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Unformatted text preview: Precalculus Precalculus
David H. Collingwood
Department of Mathematics
University of Washington K. David Prince
Minority Science and Engineering Program
College of Engineering
University of Washington Matthew M. Conroy
Department of Mathematics
University of Washington
September 2, 2011 ii Copyright c 2003 David H. Collingwood and K. David Prince. Copyright c 2011
David H. Collingwood, K. David Prince, and Matthew M. Conroy. Permission is
granted to copy, distribute and/or modify this document under the terms of the
GNU Free Documentation License, Version 1.1 or any later version published by
the Free Software Foundation; with no Invariant Sections, with no 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 experience 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. v Notation, Answers, etc.
This book is full of worked out examples. We use the the notation “Solution.” to indicate where the reasoning for a problem begins; the symbol
is used to indicate the end of the solution to a problem. There is a Table
of Contents that is useful in helping you ﬁnd a topic treated earlier in
the course. It is also a good rough outline when it comes time to study
for the ﬁnal examination. The book also includes an index at the end.
Finally, there is an appendix at the end of the text with ”answers” to most
of the problems in the text. It should be emphasized these are ”answers”
as opposed to ”solutions”. Any homework problems you may be asked to
turn in will require you include all your work; in other words, a detailed
solution. Simply writing down the answer from the back of the text would
never be sufﬁcient; the answers are intended to be a guide to help insure
you are on the right track. How to succeed in Math 120.
Most people learn mathematics by doing mathematics. That is, you learn
it by active participation; it is very unusual for someone to learn the material by simply watching their instructor perform on Monday, Wednesday, and Friday. For this reason, the homework is THE heart of the
course and more than anything else, study time is the key to success
in Math 120. We advise 15 hours of study per week, outside class.
Also, during the ﬁrst week, the number of study hours will probably be
even higher as you adjust to the viewpoint of the course and brush up
on algebra skills.
Here are some suggestions: Prior to a given class, make sure you have
looked over the reading assigned. If you can’t ﬁnish it, at least look it over
and get some idea of the topic to be discussed. Having looked over the
material ahead of time, you will get FAR MORE out of the lecture. Then,
after lecture, you will be ready to launch into the homework. If you follow
this model, it will minimize the number of times you leave class in a daze.
In addition, spread your study time out evenly over the week, rather than
waiting until the day before an assignment is due. Acknowledgments
The efforts of numerous people have led to many changes, corrections
and improvements. We want to speciﬁcally thank Laura Acuna, Patrick
˜
Averbeck, Jim Baxter, Sandi Bennett, Daniel Bjorkegren, Cindy Burton,
Michael D. Calac, Roll Jean Cheng, Jerry Folland, Dan Fox, Grant Galbraith, Peter Garﬁeld, Richard J. Golob, Joel Grus, Fred Kuczmarski,
Julie Harris, Michael Harrison, Teri Hughes, Ron Irving, Ian Jannetty,
Mark Johnson, Michael Keynes, Andrew Loveless, Don Marshall, Linda vi
Martin, Alexandra Nichifor, Patrick Perkins, Lisa Peterson, Ken Plochinski, Eric Rimbey, Tim Roberts, Aaron Schlaﬂy, David Schneider, Marilyn
Stor, Lukas Svec, Sarah Swearinger, Jennifer Taggart, Steve Tanner, Paul
Tseng, and Rebecca Tyson. I am grateful to everyone for their hard work
and dedication to making this a better product for our students.
The Minority Science and Engineering Program (MSEP) of the College of Engineering supports the development of this textbook. It is also
authoring additional materials, namely, a student study guide and an
instructor guide. MSEP actively uses these all of these materials in its
summer mathematics program for freshman 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 was used
to maintain and develop this textbook. Comments
Send comments, corrections, and ideas to [email protected]
or [email protected] Preface
Have you ever noticed this peculiar feature of mathematics: When you
don’t know what is going on, it is really hard, difﬁcult, and frustrating.
But, when you know what is going on, mathematics seems incredibly
easy, and you wonder why you had trouble with it in the ﬁrst place!
Here is another feature of learning mathematics: When you are struggling with a mathematical problem, there are times when the answer
seems to pop out at you. At ﬁrst, nothing is there, then very suddenly,
in a ﬂash, the answer is all there, and you sit wondering why you didn’t
“see” the solution sooner. We have a special name for this: It’s the “AHa!” experience. Often the difﬁculty you have in studying mathematics
is that the rate at which you are having an 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 Pictures Symbols Figure 1: Problem solving as a transformation process. actually translating information from one form to another. The arrows
also suggest that there are alternative paths you can take to get from one
form to another! This is a very, very important point: the idea that there
is more than one way to get from words to symbols.
Let’s rewind this discussion: You’re reading a story problem. But,
now, before giving any thought to what your formula is, that is, before
worrying about your symbolic information, you grab a blank sheet of paper and start drawing a picture of your problem. And, to your picture
you add symbols denoting the quantities you need in your problem. At
this point in your problem solving, you are not trying to write any equations; you are only trying to see what your problem looks like. You are
also concentrating on another extremely important step: Deciding what
symbols to use in your problem!
Now you have a good picture of your problem. It shows not only what
the problem looks like, but symbolically shows all the problem’s variables
and constants. You can start using this information to mathematically
model your problem. The process of creating a mathematical model is actually nothing more than the arrow in the diagram going from pictures to
symbols. Mathematical modeling is the jump you make from the visual
information you have created to information contained in your formulas.
Let’s summarize the problem solving process. You start with a description of a problem that is presented to you mainly in the form of
words. Instead of trying to jump directly from words to symbols, you
jump from words to pictures. Once you have a good picture, you jump
from pictures to symbols. And, all the time, you are relying on mathematical deﬁnitions as you interpret the words of your problem; on visualization techniques as you draw pictures related to your problem; and,
on your algebra skills as you are formulating the equations you need to xi
solve your problem.
There is one ﬁnal thing to notice about the diagram in this section. All
of this discussion so far deals with the situation where your direction is
from
Words =⇒ Pictures =⇒ Symbols. But when you study the diagram you see that the arrows go both ways!
So, we will leave you with this to think about: What does it mean, within
the context of problem solving, when you have
Symbols =⇒ Pictures =⇒ Words ? An Example.
Here is a worked example that is taken from a typical homework assignment for Section 1.1 of this book. See if you can recognize the multitude
of steps needed to arrive at the equations that allow us to compute a
solution. That is, try to identify the speciﬁc way in which information is
being transformed during the problem solving process.
This problem illustrates the principle used to make a good “squirt
gun”. A cylindrical tube has diameter 1 inch, then reduces to
diameter d. The tube is ﬁlled with oil and piston A moves to the
right 2 in/sec, as indicated. This will cause piston B to move to
the right m in/sec. Assume the oil does not compress; that means
the volume of the oil between the two pistons is always the same.
oil 2 in/sec
A m in/sec
B 1. If the diameter of the narrow part of the tube is
is the speed of piston B? 1
2 inch, what in
2. If B moves 11 sec , what is the diameter of the narrow part of
the tube? Solution.
The ﬁrst thing to do with any story problem is to draw a picture of the
problem. In this case, you might 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 B Piston B.
x A Piston A. Figure 2: A resketch of the original given ﬁgure. 1. Let VA and VB stand for the change in volumes as piston A moves to
the right.
2. Let dA and dB represent the diameters of each cylinder.
3. Let rA and rB represent the radii of each cylinder.
4. Let sA and sB stand for the speeds of each piston.
5. Let xA and xB stand for the distance traveled by each piston.
Now that you have some symbols to work with, you can write the given
data down this way:
inches
1. sA = 2 second . 2. dA = 1.0 inch.
After you have studied this problem for a while, you would write down
some useful relationships:
1. The volume of any cylinder is
V = πr2 h
where r is the radius of the cylinder, and h is its height or length.
From this, you can derive the volume of a cylinder in terms of its
diameter, d:
V= π2
d h.
4 xiii
2. “Distance” = “Rate” × “Time”. In terms of this problem, you would
write
x = st,
where x is the distance your piston moves, and s is the speed of the
piston’s motion.
Now you are in a position to create a mathematical model that describes what is going on:
1. From the two relationships above, you can derive the volume equations for each cylinder so that the diameters and speed of the pistons
are included:
VA =
and
VB = πd2 (sA t)
πd2 xA
A
A
=⇒ VA =
4
4 πd2 xB
πd2 (sB t)
B
B
=⇒ VB =
4
4 2. Since the oil does not compress, at each instant when piston A is
moving, you must have VA = VB , thus:
πd2 (sA t)
πd2 (sB t)
A
B
=
.
4
4
After canceling π, t, and 4, you end up with a mathematical model
describing this problem that you can use to answer all sorts of interesting questions:
d2 sA = d2 sB .
A
B
Using your model, you can compute the following solutions:
in
1. Given: dB = 1 in, dA = 1.0 in, and sA = 2 sec , ﬁnd sB . From your model,
2
you derive: d2 sA = d2 sB =⇒ sB =
A
B d2 · sA
A
d2
B from which you can compute
m = sB = 8 in
.
sec 2. Likewise, you can use your model to compute
dB =
exactly. 2
in,
11 xiv Contents
1 2 3 4 Warming Up
1.1 Units and Rates . . . . . . .
1.2 Total Change = Rate × Time
1.3 The Modeling Process . . . .
1.4 Exercises . . . . . . . . . . . .
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8 Imposing Coordinates
2.1 The Coordinate System . . . . . . . . . . . . . . .
2.1.1 Going from P to a Pair of Real Numbers.
2.2 Three Features of a Coordinate System . . . . . .
2.2.1 Scaling . . . . . . . . . . . . . . . . . . . .
2.2.2 Axes Units . . . . . . . . . . . . . . . . . .
2.3 A Key Step in all Modeling Problems . . . . . . .
2.4 Distance . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . .
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50 Three Simple Curves
3.1 The Simplest Lines . .
3.2 Circles . . . . . . . . .
3.3 Intersecting Curves I
3.4 Summary . . . . . . .
3.5 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 Uniform Linear Motion . . . . . . . . . .
4.12 Summary . . . . . . . . . . . . . . . . . .
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. CONTENTS xvi 4.13 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 6 7 8 51 Functions and Graphs
5.1 Relating Data, Plots and Equations
5.2 What is a Function? . . . . . . . . .
5.2.1 The deﬁnition of a function
(equation viewpoint) . . . .
5.2.2 The deﬁnition of a function
(conceptual viewpoint) . . .
5.3 The Graph of a Function . . . . . .
5.4 The Vertical Line Test . . . . . . . .
5.4.1 Imposed Constraints . . . .
5.5 Linear Functions . . . . . . . . . . .
5.6 Proﬁt Analysis . . . . . . . . . . . .
5.7 Exercises . . . . . . . . . . . . . . . .............
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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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7.1 Parabolas and Vertex Form . . . . . . . . . . . . .
7.1.1 First Maneuver: Shifting . . . . . . . . . .
7.1.2 Second Maneuver: Reﬂection . . . . . . .
7.1.3 Third Maneuver: Vertical Dilation . . . .
7.1.4 Conclusion . . . . . . . . . . . . . . . . .
7.2 Completing the Square . . . . . . . . . . . . . . .
7.3 Interpreting the Vertex . . . . . . . . . . . . . . .
7.4 Quadratic Modeling Problems . . . . . . . . . . .
7.4.1 How many points determine a parabola?
7.5 What’s Needed to Build a Quadratic Model? . . .
7.6 Summary . . . . . . . . . . . . . . . . . . . . . . .
7.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . .
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. 85
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101
101
103 Composition
8.1 The Formula for a Composition . . . .
8.1.1 Some notational confusion . .
8.2 Domain, Range, etc. for a Composition
8.3 Exercises . . . . . . . . . . . . . . . . . .
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. CONTENTS
9 xvii Inverse Functions
9.1 Concept of an Inverse Function . . . . . . . . .
9.1.1 An Example . . . . . . . . . . . . . . . .
9.1.2 A Second Example . . . . . . . . . . . .
9.1.3 A Third Example . . . . . . . . . . . . .
9.2 Graphical Idea of an Inverse . . . . . . . . . . .
9.2.1 Onetoone Functions . . . . . . . . . .
9.3 Inverse Functions . . . . . . . . . . . . . . . . .
9.3.1 Schematic Idea of an Inverse Function
9.3.2 Graphing Inverse Functions . . . . . .
9.4 Trying to Invert a Non onetoone Function . .
9.5 Summary . . . . . . . . . . . . . . . . . . . . . .
9.6 Exercises . . . . . . . . . . . . . . . . . . . . . . 10 Exponential Functions
10.1 Functions of Exponential Type . . . . . . .
10.1.1 Reviewing the Rules of Exponents
10.2 The Functions y = A0 bx . . . . . . . . . . .
10.2.1 The case b = 1 . . . . . . . . . . .
10.2.2 The case b > 1 . . . . . . . . . . . .
10.2.3 The case 0 < b < 1 . . . . . . . . .
10.3 Piano Frequency Range . . . . . . . . . . .
10.4 Exercises . . . . . . . . . . . . . . . . . . . .
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. 11 Exponential Modeling
11.1 The Method of Compound Interest . .
11.1.1 Two Examples . . . . . . . . . .
11.1.2 Discrete Compounding . . . .
11.2 The Number e and
the Exponential Function . . . . . . . .
11.2.1 Calculator drill . . . . . . . . .
11.2.2 Back to the original problem...
11.3 Exercises . . . . . . . . . . . . . . . . . .
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. 12 Logarithmic Functions
12.1 The Inverse Function of y = ex . . .
12.2 Alternate form for
functions of exponential type . . . .
12.3 The Inverse Function of y = bx . . .
12.4 Measuring the Loudness of Sound
12.5 Exercises . . . . . . . . . . . . . . . .
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. 133
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142 145
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. 156
157
159
163 13 Three Construction Tools
165
13.1 A Lowtech Exercise . . . . . . . . . . . . . . . . . . . . . . 165
13.2 Reﬂection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
13.3 Shifting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 CONTENTS xviii
13.4
13.5
13.6
13.7 Dilation . .
Vertex Form
Summary of
Exercises . ........
and Order of
Rules . . . .
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Operations
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. 170
173
174
178 14 Rational Functions
181
14.1 Modeling with Lineartolinear Rational Functions . . . . 185
14.2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
14.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
15 Measuring an Angle
15.1 Standard and Central Angles .
15.2 An Analogy . . . . . . . . . . .
15.3 Degree Method . . . . . . . . .
15.4 Radian Method . . . . . . . . .
15.5 Areas of Wedges . . . . . . . .
15.5.1 Chord Approximation
15.6 Great Circle Navigation . . . .
15.7 Summary . . . . . . . . . . . .
15.8 Exercises . . . . . . . . . . . .
16 Measuring Circular Motion
16.1 Different ways to measure
Cosmo’s speed . . . . . . . .
16.2 Different Ways to Measure
Circular Motion . . . . . . .
16.2.1 Three Key Formulas
16.3 Music Listening Technology
16.4 Belt and Wheel Problems . .
16.5 Exercises . . . . . . . . . . . .
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. 17 The Circular Functions
17.1 Sides and Angles of a Right Triangle . . . . . . . . . .
17.2 The Trigonometric Ratios . . . . . . . . . . . . . . . . .
17.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . .
17.4 Circular Functions . . . . . . . . . . . . . . . . . . . . .
17.4.1 Are the trigonometric ratios functions? . . . .
17.4.2 Relating circular functions and right triangles
17.5 What About Other Circles? . . . . . . . . . . . . . . . .
17.6 Other Basic Circular Function . . . . . . . . . . . . . .
17.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
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. 18 Trigonometric Functions
237
18.1 Easy Properties of Circular Functions . . . . . . . . . . . 237
18.2 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
18.3 Graphs of Circular Functions . . . . . . . . . . . . . . . . 242 CONTENTS xix 18.3.1 A matter of scaling . . . . . . . . . .
18.3.2 The sine and cosine graphs . . . . .
18.3.3 The tangent graph . . . . . . . . . .
18.4 Trigonometric Functions . . . . . . . . . . .
18.4.1 A Transition . . . . . . . . . . . . . .
18.4.2 Graphs of trigonometric functions .
18.4.3 Notation for trigonometric functions
18.5 Exercises . . . . . . . . . . . . . . . . . . . . .
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. 243
243
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247
247
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249
250 19 Sinusoidal Functions
19.1 A special class of functions . . . . . . . . . . . . . .
19.1.1 How to roughly sketch a sinusoidal graph
19.1.2 Functions not in standard sinusoidal form
19.2 Examples of sinusoidal behavior . . . . . . . . . . .
19.3 Summary . . . . . . . . . . . . . . . . . . . . . . . .
19.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
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. 267
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281 20 Inverse Circular Functions
20.1 Solving Three Equations . . . . . . .
20.2 Inverse Circular Functions . . . . . .
20.3 Applications . . . . . . . . . . . . . . .
20.4 How to solve trigonometric equations
20.5 Summary . . . . . . . . . . . . . . . .
20.6 Exercises . . . . . . . . . . . . . . . . .
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. Appendix 285 A Useful Formulas 287 B Answers 291 C GNU Free Documentation License
C.1 Applicability and Deﬁnitions . . . . .
C.2 Verbatim Copying . . . . . . . . . . .
C.3 Copying in Quantity . . . . . . . . . .
C.4 Modiﬁcations . . . . . . . . . . . . . .
C.5 Combining Documents . . . . . . . .
C.6 Collections of Documents . . . . . . .
C.7 Aggregation With Independent Works
C.8 Translation . . . . . . . . . . . . . . .
C.9 Termination . . . . . . . . . . . . . . .
C.10 Future Revisions of This License . . . 303
304
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. References 311 Index 311 xx CONTENTS Chapter 1
Warming Up
The basic theme of this book is to study precalculus within the context
of problem solving. This presents a challenge, since skill in problem
solving is as much an art or craft as it is a science. As a consequence,
the process of learning involves an active apprenticeship rather than a
passive reading of a text. We are going to start out by assembling a basic
toolkit of examples and techniques that are essential in everything that
follows. The main ideas discussed in the next couple of chapters will
surely be familiar; our perspective on their use and importance may be
new.
The process of going from equations to pictures involves the key concept of a graph , while the reverse process of going from pictures (or raw
data) to equations is called modeling . Fortunately, the study of graphing
and modeling need not take place in a theoretical vacuum. For example,
imagine you have tossed a ball from the edge of a cliff. A number of natural questions arise: Where and when does the ball reach its maximum
height? Where and when does the ball hit the ground? Where is the ball
located after t seconds?
We can attack these questions from two directions. If
we knew some basic physics, then we would have equaPath of tossed ball.
tions for the motion of the ball. Going from these equations
to the actual curved path of the ball becomes a graphing
problem; answering the questions requires that we really
Cliff.
understand the relationship between the symbolic equaGround level.
tions and the curved path. Alternatively, we could approach these questions without knowing any physics. The
idea would be to collect some data, keeping track of the
Figure 1.1: Ball toss.
height and horizontal location of the ball at various times,
then ﬁnd equations whose graphs will “best” reproduce the
collected data points; this would be a modeling approach to the problem. Modeling is typically harder than graphing, since it requires good
intuition and a lot of experience.
1 CHAPTER 1. WARMING UP 2 1.1 Units and Rates
A marathon runner passes the 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) = (constant speed) × (elapsed time)
(ft) = (ft/sec) × (sec) (1.1) To apply the formula, let t represent the elapsed time in seconds and
ﬁrst carry out a “conversion of units” using the conversion factor “5,280
ft/mile”. Recall, we can manipulate the units just like numbers, canceling common units on the top and bottom of a fraction:
mile
////· ft
mile
////
= (26.2)(5,280) ft. 26.2 mile × (5,280 ft/mile) = (26.2)(5,280)
Formula 1.1 can now be applied:
(26.2)(5,280) ft = 18 ft/sec × t
(26.2)(5,280)
18 ft
/
=t
ft/sec
/ 7,685.33 1
= 7,685.33 sec = t
1/sec So, the runner would complete the race in t = 7,685.33 seconds. If we
wanted this answer in more sensible units, we would go through yet
another units conversion:
t = 7,685.33 sec × (1 min/60 sec) × (1 hr/60 min)
7,685.33
hr
=
602
= 2.1348 hr. 1.1. UNITS AND RATES 3 The ﬁnish clock will display elapsed time in units of “hours : minutes :
seconds”. Two further conversions (see Exercise 1.5) lead to our runner
having a time of 2:08:05.33; this is a world class time!
Manipulation of units becomes especially important when we are working with the density of a substance, which is deﬁned by
def density = mass
.
volume For example, pure water has a density of 1 g/cm3 . Notice, given any two
of the quantities “density, volume or mass,” we can solve for the remaining unknown using the formula. For example, if 857 g of an unknown
substance has a volume of 2.1 liters, then the density would be
mass
volume
857 g
=
2.1 L
g
857
×
=
2.1
L
/
3
= 0.408 g/cm . d= × 1/
L
1,000 cm3 Example 1.1.1. A sphere of solid gold has a mass of 100 kg and the density of gold is 19.3 g/cm3 . What is the radius of the sphere?
Solution. This problem is more involved. To answer this, let r be the
unknown radius of the sphere in units of cm. The volume of the sphere
is V = 4 πr3 . Since the sphere is solid gold, the density of gold is the ratio
3
density of gold = mass of sphere
volume of sphere Plugging in what we know, we get the equation
19.3 g
=
cm3 100kg
4
πr3
3 = 100kg
4
πr3
3 Solving for r3 we ﬁnd
r3 = 1236.955516 cm3
from which we get
r = 1237 cm3 1/3 = 10.73457 cm 1000 g
1 kg = 105 g
4
πr3
3 CHAPTER 1. WARMING UP 4 1.2 Total Change = Rate × Time
We live in a world where things are changing as time goes by: the temperature during the day, the cost of tuition, the distance you will travel
after leaving this class, and so on. The ability to precisely describe how
a quantity is changing becomes especially important when making any
kind of experimental measurements. For this reason, let’s start with a
clear and careful deﬁnition. If a quantity is changing with respect to time
(like temperature, distance or cost), we can keep track of this using what
is called a rate (also sometimes called a rate of change ); this is deﬁned
as follows:
def rate = change in the quantity
change in time This sort of thing comes up so frequently, there is special shorthand
notation commonly used: We let the Greek letter ∆ (pronounced “delta”)
be shorthand for the phrase “change in.” With this agreement, we can
rewrite our rate deﬁnition in this way:
def rate = ∆ quantity
∆ time But, now the question becomes: How do we calculate a rate? If you think
about it, to calculate “∆ quantity” in the rate deﬁnition requires that we
compare two quantities at two different times and see how they differ
(i.e., how they have changed). The two times of comparison are usually
called the ﬁnal time and the initial time. We really need to be precise
about this, so here is what we mean:
∆ quantity = (value of quantity at ﬁnal time) −
(value of quantity at initial time)
∆ time = (ﬁnal time) − (initial time).
For example, suppose that on June 4 we measure that the temperature at 8:00 am is 65◦ F and at 10:00 am it is 71◦ F. So, the ﬁnal time is
10:00 am, the initial time is 8:00 am and the temperature is changing
according to the
∆ quantity
∆ time
ﬁnal value of quantity − initial value of quantity
=
ﬁnal time − initial time
71 − 65 degrees
=
10:00 − 8:00 hours
= 3 deg/hr. rate = As a second example, suppose on June 5 the temperature at 8:00 am is
71◦ and at 10:00 am it is 65◦ . So, the ﬁnal time is 10:00 am, the initial 1.2. TOTAL CHANGE = RATE × TIME 5 time is 8:00 am and the temperature is changing according to the
∆ quantity
∆ time
ﬁnal value of quantity − initial value of quantity
=
ﬁnal time − initial time
65 − 71 degrees
=
10:00 − 8:00 hours
= −3 deg/hr. rate = These two examples illustrate that a rate can be either a positive or a
negative number. More importantly, it highlights that we really need to
be careful when making a rate computation. In both examples, the initial
and ﬁnal times are the same and the two temperatures involved are the
same, BUT whether they occur at the initial or ﬁnal time is interchanged.
If we accidently mix this up, we will end up being off by a minus sign.
There are many situations where the rate is the same for all time
periods. In a case like this, we say we have a constant rate. For example,
imagine you are driving down the freeway at a constant speed of 60 mi/hr.
The fact that the speedometer needle indicates a steady speed of 60 mi/hr
means the rate your distance is changing is constant.
In cases when we have a constant rate, we often want to ﬁnd the total
amount of change in the quantity over a speciﬁc time period. The key
principle in the background is this:
Total Change in some Quantity = Rate × Time (1.2) It is important to mention that this formula only works when we have
a constant rate, but that will be the only situation we encounter in this
course. One of the main goals of calculus is to develop a version of (1.2)
that works for 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 cm2 /hour.
Find the surface area and dimensions of the puddle after 84 minutes.
Solution. The quantity changing is “surface area” and we are given a
“rate” and “time.” Using (1.2) with time t = 84 minutes,
Total Surface Area = Rate × Time
cm2
84
=
4
×
hr
hr
60
= 5.6 cm2 .
The formula for the area of a circular region of radius r is given at the
back of this text. Using this, the puddle has radius r =
time t = 84 minutes. 5.6
π = 1.335 cm at CHAPTER 1. WARMING UP 6 1.3 The Modeling Process
Modeling is a method used in disciplines ranging from architecture to
zoology. This mathematical technique will crop up any time we are problem solving and consciously trying to both “describe” and “predict.” Inevitably, mathematics is introduced to add structure to the model, but
the clean equations and formulas only arise after some (or typically a lot)
of preliminary work.
A model can be thought of as a caricature in that it will pick out certain features (like a nose or a face) and focus on those at the expense of
others. It takes a lot of experience to know which models are “good” and
“bad,” in the sense of isolating the right features. In the beginning, modeling will lead to frustration and confusion, but by the end of this course
our comfort level will dramatically increase. Let’s look at an illustration
of the problem solving process.
Example 1.3.1. How much time do you anticipate studying precalculus
each week?
Solution. One possible response is simply to say “a little” or “way too
much!” You might not think these answers are the result of modeling,
but they are. They are a consequence of modeling the total amount of
study time in terms of categories such as “a little,” “some,” “lots,” “way
too much,” etc. By drawing on your past experiences with math classes
and using this crude model you arrived at a preliminary answer to the
question.
Let’s put a little more effort into the problem and try to come up with
a numerical estimate. If T is the number of hours spent on precalculus a
given week, it is certainly the case that:
T = (hours in class) + (hours reading text) + (hours doing homework)
Our time in class each week is known to be 5 hours. However, the other
two terms require a little more thought. For example, if we can comfortably read and digest a page of text in (on average) 15 minutes and there
are r pages of text to read during the week, then
(hours reading text) = 15
r hours.
60 As for homework, if a typical homework problem takes (on average) 25
minutes and there are h homework problems for the week, then
(hours doing homework) = 25
h hours.
60 We now have a mathematical model for the weekly time commitment to
precalculus:
T =5+ 25
15
r + h hours.
60
60 1.3. THE MODELING PROCESS 7 Is this a good model? Well, it is certainly more informative than our
original crude model in terms of categories like “a little” or “lots.” But, the
real plus of this model is that it clearly isolates the features being used
to make our estimated time commitment and it can be easily modiﬁed as
the amount of reading or homework changes. So, this is a pretty good
model. However, it isn’t perfect; some homework problems will take a lot
more than 25 minutes! CHAPTER 1. WARMING UP 8 1.4 Exercises
Problem 1.1.
(a) Verify that 7685.33 seconds is 2 hours 8 minutes 5.33 seconds. (b) Allyson has a pace of 6 min/mile; what
is her speed? (b) Which is faster: 100 mph or 150 ft/s? (c) Adrienne and Dave are both running
a race.
Adrienne has a pace of
5.7 min/mile and Dave is running
10.3 mph. Who is running faster? (c) Gina’s salary is 1 cent/second for a
40 hour work week. Tiare’s salary is
$1400 for a 40 hour work week. Who
has a higher salary?
(d) Suppose it takes 180 credits to get a
baccalaureate degree.
You accumulate credit at the rate of one credit per
quarter for each hour that the class
meets per week. For instance, a class
that meets three hours each week of
the quarter will count for three credits.
In addition, suppose that you
spend 2.5 hours of study outside of class
for each hour in class. A quarter is
10 weeks long. How many total hours,
including time spent in class and time
spent studying out of class, must you invest to get a degree?
Problem 1.2. Sarah can bicycle a loop around
the north part of Lake Washington in 2 hours
and 40 minutes. If she could increase her average speed by 1 km/hr, it would reduce her
time around the loop by 6 minutes. How many
kilometers long is the loop?
Problem 1.3. The density of lead is
11.34 g/cm3 and the density of aluminum is
2.69 g/cm3 . Find the radius of lead and aluminum spheres each having a mass of 50 kg.
Problem 1.4. The Eiffel Tower has a mass of
7.3 million kilograms and a height of 324 meters. Its base is square with a side length of
125 meters. The steel used to make the Tower
occupies a volume of 930 cubic meters. Air
has a density of 1.225 kg per cubic meter.
Suppose the Tower was contained in a cylinder. Find the mass of the air in the cylinder. Is
this more or less than the mass of the Tower?
Problem 1.5. Marathon runners keep track
of their speed using units of pace = minutes/mile.
(a) Lee has a speed of 16 ft/sec; what is his
pace? Problem 1.6. Convert each of the following
sentences into “pseudoequations.” For example, suppose you start with the sentence:
“The cost of the book was more than $10 and
the cost of the magazine was $4.” A ﬁrst step
would be these “pseudoequations”:
(Book cost) > $10 and (Magazine cost) = $4.
(a) John’s salary is $56,000 a year and he
pays no taxes.
(b) John’s salary is at most $56,000 a year
and he pays 15% of his salary in taxes.
(c) John’s salary is at least $56,000 a year
and he pays more than 28% of his salary
in taxes.
(d) The number of students taking Math 120
at the UW is somewhere between 1500
and 1800 each year.
(e) The cost of a new red Porsche is more
than three times the cost of a new Ford
F150 pickup truck.
(f) Each week, students spend at least two
but no more than three hours studying
for each credit hour.
(g) Twice the number of happy math students exceeds ﬁve times the number of
happy chemistry students. However, all
of the happy math and chemistry students combined is less than half the total number of cheerful biology students.
(h) The difference between Cady’s high and
low midterm scores was 10%. Her ﬁnal
exam score was 97%.
(i) The vote tally for Gov. Tush was within
onehundredth of one percent of onehalf the total number of votes cast.
Problem 1.7. Which is a better deal: A 10 inch
diameter pizza for $8 or a 15 inch diameter
pizza for $16? 1.4. EXERCISES 9 Problem 1.8. The famous theory of relativity
predicts that a lot of weird things will happen when you approach the speed of light
c = 3 × 108 m/sec. For example, here is a formula that relates the mass mo (in kg) of an
object at rest and its mass when it is moving
at a speed v:
m = mo
1− v2
c2 Problem 1.11. A typical cell in the human
body contains molecules of deoxyribonucleic
acid, referred to as DNA for short. In the cell,
this DNA is all twisted together in a tight little
packet. But imagine unwinding (straightening
out) all of the DNA from a single typical cell
and laying it “endtoend”; then the sum total
length will be approximately 2 meters. . isolate DNA
from nucleus (a) Suppose the object moving is Dave, who
has a mass of mo = 66 kg at rest. What is
Dave’s mass at 90% of the speed of light?
At 99% of the speed of light? At 99.9% of
the speed of light? cell
nucleus (b) How fast should Dave be moving to have
a mass of 500 kg? Problem 1.9. During a typical evening in Seattle, Pagliacci receives phone orders for pizza
delivery at a constant rate: 18 orders in a typical 4 minute period. How many pies are sold
in 4 hours? Assume Pagliacci starts taking orders at 5 : 00 pm and the proﬁt is a constant
rate of $11 on 10 orders. When will phone order
proﬁt exceed $1,000?
Problem 1.10. Aleko’s Pizza has delivered a
beautiful 16 inch diameter pie to Lee’s dorm
room. The pie is sliced into 8 equal sized
pieces, but Lee is such a 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?
John’s part lay out
end−to−end 2m Assume the human body has 1014 cells containing DNA. How many times would the sum
total length of DNA in your body wrap around
the equator of the earth?
Problem 1.12. A water pipe mounted to the
ceiling has a leak and is dripping onto the ﬂoor
below, creating a circular puddle of water. The
area of the circular puddle is increasing at a
constant rate of 11 cm2 /hour.
(a) Find the area and radius of the puddle
after 1 minute, 92 minutes, 5 hours, 1
day.
(b) Is the radius of the puddle increasing at
a constant rate?
Problem 1.13. During the 1950s, Seattle was
dumping an average of 20 million gallons of
sewage into Lake Washington each day.
(a) How much sewage went into Lake Washington in a week? In a year? Lee’s part (b) In order to illustrate the amounts involved, imagine a rectangular prism
whose base is the size of a football
ﬁeld (100 yards × 50 yards) with height
h yards. What are the dimensions of
such a rectangular prism containing the CHAPTER 1. WARMING UP 10
sewage dumped into Lake Washington in
a single day? (Note: There are 7.5 gallons in one cubic foot. Dumping into
Lake Washington has stopped; now it
goes into the Puget Sound.)
Problem 1.14. Dave has inherited an apple
orchard on which 60 trees are planted. Under
these conditions, each tree yields 12 bushels
of apples. According to the local WSU extension agent, each time Dave removes a tree the
yield per tree will go up 0.45 bushels. Let x be
the number of trees in the orchard and N the
yield per tree.
(a) Find a formula for N in terms of the unknown x. (Hint: Make a table of data
with one column representing various
values of x and the other column the
corresponding values of N. After you
complete the ﬁrst few rows of the table,
you need to discover the pattern.)
(b) What possible reason(s) might explain
why the yield goes up when you remove
trees?
Problem 1.15. Congress is debating a proposed law to reduce tax rates. If the current
tax rate is r %, then the proposed rate after x
years is given by this formula:
r
1+ . 1
1+ 1
x Rewrite this formula as a simple fraction. Use
your formula to calculate the new tax rate af ter 1, 2, 5 and 20 years. Would tax rates increase or decrease over time? Congress claims
that this law would ultimately cut peoples’ tax
rates by 75 %. Do you believe this claim?
Problem 1.16.
(a) The temperature at 7:00
am is 44◦ F and the temperature at 10:00
am is 50◦ F. What are the initial time, the
ﬁnal time, the initial temperature and
the ﬁnal temperature? What is the rate
of change in the temperature between
7:00 am and 10:00 am?
(b) Assume it is 50◦ F at 10:00 am and
rate of change in the temperature
tween 10:00 am and 2:00 pm is
same as the rate in part (a). What is
temperature at 2:00 pm? the
bethe
the (c) The temperature at 4:30 pm is 54◦ F and
the temperature at 6:15 pm is 26◦ F.
What are the initial time, the ﬁnal time,
the initial temperature and the ﬁnal
temperature? What is the rate of change
in the temperature between 4:30 pm and
6:15 pm?
Problem 1.17.
(b) Solve for a: (a) Solve for t: 3t − 7 = 11 + t.
1
a = 3.
√
(c) Solve for x: x2 + a2 = 2a + x.
1+ (d) Solve for t: 1 − t > 4 − 2t.
(e) Write as a single fraction:
2
1
−
x x+1 Chapter 2
Imposing Coordinates
You ﬁnd yourself visiting Spangle, WA and dinner time is approaching.
A friend has recommended Tiff’s Diner, an excellent restaurant; how will
you ﬁnd it?
Of course, the solution to this simple problem amounts to locating a
“point” on a 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 ﬂow of ideas underlying twodimensional coordinate systems in such a way
Q
that it easily generalizes to threedimensions. Suppose
we start with a blank piece of paper and mark two points;
let’s label these two points “P ” and “Q.” This presents the
P
basic problem of ﬁnding a foolproof method to reconstruct
the picture.
Figure 2.1: Two points in a
The basic idea is to introduce a coordinate system for
plane.
the plane (analogous to the city map grid of streets), allowing us to catalog points in the plane using pairs of real numbers (analogous to the addresses of locations in the city).
Here are the details. Start by drawing two perpendicular lines, called
the horizontal axis and the vertical axis , each of which looks like a copy
of the real number line. We refer to the intersection point of these two
lines as the origin . Given P in the plane, the plan is to use these two axes
to obtain a pair of real numbers (x,y) that will give us the exact location of
P. With this in mind, the horizontal axis is often called the xaxis and the
vertical axis is often called the yaxis. Remember, a typical real number
line (like the xaxis or the yaxis) 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 xaxis and
yaxis. Unless we say otherwise, we will always adopt the convention that
the positive xaxis consists of those numbers to the right of the origin on
11 CHAPTER 2. IMPOSING COORDINATES 12 the xaxis and the positive yaxis consists of those numbers above the
origin on the yaxis. We have just described the xycoordinate system for
the plane:
Positive yaxis
Negative real
numbers Positive real
numbers Origin Negative xaxis Zero Positive xaxis
Negative yaxis (a) Number line. (b) xycoordinate system. Figure 2.2: Coordinates. 2.1.1 Going from P to a Pair of Real Numbers.
yaxis
P ℓ y xaxis
x
ℓ∗ Figure 2.3:
Coordinate
pairs. Imagine a coordinate system had been drawn on our piece
of paper in Figure 2.1. Let’s review the procedure of going
from a point P to a pair of real numbers:
1. First, draw two new lines passing through P, one
parallel to the xaxis and the other parallel to the
yaxis; call these ℓ and ℓ∗ , as pictured in Figure 2.3.
2. Notice that ℓ will cross the yaxis exactly once; the
point on the yaxis where these two lines cross will
be called “y.” Likewise, the line ℓ∗ will cross the
xaxis exactly once; the point on the xaxis where
these two lines cross will be called “x.”
3. If you begin with two different points, like P and Q in
Figure 2.1, you will see that the two pairs of points
you obtain will be different; i.e., if Q gives you the
pair (x∗ ,y∗ ), then either x = x∗ or y = y∗ . This shows
that two different points in the plane give two different pairs of real numbers and describes the process
of assigning a pair of real numbers to the point P. The great thing about the procedure we just described is that it is
reversible! In other words, suppose you start with a pair of real numbers,
say (x,y). Locate the number x on the xaxis and the number y on the
yaxis. Now draw two lines: a line ℓ parallel to the xaxis passing through
the number y on the yaxis and a line ℓ∗ parallel to the yaxis passing
through the number x on the xaxis. The two lines ℓ and ℓ∗ will intersect 2.2. THREE FEATURES OF A COORDINATE SYSTEM 13 in exactly one point in the plane, call it P. This procedure describes
how to go from a given pair of real numbers to a point in the plane. In
addition, if you start with two different pairs of real numbers, then the
corresponding two points in the plane are going to be different. In the
future, we will constantly be going back and forth between points in the
plane and pairs of real numbers using these ideas.
Deﬁnition 2.1.1. Coordinate System: Every point P in the xyplane corresponds to a unique pair of real numbers (x, y), where x is a number on the
horizontal xaxis and y is a number on the vertical yaxis; for this reason,
we commonly use the notation “P = (x,y).”
Having speciﬁed positive and negative directions on
the horizontal and vertical axes, we can now divide our
two dimensional plane into four quadrants . The ﬁrst
quadrant corresponds to all the points where both coordinates are positive, the second quadrant consists of
points with the ﬁrst coordinate negative and the second
coordinate positive, etc. Every point in the plane will lie
in one of these four quadrants or on one of the two axes.
This quadrant terminology is useful to give a rough sense
of location, just as we use the terminology “Northeast,
Northwest, Southwest and Southeast” when discussing
locations on a map. yaxis Second
Quadrant First
Quadrant Third
Quadrant Fourth
Quadrant Figure 2.4: Quadrants in the
xyplane. 2.2 Three Features of a Coordinate System
A coordinate system involves scaling, labeling and units on each of the
axes. 2.2.1 Scaling
Sketch two xy coordinate systems. In the ﬁrst, make the scale on each
axis the same. In the second, assume “one unit” on the x axis has the
same length as “two units” on the y axis. Plot the points (1,1), (−1,1),
39
24
11
11
24
39
4
4
− 5 , 16 , − 5 , 25 , − 5 , 25 , − 5 , 25 , (0,0), 5 , 25 , 5 , 25 , 5 , 25 , 5 , 16 , (1,1).
25
25
Both pictures illustrate how the points lie on a parabola in the xycoordinate system, but the aspect ratio has changed. The aspect ratio is
deﬁned by this fraction:
def aspect ratio = xaxis length of one unit on the vertical axis
.
length of one unit on the horizontal axis Figure 2.5(a) has aspect ratio 1, whereas Figure 2.5(b) has aspect ratio
1
. In problem solving, you will often need to make a rough assumption
2
about the relative axis scaling. This scaling will depend entirely on the CHAPTER 2. IMPOSING COORDINATES 14
1.0 yaxis 0.8
0.6 1.0 yaxis
0.8
0.6
0.4
0.2 0.4
0.2
−1.0 −0.5 0.0 0.5 xaxis
1.0 (a) Aspect ratio = 1. −1.0 −0.5 0.0 xaxis
0.5 1.0 1
(b) Aspect ratio = 2 . Figure 2.5: Coordinates. information given in the problem. Most graphing devices will allow you
to specify the aspect ratio. 2.2.2 Axes Units
Sometimes we are led to coordinate systems where each of the two axes
involve different types of units (labels). Here is a sample, that illustrates
the power of using pictures.
Example 2.2.1. As the marketing director of Turboweb software, you have
been asked to deliver a brief message at the annual stockholders meeting
on the performance of your product. Your staff has assembled this tabular collection of data; how can you convey the content of this table most
clearly?
week
1
2
3
4
5
6
7
8
9
10 sales
11.0517
12.214
13.4986
14.9182
16.4872
18.2212
20.1375
22.2554
24.596
27.1828 week
11
12
13
14
15
16
17
18
19
20 TURBOWEB SALES (in $1000’s)
sales
week
sales
week
30.0417
21
81.6617
31
33.2012
22
90.2501
32
36.693
23
99.7418
33
40.552
24
110.232
34
44.8169
25
121.825
35
49.5303
26
134.637
36
54.7395
27
148.797
37
60.4965
28
164.446
38
66.8589
29
181.741
39
73.8906
30
200.855
40 sales
221.98
245.325
271.126
299.641
331.155
365.982
404.473
447.012
494.024
545.982 week
41
42
43
44
45
46
47
48
49
50 sales
603.403
666.863
736.998
814.509
900.171
994.843
1099.47
1215.1
1342.9
1,484.131 One idea is to simply ﬂash an overhead slide of this data to the audience; this can be deadly! A better idea is to use a visual aid. Suppose
we let the variable x represent the week and the variable y represent the
gross sales (in thousands of dollars) in week x. We can then plot the
points (x,y) in the xycoordinate system; see Figure 2.6.
Notice, the units on the two axes are very different: yaxis units are
“thousands of dollars” and xaxis units are “weeks.” In addition, the
aspect ratio of this coordinate system is not 1. The beauty of this picture
is the visual impact it gives your audience. From the coordinate plot we
can get a sense of how the sales ﬁgures are dramatically increasing. In
fact, this plot is good evidence you deserve a big raise! 2.3. A KEY STEP IN ALL MODELING PROBLEMS
Mathematical modeling is all about relating concrete
phenomena and symbolic equations, so we want to embrace the idea of visualization. Most typically, visualization will involve plotting a collection of points in the plane.
This can be achieved by providing a “list” or a “prescription” for plotting the points. The material we review in the
next couple of sections makes the transition from symbolic mathematics to visual pictures go more smoothly. 15
yaxis (Thousands of Dollars)
1,400
1,200
1,000
800
600
400
200 xaxis
(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
xycoordinate system is called imposing a coordinate system : There will
often be many possible choices; it takes problem solving experience to
develop intuition for a “natural” choice. This is a key step in all modeling
problems.
Example 2.3.1. Return to the tossed ball scenario on page 1. How do we
decide where to draw a coordinate system in the picture?
Figure 2.7 on page 16 shows four natural choices of xycoordinate
system. To choose a coordinate system we must specify the origin. The
four logical choices for the origin are either the top of the cliff, the bottom
of the cliff, the launch point of the ball or the landing point of the ball.
So, which choice do we make? The answer is that any of these choices
will work, but one choice may be more natural than another. For example, Figure 2.7(b) is probably the most natural choice: in this coordinate
system, the motion of the ball takes place entirely in the ﬁrst quadrant,
so the x and y coordinates of any point on the path of the ball will be
nonnegative.
Example 2.3.2. Michael and Aaron are running toward each other, beginning at opposite ends of a 10,000 ft. airport runway, as pictured in
Figure 2.8 on page 17. Where and when will these guys collide?
Solution. This problem requires that we ﬁnd the “time” and “location” of
the collision. Our ﬁrst step is to impose a coordinate system.
We choose the coordinate system so that Michael is initially located
at the point M = (0, 0) (the origin) and Aaron is initially located at the
point A = (10,000, 0). To ﬁnd the coordinates of Michael after t seconds,
we need to think about how distance and time are related.
Since Michael is moving at the rate of 15 ft/second, then after one
second he is located 15 feet right of the origin; i.e., at the point (15, 0).
After 2 seconds, Michael has moved an additional 15 feet, for a total of
30 feet; so he is located at the point (30, 0), etc. Conclude Michael has
traveled 15t ft. to the right after t seconds; i.e., his location is the point CHAPTER 2. IMPOSING COORDINATES 16
yaxis
Path of tossed ball. yaxis
Path of tossed ball. xaxis
Cliff. Cliff.
xaxis (a) Origin at the top of the ledge. Path of tossed ball. yaxis (b) Origin at the bottom of the
ledge.
yaxis
Path of tossed ball. xaxis
Cliff. Cliff.
xaxis (c) Origin at the landing point. (d) Origin at the launch point. Figure 2.7: Choices when imposing an xycoordinate system. M(t) = (15t, 0). Similarly, Aaron is located 8 ft. left of his starting location
after 1 second (at the point (9,992, 0)), etc. Conclude Aaron has traveled
8t ft. to the left after t seconds; i.e., his location is the point A(t) =
(10,000 − 8t, 0).
The key observation required to solve the problem is that the point
of collision occurs when the coordinates of Michael and Aaron are equal.
Because we are moving along the horizontal axis, this amounts to ﬁnding
where and when the xcoordinates of M(t) and A(t) agree. This is a
straight forward algebra problem:
15t = 10,000 − 8t
23t = 10,000
t = 434.78 (2.1) To the nearest tenth of a second, the runners collide after 434.8
seconds. Plugging t = 434.78 into either expression for the position:
M(434.8) = (15(434.8), 0) = (6,522, 0). 2.4. DISTANCE
Michael: 15 17 ft
sec Aaron: 8 ft
sec yaxis
xaxis
M = (0,0) 10,000 ft (b) The xycoordinate picture. (a) The physical picture.
yaxis NOT TO SCALE!
Aaron starts here. Michael starts here. (0,0) A = (10,000, 0) M(t) = (15t, 0) A(t) = (10,000 − 8t, 0)
xaxis M after t seconds. A after t seconds. (c) Building a visual model
Michael: 10 mph. Aaron: 8 mph. 6,522 feet to collision point. (d) Michael and Aaron’s collision point.
Figure 2.8: Michael and Aaron running 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 P = (x1 ,y1 ) and Q = (x2 ,y2 ) in the xy coordinate
system, where we assume that the units on each axis are the same ;
for example, both in units of “feet.” Imagine starting at the location P
(Denver) and ﬂying to the location Q (New York) along a straight line
segment; see Figure 2.9(a). Now ask yourself this question: To what
overall extent have the x and y coordinates changed?
To answer this, we introduce visual and notational aides into this
ﬁgure. We have inserted an “arrow” pointing from the starting position P
to the ending position Q; see Figure 2.9(b). To simplify things, introduce
the notation ∆x to keep track of the change in the xcoordinate and ∆y CHAPTER 2. IMPOSING COORDINATES 18
yaxis yaxis
Q = (x2 , y2 ) Q y2 End (stop)
here. ??
?? Ending
point. d
∆y Begin (start) here.
P
xaxis
?? (a) Starting and stopping points. y1 P = (x1 , y1 )
Beginning point.
x1 xaxis x2
∆x (b) Coordinates for P and Q. Figure 2.9: The meaning of ∆x and ∆y. to keep track of the change in the ycoordinate, as we move from P to Q.
Each of these quantities can now be computed:
∆x =
=
=
∆y =
=
= change in xcoordinate going from P to Q
(2.2)
(xcoord of ending point) − (xcoord of beginning point)
x2 − x1
change in ycoordinate going from P to Q
(ycoord of ending point) − (ycoord of beginning point)
y2 − y1 . We can interpret ∆x and ∆y using the right triangle in Figure 2.9(b).
This means we can use the Pythagorean Theorem to write:
d2 = (∆x)2 + (∆y)2 ;
that is,
d= (∆x)2 + (∆y)2 , which tells us the distance d from P to Q. In other words, d is the distance
we would ﬂy if we had ﬂown along that line segment connecting the two
points. As an example, if P = (1, 1) and Q = (5, 4), then ∆x = 5 − 1 = 4,
∆y = 4 − 1 = 3 and d = 5.
There is a subtle idea behind the way we deﬁned ∆x and ∆y: You need
to specify the “beginning” and “ending” points used to do the calculation in Equations 2.2. What happens if we had reversed the choices in
Figure 2.9?
Then the quantities ∆x and ∆y will both be negative and the lengths of
the sides of the right triangle are computed by taking the absolute value 2.4. DISTANCE 19 of ∆x and ∆y. As far as a distance calculation is concerned, the previous
formula still works because of this algebra equality:
d=
= (∆x)2 + (∆y)2
(∆x)2 + (∆y)2 . We will sometimes refer to ∆x and ∆y as directed distances in the x and
y directions. The notion of directed distance becomes important in our
discussion of lines in Chapter 4 and later when you learn about vectors ;
it is also very important in calculus.
For example, if P = (5, 4) and Q = (1, 1), then ∆x =
yaxis
1 − 5 = −4, ∆y = 1 − 4 = −3 and d = 5.
Important Fact 2.4.2 (Distance formula). If P = (x1 , y1)
and Q = (x2 , y2) are two points in the plane, then the
straight line distance between the points (in the same units
as the two axes) is given by the formula
d=
= (∆x)2 + (x2 − x1 Beginning
point. d ∆y
y2 Q = (x2 , y2 )
Ending point. (∆y)2
)2 P = (x1 , y1 ) y1 x2 + (y2 − y1 )2 . (2.3) xaxis x1
∆x Figure 2.10: A different direction. If your algebra is a little rusty, a very common mistake may crop up
when you are using the distance formula. For example,
?√ = 32 + √
42 ? = 3+4
= 7. Notice, you have an impossible situation: 5 is never equal to 7. !!! 32 + 42
√
9 + 16
5 CAUTION
!!! CHAPTER 2. IMPOSING COORDINATES 20 Example 2.4.3. Two cars depart from a four way intersection at the same time, one heading East and the other
heading North. Both cars are traveling at the constant
speed of 30 ft/sec. Find the distance (in miles) between the two cars after
1 hour 12 minutes. In addition, determine when the two cars would be
exactly 1 mile apart. Solution. Begin with a picture of the situation. We have
indicated the locations of the two vehicles after t seconds
and the distance d between them at time t. By the distance formula, the distance between them is yaxis
(North)
(0,b)
d
xaxis
(East) d=
= (a,0) Figure 2.11: Two departing
cars. (a − 0)2 + (0 − b )2
a2 + b2 . This formula is a ﬁrst step; the difﬁculty is that we have
traded the mystery distance d for two new unknown numbers a and b. To ﬁnd the coordinate a for the Eastbound
car, we know the car is moving at the rate of 30 ft/sec, so
it will travel 30t feet after t seconds; i.e., a = 30t. Similarly,
we ﬁnd that b = 30t. Substituting into the formula for d
we arrive at
d= (30t)2 + (30t)2 2t2 (30)2
√
= 30t 2.
= First, we need to convert 1 hour and 12 minutes into seconds so that
our formula can be used:
1 hr 12 min = 1 + 12/60 hr
= 1.2 hr
60 min
= (1.2hr)
hr
= 4,320 sec. 60 sec
min Substituting t = 4,320 sec and recalling that 1 mile = 5,280 feet, we arrive
at
√
d = 129,600 2 feet
= 183,282 feet
= 34.71 miles.
For the second question, we specify the distance being 1 mile and want
to ﬁnd when this occurs. The idea is to set d equal to 1 mile and solve for 2.4. DISTANCE
t. However, we need to be careful, since the units for d are feet:
√
30t 2 = d
= 5,280
Solving for t:
5,280
√
30 2
= 124.45 seconds
= 2 minutes 4 seconds. t= The two cars will be 1 mile apart in 2 minutes, 4 seconds. 21 CHAPTER 2. IMPOSING COORDINATES 22 2.5 Exercises
Problem 2.1. In the following four cases, let P
be the initial (starting) point and Q the ending
point; recall Equation 2.2 and Figure 2.10 on
Page 19. Compute d = the distance from P to
Q, ∆x and ∆y. Give your answer in exact form;
√
answer, whereas 1.41 is an
eg. 2 is an exact√
approximation of 2.
(a) P = (0,0), Q = (1,1). Problem 2.4. Erik’s disabled sailboat is ﬂoating at a stationary location 3 miles East and
2 miles North of Kingston. A ferry leaves
Kingston heading due East toward Edmonds
at 12 mph. At the same time, Erik leaves the
sailboat in a dinghy heading due South at 10
ft/sec (hoping to intercept the ferry). Edmonds
is 6 miles due East of Kingston.
sailboat (b) P = (2,1), Q = (1, − 1).
(c) P = (−1,2), Q = (4, − 1). Edmonds Kingston (d) P = (1,2), Q = (1 + 3t,3 + t), where t is a
constant.
Problem 2.2. Start with two points M = (a,b)
and N = (s,t) in the xycoordinate system. Let
d be the distance between these two points.
Answer these questions and make sure you
can justify your answers:
(a) TRUE or FALSE: d = (a − s)2 + (t − b)2 . (c) TRUE or FALSE: d = (s − a)2 + (t − b)2 . Ballard
UDub (a − s)2 + (b − t)2 . (b) TRUE or FALSE: d = North (d) Suppose M is the beginning point and N
is the ending point; recall Equation 2.2
and Figure 2.10 on Page 19. What is ∆x?
What is ∆y?
(e) Suppose N is the beginning point and M
is the ending point; recall Equation 2.2
and Figure 2.10 on Page 19. What is ∆x?
What is ∆y?
(f) If ∆x=0, what can you say about the relationship between the positions of the
two points M and N ? If ∆y=0, what can
you say about the relationship between
the positions of the two points M and N?
(Hint: Use some speciﬁc values for the
coordinates and draw some pictures to
see what is going on.)
Problem 2.3. Steve and Elsie are camping in
the desert, but have decided to part ways.
Steve heads North, at 6 AM, and walks steadily
at 3 miles per hour. Elsie sleeps in, and starts
walking West at 3.5 miles per hour starting at
8 AM.
When will the distance between them be 25
miles? (a) Compute Erik’s speed in mph and the
Ferry speed in ft/sec.
(b) Impose a coordinate system and complete this table of data concerning locations (i.e., coordinates) of Erik and the
ferry. Insert into the picture the locations of the ferry and Erik after 7 minutes.
Time Ferry Erik Distance
Between 0 sec
30 sec
7 min
t hr
(c) Explain why Erik misses the ferry.
(d) After 10 minutes, a Coast Guard boat
leaves Kingston heading due East at a
speed of 25 ft/sec. Will the Coast Guard
boat catch the ferry before it reaches Edmonds? Explain.
Problem 2.5. Suppose two cars depart from
a four way intersection at the same time, one
heading north and the other heading west. The
car heading north travels at the steady speed
of 30 ft/sec and the car heading west travels
at the steady speed of 58 ft/sec. 2.5. EXERCISES 23
Brooke (a) Find an expression for the distance between the two cars after t seconds.
(b) Find the distance in miles between the
two cars after 3 hours 47 minutes. ocean
5 mi kayak reaches shore here (c) When are the two cars 1 mile apart?
shore
Problem 2.6. Allyson and Adrian have decided
to connect their ankles with a bungee cord;
one end is tied to each person’s ankle. The
cord is 30 feet long, but can stretch up to 90
feet. They both start from the same location.
Allyson moves 10 ft/sec and Adrian moves 8
ft/sec in the directions indicated. Kono’s A
6 mi (a) If she paddles along a straight line
course to the shore, ﬁnd an expression
that computes the total time to reach
lunch in terms of the location where
Brooke beaches the boat.
(b) Determine the total time to reach Kono’s
if she paddles directly to the point “A”. 20 ft (c) Determine the total time to reach Kono’s
if she paddles directly to Kono’s. Building (d) Do you think your answer to (b) or (c) is
the minimum time required for Brooke
to reach lunch? Allyson 30 ft Adrian (e) Determine the total time to reach Kono’s
if she paddles directly to a point on the
shore half way between point “A” and
Kono’s. How does this time compare to
the times in parts (b) and (c)? Do you
need to modify your answer to part (d)? start (a) Where are the two girls located after 2
seconds?
(b) After 2 seconds, will the slack in the
bungee cord be used up?
(c) Determine when the bungee cord ﬁrst
becomes tight; i.e. there is no slack
in the line. Where are the girls located
when this occurs?
(d) When will the bungee cord ﬁrst touch
the corner of the building? (Hint: Use
a fact about “similar triangles”.) Problem 2.7. Brooke is located 5 miles out
from the nearest point A along a straight
shoreline in her seakayak. Hunger strikes and
she wants to make it to Kono’s for lunch; see
picture. Brooke can paddle 2 mph and walk 4
mph. Problem 2.8. A spider is located at the position (1,2) in a coordinate system, where the
units on each axis are feet. An ant is located at
the position (15,0) in the same coordinate system. Assume the location of the spider after t
minutes is s(t) = (1 + 2t,2 + t) and the location
of the ant after t minutes is a(t) = (15 − 2t,2t).
(a) Sketch a picture of the situation, indicating the locations of the spider and ant
at times t = 0,1,2,3,4,5 minutes. Label
the locations of the bugs in your picture,
using the notation s(0), s(1),...,s(5), a(0),
a(1), ..., a(5).
(b) When will the xcoordinate of the spider
equal 5? When will the ycoordinate of
the ant equal 5?
(c) Where is the spider located when its
ycoordinate is 3?
(d) Where is each bug located when the
ycoordinate of the spider is twice as
large as the ycoordinate of the ant? 24
(e) How far apart are the bugs when their
xcoordinates coincide? Draw a picture, indicating the locations of each bug
when their xcoordinates coincide.
(f) A sugar cube is located at the position
(9,6). Explain why each bug will pass
through the position of the sugar cube.
Which bug reaches the sugar cube ﬁrst?
(g) Find the speed of each bug along its line
of motion; which bug is moving faster?
Problem 2.9. A Ferrari is heading south at a
constant speed on Broadway (a north/south
street) at the same time a Mercedes is heading
west on Aloha Avenue (an east/west street).
The Ferrari is 624 feet north of the intersection of Broadway and Aloha, at the same time
that the Mercedes is 400 feet east of the intersection. Assume the Mercedes is traveling at
the constant speed of 32 miles/hour. Find the
speed of the Ferrari so that a collision occurs
in the intersection of Broadway and Aloha.
Problem 2.10. Two planes ﬂying opposite directions (North and South) pass each other 80
miles apart at the same altitude. The Northbound plane is ﬂying 200 mph (miles per hour)
and the Southbound plane is ﬂying 150 mph.
How far apart are the planes in 20 minutes?
When are the planes 300 miles apart?
Problem 2.11. Here is a list of some algebra
problems with ”solutions.” Some of the solutions are correct and some are wrong. For
each problem, determine: (i) if the answer is
correct, (ii) if the steps are correct, (iii) identify
any incorrect steps in the solution (noting that
the answer may be correct but some steps may
not be correct). CHAPTER 2. IMPOSING COORDINATES
(a) If x = 1,
x2 − 1
x+1 x2 + (−1)1
x+1
x 2 −1
+
x
1
x−1 =
=
= (b)
(x + y)2 − (x − y)2 =
= (x2 + y2 ) − x2 − y2
0 (c) If x = 4,
9(x − 4)2
3x − 12 =
=
= 32 (x − 4)2
3x − 12
(3x − 12)2
3x − 12
3x − 12. Problem 2.12. Assume α, β are nonzero constants. Solve for x.
1
αx−β (a) αx + β =
+ 1
β = 1
x (c) α + 1
β = 1
x (b) 1
α Problem 2.13. Simplify as far as possible.
(a) (1 − t)2 + (2 + 2t)2
(b) (t + 1)2 + (−t − 1)2 − 2
(c)
(d) 1
t−1 − 1
t+ 1 (write as a single fraction) (2 + t)2 + 4t2 Chapter 3
Three Simple Curves
Before we discuss graphing, we ﬁrst want to become acyaxis
quainted with the sorts of pictures that will arise. This is
A typical curve.
surprisingly easy to accomplish: Impose an xycoordinate
system on a blank sheet of paper. Take a sharp pencil and
begin moving it around on the paper. The resulting picxaxis
ture is what we will call a curve. For example, here is a
sample of the sort of “artwork” we are trying to visualize.
A number of examples in the text will involve basic curves
in the plane. When confronted with a curve in the plane,
the fundamental question we always try to answer is this:
Figure 3.1: A typical curve.
Can we give a condition (think of it as a “test”) that will
tell us precisely when a point in the plane lies on a curve?
Typically, the kind of condition we will give involves an equation in
two variables (like x and y). We consider the three simplest situations in
this chapter: horizontal lines, vertical lines and circles. 3.1 The Simplest Lines
Undoubtedly, the simplest curves in the plane are the
horizontal and vertical lines. For example, sketch a line
parallel to the xaxis passing through 2 on the yaxis; the
result is a horizontal line ℓ, as pictured. This means the
line ℓ passes through the point (0, 2) in our coordinate system. A concise symbolic prescription for ALL of the points
on ℓ can be given using “set notation”: yaxis
(−1, 2) (0, 2) (2, 2) ℓ
xaxis Figure 3.2: The points (x, y). ℓ = {(x, 2)x is any real number}.
We read the righthand side of this expression as “the set of all points
(x, 2) where x = any real number.” Notice, the points (x, y) on the line ℓ
are EXACTLY the ones that lead to solutions of the equation y = 2; i.e.,
take any point on this line, plug the coordinates into the equation y = 2
25 (x, 2), a typical point. CHAPTER 3. THREE SIMPLE CURVES 26 yaxis
m and you get a true statement. Because the equation does not involve the
variable x and only constrains y to equal 2, we see that x can take on any
real value. In short, we see that plotting all of the solutions (x, y) to the
equation y = 2 gives the line ℓ. We usually refer to the set of all solutions
of the equation y = 2 as the graph of the equation y = 2.
As a second example, sketch the vertical line m passing through 3 on the xaxis; this means the line m passes
through the point (3, 0) in our coordinate system. A concise symbolic prescription for ALL of the points on m can
(3, y), a typical point.
be given using “set notation”:
(3, 3) m = {(3, y)y is any real number}. (3, 0) Notice, the points (x, y) on the line m are EXACTLY
the ones that lead to solutions of the equation x = 3; i.e.,
take any point on this line, plug the coordinates into the
equation x = 3 and you get a true statement. Because the
Figure 3.3: Stacked points.
equation does not involve the variable y and only speciﬁes
that x = 3, y can take on any real number value. Plotting all of the
solutions (x, y) to the equation x = 3 gives the line m. We usually refer to
the set of all solutions of the equation x = 3 as the graph of the equation
x = 3.
These two simple examples highlight our ﬁrst clear connection between a geometric ﬁgure and an equation; the link is achieved by plotting all of the solutions (x, y) of the equation in the xycoordinate system.
These observations work for any horizontal or vertical line.
xaxis (3, −2) Deﬁnition 3.1.1. Horizontal and Vertical Lines: A horizontal line ℓ
passing through k on the yaxis is precisely a plot of all solutions (x, y) of
the equation y = k; i.e., ℓ is the graph of y = k. A vertical line m passing through h on the xaxis is precisely a plot of all solutions (x, y) of the
equation x = h; i.e., m is the graph of x = h. 3.2 Circles
Another common curve in the plane is a circle. Let’s see how to relate
a circle and an equation involving the variables x and y. As a special
case of the distance formula (2.3), suppose P = (0, 0) is the origin and
Q = (x, y) is any point in the plane; then
distance from P to Q =
= (x − 0)2 + (y − 0)2
x2 + y2 . This calculation tells us that a point (x, y) is of distance r from the origin
if and only if r = x2 + y2 or, squaring each side, that x2 + y2 = r2 . This
shows
{(x, y)distance (x, y) to origin is r} = {(x, y)x2 + y2 = r2 }. (3.1) 3.2. CIRCLES 27 Pencil.
What is the lefthand side of Equation 3.1? To picr
ture all points in the plane of distance r from the origin,
fasten a pencil to one end of a nonelastic string (a string
r
that will not stretch) of length r and tack the other end to
Start.
Draw with a
the origin. Holding the string tight, the pencil point will
tight string.
locate a point of distance r from the origin. We could visuFigure 3.4: Drawing a circle.
alize all such points by simply moving the pencil around
the origin, all the while keeping the string tight.
What is the righthand side of Equation 3.1? A point (x, y) in the
righthand set is a solution to the equation x2 + y2 = r2 ; i.e., if we plug in
the coordinates we get a true statement. For example, in Figure 3.5 we
r
r
r
−
plot eight solutions (r, 0), (−r, 0), (0, r), (0, −r), A = √2 , √2 , B = √2 , √r ,
2 C= −
−
√r , √r
2
2 ,D= −
r
√r , √
2
2 , of the equation. To see that the last point is a solution, here is the sample calculation: −
√r
2 2 + r
√
2 2 = r2
2 Since the two sides of Equation 3.1 are equal, drawing
the circle of radius r is the same as plotting all of the solutions of the equation x2 + y2 = r2 . The same reasoning can
be used to show that drawing a circle of radius r centered
at a point (h, k) is the same as plotting all of the solutions
of the equation: (x − h)2 + (y − k)2 = r2 . We usually refer to
the set of all solutions of the equation as the graph of the
equation.
Deﬁnition 3.2.1 (Circles). Let (h, k) be a given point in the
xyplane and r > 0 a given positive real number. The circle
of radius r centered at (h, k) is precisely all of the solutions
(x,y) of the equation
(x − h)2 + (y − k)2 = r2 ;
i.e., the circle is the graph of this equation.
We refer to the equation in the box as the standard
form of the equation of a circle. From this equation you
know both the center and radius of the circle described. + r2
2 = r2 .
(0, r)
D A (−r, 0) (r, 0) C B
(0, −r) Figure 3.5:
Computing
points. yaxis
(x, y)
r
(h, k)
xaxis Figure 3.6: Deﬁning a circle. CHAPTER 3. THREE SIMPLE CURVES 28 Be very careful with the minus signs “−” in the standard form for a circle
equation. For example, the equation
!!!
CAUTION
!!! ( x + 3) 2 + ( y − 1) 2 = 7
is NOT in standard form. We can rewrite it in standard form:
√
(x − (−3))2 + (y − 1)2 = ( 7)2 ;
√
so, this equation describes a circle of radius 7 centered at (−3, 1).
Examples 3.2.2. Here are some of the ways we can discuss circles:
1. The circle of radius 1 centered at the origin is the graph of the equation
x2 + y2 = 1. This circle is called the unit circle and will be used
extensively.
2. A circle of radius 3 centered at the point (h, k) = (1, −1) is the graph of
the equation (x − 1)2 +(y −(−1))2 = 32 ; or, equivalently (x − 1)2 +(y + 1)2 =
32 ; or, equivalently x2 + y2 − 2x + 2y = 7.
√
3. The circle of radius 5 centered at (2, −3) does not pass through the
origin; this is because (0, 0) is not a solution of the equation (x − 2)2 +
(y + 3)2 = 5. 3.3 Intersecting Curves I
In many problem solving situations, we will have two curves in the plane
and need to determine where the curves intersect one another. Before
we discuss a general procedure, let’s make sure we really understand
the meaning of the word “intersect.” From Latin, the word “inter” means
“within or in between” and the word “sectus” means “to cut.” So, the
intersection of two curves is the place where the curves “cut into” each
other; in other words, where the two curves cross one another.
If the pictures of two curves are given to us up front, we can often
visually decide whether or not they intersect. This is one good reason for
drawing a picture of any physical problem we are trying to solve. We will
need a small bag of tricks used for ﬁnding intersections of curves. We
begin with intersections involving the curves studied in this section.
Two different horizontal lines (or two different vertical lines) will never
intersect. However, a horizontal line always intersects a vertical line exactly once; Figure 3.7(a). Given a circle and a horizontal or vertical line,
we may or may not have an intersection. Looking at Figure 3.7(b), you
can convince yourself a given horizontal or vertical line will intersect a
circle in either two points, one point or no points. This analysis is all pictorial; how do you ﬁnd the explicit coordinates of an intersection point?
Let’s look at a sample problem to isolate the procedure used. 3.3. INTERSECTING CURVES I
yaxis 29 Horizontal
line: y = k. Point:
(h, k). xaxis
Vertical
line: x = h. (a) Line equations. (b) Possible intersections.
