zotero-db/storage/CVGZ9NKA/.zotero-ft-cache

Quick Calculus
A Self-Teaching Guide
Third Edition
Daniel Kleppner
Lester Wolfe Professor of Physics Massachusetts Institute of Technology
Norman Ramsey
Higgins Professor of Physics Harvard University Nobel Prize for Physics 1989
Peter Dourmashkin
Senior Lecturer Massachusetts Institute of Technology

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COVER DESIGN: PAUL MCCARTHY

Preface
Quick Calculus is designed for you to learn the basic techniques of differential and integral calculus with a minimum of wasted effort, studying by yourself. It was created on a premise that is now widely accepted: in technical subjects such as calculus, students learn by doing rather than by listening. The book consists of a sequence of relatively short discussions, each followed by a problem whose solution is immediately available. One’s path through the book is directed by the responses. The text is aimed at newcomers to calculus, but additional topics are discussed in the final chapter for those who wish to go further.
The initial audience for Quick Calculus was composed of students entering college who did not wish to postpone physics for a semester in order to take a prerequisite in calculus. In reality, the level of calculus needed to start out in physics is not high and could readily be mastered by self-study.
The readership for Quick Calculus has grown far beyond novice physics students, encompassing people at every stage of their career. The fundamental reason is that calculus is empowering, providing the language for every physical science and for engineering, as well as tools that are crucial for economics, the social sciences, medicine, genetics, and public health, to name a few. Anyone who learns the basics of calculus will think about how things change and influence each other with a new perspective. We hope that Quick Calculus will provide a firm launching point for helping the reader to achieve this perspective.
Daniel Kleppner Peter Dourmashkin Cambridge, Massachusetts
iii

Contents

Preface

iii

Chapter One Starting Out

1

1.1 A Few Preliminaries

1

1.2 Functions

2

1.3 Graphs

5

1.4 Linear and Quadratic Functions

11

1.5 Angles and Their Measurements

19

1.6 Trigonometry

28

1.7 Exponentials and Logarithms

42

Summary of Chapter 1

51

Chapter Two Differential Calculus

57

2.1 The Limit of a Function

57

2.2 Velocity

71

2.3 Derivatives

83

2.4 Graphs of Functions and Their Derivatives

87

2.5 Differentiation

97

2.6 Some Rules for Differentiation

103

2.7 Differentiating Trigonometric Functions

114

2.8 Differentiating Logarithms and Exponentials

121

2.9 Higher-Order Derivatives

130

2.10 Maxima and Minima

134

2.11 Differentials

143

2.12 A Short Review and Some Problems

147

Conclusion to Chapter 2

164

Summary of Chapter 2

165

Chapter Three Integral Calculus

169

3.1 Antiderivative, Integration, and the Indefinite Integral

170

3.2 Some Techniques of Integration

174

v

vi Contents

3.3 Area Under a Curve and the Definite Integral

182

3.4 Some Applications of Integration

201

3.5 Multiple Integrals

211

Conclusion to Chapter 3

219

Summary of Chapter 3

219

Chapter Four Advanced Topics: Taylor Series, Numerical Integration, and

Differential Equations

223

4.1 Taylor Series

223

4.2 Numerical Integration

232

4.3 Differential Equations

235

4.4 Additional Problems for Chapter 4

244

Summary of Chapter 4

248

Conclusion (frame 449)

250

Appendix A

Derivations

251

A.1 Trigonometric Functions of Sums of Angles

251

A.2 Some Theorems on Limits

252

A.3 Exponential Function

254

A.4

dy Proof That =

1

255

dx dx∕dy

A.5 Differentiating xn

256

A.6 Differentiating Trigonometric Functions

258

A.7 Differentiating the Product of Two Functions

258

A.8 Chain Rule for Differentiating

259

A.9 Differentiating ln x

259

A.10 Differentials When Both Variables Depend on a Third

Variable

260

A.11 Proof That if Two Functions Have the Same Derivative They

Differ Only by a Constant

261

A.12 Limits Involving Trigonometric Functions

261

Appendix B Additional Topics in Differential Calculus

263

B.1 Implicit Differentiation

263

B.2 Differentiating the Inverse Trigonometric Functions

264

B.3 Partial Derivatives

267

B.4 Radial Acceleration in Circular Motion

269

B.5 Resources for Further Study

270

Frame Problems Answers

273

Answers to Selected Problems from the Text

273

Review Problems Chapter 1 Chapter 2 Chapter 3
Tables Table 1: Derivatives Table 2: Integrals
Indexes Index Index of Symbols

Contents vii
277 277 278 282
287 287 288
291 291 295

CHAPTER ONE
Starting Out

In spite of its formidable name, calculus is not a particularly difficult subject. The fundamental concepts of calculus are straightforward. Your appreciation of their value will grow as you develop the skills to use them.
After working through Quick Calculus you will be able to handle many problems and be prepared to acquire more elaborate techniques if you need them. The important word here is working, though we hope that you find that the work is enjoyable.
Quick Calculus comprises four chapters that consist of sections and subsections. We refer to the subsections as frames. Each chapter concludes with a summary. Following these chapters there are two appendixes on supplementary material and a collection of review problems with solutions.
The frames are numbered sequentially throughout the text. Working Quick Calculus involves studying the frames and doing the problems. You can check your answers immediately: they will be located at the bottom of one of the following pages or, if the solutions are longer, in a separate frame. Also a summary of frame problems answers start on page 273.
Your path through Quick Calculus will depend on your answers. The reward for a correct answer is to go on to new material. If you have difficulty, the solution will usually be explained and you may be directed to another problem.
Go on to frame 1.

1.1 A Few Preliminaries

1

Chapter 1 will review topics that are foundational for the discussions to come. These are:

• the definition of a mathematical function; • graphs of functions;

(continued)

1

2 Starting Out

Chap. 1

• the properties of the most widely used functions: linear and quadratic functions, trigonometric functions, exponentials, and logarithms.

A note about calculators: a few problems in Quick Calculus need a simple calculator. The calculator in a typical smartphone is more than adequate. If you do not happen to have access to a calculator, simply skip the problem: you can master the text without it.
Go on to frame 2.
2
Here is what’s ahead: this first chapter is a review, which will be useful later on; Chapter 2 is on differential calculus; and Chapter 3 introduces integral calculus. Chapter 4 presents some more advanced topics. At the end of each chapter there is a summary to help you review the material in that chapter. There are two appendixes—one gives proofs of a number of relations used in the book, and the other describes some supplementary topics. In addition, there is a list of extra problems with answers in the Review Problems on page 277, and a section of tables you may find useful.
First we review the definition of a function. If you are already familiar with this and with the idea of dependent and independent variables, skip to frame 14. (In fact, in this chapter there is ample opportunity for skipping if you already know the material. On the other hand, some of the material may be new to you, and a little time spent on review can be a good thing.)
A word of caution about the next few frames. Because we start with some definitions, the first section must be somewhat more formal than most other parts of the book.
Go on to frame 3.

1.2 Functions

3
The definition of a function makes use of the idea of a set. If you know what a set is, go to 4. If not, read on.
A set is a collection of objects—not necessarily material objects—described in such a way that we have no doubt as to whether a particular object does or does not belong to it. A set may be described by listing its elements. Example: 23, 7, 5, 10 is a set of numbers. Another example: Reykjavik, Ottawa, and Rome is a set of capitals.

§ 1.2

Functions 3

We can also describe a set by a rule, such as all the even positive integers (this set contains an infinite number of objects).
A particularly useful s√et is the set of all real numbers. This includes all numbers such as 5, −4, 0, 1/2, 𝜋, −3.482, 2. The set of real numbers does not include quantities involving the square root of negative numbers. Such quantities are called complex numbers; in this book we will be concerned only with real numbers.
The mathematical use of the word “set” is similar to the use of the same word in ordinary conversation, as “a set of chess pieces.”

Go to 4.

4
In the blank below, list the elements of the set that consists of all the odd integers between −10 and +10.

Elements: Go to 5 for the correct answer.
5 Here are the elements of the set of all odd integers between −10 and +10:

−9, −7, −3, −5, −1, 1, 3, 5, 7, 9.

Go to 6.

6
Now we are ready to talk about functions. Here is the definition. A function is a rule that assigns to each element in a set A one and only one element in a set B. The rule can be specified by a mathematical formula such as y = x2, or by tables of associated numbers, for instance, the temperature at each hour of the day. If x is one of the elements in set A, then the element in set B that the function f associates with x is denoted by the symbol f (x). This symbol f (x) is the value of f evaluated at the element x. It is usually read as “f of x.” The set A is called the domain of the function. The set of all possible values of f (x) as x varies over the domain is called the range of the function. The range of f need not be all of B.
(continued)

4 Starting Out

Chap. 1

In general, A and B need not be restricted to sets of real numbers. However, as mentioned in frame 3, in this book we will be concerned only with real numbers.

Go to 7. 7

For example, for the function f (x) = x2, with the domain being all real numbers, the range

is

.

Go to 8. 8

The range is all nonnegative real numbers. For an explanation, go to 9.

Otherwise, skip to 10. 9

Recall that the product of two negative numbers is positive. Thus for any real value of x positive or negative, x2 is positive. When x is 0, x2 is also 0. Therefore, the range of f (x) = x2
is all nonnegative numbers.

Go to 10. 10

Our chief interest will be in rules for evaluating functions defined by formulas. If the

domain is not specified, it will be understood that the domain is the set of all real numbers

for which the formula produces a real number, and for which it makes sense. For instance,

√

(a) f (x) = x Range =

.

(b) f (x) = 1

Range =

.

x

For the answers go to 11.

11 √
The function x is real for x nonnegative, so the answer to (a) is all nonnegative real numbers. The function 1∕x is defined for all values of x except zero, so the range in (b) is all real numbers except zero.

Go to 12.

§ 1.3

Graphs 5

12
When a function is defined by a formula such as f (x) = ax3 + b, x is called the independent variable and f (x) is called the dependent variable. One advantage of this notation is that the value of the dependent variable, say for x = 3, can be indicated by f (3).
Often, a single letter is used to represent the dependent variable, as in

y = f (x).

Here x is the independent variable, and y is the dependent variable.
Go to 13.
13
In mathematics the symbol x frequently represents an independent variable, f often represents the function, and y = f (x) usually denotes the dependent variable. However, any other symbols may be used for the function, the independent variable, and the dependent variable. For example, we might have z = H(r), which is read as “z equals H of r.” Here r is the independent variable, z is the dependent variable, and H is the function.
Now that we know what a function means, let’s describe a function with a graph instead of a formula.
Go to 14.

1.3 Graphs
14 If you know how to plot graphs of functions, skip to frame 19.
Otherwise, go to 15. 15
We start by constructing coordinate axes. In the most common cases we construct a pair of mutually perpendicular intersecting lines, one horizontal, the other vertical. The horizontal line is often called the x-axis and the vertical line the y-axis. The point of intersection is the origin, and the axes together are called the coordinate axes.
(continued)

6 Starting Out

y-axis
10

Chap. 1

5

0 –5 –4 –3 –2 –1
–5

12345

x-axis

–10
Next we select a convenient unit of length and, starting from the origin, mark off a number scale on the x-axis, positive to the right and negative to the left. In the same way, we mark off a scale along the y-axis with positive numbers going upward and negative downward. The scale of the y-axis does not need to be the same as that for the x-axis (as in the drawing). In fact, y and x can have different units, such as distance and time.
Go to 16.
16
We can represent one specific pair of values associated by the function in the following way: let a represent some particular value for the independent variable x, and let b indicate the corresponding value of y = f (x). Thus, b = f (a).
y-axis
a P
(a,b) b
x-axis
We now draw a line parallel to the y-axis at distance a from the y-axis and another line parallel to the x-axis at distance b from that axis. The point P at which these two lines intersect is designated by the pair of values (a, b) for x and y, respectively.

§ 1.3

Graphs 7

The number a is called the x-coordinate of P, and the number b is called the y-coordinate of P. (Sometimes the x-coordinate is called the abscissa, and the y-coordinate is called the ordinate.) In the designation of a typical point by the notation (a, b) we will always designate the x-coordinate first and the y-coordinate second.
As a review of this terminology, encircle the correct answers below. For the point (5, −3):

x-coordinate∶ [−5 | − 3 | 3 | 5]

y-coordinate∶ [−5 | − 3 | 3 | 5]

Go to 17. 17

The most direct way to plot the graph of a function y = f (x) is to make a table of reasonably spaced values of x and of the corresponding values of y = f (x). Then each pair of values (x, y) can be represented by a point as in the previous frame. A graph of the function is obtained by connecting the points with a smooth curve. Of course, the points on the curve may be only approximate. If we want an accurate plot, we just have to be very careful and use many points. (On the other hand, crude plots are pretty good for many purposes.)

Go to 18. 18

As an example, here is a plot of the function y = 3x2. A table of values of x and y is shown, and these points are indicated on the graph.

xy –3 27 –2 12 –1 3 00 13
2 12 3 27

y-axis 30
P
25
20 15
10
5 x-axis
–3 –2 –1 0 1 2 3

To test yourself, encircle the pair of coordinates that corresponds to the point P indicated

in the figure.

