865
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 9
Polar Coordinates; Vectors
Section 9.1
1.

The point lies in quadrant IV.

2.
2
6
9
2

 =





3.
b
a

4.
4
π
−

5. pole, polar axis
6. True
7. False
8.
cos
r
θ ; sin
r
θ
9. A
10. B
11. C
12. C
13. B
14. D
15. A
16. D
17.

18.

19.

20.

21.

22.

23.

Chapter 9: Polar Coordinates; Vectors
866
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
24.

25.

26.

27.

28.

29.

30.

31.

a.
4
0,
2
0
5,
3
r
θ
π


>
− π ≤ <
−





b.
5
0, 0
2
5,
3
r
θ
π


<
≤ < π
−





c.
8
0, 2
4
5,
3
r
θ
π


>
π ≤ < π





32.

a.
5

0,
2
0
4,
4
r
θ
π


>
− π ≤ <
−





b.
7
0, 0
2
4,
4
r
θ
π


<
≤ < π
−





c.
11
0, 2
4
4,
4
r
θ
π


>
π ≤ < π





33.

a.
(
)
0,
2
0
2, 2
r
θ
>
− π ≤ <
− π
b.
(
)
0, 0
2
2,
r
θ
<
≤ < π
− π
c.
(
)
0, 2
4
2, 2
r
θ
>
π ≤ < π
π

34.

a.
(
)
0,
2
0
3,
r
θ
>
− π ≤ <
− π
b.
(
)
0, 0
2
3, 0
r
θ
<
≤ < π
−

c.
(
)
0, 2
4
3, 3
r
θ
>
π ≤ < π
π


Section 9.1: Polar Coordinates
867
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
35.

a.
3
0,
2
0
1,
2
r
θ
π


>
− π ≤ <
−





b.
3
0, 0
2
1,
2
r
θ
π


<
≤ < π
−





c.
5
0, 2
4
1,
2
r
θ
π


>
π ≤ < π





36.

a.
(
)
0,
2
0
2,
r
θ
>
− π ≤ <
− π

b.
(
)
0, 0
2
2, 0
r
θ
<
≤ < π
−

c.
(
)
0, 2
4
2, 3
r
θ
>
π ≤ < π
π
37.

a.
5
0,
2
0
3,
4
r
θ
π


>
− π ≤ <
−





b.
7
0, 0
2
3,
4
r
θ
π


<
≤ < π
−





c.
11
(
0, 2
4
3,
4
r
θ
π


>
π ≤ < π





38.

a.
5
0,
2
0
2,
3
r
θ
π


>
− π ≤ <
−





b.
4
0, 0
2
2,
3
r
θ
π


<
≤ < π
−





c.
7
0, 2
4
2,
3
r
θ
π


>
π ≤ < π





39.
cos
3cos
3 0 0
2
x
r
θ
π
=
=
= ⋅ =

sin
3sin
3 1 3
2
y
r
θ
π
=
=
= ⋅ =

Rectangular coordinates of the point 3,
2
π






are
(
)
0, 3 .
40.
3
cos
4cos
4 0 0
2
x
r
θ
π
=
=
= ⋅ =

3
sin
4sin
4 ( 1)
4
2
y
r
θ
π
=
=
= ⋅ − = −

Rectangular coordinates of the point
3
4,
2
π






are
(
)
0, 4
−
.
41.
cos
2cos 0
2 1
2
x
r
θ
=
= −
= − ⋅ = −

sin
– 2sin 0 – 2 0 0
y
r
θ
=
=
=
⋅ =

Rectangular coordinates of the point (
)
–2, 0 are
(
)
2, 0
−
.
42.
cos
3cos
3( 1) 3
x
r
θ
=
= −
π = − − =

sin
3sin
3 0 0
y
r
θ
=
= −
π = − ⋅ =

Rectangular coordinates of the point (
)
3,
− π are
(
)
3, 0 .
43.
3
cos
6cos150º 6
3 3
2
x
r
θ


=
=
=
−
= −







1
sin
6sin150º 6
3
2
y
r
θ
=
=
= ⋅ =

Rectangular coordinates of the point (
)
6, 150º
are (
)
3 3, 3
−
.
44.
1
5
cos
5cos300º 5
2
2
x
r
θ
=
=
= ⋅ =

3
5 3
sin
5sin 300º 5
2
2
y
r
θ


=
=
=
−
= −







Chapter 9: Polar Coordinates; Vectors
868
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Rectangular coordinates of the point (
)
5, 300º
are
5
5 3
,
2
2


−






.
45.
3
2
cos
2cos
2
2
4
2
x
r
θ


π
=
= −
= −
−
=







3
2
sin
2sin
2
2
4
2
y
r
θ
π
=
= −
= − ⋅
= −

Rectangular coordinates of the point
3
2,
4
π


−





are (
)
2,
2
−
.
46.
2
1
cos
2cos
2
1
3
2
x
r
θ
π


=
= −
= − −
=





2
3
sin
2sin
2
3
3
2
y
r
θ
π
=
= −
= − ⋅
= −

Rectangular coordinates of the point
2
2,
3
π


−





are (
)
1,
3
−
.
47.
1
1
cos
1cos
1
3
2
2
x
r
θ
π


=
= −
−
= − ⋅ = −





3
3
sin
1sin
1
3
2
2
y
r
θ


π


=
= −
−
= − −
=











Rectangular coordinates of the point
1,
3
π


− −





are
1
3
,
2
2


−






.
48.
3
2
3 2
cos
3cos
3
4
2
2
x
r
θ


π


=
= −
−
= − −
=











3
2
3 2
sin
3sin
3
4
2
2
y
r
θ


π


=
= −
−
= − −
=











Rectangular coordinates of the point
3
3,
4
π


− −





are
3 2 3 2
,
2
2








.
49.
(
)
cos
2cos( 180º )
2 1
2
x
r
θ
=
= −
−
= − − =

sin
– 2sin( 180º )
– 2 0 0
y
r
θ
=
=
−
=
⋅ =

Rectangular coordinates of the point
(
)
–2, 180º
−
are (
)
2, 0 .
50.
cos
3cos( 90º )
3 0 0
x
r
θ
=
= −
−
= − ⋅ =

sin
– 3sin( 90º )
3( 1) 3
y
r
θ
=
=
−
= − − =

Rectangular coordinates of the point (
)
–3, 90º
−

are (
)
0, 3 .
51.
cos
7.5cos110º 7.5( 0.3420)
2.57
x
r
θ
=
=
≈
−
≈ −

sin
7.5sin110º 7.5(0.9397) 7.05
y
r
θ
=
=
≈
≈

Rectangular coordinates of the point (
)
7.5, 110º
are about (
)
2.57, 7.05
−
.
52.
cos
3.1cos182º
3.1( 0.9994) 3.10
x
r
θ
=
= −
≈ −
−
≈

sin
3.1sin182º
3.1( 0.0349) 0.11
y
r
θ
=
= −
≈ −
−
≈

Rectangular coordinates of the point (
)
3.1, 182º
−

are about (
)
3.10, 0.11 .
53.
(
)
cos
6.3cos 3.8
6.3( 0.7910)
4.98
x
r
θ
=
=
≈
−
≈ −

(
)
sin
6.3sin 3.8
6.3( 0.6119)
3.85
y
r
θ
=
=
≈
−
≈ −

Rectangular coordinates of the point (
)
6.3, 3.8 are
about (
)
4.98, 3.85
−
−
.
54.
(
)
cos
8.1cos 5.2
8.1(0.4685) 3.79
x
r
θ
=
=
≈
≈

(
)
sin
8.1sin 5.2
8.1( 0.8835)
7.16
y
r
θ
=
=
≈
−
≈ −

Rectangular coordinates of the point (
)
8.1, 5.2
are about (
)
3.79, 7.16
−
.
55.
2
2
2
2
1
1
1
3
0
9 3
0
tan
tan
tan 0 0
3
r
x
y
y
x
θ
−
−
−
=
+
=
+
=
=


 
=
=
=
=


 


 

Polar coordinates of the point (3, 0) are (3, 0) .
56.
2
2
2
2
0
2
4
2
r
x
y
=
+
=
+
=
=

1
1 2
tan
tan
0
y
x
θ
−
−




=
=









Since
2
0
is undefined,
2
θ
π
=

Polar coordinates of the point (0, 2) are 2,
2
π






.

Section 9.1: Polar Coordinates
869
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
57.
2
2
2
2
1
1
1
( 1)
0
1 1
0
tan
tan
tan 0 0
1
r
x
y
y
x
θ
−
−
−
=
+
=
−
+
=
=




=
=
=
=




−





The point lies on the negative x-axis, so θ = π .
Polar coordinates of the point ( 1, 0)
−
are (
)
1, π
.
58.
2
2
2
2
1
1
0
( 2)
4
2
2
tan
tan
0
r
x
y
y
x
θ
−
−
=
+
=
+ −
=
=
−




=
=









Since
2
0
−
is undefined,
2
θ
π
=
.
The point lies on the negative y-axis, so
2
θ
π
= −
.
Polar coordinates of the point (0, 2)
−
are 2,
2
π


−




.
59. The point (1, 1)
−
lies in quadrant IV.
2
2
2
2
1
1
1
1
( 1)
2
1
tan
tan
tan ( 1)
1
4
r
x
y
y
x
θ
−
−
−
=
+
=
+ −
=
−
π




=
=
=
− = −









Polar coordinates of the point (1, 1)
−
are
2,
4
π


−




.
60. The point ( 3, 3)
−
lies in quadrant II.
2
2
2
2
1
1
1
( 3)
3
3 2
3
tan
tan
tan ( 1)
3
4
r
x
y
y
x
θ
−
−
−
=
+
=
−
+
=
π




=
=
=
− = −




−





Polar coordinates of the point (
)
3, 3
−
are
3
3 2,
4
π






.
61. The point (
)
3, 1 lies in quadrant I.
(
)2
2
2
2
1
1
3
1
4
2
1
tan
tan
6
3
r
x
y
y
x
θ
−
−
=
+
=
+
=
=
π


=
=
=





Polar coordinates of the point (
)
3, 1 are 2,
6
π






.
62. The point (
)
2, 2 3
− −
lies in quadrant III.
(
)
(
)2
2
2
2
1
1
1
2
2 3
16 4
2 3
tan
tan
tan
3
2
3
r
x
y
y
x
θ
−
−
−
=
+
=
−
+ −
=
=


−
π


=
=
=
=






−





The point lies in quadrant III, so
2
3
3
π
π
θ
π
= − = −

Polar coordinates of the point (
)
2, 2 3
− −
are
2
4,
3
π


−




.
63. The point (1.3, 2.1)
−
lies in quadrant IV.
2
2
2
2
1
1
1.3
( 2.1)
6.1 2.47
2.1
tan
tan
1.02
1.3
r
x
y
y
x
θ
−
−
=
+
=
+ −
=
≈
−




=
=
≈ −









The polar coordinates of the point (
)
1.3, 2.1
−
are
(
)
2.47, 1.02
−
.
64. The point ( 0.8, 2.1)
−
−
lies in quadrant III.
2
2
2
2
1
1
( 0.8)
( 2.1)
5.05
2.25
2.1
tan
tan
1.21
0.8
r
x
y
y
x
θ
−
−
=
+
=
−
+ −
=
=


−


=
=
≈




−





Since the point lies in quadrant III,
1.21
1.93
θ
π
=
− ≈ −
.
The polar coordinates of the point (
)
0.8, 2.1
−
−
are
(
)
2.25, 1.93
−
.
65. The point (8.3, 4.2) lies in quadrant I.
2
2
2
2
1
1
8.3
4.2
86.53 9.30
4.2
tan
tan
0.47
8.3
r
x
y
y
x
θ
−
−
=
+
=
+
=
≈




=
=
≈









The polar coordinates of the point (
)
8.3, 4.2 are
(
)
9.30, 0.47 .
66. The point ( 2.3, 0.2)
−
lies in quadrant II.
2
2
2
2
1
1
( 2.3)
0.2
5.33 2.31
0.2
tan
tan
0.09
2.3
r
x
y
y
x
θ
−
−
=
+
=
−
+
=
≈




=
=
≈ −




−





Since the point lies in quadrant II,
0.09 3.05
θ = π −
≈
.
Chapter 9: Polar Coordinates; Vectors
870
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
The polar coordinates of the point (
)
2.3, 0.2
−

are (
)
2.31, 3.05 .
67.
(
)
2
2
2
2
2
2
2
2
3
2
3
2
3
3
3
6
or
2
2
2
x
y
x
y
r
r
r
+
=
+
=
=
=
=
=

68.
2
2
2
cos
cos
x
y
x
r
r
r
θ
θ
+
=
=
=

69.
(
)
2
2
2
2
4
cos
4 sin
cos
4 sin
0
x
y
r
r
r
r
θ
θ
θ
θ
=
=
−
=

70.
(
)
2
2
2
2
2
sin
2 cos
sin
2 cos
0
y
x
r
r
r
r
θ
θ
θ
θ
=
=
−
=

71.
(
)
2
2
2
1
2( cos )( sin ) 1
2sin cos
1
sin 2
1
xy
r
r
r
r
θ
θ
θ
θ
θ
=
=
=
=

72.
2
2
2
2
3
2
4
1
4( cos )
sin
1
4
cos
sin
1
1
cos
sin
4
x y
r
r
r
r
r
θ
θ
θ
θ
θ
θ
=
=
=
=

73.
4
cos
4
x
r
θ
=
=

74.
3
sin
3
y
r
θ
= −
= −

75.
2
2
2
2
2
2
2
2
2
cos
cos
0
1
1
4
4
1
1
2
4
r
r
r
x
y
x
x
x
y
x
x
y
x
y
θ
θ
=
=
+
=
− +
=
− + +
=


−
+
=





76.
2
2
2
2
2
sin
1
sin
r
r
r
r
x
y
y
x
y
θ
θ
=
+
=
+
+
= +
+

77.
(
)
(
)
(
)
2
3
3/ 2
2
3/ 2
2
2
3/ 2
2
2
cos
cos
cos
0
r
r
r
r
r
x
y
x
x
y
x
θ
θ
θ
=
=
=
+
=
+
− =

78.
2
2
2
2
2
2
2
2
2
sin
cos
sin
cos
0
1
1
1 1
4
4
4 4
1
1
1
2
2
2
r
r
r
r
x
y
y x
x
x
y
y
x
x
y
y
x
y
θ
θ
θ
θ
=
−
=
−
+
= −
+ +
− =
+ + +
− + = +




+
+
−
=









79.
2
2
2
2
4
4
r
r
x
y
=
=
+
=

80.
2
2
2
4
16
16
r
r
x
y
=
=
+
=


Section 9.1: Polar Coordinates
871
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
81.
(
)
2
2
2
2
2
2
2
2
4
1 cos
(1 cos )
4
cos
4
4
4
8
16
8
2
r
r
r
r
x
y
x
x
y
x
x
y
x
x
y
x
θ
θ
θ
=
−
−
=
−
=
+
− =
+
= +
+
=
+
+
=
+

82.


(
)
2
2
2
2
2
2
2
2
2
2
3
3 cos
(3 cos )
3
3
cos
3
3
3
3
3
9
6
9
9
9
6
9
r
r
r
r
x
y
x
x
y
x
x
y
x
x
x
y
x
x
θ
θ
θ
=
−
−
=
−
=
+
− =
+
= +
+
=
+
+
+
=
+
+

2
2
2
2
2
2
2
2
2
2
8
6
9
9 0
64
48
72
72 0
3
64
72
72
4
3
9
9
64
72
72 64
4
64
64
3
64
72
81
8
x
x
y
x
x
y
x
x
y
x
x
y
x
y
−
+
− =
−
+
−
=


−
+
=








−
+
+
=
+










−
+
=





83. a. For this application, west is a negative direction and north is positive. Therefore, the rectangular coordinate is
( 10, 36)
−
.

b. The distance r from the origin to ( 10, 36)
−
is
2
2
2
2
( 10)
(36)
1396 2 349 37.36
r
x
y
=
+
=
−
+
=
=
≈
.
Since the point ( 10, 36)
−
lies in quadrant II, we use
1
180
tan
y
x
θ
− 

=
° +




. Thus,
1
1
36
18
180
tan
180
tan
105.5
10
5
θ
−
−




=
° +
=
° +
−
≈
°




−




.
The polar coordinate of the point is
(
)
1
18
2 349, 180
tan
37.36,105.5
5
−




°+
−
≈
°








.

c. For this application, west is a negative direction and south is also negative. Therefore, the rectangular
coordinate is ( 3, 35)
− −
.

d. The distance r from the origin to ( 3, 35)
− −
is
2
2
2
2
( 3)
( 35)
1234 35.13
r
x
y
=
+
=
−
+ −
=
≈
.
Since the point ( 3, 35)
− −
lies in quadrant III, we use
1
180
tan
y
x
θ
− 

=
° +




. Thus,
1
1
35
35
180
tan
180
tan
265.1
3
3
θ
−
−


−


=
° +
=
° +
≈
°




−





The polar coordinate of the point is
(
)
1 35
1234,180
tan
35.13, 265.1
3
−




°+
≈
°








.
84. Rewrite the polar coordinates in rectangular form:
Since
(
)
1
1
1
,
P
r θ
=
and
(
)
2
2
2
,
P
r θ
=
, we have that (
)
(
)
1
1
1
1 1
1
,
cos
,
sin
x y
r
r
θ
θ
=
and
Chapter 9: Polar Coordinates; Vectors
872
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
(
)
(
)
1
1
2
2
2
2
,
cos
,
sin
x y
r
r
θ
θ
=
.
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2
2
2
1
2
1
2
2
2
2
1
1
2
2
1
1
2
2
2
2
2
2
2
2
2
2
1 2
2
1
1
1
2
2
1 2
2
1
1
1
2
2
2
2
2
2
1
1
1
2
2
2
1 2
2
1
2
1
2
2
1
2
1 2
cos
cos
sin
sin
cos
2
cos
cos
cos
sin
2
sin
sin
sin
cos
sin
cos
sin
2
cos
cos
sin
sin
2
cos
d
x
x
y
y
r
r
r
r
r
r r
r
r
r r
r
r
r
r r
r
r
r r
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
=
−
+
−
=
−
+
−
=
−
+
+
−
+
=
+
+
+
−
+
=
+
−
(
)
2
1θ
−

85.
cos
x
r
θ
=
and
sin
y
r
θ
=

86 – 87. Answers will vary.

