424
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
(c) −3 − i
Sketch a graph of −3 − i in the complex
plane.
(c)
z 3 = ⎡⎣3 (cos 50° + i sin 50°)⎤⎦
= 33 ⎡⎣ cos (3 ⋅ 50°) + i sin (3 ⋅ 50°)⎤⎦
= 27 (cos150° + i sin150°)
⎛
3 1 ⎞
27 3 27
= 27 ⎜ −
+ i⎟ = −
+ i
2
2
⎝ 2 2 ⎠
3
(d) w3 = ⎡⎣12 (cos 80° + i sin 80°)⎤⎦
= 123 ⎡⎣cos (12 ⋅ 80°) + i sin (12 ⋅ 80°)⎤⎦
= 1728 (cos 960° + i sin 960°)
⎛ 1
3 ⎞
i⎟
= 1728 ⎜ − −
⎝ 2 2 ⎠
Using a calculator, we find that the
reference angle is 18.4º. The graph shows
that θ is in quadrant III, so
θ = 180° + 18.4° = 198.4°. Therefore,
−3 − i = 10 (cos 198.4° + i sin198.4°) .
⎛1
3 ⎞
6. (a) 4 (cos 60° + i sin 60°) = 4 ⎜ +
i⎟
⎝2 2 ⎠
= 2 + 2i 3
(b) 5cis130° = 5 (cos130° + i sin130°)
= 5cos130° + 5i sin130°
= −3.2139 + 3.8302i
(c) 7 (cos 270° + i sin 270°) = 7 (0 + (−1)i )
= 0 − 7i
(d) 2 cis 0° = 2 (cos 0° + i sin 0°) = 2 (1 + 0i )
= 2 or 2 + 0i
7. w = 12 (cos 80° + i sin 80°) ,
z = 3 (cos 50° + i sin 50°)
(a) wz = 12 (cos 80° + i sin 80°)
⋅ 3 (cos 50° + i sin 50°)
⎡cos (80° + 50°)
⎤
= 12 ⋅ 3 ⎢
⎥
i
sin
80
50
+
°
+
°
(
)
⎣
⎦
= 36 (cos130° + i sin130°)
(b)
w 12 (cos 80° + i sin 80°)
=
z
3 (cos 50° + i sin 50°)
= 4 ⎣⎡cos (80° − 50°) + i sin (80° − 50°)⎤⎦
= 4 (cos 30° + i sin 30°)
⎛ 3 1 ⎞
= 4⎜
+ i ⎟ = 2 3 + 2i
⎝ 2 2 ⎠
3
=−
1728 1728 3
i = −864 − 864i 3
−
2
2
8. Find all the fourth roots of
−16 = 16 (cos180° + i sin180°) .
Since r 4 (cos 4α + i sin 4α )
= 16 (cos180° + i sin180°) , then we have
r 4 = 16 ⇒ r = 2 and 4α = 180° + 360° ⋅ k ⇒
180° + 360° ⋅ k
α=
= 45° + 90° ⋅ k , k any
4
integer.
If k = 0, then α = 45° + 0° = 45°.
If k = 1, then α = 45° + 90° = 135°.
If k = 2, then α = 45° + 180° = 225°.
If k = 3, then α = 45° + 270° = 315°.
Solution set:
2 (cos 45° + i sin 45°) ,2 (cos135° + i sin135°) ,
2 (cos 225° + i sin 225°) ,
{
}
2 (cos 315° + i sin 315°) or
{
2 + i 2, − 2 + i 2, − 2 − i 2, 2 − i 2
Section 8.5
Polar Equations and
Graphs
1. (a) II (since r > 0 and 90° < θ < 180°)
(b) I (since r > 0 and 0° < θ < 90° )
(c) IV (since r > 0 and –90° < θ < 0°)
(d) III (since r > 0 and 180° < θ < 270°)
2. (a) positive x-axis
(b) negative x-axis
Copyright © 2013 Pearson Education, Inc.