[(3, 27) | (27, 3) | none of these]

If incorrect, study frame 16 once again and then go to 19. If correct,

Go to 19.

8 Starting Out

Chap. 1

19

Here is a rather special function. It is called a constant function and assigns a single fixed number c to every value of the independent variable, x. Hence, f (x) = c.

y-axis

5 4
(0,3)
3
2 1
x-axis
–5 –4 –3 –2 –1 0 1 2 3 4 5

This is a peculiar function because the value of the dependent variable is the same for all values of the independent variable. Nevertheless, the relation f (x) = c assigns exactly one value of f (x) to each value of x as required in the definition of a function. All the values of f (x) happen to be the same.
Try to convince yourself that the graph of the constant function y = f (x) = 3 is a straight line parallel to the x-axis passing through the point (0,3) as shown in the figure.
Go to 20.
20
Another special function is the absolute value function. The absolute value of x is indicated by the symbol ∣ x ∣. The absolute value of a number x determines the size, or magnitude, of the number without regard to its sign. For example,
∣ −3 ∣ = ∣ 3 ∣ = 3

Answers: Frame 16: 5, −3 Frame 18: (3, 27)

§ 1.3

Graphs 9

Now we will define ∣ x ∣ in a general way. But first we need to recall the inequality symbols:

a > b means a is greater than b. a ≥ b means a is greater than or equal to b. a < b means a is less than b. a ≤ b means a is less than or equal to b.

With this notation we can define the absolute value function, ∣ x ∣, by the following two

rules:

{

∣x∣=

x −x

if x ≥ 0, if x < 0.

Go to 21. 21
A good way to show the behavior of a function is to plot its graph. Therefore, as an exercise, plot a graph of the function y = ∣ x ∣ in the accompanying figure.

y-axis

5 4 3 2 1
–5 –4 –3 –2 –1 0 1 2 3 4 5

x-axis

To check your answer, go to 22.

10 Starting Out 22
The graph for ∣ x ∣ is

y-axis

Chap. 1

5 4 3 2 1
–5 –4 –3 –2 –1 0 1 2 3 4 5

x-axis

This can be seen by preparing a table of x and y values as follows:

x y=∣x∣

−4

+4

−2

+2

0

0

+2

+2

+4

+4

These points may be plotted as in frames 16 and 18 and the lines drawn with the results in the above figure.
The graph and x, y coordinates described here are known as a Cartesian coordinate system. There are other coordinate systems better suited to other geometries, such as cylindrical or spherical coordinate systems, but Cartesian coordinates are the best known.
With this introduction on functions and graphs, we are now going to familiarize ourselves with some important elementary functions.
These functions are the linear, quadratic, trigonometric, exponential, and logarithmic functions.

Go to 23.

§ 1.4

Linear and Quadratic Functions 11

1.4 Linear and Quadratic Functions

23
A function defined by an equation in the form y = mx + b, where m and b are constants, is called a linear function because its graph is a straight line. This is a simple and useful function, and you need to become familiar with it.
Here is an example: Encircle the letter that identifies the graph (as labeled in the figure) of
y = 3x − 3. [A | B | C]

y-axis
5 4 3 2 1

B A

0 –5 –4 –3 –2 –1–1
–2 –3 –4 –5

12345
C

x-axis

If you missed this or if you do not feel entirely sure of the answer, go to 24. Otherwise, go to 25.

12 Starting Out 24
The table below gives a few values of x and y for the function y = 3x − 3.

xy –2 –9 –1 –6
0 –3 10 23

y-axis 5

4

3

2

1

–5 –4

–3 –2

0 ––11

x-axis 12345

–2

–3

–4

–5

Chap. 1

A few of these points are shown on the graph, and a straight line has been drawn through them. This is line B of the figure in frame 23.
Go to 25. 25
Here is the graph of a typical linear function. Let us take any two different points on the line, (x2, y2) and (x1, y1). We define the slope of the line in the following way:

slope = y2 − y1 . x2 − x1

y-axis (x1, y1)

(x2, y2)

y2 – y1

x2 – x1

x-axis

Answer: Frame 23: B

§ 1.4

Linear and Quadratic Functions 13

The idea of slope will be important in our later work, so let’s spend a little time learning more about it.
Go to 26.
26
If the x and y scales are the same, as in the figure, then the slope is the ratio of the vertical distance y2 − y1 to the horizontal distance x2 − x1 as we go from the point (x1, y1) on the line to (x2, y2). If the line is vertical, the slope is infinite (or, more strictly, undefined). Test for yourself that the slope is the same for any pair of two separate points on the line.

y-axis
5 4

(x2, y2)

vertical distance

3

= y2 – y1

2

(x1, y1)

horizontal distance

1

= x2 – x1

0

x-axis

12345

Go to 27. 27
If the vertical and horizontal scales are not the same, the slope is still defined by
slope = vertical distance , horizontal distance
but now the distance is measured using the appropriate scale. For instance, the two figures below may look similar, but the slopes are quite different. In the first figure the x and y scales are identical, and the slope is 1/2. In the second figure the y scale has been changed by a factor of 100, and the slope is 50.

y-axis 10 5
05

5
10
x-axis 10 15 20

y-axis 1000 500
05

500
10
x-axis 10 15 20

(continued)

14 Starting Out

Chap. 1

Because the slope is the ratio of two lengths, the slope is a pure number if the lengths are pure numbers. However, if the variables have different dimensions, the slope will also have a dimension.
Below is a plot of the distance traveled by a car vs. the amount of gasoline consumed.

Distance (miles)

100 80 60 40 20 0
0

½ 1 1½ 2 Gasoline (gallons)

Here the slope has the units of miles per gallon (mpg). What is the slope of the line shown?
Slope = [20 | 40 | 60 | 80] mpg
If right, go to 29. Otherwise, go to 28. 28 To evaluate the slope, let us find the coordinates of any two points on the line.

Distance (miles)

100 80 60 40 20 0
0

A
B ½ 1 1½ 2 Gasoline (gallons)

For instance, A has the coordinates (2 gallons, 80 miles) and B has the coordinates ( 1/2 gallon, 20 miles). Therefore, the slope is

(80 − 20) miles = 40 miles = 40 mpg.

(2 − 1∕2) gallon

gallon

We would have obtained the same value for the slope no matter which two points we used, because two points determine a straight line.
Go to 29.

§ 1.4

Linear and Quadratic Functions 15

29 If the line is described by an equation of the form y = mx + b, then the slope is given by slope = y2 − y1 . x2 − x1 Substituting in the above expression for y, we have

slope = (mx2 + b) − (mx1 + b) = mx2 − mx1 = m(x2 − x1) = m.

x2 − x1

x2 − x1

x2 − x1

What is the slope of y = 7x − 5?

[

]

5∕7 | 7∕5 | − 5 | − 7 | 5 | 7

If right, go to 31. Otherwise, go to 30.
30
The equation y = 7x − 5 can be written in the form y = mx + b if m = 7 and b = −5. Because slope = m, the line given has a slope of 7.

Go to 31. 31

The slope of a line can be positive (greater than 0), negative (less than 0), or 0. An example of each is shown graphically below.

y

y

y

x

x

x

Positive slope Figure 1

Negative slope Figure 2

Zero slope Figure 3

Note how a line with positive slope rises in going from left to right, a line with negative

slope falls in going from left to right, and a line of zero slope is horizontal. (It was pointed

out in frame 26 that the slope of a vertical line is not defined.)

(continued)

16 Starting Out

Chap. 1

Indicate whether the slope of the graph of each of the following equations is positive, negative, or zero by encircling your choice.

Equation

Slope

1. y = 2x – 5 { + | − | 0 }

2. y = −3x

{+|−|0}

3. p = q − 2 { + | − | 0 }

4. y = 4

{+|−|0}

The answer is in the next frame. 32
Here are the answers to the questions in frame 31. In frame 29 we saw that for a linear equation in the form y = mx + b the slope is m.

1. y = 2x − 5. Here m = 2 and the slope is 2. Clearly this is a positive number. See Figure 1 below.

2. y = −3x. Here m = −3. The slope is −3, which is negative. See Figure 2 below.

3. p = q − 2. In this equation the variables are p and q, rather than y and x. Written in the form p = mq + b, it is evident that m = 1, which is positive. See Figure 3 below.

4. y = 4. This is an example of a constant function. Here m = 0, b = 4, and the slope is 0. See Figure 4 below.

y

y

+5

+5

p

y

+5

+5

x

–5

+5 –5

x +5 –5

q

x

+5 –5

+5

–5
Positive slope y = 2x – 5 Figure 1

–5
Negative slope y = –3x Figure 2

–5
Positive slope p = q – 2 Figure 3

–5
Zero slope y = 4
Figure 4

Go to 33.

Answers: Frame 27: 40 mpg Frame 29: 7

§ 1.4

Linear and Quadratic Functions 17

33

Here is a linear equation in which the slope has a familiar meaning. The graph below shows the position S of a car on a straight road at different times. The position S = 0 means the car is at the starting point.

60

S (miles)

40

20

0
0 ½ 1 1½ 2 t (hours)

Try to guess the correct word to fill in the blank below:

The slope of the line has the same value as the car’s

.

Go to 34. 34

The slope of the line has the same value as the car’s speed (or, for this one-dimensional motion velocity).
The slope is the ratio of the distance traveled to the time required. But, by definition, the speed is also the distance traveled divided by the time. Thus the value of the slope of the line is equal to the speed.
Go to 35.
35

Now let’s look at another type of equation. An equation in the form y = ax2 + bx + c, where a, b, and c are constants (a ≠ 0), is called a quadratic function and its graph is called a parabola. Two typical parabolas are shown in the figure.

y-axis

y-axis

x1

x2 x-axis

x-axis

Go to 36.

18 Starting Out

Chap. 1

36
Roots of an Equation:
The values of x for which f (x) = 0 are called the roots of the equation. The values at y = 0, shown by x1 and x2 in the figure on the left in frame 35, correspond to values of x which satisfy ax2 + bx + c = 0 and are thus the roots of the equation. Not all quadratic equations have real roots. For example, the curve on the right represents an equation with no real value of x when y = 0.
Although you will not need to find the roots of any quadratic equation later in this book, you may want to know the formula anyway. If you would like to see a discussion of this, go to frame 37.

Otherwise, skip to frame 39.

37

The equation ax2 + bx + c = 0 has two roots. These are given by

√

√

−b + x1 =

b2 − 4ac , 2a

−b − x2 =

b2 − 4ac . 2a

The subscripts 1 and 2 serve merely to identify the roots. The two roots can be summarized

by a single equation:

√ x = −b ± b2 − 4ac .
2a

We will not prove these results, though they can be checked by substituting the values for x
in the original equation.
Here is a practice problem on finding roots: Which answer correctly gives the roots of 3x − 2x2 = 1?
( √) ( √) (a) 1∕4 3 + 17 ; 1∕4 3 − 17

(b) −1; − 1/2 (c) 1/4; − 1/4 (d) 1; 1/2

Answers: Frame 31: +, −, +, 0

§ 1.5

Angles and Their Measurements 19

Encircle the letter of the correct answer.
[a | b | c | d]
If you got the right answer, go to 39. The answer is in the following frame. 38
Here is the solution to the problem in frame 37. The equation 3x − 2x2 = 1 can be written in the standard form
2x2 − 3x + 1 = 0.

Here a = 2, b = −3, c = 1.

x=

1

[√

]

−b ± b2 − 4ac =

1

[ −(−3)

±

√ (−3)2

−

] (4)(2)(1)

2a

4

= 1 (3 ± 1). 4

x1

=

1 (3 4

+

1)

=

1.

x2

=

1 (3 4

−

1)

=

1. 2

39

Go to 39.

This ends our brief discussion of linear and quadratic functions. Perhaps you would like some more practice on these topics before continuing. If so, try working Review Problems 1–5 on page 277. At the end of this chapter there is a concise summary of the material we have had so far, which you may find useful.

Whenever you are ready, go to 40.

1.5 Angles and Their Measurements
40 Elementary features of rotations and angles:
If you are already familiar with rotations, angle, and degrees and radians, you can jump to frame 50.
(continued)

20 Starting Out

Chap. 1

B

The concept of the angle is the bedrock of trigonometry. Although

the general idea of an angle is probably familiar, it is important to agree

on the conventions and units for describing angles. For two straight-line

θ

segments OA and OB that intersect at a point O, the angle between them

O

A is a measure of how far the line segment OA must be rotated about the

point O to coincide with the line segment OB.

If the two segments initially coincide, for instance, half a revolution of either segment will

leave them pointing in opposite directions and a full revolution will bring them back to their

original positions.

The Greek letter 𝜃 (theta) symbolizes the rotation angle. The sense of rotation is shown

by the curved arrow. We will follow the convention that if the segment OA is rotated in the

counterclockwise direction to coincide with the segment OB, then the rotation is positive.

Conversely, a rotation in the clockwise direction is negative. The direction can be indicated

by a small arrowhead on the curve between the two segments. If the sense of rotation is

unimportant, the arrowhead is usually omitted.

