424 Chapter 8 Complex Numbers, Polar Equations,...

424 Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations

Copyright © 2013 Pearson Education, Inc.

(c) 3 i− −

Sketch a graph of 3 i− − in the complex

plane.

Using a calculator, we find that the

reference angle is 18.4º. The graph shows

that θ is in quadrant III, so

180 18.4 198.4 .θ = ° + ° = ° Therefore,

( )3 10 cos 198.4 sin198.4 .i i− − = ° + °

6. (a) ( ) 1 34 cos 60 sin 60 4

2 2

2 2 3

i i

i

⎛ ⎞° + ° = +⎜ ⎟⎝ ⎠

= +

(b) ( )5cis130 5 cos130 sin130

5cos130 5 sin130

3.2139 3.8302

i

i

i

° = ° + °= ° + °= − +

(c) ( ) ( )7 cos 270 sin 270 7 0 ( 1)

0 7

i i

i

° + ° = + −= −

(d) ( ) ( )2cis 0 2 cos 0 sin 0 2 1 0

2 or 2 0

i i

i

° = ° + ° = += +

7. ( )( )

12 cos80 sin 80 ,

3 cos 50 sin 50

w i

z i

= ° + °= ° + °

(a) ( )( )

( )( )

( )

12 cos80 sin 80

3 cos 50 sin 50

cos 80 5012 3

sin 80 50

36 cos130 sin130

wz i

i

i

i

= ° + °⋅ ° + °

⎡ ° + ° ⎤= ⋅ ⎢ ⎥+ ° + °⎣ ⎦= ° + °

(b) ( )( )

( ) ( )( )

12 cos80 sin 80

3 cos50 sin 50

4 cos 80 50 sin 80 50

4 cos 30 sin 30

3 14 2 3 2

2 2

iw

z i

i

i

i i

° + °=

° + °= ⎡ ° − ° + ° − ° ⎤⎣ ⎦= ° + °

⎛ ⎞= + = +⎜ ⎟⎝ ⎠

(c) ( )( ) ( )

( )

33

3

3 cos 50 sin 50

3 cos 3 50 sin 3 50

27 cos150 sin150

3 1 27 3 2727

2 2 2 2

z i

i

i

i i

= ⎡ ° + ° ⎤⎣ ⎦= ⎡ ⋅ ° + ⋅ ° ⎤⎣ ⎦= ° + °

⎛ ⎞= − + = − +⎜ ⎟⎝ ⎠

(d) ( )( ) ( )

( )

33

3

12 cos80 sin 80

12 cos 12 80 sin 12 80

1728 cos 960 sin 960

1 31728

2 2

1728 1728 3864 864 3

2 2

w i

i

i

i

i i

= ⎡ ° + ° ⎤⎣ ⎦= ⎡ ⋅ ° + ⋅ ° ⎤⎣ ⎦= ° + °

⎛ ⎞= − −⎜ ⎟⎝ ⎠

= − − = − −

8. Find all the fourth roots of

( )16 16 cos180 sin180i− = ° + ° .

Since ( )4 cos 4 sin 4r iα α+

( )16 cos180 sin180 ,i= ° + ° then we have

4 16 2r r= ⇒ = and 4 180 360 kα = ° + ° ⋅ ⇒

180 36045 90 ,

4

kkα ° + ° ⋅= = ° + ° ⋅ k any

integer.

If 0, then 45 0 45 .k α= = ° + ° = °

If 1, then 45 90 135 .k α= = ° + ° = °

If 2, then 45 180 225 .k α= = ° + ° = °

If 3, then 45 270 315 .k α= = ° + ° = °

Solution set:

( ) ( ){( )

2 cos 45 sin 45 ,2 cos135 sin135 ,

2 cos 225 sin 225 ,

i i

i

° + ° ° + °° + °

( )}2 cos 315 sin 315i° + ° or

{ }2 2, 2 2, 2 2, 2 2i i i i+ − + − − −

Section 8.5 Polar Equations and

Graphs

1. (a) II (since 0 and 90 180 )r θ> ° < < °

(b) I (since r > 0 and 0° < 90θ < ° )

(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

Section 8.5 Polar Equations and Graphs 425


(c) negative y-axis

(d) positive y-axis (since 450° – 360° = 90°)

For Exercises 3(b)−14(b), answers may vary.