}
Section 8.5 Polar Equations and Graphs
(c) negative y-axis
(d) positive y-axis (since 450° – 360° = 90°)
425
(c) Since
x = r cos θ ⇒ x = (−2) cos135° = 2 and
y = r sin θ ⇒ y = ( −2) sin135° = 2, the
For Exercises 3(b)−14(b), answers may vary.
point is
3. (a)
(
)
2, − 2 .
6. (a)
(b) Two other pairs of polar coordinates for
(1, 45°) are (1, 405°) and (–1, 225°).
2
(c) Since x = r cos θ ⇒ x = 1 ⋅ cos 45° =
2
2
,
and y = r sin θ ⇒ y = 1 ⋅ sin 45° =
2
⎛ 2 2⎞
the point is ⎜
,
⎟.
⎝ 2 2 ⎠
(b) Two other pairs of polar coordinates for
(–4, 30°) are (−4, 390°) and (4, 210°).
(c) Since
x = r cos θ ⇒ x = (−4) cos 30° = −2 3
and y = r sin θ ⇒ y = ( −4) sin 30° = −2,
(
)
the point is −2 3, −2 .
7. (a)
4. (a)
(b) Two other pairs of polar coordinates for
(3, 120°) are (3, 480°) and (–3, 300°).
(c) Since x = r cos θ ⇒ x = 3cos120 = −
and y = r sin θ ⇒ y = 3sin120° =
3
2
(b) Two other pairs of polar coordinates for
(5, – 60°) are (5, 300°) and (–5, 120°).
(c) Since x = r cos θ ⇒ x = 5 cos (−60°) =
5
2
and
3 3
,
2
y = r sin θ ⇒ y = 5sin ( −60°) = −
⎛ 3 3 3⎞
the point is ⎜ − ,
⎟.
⎝ 2 2 ⎠
5 3
,
2
⎛5 5 3⎞
the point is ⎜ , −
.
2 ⎟⎠
⎝2
5. (a)
8. (a)
(b) Two other pairs of polar coordinates for
(–2, 135°) are (–2, 495°) and (2, 315°).
(b) Two other pairs of polar coordinates for
(2, – 45°) are (2, 315°) and (–2, 135°).
Copyright © 2013 Pearson Education, Inc.
426
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
(c) Since
x = r cos θ ⇒ x = 2 cos (−45°) = 2 and
11. (a)
y = r sin θ ⇒ y = 2 sin (−45°) = − 2, the
point is
(
)
2, − 2 .
9. (a)
(b) Two other pairs of polar coordinates for
2π ⎞
⎛ 5π ⎞
⎛ 11π ⎞
⎛
⎜⎝ 3,
⎟⎠ are ⎜⎝ 3,
⎟⎠ and ⎜⎝ −3,
⎟.
3
3
3 ⎠
(c) Since x = r cos θ ⇒ x = 3cos
(b) Two other pairs of polar coordinates for
(–3, –210°) are (–3, 150°) and (3, –30°).
and y = r sin θ ⇒ y = 3sin
(c) Since
3 3
x = r cos θ ⇒ x = (−3) cos ( −210°) =
2
and
3
y = r sin θ ⇒ y = (−3) sin (−210°) = − ,
2
⎛3 3 3⎞
the point is ⎜
,− ⎟.
2⎠
⎝ 2
5π 3
=
3
2
5π
3 3
=−
,
3
2
⎛3 3 3⎞
the point is ⎜ , −
.
2 ⎟⎠
⎝2
12. (a)
10. (a)
(b) Two other pairs of polar coordinates for
π⎞
π⎞
⎛ 3π ⎞
⎛
⎛
⎜⎝ 4,
⎟ are ⎜⎝ 4, − ⎟⎠ and ⎜⎝ −4, ⎟⎠ .
2
2
2 ⎠
(c) Since x = r cos θ ⇒ x = 4 cos
(b) Two other pairs of polar coordinates for
(–1, –120°) are (–1, 240°) and (1, 60°).
y = r sin θ ⇒ y = 4 sin
3π
= 0 and
2
3π
= −4, the
2
point is (0, −4).
