ò æ O if è ðdð æ æ ò if O ò ò î î_ò í ò ò (c) ò òæ æ æò if æ O ðæ ð if æ D D if O æ if
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Answer 8.5 (c) S æ and S î î í: ³aÞ ; Answer 8.6 (a) à ð î î î Oà î í O : à ðåð O ¾ O ö ð ö à ö D O : à ðdð î î í mph when you input seconds à . (b) è ð computes O ;
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D è è ðdð , ðdð , S ðåð p l . (h) S ðåð æ p , (g) : : ðS åð S æ p , l è : S ðåS ð O . l
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APPENDIX B. ANSWERS
338
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APPENDIX B. ANSWERS
340 y
Answer 12.2 (a)
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and gives the amount the shop spent on advertising as an output value.
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(b) 1.0788 .
u p l É p u
Answer 13.2 (a) 6080 ft. (b) 29.95 mph. (c) 15.63 knots.
. (c) Answer 13.3 (a) O sq. in. (b) l u¼ ç p p p in. (e) in. ¾ ä
O in. (d) lR p ¾
Answer 13.7 (b)
ç p p ¾
sq. in. (c)
ç u
mph. (c)
p ¾
hrs.,
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rev * Answer 14.2 (a) O rad = 10.47 radians. (b) l  ¼ p  1 hour/rev = .64 hours = 38.2 minutes. (c) Using (2.2.2), ð O  l ð = 219.91 meters. ä§p
RPM.
l
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(b)
p O¥O ft.RPM. ç§uRrad ç =mph. ; ç ç ç ä (e) O ¾ p¥p u
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Answer 15.1 miles.
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î ¾ p¥p ç
ft. (b)
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with
S S S ú ð æ æ O O ð ð cos ö ² and Ú sin ö ²
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Answer 15.3 (a)
î
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Answer 15.4 290 ft.
í
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í ¥
ä ä ë lR ëp ð . ð . (b) uu ë ç ð . l ç¥ç l ¾ u
¥
à ç and ¾¼ ßp .
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(c)
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ft.
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Answer 15.9 (a) crosses the track at cation at this instant is Answer 15.10
Answer 13.8 0.685078 miles.
Answer 14.3
p
Answer 15.8
sq. in.
Answer 13.9 Middle picture: shaded area=
pO
Answer 15.7 Dam is 383 feet high.
Answer 13.5 2160 miles. (b) ¥ O l
í RPM at inch; 7í O RPM at O inches. l §p ç p uR u p u
ft. Ö (b) O O ç§l rad. (c) Answer (a)
/5 rad14.7 l¥lR p ¾ counterclockwise from .
Answer 15.6
Answer 13.4 (a) 1413.7 sq. ft. (c) 4.244 sec.
Answer 13.6 (a)
miles. »p ¾ hrs. (d) l lH¾¥ O miles.
ft/sec. (b)
Answer 14.10 (a) 70.65 mph. (b) 32.72 mph.
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Answer 14.6 (a) inches. (b)
Answer 14.9
. (b) Ú aÀ~Ô : ^ S ^ð ^ S O¥O $p O¥O . ð ( Ö ^ m ~ ö ö Ú . Domain (d) õ õ S ¢Ú : O ¥ O ¥ O O , Range . The inverse function õ lçç p S u takes the number of customers per day as an input value
Answer 12.11 (a)
í
ú rad/sec,
õ mõ RPM. ½ u
341
Answer 14.5 (b) 700 ft. (c) 70 sec. (d) 15000 sq. ft.
O
í
í
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í í
×
Answer 15.11 (a) 204.74 ft. (b) no.
î
î
Answer 15.12 If the origin is the bottom left corner of the table, then the ball should first hit the position , then the position .
Oë Ö° ð
m³ ë ORð
î
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î
î
Answer 15.13 If SeaTac is the origin, they intersect at .
æ ç l ä u ë æ p ä ä l ð
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(b) max=
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(c) min =
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Answer 16.2 (b) sinö S ð ö õ æ cos S ðdð . 342 î_ò í
APPENDIX B. ANSWERS
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6
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î
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î cî ò ò ú
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y 2 1.5 1 0.5 5
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t The graph of
Sfrag replacements
1 axis axis axis 2
O
Answer 18.1 (a1) , , not defined.