Measuring the size of rotations:
There are two conventions used for measuring the size of rotation. The first divides one revolution into 360 parts called degrees (not to be confused with degrees of temperature). The number 360 has the attraction of being large—providing good angular resolution—and possessing many divisors. The symbol for a degree of rotation is ∘; hence, a quarter turn is 90∘. The degree can be subdivided into 60 minutes (60′), and the minute subdivided into 60 seconds (60′′). Until recent years degrees, minutes, and seconds (DMS coordinates) were commonly used in map-making and navigation. With advent of GPS and laser metrology, the convention for subdividing the degree has been changed: instead of minutes and seconds, the convention is to express the fraction of a degree by a decimal number, typically with four digits.

Location on Earth: latitude and longitude:
Two numbers are needed to describe positions on a surface. Because the Earth is spherical, Cartesian coordinates are not useful for specifying a position. Instead, a system based on coordinates known as latitude and longitude is employed. One of the coordinates is based on the idea of a meridian. This is an imaginary half circle on the Earth that connects the Earth’s

Answers: Frame 37: d

§ 1.5

Angles and Their Measurements 21

South and North Poles. The prime meridian is the half circle passing through Earth’s South and North Poles and Greenwich, England. Longitude is the angle between the prime meridian and the meridian through the point of interest. Points east of the prime meridian require a clockwise rotation around the polar axis and by convention are positive; those to the west are negative. The sign of longitude reverses when passing the meridian 180 degrees east or west of the prime meridian.

North Pole

Times Square

Latitude line

60° N

60° W 30° W

40° N 20° N

Greenwich
60° E 30° E

Longitude line

Equator

0° 20° S

Prime meridian

N

W

E

S

Instead of using positive and negative values of longitude, the convention is to assign east (E) for positive values and west (W) for negative values.
Latitude lines are parallel to the equator. Latitudes north of the equator use the symbol N and latitudes south of the equator use the symbol S. For example, the latitude and longitude of Times Square in New York City are 40.7580∘N and 73.9855∘W. In old-fashioned DMS units these are 77∘ 2′ 7.008′′W and 38∘ 53′ 22.1424′′N.
Go to 41.

22 Starting Out

Chap. 1

41 As we have seen, the full circle contains 360∘, and so it follows that a semicircle contains
180∘.

B

r θ

O

A

Which of the following angles is equal to the angle 𝜃 shown in the figure? [25∘ | 45∘ | 90∘ | 180∘]

42 To find the angle 𝜃, let’s first look at a related example.

If right, go to 43. Otherwise, go to 42.

90°

The angle shown is a right angle, constructed from two perpendicular lines. (The symbol

between sides indicates a right angle.) Because there are four right angles in a full revolution,

it is apparent that the angle equals

360∘ = 90∘. 4

The angle 𝜃 shown in frame 41 is just half as big as the right angle; thus it is 45∘. Here is a circle divided into equal segments by three straight lines.

§ 1.5

Angles and Their Measurements 23

a b c

Which of the angles in the figure equals 240∘?
[a | b | c]
Go to 43. 43
The second convention for measuring the size of rotations is the radian. The symbol for the radian is rad.
B

r

s

O

θ

A

To find the value of an angle in radians, we draw a circle of radius r, about the vertex, O, of the angle so that it intersects the sides of the angle at two points, shown in the figure as A and B. The length of the arc between A and B is designated by s. Then,

𝜃 (in radians) = s = length of arc .

r

radius

Radians are used widely in scientific applications. For example, to calculate numerical values of trigonometric functions in this chapter naturally calls for radians. If no unit is given for the numerical value of an angle, the angle is in radians.
(continued)

24 Starting Out

Chap. 1

To see whether you have caught on, answer this question: There are 360 degrees in a circle; how many radians are there?

[1 | 2 | 𝜋 | 2𝜋 | 360∕𝜋]

If right, go to 45. Otherwise, go to 44.
44

s = 2πr r

s = r rθ

θ

The circumference of a circle is 𝜋d or 2𝜋r, where d is the diameter and r is the radius. The length of an arc going completely around a circle is the circumference, 2𝜋r, so the angle

enclosed is 2𝜋r∕r = 2𝜋 radians, as shown in the figure on the left. In the figure on the right the angle 𝜃 subtends an arc s = r.
Encircle the answer, which gives 𝜃.

[

]

1 rad | 1∕4 rad | 1∕2 rad | 𝜋 rad | none of these

Go to 45.

45

Because 2𝜋 rad = 360∘, the rule for converting angles from degrees to radians is

1

rad

=

360∘ 2𝜋

.

Answers: Frame 41: 45∘ Frame 42: c

§ 1.5

Angles and Their Measurements 25

Conversely,

1∘ = 2𝜋 rad . 360

Try the following problems.

60∘ = [2𝜋∕3 | 𝜋∕3 | 𝜋∕4 | 𝜋∕6] rad

𝜋∕4

=

[ 22

1∕2∘

|

45∘

|

60∘

|

90∘]

Which angle is closest to 1 rad? (Remember that 𝜋 = 3.14 … ) [30∘ | 45∘ | 60∘ | 90∘]

If correct, go to 47. If you made any mistake, go to 46.

46

Here are the solutions to the problems in frame 45. From the formulas in frame 45, one obtains

60∘ = 60 × 2𝜋 rad = 2𝜋 rad = 𝜋 rad.

360

63

𝜋 rad 4

=

𝜋 4

×

360∘ 2𝜋

=

360∘ 8

=

45∘.

Because 2𝜋 is just a little greater than 6, 1 rad is slightly less than 360∘/6 = 60∘. (A closer approximation to the radian is 57.3∘.) The figure below shows all the angles in this question.

R
90° 1 rad 60° .30° 45° R

Go to 47.

26 Starting Out 47
In the circle shown, CG is perpendicular to AE and
C D
B

Chap. 1

arc AB = arc BC = arc AH, E

O

A

arc AD = arc DF = arc FA.

H

F

G

(Arc AB means the length of the arc along the circle between A and B, going the shortest way.)
We will designate angles by three letters. For example, ∕ AOB (read as “angle AOB”) designates the angle between OA and OB.
Try the following:
∕ AOD = {60∘ | 90∘ | 120∘ | 150∘ | 180∘} ∕ FOH = {15∘ | 30∘ | 45∘ | 60∘ | 75∘ | 90 degrees}
∕ HOB = {1∕4 | 1 | 𝜋∕2 | 𝜋∕4 | 𝜋∕8}

If you did all these correctly, go to 49. If you made any mistakes, go to 48.
48 Because arc AD = arc DF = arc FA, and the sum of their angles is 360∘, ∕ AOD =
360∘∕3 = 120∘.
∕ FOA = 120∘, ∕ GOA = 90∘, ∕ GOH = 45∘.

Answers: Frame 43: 2𝜋 Frame 44: 1 rad Frame 45: 𝜋∕3, 45∘, 60∘

§ 1.5 Thus

Angles and Their Measurements 27
∕ FOH = ∕ FOG + ∕ GOH = 30∘ + 45∘ = 75∘. ∕ HOB = ∕ HOA + ∕ AOB = 45∘ + 45∘ = 90∘.

Now try the following:
90∘ = [2𝜋 | 𝜋∕6 | 𝜋∕2 | 𝜋∕8 | 1∕4] 3𝜋 = [240∘ | 360∘ | 540∘ | 720∘] 𝜋∕6 = [15∘ | 30∘ | 45∘ | 60∘ | 90∘ | 120∘]
Go to 49. 49
Rotations can be clockwise or counterclockwise. By choosing a convention for the sign of an angle, we can indicate which direction is meant. As previously explained, an angle formed by rotating in a counterclockwise direction is positive; an angle formed by moving in a clockwise direction is negative.
Here is a circle of radius r drawn with x- and y-axes, as shown:
y-axis

r

θA

θB

x-axis

We will usually choose the positive x-axis as the initial side and, in general, we will measure angles from the initial to the final or terminal side, denoted by the curved arrow. For example, the angle 𝜃A measured in the counterclockwise direction is positive and 𝜃B is negative, as shown in the figure. If there is no curved arrow associated with the angle, then we shall assume that the angle is positive.
Go to 50.

28 Starting Out

Chap. 1

1.6 Trigonometry
50
If you are not familiar with trigonometric functions, proceed with this frame. Otherwise, check yourself with frame 51, or go right to frame 52.
Our next task is to introduce the trigonometric functions. These functions relate the various sides and angles of triangles.
Do you know the general definitions of the trigonometric functions of angle 𝜃? If you do, test yourself with the quiz below. If you don’t, go right on to frame 51.
The trigonometric function√s of 𝜃 can be expressed in terms of the coordinates x and y and the radius of the circle, r = x2 + y2.

y-axis

(x, y)

r

θ

x-axis

These are shown in the figure. Try to fill in the blanks (the answers are in frame 51):

sin 𝜃 = cos 𝜃 = tan 𝜃 =

csc 𝜃 = sec 𝜃 = cot 𝜃 =

Go to frame 51 to check your answers.

Answers: Frame 47: 120∘, 75∘, 𝜋/2 Frame 48: 𝜋∕2, 540∘, 30∘

§ 1.6 51
Here are the definitions of the trigonometric functions:

Trigonometry 29

sine∶

sin 𝜃 = y , r

cotangent∶

csc 𝜃

=

1 sin 𝜃

=

r, y

cosine∶ tangent∶

cos 𝜃 = x , r
tan 𝜃 = y , x

secant∶

sec 𝜃

=

1 cos 𝜃

=

r, x

cosecant∶

cot 𝜃

=

1 tan 𝜃

=

x. y

Notice that the definitions in the right-hand equations are the reciprocal of those on the left.

y-axis

(x, y)

yr

θ

x

x-axis

√ For the angle shown in the figure, x is negative and y is positive (r = x2 + y2 and is always positive) so that cos 𝜃, tan 𝜃, cot 𝜃, and sec 𝜃 are negative, while sin 𝜃 and csc 𝜃 are positive.
Go to 52.

30 Starting Out

Chap. 1

52

Below is a circle with a radius of 5. The point shown is (−3, −4). On the basis of the

definition in the last frame, you should be able to answer the following:

[

]

sin 𝜃 = 3∕5 | 5∕3 | 3∕4 | − 4∕5 | − 3∕5 | 4∕3

[

]

cos 𝜃 = 3∕5 | 5∕3 | 3∕4 | − 4∕5 | − 3∕5 | 4∕3

[

]

tan 𝜃 = 3∕5 | 5∕3 | 3∕4 | − 4∕5 | − 3∕5 | 4∕3

y-axis

–3

θ

x-axis

(–3,–4)

–4

If all right, go to 55. Otherwise, go to 53. 53
Perhaps you had difficulty because you did not realize that x and y have different signs in different quadrants (quarters of the circle) while r, a radius, is always positive. Try this problem.

θ

θ

θ

A

B

C

Indicate whether the function specified is positive or negative, for each of the figures, by checking the correct box.

sin 𝜃 cos 𝜃 tan 𝜃

Figure A +−

Figure B +−

Figure C +−

See frame 54 for the correct answers.

§ 1.6 54
Here are the answers to the questions in frame 53.

Trigonometry 31

Figure A

Figure B

Figure C

+−

+−

+−

sin 𝜃 ✓

✓

✓

cos 𝜃 ✓

✓

✓

tan 𝜃 ✓

✓

✓

Go to 55.

55

In the figure both 𝜃 and −𝜃 are shown. The trigonometric functions for these two angles are simply related.

y-axis

θ x-axis
–θ

Can you do these problems? Encircle the correct sign. sin(−𝜃) = [+ | −] sin 𝜃 cos(−𝜃) = [+ | −] cos 𝜃 tan(−𝜃) = [+ | −] tan 𝜃

56 There are many relationships among the trigonometric functions.
y-axis

x

θ

x-axis

y r

Go to 56.
(continued)

32 Starting Out

For instance, using r2 = x2 + y2, we have

sin2𝜃 = y2 = r2 − x2 = 1 − ( x )2 = 1 − cos2𝜃.

r2

r2

r

Try these:

Chap. 1

1. sin2𝜃 + cos2 = {sec2𝜃 | 1 | tan2𝜃 | cot2𝜃} 2. 1 + tan2𝜃 = {1 | tan2𝜃 | cot2𝜃 | sec2𝜃} 3. sin2𝜃 − cos2𝜃 = {1 − 2cos2𝜃 | 1 − 2sin2𝜃 | cot2𝜃 | 1}

57 Here are the solutions to the problems in frame 56.

1. sin2𝜃 + cos2𝜃 = y2 + x2 = x2 + y2 = r2 = 1.

r2

r2

r2

r2

This is an important identity, which is worth remembering.

The other solutions are

2.

1 + tan2𝜃

=

1+

sin2𝜃 cos2𝜃

=

cos2𝜃 + sin2𝜃 cos2𝜃

=

1 cos2𝜃

=

sec2𝜃.

3. sin2𝜃 − cos2 = (1 − cos2𝜃) − cos2𝜃 = 1 − 2cos2𝜃.

If any mistakes, go to 57. Otherwise, go to 58.

