563


563
A Look Back, A Look Ahead
This chapter is in two parts: Polar Coordinates (Sections 9.1–9.3) and Vectors
(Sections 9.4–9.7). They are independent of each other and may be covered in
either order.
Sections 9.1–9.3: In Chapter 1 we introduced rectangular coordinates
(the xy-plane) and discussed the graph of an equation in two variables involving
x and y. In Sections 9.1 and 9.2, we introduce polar coordinates, an alternative to
rectangular coordinates, and discuss graphing equations that involve polar
coordinates. In Section 5.3, we discussed raising a real number to a real power.
In Section 9.3, we extend this idea by raising a complex number to a real power.
As it turns out, polar coordinates are useful for the discussion.
Sections 9.4–9.7: We have seen in many chapters that we are often
required to solve an equation to obtain a solution to applied problems. In the
last four sections of this chapter, we develop the notion of a vector and show
how it can be used to model applied problems in physics and engineering.
9
Polar Coordinates; Vectors
Outline
9.1
Polar Coordinates
9.2
Polar Equations and Graphs
9.3
The Complex Plane;
De Moivre’s Theorem
9.4
Vectors
9.5
The Dot Product
9.6
Vectors in Space
9.7
The Cross Product

Chapter Review

Chapter Test

Cumulative Review

Chapter Projects
How Airplanes Fly
Four aerodynamic forces act on an airplane in flight: lift, drag, thrust, and weight (gravity).
Drag is the resistance of air molecules hitting the airplane (the backward force),
thrust is the power of the airplane’s engine (the forward force), lift is the upward force,
and weight is the downward force. So for airplanes to fly and stay airborne, the thrust
must be greater than the drag, and the lift must be greater than the
weight (so, drag opposes thrust, and lift opposes weight).
This is certainly the case when an airplane takes off or climbs.
However, when it is in straight and level flight, the opposing
forces of lift and weight are balanced. During a descent, weight
exceeds lift, and to slow the airplane, drag has to overcome thrust.
Thrust is generated by the airplane’s engine (propeller or jet), weight is created
by the natural force of gravity acting on the airplane, and drag comes from friction
as the plane moves through air molecules. Drag is also a reaction to lift, and this lift
must be generated by the airplane in flight. This is done by the wings of the airplane.
A cross section of a typical airplane wing shows the top surface to be more curved
than the bottom surface. This shaped profile is called an airfoil (or aerofoil), and the
shape is used because an airfoil generates significantly more lift than opposing drag.
In other words, it is very efficient at generating lift.
During flight, air naturally flows over and beneath the wing and is deflected
upward over the top surface and downward beneath the lower surface. Any difference
in deflection causes a difference in air pressure (pressure gradient), and because of the
airfoil shape, the pressure of the deflected air is lower above the airfoil than below it.
As a result the wing is “pushed” upward by the higher pressure beneath, or, you can
argue, it is “sucked” upward by the lower pressure above.
Source: Adapted from Pete Carpenter. How Airplanes Fly—The Basic Principles of Flight
http://www.rc-airplane-world.com/how-airplanes-fly.html,
accessed June 2014. © rc-airplane-world.com
—See Chapter Project I—
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564 CHAPTER 9  Polar Coordinates; Vectors
So far, we have always used a system of rectangular coordinates to plot points in
the plane. Now we are ready to describe another system, called polar coordinates.
In many instances, polar coordinates offer certain advantages over rectangular
coordinates.
In a rectangular coordinate system, you will recall, a point in the plane is
represented by an ordered pair of numbers 1x, y2 , where x and y equal the signed
distances of the point from the y-axis and the x-axis, respectively. In a polar coordinate
system, we select a point, called the pole, and then a ray with vertex at the pole,
called the polar axis. See Figure 1. Comparing the rectangular and polar coordinate
systems, note that the origin in rectangular coordinates coincides with the pole in
polar coordinates, and the positive x-axis in rectangular coordinates coincides with
the polar axis in polar coordinates.
Plot Points Using Polar Coordinates
A point P in a polar coordinate system is represented by an ordered pair of numbers
1r, u2 . If r 7 0, then r is the distance of the point from the pole; u is an angle (in
degrees or radians) formed by the polar axis and a ray from the pole through the
point. We call the ordered pair 1r, u2 the polar coordinates of the point. See Figure 2.
As an example, suppose that a point P has polar coordinates a2, p
4
b . Locate P
by first drawing an angle of
p
4
radian, placing its vertex at the pole and its initial side
along the polar axis. Then go out a distance of 2 units along the terminal side of the
angle to reach the point P. See Figure 3.
1
Now Work the ‘Are You Prepared?’ problems on page 571.

ObjeCtives 1 Plot Points Using Polar Coordinates (p. 564)

2 Convert from Polar Coordinates to Rectangular Coordinates (p. 566)

3 Convert from Rectangular Coordinates to Polar Coordinates (p. 568)

4 Transform Equations between Polar and Rectangular Forms (p. 570)
9.1 Polar Coordinates
• Rectangular Coordinates (Section 1.1, pp. 2–4)
• Definition of the Trigonometric Functions
(Section 6.2, pp. 367 and 377)
• Inverse Tangent Function (Section 7.1, pp. 447–449)
• Completing the Square (Appendix A,
Section A.3, pp. A28–A29)
PRePARiNG FOR tHis seCtiON  Before getting started, review the following:
Figure 2
O Pole
Polar axis
P 5 (r, u)
r
u
Figure 3
P  2,
(
)–4
O Pole
Polar axis
2
–
4
In using polar coordinates 1r, u2 , it is possible for r to be negative. When this
happens, instead of the point being on the terminal side of u, it is on the ray from
the pole extending in the direction opposite the terminal side of u at a distance
0 r
0
units from the pole. See Figure 4 for an illustration.
For example, to plot the point a-3, 2p
3
b , use the ray in the opposite direction
of
2p
3
and go out
0 -3
0 = 3 units along that ray. See Figure 5.
Figure 1
x
O Pole
Polar axis
y
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SECTion 9.1  Polar Coordinates 565
Plotting Points Using Polar Coordinates
Plot the points with the following polar coordinates:
(a) a3, 5p
3
b


(b) a2, - p
4
b


(c) 13, 02


(d) a-2, p
4
b
Figure 6 shows the points.
ExamplE 1
Solution
Figure 4
O
P 5 (r, u), r , 0
u
_r _
Figure 5
3,
(
)
2
––
3
2
––
3
O
Figure 6
3,
5p
––
3
5p
––
3
O
D
(
2,
O
p–
4
2
p–
4
2
E
(
O
(3, 0)
F
O
22,
p–
4
p–
4
G
(
)
)
)
Now Work p r o b l e m s 1 1 , 1 9 , a n d 2 9
Recall that an angle measured counterclockwise is positive and an angle
measured clockwise is negative. This convention has some interesting consequences
related to polar coordinates.
Finding Several Polar Coordinates of a Single Point
Consider again the point P with polar coordinates a2, p
4
b , as shown in Figure 7(a).
Because
p
4
,
9p
4
, and -
7p
4
all have the same terminal side, this point P also can be
located by using the polar coordinates a2, 9p
4
b or the polar coordinates a2, - 7p
4
b ,
as shown in Figures 7(b) and (c). The point a2, p
4
b can also be represented by the
polar coordinates a-2, 5p
4
b . See Figure 7(d).
•
ExamplE 2
Figure 7
•
O
(a)
P  2, –4
–
4
(
)
O
P  2,
9
––
4
(b)
9
–––
4
(
)
P  2,
7
–––
4
O

 7
––
4
(c)
(
)
–
4
O
P  2,
5
––
4
(d)
5
––
4
(
)
2
2
2
2
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566 CHAPTER 9  Polar Coordinates; Vectors
Finding Other Polar Coordinates of a Given Point
Plot the point P with polar coordinates a3, p
6
b , and find other polar coordinates
1r, u2 of this same point for which:
(a) r 7 0, 2p … u 6 4p





(b) r 6 0, 0 … u 6 2p
(c) r 7 0, -2p … u 6 0
The point a3, p
6
b is plotted in Figure 8.
(a) Add 1 revolution 12p radians2 to the angle p
6
to get
P = a3, p
6
+ 2pb = a3, 13p
6
b .
See Figure 9.
(b) Add
1
2
revolution 1p radians2 to the angle p
6
, and replace 3 by -3 to get

P = a-3, p
6
+ pb = a-3, 7p
6
b . See Figure 10.
(c) Subtract 2p from the angle
p
6
to get P = a3, p
6
- 2pb = a3, - 11p
6
b . See
Figure 11.
ExamplE 3
Solution
Figure 9
P  3,
13
–––
6
13
–––
6
O
(
)
Figure 10
P  3,
7
––
6
7
––
6
O
(
)
Figure 11
•
P  3, 
11
–––
6
11
–––
6
O
(
)
Figure 8
P  3,
–
6
–
6
O
(
)
These examples show a major difference between rectangular coordinates and
polar coordinates. A point has exactly one pair of rectangular coordinates; however,
a point has infinitely many pairs of polar coordinates.
SummAry
A point with polar coordinates 1r, u2 , u in radians, can also be represented by either of the following:
1r, u + 2pk2 or 1- r, u + p + 2pk2 k any integer
The polar coordinates of the pole are 10, u2 , where u can be any angle.
Now Work p r o b l e m 3 3
Convert from Polar Coordinates to Rectangular Coordinates
Sometimes it is necessary to convert coordinates or equations in rectangular
form to polar form, and vice versa. To do this, recall that the origin in rectangular
coordinates is the pole in polar coordinates and that the positive x-axis in rectangular
coordinates is the polar axis in polar coordinates.
2
Conversion from Polar Coordinates to Rectangular Coordinates
If P is a point with polar coordinates 1r, u2 , the rectangular coordinates 1x, y2
of P are given by

x = r cos u y = r sin u
(1)
TheOrem
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SECTion 9.1  Polar Coordinates 567
Proof Suppose that P has the polar coordinates 1r, u2 . We seek the rectangular
coordinates 1x, y2 of P. Refer to Figure 12.
If r = 0, then, regardless of u, the point P is the pole, for which the rectangular
coordinates are 10, 02 . Formula (1) is valid for r = 0.
If r 7 0, the point P is on the terminal side of u, and r = d1O, P2 = 2x2 + y2 .
Because
cos u =
x
r
sin u =
y
r
this means
x = r cos u y = r sin u
If r 6 0 and u is in radians, the point P = 1r, u2 can be represented as
1- r, p + u2 , where - r 7 0. Because
cos1p + u2 = - cos u = x- r sin1p + u2 = - sin u =
y
- r
this means

x = r cos u y = r sin u
■
Converting from Polar Coordinates to rectangular Coordinates
Find the rectangular coordinates of the points with the following polar coordinates:
(a) a6, p
6
b



(b) a-4, - p
4
b
Use formula (1): x = r cos u and y = r sin u.
(a) Figure 13(a) shows a6, p
6
b plotted. Notice that a6, p
6
b lies in quadrant I of the
rectangular coordinate system. So both the x-coordinate and the y-coordinate
will be positive. Substituting r = 6 and u =
p
6
gives
x = r cos u = 6 cos
p
6
= 6 # 13
2
= 323
y = r sin u = 6 sin
p
6
= 6 # 1
2
= 3
The rectangular coordinates of the point a6, p
6
b are 1323, 32 , which lies in
quadrant I, as expected.
(b) Figure 13(b) shows a-4, - p
4
b plotted. Notice that a-4, - p
4
b lies in quadrant II
of the rectangular coordinate system. Substituting r = -4 and u = -
p
4
gives
x = r cos u = -4 cosa- p
4
b = -4 # 12
2
= -222
y = r sin u = -4 sina- p
4
b = -4a- 12
2
b = 222
The rectangular coordinates of the point a-4, - p
4
b are 1 -222 , 222 2 ,
which lies in quadrant II, as expected.
Now Work p r o b l e m s 4 1 a n d 5 3
ExamplE 4
Solution
•
Comment Many
calculators
have
the capability of converting from polar
coordinates to rectangular coordinates.
Consult your owner’s manual for the
proper keystrokes. In most cases this
procedure is tedious, so you will probably
find that using formula (1) is faster.
■
Figure 12
P
O
x
x
y
y
r
u
Figure 13
x
3 3
6,
6
3
–
6
–
6
(a)
y
(
)
4
x
2 2
4, 
2 2
–
4
–
4
(b)

y
(
)
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568 CHAPTER 9  Polar Coordinates; Vectors
Convert from Rectangular Coordinates to Polar Coordinates
Converting from rectangular coordinates 1x, y2 to polar coordinates 1r, u2 is a little
more complicated. Notice that each solution begins by plotting the given rectangular
coordinates.
how to Convert from rectangular Coordinates to Polar
Coordinates with the Point on a Coordinate Axis
Find polar coordinates of a point whose rectangular coordinates are 10, 32 .
3
ExamplE 5
Step-by-Step Solution
Step 1: Plot the point (x, y) and
note the quadrant the point lies in
or the coordinate axis the point lies
on.
The point 10, 32 lies on the y-axis a distance of 3 units from the origin (pole), so r = 3.
Plot the point 10, 32 in a rectangular
coordinate system. See Figure 14. The point
lies on the positive y-axis.
Step 2: Determine the distance r
from the origin to the point.
Step 3: Determine u.
A ray with vertex at the pole through 10, 32 forms an angle u = p
2
with the polar axis.
Polar coordinates for this point can be given by a3, p
2
b . Other possible
representations include a-3, -p
2
b and a3, 5p
2
b .
•
Figure 15 shows polar coordinates of points that lie on either the x-axis or the
y-axis. In each illustration, a 7 0.
Now Work p r o b l e m 5 7
how to Convert from rectangular Coordinates to Polar
Coordinates with the Point in a Quadrant
Find the polar coordinates of a point whose rectangular coordinates are 12, -22 .
ExamplE 6
Step-by-Step Solution
Step 1: Plot the point (x, y) and
note the quadrant the point lies in
or the coordinate axis the point
lies on.
Plot the point 12, -22 in a rectangular
coordinate system. See Figure 16. The point
lies in quadrant IV.
Figure 14
(x, y)  (0, 3)
x
y
3
–
2
Figure 15
x
(r, u) 5 (a, 0)
(x, y) 5 (a, 0)
a
y
D[\5aa.
p–
2
p–
2
x
(r, u) 5 a,
(x, y) 5 (0, a)
a
y
E[\5aa.
(
)
x
(r, u) 5 (a, p)
(x, y) 5 (2a, 0)
p
a
y
F[\52aa.
(r, u) 5 a,
x
a
y
G[\52aa.
(x, y)5(0,2a)
3p
–––
2
3p
–––
2
(
)
Figure 16
21
1
21
22
2
(x, y ) 5 (2, 22)
u
x
y
r
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SECTion 9.1  Polar Coordinates 569
Find u by recalling that tan u =
y
x
, so u = tan-1
y
x
, -
p
2
6 u 6
p
2
. Because 12, -22
lies in quadrant IV, this means that -
p
2
6 u 6 0. As a result,
u = tan-1
y
x
= tan-1 a -2
2
b = tan-1 1-12 = - p
4