í Answer 18.2
¾¾
¿ ¾¾¾
: ð S
y 2
æ O ¾
¾¾
O
¾¾
í
t
1 axis axis axis 2
20
u ¾ O ç , æ uH¾ O ç , O p ¾ , O ¾ ¥ç , í
µ
Ú
Ú
°î
î ,
l , O
ç . (b) Ú
±
æ lð æ S p O î if S æ ð sin pif S O l D ç Õ¾ S O sin î S æ l ð æ p if ¾ ç ä î l¥l S sin S æ l ð D p if O ç O Õ¾ S î O sin p S æ l ð æ if u ç ä l¥l S æ ð sin p S if l OD ç ç Õ¾
¨ ¨ ° ± ç O ° ¨ ± ¨ ¼ ç }ä l¥l ¨ ¨ ° ± O ç O ° ¨ ± ¨ uR ç }ä l¥l ¨ ¨ ° ± ç ç O ¨ S ¨ O
sin
Õ¾
¾
Õ¾
Õ¾
: ð is given below: S
axis
15 10
(a) 9.39, 13.63, 11.09. (b) Feb. 3, Nov. 4. PSfrag replacements
Answer 18.3 (a) Principal
; graph solution: S äR O , symmetry solution: below with these two solutions S p lH¾ graphically indicated:
O
¾¾
î
Sfrag replacements
î
¬
cos
, Answer 18.7 (a) is not sinusoidal; its rule is
x 2 1
î
hours of dry time each day.
^ ö { D O¥O ö D OO æ O¥O tan ðdð ö . ìPîcò ò î î ò î ò î_ò ò î î ò î ò Answer 18.6 The key fact to use over and over ð =M’s location after seconds = is this: O ¥O cos O O u ðåë OO sin O O u ðåð ; æ ð = T’s location after í O O íl D ðdð . í î seconds = O¥O cosí O O l D í ðdë O¥O sin
Ý
Answer 18.5 (a)
Answer 17.6 (a) A = 25, B = 5 seconds, C = 1.75, D = 28. (b) =1.75 and 4.25 seconds
ð
çz ä ¥ç í
Answer 18.4
5
axis−5
2.5
5
S
axis
7.5 10 12.5 15 17.5 20
: ð S
Ò í
344
ø ¼qèWà
î
ò
Ì
Ò í
Í
Ã àéo¾ läO läO Ò í
ð
Answer 18.8 (a) . (b) D The interior angle is 166.608 degs and so one satellite covers 46% of the circumference. Thus you need 3 (not twopointsomething) satellites to cover the earth’s The interior angle is circumference. (c) ¾ 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 . Solve arcsin = 20% D of 360 degs = 72 degs. You’ll get = 934.83 miles.
ä
ø î u ä q¼ è ò î_ò íïO î æ î ò lä ä Ö ð í í O cosí æ O D Answer 18.9 (f) í then plug in from (e).
î
ò
ò
Ì
Í
O  l ä O î ð óð ò í ðåë O siní æ O ðåð , D ë µ p0 u ëd Oë O u  . Note Answer 18.10 (a) O u O ORë ë ë } are also all valid that D u , choices for the phase shift. (b) maximum temperature=
Î5 Î5
É¸
É¸
430 . (c) minimum temperature= 400 . (d) 12.1635 minutes. (e) 14.6456 minutes. (f) 6.80907 minutes.
Answer 18.11
î_ò í
Answer 18.12
î
îcò
(³fö ² ú ² æ ç O ¥ç u ðåð ôö O
ð lR p u ð ¶
¶
ë O O ëA í Rl ç D î Answer 19.2 (a) Ú l í: í 19.3 (a1) í S Answer
í ðÁ
(b)
Ú
î
öõ
ð Á í . (d) Ú
ðÁ ö õ± h õ 
.
ð pÁ
Ê
.
:
©
p
³
O
cells. (b) True. (c) The two
í
Answer 19.7 (a) 261.31 Hz. (b) 440 Hz. (c) 27.5 Hz. (d) 16.35 Hz.