3. (a)

(b) Two other pairs of polar coordinates for (1, 45°) are (1, 405°) and (–1, 225°).

(c) Since 2

cos 1 cos 452

x r xθ= ⇒ = ⋅ ° =

and 2

sin 1 sin 45 ,2

y r yθ= ⇒ = ⋅ ° =

the point is 2 2

, .2 2

⎛ ⎞⎜ ⎟⎝ ⎠

4. (a)

(b) Two other pairs of polar coordinates for (3, 120°) are (3, 480°) and (–3, 300°).

(c) Since 3

cos 3cos1202

x r xθ= ⇒ = = −

and 3 3

sin 3sin120 ,2

y r yθ= ⇒ = ° =

the point is 3 3 3

, .2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

5. (a)

(b) Two other pairs of polar coordinates for (–2, 135°) are (–2, 495°) and (2, 315°).

(c) Since

( )cos 2 cos135 2x r xθ= ⇒ = − ° = and

( )sin 2 sin135 2,y r yθ= ⇒ = − ° = the

point is ( )2, 2 .−

6. (a)

(b) Two other pairs of polar coordinates for (–4, 30°) are (−4, 390°) and (4, 210°).

(c) Since

( )cos 4 cos30 2 3x r xθ= ⇒ = − ° = −

and ( )sin 4 sin 30 2,y r yθ= ⇒ = − ° = −

the point is ( )2 3, 2 .− −

7. (a)

(b) Two other pairs of polar coordinates for (5, – 60°) are (5, 300°) and (–5, 120°).

(c) Since ( ) 5cos 5cos 60

2x r xθ= ⇒ = − ° =

and

( ) 5 3sin 5sin 60 ,

2y r yθ= ⇒ = − ° = −

the point is 5 5 3

, .2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

8. (a)

(b) Two other pairs of polar coordinates for (2, – 45°) are (2, 315°) and (–2, 135°).



(c) Since

( )cos 2cos 45 2x r xθ= ⇒ = − ° = and

( )sin 2sin 45 2,y r yθ= ⇒ = − ° = − the

point is ( )2, 2 .−

9. (a)

(b) Two other pairs of polar coordinates for (–3, –210°) are (–3, 150°) and (3, –30°).

(c) Since

( ) ( ) 3 3cos 3 cos 210

2x r xθ= ⇒ = − − ° =

and

( ) ( ) 3sin 3 sin 210 ,

2y r yθ= ⇒ = − − ° = −

the point is 3 3 3

, .2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

10. (a)

(b) Two other pairs of polar coordinates for (–1, –120°) are (–1, 240°) and (1, 60°).

(c) Since

( ) ( ) 1cos 1 cos 120

2x r xθ= ⇒ = − − ° =

and

( ) ( ) 3sin 1 sin 120 ,

2y r yθ= ⇒ = − − ° =

the point is 1 3

, .2 2

⎛ ⎞⎜ ⎟⎝ ⎠

11. (a)

(b) Two other pairs of polar coordinates for

53,

3

π⎛ ⎞⎜ ⎟⎝ ⎠

are 11 2

3, and 3, .3 3

π π⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(c) Since 5 3

cos 3cos3 2

x r xπθ= ⇒ = =

and 5 3 3

sin 3sin ,3 2

y r yπθ= ⇒ = = −

the point is 3 3 3

, .2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

12. (a)


34,

2

π⎛ ⎞⎜ ⎟⎝ ⎠

are 4, and 4, .2 2

π π⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(c) Since 3

cos 4cos 02

x r xπθ= ⇒ = = and

3sin 4sin 4,

2y r y

πθ= ⇒ = = − the

point is (0, −4).

