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yв€’1

3= (в€—)

.

xв€’2

This may be simpliп¬Ѓed to y = 3x в€’ 5.

Math Note: The form y = 3x в€’ 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 (в€—) may be rewritten as y в€’ 1 = 3(x в€’ 2). In general, the

line with slope m that passes through the point (x0 , y0 ) can be written as y в€’ y0 =

m(x в€’ 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 (в€’3, 2)

and has slope 4.

EXAMPLE 1.11

Write the equation of the line passing through the points (в€’4, 5) and (6, 2).

SOLUTION

Let (x, y) be a variable point on the line. Using the points (x, y) and (в€’4, 5),

we may calculate the slope to be

yв€’5

m= .

x в€’ (в€’4)

On the other hand, we may use the points (в€’4, 5) and (6, 2) to calculate the

slope:

2в€’5 в€’3

m= = .

6 в€’ (в€’4) 10

Equating the two expressions for slope, we п¬Ѓnd that

yв€’5 в€’3

= .

x+4 10

Simplifying this identity, we п¬Ѓnd that the equation of our line is

в€’3

yв€’5= В· (x + 4).

10

You Try It: Find the equation of the line that passes through the points (2, в€’5)

and (в€’6, 1).

In general, the line that passes through points (x0 , y0 ) and (x1 , y1 ) has equation

y в€’ y0 y1 в€’ y0

= .

x в€’ x0 x1 в€’ x0

This is called the two-point form of the equation of a line.

EXAMPLE 1.12

Find the line perpendicular to y = 3x в€’ 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 в€’1/3.

Using the point-slope form of a line, we may immediately write the equation

of the line with slope в€’1/3 and passing through (5, 4) as

в€’1

yв€’4= В· (x в€’ 5).

3

CHAPTER 1 Basics 15

Summary: We determine the equation of a line in the plane by п¬Ѓnding 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 в€’ y0 = m(x в€’ 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 в€’ y0 y1 в€’ y0

= .

x в€’ x0 x1 в€’ 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 п¬Ѓnd 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 deп¬Ѓned

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

в€’3 9

в€’2 4

в€’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 п¬Ѓnd 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

в€’3 в€’27

в€’2 в€’8

в€’1 в€’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 п¬Ѓnd 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 в€’ 1.

18 CHAPTER 1 Basics

SOLUTION

It is convenient to make a table of values:

x y = x2 + x в€’ 1

в€’4 11

в€’3 5

в€’2 1

в€’1 в€’1

в€’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 п¬Ѓnd 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 п¬Ѓrst 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 deп¬Ѓned 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 в€’ sign).

20 CHAPTER 1 Basics

y

О±

x

positive angle

Fig. 1.26

y

x

О±

negative angle

Fig. 1.27

In degree measure, one full rotation about the unit circle is 360в—¦ ; in radian

measure, one full rotation about the circle is just the circumference of the circle or

2ПЂ . Let us use the symbol Оё to denote an angle. The principle of proportionality

now tells us that

degree measure of Оё radian measure of Оё

= .

360в—¦ 2ПЂ

In other words

ПЂ

radian measure of Оё = В· (degree measure of Оё )

180

and

180

degree measure of Оё = В· (radian measure of Оё ).

ПЂ

CHAPTER 1 Basics 21

EXAMPLE 1.16

Sketch the angle with radian measure ПЂ/6. Give its equivalent degree

measure.

SOLUTION

Since

1

ПЂ/6

=,

2ПЂ 12

the angle subtends an arc of the unit circle corresponding to 1/12 of the full cir-

cumference. Since ПЂ/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 ПЂ

В· = 30в—¦ .

6

ПЂ

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 вЂњthe angle 2ПЂ/3вЂќ 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

2ПЂвЂ”refer to Fig. 1.30.

You Try It: Sketch the angles в€’2, 1, ПЂ, 3ПЂ/2, 10вЂ”all on the same coordinate

п¬Ѓgure. 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 Оё 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 Оё. Look at Fig. 1.32. The number y is called the

sine of Оё and is written sin Оё. The number x is called the cosine of Оё and is written

cos Оё.

Since (cos Оё, sin Оё ) are coordinates of a point on the unit circle, the following

two fundamental properties are immediate:

(1) For any number Оё,

(sin Оё )2 + (cos Оё )2 = 1.

CHAPTER 1 Basics 23

y

5F/6 = 150В°

= 60В°

p/3

x

_ p = _180В°

_ 3p/4 = _135В°

Fig. 1.31

y

unit circle P = (x, y)

sin G

G

cos G x

Fig. 1.32

(2) For any number Оё,

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