58
y-axis

Go to 58.

ry θ
x

x-axis

ca θ
b

Answers: Frame 52: − 4∕5, − 3∕5, 4∕3 Frame 55: −, +, −

§ 1.6

Trigonometry 33

The trigonometric functions are particularly useful when applied to right triangles (triangles with one 90∘, or right angle). In this case 𝜃 is always acute (less than 90∘, or 𝜋∕2). You can then write the trigonometric functions in terms of the sides a and b of the right triangle shown, and its hypotenuse c. Fill in the blanks.

sin 𝜃 = cos 𝜃 = tan 𝜃 =

csc 𝜃 = sec 𝜃 = cot 𝜃 =

Check your answer in 59. 59

c a

θ b

The answers are:

sin 𝜃 = a = opposite side , csc 𝜃 = c = hypotenuse ,

c hypotenuse

a opposite side

cos 𝜃 = b = adjacent side , sec 𝜃 = c = hypotenuse ,

c hypotenuse

b adjacent side

tan 𝜃 = a = opposite side , cot 𝜃 = b = adjacent side .

b adjacent side

a opposite side

These results follow from the definitions in frame 51, providing we let a, b, and c correspond to y, x, and r, respectively. (Remember that here 𝜃 is less than 90∘.) If you are not familiar with the terms opposite side, adjacent side, and hypotenuse, they should be evident from the figure.

Go to 60.

34 Starting Out

Chap. 1

60

a

b

ϕ

θ

c

The following problems refer to the figure shown. (𝜙 is the Greek letter “phi.”) sin 𝜃 = [b∕c | a∕c | c∕a | c∕b | b∕a | a∕b]

tan 𝜙 = [b∕c | a∕c | c∕a | c∕b | b∕a | a∕b]

If all right, go to 62. Otherwise, go to 61. 61

You may have become confused because the triangle was drawn in a new position. Review the definitions in 51, and then do the following problems:

cos 𝜃 = [l∕n | n∕l | m∕n | m∕l | n∕m | l∕m] cot 𝜙 = [l∕n | n∕l | m∕n | m∕l | n∕m | l∕m]

n

θ

ϕ

m

l

If you missed either of these, you will have to put in more work learning and memorizing the definitions.
Meanwhile go to 62. 62
It is helpful to be familiar with the trigonometric functions of 30∘,45∘, and 60∘. The triangles for these angles are particularly simple.

2 45° 1 45°
1

2
30° 3

60° 1

Answer: Frame 56: 1, sec2𝜃, 1 − 2cos2𝜃

§ 1.6

Trigonometry 35

Try these problems:

cos

45∘

=

[1∕2

|

√ 1∕ 2

|

√ 22

|

2]

sin

30∘

=

[3

|

√ 3∕2

|

2∕3

|

1∕2]

sin

45∘

=

[1∕2

|

√ 1∕ 2

|

√ 2∕2

|

2]

tan

30∘

=

[1

|

√ 3

|

√ 1∕ 3

|

2]

Make sure you understand these problems. Then go to 63.

63
Many calculators provide values of trigonometric functions. With such a calculator, it is simple to plot enough points to make a good graph of the function. If you have a calculator, plot sin 𝜃 for values between 0∘ and 360∘ on the coordinate axes below, and then compare your result with frame 64. If you do not have a suitable calculator, go directly to 64 and check that sin 𝜃 has the correct values for the angles you know.

Go to 64.

36 Starting Out 64
Here is the graph of the sine function.

Chap. 1

65

1

1

1

0

π

2π

0

π

2π

0

π

2π

–1

–1

–1

(a)

(b)

(c)

1

1

1

0

π

2π

0

π

2π

0

π

2π

–1

–1

–1

(d)

(e)

(f)

Answers: Frame 60: a/c, b/a

Frame 61: m/√n, l/m

√√

Frame 62: 1∕ 2, 1/2, 1∕ 2, 1∕ 3

Go to 65.

§ 1.6

Trigonometry 37

Try to decide which graph represents each function.

cos 𝜃∶ [a | b | c | d | e | f | none of these] tan 𝜃∶ [a | b | c | d | e | f | none of these] sin(−𝜃)∶ [a | b | c | d | e | f | none of these] tan(−𝜃)∶[a | b | c | d | e | f | none of these]

If you got these all right, go to 67. Otherwise go to 66.

66

Knowing the values of the trigonometric functions at a few important points will help you identify them. Try these (∞ is the symbol for infinity, here meaning that the function is undefined):

sin 0∘ = [0 | 1 | − 1 | − ∞ | +∞]

cos 0∘ = [0 | 1 | − 1 | − ∞ | +∞]

cos

30∘

=

[1

|

1∕2

|

√ 3

|

√ 3∕2]

tan 45∘ = [0 | 1 | − 1 | − ∞ | +∞]

cos

60∘

=

[1

|

1∕2

|

√ 3

|

√ 3∕2]

sin 90∘ = [0 | 1 | − 1 | − ∞ | +∞]

cos 90∘ = [0 | 1 | − 1 | − ∞ | +∞]

Go to 67.
67
Because the angle 𝜃 + 2𝜋n, where n is any integer, is equivalent to 𝜃 as far as the trigonometric functions are concerned (i.e. for any trig function f , f (𝜃 + 2𝜋n) = f (𝜃)), we can add 2𝜋n to any angle without changing the value of the trigonometric functions. Thus, the sine and cosine (as well as its reciprocals, csc and sec) functions repeat their values whenever 𝜃 increases by 2𝜋n where n is an integer; we say that these functions are periodic in 𝜃 with a fundamental period of 2𝜋, or 360∘. (The fundamental period of the tangent and the cotangent is π.)
(continued)

38 Starting Out

Chap. 1

θ θ+2π

Using this property, you can extend the graph of sin 𝜃 in frame 64 to the following. (For variety, the angle here is in radians.)
sinθ 1

0

θ

π

π 3π

2π 5π 3π 7π 4π 9π 5π

2

2

2

2

2

–1

Go to 68. 68
It is helpful to know the sine and cosine of the sum and the difference of two angles.

ϕ
θ + ϕ θ

Answers:

Frame Frame

65: 66:

b, c, d, none of these; sin 0∘ = 0, cos 0∘ = 1,

cos 30∘

=

√ 3∕2,

tan 45∘

=

1,

cos 60∘ = 1∕2, sin 90∘ = 1, cos 90∘ = 0.

§ 1.6

Trigonometry 39

Do you happen to remember the formulas from previous studies of trigonometry? If not, go to 69. If you do, try the quiz below.

sin(𝜃 + 𝜙) =

.

cos(𝜃 + 𝜙) =

.

Go to 69 to see the correct answer. 69
Here are the formulas. They are derived in Appendix A1. sin(𝜃 + 𝜙) = sin 𝜃 cos 𝜙 + cos 𝜃 sin 𝜙, cos(𝜃 + 𝜙) = cos 𝜃 cos 𝜙 − sin 𝜃 sin 𝜙.

These formulas hold for both positive and negative values of the angles. (Note that tan(𝜃 + 𝜙) and cot(𝜃 + 𝜙) can be obtained from these formulas and the relation tan 𝜃 = sin 𝜃∕ cos 𝜃.
By using what you have already learned, circle the correct sign in each of the following:
(a) sin(𝜃 − 𝜙) = {+ | −} sin 𝜃 cos 𝜙 {+ | −} cos 𝜃 sin 𝜙 (b) cos(𝜃 − 𝜙) = {+ | −} cos 𝜃 cos 𝜙 {+ | −} sin 𝜃 sin 𝜙
If right, go to 71. If wrong, go to 70. 70
If you made a mistake in problem 69, recall from frame 55 that
sin(−𝜙) = − sin 𝜙, cos(−𝜙) = + cos 𝜙.

Then

sin(𝜃 − 𝜙) = sin 𝜃 cos(−𝜙) + cos 𝜃 sin(−𝜙) = sin 𝜃 cos 𝜙 − cos 𝜃 sin 𝜙,
cos(𝜃 − 𝜙) = cos 𝜃 cos(−𝜙) − sin 𝜃 sin(−𝜙) = cos 𝜃 cos 𝜙 + sin 𝜃 sin 𝜙.

Go to 71.

40 Starting Out

Chap. 1

71
By using the expressions for sin(𝜃 + 𝜙) and cos(𝜃 + 𝜙), one can obtain the formulas for sin(2𝜃) and cos(2𝜃). Simply let 𝜃 = 𝜙. Fill in the blanks.

sin 2𝜃 =

.

cos 2𝜃 =

.

72 Here are the answers:

See 72 for the correct answers.

sin(2𝜃) = 2 sin 𝜃 cos 𝜃, cos(2𝜃) = cos2𝜃 − sin2𝜃
= 1 − 2sin2𝜃 = 2cos2𝜃 − 1.
(Note, by convention, (sin 𝜃)2 is usually written sin2𝜃, and (cos 𝜃)2is usually written cos2𝜃.)
Go to 73
73
It is often useful to use the inverse trigonometric function. This is the value of the angle for which the trigonometric function has a specified value. The inverse sine of x is denoted by sin−1x. (Warning: This notation is standard, but it can be confusing. sin−1x always represents the inverse sine of x, not 1∕ sin x. The latter would be written (sin x)−1. An older notation for sin−1x is arcsin x.)
For example, because the sine of 30∘is 1∕2, sin−1(1∕2) = 30∘. Note, however, that the sine of 150∘is also 1∕2. Furthermore, the trigonometric functions are periodic: there is an endless sequence of angles (all differing by 360∘) having the same value for the sine, cosine, etc.

Answer: Frame 69 (a): +, −; (b): +, +

§ 1.6

Trigonometry 41

Because the definition of function (frame 6) specifies the assignment of one and only one value of f (x) for each value of x, the domain of the inverse trigonometric function must be suitably restricted.
The inverse functions are defined by

y = sin−1x Domain∶ − 1 ≤ x ≤ +1 Range∶ − 𝜋 ≤ y ≤ + 𝜋

2

2

y = cos−1x Domain∶ − 1 ≤ x ≤ +1 Range∶ 0 ≤ y ≤ 𝜋

y = tan−1x Domain∶ − ∞ < x < +∞ Range∶ − 𝜋 < y < + 𝜋

2

2

Go to 74. 74

Try these problems: √
(a) sin−1(1∕ 2) = [𝜋∕6 | 𝜋∕4 | 𝜋∕3 | 𝜋∕2]

(b) tan−1(1) = [𝜋∕6 | 𝜋∕4 | 𝜋∕3 | 𝜋]

(c) cos−1(1∕2) = [𝜋∕6 | 𝜋∕4 | 𝜋∕3 | 𝜋]

If you have a calculator with inverse trigonometric functions, try the following:

(d) sin−1(0.8) = [46.9 | 28.2 | 53.1 | 67.2] degrees (e) tan−1(12) = [0.82 | 1.49 | 1.62 | 1.83] radians (f ) cos−1(0.05) = [4.3 | 12.6 | 77.2 | 87.1] degrees

Check your answers, and then go on to the next section, which is the last one in our reviews.

Go to 75.

42 Starting Out

Chap. 1

1.7 Exponentials and Logarithms

75
Are you already familiar with exponentials? If not, go to 76. If you are, try this short quiz. a5 = [5a | 5 log a | a log 5 | none of these]
ab+c = [abac | ab + ac | cab | (b + c) log a] a f ∕a g = [( f − g) log a | a f ∕g | a f −g | none of these]
a0 = [0 | 1 | a | none of these]; a ≠ 0 (ab)c = [abac | ab+c | abc | none of these]

If any mistakes, go to 76. Otherwise, go to 77.
76 By definition am, where m is a positive integer, is the product of m factors of a. Hence, 23 = (2)(2)(2) = 8 and 102 = (10)(10) = 100.

Furthermore, by definition a−m = 1∕am. It is easy to see, then, that

aman = am+n, am = am−n, an a0 = am = 1 am
(am)n = amn,
(ab)m = ambm.

(a ≠ 0, m can be any integer)

Note that am+n is evaluated as a(m+n); the expression in the exponential is always evaluated before any other operation is carried out.
If you have not yet tried the quiz in frame 75, try it now. Otherwise,

Go to 77.

Answer: Frame 74: (a) 𝜋∕4, (b) 𝜋∕4, (c) 𝜋∕3, (d) 53.1∘, (e) 1.49, (f) 87.1∘

§ 1.7

Exponentials and Logarithms 43

77 Here are a few problems:

32 = [6 | 8 | 9 | none of these]

13

=

[ 1

|

3

|

1

|

none

of

] these

[ 2−3 = −6 |

1

3 |

] − 9 | none of these

8

43 = [48 | 4−8 | 16−1 | none of these] 45

If you did these all correctly, go to 79. If you made any mistakes, go to 78.

78

Below are the solutions to problem 77. Refer back to the rules in 76 if you have trouble understanding the solution.

Now try these:

32 = (3)(3) = 9,

13 = (1)(1)(1) = 1 (1m = 1 for any m),

2−3 = 1 = 1 , 23 8

43 45

= 43−5

=

4−2

=

1 16

=

16−1.

(3−3)3 = [1 | 3−9 | 3−27 | none of these]

52 32

=

[( 5 )2 3

||||

( 5 )−1 3

||||

5−6

||||

none

of

] these

43 = [12 | 16 | 26 | none of these]

Check your answers and try to track down any mistakes.

Then go to 79.