13. (a)


2,3

π⎛ ⎞−⎜ ⎟⎝ ⎠ are

7 42, and 2, .

3 3

π π⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



(c) Since cos 2cos 13

x r xπθ= ⇒ = − = −

and sin 2sin 3,3

y r yπθ= ⇒ = − = −

the point is ( )1, 3 .− −

14. (a)


55,

6

π⎛ ⎞−⎜ ⎟⎝ ⎠ are

115,

6

π⎛ ⎞⎜ ⎟⎝ ⎠

and

175, .

6

π⎛ ⎞−⎜ ⎟⎝ ⎠

(c) Since

5 5 3cos 5cos

6 2x r x

πθ= ⇒ = − = and

5 5sin 5sin ,

6 2y r y

πθ= ⇒ = − = − the

point is 5 3 5

, .2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

For Exercises 15(b)–26(b), answers may vary.

15. (a)

(b) ( )221 1 1 1 2r = + − = + = and

( )1 11tan tan 1 45 ,

1θ − −−⎛ ⎞= = − = − °⎜ ⎟⎝ ⎠

since

θ is in quadrant IV. Since

360 45 315 ,° − ° = ° one possibility is

( )2, 315 .° Alternatively, if 2,r = −

then 315 180 135 .θ = ° − ° = ° Thus, a

second possibility is ( )2,135 .− °

16. (a)

(b) 2 21 1 1 1 2r = + = + = and

1 11tan tan 1 45 ,

1θ − −⎛ ⎞= = = °⎜ ⎟⎝ ⎠

since θ is

in quadrant I. So, one possibility is


then 45 180 225 .θ = ° + ° = ° Thus, a

second possibility is ( )2, 225 .− °

17. (a)

(b) 2 2

0 3 0 9 9 3r = + = + = = and

90 ,θ = ° since ( )0,3 is on the positive y-

axis. So, one possibility is ( )3, 90 .°

Alternatively, if 3,r = − then

90 180 270 .θ = ° + ° = ° Thus, a second

possibility is ( )3, 270 .− °

18. (a)

(b) ( )220 3 0 9 9 3r = + − = + = = and

270 ,θ = ° since ( )0,3 is on the negative

y-axis. So, one possibility is ( )3, 270 .°


270 180 90 .θ = ° − ° = ° Thus, a second




19. (a)

(b) ( ) ( )2 2

2 2 2 2 4 2r = + = + = =

and 1 12tan tan 1 45 ,

2θ − −⎛ ⎞= = = °⎜ ⎟⎝ ⎠

since θ is in quadrant I. So, one

possibility is ( )2, 45 .° Alternatively, if

2,r = − then 45 180 225 .θ = ° + ° = °

Thus, a second possibility is ( )2, 225 .− °

20. (a)

(b) ( ) ( )2 2

2 2 2 2 4 2r = − + = + = =

and 1 2tan

2θ − ⎛ ⎞= =⎜ ⎟−⎝ ⎠

( )1tan 1 ,− − since

θ is in quadrant II we have 135 .θ = °

So, one possibility is ( )2, 135 .°


135 180 315 .θ = ° + ° = ° Thus, a second


21. (a)

(b)

2 23 3 3 9

2 2 4 4r

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

123

4= = and

3 2arctan

2 3θ ⎛ ⎞= ⋅⎜ ⎟⎝ ⎠

( )1tan 3 60 ,−= = ° since θ is in

quadrant I. So, one possibility is


then 60 180θ = ° + ° = 240 .° Thus, a

second possibility is ( )3, 240 .− °

22. (a)

(b)