Section 9.2
1. (
)
4,6
−

2. cos cos
sin
sin
A
B
A
B
+

3. (
)
(
)
(
)
(
)
2
2
2
2
2
( 2)
5
3
2
5
9
x
y
x
y
− −
+
−
=
+
+
−
=

4. odd, since sin(
)
sin
x
x
− = −
.
5.
2
2
−

6.
1
2
−

7. polar equation
8. False. They are sufficient but not necessary.
9. θ
−

10. π θ
−

11. True
12. 2n ; n
13.
4
r =

The equation is of the form
,
0
r
a a
=
>
. It is a
circle, center at the pole and radius 4. Transform
to rectangular form:
2
2
2
4
16
16
r
r
x
y
=
=
+
=




14.
2
r =

The equation is of the form
,
0
r
a a
=
>
. It is a
circle, center at the pole and radius 2. Transform
to rectangular form:
2
2
2
2
4
4
r
r
x
y
=
=
+
=


Section 9.2: Polar Equations and Graphs
873
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall



15.
3
θ
π
=

The equation is of the form θ α
=
. It is a line,
passing through the pole making an angle of
3
π

with the polar axis. Transform to rectangular
form:

3
tan
tan
3
3
3
y
x
y
x
θ
θ
π
=
π
=
=
=




16.
4
θ
π
= −

The equation is of the form θ α
=
. It is a line,
passing through the pole making an angle of
4
π
−

3
or
4
π






with the polar axis. Transform to
rectangular form:


4
tan
tan
4
1
y
x
y
x
θ
θ
π
= −
π


=
−




= −
= −




17.
sin
4
r
θ =

The equation is of the form sin
r
b
θ =
. It is a
horizontal line, 4 units above the pole.
Transform to rectangular form:
sin
4
4
r
y
θ =
=




Chapter 9: Polar Coordinates; Vectors
874
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
18.
cos
4
r
θ =

The equation is of the form cos
r
a
θ =
. It is a
vertical line, 4 units to the right of the pole.
Transform to rectangular form:


cos
4
4
r
x
θ =
=




19.
cos
2
r
θ = −

The equation is of the form cos
r
a
θ =
. It is a
vertical line, 2 units to the left of the pole.
Transform to rectangular form:
cos
2
2
r
x
θ = −
= −




20.
sin
2
r
θ = −

The equation is of the form sin
r
b
θ =
. It is a
horizontal line, 2 units below the pole.
Transform to rectangular form:
sin
2
2
r
y
θ = −
= −




21.
2cos
r
θ
=

The equation is of the form
2 cos ,
0
r
a
a
θ
= ±
>
. It is a circle, passing
through the pole, and center on the polar axis.
Transform to rectangular form:
2
2
2
2
2
2
2
2cos
2 cos
2
2
0
(
1)
1
r
r
r
x
y
x
x
x
y
x
y
θ
θ
=
=
+
=
−
+
=
−
+
=

center (1, 0) ; radius 1



22.
2sin
r
θ
=

The equation is of the form
2 sin ,
0
r
a
a
θ
= ±
>
.
It is a circle, passing through the pole, and center
on the line
2
θ
π
=
. Transform to rectangular form:
2
2
2
2
2
2
2
2sin
2 sin
2
2
0
(
1)
1
r
r
r
x
y
y
x
y
y
x
y
θ
θ
=
=
+
=
+
−
=
+
−
=

center (0, 1) ; radius 1

Section 9.2: Polar Equations and Graphs
875
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall



23.
4sin
r
θ
= −

The equation is of the form
2 sin ,
0
r
a
a
θ
= ±
>
.
It is a circle, passing through the pole, and center
on the line
2
θ
π
=
. Transform to rectangular form:
2
2
2
2
2
2
2
4sin
4 sin
4
4
0
(
2)
4
r
r
r
x
y
y
x
y
y
x
y
θ
θ
= −
= −
+
= −
+
+
=
+
+
=

center (0, 2)
−
; radius 2



24.
4cos
r
θ
= −

The equation is of the form
2 cos ,
0
r
a
a
θ
=
>
.
It is a circle, passing through the pole, and center
on the polar axis. Transform to rectangular form:
2
2
2
2
2
2
2
4cos
4 cos
4
4
0
(
2)
4
r
r
r
x
y
x
x
x
y
x
y
θ
θ
= −
= −
+
= −
+
+
=
+
+
=

center ( 2, 0)
−
; radius 2



25.
sec
4
r
θ =

The equation is a circle, passing through the
pole, center on the polar axis and radius 2.
Transform to rectangular form:
2
2
2
2
2
2
2
sec
4
1
4
cos
4cos
4 cos
4
4
0
(
2)
4
r
r
r
r
r
x
y
x
x
x
y
x
y
θ
θ
θ
θ
=
⋅
=
=
=
+
=
−
+
=
−
+
=

center (2, 0) ; radius 2



26.
csc
8
r
θ =

The equation is a circle, passing through the
pole, center on the line
2
θ
π
=
and radius 4.
Transform to rectangular form:

2
csc
8
1
8
sin
8sin
8 sin
r
r
r
r
r
θ
θ
θ
θ
=
⋅
=
=
=

Chapter 9: Polar Coordinates; Vectors
876
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
2
2
2
2
2
2
8
8
0
(
4)
16
x
y
y
x
y
y
x
y
+
=
+
−
=
+
−
=

center (0, 4) ; radius 4



27.
csc
2
r
θ = −

The equation is a circle, passing through the
pole, center on the line
2
θ
π
=
and radius 1.
Transform to rectangular form:
2
2
2
2
2
2
2
csc
2
1
2
sin
2sin
2 sin
2
2
0
(
1)
1
r
r
r
r
r
x
y
y
x
y
y
x
y
θ
θ
θ
θ
= −
⋅
= −
= −
= −
+
= −
+
+
=
+
+
=

center (0, 1)
−
; radius 1



28.
sec
4
r
θ = −

The equation is a circle, passing through the
pole, center on the polar axis and radius 2.
Transform to rectangular form:
sec
4
1
4
cos
4cos
r
r
r
θ
θ
θ
= −
⋅
= −
= −

2
2
2
2
2
2
2
4 cos
4
4
0
(
2)
4
r
r
x
y
x
x
x
y
x
y
θ
= −
+
= −
+
+
=
+
+
=

center ( 2, 0)
−
; radius 2



29. E
30. A
31. F
32. B
33. H
34. G
35. D
36. C

Section 9.2: Polar Equations and Graphs
877
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
37.
2 2cos
r
θ
= +

The graph will be a cardioid. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2 2cos(
)
2 2cos
r
θ
θ
= +
− = +
.
The graph is symmetric with respect to the polar
axis.

The line
2
θ
π
=
: Replace θ by π θ
−
.
[
]
2 2cos(
)
2 2 cos( ) cos
sin( )sin
2 2( cos
0)
2 2cos
r
θ
π
θ
π
θ
θ
θ
= +
π −
= +
+
= + −
+
= −

The test fails.

The pole: Replace by
r
r
−
.
2 2cos
r
θ
− = +
. The test fails.

Due to symmetry with respect to the polar axis,
assign values to
from 0 to
θ
π .


2 2cos
0
4
2
3 3.7
6
3
3
2
2
2
1
3
5
2
3 0.3
6
0
r
θ
θ
= +
π
+
≈
π
π
π
π
−
≈
π




38.
1 sin
r
θ
= +

The graph will be a cardioid. Check for
symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 sin(
) 1 sin
r
θ
θ
= +
− = −
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.


( )
( )
1 sin(
)
1
sin
cos
cos
sin
1 (0 sin )
1 sin
r
θ
θ
θ
θ
θ
= +
π −
= +
π
−
π




= + +
= +

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
1 sin
r
θ
− = +
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.


1 sin
0
2
3
1
0.1
3
2
1
6
2
0
1
3
6
2
3
1
1.9
3
2
2
2
r
θ
θ
π
π
π
π
π
π
= +
−
−
−
≈
−
+
≈




Chapter 9: Polar Coordinates; Vectors
878
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
39.
3 3sin
r
θ
= −

The graph will be a cardioid. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
3 3sin(
)
3 3sin
r
θ
θ
= −
− = +
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
3 3sin(
)
3 3 sin
cos
cos
sin
3 3(0 sin )
3 3sin
r
θ
θ
θ
θ
θ
= −
π −
= −
π
−
π




= −
+
= −

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
3 3sin
r
θ
− = −
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.


3 3sin
6
2
3 3
3
5.6
3
2
9
6
2
0
3
3
6
2
3 3
3
0.4
3
2
0
2
r
θ
θ
= −
π
−
π
−
+
≈
π
−
π
π
−
≈
π




40.
2 2cos
r
θ
= −

The graph will be a cardioid. Check for
symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2 2cos(
)
2 2cos
r
θ
θ
= −
− = −
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
2 2cos(
)
2 2 cos
cos
sin
sin
2 2( cos
0)
2 2cos
r
θ
θ
θ
θ
θ
= −
π −
= −
π
+
π




= − −
+
= +

The test fails.

The pole: Replace by
r
r
−
.
2 2cos
r
θ
− = −
.
The test fails.
Due to symmetry with respect to the polar axis,
assign values to
from 0 to
θ
π .


2 2cos
0
0
2
3 0.3
6
1
3
2
2
2
3
3
5
2
3 3.7
6
4
r
θ
θ
= −
π
−
≈
π
π
π
π
+
≈
π





Section 9.2: Polar Equations and Graphs
879
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
41.
2 sin
r
θ
= +

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2 sin(
)
2 sin
r
θ
θ
= +
− = −
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
2 sin(
)
2
sin
cos
cos
sin
2 (0 sin )
2 sin
r
θ
θ
θ
θ
θ
= +
π −
= +
π
−
π




= + +
= +

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
2 sin
r
θ
− = +
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.


2 sin
1
2
3
2
1.1
3
2
3
6
2
0
2
5
6
2
3
2
2.9
3
2
3
2
r
θ
θ
= +
π
−
π
−
−
≈
π
−
π
π
+
≈
π




42.
2 cos
r
θ
= −

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2 cos(
)
2 cos
r
θ
θ
= −
− = −
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
2 cos(
)
2
cos
cos
sin
sin
2 ( cos
0)
2 cos
r
θ
θ
θ
θ
θ
= −
π −
= −
π
+
π




= − −
+
= +

The test fails.

The pole: Replace by
r
r
−
.
2 cos
r
θ
− = −
.
The test fails.

Due to symmetry with respect to the polar axis,
assign values to
from 0 to
θ
π .


2 cos
0
1
3
2
1.1
6
2
3
3
2
2
2
2
5
3
2
5
3
2
2.9
6
2
3
r
θ
θ
= −
π
−
≈
π
π
π
π
+
≈
π




Chapter 9: Polar Coordinates; Vectors
880
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
43.
4 2cos
r
θ
= −

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
4 2cos(
)
4 2cos
r
θ
θ
= −
− = −
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
4 2cos(
)
4 2 cos
cos
sin
sin
4 2( cos
0)
4 2cos
r
θ
θ
θ
θ
θ
= −
π −
= −
π
+
π




= − −
+
= +

The test fails.

The pole: Replace by
r
r
−
.
4 2cos
r
θ
− = −
.
The test fails.

Due to symmetry with respect to the polar axis,
assign values to
from 0 to
θ
π .


4 2cos
0
2
4
3 2.3
6
3
3
4
2
2
5
3
5
4
3 5.7
6
6
r
θ
θ
= −
π
−
≈
π
π
π
π
+
≈
π




44.
4 2sin
r
θ
= +

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
4 2sin(
)
4 2sin
r
θ
θ
= +
− = −
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
4 2sin(
)
4 2 sin
cos
cos
sin
4 2(0 sin )
4 2sin
r
θ
θ
θ
θ
θ
= +
π −
= +
π
−
π




= +
+
= +

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
4 2sin
r
θ
− = +
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.


4 2sin
2
2
4
3 2.3
3
3
6
0
4
5
6
4
3 5.7
3
6
2
r
θ
θ
= +
π
−
π
−
−
≈
π
−
π
π
+
≈
π





Section 9.2: Polar Equations and Graphs
881
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
45.
1 2sin
r
θ
= +

The graph will be a limaçon with an inner loop.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 2sin(
) 1 2sin
r
θ
θ
= +
− = −
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
1 2sin(
)
1 2 sin
cos
cos
sin
1 2(0 sin )
1 2sin
r
θ
θ
θ
θ
θ
= +
π −
= +
π
−
π




= +
+
= +

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
1 2sin
r
θ
− = +
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.
1 2sin
1
2
1
3
0.7
3
0
6
0
1
2
6
1
3 2.7
3
3
2
r
θ
θ
= +
π
−
−
π
−
−
≈ −
π
−
π
π
+
≈
π




46.
1 2sin
r
θ
= −

The graph will be a limaçon with an inner loop.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 2sin(
) 1 2sin
r
θ
θ
= −
− = +
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
1 2sin(
)
1 2 sin
cos
cos
sin
1 2(0 sin )
1 2sin
r
θ
θ
θ
θ
θ
= −
π −
= −
π
−
π




= −
+
= −

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
1 2sin
r
θ
− = −
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.
1 2sin
3
2
1
3 2.7
3
2
6
0
1
0
6
1
3
0.7
3
1
2
r
θ
θ
= −
π
−
π
−
+
≈
π
−
π
π
−
≈ −
π
−




Chapter 9: Polar Coordinates; Vectors
882
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
47.
2 3cos
r
θ
= −

The graph will be a limaçon with an inner loop.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2 3cos(
)
2 3cos
r
θ
θ
= −
− = −
.
The graph is symmetric with respect to the polar
axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
2 3cos(
)
2 3 cos
cos
sin
sin
2 3( cos
0)
2 3cos
r
θ
θ
θ
θ
θ
= −
π −
= −
π
+
π




= − −
+
= +

The test fails.

The pole: Replace by
r
r
−
.
2 3cos
r
θ
− = −
.
The test fails.

Due to symmetry with respect to the polar axis,
assign values to
from 0 to
θ
π .
2 3cos
0
1
3 3
2
0.6
6
2
1
3
2
2
2
2
7
3
2
5
3 3
2
4.6
6
2
5
r
θ
θ
= −
−
π
−
≈ −
π
π
π
π
+
≈
π




48.
2 4cos
r
θ
= +

The graph will be a limaçon with an inner loop.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2 4cos(
)
2 4cos
r
θ
θ
= +
− = +
.
The graph is symmetric with respect to the polar
axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
2 4cos(
)
2 4 cos
cos
sin
sin
2 4( cos
0)
2 4cos
r
θ
θ
θ
θ
θ
= +
π −
= +
π
+
π




= + −
+
= −

The test fails.

The pole: Replace by
r
r
−
.
2 4cos
r
θ
− = +
.
The test fails.

Due to symmetry with respect to the polar axis,
assign values to
from 0 to
θ
π .
2 4cos
0
6
2 2 3 5.5
6
4
3
2
2
2
0
3
5
2 2 3
1.5
6
2
r
θ
θ
= +
π
+
≈
π
π
π
π
−
≈ −
π
−





Section 9.2: Polar Equations and Graphs
883
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
49.
3cos(2 )
r
θ
=

The graph will be a rose with four petals. Check
for symmetry:

Polar axis: Replace
by
θ
θ
−
.
3cos(2(
))
3cos( 2 ) 3cos(2 )
r
θ
θ
θ
=
−
=
−
=
. The
graph is symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
[
]
(
)
(
)
(
)
(
)
(
)
3cos 2(
)
3cos(2
2 )
3 cos 2
cos 2
sin 2
sin 2
3(cos 2
0)
3cos 2
r
θ
θ
θ
θ
θ
θ
=
π −
=
π −
=
π
+
π




=
+
=

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Since the graph is symmetric with
respect to both the polar axis and the line
2
θ
π
=
,
it is also symmetric with respect to the pole.

Due to symmetry, assign values to
from 0 to
2
θ
π
.
(
)
3cos 2
3
3
6
2
0
4
3
3
2
3
2
r
θ
θ
=
π
π
π
−
π
−




50.
2sin(3 )
r
θ
=

The graph will be a rose with three petals.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
[
]
(
)
2sin 3(
)
2sin( 3 )
2sin 3
r
θ
θ
θ
=
−
=
−
= −
. The
test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.


[
]
(
)
(
)
(
)
(
)
(
)
(
)
2sin 3(
)
2sin(3
3 )
2 sin 3 cos 3
cos 3
sin 3
2 0 sin 3
2sin 3
r
θ
θ
θ
θ
θ
θ
=
π −
=
π −
=
π
−
π




=
+




=

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
(
)
2sin 3
r
θ
− =
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.
(
)
(
)
2sin 3
2sin 3
2
2
2
6
0
2 1.4
3
4
2
1.4
0
4
3
2
2
6
2
0
0
r
r
θ
θ
θ
θ
=
=
π
π
−
π
π
−
≈
π
π
−
−
≈ −
π
π
−
−
−




Chapter 9: Polar Coordinates; Vectors
884
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
51.
4sin(5 )
r
θ
=

The graph will be a rose with five petals. Check
for symmetry:

Polar axis: Replace
by
θ
θ
−
.
[
]
(
)
4sin 5(
)
4sin( 5 )
4sin 5
r
θ
θ
θ
=
−
=
−
= −
.
The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
[
]
(
)
(
)
(
)
(
)
(
)
(
)
4sin 5(
)
4sin(5
5 )
4 sin 5
cos 5
cos 5
sin 5
4 0 sin 5
4sin 5
r
θ
θ
θ
θ
θ
θ
=
π −
=
π −
=
π
−
π




=
+




=

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
(
)
4sin 5
r
θ
− =
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
,
assign values to
from
to
2
2
θ
π
π
−
.


