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CALCULUS DEMYSTIFIED
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CALCULUS DEMYSTIFIED




STEVEN G. KRANTZ




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DOI: 10.1036/0071412115
To Archimedes, Pierre de Fermat, Isaac Newton, and Gottfried Wilhelm
von Leibniz, the fathers of calculus
This page intentionally left blank.
For more information about this book, click here.




CONTENTS




Preface xi

CHAPTER 1 Basics 1
1.0 Introductory Remarks 1
1.1 Number Systems 1
1.2 Coordinates in One Dimension 3
1.3 Coordinates in Two Dimensions 5
1.4 The Slope of a Line in the Plane 8
1.5 The Equation of a Line 13
1.6 Loci in the Plane 15
1.7 Trigonometry 19
1.8 Sets and Functions 30
1.8.1 Examples of Functions of a Real Variable 31
1.8.2 Graphs of Functions 33
1.8.3 Plotting the Graph of a Function 35
1.8.4 Composition of Functions 40
1.8.5 The Inverse of a Function 42
1.9 A Few Words About Logarithms and Exponentials 49


CHAPTER 2 Foundations of Calculus 57
2.1 Limits 57
2.1.1 One-Sided Limits 60
2.2 Properties of Limits 61
2.3 Continuity 64
2.4 The Derivative 66
2.5 Rules for Calculating Derivatives 71
2.5.1 The Derivative of an Inverse 76
2.6 The Derivative as a Rate of Change 76

vii
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Contents
viii

CHAPTER 3 Applications of the Derivative 81
3.1 Graphing of Functions 81
3.2 Maximum/Minimum Problems 86
3.3 Related Rates 91
3.4 Falling Bodies 94


CHAPTER 4 The Integral 99
4.0 Introduction 99
4.1 Antiderivatives and Inde¬nite Integrals 99
4.1.1 The Concept of Antiderivative 99
4.1.2 The Inde¬nite Integral 100
4.2 Area 103
4.3 Signed Area 111
4.4 The Area Between Two Curves 116
4.5 Rules of Integration 120
4.5.1 Linear Properties 120
4.5.2 Additivity 120


CHAPTER 5 Indeterminate Forms 123
5.1 l™Hôpital™s Rule 123
5.1.1 Introduction 123
5.1.2 l™Hôpital™s Rule 124
5.2 Other Indeterminate Forms 128
5.2.1 Introduction 128
5.2.2 Writing a Product as a Quotient 128
5.2.3 The Use of the Logarithm 128
5.2.4 Putting Terms Over a Common Denominator 130
5.2.5 Other Algebraic Manipulations 131
5.3 Improper Integrals: A First Look 132
5.3.1 Introduction 132
5.3.2 Integrals with In¬nite Integrands 133
5.3.3 An Application to Area 139
5.4 More on Improper Integrals 140
5.4.1 Introduction 140
5.4.2 The Integral on an In¬nite Interval 141
5.4.3 Some Applications 143
Contents ix

CHAPTER 6 Transcendental Functions 147
6.0 Introductory Remarks 147
6.1 Logarithm Basics 147
6.1.1 A New Approach to Logarithms 148
6.1.2 The Logarithm Function and the Derivative 150
6.2 Exponential Basics 154
6.2.1 Facts About the Exponential Function 155
6.2.2 Calculus Properties of the Exponential 156
6.2.3 The Number e 158
6.3 Exponentials with Arbitrary Bases 160
6.3.1 Arbitrary Powers 160
6.3.2 Logarithms with Arbitrary Bases 163
6.4 Calculus with Logs and Exponentials to Arbitrary Bases 166
6.4.1 Differentiation and Integration of loga x and a x 166
6.4.2 Graphing of Logarithmic and Exponential
Functions 168
6.4.3 Logarithmic Differentiation 170
6.5 Exponential Growth and Decay 172
6.5.1 A Differential Equation 173
6.5.2 Bacterial Growth 174
6.5.3 Radioactive Decay 176
6.5.4 Compound Interest 178
6.6 Inverse Trigonometric Functions 180
6.6.1 Introductory Remarks 180
6.6.2 Inverse Sine and Cosine 180
6.6.3 The Inverse Tangent Function 185
6.6.4 Integrals in Which Inverse Trigonometric Functions
Arise 187
6.6.5 Other Inverse Trigonometric Functions 189
6.6.6 An Example Involving Inverse Trigonometric
Functions 193