©
ìP° î
² O õ
u
Answer 19.8 (a) (d) meters.
ä
v ð
. (b)
¶
õ . (c) ³f
ä
©
O õµ¶ .
is the higher curve. (d)
fraction
60
0.8
40
0.6
20
PSfrag replacements 50
axis axis 20 axis
100
150
í
Answer 18.14 (a) Domain: . One solution:
u
S
200
250
õ æ ú³ ú õ
õ¶
D
300
350
t
0.4
PSfrag replacements 0.2 axis ; Range: axis D ; graph is below: axis
¨S¨
õ
ú³
;í æ í ä ä l æ lÕ¾Õ¾ O ç¥q¾ ¾ p ä l
î ORð
0.5
1
1.5
2
x
í
í O ¥ä ä Ò rad,
Answer 18.17 Top left triangle: rad, .
S
ä
¾
í
î
Answer 20.2 (a) ô (b)
Answer 18.15 (a) Many possible answers; for example: , , , .
î
60
80
100
p
Ü î ¥ l O p p O ç ð î í
Á ô Oî Oí .
ë î v ORS ð ð í ¥ç O O O ¾ O î Õ¾ p ð , S ð ìP ¾ S D7ç¥ O ¾ u v
Á Answer
O 20.3
O ðÁ . S ð u O ¾ ¾ u ð , Sð ä u ¥ ¾ ô ç  éo¾ ö ; no explosion. (b) Answer (a)
ô ç éo¾ 20.4 l » p p ; explosion. (c) degrees. ö ç l  u p î_ò Answer 20.5 (a) %. (b) %. îcò í î ¾¼ ä ò ä ç
Ã ð , we get and O¥OO in Answer If we use îtwo cò data20.6 ç ä ä points and a corresponding exponential model: ð uR äR ç p l¥l O ð { æ ä ç ORð . The exponential model grows ¶ Ã ð faster than the cubic model and eventually exceeds .
axis 2 1.5 1 0.5 axis 2 axis
í
¹ î ðí
40
6
î
2
O ç§u¼ç
îcò í
Answer 20.1 (a) $ . (b) $1.11. (c) It is below; should be $5.70 by the model.
PSfrag replacements
Answer 18.16 Top row: rad.
20
8
4
u
öõ
=Ï
Answer 19.9 (a)
S í O î ¥ l
í î
î í Á S æ ð öõ p . O , then ¢ ¹mð O O l ¾ ¹ ô ± ö ² mö . If Answer 19.4 If ¢ ¾ ¹ surface area ßp lb, thení î l p mö lb.. To increase byí 5%, must gain about O í í  õ Á ð is obtained by Á vertically Answer 19.5 Ú dilating Á
a is a horizontal ; it is vertically compressed. Ú Ú Á ; it is horizontally stretched. dilation of Ú
80
¨>Úm¨
î
í
sin Answer 18.13 (a) Õ¾ D . (d) The graph of when is below; you ¾ need to determine when the graph is below the horizontal axis.
äO æ
Answer 19.1 (a) 31.5443; (b) 355.1134; (c) 36.4622; (d) 0.0616; (e) 51,168; (f) 0.009794; (g)84.2; (h) D an integer.
Answer 19.6 (a) formulas are identical.
î_ò
mí
APPENDIX B. ANSWERS í
O p ç
í
lR l p l O rad,
Ó
rad,
Ð
í
í î
The OO º à ^² Á^ ð æ ¥p p O u¥l ufeet. íïî î
Answer 20.7 (c)
î í
Ú
î_ò
curve is modeled by and the minimum height is 59.46
í
î
í
î
Answer 21.1 (a) 0.6826; 2.3979; 3.3030; 3.3219;0.3010. (b) 3.555; 19.8; 0.0729. (c) log ; log ; log .
} l ð
S
ð õ Ú » í
øð
Answer 21.2 (a) Answer 21.3 (a)
Ú
»
S
O ä§p
¼ ³ l õ^ô °^¶
ð õ Ú S õ lÚ í
ð Ý . (b) meters. ¾ ¾ ç§u Ö õ Ü . (b) Ú ¼ ö ô ±a° Ü .