(
)
(
)
4sin 5
4sin 5
4
2
2
6
2 3 3.5
2 2
2.8
3
4
2 2 2.8
2 3
3.5
4
3
2
4
6
2
0
0
r
r
θ
θ
θ
θ
=
=
π
π
−
−
π
π
−
≈
−
≈ −
π
π
−
≈
−
≈ −
π
π
−
−




52.
3cos(4 )
r
θ
=

The graph will be a rose with eight petals.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
3cos(4(
))
3cos( 4 )
3cos(4 )
r
θ
θ
θ
=
−
=
−
=
. The
graph is symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
[
]
(
)
(
)
(
)
(
)
(
)
3cos 4(
)
3cos(4
4 )
3 cos 4
cos 4
sin 4
sin 4
3(cos 4
0)
3cos 4
r
θ
θ
θ
θ
θ
θ
=
π −
=
π −
=
π
+
π




=
+
=

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Since the graph is symmetric with
respect to both the polar axis and the line
2
θ
π
=
,
it is also symmetric with respect to the pole.

Due to symmetry, assign values to
from 0 to
2
θ
π
.
(
)
3cos 4
0
2
3
6
2
3
4
3
3
2
3
2
r
θ
θ
=
π
−
π
−
π
−
π





Section 9.2: Polar Equations and Graphs
885
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
53.
2
9cos(2 )
r
θ
=

The graph will be a lemniscate. Check for
symmetry:

Polar axis: Replace
by
θ
θ
−
.
2
9cos(2(
))
9cos( 2 )
9cos(2 )
r
θ
θ
θ
=
−
=
−
=
. The
graph is symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
[
]
(
)
(
)
(
)
2
9cos 2(
)
9cos(2
2 )
9 cos 2
cos 2
sin 2
sin 2
9(cos 2
0)
9cos 2
r
θ
θ
θ
θ
θ
θ
=
π −
=
π −
=
π
+
π




=
+
=

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Since the graph is symmetric with
respect to both the polar axis and the line
2
θ
π
=
,
it is also symmetric with respect to the pole.

Due to symmetry, assign values to
from 0 to
2
θ
π
.
(
)
9cos 2
0
3
3 2
2.1
6
2
0
4
undefined
3
undefined
2
r
θ
θ
= ±
±
π
±
≈ ±
π
π
π




54.
2
sin(2 )
r
θ
=

The graph will be a lemniscate. Check for
symmetry:

Polar axis: Replace
by
θ
θ
−
.
2
sin(2(
))
sin( 2 )
sin(2 )
r
θ
θ
θ
=
−
=
−
= −
. The test
fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
[
]
(
)
(
)
(
)
(
)
(
)
2
sin 2(
)
sin(2
2 )
sin 2
cos 2
cos 2
sin 2
0 sin 2
sin 2
r
θ
θ
θ
θ
θ
θ
=
π −
=
π −
=
π
−
π
= −
= −

The test fails.

The pole: Replace by
r
r
−
.
(
)
(
)
2
2
(
)
sin 2
sin 2
r
r
θ
θ
−
=
=

The graph is symmetric with respect to the pole.

Due to symmetry, assign values to
from 0 to
θ
π .
(
)
sin 2
0
0
3
6
2
3
3
2
0
2
2
undefined
3
5
undefined
6
0
r
θ
θ
= ±
π
±
π
±
π
π
π
π




Chapter 9: Polar Coordinates; Vectors
886
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
55.
2
r
θ
=

The graph will be a spiral. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
2
r
θ
−
=
. The
test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
2
r
θ
π−
=
. The test fails.

The pole: Replace by
r
r
−
.
2
r
θ
− =
. The test
fails.


2
0.1
0.3
2
0.6
4
0
1
1.7
4
3.0
2
8.8
3
26.2
2
2
77.9
r
θ
θ
=
−π
π
−
π
−
π
π
π
π
π




56.
3
r
θ
=

The graph will be a spiral. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
3
r
θ
−
=
. The
test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
3
r
θ
π−
=
.
The test fails.

The pole: Replace by
r
r
−
.
3
r
θ
− =
. The test
fails.


3
0.03
0.2
2
0.4
4
0
1
2.4
4
5.6
2
31.5
3
117.2
2
2
995
r
θ
θ
=
−π
π
−
π
−
π
π
π
π
π





Section 9.2: Polar Equations and Graphs
887
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
57.
1 cos
r
θ
= −

The graph will be a cardioid. Check for
symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 cos(
) 1 cos
r
θ
θ
= −
− = −
.
The graph is symmetric with respect to the polar
axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
1 cos(
)
1 (cos
cos
sin
sin )
1 ( cos
0)
1 cos
r
θ
θ
θ
θ
θ
= −
π −
= −
π
+
π
= − −
+
= +

The test fails.

The pole: Replace by
r
r
−
.
1 cos
r
θ
− = −
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .


1 cos
0
0
3
1
0.1
6
2
1
3
2
1
2
2
3
3
2
5
3
1
1.9
6
2
2
r
θ
θ
= −
π
−
≈
π
π
π
π
+
≈
π




58.
3 cos
r
θ
= +

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
3 cos(
)
3 cos
r
θ
θ
= +
− = +
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
3 cos(
)
3
cos
cos
sin
sin
3 ( cos
0)
3 cos
r
θ
θ
θ
θ
θ
= +
π −
= +
π
+
π




= + −
+
= −

The test fails.

The pole: Replace by
r
r
−
.
3 cos
r
θ
− = +
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .


3 cos
0
4
3
3
3.9
6
2
7
3
2
3
2
2
5
3
2
5
3
3
2.1
6
2
2
r
θ
θ
= +
π
+
≈
π
π
π
π
−
≈
π




Chapter 9: Polar Coordinates; Vectors
888
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
59.
1 3cos
r
θ
= −

The graph will be a limaçon with an inner loop.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 3cos(
) 1 3cos
r
θ
θ
= −
− = −
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
1 3cos(
)
1 3 cos
cos
sin
sin
1 3( cos
0)
1 3cos
r
θ
θ
θ
θ
θ
= −
π −
= −
π
+
π




= − −
+
= +

The test fails.

The pole: Replace by
r
r
−
.
1 3cos
r
θ
− = −
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .


1 3cos
0
2
3 3
1
1.6
6
2
1
3
2
1
2
2
5
3
2
5
3 3
1
3.6
6
2
4
r
θ
θ
= −
−
π
−
≈ −
π
−
π
π
π
+
≈
π




60.
4cos(3 )
r
θ
=

The graph will be a rose with three petals.
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
4cos(3(
))
4cos( 3 )
4cos(3 )
r
θ
θ
θ
=
−
=
−
=
. The
graph is symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
(
)
(
)
(
)
(
)
(
)
(
)
(
)
4cos 3
4cos 3
3
4 cos 3 cos3
sin 3
sin 3
4
cos3
0
4cos 3
r
θ
θ
θ
θ
θ
θ
=
π −




=
π −
=
π
+
π




=
−
+
= −

The test fails.

The pole: Replace by
r
r
−
.
(
)
4cos 3
r
θ
− =
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .


(
)
4cos 3
0
4
0
6
4
3
0
2
2
4
3
5
0
6
4
r
θ
θ
=
π
π
−
π
π
π
π
−





Section 9.2: Polar Equations and Graphs
889
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
61.
8cos
r
θ
=

The equation is of the form
2 cos ,
r
a
θ
= ±

0
a >
.
It is a circle, passing through the pole, and center
on the polar axis.


2sec
2
cos
cos
2
r
r
r
θ
θ
θ
=
=
=

The equation is of the form cos
r
a
θ =
. It is a
vertical line, 2 units to the right of the pole.




Use substitution to find the point(s) of
intersection:
2
8cos
2sec
2
8cos
cos
1
cos
4
1
cos
2
2
4
5
,
,
,
for 0
2
3 3
3
3
θ
θ
θ
θ
θ
θ
π
π
π π
θ
θ
π
=
=
=
= ±
=
≤ <

If
3
π
θ =
,
1
8cos
8
4
3
2
r
π


=
=
=




.
If
2
3
π
θ =
,
2
1
8cos
8
4
3
2
r
π


=
=
−
= −




.
If
4
3
π
θ =
,
4
1
8cos
8
4
3
2
r
π


=
=
−
= −




.
If
5
3
π
θ =
,
5
1
8cos
8
4
3
2
r
π


=
=
=




.
The points of intersection are 4,
3
π






,
2
4,
3
π


−




,
4
4,
3
π


−




, and
5
4,
3
π






.
62.
8sin
r
θ
=

The equation is of the form
2 sin ,
r
a
θ
= ±

0
a >
.
It is a circle, passing through the pole, and center
on the line
2
θ
π
=
.


4csc
4
sin
sin
4
r
r
r
θ
θ
θ
=
=
=

The equation is of the form sin
r
b
θ =
. It is a
horizontal line, 4 units above the pole.




Use substitution to find the point(s) of
intersection:
2
8sin
4csc
4
8sin
sin
1
sin
2
2
sin
2
3
5
7
,
,
,
for 0
2
4 4
4
4
θ
θ
θ
θ
θ
θ
π π π
π
θ
θ
π
=
=
=
= ±
=
≤ <

If
4
π
θ =
,
2
8sin
8
4 2
4
2
r
π


=
=
=




.
If
3
4
π
θ =
,
3
2
8sin
8
4 2
4
2
r
π


=
=
=




.
If
5
4
π
θ =
,
5
2
8sin
8
4 2
4
2
r
π


=
=
−
= −




.
If
7
4
π
θ =
,
7
2
8sin
8
4 2
4
2
r
π


=
=
−
= −




.
The points of intersection are 4 2,
4
π






,
3
4 2,
4
π






,
5
4 2,
4
π


−




, and
7
4 2,
4
π


−




.
Chapter 9: Polar Coordinates; Vectors
890
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
63.
sin
r
θ
=

The equation is of the form
2 sin ,
r
a
θ
= ±

0
a >
.
It is a circle, passing through the pole, and center
on the line
2
θ
π
=
.


1 cos
r
θ
= +

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 cos(
) 1 cos
r
θ
θ
= +
− = +
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
1 cos(
)
1
cos
cos
sin
sin
1 ( cos
0)
1 cos
r
θ
θ
θ
θ
θ
= +
π −
= +
π
+
π




= + −
+
= −

The test fails.

The pole: Replace by
r
r
−
.
1 cos
r
θ
− = +
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .


1 cos
0
2
3
1
1.9
6
2
3
3
2
1
2
2
1
3
2
5
3
1
0.1
6
2
0
r
θ
θ
= +
π
+
≈
π
π
π
π
−
≈
π





Use substitution to find the point(s) of
intersection:
(
)
( )
2
2
2
2
sin
1 cos
sin
cos
1
sin
cos
1
sin
2sin cos
cos
1
1 2sin cos
1
2sin cos
0
sin cos
0
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
= +
−
=
−
=
−
+
=
−
=
−
=
=

sin
0 or cos
0
3
0,
or
,
for 0
2
2 2
θ
θ
π π
θ
π
θ
θ
π
=
=
=
=
≤ <

If
0
θ =
,
sin 0 0
r =
=
.
If θ π
=
,
sin
0
r
π
=
=
.
If
2
π
θ =
,
sin
1
2
r
π
=
=
.
If
3
2
π
θ =
,
3
sin
1
2
r
π
=
= −
.
The points of intersection are (
)
0, 0 , (
)
0, π
,
1,
2
π






, and
3
1,
2
π


−




.
64.
3
r =

The equation is of the form
,
0
r
a a
=
>
. It is a
circle, center at the pole and radius 3.


2 2cos
r
θ
= +

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2 2cos(
)
2 2cos
r
θ
θ
= +
− = +
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
2 2cos(
)
2 2 cos
cos
sin
sin
2 2( cos
0)
2 2cos
r
θ
θ
θ
θ
θ
= +
π −
= +
π
+
π




= + −
+
= −

The test fails.

The pole: Replace by
r
r
−
.
2 2cos
r
θ
− = +
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .

Section 9.2: Polar Equations and Graphs
891
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall


2 2cos
0
4
2
3 3.7
6
3
3
2
2
2
1
3
5
2
3 0.3
6
0
r
θ
θ
= +
π
+
≈
π
π
π
π
−
≈
π





Use substitution to find the point(s) of
intersection:
3 2 2cos
1 2cos
1
cos
2
5
,
for 0
2
3 3
θ
θ
θ
π π
θ
θ
π
= +
=
=
=
≤ <

If
3
π
θ =
,
1
2 2cos
2 2
3
3
2
r
π


= +
= +
=




.
If
5
3
π
θ =
,
5
1
2 2cos
2 2
3
3
2
r
π


= +
= +
=




.
The points of intersection are 3,
3
π






and
5
3,
3
π






.
65.
1 sin
r
θ
= +

The graph will be a cardioid. Check for
symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 sin(
) 1 sin
r
θ
θ
= +
− = −
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.


[
]
1 sin(
)
1
sin( )cos
cos( )sin
1 (0 sin )
1 sin
r
θ
π
θ
π
θ
θ
θ
= +
π −
= +
−
= + +
= +

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
1 sin
r
θ
− = +
.
The test fails.

Due to symmetry with respect to the line
2
θ
π
=
, assign values to
from
to
2
2
θ
π
π
−
.


2
3
3
2
1
6
2
3
6
2
3
3
2
2
1 sin
0
1
0.1
0
1
1
1.9
2
r
π
π
π
π
π
π
θ
θ
−
−
−
= +
−
≈
+
≈



1 cos
r
θ
= +

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 cos(
) 1 cos
r
θ
θ
= +
− = +
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
1 cos(
)
1
cos
cos
sin
sin
1 ( cos
0)
1 cos
r
θ
θ
θ
θ
θ
= +
π −
= +
π
+
π




= + −
+
= −

The test fails.
Chapter 9: Polar Coordinates; Vectors
892
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

The pole: Replace by
r
r
−
.
1 cos
r
θ
− = +
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .


3
6
2
3
3
2
2
2
1
3
2
5
3
0.1
6
2
1 cos
0
2
1
1.9
1
1
0
r
θ
θ
π
π
π
π
π
≈
= +
+
≈
−
π





Use substitution to find the point(s) of
intersection:
1 sin
1 cos
sin
cos
sin
1
cos
tan
1
5
,
for 0
2
4 4
θ
θ
θ
θ
θ
θ
θ
θ
θ
π
+
= +
=
=
=
π π
=
≤ <

If
4
π
θ =
,
2
1 sin
1
1.7
4
2
r
π
= +
= +
≈
.
If
5
4
π
θ =
,
5
2
1 sin
1
0.3
4
2
r
π
= +
= −
≈
.
The points of intersection are
2
1
,
2
4
π


+




and
2 5
1
,
2
4
π


−




.
66.
1 cos
r
θ
= +

The graph will be a limaçon without an inner
loop. Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
1 cos(
) 1 cos
r
θ
θ
= +
− = +
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
( )
( )
1 cos(
)
1
cos
cos
sin
sin
1 ( cos
0)
1 cos
r
θ
θ
θ
θ
θ
= +
π −
= +
π
+
π




= + −
+
= −

The test fails.

The pole: Replace by
r
r
−
.
1 cos
r
θ
− = +
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .


1 cos
0
2
3
1
1.9
6
2
3
3
2
1
2
2
1
3
2
5
3
1
0.1
6
2
0
r
θ
θ
= +
π
+
≈
π
π
π
π
−
≈
π



3cos
r
θ
=

The equation is of the form
2 cos ,
r
a
θ
= ±

0
a >
.
It is a circle, passing through the pole, and center
on the polar axis.




Use substitution to find the point(s) of
intersection:

Section 9.2: Polar Equations and Graphs
893
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

1 cos
3cos
1 2cos
1
cos
2
5
,
for 0
2
3 3
θ
θ
θ
θ
θ
θ
π
+
=
=
=
π π
=
≤ <

If
3
π
θ =
,
1
3
1 cos
1
3
2
2
r
π
= +
= + =
.
If
5
3
π
θ =
,
5
1
3
1 cos
1
3
2
2
r
π
= +
= + =
.
The points of intersection are
3
2 3
,
π






and
3 5
2 3
,
π






.
67. The graph is a cardioid whose equation is of the
form
cos
r
a b
θ
= +
. The graph contains the point
(6,0) , so we have
6
cos 0
6
(1)
6
a b
a b
a b
= +
= +
= +

The graph contains the point
2
3,
π






, so we have
2
3
cos
3
(0)
3
a b
a b
a
π
= +
= +
=

Substituting
3
a =
into the first equation yields:
6
6 3
3
a b
b
b
= +
= +
=

Therefore, the graph has equation
3 3cos
r
θ
= +
.
68. The graph is a cardioid whose equation is of the
form
cos
r
a b
θ
= +
. The graph contains the point
(6,
)
π
, so we have
6
cos
6
( 1)
6
a b
a b
a b
π
= +
= + −
= −

The graph contains the point
2
3,
π






, so we have
2
3
cos
3
(0)
3
a b
a b
a
π
= +
= +
=

Substituting
3
a =
into the first equation yields:
6
6 3
3
a b
b
b
= −
= −
= −

Therefore, the graph has equation
3 3cos
r
θ
= −
.
69. The graph is a limaçon without inner loop whose
equation is of the form
sin
r
a b
θ
= +
, where
0 b a
< <
. The graph contains the point (
)
4,0 ,
so we have
( )
4
sin 0
4
0
4
a b
a b
a
= +
= +
=

The graph contains the point 5,
2
π






, so we have
( )
5
sin
2
5
1
5
a b
a b
a b
π
= +
= +
= +

Substituting
4
a =
into the second equation
yields:
5
5 4
1
a b
b
b
= +
= +
=

Therefore, the graph has equation
4 sin
r
θ
= +
.
70. The graph is a limaçon with inner loop whose
equation is of the form
sin
r
a b
θ
= +
, where
0 a b
< <
. The graph contains the point (
)
1,0 ,
so we have
( )
1
sin 0
1
0
1
a b
a b
a
= +
= +
=

The graph contains the point 5,
2
π






, so we have


( )
5
sin
2
5
1
5
a b
a b
a b
π
= +
= +
= +

Substituting
1
a =
into the second equation yields:
5
5 1
4
a b
b
b
= +
= +
=

Therefore, the graph has equation
1 4sin
r
θ
= +
.
71.
2
1 cos
r
θ
=
−
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
(
)
2
2
1 cos
1 cos
r
θ
θ
=
=
−
−
−
.