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 50 foot light pole, the resulting useful illuminated area is a
circular disc 120 feet in diameter. Assume the light pole is located 20 feet
east and 40 feet north of the intersection of Parkside Ave. (a north/south
street) and Wilson St. (an east/west street). What portion of each street is
illuminated? yaxis Solution. The illuminated area is a circular disc whose diameter and center are both known. Consequently, we
really need to study the intersection of this circle with the
two streets. Begin by imposing the pictured coordinate
system; we will use units of feet for each axis. The illuminated region will be a circular disc centered at the point
(20, 40) in the coordinate system; the radius of the disc
will be r = 60 feet.
We need to ﬁnd the points of intersection P, Q, R, and S
of the circle with the xaxis and the yaxis. The equation
for the circle with r = 60 and center (h, k) = (20, 40) is (Parkside) Illuminated
zone. P xaxis
(Wilson)
R S
Q Figure 3.8:
Illuminated
street. (x − 20)2 + (y − 40)2 = 3600.
To ﬁnd the circular disc intersection with the yaxis, we
have a system of two equations to work with:
(x − 20)2 + (y − 40)2 = 3,600;
x = 0.
To ﬁnd the intersection points we simultaneously solve both equations.
To do this, we replace x = 0 in the ﬁrst equation (i.e., we impose the CHAPTER 3. THREE SIMPLE CURVES 30 conditions of the second equation on the ﬁrst equation) and arrive at
(0 − 20)2 + (y − 40)2 = 3,600;
400 + (y − 40)2 = 3,600;
(y − 40)2 = 3,200;
√
(y − 40) = ± 3,200
√
y = 40 ± 3,200
= −16.57 or 96.57.
Notice, we have two solutions. This means that the circle and yaxis
intersect at the points P = (0,96.57) and Q = (0, − 16.57). Similarly, to ﬁnd
the circular disc intersection with the xaxis, we have a system of two
equations to work with:
(x − 20)2 + (y − 40)2 = 3,600;
y = 0.
Replace y = 0 in the ﬁrst equation (i.e., we impose the conditions of the
second equation on the ﬁrst equation) and arrive at
(x − 20)2 + (0 − 40)2 = 3,600;
(x − 20)2 = 2,000;
√
(x − 20) = ± 2,000;
√
x = 20 ± 2,000
= −24.72 or 64.72.
Conclude the circle and xaxis intersect at the points S = (64.72,0) and
R = (−24.72,0).
The procedure we used in the solution of Example 3.3.1 gives us a
general approach to ﬁnding the intersection points of circles with horizontal and vertical lines; this will be important in the exercises. 3.4 Summary
• Every horizontal line has equation of the form y = c.
• Every vertical line has equation of the form x = c.
• Every circle has equation of the form
(x − h)2 + (y − k)2 = r2
where (h,k) is the center of the circle, and r is the circle’s radius. 3.5. EXERCISES 31 3.5 Exercises
Problem 3.1. This exercise emphasizes the
“mechanical aspects” of circles and their equations.
(a) Find an equation whose graph is a circle
of radius 3 centered at (−3,4).
(b) Find an equation whose graph is a circle of diameter 1 centered at the point
2
(3, − 11 ).
3
(c) Find four different equations whose
graphs are circles of radius 2 through
(1,1).
(d) Consider the equation (x − 1)2 + (y + 1)2 =
4. Which of the following points lie on
the graph of √
this equation:√1,1), (1, − 1),
(
(1, − 3), (1 + 3,0), (0, − 1 − 3), (0,0).
Problem 3.2. Find the center and radius of
each of the following circles.
(a) x2 − 6x + y2 + 2y − 2 = 0
(b) x2 + 4x + y2 + 6y + 9 = 0
(c) x2 + 1 x + y2 −
3 10
3y = (d) x2 + y2 = 3 x − y +
2 127
9 35
16 Problem 3.3. Water is ﬂowing from a major
broken water main at the intersection of two
streets. The resulting puddle of water is circular and the radius r of the puddle is given by
the equation r = 5t feet, where t represents
time in seconds elapsed since the the main
broke.
(a) When the main broke, a runner was located 6 miles from the intersection. The
runner continues toward the intersection at the constant speed of 17 feet per
second. When will the runner’s feet get
wet?
(b) Suppose, instead, that when the main
broke, the runner was 6 miles east, and
5000 feet north of the intersection. The
runner runs due west at 17 feet per second. When will the runner’s feet get wet?
Problem 3.4. An amusement park Ferris
Wheel has a radius of 60 feet. The center of
the wheel is mounted on a tower 62 feet above
the ground (see picture). For these questions,
the wheel is not turning. rider
60 feet
100 feet
operator
ground level
62 ft. tower
24 feet
(a) Impose a coordinate system.
(b) Suppose a rider is located at the point in
the picture, 100 feet above the ground.
If the rider drops an ice cream cone
straight down, where will it land on the
ground?
(c) The ride operator is standing 24 feet to
one side of the support tower on the level
ground at the location in the picture.
Determine the location(s) of a rider on
the Ferris Wheel so that a dropped ice
cream cone lands on the operator. (Note:
There are two answers.)
Problem 3.5. A crawling tractor sprinkler is
located as pictured below, 100 feet South of a
sidewalk. Once the water is turned on, the
sprinkler waters a circular disc of radius 20
feet and moves North along the hose at the rate
of 1 inch/second. The hose is perpendicular
2
to the 10 ft. wide sidewalk. Assume there is
grass on both sides of the sidewalk.
N
hose
W E
S
sidewalk tractor sprinkler (a) Impose a coordinate system. Describe
the initial coordinates of the sprinkler
and ﬁnd equations of the lines forming
the North and South boundaries of the
sidewalk. CHAPTER 3. THREE SIMPLE CURVES 32
(b) When will the water ﬁrst strike the sidewalk?
(c) When will the water from the sprinkler
fall completely North of the sidewalk?
(d) Find the total amount of time water from
the sprinkler falls on the sidewalk.
(e) Sketch a picture of the situation after 33
minutes. Draw an accurate picture of
the watered portion of the sidewalk.
(f) Find the area of GRASS watered after
one hour.
Problem 3.6. Erik’s disabled sailboat is ﬂoating stationary 3 miles East and 2 miles North
of Kingston. A ferry leaves Kingston heading
toward Edmonds at 12 mph. Edmonds is 6
miles due east of Kingston. After 20 minutes
the ferry turns heading due South. Ballard is
8 miles South and 1 mile West of Edmonds.
Impose coordinates with Ballard as the origin. (e) How long does the ferry spend inside the
radar zone?
Problem 3.7. Nora spends part of her summer driving a combine during the wheat harvest. Assume she starts at the indicated position heading east at 10 ft/sec toward a circular
wheat ﬁeld of radius 200 ft. The combine cuts
a swath 20 feet wide and begins when the corner of the machine labeled “a” is 60 feet north
and 60 feet west of the westernmost edge of
the ﬁeld.
N
W E
S swath cut combine 20 ft a
center
wheat field sailboat
Kingston Edmonds (a) When does Nora’s rig ﬁrst start cutting
the wheat?
(b) When does Nora’s rig ﬁrst start cutting a
swath 20 feet wide? North
Ballard
UDub (a) Find the equations for the lines along
which the ferry is moving and draw in
these lines.
(b) The sailboat has a radar scope that will
detect any object within 3 miles of the
sailboat. Looking down from above, as
in the picture, the radar region looks
like a circular disk. The boundary is
the ”edge” or circle around this disc, the
interior is the inside of the disk, and
the exterior is everything outside of the
disk (i.e. outside of the circle). Give
a mathematical (equation) description of
the boundary, interior and exterior of
the radar zone. Sketch an accurate picture of the radar zone by determining
where the line connecting Kingston and
Edmonds would cross the radar zone.
(c) When does the ferry enter the radar
zone?
(d) Where and when does the ferry exit the
radar zone? (c) Find the total amount of time wheat is
being cut during this pass across the
ﬁeld.
(d) Estimate the area of the swath cut during this pass across the ﬁeld.
Problem 3.8. (a) Solve for x: 2 x − 2x + 1
=x−2
x+5
(b) Solve for x:
x−3
=1
x+2
(c) If x = −2, ﬁnd ALL solutions of the equation
(x + 1)2 + (y − 1)2 = 10
(d) If y = 3, ﬁnd ALL solutions of the equation
5(x + 1)2 + 2(y − 1)2 = 10 Chapter 4
Linear Modeling
Sometimes, we will begin a section by looking at a speciﬁc problem which
highlights the topic to be studied; this section offers the ﬁrst such vista.
View these problems as illustrations of precalculus in action, rather than
confusing examples. Don’t panic, the essential algebraic skills will be
reviewed once the motivation is in place. 4.1 The Earning Power Problem
The government likes to gather all kinds of data. For example, Table 4.1
contains some data on the average annual income for 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
(a) Women. YEAR
1970
1987 (b) Men. WOMEN (dollars)
$5,616
$18,531 YEAR
1970
1987 Table 4.1: Earning power data. MEN (dollars)
$9,521
$28,313 question would be to use the data in the table to construct two different
mathematical models that predict the future (or past) earning power for
women or men. In order to do that, we would need to make some kind of
initial assumption about the type of mathematical model expected. Let’s
begin by drawing two identical xycoordinate systems, where the xaxis
has units of “year” and the yaxis has units of “dollars;” see Figure 4.1. In
each coordinate system, the data in our table gives us two points to plot:
In the case of women, the data table gives us the points P = (1970, 5,616)
and Q = (1987, 18,531). Likewise, for the men, the data table gives us the
points R = (1970, 9,521) and S = (1987, 28,313).
To study the future earning power of men and women, we are going
to make an assumption: For women, if the earning power in year x is $y,
33 CHAPTER 4. LINEAR MODELING 34
PSfrag
yaxis (dollars) 30000 yaxis (dollars) 30000 S = (1987,28313)
25000 25000 20000 20000 Q = (1987,18531) 15000 15000 10000 10000 R = (1970,5616)
P = (1970,5616) 5000 5000 xaxis (year) xaxis (year)
1995 1990 1985 1980 1975 1970 1995 1990 1985 1980 1975 1970 aca (b) Data points for men.
(a) Data points for women.
Figure 4.1: Visualizing the data. then the point (x, y) lies on the line connecting P and Q. Likewise, for
men, if the earning power in year x is $y, then the point (x, y) lies on the
line connecting R and S.
In the real world, the validity of this kind of assumption would involve
a lot of statistical analysis. This kind of assumption leads us to what
is called a linear model, since we are demanding that the data points
predicted by the model (i.e., the points (x, y)) lie on a straight line in a
coordinate system. Now that we have made this assumption, our job is
to ﬁnd a way to mathematically describe when a point (x, y) lies on one
of the two lines pictured in Figure 4.2.
Our goal in the next subsection is to review the mathematics necessary to show that the lines in Figure 4.2 are the socalled graphs of
Equations 4.1 and 4.2. yaxis (dollars) 30000 yaxis (dollars) 30000 S = (1987,28313)
25000 25000
20000 Q = (1987,18531) y 15000 (x,y) on the line:
This means men
earn y dollars in
year x. 15000 (x,y) on the line: This
means women earn y
dollars in year x.
P = (1970,5616) 5000 10000 R = (1970,9521)
5000 xaxis (year) xaxis (year)
1975 1970 1995 1990 1985 1980 1975 1970 1980 x x 1995 10000 1990 y 1985 20000 (b) Linear model for men.
(a) Linear model for women.
Figure 4.2: Linear models of earning power. 4.2. RELATING LINES AND EQUATIONS 35 18,531 − 5,616
(x − 1970) + 5,616
1987 − 1970
12,915
=
(x − 1970) + 5,616
17
28,313 − 9,521
(x − 1970) + 9,521
=
1987 − 1970
18,792
=
(x − 1970) + 9,521
17 (4.1) ywomen = ymen (4.2) 4.2 Relating Lines and Equations
A systematic approach to studying equations and their graphs would
begin with the simple cases, gradually working toward the more complicated. Thinking visually, the simplest curves in the plane would be
straight lines. As we discussed in Chapter 3, a point on the vertical line
in Figure 4.3(a) will always have the same xcoordinate; we refer to this
line as the graph of the equation x = h. Likewise, a point on the horizontal line in Figure 4.3(b) will always have the same ycoordinate; we refer
to this line as the graph of the equation y = k. Figure 4.3(c) is different,
in the sense that neither the x nor the y coordinate is constant; i.e., as
you move a point along the line, both coordinates of the point are changing. It is reasonable to guess that this line is the graph of some equation
involving both x and y. The question is: What is the equation?
yaxis yaxis
graph of x = h
(h,y) is a typical
point on this line. xaxis
location h on xaxis (a) Vertical line. yaxis
(x,k) is a typical
point on this line
graph of y = k location k on yaxis xaxis (b) Horizontal line.
Figure 4.3: Lines in a plane. (x,y) is a typical
point on this line
graph
of
some
equation involving
x and y xaxis (c) Sloped line. CHAPTER 4. LINEAR MODELING 36 Here is the key geometric fact needed to model lines by mathematical
equations:
Important Fact 4.2.1. Two different points completely determine a straight
line.
This fact tells us that if you are given two different points on a line, you
can reconstruct the line in a coordinate system by simply lining a ruler
up with the two points. In our discussion, we will need to pay special
attention to the difference between vertical and nonvertical lines. 4.3 Nonvertical Lines
Assume in this section that ℓ is a nonvertical line in the plane; for example, the line in Figure 4.3(c). If we are given two points P = (x1 , y1 )
and Q = (x2 , y2 ) on a line ℓ, then Equation (2.2) on page 18 deﬁned two
quantities we can calculate:
∆x = change in x going from P to Q = x2 − x1 .
∆y = change in y going from P to Q = y2 − y1 .
We deﬁne the slope of the line ℓ to be the ratio of ∆y by ∆x, which is
usually denoted by m:
∆y
∆x
y2 − y1
=
x2 − x1
change in y
slope of ℓ =
change in x
def m= (4.3) Notice, we are using the fact that the line is nonvertical to know that
this ratio is always deﬁned; i.e., we will never have ∆x = 0 (which would
lead to illegal division by zero). There is some additional terminology that
goes along with the deﬁnition of the slope. The term ∆y is sometimes
called the rise of ℓ and ∆x is called the run of ℓ. For this reason, people
often refer to the slope of a line ℓ as “the rise over the run,” meaning
def slope of ℓ = ∆y
rise of ℓ
=
= m.
run of ℓ
∆x In addition, notice that the calculation of ∆y involves taking the difference of two numbers; likewise, the calculation of ∆x involves taking the
difference of two numbers. For this reason, the slope of a line ℓ is sometimes called a difference quotient. 4.3. NONVERTICAL LINES 37 For example, suppose P = (1, 1) and Q = (4, 5) lie on a
line ℓ. In this case, the rise = ∆y = 4 and the run = ∆x = 3,
4
so m = 3 is the slope of ℓ.
When computing ∆x, pay special attention that it is
the xcoordinate of the destination point Q minus the xcoordinate of the starting point P; likewise, when computing ∆y, it is the ycoordinate of the destination point
Q minus the ycoordinate of the starting point P. We can
reverse this order in both calculations and get the same
slope:
m= Q = (4,5)
5 P = (1,1) 1 ∆y = 4 ∆x = 3
1 4 Figure 4.4: Computing the
slope of a line. y2 − y1
−(y2 − y1 )
y1 − y2
−∆y
∆y
=
=
=
=
.
∆x
x2 − x1
−(x2 − x1 )
x1 − x2
−∆x We CANNOT reverse the order in just one of the calculations and get
the same slope:
y2 − y1
x2 − x1 = y2 − y1
x1 − x2 , and m= y2 − y1
x2 − x1 = y1 − y2
x2 − x1 It is very important to notice that the calculation of the
slope of a line does not depend on the choice of the two
points P and Q. This is a real windfall, since we are then
always at liberty to pick our favorite two points on the
line to determine the slope. The reason for this freedom
of choice is pretty easy to see by looking at a picture.
If we were to choose two other points P∗ = (x∗ , y∗ ) and
1
1
Q∗ = (x∗ , y∗ ) on ℓ, then we would get two similar right
2
2
triangles: See Figure 4.5.
Basic properties of similar triangles tell us ratios of
lengths of common sides are equal, so that
m= !!! m= CAUTION
!!! . x∗ ,y∗ = Q∗
22 (x2 ,y2 ) = Q
y2 − y1
x2 − x1 x∗ − x∗
2
1
P ∗ = x∗ ,y∗
11 Figure 4.5: Using similar triangles. y∗ − y∗
y2 − y1
2
1
=∗
;
x2 − x1
x2 − x∗
1 but this just says the calculation of the slope is the same for any pair of
distinct points on ℓ. For example, lets redo the slope calculation when
P∗ = P = (x1 , y1 ) and Q∗ = (x, y) represents an arbitrary point on the line.
Then the two ratios of lengths of common sides give us the equation
y − y1
,
x − x1
= m(x − x1 ). m=
y − y1 y∗ − y∗
2
1 (x1 ,y1 ) = P CHAPTER 4. LINEAR MODELING 38
This can be rewritten as
y = m(x − x1 ) + y1
or
y2 − y1
(x − x1 ) + y1 .
y=
x2 − x1 (4.4)
(4.5) Equation 4.4 is usually called the point slope formula for the line ℓ (since
the data required to write the equation amounts to a point (x1 ,y1 ) on
the line and the slope m), whereas Equation 4.5 is called the two point
formula for the line ℓ (since the data required amounts to the coordinates
of the points P and Q). In any event, we now see that
Important Fact 4.3.1. (x,y) lies on ℓ if and only if (x,y) is a solution to
y = m(x − x1 ) + y1 .
We can plot the collection of ALL solutions to the equation in Fact 4.3.1,
which we refer to as the graph of the equation. As a subset of the xycoordinate system, the line
Important Fact 4.3.2. ℓ = {(x, m(x − x1 ) + y1 ) x is any real number}. 1
graph of y = 4 x − 3
3 (6, 23 )
3
(4,5) (−1, − 5 )
3 (1,1)
1
(0, − 3 ) Figure 4.6: Verifying points
on a line. Example 4.3.3. Consider the line ℓ, in Figure 4.6, through
the two points P = (1,1) and Q = (4,5). Then the slope
of ℓ is m = 4/3 and ℓ consists of all pairs of points (x,y)
such that the coordinates x and y satisfy the equation y =
4
(x − 1) + 1. Letting x = 0, 1, 6 and −1, we conclude that
3
the following four points lie on the line ℓ: (0, 4 (0 − 1) + 1) =
3
4
4
(0, −1 ), (1, 3 (1 − 1) + 1) = (1,1), (6, 3 (6 − 1) + 1) = (6, 23 ) and
3
3
4
(−1, 3 (−1 − 1) + 1) = (−1, −5 ). By the same reasoning, the
3
point (0,0) does not lie on the line ℓ. As a set of points in the
plane, we have
ℓ= 4
x, (x − 1) + 1 x is any real number
3 Returning to the general situation, we can obtain a third general equation for a nonvertical line. To emphasize what is going on here, plug the
speciﬁc value x = 0 into Equation 4.4 and obtain the point R = (0,b) on
the line, where b = m(0 − x1 ) + y1 = −mx1 + y1 . But, Equation (4.4) can be
written
y = m(x − x1 ) + y1
= mx − mx1 + y1
= mx + b. 4.4. GENERAL LINES 39 The point R is important; it is precisely the point where the line ℓ
crosses the yaxis, usually called the yintercept. The slope intercept
equation of the line is the form
y = mx + b,
where the slope of the line is m and b is the yintercept of the line.
Summary 4.3.4. Nonvertical Lines: Let ℓ be a nonvertical line in the
xyplane. There are three ways to obtain an equation whose graph is ℓ,
depending on the data provided for ℓ: 1. If P = (x1 ,y1 ), Q = (x2 ,y2 ) are two different points on the line, then
y2 − y1
the twopoint formula y =
(x − x1 ) + y1 gives an equation
x2 − x1
whose graph is ℓ.
2. If P = (x1 ,y1 ) is a point on the line and m is the slope of ℓ, then
the pointslope formula y = m(x − x1 ) + y1 gives an equation whose
graph is ℓ.
3. If the line ℓ intersects the yaxis at the point (0,b) and m is the slope
of the line ℓ, then the slopeintercept formula y = mx + b gives an
equation whose graph is ℓ. 4.4 General Lines
Summarizing, any line in the plane is the graph of an equation involving
x and y and the equation always has the form
Ax + By + C = 0,
for some constants A, B, C. Equations like this are called linear equations. In general, 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 xy coordinate system, then its slope
will be the rate of change of y with respect to x:
∆y
∆x
change in y
=
change in x slope = def = rate of change of y with respect to x. 40 CHAPTER 4. LINEAR MODELING We should emphasize that this rate of change is a constant; in other
words, this rate is the same no matter where we compute the slope on
the line. The 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 y
with respect to x.
An interesting thing to notice is how the units for x and y ﬁgure into
the rate of change calculation. For example, suppose that we have the
equation y = 10,000 x + 200,000, which relates the value y of a house (in dollars) to the number of years x you own it. For example, after 5 years, x = 5
and the value of the house would be y = 10,000 (5) + 200,000 = $250,000.
In this case, the equation y = 10,000 x + 200,000 is linear and already written in slopeintercept form, so the slope can be read as m = 10,000. If
∆y
we carry along the units in the calculation of
, then the numerator
∆x
involves “dollar” units and the denominator “years” units. That means
that carrying along units, the slope is actually m = 10,000 dollars/year.
In other words, the value of the house is changing at a rate of 10,000
dollars/year.
At the other extreme, if the units for both x and y are the same, then
the units cancel out in the rate of change calculation; in other words,
the slope is a 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
s
of a moving object. Suppose an object moves along a
b
speed m
straight line at a constant speed m: See Figure 4.7.
reference
If we specify a reference point, we can let b be the startpoint
location of the obinitial location ject at time t
ing location of the moving object, which is usually called
of the object
the initial location of the object. We can write down an
Figure 4.7: Motion along a
equation relating the initial location b, the time t, the conline.
stant speed m and the location s at time t:
s = (location of object at time t)
s = (initial location of object) + (distance object travels in time t)
s = b + mt.
where t is in the same time units used to deﬁne the rate m. Notice, both
b and m would be constants given to us, so this is a linear equation
involving the variables s and t. We can graph the equation in the tscoordinate system: See Figure 4.8.
It is important to distinguish between this picture (the graph of s =
mt + b) and the path of the object in Figure 4.7. The graph of the equation
should be thought of as a visual aid attached to the equation s = mt + b.
The general idea is that using this visual aid can help answer various
questions involving the equation, which in turn will tell us things about
the motion of the object in Figure 4.7. 4.5. LINES AND RATE OF CHANGE 41 Two other comments related to this discussion are important. First, concerning notation, the speed m is often
symbolized by v to denote constant velocity and b is written as s◦ (the subscript “0” meaning “time zero”). With
these changes, the equation becomes s = s◦ + vt, which is
the form in which it would be written in a typical physics
text. As a second note, if you return to Figure 4.8, you will
notice we only drew in the positive t axis. This was because t represented time, which is always a nonnegative
quantity. saxis
graph of s = mt + b
∆s
∆t
rate of change =
b = sintercept Mookie ∆t taxis Figure 4.8: The graph of
s = mt + b. Example 4.5.1. Linda, Asia and Mookie are all playing frisbee. Mookie is
10 meters in front of Linda and always runs 5 m/sec. Asia is 34 meters
in front of Linda and always runs 4 m/sec. Linda yells “go!” and both
Mookie and Asia start running directly away from Linda to catch a tossed
frisbee. Find linear equations for the distances between Linda, Mookie and
Asia after t seconds. Linda ∆s Asia Solution. Let sM be the distance between Linda and Mookie and sA the
distance between Linda and Asia, after t seconds. An application of the
above formula tells us
sM = (initial distance between Linda and Mookie) + · · ·
· · · + (distance Mookie runs in t seconds)
sM = 10 + 5t.
Likewise,
sA = (initial distance between Linda and Asia) + · · ·
· · · + (distance Asia runs in t seconds)
sA = 34 + 4t.
If sMA is the distance between Mookie and Asia after t seconds, we compute
sMA = sA − sM = (34 + 4t) − (10 + 5t) = (24 − t) meters.
In all cases, the distances we computed are given by linear equations of
the form s = b + mt, for appropriate b and rate m. CHAPTER 4. LINEAR MODELING 42 4.6 Back to the Earning Power Problem
We now return to the motivating problem at the start of
this section. Recall the plot in Figure 4.1(b). We can
model the men’s earning power using the ﬁrst and last
data points, using the ideas we have discussed about linear equations. To do this, we should specify a “beginning
point” and an “ending point” (recall Figure 2.9) and calculate the slope: yaxis (dollars)
40000 U 35000 T
30000 S 25000
20000 (x,y) on the line:
means men earn y
dollars in year x y
15000
10000 Rbegin = (1970,9,521) R 5000 and xaxis (year) Send = (1987,28,313). 1995 1990 1985 1980 1975 1970 x Figure 4.9: Linear model of
Men’s Earning Power. We ﬁnd that ∆y ∆x m 28,313 − 9,521 1987 − 1970 ∆y
∆x 18,792 17 18,792
17 If we apply the “pointslope formula” for the equation of a line, we
arrive at the equation:
y= 18,792
(x − 1970) + 9,521.
17 (4.6) The graph of this line will pass through the two points R and S in Figure 4.2. We sketch the graph in Figure 4.9, indicating two new points T
and U.
We can use the model in (4.6) to make predictions of two different
sorts: (i) predict earnings at some date, or (ii) predict when a desired
value for earnings will occur. For example, let’s graphically discuss the
earnings in 1995:
• Draw a vertical line x = 1995 up to the graph and label the intersection point U.
• Draw a horizontal line ℓ through U. The line ℓ crosses the y axis at
the point
18,792
(1995 − 1970) + 9521 = 37,156.
17 4.7. WHAT’S NEEDED TO BUILD A LINEAR MODEL? 43 • The coordinates of the point U = (1995, 37,156).
Conclude that $37,156 is the Men’s Earning Power in 1995. For another
example, suppose we wanted to know when men’s earning power will
equal $33,000? This means we seek a data point T on the men’s earning
curve whose ycoordinate is 33,000. By (4.6), T has the form
T= x, 18,792
(x − 1970) + 9,521 .
17 We want this to be a data point of the form (x, 33,000). Setting these two
points equal and equating the second coordinates leads to an algebra
problem:
18,792
(x − 1970) + 9,521 = 33,000
17
x = 1991.24.
This means men’s earning power will be $33,000 at the end of the ﬁrst
quarter of 1991. Graphically, we interpret this reasoning as follows:
• Draw a horizontal line y = 33,000 and label the intersection point T
on the model.
• Draw a vertical line ℓ through T . The line ℓ crosses the x axis at the
point 1991.24.
• The coordinates of the point T = (1991.24,33,000).
In the exercises, you will be asked to show that the women’s earning
power model is given by the equation
y= 12,915
(x − 1970) + 5,616.
17 Using the two linear models for the earning power of men and women,
are women gaining on men? You will also be asked to think about this
question in the exercises. 4.7 What’s Needed to Build a Linear Model?
As we progress through this text, a number of different “types” of mathematical models will be discussed. We will want to think about the information needed to construct that particular kind of mathematical model.
Why would we care? For example, in a laboratory context, if we knew a
situation being studied was given by a linear model, this would effect the
amount of data collected. In the case of linear models, we can now make
this useful statement: CHAPTER 4. LINEAR MODELING 44 Important Fact 4.7.1. A linear model is completely determined by:
1. One data point and a slope (a rate of change), or
2. Two data points, or
3. An intercept and a slope (a rate of change). 4.8 Linear Application Problems
Example 4.8.1. The yearly resident tuition at the University of Washington was $1827 in 1989 and $2907 in 1995. Assume that the tuition growth at
the UW follows a linear model. What will be the tuition in the year 2000?
When will yearly tuition at the University of Washington be $10,000?
yaxis
10000 (dollars) 8000
6000
4000 xaxis
2040 2020 (year)
2010 2000 Q 2030 P
1990 2000 Figure 4.10: Linear tuition
model. Solution. If we consider a coordinate system where the
xaxis represents the year and the yaxis represents
dollars, we are given two data points: P = (1989, 1,827)
and Q = (1995, 2,907). Using the twopoint formula for the
equation of line through P and Q, we obtain the equation
y = 180(x − 1989) + 1,827. The graph of this equation gives a line through the given
points as pictured in Figure 4.10.
If we let x = 2000, we get y = $3,807, which tells us the tuition in the
year 2000. On the other hand, if we set the equation equal to $10,000, we
can solve for x:
10,000 = 180(x − 1989) + 1,827
8,173 = 180(x − 1989)
2,034.4 = x.
Conclude the tuition is $10,000 in the year 2035. 4.9 Perpendicular and Parallel Lines
Here is a useful fact to keep in mind.
Important Facts 4.9.1. Two 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. 4.10. INTERSECTING CURVES II 45 Example 4.9.2. Let ℓ be a line in the plane passing through the points
(1, 1) and (6, −1). Find a linear equation whose graph is a line parallel to
ℓ passing through 5 on the yaxis. Find a linear equation whose graph is
perpendicular to ℓ and passes through (4, 6).
Solution. Letting P = (1, 1) and Q = (6, −1), apply the “two point formula”:
−2
(x − 1) + 1
5
7
2
= − x+ .
5
5
The graph of this equation will be ℓ. This equation is in slope intercept
form and we can read off that the slope is m = −2 . The desired line a is
5
2
parallel to ℓ; it must have slope m = − 5 and yintercept 5. Plugging into
the “slope intercept form”:
y= y= −2
x + 5.
5 The desired line b is a line perpendicular to ℓ (so its slope is m ′ =
5
)
2 −1
−2
5 = and passes through the point (4, 6), so we can use the “point slope
formula”:
5
y = (x − 4) + 6.
2 4.10 Intersecting Curves II
We have already encountered problems that require us to investigate the
intersection of two curves in the plane. Ultimately, this reduces to solving
a system of two (or more) equations in the variables x and y. A useful
tool when working with equations involving squared terms (i.e., x2 or y2 ),
is the quadratic formula .
Important Fact 4.10.1. Quadratic Formula: Consider the equation
az2 + bz + c = 0, where a,b,c are constants. The solutions for this equation are given by the formula
√
−b ± b2 − 4ac
z=
.
2a
The solutions are real numbers if and only if b2 − 4ac ≥ 0.
The next example illustrates a typical application of the quadratic formula. In addition, we describe a very useful technique for ﬁnding the
shortest distance between a “line” and a “point.” CHAPTER 4. LINEAR MODELING 46 North
ﬂight path irrigated ﬁeld West East 2 miles Q P
1.5 miles crop
duster South Figure 4.11: The ﬂight path
of a crop duster. Example 4.10.2. A crop dusting airplane ﬂying a constant
speed of 120 mph is spotted 2 miles South and 1.5 miles
East of the center of a circular irrigated ﬁeld. The irrigated
ﬁeld has a radius of 1 mile. Impose a coordinate system
as pictured, with the center of the ﬁeld the origin (0,0). The
ﬂight path of the duster is a straight line passing over the
labeled points P and Q. Assume that the point Q where the
plane exits the airspace above the ﬁeld is the 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 P where the crop duster enters airspace
above the irrigated ﬁeld.
3. How much time does the duster spend ﬂying over the
irrigated ﬁeld?
4. Find the shortest distance from the ﬂight path to the
center of the irrigated ﬁeld.
Solution.
1. Take Q = (−1, 0) and S = (1.5, −2) = duster spotting point. Construct
a line through Q and S. The slope is −0.8 = m and the line equation
becomes:
y = −0.8x − 0.8. (4.7) 2. The equation of the boundary of the irrigated region is x2 + y2 = 1.
We need to solve this equation AND the line equation y = −0.8x − 0.8
simultaneously. Plugging the line equation into the unit circle equation gives:
x2 + (−0.8x − 0.8)2 = 1
x2 + 0.64x2 + 1.28x + 0.64 = 1
1.64x2 + 1.28x − 0.36 = 0
Apply the quadratic formula and ﬁnd x = −1, 0.2195. Conclude
that the x coordinate of P is 0.2195. To ﬁnd the y coordinate, plug
into the line equation and get y = −0.9756. Conclude that P =
(0.2195, −0.9756) 4.11. UNIFORM LINEAR MOTION 47 3. Find the distance from P to Q by using the distance formula:
(−1 − 0.2195)2 + (0 − (−0.9756))2
d=
= 1.562 miles
Now, 1.562 miles
120 mph = 0.01302 hours = 47 seconds. 4. The idea is to construct a line perpendicular to the ﬂight path passing through the origin of the coordinate system. This line will have
1
slope m = − −0.8 = 1.25. So this perpendicular line has equation
y = 1.25x. Intersecting this line with the ﬂight path gives us the
point closest to the center of the ﬁeld. The xcoordinate of this point
is found by setting the two line equations equal and solving:
−0.8x − 0.8 = 1.25x
x = −0.3902
This means that the closest point on the ﬂight path is (−0.39, −0.49).
Apply the distance formula and the shortest distance to the ﬂight
path is
(−0.39)2 + (−0.49)2
d=
= 0.6263. 4.11 Uniform Linear Motion
When an object moves along a line in the xyplane at a constant speed,
we say that the object exhibits uniform linear motion. Its location can
be described using a pair of linear equations involving a variable which
represents time. That is, we can ﬁnd constants a, b, c, and d such that,
at any time t, the object’s location is given by (x,y), where
x = a + bt and y = c + dt.
Such equations are called parametric equations of motion. The motion
is deﬁned in terms of the parameter t.
Since there are four constants to be determined, one needs four pieces
of information to determine these equations. Knowing the object’s location at two points in time is sufﬁcient.
Example 4.11.1. Bob is running in the xyplane. He runs in a straight line
from the point (2,3) to the point (5, − 4), taking 6 seconds to do so. Find his
equations of motion. CHAPTER 4. LINEAR MODELING 48 Solution. We begin by setting a reference for our time parameter. Let’s let
t = 0 represent the instant when Bob is at the point (2,3). In this way, t
will represent the time since Bob left the point (2,3). When t = 6, we know
he will be at the point (5, − 4). This is enough information to determine
his equations of motion.
We seek constants a, b, c, and d so that at time t, Bob’s location is
given by
x = a + bt and y = c + dt.
When t = 0, we know Bob’s location is (2,3). That is, x = 2 and y = 3.
Thus, with t = 0, we have the two equations
x = 2 = a + b(0) = a and y = 3 = c + d(0) = c.
and so a = 2 and c = 3. We’re halfway done.
When t = 6, we know Bob’s location is (5, − 4). Thus, with t = 6, we
have the two equations
x = 5 = a + b(6) = 2 + 6b and y = −4 = c + d(6) = 3 + 6d
which we can solve to ﬁnd
b= 1
7
and d = − .
2
6 So, we arrive at the equations of Bob’s motion
7
1
x = 2 + t and y = 3 − t.
2
6
Notice it is easy to check that these are correct. If we plug in t = 0, we
ﬁnd x = 2, y = 3 as required. If we plug in t = 6, we ﬁnd x = 5, y = −4, as
required. So we know we’ve done it right.
Now that we have these equations of motion, it is very easy to calculate
Bob’s location at any time. For instance, 30 seconds after leaving the
point (2,3), we can ﬁnd that he is at the point (17, − 32) since
1
7
x = 2 + (30) = 17, y = 3 − (30) = −32.
2
6 Example 4.11.2. Olga is running in the xyplane, and the coordinate are
given in meters (so, for example, the point (1,0) is one meter from the origin
(0,0)). She runs in a straight line, starting at the point (3,5) and running
along the line y = − 1 x + 6 at a speed of 7 meters per second, heading away
3
from the yaxis. What are her parametric equations of motion? 4.11. UNIFORM LINEAR MOTION 49 Solution. This example differs in some respects from the last example.
In particular, instead of knowing where the runner is at two points in
time, we only know one point, and have other information given to us
about the speed and path of the runner. One approach is to use this new
information to ﬁnd where the runner is at some other point in time: this
will then give us exactly the same sort of information as we used in the
last example, and so we may solve it in an identical manner.
We know that Olga starts at the point (3,5). Letting t = 0 represent the
time when she starts, we then know that when t = 0, x = 3 and y = 5.
To get another point (and time), we can use the fact that we know what
line she travels along, and which direction she runs. We may consider
any point on the line in the correct direction: any will do. For instance,
the point (6,4) is on the line. We then need to ﬁnd when Olga reaches
this point. To do this, we ﬁnd the distance from her starting point to the
point (6,4), and divide this by her speed. The time she takes to get to (6,4)
is thus
(6 − 3)2 + (4 − 5)2
= 0.45175395 seconds .
7
At this point, we are now in the same situation as in the last example.
We know two facts: when t = 0, x = 3 and y = 5, and when t = 0.45175395,
x = 6 and y = 4. As we saw in the last example, this is enough information
to ﬁnd the parametric equations of motion.
We seek a, b, c, and d such that Olga’s location t seconds after she
starts is (x,y) where
x = a + bt and y = c + dt.
When t = 0, x = 3, and y = 5, so
x = 3 = a + b(0) = a and y = 5 = c + d(0) = c
and so a = 3 and c = 5. Also, when t = 0.45175395, x = 6 and y = 4, so
x = 6 = a+b(0.45175395) = 3+b(0.45175395) and y = 4 = c+d(0.45175395) = 5+d(0.45175395).
Solving these equations for b and d, we ﬁnd
b= 3
−1
= 6.64078311 and d =
= −2.21359436.
0.45175395
0.45175395 Thus, Olga’s equations of motion are
x = 3 + 6.64078311t, y = 5 − 2.21359436t. CHAPTER 4. LINEAR MODELING 50 4.12 Summary
• The equation of every nonvertical line can be expressed in the form
y = m(x − h) + k (the pointslope form)
and
y = mx + b (the slopeintercept form)
• A vertical line has an equation of the form x = c.
• The shortest distance between a point, P, and a line, l, can be found
by determining a line l2 which passes through P and is perpendicular to l. Then the point at which l and l2 intersect is the point on l
which is closest to l. The distance from this point to P is the shortest
distance between P and l.
• The location of an object moving at constant speed along a line can
be described using a pair of equations (parametric equations)
x = a + bt, y = c + dt. 4.13. EXERCISES 51 4.13 Exercises
Problem 4.1. This exercise emphasizes the
“mechanical aspects” of working with linear
equations. Find the equation of a line:
(a) Passing through the points (1, − 1) and
(−2,4).
(b) Passing through the point (−1, − 2) with
slope m = 40.
(c) With yintercept b = −2 and slope m =
−2.
(d) Passing through the point (4,11) and
having slope m = 0.
(e) Perpendicular to the line in (a) and passing through (1,1).
(f) Parallel to the line in (b) and having yintercept b = −14.
(g) Having the equation 3x + 4y = 7.
(h) Crossing the xaxis at x = 1 and having
slope m = 1.
Problem 4.2. Sketch an accurate picture of
1
the line having equation y = 2 − 2 x. Let α be
an unknown constant.
(a) Find the point of intersection between
the line you have graphed and the line
y = 1 + αx; your answer will be a point in
the xy plane whose coordinates involve
the unknown α.
(b) Find α so that the intersection point in
(a) has xcoordinate 10.
(c) Find α so that the intersection point in
(a) lies on the xaxis.
Problem 4.3.
(a) What is the area of the tri1
angle determined by the lines y = − 2 x +
5, y = 6x and the yaxis?
(b) If b > 0 and m < 0, then the line y =
mx + b cuts off a triangle from the ﬁrst
quadrant. Express the area of that triangle in terms of m and b.
(c) The lines y = mx + 5, y = x and the yaxis form a triangle in the ﬁrst quadrant.
Suppose this triangle has an area of 10
square units. Find m. Problem 4.4. Complete Table 4.2 on page 52.
In many cases there may be several possible
correct answers.
Problem 4.5. The (average) sale price for
single family property in Seattle and Port
Townsend is tabulated below:
YEAR
1970
1990 SEATTLE
$38,000
$175,000 PORT TOWNSEND
$8400
$168,400 (a) Find a linear model relating the year x
and the sales price y for a single family
property in Seattle.
(b) Find a linear model relating the year x
and the sales price y for a single family
property in Port Townsend.
(c) Sketch the graph of both modeling equations in a common coordinate system;
restrict your attention to x ≥ 1970.
(d) What is the sales price in Seattle and
Port Townsend in 1983 and 1998?
(e) When will the average sales price in
Seattle and Port Townsend be equal and
what is this price?
(f) When will the average sales price in
Port Townsend be $15,000 less than the
Seattle sales price? What are the two
sales prices at this time?
(g) When will the Port Townsend sales price
be $15,000 more than the Seattle sales
price? What are the two sales prices at
this time?
(h) When will the Seattle sales price be double the Port Townsend sales price?
(i) Is the Port Townsend sales price ever
double the Seattle sales price? Problem 4.6. The cup on the 9th hole of a golf
course is located dead center in the middle of a
circular green that is 70 feet in diameter. Your
ball is located as in the picture below: CHAPTER 4. LINEAR MODELING 52
Equation Slope yintercept Point on
the line Point on
the line y = 2x + 1
(3, −4)
−2 (−1, 7) 1 1
2 (0, 1)
1,000 0
(3, 3) (3, −2) (5, −9)
Table 4.2: Linear equations table for Problem 4.4. 9
cup
green 50 feet ball ball path rough Problem 4.7. Allyson and Adrian have decided
to connect their ankles with a bungee cord;
one end is tied to each person’s ankle. The
cord is 30 feet long, but can stretch up to 90
feet. They both start from the same location.
Allyson moves 10 ft/sec and Adrian moves 8
ft/sec in the directions indicated. Adrian stops
moving at time t = 5.5 sec, but Allyson keeps
on moving 10 ft/sec in the indicated direction. 40 feet The ball follows a straight line path and exits
the green at the rightmost edge. Assume the
ball travels a constant rate of 10 ft/sec.
(a) Where does the ball enter the green? (a) Sketch an accurate picture of the situation at time t = 7 seconds. Make sure to
label the locations of Allyson and Adrian;
also, compute the length of the bungee
cord at t = 7 seconds. (b) When does the ball enter the green?
(c) How long does the ball spend inside the
green?
(d) Where is the ball located when it is
closest to the cup and when does this
occur. (b) Where is Allyson when the bungee
reaches its maximum length? 4.13. EXERCISES 53
Find the portion of the yaxis that Dave cannot see. (Hint: Let a be the xcoordinate of the
point where line of sight #1 is tangent to the
silo; compute the slope of the line using two
points (the tangent point and (12,0)). On the
other hand, compute the slope of line of sight
#1 by noting it is perpendicular to a radial line
through the tangency point. Set these two calculations of the slope equal and solve for a.) 20 ft
Building Allyson 30 ft start Adrian Problem 4.8. Dave is going to leave academia
and go into business building grain silos. A
grain silo is a cylinder with a hemispherical
top, used to store grain for farm animals. Here
is a 3D view, a crosssection, and the top view: silo
h 3Dview
r Problem 4.9. While speaking on the phone to
a friend in Oslo, Norway, you learned that the
current temperature there was −23◦ Celsius
(−23◦ C). After the phone conversation, you
wanted to convert this temperature to Fahrenheit degrees ◦ F, but you could not ﬁnd a reference with the correct formulas. You then
remembered that the relationship between ◦ F
and ◦ C is linear.
(a) Using this and the knowledge that
32◦ F = 0◦ C and 212◦ F = 100◦ C, ﬁnd an
equation that computes Celsius temperature in terms of Fahrenheit temperature; i.e., an equation of the form C=
“an expression involving only the variable F .”
(b) Likewise, ﬁnd an equation that computes Fahrenheit temperature in terms
of Celsius temperature; i.e. an equation
of the form F= “an expression involving
only the variable C .”
(c) How cold was it in Oslo in ◦ F? crosssection
yaxis blind spot line of sight #1
(12,0)
xaxis
Dave
line of sight #2
TOP VIEW If Dave is standing next to a silo of crosssectional radius r = 8 feet at the indicated position, his vision will be partially obstructed. Problem 4.10. Pam is taking a train from the
town of Rome to the town of Florence. Rome is
located 30 miles due West of the town of Paris.
Florence is 25 miles East, and 45 miles North
of Rome.
On her trip, how close does Pam get to Paris?
Problem 4.11. Angela, Mary and Tiff are all
standing near the intersection of University
and 42nd streets. Mary and Tiff do not move,
but Angela runs toward Tiff at 12 ft/sec along
a straight line, as pictured. Assume the roads
are 50 feet wide and Tiff is 60 feet north of the
nearest corner. Where is Angela located when
she is closest to Mary and when does she reach
this spot? CHAPTER 4. LINEAR MODELING 54 Problem 4.13. Margot is walking in a straight
line from a point 30 feet due east of a statue in
a park toward a point 24 feet due north of the
statue. She walks at a constant speed of 4 feet
per second. tiff
mary
42 nd St. (a) Write parametric equations for Margot’s
position t seconds after she starts walking. angela (b) Write an expression for the distance
from Margot’s position to the statue at
time t. University Way
Problem 4.12. The infamous crawling tractor
sprinkler is located as pictured below, 100 feet
South of a 10 ft. wide sidewalk; notice the hose
and sidewalk are not perpendicular. Once the
water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North
1
along the hose at the rate of 2 inch/second.
(a) Impose a coordinate system. Describe
the initial coordinates of the sprinkler
and ﬁnd the equation of the line forming
the southern boundary of the sidewalk.
(b) After 33 minutes, sketch a picture of
the wet portion of the sidewalk; ﬁnd the
length of the wet portion of the Southern
edge of the sidewalk.
(c) Find the equation of the line forming
the northern boundary of the sidewalk.
(Hint: You can use the properties of right
triangles.) W
100 ft Problem 4.14. Juliet and Mercutio are moving at constant speeds in the xyplane. They
start moving at the same time. Juliet starts at
the point (0, − 6) and heads in a straight line
toward the point (10,5), reaching it in 10 seconds. Mercutio starts at (9, − 14) and moves in
a straight line. Mercutio passes through the
same point on the x axis as Juliet, but 2 seconds after she does.
How long does it take Mercutio to reach the
yaxis?
Problem 4.15. (a) Solve for x: 1
1
−
= 3.
x x+1
(1 + t)2 + (1 − 2t)2 . (b) Solve for t: 2 = N hose (c) Find the times when Margot is 28 feet
from the statue. E
S
20 ft (c) Solve for t: 3
√
5 (d) Solve for t: 0 =
Problem 4.16. (1 + t)2 + (1 − 2t)2 . = (1 + t)2 + (1 − 2t)2 . (a) Solve for x: x4 − 4x2 + 2 = 0 100 ft. sidewalk
(b) Solve for y: circular watered zone √
y−2 y=4 Chapter 5
Functions and Graphs
Pictures are certainly important in the work of an architect, but it is
perhaps less evident that visual aids can be powerful tools for solving
mathematical problems. If we start with an equation and attach a picture, then the mathematics can come to life. This adds a new dimension
to both interpreting and solving problems. One of the real triumphs of
modern mathematics is a theory connecting pictures and equations via
the concept of a graph. This transition from “equation” to “picture” (called
graphing) and its usefulness (called graphical analysis) are the theme of
the next two sections. The importance of these ideas is HUGE and cannot be overstated. Every moment spent studying these ideas will pay
back dividends in this course and in any future mathematics, science or
engineering courses. 5.1 Relating Data, Plots and Equations
Imagine you are standing high atop an oceanside cliff and
spot a seagull hovering in the aircurrent. Assuming the
gull moves up and down along a vertical line of motion,
how can we best describe its location at time t seconds?
There are three different (but closely linked) ways to
describe the location of the gull: gull line of motion 1
0
1
0 • a table of data of the gull’s height above cliff level at
various times t;
• a plot of the data in a “time” (seconds) vs. “height”
(feet) coordinate system; ocean
Figure 5.1: Seagull’s height. • an equation relating time t (seconds) and height s
(feet).
To make sure we really understand how to pass back and forth between these three descriptive modes, imagine we have tabulated (Figure 5.2) the height of the gull above cliff level at onesecond time intervals
55 cliff level CHAPTER 5. FUNCTIONS AND GRAPHS 56 for a 10 second time period. Here, a “negative height” means the gull is
below cliff level. We can try to visualize the meaning of this data by plotting these 11 data points (t, s) in a time (sec.) vs. height (ft.) coordinate
system.
Feet
Gull Height (feet above cliff level)
s (ft)
t (sec)
s (ft)
t (sec)
20
4
10
8
6.88
5
8.12
9
2.5
6
2.5
10
8.12
7
6.88 t (sec)
0
1
2
3 50
40
30
20
10 s (ft)
20
36.88
57.5 2 4 6 8 10 Seconds −10 (b) Visual data. (a) Symbolic data. Figure 5.2: Symbolic versus visual view of data. We can improve the quality of this description by increasing the number of data points. For example, if we tabulate the height of the gull above
cliff level at 1/2 second or 1/4 second time intervals (over the same 10
second time period), we might get these two plots:
Feet Feet
50
40
30
20
10 Seconds 4 −10 2 (a) 1
2 6 8 10 second intervals. 50
40
30
20
10 Seconds 4 −10 2 (b) 1
4 6 8 10 second intervals. Figure 5.3: Shorter time intervals mean more data points. We have focused on how to go from data to a plot, but the reverse process is just as easy: A point (t, s) in any of these three plots is interpreted
to mean that the gull is s feet above cliff level at time t seconds.
Furthermore, increasing the amount of data, we see how the plotted
points are “ﬁlling in” a portion of a parabola. Of course, it is way too
tedious to create longer and longer tables of data. What we really want
is a “formula” (think of it as a prescription) that tells us how to produce
a data point for the gull’s height at any given time t. If we had such a
formula, then we could completely dispense with the tables of data and
just use the formula to crank out data points. For example, look at this
equation involving the variables t and s:
s= 15
(t − 4)2 − 10.
8 5.2. WHAT IS A FUNCTION? 57 If we plug in t = 0, 1, 2, 9, 10, then we get s = 20, 6.88, −2.5, 36.88, 57.5, respectively; this was some of our initial tabulated data. This same equation produces ALL of the data points for the other two plots, using 1/2
second and 1/4 second time intervals. (Granted, we have swept under the
rug the issue of “...where the heck the equation comes from...” ; that is
a consequence of mathematically modeling the motion of this gull. Right
now, we are focusing on how the equation relates to the data and the
plot, assuming the equation is in front of us to start with.) In addition, it
is very important to notice that having this equation produces an inﬁnite
number of data points for our gull’s location, since we can plug in any
t value between 0 and 10 and get out a corresponding height s. In other
words, the equation is A LOT more powerful than a ﬁnite (usually called
discrete ) collection of tabulated data. 5.2 What is a Function?
Our lives are chock full of examples where two changing quantities are
related to one another:
• The cost of postage is related to the weight of the item.
• The value of an investment will depend upon the time elapsed.
• The population of cells in a growth medium will be related to the
amount of time elapsed.
• The speed of a chemical reaction will be related to the temperature
of the reaction vessel.
In all such cases, it would be beneﬁcial to have a “procedure” whereby
we can assign a unique output value to any acceptable input value. For
example, given the time elapsed (an input value), we would like to predict
a unique future value of an investment (the output value). Informally,
this leads to the broadest (and hence most applicable) deﬁnition of what
we will call a function :
Deﬁnition 5.2.1. A function is a procedure for assigning a unique output
to any allowable input.
The key word here is “procedure.” Our discussion of the
hovering seagull in 5.1 highlights three ways to produce
such a “procedure” using data, plots of curves and equations. yaxis
y P = (x,y)
x xaxis Figure 5.4: Graph of a
• A table of data, by its very nature, will relate two
procedure.
columns of data: The output and input values are
listed as column entries of the table and reading across each row
is the “procedure” which relates an input with a unique output. CHAPTER 5. FUNCTIONS AND GRAPHS 58 • Given a curve in Figure 5.4, consider the “procedure”
which associates to each x on the horizontal axis the y
coordinate of the pictured point P on the curve.
• Given an equation relating two quantities x and y, plugging in a particular x value and going through the “procedure” of algebra often produces a unique output value
y. 5.2.1 The deﬁnition of a function
(equation viewpoint)
Now we focus on giving a precise deﬁnition of a function, in the situation when the “procedure” relating two quantities is actually given by an
equation. Keep in mind, this is only one of three possible ways to describe a function; we could alternatively use tables of data or the plot of
a curve. We focus on the equation viewpoint ﬁrst, since it is no doubt the
most familiar.
If we think of x and y as related physical quantities (e.g. time and distance ), then it is sometimes possible (and often desirable) to express one
of the variables in terms of the other. For example, by simple arithmetic,
the equations
3x + 2y = 4 1
x2 − x = y − 4
2 y x2 + 1 = 1, can be rewritten as equivalent equations
1
y = (4 − 3x)
2 2x2 − 2x + 8 = y y= √ 1
.
+1 x2 This leads to THE MOST IMPORTANT MATH DEFINITION IN THE WORLD:
Deﬁnition 5.2.2. A function is a package, consisting of three parts:
• An equation of the form
y = “a mathematical expression only involving the variable x,” which we usually indicate via the shorthand notation y = f(x). This
equation has the very special property that each time we plug in
an x value, it produces exactly one (a unique) y value. We call the
mathematical expression f(x) ”the rule”.
• A set D of xvalues we are allowed to plug into f(x), called the
”domain” of the function. 5.2. WHAT IS A FUNCTION? 59 • The set R of output values f(x), where x varies over the domain,
called the ”range” of the function.
Any time we have a function y = f(x), we refer to x as the independent
variable (the “input data”) and y as the dependent variable (the “output
data”). This terminology emphasizes the fact that we have freedom in
the values of x we plug in, but once we specify an x value, the y value is
uniquely determined by the rule f(x).
Examples 5.2.3.
(i) The equation y = −2x + 3 is in the
form y = f(x), where the rule is f(x) = −2x+3. Once we
specify a domain of x values, we have a function. For
example, we could let the domain be all real numbers. yaxis
Graph of y = b
b xaxis Figure 5.5: Constant func(ii) Take the same rule f(x) = −2x + 3 from (i) and let the
tion.
domain be all nonnegative real numbers. This describes a function. However, the functions f(x) = −2x+
3 on the domain of all nonnegative real numbers and f(x) = −2x + 3
on the domain of all real numbers (from (i)) are different, even though
they share the same rule; this is because their domains differ! This
example illustrates the idea of what is called a restricted domain. In
other words, we started with the function in (i) on the domain of all
real numbers, then we “restricted” to the subset of nonnegative real
numbers. (iii) The equation y = b, where b is a constant, deﬁnes a
function on the domain of all real numbers, where the
rule is f(x) = b; we call these the constant functions.
Recall, in Chapter 3, we observed that the solutions
of the equation y = b, plotted in the xy coordinate system, will give a horizontal line. For example, if b = 0,
you get the horizontal axis.
1
1
(iv) Consider the equation y = x , then the rule f(x) = x
deﬁnes a function, as long as we do not plug in x = 0.
For example, take the domain to be the nonzero real
numbers.
√
(v) Consider the equation y = 1 − x2 . Before we start
plugging in x values, we want to know the expression
under the radical symbol (square root symbol) is nonnegative; this insures the square root is a real number. This amounts to solving an inequality equation:
0 ≤ 1 − x2 ; i.e., −1 ≤ x ≤ 1. These remarks show that
√
the rule f(x) = 1 − x2 deﬁnes a function, where the
domain of x values is −1 ≤ x ≤ 1. CHAPTER 5. FUNCTIONS AND GRAPHS 60 Typically, the domain of a function y = f(x) will either be the entire
number line, an interval on the number line, or a ﬁnite union of such
intervals. We summarize the notation used to represent intervals in Table 5.1.
Common Intervals on the Number Line
Description Symbolic
Notation All numbers x between a
and b, x possibly equal to
either a or b a≤x≤b All numbers x between a
and b, x = a and x = b a<x<b All numbers x between a
and b, x = b and x possibly equal to a
All numbers x between a
and b, x = a and x possibly equal to b Picture a b a b a≤x<b a b a<x≤b a b Table 5.1: Interval Notations We can interpret a function as a “prescription” that takes a given
x value (in the domain) and produces a single unique y value (in the
range). We need to be really careful and not fall into the trap of thinking
that every equation in the world is a function. For example, if we look at
this equation
x + y2 = 1
and plug in x = 0, the equation becomes
y2 = 1.
This equation has two solutions, y = ±1, so the conclusion is that plugging in x = 0 does NOT produce a single output value. This violates one
of the conditions of our function deﬁnition, so the equation x + y2 = 1 is
NOT a function in the independent variable x. Notice, if you were to try
and solve this equation for y in terms of x, you’d ﬁrst write y2 = 1 − x
and then take a square root (to isolate y); but the square root introduces
TWO roots, which is just another way of reﬂecting the fact there can be
two y values attached to a single x value. Alternatively, you can solve the
equation for x in terms of y, getting x = 1 − y2 ; this shows the equation
does deﬁne a function x = g(y) in the independent variable y. 5.2. WHAT IS A FUNCTION? 61 5.2.2 The deﬁnition of a function
(conceptual viewpoint)
Conceptually, you can think of a function as a “process”: An allowable
input goes into a “black box” and out pops a unique new value denoted
by the symbol f(x). Compare this with the machine making “hulahoops”
in Figure 5.6. While you are problem solving, you will ﬁnd this to be a
useful viewpoint when a function is described in words.
11111111111
00000000000
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00000000000
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00000000000
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00000000000
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00000000000
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00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
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00000000000
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00000000000
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00000000000
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00000000000
11111111111
00000000000 tube in tubes out hoop f(tube) (a) A hoop machine as a
“process” taking “tubes” to
“hoops.” x
domain in out y f ( x) (b) A function as a “process”
which takes x to y. Figure 5.6: Viewing a function as a “process.” Examples 5.2.4. Here are four examples of relationships that are functions:
(i) The total amount of water used by a household since midnight
on a particular day. Let y be the total number of gallons of water
used by a household between 12:00am and a particular time t; we
will use time units of “hours.” Given a time t, the household will
have used a speciﬁc (unique) amount of water, call it S(t). Then y =
S(t) deﬁnes a function in the independent variable t with dependent
variable y. The domain would be 0 ≤ t ≤ 24 and the largest possible
value of S(t) on this domain is S(24). This tells us that the range would
be the set of values 0 ≤ y ≤ S(24).
(ii) The height of the center of a basketball as you dribble, depending on time. Let s be the height of the basketball center at time
t seconds after you start dribbling. Given a time t, if we freeze the action, the center of the ball has a single unique height above the ﬂoor,
call it h(t). So, the height of the basketball center is given by a function s = h(t). The domain would be a given interval of time you are
dribbling the ball; for example, maybe 0 ≤ t ≤ 2 (the ﬁrst 2 seconds).
In this case, the range would be all of the possible heights attained
by the center of the basketball during this 2 seconds.
(iii) The state sales tax due on a taxable item. Let T be the state
tax (in dollars) due on a taxable item that sells for z dollars. Given a
taxable item that costs z dollars, the state tax due is a single unique CHAPTER 5. FUNCTIONS AND GRAPHS 62 amount, call it W (z). So, T = W (z) is a function, where the independent variable is z. The domain could be taken to be 0 ≤ z ≤ 1,000,000,
which would cover all items costing up to onemillion dollars. The
range of the function would be the set of all values W (z), as z ranges
over the domain.
(iv) The speed of a chemical reaction depending on the temperature. Let v be the speed of a particular chemical reaction and T
the temperature in Celsius ◦ C. Given a particular temperature T , one
could experimentally measure the speed of the reaction; there will be
a unique speed, call it r(T ). So, v = r(T ) is a function, where the independent variable is T . The domain could be taken to be 0 ≤ T ≤ 100,
which would cover the range of temperatures between the freezing
and boiling points of water. The range of the function would be the
set of all speeds r(T ), as T ranges over the domain. 5.3 The Graph of a Function
Let’s start with a concrete example; the function f(x) = −2x + 3 on the
domain of all real numbers. We discussed this Example 5.2.3. Plug in
the speciﬁc x values, where x = −1, 0, 1, 2 and tabulate the resulting
y values of the function:
x
1
0
1
2
.
.
.
x y
5
3
1
1
.
.
.
−2x + 3 point (x,y)
(1,5)
(0,3)
(1,1)
(2,1)
.
.
.
(x, − 2x + 3) (a) Tabulated data. yaxis
Graph of y = −2x + 3.
xaxis (b) Visual data. Figure 5.7: Symbolic versus visual view of data. This tells us that the points (0, 3), (1, 1), (2, −1), (−1, 5) are solutions
of the equation y = −2x + 3. For example, if y = −2x + 3, x = 0, y = 3,
then 3 = −2 · 0 + 3 (which is true), or if y = −2x + 3, x = 2, y = −1, then
−1 = −2 · 2 + 3 (which is true), etc. In general, if we plug in x we get out
−2x + 3, so the point (x, − 2x + 3) is a solution to the function equation
y = f(x). We can plot all of these solutions in the xycoordinate system.
The set of points we obtain, as we vary over all x in the domain, is called
the set of solutions of the equation y = −2x + 3:
Solutions = { (x, −2x + 3)  x any real number}.
Notice that plotting these points produces a line of slope m = −2 with
yintercept 3. In other words, the graph of the function f(x) = −2x + 3 is 5.4. THE VERTICAL LINE TEST 63 the same as the graph of the equation y = −2x + 3, as we discussed in
Chapter 4.
In general, by deﬁnition, we say that a point (x,y) is a solution to the
function equation y = f(x) if plugging x and y into the equation gives a
true statement.
How can we ﬁnd ALL the solutions of the equation y = f(x)? In general,
the deﬁnition of a function is “rigged” so it is easy to describe all solutions
of the equation y = f(x): Each time we specify an x value (in the domain),
there is only one y value, namely f(x). This means the point P = (x, f(x))
is the ONLY solution to the equation y = f(x) with ﬁrst coordinate x. We
deﬁne the graph of the function y = f(x) to be the plot of all solutions of
this equation (in the xy coordinate system). It is common to refer to this
as either the “graph of f(x)” or the “graph of f.”
Graph = {(x,f(x))  x in the domain} (5.1) Important Procedure 5.3.1. Points on a graph. The description of the
graph of a function gives us a procedure to produce points on the graph
AND to test whether a given point is on the graph. On the one hand, if
you are given u in the domain of a function y = f(x), then you immediately
can plot the point (u, f(u)) on the graph. On the other hand, if someone
gives you a point (u, v), it will be on the graph only if v = f(u) is true. We
illustrate this in Example 5.3.2. Example 5.3.2. The function s = h(t) = 15 (t − 4)2 − 10 de8
ﬁnes a function in the independent variable t. If we restrict
to the domain 0 ≤ t ≤ 10, then the discussion in Chapter 7
tells us that the graph is a portion of a parabola: See Figure 5.8. Using the above procedure, you can verify that the
data points discussed in the seagull example (in §5.1) all
lie on this parabola. On the other hand, the point (0,0) is
NOT on the graph, since h(0) = 20 = 0. 60
50
40
30
20
10
−10 saxis taxis 4
2 6 8 10 Figure 5.8: s = h(t). 5.4 The Vertical Line Test
There is a pictorial aspect of the graph of a function that is very revealing:
Since (x, f(x)) is the only point on the graph with ﬁrst coordinate equal to
x, a vertical line passing through x on the xaxis (with x in the domain)
crosses the graph of y = f(x) once and only once. This gives us a decisive
way to test if a curve is the graph of a function.
Important Procedure 5.4.1. The vertical line test. Draw a curve in
the xyplane and specify a set D of xvalues. Suppose every vertical line
through a value in D intersects the curve exactly once. Then the curve is CHAPTER 5. FUNCTIONS AND GRAPHS 64 the graph of some function on the domain D. If we can ﬁnd a single vertical
line through some value in D that intersects the curve more than once, then
the curve is not the graph of a function on the domain D.
For example, draw any straight line m in the plane. By the vertical
line test, if the line m is not vertical, m is the graph of a function. On the
other hand, if the line m is vertical, then m is not the graph of a function.
These two situations are illustrated in Figure 5.9. As another example,
consider the equation x2 + y2 = 1, whose graph is the unit circle and
specify the domain D to be −1 ≤ x ≤ 1; recall Example 3.2.2. The vertical
line passing through the point 1 , 0 will intersect the unit circle twice; by
2
the vertical line test, the unit circle is not the graph of a function on the
domain −1 ≤ x ≤ 1.
yaxis yaxis yaxis l
m m
xaxis l crosses curve twice C
xaxis xaxis Figure 5.9: Applying the vertical line test. 5.4.1 Imposed Constraints
In physical problems, it might be natural to constrain (meaning to “limit”
or “restrict”) the domain. As an example, suppose the height s (in feet) of
a ball above the ground after t seconds is given by the function
s = h(t) = −16t2 + 4.
saxis
Physically interesting
portion of graph. taxis Figure 5.10: Restricting the
domain. We could look at the graph of the function in the tsplane and we will review in Chapter 7 that the graph
looks like a parabola. The physical context of this problem makes it natural to only consider the portion of the
graph in the ﬁrst quadrant; why? One way of specifying
this quadrant would be to restrict the domain of possible
1
t values to lie between 0 and 2 ; notationally, we would
write this constraint as 0 ≤ t ≤ 1 .
2 5.5 Linear Functions
A major goal of this course is to discuss several different kinds of functions. The work we did in Chapter 4 actually sets us up to describe one 5.6. PROFIT ANALYSIS 65 very useful type of function called a linear function. Back in Chapter 4,
we discussed how lines in the plane can be described using equations in
the variables x and y. One of the key conclusions was:
Important Fact 5.5.1. A nonvertical line in the plane will be the graph
of an equation y = mx + b, where m is the slope of the line and b is the
yintercept.
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 x and y we already obtained in Chapter 4: The rule f(x) = mx + b on some speciﬁed domain will
have a line of slope m and yintercept b as its graph. We call a function
of this form a linear function.
Example 5.5.2. You are driving 65 mph from the Kansas state line (mile
marker 0) to Salina (mile marker 130) along I35. Describe a linear function
that calculates mile marker after t hours. Describe another linear function
that will calculate your distance from Salina after t hours. Solution. Deﬁne a function d(t) to be the mile marker after
t hours. Using “distance=rate×time,” we conclude that
65t will be the distance traveled after t hours. Since we
started at mile marker 0, d(t) = 65t is the rule for the ﬁrst
function. A reasonable domain would be to take 0 ≤ t ≤ 2,
since it takes 2 hours to reach Salina.
For the second situation, we need to describe a different function, call it s(t), that calculates your distance
from Salina after t hours. To describe the rule of s(t) we
can use the previous work: yaxis
120
100
80
60
40
20
0.5 1 1.5 2 taxis Figure 5.11: Distance functions. s(t) = (mile marker Salina) −
(your mile marker at t hrs.)
= 130 − d(t)
= 130 − 65t.
For the rule s(t), the best domain would again be 0 ≤ t ≤ 2. We
have graphed these two functions in the same coordinate system: See
Figure 5.11 (Which function goes with which graph?). 5.6 Proﬁt Analysis
Let’s give a ﬁrst example of how to interpret the graph of a function in
the context of an application. CHAPTER 5. FUNCTIONS AND GRAPHS 66 Example 5.6.1. A software company plans to bring a new product to market. The sales price per unit is $15 and the expense to produce and market
√
x units is $100(1 + x). What is the proﬁt potential?
Two functions control the proﬁt potential of the new software. The
ﬁrst tells us the gross income, in dollars, on the sale of x units. All of
the costs involved in developing, supporting, distributing and marketing
x units are controlled by the expense equation (again in dollars):
g(x) = 15x
e(x) = 100(1 + √ (gross income function)
(expense function) x) A proﬁt will be realized on the sale of x units whenever the gross
income exceeds expenses; i.e., this occurs when g(x) > e(x). A loss occurs
on the sale of x units when expenses exceed gross income; i.e., when
e(x) > g(x). Whenever the sale of x units yields zero proﬁt (and zero loss),
we call x a breakeven point ; i.e., when e(x) = g(x).
The above approach is “symbolic.” Let’s see how to study proﬁt and
loss visually, by studying the graphs of the two functions g(x) and e(x).
To begin with, plot the graphs of the two individual functions in the xycoordinate system. We will focus on the situation when the sales ﬁgures
are between 0 to 100 units; so the domain of x values is the interval
0 ≤ x ≤ 100. Given any sales ﬁgure x, we can graphically relate three
yaxis (dollars) 1500 1500
expense gross income yaxis (dollars) 1000
(x, 15x)
500
P
20 1000
500 (x, e(x))
Q
xaxis (units sold) xaxis (units sold)
40 60 80 100 (a) Gross income graph. 20 40 60 80 100 (b) Expenses graph. Figure 5.12: Visualizing income and expenses. things:
• x on the horizontal axis;
• a point on the graph of the gross income or expense function;
• y on the vertical axis. If x = 20 units sold, there is a unique point P = (20, g(20)) = (20, 300) on
the gross income graph and a unique point Q = (20, e(20)) = (20, 547) on
the expenses graph. Since the ycoordinates of P and Q are the function
values at x = 20, the height of the point above the horizontal axis is
controlled by the function. 5.6. PROFIT ANALYSIS 67
dollars y If we plot both graphs in the same coordinate
(x, g(x))
1400
system, we can visually study the distance between
1200
points on each graph above x on the horizontal axis.