A set of polar coordinates for the point 12, -22 is a222, -p
4
b . Other possible
representations include a222, 7p
4
b and a-222, 3p
4
b .
Converting from rectangular Coordinates to Polar Coordinates
Find polar coordinates of a point whose rectangular coordinates are 1 -1, - 23 2 .
Step 1: See Figure 17. The point lies in quadrant III.
Step 2: The distance r from the origin to the point 1 -1, - 23 2 is
r = 31-122 + 1 - 23 22 = 24 = 2
Step 3: To find u, use a = tan-1
y
x
= tan-1
- 13
-1
= tan-1 23, -p
2
6 a 6
p
2
.
Since the point 1 -1, - 23 2 lies in quadrant III and the inverse tangent
function gives an angle in quadrant I, add p to the result to obtain an angle
in quadrant III.
u = p + tan-1 a - 13
-1
b = p + tan-1 23 = p + p
3
=
4p
3
A set of polar coordinates for this point is a2, 4p
3
b . Other possible
representations include a-2, p
3
b and a2, - 2p
3
b .
Figure 18 shows how to find polar coordinates of a point that lies in a quadrant
when its rectangular coordinates 1x, y2 are given.
•
ExamplE 7
Solution
•
Comment Many calculators have the
capability of converting from rectangular
coordinates
to
polar
coordinates.
Consult your owner’s manual for the
proper keystrokes.
■
Step 2: Determine the distance r
from the origin to the point using
r = 2x 2 + y 2.
r = 2x2 + y2 = 21222 + 1-222 = 28 = 222
Step 3: Determine u.
r 5 x 2 1 y 2
u 5 tan21
D
x
y
(x, y )
r 5 x 2 1 y 2
u 5 p 1 tan21
E
x
y
(x, y )
r 5 x 2 1 y 2
u 5 p 1 tan21
F
x
y
(x, y )
r 5 x 2 1 y 2
u 5 tan21
G
(x, y )
x
y
u
r
u
r
u
r
u
r
y
x
y
x
y
x
y
x
Figure 18
The preceding discussion provides the formulas

r 2 = x2 + y2 tan u =
y
x
if x ∙ 0
(2)
Figure 17
(x, y ) 5 (21, 2 3)
u
x
r
y
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570 CHAPTER 9  Polar Coordinates; Vectors
To use formula (2) effectively, follow these steps:
steps for Converting from Rectangular to Polar Coordinates
step 1: Always plot the point 1x, y2 first, as shown in Examples 5, 6, and 7.
Note the quadrant the point lies in or the coordinate axis the point
lies on.
step 2:
If x = 0 or y = 0, use your illustration to find r.

If x ∙ 0 and y ∙ 0, then r = 2x2 + y2.
step 3: Find u. If x = 0 or y = 0, use your illustration to find u.

If x ∙ 0 and y ∙ 0, note the quadrant in which the point lies.
Quadrant I or IV: u = tan-1
y
x
Quadrant II or III: u = p + tan-1
y
x
Now Work p r o b l e m 6 1
transform equations between Polar and Rectangular Forms
Formulas (1) and (2) can also be used to transform equations from polar form to
rectangular form, and vice versa. Two common techniques for transforming an
equation from polar form to rectangular form are
1. Multiplying both sides of the equation by r
2. Squaring both sides of the equation
Transforming an equation from Polar to rectangular Form
Transform the equation r = 6 cos u from polar coordinates to rectangular
coordinates, and identify the graph.
Multiplying each side by r makes it easier to apply formulas (1) and (2).
r = 6 cos u
r 2 = 6r cos u
x2 + y2 = 6x
This is the equation of a circle. Complete the square to obtain the standard form of
the equation.
x2 + y2 = 6x
1x2 - 6x2 + y2 = 0
1x2 - 6x + 92 + y2 = 9
1x - 322 + y2 = 9
This is the standard form of the equation of a circle with center 13, 02 and radius 3.
Now Work p r o b l e m 7 7
Transforming an equation from rectangular to Polar Form
Transform the equation 4xy = 9 from rectangular coordinates to polar coordinates.
4
ExamplE 8
Solution
Multiply each side by r.
r 2 = x 2 + y 2 ; x = r cos u
General form
Complete the square in x.
Factor.
•
ExamplE 9
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SECTion 9.1  Polar Coordinates 571
Use formula (1): x = r cos u and y = r sin u.
4xy = 9
41r cos u2 1r sin u2 = 9
4r 2 cos u sin u = 9
This is the polar form of the equation. It can be simplified as follows:
2r 2 12 sin u cos u2 = 9
2r 2 sin12u2 = 9
Now Work p r o b l e m 7 1
Solution
x = r cos u, y = r sin u
Factor out 2r2.
Double-angle Formula
•
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Plot the point whose rectangular coordinates are 13, -12 .
What quadrant does the point lie in? (pp. 2–4)
2. To complete the square of x2 + 6x, add
.
(pp. A28–A29)
3. If P = 1a, b2 is a point on the terminal side of the angle u

at a distance r from the origin, then tan u = .
(p. 377)
4. tan-1 1- 12 =
. (pp. 447–449)
9.1 Assess Your Understanding
Concepts and vocabulary
5. The origin in rectangular coordinates coincides with
the
in polar coordinates; the positive x-axis in
rectangular coordinates coincides with the

in
polar coordinates.
6. If P is a point with polar coordinates (r, u), the rectangular
coordinates (x, y) of P are given by x =
and
y =
.
7. For the point with polar coordinates a1, -p
2
b , which of

the following best describes the location of the point in a
rectangular coordinate system?
(a) in quadrant IV
(b) on the y-axis
(c) in quadrant II
(d) on the x-axis
8. The point a5, p
6
b can also be represented by which of the
following polar coordinates?
(a) a5, -p
6
b
(b) a-5, 13p
6
b
(c) a5, - 5p
6
b
(d) a- 5, 7p
6
b
9. True or False In the polar coordinates (r, u), r can be negative.
10. True or False The polar coordinates of a point are unique.
skill building
In Problems 11–18, match each point in polar coordinates with either A, B, C, or D on the graph.
11. a2, - 11p
6
b

12. a-2, - p
6
b

13. a-2, p
6
b

14. a2, 7p
6
b
15. a2, 5p
6
b

16. a-2, 5p
6
b

17. a-2, 7p
6
b

18. a2, 11p
6
b
In Problems 19–32, plot each point given in polar coordinates.
π
6
C
D
B
A
2
19. 13, 90°2
20. 14, 270°2
21. 1-2, 02
22. 1-3, p2
23. a6, p
6
b
24. a5, 5p
3
b
25. 1-2, 135°2
26. 1-3, 120°2
27. a4, - 2p
3
b
28. a2, - 5p
4
b
29. a-1, - p
3
b
30. a-3, - 3p
4
b
31. 1-2, -p2
32. a-3, - p
2
b
In Problems 33–40, plot each point given in polar coordinates, and find other polar coordinates 1r, u2 of the point for which:
(a) r 7 0, -2p … u 6 0

(b) r 6 0, 0 … u 6 2p

(c) r 7 0, 2p … u 6 4p
33. a5, 2p
3
b
34. a4, 3p
4
b
35. 1-2, 3p2
36. 1-3, 4p2
37. a1, p
2
b
38. 12, p2
39. a-3, - p
4
b
40. a-2, - 2p
3
b
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572 CHAPTER 9  Polar Coordinates; Vectors
In Problems 41–56, polar coordinates of a point are given. Find the rectangular coordinates of each point.
41. a3, p
2
b

42. a4, 3p
2
b

43. 1-2, 02

44. 1-3, p2
45. 16, 150°2

46. 15, 300°2

47. a-2, 3p
4
b
48. a-2, 2p
3
b
49. a- 1, - p
3
b

50. a-3, - 3p
4
b
51. 1-2, -180°2

52. 1-3, -90°2
53. 17.5, 110°2

54. 1-3.1, 182°2

55. 16.3, 3.82

56. 18.1, 5.22
In Problems 57–68, the rectangular coordinates of a point are given. Find polar coordinates for each point.
57. 13, 02

58. 10, 22

59. 1-1, 02

60. 10, -22
61. 11, - 12

62. 1-3, 32

63. 123 , 12

64. 1 - 2, - 223 2
65. 11.3, -2.12

66. 1-0.8, -2.12
67. 18.3, 4.22

68. 1-2.3, 0.22
In Problems 69–76, the letters x and y represent rectangular coordinates. Write each equation using polar coordinates 1r, u2 .
69. 2x2 + 2y2 = 3
70. x2 + y2 = x

71. x2 = 4y
72. y2 = 2x
73. 2xy = 1

74. 4x2 y = 1
75. x = 4

76. y = -3
In Problems 77–84, the letters r and u represent polar coordinates. Write each equation using rectangular coordinates 1x, y2 .
77. r = cos u
78. r = sin u + 1
79. r 2 = cos u
80. r = sin u - cos u
81. r = 2

82. r = 4

83. r =
4
1 - cos u

84. r =
3
3 - cos u
Applications and extensions
85. Chicago In Chicago, the road system is set up like a
Cartesian plane, where streets are indicated by the number
of blocks they are from Madison Street and State Street.
For example, Wrigley Field in Chicago is located at 1060
West Addison, which is 10 blocks west of State Street and
36 blocks north of Madison Street. Treat the intersection of
Madison Street and State Street as the origin of a coordinate
system, with east being the positive x-axis.
(a) Write the location of Wrigley Field using rectangular
coordinates.
(b) Write the location of Wrigley Field using polar
coordinates. Use the east direction for the polar axis.
Express u in degrees.
(c) U.S. Cellular Field, home of the White Sox, is located
at 35th and Princeton, which is 3 blocks west of State
Street and 35 blocks south of Madison. Write the
location of U.S. Cellular Field using rectangular
coordinates.
(d) Write the location of U.S. Cellular Field using polar
coordinates. Use the east direction for the polar axis.
Express u in degrees.
86. Show that the formula for the distance d between two points
P1 = 1r1 , u1 2 and P2 = 1r2 , u2 2 is
d = 2r 12 + r 22 - 2r1 r2 cos1u2 - u1 2
City of Chicago, Illinois
Addison Street
Addison Street
35th Street
1 mile
1 km
Wrigley Field
1060 West Addison
U.S. Cellular Field
35th and Princeton
Madison Street
35th Street
State Street
N
explaining Concepts: Discussion and Writing
87. In converting from polar coordinates to rectangular
coordinates, what formulas will you use?
88. Explain how to convert from rectangular coordinates to
polar coordinates.
89. Is the street system in your town based on a rectangular
coordinate system, a polar coordinate system, or some other
system? Explain.
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SECTion 9.2  Polar Equations and Graphs 573
Retain Your Knowledge
Problems 90–93 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
90. Solve: log 41x + 32 - log 41x - 12 = 2
91. Use Descartes’ Rule of Signs to determine the possible
number of positive or negative real zeros for the function
f1x2 = - 2x3 + 6x2 - 7x - 8.
92. Find the midpoint of the line segment connecting the
points 1-3, 72 and a1
2
, 2b .
93. Given that the point (3, 8) is on the graph of y = f1x2 ,
what is the corresponding point on the graph of
y = -2f1x + 32 + 5?
‘Are You Prepared?’ Answers
1. ; quadrant IV
2. 9
3.
b
a

4. -
p
4
x
y
2
2
2
2
4
(3, 1)
Now Work the ‘Are You Prepared?’ problems on page 585.

