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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,

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