1000
(x, e(x))
800
(x, e(x))
In the ﬁrst part of this plot, the expense graph is
B
600
above the income graph, showing a loss is realized;
400
(x, g(x))
200
sold units x
the exact amount of the loss will be e(x) − g(x), which
is the length of the pictured line segment. Further to
20
40
60
80 100
the right, the two graphs cross at the point labeled
Figure 5.13: Modelling proﬁt and
loss.
“B”; this is the breakeven point; i.e., expense and
income agree, so there is zero proﬁt (and zero loss).
Finally, to the right of B the income graph is above the expense graph, so
there is a proﬁt; the exact amount of the proﬁt will be g(x) − e(x), which is
the length of the rightmost line segment. Our analysis will be complete
once we pin down the breakeven point B. This amounts to solving the
equation g(x) = e(x).
√
15x = 100(1 + x)
√
15x − 100 = 100 x
225x2 − 3000x + 10000 = 10000x
225x2 − 13000x + 10000 = 0.
Applying the quadratic formula, we get two answers: x = 0.78 or 57.
Now, we face a problem: Which of these two solutions is the answer
to the original problem? We are going to argue that only the second
solution x = 57 gives us the break even point. What about the other
”solution” at x = 0.78? Try plugging x = 0.78 into the original equation:
√
15(0.78) = 100(1 + 0.78). What has happened? Well, when going from
the second to the third line, both sides of the equation were squared.
Whenever we do this, we run the risk of adding extraneous solutions.
What should you do? After solving any equation, look back at your steps
and ask yourself whether or not you may have added (or lost) solutions.
In particular, be wary when squaring or taking the square root of both
sides of an equation. Always check your ﬁnal answer in the original
equation.
We can now compute the coordinates of the breakeven point using
either function:
B = (57, g(57)) = (57, 855) = (57, e(57)). CHAPTER 5. FUNCTIONS AND GRAPHS 68 5.7 Exercises
Problem 5.1. For each of the following functions, ﬁnd the expression for
f(x + h) − f(x)
.
h
Simplify each of your expressions far enough
so that plugging in h = 0 would be allowed.
(a) f(x) = x2 − 2x.
(b) f(x) = 2x + 3
(c) f(x) = x2 − 3
(d) f(x) = 4 − x2
(e) f(x) = −πx2 − π2
√
(f) f(x) = x − 1. (Hint: Rationalize the numerator)
Problem 5.2. Here are the graphs of two linear functions on the domain 0 ≤ x ≤ 20. Find
the formula for each of the rules y = f(x) and
y = g(x). Find the formula for a NEW function v(x) that calculates the vertical distance
between the two lines at x. Explain in terms
of the picture what v(x) is calculating. What is
v(5)? What is v(20)? What are the smallest and
largest values of v(x) on the domain 0 ≤ x ≤ 20?
yaxis 60 (20,60) g(x) 40 20 (0,24) f(x) (20,20) (0,4)
xaxis 10 20 Problem 5.3. Dave leaves his ofﬁce in
Padelford Hall on his way to teach in Gould
Hall. Below are several different scenarios.
In each case, sketch a plausible (reasonable)
graph of the function s = d(t) which keeps
track of Dave’s distance s from Padelford Hall
at time t. Take distance units to be “feet” and
time units to be “minutes.” Assume Dave’s
path to Gould Hall is along a straight line
which is 2400 feet long. gould
padelford (a) Dave leaves Padelford Hall and walks at
a constant speed until he reaches Gould
Hall 10 minutes later.
(b) Dave leaves Padelford Hall and walks at
a constant speed. It takes him 6 minutes to reach the 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.
(f) Suppose you want to sketch the graph of
a new function s = g(t) that keeps track
of Dave’s distance s from Gould Hall at
time t. How would your graphs change
in (a)(e)? Problem 5.4. At 5 AM one day, a monk began a trek from his monastery by the sea to
the monastery at the top of a mountain. He
reached the mountaintop monastery at 11
AM, spent the rest of the day in meditation, 5.7. EXERCISES 69 and then slept the night there. In the morning,
at 5 AM, he began walking back to the seaside
monastery. Though walking downhill should
have been faster, he dawdled in the beautiful
sunshine, and ending up getting to the seaside
monastery at exactly 11 AM. yaxis (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?
xaxis (b) Suppose the monk walked faster on the
second day, and got back at 9 AM. What
is your answer to part (a) in this case? Recall the procedure 5.3.1 on page 63.
(a) Find the x and y intercepts of the graph. (c) Suppose the monk started later, at 10
AM, and reached the seaside monastery
at 3 PM. What is your answer to part (a)
in this case? (b) Find the exact coordinates of all points
(x,y) on the graph which have ycoordinate equal to 5.
(c) Find the coordinates of all points (x,y) on
the graph which have ycoordinate equal
to 3. Problem 5.5. Sketch a reasonable graph for
each of the following functions. Specify a reasonable domain and range and state any assumptions you are making. Finally, describe
the largest and smallest values of your function.
(a) Height of a person depending on age.
(b) Height of the top of your head as you
jump on a pogo stick for 5 seconds.
(c) The amount of postage you must put
on a ﬁrst class letter, depending on the
weight of the letter.
(d) Distance of your big toe from the ground
as you ride your bike for 10 seconds. (d) Which of these points √ on the graph:
is
√
(1, − 2), (−1,3), (2.4,8), ( 3,7 − 3 3).
(e) Find the exact coordinates of the point
√
(x,y) on the graph with x = 1 + 2.
Problem 5.7. After winning the lottery, you
decide to buy your own island. The island is
located 1 km offshore from a straight portion of
the mainland. There is currently no source of
electricity on the island, so you want to run a
cable from the mainland to the island. An electrical power 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
cable path (e) Your height above the water level in a
swimming pool after you dive off the high
board. power 1 km x
4 km
Problem 5.6. Here is a picture of the graph of
the function f(x) = 3x2 − 3x − 2. (a) Explain why we can assume the cable
follows the path indicated in the picture; CHAPTER 5. FUNCTIONS AND GRAPHS 70
i.e. explain why the path consists of
two line segments, rather than a weird
curved path AND why it is OK to assume
the cable reaches shore to the right of
the power station and the left of the island.
(b) Let x be the distance downshore from
the power substation to where the cable
reaches the land. Find a function f(x) in
the variable x that computes the cost to
lay a cable out to your island.
(c) Make a table of values of f(x), where
7
1
x = 0, 2 ,1, 3 ,2,. . . , 2 ,4. Use these calcula2
tions to estimate the installation of minimal cost. Problem 5.8. This problem deals with the
“mechanical aspects” of working with the rule
of a function. For each of the functions listed
in (a)(c), calculate: f(0), f(−2), f(x + 3), f(♥),
f(♥ + △).
1
(a) The function f(x) = 2 (x − 3) on the domain of all real numbers. (b) The function f(x) = 2x2 − 6x on the domain of all real numbers.
(c) The function f(x) = 4π2 . Problem 5.9. Which of the curves in Figure 5.14 represent the graph of a function? If
the curve is not the graph of a function, describe what goes wrong and how you might
“ﬁx it.” When you describe how to “ﬁx” the
graph, you are allowed to cut the curve into
pieces and such that each piece is the graph
of a function. Many of these problems have
more than one correct answer.
Problem 5.10. Find an EXACT answer for
each problem.
(a) Solve for x
5
30
x
+
=2
x+3 x−7
x − 4x − 21
(b) Solve for x
√
x
5x − 4 = + 2
2
(c) Solve for x
√
√
x + x − 20 = 10
(d) Solve for t
√
√
2t − 1 + 3t + 3 = 5 5.7. EXERCISES 71 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) Figure 5.14: Curves to consider for Problem 5.9 72 CHAPTER 5. FUNCTIONS AND GRAPHS Chapter 6
Graphical Analysis
We ended the previous section with an 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 63) and the
surrounding discussion. The key thing about the graph of a function
f(x) is that it keeps track of a particular set of points in the plane whose
coordinates are related by the function rule. To be precise, a point P =
(x,y) will be on the graph of the function f(x) exactly when y = f(x). 6.1.1 Visualizing the domain and range
A function is a package that consists of a rule y = f(x), a domain of
allowed xvalues and a range of output yvalues. The domain can be
visualized as a subset of the xaxis and the range as a subset of the
yaxis. 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 xaxis, then project these points to the yaxis.
The collection of all values you obtain on the yaxis will be the range of
the function. This idea of “projection” is illustrated in the two graphs be73 CHAPTER 6. GRAPHICAL ANALYSIS 74 low. We use arrows “→” to indicate going from a domain xvalue, up to the
graph, then over to the yaxis:
range = [1,9]
graph of
f (x) = −2x + 3 domain = [3,5] domain
= [−3,1] range =
[−7, − 3] graph of
f (x) = −2x + 3 (b) Domain: [−3, 1]. (a) Domain: [3, 5]. Figure 6.1: Example projections. 6.1.2 Interpreting Points on the Graph
We can visually detect where a function has positive or negative values:
Important Fact 6.1.2. The function values f(x) control the height of the
point P(x) = (x,f(x)) on the graph above the xaxis; if the function value f(x)
is negative, the point P(x) is below the xaxis. f (x) units above xaxis
f(x3 ) units above the xaxis
P (x) = (x,f (x))
P (x3 ) = (x3 ,f (x3 ))
x2
x x3 f(x) units below the xaxis xaxis P(x2 ) = (x2 ,f(x2 )) yaxis Figure 6.2: Interpreting points on a graph. In Figure 6.2 we can now divide the domain (in this case the whole number line) into segments where the function is above, below or crossing the
axis. Keeping track of this information on a number line is called a sign
plot for the function. We include a “shadow” of the graph in Figure 6.3 6.1. VISUAL ANALYSIS OF A GRAPH 75 positive positive negative negative xaxis Figure 6.3: Sign plot. to emphasize how we arrived at our “positive” and “negative” labeling of
the sign plot; in practice we would only provide a labeled number line.
By moving through a sequence of x values we can investigate how the
corresponding points on the graph move “up and down”; this then gives
us a dynamic visual sense of how the function values are changing. For
example, in Figure 6.4, suppose we let x move from 1 to 5, left to right;
we have indicated how the corresponding points on the curve will move
and how the function values will change. yaxis
f(1) P
#2: points on graph move from P to Q
f(2) #3: f(x) values move
like this
graph of y = f(x) f(3)
f(5) Q
f(4)
1 5
xaxis #1: x values move from 1 to 5 Figure 6.4: Dynamic interpretation of a graph. CHAPTER 6. GRAPHICAL ANALYSIS 76 6.1.3 Interpreting Intercepts of a Graph
The places where a graph crosses the axes are often signiﬁcant. We isolate each as an important feature to look
yintercept = (0,f(0))
for when doing graphical analysis. The graph of the func(x3 ,0)
xaxis
(x2 ,0)
tion y = f(x) crosses the yaxis at the point (0,f(0)); so,
(x1 ,0)
the yintercept of the graph is just f(0). The graph of the
xintercepts have the form “(x,0)”
function y = f(x) crosses the xaxis at points of the form
Figure 6.5: Intercepts of a
(x0 ,f(x0 )), where f(x0 ) = 0. The values x0 are called roots
graph.
or zeros of the function f(x). There can be at most one
yintercept, but there can be several xintercepts or no
xintercept: See Figure 6.5
The graph of a function y = f(x) crosses the vertical line x = h at the
point (h,f(h)). To ﬁnd where the graph of a function y = f(x) crosses the
horizontal line y = k, ﬁrst solve the equation k = f(x) for x. If the equation
k = f(x) has solutions x1 , x2 , x3 , x4 , then the points of intersection would
have the coordinates given.
yaxis yaxis upper semicircle radius
r = 2 centered at (2,1) yaxis
(h,(f(h)) y=k graph of f(x)
(x1 ,k) (x3 ,k) (x2 ,k) (x4 ,k) y=k
x1 x2 x3 x=h x4 y= xaxis
x=h (a) General curve. 1
2 xaxis (b) Semicircle. Figure 6.6: Crossing horizontal and vertical lines. As another example, the graph in Figure 6.6(b) above will cross the
horizontal line y = k twice if and only if 1 ≤ k < 3; the graph will cross the
horizontal line y = k once if and only if k = 3. The graph will not intersect
1
the line y = 2 and the graph will cross the vertical line x = h if and only if
0 ≤ h ≤ 4. 6.1.4 Interpreting Increasing and Decreasing
yaxis up hi ll xaxis
dow nh ill = local extrema Figure 6.7: Graphically interpreting increasing and decreasing. We use certain terms to describe how the function values
are changing over some domain of x values. Typically, we
want to study what is happening to the values f(x) as x
moves from “left to right” in some interval. This can be
linked graphically with the study of “uphill” and “downhill” portions of the function graph: If you were “walking
to the right” along the graph, the function values are increasing if you are walking uphill. Likewise, if you were 6.2. CIRCLES AND SEMICIRCLES 77 “walking to the right” along the graph, the function values are decreasing
if you are walking downhill.
Once we understand where the graph is moving uphill and downhill,
we can isolate the places where we change from moving uphill to downhill, or vice versa; these “peaks” and “valleys” are called local maxima
and local minima. Some folks refer to either case as a local extrema.
People have invested a lot of time (centuries!) and energy (lifetimes!) into
the study of how to ﬁnd local extrema for particular function graphs. We
will see some basic examples in this course and others will surface in
future courses once you have the tools of calculus at your disposal. Examples range from business applications that involve optimizing proﬁt to
understanding the 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 t minutes is given by the
graph in Figure 6.8: ft above gliderport
600
400
200
minutes
5 10 −200 1. When is the pilot climbing and descending? −400
−600 2. When is the pilot at the glider port elevation?
3. How much time does the pilot spend ﬂying level? Figure 6.8: Hanglider elevations. Solution.
1. Graphically, we need to determine the portions of the graph that
are increasing or decreasing. In this example, it is increasing when
0 ≤ t ≤ 2 and 7 ≤ t ≤ 9. And, it is decreasing when 3 ≤ t ≤ 5 and
9 ≤ t ≤ 10.
2. Graphically, this question amounts to asking when the elevation
is 0, which is the same as ﬁnding when the graph crosses the horizontal axis. We can read off there are four such times: t = 0, 4, 8, 10.
3. Graphically, we need to determine the portions of the graph that are
made up of horizontal line segments. This happens when 2 ≤ t ≤ 3
and 5 ≤ t ≤ 7. So, our pilot ﬂies level for a total of 3 minutes. 6.2 Circles and Semicircles
Back in Chapter 3, we discussed equations whose graphs were circles:
We found that the graph of the equation
(x − h)2 + (y − k)2 = r2 (6.1) CHAPTER 6. GRAPHICAL ANALYSIS 78 is a circle of radius r centered at the point (h, k). It is possible to manipulate this equation and become confused. We could rewrite this as
(y − k)2 = r2 − (x − h)2 , then take the square root of each side. However,
the resulting equivalent equation would be
r2 − (x − h)2 y=k± and the presence of that ± sign is tricky; it means we have two equations:
y = k+ r2 − (x − h)2 y = k− or r2 − (x − h)2 . Each of these two equations deﬁnes a function:
f(x) = k + r2 − (x − h)2 g(x) = k − or (6.2) r2 − (x − h)2 . (6.3) So, even though the Equation 6.1 is not a function, we were able to obtain two different functions f(x) and g(x) from the original equation. The
relationship between the graph of the original equation and the graphs
of the two functions in (6.2) and (6.3) is as follows: The upper semicircle
is the graph of the function f(x) and the lower semicircle is the graph of
the function g(x).
yaxis upper semicircle yaxis (h,k)
(h,k) lower semicircle
xaxis Graph of y = f(x) xaxis
Graph of y = g(x) Figure 6.9: Upper and lower semicircles. Example 6.2.1. A tunnel connecting two portions of a space station has
a circular 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. 6.3. MULTIPART FUNCTIONS Solution. Impose a coordinate system so that the origin is
at the center of the circular cross section of the tunnel; by
symmetry the walkway is centered about the origin. With
this coordinate system, the graph of the equation x2 + y2 =
152 = 225 will be the circular crosssection of the tunnel.
In the case of Deck A, we basically need to determine
how close to each edge of the tunnel a 6 foot high person
can stand without hitting his or her head on the tunnel;
a similar remark applies to Deck B. This means we are
really trying to ﬁt two sixfoothigh rectangular safe walk
zones into the picture:
Our job is to ﬁnd the coordinates of the four points
P, Q, R, and S. Let’s denote by x1 , x2 , x3 ,and x4 the xcoordinates of these four points, then P = (x1 , 6), Q =
(x2 , 6), R = (x3 , −8), and S = (x4 , −8). To ﬁnd x1 , x2 , x3 ,
and x4 , we need to ﬁnd the intersection of the circle in
Figure 6.10(b) with two horizontal lines: 79 Deck A
xaxis
Deck B r = 15ft (a) Crosssection of tunnel.
Deck A Deck B
= Safe walk zone (b) Walk zones.
Figure 6.10: Space station
tunnels. • Intersecting the the upper semicircle with the horizontal line having equation y = 6 will determine x1
and x2 ; the upper semicircle is the graph of f(x) =
√
225 − x2 .
• Intersecting the lower semicircle with the horizontal line having equation y = −8 will determine x3
and x4 ; the lower semicircle is the graph of g(x) =
√
− 225 − x2 .
For Deck A, we simultaneously solve the system of equations
√
y = 225 − x2
.
y =6
Plugging in y = 6 into the ﬁrst equation of the system gives x2 = 225 −
√
62 = 189; i.e., x = ± 189 = ±13.75. This tells us that P = (−13.75, 6) and
Q = (13.75, 6). In a similar way, for Deck B, we ﬁnd R = (−12.69, −8) and
S = (12.69, −8).
In the case of Deck A, we would paint a safety stripe 13.75 feet to the
right and left of the centerline. In the case of Deck B, we would paint a
safety stripe 12.69 feet to the right and left of the centerline. 6.3 Multipart Functions
So far, in all of our examples we have been able to write f(x) as a nice
compact expression in the variable x. Sometimes we have to work harder. CHAPTER 6. GRAPHICAL ANALYSIS 80 As an example of what we have in mind, consider the graph in Figure 6.11(a):
yaxis
1 1 2 3 4 xaxis −1 (a) Graphing a multipart
function. −1 1 −1
f(x) = 1 −1 if
if
if
if
if 0≤x<1
1≤x<2
2≤x<3
3≤x<4
x=4 (b) Writing a multipart function. Figure 6.11: A multipart function. The curve we are trying to describe in this picture is made up of ﬁve
pieces; four little line segments and a single point. The ﬁrst thing to
notice is that on the domain 0 ≤ x ≤ 4, this curve will deﬁne the graph of
some function f(x). To see why this is true, imagine a vertical line moving
from left to right within the domain 0 ≤ x ≤ 4 on the xaxis; any one of
these vertical lines will intersect the curve exactly once, so by the vertical
line test, the curve must be the graph of a function. Mathematicians use
the shorthand notation above to describe this function. Notice how the
rule for f(x) involves ﬁve cases; each of these cases corresponds to one
of the ﬁve pieces that make up the curve. Finally, notice the care with
the “open” and “closed” circles is really needed if we want to make sure
the curve deﬁnes a function; in terms of the rule, these open and closed
circles translate into strict inequalities like < or weak inequalities like ≤.
This is an example of what we call a multipart function.
The symbolic appearance of multipart functions can be somewhat
frightening. The key point is that the graph (and rule) of the function
will be broken up into a number of separate cases. To study the graph
or rule, we simply “home in” on the appropriate case. For example, in
the above illustration, suppose we wanted to compute f(3.56). First, we
would ﬁnd which of the ﬁve cases covers x = 3.56, then apply that part of
the rule to compute f(3.56) = 1.
Our ﬁrst multipart function example illustrated how to go from a
graph in the plane to a rule for f(x); we can reverse this process and
go from the rule to the graph.
Example 6.3.1. Sketch the graph of the multipart function if x ≤ −1
1 √
2 if −1 ≤ x ≤ 1
g(x) =
1+ 1−x 1
if x ≥ 1 6.3. MULTIPART FUNCTIONS 81 second part
yaxis
Solution. The graph of g(x) will consist of three pieces.
of graph
The ﬁrst case consists of the graph of the function y =
g(x) = 1 on the domain x ≤ −1, this consists of all points
xaxis
on the horizontal line y = 1 to the left of and including the
third part of graph
ﬁrst part of graph
point (−1,1). We have “lassoed” this portion of the graph
= graph of g(x)
in Figure 6.12. Likewise, the third case in the deﬁnition
yields the graph of the function y = g(x) = 1 on the domain
Figure 6.12: Multipart funcx ≥ 1; this is just all points on the horizontal line y = 1
tion g(x).
to the right of and including the point (1, 1). Finally, we
need to analyze the middle case, which means we need to
√
look at the graph of 1 + 1 − x2 on the domain −1 ≤ x ≤ 1. This is just
the upper semicircle of the circle of radius 1 centered at (0,1). If we paste
these three pieces together, we arrive at the graph of g(x). 5
4 feet Example 6.3.2. You are dribbling a basketball and the
function s = h(t) keeps track of the height of the ball’s center above the ﬂoor after t seconds. Sketch a reasonable
graph of s = h(t). 3 2
1 1 seconds 2 Solution. If we take the domain to be 0 ≤ t ≤ 2 (the ﬁrst 2
Figure 6.13: Dribbling.
seconds), a reasonable graph might look like Figure 6.13.
This is a multipart function. Three portions of the graph are decreasing
and two portions are increasing. Why doesn’t the graph touch the t axis? CHAPTER 6. GRAPHICAL ANALYSIS 82 6.4 Exercises
Problem 6.1. The absolute value function is
deﬁned by the multipart rule:
x = x
−x Problem 6.4.
(a) Let f(x) = x + 2x − 1. Find
all solutions to the equation if 0 ≤ x
if x < 0 f(x) = 8. The graph of the absolute value function is pictured below: (b) Let g(x) = 3x − 3 + x + 5. Find all values
of a which satisfy the equation yaxis g(a) = 2a + 8.
y = x (c) Let h(x) = x − 3x + 4. Find all solutions
to the equation
h(x − 1) = x − 2. xaxis Problem 6.5. Express the area of the shaded
region below as a function of x. The dimensions in the ﬁgure are centimeters. (a) Calculate: 0, 2,  − 3.
(b) Solve for x: x = 4; x = 0, x = −1.
(c) Sketch the graph of y = 1 x + 2 and y = x
2
in the same coordinate system. Find
where the two graphs intersect, label the
coordinates of these point(s), then ﬁnd
the area of the region bounded by the
two graphs. 5
Problem 6.2. For each of the following functions, graph f(x) and g(x) = f(x), and give the
multipart rule for g(x). 3 (a) f(x) = −0.5x − 1
(b) f(x) = 2x − 5
x (c) f(x) = x + 3 6
Problem 6.3. Solve each of the following equations for x.
(a) g(x) = 17, where g(x) = 3x + 5
(b) f(x) = 1.5 where
2x
4−x f(x) = if x < 3,
if x ≥ 3. (c) h(x) = −1 where
h(x) = −8 − 4x
1 + 1x
3 if x ≤ −2,
if x > −2. Problem 6.6. Pizzeria Buonapetito makes a
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 x be the
bottom length of Alice’s portion and y be the
length of the cut as shown, answer the following questions: 6.4. EXERCISES 83 20
y
x
10 20 20
y x Problem 6.8. Arthur is going for a run. From
his starting point, he runs due east at 10 feet
per second for 250 feet. He then turns, and
runs north at 12 feet per second for 400 feet.
He then turns, and runs west at 9 feet per second for 90 feet.
Express the (straightline) distance from
Arthur to his starting point as a function of
t, the number of seconds since he started. Problem 6.9. A baseball diamond is a square
with sides of length 90 ft. Assume Edgar hits a
home run and races around the bases (counterclockwise) at a speed of 18 ft/sec. Express
the distance between Edgar and home plate as
a function of time t. (Hint: This will be a multipart function.) Try to sketch a graph of this
function. (a) Find a formula for y as a multipart function of x, for 0 ≤ x ≤ 30. Sketch the graph
of this function and calculate the range.
(b) Find a formula for the area of Alice’s
portion as a multipart function of x, for
0 ≤ x ≤ 30.
(c) If Alice wants her portion to have half
the area of the pizza, where should she
make the cut? Problem 6.7. This problem deals with cars
traveling between Bellevue and Spokane,
which are 280 miles apart. Let t be the time
in hours, measured from 12:00 noon; for example, t = −1 is 11:00 am.
(a) Joan drives from Bellevue to Spokane at
a constant speed, departing from Bellevue at 11:00 am and arriving in Spokane
at 3:30 pm. Find a function j(t) that
computes her distance from Bellevue at
time t. Sketch the graph, specify the domain and determine the range.
(b) Steve drives from Spokane to Bellevue
at 70 mph, departing from Spokane at
12:00 noon. Find a function s(t) for his
distance from Bellevue at time t. Sketch
the graph, specify the domain and determine the range.
(c) Find a function d(t) that computes the
distance between Joan and Steve at
time t. 90 ft
Edgar d(t) home plate Problem 6.10. Pagliacci Pizza has designed a
cardboard delivery box from a single piece of
cardboard, as pictured.
(a) Find a polynomial function v(x) that
computes the volume of the box in terms
of x. What is the degree of v?
(b) Find a polynomial function a(x) that
computes the exposed surface area of
the closed box in terms of x. What is
the degree of a? What are the explicit dimensions if the exposed surface area of
the closed box is 600 sq. inches? CHAPTER 6. GRAPHICAL ANALYSIS 84 (a) When will the ditch be completely full? x (b) Find a multipart function that models
the vertical crosssection of the ditch. x
20 in (c) What is the width of the ﬁlled portion of
the ditch after 1 hour and 18 minutes?
(d) When will the ﬁlled portion of the ditch
be 42 feet wide? 50 feet wide? 73 feet
wide? x
x
50 in
remove shaded squares and fold to get: Problem 6.12. The graph of a function y =
g(x) on the domain −6 ≤ x ≤ 6 consistes of line
segments and semicircles of radius 2 connecting the points (−6,0),(−4,4), (0,4), (4,4), (6,0).
yaxis xaxis Problem 6.11. The vertical crosssection of a
drainage ditch is pictured below: (a) What is the range of g?
(b) Where is the function increasing? Where
is the function decreasing?
(c) Find the multipart formula for y = g(x).
(d) If we restrict the function to the smaller
domain −5 ≤ x ≤ 0, what is the range?
(e) If we restrict the function to the smaller
domain 0 ≤ x ≤ 4, what is the range? 3D−view of ditch
20 ft 20 ft
R R
R R vertical crosssection
Here, R indicates a circle of radius 10 feet and
all of the indicated circle centers lie along the
common horizontal line 10 feet above and parallel to the ditch bottom. Assume that water is
ﬂowing into the ditch so that the level above
the bottom is rising 2 inches per minute. (a) Simply as far as possible Problem 6.13.
1
1+ 1
a − a
.
a+1 (b) Find a, b, c that simultaneously satisfy
these three equations:
a+b−c =
2a − 3b + c =
a+b+c = 5
4
−1 Chapter 7
Quadratic Modeling
If you kick a ball through the air enough times, you will ﬁnd its path
tends to be parabolic. Before we can answer any detailed questions about
this situation, we need to get our hands on a precise mathematical model
for a parabolic shaped curve. This means we seek a function y = f(x)
whose graph reproduces the path of the ball. ground level Figure 7.1: Possible paths for a kicked ball are parabolic. 7.1 Parabolas and Vertex Form
OK, suppose we sit down with an xycoordinate system and draw four
random parabolas; let’s label them I, II, III, and IV: See Figure 7.2. The
relationship between these parabolas and the ﬁxed coordinate system
can vary quite a bit: The key distinction between these four curves is
that only I and IV are the graphs of functions; this follows from the vertical line test. A parabola that is the graph of a function is called a
standard parabola. We can see that any standard parabola has three
basic features:
85 CHAPTER 7. QUADRATIC MODELING 86 yaxis II
III
I IV
xaxis
not graphs of functions Figure 7.2: Relationship between a ﬁxed coordinate system and various parabolas. • the parabola will either open “upward” or “downward”;
• the graph will have either a “highest point” or “lowest point,” called
the vertex ;
• the parabola will be symmetric about some vertical line called the
axis of symmetry.
Our ﬁrst task is to describe the mathematical model for any standard
parabola. In other words, what kind of function equations y = f(x) give
us standard parabolas as their graphs? Our approach is geometric and
visual:
• Begin with one speciﬁc example, then show every other standard
parabola can be obtained from it via some speciﬁc geometric maneuvers.
• As we perform these geometric maneuvers, we keep track of how the
function equation for the curve is changing. 35
30
25
20
15
10
5
−6 −4 −2 This discussion will amount to a concrete application of a more general
set of tools developed in the following section of this chapter.
Using a graphing device, it is an easy matter to plot
yaxis
the graph of y = x2 and see we are getting the parabola
pictured in Figure 7.3. The basic idea is to describe how
we can manipulate this graph and obtain any standard
parabola. In the end, we will see that standard parabolas
xaxis
are obtained as the graphs of functions having the form
2 4 6 Figure 7.3: Graph of y = x2 . y = ax2 + bx + c, 7.1. PARABOLAS AND VERTEX FORM 87 for various constants a, b, and c, with a = 0. A function of this type is
called a quadratic function and these play a central role throughout the
course. We will divide our task into two steps.
First we show every standard parabola arises as the graph of a function having the form
y = a(x − h)2 + k,
for some constants a, h, and k, with a = 0. This is called the vertex
form of a quadratic function. Notice, if we were to algebraically expand
out this equation, we could rewrite it in the y = ax2 + bx + c form. For
example, suppose we start with the vertex form y = 2(x − 1)2 + 3, so
that a = 2, h = 1, k = 3. Then we can rewrite the equation in the form
y = ax2 + bx + c as follows:
2(x − 1)2 + 3 = 2(x2 − 2x + 1) + 3 = 2x2 − 4x + 5,
so a = 2, b = −4, c = 5. The second step is to show any quadratic function
can be written in vertex form; the underlying algebraic technique used
here is called completing the square. This is a bit more involved. For
example, if you are simply handed the quadratic function y = −3x2 + 6x − 1,
it not at all obvious why the vertex form is obtained by this equality:
−3x2 + 6x − 1 = −3(x − 1)2 + 2.
The reason behind this equality is the technique of completing the square.
In the end, we will almost always be interested in the vertex form of a
quadratic. This is because a great deal of qualitative information about
the parabolic graph can simply be “read off” from this form. 7.1.1 First Maneuver: Shifting
Suppose we start with the graph in Figure 7.3 and horiyaxis
40
zontally shift it h units to the right. To be speciﬁc, con30
sider the two cases h = 2 and h = 4. To visualize this,
20
imagine making a wire model of the graph, set in on top
of the curve, then slide the wire model h units to the right.
10
What you will obtain are the two “dashed curves” in Figxaxis
−6
−4
−2
2
4
6
ure 7.4. We will call the process just described a horizontal shift. Since the “dashed curves” are no longer the
Figure 7.4: Shift to the right.
original parabola in Figure 7.3, the corresponding function equations must have changed.
Using a graphing device, you can check that the corresponding equations for the dashed graphs would be
y = (x − 2)2
= x2 − 4x + 4, CHAPTER 7. QUADRATIC MODELING 88 which is the plot with lowest point (2, 0) and
y = (x − 4)2
= x2 − 8x + 16, 40 30 20 10 −6 −4 −2 which is the plot with lowest point (4, 0). In general, if h is positive, the
graph of the function y = (x − h)2 is the parabola obtained by shifting the
graph of y = x2 by h units to the right.
Next, if h is negative, shifting h units to the right is the
yaxis
same as shifting h units left! On the domain −6 ≤ x ≤ 6,
Figure 7.5 indicates this for the cases h = −2, − 4, using
“dashed curves” for the shifted graphs and a solid line
for the graph of y = x2 . Using a graphing device, we can
check that the corresponding equations for the dashed
xaxis
graphs would be
2 4 6 Figure 7.5: Shift to the left. y = (x − (−2))2
= (x + 2)2
= x2 + 4x + 4, which is the plot with lowest point (−2, 0) and
y = (x − (−4))2
= (x + 4)2
= x2 + 8x + 16,
which is the plot with lowest point (−4, 0). In general, if h is negative, the
graph of the function y = (x − h)2 gives the parabola obtained by shifting
the graph of y = x2 by h units to the LEFT.
The conclusion thus far is this: Begin with the graph of y = x2 in
Figure 7.3. Horizontally shifting this graph h units to the right gives a
new (standard) parabola whose equation is y = (x − h)2 .
We can also imagine vertically shifting the graph in
yaxis
Figure 7.3. This amounts to moving the graph k units
40
30
vertically upward. It turns out that this vertically shifted
20
graph corresponds to the graph of the function y = x2 + k.
10
We can work out a few special cases and use a graphing
xaxis
−4
−6
4
6
−2
2
device to illustrate what all this really means.
−10
Figure 7.6 illustrates the graphs of y = x2 + k in the
cases when k = 4, 10 and k = −4, −10, leading to vertically
Figure 7.6: Vertical shifts.
shifted graphs. Positive values of k lead to the upper two
“dashed curves” and negative values of k lead to the lower two “dashed
curves”; the plot of y = x2 is again the solid line. The equations giving
these graphs would be y = x2 − 10, y = x2 − 4, y = x2 + 4 and y = x2 + 10,
from bottom to top dashed plot.
If we combine horizontal and vertical shifting, we end up with the
graphs of functions of the form y = (x − h)2 + k. Figure 7.7(a) illustrates 7.1. PARABOLAS AND VERTEX FORM 89 the four cases with corresponding equations y = (x±2)2 ±4; as an exercise,
identify which equation goes with each curve. 7.1.2 Second Maneuver: Reﬂection
Next, we can reﬂect any of the curves y = p(x) obtained
by horizontal or vertical shifting across the xaxis. This
procedure will produce a new curve which is the graph of
the new function y = −p(x). For example, begin with the
four dashed curves in the previous ﬁgure. Here are the
reﬂected parabolas and their equations are y = −(x ± 2)2 ±
4: See Figure 7.7(b). yaxis
40
30
20
10 xaxis
−4 −6 −2 2 4 6 (a) Combined shifts. 7.1.3 Third Maneuver: Vertical Dilation yaxis
30 2 If a is a positive number, the graph of y = ax is usually
called a vertical dilation of the graph of y = x2 . There are
two cases to distinguish here: 20
10 xaxis
−6 −4 −2 2 4 6 −10
−20 • If a > 1, we have a vertically expanded graph. −30 • If 0 < a < 1, we have a vertically compressed graph. (b) Reﬂections. This is illustrated for a = 2 (upper dashed plot) and a =
1/2 (lower dashed plot): See Figure 7.7(c). yaxis
30
25
20 7.1.4 Conclusion 15
10 Starting with y = x2 in Figure 7.3, we can combine together all three of the operations: shifting, reﬂection and
dilation. This will lead to the graphs of functions that
have the form:
y = a(x − h)2 + k, 5 xaxis
−6 −4 −2 2 4 6 (c) Vertical dilations.
Figure 7.7: Shifts, reﬂections, and dilations. for some a, h and k, a = 0. If you think about it for awhile,
it seems pretty easy to believe that any standard parabola
arises from the one in Figure 7.3 using our three geometric maneuvers. In other words, what we have shown is
that any standard parabola is the graph of a quadratic equation in vertex
form. Let’s summarize.
Important Fact 7.1.1. A standard parabola is the graph of a function
y = f(x) = a(x − h)2 + k, for some constants a, h, and k and a = 0. The
vertex of the parabola is (h, k) and the axis of symmetry is the line x = h. If
a > 0, then the parabola opens upward; if a < 0, then the parabola opens
downward. CHAPTER 7. QUADRATIC MODELING 90 Example 7.1.2. Describe a sequence of geometric operations leading from
the graph of y = x2 to the graph of y = f(x) = −3(x − 1)2 + 2. reﬂect across xaxis
horizontal shift by h = 1 −3(x − 1)2 + 2
vertical shift by k = 2
vertical dilate by 3 (a) What do the symbols of an
equation mean?
20 yaxis 15
10
5 −6 −4 −2 xaxis
2 4 6 −5
−10
−15 (b) What does the equation
look like?
Figure 7.8: Interpreting an
equation. Solution. To begin with, we can make some initial conclusions about the speciﬁc shifts, reﬂections and dilations
involved, based on looking at the vertex form of the equation. In addition, by Fact 7.1.1, we know that the vertex
of the graph of y = f(x) is (1, 2), the line x = 1 is a vertical
axis of symmetry and the parabola opens downward.
We need to be a little careful about the order in which
we apply the four operations highlighted. We will illustrate a procedure that works. The full explanation for the
success of our procedure involves function compositions
and we will return to that at the end of Chapter 8. The
order in which we will apply our geometric maneuvers is
as follows:
horizontal shift ⇒ vertical dilate
⇒ reﬂect
⇒ vertical shift Figure 7.8(b) illustrates the four curves obtained by applying these successive steps, in this order. As a reference, we include the graph of y = x2 as a “dashed curve”: • A horizontal shift by h = 1 yields the graph of
y = (x − 1)2 ; this is the fat parabola opening upward
with vertex (1, 0).
• A dilation by a = 3 yields the graph of y = 3(x −
1)2 ; this is the skinny parabola opening upward with
vertex (1, 0).
• A reﬂection yields the graph of y = −3(x − 1)2 ; this is
the downward opening parabola with vertex (1, 0).
• A vertical shift by k = 2 yields the graph of y = −3(x −
1)2 + 2; this is the downward opening parabola with
vertex (1, 2). 7.2 Completing the Square
By now it is pretty clear we can say a lot about the graph of a quadratic
function which is in vertex form. We need a procedure for rewriting a
given quadratic function in vertex form. Let’s ﬁrst look at an example. 7.2. COMPLETING THE SQUARE 91 Example 7.2.1. Find the vertex form of the quadratic function y = −3x2 +
6x − 1.
Solution. Since our goal is to put the function in vertex form, we can write
down what this means, then try to solve for the unknown constants. Our
ﬁrst step would be to write
−3x2 + 6x − 1 = a(x − h)2 + k,
for some constants a, h, k. Now, expand the right hand side of this
equation and factor out coefﬁcients of x and x2 :
−3x2 + 6x − 1
−3x2 + 6x 1
−3x2 + 6x − 1
(−3)x2 + (6)x + (−1) =
=
=
= a ( x − h) 2 + k
a(x2 − 2xh + h2 ) + k
ax2 − 2xah + ah2 + k
(a)x2 + (−2ah)x + (ah2 + k). If this is an equation, then it must be the case that the coefﬁcients of
like powers of x match up on the two sides of the equation in Figure 7.9.
Now we have three equations and three unknowns (the a, h, k) and we
Equal (−3) x2 + (6) x + (−1) = (a) x2 + (−2ah) x + (ah2 + k) Equal
Equal
Figure 7.9: Balancing the coefﬁcients. can proceed to solve for these:
−3 = a
6 = −2ah
−1 = ah2 + k
The ﬁrst equation just hands us the value of a = −3. Next, we can plug
this value of a into the second equation, giving us
6 = −2ah
= −2(−3)h
= 6h, CHAPTER 7. QUADRATIC MODELING 92 so h = 1. Finally, plug the now known values of a and h into the third
equation:
−1 = ah2 + k
= −3(12 ) + k
= −3 + k,
so k = 2. Our conclusion is then
−3x2 + 6x − 1 = −3(x − 1)2 + 2.
Notice, this is the quadratic we studied in Example 7.1.2 on page 90.
The procedure used in the preceding example will always work to
rewrite a quadratic function in vertex form. We refer to this as completing
the square.
Example 7.2.2. Describe the relationship between the graphs of y = x2
and y = f(x) = −4x2 + 5x + 2. 20 yaxis 10 xaxis
−6 −4 −2 2 4 6 −10 −20 Solution. We will go through the algebra to complete the
square, then interpret what this all means in terms of
graphical maneuvers. We have
−4x2 + 5x + 2 = a(x − h)2 + k
(−4)x2 + 5x + 2 = ax2 + (−2ah)x + (ah2 + k).
This gives us three equations: Figure 7.10: Maneuvering
y = x2 . −4 = a
5 = −2ah
2 = ah2 + k. 57
We conclude that a = −4, h = 5 = 0.625 and k = 16 = 3.562. So, this tells
8
us that we can obtain the graph of y = f(x) from that of y = x2 by these
steps: • Horizontally shifting by h = 0.625 units gives y = (x − 0.625)2.
• Vertically dilate by the factor a = 4 gives y = 4(x − 0.625)2.
• Reﬂecting across the xaxis gives y = −4(x − 0.625)2.
• Vertically shifting by k = 3.562 gives y = f(x) = −4(x − 0.625)2 + 3.562.
Example 7.2.3. A drainage canal has a 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 93 20 feet centerline
Solution. Impose an xycoordinate system so that the
parabolic crosssection of the canal is symmetric about
10 feet
the yaxis and its vertex is the origin. The vertex form of
any such parabola is y = f(x) = ax2 , for some a > 0; this
is because (h, k) = (0, 0) is the vertex and the parabola
Figure 7.11:
A drainage
opens upward! The dimension information given tells us
canal.
that the points (10, 10) and (−10, 10) are on the graph of
f(x). Plugging into the expression for f, we conclude that 10 = 100a, so
a = 0.1 and f(x) = (0.1)x2. Finally, if the water is 5 feet deep, we must
√
solve the equation: 5 = (0.1)x2, leading to x = ± 50 = ±7.07. Conclude the
surface of the water is 14.14 feet wide when the water is 5 feet deep. 11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000 7.3 Interpreting the Vertex minimum
−b
value f
2a maximum
−b
value f
2a
vertex
vertex
−b
2a −b
2a Figure 7.12: The vertex as the extremum of the quadratic function. If we begin with a quadratic function y = f(x) = ax2 +bx+c, we know the
graph will be a parabola. Graphically, the vertex will correspond to either
the “highest point” or “lowest point” on the graph. If a > 0, the vertex
is the lowest point on the graph; if a < 0, the vertex is the highest point
on the graph. The maximum or minimum value of the function is the
second coordinate of the vertex and the value of the variable x for which
this extreme value is achieved is the ﬁrst coordinate of the vertex. As
we know, it is easy to read off the vertex coordinates when a quadratic
function is written in vertex form. If instead we are given a quadratic
function y = ax2 + bx + c, we can use the technique of completing the
square and arrive at a formula for the coordinates of the vertex in terms
of a, b, and c. We summarize this below and label the two situations
(upward or downward opening parabola) in the Figure 7.12. Keep in CHAPTER 7. QUADRATIC MODELING 94 mind, it is always possible to obtain this formula by simply completing
the square.
Important Fact 7.3.1. In applications involving a quadratic function
f(x) = ax2 + bx + c,
the vertex has coordinates P = −b , f −b . The second coordinate of the
2a
2a
vertex will detect the maximum or minimum value of f(x); this is often a
key step in problem solving.
Example 7.3.2. Discuss the graph of the quadratic function y = f(x) =
−2x2 + 11x − 4. −4 2 −2 4 6 8 −20 Solution. We need to place the equation y = f(x) in vertex
form. We can simply compute a = −2, h = −b = 11 and
2a
4
k = f( 11 ) = 89 , using Fact 7.3.1:
4
8
f(x) = −2x2 + 11x − 4 −40 −60 = −2 x − −80 Figure 7.13: Sketching y =
f(x). 11
4 2 + 89
.
8 This means that the graph of f(x) is a parabola opening
downward with vertex 141 , 89 and axis x = 11 ; see Fig8
4
ure 7.13. 7.4 Quadratic Modeling Problems
The real importance of quadratic functions stems from the connection
with motion problems. Imagine one of the three kicked ball scenarios in
Figure 7.1 and impose a coordinate system with the kicker located at the
origin. We can study the motion of the ball in two ways:
• Regard time t as the important variable and try to ﬁnd a function
y(t) which describes the height of the ball t seconds after the ball
is kicked; this would just be the ycoordinate of the ball at time t.
The function y(t) is a quadratic function. If we had this function in
hand, we could determine when the ball hits the ground by solving
the equation 0 = y(t), but we would not be able to determine where
the ball hits the ground.
• A second approach is to forget about the time variable and simply
try to ﬁnd a function y = f(x) whose graph models the exact path of
the ball. In particular, we could ﬁnd where the ball hits the ground
by solving 0 = f(x), but we would not be able to determine when the
ball hits the ground. 7.4. QUADRATIC MODELING PROBLEMS 95 Example 7.4.1. Figure 7.14(a) shows a ball is located on
the edge of a cliff. The ball is kicked and its height (in feet)
above the level ground is given by the function s = y(t) =
−16t2 + 48t + 50, where t represents seconds elapsed after
kicking the ball. What is the maximum height of the ball
and when is this height achieved? When does the ball hit
the ground? How high is the cliff?
Solution. The function y(t) is a quadratic function with
a negative leading coefﬁcient, so its graph in the tscoordinate system will be a downward opening parabola.
We use a graphing device to get the picture in Figure 7.14(b).
The vertex is the highest point on the graph, which
can be found by writing y(t) in standard form using
Fact 7.3.1: path of kicked ball cliff
ground level (a) What it looks like
physically. 50 −2 −1 1 2 3 4 5 −50 −100 (b) What it looks like
graphically. y(t) = −16t2 + 48t + 50
3
= −16 t −
2 2 + 86. Figure 7.14: Different views
of the ball’s trajectory. 3
The vertex of the graph of y(t) is 2 , 86 , so the maximum
height of the ball above the level ground is 86 feet, occur3
ing at time t = 2 .
The ball hits the ground when its height above the ground is zero;
using the quadratic formula: y(t) = −16t2 + 48t + 50
=0
−48 ± (48)2 − 4 (−16) (50)
t=
2 · 16
= 3.818 sec or − 0.818 sec
Conclude the ball hits the ground after 3.818 seconds. Finally, the height
of the cliff is the height of the ball zero seconds after release; i.e., y(0) = 50
feet is the height of the cliff.
Here are two items to consider carefully: 2. The function y(t) is deﬁned for all t; however, in the context of the
problem, there is no physical meaning when t < 0. !!! 1. The graph of y(t) is NOT the path followed by the ball! Finding the
actual path of the ball is not possible unless additional information
is given. Can you see why? CAUTION
!!! CHAPTER 7. QUADRATIC MODELING 96 The next example illustrates how we must be very careful to link the
question being asked with an appropriate function.
Example 7.4.2. A hot air balloon takes off from the edge of a mountain
lake. Impose a coordinate system as pictured in Figure 7.15 and assume
2
that the path of the balloon follows the graph of y = f(x) = − 2500 x2 + 4 x. The
5
land rises at a constant incline from the lake at the rate of 2 vertical feet for
each 20 horizontal feet. What is the maximum height of the balloon above
lake level? What is the maximum height of the balloon above ground level?
Where does the balloon land on the ground? Where is the balloon 50 feet
above the ground?
Solution. In the coordinate system indicated, the origin is
the takeoff point and the graph of y = f(x) is the path of
balloon path
200
the balloon. Since f(x) is a quadratic function with a negative leading coefﬁcient, its graph will be a parabola which
100
opens downward. The difﬁculty with this problem is that
at any instant during the balloon’s ﬂight, the “height of
(ft)
lake
500
1000
the balloon above the ground” and the “height of the balFigure 7.15: Visualizing.
loon above the lake level” are different! The picture in
Figure 7.16 highlights this difference; consequently, two different functions will be needed to study these two different quantities. height above lake (ft) balloon height of balloon above lake level 200 A A
100 B
lake B
500 height of balloon above ground level
height of ground above lake level 1000 Figure 7.16: The height of the balloon y as a function of x. The function y = f(x) keeps track of the height of the balloon above
lake level at a given x location on the horizontal axis. The line ℓ with slope
1
2
m = 20 = 10 passing through the origin models the ground level. This says
that the function
1
y=
x
10
keeps track of the height of the ground above lake level at a given x
location on the horizontal axis.
We can determine the maximum height of the balloon above lake level
by analyzing the parabolic graph of y = f(x). Putting f(x) in vertex form,
via Fact 7.3.1,
2
(x − 500)2 + 200.
f(x) = −
2500 7.4. QUADRATIC MODELING PROBLEMS 97 The vertex of the graph of y = f(x) is (500, 200). This just tells us that the
maximum height of the balloon above lake level is 200 feet. To ﬁnd the
landing point, we need to solve the system of equations
2
4
y = − 2500 x2 + 5 x
1
y = 10 x . As usual, plugging the second equation into the ﬁrst and solving for x,
we get
1
x
10
x2
x2 − 875x
x(x − 875) 224
x+ x
2500
5
= 875x
=0
=0 =− height above lake level (feet) vertex (high point above lake)
200 100
landing point horizontal distance
from launch (feet)
takeoff point 500 1000 Figure 7.17: Locating the takeoff and landing points. From the algebra, we see there are two solutions: x = 0 or x = 875;
these correspond to the takeoff and landing points of the balloon, which
are the two places the ﬂight path and ground coincide. (Notice, if we had
divided out x from the last equation, we would only get one solution; the
tricky point is that we can’t divide by zero!) The balloon lands at the
position where x = 875 and to ﬁnd the y coordinate of this landing point
we plug x = 875 into our function for the balloon height above lake level:
y = f(875) = 87.5 feet. So, the landing point has coordinates (875,87.5).
Next, we want to study the height of the balloon above the ground. Let
y = g(x) be the function which represents the height of the balloon above CHAPTER 7. QUADRATIC MODELING 98 the ground when the horizontal coordinate is x. We ﬁnd
height of the balloon
above lake level with
horizontal coordinate x g(x) = − elevation of ground
above lake level with
horizontal coordinate x = f(x) − g(x)
= − 224
x+ x
2500
5 (balloon above lake level) =− 1
x
10 −
minus (ground above lake level) 2
(x − 437.5)2 + 153.12,
2500 Notice that g(x) itself is a NEW quadratic function with a negative leading
coefﬁcient, so the graph of y = g(x) will be a downward opening parabola.
The vertex of this parabola will be (437.5, 153.12), so the highest elevation
of the balloon above the ground is 153.12 feet.
We can now sketch the graph of g(x) and the horizontal line determined by y = 50 in a common coordinate system, as below. Finding
where the balloon is 50 feet above the ground amounts to ﬁnding where
these two graphs intersect. We need to now solve the system of equations
2
y = − 2500 x2 +
y = 50 7
x
10 . Plug the second equation into the ﬁrst and apply the quadratic formula to get x = 796.54 or 78.46. This tells us the two possible x coordinates when the balloon is 50 feet above the ground. In terms of the
original coordinate system imposed, the two places where the balloon is
50 feet above the ground are (78.46, 57.85) and (796.54, 129.6). 7.4.1 How many points determine a parabola?
We all recall from elementary geometry that two distinct points in the
plane will uniquely determine a line; in fact, we used this to derive equations for lines in the plane. We could then ask if there is a similar characterization of parabolas.
Important Fact 7.4.3. Let P = (x1 , y1), Q = (x2 , y2 ) and R = (x3 , y3 ) be three
distinct noncollinear points in the plane such that the xcoordinates are all
different. Then there exists a unique standard parabola passing through
these three points. This parabola is the graph of a quadratic function y =
f(x) = ax2 + bx + c and we can ﬁnd these coefﬁcients by simultaneously 7.4. QUADRATIC MODELING PROBLEMS 99 vertex (high point above ground) graph of height above ground function
100 line y = 50 500 1000 places where 50 feet above ground Figure 7.18: Finding heights above the ground. solving the system of three equations and three unknowns obtained by
assuming P, Q and R are points on the graph of y = f(x): ax2 + bx1 + c = y1 1 2
ax2 + bx2 + c = y2
. ax2 + bx3 + c = y3 .
3 Example 7.4.4. Assume the value of a particular house in Seattle has
increased in value according to a quadratic function y = v(x), where the
units of y are in dollars and x represents the number of years the property
has been owned. Suppose the house was purchased on January 1, 1970
and valued at $50,000. In 1980, the value of the house on January 1
was $80,000. Finally, on January 1, 1990 the value was $200,000. Find
the value function v(x), determine the value on January 1, 1996 and ﬁnd
when the house will be valued at $1,000,000.
Solution. The goal is to explicitly ﬁnd the value of the function y = v(x).
We are going to work in a xycoordinate 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
xvariable, which represents time, will denote the number of years the
house is owned. For the yvariable, which represents value, we could
use dollars. But, instead, we will follow a typical practice in real estate
and use the units of K, where K = $1,000. For example, a house valued CHAPTER 7. QUADRATIC MODELING 100 at $235,600 would be worth 235.6 K. These will be the units we use, which
essentially saves us from drowning in a sea of zeros!
We are given three pieces of information about the value of a particular
house. This leads to three points in our coordinate system: P = (0, 50),
Q = (10, 80) and R = (20, 200). If we plot these points, they do not lie
on a common line, so we know there is a unique quadratic function
v(x) = ax2 + bx + c whose graph (which will be a parabola) passes through
these three points. In order to ﬁnd the coefﬁcients a, b, and c, we need
to solve the system of equations: = 50 a02 + b0 + c
a(10)2 + b(10) + c = 80
, a(20)2 + b(20) + c = 200 which is equivalent to c
100a + 10b + c 400a + 20b + c the system = 50 = 80
. = 200 Plugging c = 50 into the second two equations gives the system
100a + 10b = 30
400a + 20b = 150 (7.1) Solve the ﬁrst equation for a, obtaining a =
the second equation to get:
400 30−10b
,
100 then plug this into 30 − 10b
+ 20b = 150
100
120 − 40b + 20b = 150
3
b=− .
2 3
Now, plug b = − 2 into the ﬁrst equation of Equation 7.1 to get 100a +
9
10 − 3 = 30; i.e., a = 20 . We conclude that
2 92 3
x − x + 50,
20
2
keeping in mind the units here are K.
To ﬁnd the value of the house on January 1, 1996, we simply note this
9
is after x = 26 years of ownership. Plugging in, we get y = v(26) = 20 (26)2 −
3
26 + 50 = 315.2; i.e., the value of the house is $315,200. To ﬁnd when
2
the house will be worth $1,000,000, we note that $1,000,000 = 1,000 K and
need to solve the equation
y = v(x) = 92
x−
20
92
x−
0=
20 1000 = v(x) = 3
x + 50
2
3
x − 950.
2 7.5. WHAT’S NEEDED TO BUILD A QUADRATIC MODEL? 101 By the quadratic formula,
x= 3
2 −3
2 ± √ 2 −4 2 9
20 9
20 (−950) 1712.25
0.9
= 47.64 or − 44.31.
= 1.5 ± Because x represents time, we can ignore the negative solution and so
the value of the house will be $1,000,000 after approximately 47.64 years
of ownership. 7.5 What’s Needed to Build a Quadratic Model?
Back in Fact 4.7.1 on page 44, we highlighted the information required to
determine a linear model. We now describe the quadratic model analog.
Important Facts 7.5.1. A quadratic model is completely determined by:
1. Three distinct 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 f(x) has the form
f(x) = a(x − h)2 + k,
where (h, k) is the vertex. If we are given h and k, together with another
point (x0 , y0 ) on the graph, then plugging in gives this equation:
y0 = a(x0 − h)2 + k.
The only unknown in this equation is a, which we can solve for using
algebra. A couple of the exercises will depend upon these observations. 7.6 Summary
• A quadratic function is one of the form
f(x) = ax2 + bx + c.
where a = 0. CHAPTER 7. QUADRATIC MODELING 102 • The graph of a quadratic function is a parabola which is symmetric
about the vertical line through the highest (or lowest) point on the
graph. This highest (or lowest) point is known as the vertex of the
graph; its location is given by (h,k) where
h=− b
and k = f(h).
2a • If a > 0, then the vertex is the lowest (or minimum) point on the
graph, and the parabola ”opens upward”. If a < 0, then the vertex
is the highest (or maximum) point on the graph, and the parabola
”opens downward”.
• Every quadratic function can be expressed in the form
f(x) = a(x − h)2 + k
where (h,k) is the vertex of the function’s graph. 7.7. EXERCISES 103 7.7 Exercises
Problem 7.1. Write the following quadratic
functions in vertex form, ﬁnd the vertex, the
axis of symmetry and sketch a rough graph.
(a) f(x) = 2x2 − 16x + 41.
(b) f(x) = 3x2 − 15x − 77.
(c) f(x) = x2 − 3 x + 13.
7
(d) f(x) = 2x2 .
(e) f(x) = 1
2
100 x . Problem 7.2. In each case, ﬁnd a quadratic
function whose graph passes through the
given points:
(a) (0,0), (1,1) and (3, − 1).
(b) (−1,1), (1, − 2) and (3,4).
(c) (2,1), (3,2) and (5,1).
(d) (0,1), (1,1) and (1,3).
Problem 7.3.
(a) Sketch the graph of the
function f(x) = x2 − 3x + 4 on the interval
−3 ≤ x ≤ 5. What is the maximum value
of f(x) on that interval? What is the minimum value of f(x) on that interval?
(b) Sketch the graph of the function f(x) =
x2 − 3x + 4 on the interval 2 ≤ x ≤ 7. What
is the maximum value of f(x) on that interval? What is the minimum value of
f(x) on that interval?
(c) Sketch the graph of the function g(x) =
−(x + 3)2 + 3 on the interval 0 ≤ x ≤ 4.
What is the maximum value of g(x) on
that interval? What is the minimum
value of g(x) on that interval?
Problem 7.4. If the graph of the quadratic
function f(x) = x2 + dx + 3d has its vertex on
the xaxis, what are the possible values of d?
What if f(x) = x2 + 3dx − d2 + 1 ?
Problem 7.5. The initial price of buzz.com
stock is $10 per share. After 20 days the stock
price is $20 per share and after 40 days the
price is $25 per share. Assume that while the
price of the stock is not zero it can be modeled
by a quadratic function. (a) Find the multipart function s(t) giving
the stock price after t days. If you buy
1000 shares after 30 days, what is the
cost?
(b) To maximize proﬁt, when should you sell
shares? How much will the proﬁt be on
your 1000 shares purchased in (a)?
Problem
7.6. Sketch
the
graph
of
y = x2 − 2x − 3. Label the coordinates of the
x and y intercepts of the graph.
In the
same coordinate system, sketch the graph of
y = x2 − 2x − 3, give the multipart rule and
label the x and y intercepts of the graph.
Problem 7.7. A hot air balloon takes off from
the edge of a plateau. Impose a coordinate system as pictured below and assume that the
path the balloon follows is the graph of the
4
quadratic function y = f(x) = − 2500 x2 + 4 x.
5
The land drops at a constant incline from the
plateau at the rate of 1 vertical foot for each
5 horizontal feet. Answer the following questions:
height above plateau (feet)
balloon takeoff horizontal distance
from launch (feet)
ground incline (a) What is the maximum height of the balloon above plateau level?
(b) What is the maximum height of the balloon above ground level?
(c) Where does the balloon land on the
ground?
(d) Where is the balloon 50 feet above the
ground?
Problem 7.8.
(a) Suppose f(x) = 3x2 − 2.
Does the point (1,2) lie on the graph of
y = f(x)? Why or why not?
(b) If b is a constant, where does the line
y = 1 + 2b intersect the graph of y =
x2 + bx + b? 104
(c) If a is a constant, where does the
line y = 1 − a2 intersect the graph of
y = x2 − 2ax + 1?
(d) Where does the graph of y = −2x2 + 3x + 10
intersect the graph of y = x2 + x − 10?
Problem 7.9. Sylvia has an apple orchard.
One season, her 100 trees yielded 140 apples
per tree. She wants to increase her production
by adding more trees to the orchard. However,
she knows that for every 10 additional trees
she plants, she will lose 4 apples per tree (i.e.,
the yield per tree will decrease by 4 apples).
How many trees should she have in the orchard to maximize her production of apples?
Problem 7.10. Rosalie is organizing a circus
performance to raise money for a charity. She
is trying to decide how much to charge for tickets. From past experience, she knows that the
number of people who will attend is a linear
function of the price per ticket. If she charges
5 dollars, 1200 people will attend. If she
charges 7 dollars, 970 people will attend. How
much should she charge per ticket to make the
most money?
Problem 7.11. A Norman window is a rectangle with a semicircle on top. Suppose that the
perimeter of a particular Norman window is to
be 24 feet. What should its dimensions be in
order to maximize the area of the window and,
therefore, allow in as much light as possible?
Problem 7.12. Jun has 300 meters of fencing to make a rectangular enclosure. She also
wants to use some fencing to split the enclosure into two parts with a fence parallel to two
of the sides. What dimensions should the enclosure have to have the maximum possible
area?
Problem 7.13. You have $6000 with which
to build a rectangular enclosure with fencing.
The fencing material costs $20 per meter. You
also want to have two parititions across the
width of the enclosure, so that there will be
three separated spaces in the enclosure. The
material for the partitions costs $15 per meter.
What is the maximum area you can achieve for
the enclosure?
Problem 7.14. Steve likes to entertain friends
at parties with “wire tricks.” Suppose he takes CHAPTER 7. QUADRATIC MODELING
a piece of wire 60 inches long and cuts it into
two pieces. Steve takes the ﬁrst piece of wire
and bends it into the shape of a perfect circle.
He then proceeds to bend the second piece of
wire into the shape of a perfect square. Where
should Steve cut the wire so that the total area
of the circle and square combined is as small
as possible? What is this minimal area? What
should Steve do if he wants the combined area
to be as large as possible?
Problem 7.15. Two particles are moving in the
xyplane. The move along straight lines at constant speed. At time t, particle A’s position is
given by
x = t + 2, y = 1
t−3
2 and particle B’s position is given by
1
x = 12 − 2t, y = 6 − t.
3
(a) Find the equation of the line along which
particle A moves. Sketch this line, and
label A’s starting point and direction of
motion.
(b) Find the equation of the line along which
particle B moves. Sketch this line on the
same axes, and label B’s starting point
and direction of motion.
(c) Find the time (i.e., the value of t) at
which the distance between A and B is
minimal. Find the locations of particles
A and B at this time, and label them on
your graph.
Problem 7.16. Sven starts walking due south
at 5 feet per second from a point 120 feet north
of an intersection. At the same time Rudyard
starts walking due east at 4 feet per second
from a point 150 feet west of the intersection.
(a) Write an expression for the distance between Sven and Rudyard t seconds after
they start walking.
(b) When are Sven and Rudyard closest?
What is the minimum distance between
them? Problem 7.17. After a vigorous soccer match,
Tina and Michael decide to have a glass of 7.7. EXERCISES 105 their favorite refreshment. They each run in
a straight line along the indicated paths at a
speed of 10 ft/sec. Parametrize the motion of
Tina and Michael individually. Find when and
where Tina and Michael are closest to one another; also compute this minimum distance.
(200,300)
soy milk (−50,275) Find values of α that make this equation true
(your answer will involve x).
Problem 7.19. For each of the following equations, ﬁnd the value(s) of the constant α so that
the equation has exactly one solution, and determine the solution for each value.
(a) αx2 + x + 1 = 0
(b) x2 + αx + 1 = 0 beet juice (c) x2 + x + α = 0
(400,50)
Tina (d) x2 + αx + 4α + 1 = 0
Problem 7.20. (a) Solve for t Michael s = 2(t − 1)2 + 1
Problem 7.18. Consider the equation: αx2 +
2α2 x + 1 = 0. Find the values of x that make
this equation true (your answer will involve α). (b) Solve for x
y = x2 + 2x + 3 106 CHAPTER 7. QUADRATIC MODELING Chapter 8
Composition
A new home takes its shape from basic building materials and the skillful use of construction tools. Likewise, we can build new functions
from known functions through the application of analogous mathematical tools. There are ﬁve tools we want to develop: composition, reﬂection,
shifting, dilation, arithmetic. We will handle composition in this section,
then discuss the others in the following two sections.
1
To set the stage, let’s look at a simple botany experioxygen rate hr
1
ment. Imagine a plant growing under a particular steady
0.8
light source. Plants continually give off oxygen gas to
0.6
the environment at some rate; common units would be
0.4
liters/hour. If we leave this plant unbothered, we mea0.2
hours
sure that the plant puts out 1 liter/hour; so, the oxygen
2 4 6 8 10
output is a steady constant rate. However, if we apply a
ﬂash of high intensity green light at the time t = 1 and
(a) Flash at t = 1.
measure the oxygen output of the plant, we obtain the
1
plot in Figure 8.1(a).
oxygen rate hr
1
Using what we know from the previous section on
0.8
quadratic functions, we can check that a reasonable
0.6
model for the graph is this multipart function f(t) (on the
0.4
domain 0 ≤ t ≤ 10):
0.2
2 1 if t ≤ 1
22
8
t − 3 t + 3 if 1 ≤ t ≤ 3
f( t ) =
3
1
if 3 ≤ t 4 (b) Flash at t = 5.
Figure 8.1: Light ﬂashes. Suppose we want to model the oxygen consumption
when a green light pulse occurs at time t = 5 (instead of
time t = 1), what is the mathematical model? For starters,
it is pretty easy to believe that the graph for this new situation will look like the new graph in Figure 8.1(b).
But, can we somehow use the model f(t) in hand (the known function)
to build the model we want (the new function)? We will return in Exam107 6 hours
8 10 CHAPTER 8. COMPOSITION 108 ple 8.2.4 to see the answer is yes; ﬁrst, we need to develop the tool of
function composition. 8.1 The Formula for a Composition
The basic idea is to start with two functions f and g and produce a new
function called their composition. There are two basic steps in this process and we are going to focus on each separately. The ﬁrst step is fairly
mechanical, though perhaps somewhat unnatural. It involves combining
the formulas for the functions f and g together to get a new formula;
we will focus on that step in this subsection. The next step is of varying
complexity and involves analyzing how the domains and ranges of f and g
affect those of the composition; we defer that to the next subsection once
we have the mechanics down. in in out out
y x u the “f” function y = f(x) the “g” function x = g(u) in out “composed” function y = f(g(u)) Figure 8.2: Visualizing a composite function y = h(u) = f(g(u)). Here is a very common occurrence: We are handed a function y = f(x),
which means given an x value, the rule f(x) produces a new y value. In
addition, it may happen that the variable x is itself related to a third
variable u by some different function equation x = g(u). Given u, the
rule g(u) will produce a value of x; from this x we can use the rule f(x)
to produce a y value. In other words, we can regard y as a function
depending on the new independent variable u. It is important to know
the mechanics of working with this kind of setup. Abstractly, we have
just described a situation where we take two functions and build a new 8.1. THE FORMULA FOR A COMPOSITION 109 function which “composes” the original ones together; schematically the
situation looks like this:
Example 8.1.1. A pebble is tossed into a pond. The radius of the ﬁrst circular ripple is measured to increase at the constant rate of 2.3 ft/sec. What
is the area enclosed by the leading ripple after 6 seconds have elapsed?
How much time must elapse so that the area enclosed by the leading ripple
is 300 square feet? Solution. We know that an object tossed into a pond
leading ripple after t seconds
will generate a series of concentric ripples, which grow
leading ripple after 2 seconds
steadily larger. We are asked questions that relate the
area of the circular region bounded by the leading ripple
and time elapsed.
r = r(t)
leading ripple after 1 second
Let r denote the radius of the leading ripple after t secFigure 8.3: Concentric riponds; units of feet. The area A of a disc bounded by a
ples.
2
leading ripple will be A = A(r) = πr . This exhibits A as a
function in the variable r. However, the radius is changing with respect
to time:
r = r(t) = radius after t seconds = 2.3 feet
t seconds = 2.3t feet.
sec So, r = r(t) is a function of t. In the expression A = A(r), replace “r” by
“r(t),” then
A = π(2.3t)2 = 5.29πt2.
The new function a(t) = 5.29πt2 gives a precise relationship between area
and time.
To answer our ﬁrst question, a(6) = 598.3 feet2 is the area of the region
bounded by the leading ripple after 6 seconds. On the other hand, if
a(t) = 300 ft2 ;, 300 = 5.29πt2, so t = ± 300/(5.29π) = ±4.25. Since t represents time, only the positive solution t = 4.25 seconds makes sense.
We can formalize the key idea used in solving this problem, which is
familiar from previous courses. Suppose that
y = f(x)
and that additionally the independent variable x is itself a function of a
different independent variable t; i.e.,
x = g(t ).
Then we can replace every occurrence of “x” in f(x) by the expression
“g(t),” thereby obtaining y as a function in the independent variable t.
We usually denote this new function of t:
y = f(g(t)). CHAPTER 8. COMPOSITION 110 We refer to f(g(t)) as the composition of f and g or the composite function.
The process of forming the composition of two functions is a mechanical procedure. If you are handed the actual formulas for y = f(x) and
x = g(t), then Procedure 8.1.2 is what you need.
Important Procedure 8.1.2. To obtain the formula for f(g(t)), replace every occurrence of “x” in f(x) by the expression “g(t).”
Here are some examples of how to do this:
Examples 8.1.3. Use the composition procedure in each of these cases.
(i) If y = f(x) = 2 and x = g(t) = 2t, then f(g(t)) = f(2t) = 2.
(ii) If y = 3x − 7 and x = g(t) = 4, then f(g(t)) = f(4) = 3 · 4 − 7 = 5.
(iii) If y = f(x) = x2 + 1 and x = g(t) = 2t − 1, then
f(g(t)) = f(2t − 1)
= (2t − 1)2 + 1
= 4t2 − 4t + 2.