ObjeCtives 1 Identify and Graph Polar Equations by Converting to Rectangular
Equations (p. 574)

2 Test Polar Equations for Symmetry (p. 577)

3 Graph Polar Equations by Plotting Points (p. 578)
9.2 Polar equations and Graphs
• Symmetry (Section 1.2, pp. 12–14)
• Circles (Section 1.4, pp. 34–37)
• Even–Odd Properties of Trigonometric Functions
(Section 6.3, p. 393)
• Difference Formulas for Sine and Cosine
(Section 7.4, pp. 478 and 481)
• Values of the Sine and Cosine Functions at Certain
Angles (Section 6.2, pp. 369–376)
PRePARiNG FOR tHis seCtiON  Before getting started, review the following:
Just as a rectangular grid may be used to plot points given by rectangular coordinates,
such as the points (−3, 1) and (1, 2) shown in Figure 19(a), a grid consisting of concentric
circles (with centers at the pole) and rays (with vertices at the pole) can be used
to plot points given by polar coordinates, such as the points a4, 5p
4
b and a2, p
4
b
shown in Figure 19(b). Such polar grids are used to graph polar equations.
Figure 19
u 5 0
u 5 p
u 5
u 5
u 5
u 5
u 5
u 5
p–
2
3p
––
2
7p
––
4
p–
4
3p
––
4
5p
––
4
x
24 2221
23
4
2 3
1
O
y
2
4
24
1
23
3
22
(1, 2)
(23, 1)
DRectangular grid
O
r 5 5
r 5 3
r 5 1
2,
4,
(
)
(
)
EPolar grid
p–
4
5p
––
4
M09_SULL9070_10_SE_C09_pp563-637.indd 573
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574 CHAPTER 9  Polar Coordinates; Vectors
Figure 21 u =
p
4
, or y = x
x
2
1
5
4
3
O
y
p–
4
u 5 0
u 5 p
u 5p–2
u 53p
––
2
u 5 7p
––
4
u 5p–4
u 5 3p
––
4
u 5 5p
––
4
identify and Graph Polar equations by Converting
to Rectangular equations
One method that can be used to graph a polar equation is to convert the equation to
rectangular coordinates. In the following discussion, 1x, y2 represents the rectangular
coordinates of a point P, and 1r, u2 represents polar coordinates of the point P.
Identifying and Graphing a Polar equation (Circle)
Identify and graph the equation: r = 3
Convert the polar equation to a rectangular equation.
r = 3
r 2 = 9
x2 + y2 = 9
The graph of r = 3 is a circle, with center at the pole and radius 3. See Figure 20.
1
ExamplE 1
Solution
Square both sides.
r 2 = x 2 + y 2
An equation whose variables are polar coordinates is called a polar equation.
The graph of a polar equation consists of all points whose polar coordinates
satisfy the equation.
DeFInITIOn
Figure 20 r = 3, or x2 + y2 = 9
•
x
u 5 0
u 5 p
p–
2
3p
––
2
7p
––
4
p–
4
u 5
u 5
u 5
u 5
u 5
u 5
3p
––
4
5p
––
4
2
1
5
4
3
O
y
Now Work p r o b l e m 1 5
Identifying and Graphing a Polar equation (Line)
Identify and graph the equation: u =
p
4
Convert the polar equation to a rectangular equation.
u =
p
4
tan u = tan
p
4


y
x
= 1
y = x
The graph of u =
p
4
is a line passing through the pole making an angle of
p
4
with
the polar axis. See Figure 21.
Now Work p r o b l e m 1 7
ExamplE 2
Solution
Take the tangent of both sides.
tan u =
y
x
; tan
p
4
= 1
•
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SECTion 9.2  Polar Equations and Graphs 575
Identifying and Graphing a Polar equation (horizontal Line)
Identify and graph the equation: r sin u = 2
ExamplE 3
Comment A graphing utility can be
used to graph polar equations. Read
Using a Graphing Utility to Graph a Polar
Equation, Appendix B, Section B.8.
■
Figure 23 r cos u = -3, or x = -3
•
Because y = r sin u, we can write the
equation as
y = 2
Therefore, the graph of r sin u = 2 is a
horizontal line 2 units above the pole.
See Figure 22.
Solution
Figure 22 r sin u = 2, or y = 2
•
x
5
4
3
2
O
y
1
u 5 0
u 5 p
u 5p–
2
u 53p
––
2
u 5 7p
––
4
u 5p–4
u 5 3p
––
4
u 5 5p
––
4
Identifying and Graphing a Polar equation (Vertical Line)
Identify and graph the equation: r cos u = -3
ExamplE 4
Since x = r cos u, we can write the
equation as
x = -3
Therefore, the graph of r cos u = -3
is a vertical line 3 units to the left of
the pole. See Figure 23.
Solution
x
O
y
4 5
2 3
1
u 5 0
u 5 p
u 5p–2
u 53p
––
2
u 5 7p
––
4
u 5p–4
u 5 3p
––
4
u 5 5p
––
4
Examples 3 and 4 lead to the following results. (The proofs are left as exercises.
See Problems 83 and 84.)
Let a be a real number. Then the graph of the equation
r sin u = a
is a horizontal line. It lies a units above the pole if a Ú 0 and lies
0 a
0 units
below the pole if a 6 0.
The graph of the equation
r cos u = a
is a vertical line. It lies a units to the right of the pole if a Ú 0 and lies
0 a
0 units
to the left of the pole if a 6 0.
TheOrem
Now Work p r o b l e m 2 1
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576 CHAPTER 9  Polar Coordinates; Vectors
Figure 25
r = -2 cos u, or (x + 1)2 + y2 = 1
x
O
4 5
2 3
1
y
u 5 0
u 5 p
u 5p–
2
u 53p
––
2
u 5 7p
––
4
u 5p–4
u 5 3p
––
4
u 5 5p
––
4
Identifying and Graphing a Polar equation (Circle)
Identify and graph the equation: r = 4 sin u
To transform the equation to rectangular coordinates, multiply each side by r.
r 2 = 4r sin u
Now use the facts that r 2 = x2 + y2 and y = r sin u. Then
x2 + y2 = 4y
x2 + 1y2 - 4y2 = 0
x2 + 1y2 - 4y + 42 = 4
x2 + 1y - 222 = 4
This is the standard equation of a circle with center at 10, 22 in rectangular
coordinates and radius 2. See Figure 24.
Identifying and Graphing a Polar equation (Circle)
Identify and graph the equation: r = -2 cos u
To transform the equation to rectangular coordinates, multiply each side by r.
r 2 = -2r cos u
x2 + y2 = -2x

x2 + 2x + y2 = 0
1x2 + 2x + 12 + y2 = 1
1x + 122 + y2 = 1
This is the standard equation of a circle with center at 1-1, 02 in rectangular
coordinates and radius 1. See Figure 25.
ExamplE 5
Solution
Complete the square in y.
Factor.
•
ExamplE 6
Solution
Multiply both sides by r.
r 2 = x 2 + y 2 ; x = r cos u
Complete the square in x.
Factor.
•
exploration
Using a square screen, graph r1 = sin u, r2 = 2 sin u, and r3 = 3 sin u. Do you see the pattern? Clear
the screen and graph r1 = - sin u, r2 = -2 sin u, and r3 = -3 sin u. Do you see the pattern? Clear the
screen and graph r1 = cos u, r2 = 2 cos u, and r3 = 3 cos u. Do you see the pattern? Clear the screen
and graph r1 = - cos u, r2 = -2 cos u, and r3 = -3 cos u. Do you see the pattern?
Based on Examples 5 and 6 and the Exploration above, we are led to the
following results. (The proofs are left as exercises. See Problems 85–88.)
Let a be a positive real number. Then
Equation
Description
(a) r = 2a sin u
Circle: radius a; center at 10, a2 in rectangular coordinates
(b) r = -2a sin u Circle: radius a; center at 10, -a2 in rectangular coordinates
(c) r = 2a cos u
Circle: radius a; center at 1a, 02 in rectangular coordinates
(d) r = -2a cos u Circle: radius a; center at 1-a, 02 in rectangular coordinates
Each circle passes through the pole.
TheOrem
Now Work p r o b l e m 2 3
Figure 24
r = 4 sin u, or x2 + (y - 2)2 = 4
x
O
y
4 5
2 3
1
u 5 0
u 5 p
u 5p–
2
u 53p
––
2
u 5 7p
––
4
u 5p–4
u 5 3p
––
4
u 5 5p
––
4
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SECTion 9.2  Polar Equations and Graphs 577
The method of converting a polar equation to an identifiable rectangular
equation to obtain the graph is not always helpful, nor is it always necessary. Usually,
a table is created that lists several points on the graph. By checking for symmetry, it
may be possible to reduce the number of points needed to draw the graph.
test Polar equations for symmetry
In polar coordinates, the points 1r, u2 and 1r, - u2 are symmetric with respect to the
polar axis (and to the x-axis). See Figure 26(a). The points 1r, u2 and 1r, p - u2 are
symmetric with respect to the line u =
p
2
(the y-axis). See Figure 26(b). The points
1r, u2 and 1- r, u2 are symmetric with respect to the pole (the origin). See Figure 26(c).
2
D Points symmetric with
respect to the polar axis
x
u 5 0
u 5 p
u 5p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
2
1
5
4
3
O
y
u
(r, u)
(r, 2u)
p
––
2
E Points symmetric with
respect to the line u 5
x
u 5 0
u 5 p
p–
2
3p
––
2
7p
––
4
p–
4
u 5
u 5
u 5
u 5
u 5
3p
––
4
u 5 5p
––
4
2
1
5
4
3
O
y
u
(r, u)
u
(r, p 2 u)
(r, u 1 p)
u 1 p
F Points symmetric with
respect to the pole
x
u 5 0
u 5 p
u 5p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
2
1
5
4
3
y
O
u
(2r, u)
(r, u)
p 2 u
2u
Figure 26
tests for symmetry
symmetry with Respect to the Polar Axis (x-Axis)
In a polar equation, replace u by - u. If an equivalent equation results, the
graph is symmetric with respect to the polar axis.
symmetry with Respect to the Line U ∙
P
2
(y-Axis)
In a polar equation, replace u by p - u. If an equivalent equation results,
the graph is symmetric with respect to the line u =
p
2
.
symmetry with Respect to the Pole (Origin)
In a polar equation, replace r by - r or u by u + p. If an equivalent
equation results, the graph is symmetric with respect to the pole.
TheOrem
The following tests are a consequence of these observations.
The three tests for symmetry given here are sufficient conditions for symmetry,
but they are not necessary conditions. That is, an equation may fail these tests and
still have a graph that is symmetric with respect to the polar axis, the line u =
p
2
, or
the pole. For example, the graph of r = sin12u2 turns out to be symmetric with
respect to the polar axis, the line u =
p
2
, and the pole, but only the test for symmetry
with respect to the pole (replace u by u + p) works. See also Problems 89–91.
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578 CHAPTER 9  Polar Coordinates; Vectors
Graph Polar equations by Plotting Points
Graphing a Polar equation (Cardioid)
Graph the equation: r = 1 - sin u
Check for symmetry first.
Polar Axis: Replace u by - u. The result is
r = 1 - sin1- u2 = 1 + sin u
sin (- u) = - sin u
The test fails, so the graph may or may not be symmetric with respect to the polar
axis.
The Line U ∙
P
2
: Replace u by p - u. The result is
r = 1 - sin1p - u2 = 1 - 1sin p cos u - cos p sin u2
= 1 - 30 # cos u - 1-12 sin u4 = 1 - sin u
The test is satisfied, so the graph is symmetric with respect to the line u =
p
2
.
The Pole: Replace r by - r. Then the result is - r = 1 - sin u, so r = -1 + sin u.
The test fails. Replace u by u + p. The result is
r = 1 - sin(u + p)
= 1 - 3sin u cos p + cos u sin p4
= 1 - 3sin u # (-1) + cos u # 04
= 1 + sin u
This test also fails, so the graph may or may not be symmetric with respect to the
pole.
Next, identify points on the graph by assigning values to the angle u and calculating
the corresponding values of r. Due to the periodicity of the sine function and the
symmetry with respect to the line u =
p
2
, just assign values to u from -
p
2
to
p
2
, as
given in Table 1.
Now plot the points 1r, u2 from Table 1 and trace out the graph, beginning at
the point a2, - p
2
b and ending at the point a0, p
2
b . Then reflect this portion of the
graph about the line u =
p
2
(the y-axis) to obtain the complete graph. See Figure 27.
3
ExamplE 7
Solution
table 1
U
r ∙ 1 ∙ sin U
-
p
2

1 - (-1) = 2
-
p
3

1 - a- 13
2
b ≈ 1.87
-
p
6

1 - a- 1
2
b = 3
2

0
1 - 0 = 1

p
6
1 -
1
2
=
1
2

p
3
1 -
23
2
≈ 0.13

p
2
1 - 1 = 0
The curve in Figure 27 is an example of a cardioid (a heart-shaped curve).
Figure 27 r = 1 - sin u
•
x
y
2
1
(1, 0)
(0, )
(2, )
u 5 0
u 5 p
u 5 p–2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
p–
62
p–
22
(1.87, )p–32
3–
2
p–
6
p–
2
( , )
1–
2
(0.13, )p–3
( , )
exploration
Graph r1 = 1 + sin u. Clear the screen
and graph r1 = 1 - cos u. Clear the
screen and graph r1 = 1 + cos u. Do
you see a pattern?
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SECTion 9.2  Polar Equations and Graphs 579
Now Work p r o b l e m 3 9
Graphing a Polar equation (Limaçon without an Inner Loop)
Graph the equation: r = 3 + 2 cos u
Check for symmetry first.
Polar Axis: Replace u by - u. The result is
r = 3 + 2 cos1- u2 = 3 + 2 cos u cos (- u) =
cos u
The test is satisfied, so the graph is symmetric with respect to the polar axis.
The Line U ∙
P
2
: Replace u by p - u. The result is
r = 3 + 2 cos1p - u2 = 3 + 21cos p cos u + sin p sin u2
= 3 - 2 cos u
The test fails, so the graph may or may not be symmetric with respect to the
line u =
p
2
.
The Pole: Replace r by - r. The test fails, so the graph may or may not be symmetric
with respect to the pole. Replace u by u + p. The test fails, so the graph may or may
not be symmetric with respect to the pole.
Next, identify points on the graph by assigning values to the angle u and
calculating the corresponding values of r. Due to the periodicity of the cosine function
and the symmetry with respect to the polar axis, just assign values to u from 0 to p,
as given in Table 2.
Now plot the points 1r, u2 from Table 2 and trace out the graph, beginning at the
point 15, 02 and ending at the point 11, p2 . Then reflect this portion of the graph
about the polar axis (the x-axis) to obtain the complete graph. See Figure 28.
ExamplE 8
Solution
table 2
U
r ∙ 3 ∙ 2 cos U
0
3 + 2(1) = 5
p
6
3 + 2a23
2
b ≈ 4.73
p
3
3 + 2a1
2
b = 4
p
2
3 + 2(0) = 3
2p
3
3 + 2a- 1
2
b = 2
5p
6
3 + 2a- 23
2
b ≈ 1.27
p
3 + 2(-1) = 1
Cardioids are characterized by equations of the form
r = a11 + cos u2 r = a11 + sin u2
r = a11 - cos u2 r = a11 - sin u2
where a 7 0. The graph of a cardioid passes through the pole.
DeFInITIOn
Figure 28 r = 3 + 2 cos u
•
x
O
y
4
2
u = 0
u = p
u =
3
1
p–
2
u = 3p
––
2
u = 7p
––
4
u = p–4
u = 3p
––
4
u = 5p
––
4
5
(5, 0)
(2, )
(4.73, )p–6
(1.27, )
(1, p)
5p
––
6
2p
––
3
(3, )p–2
(4, )p–3
The curve in Figure 28 is an example of a limaçon (a French word for snail)
without an inner loop.
exploration
Graph r1 = 3 - 2 cos u. Clear the screen
and graph r1 = 3 + 2 sin u. Clear the
screen and graph r1 = 3 - 2 sin u. Do
you see a pattern?
M09_SULL9070_10_SE_C09_pp563-637.indd 579
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580 CHAPTER 9  Polar Coordinates; Vectors
Now Work p r o b l e m 4 5
Graphing a Polar equation (Limaçon with an Inner Loop)
Graph the equation: r = 1 + 2 cos u
First, check for symmetry.
Polar Axis: Replace u by - u. The result is
r = 1 + 2 cos1- u2 = 1 + 2 cos u
The test is satisfied, so the graph is symmetric with respect to the polar axis.
The Line U ∙
P
2
: Replace u by p - u. The result is
r = 1 + 2 cos1p - u2 = 1 + 21cos p cos u + sin p sin u2