CHAPTER 7 Methods of Integration 197
7.1 Integration by Parts 197
7.2 Partial Fractions 202
7.2.1 Introductory Remarks 202
7.2.2 Products of Linear Factors 203
7.2.3 Quadratic Factors 206
Contents
x

7.3 Substitution 207
7.4 Integrals of Trigonometric Expressions 210


CHAPTER 8 Applications of the Integral 217
8.1 Volumes by Slicing 217
8.1.0 Introduction 217
8.1.1 The Basic Strategy 217
8.1.2 Examples 219
8.2 Volumes of Solids of Revolution 224
8.2.0 Introduction 224
8.2.1 The Method of Washers 225
8.2.2 The Method of Cylindrical Shells 228
8.2.3 Different Axes 231
8.3 Work 233
8.4 Averages 237


Y
8.5 Arc Length and Surface Area 240
FL
8.5.1 Arc Length 240
8.5.2 Surface Area 243
8.6 Hydrostatic Pressure 247
AM

8.7 Numerical Methods of Integration 252
8.7.1 The Trapezoid Rule 253
8.7.2 Simpson™s Rule 256
TE



Bibliography 263

Solutions to Exercises 265

Final Exam 313

Index 339
PREFACE



Calculus is one of the milestones of Western thought. Building on ideas of
Archimedes, Fermat, Newton, Leibniz, Cauchy, and many others, the calculus is
arguably the cornerstone of modern science. Any well-educated person should
at least be acquainted with the ideas of calculus, and a scienti¬cally literate person
must know calculus solidly.
Calculus has two main aspects: differential calculus and integral calculus.
Differential calculus concerns itself with rates of change. Various types of change,
both mathematical and physical, are described by a mathematical quantity called
the derivative. Integral calculus is concerned with a generalized type of addition,
or amalgamation, of quantities. Many kinds of summation, both mathematical and
physical, are described by a mathematical quantity called the integral.
What makes the subject of calculus truly powerful and seminal is the Funda-
mental Theorem of Calculus, which shows how an integral may be calculated by
using the theory of the derivative. The Fundamental Theorem enables a number
of important conceptual breakthroughs and calculational techniques. It makes the
subject of differential equations possible (in the sense that it gives us ways to solve
these equations).
Calculus Demysti¬ed explains this panorama of ideas in a step-by-step and acces-
sible manner. The author, a renowned teacher and expositor, has a strong sense of
the level of the students who will read this book, their backgrounds and their
strengths, and can present the material in accessible morsels that the student can
study on his own. Well-chosen examples and cognate exercises will reinforce the
ideas being presented. Frequent review, assessment, and application of the ideas
will help students to retain and to internalize all the important concepts of calculus.
We envision a book that will give the student a ¬rm grounding in calculus.
The student who has mastered this book will be able to go on to study physics,
engineering, chemistry, computational biology, computer science, and other basic
scienti¬c areas that use calculus.
Calculus Demysti¬ed will be a valuable addition to the self-help literature.
Written by an accomplished and experienced teacher (the author of How to Teach
Mathematics), this book will aid the student who is working without a teacher.


xi
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
Preface
xii

It will provide encouragement and reinforcement as needed, and diagnostic exer-
cises will help the student to measure his or her progress. A comprehensive exam
at the end of the book will help the student to assess his mastery of the subject, and
will point to areas that require further work.
We expect this book to be the cornerstone of a series of elementary mathematics
books of the same tenor and utility.

Steven G. Krantz
St. Louis, Missouri
CHAPTER 1



Basics

1.0 Introductory Remarks
Calculus is one of the most important parts of mathematics. It is fundamental to all
of modern science. How could one part of mathematics be of such central impor-
tance? It is because calculus gives us the tools to study rates of change and motion.
All analytical subjects, from biology to physics to chemistry to engineering to math-
ematics, involve studying quantities that are growing or shrinking or moving”in
other words, they are changing. Astronomers study the motions of the planets,
chemists study the interaction of substances, physicists study the interactions of
physical objects. All of these involve change and motion.
In order to study calculus effectively, you must be familiar with cartesian geome-
try, with trigonometry, and with functions. We will spend this ¬rst chapter reviewing
the essential ideas. Some readers will study this chapter selectively, merely review-
ing selected sections. Others will, for completeness, wish to review all the material.
The main point is to get started on calculus (Chapter 2).