(iv) If y = f(x) = 2 + 1 + (x − 3)2 and x = g(t) = 2t2 − 1, then f(g(t)) = f(2t2 − 1)
= 2+ 1 + (2t2 − 1 − 3)2 = 2+ 4t4 − 16t2 + 17. (v) If y = f(x) = x2 and x = g(t) = t + ♥, then
f(g(t)) = f(t + ♥)
= (t + ♥)2
= t2 + 2t♥ + ♥2 .
It is natural to ask: What good is this whole business about compositions? One way to think of it is that we can use composite functions to
break complicated functions into simpler parts. For example,
y = h(x) = x2 + 1 √
can be written as the composition f(g(x)), where y = f(z) = z and z =
g(x) = x2 + 1. Each of the functions f and g is “simpler” than the original h,
which can help when studying h.
Examples 8.1.4. Here we use composite functions to “simplify” a given
function. 8.1. THE FORMULA FOR A COMPOSITION 111 (i) The function y = (x−31)2 +2 can be written as a composition y = f(g(x)),
where y = f(z) = z21 2 and z = g(x) = x − 3.
+
(ii) The upper semicircle of radius 2 centered at (1,2) is the graph of the
function y = 2 + 4 − (x − 1)2 . This function can be written as a com√
position y = f(g(x)), where y = f(z) = 2 + 4 − z2 and z = g(x) = x − 1. 8.1.1 Some notational confusion
In our discussion above, we have used different letters to represent the
domain variables of two functions we are composing. Typically, we have
been writing: If y = f(x) and x = g(t), then y = f(g(t)) is the composition.
This illustrates that the three variables t, x, and y can all be of different
types. For example, t might represent time, x could be speed and y could
be distance.
If we are given two functions that involve the same independent variable, like f(x) = x2 and g(x) = 2x + 1, then we can still form a new function
f(g(x)) by following the same prescription as in Procedure 8.1.2:
Important Procedure 8.1.5. To obtain the formula for f(g(x)), replace
every occurrence of “x” in f(x) by the expression “g(x).”
For our example, this gives us:
f(g(x)) = f(2x + 1) = (2x + 1)2 .
Here are three other examples:
√
√
• If f(x) = x, g(x) = 2x2 + 1, then f(g(x)) = 2x2 + 1.
• If f(x) = 1 , g(x) = 2x + 1, then f(g(x)) =
x 1
.
2x+1 • If f(x) = x2 , g(x) = △ − x, then f(g(x)) = △2 − 2x△ + x2 . 112 CHAPTER 8. COMPOSITION yaxis Example 8.1.6. Let f(x) = x2 , g(x) = x + 1 and h(x) = x − 1.
Find the formulas for f(g(x)), g(f(x)), f(h(x)) and h(f(x)).
Discuss the relationship between the graphs of these four
functions. 10
8
6
4
2 xaxis
−3 −2 −1 1 2 3 Figure 8.4: Sketching composite functions. Solution. If we apply Procedure 8.1.5, we obtain the composition formulas. The four graphs are given on the domain −3 ≤ x ≤ 3, together with the graph of f(x) = x2 .
f(g(x))
g(f(x))
f(h(x))
h(f(x)) =
=
=
= f( x + 1) = ( x + 1) 2
g(x2 ) = x2 + 1
f( x − 1) = ( x − 1) 2
h(x2 ) = x2 − 1. We can identify each graph by looking at its vertex:
• f(x) has vertex (0,0)
• f(g(x)) has vertex (1,0)
• g(f(x)) has vertex (0,1)
• f(h(x)) has vertex (1,0)
• h(f(x)) has vertex (0,1)
Horizontal or vertical shifting of the graph of f(x) = x2 gives the other four
graphs: See Figure 8.4. 8.2 Domain, Range, etc. for a Composition
A function is a “package” consisting of a rule, a domain of allowed input
values, and a range of output values. When we start to compose functions, we sometimes need to worry about how the domains and ranges
of the composing functions affect the composed function. First off, when
you form the composition f(g(x)) of f(x) and g(x), the range values for
g(x) must lie within the domain values for f(x). This may require that
you modify the range values of g(x) by changing its domain. The domain
values for f(g(x)) will be the domain values for g(x). 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION in domain g out in the “g” function 113 out the “f” function range of f(g(x)) range of g
domain of f Figure 8.5: What is the domain and range of a composite function? In practical terms, here is how one deals with the domain issues for a
composition. This is a reﬁnement of Procedure 8.1.5 on page 111.
Important Procedure 8.2.1. To obtain the formula for f(g(x)), replace
every occurrence of “x” in f(x) by the expression “g(x).” In addition, if
there is a condition on the domain of f that involves x, then replace every
occurrence of “x” in that condition by the expression “g(x).”
The next example illustrates how to use this principle.
Example 8.2.2. Start with the function y = f(x) = x2 on the domain
−1 ≤ x ≤ 1. Find the rule and domain of y = f(g(x)), where g(x) = x − 1.
Solution. We can apply the ﬁrst statement in Procedure 8.2.1 to ﬁnd the
rule for y = f(g(x)):
y = f(g(x))
= f( x − 1) = ( x − 1) 2
= x2 − 2x + 1.
To ﬁnd the domain of y = f(g(x)), we apply the second statement in
Procedure 8.2.1; this will require that we solve an inequality equation:
−1 ≤ g(x) ≤ 1
−1 ≤ x − 1 ≤ 1
0≤
x
≤2
The conclusion is that y = f(g(x)) = x2 − 2x + 1 on the domain 0 ≤ x ≤ 2.
√
Example 8.2.3. Let y = f(z) = z, z = g(x) = x + 1. What is the largest
possible domain so that the composition f(g(x)) makes sense? CHAPTER 8. COMPOSITION 114
y z range z
domain (a) y = z = g ( x) = x + 1 y = f(z) √
z. desired range
−1 x
required domain (b) z = x = 1. Figure 8.6: Finding the largest domain for f(g(x)). Solution. The largest possible domain for y = f(z) will consist of all nonnegative real numbers; this is also the range of the function f(z): See
Figure 8.6(a).
To ﬁnd the largest domain for the composition, we try to ﬁnd a domain
of xvalues so that the range of z = g(x) is the domain of y = f(z). So, in
this case, we want the range of g(x) to be all nonnegative real numbers,
denoted 0 ≤ z. We graph z = g(x) in the xzplane, mark the desired range
0 ≤ z on the vertical zaxis, then determine which xvalues would lead to
points on the graph with second coordinates in this zone. We ﬁnd that
the domain of all xvalues greater or equal to −1 (denoted −1 ≤ x)√
leads
to the desired range. In summary, the composition y = f(g(x)) = x + 1
is deﬁned on the domain of xvalues −1 ≤ x.
Let’s return to the botany experiment that opened this section and see
how composition of functions can be applied to the situation. Recall that
plants continually give off oxygen gas to the environment at some rate;
common units would be liters/hour.
Example 8.2.4. A plant is growing under a particular steady light source.
If we apply a ﬂash of high intensity green light at the time t = 1 and
measure the oxygen output of the plant, we obtain the plot below and the
mathematical model f(t). if t ≤ 1
1
22
8
f( t ) =
t − 3 t + 3 if 1 ≤ t ≤ 3
3
1
if 3 ≤ t Now, suppose instead we apply the ﬂash of high intensity green light at
the time t = 5. Verify that the mathematical model for this experiment is
given by f(g(t)), where g(t) = t − 4. 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION Solution. Our expectation is that the plot for this new experiment will have the “parabolic dip” shifted over to occur starting at time t = 5 instead of at time t = 1. In other
words, we expect the graph in Figure 8.7(b).
Our job is to verify that this graph is obtained from the
function f(g(t)), where g(t) = t − 4. This is a new terrain
for us, since we need to look at a composition involving a
multipart function. Here is how to proceed: When we are
calculating a composition involving a multipart function,
we need to look at each of the parts separately, so there
will be three cases to consider:
First part: f(t) = 1 when t ≤ 1. To get the formula for
f(g(t)), we now appeal to Procedure 8.2.1 and just replace
every occurrence of t in f(t) by g(t). That gives us this
NEW domain condition and function equation: 115 oxygen rate
1
0.8
0.6
0.4
0.2 hours
2 22
t
3 8
t
3 4 6 8 10 (a) Flash at t = 1.
oxygen rate
1 1 hr 0.8
0.6
0.4
0.2 hours
2 f(g(t)) = f(t − 4) = 1 when t − 4 ≤ 1
= 1 when t ≤ 5. 1 hr 4 6 8 10 (b) Flash at t = 5.
Figure 8.7: Applying light at
time t. Second part: f(t) =
−
+ 3 when 1 ≤ t ≤ 3. We
now appeal to Procedure 8.2.1 and just replace every occurrence of t in this function by g(t). That gives us this NEW domain
condition and function equation:
8
2
f(g(t)) = f(t − 4) = (t − 4)2 − (t − 4) + 3 when 1 ≤ t − 4 ≤ 3
3
3
22
73
=
t − 8t +
when 5 ≤ t ≤ 7.
3
3
Third part: f(t) = 1 when 3 ≤ t. We now appeal to Procedure 8.2.1 and
just replace every occurrence of t in this function by g(t). That gives us
this NEW domain condition and function equation:
f(g(t)) = f(t − 4) = 1 when 3 ≤ t − 4
= 1 when 7 ≤ t.
The multipart rule for this composition can now be written down and
using a graphing device you can verify its graph is the model for our
experiment. if t ≤ 5
1
22
t − 8t + 73 if 5 ≤ t ≤ 7
f(g(t)) =
3
3
1
if 7 ≤ t CHAPTER 8. COMPOSITION 116 8.3 Exercises
Problem 8.1. For this problem, f(t) = t − 1,
g(t) = −t − 1 and h(t) = t.
(a) Compute the multipart rules for h(f(t))
and h(g(t)) and sketch their graphs.
(b) Compute the multipart rules for f(h(t))
and g(h(t)) and sketch their graphs. Problem 8.5. A car leaves Seattle heading
east. The speed of the car in mph after m minutes is given by the function
C(m) = 70m2
.
10 + m2 (c) Compute the multipart rule for h(h(t)−1)
and sketch the graph. (a) Find a function m = f(s) that converts
seconds s into minutes m. Write out
the formula for the new function C(f(s));
what does this function calculate? Problem 8.2. Write each of the following functions as a composition of two simpler functions: (There is more than one correct answer.) (b) Find a function m = g(h) that converts
hours h into minutes m. Write out the
formula for the new function C(g(h));
what does this function calculate? (a) y = (x − 11)5 .
√
(b) y = 3 1 + x2 .
(c) y = 2(x − 3)5 − 5(x − 3)2 + 1 (x − 3) + 11.
2
(d) y =
(e) y = (c) Find a function z = v(s) that converts
mph s into ft/sec z. Write out the formula for the new function v(C(m); what
does this function calculate? 1
.
x2 +3 √
x + 1. (f) y = 2 − 5 − (3x − 1)2 . Problem 8.3.
(a) Let f(x) be a linear function, f(x) = ax + b for constants a and b.
Show that f(f(x)) is a linear function.
(b) Find a function g(x) such that g(g(x)) =
6x − 8.
Problem 8.4. Let f(x) = 1 x + 3.
2
(a) Sketch the graphs of f(x),f(f(x)),f(f(f(x)))
on the interval −2 ≤ x ≤ 10.
(b) Your graphs should all intersect at the
point (6,6). The value x = 6 is called
a ﬁxed point of the function f(x) since
f(6) = 6; that is, 6 is ﬁxed  it doesn’t
move when f is applied to it. Give an explanation for why 6 is a ﬁxed point for
any function f(f(f(...f(x)...))).
(c) Linear functions (with the exception of
f(x) = x) can have at most one ﬁxed
point. Quadratic functions can have at
most two. Find the ﬁxed points of the
function g(x) = x2 − 2.
(d) Give a quadratic function whose ﬁxed
points are x = −2 and x = 3. Problem 8.6. Compute the compositions
f(g(x)), f(f(x)) and g(f(x)) in each case:
(a) f(x) = x2 , g(x) = x + 3.
√
(b) f(x) = 1/x, g(x) = x.
(c) f(x) = 9x + 2, g(x) = 1 (x − 2).
9
(d) f(x) = 6x2 + 5, g(x) = x − 4.
√
(e) f(x) = 4x3 − 3, g(x) = 3 2x + 6
(f) f(x) = 2x + 1, g(x) = x3 .
(g) f(x) = 3, g(x) = 4x2 + 2x + 1.
(h) f(x) = −4, g(x) = 0.
√
4 − z2 and
Problem 8.7. Let y = f(z) =
z = g(x) = 2x + 3. Compute the composition
y = f(g(x)). Find the largest possible domain
of xvalues so that the composition y = f(g(x))
is deﬁned.
Problem 8.8. Suppose you have a function
y = f(x) such that the domain of f(x) is
1 ≤ x ≤ 6 and the range of f(x) is −3 ≤ y ≤ 5.
(a) What is the domain of f(2(x − 3)) ? (b) What is the range of f(2(x − 3)) ?
(c) What is the domain of 2f(x) − 3 ?
(d) What is the range of 2f(x) − 3 ?
(e) Can you ﬁnd constants B and C so that
the domain of f(B(x − C)) is 8 ≤ x ≤ 9? 8.3. EXERCISES
(f) Can you ﬁnd constants A and D so that
the range of Af(x) + D is 0 ≤ y ≤ 1? 117
this simpliﬁed expression:
f(x + h) − f(x)
.
h
(a) f(x) = Problem 8.9. For each of the given functions
y = f(x), simplify the following expression so
that h is no longer a factor in the denominator, then calculate the result of setting h = 0 in 1
x−1 . (b) f(x) = (2x + 1)2 .
√
(c) f(x) = 25 − x2 . 118 CHAPTER 8. COMPOSITION Chapter 9
Inverse Functions
The experimental sciences are loaded with examples of functions relating time and some measured quantity. In this case, time represents our
“input” and the quantity we are measuring is the “output.” For example,
maybe you have just mixed together some chemical reactants in a vessel. As time goes by, you measure the fraction of reactants remaining,
tabulate your results, then sketch a graph as indicated in Figure 9.1.
Viewing the input value as “time” and the output value
fraction
1
as “fraction of product,” we could ﬁnd a function y = f(t)
modeling this data. Using this function, you can easily
compute the fraction of reactants remaining at any time
in the future. However, it is probably just as interesting
to know how to predict the time when a given fraction
0
time
of reactants exists. In other words, we would like a new
Figure 9.1: Fraction of reacfunction that allows us to input a “fraction of reactants”
tants as a function of time.
and get out the “time” when this occurs. This “reverses”
the input/output roles in the original function. Is there a systematic way
to ﬁnd the new function if we know y = f(t)? The answer is yes and
depends upon the general theory of inverse functions. 9.1 Concept of an Inverse Function
Suppose you are asked to solve the following three equations for x. How
do you proceed?
(x + 2) = 64
(x + 2)2 = 64
(x + 2)3 = 64.
In the ﬁrst equation, you add “−2” to each side, then obtain x = 62. In
the third equation, you take the cube root of both sides of the equation,
giving you x + 2 = 4, then subtract 2 getting x = 2. In second equation,
you take a square root of both sides, BUT you need to remember both the
119 CHAPTER 9. INVERSE FUNCTIONS 120 positive and negative results when doing this. So, you are reduced down
to x + 2 = ±8 or that x = −10 or 6. Why is it that in two of these cases you
obtain a single solution, while in the remaining case there are two different answers? We need to sort this out, since the underlying ideas will
surface when we address the inverse circular functions in Chapter 20.
Let’s recall the conceptual idea of a function: A function is a process which takes a number x and outputs a
In
Out
x
f ( x)
new number f(x). So far, we’ve only worked with this process from “left to right;” i.e., given x, we simply put it into
The function
a symbolic rule and out pops a new number f(x). This is
Figure 9.2: A function as a
all pretty mechanical and straightforward.
process. 9.1.1 An Example
Let’s schematically interpret what happens for the speciﬁc concrete example y = f(x) = 3x − 1, when x = −1, − 1 , 0, 1 , 1, 2: See Figure 9.3.
2
2 out in
−1 in −5
2 1 −1 2 out
3x − 1 in
0 −4 3x − 1 1
−2 in
1
2 1
2 3x − 1 in out
3x − 1 out out
3x − 1 in 2 out
3x − 1 5 Figure 9.3: Function process y = 3x − 1. We could try to understand the function process in this example in
“reverse order,” going “right to left;” namely, you might ask what x value
can be run through the process so you end up with the number 11? This
is somewhat like the “Jeopardy” game show: You know what the answer
is, you want to ﬁnd the question. For our example, if we start out with
1
some given y values, then we can deﬁne a “reverse process” x = 3 (y + 1),
which returns the x value required so that f(x) = y: See Figure 9.4. 9.1. CONCEPT OF AN INVERSE FUNCTION
placements
(y+1)
3 −4 −1
2 (y+1)
3 0 (y+1)
3 −1 121 1
2 (y+1)
3 1
2 −5
2 1 (y+1)
3 2 −1 2 (y+1)
3 5 Figure 9.4: The reverse process x = 1 (y + 1).
3 9.1.2 A Second Example
If we begin with a linear function y = f(x) = mx + b, where m = 0, then we
can always ﬁnd a “reverse process” for the function. To ﬁnd it, you must
solve the equation y = f(x) for x in terms of y:
y = mx + b
y − b = mx
1
(y − b) = x
m
So, if m = 3 and b = −1, we just have the ﬁrst example above.
For another example, suppose y = −0.8x + 2; then m = −0.8 and b = 2.
In this case, the reverse process is −1.25(y − 2) = x. If we are given the
value y = 11, we simply compute that x = −11.25; i.e., f(−11.25) = 11. 9.1.3 A Third Example
The previous examples hide a subtle point that can arise when we try to
understand the “reverse process” for a given function. Suppose we begin
with the function y = f(x) = (x − 1)2 + 1. Figure 9.5 is a schematic of how
1
1
the function works when we plug in x = −1, − 2 , 0, 2 , 1, 2; what is being
illustrated is a “forward process”, in that each input generates a unique
output.
For this example, if we start out with some given y values, then we can
try to deﬁne a “reverse process” x = ??? which returns an x value required
so that f(x) = y. Unfortunately, there is no way to obtain a single formula
for this reverse process; Figure 9.6(a) shows what happens if you are
given y = 3 and you try to solve for x. CHAPTER 9. INVERSE FUNCTIONS 122 in (x − 1) + 1 −1 in
2 5 in (x − 1)2 + 1 in
13
4 2 2 (x − 1)2 + 1 1 out (x − 1)2 + 1 0 1
2 out (x − 1) + 1 −1
2 in out
2 in (x − 1)2 + 1 out
5
4 out 1 out
2 Figure 9.5: Function process y = f(x) = (x − 1)2 + 1. The conclusion is that the “reverse process” has two outputs. This
violates the rules required for a function, so this is NOT a function. The
solution is to create two new “reverse processes.”
Each of these “reverse processes” has a unique output;
√
1+ 2
reverse
in other words, each of these “reverse processes” deﬁnes
process
a function.
√
x= − y−1
3
+
So, given y √ 3, there are √
=
TWO possible x √
values,
√
1− 2
namely x = 1 ± 2, so that f(1 + 2) = 3 and f(1 − 2) = 3.
In other words, the reverse process is not given by a single
(a) Reverse process but not a
equation; there are TWO POSSIBLE reverse processes.
function.
reverse
process
1+ √
2 x =+ √ y−1 3 reverse
process
1− √
2 x =− √ y−1 3 (b) Two new reverse processes
that are functions.
Figure 9.6: What to do if a
reverse process is not a function. 9.2 Graphical Idea of an Inverse
We have seen that ﬁnding inverses is related to solving
equations. However so far, the discussion has been symbolic; we have pushed around a few equations and in
the end generated some confusion. Let’s use the tools of
Chapter 6 to visualize what is going on here. Suppose we
are given the graph of a function f(x) as in Figure 9.7(a).
What input x values result in an output value of 3? This
involves ﬁnding all x such that f(x) = 3. Graphically, this
means we are trying to ﬁnd points on the graph of f(x)
so that their ycoordinates are 3. The easiest way to to do
this is to draw the line y = 3 and ﬁnd where it intersects
the graph.
In Figure 9.7(b) we can see the points of intersection
are (−5, 3), (−1, 3), and (9, 3). That means that x = −5, −1,
9 produce the output value 3; i.e., f(−5) = f(−1) = f(9) = 3. 9.2. GRAPHICAL IDEA OF AN INVERSE 123 y = f ( x) y = f ( x) 5 5
y=3 −5 5 10 −5 (a) Given the function y = f(x). −5 5 10 −5 (b) What values of x give
f(x) = 3? Figure 9.7: Using the horizontal line y = 3 to ﬁnd values on the xaxis. This leads to our ﬁrst important fact about the “reverse process” for a
function:
Important Fact 9.2.1. Given a number c, the x values such that f(x) = c
can be found by ﬁnding the xcoordinates of the intersection points of the
graphs of y = f(x) and y = c.
Example 9.2.2. Graph y = f(x) = x2 and discuss the meaning of Fact 9.2.1
when c = 3, 1, 6. Solution. We graph y = x2 and the lines y = 1, y = 3 and
y = 6. Let’s use c = 6 as an example. We need to simultaneously solve the equations y = x2 and y = 6. Putting
√
these together, we get x2 = 6 or x = ± 6 ≈ ±2.449; i.e.,
f(±2.449) = 6. If c = 3, we get x = ±1.732; i.e., f(±1.732) = 3.
Finally, if c = 1, we get x = ±1; i.e., f(±1) = 1. y = x2
y=6
y=3
y=1 Figure 9.8: Graph of y = The pictures so far indicate another very important
f(x) = x2 .
piece of information. For any number c, we can tell exactly “how many” input x values lead to the same output value c, just by
counting the number of times the graphs of y = f(x) and y = c intersect.
Important Fact 9.2.3. For any function f(x) and any number c, the number of x values so that f(x) = c is the number of times the graphs of y = c
and y = f(x) intersect. CHAPTER 9. INVERSE FUNCTIONS 124
Examples 9.2.4. (i) If f(x) is a linear function f(x) = mx + b, m = 0, then the graph of
f(x) intersects a given horizontal line y = c EXACTLY once; i.e., the
equation c = f(x) always has a unique solution.
(ii) If f(x) = d is a constant function and c = d, then every input x value
in the domain leads to the output value c. On the other hand, if c = d,
then no input x value will lead to the output value c. For example, if
f(x) = 1 and c = 1, then every real number can be input to produce
an output of 1; if c = 2, then no input value of x will lead to an output
of 2. yaxis f(x) = mx + b yaxis y=c f ( x) = 1
xaxis xaxis
any of these inputs the only input which leads to an output of 1 leads to an output of c linear functions constant functions Figure 9.9: Does a horizontal line y = c intersect a curve once or more than once? 9.2.1 Onetoone Functions
For a speciﬁed domain, onetoone functions are functions with the property: Given any number c, there is at most one input x value in the domain so that f(x) = c. Among our examples thus far, linear functions
(degree 1 polynomials) are always onetoone. However, f(x) = x2 is not
onetoone; we’ve already seen that it can have two values for some of its
inverses. By Fact 9.2.3, we can quickly come up with what’s called the
horizontal line test.
Important Fact 9.2.5 (Horizontal Line Test). On a given domain of xvalues,
if the graph of some function f(x) has the property that every horizontal line
crosses the graph at most only once, then the function is onetoone on this
domain. 9.3. INVERSE FUNCTIONS 125 Example 9.2.6. By the horizontal line test, it is easy to
see that f(x) = x3 is onetoone on the domain of all real
numbers.
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 x values producing the
same output y value. In effect, this is saying that we can
deﬁne a “reverse process” for the function y = f(x) which
will also be a function; this is the key theme of the next
section. yaxis xaxis lines Figure 9.10: A onetoone
function f(x) = x3 . 9.3 Inverse Functions
Let’s now come face to face with the problem of ﬁnding the “reverse process” for a given function y = f(x). It is important to keep in mind that
the domain and range of the function will both play an important role
in this whole development. For example, Figure 9.11 shows the function
f(x) = x2 with three different domains speciﬁed and the corresponding
range values. range range
range domain domain horizontal domain Figure 9.11: Possible domains for a given range. These comments set the stage for a third important fact. Since the
domain and range of the function and its inverse rule are going to be
intimately related, we want to use notation that will highlight this fact.
We have been using the letters x and y for the domain (input) and range
(output) variables of f(x) and the “reverse process” is going to reverse
these roles. It then seems natural to simply write y (instead of c) for the
input values of the “reverse process” and x for its output values. CHAPTER 9. INVERSE FUNCTIONS 126 Important Fact 9.3.1. Suppose a function f(x) is onetoone on a domain
of x values. Then deﬁne a NEW FUNCTION by the rule
f−1 (y) = the x value so that f(x) = y.
The domain of y values for the function f−1 (y) is equal to the range of the
function f(x).
The rule deﬁned here is the “reverse process” for the given function. It
is referred to as the inverse function and we read f−1 (y) as “...eff inverse
of y...”.
!!!
CAUTION
!!! Both the “domain” of f(x) and the “rule” f(x) have equal inﬂuence on
whether the inverse rule is a function. Keep in mind, you do NOT get an
inverse function automatically from functions that are not onetoone! 9.3.1 Schematic Idea of an Inverse Function
Suppose that f(x) is onetoone, so that f−1 (y) is a function. As a result,
we can model f−1 (y) as a black box. What does it do? If we put in y in
the input side, we should get out the x such that f(x) = y.
Now, let’s try to unravel something very special that is
in
out
happening on a symbolic level. What would happen if we
f−1 (y)
plugged f(a) into the inverse function for some number a?
x such that
y
Then the inverse rule f−1 (f(a)) tells us that we want to
f ( x) = y
ﬁnd some x so that f(x) = f(a). But, we already know
x = a works and since f−1 (y) is a function (hence gives us
Figure 9.12: A new function
unique answers), the output of f−1 (f(a)) is just a. Symbolx = f−1 (y).
ically, this means we have Fact 9.3.2.
Important Fact 9.3.2. For every a value in the domain of f(x), we have
f−1 (f(a)) = a. (9.1) This is better shown in the black box picture of Figure 9.13. out in
a f ( x) in
f(a) out
f − 1 ( x) a Figure 9.13: Visualizing f−1 (f(a)) = a. A good way to get an idea of what an inverse function is doing is to
remember that f−1 (y) reverses the process of f(x). We can think of f−1 (y)
as a “black box” running f(x) backwards. 9.4. TRYING TO INVERT A NON ONETOONE FUNCTION 127 9.3.2 Graphing Inverse Functions
How can we get the graph of an inverse function? The idea is to manipulate the graph of our original onetoone function in some prescribed
way, ending up with the graph of f−1 (y). This isn’t as hard as it sounds,
but some confusion with the variables enters into play. Remember that
a typical point on the graph of a function y = f(x) looks like (x, f(x)). Now
let’s take a look at the inverse function x = f−1 (y). Given a number y in
the domain of f−1 (y), y = f(x) for some x in the domain of f(x); i.e., we are
using the fact that the domain of f−1 equals the range of f. The function
f−1 (y) takes the number f(x) and sends it to x, by Fact 9.3.2. So when
f(x) is the input value, x becomes the output value. Conclude a point on
the graph of f−1 (y) looks like (f(x), x). It’s similar to the graph of y = f(x),
only the x and y coordinates have reversed! What does that do to the
graph? Essentially, you reorient the picture so that the positive xaxis
and positive yaxis are interchanged. Figure 9.14 shows the process for
√
the function y = f(x) = x3 and its inverse function x = f−1 (y) = 3 y. We
place some ∗ symbols on the graph to help keep track of what is happening.
+y−axis
+x−axis *
*
+x−axis
*
rotate 900 * clockwise +y−axis
+x−axis ** +y−axis *
* ﬂip across
horiz axis Figure 9.14: Graphically ﬁnding x = f−1 (y) = * √
3 y. 9.4 Trying to Invert a Non onetoone Function
√
Suppose we blindly try to show that y is the inverse function for y = x2 ,
without worrying about all of this onetoone stuff. We’ll start out with
√
the number −7. If f−1 (y) = y, then we know that
f−1 (f(−7)) = f−1 (49) = 7.
On the other hand, the formula in Fact 9.3.2 tells us that we must have
f−1 (f(−7)) = −7,
√
so we have just shown 7 = −7! So clearly f−1 (y) = y. Even if we try
√
f−1 (y) = − y, we produce a contradiction. It seems that if you didn’t have CHAPTER 9. INVERSE FUNCTIONS 128 to worry about negative numbers, things would be all right. Then you
√
could say that f−1 (y) = y. Let’s try to see what this means graphically.
Let’s set f(x) = x2 , but only for nonnegative xvalues. That means
that we want to erase the graph to the left of the yaxis (so remember  no
negative xvalues allowed). The graph would then look like Figure 9.15.
+y
y = x2 +x
inverse function +x √ y +y domain nonnegative y domain nonnegative x Figure 9.15: Restricting the domain: No negative xvalues. This is a now a onetoone function! And now, one can see that its
√
inverse function is y. Similarly, we could have taken f(x) = x2 but only
√
for the nonpositive xvalues. In that case, f−1 (y) = − y. In effect, we
have split the graph of y = x2 into two parts, each of which is the graph
of a onetoone function; Figure 9.16.
+y
y = x2 +x
domain nonnegative y +x domain nonpositive x +y inverse function √
−y Figure 9.16: Restricting the domain: No positive xvalues. It is precisely this splitting into two cases that √
leads us to multiple
√
2
solutions of an equation like x = 5. We obtain x = 5 and x = − 5; one 9.5. SUMMARY 129 solution comes from the side of the graph to the left of the yaxis, and
the other from the right of the yaxis. This is because we have separate
inverse functions for the left and right side of the graph of y = x2 . 9.5 Summary
• Two functions f and g are inverses if
f(g(x)) = x and g(f(x)) = x
for all x in the domain of f and the domain of g.
• A function f is onetoone if every equation
f(x) = k
has at most one solution. If there is a value of k such that the
equation f(x) = k has more than one solution, then f is not onetoone.
• A function is onetoone if every horizontal line intersects the function’s graph at most once.
• A function has an inverse if the function is onetoone, and every
onetoone function has an inverse.
• The domain of a function is the range of its inverse, and the range
of a function is the domain of its inverse.
• The graph of a function and its inverse are mirror images of each
other across the line y = x. CHAPTER 9. INVERSE FUNCTIONS 130 9.6 Exercises
Problem 9.1. Let f(x) = 3x2 4 on the largest
−
domain for which the formula makes sense. A B (a) Find the domain and range of f(x), then
sketch the graph.
(b) Find the domain, range and rule for
the inverse function f−1 , then sketch its
graph. C
D Problem 9.2. Find the inverse function of
each of the following functions. Specify the domains of the inverse functions.
(a) f(x) = 1
x+8
5 5
x+3
√
(c) g(x) = 4 3 − x − 7
√
√
(d) j(x) = x + x − 1 (b) h(x) = (e) k(x) = 16 − x2 , 0 ≤ x ≤ 4 Problem 9.3. For this problem, y = f(x) =
2x2 − 3x − 1 on the domain of all real numbers.
(a) Sketch the function graph and ﬁnd the
coordinates of the vertex P = (a,b).
(b) Explain why y = f(x) does not have an
inverse function on the domain of all real
numbers.
(c) Restrict y = f(x) to the domain {a ≤
x} and ﬁnd the formula for the inverse
function f−1 (y). What are the domain
and range of the inverse function?
(d) Restrict y = f(x) to the domain {x ≤
a} and ﬁnd the formula for the inverse
function f−1 (y). What are the domain
and range of the inverse function? Problem 9.4. Which of the following graphs
are onetoone? If they are not onetoone, section the graph up into parts that are onetoone. Problem 9.5. Show that, for every value of a,
the function
f(x) = a + 1
x−a is its own inverse.
Problem 9.6. Clovis is standing at the edge
of a cliff, which slopes 4 feet downward from
him for every 1 horizontal foot. He launches
a small model rocket from where he is standing. With the origin of the coordinate system
located where he is standing, and the xaxis
extending horizontally, the path of the rocket
is described by the formula y = −2x2 + 120x.
(a) Give a function h = f(x) relating the
height h of the rocket above the sloping
ground to its xcoordinate.
(b) Find the maximum height of the rocket
above the sloping ground. What is its
xcoordinate when it is at its maximum
height?
(c) Clovis measures its height h of the
rocket above the sloping ground while it
is going up. Give a function x = g(h) relating the xcoordinate of the rocket to
h.
(d) Does this function still work when the
rocket is going down? Explain.
Problem 9.7. For each of the following functions: (1) sketch the function, (2) ﬁnd the inverse function, and (3) sketch the inverse function. In each case, indicate the correct domains and ranges. (4) Finally, make sure you 9.6. EXERCISES 131 test each of the functions you propose as an
inverse with the following compositions:
? (c) Give a function t = f−1 (w) relating the
time to the width of the surface of the
water. Make sure to specify the domain
and compute the range too. f(f−1 (x)) = x
and (b) After how many hours will the surface of
the water have width of 6 feet? ? f−1 (f(x)) = x.
(a) f(x) = 3x − 2
(b) f(x) = 1 x + 5
2 Problem 9.9. A biochemical experiment involves combining together two protein extracts. Suppose a function φ(t) monitors the
amount (nanograms) of extract A remaining at
time t (nanoseconds). Assume you know these
facts: (c) f(x) = −x2 + 3, x ≥ 0 (d) f(x) = x2 + 2x + 5, x ≤ −1
√
(e) f(x) = 4 − x2 , 0 ≤ x ≤ 2
Problem 9.8. A trough has a semicircular
cross section with a radius of 5 feet. Water
starts ﬂowing into the trough in such a way
that the depth of the water is increasing at a
rate of 2 inches per hour. 1. The function φ is invertible; i.e., it has
an inverse function.
2. φ(0) = 6, φ(1) = 5, φ(2) = 3, φ(3) = 1,
φ(4) = 0.5, φ(10) = 0.
(a) At what time do you know there will be
3 nanograms of extract A remaining? water (b) What is φ−1 (0.5) and what does it tell
you?
5 ft
crosssection of
trough (a) Give a function w = f(t) relating the
width w of the surface of the water to
the time t, in hours. Make sure to specify the domain and compute the range
too. (c) (True or False) There is exactly one time
when the amount of extract A remaining
is 4 nanograms.
(d) Calculate φ(φ−1 (1)) =
(e) Calculate φ−1 (φ(6)) =
(f) What is the domain and range of φ? 132 CHAPTER 9. INVERSE FUNCTIONS Chapter 10
Exponential Functions
If we start with a single yeast cell under favorable growth conditions, then
it will divide in one hour to form two identical “daughter cells”. In turn,
after another hour, each of these daughter cells will divide to produce
two identical cells; we now have four identical “granddaughter cells” of
the original parent cell. Under ideal conditions, we can imagine how this
“doubling effect” will continue:
TIME cells t=0 hours t=1 hours t=2 hours t=3 hours Figure 10.1: Observing cell growth. The question is this: Can we ﬁnd a function of t that
will predict (i.e. model) the number of yeast cells after t
hours? If we tabulate some data (as at right), the conclusion is that the formula
N (t ) = 2t
predicts the number of yeast cells after t hours. Now,
let’s make a very slight change. Suppose that instead
of starting with a single cell, we begin with a population
of 3 × 106 cells; a more realistic situation. If we assume
133 Total
hours Number of
yeast cells 0
1
2
3
4
5
6 1=20
2=21
4=22
8=23
16=24
32=25
64=26 Table 10.1: Cell growth data. CHAPTER 10. EXPONENTIAL FUNCTIONS 134 that the population of cells will double every hour, then
reasoning as above will lead us to conclude that the formula
N(t) = (3 × 106 )2t
gives the population of cells after t hours. Now, as long as t represents
a nonnegative integer, we know how to calculate N(t). For example, if
t = 6, then
N(t) = (3 × 106 )26 = (3 × 106 )(2 · 2 · 2 · 2 · 2 · 2)
= (3 × 106 )64
= 192 × 106. The key point is that computing N(t) only involves simple arithmetic.
But what happens if we want to know the population of cells after 6.37
hours? That would require that we work with the formula
N(t) = (3 × 106 )26.37
and the rules of arithmetic do not sufﬁce to calculate N(t). We are stuck,
since we must understand the meaning of an expression like 26.37 . In
order to proceed, we will need to review the algebra required to make
sense of raising a number (such as 2) to a noninteger power. We need to
√
understand the precise meaning of expressions like: 26.37 , 2 5 , 2−π , etc. 10.1 Functions of Exponential Type x this is a fixed positive integer this is a variable y=b b
y=x this is a fixed number Exponential Picture this is a variable Monomial Picture Figure 10.2: Viewing the difference between exponential and monomial functions. On a symbolic level, the class of functions we are trying to motivate is
easily introduced. We have already studied the monomials y = xb , where
x was our input variable and b was a ﬁxed positive integer exponent.
What happens if we turn this around, interchanging x and b, deﬁning a
new rule:
y = f(x) = bx . (10.1) 10.1. FUNCTIONS OF EXPONENTIAL TYPE 135 We refer to x as the power and b the base. An expression of this sort
is called a function of exponential type. Actually, if your algebra is a
bit rusty, it is easy to initially confuse functions of exponential type and
monomials (see Figure 10.2). 10.1.1 Reviewing the Rules of Exponents
To be completely honest, making sense of the expression
y = bx for all numbers x requires the tools of Calculus,
but it is possible to establish a reasonable comfort level
by handling the case when x is a rational number. If b ≥ 0
and n is a positive integer (i.e. n = 1, 2, 3, 4, . . . ), then we
can try to solve the equation
tn = b. (10.2) 3
2 n=2
n=4
n=6 1
0
1.5 1 0.5 0 0.5 1 1.5 1 A solution t to this equation is called an nth root of
b. This leads to complications, depending on whether n
is even or odd. In the odd case, for any real number b,
notice that the graph of y = b will always cross the graph
of y = tn exactly once, leading to one solution of (10.2).
On the other hand, if n is even and b < 0, then the
graph of y = tn will miss the graph of y = b, implying there
are no solutions to the equation in (10.2). (There will be
complex solutions to equations such as t2 = −1, involv√
ing the imaginary complex numbers ±i = ± −1, but we
are only working with real numbers in this course.) Also,
again in the case when n is even, it can happen that there
are two solutions to (10.2). We do not want to constantly
worry about this even/odd distinction, so we will henceforth assume b > 0. To eliminate possible ambiguity, we
will single out a particular nth root; we deﬁne the symbols:
√
1
n
b = b n = the largest real nth root of b. 2
3 3
2
1
0
1.5 1 0.5 0 √
q p b = √
q 1 1.5 2
3 Figure 10.3: Even and odd
monomials. (10.3) Important Facts 10.1.1 (Working with rational exponents). For all positive integers p and q, and any real number base b > 0, we have
p 0.5 1 √
Thus, whereas ±1 are both 4th roots of 1, we have deﬁned 4 1 = 1.
In order to manipulate y = bx for rational x, we need to recall some
basic facts from algebra. bq = n=3
n=5
n=7 bp . For any rational numbers r and s, and for all positive bases a and b: CHAPTER 10. EXPONENTIAL FUNCTIONS 136
n
y=t yaxis n even
= solution yaxis n
y=t n odd y=b
y=b
taxis y=b* taxis no solution or two solutions
exactly one solution Figure 10.4: How many solutions to tn = b?. 1. Product of power rule: br bs = br+s
2. Power of power rule: (br )s = brs
3. Power of product rule: (ab)r = ar br
4. Zero exponent rule: b0 = 1
5. Negative power rule: b−r = 1
br These rules have two important consequences, one theoretical and the
other more practical. On the ﬁrst count, recall that any rational number
p
r can be written in the form r = q , where p and q are integers. Consequently, using these rules, we see that the expression y = bx deﬁnes a
function of x, whenever x is a rational number. On the more practical
side of things, using the rules we can calculate and manipulate certain
expressions. For example,
2 √
3 27 5 √
3 8 27 3 =
8− 3 = 2 −5 = 32 = 9;
= 2−5 = 1
1
=.
5
2
32 The sticky point which remains is knowing that f(x) = bx actually
deﬁnes a function for all real values of x. This is not easy to verify and
we are simply going to accept it as a fact. The difﬁculty is that we need
the fundamentally new concept of a limit, which is the starting point of
a Calculus course. Once we know the expression does deﬁne a function,
we can also verify that the rules of Fact 10.1.1 carry through for all real 10.2. THE FUNCTIONS Y = A0 BX 137 exponent powers. Your calculator should have a “y to the x key”, allowing
√
you to calculate expressions such as π 2 involving nonrational powers.
Here are the key modeling functions we will work with in this Chapter.
Deﬁnition 10.1.2. A function of exponential type has the form
A(x) = A0 bx ,
for some b > 0, b = 1, and A0 = 0.
We will refer to the formula in Deﬁnition 10.1.2 as the standard exponential form. Just as with standard forms for quadratic functions, we
sometimes need to do a little calculation to put an equation in standard
form. The constant A0 is called the initial value of the exponential function; this is because if x represents time, then A(0) = A0 b0 = A0 is the
value of the function at time x = 0; i.e. the initial value of the function.
Example 10.1.3. Write the equations y = 83x and y = 7
exponential form. 1 2x−1
2 in standard Solution. In both cases, we just use the rules of exponents to maneuver
the given equation into standard form:
y = 83x
= (83 )x
= 512x
and
1
2 2x−1 y=7 2x =7 1
2 =7 1
2 = 14 1
4 −1 1
2
2 x 2
x 10.2 The Functions y = A0bx
We know f(x) = 2x deﬁnes a function of x, so we can study basic qualitative features of its graph. The data assembled in the solution of the
“Doubling Effect” beginning this Chapter, plus the rules of exponents,
produce a number of points on the graph. This graph exhibits four key
qualitative features that deserve mention: CHAPTER 10. EXPONENTIAL FUNCTIONS 138 Point on the
graph of y = 2x
.
.
.
.
.
.
.
.
.
2 1/4
(2, 1/4)
1 1/2
(1, 1/2)
0
1
(0, 1)
1
2
(1, 2)
2
4
(2, 4)
3
8
(3, 8)
.
.
.
.
.
.
.
.
.
x 2x (a) Data points from y = 2x . y=2 x (3,8) (2,4)
(−1,1/2)
(1,2)
(0,1) (−2,1/4) −1 1 (b) Graph of y = 2x . Figure 10.5: Visualizing y = 2x . • The graph is always above the horizontal axis; i.e. the function
values are always positive.
• The graph has yintercept 1 and is increasing.
• The graph becomes closer and closer to the horizontal axis as we
move left; i.e. the xaxis is a horizontal asymptote for the lefthand portion of the graph.
• The graph becomes higher and higher above the horizontal axis as
we move to the right; i.e., the graph is unbounded as we move to
the right.
The special case of y = 2x is representative of the function y = bx , but
there are a few subtle points that need to be addressed. First, recall we
are always assuming that our base b > 0. We will consider three separate
cases: b = 1, b > 1, and 0 < b < 1. 10.2.1 The case b = 1
In the case b = 1, we are working with the function y = 1x = 1; this is not
too exciting, since the graph is just a horizontal line. We will ignore this
case. 10.2.2 The case b > 1
If b > 1, the graph of the function y = bx is qualitatively similar to the
situation for b = 2, which we just considered. The only difference is the
exact amount of “concavity” in the graph, but the four features highlighted above are still valid. Figure 10.6(a) indicates how these graphs 10.2. THE FUNCTIONS Y = A0 BX 139 compare for three different values of b. Functions of this type exhibit
what is typically referred to as exponential growth ; this codiﬁes the fact
that the function values grow rapidly as we move to the right along the
xaxis.
yaxis yaxis all graphs pass through (0,1)
all graphs pass through (0,1) xaxis xaxis (a) Graph of y = bx , b > 1.
(b) Graph of y = bx , 0 < b < 1.
Figure 10.6: Visualizing cases for b. 10.2.3 The case 0 < b < 1
We can understand the remaining case 0 < b < 1, by using the remarks
above and our work in Chapter 13. First, with this condition on b, notice
1
1x
that b > 1, so the graph of y = b is of the type in Figure 10.6(a). Now,
using the rules of exponents:
y= 1
b −x = 1
b −1 x = bx .
−x 1
By the reﬂection principle, the graph of y = b
is obtained by reﬂect1x
ing the graph of y = ( b ) about the yaxis. Putting these remarks together, if 0 < b < 1, we conclude that the graph of y = bx will look like
Figure 10.6(b). Notice, the graphs in Figure 10.6(b) share qualitative features, mirroring the features outlined previously, with the “asymptote”
and “unbounded” portions of the graph interchanged. Graphs of this
sort are often said to exhibit exponential decay, in the sense that the
function values rapidly approach zero as we move to the right along the
xaxis. Important Facts 10.2.1 (Features of Exponential Type Functions). Let
b be a positive real number, not equal to 1. The graph of y = bx has these
four properties:
1. The graph is always above the horizontal axis.
2. The graph has yintercept 1.
3. If b > 1 (resp. 0 < b < 1), the graph becomes closer and closer to the
horizontal axis as we move to the left (resp. move to the right); this
says the xaxis is a horizontal asymptote for the lefthand portion of
the graph (resp. righthand portion of the graph). CHAPTER 10. EXPONENTIAL FUNCTIONS 140 4. If b > 1 (resp. 0 < b < 1), the graph becomes higher and higher above
the horizontal axis as we move to the right (resp. move to the left);
this says that the graph is unbounded as we move to the right (resp.
move to the left).
If A0 > 0, the graph of the function y = A0 bx is a vertically expanded or
compressed version of the graph of y = bx . If A0 < 0, we additionally
reﬂect about the xaxis. 10.3 Piano Frequency Range
A sound wave will cause your eardrum to move back and forth. In the
case of a socalled pure tone, this motion is modeled by a function of the
form
d(t) = A sin(2πft),
where f is called the frequency, in units of “periods/unit time”, called
“Hertz” and abbreviated “Hz”. The coefﬁcient A is related to the actual
displacement of the eardrum, which is, in turn, related to the loudness
of the sound. A person can typically perceive sounds ranging from 20 Hz
to 20,000 Hz.
A# C# D# F# G# A# C# D# F# G#A# C#D# F# G# A# C# D# F# G#A# C# D# F# G# A# C# D# F# G# A# C# D# F# G#A# ABCDE FGABCDEF GABCDEF GABCDEF GABCDEF GABCDEF GABCDEF GABC
220 Hz middle C Figure 10.7: A piano keyboard. A piano keyboard layout is shown in Figure 10.7. The white keys are
labelled A, B, C, D, E, F, and G, with the sequence running from left to
right and repeating for the length of the keyboard. The black keys ﬁt into
this sequence as “sharps”, so that the black key between A and B is “A
sharp”, denoted A# . Thus, starting at any A key, the 12 keys to the right
are A, A# , B, C, C# , D, D# , E, F, F# , G, and G# . The sequence then repeats.
Notice that between some adjacent pairs of white keys there is no black
key.
A piano keyboard is commonly tuned according to a rule requiring
that each key (white and black) has a frequency 21/12 times the frequency
of the key to its immediate left. This makes the ratio of adjacent keys
always the same (21/12 ), and it means that keys 12 keys apart have a ratio
of frequencies exactly equal to 2 (since (21/12 )12 = 2). Two such keys are 10.3. PIANO FREQUENCY RANGE 141 said to be an o ctave apart. Assuming that the key A below middle C has
a frequency of 220 Hz, we can determine the frequency of every key on
the keyboard. For instance, the A# to the right of this key has frequency
220 × 21/12 = 220 × 1.059463094... ≈ 233.08188Hz. The B to the right of this
key has frequency 233.08188 × 21/12 ≈ 246.94165Hz. CHAPTER 10. EXPONENTIAL FUNCTIONS 142 10.4 Exercises
Problem 10.1. Let’s brush up on the required
calculator skills. Use a calculator to approximate:
(a) 3π
(b) 42+ √ 5 (c) ππ
(d) 5− √
3 (d) Anja, a third member of your lab working with the same yeast cells, took these
two measurements: 7.246 × 106 cells after
4 hours; 16.504 × 106 cells after 6 hours.
Should you be worried by Anja’s results?
If Anja’s measurements are correct, does
your model over estimate or under estimate the number of yeast cells at time
t? 2 (e) 3π
√
(f) 11π−7 Problem 10.4.
middle C. (a) Find the frequency of (b) Find the frequency of A above middle C.
Problem 10.2. Put each equation in standard
exponential form:
(a) y = 3(2−x )
(b) y = 4−x/2
(c) y = ππx
(d) y = 1
(e) y = x
1 3+ 2
3 5
0.3452x−7 (f) y = 4(0.0003467)−0.4x+2 Problem 10.3. A colony of yeast cells is estimated to contain 106 cells at time t = 0. After
collecting experimental data in the lab, you decide that the total population of cells at time t
hours is given by the function
y = 106 e0.495105t . (c) What is the frequency of the lowest note
on the keyboard? Is there a way to solve
this without simply computing the frequency of every key below A220?
(d) The Bosendorfer piano is famous, due in
part, to the fact it includes additional
keys at the left hand end of the keyboard, extending to the C below the bottom A on a standard keyboard. What
is the lowest frequency produced by a
Bosendorfer?
Problem 10.5. You have a chess board as pictured, with squares numbered 1 through 64.
You also have a huge change jar with an unlimited number of dimes. On the ﬁrst square
you place one dime. On the second square you
stack 2 dimes. Then you continue, always doubling the number from the previous square. (a) How many cells are present after one
hour? (a) How many dimes will you have stacked
on the 10th square? (b) (True or False) The population of yeast
cells will double every 1.4 hours. (b) How many dimes will you have stacked
on the nth square? (c) Cherie, another member of your lab,
looks at your notebook and says : ...that
formula is wrong, my calculations predict the formula for the number of yeast
cells is given by the function (c) How many dimes will you have stacked
on the 64th square? y = 106 (2.042727)0.693147t .
Should you be worried by Cherie’s remark? (d) Assuming a dime is 1 mm thick, how
high will this last pile be?
(e) The distance from the earth to the sun
is approximately 150 million km. Relate
the height of the last pile of dimes to this
distance. 10.4. EXERCISES 143
63 64 calculates the fraction of hemoglobin saturated with oxygen at a given pressure p.
(a) The graphs of M(p) and H(p) are given
below on the domain 0 ≤ p ≤ 100; which
is which? fraction
1
0.8
0.6 10 9
0.4 1 2 3 8 0.2
20 Problem 10.6. Myoglobin and hemoglobin are
oxygen carrying molecules in the human body.
Hemoglobin is found inside red blood cells,
which ﬂow from the lungs to the muscles
through the bloodstream. Myoglobin is found
in muscle cells. The function
p
Y = M(p) =
1+p
calculates the fraction of myoglobin saturated
with oxygen at a given pressure p torrs. For
example, at a pressure of 1 torr, M(1) = 0.5,
which means half of the myoglobin (i.e. 50%)
is oxygen saturated. (Note: More precisely, you
need to use something called the “partial pressure”, but the distinction is not important for
this problem.) Likewise, the function
Y = H(p) = p2.8
262.8 + p2.8 40 60 80 100 p (b) If the pressure in the lungs is 100 torrs,
what is the level of oxygen saturation of
the hemoglobin in the lungs?
(c) The pressure in an active muscle is 20
torrs. What is the level of oxygen saturation of myoglobin in an active muscle?
What is the level of hemoglobin in an active muscle?
(d) Deﬁne the efﬁciency of oxygen transport
at a given pressure p to be M(p) − H(p).
What is the oxygen transport efﬁciency
at 20 torrs? At 40 torrs? At 60 torrs?
Sketch the graph of M(p) − H(p); are
there conditions under which transport
efﬁciency is maximized (explain)? 144 CHAPTER 10. EXPONENTIAL FUNCTIONS Chapter 11
Exponential Modeling
Example 11.0.1. A computer industry spokesperson has predicted that
the number of subscribers to geton.com, an internet provider, will grow exponentially for the ﬁrst 5 years. Assume this person is correct. If geton.com
has 100,000 subscribers after 6 months and 750,000 subscribers after 12
months, how many subscribers will there be after 5 years?
Solution. The solution to this problem offers a template for many exponential modeling applications. Since, we are assuming that the number
of subscribers N(x), where x represents years, is a function of exponential
type,
N(x) = N◦ bx ,
for some N◦ and b > 1. We are given two pieces of information about the
values of N(x):
N(0.5) = 100,000; i.e., N0 b0.5 = 100,000, and
N(1) = 750,000; i.e., N0 b = 750,000.
We can use these two equations to solve for the two unknowns N◦ and b
as follows: If we divide the second equation by the ﬁrst, we get
b1
= 7.5
b0.5
√
b0.5 = b1/2 = b = 7.5
∴ b = 56.25.
Plugging this value of b into either equation (say the ﬁrst one), we can
100,000
solve for N◦ : N◦ = (56.25)0.5 = 13,333. We conclude that the number of
geton.com subscribers will be predicted by
N(x) = 13,333(56.25)x.
In ﬁve years, we obtain N(5) = 7,508,300,000,000 subscribers, which exceeds the population of the Earth (which is between 5 and 6 billion)!
145 CHAPTER 11. EXPONENTIAL MODELING 146
Q 800000 600000 There are two important conclusions we can draw from
this problem. First, the given information provides us
with two points on the graph of the function N(x): 400000 200000 P 0.2 0.4 0.6 0.8 1 Figure 11.1: Finding the
equation for N(x) = N0 bx . P = (0.5, 100,000)
Q = (1, 750,000).
More importantly, this example illustrates a very important principal we can use when modeling with functions
of exponential type. Important Fact 11.0.2. A function of exponential type can be determined
if we are given two data points on its graph.
!!!
CAUTION
!!! When you use the above strategy to ﬁnd the base b of the exponential model, make sure to write down a lengthy decimal approximation.
As a rule of thumb, go for twice as many signiﬁcant digits as you are
otherwise using in the problem. 11.1 The Method of Compound Interest
You walk into a Bank with P0 dollars (usually called principal), wishing
to invest the money in a savings account. You expect to be rewarded by
the Bank and paid interest, so how do you compute the total value of the
account after t years?
The future value of the account is really a function of the number of
years t elapsed, so we can write this as a function P(t). Our goal is to see
that P(t) is a function of exponential type. In order to compute the future
value of the account, the Bank provides any savings account investor
with two important pieces of information:
r = annual (decimal) interest rate
n = the number of compounding periods per year
The number n tells us how many times each year the Bank will compute
the total value P(t) of the account. For example, if n = 1, the calculation
is done at oneyear intervals; if n = 12, the calculation is done each
month, etc. The bank will compute the value of your account after a
r
typical compounding period by using the periodic rate of return n . For
example, if the interest rate percentage is 12% and the compounding
period is monthly (i.e., n = 12), then the annual (decimal) interest rate is
0.12 and the periodic rate is 0.12 = 0.01.
12
!!!
CAUTION
!!! The number r always represents the decimal interest rate, which is a
decimal between 0 and 1. If you are given the interest rate percentage
(which is a positive number between 0 and 100), you need to convert
to a decimal by dividing by 100. 11.1. THE METHOD OF COMPOUND INTEREST 147 11.1.1 Two Examples
Let’s consider an example: P0 = $1,000 invested at the annual interest
percentage of 8% compounded yearly, so n = 1 and r = 0.08. To compute
the value P(1) after one year, we will have
P(1) = P0 + (periodic rate)P0
= P0 + rP0 = P0 (1 + r)
= $1,000(1 + 0.08) = $1,080.
To compute the value after two years, we need to apply the periodic
rate to the value of the account after one year:
P(2) = P(1) + (periodic rate)P(1)
= P0 (1 + r) + rP0 (1 + r) = P0 (1 + r)2
= $1,000(1 + 0.08)2 = $1,166.40.
Notice, the amount the Bank has paid after two years is $166.40, which
is slightly bigger than twice the $80 paid after one year. To compute the
value after three years, we need to apply the periodic rate to the value of
the account after two years:
P(3) = P(2) + (periodic rate)P(2)
= P0 (1 + r)2 + rP0 (1 + r)2 = P0 (1 + r)3
= $1,000(1 + 0.08)3 = $1,259.71.
Again, notice the amount the Bank is paying after three years is $259.71,
which is slightly larger than three times the $80 paid after one year.
Continuing on in this way, to ﬁnd the value after t years, we arrive at the
formula
P(t) = P0 (1 + r)t
= $1,000(1.08)t.
In particular, after 5 and 10 years, the value of the account (to the nearest
dollar) will be $1,469 and $2,159, respectively.
As a second example, suppose we begin with the same $1,000 and
the same annual interest percentage 8%, but now compound monthly,
so n = 12 and r = 0.08. The value of the account after one compounding
period is P(1/12), since a month is onetwelfth of a year. Arguing as
r
before, paying special attention that the periodic rate is now n = 0.08 , we
12
have
P(1/12) = P0 + (periodic rate)P0
.08
= P0 1 +
12
= $1,000(1 + 0.006667) = $1,006.67. CHAPTER 11. EXPONENTIAL MODELING 148 After two compounding periods, the value is P(2/12),
P(2/12) = P(1/12) + (periodic rate)P(1/12)
.08
0.08
.08
= P0 1 +
+
P0 1 +
12
12
12
2 .08
= P0 1 +
12
= $1,000(1 + 0.006667)2 = $1,013.38.
Continuing on in this way, after k compounding periods have elapsed,
k
the value will be P 12 , which is computed as
P(k/12) = P0 1 + .08
12 k . It is possible to rewrite this formula to give us the value after t years,
noting that t years will lead to 12t compounding periods; i.e., set k = 12t
in the previous formula:
P(t) = P0 .08
1+
12 12t For example, after 1, 5 and 10 years, the value of the account, to the
nearest dollar, would be $1,083, $1,490, and $2,220. 11.1.2 Discrete Compounding
The two examples above highlight a general formula for computing the
future value of an account.
Important Fact 11.1.1 (Discrete compounding). Suppose an account is
opened with P0 principal. If the decimal interest rate is r and the number of
compounding periods per year is n, then the value P(t) of the account after
t years will be
P(t) = P0 1 + r
n nt . Notice, the future value P(t) is a function of exponential type; the base
r
is the number 1 + n , which will be greater than one. Since P0 > 0, the
graph will be qualitatively similar to the ones pictured in Figure 10.6(a).
Example 11.1.2. At birth, your Uncle Hans secretly purchased a $5,000
U.S. Savings Bond for $2,500. The conditions of the bond state that the
U.S. Government will pay a minimum annual interest rate of r = 8.75%,
compounded quarterly. Your Uncle has given you the bond as a gift, subject
to the condition that you cash the bond at age 35 and buy a red Porsche. 11.2. THE NUMBER E AND THE EXPONENTIAL FUNCTION 149 On your way to the Dealer, you receive a call from your tax accountant
informing you of a 28% tax on the capital gain you realize through cashing
in the bond; the capital gain is the selling price of the bond minus the
purchase price. Before stepping onto the showroom ﬂoor, compute how
much cash will you have on hand, after the U.S. Government shares in
your proﬁts.
Solution. The value of your bond after 35 years is computed by the formula in Fact 11.1.1, using P0 = $2,500, r = 0.0875, n = 4, and t = 35.
Plugging this all in, we ﬁnd that the selling price of the bond is
P(35) = $2,500 1 + 0.0875
4 4(35) = $51,716.42. The capital gain will be $51,716.42  $2,500 = $49,216.42 and the tax
due is $(49,216.42)(0.28) = $13,780.60. You are left with $51,716.42 $13,780.60 = $37,935.82. Better make that a used Porsche! 11.2 The Number e and
the Exponential Function
What happens to the future value of an investment of P0 dollars as the
number of compounding periods is increased? For example, return to
our earlier example: P0 = $1000 and an annual interest percentage of 8%.
After 1 year, the table below indicates the value of the investment for
various compounding periods: yearly, quarterly, monthly, weekly, daily,
and hourly.
Value after 1 year
(to nearest dollar) n Compounding
Period
1 yearly
4 quarterly $1,000(1 + 0.08)1 = $1,080.00
$1,000 1 + 0.08 4
=
4 $1,082.43 12 monthly $1,000 1 + 0.08 12
12 = $1,083.00 52 weekly $1,000 1 + 0.08 52
52 = $1,083.22 0.08 365
365 = $1,083.28 0.08 8,760
8,760 =$1,083.29 365 daily
8,760 hourly $1,000 1 +
$1,000 1 + CHAPTER 11. EXPONENTIAL MODELING 150 We could continue on, considering “minute” and “second” compounding
and what we will ﬁnd is that the value will be at most $1,083.29. This
illustrates a general principal:
Important Fact 11.2.1. Initially increasing the number of compounding
periods makes a signiﬁcant difference in the future value; however, eventually there appears to be a limiting value.
Let’s see if we can understand mathematically why this is happening.
The ﬁrst step is to recall the discrete compounding formula:
r nt
.
n
If our desire is to study the effect of increasing the number of compounding periods, this means we want to see what happens to this formula as
n gets BIG. To analyze this, it is best to rewrite the expression using a
r
substitution trick: Set z = n , so that n = rz and n = 1 . Plugging in, we
r
z
have
r nt
P(t) = P0 1 +
n
rzt
1
= P0 1 +
(11.1)
z
P(t) = P0 1 + = P0 1+ 1
z z rt . So, since r is a ﬁxed number and z = n , letting n get BIG is the same
r
as letting z become BIG in (11.1). This all means we need to answer
z
this new question: What happens to the expression 1 + 1 as z becomes
z
large? On the one hand, the power in the expression is getting large; at
the same time, the base is getting close to 1. This makes it very tricky
to make quick predictions about the outcome. It is best to ﬁrst tabulate
z
some numerical data for the values of y = g(z) = 1 + 1 and look at a
z
plot of this function graph on the domain 0.01 ≤ z ≤ 100: See Figure 11.2.
z
You can see from this plot, the graph of y = 1 + 1 approaches the
z
“dashed” horizontal asymptote, as z becomes BIG. We will let the letter “e”
represent the spot where this horizontal line crosses the vertical axis and
e ≈ 2.7182818. This number is only an approximation, since e is known to
be an irrational number. What sets this irrational number apart from the
√
ones you are familiar with (e.g. 2, π, etc.) is that deﬁning the number
e requires a “limiting” process. This will be studied a lot more in your
Calculus course. The new number e is a positive number greater than 1,
so we can study the function:
y = ex . (11.2) Since e > 1, the graph will share the properties in Figure 10.6(a). This
function is usually referred to as THE exponential function. Scientiﬁc
calculators will have a key of the form “ exp(x) ” or “ ex ”. 11.2. THE NUMBER E AND
1+ z
1
2
3
4
20
100
1000
109 1
z THE EXPONENTIAL FUNCTION 151 z 2
2.25
2.37037
2.4414
2.65329
2.70481
2.71692
2.71828 2.5 2 1.5 1 0.5 20 (a) Data points for
z
1+ 1 .
z 40 60 80 (b) The graph of 1 + 100 1z
.
z Figure 11.2: What happens when z get very large? 11.2.1 Calculator drill
Plugging in x = 1, you can compute an approximation to e on your calculator; you should get e = 2.7183 to four decimal places. Make sure you
√
can compute expressions like e3 , eπ , and e = e1/2 ; to four decimal places,
you should get 20.0855, 23.1407, and 1.6487. 11.2.2 Back to the original problem...
We can now return to our future value formula (11.1) and conclude that
as the number of compounding periods increases, the future value is
approaching a limiting value:
P(t) = P0 1
1+
z z rt =⇒ P0 ert . The right hand limiting formula Q(t) = P0 ert computes the future value
using what is usually referred to as continuous compounding. From the
investors viewpoint, this is the best possible scheme for computing future
value.
Important Fact 11.2.2 (Continuous compounding). The future value of
P0 dollars principal invested at an annual decimal interest rate of r under continuous compounding after t years is Q(t) = P0 ert ; this value is alr nt
ways greater than the value of P0 1 + n , for any discrete compounding
scheme. In fact, P0 ert is the limiting value. CHAPTER 11. EXPONENTIAL MODELING 152 11.3 Exercises
Problem 11.1. In 1968, the U.S. minimum
wage was $1.60 per hour. In 1976, the minimum wage was $2.30 per hour. Assume the
minimum wage grows according to an exponential model w(t), where t represents the time
in years after 1960.
(a) Find a formula for w(t).
(b) What does the model predict for the minimum wage in 1960? (a) Sketch rough graphs of these two functions.
(b) The graph of the equation x2 − y2 = 1
is shown below; this is called the unit
hyperbola. For any value a, show that
the point (x,y) = (cosh(a), sinh(a)) is on
the unit hyperbola. (Hint: Verify that
[cosh(x)]2 − [sinh(x)]2 = 1, for all x.) y (c) If the minimum wage was $5.15 in 1996,
is this above, below or equal to what the
model predicts.
Problem 11.2. The town of Pinedale,
Wyoming, is experiencing a population boom.
In 1990, the population was 860 and ﬁve years
later it was 1210. (−1,0) (1,0) x (a) Find a linear model l(x) and an exponential model p(x) for the population of
Pinedale in the year 1990+x.
(b) What do these models estimate the population of Pinedale to be in the year
2000?
Problem 11.3. In 1989, research scientists
published a model for predicting the cumulative number of AIDS cases reported in the
United States:
a(t) = 155 t − 1980
10 3 (thousands) , where t is the year. This paper was considered
a “relief”, since there was a fear the correct
model would be of exponential type. Pick two
data points predicted by the research model
a(t) to construct a new exponential model b(t)
for the number of cumulative AIDS cases. Discuss how the two models differ and explain the
use of the word “relief”.
Problem 11.4. Deﬁne two new functions:
x y = cosh(x) =
and y = a cosh( x−h
) + C,
a for appropriate constants a, h and C.
The constant h depends on how the coordinate system is imposed. A cable for
a suspension bridge hangs from two 100
ft. high towers located 400 ft. apart. Impose a coordinate system so that the picture is symmetric about the yaxis and
the roadway coincides with the xaxis.
The hanging cable constant is a = 500
and h = 0. Find the minimum distance
from the cable to the road.
towers −x e +e
2
x (c) A hanging cable is modeled by a portion
of the graph of the function −x e −e
.
2
These are called the basic hyperbolic trigonometric functions. cable
d y = sinh(x) = 100 ft
roadway 400 ft Chapter 12
Logarithmic Functions
If we invest P0 = $1,000 at an annual rate of r = 8% compounded continuously, how long will it take for the account to have a value of $5000?
The formula P(t) = 1,000e0.08t gives the value after t years, so we need
to solve the equation:
5,000 = 1,000e0.08t
5 = e0.08t .
Unfortunately, algebraic manipulation will not lead to a further simpliﬁcation of this equation; we are stuck! The required technique involves the
theory of inverse functions. Assuming we can ﬁnd the inverse function of
f(t) = et , we can apply f−1 (t) to each side of the equation and solve for t:
f−1 (5) = f−1 (e0.08t ) = 0.08t
(12.5)f−1(5) = t
The goal in this section is to describe the function f−1 , which is usually
denoted by the symbol f−1 (t) = ln(t) and called the natural logarithm
function. On your calculator, you will ﬁnd a button dedicated to this
function and we can now compute ln(5) = 1.60944. Conclude that the
solution is t = 20.12 years. 12.1 The Inverse Function of y = ex
If we sketch a picture of the exponential function on the domain of all real
numbers and keep in mind the properties in Fact 10.2.1, then every horizontal line above the xaxis intersects the graph of y = ex exactly once:
See Figure 12.1(a). The range of the exponential function will consist
of all possible ycoordinates of points on the graph. Using the graphical
techniques of Chapter 6, we can see that the range of will be all POSITIVE
real numbers: See Figure 12.1(b).
153 CHAPTER 12. LOGARITHMIC FUNCTIONS 154
PSfrag y=e x y = ex range = all
positive numbers
graphs of y=c,
c>1 cross exponential
graph exactly once. −1 1 these horizontal
lines miss graph
of exponential function. −1 1
domain = all real numbers (a) Horizontal line test for
(b) The domain and range for
y = ex .
y = ex .
Figure 12.1: Properties needed to ﬁnd the inverse of f(x) = ex . By the horizontal line test, this means the exponential function is onetoone and the inverse rule f−1 (c) will deﬁne a function the unique solu tion of the equation , if c > 0 −1
c = ex
(12.1)
f (c ) = (undeﬁned), if c ≤ 0. y=e x reflecting line y = x y=ln (x) −1 1 Figure 12.2: Visualizing the
y = ln(x). This inverse function is called the natural logarithm
function, denoted ln(c). We can sketch the graph of the
the natural logarithm as follows: First, by Fact 9.2.1, the
domain of the function ln(y) = x is just the range of the
exponential function, which we noted is all positive numbers. Likewise, the range of the function ln(y) = x is the
domain of the exponential function, which we noted is all
real numbers. Interchanging x and y, the graph of the
natural logarithm function y = ln(x) can be obtained by
ﬂipping the graph of y = ex across the line y = x: Important Facts 12.1.1 (Graphical features of natural log). The function
y = ln(x) has these features:
• The largest domain is the set of positive numbers; e.g. ln(−1) makes
no sense.
• The graph has xintercept 1 and is increasing.
• The graph becomes closer and closer to the vertical axis as we approach x = 0; i.e. the yaxis is a vertical asymptote for the graph.
• The graph is unbounded as we move to the right.
Any time we are working with an inverse function, symbolic properties
are useful. Here are the important ones related to the natural logarithm.