The graph is symmetric with respect to the polar
axis.
Chapter 9: Polar Coordinates; Vectors
894
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
(
)
(
)
2
1 cos
2
1
cos cos
sin sin
2
1 ( cos
0)
2
1 cos
r
θ
π
θ
π
θ
θ
θ
=
−
π −
=
−
+
=
− −
+
=
+

The test fails.

The pole: Replace by
r
r
−
.
2
1 cos
r
θ
− =
−
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .
2
1 cos
0
undefined
2
14.9
6
1
3 2
4
3
2
2
2
4
3
3
5
2
1.1
6
1
3 2
1
r
θ
θ
=
−
π
≈
−
π
π
π
π
≈
+
π




72.
2
1 2cos
r
θ
=
−
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
2
2
1 2cos(
)
1 2cos
r
θ
θ
=
=
−
−
−
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
(
)
(
)
(
)
2
1 2cos
2
1 2 cos cos
sin sin
2
2
1 2
cos
0
1 2cos
r
θ
π
θ
π
θ
θ
θ
=
−
π −
=
−
+
=
=
− −
+
+

The test fails.

The pole: Replace by
r
r
−
.
2
1 2cos
r
θ
− =
−
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .
2
1 2cos
0
2
2
2.7
6
1
3
undefined
3
2
2
2
1
3
5
2
0.7
6
1
3
2
3
r
θ
θ
=
−
−
π
≈ −
−
π
π
π
π
≈
+
π





Section 9.2: Polar Equations and Graphs
895
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
73.
1
3 2cos
r
θ
=
−
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
(
)
1
1
3 2cos
3 2cos
r
θ
θ
=
=
−
−
−
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
(
)
(
)
(
)
1
3 2cos
1
3 2 cos cos
sin sin
1
1
3 2
cos
0
3 2cos
r
θ
π
θ
π
θ
θ
θ
=
−
π −
=
−
+
=
=
− −
+
+

The test fails.

The pole: Replace by
r
r
−
.
1
3 2cos
r
θ
− =
−
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .
1
3 2cos
0
1
1
0.8
6
3
3
1
3
2
1
2
3
2
1
3
4
5
1
0.2
6
3
3
1
5
r
θ
θ
=
−
π
≈
−
π
π
π
π
≈
+
π




74.
1
1 cos
r
θ
=
−
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. The result is
(
)
1
1
1 cos
1 cos
r
θ
θ
=
=
−
−
−
. The graph is
symmetric with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
(
)
(
)
(
)
1
1 cos
1
1
cos cos
sin sin
1
1
cos
0
1
1 cos
r
θ
π
θ
π
θ
θ
θ
=
−
π −
=
−
+
=
− −
+
=
+

The test fails.

The pole: Replace by
r
r
−
.
1
1 cos
r
θ
− =
−
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .
1
1 cos
0
undefined
1
7.5
6
1
3 2
2
3
2
2
2
2
3
3
5
1
0.5
6
1
3 2
1
2
r
θ
θ
=
−
π
≈
−
π
π
π
π
≈
+
π

Chapter 9: Polar Coordinates; Vectors
896
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall



75.
,
0
r θ θ
=
≥

Check for symmetry:

Polar axis: Replace
by
θ
θ
−
. r
θ
= −
.
The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
. r
θ
= π −
.
The test fails.

The pole: Replace by
r
r
−
. r θ
− =
. The test
fails.


0
0
0.5
6
6
1.0
3
3
1.6
2
2
3.1
3
3
4.7
2
2
2
2
6.3
r
θ
θ
π
π
=
π
π ≈
π
π ≈
π
π ≈
π ≈
π
π ≈
π ≈




76.
3
r
θ
=
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
3
r
θ
=
−
. The
test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
3
r
θ
=
π−
. The test fails.

The pole: Replace by
r
r
−
.
3
r
θ
− =
. The test
fails.


3
0
undefined
18
5.7
6
9
2.9
3
6
1.9
2
3
1.0
3
2
0.6
2
3
2
0.5
2
r
θ
θ
π
=
π
≈
π
π
≈
π
π
≈
π
≈
π
π
≈
π
π
≈
π





Section 9.2: Polar Equations and Graphs
897
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
77.
1
csc
2
2, 0
sin
r
θ
θ
θ
=
− =
−
< < π
Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
csc(
) 2
csc
2
r
θ
θ
=
− − = −
−
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.


(
)
(
)
csc
2
1
2
sin
1
2
sin cos
cos sin
1
2
0 cos
1 sin
1
2
sin
csc
2
r
π θ
π θ
π
θ
π
θ
θ
θ
θ
θ
=
−
−
=
−
−
=
−
−
=
−
⋅
− ⋅
=
−
=
−

The graph is symmetric with respect to the line
2
θ
π
=
.

The pole: Replace by
r
r
−
.
csc
2
r
θ
− =
−
.
The test fails.

Due to symmetry, assign values to
from 0 to
2
θ
π
.
csc
2
0
undefined
0
6
2 2
0.6
4
2 3
2
0.8
3
3
1
2
r
θ
θ
=
−
π
π
− ≈ −
π
− ≈ −
π
−




78.
sin
tan
r
θ
θ
=

Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
sin(
) tan(
)
( sin )( tan )
sin
tan
r
θ
θ
θ
θ
θ
θ
=
−
−
= −
−
=

The graph is symmetric with respect to the polar
axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
(
)
(
)
(
)
sin
tan
tan
tan
sin cos
cos sin
1 tan
tan
tan
sin
1
sin
tan
r
θ
θ
π
θ
π
θ
π
θ
π
θ
θ
θ
θ
θ
=
π −
π −
−


=
−


+


−
=
⋅
= −

The test fails.

The pole: Replace by
r
r
−
.
sin
tan
r
θ
θ
− =
.
The test fails.

Due to symmetry, assign values to
from 0 to
θ
π .
sin
tan
0
0
1
3
0.3
6
2 3
3
3
2
undefined
2
2
3
3
2
5
1
3
0.3
6
2
3
0
r
θ
θ
θ
=
π
⋅
≈
π
π
π
−


π
⋅ −
≈ −






π




Chapter 9: Polar Coordinates; Vectors
898
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
79.
tan ,
2
2
r
θ
θ
π
π
=
− < <

Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
tan(
)
tan
r
θ
θ
=
− = −
. The test fails.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.


( )
( )
tan
tan
tan
tan(
)
tan
1 tan
tan
1
r
θ
θ
θ
θ
θ
π −
−
=
π −
=
=
= −
+
π

The test fails.

The pole: Replace by
r
r
−
.
tan
r
θ
− =
. The
test fails.


tan
3
1.7
3
1
4
3
0.6
6
3
0
0
3
0.6
6
3
1
4
3 1.7
3
r
θ
θ
=
π
−
−
≈ −
π
−
−
π
−
−
≈ −
π
≈
π
π
≈




80.
cos
2
r
θ
=

Check for symmetry:

Polar axis: Replace
by
θ
θ
−
.
cos
cos
2
2
r
θ
θ


=
−
=




. The graph is symmetric
with respect to the polar axis.

The line
2
θ
π
=
: Replace
by
θ
θ
π−
.
cos
cos
2
2 2
cos
cos
sin
sin
2
2
2
2
sin
2
r
θ
θ
π
θ
π
θ
θ
π−
π




=
=
−








=
⋅
+
⋅
=

The test fails.

The pole: Replace by
r
r
−
.
cos
2
r
θ
− =
. The
test fails.

Due to symmetry, assign values to θ from 0 to π .
cos
2
0
1
0.97
6
3
0.87
3
2
2
0.71
2
2
2
1
3
2
5
0.26
6
0
r
θ
θ
=
π
π
≈
π
≈
π
π
π





Section 9.2: Polar Equations and Graphs
899
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
81. Convert the equation to rectangular form:
sin
r
a
y
a
θ =
=

The graph of sin
r
a
θ =
is a horizontal line a
units above the pole if
0
a ≥
, and |a| units below
the pole if
0
a <
.
82. Convert the equation to rectangular form:
cos
r
a
x a
θ =
=

The graph of cos
r
a
θ =
is a vertical line a units
to the right of the pole if
0
a ≥
and |a| units to
the left of the pole if
0
a <
.
83. Convert the equation to rectangular form:
2
2
2
2
2
2
2
2
2 sin ,
0
2
sin
2
2
0
(
)
r
a
a
r
a r
x
y
ay
x
y
ay
x
y a
a
θ
θ
=
>
=
+
=
+
−
=
+
−
=

Circle: radius a, center at rectangular
coordinates (0,
).
a

84. Convert the equation to rectangular form:
2
2
2
2
2
2
2
2
2 sin ,
0
2
sin
2
2
0
(
)
r
a
a
r
a r
x
y
ay
x
y
ay
x
y a
a
θ
θ
= −
>
= −
+
= −
+
+
=
+
+
=

Circle: radius a, center at rectangular
coordinates (0,
).
a
−

85. Convert the equation to rectangular form:
2
2
2
2
2
2
2
2
2 cos ,
0
2
cos
2
2
0
(
)
r
a
a
r
a r
x
y
ax
x
ax y
x a
y
a
θ
θ
=
>
=
+
=
−
+
=
−
+
=

Circle: radius a, center at rectangular
coordinates ( , 0).
a

86. Convert the equation to rectangular form:
2
2
2
2
2
2
2
2
2 cos ,
0
2
cos
2
2
0
(
)
r
a
a
r
a r
x
y
ax
x
ax y
x a
y
a
θ
θ
= −
>
= −
+
= −
+
+
=
+
+
=

Circle: radius a, center at rectangular
coordinates (
, 0)
a
−
.
87. a.
2
cos
r
θ
=
: 2
2
cos(
)
cos
r
r
θ
θ
=
π −
= −


Not equivalent; test fails.



(
)2
2
cos(
)
cos
r
r
θ
θ
−
=
−
=


New test works.

b.
2
sin
r
θ
=
:
2
2
sin(
)
sin
r
r
θ
θ
=
π −
=


Test works.



(
)2
2
sin(
)
sin
r
r
θ
θ
−
=
−
= −


Not equivalent; new test fails.
88. Answers will vary.
89. If an equation passes one or more of these tests,
then it will definitely have a graph that is
symmetric with respect to the polar axis, the line
2
π
θ =
, or the pole, depending on the test(s)
passed. However, an equation may fail these
tests and still have a graph that is symmetric with
respect to the polar axis, the line
2
π
θ =
, or the
pole.
90. Answers will vary.

Chapter 9: Polar Coordinates; Vectors
900
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 9.3
1.
4 3i
− +

2. sin cos
cos sin
A
B
A
B
+

3. cos cos
sin
sin
A
B
A
B
−

4.
3
1
,
2
2
−


5. real; imaginary
6. magnitude, modulus, argument
7.
1 2
rr ; 1
2
θ
θ
+
; 1
2
θ
θ
+

8.
n
r ; nθ ; nθ
9. three
10. True
11.
2
2
2
2
1
1
2
r
x
y
=
+
=
+
=

tan
1
45º
y
x
θ
θ
=
=
=

The polar form of
1
z
i
= +
is
(
)
(
)
cos
sin
2 cos 45º
sin 45º
z
r
i
i
θ
θ
=
+
=
+




12.
2
2
2
2
( 1)
1
2
r
x
y
=
+
=
−
+
=

tan
1
135º
y
x
θ
θ
=
= −
=

The polar form of
1
z
i
= − +
is
(
)
(
)
cos
sin
2 cos135º
sin135º
z
r
i
i
θ
θ
=
+
=
+




13.
(
)
(
)
2
2
2
2
3
1
4
2
r
x
y
=
+
=
+ −
=
=

1
3
tan
3
3
330º
y
x
θ
θ
−
=
=
= −
=

The polar form of
3
z
i
=
−
is
(
)
(
)
cos
sin
2 cos330º
sin 330º
z
r
i
i
θ
θ
=
+
=
+




14.
(
)2
2
2
2
1
3
4
2
r
x
y
=
+
=
+ −
=
=

3
tan
3
1
300º
y
x
θ
θ
−
=
=
= −
=

The polar form of
1
3
z
i
= −
is
(
)
(
)
cos
sin
2 cos300º
sin 300º
z
r
i
i
θ
θ
=
+
=
+





Section 9.3: The Complex Plane; De Moivre’s Theorem
901
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
15.
2
2
2
2
0
( 3)
9
3
r
x
y
=
+
=
+ −
=
=

3
tan
0
270º
y
x
θ
θ
−
=
=
=

The polar form of
3
z
i
= −
is
(
)
(
)
cos
sin
3 cos 270º
sin 270º
z
r
i
i
θ
θ
=
+
=
+




16.
2
2
2
2
( 2)
0
4
2
r
x
y
=
+
=
−
+
=
=

0
tan
0
2
180º
y
x
θ
θ
=
=
=
−
=

The polar form of
2
z = −
is
(
)
(
)
cos
sin
2 cos180º
sin180º
z
r
i
i
θ
θ
=
+
=
+




17.
2
2
2
2
4
( 4)
32
4 2
r
x
y
=
+
=
+ −
=
=

4
tan
1
4
315º
y
x
θ
θ
−
=
=
= −
=

The polar form of
4 4
z
i
= −
is
(
)
(
)
cos
sin
4 2 cos315º
sin 315º
z
r
i
i
θ
θ
=
+
=
+




18.
(
)2
2
2
2
9 3
9
324 18
r
x
y
=
+
=
+
=
=

9
3
tan
3
9 3
30º
y
x
θ
θ
=
=
=
=

The polar form of
9 3 9
z
i
=
+
is
(
)
(
)
cos
sin
18 cos30º
sin 30º
z
r
i
i
θ
θ
=
+
=
+
.



19.
2
2
2
2
3
( 4)
25 5
r
x
y
=
+
=
+ −
=
=

4
tan
3
306.9º
y
x
θ
θ
−
=
=
≈

The polar form of
3 4
z
i
= −
is
(
)
(
)
cos
sin
5 cos306.9º
sin 306.9º
z
r
i
i
θ
θ
=
+
=
+




Chapter 9: Polar Coordinates; Vectors
902
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
20.
(
)2
2
2
2
2
3
7
r
x
y
=
+
=
+
=

3
tan
2
40.9º
y
x
θ
θ
=
=
≈

The polar form of
2
3
z
i
= +
is
(
)
(
)
cos
sin
7 cos 40.9º
sin 40.9º
z
r
i
i
θ
θ
=
+
=
+




21.
2
2
2
2
( 2)
3
13
r
x
y
=
+
=
−
+
=

3
3
tan
2
2
123.7º
y
x
θ
θ
=
=
= −
−
≈

The polar form of
2 3
z
i
= − +
is
(
)
(
)
cos
sin
13 cos123.7º
sin123.7º
z
r
i
i
θ
θ
=
+
=
+




22.
(
)
(
)
2
2
2
2
5
1
6
r
x
y
=
+
=
+ −
=

1
5
tan
5
5
335.9º
y
x
θ
θ
−
=
=
= −
≈

The polar form of
5
z
i
=
−
is
(
)
(
)
cos
sin
6 cos335.9º
sin 335.9º
z
r
i
i
θ
θ
=
+
=
+




23.
(
)
1
3
2 cos120º
sin120º
2
2
2
1
3
i
i
i


+
=
− +




= − +

24.
(
)
3 1
3 cos 210º
sin 210º
3
2
2
3 3 3
2
2
i
i
i


+
=
−
−




= −
−

25.
7
7
2
2
4 cos
sin
4
4
4
2
2
2 2 2 2
i
i
i


π
π


+
=
−








=
−

26.
5
5
3
1
2 cos
sin
2
3
6
6
2
2
i
i
i


π
π


+
=
−
+
= −
+









27.
(
)
3
3
3 cos
sin
3 0 1
3
2
2
i
i
i
π
π


+
=
−
= −





28.
(
)
4 cos
sin
4 0 1
4
2
2
i
i
i
π
π


+
=
+
=





29.
(
)
(
)
0.2 cos100º
sin100º
0.2
0.1736 0.9848
0.035 0.197
i
i
i
+
≈
−
+
≈ −
+

30.
(
)
(
)
0.4 cos 200º
sin 200º
0.4
0.9397 0.3420
0.376 0.137
i
i
i
+
≈
−
−
≈ −
−

31.
(
)
2 cos
sin
2 0.9848 0.1736
18
18
1.970 0.347
i
i
i
π
π


+
≈
+




≈
+

32.
(
)
3 cos
sin
3 0.9511 0.3090
10
10
2.853 0.927
i
i
i
π
π


+
≈
+




≈
+


Section 9.3: The Complex Plane; De Moivre’s Theorem
903
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
33.
(
)
(
)
[
]
(
)
2 cos 40º
sin 40º 4 cos 20º
sin 20º
2 4 cos(40º 20º )
sin(40º 20º )
8 cos 60º
sin 60º
z w
i
i
i
i
=
+
⋅
+
= ⋅
+
+
+
=
+



(
)
(
)
[
]
(
)
2 cos 40º
sin 40º
4 cos 20º
sin 20º
2
cos(40º 20º )
sin(40º 20º )
4
1
cos 20º
sin 20º
2
i
z
w
i
i
i
+
=
+
=
−
+
−
=
+