Answer 21.4 (a) 1:10. (b) 7.723. (c) A 7.2 quake is about 5 times as intense as a 6.5 quake. Answer 21.5 (a)
ä ä
years. (b)
S z ç p¼ä
S x
òí
In particular, if we restrict function has rule:
æ O p
í î ð O l¥u Ü . íð OO¥O O l u ð Á . Valued at Answer 21.9 (c) S ¥ u u ½ í í î í $200,000 during ää l . í í
O (b) 1.392. (c) 0.3552. (d) Answer 21.10 (a) S . ¾(f)0.61. ¼ ¶ . (e) ð æ O ä l ;
(g) S æ O S or u l O . sin S S ¾§uç D S l ä ¾ p D Answer 21.12
î_ò
u
Ñ
î ð to O ln
arcsin
l
¨ ¨
O u , the inverse
{×~o ln { ð  ln { ö D p ö
u
D
Answer 21.14
years.
í
îcò í
O u
ä %.
Answer 21.18 Earning power will be equal at a time dur. ing the year
cî ò í _î ò í ¼ î ô õ°a¶ ö ± Ü . ¥ç ç pÖ p Ö Ou O¥O ¼ ô a Ü , ð Answer 21.20 First row: ð ò4Oí O¥O O O î ð Ü , O¥O¥O , ë ò4í ð , î ë O¥O ð . u l pu u u u ¾ ä l p l p l p î
¼ îð
Answer 21.19 (a)
Answer 22.1 (d) Graph is below with locations at times (i.e. ) and (i.e. ) indicated by ”dots”:
pëæ ð
O
Answer 21.11 (d) 8.617 weeks.
î
æ
Answer 21.17 (a)
î
î_ò í
D
O p ð , the D
{×~R? ln { ò ð  ln { ö D p ö
ln
arcsin
O ä
í
Answer 21.7 (b) 135 cm. (c) 4.84 yrs. (d) Maximum possible length of the halibut. O
îcò
æ
O p ð
¨ ¨
O î p ð
O
If we restrict to ¾FD inverse function has rule:
î 345
Answer 21.16
îcò í (a) 8.8 cm.
Answer 21.8 (a)
Ñ
òní î ð
ò
Answer 21.15 (a) 2015 (b) 2048.
l p %.
Answer 21.6 (a) A is 32.763 ft; B is 24.308 ft. (b) ft or ft. (c) After 125.64 ft. (d) 24.36 ft.
çz u
î
æ ëp ð
u
y 4
í
îcò í í ò´í 2
Á is posií zero îAnswer 21.13 òí (a)2 volts.î (b) úNever since  µ ò ð sin ö a² æ ² ð , so ë tive for all . (c) x ë a  u ë S O î l v  u .î (d)l î O u ö î ð í D . (e) O ul §l¥uFD 4 1 1 2 3 O p ð and O î_ò ¾ p luqD O ò p ð , ORë ë ë } are ALL so2 lutions. Four of these lie in the domain O Max . (f) replacements PSfrag O å ð ë ð ima have coordinates O î and minima have l D ð , where p Oë ë ë . (g) axis 4 coordinates O òí O ðdë O D u p ð O O axis If we restrict to , the inverse function has 6 axis rule: îcò u ln { ~Rî o { ¾ ln { ð ò î ö D4l arcsin ln ö Answer 22.2 (a) Plots are below; heavy plot is for the cariuî bou population. ð O O ð O O ð If we restrict to , the òí î uD p ¾FD p
¨ Ï Ñ 5¨ Î5 ¨ ¨ 5 85 ¨ ¨
Ñ
Ñ
¨ ¨
inverse function has rule:
population 300
{ ~R? ln { ò ln  { ö ð D4l arcsinî_ò ln ö O p ð D u î ð to O O l , the inverse In particular, if we restrict ò í function has rule:
Ñ
If we restrict rule:
Ñ
¨ ¨
{ ~R? ln { ln  { ö ð Dml PSfrag replacements ln ö
î_ò ò æ O p D arcsin î u òí ð O O ä , the inverse function has to ¾ æ arcsin ln {×~Rln? { ln { ö ð D p ö u
¨ ¨
250 200 150
axis axis axis Answer 22.3
òí
5
10
O ¾ç§u p
sec.