1.1 Number Systems
The number systems that we use in calculus are the natural numbers, the integers,
the rational numbers, and the real numbers. Let us describe each of these:
• The natural numbers are the system of positive counting numbers 1, 2, 3, ¦.
We denote the set of all natural numbers by N.

1
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
2 CHAPTER 1 Basics

• The integers are the positive and negative whole numbers and zero:
. . . , ’3, ’2, ’1, 0, 1, 2, 3, . . . . We denote the set of all integers by Z.
• The rational numbers are quotients of integers. Any number of the form p/q,
with p, q ∈ Z and q = 0, is a rational number. We say that p/q and r/s
represent the same rational number precisely when ps = qr. Of course you
know that in displayed mathematics we write fractions in this way:
12 7
+=.
23 6
• The real numbers are the set of all decimals, both terminating and non-
terminating. This set is rather sophisticated, and bears a little discussion. A
decimal number of the form
x = 3.16792
is actually a rational number, for it represents
316792
x = 3.16792 = .
100000
A decimal number of the form
m = 4.27519191919 . . . ,
with a group of digits that repeats itself interminably, is also a rational number.
To see this, notice that
100 · m = 427.519191919 . . .
and therefore we may subtract:
100m = 427.519191919 . . .
m = 4.275191919 . . .
Subtracting, we see that
99m = 423.244
or
423244
m= .
99000
So, as we asserted, m is a rational number or quotient of integers.
The third kind of decimal number is one which has a non-terminating dec-
imal expansion that does not keep repeating. An example is 3.14159265 . . . .
This is the decimal expansion for the number that we ordinarily call π. Such
a number is irrational, that is, it cannot be expressed as the quotient of two
integers.
CHAPTER 1 Basics 3

In summary: There are three types of real numbers: (i) terminating decimals,
(ii) non-terminating decimals that repeat, (iii) non-terminating decimals that do not
repeat. Types (i) and (ii) are rational numbers. Type (iii) are irrational numbers.
You Try It: What type of real number is 3.41287548754875 . . . ? Can you express
this number in more compact form?


1.2 Coordinates in One Dimension
We envision the real numbers as laid out on a line, and we locate real numbers from
left to right on this line. If a < b are real numbers then a will lie to the left of b on
this line. See Fig. 1.1.
_3 _2 _1 0 1 2 3 4
a b

Fig. 1.1

EXAMPLE 1.1
On a real number line, plot the numbers ’4, ’1, 2, 6. Also plot the sets
S = {x ∈ R: ’ 8 ¤ x < ’5} and T = {t ∈ R: 7 < t ¤ 9}. Label the plots.

SOLUTION
Figure 1.2 exhibits the indicated points and the two sets. These sets are called
half-open intervals because each set includes one endpoint and not the other.
_9 _6 _3 0 3 6 9



_9 _6 _3 0 3 6 9
S T

Fig. 1.2

Math Note: The notation S = {x ∈ R: ’ 8 ¤ x < ’5} is called set builder
notation. It says that S is the set of all numbers x such that x is greater than or equal
to ’8 and less than 5. We will use set builder notation throughout the book.
If an interval contains both its endpoints, then it is called a closed interval. If an
interval omits both its endpoints, then it is called an open interval. See Fig. 1.3.
closed interval open interval


Fig. 1.3
4 CHAPTER 1 Basics

EXAMPLE 1.2
Find the set of points that satisfy x ’ 2 < 4 and exhibit it on a number line.
SOLUTION
We solve the inequality to obtain x < 6. The set of points satisfying this
inequality is exhibited in Fig. 1.4.
_9 _6 _3 0 3 6 9


Fig. 1.4

EXAMPLE 1.3
Find the set of points that satisfy the condition
|x + 3| ¤ 2 (*)
and exhibit it on a number line.

SOLUTION
In case x + 3 ≥ 0 then |x + 3| = x + 3 and we may write condition (—) as
x+3¤2
or
x ¤ ’1.
Combining x + 3 ≥ 0 and x ¤ ’1 gives ’3 ¤ x ¤ ’1.
On the other hand, if x + 3 < 0 then |x + 3| = ’(x + 3). We may then write
condition (—) as
’(x + 3) ¤ 2
or
’5 ¤ x.
Combining x + 3 < 0 and ’5 ¤ x gives ’5 ¤ x < ’3.
We have found that our inequality |x + 3| ¤ 2 is true precisely when either
’3 ¤ x ¤ ’1 or ’5 ¤ x < ’3. Putting these together yields ’5 ¤ x ¤ ’1.
We display this set in Fig. 1.5.