= 1 - 2 cos u
The test fails, so the graph may or may not be symmetric with respect to the
line u =
p
2
.
The Pole: Replace r by - r. The test fails, so the graph may or may not be symmetric
with respect to the pole. Replace u by u + p. The test fails, so the graph may or may
not be symmetric with respect to the pole.
Next, identify points on the graph of r = 1 + 2 cos u by assigning values to the
angle u and calculating the corresponding values of r. Due to the periodicity of the
cosine function and the symmetry with respect to the polar axis, just assign values to u
from 0 to p, as given in Table 3.
Now plot the points 1r, u2 from Table 3, beginning at 13, 02 and ending
at 1-1, p2 . See Figure 29(a). Finally, reflect this portion of the graph about the
polar axis (the x-axis) to obtain the complete graph. See Figure 29(b).
ExamplE 9
Solution
table 3
U
r ∙ 1 ∙ 2 cos U
0
1 + 2(1) = 3
p
6
1 + 2a23
2
b ≈ 2.73
p
3
1 + 2a1
2
b = 2
p
2
1 + 2(0) = 1
2p
3
1 + 2a- 1
2
b = 0
5p
6
1 + 2a- 23
2
b ≈ -0.73
p
1 + 2(-1) = -1
Limaçons without an inner loop are characterized by equations of the form
r = a + b cos u r = a + b sin u
r = a - b cos u r = a - b sin u
where a 7 0, b 7 0, and a 7 b. The graph of a limaçon without an inner loop
does not pass through the pole.
DeFInITIOn
The curve in Figure 29(b) is an example of a limaçon with an inner loop.
x
u 5 0
u 5 p
u 5 p–2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u = 3p
––
4
u 5 5p
––
4
2
4
y
2,p–3
1,
(–1, p)
p–
2
20.73,5p
––
6
0,2p
––
3
2.73,
(3, 0)
p–
6
D
(
)
(
)
(
)
(
) (
)
Er 5 1 1 2 cos u
x
u 5 0
u 5 p
u 5 p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
2
4
y
2,p–3
1,
(21, p)
p–
2
20.73,5p
–
6
2.73,
(3, 0)
p–
6
(
)
(
)
(
)
(
)
0,2p
––
3
(
)
Figure 29
•
exploration
Graph r1 = 1 - 2 cos u. Clear the screen
and graph r1 = 1 + 2 sin u. Clear the
screen and graph r1 = 1 - 2 sin u. Do
you see a pattern?
M09_SULL9070_10_SE_C09_pp563-637.indd 580
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SECTion 9.2  Polar Equations and Graphs 581
Now Work p r o b l e m 4 7
Graphing a Polar equation (rose)
Graph the equation: r = 2 cos12u2
Check for symmetry.
Polar Axis: Replace u by - u. The result is
r = 2 cos321- u2 4 = 2 cos12u2
The test is satisfied, so the graph is symmetric with respect to the polar axis.
The Line U ∙
P
2
: Replace u by p - u. The result is
r = 2 cos321p - u2 4 = 2 cos12p - 2u2 = 2 cos12u2
The test is satisfied, so the graph is symmetric with respect to the line u =
p
2
.
The Pole: Since the graph is symmetric with respect to both the polar axis and the
line u =
p
2
, it must be symmetric with respect to the pole.
Next, construct Table 4. Because of the periodicity of the cosine function and the
symmetry with respect to the polar axis, the line u =
p
2
, and the pole, consider only
values of u from 0 to
p
2
.
Plot and connect these points as shown in Figure 30(a). Finally, because of symmetry,
reflect this portion of the graph first about the polar axis (the x-axis) and then about
the line u =
p
2
(the y-axis) to obtain the complete graph. See Figure 30(b).
ExamplE 10
Solution
table 4
U
r ∙ 2 cos(2U)
0
2(1) = 2
p
6
2a1
2
b = 1
p
4
2(0) = 0
p
3
2a- 1
2
b = -1
p
2
2(-1) = -2
Limaçons with an inner loop are characterized by equations of the form
r = a + b cos u r = a + b sin u
r = a - b cos u r = a - b sin u
where a 7 0, b 7 0, and a 6 b. The graph of a limaçon with an inner loop
passes through the pole twice.
DeFInITIOn
The curve in Figure 30(b) is called a rose with four petals.
exploration
Graph r1 = 2 cos14u2 ; clear the screen
and graph r1 = 2 cos16u2 . How many
petals did each of these graphs have?
Clear the screen and graph, in order,
each on a clear screen, r1 = 2 cos13u2 ,
r1 = 2 cos15u2 , and r1 = 2 cos17u2 .
What do you notice about the number of
petals?
D
x
u 5 0
u 5 p
u 5 p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
2
1
5
4
y
3
21,p–3
1, p–6
0, p–4
(2, 0)
(
)
(
)
(
)
Er 5 2 cos (2u)
x
u 5 0
u 5 p
u 5 p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
5
4
y
3
(2, 0)
2
1,p–6
(
)
21,p–3
(
)
22, p–2
(
)
22, p–2
(
)
Figure 30
•
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582 CHAPTER 9  Polar Coordinates; Vectors
Now Work p r o b l e m 5 1
Graphing a Polar equation (Lemniscate)
Graph the equation: r 2 = 4 sin12u2
We leave it to you to verify that the graph is symmetric with respect to the pole.
Because of the symmetry with respect to the pole, consider only those values
of u between u = 0 and u = p. Note that there are no points on the graph
for
p
2
6 u 6 p (quadrant II), since r 2 6 0 for such values. Table 5 lists points on
the graph for values of u = 0 through u =
p
2
. The points from Table 5 where r Ú 0
are plotted in Figure 31(a). The remaining points on the graph may be obtained by
using symmetry. Figure 31(b) shows the final graph drawn.
ExamplE 11
Solution
table 5
U
r2 ∙ 4 sin(2U)
r
0
4(0) = 0
0
p
6
4a13
2
b = 223 {1.9
p
4
4(1) = 4
{2
p
3
4a23
2
b = 223 {1.9
p
2
4(0) = 0
0
Rose curves are characterized by equations of the form
r = a cos1nu2 r = a sin1nu2 a ∙ 0
and have graphs that are rose shaped. If n ∙ 0 is even, the rose has 2n petals;
if n ∙ {1 is odd, the rose has n petals.
DeFInITIOn
The curve in Figure 31(b) is an example of a lemniscate (from the Greek word
for ribbon).
Figure 31
•
D
x
u = 0
u = p
u = p–
2
u = 3p
––
2
u = 7p
––
4
u = p–4
u = 3p
––
4
u = 5p
––
4
1
2
y
1.9,p–6
1.9, p–3
2, p–4
(
(
)
)
(
)
1.9,
(
)p–6
Er 2 = 4 sin (2u)
x
u = 0
u = p
u = p–
2
u = 3p
––
2
u = 7p
––
4
u = p–4
u = 3p
––
4
u = 5p
––
4
1
2
y
(
)
1.9,
(0, 0)
(0, 0)
p–
3
(
)
2, p–4
Lemniscates are characterized by equations of the form
r 2 = a2 sin12u2 r 2 = a2 cos12u2
where a ∙ 0, and have graphs that are propeller shaped.
DeFInITIOn
Now Work p r o b l e m 5 5
Graphing a Polar equation (Spiral)
Graph the equation: r = eu>5
The tests for symmetry with respect to the pole, the polar axis, and the line u =
p
2
fail. Furthermore, there is no number u for which r = 0, so the graph does not pass
through the pole. Observe that r is positive for all u, r increases as u increases, r S 0
ExamplE 12
Solution
M09_SULL9070_10_SE_C09_pp563-637.indd 582
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SECTion 9.2  Polar Equations and Graphs 583
as u S - q , and r S q as u S q . With the help of a calculator, the values in Table 6
can be obtained. See Figure 32.
table 6
U
r ∙ eU>5
-
3p
2

0.39
-p
0.53
-
p
2

0.73
-
p
4

0.85
0
1
p
4
1.17
p
2
1.37
p
1.87
3p
2
2.57
2p
3.51
table 7
Lines
Description
Line passing through the pole
making an angle a with the
polar axis
Vertical line
Horizontal line
Rectangular equation
y = (tan a)x
x = a
y = b
Polar equation
u = a
r cos u = a
r sin u = b
typical graph
Circles
Description
Center at the pole, radius a
Passing through the pole,

tangent to the line u =
p
2
,
center on the polar axis,

radius a
Passing through the pole, tangent

to the polar axis, center on the

line u =
p
2
, radius a
Rectangular equation
x2 + y2 = a2, a 7 0
x2 + y2 = {2ax, a 7 0
x2 + y2 = {2ay, a 7 0
Polar equation
r = a, a 7 0
r = {2a cos u, a 7 0
r = {2a sin u, a 7 0
typical graph
(continued)
Figure 32 r = eu/5
•
x
u 5 0
u 5 p
u 5 p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
4
y
2
1.17, p–4
(
)
(3.51, 2p)
(1, 0)
1.37, p–
2
(
2.57,
)
3p
––
2
(
)
(1.87, p)
The curve in Figure 32 is called a logarithmic spiral, since its equation may be
written as u = 5 ln r and it spirals infinitely both toward the pole and away from it.
Classification of Polar equations
The equations of some lines and circles in polar coordinates and their corresponding
equations in rectangular coordinates are given in Table 7. Also included are the
names and graphs of a few of the more frequently encountered polar equations.
y
x

y
x
y
x
y
x
a
a
y
x
a
y
x
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584 CHAPTER 9  Polar Coordinates; Vectors
sketching Quickly
If a polar equation involves only a sine (or cosine) function, you can quickly obtain
its graph by making use of Table 7, periodicity, and a short table.
Sketching the Graph of a Polar equation Quickly
Graph the equation: r = 2 + 2 sin u
You should recognize the polar equation: Its graph is a cardioid. The period of sin u
is 2p, so form a table using 0 … u … 2p, compute r, plot the points 1r, u2 , and
sketch the graph of a cardioid as u varies from 0 to 2p. See Table 8 and Figure 33.
ExamplE 13
Solution
table 8
U
r ∙ 2 ∙ 2 sin U
0
2 + 2(0) = 2
p
2
2 + 2(1) = 4
p
2 + 2(0) = 2
3p
2
2 + 2(-1) = 0
2p
2 + 2(0) = 2
table 7 (Continued)
Other equations
Name
Cardioid
Limaçon without inner loop
Limaçon with inner loop
Polar equations
r = a { a cos u, a 7 0
r = a { b cos u, 0 6 b 6 a
r = a { b cos u, 0 6 a 6 b

r = a { a sin u, a 7 0
r = a { b sin u, 0 6 b 6 a
r = a { b sin u, 0 6 a 6 b
typical graph
Name
Lemniscate
Rose with three petals
Rose with four petals
Polar equations
r2 = a2 cos(2u), a ∙ 0
r = a sin(3u), a 7 0
r = a sin(2u), a 7 0

r2 = a2 sin(2u), a ∙ 0
r = a cos(3u), a 7 0
r = a cos(2u), a 7 0
typical graph
y
x
y
x
y
x
y
x
y
x
y
x
Figure 33 r = 2 + 2 sin u
•
x
u 5 0
u 5 p
u 5 p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
y
5
4
3
2
(2, p)
(2, 0)
1
3p
––
2
(
)
0,
p
––
2
(
)
4,
Calculus Comment For those of you who are planning to study calculus, a
comment about one important role of polar equations is in order.
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SECTion 9.2  Polar Equations and Graphs 585
In rectangular coordinates, the equation x2 + y2 = 1, whose graph is the unit
circle, is not the graph of a function. In fact, it requires two functions to obtain the
graph of the unit circle:
y1 = 21 - x2 Upper semicircle

y2 = - 21 - x2 Lower semicircle
In polar coordinates, the equation r = 1, whose graph is also the unit circle, does
define a function. For each choice of u, there is only one corresponding value of r,
that is, r = 1. Since many problems in calculus require the use of functions, the
opportunity to express nonfunctions in rectangular coordinates as functions in polar
coordinates becomes extremely useful.
Note also that the vertical-line test for functions is valid only for equations in
rectangular coordinates.
Historical Feature
Polar coordinates seem to have been
invented by Jakob Bernoulli (1654–1705)
in about 1691, although, as with most
such ideas, earlier traces of the notion exist.
Early users of calculus remained committed to
rectangular coordinates, and polar coordinates
did not become widely used until the early
1800s. Even then, it was mostly geometers who
used them for describing odd curves. Finally, about the mid-1800s,
applied mathematicians realized the tremendous simplification that
polar coordinates make possible in the description of objects with
circular or cylindrical symmetry. From then on, their use became
widespread.
Jakob Bernoulli
(1654–1705)
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. If the rectangular coordinates of a point are 14, - 62 ,
the point symmetric to it with respect to the origin is