Important Facts 12.1.2 (Natural log properties). We have the following
properties: 12.1. THE INVERSE FUNCTION OF Y = EX 155 (a) For any real number x, ln(ex ) = x.
(b) For any positive number x, eln(x) = x.
(c) ln(bt ) = t ln(b), for b > 0 and t any real number;
(d) ln(ba) = ln(a) + ln(b), for all a, b > 0;
(e) ln b
a = ln(b) − ln(a), for all a, b > 0. The properties (c)(e) are related to three of the rules of exponents in
Facts 10.1.1. Here are the kinds of basic symbolic maneuvers you can
pull off using these properties:
Examples 12.1.3.
(i) ln(83 ) = 3 ln(8) = 6.2383; ln(6π) = ln(6) + ln(π) = 2.9365; ln
ln(3) − ln(5) = −0.5108.
(ii) ln √ 3
5 = 1
x = ln x1/2 = 2 ln(x); ln x2 − 1 = ln((x − 1)(x + 1)) = ln(x − 1) + ln(x + 1); ln x5
x2 +1 = ln x5 − ln x2 + 1 = 5 ln(x) − ln x2 + 1 . Examples 12.1.4.
Given the equation 3x+1 = 12, we can solve for x:
3x+1 = 12
ln 3x+1 = ln(12)
(x + 1) ln(3) = ln(12)
ln(12)
x=
− 1 = 1.2619.
ln(3)
Example 12.1.5. If $2,000 is invested in a continuously compounding
savings account and we want the value after 12 years to be $130,000,
what is the required annual interest rate? If, instead, the same $2,000
is invested in a continuously compounding savings account with r = 6.4%
annual interest, when will the exact account value be be $130,000?
Solution. In the ﬁrst scenario,
130,000 = 2,000e12r
65 = e12r
ln(65) = ln e12r
ln(65) = 12r
ln(65)
= 0.3479.
r=
12 CHAPTER 12. LOGARITHMIC FUNCTIONS 156 This gives an annual interest rate of 34.79%. In the second scenario, we
study the equation
130,000 = 2,000e(0.064)t
65 = e(0.064)t
ln(65)
= 65.22.
t=
0.064
So, it takes over 65 years to accumulate $130,000 under the second
scheme. 12.2 Alternate form for
functions of exponential type
The standard model for an exponential function is A(t) = A0 bt , for some
b > 0, b = 1, and A0 = 0. Using the properties of the natural logarithm
function,
bt = eln(b) t = et ln(b) . This means that every function as in Deﬁnition 10.1.2 can be rewritten
using the exponential function et . Another way of saying this is that you
really only need the function keys “ et ” and “ ln(t) ” on your calculator.
Important Fact 12.2.1 (Observation). A function of exponential type can
be written in the form
A(t) = A0 eat ,
for some constants A0 = 0 and a = 0.
By studying the sign of the constant a, we can determine whether the
function exhibits exponential growth or decay. For example, given the
function A(t) = eat , if a > 0 (resp. a < 0), then the function exhibits
exponential growth (resp. decay).
Examples 12.2.2.
(a) The function A(t) = 200 (2t ) exhibits exponential growth and can be
rewritten as:
A(t) = 200 et ln(2) = 200e0.69315t
(b) The function A(t) = 4e−0.2t exhibits exponential decay and can be rewritten as:
A(t) = 4e−0.2t = 4 e−0.2 t = 4 0.81873t . 12.3. THE INVERSE FUNCTION OF Y = BX 157 12.3 The Inverse Function of y = bx
For some topics in Chemistry and Physics (e.g. acid base equilibria and
acoustics) it is useful to have on hand an inverse function for y = bx ,
where b > 0 and b = 1. Just as above, we would show that f(x) = bx is
onetoone, the range is all positive numbers and obtain the graph using
ideas in Figure 12.2. We will refer to the inverse rule as the logarithm
function base b, denoted logb (x), deﬁned by the rule: logb (c) = the unique solution of the equation , if c > 0
c = bx (undeﬁned), if c ≤ 0. We will need to consider two cases, depending on the magnitude of b:
The important qualitative features of the logarithm function y = logb (x)
mirror Fact 12.1.1:
x
y=b reflecting line y = x x
y=b reflecting line y = x y=logb(x) −1 1 −1 1
y=log (x)
b The case b > 1 The case 0 < b < 1 Figure 12.3: Cases to consider for b. Important Facts 12.3.1 (Graphical features of general logs). The function y = logb (x) has these features:
• The largest domain is the set of positive numbers; e.g. logb (−1) is not
deﬁned.
• The graph has xintercept 1 and is increasing if b > 1 (resp. decreasing if 0 < b < 1).
• The graph becomes closer and closer to the vertical axis as we approach x = 0; this says the yaxis is a vertical asymptote for the
graph. CHAPTER 12. LOGARITHMIC FUNCTIONS 158 • The graph is unbounded as we move to the right.
Important Facts 12.3.2 (Log properties). Fix a positive base b, b = 1.
(a) For any real number x, logb (bx ) = x.
(b) For any positive number x, blogb (x) = x.
(c) logb (rt ) = t logb (r), for r > 0 and t any real number;
(d) logb (rs) = logb (r) + logb (s), for all r, s > 0;
r
(e) logb ( s ) = logb (r) − logb (s), for all r, s > 0. It is common to simplify terminology and refer to the function logb (x)
as the log base b function, dropping the longer phrase “logarithm”. Some
scientiﬁc calculators will have a key devoted to this function. Other calculators may have a key labeled “log(x)”, which is usually understood
to mean the log base 10. However, many calculators only have the key
“ln(x)”. This is not cause for alarm, since it is always possible to express
logb (x) in terms of the natural log function. Let’s see how to do this, since
it is a great application of the Log Properties listed in Fact 12.3.2.
Suppose we start with y = logb (x). We will rewrite this in terms of the
natural log by carrying out a sequence of algebraic steps below; make
sure you see why each step is justiﬁed.
y = logb (x)
by = x
ln(by ) = ln(x)
y ln(b) = ln(x)
ln(x)
y=
.
ln(b)
We have just veriﬁed a useful conversion formula:
Important Fact 12.3.3 (Log conversion formula). For x a positive number
and b > 0, b = 1 a base,
logb (x) = ln(x)
.
ln(b) For example,
ln(5)
= 0.699
ln(10)
ln(11)
log0.02 (11) =
= −0.613
ln(0.02)
ln( 1 )
1
2
=
log20
= −0.2314
2
ln(20)
log10 (5) = 12.4. MEASURING THE LOUDNESS OF SOUND 159 The conversion formula allows one to proceed slightly differently when
solving equations involving functions of exponential type. This is illustrated in the next example.
Example 12.3.4. Ten years ago, you purchased a house valued at $80,000.
Your plan is to sell the house at some point in the future, when the value
is at least $1,000,000. Assume that the future value of the house can
be computed using quarterly compounding and an annual interest rate of
4.8%. How soon can you sell the house?
Solution. We can use the future value formula to obtain the equation
1,000,000 = 80,000 1 + 0.048
4 4t 12.5 = (1.012)4t
Using the log base b = 1.012,
log1.012 (12.5) = log1.012 (1.012)4t
log1.012 (12.5) = 4t
ln(12.5)
t=
= 52.934.
4 ln(1.012)
Since you have already owned the house for 10 years, you would need to
wait nearly 43 years to sell at the desired price. 12.4 Measuring the Loudness of Sound
As we noted earlier, the reception of a sound wave by the ear gives rise to
a vibration of the eardrum with a deﬁnite frequency and a deﬁnite amplitude. This vibration may also be described in terms of the variation of
air pressure at the same point, which causes the eardrum to move. The
perception that rustling leaves and a jet aircraft sound different involves
two concepts: (1) the fact that the frequencies involved may differ; (2)
the intuitive notion of “loudness”. This loudness is directly related to the
force being exerted on the eardrum, which we refer to as the intensity
of the sound. We can try to measure the intensity using some sort of
scale. This becomes challenging, since the human ear is an amazing instrument, capable of hearing a large range of sound intensities. For that
reason, a logarithmic scale becomes most useful. The sound pressure
level β of a sound is deﬁned by the equation
β = 10 log10 I
I0 , (12.2) where I0 is an arbitrary reference intensity which is taken to correspond
with the average faintest sound which can be heard and I is the intensity CHAPTER 12. LOGARITHMIC FUNCTIONS 160 of the sound being measured. The units used for β are called decibels,
abbreviated “db”. (Historically, the units of loudness were called bels,
in honor of Alexander Graham Bell, referring to the quantity log10 II0 .)
Notice, in the case of sound of intensity I = I0 , we have a sound pressure
level of
β = 10 log10 I0
I0 = 10 log10 (1) = 10(0) = 0. We refer to any sound of intensity I0 as having a sound pressure level
at the threshold of hearing. At the other end of the scale, a sound of
intensity the maximum the eardrum can tolerate has an average sound
pressure level of about 120 db. The Table 12.4(a) gives a hint of the
sound pressure levels associated to some common sounds.
Source of Noise
Threshold of pain
Riveter
Busy Street Trafﬁc
Ordinary Conversation
Quiet Auto
Background Radio
Whisper
Rustle of Leaves
Threshold of Hearing Sound Pressure
Level in db
120
95
70
65
50
40
20
10
0 (a) Sources of noise levels. pain threshold
120
100 Zone of Hearing 80 db 60
40
20
0
20 100 hearing threshold 1000 10,000 20,000 Hz (b) Graphing noise levels. Figure 12.4: Considering noise levels. It turns out that the above comments on the threshold of hearing and
pain are really only averages and depend upon the frequency of the given
sound. In fact, while the threshold of pain is on average close to 120
db across all frequencies between 20 Hz and 20,000 Hz, the threshold
of hearing is much more sensitive to frequency. For example, for a tone
of 20 Hz (something like the 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 12.4(b). Such a
plot would differ from person to person and is helpful in isolating hearing
problems.
Example 12.4.1. A loudspeaker manufacturer advertises that their model
no. 801 speaker produces a sound pressure level of 87 db when a reference test tone is applied. A competing speaker company advertises that 12.4. MEASURING THE LOUDNESS OF SOUND 161 their model 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?
Solution. If we let I1 and I2 refer to the sound intensities of the two speakers reproducing the test signal, then we have two equations:
87 = 10 log10
93 = 10 log10 I1
I0
I2
I0 Using log properties, we can solve the ﬁrst equation for I1 :
I1
= 10 log10 (I1 ) − 10 log10 (I0 )
I0
log10 (I1 ) = 8.7 + log10 (I0 )
87 = 10 log10 10log10 (I1 ) = 108.7+log10 (I0 )
I1 = 108.7 10log10 (I0 ) = 108.7 I0 .
Similarly, we ﬁnd that I2 = 109.3 I0 . This means that the ratio of the intensities will be
I2
109.3 I0
= 8.7 = 100.6 = 3.98.
I1
10 I0
This means that the test signal on the X − 1 speaker produces a sound
pressure level nearly 4 times that of the same test signal on the no. 801
speaker.
To ﬁnish the problem, imagine a third speaker which produces a
sound pressure level β, which is twice that of the ﬁrst speaker. If I3
is the corresponding intensity of the sound, then as above, I3 = 10(β/10) I0 .
We are assuming that I3 = 2I1 , so this gives us the equation log10 1
I1 = I3
2
1
108.7 I0 = 10(β/10) I0
2
1 (β/10)
10
108.7 = log10
2
1
8.7 = log10
+ log10 10(β/10)
2
β
8.7 = −0.30103 +
10
90 = β 162 CHAPTER 12. LOGARITHMIC FUNCTIONS So, the test signal on the third speaker must produce a sound pressure
level of 90 db. 12.5. EXERCISES 163 12.5 Exercises
Problem 12.1. These problems will help you
develop your skills with logarithms.
(a) Compute: log5 3, loge 11, log√2 π, log2 10,
log10 2.
(b) Solve for x: 35 = ex , log3 x = e, log3 5 =
xe3 .
(c) Solve each of these equations for x in
terms of y: y = 10x , 3y = 10x , y = 103x . Problem 12.2. As light from the surface penetrates water, its intensity is diminished. In the
clear waters of the Caribbean, the intensity is
decreased by 15 percent for every 3 meters of
depth. Thus, the intensity will have the form
of a general exponential function.
(a) If the intensity of light at the water’s surface is I◦ , ﬁnd a formula for I(d), the intensity of light at a depth of d meters.
Your formula should depend on I◦ and
d.
(b) At what depth will the light intensity be
decreased to 1% of its surface intensity? Problem 12.3. Rewrite each function in the
form y = A◦ eat , for appropriate constants A◦
and a.
(a) y = 13(3t )
(b) y = 2( 1 )t
8
(c) y = −7(1.567)t−3
(d) y = −17(2.005)−t
(e) y = 3(14.24)4t Problem 12.4.
(a) If you invest Po dollars
at 7% annual interest and the future
value is computed by continuous compounding, how long will it take for your
money to double?
(b) Suppose you invest Po dollars at r% annual interest and the future value is
computed by continuous compounding.
If you want the value of the account to
double in 2 years, what is the required
interest rate? (c) A rule of thumb used by many people to
determine the length of time to double
an investment is the rule of 70. The rule
says it takes about t = 70 years to dour
ble the investment. Graphically compare
this rule to the one isolated in part b. of
this problem.
Problem 12.5. The length of some ﬁsh are
modeled by a von Bertalanffy growth function.
For Paciﬁc halibut, this function has the form
L(t) = 200 (1 − 0.956 e−0.18t )
where L(t) is the length (in centimeters) of a
ﬁsh t years old.
(a) What is the length of a 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 L(t)?
Problem 12.6. A cancerous cell lacks normal biological growth regulation and can divide continuously. Suppose a single mouse
skin cell is cancerous and its mitotic cell cycle (the time for the cell to divide once) is 20
hours. The number of cells at time t grows
according to an exponential model.
(a) Find a formula C(t) for the number of
cancerous skin cells after t hours.
(b) Assume a typical mouse skin cell is
spherical of radius 50 × 10−4 cm. Find the
combined volume of all cancerous skin
cells after t hours. When will the volume
of cancerous cells be 1 cm3 ?
Problem 12.7. Your Grandfather purchased
a house for $55,000 in 1952 and it has increased in value according to a function y =
v(x), where x is the number of years owned.
These questions probe the future value of the
house under various mathematical models. CHAPTER 12. LOGARITHMIC FUNCTIONS 164 (a) Suppose the value of the house is
$75,000 in 1962. Assume v(x) is a linear
function. Find a formula for v(x). What
is the value of the house in 1995? When
will the house be valued at $200,000?
(b) Suppose the value of the house is
$75,000 in 1962 and $120,000 in 1967.
Assume v(x) is a quadratic function.
Find a formula for v(x). What is the
value of the house in 1995? When will
the house be valued at $200,000?
(c) Suppose the value of the house is
$75,000 in 1962. Assume v(x) is a function of exponential type. Find a formula
for v(x). What is the value of the house
in 1995? When will the house be valued
at $200,000? Problem 12.8. Solve the following equations
for x:
(a) log3 (5) = log2 (x)
(b) 10log2 (x) = 3
x (c) 35 = 7
(d) log2 (ln(x)) = 3
(e) ex = 105
(f) 23x+5 = 32 Problem 12.9. A ship embarked on a long voyage. At the start of the voyage, there were 500
ants in the cargo hold of the ship. One week
into the voyage, there were 800 ants. Suppose
the population of ants is an exponential function of time.
(a) How long did it take the population to
double?
(b) How long did it take the population to
triple?
(c) When were there be 10,000 ants on
board? (d) There also was an 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?
Problem 12.10. The populations of termites
and spiders in a certain house are growing exponentially. The house contains 100 termites
the day you move in. After 4 days, the house
contains 200 termites. Three days after moving in, there are two times as many termites
as spiders. Eight days after moving in, there
were four times as many termites as spiders.
How long (in days) does it take the population of spiders to triple?
Problem 12.11. In 1987, the population of
Mexico was estimated at 82 million people,
with an annual growth rate of 2.5%. The 1987
population of the United States was estimated
at 244 million with an annual growth rate
of 0.7 %. Assume that both populations are
growing exponentially.
(a) When will Mexico double its 1987 population?
(b) When will the United States and Mexico
have the same population?
Problem 12.12. The cities of Abnarca and
Bonipto have populations that are growing exponentially. In 1980, Abnarca had a population of 25,000 people. In 1990, its population
was 29,000.
Bonipto had a population of 34,000 in 1980.
The population of Bonipto doubles every 55
years.
(a) How long does it take the population of
Abnarca to double?
(b) When will Abnarca’s population equal
that of Bonipto? Chapter 13
Three Construction Tools
Sometimes the composition of two functions can be understood by graphical manipulation. When we discussed quadratic functions and parabolas in the previous section, certain key graphical maneuvers were laid
out. In this section, we extend those graphical techniques to general
function graphs. 13.1 A Lowtech Exercise
This section is all about building new functions from ones
we already have in hand. This can be approached symboly = f ( x)
ically or graphically. Let’s begin with a simple handson
exercise involving the curve in Figure 13.1.
By the vertical line test, we know this represents the
graph of a function y = f(x). With this picture and a piece
Figure 13.1: Start with some
of bendable wire we can build an INFINITE number of new
curve.
functions from the original function. Begin by making a
“model” of this graph by bending a piece of wire to the exact shape of
the graph and place it right on top of the curve. The wire model can be
manipulated in a variety of ways: slide the model back and forth horizontally, up and down vertically, expand or compress the model horizontally
or vertically.
Another way to build new curves from old ones is to exploit the built in
symmetry of the xycoordinate system. For example, imagine reﬂecting
the graph of y = f(x) across the xaxis or the yaxis.
In all of the above cases, we moved from the original
rotate
wire model of our function graph to a new curve that (by
the vertical line test) is the graph of a new function. The
big caution in all this is that we are NOT ALLOWED to
NOT a function graph
rotate or twist the curve; this kind of maneuver does lead
Figure 13.2:
Rotating a
to a new curve, but it may not be the graph of a function:
curve.
See Figure 13.2.
The pictures in Figure 13.3 highlight most of what we have to say in
165 CHAPTER 13. THREE CONSTRUCTION TOOLS 166 this section; the hard work remaining is a symbolic reinterpretation of
these graphical operations.
left right
up
down Horizontal shift. pull pull expand horizontally push push compress horizontally Horizontal expansion.
pull Vertical shift. Horizontal compression. pull
push push
compress
vertically expand
vertically Vertical expansion. Vertical compression.
reﬂect across yaxis reﬂect across
xaxis Vertical reﬂection. Horizontal reﬂection. Figure 13.3: Shifting, dilating, and reﬂecting y = f(x). 13.2 Reﬂection
In order to illustrate the technique of reﬂection, we will use a concrete
example:
Function:
Domain:
Range: y = p(x) = 2x + 2
−2 ≤ x ≤ 2
−2 ≤ y ≤ 6 As we know, the graph of y = p(x) on the domain −2 ≤ x ≤ 2 is a line of
slope 2 with yintercept 2, as pictured in Figure 13.4(a). Now, start with 13.2. REFLECTION 167 the function equation y = p(x) = 2x + 2 and replace every occurrence of
“y” by “−y.” This produces the new equation −y = 2x + 2; or, equivalently,
y = q(x) = −2x − 2.
The domain is still −2 ≤ x ≤ 2, but the range will
change; we obtain the new range by replacing “y” by “−y”
in the original range: −2 ≤ −y ≤ 6; so 2 ≥ y ≥ −6. The
graph of this function is a DIFFERENT line; this one has
slope −2 and yintercept −2. We contrast these two curves
in Figure 13.4(b), where q(x) is graphed as the “dashed
line” in the same picture with the original p(x). Once we
do this, it is easy to see how the graph of q(x) is really just
the original line reﬂected across the xaxis.
Next, take the original function equation y = p(x) =
2x + 2 and replace every occurrence of “x” by “−x.” This
produces a new equation y = 2(−x) + 2; or, equivalently, 8
6
4
2
−2 −2
−4
−6
−8 1 8
6
4
2 2 −2 yaxis −1 −2
−4
−6
−8 1 xaxis
2 (b) Graph of q(x) = −2x − 2.
8
6
4
2
−2 −1 −2
−4
−6
−8 yaxis xaxis
1 2 (c) Graph of r(x) = −2x + 2.
Figure 13.4: Reﬂecting y =
p(x). Important Facts 13.2.1 (Reﬂection). Let y = f(x) be a
function equation.
(i) We can reﬂect the graph across the xaxis and the resulting curve is the graph of the new function obtained
by replacing “y” by “−y” in the original equation. The
domain is the same as the domain for y = f(x). If the range for y = f(x)
is c ≤ y ≤ d, then the range of −y = f(x) is c ≤ −y ≤ d. In other words,
the reﬂection across the xaxis is the graph of y = −f(x).
(ii) We can reﬂect the graph across the yaxis and the resulting curve is the graph of the new function obtained
by replacing “x” by “−x” in the original equation. The
range is the same as the range for y = f(x). If the
domain for y = f(x) is a ≤ x ≤ b, then the domain xaxis (a) Graph of p(x) = 2x + 2. y = r(x) = −2x + 2.
The domain must also be checked by replacing “x” by
“−x” in the original domain condition: −2 ≤ −x ≤ 2, so
2 ≥ x ≥ −2. It just so happens in this case, the domain
is unchanged. This is yet another DIFFERENT line; this
one has slope −2 and yintercept 2. We contrast these
two curves in Figure 13.4(c), where r(x) is graphed as the
“dashed line” in the same picture with the original p(x).
Once we do this, it is easy to see how the graph of r(x) is
really just the original curve reﬂected across the yaxis.
This example illustrates a general principle referred to
as the reﬂection principle. −1 yaxis CHAPTER 13. THREE CONSTRUCTION TOOLS 168 of y = f(−x) is a ≤ −x ≤ b. Using composition notation, the reﬂection across the yaxis is the graph of
y = f(−x).
Example 13.2.2. Consider the parallelogramshaped region R with vertices (0, 2), (0, −2), (1, 0), and (−1, 0). Use the reﬂection principle to ﬁnd
functions whose graphs bound R. P = (0,2)
l3 l 2 Q = (1,0)
xaxis
l l 4 1 Region R
yaxis Figure 13.5: The region R. yaxis
y = f(−x) −4 −2 y = f ( x) 2 4 xaxis −y = f(x) Figure 13.6: Reﬂecting the
semicircle. Solution. Here is a picture of the region R: First off, using the two point formula for the equation of a line, we
ﬁnd that the line ℓ1 passing through the points P = (0, 2)
and Q = (1, 0) is the graph of the function y = f1 (x) =
−2x + 2. By Fact 13.2.1 (i), ℓ2 is the graph of the equation −y = −2x + 2, which we can write as the function
y = f2 (x) = 2x − 2. By Fact 13.2.1 (ii) applied to ℓ2 , the line
ℓ3 is the graph of the function y = f3 (x) = −2x − 2. Finally,
by Fact 13.2.1 (i) applied to ℓ3 , the line ℓ4 is the graph of
the equation −y = −2x − 2, which we can write as the
function y = f4 (x) = 2x + 2.
Figure 13.6 illustrates the fact that we need to be
careful about the domain of the original function when
using the reﬂection principle. For example, consider
y = f(x) = 1 + 1 − (x − 3)2 . The largest possible domain of
xvalues is 2 ≤ x ≤ 4 and the graph is an upper semicircle
of radius 1 centered at the point (3, 1).
Reﬂection across the xaxis gives the graph of
y = −1 − 1 − (x − 3)2 on the same domain; reﬂection
across the yaxis gives the graph of y = 1 + 1 − (x + 3)2
on the new domain −4 ≤ x ≤ −2. 13.3 Shifting
√
Let’s start out with the function y = f(x) = 4 − x2 , which
has a largest possible domain −2 ≤ x ≤ 2. From Chap2
ter 6, the graph of this equation is an upper semicircle
1.5
1
of radius 2 centered at the origin (0, 0). Sliding the graph
back and forth horizontally or vertically (or both), never
0.5
xaxis
rotating or twisting, we are led to the “dashed curves” be−2
−1
0
1
2
low (contrasted with the original graph which is plotted
Figure 13.7: Graph of y =
with a solid curve). This describes some shifted curves on
√
4 − x2 .
a pictorial level, but what are the underlying equations?
For this example, we can use the fact that all of the shifted curves are
still semicircles and Chapter 6 tells us how to ﬁnd their equations.
2.5 yaxis 13.3. SHIFTING 169 The lower righthand dashed semicircle is of radius 2 and is centered
at (3, 0), so the corresponding equation must be y = 4 − (x − 3)2. The
upper lefthand dashed semicircle is of radius 2 and centered at (0, 3), so
√
the corresponding equation must be y = 3+ 4 − x2 . The upper righthand
dashed semicircle is of radius 2 and centered at (3, 3), so the corresponding equation must be y = 3 + 4 − (x − 3)2 .
Keeping this same example, we can continue this kind
5
of shifting more generally by thinking about the effect of
4
making the following three replacements in the equation
√
3
y = 4 − x2 :
2
1 x
y
(x and y) x−h
y−k
(x − h and y − k). −2 −1 1 2 3 4 5 Figure 13.8: Shifting the upper semicircle. These substitutions lead to three new equations, each
the equation of a semicircle:
y= 4 − ( x − h) 2 ; y−k= 4 − x2 ; and, y−k= 4 − ( x − h) 2 . ⇐ ⇐ ⇐ Upper semicircle with radius
2 and center (h, 0).
Upper semicircle with radius
2 and center (0, k).
Upper semicircle with radius
2 and center (h, k). There are three potentially confusing points with this
example:
• Be careful with the sign (i.e., ±) of h and k. In Figure 13.9, if h = 1, we horizontally shift the semicircle
1 unit to the right; whereas, if h = −1, we horizontally shift the semicircle 1 units to the right. But,
shifting −1 unit to the right is the same as shifting 1
unit to the left! In other words, if h is positive, then
a horizontal shift by h will move the graph h units
to the right; if h is negative, then a horizontal shift
by h will move the graph h units to the left.
• If k is positive, then a vertical shift by k will move
the graph k units up; if k is negative, then a vertical
shift by k will move the graph k units down. These
conventions insure that the “positivity” of h and k
match up with “rightward” and “upward” movement
of the graph. yaxis original
curve
xaxis h negative h positive curve shifted
h units left
if h is negative curve shifted
h units right
if h is positive Figure 13.9: Potentially confusing points. CHAPTER 13. THREE CONSTRUCTION TOOLS 170 • When shifting, the domain of allowed xvalues may
change.
This example illustrates an important general principle referred to as
the shifting principle.
Important Facts 13.3.1 (Shifting). Let y = f(x) be a function equation.
(i) If we replace “x” by “x − h” in the original function equation, then
the graph of the resulting new function y = f(x − h) is obtained by
horizontally shifting the graph of f(x) by h. If h is positive, the picture
shifts to the right h units; if h is negative, the picture shifts to the left
h units. If the domain of f(x) is an interval a ≤ x ≤ b, then the domain
of f(x − h) is a ≤ x − h ≤ b. The range remains unchanged under
horizontal shifting.
(ii) If we replace “y” by “y − k” in the original function equation, then
the graph of the resulting new function y = f(x) + k is obtained by
vertically shifting the graph of f(x) by k. If k is positive, the picture
shifts upward k units; if k is negative, the picture shifts downward
k units. If the range of f(x) is an interval c ≤ y ≤ d, then the range
of f(x) + k is c ≤ y − k ≤ d. The domain remains unchanged under
vertical shifting. 13.4 Dilation
1 To introduce the next graphical principle we will look at
the function
x
y = f(x) = 2
.
x +1 yaxis 0.75
0.5
0.25
−6 −4 −2 2
−0.25 4 6
xaxis Using a graphing device, we have produced a plot of the
graph on the domain −6 ≤ x ≤ 6. Figure 13.10 shows the
−1
curve has a high point H (like a “mountain peak”) and a
Figure 13.10: Graph of y =
x
low point L (like a “valley”). Using a graphing device, we
f(x) = x2 +1 .
1
can determine that the high point is H = (1, 2 ) (it lies on
1
the line with equation y = 2 ) and the low point is L = −1, − 1 (it lies on
2
1
the line with equation line y = − 1 ), so the range is − 2 ≤ y ≤ 1 . Draw two
2
2
new horizontal lines with equations y = 2 · ± 1 = ±1. Grab the high point
2
H on the curve and uniformly pull straight up, so that the high point now
lies on the horizontal line y = 1 at (1, 1). Repeat this process by pulling
L straight downward, so that the low point is now on the line y = −1
at (−1, −1). We end up with the ”stretched dashed curve” illustrated in
Figure 13.11(a). In terms of the original function equation y = x2x 1 , we
+
are simply describing the graphical effect of multiplying the ycoordinate
of every point on the curve by the positive number 2. In other words, the
2x
dashed curve is the graph of y = x2 +1 = 2 x2x 1 .
+
−0.5 −0.75 13.4. DILATION 171 1
Next, draw the two horizontal lines with equations y = 1 · ± 1 = ± 4 .
2
2
Grab the high point H on the curve (in Figure 13.10) and uniformly
push straight down, so that the high point now lies on the horizontal
1
1
line y = 4 at (1, 4 ). Repeat this process at the low point L by pushing the
1
curve straight upward, so that the low point is now on the line y = − 4
1
at (−1, − 4 ). We end up with the new ”dashed curve” illustrated in Figure 13.11(b). In terms of the original function equation y = x2x 1 , we are
+
simply describing the graphical effect of multiplying the ycoordinate of
1
every point on the curve by the positive number 2 . In other words, the
x
dashed curve is the graph of y = 2(x2 +1) .
We could repeat this process systematically:
1 yaxis 0.75
0.5 • Pick a positive number c.
• Draw the two horizontal lines y = c · ± 1 . If c > 1, then
2
c
the graph of y = 2 is parallel and above the graph of
1
y = 2 . On the other hand, if 0 < c < 1, then this new
1
line is parallel and below y = 2 .
• Uniformly deform the original graph (in Figure 13.10) so that the new curve has it’s high and
low points just touching y = ± c . This will involve
2
vertically stretching or compressing, depending on
whether 1 < c or 0 < c < 1, respectively. A number of
possibilities are pictured in Figure 13.11(c).
We refer to each new dashed curve as a vertical dilation
of the original (solid) curve. This example illustrates an
important principle. 0.25
−6 −4 −2 2
−0.25 y
c (i) If we replace “ y ” by “ ” in the original equation, then
the graph of the resulting new equation is obtained by
vertical dilation of the graph of y = f(x). The domain
of xvalues is not affected.
(ii) If c > 1, then the graph of y = f(x) (i.e., y = cf(x)) is a
c
vertically stretched version of the original graph. 6 xaxis −0.75
−1 (a) Vertical expansion.
yaxis 1
0.75
0.5 11 11
00 00
11 11
00 00 11 11
00 00
−6
−4 1111
0000
1111
0000
0.25
1111
0000
2
111
000
111
000
−0.25
111
000 −2 4 6 xaxis −0.5 111
000
111
000
111
000 −0.75
−1 (b) Vertical compression.
1
0.75 Important Facts 13.4.1 (Vertical dilation). Let c > 0 be a
positive number and y = f(x) a function equation. 4 −0.5 1111
0000
1111
0000
yaxis
1111
0000 0.5
0.25 11
00
111 11
000 00
11
00 −4 00
111 −2
−6 000 11
11
00
111 11
000 00 2
−0.25 4 6 xaxis −0.5
−0.75
−1 (c) Many possibilities.
Figure 13.11: Dilating y =
f(x). (iii) If 0 < c < 1, then the graph of y = f(x) (i.e., y = cf(x)) is
c
a vertically compressed version of the original graph.
If we combine dilation with reﬂection across the xaxis,
we can determine the graphical relationship between y =
f(x) and y = cf(x), for any constant c. The key observation
is that reﬂection across the xaxis corresponds to the case c = −1. CHAPTER 13. THREE CONSTRUCTION TOOLS 172 Example 13.4.2. Describe the relationship between the graphs of
y = f(x) = 1 − (x + 1)2 , y = −f(x) = − 1 − (x + 1)2, and
y = −4f(x) = −4 1 − (x + 1)2 . Solution. The graph of y = f(x) is an upper semicircle of
radius 1 centered at the point (−1,0). To obtain the picture of the graph of y = −4f(x), we ﬁrst reﬂect y = f(x)
across the xaxis; this gives us the graph of y = −f(x).
Then, we vertically dilate this picture by a factor of c = 4
to get the graph of y = −f(x), which is the same as the
4
graph of the equation y = −4f(x). See Figure 13.12. step 1: start with upper
semicircle xaxis
step 2: reﬂect across
step 3: stretch curve from
step 2: to get
y = −4f(x) Figure 13.12: Reﬂecting and
dilating a lower semicircle. Let’s return to the original example y = x2x 1 and in+
vestigate a different type of dilation where the action is
taking place in the horizontal direction (whereas it was in
the vertical direction before). Grab the righthand end of the graph (in
Figure 13.10) 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
1
1
high point H = 1, 2 of the original curve moves to the new location 2, 2
1
1
and the low point L = −1, − 2 moves to the new location −2, − 2 .
1 yaxis 1 0.75
0.5 −4 0.5 0.25 111
000
−6
111
000 yaxis 0.75
0.25 −2 2
−0.25 4 6 xaxis 11
00 −4
111
−6 000
11
00
111
000 −2 2
−0.25 −0.5 −0.75 −1 6 xaxis −0.5 −0.75 4 −1 (a) Stretching.
(b) Compressing.
Figure 13.13: Horizontally dilating y = x2x 1 .
+ The result will be somewhat analogous to stretching a spring. By the
same token, we could push the 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 H = 1, 1 of the original curve moves
2
11
to the new location 2 , 2 and the low point L = −1, − 1 moves to the new
2
1
location − 1 , − 2 . These two situations are indicated in Figure 13.13. We
2
refer to each of the dashed curves as a horizontal dilation of the original
(solid) curve. 13.5. VERTEX FORM AND ORDER OF OPERATIONS 173 The tricky point is to understand what happens on the level of the
original equation. In the case of the stretched graph in Figure 13.13(a),
you can use a graphing device to verify that this looks like the graph
x/2
of y = (x/2)2 +1 ; in other words, we replaced “x” by “x/2” in the original
equation. In the case of the compressed graph in Figure 13.13(b), you can
use a graphing device to verify that this looks like the graph of y = (2x2x+1 ;
)2
x
in other words, we replaced “x” by “ 1/2 = 2x” in the original equation.
The process just described leads to a general principle.
Important Facts 13.4.3 (Horizontal dilation). Let c > 0 be a positive
number and y = f(x) a function equation.
(i) If we replace “ x” by “ x ” in the original function equation, then the
c
graph of the resulting new function y = f x is obtained by a horizonc
tal dilation of the graph of y = f(x). If the domain of f(x) is a ≤ x ≤ b,
then the domain of y = f x is a ≤ x ≤ b.
c
c
(ii) If c > 1, then the graph of y = f x
c (iii) If 0 < c < 1, then the graph of y = f is a horizontal stretch.
x
c is a horizontal compression. 13.5 Vertex Form and Order of Operations
Using the language of function compositions we can clarify our discussion in Example 7.1.2. Let’s revisit that example:
Example 13.5.1. The problem is to describe a sequence of geometric maneuvers that transform the graph of y = x2 into the graph of y = −3(x − 1)2 + 2.
Solution. The idea is to rewrite y = −3(x − 1)2 + 2 as a composition of y = x2
with four other functions, each of which corresponds to a horizontal shift,
vertical shift, reﬂection or dilation. Once we have done this, we can
read off the order of geometric operations using the order of composition.
Along the way, pay special attention to the exact order in which we will
be composing our functions; this will make a big difference.
To begin with, we can isolate four key numbers in the equation:
Horizontal shift by h = 1 Reﬂect − 3
Dilate by 3 (x − 1 )2 + 2
Vertical shift by k = 2 We want to use each number to deﬁne a new function, then compose
these in the correct order. We will also give our starting function y = x2 a
speciﬁc name to make things deﬁnite:
f(x) = x2
v(x) = x + 2
d(x) = 3x. h(x) = x − 1
r(x) = −x CHAPTER 13. THREE CONSTRUCTION TOOLS 174
Now, verify that v r d f h(x) = v r d f[x − 1] = v r d{(x − 1)2} = v r(3(x − 1)2 )
= v[−3(x − 1)2 ]
= −3(x − 1)2 + 2
= −3x2 + 6x − 1. 13.6 Summary of Rules 2 −2 2 Figure 13.14: A multipart
function. For quick reference, we summarize the consequence of
shifting and expanding symbolically and pictorially. The
running example for Tables 13.1, 13.2, and 13.3 will be
a multipart function y = f(x) whose graph, seen in Figure 13.14, consists of a line segment and a quarter circle
on the domain −2 ≤ x ≤ 2:
f(x) = x+2
if −2 ≤ x ≤ 0
√
2 if 0 ≤ x ≤ 2
4−x 13.6. SUMMARY OF RULES 175 Reﬂection
Symbolic
Change Replace x with
−x. Replace
f(x)
with −f(x). New Equation Graphical
Consequence y = f(−x) A reﬂection
across the
yaxis. y = −f(x) A reﬂection
across the
xaxis. Table 13.1: Reﬂecting y = f(x). Picture 2 −2 2 −2 2 −2 CHAPTER 13. THREE CONSTRUCTION TOOLS 176 Shifting (Assume c > 0 )
Symbolic
Change New Equation Graphical
Consequence y = f( x − c ) A shift to the
right c units. Picture
2 Replace x with
(x − c ). 2 Replace x with
(x + c ). y = f( x + c ) 4 2 A shift to the
left c units.
−4 −2 4 Replace
f(x)
with (f(x) + c). y = f(x) + c A shift up c
units. 2 −2 Replace
f(x)
with (f(x) − c). y = f(x) − c A shift down c
units. 2 −2 2
2 Table 13.2: Shifting y = f(x). 13.6. SUMMARY OF RULES 177 Dilation
Symbolic
Change New Equation Graphical
Consequence Picture c=2 If c > 1, replace
x with x .
c y=f x
c A horizontal
expansion.
−4 −3 −2 −1 c= If 0 < c < 1, replace x with x .
c y=f x
c A horizontal
compression. 1
2 4
3
2
1
01234 4
3
2
1 11
00
−4 −3 −2 −1 00 1
0
11 c=2 If c > 1, replace
f(x) with (cf(x)). y = cf(x) A vertical
expansion.
−4 −3 −2 −1 If 0 < c < 1, replace f(x) with
(cf(x)). c= y = cf(x) A vertical
compression. 1
2 23 4 111
000
111
000
4
111
000
111
000
111
000 3
2
1 11
00
0
11
00 1234 4
3
2
1
0
1
1
0
1
0 111
−4 −3 −2 −1 000 1 2 3 4
0
111
000 Table 13.3: Dilating y = f(x). CHAPTER 13. THREE CONSTRUCTION TOOLS 178 13.7 Exercises
Problem 13.1. On a single set of axes, sketch
a picture of the graphs of the √
following four
√
√
equations: y = −x + 2, y = −x − 2, y = x + 2,
√
and y = x − 2. These equations determine
lines, which in turn bound a diamond shaped
region in the plane.
(a) Show that the unit circle sits inside this
diamond tangentially; i.e. show that the
unit circle intersects each of the four
lines exactly once.
(b) Find the intersection points between the
unit circle and each of the four lines.
(c) Construct a diamond shaped region in
which the circle of radius 1 centered at
(−2, − 1) sits tangentially. Use the techniques of this section to help.
Problem 13.2. The graph of a function y = f(x)
is pictured with domain −2.5 ≤ x ≤ 3.5. y
π
f(x) π/2 −1 1 x Problem 13.4.
(a) Each of the six functions
y = f(x) below can be written in the
“standard form”
y = AB(x − C) + D,
for some constants A,B,C,D. Find these
constants, describe the precise order of
graphical operations involved in going
from the graph of y = x to the graph of
y = f(x) (paying close attention to the order), write out the multipart rule, sketch
the graph and calculate the coordinates
of the “vertex” of the graph.
(a1) f(x) = x − 2
(a2) f(x) = 2x + 3 Sketch the graph of each of the new functions listed below. (a3) f(x) = 2x − 1 (a) g(x) = 2f(x + 1) (a4) f(x) = 2(x − 1) (b) h(x) = 1 f(2x − 1)
2 (a5) f(x) = 32x − 1 + 5 (c) j(x) = 5f( 1 x + 2) − 2
3 (a6) f(x) = −2x + 3 − 1 Problem 13.3. The graph of a function y = f(x)
is pictured with domain −1 ≤ x ≤ 1. Sketch the
graph of the new function
y = g(x) = 1
f(3x) − 0.5.
π Find the largest possible domain of the function y = g(x). (b) Solve the following inequalities using
your work in the previous part of this
problem:
(b1) x − 2 ≤ 3
(b2) 1 ≤ 2x + 3 ≤ 5
(b3) y = 32x − 1 + 5 ≥ 10 13.7. EXERCISES
(c) The graphs of y = 32x − 1 + 5
and y = −x − 3 + 10 intersect to form a
bounded region of the plane. Find the
vertices of this region and sketch a picture.
Problem 13.5. Consider the function y = f(x)
with multipart deﬁnition if x ≤ −1
0 2x + 2 if −1 ≤ x ≤ 0
f(x) = −x + 2 if 0 ≤ x ≤ 2 0
if x ≥ 2 179
(c) The graph of z = a(x) from part (a) is
given below. Sketch the graph and ﬁnd
the rule for the function z = 2a(3x + 3)+ 1;
make sure to specify the domain and
range of this new function.
10 yaxis 8
6
4
2
xaxis (a) Sketch the graph of y = f(x). (b) Is y = f(x) an even function? Is y = f(x)
an odd function? (A function y = f(x) is
called even if f(x) = f(−x) for all x in the
domain. A function y = f(x) is called odd
if f(−x) = −f(x) for all x in the domain.)
(c) Sketch the reﬂection of the graph across
the xaxis and yaxis. Obtain the resulting multipart equations for these reﬂected curves.
(d) Sketch the vertical dilations y = 2f(x)
1
and y = 2 f(x).
(e) Sketch the horizontal dilations y = f(2x)
and y = f( 1 x).
2
(f) Find a number c > 0 so that the highest
point on the graph of the vertical dilation
y = cf(x) has ycoordinate 11.
(g) Using horizontal dilation, ﬁnd a number
c > 0 so that the function values f( x ) are
c
5
nonzero for all − 2 < x < 5.
(h) Using horizontal dilation, ﬁnd positive
numbers c,d > 0 so that the function
values f( 1 (x − d)) are nonzero precisely
c
when 0 < x < 1.
Problem 13.6. An isosceles triangle has sides
of length x, x and y. In addition, assume the
triangle has perimeter 12. 2 4 6 8 10 Problem 13.7. Describe how each graph differs from that of y = x2 .
(a) y = 2x2
(b) y = x2 − 5
(c) y = (x − 4)2
(d) y = (3x − 12)2
(e) y = 2(3x − 12)2 − 5
Problem 13.8. In each case, start with the
function y = x and perform the operations
described to the graph, in the order speciﬁed.
Write out the resulting rule for the function
and sketch the ﬁnal graph you obtain.
(a) (1) horizontally compress by a factor of
2; (2) horizontally shift to the left by 2;
(3) vertically stretch by a factor of 7; (4)
vertically shift up 2 units.
(b) (1) horizontally stretch by a factor of 2;
(2) horizontally shift to the right by 2; (3)
vertically compress by a factor of 7; (4)
vertically shift down 2 units.
(c) (1) horizontally shift to the right 2 units;
(2) horizontally compress by a factor of
3. (a) Find the rule for a function that computes the area of the triangle as a function of x. Describe the largest possible
domain of this function. Problem 13.9.
(a) Begin with the function
y = f(x) = 2x . (b) Assume that the maximum value of the
function a(x) in (a) occurs when x = 4.
Find the maximum value of z = a(x) and
z = 2a(3x + 3) + 1. (a1) Rewrite each of the following functions in standard exponential form:
f(2x), f(x − 1), f(2x − 1), f(2(x − 1)),
3f(x), 3f(2(x − 1)). 180 CHAPTER 13. THREE CONSTRUCTION TOOLS
(a2) Is the function 3f(2(x − 1))+ 1 a function of exponential type? cal dilation, horizontal shifting, horizontal dilation) to the standard exponential
model y = Ao bx . For which of the four
operations is the resulting function still
a standard exponential model? (a3) Sketch the graphs of f(x), f(2x), f(2(x−
1)), 3f(2(x − 1)) and 3f(2(x − 1)) + 1
in the same coordinate system
and explain which graphical operation(s) (vertical shifting, vertical
Problem 13.10. Begin with a sketch of the
dilation, horizontal shifting, horigraph of the function y = 2x on the domain
zontal dilation) have been carried
of all real numbers. Describe how to use the
out.
“four tools” of Chapter 13 to obtain the graphs
of these functions: y = −2x , y = 2−x , y = 3(2x ),
(b) In general, explain what happens when
y = 1 (2x ), y = 3 + 2x , y = 2x − 2, y = 2x−2 ,
you apply the four construction tools
3
y = 2x+2 , y = 23x , y = 2x/3 .
of Chapter 13 (vertical shifting, verti Chapter 14
Rational Functions
p
A rational function is a function of the form f(x) = q(x) where p(x) and q(x)
( x)
are polynomials. For example, the following are all rational functions. f(x) = x
x2 + 1
4x5 − 4x2 − 8
x6
g(x) =
h(x) = 3
j(x) = 8
3x + 4
3x − 5
x + x2 − x + 1
x + 5x − 1
2 There is a huge variety of rational functions. In this course, we will
concentrate our efforts exclusively toward understanding the simplest
type of rational functions: lineartolinear rational functions. Lineartolinear rational functions are rational functions in which the numerator and the denominator are both linear polynomials. The following are
lineartolinearrational functions.
k(x) = 4x + 3
5x − 6
0.34x − 0.113
x
4
m(x) =
n(x) =
p(x) = 5
3x + 4
2x + 1
x−1
x − 1.117
8 The simplest example of a lineartolinear rational
1
function is f(x) = x whose graph is shown in Figure 14.1.
This is an important example for the study of this class of
functions, as we shall see.
1
Let’s consider the graph of this function, f(x) = x . We
ﬁrst begin by considering the domain of f. Since the nu1
merator of x is a constant, and the denominator is just x,
the only way we can run into difﬁculty when evaluating
this function is if we try to divide by zero; that is, the only
value of x not in the domain of this function is zero. The
function is deﬁned for all x except x = 0. At x = 0, there
must be a gap, or hole, in the graph.
Figure 14.1: The graph of
1
To get the graph started we might simply give ourselves
f(x) = x .
a point on the graph. For instance, we see that f(1) =
1/1 = 1, so the point (1,1) is on the graph. Then, if we use larger values of
x, we see that 1/x becomes smaller as x grows. For instance, f(2) = 1/2,
f(10) = 1/10, and f(1000) = 1/1000. In addition, we see that, other than
the fact that 1/x > 0 for positive x, there is nothing preventing us from
10 5 0 10 5 0 5 10 181 5 10 CHAPTER 14. RATIONAL FUNCTIONS 182 making 1/x as small as we want simply by taking x big enough. Want 1/x
to be less than 0.001? Just pick x bigger than 1000. Want 1/x to be less
than 0.000001? Just use x bigger than 1,000,000.
What this means graphically is that as x gets bigger (starting from
x = 1), the curve y = 1/x gets closer and closer to the xaxis. As a result,
we say that the xaxis is a horizontal asymptote for this function.
We see the same behavior for negative values of x. If x is large, and
negative (think 1000, or 1000000), then 1/x is very small (i.e., close to
zero), and it gets smaller the larger x becomes. Graphically, this means
that as x gets large in the negative direction, the curve y = 1/x gets closer
and closer to the xaxis. We say that, in both the positive and negative
directions, y = 1/x is asymptotic to the xaxis.
A similar thing happens when we consider x near zero. If x is a small
positive number (think 1/2, or 1/10, or 1/10000), then 1/x is a large
positive number. What’s more, if we think of x as getting closer to zero,
1/x gets bigger and bigger. Plus, there is no bound on how big we can
make 1/x simply by taking x as close to zero as we need to.
For instance, can 1/x be as big as 10000? All you need to do is pick a
positive x smaller than 1/10000.
Graphically, what this means is that as x approaches zero from the
positive side, the y value gets larger and larger. As a result, the curve
approaches the yaxis as y gets larger. We say that the yaxis is a vertical
asymptote for the curve y = 1/x.
We see the same phenomenon as x approaches zero from the negative
side: y = 1/x gets larger in the negative direction (i.e., it gets more and
more negative). The curve gets closer and closer to the negative yaxis as
y becomes more and more negative. Again, we say that the y − axis is a
vertical asymptote for the curve y = 1/x.
It turns out that every lineartolinear rational function has a graph
that looks essentially the same as the graph of y = 1/x. Let’s see why.
+b
Consider the lineartolinear rational function f(x) = ax+d . If we divide
cx
cx + d into ax + b, the result is
f(x) = ax + b
a b − ad
c
=+
cx + d
c cx + d which we can rewrite as
f(x) = a
1
1
1
ad
a
ad
a
bc − ad
·
·
.
·
+ b−
= + b−
=+
d
2
c
c
cx + d
c
c
c
c
c (x + c )
x+ d
c If we now let
A= bc − ad
c2 a
,B=
c then we can write
f(x) = A + B 1
.
x+C , and C = d
,
c 183
If we let g(x) = 1/x then we have shown that f(x) = A + Bg(x + C), and so
the graph of the function f is just a horizontally shifted, vertically shifted
and vertically dilated version of the graph of g. Also, if B turned out to be
negative the graph would be vertically ﬂipped, too.
Why is that useful? It means that the graph of a lineartolinear rational function can only take one of two forms. Either it looks like this: Or like this: We can sketch an accurate graph of a lineartolinear rational function
by sketching the asymptotes and then sketching in just one point on the
graph. That will be enough information to nail down a decent sketch. We
can always plot more points to give us more precision, but one point is
enough to capture the essence of the graph.
Given a lineartolinear function that we wish to graph, we must ﬁrst
determine the asymptotes. There will be one horizontal asymptote and
one vertical asymptote.
The vertical asymptote will be a vertical line with equation x = k where
k is the one x value which is not in the domain of the function. That is,
ﬁnd the value of x which makes the denominator zero and that will tell
you the vertical asymptote.
The horizontal asymptote is a little more involved. However, we can
quickly get to a shortcut. The essence of a horizontal asymptote is that it
describes what value the function is approximately equal to for very large
values of x. To study what a lineartolinear function is like when x is
very large, we can perform the following algebraic manipulation:
f(x) = a+ b
ax + b
ax + b 1/x
x
=
·
=
cx + d
cx + d 1/x
c+ d
x While dividing by x is troublesome if x equals zero, here we are assuming
x is very large, so it is certainly not zero.
Now, consider this last expression in the above equation. If x is very
large, then
b
≈0
x CHAPTER 14. RATIONAL FUNCTIONS 184 (where ≈ means ”is approximately”) and likewise
d
≈ 0.
x
Hence, when x is very large,
f(x) ≈ a
a+0
=.
c+0
c We can interpret this by saying that when x is very large, the function
f(x) is is close to a constant, and that constant is a . Thus, the horizontal
c
ax + b
is the horizontal line y = a .
asymptote of f(x) =
c
cx + d
Example 14.0.1. Sketch the graph of the function f(x) = 3x−1
.
2x+7 Solution. We begin by ﬁnding the asymptotes of f.
The denominator is equal to zero when 2x + 7 = 0, i.e., when x = −7/2.
As a result, the vertical asymptote for this function is the vertical line
x = −7/2.
By taking the ratio of the coefﬁcients of x in the numerator and denominator, we can ﬁnd that the horizontal asymptote is the horizontal
line
3
y= .
2
10 5 0
10 5 0 5 10 5 10 Figure 14.2: The graph of f(x) = 3x−1
2x+7 . We then sketch these two asymptotes. The last thing we need is a
single point. For instance, we may evaluate f(0):
f( 0) = −1
7 14.1. MODELING WITH LINEARTOLINEAR RATIONAL FUNCTIONS185
and so the point (0, −1/7) is on the graph. With this information, we know
that the curve lies below the horizontal asymptote to the right of the
vertical asymptote, and consequently the curve lies above the horizontal
asymptote to the left of the vertical asymptote.
We graph the result in Figure 14.2. 14.1 Modeling with Lineartolinear Rational
Functions
As we have done with other sorts of functions, such as linear and quadratic,
we can also model using lineartolinear rational functions. One reason
for using this type of function is their asymptotic nature. Many changing quantities in the world continually increase or decrease, but with
bounds on how large or small they can get. For instance, a population
may steadily decrease, but a population can never be negative. Conversely, a population may steadily increase, but due to environmental
and other factors we may hypothesize that the population will always
stay below some upper bound. As a result, the population may ”level
off”. This leveling off behavior is exempliﬁed by the asymptotic nature of
the lineartolinear rational functions, and so this type of function provides a way to model such behavior.
Given any lineartolinear rational function, we can always divide the
numerator and the denominator by the coefﬁcient of x in the denominator. In this way, we can always assume that the coefﬁcient of x in the
denominator of a function we seek is 1. This is illustrated in the next
example.
Example 14.1.1. Let f(x) =
f(x) =
= 2x + 3
·
5x − 7 2x+3
.
5x−7 Then 1
5
1
5 3
+5
.
x− 7
5 2
x
5 Thus, in general, when we seek a lineartolinear rational function,
we will be looking for a function of the form
f(x) = ax + b
x+c and thus there are three parameters we need to determine.
Note that for a function of this form, the horizontal asymptote is y = a
and the vertical asymptote is x = −c.
Since these functions have three parameters (i.e., a, b and c), we will
need three pieces of information to nail down the function. CHAPTER 14. RATIONAL FUNCTIONS 186 There are essentially three types of modeling problems that require the
determination of a lineartolinear function. The three types are based
on the kind of information given about the function. The three types are:
1. You know three points the the graph of the function passes through;
2. You know one of the function’s asymptotes and two points the graph
passes through;
3. You know both asymptotes and one point the graph passes through.
Notice that in all cases you know three pieces of information. Since
a lineartolinear function is determined by three parameters, this is exactly the amount of information needed to determine the function.
The worst case, in terms of the amount of algebra you need to do, is
the ﬁrst case. Let’s look at an example of the algebra involved with this
sort.
Example 14.1.2. Find the lineartolinear rational function f(x) such that
f(10) = 20, f(20) = 32 and f(25) = 36.
Solution. Since f(x) is a lineartolinear rational function, we know
f(x) = ax + b
x+c for constants a, b, and c. We need to ﬁnd a, b and c.
We know three things.
First, f(10) = 20. So
f(10) = 10a + b
= 20,
10 + c which we can rewrite as
10a + b = 200 + 20c. (14.1) Second, f(20) = 32. So
f(20) = 20a + b
= 32,
20 + c which we can rewrite as
20a + b = 640 + 32c.
Third, f(25) = 36. So
f(25) = 25a + b
= 36,
25 + c (14.2) 14.1. MODELING WITH LINEARTOLINEAR RATIONAL FUNCTIONS187
which we can rewrite as
25a + b = 900 + 36c. (14.3) These three numbered equations are enough algebraic material to
solve for a, b, and c. Here is one way to do that.
Subtract equation 14.1 from equation 14.2 to get
10a = 440 + 12c (14.4) and subtract equation 14.2 from equation 14.3 to get
5a = 260 + 4c (14.5) Note that we’ve eliminated b. Now multiply this last equation by 2 to
get
10a = 520 + 8c
Subtract equation 14.4 from this to get
0 = 80 − 4c
which easily give us c = 20.
Plugging this value into equation 14.4, we can ﬁnd a = 68, and then
we can ﬁnd b = −80.
Thus,
f(x) = 68x − 80
.
x + 20 We can check that we have done the algebra correctly by evaluating
f(x) at x = 10, x = 20 and x = 25. If we get f(10) = 20, f(20) = 32 and
f(25) = 36, then we’ll know our work is correct.
Algebraically, this was the worst situation of the three, since it required the most algebra. If, instead of knowing three points, we know
one or both of the asymptotes, then we can easily ﬁnd a and/or c, and
so cut down on the amount of algebra needed. However, the method is
essentially identical.
Let’s now apply these ideas to a real world problem.
Example 14.1.3. Clyde makes extra money selling tickets in front of the
Safeco Field. The amount he charges for a ticket depends on how many he
has. If he only has one ticket, he charges $100 for it. If he has 10 tickets,
he charges $80 a piece. But if he has a large number of tickets, he will
sell them for $50 each. How much will he charge for a ticket if he holds 20
tickets? CHAPTER 14. RATIONAL FUNCTIONS 188 Solution. We want to give a lineartolinear rational function relating the
price of a ticket y to the number of tickets x that Clyde is holding. As we
saw above, we can assume the function is of the form
y= ax + b
x+c where a, b and c are numbers. Note that y = a is the horizontal asymptote. When x is very large, y is close to 50. This means the line y = 50 is
a horizontal asymptote. Thus a = 50 and
y= 50x + b
.
x+c Next we plug in the point (1,100) to get a linear equation in b and c.
50 · 1 + b
1+c
100 · (1 + c) = 50 + b
50 = b − 100c
100 = Similarly, plugging in (10,80) and doing a little algebra (do it now!)
gives another linear equation 300 = b − 80c. Solving these two linear
equations simultaneously gives c = 12.5 and b = 1300. Thus our function
is
y= 50x + 1300
x + 12.5 and, if Clyde holds 20 tickets, he will charge
y= 50 · 20 + 1300
= $70.77
20 + 12.5 per ticket. 14.2 Summary
• Every lineartolinear rational function has a graph which is a shifted,
scaled version of the curve y = 1/x. As a result, they have one vertical asymptote, and one horizontal asymptote.
• Every lineartolinear rational function f can be expressed in the
form
f(x) = ax + b
.
x+c This function has horizontal asymptote y = a and vertical asymptote x = −c. 14.3. EXERCISES 189 14.3 Exercises
Problem 14.1. Give the domain of each of
the following functions. Find the x and yintercepts of each function. Sketch a graph
and indicate any vertical or horizontal asymptotes. Give equations for the asymptotes.
(a) f(x) =
(c) h(x) =
(e) k(x) = 2x
x−1
x+1
x−2
8x+16
5x− 1
2 3x+2
2x−5
(d) j(x) = 4x−12
x+8
9x+24
(f) m(x) = 35x−100 (b) g(x) = Problem 14.2. Oscar is hunting magnetic
ﬁelds with his gauss meter, a device for measuring the strength and polarity of magnetic
ﬁelds. The reading on the meter will increase
as Oscar gets closer to a magnet. Oscar is in
a long hallway at the end of which is a room
containing an extremely strong magnet. When
he is far down the hallway from the room, the
meter reads a level of 0.2. He then walks down
the hallway and enters the room. When he has
gone 6 feet into the room, the meter reads 2.3.
Eight feet into the room, the meter reads 4.4.
(a) Give a lineartolinear rational model relating the meter reading y to how many
feet x Oscar has gone into the room.
(b) How far must he go for the meter to
reach 10? 100?
(c) Considering your function from part (a)
and the results of part (b), how far into
the room do you think the magnet is?
Problem 14.3. In 1975 I bought an old Martin ukulele for $200. In 1995 a similar uke
was selling for $900. In 1980 I bought a new
Kamaka uke for $100. In 1990 I sold it for
$400.
(a) Give a linear model relating the price p of
the Martin uke to the year t. Take t = 0
in 1975.
(b) Give a linear model relating the price q
of the Kamaka uke to the year t. Again
take t = 0 in 1975.
(c) When is the value of the Martin twice the
value of the Kamaka?
(d) Give a function f(t) which gives the ratio
of the price of the Martin to the price of
the Kamaka. (e) In the long run, what will be the ratio of
the prices of the ukuleles?
Problem 14.4. Isobel is producing and selling
casette tapes of her rock band. When she had
sold 10 tapes, her net proﬁt was $6. When she
had sold 20 tapes, however, her net proﬁt had
shrunk to $4 due to increased production expenses. But when she had sold 30 tapes, her
net proﬁt had rebounded to $8.
(a) Give a quadratic model relating Isobel’s
net proﬁt y to the number of tapes sold
x.
(b) Divide the proﬁt function in part (a) by
the number of tapes sold x to get a model
relating average proﬁt w per tape to the
number of tapes sold.
(c) How many tapes must she sell in order
to make $1.20 per tape in net proﬁt?
Problem 14.5. Find the lineartolinear function whose graph passes through the points
(1,1), (5,2) and (20,3). What is its horizontal
asymptote?
Problem 14.6. Find the lineartolinear function whose graph has y = 6 as a horizontal
asymptote and passes through (0,10) and (3,7).
Problem 14.7. The more you study for a certain exam, the better your performance on it.
If you study for 10 hours, your score will be
65%. If you study for 20 hours, your score will
be 95%. You can get as close as you want to
a perfect score just by studying long enough.
Assume your percentage score is a 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 14.8. A street light is 10 feet above a
straight bike path. Olav is bicycling down the
path at a rate of 15 MPH. At midnight, Olav is
33 feet from the point on the bike path directly
below the street light. (See the picture). The
relationship between the intensity C of light (in
candlepower) and the distance d (in feet) from
k
the light source is given by C = d2 , where k is
a constant depending on the light source. CHAPTER 14. RATIONAL FUNCTIONS 190
(a) From 20 feet away, the street light has
an intensity of 1 candle. What is k?
(b) Find a function which gives the intensity
of the light shining on Olav as a function
of time, in seconds.
(c) When will the light on Olav have maximum intensity?
(d) When will the intensity of the light be 2
candles? 10ft
olav
path Problem 14.10. The number of customers
in a local dive shop depends on the amount
of money spent on advertising. If the shop
spends nothing on advertising, there will be
100 customers/day.
If the shop spends
$100, there will be 200 customers/day. As
the amount spent on advertising increases,
the number of customers/day increases and
approaches (but never exceeds) 400 customers/day.
(a) Find a linear to linear rational function
y = f(x) that calculates the number y of
customers/day if $x is spent on advertising. 33ft (b) How much must the shop spend on advertising to have 300 customers/day.
Problem 14.9. For each of the following ﬁnd
the linear to linear function f(x) satisfying the
given requirements:
(a) f(0) = 0, f(10) = 10, f(20) = 15
(b) f(0) = 10, f(5) = 4, f(20) = 3
(c) f(10) = 20, f(30) = 25, and the graph of
f(x) has y = 30 as its horizontal asymptote (c) Sketch the graph of the function y = f(x)
on the domain 0 ≤ x ≤ 5000.
(d) Find the rule, domain and range for the
inverse function from part (c). Explain
in words what the inverse function calculates. Chapter 15
Measuring an Angle
So far, the equations we have studied have an algebraic
character, involving the variables x and y, arithmetic operations and maybe extraction of roots. Restricting our
attention to such equations would limit our ability to describe certain natural phenomena. An important example
involves understanding motion around a circle, and it can
be motivated by analyzing a very simple scenario: Cosmo
the dog, tied by a 20 foot long tether to a post, begins
walking around a circle. A number of very natural questions arise: S motors
around
the Q circle
P 20 feet R Figure 15.1: Cosmo the dog
walking a circular path. Natural Questions 15.0.1. How can we measure the angles ∠SPR, ∠QPR, and
∠QPS? How can we measure the arc lengths arc(RS), arc(SQ) and arc(RQ)? How
can we measure the rate Cosmo is moving around the circle? If we know how to
measure angles, can we compute the coordinates of R, S, and Q? Turning this
around, if we know how to compute the coordinates of R, S, and Q, can we then
measure the angles ∠SPR, ∠QPR, and ∠QPS ? Finally, how can we specify the
direction Cosmo is traveling? We will answer all of these questions and see how the theory which
evolves can be applied to a variety of problems. The deﬁnition and basic
properties of the circular functions will emerge as a central theme in this
Chapter. The full problemsolving power of these functions will become
apparent in our discussion of sinusoidal functions in Chapter 19.
The xycoordinate 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.
191 Cosmo CHAPTER 15. MEASURING AN ANGLE 192 15.1 Standard and Central Angles
An angle is the union of two rays emanating from a common point called
the vertex of the angle. A typical angle can be dynamically generated
by rotating a single ray from one position to another, sweeping counterclockwise or clockwise : See Figure 15.2. We often insert a curved arrow
to indicate the direction in which we are sweeping out the angle. The ray
ℓ1 is called the initial side and ℓ2 the terminal side of the angle ∠AOB. (terminal side)
l2
B
SWEEP CLOCKWISE O A l1 vertex
O vertex
l1
A (initial side) O A (initial side)
l1
(terminal side) START SWEEP COUNTERCLOCKWISE B l2 Figure 15.2: Angle ∠AOB. Working with angles, we need to agree on a standard frame of reference for viewing them. Within the usual xycoordinate system, imagine a
model of ∠AOB in Figure 15.2 constructed from two pieces of rigid wire,
welded at the vertex. Sliding this model around inside the xyplane 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 xaxis and the vertex is the origin. Whenever we do
this, we say the angle is in standard position. Once an angle is in standard position, we can construct a circle centered at the origin and view
our standard angle as cutting out a particular “pie shaped wedge” of the
corresponding disc.
Notice, the shaded regions in Figure 15.3 depend on whether we sweep
the angle counterclockwise or clockwise from the initial side. The portion
of the “pie wedge” along the circle edge, which is an arc, is called the arc
subtended by the angle. We can keep track of this arc using the notation
arc(AB). A central angle is any angle with vertex at the center of a circle,
but its initial side may or may not be the positive xaxis. For example,
∠QPS in the Figure beginning this Chapter is a central angle which is not
in standard position. 15.2. AN ANALOGY 193 l2 yaxis yaxis B
arc subtended l2
B arc(AB) arc(AB) vertex
O A l1 xaxis COUNTERCLOCKWISE vertex
O xaxis
A l1 CLOCKWISE Figure 15.3: Standard angles and arcs. 15.2 An Analogy
To measure the dimensions of a box you would use a ruler. In other
words, you use an instrument (the ruler) as a standard against which
you measure the box. The ruler would most likely be divided up into
either English units (inches) or metric units (centimeters), so we could
express the dimensions in a couple of different ways, depending on the
units desired.
By analogy, to measure the size of an angle, we need a standard
against which any angle can be compared. In this section, we will describe two standards commonly used: the degree method and the radian
method of angle measurement. The key idea is this: Beginning with a
circular region, describe how to construct a “basic” pie shaped wedge
whose interior angle becomes the standard unit of angle measurement. 15.3 Degree Method
Begin by drawing a circle of radius r, call it Cr , centered at the origin.
Divide this circle into 360 equal sized pie shaped wedges, beginning with
the the point (r,0) on the circle; i.e. the place where the circle crosses the
xaxis.
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 CHAPTER 15. MEASURING AN ANGLE 194 circle Cr
etc.
a total of 360
equal sized
pie shaped
wedges inside
this disk r
(r,0) typical
wedge r this angle is
DEFINED to
have measure
1 degree etc. ***NOT TO SCALE*** Figure 15.4: Wedges as 1◦ arcs. deﬁned to have a measure of one degree, denoted 1◦ .
We can now use onedegree arcs to measure any angle: Begin by sliding the angle ∠AOB into standard central position, as in Figure 15.3.
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 ∠AOB to the nearest degree. If
we are allowed to divide a onedegree arc into a fractional portion, then
we could precisely determine the number m of onedegree arcs which
consecutively ﬁt together into the given arc. If we are sweeping counterclockwise from the initial side of the angle, m is deﬁned to be the degree
measure of the angle. If we sweep in a clockwise direction, then −m is
deﬁned to be the degree measure of the angle. So, in Figure 15.3, the
lefthand angle has positive degree measure while the righthand angle
has negative degree measure. Simple examples would be angles like the
ones in Figure 15.5.
Notice, with our conventions, the rays determining an angle with measure −135◦ sit inside the circle in the same position as those for an angle
of measure 225◦; the minus sign keeps track of sweeping the positive
xaxis 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 1 ′ and one second, denoted 1 ′′ :
1◦ = 60 minutes
= 3600 seconds. 15.3. DEGREE METHOD 195 180◦ 90◦ 270◦ 45◦ 315◦
−135◦ Figure 15.5: Examples of common angles. For example, an angle of measure θ = 5 degrees 23 minutes 18 seconds is usually denoted 5◦ 23 ′ 18 ′′ . We could express this as a decimal of
degrees:
◦ ′ 5 23 18 ′′ 23
18
=
5+
+
60 3600
= 5.3883◦. ◦ ←In degrees! As another example, suppose we have an angle with measure 75.456◦ and
we wish to convert this into degree/minute/second units. First, since
75.456◦ = 75◦ + 0.456◦, we need to write 0.456◦ in minutes by the calculation:
0.456 degree × 60 minutes
= 27.36 ′.
degree This tells us that 75.456◦ = 75◦ 27.36 ′ = 75◦ 27 ′ + 0.36 ′. Now we need to write
0.36 ′ in seconds via the calculation:
0.36 minutes × 60 seconds/minute = 21.6 ′′ .
In other words, 75.456◦ = 75◦ 27 ′ 21.6 ′′.
Degree measurement of an angle is very closely tied to direction in the
plane, explaining its use in map navigation. With some additional work,
it is also possible to relate degree measure and lengths of circular arcs.