2 23 1 3 1

2 2 4 4r

⎛ ⎞ ⎛ ⎞= − + − = +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

41

4= = and

1 2arctan

2 3θ

⎛ ⎞⎛ ⎞− −= ⋅ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1 1tan

3

− ⎛ ⎞= =⎜ ⎟⎝ ⎠

1 3tan 210

3

− ⎛ ⎞= °⎜ ⎟⎝ ⎠

since

θ is in quadrant III. So, one possibility is

( )1, 210 .° Alternatively, if 1,r = − then

210 180 30 .θ = ° − ° = ° Thus, a second


23. (a)

(b) 2 2

3 0 9 0 9 3r = + = + = = and

0 ,θ = ° since ( )3,0 is on the positive

x-axis. So, one possibility is ( )3, 0 .°


0 180 180 .θ = ° + ° = ° Thus, a second




24. (a)

(b) ( )2 22 0 4 0 4 2r = − + = + = = and

180 ,θ = ° since (−2, 0) is on the negative

x-axis. So, one possibility is (2, 180°). Alternatively, if r = −2, then

180 180 0 .θ = ° − ° = ° Thus, a second

possibility is (−2, 0°).

25. (a)

(b)

223 3 3 9 27

2 2 4 4

363

4

r⎛ ⎞⎛ ⎞= − + − = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= =

and

3 3

1 1232

tan tan 3.θ − −⎛ ⎞−

= =⎜ ⎟⎜ ⎟−⎝ ⎠

Since

θ is in quadrant III we have 240 .θ = °

So, one possibility is (3, 240°). Alternatively, if r = −3, then

240 180 60 .θ = ° − ° = ° Thus, a second


26. (a)

(b)

221 3 1 3

12 2 4 4

r⎛ ⎞⎛ ⎞= + − = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and ( )3

1 1212

tan tan 3 .θ − −⎛ ⎞−

= = −⎜ ⎟⎜ ⎟⎝ ⎠

Since θ is in quadrant IV we have

300 .θ = ° So, one possibility is (1, 300°).

Alternatively, if r = −1, then

300 180 120 .θ = ° − ° = ° Thus, a second


27. 4x y− =

Using the general form for the polar equation

of a line, ,cos sin

cr

a bθ θ=

+ with

1, 1, and 4,a b c= = − = the polar equation is

4.

cos sinr

θ θ=

−

28. 7x y+ = −


of a line, ,cos sin

cr

a bθ θ=

+ with

1, 1, and 7,a b c= = = − the polar equation is

7.

cos sinr

θ θ−=+



29. 2 2 216 16 4x y r r+ = ⇒ = ⇒ = ±

The equation of the circle in polar form is r = 4 or r = −4.

30. 2 2 29 9 3x y r r+ = ⇒ = ⇒ = ±

The equation of the circle in polar form is r = 3 or r = −3.

31. 2 5x y+ =


of a line, ,cos sin

cr

a bθ θ=

+with

2, 1, and 5,a b c= = = the polar equation is

5.

2cos sinr

θ θ=

+

32. 3 2 6x y− =


of a line, ,cos sin

cr

a bθ θ=

+with

3, 2, and 6,a b c= = − = the polar equation is

6.

3cos 2sinr

θ θ=

−

33. sinr kθ = 34. sin

kr

θ=

35. sin

kr

θ= ⇒ cscr k θ=

36. y = 3

37. cosr kθ = 38. cos

kr

θ=

39. seccos

kr r k θ

θ= ⇒ =

40. x = 3

41. r = 3 represents the set of all points 3 units from the pole. The correct choice is C.

42. cos 3r θ= is a rose curve with 3 petals. The

correct choice is D.

43. cos 2r θ= is a rose curve with 2 2 4⋅ =

petals. The correct choice is A



44. The general form for the polar equation of a line is ,cos sin

cr

a bθ θ=

+ where the standard form of a line ax

+ by = c. 2

cos sinr

θ θ=

+ is a line. The correct choice is B.