_9 _6 _3 0 3 6 9

Fig. 1.5

You Try It: Solve the inequality |x’4| > 1. Exhibit your answer on a number line.
You Try It: On a real number line, sketch the set {x: x 2 ’ 1 < 3}.
CHAPTER 1 Basics 5

1.3 Coordinates in Two Dimensions
We locate points in the plane by using two coordinate lines (instead of the single
line that we used in one dimension). Refer to Fig. 1.6. We determine the coordinates
of the given point P by ¬rst determining the x-displacement, or (signed) distance
from the y-axis and then determining the y-displacement, or (signed) distance from
the x-axis. We refer to this coordinate system as (x, y)-coordinates or Cartesian
coordinates. The idea is best understood by way of some examples.

y



P



x



Fig. 1.6



EXAMPLE 1.4
Plot the points P = (3, ’2), Q = (’4, 6), R = (2, 5), S = (’5, ’3).

SOLUTION
The ¬rst coordinate 3 of the point P tells us that the point is located 3 units
to the right of the y-axis (because 3 is positive). The second coordinate ’2 of
the point P tells us that the point is located 2 units below the x-axis (because
’2 is negative). See Fig. 1.7.
The ¬rst coordinate ’4 of the point Q tells us that the point is located 4 units
to the left of the y-axis (because ’4 is negative). The second coordinate 6 of
the point Q tells us that the point is located 6 units above the x-axis (because
6 is positive). See Fig. 1.7.
The ¬rst coordinate 2 of the point R tells us that the point is located 2 units
to the right of the y-axis (because 2 is positive). The second coordinate 5 of the
point R tells us that the point is located 5 units above the x-axis (because 5 is
positive). See Fig. 1.7.
The ¬rst coordinate ’5 of the point S tells us that the point is located 5 units
to the left of the y-axis (because ’5 is negative). The second coordinate ’3 of
the point S tells us that the point is located 3 units below the x-axis (because
’3 is negative). See Fig. 1.7.
6 CHAPTER 1 Basics

y


Q
R
4



1

x
1 4

S P




Fig. 1.7

EXAMPLE 1.5
Give the coordinates of the points X, Y, Z, W exhibited in Fig. 1.8.

y




Z



Y

x

X

W




Fig. 1.8


SOLUTION
The point X is 1 unit to the right of the y-axis and 3 units below the x-axis.
Therefore its coordinates are (1, ’3).
The point Y is 2 units to the left of the y-axis and 1 unit above the x-axis.
Therefore its coordinates are (’2, 1).
CHAPTER 1 Basics 7

The point Z is 5 units to the right of the y-axis and 4 units above the x-axis.
Therefore its coordinates are (5, 4).
The point W is 6 units to the left of the y-axis and 5 units below the x-axis.
Therefore its coordinates are (’6, ’5).
You Try It: Sketch the points (3, ’5), (2, 4), (π, π/3) on a set of axes. Sketch
the set {(x, y): x = 3} on another set of axes.
EXAMPLE 1.6
= {(x, y): y = 3}. Sketch the set of points k =
Sketch the set of points
{(x, y): x = ’4}.
SOLUTION
The set consists of all points with y-coordinate equal to 3. This is the set
of all points that lie 3 units above the x-axis. We exhibit in Fig. 1.9. It is a
horizontal line.




l




Fig. 1.9

The set k consists of all points with x-coordinate equal to ’4. This is the set
of all points that lie 4 units to the left of the y-axis. We exhibit k in Fig. 1.10.
It is a vertical line.

EXAMPLE 1.7
Sketch the set of points S = {(x, y): x > 2} on a pair of coordinate axes.
SOLUTION
Notice that the set S contains all points with x-coordinate greater than 2.
These will be all points to the right of the vertical line x = 2. That set is
exhibited in Fig. 1.11.