To do this carefully, ﬁrst go back to Figure 15.3 and recall the situation CHAPTER 15. MEASURING AN ANGLE 196 where an arc arc(AB) is subtended by the central angle ∠AOB. In this
situation, the arc length of arc(AB), commonly denoted by the letter s, is
the distance from A to B computed along the circular arc; keep in mind,
this is NOT the same as the straight line distance between the points A
and B.
For example, consider the six angles pictured above, of measures 90◦ ,
180◦ , 270◦, 45◦ , −135◦ , and 315◦ . If the circle is of radius r and we want to
compute the lengths of the arcs subtended by these six angles, then this
can be done using the formula for the circumference of a circle (on the
back of this text) and the following general principle:
Important Fact 15.3.1.
(length of a part) = (fraction of the part) × (length of the whole)
For example, the circumference of the entire circle of radius r is 2πr;
this is the “length of the whole” in the general principle. The arc sub90
1
tended by a 90◦ angle is 360 = 4 of the entire circumference; this is the
“fraction of the part” in the general principle. The boxed formula implies:
s = arc length of the 90◦ arc =
s = distance
along the arc
r
θ degrees πr
1
2πr =
.
4
2 Similarly, a 180◦ angle subtends an arc of length s = πr,
315
a 315◦ angle subtends an arc of length s = 360 2πr = 7πr ,
4
etc. In general, we arrive at this formula:
Important Fact 15.3.2 (Arc length in degrees). Start with
a central angle of measure θ degrees inside a circle of radius r. Then this angle will subtend an arc of length r Cr Figure 15.6: The deﬁnition
of arc length. s= 2π
rθ
360 15.4 Radian Method
The key to understanding degree measurement was the description of a
“basic wedge” which contained an interior angle of measure 1◦ ; this was
straightforward and familiar to all of us. Once this was done, we could
proceed to measure any angle in degrees and compute arc lengths as in
Fact 15.3.2. However, the formula for the length of an arc subtended by
an angle measured in degrees is sort of cumbersome, involving the curi2π
ous factor 360 . Our next goal is to introduce an alternate angle measurement scheme called radian measure that begins with a different “basic
wedge”. As will become apparent, a big selling point of radian measure is
that arc length calculations become easy. 15.4. RADIAN METHOD 197 r r r
r equilateral
wedge
r (r,0) circle Cr r
this angle is
DEFINED to
have
measure 1
radian Figure 15.7: Constructing an equilateral wedge. As before, begin with a circle Cr of radius r. Construct an equilateral
wedge with all three sides of equal length r; see Figure 15.7. We deﬁne
the measure of the interior angle of this wedge to be 1 radian.
Once we have deﬁned an angle of measure 1 radian, we can deﬁne an
angle of measure 2 radians by putting together two equilateral wedges.
Likewise, an angle of measure 1 radian is obtained by symmetrically di2
viding an equilateral wedge in half, etc.
Reasoning in this way, we can piece together equilateral wedges or
fractions of such to compute the radian measure of any angle. It is important to notice an important relationship between the radian measure
of an angle and arc length calculations. In the ﬁve angles pictured above,
1
1 radian, 2 radian, 3 radian, 2 radian and 1 radian, the length of the arcs
4
1
1
subtended by these angles θ are r, 2r, 3r, 2 r, and 4 r. In other words, a
pattern emerges that gives a very simple relationship between the length
s of an arc and the radian measure of the subtended angle:
Important Fact 15.4.1 (Arc length in radians).
Start with a central angle of measure θ radians inside a
circle of radius r. Then this angle will subtend an arc of
length s = θr.
These remarks allow us to summarize the deﬁnition of
the radian measure θ of ∠AOB inside a circle of radius r
by the formula:
θ= s
r if angle is swept counterclockwise
− s if angle is swept clockwise
r s = distance
along the arc
r
θ radians r Cr Figure 15.9: Deﬁning arc
length when angles are measured in radians. CHAPTER 15. MEASURING AN ANGLE 198 3r 2r r 3 radians = θ 2 radians = θ r r r r r 1 radian = θ
r 1
2 r 1
4 radian = θ radian = θ
r
2 r r r
4 r Figure 15.8: Measuring angles in radians. The units of θ are sometimes abbreviated as rad. It is
important to appreciate the role of the radius of the circle
B
Cr when using radian measure of an angle: An angle of
radian measure θ will subtend an arc of length θ on the
θ radian
unit circle. In other words, radian measure of angles is
O
A
xaxis
exactly the same as arc length on the unit circle; we
couldn’t hope for a better connection!
circle radius r
The difﬁculty with radian measure versus degree measure is really one of familiarity. Let’s view a few common
Figure 15.10: Arc length
angles in radian measure. It is easiest to start with the
after imposing a coordinate
system.
case of angles in central standard position within the unit
circle. Examples of basic angles would be fractional parts
1
of one complete revolution around the unit circle; for example, 12 revolu1
3
tion, 1 revolution, 6 revolution, 1 revolution, 1 revolution and 4 revolution.
8
4
2
One revolution around the unit circle describes an arc of length 2π and
so the subtended angle (1 revolution) is 2π radians. We can now easily
1
ﬁnd the radian measure of these six angles. For example, 12 revolution
1
would describe an angle of measure ( 12 )2π rad= π rad. Similarly, the other
6
ﬁve angles pictured below have measures π rad, π rad, π rad, π rad and
4
3
2
3π
rad.
2
All of these examples have positive radian measure. For an angle with
negative radian measure, such as θ = − π radians, we would locate B
2
1
by rotating 4 revolution clockwise, etc. From these calculations and our
yaxis length of
arc(AB) = s 15.5. AREAS OF WEDGES 199 placements π
4 π
6 1
12 π
2 1
4 1
8 revolution 1
6 revolution revolution 3
4 revolution 3π
2 π revolution π
3 1
2 revolution Figure 15.11: Common angles measured in radians. previous examples of degree measure we ﬁnd that
180 degrees = π radians. (15.1) Solving this equation for degrees or radians will provide conversion
formulas relating the two types of angle measurement. The formula
π
2π
also helps explain the origin of the curious conversion factor 180 = 360
in Fact 15.3.2. 15.5 Areas of Wedges
The beauty of radian measure is that it is rigged so that we can easily
compute lengths of arcs and areas of circular sectors (i.e. “pieshaped
regions”). This is a key reason why we will almost always prefer to work
with radian measure.
Example 15.5.1. If a 16 inch pizza is cut into 12 equal slices, what is the
area of a single slice?
This can be solved using a general principle:
(Area of a part) = (area of the whole) × (fraction of the part) CHAPTER 15. MEASURING AN ANGLE 200
So, for our pizza: O θ (area one slice) = (area whole pie) × (fraction of pie)
1
= (82 π)
12
16π
.
=
3
Let’s apply the same reasoning to ﬁnd the area of a
circular sector. We know the area of the circular disc
B
bounded by a circle of radius r is πr2 . Let Rθ be the “pie
shaped wedge” cut out by an angle ∠AOB with positive
Rθ
measure θ radians. Using the above principle
A Cr Figure 15.12: Finding the
area of a “pie shaped wedge”. area(Rθ ) = (area of disc bounded by Cr )
× (portion of disc accounted for by Rθ )
1
θ
= r2 θ.
= (πr2 )
2π
2
For example, if r = 3 in. and θ =
9
the pie shaped wedge is 8 π sq. in. π
4 rad, then the area of Important Fact 15.5.2 (Wedge area). Start with a central angle with positive measure θ radians inside a circle of radius r. The area of the “pie
shaped region” bounded by the angle is 1 r2 θ.
2
Example 15.5.3. A water drip irrigation arm 1200 feet long rotates around
a pivot P once every 12 hours. How much area is covered by the arm in
one hour? in 37 minutes? How much time is required to drip irrigate 1000
square feet?
Solution. The irrigation arm will complete one revolution in 12 hours.
The angle swept out by one complete revolution is 2π radians, so after t
hours the arm sweeps out an angle θ(t) given by
θ(t) = π
2π radians
× t hours = t radians.
12 hours
6 Consequently, by Important Fact 15.5.2 the area A(t) of the irrigated
region after t hours is
1
1
π
A(t) = (1200)2θ(t) = (1200)2 t = 120,000πt square feet.
2
2
6
After 1 hour, the irrigated area is A(1) = 120,000π = 376,991 sq. ft. Like37
wise, after 37 minutes, which is 60 hours, the area of the irrigated region
37
37
is A( 60 ) = 120,000π( 60 ) = 232,500 square feet. To answer the ﬁnal question, we need to solve the equation A(t) = 1000; i.e., 120,000πt = 1000, so
3600 seconds
1
= 9.55 seconds.
hours ×
t=
120π
hour 15.5. AREAS OF WEDGES 201 15.5.1 Chord Approximation
Our ability to compute arc lengths can be used as an estimating tool for
distances between two points. Let’s return to the situation posed at the
beginning of this section: Cosmo the dog, tied by a 20 foot long tether
to a post in the ground, begins at location R and walks counterclockwise
to location S. Furthermore, let’s suppose you are standing at the center
of the circle determined by the tether and you measure the angle from R
to S to be 5◦ ; see the lefthand ﬁgure. Because the angle is small, notice
that the straight line distance d from R to S is approximately the same as
the arc length s subtended by the angle ∠RPS; the righthand picture in
Figure 15.13 is a blowup: S
S
5◦
P 20 feet d R s 5◦
P 20 feet R Figure 15.13: Using the arc length s to approximate the chord d. Example 15.5.4. Estimate the distance from R to S.
Solution. We ﬁrst convert the angle into radian measure via (15.1): 5◦ =
0.0873 radians. By Fact 15.3.2, the arc s has length 1.745 feet = 20.94 inches.
This is approximately equal to the distance from R to S, since the angle
is small. CHAPTER 15. MEASURING AN ANGLE 202
S
s
chord
R
O We call a line segment connecting two points on a circle a chord of the circle. The above example illustrates
a general principal for approximating the length of any
chord. A smaller angle will improve the accuracy of the
arc length approximation.
Important Fact 15.5.5 (Chord Approximation). In Figure 15.14, if the central angle is small, then s ≈ RS. Figure 15.14: Chord approximation. 15.6 Great Circle Navigation A basic problem is to ﬁnd the shortest route between any two locations
on the earth. We will review how to coordinatize the surface of the earth
and recall the fact that the shortest path between two points is measured
along a great circle.
View the earth as a sphere of radius r = 3,960 miles. We could slice
the earth with a twodimensional plane P0 which is both perpendicular
to a line connecting the North and South poles and passes through the
center of the earth. Of course, the resulting intersection will trace out a
circle of radius r = 3,960 miles on the surface of the earth, which we call
the equator. We call the plane P0 the equatorial plane. Slicing the earth
with any other plane P parallel to P0 , we can consider the right triangle
pictured below and the angle θ:
line of latitude North Pole North Pole b b
θ
r r
r θ equatorial plane P0 r 90◦ − θ◦ equator
equator
center of earth
South Pole center of earth South Pole Figure 15.15: Measuring latitude. Essentially two cases arise, depending on whether or not the plane P
is above or below the equatorial plane. The plane P slices the surface of
the earth in a circle, which we call a line of latitude. This terminology
is somewhat incorrect, since these lines of latitude are actually circles
on the surface of the earth, but the terminology is by now standard. Depending on whether this line of latitude lies above or below the equatorial
plane, we refer to it as the θ◦ North line of latitude (denoted θ◦ N) or the
θ◦ South line of latitude (denoted θ◦ S). Notice, the radius b of a line of
latitude can vary from a maximum of 3,960 miles (in the case of θ = 0◦ ), 15.6. GREAT CIRCLE NAVIGATION 203 to a minimum of 0 miles, (when θ = 90◦ ). When b = 0, we are at the North
or South poles on the earth.
In a similar spirit, we could imagine slicing the earth with a plane Q
which is perpendicular to the equatorial plane and passes through the
center of the earth. The resulting intersection will trace out a circle of
radius 3,960 miles on the surface of the earth, which is called a line of
longitude. Half of a line of longitude from the North Pole to the South
Pole is called a meridian. We distinguish one such meridian; the one
which passes through Greenwich, England as the Greenwich meridian.
Longitudes are measured using angles East or West of Greenwich. Pictured below, the longitude of A is θ. Because θ is east of Greenwich, θ
measures longitude East, typically written θ◦ E; west longitudes would
be denoted as θ◦ W. All longitudes are between 0◦ and 180◦ . The meridian
which is 180◦ West (and 180◦ E) is called the International Date Line.
Introducing the grid of latitude and longitude lines on
Greenwich, England
the earth amounts to imposing a coordinate system. In
North Pole
line of longitude
other words, any position on the earth can be determined
A
Greenwich
International
meridian
Date Line
by providing the longitude and latitude of the point. The
r
θ
r
usual convention is to list longitude ﬁrst. For example,
Equator
Seattle has coordinates 122.0333◦ W, 47.6◦ N. Since the lacenter of earth South Pole
bels “N and S” are attached to latitudes and the labels “E
and W” are attached to longitudes, there is no ambiguity
Figure 15.16: The International Date Line.
here. This means that Seattle is on the line of longitude
122.0333◦ West of the Greenwich meridian and on the line
of latitude 47.6◦ North of the equator. In the ﬁgure below, we indicate the
key angles ψ = 47.6◦ and θ = 122.0333◦ by inserting the three indicated
radial line segments. not great circle
N
great circles Seattle, WA Ψ r
θ
Greenwich
Meridian
S
not great circle Figure 15.17: Distances along great circles. CHAPTER 15. MEASURING AN ANGLE 204 Now that we have imposed a coordinate system on the earth, it is
natural to study the distance between two locations. A great circle of a
sphere is deﬁned to be a circle lying on the sphere with the same center
as the sphere. For example, the equator and any line of longitude are
great circles. However, lines of latitude are not great circles (except the
special case of the equator). Great circles are very important because
they are used to ﬁnd the shortest distance between two points on the
earth. The important fact from geometry is summarized below.
Important Fact 15.6.1 (Great Circles). The shortest distance between
two points on the earth is measured along a great circle connecting them.
Example 15.6.2. What is the shortest distance from the North Pole to Seattle, WA ? Solution. The line of longitude 122.0333◦ W is a great circle
connecting the North Pole and Seattle. So, the shortest
distance will be the arc length s subtended by the angle
∠NOW pictured in Figure 15.18. Since the latitude of
Seattle is 47.6◦, the angle ∠EOW has measure 47.6◦ . Since
∠EON is a right angle (i.e., 90◦ ), ∠NOW has measure 42.4◦ .
By Fact 15.4.1 and Equation 15.1, N W O
E equator S s = (3960 miles)(42.4◦)(0.01745 radians/degree)
= 2943.7 miles, Greenwich
Meridian Figure 15.18: Distance between the North Pole and
Seattle, Washington. which is the shortest distance from the pole to Seattle. 15.7 Summary
• 360◦ = 2π radians
• A circular arc with radius r and angle θ has length s, with
s = rθ
when θ is measured in radians.
• A circular wedge with radius r and angle θ has area A, with
1
A = r2 θ
2
when θ is measured in radians. 15.8. EXERCISES 205 15.8 Exercises
Problem 15.1. Let ∠AOB be an angle of measure θ.
(a) Convert θ = 13.4o into degrees/ minutes/ seconds and into radians.
(b) Convert θ = 1o 4 ′ 44 ′′ into degrees and radians.
(c) Convert θ = 0.1 radian into degrees and
degrees/ minutes/ seconds.
Problem 15.2. A nautical mile is a unit of distance frequently used in ocean navigation. It is
deﬁned as the length of an arc s along a great
circle on the earth when the subtending angle
has measure 1 ′ = “one minute” = 1/60 of
one degree. Assume the radius of the earth is
3,960 miles.
(a) Find the length of one nautical mile to
the nearest 10 feet.
(b) A vessel which travels one nautical mile
in one hours time is said to have the
speed of one knot ; this is the usual navigational measure of speed. If a vessel is
traveling 26 knots, what is the speed in
mph (miles per hour)?
(c) If a vessel is traveling 18 mph, what is
the speed in knots?
Problem 15.3. The rear window wiper blade
on a station wagon has a length of 16 inches.
The wiper blade is mounted on a 22 inch arm,
6 inches from the pivot point. 6"
16" (a) If the wiper turns through an angle of
110◦ , how much area is swept clean?
(b) Through how much of an angle would
the wiper sweep if the area cleaned was
10 square inches? (c) Suppose bug A lands on the end of the
blade farthest from the pivot. Assume
the wiper turns through an angle of 110◦ .
In one cycle (back and forth) of the wiper
blade, how far has the bug traveled?
(d) Suppose bug B lands on the end of the
wiper blade closest to the pivot. Assume
the wiper turns through an angle of 110◦ .
In one cycle of the wiper blade, how far
has the bug traveled?
(e) Suppose bug C lands on an intermediate
location of the wiper blade. Assume the
wiper turns through an angle of 110◦ . If
bug C travels 28 inches after one cycle
of the wiper blade, determine the location of bug C on the wiper blade. Problem 15.4. A water treatment facility operates by dripping water from a 60 foot long arm
whose end is mounted to a central pivot. The
water then ﬁlters through a layer of charcoal.
The arm rotates once every 8 minutes.
(a) Find the area of charcoal covered with
water after 1 minute.
(b) Find the area of charcoal covered with
water after 1 second.
(c) How long would it take to cover 100
square feet of charcoal with water?
(d) How long would it take to cover 3245
square feet of charcoal with water? Problem 15.5. Astronomical measurements
are often made by computing the small angle
formed by the extremities of a distant object
and using the estimating technique in 15.5.1.
In the picture below, the full moon is shown
1o
to form an angle of 2 when the distance indicated is 248,000 miles. Estimate the diameter
of the moon. CHAPTER 15. MEASURING AN ANGLE 206
moon (a) θ = 45◦
(b) θ = 80o 248,000 miles (c) θ = 3 radians
(d) θ = 2.46 radians
(e) θ = 97o 23 ′ 3 ′′ o
1/2 (f) θ = 35o 24 ′ 2 ′′
earth Problem 15.6. An aircraft is ﬂying at the
speed of 500 mph at an elevation of 10
miles above the earth, beginning at the North
pole and heading South along the Greenwich
meridian. A spy satellite is orbiting the earth
at an elevation of 4800 miles above the earth
in a circular orbit in the same plane as the
Greenwich meridian. Miraculously, the plane
and satellite always lie on the same radial line
from the center of the earth. Assume the radius of the earth is 3960 miles. Problem 15.8. Matilda is planning a walk
around the perimeter of Wedge Park, which is
shaped like a circular wedge, as shown below.
The walk around the park is 2.1 miles, and the
park has an area of 0.25 square miles.
If θ is less than 90 degrees, what is the
value of the radius, r?
r θ r
satellite
plane earth Problem 15.9. Let C6 be the circle of radius
6 inches centered at the origin in the xycoordinate system. Compute the areas of the
shaded regions in the picture below; the inner
circle in the rightmost picture is the unit circle:
y=x
y=−x (a) When is the plane directly over a location with latitude 74◦ 30 ′ 18 ′′ N for the ﬁrst
time? C C 6 (b) How fast is the satellite moving? 6
y=x (c) When is the plane directly over the equator and how far has it traveled?
(d) How far has the satellite traveled when
the plane is directly over the equator?
y=−(1/4)x + 2 Problem 15.7. Find the area of the sector of a
circle of radius 11 inches if the measure θ of a
central angle of this sector is: C 6 Chapter 16
Measuring Circular Motion
If Cosmo begins at location R and walks counterclockwise,
always maintaining a tight tether, how can we measure
Cosmo’s speed?
This is a “dynamic question” and requires that we discuss ways of measuring circular motion. In contrast, if
we take a snapshot and ask to measure the speciﬁc angle
∠RPS, this is a “static question”, which we answered in
the previous section. Cosmo moves
counterclockwise,
maintaining
S
a tight tether. R
P Figure 16.1: How fast is
Cosmo moving? 16.1 Different ways to measure
Cosmo’s speed
If Cosmo starts at location R and arrives at location S after some amount
of time, we could study
ω= measure ∠RPS
.
time required to go from R to S The funny Greek letter “ω” on the left of side of the equation is pronounced “ohmega”. We will refer to this as an angular speed. Typical
units are “ degrees ”, “ degrees ”, “ radians ”, etc. For example, if the angle
minute
second
minute
swept out by Cosmo after 8 seconds is 40◦ , then Cosmo’s angular speed
40◦
5◦
is 8 seconds = sec . Using (15.1), we can convert to radian units and get
π
5
rad
ω = 36 rad = 3 π min . This is a new example of a rate and we can ask to
sec
ﬁnd the total change, in the spirit of (1.2). If we are given ω in units of
deg
rad
“ time ” or “ time ”, we have
θ = ωt,
which computes the measure of the angle θ swept out after time t (i.e.
the total change in the angle). Angular speed places emphasis upon the
207 20 feet CHAPTER 16. MEASURING CIRCULAR MOTION 208 “size of the angle being swept out per unit time” by the moving object,
starting from some initial position. We need to somehow indicate the
direction in which the angle is being swept out. This can be done by indicating “clockwise” our “counterclockwise”. Alternatively, we can adopt
the convention that the positive rotational direction is counterclockwise,
then insert a minus sign to indicate rotation clockwise. For example,
saying that Cosmo is moving at an angular speed of ω = − π rad means he
2 sec
is moving clockwise π rad .
2 sec
Another way to study the rate of a circular motion is to count the
number of complete circuits of the circle per unit time. This sort of rate
has the form
Number of Revolutions
;
Unit of Time
we will also view this as an angular speed. If we take “minutes” to be
the preferred unit of time, we arrive at the common measurement called
revolutions per minute, usually denoted RPM or rev/min. For example, if
Cosmo completes one trip around the circle every 2 minutes, then Cosmo
is moving at a rate of 1 RPM. If instead, Cosmo completes one trip around
2
the circle every 12 seconds, then we could ﬁrst express Cosmo’s speed in
1
units of revolutions/second as 12 rev/second, then convert to RPM units:
1 rev
12 sec 60 sec
= 5 RPM.
min As a variation, if we measure that Cosmo completed 3 of a revolution in
7
2 minutes, then Cosmo’s angular speed is computed by
3
rev
7 2 min = 3
RPM.
14 The only possible ambiguity involves the direction of revolution: the object can move clockwise or counterclockwise.
The one shortcoming of using angular speed is that we are not directly
keeping track of the distance the object is traveling. This is fairly easy
to remedy. Returning to Figure 16.1, the circumference of the circle of
motion is 2π(20) = 40π feet. This is the distance traveled per revolution,
so we can now make conversions of angular speed into “distance traveled
per unit time”; this is called the linear speed. If Cosmo is moving 1 RPM,
2
then he has a linear speed of
v= 1 rev
2 min 40π ft
rev = 20π Likewise, if Cosmo is moving
v= π rad
7 sec 1 rev
2π rad ft
.
min π rad
,
7 sec 40π ft
rev then
= 20π ft
.
7 sec 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION 209 Important Fact 16.1.1. This discussion is an example of what is usually
called “units analysis”. The key idea we have illustrated is how to convert
between two different types of units:
rev
min converts to −→ ft
min 16.2 Different Ways to Measure
Circular Motion
The discussion of Cosmo applies to circular motion of any object. As a
matter of convention, we usually use the Greek letter ω to denote angular
speed and v for linear speed. If an object is moving around a circle of
radius r at a constant rate, then we can measure it’s speed in two ways:
• The angular speed
ω= “revolutions”
“degrees swept”
“radians swept”
or
or
.
“unit time”
“unit time”
“unit time” • The linear speed
v= “distance traveled”
.
“per unit time” Important Facts 16.2.1 (Measuring and converting). We can convert between angular and linear speeds using these facts:
• 1 revolution = 360◦ = 2π radians;
• The circumference of a circle of radius r units is 2πr units. 16.2.1 Three Key Formulas
If an object begins moving around a circle, there are a number of quantities we can try to relate. Some of these are “static quantities”: Take
a visual “snapshot” of the situation after a certain amount of time has
elapsed, then we can measure the radius, angle swept, arc length and
time elapsed. Other quantities of interest are “dynamic quantities”: This
means something is CHANGING with respect to time; in our case, the linear speed (which measures distance traveled per unit time) and angular
speed (which measures angle swept per unit time) fall into this category. CHAPTER 16. MEASURING CIRCULAR MOTION 210 STATIC QUANTITIES DYNAMIC QUANTITIES S S 1. arc length s 5. angular speed ω 2. angle swept in time r 6. linear speed v θ
P P R R 3. radius r 4. elapsed time t ...take a “snapshot” after time t... ...see what happens per unit time... Figure 16.2: Measuring linear and angular speed. We now know two general relationships for circular motion:
(i) s = rθ, where s=arclength (a linear distance), r=radius of the circular
path and θ=angle swept in RADIAN measure; this was the content
of Fact 15.4.1 on page 197.
(ii) θ = ωt, where θ is the measure of an angle swept, ω= angular speed
and t represents time elapsed. This is really just a consequence of
units manipulation.
Notice how the units work in these formulas. If r=20 feet and θ = 1.3
radians, then the arc length s = 20(1.3) feet= 26 feet; this is the length
of the arc of radius 20 feet that is subtending the angle θ. If ω = 3
rad
rad/second and t = 5 seconds, then θ = 3 seconds × 5 seconds = 15 radians.
If we replace “θ” in s = rθ of (i) with θ = ωt in (ii), then we get
s = rωt.
This gives us a relationship between arclength s (a distance) and time t.
Plug in the fact that the linear speed is deﬁned to be v = “distance” and we
t
get
s
rωt
=
= rω.
t
t
All of these observations are summarized below.
v= Important Facts 16.2.2 (Three really useful formulas). If we measure
angles θ in RADIANS and ω in units of radians per unit time, we have
these three formulas:
s = rθ
θ = ωt
v = rω (16.1)
(16.2)
(16.3) 16.2. DIFFERENT WAYS TO MEASURE CIRCULAR MOTION Example 16.2.3. You are riding a stationary exercise bike
and the speedometer reads a steady speed of 40 MPH
(miles per hour). If the rear wheel is 28 inches in diameter, determine the angular speed of a location on the rear
tire. A pebble becomes stuck to the tread of the rear tire.
Describe the location of the pebble after 1 second and 0.1
second. 211 *
pebble sticks to tread here Figure 16.3: Where is the
pebble after t seconds? Solution. The tires will be rotating in a counterclockwise
direction and the radius r = 1 28 = 14 inches. The other given quantity,
2
“40 MPH”, involves miles, so we need to decide which common units to
work with. Either will work, but since the problem is focused on the
wheel, we will utilize inches.
If the speedometer reads 40 MPH, this is the linear speed of a speciﬁed
location on the rear tire. We need to convert this into an angular speed,
using unit conversion formulas. First, the linear speed of the wheel is
miles
hr
in
= 704
.
sec v= 5280 40 ft
mile 12 in
ft 1 hr
60 min 1 min
60 sec Now, the angular speed ω of the wheel will be
inches
704 second ω= inches
2(14)π revolution
revolution
=8
second
= 480 RPM It is then an easy matter to convert this to
revolution
second
degrees
= 2,880
.
second ω= 8 360 degrees
revolution If the pebble begins at the “6 o’clock” position (the place the tire touches
the ground on the wheel), then after 1 second the pebble will go through 8
revolutions, so will be in the “6 o’clock” position again. After 0.1 seconds,
the pebble will go through
8 rev
(0.1 sec) = 0.8 rev
sec
= (0.8 rev) 360
= 288◦ . deg
rev CHAPTER 16. MEASURING CIRCULAR MOTION 212 Keeping in mind that the rotation is counterclockwise, we can view the
location of the pebble after 0.1 seconds as pictured below: 288◦
counterclockwise rotation after 0.1 second *
located here at
time = 0.1 sec starts
here *
pebble sticks to tread in 6 o’clock position Figure 16.4: Computing the pebble’s position after t = 0.1 sec. We solved the previous problem using the “unit conversion method”.
There is an alternate approach available, which uses one of the formulas
in Fact 16.2.2. Here is how you could proceed: First, as above, we know
the linear speed is v = 704 in/sec. Using the “v = ωr” formula, we have
704 in
= ω(14 in)
sec
rad
ω = 50.28
.
sec Notice how the units worked out in the calculation: the “time” unit comes
from v and the “angular” unit will always be radians. As a comparison
with the solution above, we can convert ω into RPM units:
ω=
=8 50.28 rad
sec 1 rev
2π rad rev
.
sec All of the problems in this section can be worked using either the “unit
conversion method” or the “v = ωr method”. 16.3 Music Listening Technology
The technology of reproducing music has gone through a revolution since
the early 1980’s. The “old” stereo long playing record (the LP ) and the 16.3. MUSIC LISTENING TECHNOLOGY 213 “new” digital compact disc (the CD ) are two methods of storing musical
data for later reproduction in a home stereo system. These two technologies adopt different perspectives as to which notion of circular speed is
best to work with.
Long playing stereo records are thin vinyl plastic discs of radius 6
inches onto which small spiral grooves are etched into the surface; we
can approximately view this groove as a circle. The LP is placed on a
ﬂat 12 inch diameter platter which turns at a constant angular speed of
1
33 3 RPM. An arm on a pivot (called the tone arm) has a needle mounted
on the end (called the cartridge), which is placed in the groove on the
outside edge of the record. Because the grooves wobble microscopically
from 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! amp LP turning at 33 1 RPM
3 speakers tonearm needle Figure 16.5: Reproducing music using analogue technology. This is known as analogue technology and is based upon the idea of
1
maintaining a constant angular speed of 33 3 RPM for the storage medium
(our LP). (Older analogue technologies used 45 RPM and 78 RPM records.
However, 33 1 RPM became the consumer standard for stereo music.) With
3
an LP, the beginning of the record (the 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 33 1 RPM, the linear speed at the
3
needle can vary quite a bit, depending on the needle location. 214 CHAPTER 16. MEASURING CIRCULAR MOTION 1 Example 16.3.1 (Analogue LP’s). The “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. 6 Figure 16.6: Leadin and
exit grooves. Solution. This is a straightforward application of Fact 16.2.1.
Let v6 (resp. v1 ) be the linear speed at the lead in groove (resp. exit groove);
the subscript keeps track of the needle radial location. Since the groove
is approximately a circle,
1 rev
inches
2(6)π
3 min
rev
in
= 1257
min
min
in
1257 min 60 hour
=
ft
5280 mile 12 in
ft
= 1.19 MPH v6 = 33 Similarly, v1 = 0.2 MPH. To answer the remaining question, let r be the
radial distance from the center of the LP to the needle location on the
record. If vr = 1 MPH:
mile
1
= vr
hour
1 rev
in
min
1 ft
1 mile
=
33
2rπ
60
3 min
rev
hr
12 in
5280 ft
So, when the needle is r = 5.04 inches from the center, the linear speed is
1 MPH.
laser laser support arm
moves back and forth spinning CD Figure 16.7: Reproducing
music using digital technology. In the early 1980’s, a new method of storing and reproducing music was introduced; this medium is called the
digital compact disc, referred to as a CD for short. This is
a thin plastic disc of diameter 4.5 inches, which appears
to the naked eye to have a shiny silver coating on one side.
Upon microscopic examination one would ﬁnd concentric
circles of pits in the silver coating. This disc is placed in
a CD player, which spins the disc. A laser located above
the spinning disc will project onto the spinning disc. The
pits in the silver coating will cause the reﬂected laser light
to vary in intensity. A sensor detects this variation, converting it to a digital signal (the analogue to digital or AD
conversion ). This is fed into a digital to analogue or DA
conversion device, which sends a signal to your stereo,
again producing music. 16.4. BELT AND WHEEL PROBLEMS 215 The technology of CD ′ s differs from that of LP ′ s in two crucial ways.
First, the circular motion of the spinning CD is controlled so that the
target on the disc below the laser is always moving at a constant linear
inches
speed of 1.2 meters = 2,835 minute . Secondly, the beginning location of the
sec
laser will be on the inside portion of the disc, working its way outward to
the end. In this context, it makes sense to study how the angular speed
of the CD is changing, as the laser position changes.
Example 16.3.2 (Digital CD’s). What is the angular speed
(in RPM) of a CD if the laser is at the beginning, located 3
4
inches from the center of the disc ? What is the angular
speed (in RPM) of a CD if the laser is at the end, located 2
inches from the center of the disc ? Find the location of the
laser if the angular speed is 350 RPM.
Solution. This is an application of Fact 16.2.1. Let ω3/4 be
the angular speed at the start and ω2 the angular speed
at the end of the CD ; the subscript is keeping track of the
laser distance from the CD center.
(2835 inches/min)
ω2 =
= 225.6 RPM
(2(2)π inches/rev) 2 ′′ 3 ′′
4 Start of CD End of CD Figure 16.8: Computing the
angular speed of a CD. 2835 inches/min)
= 601.6 RPM
(2(0.75)π inches/rev)
To answer the remaining question, let r be the radial distance from
the center of the CD to the laser location on the CD. If the angular speed
ωr at this location is 350 RPM, we have the equation
ω3/4 = 350 RPM = ωr
inches
2,835 minute
=
inches
2rπ revolution
1.289 inches = r.
So, when the laser is 1.289 inches from the center, the CD is moving
350 RPM. 16.4 Belt and Wheel Problems
The industrial revolution spawned a number of elaborate machines involving systems of belts and wheels. Computing the speed of various
belts and wheels in such a system may seem complicated at ﬁrst glance.
The situation can range from a simple system of two wheels with a belt
connecting them, to more elaborate designs. We call problems of this sort
belt and wheel problems, or more generally, connected wheel problems.
Solving problems of this type always uses the same strategy, which we
will ﬁrst highlight by way of an example. CHAPTER 16. MEASURING CIRCULAR MOTION 216 J
E
C D G
F A H B I Figure 16.9: Two typical connected wheel scenarios. front sprocket radius = 5 inches rear sprocket radius = 2 inches (a) A stationary exercise bike.
radius A = 14 inches
A
C radius C = 5 inches B
radius B =
2 inches (b) A model of the bike’s
connected wheels.
Figure 16.10: Visualizing
the connected wheels of an
exercise bike. Example 16.4.1. You are riding a stationary exercise bike.
Assume the rear wheel is 28 inches in diameter, the rear
sprocket has radius 2 inches and the front sprocket has
radius 5 inches. How many revolutions per minute of the
front sprocket produces a forward speed of 40 MPH on the
bike (miles per hour)?
Solution. There are 3 wheels involved with a belt (the bicycle chain) connecting two of the wheels. In this problem,
we are provided with the linear speed of wheel A (which is
40 MPH) and we need to ﬁnd the angular speed of wheel
C=front sprocket.
Denote by vA , vB , and vC the linear speeds of each of
the wheels A, B, and C, respectively. Likewise, let ωA , ωB ,
and ωC denote the angular speeds of each of the wheels A,
B, and C, respectively. In addition, the chain connecting
the wheels B and C will have a linear speed, which we will
denote by vchain . The strategy is broken into a sequence of
steps which leads us from the known linear speed vA to
the angular speed ωC of wheel C:
• Step 1: Given vA , ﬁnd ωA . Use the fact ωA = vA
.
rA • Step 2: Observe ωA = ωB ; this is because the wheel
and rear sprocket are both rigidly mounted on a
common axis of rotation.
• Step 3: Given ωB , ﬁnd vB . Use the fact vB = rB ωB =
rB
rB ωA = rA vA .
• Step 4: Observe vB = vchain = vC ; this is because the
chain is directly connecting the two sprockets and
assumed not to slip. 16.4. BELT AND WHEEL PROBLEMS
• Step 5: Given vC , ﬁnd ωC . Use the fact ωC =
vB
rC = rB
rA rC 217
vC
rC = vA . Saying that the speedometer reads 40 MPH is the same as saying that
the linear speed of a location on the rear wheel is vA = 40 MPH. Converting
this into angular speed was carried out in our solution to Example 16.2.3
above; we found that ωA = 480 RPM. This completes Step 1 and so by Step
2, ωA = ωB = 480 RPM. For Step 3, we convert ωB = 480 RPM into linear
speed following Fact 16.2.1:
revolution
inches
(2(2)π)
minute
revolution
inches
= 6,032
.
minute
By Step 4, conclude that the linear speed of wheel C is vC = 6,032 inches/min.
Finally, to carry out Step 5, we convert the linear speed into angular
speed:
vB = ωC = 480 6,032 inches
min 2(5)π inches
rev
= 192 RPM
rev
.
= 3.2
sec
In conclusion, the bike rider must pedal the front sprocket at the rate of
rev
3.2 sec .
This example indicates the basic strategy used in all belt/wheel problems.
Important Facts 16.4.2 (Belt and Wheel Strategy). Three basic facts are
used in all such problems:
• Using “unit conversion” or Fact 16.2.2 allows us to go from linear
speed v to angular speed ω, and vice versa.
• If two wheels are fastened rigidly to a common axle, then they have
the same angular speed. (Caution: two wheels fastened to a common
axle typically do not have the same linear speed!)
• If two wheels are connected by a belt (or chain), the linear speed of
the belt coincides with the linear speed of each wheel. CHAPTER 16. MEASURING CIRCULAR MOTION 218 16.5 Exercises
Problem 16.1. The restaurant in the Space
Needle in Seattle rotates at the rate of one revolution per hour.
(a) Through how many radians does it turn
in 100 minutes?
(b) How long does it take the restaurant to
rotate through 4 radians?
(c) How far does a person sitting by the window move in 100 minutes if the radius of
the restaurant is 21 meters?
Problem 16.2. You are riding a bicycle along
a level road. Assume each wheel is 26 inches
in diameter, the rear sprocket has a radius of
3 inches and the front sprocket has a radius
of 7 inches. How fast do you need to pedal (in
revolutions per minute) to achieve a speed of
35 mph?
front wheel
rear wheel rear sprocket Problem 16.4. Lee is running around the
perimeter of a circular track at a rate of 10
ft/sec. The track has a radius of 100 yards.
After 10 seconds, Lee turns and runs along a
radial line to the center of the circle. Once he
reaches the center, he turns and runs along a
radial line to his starting point on the perimeter. Assume Lee does not slow down when he
makes these two turns.
(a) Sketch a picture of the situation.
(b) How far has Lee traveled once he returns
to his starting position?
(c) How much time will elapse during Lee’s
circuit?
(d) Find the area of the pie shaped sector
enclosed by Lee’s path.
Problem 16.5. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the worlds
greatest ferris wheel. Tiff isn’t into math; she
simply has a vision and has told John these
constraints on her dream: (i) the wheel should
rotate counterclockwise with an angular speed
of 12 RPM; (ii) the linear speed of a rider should
be 200 mph; (iii) the lowest point on the ride
should be 4 feet above the level ground. front sprocket
12 RPM Problem 16.3. Answer the following angular
speed questions.
(a) A wheel of radius 22 ft. is rotating 11
RPM counterclockwise. Considering a
point on the rim of the rotating wheel,
what is the angular speed ω in rad/sec
and the linear speed v in ft/sec?
(b) A wheel of radius 8 in. is rotating
15o /sec. What is the linear speed v, the
angular speed in RPM and the angular
speed in rad/sec? θ
P 4 feet (a) Find the radius of the ferris wheel. (c) You are standing on the equator of the
earth (radius 3960 miles). What is your
linear and angular speed? (b) Once the wheel is built, John suggests
that Tiff should take the ﬁrst ride. The
wheel starts turning when Tiff is at the
location P, which makes an angle θ with
the horizontal, as pictured. It takes her
1.3 seconds to reach the top of the ride.
Find the angle θ. (d) An auto tire has radius 12 inches. If you
are driving 65 mph, what is the angular
speed in rad/sec and the angular speed
in RPM? (c) Poor engineering causes Tiff’s seat to ﬂy
off in 6 seconds. Describe where Tiff is
located (an angle description) the instant
she becomes a human missile. 16.5. EXERCISES 219 Problem 16.6. Michael and Aaron are on the
“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 of the ride
to a drive wheel of radius 3 feet: (e) Assume Michael has traveled 88 feet
from the position P to a new position Q.
How many seconds will this take? What
will be the angle swept out by Michael? length of arc (PQ) is 88 ft Aaron
Drive wheel
radius 3 ft
Michael
O P O main ride
radius 20 ft drive belt (a) Assume Michael is seated on the edge of
the ride, as pictured. What is Michael’s
linear speed in mph and ft/sec?
(b) What is the angular speed of the drive
wheel in RPM?
(c) Suppose Aaron is seated 16 feet from the
center of the ride. What is the angular
speed of Aaron in RPM? What is the linear speed of Aaron in ft/sec?
(d) After 0.23 seconds Michael will be located at S as pictured. What is the angle ∠POS in degrees? What is the angle
∠POS in radians? How many feet has
Michael traveled? S P Q Problem 16.7. You are riding a bicycle along a
level road. Assume each wheel is 28 inches in
diameter, the rear sprocket has radius 3 inches
and the front sprocket has radius r inches.
Suppose you are pedaling the front sprocket
rev
at the rate of 1.5 sec and your forward speed is
11 mph on the bike. What is the radius of the
front sprocket?
Problem 16.8. You are designing a system of
wheels and belts as pictured below. You want
wheel A to rotate 20 RPM while wheel B rotates
42 RPM. Wheel A has a radius of 6 inches,
wheel B has a radius of 7 inches and wheel C
has a radius of 1 inch. Assume wheels C and
D are rigidly fastened to the same axle. What
is the radius r of wheel D? θ
O P D
A C B 220 CHAPTER 16. MEASURING CIRCULAR MOTION Chapter 17
The Circular Functions
Suppose Cosmo begins at location R and walks in a counterclockwise direction, always maintaining a tight 20 ft
long tether. As Cosmo moves around the circle, how can
we describe his location at any given instant?
In one sense, we have already answered this question:
The measure of ∠RPS1 exactly pins down a location on
the circle of radius 20 feet. But, we really might prefer
a description of the horizontal and vertical coordinates of
Cosmo; this would tie in better with the coordinate system
we typically use. Solving this problem will require NEW
functions, called the circular functions. S1
S2 P 20 feet S3 R S4 Figure 17.1: Cosmo moves
counterclockwise maintaining a tight tether. Where’s
Cosmo? 17.1 Sides and Angles of a Right Triangle
Example 17.1.1. You are preparing to make your ﬁnal
shot at the British Pocket Billiard World Championships.
The position of your ball is as in Figure 17.2, and you must
play the ball off the left cushion into the 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.
4 ft ﬁnd
this
location 5 ft this pocket
for the
big money 6 ft 12 ft Solution. This problem depends on two basic facts. First,
the angles of entry and exit between the path the cushion
will be equal. Secondly, the two obvious right triangles
in this picture are similar triangles. Let x represent the
distance from the bottom left corner to the impact point
of the ball’s path:
Properties of similar triangles tell us that the ratios of
4
common sides are equal: 5−x = 12 . If we solve this equation
x
for x, we obtain x = 15 = 3.75 feet.
4
221 4
5−x θ
θ x Mathmatically modeling the bank shot. Figure 17.2: A pocket billiard banking problem. CHAPTER 17. THE CIRCULAR FUNCTIONS 222 This discussion is enough to win the tourney. But, of course, there
are still other questions we can ask about this simple example: What is
the angle θ? That is going to require substantially more work; indeed the
bulk of this Chapter! It turns out, there is a lot of mathematical mileage
in the idea of studying ratios of sides of right triangles. The ﬁrst step,
which will get the ball rolling, is to introduce new functions whose very
deﬁnition involves relating sides and angles of right triangles. 17.2 The Trigonometric Ratios
From elementary geometry, the sum of the angles of any
triangle will equal 180◦ . Given a right triangle △ABC, since
one of the angles is 90◦ , the remaining two angles must be
acute angles ; i.e., angles of measure between 0◦ and 90◦ .
If we specify one of the acute angles in a right triangle
△ABC, say angle θ, we can label the three sides using
this terminology. We then consider the following three
ratios of side lengths, referred to as trigonometric ratios : B
hypotenuse
side opposite θ
A θ
C
side adjacent θ Figure 17.3: Labeling the
sides of a right triangle. def sin(θ) = def cos(θ) = def tan(θ) = length of side opposite θ
length of hypotenuse (17.1) length of side adjacent θ
length of hypotenuse (17.2) length of side opposite θ
.
length of side adjacent to θ (17.3) For example, we have three right triangles in Figure 17.4; you can
verify that the Pythagorean Theorem holds in each of the cases. In the
5
5
lefthand triangle, sin(θ) = 13 , cos(θ) = 12 , tan(θ) = 12 . In the middle
13
1
1
triangle, sin(θ) = √2 , cos(θ) = √2 , tan(θ) = 1. In the righthand triangle,
√ 1
sin(θ) = 1 , cos(θ) = 23 , tan(θ) = √3 . The symbols “sin”, “cos”, and “tan”
2
are abbreviations for the words sine, cosine and tangent, respectively.
As we have deﬁned them, the trigonometric ratios depend on the dimensions of the triangle. However, the same ratios are obtained for any right
triangle with acute angle θ. This follows from the properties of similar
triangles. Consider Figure 17.5. Notice △ABC and △ADE are similar. If
we use △ABC to compute cos(θ), then we ﬁnd cos(θ) = AC . On the other
AB hand, if we use △ADE, we obtain cos(θ) = AE
.
AD Since the ratios of com mon sides of similar triangles must agree, we have cos(θ) = AC
AB = AE
,
AD 17.2. THE TRIGONOMETRIC RATIOS √ 13 223 2 2 5 1 θ 1 θ
12 θ
1 √ 3 Figure 17.4: Computing trigonometric ratios for selected right triangles. which is what we wanted to be true. The same argument can be used
to show that sin(θ) and tan(θ) can be computed using any right triangle
with acute angle θ.
Except for some “rigged” right triangles, it is not easy
D
to calculate the trigonometric ratios. Before the 1970’s,
B
approximate values of sin(θ), cos(θ), tan(θ) were listed
in long tables or calculated using a slide rule. Today, a
scientiﬁc calculator saves the day on these computations.
θ
Most scientiﬁc calculators will give an approximation for
A
C
E
the values of the trigonometric ratios. However, it is good
Figure 17.5:
Applying
to keep in mind we can compute the EXACT values of the
π
π
π
π
trigonometric ratios to any
trigonometric ratios when θ = 0, 6 , 4 , 3 , 2 radians or,
right triangle.
equivalently, when θ = 0◦ , 30◦ , 45◦ , 60◦ , 90◦ .
Angle θ Trigonometric Ratio Deg Rad sin(θ) cos(θ) tan(θ) 0◦ 0 0 1 0 30◦ π
6 1
2 √
3
2 1
√
3 45◦ π
4 √
2
2 1 60◦ π
3 90◦ π
2 √ 2
2 √ 3
2 1 0 √ 1
2 3 Undeﬁned Table 17.1: Exact Trigonometric Ratios CHAPTER 17. THE CIRCULAR FUNCTIONS 224 Some people make a big deal of “approximate” vs. “exact” answers; we
won’t worry about it here, unless we are speciﬁcally asked for an exact
answer. However, here is something we will make a big deal about:
!!! When computing values of cos(θ), sin(θ), and tan(θ) on your calculator, make sure you are using the correct “angle mode” when entering
θ; i.e. “degrees” or “radians”. CAUTION
!!! For example, if θ = 1◦ , then cos(1◦ ) = 0.9998, sin(1◦ ) = 0.0175, and
tan(1◦ ) = 0.0175. In contrast, if θ = 1 radians, then cos(1) = 0.5403,
sin(1) = 0.8415, and tan(1) = 1.5574. 17.3 Applications h
h sin(θ)
θ a tan(θ)
θ h cos(θ) When confronted with a situation involving a right triangle where the measure of one acute angle θ and one side
are known, we can solve for the remaining sides using the
appropriate trigonometric ratios. Here is the key picture
to keep in mind: a Figure 17.6: What do these
ratios mean? Important Facts 17.3.1 (Trigonometric ratios).
Given a right triangle, the trigonometric ratios relate the
lengths of the sides as shown in Figure 17.6. Example 17.3.2. To measure the distance across a river for a new bridge,
surveyors placed poles at locations A, B and C. The length AB = 100 feet
and the measure of the angle ∠ABC is 31◦ 18 ′ . Find the distance to span the
river. If the measurement of the angle ∠ABC is only accurate within ±2 ′ ,
ﬁnd the possible error in AC. C Solution. The trigonometric ratio relating these two sides
would be the tangent and we can convert θ into decimal
form, arriving at: d
B 100 A tan(31◦18 ′ ) = tan(31.3◦) = 310 18 ′ Figure 17.7: The distance
spanning a river. AC
d
=
100
BA therefore d = 60.8 feet. This tells us that the bridge needs to span a gap of
60.8 feet. If the measurement of the angle was in error by +2 ′ , then
tan(31◦20 ′ ) = tan(31.3333◦) = 0.6088 and the span is 60.88 ft. On the
other hand, if the measurement of the angle was in error by −2 ′ , then
tan(31◦16 ′ ) = tan(31.2667◦) = 0.6072 and the span is 60.72 ft. 17.3. APPLICATIONS 225 Example 17.3.3. A plane is ﬂying 2000 feet above sea
level toward a mountain. The pilot observes the top of the
mountain to be 18◦ above the horizontal, then immediately
ﬂies the plane at an angle of 20◦ above horizontal. The
airspeed of the plane is 100 mph. After 5 minutes, the
plane is directly above the top of the mountain. How high
is the plane above the top of the mountain (when it passes
over)? What is the height of the mountain? T
E
P
2000 ft L 111
000
S sealevel
111
000
111
000 Figure 17.8: Flying toward a
mountain. Solution. We can compute the hypotenuse of △LPT by using the speed and time information about the plane:
PT  = (100 mph)(5 minutes)(1 hour/60 minutes) = 25
miles.
3 The deﬁnitions of the trigonometric ratios show:
25
sin(20◦ ) = 2.850 miles, and
3
25
cos(20◦ ) = 7.831 miles.
PL =
3
TL = With this data, we can now ﬁnd EL:
EL = PL tan(18◦ ) = 2.544 miles.
The height of the plane above the peak is TE = TL − EL = 2.850 − 2.544 =
0.306 miles = 1,616 feet. The elevation of the peak above sea level is
given by: Peak elevation = plane altitude + EL = SP + EL = 2,000 +
(2.544)(5,280) = 15,432 feet. Example 17.3.4. A Forest Service helicopter needs to determine the width of a deep canyon. While hovering, they
measure the angle γ = 48◦ at position B (see picture), then
descend 400 feet to position A and make two measurements of α = 13◦ (the measure of ∠EAD), β = 53◦ (the measure of ∠CAD). Determine the width of the canyon to the
nearest foot. B
400 ft C γ
A
β α
E 111
000
111
000
canyon
111
000 D Figure 17.9: Finding the
width of a canyon. Solution. We will need to exploit three right triangles in
the picture: △BCD, △ACD, and △ACE. Our goal is to compute ED =
CD − CE, which suggests more than one right triangle will come into
play. CHAPTER 17. THE CIRCULAR FUNCTIONS 226 The ﬁrst step is to use △BCD and △ACD to obtain a system of two
equations and two unknowns involving some of the side lengths; we will
then solve the system. From the deﬁnitions of the trigonometric ratios,
CD = (400 + AC) tan(48◦)
CD = AC tan(53◦ ).
Plugging the second equation into the ﬁrst and rearranging we get
AC = 400 tan(48◦ )
= 2,053 feet.
tan(53◦) − tan(48◦ ) Plugging this back into the second equation of the system gives
CD = (2053) tan(53◦) = 2724 feet.
The next step is to relate △ACD and △ACE, which can now be done
in an effective way using the calculations above. Notice that the measure
of ∠CAE is β − α = 40◦ . We have
CE = AC tan(40◦ ) = (2053) tan(40◦ ) = 1,723 feet.
As noted above, ED = CD − CE = 2,724 − 1,723 = 1,001 feet is the width
of the canyon. 17.4 Circular Functions
S = (x,y) 20
y
θ
P x R If Cosmo is located somewhere in the ﬁrst quadrant of
Figure 17.1, represented by the location S, we can use the
trigonometric ratios to describe his coordinates. Impose
the indicated xycoordinate system with origin at P and
extract the pictured right triangle with vertices at P and
S. The radius is 20 ft. and applying Fact 17.3.1 gives
S = (x, y) = (20 cos(θ), 20 sin(θ)). Figure 17.10: Cosmo on a
circular path. Unfortunately, we run into a snag if we allow Cosmo to
wander into the second, third or fourth quadrant, since
then the angle θ is no longer acute. 17.4. CIRCULAR FUNCTIONS 227
17.4.1 Are the trigonometric ratios functions?
Recall that sin(θ), cos(θ), and tan(θ) are deﬁned for acute
A unit circle with
angles θ inside a right triangle. We would like to say that
P
radius = 1.
these three equations actually deﬁne functions where the
1
variable is an angle θ. Having said this, it is natural to ask
θ
if these three equations can be extended to be deﬁned for
π
O
R
ANY angle θ. For example, we need to explain how sin 23
1
is deﬁned.
To start, we begin with the unit circle pictured in the
xycoordinate system. Let θ = ∠ROP be the angle in stanFigure 17.11: Coordinates
dard central position shown in Figure 17.11. If θ is posof points on the unit circle.
itive (resp. negative), we adopt the convention that θ is
swept out by counterclockwise (resp. clockwise) rotation
of the initial side OR. The objective is to ﬁnd the coordinates of the point
P in this ﬁgure. Notice that each coordinate of P (the xcoordinate and
the ycoordinate) will depend on the given angle θ. For this reason, we
need to introduce two new functions involving the variable θ.
Deﬁnition 17.4.1. Let θ be an angle in standard central
position inside the unit circle, as in Figure 17.11. This angle
determines a point P on the unit circle. Deﬁne two new
functions, cos(θ) and sin(θ), on the domain of all θ values
as follows: 11
00
11
00 1111 11
1111
0000 11
r 0000 00
= 1 kilometer
1111 00
0000 00
1111
0000 11 0.025 rad
sec
Michael
starts here def cos(θ) = horizontal xcoordinate of P on unit circle
def sin(θ) = vertical ycoordinate of P on unit circle.
Figure 17.12: A circular We refer to sin(θ) and cos(θ) as the basic circular funcdriving track.
tions. Keep in mind that these functions have variables
which are angles (either in degree or radian measure). These functions
will be on your calculator. Again, BE CAREFUL to check the angle mode
setting on your calculator (“degrees” or “radians”) before doing a calculation.
Example 17.4.2. Michael is test driving a vehicle counterclockwise around a desert test track which is circular of
radius 1 kilometer. He starts at the location pictured, traveling 0.025 rad . Impose coordinates as pictured. Where is
sec
Michael located (in xycoordinates) after 18 seconds?
Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds and θ(t) the
angle swept out the by Michael after t seconds. Since we
are given the angular speed, we get
θ(t) = 0.025t radians. yaxis
M(t) = (x(t),y(t))
0.025
θ(t ) rad
sec xaxis
Michael
starts here Figure 17.13:
Modeling
Michael’s location. CHAPTER 17. THE CIRCULAR FUNCTIONS 228 Since the angle θ(t) is in central standard position, we get
M(t) = (cos(θ(t)), sin(θ(t))) = (cos(0.025t), sin(0.025t)).
So, after 18 seconds Michael’s location will be M(18) = (0.9004, 0.4350).
!!! Interpreting the coordinates of the point P = (cos(θ), sin(θ)) in Figure 17.11 only works if the angle θ is viewed in central standard position. You must do some additional work if the angle is placed in a
different position; see the next Example. CAUTION
!!! yaxis Angela
starts here 0.025 rad
sec r = 1 kilometer xaxis
Michael
starts here 0.03 rad
sec (a) Angela and Michael on the
same test track.
yaxis
M(t)
Angela
starts here β(t)
α(t) 0.03 rad
sec 0.025 rad
sec
θ(t) xaxis
Michael
starts here A(t) (b) Modeling the motion of
Angela and Michael.
Figure 17.14: Visualizing
motion on a circular track. Example 17.4.3. Both Angela and Michael are test driving
vehicles counterclockwise around a desert test track which
is circular of radius 1 kilometer. They start at the locations shown in Figure 17.14(a). Michael is traveling 0.025
rad/sec and Angela is traveling 0.03 rad/sec. Impose coordinates as pictured. Where are the drivers located (in
xycoordinates) after 18 seconds?
Solution. Let M(t) be the point on the circle of motion representing Michael’s location after t seconds. Likewise, let
A(t) be the point on the circle of motion representing Angela’s location after t seconds. Let θ(t) be the angle swept
out the by Michael and α(t) the angle swept out by Angela
after t seconds.
Since we are given the angular speeds, we get
θ(t) = 0.025t radians, and
α(t) = 0.03t radians.
From the previous Example 17.4.2,
M(t) = (cos(0.025t), sin(0.025t)), and
M(18) = (0.9004, 0.4350).
Angela’s angle α(t) is NOT in central standard position, so
we must observe that α(t) + π = β(t), where β(t) is in central standard position: See Figure 17.14(b). We conclude
that
A(t) = (cos(β(t)), sin(β(t)))
= (cos(π + 0.03t), sin(π + 0.03t)). So, after 18 seconds Angela’s location will be A(18) =
(−0.8577, −0.5141). 17.4. CIRCULAR FUNCTIONS 229 17.4.2 Relating circular functions and right triangles
If the point P on the unit circle is located in the ﬁrst
quadrant, then we can compute cos(θ) and sin(θ) using
trigonometric ratios. In general, it’s useful to relate right
triangles, the unit circle and the circular functions. To describe this connection, given θ we place it in central standard position in the unit circle, where ∠ROP = θ. Draw
a line through P perpendicular to the xaxis, obtaining
an inscribed right triangle. Such a right triangle has hypotenuse of length 1, vertical side of length labeled b and
horizontal side of length labeled a. There are four cases:
See Figure 17.16.
P O θ
a b sin(θ)
θ
O R 1
0
1
0
a
1
0
O Figure 17.15: The point P in
the ﬁrst quadrant. θ θ a
R b 11
00
O
11
00
11
00 R θ 11
00
11 11
00 00 11 11
00 00
a
O
bR
P P
CASE I CASE II CASE III CASE IV Figure 17.16: Possible positions of θ on the unit circle. Case I has already been discussed, arriving at cos(θ) = a and sin(θ) =
b. In Case II , we can interpret cos(θ) = −a, sin(θ) = b. We can reason
similarly in the other Cases III and IV, using Figure 17.16, and we arrive
at this conclusion:
Important Facts 17.4.4 (Circular functions and triangles). View θ as in
Figure 17.16 and form the pictured inscribed right triangles. Then we can
interpret cos(θ) and sin(θ) in terms of these right triangles as follows:
Case I :
Case II :
Case III :
Case IV : cos(θ) = a,
cos(θ) = −a,
cos(θ) = −a,
cos(θ) = a, sin(θ) = b
sin(θ) = b
sin(θ) = −b
sin(θ) = −b R cos(θ) P b unit circle (radius = 1)
P CHAPTER 17. THE CIRCULAR FUNCTIONS 230 17.5 What About Other Circles?
T
P
θ
O R Cr S unit circle Figure 17.17:
Points on
other circles. What happens if we begin with a circle Cr with radius r
(possibly different than 1) and want to compute the coordinates of points on this circle?
The circular functions can be used to answer this more
general question. Picture our circle Cr centered at the
origin in the same picture with unit circle C1 and the angle
θ in standard central position for each circle. As pictured,
we can view θ = ∠ROP = ∠SOT . If P = (x,y) is our point
on the unit circle corresponding to the angle θ, then the
calculation below shows how to compute coordinates on general circles:
P = (x,y)
= (cos(θ), sin(θ)) ∈ C1 ⇔
⇔ ⇔
⇔ x2 + y2 = 1
r2 x2 + r2 y2 = r2
(rx)2 + (ry)2 = r2
T = (rx, ry)
= (r cos(θ), r sin(θ)) ∈ Cr . Important Fact 17.5.1. Let Cr be a circle of radius r centered at the origin
and θ = ∠SOT an angle in standard central position for this circle, as in
Figure 17.17. Then the coordinates of T = (r cos(θ), r sin(θ)). yaxis
β = π − α = 2.9416
R
U
α
B T Q
S P
O Examples 17.5.2. Consider the picture below, with θ = 0.8
radians and α = 0.2 radians. What are the coordinates of
the labeled points? θ
A xaxis
circle radius = 1
circle radius = 2
circle radius = 3 Figure 17.18: Coordinates
of points on circles. Solution. The angle θ is in standard central position; α is
a central angle, but it is not in standard position. Notice,
β = π − α = 2.9416 is an angle in standard central position which locates the same points U, T, S as the angle α.
Applying Deﬁnition 17.4.1 on page 227:
P = (cos(0.8), sin(0.8))
= (0.6967, 0.7174)
Q = (2 cos(0.8), 2 sin(0.8))
= (1.3934, 1.4347)
R = (3 cos(0.8), 3 sin(0.8))
= (2.0901, 2.1521)
S = (cos(2.9416), sin(2.9416))
= (−0.9801, 0.1987)
T = (2 cos(2.9416), 2 sin(2.9416)) = (−1.9602, 0.3973)
U = (3 cos(2.9416), 3 sin(2.9416)) = (−2.9403, 0.5961). 17.6. OTHER BASIC CIRCULAR FUNCTION 231 Example 17.5.3. Suppose Cosmo begins at the position R
in the ﬁgure, walking around the circle of radius 20 feet
4
with an angular speed of 5 RPM counterclockwise. After 3
minutes have elapsed, describe Cosmo’s precise location.
4
Solution. Cosmo has traveled 3 5 = 12 revolutions. If θ is
5
the angle traveled after 3 minutes, θ = 152 rev 2π radians =
rev
24π
radians = 15.08 radians. By (15.5.1), we have x =
5
20 cos 24π rad = −16.18 feet and y = 20 sin 24π rad =
5
5
11.76 feet. Conclude that Cosmo is located at the point
S = (−16.18, 11.76). Using (15.1), θ = 864◦ = 2(360◦) + 144◦ ;
this means that Cosmo walks counterclockwise around
the circle two complete revolutions, plus 144◦. S P Given any angle θ, our constructions offer a concrete link between the
cosine and sine functions and right triangles inscribed inside the unit
circle: See Figure 17.20. P
θ θ
O R O θ
R O R
R θO
P P
CASE I CASE II CASE III CASE IV Figure 17.20: Computing the slope of a line using the function tan(θ). The slope of the hypotenuse of these inscribed triangles is just the
slope of the line through OP. Since P = (cos(θ), sin(θ)) and O = (0, 0):
Slope = sin(θ)
∆y
=
;
∆x
cos(θ) this would be valid as long as cos(θ) = 0. This calculation motivates a
new circular function called the tangent of θ by the rule
tan(θ) = sin(θ)
,
cos(θ) provided cos(θ) = 0. R Figure 17.19:
Where is
Cosmo after 3 minutes? 17.6 Other Basic Circular Function P 20 feet CHAPTER 17. THE CIRCULAR FUNCTIONS 232 The only time cos(θ) = 0 is when the corresponding point P on the
unit circle has xcoordinate 0. But, this only happens at the positions
(0, 1) and (0, −1) on the unit circle, corresponding to angles of the form
θ = ± π , ± 3π , ± 5π , · · · . These are the cases when the inscribed right tri2
2
2
angle would “degenerate” to having zero width and the line segment OP
becomes vertical. In summary, we then have this general idea to keep in
mind:
Important Fact 17.6.1. The slope of a line = tan(θ), where θ is the
angle the line makes with the xaxis (or any other horizontal line)
Three other commonly used circular functions come up from time to
time. The cotangent function y = cot(θ), the secant function y = sec(θ)
and the cosecant function y = csc(θ) are deﬁned by the formulas:
def sec(θ) = 1
,
cos(θ) def csc(θ) = 1
,
sin(θ) def cot(θ) = 1
.
tan(θ) Just as with the tangent function, one needs to worry about the values
of θ for which these functions are undeﬁned (due to division by zero). We
will not need these functions in this text. North
Alaska Northwest 500
1150
West East
200 SeaTac
South Delta (a) The ﬂight paths of three
airplanes.
x = 30 Alaska
Q Northwest N P y = 20 W E
R Delta x = −50
S (b) Modeling the paths of each
ﬂight.
Figure 17.21: Visualizing
and modeling departing airplanes. Example 17.6.2. Three airplanes depart SeaTac Airport.
A NorthWest ﬂight is heading in a direction 50◦ counterclockwise from East, an Alaska ﬂight is heading 115◦ counterclockwise from East and a Delta ﬂight is heading 20◦
clockwise from East. Find the location of the Northwest
ﬂight when it is 20 miles North of SeaTac. Find the location of the Alaska ﬂight when it is 50 miles West of SeaTac.
Find the location of the Delta ﬂight when it is 30 miles East
of SeaTac.
Solution. We impose a coordinate system in Figure 17.21(a), where “East” (resp. “North”) points along the
positive xaxis (resp. positive yaxis). To solve the problem, we will ﬁnd the equation of the three lines representing the ﬂight paths, then determine where they intersect
the appropriate horizontal or vertical line. The Northwest
and Alaska directions of ﬂight are angles in standard central position; the Delta ﬂight direction will be −20◦ . We can
imagine right triangles with their hypotenuses along the
directions of ﬂight, then using the tangent function, we
have these three immediate conclusions:
slope NW line = tan(50◦ ) = 1.19,
slope Alaska line = tan(115◦ ) = −2.14, and
slope Delta line = tan(−20◦ ) = −0.364. 17.6. OTHER BASIC CIRCULAR FUNCTION 233 All three ﬂight paths pass through the origin (0,0) of our coordinate
system, so the equations of the lines through the ﬂight paths will be:
NW ﬂight : y = 1.19x,
Alaska ﬂight : y = −2.14x,
Delta ﬂight : y = −0.364x.
The Northwest ﬂight is 20 miles North of SeaTac when y = 20; plugging
into the equation of the line of ﬂight gives 20 = 1.19x, so x = 16.81 and
the plane location will be P = (16.81, 20). Similarly, the Alaska ﬂight is
50 miles West of SeaTac when x = −50; plugging into the equation of the
line of ﬂight gives y = (−2.14)(−50) = 107 and the plane location will be
Q = (−50, 107). Finally, check that the Delta ﬂight is at R = (30, −10.92)
when it is 30 miles East of SeaTac. CHAPTER 17. THE CIRCULAR FUNCTIONS 234 17.7 Exercises
Problem 17.1. John has been hired to design an exciting carnival ride. Tiff, the carnival owner, has decided to create the world’s
greatest ferris wheel. Tiff isn’t into math; she
simply has a vision and has told John these
constraints on her dream: (i) the wheel should
rotate counterclockwise with an angular speed
of 12 RPM; (ii) the linear speed of a rider should
be 200 mph; (iii) the lowest point on the ride
should be 4 feet above the level ground. Recall,
we worked on this in Exercise 16.5. forest) as a potential landing site, but are uncertain whether it is wide enough. They make
two measurements from A (see picture) ﬁnding
α = 25o and β = 54o . They rise vertically 100
feet to B and measure γ = 47o . Determine the
width of the clearing to the nearest foot.
B
γ 100 feet
A
β
C α
E clearing D 12 RPM
θ
P 4 feet (a) Impose a coordinate system and ﬁnd the
coordinates T (t) = (x(t),y(t)) of Tiff at
time t seconds after she starts the ride.
(b) Tiff becomes a human missile after 6
seconds on the ride. Find Tiff’s coordinates the instant she becomes a human
missile.
(c) Find the equation of the tangential line
along which Tiff travels the instant she
becomes a human missile. Sketch a picture indicating this line and her initial
direction of motion along it when the
seat detaches. Problem 17.2.
(a) Find the equation of a
line passing through the point (1,2) and
making an angle of 13o with the xaxis.
(Note: There are two answers; ﬁnd them
both.)
(b) Find the equation of a line making an
angle of 8o with the yaxis and passing
through the point (1,1). (Note: There are
two answers; ﬁnd them both.)
Problem 17.3. The crew of a helicopter needs
to land temporarily in a forest and spot a ﬂat
horizontal piece of ground (a clearing in the Problem 17.4. Marla is running clockwise
around a circular track. She runs at a constant speed of 3 meters per second. She takes
46 seconds to complete one lap of the track.
From her starting point, it takes her 12 seconds to reach the northermost point of the
track.
Impose a coordinate system with the center of the track at the origin, and the northernmost point on the positive yaxis.
(a) Give Marla’s coordinates at her starting
point.
(b) Give Marla’s coordinates when she has
been running for 10 seconds.
(c) Give Marla’s coordinates when she has
been running for 901.3 seconds.
Problem 17.5. A 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. rotating 3 RPM θ jump on here 17.7. EXERCISES 235 (a) If θ = 34o , then what are your xycoordinates after 4 minutes?
(b) If θ = 20o , then what are your xycoordinates after 45 minutes?
(c) If θ = −14o , then what are your xycoordinates after 6 seconds? Draw an
accurate picture of the situation.
(d) If θ = −2.1 rad, then what are your
xycoordinates after 2 hours and 7 seconds? Draw an accurate picture of the
situation.
(e) If θ = 2.1 rad, then what are your xycoordinates after 5 seconds? Draw an
accurate picture of the situation.
Problem 17.6. Shirley is on a ferris wheel
which spins at the rate of 3.2 revolutions per
minute. The wheel has a radius of 45 feet, and
the center of the wheel is 59 feet above the
ground. After the wheel starts moving, Shirley
takes 16 seconds to reach the top of the wheel.