(c) Since
x = r cos θ ⇒ x = (−1) cos (−120°) =
1
2
13. (a)
and
y = r sin θ ⇒ y = ( −1) sin (−120°) =
⎛1 3 ⎞
the point is ⎜ ,
⎟.
⎝2 2 ⎠
3
,
2
(b) Two other pairs of polar coordinates for
π⎞
7π ⎞
⎛
⎛
⎛ 4π ⎞
⎜⎝ −2, ⎟⎠ are ⎜⎝ −2,
⎟⎠ and ⎜⎝ 2,
⎟.
3
3
3 ⎠
Copyright © 2013 Pearson Education, Inc.
Section 8.5 Polar Equations and Graphs
(c) Since x = r cos θ ⇒ x = −2 cos
and y = r sin θ ⇒ y = −2 sin
(
)
π
3
π
3
= −1
427
16. (a)
= − 3,
the point is −1, − 3 .
14. (a)
(b) r = 12 + 12 = 1 + 1 = 2 and
⎛1 ⎞
θ = tan −1 ⎜ ⎟ = tan −1 1 = 45°, since θ is
⎝1 ⎠
in quadrant I. So, one possibility is
(
(b) Two other pairs of polar coordinates for
5π ⎞
⎛
⎛ 11π ⎞
⎟ and
⎜⎝ −5,
⎟⎠ are ⎜⎝ 5,
6
6 ⎠
17π ⎞
⎛
⎜⎝ −5,
⎟.
6 ⎠
)
2, 45° . Alternatively, if r = − 2,
then θ = 45° + 180° = 225°. Thus, a
(
)
second possibility is − 2, 225° .
17. (a)
(c) Since
5π 5 3
=
and
6
2
5π
5
= − , the
y = r sin θ ⇒ y = −5sin
6
2
⎛5 3
5⎞
point is ⎜
, − ⎟.
2⎠
⎝ 2
x = r cos θ ⇒ x = −5 cos
(b) r = 02 + 32 = 0 + 9 = 9 = 3 and
θ = 90°, since (0, 3) is on the positive y-
For Exercises 15(b)–26(b), answers may vary.
axis. So, one possibility is (3, 90°) .
15. (a)
Alternatively, if r = −3, then
θ = 90° + 180° = 270°. Thus, a second
possibility is (−3, 270°) .
18. (a)
(b) r = 12 + (−1) = 1 + 1 = 2 and
2
⎛ −1 ⎞
= tan −1 ( −1) = −45°, since
⎝ 1 ⎟⎠
θ is in quadrant IV. Since
360° − 45° = 315°, one possibility is
θ = tan −1 ⎜
(
)
2, 315° . Alternatively, if r = − 2,
(b) r = 02 + ( −3) = 0 + 9 = 9 = 3 and
2
θ = 270°, since (0, 3) is on the negative
then θ = 315° − 180° = 135°. Thus, a
y-axis. So, one possibility is (3, 270°) .
second possibility is − 2,135° .
Alternatively, if r = −3, then
θ = 270° − 180° = 90°. Thus, a second
possibility is (−3, 90°) .
(
)
Copyright © 2013 Pearson Education, Inc.
428
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
19. (a)
2
⎛ 3 ⎞ ⎛ 3 ⎞2
(b) r = ⎜
⎟ +⎜ ⎟ =
⎝ 2 ⎠ ⎝2⎠
=
3 9
+
4 4
⎛3 2 ⎞
12
= 3 and θ = arctan ⎜ ⋅
⎝ 2 3 ⎟⎠
4
( )
= tan −1 3 = 60°, since θ is in
quadrant I. So, one possibility is
(b) r =
( 2) + ( 2)
2
2
(
= 2+2 = 4 = 2
2⎞
−1
and θ = tan ⎜
⎟ = tan 1 = 45°,
⎝ 2⎠
since θ is in quadrant I. So, one
possibility is (2, 45°) . Alternatively, if
)
3, 60° . Alternatively, if r = − 3,
then θ = 60° + 180° = 240°. Thus, a
(
−1 ⎛
)
second possibility is − 3, 240° .
22. (a)
r = −2, then θ = 45° + 180° = 225°.