45. 2 2cosr θ= + (cardioid)

θ 0° 30° 60° 90° 120° 150°

cos θ 1 0.9 0.5 0 –0.5 –0.9

2 2cosr 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

2 2cosr θ= + 0 0.3 1 2 3 3.7

46. 8 6cos (limaçon)r θ= +

θ 0° 30° 60° 90° 120° 150°

cos θ 1 0.9 0.5 0 –0.5 –0.9

8 6cos r 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

8 6cos r θ= +

2 2.8 5 8 11 13.2

47. 3 cos r θ= + (limaçon)

θ 0° 30° 60° 90° 120° 150°

3 cos r θ= + 4 3.9 3.5 3 2.5 2.1

θ 180° 210° 240° 270° 300° 330°

3 cos r θ= + 2 2.1 2.5 3 3.5 3.9

48. 2 cos r θ= − (limaçon)

θ 0° 30° 60° 90° 135° 180° 225° 270° 315°

2 cos r θ= − 1 1.1 1.5 2 2.7 3 2.7 2 1.3



49. 4cos 2 r θ= (four-leaved rose)

θ 0° 30° 45° 60° 90° 120° 135° 150°

4cos 2 r θ=

4 2 0 –2 –4 –2 0 2

θ 180° 210° 225° 240° 270° 300° 315° 330°

4cos 2 r θ=

4 2 0 –2 –4 –2 0 2

50. 3cos5 r θ= (five-leaved rose)

r = 0 when cos 5 0,θ = or 5 90 360 18 72 , where is an integer, ork k kθ = ° + ° ⋅ = ° + ° ⋅ 18 , 90 , 162 , 234 .θ = ° ° ° °

θ 0° 18° 36° 54° 72° 90° 108° 162°

3cos5 r θ= 3

0 –3 0 3 0 –3 0

Pattern 3, 0, –3, 0, 3 continues for every 18°.

51. 2 4cos 2 2 cos 2 (lemniscate)r rθ θ= ⇒ = ±

Graph only exists for [0°, 45°], [135°, 225°], and [315°, 360°] because cos 2θ must be positive.

θ 0° 30° 45° 135° 150°

2 cos 2r θ= ± ±2 ±1.4 0 0 ±1.4

θ 180° 210° 225° 315° 330°

2 cos 2r θ= ± ±2 ±1.4 0 0 ±1.4

52. 2 4sin 2 2 sin 2 (lemniscate)r rθ θ= ⇒ = ±

Graph only exists for [0°, 90°] and [180°, 270°] because sin 2θ must be positive.

θ 0° 30° 45° 60° 90° 180° 225° 270°

2 sin 2r θ= ± 0 ±1.86 ±2 ±1.86 0 0 ±2 0



53. 4 4cosr θ= − (cardioid)

θ 0° 30° 60° 90° 120° 150°

4 4cosr θ= − 0 0.5 2 4 6 7.5

θ 180° 210° 240° 270° 300° 330°

4 4cosr θ= − 8 7.5 6 4 2 0.5

54. 6 3cos (limaçon)r θ= −

θ 0° 45° 90° 135° 180° 270° 360°

6 3cosr θ= − 3 3.9 6 8.1 9 6 3

55. 2sin tanr θ θ= (cissoid)

r is undefined at θ = 90° and θ = 270°.

θ 0° 30° 45° 60° 90° 120° 135° 150° 180°

2sin tanr θ 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°).

56. cos 2

(cissoid with a loop)cos

rθθ

=

r is undefined at 90θ = ° and θ = 270° and r = 0 at 45°, 135°, 225°, and 315°.

θ 0° 45° 60° 70° 80°

cos 2

cosr

θθ

= 1 0 –1 –2.2 –5.4

θ 90° 100° 110° 135° 180°

cos 2

cosr

θθ

= undefined 5.4 2.2 0 –1

Notice that for [180°, 360°), the graph retraces the path traced for [0°, 180°).