44 Starting Out

Chap. 1

79

Here are a few more problems. 100 = [0 | 1 | 10]

10−1 = [−1 | 1 | 0.1]

0.00003

=

[1

×

10−3

|

10−3

|

3

×

] 10−5

3

0.4 × 10−4 = [4 × 10−5 | 4 × 10−3 | 2.5 × 10−5]

3 6

× ×

10−7 10−3

=

[ 1 2

×

1010

||||

5

×

104

||||

0.5

×

] 10−4

If these were all correct, go to 81. If you made any mistakes, go to 80.

80

Here are the solutions to the problems in 79:
100 = 10 = 1 10
10−1 = 1 = 0.1, 10
0.00003 = 0.00001 × 3 = 3 × 10−5,

0.4 × 10−4 = (4 × 10−1) × 10−4 = 4 × 10−5,

3 × 10−7 6 × 10−3

=

3 6

×

10−7 10−3

=

1 2

× 10−7+3

=

0.5 × 10−4.

Go to 81.

81

Let’s introduce the idea of fractional exponents. If bn = a, then b is called the nth root of a and is written b = a1∕n. Hence 161∕4 = (fourth root of 16) = 2. That is, 24 = 16.

Answers: Frame 75: a5 = none of these, ab+c = abac, a f ∕ag = a f −g, a0 = 1, (ab)c = abc Frame 77: 9, 1, 1/8, 16−1
Frame 78: 3−9, ( 5∕3)2, 26

§ 1.7

Exponentials and Logarithms 45

If y = am∕n, where m and n are integers, then y = (a1∕n)m. For instance

82∕3 = (81∕3)2 = 22 = 4.

Try these:

[

]

27−2∕3 = 1∕18 | 1∕18 | 1∕9 | − 18 | none of these

163∕4 = [12 | 8 | 6 | 64]

82 The answers are:

27−2∕3 = (271∕3)−2 = 3−2 = 1∕9, 163∕4 = (161∕4)3 = 23 = 8.

If right, go to 84. If wrong, go to 82.

Do these problems:
253∕2 = [125 | 5 | 15 | none of these] (0.00001)−3∕5 = [0.001 | 1000 | 10−15 | 10−25]

83 Here are the solutions to the problems in 82.

If your answers were correct, go to 84. Otherwise, go to 83.

253∕2 = (251∕2)3 = 53 = 125, (0.00001)−3∕5 = (10−5)−3∕5 = 1015∕5 = 103 = 1000.

Here are a few more problems. Encircle the correct answers.

(

27∕64 × (49 ×

10−6)1∕3 10−4)1∕4

= =

[[3√∕4070∕||103∕1||||6(×101×0−72)||−29∕||||64√×71∕01−040]0,]

.

Go to 84 after checking your answers.

46 Starting Out

Chap. 1

84

Although our original definition of am applied only to integral values of m, we have also

defined (am)1∕n = am∕n, where both m and n are integers. Thus we have a meaning for ap,

where p is either an integer or a fraction (ratio of integers).

√ As yet we do not know how to evaluate ap if p is an irrational number, such as 𝜋 or

2. However, we can approximate an irrational number as closely as we desire by a fraction.

For instance, 𝜋 is approximately 31,416/10,000. This is in the form m∕n, where m and n are

integers, and we know how to evaluate it. Therefore, y = ax, where x is any real number, is a

meaningful expression in the sense that we can evaluate it as accurately as we please. (A more

rigorous treatment of irrational exponents can be based on the properties of suitably defined

logarithms.)

Try the following problem.

a𝜋 ax a3

=

[a𝜋x∕3

||

a𝜋+x−3

||

a3𝜋x

||

a(𝜋+x)∕3]

If right, go to 86.

If wrong, go to 85.

85

The rules given in frame 76 apply here as if all exponents were integers. Hence

a𝜋ax = a𝜋+x−3. a3

Here is another problem:

(𝜋 2 )(2𝜋 )

=

[ 1

||

(2𝜋)2𝜋

||

2𝜋2+𝜋

||

none

of

] these

If right, go to 87. If wrong, go to 86.

Answers: Frame 79: 1, 0.1, 3 × 10−5, 4 × 10−5, 0.5 × 10−4 Frame 81: 1∕9, 8 Frame 82: 125, 10√00 Frame 83: 3/400, 7∕10

§ 1.7

Exponentials and Logarithms 47

86
The quantity (𝜋2)(2𝜋) is the product of two different numbers raised to two different exponents. None of our rules apply to this and, in fact, there is no way to simplify this expression.
Now go to 87.
87

If you do not clearly remember logarithms, go to 88.

If you do, try the following test. Let x be any positive number, and let log x represent the

log of x to the base 10. Then:

10log x =

.

Go to 88 for the correct answer. 88
The answer to 87 is x; in fact we will take the logarithm of x to the base 10 to be defined by
10log x = x.

That is, the logarithm of a number x is the power to which 10 must be raised to produce the number x itself. This definition only applies for x > 0. Here are two examples:
100 = 102, therefore log 100 = 2; 0.001 = 10−3, therefore log 0.001 = −3.

Now try these problems: log 1,000,000 = [1,000,000 | 6 ∣ 60 ∣ 600] log 1 = [0 | 1 | 10 | 100]

If right, go to 90. If wrong, go to 89.

89

Here are the answers: log 1,000,000 = log(106) = 6 log 1 = log(100) = 0

(check, 106 = 1,000,000), (check, 100 = 1).

(continued)

48 Starting Out

Chap. 1

Try the following: log(104∕10−3) = [107 | 1 | 10 | 7 | 70] log(10n) = [10n | n | 10n | 10∕n] log(10−n) = [−10n | − n | − 10n | − 10∕n]

If you had trouble with these, carefully review the material in this section. Then go to 90.
90
Here are three important relations for manipulating logarithms, a and b are any positive numbers:
log(ab) = log a + log b, log(a∕b) = log a − log b,
log(an) = n log a.

If you are familiar with these rules, go to 92. If you want to see how they are derived, Go to 91.
91
We can derive the required rules as follows. From the definition of log x, a = 10log a and b = 10log b. Consequently, from the properties of exponentials,
ab = (10log a)(10log b) = 10log a+log b.
Taking the log of both sides, and again using log 10x = x gives log (ab) = log 10log a+log b = log a + log b.

Similarly,

a∕b = 10log a10− log b = 10log a−log b. log(a∕b) = log a − log b

Answers: Frame 84: a𝜋+x−3 Frame 85: None of these Frame 88: 6, 0

§ 1.7

Exponentials and Logarithms 49

Likewise,

an = (10log a)n = 10n log a,

so that

log(an) = n log a.

92

Try these problems:

[

]

If log n = −3, n = 1∕3 | 1∕300 | 1∕1000

10log 100 = [1010 | 20 | 100 | none of these]

log 1000 [

]

log 100 = 3∕2 | 1 | − 1 | 10

93 The answers are: 10log n = n, so if log n = −3, n = 10−3 = 1∕1000.

Go to 92.
If right, go to 94. If wrong, go to 93.

For the same reason,

10log 100 = 100. log 1000 = log 103 = 3 . log 100 log 102 2

Try these problems: 1∕2 log 16 = [2 | 4 | 8 | log 2 | log 4]
log(log 10) = [10 | 1 | 0 | − 1 | − 10]

94

Go to 94.

In this section we have discussed only logarithms to the base 10. However, any positive number except 1 can be used as a base. Bases other than 10 are usually indicated by a subscript.
(continued)

50 Starting Out

Chap. 1

For instance, the logarithm of 8 to the base 2 is written log28, and is equal to 3 because 23 = 8. If our base is denoted by r, then the defining equation for logrx is

rlogrx = x.

All the relations explained in frame 91 are true for logarithms to any base (provided, of course, that the same base is used for all the logarithms in each equation).
There is a special base number

e = 2.71828 … ,

called Euler’s number, that is used to define natural logarithms that are usually designated by the

symbol ln x = logex. Euler’s number is an irrational number, and the three dots, known as an ellipsis, indicate the indefinite continuation of that number. The defining equation for natural

logarithms is then

eln x = x.

From the defining equation, set x = e, then eln e = e, thus

ln e = 1.

The significance of this special property will be described in Chapter 2.
Go to 95. 95
From the definition of logarithm in the last frame we can obtain the rule for changing logarithms from one base to another, for instance from base 10 to the base e. (Many calculators give both log x, i.e. log10x, and ln x.) Take log10 of both sides of the defining equation eln x = x,
log(eln x) = log x.

Because log xn = n log x (frame 91), this gives ln x log e = log x or ln x = log x . log e

Answers: Frame 89: 7, n, −n Frame 92: 1∕1000, 100, 3∕2 Frame 93: log 4, 0

Summary of Chapter 1 51
The numerical value of log e is 1∕(2.303 … ) so ln x = (2.303) log x.
If you have a calculator which evaluates both ln x and log x, check this relation for a few values of x.
The ln x satisfies the same properties as log x as listed in frame 90,
ln(ab) = ln a + ln b, ln(a∕b) = ln a − ln b,
ln(an) = n ln a.
Go to 96. 96
Before concluding Chapter 1 it is worth commenting on how to find the values of the functions in this chapter. In former times one had to consult bulky books of tables. Today the values are essentially instantly generated on simple and inexpensive calculators. The technique for doing this will be explained in Chapter 4 in the section on Taylor’s formula. This technique requires differential calculus, which is introduced in the next chapter.
On page 277, following the appendices and the solutions to the problems, there is a collection of review problems with answers, an index to the symbols, and an index to the text.
Before going on, here is a summary of Chapter 1 to help you review what you have learned. Take a look if you feel that this would be helpful.
As soon as you are ready, go to Chapter 2.
Summary of Chapter 1
1.2 Functions (frames 3–13)
A set is a collection of objects—not necessarily material objects—described in such a way that we have no doubt as to whether a particular object does or does not belong to the set. A set may be described by listing its elements or by a rule.
A function is a rule that assigns to each element in a set A one and only one element in a set B. The rule can be specified by a mathematical formula such as y = x2, or by tables of associated numbers. If x is one of the elements of set A, then the element in set B that the function f associates with x is denoted by the symbol f (x), which is usually read as “f of x.”

52 Starting Out

Chap. 1

The set A is called the domain of the function. The set of all possible values of f (x) as x varies over the domain is called the range of the function. The range of f need not be all of B.
When a function is defined by a formula such as f (x) = ax3 + b, then x is often called the independent variable and f (x) is called the dependent variable. Often, however, a single letter is used to represent the single variable as in y = f (x).
Here x is the independent variable and y is the dependent variable. In mathematics the symbol x frequently represents an independent variable, f often represents the function, and y = f (x) usually denotes the dependent variable. However, any other symbols may be used for the function, the independent variable, and the dependent variable, for example, x = H(r).

1.3 Graphs (frames 14–22)

A convenient way to represent a function is to plot a graph as described in frames 15–18.

The mutually perpendicular coordinate axes intersect at the origin. The axis that runs hor-

izontally is called the horizontal axis, or x-axis. The axis that runs vertically is called the

vertical axis, or y-axis. Sometimes the value of the x-coordinate of a point is called the

abscissa, and the value of the y-coordinate is called the ordinate. In the designation of a typ-

ical point by the notation (a, b), we will always designate the x-coordinate first and the

y-coordinate second.

The constant function assigns a single fixed number c to each value of the independent

variable x. The absolute value function ∣ x ∣ is defined by

{

∣x∣=

x if x ≥ 0, −x if x < 0.

1.4 Linear and Quadratic Functions (frames 23–39)
An equation of the form y = mx + b where m and b are constants is called linear because its graph is a straight line. The slope of a linear function is defined by
Slope = y2 − y1 = y1 − y2 . x2 − x1 x1 − x2
From the definition it is easy to see (frame 29) that the slope of the above linear equation is m.
An equation of the form y = ax2 + bx + c, where a, b, and c, are constants (and a ≠ 0), is called a quadratic equation. Its graph is called a parabola. The values of x at y = 0 satisfy ax2 + bx + c = 0 and are called the roots of the equation. Not all quadratic equations have real roots. The equation ax2 + bx + c = 0 has two roots given by
√ x = −b ± b2 − 4ac .
2a

Summary of Chapter 1 53

1.5–1.6 Angles and Their Measurements; Trigonometry (frames 40 – 74)

Angles are measured in either degrees or radians. A circle is divided into 360 equal degrees. The number of radians in an angle is equal to the length of the subtending arc divided by the length of the radius (frame 42). The relation between degrees and radians is

1

rad

=

360∘ 2𝜋

.

Rotations can be clockwise or counterclockwise. An angle formed by rotating in a coun-

terclockwise direction is taken to be positive.

The trigonometric functions are defined in conjunction with the figure.

The definitions are

sin 𝜃 = y , r

cos 𝜃 = x , r

tan 𝜃 = y , x

cot 𝜃

=

1 tan 𝜃

=

x, y

sec 𝜃

=

1 cos 𝜃

=

r, x

csc 𝜃

=

1 sin 𝜃

=

r. y

(x, y)

y-axis

yr

θ

x

x-axis

√ Although r = x2 + y2 is always positive, x and y can be either positive or negative and the above quantities may be positive or negative depending on the value of 𝜃. From the Pythagorean theorem it is easy to see (frame 56) that
sin2𝜃 + cos2𝜃 = 1.
The sines and cosines for the sum of two angles are given by:
sin(𝜃 + 𝜙) = sin 𝜃 cos 𝜙 + cos 𝜃 sin 𝜙, cos(𝜃 + 𝜙) = cos 𝜃 cos 𝜙 − sin 𝜃 sin 𝜙.