Thus, a second possibility is (−2, 225°) .
20. (a)
2
2
⎛
3 ⎞ ⎛ 1⎞
(b) r = ⎜ −
⎟ + ⎜− ⎟ =
⎝ 2 ⎠ ⎝ 2⎠
=
(b) r =
(− 2 ) + ( 2 )
2
2
⎛ −1 ⎛ −2 ⎞ ⎞
4
= 1 and θ = arctan ⎜ ⋅ ⎜
⎟⎟
4
⎝ 2 ⎝ 3 ⎠⎠
⎛ 3⎞
⎛ 1 ⎞
= tan −1 ⎜
= tan −1 ⎜
⎟ = 210° since
⎝ 3 ⎟⎠
⎝ 3 ⎠
θ is in quadrant III. So, one possibility is
(1, 210°) . Alternatively, if r = −1, then
= 2+2 = 4 = 2
⎛ 2 ⎞
−1
and θ = tan −1 ⎜
⎟ = tan (−1) , since
⎝− 2 ⎠
θ is in quadrant II we have θ = 135°.
So, one possibility is (2, 135°) .
Alternatively, if r = −2, then
θ = 135° + 180° = 315°. Thus, a second
possibility is (−2, 315°) .
3 1
+
4 4
θ = 210° − 180° = 30°. Thus, a second
possibility is (−1, 30°) .
23. (a)
21. (a)
(b) r = 32 + 02 = 9 + 0 = 9 = 3 and
θ = 0°, since (3, 0) is on the positive
x-axis. So, one possibility is (3, 0°) .
Alternatively, if r = −3, then
θ = 0° + 180° = 180°. Thus, a second
possibility is (−3, 180°) .
Copyright © 2013 Pearson Education, Inc.
Section 8.5 Polar Equations and Graphs
24. (a)
2
2
3⎞
⎛1⎞ ⎛
=
(b) r = ⎜ ⎟ + ⎜ −
⎝ 2 ⎠ ⎝ 2 ⎟⎠
1 3
+ =1
4 4
⎛− 3 ⎞
and θ = tan −1 ⎜ 12 ⎟ = tan −1 − 3 .
⎜⎝ 2 ⎟⎠
Since θ is in quadrant IV we have
θ = 300°. So, one possibility is (1, 300°).
Alternatively, if r = −1, then
θ = 300° − 180° = 120°. Thus, a second
possibility is (−1, 120°).
(
(b) r =
( − 2 )2 + 0 2
= 4 + 0 = 4 = 2 and
θ = 180°, since (−2, 0) is on the negative
x-axis. So, one possibility is (2, 180°).
Alternatively, if r = −2, then
θ = 180° − 180° = 0°. Thus, a second
possibility is (−2, 0°).
25. (a)
2
=
4
.
cos θ − sin θ
9 27
+
4 4
36
=3
4
⎛− 3 3 ⎞
and θ = tan ⎜ 23 ⎟ = tan −1 3. Since
⎜⎝ − 2 ⎟⎠
θ is in quadrant III we have θ = 240°.
So, one possibility is (3, 240°).
Alternatively, if r = −3, then
θ = 240° − 180° = 60°. Thus, a second
possibility is (−3, 60°).
−1
)
27. x − y = 4
Using the general form for the polar equation
c
, with
of a line, r =
a cos θ + b sin θ
a = 1, b = −1, and c = 4, the polar equation is
r=
2
⎛ 3⎞ ⎛ 3 3 ⎞
=
(b) r = ⎜ − ⎟ + ⎜ −
⎝ 2⎠ ⎝
2 ⎟⎠
429
28. x + y = −7
Using the general form for the polar equation
c
, with
of a line, r =
a cos θ + b sin θ
a = 1, b = 1, and c = −7, the polar equation is
r=
−7
.
cos θ + sin θ
26. (a)
Copyright © 2013 Pearson Education, Inc.
430
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
29. x 2 + y 2 = 16 ⇒ r 2 = 16 ⇒ r = ±4
The equation of the circle in polar form is
r = 4 or r = −4.