You Try It: Sketch the set {(x, y): x + y < 4}.
You Try It: Identify the set (using set builder notation) that is shown in Fig. 1.12.
8 CHAPTER 1 Basics




k




Fig. 1.10

y




Y
FL
AM

x
TE




Fig. 1.11



1.4 The Slope of a Line in the Plane
A line in the plane may rise gradually from left to right, or it may rise quite steeply
from left to right (Fig. 1.13). Likewise, it could fall gradually from left to right,
or it could fall quite steeply from left to right (Fig. 1.14). The number “slope”
differentiates among these different rates of rise or fall.
Look at Fig. 1.15. We use the two points P = (p1 , p2 ) and Q = (q1 , q2 ) to
calculate the slope. It is
q 2 ’ p2
m= .
q1 ’ p1
CHAPTER 1 Basics 9

y




1

x
1




Fig. 1.12

y




x




Fig. 1.13

It turns out that, no matter which two points we may choose on a given line, this
calculation will always give the same answer for slope.
EXAMPLE 1.8
Calculate the slope of the line in Fig. 1.16.

SOLUTION
We use the points P = (’1, 0) and Q = (1, 3) to calculate the slope of
this line:
3’0 3
m= =.
1 ’ (’1) 2
10 CHAPTER 1 Basics

y




x




Fig. 1.14

y




Q



P

x




Fig. 1.15


We could just as easily have used the points P = (’1, 0) and R = (3, 6) to
calculate the slope:
6’0 6 3
m= ==.
3 ’ (’1) 4 2
If a line has slope m, then, for each unit of motion from left to right, the line
rises m units. In the last example, the line rises 3/2 units for each unit of motion to
the right. Or one could say that the line rises 3 units for each 2 units of motion to
the right.
CHAPTER 1 Basics 11

y


R = (3,6)



Q = (1,3)

P = (_ 1,0)
x




Fig. 1.16

y



R = (_ 2,10) 10
8
6
S = (_ 1,5)
4
2

x
246

T = (1,_ 5)




Fig. 1.17


EXAMPLE 1.9
Calculate the slope of the line in Fig. 1.17.

SOLUTION
We use the points R = (’2, 10) and T = (1, ’5) to calculate the slope of
this line:
10 ’ (’5)
m= = ’5.
(’2) ’ 1
12 CHAPTER 1 Basics

We could just as easily have used the points S = (’1, 5) and T = (1, ’5):
5 ’ (’5)
m= = ’5.
’1 ’ 1
In this example, the line falls 5 units for each 1 unit of left-to-right motion. The
negativity of the slope indicates that the line is falling.
The concept of slope is unde¬ned for a vertical line. Such a line will have any
two points with the same x-coordinate, and calculation of slope would result in
division by 0.

You Try It: What is the slope of the line y = 2x + 8?

You Try It: What is the slope of the line y = 5? What is the slope of the line
x = 3?
Two lines are perpendicular precisely when their slopes are negative reciprocals.
This makes sense: If one line has slope 5 and the other has slope ’1/5 then we
see that the ¬rst line rises 5 units for each unit of left-to-right motion while the
second line falls 1 unit for each 5 units of left-to-right motion. So the lines must be
perpendicular. See Fig. 1.18(a).

y




x




Fig. 1.18(a)


You Try It: Sketch the line that is perpendicular to x +2y = 7 and passes through
(1, 4).
Note also that two lines are parallel precisely when they have the same slope.
See Fig. 1.18(b).
CHAPTER 1 Basics 13

y




x


Fig. 1.18(b)



1.5 The Equation of a Line
The equation of a line in the plane will describe”in compact form”all the points
that lie on that line. We determine the equation of a given line by writing its
slope in two different ways and then equating them. Some examples best illustrate
the idea.
EXAMPLE 1.10
Determine the equation of the line with slope 3 that passes through the point
(2, 1).

SOLUTION
Let (x, y) be a variable point on the line. Then we can use that variable point
together with (2, 1) to calculate the slope:
y’1
m= .
x’2
On the other hand, we are given that the slope is m = 3. We may equate the
two expressions for slope to obtain
y’1
3= (—)
.
x’2
This may be simpli¬ed to y = 3x ’ 5.

Math Note: The form y = 3x ’ 5 for the equation of a line is called the slope-
intercept form. The slope is 3 and the line passes through (0, 5) (its y-intercept).