15
20
25
30
t
346
òÒí
ò í
O u¥í u¥l î u u í O ¥¥ ë p ç§l ò Ú S ð äz }S æ O O äz lç S ö î î ò í î_ò í î ò O, Answer 22.5 (c) On domain ð cos ð D l æ p sin ðdð ö í î(e)cò O í O , O î O . ò î_ò í î ò l l O ¾¥ p rad/sec. mph. (b) Answer 22.6 (a) î O½ = 18.85 ð ¥ O O O å ð ë ð ¥ O O cos Ú sin SO ð . (c) 200l ft. l  D  D ¾¥ ò p l D ¾ p îcò
ë O ð . (c)698.1 Answer 22.7 (a) 0.9 minutes. (b) æ uR ä ò range is çzî and O O ft. í ø domain î ð O¥O î is ò î ç seconds
O¥îcòO (d1) p . (d2) ø ð ðåð ö D l O¥O sin l æ ð æ OORð ö Ol ¥O cos l æ u u p ò p ò î_ò í î î î î
Answer 22.4 (a) 6 feet. (b) seconds. (c) No. (d) At times seconds it is 22 feet Õ¾ ¾Õ¾ above the floor. (e) Maximum height: is 24.5675 feet. (f) Allyson teaches Lee a lesson. (g) D ¾Õ¾ .
ª
®
Ë
¨ ¨
¨ ÓÒ ¨
(e)
Ò
î_ò í î ò øì ð î ò l O¥O cos l æ î_ò åð ð ö í D l O O sin î l ò æ îcò í î ò î up Öú
ð Answer 22.8 upper left: R Ö ú °
sin Ö ú ð . lower right: S ð cos cos RÖ ú S ° D Ú ð ° sin ° D æ O u ðåð .
_î ò Ã í î ð æ ðö up ð, Ú ð æ O ðåð , u
Answer 23.1 Answer 23.2
í Answer 23.3
î_ò í
ò
î_ò í
æ O ç , Ú ð lD O p ð O p ½ S O ç ä§p . (c) 5.814 feet.
Answer 23.4 (a) , ,
æ O ç ~ ½
í
î ò í î APPENDIX B. ANSWERS î î ò
ð ,Ú Answer 23.5 (a) 78.79 feet. (b) = æ
æ ¾ D . (c) S lR u ð = æ lR l¥l¥l S ft andl Ú D lR u ð  l= æ l O ft òâí ft. (d) crosses î intoíZgreen î so the distance is 32.83 at æ »p u ë æ l ä ç p ð at time = 3.3347 seconds. òmí î í Ö î ò î ò, location í òí Answer 23.6 At time is ð; ð and ç ë p disdist(origin,P(1))=5.385 ft. Closest when u î 6 feetò ø ð î_ò O í . Distance î ò from î_ò í origin tance is u ¾ ¥ p p
æ ò O ð O ð when D ç ö D u p ö ; i.e. ò when lR u . î î ð îcò l í O ò æ l î O  î_ò ¾ ð ë Ú ð î î_p ò æ p¼ ¾ ðí Answer 23.7 (a) S (b) Enters green when 2.08 sec at
î O 11.78ë ft.ò ð .(c)î Leaves ò green when == 11.92 ò sec at æ ç§O Hu ç§¾ Hu ¾ zë æ ç¥ ä ç¥ ä ð . (d) = 6.8862ò sec, S = 0.4878, ø ð ðö D Ú ð D p ðö Ú = 0.3902 . (e) S (f)ð æ Two p
æ ð O ð O answers: = ßp ¾  p¼ä ö ä p¼ ¾ D »p 5.849 sec, so = 4.932, Ú = 3.946; = 7.923, so = S î S 3.956, Ú = 3.165. ò
í ò
½
. (b)
Á
¬ ¬
¬
¬
î_ò í
¸
î_ò
ò
ò
ò RO ë ð Rl ¾§u
Answer 23.8 (a) Impose coordinates with the bottom left pocket the origin; by 15.1.1 ball hits cushion at .