. (pp. 12–14)
2. The difference formula for cosine is cos1A - B2 = .
(p. 478)
3. The standard equation of a circle with center at 1-2, 52 and
radius 3 is . (pp. 34–37)
4. Is the sine function even, odd, or neither? (p. 393)
5. sin
5p
4
=
. (pp. 369–376)
6. cos
2p
3
=
. (pp. 369–376)
9.2 Assess your understanding
Concepts and vocabulary
7. An equation whose variables are polar coordinates is called
a(n)

.
8. True or False The tests for symmetry in polar coordinates
are always conclusive.
9. To test whether the graph of a polar equation may be
symmetric with respect to the polar axis, replace u by
.
10. To test whether the graph of a polar equation may be symmetric

with respect to the line u =
p
2
, replace u by
.
11. True or False A cardioid passes through the pole.
12. Rose curves are characterized by equations of the form
r = a cos (nu) or r = a sin (nu), a ∙ 0. If n ∙ 0 is even,
the rose has
petals; if n ∙ {1 is odd, the rose has

petals.
13. For a positive real number a, the graph of which of the
following polar equations is a circle with radius a and center
at (a, 0) in rectangular coordinates?
(a) r = 2a sin u
(b) r = -2a sin u
(c) r = 2a cos u
(d) r = -2a cos u
14. In polar coordinates, the points 1r, u2 and 1- r, u2 are
symmetric with respect to which of the following?
(a) the polar axis (or x-axis)
(b) the pole (or origin)
(c) the line u =
p
2
(or y-axis)
(d) the line u =
p
4





(or y = x)
skill building
In Problems 15–30, transform each polar equation to an equation in rectangular coordinates. Then identify and graph the equation.
15. r = 4
16. r = 2
17. u =
p
3

18. u = -
p
4
19. r sin u = 4
20. r cos u = 4
21. r cos u = -2
22. r sin u = - 2
M09_SULL9070_10_SE_C09_pp563-637.indd 585
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586 CHAPTER 9  Polar Coordinates; Vectors
23. r = 2 cos u
24. r = 2 sin u
25. r = -4 sin u
26. r = -4 cos u
27. r sec u = 4
28. r csc u = 8
29. r csc u = -2
30. r sec u = - 4
In Problems 31–38, match each of the graphs (A) through (H) to one of the following polar equations.
31. r = 2

32. u =
p
4


33. r = 2 cos u

34. r cos u = 2
35. r = 1 + cos u

36. r = 2 sin u

37. u =
3p
4


38. r sin u = 2
x
O
y
2
u 5 0
u 5 p
u 5 p–2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
$
x
O
y
u 5 0
u 5 p
u 5
3
1
p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
%
x
O
y
2
u 5 0
u 5 p
u 5 p–2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
&
x
O
y
2
4
u 5 0
u 5 p
u 5 p–2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
'
x
O
y
u 5 0
u 5 p
u 5
3
1
p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
(
x
O
y
u 5 0
u 5 p
u 5
3
1
p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
)
x
O
y
4
u 5 0
u 5 p
u 5
2
p–
2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
*
x
O
y
2 u 5 0
u 5 p
u 5 p–2
u 5 3p
––
2
u 5 7p
––
4
u 5 p–4
u 5 3p
––
4
u 5 5p
––
4
+
In Problems 39–62, identify and graph each polar equation.
39. r = 2 + 2 cos u
40. r = 1 + sin u
41. r = 3 - 3 sin u
42. r = 2 - 2 cos u
43. r = 2 + sin u
44. r = 2 - cos u
45. r = 4 - 2 cos u
46. r = 4 + 2 sin u
47. r = 1 + 2 sin u
48. r = 1 - 2 sin u
49. r = 2 - 3 cos u
50. r = 2 + 4 cos u
51. r = 3 cos12u2
52. r = 2 sin13u2
53. r = 4 sin15u2
54. r = 3 cos14u2
55. r 2 = 9 cos12u2
56. r 2 = sin12u2
57. r = 2u
58. r = 3u
59. r = 1 - cos u
60. r = 3 + cos u
61. r = 1 - 3 cos u
62. r = 4 cos13u2
Mixed Practice
In Problems 63–68, graph each pair of polar equations on the same polar grid. Find the polar coordinates of the point(s) of intersection
and label the point(s) on the graph.
63. r = 8 cos u; r = 2 sec u
64. r = 8 sin u; r = 4 csc u
65. r = sin u; r = 1 + cos u
66. r = 3; r = 2 + 2 cos u
67. r = 1 + sin u; r = 1 + cos u
68. r = 1 + cos u; r = 3 cos u
Applications and extensions
In Problems 69–72, the polar equation for each graph is either r = a + b cos u or r = a + b sin u, a 7 0. Select the correct equation and
find the values of a and b.
69.

70.
x
y
u5
u5
u5
u5 p
u5 0
u5
u5
u5
2
0
4 6 8
3,
(6, 0)
10
(
)
5p
²²
4
7p
²²
4
3p
²²
2
3p
²²
4
p²
2
p²
2
p²
4
x
y
u5
u5
u5
u5 p
u5 0
u5
u5
u5
2
0
4 6 8
3,
(6, p)
10
(
)
5p
²²
4
7p
²²
4
3p
²²
4
3p
²²
2
p²
2
p²
2
p²
4
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SECTion 9.2  Polar Equations and Graphs 587
71.

72.
In Problems 73–82, graph each polar equation.
73. r =
2
1 - cos u

(parabola)
74. r =
2
1 - 2 cos u

(hyperbola)
75. r =
1
3 - 2 cos u

(ellipse)
76. r =
1
1 - cos u

(parabola)
77. r = u, u Ú 0 (spiral of Archimedes)
78. r =
3
u

(reciprocal spiral)
79. r = csc u - 2, 0 6 u 6 p
(conchoid)
80. r = sin u tan u
(cissoid)
81. r = tan u, -
p
2
6 u 6
p
2
(kappa curve)
82. r = cos
u
2
x
y
u5
u5
u5
u5 p
u5 0
u5
u5
u5
1
0
2 3 4
5,
(4, 0)
5
(
)
5p
²²
4
3p
²²
2
7p
²²
4
3p
²²
4
p²
2
p²
2
p²
4
x
y
u5
u5
u5 p
u5 0
u5
u5
u5
1
0
2 3 4
5,
(1, 0)
5
(
)
5p
²²
4
7p
²²
4
3p
²²
2
3p
²²
4
p²
2
p²
2
u5p²4
83. Show that the graph of the equation r sin u = a is a
horizontal line a units above the pole if a Ú 0 and
0 a
0 units
below the pole if a 6 0.
84. Show that the graph of the equation r cos u = a is a vertical
line a units to the right of the pole if a Ú 0 and
0 a
0 units to
the left of the pole if a 6 0.
85. Show that the graph of the equation r = 2a sin u, a 7 0, is a
circle of radius a with center at 10, a2 in rectangular
coordinates.
86. Show that the graph of the equation r = -2a sin u, a 7 0, is a
circle of radius a with center at 10, -a2 in rectangular
coordinates.
87. Show that the graph of the equation r = 2a cos u, a 7 0, is
a circle of radius a with center at 1a, 02 in rectangular
coordinates.
88. Show that the graph of the equation r = -2a cos u, a 7 0, is
a circle of radius a with center at 1-a, 02 in rectangular
coordinates.
explaining Concepts: Discussion and Writing
89. Explain why the following test for symmetry is valid:
Replace r by - r and u by - u in a polar equation. If an
equivalent equation results, the graph is symmetric with

respect to the line u =
p
2
(y-axis).
(a) Show that the test on page 577 fails for r 2 = cos u, yet
this new test works.
(b) Show that the test on page 577 works for r 2 = sin u, yet
this new test fails.
90. Write down two different tests for symmetry with respect to the
polar axis. Find examples in which one test works and the other
fails. Which test do you prefer to use? Justify your answer.
91. The tests for symmetry given on page 577 are sufficient, but
not necessary. Explain what this means.
92. Explain why the vertical-line test used to identify functions
in rectangular coordinates does not work for equations
expressed in polar coordinates.
Retain Your Knowledge
Problems 93–96 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
93. Solve:
5
x - 3
Ú 1
94. Convert
7p
3
radians to degrees.
95. Determine the amplitude and period of y = -2 sin (5x)
without graphing.
96. Find any asymptotes for the graph of
R1x2 =
x + 3
x2 - x - 12
.

‘Are You Prepared?’ Answers
1. 1-4, 62
2. cos A cos B + sin A sin B
3. 1x + 222 + 1y - 522 = 9
4. Odd 5. -
12
2


6. -
1
2
M09_SULL9070_10_SE_C09_pp563-637.indd 587
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588 CHAPTER 9  Polar Coordinates; Vectors
Figure 34 Complex plane
y
Imaginary
axis
Real
axis
x
z 5 x 1 yi
O
Plot Points in the Complex Plane
Complex numbers are discussed in Appendix A, Section A.7. In that discussion, we
were not prepared to give a geometric interpretation of a complex number. Now
we are ready.
A complex number z = x + yi can be interpreted geometrically as the point
1x, y2 in the xy-plane. Each point in the plane corresponds to a complex number, and
conversely, each complex number corresponds to a point in the plane. The collection
of such points is referred to as the complex plane. The x-axis is referred to as the real
axis, because any point that lies on the real axis is of the form z = x + 0i = x, a real
number. The y-axis is called the imaginary axis, because any point that lies on it is of
the form z = 0 + yi = yi, a pure imaginary number. See Figure 34.
Plotting a Point in the Complex Plane
Plot the point corresponding to z = 13 - i in the complex plane.
1
ExamplE 1
Now Work the ‘Are You Prepared?’ problems on page 594.

ObjeCtives 1 Plot Points in the Complex Plane (p. 588)

2 Convert a Complex Number between Rectangular Form and Polar
Form (p. 589)

3 Find Products and Quotients of Complex Numbers in Polar Form (p. 590)

4 Use De Moivre’s Theorem (p. 591)

5 Find Complex Roots (p. 592)
9.3 the Complex Plane; De Moivre’s theorem
• Complex Numbers (Appendix A, Section A.7,
pp. A53–A58)
• Values of the Sine and Cosine Functions at Certain
Angles (Section 6.2, pp. 369–376)
• Sum and Difference Formulas for Sine and Cosine
(Section 7.4, pp. 478 and 481)
PRePARiNG FOR tHis seCtiON  Before getting started, review the following:
The point corresponding to z = 23 - i has the
rectangular coordinates 123, -12 . This point,
located in quadrant IV, is plotted in Figure 35.
Solution
Figure 35
•
Imaginary
axis
Real
axis
22
2
2
22
z 5 3 2 i
O
Let z = x + yi be a complex number. The magnitude or modulus of z, denoted
by
0 z
0 , is defined as the distance from the origin to the point 1x, y2 . That is,

0 z
0 = 2x2 + y2
(1)
DeFInITIOn
See Figure 36 for an illustration.
This definition for
0 z
0 is consistent with the definition for the absolute value of
a real number: If z = x + yi is real, then z = x + 0i and
0 z
0 = 2x2 + 02 = 2x2 =
0 x
0
For this reason, the magnitude of z is sometimes called the absolute value of z.
Figure 36
y
Imaginary
axis
Real
axis
x
z 5 x 1 yi
O
_ z _
5
x
2 1
y
2
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SECTion 9.3  The Complex Plane; De Moivre’s Theorem 589
Recall that if z = x + yi, then its conjugate, denoted by z, is z = x - yi.
Because zz = x2 + y2, which is a nonnegative real number, it follows from
equation (1) that the magnitude of z can be written as

0 z
0 = 2zz
(2)
Convert a Complex Number between Rectangular
Form and Polar Form
When a complex number is written in the standard form z = x + yi, it is in
rectangular, or Cartesian, form, because 1x, y2 are the rectangular coordinates
of the corresponding point in the complex plane. Suppose that 1r, u2 are polar
coordinates of this point. Then

x = r cos u y = r sin u
(3)
2
If r Ú 0 and 0 … u 6 2p, the complex number z = x + yi may be written in
polar form as
z = x + yi = 1r cos u2 + 1r sin u2 i = r1cos u + i sin u2
(4)
DeFInITIOn
See Figure 37.
If z = r1cos u + i sin u2 is the polar form of a complex number,* the angle u,
0 … u 6 2p, is called the argument of z.
Also, because r Ú 0, we have r = 2x2 + y2 . From equation (1), it follows that
the magnitude of z = r1cos u + i sin u2 is
0 z
0 = r
Writing a Complex number in Polar Form
Write an expression for z = 23 - i in polar form.
The point, located in quadrant IV, is plotted in Figure 35. Because x = 23 and
y = -1, it follows that
r = 2x2 + y2 = 3123 22 + 1-122 = 24 = 2
so
sin u =
y
r
=
-1
2
cos u =
x
r
=
23
2
0 … u 6 2p
The angle u, 0 … u 6 2p, that satisfies both equations is u =
11p
6
. With
u =
11p
6
and r = 2, the polar form of z = 23 - i is
z = r1cos u + i sin u2 = 2acos 11p
6
+ i sin
11p
6
b
Now Work p r o b l e m 1 3
Plotting a Point in the Complex Plane and Converting
from Polar to rectangular Form
Plot the point corresponding to z = 21cos 30° + i sin 30°2 in the complex plane,
and write an expression for z in rectangular form.
ExamplE 2
Solution
•
ExamplE 3
*Some texts abbreviate the polar form using z = r(cos u + i sin u) = r cis u.
Figure 37
y
r
z
Imaginary
axis
Real
axis
x
z 5 x 1 yi 5 r (cos u 1 i sin u),
r ≥ 0, 0 ≤ u , 2p
O
u
M09_SULL9070_10_SE_C09_pp563-637.indd 589
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590 CHAPTER 9  Polar Coordinates; Vectors
Figure 38 z = 23 + i
Imaginary
axis
Real
axis
22
2
2
2
z 5 2(cos 308 1 i sin 308)
O
308
To plot the complex number z = 21cos 30° + i sin 30°2 , plot the point whose polar
coordinates are 1r, u2 = 12, 30°2 , as shown in Figure 38. In rectangular form,
z = 21cos 30° + i sin 30°2 = 2a 13
2
+
1
2
ib = 23 + i
Now Work p r o b l e m 2 5
Find Products and Quotients of Complex Numbers
in Polar Form
The polar form of a complex number provides an alternative method for finding
products and quotients of complex numbers.
Solution
•
3
In Words
The magnitude of a complex
number z is r, and its argument
is u, so when
z = r (cos u +
i sin u)
the magnitude of the product
(quotient) of two complex
numbers equals the product
(quotient) of their magnitudes;
the argument of the product
(quotient) of two complex
numbers is determined by the
sum (difference) of their
arguments.
Let z1 = r11cos u1 + i sin u12 and z2 = r21cos u2 + i sin u22 be two complex
numbers. Then

z1 z2 = r1r2 3cos1u1 + u22 + i sin1u1 + u22 4
(5)
If z2 ∙ 0, then

z1
z2
=
r1
r2
3cos1u1 - u22 + i sin1u1 - u22 4
(6)
TheOrem
Proof We will prove formula (5). The proof of formula (6) is left as an exercise
(see Problem 68).
z1 z2 = 3r11cos u1 + i sin u12 4 3r21cos u2 + i sin u22 4
= r1r23 1cos u1 + i sin u12 1cos u2 + i sin u22 4
= r1 r23 1cos u1 cos u2 - sin u1 sin u22 + i1sin u1 cos u2 + cos u1 sin u22 4
= r1 r23cos1u1 + u22 + i sin1u1 + u22 4
Let’s look at an example of how this theorem can be used.
Finding Products and Quotients of Complex numbers in Polar Form
If z = 31cos 20° + i sin 20°2 and w = 51cos 100° + i sin 100°2 , find the following
(leave your answers in polar form).
(a) zw