34.
(
) (
)
cos120º
sin120º
cos100º
sin100º
cos(120º 100º )
sin(120º 100º )
cos 220º
sin 220º
z w
i
i
i
i
=
+
⋅
+
=
+
+
+
=
+



(
)
(
)
cos120º
sin120º
cos100º
sin100º
cos(120º 100º )
sin(120º 100º )
cos 20º
sin 20º
i
z
w
i
i
i
+
=
+
=
−
+
−
=
+

35.
(
)
(
)
[
]
(
)
[
]
(
)
3 cos130º
sin130º 4 cos270º
sin 270º
3 4 cos(130º 270º )
sin(130º 270º )
12 cos400º
sin 400º
12 cos(400º 360º)
sin(400º 360º )
12 cos40º
sin 40º
zw
i
i
i
i
i
i
=
+
⋅
+
= ⋅
+
+
+
=
+
=
−
+
−
=
+



(
)
(
)
[
]
(
)
[
]
(
)
3 cos130º
sin130º
4 cos270º
sin 270º
3
cos(130º 270º)
sin(130º 270º )
4
3
cos( 140º )
sin( 140º )
4
3
cos(360º 140º)
sin(360º 140º )
4
3
cos220º
sin 220º
4
i
z
w
i
i
i
i
i
+
=
+
=
−
+
−
=
−
+
−
=
−
+
−
=
+

36.
(
)
(
)
[
]
(
)
2 cos80º
sin 80º 6 cos 200º
sin 200º
2 6 cos(80º 200º )
sin(80º 200º )
12 cos 280º
sin 280º
z w
i
i
i
i
=
+
⋅
+
= ⋅
+
+
+
=
+



(
)
(
)
[
]
[
]
[
]
(
)
2 cos80º
sin80º
6 cos 200º
sin 200º
2
cos(80º 200º )
sin(80º 200º )
6
1
cos( 120º )
sin( 120º )
3
1
cos(360º 120º )
sin(360º 120º )
3
1
cos 240º
sin 240º
3
i
z
w
i
i
i
i
i
+
=
+
=
−
+
−
=
−
+
−
=
−
+
−
=
+

37.
2 cos
sin
2 cos
sin
8
8
10
10
2 2 cos
sin
8 10
8 10
9
9
4 cos
sin
40
40
zw
i
i
i
i
π
π
π
π




=
+
⋅
+









π
π
π
π 




= ⋅
+
+
+












π
π


=
+







2 cos
sin
8
8
2 cos
sin
10
10
2
cos
sin
2
8 10
8 10
cos
sin
40
40
i
z
w
i
i
i
π
π


+




=
π
π


+





π
π
π
π 




=
−
+
−












π
π
=
+

38.
3
3
9
9
4 cos
sin
2 cos
sin
8
8
16
16
3
9
3
9
4 2 cos
sin
8
16
8
16
15
15
8 cos
sin
16
16
z w
i
i
i
i
π
π
π
π




=
+
⋅
+









π
π
π
π 




= ⋅
+
+
+












π
π


=
+







3
3
4 cos
sin
8
8
9
9
2 cos
sin
16
16
4
3
9
3
9
cos
sin
2
8
16
8
16
3
3
2 cos
sin
16
16
3
3
2 cos 2
sin 2
16
16
29
29
2 cos
sin
16
16
i
z
w
i
i
i
i
i
π
π
π
π


+




=
π
π


+





π
π
π
π 




=
−
+
−













π
π 




=
−
+
−













π
π 




=
−
+
−












π
π


=
+





Chapter 9: Polar Coordinates; Vectors
904
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
39.
2
2
2 2
2
2
8 2 2
z
i
r
= +
=
+
=
=

2
tan
1
2
45º
θ
θ
= =
=

(
)
2 2 cos 45º
sin 45º
z
i
=
+



(
)
(
)
2
2
3
3
1
4 2
w
i
r
=
−
=
+ −
=
=



1
3
tan
3
3
330º
θ
θ
−
=
= −
=

(
)
2 cos330º
sin 330º
w
i
=
+



(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 2 cos45º
sin 45º 2 cos330º
sin330º
2 2 2 cos 45º 330º
sin 45º 330º
4 2 cos375º
sin375º
4 2 cos 375º 360º
sin 375º 360º
4 2 cos15º
sin15º
zw
i
i
i
i
i
i
=
+
⋅
+
=
⋅
+
+
+




=
+
=
−
+
−




=
+



(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 2 cos 45º
sin 45º
2 cos330º
sin 330º
2 2
cos 45º 330º
sin 45º 330º
2
2 cos
285º
sin
285º
2 cos 360º 285º
sin 360º 285º
2 cos 75º
sin 75º
i
z
w
i
i
i
i
i
+
=
+
=
−
+
−




=
−
+
−




=
−
+
−




=
+

40.
2
2
1
1
( 1)
2
z
i
r
= −
=
+ −
=

1
tan
1
1
315º
θ
θ
−
=
= −
=

(
)
2 cos315º
sin 315º
z
i
=
+



(
)2
2
1
3
1
3
4
2
w
i
r
= −
=
+ −
=
=

3
tan
3
1
300º
θ
θ
−
=
= −
=

(
)
2 cos300º
sin 300º
w
i
=
+



(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 cos315º
sin315º 2 cos300º
sin300º
2 2 cos 315º 300º
sin 315º 300º
2 2 cos615º
sin 615º
2 2 cos 615º 360º
sin 615º 360º
2 2 cos255º
sin 255º
zw
i
i
i
i
i
i
=
+
⋅
+
=
⋅
+
+
+




=
+
=
−
+
−




=
+



(
)
(
)
(
)
(
)
(
)
2 cos315º
sin315º
2 cos300º
sin300º
2
cos 315º 300º
sin 315º 300º
2
2
cos15º
sin15º
2
i
z
w
i
i
i
+
=
+
=
−
+
−




=
+

41.
(
)
(
)
(
)
(
)
3
3
4 cos 40º
sin 40º
4 cos 3 40º
sin 3 40º
64 cos120º
sin120º
1
3
64
2
2
32 32 3
i
i
i
i
i
+




=
⋅
+
⋅




=
+


=
− +




= − +

42.
(
)
[
]
(
)
3
3
3 cos80º
sin80º
3 cos(3 80º )
sin(3 80º )
27 cos 240º
sin 240º
1
3
27
2
2
27 27 3
2
2
i
i
i
i
i
+




=
⋅
+
⋅
=
+


=
− −




= −
−

43.
(
)
5
5
2 cos
sin
10
10
2 cos 5
sin 5
10
10
32 cos
sin
2
2
32 0 1
32
i
i
i
i
i

π
π 


+









π
π 




=
⋅
+
⋅












π
π


=
+




=
+
=


Section 9.3: The Complex Plane; De Moivre’s Theorem
905
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
44.
(
)
4
4
5
5
2 cos
sin
16
16
5
5
2
cos 4
sin 4
16
16
5
5
4 cos
sin
4
4
2
2
4
2
2
2 2 2 2
i
i
i
i
i

π
π 


+









π
π 




=
⋅
+
⋅












π
π


=
+






=
−
−




= −
−

45.
(
)
(
) [
]
(
)
6
6
3 cos10º
sin10º
3
cos(6 10º )
sin(6 10º )
27 cos 60º
sin 60º
1
3
27
2
2
27 27 3
2
2
i
i
i
i
i


+


=
⋅
+
⋅
=
+


=
+




=
+

46.
(
)
[
]
(
)
(
)
5
5
1
cos 72º
sin 72º
2
1
cos(5 72º )
sin(5 72º )
2
1
cos360º
sin 360º
32
1
1 0
32
1
32
i
i
i
i


⋅
+






=
⋅
+
⋅




=
⋅
+
=
⋅ +
=

47.
4
3
3
5 cos
sin
16
16
i

π
π 


+









(
)4
3
3
5
cos 4
sin 4
16
16
3
3
25 cos
sin
4
4
2
2
25
2
2
25 2
25 2
2
2
i
i
i
i

π
π 




=
⋅
+
⋅












π
π


=
+






=
−
+




= −
+

48.
(
)
6
6
5
5
3 cos
sin
18
18
5
5
3
cos 6
sin 6
18
18
i
i

π
π 


+









π
π 




=
⋅
+
⋅













5
5
27 cos
sin
3
3
1
3
27
2
2
27 27 3
2
2
i
i
i
π
π


=
+






=
−






=
−

49.
2
2
1
1
( 1)
2
i
r
−
=
+ −
=

1
tan
1
1
7
4
θ
θ
−
=
= −
π
=

7
7
1
2 cos
sin
4
4
i
i
π
π


− =
+







(
)
5
5
5
7
7
(1
)
2 cos
sin
4
4
7
7
2
cos 5
sin 5
4
4
35
35
4 2 cos
sin
4
4
2
2
4 2
2
2
4 4
i
i
i
i
i
i

π
π 


−
=
+









π
π 




=
⋅
+
⋅












π
π


=
+






=
−
+






= − +

50.
(
)2
2
3
3
( 1)
4
2
i
r
−
=
+ −
=
=

1
3
tan
3
3
330º
θ
θ
−
=
= −
=

(
)
3
2 cos330º
sin 330º
i
i
− =
+

(
)
(
)
(
)
(
)
(
)
(
)
6
6
6
3
2 cos330º
sin 330º
2
cos 6 330º
sin 6 330º
64 cos1980º
sin1980º
64 1 0
64
i
i
i
i
i
−
=
+




=
⋅
+
⋅




=
+
=
− +
= −

Chapter 9: Polar Coordinates; Vectors
906
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
51.
(
)2
2
2
2
( 1)
3
i
r
−
=
+ −
=

1
2
tan
2
2
324.736º
θ
θ
−
=
= −
≈

(
)
2
3 cos324.736º
sin 324.736º
i
i
− ≈
+



(
)
(
)
(
)
(
)
(
)
(
)
(
)
6
6
6
2
3 cos324.736º
sin 324.736º
3
cos 6 324.736º
sin 6 324.736º
27 cos1948.416
sin1948.416
27 0.8519 0.5237
23 14.142
i
i
i
i
i
i


−
≈
+


=
⋅
+
⋅




=
° +
°
≈
−
+
≈ − +

52.
(
)2
2
1
5
1
5
6
i
r
−
=
+ −
=

5
tan
5
1
294.095º
θ
θ
−
=
= −
≈

(
)
1
5
6 cos 294.095º
sin 294.095º
i
i
−
≈
+

(
)
(
)
(
)
(
)
[
]
(
)
8
8
1
5
6
cos 8 294.095º
sin 8 294.095º
1296 cos 2352.76º
sin 2352.76º
1296 0.9753 0.2208
1264 286.217
i
i
i
i
i
−
≈
⋅
+
⋅




=
+
≈
−
−
≈ −
−

53.
2
2
1
1
1
2
i
r
+
=
+
=

1
tan
1
1
45º
θ
θ
= =
=

(
)
1
2 cos 45º
sin 45º
i
i
+ =
+

The three complex cube roots of
(
)
1
2 cos 45º
sin 45º
i
i
+ =
+
are:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
3
6
6
0
6
6
1
6
2
45º
360º
45º
360º
2 cos
sin
3
3
3
3
2 cos 15º 120º
sin 15º 120º
2 cos 15º 120º 0
sin 15º 120º 0
2 cos15º
sin15º
2 cos 15º 120º 1
sin 15º 120º 1
2 cos135º
sin135º
2
k
k
k
z
i
k
i
k
z
i
i
z
i
i
z






=
+
+
+












=
+
+
+




=
+
⋅
+
+
⋅




=
+
=
+
⋅ +
+
⋅




=
+
=
(
)
(
)
(
)
6
6
cos 15º 120º 2
sin 15º 120º 2
2 cos 255º
sin 255º
i
i
+
⋅
+
+
⋅




=
+

54.
(
)
(
)
2
2
3
3
1
4
2
i
r
−
=
+ −
=
=

1
3
tan
3
3
330º
θ
θ
−
=
= −
=

(
)
3
2 cos330º
sin 330º
i
i
− =
+


The four complex fourth roots of
(
)
3
2 cos330º
sin 330º
i
i
− =
+
are:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
4
4
4
0
4
4
1
4
330º
360º
330º
360º
2 cos
sin
4
4
4
4
2 cos 82.5º 90º
sin 82.5º 90º
2 cos 82.5º 90º 0
sin 82.5º 90º 0
2 cos 82.5º
sin 82.5º
2 cos 82.5º 90º 1
sin 82.5º 90º 1
2 cos 1
k
k
k
z
i
k
i
k
z
i
i
z
i






=
+
+
+












=
+
+
+




=
+
⋅
+
+
⋅




=
+




=
+
⋅ +
+
⋅




=
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
4
2
4
4
3
4
72.5º
sin 172.5º
2 cos 82.5º 90º 2
sin 82.5º 90º 2
2 cos 262.5º
sin 262.5º
2 cos 82.5º 90º 3
sin 82.5º 90º 3
2 cos 352.5º
sin 352.5º
i
z
i
i
z
i
i
+




=
+
⋅
+
+
⋅




=
+




=
+
⋅
+
+
⋅




=
+





55.
(
)2
2
4 4 3
4
4 3
64 8
i
r
−
=
+ −
=
=

4 3
tan
3
4
300º
θ
θ
−
=
= −
=

(
)
4 4 3
8 cos300º
sin 300º
i
i
−
=
+

The four complex fourth roots of

Section 9.3: The Complex Plane; De Moivre’s Theorem
907
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
(
)
4 4 3
8 cos300º
sin 300º
i
i
−
=
+
are:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
4
4
4
0
4
4
1
4
4
2
300º
360º
300º
360º
8 cos
sin
4
4
4
4
8 cos 75º 90º
sin 75º 90º
8 cos 75º 90º 0
sin 75º 90º 0
8 cos75º
sin 75º
8 cos 75º 90º 1
sin 75º 90º 1
8 cos165º
sin165º
8 cos
k
k
k
z
i
k
i
k
z
i
i
z
i
i
z






=
+
+
+












=
+
+
+




=
+
⋅
+
+
⋅




=
+
=
+
⋅ +
+
⋅




=
+
=
(
)
(
)
(
)
(
)
(
)
(
)
4
4
3
4
75º 90º 2
sin 75º 90º 2
8 cos 255º
sin 255º
8 cos 75º 90º 3
sin 75º 90º 3
8 cos345º
sin 345º
i
i
z
i
i
+
⋅
+
+
⋅




=
+
=
+
⋅
+
+
⋅




=
+

56.
2
2
8 8
( 8)
( 8)
8 2
i
r
− −
=
−
+ −
=

8
tan
1
8
225º
θ
θ
−
=
=
−
=

(
)
8 8
8 2 cos 225º
sin 225º
i
i
− − =
+

The three complex cube roots of
(
)
8 8
8 2 cos 225º
sin 225º
i
i
− − =
+
are:
(
)
(
)
3
6
225º
360º
225º
360º
3
3
3
3
8 2 cos
sin
2 2 cos 75º 120º
sin 75º 120º
k
k
k
z
i
k
i
k






=
+
+
+












=
+
+
+





(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
6
0
6
6
1
6
6
2
6
2 2 cos 75º 120º 0
sin 75º 120º 0
2 2 cos75º
sin 75º
2 2 cos 75º 120º 1
sin 75º 120º 1
2 2 cos195º
sin195º
2 2 cos 75º 120º 2
sin 75º 120º 2
2 2 cos315º
sin 315º
z
i
i
z
i
i
z
i
i
=
+
⋅
+
+
⋅




=
+
=
+
⋅ +
+
⋅




=
+
=
+
⋅
+
+
⋅




=
+

57.
(
)2
2
16
0
16
256 16
i
r
−
=
+ −
=
=

16
tan
0
270º
θ
θ
−
=
=

(
)
16
16 cos 270º
sin 270º
i
i
−
=
+

The four complex fourth roots of
(
)
16
16 cos 270º
sin 270º
i
i
−
=
+
are:
(
)
(
)
4
270º
360º
270º
360º
4
4
4
4
16 cos
sin
2 cos 67.5º 90º
sin 67.5º 90º
k
k
k
z
i
k
i
k






=
+
+
+












=
+
+
+





(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
1
2
3
2 cos 67.5º 90º 0
sin 67.5º 90º 0
2 cos 67.5º
sin 67.5º
2 cos 67.5º 90º 1
sin 67.5º 90º 1
2 cos157.5º
sin157.5º
2 cos 67.5º 90º 2
sin 67.5º 90º 2
2 cos 247.5º
sin 247.5º
2 cos 67.5º 90º 3
si
z
i
i
z
i
i
z
i
i
z
i
=
+
⋅
+
+
⋅




=
+
=
+
⋅ +
+
⋅




=
+
=
+
⋅
+
+
⋅




=
+
=
+
⋅
+
(
)
(
)
n 67.5º 90º 3
2 cos337.5º
sin 337.5º
i
+
⋅




=
+

58.
2
2
8
( 8)
0
8
r
−
=
−
+
=

0
tan
0
8
180º
θ
θ
=
=
−
=

(
)
8 8 cos180º
sin180º
i
− =
+

The three complex cube roots of
(
)
8 8 cos180º
sin180º
i
− =
+
are:
(
)
(
)
3
180º
360º
180º
360º
8 cos
sin
3
3
3
3
2 cos 60º 120º
sin 60º 120º
k
k
k
z
i
k
i
k






=
+
+
+












=
+
+
+





(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
1
2
2 cos 60º 120º 0
sin 60º 120º 0
2 cos 60º
sin 60º
2 cos 60º 120º 1
sin 60º 120º 1
2 cos180º
sin180º
2 cos 60º 120º 2
sin 60º 120º 2
2 cos300º
sin 300º
z
i
i
z
i
i
z
i
i
=
+
⋅
+
+
⋅