Math Note: Equation (—) may be rewritten as y ’ 1 = 3(x ’ 2). In general, the
line with slope m that passes through the point (x0 , y0 ) can be written as y ’ y0 =
m(x ’ x0 ). This is called the point-slope form of the equation of a line.
14 CHAPTER 1 Basics

You Try It: Write the equation of the line that passes through the point (’3, 2)
and has slope 4.
EXAMPLE 1.11
Write the equation of the line passing through the points (’4, 5) and (6, 2).
SOLUTION
Let (x, y) be a variable point on the line. Using the points (x, y) and (’4, 5),
we may calculate the slope to be
y’5
m= .
x ’ (’4)
On the other hand, we may use the points (’4, 5) and (6, 2) to calculate the
slope:
2’5 ’3
m= = .
6 ’ (’4) 10
Equating the two expressions for slope, we ¬nd that
y’5 ’3
= .
x+4 10
Simplifying this identity, we ¬nd that the equation of our line is
’3
y’5= · (x + 4).
10
You Try It: Find the equation of the line that passes through the points (2, ’5)
and (’6, 1).
In general, the line that passes through points (x0 , y0 ) and (x1 , y1 ) has equation
y ’ y0 y1 ’ y0
= .
x ’ x0 x1 ’ x0
This is called the two-point form of the equation of a line.
EXAMPLE 1.12
Find the line perpendicular to y = 3x ’ 6 that passes through the point
(5, 4).
SOLUTION
We know from the Math Note immediately after Example 1.10 that the given
line has slope 3. Thus the line we seek (the perpendicular line) has slope ’1/3.
Using the point-slope form of a line, we may immediately write the equation
of the line with slope ’1/3 and passing through (5, 4) as
’1
y’4= · (x ’ 5).
3
CHAPTER 1 Basics 15

Summary: We determine the equation of a line in the plane by ¬nding two
expressions for the slope and equating them.
If a line has slope m and passes through the point (x0 , y0 ) then it has equation
y ’ y0 = m(x ’ x0 ).
This is the point-slope form of a line.
If a line passes through the points (x0 , y0 ) and (x1 , y1 ) then it has equation
y ’ y0 y1 ’ y0
= .
x ’ x0 x1 ’ x0
This is the two-point form of a line.
You Try It: Find the line perpendicular to 2x + 5y = 10 that passes through
the point (1, 1). Now ¬nd the line that is parallel to the given line and passes
through (1, 1).


1.6 Loci in the Plane
The most interesting sets of points to graph are collections of points that are de¬ned
by an equation. We call such a graph the locus of the equation. We cannot give all
the theory of loci here, but instead consider a few examples. See [SCH2] for more
on this matter.
EXAMPLE 1.13
Sketch the graph of {(x, y): y = x 2 }.
SOLUTION
It is convenient to make a table of values:

x y = x2
’3 9
’2 4
’1 1
0 0
1 1
2 4
3 9

We plot these points on a single set of axes (Fig. 1.19). Supposing that the
curve we seek to draw is a smooth interpolation of these points (calculus will
later show us that this supposition is correct), we ¬nd that our curve is as shown
in Fig. 1.20. This curve is called a parabola.
16 CHAPTER 1 Basics

y




x



Fig. 1.19




Fig. 1.20

EXAMPLE 1.14
Sketch the graph of the curve {(x, y): y = x 3 }.

SOLUTION
It is convenient to make a table of values:

x y = x3
’3 ’27
’2 ’8
’1 ’1
0 0
1 1
2 8
3 27
We plot these points on a single set of axes (Fig. 1.21). Supposing that the
curve we seek to draw is a smooth interpolation of these points (calculus will
later show us that this supposition is correct), we ¬nd that our curve is as shown
in Fig. 1.22. This curve is called a cubic.
CHAPTER 1 Basics 17

y




6
3

x
36




Fig. 1.21




Fig. 1.22


You Try It: Sketch the graph of the locus |x| = |y|.
EXAMPLE 1.15
Sketch the graph of the curve y = x 2 + x ’ 1.
18 CHAPTER 1 Basics

SOLUTION
It is convenient to make a table of values:


x y = x2 + x ’ 1
’4 11
’3 5
’2 1
’1 ’1
’1
0
1 1
2 5
3 11

We plot these points on a single set of axes (Fig. 1.23). Supposing that the
curve we seek to draw is a smooth interpolation of these points (calculus will


Y
later show us that this supposition is correct), we ¬nd that our curve is as shown
in Fig. 1.24. This is another example of a parabola.
FL
y
AM
TE




x




Fig. 1.23


You Try It: Sketch the locus y 2 = x 3 + x + 1 on a set of axes.
The reader unfamiliar with cartesian geometry and the theory of loci would do
well to consult [SCH2].
CHAPTER 1 Basics 19




Fig. 1.24


1.7 Trigonometry
Here we give a whirlwind review of basic ideas of trigonometry. The reader who
needs a more extensive review should consult [SCH1].
When we ¬rst learn trigonometry, we do so by studying right triangles and
measuring angles in degrees. Look at Fig. 1.25. In calculus, however, it is convenient
to study trigonometry in a more general setting, and to measure angles in radians.