How high above the ground is she when
the wheel has been moving for 9 minutes?
Problem 17.7. The top of the Boulder Dam
has an angle of elevation of 1.2 radians from
a point on the Colorado River. Measuring the
angle of elevation to the top of the dam from
a point 155 feet farther down river is 0.9 radians; assume the two angle measurements are
taken at the same elevation above sea level.
How high is the dam? downriver
dam
0.9 1.2 relatively ﬂat area nearby the tower (not necessarily the same altitude as the bottom of the
tower), and standing some unknown distance
away from the tower, you make three measurements all at the same height above sea
level. You observe that the top of the old tower
makes an angle of 39◦ above level. You move
110 feet away from the original measurement
and observe that the old top of the tower now
makes an angle of 34◦ above level. Finally, after the new construction is complete, you observe that the new top of the tower, from the
same point as the second measurement was
made, makes an angle of 40◦ above the horizontal. All three measurements are made at
the same height above sea level and are in line
with the tower. Find the height of the addition
to the tower, to the nearest foot.
Problem 17.9. Charlie and Alexandra are
running around a circular track with radius
60 meters. Charlie started at the westernmost point of the track, and, at the same time,
Alexandra started at the northernmost point.
They both run counterclockwise. Alexandra
runs at 4 meters per second, and will take exactly 2 minutes to catch up to Charlie.
Impose a coordinate system, and give the
x and ycoordinates of Charlie after one
minute of running.
Problem 17.10. George and Paula are running around a circular track. George starts at
the westernmost point of the track, and Paula
starts at the easternmost point. The illustration below shows their starting positions and
running directions. They start running toward
each other at constant speeds. George runs at
9 feet per second. Paula takes 50 seconds to
run a lap of the track. George and Paula pass
each other after 11 seconds. N 155 ft a Problem 17.8. A radio station obtains a permit to increase the height of their radio tower
on Queen Anne Hill by no more than 100 feet.
You are the head of the Queen Anne Community Group and one of your members asks you
to make sure that the radio station does not
exceed the limits of the permit. After ﬁnding a George Paula After running for 3 minutes, how far east
of his starting point is George? CHAPTER 17. THE CIRCULAR FUNCTIONS 236 Problem 17.11. A kite is attached to 300 feet
of string, which makes a 42 degree angle with
the level ground. The kite pilot is holding the
string 4 feet above the ground. (d) What are the coordinates of the bug after 1 second? After 0 seconds? After 3
seconds? After 22 seconds? ω=4π/9 rad/sec
bug lands here kite 1.2 rad o
42
4 feet 2 in ground level (a) How high above the ground is the kite?
(b) Suppose that power lines are located
250 feet in front of the kite ﬂyer. Is
any portion of the kite or string over the
power lines?
bug lands here Problem 17.12. In the pictures below, a bug
has landed on the rim of a jelly jar and is moving around the rim. The location where the
bug initially lands is described and its angular speed is given. Impose a coordinate system with the origin at the center of the circle
of motion. In each of the cases, answer these
questions:
(a) Find an angle θ0 in standard central position that gives the bugs initial location.
(In some cases, this is the angle given in
the picture; in other cases, you will need
to do something.)
(b) The location angle of the bug at time t is
given by the formula θ(t) = θ0 + ωt. Plug
in the values for θ0 and ω to explicitly
obtain a formula for θ(t).
(c) Find the coordinates of the bug at time
t. 2 in ω=4π/9rad/sec ω= 4π/9 rad/sec
bug lands
here
0.5 rad 2 in Chapter 18
Trigonometric Functions
Our deﬁnitions of the circular functions are based upon the unit circle.
This makes it easy to visualize many of their properties. 18.1 Easy Properties of Circular Functions
How can we determine the range of function values for
cos(θ) and sin(θ)?
To begin with, recall the abstract
deﬁnition for the range of a function f(θ): yaxis
(0,1) P (θ) sin(θ)
1 θ (−1,0) (1,0) cos(θ) Range of f = {f(θ) : θ is in the domain}. ball moves
counterclockwise xaxis UNIT CIRCLE
(0, − 1) Using the unit circle constructions of the basic circular
Figure 18.1: Visualizing the
functions, it is easy to visualize the range of cos(θ) and
range of sin(θ) and cos(θ).
sin(θ). Beginning at the position (1, 0), imagine a ball
moving counterclockwise around the unit circle. If we
“freeze” the motion at any point in time, we will have swept out an angle
θ and the corresponding position P(θ) on the circle will have coordinates
P(θ) = (cos(θ), sin(θ)).
yaxis light source ball moves from 0 to
π radians around
2 ball moves from 0
to π radians
2 unit
circle xaxis around unit circle light
source xaxis yaxis (b) What do you see on the
(a) What do you see on the
xaxis?
yaxis?
Figure 18.2: Projecting the coordinates of points onto the yaxis and the xaxis. By studying the coordinates of the ball as it moves in the ﬁrst quadrant, we will be studying cos(θ) and sin(θ), for 0 ≤ θ ≤ π/2 radians.
237 CHAPTER 18. TRIGONOMETRIC FUNCTIONS 238 We can visualize this very concretely. Imagine a light source as in Figure 18.2(a); then a shadow projects onto the vertical yaxis. The shadow
locations you would see on the yaxis are precisely the values sin(θ), for
0 ≤ θ ≤ π/2 radians. Similarly, imagine a light source as in Figure 18.2(b);
then a shadow projects onto the horizontal xaxis. The shadow locations
you would see on the xaxis are precisely the values cos(θ), for 0 ≤ θ ≤ π/2
radians.
There are two visual conclusions: First, the function values of sin(θ)
vary from 0 to 1 as θ varies from 0 to π/2. Secondly, the function values
of cos(θ) vary from 1 to 0 as θ varies from 0 to π/2. Of course, we can go
ahead and continue analyzing the motion as the ball moves into the second, third and fourth quadrant, ending up back at the starting position
(1, 0). See Figure 18.3.
light source yaxis
#2 #1 ball moves from
0 to 2π radians
around unit
circle xaxis
#3
#4 ball moves from 0
#1 #2
#3 #4 to 2π radians
around nit circle xaxis light
source (a) What do you see on the
yaxis? (b) What do you see on the
xaxis? Figure 18.3: Analyzing the values of the sine and cosine functions. The conclusion is that after one complete counterclockwise rotation,
the values of sin(θ) and cos(θ) range over the interval [−1, 1]. As the
ball moves through the four quadrants, we have indicated the “order” in
which these function values are assumed by labeling arrows #1 — #4:
For example, for the sine function, look at Figure 18.3(a). The values of
the sine function vary from 0 up to 1 while the ball moves through the
ﬁrst quadrant (arrow labeled #1), then from 1 down to 0 (arrow labeled
#2), then from 0 down to −1 (arrow labeled #3), then from −1 up to 0
(arrow labeled #4).
What about the tangent function? We have seen that the tangent
function computes the slope of the hypotenuse of an inscribed triangle.
This means we can determine the range of values of tan(θ) by investigating the possible slopes for these inscribed triangles. We will maintain the
above model of a ball moving around the unit circle.
We look at two cases, each starting at (1, 0). In the ﬁrst quadrant, the
ball moves counterclockwise and in the fourth quadrant it moves clockwise: In the ﬁrst quadrant, we notice that these hypotenuse slopes are
always nonnegative, beginning with slope 0 (the degenerate right triangle
when θ = 0) then increasing. In fact, as the angle θ approaches π/2 radians, the ball is getting closer to the position (0, 1) and the hypotenuse 18.1. EASY PROPERTIES OF CIRCULAR FUNCTIONS
yaxis yaxis
8 7 239 6
5
4
3
2
1 xaxis 1 xaxis 2
3
4
5
6
7 (a) What happens to the
slopes of these triangles? (b) What happens to the
slopes of these triangles? Figure 18.4: Analyzing the values of the tangent function. is approaching a vertical line. This tells us that as θ varies from 0 to
π/2 (but not equal to π/2), these slopes attain all possible nonnegative
values. In other words, the range of values for tan(θ) on the domain
0 ≤ θ < π/2 will be 0 ≤ z < ∞. Similar reasoning shows that as the ball
moves in the fourth quadrant, the slopes of the hypotenuses of the triangles are always nonpositive, varying from 0 to ANY negative value. In
other words, the range of values for tan(θ) on the domain −π/2 < θ ≤ 0
will be −∞ < z ≤ 0.
On your calculator, you can verify the visual conclusions we just established by studying the values of tan(θ) for θ close (but not equal) to π
2
radians = 90◦ :
tan(89◦ ) = 57.29
tan(89.9◦) = 572.96
tan(89.99◦) = 5729.58
.
.
. tan(−89◦ ) = −57.29
tan(−89.9◦) = −572.96
tan(−89.99◦) = −5729.58
.
.
. The fact that the values of the tangent function become arbitrarily large
as we get close to ±π/2 radians means the function output values are
unbounded.
Important Fact 18.1.1 (Circular function values). For any angle θ, we
always have −1 ≤ cos(θ) ≤ 1 and −1 ≤ sin(θ) ≤ 1. On domain 0 ≤ θ ≤ 2π,
the range of both cos(θ) and sin(θ) is −1 ≤ z ≤ 1. In contrast, on the
domain of all θ values for which tangent is deﬁned, the range of tan(θ) is
all real numbers.
For the sine and cosine functions, if the domain is not 0 ≤ θ ≤ 2π,
then we need to consider the “periodic qualities” of the circular functions
to determine the range. This is discussed below. 240 CHAPTER 18. TRIGONOMETRIC FUNCTIONS 18.2 Identities
There are dozens of formulas that relate the values of two or more circular functions; these are usually lumped under the heading of Trigonometric Identities. In this course, we only need a couple frequently used
identities.
If we take the point P = (cos(θ), sin(θ)) on the unit circle, corresponding to the standard central position angle θ, then recall the equation of
the unit circle tells us x2 + y2 = 1. But, since the x coordinate is cos(θ) and
the y coordinate is sin(θ), we have (cos(θ))2 + (sin(θ))2 = 1. It is common
notational practice to write (cos(θ))2 = cos2 (θ) and (sin(θ))2 = sin2 (θ).
This leads to the most important of all trigonometric identities:
Important Fact 18.2.1 (Trigonometric identity). For any angle θ, we
have the identity cos2 (θ) + sin2 (θ) = 1.
Adding any multiple of 2π radians (or 360◦ ) to an angle will not change
the values of the circular functions. If we focus on radians for a moment,
this says that knowing the values of cos(θ) and sin(θ) on the domain
0 ≤ θ ≤ 2π determines the values for any other possible angle.
There is something very general going on here, so let’s pause a moment to make a deﬁnition and then an observation.
Deﬁnition 18.2.2 (Periodic function). For c > 0, a function f(θ) is called
cperiodic if two things are true:
(i) f(θ + c) = f(θ) holds for all θ;
(ii) There is no smaller d, 0 < d < c, such that f(θ + d) = f(θ) holds for all
θ.
We usually call c the period of the function.
Using this new terminology, we conclude that the sine and cosine
circular functions are 2πperiodic. In the case of the tangent circular
function, it is also true that tan(θ) = tan(θ + 2πn), for every integer n.
However, referring back to the unit circle deﬁnitions of the circular functions, we have tan(θ) = tan(θ + nπ), for all integers n. If you take n = 1,
then this tells us that the tangent circular function is πperiodic. We
summarize this information below.
Important Fact 18.2.3 (Periodicity identity). For any angle θ and any
integer n = 0, ±1, ±2, ±3, . . . , we have cos(θ) = cos(θ + 2πn), sin(θ) = sin(θ +
2πn), and tan(θ) = tan(θ + nπ). 18.2. IDENTITIES
Next, we draw an angle θ and its negative in the same
unit circle picture in standard central position. We have
indicated the points Pθ and P−θ used to deﬁne the circular functions. It is clear from the picture in Figure 18.5
that Pθ and P−θ have the same xcoordinate, but the ycoordinates are negatives of one another. This gives the
next identity: 241
unit
circle (cos(θ), sin(θ)) = Pθ
θ
O R
−θ
(cos(−θ), sin(−θ)) = P−θ Figure 18.5: Visualizing a
trigonometric identity. Important Fact 18.2.4 (Even/Odd identity). For any angle θ, sin(−θ) =
− sin(θ), and cos(−θ) = cos(θ). We can use the terminology of even and odd functions here. In this
language, this result says that the cosine function is an even function
and the sine function is an odd function.
Next, draw the angles θ and θ + π in the same unit
circle picture in standard central position. We have indicated the corresponding points Pθ and Pθ+π on the unit
circle and their coordinates in terms of the circular functions: From the picture in Figure 18.6, the xcoordinate of
Pθ must be the “negative” of the xcoordinate of Pθ+π and
similarly, the ycoordinate of Pθ must be the “negative” of
the ycoordinate of Pθ+π . This gives us the next identity: Pθ = (cos(θ), sin(θ))
θ
R θ+π
unit
circle Figure Pθ+π =
(cos(θ + π), (sin(θ + π)) 18.6:
Visualizing
Fact 18.2.5. Important Fact 18.2.5 (Plus π identity). For any angle θ, we have sin(θ +
π) = − sin(θ), and cos(θ + π) = − cos(θ). Important Fact 18.2.6. For any angle θ, we have sin(π − θ) = sin(θ) and
cos(π − θ) = − cos(θ). π
1
For example, we have sin 56 = sin π = 2 . This calculation leads to a
6
computational observation: Combining Table 17.1 with the previous two
identities we can compute the EXACT value of cos(θ), sin(θ), and tan(θ)
at an angle θ which is a multiple of 30◦ = π radians or 45◦ = π radians.
6
4
Here are some sample calculations together with a reference as to “why”
each equality is valid: CHAPTER 18. TRIGONOMETRIC FUNCTIONS 242
Example 18.2.7.
(i) (ii) cos(−45◦ ) = cos(45◦ )
√
2
=
2 Fact 18.2.4 on page 241
Table 17.1 on page 223 sin(225◦ ) = sin(45◦ + 180◦ ) Fact 18.2.3 on page 240
= − sin(45◦ )
Fact 18.2.5 on page 241
√
2
Table 17.1 on page 223
=−
2 (iii) cos 2π
3 π
+π
3
π
= − cos −
3
π
= − cos
3
1
=−
2
= cos − Fact 18.2.3 on page 240
Fact 18.2.5 on page 241
Fact 18.2.4 on page 241
Table 17.1 on page 223 18.3 Graphs of Circular Functions
We have introduced three new functions of the variable θ and it is important to understand and interpret the pictures of their graphs. To do this,
we need to settle on a coordinate system in which to work. The horizontal
axis will correspond to the independent variable, so this should be the θaxis. We will label the vertical axis, which corresponds to the dependent
variable, the zaxis. With these conventions, beginning with any of the
circular functions z = sin(θ), z = cos(θ), or z = tan(θ), the graph will be
a subset of the θzcoordinate system. Precisely, given a circular function
z = f(θ), the graph consists of all pairs (θ, f(θ)), where θ varies over a domain of allowed values. We will record and discuss these graphs below;
a graphing device will painlessly produce these for us!
There is a point of possible confusion that needs attention. We purposely did not use the letter “y” for the dependent variable of the circular
functions. This is to avoid possible confusion with our construction of
the sine and cosine functions using the unit circle. Since we viewed the
unit circle inside the xycoordinate system, the xcoordinates (resp. ycoordinates) of points on the unit circle are computed by cos(θ) (resp.
sin(θ)). 18.3. GRAPHS OF CIRCULAR FUNCTIONS z 243 y
P = (cos(θ), sin(θ)) θ
x θ Coordinate system used to
GRAPH the circular functions. Coordinate system used to DEFINE the circular functions. Figure 18.7: The zθ versus xy coordinate systems. 18.3.1 A matter of scaling
The ﬁrst issue concerns scaling of the axes used in graphing the circular
functions. As we know, the deﬁnition of radian measure is directly tied
to the lengths of arcs subtended by angles in the unit circle:
Important Fact 18.3.1. An angle of measure 1 radian inside the unit
circle will subtend an arc of length 1.
Since length is a good intuitive scaling quantity, it is natural to scale
the θaxis so that the length of 1 radian on the θaxis (horizontal axis)
is the same length as 1 unit on the vertical axis. For this reason, we
will work primarily with radian measure when sketching the graphs of
circular functions. If we need to work explicitly with degree measure for
angles, then we can always convert radians to degrees using the fact:
360◦ = 2π radians. 18.3.2 The sine and cosine graphs
Using Fact 18.1.1, we know that −1 ≤ sin(θ) ≤ 1 and −1 ≤ cos(θ) ≤ 1.
Pictorially, this tells us that the graphs of z = sin(θ) and z = cos(θ) lie
between the horizontal lines z = 1 and z = −1; i.e. the graphs lie inside
the darkened band pictured in Figure 18.8.
By Fact 18.2.3, we know that the values of the sine and cosine repeat
themselves every 2π radians. Consequently, if we know the graphs of the
sine and cosine on the domain 0 ≤ θ ≤ 2π, then the picture will repeat for
the interval 2π ≤ θ ≤ 4π, −2π ≤ θ ≤ 0, etc. CHAPTER 18. TRIGONOMETRIC FUNCTIONS 244 zaxis
z=1 θaxis
z = −1 Figure 18.8: Visualizing the range of sin(θ) and cos(θ). y=1
Repeat Repeat Repeat
0 −2π −4π zaxis
Repeat 2π θaxis 4π 6π y = −1
picture in here repeats each 2π units Figure 18.9: On what intervals will the graph repeat? Sketching the graph of z = sin (θ) for 0 ≤ θ ≤ 2π can be roughly
√
√
1
achieved by plotting points. For example, π , 2 , π , 22 , π , 23 , and
6
4
3
√ π
,1
2 √ π
π
π
lie on the graph, as do 32 , −1 , 53 , − 23 , 74 , − 22 , 11π , − 1 , and
6
2
(2π, 0), etc. If we return to our analysis of the range of values for the
sine function in Figure 18.2, it is easy to see where sin(θ) is positive or
negative; combined with Chapter 4, this tells us where the graph is above
and below the horizontal axis (see Figure 18.10). zaxis θaxis
−2π −π π 2π 3π Figure 18.10: Where is the graph positive or negative? We now include a software plot of the graph of sine function, observing
the three qualitative features just isolated: bounding, periodicity and
sign properties (see Figure 18.11). 18.3. GRAPHS OF CIRCULAR FUNCTIONS 245 zaxis
1 θaxis
−2π −π π 2π 3π −1 one period Figure 18.11: The graph of z = sin(θ). We could repeat this analysis to arrive at the graph
of the cosine. Instead, we will utilize an identity. Given
an angle θ, place it in central standard position in the
unit circle, as one of the four cases of Figure 17.16. For
example, we have pictured Case I in this ﬁgure. Since
the sum of the angles in a triangle is 180◦ = π radians,
we know that θ, π/2, and π − θ are the three angles of the
2
inscribed right triangle. From the picture in Figure 18.12,
it then follows that π
2 θ
unit
circle π
2 Figure 18.12: Visualizing
the conversion identity. side adjacent to θ
hypotenuse
side opposite to π − θ
2
=
hypotenuse
π
−θ .
= sin
2 cos(θ) = Using the same reasoning this identity is valid for all θ. This gives us
another useful identity:
Important Fact 18.3.2 (Conversion identity). For any angle θ, cos(θ) =
sin π − θ , and sin(θ) = cos π − θ .
2
2
This identity can be used to sketch the graph of the cosine function.
First, we do a calculation using our new identity:
cos(θ) = cos(−θ)
π
− (−θ)
= sin
2
π
= sin θ − −
2 −θ Fact 18.2.4 on page 241
Fact 18.3.2 (above)
Since: (a + b) = (a − (−b)) = (b − (−a)) By the horizontal shifting principle in Fact 13.3.1 on page 170, the graph
of z = cos(θ) is obtained by horizontally shifting the graph of z = sin(θ)
by π units to the left. Here is a plot of the graph of the cosine function:
2
See Figure 18.13. CHAPTER 18. TRIGONOMETRIC FUNCTIONS 246 zaxis
1 θaxis
−2π −π π 2π 3π −1 one period Figure 18.13: The graph of z = cos(θ). 18.3.3 The tangent graph
θ = −π
2 graph heads this
way, getting close
to vertical line as θ
gets close to π
2 −π
2 π
2 zaxis θaxis
graph heads this
way, getting close
to vertical line as θ
gets close to − π
2 θ= π
2 Figure 18.14: The behavior of tan(θ) as θ approaches
asymptotes. As we have already seen, unlike the sine and cosine circular functions, the tangent function is NOT deﬁned for all
sin
values of θ. Since tan(θ) = cos(θ) , here are some properties
(θ)
we can immediately deduce:
• The function z = tan(θ) is undeﬁned if and only if
θ = π + kπ, where k = 0, ±1, ±2, ±3, · · · .
2
• The function z = tan(θ) = 0 if and only if θ = kπ,
where k = 0, ±1, ±2, ±3, · · · .
• By Fact 18.2.3, the tangent function is πperiodic,
so the picture of the graph will repeat itself every πunits and it is enough to understand the graph when
−π < θ < π.
2
2
• On the domain 0 ≤ θ < π , tan(θ) ≥ 0; on the domain
2
− π < θ ≤ 0, tan(θ) ≤ 0.
2 In the θzcoordinate system, the vertical lines θ = π + kπ, where
2
k = 0, ±1, ±2, · · · will be vertical asymptotes for the graph of the tangent
function. Using our slope interpretation in Figure 18.4, what becomes
clear is this: As the values of θ get close to π , the graph is getting close
2
to the vertical line θ = π AND becoming farther and farther away from
2
the horizontal axis: To understand this numerically, ﬁrst suppose θ is
slightly smaller than π , say θ = π − 0.1 , π − 0.01 , and π − 0.001 . Then
2
2
2
2
the calculation of tan(θ) involves dividing a number very close to 1 by a
very small positive number:
π
− 0.1
2
π
− 0.01
tan
2
π
tan
− 0.001
2
tan = 9.9666,
= 99.9967, and
= 1000. 18.4. TRIGONOMETRIC FUNCTIONS
Conclude that as θ “approaches π from below”, the val2
ues of tan(θ) are becoming larger and larger. This says
that the function values become “unbounded”. Likewise,
imagine the case when θ is slightly bigger than − π , say
2
θ = − π + 0.1 , − π + 0.01 , and − π + 0.001 . Then the cal2
2
2
culation of tan(θ) involves dividing a number very close to
−1 by a very small positive number:
π
tan − + 0.1
2
π
tan − + 0.01
2
π
tan − + 0.001
2 247
θ = − π zaxis
2 one period
θ = 5π
2
etc. etc. θaxis
−2π −π π θ = − 3π
2 θ= π
2 2π 3π θ = 3π
2 Figure 18.15: The graph of
z = tan(θ). = −9.9666,
= −99.9967, and
= −1000. Conclude that as θ “approaches − π from above”, the values of tan(θ) are
2
becoming negative numbers of increasingly larger magnitude:
Again, this tells us the function values are becoming “unbounded”.
The graph of z = tan(θ) for − π < θ < π can be roughly achieved by com2
2
bining the calculations as in Example 18.2.7 and the qualitative features
highlighted. Figure 18.15 shows a software plot. 18.4 Trigonometric Functions
To become successful mathematical modelers, we must have wide variety
of functions in our toolkit. As an illustration, the graph below might
represent the height of the tide above some reference level over the course
of several days. The curve drawn is clearly illustrating that the height of
the tide is “periodic” as a function of time t; in other words, the behavior
of the tide repeats itself as time goes by. However, if we try to model this
periodic behavior, the only weapon at our disposal would be the circular
functions and these require an angle variable, not a time variable such
as t; we are stuck!
Modeling the tide graph requires the trigonometric functions, which lie
at the heart of studying all kinds of periodic behavior. We have no desire
to “reinvent the wheel”, so let’s use our previous work on the circular
functions to deﬁne the trigonometric functions. 18.4.1 A Transition
Given a real number t, is there a sensible way to deﬁne cos(t) and sin(t)?
The answer is yes and depends on the ideas surrounding radian measure
of angles. Given the positive real number t, we can certainly imagine an
angle of measure t radians inside the unit circle (in standard position)
and we know the arc subtended by this angle has length t (this is why
we use the unit circle). CHAPTER 18. TRIGONOMETRIC FUNCTIONS 248
feet +20
+15
+10
+5
0
t (time) −5
−10
−15
−20 Figure 18.16: A periodic function with input variable t. P (t) = (x,y) We already know that cos(t radians) and sin(t radians)
compute the x and y coordinates of the point P(t). In
effect, we are just using the measure of the angle t to
help us locate the point P(t). An alternate way to locate
P(t) is to move along the circumference counterclockwise,
beginning at (1, 0), until we have an arc of length t; that
again puts us at the point P(t). In the case of an angle
of measure −t radians, the point P(−t) can be located by
moving along the circumference clockwise, beginning at
(1, 0), until we have an arc of length t. arc length = t
t rads
(1,0) unit circle Figure 18.17: The circular
functions with input variable
t. Deﬁnition 18.4.1 (Trigonometric functions). Let t be a real number. We
DEFINE the sine function y = sin(t), the cosine function y = cos(t) and the
tangent function y = tan(t) by the rules
def sin(t) = ycoordinate of P(t) = sin(t radians)
def cos(t) = xcoordinate of P(t) = cos(t radians)
def sin(t)
tan(t) =
= tan(t radians)
cos(t) We refer to these as the basic trigonometric functions. If we are working with radian measure and t is a real number, then there is no difference between evaluating a trigonometric function at the real number t
and evaluating the corresponding circular function at the angle of measure t radians.
Example 18.4.2. Assume that the number of hours of daylight in Seattle
2π
during 1994 is given by the function d(t) = 3.7sin 366 (t − 80.5) + 12, where 18.4. TRIGONOMETRIC FUNCTIONS 249 t represents the day of the year and t = 0 corresponds to January 1. How
many hours of daylight will there be on May 11?
Solution. To solve the problem, you need to consult a calendar, ﬁnding
every month has 31 days, except: February has 28 days and April, June,
September and November have 30 days. May 11 is the 31+28+31+30+11 =
131st day of the year. So, there will be d(131) = 3.7 sin(2(50.5)π/366) + 12 =
14.82 hours of daylight on May 11. 18.4.2 Graphs of trigonometric functions
The graphs of the trigonometric functions y = sin(t), y = cos(t), and
y = tan(t) will look just like Figures 18.11, 18.13, and 18.15, except that
the horizontal axis becomes the taxis and the vertical axis becomes the
yaxis. 18.4.3 Notation for trigonometric functions
In many texts, you will ﬁnd the sine function written as y = sin t; i.e. the
parenthesis around the “t” are omitted. A similar comment applies to all
of the trigonometric functions. We will never do this and the reasoning
is simply this: Maintaining the parenthesis, as in y = sin(t), emphasizes
the fact that we are dealing with a function and the “input values” are
located between the parenthesis. For example, if we write the function
y = sin(t2 + 2t + 1), it is crystal clear that the sine function is applied to the
expression “t2 + 2t + 1”; using the alternate notation yields the expression
y = sin t2 + 2t + 1, which is interpreted to mean y = (sin t2 ) + (2t + 1).
!!! As a rule, whenever you see an expression involving sin(· · · ), cos(· · · ),
or tan(· · · ), we assume “ · · · ” is in units of RADIANS, unless otherwise
noted. When computing values on your calculator, MAKE SURE YOU
ARE USING RADIAN MODE! CAUTION
!!! CHAPTER 18. TRIGONOMETRIC FUNCTIONS 250 18.5 Exercises
Problem 18.1. Work the following problems
without using ANY calculators.
(a) Sketch y = sin(x). (b) If sin(θ) = −0.8 and θ is in the third
quadrant of the xy plane, what is cos(θ)?
(c) If sin(θ) = 3 , what is sin( π − θ)?
7
2 (b) Sketch y = sin2 (x).
(c) Sketch y = 1
.
1 + sin2 (x) Problem 18.4. These graphs represent periodic functions. Describe the period in each
case. Problem 18.2. Sketch the graphs of these
functions:
(a) f(t) =  sin(t).
(b) f(t) =  cos(t).
(c) f(t) =  tan(t).
Problem 18.3. Solve the following:
24
(a) If cos(θ) = 25 , what are the two possible
values of sin(θ)? Problem 18.5. Start with the equation
sin(θ) = cos(θ). Use the unit circle interpretation of the circular functions to ﬁnd the solutions of this equation; make sure to describe
your reasoning. Chapter 19
Sinusoidal Functions depth 5
A migrating salmon is heading up a portion of the
4
Columbia River. It’s depth d(t) (in feet) below the water
3
surface is measured and plotted over a 30 minute period,
2
as a function of time t (minutes). What is the formula for
1
d(t)?
5
10 15 20
25 30
time
In order to answer the question, we need to introduce
Figure 19.1: The depth of a
an important new family of functions called the sinusoidal
salmon as a function of time.
functions. These functions will play a central role in modeling any kind of periodic phenomena. The amazing fact
is that almost any function you will encounter can be approximated by a
sum of sinusoidal functions; a result that has farreaching implications
in all of our lives. 19.1 A special class of functions
Beginning with the trigonometric function y = sin(x), what is the most
general function we can build using the graphical techniques of shifting
and stretching?
yaxis
1 y = sin(x) horizontally shift vertically
xaxis
horizontally dilate −π π
2 −π
2 π vertically dilate shift −1 Figure 19.2: Visualizing the geometric operations available for curve sketching. The graph of y = sin(x) can be manipulated in four basic ways: horizontally shift, vertically shift, horizontally dilate or vertically dilate. Each
of these “geometric operations” corresponds to a simple change in the
“symbolic formula” for the function, as discussed in Chapter 13.
251 CHAPTER 19. SINUSOIDAL FUNCTIONS 252 If we vertically shift the graph by D units upward, the resulting curve
would be the graph of the function y = sin(x)+D; see Facts 13.3.1. Recall,
the effect of the sign of D: If D is negative, the effect of shifting D units
upward is the same as shifting D units downward. Notice, the function
y = sin(x) + D is still a periodic function, having the same period 2π as
y = sin(x). Notice, whereas the graph of the function y = sin(x) oscillates
between the horizontal lines y = ±1, the graph of y = sin(x) + D oscillates
between y = D ± 1. For this reason, we sometimes refer to the constant
D as the mean of the function y = sin(x) + D. In Figure 19.3, notice that
the graph of y = sin(x) + D is symmetrically split by the horizontal “mean”
line y = D.
yaxis y = sin(x) + D yaxis
y = sin(x)
y=D
D
xaxis xaxis
shift D units Figure 19.3: Interpreting the mean. Next, consider the effect of horizontally shifting the graph of y = sin(x)
by C units to the right. By Facts 13.3.1, the new curve is the graph of
the function y = sin(x − C). Also, recall the effect of the sign of C: If
C is negative, the effect of shifting C units right is the same as shifting
C units left. If the domain of sin(x) is 0 ≤ x ≤ 2π, then the domain of
sin(x − C) is 0 ≤ x − C ≤ 2π, again by Facts 13.3.1. Rewriting this, the
domain of sin(x − C) is C ≤ x ≤ 2π + C and the graph will go through
precisely one period on this domain. In other words, the new function
sin(x − C) is still 2πperiodic. The constant C is usually called the phase
shift of y = sin(x − C). Looking at Figure 19.4, it is possible to interpret C
graphically: C will be a point where the graph crosses the horizontal axis
on its way up from a minimum to a maximum.
yaxis yaxis C
y = sin(x − C) y = sin(x) xaxis xaxis
shift C units Figure 19.4: Interpreting the phase shift. Vertically dilating the graph, either by vertical expansion or compression, leads to a new curve. The graph of this vertically dilated curve
is y = A sin(x), for some positive constant A. Furthermore, if A > 1, 19.1. A SPECIAL CLASS OF FUNCTIONS 253 the graph of y = A sin(x) is a vertically expanded version of y = sin(x),
whereas, if 0 < A < 1, then the graph of y = A sin(x) is a vertically compressed version of y = sin(x). Notice, the function y = A sin(x) is still 2πperiodic. What has changed is the band of oscillation: whereas the graph
of the function y = sin(x) stays between the horizontal lines y = ±1, the
graph of y = A sin(x) oscillates between the horizontal lines y = ±A. We
usually refer to A as the amplitude of the function y = A sin(x).
y = A sin(x) yaxis yaxis
y = sin(x) A
xaxis
xaxis
stretch A units Figure 19.5: Interpreting the amplitude. Finally, horizontally dilating the graph, either by horizontal expansion
or compression, leads to a new curve. The equation of this horizontally
dilated curve is y = sin(cx), for some constant c > 0. We know that y =
sin(x) is a 2πperiodic function and observe that horizontally dilation still
results in a periodic function, but the period will typically NOT be 2π. For
future purposes, it is useful to rewrite the equation for the horizontally
stretched curve in a way more directly highlighting the period. To begin
with, once the horizontal stretching factor c is known, we could rewrite
c= 2π
,
B for some B = 0. yaxis yaxis
y = sin(x) y = sin(( 2π )x)
B xaxis
xaxis stretch Figure 19.6: Interpreting the period. Here is the point of this yoga with the horizontal dilating constant: If
we let the values of x range over the interval [0, B], then 2π x will range over
B
π
the interval [0,2π]. In other words, the function y = sin 2B x is Bperiodic
2π
and we can read off the period of y = sin B x by viewing the constant
in this mysterious way. The four constructions outlined lead to a new
family of functions. CHAPTER 19. SINUSOIDAL FUNCTIONS 254 Deﬁnition 19.1.1 (The Sinusoidal Function). Let A, B, C and D be ﬁxed
constants, where A and B are both positive. Then we can form the new
function
y = A sin 2π
(x − C) + D,
B which is called a sinusoidal function. The four constants can be interpreted
graphically as indicated: B
yaxis yaxis A
D
xaxis xaxis all four
y = sin(x) C operations
y = A sin(( 2π )(x − c)) + D
B Figure 19.7: Putting it all together for the sinusoidal function. 19.1.1 How to roughly sketch a sinusoidal graph
Important Procedure 19.1.2. Given a sinusoidal function in the standard
form
y = A sin 2π
(x − C) + D,
B once the constants A, B, C, and D are speciﬁed, any graphing device can
produce an accurate graph. However, it is pretty straightforward to sketch
a rough graph by hand and the process will help reinforce the graphical
meaning of the constants A, B, C, and D. Here is a “ﬁve step procedure”
one can follow, assuming we are given A, B, C, and D. It is a good idea to
follow Example 19.1.3 as you read this procedure; that way it will seem a
lot less abstract.
1. Draw the horizontal line given by the equation y = D; this line will
π
“split” the graph of y = A sin 2B (x − C) + D into symmetrical upper
and lower halves.
2. Draw the two horizontal lines given by the equations y = D ± A. These
two lines determine a horizontal strip inside which the graph of the 19.1. A SPECIAL CLASS OF FUNCTIONS 255 sinusoidal function will oscillate. Notice, the points where the sinusoidal function has a maximum value lie on the line y = D + A. Likewise, the points where the sinusoidal function has a minimum value
lie on the line y = D − A. Of course, we do not yet have a prescription
that tells us where these maxima (peaks) and minima (valleys) are
located; that will come out of the next steps.
3. Since we are given the period B, we know these important facts: (1)
The period B is the horizontal distance between two successive maxima (peaks) in the graph. Likewise, the period B is the horizontal
distance between two successive minima (valleys) in the graph. (2)
The horizontal distance between a maxima (peak) and the successive
minima (valley) is 1 B.
2
4. Plot the point (C, D). This will be a place where the graph of the
sinusoidal function will cross the mean line y = D on its way up
from a minima to a maxima. This is not the only place where the
graph crosses the mean line; it will also cross at the points obtained
1
from (C, D) by horizontally shifting by any integer multiple of 2 B.
For example, here are three places the graph crosses the mean line:
1
(C, D), (C + 2 B, D), (C + B, D)
1
5. Finally, midway between (C, D) and (C+ 2 B, D) there will be a maxima
1
(peak); i.e. at the point (C + 4 B, D + A). Likewise, midway between
(C + 1 B, D) and (C + B, D) there will be a minima (valley); i.e. at the
2
3
point (C + 4 B,D − A). It is now possible to roughly sketch the graph on
the domain C ≤ x ≤ C + B by connecting the points described. Once
this portion of the graph is known, the fact that the function is periodic
tells us to simply repeat the picture in the intervals C + B ≤ x ≤ C + 2B,
C − B ≤ x ≤ C, etc. To make sense of this procedure, let’s do an explicit example to see
how these ﬁve steps produce a rough sketch. CHAPTER 19. SINUSOIDAL FUNCTIONS 256 eplacements
Example 19.1.3. The temperature (in ◦ C) of AdriN’s dorm room varies
π
during the day according to the sinusoidal function d(t) = 6 sin 12 (t − 11) +
19, where t represents hours after midnight. Roughly sketch the graph of
d(t) over a 24 hour period.. What is the temperature of the room at 2:00
pm? What is the maximum and minimum temperature of the room?
Solution. We begin with the rough sketch. Start by taking an inventory
of the constants in this sinusoidal function:
d(t) = 6 sin π
2π
(t − 11) + 19 = A sin
(t − C) + D.
12
B Conclude that A = 6, B = 24, C = 11, D = 19. Following the ﬁrst four
steps of the procedure outlined, we can sketch the lines y = D = 19,
y = D ± A = 19 ± 6 and three points where the graph crosses the mean
line (see Figure19.8). 30 d(t) graph
will oscillate
inside this
strip 25 y = 25 20
y = 19
(11, 19) (23, 19) (35, 19) 15 y = 13 10 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.8: Sketching the mean D and amplitude A. According to the ﬁfth step in the sketching procedure, we can plot the
3
maxima (C + 1 B, D + A) = (17, 25) and the minima (C + 4 B, D − A) = (29, 13).
4
We then “connect the dots” to get a rough sketch on the domain 11 ≤ t ≤
35.
30 d(t) graph
will oscillate
inside this maxima
(17, 25) 25 y = 25
(23, 19) 20 y = 19
(11, 19) strip (35, 19) 15 y = 13
(29, 13)
minima 10 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.9: Visualizing the maximum and minimum over one period. 19.1. A SPECIAL CLASS OF FUNCTIONS 257 Finally, we can use the fact the function has period 24 to sketch the
graph to the right and left by simply repeating the picture every 24 horizontal units.
yaxis
30 maxima
(−7,25) maxima
(17,25) maxima
(41,25) y = 25 25
(11,19) 20
(−13,19) (23,19) (35,19)
y = 19 (−1,19)
15 (47,19) (59,19)
y = 13 (5,13)
minima 10 (29,13)
minima (53,13)
minima
taxis −12 −8
−4
−14 −10 −6 −2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.10: Repeat sketch for every full period. We restrict the picture to the domain 0 ≤ t ≤ 24 and obtain the computer generated graph pictured in Figure 19.11; as you can see, our
rough graph is very accurate. The temperature at 2:00 p.m. is just
d(14) = 23.24◦ C. From the graph, the maximum value of the function will
be D + A = 25◦ C and the minimum value will be D − A = 13◦ C. temp (C) 25
20
15
10
5 10 15 20 t (hours) Figure 19.11: The computer generated solution. 19.1.2 Functions not in standard sinusoidal form
Any time we are given a trigonometric function written in the standard
form
y = A sin 2π
(x − C) + D,
B for constants A, B, C, and D (with A and B positive), the summary in
Deﬁnition 19.1.1 tells us everything we could possibly want to know CHAPTER 19. SINUSOIDAL FUNCTIONS 258 about the graph. But, there are two ways in which we might encounter a
trigonometric type function that is not in this standard form:
• The constants A or B might be negative. For example, y = −2 sin(2x−
7) − 3 and y = 3 sin − 1 x + 1 + 4 are examples that fail to be in
2
standard form.
• We might use the cosine function in place of the sine function. For
example, something like y = 2 cos(3x + 1) − 2 fails to be in standard
sinusoidal form.
Now what do we do? Does this mean we need to repeat the analysis that
led to Deﬁnition 19.1.1? It turns out that if we use our trig identities
just right, then we can move any such equation into standard form and
read off the amplitude, period, phase shift and mean. In other words,
equations that fail to be in standard sinusoidal form for either of these
two reasons will still deﬁne sinusoidal functions. We illustrate how this
is done by way of some examples:
Examples 19.1.4.
(i) Start with y = −2 sin(2x − 7) − 3, then here are the steps with reference
to the required identities to put the equation in standard form:
y = −2 sin(2x − 7) − 3
= 2 (− sin(2x − 7)) − 3
= 2 sin(2x − 7 + π) + (−3)
2π
7−π
= 2 sin
x−
π
2 Fact 18.2.5 on page 241
+ (−3). This function is now in the standard form of Deﬁnition 19.1.1, so it is
a sinusoidal function with phase shift C = 7−π = 1.93, mean D = −3,
2
amplitude A = 2 and period B = π.
1
(ii) Start with y = 3 sin(− 2 x + 1) + 4, then here are the steps with reference
to the required identities to put the equation in standard form: 1
y = 3 sin − x + 1 + 4
2
1
x−1
+4
= 3 sin −
2
1
= 3 − sin
x−1
+4
Fact 18.2.4 on page 241
2
1
x−1+π +4
Fact 18.2.5 on page 241
= 3 sin
2
2π
(x − [2 − 2π]) + 4
= 3 sin
4π 19.2. EXAMPLES OF SINUSOIDAL BEHAVIOR 259 This function is now in the standard form of Deﬁnition 19.1.1, so it
is a sinusoidal function with phase shift C = 2 − 2π, mean D = 4,
amplitude A = 3 and period B = 4π.
(iii) Start with y = 2 cos(3x + 1) − 2, then here are the steps to put the equation in standard form. A key simplifying step is to use the identity:
cos(t) = sin( π + t).
2
y = 2 cos(3x + 1) − 2
π
+ 3x + 1 − 2
= 2 sin
2
π
= 2 sin 3x − −1 −
+ (−2)
2
2π
1
π
= 2 sin
x−
−1 −
2π
3
2
3 + (−2) This function is now in the standard form of Deﬁnition 19.1.1, so it is
a sinusoidal function with phase shift C = 1 [−1 − π ], mean D = −2,
3
2
amplitude A = 2 and period B = 2π .
3 19.2 Examples of sinusoidal behavior
Problems involving sinusoidal behavior come in two basic ﬂavors. On the
one hand, we could be handed an explicit sinusoidal function
y = A sin 2π
(x − C) + D
B and asked various questions. The answers typically require either direct
calculation or interpretation of the constants. Example 19.1.3 is typical
of this kind of problem. On the other hand, we might be told a particular
situation is described by a sinusoidal function and provided some data
or a graph. In order to further analyze the problem, we need a “formula”,
which means ﬁnding the constants A, B, C, and D. This is a typical
scenario in a “mathematical modeling problem”: the process of observing
data, THEN obtaining a mathematical formula. To ﬁnd A, take half the
difference between the largest and smallest values of f(x). The period B
is most easily found by measuring the distance between two successive
maxima (peaks) or minima (valleys) in the graph. The mean D is the
average of the largest and smallest values of f(x). The phase shift C
(which is usually the most tricky quantity to get your hands on) is found
by locating a “reference point”. This “reference point” is a location where
the graph crosses the mean line y = D on its way up from a minimum
to a maximum. The funny thing is that the phase shift C is NOT unique;
there are an inﬁnite number of correct choices. One choice that will work CHAPTER 19. SINUSOIDAL FUNCTIONS 260 is C = (xcoordinate of a maximum) − B . Any other choice of C will differ
4
from this one by a multiple of the period B.
max value − min value
2
B = distance between two successive peaks (or valleys)
B
C = xcoordinate of a maximum −
4
max value + min value
D=
.
2
A= daxis (hours of daylight) Example 19.2.1. Assume that the number of hours of daylight in Seattle
is given by a sinusoidal function d(t) of time. During 1994, assume the
longest day of the year is June 21 with 15.7 hours of daylight and the
shortest day is December 21 with 8.3 hours of daylight. Find a formula
d(t) for the number of hours of daylight on the tth day of the year. 15
12.5
10
7.5
5
50 100 150 200 250 300 350
taxis (days) Solution. Because the function d(t) is assumed to be siπ
nusoidal, it has the form y = A sin 2B (t − C) + D, for constants A, B, C, and D. We simply need to use the given
information to ﬁnd these constants. The largest value of
the function is 15.7 and the smallest value is 8.3. Knowing this, from the above discussion we can read off : Figure 19.12: Hours of daylight in Seattle in 1994. D= 15.7 + 8.3
= 12
2 A= 15.7 − 8.3
= 3.7.
2 To ﬁnd the period, we need to compute the time between two successive
maximum values of d(t). To ﬁnd this, we can simply double the time
length of onehalf period, which would be the length of time between
successive maximum and minimum values of d(t). This gives us the
equation
B = 2(days between June 21 and December 21) = 2(183) = 366.
Locating the ﬁnal constant C requires the most thought. Recall, the
longest day of the year is June 21, which is day 172 of the year, so
C = (day with max daylight) − B
366
= 172 −
= 80.5.
4
4 In summary, this shows that
d(t) = 3.7 sin 2π
(t − 80.5) + 12.
366 A rough sketch, following the procedure outlined above, gives this graph
on the domain 0 ≤ t ≤ 366; we have included the mean line y = 12 for
reference. 19.2. EXAMPLES OF SINUSOIDAL BEHAVIOR 261 We close with the example that started this section.
Example 19.2.2. The depth of a migrating salmon below the water surface changes according to a sinusoidal function of time. The ﬁsh varies
between 1 and 5 feet below the surface of the water. It takes the ﬁsh
1.571 minutes to move from its minimum depth to its successive maximum
depth. It is located at a maximum depth when t = 4.285 minutes. What
is the formula for the function d(t) that predicts the depth of the ﬁsh after
t minutes? What was the depth of the salmon when it was ﬁrst spotted?
During the ﬁrst 10 minutes, how many times will the salmon be exactly 4
feet below the surface of the water? max depth − min depth
5−1
A=
=
=2
2
2
5+1
max depth + min depth
=
= 3.
D=
2
2 5
depth Solution. We know that d(t) = A sin( 2π (x − C)) + D, for apB
propriate constants A, B, C, and D. We need to use the
given information to extract these four constants. The
amplitude and mean are easily found using the above formulas: 4
3
2
1
2 4 B = 2(1.571) = 3.142
Finally, to ﬁnd C we
B
3.142
= 4.285 −
= 3.50.
4
4 The formula is now
d(t) = 2 sin 8 10 Figure 19.13: Depth of a migrating salmon. The period can be found by noting that the information about the time
between a successive minimum and maximum depth will be half of a
period (look at the picture in Figure 19.13): C = (time of maximum depth) − 6
time 2π
(t − 3.5) + 3 = 2 sin(2t − 7) + 3
3.142 The depth of the salmon when it was ﬁrst spotted is just
d(0) = 2 sin(−7) + 3 = 1.686 feet.
Finally, graphically, the last question amounts to determining how many
times the graph of d(t) crosses the line y = 4 on the domain [0,10].
This can be done using Figure 19.13. A simultaneous picture of the
two graphs is given, from which we can see the salmon is exactly 4 feet
below the surface of the water six times during the ﬁrst 10 minutes. CHAPTER 19. SINUSOIDAL FUNCTIONS 262 19.3 Summary
• A sinusoidal function is one of the form
f(t) = A sin 2π
(t − C) + D
B where A, B, C, and D are constants.
– A is the amplitude of the function; this is half the vertical distance between a high point and a low point on its graph.
– B is the period of the function; this is the horizontal distance between two consecutive high points (or low points) on its graph.
– C is the phase shift of the function; it is multivalued, but one
choice for C is a value of t at which the function is increasing
and equal to D.
– D is the mean value of the function; it is the yvalue of the horizontal line about which the graph of the function is balanced.
• The graph of a sinusoidal function is a shifted, scaled version of the
graph of y = sin t. 19.4. EXERCISES 263 19.4 Exercises
Problem 19.1. Find the amplitude, period, a
phase shift and the mean of the following sinusoidal functions.
(a) y = sin(2x − π) + 1
(b) y = 6 sin(πx) − 1
(c) y = 3 sin(x + 2.7) + 5.2
(d) y = 5.6 sin
(e) y = 2.1 sin 2
3x
x
π − 7 − 12.1 + 44.3 − 9.8 (f) y = 3.9 (sin(22.34(x + 18)) − 11)
(g) y = 11.2 sin 5
π (x − 9.2) + 8.3 Problem 19.2. A weight is attached to a spring
suspended from a beam. At time t = 0, it
is pulled down to a point 10 cm above the
ground and released. After that, it bounces up
and down between its minimum height of 10
cm and a maximum height of 26 cm, and its
height h(t) is a sinusoidal function of time t.
It ﬁrst reaches a maximum height 0.6 seconds
after starting.
(a) Follow the procedure outlined in this
section to sketch a rough graph of h(t).
Draw at least two complete cycles of the
oscillation, indicating where the maxima
and minima occur.
(b) What are the mean, amplitude, phase
shift and period for this function?
(c) Give four different possible values for
the phase shift. machine begins recording a plot of volume per
breath versus time (in seconds). Let b(t) be a
function of time t that tells us the volume (in
liters) of a breath that starts at time t. During the test, the smallest volume per breath
is 0.6 liters and this ﬁrst occurs for a breath
that starts 5 seconds into the test. The largest
volume per breath is 1.8 liters and this ﬁrst
occurs for a breath beginning 55 seconds into
the test.
(a) Find a formula for the function b(t)
whose graph will model the test data for
this patient.
(b) If the patient begins a breath every 5
seconds, what are the breath volumes
during the ﬁrst minute of the test?
Problem 19.4. Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. at which
time the water is 10 feet above the height of
low tide. Low tides occur 6 hours after high
tides. Suppose there are two high tides and
two low tides every day and the height of the
tide varies sinusoidally.
(a) Find a formula for the function y = h(t)
that computes the height of the tide
above low tide at time t. (In other words,
y = 0 corresponds to low tide.)
(b) What is the tide height at 11:00 a.m.?
Problem 19.5. Your seat on a Ferris Wheel is
at the indicated position at time t = 0. (d) Write down a formula for the function
h(t) in standard sinusoidal form; i.e. as
in 19.1.1 on Page 254.
(e) What is the height of the weight after
0.18 seconds?
(f) During the ﬁrst 10 seconds, how many
times will the weight be exactly 22 cm
above the ﬂoor? (Note: This problem
does not require inverse trigonometry.)
Problem 19.3. A respiratory ailment called
“CheyneStokes Respiration” causes the volume per breath to increase and decrease in a
sinusoidal manner, as a function of time. For
one particular patient with this condition, a Start 53 feet Let t be the number of seconds elapsed after
the wheel begins rotating counterclockwise.
You ﬁnd it takes 3 seconds to reach the top,
which is 53 feet above the ground. The wheel
is rotating 12 RPM and the diameter of the
wheel is 50 feet. Let d(t) be your height above
the ground at time t. 264 CHAPTER 19. SINUSOIDAL FUNCTIONS
ω=4π/9 rad/sec (a) Argue that d(t) is a sinusoidal function,
describing the amplitude, phase shift,
period and mean. bug lands here 1.2 rad
2 in (b) When are the ﬁrst and second times you
are exactly 28 feet above the ground? bug lands here (c) After 29 seconds, how many times will
you have been exactly 28 feet above the
ground? 2 in ω=4π/9rad/sec ω= 4π/9 rad/sec
bug lands
here
0.5 rad 2 in Problem 19.6. In Exercise 17.12, we studied
the situation below: A bug has landed on the
rim of a jelly jar and is moving around the rim.
The location where the bug initially lands is
described and its angular speed is given. Impose a coordinate system with the origin at the
center of the circle of motion. In each of the
cases, the earlier exercise found the coordinates P(t) of the bug at time t. For each of the
scenarios below, answer these two questions: Problem 19.7. The voltage output(in volts) of
an electrical circuit at time t seconds is given
by the function
V (t) = 23 sin(5πt−3π)+1 .
(a) What is the initial voltage output of the
circuit? (a) Both coordinates of P(t) = (x(t),y(t)) are
sinusoidal functions in the variable t.
Sketch a rough graph of the functions
x(t) and y(t) on the domain 0 ≤ t ≤ 9. (b) Use the graph sketches to help you ﬁnd
the the amplitude, mean, period and
phase shift for each function. Write x(t)
and y(t) in standard sinusoidal form. (b) Is the voltage output of the circuit ever
equal to zero? Explain.
(c) The function V (t) = 2p(t) , where p(t) =
3 sin(5πt − 3π) + 1. Put the sinusoidal
function p(t) in standard form and
sketch the graph for 0 ≤ t ≤ 1. Label the
coordinates of the extrema on the graph.
(d) Calculate the maximum and minimum
voltage output of the circuit.
(e) During the ﬁrst second, determine when
the voltage output of the circuit is 10
volts. 19.4. EXERCISES 265 (f) A picture of the graph of y = V (t) on the
domain 0 ≤ t ≤ 1 is given; label the coordinates of the extrema on the graph. free to move back and forth along the xaxis.
The point A is at (2,0) at time t = 0, and the
wheel rotates counterclockwise at 3 rev/sec.
2
A volts A
2 15 2 B B
4 6 8 2 4 6 8 2 12.5 time t > 0
time t=0 10 (a) As the point A makes one complete revolution, indicate in the picture the direction and range of motion of the point B. 7.5
5
2.5
0.2 0.4 0.6 0.8 1 t axis (g) Restrict the function V (t) to the domain
0.1 ≤ t ≤ 0.3; explain why this function
has an inverse and ﬁnd the formula for
the inverse rule. Restrict the function
V (t) to the domain 0.3 ≤ t ≤ 0.5; explain why this function has an inverse
and ﬁnd the formula for the inverse rule.
Problem 19.8. A six foot long rod is attached
at one end A to a point on a wheel of radius 2
feet, centered at the origin. The other end B is (b) Find the coordinates of the point A as a
function of time t.
(c) Find the coordinates of the point B as a
function of time t.
(d) What is the xcoordinate of the point B
when t = 1? You should be able to ﬁnd
this two ways: with your function from
part (c), and using some common sense
(where is point A after one second?).
(e) Is the function you found in (c) a sinusoidal function? Explain. 266 CHAPTER 19. SINUSOIDAL FUNCTIONS Chapter 20 An aircraft is ﬂying at an altitude 10 miles above the elevation of an airport. If the airplane begins a steady descent 100 miles from the airport, what is the angle θ of
descent?
The only natural circular function we can use is z =
tan(θ), leading to the equation:
tan(θ) = 10 miles Inverse Circular Functions θ
100 miles Figure 20.1: An aircraft decending toward an airport. 10
1
=.
100
10 The problem is that this equation does not tell us the value of θ. Moreover, none of the equation solving techniques at our disposal (which all
amount to algebraic manipulations) will help us solve the equation for θ.
What we need is an inverse function θ = f−1 (z); then we could use the
fact that tan−1 (tan(θ)) = θ and obtain:
θ = tan−1 (tan(θ)) = tan−1 1
10 . Computationally, without even thinking about what is going on, any scientiﬁc calculator will allow us to compute values of an inverse circular
function and leads to a solution of our problem. In this example, you will
1
ﬁnd θ = tan−1 10 = 5.71◦. Punch this into your calculator and verify it! 20.1 Solving Three Equations
Example 20.1.1. Find all values of θ (an angle) that make this equation
1
true: sin(θ) = 2 .
Solution. We begin with a graphical reinterpretation: the solutions correspond to the places where the graphs of z = sin(θ) and z = 1 intersect in
2
the θzcoordinate system. Recalling Figure 18.11, we can picture these
two graphs simultaneously as below:
267 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 268
cross cross cross zaxis cross 1 Graph of z =
A B 1
2 θaxis
−2π −π π 2π 3π −1 Graph of z = sin(θ) one period Figure 20.2: Where does sin(θ) cross z = 1
2 ? The ﬁrst thing to notice is that these two graphs will cross an inﬁnite number of times, so there are inﬁnitely many solutions to Example 20.1.1! However, notice there is a predictable spacing of the crossing
points, which is just a manifestation of the periodicity of the sine function. In fact, if we can ﬁnd the two crossing points labeled “A” and “B”,
then all other crossing points are obtained by adding multiples of 2π to
either “A” or “B”. By Table 17.1, θ = π radians is a special angle where we
6
computed sin π = 1 , which tells us that the crossing point labeled “A”
6
2
is the point ( π , 1 ). Using the identities in Facts 18.2.4 and 18.2.5, notice
62
that
sin 5π
6 π
+π
6
π
= − sin −
6
π
= − − sin
6
π
= sin
6
1
=.
2
= sin − So, θ = 5π is the only other angle θ between 0 and 2π such that Ex6
ample 20.1.1 holds. This corresponds to the crossing point labeled “B”,
π
which has coordinates 56 , 1 . In view of the remarks above, the crossing
2
points come in two ﬂavors:
π
1
, k = 0, ±1, ±2, ±3, · · · , and
+ 2kπ,
6
2
5π
1
, k = 0, ±1, ±2, ±3, · · · .
+ 2kπ,
6
2 Taking this example as a model, we can tackle the more general problem: For a ﬁxed real number c, describe the solution(s) of the equation 20.1. SOLVING THREE EQUATIONS 269 c = f(θ) for each of the circular functions z = f(θ). Studying solutions
of these equations will force us to come to grips with three important
issues:
• For what values of c does f(θ) = c have a solution?
• For a given value of c, how many solutions does f(θ) = c have?
• Can we restrict the domain so that the resulting function is 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 20.1.2. None of the circular functions is onetoone on
the domain of all θ values. The equations c = sin(θ) and c = cos(θ) have
a solution if and only if −1 ≤ c ≤ 1; if c is in this range, there are inﬁnitely
many solutions. The equation c = tan(θ) has a solution for any value of c
and there are inﬁnitely many solutions. Example 20.1.3. If two sides of a right triangle have
√
lengths 1 and 3 as pictured below, what are the acute
angles α and β?
β Solution. By the Pythagorean Theorem the remaining side
has length
1+ √ 2 3 √
3 α = 2. 1
√
Since tan(α) = 3, we need to solve this equation for α.
Figure 20.3: What are the
√
Graphically, we need to determine where z = 3 crosses
values for α and β?
the graph of the tangent function:
From Fact 20.1.2, there will be inﬁnitely many solutions to our equation, but notice that there is exactly one solution in the interval − π , π
22
and we can ﬁnd it using Table 17.1: ◦ π
sin(60 )
tan
= tan(60◦ ) =
=
3
cos(60◦) √ 3
2
1
2 = √ 3. So, α = π radians = 60◦ is the only acute angle solution and β = 180◦ −
3
60◦ − 90◦ = 30◦ . CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 270 zaxis θ = −π
2 one period
θ= 5π
2 only place graphs
cross √ this period
in
is ( π , 3)
3
etc. etc. θaxis
−2π θ = − 3π
2 −π π θ= π
2 2π θ= Figure 20.4: Where does the line z = 3π 3π
2 √
3 cross z = tan θ? 20.2 Inverse Circular Functions
Except for specially chosen angles, we have not addressed the serious
problem of FINDING values of the inverse rules attached to the circular
function equations. (Our previous examples were “rigged”, so that we
could use Table 17.1.) To proceed computationally, we need to obtain
the inverse circular functions. If we were to proceed in a sloppy manner,
then a ﬁrst attempt at deﬁning the inverse circular functions would be to
write
$ sin−1 (z) = solutions θ of the equation z = sin(θ).
$ cos−1 (z) = solutions θ of the equation z = cos(θ). (20.1) $ tan−1 (z) = solutions θ of the equation z = tan(θ).
There are two main problems with these rules as they stand. First, to
have a solution θ in the case of sin−1 (z) and cos−1 (z), we need to restrict z
so that −1 ≤ z ≤ 1. Secondly, having made this restriction on z in the ﬁrst
two cases, there is no unique solution; rather, there are an inﬁnite number of solutions. This means that the rules sin−1 , cos−1 , and tan−1 as they
now stand do not deﬁne functions. Given what we have reviewed about
inverse functions, the only way to proceed is to restrict each circular
function to a domain of θ values on which it becomes onetoone, then
we can appeal to Fact 9.3.1 and conclude the inverse function makes
sense. 20.2. INVERSE CIRCULAR FUNCTIONS 271 At this stage a lot of choice (ﬂexibility) enters into determining the
domain on which we should try to invert each circular function. In effect,
there are an inﬁnite number of possible choices. If z = f(θ) denotes one
of the three circular functions, there are three natural criteria we use
to guide the choice of a restricted domain, which we will call a principal
domain :
• The domain of f(θ) should include the angles between 0 and π , since
2
these are the possible acute angles in a right triangle.
• On the restricted domain, the function f(θ) should take on all possible values in the range of f(θ). In addition, the function should be
onetoone on this restricted domain.
• The function f(θ) should be “continuous” on this restricted domain;
i.e. the graph on this domain could be traced with a pencil, without
lifting it off the paper.
In the case of z = sin(θ), the principal domain − π ≤ θ ≤ π satisﬁes our
2
2
criteria and the picture is given below. Notice, we would not want to take
the interval 0 ≤ θ ≤ π, since z = sin(θ) doesn’t achieve negative values on
this domain; in addition, it’s not onetoone there.
zaxis
1 yaxis principal domain
on the unit circle θ
−π π
2 −π
2 −1 θaxis
π xaxis sine restricted to
principal domain Figure 20.5: Principal domain for sin(θ). In the case of z = cos(θ), the principal domain 0 ≤ θ ≤ π satisﬁes our
criteria and the picture is given below. Notice, we would not want to take
the interval − π ≤ θ ≤ π , since z = cos(θ) doesn’t achieve negative values
2
2
on this domain; in addition, it’s not onetoone there.
In the case of z = tan(θ), the principal domain − π < θ < π satisﬁes
2
2
our criteria and the picture is given below. Notice, we would not want to
take the interval 0 ≤ θ ≤ π, since z = tan(θ) does not have a continuous
graph on this interval; in other words, we do not include the endpoints
since tan(θ) is undeﬁned there.
Important Facts 20.2.1 (Inverse circular functions). Restricting each circular function to its principal domain, its inverse rule f−1 (z) = θ will deﬁne
a function. CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 272
zaxis
1 yaxis principal domain
on the unit circle θ
−π π
2 −π
2 −1 π 3π
2 θaxis xaxis cosine restricted to
principal domain Figure 20.6: Principal domain for cos(θ). (i) If −1 ≤ z ≤ 1, then sin−1 (z) is the unique angle θ in the principal
domain − π ≤ θ ≤ π with the property that sin(θ) = z.
2
2
(ii) If −1 ≤ z ≤ 1, then cos−1 (z) is the unique angle θ in the principal
domain 0 ≤ θ ≤ π with the property that cos(θ) = z.
(iii) For any real number z, tan−1 (z) is the unique angle θ in the principal
domain − π < θ < π with the property that tan(θ) = z.
2
2
We refer to the functions deﬁned above as the inverse circular functions. These are sometimes referred to as the “arcsine”, “arccosine” and
“arctangent” functions, though we will not use that terminology. The inverse circular functions give us one solution for each of these equations:
c = cos(θ)
c = sin(θ)
c = tan(θ);
these are called the principal solutions. We also can refer to these as the
principal values of the inverse circular function rules θ = f−1 (z). !!!
CAUTION
!!! As usual, be careful with “radian mode” and “degree mode” when
making calculations. For example, if your calculator is in “degree”
mode and you type in “tan−1 (18)”, the answer given is “86.82”. This
means that an angle of measure θ = 86.82◦ has tan(86.82) = 18. If your
calculator is in “radian” mode and you type in “sin−1 (0.9)”, the answer
given is “1.12”. This means that an angle of measure θ = 1.12 radians
has sin(1.12) = 0.9.
There is a key property of the inverse circular functions which is useful in equation solving; it is just a direct translation of Fact 9.3.2 into our
current context:
Important Facts 20.2.2 (Composition identities). We have the following
equalities involving compositions of circular functions and their inverses: 20.3. APPLICATIONS 273 zaxis
yaxis 1
−π −π
2 −1 principal domain
on the unit circle θ
π
2 θaxis xaxis π tangent function
restricted to principal domain Figure 20.7: Principal domain for tan(θ). (a) If − π ≤ θ ≤ π , then sin−1 (sin(θ)) = θ.
2
2
(b) If 0 ≤ θ ≤ π, then cos−1 (cos(θ)) = θ.
(c) If − π < θ < π , then tan−1 (tan(θ)) = θ.
2
2
We have been very explicit about the allowed θ values for the equations
in Fact 20.2.2. This is important and an Exercise will touch on this issue. 20.3 Applications
As a simple application of Fact 20.2.2, we can return to the beginning of
this section and justify the reasoning used to ﬁnd the angle of descent of
the aircraft:
θ = tan−1 ((tan(θ)) = tan−1 1
10 = 0.09967 rad = 5.71◦. Let’s look at some other applications.
Example 20.3.1. Find two acute angles θ so that the following equation
is satisﬁed:
252
9
1 − cos2 (θ) .
=
4 cos2 (θ)
16
Solution. Begin by multiplying each side of the equation by cos2 (θ) and
rearranging terms:
9
252
1 − cos2 (θ) cos2 (θ)
=
4
16
252
252
9
0=
cos4 (θ) −
cos2 (θ) + .
16
16
4 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 274 To solve this equation for θ, we use what is called the technique of substitution. The central idea is to bring the quadratic formula into the picture
by making the substitution z = cos2 (θ):
9
252 2 252
z−
z+ .
16
16
4
Applying the quadratic formula, we obtain
0= z= 252
16 ± 2 252
16 −4 2(252 )
16 252
16 9
4 = 0.9386 or 0.06136. We now use the fact that z = cos2 (θ) and note the cosine of an acute angle
is nonnegative to conclude that
cos2 (θ) = 0.9386 or
cos(θ) = 0.9688 or cos2 (θ) = 0.06136
cos(θ) = 0.2477. Finally, we use the inverse cosine function to arrive at our two acute
angle solutions:
cos(θ) = 0.9688
cos(θ) = 0.2477 11111111111111111111111111
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11111111111111111111111111
00000000000000000000000000
11111111111111111111111111
00000000000000000000000000
11111111111111111111111111
00000000000000000000000000
11111111111111111111111111
00000000000000000000000000
11111111111111111111111111
00000000000000000000000000
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00000000000000000000000000 111
000
111
000
111
000
111
000 h α 32 d ⇒ θ = cos−1 (0.9688) = 14.35◦ ⇒ θ = cos−1 (0.2477) = 75.66◦ Example 20.3.2. A 32 ft ladder leans against a building
(as shown below) making an angle α with the wall. OSHA
(Occupational Safety and Health Administration) speciﬁes
a “safety range” for the angle α to be 15◦ ≤ α ≤ 44◦ . If the
base of the ladder is d = 10 feet from the house, is this a
safe placement? Find the highest and lowest points safely
accessible. Solution. If d = 10, then sin(α) = 10 , so the principal so32
10
lution is α = sin−1 32 = 18.21◦; this lies within the safety
zone. From the picture, it is clear that the highest point safely reached
will occur precisely when α = 15◦ and as this angle increases, the height
h decreases until we reach the lowest safe height when α = 44◦ . We need
to solve two right triangles. If α = 15◦ , then h = 32 cos(15◦ ) = 30.9 ft. If
α = 44◦ , then h = 32 cos(44◦) = 23.02 ft. Figure 20.8: A ladder problem. Example 20.3.3. A Coast Guard jet pilot makes contact with a small
unidentiﬁed propeller plane 15 miles away at the same altitude in a direction 0.5 radians counterclockwise from East. The prop plane ﬂies in
the direction 1.0 radians counterclockwise from East. The jet has been instructed to allow the prop plane to ﬂy 10 miles before intercepting. In what
direction should the jet ﬂy to intercept the prop plane? If the prop plane is
ﬂying 200 mph, how fast should the jet be ﬂying to intercept? 20.3. APPLICATIONS 275 Intercept point North Solution. A picture of the situation is shown in Figure 20.9(a). After a look at the picture, three right triangles pop out and beg to be exploited. We highlight these in
the Figure 20.9(b), by imposing a coordinate system and
labeling the various sides of our triangles. We will work
in radian units and label θ to be the required intercept
heading.
We will ﬁrst determine the sides x + y and u + w of the
large right triangle. To do this, we have ﬁnd speed to
intercept Jet
West 10 miles
1 rad
East
prop
plane
spotted here
15 miles East
0.5 rad
θ = intercept heading
South (a) The physical layout.
Intercept point yaxis
1 rad u x = 15 cos(0.5) = 13.164 miles,
y = 10 cos(1.0) = 5.403 miles,
w = 15 sin(0.5) = 7.191 miles, and
u = 10 sin(1.0) = 8.415 miles. d u+w
prop plane
w spotted
here 15 0.5 rad
Jet 10 θ
x y xaxis x+y (b) Modeling the problem. w+u
x+y We now have tan(θ) =
= 0.8405, so the principal soluFigure 20.9: Visualizing the
tion is θ = 0.699 radians, which is about 40.05◦. This is the
Coast Guard problem.
only acute angle solution, so we have found the required
intercept heading.