=
+
=
+
⋅ +
+
⋅




=
+
=
+
⋅
+
+
⋅




=
+

59.
2
2
0
1
1 1
i
r =
+
=
=

1
tan
0
90º
θ
θ
=
=



1(cos90º
sin 90º )
i
i
=
+

The five complex fifth roots of
(
)
1 cos90º
sin 90º
i
i
=
+
are:
Chapter 9: Polar Coordinates; Vectors
908
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
[
]
5
90º
360º
90º
360º
1 cos
sin
5
5
5
5
1 cos(18º 72º )
sin(18º 72º )
k
k
k
z
i
k
i
k






=
+
+
+












=
+
+
+

(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
1
2
3
1 cos 18º 72º 0
sin 18º 72º 0
cos18º
sin18º
1 cos 18º 72º 1
sin 18º 72º 1
cos90º
sin 90º
1 cos 18º 72º 2
sin 18º 72º 2
cos162º
sin162º
1 cos 18º 72º 3
sin 18º 72º 3
cos 234º
sin 234º
z
i
i
z
i
i
z
i
i
z
i
i
=
+
⋅
+
+
⋅




=
+
=
+
⋅ +
+
⋅




=
+
=
+
⋅
+
+
⋅




=
+
=
+
⋅
+
+
⋅




=
+
(
)
(
)
4
1 cos 18º 72º 4
sin 18º 72º 4
cos306º
sin 306º
z
i
i
=
+
⋅
+
+
⋅




=
+

60.
2
2
0
( 1)
1 1
i
r
−
=
+ −
=
=

1
tan


270º
0
θ
θ
−
= 
=

1(cos 270º
sin 270º )
i
i
− =
+

The five complex fifth roots of
(
)
1 cos 270º
sin 270º
i
i
− =
+
are:
[
]
5
270º
360º
270º
360º
1 cos
sin
5
5
5
5
1 cos(54º 72º )
sin(54º 72º )
k
k
k
z
i
k
i
k






=
+
+
+












=
+
+
+

[
]
[
]
0
1
1 cos(54º 72º 0)
sin(54º 72º 0)
cos54º
sin 54º
1 cos(54º 72º 1)
sin(54º 72º 1)
cos126º
sin126º
z
i
i
z
i
i
=
+
⋅
+
+
⋅
=
+
=
+
⋅ +
+
⋅
=
+



[
]
[
]
2
3
1 cos(54º 72º 2)
sin(54º 72º 2)
cos198º
sin198º
1 cos(54º 72º 3)
sin(54º 72º 3)
cos 270º
sin 270º
z
i
i
z
i
i
=
+
⋅
+
+
⋅
=
+
=
+
⋅
+
+
⋅
=
+



[
]
4
1 cos(54º 72º 4)
sin(54º 72º 4)
cos342º
sin 342º
z
i
i
=
+
⋅
+
+
⋅
=
+

61.
2
2
1 1 0
1
0
1 1
i
r
= +
=
+
=
=

0
tan
0
1
0º
θ
θ
= =
=

(
)
1 0
1 cos 0º
sin 0º
i
i
+
=
+

The four complex fourth roots of unity are:
(
)
(
)
4
0º
360º
0º
360º
1 cos
sin
4
4
4
4
1 cos 90º
sin 90º
k
k
k
z
i
k
i
k






=
+
+
+












=
+





(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
1
2
3
cos 90º 0
sin 90º 0
cos 0º
sin 0º 1 0
1
cos 90º 1
sin 90º 1
cos90º
sin 90º 0 1
cos 90º 2
sin 90º 2
cos180º
sin180º
1 0
1
cos 90º 3
sin 90º 3
cos 270º
sin 270º 0 1
z
i
i
i
z
i
i
i
i
z
i
i
i
z
i
i
i
i
=
⋅
+
⋅
=
+
= +
=
=
⋅ +
⋅
=
+
= + =
=
⋅
+
⋅
=
+
= − +
= −
=
⋅
+
⋅
=
+
= − = −

The complex fourth roots of unity are:
1, , 1,
i
i
− −
.



62.
2
2
1 1 0
1
0
1 1
i
r
= +
=
+
=
=

0
tan
0
1
0º
θ
θ
= =
=

(
)
1 0
1 cos 0º
sin 0º
i
i
+
=
+

The six complex sixth roots of unity are:
(
)
(
)
6
0º
360º
0º
360º
1 cos
sin
6
6
6
6
1 cos 60º
sin 60º
k
k
k
z
i
k
i
k






=
+
+
+












=
+






Section 9.3: The Complex Plane; De Moivre’s Theorem
909
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
1
2
3
4
cos 60º 0
sin 60º 0
cos 0º
sin 0º 1 0
1
cos 60º 1
sin 60º 1
1
3
cos 60º
sin 60º
2
2
cos 60º 2
sin 60º 2
1
3
cos120º
sin120º
2
2
cos 60º 3
sin 60º 3
cos180º
sin180º
1 0
1
cos 60º 4
sin 60º 4
c
z
i
i
i
z
i
i
i
z
i
i
i
z
i
i
i
z
i
=
⋅
+
⋅
=
+
= +
=
=
⋅ +
⋅
=
+
= +
=
⋅
+
⋅
=
+
= − +
=
⋅
+
⋅
=
+
= − +
= −
=
⋅
+
⋅
=
(
)
(
)
5
1
3
os 240º
sin 240º
2
2
cos 60º 5
sin 60º 5
1
3
cos300º
sin 300º
2
2
i
i
z
i
i
i
+
= − −
=
⋅
+
⋅
=
+
= −

The complex sixth roots of unity are:
1
3
1
3
1
3 1
3
1,
,
,
1,
,
2
2
2
2
2
2
2
2
i
i
i
i
+
− +
− − −
−
.




63. Let
(
)
cos
sin
w r
i
θ
θ
=
+
be a complex number.
If
0
w ≠
, there are n distinct nth roots of w,
given by the formula:
2
2
cos
sin
,
where
0, 1, 2, ... ,
1
for all
n
k
n
k
k
k
z
r
i
n
n
n
n
k
n
z
r
k
θ
θ

π
π 




=
+
+
+












=
−
=

64. Since
n
k
z
r
=
for all k, each of the complex
nth roots lies on a circle with center at the origin
and radius
n
n w
r
=
, where w is the original
complex number.
65. Examining the formula for the distinct complex nth roots of the complex number
(
)
cos
sin
w r
i
θ
θ
=
+
,
2
2
cos
sin
,
n
k
k
k
z
r
i
n
n
n
n
θ
θ

π
π 




=
+
+
+












where
0, 1, 2, ... ,
1
k
n
=
−
, we see that the k
z
are spaced apart by an
angle of
2
n
π
.
66. Let
(
)
1
1
1
1
cos
sin
z
r
i
θ
θ
=
+
and
(
)
2
2
2
2
cos
sin
z
r
i
θ
θ
=
+
. Then
(
)
(
)
(
)
(
)
(
)
(
)
1
1
1
1
2
2
2
2
1
1
1
2
2
2
2
2
2
2
1
1
2
1
2
1
2
1
2
2
2
2
2
2
1
2
1
2
1
2
1
2
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
cos
cos
cos
sin
sin
cos
sin
sin
cos
sin
cos
cos
sin
sin
sin
cos
co
r
i
z
z
r
i
r
i
i
r
i
i
r
i
i
r
i
r
r
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
+
=
+
+
−
=
⋅
+
−
⋅
−
⋅
+
⋅
+
⋅
=
⋅
+
⋅
+
⋅
+
⋅
−
=
⋅
(
)
(
)
(
)
1
2
1
1
2
1
2
2
s
sin
1
cos
sin
r
i
r
θ
θ
θ
θ
θ
θ
⋅
=
−
+
−





Chapter 9: Polar Coordinates; Vectors
910
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
67. a.
0
1
2
3
4
5
6
0.1 0.4
0.05 0.48
0.13 0.35
0.01 0.31
0.01 0.35
0.2 0.41
0.07 0.42
0.5 0.8
0.11 1.6
2.05 1.15
3.37 3.92
3.52 25.6
641.9 181.1
379290 232509
0.9 0.7
0.58 0.56
0.87 1.35
1.95 1.
z
a
a
a
a
a
a
a
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
=
−
−
−
−
−
−
−
−
−
−
+
+
−
+
−
−
−
−
+
−
−
+
−
−
+
−
− 67
0.13 7.21
52.88 2.56
2788.5 269.6
1.1 0.1
0.1 0.12
1.10 0.76 0.11 0.068
1.09 0.085
0.085 0.85
1.10 0.086
0 1.3
1.69 1.3
1.17 3.09
8.21 5.92 32.46 98.47
8643.6 6393.7
33833744 110529134.4
1
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
−
+
−
−
+
−
−
+
−
−
+
−
−
+
−
−
−
+
−
+
−
−
−
+
+
15
14
1
1 3
7 7
1 97
9407 193
88454401 3631103
7.8 10
6.4 10
i
i
i
i
i
i
i
+
− +
−
−
−
+
×
+
×


b.
1z and 4
z
are in the Mandlebrot set.
6a for the complex numbers not in the set have very large components.

c.
6
15
0.1 0.4
0.4
0.4
0.5 0.8
0.9
444884
0.9 0.7 1.1
2802
1.1 0.1 1.1
1.1
0 1.3
1.3 115591573
1 1
1.4
7.8 10
z
z
a
i
i
i
i
i
i
−
+
−
+
−
+
−
+
×

The numbers which are not in the Mandlebrot set satisfy this condition. The numbers which are in the
Mandlebrot set satisfy the condition
2
n
a ≤
.


Section 9.4
1. vector
2. 0
3. unit
4. position
5. horizontal, vertical
6. resultant
7. True
8. False
9.
+
v w

10.
+
u v



11. 3v



12. 4w


Section 9.4: Vectors
911
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
13.
−
v w



14.
−
u v



15. 3
2
+ −
v u
w

16. 2
3
−
+
u
v w



17. True
18. False
+ = −
K G
F
19. False
= − + −
C
F E D
20. True
21. False
− = +
D E H G
22. False
+
= − −
C H G F
23. True
24. True
25. If
4
=
v
, then 3
3
3 4 12
=
⋅
= ⋅ =
v
v
.
26. If
2
=
v
, then
4
4
4 2 8
−
= −
⋅
=
⋅ =
v
v

27.
(0, 0),
(3, 4)
(3 0)
(4 0)
3
4
P
Q
=
=
= −
+ −
= +
v
i
j
i
j

28.
(0, 0),
( 3, 5)
( 3 0)
( 5 0)
3
5
P
Q
=
= − −
= − −
+ − −
= − −
v
i
j
i
j

29.
(3, 2),
(5, 6)
(5 3)
(6 2)
2
4
P
Q
=
=
= −
+ −
= +
v
i
j
i
j

30.
[
]
( 3, 2),
(6, 5)
6 ( 3)
(5 2)
9
3
P
Q
= −
=
= − −
+ −
= +
v
i
j
i
j

31.
[
]
[
]
( 2, 1),
(6, 2)
6 ( 2)
2 ( 1)
8
P
Q
= − −
=
−
= − −
+ − − −
= −
v
i
j
i
j

32.
[
]
( 1, 4),
(6, 2)
6 ( 1)
(2 4)
7
2
P
Q
= −
=
= − −
+ −
= −
v
i
j
i
j

33.
(1, 0),
(0, 1)
(0 1)
(1 0)
P
Q
=
=
=
−
+ −
= − +
v
i
j
i
j

34.
(1, 1),
(2, 2)
(2 1)
(2 1)
P
Q
=
=
=
−
+ −
= +
v
i
j
i
j

35. For
3
4
= −
v
i
j ,
2
2
3
( 4)
25 5
=
+ −
=
=
v
.
36. For
5
12
= − +
v
i
j ,
2
2
( 5)
12
169 13
=
−
+
=
=
v
.
37. For = −
v
i
j ,
2
2
1
( 1)
2
=
+ −
=
v
.
38. For = − −
v
i
j ,
2
2
( 1)
( 1)
2
=
−
+ −
=
v
.
39. For
2
3
= − +
v
i
j ,
2
2
( 2)
3
13
=
−
+
=
v
.
40. For
6
2
=
+
v
i
j ,
2
2
6
2
40
2 10
=
+
=
=
v
.
41.
(
)
(
)
2
3
2 3
5
3
2
3
6 10
6
9
+
=
−
+ − +
= −
− +
= −
v
w
i
j
i
j
i
j
i
j
j

Chapter 9: Polar Coordinates; Vectors
912
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
42.
(
)
(
)
3
2
3 3
5
2
2
3
9 15
4
6
13
21
−
=
−
− − +
= −
+ −
=
−
v
w
i
j
i
j
i
j
i
j
i
j

43.
(
)
(
)
2
2
3
5
2
3
5
8
5
( 8)
89
−
=
−
− − +
=
−
=
+ −
=
v w
i
j
i
j
i
j

44.
(
)
(
)
2
2
3
5
2
3
2
1
( 2)
5
+
=
−
+ − +
= −
=
+ −
=
v w
i
j
i
j
i
j

45.
2
2
2
2
3
5
2
3
3
( 5)
( 2)
3
34
13
−
=
−
− − +
=
+ −
− −
+
=
−
v
w
i
j
i
j

46.
2
2
2
2
3
5
2
3
3
( 5)
( 2)
3
34
13
+
=
−
+ − +
=
+ −
+ −
+
=
+
v
w
i
j
i
j

47.
5
5
5
5
5
25 0
=
=
=
=
=
+
v
i
i
i
u
i
v
i

48.
3
3
3
3
3
0 9
−
−
−
=
=
=
=
= −
−
+
v
j
j
j
u
j
v
j

49.
3
4
3
4
−
=
=
−
v
i
j
u
v
i
j
2
2
3
4
3
( 4)
3
4
25
3
4
5
3
4
5
5
−
=
+ −
−
=
−
=
=
−
i
j
i
j
i
j
i
j

50.
5
12
5 12
− +
=
=
− +
v
i
j
u
v
i
j
2
2
5
12
( 5)
12
5 12
169
5 12
13
5
12
13
13
− +
=
−
+
− +
=
− +
=
= −
+
i
j
i
j
i
j
i
j

51.
2
2
1
( 1)
−
−
=
=
=
−
+ −
v
i
j
i
j
u
v
i
j
2
1
1
2
2
2
2
2
2
−
=
=
−
=
−
i
j
i
j
i
j

52.
2
2
2
2
2
2
( 1)
−
−
=
=
=
−
+ −
v
i
j
i
j
u
v
i
j
2
5
2
1
5
5
2 5
5
5
5
−
=
=
−
=
−
i
j
i
j
i
j

53. Let
a
b
=
+
v
i
j . We want
4
=
v
and
2
a
b
=
.
(
)2
2
2
2
2
2
5
a
b
b
b
b
=
+
=
+
=
v

2
2
2
5
4
5
16
16
5
16
4
4 5
5
5
5
b
b
b
b
=
=
=
= ±
= ±
= ±

4 5
8 5
2
2
5
5
a
b


=
=
±
= ±







8 5
4 5
8 5
4 5
or
5
5
5
5
=
+
= −
−
v
i
j
v
i
j
54. Let
a
b
=
+
v
i
j . We want
3
=
v
and a b
=
.
2
2
2
2
2
2
a
b
b
b
b
=
+
=
+
=
v

2
2
2
2
3
2
9
9
2
9
3
3 2
2
2
2
b
b
b
b
=
=
=
= ±
= ±
= ±

3 2
2
a b
= = ±

3 2
3 2
3 2
3 2
or
2
2
2
2
=
+
= −
−
v
i
j
v
i
j

Section 9.4: Vectors
913
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
55.
2
,
3 ,
5
x
=
−
= +
+
=
v
i
j w
i
j
v w

2
2
2
2
2
3
(2
)
2
(2
)
2
4
4 4
4
8
x
x
x
x
x
x
x
+
=
− + +
=
+
+
=
+
+
=
+
+ +
=
+
+
v w
i
j
i
j
i
j


Solve for x:
2
2
2
4
8 5
4
8 25
4
17 0
x
x
x
x
x
x
+
+ =
+
+ =
+
−
=

4
16 4(1)( 17)
2(1)
4
84
2
4 2 21
2
2
21
x
− ±
−
−
=
− ±
=
− ±
=
= − ±

The solution set is {
}
2
21, 2
21
− +
− −
.
56.
[
]
( 3,1),
( , 4),
( 3)
(4 1)
(
3)
3
P
Q
x
x
x
= −
=
=
− −
+ −
=
+
+
v
i
j
i
j

2
2
2
2
(
3)
3
6
9 9
6
18
x
x
x
x
x
=
+
+
=
+
+ +
=
+
+
v

Solve for x:
2
2
2
6
18 5
6
18 25
6
7 0
(
7)(
1)
0
x
x
x
x
x
x
x
x
+
+
=
+
+
=
+
− =
+
− =

7 or
1
x
x
= −
=

The solution set is { 7, 1}.
−

57.
5,
60º
α
=
=
v

(
)
(
)
(
)
(
)
cos
sin
5 cos 60º
sin 60º
1
3
5
2
2
5
5 3
2
2
α
α
=
+
=
+


=
+






=
+
v
v
i
j
i
j
i
j
i
j

58.
8,
45º
α
=
=
v

(
)
(
)
(
)
cos
sin
8 cos 45º
sin 45º
2
2
8
2
2
4 2
4 2
α
α
=
+
=
+






=
+






=
+
v
v
i
j
i
j
i
j
i
j

59.
14,
120º
α
=
=
v

(
)
(
)
(
)
(
)
cos
sin
14 cos 120º
sin 120º
1
3
14
2
2
7
7 3
α
α
=
+
=
+


=
−
+






= − +
v
v
i
j
i
j
i
j
i
j

60.
3,
240º
α
=
=
v

(
)
(
)
(
)
cos
sin
3 cos 240º
sin 240º
1
3
3
2
2
3
3 3
2
2
α
α
=
+
=
+






=
−
−






= −
−
v
v
i
j
i
j
i
j
i
j

61.
25,
330º
α
=
=
v

(
)
(
)
(
)
cos
sin
25 cos 330º
sin 330º
3
1
25
2
2
25 3
25
2
2
α
α
=
+
=
+






=
−






=
−
v
v
i
j
i
j
i
j
i
j

62.
15,
315º
α
=
=
v

(
)
(
)
cos
sin
15 cos315º
sin 315º
2
2
15 2
15 2
15
2
2
2
2
α
α
=
+
=
+


=
−
=
−






v
v
i
j
i
j
i
j
i
j

63.
(
)
(
)
3
3
3
cos
sin
α
α
= + =
+
=
+
v
i
j
i
j
v
i
j



tan
1
α =

( )
1
tan
1
45
α
−
=
=
°
Chapter 9: Polar Coordinates; Vectors
914
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