54 Starting Out

Chap. 1

The inverse trigonometric function designates the angle for which the trigonometric
function has the specified value. Thus the inverse trigonometric function to x = sin 𝜃 is 𝜃 = sin−1x and similar definitions apply to cos−1x, tan−1x, etc. [Warning: This notation is standard, but it can be confusing: sin−1x ≠ (sin x)−1. An older notation for sin−1x is arc sin x.]

1.7 Exponentials and Logarithms (frames 75–95)
If a is multiplied by itself as aaa · · · with m factors, the product is written as am. Furthermore, by definition, a−m = 1∕am. From this it follows that

aman = am+n,

am an

=

am−n,

a0 = am = 1, am

(am)n = amn,

(ab)m = ambm.

If bn = a, b is called the nth root of a and is written as b = a1∕n. If m and n are integers, am∕n = (a1∕n)m.

The meaning of exponents can be extended to irrational numbers (frame 84) and the above relations also apply with irrational exponents, so (ax)b = axb, etc.
The definition of log x (the logarithm of x to the base 10) is
x = 10log x.

The following important relations can easily be seen to apply to logarithms (frame 91):

log(ab) = log a + log b, log (a∕b) = log a − log b,
log(an) = n log a.

The logarithm of x to another base r is written as logrx and is defined by x = rlogrx.

The above three relations for logarithms of a and b are correct for logarithms to any base provided the same base is used for all the logarithms in each equation.

Summary of Chapter 1 55 A particular important base is r = e = 2.71828 … as defined in frame 109. Logarithms to the base e are so important in calculus that they are given a different name; they are called natural logarithms and written as ln. With this notation the natural logarithm of x is defined by
eln x = x.
If we take the logarithm to base 10 of both sides of the equation,
log eln x = log x, ln x log e = log x,
ln x = log x . log e
Because the numerical value of 1∕ log e = 2.303 … ,
ln x = (2.303) log x.
The special value of e and the importance that ln e = 1 will be discussed in Chapter 2. Continue to Chapter 2.

CHAPTER TWO
Differential Calculus
In this chapter you will learn about • The concept of the limit of a function; • What is meant by the derivative of a function; • Interpreting derivatives graphically; • Shortcuts for finding derivatives; • How to recognize derivatives of some common functions; • Finding the maximum or minimum values of functions; • Applying differential calculus to a variety of problems.
2.1 The Limit of a Function
97 Before diving into differential calculus, it is essential to understand the concept of the limit
of a function. The idea of a limit may be new to you, but it is at the heart of calculus, and it is essential to understand the material in this section before going on. Once you understand the concept of limits, you should be able to grasp the ideas of differential calculus quite readily.
Limits are so important in calculus that we will discuss them from two different points of view. First, we will discuss limits from an intuitive point of view. Then, we will give a precise mathematical definition.
Go to 98.
57

58 Differential Calculus

Chap. 2

98

Here is a little bit of mathematical shorthand, which will be useful in this section. Suppose a variable x has values lying in an interval with the following properties:

1. The interval surrounds some number a. 2. The difference between x and a is less than another number B, where B is any number
that you choose. 3. x does not take the particular value a. (We will see later why this point is excluded.)
The above three statements can be summarized by the following:

|x − a| > 0 |x − a| < B

(This statement means x cannot have the value a.) (The magnitude of the difference between x and a is less than the arbitrary number B.)

These relations can be combined in the single statement:

0 < |x − a| < B.

(If you need to review the symbols used here, see frame 20.) The values of x which satisfy 0 < | x − a | < B are indicated by the interval along the x-axis
shown in the figure.

allowed values of x (x = a excluded)

B

B

a–B a a+B

x-axis

Go to 99.

§ 2.1

The Limit of a Function 59

99

We begin our discussion of limits with an example. We are going to work with the equation y = f (x) = x2, as shown in the graph. P is the point on the curve corresponding
to x = 3, y = 9.

y-axis

25

y = 25

20

15

10

P

A′

C

5

B

A

y = 1 x-axis

0 12345

Let us concentrate on the behavior of y for values of x in an interval about x = 3. For
reasons we shall see shortly, it is important to exclude the particular point of interest P, and to
remind us of this, the point is encircled on the curve.
We start by considering values of y corresponding to values of x in an interval about x = 3,
lying between x = 1 and x = 5. With the notation of the last frame, this can be written as 0 < | x − 3 | < 2. This interval for x is shown by line A in the figure. The corresponding interval for y is shown by line A′ and includes points between y = 1 and y = 25, except y = 9.
A smaller interval for x is shown by line B. Here 0 < | x − 3 | < 1, and the corresponding interval for y is 4 < y < 16, with y = 9 excluded.
The interval for x shown by the line C is given by 0 < | x − 3 | < 0.5. Write the corresponding interval for y in the blank below, assuming y = 9 is excluded.

In order to find the correct answer, go to 100. 100
The interval for y which corresponds to 0 < | x − 3 | < 0.5 is 6.25 < y < 12.25,
(continued)

60 Differential Calculus

Chap. 2

which you can check by substituting the values 2.5 and 3.5 for x in y = x2 in order to find the values of y at either end point.
So far we have considered three successively smaller intervals of x about x = 3 and the corresponding intervals of y. Suppose we continue the process. The drawing shows the plot y = x2 for values of x between 2.9 and 3.1. (This is an enlarged piece of the graph in frame 99. Over the short distance shown the parabola looks practically straight.)

9.5

9.0

P

8.5

8.0

2.9

3.0

3.1

Three small intervals of x around x = 3 are shown along with the corresponding interval in y. The table below shows the values of y, corresponding to the boundaries of x at either end of the interval. (The last entry is for an interval too small to show on the drawing.)

Interval of x
1–5 2–4 2.5 – 3.5 2.9 – 3.1 2.95 – 3.05 2.99 – 3.01 2.999 – 3.001

Corresponding interval of y
1 – 25 4 – 16 6.25 – 12.25 8.41 – 9.61 8.70 – 9.30 8.94 – 9.06 8.994 – 9.006

Go to 101.

§ 2.1

The Limit of a Function 61

101
We hope it is apparent from the discussion in the last two frames that as we diminish the interval for x around x = 3, the values for y = x2 cluster more and more closely about y = 9. In fact, it appears that we can make the values for y cluster as closely as we please about y = 9 by merely limiting x to a sufficiently small interval about x = 3. Because this is true, we say that the limit of x2, as x approaches 3, is 9. We write this as

lim x2 = 9.
x→3

Let’s put this in more general terms. If a function f (x) is defined for values of x about some fixed number a, and if, as x is confined to smaller and smaller intervals about a, the values of f (x) cluster more and more closely about some specific number L, the number L is called the limit of f (x) as x approaches a. The statement that “the limit of f (x) as x approaches a is L” is customarily abbreviated by
lim f (x) = L.
x→a

In the example at the top of the page f (x) = x2, a = 3, and L = 9. The important idea in the definition is that the intervals we use lie on either side of the point of interest a, but that the point itself is not included. The value of the function f (a) when x = a may be different from lim f (x), as we shall see.
x→a
To summarize the mathematical argument in more familiar language: in the spirit of “Anything you can do I can do better!” the challenge to an opponent is “Pick a point as close as you want to L as you please, and I can find a point close to a for which f (x) will be closer to L than the point that you chose.”

Go to 102.

102

You may be wondering why we have been giving such a complicated discussion of an apparently simple problem. Why bother with lim x2 = 9 when it is obvious
x→3
that x2 = 9 for x = 3? The reason is that the value of a function for a particular x = a may not be defined, whereas
the limit as x approaches a is perfectly well defined. For instance at 𝜃 = 0 the function sin 𝜃/𝜃
has the value 0/0, which is not defined. Nevertheless

lim
𝜃→0

sin 𝜃

𝜃

=

1.

(continued)

62 Differential Calculus

Chap. 2

You can see that this result is reasonable by graphing the function sin 𝜃/𝜃 as shown below. If you have a calculator, explore for yourself values of sin 𝜃/𝜃 as 𝜃 approaches zero. If you try to evaluate the function at 𝜃 = 0, most calculators will indicate an error. This is as it should be because the function is not defined at 𝜃 = 0. Nevertheless, its limit is well defined and has
the value 1. (This is formally proved in Appendix A12.)

sin θ θ
1.0

0.5

–2π –3π 2

–π

–

π 2

π

π 3π 2π θ (radians)

2

2

The actual procedure for finding a limit varies from problem to problem. For those interested in learning more, there are a number of theorems for finding the limits of simple functions in Appendix A2. As another illustration consider

f (x) = x2 − 1 . x−1

For x = 1, f (1) = 1−1 = 0 , which is not defined. However, we can divide through by x − 1
1−1 0
provided x is not equal to 1, and we obtain

f (x) = x2 − 1 = (x + 1)(x − 1) . = x + 1.

x−1

x−1

Therefore, even though f (1) is not defined,

lim f (x) = lim(x + 1) = 2.

x→1

x→1

Formal justification of these steps is given in Appendix A2, along with a number of rules for handling limits. (There is no need to read the appendix now unless you are really interested.)
We could also have obtained the above result graphically by studying the graph of the function in the neighborhood of x = 1 as we did in frame 99.

Go to 103.

§ 2.1

The Limit of a Function 63

103

To see whether you have caught on, find the limit of the following slightly more complicated functions by procedures similar to the ones described in frame 102. (You will probably have to work these out on paper. Both of them involve a little algebraic manipulation.)

(a) lim (1 + x)2 − 1 = {1 | x | −1| 2}

x→0

x

(b)

1− lim

(1

+ x)3

=

{1

|

x

|

3

|

−3}

x→0

x

If right, go to 105. Otherwise, go to 104.

104

Here are the solutions to the problems in frame 103:

(1 + x)2 − 1

(1 + 2x + x2) − 1

(a) lim

= lim

x→0

x

x→0

x

= lim 2x + x2 = lim(2 + x) = lim 2 + lim x = 2 + 0 = 2.

x→0 x

x→0

x→0

x→0

1 − (1 + x)3

1 − (1 + x)(1 + x)(1 + x)

lim

= lim

x→0

x

x→0

x

(b)

=

1 lim

−

(1

+

3x

+

3x2

+

x3)

=

lim(−3

−

3x

−

x2)

x→0

x

x→0

= lim(−3) + lim(−3x) + lim(−x2) = −3.

x→0

x→0

x→0

(These steps are formally justified in Appendix A2.)

Go to 105.
105
So far we have discussed limits using expressions such as “confined to a smaller and smaller interval” and “clustering more and more closely.” These expressions convey the intuitive meaning of a limit, but they are not precise mathematical statements. Now we are ready for a
(continued)

64 Differential Calculus

Chap. 2

precise definition of a limit. (Because it is an almost universal custom, in the definition of a limit we will use the Greek letters 𝛿 (delta) and 𝜀 (epsilon).)
Definition of a Limit:
Let f (x) be defined for all values of x in an interval centered about x = a but not necessarily at x = a. If there is a number L such that to each positive number 𝜀 there corresponds a positive number 𝛿 such that
| f (x) − L| < 𝜀 provided 0 < |x − a| < 𝛿,

we say that L is the limit of f (x) as x approaches a, and write
lim f (x) = L.
x→a
To see how to apply this definition,
Go to 106.
106
Suppose we assert that lim f (x) = L, and an opponent disagrees. The formal definition
x→a
of a limit in frame 105 provides a clear basis for settling the dispute as to whether the limit exists and is L. As a first step, we tell the opponent to pick any positive number 𝜀 no matter how small, say 0.001, or if the opponent wants to be difficult, 0.00001. Our task is to find some other number 𝛿, such that for all x in the interval 0 < | x − a | < 𝛿, the difference between f (x) and L is smaller than 𝜀. If we can always do this, we win the argument—the limit exists and is L. These steps are illustrated for a particular function in the drawings below.

f (x)

f (x)

L

2ε

a

x

Our opponent has challenged us to find a δ to fit this ε.

L

2ε

2δ

a

x

Here is one choice of δ. Obviously, for all values of x in the interval shown, f (x) will satisfy | f (x) – L | < ε.

Answer: Frame 103: 2, −3

§ 2.1

The Limit of a Function 65

It may be that our opponent can find an 𝜀 such that we can never find a 𝛿, no matter how small, that satisfies our requirement. In this case, she wins and f (x) does not have the limit L. (In frame 114 we will come to an example of a function that does not have a limit.)

Go to 107. 107
In the examples we have studied so far, the function has been expressed by a single equation. However, this is not necessarily the case. Here is an example:

f (x) = 1 for x ≠ 2, f (x) = 3 for x = 2.