(b)
z
w
(a) zw = 331cos 20° + i sin 20°2 4 351cos 100° + i sin 100°2 4

= 13 # 52 3cos120° + 100°2 + i sin120° + 100°2 4

= 151cos 120° + i sin 120°2
(b)
z
w
=
31cos 20° + i sin 20°2
51cos 100° + i sin 100°2

=
3
5
3cos120° - 100°2 + i sin120° - 100°2 4

=
3
5
3cos1-80°2 + i sin1-80°2 4

=
3
5
1cos 280° + i sin 280°2
Now Work p r o b l e m 3 5
■
ExamplE 4
Solution
Apply equation (5).
Apply equation (6).
•
The argument must lie between
0° and 360°.
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SECTion 9.3  The Complex Plane; De Moivre’s Theorem 591
Use De Moivre’s theorem
De Moivre’s Theorem, stated by Abraham De Moivre (1667–1754) in 1730, but
already known to many people by 1710, is important for the following reason:
The fundamental processes of algebra are the four operations of addition,
subtraction, multiplication, and division, together with powers and the extraction
of roots. De Moivre’s Theorem allows the last two fundamental algebraic operations
to be applied to complex numbers.
De Moivre’s Theorem, in its most basic form, is a formula for raising a complex
number z to the power n, where n Ú 1 is a positive integer. Let’s try to conjecture
the form of the result.
Let z = r1cos u + i sin u2 be a complex number. Then equation (5) yields
n = 2: z2 = r 2 3cos12u2 + i sin12u2 4
n = 3: z3 = z2 # z

= 5r 2 3cos12u2 + i sin12u2 4 6 3r1cos u + i sin u2 4

= r 3 3cos13u2 + i sin13u2 4
n = 4: z4 = z3 # z

= 5r 3 3cos13u2 + i sin13u2 4 6 3r1cos u + i sin u2 4

= r 4 3cos14u2 + i sin14u2 4
Do you see the pattern?
4
Equation (5)
Equation (5)
Equation (5)
De Moivre’s theorem
If z = r1cos u + i sin u2 is a complex number, then

zn = r n 3cos1nu2 + i sin1nu2 4
(7)
where n Ú 1 is a positive integer.
TheOrem
The proof of De Moivre’s Theorem requires mathematical induction (which is
not discussed until Section 12.4), so it is omitted here. The theorem is actually true
for all integers, n. You are asked to prove this in Problem 69.
Using De moivre’s Theorem
Write 321cos 20° + i sin 20°2 43 in the standard form a + bi.
321cos 20° + i sin 20°2 43 = 233cos13 # 20°2 + i sin13 # 20°24
= 81cos 60° + i sin 60°2
= 8a1
2
+
23
2
ib = 4 + 423i
Now Work p r o b l e m 4 3
Using De moivre’s Theorem
Write 11 + i25 in the standard form a + bi.
To apply De Moivre’s Theorem, first write the complex number in polar form. Since
the magnitude of 1 + i is 212 + 12 = 12, begin by writing
1 + i = 22 a 112 + 112 ib = 22 acos p4 + i sin p4 b
ExamplE 5
Solution
Apply De Moivre’s Theorem.
•
ExamplE 6
Solution
note In the solution of Example 6, the
approach used in Example 2 could also be
used to write 1 + i in polar form.
■
M09_SULL9070_10_SE_C09_pp563-637.indd 591
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592 CHAPTER 9  Polar Coordinates; Vectors
Now
11 + i25 = c 22 acos p
4
+ i sin
p
4
b d
5
= 122 25 c cosa5 # p
4
b + i sina5 # p
4
b d
= 422 acos 5p
4
+ i sin
5p
4
b
= 422 c - 112 + a- 112 b i d = -4 - 4i
Find Complex Roots
Let w be a given complex number, and let n Ú 2 denote a positive integer. Any
complex number z that satisfies the equation
zn = w
is a complex nth root of w. In keeping with previous usage, if n = 2, the solutions
of the equation z2 = w are called complex square roots of w, and if n = 3, the
solutions of the equation z3 = w are called complex cube roots of w.
•
5
Finding Complex Roots
Let w = r1cos u0 + i sin u02 be a complex number, and let n Ú 2 be an integer.
If w ∙ 0, there are n distinct complex nth roots of w, given by the formula

zk = 2
n
r Jcos¢ u0
n
+
2kp
n
≤ + i sin¢ u0
n
+
2kp
n
≤ R
(8)
where k = 0, 1, 2,c, n - 1.
TheOrem
Proof (Outline) We will not prove this result in its entirety. Instead, we shall show
only that each zk in equation (8) satisfies the equation zk
n = w, proving that each zk
is a complex nth root of w.
znk = b2
n
r Jcos¢ u0
n
+
2kp
n
≤ + i sin¢ u0
n
+
2kp
n
≤R rn
= 12
n
r 2n bcosJn¢u0
n
+
2kp
n
≤R + i sinJn¢u0
n
+
2kp
n
≤R r
= r3cos1u0 + 2kp2 + i sin1u0 + 2kp2 4
= r1cos u0 + i sin u02 = w
So each zk , k = 0, 1, c, n - 1, is a complex nth root of w. To complete the proof,
we would need to show that each zk , k = 0, 1, c, n - 1, is, in fact, distinct and that
there are no complex nth roots of w other than those given by equation (8).
■
Finding Complex Cube roots
Find the complex cube roots of -1 + 13i . Leave your answers in polar form, with
the argument in degrees.
First, express -1 + 13i in polar form using degrees.
-1 + 23i = 2a- 1
2
+
13
2
ib = 21cos 120° + i sin 120°2
Apply De Moivre’s Theorem.
Simplify.
Periodic Property
ExamplE 7
Solution
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Section 9.3  The Complex Plane; De Moivre’s Theorem 593
The three complex cube roots of -1 + 13i = 21cos 120° + i sin 120°2 are
zk = 2
3 2 c cosa120°
3
+
360°k
3
b + i sina120°
3
+
360°k
3
b d

= 2
3 2 3cos140° + 120°k2 + i sin140° + 120°k2 4 k = 0, 1, 2
so
z0 = 2
3 2 3cos140° + 120° # 02 + i sin140° + 120° # 02 4 = 2
3 2 1cos 40° + i sin 40°2
z1 = 2
3 2 3cos140° + 120° # 12 + i sin140° + 120° # 12 4 = 2
3 2 1cos 160° + i sin 160°2
z2 = 2
3 2 3cos140° + 120° # 22 + i sin140° + 120° # 224 = 2
3 2 1cos 280° + i sin 280°2
Notice that all of the three complex cube roots of -1 + 13i have the same
magnitude, 2
3 2. This means that the points corresponding to each cube root lie the
same distance from the origin; that is, the three points lie on a circle with center at
the origin and radius 2
3 2. Furthermore, the arguments of these cube roots are 40°, 160°,
and 280°, the difference of consecutive pairs being 120° =
360°
3
. This means that the
three points are equally spaced on the circle, as shown in Figure 39. These results are
not coincidental. In fact, you are asked to show that these results hold for complex
nth roots in Problems 65 through 67.
•
Warning Most graphing utilities
will provide only the answer z 0 to the
calculation (- 1 + 13 i ) ¿ (1/3). The
paragraph following Example 7 explains
how to obtain z1 and z2 from z 0 .
■
Figure 39
Imaginary
axis
Real
axis
1
21
22
1
21
22
2
2
z0 5
3 2(cos 408 1 i sin 408)
O
408
z1 5
3 2(cos 1608 1 i sin 1608)
z2 5
3 2(cos 2808 1 i sin 2808)
3 2)2
x2 1 y2 5 (
1208
1208
1208
Now Work p r o b l e m 5 5
Historical Feature
T he Babylonians, Greeks,
and Arabs
considered square roots of negative
quantities to be impossible and equations
with complex solutions to be unsolvable. The
first hint that there was some connection
between real solutions of equations and complex
numbers
came when Girolamo Cardano
(1501–1576) and Tartaglia (1499–1557) found
real roots of cubic equations by taking cube roots
of complex quantities. For centuries thereafter,
mathematicians worked with complex numbers without much belief
in their actual existence. In 1673, John Wallis appears to have been
the first to suggest the graphical representation of complex numbers,
a truly significant idea that was not pursued further until about 1800.
Several people, including Karl Friedrich Gauss (1777–1855), then
rediscovered the idea, and graphical representation helped to establish
complex numbers as equal members of the number family. In practical
applications, complex numbers have found their greatest uses in the
study of alternating current, where they are a commonplace tool, and in
the field of subatomic physics.
John Wallis
Historical Problems
1. The quadratic formula works perfectly well if the coefficients are complex numbers. Solve the following.
(a) z2 - (2 + 5i)z - 3 + 5i = 0




(b) z2 - (1 + i)z - 2 - i = 0
M09_SULL9070_10_SE_C09_pp563-637.indd 593
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594 CHAPTER 9  Polar Coordinates; Vectors
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. The conjugate of -4 - 3i is
. (pp. A53–A58)
2. The sum formula for the sine function is
sin1A + B2 = . (p. 481)
3. The sum formula for the cosine function is

cos1A + B2 = . (p. 478)
4. sin 120° = ; cos 240° = . (pp. 369–376)
9.3 Assess Your Understanding
Concepts and vocabulary
5. In the complex plane, the x-axis is referred to as the

axis, and the y-axis is called the
axis.
6. When a complex number z is written in the polar form
z = r1cos u + i sin u2 ,
the nonnegative number r
is
the
or
of z, and the angle u,
0 … u 6 2p, is the
of z.
7. Let z1 = r11cos u1 + i sin u12 and z2 = r21cos u2 + i sin u22
be two complex numbers. Then
z1z2 =
3cos 1
2 + i sin 1
24 .
8. If
z = r(cos u + i sin u)
is
a
complex
number,
then zn =
3cos1
2 + i sin1
2 4 .
9. Every nonzero complex number will have exactly

distinct complex cube roots.
10. True or False The polar form of a nonzero complex number
is unique.
11. If z = x + yi is a complex number, then z  equals which of
the following?
(a) x2 + y2

(b) x  +
y 
(c) 2x2 + y2
(d) 2 x  +
y 
12. If z1 = r1( cos u1 + i sin u1) and z2 = r2( cos u2 + i sin u2)

are complex numbers, then
z1
z2
, z2 ≠ 0, equals which of the
following?
(a)
r1
r2
3cos (u1 - u2) + i sin (u1 - u2)4
(b)
r1
r2
c cos au1
u2
b + i sin au1
u2
b d
(c)
r1
r2
3cos (u1 + u2) - i sin (u1 + u2)4
(d)
r1
r2
c cos au1
u2
b - i sin au1
u2
b d
skill building
In Problems 13–24, plot each complex number in the complex plane and write it in polar form. Express the argument in degrees.
13. 1 + i
14. -1 + i
15. 23 - i
16. 1 - 23i
17. -3i
18. -2
19. 4 - 4i
20. 923 + 9i
21. 3 - 4i
22. 2 + 23i
23. -2 + 3i
24. 25 - i
In Problems 25–34, write each complex number in rectangular form.
25. 21cos 120° + i sin 120°2

26. 31cos 210° + i sin 210°2

27. 4acos 7p
4
+ i sin
7p
4
b
28. 2acos 5p
6
+ i sin
5p
6
b

29. 3acos 3p
2
+ i sin
3p
2
b

30. 4acos p
2
+ i sin
p
2
b
31. 0.21cos 100° + i sin 100°2
32. 0.41cos 200° + i sin 200°2
33. 2acos p
18
+ i sin
p
18
b

34. 3acos p
10
+ i sin
p
10
b
In Problems 35–42, find zw and
z
w
. Leave your answers in polar form.
35. z = 21cos 40° + i sin 40°2
36. z = cos 120° + i sin 120°
37. z = 31cos 130° + i sin 130°2

w = 41cos 20° + i sin 20°2
w = cos 100° + i sin 100°
w = 41cos 270° + i sin 270°2
38. z = 21cos 80° + i sin 80°2
39. z = 2acos p
8
+ i sin
p
8
b
40. z = 4acos 3p
8
+ i sin
3p
8
b


w = 61cos 200° + i sin 200°2

w = 2acos p
10
+ i sin
p
10
b
w = 2acos 9p
16
+ i sin
9p
16
b
41. z = 2 + 2i
42. z = 1 - i

w = 23 - i
w = 1 - 23i
In Problems 43–54, write each expression in the standard form a + bi.
43. 341cos 40° + i sin 40°2 43
44. 331cos 80° + i sin 80°2 43
45. c 2acos p
10
+ i sin
p
10
b d
5