57. r = 2 sin θ Multiply both sides by r to obtain

2 2 sin .r r θ= Since 2 2 2 and sin ,r x y y r θ= + = 2 2 2x y y+ = .

Complete the square on y to obtain

( )22 2 22 1 1 1 1.x y y x y+ − + = ⇒ + − =

The graph is a circle with center at (0, 1) and radius 1.

58. r = 2 cos θ Multiply both sides by r to obtain

2 2 cos .r r θ= 2 2 2Since and cos ,r x y x r θ= + =

2 2 2 .x y x+ = Complete the square on x to get

the equation of a circle to obtain

( )22 2 22 0 1 1.x x y x y− + = ⇒ − + =

The graph is a circle with center at (1, 0) and radius 1.

59. 2

1 cosr

θ=

−

Multiply both sides by 1 – cosθ to obtain

r – r cosθ = 2. Substitute 2 2 r x y= + to

obtain

( )( )

2 2 2 2

22 2 2 2 2

2

2 2

2 4 4

4 1

x y x x y x

x y x x y x x

y x

+ − = ⇒ + = + ⇒

+ = + ⇒ + = + + ⇒= +

The graph is a parabola with vertex at (−1, 0) and axis y = 0.

60. 3

1 sinr

θ=

−

2 2

2 2 2 2

2

3sin 3

1 sin

sin 3 3

6 9 6 9

36

2

r r r

r r x y y

x y y y x y

x y

θθ

θ

= ⇒ − = ⇒−

= + ⇒ + = ++ = + + ⇒ = + ⇒

⎛ ⎞= +⎜ ⎟⎝ ⎠

The graph is a parabola with axis x = 0 and

vertex 3

0, .2

⎛ ⎞−⎜ ⎟⎝ ⎠

61. 2 cos 2sinr θ θ+ = −

( ) ( )

2 2 2

2 2

2 2

2 2

2cos 2sin

2 sin 2 cos 2 2

2 2 0

2 1 2 1 2

1 1 2

r

r r r x y y x

x x y y

x x y y

x y

θ θθ θ

+ = − ⇒= − − ⇒ + = − −+ + + = ⇒+ + + + + = ⇒+ + + =

The graph is a circle with center (−1, −1) and

radius 2.



62. 3

4cos sinr

θ θ=

−


of a line, ,cos sin

cr

a bθ θ=

+with

4, 1,a b= = − and 3,c = we have

4 3.x y− = The graph is a line with intercepts

( )0, 3− and ( )34

, 0 .

63. 2secr θ=

22sec cos 2 2

cosr r r xθ θ

θ= ⇒ = ⇒ = ⇒ =

The graph is a vertical line, intercepting the x-axis at 2.

64. 5cscr θ= −

55csc

sinsin 5 5

r r

r y

θθ

θ

= − ⇒ = − ⇒

= − ⇒ = −

The graph is a horizontal line, intercepting the y-axis at −5.

65. 2

cos sinr

θ θ=

+


of a line, ,cos sin

cr

a bθ θ=

+with

1, 1,a b= = and 2,c = we have 2.x y+ =

The graph is a line with intercepts (0, 2) and (2, 0).

66. 2

2cos sinr

θ θ=

+


of a line, ,cos sin

cr

a bθ θ=

+with

2, 1,a b= = and 2,c = we have 2 2.x y+ =

The graph is a line with intercepts (0, 2) and (1, 0).



67. Graph r = θ , a spiral of Archimedes.

θ –360° –270° –180° –90° 0°

θ

(radians) –6.3 –4.7 –3.1 –1.6 0

r θ= –6.3 –4.7 –3.1 –1.6 0

θ 90° 180° 270° 360°

θ

(radians) 1.6 3.1 4.7 6.3

r θ= 1.6 3.1 4.7 6.3

68.