The angle is in quadrant I, thus,
45
α =
° .
64.
(
)
3
cos
sin
α
α
= +
=
+
v
i
j
v
i
j



tan
3
α =

(
)
1
tan
3
60
α
−
=
=
°

The angle is in quadrant I, thus,
60
α =
° .
65.
(
)
(
)
3 3
3
3
3
cos
sin
α
α
= −
+ = −
+
=
+
v
i
j
i
j
v
i
j



1
tan
3
α = −

1
1
tan
30
3
α
− 

=
−
= − °






The angle is in quadrant II, thus,
150
α =
° .
66.
(
)
(
)
5
5
5
cos
sin
α
α
= − − = − −
=
+
v
i
j
i
j
v
i
j



tan
1
α =

( )
1
tan
1
45
α
−
=
=
°

The angle is in quadrant III, thus,
225
α =
° .
67.
(
)
(
)
4
2
2 2
cos
sin
α
α
= −
=
−
=
+
v
i
j
i
j
v
i
j



1
tan
2
α = −

1
1
tan
26.6
2
α
− 

=
−
= −
°






The angle is in quadrant IV, thus,
333.4
α =
° .
68.
(
)
(
)
6
4
2 3
2
cos
sin
α
α
= −
=
−
=
+
v
i
j
i
j
v
i
j



2
tan
3
α = −

1
2
tan
33.7
3
α
− 

=
−
= −
°






The angle is in quadrant IV, thus,
326.3
α =
° .
69.
(
)
5
cos
sin
α
α
= − −
=
+
v
i
j
v
i
j



tan
5
α =

( )
1
tan
5
78.7
α
−
=
=
°

The angle is in quadrant III, thus,
258.7
α =
° .
70.
(
)
3
cos
sin
α
α
= − +
=
+
v
i
j
v
i
j



tan
3
α = −

(
)
1
tan
3
71.6
α
−
=
− = −
°

The angle is in quadrant II, thus,
108.4
α =
° .
71. a. Let
3, 1
=
−
u
. Then
3, 1
4,5
1, 4
= + =
− + −
= −
u' u v
.
The new coordinate will be ( 1, 4).
−


b.
(−4, 5)

72. a. Let
3,0
= −
a
,
1, 2
= − −
b
,
3,1
=
c
, and
1,3
=
d
. Then
3,0
3, 2
0, 2
= + = −
+
− =
−
a' a v
,
1, 2
3, 2
2, 4
= + = − − +
− =
−
b' b v
,
3, 1
3, 2
6, 1
= + =
+
− =
−
c'
c v
, and
1,3
3, 2
4,1
= + =
+
− =
d' d v
.
The vertices of the new parallelogram
'
'
'
'
A B C D are (0, 2),
−
(2, 4),
−
(6, 1),
−

and (4, 1).

b.
1
1
3
3, 2
,1
2
2
2
−
= −
− = −
v
. Then
1
3
9
3,0
,1
,1
2
2
2
= −
= −
+ −
= −
a' a
v
,
1
3
5
1, 2
,1
, 1
2
2
2
= −
= − − + −
= −
−
b' b
v
,
1
3
3
3,1
,1
,2
2
2
2
= −
=
+ −
=
c'
c
v
, and
1
3
1
1,3
,1
,4
2
2
2
= −
=
+ −
= −
d' d
v
.
The vertices of the new parallelogram
'
'
'
'
A B C D are
9
,1 ,
2


−





5
, 1 ,
2


−
−




3
, 2 ,
2







and
1
, 4 .
2


−





73.
(
)
(
)
40 cos 30º
sin 30º
3
1
40
2
2
20 3
20
=
+






=
+






=
+
F
i
j
i
j
i
j

74.
(
)
(
)
100 cos 20º
sin 20º
=
+




F
i
j


Section 9.4: Vectors
915
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
75.
[
]
1
40 cos(30º )
sin(30º )
3
1
40
20 3
20
2
2
=
+


=
+
=
+




F
i
j
i
j
i
j



[
]
2
60 cos( 45º )
sin( 45º )
2
2
60
30 2
30 2
2
2
=
−
+
−


=
−
=
−






F
i
j
i
j
i
j



(
)
(
)
1
20 3
20
30 2
30 2

20 3 30 2
20 30 2
=
+
=
+
+
−
=
+
+
−
2
F F F
i
j
i
j
i
j

76.
(
)
(
)
1
30 cos 45º
sin 45º
2
2
30
15 2
15 2
2
2
=
+






=
+
=
+






F
i
j
i
j
i
j



(
)
(
)
2
70 cos 120º
sin 120º
1
3
70
35
35 3
2
2
=
+






=
−
+
= −
+






F
i
j
i
j
i
j



(
)
(
)
1
15 2
15 2
( 35)
35 3

15 2 35
15 2 35 3
=
+
=
+
+ −
+
=
−
+
+
2
F F F
i
j
i
j
i
j

77. Let a
v
= the velocity of the plane in still air,
w
v
= the velocity of the wind, and g
v
= the
velocity of the plane relative to the ground.

a.
550
100(cos 45
sin 45 )
2
2
100
2
2
50 2
50 2
a
w
=
=
° +
°


=
+






=
+
v
j
v
i
j
i
j
i
j


b.

(
)
550
50 2
50 2
50 2
550 50 2
g
a
w
=
+
=
+
+
=
+
+
v
v
v
j
i
j
i
j


c. The speed of the plane relative to the ground
is:



(
)
(
)
2
2
50 2
550 50 2
390281.7459 624.7
g =
+
+
=
≈
v



To find the direction, find the angle between
g
v
and the x-axis.



550 50 2
tan
50 2
83.5
α
α
+
=
=
°



The plane is traveling with a ground speed of
624.7 mph in an approximate direction of
6.5° degrees east of north (
6.5
N
E
°
).
78. Let a
v
= the velocity of the plane in still air,
w
v
= the velocity of the wind, and g
v
= the
velocity of the plane relative to the ground.

a.
500
100(cos315
sin 315 )
2
2
100
2
2
50 2
50 2
a
w
= −
=
° +
°


=
−






=
−
v
i
v
i
j
i
j
i
j


b.
(
)
500
50 2
50 2
50 2 500
50 2
g
a
w
=
+
= −
+
−
=
−
−
v
v
v
i
i
j
i
j


c. The speed of the plane relative to the ground
is:



(
)
(
)
2
2
50 2 500
50 2
189289.3219
435.1
g =
−
+ −
=
≈
v



To find the direction, find the angle between
g
v
and the x-axis and consider a convenient
vector such as due south.



50 2
tan
50 2 500
9.4
α
α
−
=
−
=
°



The plane is traveling with a ground speed of
435.1 mph in an approximate direction of
80.6° degrees west of south (
80.6
S
W
°
).
79. Let a
v
= the velocity of the plane in still air, w
v

= the velocity of the wind, and g
v
= the velocity
of the plane relative to the ground.
=
+
g
a
w
v
v
v

Chapter 9: Polar Coordinates; Vectors
916
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall


(
)
(
)
(
)
(
)
500 cos 45º
sin 45º
2
2
500
2
2
250 2
250 2
60 cos 120º
sin 120º
1
3
60
2
2
30
30 3
=
+






=
+






=
+
=
+






=
−
+






= −
+
a
w
v
i
j
i
j
i
j
v
i
j
i
j
i
j



(
)
(
)
250 2
250 2
30
30 3
250 2 30
250 2 30 3
=
+
=
+
−
+
=
−
+
+
g
a
w
v
v
v
i
j
i
j
i
j


The speed of the plane relative to the ground is:


(
)
(
)
2
2
30 250 2
250 2 30 3
269,129.1 518.8 km/hr
=
− +
+
+
≈
≈
g
v


To find the direction, find the angle between g
v

and a convenient vector such as due east, i .
(
)
(
)

tan

250 2 30 3
250 2 30
1.2533
51.5º
component
component
θ
θ
=
+
=
−
≈
≈
j
i

The plane is traveling with a ground speed of
about 518.8 km/hr in a direction of 38.6º east of
north (
)
N38.6 E
°
.



80. Let a
v
= the velocity of the plane in still air, w
v

= the velocity of the wind, and g
v
= the velocity
of the plane relative to the ground.
=
+
g
a
w
v
v
v

(
)
(
)
600 cos 60º
sin 60º
1
3
600
300
300 3
2
2
=
−






=
−
=
−






a
v
i
j
i
j
i
j

(
)
(
)
40 cos 45º
sin 45º
2
2
40
2
2
20 2
20 2
=
−






=
−






=
−
w
v
i
j
i
j
i
j

(
)
(
)
300
300 3
20 2
20 2
300 20 2
300 3 20 2
=
+
=
−
+
−
=
+
−
+
g
a
w
v
v
v
i
j
i
j
i
j

The speed of the plane relative to the ground is:
(
)
(
)
2
2
300 20 2
300 3 20 2
407,964 638.7 km/hr
=
+
+ −
−
≈
≈
g
v


To find the direction, find the angle between g
v

and a convenient vector such as due east, i .
(
)
(
)

tan

300 3 20 2
300 20 2
1.6689
59.1º
component
component
θ
θ
=
−
−
=
+
≈ −
≈ −
j
i

The plane is traveling with a ground speed of
about 638.7 km/hr in a direction of about 30.9°
degrees east of south ( S30.9 E
°
).

81. Let 1F be the force of gravity and 2F be the
force required to hold the weight on the ramp.
Then
2
1
1
1
sin10
700
sin10
700
sin10
4031
° =
° =
=
°
=
F
F
F
F


So the combined weight of the boat and its trailer
is 4031 lbs.
82. Let 1F be the force of gravity and 2F be the
force required to hold the weight on the ramp.
Then
2
1
1
1
sin15
1200
sin15
1200
sin15
4636
° =
° =
=
°
=
F
F
F
F


So the weight of the car is 4636 lbs.

Section 9.4: Vectors
917
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
83. Let the positive x-axis point downstream, so that
the velocity of the current is
3
=
c
v
i . Let w
v
=
the velocity of the boat in the water, and g
v
=
the velocity of the boat relative to the land. Then
=
+
g
w
c
v
v
v and
=
g
v
k since the boat is going
directly across the river. The speed of the boat is
20
=
w
v
; we need to find the direction. Let
a
b
=
+
w
v
i
j , so
2
2
2
2
20
400
a
b
a
b
=
+
=
+
=
w
v


Since
=
+
g
w
c
v
v
v ,
3
(
3)
k
a
b
a
b
=
+
+ =
+
+
j
i
j
i
i
j
3 0
3
a
a
+ =
= −

2
2
2
2
400
9
400
391
391 19.8
k
b
a
b
b
b
k b
=
+
=
+
=
=
= =
≈

3
391
= − +
w
v
i
j and
391
=
g
v
j

Find the angle between w
v
and j :
( )
2
2
cos
3 0
391(1)
391
0.9887
20
20 0
1
8.6º
θ
θ
⋅
=
−
+
=
=
=
+
≈
w
w
v
j
v
j

The heading of the boat needs to be about 8.6º
upstream. The velocity of the boat directly
across the river is about 19.8 kilometers per
hour. The time to cross the river is:
0.5
0.025 hours
19.8
t =
≈
or
0.5
60 1.52 minutes
19.8
t =
⋅
≈
.
84. Let a
v
= the velocity of the plane in still air,
w
v
= the velocity of the wind, and g
v
= the
velocity of the plane relative to the ground.
=
+
g
a
w
v
v
v

(
)
250 cos
sin
α
α
=
+
a
v
i
j

(
)
(
)
40
40
40
20 2
1 1
2


−
=




−


−


=
=
− =
−


+


w
i
j
v
i
j
i
j
i
j
i
j

250cos
250sin
20 2
20 2
a
α
α
=
+
=
+
+
−
=
g
a
w
v
v
v
i
j
i
j
i

Examining the j components:
250sin
20 2 0
250sin
20 2
20 2
sin
0.11314
250
6.5
α
α
α
α
−
=
=
=
≈
≈
°

The heading of the plane should be about
N83.5˚E, that is, about 6.5° north of east.

Examining the i components:
(
)
250cos 6.5º
20 2
276.7
a
a
+
=
≈
i
i
i
The speed of the plane relative to the ground is
about 276.7 miles per hour.
85. Let 1F be the tension on the left cable and 2
F be
the tension on the right cable. Let 3
F represent
the force of the weight of the box.
(
)
(
)
(
)
(
)
(
)
(
)
cos 155º
sin 155º
0.9063
0.4226
cos 40º
sin 40º
0.7660
0.6428
1000
=
+




≈
−
+
=
+




≈
+
= −
1
1
1
2
2
2
3
F
F
i
j
F
i
j
F
F
i
j
F
i
j
F
j

For equilibrium, the sum of the force vectors
must be zero.
(
)
(
)
0.9063
0.4226

0.7660
0.6428
1000
0.9063
0.7660

0.4226
0.6428
1000
0
+
+
= −
+
+
+
−
= −
+
+
+
−
=
1
2
3
1
1
2
2
1
2
1
2
F
F
F
F i
F
j
F
i
F
j
j
F
F
i
F
F
j

Set the and
i
j components equal to zero and
solve:
0.9063
0.7660
0
0.4226
0.6428
1000 0
−
+
=


+
−
=

1
2
1
2
F
F
F
F

Solve the first equation for
2
F
and substitute
the result into the second equation to solve the
Chapter 9: Polar Coordinates; Vectors
918
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
system:
0.9063
1.1832
0.7660
=
≈
2
1
1
F
F
F

(
)
0.4226
0.6428 1.1832
1000 0
1.1832
1000
845.2
+
−
=
=
≈
1
1
1
1
F
F
F
F

(
)
1.1832 845.2
1000
≈
≈
2
F

The tension in the left cable is about 845.2
pounds and the tension in the right cable is about
1000 pounds.
86. Let 1F be the tension on the left cable and 2
F be
the tension on the right cable. Let 3
F represent the
force of the weight of the box.
(
)
(
)
(
)
(
)
(
)
(
)
cos 145º
sin 145º
0.8192
0.5736
cos 50º
sin 50º
0.6428
0.7660
800
=
+




≈
−
+
=
+




≈
+
= −
1
1
1
2
2
2
3
F
F
i
j
F
i
j
F
F
i
j
F
i
j
F
j

For equilibrium, the sum of the force vectors must
be zero.
(
)
(
)
2
2
2
2
0.8192
0.5736

0.6428
0.7660
800
0.8192
0.6428

0.5736
0.7660
800
0
+
+
= −
+
+
+
−
= −
+
+
+
−
=
1
2
3
1
1
1
1
F
F
F
F i
F
j
F
i
F
j
j
F
F
i
F
F
j

Set the and
i
j components equal to zero and
solve:
2
2
0.8192
0.6428
0
0.5736
0.7660
800 0
−
+
=


+
−
=

1
1
F
F
F
F

Solve the first equation for
2F
and substitute the
result into the second equation to solve the system:
2
0.8192
1.2744
0.6428
=
≈
1
1
F
F
F

(
)
0.5736
0.7660 1.2744
800 0
1.5498
800
516.2
+
−
=
=
≈
1
1
1
1
F
F
F
F

(
)
2
1.2744 516.2
657.8
≈
≈
F

The tension in the left cable is about 516.2 pounds
and the tension in the right cable is about 657.8
pounds.
87. Let 1F be the tension on the left end of the rope
and 2F be the tension on the right end of the rope.
Let 3
F represent the force of the weight of the
tightrope walker.
(
)
(
)
(
)
(
)
(
)
(
)
2
2
2
3
cos 175.8º
sin 175.8º
0.99731
0.07324
cos 3.7º
sin 3.7º
0.99792
0.06453
150
=
+




≈
−
+
=
+




≈
+
= −
1
1
1
F
F
i
j
F
i
j
F
F
i
j
F
i
j
F
j


For equilibrium, the sum of the force vectors must
be zero.