(The symbol ≠ means “not equal.”) A sketch of this peculiar function is shown. You should be able to convince yourself that
lim f (x) = 1, whereas f (2) = 3.
x→2

f (x)
3

2

1

–1 0

x 12345

If you would like further explanation of this, go to 108; otherwise, go to 109.
Go to 108. 108
For every value of x except x = 2, the value of f (x) = 1. Consequently, f (x) − 1 = 0 for all x except x = 2. Because 0 is less than the smallest positive number 𝜀 that your opponent could select, it follows from the definition of a limit that lim f (x) = 1, even though f (2) = 3.
x→2
Continue with 109.

66 Differential Calculus

Chap. 2

109

Calculator Problem

Here is another function which has a well-defined limit at a point but which can’t be evaluated at that point: f (x) = (1 + x)1/x. Although the value of f (x) at x = 0 is puzzling, it is possible to find lim(1 + x)1∕x.
x→0
Most calculators have the function yx. If you have such a calculator, determine the values in Table 1:

Table 1 x 1 0.1 0.01 0.001 0.0001 0.00001

f (x) = (1 + x)1/x

The limit of (1 + x)1/x as x → 0 will play a key role in our study of logarithms. Its value is given a special symbol, e. Like 𝜋, e is an unending and unrepeating decimal; it is irrational. The value of e is 2.7182818 . . . . If you tried evaluating e with a calculator, the last entry in the table should give correct values for the first four digits after the decimal point. A proof that lim(1 + x)1∕x = e is presented in Appendix A4.
x→0
Go to 110.
110

The actual procedure for finding a limit varies from problem to problem. For those interested in learning more, there are a number of theorems for finding the limits of simple functions in Appendix A2. The result mentioned earlier,

lim
𝜃→0

sin 𝜃

𝜃

=

1,

is proved in Appendix 12.

§ 2.1

sin θ θ
1.0

The Limit of a Function 67

0.5

–2π

–3π 2

–π

–

π 2

π

π

3π

θ (radians)
2π

2

2

You can see that this result is reasonable by graphing the function sin 𝜃/𝜃 as shown above. If you have a calculator, explore for yourself values of sin 𝜃/𝜃 as 𝜃 approaches zero. The function is not defined at 𝜃 = 0, but its limit is well defined and has the value 1.

111 Continuous Function

Go to 111.

So far in most of our discussion of limits we have been careful to exclude the actual value of f (x) at the point of interest, a. In fact, f (a) does not even need to be defined for the limit to exist (as in the last frame). However, frequently f (a) is defined. If this is so, and if in addition

lim f (x) = f (a),
x→a

then the function is said to be continuous at a. To summarize, fill in the blanks:

A function f (x) is continuous at x = a if

1. f (a) is

.

2. lim f (x) =

.

x→a

Check your answers in frame 112.

68 Differential Calculus 112
Here are the answers: A function f (x) is continuous at x = a if

Chap. 2

1. f (a) is defined.

2. lim f (x) = f (a).
x→a

A more picturesque description of a continuous function is that it is a function you can graph without lifting your pencil from the paper in the region of interest.
Try to determine whether each of the following functions is continuous or discontinuous (not continuous) at the point indicated.

1. f (x) = x2 + 3 . 9 − x2

At x = 3, f (x) is

{

2. f (x) =

1, x ≥ 0, 0, x < 0.

{continuous | discontinuous}

At x = 1, f (x) is {continuous | discontinuous} 3. f (x) = |x|.

At x = 0, f (x) is 4. f (x) = sin x .
x

{continuous | discontinuous}

At x = 0, f (x) is {continuous | discontinuous}

If you made any mistakes, or want more explanation, go to 113. Otherwise, skip on to 114.
113
Here are the explanations of the problems in frame 112.

1. At x = 3, f (x) = x2+3 = 12 . This is an undefined expression and, therefore, the function

9−x2

0

is not continuous at x = 3.

§ 2.1 2. Here is a plot of the function given.

The Limit of a Function 69

f (x) 2

1

–3 –2 –1 0

x 123

This function satisfies both conditions for continuity at x = 1, and is thus continuous there. (It is, however, discontinuous at x = 0.) 3. Here is a plot of f (x) = | x |.
f (x) = x

x
This function is continuous at x = 0 because it satisfies all the formal requirements. 4. As discussed in frame 110, sin x/x is not defined at x = 0, and so it is discontinuous at this
point. (It is, however, continuous for all other values of x.)
Go to 114. 114
Before leaving the subject of limits, it is worth looking at some examples of functions that somewhere have no limit. One such function is presented in problem 2 of the previous frame. The graph of the function is shown in the figure. We can prove that this function has no limit at x = 0 by following the procedure described in the definition of a limit.
(continued)

70 Differential Calculus

f (x) 2

1

–3 –2 –1 0

x 123

Chap. 2

For purposes of illustration, suppose we guess that lim f (x) = 1. Next, our opponent
x→0
chooses a value for 𝜀, say 1/4. Now, for |x − 0 | < 𝛿, where 𝛿 is any positive number, {|1 − 1 | = 0 if x > 0,
| f (x) − 1| = |0 − 1 | = 1 if x < 0.

Therefore, for all negative values of x in the interval, | f (x) − 1| = 1, which is greater than 𝜀 = 1/4. Thus 1 is not the limit. You should be able to convince yourself that there is no number L, which satisfies the criterion because f (x) changes by 1 when x jumps from negative to positive values.
Go to 115.
115
Here is another example of a function that has no limit at a particular point. From the graph it is obvious that cot 𝜃 has no limit as 𝜃 → 0. Instead of clustering more and more closely to any number, L, the value of the function gets increasingly larger as 𝜃 → 0 in the direction shown by A, and increasingly more negative as 𝜃 → 0 in the direction shown by B.

cot(θ)

4.0 3.0 2.0
B 1.0 A

–π

–

π 2

–1.0

π 2

–2.0

–3.0

–4.0

πθ

Answer: Frame 112: (1) discontinuous, (2) continuous, (3) continuous, (4) discontinuous

§ 2.2

Velocity 71

This concludes our study of the limit of a function for the present. If you want some more
practice with limits, see review problems 21–28 start on page 284. Now we are ready to go
on to the next section, a discussion about velocity. Go to 116.

2.2 Velocity

116
Our discussion has become a little abstract, so before we go on to differential calculus, let’s talk about something down to earth: motion. As a matter of fact, Leibniz and Newton invented calculus because they were concerned with problems of motion, so it is a good place to start. Besides, you already know quite a bit about motion.
Go to 117.

117

In this chapter, we will only consider motion along a straight line. Here is a warm-up
problem. A train travels away at a velocity v mph (miles per hour). At t = 0, it is distance S(0) = S0
from us. (The subscript on S0 is to avoid confusion. S0 is a particular distance and is a constant; S(t) is a function that gives the distance the train is from us at time t.) Write the equation
for S(t) in terms of time t. (Take the unit of t to be hours.)

S(t) =

.

Go to 118 for the answer.

118
If you wrote S(t) = S0 + vt, you are correct. Go on to frame 119. If your answer was not equivalent to the above, try to convince yourself that this answer is correct. Note that it yields S0 when t = 0, as required. The equation is that of a straight line, and it might be worthwhile reviewing the section on linear functions, frames 23–39, before continuing. Whenever you are satisfied with this result,
Go to 119.

72 Differential Calculus
119
400

Chap. 2

300

S (miles)

200

100

0

0

2

4

6

8

t (hours)

Here is a plot of the positions at different times of a train going in a straight line. Obviously,

this represents a linear equation. Write the equation for the position of this train (in miles)

in terms of time (in hours).

S(t) =

.

Find the velocity of the train from your equation.

v=

.

Go to 120 for the correct answers. 120
Here are the answers to the questions in frame 119.

S(t) = −60t + 300 mi, v = −60 mph.

The velocity is negative because S(t) decreases as time increases. (Note that the velocity along a straight line is positive or negative depending on the direction of motion. The speed, which is the magnitude of the velocity, is always positive.) If you would like further discussion, review frames 33 and 34.

Go to 121.

§ 2.2 121
Here is another plot of position of a train traveling in a straight line.
S (t)

Velocity 73

t

The property of the line that represents the velocity of the train is the of the line.
Go to 122 for the answer.
122
The property of the line that represents the velocity of the train is the slope of the line. If you wrote this, go right on to 123. If you wrote anything else, or nothing at all, then you may have forgotten what we reviewed back in frames 23–39. It would be worthwhile going over that section once again (particularly frames 33 and 34) and think about this problem before going on. At least convince yourself that the slope really represents the velocity.
Go to 123.
123
In the figure below are plots of the positions vs. time of six objects moving along straight lines. Which plot corresponds to the object that

Has the greatest velocity forward? {a | b | c | d | e | f }

Is moving backward most rapidly? {a | b | c | d | e | f }

Is at rest?

{a | b | c | d | e | f }

(continued)

74 Differential Calculus

S (t)

e

a

c

bd

f

t

Chap. 2

If all right, go to 125. If any wrong, go to 124.
124
The velocity of the object is given by the slope of the plot of its position against time. Don’t confuse the slope of a line with the line’s location.

S (t)

S (t)

d

c

b

t

a

t

All the above lines have the same slope.

All these lines have different slopes.

A positive slope means that position is increasing with time, which corresponds to a positive velocity. Likewise, a negative slope means that position is decreasing in time, which means the velocity is negative. If you need to review the idea of slope, look at frames 25–27 before continuing.
Which line in the figure above on the right has

Negative slope?

{a | b | c | d}

Greatest positive slope? {a | b | c | d}

Go to 125.

§ 2.2

Velocity 75

125

So far, the velocities we have considered have all been constant in time. But what if the velocity changes?
S (t)

S2

(t2, S2)

S1

(t1, S1)

t

t1

t2

Here is a plot of the position of a car whose velocity is varying while it moves along a

straight line. In order to describe this, we introduce the average velocity v (read as “v bar”).

This is the ratio of the change in position to the time taken. The change in position is called

the displacement. For example, between the times t1 and t2 the displacement of the car is S2 − S1,

so (S2 − S1)/(t2 − t1) is its

during the time.

Go to 126.

126

The answer to frame 125 is
(S2 − S1)∕(t2 − t1) is its average velocity during the time interval t2 − t1.
(The single word “velocity” is not a correct answer because the velocity was changing during this interval.)
Go to 127. 127

In addition to defining the average velocity v algebraically,
v = S2 − S1 , t2 − t1
we can interpret v graphically. If we draw a straight line between the points (t1, S1) and (t2, S2), then the average velocity is simply the slope of that line.
S (t)

S2

S1

t

t1

t2

Go to 128.

76 Differential Calculus 128
During which interval in the figure was the average velocity

Chap. 2

Closest to 0?

{1 | 2 | 3}

Largest forward? {1 | 2 | 3}

Largest backward? {1 | 2 | 3}

S (t)

2

3

B C

A t

1

129 Let us analyze the last problem in detail.

If right, go to 130. If wrong, go to 129.

S (t)

B

II III

C

A

I

t

Here are straight lines drawn through the points A, B, C. Line I has a small positive slope and corresponds to almost 0 velocity. Line II has positive slope, and line III has negative slope, corresponding to positive and negative average velocities, respectively.
Go to 130.

Answers: Frame 123: d, b, e Frame 124: d, a

§ 2.2

Velocity 77

130
We now extend our idea of velocity in a very important manner: instead of asking, “What is the average velocity between time and t1 and t2?” let us ask, “What is the velocity at time t1?” The velocity at a particular time is called the instantaneous velocity. This is a new term, and we will give it a precise definition shortly even though it may already be familiar to you.

Go to 131. 131

B S(t)
l
A (t1, S1)

(t2, S2) t

We can give a graphical meaning to the idea of instantaneous velocity. The average velocity is the slope of a straight line joining two points on the curve, (t1, S1) and (t2, S2). To find the instantaneous velocity, we want t2 to be very close to t1. As we let point B on the curve approach point A (i.e. as we consider intervals of time starting at t1 that become shorter and shorter), the slope of the line joining A and B approaches the slope of the line, which is labeled l. The instantaneous velocity is then the slope of line l. In a sense, then, the straight line l has the same slope as the curve at the point A. Line l is called a tangent to the curve at A.
Go to 132.
132
Here is where the idea of a limit becomes very important. If we draw a straight line through the given point A on the curve and some other point on the curve B and then let B get closer and closer to A, the slope of the straight line approaches a unique value and can be identified with the slope of the curve at A. What we must do is consider the limit of the slope of the line through A and B as B → A.
Now, go to 133.

78 Differential Calculus

Chap. 2

133
We can now give a precise meaning to the intuitive idea of instantaneous velocity as the slope of a curve at a point. We start by considering the average velocity:

v = (S2 − S1)∕(t2 − t1) = the slope of the line connecting points 1 and 2.

S (t)
S2 v
S1 (t1, S1)

(t2, S2) S2 S1

t2 t1

t

t1

t2

S (t)
S2
S1 (t1, S1)
t1

(t2, S2)

v(t1)

S2 S1

t2 t1

t

t2

As t2 → t1, the average velocity approaches the instantaneous velocity, that is, v → v(t1)

as t2 → t1, or

v(t1)

=

lim
t2→t1

S2 t2

− −

S1 t1

.

Go to 134. 134

Because the ideas presented in the last few frames are important, let’s review them. If a point moves from S1 to S2 during the time t1 to t2, then

(S2 − S1)∕(t2 − t1)

is the

, v.