To ﬁnd the intercept speed, ﬁrst compute your
distance to the intercept point, which is the length
of the hypotenuse of the big right triangle: d =
(18.567)2 + (15.606)2 = 24.254 miles. You need to travel
this distance in the same amount of time T it takes the
10
prop plane to travel 10 miles at 200 mph; i.e. T = 200 = 0.05 hours. Thus,
the intercept speed s is
s= distance traveled 24.254
=
= 485 mph.
time T elapsed
0.05 Later you will use an alternative approach to this problem using velocity vectors.
In certain applications, knowledge of the principal solutions for the
equations c = cos(θ), c = sin(θ), and c = tan(θ) is not sufﬁcient. Here is
a typical example of this, illustrating the reasoning required.
Example 20.3.4. A rigid 14 ft pole is used to vault. The vaulter leaves
and returns to the ground when the tip is 6 feet high, as indicated. What
are the angles of the pole with the ground on takeoff and landing? CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 276 14 ft
α 6 ft leaving ground β 6 ft
airborne returning to ground Figure 20.10: Various angles of a vaulter’s pole. yaxis
1
Q y= 3
7 P β
α
1 xaxis Solution. From the obvious right triangles in the picture,
6
3
we are interested in ﬁnding angles θ where sin(θ) = 14 = 7 .
The idea is to proceed in three steps:
3
• Find the principal solution of the equation sin(θ) = 7 ; • Find all solutions of the equation sin(θ) = 3 ;
7
Figure 20.11: Modeling the
problem with a unit circle. • Use the constraints of the problem to ﬁnd α and β
among the set of all solutions. 3
Solving the equation sin(θ) = 7 involves ﬁnding the points
3
on the unit circle with ycoordinate equal to 7 . From the
picture, we see there are two such points, labeled P and Q.
3
The coordinates of these points will be P = cos(α), 3 and Q = cos(β), 7 .
7
Notice, α is the principal solution of our equation sin(θ) = 3 , since
7
0 ≤ α ≤ 90◦ ; so α = sin−1 3 = 25.38◦. In general, the solutions come
7
in two basic ﬂavors: θ = α + 2k(180◦ )
= 25.38◦ + 2k(180◦ ),
or
θ = β + 2k(180◦),
where k = 0, ± 1, ± 2, ± 3, . . . . To ﬁnd the angle β, we can use basic
properties of the circular functions:
sin(β) = sin(α) = − sin(−α) = sin(180◦ − α) = sin(154.62◦).
This tells us β = 154.62◦. 20.4 How to solve trigonometric equations
So far, our serious use of the inverse trigonometric functions has focused
on situations that ultimately involve triangles. However, many trigonometric modeling problems have nothing to do with triangles and so we 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 277 need to free ourselves from the necessity of relying on such a geometric picture. There are two general strategies for ﬁnding solutions to the
equations c = sin(θ), c = cos(θ), and c = tan(θ):
• The ﬁrst strategy is summarized in Procedure 20.4.1. This method
has the advantage of offering a “prescription” for solving the equations; the disadvantage is you can lose intuition toward interpreting
your answers.
• The second strategy is graphical in nature and is illustrated in Example 20.4.2 below. This method usually clariﬁes interpretation of
the answers, but it does require more work since an essential step is
to roughly sketch the graph of the trigonometric function (following
the procedure of Chapter 19 or using a graphing device).
Each approach has its merits as you will see in the exercises.
Important Procedure 20.4.1. To ﬁnd ALL solutions to the equations c =
sin(θ), c = cos(θ), and c = tan(θ), we can lay out a foolproof strategy.
Step Sine case Cosine case Tangent case 1. Find principal solution θ = sin−1 (c) θ = cos−1 (c) θ = tan−1 (c) 2. Find symmetry solution θ = − sin−1 (c) + π θ = − cos−1 (c) not applicable 3. Write out multiples of
period k = 0, ± 1, ± 2, · · · 2kπ 2kπ kπ 4. Obtain general principal solutions θ = sin−1 (c) + 2kπ θ = cos−1 (c) + 2kπ θ = tan−1 (c) + kπ 5. Obtain general symmetry solutions θ = − sin−1 (c) + π + 2kπ θ = − cos−1 (c) + 2kπ not applicable Example 20.4.2. Assume that the number of hours of daylight in your
2π
hometown during 1994 is given by the function d(t) = 3.7 sin 366 (t − 80.5) +
12, where t represents the day of the year. Find the days of the year during
which there will be approximately 14 hours of daylight?1
Solution. To begin, we want to roughly sketch the graph of y = d(t) on
the domain 0 ≤ t ≤ 366. If you apply the graphing procedure discussed
in Chapter 19, you obtain the sinusoidal graph below on the larger domain −366 ≤ t ≤ 732. (The reason we use a larger domain is so that
the “pattern” that will arise in the strategy described below is more evident. Ultimately, we will restrict our attention to the smaller domain
0 ≤ t ≤ 366.) To determine when there will be 14 hours of daylight, we
need to solve the equation 14 = d(t). Graphically, this amounts to ﬁnding CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 278 yaxis
15
12.5
10
7.5 taxis −200 200 400 600 Figure 20.12: Where does the sinusoidal function d(t) cross the line y = 14. the places where the line y = 14 intersects the graph of d(t). As can be
seen in Figure 20.12, there are several such intersection points.
We now outline a “three step strategy” to ﬁnd all of these intersection
points (which amounts to solving the equation 14 = d(t)):
1. Principal Solution. We will ﬁnd one solution by using the inverse
sine function. If we start with the function y = sin(t) on its principal
domain −π ≤ t ≤ π , then we can compute the domain of d(t) =
2
2
2π
3.7 sin 366 (t − 80.5) + 12:
2π
π
π
≤
(t − 80.5) ≤
2
366
2
−11 ≤ t ≤ 172.
− Now, using the inverse sine function we can ﬁnd the principal solution to the equation 14 = d(t):
14 = 3.7 sin
0.54054 = sin 2π
(t − 80.5) + 12
366 (20.2) 2π
(t − 80.5)
366 0.57108 = sin−1 (0.54054) = π
(t − 80.5)
183 t = 113.8
Notice, this answer is in the domain −11 ≤ t ≤ 172. In effect, we
have found THE ONLY SOLUTION on this domain. Conclude that
there will be about 14 hours of daylight on the 114th day of the year.
1 You can get the actual data from the naval observatory at this world wide web
address: <http://tycho.usno.navy.mil/time.html> 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 279 2. Symmetry Solution. To ﬁnd another solution to the equation 14 =
d(t), we will use symmetry properties of the graph of y = d(t). This
is where having the graph of y = d(t) is most useful. We know the
a maxima on the graph occurs at the point M = (172, 15.7); review
Example 19.2.1 for a discussion of why this is the case. From the
graph, we can see there are two symmetrically located intersection
points on either side of M. The principal solution gives the intersection point (113.8, 14). This point is 58.2 horizontal units to the left of
M; see the picture below. So a symmetrically positioned intersection
point will be (172 + 58.2, 14) = (230.2, 14). yaxis M 15
12.5
10
7.5
taxis
−200 200 400 600 Figure 20.13: Finding the symetry solution. In other words, t = 230.2 is a second solution to the equation 14 =
d(t). We call this the symmetry solution.
3. Other Solutions. To ﬁnd all other solutions of 14 = d(t), we add
integer multiples of the period B = 366 to the tcoordinates of the
principal and symmetric intersection points. On the domain −366 ≤
t ≤ 732 we get these six intersection points; refer to the picture of
the graph:
(−252.2, 14), (113.8, 14), (479.8, 14),
(−135.8, 14), (230.2, 14), (596.2, 14).
So, on the domain −366 ≤ t ≤ 732 we have these six solutions to the
equation 14 = d(t):
t = −252.2, −135.8, 113.8, 230.2, 479.8, 596.2.
To conclude the problem, we only are interested in solutions in the
domain 0 ≤ t ≤ 366, so the answers are t = 113.8, 230.2; i.e. on days 114
and 230 there will be about 14 hours of daylight. CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 280 20.5 Summary
• The inverse sine function, sin−1 x, is deﬁned as the inverse of the sine
function, sin x, restricted to the domain −π/2 ≤ x ≤ π/2.
• Inverse cosine and inverse tangent are deﬁned similarly.
• To ﬁnd the solutions to an equation of the form
f(t) = A sin 2π
(t − C) + D = k,
B proceed as follows:
1. Sketch the graph of the function f(t), including a few periods
and t = C.
2. Use algebra and the inverse sine function (sin−1 ) to ﬁnd one
solution. This is the principal solution; it is the solution nearest
to t = C. Call this solution P.
3. Use the principal solution and your knowledge of the graph of
the function to ﬁnd the symmetry solution. Call this solution S.
4. All solutions are then of the form
P, P ± B, P ± 2B, P ± 3B, . . .
or
S, S ± B, S ± 2B, S ± 3B, . . . . 20.6. EXERCISES 281 20.6 Exercises
Problem 20.1. Let’s make sure we can handle
the symbolic and mechanical aspects of working with the inverse trigonometric functions:
(a) Set your calculator to “radian mode” and
compute to four decimal places:
(a1) sin−1 (x), for x = 0,1,−1,
−3
11 ,2.
(a2) cos−1 (x), for x =
−3
11 ,2. √
3
2 ,0.657, √
0,1,−1, 23 ,0.657, (a3) tan−1 (x), for x = 0,1,−1,
−3
11 ,2. √
3
2 ,0.657, (b) Redo part (a) with your calculator set in
“degree mode”.
(c) Find four values of x that satisfy the
equation 5 sin(2x2 + x − 1) = 2.
(d) Find four solutions to the equation
5 tan(2x2 + x − 1) = 2
Problem 20.2. For each part of the problem
below:
• Sketch the graphs of f(x) and g(x) on the
same set of axes.
• Set f(x) = g(x) and ﬁnd the principal and
symmetry solutions.
• Calculate at least two other solutions to
f(x) = g(x). Indicate them on your graph.
(a) f(x) = sin x − π
2 1
, g(x) = 3 . (b) f(x) = sin x + π
6 , g(x) = −1. (c) f(x) = sin (2x − 1), g(x) = 1 .
4
(d) f(x) = 10 cos (2x + 1) − 5, g(x) = −1. Problem 20.3. Assume that the number of
hours of daylight in New Orleans in 1994 is
7
2π
given by the function D(x) = 3 sin 365 x + 35 ,
3
where x represents the number of days after
March 21.
(a) Find the number of hours of daylight on
January 1, May 18 and October 5.
(b) On what days of the year will there be
approximately 10 hours of daylight? Problem 20.4. Hugo bakes world famous
scones. The key to his success is a special
oven whose temperature varies according to a
sinusoidal function; assume the temperature
(in degrees Fahrenheit) of the oven t minutes
after inserting the scones is given by
y = s(t) = 15 sin 3π
π
t−
5
2 + 415 (a) Find the amplitude, phase shift, period
and mean for s(t), then sketch the graph
on the domain 0 ≤ t ≤ 20 minutes. (b) What is the maximum temperature of
the oven?
Give all times when the
oven achieves this maximum temperature during the ﬁrst 20 minutes.
(c) What is the minimum temperature of
the oven?
Give all times when the
oven achieves this minimum temperature during the ﬁrst 20 minutes.
(d) During the ﬁrst 20 minutes of baking,
calculate the total amount of time the
oven temperature is at least 410o F.
(e) During the ﬁrst 20 minutes of baking,
calculate the total amount of time the
oven temperature is at most 425o F.
(f) During the ﬁrst 20 minutes of baking,
calculate the total amount of time the
oven temperature is between 410o F and
425o F.
Problem 20.5. The temperature in Gavin’s
oven is a sinusoidal function of time. Gavin
sets his oven so that it has a maximum temperature of 300◦ F and a minimum temperature of 240◦ . Once the temperature hits 300◦ ,
it takes 20 minutes before it is 300◦ again.
Gavin’s cake needs to be in the oven for 30
minutes at temperatures at or above 280 ◦ . He
puts the cake into the oven when it is at 270◦
and rising. How long will Gavin need to leave
the cake in the oven?
Problem 20.6. Maria started observing Elasticman’s height at midnight. At 3 AM, he was
at his shortest: only 5 feet tall. At 9 AM, he
was at his tallest: 11 feet tall.
Elasticman’s height is a sinusoidal function of
time. CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 282 In the 24 hours after Maria began observing
Elasticman, how much of the time will Elasticman be less than 6 feet tall?
Problem 20.7. Suppose
T (t) = 23 sin 2π
(t − 7) + 66
24 is the temperature (in degrees Fahrenheit) at
time t, where t is measured in hours after midnight on Sunday. You paint the exterior door
to your house at 5 p.m. on Monday. The paint
information states that 48 hours of 75◦ F drying time is required; i.e., you can only count
time periods when the temperature is at least
75◦ F. When will the door be dry?
Problem 20.8. Tiffany and Michael begin running around a circular track of radius 100
yards. They start at the locations pictured.
Michael is running 0.025 rad/sec counterclockwise and Tiffany is running 0.03 rad/sec
counterclockwise. Impose coordinates as pictured. r= 100 yards 0.025 rad/sec (i) Find where Tiffany passes Michael the
ﬁrst time.
(j) Find when Tiffany passes Michael the
second time.
(k) Find where Tiffany passes Michael the
second time.
Problem 20.9. A communications satellite orbits the earth t miles above the surface. Assume the radius of the earth is 3,960 miles.
The satellite can only “see” a portion of the
earth’s surface, bounded by what is called a
horizon circle. This leads to a twodimensional
crosssectional picture we can use to study the
size of the horizon slice:
satellite
Earth horizon circle
center of Earth Earth satellite α
α Tiff starts here t Michael starts here CROSS−SECTION
0.03 rad/sec (a) Find a formula for α in terms of t.
(a) Where is each runner located (in
xycoordinates) after 8 seconds?
(b) How far has each runner traveled after
8 seconds?
(c) Find the angle swept out by Michael
after t seconds.
(d) Find the angle swept out by Tiffany
after t seconds.
(e) Find the xycoordinates of Michael and
Tiffany after t seconds.
(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. (b) If t = 30,000 miles, what is alpha? What
percentage of the circumference of the
earth is covered by the satellite? What
would be the minimum number of such
satellites required to cover the circumference?
(c) If t = 1,000 miles, what is alpha? What
percentage of the circumference of the
earth is covered by the satellite? What
would be the minimum number of such
satellites required to cover the circumference?
(d) Suppose you wish to place a satellite
into orbit so that 20% of the circumference is covered by the satellite. What is
the required distance t?
Problem 20.10. Answer the following questions: 20.6. EXERCISES
(a) If y = sin(x) on the domain − π ≤ x ≤ π ,
2
2
what is the domain D and range R of
y = 2 sin(3x − 1) + 3? How many solutions
does the equation 4 = 2 sin(3x−1)+3 have
on the domain D and what are they?
(b) If y = sin(t) on the domain − π ≤ t ≤ π ,
2
2
what is the domain D and range R of
2π
y = 8 sin( 1.2 (t−0.3))+18. How many solu2π
tions does the equation 22 = 8 sin( 1.2 (t −
0.3))+ 18 have on the domain D and what
are they?
(c) If y = sin(t) on the domain − π ≤ t ≤
2
π
2 , what is the domain D and range R
2π
of y = 27 sin( 366 (t − 80.5)) + 45. How
many solutions does the equation 40 =
2π
27 sin( 366 (t − 80.5)) + 45 have on the domain D and what are they?
(d) If y = cos(x) on the domain 0 ≤ x ≤ π,
what is the domain D and range R of
y = 4 cos(2x + 1)− 3? How many solutions
does the equation −1 = 4 cos(2x + 1) − 3
have on the domain D and what are
they?
(e) If y = tan(x) on the domain − π < x < π ,
2
2
what is the domain D and range R of y =
2 tan(−x + 5) + 13? How many solutions
does the equation 100 = 2 tan(−x + 5) + 13
have on the domain D and what are
they?
Problem 20.11. Tiffany is a model rocket enthusiast. She has been working on a pressurized rocket ﬁlled with laughing gas. Ac 283
cording to her design, if the atmospheric pressure exerted on the rocket is less than 10
pounds/sq.in., the laughing gas chamber inside the rocket will explode. Tiff worked from
a formula p = (14.7)e−h/10 pounds/sq.in. for
the atmospheric pressure h miles above sea
level. Assume that the rocket is launched at
an angle of α above level ground at sea level
with an initial speed of 1400 feet/sec. Also,
assume the height (in feet) of the rocket at
time t seconds is given by the equation y(t) =
−16t2 + 1400 sin(α)t.
(a) At what altitude will the rocket explode?
(b) If the angle of launch is α = 12o , determine the minimum atmospheric pressure exerted on the rocket during its
ﬂight. Will the rocket explode in midair?
(c) If the angle of launch is α = 82o , determine the minimum atmospheric pressure exerted on the rocket during its
ﬂight. Will the rocket explode in midair?
(d) Find the largest launch angle α so that
the rocket will not explode.
Problem 20.12. Let’s make sure we can handle the symbolic and mechanical aspects of
working with the inverse trigonometric functions:
(a) Find four solutions of tan(2x2 + x − 1) = 5.
(b) Solve for x: tan−1 (2x2 + x − 1) = 0.5 284 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix 285 286 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix A
Useful Formulas
Abbreviations
inch = in liter = L feet = ft Joule = J yard = yd calorie = cal mile = mi atmosphere = atm millimeter = mm Coulomb = C centimeter = cm radian = rad meter = m
degree = deg
=◦ kilometer = km
second = sec = s
minute = min miles per hour = mi/hr
= mph hour = hr
year = yr feet per second = ft/sec
= ft/s ounce = oz
pound = lb
gram = g meters per second = m/sec
= m/s kilogram = kg
quart = qt revolutions
= rev/min
per minute
= RPM gallon = gal
milliliter = ml 287 APPENDIX A. USEFUL FORMULAS 288 Conversion Factors
Length Energy 1 in = 2.54 cm 1 J = 1 kg m2 /s2 1 ft = 0.3048 m 1 cal = 4.184 J 1 mi = 1.609344 km Mass Volume
1 gal = 3.7854 L 1 oz = 28.3495 g 1 qt = 0.946353 L 1 lb = 0.453592 kg Formulas from Plane and Solid Geometry
Rectangle
r • Perimeter = 2ℓ + 2w
• Area = ℓw w Rectanglular prism
• Surface Area = 2(ℓw + ℓh + wh) ℓ • Volume = ℓhw Triangle h • Perimeter = a + b + c
• Area = w 1
bh
2 ℓ a c
h b Right circular cylinder
• Surface Area = 2πr2 + 2πrh
• Volume = πr2 h Circle
• Perimeter = 2πr
2 • Area = πr r
h 289 Sphere Right circular cone • Surface Area = 4πr2 • Surface Area = πr2 + πrs 4
• Volume = 3 πr3 • Volume = 1 πr2 h
3 r r Constants
Avogadro’s number = N = 6.022142 × 1023
speed of light = c = 2.99792 × 108 m/s2
density of water = 1 g/cm3 mass of earth = 5.9736 × 1024 kg earth’s equatorial radius = 3,960 mi = 6.38 × 106 m acceleration of gravity at earth’s surface = 32 ft/sec2 = 9.8 m/s2 Algebra
• ax ay = ax+y
• (ax )y = axy
• a0 = 1
• a
b x = ax
bx ax
= ax−y
ay
√
• n a = a1/n • • (a + b)(c + d) = ac + ad + bc + bd
• (a + b)2 = a2 + 2ab + b2
• (a + b)3 = a3 + 3a2 b + 3ab2 + b3
2 • Quadratic Formula: If ax + bx + c = 0, then x = −b ± √ b2 − 4ac
2a APPENDIX A. USEFUL FORMULAS 290 b
• Completing the Square: ax + bx + c = a x +
2a
2 2 − b2
+c
4a Trigonometry
• sin(−x) = − sin x • cos(π − x) = − cos x • cos(−x) = cos x • sin(π + x) = − sin x • sin = cos x • cos(π + x) = − cos x = sin x • sin(x + y) = sin x cos y + cos x sin y π
−x
2
π
• cos
−x
2
π
+x
• sin
2
π
• cos
+x
2 = cos x
= − sin x • sin(π − x) = sin x • cos(x + y) = cos x cos y − sin x sin y
• sin2 x + cos2 x = 1
• sin 2x = 2 sin x cos x
• cos 2x = cos2 x − sin2 x Appendix B
Answers
Answer 1.1 (b) 150 ft/sec. (c) Gina. (d) 6300 hours. Answer 1.14 (a) N = 12 + 0.45(60 − x). (b) Reduced competition for resources (eg. water, nutrients, etc.). Answer 1.2 68.4444 km.
Answer 1.3 r = 10.172 cm for lead; r = 16.433 cm for aluminum.
Answer 1.4 The mass of the air is 9.740 million kg, more
than the mass of the Tower.
Answer 1.5 (a) 5.5 min/mi = 5:30 pace. (b) 14 2 ft/sec. (c)
3
Adrienne.
Answer 1.6 (a) (John’s Salary) = $56000 and (taxes) = $0.
(b) (John’s Salary) ≤$56000 and (taxes) = 0.15× (John’s
Salary). (c) (John’s Salary) ≥ $56000 and (John’s taxes)
> 0.28× (John’s Salary). (d) 1500 ≤(number of 120 students each year) ≤ 1800. (e) (cost red Porsche) > 3× (cost
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) < 1 × (number of
2
cheerful biology students). (h) (Cady’s high score)  (Cady’s
low score) = 10%. (Cady’s ﬁnal exam score) = 97%. (i)
1
(0.9999) × 2 (total votes cast)≤ (Tush votes) ≤ (1.0001) × 1
2
(total votes cast). (
Answer 1.15 Formula simpliﬁes to r11+x) . Rates are
+2x
about 0.67r, 0.6r, 0.54r and 0.51r at the indicated times.
Rates decrease over time, but will never be less than 0.5r. Answer 1.16 (a) Initial time = 7am, Initial temperature =
44◦ F, Final time = 10am, Final temperature = 50◦ F, rate of
change = 2◦ F/hr. (b) 58◦ F. (c) Initial time = 4.5pm, Initial
temperature = 54◦ F, Final time = 6.25pm, Final temperature = 26◦ F, rate of change = 16◦ F/hr.
Answer 1.17 (a) t = 9. (b) a = 1/8. (c) x = −3a/4. (d) t > 3.
x+ 2
(e) x(x+1) .
√
√
Answer 2.1 (a) d = 2, ∆x = 1 = ∆y = 1. (b) d = 5,
√
∆x = −1, ∆y = −2. (c) d = 34, ∆x = 5, ∆y = −3. (d)
√
d = 10t2 + 2t + 1, ∆x = 3t, ∆y = 1 + t.
Answer 2.2 (ac) True. (d) ∆x = s − a, ∆y = t − b. (e)
∆x = a − s, ∆y = b − t. (f) ∆x = 0 means the points line on
the same vertical line; ∆y = 0 means the points line on the
same horizontal line.
Answer 2.3 Just after 12:29 PM that afternoon. Answer 1.10 Lee has 2.265 in2 more pie. Answer 2.4 (a) Erik= 6.818 mph, Ferry = 17.6 ft/sec. (b)
Impose coordinates with Kingston the origin and units of
miles on each axis; then Edmonds is located at (6,0) and
Erik’s sailboat is at (3,2). The table rows have these entries:
(0,0), (0.1,0), (1.4,0), (12t,0).
(3,2), (3,1.9432), (3,1.2045), (3,2 − 6.818t)
3.606, 3.491, 2.003,
(12t − 3)2 + (2 − 6.818t)2 (c) Use
coordinates as in (b), then when the ferry reaches (3,0),
Erik is at (3,0.296). (d) CG vessel does not catch the ferry
before Edmonds. Answer 1.11 About 5 × 106 times around the equator. Answer 2.5 (a) d(t) = (65.3)t (b) 227 minutes, 168.4 miles
(c) t = 80.86 seconds. Answer 1.7 Go for the 15 inch pie.
Answer 1.8 (a) m90 = 151kg,m99 = 468kg,m99.9 = 1476kg.
m
(b) v = 2.974 × 108 sec .
Answer 1.9 1080 pizzas sold in 4 hours. Proﬁt reaches
$1000 at about 8 : 22 pm. Answer 1.12 (a) Radii are about r = 0.2416, 2.317, 4.184
and 9.167 cm at the indicated times. (b) No.
Answer 1.13 (a) 140 million gallons per week; about 7300
million gallons per year. (b) 100 yds × 50 yds ×20 yds. Answer 2.6 (a) Allyson’s coordinate position: (0 ft,20 ft).
Adrienne’s coordinate position: (−16 ft,0 ft). (b) After 2
seconds there will still be slack in the bungee cord. (c)
t ≈ 2.34. Use this time to ﬁnd where Allyson and Adrienne
are located. (d) t = 5.5 seconds. 291 APPENDIX B. ANSWERS 292
Answer 2.7 Impose a coordinate system with origin at
A and the shore the horizontal x axis; let x be the location on the shore where Brooke beaches. Equation
√
1
25 + x2 + 1 (6 − x) gives time to reach Kono’s. T (0) = 4
2
4
hr, T (6) = 3.9 hr. Neither time will be the minimum time.
7
Answer 2.8 (b) t = 2; t = 5 . (c) (3,3) (d) spider=( 3 , 8 ),
2
3
ant=( 41 , 4 ). (e) 1.5 feet. (f) Spider reaches (9,6) when t = 4;
33
√
ant reaches (9,6) when t = 3. (g) spider speed is 5 ft/sec;
√
ant speed is 8 ft/sec. Answer 3.7 (a) 6.92 seconds. (b) 7.67 seconds. (c) 38.16
seconds. (d) 7332 ft2 < area < 7632 ft2 .
Answer 3.8 (a) x = 11/5. (b) no solutions. (c) (−2, − 2) and
(−2,4). (d) (−1 ± 2/5,3).
5
Answer 4.1 (a) y = − 3 (x − 1) − 1 (b) y = 40(x + 1) − 2
(c) y = −2x − 2 (d) y = 11 (e) m = 3 , y = 3 (x − 1) + 1 (f)
5
5
y = 40x − 14 (g) y = − 3 x + 7 (h) y = x − 1
4
4 Answer 2.9 49.92 mph (exactly).
Answer 2.10 141.46 miles. They are 300 miles apart at
time 0.826 hr = 49.6 minutes.
Answer 2.11 (a) Final answer is correct, but second
equality is wrong. (b) Final answer should be 4xy; key
fact is that (x + y)2 = x2 + 2xy + y2 , etc. (c) Answer and
steps correct.
Answer 2.12 (a) x = ± 1+β2
α2 (b) x = αβ
α+ β 1+4α
2
Answer 4.2 (a)( 1+2α , 1+2α ). (b) α = −2/5. (c) α = −1/4. Answer 4.3 (a) area = 25
13 2 b
(b) area = − 2m (c) m = − 1
4 Answer 4.4 In some cases, answers in this problem are
unique. In other cases, the answers are not unique. Here
is a possible solution set: β
αβ+1 (c) x = Eqn Answer 2.13 (a) 5t2 + 6t + 5 (b) 2t2 + 4t (c)
√
5t2 + 4t + 4 2
t2 −1 Answer 3.1 (a) (x+3)2 +(y−4)2 = 9 (b) (x−3)2 + y + 11
3 2 = (c) Draw a vertical and horizontal line through (1, 1). On
the vertical line, a circle of radius 2 will have a center at either (1, 3) or (1, −1). Likewise, the horizontal line will have
circles at either (−1, 1) or (3, 1).
C(h,k) 1.
2.
3.
4. (3, 1)
(1, 3)
(−1, 1)
(1, −1) (d) (1,1), (1, − 3), (1 + √ (x − h)2 + (y − k)2 = r2
(x − 3)2
(x − 1)2
(x + 1)2
(x − 1)2 + (y − 1)2
+ (y − 3)2
+ (y − 1)2
+ (y + 1)2 3,0), (0, − 1 − √ yint Pt on line Pt on line y = 2x + 1 1
16 Eqn Slope
2 1 (0, 1) (− 1 , 0)
2 y = − 11 x + 17
4
4 − 11
4 17
4 (3, − 4) (−1,7) y = −2x + 1 −2 1 (0, 1) ( 1 , 0)
2 y = 1x + 1
2 1
2 1 (0, 1) (−2, 0) y = 1,000 0 1,000 (0, 1,000) (0, 1,000) y=0 0 0 (0, 0) (1, 0) x=3 Undef None (3, 3) (3, −2) y = x − 14 1 −14 (5, − 9) (0, −14) (d) =4
=4
=4
=4 3) √
Answer 3.2 (a) center (3,1), radius 2 3 (b) center (2,3),
radius 2 (c) center (1/6,5/3), radius 203/12 (d) center
√
(3/4, 1/2), radius 3
Answer 3.3 (a) Wet in 24.0 minutes.
25.4154041 minutes. Answer 4.5 (a) y = 6,850(x − 1970) + 38,000 (b) y = 8,000(x −
1970) + 8,400 (d) Tabularize your answer: (b) Wet in Answer 3.4 (a) Imposing with xaxis along ground and yaxis along tower, wheel modeled by x2 + (y − 62)2 = 602 . (b)
46.43 ft. to right of tower. (c) (−24,7) and (−24,117).
Answer 3.5 (a) Impose a coordinate system so that the
tractor is at the origin at t = 0 seconds. With this coordinate system, the south edge of the sidewalk is modeled by
ys = 100; the north edge is modeled by, yn = 110. (b) t = 32
minutes. (c) t = 52 minutes. (d) 20 minutes.
Answer 3.6 (a)] The equation for eastward travel from
Kingston is y = 8. Southward travel along x = −1. (b)
Boundary: (x +2)2 +(y−10)2 = 9; Interior:(x +2)2 +(y−10)2 <
9; Exterior:(x + 2)2 + (y − 10)2 > 9. (c) 3.82 minutes. (d)
Plug x = −1 into circle equation to ﬁnd exit point. Use
this to ﬁnd when the ferry exits the radar zone. Be careful
to reference all times relative to when the ferry departed
Kingston. (e) 20.32 minutes. Year Seattle Port
Townsend 1983
1998 $127,050
$229,800 $112,400
$232,400 (e) 1995.74, $214,313 (f) 1982.70, Sea = $124,965 (g) 2008.78,
Sea = $303,661 (h) 1972.32, Sea = $53,892 (i) No.
Answer 4.6 For these answers, we take the hole to be the
origin of our coordinate system. Then, the ball’s line of
travel is y = (0.667)(x − 35). (a) (−13.46, −32.31) (b) 3.19 seconds (c) 5.82 seconds (d) (10.78, −16.17) at about t = 6.1 sec.
Answer 4.7 (a) Allyson at (0,70); Adrian at (−44,0).
Bungee is 83.1401 ft. long. (b) Occurs at time t = 7.7546
seconds and Allyson is at (0,77.546). Allyson’s ﬁnal location
is 77.546 ft. from her starting point. 293
Answer 4.8 The lines of tangency have equations y =
2
± √ (x − 12). The nonvisible portion of the yaxis is
5
√ − 245 5 ≤y≤ √
24 5
.
5 Answer 4.9 C
temperature is = 5 (F −
9
−9.4◦ F. 32) and F = Answer 4.10 At the closest
26.22471828 miles from Paris. 9
C
5 + 32. In Oslo, the point, she will 5
1
Answer 5.8 (a) − 3 ,− 2 , x , 2 (♥ − 3). (b) 0, 20, 6x + 2x2 ,
2
2
2 − 6♥, 2(♥ + △ − 3)(♥ + △). (c) Always 4π2 .
2♥ Answer 5.9 Use the vertical line test. For example, (a) is
not a function, by the vertical line test; you can split it
into two function graphs by slicing the ellipse symmetrically into upper and lower halves. On the other hand, (o)
is a function, by the vertical line test; etc. be
Answer 5.10 (a) x = 5.
√
242− Answer 4.11 Impose coordinates with Angela’s initial location as the origin.
Angela is closest to Mary at
(18.8356,41.4383); this takes approximately 3.8 seconds.
Answer 4.12 (a) Impose coordinates with sprinkler initial location as the origin (0,0). Line equation becomes
1
y = − 5 x + 100. (b) Sprinkler is located at (0,82.5) at time
t = 33 minutes. Circular boundary of watered zone hits
southern edge of sidewalk at the points (−6.708,101.3416)
1
and (13.4386,97.3123). (c) y = − 5 x + 110.198.  − 0.5x − 1 = if x ≤ −2
if x > −2 −0.5x − 1
0.5x + 1
y
3
2.5
2
1.5
1
0.5 √
Answer 4.15 (a) (−1 ± 7/3)/2 = (−3 ± 21)/6. (b) t =
−
√0.463325, 0.863325; or the exact answer would be (1 ±
11)/5. (c) t = 0.2. (d) No real number solutions.
2+ √ 2 Answer 5.1 (a) −2 + h + 2x. (b) 2. (c) h + 2x. (d) −h − 2x. (e)
−π(h + 2x) (f) √h+x−1 +√x−1 .
1 4 Answer 5.3 For example, in (a), suppose Dave has constant speed v ft/min. Then the function s = d(t) = vt
will compute the distance Dave travels in t minutes. The
graph would be a line with sintercept 0 and slope v; the
domain would be 0 ≤ t ≤ 2400 ; etc.
v
Answer 5.4 2 4 2 x Answer 6.3 (a) x = 4 and x = − 22 (b) x = 0.75 (c) the
3
equation has no solutions.
Answer 6.4 (a) x = −7 and x = 3 (b) a = 3 (c) x = 8/3
Answer 6.5 For 0 ≤ x ≤ 6, the area is 1
x( x
6 + 18). Answer 6.6 (a) The rule is
2x
−x + 30 y= Answer 5.2 g(x) = 9 x + 24, f(x) = 4 x + 4, v(x) = x + 20,
5
5
v(5) = 25, minimum v(x) = 20, maximum v(x) = 40. if 0 ≤ x ≤ 10
if 10 ≤ x ≤ 30 and the range is 0 ≤ y ≤ 20.
(b) The rule for the area function a(x) is
a(x) = 1
− 2 x2 x2
+ 30x − 150 if 0 ≤ x ≤ 10
if 10 ≤ x ≤ 30 (c) x = 12.6795 inches.
Answer 6.7 (a) Answer 5.5 Several possible answers for each one.
j( t ) = √ Answer 5.6 (a) xintercepts= 1 (3 ±
√
√6
1
(b) ( 6 (3 − 93),5) and ( 1 (3 + 93),5).
6
1+ √ 2, − 2 − 3 (d) Answer 6.2 (a) Answer 4.14 18.923076923 seconds No, Yes. (e) ( (c) x = 36. Answer 6.1 (a) 0, 2, 3. (b) x = ±4, x = 0, no solution. (c)
Intersect at (4,4) and (− 4 , 4 ). Area is 16 .
33
3 Answer 4.13 (a) With the origin set at the statue, x =
30 − 3.123475t, y = 2.49878t. (b) The distance to the statue is
√
16t2 − 187.4085t + 900 feet. (c) Margot will be 28 feet from
the statue after 0.655672 seconds, and after 11.057360
seconds (assuming she continues past the point due north
of the statue). Answer 4.16 (a) There are four answers: x = ±
√
√
and x = ± 2 − 2 (b) y = 6 + 2 5. (b) x = 4, 8. 56400
2 33). yintercept = −2.
(c) None. (d) Yes, No,
√
√
1 + 2 + 3(1 + 2)). Answer 5.7 (b) f(x) = 30000x + 50000 (4 − x)2 + 1. (c) The
table of values (x,f(x)) will be: (0., 206155), (0.5, 197003),
(1, 188114), (1.5, 179629), (2,171803), (2.5, 165139), (3., 160711),
(3.5, 160902), (4,170000). Minimal cost occurs for some
2.5 < x < 3.5; the exact answer is x = 13 , but we can4
not solve this in our class since it requires the tools of
Calculus. 0
62.22(t + 1) if t < −1
if −1 ≤ t ≤ 3.5 (b)
s(t) = 280
280 − 70t if t < 0
if 0 ≤ t ≤ 4 (c) 280 280 − 62.22(t + 1) 217.78 − 132.22t
d(t) = 132.22t − 217.78 70t if
if
if
if
if t < −1
−1 ≤ t < 0
0 ≤ t < 1.6471
1.6471 ≤ t < 3.5
3.5 ≤ t < 4 APPENDIX B. ANSWERS 294
Answer 6.8 The distance to his starting point t seconds
after he starts is 10t 2502 + (12(t − 25))2
d(t) = 4002 + (250 − 9(t − 175 ))2
3 if 0 ≤ t ≤ 25,
if 25 ≤ t ≤ 175 ,
3 if 175
3 ≤t≤ 205
.
3 Answer 6.9 y = d(t) = if
if
if
if 18t
− 5)2
902 + (90 − 18(t − 10))2
90 − 18(t − 15)
902 + 182 (t 0≤x≤5
5 ≤ x ≤ 10
10 ≤ x ≤ 15
15 ≤ x ≤ 20 Here is the graph of d(t) 2
Answer 7.2 (a) y = − 3 x2 + 5 x. (b) y = 1.125x2 − 1.5x − 1.625.
3
2 + 3.5x − 4. (d) No solution.
(c) y = −0.5x Answer 7.3 (a) maximum value is 22; minimum value is
1.75 (b) maximum value is 32; minimum value is 2 (c)
maximum value is 6; minimum value is 46
Answer 7.4 In the ﬁrst case, d = 0 or d = 12. In the second case, d = ± √2 .
13 5
1
Answer 7.5 (a) Multipart function: s(t) = − 160 t2 + 8 t + 10
for 0 ≤ t ≤ 114.031, s(t) = 0 for t ≥ 114.031.; $23,125. (b)
Sell at time t = 50 days; $2500. Answer 7.6 The parabola has x intercepts at −1, 3 and y
intercept at 3. The vertex of the parabola is (1, − 4). ft
140
120
100
80
60
40
20 2 x − 2x − 3 = x2 − 2x − 3
−x2 + 2x + 3 x2 − 2x − 3 if x ≤ −1
if −1 < x < 3
if x ≥ 3 Answer 7.7 (a) 100 ft. (b) 156.25 ft. (c) (625, − 125). (d)
When x = 54.81 ft. or x = 570.19 ft.; i.e. at (54.81,39.04)
and (570.19, − 64.04).
5 15 10 20 t sec Answer 6.10 (a) v(x) = 2x3 − 70x2 + 500x; degree 3. (b)
a(x) = 1000 − 50x − 2x2 , degree 2. To get 600 sq. in. dimensions are: 6.3746 × 7.2508× 18.6254.
Answer 6.11 (a) 2 hours. (b) Impose coordinates with x
axis the bottom of the ditch and the y axis the pictured
centerline. y= 10 + 10 − 10 − 10 + if
if
if
if
if
if
if x ≤ −40
−40 ≤ x ≤ −30
−30 ≤ x ≤ −20
−20 ≤ x ≤ 20
20 ≤ x ≤ 30
30 ≤ x ≤ 40
40 ≤ x Answer 6.12 (a) 0 ≤ y ≤ 6. (b) Increasing: −6 ≤ x ≤ −2
and 2 ≤ x ≤ 4. Decreasing: −2 ≤ x ≤ 2 and 4 ≤ x ≤ 6.
(c) 4+
y= 4− Answer 7.9 She should have 225 trees in the orchard.
Answer 7.10 She should charge $7.72 to make the most
money.
Answer 7.11 The radius of the circular part should be 20
100 − (x + 40)2
100 − (x + 20)2
0
100 − (x − 20)2
100 − (x − 40)2
20 √
(c) 2(40 − 91) feet. (d) 42 ft. wide: 0.3008 minutes. 50 ft.
wide: 8.038 minutes. 73 ft. wide: 116.205 minutes. Answer 7.8 (a) No, since f(1) = 1 = 2. (b) The points
(1,1 + 2b) and (−(1 + b),1 + 2b). (c) Only the point (a,1 − a2 ).
(d) The points (−2.2701, − 7.1168) and (2.9368,1.5612). 2x + 12
4 − (x + 2)2
4 − (x − 2)2
−2x + 12 if
if
if
if −6 ≤ x ≤ −4
−4 ≤ x ≤ 0
0≤x≤4
4≤x≤6 (d) 2 ≤ y ≤ 6. (e) 2 ≤ y ≤ 4.
Answer 6.13 (a) 0. (b) a = 13
,
5 b = − 3 , c = −3.
5 Answer 7.1 (a) 2(x − 4)2 + 9, Vertex: (4, 9), Axis: x = 4. (b)
3(x − 5/2)2 − 383/4, Vertex: (5/2, 383/4), Axis: x = 5/2.
(c) (x − 3/14)2 + 2539/196, Vertex: (3/14, 2539/196), Axis:
x = 3/14. (d)2(x − 0)2 − 0, Vertex: (0, 0), Axis: x = 0. (e)
(1/100)(x − 0)2 − 0, Vertex: (0, 0), Axis: x = 0. 24
≈ 3.36059492 feet.
4+π
The short side of the rectangular part should also be equal
to this.
Answer 7.12 The enclosure should be 50 meters by 75
meters.
Answer 7.13 The maximum possible area is 4285.71 m2 .
Answer 7.14 To achieve the minimal combined area, cut
the wire so that the pieces have lengths 26.394 in. and
33.606 in.; bend the 26.394 inch piece into a circle.
Answer 7.15 (a) y =
3.8681948. 1
x−4
2 1
(b) y = 4 + 6 x (c) t = 1350/349 ≈ Answer 7.16 (a) The distance between Sven and Rudyard
is
41t2 − 2400t + 36900 feet, where t is the number of
seconds after they start moving. (b) They will be closest
together after they have been moving for 29.268292 seconds, and will be 42.166916 feet apart at that time.
Answer 7.17 Michael M(t) = (5.547t,8.32t), Tina T (t) =
(400 − 8.944t,50 + 4.472t). Study the distance SQUARED
from M(t) to T (t). Michael and Tina will be closest when
t = 26.641 sec; they are 54.3 ft at that instant. 295
Answer 8.5 (a) f(s) = Answer 7.18
x= −2α2 ± α= −x2 ± √ 4α4 − 4α
2α C(f(s)) computes mph when you input seconds s.
g(h) = 60h; x4 − 8x
4x Answer 7.19 (d) There are two possible values for α: If
√
,
α = 8 + 2 17√then the unique solution of the equation will
√
, then the unique solution
be x = −4 − 17. If α = 8 − 2 17√
of the equation will be x = −4 + 17.
1
(2 ±
2 70s2
10(602 ) + s2 C(f(s)) = √ Answer 7.20 (a) t = s
;
60 (b) 252000h2
10 + 3600h2 C(g(h)) = C(g(h)) computes mph when you input hours h. (c) v(s) =
22
s;
15 √√
√
2 s − 1). (b) x = −1 ± y − 2. 308m2
30 + 3m2 v(C(m)) = v(C(m)) computes ft/sec when you input minutes m. Answer 8.1 (a) Answer 8.6 (a) f(g(x)) = (x + 3)2 , f(f(x)) = x4 , g(f(x)) = x2 +
1
1
3. (b) f(g(x)) = √x , f(f(x)) = x, g(f(x)) = √x . (c) f(g(x)) = x, h(f(t)) = −t + 1
t−1 if t ≤ 1
if 1 ≤ t h(g(t)) = −t − 1
t+1 if t ≤ −1
if −1 ≤ t f(f(x)) = 81x + 20, g(f(x)) = x. (d) f(g(x)) = 6(x − 4)2 + 5,
f(f(x)) = 6(6x2 + 5)2 + 5, g(f(x)) = 6x2 + 1. (e) f(g(x)) = 8x + 21,
f(f(x)) = 4(4x3 − 3)3 − 3, g(f(x)) = 2x. (f) f(g(x)) = 2x3 + 1,
f(f(x)) = 4x + 3, g(f(x)) = (2x + 1)3 . (g) f(g(x)) = 3, f(f(x)) = 3,
g(f(x)) = 43. (h) f(g(x)) = −4, f(f(x)) = −4, g(f(x)) = 0. (b) Answer 8.7 −5/2 ≤ x ≤ −1/2
f(h(t)) = −t − 1
t−1 if t ≤ 0
if 0 ≤ t g(h(t)) = t−1
−t − 1 if t ≤ 0
if 0 ≤ t 7
Answer 8.8 (a) 2 ≤ x ≤ 6. (b) −3 ≤ y ≤ 5. (c) 1 ≤ x ≤ 6.
3
(d) −9 ≤ y ≤ 7. (e) B = 5, C = 39 . (f) A = 1 , D = 8 .
5
8 Answer 8.9 (a) −1
,
(x−1)(x+h−1) 4 + 4h + 8x, set h = 0 to get 4 + 8x. (c) (c) −1
(b)
(x−1)2
−h−2x
√
√
,
25−x2 + 25−(x+h)2 set h = 0 to get −x
.
25−x2 set h = 0 to get √ −t − 1 t+1
h(h(t) − 1) = −t + 1 t−1 if
if
if
if t ≤ −1
−1 ≤ t ≤ 0
0≤t≤1
1≤t Answer 9.1 (a) domain of f ={xx =
(b) Answer 9.3 (c) f−1 (y) =
− 17 }.
8 2.5 range=f= {yy = 0}. 3+ √ 17+8y
4 on the domain {yy ≥ Answer 9.5 1.5
1 Answer 9.6 (a) h = f(x) = −2x2 + 124x. (c) x = g(h) =
√
1
31 − 2 3844 − 2h. 0.5
1 4
};
3 Answer 9.4 Only (B) is onetoone on the entire domain. 2 2 = 2+4y
.
3y Answer 9.2 y
3 3 f−1 (y) 1 2 3 x Answer 9.7 (a) (1) 0.5 y 1
Answer 8.2 (a) y = f(g(x)), if f(x) = x5 , g(x) = x − 11. (b)
√
y = f(g(x)), if f(x) = 3 x, g(x) = 1 + x2 . (c) y = f(g(x)), if
5 − 5x2 + (1/2)x + 11, g(x) = x − 3. (d) y = f(g(x)), if
f(x) = 2x
√
f(x) = (1/x), g(x) = x2 + 3. (e) y = f(g(x)), if f(x√ = x,
)
√
g ( x) =
x + 1. (f) y = f(g(x)), if f(x) = 2 − 5 − x2 ,
g(x) = 3x − 1.
Answer 8.4 (c) x = −1 and x = 2 x (2) x = f−1 (y) =
bers. (3) y+2
3 has domain and range all real num APPENDIX B. ANSWERS 296
x Answer 11.4 (c) The curve is modeled by y
=
x
500cosh( 500 ) − 440.536 and the minimum height is 59.46
feet.
Answer 12.1 (a) 0.6826; 2.3979; 3.3030; 3.3219;0.3010.
(b) 3.555; 19.8; 0.0729. (c) x = log10 y; x = log10 (3y);
x = (1/3) log10 (y). y Answer 12.2 (a) I(d) = I◦ (0.94727)d . (b) 85 meters.
(4) f(f−1 (y)) = f( y+2 ) = 3( y+2 ) − 2 = y + 2 − 2 = y;
3
3
f−1 (f(x)) = f−1 (3x − 2) = 3x−2+2 = x.
3
t
Answer 9.8 (a) w = f(t) = 2 25 − ( 6 − 5)2 . Domain:
0 ≤ t ≤ 30 hours; Range: 0 ≤ w ≤ 10 ft. (b) t = 6 hours. (c) t = f−1 (w) = 30 − 6 25 − ( w )2 . Domain: 0 ≤ w ≤ 10 ft.;
2
Range: 0 ≤ t ≤ 30 hours. Answer 9.9 (a) 2 nanoseconds. (f) domain φ={0 ≤ t ≤ 10};
range φ={0 ≤ y ≤ 6}.
Answer 10.1 (a) 31.5443; (b) 355.1134; (c) 36.4622; (d)
0.0616; (e) 51,168; (f) 0.009794;
Answer 10.2 (a) y = 3( 1 )x (b) y = ( 1 )x . (d) y =
2
2 1√x
(1).
27
3 Answer 10.3 (a) 1.64 × 106 cells. (b) True. (c) The two
formulas are identical.
Answer 10.4 (a) 261.31 Hz. (b) 440 Hz. (c) 27.5 Hz. (d)
16.35 Hz.
Answer 10.5 (a) 29 = 512. (b) 2n−1 . (c) 263 = 9.2 × 1018 . (d)
9.2 × 1015 meters.
Answer 10.6 (a) M(p) is the higher curve. (d) Answer 12.3 (a) y = 13e1.09861t . (b) y = 2e−2.0794t .
Answer 12.4 (a) 9.9 years. (b) 34.66%.
Answer 12.5 (a) 8.8 cm. (b) 135 cm. (c) 4.84 yrs. (d)
Maximum possible length of the halibut.
Answer 12.6 (a) C(t) = 1.03526t .
Answer 12.7 (c) v(x)
$200,000 during 1993. 55000(1.0315)x . = Valued at Answer 12.8 (a) x = 2.7606. (b) 1.392. (c) 0.3552. (d)
x = e8 . (e) x = 11.513. (f)0.61. (g) sin(x) = −0.6931;
x = −0.7658 + 2kπ or x = 3.9074 + 2kπ.
Answer 12.9 (d) 8.617 weeks.
Answer 12.10 31.699250014 days
Answer 12.11 (a) 2015 (b) 2048.
Answer 12.12 (a) 46.701735 years (b) 137.3113631016
years after 1980
Answer 13.1 The graph is given below: fraction
0.8 y
2 0.6 1.5 0.4 1 0.2
20 40 60 80 $1.113(1.046408)t . Answer 11.1 (a) w(t) =
is below; should be $5.70 by the model. 100 0.5 p 2 1.5 1 0.5 0.5 (b) $1.11. (c) It 0.5 Answer 11.2 (a) p(x) = 860(1.070674)x , l(x) = 70x + 860. (b)
p(10) = 1702, l(10) = 1560. 1
1.5 Answer 11.3 If we use 1989 and 2000 in a(t), we get
two data points and a corresponding exponential model:
E(t) = 15.918(1.243301)( t − 1980). The exponential model
grows faster than the cubic model and eventually exceeds
a(t). 2
Answer 13.2 (a) 1 1.5 2 x 297
y
30 6 25
4 20
2 15
10 0
3 2 1 0 1 2 5
2 1 2 1 2 4 3 x 4 Answer 13.5 (a) 6 y
4 (b) 3
2 1.5 1
1 4 3 2 1 1 2 4 3 x 0.5 1
0
0.5 0 0.5 1 1.5 2 0.5 (b) No. (c) 0 x+2
y = f(−x) = −2x + 2 0 1 1.5 if
if
if
if x ≤ −2
−2 ≤ x ≤ 0
0≤x≤1
x≥1 y
4 (c) 3
2
1
10 4 3 2 1 5 10 8 6 4 2 0 2 4 5 0 −2x − 2
y = −f(x) x−2 0 4 Answer 13.3 3 4 x if
if
if
if x ≤ −1
−1 ≤ x ≤ 0
0≤x≤2
x≥2 y
1 10 −1
3 2 1 0
12 1 ≤ x ≤ 0. 3 2 1 1 2 3 4 x 1
2 Answer 13.4 (a5) (1) horizontally dilate (compress) by a
1
factor of 2; (2) horizontal shift right by 2 ; (3) vertical dilate
(expand) by a factor of 3; vertical shift up by 5. f ( x) = −6x + 8
6x + 2 if x <
if x ≥ 1
2
1
2 3
4 (d) Here are the graphs of y = 2f(x) and y =
tively: 1
f ( x) ,
2 respec APPENDIX B. ANSWERS 298
y
4 Answer 13.10 y = 1 (2x ) is obtained by vertically dilating
3
y = 2x ; it is vertically compressed. y = 2x/3 is a horizontal
x ; it is horizontally stretched.
dilation of y = 2 3
2 Answer 14.1 (a) domain={xx = 1}; range={yy = 2}; zero
at x = 0; horizontal asymptote y = 2; vertical asymptote
x = 1; graph below: 1
4 3 2 1 1 2 3 4 x y
15 1
10
y
4 5
4 2 3 5 2 3 2 10 1
4 15 1 1 2 3 4 x 1
1
(e) Here are the graphs of y = f(2x) and y = f( 2 x), respectively: Answer 14.2 (a) y =
ft. (c) 10 ft. Answer 14.5 f(x) =
2 asymptote is y =
1
2 1 (b) x = 9.143 ft.; x = 9.916 Answer 14.3 (a) m(t) = 35t + 200. (b) k(t) = 30t − 50. (c)
1987. (d) r(t) = 35t+200 . (e) 7 .
30t−50
6 3 3 0.2x−10.4
.
x−10 Answer 14.4 (a) y = 0.03x2 − 1.1x + 14 (b) w = 0.03x − 1.1 + 14
x
(c) 70 or 6 2
3 y
4 4 x 4 2 1 2 3 4 x 41
.
11 Answer 14.6 f(x) = 41
x
11 35
11
65
11 + x+ = 41x + 35
. The horizontal
11x + 65 6x + 10
x+1 1 Answer 14.7 You should study for 11.25 hours.
y
4 Answer 14.8 (a) k = 400 (b) I(t) =
t = 1.5 (d) t = 1.05 and 1.95. 400
484t2 −1452t+1189 (c) 3 Answer 14.9 2 Answer 14.10 (a)y = f(x) = 1 (d)x =
4 2 2 4 x 1 (f) c = 11
.
2 (g) c = 5
.
2 (h) c = d = 1
.
3 √ Answer 13.6 (a) a(x) = 2(6 − x) 3x − 9 with domain 3 ≤
√
x√ 6. (b) Max of a(x) is 4 3 and max of 2a(3x + 3) + 1 is
≤
√
1
8 3 + 1 (when x = 3 ). (c) 1 + 36(1 − x) x on the domain
√
0 ≤ x ≤ 1. The range is all y values between 1 and 8 3 + 1.
Answer 13.7 (c) Horizontally shift the graph of y = x2 to
the right 4 units. 20000
= 200y−−y .
400
−1 = {0 ≤ x ≤
f 400x+20000
.
x+200
Domain f−1 (b) x = $400.
= {100 ≤ y ≤ 388.46}, Range
5000}. The inverse function
takes the number of customers per day as an input value
and gives the amount the shop spent on advertising as an
output value.
Answer 15.1 (a) 13o 24 ′ or 0.233874 rads. (b) 1.0788 degs
or .01882 rads. (c) 5.7296 degs or 5o 43 ′ 46.5".
Answer 15.2 (a) 6080 ft. (b) 29.95 mph. (c) 15.63 knots.
Answer 15.3 (a) 430 sq. in. (b) 2.56◦ . (c) 84.47 in. (d) 23.04
in. (e) 7.29 in.
Answer 15.4 (a) 1413.7 sq. ft. (c) 4.244 sec.
Answer 15.5 2164.208272472 miles. Answer 13.8 (a) y = 72x + 4 + 2.
Answer 13.9 (a1) f(2x) = 4x . f(2x − 1) = f−1 (y) 1x
4.
2 Answer 15.6 (a) 2.147 hrs. (b) 1103 mph. (c) 12.47 hrs.,
6236 miles. (d) 13760 miles. 299
Answer 17.12 Top right scenario: (a) θ◦ = 1.2 rad.
(b)θ(t) = 1.2 + 4πt (c) b(t) = (2 cos(1.2 + 4πt ),2 sin(1.2 + 4πt )).
9
9
9
(d) b(1) = (−1.710,1.037). b(0) = (0.725,1.864). b(3) =
(1.252, − 1.560). b(22) = (1.753,0.962). Answer 15.7 (b) 84.47 sq. in. (c) 181.5 sq. in.
Answer 15.8 0.685078 miles.
Answer 15.9 Middle picture: shaded area= 12.537 sq. in. Answer 18.1 (b) sin2 (x) =
Answer 16.1 (a) 10π/3 rad = 10.47 radians. (b) 4/2π rev *
1 hour/rev = .64 hours = 38.2 minutes. (c) Using (2.2.2),
(21)(10π/3) = 219.91 meters. 1
(1
2 − cos(2x)). 2
1.5
1 Answer 16.2 194 RPM. 0.5
11π
30 Answer 16.3 (a) ω =
rad/sec, v =
v = 2.094 in/sec, ω = 2.5 RPM. 121π
15 ft/sec. (b) 6 4 2 4 6 2 4 6 2 4 6 2 2 4 6 0.5
1 Answer 16.4 (b) 700 ft. (c) 70 sec. (d) 15000 sq. ft. 1.5
2 Answer 16.5 (a) r = 233.427 ft. (b) θ = 0.06283 rad. (c) 2π/5
rad counterclockwise from P. (c) Answer 16.6 (a) 40π ft/sec; 85.68 mph. (b) 400 RPM. (c) 60
RPM; 32π ft/sec. (d) 1.445 rad = 82.8◦ ; 28.9 ft. (e) 0.7 sec;
1.4π rad. 1.5 Answer 16.7 4.4 inches. 0.5 2 1 6 Answer 16.8 r = 2.45 inches. 4 2
0.5
1 Answer 17.1 (a) If you impose coordinates with the center of the wheel at (0,237.427), then ground level coincides with the xaxis. (a) T (t) = (x(t),y(t)), where x(t) =
233.427 cos( 2π t − 0.06283) and y(t) = 233.427 sin( 2π t −
5
5
0.06283) + 237.427. (b) T (6) = (85.93,454.45). (c) First nd
the slope of a radial line from the wheel center out to Tiff’s
launch point. 1.5
2 Answer 18.2 (a)
2 Answer 17.2 (a) y = ±0.2309(x + 1) + 2 1.5 Answer 17.3 290 ft. 0.5 1 6 Answer 17.4 (a) (21.91218, 1.498834) (b) (5.92564,
21.14892) (c) (19.07064, 10.89497)
Answer 17.5 (a) (19.9,13.42).
(−1.674,23.942). (d) (23.882,2.375). (b) (22.55,8.21). 4 2
0.5
1
1.5 (c) Answer 17.6 101.496936 feet above the ground. 2 (b) Answer 17.7 The dam is 383 feet high. 2
1.5 Answer 17.8 108 ft. 1
0.5 Answer 17.9 With the center of the track at the origin, and the northernmost point on the positive yaxis, Charlie’s location after one minue of running is
(59.84016,4.37666). 6 4 2
0.5
1
1.5 Answer 17.10 105.2718216 feet
Answer 17.11 (a) 204.74 ft. (b) no. 2 (c) APPENDIX B. ANSWERS 300
x y
2
15 1 4 2 6 t 8 10 1
2 5 0
6 4 2 0 2 4 Answer 18.3 (a) 7/25 or 7/25. (b) 0.6. (c) ± √
40
.
7 Answer 18.4 In the ﬁrst case, 6 Answer 19.7 (a)2 volts. (b) Never zero since 2x is positive
3
2π
for all x. (c)p(t) = 3 sin( 2/5 (t − 5 ) + 1, so A = 3, D = 1, B =
2/5, C = 3/5. (d) 0.25 ≤ V (t) ≤ 16. (e)t = 0.65635 + k(0.4) and
t = 0.74365 + k(0.4), k = 0, ± 1, ±2, . . . are ALL solutions.
Four of these lie in the domain 0 ≤ t ≤ 1. (f) Maxima have
coordinates (0.3 + k(0.4),16) and minima have coordinates
(0.1 + k(0.4),0.25), where k = 0, ± 1, ±2, . . . . (g) If we restrict
V (t) to 0.5 ≤ t ≤ 0.7, the inverse function has rule:
t= y
arcsin( ln(3 )−ln)(2) ) + 3π
ln(2 5π . If we restrict V (t) to 0.5 + k(0.4) ≤ t ≤ 0.7 + k(0.4), the inverse
function has rule:
one period t = k(0.4) + arcsin( ln(y)−ln(2)
)
3 ln(2) + 3π
. 5π In particular, if we restrict V (t) to 0.1 ≤ t ≤ 0.3, the inverse
function has rule:
Answer 18.5 π
4 + 2kπ and Answer 19.1 (a) 1, π, π
,
2 5π
4 + 2kπ, k = 0, ±1, ±2, ±3, . . . . t = −0.4 + y
arcsin( ln(3 )−ln)(2) ) + 3π
ln(2 5π 1. (b) 6, 2, 0, 1. . If we restrict V (t) to 0.7 ≤ t ≤ 0.9, the inverse function has
rule:
ln(y)−ln(2)
− arcsin( 3 ln(2) ) + 4π
t=
.
5π 2π
Answer 19.2 (d) h(t) = 8 sin( 1.2 (t − 0.3)) + 18.
2π
Answer 19.3 (a) b(t) = 0.6 sin( 100 (t − 30)) + 1.2. Answer 19.4 (a) h(t) = 5 sin( π (t −10)+5, where t indicates
6
hours after midnight. (b) 7.5 ft. above low tide.
Answer 19.5 (a) A = 25, B = 5 seconds, C = 1.75, D = 28.
(b) t=1.75 and 4.25 seconds
2π
Answer 19.6 First scenario: x(t) = 2 sin( 4.5 (t − Cx )),
π
9
2π
where Cx = −(1.2 + 2 )( 4π ); y(t) = 2 sin( 4.5 (t − Cy )), where
9
Cy = −(1.2)( 4π ). Plots are below: If we restrict V (t) to 0.7 + k(0.4) ≤ t ≤ 0.9 + k(0.4), the inverse
function has rule: t = k(0.4) + ln(y)−ln(2)
)
3 ln(2) + 4π
. 5π In particular, if we restrict V (t) to 0.3 ≤ t ≤ 0.5, the inverse
function has rule: t = −0.4 + x
2 − arcsin( y
− arcsin( ln(3 )−ln)(2) ) + 4π
ln(2 5π . Answer 19.8 (c) On domain t ≥ 0, 1 B = B(t) = 2 cos(6πt) + 2 4 6 8 36 − 4(sin(6πt))2 . t
(e) 0.1023, 0.2310. 1
2 Answer 20.1 (a1) 0, 1.5708, −1.5708, 1.0472, 0.7168, −0.2762,
not deﬁned. 301
Answer 20.2 (a) Principal solution: x = 1.9106, symmetry
solution: x = 4.3726; graph below with these two solutions
graphically indicated: y
2
1.5
1
0.5
10 5 5 10 Answer 20.9 (a)α = arcsin[3960/(3960 + t)]. (b) α =
6.696degs. The interior angle is 166.608 degs and so one
satellite covers 46% of the circumference. Thus you need
3 (not twopointsomething) satellites to cover the earth’s
circumference. (c)α = 52.976deg. The interior angle is
74.047 degs and so one satellite covers 20.57% of the circumference. Thus you need 5 satellites to cover the earth’s
circumference. (d) Get an equation for the interior angle
in terms of t. Solve 2(90 − arcsin[3960/(3960 + t)]) = 20% of
360 degs = 72 degs. You’ll get t = 934.83 miles. x 0.5
1
1.5 Answer 20.10 (a) Domain:
1 ≤ y ≤ 5. One solution: x = 1
3
1
3 1
− π ≤ x ≤ 3 + π ; Range:
6
6
π
+ 18 ; graph is below: 2
8
Answer 20.3 (a) 9.39, 13.63, 11.09. (b) Feb. 3, Nov. 4. 6 Answer 20.4 (a) A = 15, D = 415, B = 10, C = 15/2. Note
that C = 10k + 15/2, k = 0, ±1, ±2, . . . are also all valid
choices for the phase shift. (b) maximum temperature=
430o F. (c) minimum temperature= 400o F. (d) 12.1635
minutes. (e) 14.6456 minutes. (f) 6.80907 minutes.
Answer 20.5 The cake should be in the oven for
67.572141357 minutes. 4
2
2 1.5 1 0.5 0.5 1 1.5 2 x 2
Answer 20.6 6.42529 hours
Answer 20.7 8.9286 hours of dry time each day.
Answer 20.8 The key fact to use over and over
is this:
M(t)=M’s location after t seconds =
(100 cos(0.025t),100 sin(0.025t)); T (t)= T’s location after t
seconds = (100 cos(0.03t + π),100 sin(0.03t + π)). Answer 20.11 (a) 3.852624 miles (a) 14.34lb/in2 ; no explosion (b) 8.32lb/in2 ; explosion (c) 54.6 degrees Answer 20.12 (a) Many possible answers; for example:
−1.9293, −1.3677, 0.8677, 1.4293. 302 APPENDIX B. ANSWERS Appendix C
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Foundation. If the Document does not specify a version number of this
License, you may choose any version ever published (not as a draft) by
the Free Software Foundation. Index
xaxis, 11
xaxis,positive, 11
xycoordinate system, 11, 12
yaxis, 11
yaxis,positive, 12
yintercept, 39
nth root, 135
adjacent, 222
amplitude, 253
analogue LP’s, 213
angle, 192
angle,central, 192
angle,initial side, 192
angle,standard position, 192
angle,terminal side, 192
angle,vertex, 192
angular speed, 207
arc,length, 199
arc,subtended, 192
arccosine function, 272
arcsine function, 272
arctangent function, 272
area,sector, 199
aspect ratio, 13
axis scaling, 13
axis units, 14
axis,horizontal, 11
axis,vertical, 11
belt/wheel problems, 215
CD’s, 215
central angle, 192
chord, 202
circle, 25
circles, 26, 77
circles,circular function, 230 circles,great, 204
circles,point coordinates on, 230
circles,unit, 28
circular function, 229
circular function,triangles, 229
circular function, 191, 221, 226,
227
circular function,circles, 230
circular function,inverse, 267
circular function,special values, 223
circular motion, 209
compound interest, 146, 148
compounding periods, 146
continuous compounding, 151
converting units, 1
coordinates,imposing, 11
cosecant function, 232
cosine function, 222, 248
cotangent function, 232
curves,intersecting, 28
db, 160
decibel, 160
decreasing function, 76
degree, 194
degree method, 193
degree,minute, 194
degree,second, 194
density, 3
dependent variable, 59
difference quotient, 36
digital compact disc, 215
dilation, 170
dilation,horizontal, 173
dilation,vertical, 171
directed distance, 19
distance, directed, 19
distance,between two points, 17, 19
311 INDEX 312
domain, 58
e, 149, 150
envelope of hearing, 160
equation,quadratic, 45
equatorial plane, 202
even function, 241
exponential decay, 139
exponential function, 149
exponential growth, 139
exponential modeling, 145
exponential type, 139 graphing, 1
great circle, 202, 204
horizon circle, 282
horizontal line, 26
horizontal axis, 11 identity,composition, 272
identity,even/odd, 241
identity,key, 240
identity,periodicity, 240
imposing coordinates, 11, 15
independent variable, 59
interest, 146
function, 58
function,circular, 191, 221, 226, 227, intersecting curves, 28
intervals, 60
229
inverse circular function, 267, 270,
function,cosine, 222, 248
271
function,decreasing, 76
inverse function, 267
function,even, 241
function,exponential, 149
function,exponential type, 139
function,logarithm base b, 157
function,logarithmic, 153
function,multipart, 79
function,natural logarithm, 154
function,odd, 241
function,periodic, 240
function,picturing, 55, 57
function,rational, 181
function,sine, 222, 248
function,sinusoidal, 191, 247, 251
function,tangent, 222, 248
function,trigonometric, 247
function,cos(θ), 222
function,cos(x), 248
function,cos−1(z), 271
function,sin(θ), 222
function,sin(x), 248
function,sin−1 (z), 271
function,tan(theta), 222
function,tan(x), 248
function,tan−1(z), 271
graph, 1, 26, 27, 38
graph,circular function, 242
graph,sin(θ), 244 knot, 205
latitude, 202
line,horizontal, 26
line,vertical, 26
linear speed, 208
linear functions, 64
linear modeling, 33
lines, 33, 39
lines,horizontal, 25
lines,parallel, 44
lines,perpendicular, 44
lines,point slope formula, 38
lines,slope intercept formula, 39
lines,two point formula, 38
lines,vertical, 25
logarithm conversion formula, 158
logarithm function base b, 157
logarithmic function, 153
longitude, 203
loudness of sound, 159
LP’s, 213
mean, 252
meridian, 203
meridian,Greenwich, 203
modeling, 1 INDEX
modeling,exponential, 145
modeling,linear, 33
modeling,sinusoidal, 251
motion,circular, 209
mulitpart function, 80
multipart functions, 79
natural logarithm, 153
natural logarithm function, 154
natural logarithm function, properties, 154
nautical mile, 205
navigation, 202
odd function, 241
origin, 11
parabola,three points determine, 98
parametric equations, 47
period, 253
periodic, 240, 247
periodic rate, 146
phase shift, 252
piano frequency range, 140
picturing a function, 55, 57
positive xaxis, 11
positive yaxis, 12
principal, 146
principal domain, 271
principal domain, cosine, 271
principal domain, sine, 271
principal domain, tangent, 271
principal solution, 270
Pythagorean Theorem, 18
quadrants, 13
quadratic formula, 45
radian, 198
radian method, 196
range, 59
rate, 4, 39
rate of change, 4
rational function, 181
reﬂection, 166
restricted domain, 59
right triangles, 229 313
RPM, 208
rules of exponents, 135
scaling, 13
secant function, 232
sector,area, 199
semicircles, 77
shifting, 168
shifting,principle, 170
sign plot, 75
sine function, 222, 248
sinusoidal function, 247
sinusoidal function, 191, 251
sinusoidal modeling, 251
slope, 36
solve the triangle, 267
sound pressure level, 159
speed,angular, 207
speed,circular, 209
speed,linear, 208
standard position, 192
standard angle, 192
standard form, 27
tangent function, 222, 248
triangle,sides, 221
trigonometric function, 247
trigonometric ratios, 223
uniform linear motion, 47
unit circle, 28
units, 1
vertical axis, 11
vertical line test, 63
vertical lines, 26 ...
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