33. r sin θ = k
30. x 2 + y 2 = 9 ⇒ r 2 = 9 ⇒ r = ±3
The equation of the circle in polar form is
r = 3 or r = −3.
35. r =
r=
r=
k
sin θ
r=
k
cos θ
k
⇒ r = k csc θ
sin θ
36. y = 3
37. r cos θ = k
31. 2 x + y = 5
Using the general form for the polar equation
c
, with
of a line, r =
a cos θ + b sin θ
a = 2, b = 1, and c = 5, the polar equation is
34.
39. r =
38.
k
⇒ r = k sec θ
cos θ
40. x = 3
5
.
2 cos θ + sin θ
41. r = 3 represents the set of all points 3 units
from the pole. The correct choice is C.
42. r = cos 3θ is a rose curve with 3 petals. The
correct choice is D.
32. 3 x − 2 y = 6
Using the general form for the polar equation
c
, with
of a line, r =
a cos θ + b sin θ
a = 3, b = −2, and c = 6, the polar equation is
r=
43. r = cos 2θ is a rose curve with 2 ⋅ 2 = 4
petals. The correct choice is A
6
.
3cos θ − 2sin θ
Copyright © 2013 Pearson Education, Inc.
Section 8.5 Polar Equations and Graphs
44. The general form for the polar equation of a line is r =
+ by = c. r =
431
c
, where the standard form of a line ax
a cos θ + b sin θ
2
is a line. The correct choice is B.
cos θ + sin θ
45. r = 2 + 2 cos θ (cardioid)
θ
0°
30°
60°
90°
120°
150°
cos θ
1
0.9
0.5
0
–0.5
–0.9
r = 2 + 2 cos θ
4
3.7
3
2
1
0.3
θ
180°
210°
240°
270°
300°
330°
cos θ
–1
–0.9
–0.5
0
0.5
0.9
r = 2 + 2 cos θ
0
0.3
1
2
3
3.7
46. r = 8 + 6 cos θ (limaçon)
θ
0°
30°
60°
90°
120°
150°
cos θ
1
0.9
0.5
0
–0.5
–0.9
r = 8 + 6 cos θ
14
13.2
11
8
5
2.8
θ
180°
210°
240°
270°
300°
330°
cos θ
–1
–0.9
–0.5
0
0.5
0.9
r = 8 + 6 cos θ
2
2.8
5
8
11
13.2
47. r = 3 + cos θ (limaçon)
θ
0°
30°
60°
90°
120°
150°
r = 3 + cos θ
4
3.9
3.5
3
2.5
2.1
θ
180°
210°
240°
270°
300°
330°
r = 3 + cos θ
2
2.1
2.5
3
3.5
3.9
48. r = 2 − cos θ (limaçon)
θ
0°
30°
60°
90°
135°
180°
225°
270°
315°
r = 2 − cos θ
1
1.1
1.5
2
2.7
3
2.7
2
1.3
Copyright © 2013 Pearson Education, Inc.
432
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
49. r = 4 cos 2θ (four-leaved rose)
θ
0°
30°
45°
60°
90°
120°
135°
150°
r = 4 cos 2θ
4
2
0
–2
–4
–2
0
2
θ
180°
210°
225°
240°
270°
300°
315°
330°
r = 4 cos 2θ
4
2
0
–2
–4
–2
0
2
50. r = 3cos 5θ (five-leaved rose)
r = 0 when cos 5θ = 0, or 5θ = 90° + 360° ⋅ k = 18° + 72° ⋅ k , where k is an integer, or θ = 18°, 90°, 162°, 234°.
θ
0°
18°
36°
54°
72°
90°
108°
162°
r = 3cos 5θ
3
0
–3
0
3
0
–3
0
Pattern 3, 0, –3, 0, 3 continues for every 18°.
51. r 2 = 4 cos 2θ ⇒ r = ±2 cos 2θ (lemniscate)
Graph only exists for [0°, 45°], [135°, 225°], and [315°, 360°] because cos 2θ must be positive.