46. c 22 acos 5p
16
+ i sin
5p
16
b d
4

47. 323 1cos 10° + i sin 10°2 46
48. c 1
2
1cos 72° + i sin 72°2 d
5

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06/01/15 3:29 pm
SECTion 9.3  The Complex Plane; De Moivre’s Theorem 595
49. c 25 acos 3p
16
+ i sin
3p
16
b d
4

50. c 23 acos 5p
18
+ i sin
5p
18
b d
6

51. 11 - i25
52. 123 - i26

53. 122 - i26

54. 11 - 25i28
In Problems 55–62, find all the complex roots. Leave your answers in polar form with the argument in degrees.
55. The complex cube roots of 1 + i
56. The complex fourth roots of 23 - i
57. The complex fourth roots of 4 - 423i
58. The complex cube roots of -8 - 8i
59. The complex fourth roots of -16i
60. The complex cube roots of -8
61. The complex fifth roots of i
62. The complex fifth roots of - i
Applications and extensions
63. Find the four complex fourth roots of unity (1) and plot them.
64. Find the six complex sixth roots of unity (1) and plot them.
65. Show that each complex nth root of a nonzero complex
number w has the same magnitude.
66. Use the result of Problem 65 to draw the conclusion that
each complex nth root lies on a circle with center at the
origin. What is the radius of this circle?
67. Refer to Problem 66. Show that the complex nth roots of a
nonzero complex number w are equally spaced on the circle.
68. Prove formula (6).
69. Prove that De Moivre’s Theorem is true for all integers n by
assuming it is true for integers n Ú 1 and then showing it is
true for 0 and for negative integers.
Hint: Multiply the numerator and the denominator by the
conjugate of the denominator, and use even-odd properties.
70. Mandelbrot Sets
(a) Consider the expression an = 1an-1 22 + z, where z is
some complex number (called the seed) and a0 = z.
Compute a11=a02 + z2 , a21=a12 + z2 , a31=a22 + z2, a4, a5,
and a6 for the following seeds: z1 = 0.1 - 0.4i,
z2 = 0.5 + 0.8i, z3 = -0.9 + 0.7i, z4 = -1.1 + 0.1i,
z5 = 0 - 1.3i, and z6 = 1 + 1i.
(b) The dark portion of the graph represents the set of
all values z = x + yi that are in the Mandelbrot set.

Determine which complex numbers in part (a) are in this

set by plotting them on the graph. Do the complex
numbers that are not in the Mandelbrot set have any
common characteristics regarding the values of a6 found
in part (a)?
(c) Compute
0 z
0 = 2x2 + y2 for each of the complex
numbers in part (a). Now compute
0 a6
0 for each of
the complex numbers in part (a). For which complex
numbers is
0 a6
0 …
0 z
0 and
0 z
0 … 2? Conclude that the
criterion for a complex number to be in the Mandelbrot
set is that
0 an
0 …
0 z
0 and
0 z
0 … 2.
–2
–1
y
1
Imaginary axis
x
1
Real axis
Retain Your Knowledge
Problems 71–74 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
71. Find the area of the triangle with a = 8, b = 11,
and C = 113°.
72. Convert 240° to radians. Express your answer as a
multiple of p.
73. Find the exact distance between the points (- 3, 4) and
(2, - 1).
74. Determine whether f(x) = 5x2 - 12x + 4 has a maximum
value or a minimum value, and then find the value.
‘Are You Prepared?’ Answers
1. - 4 + 3i

2. sin A cos B + cos A sin B

3. cos A cos B - sin A sin B


4.
13
2
; -
1
2
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596 CHAPTER 9  Polar Coordinates; Vectors
In simple terms, a vector (derived from the Latin vehere, meaning “to carry”) is a
quantity that has both magnitude and direction. It is customary to represent a vector
by using an arrow. The length of the arrow represents the magnitude of the vector,
and the arrowhead indicates the direction of the vector.
Many quantities in physics can be represented by vectors. For example, the
velocity of an aircraft can be represented by an arrow that points in the direction of
movement; the length of the arrow represents the speed. If the aircraft speeds up,
we lengthen the arrow; if the aircraft changes direction, we introduce an arrow in the
new direction. See Figure 40. Based on this representation, it is not surprising that
vectors and directed line segments are somehow related.
Geometric vectors
If P and Q are two distinct points in the xy-plane, there is exactly one line containing
both P and Q [Figure 41(a)]. The points on that part of the line that joins P to Q,
including P and Q, form what is called the line segment PQ [ Figure 41(b) ]. Ordering
the points so that they proceed from P to Q results in a directed line segment from
P to Q, or a geometric vector, which is denoted by PQ
>
. In a directed line segment
PQ
>
, P is called the initial point and Q the terminal point, as indicated in Figure 41(c).
9.4 vectors

ObjeCtives 1 Graph Vectors (p. 598)

2 Find a Position Vector (p. 598)

3 Add and Subtract Vectors Algebraically (p. 600)

4 Find a Scalar Multiple and the Magnitude of a Vector (p. 601)

5 Find a Unit Vector (p. 601)

6 Find a Vector from Its Direction and Magnitude (p. 602)

7 Model with Vectors (p. 603)
*Boldface letters will be used to denote vectors, to distinguish them from numbers. For handwritten
work, an arrow is placed over the letter to signify a vector. For example, write a vector by hand as v
>
.
Figure 40
Figure 41
P
Q
(a) Line containing P and Q
(b) Line segment PQ
P
Q
Terminal
point
Initial
point
(c) Directed line segment PQ
P
Q
The magnitude of the directed line segment PQ
>
is the distance from the point P
to the point Q; that is, it is the length of the line segment. The direction of PQ
>
is
from P to Q. If a vector v* has the same magnitude and the same direction as the
directed line segment PQ
>
, write
v = PQ
>
The vector v whose magnitude is 0 is called the zero vector, 0. The zero vector
is assigned no direction.
Two vectors v and w are equal, written
v = w
if they have the same magnitude and the same direction.
For example, the three vectors shown in Figure 42 have the same magnitude and
the same direction, so they are equal, even though they have different initial points
and different terminal points. As a result, it is useful to think of a vector simply as
an arrow, keeping in mind that two arrows (vectors) are equal if they have the same
direction and the same magnitude (length).
Figure 42 Equal vectors
R
S
P
Q
T
U
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SECTion 9.4  Vectors 597
Adding vectors Geometrically
The sum v + w of two vectors is defined as follows: Position the vectors v and w so
that the terminal point of v coincides with the initial point of w, as shown in Figure 43.
The vector v + w is then the unique vector whose initial point coincides with the
initial point of v and whose terminal point coincides with the terminal point of w.
Vector addition is commutative. That is, if v and w are any two vectors, then
v + w = w + v
Figure 44 illustrates this fact. (Observe that the commutative property is
another way of saying that opposite sides of a parallelogram are equal and parallel.)
Vector addition is also associative. That is, if u, v, and w are vectors, then
u + 1v + w2 = 1u + v2 + w
Figure 45 illustrates the associative property for vectors.
The zero vector 0 has the property that
v + 0 = 0 + v = v
for any vector v.
If v is a vector, then -v is the vector that has the same magnitude as v, but
whose direction is opposite to v, as shown in Figure 46.
Furthermore,
v + 1-v2 = 0
If v and w are two vectors, then the difference v - w is defined as
v - w = v + 1-w2
Figure 47 illustrates the relationships among v, w, v + w, and v - w.
Multiplying vectors by Numbers Geometrically
When dealing with vectors, real numbers are referred to as scalars. Scalars are
quantities that have only magnitude. Examples of scalar quantities from physics are
temperature, speed, and time. We now define how to multiply a vector by a scalar.
If a is a scalar and v is a vector, the scalar multiple av is defined as follows:
1. If a 7 0, av is the vector whose magnitude is a times the magnitude of v
and whose direction is the same as that of v.
2. If a 6 0, av is the vector whose magnitude is
0a
0 times the magnitude of v
and whose direction is opposite that of v.
3. If a = 0 or if v = 0, then av = 0.
DeFInITIOn
See Figure 48 for some illustrations.
For example, if a is the acceleration of an object of mass m due to a force F
being exerted on it, then, by Newton’s second law of motion, F = ma. Here, ma is
the product of the scalar m and the vector a.
Figure 48 Scalar multiples
1v
2v
v
Figure 44 v + w = w + v
v 
w
w 
v
w
v
v
w
Figure 43 Adding vectors
v  w
Initial point of v
Terminal point of w
w
v
Figure 45
(u + v) + w = u + (v + w)
w
u  v
v  w
u
v
Figure 46 Opposite vectors
v
 v
Figure 47
v  w
v 
w
w
 w
v
v
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598 CHAPTER 9  Polar Coordinates; Vectors
Scalar multiples have the following properties:
0v = 0 1v = v
-1v = -v
1a + b2v = av + bv a1v + w2 = av + aw
a1bv2 = 1ab2v
Graph vectors
Graphing Vectors
Use the vectors illustrated in Figure 49 to graph each of the following vectors:
(a) v - w


(b) 2v + 3w


(c) 2v - w + u
Solution Figure 50 shows each graph.
1
ExamplE 1
Figure 50
•
v  w
(a) v  w
2v  3w
(b) 2v  3w
2v  w  u
(c) 2v  w  u
 w
v
2v
3w
u
 w
2v
Now Work p r o b l e m s 1 1 a n d 1 3
Magnitude of vectors
The symbol 7v 7 represents the magnitude of a vector v. Since 7v 7 equals the length
of a directed line segment, it follows that 7v 7 has the following properties:
Properties of 7 v 7
If v is a vector and if a is a scalar, then
(a) 7v 7 Ú 0
(b) 7v 7 = 0 if and only if v = 0
(c) 7 -v 7 =
7v 7
(d) 7av 7 =
0a
0
7v 7
TheOrem
Property (a) is a consequence of the fact that distance is a nonnegative number.
Property (b) follows because the length of the directed line segment PQ
>
is positive
unless P and Q are the same point, in which case the length is 0. Property (c) follows
because the length of the line segment PQ equals the length of the line segment QP.
Property (d) is a direct consequence of the definition of a scalar multiple.
A vector u for which 7u 7 = 1 is called a unit vector.
DeFInITIOn
Find a Position vector
To compute the magnitude and direction of a vector, an algebraic way of representing
vectors is needed.
2
An algebraic vector v is represented as
v = 8a, b9
where a and b are real numbers (scalars) called the components of the vector v.
DeFInITIOn
Figure 49
w
u
v
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Section 9.4  Vectors 599
A rectangular coordinate system is used to represent algebraic vectors in the
plane. If v = 8a, b9 is an algebraic vector whose initial point is at the origin, then v
is called a position vector. See Figure 51. Notice that the terminal point of the
position vector v = 8a, b9 is P = 1a, b2 .
The next result states that any vector whose initial point is not at the origin is
equal to a unique position vector.
Figure 53
Suppose that v is a vector with initial point P1 = 1x1, y12, not necessarily the
origin, and terminal point P2 = 1x2 , y22 . If v = P1P2
>
, then v is equal to the
position vector

v = 8x2 - x1, y2 - y19
(1)
Theorem
To see why this is true, look at Figure 52.
Figure 52 v = 8a, b9 = 8x2 - x1 , y2 - y19
a
b
v
P 5 (a, b)
a
A
v
P2 5 (x2, y2)
P1 5 (x1, y1)
x2 2 x1
y2 2 y1
Q
O
b
x
y
Triangle OPA and triangle P1P2Q are congruent. [Do you see why? The line
segments have the same magnitude, so d1O, P2 = d1P1 , P22 ; and they have the
same direction, so ∠POA = ∠P2 P1Q. Since the triangles are right triangles, we
have angle–side–angle.] It follows that corresponding sides are equal. As a result,
x2 - x1 = a and y2 - y1 = b, so v may be written as
v = 8a, b9 = 8x2 - x1 , y2 - y19
Because of this result, any algebraic vector can be replaced by a unique position
vector, and vice versa. This flexibility is one of the main reasons for the wide use of
vectors.
Finding a Position Vector
Find the position vector of the vector v = P1P2
>
if P1 = 1-1, 22 and P2 = 14, 62 .
Solution By equation (1), the position vector equal to v is
v = 84 - 1-12 , 6 - 29 = 85, 49
See Figure 53.
ExamplE 2
•
y
v  <5, 4>
(5, 4)
P2  (4, 6)
P1  (1, 2)
O
5
5
x
Figure 51 Position vector v
y
P  (a, b)
x
v =
, b
>
O
in Words
An algebraic vector represents
“driving directions” to get from the
initial point to the terminal point
of a vector. So if v = 85, 49 ,
travel 5 units right and 4 units up
from the initial point to arrive at
the terminal point.
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600 CHAPTER 9  Polar Coordinates; Vectors
Two position vectors v and w are equal if and only if the terminal point of v is
the same as the terminal point of w. This leads to the following result:
equality of vectors
Two vectors v and w are equal if and only if their corresponding components
are equal. That is,
If v = 8a1 , b19 and w = 8a2 , b29
then v = w if and only if a1 = a2
and b1 = b2 .
TheOrem
We now present an alternative representation of a vector in the plane that is
common in the physical sciences. Let i denote the unit vector whose direction is
along the positive x-axis; let j denote the unit vector whose direction is along the
positive y-axis. Then i = 81, 09 and j = 80, 19 , as shown in Figure 54. Any vector
v = 8a, b9 can be written using the unit vectors i and j as follows:
v = 8a, b9 = a81, 09 + b80, 19 = ai + bj
The quantities a and b are called the horizontal and vertical components of v,
respectively. For example, if v = 85, 49 = 5i + 4j, then 5 is the horizontal
component and 4 is the vertical component.
Now Work p r o b l e m 3 1
Add and subtract vectors Algebraically
The sum, difference, scalar multiple, and magnitude of algebraic vectors are defined
in terms of their components.
3
Let v = a1i + b1j = 8a1 , b19 and w = a2i + b2j = 8a2 , b29 be
two

vectors, and let a be a scalar. Then
v + w = 1a1 + a22 i + 1b1 + b22 j = 8a1 + a2 , b1 + b29
(2)
v - w = 1a1 - a22 i + 1b1 - b22 j = 8a1 - a2 , b1 - b29
(3)

av = 1aa12 i + 1ab12 j = 8aa1 , ab19
(4)