¨ ¨ ¨ ¨ ¨ ¨
x u ßp p æ æ u ä O ç ð ifif O O u ßp ò u ßp ò u ä ç O O uH¾ î_ò í î_ò ò
x u ßp î Ú î_ò ð í ò æ î_O ò í u æ ò æ O p¼ä ¾ ð ifif O u ßp ò u ßp O O u ¾ î_ò í l ¾ u ò p¼îcä ò ¾ í î_ò í òAí = Oë ORð . ò ò Answer 23.9 (a) If we impose coordinates with ò í ð , Ú ð moves u O ft/sec. S ð O u æ O , Ú ð p u O, puck S (d) øu ð l O O æ ìP îcò O¥D O í¢ p O î . O¥O ò . (e) ò O O¥O î_òandí ö p l u ò u D î §lu . (f)ò Never. ìPî_ò î_ò òí ð uR u p ¾ ë ç l ð , Tina æ ð Answer 23.10 Michael
æ O D p p ¾ ð . Study distance SQUARED from p O¥O ð toç æ ä§p¥p ð .ë u Closest when z p0 sec and distance is 54.3 ft. S ð
¹¸
Appendix C GNU Free Documentation License Version 1.1, March 2000
Ô
Copyright c 2000 Free Software Foundation, Inc. 59 Temple Place, Suite 330, Boston, MA 021111307 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. This License is a kind of “copyleft”, which means that derivative works of the document must themselves be free in the same sense. It complements the GNU General Public License, which is a copyleft license designed for free software. We have designed this License in order to use it for manuals for free software, because free software needs free documentation: a free program should come with manuals providing the same freedoms that the software does. But this License is not limited to software manuals; it can be used for any textual work, regardless of subject matter or whether it is published as a printed book. We recommend this License principally for works whose purpose is instruction or reference. 347
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APPENDIX C. GNU FREE DOCUMENTATION LICENSE
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APPENDIX C. GNU FREE DOCUMENTATION LICENSE
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Index axis, 13
circles,great, 181 circles,point coordinates on, 208 circles,unit, 32 circular function, 207 circular function,triangles, 207 circular motion, 308 circular function, 169, 199, 204, 205 circular function,circles, 208 circular function,inverse, 247 circular function,special values, 201 circular motion, 187, 308, 309 compound interest, 278, 280 compounding periods, 278 continuous compounding, 283 converting units, 1 coordinates,imposing, 13 cosecant function, 210 cosine function, 200, 228 cotangent function, 210 curves,intersecting, 32
axis,positive, 13 M coordinate system, 13, 14
M axis, 13 14 M axis,positive, ò M intercept, 45
Ä
root, 267
adjacent, 200 amplitude, 235 analogue LP’s, 191 angle, 170 angle,central, 170 angle,initial side, 170 angle,standard position, 170 angle,terminal side, 170 angle,vertex, 170 angular speed, 185 arc,length, 177 arc,subtended, 170 arccosine function, 252 arcsine function, 252 arctangent function, 252 area,sector, 177 aspect ratio, 15 axis scaling, 15 axis units, 16 axis,horizontal, 13 axis,vertical, 13
db, 295 decibel, 295 decreasing function, 82 degree, 89, 172 degree method, 171 degree,minute, 172 degree,second, 172 density, 3 dependent variable, 63 difference quotient, 42 digital compact disc, 193 dilation, 128 dilation,horizontal, 131 dilation,vertical, 129 directed distance, 21, 316 directed quantity, 318
belt/wheel problems, 193 CD’s, 193 central angle, 170 chord, 179 circle, 29 circles, 30, 84 circles,circular function, 208 355
INDEX