If we consider the limit of the average velocity as the averaging time goes to zero, the result

is called the

, v.

Now let’s try to present these ideas in a neater form. If you can, write a formal definition

of v in the blank space. v=

Go to frame 135 for the answers.

Answer: Frame 128: 1, 2, 3

§ 2.2

Velocity 79

135

The correct answers to frame 134 are the following: If a point moves from S1 to S2 during the time t1 to t2, then (S2 − S1)∕(t2 − t1) is the average velocity, v. If we consider the limit of the average velocity as the averaging time goes to zero, the result is called the instantaneous velocity, v.
v = lim S2 − S1 . t2→t1 t2 − t1

If you wrote this, congratulations! Go on to 136. If you wrote something different, go back to frame 133 and work your way to this frame once more.

Then go on to 136. 136

The Greek capital letter Δ (“delta”) is often used to indicate the change in a variable. Thus, to make the notation more succinct, we can write ΔS = S2 − S1, and Δt = t2 − t1. (ΔS is a single symbol read as “delta S”; it does not mean Δ × S.) Although this notation may
be new, it saves lots of writing and is worth the effort to get used to.

S (t)
S2 S1 (t1, S1)
t1

(t2, S2)
ΔS
Δt t2

t-axis

With this notation, our definition of instantaneous velocity is

v=

.

Go to 137 to find the correct answer.

80 Differential Calculus

137

If you wrote

v = lim S2 − S1 Δt→0 t2 − t1

or v = lim ΔS , Δt→0 Δt

go ahead to frame 138. If you missed this, review frames 134–136 before going to 138.

Chap. 2

138
Now we are going to calculate an instantaneous velocity by analyzing an example step by step. Later on we will find shortcuts for doing this.
Suppose that we are given the following expression relating position and time:
S(t) = k t2 (k is a constant).

The goal is to find ΔS = S(t + Δt) − S(t), for any Δt, and then to evaluate the limit ΔS/Δt as Δt → 0.
Here are the steps

ΔS = S(t + Δt) − S(t) = k(t + Δt)2 − kt2

= k[t2 + 2t Δt + (Δt)2] − kt2

= k[2t Δt + (Δt)2],

ΔS

=

k[2t

Δt + (Δt)2] = 2kt + k

Δt,

Δt

Δt

v = lim ΔS = lim (2kt + k Δt) = 2kt. Δt→0 Δt Δt→0

A simpler problem for you to try is in the next frame. 139

Go to 139.

Suppose we are given that S(t) = v0t + S0. The problem is to find the instantaneous velocity from our definition.
In time Δt the point moves distance ΔS.

ΔS =

.

v = lim ΔS = Δt→0 Δt

. Write in the answers and go to 140.

§ 2.2

Velocity 81

140

If

you

wrote

ΔS

=

v0

Δt

and

v

=

lim ΔS
Δt→0 Δt

=

lim v0Δt
Δt→0 Δt

=

v0,

you are correct and can skip on to frame 142.

If you wrote something different, study the detailed explanation in frame 141.

141

Here is the correct procedure. Because S(t) = v0t + S0, ΔS = S(t + Δt) − S(t)

= v0(t + Δt) + S0 − (v0t + S0)

= v0 Δt,

lim ΔS Δt→0 Δt

= lim v0 Δt Δt→0 Δt

=

Δlitm→0v0

=

v0.

The instantaneous velocity and the average velocity are the same in this case because the velocity is a constant, v0.
Go to frame 142.
142

Here is a problem for you to work out. Suppose the position of an object is given by

S(t) = k t2 + bt + S0,

where k, b, and S0 are constants. Find v(t).

v(t) = lim ΔS = Δt→0 Δt

. To check your answer, go to 143.

82 Differential Calculus

Chap. 2

143

The answer is v(t) = 2kt + b. If you obtained this result and are ready to move on to the next section, go to frame 146.
Otherwise, go to 144.
144

Here is the solution to the problem in frame 142.

S(t) = kt2 + bt + S0,

S(t + Δt) = k(t + Δt)2 + b(t + Δt) + S0

= k[t2 + 2t Δt + (Δt)2] + b(t + Δt) + S0,

ΔS = S(t + Δt) − S(t) = k[2t Δt + (Δt)2] + b Δt,

{

}

v(t) = lim ΔS = lim k[2t Δt + (Δt)2] + b Δt

Δt→0 Δt Δt→0

Δt

= lim [k(2t + Δt) + b] = 2kt + b.
Δt→0

Now try this problem: If S(t) = At3, where A is a constant, find v(t).

Answer:
To check your solution, go to 145. 145
Here is the answer: v(t) = 3At2. Go right on to frame 146 unless you would like to see the solution, in which case continue here.
S(t) = At3, ΔS = S(t + Δt)3 − St3
= A[t3 + 3t2 Δt + 3t(Δt)2 + (Δt)3] − At3 = 3At2 Δt + 3At(Δt)2 + A(Δt)3, v(t) = lim ΔS = lim [3At2 + 3At Δt + A(Δt)2] = 3At2.
Δt→0 Δt Δt→0 Go to frame 146.

§ 2.3

Derivatives 83

2.3 Derivatives

146

In this section we will generalize our results on velocity. This will lead us to the idea of the derivative of a function, which is at the heart of differential calculus.

Go to 147. 147

To launch the discussion, let’s start with a couple of review questions. When we write S(t) we are stating that position depends on time.

Here position is the dependent variable and time is the

variable.

The velocity is the rate of change of position with respect to time. By this we mean that velocity is (give the formal definition again):

v(t) =

Go to frame 148 for the correct answers. 148
In the last frame you should have written … time is the independent variable, and v(t) = lim ΔS .
Δt → 0 Δt
Go on to 149.
149
Let us consider any continuous function defined by y = f (x). Here y is our dependent variable, and x is our independent variable. If we ask “At what rate does y change as x changes?,” we can find the answer by taking the following limit:

Rate of change of y with respect to x = lim Δy . Δx→0 Δx

Go on to 150.

84 Differential Calculus

Chap. 2

150

You can give a geometrical meaning to lim Δy , where y = f (x). To do so, fill in the blanks:

Δx→0 Δx

Geometrically, lim Δy can be found by drawing a straight line through the point (x, y)

Δx→0 Δx

and the point ( ,

) as shown. The slope of that line is given by Δy , and lim Δy is

Δx

Δx→0 Δx

the

of the tangent line to the curve at (x, y).

y (x)
y + Δy

y

x

x

x + Δx

Go on to 151. 151

The correct insertions for frame 150 are (x + Δx,y + Δy) and lim Δy . Δx→0 Δx

For brevity, the slope of the tangent to a curve is usually called the slope of the curve. (If you would like to see a discussion of this, review frame 131 before continuing.)

Go on to 152. 152

Another way of writing Δy is
Δx

Δy =

y2 − y1

=

y(x2) − y(x1) .

Δx x2 − x1

x2 − x1

If the notation used here still seems unfamiliar, review frame 136 before proceeding. Go on to 153.

§ 2.3

Derivatives 85

153

Let’s review just once more. Fill in the blank below. If we want to know how y changes as x changes, we find out by calculating the following limit:

Go on to 154. 154

The correct answer to frame 153 is

Δy lim

or

lim y2 − y1 .

Δx→0 Δx

x2→x1 x2 − x1

If you were correct, go on to 155. If you missed this, go back to 149.

155

Because the quantity lim Δy is so useful, we give it a special name and a special sym-
Δx→0 Δx
bol: lim Δy is called the derivative of y with respect to x, and it is often written with the
Δx→0 Δx
symbol dy ,
dx
dy = lim Δy , dx Δx→0 Δx

where Δy = y(x + Δx) − y(x).

Once again: dy is the
dx

of

with respect to

.

Go to 156 for the correct answer. 156

The answer is

dy is the derivative of y with respect to x. dx

This symbol is read as “dee y by dee x.” The derivative is frequently written in another

form:

dy = y′. dx

(continued)

86 Differential Calculus

Chap. 2

(The symbol y′ is read as “y prime.”) y′ and dy mean the same thing:
dx

y′ = dy = lim Δy . dx Δx→0 Δx

(Another symbol sometimes used for the derivative operator is D. Thus Dy = y′. However,

we will not use the “D” symbol.)

Having two separate symbols for the derivative may look confusing at first, but they should

both quickly become familiar. Each has its advantages. The symbol dy leaves no doubt that

the

independent

variable

is

x,

whereas

y′

might

be

ambiguous:

dx
because

y

could

be

a

func-

tion of some other variable, z. (To avoid confusion, the “prime” form is sometimes written as y′(x).) On the other hand, the symbol dy can be cumbersome to write. More seriously,
dx
in the form dy the derivative looks like the simple ratio of two quantities, dy and dx, which it
dx
is not.

We can apply the idea of a derivative to the notion of velocity, which we discussed earlier.

Instantaneous velocity is the rate of change of position with respect to time, in other words,

instantaneous velocity is the derivative of position with respect to time. Unless otherwise

specified, velocity will be used as the common meaning of instantaneous velocity.

Go to 157. 157

Let’s state the definition of a derivative using different variables. Suppose z is some

independent variable, and q depends on z. Then the derivative of q with respect to z is

defined by

dq =

.

dz

For the right answer, go to 158.

158

The correct answer is

dq = lim Δq . dz Δz→0 Δz

If correct, go to 159. If not, review to frame 155 and try again.

§ 2.4

Graphs of Functions and Their Derivatives 87

159
We have established the definition of df /dx, but there is more to explore. The symbol df /dx can be thought of as a derivative operator d , operating on the function f .
dx
If f (x) = x3 + 3, then the derivative can be written in any of the following forms:

df

d(x3 + 3)

=

=

d (x3 + 3).

dx

dx

dx

Similarly, if f (𝜃) = 𝜃2 sin 𝜃, then

d(𝜃2sin 𝜃) d𝜃

=

d d𝜃

(𝜃

2

sin

𝜃).

(Here, 𝜃 is merely another variable.)

Thus d ( ) means differentiate with respect to x whatever function f (x) happens to be

in

the

dx
parentheses.

For

functions

such

as

sin

𝜃

that

are

not

products,

the

derivative

will

be

written as d sin 𝜃 with no parentheses. The symbol df means that one should obtain an

dx

dx

expression for

Δf = f (x + Δ x) − f (x),

and then use it to evaluate

df = lim Δ f . dx Δx→0 Δx

However, as we shall see, there are lots of shortcuts for calculating derivatives.

Go to 160.

2.4 Graphs of Functions and Their Derivatives
160 We have just learned the formal definition of a derivative. Graphically, the derivative of a
function f (x) at some value of x is equivalent to the slope of the straight line that is tangent to the graph of the function at that point. Our chief concern in the rest of this chapter will
(continued)

88 Differential Calculus

Chap. 2

be to find methods for evaluating derivatives of different functions. In doing this it is helpful to have some intuitive idea of how the derivative behaves, and we can obtain this by looking at the graph of the function. If the graph has a steep positive slope, the derivative is large and positive. If the graph has a slight slope downward, the derivative is small and negative. In this section we will get some practice putting to use such qualitative ideas as these, and in the following sections we will learn how to obtain derivatives precisely.

Go to 161.

161

Here is a plot of the simple function y = x. We have plotted y′ = dy . Because the slope

of y is positive and constant, y′ is a positive constant.

dx

y

3

2

1

0 –3 –2 –1

x
123

–1

–2

y′

3 2 1 0 –3 –2 –1 –1 –2

x
123

The graph indicates that d x = 1. Can you prove this?
dx

Go to 162.

§ 2.4

Graphs of Functions and Their Derivatives 89

162

To prove that d x = 1, let y(x) = x. Then
dx
Δy = y(x + Δx) − y(x) = x + Δx − x = Δx.

Hence,

dy = lim Δy = lim Δx = 1. dx Δx→0 Δx Δx→0 Δx

Here is a plot of y = |x|. (If you have forgotten the definition of |x|, see frame 20.)

y

3 2 1 0 –3 –2 –1 –1 –2

x
123

On the coordinates below, sketch y′.

y′

2

1

0 –3 –2 –1
–1

x
123

–2

For the correct answer, go to 163.

90 Differential Calculus

Chap. 2

163
Here are sketches of y = | x| and y′. If you drew this correctly, go on to 164. If you made a mistake or want further explanation, continue here.

y

3 2 1 0 –3 –2 –1 –1 –2

x
123

y′

3 2 1 0 –3 –2 –1 –1 –2

x
123

As you can see from the graph, y = | x | = x for x > 0. So for x > 0 the problem is identical to that in frame 161, and y′ = 1. However, for x < 0, the slope of |x| is negative and is easily seen to be −1. At x = 0, the slope is undefined, for it has the value +1 if we approach 0 along the positive x-axis and has the value −1 if we approach 0 along the negative x-axis. Therefore, d |x| is discontinuous at x = 0. (The function |x| is continuous at this point, but the break
dx
in its slope at x = 0 causes a discontinuity in the derivative.)
Go to 164.
164
Here is the graph of a function y = f (x). Sketch its derivative in the space provided below. (The sketch does not need to be exact—just show the general features of y′.)