7v 7 = 2a21 + b21
(5)
DeFInITIOn
These definitions are compatible with the geometric definitions given earlier in
this section. See Figure 55.
Figure 55
a2
a2
b1
b2
b2
a1
y
v 1
w
(a2, b2)
(a1, b1)
(a1 1 a2, b1 1 b2)
O
x
DIllustration of property (2)
aa1
b1
ab1
a1
y
av
(a1, b1)
(aa1, ab1)
O
x
EIllustration of property (4), a . 0
b1
b1
y
P1 5 (a1, b1)
O
x
F
a21 1 b
2
1
a1
Illustration of property (5):
|| v || 5 Distance from O to P1
|| v || 5
w
v
v
v
In Words
To add two vectors, add
corresponding components. To
subtract two vectors, subtract
corresponding components.
Figure 54 Unit vectors i and j
y
j
i
(0, 1)
(1, 0)
x
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SECTion 9.4  Vectors 601
Adding and Subtracting Vectors
If v = 2i + 3j = 82, 39 and w = 3i - 4j = 83, -49 , find:
(a) v + w




(b) v - w
(a) v + w = 12i + 3j2 + 13i - 4j2 = 12 + 32 i + 13 - 42 j = 5i - j
or

v + w = 82, 39 + 83, -49 = 82 + 3, 3 + 1-42 9 = 85, -19
(b) v - w = 12i + 3j2 - 13i - 4j2 = 12 - 32 i + 33 - 1-42 4 j = - i + 7j
or

v - w = 82, 39 - 83, -49 = 82 - 3, 3 - 1-42 9 = 8 -1, 79
Find a scalar Multiple and the Magnitude of a vector
Finding Scalar multiples and magnitudes of Vectors
If v = 2i + 3j = 82, 39 and w = 3i - 4j = 83, -49 , find:
(a) 3v



(b) 2v - 3w


(c) ‘v ‘
(a) 3v = 312i + 3j2 = 6i + 9j
or

3v = 382, 39 = 86, 99
(b) 2v - 3w = 212i + 3j2 - 313i - 4j2 = 4i + 6j - 9i + 12j

= -5i + 18j
or

2v - 3w = 282, 39 - 383, -49 = 84, 69 - 89, -129

= 84 - 9, 6 - 1-122 9 = 8 -5, 189
(c)
7v 7 =
72i + 3j 7 = 222 + 32 = 213
Now Work p r o b l e m s 3 7 a n d 4 3
For the remainder of the section, we will express a vector v in the form ai + bj.
Find a Unit vector
Recall that a unit vector u is a vector for which 7u 7 = 1. In many applications, it is
useful to be able to find a unit vector u that has the same direction as a given vector v.
ExamplE 3
Solution
•
4
ExamplE 4
Solution
•
5
Unit vector in the Direction of v
For any nonzero vector v, the vector

u =
v
7 v 7
(6)
is a unit vector that has the same direction as v.
TheOrem
Proof Let v = ai + bj. Then 7v 7 = 2a2 + b2 and
u =
v
7 v 7 =
ai + bj
2a2 + b2 =
a
2a2 + b2 i +
b
2a2 + b2 j
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602 CHAPTER 9  Polar Coordinates; Vectors
The vector u is in the same direction as v, since 7v 7 7 0. Furthermore,
7u 7 = B
a2
a2 + b2
+
b2
a2 + b2
= Ba2 + b2
a2 + b2
= 1
That is, u is a unit vector in the direction of v.
■
As a consequence of this theorem, if u is a unit vector in the same direction as
a vector v, then v may be expressed as

v =
7v 7u
(7)
This way of expressing a vector is useful in many applications.
Finding a Unit Vector
Find a unit vector in the same direction as v = 4i - 3j.
Find 7v ‘ first.
7v 7 =
74i - 3j 7 = 216 + 9 = 5
Now multiply v by the scalar
1
7 v 7 =
1
5
. A unit vector in the same direction as v is
v
7 v 7 =
4i - 3j
5
=
4
5
i -
3
5
j
Check: This vector is indeed a unit vector because
" v
7v 7
" = Ba45 b2 + a- 35 b2 = B16
25
+
9
25
= B25
25
= 1
Now Work p r o b l e m 5 3
Find a vector from its Direction and Magnitude
If a vector represents the speed and direction of an object, it is called a velocity
vector. If a vector represents the direction and amount of a force acting on an object,
it is called a force vector. In many applications, a vector is described in terms of its
magnitude and direction, rather than in terms of its components. For example, a ball
thrown with an initial speed of 25 miles per hour at an angle of 30° to the horizontal
is a velocity vector.
Suppose that we are given the magnitude ‘v ‘ of a nonzero vector v and the
direction angle a, 0° … a 6 360°, between v and i. To express v in terms of ‘v ‘
and a, first find the unit vector u having the same direction as v.
Look at Figure 56. The coordinates of the terminal point of u are 1cos a, sin a2 .
Then u = cos a i + sin a j and, from equation (7),
ExamplE 5
Solution
•
6

v = ‘v ‘ 1cos a i + sin a j2
(8)
where a is the direction angle between v and i.
Figure 56 v =
7v 7 (cos a i + sin a j)
1
1
y
x

i
j
(cos , sin )
u
v
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SECTion 9.4  Vectors 603
Figure 57
y
x
12.5 j
21.65 i
12.5
21.65
v = 25(cos 30°i + sin 30°j)
25
30°
Finding a Vector When Its magnitude and Direction Are Given
A ball is thrown with an initial speed of 25 miles per hour in a direction that makes
an angle of 30° with the positive x-axis. Express the velocity vector v in terms of i
and j. What is the initial speed in the horizontal direction? What is the initial speed
in the vertical direction?
The magnitude of v is 7v 7 = 25 miles per hour, and the angle between the direction
of v and i, the positive x-axis, is a = 30°. By equation (8),
ExamplE 6
Solution
v =
7v 7 1cos ai + sin aj2 = 251cos 30°i + sin 30°j2
= 25a 13
2
i +
1
2
jb = 2513
2
i +
25
2
j
The initial speed of the ball in the horizontal direction is the horizontal component
of v,
2523
2
≈ 21.65 miles per hour. The initial speed in the vertical direction is the
vertical component of v,
25
2
= 12.5 miles per hour. See Figure 57.
Now Work p r o b l e m 5 9
Finding the Direction Angle of a Vector
Find the direction angle a of v = 4i - 4j.
See Figure 58. The direction angle a of v = 4i - 4j can be found by solving
tan a =
-4
4
= -1
Because 0° … a 6 360°, the direction angle is a = 315°.
Now Work p r o b l e m 6 5
Model with vectors
Because forces can be represented by vectors, two forces “combine” the way that
vectors “add.” If F1 and F2 are two forces simultaneously acting on an object, the
vector sum F1 + F2 is the resultant force. The resultant force produces the same
effect on the object as that obtained when the two forces F1 and F2 act on the object.
See Figure 59.
Finding the Actual Speed and Direction of an Aircraft
A Boeing 737 aircraft maintains a constant airspeed of 500 miles per hour headed
due south. The jet stream is 80 miles per hour in the northeasterly direction.
(a) Express the velocity va of the 737 relative to the air and the velocity vw of the jet
stream in terms of i and j.
(b) Find the velocity of the 737 relative to the ground.
(c) Find the actual speed and direction of the 737 relative to the ground.
•
ExamplE 7
Solution
•
7
ExamplE 8
Figure 58
x

v  4i  4j
(4,4)
4
4
Figure 59 Resultant force
F1 + F2
Resultant
F1
F2
Orlando
Naples
Miami
Wind
N
S
W
E
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604 cHAPteR 9  Polar Coordinates; Vectors
(a) Set up a coordinate system in which north (N) is along the positive y-axis. See
Figure 60. The velocity of the 737 relative to the air is va = -500j. The velocity
of the jet stream vw has magnitude 80 and direction NE (northeast), so the angle
between vw and i is 45°. Express vw in terms of i and j as
vw = 801cos 45° i + sin 45° j2 = 80a 22
2
i +
22
2
jb = 4022 1i + j2
(b) The velocity of the 737 relative to the ground vg is
vg = va + vw = -500j + 40221i + j2 = 4022 i + 14022 - 5002j
(c) The actual speed of the 737 is
7vg 7 = 31402222 + 14022 - 50022 ≈ 447 miles per hour
To find the actual direction of the 737 relative to the ground, determine the
direction angle of vg. The direction angle is found by solving
tan a =
4022 - 500
4022
Then a ≈ -82.7°. The 737 is traveling S 7.3° E.
Now Work p r o b l e m 7 7
Finding the Weight of a Piano
Two movers require a magnitude of force of 300 pounds to push a piano up a ramp
inclined at an angle 20° from the horizontal. How much does the piano weigh?
Let F1 represent the force of gravity, F2 represent the force required to move the
piano up the ramp, and F3 represent the force of the piano against the ramp. See
Figure 61. The angle between the ground and the ramp is the same as the angle between
F1 and F3 because triangles ABC and BDE are similar, so ∠BAC = ∠DBE = 20°.
To find the magnitude of F1 (the weight of the piano), calculate
sin 20° =
7 F2 7
7 F1 7 =
300
7 F1 7
7F1 7 = 300 lb
sin 20°
≈ 877 lb
The piano weighs approximately 877 pounds.
An object is said to be in static equilibrium if the object is at rest and
the sum of all forces acting on the object is zero—that is, if the resultant force is 0.
Analyzing an object in Static equilibrium
A box of supplies that weighs 1200 pounds is suspended by two cables attached to
the ceiling, as shown in Figure 62. What are the tensions in the two cables?
Draw a force diagram using the vectors as shown in Figure 63. The tensions in
the cables are the magnitudes 7F1 7 and 7F2 7 of the force vectors F1 and F2 . The
magnitude of the force vector F3 equals 1200 pounds, the weight of the box. Now write
each force vector in terms of the unit vectors i and j. For F1 and F2 , use equation (8).
Remember that a is the angle between the vector and the positive x-axis.
Solution
•
ExamplE 9
Solution
•
ExamplE 10
Solution
Figure 60
W
N
S
vg
500
E
y
x
500
vw
va  500j
Figure 61
20°
20°
A
B
D
C
E
F3
F1
F2
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SECTion 9.4  Vectors 605
F1 =
7F1 7 1cos 150°i + sin 150°j2 =
7F1 7 a- 13
2
i +
1
2
jb = - 13
2
7F1 7i + 12 7F1 7 j
F2 =
7F2 7 1cos 45°i + sin 45°j2 =
7F2 7 a 12
2
i +
12
2
jb = 12
2
7F2 7 i + 12
2
7F2 7 j
F3 = -1200j
For static equilibrium, the sum of the force vectors must equal zero.
F1 + F2 + F3 = -
13
2
7F1 7 i + 12 7F1 7 j +
12
2
7F2 7 i + 12
2
7F2 7 j - 1200j = 0
The i component and j component will each equal zero. This results in the two equations

-
13
2
7F1 7 + 12
2
7F2 7 = 0
(9)


1
2
7F1 7 + 12
2
7F2 7 - 1200 = 0
(10)
Solve equation (9) for 7F2 7 to obtain

7F2 7 = 2312 7F1 7
(11)
Substituting into equation (10) and solving for 7F1 7 yields

1
2
7F1 7 + 12
2
a 1312 7F1 7 b - 1200 = 0

1
2
7F1 7 + 13
2
7F1 7 - 1200 = 0

1 + 13
2
7F1 7 = 1200
7F1 7 =
2400
1 + 13 ≈ 878.5 pounds
Substituting this value into equation (11) gives 7F2 7 .
7F2 7 = 1312 7F1 7 = 1312 # 2400
1 + 13 ≈ 1075.9 pounds
The left cable has tension of approximately 878.5 pounds, and the right cable has
tension of approximately 1075.9 pounds.
Now Work p r o b l e m 8 7
•
Historical Feature
T he history of vectors
is surprisingly
complicated for such a natural concept.
In the xy-plane, complex numbers do
a good job of imitating vectors. About 1840,
mathematicians became interested in finding a
system that would do for three dimensions what
the complex numbers do for two dimensions.
Hermann Grassmann (1809–1877), in Germany,
and William Rowan Hamilton (1805–1865), in
Ireland, both attempted to find solutions.
Hamilton’s system was the quaternions, which are best thought
of as a real number plus a vector; they do for four dimensions
what complex numbers do for two dimensions. In this system the
order of multiplication matters; that is, ab 3 ba. Also, two products of
vectors emerged, the scalar product (or dot product) and the vector
product (or cross product).
Grassmann’s abstract style, although easily read today, was almost
impenetrable during the nineteenth century, and only a few of his ideas
were appreciated. Among those few were the same scalar and vector
products that Hamilton had found.
About 1880,
the American physicist Josiah Willard Gibbs
(1839–1903) worked out an algebra involving only the simplest
concepts: the vectors and the two products. He then added some
calculus, and the resulting system was simple, flexible, and well adapted
to expressing a large number of physical laws. This system remains in
use essentially unchanged. Hamilton’s and Grassmann’s more extensive
systems each gave birth to much interesting mathematics, but little
of it is seen at elementary levels.
Josiah Gibbs
(1839–1903)
Figure 62
30°
30°
45°
45°
1200
pounds
Figure 63 Force diagram
y
30°
x
F1
F2
F3
45°
150°
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