=

= measured in degrees
Fig. 1.25

Angles will be measured by rotation along the unit circle in the plane, beginning
at the positive x-axis. See Fig. 1.26. Counterclockwise rotation corresponds to
positive angles, and clockwise rotation corresponds to negative angles. Refer to
Fig. 1.27. The radian measure of an angle is de¬ned to be the length of the arc
of the unit circle that the angle subtends with the positive x-axis (together with an
appropriate + or ’ sign).
20 CHAPTER 1 Basics

y




±
x




positive angle



Fig. 1.26

y




x
±




negative angle



Fig. 1.27

In degree measure, one full rotation about the unit circle is 360—¦ ; in radian
measure, one full rotation about the circle is just the circumference of the circle or
2π . Let us use the symbol θ to denote an angle. The principle of proportionality
now tells us that
degree measure of θ radian measure of θ
= .
360—¦ 2π
In other words
π
radian measure of θ = · (degree measure of θ )
180
and
180
degree measure of θ = · (radian measure of θ ).
π
CHAPTER 1 Basics 21

EXAMPLE 1.16
Sketch the angle with radian measure π/6. Give its equivalent degree
measure.

SOLUTION
Since
1
π/6
=,
2π 12
the angle subtends an arc of the unit circle corresponding to 1/12 of the full cir-
cumference. Since π/6 > 0, the angle represents a counterclockwise rotation.
It is illustrated in Fig. 1.28.
y




p/6
x




Fig. 1.28

The degree measure of this angle is
180 π
· = 30—¦ .
6
π
Math Note: In this book we always use radian measure for angles. (The reason
is that it makes the formulas of calculus turn out to be simpler.) Thus, for example,
if we refer to “the angle 2π/3” then it should be understood that this is an angle in
radian measure. See Fig. 1.29.
Likewise, if we refer to the angle 3 it is also understood to be radian measure.
We sketch this last angle by noting that 3 is approximately 0.477 of a full rotation
2π”refer to Fig. 1.30.

You Try It: Sketch the angles ’2, 1, π, 3π/2, 10”all on the same coordinate
¬gure. Of course use radian measure.
22 CHAPTER 1 Basics

y




2F/3



x




Fig. 1.29

y




3


x




Fig. 1.30

EXAMPLE 1.17
Several angles are sketched in Fig. 1.31, and both their radian and degree
measures given.
If θ is an angle, let (x, y) be the coordinates of the terminal point of the corre-
sponding radius (called the terminal radius) on the unit circle. We call P = (x, y)
the terminal point corresponding to θ. Look at Fig. 1.32. The number y is called the
sine of θ and is written sin θ. The number x is called the cosine of θ and is written
cos θ.
Since (cos θ, sin θ ) are coordinates of a point on the unit circle, the following
two fundamental properties are immediate:
(1) For any number θ,
(sin θ )2 + (cos θ )2 = 1.
CHAPTER 1 Basics 23

y




5F/6 = 150°
= 60°
p/3




x
_ p = _180°
_ 3p/4 = _135°




Fig. 1.31

y


unit circle P = (x, y)

sin G
G
cos G x




Fig. 1.32


(2) For any number θ,

’1 ¤ cos θ ¤ 1 ’ 1 ¤ sin θ ¤ 1.
and


Math Note: It is common to write sin2 θ to mean (sin θ )2 and cos2 θ to mean
(cos θ )2 .

EXAMPLE 1.18
Compute the sine and cosine of π/3.
24 CHAPTER 1 Basics

SOLUTION
We sketch the terminal radius and associated triangle (see Fig. 1.33). This is

a 30“60“90 triangle whose sides have ratios 1 : 3 : 2. Thus
1 1
=2 x= .
or
2
x
Likewise,

√ √ 3
y
=3 y = 3x =
or .
2
x
It follows that

3
π
sin =
3 2
and
1
π
=.
cos
3 2

y




unit circle
√3
2
F/3
x
1
2




Fig. 1.33


You Try It: The cosine of a certain angle is 2/3. The angle lies in the fourth
quadrant. What is the sine of the angle?