(
)
(
)
2
2
2
2
0.99731
0.07324

0.99792
0.06453
150
0.99731
0.99792

0.07324
0.06453
150
0
+
+
= −
+
+
+
−
= −
+
+
+
−
=
1
2
3
1
1
1
1
F F
F
F i
F
j
F
i
F
j
j
F
F
i
F
F
j

Set the and
i
j components equal to zero and
solve:
2
2
0.99731
0.99792
0
0.07324
0.06453
150 0
−
+
=


+
−
=

1
1
F
F
F
F

Solve the first equation for
2F
and substitute the
result into the second equation to solve the system:
2
0.99731
0.99939
0.99792
=
≈
1
1
F
F
F

(
)
0.07324
0.06453 0.99939
150 0
+
−
=
1
1
F
F


0.13773
150
1089.1
=
=
1
1
F
F

2
0.99939(1089.1) 1088.4
=
≈
F

The tension in the left end of the rope is about
1089.1 pounds and the tension in the right end of
the rope is about 1088.4 pounds.
88. Let 1F be the tension on the left end of the rope
and 2F be the tension on the right end of the rope.
Let 3
F represent the force of the weight of the
tightrope walker.
(
)
(
)
(
)
cos 176.2º
sin 176.2º
0.99780
0.06627
=
+




≈
−
+
1
1
1
F
F
i
j
F
i
j

(
)
(
)
(
)
2
2
2
3
cos 2.6º
sin 2.6º
0.99897
0.04536
135
=
+




≈
+
= −
F
F
i
j
F
i
j
F
j


Section 9.4: Vectors
919
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
For equilibrium, the sum of the force vectors must
be zero.
(
)
(
)
2
2
2
2
0.99780
0.06627

0.99897
0.04536
135
0.99780
0.99897

0.06627
0.04536
135
0
+
+
= −
+
+
+
−
= −
+
+
+
−
=
1
2
3
1
1
1
1
F
F
F
F i
F
j
F
i
F
j
j
F
F
i
F
F
j


Set the and
i
j components equal to zero and
solve:
2
2
0.99780
0.99897
0
0.06627
0.04536
135 0
−
+
=


+
−
=

1
1
F
F
F
F

Solve the first equation for
2F
and substitute the
result into the second equation to solve the system:
2
0.99780
0.99883
0.99897
=
≈
1
1
F
F
F

(
)
0.06627
0.04536 0.99883
135 0
+
−
=
1
1
F
F

0.11158
135
1209.9
=
≈
1
1
F
F

2
0.99883(1209.9) 1208.4
=
=
F

The tension in the left end of the rope is about
1209.9 pounds and the tension in the right end of
the rope is about 1208.4 pounds.
89.
1
3000
=
F
i


(
)
(
)
2
2000 cos 45º
sin 45º
2
2
2000
2
2
1000 2
1000 2
=
+






=
+






=
+
F
i
j
i
j
i
j



(
)
1
3000 1000 2
1000 2

3000 1000 2
1000 2
=
+
=
+
+
=
+
+
2
F F F
i
i
j
i
j

(
)
(
)
2
2
3000 1000 2
1000 2
4635.2
=
+
+
≈
F

The monster truck must pull with a force of
approximately 4635.2 pounds in order to remain
unmoved.
90. a.
1
7000
=
F
i



(
)
(
)
(
)
2
5500 cos 40º
sin 40º
5500 0.766044
0.642788
4213.24
3535.33
=
+




≈
+
≈
+
F
i
j
i
j
i
j




1
7000
4213.24
3535.33

11,213.24
3535.33
=
+
≈
+
+
=
+
2
F F F
i
i
j
i
j

2
2
(11,213.24)
(3535.33)
11,757.4
=
+
≈
F

The farmer will not be successful in
removing the stump. The two tractors will
have a combined pull of only about 11,757.4
pounds, which is less than the 6 tons needed.

b.
1
7000
=
F
i



(
)
(
)
(
)
2
5500 cos 25º
sin 25º
5500 0.906308
0.422618
4984.69
2324.40
=
+




≈
+
≈
+
F
i
j
i
j
i
j




1
7000
4984.69
2324.40

11,984.69
2324.40
=
+
≈
+
+
=
+
2
F F F
i
i
j
i
j

2
2
(11,984.69)
(2324.40)
12,208.0
=
+
≈
F

The farmer will be successful in removing
the stump. The two tractors will have a
combined pull of about 12,208 pounds,
which is more than the 6 tons needed.
91. The given forces are:
3
4
3 ;
4 ;
4
2 ;
4
= −
= − +
=
−
= −
1
2
F
i F
i
j F
i
j F
j

A vector
a
b
=
+
v
i
j needs to be added for
equilibrium. Find vector
a
b
=
+
v
i
j :
0
3
(
4 )
(4
2 )
( 4 )
(
)
0
0
2
(
)
0
( 2
)
0
a
b
a
b
a
b
+
+
+
+ =
− + − +
+
−
+ −
+
+
=
+ − +
+
=
+ − +
=
1
2
3
4
F
F
F F
v
i
i
j
i
j
j
i
j
i
j
i
j
i
j


0;
2
0
2
a
b
b
=
− + =
=


Therefore,
2
=
v
j .
Chapter 9: Polar Coordinates; Vectors
920
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall



92 – 93. Interactive Exercises.

94 – 96. Answers will vary.

Section 9.5
1.
2
2
2
2
cos
c
a
b
ab
C
=
+
−


2. dot product
3. orthogonal
4. parallel
5. True
6. False
7.
= −
= +
v
i
j, w
i
j
a.
1(1)
( 1)(1) 1 1 0
⋅
=
+ −
= − =
v w


b.
2
2
2
2
0
cos
0
1
( 1)
1
1
90º
θ
θ
⋅
=
=
=
+ −
+
=
v w
v w


c. The vectors are orthogonal.
8.
= +
= − +
v
i
j, w
i
j
a.
1( 1) 1(1)
1 1 0
⋅
= − +
= − + =
v w


b.
2
2
2
2
0
cos
0
1
1
( 1)
1
90º
θ
θ
⋅
=
=
=
+
−
+
=
v w
v w


c. The vectors are orthogonal.
9.
2
2
=
+
= −
v
i
j, w
i
j
a.
2(1) 1( 2)
2 2 0
⋅
=
+ − = − =
v w


b.
2
2
2
2
0
cos
2
1 1
( 2)
0
0
0
5
5 5
90º
θ
θ
⋅
=
=
+
+ −
=
= =
=
v w
v w


c. The vectors are orthogonal.
10.
2
2
2
=
+
= +
v
i
j, w
i
j
a.
2(1) 2(2)
2 4 6
⋅
=
+
= + =
v w


b.
2
2
2
2
6
cos
2
2
1
2
6
3
3 10
10
2 2 5
10
18.4º
θ
θ
⋅
=
=
+
+
=
=
=
≈
v w
v w


c. The vectors are neither parallel nor
orthogonal.
11.
3
=
−
= +
v
i
j, w
i
j
a.
3(1)
( 1)(1)
3 1
⋅
=
+ −
=
−
v w


b.
(
)2
2
2
2
o
3 1
cos
3
( 1)
1
1
3 1
3 1
6
2
4
4 2
2 2
75
θ
θ
⋅
−
=
=
+ −
+
−
−
−
=
=
=
=
v w
v w


c. The vectors are neither parallel nor
orthogonal.
12.
3
= +
= −
v
i
j, w
i
j
a.
1(1)
3( 1) 1
3
⋅
=
+
− = −
v w


b.
(
)2
2
2
2
1
3
cos
1
3
1
( 1)
1
3
1
3
2
6
4
4 2
2 2
105º
θ
θ
⋅
−
=
=
+
+ −
−
−
−
=
=
=
=
v w
v w


c. The vectors are neither parallel nor
orthogonal.

Section 9.5: The Dot Product
921
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
13.
3
4
6
8
= +
= − −
v
i
j, w
i
j
a.
3( 6) 4( 8)
18 32
50
⋅
= − + − = − −
= −
v w


b.
2
2
2
2
50
cos
3
4
( 6)
( 8)
50
50
1
50
25 100
180º
θ
θ
⋅
−
=
=
+
−
+ −
−
−
=
=
= −
=
v w
v w


c. Note that
1
2
= −
v
w and
180º
θ =
, so the
vectors are parallel.
14.
3
4
9 12
= −
= −
v
i
j, w
i
j
a.
3(9)
( 4)( 12)
27 48 75
⋅
=
+ −
−
=
+
=
v w


b.
2
2
2
2
cos
75
3
( 4)
9
( 12)
75
75
1
75
25 225
0º
θ
θ
⋅
=
=
+ −
+ −
=
=
=
=
v w
v w


c. Note that
1
3
=
v
w and
0º
θ =
, so the
vectors are parallel.
15.
4
=
=
v
i , w
j
a.
4(0) 0(1)
0 0 0
⋅
=
+
= + =
v w


b.
2
2
2
2
0
cos
4
0
0
1
0
0
0
4 1 4
90º
θ
θ
⋅
=
=
+
+
=
= =
⋅
=
v w
v w


c. The vectors are orthogonal.
16.
3
=
= −
v
i , w
j
a.
1(0) 0( 3)
0 0 0
⋅
=
+ − = + =
v w


b.
2
2
2
2
cos
0
0
0
0
1 3
3
1
0
0
( 3)
90º
θ
θ
⋅
=
=
=
= =
⋅
+
+ −
=
v w
v w


c. The vectors are orthogonal.
17.

2
3
a
= −
=
+
v
i
j, w
i
j
Two vectors are orthogonal if the dot product is
zero. Solve for a:
0
1(2)
(
)(3)
0
2 3
0
3
2
2
3
a
a
a
a
⋅
=
+ −
=
−
=
=
=
v w

18.
b
= +
= +
v
i
j, w
i
j
Two vectors are orthogonal if the dot product is
zero. Solve for b:
0
1(1) 1( )
0
1
0
1
b
b
b
⋅
=
+
=
+ =
= −
v w

19.
2
3
=
−
= −
v
i
j, w
i
j
(
)
(
)
(
)
(
)
1
2
2
2
2
2(1)
( 3)( 1)
1
( 1)
5
5
5
2
2
2
⋅
+ −
−
=
=
−
+ −
=
− =
−
v w
v
w
i
j
w
i
j
i
j

(
)
5
5
1
1
2
3
2
2
2
2


= −
=
−
−
−
= −
−




2
1
v
v v
i
j
i
j
i
j
20.
3
2
2
= − +
=
+
v
i
j, w
i
j
(
)
(
)
(
)
(
)
1
2
2
2
2
3(2) 2(1)
2
2
1
4
8
4
2
5
5
5
⋅
−
+
=
=
+
+
= −
+ = −
−
v w
v
w
i
j
w
i
j
i
j

(
)
8
4
3
2
5
5
7
14
5
5


= −
= − +
− −
−




= −
+
2
1
v
v v
i
j
i
j
i
j

21.
2
= −
= − −
v
i
j, w
i
j
(
)
(
)
(
)
(
)
(
)
(
)
1
2
2
2
2
1( 1)
( 1)( 2)
2
1
2
1
1
2
2
5
5
5
1
2
6
3
5
5
5
5
⋅
− + − −
=
=
− −


−
+ −




= −
+
= −
−


= −
= − − −
−
=
−




2
1
v w
v
w
i
j
w
i
j
i
j
v
v v
i
j
i
j
i
j

Chapter 9: Polar Coordinates; Vectors
922
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
22.
2
2
=
−
= −
v
i
j, w
i
j
(
)
(
)
(
)
(
)
(
)
1
2
2
2
2
2(1)
( 1)( 2)
2
1
( 2)
4
4
8
2
5
5
5
4
8
6
3
2
5
5
5
5
⋅
+ − −
=
=
−
+ −
=
−
=
−


= −
=
− −
−
=
+




2
1
v w
v
w
i
j
w
i
j
i
j
v
v v
i
j
i
j
i
j

23.
3
2
= +
= − −
v
i
j, w
i
j
(
)
(
)
(
)
(
)
(
)
1
2
2
2
2
3( 2) 1( 1)
2
( 2)
( 1)
7
14
7
2
5
5
5
14
7
1
2
3
5
5
5
5
⋅
−
+ −
=
=
− −
−
+ −
= −
− − =
+


= −
=
+ −
+
=
−




2
1
v w
v
w
i
j
w
i
j
i
j
v
v v
i
j
i
j
i
j

24.
3
4
= −
=
−
v
i
j, w
i
j
(
)
(
)
(
)
(
)
(
)
1
2
2
2
2
1(4)
( 3)( 1)
4
4
( 1)
7
28
7
4
17
17
17
28
7
3
17
17
11
44
17
17
⋅
+ −
−
=
=
−
+ −
=
− =
−


= −
= −
−
−




= −
−
2
1
v w
v
w
i
j
w
i
j
i
j
v
v v
i
j
i
j
i
j

25.
(
)
(
)
3 cos 60º
sin 60º
1
3
3
3 3
3
2
2
2
2
=
+






=
+
=
+






F
i
j
i
j
i
j

( )
3
3 3
6
2
2
3
3 3
(6)
0
9 ft-lb
2
2
W
AB


=
⋅
=
+
⋅






=
+
=
F
i
j
i


26.
(
)
(
)
20 cos 30º
sin 30º
3
1
20
10 3 10
2
2
=
+






=
+
=
+






F
i
j
i
j
i
j

(
)
( )
10 3 10 100
10 3(100) 10 0
1000 3 1732 ft-lb
W
AB
=
⋅
=
+
⋅
=
+
=
≈
F
i
j
i


27. a.
2
2
( 0.02)
( 0.01)
0.0005 0.022
=
−
+ −
=
≈
I

The intensity of the sun’s rays is about
0.022 watts per square centimeter.



2
2
(300)
(400)
250,000 500
=
+
=
=
A

The area of the solar panel is 500 square
centimeter.

b.
( 0.02
0.01 ) (300
400 )
( 0.02)(300)
( 0.01)(400)
6 ( 4)
10
10
W =
⋅
= −
−
⋅
+
= −
+ −
= − + −
= −
=
I A
i
j
i
j

This means 10 watts of energy is collected.

c. To collect the maximum number of watts, I
and A should be parallel with the solar
panels facing the sun.
28. a.
2
2
(0.75)
( 1.75)
3.625 1.90
=
+ −
=
≈
R

About 1.90 inches of rain fell.



2
2
(0.3)
(1)
1.09 1.04
=
+
=
=
A

The area of the opening of the gauge is
about 1.04 square inches.

b.
(0.75 1.75 ) (0.3
)
(0.75)(0.3)
( 1.75)(1)
0.225 ( 1.75)
1.525 1.525
V =
⋅
=
−
⋅
+
=
+ −
=
+ −
= −
=
R A
i
j
i
j

This means the gauge collected 1.525 cubic
inches of rain.

c. To collect the maximum volume of rain, R
and A should be parallel and oriented in
opposite directions.
29. Split the force into the components going down the
hill and perpendicular to the hill.
d
F
8º
p
F
F

sin8º 5300sin8º 737.6
cos8º 5300cos8º 5248.4
=
=
≈
=
=
≈
d
p
F
F
F
F

The force required to keep the Sienna from rolling
down the hill is about 737.6 pounds. The force
perpendicular to the hill is approximately 5248.4
pounds.

Section 9.5: The Dot Product
923
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
30. Split the force into the components going down the
hill and perpendicular to the hill.
10º
d
F
pF
F

sin10º 4500sin10º 781.4
cos10º 4500cos10º 4431.6
=
=
≈
=
=
≈
d
p
F
F
F
F

The force required to keep the Bonneville from
rolling down the hill is about 781.4 pounds. The
force perpendicular to the hill is approximately
4431.6 pounds.
31. We must determine the component force going
down the ramp.
sin20º 250sin20º 85.5
=
=
≈
d
F
F

Timmy must exert about 85.5 pounds of force to
hold the piano in position.
32. We must determine the angle θ if the force of the
boulder is
5000
=
F
pounds and the component
force going down the hill is
1000
=
d
F
pounds.
1
sin
1000 5000sin
1000
sin
0.2
5000
sin (0.2) 11.5
θ
θ
θ
θ
−
=
=
=
=
=
≈
°
d
F
F

The angle of inclination of the hill is about 11.5 .°
33.
2,
4
W
AB W
AB
=
⋅
=
=
F
i



(
)
cos
sin
2
cos
sin
4
2 4cos
1
cos
2
60º
α
α
α
α
α
α
α
=
−
=
−
⋅
=
=
=
F
i
j
i
j
i

34. Let
1
1
a
b
=
+
u
i
j ,
2
2
a
b
=
+
v
i
j , and
3
3
a
b
=
+
w
i
j .
(
)
(
)(
)
(
)(
)
(
)[
]
(
)(
)
(
)(
)
1
1
2
2
3
3
1
1
2
3
2
3
1
1
2
3
2
3
1
2
3
1
2
3
1 2
1 3
1 2
1 3
1 2
1 2
1 3
1 3
1
1
2
2
1
1
3
3
(
)
(
)
(
)
(
)
a
b
a
b
a
b
a
b
a
a
b
b
a
b
a
a
b
b
a a
a
b b
b
a a
a a
b b
b b
a a
b b
a a
b b
a
b
a
b
a
b
a
b
⋅
+
=
+
+
+
+
=
+
+
+
+
=
+
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
=
+
+
+
+
+
= ⋅ +
⋅
u v w
i
j
i
j
i
j
i
j
i
i
j
j
i
j
i
j
i
j
i
j
i
j
i
j
u v u w

35. Since
0
0
=
+
0
i
j and = +
a b
v
i
j , we have that
0
0
0
a
b
⋅
=
⋅ +
⋅ =
0 v
.
36. Let
x
y
=
+
v
i
j . Since v is a unit vector, we
have that
2
2
1
x
y
=
+
=
v
, or
2
2

1
x
y
+
=
.
If α is the angle between v and i , then
(
)
cos
1 1
x
y
x
α
+
⋅
⋅
=
=
=
⋅
i
j
i
v i
v
i
. Now,
2
2
2
2
2
2
2
2
1
cos
1
1 cos
sin
sin
x
y
y
y
y
y
α
α
α
α
+
=
+
=
= −
=
=

Thus,
cos
sin
α
α
=
+
v
i
j .
37. If
1
1
cos
sin
a
b
α
α
=
+
=
+
v
i
j
i
j and
2
2
cos
sin
a
b
β
β
=
+
=
+
w
i
j
i
j , then
1 2
1 2
cos(
)
cos cos
sin
sin
a a
b b
α β
α
β
α
β
−
= ⋅
=
+
=
+
v w

38. Let
a
b
=
+
v
i
j . The projection of v onto i is
(
)
(
)
(
)
2
2
2
2
2
the projection of onto
(1)
(0)
1
1
0
a
b
a
b
a
=
+
⋅
⋅
+
=
=
=
=
+
1
1
v
v
i
i
i
i
v
i
i
i
i
i
i

2
1 a
b
a
b
= −
=
+ −
=
v
v v
i
j
i
j
Since
(
)
( )( )
( )( )
1
0
a
b
a
b
a
⋅ =
+
⋅ =
+
=
v i
i
j
i
and
(
)
( )( )
( )( )
0
1
a
b
a
b
b
⋅ =
+
⋅ =
+
=
v j
i
j
j
,
(
)
(
)
1
2
.
=
+
=
⋅
+
⋅
v v
v
v i i
v j j