θ
0°
30°
45°
135°
150°
r = ±2 cos 2θ
±2
±1.4
0
0
±1.4
θ
180°
210°
225°
315°
330°
r = ±2 cos 2θ
±2
±1.4
0
0
±1.4
52. r 2 = 4 sin 2θ ⇒ r = ±2 sin 2θ (lemniscate)
Graph only exists for [0°, 90°] and [180°, 270°] because sin 2θ must be positive.
θ
0°
30°
45°
60°
90°
180°
225°
270°
r = ±2 sin 2θ
0
±1.86
±2
±1.86
0
0
±2
0
Copyright © 2013 Pearson Education, Inc.
Section 8.5 Polar Equations and Graphs
53. r = 4 − 4 cos θ (cardioid)
θ
r = 4 − 4 cos θ
0°
30°
60°
90°
120°
150°
0
0.5
2
4
6
7.5
θ
180°
210°
240°
270°
300°
330°
r = 4 − 4 cos θ
8
7.5
6
4
2
0.5
54. r = 6 − 3cos θ (limaçon)
θ
r = 6 − 3cos θ
0°
45°
90°
135°
180°
270°
360°
3
3.9
6
8.1
9
6
3
55. r = 2sin θ tan θ (cissoid)
r is undefined at θ = 90° and θ = 270°.
θ
0°
30°
45°
60°
90°
120°
135°
150°
180°
r = 2sin θ tan θ
0
0.6
1.4
3
undefined
–3
–1.4
–0.6
0
Notice that for [180°, 360°), the graph retraces the path traced for [0°, 180°).
cos 2θ
(cissoid with a loop)
cos θ
r is undefined at θ = 90° and θ = 270° and r = 0 at 45°, 135°, 225°, and 315°.
56. r =
r=
θ
0°
45°
60°
70°
80°
cos 2θ
cos θ
1
0
–1
–2.2
–5.4
90°
100°
110°
135°
180°
undefined
5.4
2.2
0
–1
θ
r=
cos 2θ
cos θ
Notice that for [180°, 360°), the graph retraces the path traced for [0°, 180°).
Copyright © 2013 Pearson Education, Inc.
433
434
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
57. r = 2 sin θ
Multiply both sides by r to obtain
r 2 = 2r sin θ . Since
r 2 = x 2 + y 2 and y = r sin θ , x 2 + y 2 = 2 y .
Complete the square on y to obtain
x 2 + y 2 − 2 y + 1 = 1 ⇒ x 2 + ( y − 1) = 1.
The graph is a circle with center at (0, 1) and
radius 1.
2
3
1 − sin θ
3
r=
⇒ r − r sin θ = 3 ⇒
1 − sin θ
r = r sin θ + 3 ⇒ x 2 + y 2 = y + 3
60. r =
58. r = 2 cos θ
Multiply both sides by r to obtain
r 2 = 2r cos θ .
x2 + y2 = y2 + 6 y + 9 ⇒ x2 = 6 y + 9 ⇒
3⎞
⎛
x2 = 6 ⎜ y + ⎟
⎝
2⎠
The graph is a parabola with axis x = 0 and
3⎞
⎛
vertex ⎜ 0, − ⎟ .
⎝
2⎠
Since r 2 = x 2 + y 2 and x = r cos θ ,
x 2 + y 2 = 2 x. Complete the square on x to get
the equation of a circle to obtain
x 2 − 2 x + y 2 = 0 ⇒ ( x − 1) + y 2 = 1.
The graph is a circle with center at (1, 0) and
radius 1.
2
61. r + 2 cos θ = −2 sin θ
r + 2 cos θ = −2 sin θ ⇒
r 2 = −2r sin θ − 2r cos θ ⇒ x 2 + y 2 = −2 y − 2 x
x2 + 2 x + y 2 + 2 y = 0 ⇒
x2 + 2 x + 1 + y2 + 2 y + 1 = 2 ⇒
( x + 1)2 + ( y + 1)2 = 2
2
1 − cos θ
Multiply both sides by 1 – cos θ to obtain
59. r =
The graph is a circle with center (−1, −1) and
radius 2.
r – r cos θ = 2. Substitute r = x 2 + y 2 to
obtain
x2 + y2 − x = 2 ⇒ x2 + y2 = 2 + x ⇒
x 2 + y 2 = (2 + x ) ⇒ x 2 + y 2 = 4 + 4 x + x 2 ⇒
2
y 2 = 4 (1 + x )
The graph is a parabola with vertex at (−1, 0)
and axis y = 0.