69. In rectangular coordinates, the line passes

through ( )1,0 and ( )0, 2 . So

2 0 22

0 1 1m

−= = = −− −

and

( ) ( )0 2 1 2 2y x y x− = − − ⇒ = − + ⇒

2 2.x y+ = Converting to polar form

cos sin

cr

a bθ θ=

+, we have:

2.

2cos sinr

θ θ=

+

70. Answers will vary.

71. (a) ( ),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

(e) π θ+ (f) the polar axis

(g) the line 2

πθ =

73.

74.

75.



76.

77. r = 4 sinθ , r = 1 + 2 sinθ , 0 ≤ θ < 2π

4sin 1 2sin 2sin 1

1 5sin or

2 6 6

θ θ θπ πθ θ

= + ⇒ = ⇒

= ⇒ =

The points of intersection are

4sin , 2, 6 6 6

π π π⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ and

5 5 54sin , 2, .

6 6 6

π π π⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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

4sin 1 2sin .θ θ= +

78. 3, 2 2 cos ; 0 360r r θ θ= = + ° ≤ < °

3 2 2 cos 1 2 cos

1cos 60 or 300

2

θ θ

θ θ

= + ⇒ = ⇒

= ⇒ = ° °

The points of intersection are (3, 60°) and (3, 300°).

79. r = 2 + sinθ , r = 2 + cosθ , 0 ≤ θ < 2π

2 sin 2 cos sin cos

5 or

4 4

θ θ θ θπ πθ

+ = + ⇒ = ⇒

=

2 4 2

2 sin 24 2 2

rπ += + = + = and

5 2 4 22 sin 2

4 2 2r

π −= + = − =

The points of intersection are

4 2 4 2 5, and , .

2 4 2 4

π π⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

80. sin 2 , 2cos , 0r rθ θ θ= = ≤ < π

( )

sin 2 2 cos 2sin cos 2 cos

2sin cos 2 cos 0

cos 2sin 2 0

cos 0 or 2sin 2 0

22sin 2 sin

2

θ θ θ θ θθ θ θθ θ

θ θ

θ θ

= ⇒ = ⇒− = ⇒− =

= − = ⇒

= ⇒ =

Thus, or 2

πθ = 3 or .

4 4

π πθ = The points

of intersection are sin 2 , 0, ,2 2 2

π π π⎛ ⎞ ⎛ ⎞⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sin 2 , 1, ,4 4 4

π π π⎛ ⎞ ⎛ ⎞⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ and

3 3 3sin 2 , 1, .

4 4 4

π π π⎛ ⎞ ⎛ ⎞⋅ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

81. (a) Plot the following polar equations on the same polar axis in radian mode:

Mercury: 20.39(1 0.206 )

;1 0.206cos

rθ

−=+

Venus: 20.78(1 0.007 )

;1 0.007 cos

rθ

−=+

Earth: 21(1 0.017 )

1 0.017 cosr

θ−=

+;

Mars: 21.52(1 0.093 )

.1 0.093cos

rθ

−=+

(continued on next page)



(continued)

Plot the following polar equations on the same polar axis:

Earth: 21(1 0.017 )

1 0.017 cosr

θ−=

+;

Jupiter: 25.2(1 0.048 )

1 0.048cosr

θ−=

+;

Uranus: 219.2(1 0.047 )

1 0.047 cosr

θ−=

+;

Pluto: 239.4(1 0.249 )

1 0.249cosr

θ−=

+.

(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.

Neptune: 230.1(1 0.009 )

;1 0.009cos

rθ

−=+

Pluto: 239.4(1 0.249 )

1 0.249cosr

θ−=

+

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.)

82. (a) In degree mode, graph 2

40,000cos 2 .r 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 2

22,500sin 2 .r θ=

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.

424 Chapter 8 Complex Numbers, Polar Equations,...

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Transcript of 424 Chapter 8 Complex Numbers, Polar Equations,...