356 distance, directed, 21 distance,between two points, 19, 21 distance,directed, 316 domain, 63 e, 281, 282 effective yield, 284 envelope of hearing, 296 equation,quadratic, 51 equatorial plane, 180 even function, 221 exponential decay, 271 exponential function, 281 exponential growth, 271 exponential modeling, 277 exponential type, 271 function, 62 function,circular, 169, 199, 204, 205, 207 function,cosine, 200, 228 function,decreasing, 82 function,even, 221 function,exponential, 281 function,exponential type, 271 function,logarithm base b, 291 function,logarithmic, 287 function,multipart, 86 function,natural logarithm, 288 function,odd, 221 function,periodic, 220 function,picturing, 59, 61 function,rational, 163 function,sine, 200, 228 function,sinusoidal, 169, 227, 233 function,step, 143 function,tangent, 200, 228 function,trigonometric, 227 function, ï T 6 A9 , 200 function, ï T 6 @9 , 228 æ function, ï T 6µî 9 , 251 function, Ä¢6 A9 , 200 function, Ä¢6 9 , 228 æ function, Ä 6* î 9 , 251 function,/ 'ÅÄ¢6 / \e/ ' 9 , 200
function,/ Å ' Ä¢6 @9 , 228 æ function,/ Å ' Ä 6µî 9 , 251 graph, 1, 30, 31, 44 graph,circular function, 222 graph, Ä¢6 A9 , 224 graphing, 1 great circle, 180, 181
horizon circle, 261 horizontal line, 30 horizontal axis, 13 horizontal speed, 315, 318 identity,composition, 252 identity,even/odd, 221 identity,key, 220 identity,periodicity, 220 imposing coordinates, 13, 17 independent variable, 63 interest, 278 intersecting curves, 32 intervals, 64 inverse circular function, 247, 250, 251 inverse function, 247 knot, 183 latitude, 180 line,horizontal, 30 line,vertical, 30 linear speed, 186 linear functions, 69 linear modeling, 39 linear motion, 315, 319 lines, 39, 45 lines,horizontal, 29 lines,parallel, 51 lines,perpendicular, 51 lines,point slope formula, 44 lines,slope intercept formula, 45 lines,two point formula, 44 lines,vertical, 29 logarithm conversion formula, 292 logarithm function base b, 291 logarithmic function, 287
INDEX
357
longitude, 180 loudness of sound, 294 LP’s, 191
principal domain, tangent, 251 principal solution, 250 Pythagorean Theorem, 20
mean, 234 meridian, 180 meridian,Greenwich, 180 modeling, 1 modeling,exponential, 277 modeling,linear, 39 modeling,sinusoidal, 233 motion,circular, 187, 308, 309 motion,in the coordinates, 315 motion,linear, 315, 319 mulitpart function, 86 multipart functions, 86
quadrants, 15 quadratic formula, 51
natural logarithm, 287 natural logarithm function, 288 natural logarithm function, properties, 288 nautical mile, 183 navigation, 180
radian, 176 radian method, 174 range, 63 rate, 4, 45 rate of change, 4 rational function, 163 reflection, 124 restricted domain, 63 right triangles, 207 RPM, 186 rules of exponents, 267
scaling, 15 secant function, 210 sector,area, 177 semicircles, 84 shifting, 126 odd function, 221 shifting,principle, 127 origin, 13 sign plot, 81 sine function, 200, 228 parabola,three points determine, sinusoidal function, 227 105 sinusoidal function, 169, 233 parametric equation, 305 sinusoidal modeling, 233 parametrized curves, 305, 306 slope, 42 parametrized curves,as function graph, solve the triangle, 247 307 sound pressure level, 295 parametrized curves,motivation, 304 speed,angular, 185 period, 235 speed,circular, 187 periodic, 220, 227 speed,horizontal, 315 periodic rate, 278 speed,linear, 186 phase shift, 234 speed,vertical, 315 piano frequency range, 272 standard position, 170 picturing a function, 59, 61 standard angle, 170 polynomial,degree, 89 standard form, 31 positive axis, 13 step function, 143 positive M axis, 14 principal, 278 tangent function, 200, 228 principal domain, 251 triangle,sides, 199 principal domain, cosine, 251 trigonometric function, 227 principal domain, sine, 251 trigonometric ratios, 201
INDEX
358 unit circle, 32 units, 1 vertical vertical vertical vertical
axis, 13 line test, 68 lines, 30 speed, 315, 318