Copyright © 2013 Pearson Education, Inc.
Section 8.5 Polar Equations and Graphs
3
4 cos θ − sin θ
Using the general form for the polar equation
c
, with
of a line, r =
a cos θ + b sin θ
a = 4, b = −1, and c = 3, we have
4 x − y = 3. The graph is a line with intercepts
62. r =
(0, −3)
and
( 34 , 0) .
63. r = 2 sec θ
2
r = 2 sec θ ⇒ r =
⇒ r cos θ = 2 ⇒ x = 2
cos θ
The graph is a vertical line, intercepting the xaxis at 2.
435
2
cos θ + sin θ
Using the general form for the polar equation
c
, with
of a line, r =
a cos θ + b sin θ
a = 1, b = 1, and c = 2, we have x + y = 2.
The graph is a line with intercepts (0, 2) and
(2, 0).
65. r =
2
2 cos θ + sin θ
Using the general form for the polar equation
c
, with
of a line, r =
a cos θ + b sin θ
a = 2, b = 1, and c = 2, we have 2 x + y = 2.
The graph is a line with intercepts (0, 2) and
(1, 0).
66. r =
64. r = −5 csc θ
5
⇒
sin θ
r sin θ = −5 ⇒ y = −5
The graph is a horizontal line, intercepting the
y-axis at −5.
r = −5 csc θ ⇒ r = −
Copyright © 2013 Pearson Education, Inc.
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
436
67. Graph r = θ , a spiral of Archimedes.
θ
θ
71. (a)
–360°
–270°
–180°
–90°
0°
–6.3
–4.7
–3.1
–1.6
0
r =θ
–6.3
–4.7
–3.1
–1.6
0
θ
θ
90°
180°
270°
360°
1.6
3.1
4.7
1.6
3.1
4.7
(radians)
(radians)
r =θ
(r , −θ )
(b)
(r , π − θ )
or (− r , −θ )
(c)
(r , π + θ )
or (− r , θ )
72. (a) −θ
(b)
π −θ
(c) –r; – θ
(d)
–r
6.3
(e) π + θ
(f)
the polar axis
6.3
(g) the line θ =
73.
68.
74.
75.
69. In rectangular coordinates, the line passes
through (1, 0) and (0, 2) . So
2−0 2
=
= −2 and
0 − 1 −1
( y − 0) = −2 ( x − 1) ⇒ y = −2 x + 2 ⇒
2 x + y = 2. Converting to polar form
m=
c
, we have:
a cos θ + b sin θ
2
.
r=
2 cos θ + sin θ
r=
70. Answers will vary.
Copyright © 2013 Pearson Education, Inc.
π
2
Section 8.5 Polar Equations and Graphs
The points of intersection are
⎛4+ 2 π ⎞
⎛ 4 − 2 5π
⎜ 2 , 4 ⎟ and ⎜ 2 , 4
⎝
⎠
⎝
76.
437
⎞
⎟.
⎠
80. r = sin 2θ , r = 2cos θ , 0 ≤ θ < π
sin 2θ = 2 cos θ ⇒ 2 sin θ cos θ = 2 cos θ ⇒
2 sin θ cos θ − 2 cos θ = 0 ⇒
(
)
cos θ 2sin θ − 2 = 0
77. r = 4 sin θ , r = 1 + 2 sin θ , 0 ≤ θ < 2π
4 sin θ = 1 + 2 sin θ ⇒ 2 sin θ = 1 ⇒
π
1
5π
sin θ = ⇒ θ =
or
2
6
6
The points of intersection are
π π⎞ ⎛ π⎞
⎛
⎜⎝ 4 sin , ⎟⎠ = ⎜⎝ 2, ⎟⎠ and
6 6
6
5π 5π ⎞ ⎛ 5π ⎞
⎛
4 sin
,
⎟ = ⎜ 2,
⎟.