Math Note: Notice that if θ is an angle then θ and θ + 2π have the same terminal
radius and the same terminal point (for adding 2π just adds one more trip around
the circle”look at Fig. 1.34).
As a result,
sin θ = x = sin(θ + 2π )
CHAPTER 1 Basics 25

y




sin G
G
cos G x

unit circle




Fig. 1.34

and
cos θ = y = cos(θ + 2π ).
We say that the sine and cosine functions have period 2π : the functions repeat
themselves every 2π units.
In practice, when we calculate the trigonometric functions of an angle θ, we
reduce it by multiples of 2π so that we can consider an equivalent angle θ , called
the associated principal angle, satisfying 0 ¤ θ < 2π . For instance,
15π/2 has associated principal angle
(since 15π/2 ’ 3π/2 = 3 · 2π )
3π/2
and
’10π/3 has associated principal angle
(since ’ 10π/3 ’ 2π/3 = ’12π/3 = ’2 · 2π ).
2π/3
You Try It: What are the principal angles associated with 7π , 11π/2, 8π/3,
’14π/5, ’16π/7?
What does the concept of angle and sine and cosine that we have presented here
have to do with the classical notion using triangles? Notice that any angle θ such
that 0 ¤ θ < π/2 has associated to it a right triangle in the ¬rst quadrant, with
vertex on the unit circle, such that the base is the segment connecting (0, 0) to (x, 0)
and the height is the segment connecting (x, 0) to (x, y). See Fig. 1.35.
26 CHAPTER 1 Basics

y

unit circle (x, y)
opposite
side
G
x
G
adjacent side




Fig. 1.35

Then
opposite side of triangle
y
sin θ = y = =
1 hypotenuse
and
adjacent side of triangle
x
cos θ = x = = .
1 hypotenuse
Thus, for angles θ between 0 and π/2, the new de¬nition of sine and cosine using
the unit circle is clearly equivalent to the classical de¬nition using adjacent and
opposite sides and the hypotenuse. For other angles θ, the classical approach is to
reduce to this special case by subtracting multiples of π/2. Our approach using the
unit circle is considerably clearer because it makes the signatures of sine and cosine
obvious.
Besides sine and cosine, there are four other trigonometric functions:
sin θ
y
tan θ = =
cos θ
x
cos θ
x
cot θ = =
sin θ
y
1 1
sec θ = =
cos θ
x
1 1
csc θ = = .
sin θ
y
Whereas sine and cosine have domain the entire real line, we notice that tan θ and
sec θ are unde¬ned at odd multiples of π/2 (because cosine will vanish there) and
cot θ and csc θ are unde¬ned at even multiples of π/2 (because sine will vanish
there). The graphs of the six trigonometric functions are shown in Fig. 1.36.
EXAMPLE 1.19
Compute all the trigonometric functions for the angle θ = 11π/4.
CHAPTER 1 Basics 27

1 1


0.5 0.5


_6 _4 _2 _6 _4 _2
2 4 6 2 4 6

_ 0.5 _ 0.5


_1 _1

Graphs of y = sin x and y = cos x.
Fig. 1.36(a)



30 30

20 20

10 10

_6 _4 _2 _6 _4 _2
2 4 6 2 4 6
_ 10 _ 10

_ 20 _ 20

_ 30 _ 30


Graphs of y = tan x and y = cot x.
Fig. 1.36(b)



15 15

10 10

5 5
_4 _2 _2 6
2 4 4
_4
_6 _6
6 2
_5 _5

_10 _10

_15 _15


Graphs of y = sec x and y = csc x.
Fig. 1.36(c)



SOLUTION
We ¬rst notice that the principal associated angle is 3π/4, so we deal with
that angle. Figure 1.37 shows that the triangle associated to this angle is an
isosceles right triangle with hypotenuse 1.
28 CHAPTER 1 Basics

y




11F/4
1
√2

x
1
√2

unit circle




Y Fig. 1.37
FL
AM

√ √
Therefore x = ’1/ 2 and y = 1/ 2. It follows that
1
sin θ = y = √
TE



2
1
cos θ = x = ’ √
2
y
tan θ = = ’1

. 1
( 11)



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