⎝⎜
6 6 ⎠ ⎝ 6 ⎠
Using a graphing calculator, we see that the
pole, (0, 0) is also an intersection. However, it
is not reached at the same value of theta for
each equation, which is why it does not appear
as a solution of the equation
4 sin θ = 1 + 2 sin θ .
cos θ = 0 or 2 sin θ − 2 = 0 ⇒
2
2 sin θ = 2 ⇒ sin θ =
2
π
π
3π
Thus, θ =
or
. The points
or θ =
4
4
2
π π⎞ ⎛ π⎞
⎛
of intersection are ⎜ sin 2 ⋅ , ⎟ = ⎜ 0, ⎟ ,
⎝
2 2⎠ ⎝ 2⎠
π π⎞ ⎛ π⎞
⎛
⎜⎝ sin 2 ⋅ , ⎟⎠ = ⎜⎝1, ⎟⎠ , and
4 4
4
3π 3π ⎞ ⎛
3π ⎞
⎛
,
⎜⎝ sin 2 ⋅
⎟⎠ = ⎜⎝ −1,
⎟.
4 4
4 ⎠
81. (a) Plot the following polar equations on the
same polar axis in radian mode:
Mercury: r =
Venus: r =
0.39(1 − 0.2062 )
;
1 + 0.206 cos θ
0.78(1 − 0.0072 )
;
1 + 0.007 cos θ
Earth: r =
1(1 − 0.0172 )
;
1 + 0.017 cos θ
Mars: r =
1.52(1 − 0.0932 )
.
1 + 0.093cos θ
78. r = 3, r = 2 + 2 cos θ ; 0° ≤ θ < 360°
3 = 2 + 2 cos θ ⇒ 1 = 2 cos θ ⇒
1
cos θ = ⇒ θ = 60° or 300°
2
The points of intersection are (3, 60°) and
(3, 300°).
79. r = 2 + sin θ , r = 2 + cos θ , 0 ≤ θ < 2π
2 + sin θ = 2 + cos θ ⇒ sin θ = cos θ ⇒
π
5π
θ = or
4
4
π
2 4+ 2
r = 2 + sin = 2 +
=
and
4
2
2
r = 2 + sin
5π
2 4− 2
= 2−
=
4
2
2
(continued on next page)
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438
Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations
(continued)
Plot the following polar equations on the
same polar axis:
Earth: r =
1(1 − 0.0172 )
;
1 + 0.017 cos θ
Jupiter: r =
5.2(1 − 0.0482 )
;
1 + 0.048 cos θ
Uranus: r =
19.2(1 − 0.047 2 )
;
1 + 0.047 cos θ
Pluto: r =
The graph shows that their orbits are very
close near the polar axis. Use ZOOM or
change your window to see that the orbit
of Pluto does indeed pass inside the orbit
of Neptune. Therefore, there are times
when Neptune, not Pluto, is the farthest
planet from the sun. (However, Pluto’s
average distance from the sun is
considerably greater than Neptune’s
average distance.)
39.4(1 − 0.2492 )
.
1 + 0.249 cos θ
82. (a) In degree mode, graph
r 2 = 40, 000 cos 2θ .
(c) We must determine if the orbit of Pluto is
always outside the orbits of the other
planets. Since Neptune is closest to Pluto,
plot the orbits of Neptune and Pluto on the
same polar axes.
30.1(1 − 0.0092 )
Neptune: r =
;
1 + 0.009 cos θ
Pluto: r =
39.4(1 − 0.2492 )
1 + 0.249 cos θ
Inside the “figure eight” the radio signal
can be received. This region is generally
in an east-west direction from the two
radio towers with a maximum distance of
200 mi.
(b) In degree mode, graph
r 2 = 22, 500 sin 2θ .
Inside the “figure eight” the radio signal
can be received. This region is generally
in a northeast-southwest direction from
the two radio towers with a maximum
distance of 150 mi.
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